Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Converse to Jensen's inequality for $1/x$ on a positive bounded interval? Consider the function $f(x) = 1/x$ on the interval $I = [a, b]$, where $0 < a \leq b$.
By Jensen's inequality, we have for any $\{x_j \}_{j=1}^n \subset I$,
$$
f(\overline{x}) \leq \frac{1}{n} \sum_{i=1}^n f(x_i).
$$
Above, $\overline{x} = (1/n)... | Yes, there is a bound available.
Let $t_i := \max \Big\{
\tfrac{(x_i - \overline{x})^2}{x_i^3},
\tfrac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}$.
We have that the remainder is the average $\frac{1}{n} \sum_{j=1}^n t_j$.
Consider the two cases separately. First suppose $x_i > \overline{x}$.
Then
$$
t_i = \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact ... | I will directly follow your step.
$$
\begin{aligned}
\frac{\operatorname{d}y}{\operatorname{d}x}&=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}\\
&=-\frac{1}{2}\frac{1}{\sqrt{1+x}}\Big(\frac{1}{\sqrt{1-x}}+\frac{\sqrt{1-x}}{1+x}\Big)\\
&=-\frac{1}{2\sqrt{1+x}}\frac{2}{(1+x)\sqrt{1-x}}\\
&=-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find all bounded infinite sequences of positive integers $(a_n)$ with $a_n = (a_{n-1}+a_{n-2})/(\gcd(a_{n-1},a_{n-2}))$ for $n\ge 2$.
Find all bounded infinite sequences of positive integers $(a_n)$ with $a_n = (a_{n-1}+a_{n-2})/(\gcd(a_{n-1},a_{n-2}))$ for $n\ge 2$.
I think the only possible sequence is the constant... | $\textbf{Lemma:}$ when $j \in \mathbb{N}$ we have that $\text{gcd}(a_{j+1},a_{j+2})$ divides $\text{gcd}(a_{j},a_{j+1})$
$\textbf{Proof:} $ Put $\text{gcd}(a_{j},a_{j+1}) = A_{j}$ and write $a_{j} = A_{j}v_{j}, a_{j+1} = A_{j+1}v_{j+1}$ where $\text{gcd}(v_{j}, v_{j+1}) = 1$. We have $a_{j+2} = v_{j}+v_{j+1}$ and $\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations... | If we call $ \ b \ $ and $ \ d \ $ the roots of a quadratic equation, the Viete relations tell us that the equation is $ \ x^2 + (6 - c^2)x + 6 \ = 0 \ \ \ $ (from the sum and product of the roots). The difference of the zeroes of a monic quadratic polynomial is equal to the square-root of its discriminant $ \ ( \ b -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
There is a disk of radius $r(x)=2-\frac{2}{5}|x|$ at every point on $[-5, 5]$. Find equation of a union of all disks Modeling shows that the figure look like this:
But I don't know how to get an equation for it analytically
Looks like for some $0<x_0<5$ for $(x_0, 5]$ and $[-5, x_0)$ correspoing disks touch final shap... | Based on the diagram and the comment from @ajotatxe, I will assume that you meant to write $r(x) = 2 - \frac{2}{5}|x|$. Fix a point $a$ on the interval $[0,5]$ and consider the disk
$$(x-a)^2 + y^2 = r(a)^2 = \left(2 - \frac{2}{5}a\right)^2 = 4 - \frac{8}{5}a + \frac{4}{25}a^2$$
centered at $a$. There is a unique coor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation of a line orthogonal to two other lines Consider the two lines $\ell_1:\mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+s\begin{pmatrix}-1\\2\\-2\end{pmatrix}$ and $\ell_2:\mathbf{r}=\begin{pmatrix}-2\\3\\-2\end{pmatrix}+t\begin{pmatrix}0\\-1\\2\end{pmatrix}$. I need to determine the equation of the line $\ell_... | If
\begin{align}
\mathbf{r}_1&=\begin{pmatrix}1\\1\\0\end{pmatrix}+t_1\begin{pmatrix}-1\\2\\-2\end{pmatrix}\ \text{and}\\
\mathbf{r}_2&=\begin{pmatrix}-2\\3\\-2\end{pmatrix}+t_2\begin{pmatrix}0\\-1\\2\end{pmatrix}
\end{align}
are the points of intersection of $\ \ell_1\ $ and $\ \ell_2\ $ with $\ \ell_3\ $, respectivel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Everywhere differentiable function whose derivative is not locally integrable My question is all in the title: I can’t come up with an example of a function $u \colon \mathbb R \to \mathbb R$ such that
*
*$u$ is differentiable at every $x \in \mathbb R$
*$u’ \notin L^1_{\text{loc}}(\mathbb R)$
Any ideas?
| To combine all the ideas from the comments, define $$f(x)=\begin{cases} x^2\sin(\frac{1}{x^2}) & x\not=0 \\ 0 &x=0\end{cases}$$
Then it's easy to see $f(x)$ is differentiable everywhere, and whenever $x\not=0$, $$f'(x)=2x\sin(\frac{1}{x^2})+x^2\cos(\frac{1}{x^2})\frac{-2}{x^3}=2x\sin(\frac{1}{x^2})+\frac{-2\cos(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried r... | So your sum looks like
$$
s(x) = 2 + \sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k.
$$
One approach is to expand the sum to
$$
\begin{split}
\sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k
&= \sum_{k=1}^\infty \frac{k^2+2k+1}{k} x^k \\
&= \sum_{k=1}^\infty kx^k
+ 2 \sum_{k=1}^\infty x^k
+ \sum_{k=1}^\infty \frac{x^k}{k}.
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
At which points does the graph of the $f(x)$ have horizontal tangent line? $(2x^2-2x-1)/(x^4+1)$
I tried to solve with the derivative method. But came across a fifth degree Polynomial involving $5$ terms. After that my math failed to factorize it. This is the derivative $2x^5 - 3x^4-2x^3-2x+1$. Kindly help.
| With derivative method, we get
$$f'(x) = \frac{(4x-2)(x^4+1)-4x^3(2x^2-2x-1)}{(x^4+1)^2} = 0.$$
We have to find $x$ so that the equality can be fulfilled.
We obtain
$$(4x-2)(x^4+1) - 4x^3(2x^2-2x-1) = 0$$
$$ \iff (4x^5-2x^4 + 4x-2)- 8x^5+8x^4+4x^3 = 0 $$
$$\iff -4x^5 + 6x^4+4x^3+4x-2 = 0 $$
$$\iff 2x^5 - 3x^4 -2x^3-2x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
show that all the roots of a given equation are in $[-1,1]$
*
*Suppose $c\in \mathbb{R}$ is such that one of the roots of the equation $x^3 - \dfrac{3}4 x+c=0$ belongs to $[-1,1]$.
*Show that all the roots belong to $[-1,1]$.
*The equation is equivalent to $p(x)=0$ where $p(x) = 4x^3 - 3x+4c.$
*
*Observe that $... | Without calculus:
Let $r$ be a root of the equation in $[-1, 1]$. We have $c = -r^3 + \frac34 r$.
The equation becomes
$$x^3 - \frac34x - r^3 + \frac34 r = 0$$
or
$$\frac14(x-r)(4x^2 + 4rx + 4r^2 - 3) = 0$$
which has exactly three real roots
$$x_1 = r,\quad x_2 = -\frac12 r + \frac12\sqrt{3 - 3r^2}, \quad
x_3 = -\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let $x,y,z\in\mathbb{R}^{+}$ such that $x+y+z = 1.$ T/F: $\frac{x^2 y^2}{(x+y)^2}+\frac{y^2 z^2}{(y+z)^2}+\frac{z^2 x^2}{(z+x)^2}\geq\frac{9}{4}xyz$ In Springer's "Inequalities" book, Exercise $2.5$ is the following:
Let $x,y,z\in\mathbb{R}^{+}\ $ such that $\ x+y+z = 1.\ $ Then:
$$ xy + yz + zx \geq 9xyz.\qquad (1)$$
... | $$\sum_{cyc}\frac{ 4x^2 y^2 }{ (x+y)^2 } \ge 9xyz$$
$$\sum_{cyc}\frac{ 4x y z(x+y+z)}{ (x+y)^2z^2 } \ge 9$$
define $a=yz,b=xz,c=xy$
$$\sum_{cyc}\frac{ 4(ab+bc+ac)}{ (a+b)^2 } \ge 9$$
$$\sum_{cyc}\frac{4}{ (a+b)^2 } \ge \frac{9}{ab+bc+ac}$$
which is well known prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Write $ z = \frac{(1-i)^3(√3+i)}{4i}$ to polar form Write the complex number in polar form:
$$ z = \frac{(1-i)^3(\sqrt 3+i)}{4i}$$
So my try goes as follows:
\begin{align}
\frac{(1−i)^3(\sqrt 3+i)}{4i} &=
\frac{(1−i)^3(\sqrt 3+i) \times -4i}{16}\\& =
\frac{(1-3i-3+i)(\sqrt 3+i)\times-4i}{16}\\& =
\frac{(-2-2i)(\sqrt 3... | Problems
The absolute value $r$ is correct, but the angle is not!
