Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$?
What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ?
My trial:
from given equation: $10x^2+14xy+10y^2-7=0$
$$10x^2+14xy+10y^2=7$$
$$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$
I know the standard form of ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
But the term of $xy$ is a bottleneck.
From here I can't proceed.
Can somebody please help me solve this problem?
Thank you.
|
The ellipse we have is in the form $S(x, y) = ax^2 + by^2 + 2hxy + c$. To remove the $xy$ term we rotate the ellipse by the angle $\tan (2\theta) = \dfrac{2h}{a - b}$. Which is same as rotating coordinate system by $-\theta$.
So if $(X', Y')$ is coordinates in new coordinate system then we can use
$${x \choose y} = \left[\begin{matrix}\cos\theta & -\sin \theta \\ \sin \theta& \cos \theta \end{matrix}\right] {X^\prime \choose Y^\prime}$$
to find relation between old coordinate system $(x,y)$ and new coordinate system $(X', Y')$.
Plugging $ \theta = -\pi/4$
$${x \choose y} = \dfrac{1}{\sqrt{2}}\left[\begin{matrix}1 & 1 \\ -1& 1 \end{matrix}\right] {X^\prime \choose Y^\prime} = \dfrac{1}{\sqrt{2}}{X' +Y'\choose Y'-X'}$$
So, $$10x^2+14xy+10y^2-7= 3 X'^2 + 17 Y'^2 - 7 \implies \dfrac{X'^2}{7/3} + \dfrac{Y'^2}{7/17} = 1$$
The major axis is $2\sqrt{\dfrac73}$ and minor axis is $2\sqrt{\dfrac7{17}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2615815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Calculate the triple integral: $\iiint_W(x^2+y^2)z\ dx\,dy\,dz$ I am trying to solve the following triple integral:
$$
\iiint_W(x^2+y^2)z\,dx\,dy\,dz \\
W=\{(x,y,z) \in \mathbb{R}: x^2+y^2+z^2 \le 9; x^2+y^2\le 1; x \ge 0; y \ge 0; z \ge 0\}
$$
From which I know that there are two surfaces:
$$
x^2+y^2+z^3 = 9 \rightarrow \text{Sphere of radius 3} \\
x^2+y^2=1 \rightarrow \text{Cylinder with a bade radius of 1}
$$
Converting this to Cylindrical Coordinates:
$$
\rho^2+z^2 \le 9 \implies \rho^2 \le 9-z^2 \implies \rho \le \sqrt{9-z^2} \\
\rho^2\le 1
$$
Where do I go from here?
|
The integral is getting calculated in 1st octant therefore:
$$\iiint_W(x^2+y^2)z\,dx\,dy\,dz=\iiint_W R^5\sin^{3}\theta\cos\theta dR d\theta d \phi$$also we know that for $0<\theta<\theta_0$ where $\theta_0=\sin^{-1}\frac{1}{3}$ we have $0<R<3$ and for $\frac{\pi}{2}>\theta>\theta_0$ we have $0<R<\frac{1}{\sin\theta}$. So we can split the integral by two:$$I=\int_{0}^{\frac{\pi}{2}}\int_{\theta_0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin\theta}} R^5\sin^{3}\theta\cos\theta dR d\theta d \phi+\int_{0}^{\frac{\pi}{2}}\int_{0}^{\theta_0}\int_{0}^{3} R^5\sin^{3}\theta\cos\theta dR d\theta d \phi$$$$=\frac{1}{6}\int_{0}^{\frac{\pi}{2}}\int_{\theta_0}^{\frac{\pi}{2}}\frac{1}{\sin^{3}\theta}\cos\theta d\theta d \phi+\frac{243}{2}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\theta_0}\sin^{3}\theta\cos\theta d\theta d \phi=\frac{3\pi}{16}+\frac{\pi}{3}=\frac{25\pi}{48}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2616263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is this a correct derivation of completing the square? $x^2 + bx$
$=x^2 + bx + c - c$
$=(x + k)^2 - c$
$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$
This implies:
$2k = b$, so $k = b/2$, and:
$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$
So to complete the square we are making the transformation:
$x^2 + bx \implies (x + b/2)^2 - (b/2)^2$
|
Okay. You are trying to derive.
As you want to derive $x^2 + bx = (x + k)^2 + c$ I'd start with that line:
"we want $x^2 + bx = (x+k)^2 +c$ and we need to find the $c$ and $k$ that will make it so.
You started instead with
$x^2 + bx = x^2 + bx + c - c = (x+k)^2 -c$ with, I guess (????) then assumption that $(x+k)^2 = x^2 + bx + c$. But without any reason I don't get what you are doing. What is that $c$. It just came out of nowhere. Why do you think that $x + bx + c = (x + k)^2$ where did the $k$ come from? What are you doing?
It doesnt seem clear.
As you want to solve $x^2 + bx = (x + k)^2 -c$ then start there. It's almost as though you are afraid to ask the right question so you ask the wrong question $x^2 + bx + c -c = (x + k)^2 - c$.
But if you say:
"$x^2 + bx = (x+k)^2 -c $; what are the $k$ and $c$ we need to make the happen?"
The solution is clear.
$(x+k)^2 - c = x^2 + 2kx + k^2 -c$ and we want $x^2 + bx = x^2 + 2kx +k^2 - c$ so $k = \frac b2$ and $k^2 - c = \frac {b^2}4 - c = 0$. So $c = -\frac {b^2}4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2616753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How can I calculate the remainder of $3^{2012}$ modulo 17? So far this is what I can do:
Using Fermat's Little Theorem I know that $3^{16}\equiv 1 \pmod {17} $
Also: $3^{2012} = (3^{16})^{125}*3^{12} \pmod{17}$
So I am left with $3^{12}\pmod{17}$.
Again I'm going to use fermat's theorem so: $ 3^{12} = \frac{3^{16}}{3^{4}} \pmod{17}$
Here I am stuck because I get $3^{-4} \pmod{17}$ and I don't know how to calculate this because I don't know what $\frac{1}{81} \pmod{17}$ is.
I know $81 = 13 \pmod{17}$
But I know the answer is 4. What did I do wrong?
|
Just do it.
$3^4 = 81 \equiv -4$.
$3^{12} \equiv (3^4)^3 = (-4)^3 \equiv -81 \equiv 4 \mod 17$.
For insight:
You know $3^{16}\equiv 1 \mod 17$ so $3^{8}\equiv \pm 4$ so $3^4 \equiv \pm 1, \pm \sqrt{-1}$. So $-1 \equiv 16$ one of the $\sqrt {-1} \equiv 4\mod 17$. (the other is $13$). This should tell you to try to find $3^{12}$ via iterations $3^4$.
Also: $81 \equiv 13 \equiv - 4 \mod 17$. So $\frac 1{81} \equiv -\frac 14$. And figuring $\frac 14$ shouldn't be hard $1 \equiv 18$ so $\frac 12 \equiv 9 \mod 17$ and $9 \equiv 26$ so $\frac 14 \equiv 13\equiv -4$. So $-\frac 14 = 4$. And that makes sense. $(-4)*4 = -16 \equiv 1 \mod 17$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2617207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
}
|
Matrix Row Reduction - What am I doing wrong? In this homework problem, I'm asked to find the solution to the following augmented matrix that represents a system of linear equations. Extracting the solution from the matrix isn't a problem, so I'll just show you guys how I got to my reduced echelon form.
$\left[\begin{array}{c c c c c} 3 & 3 & 6 & 12 \\ -6 & -6 & -12 & -24 \\ 0 & -4 & -12 & 16 \\ \end{array}\right]$
Here is my thought process:
I see that the 2nd row is simply a scalar multiple of the first, so that will turn into a row of zeros and move to the bottom. Further, the first and third rows (in the original matrix) can be divided by scalars $3$ and $4$, respectively. So far, we have
$\left[\begin{array}{c c c c c} \color{red}3 & \color{red} 3 & \color{red} 6 & \color{red}{12} \\ \color{blue}{-6} & \color{blue}{-6} & \color{blue}{-12} & \color{blue}{-24} \\ \color{green}{0} & \color{green}{-4} & \color{green}{-12} & \color{green}{16} \\ \end{array}\right] \sim \left[\begin{array}{c c c c c} \color{red}1 & \color{red}{1} & \color{red}{2} & \color{red}{4} \\ \color{green}{0} & \color{green}{-1} & \color{green}{-3} & \color{green}{4} \\ \color{blue}0 & \color{blue}0 & \color{blue}0 & \color{blue}0 \end{array}\right]$
Now we replace row 2 with the sum of rows 1 and 2, and get
$\left[\begin{array}{c c c c c} 1 & 1 & 2 & 4 \\ 0 & 0 & -1 & 8 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$
Finally, we add replace row 1 with sum of twice row 2 and row 1 to get the following reduced echelon form:
$\left[\begin{array}{c c c c c} 1 & 1 & 0 & 20 \\ 0 & 0 & -1 & 8 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$
Apparently, the correct reduced echelon form is
$\left[\begin{array}{c c c c c} 1 & 0 & -1 & 8 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$
I can see that the discrepancy results from my second step: instead of replacing row 2 with the sum of rows 1 and 2, it seems it is correct to replace row 1 instead with the same row operation. But why is what I did wrong?
|
As you replace row $2$ with sum of row $1$ and row $2$, you have forgotten about the first entry which is non-zero.
You should have added row $2$ to row $1$ instead.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2620198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.
I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)
|
Here is an alternative approach that uses a triple integral.
We begin by noting that
$$\frac{1}{n} = \int_0^1 x^{n - 1} \, dx, \quad \frac{1}{2n + 1} = \int_0^1 y^{2n} \, dy, \quad \frac{1}{n + 1} = \int_0^1 z^n \, dz.$$
The sum can therefore be written as
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{2n(2n + 1)(2n + 2)} &= \frac{1}{4} \sum_{n = 1}^\infty \frac{1}{n(2n + 1)(n + 1)}\\
&= \frac{1}{4} \sum_{n = 1}^\infty \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{2n} z^n \, dx dy dz\\
&= \frac{1}{4} \int_0^1 \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty (xy^2 z)^n \, dx dy dz \tag1\\
&= \frac{1}{4} \int_0^1 \int_0^1 \int_0^1 \frac{1}{x} \cdot \frac{xy^2 z}{1 - xy^2 z} \, dx dy dz \tag2\\
&= \frac{1}{4} \int_0^1 \int_0^1 \int_0^1 \frac{y^2 z}{1 - xy^2 z} \, dx dy dz\\
&= -\frac{1}{4} \int_0^1 \int_0^1 \Big{[} \ln (1 -x y^2 z) \Big{]}_0^1 \, dy dz\\
&= -\frac{1}{4} \int_0^1 \int_0^1 \ln (1 -y^2 z) \, dz dy \tag3 \\
&= -\frac{1}{4} \int_0^1 \left [\frac{(y^2 z - 1)[\ln (1 - y^2 z) - 1]}{y^2} \right ]_0^1 \, dy\\
&= \frac{1}{4} \int_0^1 dy - \frac{1}{4} \int_0^1 \frac{y^2 - 1}{y^2} \ln (1 - y^2) \, dy\\
&= \frac{1}{4} - \frac{1}{4} \left (2 + 2 \int_0^1 \ln (1 - y^2) \, dy \right ) \tag4\\
&= -\frac{1}{4} -\frac{1}{2} \int_0^1 \ln (1 - y^2) \, dy\\
&= -\frac{1}{4} + \int_0^1 \frac{y(1 - y)}{1 - y^2} \, dy \tag5\\
&= -\frac{1}{4} + \int_0^1 \frac{y}{1 + y} \, dy\\
&= -\frac{1}{4} + \int_0^1 \frac{(1 + y) - 1}{1 + y} \, dy\\
&= -\frac{1}{4} +\int_0^1 dy - \int_0^1 \frac{dy}{1 + y}\\
&= -\frac{1}{4} + 1 - \ln (2)\\
&= \frac{3}{4} - \ln (2).
\end{align*}
Explanation
(1) Interchanging the summation with the triple integration.
(2) Summing the series which is geometric.
(3) Interchanging the order of integration.
(4) Integrating by parts.
(5) Integrating by parts.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2621718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$ whose slope is $$m_1=\frac{-4 \cot a}{5}$$
Also slope of $AP$ is given by
$$m_2=\frac{4 \sin a+5}{5 \cos a-3}$$
So both slopes are equal , with that we get
$$12 \cos a-25 \sin a=20 \tag{1}$$
Now distance $AP$ is given by
$$AP=\sqrt{(3-5 \cos a)^2+(5+4 \sin a)^2}=\sqrt{75-30 \cos a+40 \sin a} \tag{2}$$
Now using $(1)$ we have to find $\cos a$ and $\sin a$ and then substitute in $(2)$ which becomes lengthy.
Any better way?
|
The polar of $(3,-5)$ w.r.t. the ellipse is $$\frac{3x}{25}+\frac{-5y}{16}-1=0.$$ Intersecting the polar and the ellipse you get the tangent points of the two tangents through $(3,-5).$
They are $(\frac{375\sqrt{41}+1200}{769},\frac{144\sqrt{41}-2000}{769})$ and $(\frac{-375\sqrt{41}+1200}{769},\frac{-144\sqrt{41}-2000}{769})$. The rest is just the distance between these points and $(3,-5)$ and the answer is the two lengths $$\frac{3}{769}\sqrt{1249475\pm 33210\sqrt{41}}.$$
Aside: If you want the equations of the tangents, you can look at the dual conic $25X^2+16Y^2-1=0.$ It intersects the dual line $3X-5Y+1=0$ in two points $$(-\frac{15\sqrt{41}+48}{769},-\frac{9\sqrt{41}-125}{769}),
(\frac{15\sqrt{41}-48}{769},\frac{9\sqrt{41}+125}{769})$$ that have duals the tangent lines: $$-\frac{15\sqrt{41}+48}{769}x-\frac{9\sqrt{41}-125}{769}y+1=0$$ and $$\frac{15\sqrt{41}-48}{769}x+\frac{9\sqrt{41}+125}{769}y+1=0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2623243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Region(area) enclosed between line and parabola The problem is to find the area of the region enclosed by the given parabola and line:
$y=-x^2+3x-1$
$y=x-2$
I tried solving the definite integral, but the roots (points of intersection of the parabola and line) on the graph were too weird and I ended up with the wrong solution. Does anyone know a more efficient method?
|
This is a method where you do not need to find the actual roots from the graph, from which the problem of "weird points of intersection" will be completely eradicated. Trust me.
From $-x^2+3x-1=x-2$, $x^2-2x-1=0$.
Letting $a$ and $b$ (where $a<b$) be the solutions of $x^2-2x-1=0$, $a+b=-2$ and $ab=1$, due to the root-coefficient relationship.
Therefore, the area, $S = \int_a^b\left[-x^2+3x-1-\left(x-2\right)\right]dx = -\int_a^b\left(x^2-2x-1\right)dx = -\int_a^b\left(x-a\right)\left(x-b\right)dx = \frac{1}{6}\left(b-a\right)^3$
From $(b-a)^2=(a+b)^2-4ab=8$, $b-a=2\sqrt{2}$
Therefore, $\left(b-a\right)^3=16\sqrt{2}$.
Thus, the area, $S=\frac{8\sqrt{2}}{3}$.
And this answer was able to be posted quickly due to my MathJax friend, Desmos.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2625152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sqrt{n})(\sqrt{n+\sqrt{n}}+\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}-n)}{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}+\sqrt{n})} \right) \\
&= \lim_{n \to \infty}\left( \frac{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}-\frac{n}{\sqrt{n}})}{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}+ \frac{\sqrt{n}}{\sqrt{n}})} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{n}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{\sqrt{n}\sqrt{n}}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\sqrt{n}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{1}{2} - \frac{\sqrt{n}}{2} \right) \\
\end{align}$
Which is wrong.
Where could be my mistake?
|
To simplify the derivation you can use the binomial series and note that
$$\sqrt{n+\sqrt{n}}= \sqrt{n}\left( \sqrt{1+\frac1{\sqrt{n}}} \right)=\sqrt{n}\left( 1+\frac1{2\sqrt{n}}+o\left(\frac1{\sqrt{n}}\right) \right)=\sqrt{n}+\frac12+o(1)$$
thus
$$\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)=\sqrt{n}+\frac12+o(1)-\sqrt{n}=\frac12+o(1)\to\frac12$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2626347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Why is major axis of this ellipse is along what appears to be minor axis? Given any ellipse in standard form, like
$$x^2/5+y^2/10=1$$
is always elongated towards $y$ as its denominator is larger or $b>a$.
But consider this ellipse
$$\frac{\left(x+2y\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1$$
This is elongated towards $x+2y=0$, even though it has less denominator.
Whys is that? Can't this ellipse be seen as one with axes $x+2y=0$ and $2x-y=0$
Sketch of this ellipse is here:
|
The two cases are similar
*
*for $x^2/5+y^2/10=1$ when $x=0$ we have the max elongation towards $y$
(that is $x=0$)
*for $\frac{\left(x+2y\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1
$ when $x+2y=0$ we have the max elongation towards $2x-y$
(that is $x+2y=0$)
Indeed note that
$$x+2y=0\implies \frac{\left(0\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1
\implies \left(2x-y\right)^2=20 \implies25y^2=20\\\implies y=\pm\frac{2\sqrt{5}}{5}\quad x=\mp\frac{4\sqrt{5}}{5}\implies\rho_1=4\frac{\sqrt{6}}5$$
with $\rho_1$ along the $x+2y=0$ axis.
$$2x-y=0\implies \frac{\left(x+2y\right)^2}{5}+\frac{\left(0\right)^2}{20}=1
\implies \left(x+2y\right)^2=5 \implies25x^2=5\\\implies x=\pm\frac{\sqrt{5}}{5}\quad y=\pm \frac{2\sqrt{5}}{5} \implies\rho_2=1$$
with $\rho_2$ along the $2x-y=0$ axis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2628166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Compute the 100th power of a given matrix So, I was asked this question in an interview.
Given a matrix
$$M = \begin{bmatrix} 0 & 1& 1 \\ 1& 0& 1 \\ 1&1& 0\end{bmatrix},$$
find $M^{100}$.
How does one approach this question using pen and paper only?
|
Here's another trick. Denote
$$
\mathbf 1 = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1& 1 & 1\end{bmatrix}, $$
and let $I=\mathrm{diag}(1,1,1)$ be the identity matrix. Then by the binomial formula
$$
M^{100}=(\mathbf 1 - I)^{100}=\sum_{k=0}^{100} \binom{100}{k}(-1)^k \mathbf 1^{100-k}, $$
so we are led to considering powers of $\mathbf 1$, which are easy to compute:
$$\mathbf 1^{j}=\begin{cases} 3^{j-1}\mathbf 1, & j\ge 1 \\ I & j=0.\end{cases}$$
We conclude that $M^{100} = \big(\sum_{k=0}^{99} \binom{100}{k}(-1)^k 3^{100-k-1}\big)\mathbf 1 + I, $ and the binomial sum can be computed by observing that
$$
(3-1)^{100}=\sum_{k=0}^{100} (-1)^k \binom{100}{k}3^{100-k}.$$
We conclude that
$$
M^{100}=\begin{bmatrix} \frac{2^{100}+2}3 & \frac{2^{100}-1}3 & \frac{2^{100}-1}3 \\ \frac{2^{100}-1}3 & \frac{2^{100}+2}3 & \frac{2^{100}-1}3 \\ \frac{2^{100}-1}3 & \frac{2^{100}-1}3 & \frac{2^{100}+2}3 \end{bmatrix}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2628253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Finding all solutions to a trigonometric equation I was solving a simple trigonometric equation for my brother, going this path:
$16\sin^2(2x)\cos^2(2x) = 3$
${[4\sin(2x)\cos(2x)]}^2 = 3$
Applied the formula for $\sin(2a) = 2\sin(a)\cos(a)$ "backwards"
$[2\sin(4x)]^2 = 3$
$\sin(4x) = \frac{\sqrt{3}}{2}$
Then the solutions are easily found:
$\frac{\pi}{12} + k\frac{\pi}{2}$ and $\frac{\pi}{6} + k\frac{\pi}{2}$
However the book lists 4 different solutions. I can find them with little effort, but very arbitrarily. How can I know in advance how many solutions there are to such an equation? And to a generic $f(x) = 0$?
|
Due to taking the square root, $$\sin(4x)=\pm\frac{\sqrt3}2\implies 4x=\frac\pi3, \frac{2\pi}3, \frac{4\pi}3, \frac{5\pi}3, \frac{7\pi}3, \frac{8\pi}3, \frac{10\pi}3, \frac{11\pi}3$$ so $$\boxed{x=\frac\pi{12}, \frac\pi6, \frac\pi3, \frac{5\pi}{12} \frac{7\pi}{12},\frac{2\pi}3, \frac{5\pi}6, \frac{11\pi}{12}}$$ are the eight solutions for $0<x<2\pi$.
As commented below an infinite set of solutions can be found by adding $\dfrac{k\pi}4$ to each solution for $k\in\mathbb{Z}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2631771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Prove the slope of the asymptote of a hyperbola Consider a hyperbola that opens horizontally (left to right):
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
How do we prove that the slopes of the asymptotes are $\pm \frac{b}{a}$?
|
Trying to solve it myself:
Rearranging for $y$ I get:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
$$a^2(y-k)^2 = b^2(x-h)^2 - a^2b^2$$
$$y = k \pm \sqrt{\frac{b^2(x-h)^2 - a^2b^2}{a^2}}$$
$$y = f(x) = k \pm \frac{b \sqrt{-a^2 + h^2 + x^2 - 2 h x}}{a}$$
So then the slope is
$$\lim_{x\to\infty}\frac{f(x)}{x}$$
or
$$\lim_{x\to\infty}\frac{k}{x} \pm \frac{b \sqrt{-a^2 + h^2 + x^2 - 2 h x}}{ax}$$
$$\lim_{x\to\infty}\frac{k}{x} \pm \frac{b \sqrt{\frac{-a^2 + h^2 + x^2 - 2 h x}{x^2}}}{a}$$
$$\lim_{x\to\infty}\frac{k}{x} \pm \frac{b \sqrt{-\frac{a^2}{x^2} + \frac{h^2}{x^2} + \frac{1}{1} - \frac{2 h}{x}}}{a}$$
$$\lim_{x\to\infty}\frac{k}{\infty} \pm \frac{b \sqrt{-\frac{a^2}{\infty^2} + \frac{h^2}{\infty^2} + \frac{1}{1} - \frac{2 h}{\infty}}}{a}$$
$$\lim_{x\to\infty}0 \pm \frac{b \sqrt{0 + 0 + 1 - 0}}{a}$$
$$\lim_{x\to\infty}\pm \frac{b}{a}$$
The result seems to be right but is my work here correct?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2632343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Eigenvalues of symmetric $\mathbb{R}^{p\times p}$ matrix I want to prove that
$$A^{(p)}
=
\begin{pmatrix}
a & 1 & 1 & \dots & 1 & 1 & 1 \\
1 & a & 1 & \dots & 1 & 1 & 1 \\
1 & 1 & a & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a & 1 \\
1 & 1 & 1 & \dots & 1 & 1 & a
\end{pmatrix}
\in\mathbb{R}^{p\times p}$$
has eigenvalues $\lambda_+=a+p-1$ and $\lambda_-=a-1$ with degeneracy $p-1$ by using mathematical induction. Evidence that these eigenvalues could be correct were found 'empiricaly' by Mathematica.
