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How to solve $2 \tan x / (1 - (\tan x)^2) = (\sin 2x)^2$? $$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$
I tried a lot but I get nowhere
| $\frac{2\tan x}{1-\tan^2x}=(\sin 2x)^2$
$\tan 2x=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}=\sin^2 2x$
$\frac{\sin 2x}{\cos 2x}-\sin^2 2x=0$
$\frac{\sin 2x}{\cos 2x}(1-\sin 2x \cos 2x)=0$
either $\sin 2x=0$ or $1-\sin 2x \cos 2x=0$
$\sin 2x=0$ gives $x=\frac{n\pi}{2}$
$1-\sin 2x \cos 2x=0\Rightarrow\sin 4x=2$ which is not possible as $\sin$ cannot be greater than 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ways to write $2010$?
Let $ N$ be the number of ways to write $ 2010$ in the form $ 2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $ a_i$'s are integers, and $ 0 \le a_i \le 99$. An example of such a representation is $ 1\cdot10^3 + 3\cdot10^2 + 67\cdot10^1 + 40\cdot10^0$. Find $ N$.
I picked the biggest $a_1$ so: $a_1 = 2$, there are only two ways to form $2010$.
Take $a_1 = 1$ now. This opens up to a lot of possibilities.
Specific Casework should work:
Cases 1-1: $a_2 = 10$, then possibilities are: $a_1 = 1, a_0 = 0$ or $a_1 = 0, a_0 = 10$
Actually, I think a number-theoretic way is easier.
But still.
Case 1: $a_1 = 1$ then we must solve:
$100x + 10y + z = 1010$.
Since $0 \le x \le 10$, we can casework $x$ so that:
Case 1-1:$x = 0$. So that:
$10y + z = 1010 \implies z \equiv 0 \pmod{10}, z = 10k$ and $y = 101 - k$.
Hence, $(0, 101 - k, 10k)$. $\min{k} = 0 $ and we need to find the max of $k$. We must have, $101 - k \le 99$ and $10k \le 99$. This suggests, $k \le 9$.
Cases 1-2: $x=1$. So that:
$10y + z = 910 \implies z \equiv 0 \pmod{10}$ Again, $z = 10k$ and $y = 91 - k$. Giving a set of $(1, 91 - k, 10k)$.Here again, $\min{k} = 0$ and $10k \le 99$ so $k \le 9$.
I am conjecturing that since we are always increasing $x$ values, the value on the RHS will always be divisible by $10$.
$x = 9$ so that:
$10y + z = 110 \implies z = 10k$ and $y = 11 - k$, which again there are $9$ values.
Except if $x=10$ then there is: $10y + z = 10$ then $z = 10k$ and $y = 1 - k$. Then $k$ must be $1$.
So there are: $10(9) + 1 + 2 = 93$ solutions total.
This is just an attempt!
Bump: anybody have anything?
| It's easier to start from the other end. We observe first that $a_0$ can be any multiple of $10$ from $0$ through $90$. There are, of course, ten such values.
Then $a_1$ can be any one- or two-digit value equivalent to $11-(a_0/10) \bmod 10$. Again, there are ten such values.
The choices of $a_2$ are much more limited by the fact that the target sum is only $2010$. In which cases are there two different values? In which cases are there three?
Once $a_0, a_1, a_2$ have been decided, there is only one usable value of $a_3$.
| {
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Into how many equivalences classes does $R$ partition $\mathbb{Z}$? Let $R= \{ (a,b) \in\mathbb{Z}\times\mathbb{Z} \mid a^2\equiv b^2 \bmod 7\}$.
Into how many equivalences classes does $R$ partition $\mathbb{Z}$?
My best guess is that there are $7$ equivalence classes: $a^2-b^2$ has remainder $0 \bmod 7$, $a^2-b^2$ has remainder $1 \bmod 7$, $a^2-b^2$ has remainder $2 \bmod 7, \ldots, a^2-b^2$ has remainder $6 \bmod 7$.
Am I misunderstanding equivalence class partitions?
| The fact that $7$ is prime is relevant here. A number can have more than two square roots modulo a composite number, but modulo a prime number, each number can have only two square roots. Thus $a\equiv b\bmod 7$ if and only if $a\equiv\pm b\bmod 7$.
\begin{align}
0^2 & \equiv (-0)^2 \equiv 0 \\
1^2 & \equiv (-1)^2 \equiv 1 \\
2^2 & \equiv (-2)^2 \equiv 4 \\
3^2 & \equiv (-3)^2 \equiv 2
\end{align}
Since the only congruence classes modulo $7$ are $0,\pm1,\pm2,\pm3$, this list is complete. There are $7-1$ non-zero congruence classes and hence $(7-1)/2$ non-zero squares. Each non-zero square coresponds to one of the equivalence classes you're looking for, $0$ corresponds to another.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve$\frac {1}{\sin {x}} + \frac {\sqrt {3}}{\cos {x}} = 4$ $$\frac {1}{\sin {x}} + \frac {\sqrt {3}}{\cos {x}} = 4$$
Can you help me solve this?
| We may assume $x\not\in \frac{\pi}{2}\mathbb{Z}$, hence your equation is equivalent to:
$$ \cos(x)+\sqrt{3}\sin(x) = 4\sin(x)\cos(x) = 2\sin(2x) $$
or to:
$$ \sin\left(x+\frac{\pi}{6}\right) = \sin(2x) $$
whose fundamental solutions are given by $2x=x+\frac{\pi}{6}$, leading to $x=\frac{\pi}{6}$, and by $x+\frac{\pi}{6}=(2k+1)\pi-2x$, leading to $x\in\left\{\frac{5\pi}{18},\frac{17\pi}{18},\frac{29\pi}{18}\right\}$. So we get:
$$ x \in \left\{\frac{\pi}{6},\frac{5\pi}{18},\frac{17\pi}{18},\frac{29\pi}{18}\right\}+2\pi\mathbb{Z}.$$
| {
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Dihedral angle Finding I meet a wall while some problem solving.
