Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Choosing convenient limits of integration on Basel problem I have recently discovered $$\sum_{k=1}^{\infty}\frac{\cos\left(k\alpha\right)}{k^{2}}-\sum_{k=1}^{\infty}\frac{\cos\left(k\beta\right)}{k^{2}}=\int_{\alpha}^{\beta}\tan^{-1}\left(\cot\left(\theta/2\right)\right)\,\mathrm{d}\theta$$ which seems to be a nice way to attack Basel problem. The integral nicely reduces to $$\int_{0}^{\beta}\frac{\pi}{2}-\frac{\theta}{2}\,\mathrm{d}\theta=\frac{2\pi\beta-\beta^{2}-2\pi\alpha+\alpha^{2}}{4}=\frac{1}{4}\left(\alpha-\beta\right)\left(\alpha+\beta-2\pi\right)$$ when $\alpha,\,\beta\in\left[0,\,2\pi\right]$. The question: how do I choose $\alpha$ and $\beta$ such that $\cos\left(k\alpha\right)-\cos\left(k\beta\right)=1$ for all $k\in\mathbb{N}$? Does this $k$ even exist?
| Sadly, there's no such $\alpha$ and $\beta$. We'll try and solve the more general equation - $\cos(k\alpha)-\cos(k\beta)$ is a non-zero constant, say $C$.
Assume there are such $\alpha, \beta$. Let $x=\cos \alpha, y = \cos \beta$.
Note that $\cos(2t) = 2\cos(t)^2 - 1, \cos(3t) = 4\cos(t)^{3} - 3\cos(t)$. Then the following 2 equalities hold:
$$x-y = C$$
$$2x^2-2y^2 = C$$
Factoring $2x^2-2y^2$ as $2(x-y)(x+y)$ and plugging the first equation, we find that $x+y = 0.5$ This leads to $x = \frac{C+0.5}{2}, y=\frac{0.5 - C}{2}$.
Now we choose $k=3$:
$$4x^3 - 3x - (4y^3 - 3y) = C$$
The LHS factors as $(x-y)(4(x^2+xy+y^2)-3)=(x-y)(4(x+y)^2-4xy-3)$, which is
$$C(4(0.5)^2-4(\frac{1}{16} - \frac{C^2}{4})-3)=C$$
Cancelling the $C$s, it becomes
$$C^2 = 3.25$$
So $C\neq 1$!
Now we note that $\cos(4t) = 8(\cos(t)^4-\cos(t)^2)+1$, which gives the following equation:
$$8(x^4-x^2-(y^4-y^2)) = 1$$
The LHS factors as $8(x^2-y^2)(x^2+y^2-1)=1$. We can plug $2x^2-2y^2 = C$ and this becomes
$$x^2+y^2 = \frac{1}{4C}+1$$
But $x^2+y^2 = \frac{C^2}{2}+\frac{1}{8} = 1.75$, so it follows that $C=\frac{1}{3}$, a contradiction to $C^2 = 3.25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/277147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Exponent questions algebra How would I solve the following two exponent questions?
(1) The first question is
$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}$$
I got $\quad \displaystyle \frac{-xy^{-2}}{x^2+y},\;\;$but this does not seem to be correct.
(2) My second question is
$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}$$
I got $\quad\displaystyle \frac{A^6B^4}{1/9},\quad$ but my book's answer is $\quad\displaystyle \frac{1}{9A^6B^4}$
| You got
$$\frac{-xy^{-2}}{x^2+y}$$
Here's what I got:
$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}=\left(\frac{xy^2}{x^{-2}+y^{-1}}\right)=\left(\frac{xy^2}{\frac{1}{x^2}+\frac{1}{y}} \right)=\left(\frac{xy^2}{\frac{y+x^2}{yx^2}} \right)=xy^2\cdot\left(\frac{yx^2}{y+x^2} \right)=\frac{x^3y^3}{y+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/279637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding all solutions of the equation $(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$. Find all solutions of the equation
$(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$.
not getting any clue.can somebody help me please.thanks.
| If $x,y,z$ are real, we can set $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$
Now, $$\cos A\cos B+\cos A\sin B+\sin A=\cos A\sqrt2\cos(B-\frac\pi4)+\sin A\le \sqrt2\cos A+\sin A$$
Let $\sqrt2=R\cos C,1=R\sin C$ where $R>0\implies R^2=2+1=3,R=\sqrt3, \cos C=\sqrt\frac23$
$\implies \sqrt2\cos A+\sin A=\sqrt3\cos(A-\arccos\sqrt\frac23)\le \sqrt 3$
So, $x+y+z\le \sqrt3$ which occurs when $A-\arccos\sqrt\frac23=2n\pi\implies \cos A=\sqrt\frac23$ and $B=\frac \pi4$
So there is no real solution for $x+y+z=3$ if $x^2+y^2+z^2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/280434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the angle of rotation of an ellipse from its general equation and the other way around The general equation for an ellipse is $Ax^2+Bxy+Cy^2+D=0$. How do I find the angle of rotation, the dimensions, and the coordinates of the center of the ellipse from the general equation and vice versa? Please avoid using matrices or parametric equations. I'd like all-in-one equations for each parameter.
| For the vice versa part of the question: the parameterization for the ellipse centered at the origin and with semi major axis equal to $a$ (parallel to the frame of reference) and semi minor axis equal to $b$ is
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}\qquad\varphi\in[0,2\pi)$$
If I rotate the ellipse of the angle $\theta$ with the rotation matrix $R=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ I will get:
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$
If I translate the center from $(0,0)$ to $(x_c,y_c)$ I will get:
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}+\begin{pmatrix}x_c\\y_c\end{pmatrix}$$
Now I would like to isolate $\cos\varphi$ and $\sin\varphi$, then square them and finally get the quadratic form.
$$\begin{pmatrix}x-x_c\\y-y_c\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$
Since $R$ is a rotation matrix, it holds $R^{-1}=R^T$ and so I can multiply both sides by $R^T$:
$$\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}x-x_c\\y-y_c\end{pmatrix}=\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$
Now I do the matrix vector product:
$$\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)=a\cos\varphi$$
$$-\sin(\theta)(x-x_c)+\cos(\theta)(y-y_c)=b\sin\varphi$$
and I isolate $\cos\varphi$ and $\sin\varphi$:
$$\frac{\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)}{a}=\cos\varphi$$
$$\frac{-\sin(\theta)(x-x_c)+\cos(\theta)(y-y_c)}{b}=\sin\varphi$$
Now, since $\cos^2\varphi+\sin^2\varphi=1$, I get
$$\left(\frac{\cos(\theta)(x-x_c)+\sin(\theta)(y-y_c)}{a}\right)^2+
\left(\frac{\cos(\theta)(y-y_c)-\sin(\theta)(x-x_c)}{b}\right)^2=1$$
I will now use sympy to extract the coefficients $A$, $B$, $C$, $D$, $E$, $F$ of the quadratic form $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$
from sympy import *
theta,x_c,y_c,a,b,x,y=symbols('theta x_c y_c a b x y')
exp_1=((cos(theta)*(x-x_c)+sin(theta)*(y-y_c))/a)**2+((cos(theta)*(y-y_c)-sin(theta)*(x-x_c))/b)**2-1
exp_2=expand(exp_1)
P=poly(exp_2,x,y)
A,B,C,D,E,F=symbols('A B C D E F')
A_=P.coeff_monomial(x**2)
B_=P.coeff_monomial(y**2)
C_=P.coeff_monomial(x*y)
D_=P.coeff_monomial(x)
E_=P.coeff_monomial(y)
F_=P.coeff_monomial(1)
print(latex(relational.Eq(A, A_)))
print(latex(relational.Eq(B, B_)))
print(latex(relational.Eq(C, C_)))
print(latex(relational.Eq(D, D_)))
print(latex(relational.Eq(E, E_)))
print(latex(relational.Eq(F, F_)))
$$A = \frac{1}{b^{2}} \sin^{2}{\left (\theta \right )} + \frac{1}{a^{2}} \cos^{2}{\left (\theta \right )}$$
$$B = \frac{1}{b^{2}} \cos^{2}{\left (\theta \right )} + \frac{1}{a^{2}} \sin^{2}{\left (\theta \right )}$$
$$C = - \frac{2}{b^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} + \frac{2}{a^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )}$$
$$D = - \frac{2 x_{c}}{b^{2}} \sin^{2}{\left (\theta \right )} + \frac{2 y_{c}}{b^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} - \frac{2 x_{c}}{a^{2}} \cos^{2}{\left (\theta \right )} - \frac{2 y_{c}}{a^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )}$$
$$E = \frac{2 x_{c}}{b^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} - \frac{2 y_{c}}{b^{2}} \cos^{2}{\left (\theta \right )} - \frac{2 x_{c}}{a^{2}} \sin{\left (\theta \right )} \cos{\left (\theta \right )} - \frac{2 y_{c}}{a^{2}} \sin^{2}{\left (\theta \right )}$$
$$F = -1 + \frac{x_{c}^{2}}{b^{2}} \sin^{2}{\left (\theta \right )} - \frac{2 x_{c}}{b^{2}} y_{c} \sin{\left (\theta \right )} \cos{\left (\theta \right )} + \frac{y_{c}^{2}}{b^{2}} \cos^{2}{\left (\theta \right )} + \frac{x_{c}^{2}}{a^{2}} \cos^{2}{\left (\theta \right )} + \frac{2 x_{c}}{a^{2}} y_{c} \sin{\left (\theta \right )} \cos{\left (\theta \right )} + \frac{y_{c}^{2}}{a^{2}} \sin^{2}{\left (\theta \right )}$$
Inspired by Compute center, axes and rotation from equation of ellipse
See also https://en.wikipedia.org/wiki/Ellipse#General_ellipse
| {
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"url": "https://math.stackexchange.com/questions/280937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Why a closed bounded convex set in $\mathbb{R}^{n}$ always has an extreme point? Let $X\subseteq \mathbb{R}^{n}$ is closed, bounded convex set. How to prove that $X$ contains such point $x$ that we can't represent as $x=\frac{1}{2}x_{1}+\frac{1}{2}x_{2}$ where $x_1\in X$ and $x_2\in X$ and $x_1 \neq x_2$?
For example closed square in $\mathbb{R}^{2}$ has $4$ such points.
| Let $R$ be the diameter of $X$. Choose any $x, y \in X$ such that $\lVert x - y\rVert = R$; prove by contradiction that $x$ is the desired point.
Basically, suppose that $x = \frac{1}{2}x_1 + \frac{1}{2}x_2$ for some distinct $x_1, x_2 \in X$. Note that either $\lVert x_1 - y\rVert$ or $\lVert x_2 - y\rVert$ must exceed $R$.
Detailed Proof:
Suppose that $x = \frac{1}{2}x_1 + \frac{1}{2}x_2$ for some distinct $x_1, x_2 \in X$. Let $x_1 = x + v$ and $x_2 = x - v$. Since $x_1 \neq x_2$, $v \neq 0$.
$$
\begin{align}
\lVert x_2 - y \rVert &\leq R\\
\implies \lVert x_2 - y \rVert^2 &= ((x - y) - v) \cdot ((x - y) - v)\\
&= R^2 + \lVert v \rVert^2 - 2v \cdot (x - y)\\
&\leq R^2
\end{align}
$$
Hence, $2v \cdot (x - y) \geq \lVert v \rVert^2 \geq 0$ and thus $R^2 + \lVert v \rVert^2 + 2v \cdot (x - y) \geq R^2$ ($\ast$).
Similarly,
$$
\begin{align}
\lVert x_1 - y \rVert &\leq R\\
\implies \lVert x_1 - y \rVert^2 &= ((x - y) + v) \cdot ((x - y) + v)\\
&= R^2 + \lVert v \rVert^2 + 2v \cdot (x - y)\\
&\leq R^2
\end{align}
$$
Together with ($\ast$), this gives us $R^2 + \lVert v \rVert^2 + 2v \cdot (x - y) = R^2$, which implies that $\lVert x_2 - y \rVert = \lVert x_2 - y \rVert = R$.
Here comes the part where the Euclidean metric is crucial. Since $x_1$ and $x_2$ lie on the sphere of radius $R$ and centered at $y$, $x$ is the midpoint of the chord joining $x_1$ and $x_2$, so $x - y$ is perpendicular to $v$. The Pythagorean Theorem gives us $R^2 = \lVert x - y\rVert^2 = R^2 - \lVert v\rVert^2 < R^2$. Contradiction!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281982",
"timestamp": "2023-03-29T00:00:00",
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Calculate $\dim W+V$ and $W\cap V$ This is a task from an old exam.
Let define:
$V_{t} = \text{lin}((1,2,2,1),(1,1,-1,t))$
$W = \cases{x_1-x_2=0\\x_3-x_4 =0}$
Calculate $\dim W+V_{t}$ and $\dim W\cap V_{t}$
Please verify answer below
| basis of $W$ = $\text{lin} ((1,1,0,0),(0,0,1,1))$
$ W+V_{t}=\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
1 & 1 & -1 & t\\
1 & 1 & 0 & 0\\
0 & 0 & 1 & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
0 & -1 & -3 & t-1\\
0 & -1 & -2 & -1\\
0 & 0 & 1 & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
0 & 0 & -1 & t\\
0 & -1 & -2 & -1\\
0 & 0 & 1 & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
0 & -1 & -2 & -1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & t+1
\end{array}\right]
$
$\dim W+V_{t}=3$ for $t=-1$ and $4$ in other case.
$V_{t}:$
$\left[\begin{array}{ccc}
1 & 1 & x_{1}\\
2 & 1 & x_{2}\\
2 & -1 & x_{3}\\
1 & t & x_{4}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 1 & x_{1}\\
0 & 0 & x_{2}-2x_{1}\\
0 & -2 & x_{3}-2x_{1}\\
0 & t & x_{4}-2x_{1}
\end{array}\right]\implies V_{1}:\begin{cases}
x_{2}-2x_{1}=0\\
x_{4}-2x_{1}=0
\end{cases}
$
$W\cap V_{t}:\begin{cases}
x_{1}-x_{2}=0\\
x_{3}-x_{4}=0\\
x_{2}-2x_{1}=0\\
x_{4}-2x_{1}=0
\end{cases}
$
$\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
& & 1 & -1\\
-2 & 1\\
-2 & & & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
& & 1 & -1\\
0 & -1\\
0 & -2 & & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
0 & -1\\
& & 1 & -1\\
0 & -2 & & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
0 & -1\\
& & 1 & -1\\
0 & 0 & & 1
\end{array}\right]
$
$\dim W\cap V_{t}=0
$
| {
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"url": "https://math.stackexchange.com/questions/283758",
"timestamp": "2023-03-29T00:00:00",
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How can I show the convergence and find limit.
Let $a_1=0$ and $$a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}~,~n \geq 1$$ . Show that $\{\ a_n \}$ converges and find its limit.
Trial: Here $$\begin{align} a_1 &=0\\ \text{and}~ a_2 &=\frac{3}{2} \\ \text{and}~ a_3 &=\frac{21}{20} \end{align}$$ . Here I find no relationship between $a_n ~\text{and}~ a_{n+1}$. Please help.
| Okay what do we need for conergence? If we can show that the sequence is bounded and monotone we know that it converges.
*
*bounded: Obviously we have $0\leq a_n$. I will show that $a_n<10$ (which is rather arbitary) by induction. We have $a_0<10$. Now we make the step $n\rightarrow n+1$.
\begin{align} a_{n+1} &= \frac{a_n^2+3}{2(a_n+1)} <\frac{a_n^2+2a_n+3}{2(a_n+1)}
\\ &=\frac{(a_n+1)^2+2}{2(a_n+1)} = \frac{(a_n+1)^2}{2(a_n+1)}+\frac{2}{2(a_n+1)}
\\ &= \frac{a_n+1}{2} + \frac{2}{2(a_n+1)} \leq^* \frac{10+1}{2}+\frac{2}{2(0+1)}
\\ &= 5.5 +1 <10\end{align}
Where we use $0\leq a_n<10$ at $^ *$.
*monotone: here we show, that $\forall n > 0$ we have $\displaystyle \frac{a_{n+1}}{a_n}<1$. I think this is something you should proof.
For the last step: We proofed that our sequence converges.
So there exists an $a=\lim_{n\rightarrow \infty} a_n$ Knowing this, you can solve
$a_{\infty} = \frac{a_{\infty}^2+3}{2(a_{\infty}+1)}$ for $a_{\infty}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/284209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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polynomial division problem ( find the remainder) How to find out what remainder will $\,(x-1)^{2013}\,$ have upon division by $\,x^2-2x+2\,?\;$
I've never solved anything like this before, so I have no ideas at all..
Thanks in advance!
| Hint: Upon division by $x^2 - 2x + \color{blue}{\bf{1}}$, there is no remainder: that is $$\frac{(x-1)^{2013}}{x^2 - 2x + 1} = \frac{(x - 1)^{2013}}{(x-1)^2} = (x - 1)^{2011}$$
Note: $\quad(x-1)^2 + 1 \equiv 0\pmod{x^2-2x+2}\;\implies\; (x-1)^2 \equiv -1 \pmod {x^2-2x+2}$
So what does this mean for
$
\begin{align}
(x - 1)^{2013}\pmod{x^2 - 2x +2} &= (x-1)(x-1)^{2012}\pmod{x^2 - 2x +2} \\ &= (x-1)[(x-1)^2]^{2012/2}\pmod{x^2 - 2x +2}\\ &= (x-1)[(x-1)^2]^{1006}\pmod{x^2 - 2x + 1}\quad?\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A limit with infinite product $\lim\limits_{n\to\infty}\frac1{n^4}\prod_{i=1}^{2n}\left(n^2+i^2\right)^{\frac1n}$ How to calculate :
$$\lim_{n\to\infty}\frac{1}{n^4}\prod_{i=1}^{2n}\left(n^2+i^2\right)^{\frac{1}{n}}$$
| Pull the $n$ out of the product to get
$$L = \lim_{n \rightarrow \infty} \prod_{i=1}^{2 n} \left ( 1+\frac{i^2}{n^2} \right )^\frac{1}{n} $$
Rewrite as
$$\log{L} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{2 n} \left ( 1+\frac{i^2}{n^2} \right )$$
This is a Riemann sum whose limit is
$$\begin{align}\log{L} &= \int_0^2 dx \: \log{(1+x^2)}\\ &= [x \log{(1+x^2)}]_0^2 - 2 \int_0^2 dx \: \frac{x^2}{1+x^2} \\ &= 2 \log{5} + 2 \arctan{2} - 4 \end{align}$$
Then the limit $L$ is
$$L = 25 e^{-2 (2-\arctan{2})} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Stuck on this linear equation $y'=(3x^2-e^x)/2y-5$ Given this separable equation: $y'=(3x^2-e^x)/2y-5$.
I did integration, I found my constant $c=-3$.
And now I am stuck on this part: $y^2-5y=x^3-e^x-3$.