Induction Base Case $(p=2)$
$$\det(A^{(2)}-\lambda E_2)
=
\det\begin{pmatrix}a-\lambda & 1\\1 & a-\lambda\end{pmatrix}
=(\lambda-(a-1))(\lambda-(a+1)) $$
Induction Hypothesis $(p\in\mathbb{N})$
$$\det(A^{(p)}-\lambda E_p)
=(\lambda-(a-1))^{p-1}(\lambda-(a+p-1))$$
Induction Step $(p\to p+1)$
$$\det(A^{(p+1)}-\lambda E_{p+1})
=
\det\begin{pmatrix}
a-\lambda & 1 & 1 & \dots & 1 & 1 & 1 \\
1 & a-\lambda & 1 & \dots & 1 & 1 & 1 \\
1 & 1 & a-\lambda & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a-\lambda & 1 \\
1 & 1 & 1 & \dots & 1 & 1 & a-\lambda
\end{pmatrix}$$
We can subtract the last row from the first row
$$\det\begin{pmatrix}
a-\lambda-1 & 0 & 0 & \dots & 0 & 0 & 1-a+\lambda \\
1 & a-\lambda & 1 & \dots & 1 & 1 & 1 \\
1 & 1 & a-\lambda & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a-\lambda & 1 \\
1 & 1 & 1 & \dots & 1 & 1 & a-\lambda
\end{pmatrix}$$
and apply Laplace's formula to the first row. The first entry (upper left corner) gives
$$(a-\lambda-1)
\det\begin{pmatrix}
a-\lambda & 1 & \dots & 1 & 1 & 1 \\
1 & a-\lambda & \dots & 1 & 1 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
1 & 1 & \dots & 1 & a-\lambda & 1 \\
1 & 1 & \dots & 1 & 1 & a-\lambda
\end{pmatrix}$$
which is just the case of our induction hypothesis, thus
$$-(\lambda-(a-1))^p(\lambda-(a+p-1)) $$
is the first contribution to Laplace's formula. The second contribution reads
$$(1-a+\lambda) (-1)^{p+2}
\det\begin{pmatrix}
1 & a-\lambda & 1 & \dots & 1 & 1 \\
1 & 1 & a-\lambda & \dots & 1 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & 1 & 1 & \dots & 1 & a-\lambda\\
1 & 1 & 1 & \dots & 1 & 1
\end{pmatrix}$$
and we can subtract the first column from all other columns yielding
$$\det\begin{pmatrix}
0 & a-\lambda-1 & 0 & \dots & 0 & 0 \\
0 & 0 & a-\lambda-1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & 0 & a-\lambda-1\\
1 & 1 & 1 & \dots & 1 & 1
\end{pmatrix}$$
on which we can reapply Laplace's formula and use that the determinant of a diagonal matrix equals the product of the diagonal entries, hence
$$(-1)^{p+1}(\lambda-(a-1))^{p+1} $$
is the second contribution to the first application of Laplace's formula. Now combining both I find
$$\det(A^{(p+1)}-\lambda E_{p+1})
=
(-1)(\lambda-(a-1))^p(\lambda-(a+p-1))+(-1)^{p+1}(\lambda-(a-1))^{p+1}
=(-1)(\lambda-(a-1))^p(\lambda-a-p+1+(-1)^p(\lambda-a+1)) $$
which seems faulty: on the one hand side the factor $(-1)^{p}$ should not exist as for odd $p$ the characteristic polynomial should varnish which has been shown to be not the case for i.e. $p=3$ (see Mathematica). On the other hand even if that factor would be wrong the final expression would yield
$$(-2)(\lambda-(a-1))^p(\lambda-(a-p/2+1))$$
which would contradict the induction hypothesis.
Did I made a mistake in evaluating the determinant or is the induction hypothesis wrong?
|
The last diagonal matrix you are considering will be $(p-1)\times(p-1)$, because it results from applying Laplace expansion twice on a $(p+1)\times (p+1)$ matrix. Looking at the exponent of $(\lambda-a+1)$, you do not seem to have taken it into account.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2632671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by parts implies $$\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{(x^2+a^2)-a^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{1}{(x^2+a^2)^{n}}-2na^2\int\frac{dx}{(x^2+a^2)^{n+1}}$$ but I don't think I'm doing this correctly since the power of the denominator $(x^2+a^2)$ is not decreasing.
|
hint
$$\frac {1}{(x^2+a^2)^n}=\frac {1}{a^2}\frac {a^2+x^2-\frac {2x}{2}x}{(a^2+x^2)^n} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2633107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the first element of every individual term makes a sequence $1, 2, 4, 7...$. $\therefore$ the first element of each term is $(1+\frac{n(n-1)}2)^3$ and this the last element is $(\frac{n(n-1)}2 + 1 + (n-1))^3$. But I don't understand what to do next.
|
Hint: The sum is equal to
$$S_n=1^3+2^3+3^3+...+n^3=\dfrac{n^2(n+1)^2}{4}$$
The $n^\text{th}$ term in your representation is given as
$$S_{n(n+1)/2}-S_{n(n-1)/2}.$$
The indices are just the partial sums for
$$1+2+...+n = \dfrac{n(n+1)}{2}$$
$$1+2+...+(n-1)=\dfrac{n(n-1)}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2634040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Inequality : $ \sqrt{\frac{a}{b+1}} + \sqrt{\frac{b}{c+1}} +\sqrt{\frac{c}{a+1}} \le \frac{3}{2} $ $a+b+c =1$ , $a, b, c>0 $
Prove
$$ \sqrt{\frac{a}{b+1}} + \sqrt{\frac{b}{c+1}} + \sqrt{\frac{c}{a+1}} \le \frac{3}{2}$$
When I meet the inequality with square root symbols, I don't have any idea. It's my tragedy. I know the triangular inequality : $$\cos A + \cos B + \cos C \le \frac{3}{2} $$
But I failed to find a link between the two inequalities. I want some another hints. Thank you.
|
In eliminating square roots, a well chosen CS is your friend, (and for higher roots, often Holder). Note by Cauchy-Schwarz inequality,
$$\sum_{cyc} (a+1) \cdot\sum_{cyc} \frac{a}{(a+1)(b+1)}\geqslant \left(\sqrt{\frac{a}{b+1}}+\sqrt{\frac{b}{c+1}}+\sqrt{\frac{c}{a+1}} \right)^2$$
Hence it is enough to show that
$$\sum_{cyc} \frac{a}{(a+1)(b+1)} \leqslant \frac9{16}$$
$$\iff 16\sum_{cyc} a(c+1) \leqslant 9\prod_{cyc} (a+1)$$
$$\iff 9abc+2 \geqslant 7(ab+bc+ca)$$
In symmetric form this is:
$$2(a+b+c)^3 + 9abc \geqslant 7(a+b+c)(ab+bc+ca)$$
$$\iff 2(a^3+b^3+c^3) \geqslant \sum_{cyc} ab(a+b)$$
which is true by Muirhead’s theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2634120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
$$x^2+y^2+z^2+2xyz=1$$
My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1-x^2})+\sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2} + y\sqrt {1-x^2})=\dfrac {\pi}{2} - \sin^{-1} (z)$$
|
Put differently, if $\alpha+\beta+\gamma=\frac \pi2$, then
$$ \tag1\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=1.$$
Let's investigate how the left hand side aries if we keep $\gamma $ fixed, that is, we consider
$$\tag2 \sin^2(\alpha+t)+\sin^2(\beta-t)+\sin^2\gamma+2\sin(\alpha+t)\sin(\beta-t)\sin\gamma$$
and compute its derivative $\frac d{dt}$ at $t=0$:
$$ \begin{align}&2\sin \alpha\cos \alpha-2\sin \beta\cos\beta+2(\cos\alpha\sin\beta-\sin\alpha\cos\beta)\sin\gamma\\
={}&\sin2\alpha-\sin 2\beta-2\sin(\alpha-\beta)\sin\gamma\\
{}={}&2\cos(\alpha+\beta)\sin(\alpha-\beta)-2\sin(\alpha-\beta)\sin\gamma\\
={}&2\sin\gamma\sin(\alpha-\beta)-2\sin(\alpha-\beta)\sin\gamma\\={}&0\end{align}$$
We conclude that $(2)$ is constant as a function of $t$. Therefore, it suffices to show $(1)$ for the case $\beta=0$ (i.e., $t=\beta$), i.e., we are reduced to showing
If $\alpha+\gamma=\frac\pi2$ then $\sin^2\alpha+\sin^2\gamma=1$.
But that is of course true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2634292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
How to solve $\int \frac{dx}{x+x^2}$ without using partial fractions? I can easily rewrite it as $\frac{1}{x} - \frac{1}{x+1}$, but is it possible to solve this without partial fractions?
|
Let $\displaystyle y=1+\frac{1}{x}$. Then $\displaystyle \frac{dy}{dx}=-\frac{1}{x^2}$.
\begin{align*}
\int\frac{dx}{x+x^2}&=-\int\frac{\frac{-1}{x^2}dx}{1+\frac{1}{x}}\\
&=-\ln\left(1+\frac{1}{x}\right)+C
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2635601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Which one is correct $\sum_{n=1}^{k} 2^{n}× n^2$ or $\sum_{n=1}^{k} (2^{n}× n^2)$? I have $2$ questions about notation.
Question 1.
Suppose, numbers are as follows.
$\left\{3\right\}\longrightarrow7$
$\left\{3,6\right\}\longrightarrow9$
$\left\{3,6,7\right\}\longrightarrow11$
$\left\{3,6,7,5\right\}\longrightarrow15$
$...............$
$\left\{p_1,p_2,p_3,...p_n\right\}\longrightarrow q_n$
And $p_1,p_2,...,p_{n}$ are arbitrary natural numbers.But, the value $q_n$ depend on the values of ${p_1,p_2,...,p_n}.$
I want to write a notation.
I think, I can write, for example,
$$f(p_1,p_2,...p_n)=q_n$$ like a function.Is it correct or ıs there an other notation?
Question 2.
Which notation is correct?
$\sum_{n=1}^{k} 2^{n}× n^2$ or $\sum_{n=1}^{k} (2^{n}× n^2)$
and
$\sum_{n=1}^{k} 2^{n}+n^2$ or $\sum_{n=1}^{k} (2^{n}+n^2)$
|
A word of warning to question 2: The subscript $n$ in
\begin{align*}
\sum_{n=1}^k 2^n+n^2
\end{align*}
does not tell anything about the scope of the sigma operator $\Sigma$.
*
*The scope of the sigma operator $\Sigma$ is solely defined via arithmetic precedence rules. It is given by the expression that follows immediately the $\Sigma$ and is valid respecting the arithmetic precedence rules up to an operator with precedence level equal to '$+$' or up to the end if no such operator follows.
This implies that
\begin{align*}
\color{blue}{\sum_{n=1}^k 2^n}+n^2&=\color{blue}{\sum_{n=1}^k \left(2^n\right)}+n^2=\color{blue}{\left(\sum_{n=1}^k 2^n\right)}+n^2\\
&=\color{blue}{2^1+2^2+\cdots+2^k}+n^2
\end{align*}
The scope of the Sigma symbol does not extend to $n^2$, which is a variable by its own independent of the index $n$ (and a bad naming style).
On the other hand according to the arithmetic precedence rules we have
\begin{align*}
\sum_{n=1}^k \left(2^n+n^2\right)&=(2^1+1^2)+(2^2+2^2)\cdots (2^k+k^2)\\
&=2^1+2^2+\cdots+2^k+1^2+2^2+\cdots+k^2\\
&=\sum_{n=1}^k 2^n +\sum_{n=1}^k n^2
\end{align*}
We conclude in general we have
\begin{align*}
\sum_{n=1}^k 2^n+n^2\ne\sum_{n=1}^k \left(2^n+n^2\right)
\end{align*}
Since multiplication has higher precedence level than addition we obtain
\begin{align*}
\sum_{n=1}^k2^n\cdot n^2&=\sum_{n=1}^k\left(2^n\cdot n^2\right)
=2^1\cdot1^2+2^2\cdot 2^2+\cdots 2^k\cdot k^2
\end{align*}
Hints to question 1:
The representation
\begin{align*}
f(p_1,p_2,\ldots,p_n)=q_n
\end{align*}
indicates a function $f:\mathbb{N}^n\rightarrow \mathbb{N}$. Here the argument of $f$ is an $n$-tuple $(p_1,p_2,\ldots,p_n)$ where order matters. This does not match the mapping of
\begin{align*}
\{p_1,p_2,\ldots,p_n\}\rightarrow q_n
\end{align*}
where the left-hand side is an unordered set.
We could write
\begin{align*}
&f:\mathcal{P}(\mathbb{N})\setminus\emptyset\rightarrow \mathbb{N}\\
&f(\{p_1,p_2,\ldots,p_n\})=q_n
\end{align*}
Here the argument is a set consisting of $n$ natural numbers. In order to avoid ambiguities we also have to specify in this case that $p_1,p_2,\ldots,p_n$ are pairwise different.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2636163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
FInd the limit without the l’Hospital’s rule FInd the limit without the l’Hospital’s rule:
$$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x}-\sqrt{x+2}}$$
I have tried multiplying the equation with:
$\frac{\sqrt{2x}+\sqrt{x+2}}{\sqrt{2x}+\sqrt{x+2}}$ and got stuck with
$\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{2x-(x+2)}$
$\implies$ $\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{x-2}$
$\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}$ and got stuck with
$\frac{3^2-(x^2+5)}{(\sqrt{2x}-\sqrt{x+2})(3+\sqrt{x^2+5})}$
$\implies$ $\frac{14-x^2}{(\sqrt{2x}-\sqrt{x+2})(3+\sqrt{x^2+5})}$
I tried to solve the denominator in the second one (which is the same as the nominator in the first but with different signs) which ended up with:
$\frac{14-x^2}{3\sqrt{2x}+\sqrt{2x(x^2+5)}-3\sqrt{x+2}-\sqrt{(x+2)(x^2+5)}}$
What can I do after that?
The answer with the l’Hospital’s rule (using Symbolab) $-\frac{8}{3}$
|
You've made some errors in arithmetic. The solution involves doing both of the things you tried:
$$\begin{align*}
\frac{3-\sqrt{x^2+5}}{\sqrt{2x} - \sqrt{x+2}} &= \frac{(3^2 - (x^2+5))( \sqrt{2x} + \sqrt{x+2})}{(2x - (x+2))(3 + \sqrt{x^2+5})} \\ &= \frac{(4-x^2)(\sqrt{2x} + \sqrt{x+2})}{(x-2)(3 + \sqrt{x^2+5})} \\
&= -(x+2) \frac{\sqrt{2x} + \sqrt{x+2}}{3 + \sqrt{x^2+5}}. \end{align*}$$ Now all expressions are nonzero when $x = 2$, giving $$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x} - \sqrt{x+2}} = -(2+2) \frac{\sqrt{4} + \sqrt{4}}{3 + \sqrt{9}} = -\frac{8}{3}.$$
As you can see, the reason why you need to do both is because there is a factor of $x-2$ that appears in both the numerator and denominator, and this is what needs to cancel out in order for the limit to be computed. You did each method by itself but without doing both, that factor cannot be removed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2636567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
A proof in combinatorics (prove by any method) I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following by using any method?
$$\left( \begin{array}{c} n \\ 1\ \end{array} \right) + 3 \left( \begin{array}{c} n \\ 3\ \end{array} \right) +5 \left( \begin{array}{c} n \\ 5\ \end{array} \right) +... = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + 4 \left( \begin{array}{c} n \\ 4\ \end{array} \right)+... $$
Could use some help as I work through possible exam questions.
|
Putting the right-hand side to the left we have
\begin{align*}
\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-\cdots=
\begin{cases}
1\qquad n=1\\
0\qquad n>1
\end{cases}
\end{align*}
or in more compact notation using Iverson brackets:
\begin{align*}
\sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}=[[n=1]]
\end{align*}
We obtain for integer $n>0$
\begin{align*}
\color{blue}{\sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}}
&=n\sum_{k=1}^n(-1)^{k+1}\binom{n-1}{k-1}\tag{1}\\
&=n\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\tag{2}\\
&=n(1-1)^{n-1}\tag{3}\\
&\color{blue}{=[[n=1]]}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we apply the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$.
*In (2) we shift the index $k$ to start with $k=0$.
*In (3) we apply the binomial theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2637101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0$$
Now what do I do?
|
The condition $f(-x)=-f(x)$, i.e.,
$$|-x+a|-|-x+2|+b|-x+5|=-(|x+a|-|x+2|+b|x+5|)$$
is equivalent to
$$|x+a|+|-x+a|+b|x+5|+b|-x+5|=|x+2|+|-x+2|$$
which can be rewritten as
$$(|x+|a||+|x-|a||)+b(|5+x|+|5-x|)=|x+2|+|x-2|$$
Now think of each side of this equation as the formula for a continuous function whose derivative fails to exist only at values of $x$ where the linear expression inside an absolute value sign takes the value $0$. For the function on the right hand side, the derivative fails to exist at $x=\pm2$ (and exists for all other values of $x$). Since $5\pm2\not=0$, the only way that the derivative of the left hand function can fail to exist at $x=\pm2$ is if $a=\pm2$. But if $b\not=0$, the derivative of the left hand function will also fail to exist at $x=\pm5$, where the right hand function has a well defined derivative. Thus $b=0$ and $a=\pm2$, hence $a+b=\pm2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2638472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Mathematical proof that $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ (Table proof provided) My question is how to express my solution to $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ as a mathematical proof.
I understand that to prove by contradiction, the result is to be a cube of an integer:
\begin{align}
m^4 + 8 & = n^3 \\
m^4 &= n^3 - 8 \\
\therefore m^4 &\equiv n^3-8\pmod{13}\\
\\
\end{align}
That equivalence is however impossible, because of the following table:
We see that $m^4 \equiv n^3-8\pmod{13}$ can only occur when m $\equiv 0 \pmod{13}$ and n = 2, 5 or 6.
This shows that since $13$ can not divide $m$, $m^4+8$ is not a cube of an integer. I am now sure how to express this mathematically.
Any help would be appreciated!
|
Your method is perfectly fine for this. Just for fun, I'm giving an alternative method in which you can make do with only a table of squares mod 13.
n 0 1 2 3 4 5 6 7 8 9 10 11 12
n^2 0 1 4 9 3 12 10 10 12 3 9 4 1
$$m^4+8 \equiv n^3 \mod 13\\
(m^4+8)^4 \equiv n^{12} \mod 13$$
From Fermat's little theorem we know that $n^{12}\equiv0,1 \mod 13$, so we then get:
$$(m^4+8)^4 \equiv 0,1 \mod 13\\
(m^4+8)^2 \equiv 0,1,12 \mod 13\\
m^4+8 \equiv 0,1,5,8,12 \mod 13\\
m^4 \equiv 0,4,5,6,10 \mod 13$$
Note that $5$ and $6$ are not squares, so they are certainly not fourth powers.
$$m^4 \equiv 0,4,10 \mod 13\\
m^2 \equiv 0,2,6,7,11 \mod 13$$
Note that the only one of these that is a square is $0$ so we are just left with
$$m^2 \equiv 0 \mod 13\\
m \equiv 0 \mod 13$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2642805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Solve the equation $ { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $ Prove that one root of the Equation
$${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$
is
$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$
My progress:
After simplyfying the given Equation I got
$$ x^8-16x^6+88x^4-192x^2+x+140=0$$
Then I tried to factorise it, got this
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^5-x^4-9x^3+10x^2+17x-20)$
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^3-2x^2-3x+5)(x^2+x-4)$
Should I now solve cubic?
|
I created this equation by the following way.
$$3^n\equiv1,3,9,8,5,15,7,2,6,...\!\pmod{19}$$
Now, let
$$x_1=2\left(\cos\frac{2\cdot1\pi}{19}+\cos\frac{2\cdot8\pi}{19}+\cos\frac{2\cdot7\pi}{19}\right),$$
$$x_2=2\left(\cos\frac{2\cdot3\pi}{19}+\cos\frac{2\cdot5\pi}{19}+\cos\frac{2\cdot2\pi}{19}\right)$$ and
$$x_3=2\left(\cos\frac{2\cdot9\pi}{19}+\cos\frac{2\cdot15\pi}{19}+\cos\frac{2\cdot6\pi}{19}\right).$$
Thus, $x_1$, $x_2$ and $x_3$ they are roots of the cubic equation with rational coefficients.
Easy to show that it's $$x^3+x^2-6x-7=0.$$
Also, easy to show that $$-x_1=\sqrt{4-x_2},$$
$$x_2=\sqrt{4-x_3}$$ and $$-x_3=\sqrt{4-x_1}$$
which gives an equation with radicals.
I posted this equation here: https://artofproblemsolving.com/community/c6h69475
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2643622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem, i applied AM-GM inequality for $(x \;, \;6-x)$
$$\Biggl (\frac{(x+(6-x)}{2}\Biggr) \ge \sqrt {x(x-6)}$$
Expanding out we get $$(x-3)^2\ge0$$
Holding the equality when $x=3$
We can do the same with $(y \;, \; 6-y)$ and $(z \;, \; 6-z)$ getting $(y-3)^2\ge0$ and $(z-3)^2\ge0$ holding when $y,z =3$ and getting that one solution for the system is $x=y=z=3$ but i don't know if this is enough for proving that those are the only solutions.
|
Replace $x$ with $3+e$
If $x = 3 + e$ then $z(3-e) = 9; z=\frac 9{3-e}$ and $(3+e)(6-y)=9$ while $y(6- \frac 9{3-e}) = 9$ or $y(2-\frac 3{3-e}) = 3$.
So $y = \frac 3{2-\frac 3{3-e}} = 6-\frac 9{3+e}$
Let solve for $e$
$ \frac 3{2-\frac 3{3-e}}= \frac {3(3-e)}{2(3-e) - 3}=3\frac {3-e}{3-2e}=6-\frac9{3+e}=3(2 - \frac 3{3+e})$
$\frac{(3-e)(3+e)}{3-2e}=(2(3+e) - 3)=3+2e$
$(3-e)(3+e) = (3+2e)(3- 2e)$
$9 - e^2 = 9 - 4e^2$ so $e = 0$
So $x = 3+0 = 3$ and $z=\frac 9{3-0} = 3$ and $y = \frac 3{2-\frac 3{3-0}} = 6-\frac 9{3+0} = 3$
So, unless I made an error, $x = y = z = 3$ is the only solution.