The wall is following question.
There is a triangle ABC, The vertice A touch bottom plane and the distance from B, C to the bottom : BE = b, CD = c .
When ED= l_1 , DA=l_2 , AE = l_3 , What can I find the dihedral angle between plane ABC and plane AED ?
I want to break this wall.
Please some help to me !
| Let $A = O = (0, 0, 0)$ - is origin. $\overrightarrow{Oz}$ axis directed away from $A$ and $\overrightarrow{Oz} \perp AED$. $\overrightarrow{Ox}$ is codirectional to $\overrightarrow{AE}$, $\Rightarrow E = (l_3, 0, 0)$ and $B = (l_3, 0, b)$. What are $x$ and $y$ coordinates of $D$ and $C$?
$$
\begin{array}{lr}
p = \frac{l_1 + l_2 + l_3}{2} & \\
y = \frac{2}{l_3} \cdot \sqrt{p \cdot (p - l_1) \cdot (p - l_2) \cdot (p - l_3)} &
\text{($y$ is height of the $EAD$ triangle from $D$ to $EA$)} \\
x = \sqrt{l_2^2 - y^2} &
\text{$x$ is $x$-coordinate of mentioned above height}
\end{array}
$$
$\Rightarrow$ $D = (x, y, 0)$ and $C = (x, y, c)$.
It is well-known that normalized normal hyperplane equation for hyperplane supports the simplex spanned on $p_i, i \in \{1, 2, \dots, D\}$ points is:
$$
\sum \limits_{i = 1}^{D}\left(x_i \cdot \frac{n_i}{|\overrightarrow{n}|}\right) - \frac{d}{|\overrightarrow{n}|} = 0
$$
where $\overrightarrow{n} = (n_1, n_2, \dots, n_D)$ — normal to the hyperplane and $\frac{d}{|\overrightarrow{n}|}$ is a distance from the origin to the hyperplane:
$$
n_i =
\begin{vmatrix}
p_1^1 & p_1^2 & \dots & p_1^{i - 1} & 1 & p_1^{i + 1} & \dots & p_1^D \\
p_2^1 & p_2^2 & \dots & p_2^{i - 1} & 1 & p_2^{i + 1} & \dots & p_2^D \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
p_D^1 & p_D^2 & \dots & p_D^{i - 1} & 1 & p_D^{i + 1} & \dots & p_D^D \notag
\end{vmatrix} \\
d =
\begin{vmatrix}
p_1^1 & p_1^2 & \dots & p_1^D \\
p_2^1 & p_2^2 & \dots & p_2^D \\
\vdots & \vdots & \ddots & \vdots \\
p_D^1 & p_D^2 & \dots & p_D^D \notag
\end{vmatrix}
$$
In our case $D = 3$ and above huge formulas became much simplier. All we need are just a coordinates of the normal $\overrightarrow{n}$ to each hyperplane (id est to triangles $ABC$ and $AED$).
$$
n_{ABC} =
\left(
\begin{vmatrix}
1 & 0 & 0 \\
1 & 0 & b \\
1 & y & c
\end{vmatrix},
\begin{vmatrix}
0 & 1 & 0 \\
l_3 & 1 & b \\
x & 1 & c
\end{vmatrix},
\begin{vmatrix}
0 & 0 & 1 \\
l_3 & 0 & 1 \\
x & y & 1
\end{vmatrix}
\right) = (-b \cdot y, b \cdot x - l_3 \cdot c, l_3 \cdot y)
$$
$$
n_{AED} =
\left(
\begin{vmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
1 & y & 0
\end{vmatrix},
\begin{vmatrix}
0 & 1 & 0 \\
l_3 & 1 & 0 \\
x & 1 & 0
\end{vmatrix},
\begin{vmatrix}
0 & 0 & 1 \\
l_3 & 0 & 1 \\
x & y & 1
\end{vmatrix}
\right) = (0, 0, l_3 \cdot y)
$$
Finally, dihedral angle between $ABC$ and $AED$ is:
$$
\angle (ABC, AED) = \arccos\left(\frac{(\overrightarrow{n_{ABC}} \cdot \overrightarrow{n_{AED}})}{|\overrightarrow{n_{ABC}}| \cdot |\overrightarrow{n_{AED}}|}\right) = \arccos\left(\frac{l_3 \cdot y}{\sqrt{(b \cdot y)^2 + (b \cdot x - l_3 \cdot c)^2 + (l_3 \cdot y)^2}}\right)
$$
is an $\arccos(\cdot)$ of dot product of normalized normals to planes, supports the triangles.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I find the $\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}$ without using L’Hopital’s? $$\Large{\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}}$$
I know how to find this Limit only by using L’Hopital’s. Here's How:
$$={\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}}$$
$$={\lim_{x\to 0}\frac{21x^2-8x}{6x\cos(3x^2)}}$$
$$={\lim_{x\to 0}\frac{x(21x-8)}{6x\cos(3x^2)}}$$
$$={\lim_{x\to 0}\frac{(21x-8)}{6\cos(3x^2)}}$$
$$=\frac{21\cdot 0 - 8}{6 \cos 0}$$
$$=\frac{-4}{3}$$
My Question is, is there a way to do this without using the L’Hopital’s? If so, How?
