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Does $\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + cs}$ have a closed-form expression?
I'm wondering whether
$$
\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + c \cdot s},
$$
with $\delta \in (0,1)$ and $c\ge 0$ has a closed-form expression.
While I don't know of a general closed-form I can think if a special case:
Setting $c=0$ we obtain
$$
\sum^\infty_{s=1}\delta^{s-1}=\frac{1}{1-\delta},
$$
which is the geometric series. As for the more general case $c\neq 0$ I'm not sure.
|
For $|x^c|< 1$, consider the geometric series $\sum\limits_{n=0}^{\infty}x^{nc}$. Then we know
$$\frac{1}{1-x^c}=\sum\limits_{n=0}^{\infty}x^{nc}.$$
So
\begin{align*}
\frac{x^c}{1-x^c}& =\sum_{n=0}^{\infty}x^{nc+c}\\
\int\frac{x^c}{1-x^c} \, dx& =\sum_{n=0}^{\infty}\frac{x^{nc+c+1}}{1+c(n+1)}\\
\int\frac{x^c}{1-x^c} \, dx& =x^{c+1}\sum_{n=0}^{\infty}\frac{x^{nc}}{1+c(n+1)}.
\end{align*}
Now let $\delta=x^c$ (with $c \neq 0$), so we have
\begin{align*}
\frac{1}{c}\int\frac{\delta^{1/c}}{1-\delta} \, d\delta& =\delta^{\frac{c+1}{c}}\sum_{n=0}^{\infty}\frac{\delta^{n}}{1+c(n+1)}\\
\color{red}{\frac{1}{c\delta^{\frac{c+1}{c}}}\int\frac{\delta^{1/c}}{1-\delta} \, d\delta}& =\sum_{n=0}^{\infty}\frac{\delta^{n}}{1+c(n+1)}.
\end{align*}
So if we can solve the integral then we can get some closed form. However solving this integral may not be that straightforward.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4187156",
"timestamp": "2023-03-29T00:00:00",
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|
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $abc\geq ab+2c\geq ab+c+c\geq a+b+c+c\geq a+b+c+2$.
The main problem I face is to justify that $$abc=a+b+c+2\implies a=b=c=2.$$
My idea is to assume that $a>2$ and try to get a contradiction.
|
WLOG we assume $c \geq b \geq a \geq 2$
Then $3c \geq a + b + c$
We have,
$(ab-3) c \geq 2$
$abc \geq 2 + 3c \geq 2 + a + b + c$
For the equality case,
we are given, $abc = a + b + c +2 \leq 2 + 3 c$
$ \implies (ab-3) c \leq 2$
$ab - 3 \leq \dfrac{2}{c} \leq 1 \ ( \text {as } c \geq 2)$
That leads to $ab = 4, c = 2$.
i.e. $a = b = c = 2$
|
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|
Is $R$ a linear transformation from $\Bbb R^2 $ to $\Bbb R^2$? Question :
Let $R$ be the rotation around the origin clockwise by $\frac{ \pi}{3}$. Is $R$ a linear transformation from $\Bbb R^2 $ to $\Bbb R^2$? Write down the corresponding matrix $Au = R(u)$. Compute $R(u)$ with $u=[1,3]$
My approach:
$$R(u)=\underbrace{\begin{pmatrix}\cos \theta&\sin \theta\\-\sin \theta &\cos \theta\end{pmatrix}}_{\text { rotation matrix for clockwise direction}}\vec u$$
$$ R
\begin{bmatrix}
1 \\ 3
\end{bmatrix} = \begin{pmatrix}\cos \frac{ \pi}{3} &\sin \frac{ \pi}{3}\\-\sin \frac{ \pi}{3} &\cos \frac{ \pi}{3}\end{pmatrix} \begin{bmatrix}
1 = \vec u_{1} \\ 3 = \vec u_{2}
\end{bmatrix} $$
Product of matrices :
$$ R \begin{bmatrix}
1 \\ 3
\end{bmatrix} = \begin{pmatrix}\cos \frac{ \pi}{3} \cdot 1 ~~~+&\sin \frac{ \pi}{3} \cdot 3\\-\sin \frac{ \pi}{3} \cdot 1 ~~~+&\cos \frac{ \pi}{3} \cdot 3\end{pmatrix}$$
$$ \underbrace{R
\begin{bmatrix}
1 \\ 3
\end{bmatrix}}_{\text{R(u)}}= \underbrace{\begin{pmatrix} \frac{1+3 \sqrt{3}}{2}\\ \frac{3-\sqrt{3}}{2}\end{pmatrix}}_{\text{A x u}} $$
Conclusion :
$R $ is a linear transformation.
I feel I understand the Question, but I was hesitant on following wording
— “is R a linear transformation ?"
Could guys please validate the my answer and elaborate a little on the Wording.
Edit :
$ \vec u = \begin{bmatrix}
x_1\\ x_2
\end{bmatrix} $ , then a clockwise rotation by $\frac{\pi}{2}$ gives $\vec w = \begin{bmatrix}
x_2\\ -x_1
\end{bmatrix}$
Using trigonometry:
$$ R(\vec u) = \cos \left(\frac{\pi}{2}\right) ~\vec u + \sin \left(\frac{\pi}{2}\right)~\vec w$$
Thus, a general form would be :
$$ R(\vec u) = (\cos \theta) ~\vec u + (\sin \theta)~\vec w$$
Using $\theta = \frac{\pi}{3}$ in clockwise direction :
$$ = \cos \left(\frac{\pi}{3}\right) \begin{bmatrix}
x_1 \\ x_2
\end{bmatrix} + \sin \left(\frac{\pi}{3}\right) \begin{bmatrix}
x_2 \\ -x_1
\end{bmatrix} $$
$$ = \begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) \cdot x_1 + \sin \left(\frac{\pi}{3}\right) \cdot x_2\\ \cos \left(\frac{\pi}{3}\right) \cdot x_2 - \sin \left(\frac{\pi}{3}\right) \cdot x_1
\end{bmatrix} $$
Rearranging :
$$ = \begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) \cdot x_1 + \sin \left(\frac{\pi}{3}\right) \cdot x_2\\ - \sin \left(\frac{\pi}{3}\right) \cdot x_1 +
\cos \left(\frac{\pi}{3}\right) \cdot x_2 \end{bmatrix} $$
$$ = \begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) & \sin \left(\frac{\pi}{3}\right) \\ - \sin \left(\frac{\pi}{3}\right) & \cos \left(\frac{\pi}{3}\right)
\end{bmatrix} ~ \begin{bmatrix}
x_1\\ x_2
\end{bmatrix} $$
$$ R(\vec u) = \underbrace{\begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) & \sin \left(\frac{\pi}{3}\right) \\ - \sin \left(\frac{\pi}{3}\right) & \cos \left(\frac{\pi}{3}\right)
\end{bmatrix}}_{\text { rotation matrix}}~ \vec u$$
Hence proved it’s linear because it gives the form
$$ R(\vec u) = A \cdot ~\vec u$$
Lastly :
$$ \underbrace{R
\begin{bmatrix}
1 \\ 3
\end{bmatrix}}_{\text{R(u)}}= \underbrace{\begin{pmatrix} \frac{1+3 \sqrt{3}}{2}\\ \frac{3-\sqrt{3}}{2}\end{pmatrix}}_{\text{A x u}} $$
|
Your approach is correct for the last two questions. However, you have not proven that $R$ is a linear transformation. I will answer your question with detail.
A linear transformation from a $K$-vector space $V$ to another $K$-vector space $W$ is a function $R:V\to W$ such that the following two properties hold:
*
*$R(v+w)=R(v) + R(w)$ for all $v,w\in V$
*$R(\lambda v) = \lambda R(v)$ for all $\lambda\in K$ and $v\in V$
In your case, in order to prove that $R\colon \mathbb{R}^2\to\mathbb{R}^2$ is a linear transformation, you just have to prove that the properties 1. and 2. hold with $V,W=\mathbb{R}^2$, $K=\mathbb{R}$ and arbitrary vectors $v,w$ (as @Salihcyilmaz has pointed out in a comment). If you have already proven that $R(v)=Av$, then said properties are immediate to prove, since:
*
*$R(v+w) = A(v+w) = Av+Aw = R(v)+R(w)$ for any two $v,w\in\mathbb{R}^2$
*$R(\lambda v) = A(\lambda v) = \lambda Av = \lambda R(v)$ for any $\lambda\in \mathbb{R}$, $v\in\mathbb{R}^2$
P.S.: Note that we have only made use of the fact that there exists a fixed matrix $A$ such that $R(v)=Av$ for all $v\in\mathbb{R}^2$. Therefore, any transformation which can be represented by a matrix (in this sense) is a linear transformation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Are there positive integers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers? Are there positive intgers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers?
My research
We have $2a(b^c+1)-2a(b^c-1)=4a$, so I looked at perfect powers which have a multiple of $4$ as a difference. I found this question which talks about differences of powers, and I found:
$5^3-11^2=4$
$47^2-13^3=12$
$312^2-46^3=8$
$2^{17}-362^2=28$
and a few more, but none of them seemed to be in our form.
I also found On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ which is kind of similar to what are talking about. They say we don't have a solution to their problem in the special case due to Catalan's conjecture, so I guess we also don't have one for our problem, but couldn't prove it.
|
If $b$ is even then $b^c+1$ and $b^c-1$ are coprime, and so we have
$$b^c+1=\prod p^{u_p}\qquad\text{ and }\qquad b^c-1=\prod q^{v_q},$$
where the $p$ and $q$ are all distinct prime numbers, and the $u_p$ and $v_q$ are distinct positive integers. To find a positive integer $a$ such that $2a(b^c+1)$ and $2a(b^c-1)$ are both perfect powers, it suffices to find nonnegative integers $s_p$ and $t_q$ such that
$$u_p+s_p\equiv t_q\equiv0\pmod{m}\qquad\text{ and }\qquad v_q+t_q\equiv s_p\equiv0\pmod{n},$$
for some integers $m,n>1$, as then $a:=2^{mn-1}\prod p^{s_p}\prod q^{t_q}$ will make the two products $2a(b^c+1)$ and $2a(b^c-1)$ perfect $m$-th and $n$-th powers, respectively. By the Chinese remainder theorem, this can always be done by choosing $m$ coprime to the $v_q$, and $n$ coprime to the $u_p$, and $m$ and $n$ coprime to eachother.
For a concrete example, take $b=6$ and $c=3$ so that
$$b^c+1=217=7\times31\qquad\text{ and }\qquad b^c-1=215=5\times43.$$
Then $u_7=u_{31}=1$ and $v_5=v_{43}=1$. We may simply take $m=2$ and $n=3$, for which the smallest $s_p$ and $t_q$ are
$$s_7=3,\quad s_{31}=3,\quad t_5=2,\quad t_{43}=2,$$
corresponding to
$$a=2^{2\times3-1}\times(7^3\times31^3)\times(5^2\times43^2)=15114928589600,$$
with
\begin{eqnarray*}
2a(b^c+1)&=&80993080^2,\\
2a(b^c-1)&=&186620^3.
\end{eqnarray*}
Easy exercise: Can you find an appropriate value for $a$ given $(b,c)=(2,21)$?
Harder exercise: Can you extend this construction to odd $b$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a matrix for a unitary transform between matrices or prove that there is none I have hermitian matrices $A,\,B$ and would like to find a unitary matrix $U$ such that
$$UAU^\dagger=B$$
or show that there is no such matrix.
Example:
For
$$
A=\begin{pmatrix}
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & -1\\
0 & 0 & -1 & 0
\end{pmatrix},\,
B=\begin{pmatrix}
0 & 0 & -1 & 0\\
0 & 0 & 0 & 1\\
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0
\end{pmatrix}
$$
we find
$$
U=\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & -1 & 0 \\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}
$$
by guesswork. A constructive method would be useful.
Next example, where I assume that no such $U$ exists, but do not know how to show it:
$$
A=\begin{pmatrix}
b & d & c & a\\
d & - b & a & - c\\
c & a & - b & - d\\
a & - c & - d & b
\end{pmatrix}
=a\cdot\sigma_x\otimes\tau_x + b\cdot\sigma_z\otimes\tau_z
+ c\cdot\sigma_x\otimes\tau_z + d\cdot\sigma_z\otimes\tau_x\\
B=\begin{pmatrix}
- b & d & - c & - a\\
d & b & - a & c\\
- c & - a & b & - d
\\- a & c & - d & - b\end{pmatrix}
=-a\cdot\sigma_x\otimes\tau_x - b\cdot\sigma_z\otimes\tau_z
- c\cdot\sigma_x\otimes\tau_z + d\cdot\sigma_z\otimes\tau_x
$$
where $\sigma_x,\,\sigma_z,\,\tau_x,\,\tau_z$ denote the Pauli matrices
$$
\sigma_0 = \tau_0 =
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix},\,
\sigma_x = \tau_x =
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix},\,
\sigma_y = \tau_y =
\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix},\,
\sigma_z = \tau_z =
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
$$
Note: I am mostly interested in $4\times4$ matrices, but $8\times8$ would be nice, too.
EDIT: Thanks to the answer of Kurt G.. Assume $a,\,b,\,c,\,d\in\mathbb{R}\backslash\{0\}$. I was aware that a solution for $d=0$ exists. I specifically included $d$ to avoid a simple solution of the form $\sigma_a\otimes\tau_b$. When $d=0$, then a solution exists independently of the other values, for example $U=\sigma_0\otimes\tau_y$
|
In the second problem, the two matrices are unitarily similar when $abcd=0$. When $abcd\ne0$, one can verify that
\begin{aligned}
\operatorname{tr}(A^4)&=4\left[(a^2+b^2+c^2+d^2)^2+4(ab-cd)^2\right],\\
\operatorname{tr}(B^4)&=4\left[(a^2+b^2+c^2+d^2)^2+4(ab+cd)^2\right].
\end{aligned}
Hence $\operatorname{tr}(A^4)\ne\operatorname{tr}(B^4)$ in this case and $A,B$ are not similar.
|
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|
Min and max of $f(x,y)=e^{-xy}$ where $x^2+4y^2 \leq 5$ I am trying to use Lagrange multipliers to find the maximum and minimum values of the function $$f(x,y)=e^{-xy}$$
constrained as $$x^2+4y^2=5$$
I began this problem by setting up the Lagrangian:
$$f(x,y) = e^{-xy}$$
$$g(x,y) = x^2+4y^2-5$$
$$L(x,y) = f(x,y) - \lambda g(x,y) = e^{-xy} - \lambda (x^2+4y^2-5)$$
So our equations are:
$$-ye^{-xy} - 2\lambda x = 0$$
$$-xe^{xy} -8\lambda y = 0$$
$$x^2+4y^2-5 = 0$$
Now from the first equation, $e^{-xy} = 2\lambda x/y$. Substituting into the second equation yields: $2x^2 \lambda / y = 8 \lambda y$ or $x^2+4y^2 = 0$. This clearly violates the third equation, meaning this system has no solution. Doe this mean there are no local maxima or minima?
Any guidance is greatly appreciated!
|
The function $ \ f(x \ , \ y) \ = \ e^{\ -xy} \ $ to be extremized has "diagonal symmetry", which is to say that $ \ f(-x \ , \ -y) \ = \ f(x \ , \ y) \ \ ; $ it also has an anti-symmetric property in that $ \ f(-x \ , \ y) \ = \ f(x \ , \ -y) \ = \ \frac{1}{f(x \ , \ y)} \ \ . $ The elliptical region within which we wish to locate extrema, $ \ x^2 + 4y^2 \ \le \ 5 \ \ , $ has "four-fold" symmetry about the origin, so we expect to find the minima of $ \ f(x \ , \ y) \ $ in the first and third quadrants at the points $ \ ( \pm x \ , \ \pm y ) \ $ and its maxima in the second and fourth quadrants at $ \ ( \pm x \ , \ \mp y ) \ \ . $
While we can perform the usual critical-point analysis in this elliptical region, we may also examine the values of our function on the concentric ellipses $ \ x^2 + 4y^2 \ = \ c^2 \ \ , \ $ with $ \ 0 \ \le \ c^2 \ \le \ 5 \ $ to find the value of $ \ c \ $ where the extrema reach their greatest and smallest values. Due to the symmetries noted, we can just look at each ellipse in the first quadrant, for which $ \ f(x \ , \ y) \ = \ \large{e^{\ -\sqrt{c^2 - 4y^2} \ · \ y }} \ \ . $ This plainly takes on its largest value at the origin, where $ \ c \ = \ 0 \ \Rightarrow \ f(0 \ , \ 0) \ = \ e^{\ 0} \ = \ 1 \ \ . $ We have, for $ \ c \ \neq \ 0 \ \ , $ the derivative
$$ \frac{d}{dy} \ [ \ e^{\ -\sqrt{c^2 - 4y^2} \ · \ y} \ ] \ \ = \ \ e^{\ -\sqrt{c^2 - 4y^2} \ · \ y} \ · \ \frac{d}{dy} \ [ \ -\sqrt{c^2 - 4y^2} \ · \ y \ ] $$
and since $ \ f(x \ , \ y) \ \neq \ 0 \ \ $ for real values of the coordinate variables, we find the minimum value of our function from
$$ \frac{d}{dy} \ [ \ -\sqrt{c^2 - 4y^2} \ · \ y \ ] \ \ = \ \ -\frac12·\frac{1}{\sqrt{c^2 - 4y^2}}·(-8y) \ · \ y \ + \ -\sqrt{c^2 - 4y^2} \ · \ 1 \ \ = \ \ 0 $$
$$ \Rightarrow \ \ -4y^2 \ \ = \ \ c^2 - 4y^2 \ \ \Rightarrow \ \ y^2 \ \ = \ \ \frac{c^2}{8} \ \ \Rightarrow \ \ x^2 \ \ = \ \ c^2 \ - \ 4·\left(\frac{c^2}{8} \right) \ \ = \ \ \frac{c^2}{2} \ \ . $$
The minima on the concentric ellipses are therefore $ \ f\left(\pm \sqrt{\frac{c^2}{2}} \ , \ \pm \sqrt{\frac{c^2}{8}} \right) \ = \ \large{e^{\ - \sqrt{\frac{c^2}{2}} \ · \ \sqrt{\frac{c^2}{8}} }} \ = \ \normalsize{e^{\ - c^2/4} } \ \le \ 1 \ \ $ and the maxima are
$ \ f\left(\pm \sqrt{\frac{c^2}{2}} \ , \ \mp \sqrt{\frac{c^2}{8}} \right) \ = \ \large{e^{\ \sqrt{\frac{c^2}{2}} \ · \ \sqrt{\frac{c^2}{8}} }} \ = \ \normalsize{e^{\ c^2/4} } \ \ge \ 1 \ \ . $
The global extrema are attained by making $ \ c^2 \ $ as large as is permitted on the elliptical region, that is, at its boundary:
$$ \mathbf{\text{global minima:}} \ \ \ f\left(\pm \sqrt{\frac{5}{2}} \ , \ \pm \sqrt{\frac{5}{8}} \right) \ \ = \ \ e^{\ - 5/4} \ \ ; $$
$$ \mathbf{\text{global maxima:}} \ \ \ f\left(\pm \sqrt{\frac{5}{2}} \ , \ \mp \sqrt{\frac{5}{8}} \right) \ \ = \ \ e^{\ 5/4} \ \ . $$
|
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|
Finding the surface area of a sphere within a cylinder and above the $xy$-plane using double integral I'm trying to find the surface area of the part of the sphere $x^2+y^2+z^2=a^2$ that lies within the cylinder $x^2+y^2=ax$ and above the xy-plane.
I want to do this using $$A(S)=\iint_D \sqrt{[f_{x}(x,y)]^2+[f_{y}(x,y)]^2+1} \; dA.$$
My understanding is that $f=z=\sqrt{a^2-x^2-y^2}$, $z_x=-x(a^2-x^2-y^2)^{-1/2}$, $z_y=-y(a^2-x^2-y^2)^{-1/2}$.
What I don't understand is what the bounds for the integrals should be. I think it has to be done using polar coordinates, but I'm lost as to what the radius and angle bounds would be -- I'm not visualizing the cylinder and sphere very well.
Any help would be appreciated.
|
Yes it is easier to evaluate using polar coordinates. Based on your working,
$\displaystyle \sqrt{1+z_x^2 + z_y^2} = \cfrac{a}{\sqrt{a^2-x^2-y^2}}$
$\displaystyle dS = \cfrac{a}{\sqrt{a^2-x^2-y^2}} \ dx \ dy$
So integral to find surface area of sphere inside the cylinder above xy-plane is,
$\displaystyle \int_{x^2+y^2 \leq ax} \cfrac{a}{\sqrt{a^2-x^2-y^2}} \ dx \ dy$
Converting to polar coordinates,
$x = r \cos\theta, y = r\sin\theta, x^2 + y^2 = r^2$
$x^2 + y^2 \leq ax \implies r \leq a \cos\theta, - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
[please note the bounds of $\theta$. As we are using polar coordinates centered at the origin, the circle $r = a \cos\theta$ forms for $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) \ $]
So the integral is,
$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{a\cos\theta} \cfrac{a}{\sqrt{a^2-r^2}} \ r \ dr \ d\theta$
|
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|
$p$-adic structure of Pell-type equations As an example, consider an integer solution of $ x ^ 2-3y ^ 2 = 13 $.
$ y $ that satisfies this equation is
$y_k = \frac {(4+ \sqrt {3}) (2+ \sqrt {3}) ^ k-(4- \sqrt {3}) (2- \sqrt {3}) ^ k} { 2 \sqrt {3}} $
for any integer $ k $.
By calculation
$ 3 | y_k \Leftrightarrow k \equiv1 \mod 3 $
$ 3 ^ 2 | y_k \Leftrightarrow k \equiv7 \mod 3 ^ 2 $
$ 3 ^ 3 | y_k \Leftrightarrow k \equiv7 \mod 3 ^ 3 $
$ 3 ^ 4 | y_k \Leftrightarrow k \equiv7 \mod 3 ^ 4 $
$ 3 ^ 5 | y_k \Leftrightarrow k \equiv7+2\cdot 3^4 \mod 3 ^ 5 $
$ 3 ^ 6 | y_k \Leftrightarrow k \equiv7+2\cdot 3^4+2\cdot 3^5 \mod 3 ^ 6 $
You can see that it has a regular structure.
In other words, a solution divisible by $3^d$ appear at intervals of $3$ powers and do not appear anywhere else.
I tried similar calculations for other Pell-type equations, and I was convinced that they had a similar structure.
Can you prove this structure in general?
We consider the Pell-type equation $ x ^ 2-py ^ 2 = N $ with a solution. For any $ d $
$ p ^ d | y_k \Leftrightarrow k \equiv r \mod p^{d+e}. $
$ r $ should be uniquely determined as $ p $-adic number.
|
As $d$ gets larger and $p^d|y_k$, this can be thought of as taking the limit as $y \to 0$ in the p-adics, since $\lim_{d \to \infty}|p^d|_p=0$. So in your equation we're looking at solving just $x^2=N$.
Since in your specific problem you're interested in $k$, we can set $y_k=0$ and solve for $k$,
$$0 = \frac {(4+ \sqrt {3}) (2+ \sqrt {3}) ^ k-(4- \sqrt {3}) (2- \sqrt {3}) ^ k} { 2 \sqrt {3}} $$
$$k=\frac{\log \frac{4- \sqrt {3}}{4+ \sqrt {3}}}{\log \frac{2+ \sqrt {3}}{2- \sqrt {3}}}$$
This gets us exactly the digits you have so far (rewriting $7=1+2\cdot 3$ of course)
$$k = 1 + 2\cdot 3 + 2\cdot 3^4 + 2\cdot 3^5 + 2\cdot 3^6 + 3^7 + 3^8 + \dots$$
For your convenience, here is the sage code I used to compute it almost exactly copied from here:
R = Zp(3, 10)
S.<x> = ZZ[]
f = x^2 - 3
W.<w> = R.ext(f)
log((4-w)/(4+w))/log((2+w)/(2-w))
The general case follows identically, and I leave it to you.
|
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|
To prove $f=x^4+x^3+x^2+x^1+1$ is irreducible over the $\mathbb{Q}$ My line of proof is as follows:
*
*$\pm1$ are the only candidates for being rational roots (Rational Root Theorem)
*Since $f(1)=5$ and $f(-1)=1$, none is a root
And hence the given polynomial is irreducible over the rationals $\mathbb{Q}$.
Is this proof correct? I have doubt because the solution given in the book uses Eisenstein's criterion by modifying the polynomial to $f(x+1)=\frac{(x+1)^5-1}{x}$ and so on... which seems complicated to me.
|
(Alternative proof, not using Eisenstein's criterion.)
The polynomial is reciprocal and can be easily factored over the reals:
$$
\begin{align}
x^4+x^3+x^2+x+1 &= x^2 \left(\left(x^2+\frac{1}{x^2}\right)+\left(x+\frac{1}{x}\right)+1\right)
\\ &= x^2 \left(\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-1\right)
\\ &= x^2\left(x+\frac{1}{x}-\frac{-1+\sqrt{5}}{2}\right)\left(x+\frac{1}{x}-\frac{-1-\sqrt{5}}{2}\right)
\\ &= \left(x^2+\frac{1-\sqrt{5}}{2}x + 1\right)\left(x^2+\frac{1+\sqrt{5}}{2}x + 1\right)
\end{align}
$$
Since neither quadratic has real roots this is the unique irreducible factorization over $\mathbb R$, and since the coefficients are not rational the polynomial is irreducible over $\mathbb Q$.
|
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|
Solutions to $2^a3^b+1=2^c+3^d$
Find all $a,b,c,d$ positive integer such that:
$2^a3^b+1=2^c+3^d$
My progress:
One solution satisfying is $$\boxed{a=1,b=1,c=2,d=1} $$
We first take $\mod 3$ which gives $$ L.H.S\equiv 1\mod 3,~~R.H.S\equiv 2^c\mod 3$$
Hence we get $c$ even. So let $c=2k.$
We get $$2^a3^b-3^d=2^{2k}-1=(2^k-1)(2^k+1). $$
Now note that $2^k-1,2^k+1$ are relatively prime. Because, if not then let $d$ be the common divisor.
Then $$d|2^k-1,~~d|2^k+1\implies d|(2^k+1)-(2^k-1)=2\implies d|2^k\implies d|2^k+1-2^k=1.$$
Now there are two cases. So using the fact that $2^k-1,2^k+1$ are relatively prime and then for odd k $3|2^k+1$ and for even $k$ $3|2^k-1$
Case 1: When $d<b$ then $$2^a3^b-3^d=3^d(2^a3^{b-d}-1)=(2^k-1)(2^k+1)$$
*
*$K$ is odd $$\implies v_3(2^k+1)=d,~~3\nmid 2^k-1. $$
*$K$ is even $$\implies v_3(2^k-1)=d,~~3\nmid 2^k+1. $$
Case 2: When $d>b$ then $$2^a3^b-3^d=3^b(2^a -3^{d-b})=(2^k-1)(2^k+1)$$
*
*$K$ is odd $$\implies v_3(2^k+1)=b,~~3\nmid 2^k-1. $$
*$K$ is even $$\implies v_3(2^k-1)=b,~~~~3\nmid 2^k+1. $$
P.S. This is my 100th post in MSE. The other solutions are there in the chat. Any elementary method?
|
There might not be an easy elementary solution to this equation. It was solved (there are $12$ nontrivial solutions in integers, where trivial solutions have at least $2$ of the exponents equal to $0$) by Tijdeman and Wang in a paper in the Pacific Journal in 1988; their argument uses bounds for linear forms in logarithms. The nontrivial solutions are as noted by Robert Israel, as well as $(a,b,c,d)$ in the following list :
$$
(2,0,1,1),\; (4,0,3,2), \;(-2,2,-2,1),\; (2,-1,1,-1).
$$
|
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|
Conditional inequality $2a^3+b^3≥3$
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?
So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{2}+\lambda(5 a^{4}+5 a^{4} b^{5})=0 \ldots$$
$$\tag2 \frac{\partial f}{\partial b}=3 b^{2}+\lambda(5 a^{5} b^{4})=0 \ldots$$
$$\left\{\begin{array}{c} 6+\lambda(5 a^{2}+5 a^{2} b^{5})=0 \\ 3+\lambda(5 a^{3} b^{4})=0 \end{array}\right. $$
$$ \lambda=-\frac{6}{5 a^{2}+5 a^{2} b^{5}}=-\frac{3}{5 a^{3} b^{4}} $$
Multiply by $a^3$, $\,2 a^{6} b^{4}=a^{5}+a^{5} b^{5}=2$.
$$ a^{6} b^{4}=1,\, a^{3} b^{2}=1 \Rightarrow b=\frac{1}{a^{\frac{3}{2}}}\quad a, b \geq 0 $$
$$ \begin{gathered} a^{5}+a^{5} b^{5}=2 \Rightarrow a^{5}+\frac{a^{5}}{a^{\frac{15}{2}}}=2 \Rightarrow a^{5}+\frac{1}{a^{\frac{5}{2}}}=2 \Rightarrow a^{5}-2 a^{\frac{5}{2}}+1=0 \\ \Rightarrow\left(a^{\frac{5}{2}}-1\right)^{2}=0 \Rightarrow a=1 \end{gathered} $$
And the minimum is at $(a,b)=(1,1)$.
I'm not sure I solved this inequality correctly, I would like to see a more beautiful way.
|
Remarks: Without calculus.
We have $a^5 = \frac{2}{1 + b^5}$.
We need to prove that $2a^3 \ge 3 - b^3$.
WLOG, assume that $3 - b^3 \ge 0$.
It suffices to prove that
$$2^5 a^{15} \ge (3 - b^3)^5$$
or
$$2^5 \left(\frac{2}{1 + b^5}\right)^3 \ge (3 - b^3)^5$$
or
$$2^9 \ge (1 + b^5)^3 \cdot (3 - b^3)^3 \cdot 2(3 - b^3)^2.$$
By AM-GM, we have
$$b^5 \le \frac{b^3 + b^6 + b^6}{3}$$
and
$$(3 - b^3)\cdot (3 - b^3)\cdot 2
\le \left(\frac{3 - b^3 + 3 - b^3 + 2}{3}\right)^3.$$
Thus, it suffices to prove that
$$2^9 \ge \left(1 + \frac{b^3 + b^6 + b^6}{3}\right)^3\cdot (3 - b^3)^3 \cdot \left(\frac{3 - b^3 + 3 - b^3 + 2}{3}\right)^3$$
or
$$2^3 \ge \left(1 + \frac{b^3 + b^6 + b^6}{3}\right)\cdot (3 - b^3) \cdot \left(\frac{3 - b^3 + 3 - b^3 + 2}{3}\right)$$
or
$$\frac{2}{9}b^3(9 - 2b^3)(b^3 - 1)^2 \ge 0$$
which is true.
We are done.
|
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|
Prove that $(a+b)^c\cdot(b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot (a+b+c)\right]^{a+b+c}$ where $a,b,c\in \mathbb{Q}^{+}$ unless $a=b=c$.