You used the wrong formula for the angle...
If you got a complex number $z = a + b \cdot \mathrm{i} = r \cdot \operatorname{cis}(\theta)$ then your angle is $\theta = \operatorname{arctan2}(b, ~a)$. But you used another formula.
If you use $\theta = \ope... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the system $ \left\lbrace \begin{array}{ccc} \sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\ x+y &=& 2xy \end{array}\right. $ I have to solve the following system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\
x+y &=& 2xy
\end{array}\right.
$$
I've showed that it is equivalent to
$... | We can still solve the system from its original form. From the first equation you get: $y \le 1$, and $y - x \ge 0$ or $y \ge x$. This leads to: $2x \le x+y = 2xy \le 2x\implies 2x \le 2xy \le 2x\implies 2x = 2xy \implies 2x(1-y) = 0\implies x = 0$ or $y = 1$ or both. If $x = 0$ then $y = 0$ from the second equation, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$ Let $x \in \mathbb{R}^*$ and $y \geq \frac{1}{4}$.
Show that if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$
I tried the following idea:
$$
2\sqrt{y-\frac{1}{4}}\leq \sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y
$$
which leads to:
$$
y... | Claim: $y \ge 1$.
Proof: For $y \ge \dfrac{1}{4}$ we have: $y = \sqrt{x^2+x+y} + \sqrt{x^2-x+y}\ge \sqrt{x^2+x+\dfrac{1}{4}}+\sqrt{x^2-x+\dfrac{1}{4}}=\left|x+\dfrac{1}{2}\right|+\left|x-\dfrac{1}{2}\right|$. So if $x \le -\dfrac{1}{2}$, then $y \ge -x-\dfrac{1}{2}-x+\dfrac{1}{2}=-2x \ge 1$. If $-\dfrac{1}{2} \le x \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How many natural numbers have $ n$ digits such that the sum of its digits is $ m.$ How many n-digit numbers such that the sum of its digits is $ m.$
My attempt :
We have the gf:
$\begin {align*}
f(x)&=x(1-x^9)(1-x^{10})^{n-1}(1-x)^{-n}\\&=(x-x^{10})(1-x^{10})^{n-1}(1-x)^{-n}
\end{align*}$
Extracting the coefficients of... | We obtain
\begin{align*}
\color{blue}{[x^m]}&\color{blue}{\left(x-x^{10}\right)\left(1-x^{10}\right)^{n-1}(1-x)^{-n}}\\
&=\left([x^{m-1}]-[x^{m-10}]\right)\sum_{j=0}^{\infty}\binom{-n}{j}(-x)^j\left(1-x^{10}\right)^{n-1}\tag{1}\\
&=\left([x^{m-1}]-[x^{m-10}]\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^j\left(1-x^{10}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{x\to 0}\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}$ How to evaluate the following limit without L'Hôpital's rule
?$$\lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)$$
My attempt:
$$\begin{align}L &= \lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)\tag1\\& = \lim_{x\to 0}\... | Using $$ \sin^{-1}(x) = x + \frac{x^3}{6} + \frac{3 \, x^5}{40} + \mathcal{O}(x^7) $$ then $$\frac{1}{(\sin^{-1}(x))^2} = \frac{1}{x^2} - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6). $$ This gives $$ \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} = - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int{\sqrt{x^2+4x+13}\, dx}$ I was trying to calculate with $\sinh$ :
$$\begin{align}I&=\int{\sqrt{x^2+4x+13}\, dx}\\&=\int{\sqrt{(x+2)^2+9}\, dx}\end{align}$$
Now with $x+2=3\sinh(u)$, $dx=3\cosh(u)\,du$
$$\begin{align}I&=3\int{\left(\sqrt{9\sinh^2(u)+9}\, \right)\cosh(u)\, du}\\&=9\int{\cosh(u)\sqrt{\sinh... | Letting $x+2=3\tan \theta$ changes the integral into
$$
I=9 \int \sec ^3 \theta d \theta
$$
$$
\begin{aligned}
\int \sec ^3 \theta d \theta &=\int \sec \theta d(\tan \theta) \\
&=\sec \theta \tan \theta-\int \sec \theta \tan ^2 \theta d \theta \\
&=\sec \theta \tan \theta-\int \sec \theta\left(\sec ^2 \theta-1\right) d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$ The question goes:
Evaluate $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$.
I have tried by parts, substitutions and other techniques I have learned from first year...but I didn't get any thing look good...
| The only viable solution using first-year knowledge is a series solution. We start by replacing the sine function with its respective Taylor series. Then we can integrate term by term
$$I=\int_1^{\frac{\pi}{2}}\frac{\ln{x}\sin{x}}{x^2}dx=\int_1^{\frac{\pi}{2}}\frac{\ln{x}}{x^2}\left(x+\sum_{k=1}^\infty\frac{(-1)^kx^{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$... | The following method has the advantage that it can be generalized
for similar kinds of problems with more variables.
Consider the equation satisfied by $\;z=x\;$ and $\;z=y\;$
$$ z^2 - (x+y)z + (xy) = (z-x)(z-y) = 0. \tag1 $$
Then for all integer $n\ge 0$ it is true that
$$ z^{n+2} - (x+y)z^{n+1} + (xy)z^n = 0. \tag2 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Where is my mistake when I am trying to solve $| \int_{\Gamma} \frac{1} {(z^2 +1)^2} dz | $
Find an upper bound for
$$ \bigg| \int_{\Gamma} \frac{1} {(z^2 +1)^2} dz \bigg|.$$
where Γ is the upper half-circle |z| = a with radius a > 1 traversed once in the counter clockwise direction.
I know the correct answer but I d... | There is a classic bound, best seen by geometric reasoning: $$\left|\int_\Gamma\frac{1}{(z^2+1)^2}\,\mathrm{d}z\right|<\frac{\pi a}{(a^2-1)^2}$$
Which is slightly sharper than yours, $a>1$. The explanation: $\Gamma$ traces a half-circle - the locus of $z^2,\,z\in\Gamma$ traces a full circle (circle wraps around, argume... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $? I need to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $. I've checked numerically that this is true, but I haven't been able to prove it.
I've tried trigonometric substitutions. Let $\tan u= t:$
$$\int_0^1 (1+t^2)^{\frac 7 2} dt = \int_0^{\frac... | Since $t^2 \in [0,1],\ $ we may use the Binomial expansion,
$$ \left( 1+ t^2 \right)^{7/2} = 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8 + (\text{ alternating sequence of decreasing terms with negative leading term }),$$
so,
$$ \left( 1+ t^2 \right)^{7/2} < 1 + \frac{7}{2}t^2 + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I solve this comparing of coefficient in this differential equation?
Solve $y'' + 2y' = \cos \pi x$
Homogeneous equation: $y= C_1 + C_2e^{-2x}$
Particular solution:
$y_p = A \cos \pi x + B\sin \pi x$
$y'=-A \pi \sin \pi x + B\pi \cos \pi x$
$y'' = -A \pi^2 \cos \pi x - B\pi^2 \sin \pi x$
Substituting back into... | From the second equation
$$A = -\dfrac{B \pi}{2}$$
Substituting into the first
$$B \pi^3 + 4 B \pi = 2$$
Solving for $B$
$$B = \dfrac{2}{\pi(\pi^2+4)}$$
Can you finish it off?
There are other approaches to finding $A$ and $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Parametric equation appears to generate circle and violate $x(t)^2+y(t)^2={R+}$ The context of the equations is from plotting the real against the imaginary impedance of an RC circuit however the problem is entirely mathematical.