Also Note: We have not been taught the series expansion of $\sin x$.
| So we have $$\lim_{x\to 0} \frac{7x^3 - 4x^2}{\sin(3x^2)}.$$
First, we have to factor the numerator so $x^3 - 4x^2 = x^2(7x - 4)$.
Therefore, we have $$\lim_{x\to 0} \frac{x^2(7x - 4)}{\sin(3x^2)}.$$
After this, we multiply the numerator and the denominator by $3x^2$.
Note $\frac{3x^2}{3x^2}=1$ so we aren't doing anything arbitrarily and it is still equivalent. We know that $$\lim_{x\to 0} \frac{\sin(x)}{x}=1 \text{ and } \lim_{x\to 0}\frac{x}{\sin(x)}=1$$ (you have to use a calculator or L'Hopital's theorem to check this). By doing this multiplication, we get $$\lim_{x\to 0} \frac{3x^2 x^2 (7x - 4)}{3x^2 \sin (3x^2)}.$$ And since this is similar to $x / \sin(x)$ because $x / \sin(x)$ is essentially $1 \cdot 1 / (1 \cdot \sin(1 \cdot x))$, the limit as $x$ approaches $0$ of $(x / \sin(x)) = (3x^2 / 3x^2 \sin(3x^2)$ must be true due to proportionality or similarity. After this, we can simplify the limit as $x$ approaches $0$ of $(3x^2 \cdot x^2 \cdot (7x - 4))/(3x^2\sin(3x^2))$ to the limit as $x$ approaches $0$ of $(x^2(7x - 4)/x^2)$. After this, we simplify the $x^2$ in the numerator and the denominator thus leaving the limit as $x$ approaches $0$ of $(7x -4)$ which is $-4/3$. I hope this makes sense though!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the limit of $\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}$? $$\lim_{n\to \infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)=? $$
I tried with the squeeze theorem, though I got the upper bound, but I couldn't find the lower bound. I also tried to solve it with the order limit theorem, but without any success. I guessed the result should be $\frac{1}{\sqrt{2}}$. How can I do this?
| If you don't spot that the series telescopes (and this is always worth a check - usually careful inspection of the first few terms will suffice) then here is a solution using the Stolz–Cesàro theorem.
Write the desired series as a fraction:
$$S_n = \frac{\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}}{\sqrt{n}} = \frac{a_n}{b_n} $$
Now $b_n = \sqrt{n}$ is strictly monotone and divergent (since it is strictly increasing and $b_n \to +\infty$). We need this condition for Stolz–Cesàro to apply.
Now consider the "fraction of differences" (does it have a more technical name?):
$$\frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \frac{\left(\sqrt{2n+1}+\sqrt{2n-1}\right)^{-1}}{\sqrt{n}-\sqrt{n-1}} = \frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac{1+\sqrt{1-\frac{1}{n}}}{\sqrt{2+\frac{1}{n}}+\sqrt{2-\frac{1}{n}}}$$
The numerator $1+\sqrt{1-\frac{1}{n}} \to 2$ and the denominator $\sqrt{2+\frac{1}{n}}+\sqrt{2-\frac{1}{n}} \to 2\sqrt{2}$ so the fraction tends to $\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$.
Then since $b_n$ was strictly monotone and divergent, and the limit of $\frac{a_n - a_{n-1}}{b_n - b_{n-1}}$ exists and equals $\frac{1}{\sqrt{2}}$, we can conclude from the Stolz–Cesàro theorem that the limit of $\frac{a_n}{b_n}$ also exists and also equals $\frac{1}{\sqrt{2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why the following series doesn't absolute convergence $\sum_{k=1}^{\infty} (-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})$ I have the following series, I don't understand why it doesn't converge absolutely.
$$\sum_{k=1}^{\infty} (-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})$$
This is what I did:
$$\sum_{k=1}^{\infty} |(-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})|=\sum_{k=1}^{\infty} |(\sqrt{k^2+1}-\sqrt{k^2-1})*\frac{\sqrt{k^2+1}+\sqrt{k^2-1}}{\sqrt{k^2+1}+\sqrt{k^2-1}}|=\sum_{k=1}^{\infty}\frac{2}{\sqrt{k^2+1}+\sqrt{k^2-1}} \geq \sum_{k=1}^{\infty} \frac{1}{\sqrt{k^2+1}} \geq\sum_{k=1}^{\infty} \frac{1}{2k^2}$$
And $\sum_{k=1}^{\infty} \frac{1}{2k^2}$ convergences therefore $\sum_{k=1}^{\infty} (-1)^k(\sqrt{k^2+1}-\sqrt{k^2-1})$ is absolutely convergent.
I don't understand why it's not true.
Any ideas? Thank you!
| There are three points I want to make.
Firstly, your inequality is incorrect. Let me show you the correct one:
\begin{align*}
\sum_{k=1}^n(\sqrt{k^2+1}-\sqrt{k^2-1})={}&\sum_{k=1}^n(\sqrt{k^2+1}-\sqrt{k^2-1})\frac{\sqrt{k^2+1}+\sqrt{k^2-1}}{\sqrt{k^2+1}+\sqrt{k^2-1}}={} \\
{}={}&\sum_{k=1}^n\frac{2}{\sqrt{k^2+1}+\sqrt{k^2-1}}\leq\sum_{k=1}^n\frac{2}{\sqrt{k^2+1}},
\end{align*}
because when you reduce a denominator, the fraction increases.