If $a,b,c$ be positive rational numbers, prove that
$$(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot
(a+b+c)\right]^{a+b+c}$$ unless $a=b=c$
My try :
Consider $(a+b),(b+c),(c+a)$ be three numbers with associated weights $c,a,b$
By weighted AM-GM inequality,
$$\frac{c\cdot (a+b)+a\cdot (b+c)+b\cdot (c+a)}{a+b+c}\ge \biggl[(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\biggl]^\frac{1}{a+b+c}$$
the equality occurs when $(a+b)=(b+c)=(c+a)$ that is $a=b=c$
Therefore
$$\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$$
unless $a=b=c$
Now $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
Therefore $$(a+b+c)^2\gt 2(ab+bc+ca)-------(2)$$
using equation $(1)$ and $(2)$, we get,
$$\left(a+b+c\right)^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b$$
I am stuck. please give me some hint. Thanks in advance.
|
We need to prove that:
$$\sum_{cyc}\frac{c}{a+b+c}\ln(a+b)\leq\ln\left(\frac{2}{3}(a+b+c)\right).$$
Now, $\sum\limits_{cyc}\frac{c}{a+b+c}=1$ and we can use Jensen for the concave function $\ln$.
Indeed, $$\sum_{cyc}\frac{c}{a+b+c}\ln(a+b)\leq\ln\left(\sum_{cyc}\frac{c(a+b)}{a+b+c}\right)$$ and it's enough to prove that
$$2(a+b+c)^2\geq3\sum_{cyc}c(a+b),$$ which is $$\sum_{cyc}(a-b)^2\geq0.$$
|
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|
Why is AM-GM giving 2 contradictory results for the same problem? The problem is the following:
given x and y such that $x+y=1$ and $x,y \geq 0$, find the maximum value of $x^2y$.
The point is, I tried to solve it in 2 different ways, which are shown below, but the second method I used gives a wrong result, and I can't understand where the mistake is.
First Method
Using AM-GM inequality you can say
$ \displaystyle{\left[{\frac{x}{2}\cdot \frac{x}{2} \cdot y}\right]^{\frac{1}{3}} \leq \frac{\frac{x}{2}+\frac{x}{2}+y}{3}}$
But we also have
$\displaystyle{\frac{x}{2}+\frac{x}{2}+y = x+y=1}$
Hence
$\displaystyle{\frac{x^2y}{4} \leq \left(\frac{1}{3}\right)^3 \quad {\LARGE\Rightarrow} \quad x^2y \leq \frac{4}{27}}$
So the solution is
$ max=\frac{4}{27} $, where $x= \frac {2}{3}$ and $y=\frac{1}{3}$
Second Method
Using AM-GM inequality again, we have
$ \displaystyle{\left[x\cdot x \cdot y\right]^{\frac{1}{3}} } \leq \frac{x+x+y}{3}$
Now if you substitute $x+y=1$, you get
$\displaystyle{\left[x\cdot x \cdot y\right]^{\frac{1}{3}} } \leq \frac{x+1}{3}$
Cubing both sides of the inequality, it becomes
$\displaystyle {x^2y\leq \frac{(x+1)^3}{27} }$
It follows that $x^2y$ is maximum when it is equal to the other side of the inequality, so we have:
$\displaystyle {x^2y - \frac{(x+1)^3}{27}= 0 }$
As it is given that $x+y =1$, then $ y=1-x$ and we have a cubic equation which is:
$\displaystyle {x^2(1-x) - \frac{(x+1)^3}{27}= 0 }$
I expected that its solution would be $ x= \frac{2}{3} $ as above, but instead, you can check that its only real solution is $x=y=\frac{1}{2}$
Clearly the correct solution is given by the first method that I showed, as it can also be confirmed by looking for the maximum using the derivative. I just don't know why this happens, apparently I don't see any mistake in the second method.
If anyone is able to come up with an explanation, I would appreciate it, thanks.
|
Your mistake is in this assumption in your second approach:
It follows that $x^2y$ is maximum when it is equal to the other side of the inequality
This is not true; while at $x = \frac12$,
$$x^2y = \frac{(x+1)^3}{27},$$
the right-hand side is not a constant upper bound, but increasing.
Then it turns out (from the first approach) that the maximum value of $x^2y$ is at $x=\frac23$, when the right-hand side upper bound is larger than the value of $x^2y$.
The graph of the two sides of the (in)equality demonstrates better than my words.
From just the condition $0\le x \le 1$ and the inequality
$$x^2y \le \frac{(x+1)^3}{27},$$
the right-hand side has the global maximum $\frac 8{27}$ at $x=1$, so a rough constant upper bound from these would be
$$x^2y \le \frac{(x+1)^3}{27} \le \frac{8}{27}.$$
But the two equalities happen at different $x$, so overall it's a strict inequality
$$x^2y < \frac{8}{27}$$
and $\frac{8}{27}$ is not the maximum value of $x^2y$.
|
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|
How many different ordered triples are there such that $a+b+c = 50$ and $a\geq b\geq c\geq 0$? What if $a>b>c>0$? How many different ordered triples are there such that $a+b+c = 50$ and $a\geq b\geq c\geq 0$? What if $a>b>c>0$?
a,b, and c are all integers
I found this question from a textbook, but the author didn't give the answer so I don't know if my answers are correct, can anyone help me with this problem?
My answers:
if $a\geq b\geq c\geq 0$: $221$ possible triples
if $a>b>c>0$: $196$ possible triples
Thanks!
|
Using generating functions:
First rewrite the problem by using the substitutions
$$d = a-b, \qquad e = b-c, \qquad f = c.$$
so we have $d,e,f \geq 0$ are integers such that $d + 2e + 3f = 50$.
The number of such triplets can be found as the coefficient of $x^{50}$ in the generating function
$$\frac{1}{(1-x)(1-x^2)(1-x^3)}$$
By using partial fractions, we may find,
$$\begin{align*}\frac{1}{(1-x)(1-x^2)(1-x^3)} &= \frac{17x^2-52x+47}{72(1-x)^3} + \frac{1}{8(1+x)} + \frac{1}{3(1-x^3)} - \frac{1}{9(1-x)} \\ &= \frac{17x^2-52x+47}{72}\sum_{n=0}^\infty \binom{n+2}{2}x^n + \frac{1}{8}\sum_{n=0}^\infty (-1)^nx^n + \frac{1}{3}\sum_{n=0}^\infty x^{3n} - \frac{1}{9}\sum_{n=0}^\infty x^n \end{align*}$$
giving an answer of
$$\frac{17}{72}\binom{50}{2} - \frac{52}{72}\binom{51}{2} + \frac{47}{72}\binom{52}{2} + \frac{1}{8} - \frac{1}{9} = 234$$
for the first problem.
For the second problem, you would have $d,e,f > 0$, or by another substitution of $d' = d-1, e'=e-1, f'=f-1$ we'd be counting the number of triples $d',e',f' \geq 0$ such that $d' + 2e' + 3f' = 44$, so it's just the coefficient of $x^{44}$ of the same generating function. This gives
$$\frac{17}{72}\binom{44}{2} - \frac{52}{72}\binom{45}{2} + \frac{47}{72}\binom{46}{2} + \frac{1}{8} - \frac{1}{9} = 184$$
|
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|
Exercise on limit of indeterminate form: $\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt2$ I'm not able to show that
$\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt{2}$
I proceeded as follows:
Substituting 1 to $x$ gives the indeterminate form $\frac{0}{0}$.
I rationalized by multiplying the numerator and the denominator by $\sqrt{1+x} + \sqrt{2}$ but I was not able to prosecute, since I obtained again an indetermiante form $\frac{0}{0}$.
I used
$y = x -1$
$x = y + 1$
and computed the limit for $y \rightarrow 0$ and then I rationalized the denominator, but I was again stuck with the indeterminate form $\frac{0}{0}$.
The exercise is taken from a high school book and the chapter comes before the one about de l'Hopital rule, so it must be solved without using it.
|
If you rationalize both the numerator and the denominator you will end up with $\frac {2(\sqrt {1+x}+\sqrt 2)} {\sqrt {x+3}+\sqrt {5-x}}$ so the limit is $\frac {2(\sqrt2 +\sqrt 2)} {\sqrt 4 +\sqrt 4}=\sqrt {2}$.
|
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Function corresponds to $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}$ using the Fourier expansion
Find the function corresponds to $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}$ using the Fourier expansion
$$
f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty (a_n\cos(nx)+b_n\sin(nx))
$$
where $$\frac{a_0}{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)dx,\\{a_n}=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(n x)dx,\\{b_n}=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)dx$$
$$
\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}=\sum_{n=1}^\infty \frac{\cos(n\pi x)}{n^2+\alpha^2}
$$
How do I proceed further ?
|
You may use the Poisson summation formula:
If $f\in L_1(\mathbb{R})$ and $\widehat{f}\in L_1(\mathbb{R})$, then $Pf(x)=\sum_{n\in\mathbb{Z}}f(x+n)$ converges uniformly, $f\in\mathcal{C}(\mathbb{S}^1)$, and
$$Pf(x)=\sum_{n\in\mathbb{Z}} \widehat{f}(n) e^{2\pi kx}$$
where $\widehat{f}$ is the Fourier transform of $f$.
Using the fact that $\frac12 e^{-|2\pi t|}=\frac{1}{2\pi}\int^\infty_{-\infty} \frac{e^{-i2\pi tx}}{1+x^2}\,dx$, (this requires a little bit of Fourier analysis on $\mathbb{R}$, or some knowledge on characteristic functions of certain probability distributions), we have that
$$\sum_{n\in\mathbb{Z}} e^{-2\pi a |n|}e^{-2\pi kx}=Pf(x)=\frac{1}{\pi}\sum_{n\in\mathbb{Z}}\frac{a}{a^2+|x+n|^2}$$
At $x=0$ one gets
\begin{align}
\frac{\pi}{2a}\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}}&=\sum^\infty_{n=1}\frac{1}{a^2+n^2} +\frac{1}{2a^2}=\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2}+\frac{1}{2a^2} +\frac14\sum^{\infty}_{n=1}\frac{1}{\big(\tfrac{a}{2}\big)^2+ n^2}\\
&=\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2} +\frac{1}{2a^2} + \frac{1}{4}\Big(\frac{\pi}{a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}-\frac{2}{a^2} \Big)\\
&=\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2}+\frac{\pi}{4a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}
\end{align}
Hence
$$
\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2} =\frac{\pi}{2a}\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} -\frac{\pi}{4a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}
$$
The series in the OP can be then expressed as
$$\begin{align}
\sum^\infty_{n=1}\frac{(-1)^n}{a^2+n^2}&=-\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2}+\frac14\sum^{\infty}_{n=1}\frac{1}{\big(\tfrac{a}{2}\big)^2+ n^2}\\
&=-\frac{\pi}{2a}\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} +\frac{\pi}{2a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}-\frac{1}{2a^2} \\
&=\frac{\pi}{2a}\Big(\frac{1+e^{-\pi a}}{1-e^{-\pi a}}-\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}}\Big)-\frac{1}{2a^2} \\
&=\frac{\pi}{2a}\operatorname{csch}(\pi a)-\frac{1}{2a^2}
\end{align}
$$
|
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Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\log a_{n} =n\log\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right) =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{2n}{n^{\frac{1}{n}}} -\frac{n}{n^{\frac{2}{n}}} -n\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{n^{\frac{1}{n}} -1}{n^{\frac{1}{n}}}\right)^{2} .( -n)
\end{array}$
The first term on RHS has limit equal to $\displaystyle 1\ $but the second term is giving me problem.
Please help. Thanks.
|
Use these two results:
We can use that for $|\alpha x|>>1$, $$(1+\alpha)^x \sim e^{\alpha x}
\tag{1}$$
We have the basic limit for logarithm as; $$\lim_{n \rightarrow
\infty}n\left(a^{1/n}-1\right)=\log a \hspace{5px}\text{, for all
positive } a$$ for $a =n$ itself we have: $$\lim_{n \rightarrow
\infty}n\left(n^{1/n}-1\right)=\log n \tag{2}$$
$$\begin{align}\lim_{n\rightarrow \infty}\frac{\left(2n^{1/n}-1\right)^{n}}{n^{2}}&=\lim_{n\rightarrow \infty}\frac{\left(1+2\left(n^{1/n}-1\right)\right)^{n}}{n^{2}} \\ &=\lim_{n\rightarrow \infty}\frac{\exp\left(2n\left(n^{1/n}-1\right)\right)}{n^{2}} \tag{By Eq. 1 & 2}\\ &=\lim_{n\rightarrow \infty}\frac{\exp\left(2\log n\right)}{n^{2}} \tag{By Eq. 2}\\ &=1\end{align}$$
|
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|
Complicated Integration involving Bessel function and exponential function Sir,
while studying the quantum scattering for various cases, I have got the following integrals:
$$\int_{t_0=-\infty}^{t-R/c}\Big[\frac{\exp{(-i\omega t_0)}}{(t-t_0)}\Big]J_0\big(\sqrt{(t-t_0)^2-(R/c)^2}\big)\,dt_0$$ and similarly,
$$\int_{t_0=-\infty}^{t-R/c}\Big[-\frac{\exp{(-i\omega t_0)}}{(t-t_0)^2}\Big]J_0\big(\sqrt{(t-t_0)^2-(R/c)^2}\big)\,dt_0$$
where, $t$ is observation time, $t_0$ is the retarded time, $J_0$ is the Bessel function of first kind and $R$ is the distance between the observation point and the point on scatterer such that $R=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ , where, $(x,y,z)$ is the point of observation.$c$ is the speed of light and taken as constant. Further, the causality condition is followed such that $t-t_0\geq R/c$ or $t_0\leq t- R/c$. According to the physical resctriction imposed on the problem, it is also true that in the limit $t_0\rightarrow -\infty$, the integrand should be zero.
I am trying to find out the closed form answer of the integrals (if one can be solved, the other can be solved by the same technique).I searched through the "Table of integrals" by Gradshteyn but somehow, I could not find out the exact match.
Would you kindly suggest me any method or any relevant texts from where I can get some help.
|
$$
\int_{-\infty}^{t-a}{\frac{e^{-i\omega u}}{t-u}J_0\left( \sqrt{\left( t-u \right) ^2-a^2} \right) du}
\\
u=t-r,=e^{-i\omega t}\int_a^{\infty}{\frac{e^{i\omega r}}{r}J_0\left( \sqrt{r^2-a^2} \right) dr}
\\
I\left( \omega \right) =\int_a^{\infty}{\frac{e^{i\omega r}}{r}J_0\left( \sqrt{r^2-a^2} \right) dr}
\\
I'\left( \omega \right) =i\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( n! \right) ^2}}\left( \frac{1}{2} \right) ^{2n}\int_a^{\infty}{e^{i\omega r}\left( r^2-a^2 \right) ^ndr}
\\
r=ax,=i\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( n! \right) ^2}}\left( \frac{a}{2} \right) ^{2n}\int_1^{\infty}{e^{i\omega ax}\left( x^2-1 \right) ^ndx}
\\
{H_{\nu}}^{\left( 1 \right)}\left( x \right) =-\frac{2i\left( \frac{x}{2} \right) ^{-\nu}}{\sqrt{\pi}\Gamma \left( \frac{1}{2}-\nu \right)}\int_1^{\infty}{\frac{e^{ixt}}{\left( t^2-1 \right) ^{\nu +\frac{1}{2}}}dt}
\\
{H_{-\frac{1}{2}-n}}^{\left( 1 \right)}\left( \omega a \right) =-\frac{2i\left( \frac{\omega a}{2} \right) ^{\frac{1}{2}+n}}{\sqrt{\pi}n!}\int_1^{\infty}{e^{i\omega at}\left( t^2-1 \right) ^ndt}
\\
I'\left( \omega \right) =i\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( n! \right) ^2}}\left( \frac{a}{2} \right) ^{2n}\int_1^{\infty}{e^{i\omega ax}\left( x^2-1 \right) ^ndx}
\\
=-\frac{\sqrt{\pi}}{2\omega}\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{n!}}\left( \frac{a}{2\omega} \right) ^{n-\frac{1}{2}}{H_{-\frac{1}{2}-n}}^{\left( 1 \right)}\left( \omega a \right)
\\
=-\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( -\frac{a}{2\omega} \right) ^n\left( J_{-\frac{1}{2}-n}\left( \omega a \right) +iY_{-\frac{1}{2}-n}\left( \omega a \right) \right)
\\
=-\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( -\frac{a}{2\omega} \right) ^n\left( \left( -1 \right) ^{n-1}Y_{\frac{1}{2}+n}\left( \omega a \right) +i\left( -1 \right) ^nJ_{\frac{1}{2}+n}\left( \omega a \right) \right)
\\
=\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{Y_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n-i\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{J_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n
$$
The spherical Bessel functions have the generating functions
$$
\frac{\cos \left( \sqrt{z^2-2zt} \right)}{z}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}J_{n-1/2}(z)
\\
\frac{\sin \left( \sqrt{z^2-2zt} \right)}{z}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}Y_{n-1/2}(z)
\\
\therefore \frac{\sin \left( \sqrt{z^2-2zt} \right)}{\sqrt{z^2-2zt}}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}J_{n+1/2}(z)
\\
-\frac{\cos \left( \sqrt{z^2-2zt} \right)}{\sqrt{z^2-2zt}}=\sum_{n=0}^{\infty}{\frac{t^n}{n!}}\sqrt{\frac{\pi}{2z}}Y_{n+1/2}(z)
\\
\therefore \frac{\sin \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}=\sum_{n=0}^{\infty}{\frac{\left( \frac{a}{2\omega} \right) ^n}{n!}}\sqrt{\frac{\pi}{2\omega a}}J_{n+1/2}(\omega a)
\\
-\frac{\cos \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}=\sum_{n=0}^{\infty}{\frac{\left( \frac{a}{2\omega} \right) ^n}{n!}}\sqrt{\frac{\pi}{2\omega a}}Y_{n+1/2}(\omega a)
\\
I'\left( \omega \right) =\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{Y_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n-i\sqrt{\frac{\pi}{2\omega a}}\sum_{n=0}^{\infty}{\frac{J_{\frac{1}{2}+n}\left( \omega a \right)}{n!}}\left( \frac{a}{2\omega} \right) ^n
\\
=-\frac{\cos \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}-i\frac{\sin \left( a\sqrt{\omega ^2-1} \right)}{a\sqrt{\omega ^2-1}}
\\
I\left( \omega \right) =-\frac{1}{a}\int_{\infty}^{\omega}{\left( \frac{\cos \left( a\sqrt{t^2-1} \right)}{\sqrt{t^2-1}}+i\frac{\sin \left( a\sqrt{t^2-1} \right)}{\sqrt{t^2-1}} \right) dt}
\\
t=\cosh u,=\frac{1}{a}\int_{\mathrm{arc}\cosh \omega}^{\infty}{e^{ia\sinh u}du}
\\
\int_{-\infty}^{t-a}{\frac{e^{-i\omega u}}{t-u}J_0\left( \sqrt{\left( t-u \right) ^2-a^2} \right) du}=e^{-i\omega t}I\left( \omega \right)
\\
=\frac{1}{a}e^{-i\omega t}\int_{\mathrm{arc}\cosh \omega}^{\infty}{e^{ia\sinh u}du}
$$
I think this integral may not have a closed-form solution.
|
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|
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$ The question is
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$.
My attempt:
This reduces to finding primes $p,q$ such that p divides $q^3+1=(q+1)(q^2-q+1)$ and q divides $p^3+1=(p+1)(p^2-p+1)$.
Now we have 4 cases: If p divides $q+1$ and q divides $p+1$, then $\{p,q\}=\{2,3\}$.
I don't know how to deal with the other cases. For example, what happens if p divides $q+1$ and q divides $p^2-p+1$?
|
If $p \mid q+1$ and $q \mid p^2-p+1$, then $\frac{p^2-p+1}q$ must be $-1$ modulo $p$, since $q\equiv -1\pmod p$. As a result, there exist positive $m,n$ for which
$$(mp-1)(np-1)=p^2-p+1.$$
We clearly can't have $m=n=1$, so we must have
$$(p-1)(2p-1)\leq (mp-1)(np-1)=p^2-p+1\implies p\leq 2,$$
and thus $(2,3)$ is the only solution here as well. The same goes for if $p$ and $q$ are swapped.
For the final case, we claim that $a \mid b^2-b+1$ and $b \mid a^2-a+1$ has no solutions in positive integers besides $(a,b)=(1,1)$. To show this, assume for the sake of contradiction that there exists such a pair with minimal sum, and without loss of generality assume $a>b$. Now, consider
$$c=\frac{b^2-b+1}a.$$
Since $b>1$ (as otherwise $a=b=1$), $c<a$, so $b+c<a+b$. However, $c \mid b^2-b+1$, and
$$c^2-c+1\equiv \frac1{a^2}-\frac1a+1\equiv \frac{a^2-a+1}{a^2}\equiv 0\pmod b$$
(using that $\gcd(a,b)=1$), so $b \mid c^2-c+1$ as well. Then $(b,c)$ is a pair satisfying the conditions with smaller sum than $(a,b)$, a contradiction. So, not only can there not exist primes $(p,q)$ with $p \mid q^2-q+1$ and $q \mid p^2-p+1$, there can't exist any such pairs of positive integers.
|
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|
Determining convergence of sequence $a_n=a_{n-1}^{-1}+a_{n-2}^{-1}$ I'm trying to prove convergence for the sequence
$$a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}$$
with $a_0=a_1=1$. If it does converge then it converges to $\sqrt{2}$, which agrees with numerical tests. $$$$I've managed to prove two facts about the sequence, although I'm not sure they're relevant to the question or not. Firstly, for every set of three consecutive terms in the sequence, at least one is bigger than $\sqrt{2}$ and at least one is smaller than $\sqrt{2}$. Secondly, if two consecutive terms are in the interval $(\alpha,\,2\alpha^{-1})$ for some $\alpha\in[1,\,\sqrt{2})$, then all terms that follow are in this interval as well. $$$$Does anyone have an idea for how to prove convergence, with or without these facts?
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By induction, we can easily show that $1 \leq a_n \leq 2$ for all $n$. Next, note that the recurrence relation can be recast as
$$ a_n - \sqrt{2} = -\frac{a_{n-1} - \sqrt{2}}{\sqrt{2}a_{n-1}} - \frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}}. $$
Applying this relation to the factor $a_{n-1} - \sqrt{2}$ on the right-hand side again,
\begin{align*}
a_n - \sqrt{2}
&= -\frac{1}{\sqrt{2}a_{n-1}}\left( -\frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}} - \frac{a_{n-3} - \sqrt{2}}{\sqrt{2}a_{n-3}} \right) - \frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}} \\
&= - \frac{\sqrt{2} - a_{n-1}^{-1}}{2a_{n-2}} (a_{n-2} - \sqrt{2}) + \frac{1}{2a_{n-1}a_{n-3}}(a_{n-3} - \sqrt{2}).
\end{align*}
From this and the bounds for $(a_n)$ altogether, we get
$$ \left|a_n - \sqrt{2}\right| \leq \frac{\sqrt{2}-\frac{1}{2}}{2}\left|a_{n-2}-\sqrt{2}\right| + \frac{1}{2}\left|a_{n-3} - \sqrt{2}\right|. $$
From this, we find that $\varepsilon := \limsup_{n\to\infty} |a_n - \sqrt{2}|$ satisfies
$$ \varepsilon \leq \frac{\sqrt{2}-\frac{1}{2}}{2} \varepsilon + \frac{1}{2} \varepsilon = \frac{2\sqrt{2}+1}{4} \varepsilon. $$
Since $\frac{2\sqrt{2}+1}{4} < 1$, this inequality is satisfied only when $\varepsilon = 0$, and therefore the claim follows.
|
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|
Evaluate $\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$ Evaluate
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$$
I transformed it into:
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2(1-\cos^2(\theta)) + b^2\cos^2(\theta)} \ d\theta\\ \rightarrow \frac{b}{\sqrt{b^2-a^2}}\int_{0}^{2\pi} |\sin(\theta)| \sqrt{\frac{a^2}{b^2-a^2} + \cos^2(\theta)} \ d\theta$$
Using $u = \cos(\theta)$ I can make the substitution:
$$\frac{-b}{\sqrt{b^2-a^2}}\int_{1}^{0} \sqrt{\frac{a^2}{b^2-a^2} + u^2} \ du$$
Now I want to use the formula for $\int \sqrt{x^2 + a^2} dx$ but that will give me an expression of $\ln(x)$ I somehow need to end up with :
$$b^2 +\frac{a^2b}{\sqrt{a^2-b^2}}\arcsin(\frac{\sqrt{a^2-b^2}}{a})$$
Does anyone see a route to get this?
|
Let $$F(a,b)=\int_0^{2\pi}|\sin(x)|\sqrt{a^2\sin^2(x)+b^2\cos^2(x)}\,dx.$$
Symmetry of the integrand gives
$$F(a,b)=4\int_0^{\pi/2}|\sin(x)|\sqrt{a^2\sin^2(x)+b^2\cos^2(x)}\,dx.$$
For $x\in[0,\pi/2]$, $|\sin x|=\sin x$, so the integral is
$$F(a,b)=4\int_0^{\pi/2}\sin(x)\sqrt{a^2\sin^2(x)+b^2\cos^2(x)}\,dx.$$
Then the trick is to assume $a^2>b^2$, then we have
$$F(a,b)=4\int_0^{\pi/2}\sin(x)\sqrt{a^2-(a^2-b^2)\cos^2(x)}\,dx,$$
which is
$$F(a,b)=\frac{4}{\sqrt{a^2-b^2}}\int_0^{\pi/2}\sin(x)\sqrt{q^2-\cos^2(x)}\,dx,$$
where $$q=\frac{a}{\sqrt{a^2-b^2}}.$$
Take $u=\cos x$,
$$F(a,b)=\frac{4q}{a}\int_0^1\sqrt{q^2-u^2}\,du.$$
Can you take it from here?
|
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|
Or/Not Conditions in Number Theory How many positive integers, not exceeding 100, are multiples of 2 or 3 but not 4?
Hello, I was wondering how to solve problems like these. My thoughts are to count the number that are multiples of 2 but not 4, multiples of 3 but not 4, and then multiples of 6 but not 4, then use Principle of Inclusion and Exclusion as follows:
$$\left\lfloor \frac{100}{2} \times \frac{1}{2} \right\rfloor + \left\lfloor \frac{100}{3} \times \frac{3}{4} \right\rfloor - \left\lfloor \frac{100}{6} \times \frac{1}{2} \right\rfloor = 42,$$
but I am unsure as to if this is a valid solution. Can anybody confirm/deny that I am using Principle of Inclusion and Exclusion properly? Thanks!
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What you did is not quite correct, although it does give the correct result. This is because each floor function in inclusion-exclusion is for just one set of divisions only, not multiple ones combined as you did, although combining them still sometimes gives the same values anyway, as it does in your case. An example where it doesn't work is if $102$ were used instead, since there are $26$ multiples of $2$ but not $4$, but your technique gives $\left\lfloor \frac{102}{4} \right\rfloor = 25$. Instead, as shown in \eqref{eq1A} below, using inclusion-exclusion appropriately gives $\left\lfloor \frac{102}{2} \right\rfloor - \left\lfloor \frac{102}{4} \right\rfloor = 51 - 25 = 26$.
The procedure you seem to be following is to start with the number of multiples of $2$, but not $4$. Next, add the number of multiples of $3$, but not $12$. Finally, since you double counted these, subtract the number of multiples of $6$, but not $12$. The principle of inclusion and exclusion then gives
$$\begin{equation}\begin{aligned}
& \left(\left\lfloor \frac{100}{2} \right\rfloor - \left\lfloor \frac{100}{4} \right\rfloor \right) + \left(\left\lfloor \frac{100}{3} \right\rfloor - \left\lfloor \frac{100}{12} \right\rfloor \right) - \left(\left\lfloor \frac{100}{6} \right\rfloor - \left\lfloor \frac{100}{12} \right\rfloor \right) \\
& = (50 - 25) + (33 - 8) - (16 - 8) \\
& = 25 + 25 - 8 \\
& = 42
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
FYI, here's a bit simpler way to solve the problem. As you did, start with those numbers which are multiples of $2$ but not $4$, i.e., $2x$ with $x \in \{1, 3, \ldots, 49\}$, of which there are $25$ integers. Next, since it's a disjoint subset, you can just add the number of odd multiples of $3$, i.e., $3x$ with $x \in \{1, 3, \ldots, 33\}$, so there's $17$ of them. This then gives the same result as yours, i.e.,
$$25 + 17 = 42 \tag{2}\label{eq2A}$$
Note you can also get the same thing using inclusion-exclusion. This is done by, after dealing with the multiples of $2$ but not $4$, then adding the number of multiples of $3$ but not $6$, giving
$$\begin{equation}\begin{aligned}
& \left(\left\lfloor \frac{100}{2} \right\rfloor - \left\lfloor \frac{100}{4} \right\rfloor \right) + \left(\left\lfloor \frac{100}{3} \right\rfloor - \left\lfloor \frac{100}{6} \right\rfloor \right) \\
& = (50 - 25) + (33 - 16) \\
& = 25 + 17 \\
& = 42
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Also note that \eqref{eq1A} is basically the same as \eqref{eq3A} since the two $\left\lfloor \frac{100}{12} \right\rfloor = 8$ terms cancel each other out.
|
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|
Given a polynomial $W(x)$. Find all pairs of integers $a,b$ that satisfy $W(a)=W(b)$ Given a polynomial $W(x) = x^4-3x^3+5x^2-9x$. Find all pairs of distinct integers $a,b$ that satisfy $W(a)=W(b)$. My approach was to factor the polynomial.
$$\begin{align*}
x^4-3x^3+5x^2-9x &= (x^4-3x^3+2x^2)+(3x^2-9x+6)-6\\
&= x^2(x^2-3x+2)+3(x^2-3x+2)-6\\
&=(x-1)(x-2)(x^2+3)-6
\end{align*}$$
As $-6$ is a constant we can now consider the following expression $$\\(x-1)(x-2)(x^2+3)$$ form this form we see that $(a,b)=(1,2)$ and $(a,b)=(2,1)$ satisfy the condition. We can also graph this function and see that $(1)$WLOG $a\ge 2$ and $b\le 1$ but as $1$ and $2$ have been used we get $a\ge 3$ and $b\le 0$. Now, if $|b|>|a|$ the following holds $$\\(b-1)(b-2)(b^2+3)>(a-1)(a-2)(a^2+3)$$
So we get that $|a|>|b|$ and $(1)$. How do I proceed? Any help appreciated.
|
Some calculation of $W(a)-W(b) = 0$ give us :
$$(a+b)\Big((a+b)^2-2ab\Big) - 3\Big((a+b)^2-ab\Big) +5(a+b)-9=0$$
Let $x=a+b$ and $y=ab$ and we get $$(2x-3)y = x^3-3x^2+5x-9$$
Let $n=2x-3$ so $n$ is odd, then we have $ x\equiv_n {3\over 2}$ so $$0 \equiv_n x^3-3x^2+5x-9 \implies 39\equiv_n 0 $$ so $$2x-1\mid 39 \implies x\in \{-1, 0,1, 2,-6,7,-19,20 \}$$
Now it is not difficult to get $y$ and then solve each of $8$ systems.
|
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|
Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$? Let $P(z)\equiv 1($ mod $ \ z) $ be a polynomial of degree $n>3$ with integer coefficients. Are there infinitely many positive integers $x, y, z$ such that $(xz+1)(yz+1)=P(z)$?