(Actually, I believe this shows that $x=y=z=3$ are the only solutions in $\mathbb R$ whether positive or negative.)
======
If you want to use A.M-G.M (and be perverse)
Note $a + b \ge 2\sqrt ab$ with equality holding if and only if $a=b$. ($a- 2\sqrt ab + b = (\sqrt a - \sqrt b)^2 \ge 0$ with equality holdig if and only if $a = b$.)
So suppose $x = 3+e; e > 0$
then $x(3-e) = (3+e)(3-e) = 9-e^2 < 9=x(6-y)$ so $3-e < 6-y; y< 3+e$. But if $x > 3$ then $6-y < 3$ so $y > 3$.
So $y = 3 + d; d < e$.
But the same argument $y(3-d) = (3+d)(3-d) = 9 -d^2 < 9 = y(6-z)$ so $3-d < 6-z; z< 3+d$ but $z > 3$ so
$z = 3+ c; c < d$.
And by the same argument as above $z(3-c) = (3+c)(3-c) = 9 - c^2 < 9 z(6-x)$ so $3-c < 6-x so $x < 3 + c < 3 + d < 3+ e = x$ which is a contradiction.
THe same argument would apply if we had assume $x < 3$ so $x = 3-e$ or if we had begun with any variable.
So $x = y = z = 3$ is the only possible solution. (If there are any solution; which obviously $x=y=z=3$ is one.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2646322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
$a,b,c$ are digits. İf $a+2b+3c=40$ then what is the largest value for $a+b+c$? I am finding it hard to understand the method and the logic behind this.
I read that we have to give the one with the smallest coefficient the biggest value ,why is this . ?
|
Assume you have found an optimal triple $(a,b,c)$ with $a,b,c\in\{0,\ldots,9\}$.
Note that $a+2b+3c=40$ implies $3c=40-a-2b\ge 40-9-18=13$ and so $c>4$.
Also, $2b=40-a-3c\ge 40-9-27=4$ and so $b\ge 2$.
If $a<8$ in our optimal solution, then (as $b\ge 2$ and so $b-1$ is still a digit), then $(a+2,b-1,c)$ is a valid triple and has larger target sum, contradicting optimality.
We conclude that $a\ge 8$ in any optimal triple.
Similarly, if $b<7$, then $(a,b+3,c-2)$ is an improvement. We conclude that $b\ge 7$.
In particular, $a$ and $b$ are not far apart: $|a-b|\le 2$. then from $a-b=40-3b-3c=1+3\cdot(13-b-c)$, we conclude that $a-b=1$ and also that $13-b-c=0$. This makes $a+b+c=13+a$, so we certainly want to make $a$ as large as possible, namely $a=9$, leading to the maximal target sum $a+b+c=13+a=22$, where by the way $b=a-1=8$ and $c=13-b=5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2647795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Let $n$ be a 6-digit number, perfect square and perfect cube. If $n-6$ is not even or a multiple of 3, find $n$ Let $n$ be a 6-digit number, perfect square and perfect cube. If $n-6$ is not even or a multiple of 3, find $n$.
My try
Playing with the first ten perfect squares and cubes I ended with:
The last digit of $n \in (1,5,9)$
If $n$ last digit is $9$, then the cube ends in $9$, Ex: if $n$ was $729$, the cube is $9^3$ (ends in $9$) and the square ends in $3$ or $7$
If $n$ last digit is 5, then the cube ends in 5 and the square ends in 5
If $n$ last digit is 1, then the cube ends in 1 and the square ends in 1
By brute force I saw that from $47^3$ onwards, the cubes are 6-digit, so I tried some cubes (luckily for me not for long) and $49^3 = 343^2 = 117649$ worked.
So I found $n=117649$ but I want to know what is the elegant or without brute force method to find this number because my method isn't very good, just pure luck maybe.
|
Note that the required number is both a square and a cube, so it must be a sixth power. Already $10^6=1000000$ has seven digits and $5^6=15625$ has only five digits, so that leaves us with $6^6,7^6,8^6,9^6$ to test.
Furthermore, we are given that $n-6$ is not even and not a multiple of 3, which implies that $n$ itself is also not even and not a multiple of 3. This eliminates $6^6,8^6$ and $9^6$ immediately, leaving $7^6$ as the only possible answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2651165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Joukowski mapping a circle onto an ellipse problem The Joukowski mapping is defined by
$$\displaystyle w = J(z) = \frac{1}{2} \left(z + \frac{1}{z}\right)$$
where $ z = x + yi $
Show that
$J$ maps the circle $|z| = r (r > 0, r \neq 1)$ onto the ellipse
$$ \frac{u^2}{\left[\frac{1}{2}\left(r + \frac{1}{r}\right)\right]^2} +\frac{v^2}{\left[\frac{1}{2}\left(r - \frac{1}{r}\right)\right]^2} = 1 $$
which has a foci at $\pm 1$.
Would I have make use of the fact that $z = re^{i\theta} = r(\cos \theta + i\sin \theta)$ and $ u + iv = w = J(z) = \frac{1}{z} \left(z + \frac{1}{z}\right) = \frac{1}{z} \left(re^{i\theta} + \frac{1}{r} e^{-i\theta}\right)$?
|
HINT...You are almost there: you have $$u+iv=\frac 12\left(r(\cos\theta+i\sin\theta)+\frac{1}{r(\cos\theta+i\sin\theta)}\right)$$
$$=\frac 12\left(r(\cos\theta+i\sin\theta)+\frac 1r(\cos\theta-i\sin\theta)\right)$$
So now you can get $\cos\theta$ and $\sin\theta$ in terms of $u$ and $v$ and use the Pythagorean identity to eliminate $\theta$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2651865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equation, we get $\cos x = \frac{1}{3}$
The general solution of the equation is given by $\cos x = 2n\pi \pm \cos^{-1}\frac{1}{3}$
For having $7$ distinct solutions, $n$ can have value = 0,1,2,3
So, from here we can conclude that $n$ is anything but greater than $6$. So, according to the options given in the questions, the greatest value of $n$ should be $13$. But the answer given is $14$. Can anyone justify?
|
simplifying the given equation we get
$$-5\cos(x)^2-5\cos(x)+4=0$$
with $$\cos(x)=t$$ we get the quadratic equation
$$-5t^2-5t+4=0$$
solving this we get
$$t_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{105}}{10}$$
can you finish?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2654538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Partial fraction decomposition of $\frac{5z^4 + 3z^2 + 1}{2z^2 + 3z + 1}$ So I'm working on this problem:
$$\frac{5z^4 + 3z^2 + 1}{2z^2 + 3z + 1}$$ where $z = x + yi$
It might be that my brain is just blanking right now but, how would one find the partial fraction decomposition of this expression? I tried long division but that didn't simplify out too nicely.
|
First of all, you should divide $5z^4+3z^2+1$ by $2z^2+3z+1$, getting$$5z^4+3z^2+1=\left(\frac52z^2-\frac{15}4z+\frac{47}8\right)\left(2z^2+3z+1\right)-\frac{111}8z-\frac{39}8.$$So$$\frac{5z^4+3z^2+1}{2z^2+3z+1}=\frac52z^2-\frac{15}4z+\frac{47}8-\frac{\frac{111}8z+\frac{39}8}{(2z+1)(z+1)}.$$Now, find numbers $a$ and $b$ such that$$\frac{\frac{111}8z+\frac{39}8}{(2z+1)(z+1)}=\frac a{2z+1}+\frac b{z+1}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2655834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Prove polynomial identity $x^k-1=(x-1)(x^{k-1}+x^{k-2}+\ldots+1).$ I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below:
$(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$
|
This one cries out for a simple proof by induction:
If $k = 1$ we evidently have
$x^1 - 1 = (x^1 - 1)(1), \tag 1$
and if $k = 2$:
$x^2 - 1 = (x -1)(x + 1) = (x - 1)\left ( \displaystyle \sum_0^1 x^i \right ); \tag 2$
if we now suppose that the formula holds for some positive $m \in \Bbb Z$,
$x^m - 1 = (x - 1) \left ( \displaystyle \sum_0^{m - 1} x^i \right ), \tag 3$
then
$x^{m + 1} - 1 = x^{m + 1} - x^m + x^m - 1$
$= x^m(x - 1) + (x - 1) \left ( \displaystyle \sum_0^{m - 1} x^i \right ) = (x - 1) \left ( x^m + \displaystyle \sum_0^{m - 1} x^i \right ) = (x - 1) \left ( \displaystyle \sum_0^m x^i \right ), \tag 4$
which shows that the formula
$x^k - 1 = (x - 1) \left ( \displaystyle \sum_0^{k -1} x^i \right ) \tag 5$
is valid for all positive integers $k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2657227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}
Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake?
Thank you.
|
Prove that $n^2-2n+7$ is even then $n+1$ is even.
If $n^2-2n+7$ is even, then $n^2-2n$ is odd.
For all even numbers, this does not work since $(even)^2-2(even)$ always results in an even number.
Therefore, the number $n$ must be an odd number. Since $n$ is odd, $n+1$ is even.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2662554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
Different ways of solving inequation - different solutions? After 'solving' the following exercise, I entered the solution in a website that gave me the graph and I noticed that It wasn't the same as the solution it gave me for the 'original' inequation.
The inequation is:
$5-\frac{4-5x}{2}<3\left(2x-1\right)$
And I solved it like:
1)
$\frac{10-4+5x}{2}*\frac{1}{2x-1}<3$
2)
$\frac{6+5x}{4x-2}-\frac{3(4x-2)}{4x-2}<0$
3)
$\frac{12-7x}{4x-2}<0$
Then I solved it and I got:
Solution= $]-\infty ,\frac{1}{2}[ U [\frac{12}{7},+\infty[$
But, it's wrong, it should be: $[\frac{12}{7},+\infty[$
I then did the exercise in a different way:
1) $\frac{6+5x}{2}-3(2x-1)<0$
2)$\frac{6+5x}{2}-\frac{2*3(2x-1)}{2}<0$
3)$\frac{12-7x}{2}<0$
And the result is now the correct one: $[\frac{12}{7},+\infty[$
Can someone explain me why there are these differences between the first way of solving and the second one?
|
With inequalities you need to be careful because when you multiply for negative value inequality reverses. For this reason the first attempt gives a wrong result and the second was correct.
Note that you can also use first method but you need to distiguish the cases, notably
*
*for $2x-1>0$
$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies \frac{10-4+5x}{2}\cdot\frac{1}{2x-1}<3$$
*
*for $2x-1<0$
$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies \frac{10-4+5x}{2}\cdot\frac{1}{2x-1}>3$$
*
*for $2x-1=0$
$$5-\frac{4-5x}{2}<3\left(2x-1\right) \implies 5-\frac{4-5\frac12}{2}<0\implies 5+\frac34<0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2663299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Binomial Harmonic Numbers Prove this equation for $0 \leq m \leq n$:
$$
\frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n - H_{n-m}
$$
where $H_k$ denotes the k-th harmonic number $\left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right)$.
Tried to use Abels partial summation $\big(\sum_{k=1}^m a_k b_k = a_m \sum_{k=1}^m - \sum_{k=1}^{m-1} (a_{k+1}-a_k)\sum_{i=1}^k b_i \big)$, but it leads to nowhere.
|
This solution is similar to an inductive proof with respect to $m$.
Let the sum of interest be
$$s_{n,m}=\frac{1}{\binom{n}{m}} \sum _{k=1}^m \frac{1}{k} \binom{n-k}{n-m}$$
For $m=1$ we find
$$s_{n,1}=\frac{1}{n}$$
The difference of $s$ with respect to $m$ turns out to be surprisingly simple:
$$s_{n,{m+1}}-s_{n,m}=\frac{1}{n-m}\tag{1}$$
so that
$$s_{n,2}=\frac{1}{n}+\frac{1}{n-1}$$
$$s_{n,3}=\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}$$
$$...$$
$$s_{n,m}=\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+...+\frac{1}{n-m+1}$$
The last expression is easily identified as the difference between two harmonic numbers. Hence
$$s_{n,m}=H_{n}-H_{n-m}$$
as requested.
And now we prove the recurence relation (1) using the definition
$$\binom{n}{m}=\frac{n!}{m!\;(n-m)!}$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}
s_{n,{m+1}}-s_{n,m}
&=& \frac{1}{\binom{n}{m+1}} \sum_{k=1}^{m+1} \frac{1}{k}\binom{n-k}{n-m-1} - \frac{1}{\binom{n}{m}} \sum_{k=1}^{m} \frac{1}{k}\binom{n-k}{n-m}\\
&=& \frac{1}{\binom{n}{m}}\left( \frac{m+1}{n-m}\left( \sum_{k=1}^m \frac{1}{k} \binom{n-k}{n-m-1} +\frac{1}{m+1}\right) -\sum_{k=1}^m \frac{1}{k} \binom{n-k}{n-m}\right) \\
&=& \frac{1}{\binom{n}{m}}\left( \frac{1}{n-m}+ \sum_{k=1}^m \frac{1}{k}\binom{n-k}{n-m} \left( \frac{m+1}{m+1-k} -1 \right) \right) \\
&=& \frac{1}{\binom{n}{m}}\left( \frac{1}{n-m}+ \sum_{k=1}^m \frac{1}{m+1-k}\binom{n-k}{n-m} \right) \\
&=& \frac{1}{n-m}\frac{1}{\binom{n}{m}}\left( 1+\sum_{k=1}^m \frac{n-m}{m+1-k}\binom{n-k}{n-m} \right) \\
&=& \frac{1}{n-m}\frac{1}{\binom{n}{m}}\left( 1+\sum_{k=1}^m \binom{n-k}{m-k+1} \right) =\frac{1}{n-m} \\
\end{eqnarray*}$
In the last step we have used the identity
$\begin{eqnarray*}
\binom{n}{m}
&=& \sum_{k=1}^m \binom{n-k}{m-k+1} +1 = \sum_{k=1}^{m+1} \binom{n-k}{m-k+1} \\
\end{eqnarray*}$
which in turn results from iterating the basic recursion of the binomial coefficients
$\begin{eqnarray*}
\binom{n}{m}
&=& \binom{n-1}{m}+\binom{n-1}{m-1} \\
&=& \binom{n-1}{m}+\binom{n-2}{m-1}+\binom{n-2}{m-2} \\
&=& \binom{n-1}{m}+\binom{n-2}{m-1}+...+\binom{n-m}{1} +\binom{n-m-1}{0}\tag{2}\\
\end{eqnarray*}$
This completes the proof of (1) and hence the solution.
Comments
1) I feel that a much shorter proof should be possible starting from (2), but I didn't find it.
2) Interesting related formula (Derivative of binomial coefficients)
$$\frac{d}{dn}\ln\binom{n}{k} = H_{n} - H_{n-k}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2668419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$ $A=\begin{bmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{bmatrix}$
here after row reduction
$\begin{bmatrix}1&0&2&1\\0&1&-1&2\\0&0&1&2\\0&0&0&0\end{bmatrix}$
clearly determinant is zero
but how can I find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$
$A^2=\begin{bmatrix}6&3&4&2\\0&0&0&0\\4&0&6&0\\2&0&3&0\end{bmatrix}$
so I try to row reduce $A^2$ using gauss elimination
$A^2=\begin{bmatrix}1&0&\frac{3}{2}&0\\0&1&\frac{-5}{3}&\frac{2}{3}\\0&0&0&0\\0&0&0&0\end{bmatrix}$
so $x_3$ and $x_4$ is free variable, but in this question I'm not sure what I need to find and should I find Eigen value first?
|
HINT
Take a vector in $\ker(A^2)$ in the general form $v=sv_1+tv_2$ then solve for $Av\neq 0 $ to find coefficients $s$ and $t$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2668782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How to construct a triangle given two sides and their bisector? Suppose I have two triangle sides $AB$ and $AC$, and the length of the angle bisector of $A$. How can I construct (straightedge and compass) the triangle? (This question is from one of the earlier Moscow Math Olympiads.)
|
Let
the length
of the angle bisector of $A$
$|AD|=d$
and the side lengths
$|BC|=a,|AC|=b,|AB|=c$,
$|BD|=m$.
Then according to
Stewart's Theorem
for $\triangle ABC$,
\begin{align}
c^2(a-m)+b^2m
-a(d^2+(a-m)m)
&=0
\tag{1}\label{1}
.
\end{align}
By the law of sines,
\begin{align}
\triangle ABD:\quad
\frac{m}{\sin\tfrac\alpha2}
&=
\frac{c}{\sin\delta}
,\\
\triangle ADC:\quad
\frac{a-m}{\sin\tfrac\alpha2}
&=
\frac{b}{\sin\delta}
,\\
\text{hence, }\quad
\frac{m}{a-m}&=\frac{c}{b}
,\\
a&=\frac{m(c+b)}{c}
,
\end{align}
and \eqref{1} becomes
\begin{align}
\frac{m(c+b)(c^2b-m^2b-d^2c)}{c^2}
&=0
,\\
\end{align}
\begin{align}
m^2&=
\frac{c(bc-d^2)}{b}
.
\end{align}
Given $m$ we can found that
\begin{align}
a^2&=\frac{(bc-d^2)(b+c)^2}{bc}
,
\end{align}
hence, the length of the missing side $a$
can be constructed from known values $b,c,d$,
for example, using
Intersecting_chords_theorem
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2677084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Adding Absolute value to a complex number: $ z+| z|=2+8i$ I would like to know my error in this problem.
Find the complex number such that:
$$ z+|z|=2+8i$$
So far, I have:
$$
\begin{split}
a+bi+\sqrt{a^2+b^2} &= 2 + 8i\\
a^2-b^2+a^2+b^2&=4-64\\
2a^2 -b^2 + b^2&=-60\\
a^2&=-30
\end{split}
$$
But I should end up with
$$a^2=-15$$
No matter how hard I try, I can't seem to find what I did wrong. Any suggestions?
|
$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$
$$a^2-b^2+a^2+b^2=4-64$$
This is where you went wrong, it looks like you tried to square both sides and take the real part but you made the wrong assumption that all the cross-multiplication terms would be imaginary.
$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$
$$a^2-b^2+(a^2+b^2) + 2abi + 2a\sqrt{a^2+b^2} + 2b\sqrt{a^2+b^2}i = 4 - 64 +16i$$
$$a^2-b^2+a^2+b^2 + 2a\sqrt{a^2+b^2} = 4 - 64 $$
$$2ab + 2b\sqrt{a^2+b^2} = 16$$
Which frankly doesn't look much like progress towards a soloution. We have a pair of equations but both of them still contain our problematic square root.
To eliminate the square root we need to move it to it's own side of the equation before squaring.
$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$
$$\sqrt{a^2+b^2} = (2 - a)+ (8- b)i$$
$$a^2+b^2 =(2 - a)^2 - (8- b)^2 + 2(2 - a)(8- b)i $$
$$a^2+b^2 = a^2 - 4a + 4 - b^2 +8b -64 $$
$$2b^2 =- 4a + 4 +8b -60 $$
$$0 = (2 - a)(8- b)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2677177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
}
|
A difficult improper integral
What is the value of the $$\int_{1}^{\infty}\frac{dx}{ x^2(e^x+1)}?$$
I got $2$ as a result. Wolfram has given $0.11111$ as a result. I am confused.
Is it correct? Or incorrect ? Please tell me...
Thanks in advance
|
As an alternative expression, we have
$$ \int_{1}^{\infty} \frac{dx}{x^2(e^x + 1)}
= C + \int_{0}^{1} \left( \frac{1}{e^x + 1} \right)'' \log x \, dx $$
where $C$ is given by
\begin{align*}
C
&= \frac{1}{4} + \frac{1}{1+e} + \frac{\gamma}{4} + \frac{\log 2}{3} - 3\log A \\
&\approx 0.14803096668067398473\cdots.
\end{align*}
and $\gamma$ is the Euler-Mascheroni constant and $A$ is the Glaisher-Kinkelin constant. So if we plug the Taylor expansion of $\frac{1}{e^x+1} = \frac{1}{2}(1-\tanh(x/2))$, we obtain
$$ \int_{1}^{\infty} \frac{dx}{x^2(e^x + 1)}
= C - \sum_{n=2}^{\infty} (-1)^n \frac{2n-1}{n-1} \cdot \frac{2^{2n}-1}{(2\pi)^{2n}} \zeta(2n). $$
Finally, expanding $\zeta(2n) = \sum_{k=1}^{\infty} \frac{1}{k^{2n}} $ and interchanging the order of summation, we find that
\begin{align*}
\sum_{n=2}^{\infty} (-1)^n \frac{2n-1}{n-1} \cdot \frac{2^{2n}-1}{(2\pi)^{2n}} \zeta(2n)
&= \frac{3-e}{4(1+e)} + \sum_{k=1}^{\infty} \frac{1}{(\pi k)^2} \log\left(1 + \frac{1}{(\pi k)^2} \right) \\
&\hspace{6em} - \sum_{k=1}^{\infty} \frac{1}{(2\pi k)^2} \log\left(1 + \frac{1}{(2\pi k)^2} \right)
\end{align*}
which might be better suited for studying this quantity.
Addendum. Here is a numerical computation:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2677552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Simplification algebraic of a cube root. I am trying to simplify this:
$$\frac{1000}{\pi \cdot (\frac{500}{\pi})^{\frac{2}{3}}}$$
and I think it becomes:
$$2 \cdot \sqrt[3]{\frac{500}{\pi}}$$
I basically thought we cube root the $\frac{500}{\pi}$ and then multiply the denominator by $\frac{500}{\pi}$ which could cancel out some stuff. This is how I thought about it:
$$\frac{1000}{\pi \cdot \sqrt[3]{\frac{500}{\pi}} \cdot \frac{500}{\pi}}$$
$$ = \frac{1000 \cdot \pi}{\pi \cdot \sqrt[3]{\frac{500}{\pi}} \cdot 500}$$
Is there a better way to do this cancellation?
|
I would do it like this:
$$\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{500^2}}{\sqrt[3]{\pi^2}}}=\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{500^2}}{\sqrt[3]{\pi^2}}\cdot\dfrac{\sqrt[3]\pi}{\sqrt[3]{\pi}}}=\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{250000\pi}}{\pi}}=\dfrac{1000}{\sqrt[3]{250000\pi}}=\dfrac{1000}{50\sqrt[3]{2\pi}}=\dfrac{20}{\sqrt[3]{2\pi}}\cdot\dfrac{\sqrt[3]{4\pi^2}}{\sqrt[3]{4\pi^2}}$$
$$=\dfrac{20\sqrt[3]{4\pi^2}}{2\pi}=\dfrac{10\sqrt[3]{4\pi^2}}{\pi}$$
The part where you are going wrong is where you are disregarding the square inside the cube root of $\frac {500}{\pi}$, and you are just "taking it out" of the cube root to the denominator.