Secondly, the above majoring series doesn't converge anyway, because it is asymptotic to $\sum\frac{1}{k}$, the harmonic series, which is well-known not to converge.
Thirdly, the starting series does not converge, because:
\begin{align*}
\frac{2}{\sqrt{k^2+1}+\sqrt{k^2-1}}={}&\frac{2}{k}\frac{1}{\sqrt{1+\frac{1}{k^2}}+\sqrt{1-\frac{1}{k^2}}}={} \\
{}={}&\frac{2}{k}\frac{1}{1+\frac{2}{k^2}+o\left(\frac{1}{k^2}\right)+1-\frac{2}{k^2}+o\left(\frac{1}{k^2}\right)}={} \\
{}={}&\frac{2}{2k}\frac{1}{1+o\left(\frac{1}{k^2}\right)}\sim\frac{1}{k},
\end{align*}
the Harmonic series once more.
Note: I assumed you knew that $x_n\sim y_n$ for $n\to\infty$ implies the convergence of $\sum x_n$ and $\sum y_n$ is the same. This is proved by noting asymptotic means there are constants $a,b$ such that eventually $ax_n\leq y_n\leq bx_n$, so if $\sum x_n$ converges so does $\sum bx_n$ and thus so does $\sum y_n$, whereas if $\sum x_n$ doesn't converge then neither does $\sum ax_n$ and thus neither does $\sum y_n$. That works at least with positive terms as in the case where I used it. With alternating signs there might be extra cancellations made possible by a majoring.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving limit of radicals without L'Hopital $\lim_{x\to 64} \frac{\sqrt x - 8}{\sqrt[3] x - 4} $ I've been trying to solve this for some time now, but I cannot get any closer to the solution.
I need to solve this limit without using L'Hopital's rule.
$$\lim_{x\to 64} \dfrac{\sqrt x - 8}{\sqrt[3] x - 4} $$
By using L'Hopital I know the result should be 3, but I cannot get to it any other way...
How would I simplify this equation?
| It isn't mentioned much in calculus courses, but you could rationalize the cube root in the denominator using $(A-B)(A^2 + AB + B^2) = A^3 - B^3$, to wit:
$$\frac{\sqrt{x}-8}{\sqrt[3]{x}-4} = \frac{(\sqrt{x}-8)(\sqrt x + 8)(\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16)}{(\sqrt[3]{x}-4)(\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16)(\sqrt x + 8)} = \frac{(x-64)(\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16)}{(x-64)(\sqrt x + 8)}$$ to get $$\lim_{x \to 64} \frac{\sqrt{x}-8}{\sqrt[3]{x}-4} = \lim_{x \to 64} \frac{\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16}{\sqrt x + 8} = 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the discriminant of a second order equation related to the graph of $ax^2+bxy+cy^2+dx+ey+f=0$? Is the discriminant of a second order equation related to the graph of $ax^2+bxy+cy^2+dx+ey+f=0$?
Most people who took precalculus know that $ax^2+bxy+cy^2+dx+ey+f=0$ is the graph of:
*
*An ellipse if $b^2-4ac\lt0$
*A parabola if $b^2-4ac=0$
*A hyperbola if $b^2-4ac\gt0$
Is this related to the fact that the discriminant of $ax^2+bx+c=0$ is $\Delta=b^2-4ac$, which tell us that the equation has 2 solutions if $b^2-4ac\gt0$, one if $b^2-4ac=0$, and none (in the reals) $b^2-4ac\lt0$?
| Both solving a quadratic and diagonalising a quadratic form involve completing the square:
$$ 0 = ax^2+bx+c = a\left( x+\frac{b}{2a}\right)^2- \frac{1}{4a}(b^2-4ac) \\
ax^2+bxy+cy^2 = a\left( x+\frac{b}{2a}y\right)^2- \frac{1}{4a}(b^2-4ac)y^2,
$$
so in both cases the sign of the quantity $b^2-4ac$ determines the nature of solutions.
Geometrically, $ax^2+bxy+cy^2+dx+ey+f=0$ is the intersection of the conic $z=ax^2+bxy+cy^2$ and the plane $ dx+ey+z+f=0 $. Choosing the plane as $z=0$, we can then ask the question: what does the conic
$$ ax^2+bxy+cy^2 = 0 $$
look like? Clearly the point $x=y=0$ is part of it. Does it contain any other points? If $a=0$, it includes the line $y=0$. Now suppose $a \neq 0$. Then there are clearly no points with $y=0$ other than $x=y=0$.
$$ a\frac{x^2}{y^2} + b \frac{x}{y}+c=0 $$
is a quadratic in $x/y$, and has real solutions if and only if $b^2-4ac>0$. Therefore the conclusion is:
*
*If $a=0$, the conic $ ax^2+bxy+cy^2 = 0 $ consists of the lines $y=0$ and $bx+cy=0$.
*If $a \neq 0$, the conic $ ax^2+bxy+cy^2 = 0 $ consists of the lines $x-\alpha_i y=0$, where $\alpha_i$ are the roots of the quadratic $az^2+bz+c=0$.
Therefore the conclusion is that if $b^2-4ac>0$, $ ax^2+bxy+cy^2 = 0 $ consists of 2 lines, if $b^2-4ac=0$, it is one line, and if $b^2-4ac<0$, it is only the point $x=y=0$.