If $P(z) = a_nz^n+1$, it has be proven that the Diophantine equation has infinitely many solutions in positive integers $x, y, z$ Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers..
From experiment, it appears the assertion is true for all polynomials $P(z)\equiv 1$(mod$ \ z)$ of degree $n>3$. How do we go about proving this?
Note if $n=3$, it has been shown that the Diophantine equation has a finite number of solutions
https://mathoverflow.net/questions/392002/is-xz1-a-proper-divisor-of-a-3z3a-2z2a-1z1-finitely-often/392018#392018
|
your first recipe, 5,20, 51, 104.. works
$$ z = n^3 + 2n^2 + 2n $$
$$x = n+1 $$
$$ 1+xz = n^2 + 3 n^3 + 4 n^2 +2n+1 $$
$$ y = n^5 +3n^4 +5n^3 + 4 n^2 + 1 $$
$$ 1 + yz = n^8 +5n^7 +13n^6 +20n^5 + 19n^4 + 10n^3 + 2 n^2 + 1$$
$$ ( 1+xz)(1+yz) = n^{12} + 8n^{11} + 32n^{10} + 81n^9 + 142n^8 + 178n^7 + 161n^6 + 104n^5 + 48n^4 + 17n^3 + 6n^2 + 2n + 1 $$
which is the same as $z^4 + z^3 + z^2 + z + 1.$
|
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|
Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :-
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is clearly rejected as $x,n$ are both positive.
CASE $2$ : $mn<0$ and $m^3+y^3=8+2my$
For this we apply AM - GM to get,
$$9 + 2my = m^3 + y^3 + 1 \geq 3my \implies my \le 9 $$
Case bash (sadly) gives that the solutions to this case are
$$(m,y)=(0,2), \enspace (2,0), \enspace (2,2)$$
Now we are left with two cases
CASE $3$ : $mn>0$ and $m^3-n^3=8-2mn$
CASE $4$ : $mn>0$ and $x^3-y^3=2xy-8$
Now how to finish off these last $2$ cases?
Thanks in advance!
|
With a little help from ultralegend5385's hint.
Let $m, n \in \mathbb{Z}$. Factoring the LHS, we obtain $$m^3-n^3 = (m-n)(m^2+mn+n^2)$$
Then our equation becomes
$$ m^3-n^3=(m-n)(m^2+mn+n^2)=2mn+8=2(mn+4)$$
$$ (m-n)(m^2+mn+n^2)=2(mn+4)$$
We can split this equation into two cases:
Case 1: $$(m-n) = 2 \tag{1}\label{eq1}$$
$$(m^2+mn+n^2)=(mn+4)\tag{2}\label{eq2}$$
Equation \eqref{eq1} becomes $m=n+2$, then substituting into equation \eqref{eq2}, we get
$$ (n+2)^2 + (n+2)n + n^2 = (n+2)n+4$$
$$ (n^2 + 4n + 4) + n^2+2n + n^2 = n^2 + 2n+ 4$$
$$2n^2+4n+4=4$$
$$n^2+2n=0$$
$$n(n+2)=0$$
So then $n=0, -2$ satisfy the equation. Then $n=0\implies m=2$ and $n=-2 \implies m=0$. Two solutions to our original equation are $(m,n)=(2,0)$ and $(m,n)=(0, -2)$.
Case 2: $$(m-n) = (mn+4) \tag{3}\label{eq3}$$
$$(m^2+mn+n^2)= 2\tag{4}\label{eq4}$$
From Equation \eqref{eq3}, we get that $(m-n)$ must be even. If m and n are of opposite parity, their product must be even so the RH side will be even but the LH side will be odd which is impossible. If they are both odd then their product is odd so the RH side will be odd but their difference will be even again impossible. Therefore m and n must both be even.
Let $k\in \mathbb{Z}$
$$ 2k=mn+4$$
$$ 2k-4=mn$$
$$2(k-2)=mn \implies$$ $$m=2 \text{ and } n=(k-2) \tag{5}\label{eq5}$$ or
$$m=(k-2) \text{ and } n=2 \tag{6}\label{eq6}$$
Let's substitute equations \eqref{eq5} into equations \eqref{eq4}, so that we can attempt to solve for k
$$ (2^2 + 2(k-2)+(k-2)^2)=2$$
$$ 4 + 2k - 4 + (k^2 - 4k + 4) = 2$$
$$ k^2-2k+2=0\tag{7}\label{eq7}$$
Equation \eqref{eq7} is in the form of a quadratic equation, so we can apply the quadratic formula.
When we do that we get imaginary roots, so there is no solution for k in the integers.
We can observe that equation \eqref{eq4} is symmetric in terms of m and n, so equations \eqref{eq6} yield the same result.
Therefore, our only solutions for the equation $m^3-n^3=2mn+8$ is
$(m,n)=(2,0)$ and
$(m,n)=(0, -2)$.
P.S. Sorry for the formatting this is my first longer post
|
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|
Solve: $y''=(1+(y')^2)^{3/2}$ / $y(1)=\frac 15, y'(1)=\frac 34$. My attempt:
$y''=(1+(y')^2)^{3/2}$
$\displaystyle\implies \frac {y''}{(1+(y')^2)^{3/2}}=1$
$\displaystyle\implies \int_{}^{} \frac {y''}{(1+(y')^2)^{3/2}} dx=x+c$
$\displaystyle\implies \int\limits_{}^{}\frac{1}{(1+u^2)\sqrt{1+u^2}}du=x+c,$ where $ u=y'$
$\displaystyle\implies \int\limits_{}^{}\frac{\sqrt{1+u^2}}{(1+u^2)^2}du=x+c$
I am stuck here, is my approach correct and can solve the problem? Thanks for your help.
|
You have found that
$\displaystyle\int\frac{du}{(1+u^2)^{\frac{3}{2}}}=x+c$
Let $u=\tan\theta$ , therefore $du=\sec^2\theta d\theta$
$\displaystyle\int\frac{\sec^2\theta d\theta}{\sec^3\theta}=x+c$
$\displaystyle\int\cos\theta=x+c$
$\sin\theta=x+c$
$\displaystyle \frac{u}{\sqrt{1+u^2}}=x+c$
where $u=y'$
We can find $c$ by the fact the $u(1)=\frac{3}{4}$
Solving we get
$\displaystyle \frac{u}{\sqrt{1+u^2}}=x-\frac{2}{5}$
Now $u=\frac{dy}{dx}$ and then our integral becomes
$\displaystyle\int dy =\int \frac{x-\frac{2}{5}}{\sqrt{1-(x-\frac{2}{5}})^2}dx$
which can be easily solved using substitution.
The final answer would be
$(y-1)^2+(x-\frac{2}{5})^2=1$
|
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|
What is the value of the PL segment if QL = 12? For reference:
My progress...
the idea would be to demonstrate that $triangle AQL \cong \triangle LQC$
$\triangle PAB \sim QCB (AA)\\
\measuredangle PBA = \measuredangle QCB=90-\theta$
Could I say that $\triangle ABC$ is isoscelels without showing that BL is perpendicular to AC?
with the answer , the drawing to scale would look like this...
|
This is an interesting question. Please note your mistake is in trying to prove that $\triangle ABC$ is isosceles. $PL = QL$ holds regardless of whether $\triangle ABC$ is isosceles or not.
All we are given is that $\angle BAP = \angle BCQ = \theta$, and $BP$ and $BQ$ are perpendicular drawn to $AP$ and $CQ$ respectively. $L$ is the midpoint of $AC$.
We draw perp $BH$ from $B$ to $AC$. $LH = x$, $AL = LC = d$ (say).
As $\triangle BAP \sim \triangle BCQ$, $ \frac{BP}{BQ} = \frac{AP}{CQ} = \frac{AB}{BC} = k \ $ (say)
As quadrilateral $AHBP$ and $CHBQ$ are cyclic, we can apply Ptolemy's theorem.
In $AHBP$, $ \ AH \cdot BP + BH \cdot AP = AB \cdot HP$
$(d + x) \cdot k \cdot BQ + BH \cdot k \cdot CQ = k \cdot BC \cdot HP$
$(d + x) \cdot BQ + BH \cdot CQ = BC \cdot HP \tag1$
In $CHBQ$, $ \ CH \cdot BQ + BH \cdot CQ = BC \cdot HQ$
$(d - x) \cdot BQ + BH \cdot CQ = BC \cdot HQ \tag2$
Subtracting $(2)$ from $(1)$,
$2x \cdot BQ = BC \cdot (HP - HQ)$
$2x \sin \theta = HP - HQ \tag3$
Now $\angle BHP = \angle BAP = \theta$, so $\angle AHP = 90^\circ - \theta$. Similarly, $\angle BHQ = \angle BCQ = \theta$ and $\angle AHQ = 90^\circ - \theta$.
By law of cosine in $\triangle PHL$,
$PL^2 = x^2 + HP^2 - 2 x \cdot HP \cos(90^\circ - \theta)$
$PL^2 = x^2 + HP^2 - HP (HP - HQ) = x^2 + HP \cdot HQ\ $ (using $3$)
By law of cosine in $\triangle QHL$,
$QL^2 = x^2 + HQ^2 - 2 x \cdot HQ \cos(90^\circ + \theta)$
$QL^2 = x^2 + HQ^2 + HQ (HP - HQ) = x^2 + HP \cdot HQ\ $
So, $PL = QL = 12$
|
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|
Simplifiying $A \cos(x-\phi)$ I have to proof that we can transform this expression : $$a \cos x + b \sin x$$
to this one : $$A \cos (x-\phi)$$
Indeed, I can do the reverse path using the formula of $\cos(x-y)= \cos x \cos y + \sin x \sin y$
but I get stuck on the initial path.
Which trig formula should I use?
Thanks
|
You can proceed with your initial path as follows
$$a\cos x +b\sin x$$
Multiply and divide by $\displaystyle\frac{1}{\sqrt{a^2+b^2}}$
$$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos x +\frac{b}{\sqrt{a^2+b^2}}\sin x)$$
Now $(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1$ and both terms are less than $1$
Therefore if we let one as $\cos \phi $ other can be let us $\sin \phi$
Therefore we have
$$\sqrt{a^2+b^2}(\cos\phi\cos x + \sin\phi\sin x)$$
$$A\cos (\phi-x)$$
Thus there exists two reals $A=\sqrt{a^2+b^2}$ and $\phi= \cos^{-1}(\frac{a}{\sqrt{a^2+b^2}})$ for which $a\cos x +b\sin x$ is identically equal to $A\cos (\phi-x)$
|
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Why is $\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$? I found the relation for $n\geq3$ $$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$$But despite my best efforts, I still have no idea as to how to prove it.
Things I've tried:
*
*Adding $\cos^2$ terms. Of course, $\sin^2x+\cos^2x$ and $\cos^2x-\sin^2x$ are both simplifiable, so I thought about adding $\cos^2$ terms to make the sum into $\frac n2$ and hope that the $\sin^2$ and $\cos^2$ terms sum to equal amounts. They don't in the $n$ odd case, so I didn't know what to do with this approach.
*Adding more $\sin^2$ terms: Since $\sin^2 x=\sin^2(\pi-x)$, we can add terms to this sum, but honestly, it didn't make the sum any easier to evaluate.
Any help here?
Edit:
Based on the comments, here are my attempts.
$$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac12\cdot\left\lfloor\frac n2\right\rfloor-\frac12\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\cos\left((2k-1)\frac{2\pi}n\right)$$$$=\frac12\cdot\left\lfloor\frac n2\right\rfloor-\frac12\mathfrak{Re}\left(\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)\right)$$How do I finish?
|
Following your way, by geometric series we have that for $n=2N$
$$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)=\sum\limits_{k=1}^{N}\exp\left((2k-1)\frac{i\pi}N\right)=\\=e^{-\frac{i\pi }N}\sum\limits_{k=1}^{N}\left(e^{\frac{i2\pi}N}\right)^k=e^{-\frac{i\pi }N}\frac{e^{\frac{i2\pi}N}-e^{\frac{i2\pi(N+1)}N}}{1-e^{\frac{i2\pi}N}}=e^{\frac{i\pi }N}\frac{1-e^{i2\pi}}{1-e^{\frac{i2\pi}N}}=0$$
then
$$\frac12\cdot\left\lfloor\frac n2\right\rfloor-\frac12\mathfrak{Re}\left(\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)\right)=\frac12 N+0 = \frac n 4$$
For $n=2N+1$
$$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)=\sum\limits_{k=1}^{N}\exp\left((2k-1)\frac{i\pi}{2N+1}\right)=\\
=e^{-\frac{i2\pi }{2N+1}}\sum\limits_{k=1}^{N}\left(e^{\frac{i4\pi}{2N+1}}\right)^k
=e^{-\frac{i2\pi }{2N+1}}\frac{e^{\frac{i4\pi}{2N+1}}-e^{\frac{i4\pi(N+1)}{2N+1}}}{1-e^{\frac{i4\pi}{2N+1}}}
=\frac{e^{\frac{i2\pi}{2N+1}}-1}{1-e^{\frac{i4\pi}{2N+1}}}=-\frac1{1+e^{\frac{i2\pi}{2N+1}}}$$
and we also have
$$w=-\frac1{1+e^{\frac{i2\pi}{2N+1}}} \implies w+\bar w=-1 \implies \Re(w)=-\frac12$$
indeed
$$w+\bar w=-\frac1{1+e^{\frac{i2\pi}{2N+1}}}-\frac1{1+e^{\frac{-i2\pi}{2N+1}}}=-\frac1{1+e^{\frac{i2\pi}{2N+1}}}-\frac{e^{\frac{i2\pi}{2N+1}}}{1+e^{\frac{i2\pi}{2N+1}}}=-1$$
then
$$\frac12\cdot\left\lfloor\frac n2\right\rfloor-\frac12\mathfrak{Re}\left(\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)\right)=\frac12 N+\frac14 = \frac{2N+1}{4}=\frac n 4$$
|
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|
Integrating inverse of a polynomial using valid method gives wrong result The integral is:
$$\int \frac{1}{x^2+(x-1)^2}dx$$
One way to integrate inverses of polynomials is to find their roots and find the fractions which when added together, give the original fraction (with the roots being the denominators). For example:
$$\int \frac{1}{x^2-x-2}dx = \int \frac{1}{(x-2)(x+1)}dx\rightarrow \frac{a}{x-2}+\frac{b}{x+1}=\frac{a(x+1)}{(x-2)(x+1)}+\
\frac{b(x-2)}{(x-2)(x+1)}$$
$$\rightarrow ax+a+bx-2b=1 \rightarrow \left\{
\begin{array}{c}
ax+bx=0x \\
a-2b=1 \\
\end{array}
\right. \rightarrow a=\frac{1}{3}, b=-\frac{1}{3}$$
$$\rightarrow \int \frac{1}{x^2-x-2}dx=\int (\frac{1/3}{x-2}-\frac{1/3}{x+1})dx$$
And then solve.
I know this is known, I just wanted to show what I've been taught exactly, no ambiguation.
So about the question. We can open up the squares to get:
$$\int \frac{1}{x^2+(x-1)^2}dx = \int \frac{1}{2x^2-2x+1}dx=\int \frac{1}{2(x-0.5-0.5i)(x-0.5+0.5i)}dx$$
$$\rightarrow \frac{a}{x-0.5-0.5i}+\frac{b}{x-0.5+0.5i}=\frac{1/2}{(x-0.5-0.5i)(x-0.5+0.5i)}$$
I'll spare the details:
$$a=-\frac{i}{2},b=\frac{i}{2}$$
$$\rightarrow \int \frac{1}{x^2+(x-1)^2}dx=\int (\frac{i/2}{x-0.5+0.5i}-\frac{i/2}{x-0.5-0.5i})dx=i(\int \frac{1}{2x-1+i}dx-\int \frac{1}{2x-1-i}dx)$$
Once again sparing the details of some simple u substitution:
$$=\frac{i}{2}(ln\lvert 2x-1+i\rvert-ln\lvert 2x-1-i\rvert)$$
Absolute value (which's added from integrating something to the (-1)) always gives out a real number, and the entire "ln" expression is multiplied by an imaginary number, so the output is always imaginary.
The actual video I took this from contained a completely different solution. It's not even close to mine, which's always imaginary. Why? How did this happen?
Thanks.
|
You did it as if$$\int\frac1{x-i}\,\mathrm dx=\log|x-i|.$$Actually, not only this is not true as it doesn't make sense, since $\log|x-i|$ is not differentiable, in the sense of Complex Analysis. You should work with$$\int\frac1{x-i}\,\mathrm dx=\log(x-i),$$with a carefully chosen branch of the logarithm.
|
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|
Showing a solution to 3rd order differential equations forms a subspace Let $S$ denote the set of all solutions of the following differential equation defined on $C^3[0,\infty)$;
$$
\begin{align}
\frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx = 0
\end{align}
$$
Show that $S$ is a linear subspace of $C^3[0, \infty)$
I know that subspaces are closed under addition and scalar multiplication and also contan the zero vector, but I'm not sure how to show that all solutions form a linear subspace. (I'm not sure how to solve this equation either)
Final Answer
Note that:
$$
\begin{align}
0 &= \frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx \\
&= \alpha \left(\frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx\right)
\end{align}
$$
Where $\alpha$ is a constant.
Consider arbitrary functions $f$ and $g$ and scalar $\alpha$. By linearity of operators we have:
$$
\begin{align}
\frac{d^3}{dt^3}(\alpha f + g ) &= \alpha \frac{d^3f}{dt^3} + \frac{d^3g}{dt^3} \\
\frac{d^2}{dt^2}(\alpha f + g ) &= \alpha \frac{d^2f}{dt^2} + \frac{d^2g}{dt^2} \\
\frac{d}{dt}(\alpha f + g ) &= \alpha \frac{df}{dt} + \frac{dg}{dt} \\
d(\alpha f + g) &= \alpha df + dg
\end{align}
$$
Now assign $f$ and $g$ to arbitrary individual solutions to our differential equation, we have:
$$
\begin{align}
0 &= \frac{d^3}{dt^3}(\alpha f + g) + b \frac{d^2}{dt^2}(\alpha f + g) + c \frac{d}{dt}(\alpha f + g) + d(\alpha f + g) \\
&= \alpha(\frac{d^3}{dt^3}f + b \frac{d^2}{dt^2}f + c \frac{d}{dt}f + df ) + (\frac{d^3}{dt^3}g + b \frac{d^2}{dt^2}g + c \frac{d}{dt}g + dg) \\
&= \alpha \pmb f + \pmb g
\end{align}
$$
This shows that $\alpha \pmb f + \pmb g$ solves the ODE which proves that $\alpha \pmb f + \pmb g \in \mathcal S$
The addition holds because the sum that describes $\pmb f$ sums to $0$. Likewise for $\pmb g$.
Therefore, all arbitrary combinations of $\pmb f$ and $\pmb g$ are in $\mathcal S$ (This can be seen by also allowing $\pmb f$ be $\pmb g$ and $\pmb g$ be $\pmb f$).
Finally, $\pmb 0$ is also in the set of solutions.
Thus, the set of all solutions forms a subspace.
|
It is easy to see that $S\subset \mathcal{C}^3[0,\infty)$.
All you need do is show that if $f$ and $g$ are in $S$ and $\alpha$ is a scalar then $\alpha f+g$ is a solution of the ODE. In other words, $\alpha f+g \in S$.
Clearly $\frac{d^3}{dt^3}(\alpha f+g) = \alpha \frac{d^3f}{dt^3} + \frac{d^3g}{dt^3}$, $\frac{d^2}{dt^2}(\alpha f+g) = \alpha \frac{d^2f}{dt^2} + \frac{d^2g}{dt^2}$, and
$\frac{d}{dt}(\alpha f+g) = \alpha \frac{df}{dt} + \frac{dg}{dt}$.
So,
$\frac{d^3}{dt^3}(\alpha f+g) + \frac{d^2}{dt^2}(\alpha f+g) + \frac{d}{dt}(\alpha f+g) +d(\alpha f+g)$
$ = \alpha \bigg(\frac{d^3f}{dt^3} + \frac{d^2f}{dt^2}+\frac{df}{dt} + df\bigg) +\frac{d^3g}{dt^3}+ \frac{d^2g}{dt^2}+ \frac{dg}{dt} + dg = 0$
|
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|
Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?
|
If $A=0$, then the equation $Ax^2+Bx+C=0$ is a linear equation, not a quadratic equation. Hence, it only has one root. Since $Bx+C=0$ is not a quadratic equation, there is no reason to think that applying the quadratic formula would produce the correct roots.
If you consider the derivation of the quadratic formula, in the very first line we rewrite the equation $Ax^2+Bx+C=0$ as $x^2+\frac{B}{A}x+\frac{C}{A}=0$. This step is only valid if $A\neq0$. Indeed, the entire derivation is premised on the assumption that $A\neq0$. Therefore, it is unsurprising that trying to apply the formula to the case $A=0$ does not work. The expression $\frac{-B\pm\sqrt{B^2-4AC}}{2A}$ does not even make sense when $A=0$.
|
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|
What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \frac{\sqrt3}{BD}\therefore BD = 2\sqrt3\\
cos 30 = \frac{BE}{BD}\rightarrow \frac{\sqrt3}{2}=\frac{BE}{2\sqrt3}\therefore BE = 3\implies BN = 3+\sqrt3\\
\triangle BNG:cos 30 = \frac{BN}{BG}\rightarrow \frac{\sqrt3}{2}=\frac{3+\sqrt3}{BG}\rightarrow BG = \frac{6+2\sqrt3}{\sqrt3}=2\sqrt3+2\\
BG = \frac{2BP}{3}\rightarrow BP = 3\sqrt3 + 3\\
\triangle BPC: tg30 = \frac{PC}{BP}\rightarrow \frac{\sqrt3}{3} = \frac{PC}{3\sqrt3+3} \implies \boxed{ PC = 3+\sqrt3}\\
\therefore \boxed{\color{red}AC = 2(3+\sqrt3) = 6+2\sqrt3}$
My question...only the equilateral triangle meets the conditions? Why if the quadrilateral is indescribable $\measuredangle MGN=120^o$
|
Given $ \small MBNG$ is a tangential quadrilateral, it is easy to see that,
$ \small BN + MG = BM + GN$
[How? $ \small BE + EN = BF + NH, MK + KG = MF + GH$. In fact there is a theorem called Pitot Theorem that states the same.]
So if $ \small AN = d, CM = e$, we have $ \small MG = \frac{e}{3}, \small GN = \frac{d}{3}$ and we rewrite $\small BN + MG = BM + GN$ as,
$\frac{c}{2} + \frac{d}{3} = \frac{a}{2} + \frac{e}{3}$
$ d + \frac{3c}{2} = e + \frac{3a}{2} \tag1$
Now given the circle is also the incircle of $\triangle BCM$ and $\triangle ABN$, AND the area of both $\triangle ABN$ and $\triangle BCM$ are same, which is half the area of $\triangle ABC$, we conclude they must have the same perimeter.
$c + d + \frac{a}{2} = a + e + \frac{c}{2}$
$d + \frac{c}{2} = e + \frac{a}{2} \tag2$
Subtracting $(2)$ from $(1)$, $c = a$.
As $a = c ~$ and angle between the sides is $60^\circ$, $\triangle ABC$ must be an equilateral triangle.
So we easily find that,
$ \small AC = BC = 2 BN = 2 (BE + EN) = 2 (3 + \sqrt3)$
|
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|
Can pmf of discrete random variables vary? Can probability mass function be in different form even though when we plug in $k =1,2,3,\cdots$ for $P(X=k)$, the probability for each $k$ between pmf A and pmf B would still be the same. For example :
$1.6.2$. Let a bowl contain $10$ chips of the same size and shape. One and only one of these chips is red. Continue to draw chips from the bowl, one at a time and at random and without replacement, until the red chip is drawn.
(a) Find the pmf of $X$, the number of trials needed to draw the red chip. (b) Compute $P(X \leq 4)$. (from Introduction to Mathematical Statistics by C. Hogg)
The key answer would be $p(x) = \dfrac{\binom{9}{x-1}}{\binom{10}{x-1}}\times\dfrac1{11-x}$ for $x = 1,2,...,10$. What if i write $p(x) = \dfrac{\binom{x}{1}}{\binom{10}{1}} \times\frac1x$ for $x = 1,2,...,10$?
|
Note that \begin{align*}
\frac{\binom9{x-1}}{\binom{10}{x-1}}\times \frac{1}{11-x}&=\frac{\frac{9!}{(x-1)!(10-x)!}}{\frac{10!}{(x-1)!(11-x)!}}\times \frac{1}{11-x}=\frac{9!}{10!}\times \frac{(11-x)!}{(10-x)!}\times \frac{1}{11-x}\\
&= \frac{1}{10}\times \frac{(10-x)!}{(10-x)!}=\frac{1}{10}
\end{align*}
and your answer is equal to $$\frac{\binom x1}{\binom{10}{1}}\times \frac1x=\frac x{10}\times \frac 1x=\frac{1}{10} $$
The two expressions are equivalent.
|
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Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}.$$
I solved this problem by Hölder:
$$\left(\sum_{cyc}\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}\right)^2\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2}\geq\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3$$
and it remains to prove that
$$(ab+ac+bc)\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3\geq3\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2},$$ which is true by BW and by using computer.
In this topic https://artofproblemsolving.com/community/c6h542992 there is a proof (from gxggs), but it's very very complicated.
I found another way, a smooth enough, but it's still a very hard solution.
I am looking for a nice proof by hand, for which there is a possibility to find this proof during a competition.
Thank you!
|
My attempt for this very hard inequality as conjecture :
Conjecture :
Let $x\geq z$ and $z\leq y\leq 1.5z$ and $z\geq 1$ then it seems we have :
$$0\le f\left(x\right)+f\left(y\right)-2f\left(\frac{2}{\frac{1}{x}+\frac{1}{y}}\right)$$
Where : $$f\left(a\right)=\frac{1}{\sqrt{2a^{2}+z\left(a-z\right)}}$$
The trick here is to use $a+b=u$ and $ab=b(u-b)$
Next it seems easier to show for $x,b,c>0$:
$$g\left(x\right)=\frac{2}{\sqrt{2\left(\frac{2}{\frac{1}{x+b}+\frac{1}{b+c}}\right)^{2}+b\left(\frac{2}{\frac{1}{x+b}+\frac{1}{b+c}}-b\right)}}+\frac{1}{\sqrt{2x^{2}+2c^{2}+5xc}}-\sqrt{\frac{3}{xb+bc+cx}}\geq 0$$
Hope it helps !
|
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|
Integrate $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx} = $ Solve $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx} = \_\_\_\_\_$ where [.] represent greatest integer function.
My approch is as follow
$\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{\left[ {{{\left( {x - 4} \right)}^2}} \right] + \left[ {{x^2}} \right]}}dx} $
$y=(x-4)^2$ represent a quadratic equation
Refer to the image below
We see that for some x value y will have values of 9,8,7,....,1
SO how we will proceed this
I presume it will be a telescopic function
|
$$I=\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {(x-4)^2} \right]}} + \left[ {{x^2}} \right]}}dx}!$$
Use $$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$$, then
$$I=\int\limits_1^3 {\frac{{\left[ {{(x-4)^2}} \right]}}{{{{\left[ {x^2} \right]}} + \left[ {{(x-4)^2}} \right]}}dx}$$
Adding the two we get $$2I=\int_{1}^{3} dx \implies I=1.$$
|
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Finding all solutions to quartic Diophantine equation Consider the Diophantine equation $6x^2 = y^2(2y-1)(y-1)$. I am interested in finding all solutions to this such that $y$ is a positive integer -- or at the very least knowing whether there are infinitely many. Certainly there are some; the smallest being $(x,y) = (0,1)$, and then (less trivially) $(x,y) = (350,25)$. Generating more is possible.
This Diophantine equation arises in my research on certain determinants; I am not fluent enough in this area to see any immediate ways this question might be resolved. I had a look in Mordell, but could not see anything that could be shaped to give any direct answer (though perhaps I am wrong!).
|
*
*Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
*Completing the Square on RHS gives us $ 48z^2+1 = (4y-3)^2 $.
This is a Pell's equation with solutions $$ z = \frac{ ( 7+ 4\sqrt{3}
)^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }, 4y-3 =
\frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] .$$
*Show that $y$ is an integer iff $n $ is even.
*Hence, conclude that the solutions are .... (Just backtrack. I'm too lazy to paste the expressions.)
The first non-trivial solution is $ n = 2, 4y-3 = 97, z = 14, y = 25, x = 350$.
The next solution is $ n = 4, 4y-3 = 18817, z = 2716, y
= 4705, x = 12778780$.
Previous version:
*
*Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
*Treating this as a quadratic in $y$, we have $y = \frac{1}{4} ( 3 \pm \sqrt{ 48z^2 + 1 } )$. We have a solution if the expression is a perfect square.
*Finally, Pell's equation on $a^2 = 48z^2 + 1$ gives us $ z = \frac{ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }$ and $ a = \frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] $
*It remains to backtrack (and check for positivity / integer-ness).
|
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|
Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}{3a^2+ab}+\dfrac{b^2}{3b^2+bc}+\dfrac{c^2}{3c^2+ca}\ge\dfrac{(a+b+c)^2}{3(a^2+b^2+c^2)+ab+bc+ca}=\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}$
$\bullet$ $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}$ so we have $\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}\le \dfrac{3}{4}$ but we need to prove $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} \ge \dfrac{3}{4}$
Can you give me some hint ?
|
Another way.
We know that the expression is closed to $\frac{1}{3}$, but by C-S
$$\sum_{cyc}\frac{a}{b+3a}=\frac{a^2}{ab+3a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3a^2+ab)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3a^2+6ab)}=\frac{1}{3},$$ which says that $\frac{1}{3}$ is an infimum.
|
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|
Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM
*
*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$
*$\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)\le \frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Square both sides)
*$\sqrt{x}\sqrt{y}\le \:\frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Simplify Left Side)
*$4xy\sqrt{x}\sqrt{y}\le x^3+2xy\sqrt{x}\sqrt{y}+y^3$ (Move 4xy to the other side and Square remaining)
*But now I see that this will not end up in $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$
Please help????
|
Let $u = \sqrt{x/y}$. The problem is equivalent to showing
$$
u^3 + 1 \ge u^2 + u,
$$
which follows immediately from
$$
(u-1)^2(u+1) \ge 0
$$
for all $u \ge 0$.
|
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|
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$
and I wanted to try and prove it.