No matter what I do, I cannot come to this correct answer, according to the textbook, $y=-5/2-sqrt(x^3-e^x+13/4)$.
| You did it right. Just see that : $$y^2-5y+\frac{25}{4}=x^3-e^x-3+\frac{25}{4}$$ so $$\left(y+\frac{5}{2}\right)^2=x^3-e^x-3+\frac{25}{4}=x^3-e^x+\frac{13}{4}$$ so one solution may arise of a form: $$y=-\frac{5}{2}+\sqrt{x^3-e^x+\frac{13}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Computing with Cauchy Residue theorem how do I calculate $$\operatorname{Res}\left(\frac{1}{z^2 \cdot \sin(z))}, 0\right)$$ What is the order of the pole? $3$?
| Yes, the pole is of order 3. Thus
$$\begin{align}\mathrm{Res}_{z=0} \frac{1}{z^2 \sin{z}} &= \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left [ z^3 \frac{1}{z^2 \sin{z}} \right ] \\ &= \frac{1}{2}\lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left ( \frac{z}{\sin{z}} \right )\\ &= \frac{1}{2} \lim_{z \rightarrow 0} [z (\csc^3{z} + \cot^2{z} \csc{z}) - 2 \cot{z} \csc{z}]\\ &= \lim_{z \rightarrow 0}\frac{1}{2} \left (\frac{2}{z^2}+\frac{1}{3}-\frac{2}{z^2} \right )\\ &= \frac{1}{6} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\}$ Evaluate
$$\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\{\sqrt{k^2+2k+2}\}$$
where $\{x\}$ is the fractional part of $x$. Some suggestions here? Thanks!
| Note that $\{\sqrt{k^2+2k+2}\}=\sqrt{k^2+2k+2}-(k+1)=\frac{1}{\sqrt{k^2+2k+2}+(k+1)}=\frac{1}{2k}+O\left(\frac{1}{k(k+1)}\right)$.
\begin{align}
& \lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\binom{n-1}{k-1}\{\sqrt{k^2+2k+2}\} \\
& =\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{1}{2k}+O\left(\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\dbinom{n-1}{k-1}\frac{1}{k(k+1)}\right) \\
& =\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\binom{n}{k}\frac{1}{2n}+O\left(\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^{n}\binom{n+1}{k+1}\frac{1}{n(n+1)}\right) \\
&=\lim_{n\to\infty}\frac{2^n-1}{2^{n+1}}+O\left(\lim_{n\to\infty}\frac{2^{n+1}-1-(n+1)}{2^n(n+1)}\right) \\
&=\frac{1}{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Upper bound for $\displaystyle\int_1^\infty \frac{\lfloor x\rfloor }{x^3} dx $ How do I find $f(x) > \lfloor x \rfloor $ in the interval $[1, \infty)$ such that
$$\int_1^\infty \frac{2\lfloor x\rfloor }{x^3 }dx < \int_1^\infty \frac{2f(x)}{x^3} dx = \frac {10}{6}$$
Actually I am trying to show that $\large\frac {10}{6}$ is upper bound for $\zeta(2)$ using Euler Summation formula.
| $\displaystyle \int_1^{\infty} \frac{2\lfloor x \rfloor }{x^3}\ \mathrm{d}x =\sum_{n=1}^{\infty} \int_n^{n+1} \frac{2n}{x^3}\ \mathrm{d}x=\sum_{n=1}^{\infty}\frac{2 n+1}{n (n+1)^2}=\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+1}+\frac{1}{(n+1)^2}=\frac{\pi^2}{6}$
to prove that : $\displaystyle \frac{\pi^2}{6} \leq \frac{5}{3} $ :
$S=\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} =\sum_{n=1}^{5}\frac{1}{n^2}+\sum_{n=6}^{\infty} \frac{1}{n^2} \leq \sum_{n=1}^{5}\frac{1}{n^2}+\sum_{n=6}^{\infty} \frac{1}{n(n-1)}$
$\displaystyle S=\sum_{n=1}^{5}\frac{1}{n^2}+\sum_{n=6}^{\infty} \frac{1}{n}-\frac{1}{n-1}=\frac{5989}{3600}\leq \frac{10}{6}$
| {
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A limit with Product and Cosine Can anyone help me please to find the limit of this series:
$$\lim_{x \to 0}\frac{\displaystyle\prod_{k=1}^{n}\left(1-\sqrt[k]{\cos x}\right)}{x^{2n-2}}$$
Thanks for all :D
| As written, the limit is zero.
$$1-(\cos{x})^{1/k} = \frac{x^2}{2 k} + O(x^4)$$
so that the product is some factor times $x^{2 n}$ in this limit.
However,
$$\lim_{x \to 0}\frac{\prod_{k=1}^{k=n}1-\sqrt[k]{\cos x}}{x^{2n}} = \prod_{k=1}^{n} \frac{1}{2 k} = \frac{1}{2^n n!}$$
EDIT
Justifying the first step:
$$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$$
$$\begin{align}(\cos{x})^{1/k} &= \left ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \right )^{1/k} \end{align}$$
Now use the fact that $(1-z)^{1/k} = 1-\frac{z}{k}+\ldots$ and get
$$\begin{align}(\cos{x})^{1/k} &= \left ( 1 - \frac{x^2}{2!k} +\ldots \right )\end{align}$$
Note that the first term that I left out starts with $x^4$. Then we write, in place of the ellipses, $O(x^4)$ to represent the error which goes to zero as $x \rightarrow 0$. The first step then follows.
| {
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"source": "stackexchange",
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Induction Proof: Fibonacci Numbers Identity with Sum of Two Squares Using induction, how can I show the following identity about the fibonacci numbers? I'm having trouble with simplification when doing the induction step.
Identity: $$f_n^2 + f_{n+1}^2 = f_{2n+1}$$
I get to:
$$f_{n+1}^2 + f_{n+2}^2$$
Should I replace $f_{n+2}$ using the recursion? When I do that, I end up with the product of terms, and that just doesn't seem right. Any guidance on how to get manipulate during the induction step?
Thanks!
| Since fibonacci numbers are a linear recurrence - and the initial conditions are special - we can express them by a matrix $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$$ this is easy to prove by induction:
*
*$$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^1 = \begin{pmatrix} F_{2} & F_1 \\ F_1 & F_{0} \end{pmatrix}$$
*$$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_n + F_{n+1} & F_{n+1} \\ F_n + F_{n-1} & F_{n} \end{pmatrix} = \begin{pmatrix} F_{n+2} & F_{n+1} \\ F_{n+1} & F_{n} \end{pmatrix}$$
Now your theorem follows from $$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}^2 = \begin{pmatrix} F_{n+1}^2 + F_n^2 & - \\ - & - \end{pmatrix} = \begin{pmatrix} F_{2n+1} & F_{2n} \\ F_{2n} & F_{2n-1} \end{pmatrix}$$
https://en.wikipedia.org/wiki/Fibonacci_number
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$
| $$\sum_{i=1}^{n}\frac{1}{i(i+1)}=\sum_{i=1}^{n}\Big(\frac{1}{i}-\frac{1}{i+1}\Big)=$$
$$=1+\sum_{i=2}^{n}\frac{1}{i}-\Big(\sum_{j=2}^{n}\frac{1}{j}+\frac{1}{n+1}\Big)=1-\frac{1}{n+1}$$
| {
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Can the distance from the vertices of a square of integer width to an inscribed circle all be integer? I'm looking for solutions to the following British Mathematical Olympiad question:
Suppose that $ABCD$ is a square and that $P$ is a point which is on the circle inscribed in the square. Determine whether or not it is possible that $PA$, $PB$, $PC$, $PD$ and $AB$ are all integers.
I first assumed that $AB=2$, Therefore, the radius of our circle $r=1$, so we have the Cartesian equation of the circle:
$$x^{2}+y^{2}=1$$
Which gives us the parametric equations:
$$x=\cos{\theta} \\ y=\sin{\theta}$$
Therefore, any arbitrary point $P$ on the circle can be written $P(\cos{\theta},\sin{\theta})$, for some $\theta\in[0,2\pi]$. As we know the position vectors for each of the points $A(1,1)$, $B(1,-1)$, $C(-1,-1)$ and $D(-1,1)$, we can find the vectors:
$$\overrightarrow{AP}=\begin{pmatrix}\cos{\theta}-1 \\ \sin{\theta}-1\end{pmatrix} \\ \overrightarrow{BP}=\begin{pmatrix}\cos{\theta}-1 \\ \sin{\theta}+1\end{pmatrix} \\ \overrightarrow{CP}=\begin{pmatrix}\cos{\theta}+1 \\ \sin{\theta}+1\end{pmatrix} \\ \overrightarrow{DP}=\begin{pmatrix}\cos{\theta}+1 \\ \sin{\theta}-1\end{pmatrix}$$
And therefore, we have a general expression for the Euclidean norm of these vectors:
$$\|\overrightarrow{AP}\|=\sqrt{3-2\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)} \\ \|\overrightarrow{BP}\|=\sqrt{3-2\sqrt{2}\sin\left(\frac{\pi}{4}-\theta\right)} \\ \|\overrightarrow{CP}\|=\sqrt{3+2\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)} \\ \|\overrightarrow{DP}\|=\sqrt{3+2\sqrt{2}\sin\left(\frac{\pi}{4}-\theta\right)}$$
From this we can see $\not\exists \theta \in [0,2\pi]$ such that $\|\overrightarrow{AP}\|$,$\|\overrightarrow{BP}\|$,$\|\overrightarrow{CP}\|$ and $\|\overrightarrow{DP}\|$ are all integral. However, I cannot see how to prove that there is no $\theta$ such that they are all rational, and so cannot finish the proof, so is this proof salvageable?
| We can rescale the square so that $AB = 2$, up to allowing the other distances to be rational.
Let $P(x,y)$ be a point on your circle (i.e. $x^2 + y^2 = 1$).
Then $AP^2 + CP^2 = (x-1)^2 + (y-1)^2 + (x+1)^2 + (y+1)^2 = 2x^2+2y^2+4 = 6$.
But $6$ is not the sum of two rational squares :
Suppose $a^2 + b^2 = 6c^2$ with $a,b,c$ integers with $c \neq 0$. Looking at this modulo $3$, we quickly get that $a \equiv b \equiv 0 \pmod 3$, and then that $c \equiv 0 \pmod 3$, etc. Thus $a,b,c$ are infinitely divisible by $3$, and so $a=b=c=0$ is the only solution.
| {
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How do I calculate $\sum_{n\geq1}\frac{1}{n^4+1}$? How do I calculate the following sum $$\sum_{n\geq1}\frac{1}{n^4+1}$$
| $$\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\frac{1}{x^5}..$$
$$\sum_{n=1}^\infty \frac{1}{n^4+1}=\zeta(4)-\zeta(8)+\zeta(12)-\zeta(16)+\zeta(20)...$$
$$\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2} = \zeta(2) t^2 - \zeta(4) t^4 + \zeta(6) t^6 - \zeta(8)t^8+\zeta(10)t^{10}\cdots.$$
$$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} = -\zeta(2) t^2 - \zeta(4) t^4 -\zeta(6) t^6 -\zeta(8)t^8-\zeta(10)t^{10} \cdots.$$
$$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} +\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2}=-2( \zeta(4)t^4+\zeta(8)t^8+\zeta(12)t^{12}+\zeta(16)t^{16}..)$$
I think you can figure out the rest...
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple proof of showing the Harmonic number $H_n = \Theta (\log n)$ Consider the partial sum of the divergent Harmonic series $H_n = \sum\limits_{k = 1}^{n}\frac{1}{k}$. I recently saw a question which required finding out the asymptotic bounds of $H_n$. Now, I could not find any bound closer than $O(n)$. So I looked up the series and found this Wikipedia article which says that $H_n - \ln n$ approaches a constant (the Euler–Mascheroni constant). This clearly shows that $H_n = \Theta(\log n)$.
However, considering the question was in an beginners undergraduate algorithms course in CS, I was wondering if there is a simpler proof of this? Is there some clever proof requiring elementary mathematics to show $H_n = \Theta(\log n)$?
PS: This is not homework in the sense that it is not my homework, and the actual question does not ask for a proof, but rather asks a lot of functions to be arranged asymptotically.
| Here's my favorite "calculus-free" proof that $H_n = \Theta(\log n)$. This proof is essentially an extension of the calculus-free proof that the harmonic series diverges.
Start with the powers of 2, $n = 2^k$, and break up $H_{2^k}$ into $k$ groups, each one twice as large as the previous:
$$\begin{align*}
H_{2^k} = &\sum_{j = 1}^{2^k} \frac{1}{j} \\
= & 1 + \\
&\frac{1}{2} + \frac{1}{3} + \\
&\frac{1}{4}+\frac{1}{5} + \frac{1}{6} + \frac{1}{7}\\
& \cdots\\
& \frac{1}{2^{k-1}} + \frac{1}{2^{k-1} + 1} + \cdots + \frac{1}{2^k - 1}
\end{align*}$$
There's also one extra term, $\frac{1}{2^k}$.
We can bound each group above and below by a constant by bounding each term by the power of 2 above and below it.
\begin{array}{lll}
\frac{1}{2}& \leq 1 &\leq 1\\
\frac{1}{4}+\frac{1}{4}& \leq \frac{1}{2} + \frac{1}{3} &\leq \frac{1}{2} + \frac{1}{2}\\
\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}& \leq \frac{1}{4}+\frac{1}{5} + \frac{1}{6} + \frac{1}{7} & \leq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}\\
& \cdots &\\
\frac{1}{2^k} + \frac{1}{2^k} + \cdots + \frac{1}{2^k} & \leq \frac{1}{2^{k-1}} + \frac{1}{2^{k-1} + 1} + \cdots + \frac{1}{2^k - 1} & \leq \frac{1}{2^{k-1}} + \frac{1}{2^{k-1}} + \cdots + \frac{1}{2^{k-1}}
\end{array}
Each of the left-hand sums is $\frac{1}{2}$. Each of the right-hand sums is 1. Together we have $\frac{1}{2}k +\frac{1}{2^k}\leq H_{2^k} \leq k + \frac{1}{2^k}$. Written in terms of $n$,
$$ \frac{1}{2} \log_2 n \leq H_n \leq \log_2n + 1$$
This is essentially the whole proof: it says when $n = 2^k$, $H_n = \Theta( \log n)$. For $n$ that aren't powers of 2, letting $n_-$ and $n_+$ be the surrounding powers of 2 of $n$, by monotonicity of $H_n$ and the observation that
$$n/2 < n_- < n < n_+ < 2n$$
we have
$$\frac{1}{2}\log_2(n/2) < \frac{1}{2}\log_2 n_- < H_{n_-} < H_n < H_{n_+} < \log_2 n_+ +1< \log_2 (2n)+1$$
$$\frac{1}{2} \log_2n - \frac{1}{2} < H_n < \log_2 n + 2$$
And so $H_n = \Theta(\log n)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we show that $x^5y^3 + x^4y^4 + x^3y^5$ is $\Omega(x^3y^3)$ Basically I'm wondering how I can show that $x^5y^3 + x^4y^4 + x^3y^5$ is $\Omega(x^3y^3)$. Any ideas? Thanks a lot!
| Denote the expression in the OP $F(x,y)$. For any $x,y >0 \ F(x,y) \geq 3 x^3 y^3 = \Omega (x^3 y^3)$
EDIT: $F(x,y) \geq x^3y^3 + x^3 y^3 + x^3 y^3 = 3 x^3 y^3 = \Omega(x^3 y^3)$ since $x^5>x^4>x^3$. Same with $y$.
| {
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The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is
(a) 0 (b) 2
(c) 4 (d) none of these
Solution: put $y = x + (3+5)/2 = x+4$
the equation becomes
=> $(y-1)^4 + (y+1)^4 = 16$ ---- (i)
=> $2{y^4 + 6(y)^2 + 1 } = 16$ --------(ii)
My question is how was (i) converted to (ii)? I just couldn't get it. Please help?
| Multiply: $(y-1)^4-(y+1)^4=(y^4-4y^3+6y^2-4y+1)+(y^4+4y^3+6y^2+4y+1)$. Add.
| {
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Method to find $\sin (2\pi/7)$ I just thought a way to find $\sin\frac{2π}{7}$.
Considering the equation $x^7=1$
$⇒(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0$
$⇒(x-1)[(x+\frac1 x)^3+(x+\frac1 x)^2-2(x+\frac1 x)-1]=0$
We can then get the 7 solutions of x, but the steps will be very complicated, especially when solving cubic equation, and expressing x as a+bi. The imaginary part of the second root of x will be $\sin\frac{2π}{7}$.
Besides this troublesome way, are any other approach? Thank you.
| Using this or Point#$24$ of this ,
$\sin 7x=7s-56s^3+112s^5-64s^7$ where $s=\sin x$
If $\sin 7x=0,7x=n\pi$ where $n$ is any integer.
So, $x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$
So, the roots of $7s-56s^3+112s^5-64s^7=0--->(1)$ are $\sin\frac{n\pi}7$ where $n=0,1,2,\cdots 5,6$
So, the roots of $64s^6-112s^4+56s^2-7=0--->(2)$ are $\sin\frac{n\pi}7$ where $n=1,2,\cdots 5,6$
So, the roots of $64t^3-112t^2+56t-7=0 --->(3)$ are $\sin^2\frac{n\pi}7$ where $n=1,2,4$ or $3,5,6$ as $\sin \frac{(7-r)\pi}7=\sin(\pi-\frac{r\pi}7)=\sin\frac{r\pi}7$
If we choose $n=1,2,4$ observe that $\sin^2\frac{4\pi}7-\sin^2\frac{2\pi}7=2\sin\frac{\pi}7\cos\frac{3\pi}7>0$ (Using $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$)
Similarly, $\sin^2\frac{2\pi}7-\sin^2\frac{\pi}7>0$
So, $\sin^2\frac{4\pi}7>\sin^2\frac{2\pi}7>\sin^2\frac{\pi}7$
Using Cardano's method, we can solve the Cubic equation $(3)$
| {
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Arithmetic progression of logarithms
If $\log_{\sqrt{2}}a$, $\log_{\sqrt{2}}(2a^2)$, $\log_{\sqrt{2}}(a^3+4)$ are in A.P, find the value of "$a$".
I tried solving this and I am getting $a^3 = \frac{4}{3}$ is it correct. Please guide.
| If $$\log_{\sqrt{2}}a, \log_{\sqrt{2}}(2a^2), \log_{\sqrt{2}}(a^3+4)$$ are in A.P,
$$\log_{\sqrt{2}}(2a^2)-\log_{\sqrt{2}}a=\log_{\sqrt{2}}(a^3+4)-\log_{\sqrt{2}}(2a^2)$$
$$\implies \log_{\sqrt{2}}a+\log_{\sqrt{2}}(a^3+4)=2\log_{\sqrt{2}}(2a^2)$$
$$\implies \log_{\sqrt{2}}a(a^3+4)=\log_{\sqrt{2}}(2a^2)^2$$
$$\implies a(a^3+4)=(2a^2)^2\implies 3a^4=4a\implies 3a^3=4\text { as } a\ne0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A two variable Inequality Show that $(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2$
Can we go like this : $(a^2+b^2)=(a+b)^2-2ab \& (a^4+b^4) = (a^2+b^2)^2-2a^2b^2$ and right hand side of the inequality : $[(a+b)^3-3ab(a+b)]^2$
| $(a^2+b^2)(a^4+b^4)-(a^3+b^3)^2$
$=(a^6+a^4b^2+a^2b^4+b^6)-(a^6+2a^3b^3+b^6)$
$=a^4b^2+a^2b^4-2a^3b^3$
$=a^2b^2(a^2+b^2-2ab)$
$=a^2b^2(a-b)^2$
$\geq 0$
$∴(a^2+b^2)(a^4+b^4)\geq(a^3+b^3)^2$
Q.E.D.
| {
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Solve $x^2+x+7 \equiv 0\pmod{81}$ Solve $x^2+x+7\equiv 0 \pmod{81}$
My work:
Prime factorization $81 = 9^2 = 3^4$
Test the value $x\equiv0,1,2$ for $x^2+x+7\equiv0\mod{3}$
we have $x\equiv1\mod{3}$ works.
Now life this to $\mod{3^2} = \mod{9}$
Let $x=1+3k$ for some integer$k$
$(1+3k)^2 + (1+3k) +7 \equiv0\mod{9}$
$1+6k+0+1+3k+7\equiv0\mod{9}$
$9+9k\equiv0\mod9$
$1+k\equiv0\mod9$
$k\equiv-1\mod9$
$k=-1+9m$ for some integer m
Lift again to $\mod81$
$(-1+9m)^2 + (-1+9m)+7\equiv0\mod81$
$-9m+7\equiv0\mod81$
I cant continue... how can I make it work???
thanks!!
| Let us do the lifting, patiently. As was pointed out, the solution $x\equiv 1\pmod{3}$ lifts to three solutions modulo $3^2$, namely $1$, $4$, and $7$.
Let us try to now lift to $3^3$. We do only the case $x\equiv 1\pmod 3^2$. Let $x=1+9k$ for some $k$. Then $x^2+x+7\equiv 1+18k+1+9k+7\equiv 27k+9\pmod{3^3}$. It is clear that there is no $k$ that works. A similar calculation shows that the solutions $x\equiv 4\pmod{3^2}$ and $x\equiv 7\pmod{3^2}$ do not lift to $3^3$.
| {
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} |
Solution of $(E):z^2-2mz+1=0$ in $\mathbb{C}$ Suppose the equation $(E):z^2-2mz+1=0 \quad / m\in \mathbb{C}\quad z\in\mathbb{C}$ and we suppose $z_{1}$ and $z_{2}$ are the two solution of this equation.