Those are the connections, as far as I can see.
| {
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$y=\sqrt{\cot 2x} \;$ revolved around x-axis between $x = \frac{\pi }{12} \;$ and $x=\frac{\pi }{4} \;$ find volume The region between the curve $y=\sqrt{\cot 2x} \;\;$ and the x-axis, from $x=\frac{\pi }{12} \;\;$ to $x=\frac{\pi }{4} \;\;$ is revolved around the x-axis to generate a solid. Find the volume of the solid.
I tried finding critical points, by taking derivative, but it seems y' ≠ 0
so I just plugged in $\frac{\pi }{12}$ and $\frac{\pi }{4}$ into $\sqrt{\cot 2x}$ and it seems that at $\frac{\pi }{12}$ function has the maximum y-value of $3^\frac{1}{4}$.
My radius will vary, r = $(3^\frac{1}{4} - \sqrt{\cot2x})$
Area = $\pi \cdot r^2$
Volume = $\pi \cdot r^2 \cdot dx$
Integration:
$$ \int^{3^\frac{1}{4}}_{0} \pi (\sqrt{3} - 2\cdot3^\frac{1}{4} \cdot \sqrt{\cot2x} + \cot2x) $$
but wolfram can't integrate it...
is it even correct?
| HINT: The volume of the solid generated by rotating the curve $y=f(x)$ about the x-axis between $x=a$ & $x=b$ is $$=\color{red}{\int_{a}^{b}\pi y^2dx}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
$\bf{My\; try::}$ We can write $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldots ) = \frac{(1-x^3)(1)}{1-x^5}$$
So we can write it as $\displaystyle \frac{(1-x)(x^2+x+1)}{(1-x)(x^4+x^3+x^2+x+1)}$
So our Integral Convert into $\displaystyle \int_{0}^{1}\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$
Now How can I solve it, Help me
Thanks
| Here is a sketch of a way that does not use the digamma function.
If we start with your own favourite(?) change of variable,
$$
u=\frac{1}{x}-x,
$$
then, we end up with
$$
\int_0^{+\infty}\frac{2+u^2}{5+5u^2+u^4}\,du-\int_0^{+\infty}\frac{1}{\sqrt{4+u^2}(5+5u^2+u^4)}\,du
$$
In the first integral, it is easier (than in the original one) to do partial fraction decomposition. In the second integral, we let
$$
t=\frac{u}{\sqrt{4+u^2}},
$$
which will give us the integral
$$
\int_0^1\frac{1-t^2}{5+10t^2+t^4}\,dt,
$$
which also can be done using partial fraction decomposition (don't forget that it should come with a minus). I leave the funny calculations to you...
The final result (as already mentioned in several places before) turns out to be
$$
\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}.
$$
Probably there are smarter substitutions, leading to an even simpler expression to integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Integral by the given $u$-substitution Original equation $$\int\sqrt{e^t-5}dt \qquad u=\sqrt{e^t-5}$$
This is what I've done so far$$\int{u}dt$$ $$du=\frac{e^t}{2\sqrt{e^t-5}}dt$$$$dt=\frac{2\sqrt{e^t-5}}{e^t}du$$$$\int{u}{\frac{2\sqrt{e^t-5}}{e^t}}du$$ From this point on what would i need to do?
| Notice, let $e^t-5=u^2\implies e^tdt=2udu$ $$dx=\frac{2udu}{u^2+5}$$
$$\int\sqrt{e^t-5}dt=\int u\frac{2udu}{u^2+5}$$
$$=2\int \frac{u^2du}{u^2+5}$$ $$=2\int \frac{u^2+5-5du}{u^2+5}$$
$$=2\int \frac{u^2+5du}{u^2+5}-10\int \frac{du}{u^2+5}$$
$$=2\int du-10\int \frac{du}{u^2+5}$$ $$=2u-10\frac{1}{\sqrt 5}\tan^{-1}\left(\frac{u}{\sqrt 5}\right)+C$$
$$=2\sqrt{e^t-5}-2\sqrt 5\tan^{-1}\left(\sqrt{\frac{e^t-5}{5}}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
prove $2\sin^4(x)+2\cos^4(x)+\sin^2(2x)=2$ \begin{align}2\sin^4(x) + 2\sin^2(x)\cos^2(x) + 2\cos^4(x)&=2(\sin^4(x) + \sin^2(x)\cos^2(x) + \cos^4(x))\\
&=2(\sin^2(x) + \cos^2(x))^2 -\sin^2(x)\cos^2(x)\\
&=2(1) -\sin^2(x)\cos^2(x) \\
&=2-\sin^2(x)\cos^2(x)
\end{align}
And I'm stuck
| $$\sin^2 2x = (\sin 2x)^2 = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x,$$ therefore, the LHS is $$2 \sin^4 x + 4 \sin^2 x \cos^2 x + 2 \cos^4 x,$$ which immediately factors as $$2 (\sin^2 x + \cos^2 x)^2 = 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\alpha:\mathbb{R}\to\mathbb{R}$ where $\alpha(x)=\frac{x^{3}}{x^{2}+1}$ is bijective Using the usual method that I was taught, in order to find that something is injective, I assume $f(x)=f(y)$ for some $x,y\in\mathbb{R}$, and then I demonstrate that $x=y$. This is not working out so well for me, using the following steps:
\begin{align}
\frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
x^3(y^2+1)&=y^3(x^2+1)\\
x^3y^2+x^3&=x^2y^3+y^3\\
x^3-y^3&=x^2y^3-x^3y^2\\
(x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
(x^2+xy+y^2)&=-x^2y^2\\
x^2+y^2&=-x^2y^2-xy\\
x^2+y^2&=-xy(xy+1)
\end{align}
And here is where I'm stuck. Am I assuming correctly that I should find that $x=y$ somehow?