Inspired by this answer by Jack D'Aurizio, I took the Weierstrass product for $\cosh(x)$ to obtain
$$
\cosh\left(\frac{\pi x}{2} \right) = \prod_{n \ge 1}\left(1 + \frac{x^2}{(2n-1)^2} \right)
$$
And by logarithmically differentiating twice we get
$$
\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi x}{2} \right) = \sum_{n \ge 1} \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}
$$
Which means we get
\begin{align*}
\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx & =\frac{4}{\pi^2}\sum_{n\ge 1} \int_{0}^{\infty} \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}\right)\, dx
\end{align*}
However, after this, I couldn't figure out how to evaluate the resulting integral.
Does anyone know how I could continue this method? Or alternatively, does anyone know another way in which the result can be proven? Thank you very much!!
Edit:
Per jimjim's request, I'll add that I found this integral on the Wikipedia article for $\zeta(3)$. I believe the reference is to this text where the following formula is given
$$
(s-1) \zeta(s) = 2\pi \int_{\mathbb{R}}\frac{\left(\frac{1}{2} + xi \right)^{1-s}}{\left(e^{\pi x} +e^{-\pi x} \right)^2}\, dx
$$
which for the case of $s=3$ reduces to the surprisingly concise
$$
\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx = \frac{\zeta(3)}{\pi}
$$
And I presume that one can modify the previous equation to get to the original integral from the question, but it is not apparent to me how this may be done.
Edit 2:
Random Variable has kindly posted in the comments how to go from $\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx$ to $ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx$. Thank you very much!
|
$$\int_0^\infty \frac{1-x^2}{(1+x^2)^2}\operatorname{sech}^2 \left(\frac{\pi x}{2}\right) dx\overset{IBP}=\pi\int_{0}^\infty \frac{x}{1+x^2}\operatorname{sech}^2 \left(\frac{\pi x}{2}\right)\tanh\left(\frac{\pi x}{2}\right)dx$$
$$=2\pi\int_{-\infty}^\infty \frac{x}{1+x^2}\color{blue}{\left(\frac{e^{\pi x/2}}{\left(e^{\pi x/2}+e^{-\pi x/2}\right)^3}-\frac{e^{-\pi x/2}}{\left(e^{\pi x/2}+e^{-\pi x/2}\right)^3}\right)}dx$$
$$\overset{x\to -x}=\color{blue}{2}\cdot 2\pi\int_{-\infty}^\infty \frac{x}{1+x^2}\color{blue}{\frac{e^{\pi x/2}}{\left(e^{\pi x/2}+e^{-\pi x/2}\right)^3}}dx\overset{\pi x\to x}=4\pi\int_{-\infty}^\infty
\frac{x}{\pi^2+x^2}\frac{e^{2 x}}{(1+e^{ x})^3}dx$$
$$=4\pi\int_{-\infty}^\infty \Im\left(-\color{red}{\frac{1}{\pi+ix}}\right)\frac{e^{2 x}}{(1+e^{ x})^3}dx=-4\pi\Im\int_{-\infty}^\infty \color{red}{\int_0^\infty e^{-(\pi+ix)t}}\frac{e^{2 x}}{(1+e^{ x})^3}\color{red}{dt}dx$$
$$\small \overset{\large e^x\to x}=-4\pi\Im\int_0^\infty e^{-\pi t}\left(\lim_{a\to 1-it}\int_{0}^\infty\frac{x^a}{(1+x)^3}dx\right)dt=-4\pi\Im\int_0^\infty e^{-\pi t}\left(\lim_{a\to 1-it}\frac{\pi}{2}\frac{ a(1-a)}{\sin(\pi a)}\right)dt$$
$$=2\pi^2\int_0^\infty e^{-\pi t}\frac{t^2}{\sinh(\pi t)}dt\overset{\pi t =x}=\frac{4}{\pi} \int_0^\infty \frac{e^{-2x }x^2}{1-e^{-2x}}dx\overset{\large e^{-x}\to x}=\frac{4}{\pi}\int_0^1 \frac{x\ln^2 x}{1-x^2}dx$$
$$\overset{x^2\to x}=\frac{2}{\pi}\int_0^1\frac{\ln^2 x}{1-x}dx=\frac{2}{\pi}\sum_{n=1}^\infty \int_0^1 x^{n-1}\ln^2 x\, dx=\frac{2}{\pi}\sum_{n=1}^\infty\frac{1}{2n^3}=\frac{\zeta(3)}{\pi}$$
|
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Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11, x>3,$ which doesn't make sense.
|
Edited for clarity.
Expanding on the comment of sonicsid, when you are faced with the necessity of clearing the denominator of $(x + 4)$, since $x + 4 = 0$ is disallowed, you must break the analysis into two cases:
Case 1: $(x + 4) > 0.$
$-1 < \frac{7}{x+4} < 1.$
$ (-1) (x+4) < 7 < (x+4).$
$(x + 4) > -7$ and $(x + 4) > 7.$
This means that all three of the following constraints must be satisfied for Case 1:
*
*$x > -4.$
*$x > -11.$
*$x > 3.$
Therefore, the Case 1 solutions resolve to
$x > 3.$
Case 2: $(x + 4) < 0.$
$-1 < \frac{7}{x+4} < 1.$
$(-1) (x + 4) > 7$ and $7 > (x + 4).$
This means that all three of the following constraints must be satisfied for Case 2:
*
*$x < -4.$
*$x < -11.$
*$x < 3.$
Therefore, the Case 2 solutions resolve to
$x < -11.$
Putting Case 1 and Case 2 together:
Either $x > 3$ or $x < -11.$
|
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Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the following lines:
Since $\gcd(6,5) = 1$ this means that if we multiply $5$ with all the numbers from $1$ to $5$ we have to get all the numbers in different order.
I.e.
The numbers less than $6$: $1,2,3,4,5$.
Multiply the above by $5$: $5,4,3,2,1$ i.e. we get the same set of numbers but in different order.
Since: $1\cdot 2\cdot 3\cdot 4\cdot 5 = 5\cdot 4\cdot 3\cdot 2\cdot 1$ we also have:
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$
Which is actually wrong.
What is the problem in my approach and thought process?
|
A most immediate thing to note is that $(m-1)\equiv -1 \pmod m$ always.
So $(m-1)^{m-1} \equiv (-1)^{m-1}\pmod m$. And $(-1)^{m-1} = -1$ if $m-1$ is odd and $(-1)^{m-1} = 1$ if $m-1$ is even. And as $-1 \equiv m-1\pmod m$ and $1\equiv m+ 1\pmod m$ we realize what you probably heard was
$(m-1)^{m-1} \equiv (-1)^{m-1} \equiv \begin {cases}1\equiv m+1 \pmod m& m\text{ is odd}\\-1\equiv m-1 \pmod m& m\text{ is even}\end{cases}$
"The numbers less than 6: 1,2,3,4,5.
Multiply the above by 5: 5,4,3,2,1 i.e. we get the same set of numbers but in different order."
Note: That is because $5\equiv -1$ so $1,2,3,4,5$ times $5$ will be $-1,-2,-3,-4,-5$ which are $5,4,3,2,1$.
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$ Which is actually wrong.
But $1\cdot 2\cdot 3\cdot 4\cdot 5\not \equiv 1$.
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 0$.
And you od have $1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 0 \equiv 5^5\cdot 0$
Ah.... I see the issue...
Wilson's theorem: $(p-1)! \equiv -1 \pmod p$ !!!IF $p$ IS PRIME!!!
But if $m$ is composite that the theorem does not hold $(m-1)! \not \equiv 1 \pmod m$.
In fact:
If $m> 4$ is composite then $(m-1)! \equiv 0 \pmod m$.
Pf: If $m$ is composite then $m = dk$ for some $d,k$ which are neither $1$ nor $m$ and so are components of $(m-1)!$. So $m|(m-1)!$ if $m$ is composite and $(m-1)!\equiv 0 \pmod m$.
....unless!..... $m = p^2$ for some prime. Then $p$ only occurs as a component once. But then $p < 2p < .....< p^2$ and $p$ and $2p$ are components. In which case $p\cdot 2p = 2p^2 = 2m|(m-1)!$ and so $m|(m-1)!$ and $(m-1)! \equiv 0 \pmod m$.
....unless! .... $p=2$ and $m =2^2 = 4$. But in that case $(4-1)! = 6$ and $6\equiv 2\pmod 4$. That is the exception. But if $m > 4$ there is no exception.
=====
Oh.... You weren't doing Wilsons theorem... which would have been negative anyway....
You were assuming $ac \equiv bc \pmod m \implies a \equiv b\pmod m$.
No. Division is one of the operations you can't do via modular arithmetic (unless $m$ is prime).
If $ac \equiv bc \pmod m$ then $m|ac -bc=(a-b)c$. But that does not mean $m|a-b$ unless $\gcd(m,c) =1$. For example if we let $a= 5$ and $b=3$ and $c= 3$ and $m = 6$ we have $15 \equiv 9\pmod 6$ and $6|15-9 = (5-3)3 = 2\times 3$ but we do not have $6|2$. So we do not have $5\equiv 3 \pmod 6$.
So we can't do division. (Unless $m$ is prime)
|
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|
Choosing the "right order" of approximation in Taylor Series I was evaluating the following limit using Taylor series, $\lim_{x \rightarrow 0}(2 \frac{e^x-1-x}{x^3}-1/x)$, and I wrote the exponential function as $e^x=1+x+\frac{x^2}{2}+ \frac{x^3}{6}+o(x^4)$ and got that the limit is $\frac13$. But, if I had written $e^x=1+x+\frac{x^2}{2}+o(x^3)$, I would have obtained a different result, so how do I choose correctly at which order I'm supposed to stop when writing a Taylor series to evaluate a limit?
|
$$\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}=\lim_{x\to0}\frac{1}{x}+\frac{1}{3}+\frac{1}{12}x+\cdots-\frac{1}{x}=\frac{1}{3}$$
However, if you choose to use little-o notation, recall that the definition of little-0 is that:
$$f(x)\in\omicron(g(x)),\,x\to0\iff\lim_{x\to0}\frac{f(x)}{g(x)}=0$$
Whereas the definition of big-O notation is:
$$f(x)\in O(g(x)),\,x\to0\iff\lim_{x\to0}\frac{f(x)}{g(x)}\lt\infty$$
So we notice that:
$$\frac{x^3}{6}+\frac{x^4}{24}+\cdots\in O(x^3)$$
But is not little-o of $x^3$ - this is because the remainder terms divided by $x^3$ leave you with $\frac{1}{6}$ and some other terms that go to zero, leaving $\frac{1}{6}\neq0$ as the limit.
So in fact $e^x=1+x+\frac{x^2}{2}+O(x^3)$, not little-0!
Now:
$$\begin{align}\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}&=\lim_{x\to0}2\frac{\frac{x^2}{2}+O(x^3)}{x^3}-\frac{1}{x}\\&=\lim_{x\to0}2\cdot\frac{O(x^3)}{x^3}\end{align}$$
Which is an ambiguous expression that is not necessarily (and not actually, in this case) equal to $0$. If we replaced $O(x^3)$ with: $\frac{1}{6}x^3+o(x^3)$ however, that would be correct, as the remainder terms when divided by $x^3$ are $\frac{1}{24}x+\frac{1}{120}x^2+\cdots$ which do go to zero as $x$ does.
To demonstrate the proper use of these notations, this is what you should be doing:
$$\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}=\cdots=\lim_{x\to0}\frac{1}{3}+\frac{o(x^3)}{x^3}=\frac{1}{3}$$
Since $o(x^3)/x^3\to0$ by definition.
|
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Proof that cube root of product of 2 prime numbers are irrational I am stuck with this problem from my son's homework:
Given $p$ and $q$ are prime numbers, prove that $\sqrt[3]{pq}$ is irrational
Could someone please shed some light? Thanks!
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(You don't explicitly state $p \neq q$. I assume this here. A similar argument can be made for the $p = q$ case.)
Suppose $\sqrt[3]{pq} = \frac{a}{b}$ is rational in lowest terms (so $a$ and $b$ are integers, $b > 0$, and $\gcd(a,b) = 1$). Then \begin{align*}
pq &= \frac{a^3}{b^3} \\
b^3pq &= a^3
\end{align*}
Since $p$ is prime and $p$ divides the left-hand side, $p$ divides the right-hand side. A prime that divides $a^3$ must be a prime divisor of $a$, so $a = p a_1$ for some integer $a_1$. We have
\begin{align*}
b^3pq &= (pa_1)^3 \\
b^3pq &= p^3a_1^3 \\
b^3 &= p^2a_1^3 \\
\end{align*}
By a similar argument as used previously, since prime $p$ divides the right-hand side, it divides $b$. We have $p$ divides $a$ and $p$ divides $b$, so $\gcd(a,b) \geq p > 1$. This contradicts that $a/b$ is in lowest terms. Therefore, $\sqrt[3]{ab}$ is irrational.
|
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|
Solve integral $\int_0^{2\pi}\frac{\sin^2\frac{(N+1)x}{2}}{2\pi(N+1)\sin^2\frac x2} dx $ My question is to solve
$$\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{2\pi(N+1)\sin^2(x/2)} dx $$
Up to $1000$ or more, Wolfram-Alpha solves this integral and gives the value $1$ for every integer. But, I couldn't prove that this is the case for any $N\in\mathbb{N}$.
I know if we didn't have the $N+1$ part and squares, this integral should be $1$. We also have
$$ \frac{\sin^2\left(\frac{N+1}{2}x\right)}{\sin^2(x/2)}=\frac{\cos((N+1)x)-1}{\cos(x)-1}=\sum_{n=0}^N\left(\sum_{k=-n}^n e^{inx} \right)$$
and
$$ \frac{\sin\left((N+\frac{1}{2})x\right)}{\sin(x/2)}=\sum_{n=-N}^Ne^{inx}= \sum_{k=1}^N2\cos(kx)+1$$
|
\begin{align}
\frac{1}{2\pi(N+1)}\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{\sin^2\left(\frac{x}{2}\right)} dx &\overset{\color{blue}{u = \frac{x}{2}}, \, \color{green}{n=N+1}}{=}\frac{1}{2\pi n}\color{blue}{2}\int_0^{\color{blue}{\pi}}\frac{\sin^2(nu)}{\sin^2(u)} du\\
& \overset{(\color{purple}{\text{Even})}}{=} \frac{1}{2\pi n}\int_{\color{purple}{-\pi}}^{\pi}\frac{\sin^2(nu)}{\sin^2(u)} du\\
& \overset{\color{blue}{s = u +\pi}}{=} \frac{1}{2\pi n}\int_{\color{blue}{0}}^{\color{blue}{2\pi}}\frac{(-1)^{2n}\sin^2(ns)}{(-1)^{2n}\sin^2(s)} ds\\
& = \frac{1}{2\pi n}2n\pi =1
\end{align}
where on the last step you use $\int_{0}^{2\pi} \frac{\sin^2(nx)}{\sin^2(x)} dx=2n\pi$ with $n \in \mathbb{N}$.
|
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|
Isn't the book wrongly taking $\sin^{-1}$ on both sides here? Question:
If $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$, then show that $\theta=\pm\frac{1}{2}\sin^{-1}\frac{3}{4}$.
My book's solution:
$$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$$
$$\sin(\pi\cos\theta)=\sin(\frac{\pi}{2}\pm\pi\sin\theta)\ [\text{Formula:}\cos\theta=\sin(\frac{\pi}{2}\pm\theta)]$$
$$\sin^{-1}(\sin(\pi\cos\theta))=\sin^{-1}(\sin(\frac{\pi}{2}\pm\pi\sin\theta))...(i)$$
$$\pi\cos\theta=\frac{\pi}{2}\pm\pi\sin\theta...(ii)$$
$$\cos\theta=\frac{1}{2}\pm\sin\theta$$
$$\cos\theta\pm\sin\theta=\frac{1}{2}$$
$$\cos^{2}\theta\pm2\sin\theta\cos\theta+\sin^{2}\theta=\frac{1}{4}$$
$$1\pm\sin2\theta=\frac{1}{4}$$
$$\sin2\theta=\pm\frac{3}{4}$$
$$\sin^{-1}(\sin2\theta)=\sin^{-1}(\pm\frac{3}{4})$$
$$2\theta=\pm\sin^{-1}(\frac{3}{4})$$
$$\theta=\pm\frac{1}{2}\sin^{-1}(\frac{3}{4})\ \text{(showed)}$$
This solution is good and all, but if we input the value of $\theta$ in line (i) we will see something interesting:-
$$\sin^{-1}(\sin(\pi\cos\theta))=\sin^{-1}(\sin(\frac{\pi}{2}\pm\pi\sin\theta))$$
$$[\text{Let's input $\theta=\frac{1}{2}\sin^{-1}(\frac{3}{4})$}]$$
$$\sin^{-1}(\sin(164.058809^{\circ}))=\sin^{-1}(\sin(164.058809^{\circ})$$
$$164.058809^{\circ}=164.058809^{\circ}$$
This is what my book did essentially. However, isn't $164.058809^{\circ}$ outside the restricted range of $\sin^{-1}(x)$: $[\frac{\pi}{2},-\frac{\pi}{2}]$? So, is the line (ii) in the solution of the book valid?
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The reasoning shown in the book is incorrect. Indeed, were it correct, then we could make the following deduction.
$$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$$
$$\sin(\pi\cos\theta)=\sin(\frac{\pi}{2}+\pi\sin\theta)$$
This is exactly the same as your step (i), but with a $+$, rather than a $\pm$. That's perfectly fine, since $\cos\theta=\sin(\frac{\pi}{2}+\theta)$ is a correct identity for any $\theta$.
$$\sin^{-1}(\sin(\pi\cos\theta))=\sin^{-1}(\sin(\frac{\pi}{2}+\pi\sin\theta))$$
$$\pi\cos\theta=\frac{\pi}{2}+\pi\sin\theta$$
This is the dodgy step. In your example, it works out for the author, but now the statement is genuinely incorrect.
$$\cos\theta=\frac{1}{2}+\sin\theta$$
$$\cos\theta-\sin\theta=\frac{1}{2}$$
$$\cos^{2}\theta-2\sin\theta\cos\theta+\sin^{2}\theta=\frac{1}{4}$$
$$1-\sin2\theta=\frac{1}{4}$$
$$\sin2\theta=\frac{3}{4}$$
$$\sin^{-1}(\sin2\theta)=\sin^{-1}(\frac{3}{4})$$
(assuming that $\theta$ lies in $[-\pi,\pi]$)
$$2\theta=\sin^{-1}(\frac{3}{4})$$
$$\theta=\frac{1}{2}\sin^{-1}(\frac{3}{4})\ \Box$$
But this is now incorrect, since $\theta = -\frac12\sin^{-1}\frac34$ is also a solution.
|
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|
Prove that $\lim\limits_{n\to\infty} \frac{\sin\left(\log n + n^5 + e^{n^2}\right)}{n} = 0$. Prove that $\lim\limits_{n\to\infty}x_n = 0$, where $x_n = \frac{\sin(\log n + n^5 + e^{n^2})}{n}$.
Demostración: Suppose $\epsilon > 0$. We first note the $\left|\sin(x)\right| \leq 1$ for $x\in\mathbb{R}$. Thus, $\left|\sin\left(\log n + n^5 + e^{n^2}\right)\right|\leq 1$. Pick $N > \frac{1}{\epsilon} \implies \frac{1}{N} < \epsilon$. For $n\geq N $, we have $\frac{1}{n} \leq \frac{1}{N} < \epsilon$. Second note that $\frac{1}{n} > -\epsilon $. Thus, $\left|\frac{1}{n}\right| < \epsilon$. Finally,
$$\left|\frac{1}{n}\cdot \sin\left(\log n + n^5 + e^{n^2}\right)\right| < \epsilon.\text{ QED}$$
Feedback is appreciated!
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HINT
We can use
$$-\frac1n \le \frac{\sin\left(\log n + n^5 + e^{n^2}\right)}{n}\le \frac1n$$
|
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|
Is $g(x, y) = \frac{f(x, y)^{y+1}-1}{(f(x,y)-1)(xf(x,y)+1)} $ always an integer? Let \begin{equation} f(x, y) = \frac{x^y-1}{x+1} \end{equation}
And \begin{equation}
g(x, y) = \frac{f(x, y)^{y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}
\end{equation}
where $x, y $ are positive integers with $y$ even, $f(x, y) \not = 1$.
It appears $g(x, y) $ is always an integer.
To prove this, I tried fixing $x$ and doing induction on $y$ but got stuck in the induction step. (An elementary proof would be excellent)
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Here is an elementary proof:
Let us change the notation so that
$$ f(x,y):= \frac{x^{2y}-1}{x+1}$$ and
$$ g(x,y):= \frac{f(x,y)^{2y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}.$$
We first show that $f(x,y)-1$ and $xf(x,y)+1$ are coprime. Suppose that there exist $p$ a prime which divides both $f(x,y)-1$ and $xf(x,y)+1$ then $p$ divides $(x+1)f(x,y)$ then $p$ divides $x+1$.
Then if $p$ divides $f(x,y)-1$ then $p^2$ divide $x^{2y+1}-x-2$, and if $p$ divides $xf(x,y)+1= \frac{x^{2y+1}+1}{x+1}$ then $p^2$ divides $x^{2y+1}+1$. Then $p^2$ divides $x+3$.
Then $p$ must be equal to $2$ and $x$ must be odd, but then $xf(x,y)+1= 1-x+x^2- \cdot\cdot +x^{2y}$ must be odd, which is contradiction.
Now if $q$ is prime and $q^h$ , $h \gt 0$, is the largest power of $q$ which divides $f(x,y)-1$ then since $f(x,y)-1$ divides $f(x,y)^{2y+1}-1$ ,then $q^h$ also divides $f(x,y)^{2y+1}-1$.
And if $q^h$ , $h\gt 0$, is the largest power of $q$ which divides $xf(x,y)+1$ then $-xf(x,y) \equiv 1 \bmod q^h$, then neither $x$ nor $f(x,y)$ is divisible by $q$ and $f(x,y)^{2y+1}-1 \equiv -\frac{1}{x^{2y+1}}-1 \equiv -\frac{x^{2y+1}+1}{x^{2y+1}} \bmod q^h$. That is
$$ x^{2y+1} (f(x,y)^{2y+1}-1) \equiv - (x+1) (xf(x,y)+1) \equiv 0 \bmod q^h$$ then $q^h$ divides $f(x,y)^{2y+1}-1$.
|
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|
Euler substitution inconsistency in evaluating $\int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}}$
$$I = \int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}} = 4\sqrt2-2-\pi$$
Euler's substitutions immediately come to mind. But as is, the integrand is free of $\sqrt{ax^2+bx+c}$ (with $a\neq0$). So, I considered rewriting the integrand as
$$\frac{1}{\sqrt{1-x}+2+\sqrt{1+x}} = \dfrac{-\sqrt{1-x}+2-\sqrt{1+x}}{-(1-x)-2\sqrt{1-x^2}+4-(1+x)} = \frac{\sqrt{1-x}-2+\sqrt{1+x}}{2\sqrt{1-x^2}-2}$$
That is, if the denominator is $a+b+c$, I multiplied by $-a+b-c$. Now I can set up the second Euler substitution:
$$\sqrt{1-x^2} = xt-1 \implies 1-x^2 = x^2t^2-2xt+1 \implies x=\frac{2t}{t^2+1}$$
Then $\mathrm dx = \frac{2(1-t^2)}{(1+t^2)^2}\,\mathrm dt$. The integration range remains unchanged.
In terms of $t$, the square roots are
$$\begin{cases}\sqrt{1-x^2} = xt-1 = \frac{2t^2}{1+t^2}-1 = \frac{t^2-1}{1+t^2} & \color{red}{(1)}\\[1ex] \sqrt{1\pm x} = \sqrt{1\pm\frac{2t}{1+t^2}} = \frac{|t\pm1|}{\sqrt{1+t^2}}\end{cases}$$
Since $|t|<1$, we have $|t+1|=t+1$ and $|t-1|=1-t$. In the integral,
$$\begin{align}
I &= \frac12 \int_{-1}^1 \frac{\sqrt{1-x}-2+\sqrt{1+x}}{\sqrt{1-x^2}-1} \,\mathrm dx \\[1ex]
&= \int_{-1}^1 \frac{1-t^2}{1+t^2} - \frac{1-t^2}{(1+t^2)^{3/2}}\,\mathrm dt
\end{align}$$
But this leads to a different value of $-2-2\sqrt2+\pi+2\sinh^{-1}(1)$.
It seems the problem lies within the substitution:
$$\sqrt{1-x^2} = \sqrt{1-\frac{4t^2}{(t^2+1)^2}} = \sqrt{\frac{(t^2-1)^2}{(t^2+1)^2}} = \frac{|t^2-1|}{t^2+1} = \frac{1-t^2}{t^2+1} \quad \color{red}{(2)}$$
since $|t|<1$. But shouldn't $\color{red}{(1)}$ and $\color{red}{(2)}$ be in agreement?
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The alternative substitution $x=\sin2t$ may be simpler
$$\begin{align}
& \hspace{5mm}\int_{-1}^1 \frac{1}{\sqrt{1-x}+2+\sqrt{1+x}} \ dx\\
& = \int_0^{\pi/4}\frac{2\cos2t}{1+\cos t}dt =\int_0^{\pi/4}\left( \sec^2 \frac t2-8\sin^2\frac t2\right)dt = 4\sqrt2-2-\pi
\end{align}$$
|
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|
Is this matrix of maximal rank? Let $(A_0,A_1)\in (M_{p\times r}(\mathbb{R}))^2$, $(B_0,B_1)\in (M_{p\times q}(\mathbb{R}))^2$, and suppose that the matrix
$$\begin{pmatrix}A_1&B_0\\A_0&B_1\end{pmatrix}$$
has rank $2p$. Has the matrix
$$\begin{pmatrix}A_1&B_0&&&\\
&B_1&A_0&&\\
&&A_1&B_0&\\
&&&B_1&A_0\end{pmatrix},$$
where blank spaces are all zero, rank $4p$? Some numerical experiments seems to say so, but trying to make appear the matrix from the hypothesis and doing Gaussian elimination like in
$$\begin{pmatrix}A_1&B_0&&&\\
&B_1&A_0&&\\
&&A_1&B_0&\\
&&&B_1&A_0\end{pmatrix}\sim\begin{pmatrix}A_1&B_0&&&\\
A_0&B_1&A_0&&\\
A_1&&A_1&B_0&\\
&&A_0&B_1&A_0\end{pmatrix}$$
won't directly work, due to terms like the $A_1$ in "$(3,1)$" position here, which prevent me to put the matrix in canonical form. I would also be interested if there was a similar argument for matrices of the form
$$\begin{pmatrix}A_1&B_0&&&&\\
&B_1&A_0&&&\\
&&A_1&B_0&&\\
&&&B_1&A_0&\\
&&&&A_1&\dots\end{pmatrix},$$
where eventually you stop at some $A_0$ or $B_0$. Thanks in advance for any ideas!
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For simplicity, set $$M = \begin{pmatrix} A_{1} & B_{0}\\ A_{0} & B_{1} \end{pmatrix} \in M_{2 p, q +r}(\mathbb{R}) \quad \text{and} \quad N = \begin{pmatrix} A_{1} & B_{0} & & & \\ & B_{1} & A_{0} & & \\ & & A_{1} & B_{0} & \\ & & & B_{1} & A_{0} \end{pmatrix} \in M_{4 p, 2 q +3 r}(\mathbb{R}) \, \text{.}$$ To prove that $N$ has rank $4 p$, it suffices to show that, if $X \in M_{1, 4 p}(\mathbb{R})$ satisfies $X N = 0$, then $X = 0$. Suppose that $X \in M_{1, 4 p}(\mathbb{R})$ is such that $X N = 0$. Writing $$X = \left( X_{1} \vert X_{2} \vert X_{3} \vert X_{4} \right) \quad \text{with} \quad X_{1}, X_{2}, X_{3}, X_{4} \in M_{1, p}(\mathbb{R}) \, \text{,}$$ we obtain the equations $$\left\lbrace \begin{array}{lr} X_{1} A_{1} = 0 & (1)\\ X_{1} B_{0} +X_{2} B_{1} = 0 & (2)\\ X_{2} A_{0} +X_{3} A_{1} = 0 & (3)\\ X_{3} B_{0} +X_{4} B_{1} = 0 & (4)\\ X_{4} A_{0} = 0 & (5) \end{array} \right. \, \text{.}$$ The equalities $(1) +(3) +(5)$ and $(2) +(4)$ can be rewritten as $$\left( X_{1} +X_{3} \vert X_{2} +X_{4} \right) M = 0 \, \text{.}$$ Since $M$ has rank $2 p$ by hypothesis, it follows that $X_{3} = -X_{1}$ and $X_{4} = -X_{2}$. In particular, we have $$\left\lbrace \begin{array}{lr} X_{1} A_{1} = 0 & (i)\\ X_{1} B_{0} +X_{2} B_{1} = 0 & (ii)\\ X_{2} A_{0} = 0 & (iii) \end{array} \right. $$ by the equations $(1)$, $(2)$ and $(5)$. The equalities $(i) +(iii)$ and $(ii)$ can be rewritten as $$\left( X_{1} \vert X_{2} \right) M = 0 \, \text{.}$$ Therefore, we have $X_{1} = 0$ and $X_{2} = 0$ since $M$ has rank $2 p$. Thus, we have $X = 0$, and the desired result is proved.
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|
Integrating $\int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$ I found the following integral and wanted to know if there is a nice closed form solution in terms of elementary or some special functions (Polylogarithm, Clausen, etc).
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$$
I know that the integral converges numerically to $\approx 0.403926$
Here is my try using integration by parts:
Let
$$ du = \frac{\arctan(x)}{x^2} \Longrightarrow u = -\frac{1}{2}\ln(1+x^2) + \ln(x) - \frac{\arctan(x)}{x} $$
$$ v = \arctan(x^2) \Longrightarrow dv = \frac{2x}{x^4+1}$$
Hence
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx \stackrel{IBP}{=} -\frac{\pi^2}{16} - \frac{1}{8}\pi \ln(2) -2\underbrace{\int_{0}^{1} \frac{x\ln(x)}{x^4+1} dx}_{I_{1}} + 2\underbrace{\int_{0}^{1}\frac{\arctan(x)}{x^4+1}dx}_{I_{2}} + \underbrace{\int_{0}^{1} \frac{x\ln(1+x^2)}{x^4+1}dx}_{I_{3}} $$
Can be proven that
$$I_{1} = -\frac{C}{4}$$
where $C$ is the Catalan constant and
$$I_{3} = \frac{1}{16} \pi \ln(2) $$
but I'm stuck with $I_{2}$
Another way could be:
Define
$$\Psi(a) = \int_{0}^{1} \frac{\arctan(ax)\arctan(x^2)}{x^2} dx$$
Hence
$$\Psi'(a) = \int_{0}^{1} \frac{\arctan(x^2)}{x(a^2x^2+1)}dx = \frac{1}{2}\int_{0}^{1} \frac{\arctan(w)}{w(a^2w+1)}dx$$
Using integration by parts, we have:
$$du = \frac{\arctan(w)}{1+a^2w} \Longrightarrow u= \frac{1}{1+a^4}\ln\left( \frac{1+a^2w}{\sqrt{1+w^2}} \right) - \frac{a^2-w}{(1+a^4)(1+a^2w)} \arctan(w) $$
$$ v = \frac{1}{w} \Longrightarrow dv = \ln(w) $$
However, this path seems even more rugged that the other.