How can I prove that $|z_1|+|z_2| = |m-1|+|m+1|$?
| This is not very elegant but you can do it explicitly: Set $m=x+iy$ and choose a squareroot of $m^2-1$, $a+bi=\sqrt{m^2-1}$. Then $z_{1,2}= x+iy\pm (a+bi)$. We want to prove $|z_1|+|z_2|=|m-1|+|m+1|$ or equivalently (after squaring and remembering $z_1z_2=1$):
$$
|z_1|^2+2+|z_2|^2=|m-1|^2+|m+1|^2+2|m^2-1|
$$
The left hand side is
$$
(x+a)^2+(y+b)^2+2+(x-a)^2+(y-b)^2=2(x^2+y^2+1+a^2+b^2).
$$
The right hand side equals
$$
2(x^2+y^2+1+|m^2-1|),
$$
so it remains to show $a^2+b^2=|m^2-1|$. This is true since $a+bi=\sqrt{m^2-1}$ and so
$|m^2-1|=|\sqrt{m^2-1}|^2=a^2+b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
How do you evaluate this limit? $\lim\limits_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$ $$\lim_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$$
This is what I have tried so far, $$\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10}+x^2)^{\frac{1}{2}}}$$
$$\begin{align}
&\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10})^{\frac{1}{2}}+(x)^{\frac{1}{2}}}\\
&\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{5})+(x)}\\
&\lim_{x\to-\infty}\frac{x^2}{x}\\
&\lim_{x\to-\infty}x = -\infty\\
\end{align}$$
Is this the correct approach to finding the limit?
| Well, if you can't get rid of a square root, put everything in a square root. Notice that for $x$ sufficiently negative--in particular, for $x<-\frac1{\sqrt[3]2}$--we have $2x^5+x^2<0,$ so that $$2x^5+x^2=-|2x^5+x^2|=-\sqrt{(2x^5+x^2)^2}$$ for sufficiently negative $x$-values. Thus, for such $x$-values, we have
$$\begin{align}
\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}
&= \frac{-\sqrt{(2x^5+x^2)^2}}{\sqrt{2x^{10}+x^2}} \\
&= -\sqrt{\frac{(2x^5+x^2)^2}{2x^{10}+x^2}} \\
&= -\sqrt{\frac{4x^{10}+4x^7+x^4}{2x^{10}+x^2}} \\
&= -\sqrt{2+\frac{4x^{10}+4x^7+x^4-2(2x^{10}+x^2)}{2x^{10}+x^2}} \\
&= -\sqrt{2+\frac{4x^7+x^4-2x^2}{2x^{10}+x^2}} \\
&= -\sqrt{2}\sqrt{1+\frac{4x^7+x^4-2x^2}{4x^{10}+2x^2}} \\
\end{align}
$$
The fraction inside the right-hand square root is about
$1/x^3$,
so the right-hand square root goes to $1$ as $x\to-\infty$,
so the whole thing goes to
$-\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to simplify $42\sqrt{45} \over 7\sqrt{35}$? The problem is $$42\sqrt{45} \over 7\sqrt{35}$$ HELP! My daughter's math sheet shows how to reduce the squareroots, but the examples all use the same square-root; the problems show two different numbers square-rooted. Can you please help work this problem so I know how to help her with this assignment?
Thanks!
| $$\frac{42\sqrt{45}}{7\sqrt{35}} = \frac{6\times 7\sqrt{9\times 5}}{7\sqrt{7\times 5}} $$
$$ = \frac{6\sqrt 9 \times \sqrt 5}{\sqrt 7 \times \sqrt 5} = \frac{6\times \sqrt 9}{\sqrt 7}$$
$$= \frac{6 \times 3}{\sqrt 7} \times \frac{\sqrt 7}{\sqrt 7} = \frac{18\sqrt 7}{ 7}$$
Cancel common factors, then multiply denominator to clear the square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Finding the maximum value If $a^2+b^2+c^2+d^2=4$ are real numbers, then how to find the maximum value for:$$a^3+b^3+c^3+d^3$$
| To maximize/minimize $a^3+b^3+c^3+d^3$ under the constraint that $a^2+b^2+c^2+d^2=4$ means we need
$$
a^2\delta a+b^2\delta b+c^2\delta c+d^2\delta d=0\tag{1}
$$
for any deltas so that
$$
a\,\delta a+b\,\delta b+c\,\delta c+d\,\delta d=0\tag{2}
$$
Standard orthogonality arguments say that there must be a $\lambda$ so that
$$
(a^2,b^2,c^2,d^2)=\lambda\,(a,b,c,d)\tag{3}
$$
That is, $(a^2,b^2,c^2,d^2)$ must be perpendicular to any delta which is perpendicular to $(a,b,c,d)$.
$(3)$ says that $a^2-\lambda a=0$. Thus, either $a=0$ or $a=\lambda$. The same is true for $b$, $c$, and $d$.
There are a finite number of possibilities, each depending on how many of $b$, $c$, and $d$ are zero and whether $\lambda$ is positive or negative.
$0$ zeroes: $\lambda=\pm1$ and $a^3+b^3+c^3+d^3=\pm4$
$1$ zero: $\lambda=\pm\sqrt{\frac43}$ and $a^3+b^3+c^3+d^3=\pm4\sqrt{\frac43}$
$2$ zeroes: $\lambda=\pm\sqrt{2}$ and $a^3+b^3+c^3+d^3=\pm4\sqrt{2}$
$3$ zeroes: $\lambda=\pm2$ and $a^3+b^3+c^3+d^3=\pm8$
Thus, the maximum of $a^3+b^3+c^3+d^3$ is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
| Consider a rectangle with dimension (x, y) of unit area, the half perimeter is defined as $a+b$
a square is a one and only one special form of rectangle with equal dimension should either be a maxima or minima
Now lets calculate the half perimeter of a square
$x.y = 1$
as $x = y$
we have
$x.x = 1$
so $x = 1$
which gives us
$x + y = 2$
Consider any one rectangular figure
$x = 2$
and
$y = \frac{1}{2}$
this gives the perimeter as
$x + y = 2 + 1/2 = 2.5$
$2.5(rectangle) > 2 (square)$
which makes us to believe
a square will have the minimum perimeter
thus we can conclude
the minimum value of $x + y$ is $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 13,
"answer_id": 11
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How to find all the solutions to $-8x^4-18x^2-11+\frac{1}{x^4}=0$? How to know all the roots for fourth degree equation like this :
$$-8x^4-18x^2-11+\frac{1}{x^4}=0$$
| Just by the way, most people would not call this a degree 4 equation, because it also involves the reciprocal of $x$... only something like $4x^4-2x^3+x-5$ would be degree 4.
To answer your question:
First, note that the equation only involves even powers of $x$. Thus, if we let $t=x^2$, we can reduce the (apparent) complexity of the problem, and then once we've solved for $t$, the values of $x$ which are solutions will just be the various square roots of the values of $t$ which were solutions to the modified problem.
Letting $t=x^2$, we now want to solve for $t$ in
$$-8t^2-18t-11+\frac{1}{t^2}=0.$$
Now note that $t=0$ cannot possibly be a solution, because the left side involves a $\frac{1}{t}$. We have to be careful when multiplying both sides of an equation by a quantity involving the variable we're solving for, because sometimes it introduces fake solutions (as an illustration: if we had the equation $2=3$, which is false, multiplying both sides by $0$ gives $0=0$, which is true). Thus, even if $t=0$ appears to be a solution later, we know to throw it out.
Multiplying both sides by $t^2$ produces
$$-8t^4-18t^3-11t^2+1=0.$$
Now use the rational root theorem to check for possible solutions that are rational numbers. This won't work for every polynomial, but presumably, if these questions are being given to you in class, they will be designed so that the rational root theorem will help you.
In our problem, the rational root theorem tells us that, if there is some rational number $\frac{p}{q}$ (written in lowest terms, i.e. $p$ and $q$ have no common factors) such that $t=\frac{p}{q}$ is a solution, then we would have to have that
$$p\in\{\pm 1\},\quad q\in\{\pm 1,\pm 2,\pm 4,\pm 8\}.$$
Checking each of the numbers
$$\pm 1,\quad \pm \frac{1}{2},\quad \pm \frac{1}{4},\quad\pm \frac{1}{8}$$
to see whether they are solutions, we see that there are 3 rational number solutions to
$$-8t^4-18t^3-11t^2+1=0,$$
namely $t=-1$, $t=-\frac{1}{2}$, and $t=\frac{1}{4}$. Now, we know that a degree $n$ polynomial can have at most $n$ distinct roots; here, we have a degree 4 polynomial, and we found 3 roots that were rational numbers. Perhaps there is another root that is an irrational number? In fact, this is not the case here. The root $t=-1$ is a double root. You can check this by using whatever polynomial division algorithm you've learned to successively divide the polynomial $-8t^4-18t^3-11t^2+1$ by $t-(-1)$, $t-(-\frac{1}{2})$, and $t-(\frac{1}{4})$, and then noting that $-1$ is a root of the degree 1 polynomial that's left over.
Thus, we've determined that the only solutions to
$$-8t^4-18t^3-11t^2+1=0$$
are $t=-1$, $t=-\frac{1}{2}$, and $t=\frac{1}{4}$. Now you need to go back to the original question, namely, finding the solutions to
$$-8x^8-18x^6-11x^4+1=0.$$
Remember, since we set $t=x^2$, if $t$ is a solution to the equation involving $t$, then the two different square roots of $t$, namely $\sqrt{t}$ and $-\sqrt{t}$, will both be solutions for $x$ in the equation involving $x$.
Therefore, the solutions to
$$-8x^8-18x^6-11x^4+1=0$$
are
$$\begin{align*}
x&=\sqrt{-1}=i, & x&=\sqrt{-\frac{1}{2}}=\frac{i\sqrt{2}}{2},& x=\sqrt{\frac{1}{4}}=\frac{1}{2},\\\\\\
x&=-\sqrt{-1}=-i, & x&=-\sqrt{-\frac{1}{2}}=-\frac{i\sqrt{2}}{2}, & x=-\sqrt{\frac{1}{4}}=-\frac{1}{2}
\end{align*}$$
You should note that because $t=-1$ was a double root of $-8t^4-18t^3-11t^2+1=0$, the solutions for $x$ that arose from it, namely $x=i$ and $x=-i$, will each be double roots of $-8x^8-18x^6-11x^4+1=0$. Thus, we have found a total of 8 (counting with muliplicity) solutions to a degree 8 equation, and therefore we must have got all of them. We're done :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Alternative solutions to $\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$ Here is a limit that can be computed directly by performing the integration and then taking
the limit, but the way is rather ugly. What else can we do? Might we avoid the integration?
$$\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$$
| Not strictly avoiding the integration, but expanding the integrand makes it pretty straightforward:
$$\frac{\ln(1+x)}{x^3}=\frac{1}{x^2}-\frac{1}{2x}+\frac{1}{3}-\frac{x}{4}+\cdots$$
$$\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=\left[-\frac{1}{x}-\frac{\ln x}{2}+\frac{x}{3}-\cdots\right]_{1/\sqrt{n}}^1=-\frac{3}{4}+\sqrt{n}+\frac{\ln \sqrt{n}}{2}-\frac{1}{3\sqrt{n}}+\cdots$$
$$\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=1-\frac{3}{4\sqrt{n}}+\frac{\ln \sqrt{n}}{2\sqrt{n}}-\frac{1}{3n}+\cdots\to 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 2
} |
Solving for $b$ in $25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$ What are the steps to get from:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
to:
$$b = \frac{\sqrt{5}}{5} + \frac{2\sqrt{5} - \sqrt{10}}{10}$$
Thanks.
| $$
\begin{align}
25\left(\dfrac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b &= \sqrt{5} \\
5b &= \sqrt{5} - \left(\dfrac{\sqrt{10}-2\sqrt{5}}{2}\right) \\
5b &= \sqrt{5} - \left(\dfrac{- ( - \sqrt{10} + 2\sqrt{5} )}{2}\right) \\
5b &= \sqrt{5} + \left(\dfrac{ - \sqrt{10} + 2\sqrt{5}}{2}\right) \\
5b &= \sqrt{5} + \left(\dfrac{ 2\sqrt{5} - \sqrt{10} }{2}\right) \\
b &= \dfrac{1}{5} \cdot \left( \sqrt{5} + \left(\dfrac{ 2\sqrt{5} - \sqrt{10} }{2}\right) \right) \\
b &= \dfrac{\sqrt{5}}{5} + \dfrac{2\sqrt{5} - \sqrt{10}}{10}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Trouble with factorising a polynomial I'm supposed to show that:
$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$
and the required answers are: $$ P=5, Q=4, R=-4 $$
I tried to solve this with partial fractions like so:
$$5(x-1)(x+2) = A(x+3) + B(x-2)$$
$\implies$ $A$=4, $B$=-4
$\implies$ $Q$=4, R=-4
But where does $P$=5 come from?
Or should I have first multiplied out the numerator and denominator and then used long division to solve?
| You don't need to do any long division. In the expression
$$y = \frac{N(x)}{D(x)} = \frac{5(x-1)(x+2)}{(x-2)(x+3)}$$
The numerator $N(x) = 5(x-1)(x+2)$ and denominator $D(x) = (x-2)(x+3)$
are polynomials of degree $2$. Since the roots of $D(x)$ are simple,
$y$ can be rewritten in the form:
$$y = P(x) + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$
where $P(x)$ is a polynomial of degree $\deg N(x) - \deg D(x) = 2 - 2 = 0$, i.e. a constant. To evaluate the 3 coefficients, you can evaluate both side at 3 different
values of $x$: the two roots of $D(x)$ and $\infty$:
$$\begin{align}
P &= \lim_{x\to\infty} \frac{N(x)}{D(x)} = 5\\
Q &= \lim_{x\to 2} (x-2)\frac{N(x)}{D(x)} = \lim_{x\to 2}\frac{5(x - 1)(x+2)}{x+3} = \frac{5(2 - 1)(2+2)}{2+3} = 4\\
R &= \lim_{x\to -3}(x+3)\frac{N(x)}{D(x)} = \lim_{x\to-3}\frac{5(x-1)(x+2)}{x-2} = \frac{5(-3-1)(-3+2)}{-3-2} = -4
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
With $xy+yz+zx=-1$, proving: $x^2+2y^2+2z^2 .....$ Assuming $xy+yz+zx=-1$, prove that :
$$x^2+2y^2+2z^2 \geq \frac{1+\sqrt{17}}{2}$$
| We need to prove that
$$x^2+2y^2+2z^2+\frac{1+\sqrt{17}}{2}(xy+xz+yz)\geq0$$ or
$$x^2+\frac{1+\sqrt{17}}{2}(y+z)x+2y^2+2z^2+\frac{1+\sqrt{17}}{2}yz\geq0,$$
for which it's enough to prove that
$$\left(\frac{1+\sqrt{17}}{2}(y+z)\right)^2-4\left(2y^2+2z^2+\frac{1+\sqrt{17}}{2}yz\right)\leq0,$$
which is $(y-z)^2\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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A conjecture from Cauchy–Schwarz inequality we know the Cauchy–Schwarz inequality in $R^3$:
$$(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)\geq(x_1y_1+x_2y_2+x_3y_3)^2$$
$$(x_1+x_2+x_3)(y_1+y_2+y_3)\geq(\sqrt{x_1y_1}+\sqrt{x_2y_2}+\sqrt{x_3y_3})^2$$
i guess the following inequlity exists too:
$$(x_1+x_2+x_3)(y_1+y_2+y_3)(z_1+z_2+z_3)\geq(\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2}+\sqrt[3]{x_3y_3z_3})^3$$
is it true? how to prove?
| Lets look at a simple proof of Cauchy Schwarz inequality and see if we can extend in the direction you want.
Let $A^2 = \sum_{i=1}^n {a_i ^2}, \quad B^2 = \sum_{i=1}^n {b_i^2}$
Then $\sum_{i=1}^n \dfrac{a_i b_i}{AB} \le \sum_{i=1}^n \dfrac{1}{2}\left( \dfrac{a_i^2}{A^2} + \dfrac{b_i^2}{B^2} \right) = 1$
So $\sum_{i=1}^n {a_i b_i} \le AB$, or if you prefer $ A^2 B^2 \ge \left(\sum_{i=1}^n {a_i b_i} \right)^2$ which is Cauchy Schwarz.
Extending this, we have:
Let $A^3 = \sum_{i=1}^n {a_i ^3}, \quad B^3 = \sum_{i=1}^n {b_i^3}, \quad C^3 = \sum_{i=1}^n {c_i^3}$
Then $\sum_{i=1}^n \dfrac{a_i b_i c_i}{ABC} \le \sum_{i=1}^n \dfrac{1}{3} \left( \dfrac{a_i^3}{A^3} + \dfrac{b_i^3}{B^3} + \dfrac{c_i^3}{C^3} \right) = 1$
So $\sum_{i=1}^n {a_i b_i c_i} \le ABC$, or if you prefer, $A^3B^3C^3 \ge \left(\sum_{i=1}^n {a_i b_i c_i} \right)^3$ is the extended version. Essentially your RHS needs to be cubed, not squared.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Range of $\sin^2x-\sin x +1$ How can we find the range of $f(x) =\sin^2x-\sin x +1$?
The function can be written as $(\sin x-\frac{1}{2})^2+\frac{3}{4}$.
Range of $\sin x$ function is given by : $-1 \leq \sin x \leq 1$. Please guide how to get the result.
| $-1 \leq sinx \leq 1 \Rightarrow -\frac{3}{2} \leq sinx-\frac{1}{2} \leq \frac{1}{2} \Rightarrow 0 \leq (sinx - \frac{1}{2})^2 \leq \frac{9}{4} \Rightarrow \frac{3}{4}\leq(sinx-\frac{1}{2})^2 + \frac{3}{4}\leq 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Is $\phi : \mathbb R^2 \rightarrow \mathbb R$ a differentiable function? We have the following example on the book "Matrix Differential Calculus with Applications in Statistics and Econometrics", 3rd edition, p. 100.
Let $\phi : \mathbb R^2 \rightarrow \mathbb R$ be a real-valued function defined by
$$
\phi(x,y) = \left\{
\begin{array}{ll}
x^2[y+\text{sin}(1/x)] & \text{if } x\neq 0 \\
0& \text{if } x= 0
\end{array}
\right.
$$
Then $\phi$ is differentiable at every point in $\mathbb R^2$ with partial derivatives
$$
\mathrm D_1\phi(x,y) = \left\{
\begin{array}{ll}
2x[y+\text{sin}(1/x)] - \text{cos}(1/x) & \text{if } x\neq 0\\
0 & \text{if } x= 0
\end{array}
\right.
$$
and $\mathrm D_2\phi(x,y) = x^2$. We see that $\mathrm D_1\phi$ is not continuous at any point on the $y$-axis since cos(1/x) in $\mathrm D_1\phi(x,y)$ does not tend to a limit as $x\rightarrow 0$.
Letting $c_1\neq 0$ and considering the open ball in $\mathbb R^2$ where $\sqrt{x^2+y^2}<c_1$ (to make $x\neq -c_1$), we have
\begin{align}
\phi(c_1+x,c_2+y) &= \phi(c_1,c_2) + [\mathrm D_1\phi(c_1,c_2) \quad \mathrm D_2\phi(c_1,c_2)] \left[\begin{array}{c}x\\
y
\end{array}\right] + r(x,y)\\
(c_1+x)^2(c_2+y+\sin\frac{1}{c_1+x}) &= c_1^2(c_2+\sin\frac{1}{c_1}) + x(2c_1(c_2+\sin\frac{1}{c_1})-\cos\frac{1}{c_1})+yc_1^2+r(x,y).
\end{align}
Thus,
$$
r(x,y) = (c_1+x)^2(c_2+y+\sin\frac{1}{c_1+x}) - c_1^2(c_2+\sin\frac{1}{c_1})-x(2c_1(c_2+\sin\frac{1}{c_1})-\cos\frac{1}{c_1})-yc_1^2.
$$
However, it seems that
$$
\lim_{(x,y)\rightarrow (0,0)} \frac{r(x,y)}{\sqrt{x^2+y^2}}=0
$$
is not true, which is required to $\phi$ be differentiable everywhere as stated by the above example. How to verify the differentiability of $\phi$?
| The partial derivatives exist at every point and are continuous for any $(c_1,c_2)$ such that $c_1\neq 0$. Thus, $\phi$ is differentiable at every $(c_1,c_2)$ such that $c_1\neq 0$.
Let $(c_1,c_2)$ be such that $c_1= 0$. From
\begin{align}
\phi(c_1+x,c_2+y) &= \phi(c_1,c_2) + [\mathrm D_1\phi(c_1,c_2) \quad \mathrm D_2\phi(c_1,c_2)] \left[\begin{array}{c}x\\
y
\end{array}\right] + r_{c_1,c_2}(x,y)\\
\phi(x,c_2+y) &=r_{c_1,c_2}(x,y),
\end{align}
we have
\begin{align}
r_{c_1,c_2}(x,y) = \left\{
\begin{array}{ll}
x^2[y+c_2+\sin(1/x)] & \text{if } x\neq 0\\
0 & \text{if } x= 0
\end{array}
\right.
\end{align}
Consider
$$
\lim_{(x,y)\rightarrow (0,0)} \frac{r_{c_1,c_2}(x,y)}{\sqrt{x^2+y^2}}=\lim_{(x,y)\rightarrow (0,0)}\frac{x^2y}{\sqrt{x^2+y^2}} + \lim_{(x,y)\rightarrow (0,0)}\frac{x^2c_2}{\sqrt{x^2+y^2}} + \lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sin(1/x)}{\sqrt{x^2+y^2}}.
$$
Note that $0\leq\frac{x^2}{\sqrt{x^2+y^2}}\le\frac{x^2}{\sqrt{x^2}}=|x|$, where $|x|\rightarrow 0$ as $(x,y)\rightarrow (0,0)$. Thus, $\lim_{(x,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}}=0$, $\lim_{(x,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} \lim_{(x,y)\rightarrow (0,0)}y =0$, and $c_2\lim_{(x,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} =0$.