This is the step where you have gone wrong:$$\dfrac{1000}{\pi\cdot \sqrt[3]{\frac{500}{\pi}}\cdot \frac {500}{\pi}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2683213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}=-ie^{
\frac{i2 \pi k}{11}}$$
so,$$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=-i \sum_{k=1}^{10}e^{
\frac{i2 \pi k}{11}}=-i \times e^{
\frac{i2 \pi }{11}} \times \frac{{\{e^{
\frac{i2 \pi }{11}}\}}^{10} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
$$=-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
now how can i proceed and simplify the result?
Answer $=i$
|
Denote the required sum value by $S$. Multiplying by $i$ we get
$$iS = \sum_{k=1}^{10}\bigg(\cos \frac{2\pi k}{11} + i\sin \frac{2\pi k}{11}\bigg)$$
Now in the summation introduce one extra term corrsponding to $k=11$ whose value will be 1. We get LHS to be $iS+1$. The RHS now is a 11-term summation which is actually the sum of all the 11th roots of unity which is zero.
So we get $iS+1=0$. So it follows that $S=i$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2683421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Find the smallest distance between point and ellipsoid
Find the smallest distance between the points on the ellipsoid $x^2+2y^2+z^2=16$ and the point $(0,0,1)$.
Answer:
Let $(x,y,z)$ be the closest point of the ellipsoid. The distance is
$$\sqrt{x^2+y^2+(z-1)^2}$$
Let $f(x,y,z)=x^2+y^2+(z-1)^2$. Construct the Lagrangian:
$$L(x,y,z)=x^2+y^2+(z-1)^2+\lambda (x^2+2y^2+z^2-16)$$
For smallest distance,
$$L_x=0 \Rightarrow x(\lambda+1)=0\\
L_y=0 \Rightarrow y(2 \lambda+1)=0\\
L_z=0 \Rightarrow z(\lambda+1)=1$$
The 3rd equation implies that $\lambda\ne-1$. Thus we have $x=0$. Now 2nd and 3rd equation solution gives $y=0$ or $\lambda=-\frac12$ and $z=\frac1{1+\lambda}$.
But I can not find the final solution .
Help me out
|
Alternative way:
We have found that $x=0$, and we need $|z| \le 4$,
\begin{align}x^2+y^2+(z-1)^2&=\frac{16-z^2}2+(z-1)^2 \\
&=\frac{16+z^2-4z+2}{2}\\
&=\frac{(z-2)^2-4+16+2}{2}\\
&=\frac{(z-2)^2}{2}+7\end{align}
Note that $|2| \le 4$, and hence the smallest distance is $\sqrt{7}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2684560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?
So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$
$$128x\left(1-\frac{2x^2}{9}\right)$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals $0$ when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.
Is this valid? If so why? Does squaring not cause any problems?
|
We all know that the quadrangle of maximal area inscribed in a disc is a square, and that this square covers ${2\over\pi}$ of the area of the disc. Your ellipse has semiaxes $2$ and $3$, hence area $6\pi$. The largest rectangle inscribed in this ellipse therefore has area
$$\leq{2\over\pi}\cdot 6\pi=12\ ,$$
and this value is attained for a suitable axis-aligned rectangle.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
\end{align*}
and I'm stuck, should I divide $63$ by $9$ to get $7$? but it does not comply to the given equation as when $x=7$, it would become $16 \pmod{47}$.
|
Using the Extended Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&5&4&2\\\hline
1&0&1&-4&9\\
0&1&-5&21&-47\\
47&9&2&1&0\\
\end{array}
$$
shows that $21\cdot9-4\cdot47=1$ so we want $63\cdot9-12\cdot47=3$ which can be reduced, by subtracting $47$ from $63$ and $9$ from $12$, to
$$
16\cdot9-3\cdot47=3
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2687806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$ How can we find the following limit? $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}\qquad \alpha>0$$
By using the polar coordinate, we get
$$\lim_{r\to 0}r^{\alpha+2}\frac{\cos^\alpha\theta \sin^4\theta}{\cos^2\theta+r^2\sin^4{\theta}}=0$$ if $\theta\notin\{\frac{\pi}{2}+\pi k:k\in\Bbb Z\}$.
Now, if $\theta = \frac{\pi}{2}+\pi k$ for some $k\in\Bbb Z$, then we get $$\lim_{(0,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0.$$
Can we conclude that $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}=0?$$
|
From the AM-GM inequality we have $$\left|\frac{x^ay^b}{x^2+y^4}\right|\le \frac12 |x|^{a-1}|y|^{b-2}$$
where we assume that either $a$ is such that $x^a\in \mathbb{R}$ for $x$ in a neighborhood of $0$ or that the limit is taken as $(x,y)\to (0^+,0)$.
And equality holds when $x=\pm y^2$. Hence, we find
$$\lim_{(x,y)\to(0,0)}\frac{x^ay^b}{x^2+y^4}=0$$
whenever we have $2a+b>4$. The limit fails to exist otherwise.
In the case at hand, $a=\alpha$ and $b=4$ and we see that the limit is $0$ for $\alpha>0$ and fails to exist otherwise.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2688350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
|
You should simplify a 2, a 3, and a 5. This is equivalent to find the number of positive divisors of
$$
\frac{2079000}{30} = 69300 = 2^2 \cdot 3^2\cdot 5^2\cdot 7\cdot 11
$$
which is
$
3^3 \cdot 2^2.
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
}
|
The logarithmic inequality $$ \log_{3x+7}{(9+12x+4x^2)}+ \log_{2x+3}{(6x^2+23x+21)} \ge 4$$
The logarithmic inequality is defined for: $ x \in (-3/2, -1) \cup(-1, \infty)$.
First, I supposed that my solutions are in the interval $(-3/2, -1)$. Following this interval for which the logarithmic function $t$ are decreasing and multiplying both sides of the inequality by $\log_{2x+3}{(3x+7)}$, I obtained:
$$ 2+ [1+\log_{2x+3}{(3x+7)}-\log_{2x+3}{6}]\cdot\log_{2x+3}{(3x+7)} \le4\log_{2x+3}{(3x+7)}\\ \log_{2x+3}{(3x+7)}=t<0\\t^2-(3+\log_{2x+3}{6})t+2 \le0$$
And here I got stuck. This inequailty is intended to be solved witouth a calculator, is it possible to continuie doing the inequality withouth it? And what I did wrong?
|
Hint:
$$9+12x+4x^2=(2x+3)^2$$
$$ 6x^2+23x+21=(2x+3)(3x+7)$$
$$4\le\log_{3x+7}(2x+3)^2+\log_{2x+3}(2x+3)(3x+7)=2\log_{3x+7}(2x+3)+1+\log_{2x+3}(3x+7)$$
As $\log(y)$ is real for $y>0,$
we need $2x+3, 3x+7>0\implies x>$max$(-3/2,-7/3)=-3/2$
Now if $\log_{3x+7}(2x+3)=a,$
$$4\le2a+1+\dfrac1a$$
$$\iff0\le2a+\dfrac1a-3=\dfrac{2a^2-3a+1}a=\dfrac{(2a-1)(a-1)}a$$
If $\dfrac{(2a-1)(a-1)}a=0, a=1$ or $\dfrac12$
Else we need $a(2a-1)(a-1)>0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2690411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How I can solve $s(n)=n+s(n-1)$ by iteration method? $$
s(n)=
\begin{cases}
0, \text{if $n=0$}\\
n+s(n-1), \text{if $n>0$}\\
\end{cases}
$$
Using the relation
\begin{align}
s(n) &= n+s(n-1) \\
&= n+n-1+s(n-2) \\
&= n+n-1+n-2+s(n-3) \\
&= \dots \\
&= n+n-1+n-2+\dots+1 \\
&= \dfrac{n^2+n}2 \\
&= \theta(n^2)
\end{align}
Is it right?
|
$s(n)=n+s(n-1)$
$s(n-1)=n-1 + s(n-2)$
$s(n-2)=n-2 + s(n-3)$
$s(n-3)=n-3 + s(n-4)$
.
.
.
$s[n-(n-1)]=n-(n-1) + s(0)$
⇒ $s(n)=n\times n -\frac{(n-1)(n-1+1)}{2}=n^2-\frac{(n)(n-1)}{2}=\frac{n^2+n}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2691632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
is $\ c= \big( \overline{x+yi} \big)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $? is $\ c= \bigg( \overline{x+yi} \bigg)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $?
EDIT
$c$ is complex, but not real.
|
You have $c=a+ib$, then $c^{\frac 12}=[r(\cos x+i \sin x)]^{\frac 12}$, where $r=\sqrt{a^2+b^2}$ and $x=arctan(\frac{b}{a})$
Now by Moivre's formula
$c^{\frac 12}=
r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin
\left( \dfrac {x+2\pi k}{2}\right )\right ],k=0,1$.
Finally when calculating the conjugate you will have to
$\overline{c^{\frac 12}}=
\overline{r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin
\left( \dfrac {x+2\pi k}{2}\right )\right ]} ,k=0,1$.
$\overline{c^{\frac 12}}=
\overline{r^{\frac 12}} \overline{\left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin
\left( \dfrac {x+2\pi k}{2}\right )\right ]} ,k=0,1$.
$\overline{c^{\frac 12}}=
r^{\frac 12} \overline{\left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin
\left( \dfrac {x+2\pi k}{2}\right )\right ]} ,k=0,1$.
$\overline{c^{\frac 12}}=
r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )-i \sin
\left( \dfrac {x+2\pi k}{2}\right )\right ] ,k=0,1$.
On the other hand
$c=\overline{(x+iy)}^{2}=(x-iy)^2=x^2 -2ixy+i^2 y=x^2-y^2-2xyi$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2691745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Seating $m$ people in $m^2$ chairs at a round table I am having difficulty with the following counting problem.
Suppose there are $m^2$ indistinguishable chairs at a round table. There are $m$ people to be seated. Each arrangement is equally likely. Let $X_m$ denote the number of pairs of people sitting adjacent to one another (i.e., no seats between them). Compute the following limits:
$$\lim_{m\rightarrow\infty}E[X_m], \qquad \lim_{m\rightarrow\infty}\operatorname{Var}(X_m).$$
The total number of arrangements can be reasoned out to be
$$(m-1)!\binom{m^2-1}{m-1} = \frac{(m^2-1)!}{(m^2-m)!}.$$
However, I am not sure how to count the number of ways that $X_m = 0,1,\cdots,\dfrac{m}{2}$. Doing this for $m = 1$ and $m = 2$ are pretty simple, i.e.
$$P(X_1 = 0) = 1, \qquad E[X_1] = 0.\\
P(X_2 = k) = \begin{cases}\dfrac{1}{3}; & k = 0 \\ \dfrac{2}{3}; & k = 1\end{cases}, \qquad E[X_2] = \frac{2}{3}.$$
EDIT: Could I possibly conjecture that $E[X_m] = \dfrac{m}{m+1}$?
|
First, it makes no difference whether the chairs are distinguishable or not because each configuration in the scenario of indistinguishable chairs corresponds to exactly $m^2$ configurations in the distinguishable scenario (Note that there are $m^2$ cyclic permutations from $\{1, \cdots, m^2\}$ to itself). Thus in the derivation below, chairs are clockwise numbered as $C_1, \cdots, C_{m^2}$, and $C_{m^2 + 1} := C_1$.
Now for $1 \leqslant k \leqslant m^2$, define$$
Y_k = \begin{cases}
1; & \text{both } C_k \text{ and } C_{k + 1} \text{ are occupied}\\
0; & \text{otherwise}
\end{cases}.
$$
Note that $X = \sum\limits_{k = 1}^{m^2} Y_k$. For any $a \geqslant b \geqslant 0$, denote $P(a, b) = \dfrac{a!}{(a - b)!}$. For $1 \leqslant k \leqslant m^2$,$$
P(Y_k = 1) = \frac{P(m, 2) P(m^2 - 2, m - 2)}{P(m^2, m)} = \frac{1}{m(m + 1)}.
$$
For $1 \leqslant j < k \leqslant m^2$, if $k = j + 1$ or $(j, k) = (1, m^2)$, then$$
P(Y_j = Y_k = 1) = \frac{P(m, 3) P(m^2 - 3, m - 3)}{P(m^2, m)} = \frac{m - 2}{m(m + 1)(m^2 - 2)}.
$$
Otherwise,$$
P(Y_j = Y_k = 1) = \frac{P(m, 4) P(m^2 - 4, m - 4)}{P(m^2, m)} = \frac{(m - 2)(m - 3)}{m(m + 1)(m^2 - 2)(m^2 - 3)}.
$$
Therefore,$$
E(X) = \sum_{k = 1}^{m^2} E(Y_k) = \sum_{k = 1}^{m^2} P(Y_k = 1) = \frac{m}{m + 1},
$$\begin{align*}
E(X^2) &= E\left( \sum_{k = 1}^{m^2} Y_k^2 + 2 \sum_{1 \leqslant j < k \leqslant m^2} Y_j Y_k \right) = \sum_{k = 1}^{m^2} E(Y_k^2) + 2 \sum_{1 \leqslant j < k \leqslant m^2} E(Y_j Y_k)\\
&= \sum_{k = 1}^{m^2} P(Y_k = 1) + 2 \sum_{1 \leqslant j < k \leqslant m^2} P(Y_j = Y_k = 1)\\
&= \frac{1}{m(m + 1)} · m^2 + 2 \Biggl( \frac{m - 2}{m(m + 1)(m^2 - 2)} · m^2\\
&\quad + \frac{(m - 2)(m - 3)}{m(m + 1)(m^2 - 2)(m^2 - 3)} · \left( \frac{1}{2} m^2 (m^2 - 1) - m^2 \right) \Biggr)\\
&= \frac{m^2 (2m - 3)}{(m + 1)(m^2 - 2)},
\end{align*}$$
D(X) = E(X^2) - (E(X))^2 = \frac{m^2 (m^2 - m - 1)}{(m + 1)^2 (m^2 - 2)},
$$
and$$
\lim_{m → ∞} E(X) = 1, \quad \lim_{m → ∞} D(X) = 1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2692804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Significance of the matrix $A^2$? Say you have destinations/places, $A,B,C,D$ and $E$ that are points that you can travel to and from. The only ways to travel between this is by the following schematic:
$A \rightarrow B \\ A \rightarrow E \\ B \rightarrow C \\ B \rightarrow D \\ C \rightarrow A \\ D \rightarrow A \\ D \rightarrow E \\ E \rightarrow B \\ E \rightarrow D$
The links are the arrows and the nodes are the destinations. The connection-matrix is created by the rule
$$a_{ij}=\left\{
\begin{array}{rcr}
1&,& \text{if there is a link from $i$ to $j$.} \\
0&,& \text{else.} \\
\end{array}
\right.$$
In this case, we get the connection matrix
$$A_c=\pmatrix{0&1&0&0&1 \\0&0&1&1&0 \\ 1&0&0&0&0 \\ 1&0&0&0&1 \\ 0&1&0&1&0}.$$
Computing $(A_c)^2$ and $A+(A_c)^2$ I get
$$(A_c)^2=\pmatrix{0&1&1&2&0 \\2&0&0&0&1 \\ 0&1&0&0&1 \\ 0&2&0&1&1 \\ 1&0&1&1&1}, \quad A+(A_c)^2=\pmatrix{0&2&1&2&1 \\2&0&1&1&1 \\ 1&1&0&0&1 \\ 1&1&1&2&1 \\ 0&1&0&1&0}.$$
Questions: What is the explanation/significance of the elements in $(A_c)^2$ and $A+(A_c)^2$? What do they intuitivley mean in terms of flights between the destinations?
How should I think and what should I do in order to derive/realize the answer to the above?
|
The square of an adjacency matrix like $A_c$ can be interpreted as the number of paths of length $2$. For example, the $2$ in the $2$nd row and $1$st column shows that there are two paths of length exactly $2$ from $B$ to $A$ ($B \to C \to A$ and $B \to D \to A$). Correspondingly $A_c^2 + A_c$ gives you all the paths of at most length $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2696090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Showing a function grows like a certain power of $n$ Consider $f(n) = \frac{1}{n} {n \choose (n+1)/2} \frac{1}{2^n}$ for odd $n$. I suspect that $\sum_{n \geq 1} n^{\beta}f(n)= \infty$ for $\beta \geq 1/2$. For this purpose, it would be enough to show that $f(n) \sim n^{-3/2}$. How can I show this last statement?
|
Convert the polynomial in the representation with Gamma function
$$
\eqalign{
& b(n) = \left( \matrix{
n \cr
n/2 + 1/2 \cr} \right) = {{\Gamma \left( {n + 1} \right)} \over {\Gamma \left( {n/2 + 3/2} \right)\Gamma \left( {n/2 + 1/2} \right)}} = \cr
& = {{\Gamma \left( {n + 1} \right)} \over {\left( {n/2 + 1/2} \right)\Gamma \left( {n/2 + 1/2} \right)^{\,2} }} = {{\Gamma \left( {2\left( {n/2 + 1/2} \right)} \right)} \over {\left( {n/2 + 1/2} \right)\Gamma \left( {n/2 + 1/2} \right)^{\,2} }} \cr}
$$
Apply the duplication formula
$$
\Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
$$
to get
$$
b(n) = {{{{2^{\,n} } \over {\sqrt \pi }}\Gamma \left( {n/2 + 1} \right)} \over {\left( {n/2 + 1/2} \right)\Gamma \left( {n/2 + 1/2} \right)}} = {{2^{\,n} } \over {\left( {n/2 + 1/2} \right)\sqrt \pi }}{{\Gamma \left( {n/2 + 1} \right)} \over {\Gamma \left( {n/2 + 1/2} \right)}}
$$
Now apply the Stirling approximation
$$
\Gamma (z) \approx \sqrt {\,{{2\,\pi } \over z}\,} \left( {{z \over e}} \right)^{\,z}
$$
to arrive to
$$
\eqalign{
& b(n) \approx {{2^{\,n} } \over {\left( {n/2 + 1/2} \right)\sqrt {\pi \,e} }}\sqrt {\,{{n/2 + 1/2} \over {n/2 + 1}}\,} {{\left( {n/2 + 1} \right)^{\,n/2 + 1} } \over {\left( {n/2 + 1/2} \right)^{\,n/2 + 1/2} }} \approx \cr
& \approx {{2^{\,n} } \over {\sqrt {\left( {n/2 + 1/2} \right)} \sqrt {\pi \,e} }}\left( {{{n/2 + 1} \over {n/2 + 1/2}}} \right)^{\,\left( {n + 1} \right)/2} \approx \cr
& \approx {{2^{\,n + 1/2} } \over {\sqrt {\left( {n + 1} \right)} \sqrt {\pi \,e} }}\left( {1 + {1 \over {n + 1}}} \right)^{\,\left( {n + 1} \right)/2} \approx \cr
& \approx {{2^{\,n + 1/2} } \over {\sqrt {\left( {n + 1} \right)} \sqrt \pi }} \cr}
$$
Inserting the other terms, finally you get
$$
f(n) = {1 \over {n\,2^{\,\,n} }}b(n) \approx \sqrt {{2 \over \pi }} {1 \over {n\sqrt {n + 1} }} \approx \sqrt {{2 \over \pi }} \;n^{\, - 3/2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2697091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Derive Newton QUADRATE (high precision) of the form $\int_a^b f(x)dx ≈ w0 f(a)+w1 f(b)+w2 f′(a)+w3 f′(b)$ Derive a formula of the form
$$\int_a^b f(x)dx ≈ w0 f(a)+w1 f(b)+w2 f′(a)+w3 f′(b)$$
that is exact for polynomials of the highest degree possible.
For a simpler version like $$\int_0^1 f(x) dx ≈ w1f(0) + w2f(x1)$$ i know how to solve: i would need to solve a system of equations
$$f(x) = x0 \int_0^1 1 dx = w1 · 1 + w2 · 1$$
$$f(x) = x1 \int_0^1 x dx = w1 · 0 + w2 · x1$$
$$f(x) = x2 \int_0^1 x^2 dx=w1 · 0 + w2 · x1^2$$
would anyone please explain briefly (or hint at) what i wouold do for this problem. thanks.
|
You know that the formula can't be exact for $4^{\text{th}}$ degree polynomials because
$$\int_a^b(x-a)^2(x-b)^2dx\ne0$$
Because the integrand is positive in $(a,b)$, but your quadrature formula predicts it is $0$. Thus all you need is a basis for $\mathcal{P}_3$, the vector space of polynomials of degree at most $3$. Apply your quadrature formula to each basis polynomial and you get $4$ equations in $4$ unknowns. There are $2$ schools of thought here, first is to use a simple basis:
$$\begin{align}\int_a^b1\,dx&=w_0+w_1=2\left(\frac{b-a}2\right)\label{a}\tag{1}\\
\int_a^b\left(x-\frac{b+a}2\right)\,dx&=-\left(\frac{b-a}2\right)w_0+\left(\frac{b-a}2\right)w_1+w_2+w_3=0\label{b}\tag{2}\\
\int_a^b\left(x-\frac{b+a}2\right)^2\,dx&=\left(\frac{b-a}2\right)^2w_0+\left(\frac{b-a}2\right)^2w_1-2\left(\frac{b-a}2\right)w_2+2\left(\frac{b-a}2\right)w_3=\frac23\left(\frac{b-a}2\right)^3\label{c}\tag{3}\\
\int_a^b\left(x-\frac{b+a}2\right)^3\,dx&=-\left(\frac{b-a}2\right)^3w_0+\left(\frac{b-a}2\right)^3w_1+3\left(\frac{b-a}2\right)^2w_2+3\left(\frac{b-a}2\right)^2w_3=0\label{d}\tag{4}\end{align}$$
If we multiply eq$(\ref{b})$ by $\left(\frac{b-a}2\right)^2$ and subtract from eq$(\ref{d})$ we see that
$$2\left(\frac{b-a}2\right)^2w_2+2\left(\frac{b-a}2\right)^2w_3=0$$
So $w_2=-w_3$. Using this result in eq$(\ref{b})$ we have
$$-\left(\frac{b-a}2\right)w_0+\left(\frac{b-a}2\right)w_1=0$$
So $w_0=w_1$. Multiply eq$(\ref{a})$ by $\left(\frac{b-a}2\right)^2$ and subtract from eq$(\ref{c})$ to show that
$$-2\left(\frac{b-a}2\right)w_2+2\left(\frac{b-a}2\right)w_3=-\frac43\left(\frac{b-a}2\right)^3$$
And we can conclude that $w_2=-w_3=\frac{(b-a)^2}{12}$. And from eq$(\ref{a})$, $w_0=w_1=\left(\frac{b-a}2\right)$.