One last hint could be the following integral:
$$\int_{0}^{1} \frac{\arctan(x) \arctan(x^3)}{x} dx = \frac{7}{72}\zeta(3) + \frac{\pi}{3}C - \frac{5\pi}{12}\operatorname{Cl}_{2}\left(\frac{2\pi}{3} \right)$$
where $\operatorname{Cl}_{2}$ is the Clausen function of order 2. However, I do not know the proof of this result either.
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Beside @Zacky's solution, using partial fraction, you can write
$$\frac{\tan ^{-1}(x)}{x^4+1}=\frac i 2 \left(\frac{\tan ^{-1}(x)}{x^2+i}-\frac{\tan ^{-1}(x)}{x^2-i}\right)$$ and face two integrals
$$I(k)=\int \frac{\tan ^{-1}(x)}{x^2+k}\,dx$$ which can be integrated (have a look here).
The problem is that the result for the definite integral is a real monster.
|
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|
Solve below somewhat symmetric equations Solve $x,y,z$ subject to
$$x^2+y^2 - xy = 3$$
$$(x-z)^2+(y-z)^2 - (y-z)(x-z) = 4$$
$$(x-z)^2+y^2 - y(x-z) = 1$$
$$x,y,z \in R^{+}$$
My attempts:
$(x-z)^2+(y-z)^2 - (y-z)(x-z) - ((x-z)^2+y^2 - y(x-z)) = xz - 2yz = (x-2y)z = 3$
$x^2+y^2 - xy - ((x-z)^2+y^2 - y(x-z)) = 2xz - yz - z^2 = (2x-y-z)z = 2$
divide above, we have $\frac{2x-y-z}{x-2y} = \frac{2}{3}$, or $3z = 4x+y$
Then the rest is easy. But I start to get interested in if there are some geometric solution. Like the comment says, they are the elliptical cylinders in the positive octant
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We can take a quadrilateral $ABCD$, for which:
$AC\cap BD=\{E\},$ $AC=BD=z$,$ED=x$, $EC=y$, $\measuredangle AEB=60^{\circ},$$AB=2$, $BC=1$ and $CD=\sqrt3$.
Thus, we obtain your conditions:
$$x^2+y^2-xy=ED^2+EC^2-2ED\cdot EC\cos60^{\circ}=CD^2=3,$$
$$(x-z)^2+(y-z)^2-(y-z)(x-z)=(z-x)^2+(z-y)^2-(z-x)(z-y)=$$
$$=EB^2+EA^2-2EB\cdot EA\cos60^{\circ}=AB^2=4$$ and
$$(x-z)^2+y^2-y(x-z)=(z-x)^2+y^2+y(z-x)=$$
$$=EB^2+EC^2-2EB\cdot EC\cos120^{\circ}=1.$$
Now, let $BFCD$ be a parallelogram.
Thus, $$AC=z=BD=FC$$ and
$$\measuredangle ACF=\measuredangle DEC=60^{\circ},$$ which gives that $\Delta AFC$ is equilateral triangle, $BA=2$, $BF=CD=\sqrt3$ and $BC=1.$
Now, let $R$ be a rotation around $C$ by $-60^{\circ}$ and $R\left(\left\{B\right\}\right)=\left\{G\right\}$ thus, since $\Delta BGC$ is an equilateral triangle, we obtain:
$BG=1$, $FG=R(AB)=2$ and $$FG^2=2^2=(\sqrt3)^2+1^2=BF^2+BC^2,$$ which says $$\measuredangle FBG=90^{\circ},$$ $$\measuredangle FBC=\measuredangle FBC+\measuredangle GBC=90^{\circ}+60^{\circ}=150^{\circ},$$ which gives $$z^2=AC^2=FC^2=FB^2+BC^2-2FB\cdot BC\cos150^{\circ}=3+1+3=7,$$
which gives $$z=\sqrt7$$ and from here we get the answer:
$$(x,y,z)=\left(\frac{5}{\sqrt7},\frac{1}{\sqrt7},\sqrt7\right).$$
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|
Proving $\sum_{k=2}^n \frac{(k-2){n-k+2\choose k-1}+k{n-k+1\choose k-1}}{k{n\choose k}}=1$ for $n\geq 2$ My friend showed me a difficult combinatorial identity I cannot solve.
Prove that:
$$\sum_{k=2}^n \frac{(k-2){n-k+2\choose k-1}+k{n-k+1\choose k-1}}{k{n\choose k}}=1$$ for all $n\ge 2$.
How do I prove this?
Edit: I would like to bump this again, due to lack of attention.
|
This can be turned into a telescoping sum.
Applying the Pascal's triangle recurrence to the first binomial coefficient in the numerator yields
$$
\binom{n-k+2}{k-1}=\binom{n-k+1}{k-2}+\binom{n-k+1}{k-1},
$$
which allows us to rewrite the identity as
$$
D_n=\sum_{k\ge2}\frac{(k-2)\binom{n-k+1}{k-2}+2(k-1)\binom{n-k+1}{k-1}}{k\binom{n}{k}}=1.\tag{1}
$$
We have named the left side of the identity $D_n$ and observed that the upper limit can be taken to be $\infty$ since the summand becomes $0$ after $k\approx\frac{n}{2}$.
The sum now telescopes by combining the second term of the $k^\text{th}$ summand with the first term of the $(k+1)^\text{st}$ summand:
\begin{align}
D_n&=\frac{(2-2)\binom{n-2+1}{2-2}}{2\binom{n}{2}}+\sum_{k\ge2}\left[\frac{2(k-1)\binom{n-k+1}{k-1}}{k\binom{n}{k}}+\frac{(k-1)\binom{n-k}{k-1}}{(k+1)\binom{n}{k+1}}\right]\\
&=\sum_{k\ge2}\left[\frac{2(k-1)(n-k)\binom{n-k+1}{k-1}}{kn\binom{n-1}{k}}+\frac{(k-1)\binom{n-k}{k-1}}{n\binom{n-1}{k}}\right]\\
&=\sum_{k\ge2}\left[\frac{2(k-1)(n-k)\left[\binom{n-k}{k-2}+\binom{n-k}{k-1}\right]}{kn\binom{n-1}{k}}+\frac{(k-1)\binom{n-k}{k-1}}{n\binom{n-1}{k}}\right]\\
&=\sum_{k\ge2}\left[\frac{(k-2)n\binom{n-k}{k-2}+k(n-2k+2)\binom{n-k}{k-2}+2(k-1)(n-k)\binom{n-k}{k-1}}{kn\binom{n-1}{k}}+\frac{(k-1)\binom{n-k}{k-1}}{n\binom{n-1}{k}}\right]\\
&=\sum_{k\ge2}\frac{(k-2)n\binom{n-k}{k-2}+k(k-1)\binom{n-k}{k-1}+2n(k-1)\binom{n-k}{k-1}-2k(k-1)\binom{n-k}{k-1}+k(k-1)\binom{n-k}{k-1}}{kn\binom{n-1}{k}}\\
&=\sum_{k\ge2}\frac{(k-2)\binom{n-k}{k-2}+2(k-1)\binom{n-k}{k-1}}{k\binom{n-1}{k}}\\
&=D_{n-1}.
\end{align}
In the second line we have used $\binom{n}{r}=\frac{n}{n-r}\binom{n-1}{r}$ to rewrite the first denominator and $r\binom{n}{r}=n\binom{n-1}{r-1}$ to rewrite the second denominator. In the third line we have used the Pascal's triangle recurrence in the first numerator. Since $D_n=D_{n-1}$, the result follows by induction.
The denominator of the summand of $(1)$ is the number of selections of $k$ elements from $n$ with one marked element. Let's say that the $n$ elements are members of an organization ranked by seniority, with $1$ having highest seniority and $n$ having lowest, and that a committee of $k$ members is to be chosen with one member designated as chairperson. Then the numerator of the summand may be interpreted as the number of such committees in which the highest seniority member has rank $k-2$ or $k-1$ and in which the chairperson has seniority lower than either of those members. It might be interesting to try to understand the manipulations above in terms of this interpretation.
|
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|
Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$
I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y = 5.$ (I found $5$ as a solution of the equation by hit and trial method).
Therefore, $x = 0$ or $3.$
I'm wondering if there's any another method to solve it as the repeated squaring step seems to be somewhat absurd.
|
It's not so hard to imagine someone guessing the solutions $0$ and $3$. The solution $0$ is something one can see from the positions of all of the "$x$". The solution $3$ might be inspired by asking what would make $\sqrt{3x}$ rational.
Now the idea is to prove that there can be no more than two solutions by showing that the left side is concave down. If $f(x)$ is positive and concave down, then first of all $2\sqrt{x+f(x)}$ is also positive. But:
$$\begin{align}
\frac{d^2}{dx^2}2\sqrt{x+f(x)}
&=\frac{d}{dx}\frac{1+f'(x)}{\sqrt{x+f(x)}}\\
&=\frac{f''(x)\sqrt{x+f(x)}-\frac{(1+f'(x))^2}{2\sqrt{x+f(x)}}}{(x+f(x))}\\
&=\frac{f''(x)(x+f(x))-\frac12(1+f'(x))^2}{(x+f(x))\sqrt{x+f(x)}}
\end{align}$$
By assumption, $f''(x)$ is negative and $f(x)$ is positive, so this expression is also negative. So $2\sqrt{x+f(x)}$ is also concave down.
So since $2\sqrt{3x}$ is positive and concave down, so is $2\sqrt{x+2\sqrt{3x}}$. And so is $2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}$. And so is $2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}$.
Since that last expression $F(x)$ is concave down, there can be at most two solutions to $\frac12F(x)=x$.
|
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|
Solving $x^3-7x^2+14x-8-\frac12\sin x=0$ With the given problem:
$$x^3-7x^2+14x-8-\frac12\sin x=0$$
I factorized the cubic part:
$$x^3-7x^2+14x-8=0$$
where we test the following solutions of the fraction of the last coefficient divided by the first coefficient and its even multiplicates:
$$\pm\frac{8}{1}, \pm\frac{8}{2}, \pm\frac{8}{4}, \pm\frac{8}{6}, \pm\frac{8}{8}$$
Here, we see that positive value of the second, the third and the fifth fractions are solutions to the problem in eqn. 19. We therefore rewrite the original function to
\begin{equation}
(x-4)(x-2)(x-1)=\frac{1}{2}\sin x
\end{equation}
But now, the situation looks only nicer, but there is no chance of solving this with any regular means. I tried to change the trigonometric part into a Taylor series, but it gets messy and I get
\begin{equation}
(x-4)(x-2)(x-1)=\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n x^{1 + 2 n}}{(1 + 2 n)!}
\end{equation}
Any ideas what to do here?
I take a guess that the domain of solutions is 0-5, and reduce the nasty part on the right to something even nastier:
\begin{equation}
\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n x^{1 + 2 n}}{(1 + 2 n)!}=-\frac{x^{11}}{39916800}-\frac{x^9}{362880}-\frac{x^7}{5040}-\frac{x^5}{120}-\frac{x^3}{6}+x
\end{equation}
which gives on the domain $x\in$(0,5)
\begin{equation}
2(x-4)(x-2)(x-1)=-\frac{x^{11}}{39916800}-\frac{x^9}{362880}-\frac{x^7}{5040}-\frac{x^5}{120}-\frac{x^3}{6}+x
\end{equation}
This leads nowhere also.
|
Here is an approximation to the zero points.
*
*We put $f_1=x^3−7x^2+14x−8$ and $f_2=\frac{1}{2}\sin x$, so that $f_1(x)=f(x)-f_2(x)$.
We then calculate the values:
$f(0)=-8\leqslant0$, $f(1.5) = 0.1263 > 0$ , $f(3) = −2.0706 < 0$, $f(5) = 12.4795 > 0$
By the mean-value theorem of continuous functions we know that $f(x) = x^3 −7x^2 + 14x −8 −0.5 \sin x = f_1(x)−f_2(x)$ has zeros in the intervals (a,b)=(0,1.5),(1.5,3) and (3,5) and no zeros if $x > 5$. See plot
*We have that $f_1(x) = x^3 −7x^2 + 14x −8 = (x −1)(x2 −6x + 8) = (x −1)(x −2)(x −4)$, that is, $f_1(x) \longrightarrow-\infty,\ x\longrightarrow-\infty,\ f_1(x)\longrightarrow-\infty,\ x\longrightarrow-\infty$.
Then, $\underset{max}{0\leqslant x\leqslant2}$ and
$f_1(x) = f_1(1.5) = 0.625 > 0.5 \sin(1.5)$ for $0.5 \sin(1.5) < 0.5$ and $\underset{2<x<4}{min}$ $f_1(x) = f_1(3) = −2 < 0.5 \sin(3)$ for $0.5 \sin(1.5) > 0 (3 < \pi)$.
Note that $f_1(x) = x^3 −7x^2 + 14x −8$ has three real zeros, x= 1, 2
and 4, and then it can be assumed that its two local extrema are precisely between the zeros 1.5 and 3.
These properties of $f_1(x) = x^3 −7x^2 + 14x −8$ and $f_2(x) = 0.5 \sin x$ show again that the function
$f(x) = x^3 −7x^2 + 14x −8 −0.5 \sin x = f_1(x) −f_2(x)$ has a zero on each interval (a,b)= (0,1.5),
(1.5,3) and (3,5) and no zeros if $ x < 0$ and $x > 5$.
|
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|
Prove the binomial sum $\sum_{k=0}^n k\binom{2n}{k} = \frac{n4^n}2$. Prove $\displaystyle \sum_{k=0}^n k\binom{2n}{k} = \frac{n4^n}2$.
This is from section 0.241 of Gradshteyn and Ryzhik's table of integrals.
I have managed to find a proof for this using the identity $k\binom nk = n \binom{n-1}{k-1}$.
Hence \begin{align*}
& \binom{2n}1+2\binom{2n}2+3\binom{2n}3+\cdots + n\binom{2n}n\\
&= 2n\binom{2n-1}0 + 2n\binom{2n-1}1 + 2n\binom{2n-1}2 + \cdots + 2n\binom{2n-1}{n-1}\\
&= 2n\bigg[\binom{2n-1}0 + \binom{2n-1}1 + \binom{2n-1}2 + \cdots + \binom{2n-1}{n-1}\bigg]\\
&= 2n\left(\frac{2^{2n-1}}2\right)\\
&= n \cdot 2^{2n-1}
\end{align*}
I was wondering if there was another method to do this (such as using calculus as the coefficients suggest this may work). I was unable to find a calculus solution as I don't think I can use symmetry to remove the second half of the terms.
|
Too complex solution based on algebra.
Consider
$$k \binom{2 n}{k} x^{k-1}=\Bigg[ \binom{2 n}{k} x^k\Bigg]'$$
$$S_n=\sum_{k=0}^n \binom{2 n}{k} x^k=(x+1)^{2 n}-\binom{2 n}{n+1} x^{n+1} \, _2F_1(1,1-n;n+2;-x)$$ where appears the gaussian hypergeometric function.
Computing the derivative with respect to $x$, we have
$$S'_n=2 n (x+1)^{2 n-1}-\frac{(n-1) \binom{2 n}{n+1} x^{n+1} \, _2F_1(2,2-n;n+3;-x)}{n+2}-$$ $$(n+1) \binom{2
n}{n+1} x^n \, _2F_1(1,1-n;n+2;-x)$$
Now, making $x=1$ and using the properties of the gaussian hypergeometric function
$$S'_n=\sum_{k=0}^n k \binom{2 n}{k} =n\,4^n-n\,2^{2 n-1} =n\,2^{2 n-1} $$
|
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|
Prove $\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1 $ Prove $$\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1,
$$
where $a,b,c \in (0,1)$.
I tried to solve it in this way:
$$\log_{a}{(\frac{b^2}{ac}-b+ac})\geq1$$
$$\frac{b^2}{ac}-b+ac\geq a$$
$$\frac{(b-ac)^2+abc}{ac}\geq a$$
$$\frac{(b-ac)^2}{ac}+b\geq a$$
We ca affirm that
$$\frac{(b-ac)^2}{ac}\geq 1,b \geq1, a\geq 1$$ then the product
$\frac{(b-ac)^2}{ac}+b\geq a$ is true.
Analogous for $\log_{b}{(\frac{c^2}{ab}-c+ab})$ and $\log_{c}{(\frac{a^2}{bc}-a+bc})$.
The question is whether this solution is properly addressed?
|
Another attempt:
At least of logarithms must be $\geq1$. We suppose the first:
$$\frac{b^2}{ac}-b+ac=ac\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)$$
$$ac\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)\geq a\leftrightarrow c\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)-1\geq0$$
Let
$\frac{b}{ac}=x$:
the function $f(x)=c(x^2-x+1)-1=cx^2-cx+c-1$ have the minimum $f_{min}=\frac{3}{4}c-1$, for $x=\frac{b}{ac}=\frac{1}{2}$.
$$f_{min}\geq0\rightarrow \frac{3}{4}c\geq1\rightarrow c\geq\frac{4}{3}>1$$
hence $c\notin(0,1)$!
So, I think, the inequality is not true.
|
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|
Prove: $\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\sum_{cyc}{\sqrt[4]{\frac{5ab}{c}+4a}}$
Prove that the following inequality :$$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\left(\sqrt[4]{\frac{5ab}{c}+4a}+\sqrt[4]{\frac{5bc}{a}+4b}+\sqrt[4]{\frac{5ca}{b}+4c}\right)$$ holds for all positive real numbers such that: $abc=1$.
I saw problem on: AoPS. I tried to continue Arqady comment:
By Holder $$\sqrt3\sum_{cyc}\sqrt[4]{\frac{5ab}{c}+4a}=\sqrt3\sum_{cyc}\sqrt[4]{ab(5ab+4ac)}\leq$$
$$\leq\sqrt3\sqrt[4]{\left(\sum\limits_{cyc}\sqrt[4]{ab}\right)^3\sum_{cyc}\sqrt[4]{ab}(5ab+4ac)}\leq(\sqrt a+\sqrt b+\sqrt c)^2.$$
We have: $$\sqrt[4]{ab}+\sqrt[4]{bc}+\sqrt[4]{ca}\le\sqrt{a}+\sqrt{b}+\sqrt{c}$$
The rest is prove the following inequality: $$9\sum_{cyc}{\sqrt[4]{ab}(5ab+4ac)}\leq(\sqrt{a}+\sqrt{b}+\sqrt{c})^5$$
However, it is not true by acountable example. I hope we can find a good solution.
Thank you!
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We need to prove that $$(a+b+c)^2\geq\sum_{cyc}\sqrt[4]{9a^4b^2(5b^2+4c^2)}.$$
Now, by Holder and Rearrangement we obtain:
$$\sum_{cyc}\sqrt[4]{9a^4b^2(5b^2+4c^2)}=\sum_{cyc}\sqrt[4]{9a^2b^2\cdot a(5b^2a+4c^2a)}\leq$$
$$\leq\sqrt[4]{9(ab+ac+bc)^2(a+b+c)\sum_{cyc}(5b^2a+4c^2a)}\leq$$
$$\leq\sqrt[4]{9(ab+ac+bc)^2(a+b+c)\left(4\sum_{cyc}(a^2b+a^2c)+\frac{4(a+b+c)^3}{27}-abc\right)}.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u^2\geq v^2$ and we need to prove that:
$$(3u)^7\geq9(3v^2)^2(4(9uv^2-3w^3)+4u^3-w^3)$$ or
$$27u^7-4u^3v^4-36uv^6+13v^4w^3\geq0$$ and since by Schur $w^3\geq4uv^2-3u^3,$ it's enough to prove that:
$$27u^7-4u^3v^4-36uv^6+13v^4(4uv^2-3u^3)\geq0$$ or
$$u(u^2-v^2)(27u^4+27u^2v^2-16v^4)\geq0,$$ which is obvious.
|
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|
$\sqrt{6}+\sqrt{3}$ is not rational proof I want to prove $\sqrt{6}+\sqrt{3}$ is not rational; here is my attempt:
Assume for the sake of contradiction that $\sqrt{6}+\sqrt{3}$ is rational. Then $(\sqrt{6}+\sqrt{3})^2$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=9+6\sqrt2,$$
we see $9+6\sqrt2$ must be rational. But, since $\sqrt{2}$ is not rational we have a contradiction and hence $\sqrt{6}+\sqrt{3}$ is not rational.
Is it correct?
|
Looks more or less ok, I would just add one more step.
You correctly prove that if $\sqrt 6+\sqrt 3$ is rational, then $9+6\sqrt2$ is also rational.
I would now add one more step and say that this also means that because $\sqrt{2} = \frac{9+6\sqrt{2} - 9}{6}$, that therefore, $\sqrt{2}$ is also rational, and only then move on to the conclusion.
It's a minor thing, but I would rather be a little too clear than skip an important step. Also, you can then completely explicitly write that
$$\sqrt{2} = \frac{(\sqrt{6} + \sqrt{3})^2 - 9}{6}$$
and make the connection between the two numbers even more explicit.
|
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|
Probability of randomly assigning elements to groups
10 students are tested in an exam with 4 different versions. Each
student is randomly assigned to one of the versions. What is the
probability that there are exactly $i$ versions in which exactly > $3$ students were assigned. Answer separately for $i=2, i = 3$.
I've just started my probability course and it's been a while since I've touched combinatorics (and admittedly I was never good at it), and I'm having trouble getting it right..
My attempt:
Note that as we choose the cards randomly we have a symmetric probability space.
We have $10$ students, each student "chooses" one of $4$ versions. Order matters and repetition allowed, therefore
$|\Omega| = 4^{10}$
For $i = 2$ we first choose two versions which will have exactly $3$ students
${4 \choose 2}$
Then we choose $6$ students to be assigned to the versions we've chosen
${10 \choose 6}$
Then we count the number of different combinations for $6$ students in $2$ versions, where there are exactly $3$ students in each version. We choose $3$ students out of $6$ to place in one version and the remaining students will be in the second version.
${6 \choose 3}$
Now we are left with $4$ students to assign between the remaining two versions. We either have two versions with $2$ students each, or one version with $4$ students. We calculate total options to allocate students and subtract the options in which there are $3$ students in the same version.
$3$ Students in $1$ version - We choose the version ${2 \choose 1}$, allocate $3$ out of $4$ students to it with the remaining student in the last version $\frac{4!}{\left(4-3\right)!}$, so in total
${2 \choose 1}\frac{4!}{\left(4-3\right)!} = 8$
Total options to allocate $4$ students into $2$ versions without restrictions $2^{4} = 16$
Therefore in total $|A| = {4 \choose 2}{10 \choose 6}{6 \choose 3}(16-8)={4 \choose 2}{10 \choose 6}{6 \choose 3}8$ and we have
$P(A) = \frac{\left|A\right|}{\left|Ω\right|} = \frac{{4 \choose 2}{10 \choose 6}{6 \choose 3}8}{4^{10}} = 0.384$
For $i = 3$ we choose $3$ versions to have $3$ students.
${4 \choose 3}$
We choose $9$ students to allocate to these versions, with the remaining one on the other version.
${10 \choose 9}$
Total options to organize $9$ students in $3$ versions.
$3^{9}$
Therefore $|A| = {4 \choose 3}{10 \choose 9}3^{9}$ and we have
$P(A) = \frac{\left|A\right|}{\left|Ω\right|} = \frac{{4 \choose 3}{10 \choose 9}3^{9}}{4^{10}} = 0.750$
I can't seem to find any mistakes, but I would also be really grateful if you can comment whether my way of thinking is correct and maybe give a few tips on how to approach these types of questions!
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For another approach and showing the mistake :
$1-)$ For $i=2$ , By your reasoning select firstly $2$ case which take exactly $3$ sutdents by $C(4,2)$ ,after that , choose $3$ students for the first and the second by $C(10,3) \times C(7,3)$ ,respectively .Now , we have $4$ students to disperse to $2$ versions , but they cannot have $3$ students. Lets find that case by exponential generating functions such that the exponential generating function for each version is $$\bigg(1 + x+ \frac{x^2}{2!}+\frac{x^4}{4!}\bigg)$$
Then , find $[x^4]$ in the expasion of $$\bigg(1 + x+ \frac{x^2}{2!}+\frac{x^4}{4!}\bigg)^2$$
So , $[x^4]=8$
Calculation of E.G.F :
Then , $$\frac{C(4,2) \times C(10,3) \times C(7,3) \times 8 }{4^{10}} = \frac{6 \times 120 \times 35 \times 8}{4^{10}}= 0,1922...$$
I guess you made mistake in $C(4,2)$ , you might have counted it $12$ instead of $6$
$2-)$For , $i=3$ , Select $3$ versions taking exactly $3$ students by $C(4,3)$ , select $3$ students for the selectd versions by $C(10,3) \times C(7,3) \times C(4,3)$. The remainig will go to the remaining version automatically. Then , $$\frac{C(4,3) \times C(10,3) \times C(7,3) \times C(4,3) }{4^{10}}= \frac{4 \times 120 \times 35 \times 4}{4^{10}}=0,06408..$$
When you write $3^9$ in your answer , you do not ensure that the selected version will have exactly $3$ students , it is the mistake..
|
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|
$x(t)= a\cos(t)$ , $y(t)= b\sin(t)$ in terms of the arc length $S$ I'm trying to parameterize the ellipse $x(t)= a\cos(t)$ , $y(t)= b\sin(t)$ in terms of the arc length $S$ but I don't know how to do it.
Supposing that $\gamma:[a,b]\to \mathbb{R}$ is a smooth curve with $\gamma'(t)\neq 0$ for $t\in [a,b]$ , I know that $s(t)$= $\int_{a}^{t}\left\| \gamma'(\psi)\right\|d\psi$ for $t\in [a,b]$ then I find the inverse funtion of $s$.
Can anybody help me find a way to express the parameterization for the ellipse? I’m looking for a solution in terms of sine amplitude and cosine amplitude.
|
There is no standard closed form but series solutions only
*
*Local canonical form starting from minor axis (clockwise convention):
\begin{align}
k &= \sqrt{1-\frac{b^2}{a^2}} \\
s(t) &= \int_0^t \sqrt{a^2\cos^2 \theta+b^2\sin^2 \theta} \, d\theta \\
&= aE(t,k) \\
t &= E^{-1} \left( \frac{s}{a},k \right) \\
\begin{pmatrix} x \\ y \end{pmatrix} &=
\begin{pmatrix} a\sin t \\ b\cos t \end{pmatrix} \\
&=
\begin{pmatrix}
s-\frac{b^2 s^3}{6a^4}+\frac{b^2(13b^2-12a^2)s^5}{120a^8}+\ldots \\
b-\frac{b s^2}{2a^2}+\frac{b(4b^2-3a^2)s^4}{24a^6}+\ldots
\end{pmatrix} \\
\kappa &= -\frac{b}{a^2}+\frac{3b(b^2-a^2)s^2}{2a^6}+
\frac{b(b^2-a^2)(15a^2-19b^2)s^4}{8a^{10}}+\ldots \\
\end{align}
*
*Local canonical form starting from major axis (anti-clockwise convention):
\begin{align}
k' &= \frac{b}{a} \\
s(u) &= \int_0^u \sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta} \, d\theta \\
&= bE\left( u,\frac{ik}{k'} \right) \\
u &= E^{-1}\left( \frac{s}{b},\frac{ik}{k'} \right) \\
\begin{pmatrix} x \\ y \end{pmatrix} &=
\begin{pmatrix} a\cos u \\ b\sin u \end{pmatrix} \\
&=
\begin{pmatrix}
a-\frac{a s^2}{2b^2}+\frac{a(4a^2-3b^2)s^4}{24b^6}+\ldots\\
s-\frac{a^2 s^3}{6b^4}+\frac{a^2(13a^2-12b^2)s^5}{120b^8}+\ldots
\end{pmatrix} \\
\kappa &= \frac{a}{b^2}+\frac{3a(b^2-a^2)s^2}{2b^6}+
\frac{a(a^2-b^2)(19a^2-15b^2)s^4}{8b^{10}}+\ldots \\
\end{align}
*
*Plots of the partial sums for $s\in [0,3]$:
*
*Fourier series for eccentric angle:
\begin{align}
ds &= a\sqrt{1-k^2\sin^2 t} \, dt \\
t &= \frac{1}{2aE(k)} \sum_{n=1}^{\infty}
\sin \frac{n\pi s}{2aE(k)} \int_{-2aE}^{2aE} E^{-1}
\left( \frac{s'}{a}, k \right) \sin \frac{n\pi s'}{2aE(k)} ds' \\
&= \sum_{n=1}^{\infty} \sin \frac{n\pi s}{2aE(k)} \int_{-\pi}^{\pi}
\underbrace{
\frac{\tau \sqrt{1-k^2\sin^2 \tau}}{2E(k)}
\sin \frac{n\pi E(\tau,k)}{2E(k)}
}_{\text{expand as Maclaurin series in }k} \, d\tau \\
&= \frac{\pi s}{2aE(k)}-
\left( \frac{k^2}{8}+\frac{k^4}{16}+\ldots \right)
\sin \frac{\pi s}{aE(k)}+\left( \frac{5k^4}{256}+\ldots \right)
\sin \frac{2\pi s}{aE(k)}-\ldots \\
\end{align}
*
*The first term comes from sawtooth waveform:
$$\frac{\pi s}{2aE(k)}=\phi=2
\left(
\sin \phi-\frac{\sin 2\phi}{2}+\frac{\sin 3\phi}{3}-\ldots
\right)$$
*
*Compare with the Jacobi amplitude here.
*See also the anti-clockwise version in my older post here.
|
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|
Integration of complicated equation I was looking at a Wolfram Mathworld article about bean curves: https://mathworld.wolfram.com/BeanCurve.html
And it states that the area enclosed by the curve of
$$x^4+x^2y^2+y^4=ax(x^2+y^2)$$ is
$$A=\sqrt{2}a^2\int_0^1\sqrt{x\left(1-x+\sqrt{1+(2-3x)x}\right)}\text{d}x=\frac{7\pi a^2}{12\sqrt{3}}$$
Although a brief calculation shows that the area enclosed by the curve is $$A=\sqrt{2}a^2\int_0^1\sqrt{x\left(1-x+\sqrt{1+(2-3x)x}\right)}\text{d}x$$ I have no clue how the integration, in which $$A=\frac{7\pi a^2}{12\sqrt{3}}$$
, works.