| Here is an explicit inverse.
$$x=\frac{\sqrt[3]{2y}}{\sqrt[3]{1+\sqrt{1+\frac{4}{27}y^2}}+\sqrt[3]{1-\sqrt{1+\frac{4}{27}y^2}}}$$
To find this, start with $y=\frac{x^3}{x^2+1}$, and solve for $x$ using Cardano's cubic solution. I subbed $z=x^{-1}$ to get a cubic that was already missing its quadratic term, which simplified the Cardano method a lot.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to Compute $\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}$
How to compute
$$\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=-\dfrac{3}{2}$$
I'm interested in more ways of computing limit for this expression.
My Thoughts
at least i tried to use L'Hospital's rule but with no luck
\begin{align}
&\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\lim _{x\to \:0}\dfrac{2x\left(\dfrac{\cos \left(x^2\right)}{\sin \left(x^2\right)+1}-1\right)}{\dfrac{2\arcsin \left(x\right)}{\sqrt{1-x^2}}-2x}\\
&=\lim _{x\to \:0}\dfrac{2\left(\dfrac{\cos \left(x^2\right)\left(\sin \left(x^2\right)+1\right)-2x^2\left(\cos ^2\left(x^2\right)+\sin ^2\left(x^2\right)+\sin \left(x^2\right)\right)}{\left(\sin \left(x^2\right)+1\right)^2}-1\right)}{2\left(\dfrac{x\arcsin \left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{1}{1-x^2}-1\right)}
\end{align}
Note this limit was taken from competition mathematics so i can't go with L'Hopital because each time i use it i got big terms that i have to derivative for the next time
| We will use the following results
$$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\text{ (via LHR)}$$ and $$\lim_{x \to 0}\frac{\arcsin x}{x} = \lim_{t \to 0}\frac{t}{\sin t} = 1\text{ (by putting }t = \arcsin x)$$
We can then proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - x^{2}}{(\arcsin x)^{2} - x^{2}}\notag\\
&= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - x^{2}}{x^{4}}\cdot\frac{x^{4}}{(\arcsin x)^{2} - x^{2}}\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - z}{z^{2}}\cdot\lim_{x \to 0}\frac{x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\text{ (putting }z = x^{2})\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z + \sin z - z}{z^{2}}\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{x}{\arcsin x}\cdot\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\left(\frac{\log(1 + \sin z) - \sin z}{z^{2}} - z\cdot\frac{z - \sin z}{z^{3}}\right)\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\left(\frac{\log(1 + \sin z) - \sin z}{z^{2}} - 0\cdot\frac{1}{6}\right)\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{(\arcsin x)^{3}}\cdot\frac{(\arcsin x)^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z}{\sin^{2}z}\cdot\frac{\sin^{2}z}{z^{2}}\notag\\
&\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{(\arcsin x)^{3}}{\arcsin x - x}\notag\\
&= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z}{\sin^{2}z}\cdot\lim_{t \to 0}\frac{t}{t + \sin t}\cdot\frac{t^{3}}{t - \sin t}\text{ (putting }t = \arcsin x)\notag\\
&= 6\lim_{u \to 0}\frac{\log(1 + u) - u}{u^{2}}\cdot\lim_{t \to 0}\dfrac{1}{1 + \dfrac{\sin t}{t}}\text{ (putting }u = \sin z)\notag\\
&= 3\lim_{u \to 0}\frac{\log(1 + u) - u}{u^{2}}\notag\\
&= 3\lim_{u \to 0}\dfrac{\dfrac{1}{1 + u} - 1}{2u}\text{ (via LHR)}\notag\\
&= -\frac{3}{2}\lim_{u \to 0}\frac{1}{1 + u}\notag\\
&= -\frac{3}{2}\notag
\end{align}
This involves 2 applications of LHR. The use of LHR can be replaced with Taylor series also. Note that both LHR and Taylor series are powerful tools to evaluate limits but they should always be used along with algebraic simplification of limit expression (and use of standard limits) otherwise their usage is a bit complicated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Identifying when vectors are linearly dependent Let u = (1, 2, 1), v = (0, 1, s) and w = (2, 0, t). Find the condition ons and t which makes the set {u, v, w} linearly dependent.
I set up the following matrix, and started to row reduce said matrix:
$ \left[
\begin{array}{ccc|c}
1&0&2&0\\
2&1&0&0\\
1&s&t&0
\end{array}
\right] $
$\rightarrow$$ \left[
\begin{array}{ccc|c}
1&0&2&0\\
0&1&-4&0\\
0&s&t-2&0
\end{array}
\right] $
$\rightarrow$ $ \left[
\begin{array}{ccc|c}
1&0&2&0\\
0&1&-4&0\\
0&s&t-2&0
\end{array}
\right] $
$\rightarrow$ $ \left[
\begin{array}{ccc|c}
1&0&2&0\\
0&1&-4&0\\
0&0&4s+(t-2)&0
\end{array}
\right] $
I then got confused as to where to go. I divided the last row by a factor of (4s+(t-2)) so that the matrix became
$ \left[
\begin{array}{ccc|c}
1&0&2&0\\
0&1&-4&0\\
0&0&1&0
\end{array}
\right] $
This of course would reduce to the identity, giving only the trivial solutions. That would suggest that irrespective of what s and t are the vectors could not be linearly dependent. Have I gone wrong some where, as the questions seems to suggest that there probably should be some case where the vectors are linear dependent
Thanks in advance!