Given the two equations:
$x(w) = \frac{R}{C^2 R^2 w^2 + 1}$
$y(w) = \frac{C R^2 w}{C^2 R^2... | Let's assume your parametric equation describes (at least part of) a circle, with unknown center $(a,b)$ and radius $r$. Hence, for all $w$,
$$(x-a)^2+(y-b)^2-r^2=0$$
The first term is
$$\left(\frac{R}{R^2C^2w^2+1}-a\right)^2=\frac{(R-aR^2C^2w^2-a)^2}{(R^2C^2w^2+1)^2}=\frac{a^2R^4C^4w^4+(R-a)^2-2a(R-a)R^2C^2w^2}{(R^2C^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int \frac{\cos 4x}{4 \sin 2x} dx$ $\int \frac{\cos 4x}{4 \sin 2x} dx$
Let $u=2x$, $dx = 1/2 du$
$\int \frac{\cos 2u}{4 \sin u} \frac{1}{2} du = \frac{1}{8} \int \frac{1-2\sin^2 u}{\sin u}du \frac{1}{8} \int \frac{1}{\sin u} du - \frac{1}{8} \int 2 \sin u$
How do I integrate $\int \frac{1}{\sin u} du$ to get $\ln (\ta... | Hint Rewrite the standard integration formula
$$\int \cot u \,du = - \log\left\vert\csc u + \cot u\right\vert + C$$
using a half-angle identity for cotangent,
$$\csc u + \cot u = \frac{1 + \cos u}{\sin u} = \cot \frac{u}{2} = \cot x .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Minimize function $\sqrt{x^2-4x+5}+\sqrt{4+x^2}$ Suppose I want to find
$$
\min f(x) = \min(\sqrt{x^2-4x+5}+\sqrt{4+x^2})
$$
I'd start with computing derivative and set it to $0$
$$
f'(x) = \frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}}
$$
Then
$$
\frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}} = 0
$$
$$
(x-... | Squaring both sides gives you the solutions to $(2-x)\sqrt{4+x^2} = x\sqrt{x^2-4x+5}$ and to $(2-x)\sqrt{4+x^2} = -x\sqrt{x^2-4x+5}$. So if our possible candidates are $x = 4/3, 4$, we can substitute these values into the equation before squaring.
We find that only $x_1 = 4/3$ satisfies the original equation.
Of course... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead... | If the Law of cosines works the Law of sines probably works. From $\frac{\sin A}{5}=\frac{\sin B}{4}=k$ and $\cos A\cos B+\sin A\sin B=\cos(A-B)$ we get the equation $$\sqrt{1-25k^2}\sqrt{1-16k^2}+20k^2=\frac{7}{8}$$
and $k=\frac{\sqrt{5}}{8\sqrt{2}}$. Hence, $\sin A=\frac{5\sqrt{5}}{8\sqrt{2}}$, $\sin B=\frac{4\sqrt{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Find the solution to a system of equations Initially I'd like to solve the following problem:
Find $x$: $$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$$
I'm already aware of one approach to solve it, but now I wonder what if I set $y=\sqrt{2+\sqrt{2+x}}$
?
That is, to solve the following system:
$$ \begin{cases} x = \sqrt... | I've just got an idea.
Let's suppose $x>y$, then $x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt{2+\sqrt{2+y}}<\sqrt{2+\sqrt{2+x}}=y$, which is impossible. When $x<y$ it's the same. Hence $x=y$.
As a result, $x=\sqrt{2+\sqrt2+x}\Rightarrow x^2=2+\sqrt{2+x}\Rightarrow x^4-4x^2+4=x+2\Rightarrow x^4-4x^2-x+2=0\Rightarrow$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ In a Calculus textbook, I was faced with the following
Problem: If $n$ is a natural number with $n \ge 1$, proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ for all $k \in \mathbb{N}$.
The book presented a different solution than the one, I came up with. So ... | There is a mistake in your derivation: $$\binom{n}{k+1} = \frac{n!}{(k+1)!(n-k\color{red}-1)!} \color{blue}{= \frac{n!}{k!(n-k)!} \cdot \frac{n-k}{k+1} = \binom{n}{k} \cdot \frac{n-k}{k+1}}.$$
So, the final step of your proof should be the following:
\begin{align}
\binom{n}{k+1} \frac{1}{n^{k+1}} & = \binom{n}{k} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $T(n) = T(\lfloor n/2 \rfloor) * T(\lfloor n/2 \rfloor) \leq 2^n$ I am stuck on the below problem.
Here is what I have so far, and I have highlighted the area that I am not understanding, particularly in the induction step when we start evaluating for k+1.
$T(n) = T(\lfloor \frac{n}{2} \rfloor) * T(\lfloor \frac{... | You've begun well with your base step, but there are a couple of issues with your induction step. First, the problem uses the floor function for the RHS of the function definition, i.e.,
$$T(n) = T\left(\left\lfloor\frac{n}{2}\right\rfloor\right)\times T\left(\left\lfloor\frac{n}{2}\right\rfloor\right) \tag{1}\label{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\frac{1}{x^3+y^3} + \frac{1}{x^6+y^6}$ Simplify: $$\frac{1}{x^3+y^3} + \frac{1}{x^6+y^6}$$
I know $\frac {na}{nb} = \frac {a}{b}$.
But I also know $(\frac{a}{b})^2 = \frac {a^2}{b^2} \neq \frac{a}{b}$
So how do I get the same denominator (go from $x^3+y^3$ to $x^6+y^6$) if I can't just square it, since t... | How about if you let $x^3 = a$ and $y^3 = b$?
Then $x^3 + y^3 = a + b$ and $x^6 + y^6 = a^2 + b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a better method to factoring large polynomials such as $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$? The question is as such:
Express $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ in irreducible factors $\in \mathbb{R}$
My initial thoughts to this were to use grouping of factors but this method was to no avail. Instead, I ... | Another well-known method is to rewrite the polynomial as follows.
$\begin{align}
&1+z+z^2+z^3+z^4+z^5+z^6=\\
&=z^3\left(\dfrac{1}{z^3}+\dfrac{1}{z^2}+\dfrac{1}{z}+1+z+z^2+z^3\right)=\\
&=z^3\left[\left(\dfrac{1}{z^3}+z^3\right)+\left(\dfrac{1}{z^2}+z^2\right)+\left(\dfrac{1}{z}+z\right)+1\right]=\\
&=z^3\left[\left(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove the inequality if $a,b,c>0$, $a \geq b+c$. Prove or disprove the inequality
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 7abc$$ if $$a,b,c>0, a \geq b+c.$$
I thought to use this evaluation:
$$a^2b+b^2c+c^2a \geq 3abc.$$
So we have:
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 3abc+3abc=6abc,$$ which is obvious that $... | Since $$b+c\geq\frac{7bc-b^2-c^2}{2(b+c)},$$ we obtain:
$$\sum_{cyc}(a^2b+a^2c)-7abc=(b+c)a^2+(b^2+c^2-7bc)a+bc(b+c)\geq$$
$$\geq(b+c)^3+(b^2+c^2-7bc)(b+c)+bc(b+c)=2(b+c)(b-c)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $3 = x^2 + y^2 + z^2 - xy - yz - zx$ for integer $x$, $y$, $z$ I've been thinking about a solution for the following equation for integers $x, y, z$:
$$3 = x^2 + y^2 + z^2 - xy - yz - zx$$
A possible approach would probably be to transform the original equation to the following one:
$$6 = (x-y)^2 + (y-z)^2 + (x... | You have made great Progress to get :
$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$
The Squares $S1,S2,S3$ must be (Case 1) all 3 even or (Case 2) 1 even with 2 odd.
Let $S1 \le S2 \le S3$ , without loss of generality.
Case 1 : This implies that the Squares are (1A) $0,0,6$ (1B) $0,2,4$ (1C) $2,2,2$
We can then see that no Integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Getting different values of the integral when integrating with different order I want to determine the volume of the shape given by the following
$$ K = \{(x,y,z): x^2+y^2+z^2 \leq 2 \, \, , x+y>0 \, \, , \, z\leq 1\} $$
I thought that i can integrate with respect to $xy$-plane first by using polar coordinates
$$ \int_... | The first integral is correct. As regards the second integral note that $z=\sqrt{2-x^2-y^2}\leq 1$ for $1\leq x^2+y^2\leq \sqrt{2}$.