I would like to know how!
|
use polar coordinates to evaluate the area.
$$r^4(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))=ar^3\cos(\theta)$$
$$r=\frac{a\cos(\theta)}{\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta)}$$
$$\frac{A}{a^2}=\frac{1}{2a^2}\int_{-\pi/2}^{\pi/2}r^2 d\theta$$
$$=\frac{1}{2}\int_{-\pi/2}^{\pi/2} \frac{\cos^2(\theta)}{(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))^2} d\theta$$
$$=\int_{0}^{\pi/2} \frac{\cos^2(\theta)}{(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))^2} d\theta$$
$$=\int_{0}^{\pi/2} \frac{\cos^2(\theta)}{(1-\frac{1}{4}\sin^2(2\theta))^2} d\theta$$
$$=8\int_{0}^{\pi/2} \frac{2\cos^2(\theta)}{(4-sin^2(2\theta))^2} d\theta$$
$$=8\int_{0}^{\pi/2} \frac{1+\cos(2\theta)}{(4-\sin^2(2\theta))^2} d\theta$$
$$=4\int_{0}^{\pi} \frac{1+\cos(\phi)}{(4-\sin^2(\phi))^2} d\phi$$
$$=4\int_{0}^{\pi} \frac{1}{(4-\sin^2(\phi))^2} d\phi$$
$$=16\int_{0}^{\pi} \frac{1}{(8-2\sin^2(\phi))^2} d\phi$$
$$=16\int_{0}^{\pi} \frac{1}{(7+\cos(2\phi))^2} d\phi$$
$$=8\int_{0}^{2\pi} \frac{1}{(7+\cos(\alpha))^2} d\alpha$$
$$=16\int_{0}^{\pi} \frac{1}{(7+\cos(\alpha))^2} d\alpha$$
$$=16*\frac{7\sqrt3}{288}*\tan^{-1}(\frac{\sqrt3}{2}\tan(\frac{\alpha}{2}))|_0^{\pi}$$
$$=\frac{7\sqrt3}{18}*\frac{\pi}{2}$$
$$A=\frac{7\pi a^2}{12\sqrt3}$$
change of variables: $\phi=2\theta$ and $\alpha=2\phi$
|
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|
How do you prove by induction that $\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$? For $n=1$ this is true because $\frac{1}{2^{1}}=2-\frac{1+2}{2^{1}}=\frac{1}{2}$. Further, it is a little more complicated, can we now assume that this is true up to the number $n-1$? Then do the induction step from $n-1$ to $n$.
So what I've tried:
$$\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{n-1}{2^{n-1}}+\frac{n}{2^{n}}=$$
$$=2-\frac{n+1}{2^{n-1}}+\frac{n}{2^{n}}=$$
|
You can write this sum using sigma notation, for clarity.
$\frac{1}{2} +\frac{2}{2^2} + \dots+\frac{n}{2^n} = \sum_{i=1}^{n} \frac{i}{2^i} = 2-\frac{n+2}{2^n}.$ (*)
Let our induction hypothesis $P(n)$ be the equation (*).
Let's apply mathematical induction now.
Base case $(n=1):$ $$\sum_{i=1}^{1} \frac{i}{2^i} = \frac{1}{2^1} = \frac{1}{2} = 2-\frac{1+2}{2^1} = 2-\frac{3}{2} = \frac{1}{2}$$ $P(1)$ holds. $\blacksquare$
Inductive step: Assume $P(n)$ holds for some $n \geq 1.$ Show that $P(n+1)$ holds.
Since $P(n)$ holds, we can write
$$\sum_{i=1}^{n} \frac{i}{2^i} = 2-\frac{n+2}{2^n}$$
$$\sum_{i=1}^{n} \frac{i}{2^i} + \frac{n+1}{2^{n+1}}= 2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}$$
$$\sum_{i=1}^{n+1} \frac{i}{2^i} = 2+\frac{(-2)\cdot(n+2) + (n+1)}{{2}^{n+1}}$$
$$\sum_{i=1}^{n+1} \frac{i}{2^i} = 2+\frac{-n-3}{{2}^{n+1}}$$
$$\sum_{i=1}^{n+1} \frac{i}{2^i} = 2-\frac{n+3}{{2}^{n+1}}$$
$$\sum_{i=1}^{n+1} \frac{i}{2^i} = 2-\frac{(n+1)+2}{{2}^{n+1}}$$
Since $P(n)$ holds for $n+1$, our induction hypothesis is true $\forall n \geq 1.$ $\blacksquare$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can we approximate $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$? It seems like from the graph $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$ is somehow alike to the graph $e^{-x^2}$, the main problem is that the limitations of the software makes it hard to graph for large numbers. Is it possible to do so or it diverges?
|
The partial product $$P_m=\prod _{n=1}^{m } \left(1-\frac{(2 n+1) }{n^2\pi ^2 }x^2\right)$$ effectively looks like a gaussian (tested for $m=10^5$). Using Pochhammer symbols
$$P_m=\frac{\left(\frac{\pi^2-x^2-\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi
^2}\right){}_m \left(\frac{\pi^2-x^2+\sqrt{x^2 \left(x^2+\pi ^2\right)} }{\pi
^2}\right){}_m}{(m!)^2}$$
Taking logarithms and expanding as series for large values of $m$, we have
$$\log(P_m)\sim-\frac{2 x^2 \log (m)}{\pi ^2}\implies P_m\sim \exp\Bigg[-\frac{2 x^2 }{\pi ^2} \log (m)\Bigg]=m^{-\frac{2 }{\pi ^2}x^2}$$
Try it and let us know.
Edit
We can even do much better using the next term of the expansion which gives
$$\log(P_m)\sim-\frac{2 x^2 \log (m)}{\pi ^2}-$$ $$\log \left(\Gamma \left(\frac{\pi^2-x^2+\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi
^2}\right) \Gamma \left(\frac{\pi^2-x^2-\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi ^2}\right)\right)$$ Expanding for small values of $x$, this gives
$$ P_m\sim \exp\Bigg[-\frac{\left(12 \log (m)+\pi ^2+12 \gamma \right)}{6 \pi ^2}\,x^2\Bigg]$$ which is much better.
We could even go further introducing terms in $x^4$ but they will include some polygamma terms.
|
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|
Where is the mistake? Finding an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
Find an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
We consider the foci in the coordinate system $XY$ such that $X=x-2$ and $Y=y-1-x$, the coordinates of the foci in this system are $(-1,0)$ and $(1,0)$, furthermore $2a=5$, the equation of the ellipse in $XY$ is
\begin{equation}
\left( \frac{X}{2.5}\right)^2 + \left( \frac{Y}{\sqrt{5.25}} \right)^2 = 1
\end{equation}
and this can be expressed in $xy$ as
$$\left( \frac{x-2}{2.5} \right) + \left( \frac{y-1-x}{\sqrt{5.25}}\right) = 1$$
I have made the graph of the last equation and it is not the case that foci are $(1,2)$ and $(3,4)$, so, can anyone help me to see the mistake please?
|
You just need to write it like what it is, the sum of the distance to the foci is 5:
$$\sqrt{(x-1)^2+(y-2)^2} + \sqrt{(x-3)^2+(y-4)^2} = 5$$
|
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|
Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
|
Treat $4x^2 - 2xy - 4x + 3y - 3$ as a quadratic in $x$.
Set $4x^2 - 2x(y + 2) + 3(y - 1) = 0$, then solve for $x$ (using the Quadratic Formula), to factor the LHS:
$$x = \dfrac{2(y+2) \pm \sqrt{(2(y+2))^2 - 4(4)(3)(y-1)}}{8} = \dfrac{y + 2 \pm \sqrt{y^2 + 4y + 4 - 12y + 12}}{4} = \dfrac{y + 2 \pm \sqrt{y^2 - 8y + 16}}{4} = \dfrac{y + 2 \pm \sqrt{(y - 4)^2}}{4} = \dfrac{(y + 2) \pm (y - 4)}{4}$$
so that the LHS factors as
$$4x^2 - 2xy - 4x + 3y - 3 = 4\Bigg(x - \dfrac{y - 1}{2}\Bigg)\Bigg(x - \dfrac{3}{2}\Bigg) = (2x - y + 1)(2x - 3).$$
|
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|
Minimize the ratio involving the ellipse
Let $P$ be any point on the curve $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$,
and $A,B$ be two fixed points $\left(\dfrac{1}{2},0\right)$ and
$(1,1)$ respectively. Find the minimum value of
$\dfrac{|PA|^2}{|PB|}$.
Assume $x=2\cos\theta,y=\sqrt{3}\sin\theta$, then
$$\dfrac{|PA|^2}{|PB|}=\frac{\left(2\cos\theta-\frac{1}{2}\right)^2+(\sqrt{3}\sin\theta-0)^2}{\sqrt{\left(2\cos\theta-1\right)^2+(\sqrt{3}\sin\theta-1)^2}},$$
but which is not so easy to tackle.
WolframeAlpha gives the result $1$.
|
Denote $c = \cos \theta$ and $s = \sin\theta$.
We have
$$|\mathrm{PA}|^2 = (2c - 1/2)^2 + (s\sqrt 3 - 0)^2= c^2 - 2c + \frac{13}{4}$$
and
$$|\mathrm{PB}|^2 = (2c - 1)^2 + (s\sqrt 3 - 1)^2
= c^2 - 4c + 5 - 2\sqrt 3\, s.$$
We have
\begin{align*}
|\mathrm{PA}|^4 - |\mathrm{PB}|^2
&= (c^2 - 2c + 13/4)^2 - (c^2 - 4c + 5 - 2\sqrt 3\, s)\\
&= (c^2 - 2c + 13/4)^2 - c^2 + 4c - 5 + 2\sqrt 3\, s\\
&\ge 0
\end{align*}
where we have used
$(c^2 - 2c + 13/4)^2 - c^2 + 4c - 5 > 0$ and
\begin{align*}
&[(c^2 - 2c + 13/4)^2 - c^2 + 4c - 5]^2 - (2\sqrt 3\, s)^2\\
=\,& [(c^2 - 2c + 13/4)^2 - c^2 + 4c - 5]^2 - 12(1-c^2)\\
=\,&\frac{1}{256}(64c^6 - 448c^5 + 1776c^4 - 4128c^3 + 6524c^2 - 6236c + 4849)(2c-1)^2\\
\ge\,& 0.
\end{align*}
On the other hand, when $\theta = \frac{5\pi}{3}$,
we have $|\mathrm{PA}|^4 - |\mathrm{PB}|^2 = 0$ and $\frac{|\mathrm{PA}|^2}{|\mathrm{PB}|} = 1$.
Thus, the minimum of $\frac{|\mathrm{PA}|^2}{|\mathrm{PB}|}$ is $1$.
|
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|
Let $p, q$ be different primes. Then $p^2+q^2-pq$ is not a perfect square. This question originally comes from the following problem:
Let $ABC$ be a triangle with integer side lengths with $\angle ABC = 60^{\circ}$. Suppose length $\overline{AB}$ and $\overline{BC}$ are prime numbers. Determine and prove what kind of triangle $ABC$ is.
I suspect $ABC$ must be an isosceles triangle where $\overline{AB} = \overline{BC}$. By the law of cosine we have
$$ \overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 - 2\overline{AB}\overline{BC}\cos 60^{\circ}.$$
Let $p = \overline{AB}$ and $q = \overline{BC}$ the above statement is equivalent to $p^2 + q^2 - pq$ being a perfect square. My job will be proving that this statement holds for $p \neq q$(where $p=q$ corresponds to the case where $\overline{AB} = \overline{BC}$ i.e. $ABC$ being isoceles triangle).
I tried factorizing and discussing case by case and did not work out for me. How can I approach this problem?
|
We have: $$(p-q)^2<p^2-pq+q^2<(p+q)^2.$$
If $p>q>0$ and $p^2-pq+q^2=c^2,$ for $c>0,$ then then $q^2\equiv c^2\pmod p,$ so $$c\equiv \pm q\pmod p$$ Now, since $0<p-q< c<p+q,$ this means our options are $c=q$ or $c=2p-q.$ In both cases, $c$ is in the range only if $p<2q.$
If $c=q,$ then $p^2-qp=0,$ so $p=0$ or $p=q.$
If $c=2p-q$ then $$p^2-pq+q^2=4p^2-4pq+q^2\\\implies3p^2-3pq=0$$ and again $p=q$ or $p=0.$
So we’ve only really used that the larger of $p,q$ is prime (to conclude $c\equiv\pm q\pmod p.$) Indeed, we could restrict it to the larger of the two being a power of a prime.
So we get:
If $1<m<n$ are relatively prime, and $n$ is the power of an odd prime, or $n=2,4$ then $m^2-mn+n^2$ is not a perfect square.
We can’t use other powers of $2$ because $1$ has more than $2$ square roots modulo $2^n$ for $n>2.$
Not sure if it is possible for $m$ to be an odd prime power if $q$ is not.
We can apply the usual technique to find a formula for all rational solutions to $$x^2-xy+y^2=1.$$
Pick one rational solution $(x,y)=(1,1).$ Every other rational solution is on a line $(1+at,1+bt)$ for some $a,b$ integers and $t\neq 0.$ But given $a,b$ you get:
$$
\begin{align}
1&=(1+at)^2-(1+at)(1+bt)+(1+bt)^2\\&=1+(2a-(a+b)+2b)t+(a^2-ab+b^2)t^2\\
t&=-\frac{a+b}{a^2-ab+b^2}
\end{align} $$
Now $$\gcd(a+b,a^2-ab+b^2)=1,3$$ with $3$ if $a\equiv $b\pmod 3,$ and $1$ otherwise.
So $$x=1+at=\frac{b^2-2ab}{a^2-ab+b^2},\\
y=1+bt=\frac{a^2-2ab}{a^2-ab+b^2}.$$
So, at least for relatively prime $(m,n)$ if $m^2-nm+n^2$ is a perfect square, then:$$m=b^2-2ab\\n=a^2-2ab$$ for some relatively prime $a,b,$ with $a+b\not\equiv 0\pmod 3,$ or $$m=\frac{b^2-2ab}3\\n=\frac{a^2-2ab}{3}$$
for some relatively prime $a,b$ with $a+b\equiv 0\pmod 3.$
This shows why it is hard for $m,n$ to be prime.
|
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|
How should I evaluate $\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$? $$\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$$
$$\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{1}{\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}x$$
$$\frac1{\pi}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{x^2}{x^4+x^2+1}\ \mathrm{d}x$$
then substituting $t^3=x$
$$\frac3{\pi}\int_0^{\infty}\frac{t^9}{(1+t)(t^{12}+t^6+1)}\ \mathrm{d}t$$
what should i do after this should i write $\frac{1}{1+t}$ as $\sum(-1)^kt^k$?
|
If, as @RAHUL commented, you use partial fraction decomposition, you should be able to prove that the result is the smallest positive root of
$$19683 y^6-94041 y^4+105786 y^2-2809=0$$
Solving the cubic equation in $y^2$ and taking the square root of it, the result is
$$\sqrt{\frac{1}{27} \left(43-2 \sqrt{543} \cos \left(\frac{1}{3} \cos
^{-1}\left(\frac{739}{362}\sqrt{\frac{3}{181}}\right)\right)\right)}=0.164948\cdots$$ which is confirmed by numerical integration.
Now, just prove it.
The partial fraction decomposition
$$\frac{t^9}{(t+1)(t^{12}+t^6+1)}=\frac{t^3}{1+t}\times\frac{t^6}{(t^{12}+t^6+1)}$$
$$\frac{t^6}{(t^{12}+t^6+1)}=\frac 1{a-b}\Big[\frac{a}{t^6-a}-\frac{b}{t^6-b} \Big]$$ where
$$a=-\frac{1}{2}-i\frac{ \sqrt{3}}{2}\qquad \text{and}\qquad b=-\frac{1}{2}+i\frac{ \sqrt{3}}{2}$$
So now, we face two terms looking like
$$\frac{t^3}{(t+1)(t^6-c)}=\frac 1{c-1}\Bigg[\frac 1{t+1}+\frac{c t^2-c t+c-t^5+t^4-t^3}{t^6-c} \Bigg]$$
$$I_k=\int \frac {t^k}{t^6-c}\,dt=-\frac{t^{k+1} }{c (k+1)}\,
_2F_1\left(1,\frac{k+1}{6};\frac{k+7}{6};\frac{t^6}{c}\right)$$ All these integrals are easily computed. The simplest are
$$I_2=-\frac{1}{3 \sqrt{c}}\tanh ^{-1}\left(\frac{t^3}{\sqrt{c}}\right)\qquad \qquad I_5=\frac{1}{6} \log \left(1-\frac{t^6}{c}\right)$$ The other ones lead to sums of logarithms.
On the other side
$$J_k=\int_0^\infty \frac {t^k}{t^6-c}\,dt=\frac{\pi}{6} \left(-\frac{1}{c}\right)^{\frac{5-k}{6}} \sec
\left(\frac{(k-2)\pi}{6} \right)$$
To make the link with @Laxmi Narayan Bhandari's answer, notice that
$$a^{1/6}=\cos \left(\frac{\pi }{9}\right)-i \sin \left(\frac{\pi }{9}\right)\qquad \qquad b^{1/6}=\cos \left(\frac{\pi }{9}\right)+i \sin \left(\frac{\pi }{9}\right)$$
|
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|
$\lim_{n \rightarrow \infty } \int_0^1 \frac{nf(x)}{1+ n^{2} x^{2} }dx = \frac{ \pi }{2} f(0)$ for continuous $f$. If f is a continuous function on $[0,1]$, then show that $$\lim_{n \rightarrow \infty } \int_0^1 \frac{nf(x)}{1+ n^{2} x^{2} }dx = \frac{ \pi }{2} f(0)$$
Can anybody help me to solve this? I tried but i have no idea about how to prove this.
Thanks in advance!
|
Put $I = \int_{0}^1 \frac{nf(x)}{1+n^2x^2} dx = I_1 + I_2$ where
$I_1 = \int_{0}^{n^{-\frac13} } \frac{nf(x)}{1+n^2x^2} dx$ and $I_2 = \int_{n^{-\frac13} }^1 \frac{nf(x)}{1+n^2x^2} dx$.
Further $|f(x)| \le M$ because $f$ is continious. We have
$$|I_2| \le \int_{n^{-\frac13} }^1 \bigg|\frac{nf(x)}{1+n^2x^2} \bigg| dx \le \int_{n^{-\frac13} }^1 \frac{nM}{1+n^2x^2} dx \le \int_{n^{-\frac13} }^1 \frac{nM}{1+n^2(n^{-\frac13})^2} dx $$
$$=\frac{nM}{1+n^2(n^{-\frac13})^2} \cdot (1 - n^{-\frac13}) = o(1). $$
Put $I_3 = \int_{0}^{n^{-\frac13} } \frac{nf(0)}{1+n^2x^2} dx$. We have
$$|I_1 - I_3| = \bigg| \int_{0}^{n^{-\frac13} } \frac{n(f(x)-f(0))}{1+n^2x^2} dx \bigg| \le \int_{0}^{n^{-\frac13} } \frac{n |f(x)-f(0)|}{1+n^2x^2} dx $$
$$ \le \sup_{ x \in [0, n^{-\frac13}]} |f(x)-f(0)| \cdot \int_{0}^{n^{-\frac13} } \frac{n}{1+n^2x^2} dx = o(1) \cdot \int_{0}^{n^{-\frac13} } \frac{d(nx)}{1+(nx)^2} $$
$$ = o(1) \cdot \arctan (n \cdot n^{-\frac13}) = o(1) \cdot O(1) = o(1). $$
Hence $$I = I_1 + I_2 = I_3 + (I_1 - I_3) + I_2 = I_3 + o(1) + o(1)$$ where
$$I_3 = f(0) \int_{0}^{n^{-\frac13} } \frac{d(nx)}{1+n^2x^2} = f(0) \cdot \arctan (n \cdot n^{-\frac13}) = f(0) \frac{\pi}2 +o(1). $$
So $I_1 \to f(0) \frac{\pi}2$ as $n \to \infty$.
Addition:
As Oolong milk tea said above, we may solve the problem much easier if it's allowed to use dominated convergence theorem: $I = \int_{[0,n]} \frac{f(\frac{y}n)}{1+y^2}dy = o(1) + \int_{[0,\infty]} \frac{f(\frac{y}n)}{1+y^2}dy = o(1) + \int_{[0,\infty]} \frac{f(0)}{1+y^2}dy$ where $f(x) = f(1)$ for $x > 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of $x^2dx$ by definition - what am I doing wrong? $$\int_a^bx^2dx$$
By definition
$$b-a=nh$$
$$\lim_{h\to0}a^2h+(a+h)^2h+(a+2h)^2h+...+(a+(n-1)h)^2h$$
$$\lim_{h\to0}h(a^2+(a+h)^2+(a+2h)^2+...+(a+(n-1)h)^2)$$
$$\lim_{h\to0}h(a^2+(a^2+1^2h^2+2*1h)+(a^2+2^2h^2+2*2h)+...+(a^2+(n-1)^2h^2+2*(n-1)h))$$
$$\lim_{h\to0}h(na^2 + (1^2+2^2+...+(n-1)^2)h^2+(2*1+2*2+...+2*(n-1))h)$$
$$\lim_{h\to0}nha^2+\lim_{h\to0}h^3(1^2+2^2+...+(n-1)^2)+\lim_{h\to0}2h^2(1+2+...+(n-1))$$
$$(b-a)a^2+\lim_{h\to0}h^3n\frac{(n-1)(2n-1)}{6}+\lim_{h\to0}h^2n(n-1)$$
$$(b-a)a^2+\frac{1}{6}\lim_{h\to0}nh\lim_{h\to0}h^2(2n^2+3n+1)+\lim_{h\to0}nh\lim_{h\to0}nh-h$$
$$(b-a)a^2+\frac{1}{6}(b-a)\lim_{h\to0}(2(nh)^2+3nh^2+h^2)+(b-a)(\lim_{h\to0}nh-\lim_{h\to0}h)$$
$$(b-a)(a^2+\frac{1}{6}\lim_{h\to0}(2(nh)^2+3nh^2+h^2)+(\lim_{h\to0}nh-\lim_{h\to0}h))$$
$$(b-a)(a^2+\frac{1}{6}(2\lim_{h\to0}(nh)^2+3\lim_{h\to0}nh\lim_{h\to0}h+\lim_{h\to0}h^2)+((b-a)-0))$$
$$(b-a)(a^2+\frac{1}{6}(2(b-a)^2+3(b-a)*0 + 0)+(b-a))$$
$$(b-a)(a^2+\frac{1}{6}2(b-a)^2+(b-a))$$
$$(b-a)(a^2+\frac{1}{3}(b-a)^2+(b-a))$$
$$(b-a)(\frac{3a^2+(b-a)^2+(b-a)}{3})$$
$$(b-a)(\frac{3a^2+b^2+a^2-2ab+b-a}{3})$$
$$\frac{(b-a)}{3}(4a^2+b^2-2ab+b-a)$$
which is not equal to $$\frac{b^3-a^3}{3}$$ for this to happen the fraction should have been $$\frac{(b-a)}{3}(b^2+a^2+ab)$$
Please tell me where did I go wrong
|
Please tell me where did I go wrong
$$\lim_{h\to0}a^2h+(a+h)^2h+(a+2h)^2h+...+(a+(n-1)h)^2h$$
$$\lim_{h\to0}h(a^2+(a+h)^2+(a+2h)^2+...+(a+(n-1)h)^2)$$
$$\lim_{h\to0}h(a^2+(a^2+1^2h^2+2*1h\color{red}{*a})+(a^2+2^2h^2+2*2h\color{red}{*a})+...+(a^2+(n-1)^2h^2+2*(n-1)h\color{red}{*a}))$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence or Divergence of $ 1 - 1 + \frac{1}{2} - . . .$ and another series. I'm testing ideas about the convergence and divergence of series:
Say I have a series $$ 1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + . . .$$
I believe that this series would converge to $0$ as every odd term in the series is canceled by the following term. (In addition, the individual terms are decreasing to 0 from 1 and -1.) In other words, the sequence of partial sums of this series tend to a limit, $0$.
Say I have a different series $$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{2^3} + \frac{1}{4} - \frac{1}{2^4} + . . .$$
My assumption would be that the sum of odd terms is not canceled out by the sum of even terms. In other words, this series would increase to infinity. The partial sum of the first $i$ terms would be less than the partial sum of the first $i+2$ terms, and so forth.
How do these series converge or diverge? Am I looking at this the right way?
|
$$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{2^3} + \frac{1}{4} - \frac{1}{2^4} + . . .$$
can be coded as $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big)$. If this series would converge, as the sum of convergent series is a convergent series, then $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big) +\sum_{n\geq1}\frac{1}{2^n}=\sum_{n\geq1}\frac{1}{n}$ would converge. This is a absurd, therefore the series diverges.
The reasoning behind your conclusion - "The partial sum of the first $i$ terms would be less than the partial sum of the first $i+2$ terms, and so forth" might be wrong... take for example $e=\sum_{n\geq0}\frac{1}{n!}$.
|
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|
Prove that the sequence $(a_n)$ is Cauchy and find the limit. Let us define a sequence $(a_n)$ as follows:
$$a_1 = 1, a_2 = 2 \text{ and } a_{n} = \frac14 a_{n-2} + \frac34 a_{n-1}$$
Prove that the sequence $(a_n)$ is Cauchy and find the limit.
I have proved that the sequence $(a_n)$ is Cauchy. But unable to find the limit. I have observed that the sequence $(a_n)$ is decreasing for $n \ge 2$.
|
Rewrite $a_n$ as $$a_1=1,\ a_2=2,\ a_{n+2}=\dfrac{3}{4}a_{n+1}+\dfrac{1}{4}a_n \mathrm{\ for\ } n\geqq 1.$$
We can get
\begin{align}
&a_{n+2}-a_{n+1}=-\dfrac{1}{4}(a_{n+1}-a_{n}) \cdots (A)\\
&a_{n+2}+\dfrac{1}{4}a_{n+1}=a_{n+1}+\dfrac{1}{4}a_{n} \cdots (B)
\end{align}
Letting $b_n=a_{n+1}-a_n$, we get $b_{n+1}=-\dfrac{1}{4}b_n$ from $(A)$, thus $\{b_n \}$ is a geometric progression of ratio $-\dfrac{1}{4}$. Thus $b_n=b_1 \cdot (-\frac{1}{4})^{n-1}=(-\frac{1}{4})^{n-1}$. Therefore $$a_{n+1}-a_n=\left(-\frac{1}{4}\right)^{n-1} \cdots (C)$$
Next, letting $c_n=a_{n+1}+\dfrac{1}{4}a_n$, we get $c_{n+1}=c_n$ from $(B)$. This means that all terms of $\{c_n \}$ are equal, so $c_n=c_1=\dfrac{9}{4}$. Thus $$a_{n+1}+\dfrac{1}{4}a_n=\dfrac{9}{4} \cdots (D)$$
Calculating $(D)-(C)$, we get $\dfrac{5}{4}a_n=\dfrac{9}{4}-\left(-\dfrac{1}{4}\right)^{n-1},$ i.e., $a_n=\dfrac{9}{5}-\dfrac{4}{5}\left(-\dfrac{1}{4}\right)^{n-1}$.
Letting $n\to \infty,$ we get $\displaystyle\lim_{n\to \infty} a_n =\dfrac{9}{5}.$
|
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|
The Operator Norm of Basis Change Matrices for a more explicit Gelfand Formula Gelfand's formula states that the operator norm of a matrix exponent $A^n$ is in some sense equal to the exponent of the spectral radius $\rho(A)^n$ in the sense that
$$
\lim_{n\to\infty} \sqrt[n]{\|A^n\|} =
\inf_{n\in\mathbb{N}} \sqrt[n]{\|A^n\|} = \rho(A).
$$
To obtain a more explicit version for my particular problem I used
$$
A^n = UJ^nU^{-1}
$$
where $U$ is a basis change matrix and $J$ is the jordan form. This resulted in
$$
\|A^n\|\le \|U\|\|U^{-1}\| (n+\rho(A))\rho(A)^{n-1}
$$
in my particular case. This is in some sense more explicit, but due to the constant in front it is not really.
So the question is whether or not there are methods to bound basis change matrices. I am mostly interested in 2 dimensional matrices.
Kozyakin (2009) showed something similar using the constant
$$
\frac{\|A\|^d}{\|A^d\|}
$$
directly where $A$ is a $d\times d$ matrix, instead of using a basis change. So in my case $d=2$. It is very likely that there is no good bound for the general case. In that case I would be glad to hear about special cases which might be helpful.
The particular matrix I am interested in looks like this
$$
\begin{pmatrix}
0 & 1 \\
-\beta & 1+\beta - a
\end{pmatrix}
$$
where $0\le\beta\le1$ and $0\le a\le (1-\sqrt{\beta})^2$.
|
Instead of using the Jordan decomposition one should use the Schur Decomposition in this case as Unitary matrices are isometries in euclidean norms and therefore isometries on the induced operator norms. This means we do not have to estimate them at all.
Lemma: Explicit Schur Decomposition for Specific Matrix
For a real matrix of the form
$$
R = \begin{pmatrix}
0 & 1 \\
-\beta & \xi
\end{pmatrix},
$$
the unitary operator
$$
Q:=\frac{1}{\sqrt{1+|r_1|^2}} \begin{pmatrix}
1 & \overline{r}_1 \\
r_1 & -1
\end{pmatrix}
$$
where
$
r_{1/2}
= \tfrac12 \left(
\xi \pm \sqrt{\xi^2 - 4\beta}
\right)
$
are the eigenvalues of (R), results in the Schur Decomposition
\begin{align*}
Q^* R Q
&=\begin{pmatrix}
r_1
& -\frac{(1+\beta)(1+\overline{r}_1^2)}{1+|r_1|^2} \\
0 & \frac{\beta\overline{r}_1 + r_2}{1+|r_1|^2}
\end{pmatrix}.
\end{align*}
For complex eigenvalues, more specifically $4\beta\ge\xi^2$, we have
\begin{align*}
Q^* R Q
&=\begin{pmatrix}
r_1
& -(1+\overline{r}_1^2) \\
0 & \overline{r}_1
\end{pmatrix}.
\end{align*}
Note that $r_1$ and $r_2$ could always be swapped to achieve a more favorable result.
Proof
Since
\begin{align*}
\frac{1}{\sqrt{1+|r_1|^2}} \begin{pmatrix}
1 \\
r_1
\end{pmatrix}
\end{align*}
is a normalized eigenvector for eigenvalue (r_1) of (R), there are
only two options for an orthonormal second vector, which results in (Q).
Recall that any eigenvalue of $R$, in particular $r_1$ and
$\overline{r}_1\in\{r_1, r_2\}$, is a root of the characteristic polynomial
\begin{align}\label{eq: property of eigenvalues of R}
\det (r\mathbb{I} - R) = r^2 - r\xi + \beta = 0.