The other school of thought is to construct a fancy basis that avoids systems of equations.
$$\begin{align}w_0&=\int_a^b\frac{2x^3-3(a+b)x^2+6abx+b^2(b-3a)}{(b-a)^3}dx\\
&=\frac1{(b-a)^3}\left(\frac12(b^4-a^4)-(a+b)(b^3-a^3)+3ab(b^2-a^2)+b^2(b-3a)(b-a)\right)\\
&=\frac{b-a}2\end{align}$$
$$\begin{align}w_1&=\int_a^b\frac{2x^3-3(a+b)x^2+6abx+a^2(a-3b)}{(a-b)^3}\\
&=\frac1{(a-b)^3}\left(\frac12(b^4-a^4)-(a+b)(b^3-a^3)+3ab(b^2-a^2)+a^2(a-3b)(b-a)\right)\\
&=\frac{b-a}2\end{align}$$
$$\begin{align}w_2&=\int_a^b\frac{(x-a)(x-b)^2}{(b-a)^2}dx\\
&=\frac1{(b-a)^2}\left(\frac14(b^4-a^4)-\frac13(2b+a)(b^3-a^3)+\frac12(b^2+2ab)(b^2-a^2)-ab^2(b-a)\right)\\
&=\frac{(b-a)^2}{12}\end{align}$$
$$\begin{align}w_3&=\int_a^b\frac{(x-a)^2(x-b)}{(b-a)^2}dx\\
&=\frac1{(b-a)^2}\left(\frac14(b^4-a^4)-\frac13(2a+b)(b^3-a^3)+\frac12(a^2+2ab)(b^2-a^2)-a^2b(b-a)\right)\\
&=-\frac{(b-a)^2}{12}\end{align}$$
So either way seems to work out OK.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2698314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$
If $\alpha,\beta,\gamma$ be a variable and $k$ be a constant such that $a\tan\alpha+b\tan\beta+c\tan\gamma=k$.Then find minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$ is
Try: Using Cauchy Schwarz Inequality:
$(a^2+b^2+c^2)(\tan^2\alpha+\tan^2\beta+\tan^2\gamma)\geq (a\tan\alpha+b\tan\beta+c\tan\gamma)^2$
So we have $$\tan^2\alpha+\tan^2\beta+\tan^2\gamma)\geq \frac{k^2}{a^2+b^2+c^2}$$
Could some help me to solve it without Cauchy Schwarz Inequity .
Please explain, thanks
|
A bit of Geometry.
$x=\tan \alpha$; $ y= \tan \beta$; $z=\tan \gamma$.
Plane: $ax+by +cz=k.$
$d^2 := x^2+y^2+z^2$ is the squared distance from the origin to a point $(x,y,z)$.
Need to find the (perpendicular= minimal) distance of the plane from the origin.
$\vec n = \dfrac{1}{a^2+b^2+c^2}(a,b,c)$ is the unit normal of the plane.
$\vec r(t) = t \vec n.$
Plane:
$\vec n\cdot(\vec r(t) -\vec r_0)=0.$
Choose: $\vec r_0 = (0,0,k/c).$
Find $t:$
$\vec n \cdot (t\vec n -\vec r_0)=0$, or
$t - \vec n \cdot \vec r_0=0.$
Hence:
$t= \dfrac{k}{\sqrt{a^2+b^2+c^2}}.$
Since $\vec n$ is a unit vector $t$ is the distance from the origin to the plane .
We are looking for the squared distance:
$t^2 = \dfrac{k^2}{a^2+b^2+c^2}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2699975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Catalan's constant I am studying the integral $$\int_1^\infty\!\ \frac{\ln(x)}{1+x^{2}}$$ and ran across this representation of :
$$ \frac{1}{1+x^{2}}=\frac{1}{\frac{1}{x^{2}}(1-\frac{1}{x^{2}}+\frac{1}{x^{4}}-\frac{1}{x^{6}}+\frac{1}{x^{8}}...)}$$
where the denominator in the original integral has been rewritten to turn the integral into the following expression :
$$\int_1^\infty\!\frac{\ln(x)}{x^{2}}-\int_1^\infty\frac{\ln(x)}{x^{4}}+\int_1^\infty\frac{\ln(x)}{x^{6}} ...$$
which can then be evaluated by integrating by parts. I can do the integral but am stumped as to how the representation of $$\frac{1}{1+{x^{2}}}= \frac{1}{\frac{1}{x^{2}}(1-\frac{1}{x^{2}}+\frac{1}{x^{4}}-\frac{1}{x^{6}}+\frac{1}{x^{8}}...)}$$
was derived, and am assuming that the purpose was to get powers of $x$ into the denominator as opposed to having $\ln(x)$ multiplying the powers of $x$. Any help would be appreciated.
|
The correct representation is:
$$
\frac{1}{1+x^2}=\frac{1}{x^2}\frac{1}{1+x^{-2}}=\frac{1}{x^2}\sum_{k=0}^\infty\left(-\frac{1}{x^2}\right)^k,
$$
which you incorrectly put in the denominator of RHS.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2700796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Simplifying calculations for hitting probabilities using communicating classes When given a rather elaborate transition matrix for 6 state Markov chain, for example
$$\begin{bmatrix} 0&0&\frac{1}{2}&0&0&\frac{1}{2} \\ \frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&0\\ \frac{1}{3}&0&\frac{1}{3}&0&0&\frac{1}{3} \\ \frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6} \\ 0&0&0&0&1&0 \\ \frac{1}{4}&0&\frac{1}{2}&0&0&\frac{1}{4} \end{bmatrix}, $$
if we wanted to find say the hitting probability that we eventually land at state 6 given that we start at state 2, it seems to me that to calculate this manually using the standard method of finding the lowest non-negative vector solution to
$$ h_{i}^{6} = \begin{cases} 1 & i = 6 \\ \sum_{j} h_{j}^{6}p_{ij} & i \neq 6
\end{cases}$$
where $h_{i}^{6} = \Pr(\text{hitting time to 6 is finite} |X_{0} =i)$, and then setting $i = 2$ seems quite cumbersome and error prone.
Would it be possible to instead recognise that $\{1, 3, 6 \}$ is a closed communicating class and then merge the states together to form a simplified transition matrix? Reason being since the class is closed we merely need to find the probability that $X_{n} \in \{1, 3, 6 \} $ and we are guaranteed that $(X)_{n} = 6$ eventually?
For example the simplified transition matrix inscribing states (1,3,6), (2), (4) and (5) would then become
$$\begin{bmatrix} 1&0&0&0 \\ \frac{2}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5} \\ \frac{1}{2}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6} \\ 0&0&0&1 \end{bmatrix}. $$
Here we can immediately read of that $h_{1,3,6}^{1,3,6} = 1$, $h_{5}^{1,3,6} = 0 $ and solving for the other states becomes a lot easier. Is this a valid strategy to adopt? Is there anything wrong with the reasoning that it is sufficient to find that probability that we enter the closed communicating class to find the probability that we eventually reach a state in that class? Thanks.
|
Partition the state space $S=\{1,2,3,4,5,6\}$ into $S=A\cup B\cup C$ with $A=\{1,3,6\}$, $B=\{2,4\}$, $C=\{5\}$. Then $\mathbb P(X_{n+1}\in A\mid X_n\in A) = 1$ and $\mathbb P(X_{n+1}\in C\mid X_n\in C)=1$, so $A$ and $C$ are closed communicating classes, and since $\mathbb P(X_{n+1}\in A\mid X_n\in B)>0$ and $\mathbb P(X_{n+1}\in C\mid X_n\in B)>0$, $B$ is the class of transient states. For $i\in B$, let
\begin{align}
\tau_{i,A} &= \inf\{n>0:X_n\in A\mid X_0 = i \}\\
\tau_{i,C} &= \inf\{n>0:X_n\in C\mid X_0 = i \}.
\end{align}
Denote $\rho_{i,A}:=\mathbb P(\tau_{i,A}<\infty)$ and $\rho_{i,C}:=\mathbb P(\tau_{i,C}<\infty)$, then
\begin{align}
\rho_{i,A}= \sum_{j\in A}p_{ij} + \sum_{j\in B}p_{ij}\rho_{i,A},
\end{align}
and $\rho_{i,C} = 1 - \rho_{i,A}$. It follows that
\begin{align}
\rho_{2,A} &= \frac 25 +\frac15\rho_{2,A} + \frac15\rho_{4,A}\\
\rho_{4,A} &= \frac 12 + \frac16\rho_{2,A} + \frac16\rho_{4,A},
\end{align}
and hence
$$
\rho_{2,A} = \frac{13}{19},\ \rho_{4,A} = \frac{14}{19},\ \rho_{2,C} = \frac6{19},\ \rho_{4,C} = \frac5{19}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2701368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Stuck on this square root conjugate problem I have to multiply the following:
$((x+h)\sqrt{x+h} - x\sqrt{x}) * ((x+h)\sqrt{x+h} + x\sqrt{x})$
I tried to use $(a - b)(a + b) = a^2 + b^2$ so:
$((x+h)\sqrt{x+h} - x\sqrt{x})((x+h)\sqrt{x+h} + x\sqrt{x}) = (x^2 + h^2)(x + h) - x^2(x)$
and then:
$x^3 + x^2h + h^2x + h^3 - x^3 $
and then:
$x^2h + h^2x + h^3$
But the book is showing the answer as $3x^2h + 3xh^2 + h^3$
I don't know where they are getting those $3$'s from. Can someone please tell me what I'm doing wrong?
|
Your error is in the first line.
You should have (take $a = (x+h)\sqrt{x+h}$ and $b = x\sqrt{x}$ and apply $(a - b)(a + b) = a^2 - b^2$):
\begin{align}
\big((x+h)\sqrt{x+h} - x\sqrt{x}\big)\big((x+h)\sqrt{x+h} + x\sqrt{x}\big) &= \color{blue}{(x + h)^2}(x + h) - x^2(x) \\
&= (x+h)^3 - x^3 \tag{1}
\end{align}
and NOT:
$\big((x+h)\sqrt{x+h} - x\sqrt{x}\big)\big((x+h)\sqrt{x+h} + x\sqrt{x}\big) = \color{blue}{(x^2 + h^2)}(x + h) - x^2(x)$.
Since $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$, you can rewrite $(1)$ as follows:
\begin{align}
(x+h)^3 - x^3 = 3x^2h + 3xh^2 + h^3
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2701634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Proof of sum formula, no induction $$\sum_{k=1}^n k=\frac{n(n+1)}2$$
So I was trying to prove this sum formula without induction. I got some tips from my textbook and got this.
Let $S=1+2+\cdots+n-1+n$ be the sum of integers and $S=n+(n+1)+\cdots+2+1$ written backwards. If I add these $2$ equations I get $2S=(1+n)+(1+n)\cdots(1+n)+(1+n)$ $n$ times.
This gives me $2S=n(n+1) \Rightarrow S=\frac{n(n+1)}2$ as wanted.
However if I changed this proof so that n was strictly odd or strictly even, how might I got about this. I realize even means n must be $n/2$. But I haven't been able to implement this in the proof correctly.
Edit: error in question fixed, also by $n/2$ I mean should I implement this idea somewhere in the proof, cause even means divisible by $2$.
|
Method 1: (requires you to consider whether $n$ is odd or even.)
$S = 1 + 2 + ...... + n$.
Join up the first to term to the last term and second to second to last and so on.
$S = \underbrace{1 + \underbrace{2 + \underbrace{3 +....+(n-2)} + (n-1)} + n}$.
$= (n+1) + (n+1) + .....$.
If $n$ is even then:
$S = \underbrace{1 + \underbrace{2 + \underbrace{3 +..+\underbrace{\frac n2 + (\frac n2 + 1)}+..+(n-2)} + (n-1)} + n}$
And you have $\frac n2$ pairs that add up to $n+1$. So the sum is $S= \frac n2(n+1)$.
If $n$ is odd then:
$S = \underbrace{1 + \underbrace{2 + \underbrace{3 +..+\underbrace{\frac {n-1}2 + [\frac {n+1}2] + (\frac {n+1}2 + 1)}+..+(n-2)} + (n-1)} + n}$
And you have $\frac {n-1}2$ pairs that also add up to $n+1$ and one extra number $\frac {n+1}2$ which didn't fit into any pair. So the sum is $\frac {n-1}2(n+1) + \frac {n+1}2 =(n-1)\frac {n+1}2 + \frac {n+1}2 = (n-1 + 1)\frac {n+1}2n=n\frac {n+1}2$.
Method 1$\frac 12$ (Same as above but waves hands over doing tso cases).
$S = average*\text{number of terms} = average*n$.
Now the average of $1$ and $n$ is $\frac {n+1}2$ and the average of $2$ and $n-1$ is $\frac {n+1}2$ and so on. So the average of all of them together is $\frac {n+1}2$. So $S = \frac {n+1}2n$.
Method 2: (doesn't require considering whether $n$ is odd or even).
$S = 1 + 2 + 3 + ...... + n$
$S = n + (n-1) + (n-2) + ...... + 1$.
$2S = S+S = (n+ 1) + (n+1) + ..... + (n+1) = n(n+1)$>
$S = \frac {n(n+1)}2$.
Note that by adding $S$ to itself this doesn't matter whether $n$ is even or odd.
And lest you are wondering why can we be so sure that $n(n+1)$ must be even (we constructed it so it must be true... but why?) we simply note that one of $n$ or $n+1$ must be even.
So no problem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2702057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Is this sequence of functions uniformly convergent on [0, 2] ?? Define a sequence of functions $f_n : [0,2] \to \Bbb R$ as:
$$f_n(x) = \frac {1-x} {1+x^n}$$
Is this sequence of functions uniformly convergent on $[0,2]$?
|
Let $f_n(x)=\frac{1-x}{1+x^n}$ and $f(x)=\begin{cases}1-x&,0\le x\le 1\\\\0&,1\le x\le 2\end{cases}$
Clearly we have
$$\lim_{n\to \infty}f_n(x)=f(x)$$
Furthermore, we see that
$$|f_n(x)-f(x)|=\begin{cases}\frac{(1-x)x^n}{1+x^n}&,0\le x\le 1\\\\\frac{x-1}{1+x^n}&,1\le x\le 2\end{cases}$$
Next, we have the following estimates for $x\in [0,1]$
$$\begin{align}
\frac{(1-x)x^n}{1+x^n}&\le (1-x)x^n\\\\
&\le \left(\frac{1}{n+1}\right)\left(\frac{n}{n+1}\right)^n\\\\
&<\frac{1}{n+1}\\\\
&<\frac{1}{n-1}\\\\
&<\epsilon
\end{align}$$
whenever $n>1+\frac1\epsilon$.
Similarly, we have the following estimates for $x\in[1,2]$
$$\begin{align}
\frac{x-1}{1+x^n}&\le (x-1)x^{-n}\\\\
&\le \left(\frac{1}{n-1}\right)\left(\frac{n-1}{n}\right)^n\\\\
&<\frac{1}{n-1}\\\\
&<\epsilon
\end{align}$$
whenever $n>1+\frac1\epsilon$.
Putting it all together, we see that for all $\epsilon>0$
$$|f_n(x)-f(x)|<\epsilon$$
whenever $n>1+\frac1\epsilon$ for all $x\in [0,2]$.
The convergence is uniform.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2702759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Calculus, how to find the value of delta in $\epsilon$-$\delta$ limit proof
Problem: Show that $\lim_{x\to 4}x^3=64$.
My steps:
$|f(x)-L|<\epsilon$ so $|x^3-64|<\epsilon$
$|x^3|<64+\epsilon \rightarrow 64-\epsilon < x^3 < 64+\epsilon$
$\sqrt[3]{64-\epsilon}<x<\sqrt[3]{64+\epsilon}$
$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$
So $\delta<\sqrt[3]{64+\epsilon}-4$
Let $\epsilon>0$. Pick $\delta$ so that $\delta<\sqrt[3]{64+\epsilon}-4$.
Suppose $0<|x-4|<\delta$. Then $|x-4|<\sqrt[3]{64+\epsilon}-4$.
Thus, $|x|<\sqrt[3]{64+\epsilon}$ and cube to get $|x^3|<64+\epsilon$.
Then $|x^3-64|<\epsilon$ which completes the proof.
But this is what the answer said:
Let $\epsilon>0$. Pick $\delta$ so that $\delta<1$ and $\delta<\frac{\epsilon}{61}$.
Suppose $0<|x−4|<\delta$.
Then $4−\delta<x<4+\delta$.
Cube to get $(4−\delta)^3 <x^3 <(4+\delta)^3$.
Expanding the right-side inequality, we get $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64$ $=64+\epsilon$.
The other inequality is similar.
So my questions are:
*
*How can I get $\delta<\frac{\epsilon}{61}$?
*Why $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64=64+\epsilon$?
*When we need to pick $\delta=\min(1,\frac{\epsilon}{61}$) in this case?
*Does my proof has problems ?
I am just a newbie to calculus... Thanks in advance.
|
Note that in you proof is ok from here
$$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$$
you should set
$$\delta=min\{4-\sqrt[3]{64-\epsilon},\sqrt[3]{64+\epsilon}-4\}=\sqrt[3]{64+\epsilon}-4$$
and what you find is the optimal value for $\delta$.
Note that by binomial expansion
$$\sqrt[3]{64+\epsilon}-4=4\sqrt[3]{1+\epsilon/64}-4\approx\frac{\epsilon}{48}$$
then for $\epsilon$ small also $\frac{\epsilon}{61}<\frac{\epsilon}{48}$ fulfills the condition.
The other limit is obtained by
$$|x^3-64|<\epsilon\iff |x-4||x^2+4x+16|<\epsilon\iff |x-4|<\frac{\epsilon}{ |x^2+4x+16|}=\delta$$
thus for $0<|x-4|<1 \implies3< x<5 \quad x\ne 4$ the minimum value for $\delta$ is attained for $x=5$ that is
$$ \delta_{min}=\frac{\epsilon}{ |5^2+4\cdot5+16|}=\frac{\epsilon}{61} $$
then the condition is satisfy for $\delta=\min\left(1,\frac{\epsilon}{61}\right)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2703393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am not sure how this is useful.
|
You know that $2^{36}\equiv 1$, since $2$ is a primitive root. Hence you may conclude that $(2^{12})^3\equiv 1$, and also $(2^{24})^3\equiv (2^{36})^2\equiv 1^2$. These correspond to $26$ and $10$ respectively. Of course $2^{36}=1$.
Now, suppose that $(2^a)^3\equiv 1$. Then $2^{3a}\equiv 1$, so $36|3a$, or $12|a$. That doesn't leave many choices for $a$, since $1\le a\le 36$...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2708291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
}
|
If $\frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
If $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
I tried$$
\sin(\alpha+\beta)=-\sin \beta \cos \beta,\\
2\sin(\alpha+\beta)=-\sin (2\beta),$$
and $$\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha} =\frac{\sin\alpha\cos^3\beta+\sin^3\beta \cos \alpha}{\sin \alpha \sin \alpha},$$
but unable to find that ratio.
Any help, please.
|
$\def\peq{\mathrel{\phantom{=}}{}}$Denote $γ = α + β$, then$$
\frac{\cos α}{\cos β} + \frac{\sin α}{\sin β} = -1 \Longrightarrow \sin γ = \sin(α + β) = -\sin β\cos β.
$$
Thus,$$
\sin(γ - β) = \sin γ\cos β - \cos γ\sin β = -\sin β\cos^2 β - \cos γ\sin β,\\
\cos(γ - β) = \cos γ\cos β + \sin γ\sin β = \cos γ \cos β - \sin^2 β\cos β.
$$
Now, note that$$
1 = (\sin^2 β + \cos^2 β)^2 \Longrightarrow \sin^4 β + \cos^4 β = 1 - 2\sin^2 β\cos^2 β,
$$
then\begin{align*}
&\peq \frac{\cos^3 β}{\cos α} + \frac{\sin^3 β}{\sin α} = \frac{\cos^3 β}{\cos(γ - β)} + \frac{\sin^3 β}{\sin(γ - β)}\\
&= \frac{\cos^3 β}{\cos γ \cos β - \sin^2 β\cos β} + \frac{\sin^3 β}{-\sin β\cos^2 β - \cos γ\sin β}\\
&= \frac{\cos^2 β}{\cos γ - \sin^2 β} - \frac{\sin^2 β}{\cos γ + \cos^2 β}\\
&= \frac{\cos^2 β(\cos γ + \cos^2 β) - \sin^2 β(\cos γ - \sin^2 β)}{(\cos γ - \sin^2 β)(\cos γ + \cos^2 β)}. \tag{1}
\end{align*}
Because\begin{align*}
&\peq \cos^2 β(\cos γ + \cos^2 β) - \sin^2 β(\cos γ - \sin^2 β)\\
&= (\cos^2 β - \sin^2 β)\cos γ + \sin^4 β + \cos^4 β\\
&= \cos 2β\cos γ + 1 - 2\sin^2 β\cos^2 β,
\end{align*}
and\begin{align*}
&\peq (\cos γ - \sin^2 β)(\cos γ + \cos^2 β)\\
&= \cos^2 γ + (\cos^2 β - \sin^2 β)\cos γ - \sin^2 β\cos^2 β\\
&= (1 - \sin^2 γ) + \cos 2β\cos γ - \sin^2 β\cos^2 β\\
&= (1 - \sin^2 β\cos^2 β) + \cos 2β\cos γ - \sin^2 β\cos^2 β\\
&= \cos 2β\cos γ + 1 - 2\sin^2 β\cos^2 β,
\end{align*}
then $(1) = 1$, i.e.$$
\frac{\cos^3 β}{\cos α} + \frac{\sin^3 β}{\sin α} = 1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2708677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Find limit of $\frac{2x+7}{\sqrt{x^2+2x-1}}$ (check my steps please..)
Compute the limit $$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}$$
Here are my steps:
$$\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}=\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}\cdot\frac{-1/x}{-1/x}=\frac{\displaystyle \lim_{x\to-\infty}-2-\frac{7}{x}}{\displaystyle \lim_{x\to-\infty}\sqrt{\frac{x^2+2x-1}{x^2}}}=\frac{-2}{1}=-2$$
But I am not confident about $\displaystyle \lim_{x\to-\infty}{\sqrt\frac{x^2+2x-1}{x^2}}$. I just suppose that $x$ can be 'large enough' to ignore $+2x-1$.
|
You are right. Note that\begin{align}\lim_{x\to-\infty}\sqrt{\frac{x^2+2x-1}{x^2}}&=\sqrt{\lim_{x\to-\infty}\left(1+\frac2x-\frac1{x^2}\right)}\\&=\sqrt{1}\\&=1.\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2708828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\frac{1}{2} d\ln(1+x^2)=\frac{1}{2}d\ln(1+y^2)(1+x^2)=0$$
Let $c=(1+y^2)(1+x^2)$, so our equation becomes:
$$
d\ln c=0
$$
So what should I do here, should I integrate, or should I divide by $dx$?
If I divide by dx I get the expression $2x+2yy'+2xy^2+2x^2yy'=0$ which has $x$, $y$ and $y'$ and doesn't help me get anywhere.