Therefore we need to split the $xy$-domain $D$ in two parts:
$$V=\iint_{D_1} \biggl( \int_{-\sqrt{2-(x^2+y^2)}}^1 \, dz \biggr) \, dxdy
+\iint_{D_2} \biggl( \int_{-\sqrt{2-(x^2+y^2)}}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after... | By using the symmetry in quartic,
\begin{array}
.x^4+3x^3+9x^2+3x+1&=x^2(x^2+3x+\frac{9}{2}+\frac{9}{2}+3x^{-1}+x^{-2})\\
&=x^2\large\left((x+\frac{3}{2})^2+(x^{-1}+\frac{3}{2})^2+\frac{9}{2}\large\right)\geq0
\end{array}
and at $x=0$ it is $1\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, $\frac{\beta^2+\gamma^2}{\alpha^2}$,$\frac{\alpha^2+\gamma^2}{\beta^2}$,$\frac{\beta^2+\alpha^2}{\gamma^2}$.
My solution g... | It’s not clear why you have written $$\frac{\gamma^2-2\alpha\beta}{\gamma^2}=\frac{1-2\alpha\beta\gamma}{\gamma^2\gamma}$$ This would mean that $\gamma^3=1$, which is not true.
It should be obvious to you that your answer is incorrect as you have arrived at a quadratic polynomial, not a cubic.
HINT…If you want a straig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to convert Quaternions into Polar form? I would like to know how to write quaternions as polar form. Because I heard that if $A$ and $B$ are elements of $C$, this can be done with the form $A \cdot e^{B \cdot j}$.
But how can I do that?
Can I do it like this
$
\begin{align*}
a + b \cdot i + c \cdot j + d \cdot k &=... |
How to convert Quaternions into Polar form?
In general if you have a quaternion $q$ with $ q = q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} $, you can write it as $ q = \left| q \right| \cdot \exp\left( \theta \cdot \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tricky integral in a complex plane How to integrate using residues:
$$\oint\limits_{|z|=2}\frac{1}{(z^6-1)(z-3)} dz $$
if the idea probably requires to change the sum of 6 residues to aditive inverse of residue at infinity.
I believe that the sum of 6 residues is:
$$2 \pi i\sum_{i=1}^6 Res_i \frac{1}{(z^6-1)(z-3)}=-2\p... | $$Res_{\infty}f(z)=Res_{z=0}-\frac{1}{z^2}f(\frac1z)=Res_{z=0}\frac{z^5}{(z^6-1)(1-3z)}=0$$
$$Res_{z=3}f(z)=Res_{z=3}\frac{1}{(z^6-1)(z-3)}=\left.\frac{1}{z^6-1}\right\rvert_{z=3}=\frac{1}{728}.$$
$$I=-2\pi i(Res_{\infty}f(z)+Res_{z=3}f(z))=-\frac{\pi i}{364}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Does a vector space need a quadratic form in order to define a wedge product? Geometric (Clifford) algebras require the vector space to be endowed with a quadratic form in order to define the geometric product.
Meanwhile, an exterior algebra has the wedge product as its defining product. It is not clear to me if a quad... | The wedge product in geometric algebra is typically defined in one of two ways, the first as the completely antisymmetrized product
$$\mathbf{a} \wedge \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right),$$
and the second, as the grade-2 selection of a product of two vectors
$$\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\mathbb{E}[X \mid X \geq n/2]$ when $X \sim \text{Bin}(n, 0.5)$ I'm wondering if there's a closed form for the conditional expectation $\mathbb{E}[X \mid X \geq n/2]$ when $X \sim \text{Bin}(n, 0.5)$ is drawn from the Binomial distribution with parameters $n$ and $0.5$. Thanks!
| @xzm is correct that
$$
\begin{align}
E[X|X \ge \frac n2] &= \frac{\sum_{k=\lfloor\frac{n+1}2\rfloor}^n k{n\choose k}}{\sum_{k=\lfloor\frac{n+1}2\rfloor}^n {n\choose k}}
\end{align}
$$
but it is possible to obtain this in closed form.
First note that $k{n\choose k}=n{n-1 \choose k-1}$. Then consider odd and even $n$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$ Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$. I think this is supposed be solved using the maclaurin series.
Let $f(x)=\tan x$ then $f'=1/\cos ^2x$ and $f''=\frac{-2\cos x \sin x}{\cos^4 x}=\frac{-\sin 2x}{\cos^4 x}$. Since $f(0)=0$ and $f'(0)=1$ we have that
$|x... | Let $f(x) = \tan x - x$ for all $x$ such that $|x| < \pi/2.$
Then $f(-x) = -f(x)$ for all $x,$ so in order to prove that
$|f(x)| \leqslant 8x^2$ for all $x$ such that $|x| \leqslant \pi/3,$
it is enough to prove it for all $x$ such that
$0 < x \leqslant \pi/3.$
We have $f'(x) = \sec^2 x - 1 > 0$ for all $x$ such that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so... | A visual demonstration of sorts:
Starting from $ \ x \ = \ \sqrt3 - 1 \ \ , \ $ we have
$$ x \ + \ 1 \ \ = \ \ \sqrt3 \ \ \Rightarrow \ \ x^2 \ + \ 2x \ + \ 1 \ \ = \ \ 3 \ \ . \ $$ So $ \ \sqrt3 - 1 \ $ is the larger of the two zeroes of $ \ x^2 + 2x - 2 \ \ . \ $ As for $ \ x \ = \ \sqrt5 \ - \ \sqrt2 \ \ , \ $ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Probability of subsets with different constraints Let $U = \{1, 2, \dots, 1000\}$. We pick a subset from U uniformly at random. That is, every subset has an equal chance of showing up.
*
*What is the probability that the empty set is chosen?
*What is the probability that the subset has at most ten elements?
*What i... | 1 and 2 - OK.
*$(1 - \frac{2^{1000-1}}{2^{1000}}) + (1 - \frac{2^{1000-1}}{2^{1000}}) = (1 - \frac12) + ( 1 -\frac12) = 1$, but the answer in obviously less than $1$. So no all is good in 4.
In 5 all is OK but there's a closed form.
We know that $(1+x)^n = \sum_{k=0}^{1000} \dbinom{1000}{k}x^k$. Put $x=\pm 1$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\frac {\cos^4 x}{\cos^2 y}＋\frac {\sin^4 x}{\sin^2 y}＝1$ How do you simplify the following?
$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$
What I've tried:
$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$
$$\sin^2(x)+\cos^2(x)=1$$
$$\implies\frac {\cos^4(x)}{\cos^2(y)}＋\frac... | \begin{eqnarray}\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}&=&1\\
\cos^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-\sin^2x)^2\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x+\sin^4x)\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x)\sin^2y+\sin^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simplifying partial derivative I have trouble understanding how this simplification was done. Can anyone explain please?
$\frac{\partial V}{\partial M}=\frac{5}{6}\frac{2^{1/3}3^{1/2}}{5^{5/6}}\frac{M^{-1/6}}{p_1^{1/3}p_2^{1/2}}=\frac{5^{1/6}}{2^{2/3}3^{1/2}p_1^{1/3}p_2^{1/2}}M^{-1/6}$
Thanks for your help.
Stan
| Besides some rearrangements, all the simplification comes in the numeric terms through factorisation and applying exponent rules:
$\begin{eqnarray}\frac{5}{6} \frac{2^{1/3} 3^{1/2}}{5^{5/6}} & = & \frac{5}{2 \times 3} \frac{2^{1/3} 3^{1/2}}{5^{5/6}} \\
& = & \frac{5^{6/6}}{5^{1/6}} \frac{2^{1/3}}{2^{3/3}} \frac{3^{1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Imaginary part of this integral I'm reading a physics paper and there is one step involving the imaginary part of an integral that I cannot reproduce (eq 23). The integral is
$$I=\int_0^1 dx\frac{1}{\Big[xa^2+(1-x)b^2-x(1-x)c^2-i\epsilon\Big]^{3/2}},$$
where all parameters are real and $a,\,b,\,\epsilon>0$. They claim ... | You can just plug this integral into some computer algebra system and get
$$I = \frac{2}{a^4 + b^4 + c^4 - 2(a^2 b^2 + a^2 c^2 + b^2 c^2) + 4i\epsilon c^2}\left[\frac{b^2-a^2-c^2}{\sqrt{a^2-i\epsilon}} + \frac{a^2-b^2-c^2}{\sqrt{b^2-i\epsilon}}\right].$$
Since $a$ and $b$ are nonzero and $\epsilon$ is infinitesimal, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
One-variable inequality Let $t \in [0;1]$. Prove that
$$\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}} \ge \frac{{\sqrt[3]{{14}}}}{7}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right).$$
Direction 1: I consider two functions $f,g$ as follows
$$f\left( t \right) = \... | The desired inequality is written as
$$\sqrt[3]{a} - \sqrt[3]{b} \ge \sqrt[3]{c} - \sqrt[3]{d}$$
or
$$\frac{a - b}{a^{2/3} + a^{1/3}b^{1/3} + b^{2/3}}
\ge \frac{c - d}{c^{2/3} + c^{1/3}d^{1/3} + d^{2/3}} \tag{1}$$
where
$a = t(t + 1)^2 + 4(t + 1)$,
$b = 12t$,
$c = \frac{14}{7^3}\cdot 2(t + 7)^2(3t + 1)$, and
$d = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Please explain how to solve limit. I know the answer but how to explain it? Problem:
$a@b = \frac{a+b}{ab+1}$. Solve limit:
$\lim_{n \to \infty}(2@3@...@n)$.