Since
$$
\frac{-x^2}{\sqrt{x^2+y^2}}\le\frac{x^2\sin(1/x)}{\sqrt{x^2+y^2}}\le \frac{x^2}{\sqrt{x^2+y^2}},
$$
we also have $\lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sin(1/x)}{\sqrt{x^2+y^2}}=0$. Thus, $\lim_{(x,y)\rightarrow (0,0)} \frac{r_{c_1,c_2}(x,y)}{\sqrt{x^2+y^2}}=0$ and $\phi$ is differentiable also at $(0,c_2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve for system of diophantine equations $\cases{x+1=a^2 \cr x^3-x^2+1=b^2}$
I just can found a trivial solution $x=0$. Is there any other ?
| Here's an elementary solution. First note that we may assume $a,b \ge 0$ since negative values yield exactly the same $x$. It's easy to see that $a=0$ doesn't work and $a=1$ does, so let's assume $a \ge 2$.
The two equations are obviously equivalent to $(a^2-1)^3 - (a^2-1)^2 + 1 = b^2$. The LHS expands to $a^6 - 4a^4 + 5a^2 - 1 = (a^3-2a)^2 + (a^2-1)$.
Since $a \ge 2$, both $a^3-2a$ and $a^2-1$ are positive. In general, whenever $n^2+m$ is a square ($n,m$ positive integers) we must have $m \ge 2n+1$ by considering the smallest square after $n^2$.
Therefore, if $(a^3-2a)^2 + (a^2-1)$ is a square, $a^2-1 \ge 2(a^3-2a)$. But it's easy to check this is impossible for $a \ge 2$, so there are no further solutions.
This simple technique is often very effective for detecting square values of monic even-degree polynomials in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Estimating the limit $\lim_{n\to\infty}\frac{n _2F_1[1 - n, 1 + 2^n n; 2 + 2^n n; -1]}{1 + 2^n n}$ Can you please help me solve the limit below?
$$\lim_{n\to\infty}\frac{n _2F_1[1 - n, 1 + 2^n n; 2 + 2^n n; -1]}{1 +
2^n n}$$
where $_2F_1(a,b;c;z)$ - hypergeometric function
| The hypergeometric function is a polynomial:
\begin{eqnarray*}
& & {}_2F_1[1 - n, 1 + 2^n n; 2 + 2^n n;z]\\
&=&\sum_{k\ge 0} \frac{(1-n)_k (1+2^n n)_k}{(2+2^n n)_k k!} z^k\\
& =& \sum_{0\le k\le n-1} \frac{(-1)^k(n-1)\cdots (n-k) (2^n n+1)}{(2^n n + k + 1) k!} z^k\\
&=& \sum_{0\le k\le n-1} \binom{n-1}{k} \left(1-\frac{k}{2^n n + k + 1}\right)(-z)^k
\end{eqnarray*}
so
\begin{eqnarray*}
&&
n {}_2F_1[1 - n, 1 + 2^n n; 2 + 2^n n;-1]\\
&=& n \sum_{0\le k\le n-1} \binom{n-1}{k} \left(1-\frac{k}{2^n n + k + 1}\right)\\
&=& 2^{n-1}n - n \sum_{0\le k\le n-1} \frac{k}{2^n n + k + 1}\binom{n-1}{k}\\
&=& 2^{n-1}n + O(n),
\end{eqnarray*}
meaning that the limit is $\frac 12 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the antiderivative of $\sin^6x\cos^2x$ I need to find the antiderivative of
$$\int\sin^6x\cos^2x \mathrm{d}x.$$ I tried symbolizing $u$ as squared $\sin$ or $\cos$ but that doesn't work. Also I tried using the identity of $1-\cos^2 x = \sin^2 x$ and again if I symbolize $t = \sin^2 x$ I'm stuck with its derivative in the $dt$.
Can I be given a hint?
| $$\text{ As }\cos2y=2\cos^2y-1=1-2\sin^2y$$
$$\sin^6x\cos^2x=\left(\frac{1-\cos2x}2\right)^3\left(\frac{1+\cos2x}2\right)$$
$$16\sin^6x\cos^2x=(1-3\cos2x+3\cos^2x-\cos^32x)(1+\cos2x)$$
$$=\left(1-3\cos2x+3\frac{(1+\cos4x)}2-\frac{(\cos6x+3\cos2x)}4\right)(1+\cos2x)$$ (applying $\cos3y=4\cos^3y-3\cos y$)
$$64\sin^6x\cos^2x=(10-15\cos2x+6\cos4x-\cos6x)(1+\cos2x)$$
$$=10-15\cos2x+6\cos4x-\cos6x+10\cos2x-15\cos^22x+6\cos4x\cos2x-\cos6x\cos2x$$
$$=10-5\cos2x+6\cos4x-\cos6x+10\cos2x-15\frac{(1+\cos4x)}2+6\frac{(\cos2x+\cos6x)}2-\frac{(\cos4x+\cos8x)}2$$ (applying $2\cos A\cos B=\cos(A-B)+\cos(A+B)$)
$$\text{ So, }128\sin^6x\cos^2x=5-4\cos2x-4\cos4x+4\cos6x-\cos8x$$
Alternatively
as we know, $e^{ix}=\cos x+i\sin x,e^{-ix}=\cos x-i\sin x\implies \cos x=\frac{e^{ix}+e^{-ix}}2,\sin x=\frac{e^{ix}-e^{-ix}}{2i}$
So, $$\sin^6x\cos^2x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^6\left(\frac{e^{ix}+e^{-ix}}2\right)^2$$
$$=\frac{\left(e^{6ix}+e^{-6ix}-\binom61(e^{4ix}+e^{-4ix})+\binom62(e^{2ix}+e^{-2ix})-\binom63\right)}{-2^6}$$
$$\cdot\frac{\left(e^{2ix}+e^{-2ix}+2\right)}{2^2}$$
$$=\frac{e^{8ix}+e^{-8ix}-(6-2)(e^{6ix}+e^{-6ix})+(1+\binom62-2\cdot\binom61)(e^{4ix}+e^{-4ix})-(\binom63-2\cdot\binom62+\binom61)(e^{2ix}+e^{-2ix})+2\binom62-2\binom63}{-2^8}$$
$$=\frac{2\cos8x-4\cdot2\cos6x+4\cdot2\cos4x+4\cdot2\cos2x-10}{-256}\text{ as }e^{ix}+e^{-ix}=2\cos x$$
Now, simplify and use $\int\cos mxdx=\frac{\sin mx}m+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Finding an inflection point Can anyone help me find an inflection point for the following function without using graphing calculator.
Determine the intervals of concavity and points of inflection for the function
$f(x)=x\sqrt{25-x^2}$
for my first derivative I did
$f'(x)=1\sqrt{25-x^2}+(x)\frac{1}{2\sqrt{25-x^2}}(-2x)$
for my common denominator I got
$\frac{50-4x^2}{2\sqrt{25-x^2}}$
for my second derivative I got
$f''(x)=\frac{(-4x)\sqrt{25-x^2}-(25-2x^2)(1/2)\sqrt{25-x^2}(-2x)}{\sqrt{(25-x^2)^2}}$
But I am having trobule finding the inflection point.
| You have miscalculated your second derivative. First, I'd rewrite $$f'(x)=\sqrt{25-x^2}+\frac{-x^2}{\sqrt{25-x^2}}=\frac{25-2x^2}{\sqrt{25-x^2}},$$ from which we see that $$\begin{align}f''(x) &= \cfrac{(-4x)\sqrt{25-x^2}-(25-2x^2)\frac{-x}{\sqrt{25-x^2}}}{25-x^2}\\ &= \cfrac{(-4x)(25-x^2)-(25-2x^2)(-x)}{(25-x^2)\sqrt{25-x^2}}\\ &= \frac{-75x-2x^3}{(25-x^2)^\frac32}\\ &= \frac{x(x-5)(x+5)}{(25-x^2)^\frac32}.\end{align}$$ At which point(s), if any, does $f''(x)$ change signs? Such points are the inflection points. Note in particular that the denominator of $f''(x)$ is never negative--and always positive where $f''(x)$ is defined. The problem, then, amounts to finding the odd-multiplicity zeroes of the numerator of $f''(x)$ that are in the domain of $f''(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Partial Fraction Decomposition Problem... half answered... $$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$
I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:
$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{Cx+D}{x^2+3}$$
Multiplying by the LCD : $(x+3)^2(x^2+3)$
I am left with :
$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2$$
By setting $x=-3:B=-4$
Now is where I am running into trouble. Now that I can substitute B into the original decomposition equation, There is no value of x that will leave only one variable to solve for. Please lend me a hand you guys(and girls). Thanks!
| \begin{align}
5x^3+19x^2+27x-3 &= A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2
&\text{Let $x = -3$}\\
-48&=12B \\
B &= -4 & \text{Go back and let $B = -4$}\\
\hline
5x^3+19x^2+27x-3 &= A(x+3)(x^2+3)-4(x^2+3)+(Cx+D)(x+3)^2 \\
5x^3 + 23x^2 + 27x + 9 &= A(x+3)(x^2+3)+(Cx+D)(x+3)^2
&\text{Divide both sides by $(x+3)$}\\
\hline
5x^2 + 8x + 3 &= A(x^2+3)+(Cx+D)(x+3) &\text{Let $x=-3$} \\
24 &= 12A \\
A &= 2 &\text{Go back and let A = 2} \\
\hline
5x^2 + 8x + 3 &= 2(x^2+3)+(Cx+D)(x+3) \\
3x^2 + 8x - 3 &= (Cx+D)(x+3)
&\text{Divide both sides by $(x+3)$} \\
3x-1 &= Cx+D \\
C &= 3 \\
D &= -1
\end{align}
Computational note.
The quotient $(5x^3 + 23x^2 + 27x + 9) \div (x+3)$ and the value of
that quotient at $x=-3$ can be accomplished with synthetic division.
\begin{array}{r|rrrrr}
& 5 & 23 & 27 & 9 \\
-3 & 0 & -15 & -24 & -9 \\
\hline
& 5 & 8 & 3 \\
-3 & 0 & -15 & 21\\
\hline
& 5 & -7 & 24
\end{array}
Hence $(5x^3 + 23x^2 + 27x + 9) \div (x+3) = 5x^2+8x+3$ and the value of
$5x^2+8x+3$ when $x=-3$ is $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove or disprove inequality: $2a^2 + 2b^2 + 3c^2 \ge 16P$ Let $a,b,c$ be the lengths of the sides of a triangle with area $P$
prove or disprove inequality: $2a^2 + 2b^2 + 3c^2 \ge 16P$
|
$$2a^2 + 2b^2 + 3c^2 \overset{?}{\ge} 16P\tag{1}$$
Hint: It will be helpful to know Heron's formula that provides a formula for the area of a triangle with sides length $a, b, c$:
$$P = \sqrt{s(s-a)(s-b)(s-c)}$$
where $ s=\dfrac{a+b+c}{2}$. This gives us
$$\text{Area}\;=\;\;P = \frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}$$
where $a, b, c$ are the side lengths of a triangle
Substituting $P$ into your original inequality, you can work with proving/disproving the following inequality:
$$2a^2 + 2b^2 + 3c^2 \geq 4\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}\tag{2}$$
*
*Start with squaring each side of the inequality $(2)$. (This is fine to do, without needing to consider a change in the direction of the inequality, because both sides are non negative.)
$$(2a^2 + 2c^2 + 3c^2)^2 \geq 16(a^2 + b^2 + c^2)^2 - 32(a^4 + b^4 + c^4)\tag{3}$$
*If you need it, you also have the triangle inequality to work with to put additional restrictions on the inequality: $a + b\geq c,; b + c\geq a,\; a + c\geq b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Calculate:$\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx$ Calculate following integration
$$\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx$$
| We have
$$I=\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx=\int\frac{dx}{\left(\frac{x+1}{x+2}\right)^{3/4}(x+2)^2}$$
Now we pose $t=\frac{x+1}{x+2}$ then $x=\frac{1-2t}{t-1}$ and $dx=\frac{dt}{(t-1)^2}$, so
$$I=\int t^{-3/4}dt=4t^{1/4}+C,$$
hence
$$I=4\left(\frac{x+1}{x+2}\right)^{1/4}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplify : $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$? Just out of curiosity, is $$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},a \gt0\quad?$$
Thanks
| Take $a=8b^2,$ where $b >0.$
Now $$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = 2\sqrt\frac {8b^2}{2}=4b > 2\sqrt 2 b=\sqrt a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
The generating function for the Fibonacci numbers Prove that $$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...=\frac{1}{1-(z+z^2)}$$
The coefficients are Fibonacci numbers, i.e., the sequence $\left\{1,1,2,3,5,8,13,21,...\right\}$.
| A related technique. What you have is the ordinary generating function of Fibonacci numbers. Use the recurrence relation of the Fibonacci numbers
$$ F_{n+2} = F_{n+1} + F_{n} $$
to get the generating function. See here for a related problem.
Added: We will derive the ordinary generating function. Let $g(z)=\sum_{n=0}^{\infty} F_n z^n $, $F_0=F_1=1$, then
$$\sum_{n=0}^{\infty} F_{n+2} z^n = \sum_{n=0}^{\infty} F_{n+1} z^n + \sum_{n=0}^{\infty} F_{n} z^n $$
$$\implies \sum_{n=2}^{\infty} F_{n} z^{n-2} = \sum_{n=1}^{\infty} F_{n} z^{n-1} + g(z) $$
$$\implies \frac{1}{z^2}\sum_{n=2}^{\infty} F_{n} z^{n} =\frac{1}{z} \sum_{n=1}^{\infty} F_{n} z^{n} + g(z) $$
$$ \implies \frac{1}{z^2}\sum_{n=0}^{\infty} F_{n} z^{n}-\frac{F_0}{z^2}-\frac{F_1}{z}= \frac{1}{z}\sum_{n=0}^{\infty} F_{n} z^{n}-\frac{F_0}{z} + g(z) $$
$$ \implies \frac{g(z)}{z^2}-\frac{1}{z^2}-\frac{1}{z} = \frac{1}{z}g(z)-\frac{1}{z} + g(z) $$
$$ \implies g(z) = \frac{1}{1-(z+z^2)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
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"answer_id": 2
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$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $ Let be a circle inscribed in the triangle $\triangle ABC$ wiht the center $I$. The intersection of the circle with $AI$ is $A'$, with $BI$ is $B'$ and with $CI$ is $C'$.
Prove that:
$$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $$
thanks. seems to be to hard for me.
| $\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}} \\ =\dfrac{2\cos \left(\dfrac{π+C}{4}\right)}{\cos \left(\dfrac{A-B}{2}\right)-\cos \left(\dfrac{A+B}{2}\right)} \\ \ge \dfrac{2\cos \left(\dfrac{π+C}{4}\right)}{1-2\sin \left(\dfrac{C}{2}\right)} \\ =\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\left(\dfrac{1}{\sqrt{2}}\cos \left(\dfrac{C}{4}\right)-\dfrac{1}{\sqrt{2}}\sin \left(\dfrac{C}{4}\right)\right)^2} \\ =\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\left(\cos \left(\dfrac{π+C}{4}\right)\right)^2}=\dfrac{1}{\cos \left(\dfrac{π+C}{4}\right)}$
$f\left(x\right)=\dfrac{1}{\cos \left(\dfrac{π+x}{4}\right)}-4\sin \dfrac{x}{2},f''\left(x\right)=\sin \dfrac{x}{2}+\dfrac{1}{16\cos \dfrac{π+x}{4}}+\dfrac{\sin ^2 \dfrac{π+x}{4}}{8\cos ^3 \dfrac{π+x}{4}} > 0 \implies $
$f\left(x\right) $ is convex function $\implies \\ ∑_{\rm cyc} \left(\dfrac{1}{\cos \left(\dfrac{π+A}{4}\right)}-4\sin \dfrac{A}{2} \right)\ge 3 f\left(\dfrac{A+B+C}{3}\right)=3f\left(\dfrac{π}{3}\right)=0$
QED.
| {
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"url": "https://math.stackexchange.com/questions/342613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Maximum value of $Z$ How to find maximum value of $| Z| $ if: $$ \Big| Z-\dfrac{4}{Z} \Big|=2; $$
Where $Z$ is a complex mumber
| Setting $z=Z/2$ we have
$$
|Z-4/Z|=2 \iff |z-z^{-1}|=1.
$$
So the problem is to find
$$
R=2\max_{z \in M}|z| \ \text{ where }\ M:=\{z \in \mathbb{C}:\ |z-z^{-1}|=1\}.
$$
We have
$$
M=\{re^{i\theta}:\ r>0, \ -\pi \le \theta \le\pi,\ r^4-(1+2\cos\theta)r^2+1=0\}.
$$
Solving
$$
r^4-(1+2\cos\theta)r^2+1=0,\ r>0,\ -\pi \le \theta \le\pi,
$$
we obtain
\begin{eqnarray}
r_1^2(\theta)&=&\frac{1}{2}\left(1+2\cos\theta+\sqrt{(1+2\cos\theta)^2-1}\right),\ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2},\\
r_2^2(\theta)&=&\frac{1}{2}\left(1+2\cos\theta-\sqrt{(1+2\cos\theta)^2-1}\right),\ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}.
\end{eqnarray}
Since
$$
r_1^2(\theta)\ge r_2^2(\theta) \quad \forall \theta \in [-\pi/2,\pi/2],
$$
it follows that
$$
R=2\max_{-\pi/2\le \theta \le \pi/2}r_1(\theta)=\max_{1 \le t \le 3}\sqrt{2[t+\sqrt{t^2-1}]}=\sqrt{2(3+2\sqrt{2})}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.
I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this.
Thank you!
Sorry, my approach:
$$\begin{align}
\frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\
\sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\
\theta + 45^\circ &= 45^\circ,\ 135^\circ \\
\theta &= 0^\circ, \ 90^\circ
\end{align}$$
| Are you familiar with the unit circle? If so, then take that approach to see that
$\sin \theta + \cos \theta =1$ is only satisfied when either $\theta = 90^{\circ}$ or
$\theta = 0^{\circ}$, or equivalently, $\theta = 360^{\circ}$. This comes from the fact that the unit circle has points $(1,0)$ at $0^{\circ}$, and (0,1) at $90^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 3
} |
$5 \frac {3}{*} \times 3 \frac {*}{2}=19$? One of my friends gave me this apparently easy-looking problem which I do not know how to crack. The problem is to find the values of "*" where
$$5 \frac {3}{*} \times 3 \frac {*}{2}=19\text{ ?}$$
I can rearrange the problem as $5 \frac {3}{x} \times 3 \frac {y}{2}=19$ and I have to find the values of $x,y.$ Now $5 \frac {3}{x} \times 3 \frac {y}{2}=19 \implies \frac {5x+3}{x} \times \frac {6+y}{2}=19.$ Now, I do not know which way to go?
Can someone point me in the right direction? Thanks in advance for your time.
EDIT: Here "*"-s are not same. Infact I know the answer but do not know how to get it. Here $x=7,y=1.$
| On simplification, $y=\frac{8x-18}{5x+3}$
Now, $$8(5x+3)-5(8x-18)=114$$
So, $5x+3$ must divide $114$ and as $x>0, 5x+3>3$
Now the factors of $114$ are $1,2,3,6,19,38,57,114$
So, $5x+3$ can be $6,19,38,57,114$
$6\equiv1\pmod 5,19\equiv4, 38\equiv3,57\equiv2,114\equiv4$
So, $5x+3$ can be $38\implies x=7,y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
How find this maximum $f=\dfrac{8a^2-6ab+b^2}{4a^2-2ab+ac}$ The quadratic equation $ax^2+bx+c=0$ has two roots in the interval $[0,2]$,Find the maximum of
$$f=\dfrac{8a^2-6ab+b^2}{4a^2-2ab+ac}$$
my idea:we have
(1):if $a>0$,then let $g(x)=ax^2+bx+c$
$$\Delta=b^2-4ac>0,g(0)\ge 0,g(2)\ge 0$$
then we have $$b^2>4ac,c\ge0,4a+2b+c\ge0$$
(2):if $a<0$,then we have
$$b^2-4ac>0,c\le 0,4a+2b+c\le 0$$.
so I think this method is very ugly. can someone have nice methods? Thank you
by the @Yimin hint:we have
$x_{1}+x_{2}=-\dfrac{b}{a},x_{1}x_{2}=\dfrac{c}{a}$,then we have $$f=\dfrac{8a^2-6ab+b^2}{4a^2-2ab+ac}=\dfrac{8-\dfrac{b}{a}+\left(\dfrac{b}{a}\right)^2}{4-2\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{8+x_{1}+x_{2}+(x_{1}+x_{2})^2}{4+2(x_{1}+x_{2})+x_{1}x_{2}},0\le x_{1},x_{2}\le 2$$
| Suppose the roots are $x_1,x_2$, then represent the objective function with the roots.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Finding the indefinite integral $\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$ This is a homework question, I tried many subtitutions but nothing worked for me...