\end{align}
Now we can calculate the Schur decomposition of $R$:
\begin{align*}
Q^* R Q
&= \frac{1}{1+|r_1|^2} \begin{pmatrix}
1 & \overline{r}_1 \\
r_{1} & -1
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
-\beta & \xi
\end{pmatrix}
\begin{pmatrix}
1 & \overline{r}_1 \\
r_1 & -1
\end{pmatrix}\\
&= \frac{1}{1+|r_{1}|^2} \begin{pmatrix}
1 & \overline{r}_1 \\
r_1 & -1
\end{pmatrix}
\begin{pmatrix}
r_1 & -1 \\
\smash{\underbrace{-\beta + \xi r_1}_{
=r_1^2
}}
& -\beta \overline{r}_1 - \xi
\end{pmatrix}
\vphantom{\underbrace{\begin{pmatrix}1\\1\end{pmatrix}}_{=2}}
\\
&=\frac{1}{1+|r_{1}|^2}\begin{pmatrix}
r_1(1+|r_1|^2)
& -\beta \overline{r}_1^2 + (1+\xi \overline{r}_1) \\
0 & \beta\overline{r}_1 + \xi - r_1
\end{pmatrix}\\
&\le\begin{pmatrix}
r_1
& -\frac{\beta \overline{r}_1^2 + 1+ (\overline{r}_1^2 + \beta)}{1+|r_1|^2} \\
0 & \frac{\beta\overline{r}_1 + \xi - r_1}{1+|r_1|^2}
\end{pmatrix}\\
&=\begin{pmatrix}
r_1
& -\frac{(1+\beta)(1+\overline{r}_1^2)}{1+|r_1|^2} \\
0 & \frac{\beta\overline{r}_1 + r_2}{1+|r_1|^2}
\end{pmatrix}
\end{align*}
where we have used $\xi - r_1 = r_2$ in the last equation. Now in the complex
case (and the case $4\beta = \xi^2$) we have
$|r_1|=|r_2|=\sqrt{\beta}$ and $\overline{r_1} = r_2$ which results
in
\begin{align*}
Q^* R Q
&=\begin{pmatrix}
r_1
& -(1+\overline{r}_1^2) \\
0 & \overline{r}_1
\end{pmatrix}.
\end{align*}
|
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|
What's the measure of the segment $EF$ in the triangle below? For reference:
The triangle $ABC$ where $AB = 7$, $BC = 8$ and $AC = 9$ is inscribed in a circle. Calculate the measure of the arrow ($EF$) of the side $AC$. (Answer: $\frac{3\sqrt5}{2}$)
My progress:
Here are the relations I found:
$EF = R - OE \\
\triangle AEF: EF^2 + OE^2 = AF^2\implies\\
(R-OE)^2 + (\frac{9}{2})^2 = AF^2 \implies:
(R-OE)^2 + \frac{81}{4} = AF^2\\
\triangle AOE: OE^2+AE^2 = AO^2 \implies OE^2 +(\frac{9}{2})^2 =R^2 \implies OE^2+ \frac{81}{4} = R^2\\
\triangle AOF: AF^2 = R^2 +OF^2-2OF.OE \implies\\
AF^2 = R^2 +R^2-2R.OE \therefore AF^2 = 2R^2-2R.OE = 2R\underbrace{(R-OE)}_{=EF}$
...??
|
Draw altitude $AD$ and call $BD=x$.
From Pythagoras' theorem, $$7^2-x^2=9^2-(8-x)^2\implies x=2$$
It follows that $AD=3\sqrt5$.
We see $\triangle ABD\sim\triangle AOE$. Therefore, $$\frac{AO}7=\frac{OE}2=\frac{9/2}{3\sqrt5}\\\implies AO=\frac{3\cdot7}{2\sqrt5},\: OE=\frac{3\cdot2}{2\sqrt5}$$
Thus $$EF=\underbrace{OF}_{=AO}-OE=\frac{3\cdot5}{2\sqrt5}=\frac{3\sqrt5}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $ \frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;. $ To find the value of $$
\frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;.
$$
I presented it as $$
\sum\limits_{n = 2}^\infty {\frac{1}{{n(n^2 - 1)}}} .
$$
Then using partial fractions method
I wrote the sum as
$$
\sum\limits_{n = 2}^\infty {\left( { - \frac{1}{n} + \frac{1}{{2(n + 1)}} + \frac{1}{{2(n - 1)}}} \right)} .
$$
Then I wrote down the individual terms to see if most of the terms cancel out like they do sometimes.
$$
- \frac{1}{2} + \frac{1}{6} + \frac{1}{2} - \frac{1}{3} + \frac{1}{8} + \frac{1}{4} - \frac{1}{4} + \frac{1}{{10}} + \frac{1}{6} - \frac{1}{5} + \frac{1}{{12}} + \frac{1}{8} - \frac{1}{6} + \frac{1}{{14}} + \frac{1}{{10}} + \cdots \; .
$$
Unfortunately they don't seem to cancel out. Is there a way to find out the sum in this method or any other method?
|
Note : $$\frac{1}{a.b.c}=\frac{1}{c-a}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$
and $$\frac{1}{a.b}=\frac{1}{b-a}\left(\frac{1}{a}-\frac{1}{b}\right)$$
Use this in $$\sum_{n\geq2}\frac{1}{(n-1)(n)(n+1)}$$
Here's what you'll get : $$\sum_{n\geq2}\frac{1}{2}\left(\frac{1}{(n-1)(n)}-\frac{1}{(n)(n+1)}\right)\implies\frac{1}{2}\left(\color{red}{\sum_{n\geq2}\frac{1}{(n-1)(n)}}-\color{blue}{\sum_{n\geq2}\frac{1}{(n)(n+1)}}\right)\tag{*}$$
So , the first sum is $$\color{red}{\sum_{n\geq2}\frac{1}{(n-1)(n)}}=\sum_{n\geq2}\left(\frac{1}{(n-1)}-\frac{1}{n}\right)=1\tag1$$
and the secong sum will be $$\color{blue}{\sum_{n\geq2}\frac{1}{(n)(n+1)}}=\sum_{n\geq2}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}\tag2$$
Using the values of sums from $(1)$ and $(2)$ in $(*)$ $$\sum_{n\geq2}\frac{1}{n(n^2-1)}=\frac{1}{2}\left(\color{red}{1}-\color{blue}{\frac{1}{2}}\right)\implies\frac{1}{4}$$
|
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|
What's the the value of $x$ in the circumference below? For reference:
In the figure; calculate $x$, if $r =\sqrt2$.
(Answer: $x = \sqrt2$)
My progress:
Draw $PO_1\perp HG\:(O_1 \in HG).$
Let $O$ be the center of the largest circle.
Using Euclid's Theorem:
$\triangle OPF:OP^2 = OQ^2+PQ^2-2\cdot OQ\cdot FQ$
$\implies ((R-x)^2 =(R-r_2)^2+(r_2+x)^2-2(R-r_2)((r_2-x)$
$\implies R^2 -2Rx+x^2 = R^2-2Rr_2+r_2^2 +r_2^2+2r_2x+x^2 -2Rr_2+2Rx+2r_2^2-2r_2x$
$\therefore\boxed{
r_2^2-r_2R-Rx = 0}$
$\triangle MJR: ((r_1+r)^2 = IH^2 +(r_1-x)^2$
$\implies r_1^2+2r_1r+r^2=IH^2+r_1^2-2r_1x+x^2$
$\therefore \boxed{2r_1(r+x)-x^2 = IH^2}$
$\triangle PFQ: PQ^2=PF^2+FQ^2 $
$\implies (r_2+x)^2=PF^2 + (r_2-x)^2 $
$\implies r_2^2+2r_2x+x^2=PF^2+r_2^2-2r_2x+x^2$
$\therefore \boxed{4r_2x = PF^2}$
...?
|
In fact, the two small circles here are called 'Twin circles'. As the name suggests, they are congruent. Here's a proof:
Following the image, $EH\parallel AB$ leads to $FEA,$ $FHB,$ $EDC,$ $ADH$ are straight lines. Also $AF$ is extended to meet $CH$ at $G$.
Since $\angle AFB=\angle ACG=90^\circ$, we can say that $H$ is the orthocenter of $\triangle ABG$. That implies $AI\perp BG$.
Therefore $\angle AIB=\angle ADC=90^\circ\implies BG\parallel CE.$
$\displaystyle \therefore \frac{EH}{AC}=\frac{EG}{AG}=\frac{CB}{AB}\implies EH=\frac{AC\cdot CB}{AB}.$
Similarly, we can prove that the diameter of the other circle is also equal to $\frac{AC\cdot CB}{AB}.$
Thus, you get the answer to your question from here.
|
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|
Find the sum of the squares of the medians of triangle APC below For reference: The diameter of a circle measures 8m. On this diameter are located the points A and B equidistant 1 m from the center; through B draw any chord PC , determine the sum of the squares of the medians of triangle APC.(Answer: $73,5$)
My progress:
sum of the squares of the median
$4(PE^2+CD^2+AB^2) = 3(AC^2+AP^2+CP^2)\therefore\\
PE^2+CD^2+AB^2=\frac{3}{4}(AC^2+AP^2+CP^2)\\
AC = AP \implies PE^2+CD^2+AB^2=\frac{3}{4}(2AC^2+CP^2)\\
CB=\frac{CP}{2} \\
\triangle ABC: 4+(\frac{CP}{2})^2=AC^2\implies \frac{16+CP^2}{4} =AC^2\\
\therefore PE^2+CD^2+AB^2 = \frac{3}{8}(16+3CP^2)
$
...??
|
Here's a proof for the general case where $\small PC$ is not necessarily perpendicular to $\small AB$:
Given that we have already proven $$\mathsf{4.(sum\ of\ squares\ of\ medians)=3.(sum\ of\ squares\ of\ sides)}$$ using Appollonius theorem, continue from there.
So let's calculate $\small AP^2+AC^2+PC^2\tag{*}$
Drop the perp from $\small A$ to $\small PC$ and the foot $\small I$ must lie on the small circle (angle of half circle=$\small 90^\circ$).
Now using power of point,
$\small \begin{align}PI\cdot PB&=3\cdot 5\\ PI\cdot(PI+IB)&=15\\PI^2+PI\cdot IB&=15\end{align}$
See $\small PI=BC$.
Also note that $\small PI\cdot IC=PI\cdot(PI+IB)$.
From Pythagoras' theorem $\small AP^2=AI^2+PI^2$, $\small AC^2=AI^2+IC^2$ and $\small AB^2=AI^2+IB^2$.
Now substitute these in (*).
$\small \begin{align}&=AP^2+AC^2+PI^2+IC^2+2\cdot PI\cdot IC\\ &=2AI^2+2PI^2+2IC^2+30\\&=2(AB^2-IB^2+PI^2+IC^2+15)\\&=2(2^2+PI^2+(IC^2-IB^2)+15)\\&=2(4+PI^2+(IC-IB)(IC+IB)+15)\\&=2(4+PI^2+PI(PI+2IB)+15)\\&=2(4+2(PI^2+PI\cdot IB)+15)\\&=2(4+2\cdot15+15)\\&=98\end{align}$
Now using the known identity, sum of squares of medians
$=98\cdot\dfrac34=73.5$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can a ratio of sinusoidal functions with the same frequency always be written as a tangent function? In general, two sinusoidal functions $y_1$ and $y_2$ with angular frequency $\omega$ can be written as
*
*$y_1 = A\sin(\omega x) + B \cos(\omega x)$,
*$y_2 = C\sin(\omega x) + D \cos(\omega x)$.
Where $A, B, C, D$ are real numbers and it is not the case that $A=B=0$ or $C=D=0$.
Can the ratio $\frac{y_1}{y_2}$ always be written in the form
$$ \frac{y_1}{y_2} = E \tan(\omega x + h) + k$$
for some constants $E, h$, and $k$?
Edit:
For example
$$\frac{12 \sin (x) + 3 \cos(x)}{ 4 \sin(x) + 2 \cos(x)}$$
has a graph like this:
Which looks like the tangent function transformed by scaling and shifting.
|
The answer is yes.
Claim 1 : If $D=0$, then
$$\dfrac{y_1}{y_2}=-\frac BC \tan\bigg(\omega x -\frac{\pi}{2}\bigg) + \frac AC$$
Claim 2 : If $D\not=0$, then $$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$
Claim 1 : If $D=0$, then
$$\dfrac{y_1}{y_2}=-\frac BC \tan\bigg(\omega x - \frac{\pi}{2}\bigg) + \frac AC$$
Proof :
One has
$$\frac{y_1}{y_2}=\frac BC\cdot\frac{\cos(\omega x)}{\sin(\omega x)}+\frac AC=-\frac BC \tan\bigg(\omega x - \frac{\pi}{2}\bigg) + \frac AC\ .\quad\blacksquare$$
Claim 2 : If $D\not=0$, then $$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$
Proof :
If $A=\dfrac{BC}{D}$, then
$$\frac{y_1}{y_2}=\frac{B(C\sin(\omega x)+D\cos(\omega x))}{D(C\sin(\omega x)+D\cos(\omega x))}=\frac BD$$
In the following, $A\not=\dfrac{BC}{D}$.
For $x$ such that $\cos(\omega x)\not=0$, one has
$$\frac{y_1}{y_2}=\frac{ A\dfrac{\sin(\omega x)}{\cos(\omega x)} + B}{C\dfrac{\sin(\omega x)}{\cos(\omega x)} + D}=\frac{A\tan(\omega x)+B}{C\tan(\omega x)+D}=\frac{\dfrac AD\tan(\omega x)+\dfrac BD}{\dfrac CD\tan(\omega x)+1}$$
Now,
$$\begin{align}G\bigg(\frac{y_1}{y_2}+F\bigg)&=\frac{G\dfrac AD\tan(\omega x)+G\dfrac BD}{\dfrac CD\tan(\omega x)+1}+GF
\\\\&=\frac{G\dfrac AD\tan(\omega x)+G\dfrac BD+GF\bigg(\dfrac CD\tan(\omega x)+1\bigg)}{\dfrac CD\tan(\omega x)+1}
\\\\&=\frac{G\bigg(\dfrac AD+\dfrac{CF}{D}\bigg)\tan(\omega x)-G\bigg(-\dfrac BD-F\bigg)}{1+\dfrac CD\tan(\omega x)}\end{align}$$
Solving the system
$$\begin{cases}G\bigg(\dfrac AD+\dfrac{CF}{D}\bigg)=1
\\G\bigg(-\dfrac BD-F\bigg)=\dfrac CD\end{cases}$$
for $F$ and $G$, one has
$$G=\frac{C^2+D^2}{DA-BC},\qquad F=\frac{-CA-BD}{C^2+D^2}$$
So, taking $(F,G)=\bigg(\dfrac{-CA-BD}{C^2+D^2},\dfrac{C^2+D^2}{DA-BC}\bigg)$, one has
$$G\bigg(\frac{y_1}{y_2}+F\bigg)=\frac{\tan(\omega x)-\dfrac CD}{1+\dfrac CD\tan(\omega x)}=\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)$$
so one gets
$$\frac {y_1}{y_2}=\frac 1G\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)-F,$$
i.e.
$$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$
which holds even for $x$ such that $\cos(\omega x)=0$ since for $x=\dfrac{1}{\omega}\bigg(\dfrac{\pi}{2}+n\pi\bigg)$ where $n$ is an integer, one has
$$\begin{align}\text{RHS}&=\frac{DA-BC}{C^2+D^2}\tan\bigg(\dfrac{\pi}{2}+n\pi-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}
\\\\&=\frac{DA-BC}{C^2+D^2}\cdot\frac{\cos\bigg(\arctan\bigg(\dfrac CD\bigg)\bigg)}{\sin\bigg(\arctan\bigg(\dfrac CD\bigg)\bigg)}+\frac{CA+BD}{C^2+D^2}
\\\\&=\frac{DA-BC}{C^2+D^2}\cdot\frac{1}{\dfrac CD}+\frac{CA+BD}{C^2+D^2}
\\\\&=\dfrac AC
\\\\&=\dfrac{y_1}{y_2}\end{align}$$
Therefore, one finally can say that if $D\not=0$, then
$$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}\ .\quad\blacksquare$$
Using this, your example can be written as
$$\frac{12 \sin (x) + 3 \cos(x)}{ 4 \sin(x) + 2 \cos(x)}=\frac{3}{5}\tan\bigg(x-\arctan(2)\bigg)+\frac{27}{10}$$
|
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|
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x+2}=0\\(x+1)(x+2)-2x\sqrt{3x+2}=0$$ What next? Thank you!
|
Multiply by $\sqrt{3x+2}$, assuming $3x+2\ge 0$ i.e. $x\ge-2/3$:
$$\left(\sqrt{3x+2}\right)^2-2x\sqrt{3x+2}+x^2=0$$
$$\left(\sqrt{3x+2}-x\right)^2=0$$
$$\sqrt{3x+2}=x$$
$$3x+2=x^2$$
Solve the quadratic equation to find $x=\frac{3\pm\sqrt{17}}{2}$. As the final equation is a consequence of the original one, but not necessarily equivalent to it, you now need to try those two solutions in the original equation. The only real solution will turn out to be $x=\frac{3+\sqrt{17}}{2}$ : the other one will be spurious and needs to be rejected.
|
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|
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$
Here's what I've done so far:
$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
Not sure where to go from here, any help's appreciated. I think the $AM-GM$ inequality should be used here in some way.
|
The key is to write:
$$\frac ab+\frac ba+\frac bc+\frac cb+\frac ca+\frac ac=\frac{b+c}a+\frac{c+a}b+\frac{a+b}c=\frac{1-a}a+\frac{1-b}b+\frac{1-c}c$$
and
$$\frac{1-a}a+2\ge2\sqrt2\sqrt{\frac{1-a}a}$$
by AM-GM.
Summing similar inequalities yields the result.
|
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|
For positive real numbers $a,b,c$ prove that $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$ For positive real numbers $a,b,c$ prove that $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
I think the $A.M-G.M$ inequality needs to be used here. Pretty sure we have to use the fact that
$\frac{a^6+b^6+c^6}{3}\ge a^2b^2c^2$
and that
$\frac{x^2+y^2}{2}\ge xy$
But I'm not sure in what way to use it exactly, I can't seem to get the sum of two squares in any way, so any help is appreciated.
|
$2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
$12+2a^6+2b^6+2c^6\ge 6ab+6ac+6bc$
We can write this as
$a^6+b^6+1+1+1+1+a^6+c^6+1+1+1+1+b^6+c^6+1+1+1+1\ge 6ab+6ac+6bc$
We also know that, from the $A.M-G.M$ inequality
$a^6+b^6+1+1+1+1\ge6\sqrt[6]{a^6b^6}=6ab$
$a^6+c^6+1+1+1+1\ge6\sqrt[6]{a^6c^6}=6ac$
$b^6+c^6+1+1+1+1\ge6\sqrt[6]{b^6c^6}=6bc$
Summing the three inequalities above we get
$a^6+b^6+1+1+1+1+a^6+c^6+1+1+1+1+b^6+c^6+1+1+1+1\ge 6ab+6ac+6bc$
$=12+2a^6+2b^6+2c^6\ge 6ab+6ac+6bc$
$=2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
So we've proved the original inequality.
|
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|
Let $A,B$ be $2\times2$ matrices. Given that we know $\text{Tr}(A)$, $\text{Tr}(B),\text{Tr}(AB)$, how do I find $A$ and $B$ that have those traces? Let $A,B$ be $2\times 2$ matrices. Given that we know $\operatorname{Tr}(A)$, $\operatorname{Tr}(B),\operatorname{Tr}(AB)$ how do I find $A$ and $B$ that have those traces?
Naively, I would let $A$ be the diagonal matrix with entries $\operatorname{Tr}(A)/2$ and the diagonal entries of $B$ be $\operatorname{Tr}(B)/2$ but this just doesn't work.
(I know those traces do not uniquely determine $A,B$ but I just need one instance of $A,B$)
|
Edit: I've revamped the answer to include the determinate of $1$ condition.
For an elementary approach: Let $Tr(A)=x$, $Tr(B)=y$, and $Tr(AB)=z$. For some $a,b$, we have
\begin{align*}
A&=\begin{pmatrix}a&a_1\\a_2&x-a\end{pmatrix}&
B&=\begin{pmatrix}b&b_1\\b_2&y-b\end{pmatrix}
\end{align*}
This gives us
$$AB=\begin{pmatrix}ab+a_1b_2&\square\\\square&a_2b_1+(x-a)(y-b)\end{pmatrix}$$
Therefore, $$z=ab+a_1b_2+a_2b_1+(x-a)(y-b)\implies a_1b_2+a_2b_1=z-ab-(x-a)(y-b).$$
Now, if we want the added condition that $|A|=1$ and $|B|=1$, then
\begin{align*}
a(x-a)-a_1a_2=1&\implies a^2-ax+a_1a_2+1=0\\
b(y-b)-b_1b_2=1&\implies b^2-by+b_1b_2+1=0
\end{align*}
Hence,
$$a=\frac{x\pm\sqrt{x^2-4a_1a_2-4}}{2}\quad\text{ and }\quad b=\frac{y\pm\sqrt{y^2-4b_1b_2-4}}{2}.$$
As an example, say $Tr(A)=7$, $Tr(B)=-2$, and $Tr(AB)=4$. Then
\begin{align*}
a_1b_2+a_2b_1&=4-ab-(7-a)(-2-b)\\
&=18-2ab+7b-2a\\
\end{align*}
We have the conditions that
$$a=\frac{7\pm\sqrt{45-4a_1a_2}}{2}\quad\text{ and }\quad b=-1\pm\sqrt{-b_1b_2}.$$
We will just take the "$+$" solution. Therefore, we can substitute these values and simplify (quite a bit) to get
\begin{align*}
a_1b_2+a_2b_1=11-\sqrt{4a_1a_2b_1b_2-45b_1b_2}
\end{align*}
At this point, we have many choices, so let $a_2=0$, $b_1=0$, $b_2=-1$. Then we have
\begin{align*}
a_1=-11
\end{align*}
We can now solve for $a$ and $b$:
$$a=\frac{7+3\sqrt{5}}{2}\quad\text{ and }\quad b=-1.$$
Therefore, we have the matrices
\begin{align*}
A&=\begin{pmatrix}\frac{7+3\sqrt{5}}{2}&-11\\0&7-\frac{7+3\sqrt{5}}{2}\end{pmatrix}&
B&=\begin{pmatrix}-1&0\\-1&-1\end{pmatrix}
\end{align*}
Indeed, both of the matrices have a determinate of $1$. Moreover,
$$AB=\begin{pmatrix}-\frac{7+3\sqrt{5}}{2}+11&\square\\\square&-7+\frac{7+3\sqrt{5}}{2}\end{pmatrix},$$
giving us a trace of $4$.
|
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|
Find the minimum value of $2a+ (1/a) + (1/2b) + b$, where a, b > 0 Find the minimum value of $2a+ (1/a) + (1/2b) + b$, where a, b > 0
My approach:-
Since a and b are positive numbers, I applied AM-GM inequality
$(2a+(1/a)+(1/2b)+b) /4$ ≥ $(2a\cdot (1/a)\cdot (1/2b)\cdot b) ^ {1/4} $
Giving me the answer as $4$
but the correct answer given in my textbook is $3\cdot \sqrt(2)$
I guess I am applying the inequality correctly, then what is the issue ?
|
AM-GM inequality applied to $(2a,1/a,1/2b,b)$ gives
$$2a+\frac{1}{a}+\frac{1}{2b}+b \ge 4 \tag{1}$$
where equality is to be achieved for $2a=1/a=1/2b=b$. Solving this gives $a=1/\sqrt{2}$ but does not give a unique/consistent value for $b$.
Conclusion is, AM-GM cannot be applied to all four quantities at once. However, since $a,b$ are independent positive quantities, one can safely apply AM-GM separately to $(2a,1/a)$ and $(1/2b,b)$.
$$2a+\frac{1}{a}\ge 2\sqrt{2}$$
where equality is achieved for $2a=1/a \Rightarrow a=1/\sqrt{2}$. Similarly
$$\frac{1}{2b}+b\ge \sqrt{2}$$
where equality is achieved for $1/2b=b \Rightarrow b=1/\sqrt{2}$.
Hence $2a+1/a + 1/2b+b$ is minimized for $a=1/\sqrt{2}$ and $b=1/\sqrt{2}$. At these values, the expression takes the value of
$$2\sqrt{2}+\sqrt{2}=3\sqrt{2}$$
Indeed $3\sqrt{2}$ is larger than $4$ because our first inequation, $(1)$, still holds true.
|
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|
Conver the Equation $r=a \sin x+b\cos x$ into Cartesian Form Hello :)) Wanted to convert the equation $r=a \sin x+b\cos x$, where $a, b \in \mathbb{R}$. Knowing that $x = r \cos x$ and $y= = r \sin x$, I can make $x = (a \sin x + b \cos x) \cos x$. But this is kinda a dead end because I can't simplify this further. Also, knowing that $r = \sqrt{x^2 + y ^2}$, I can make $\sqrt{x^2 + y^2}= a \sin x + b \cos x$. This also seems like a dead end...so, any suggestions?
|
If you need to solve fpr $x$ you can easily divide both side by $\sqrt{a^2+b^2}$. Then, we have:
$$\frac{r}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)$$
Let:
$$\theta=\cos^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)$$
We have:
$$\sin(x+\theta)=\frac{r}{\sqrt{a^2+b^2}}$$
Can you finish it now?
|
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|
Coefficient extraction I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\alpha + k}}
\end{equation}
So far I got
\begin{equation*}
\frac{1}{(1 - z)^{\alpha + 1}} = (1-z)^{-(\alpha + 1)} = \sum_{n \geq 0}{\binom{-\alpha - 1}{k}(-1)^k z^k} = \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n}
\end{equation*}
where I used $\binom{-\alpha}{k} = (-1)^k \binom{\alpha + k - 1}{k}$and
\begin{equation*}
\log \frac{1}{1 - z} = - \log 1- z = \sum_{n \geq 1}\frac{z^n}{n} = z\sum_{n \geq 0}\frac{z^{n}}{n+1}
\end{equation*}
therefore
\begin{align*}
\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} &= z\left( \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n} \right) \left( \sum_{n \geq 0}\frac{z^{n}}{n+1} \right) \\
&= z \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n} \\
&= \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n + 1}.
\end{align*}
Now the $n$-th coefficient is
\begin{align*}
\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\frac{n!}{(n-k)! k!} \frac{1}{\binom{\alpha + n}{n - k}} \frac{1}{n - k + 1}} + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k} \frac{1}{\binom{\alpha + n}{n - k}} \frac{1}{n - k + 1}} + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k - 1} \frac{1}{\binom{\alpha + n}{n - k - 1}} \frac{k-1}{n - k -1} \frac{1}{\alpha + k + 1} } + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^{n+1}_{k = 2}{\binom{n}{k - 2} \frac{1}{\binom{\alpha + n}{n - k}} \frac{k- 2 }{n - k} \frac{1}{\alpha + k} } + \frac{1}{n+1}
\end{align*}
but from now on I can't see how to proceed.
I also know that
\begin{equation}
[z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n
\end{equation}
somehow I also tried to use some sort of transformation law for
coefficient extraction, but I am not aware of any kind of transformation law for coefficient extraction. Using $1-u = (1-z)^{\alpha + 1}$ or $z = 1 - (1-u)^{1/(\alpha + 1)}$gives
\begin{equation}
\frac{1}{\alpha + 1}\frac{1}{1-u} \log \frac{1}{1 - u}
\end{equation}
and for this I get the coefficient $\frac{1}{\alpha + 1} H_n$.
Another approach I did was the following:
Since
\begin{equation*}
(1-z)^{-m} = \exp(-m \ln(1-z))
\end{equation*}
I get
\begin{equation*}
\frac{\partial }{\partial m} \exp(-m \ln(1-z)) = -\ln(1-z) \exp(- m \ln(1-z)) = - \ln(1-z) \frac{1}{(1-z)^m} = \frac{1}{(1-z)^m} \ln \frac{1}{1-z}
\end{equation*}
therefore
\begin{align*}
\frac{\partial }{\partial m} (1-z)^{-m} &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \binom{m + n - 1}{n} z^n} \\
&= \sum_{n \geq 0}{\frac{\partial }{\partial m} \frac{\Gamma(m + n)}{n!\Gamma(m)} z^n}
\end{align*}
But I don't know any identities for the derivative of the gamma function.
|
We will need an auxiliary identity before we can move on to the subject
by OP, which is
$$[z^n] \frac{1}{(1-z)^{\alpha+1}} \log\frac{1}{1-z}
= {n+\alpha\choose n} (H_{n+\alpha} - H_\alpha).$$
with $\alpha$ a non-negative integer.
A binomial identity
Introduce with $q\ge 1$
$$f(z) = n! (-1)^n \frac{1}{z+q} \prod_{p=0}^n \frac{1}{z-p}.$$
This has the property that for $0\le r\le n$
$$\mathrm{Res}_{z=r} f(z)
= n! (-1)^n \frac{1}{r+q}
\prod_{p=0}^{r-1} \frac{1}{r-p}
\prod_{p=r+1}^n \frac{1}{r-p}
\\ = n! (-1)^n \frac{1}{r+q}
\frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!}
= {n\choose r} \frac{(-1)^r}{r+q}.$$
With the residue at infinity being zero by inspection we obtain
$$\sum_{r=0}^n {n\choose r} \frac{(-1)^r}{r+q}
= - \mathrm{Res}_{z=-q} f(z)
\\ = - n! (-1)^n \prod_{p=0}^n \frac{1}{-q-p}
= n! \prod_{p=0}^n \frac{1}{q+p}
= n! \frac{(q-1)!}{(q+n)!}
= \frac{1}{q} {n+q\choose q}^{-1}.$$
Therefore with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= \sum_{r=0}^{n-k} {n-k\choose r} \frac{(-1)^r}{r+k}
= (-1)^{n-k} \sum_{r=0}^{n-k} {n-k\choose r} \frac{(-1)^r}{n-r}
\\ = [z^n] \log\frac{1}{1-z}
(-1)^{n-k} \sum_{r=0}^{n-k} {n-k\choose r} (-1)^r z^r
\\ = [z^n] \log\frac{1}{1-z}
(-1)^{n-k} (1-z)^{n-k}.$$
Main identity
We get for the LHS from first principles that it is (apply identity
setting $n$ to $n+\alpha$ and $k$ to $q$)
$$\sum_{q=1}^n {n-q+\alpha\choose n-q} \frac{1}{q}
\\ = [z^{n+\alpha}] \log\frac{1}{1-z}
\sum_{q=1}^n {n+\alpha\choose q} {n-q+\alpha\choose \alpha}
(-1)^{n+\alpha-q} (1-z)^{n+\alpha-q}.$$
Note that for $q=0$ we get
$${n+\alpha\choose \alpha}
[z^{n+\alpha}] \log\frac{1}{1-z}
(-1)^{n+\alpha} (1-z)^{n+\alpha}.$$
This will be our first piece. We include it in our sum at this time. Next
observe that
$${n+\alpha\choose q} {n-q+\alpha\choose \alpha}
= \frac{(n+\alpha)!}{q! \times \alpha! \times (n-q)!}
= {n+\alpha\choose \alpha} {n\choose q}.$$
We have for the augmented sum without the binomial scalar in front
$$[z^{n+\alpha}] \log\frac{1}{1-z}
\sum_{q=0}^n {n\choose q} (z-1)^{n+\alpha-q}
\\ = [z^{n+\alpha}] \log\frac{1}{1-z}
(z-1)^{n+\alpha} \left[1+\frac{1}{z-1}\right]^n
\\ = [z^{n+\alpha}] \log\frac{1}{1-z}
(z-1)^\alpha z^n
= [z^\alpha] \log\frac{1}{1-z} (z-1)^\alpha.$$
This is the second piece. Now to evaluate these two pieces
we evidently require
$$\;\underset{z}{\mathrm{res}}\;
\frac{1}{z^{m+1}} \log\frac{1}{1-z} (-1)^m (1-z)^m.$$
We put $z/(1-z) = w$ so that $z=w/(1+w)$ and $dz = 1/(1+w)^2 \; dw$ to
obtain
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{m+1}} (1+w)
\log\frac{1}{1-w/(1+w)} (-1)^m \frac{1}{(1+w)^2}
\\ = (-1)^m \;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{m+1}} \frac{1}{1+w} \log(1+w)
= (-1)^m [w^m] \frac{1}{1+w} \log(1+w)
\\ = [w^m] \frac{1}{1-w} \log(1-w)
= - [w^m] \frac{1}{1-w} \log\frac{1}{1-w}
= - H_m.$$
Hence our first piece is $- {n+\alpha\choose\alpha} H_{n+\alpha}$ while
the second is $- {n+\alpha\choose\alpha} H_{\alpha}$. Subtract the
first from the second to obtain our claim,
$${n+\alpha\choose n} (H_{n+\alpha} - H_\alpha).$$
|
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How do we calculate the expected value of $Y$? Let $\Omega = \{−1, 0, 1, 3\}$ with the probability function $p$ given by
I want to calculate the expected value and the variance of $X(\omega)=\omega$, $Y(\omega)=5\omega-3$, $Z(\omega)=(\omega-1)^2$.