I've tried to solve this problem by just calculating:
$$2 @ 3 = 0.714$$
$$2 @ 3 @ 4 = 1.222$$
$$2 @ 3 @ 4 @ 5 = 0.875$$
$$2 @ 3 @ 4 @ 5 @ 6 = 1.1$$
I found the pa... | It holds that $\lim x_n = x$ if and only if every subsequence of $x_n$ contains, in turn, a subsequence that converges to $x$.
Check that the subsequence of even partial terms is strictly increasing and of odd partial terms is strictly decreasing.
Take any subsequence of the initial sequence. If it contains odd partial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Range of Trigonometric function having square root
Finding range of function
$\displaystyle f(x)=\cos(x)\sin(x)+\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}$
I have use Algebric inequality
$\displaystyle -(a^2+b^2)\leq 2ab\leq (a^2+b^2)$
$\displaystyle (\cos^2(x)+\sin^2(x))\leq 2\cos(x)\sin(x)\leq \cos^2(x)+\sin^2(x)\cdo... | Let $p=\sin^2\alpha,\; y=\cos 2x,\;$ then
$$\cos x \sin x+\cos x\sqrt{\sin^2x+\sin^2\alpha\mathstrut}
=\dfrac{\pm\sqrt{1-y^2}\pm\sqrt{1-y^2+2p(1+y)\mathstrut}}2 = f(y),$$
$$2f'(y) = \mp\dfrac y{\sqrt{1-y^2}}\pm\dfrac{p-y}{\sqrt{1-y^2+2p(1+y)}}.$$
The set of possible extremes of $f(y)$ corresponds with the expression
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n≥x^n+y^n$ for $n>1$? Let $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ where $n=2$.
Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ for $n \ge 1$?
I have just been che... | Assuming that the inequality holds for $n - 1$ and $n - 2$, we can use induction.
\begin{align}
a^n + b^n &= a^{n - 1}(a + b) - a^{n - 1}b + b^{n - 1}(a + b) - ab^{n - 1} \\
&=(a + b)(a^{n - 1} + b^{n - 1}) - ab(a^{n - 2} + b^{n - 2}) \\
&\geq (x + y)(x^{n - 1} + y^{n - 1}) - xy(a^{n - 2} + b^{n - 2}) \\
&= x^n + y^n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the following integral $ \int\frac{2x+1}{x^{n+2}(x+1)^{n+2}}\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)dx$ Hello I am trying to solve a pretty complicated integral. It is a from a set of problems, published in a monthly journal for high school students and they are exercises in preparation for a c... | Integrate by parts:
$$\int \frac{2x+1}{x^{n+2}(x+1)^{n+2}} \ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2} + \frac7{16}\right) \, dx \\
= -\frac1{n+1} \frac{\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)}{x^{n+1}(x+1)^{n+1}} - \frac{32}{n+1} \int \frac{2x^3+3x^2+3x+1}{x^{n+2}(x+1)^{n+2} \left(7x^4+14x^3+39x^2+32x+16\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{al... | It is known that $\cos(50^o)=\frac{\phi}2$. So, we need to find the floor of $\phi+\sqrt{3}$. $\sqrt{3}$ is less than $2$ since it is less than $\sqrt{4}$. $\phi$ is less than $2$ since $(1+\sqrt{5})/2<(1+\sqrt{9})/2=2$. We could do a similar proof that $\sqrt{3},\phi>1$. So, the answer must be between $\lfloor4\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Solving for x in logarithmic equation $\log_4(2x) = \frac{1}{2}x^2 - 1$ I am trying to solve for $x$ in the equation $\log_4(2x) = \frac{1}{2}x^2 - 1$. I have tried converting the logarithmic expression to exponential form, but I am not able to isolate $x$ in the resulting equation.
This is what I have tried as of now:... | $$\log_4{(2x)}=\frac{x^2}{2}-1$$
As stated already in a previous answer, try $x=2$:
$$x=2: \log_4{(2\cdot 2)}=\frac{2^2}{2}-1\Leftrightarrow 1=1$$
So $x=2$ is a solution. So far nothing new (from previous answer).
Next I would transform the equation into an exponential one:
$$\begin{align}\log_4{(2x)}&=\frac{1}{2}(1+\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Determining whether $\frac{1}{2^{2}} + \frac{2}{3{^2}} + ... +\frac{n}{(n+1)^{2}}$ is a Cauchy sequence. My attempt so far: If $n \leq m$, then
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} < \frac{n+1}{(n+2)(n+1)}+\frac{n+2}{(n+3)(n+2)}+...+\frac{m}{(m+1)m} = \frac{1}{(n+2)}+\frac{1}{(... | Put $m=2n$ and check that $|a_m-a_n|$ does not go to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Generatingfunctionology Chapter 1 Exercise 1.c Regarding the exercises of the Generatingfunctionology book available at (https://www2.math.upenn.edu/~wilf/DownldGF.html).
In particular Chapter 1, Exercise 1.(c), which asks to find the ordinary power series generating function of
$$a_n = n^2.$$
Following the method expl... | $\left(x^2 \frac{d^2}{dx^2} + x \frac{d}{dx}\right)\left(\frac{1}{1-x}\right)$
gives
$2x^2\left(\frac{1}{1-x}\right)^3+x\left(\frac{1}{1-x}\right)^2$
$(xD)^2$ differentiates in two stages.
$(xD)(xD)\left(\frac{1}{1-x}\right)$
$=(xD)x\left(\frac{1}{1-x}\right)^2$
$=x\left(2x\left(\frac{1}{1-x}\right)^3+\left(\frac{1}{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4637742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
| $$\large{\int} \normalsize{\dfrac{1}{x}\cdot\sqrt{\dfrac{1+x^2}{1-x^2}}} \space \mathrm{dx} \quad \xrightarrow{\large{x \space = \space \tan{u}}} \quad \large{\int} \normalsize{\dfrac{\cos{u}}{\sin{u}}\cdot\sqrt{\dfrac{1+\tan^2{u}}{1-\tan^2{u}}}} \cdot \sec^2{u} \space \mathrm{du} \\ \quad \\$$
$$\require{\cancel} = \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Given $a + b + c = 20$, find $\max(ab + ac + bc)$ Given that $a + b + c = 20$, what's the maximum possible value for $ab +ac + bc$?
$$\left(a, b, c \in \mathbb{N}\right)$$
I tried following this post, Evaluating max(ab+bc+ac), which lead to:
$ab + bc + ac = 200 - (b - 10)^2 - (c - 10)^2 - bc$, but I couldn't get anyw... | Here is a Solution :
$a+b+c=20$
$(a+b+c)^2=400$
$a^2+b^2+c^2+2(ab+bc+ca)=400$
$X=(ab+bc+ca)=200-(a^2+b^2+c^2)/2$
We want to maximize X which occurs when $Y=(a^2+b^2+c^2)$ is minimized.
When $a<b$ , we can increase $a$ by $\delta$ while decreasing $b$ by $\delta$ (keeping total sum Constant) , which decreases $Y$.
When ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proof that $3 \mid 10^{n+2} - 2*10^n + 7, \forall n \in \mathbb{Z}^+$. This is what I have so far.
Proof by Induction. Let $n \in \mathbb{Z}^+$ Let $P(n)$ be the statement that $10^{n+2} - 2*10^n + 7$ is divisible by 3.