$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$
Any clue will help.
Thanks.
| Hint::
$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$
$$=\int\frac{\sqrt{x+8}(\sqrt{x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3}+\sqrt{x+3})}dx$$
$$=\int\frac{\sqrt{x+8}\sqrt{x-3}}{6}dx+\int\frac{\sqrt{x+8}\sqrt{x+3}}{6}dx$$
$$=\int\frac{\sqrt{x^2+5x-24}}{6}dx+\int\frac{\sqrt{x^2+5x+24}}{6}dx$$
$$=\frac{I_1+I_2}{6}$$
$$I_1 = \int\sqrt{x^2+5x-24}$$
$$I_1= \int\sqrt{\left(x+\frac{5}{2}\right)^2-\frac{121}{4}}dx$$
Substitute $u = \left(x+\frac{5}{2}\right)dx$
$$I_1= \int\sqrt{u^2-\frac{121}{4}}du$$
Substitute $\frac{11\sec\theta}{2} = u;du = \frac{11\sec\theta\tan \theta}{2}d\theta$
$$I_1= \frac{121}{4}\int\tan ^2\theta \sec \theta d\theta$$
$$I_1= \frac{121}{4}\left[\int\sec^3 \theta d\theta - \int \sec \theta d\theta\right]$$
Using reduction formula
$$\int{sec^m \theta} = \frac{\sin \theta \sec ^{m-1} \theta}{m-1} + \frac{m-2}{m-1}\int\sec^{m-2}\theta d\theta$$
Substitute and Solve
| {
"language": "en",
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"answer_id": 0
} |
Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
| $S_n = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$
| {
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"answer_id": 4
} |
Solve $x^2 \equiv 29\pmod {35}$ What are the solutions of the congruence $x^2 \equiv 29\pmod {35}$. Use congruence modulo $5$ and modulo $7$.
| Solving modulo the primes $\,5,7,\,$ then combining solutions by CRT (Chinese Remainder),
$\rm\quad x^2\equiv 4\ \ (mod\ 5)\:\Rightarrow\: 5\mid(x-2)(x+2)\:\Rightarrow\:5\mid x\!-\!2\ \ or\ \ 5\mid x\!+\!2\:\Rightarrow\:x\equiv \color{#0A0}2,\color{blue}{-2}\ \ (mod\ 5)$
$\rm\quad x^2\equiv 1\ \ (mod\ 7)\:\Rightarrow\: 7\mid(x-1)(x+1)\:\Rightarrow\:7\mid x\!-\!1\ \ or\ \ 7\mid x\!+\!1\:\Rightarrow\:x\equiv \color{#C00}1,-1\ \ (mod\ 7)$
Now combine solutions by CRT. If $\rm\:x\equiv a\pmod 5,\ \ x\equiv b\pmod 7\:$ then $\rm\: x = b + 7n,\:$ so $\rm\: mod\ 5\!:\ a \equiv x\equiv b + 7n\equiv b+2n\!\iff\! \color{#90f}{2n\equiv a-b}.$
Case $\rm1\!:\,\ a\equiv \color{#0A0}2,\ b\equiv \color{#C00}1,\,\ $ hence $\rm\,\ mod\ 5\!:\ \color{#90f}{2n\equiv a-b} \equiv 1\equiv 6\:\Rightarrow\:n\equiv 3,\ $ so $\rm\ n = 3+5k,\ $ so $\rm\ x = b+7n = 1+7(3+5k) = 22+35k.\:$
Case $\rm2\!:\,\ a\equiv \color{#0A0}2,\ b\equiv -1,\, $ hence $\rm\, mod\ 5\!:\ \color{#90f}{2n\equiv a-b} \equiv 3\equiv 8\:\Rightarrow\:n\equiv 4,\ $ so $\rm\ n = 4+5k,\ $ so $\rm\ x = b+7n = -1+7(4+5k) = 27+35k.\:$
The other $\,2\,$ cases $\rm\: (a,b)\equiv (\color{blue}{-2},\color{#C00}1),\ (\color{blue}{-2},-1)\:$ are solved similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
$c$.
I have no idea how to solve these types of problems. Can anybody help me please?
| We shall prove this by induction on $n$.
Base Case: For $n = 1$, we note that any $3$-digit integer with $3$ identical digits is divisible by $3$.
Since, for any $k \in \{1, \dots, 9\}$, $kkk = k \cdot (111)$. Further, $111$ is divisible by $3$. Therefore, $kkk$ is
divisible by $3$.
Hypothesis: Assume that the statement is true for $n = k$, $k \geq 1$.
Induction Step: For $n = k + 1$, $k \geq 1$. Let $x$ be an integer composed of $3^{k+1}$ identical digits.
We note that $x$ can be written as
$x = y \times z$
where $y$ is an integer composed of $3^k$ identical digits, and $z = 10^{2\cdot 3^k}
+ 10^{3^k}+ 1$.
For example, $x = 666666666 = 666 \times 1001001 = y × (10^{2\cdot 3}+ 10^3+ 1)$. $y$ is divisible by $3^k$ by the hypothesis and $z$ is divisible by $3$ (sum of the digits is divisible by $3$). Thus $x$ is divisible by
$3^{k+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Integrating $\int \sqrt{\frac{1+x}{x}}dx$ Integrating $\int\sqrt{\frac{1+x}{x}}dx$
Let, $x=\tan^{2}\theta$
$dx=2\tan\theta \sec^{2}\theta d\theta$
Integral = $\int \frac{\sec\theta}{\tan\theta}{2\tan\theta\sec^{2}\theta}d\theta$
Integral = $\int {2\sec^{3}\theta}d\theta$
| Let, $\sqrt\frac{x+1}{x}=t$
$\frac{x+1}{x}=t^{2}$
$1+\frac{1}{x}=t^{2}$
$\frac{1}{x}=t^{2}-1$
$\frac{1}{t^{2}-1}=x$
$dx=-\frac{2t}{(t^2-1)^{2}}dt$
Integral = -$\int\frac{2t^2}{(t^2-1)^{2}}dt$
Integral = -$\int\frac{2(t^2-1)+2}{(t^2-1)^{2}}dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{2}{(t^2-1)^{2}})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{2}{(t+1)^{2}(t-1)^{2}})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +2(\frac{1}{(t+1)(t-1)})^{2})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{1}{2}(\frac{1}{(t+1)}-\frac{1}{(t-1)})^{2})dt$
Integral = -$\int(\frac{2}{(t^2-1)} +\frac{1}{2}(\frac{1}{(t+1)^2}+\frac{1}{(t-1)^2}-\frac{2}{(t^2-1)}))dt$
Integral = -$\int(\frac{1}{(t^2-1)} +\frac{1}{2}(\frac{1}{(t+1)^2}+\frac{1}{(t-1)^2}))dt$
Integral = $\sin^{-1}{t}+\frac{t}{t^2-1}+c$
Integral = $\sin^{-1}{\sqrt{\frac{x+1}{x}}}+{\sqrt{x^2+x}}+c$
| {
"language": "en",
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Show that $x^4 +1$ is reducible over $\mathbb{Z}_{11}[x]$ and splits over $\mathbb{Z}_{17}[x]$. Reduction into linear factors $\mathbb{Z}_{17}[x]$:
This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so
$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv x^4+1$ mod 17.
Reduction over $\mathbb{Z}_{11}[x]$:
This one doesn't have such an easy solution, as neither of $y^2 \equiv -1$ mod 11 or $x^4 \equiv -1$ mod 11 have solutions.
I've tried $x^4+1 = (x^2- \sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ but $x^2 \equiv 2$ mod 11 also has no solutions so there's no easy substitution here.
I think that this approach is not going to work here, so I need something new. Any suggestions?
| An abstract algebra argument, without much computation.
Claim: The polynomial $x^4+1$ splits over $\mathbb F_{p^2}$ for any odd (*) prime $p$.
We can see this because $F_{p^2}^\times$ is a cyclic group of order $p^2-1$, a multiple of $8$. If $g$ is a generator, then $g^{(1+2k)\frac{p^2-1}{8}}$ gives distinct roots for $k=0,1,2,3$.
This means that $x^4+1$ can't be irreducible in $\mathbb Z_p$, because if it was, the splitting field would have $p^4$ elements.
(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.
Specific solutions:
We can make this explicit by looking at the "standard" complex $4$th roots of $-1$:
$$\pm\frac{\sqrt 2}2 \pm\frac{\sqrt{2}}{2}i=\frac{1}{2}(\pm \sqrt{2}\pm\sqrt{-2})$$
If $p\equiv 1\pmod 8$ then $2$ and $-2$ have square roots in $\mathbb Z_p$, so $x^4+1$ has four roots.
If $p\equiv -1\pmod 8$ then $2$ is a square, but $-2$ is not. Write $a^2\equiv 2\pmod p$. Then write:
$$x^4+1 = \left(x^2 -ax + 1\right)\left(x^2+ax+1\right)$$
If $p\equiv 3\pmod 8$ then $-2$ is a square and $2$ is not. Letting $b^2\equiv -2\pmod p$, we get:
$$x^4+1 = \left(x^2-bx-1\right)\left(x^2+bx-1\right)$$
Finally, if $p\equiv 5\pmod 8$ then neither of $\pm 2$ is a square, but $-1$ is a square. Letting $c^2\equiv -1\pmod p$, we see that the roots are $(\pm 1 \pm c)\sqrt{2}/2$ and the result is:
$$x^4+1 = (x^2-c)(x^2+c)$$
| {
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"source": "stackexchange",
"question_score": "5",
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Proving that $T(n) = 3T\left(\frac n3\right) + \sqrt n = \Theta(n)$
Show that $T(n)$ is bounded both above and below by $n$ (abusing the Big O notation) for some positive constants $c_1$ and $c_2$:
$$
T(n) = 3T\left(\frac n3\right) + \sqrt n = \Theta(n)
$$
Let's do some substutution for proving first that is $O(n)$. Because $T\left(\frac n3\right) \le c\frac n3$:
$$
\begin{align}
T(n) &= 3T\left(\frac n3\right) + \sqrt n = 3c\frac n3 + \sqrt n\\
&= cn + \sqrt n \le cn
\end{align}
$$
So $cn + \sqrt n \le cn$: where we are adding $\sqrt n$ on the left and pretending that we get something bigger on the right. I would say that $T(n) = \Omega(n)$ for $c \ge 1$, but not $T(n) = O(n)$!
I'm sure that I'm wrong (I have the solution sheet), can you help me find out why?
EDIT: actually I've found a lot of examples where induction does not work, like:
$$
T(n) = 2T\left(\frac n2\right) +1
$$
The technique to solve this (subtract a lower order term from the right side
of the inequality we wish to prove) is explained by Hagen von Eitzen answer. The question remains: why does induction not work in this case?
| Suppose the base three representation of $n$ is given by
$$ n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k$$
and we let the base case be $T(0) = 0.$
Then it is not difficult to see that
$$ T(n) = \sum_{j=0}^{\lfloor \log_3 n \rfloor} 3^j
\sqrt{\sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^{k-j}}.$$
Now to get an upper bound on this consider the case of all digits being two, giving
$$ T(n)\le \sum_{j=0}^{\lfloor \log_3 n \rfloor} 3^j
\sqrt{\sum_{k=j}^{\lfloor \log_3 n \rfloor} 2\times 3^{k-j}} =
\sum_{j=0}^{\lfloor \log_3 n \rfloor} 3^j
\sqrt{3^{\lfloor \log_3 n \rfloor -j+1}-1} \\ \le
\sqrt{3}^{\lfloor \log_3 n \rfloor+1}
\sum_{j=0}^{\lfloor \log_3 n \rfloor} 3^j \sqrt{3}^{-j} =
\sqrt{3}^{\lfloor \log_3 n \rfloor+1}
\sum_{j=0}^{\lfloor \log_3 n \rfloor}\sqrt{3}^j =
\sqrt{3}^{\lfloor \log_3 n \rfloor+1}
\frac{\sqrt{3}^{\lfloor \log_3 n \rfloor+1}-1}{\sqrt{3}-1}.$$
For a lower bound consider the case of all digits being zero except the first one, giving
$$ T(n)\ge \sum_{j=0}^{\lfloor \log_3 n \rfloor} 3^j
\sqrt{3}^{\lfloor \log_3 n \rfloor-j} =
\sqrt{3}^{\lfloor \log_3 n \rfloor}
\sum_{j=0}^{\lfloor \log_3 n \rfloor} 3^j \sqrt{3}^{-j} =
\sqrt{3}^{\lfloor \log_3 n \rfloor}
\sum_{j=0}^{\lfloor \log_3 n \rfloor} \sqrt{3}^j \\=
\sqrt{3}^{\lfloor \log_3 n \rfloor}
\frac{\sqrt{3}^{\lfloor \log_3 n \rfloor+1}-1}{\sqrt{3}-1}$$
Finally, taking the leading terms of the upper and the lower bound together, we have shown that
$$ T(n) \in \Theta\left(\sqrt{3}^{\lfloor \log_3 n \rfloor} \times \sqrt{3}^{\lfloor \log_3 n \rfloor}\right) = \Theta\left(3^{\lfloor \log_3 n \rfloor} \right) =
\Theta(n).$$
The next term in the asymptotic expansion is $\sqrt{3}^{\lfloor \log_3 n \rfloor}\sim\sqrt{n}.$
A better upper bound can be found by using the asymptotic expansion of $\sqrt{1-x}$ in a neighborhood of zero, using
$$ \sqrt{3^{\lfloor \log_3 n \rfloor -j+1}-1} \\=
\sqrt{3^{\lfloor \log_3 n \rfloor -j+1}}
\left(1 - \frac{1}{2} \left(\frac{1}{3^{\lfloor \log_3 n \rfloor -j+1}}\right)
- \frac{1}{8} \left(\frac{1}{3^{\lfloor \log_3 n \rfloor -j+1}}\right)^2
- \frac{1}{16} \left(\frac{1}{3^{\lfloor \log_3 n \rfloor -j+1}}\right)^3
- \cdots\right).$$
| {
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"url": "https://math.stackexchange.com/questions/353075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding surface area of a cone I will describe the problem then show what I tried to solve it.
I need to find the area of the cone defined as follows:
$$z^2=a^2(x^2+y^2)$$
$$0\leq z\leq bx+c$$
where $a,b,c>0$ and $b<a$.
For this I considered the parametrization $x=r\rm{cos}\theta$, $y=r\rm{sin}\theta$, $z=ar$, with $0\leq\theta\leq2\pi$. To find the interval which $r$ varies, I used the fact that $0\leq z\leq bx+c$, then $0\leq ar\Rightarrow 0\leq r$, and $ar\leq br\rm{cos}\theta+c\Rightarrow r\leq c/(a-b\rm{cos}\theta)$, therefore $0\leq r\leq c/(a-b\rm{cos}\theta)$.
The area of this surface is $$\int_0^{2\pi}\int_0^{\frac{c}{a-b\rm{cos}\theta}}\bigg\lvert(\rm{cos}\theta, \rm{sin}\theta, a)\times(-r\rm{sin}\theta,r\rm{cos}\theta, 0)\bigg\lvert dr \ d\theta=$$
$$=\int_0^{2\pi}\int_0^{\frac{c}{a-b\rm{cos}\theta}}r\sqrt{a^2+1} dr \ d\theta=\sqrt{a^2+1}\int_0^{2\pi}\bigg[\frac{r^2}{2}\bigg]_0^{\frac{c}{a-b\rm{cos}\theta}} \ d\theta=$$
$$=\sqrt{a^2+1}\int_0^{2\pi}\frac{c^2}{2(a-b\rm{cos}\theta)^2} \ d\theta=\frac{c^2\sqrt{a^2+1}}{2}\int_0^{2\pi}\frac{1}{(a-b\rm{cos}\theta)^2} \ d\theta .$$
Well, this last integral is quite difficult, in fact i can't integrate this, and wolframaplha showed me a terrible solution. Now I think I've made some mistake but don't know where. Any help is very welcome. Thanks.
PS: I tried to find something like a^2(x^2+y^2) or a^2\cdot(x^2+y^2) in the search but that was useless. Later I did some tests and verified that this search is not that good for TeX search, is there some better way to better searching when using TeX language? Again, thanks.
| I have not looked at the complete problem in detail yet, but you did mention difficulty with the integral
$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2}$$
Evaluation of this integral is not too bad using the Residue theorem. Let $z=e^{i \theta}$, $d\theta = dz/(i z)$, $\cos{\theta} = (z + z^{-1})/2$, and the integral becomes
$$\oint_{|z|=1} \frac{dz}{i z} \frac{1}{(a-b(z+z^{-1})/2)^2} = -i \frac{4}{b^2} \oint_{|z|=1} dz \: \frac{z}{(z^2 - 2 (a/b) z + 1)^2}$$
By the residue theorem, this integral is $i 2 \pi$ times the sum of the residues of the poles inside the unit circle. The poles of the integrand are at
$$z_{\pm} = \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b} \right )^2 - 1}$$
Note that only $z_-$ is inside the unit circle, so we need only evaluate the residue at that pole to get the value of the integral. Note also that this is a double pole, so the residue calculation looks like
$$\begin{align}\text{Res}_{z=z_-} \frac{z}{(z^2 - 2 (a/b) z + 1)^2} &= \lim_{z \rightarrow z_-} \frac{d}{dz} \left [(z-z_-)^2 \frac{z}{(z^2 - 2 (a/b) z + 1)^2} \right ] \\ &= \left[\frac{d}{dz} \frac{z}{(z-z_+)^2}\right]_{z=z_-} \\ &= \frac{1}{(z_- -z_+)^2} - \frac{2 z_-}{(z_--z_+)^3}\\ &= \frac{z_++z_-}{(z_+-z_-)^3}\\ &=\frac{2 (a/b)}{(2 \sqrt{(a/b)^2-1})^3}\\ &= \frac{1}{4} \frac{(a/b)}{[(a/b)^2-1]^{3/2}}\end{align}$$
We then multiply this residue by $(i 2 \pi)(-i 4/b^2)$ and the result is
$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2} = 2 \pi \, a \, (a^2-b^2)^{-3/2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing a function is differentiable using definition of derivative Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $f(x,y)=(x^2-xy, x+y^2)$. Use the definition of the derivative of a function to show that $f$ is differentiable at the point $p=(1,-1)$.
My attempt is as follows:
Definition of derivative: $\lim\limits_{x \to p}\frac{|f(x)-f(p)-df_p(x-p)|}{|x-p|}$. So,
$$\lim\limits_{(x,y)\to(1,-1)}\frac{|f((x,y))-f((1,-1))-df|_{(1,-1)}((x,y)-(1,-1)}{|(x,y)-(1,-1)|}.$$
Where $df|_{(1,-1)}=\begin{bmatrix}
3 & -1 \\
1 & -2
\end{bmatrix}.$
Then, we have:
$$\lim\limits_{(x,y) \to (1,-1)} \frac{|<x^2-xy, x+y^2>-<2,2>-<3x-y-4, x-2y-3>|}{|<x-1, y+1>|}$$
Simplified a bit:
$$\lim\limits_{(x,y) \to (1,-1)} \frac{|<x^2-xy-3x+y+2, x+y^2-x+2y+1>|}{|<x-1, y+1>|} $$
$$\lim\limits_{(x,y) \to (1,-1)} \frac{\sqrt{(x^2-xy-3x+y+2)^2+(x+y^2-x+2y+1)^2}}{\sqrt{(x-1)^2+(y+1)^2}}$$
I still noticed that the bottom goes to zero. Is there something that I am messing up computationally?
| It's easier to show that
$$
\lim_{h \to 0}\frac{\|f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\|}{\|h\|} = 0.
$$
Indeed, for $h = (h_1,h_2) \in \mathbb{R}^2$,
$$
f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\\
= ((1+h_1)^2-(1+h_1)(-1+h_2),(1+h_1)+(-1+h_2)^2) - (1^2-1(-1),1+(-1)^2) - (3h_1 -h_2,h_1-2h_2)\\
= (h_1^2 -h_1 h_2+3h_1-h_2+2,h_2^2+h_1-2h_2+2) -(2,2)-(3h_1-h_2,h_1-2h_2)\\
= (h_1^2-h_1h_2,h_2^2) = (h_1(h_1-h_2),h_2^2),
$$
so that
$$
\|f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\|^2 = \|(h_1(h_1-h_2)h_2,h_2^2)\|^2\\
= h_1^2(h_1-h_2)^2 + h_2^4 = h_1^2(h_1^2-2h_1h_2+h_2^2) + h_2^4\\
\leq h_1^2(h_1^2 + (h_1^2+h_2^2)+h_2^2) + h_2^4\\
= 2h_1^4 + 2h_1^2h_2^2 + h_2^4 \leq 2(h_1^2+h_2^2)^2 = 2\|h\|^4,
$$
and hence
$$
\lim_{h \to 0}\frac{\|f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\|}{\|h\|}\\
\leq \lim_{h \to 0} \frac{\sqrt{2}\|h\|^2}{\|h\|} = \sqrt{2}\lim_{h\to 0}\|h\| = 0,
$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the improper integral?
| Another method!
$$t=\frac{1}{1+x^3}:$$
$$\begin{aligned}\int_0^{\infty} \frac{dx}{1+x^3}&=\frac{1}{3}\int_0^1 \frac{1}{t}\left(\frac{1}{t}-1\right)^{-2/3} dt\\[7pt]&=\frac{1}{3}\int_0^1 t^{-1/3}\left(1-t\right)^{-2/3}\,dt\\[7pt]&= \frac{1}{3}\text{B}\left(\frac{2}{3},\frac{1}{3}\right)=\frac{1}{3}\Gamma\left(1-\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)\\[7pt]&=\frac{\pi}{3}\csc\frac{\pi}{3}\\[7pt]&=\frac{2\pi}{3\sqrt{3}}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Area of a Regular Polygon I was trying to find the area of a regular polygon in terms of n, the side length and s, the number of sides.