| The set $\{u,v,w\}$ is linearly independent if and only if there exist three not all zero real coefficients $\alpha,\beta,\gamma$ such that
$$\alpha u+\beta v= w$$
Since the first component of $v$ is zero, we can easily conclude that $\alpha=2$. Knowing this and observing that the second coefficient of $w$ is zero, we have $\beta=-4$. Now,
$$\alpha \begin{pmatrix} 1 \\ 2 \\1 \end{pmatrix}+\beta \begin{pmatrix} 0 \\ 1 \\s \end{pmatrix} =\begin{pmatrix} 2 \\ 4 \\2 \end{pmatrix}+\begin{pmatrix} 0 \\ -4 \\-4s \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \\t \end{pmatrix}$$
$$\implies \begin{pmatrix} 2 \\ 0 \\2-4s \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \\t \end{pmatrix}$$
For the vectors to be linearly dependent, we must have $t=2-4s$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the cube roots of $-11-2i$. How do I find the roots of $\sqrt[3]{ - 11 - 2i}$ ?
Tried to use Moivre's theorem, but can not find the solutions by using the polar form:
$z_k=\sqrt{5}[\cos(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})+i.\sin(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})]$, for $k=0, 1$ and $2$.
Also resorted to $(a+bi)^3=-11-2i$ and did not have success, because I could not solve
$a³-3ab² = -11$ and $3a²b - b³ = -2$.
Solutions at the end of the book are $1+2i$ and $-\frac{1+2\sqrt{3}}{2}+\frac{\sqrt{3}-2}{2}i$ and $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.
| The argument of $-11-2i$ is not $\tan^{-1}\frac{2}{11}$ since the complex vector lies in the third quadrant. So, we have as principal value
$$\arg(-11-2i)=\tan^{-1}\frac{2}{11}-\pi$$
Now, for $k=1$ we have
\begin{align}
z_1&=\sqrt{5}\left[\cos\left(\frac{\tan^{-1}\frac{2}{11}-\pi+2\pi}{3}\right)+i\sin \left(\frac{\tan^{-1}\frac{2}{11}-\pi+2\pi}{3}\right)\right]\\
&=\sqrt{5}\left[\cos\left(\frac{\tan^{-1}\frac{2}{11}+\pi}{3}\right)+i\sin \left(\frac{\tan^{-1}\frac{2}{11}+\pi}{3}\right)\right]\\
\end{align}
Calculator give us
$$z_1=1+2i$$
The another two roots can be found by multiplying $z_1$ by $\omega$ and $\omega^2$, where $\omega$ is a cubic root of $1$, $\omega\neq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit of $x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$ when $x\to+\infty$
Find $$\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$$
I calculated $\lim\limits_{x\to\infty}\left(\frac{x+2}{x+1}\right)^x=e$ but then the limit in question becomes $0\times \infty $ form, and further solution becomes messy.
Please tell a solution without the use of series expansions because I have no knowledge of these.
| \begin{align}
\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)&=\lim_{x\to\infty}\frac{e-\left(\frac{x+2}{x+1}\right)^x}{\frac1x}\\
&=\lim_{x\to\infty}\frac{\frac{\left(\frac{1}{x+1}+1\right)^x \left(x-(x+1) (x+2) \log
\left(\frac{1}{x+1}+1\right)\right)}{(x+1) (x+2)}}{-\frac1{x^2}}\\
&=-e\lim_{x\to\infty}\left(x-(x+1) (x+2) \log
\left(\frac{1}{x+1}+1\right)\right)\\
&=-e\lim_{x\to0}\frac{x-(x+1) (2 x+1) \left(x-\frac{3 x^2}{2}+\frac{7 x^3}{3}-\frac{15 x^4}{4}+o(x^4)\right)}{x^2}\\
&=-e \lim_{x\to 0}(\frac{15 x^4}{2}+\frac{79
x^3}{12}-\frac{x^2}{4}+\frac{x}{6}-\frac{3}{2}+o(x^2))\\
&=\frac32e
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Shortest method for $\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$ I don't want to solve by expanding it and all, I tried corollary but denominator becomes messy, also in the options there is $\pi$ so tried several trigonometric substitutions too
| $f(z)=\frac{z^4(1-z)^4}{1+z^2}$ is a meromorphic function with simple poles at $z=\pm i$. Since
$$\text{Res}_{z=\pm i}\frac{z^4(1-z)^4}{(z+i)(z-i)}=\lim_{z\to \pm i}\frac{z^4(1-z)^4}{z\pm i}=\pm 2i \tag{1}$$
we have that
$$ f(z)-\left(\frac{2i}{z-i}+\frac{-2i}{z+i}\right) = f(z)+\frac{4}{z^2+1}\tag{2} $$
is an entire function: a polynomial, in particular. It follows that $z^4(1-z)^4+4$ is a multiple of $z^2+1$. Not surprising, since by Sophie Germain's identity
$$ w^4+4 = (w^2-2w+2)(w^2+2w+2) \tag{3}$$
hence:
$$ z^4(1-z)^4+4 = (1+z^2)(2-2z+z^2)(2+2 z-z^2-2 z^3+z^4)\tag{4} $$
as well as:
$$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx = -\pi+\int_{0}^{1}(2-2z+z^2)(2+2 z-z^2-2 z^3+z^4)\,dz.\tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\frac{1}{2}(x-y)$ is prime where $x,y$ are distinct odd primes and $y^2 + y$ divides $x^2+x$ I need help with the following exercise from the Polish XI Olimpiada Matematyczna Gimnazjalistów (question 6 from this document. The submission deadline of 12 October 2015 has passed.