($\textit{Base Case}$): Let $n = 1$.
$$
10^{1+2} - 2*10^1 + 7 = 1000 - 20 + 7 = 987
$$
$3 \mid 987$... | With the inductive step note that we have:
\begin{align*}
& 10^{k+2} -2\cdot10^{k} + 7 = 3m\quad\text{for some }m\in \mathbb{Z}\\
\implies & 10^{k+2} -2\cdot10^{k} = 3m-7 \qquad(*)
\end{align*}
Now for the case P(k+1):
\begin{align*}
10^{(k+1)+2} -2\cdot10^{k+1} + 7 & = 10\bigg( 10^{k+2} -2\cdot 10^{k}\bigg) + 7\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,... | In this answer, I used only Bernoulli's inequality to show that
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
\le\frac{2n+1}{n+1}\tag{1}
$$
The squeeze theorem and $(1)$, show that
$$
\exp\left[\lim\limits_{n\to\infty}\left(\frac1{n+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 12,
"answer_id": 6
} |
Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:
$$x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$
For cubic equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three ... | Contrary to common opinion, the resolution of the quartic with real coeficients is not so difficult.
I take for granted that depletion and reduction of the equation by a linear change of variable
$$\alpha x^4+\beta x^3+\gamma x^2+\delta x+\epsilon=0\to x^4+px^2+qx+r=0$$ are known.
Then we factor this depleted quadrinom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "165",
"answer_count": 12,
"answer_id": 1
} |
How do I count the subsets of a set whose number of elements is divisible by 3? 4? Let $S$ be a set of size $n$. There is an easy way to count the number of subsets with an even number of elements. Algebraically, it comes from the fact that
$\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n$
while
$\displaystyle... | I try to add details with simple calculation:
for example, I show $\binom{n}{1}+\binom{n}{3}+\binom{n}{6}+\cdots=\frac{1}{3}[2^n+2\cos(\frac{n\pi}{3})]$
we know $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n$, so if in this identity we put $1,\alpha,\alpha^2$ where $\alpha=\cos\frac{2\pi}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 3,
"answer_id": 2
} |
Help in getting the Quadratic Equation I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.
$$ x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$
So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation
$$ax+by+c = 0.$$
The solutio... | You factored 2a out of the square root and put it in the denominator without factoring it out of -(b/2a).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$
G... | Consider the function
$$f(a) = \sum_{k=0}^{\infty} \frac{1\cdot3\cdots(2k-1)}{2\cdot4\cdots(2k)}\cdot a^{2n+2k}$$
$$|a| \leq 1$$
What you want is $$\int_{0}^{1} f(a) da$$
The $k^{th}$ coefficient can be written as
$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}{\sin^{2k}x}dx$$
(again Wallis's like product as in the other quest... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $. How does one prove the following limit?
$$
\lim_{n \to \infty}
\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}}
= 3.
$$
| This is Ramanujan's famous nested radical.
More information can be found here: http://www.isibang.ac.in/~sury/ramanujanday.pdf
See Also: http://mathworld.wolfram.com/NestedRadical.html (number 26).
Apparently, this is how he came up with it (sorry, no reference for this claim).
Start with
$$3 = \sqrt{9} = \sqrt{1 + 8} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 3,
"answer_id": 0
} |
How to solve this non-linear differential equation? I'm trying to solve this non-linear differential equation using substitution $\dfrac{y}{x}$ to $t$. However, I can't solve this equation.
$$
\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y
$$
How to solve this equation and what's the general solution?
Thanks.
| Here is my solution. Is this right?
\begin{align}
\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y \\ \\
\frac{y}{x}=u, \\
y = \text{ux}, \\
\text{y$\prime $} = \text{u$\prime $x} + u \\ \\
x^2\text{u$\prime $} = u^3 \\
\int \frac{1}{u^3} \, du =\int \frac{1}{x^2} \, dx \\
-\frac{1}{2u^2} = -\frac{1}{x} + c \\
u^{2 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $ I am stuck while proving this identity I verified it using induction but like the other two (1) (2), I am seeking a more of a general way (algebraic will be much appreciated)
$$\frac {1}{(n-1)!... | Seems easy is $n$ is odd. Multiplying both sides by $n!$ we have
$$ {n \choose 1} + {n \choose 3} + ... {n \choose n} = \frac{1}{2} \left[ {n \choose 0} + {n \choose 1} + ... {n \choose n} \right] = 2^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/7835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | Another variation. We make use of the following identity (proved at the bottom of this note):
$$\sum_{k=1}^n \cot^2 \left( \frac {2k-1}{2n} \frac{\pi}{2} \right) = 2n^2 – n. \quad (1)$$
Now $1/\theta > \cot \theta > 1/\theta - \theta/3 > 0$ for $0< \theta< \pi/2 < \sqrt{3}$ and so
$$ 1/\theta^2 – 2/3 < \cot^2 \theta < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 13
} |
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ Using $\text{n}^{\text{th}}$ root of unity
$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$
Prove that
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
| Consider $z^n=1$, each root is
$$\xi_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} = e^{i\frac{2k\pi}{n}}, k=0,1,2,...,n-1 $$
So, we have
$$ z^n -1 = \prod_{k=0}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-\xi_0)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-1)\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "76",
"answer_count": 3,
"answer_id": 1
} |
Eigenvalues of companion matrix of $4x^3 - 3x^2 + 9x - 1$ I want to find all the roots of a polynomial and decided to compute the eigenvalues of its companion matrix.
How do I do that?
For example, if I have this polynomial: $4x^3 - 3x^2 + 9x - 1$, I compute the companion matrix:
$$\begin{bmatrix} 0&0&\frac{3}{4} \\ 1&... | Hey There, so if I am assuming correctly for your case, you want to find eigenvalues for this matrix, which is essentially solving for your roots of the characteristic polynomial of the matrix after doing the determinant operation on it. So to go off from Robert idea, we want to use the equation,
det(A$-\lambda$ I) = $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Expressions of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ I am trying to understand the interpretation of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ from my book, here is how it is given :
We have $ \bigl( \sin \frac{A}{2} + \cos \frac{A}{2} \bigr)^{2} = 1 + \sin A $ and $ \bigl... | We have that
$$\cos A=\cos \left(2\cdot\frac{A}{2}\right) = \cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}$$
and
$$1 = \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}}.$$
Therefore
\begin{align*}
1+\cos A & = \cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}} + \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} \\
& = 2\cos^2{\frac{A}{2}} \\
\Rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However... | Tangent half-angle substitution
$$\displaystyle \int \frac{1}{\sin x\cos x} dx=\displaystyle \int \frac{(1+t^2)}{t(1-t^2)} dt=\displaystyle \int \frac{1}{t} dt-\displaystyle \int \frac{1}{1-t} dt-\displaystyle \int \frac{1}{1+t} dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/9075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 4
} |
Calculating the integral $\int_0^\infty \frac{\cos x}{1+x^2}\, \mathrm{d}x$ without using complex analysis Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form?
$$\int_0^\infty\frac{\cos x}{1+x^2}\,\mathrm{d}x$$
| The method of brackets, which is an extension of Ramanujan's master theorem, can be used to evaluate this classic integral.
The term bracket refers to the assignment of the symbol $\langle a \rangle$ to the divergent integral $\int_{0}^{\infty} x^{a -1} \, \mathrm{d}x$.
You can read about the method in the following pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/9402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "111",
"answer_count": 11,
"answer_id": 0
} |
$x^y = y^x$ for integers $x$ and $y$ We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
| For every integer $n$, $x = y = n$ is a solution. So assume $x \neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x \mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/9505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "92",
"answer_count": 6,
"answer_id": 1
} |
How to calculate the expected number of distinct items when drawing pairs? Suppose I have a set $\mathcal{S}$ of $N$ distinct items. Now consider the set $\mathcal{P}$ of all possible pairs that I can draw from $S$. Naturally, $|\mathcal{P}| = \binom{N}{2}$. Now when I draw $k$ items (pairs) from $\mathcal{P}$ with a u... | For choosing without replacement, here is an exact answer. Assuming $n \geq 2$, so that there is at least one pair, and $1 \leq k \leq \binom{n}{2}$, so that you're choosing at least one pair but not more than the total number of pairs, the expected value is
$$n - \left(\frac{n^2 - 3n - 2k + 4}{n-1}\right) \frac{\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Another limit task, I've multiplied by the conjugate, now what? The task:
$\lim_{x\to\infty} \sqrt{x^2+1} -x $
I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this
$\lim_{x\to\infty} \frac{1}{\sqrt{x^2+1} +x} $
Is this correct so far? And what would be the next step?
| We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
Possible Duplicate:
On the sequence $x_{n+1} = \sqrt{c+x_n}$
Where does this sequence converge?