Because there are $s$ sides number of isosceles triangles in a regular polygon, I decided to work out the area of an isosceles triangle in terms of $A$, the unique angle and $a$, the unique side:
$$Area=\frac{1}{2}ab\sin{C}$$
$$b=\frac{\sin{C}\times a}{\sin{A}}$$
(Sine rule, $C=B$)
$$b=\frac{\sin{\frac{180-A}{2}}\times a}{\sin{A}}$$
$$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$
$$C=\frac{180-A}{2}=90-\frac{A}{2}$$
$$Area=\frac{{(\sin({90-\frac{A}{2}})\times{a})}^{2}}{\sin{A}}$$
And that was where I got to in finding the area of an isosceles triangle. Then I tried to find the area of the whole regular polygon:
$$Area=s\times\frac{{(\sin({90-\frac{A}{2})}\times{n})}^{2}}{\sin{A}}$$
Where $s$ is the number of sides and $n$ replaces $a$
$$A=\frac{360}{s}$$
$$Area=s\times\frac{{(\sin({90-\frac{\frac{360}{s}}{2}})\times{n})}^{2}}{\sin{\frac{360}{s}}}=s\times\frac{\sin^{2}({90-\frac{180}{s}})\times{n}^{2}}{\sin{\frac{360}{s}}}$$
$$Area=\frac{s{n}^{2}\sin^{2}({90-\frac{180}{s}})}{\sin{\frac{360}{s}}}$$
However, when I tested this formula it was wrong. Can someone tell me where I've gone wrong?
|
Here's a kind of different approach:
Let $n$ be number of sides.
$\angle DAC= \dfrac{180}{n}$
$\tan \angle DAC=\dfrac{x/2}{AD} \implies AD= \dfrac{\frac{x}{2}}{\tan \frac{180}{n}} \implies AD=\dfrac{x}{2 \tan \frac{180}{n}}$
Call $\dfrac{180}{n}=p$
Area of $\triangle ABC= \dfrac{x^2}{4 \tan p}$, you will have $n$ triangles in $n$-sided regular polygon. Area of polygon=$\dfrac{x^2 \times n}{4 \tan p}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\frac{1}{1-x}$ series expansion How do I know that the expression:
$$\frac{1}{1-x}$$
Is equal to the infinite sum:
$$-\left(\frac{1}{x}\right)-\left(\frac{1}{x}\right)^2-\left(\frac{1}{x}\right)^3-\left(\frac{1}{x}\right)^4+...$$
Thanks!
| \begin{align}
\frac{1}{1-x}
=
&
\frac{1}{x\left[\frac{1}{x}-1\right]}
\\
=
&
-\left(\frac{1}{x}\right)\frac{1}{1-\left(\frac{1}{x}\right)}
\\
=
&
-\left(\frac{1}{x}\right)\left[ 1+\left(\frac{1}{x}\right)^1+\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3+\left(\frac{1}{x}\right)^4+\cdots\right]
\\
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$ I need to factor the following:$$\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \left(\dfrac{3}{4}z -\dfrac{5}{3}y\right)^3 - \left(\dfrac{3}{4}z + \dfrac{2}{3}x\right)^3$$A friend of mine suggested that $(a + b)^3 +(c - b)^3 - (c + b)^3 = 3(a+b)(a+c)(b +a)$, and it's right. But this is just a normal exam question in 9th grade... so it must have a “normal” way to be done. I tried factoring $(a+b)^3 + (c - b)^3$ by using $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ but all hell breaks loose.
| You have $\left(\dfrac{2}{3}x + \dfrac{5}{3}y\right)^3 + \left(\dfrac{3}{4}z -\dfrac{5}{3}y\right)^3 - \left(\dfrac{3}{4}z + \dfrac{2}{3}x\right)^3$, which (as you observed) is of the form $(a+b)^3 + (c-b)^3 - (c+a)^3$.
A short cut is to simply observe that if $a = -b$ or $-c$, the expression is $0$. Similarly, you should be able to easily observe that $b=c$ also leads to the expression being $0$. This gives you $(a+b)(a+c)(b-c)$ as factors immediately. Hence
$(a+b)^3 + (c-b)^3 - (c+a)^3 = k(a+b)(a+c)(b-c)$, for some scalar $k$.
To find $k$, you could compare coefficients of some power or test a suitable value. For e.g. let $a = 0, b = 1, c =-1$. Then LHS $= 1 - 8 + 1 = -6,$ and RHS $= -2k$, so $k=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
expressing $\int tan^{-1}tdt$ as a power series I have a question which asks me to state $\int_0^x\tan^{-1}(t)dt$ as a power series in $x$ and then use that result to show that $\frac{\pi}{4}-\log\sqrt{2}= 1-1/2-1/3+1/4+1/5--++\cdots$
For the first part:
$$\tan^{-1}t= \int\sum_{n=0}^\infty(-1)^nx^{2n}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\cdots$$6
then integrating from $0$ to $x$ we get
$$\int_0^x\tan^{-1}(t)dt= \frac{x^2}{2}-\frac{x^4}{3\cdot4}+\frac{x^5}{5\cdot6}-\frac{x^8}{7\cdot8}\cdots=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{n(n-1)}$$
But how can I use this to show the second part?
| You can use the integration-by-parts formula to get the second:
\begin{eqnarray*}
\int_0^x\tan^{-1}tdt&=&t\tan^{-1}t\big|_0^x-\int_0^x\frac{t}{t^2+1}dt\\
&=&t\tan^{-1}t\big|_0^x-\frac{1}{2}\ln(t^2+1)\big|_0^x\\
&=&x\tan^{-1}x-\frac{1}{2}\ln(x^2+1).
\end{eqnarray*}
So
$$ \int_0^1\tan^{-1}tdt=\frac{\pi}{4}-\frac{1}{2}\ln 2=\cdots. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/363918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding the residue at $z=0$ for complex function $1/(z^2\sin z$) Find the residue at $z=0$ for complex function $1/(z^2\sin z$)
I know $z=0$ is a pole of order 3 but can't seem to calculate the residue value for it.
| You can reference the result for a pole of order $3$ at $z=0$:
$$\text{Res}_{z=0} \frac{1}{z^2 \sin{z}} = \frac{1}{2!} \lim_{z \rightarrow 0} \left[\frac{d^2}{dz^2} \frac{z}{\sin{z}}\right]$$
$$\frac{d}{dz} \frac{z}{\sin{z}} = \frac{\sin{z} - z \cos{z}}{\sin^2{z}} $$
$$\begin{align}\frac{d^2}{dz^2} \frac{z}{\sin{z}} &= z \left(\csc ^3{z}+\cot ^2{z} \csc{z}\right)-2 \cot{z} \csc{z}\\ &=\left (\frac{1}{z^3} + \frac{1}{2} + \frac{1}{z^3} - \frac{1}{2}\right ) - \frac{2}{z^2} + \frac{1}{3} + O(z)\\ &= \frac{1}{3}\end{align}$$
Therefore the residue is $1/6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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How to do this integration? I'm trying to integrate $$\frac{8x^2+3x+1}{x(2x+1)^2}$$
I did a partial fraction expansion: $$\frac{8x^2+3x+1}{x(2x+1)^2}= \frac{1}{x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}$$
and now I'm left with $$\int\left(\frac{1}{x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}\right)dx$$
I would get $\ln(x) + \ln(2x+1)$ from integrating $\frac{1}{x}+\frac{2}{2x+1}$ but I do not quite know how to integrate
$$ \frac{3}{(2x+1)^2} $$
It would be great if someone can teach me what steps I need to take to integrate $\frac{3}{(2x+1)^2}$, thanks.
| Substitute $u=2x+1 \implies du=2dx $. Then the integral is $$\int\frac{3}{2}\frac{1}{u^2}du=\frac{-3}{2u}+C'=\frac{-3}{4x+2}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
| You can work in $\mathbb{Z}[\omega],$ where $\omega = e^{\frac{2 \pi i}{3}}.$
There are two essential cases to deal with. In both cases, you can easily reduce to considering the case that $x,y,z$ are pairwise coprime ( you can afterwards multiply $x$ and $y$ by $d^{2}$ and $z$ by $d^{3}$ for some integer $d$). The case when $3$ does not divide $z$ is easier.
In that case, you can assume that $x + y = a^{2}$ for some integer $a$ and that $x + \omega y = (b + \omega c)^{2}$ for integers $b$ and $c.$ Then you can see that
$x^{3} + y^{3} = (a(b^{2}-bc +c^{2}))^{2}.$ Note that the second equation gives
$x = b^{2}-c^{2}$ and $y = 2bc - c^{2}$ since $1 + \omega + \omega^{2} = 0.$
Hence we additionally need the integers $a,b,c$ to satisfy $a^{2} + 3c^{2} = (b+c)^{2},$ but this is easy to arrange.
In the (still pairwise coprime case that $3$ divides $z,$ you must have $x + y = 3^{2m-1}a^{2}$ for some integer $a$ coprime to $3$, and $x + \omega y = (1- \omega)(b + \omega c)^{2}$ for some integers $b$ and $c$ with $b+c$ not divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 12,
"answer_id": 8
} |
Series $\sum \frac{\sin(n)}{n} \cdot \left(1+\cdots +\frac{1}{n}\right)$ convergence question I want to test the convergence of the series. $$ \sum_{n=1}^{\infty} \frac{\sin(n)}{n} \cdot \left(1+\frac{1}{2} + \cdots + \frac{1}{n}\right)$$ My guess is this should diverge, and the below I provide the details
Note that
*
*$\displaystyle 1+\frac{1}{2} + \cdots + \frac{1}{n} \geq 1 +\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-1}} = 2 \cdot \left(1-\frac{1}{2^{n}}\right)$
*So from above I conclude that $$\sum_{n=1}^{\infty} \frac{\sin(n)}{n} \cdot \left(1+\frac{1}{2}+\cdots + \frac{1}{n}\right) \geq 2 \cdot \sum_{n=1}^{\infty} \frac{\sin(n)}{n} - \cdot \underbrace{\sum_{n=1}^{\infty} \frac{\sin(n)}{n \cdot 2^{n}}}_{A}$$
*Now $A$ converges and I think $\displaystyle \sum \frac{\sin(n)}{n}$ diverges hence my given series should diverge. Am I correct in my reasoning?
| The series is convergent and, for fun, I'll provide a derivation.
Let's first rewrite your sum as ($H_n$ is the $n$-th harmonic number) :
$$\tag{1}S:=\sum_{n=1}^{\infty} H_n\frac{\sin(n)}{n}=\Im\left(\sum_{n=1}^{\infty} H_n\frac{e^{in}}{n}\right)$$
Since $H_n\sim \log(n)\,$ as $\,n\to\infty$ (because the 'harmonic integral' corresponding to the arithmetic sum is the logarithm)
To evaluate $S$ let's use generating functions and suppose that the derivative of $f$ is defined by :
$$\tag{2}f'(x):=\sum_{n=1}^{\infty} H_n\;x^n=-\frac 1{1-x}\ln(1-x)$$
(expand $\frac 1{1-x}=1+x+x^2+\cdots\ $ and $\ -\ln(1-x)=x+\frac {x^2}2+\frac{x^3}3+\cdots$ and multiply !)
Integrating $(2)$ we get :
$$\tag{3}f(x)=\sum_{n=1}^{\infty} H_n \frac{x^{n+1}}{n+1}=-\int\frac{\ln(1-x)}{1-x} dx=\frac 12(\ln(1-x))^2$$
so that (using $\;H_n=H_{n+1}-\frac 1{n+1}\;$ and setting $\,k:=n+1\;$) :
$$\sum_{k=2}^{\infty} H_k \frac{x^k}k-\sum_{k=2}^{\infty} \frac{x^k}{k^2}=\sum_{k=1}^{\infty} H_k \frac{x^k}k-\sum_{k=1}^{\infty} \frac{x^k}{k^2}=\frac 12(\ln(1-x))^2$$
$\operatorname{Li}_2(x):=\sum_{k=1}^{\infty} \frac{x^k}{k^2}$ is the dilogarithm but we prefer to use directly :
$$\tag{4}\sum_{k=1}^{\infty} H_k \frac{x^k}k=\sum_{k=1}^{\infty} \frac{x^k}{k^2}+\frac 12(\ln(1-x))^2$$
For $x=e^i$ in $(1)$ we get following expression of $S$ :
\begin{align}
S&=\Im\left(\sum_{k=1}^{\infty} H_k \frac{e^{ik}}k\right)\\
S&=\sum_{k=1}^{\infty} \frac{\sin(k)}{k^2}+\frac 12\Im\left(\ln\bigl(1-e^i\bigr)\right)^2\\
S&=\boxed{\displaystyle\operatorname{Cl}_2(1)-\ln\left(2\sin\left(\frac 12\right)\right)\frac{\pi-1}2}\\
S&\approx 1.05895346485231034922735
\end{align}
with $\operatorname{Cl}_2$ the Clausen function : $\displaystyle\operatorname{Cl}_2(x)=-\int_0^x \ln\left(2\sin\left(\frac t2\right)\right)dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/370402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Parametrization of a curve in polar coordinates I'm trying to change this parametrics equations to polar coordinates
$$ X(t) = 2\cos(t) - \sin(2t) \\
Y(t) = 2\sin(t) - \cos(2t) $$
What i tryed to do was raise the two equations squared, sum then and make some algebric manipulations.
$$ X^2 = (2\cos(t) - \sin(2t) )^2 = 4\cos^2(t) -4\cos(t)\sin(2t) + \sin^2(2t) \\
Y^2 = (2\sin(t) - \cos(2t))^2 = 4\sin^2(t) -4\sin(t)\cos(2t) + \cos^2(2t) \\
X^2 + Y^2 = 4(\sin(t) + \cos(t))^2 -4(\sin(t)\cos(2t) +\sin(2t)\cos(t))+(\sin(2t)+\cos(2t))^2
\\ \to X^2 + Y^2 = 5 - 4[ \sin(t)(2\cos^2(t) -1) +2\sin(t)\cos^2(t) ]
\\ \to X^2 + Y^2 = 5 - 4\sin(t)(4(\cos(t))^2 -1)
$$
For last, we can obtain:
$$ X^2 + Y^2 = 5 - 4\sin(3t) $$
Considering that $R = \sqrt{X^2 + Y^2} $ and $ \theta(t) = \arctan(\frac{Y}{X}) $ ,
what i can do to replace the right hand side of the equation for polar?
EDIT:
Using the help of the Lord_Farin, I derivated the main equation and now i'm trying to found a relation between $\frac{X}{Y} $
and $\sin(3t) $ but i don't see a simplification in my equations.
$$
\frac{d}{dt} (X^2 + Y^2) = \frac{d}{dt} (5 -4\sin(3t)) \\
2\dot{X}X + 2\dot{Y}Y = -4(3\cos(3t)) \to \\
2\dot{X}\frac{X}{Y} + \frac{2\dot{Y}Y}{Y} = \frac{-12\cos(3t)}{Y} \to \\
\frac{X}{Y} = \left( \frac {-12\cos(3t)}{Y} -2\dot{Y} \right).\frac{1}{2\dot{X}}
$$
where $$
\dot{X} = -2\sin(t) -2\cos(2t) \\
\dot{Y}= +2\cos(t) +2\sin(2t)$$
so
$$ \frac{X}{Y} = \left( \frac {-12\cos(3t)} {2\cos(t) +2\sin(2t)} -2(2\cos(t) +2\sin(2t)) \right).\frac{1}{2(-2\sin(t) -2\cos(2t))} $$
I manipulated the values and didn't found nothing that could be replaced by $\sin(3t)$. I would be grateful if someone find a relationship.
| You may manipulate the relations $$x(t) = 2\cos(t) - \sin(2t) \\
y(t) = 2\sin(t) - \cos(2t)$$ many times to get the right answer, but plotting it by Maple:
and having the whole shape in my mind I could find the right connecting relation here. Note that we need a proper transformation to make the plot skew.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Equation with Logarithm I want to solve the following equation:
$$3^x3^{x-1} = 243.$$
My approach is the following:
$3^{2x-1} = 243$ then:
$(2x-1)\cdot\log3 = \log 243$ and then:
$x = (\frac{\log243}{\log3}+1)/2$
Is this correct?
| This is correct. Alternatively you could use the base $3$ logarithm, in which case you obtain $$x = \frac{1}{2}(\log_3 243+1).$$ Now note that $243 = 3^5$ so this reduces to $x = \frac{1}{2}(5 + 1) = 3$.
If instead you use your expression for $x$, then you have $$x = \frac{1}{2}\left(\frac{\log 243}{\log 3} + 1\right) = \frac{1}{2}\left(\frac{\log 3^5}{\log 3} + 1\right) = \frac{1}{2}\left(\frac{5\log 3}{\log 3} + 1\right) = \frac{1}{2}(5 + 1) = 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{x \to +\infty} \left(1+\frac{\cos x}{2\sqrt{x}}\right)$ Let $f(x)=x+\sin(\sqrt{x})$. I want to find $\lim_{x \to +\infty} f'(x)$.
Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \rightarrow \infty$, hence $\frac{1}{2\sqrt{x}} \rightarrow 0$. Then $1 + \frac{1}{2\sqrt{x}} \rightarrow 1$. Therefore by the Sandwich Theorem $f'(x) \rightarrow 1$.
Lemma
$\lim_{x \to \infty} g(x)=l$ if and only if $\lim_{n \to \infty} g(x_{n})=l$ for all sequences $(x_{n}) \subset E$ with $\lim_{n \to \infty} x_{n} = \infty$, where $E$ is the domain of $g$.
Attempt 2
$f'(x)=1+\frac{\cos x}{2\sqrt{x}}$. Now let $g(x)=f'(x)$ and $x_{n}=n^2$. Then $\lim_{n \to \infty} (x_{n})=\infty$. We have $g(x_{n})=1+\frac{\cos(n)}{2n}$. Then as $n \rightarrow \infty$, $g(x_{n}) \rightarrow 1.$ Therefore by the Lemma above, $\lim_{x \to +\infty} g(x)=1$.
Question
Are the attempts above correct?
Thank you for your time.
Edited Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ For all $x \geq 0$, $1+\frac{\cos x}{2\sqrt{x}} \geq 0$. So $$0 \leq 1+\frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \rightarrow \infty$, hence $\frac{1}{2\sqrt{x}} \rightarrow 0$. Then $1 + \frac{1}{2\sqrt{x}} \rightarrow 1$. Therefore by the Sandwich Theorem $f'(x) \rightarrow 1$.