Thanks in advance.
|
$$d\ln c=0 \implies \ln(c)=K$$
Another Hint
$$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$
$$(x + xy^2)dx + (y + yx^2)dy=0$$
It's an exact differential...
$$\frac {\partial P}{\partial y}=\frac {\partial Q}{\partial x} \implies 2xy=2xy$$
$$
\begin{cases}
f(x,y)=\int x+xy^2dx \\
f(x,y)=\int y+yx^2dy
\end{cases}
$$
Therefore
$$\boxed{x^2+y^2+y^2x^2=K}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2710438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
}
|
Find $\tan a \cdot\tan b$ given equation involving $\sin x$ and $\cos x$
Given:
$$(9\sin a + 44\cos a) \cdot (9\sin b + 44\cos b) = 2017$$
Find $\tan a \cdot \tan b$.
I factored the given equation, but I don't know how to proceed. Note that $9^2 + 44^2 = 2017$, that should be helpful in some way.
|
$$
(9\sin a + 44\cos a) (9\sin b + 44\cos b) = 2017
$$
Since $9^2+44^2 = 2017,$ you have
$$
\left(\frac 9 {\sqrt{2017}} \right)^2 + \left( \frac{44}{\sqrt{2017}} \right)^2 = 1
$$
and so for some value of $\theta$ you have
$$
\cos\theta = \frac 9 {\sqrt{2017}}, \quad \text{and} \quad \sin\theta = \frac{44}{\sqrt{2017}}.
$$
So
$$
(\cos\theta\sin a+ \sin\theta\cos a) (\cos\theta\sin b + \sin\theta\cos b) = 1
$$
$$
\sin(a+\theta) \sin(b+\theta) = 1
$$
Assuming all numbers are real, that can happen only if both sines are $+1$ or both are $-1.$
If $a+\theta$ and $b+\theta$ are both $\pi/2,$ modulo $2\pi,$ then $a$ and $b$ are the same modulo $2\pi.$
Since $\tan\theta = \dfrac{44}9,$ its complementary angle has tangent $\dfrac 9 {44}.$ Thus $\tan a\tan b = (\tan a)^2 = 9^2/44^2.$
Do something similar if both sines are $-1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2711173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Twin primes of the form $n^2+1$ and $N^2+3$? Assume that there are infinity many primes of the form $n^2+1$ and there are infinity many primes of the form $N^2+3$ , Then could we show that there are infinity primes of the form $n^2+1$ and $N^2+3$ (twin primes ) ?
Edit: I have edit the question just to show that $n$ and $N$ are not the same
|
A (trivial) insight...It's a lot easier to show which primes $n^2+1$ can't produce. Consider: $n^2+1 = n^2-1+2 = (n-1)(n+1)+2$; if $n^2+1 = p$, $p$ prime, then $(n-1)(n+1)+2 = p$, or $(n-1)(n+1) = p - 2$. Clearly $p$ can never be the larger of any two twin prime pairs because $p-2$ has factors $(n-1)$ and $(n+1)$. Further, $n^2+1 = (n^2-a^2)+(a^2+1)$. If $(n^2+1)$ is prime, then $p-a^2-1$ cannot be for all $0<a<n-1$. For a even, $a^2+1$ is odd and $p-a^2-1$ is even, so only $a$ odd is of interest. Conclude: Given two primes $p_1$ and $p_2$, where $p_2>p_1$, then if $p_2=n^2+1$ for certain $n>0$, then $p_2-p_1$ cannot equal $a^2+1$, $0<a<n-1$.
So the primes $n^2+1$ can produce are somewhat reduced in number because of the growing number of gap conditions that need to be satisfied. And the further one goes out, the more conditions there are to satisfy. So another form of the $n^2+1$ conjecture might be to ask whether it is possible to satisfy the conditions as $n$ increases. -FW
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2713410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Radius of a circle inscribed in a right triangle is given, the hypotenuse is given, find the legs? The radius of the circle inscribed in the right triangle is 4, the hypotenuse is 20cm. How do I find the legs? Any help would be appreciated!
|
For the right triangle $ABC$
with
the hypotenuse $|AB|=c=20$ and
the radius of inscribed circle $r=4$,
the radius of circumscribed circle is known to be
found as
\begin{align}
R&=\tfrac{c}2=10
,
\end{align}
and the semiperimeter
\begin{align}
\rho&=\tfrac12(a+b+c)
=r+c
,
\end{align}
Considering the side lengths of the triangle
as the roots of a general cubic polynomial
in terms of semiperimeter $\rho$,
inradius $r$ and circumradius $R$,
\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R
,
\end{align}
we can first simplify this cubic using the information given to
\begin{align}
x^3-2(r+c) x^2+((r+c)^2+r^2+2r c) x-2(r+c)r c
&=
(x-c)(x^2-(2r+c)x+2(r+c)r)
\\
\text{and find the legs as }\quad
a,b&=r+\tfrac{c}2\pm\tfrac12\sqrt{c^2-4r(r+c)}
=16,12
.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2714521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8...}}}}$. This can be written more generally as $f(x)=\sqrt{x+(x+1)f(x+2)}$ where $x=1$. This is where the problem arises from. If anybody can find an expression for the nested radical or find $f(x)$ I would be very happy!
|
The functional equation for $f(x)$ actually implies functional
relations also for all its derivatives
$$
\left\{ \matrix{
f(x)^{\,2} = x + \left( {x + 1} \right)f(x + 2) \hfill \cr
2f(x)f'(x) = 1 + f(x + 2) + \left( {x + 1} \right)f'(x + 2) \hfill \cr
2f'(x)^{\,2} + 2f(x)f''(x) = 2f'(x + 2) + \left( {x + 1} \right)f''(x + 2) \hfill \cr
\quad \quad \vdots \hfill \cr} \right.
$$
so that
$$
\left\{ \begin{gathered}
f(2) = f(0)^{\,2} \hfill \\
f'(2) = 2f(0)f'(0) - f(0)^{\,2} - 1 \hfill \\
f''(2) = 2f'(0)^{\,2} + 2f(0)f''(0) - 4f(0)f'(0) + 2f(0)^{\,2} + 2 \hfill \\
\quad \quad \vdots \hfill \\
\end{gathered} \right.
$$
Therefore, being $f(x)$ continuous, we are not free to fix $f(x)\quad |\;0\le x < 2$
equal to whatever continuous function respecting only $f(2)=f(0)^2$.
Instead it shall be such as to respect the functional relation, at $x$ and $x+2$, for all the derivatives.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2715665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
}
|
Proof by induction of series summation I understand the core principles of how to prove by induction and how series summations work. However I am struggling to rearrange the equation during the final (induction step).
Prove by induction for all positive integers n,
$$\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$$
After both proving for $n=1$ and assuming it holds true for $n=k$:
$$\sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k+1)^2+(k+1)^3$$
However I am unsure of how to proceed from here, the textbook says that the next step is to rearrange to give:
$$\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k+1)^2(k^2+4(k+1))$$
However I don't understand how they did this, can someone please clarify what they have done or suggest an alternative method to rearrange this equation to prove that the statement holds true for $k+1$ to give:
$$\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k+1)^2((k+1)+1)^2$$
|
$$\frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^2+(k+1)(k+1)^2=(k+1)^2\left(\frac14k^2+k+1\right)=(k+1)^2\left(\frac14k^2+\frac44(k+1)\right)=(k+1)^2\left(\frac14(k^2+4(k+1))\right)=\frac14(k+1)^2\left(k^2+4(k+1))\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2716363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove the relation is an equivalence relation. Problem
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \vert(a^2 + b) $. Prove that $R$ is an equivalence relation.
This is what I have so far.
Claim:
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \mid(a^2 + b) $. The relation $R$ is an equivalence relation.
Proof:
Part 1 (Reflixivity):
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that $b \in \Bbb{N}$.
Then, for some integer $k$:
$$\require{enclose}
\enclose{downdiagonalstrike}{\begin{align}
2 \mid b^2 +b \iff 2k &= b^2 +b \\
& = b^2 + b - (b^2 + b) + (b^2 + b) \\
& = 2b^2 + 2b - (b^2 + b) \\
& = -2b^2 - 2b + (b^2 + b)
\end{align}}$$
Therefore,
$$\enclose{downdiagonalstrike}{\begin{align}
2 \mid b^2 +b \iff b^2 + b &= 2k -2b^2 - 2b \\
& = 2(k - b^2 - b)\\
&
\end{align}}$$
$\enclose{horizontalstrike}{\text{Thus, $2 \mid b^2 +b$ for some integer $(k - b^2 - b)$. Which implies that $R$ is Reflexive.}} $
EDIT: Thanks to some positive feed back I have been let known that this is not showing Reflexivity.
Part 2 (Symmetry)
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that, for any $a,b \in \Bbb{N}$, $ a\mathbf{R}b \leftrightarrow b\mathbf{R}a.$
Then, for some integer $k$:
$$\enclose{updiagonalstrike}{\begin{align}
2 \mid a^2 + b & \iff 2k = a^2 + b \\
& \iff 2k + (a + b^2) = (a^2 + b) + (a + b^2)\\
& \iff (a + b^2) = (a + b) + (a^2 + b^2) - 2k \\
&
\end{align}}$$
$\enclose{horizontalstrike}{\text{Since, the relation $R$ is proven to be Reflexive, let the integer $ m = a = b $ and let the integer $ n = a^2 = b^2 $. Then, }}$
EDIT: This is not a valid way to show Symmetry, since Part 1 (Reflexivity) has not been proven.
$$\enclose{updiagonalstrike}{\begin{align}
a + b^2 = (a + b) + (a^2 + b^2) - 2k & \iff (a + b^2) = 2m + 2n - 2k \\
& \iff (a + b^2) = 2(m + n -k)\\
&
\end{align}}$$
$\enclose{horizontalstrike}{\text{Thus, $2 \mid b^2 + a$ which implies that $R$ is Symmetric since $ a\mathbf{R}b \leftrightarrow b\mathbf{R}a $.}}$
Part 3 (Transitivity)
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that, for any $a,b,c \in \Bbb{N}$, $ a\mathbf{R}b \text{ and } b\mathbf{R}c.$ Then, let $k$ and $h$ be some integers:
\begin{align}
2k = a^2 + b \text{ and } 2h = b^2 + c & \implies 2(k + h) = (a^2 + b) + (b^2 + c) \\
& \\
& \\
&
\end{align}
Comment: And, this is where I get stuck. I am struggling to find a way to show that $ (2 \mid a^2 + b) \land (2 \mid b^2 + c) \implies 2 \mid a^2 + c$. I've also tried eliminating $b$ like so, $2h = (2k - a^2)^2 + c$, but this doesn't seem to get me any where. I feel like I'm running in circles here.
My Question
Can you argue that $R$ is transitive since it has already been shown that $2 \mid b^2 + b$?
This would imply something like "$2(k - b^2 - b) + 2h = (a^2 + c) +(b^2 + b)$ is logically equivalent to $(2 \mid b^2 + b) \land (2 \mid a^2 + c)$." And, this simplifies to just $2 \mid a^2 + c$ by the inference rule of simplification [$(p \land q) \to p$]. Which ultimatily I believe gets me to my goal, but I'm not sure if it is two far of a leep to go from $2(k - b^2 - b) + 2h = (a^2 + c) +(b^2 + b) \implies (2 \mid b^2 + b) \land (2 \mid a^2 + c)$.
I hope my question was specific enough. Otherwise, I would much appreciate some guidance on showing how this relation is transitive if anyone is feeling generous. Thanks!
|
If $2\mid a^2+b$ and $2\mid b^2+c$, then $2\mid a^2+b+b^2+c$. But $b+b^2=b(b+1)$, which is the product of two consecutive natural numbers, and therefore it's an even number. So, since $2\mid a^2+b+b^2+c$ and since $2\mid b^2+b$, $2\mid a^2+c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2716634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Minimum of $\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}$ Let $0\leq x,y,z<1$ and $x^2+y^2+z^2=1$. What is the minimum value of
$$\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}?$$
From the condition, the point $(x,y,z)$ lies on a sphere with radius $1$. At the equality point, all three variables are equal to $1/\sqrt{3}$, and the value taken is $3/\sqrt{2}$. If one variable is $0$ and the other two are equal, the two variables are equal to $1/\sqrt{2}$, and the value taken is $2$.
|
You may convert it to become a function of two variables, the find the minimum value of that function. To do this, use
$$
z = \sqrt{1-(x^{2}+y^{2})}
$$
And substitute this in
$$ \frac{x}{\sqrt{1-x^{2}}} + \frac{y}{\sqrt{1-y^{2}}} +
\frac{z}{\sqrt{1-z^{2}}}$$
so the function will be
$$
f(x,y) = \frac{x}{\sqrt{1-x^{2}}} + \frac{y}{\sqrt{1-y^{2}}} +
\frac{ \sqrt{1-(x^{2}+y^{2})} }{ \sqrt{x^{2}+y^{2}} }
$$
Use partial derivatives to find the minimum value..?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2717109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\left ( b+ c \right )^{2}\geq 4bc$$
$$\left ( c+ a \right )^{2}\geq 4ca$$
So I want to prove $abc\geq 1$. I need to the help. Thanks!
|
Let $$\text{P}=27\, \left( a+b \right) ^{2} \left( b+c \right) ^{2} \left( c+a
\right) ^{2}-64\,abc \left( a+b+c \right) ^{3}$$
We have$:$ $$\text{P} =\sum\limits_{cyc}c \left( 44\,a{b}^{2}+19\,abc+49\,{c}^{2}a+5\,b{c}^{2}+27\,{c}^{3}
\right) \left( a-b \right) ^{2} \geqq 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2721251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
}
|
interval in which $h(x)$ is increasing If $h(x)=2g(2x^3-3x^2)+g(6x^2-4x^3-3)\forall x\in\mathbb{R}$ and $g''(x)>0\forall x\in\mathbb{R}$. Then find values of $x$ for which $h(x)$ is increasing
solution i try
$h(x)=2g'(2x^3-3x^2)(6x^2-6x)-g'(6x^2-4x^3-3)(12x^2-12x)$
From $g''(x)>0$ implies $g'(x)$ is increasing function.
Not be able to solve ahead Help required
|
Jacky.
$$ h(x) = 2g(2x^3−3x^2)+g(6x^2−4x^3−3) $$
Now let $g(2x^3−3x^2) = g(a)$ and $g(6x^2−4x^3−3)=g(b).$ So the 1st deriv of $h(x)$ is :
$$h'(x) = 12 \left[ g'(a)(x^2 - x) + g'(b)(x-x^2) \right] $$
$$ = 12 (x^2-x)\left[ g'(a) - g'(b) \right] $$
Now $h'(x)$ will be positive when
$$ x^2-x > 0 \:\: \text{and} \:\: g'(a)>g'(b) $$
$$ \cup $$
$$ x^2-x < 0 \:\: \text{and} \:\: g'(a) < g'(b) $$
Since we know that $g'(x)$ is increasing, then $g'(a) > g'(b)$ is achieved when $a > b$, or $2x^3-3x^2 > 6x^2-4x^3-3$.
so now $h'(x)$ will be positive when
$$ x^2-x > 0 \:\: \text{and} \:\: 2x^3-3x^3 > 6x^2-4x^3-3 $$
$$ \cup $$
$$ x^2-x < 0 \:\: \text{and} \:\: 2x^3-3x^3 < 6x^2-4x^3-3 $$
Can you continue from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2721505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\
\frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}
$$
How do I poceed further and find the derivative ?
|
HINT :
There is no parameter $a$ in the formula to be proved. So, first transform the initial equation into an equation where $a$ will be immediately eliminated by differentiation:
$$\frac{\sqrt{1-x^2}+\sqrt{1-y^2}}{x-y}=a$$
Differentiate and simplify.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2723169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
}
|
If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$b=2R \sin B$$
$$c=2R \sin C$$
Then,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
|
If $8R^2=a^2+b^2+c^2$ then putting $$\begin{cases}R=\dfrac{r^2+s^2}{2}\\a=r^2-s^2\\b=2rs\\c=r^2+s^2\end{cases}$$ it is verified the identity
$$8(\dfrac{r^2+s^2}{2})^2=(r^2-s)^2+(2rs)^2+(r^2+s^2)^2\iff2(r^2+s^2)^2=2(r^2+s^2)^2$$
This show that $a,b,c$ satisfy the well known parametrics of the Pythagorean triples (when $c$ is the diameter of the circumcircle i.e. $2R=r^2+s^2$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
solving for positive integer solutions $x^2 = yz,$
$x + y + z = 61$
positive integral solutions.
On wolframalpha, i could find the solution, but not how to solve.
I tried by taking a single equation
$xz + x^2 +z^2 - 61z = 0$
Tried solving for both x & z separately but did not find a way.
Please help.
EDIT: This is a high school problem in numerical analysis. Diophantine Equations is not yet in course.
|
I wanted to see that a more algebraic approach looked like
$$\begin{align}
&\left[x^2=yz \quad | \quad z=61-x-y\right]\implies x^2=61y-xy-y^2 \\
&\qquad x^2+xy+\frac{y^2}{4}-\frac{y^2}{4}=61y-y^2 \\
&\qquad \left(x+\frac{y}{2}\right)^2=61y-\frac{3}{4}y^2 \\
&\qquad \left(2x+y\right)^2=244y-3y^2 \\
&\qquad -\frac{1}{3}\left(2x+y\right)^2=-\frac{122^2}{9}+\frac{122^2}{9}-\frac{244}{3}y+y^2 \\
&\qquad -\frac{1}{3}\left(2x+y\right)^2=-\frac{122^2}{9}+\left(\frac{122}{3}-y\right)^2 \\
&\qquad -3(2x+y)^2=-122^2+(122-3y)^2 \\
&\qquad (3y-122)^2+3(2x+y)^2=122^2=14884=2^2 \cdot 61^2 \implies\\
&\left[\alpha^2+3 \beta ^2=14884\right] \quad | \quad \alpha=3y-122 \ \ \land \ \ \beta= 2x+y
\end{align}$$
Whatever $\alpha,\beta$ that might solve the above
$$\begin{align}y=\frac{\alpha+122}{3} \ \ \land \ \ x&=\frac{\beta-y}{2} \\
&=\frac{\beta-\frac{\alpha+122}{3}}{2} \\
&=\frac{3\beta-\alpha -122}{6}\end{align}$$
$$ \underline{(\alpha_i, \beta_i) \to (x_i, y_i, z_i)} \\
(-74,56) \to (20,16,25) \\
(-74,-56) \to (-36,16,81) \\
(-47,65) \to (20,25,16) \\
(-47,-65) \to (-45, 25,81) \\
(121,9) \to (-36,81,16) \\
(121,-9) \to (-45,81,25) \\
$$
$(\alpha, \beta)$ Solutions obtained through the help of Darios online equation solver: https://www.alpertron.com.ar/QUAD.HTM
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2724579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Finding Pointwise Limit Of These Functions Sequences of functions $f_n, g_n : [0, \infty) \to \Bbb R$, defined as
$f_n(x) = \dfrac{x}{1+x^n}$ and
$g_n(x) =
\begin{cases}
1
& \text{if } x > \frac{1}{n} \\
nx & \text{$0\leq x\leq1/n$}
\end{cases}
$
Find the Pointwise Limit.
Solution Attempt
Pointwise limit for $f_n(x)$ =
\begin{cases}
0
& \text{if } 0\leq x <1 \\
\frac{1}{2}
& \text{if } x = 1 \\
0 &\text{if } x > 1
\end{cases}
Pointwise limit for $g_n(x)$ =
\begin{cases}
0
& \text{if } x = 0 \\
1
& \text{if } x > 0 \\
\end{cases}
|
Your result for $g_n$ is correct.
The first comment suggests that $f_n$ is not correct. OP has the correct pointwise limit for $x \ge 1$. To fix the solution, first observe that $f_n(0) = 0$. When $0 < x < 1$,
$$\frac{x}{1+x^n} = \frac{1}{\dfrac1x + x^{n-1}} \xrightarrow[n\to\infty]{} x,$$
so the pointwise limit of $(f_n)$ is
\begin{cases}
\color{red}{x}
& \text{if } 0\leq x <1 \\
\dfrac{1}{2}
& \text{if } x = 1 \\
0 &\text{if } x > 1.
\end{cases}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2725603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Where’s the error in my factorial proof? On one of our tests, the extra credit was to find which number you would take out from the set $\{1!,2!,3!,...(N-1)!,N!\}$ such that the product of the set is a perfect square, for even $N$ My answer was as follows:
Assume $N$ is even. First note that $(n!)=(n-1)!\cdot n$.
Apply this to the odd numbers to get the product:
$$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots ((N-2)!)((N-2)!)\cdot (N-1)\cdot N!$$
Let $ 2!4!6!...(N-2)!=E$. Then our equation is equal to:
$$E^23\cdot 5\cdot 7\cdots (N-1)\cdot (N!)$$
Expand $N!$:
$$E^2\cdot 3\cdot 5\cdot 7\cdots (N-1)\cdot 2\cdot 3\cdot 4\cdots N$$
Group the odd terms together:
$$E^2\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7 \cdot 7\cdots (N-1)\cdot 2\cdot 4\cdot 6\cdots N$$
Let $O=1\cdot3\cdot5...\cdot (N-1)$:
$$E^2O^2 2\cdot 4\cdot 6\cdots N = E^2O^2\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)\cdots (2\cdot (N/2))$$
Group together the $2$'s:
$$E^2O^22^{(N/2)}1\cdot2\cdot3...(N/2)=E^2O^22^{(N/2)}\cdot(N/2)!$$
So, if $N/2$ is even, it can be expressed as $2m$ for some $m$. So we have:
$$E^2O^22^{2m}(N/2)!=(EO2^m)^2(N/2)!$$
Therefore, if $N$ is even, the number missing is $(N/2)!$ if $N/2$ is even. For example, for $N=4$, $2!$ is missing, $N=6$ is impossible ($3$ is odd), and for $N=100$, $50!$ is missing.
However, this turned out not to be correct - indeed, for $N=8$ we have solutions of $3!$ and $4!$. So where did I go wrong in my proof, or what did I leave out?
|
On this line:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2))$$
I believe you missed the first $2$. It should be:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2 \cdot 2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2))$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2728769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
$1-\cos (x) \leq \frac{x^2}{2} + \frac{x^3}{6}$, for $x > 0$, using Taylor expansion I want to solve this problem using Taylor expansions.
I tried
\begin{align*}
1 - \cos (x) = 1 - \left( \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} +R_{2n}(x) \right) \;,
\end{align*}
where $R_{2n}(x)$ is the remainder function.
For $n=2$, this yields
\begin{align*}
1 - \cos (x) = 1 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} + R_4(x) \right) = \frac{x^2}{2} - \frac{x^4}{24} - o(x^4) \;.
\end{align*}
Is it now possible to just state
\begin{align*}
|o(x^4)| \leq \frac{x^4}{24} \; ,
\end{align*}
and hence
\begin{align*}
\frac{x^2}{2}-\frac{x^4}{24}-o(x^4)\leq \frac{x^2}{2} \leq \frac{x^2}{2} + \frac{x^3}{6} \; ?