$\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...
| For a proof of convergence,
Define the sequence as
$\displaystyle x_{0} = 0$
$\displaystyle x_{n+1} =\sqrt{7 + x_n}$
Note that $\displaystyle x_n \geq 0 \ \ \forall n$.
Notice that $\displaystyle x^2 - x - 7 = (x-a)(x-b)$ where $\displaystyle a \lt 0$ and $\displaystyle b \gt 0$.
We claim the following:
i) $\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
} |
Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$ Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$.
A problem from Mark Krusemeyer. Should one be using the s... | It is possible to use the symmetry as a slight shortcut to Simon's answer: Given the graph and conditions as laid out there, the relevant point of intersection must be at $x=y=1$. Because of the symmetry, the angles formed by the tangent lines to the curves at that point and the line $y=x$ must be $\frac{\pi}{8}$, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/17177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can my proof be simplified and is it valid? $n(n+2) = (n+1)^2 - 1$ for all integers I've decided to learn the basics of proofs and here is my first attempt. Could I improve or simplify my proof in any way? Is my formal language correct? Thanks!
Let $n$ be any integer. $$n(n+2) = (n+1)^2 - 1$$ I will prove this identify... | Using $a^2 - b^2 = (a-b)(a+b)$ we have $$(n+1)^2 - 1 = (n+1)^2 - 1^2 = (n+1-1)(n+1+1) = n(n+2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/19554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$ $$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$
Complete the square?
$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$
Not sure what do do next, or if I should try something else?
Big help if you can show as "step-by-step" possible as you can.
Thanks in advan... | Hint. $(2x^{2}+4x-2) = 2(x^{2}+2x +1) -4 = 2(x+1)^{2}-4$. Now subsitute $x+1 = \sqrt{2} \sec{t}$. Then you have $dx = \sqrt{2} \sec{t} \cdot \tan{t} \ dt$. Now, $$2 \cdot \Bigl[(x+1)^{2} -2 \Bigr]= 2 \cdot \Bigl[ 2 \cdot (\sec^{2}{t}-1)\Bigr] = 4 \cdot (\sec^{2}{t}-1)$$.
Now what will you get if you power them by $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Show that 13 divides $2^{70}+3^{70}$
Show that $13$ divides $2^{70} + 3^{70}$.
My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?
Thanks!
| So by FlT, you know that $2^{12}\equiv 1\pmod{13}$ and $3^{12}\equiv 1\pmod{13}$. So
$$
2^{70}+3^{70}\equiv (2^{12})^5\cdot 2^{10}+(3^{12})^5\cdot 3^{10}\equiv 2^{10}+3^{10}\pmod{13}.
$$
However, again by FlT,
$$
2^{12}\equiv 2^{10}\cdot 2^{2}\equiv 1\pmod{13}\implies 2^{10}\equiv 2^{-2}\pmod{13}.
$$
That is to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/25701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? My attempt was:
$x^3 + 2x + 2 \equiv 0 \pmod{25}$
By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:
$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 125k^3 + 75k^2 + 25k ... | For this, I think the quickest way is simply to work out the 25 possibilities.
For 0,1,2,...24 you have $x^3+2x+2$ giving 2, 5, 14, 35, 74, 137, 230, 359, 530, 749, 1022, 1355, 1754, 2225, 2774, 3407, 4130, 4949, 5870, 6899, 8042, 9305, 10694, 12215, 13874 equivalent modulo 25 to 2, 5, 14, 10, 24, 12, 5, 9, 5, 24, 22... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/26339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Diophantine equation from Carmichael: $2x^4 - 2y^4 = z^2$ I would to solve the Diophantine equation $2x^4 - 2y^4 = z^2$ by descent. This is an exercise from Carmichael Diophantine Analysis but I cannot do it.
Since the LHS is even $2|z$, let $z = 2z'$ and divide both sides by $2$ to get $$(x-y)(x+y)(x^2+y^2)=2z'^2.$$ C... | If $p$ is a prime and it divides both $x$ and $y$, then it divides $z$: if $p$ is odd, then $p^4|z^2$; if $p=2$, then $32|z^2$. Either way, you can factor out the common prime from all of $x$, $y$, and $z$.
Thus, we may assume that $\gcd(x,y)=1$. Since you also know that $x\equiv y\pmod{2}$, you can reduce to the case ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/30335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the reduction formula The question is to "prove the reduction formula"
$$ \int{ \frac{ x^2 }{ \left(a^2 + x^2\right)^n } dx } = \frac{ 1 }{ 2n-2 } \left( -\frac{x}{ \left( a^2+x^2 \right)^{n-1} } + \int{ \frac{dx}{ \left( a^2 + x^2 \right)^{n-1} } } \right) $$
What I got is
Set
$ u = x $
$ du = dx $
$\di... | You went wrong when you integrated $dv$.
You have $dv = x(a^2+x^2)^{-n}\,dx$. When you integrate, you add one to the exponent. But adding one to $-n$ gives $-n+1 = -(n-1)$. So
$$v = \frac{1}{2(-n+1)}(a^2+x^2)^{-n+1} = \frac{1}{2(1-n)(a^2+x^2)^{n-1}}.$$
The minus sign from integration by parts can be cancelled out by sw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/30595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Coloring of an $1\times n$ board using 4 colors? How can I find the number of ways to color an $1\times n$ board using the colors red, blue, green and orange if:
# of red squares is even
# of green squares is even
We did the tilings of a $1\times n$ board using squares and dominos in class, but I'm not sure how to do t... | I just want to point out that exponential generating functions serve as a good approach here. Let $g(x)$ be the exponential generating function of the sequence $h_0, h_1, h_2,\ldots$, where $h_n$ is the number of acceptable colorings of a $1\times n$ board. So $h_n$ is the coefficient of $x^n$ in $g(x)$.
Then
$$
g(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/35063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Trig substitution for a triple integral This problem involves calculating the triple integral of the following fraction, first with respect to $p$:
$$
\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^{2} \frac{p^2\sin(\phi)}{\sqrt{p^2 + 3}} dp d\phi d\theta
$$
This involves trig-substitution (I believe), and I a... | Update: I don't use a trig-substitution because I don't see which one could work. To integrate $p^{2}/\sqrt{p^{2}+3}$ I applied a technique derived from this more general one.
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$$\int \frac{P(x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/37756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Complete induction of $10^n \equiv (-1)^n \pmod{11}$ To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.
It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$
For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$
$n\Rightarrow n+1$
$$\begin{align*}
(-1) ^{n+1} - (10) ^{n+... | You are not setting up your induction very well. You should not start with the equality you want to establish, namely that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$. Instead, you should start with the Induction Hypothesis, which is that $(-1)^n - 10^n$ is a multiple of $11$.
So: the Inductive Step is to show that if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/39882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 7
} |
Explaining an algebra step in $ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4)$ I have encountered this step in my textbook and I do not understand it, could someone please list the intermediate steps?
$$ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4). $$
Thanks,
| One way to see it is to factor out the $(n+1)^2/4$. Then
$$
\begin{align*}
\frac{n^2(n+1)^2}{4} + (n+1)^3 &=\frac{(n+1)^2}{4}\bigl(n^2+4(n+1)\bigr) \\
&= \frac{(n+1)^2}{4}(n^2+4n+4)
\end{align*}
$$
Since you factor out a $1/4$, you have to keep a $4$ in the numerator in the second term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/42566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain ... | A High School Proof:
$$S_n = 1^2 + 2^2 + 3^2 +\dots+ n^2$$
We know,
$$r^3 - 3r^2 + 3r - 1 = (r-1)^3 $$
$$r^3 - (r-1)^3 = 3r^2 - 3r + 1$$
When $r=1, 2, 3,\dots, n$
$$1^3 - 0^3 = 3*1^2 - 3*1 + 1\qquad(1)$$
$$2^3 - 1^3 = 3*2^2 - 3*2 + 1\qquad(2)$$
$$3^3 - 2^3 = 3*3^2 - 3*3 + 1\qquad(3)$$
$$\dots\dots\dots\dots\dots\dots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "145",
"answer_count": 32,
"answer_id": 12
} |
How do I resolve a recurrence relation when the characteristic equation has fewer roots than terms? I know how to solve "simple" recurrence relations. For instance, say you have:
$$c_0 = 20$$
$$c_1 = 30$$
$$c_n = 3 c_{n-1} - 2 c_{n-2}$$
We can write the characteristic equation as:
$$3x^{n-1} - 2x^{n-2} = x^n$$
Solving ... | Use generating function directly: define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence with no subtractions in indices:
$$
a_{n + 3} = 9 a_{n + 2} - 15 a_{n + 1} - 25 a_n
$$
Multiply by $z^n$ and sum over $n \ge 0$, recognize:
$$
\sum_{n \ge 0} a_{n + k} z^n
= \frac{A(z) - a_0 - a_1 z - \ldots - a_{k - 1} z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/54862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:
Find the largest number $n$ such that $(2004!)!$ is divisible by
$((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?