Question
Is this correct? Also, I would like to know if it is necessary to show that $$1-\frac{1}{2\sqrt{x}} \leq 1 + \frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}} \tag{1}$$
instead of $$0 \leq 1+\frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}}. \tag{2}$$
In my attempt to show inequality $(1)$, I got as far as $$1-\left|\frac{\cos x}{2\sqrt{x}}\right| \le \left|1 -\left(-\frac{\cos x}{2\sqrt{x}}\right)\right| \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ Could you please help me show that $$1-\frac{1}{2\sqrt{x}} \leq 1 + \frac{\cos x}{2\sqrt{x}}.$$
Thank you.
| I might be wrong here, but why not just use the squeeze lemma by using upper and lower bounds on $\cos x$? Obviously
$$
1= \lim_{x \to \infty} (1-\frac{1}{2 \sqrt{x}}) \leq L \leq \lim_{x \to \infty} (1 + \frac{1}{2 \sqrt{x}}) = 1
$$
Hence $L=1$
| {
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What is $(26+15(3)^{1/2})^{1/3}+(26-15(3)^{1/2})^{1/3}$? $$(26+15\cdot\sqrt3)^{1/3}+(26-15\cdot\sqrt3)^{1/3}$$
I'm trying to get the result of this number. Through some algebra I found that it is close to $52^{1/3}$. Through some observation I found that it is a root of this cubic equation $x^3-3x-52=0$ and I found that the only real solution of that eq. is $4$ so now I know that the number I'm looking for is $4$ (close to my first try $52^{1/3}$). My question is, is there any algebraic process to get from the original expression and simplify it to $4$?
Thank you all very much in advance.
| $26 + 15 \sqrt{3} = 8 + 12 \sqrt{3} + 18 + 3 \sqrt{3} = 2^3 + 3 \cdot 2^2 \sqrt{3} + 3 \cdot 2 \sqrt{3}^2 + \sqrt{3}^3 = (2 + \sqrt{3})^3.$
$26 - 15 \sqrt{3} = 8 - 12 \sqrt{3} + 1 8- 3 \sqrt{3} = 2^3 - 3 \cdot 2^2 \sqrt{3} + 3 \cdot 2 \sqrt{3}^2 - \sqrt{3}^3 = (2 - \sqrt{3})^3.$
$(26 + 15 \sqrt{3})^{\frac{1}{3}} + (26 - 15 \sqrt{3})^{\frac{1}{3}} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral Question - $\int\frac{1}{x^2-6x}\,\mathrm dx$ How I can evaluate the indefinite integral? :
$$\int\frac{1}{x^2-6x}\,\mathrm dx$$
Do I need to bring it to this format? : $\displaystyle \int\frac{1}{x^2-a^2}\,\mathrm dx\;$?
Thanks!
| $\int\frac{1}{x^2-a^2}\,\mathrm dx=\frac{1}{2a}\int\frac{1}{x-a}-\frac{1}{x+a}dx=\frac{1}{2a}(ln(x-a) -ln(x+a))$
and
about $\int \frac{1}{x^2-6x}=\frac{-1}{6}\int \frac{1}{x}-\frac{1}{x-6}dx=\frac{-1}{6}(lnx-ln (x-6))$
| {
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Integral question - $\int\frac{(4-x)\,dx}{x^2+4x+8}$ Integral question - $$\int\frac{(4-x)\,dx}{x^2+4x+8}$$
To solve it I need to bring the numerator to be the derivative of the dominator right?
I need to do the trick that not change the integral any ideas?
$$\frac{1}{2}\int\frac{(2x+4) \, dx}{x^2+4x+8}$$
Thanks
| We have
$$4-x=-\frac{1}{2}(2x+4)+6$$
and
$$x^2+4x+8=(x+2)^2+4=4((\frac{1}{2}(x+2))^2+1)=4(u^2+1)$$
hence
$$\int\frac{(4-x)\,dx}{x^2+4x+8}=-\frac{1}{2}\int\frac{(2x+4)}{x^2+4x+8}dx+6\int\frac{dx}{x^2+4x+8}\\=-\frac{1}{2}\log|2x^2+4x+8|+{3}\int\frac{du}{u^2+1}=-\frac{1}{2}\log|2x^2+4x+8|+3\arctan(\frac{x+2}{2})+C$$
| {
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Does there always exist an odd number of elements? Given a nonzero integer $k$, does there always exist a positive integer $n$ such that there are exactly an odd number of elements $i\in\{0,1,...,n-1\}$ with $\frac{2^n-1}4 < 2^ik \mod{2^n-1} < \frac{3(2^n-1)}4$? Here $a\mod b\in\{0,1,...,b-1\}$.
| Taking $k = 2$ seems to be an exception. For $n \geq 3$ we have the following:
*
*For $0 \leq i < n-3$ we have $2^i k = 2^{i+1} < 2^{n-3} < \frac{1}{4}(2^n - 1)$ and hence $2^i k$ is not in the desired range.
*For $i = n-3$ we have $2^i k = 2^{n-2} = \frac{1}{4}2^n > \frac{1}{4}(2^n - 1)$ which is just in the desired range.
*For $i = n-2$ we have $2^i k = 2^{n-1} = \frac{1}{2}2^n$ which is also in the desired range.
*For $i = n-1$ we have $2^i k = 2^n \equiv 1$ which is not in the desired range.
So for any $n \geq 3$, the number of elements $i \in \{0, \ldots, n-1\}$ satisfying $$\frac 14 (2^n-1) < \left(2^ik \mod{2^n-1}\right) < \frac 34 (2^n-1)$$ is exactly $2$, which is even.
For $n = 2$ we have a lower bound of $3/4 < 1$ and upper bound of $9/4 > 2$, in which case both $i = 0$ and $i = 1$ satisfy the equation. So again, the number of elements is even.
Finally, for $n = 1$ we have a lower bound of $1/4$ and upper bound of $3/4$, and no integer is in this range. So again, the number of elements satisfying the requirement is even.
Concluding, $k = 2$ seems to be a counterexample.
| {
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Limits with sums and integrals It's one of my homework exercises that is rather problematic to me. Apparently the last thing to do is to squeeze it but I don't see yet how to do that. Could you help?
$$\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^{n}\frac{2n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2-x+1}dx}{\displaystyle\sum_{k=1}^{n}\frac{n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2+x+1}dx}$$
| We can prove easily by change of variable that
$$2\int_0^1\frac{dx}{x^2+x+1}=\int_0^1\frac{dx}{x^2-x+1}=\frac{2\sqrt{3}\pi}{9}=2\alpha$$
so we have
$$\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^{n}\frac{2n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2-x+1}dx}{\displaystyle\sum_{k=1}^{n}\frac{n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2+x+1}dx}=\lim_{n\to\infty}\frac{2S_n-2\alpha}{S_n-\alpha}=2$$
Added
\begin{align}\int_0^1\frac{dx}{x^2+x+1}&=\int_0^1\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}=\frac{4}{3}\int_0^1\frac{dx}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1}=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^\sqrt{3}\frac{du}{u^2+1}\\&=\frac{2}{\sqrt{3}}\left[\arctan u\right]_{\frac{1}{\sqrt{3}}}^\sqrt{3}=\frac{\pi\sqrt{3}}{9}\end{align}
We can calculate the other integral by the same method.
| {
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Integrating the arithmetic series to get the sum of first $n$ squares? I only learnt today that series can be differentiated and integrated and I wondered if it is possible to apply the techniques of differential calculus to the finite arithmetic series
$$\sum_{x=0}^n x = n(n+1)/2$$
to obtain the sum of the first $n$ squares:
$$\sum_{x=0}^n x^2 = \frac{n(n+1)(2n+1)}{6}$$
| Let's say we have, $$f(x)=1+x+x^2+x^3+\cdots+x^n$$
$$\frac{x^{n+1}-1}{x-1} = 1+x+x^2+x^3+\cdots+x^n$$
Differentiating,
$$\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} = 1+2x+3x^2+\cdots+nx^{n-1}$$
Multiplying both sides by $x$
$$\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2} = x+2x^2+3x^3+\cdots+nx^{n}$$
Differentiating, again
$$\frac{(-2n^2-2n+1)x^{n+1} +n^2x^{n+2} +(n+1)^2x^n-x-1 }{(x-1)^3} = 1+2^2x+3^2x^2+\cdots+n^2x^{n-1}$$
Then take the limit at $x\to1$,
and we're done.
Disclaimer: This is not at all advisable, and was done just for curiosity. I had to use WA for the differentiation, as I lack the patience
| {
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Can you prove if $5^n \mid r$ and $r$ odd then $5^{n+1} \mid 2^r+3^r$? I've looked around at this, but been unable to find it. After generating prime factorizations in Maple of $2^r+3^r$ for $r>0$ it seems that this holds. It's easy enough to prove case by case. The case $n=0$ is trivial since for $r$ odd $x+y|x^r+y^r$ (I'm pretty sure it works by factoring by grouping).
For $n=1$ suppose $r>0$ is odd and divisible by 5. Then $r=20k+5$ or $r=20k+15$ for some $k\in\mathbf{N}$.
Case 1: $2^{20k+5}+3^{20k+5}=2^5(2^k)^{20}+3^5(3^k)^{20}\equiv 7(1)+18(1)$ (mod 25) $\equiv$ 0 (mod 25)
Case2: $2^{20k+15}+3^{20k+15}=2^{15}(2^k)^{20}+3^{15}(3^k)^{20}\equiv 18(1)+7(1)$ (mod 25) $\equiv$ 0 (mod 25)
This works nicely because 20 is $\phi(25)$. In fact for any power of 5, $\phi(5^{n+1})=5^n(4)$ so $r$ can be broken into two cases $5^{n}(4)k+5^{n}$ or $5^nk(4)+3(5^n)$. Since a power of 2 or 3 can't be congruent to a power of 5, raising $2^k$ and $3^k$ to $\phi(5^{n+1})$ gives 1 modulo $5^{n+1}$. But here's is where I'm stuck, showing that $2^{5^n}+3^{5^n}\equiv$ $0$ (mod $5^{n+1}$). Or maybe I'm just approaching it the wrong way? It seems that by looking at Maple data this holds for any $x$, $y$, and $x+y$ where $x+y$ is odd.
| Here's another proof.
Lemma: If $a\equiv b \pmod {p^k}$ then $a^p \equiv b^p \pmod {p^{k+1}}$.
Proof: Suppose $a\equiv b \pmod {p^k}$. Then $a = b + np^k$ whence we have
$$a^p = \sum_{i=0}^p\binom{p}{i}b^{p-i}n^ip^{ki}$$
Taking the above modulo $p^{k+1}$, all summands with $i>1$ are eliminated by the term $p^{ki}$. The term with $i=1$ is eliminated since $p^{k+1}\mid \binom{p}{1}p^k$ leaving only the $i=0$ term where we have $$a^p\equiv b^p\pmod{p^{k+1}}$$
which shows the result. $\square$
Now note that since $r$ is odd and $5^n\mid r$ we must have $r=(4k\pm 1)5^n$. Then
$$2^r + 3^r \equiv 2^{4\cdot 5^n \pm 5^n} + 3^{4\cdot 5^n \pm 5^n} \equiv 2^{\pm 5^n}+3^{\pm 5^n}\pmod{5^{n+1}}$$
Note that negative powers in this case simply denotes the multiplicative inverse.
Now $3\equiv -2 \pmod 5$ which implies $3^5 \equiv (-2)^5 \equiv -2^5 \pmod {5^2}$ and inductively,
$$3^{5^n} \equiv -2^{5^n}\pmod {5^{n+1}}$$
and similarly
$$3^{-5^n} \equiv -2^{-5^n}\pmod {5^{n+1}}$$
which is the desired result.
| {
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Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$ How do I simplify $\sqrt{(4+2\sqrt{3})}+\sqrt{(4-2\sqrt{3})}$?
I've tried to make it $x$ and square both sides but I got something extremely complicated and it didn't look right.
I got $2\sqrt{3}$ on wolframalpha, but I'm not sure how is it possible?
Help would be appreciated! Thanks!
| \begin{align}
&\ \ \ \sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}
\\
\\
&=\sqrt{(\sqrt{3}+1)^2}+\sqrt{(\sqrt{3}-1)^2}\
\\
\\
&=\sqrt{3}+1+\sqrt{3}-1
\\
\\
&=\boxed{2\sqrt{3}}
\end{align}
| {
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Calculate the integral, with the help of Euler's integral Calculate the follwing integral:
$$\int_0^1 \sqrt{\frac{1-x^2}{x+2x^3+x^5}}dx$$
| Hint:
$$
x+2x^3+x^5 = x(1+x^2)^2
$$
Now find a substitution based on this that makes the integral look more like the Beta function.
Substitute $x=\sqrt{u}$, giving
$$
\int_0^1 \sqrt{\frac{1-u}{(1+u)^2}}\frac{du}{2u^{3/4}} = \frac12\int_0^1 \frac{\sqrt{1-u}}{(1+u)u^{3/4}}du
$$
You will notice that this is already quite close to being in the form of the Beta function. From here, you can use a taylor series on $\frac1{1+u}$ (which will converge correctly for all points in the integral) to get a sum of beta functions. This is how one would calculate it using Euler's Beta function.
It can be integrated directly to obtain a beta function, but this requires some more clever work. Let $\frac{1-x^2}{x}=2A^2$, so $x=\frac{\sqrt{A^4+4}-A^2}{2}$ and $\frac{dx}{1+x^2} = -\frac{2A}{A^4+4}dA$
$$
\int_0^1 \sqrt{\frac{1-x^2}{x}}\frac{dx}{1+x^2} = \sqrt{2}\int_0^\infty \frac{A^2}{A^4+1}dA
$$
Now, let $A^4=B$, so $A=\sqrt[4]{B}$ and $dA=\frac{dB}{4B^{3/4}}$, giving
$$
\sqrt{2}\int_0^\infty \frac{1}{B+1}\frac{dB}{4B^{1/4}}
$$
Now, let $B=\frac1C-1$, do $dB=-\frac{dC}{C^2}$. This gives
\begin{align}
\sqrt{2}\int_0^1 \frac{C}{4(\frac1C-1)^{1/4}}\frac{dC}{C^2}&=\sqrt{2}\int_0^1\frac{1}{4(1-C)^{1/4}C^{3/4}}dC\\
&=\frac{1}{2^{3/2}}B(\frac14,\frac34)
\end{align}
And if you look into it, $B(\frac14,\frac34)=\sqrt{2}\pi$. Therefore, the integral is equal to $\frac\pi2$ (which can be verified by numerical integration).
Note: the overall substitution works out to be $\frac{1-x^2}{x}=2\sqrt{\frac{1-C}C}$, or $x=\frac{1-\sqrt{1-C}}{\sqrt{C}}$.
| {
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Find the rank of the following matrix. $A= \left[ \begin{array}{ccc}
3 & -1 & 2 \\
-6 & 2 & 4 \\
-3 & 1 & 2 \end{array} \right]$
Applying, $R_{3}-\frac{1}{2}R_{2}$
~ $A= \left[ \begin{array}{ccc}
3 & -1 & 2 \\
-6 & 2 & 4 \\
0 & 0 & 0 \end{array} \right]$
Applying, $R_{2}+2R_{1}$
~ $A= \left[ \begin{array}{ccc}
3 & -1 & 2 \\
0 & 0 & 8 \\
0 & 0 & 0 \end{array} \right]$
| Row reduce the matrix and count the nonzero rows that remain.
| {
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Radius of convergence of a power series with Bernoulli numbers Say, we use the definition:
Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$
and then derive power series representations of the functions:
\begin{align}
\\ x \coth(x) &= \sum_{n=0}^\infty \frac{2^{2n}}{(2n)!}B_{2n}x^{2n}
\\ x \cot(x) &= \sum_{n=0}^\infty (-1)^n\frac{2^{2n}}{(2n)!}B_{2n}x^{2n}
\\ \tan(x) &= \sum_{n=1}^\infty (-1)^{n-1}\frac{2^{2n}(2^{2n}-1)}{(2n)!}B_{2n}x^{2n-1}
\end{align}
If we take for granted that $\frac{x}{e^x-1}$ is analytic in $x_0=0$, can we conclude the following?
*
*$x \coth(x) = \frac{x}{2} + \frac{x}{e^x-1}$, therefore $x \coth(x)$ is also analytic in $x_0=0$ (as a sum of analytic functions).
*$x \cot(x) = (ix)\coth(ix)$, therefore $x \cot(x)$ is also analytic in $x_0=0$ (as a composition of analytic functions).
*$\tan(x) = \cot(x) - 2\cot(2x)$, therefore $x \tan(x)$ is also analytic in $x_0=0$ (as a sum of analytic functions).
This (if correct) gives us that the convergence radiuses are greater than zero.
But what else can be said about their radiuses of convergence? How can we calculate them? The Bernoulli numbers scare me away from applying the usual root/ratio tests.
| The radius of convergence of the first shown power series centered at the origin is the distance to the nearest singularity in the complex plane, namely at $\pm 2\pi i$. So radius of convergence is $2 \pi$.
Note that the denominator of $x/(e^x -1)$ gives a simple pole at every nonzero integer multiple of $2 \pi i$ owing to periodicity of the exponential function.
We'd arrive at the same radius of convergence by applying the ratio test, using an asymptotic approximation for the Bernoulli numbers:
$$ |B_{2k}| \approx 4 \sqrt{\pi k} (\frac{k}{\pi e})^{2k} $$
Since all the odd index Bernoulli numbers after the first are zero:
$$ \sum_{k=0}^\infty \frac{B_k x^k}{k!} = \frac{-x}{2}
+ \sum_{k=0}^\infty \frac{B_{2k} x^{2k}}{(2k)!} $$
we should apply the ratio test to consecutive even (nonzero) terms of the series:
$$ \lim_{k \rightarrow \infty} |\frac{B_{2k+2}}{B_{2k}} \frac{x^2}{(2k+2)(2k+1)}| \lt 1 $$
Replacing $B_{2k},B_{2k+2}$ by the asymptotic approximations gives:
$$ |\frac{B_{2k+2}}{B_{2k}}| \approx \sqrt{1+\frac{1}{k}} (1+\frac{1}{k})^{2k}
\frac{(k+1)^2}{(\pi e)^2} $$
Combining this approximation with the rest of the ratio test expression and taking the limit leads to:
$$ \lim_{k \rightarrow \infty} \sqrt{1+\frac{1}{k}} (1+\frac{1}{k})^{2k}
\frac{(k+1)^2}{(2k+2)(2k+1)} \frac{x^2}{(\pi e)^2} = e^2 \cdot \frac{1}{4}
\cdot \frac{x^2}{\pi^2 e^2} = \frac{x^2}{4\pi^2} $$
This ratio test limit is less than one precisely when $|x| \lt 2\pi$, and that again gives the radius of convergence as $2\pi$.
| {
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How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
| I'll go step by step.
$${1 - \cos A \over \sin A} + {\sin A \over 1 - \cos A}$$
$${(1 - \cos A)^2 \over \sin A (1 - \cos A)} + {\sin^2 A \over \sin A(1 - \cos A)}$$
$$(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A)$$
Expanding $(1 - \cos A)^2$ yields:
$$1 - 2\cos A + \cos^2 A + \sin^2 A \over \sin A(1 - \cos A)$$
Knowing that $\cos^2 x + \sin^2 x = 1$, we can change the expression to the following:
$$1 - 2 \cos A + 1 \over \sin A(1 - \cos A)$$
$$2 - 2 \cos A \over \sin A(1 - \cos A)$$
We factorize using $2$:
$$2(1 - \cos A) \over \sin A(1 - \cos A)$$
$$2 \over \sin A$$
$$2 \left ({1 \over \sin A} \right)$$
Finally:
$$ 2 \csc A$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the following integral: Find $$\int \sqrt{\tan x}dx$$
My attempt:
$$\text{Let}\ I=\int \sqrt{\tan(x)}dx$$
$$\text{Let}\ u=\tan(x), du=(1+\tan^{2}(x))dx$$
$$I=\int \frac{\sqrt{u}}{u^{2}+1}$$
$$\text{Let}\ v=\sqrt{u}, dv=\frac{du}{2\sqrt{u}}$$
$$I=2\int \frac{v^{2}}{v^{4}+1}$$
$$\int_0^\infty\frac{x^2}{1+x^4}dx$$
$$\text{Let}\ t=\frac{1}{v} \therefore dt=\frac{-dv}{v^2}$$
$$\therefore I=\int \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$
$$I=-\int \frac{dt}{1+t^4}$$
Where do I go from here?
| First let us compute the following $$\int(\sqrt{\tan x}+\sqrt{\cot x}) dx=\int\frac{\sin x +\cos x}{\sqrt{\sin x\cdot \cos x}}dx=\sqrt 2\int\frac{d(\sin x - \cos x)}{\sqrt{1-(\sin x -\cos x)^2}}$$ which is same as the $$\sqrt 2\int\frac{dz}{\sqrt{1-z^2}}=\sqrt 2\sin ^{-1}z+c$$ Again compute $$\int(\sqrt{\tan x}-\sqrt{\cot x})dx=\int\frac{\sin x -\cos x}{\sqrt{\sin x\cdot \cos x}}dx=\sqrt 2\int\frac{-d(\sin x+\cos x)}{\sqrt{(\sin x+\cos x)^2 -1} } dx$$ and this is same as $$-\sqrt 2\int \frac {dw}{\sqrt{w^2 -1}}=-\sqrt 2\int\frac{\sec u\tan u}{\tan u} du=-\sqrt 2\ln(\sec u +\tan u)+C$$ where $w=\sec u$. Now add both the integrals to obtain the result.
| {
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A sum with binomial coefficients Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$
| Suppose we seek to evaluate
$$\sum_{k=0}^n {n\choose k} (-1)^k (n-2k)^{n+2}.$$
Introduce
$$(n-2k)^{n+2}
= \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+3}} \exp((n-2k)z) \; dz.$$
We thus get for the sum
$$\frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+3}}
\sum_{k=0}^n {n\choose k} (-1)^k \exp((n-2k)z) \; dz
\\ = \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+3}} \exp(nz)
\sum_{k=0}^n {n\choose k} (-1)^k \exp(-2kz) \; dz
\\ = \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+3}} \exp(nz)
\left(1-\exp(-2z)\right)^n \; dz.$$
Categorizing the contributions to the residue according to the term
from the exponential raised to the power $n$ and using the series
expansion
$$1-\exp(-2z) = 2z - 2z^2 + \frac{4}{3}z^3 - \cdots$$
we get
First, with $z^n$, $$\frac{1}{2} n^2 \times 2^n.$$
Second, with $z^{n+1}$, $$n \times - {n\choose 1} 2^n.$$
Third, with $z^{n+2}$, $$1 \times {n\choose 2} 2^n
+ 1 \times {n\choose 1} 2^{n-1} \times \frac{4}{3}.$$
Collecting these yields
$$\left(\frac{1}{2} n^2 - n^2 + \frac{1}{2} n(n-1)
+ \frac{2}{3} n\right) \times 2^n.$$
This is
$$\left(\frac{2}{3}-\frac{1}{2}\right)n\times 2^n,$$
for a final answer of
$$\frac{1}{6} n \times 2^n \times (n+2)!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
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Absolute values in $\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$ in my math class we were given a list of indefinite integrals, and one of them was:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$$
My working:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}=\int \frac{dx}{(x+2)\sqrt{(x+2)^2-1}}$$
Then I used the substitution $x+2=\sec t$ to get:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int \frac{\tan t}{|\tan t|}dt= t\,\text{sgn}\, (\tan t)+C...$$
Then I checked the answer sheet, and this is what they did:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int dt=t+C=\text{arcsec}(x+2)+C$$
What I don't understand is, why are they allowed to say $\sqrt{\sec^2 t-1}=\tan t?$ I tried to put some values in and I have found that:
$$\int_{\sec \left(\frac{8}{5}\right)-2}^{\sec \left(\frac{9}{5}\right)-2} \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}<0$$ but according to the answer sheet I would get $\dfrac{1}{5}$
My answer looks wrong, I would be happy if someone could explain what the problem is, and also why we are allowed to simplify like they did.
| The answer of the book looks wrong. Since the principal range of $\sec x$ is:
*
*$\left[ 0; \frac{\pi}{2} \right)$ if $x \ge 1$, and on this domain, $\tan t$ is positive.