\end{align*}
|
For $x>0$,
$$\cos x > 1 -\frac{x^2}2$$
$$\begin{aligned}1-(\cos x) &< 1-\left(1 -\frac{x^2}2\right)\\
1-\cos x&< \frac{x^2}2 +\frac{x^3}6\end{aligned}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2730985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Two disjoint random events You roll twice with four-sided die in which the numbers one and two occur with probability $\frac{1}{3}$, and the numbers three and four each with probability $\frac{1}{6}$. Let X be the number of singles and Y the number of fours that occurred after two throws.
How do I create a table of probability function $p_{x,y}(x,y)=P\left \{X=x \wedge Y=y\right \}$?
This symbols at the end of this quations are little bit confusing to me.
$P(X=1)=\frac{1}{3}$, $P(X=2)=\frac{1}{3}$, $P(X=3)=\frac{1}{6}$ and $P(X=4)=\frac{1}{6}$.
$P(X,Y)=P(x)P(Y)=\frac{1}{3}\frac{1}{6}=\frac{1}{18}$
So do I just write Table:
$x_i$ $P(X=x_i)$
1 $2*1/3$
4 $2*1/6$
Because there are two throws or?
How do I calculate $P\left \{X+Y>0\right \}$?
Do I just add them?
$P\left \{X+Y>0\right \}=\frac{1}{3}+\frac{1}{6}$
|
If X is the number of ones seen in two rolls.
Each roll has a $\frac 13$ chance of seeing a $1$
$P(X = 2) = (\frac {1}{3})^2\\
P(X = 0) = (1-\frac {1}{3})^2\\
P(X = 1)= 2(\frac 13)(\frac 23)$
If Y is the number of $4's$ seen in two rolls.
Each roll has a $\frac 16$ chance of seeing a $4$
$P(Y = 2) = (\frac {1}{6})^2\\
P(Y = 0) = (1-\frac {1}{6})^2\\
P(Y = 1)= 2(\frac 16)(\frac 56)$
Regarding $P(X+Y>0)$
Seeing a 4 and seeing a 1 on any roll are mutually exclusive events. I think it is easier to look at the possibility that you roll a 4 or a 1. The probability is $\frac 12$ on any given roll.
$P(X+Y>0) = 1-\frac 14$
Covariance:
$var(X) = \sum P(X) X^2 - E[X]^2\\
2^2(\frac 19) + 1^2(\frac 49) - (2(\frac 19) + 1(\frac 49))^2\\
\frac {4}{9}$
$var(Y) = 2^2(\frac 1{36}) + 1^2(\frac {10}{36}) - (2(\frac 1{36}) + 1(\frac {10}{36}))^2\\
\frac {10}{36}$
$var (X+Y)= 2^2(\frac 1{4}) + 1^2(\frac {1}{2}) - (2(\frac 1{4}) + 1(\frac {1}{2}))^2\\
\frac {1}{2}$
$var (X+Y) = var(X) + var(Y) + 2cov(X,Y)\\
\frac {1}{2} = \frac {4}{9} + \frac {5}{36} + 2cov(X,Y)\\
cov(X,Y) = -\frac {1}{9}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2731490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Conditions on coefficients of positive semidefinite matrix with certain symmetries I have a real matrix, with certain symmetries, defined as
$
A = \left( {\begin{array}{*{20}{c}}
1-x&a&b&c\\
a&x&d&b\\
b&d&x&a\\
c&b&a&1-x
\end{array}} \right),
$
with $x,a,b,c,d \in \mathbb{R},{\rm{~ }}0 \le x \le 1$.
I want to obtain conditions on coefficients $x,a,b,c,d$ for the matrix to be positive semidefinite.
For the particular case $x=0$, I obtain with the Mathematica Reduce command the following conditions for the eigenvalues to be nonnegative:
$a = b = 0,{\rm{~}} -1 \le c \le 1,{\rm{ ~}}d = 0$.
In[1]:= A = {{(1 - x), a, b, c}, {a, x, d, b}, {b, d, x, a}, {c, b,
a, (1 - x)}}
Out[1]= {{1 - x, a, b, c}, {a, x, d, b}, {b, d, x, a}, {c, b, a,
1 - x}}
In[2]:= FullSimplify[Eigenvalues[A /. x -> 0]]
Out[2]= {1/2 (1 - c - d - Sqrt[4 (a - b)^2 + (1 - c + d)^2]),
1/2 (1 - c - d + Sqrt[4 (a - b)^2 + (1 - c + d)^2]),
1/2 (1 + c - Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d),
1/2 (1 + c + Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d)}
In[3]:= Reduce[
1/2 (1 - c - d - Sqrt[4 (a - b)^2 + (1 - c + d)^2]) >= 0 &&
1/2 (1 - c - d + Sqrt[4 (a - b)^2 + (1 - c + d)^2]) >= 0 &&
1/2 (1 + c - Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d) >= 0 &&
1/2 (1 + c + Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d) >= 0]
Out[3]= b == 0 && a == 0 && d == 0 && -1 <= c <= 1
Similarly, for the particular case $x=1~$ I get
$~a = b = 0,{\rm{~}} -1 \le d \le 1,{\rm{ ~}}c = 0$.
However, for a general $~0\le x \le1$, Mathematica takes weeks without giving an answer. I suspect that for $~0< x <1~$ the condition $a = b = 0~$ must be satisfied.
Do you think there is a way to obtain some conditions for the general case $~0\le x \le1$? The general eigenvalues of matrix $A$ have the following expressions:
$
\frac{1}{2} \left(-\sqrt{4 (a-b)^2+(-c+d-2 x+1)^2}-c-d+1\right),\frac{1}{2} \left(\sqrt{4 (a-b)^2+(-c+d-2 x+1)^2}-c-d+1\right),\frac{1}{2} \left(-\sqrt{4 (a+b)^2+(c-d-2 x+1)^2}+c+d+1\right),\frac{1}{2} \left(\sqrt{4 (a+b)^2+(c-d-2 x+1)^2}+c+d+1\right).
$
|
Essentially the same approach:
the real symmetric matrix $A$ is positive if and only if the polynomial in $\lambda$
$$\det(\lambda I +A)$$ has all coefficients $\ge 0$. The nice thing in this case is that this polynomial of degree $4$ factors into two polynomials of degree $2$. The equivalent condition is that both these two polynomials have positive coefficients. We get the conditions
$$1-(c+d)\ge 0\\
(1-x-c)(x-d)\ge (a-b)^2\\
1+c+d \ge 0\\
(1-x+c)(x+d)\ge (a+b)^2$$
They appear weaker than the ones obtained by @Delta-u:, but in fact they are equivalent.
Since the characteristic polynomial factors in this way, our matrix should be similar (by an orthogonal matrix) to the block matrix
$$\left( \begin{matrix} 1-x -c & a-b& 0& 0\\
a-b& x-d & 0 & 0 \\
0 & 0 & 1- x + c & a+b \\
0& 0& a+b & x + d \end{matrix} \right)$$
I don't see right now a simple transformation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2732005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
prove this inequality by $abc=1$ Let $a,b,c>0$ and $abc=1$,show that
$$(a^{10}+b^{10}+c^{10})^2\ge 3(a^{14}+b^{14}+c^{14})$$
since
$$LHS=\sum \left(a^{20}+\dfrac{2}{a^{10}}\right)$$
it is prove
$$\sum_{cyc}\left(a^{20}+\dfrac{2}{a^{10}}-3a^{14}\right)\ge 0$$ not easy,and I found $x^{20}+2x^{-10}-3x^{14}$http://www.wolframalpha.com/input/?i=x%5E(20)%2B2x%5E(-10)-3x%5E(14)
|
We start with two identities that we can call Lamé-type identites:
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)=P$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))=Q$$
Furthermore we have :
$$(a+b)(b+c)(c+a)=\frac{(a+b+c)^3-a^3-b^3-c^3}{3}$$
Or
With the identity of Gauss we have :
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
So we deduce this :
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})(x^2+y^2+z^2+xy+xz+yz)$$
And
$$(x+y+z)^7 - (x^7+y^7+z^7)=7(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$
your inequality is equivalent to with ($xyz=1$) :
$$(x^5+y^5+z^5)^2\ge 3(x^7+y^7+z^7)$$
Or :
$$(x^5+y^5+z^5-(x+y+z)^5+(x+y+z)^5)^2\ge 3(x^7+y^7+z^7-(x+y+z)^7+(x+y+z)^7)$$
Or :
$$(-P+(x+y+z)^5)^2\geq 3(-Q+(x+y+z)^7)$$
Or :
$$-P\geq (3(-Q+(x+y+z)^7))^{0.5}-(x+y+z)^5$$
But :
$$-P=-(x+y+z)^5 + (x^5+y^5+z^5) = 5(\frac{-(x+y+z)^3+(x+y+z)(-xz-yz-yz+x^2+y^2+z^2)+3}{3})((x+y+z)^2-(xy+xz+yz))$$
$$-Q=-(x+y+z)^7 + (x^7+y^7+z^7)=7(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})(-(-(x+y+z)^2+xy+zx+yz)^2-xyz(x+y+z))$$
Now we remark that if we put $x+y+z=\alpha$$\quad$ so $xy+yz+zx$ is maximal for $x=y=z=\frac{\alpha}{3}$
Furthermore in $-Q$ all the $xy+yz+zx$ are positive and in $-P$ there are negative so we can minimizing and maximazing the LHS and the RHS respectively
So we can reduce the problem to a one variable problem wich is easily solvable if we remark that :
$$xyz=1 \implies x+y+z\geq 3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2732128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
How do I determine this integral? $\int_{0}^{+\infty}\sin^2(1/x)\frac{dx}{(4+x^2)^2}$ $$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}$$
$${\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1-\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1\over 2(4a^2+x^2)^2}-{\cos\left({2a\over x}\right)\over 2(4a^2+x^2)^2}$$
$${1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
$${\pi\over 8(2a)^3}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
$$\int \mathrm dx{1\over (b^2+x^2)^2}={x\over 2b^2(b^2+x^2)}+{1\over 2b^3}\arctan\left({x\over b}\right)+K$$
$$\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
Enforcing a substitution of $u=\dfrac{2a}{x}$
$${1\over (2a)^3}\int_{0}^{\infty}\mathrm du {u^2\cos(u)\over (1+u^2)^2}$$
Now this integral is more harder than the original due to the extra $u^2$ at the numerator.
This is an even function, so can be expressed as
$${1\over 2(2a)^3}\int_{-\infty}^{\infty}\mathrm du {u^2\cos(u)\over (1+u^2)^2}$$
Decomposition of fraction
$${u^2\over (1+u^2)^2}={Au+B\over 1+u^2}+{Cu+B\over (1+u^2)^2}$$
This look like a nightmare, so how do I determine this integral?
|
What I suggest is to write
$$
I\left(a\right)=\int_{0}^{+\infty}\frac{\sin^2\left(\displaystyle \frac{a}{x}\right)}{\left(4a^2+x^2\right)^2}\text{d}x \text{ and }J\left(a\right)=\int_{0}^{+\infty}\frac{\cos^2\left(\displaystyle\frac{a}{x}\right)}{\left(4a^2+x^2\right)^2}\text{d}x
$$
Then you have
$$
I\left(a\right)+J\left(a\right)=\int_{0}^{+\infty}\frac{\text{d}x}{\left(4a^2+x^2\right)^2}
$$
which can be calculated by using integration by part on
$$
\int_{0}^{+\infty}\frac{\text{d}x}{\left(4a^2+x^2\right)}
$$
which gives you
$$
\int_{0}^{+\infty}\frac{\text{d}x}{\left(4a^2+x^2\right)^2}=\frac{\pi}{32a^3}
$$
Then you can calculate
$$I(a)-J(a)=\int_{0}^{+\infty}\frac{\cos\left(\displaystyle \frac{2a}{x}\right)}{\left(4a^2+x^2\right)^2}\text{d}x=0$$
Hence you have
$\displaystyle 2I\left(a\right)=\frac{\pi}{32a^3}$ so you can conclude that
$$
I\left(a\right)=\int_{0}^{+\infty}\frac{\sin^2\left(\displaystyle \frac{a}{x}\right)}{\left(4a^2+x^2\right)^2}\text{d}x=\frac{\pi}{64a^3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2733663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
I have omitted the Base Case due to lack of specific relevance. An excerpt from the book's page:
Assume the assertion has been proved for a specific value of $n$, say $n = k$. That is, assume we have proved
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 \lt \frac{k^3}{3}$$
for a fixed $k \ge 1$. Now using this, we shall deduce the corresponding result for $k + 1:$
$$A(k + 1): 1^2 + 2^2 + \cdots + k^2 \lt \frac{(k + 1)^3}{3}.$$
Start with $A(k)$ and add $k^2$ to both sides. This gives the inequality
$$1^2 + 2^2 + \cdots + k^2 \lt \frac{k^3}{3} + k^2.$$
To obtain $A(k + 1)$ as a consequence of this, it suffices to show that
$$\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}.$$
But this follows at once from the equation
$$\frac{(k + 1)^3}{3} = \frac{k^2 + 3k^2 + 3k + 1}{3} = \frac{k^3}{3} + k^2 + k + \frac13.$$
Therefore, we have shown that $A(k + 1)$ directly follows from $A(k)$.
I can not understand the last two steps. Why do the expressions seem to flip on the inequality in step 4 such that $\frac{k^3}{3} + k^2$ is now on the other side of the less than symbol, and how does the final equation prove the assertion?
Thank you.
|
you are given:
$$ 1^2 + 2^2 + ... + (k-1)^2 < \frac{k^3}{3} $$
by induction hypothesis. Now, what he is doing is
$$ 1^2 + 2^2 + ... + (k-1)^2 {\color{red} + } \color{red}{k^2} < \frac{k^3}{3} {\color{red} + } \color{red}{k^2} $$
Now, you want to prove that
$$ 1^2 + 2^2 + ... + (k-1)^2 + k^2 < \frac{(k+1)^3}{3} $$
But,
by the second line, you might as well just prove that
$$ \frac{k^3}{3} {\color{red} + } \color{red}{k^2} < \frac{ (k+1)^3}{3} $$
also, obviously,
$$ k + \frac{1}{3} > 0 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2736318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Find the limit $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$ and prove it. Find $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$.
I claim that $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$. To prove this, for given $\varepsilon >0$, I have to find $M\in N$ such that $|\frac {2n^2+10n+5}{n^2}-2|<\varepsilon$ for $n \ge M$.
By Archimedean property, we can find $M \in N$ such that $\frac {15}M<\varepsilon$, and note that $n\ge M \rightarrow \frac 1n \le \frac 1M \rightarrow \frac {15}n \le \frac {15}M$.
Then, for $n \ge M$, we have that $|\frac {2n^2+10n+5}{n^2}-2|=|\frac {10n+5}{n^2}| < |\frac {15n}{n^2}|$ (since $n \ge M \in N$) $<\frac {15}n\le \frac {15}M<\varepsilon$.
Therefore, by definition of convergence, $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$.
Should I say $M\in Z^+$ (because I am worried about the case where $M=0$)? Can you find any mistakes in this proof?
Thank you in advance.
|
Your proof is fine, to simplify note that
$$\frac {2n^2+10n+5}{n^2}=2+\frac{10}{n}+\frac5{n^2}$$
then it suffices to prove that $\frac1n \to 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2737895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{2\pi}{11},2\cos\frac{8\pi}{11},2\cos\frac{14\pi}{11}$$
I tried to eliminate options using the sum and product of roots but I can't figure out if $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$$
or
$$\cos\frac{2\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{14\pi}{11}=0$$
|
Let $x=a\cos t$
Comparing with $\cos3t$ formula $$\dfrac{a^3}4=\dfrac{3a}3, a=?$$
$$a\cos3t=-1$$
$$3t:360^\circ n\pm120^\circ$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2739299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Fourier Series of Absolute Value I think I understand how to evaluate Fourier series, but I would love to have someone check this to make sure I am doing this correctly.
Find the Fourier series of the function $f(x) = 11 + \lvert 6x \rvert$.
First, we find $a_0$.
\begin{align}
a_0 &= \bigg\langle f,\frac{1}{\sqrt{2}} \bigg\rangle \\
&= \frac{1}{\pi} \int_{-\pi}^{\pi}\frac{1}{\sqrt{2}} f(x) dx \\
&= \frac{1}{\pi} \int_{-\pi}^{\pi}\frac{1}{\sqrt{2}} (11 + \lvert 6x \rvert)dx
\end{align}
\begin{align}
a_k &= \bigg \langle f, \cos(kx) \bigg \rangle \\
&= \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(kx) f(x) dx \\
&= \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(kx)( 11 + \lvert 6x \rvert)dx
\end{align}
\begin{align}
b_k &= \bigg \langle f, \sin(kx) \bigg \rangle \\
&= \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(kx) f(x) dx \\
&= \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(kx)(11 + \lvert 6x \rvert)dx
\end{align}
So, once you find these constants, you plug them into the following formula:
$$f \sim \frac{a_{0}}{\sqrt{2}} + \sum_{k=1}^{\infty} a_k \cos(kx) + b_{k} \sin(kx)$$
Is this right? If so, how do I solve the integral including the absolute value term?
|
Guide:
Note that in the $a_0$, don't forget the braces.
$$a_0 = \frac1{\pi}\int_{-\pi}^\pi \frac1{\sqrt2} \color{blue}(11+ |6x|\color{blue}) \, dx$$
Also, note that $11+|6x|$ is an even function, hence $b_k=0$.
For $a_k$,
$$a_k = \frac{2}{\pi} \int_0^\pi \cos (kx) (11+6x)\, dx,$$
I think integration by parts should work.
Edit:
\begin{align}a_k &= \frac{2}{\pi} \int_0^\pi \cos (kx) (11+6x)\, dx,\\
&= \frac2\pi \left( \left.(11+6x)\frac{\sin(kx)}{k}\right|_{x=0}^{x=\pi}-6\int_0^\pi \frac{\sin(kx)}{k} \, dx\right)\\
&=\frac{-12}{\pi k}\int_0^\pi \sin(kx) \, dx\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2742274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve the equation $ \log_2(9x+2)=\log_3(16x+3). $
Solve the equation
$$
\log_2(9x+2)=\log_3(16x+3).
$$
It is easy to see that $x=0$ is a solution but how to prove that there are no more solutions?
My idea was prove that the function $f(x)=\log_2(9x+2)-\log_3(16x+3)$ is monotone for $x \geq 0$. We have
$$
f'(x)=\frac9{ \left( 9\,x+2 \right) \ln \left( 2 \right) }-
\frac{16}{ \left( 16\,x+3 \right) \ln \left( 3 \right) },
$$
and the inequality $f'(x)>0$ has the solution
$$
\left( -2/9, -3/16 \right) \cup
\left( a
,\infty \right),
$$
where $$a=-{\frac {1}{144}}\,{\frac {32\,\ln \left( 2 \right) -27\,\ln
\left( 3 \right) }{\ln \left( 2 \right) -\ln \left( 3 \right) }}$$
so $f'(x)>0$ for $x>0$. Thus $f(x)$ is there monotone and $f(x)=0$ has only one solution for $x \geq 0$.
My question Is there a solution without using the derivative?
|
YES, we can solve this equation without using derivative.
here I solved this equation by a method without using derivative.
$\log_2(9x+2)=\log_3(16x+3)=y $(say)
$\log_2(9x+2)=y \implies2^y=9x+2$
multiply this equation by 16$\implies2^{y+4}=144x+32$.........(1)
second part of given equation is $\log_3(16x+3)=y\implies 3^y=16x+3$
multiply this equation by 9 $\implies3^{y+2}=144x+27$.........(2)
$(1)-(2)\implies 2^{y+4}-3^{y+2}=5$.......(3)
obviously $y=1,-1$ is a solution
if $y=1\implies\log_2(9x+2)=\log_3(16x+3)=1$.By using the formula: if,$log_a(b)=1\implies b=a.$
similarly $y=-1$ we find $x=-1/6$
$9x+2=2\implies x=0$.
similarly, $16x+3=3\implies x=0$
$x=0,-1/6$ is a solution . we want to prove $x=0,-1/6$ is only two solution
from (3), $2^{y+4}-3^{y+2}=5$,
if $y$ is greater than $1$, $2^{y+4}-3^{y+2}$,is negative.i.e,$2^{y+4}\lt3^{y+2}$.we prove it by induction
it is true for $y=1$,let it is true for $y=k$.
i.e,$2^{k+4}\lt3^{k+2}$ we know that $2\lt3$
multiply these two inequalities,we get $2^{(k+1)+4}\lt3^{(k+1)+2}$, by $PMI$ it is true for all natural numbers greater than $1$.
for $y\gt1$,there is no solution for$2^{y+4}-3^{y+2}=5$,since , $2^{y+4}-3^{y+2}$ is negative for $y\gt1$.
therefore, there is no corresponding value for $x$.
Therefore $\log_2(9x+2)-\log_3(16x+3)$ has no solution other than $x=0,-1/6$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2743506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Short way for upper triangularization
We are given a matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and we are asked to find a matrix $P$ such that $P^{-1}AP$ is upper triangular.
Here, we first find one eigenvalue as $\lambda= 4$. Then the matrix $$
A-4I =
\begin{bmatrix}
-1 & 0 & 1 \\
-1 & 0 & -3 \\
-1 & 0 & 1 \\
\end{bmatrix}$$
has basis formed from $f_1 = (-1,-1,-1)^T$, $f_2 = (1,-3,1)^T$. We extend this to a basis of the whole space by adjoining $f_3 = (1,0,0)^T$, and so we have base-change matrix
$$ P =
\begin{bmatrix}
-1 & 1 & 1 \\
-1 & -3 & 0 \\
-1 & 1 & 0 \\
\end{bmatrix}.$$
Then, by using some computational tools, we find that
$$ P^{-1}AP =
\begin{bmatrix}
3 & 1 & 1 \\
-1 & 5 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$$
Now, we need to look at $$B = \begin{bmatrix}
3 & 1 \\
-1 & 5 \\
\end{bmatrix},$$ which has eigenvalue $\lambda = 4$. So we have $$B-4I = \begin{bmatrix}
-1 & 1 \\
-1 & 1 \\
\end{bmatrix}.$$
So basis for the image of this is $(1,1)^T$. We extend this to the basis $(1,1)^T$, $(1,0)^T$ of $\mathbb{R}^2$. Now, going back to $\mathbb{R}^3$, we have the matrix $$Q = \begin{bmatrix}
1 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}.$$
Then, by using calculation tools, we get
$$Q^{-1}P^{-1}APQ = \begin{bmatrix}
4 & -1 & 0 \\
0 & 4 & 1 \\
0 & 0 & 4 \\
\end{bmatrix},$$
which is in upper triangular form.