For $X$ I have done the following :
We have that $X\in \{-1,0,1,3\}$.
The expected value is \begin{equation*}E[X]=\sum_{x\in X(\Omega)}xP[X=x]=(-1)\cdot \frac{1}{10}+0\cdot \frac{3}{10}+1\cdot \frac{2}{10}+3\cdot \frac{4}{10}= \frac{13}{10}\end{equation*}
The variance is \begin{align*}Var[X]&=\sum_{x\in X(\Omega)}(x-E[X])P[X=x]\\ & =\left (-1-\frac{13}{10}\right )\cdot \frac{1}{10}+\left (0-\frac{13}{10}\right )\cdot \frac{3}{10}+\left (1-\frac{13}{10}\right )\cdot \frac{2}{10}+\left (3-\frac{13}{10}\right )\cdot \frac{4}{10}\\ & =0\end{align*}
Doe $Y$ we have $Y=5\omega-3\in \{-8,-3,2,12\}$, or not? But then we don't knowthe probabilities.
Or do weuse the property $E[aX+b]=aE[X]+b$ ?
|
Your calculation of $E[X]$ is correct but the calculation of $\text{Var}[X]$ is not correct.
$ \text {Var}[X] = \sum(x - E[X])^2 \cdot P(X = x)$ which translates to,
$ \text {Var}[X] = E[X^2] - (E[X])^2 $
Now, $E[X^2] = (-1)^2 \cdot \dfrac{1}{10} + 1^2 \cdot \dfrac{2}{10} + 3^2 \cdot \dfrac{4}{10} = \dfrac{39}{10}$
So, $ \displaystyle \text {Var}[X] = \frac{39}{10} - \left(\frac{13}{10}\right)^2 = \frac{221}{100}$
Yes you can use the formula $~E[aX+b] = aE[X] + b$
Also note that $~ \text{Var} [aX+b] = a^2 \text{Var}[X]$
|
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|
Finding the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$ I need to find the the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$. My issue is that just looking at the graph of $\arg{(z-2i)}=\frac{\pi}{6}$ (which is a ray from $(0,2)$ on the Argand diagram) and $|z-3+i|$ (a circle with centre at $(3,-1)$) wouldn't the lowest value just be the distance between $(0,2)$ and $(3,-1)$? However the question is worth $4$ marks and this isn't $4$ marks of working so I feel like I'm seriously overlooking something. Could someone clarify what I've missed or is the question that simple?
|
Alternative approach:
$\displaystyle \tan(\pi/6) = \frac{1}{\sqrt{3}}$.
Therefore, with $z = x + iy$, two constraints must be satisfied:
*
*$\displaystyle \frac{y-2}{x} = \frac{1}{\sqrt{3}}.$
*Both $(x)$ and $(y-2)$ must be positive (i.e. in the 1st quadrant), rather than both being negative.
The above constraints imply that $0 < x = (y-2)\sqrt{3}$.
You can minimize the absolute value of a complex number by minimizing the square of the absolute value.
Therefore, you want to minimize
$|z - 3 + i|^2 = (x - 3)^2 + (y + 1)^2$
$ = D(y) = [(y - 2)\sqrt{3} - 3]^2 + (y + 1)^2.$
Examining derivatives:
$D'(y) = 2[(y - 2)\sqrt{3} - 3]\sqrt{3} + 2(y + 1)$.
This simplifies to $(y)[8] + [(-12) + (-6\sqrt{3}) + (2)]
= 8y - 10 - 6\sqrt{3}$.
So, $~\displaystyle D'(y) = 0 \iff y = \frac{5 + 3\sqrt{3}}{4} \implies (y - 2) > 0.$
Further, $D''(y) = 8 > 0.$
Therefore, $D$ is minimized at
$\displaystyle y = \frac{5 + 3\sqrt{3}}{4}.$
At that value for $y$, you have that
$\displaystyle D(y) = \left[\frac{3\sqrt{3} - 3}{4}\sqrt{3} - 3\right]^2 + \left[\frac{9 + 3\sqrt{3}}{4}\right]^2$
$\displaystyle = \left[\frac{- 3 - 3\sqrt{3}}{4}\right]^2 + \left[\frac{9 + 3\sqrt{3}}{4}\right]^2$
$\displaystyle = \frac{1}{16} ~~\times
~~\left\{ ~\left[9 + 27 + 18\sqrt{3}\right]
~+ \left[81 + 27 + 54\sqrt{3}\right] ~\right\}$
$\displaystyle = \frac{1}{16} ~~\times
~~\left\{ ~\left[144 + 72\sqrt{3}\right]
~\right\} = \frac{1}{4} ~~\times ~~\left[36 + 18\sqrt{3}\right].$
In order to compute $\sqrt{D(y)}$, you must find $a,b \in \Bbb{R}$ such that
*
*$\left(a + b\sqrt{3}\right) > 0$
*$a^2 + 3b^2 = 36$
*$(2ab) = 18$.
Note that the $3$rd constraint above implies that $a$ and $b$ are both positive or both negative. Then, the first constraint above implies that $a,b$ are both positive.
You can guess that $a = b = 3$, or (more formally)
Setting $\displaystyle a = \frac{9}{b}$ leads to
$$\frac{81}{b^2} + 3b^2 = 36 \implies
3b^4 - 36b^2 + 81 = 0.\tag1 $$
Regarding (1) above as a quadratic in $b^2$ gives:
$\displaystyle b^2 = \frac{1}{6} \left[36 \pm \sqrt{1296 - (972)}\right] = \frac{1}{6} \left[36 \pm 18\right].$
The convenient try is $\displaystyle b^2 = \frac{36 + 18}{6} = 9 \implies b = 3.$
Therefore, the minimum value for $\sqrt{D(y)}$ is
$$\sqrt{\frac{1}{4} \times \left[36 + 18\sqrt{3}\right]}
~= ~\frac{1}{2} \left[3 + 3\sqrt{3}\right].$$
|
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|
Find the roots of the polynomial $x^4+2x^3-x-1=0$. Problem
Find all $4$ roots of the polynomial: $$f(x)=x^4+2x^3-x-1.$$
My Attempt
Observe,
$$f(-2)=1,\; f(-1)=-1,\; f(0)=-1,\; f(1)=1.$$
Therefore, $f(x)$ has a root between $-2$ and $-1$, another root between $0$ and $1$.
Since $f(x)$ does not contain a second degree term, a clever substitution might change $f(x)$ into a polynomial that is easier to deal with. For example $(x\to x-1)$,
$$f(x-1)=x^4-2x^3+x-1=f(-x)$$
Also, I am avoiding computational or graphical assistance.
|
With the substitution $x=t-b/na$, you eliminate the second highest order term of a polynomial $ax^n + bx^{n-1} + \cdots .$
If you substitute $x=t-2/4 = t-1/2$ your polynomial becomes
$$t^4 -\frac{3}{2}t^2 -\frac{11}{16}.$$
This is quadratic in $t^2$ so you can solve to get
$$t^2 = \frac{3\pm 2\sqrt{5}}{4}.$$
So the four roots of your polynomial are plus and minus the square roots of those two solutions (plus $1/2$ because of the original substitution.) We were lucky that the first order term also disappeared.
|
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$p$ prime and $m_p$ a proper divisor of $p−1$. Is it $\sigma(m_p)-\nu(m_p)(This is a sharper version of this other question of mine.)
Along the problem I'm facing, I've come to the following lemma (if it is true):
Let $p$ be a prime and $m_p$ a proper divisor of $p-1$. Then the difference between the sum of all the divisors of $m_p$, say $\sigma(m_p)$, and their number, say $\nu(m_p)$, is less than $p-2$.
Examples:
*
*$p=13$; then, for $m_{13}=6$: $1+2+3+6-4=8<11$, and likewise for $m_{13}=4,3,2,1$;
*$p=31$; then for $m_{31}=15$ we have: $1+3+5+15-4=20<29$, or for $m_{31}=10$: $1+2+5+10-4=14<29$, and likewise for $m_{31}=5,3,2,1$;
*$p=37$; then for $m_{37}=18$ we have: $1+2+3+6+9+18-6=33<35$, and likewise for $m_{37}=9,6,3,2,1$;
*$p=101$; then for $m_{101}=50$ we have: $1+2+5+10+25+50-6=87<99$, and likewise for $m_{101}=25,20,10,5,2,1$.
Is it true for every $p$ and $m_p$?
|
By computer search, $p = 61$ and $m_p = 30$ seems to be the smallest counterexample. We get $\sigma(m_p) = 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72$ and $\nu(m_p) = 8$, so $\sigma(m_p) - \nu(m_p) = 72 - 8 = 64$.
|
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Integrating $\int \frac{dx}{x^{11}\sqrt{1+x^4}}$
Find $$I=\int \frac{dx}{x^{11}\sqrt{1+x^4}}$$
With $x^2=t$, we get $I=\int _{ }^{ }\frac{d\ \sqrt{t}}{\left(\sqrt{t}\right)^{11}.\sqrt{1+t^2}}=\frac{1}{2}\int _{ }^{ }\frac{dt}{t^6\sqrt{1+t^2}}$
Now, with $t= \tan k$ we get $I=\frac{1}{2}\int _{ }^{ }\frac{\cos ^5k.\ dk}{\sin ^6k}$ .
At this step, I can not do anything more. Can you show me the way to solve this?
I try to go with this way but it doesn't work.
|
After substituting $t = \frac{1}{x^4}$:
$$I = -\frac{1}{4} \int \frac{x^5 \ dt}{x^{11} \sqrt{1+x^4}} = -\frac{1}{4} \int \frac{t^2 x^2 \ dt}{\sqrt{x^4(t+1)}}= -\frac{1}{4}\int \frac{t^2 \ dt}{\sqrt{1+t}}.$$
Now further substitute $u = 1 + t$ and expand $t^2 = (u - 1)^2$.
|
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multivariable calculus divergence theorem help I am stuck on a problem:
Use the Divergence Theorem to evaluate $\iint \mathbf{F} \cdot d\mathbf{S}$, where $$\mathbf{F}(x,y,z)=z^2x\mathbf{i}+(\frac{1}{3}y^3+\tan(z))\mathbf{j}+(x^2z+y^2)\mathbf{k}$$ and $S$ is the top half of the sphere $x^2+y^2+z^2=1$.
[Hint: Note that $S$ is not a closed surface. First compute integrals over $S_1$ and $S_2$, where $S_1$ is the disk $x^2+y^2\le 1$, oriented downward, and $S_2 = S \cup S_1$.]
my working process:
$\iint \mathbf{F} \cdot d\mathbf{S} = \iiint\limits_E div\mathbf{F} \cdot d\mathbf{V}
= \iiint\limits_E (x^2+y^2+z^2) d\mathbf{V}$.
for $S_2$, parametrise $x=r\sin(\phi)\cos(\theta)$, $y=r\sin(\phi)\sin(\theta)$, $z=r\cos(\theta)$ and $0\le r \le 1,\;\; 0\le \theta \le 2\pi,\;\; 0\le \phi \le \pi.$
=$\iiint\limits_{S_2} (r^2\cos^2(\theta)+r^2\sin^2(\phi)) dV+\iint\limits_{S_1} \mathbf{F} \cdot d\mathbf{S}$
=$\frac{14\pi}{15}$+$\iint\limits_{S_1} \mathbf{F} \cdot d\mathbf{S}$
(the correct answer is $\frac{13\pi}{20}$)
could you point out the mistake? Thank you very much!!!
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Applying divergence theorem, you get flux over closed surface $S_2$.
In spherical coordinates, $x^2 + y^2 + z^2 = \rho^2$
where $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi. z = \rho \cos\phi$
As it is part of the sphere above $z = 0, \phi \leq \pi/2$.
So the volume integral is,
$ \displaystyle \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^4 \sin\phi ~d\rho ~d\phi ~d\theta = \frac{2\pi}{5}$
Now the unit normal vector to the disk surface $S1: x^2 + y^2 \leq 1, z = 0 ~$ is $~\hat n = (0, 0, -1)$
$\vec F \cdot \hat n = - y^2$
Using polar coordinates, the surface integral over disk surface is,
$ \displaystyle \int_0^{2\pi} \int_0^1 - r^3\sin^2\theta ~ dr ~ d\theta = - \frac{\pi}{4}$
So flux over spherical surface $S$ is,
$ \displaystyle \frac{2\pi}{5} - \left(- \frac{\pi}{4}\right) = \frac{13\pi}{20}$
|
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Calculate the ratio based on the lengths of edges of triangle The problem
My problem is,
Let $\triangle ABC$ be an acute triangle, circumscribed in $(O)$ and has orthocenter $H$. Let $HO$ intersect $(O)$ at $E$ and $F$, as shown in the image. $AE$ cuts $BC$ at $K$ and $AF$ cuts $BC$ at $L$. $AO$ cuts $BC$ at $T$. Calculate the value of $\frac{TK}{TL}$ according to the side-lengths and angles of $\triangle ABC$
My approach is that I used the Anti-Steiner point and then use Menelaus Theorem. But that does not actually relates to $a,b,c$ (which are the side lengths of $BC$, $CA$, $AB$, respectively).
Any help is appreciated!
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triangle ABC with sides a. b, and c.
$\begin{array}{} A=(\frac{a^2+b^2-c^2}{2a},\frac{2S}{a}) & B=(a,0) & C=(0,0) & S=Δ(ABC) \end{array}$
$\begin{array}{} \text{Euler line (OH)} & l·x+m·y+n=0 & l=\frac{2a^4-(b^2-c^2)^2-a^2(b^2+c^2)}{2a} & m=\frac{2S(c^2-b^2)}{a} \end{array}$
$\begin{array}{} \left\{ D,E \right\}=OH∩Circle(O,r=R) & R=\frac{abc}{4S} & . \\ K=AD∩BC & L=AE∩BC & T=AO∩BC \\ \end{array}$
$\begin{array}{} m_{OH}=tan(ε)=\frac{-l}{m} & ε+θ=π & tan(θ)=\frac{l}{m} \end{array}$
$\begin{array}{} \text{triangle ODP} & OP=Rcos(θ) & DP=Rsin(θ) \\ x_{D}=x_{O}-Rcos(θ) & y_{D}=y_{O}+Rsin(θ) & ε+θ=π \\ cos(ϕ)=\frac{±1}{\sqrt{1+tan(ϕ)^2}} & cos(θ)=\frac{m}{ρ} & sin(θ)=\frac{l}{ρ} \\ ρ=\sqrt{l^2+m^2} & x_{O}=\frac{a}{2} & y_{O}=\frac{a(-a^2+b^2+c^2)}{8S} \\ x_{D}=\frac{a}{2}-\frac{Rm}{ρ} &y_{D}=Rcos(α)+\frac{Rl}{ρ} & D+E=2O \\ x_{E}=\frac{a}{2}+\frac{Rm}{ρ} & y_{E}=Rcos(α)-\frac{Rl}{ρ} & cos(α)=\frac{-a^2+b^2+c^2}{2bc}\end{array}$
$\begin{array}{} \text{points: K,L,T } & \left| \begin{array}{} x_{A} & y_{A} & 1 \\ x_{D} & y_{D} & 1 \\ x_{K} & 0 & 1 \\ \end{array} \right| =0 & \left| \begin{array}{} x_{A} & y_{A} & 1 \\ x_{E} & y_{E} & 1 \\ x_{L} & 0 & 1 \\ \end{array} \right|=0 & \left| \begin{array}{} x_{A} & y_{A} & 1 \\ x_{O} & y_{O} & 1 \\ x_{T} & 0 & 1 \\ \end{array} \right| =0 \end{array}$
replace everything: $x_{A}$,$y_{A}$,...$R$, $l$,$m$ and $S$. Solve $x_{K}$, $x_{L}$ and $x_{T}$. Simplify
$\begin{array}{} S=\sqrt{s(s-a)(s-b)(s-c)} & s=\dfrac{a+b+c}{2} \end{array}$
$x_{K}=\dfrac{b (a² - b² + c²) ρ - a c [(a² - c²)² + b² (a² + c²) - 2b⁴]}{a b c [(b² - c²)² + a² (b² + c²) - 2a⁴] - [(b² - c²)² - a² (b² + c²)] ρ} a b$
$x_{L}=\dfrac{b (a² - b² + c²) ρ + a c [(a² - c²)² + b² (a² + c²) - 2b⁴]}{a b c [(b² - c²)² + a² (b² + c²) - 2a⁴] + [(b² - c²)² - a² (b² + c²)] ρ} (-a b)$
$x_{T}=\dfrac{a b² (a² - b² + c²)}{a² (b² + c²) - (b² - c²)²}$
$\begin{array}{} r=\dfrac{TK}{TL} & r=\dfrac{x_{T}-x_{K}}{x_{T}-x_{L}} & r=\dfrac{1-p}{1+p} & \text{where:} \end{array}$
$p=\dfrac{[a² (b² + c²) - (b² - c²)²] ρ}{ a b c [(b² - c²)² + a² (b² + c²) - 2a⁴]}$
$ρ=\sqrt{(abc)^2-(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}$
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Find radius that minimizes the surface of the solid I have solved the following problem:
"A solid, with a volume of $8cm^3$, consists of a cylinder and two equilateral cones, external to the cylinder and each with a base in common with the cylinder itself.
Find the base radius so that the surface of the solid is minimal."
but the result doesn't agree with the result of the book, which is $\sqrt[3]{\frac{3+\sqrt{3}}{\pi}}$. I don't see where my mistake is, so I would appreciate some feedback on my solution, thanks.
My solution:
Since the cones are equilateral their lateral surface is $\pi r\cdot 2r=2\pi r^2$; the lateral surface of the cyilinder is $2\pi rh$, where $h$ is the height of the cylinder so the total lateral surface is $S=2\pi r^2+2\pi r h$.
From $V_{tot}=8$ we can find $h$ as a function of $r$ by noting that $V_{cone}=\frac{1}{3}\pi r^2\cdot \sqrt{3}r$, $V_{cyl}=\pi r^2h$ so $V_{tot}=2V_{cone}+V_{cyl}=2\left( \frac{1}{3}\pi r^2\cdot \sqrt{3}r\right)+\pi r^2h=8$ which implies that $h=\frac{8}{\pi r^2}-\frac{2}{\sqrt{3}}r$.
So, $S=2\pi r^2+2\pi r\left(\frac{8}{\pi r^2}-\frac{2}{\sqrt{3}}r \right)=2\pi r^2(1-\frac{2}{\sqrt{3}})+\frac{16}{r}$ hence $S'(r)=4\pi r(1-\frac{2}{\sqrt{3}})-\frac{16}{r^2}=0\Leftrightarrow\fbox{$r=\sqrt[3]{\frac{4\sqrt{3}}{\pi (\sqrt{3}-2)}}$}$.
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The surface area of one cone will be $2\pi r^2$ (apothem will be $2r$) so two cones will give you $4\pi r^2$. Thus, $S'(r)=8\pi r(1-\frac{1}{\sqrt{3}})-\frac{16}{r^2}=0\implies r=\sqrt[3]{\frac{2\sqrt{3}}{\pi (\sqrt{3}-1)}}=\sqrt[3]{\frac{3+\sqrt{3}}{\pi }}$
|
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About the inequality $\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}$ Hi It's a generalization found on Aops starting from this question Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$ (see comments) :
Let $x_i>0$, $1\leq i\leq n$,$n\geq 3$ such that $x_1=x_{n+1}$ then we have :
$$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}\tag{I}$$
At first glance it seems that Am-Gm is too weak so as in my answer I use Radon's inequality (see vivid edit) we need to show :
$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_{i+1}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq n\sqrt{2}$$
Then I used Minkowski's inequality with $p=\frac{2}{3}$ we need to show :
$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}+\left(\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq n\sqrt{2}$$
Edit :
We can apply Jensen's inequality and we need to show :
$$\frac{2^{-\frac{1}{2}}\left(\sum_{i=1}^{n}\left(\frac{1}{x_{i}}\right)^{\frac{2}{3}}+\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_{i}}+x_{i}\right)}}-n\sqrt{2}\geq 0$$
Question :
How to achieve my work (if true) or show $(I)$?
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Lemma 1: For $a,b>0$,
$$2\sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
Proof:
$$\left(\frac{1}{\sqrt{a}}-\sqrt{b}\right)^2 \geqslant 0$$
$$\frac{1}a + b \geqslant 2\frac{\sqrt{b}}{\sqrt{a}}$$
$$\left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)^2= \frac{1}a + b + 2\frac{\sqrt{b}}{\sqrt{a}} \leqslant 2\left(\frac{1}a + b\right) $$
and the result follows.
Lemma 2: For $a>0$,
$$\sqrt{\frac{1}{a}+a} \leqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{a}-1\right)$$
Proof:
$$(\sqrt{a}-1)^4\geqslant 0$$
$$a^2-4\sqrt{a}a+6a-4\sqrt{a}+1\geqslant 0$$
$$2(a^2-2\sqrt{a}a+3a-2\sqrt{a}+1)\geqslant a^2+1$$
$$2(a-\sqrt{a}+1)^2\geqslant a^2+1$$
$$2\left(\sqrt{a}-1+\frac{1}{\sqrt{a}}\right)^2\geqslant a+\frac{1}{a}$$
and the result follows.
Lemma 3:
For $a,b>0$,
$$\frac{{\frac{1}{a}+b}}{\sqrt{\frac{1}{a}+a}} \geqslant \sqrt2 (\sqrt{b}-\sqrt{a}+1)$$
Proof:
By AM-GM and then Lemma 1:
$$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\sqrt{\frac{1}{a}+a} \geqslant 2 \sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
Combining with Lemma 2:
$$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+ \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{a}-1\right) \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
and the result follows.
Theorem: For $x_i>0$,
$$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}$$
Proof:
Let $a=x_i$ and $b=x_{i+1}$ in Lemma 3 and sum.
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|
Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a multiple of 11?
Dont know where to progress after this. All help appreciated.
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Hint: First, dividing your second line by $2$ gives
$$xy + 11x + 11y = 0 \tag{1}\label{eq1A}$$
Next, using Simon's favorite factoring trick gives
$$(x + 11)(y + 11) = xy + 11x + 11y + 121 \tag{2}\label{eq2A}$$
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|
$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.
I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second function, but it did not work.
Need some help to compute it.
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The simple Mathematica result given by Varun Vejalla in the comments is indeed correct. In order to make the final result look a bit nicer, we can define the function $f \colon (-1,1) \to \mathbb{R}$,
$$ f(a) = \int \limits_{-1}^1 \frac{\operatorname{artanh}(x)}{1-ax} \, \mathrm{d} x \, . $$
Since $\operatorname{artanh}(x) = \frac{1}{2} \log\left(\frac{1+x}{1-x}\right)$, your integral is $2f(a)$.
We first let $x = \frac{1-t}{1+t} ~ \Leftrightarrow ~ t = \frac{1-x}{1+x}$ to obtain
$$ f(a) = \frac{1}{1+a} \int \limits_0^\infty \frac{-\log(t)}{(1+t) \left(\frac{1-a}{1+a} +t\right)} \, \mathrm{d} t \, .$$
Under $t \mapsto \frac{1-a}{1+a} t^{-1}$ this expression transforms into
$$ f(a) = \frac{1}{1+a} \int \limits_0^\infty \frac{\log(t) + 2 \operatorname{artanh}(a)}{(1+t) \left(\frac{1-a}{1+a} +t\right)} \, \mathrm{d} t \, .$$
Averaging these two equations we find (the final integral is easy to compute using partial fractions)
$$ f(a) = \frac{\operatorname{artanh}(a)}{1+a} \int \limits_0^\infty \frac{\mathrm{d} t}{(1+t) \left(\frac{1-a}{1+a} +t\right)} = \frac{\operatorname{artanh}(a)}{1+a} \frac{\log\left(\frac{1-a}{1+a}\right)}{\frac{1-a}{1+a} - 1} = \frac{\operatorname{artanh}^2(a)}{a} $$
for $a \in (-1,1)$ (with $f(0) = 0$).
In particular, $2 f\!\left(\frac{1}{2}\right) = \log^2(3)$, which agrees with Eevee Trainer's result after some dilogarithm identities are used.
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|
Extension of Euclid's formula for Pythagorean triples to higher degrees I recently discovered that you can get all the primitive pythagorean triples with this neat formula known as Euclid's formula
$$a = m^2 - n^2, b = 2mn, c = m^2 + n^2$$
Where, m and n are co-prime and each having a different parity, that is one is odd and the other is even.
PrimitivePythagoreanTriples
I am curious to know if it is possible to find such a formula for cubes that is an expression with 2 numbers or multiple numbers $(x_1, x_2, x_3, ...)$ such that setting $ a, b, c, d $ respectively with the expressions satisfies $a^3 + b^3 + c^3 = d^3$
if it does exist can such expressions exist for any integral power n,
such that $ a_1^n + a_2^n + a_3^n + ... + a_n^n = b^n$.
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The fourth power case is the Jacobi Madden equation: $a^4+b^4+c^4+d^4=(a+b+c+d)^4$ Jacobi Madden equation.
It gives infinitely many ineteger solutions for $a^4+b^4+c^4+d^4=e^4$ using elliptic curve.
Unfortunately, the parametric solution is not known.
Some small solutions are
$955^4 + 1770^4 + (-2634)^4 + 5400^4 = 5491^4$
$48150^4 + (-31764)^4 + 27385^4 + 7590^4 = 51361^4$
$561760^4 + 1493309^4 + 3597130^4 + (-1953890)^4 = 3698309^4$
For $n=5,$ see Diophantine equation 5th power.
Sastry found $(75v^5-u^5)^5+(u^5+25v^5)^5+(u^5-25v^5)^5+(10u^3v^2)^5+(50uv^4)^5 =(u^5+75v^5)^5$
Other parametric solutions are here.
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|
Integrate $\sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1)$ I'm trying to learn calculus through self study and I happened upon the following exercise:
$$ \int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx $$
Seeing the sgn I thought: well this is easy and concluded that since $ \int_{0}^{2} \operatorname{sgn}(x-1) \,dx = 0$, the answer should be zero. But of course then I remembered that $ \int f(x)g(x) \,dx \neq \int f(x) \,dx \int g(x) \,dx$ and got lost.
I know how to solve $ \int_{0}^{2} \sqrt{4 - x^2}dx$, since that geometrically corresponds to a quarter circle of radius $2$ it's just $\pi$, but how do I approach the product of these things?
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To expand on @José Carlos Santos' answer, you can compute $\int \sqrt{4-x^2}\; dx$ by trigonometric substitution: let $x=2\sin(t)$ with $-\frac{\pi}{2}\le t\le \frac{\pi}{2}$ . Then $\sqrt{4-x^2}=2\cos(t)$ and $dx=2\cos(t)\; dt$ so
$$\int \sqrt{4-x^2}\; dx = \int 4\cos^2(t)\; dt = 2 \int 1+\cos(2t)\; dt
=
2\left(t + \frac{\sin(2t)}{2} \right)+ C$$
Substituting back, we get
$$\int \sqrt{4-x^2}\; dx = 2 \arcsin\left(\frac x 2\right) + \frac{x\sqrt{4-x^2}}{2} + C$$
Therefore,
$$\int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx
=
-\int_0^1\sqrt{4-x^2}\,\mathrm dx+\int_1^2\sqrt{4-x^2}\,\mathrm dx
=
-\left(2\arcsin(1/2)+\frac{\sqrt{3}}{2}
\right)
+
\left(2\arcsin(1)-2\arcsin(1/2)-\frac{\sqrt{3}}{2}
\right)
=\frac{\pi}{3}-\sqrt{3}$$
|
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|
Comparing $\frac {9}{\sqrt{11} - \sqrt{2}}$ and $\frac {6}{3 - \sqrt{3}}$ (without calculator) We want to compare the following two numbers:
$$x = \frac {9}{\sqrt{11} - \sqrt{2}} \quad\text{and}\quad y = \frac {6}{3 - \sqrt{3}}$$
My attempts so far:
I multiply both numerator and denominator of $x$ by $\sqrt{11} + \sqrt{2}$ so I get:
$$x = \frac {9(\sqrt{11} + \sqrt{2})}{(\sqrt{11} - \sqrt{2})\cdot(\sqrt{11} + \sqrt{2})}$$ so
$$x = \frac {9(\sqrt{11} + \sqrt{2})}{(11 - 2)} = \sqrt{11} + \sqrt{2}$$
Similarly, $y = 3 + \sqrt{3}$.
But how do I take it from this point forward?
Of course $y>x$ but I must prove it.
I also tried to compare $\sqrt 11$ with $\sqrt 12$ which equals $2 \sqrt3$ but again I am not getting anywhere.
Thank you.
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Render
$[(\sqrt{11}+\sqrt2)-(3+\sqrt3)][(\sqrt{11}+\sqrt2)-(3+\sqrt3)]=(\sqrt{11}+\sqrt2)^2-(3+\sqrt3)^3=(13+2\sqrt{22})-(12+6\sqrt3)=1+2\sqrt{22}-6\sqrt3$
Then
$6\sqrt3-2\sqrt{22}=2(3\sqrt3-\sqrt{22})=\dfrac{2(27-22)}{3\sqrt3+\sqrt{22}}=\dfrac{10}{3\sqrt3+\sqrt{22}}$
And finally
$(3\sqrt3+\sqrt{22})^2=49+6\sqrt{66}<49+6\sqrt{8×9}\overset{AM-GM}{<}49+(6×8.5)=100$
So (parentheses indicate calculator results):
$3\sqrt3+\sqrt{22}<10 (9.887)$
$6\sqrt3-2\sqrt{22}>1 (1.011)$
$1+2\sqrt{22}-6\sqrt3=1-(6\sqrt3-2\sqrt{22})<0 (-0.011)$
$\sqrt11+\sqrt2<3+\sqrt3 (4.731<4.732)$
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|
Subsets and Splits
Fractions in Questions and Answers
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