For the second one the suggested solution is like this : $ 2^8 + 2^{11... | A different way for the second question:
First consider $2^8+2^{11}+2^n$ modulo $3$. Because $2^2\equiv 1\pmod 3$, we have
$$ 2^8 + 2^{11} + 2^n \equiv 2^0+2^1+2^{n\bmod 2} \equiv 2^{n\bmod 2} \bmod 3$$
and since $2^1$ is not a square modulo $3$, $n$ must be even. Similarly, $2^3\equiv 1\bmod 7$, so
$$ 2^8+2^{11} + 2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/57177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Limit of $(x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$ when $x\to \infty$ Okay, this is the last limit I have to solve but it's not that easy ;)
$$\lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$$
| I know the following method is not valid for the purpose of this exercise,
but I would like to post it. I computed in the Computer Algebra System included in SWP the following power series expansions:
$$\begin{eqnarray*}
(t^{-1}+3)^{1+t} &=&t^{-1}+\left( 3-\ln t\right) +\left( -\frac{3}{2}+\frac{1
}{2}\left( 3-\ln t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Question about computing a Fourier transform of an integral transform related to fractional Brownian motion I am trying to show an integral transform has a fixed point.
Let $H \in (0,1)$ and consider the following integral transform whose kernel is the density of fractional Brownian motion: $$T_H f(x) = 2 \int_{0}^{\... | It is rather easy to verify numerically, that $\phi(x) = \exp(-c \vert x \vert^{\frac{1}{2(1-H)}} )$ does not satisfy $T_H \phi = \phi$.
To this end it is enough to observe that $(T_H \phi_c)(x) = c^{-2 (1-h)^2} (T_H \phi_1)(x c^{2 h(1-h)})$, which is obtained via changing variables $y \to c^{-2(1-h)} y$. This scaling... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding limits of rational functions as $x\to\infty$
Possible Duplicate:
Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials
$$\displaystyle \lim_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}.$$
I do not know where to start on this, I tried multiplying it out and that didn't help really. It seems ... | Hint: Divide the numerator and the denominator by the highest power of $x$ that occurs (in this case, $x^4$); distribute it so that it shows you exactly what is going on. For example,
$$\begin{align*}
\frac{1}{x^4}(x-1)^2(x^2+x) &=\frac{1}{x^2}\times\frac{1}{x^2}\times(x-1)^2\times(x^2+x) &&\text{(factor }\frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
How to solve this equation: $\frac{\sqrt[5]{x^3 \sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$ Please, help me to solve this equation:
$$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$$
I tried to shorten fraction, but I get very weird numbers like
$$\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}... | If $x\gt 0$, then:
*
*$\sqrt[3]{x^{-2}} = x^{-2/3}$.
*$x\sqrt[3]{x^{-2}} = xx^{-2/3} = x^{1/3}$.
*$\sqrt{x\sqrt[3]{x^{-2}}} = x^{1/6}$.
*$x^3\sqrt{x\sqrt[3]{x^{-2}}} = x^{19/6}$.
*$\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}} = x^{19/30}$.
*$x\sqrt[3]{x} = xx^{1/3} = x^{4/3}$.
*$\sqrt[4]{x\sqrt[3]{x}} = x^{1/3}$.
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/63686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $ How could we prove that this inequality holds
$$ \left(1+\frac{1}{n+1}\right)^{n+1} \gt \left(1+\frac{1}{n} \right)^{n} $$
where $n \in \mathbb{N}$, I think we could use the AM-GM inequality for this but not getting how?
| No AM-GM inequality - just simple computation:
$$\begin{align}
\frac{(1+\frac{x}{n+1})^{n+1}}{(1+\frac{x}{n})^n}
&= (1+\frac{x}{n})\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1} \\\\
&= (1+\frac{x}{n})\left(\frac{n(n+1)+nx}{(n+1)(n+x)}\right)^{n+1} \\\\
&= (1+\frac{x}{n})\left(\frac{(n+1)(n+x)-x}{(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 6,
"answer_id": 0
} |
Demonstrate inequality $$15\le(3+\sin^2x)(4+\cos^2x)\le16 \mbox{ for any }x \in \mathbb{R}$$
I've wrote everything using $\sin$:
\begin{align*}15\le(3+\sin^2x)(4+1-\sin^2x)\le16&\Rightarrow
15\le(3+\sin^2x)(5-\sin^2x)\le16\\
&\Rightarrow 15\le15+2\sin^2x-\sin^4x\le16 \mid-15\\
&\Rightarrow 0\le2\sin^2x-\sin^4x\le1.
\e... | Welcome to math.stackexchange Daniel!
Just expand and simplify the middle term as such: $$ (3 + \sin^2 x)(4+\cos^2 x) = 12 + 3\cos^2 x + 4\sin^2 x + \sin^2 x \cos^2 x $$
Now, since $\sin^2 x + \cos^2 x = 1$ we can write that term as $$ 15 + \sin^2x + \sin^2 x\cos^2 x$$
Thus, this reduces our problem to showing
$$ 0 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/66058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find M, since $\log_5 M = 2\log_5 A - \log_5 B+2$ Find M, since $\log_5 M = 2\log_5 A - \log_5 B+2$
I tried this:
The answer is in function of A and B.
$\frac{\log_M M}{\log_M 5} = 2\frac{\log_M A}{\log_M 5} - \frac{\log_M B+2}{\log_M 5}$
$1=2\log_M A - \log_M B+2$
$\log_M A^2 = \log_M B+2$
$A^2=B+2$
$\log_5 M = 2\log_... | $$\log_5 M = 2 \log_5 A − \log_5 B + 2 = \log_5 A^2 + \log_5 \frac{1}{B} + \log_5 25 = \log_5 \left(\frac{25 A^2}{B}\right) \quad \Longrightarrow \quad M = \frac{25A^2}{B}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse Laplace Transform -s domain How can I find the inverse Laplace transforms of the following function?
$$ G\left(s\right)=\frac{2(s+1)}{s(s^2+s+2)} $$
I solved so far. After that, how do I do?
$$ \frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}=G\left( s \right)$$
| Use:
$$
\mathcal{LT}_s\left( \sin(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \sin(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{\alpha}{(s+b)^2 + \alpha^2}
$$
$$
\mathcal{LT}_s\left( \cos(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \cos(\alpha x) \mathrm{e}^{-b x} \mathrm{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/68991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Parametric form of an ellipse given by $ax^2 + by^2 + cxy = d$ If $c = 0$, the parametric form is obviously $x = \sqrt{\frac{d}{a}} \cos(t), y = \sqrt{\frac{d}{b}} \sin(t)$.
When $c \neq 0$ the sine and cosine should be phase shifted from each other. How do I find the angular shift and from there how do I adjust the f... | If the parametric ellipse coordinates are $\left(x(t),y(t)\right) = (X \cos\varphi \cos(t)-Y \sin\varphi \sin(t), Y \cos\varphi \sin(t)+X \sin\varphi \cos(t)) $
Then the parameters $ X,Y,\varphi $ are
$$ X =\pm \sqrt{\frac{2 d}{a+b+\sqrt{(a-b)^2+c^2}}} $$
$$ Y =\pm \sqrt{\frac{2 d}{a+b-\sqrt{(a-b)^2+c^2}}} $$
$$ \varph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/70069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Simplifying fractions How to transform $$\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m$$ into $$\left({b\over a}\right)^n -1\over \left({b\over a}\right)^m -1$$?
Thanks.
Added: As per comments, there is an... | With the added information, which is key (otherwise, the two expressions are just not equal in general, as noted by Ross):
If $b=1-a$ and $b\gt a$, then $1-a\gt a$ so $1\gt 2a$. Then
$$1 -4ab = 1-4a(1-a) = 1-4a+4a^2 = (1-2a)^2,$$
and since $1-2a\gt 0$, we have
$$\sqrt{1-4ab} = \sqrt{(1-2a)^2} = |1-2a| = 1-2a.$$
So
$$\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/72289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.