*$\left( \frac{\pi}{2} ; \pi \right]$ if $x \le -1$, whereas on this domain, $\tan t$ is negative.
So, in fact, the solution to that integral should be split in to 2 different parts:
$$\int\limits \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}} = \left\{ \begin{array}{ll} -\mbox{arcsec}(x + 2) + C_1 &, \mbox{for }x < -3 \\
\mbox{arcsec}(x + 2) + C_2 &, \mbox{for }x > -1 \end{array} \right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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} |
$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$? According to an example in my text book:
$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$$
I don't see how this works. The closest I can get is:
$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2\sqrt{x}}{\sqrt{x}(\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1})} = \frac{2}{\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1}}$$
Which is slightly off. But I'm not even sure if I calculated that correctly. What am I missing, what's the way to think?
| You have
$$
\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}} = \sqrt{x(1 + \frac{1}{\sqrt{x}})} + \sqrt{x( 1 - \frac{1}{\sqrt{x}})}
$$
It looks like you only factored out a $\sqrt{x}$ from last term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$ I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$
But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$
I ran out of imagination as what to try now
| $$\int_0^\infty \frac{dx}{x^4+1} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^4+1}$$
The latter integral is trivial by means of contour integration.
Let $C$ be the canonical semicircle contour, along the real axis from $-R$ to $-R$ and around the semicircle $Re^{i\theta}$ for $\theta \in [0,\pi]$. Letting $R \to \infty$, we consider the function on the arc ($|z| = R$):
$$\left|\frac{1}{x^4+1}\right|\le \frac{1}{|x^4+1|} \le \frac{1}{|x^4|-1} \le \frac{2}{R^4} \to 0$$
so by the estimation lemma, the integral around the arc disappears.
$$\oint_C \frac{dz}{z^4+1} = \int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i \sum \operatorname*{Res}\frac{1}{z^4+1}$$
where the residues are of poles in the upper half plane. These poles are $z_1=e^{i\pi/4}$ and $z_2=e^{3i\pi/4}$. It follows that
$$b_1=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{i\pi/4}$$
$$b_2=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{3 i\pi/4}$$
Then
$$\int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i (b_1+b_2) = \frac{\pi}{\sqrt{2}}$$
and finally
$$\int_0^\infty \frac{dx}{x^4+1} = \frac{\pi}{2\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even I want to know for what valuyes this function is even
I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!
| $$f(x)=f(-x)\\
x^6+ax^3-2x^3-2x^2+1=x^6-ax^3+2x^3-2x^2+1\\
(a-2)x^3=(2-a)x^3
$$
So for what values of $a$ is this equality satisfied?
$$a-2=2-a\iff 2a=4\iff a=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solve equation $\sqrt{4t + 1} = 3 - 3t$ Solve equation $\sqrt{4t + 1} = 3-3t$
→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$
→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$
I'm stuck at that point. Can someone tell me what I am doing wrong?
| For starters, squaring $\sqrt{4t+1}=3-3t$ gives $4t+1=9-18t+9t^2$. A little rearranging and we have $9t^2-22t+8=0$.
With a bit of insight we can see that this is $(9x-4)(x-2)$.
However, I'm pretty crap at factoring so lets use the quadratic formula:
$$ax^2+bx+c=0$$ implies
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
So we take $$x=\frac{-(-22)\pm\sqrt{22^2-4(9)(8)}}{2(9)}$$ now step-by-step we have
$$x=\frac{22\pm\sqrt{484-288}}{18}=\frac{22\pm\sqrt{196}}{18}=\frac{22\pm14}{18}
=\frac{11\pm7}{9}$$
So for "$+$" we have $x=\frac{18}{9}=2$ and for "$-$" we have $x=\frac 4 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Complete the square for $f(x) = 2x^2 + 4x - 6$ I'm studying for a math test. This is the question:
$f(x) = 2x^2 + 4x - 6$. complete the square.
This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$
But I get stuck here. Can someone complete it and explain?
I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$
| Given:
$f(x) = 2x^2 + 4x - 6$. complete the square.
$f(x)=2x^2+4x-6$
$f(x)=2x^2+4x-6 $
$f(x)=2(x^2+2x)-6$
$f(x)=2[x^2+\dfrac{2x}{2}+(1)^2]-6$
$f(x)=[2(x+1)^2 -1]-6$
$f(x)=2(x+1)^2-8$ which is the same as: $f(x) = 2x^2 + 4x - 6$
This can be confirmed by using a graphing calculator such as: https://www.desmos.com/
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Integrate $\int {{{\left( {\cot x - \tan x} \right)}^2}dx} $ $\eqalign{
& \int {{{\left( {\cot x - \tan x} \right)}^2}dx} \cr
& = {\int {\left( {{{\cos x} \over {\sin x}} - {{\sin x} \over {\cos x}}} \right)} ^2}dx \cr
& = {\int {\left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {\sin x\cos x}}} \right)} ^2}dx \cr
& = \int {{{\left( {{{\cos 2x} \over {{1 \over 2}\sin 2x}}} \right)}^2}dx} \cr
& = \int {{{\left( {2\cot 2x} \right)}^2}} \cr
& = \int {4{{\cot }^2}2xdx} \cr
& = \int {4\left( {{{\csc }^2}2x - 1} \right)dx} \cr
& = \int {\left(4{{\csc }^2}2x - 4\right)dx} \cr
& = 4 \times {{ - 1} \over 2}\cot 2x - 4x + C \cr
& = - 2\cot 2x - 4x + C \cr} $
Where have I gone wrong? I've tried to spot an error so many times yet I can't find it, I need another pair of eyes.. Thanks.
| Why do you think that you've gone wrong? Your answer is correct.
Edit. To show why the solution is the same as the one in your answers, we just need to show that $\tan x-\cot x=-2\cot 2x$.
$$
\begin{align*}
&\tan x-\cot x+2\cot 2x \\
&=\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}+2\frac{\cos 2x}{\sin 2x} \\
&=\frac{2\cos x\cos 2x\sin x-\cos^2 x\sin 2x+\sin^2 x\sin 2x}{\cos x\sin x\sin 2x} \\
&=\frac{2\cos x(\cos^2 x-\sin^2 x)\sin x-\cos^2 x(2\sin x\cos x)+\sin^2 x(2\sin x\cos x)}{\cos x\sin x\sin 2x} \\
&=\frac{2\cos^3 x\sin x-2\cos x\sin^3 x-2\sin x\cos^3 x+2\cos x\sin^3 x}{\cos x\sin x\sin 2x} \\
&=0
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evalutate $\int\frac{dx}{x\sqrt{1-\frac{4}{3}x^4}}$ How to integrate
$$\int\frac{dx}{x\sqrt{1-\frac{4}{3}x^4}}$$
Mathematica found
$$\ln x-\frac{1}{2}\ln\left(1+\sqrt{1-\frac{4}{3}x^4}\right)$$
but I can't find a method to arrive at this solution.
| Try $u=\sqrt{1-(4/3)x^4}$, $u^2=1-(4/3)x^4$, $2u\,du=-(16/3)x^3\,dx$, together with $dx/x=x^3\,dx/x^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Probability assignment There are two boxes. In the first one there are, 4 black and 3 white balls, and in the second one there are 3 black and 5 white balls. We take two balls from the first box, and one from the second one and we put then in a third box. If we take out one ball from the third box, what is the probability that the ball we took out is white?
Can someone help me with this?
| Let $A$ be the event that a white ball is taken from box 3. Also let $X_k$ be the event that $k$ white balls are in box 3.
Using the law of total probability we get,
$P(A) = \sum_{k=0}^3P(A|X_k)P(X_k)$.
$P(A|X_k)$ are easy to calculate. If there are NO white balls in box 3, that is $k=0$, then $P(A|X_0) = 0$. If there are 1 white balls in box 3 ($k=1$), then $P(A|X_1)=1/3$. Similar argument shows that $P(A|X_2) = 2/3$ and $P(A|X_3) = 1$.
Next we calculate $P(X_k)$.
$P(X_3)$ = $P$(2 white balls from box 1 AND 1 white ball from box 2). Since taking ball(s) from box 1 and box 2 are independent events we have
$P$(2 white balls from box 1 AND 1 white ball from box 2) = $P$(2 white balls from box 1)$P$(1 white ball from box 2)
$P$(2 white balls from box 1) = $\frac{3\choose2}{7\choose2}$
$P$(1 white ball from box 2) = $5/8$.
Hence $P(X_3) = \frac{3\choose2}{7\choose2}\frac{5}{8} = \frac{5}{56}$
Similar arguement shows that
$P(X_2) = \frac{3\choose2}{7\choose2}\frac{3}{8} + \frac{{3\choose1}{4\choose1}}{7\choose2}\frac{5}{8} = \frac{23}{56}$.
$P(X_1)= \frac{4\choose2}{7\choose2}\frac{5}{8} + \frac{{3\choose1}{4\choose1}}{7\choose2}\frac{3}{8} = \frac{22}{56}$.
We do not have to calculate $P(X_0)$ since it will be cancelled out by $P(A|X_0)$ which equals $0$.
Finally, putting it all together,
$P(A) = 0 + \frac{1}{3}\frac{22}{56} + \frac{2}{3}\frac{23}{56} + \frac{5}{56}$
which is approximately $0.49404$
| {
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"timestamp": "2023-03-29T00:00:00",
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How is "n+n/2+n/4....1" equal to "2n-1" using the formula for geometric series? I never knew not having good knowledge of basic maths will be so crippling!! So please help me out this time. I'll be working on my maths from today on.
I was discussing about complexity of an algorithm on StackOverflow and I was told that the series $n+n/2+n/4 + \dots + 1$ evaluates to $2n-1$ and I was linked to the following formula on Wikipedia:
Even after trying hard, I regret to say I still don't get it how using this formula I can conclude that my series evaluates to $2n-1$. Please help me out as I am sure it will take only a few seconds for you.
https://stackoverflow.com/questions/16748454/complexity-for-nested-loops-diving-by-2?noredirect=1#comment24124682_16748454
| The formula provided on the wikipedia page works if you know how many terms you are adding. For example, if you want to calculate the sum $1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \frac{1}{2^9}$, you know you are adding 10 terms (so $n = 10$). Also, $a = 1$ (the number you begin with) and $r = \frac{1}{2}$ (the common ratio). Now you could use the provided formula to calculate the sum:
\begin{align*}
a \frac{1-r^n}{1-r} &= 1 \frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}}\\
&= \frac{1 - \frac{1}{2^{10}}}{\frac{1}{2}}\\
&= 2(1 - \frac{1}{2^{10}})\\
&= 2 - \frac{2}{2^{10}}\\
&= 2 - \frac{1}{2^9}
\end{align*}
You are talking abouth the sum $a + \frac{a}{2} + \frac{a}{4} + \dots + 1$, so the last term is 1, instead of $ar^{n-1}$. Note that $r = \frac{1}{2}$ in your case, as each term is twice as small as the term before. The formula could still solve this problem, but you will have to set $a\left(\frac{1}{2}\right)^{n-1}$ equal to 1:
\begin{align*}
a\left(\frac{1}{2}\right)^{n-1} &= 1\\
a &= \frac{1}{\left(\frac{1}{2}\right)^{n-1}} = \frac{1}{\frac{1}{2^{n-1}}} = 2^{n-1}
\end{align*}
So, your formula will only work when $a$ is a power of two. For example, if $a = 4$:
$$
4 + 2 + 1 = 7 = 8 - 1 = 2*4 - 1
$$
So for $a=4$, your formula works. Let's try $a = 16$:
$$
16 + 8 + 4 + 2 + 1 = 31 = 32 - 1 = 2 * 16 - 1
$$
Again, your formula works. How would we prove this? The formula you found on wikipedia has a $r^n$ term that we would like to get rid of:
\begin{align*}
a\left(\frac{1}{2}\right)^{n-1} &= 1\\
\left(\frac{1}{2}\right)^{n-1} &= \frac{1}{a}\\
\left(\frac{1}{2}\right)^n &= \frac{1}{a} * \frac{1}{2} = \frac{1}{2a}
\end{align*}
So, $r = \frac{1}{2}$ and $r^n = \left(\frac{1}{2}\right)^n = \frac{1}{2a}$. Now, let's plug that in:
\begin{align*}
a \frac{1-r^n}{1-r} &= a \frac{1 - \frac{1}{2a}}{1 - \frac{1}{2}}\\
&= a \frac{1 - \frac{1}{2a}}{\frac{1}{2}}\\
&= 2a(1 - \frac{1}{2a})\\
&= 2a - \frac{2a}{2a}\\
&= 2a - 1
\end{align*}
So there you have it: for every $a = 2^k$ (with $k \in \mathbb{N}$):
$$
a + \frac{a}{2} + \frac{a}{4} + \dots + 1 = 2a - 1
$$
| {
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"url": "https://math.stackexchange.com/questions/401937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Radical equation $x- 5\sqrt{x} + 4 = 0$ Solve: Radical equation $x- 5\sqrt{x} + 4 = 0$
I think I understand what I had did wrong. Instead of squaring a product of a number I was squaring the equation.
| $$x-5\sqrt{x}+4=0$$
$$x+4=5\sqrt{x}$$
$$x^2+16+8x=25x$$
$$x^2-17x+16=0$$
$$x_{1,2}=\frac{17\pm\sqrt{17^2-16\cdot 4}}{2}$$
$$x_1=16\qquad;\qquad x_2=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$ What options do I have for this series? No idea how to do it.
$$\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$$
| Express the sum as follows:
$$\sum_{k=1}^{\infty} \frac{\cos^2{k}}{4 k^2-1} = \frac12 \sum_{k=1}^{\infty} \frac{1}{4 k^2-1} + \frac12 \sum_{k=1}^{\infty} \frac{\cos{2 k}}{4 k^2-1}$$
Now, the first sum on the RHS is equal to $1/2$. This may be shown using residue theory very easily:
$$\sum_{k=-\infty}^{\infty} \frac{1}{4 k^2-1} = -\text{Res}_{z=\pm 1/2} \frac{\pi \cot{\pi z}}{4 z^2-1} = 0$$
which means that
$$2 \sum_{k=1}^{\infty} \frac{1}{4 k^2-1} - 1 = 0$$
For the other sum, break into partial fractions and reorganize. You end up with
$$\begin{align}\sum_{k=1}^{\infty} \frac{\cos{2 k}}{4 k^2-1} &= \cos{2} + \sum_{k=1}^{\infty} \frac{\cos{2 (k+1)}-\cos{2 k}}{2 k+1}\\ &= \cos{2} - \sin{1} \sum_{k=1}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1}\end{align}$$
That last sum may be derived from the well-known sum
$$\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} = \frac{\pi}{4}$$
Therefore, I get as the sum
$$\begin{align}\sum_{k=1}^{\infty} \frac{\cos^2{k}}{4 k^2-1} &= \frac14 + \frac14 \left (\cos{2} - \frac{\pi}{4}\sin{1} + 2 \sin^2{1} \right )\\ &= \frac{1}{2} - \frac{\pi}{8} \sin{1}\end{align}$$
You may verify that this checks out numerically in Mathematica or WA.
ADDENDUM
The "well-known" sum may be evaluated by considering the series expansion of the function $\text{arctanh}{z}$:
$$\text{arctanh}{z} = \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$
so that
$$
\begin{align}\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} &= \Im{\left [\sum_{k=0}^{\infty} \frac{e^{i(2 k+1)}}{2 k+1}\right]}\\ &= \frac12 \Im{\left[\text{arctanh}{e^i}\right]}\\ &= \frac12 \Im{\left[\log{\left (\frac{1+e^i}{1-e^i} \right )}\right]}\\ &= \frac12 \Im{\left[\log{\left (i \frac{\cos{1/2}}{\sin{1/2}} \right )}\right]}\\ &= \frac{\pi}{4} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine whether this relation is reflexive, symmetric... Determine whether this relation $R$ on the set of all integers is reflexive, symmetric, anti-symmetric and/or transitive where $x\,R\,y$ iff
$x = y + 1$ or $x = y-1$
*
*It is not reflexive:
Let $x = 2$: $2\neq 2 + 1$ and $2 \neq 2 - 1$.
*It is symmetric:
If $x = y + 1$ then $y = x - 1$
and if $x = y - 1$ then $y = x + 1$.
*It is not anti-symmetric:
Let $x = 3$ and $y = 2$;
then $3 = 2 + 1$ ($x\,R\,y$) and $2 = 3 - 1$ ($y\,R\,x$)
And let $x = 2$ and $y = 3$;
then $2 = 3 - 1$ ($x\,R\,y$) and $3 = 2 + 1$ ($y\,R\,x$)
but $3\neq 2$.
Can anyone prove whether this relation is transitive or not?
thanks.
| You did fine with reflexivity, and with symmetry and antisymmetry.
Now, let's look at transitivity:
We can summarize the relation as follows: $xRy$ if and only if $x$ and $y$ differ by $1$.
So, suppose $xRy$ ($x$ and $y$ differ by one) and $yRz$ ($y$ and $z$ differ by one),
What may be the case about the difference between $x$ and $z$?
*
*(Suspect a counterexample exists: just find $x, y, z$ such that $x = y - 1, y = z - 1 \implies x = z - 2$.
Or, vice versa, $x = y+1, y = z+1 \implies x = z+2)$
Let $x = 0$, $y = 1$, and $z = 2$, so we certainly have $x, y, z \in \mathbb Z$
*
*Clearly, $x = y - 1$ since $0 = 1-1$, so $x R y$,
*And $y = z - 1$, since $1 = 2 - 1$, so $y R z$.
*But it is not true that $x = z + 1 $, since $0 \neq 2+1 = 3$ nor
does $x = z - 1$, since $0 \neq 2 - 1 = 1$.
Hence, $x$ is not related to $z$, and transitivity fails.
All we need is one counterexample to prove that a relation is non-transitive, and we've just found one such couterexample
| {
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"url": "https://math.stackexchange.com/questions/404196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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