Now, what I wanted to ask is that is there a way to directly find the matrix $R = PQ$ such that $R^{-1}AR$ is upper triangular, without going through these steps?
|
$\newenvironment{gmatrix}{\left\lgroup\begin{matrix}}{\end{matrix}\right\rgroup}$Here it is a general way of finding Jordan normal forms. It applies to your question because all eigenvalues of $A$ are real.
Step 1: Computing the invariant factors of $λI - A$ to get its elementary divisors.
The monic greatest common divisor of all $1 × 1$ minors of $λI - A$ is $D_1(λ) = 1$, and that of all $2 × 2$ minors is $D_2(λ) = 1$, and that of the $3 × 3$ minors is$$
D_3(λ) = |λI - A| = (λ - 4)(λ - 4 + \sqrt{2})(λ - 4 - \sqrt{2}).
$$
Thus the invariant factors of $λI - A$ are$$
d_1(λ) = D_1(λ) = 1,\quad d_2(λ) = \frac{D_1(λ)}{D_2(λ)} = 1,\\
d_3(λ) = \frac{D_3(λ)}{D_2(λ)} = (λ - 4)(λ - 4 + \sqrt{2})(λ - 4 - \sqrt{2}),
$$
and the elementary divisors of $λI - A$ are$$
λ - 4,\ λ - 4 + \sqrt{2},\ λ - 4 - \sqrt{2}.
$$
Step 2: Writing down the Jordan normal form of $A$ based on the elementary divisors of $λI - A$.
Since the elementary divisors of $λI - A$ are $λ - 4$, $λ - 4 + \sqrt{2}$, $λ - 4 - \sqrt{2}$, the Jordan normal form of $A$ is$$
J = \begin{gmatrix}
4 &&\\
& 4 - \sqrt{2} &\\
&& 4 + \sqrt{2}
\end{gmatrix}.
$$
Step 3: Finding all the Jordan chains and the $P$.
In this case, it is easy to get that$$
x_1 = \begin{gmatrix}
0 \\ 1 \\ 0
\end{gmatrix}, x_2 = \begin{gmatrix}
1 + \sqrt{2} \\ 1 + 2\sqrt{2} \\ 1
\end{gmatrix}, x_3 = \begin{gmatrix}
1 - \sqrt{2} \\ 1 - 2\sqrt{2} \\ 1
\end{gmatrix}
$$
are the three Jordan chains (here eigenvectors in fact) of $λ - 4$, $λ - 4 + \sqrt{2}$, $λ - 4 - \sqrt{2}$, respectively. Thus$$
P = (x_1, x_2, x_3) = \begin{gmatrix}
0 & 1 + \sqrt{2} & 1 - \sqrt{2}\\
1 & 1 + 2\sqrt{2} & 1 - 2\sqrt{2}\\
0 & 1 & 1
\end{gmatrix}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2744829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How many possibilities are there for at least $k$ consecutive heads to show up out of $n$ tosses? Consider $n$ coin tosses. In how many ways can we have at least $k$ consecutive heads? Call this number $f(n,k)$. Is there a general expression for it? Or at least tight upper and lower bounds?
For example take $n=5,k=2$. Then the possibilities are:
HHTTT, THHTT, TTHHT,TTTHH,
HHHTT, HHTHT, HHTTH, THHHT, THTHH, THHTH, HTHHT, TTHHH, HTTHH,
HHHHT, HHHTH, HHTHH, HTHHH, THHHH,
HHHHH.
that is 19 ways
|
We consider the binary alphabet $V=\{H,T\}$. We are looking for the number $g(n,k)$ of strings of length $n$ having runs of $H$ at most length $k-1$. The wanted number is $$f(n,k)=2^n-g(n,k)$$
Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.
A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left(1-\frac{2z}{1+z}\right)^{-1}\tag{1}
\end{align*}
Replacing occurrences of $H$ in a Smirnov word by one up to $k-1$ $H$ generates words having runs of $H$ with length less than $k$.
\begin{align*}\
z\longrightarrow z+z^2+\cdots+z^{k-1}=\frac{z\left(1-z^{k-1}\right)}{1-z}\tag{2}
\end{align*}
Since there are no restrictions to the length of runs of $T$'s we replace occurrences of $T$ in a Smirnov word by one or more $T$s.
\begin{align*}\
z\longrightarrow z+z^2+\cdots=\frac{z}{1-z}\tag{3}
\end{align*}
The resulting generating function is by substituting (2) and (3) in (1)
\begin{align*}
\left(1- \frac{\frac{z\left(1-z^{k-1}\right)}{1-z}}{1+\frac{z\left(1-z^{k-1}\right)}{1-z}}-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}\right)^{-1}
&=\frac{1-z^k}{1-2z+z^{k+1}}
\end{align*}
Denoting with $[z^n]$ the coefficient of $z^n$ in a series we obtain the number of wanted words of length $n$ as
\begin{align*}
\color{blue}{f(n,k)}&=2^n-g(n,k)\\
&\color{blue}{=[z^n]\left(\frac{1}{1-2z}-\frac{1-z^k}{1-2z+z^{k+1}}\right)}
\end{align*}
Example: Let's look at OPs example. We take $k=2$. We obtain with some help of Wolfram Alpha
\begin{align*}
\frac{1}{1-2z}-\frac{1-z^2}{1-2z+z^{3}}=z^2+3 z^3 + 8z^4+\color{blue}{19} z^5 + 43 z^6 +\cdots
\end{align*}
The blue colored coefficient of $z^5$ shows there are $\color{blue}{19}$ words of length $5$ built from characters $\{H,T\}$ and runs of $H$ with length at least $k=2$ in accordance with OP's result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2746110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \in D^3$ it is proved that a + b + c = 0 => a x b = b x c = c x a If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \in D^3$ it is proved that $ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \Rightarrow \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a} $
So taking the first part with cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively
$$
\left\{\begin{array}{test}
\overrightarrow{a} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = 0 \\
\overrightarrow{b} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = 0
\end{array}\right\} \Rightarrow \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a}
$$
Is this way of thinking right, or the two equations have infinite solutions? I mean, no matter which vector we multiply with $( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} )$ it's going to be zero anyway.
So is this a gap in our reasoning?
|
It looks from the comments that you did not understand the details
that are missing from your proof, so here is a start.
$\overrightarrow{a} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = \overrightarrow0$, hence
$\overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{c} = \overrightarrow0$.
Use that $\overrightarrow{a} \times \overrightarrow{a}=\overrightarrow0$, and that $\overrightarrow{a} \times \overrightarrow{c}=
-\overrightarrow{c} \times \overrightarrow{a}$ to obtain
$\overrightarrow{a} \times \overrightarrow{b} - \overrightarrow{c} \times \overrightarrow{a} = \overrightarrow0$, and finally
$\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{a}
$.
Then do a similar argument at least once more.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2746352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Diagonalization of a Matrix. Having Trouble.. This is for Continuous-time Markov chains but I'm having trouble with the linear algebra.
Formal Definitions:
If Q is diagonalizable, then so is $e^{tQ}$, and the transition function can be expressed in terms of the eigenvalues and eigenvectors of $Q$. Write $Q=SDS^{-1}$, where $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $Q$, and $S$ is an invertible matrix whose columns are the corresponding eigenvectors. This gives,
$$e^{tQ} = Se^{tD}S^{-1}$$
The Problem:
A Markov Chain has generator matrix,
$$Q=
\begin{pmatrix}
-1 & 1 & 0 \\
0 & -2 & 2 \\
3 & 0 & -3 \\
\end{pmatrix}
$$
Find the transition function by diagonalizing the generator and finding the matrix exponential.
My pr0fessor seems to be getting eigenvalues $\lambda = -4, \lambda = -2, \lambda = 0$.
I, on the other hand, seem to be getting $\lambda = -3, \lambda = 0$.
Edit: I recalculated but still am getting $\lambda = 0$.
This gets him a completely different set of eigenvectors.. Am I miscalculating something?
EDIT:
In the back of the book I have,
$$P(t)=
\begin{pmatrix}
-1 & 0 & 1 \\
-3 & 1 & 1 \\
3 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
e^{-4t} & 0 & 0 \\
0 & e^{-2t} & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-1/4 & 0 & 1/4 \\
-3/2 & 1 & 1/2 \\
3/4 & 0 & 1/4 \\
\end{pmatrix}
$$
Where $P(t)$ is the transition function asked for.
|
There's a typo. Reverse engineering the given answer we have:
$$
\begin{align}
Q&=PDP^{-1}\\
&=\begin{pmatrix}
-1 & 0 & 1 \\
-3 & 1 & 1 \\
3 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-4 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
-1/4 & 0 & 1/4 \\
-3/2 & 1 & 1/2 \\
3/4 & 0 & 1/4 \\
\end{pmatrix}\\
&=\begin{pmatrix}
-1 & 0 & 1 \\
0 & -2 & 2 \\
3 & 0 & -3 \\
\end{pmatrix}
\end{align}
$$
Now knowing that
$$
Q=\begin{pmatrix}
-1 & 0 & 1 \\
0 & -2 & 2 \\
3 & 0 & -3 \\
\end{pmatrix}
$$
You can find your characteristic polynomial, e.g. via cofactor expansion down the second column to get $$p(\lambda)=-\lambda(\lambda+2)(\lambda+4)$$ and proceed from there.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2751710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Real roots of an equation A friend gave me this equation which I have trouble finding the real roots. $$x^9+3x^6+3x^3-16x+9=0$$ One can easily see that 1 is a root then with the help of Horner's method this can be simplified. However I am looking for an elegant solution if possible not just to use a computer to do that madness calculation. I reduced it to $$x^3+1=2(2x-1)^{\frac{1}{3}}$$ I don't have any useful attempts.
EDIT: After several tries I think I found something useful. After getting to this form $x^3-2(2x-1)^{\frac{1}{3}}+1=0$ we might notice that it's similar to the equation that you provided me in the answers namely$$(x-1)(x^2 +x-1)=x^3 - 2x+1=0$$ Now in order that this to give me the solution, we must have that $-2(2x-1)^{\frac{1}{3}}=-2x\, $ If we cube both sides we return to $x^3 - 2x+1=0$ (the solutions) Is this a repeatedly loop in the equation? Can this be useful?
|
Let $f(x)=(x^3+1)/2$. Then $f$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$ with inverse $g(x)=(2x-1)^{1/3}$. As you already remarked, the original equation can be written as
$$((x^3+1)=2(2x-1)^{1/3}.$$
This equation is equivalent to $f(x)=g(x)$ or $f(f(x))=x.$
Now note that if $f(x)=x$, that is $x^3-2x+1=0$ then $f(f(x))=x$ which implies that $x^3-2x+1$ divides the given polynomial. After the division, we obtain the factorization
$$(x^3-2x+1)(\underbrace{(x^3+1)^2 + 2 x^4+3x^2+(x+1)^2}_{\geq 0} + 7) \tag{*}.$$
So it remains to solve
$$x^3-2x+1=(x - 1) (x^2 + x - 1)=0$$
and the desired real roots are $x_1=1$, $x_2=(-1-\sqrt{5})/2$, and $x_3=(-1+\sqrt{5})/2$ .
P.S. In my first answer I used maple to find the factorization $(*)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2753587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
|
In your first reformulation, you should get $\sqrt[15]{2^{4 \cdot 3} \cdot 16}$ and in the last part note that $\sqrt[a]{c^b} = c^{\frac{b}{a}}$ and not $c^{\frac{a}{b}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2756282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
}
|
$ Q(x_{1}) = x_{2}+x_{3} $ , $ Q(x_{2}) = x_{1}+x_{3} $ , $ Q(x_{3}) = x_{1}+x_{2} $ , $ Q(x_{1}x_{2}x_{3}) = 6 $ Let $ P(X) = X^3+7X^2+3X+1 $, with the roots $ x_{1},x_{2},x_{3} \in \mathbb{C} $
Let $ Q $ be a third grade polynomial with the following properties :
$ Q(x_{1}) = x_{2}+x_{3} $
$ Q(x_{2}) = x_{1}+x_{3} $
$ Q(x_{3}) = x_{1}+x_{2} $
$ Q(x_{1}x_{2}x_{3}) = 6 $
Let $ y_{1},y_{2},y_{3} \in \mathbb{C} $ be the roots of $ Q $
$ 1) $ Calculate $y_{1}+y_{2}+y_{3}$
$ 2) $ Calculate $P(y_{1})+P(y_{2})+P(y_{3})$
Both problems could be solved easily with Viete's formulas if we know the coefficients of $ Q $
One easy thing to deduce is that $Q(x_{1}x_{2}x_{3}) = Q(-1)=6 $
But I don't get enough relations to find $a,b,c,d$ in $aX^3+bX^2+cX+d$
Any other ideas?
|
We have $\sum_i x_i=-7,\,\prod_i x_i = -1$ so $Q(x_i)=-7-x_i,\,Q(-1)=6$. Define $R(x):=Q(x)+x+7$ so $R(x_i)=0,\,R(-1)=12$. Since $P(-1)=4$, $R=3P$ and $Q(x)=3P(x)-x-7=3x^3+21x^2+8x-4$. Thus $\sum_i y_i = -7$ and $\sum_i P(y_i)=\sum_i \frac{y_i+7}{3}=7+\frac{1}{3}\sum_i y_i=\frac{14}{3}$, since each root of $Q$ satisfies $P(y)=\frac{y+7}{3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2757403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probability that the ball you toss lands in any one of the bins. What is the expected number of tosses?
Answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1}) \\
p_3 &=& 1 - (\frac{n-1}{n})(\frac{n-1-1}{n-1}) \\
p_3 &=& 1 - (\frac{n-2}{n}) = \frac{2}{n} \\
p_4 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1})(1 - \frac{1}{n-2}) \\
p_4 &=& 1 - ( \frac{n-1}{n} )( \frac{n-2}{n-1} )( \frac{n - 2 -1}{n - 2} ) \\
p_4 &=& 1 - \frac{n-3}{n} = \frac{3}{n} \\
\end{eqnarray*}
Now for $1 <= i <= n$ we have: $p_i = \frac{i-1}{n}$.
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
E &=& \sum_{i = 1}^{n} \frac{i(i+1)}{n} = \frac{1}{2n} \sum_{i=1}^{n} i^2 + i \\
E &=& \frac{1}{2n}(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ) \\
E &=& \frac{n+1}{4n} ( \frac{2n+1}{3} + 1 ) \\
\end{eqnarray*}
Here is an update to my answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\newline
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (\frac{n-1}{n})( \frac{n-2}{n}) \\
p_3 &=& 1 - \frac{(n-1)(n-2)}{n^2} = \frac{n^2 - (n^2 - 3n + 2)}{n^2} \\
p_3 &=& \frac{3n-2}{n^2} \\
p_4 &=& 1 - (\frac{n-1}{n})(\frac{n-2}{n})(\frac{n-3}{n}) \\
p_4 &=& 1 - \frac{(n^2-3n+2)(n-3)}{n^3}\\
p_4 &=& 1 - \frac{n^3-3n^2+2n - 3n^2 +9n - 6}{n^3}\\
p_4 &=& \frac{3n^2-2n + 3n^2 - 9n + 6}{n^3}\\
p_4 &=& \frac{3n^2 + 3n^2 - 11n + 6}{n^3}\\
\end{eqnarray*}
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
\end{eqnarray*}
Now, am on the right track? That is, is what I have so far correct?
Thanks,
Bob
|
Let $T$ be the number of tosses it takes to get $2$ balls in $1$ of the bins. If you are still tossing balls after $i$ tosses it is because none of the bins has more than $1$ ball in it, so there is $1$ ball in $i$ bins
$$ \Pr(T=i+1|T>i) = \frac{i}{n} .$$
Next, using Bayes' rule
$$ \Pr(T=i+1|T>i) = \frac{\Pr(T=i+1)}{\Pr(T>i)} ,$$
which implies that
$$ \Pr(T=i+1) = \Pr(T>i)\Pr(T=i+1|T>i) .$$
Since
$$ \Pr(T>i) = 1 - \sum_{k=1}^i\Pr(T=k) ,$$
and letting $p_i =\Pr(T=i+1)$, it follows that
$$p_2 = \frac{1}{n} \quad\text{and}\quad p_{i+1}=\frac{(n+1-i)i}{n(i-1)}p_{i} \quad\text{for}\; i\ge 3. $$
Then,
$$\mathbb E[T] = \sum_{i=2}^n p_i i,$$
which I have not been able to solve analytically but have written the following Matlab code to compute it:
N = 100;
E = zeros(N,1);
for n = 2:N
p = zeros(n+1,1);
p(2) = 1/n;
for i = 2:n
p(i+1) = p(i)*(n+1-i)*i/(n*(i-1));
end
E(n) = sum(p'*(1:n+1)');
end
plot(2:N,E(2:N,1))
The code generates the following plot with $n$ in the $x$-axis and $\mathbb E[T]$ in the $y$-axis:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2758092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
Why does this relation of binomials hold? Why does $\sum_{m=n}^N m\binom {m-1}{n-1}=\sum_{m=n}^N n \binom mn=n \binom{N+1}{n+1}$ is true? Is there some special formula for it?
|
The first equality is due to the equality in @OlivierOloa's answer. The second one is a bit trick to obtain.
*
*Change $\binom nn$ to $\binom {n+1}{n+1}$.
*Group the first two term.
*Apply the equality $\binom nk + \binom n{k+1} = \binom {n+1}{k+1}$ to condense the leftmost two terms into one.
*Repeat (3) until only one term left.
\begin{align}
&\binom nn + \binom {n+1}n + \binom {n+2}n + \dots + \binom Nn \\
&= \left[\binom {n+1}{n+1} + \binom {n+1}n \right] + \binom {n+2}n + \dots + \binom Nn \\
&= \left[\binom {n+2}{n+1} + \binom {n+2}n\right] + \dots + \binom Nn \\
&= \cdots \\
&= \binom{N+1}{n+1}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2759256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Infinite positive integer solutions of the equation: $x^2+x+1=(y^2+y+1)(n^2+n+1)$ Could anybody help me some ideas on the below problem:
Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that
$$x^2+x+1=(y^2+y+1)(n^2+n+1).$$
Thanks in advance.
|
To solve the Diophantine equation.
$$x^2+x+1=(z^2+z+1)(y^2+y+1)$$
It is necessary to use the solutions of the Pell equation. $p^2-(z^2+z+1)s^2=\pm1$
Then the solutions can be written as follows.
$$x=\mp((z+1)p^2+(z^2+z+1)(zs-p)s)$$
$$y=\mp((2z+1)p-(z^2+z+1)s)s$$
For positive you need to take decisions at $-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2761139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Which $z\in\mathbb{C}$ satisfy the equation $|z-i|=\sqrt{2}|\bar{z}+1|$? Let $S$ be the set of all complex numbers $z$ satisfying the rule $$|z-i|=\sqrt{2}|\bar{z}+1|$$
Show that $S$ contains points on a circle.
My attempt,
By substituting $z = x + yi$, and squaring both sides. But I can't get the circle form.
|
Let $z=x+iy$. Note that $\overline{z}=x-iy$.
$$\begin{align}
|z-i|&=\sqrt{2}|\overline{z}+1|\\
|z-i|^2&=2|\overline{z}+1|^2\\
|x+i(y-1)|^2&=2|(x+1)-iy|^2\\
x^2+(y-1)^2&=2(x+1)^2+y^2)\\
x^2+4x+2+y^2+2y-1&=0\\
(x+2)^2+(y+1)^2&=2^2
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2764586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
In a triangle ABC if $B-C = \frac\pi 4$ and .......... What will be the value of $\frac {a+b+c}{a}$
In a triangle ABC if $B-C = \frac\pi 4$ . Consider the following determinant.
|-2 cosC cosB |
| |
|cosC -1 cosA | = P
| |
|cosB cosA -1 |
If $ P + \cos ^2 A = 0$ then what is the value of $\frac {a+b+c}{a}$?
(Where $A$,$B$,$C$ represent the angles opposite to the sides $a$,$b$,$c$)
I tried to expand the determinant and got $\cos ^2 B +2\cos ^2 A +\cos ^2 C + 2\cos A\cos B\cos C - 2$. I don't think that helps?
Just to solve it I assumed the triangle ABC with $B = 90^0$, $A = 45^0$, $C = 45^0$. And lucky me, the triangle satisfied the given equation. But I wouldn't call that a real method.
All help will be appreciated.
|
►The function $f(x)$ where $x=C$ solution of your problem can be find out without using your determinant. The variable $x$ should be such that $0\lt x\lt\dfrac{3\pi}{8}(=135^{\circ})$, as you can verify by drawing triangles appropriate to the problem, and the function is increasing from $2$ to $\infty$. You have
$$f(x)=\frac{\sin(x+\frac{\pi}{4})+\sin( x)}{\sin(\frac{3\pi}{4}-2x)}+1 \left(\text {this is }f(x)=\frac {a+b+c}{a}\right)$$ and its (blue) graphic where figure the red point $\left(\dfrac{\pi}{4},2+\sqrt2\right)$ corresponding to your verification with $C=45^{\circ}$.
► Another thing (more difficult) is to use the $p$ value of your determinant.
We have to calculate $\dfrac {b+c}{a}+1$.
One has $$\frac {b+c}{a}=\frac{\sin B+\sin C}{\sin A}\space\space \qquad (1)$$
and the determinant equal to $p$ above is equal to
$$p=2\cos A\cos B\cos C-2+2\cos^2A+\cos^2B+\cos^2C$$ or $$\\2\cos A\cos B\cos C-\sin^2B-\sin^2C=3p\qquad (2)$$
Noting $C=x$ we have $B=x+\dfrac{\pi}{4}$ and $A=\dfrac{3\pi}{4}-2x$ it follows
$\sin A=\dfrac{\sqrt2}{2}(\sin(2x)+\cos(2x))\Rightarrow \sin^2 A=\dfrac 12(1+\sin 4x)=P$ where $0\lt P=-p$.
Therefore $\sin 4x=2P-1\Rightarrow \boxed{x=\dfrac{\arcsin(2P-1)}{4}}$. This solves already the problem because this is to find a relationship between $ x $ and $ P $ but if you want to use $(2)$ for the purpose of finding a simpler relationship between $x$ and $P$, you have
$$2\cos A\cos B\cos C=(\sin 2x-\cos 2x)(\sin x-\cos x)\cos x=(\cos x-\sin 3x)\cos x\\\sin^2B=\frac 12(1+\sin 2x)$$
from which you get in $(2)$ $$(\cos x-\sin 3x)\cos x-\frac 12(1+\sin 2x)-\sin^2x=-3P\\$$
I stop here. Whatever you can deduce, you will find at the end an expression for the angle x equal to the boxed above. I add an easy exercise to find algebraic expressions for trigonometric functions of the variable $x$. In the attached figure is drawn a right triangle in which a construction of the angle $ 4x $ is presented. By the geometric theorem of the bisector you can find expressions for trigonometric functions of $ x $ and $2x$ as a function of $ P $. The result, if you get it, will be a quite complicated algebraic expression for $ f (x) $.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2766487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.