Assume that $x≠y$ are both prime, odd and that $y^2+y$ divides $x^2+x$. Prove that $\frac{1}{2}(x−y)$ is composite.
I think I'm really close to find a solution. My approach takes advantage of the Euclidean algorithm:
$\gcd(x^2+x, y^2+y) = \gcd(x^2-y^2+x-y, y^2+y)=\gcd((x-y)(x+y)+(x-y), y^2+y)=\gcd((x-y)(x+y+1),y^2+y)=y^2+y$
Now it's true, that:
$\frac{(x-y)(x+y+1)}{(y^2+y)}=a$, where $a$ is natural greater than $1$
$(x-y)(x+y+1)=a(y^2+y)$
$\frac{1}{2}(x-y)=\frac{a(y^2+y)}{2(x+y+1)}$
I think now I should prove, that $\frac{a(y^2+y)}{2(x+y+1)}$ is composite, but I don't know how to.
| You are given
$$y(y+1)\mid x(x+1) \tag{1}$$
so the fact that $y\ge3$ and $y\ne x$ implies that $y<x$. Since $x,y$ are both odd primes $y\le x-2$ and $y\ge3$.
The since $x$ is prime and $y<y+1<x$ we must have
$$\begin{array}{ll}
\gcd(y,x)=1\text{ and }\gcd(y+1,x)=1 \\
\implies \color{blue}{y(y+1)\mid x+1} &(\text{from (1)}) \\
\implies x+1=ky(y+1),\quad k\ge1,k\in\mathbb{Z}
\end{array}$$
So $\frac{1}{2}(x-y)=\frac{1}{2}(ky-1)(y+1)=(ky-1)(\frac{y+1}{2})$. Since $k\ge1,y\ge3$ have that $ky-1\ge2$ is an integer, and $y\ge3$ and odd means that $\frac{y+1}{2}\ge2$ is an integer, so $\frac{1}{2}(x-y)$ has two integer factors of at least two, so cannot be a prime.
A solution to (1) is $x=11,y=3$ from which we see that $y(y+1)=12,x+1=12 \implies k=1$, so then $ky-1=2$ and $\frac{y+1}{2}=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^2 + 2^n$ is composite if $n\not\equiv3\pmod{6}$.
If $n>1$ is an integer not of the form $6k+3$, Prove that $n^2 + 2^n$ is composite.
Any idea of how to think about this problem? I have been thinking about it a lot and yet I was not able to come up with anything.
| If $n$ is even, the proof is trivial. Let $n=2m$, then $n^2+2^n=4m^2+4^m$ which is divisible by 4.
If $n$ is odd, consider $n=6k+1$.
When $k=1, n^2+2^n=177$
When $k=2, n^2+2^n=8361$
When $k=3, n^2+2^n=52469$
We wish to show that for all $k>0$, $(6k+1)^2+2^{6k+1} \equiv 0\pmod 3$. Let $a_k=(6k+1)^2+2^{6k+1}$. Assume the case $n=k$ is true. When $n=k+1$
$$a_{k+1}=(6(k+1)+1)^2+2^{6(k+1)+1}=(6k+1)^2+36+12(6k+1)+64(2^{6k+1})\equiv (6k+1)^2+2^{6k+1}=a_k$$
Thus, this means $$a_k\equiv a_{k-1} \equiv a_{k-2}\equiv...\equiv a_1\equiv 0\pmod 3$$
Our inductive proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
What is the 1393th number? What is the 1393th number in this string??
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5........
Could you explain me how to answer this kind of questions .
| It will be a bit easier if we start counting at 0.
start 0 1 3 6 10 15
index 0, 1 2 3 4 5 6 7 8 9 10 11 12 12 14 15 16
number 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6,........
The number 1 starts at index 0 = 0
The number 2 starts at index 1 = 0+1.
The number 3 starts at index 3 = 0+1+2
The number 4 starts at index 6 = 0+1+2+3
The number 5 starts at index 10 = 0+1+2+3+4
The number 6 starts at index 15 = 0+1+2+3+4+5
The number n starts at index n(n-1)/2.
So we need to find the largest n such that
n(n-1)/2 ≤ 1392 (because we started counting at 0 instead of 1)
n(n-1) ≤ 2784
Since $n(n-1) = 2784$ is about $n^2 = 2784$, a good guess for $n$ would be $\lfloor \sqrt{2784} \rfloor = 52$
We compute
\begin{align}
52\times 51 &= 2652\\
53 \times 52 &= 2756\\
54 \times 53 &= 2862\\
\end{align}
So the number 53 starts at index 53(53-1)/2 = 1378
and the number 54 starts at index 54(54-1)/2 = 1431.
That means that the 1379th number is 53
and the 1432th number is 54.
So the 1393th number is 53.
| {
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"url": "https://math.stackexchange.com/questions/1454847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
| Hint: Prove that $\bigg(1+\dfrac13-\dfrac15-\dfrac17+\dfrac19+\dfrac1{11}\pm\cdots\bigg)^2=1+\dfrac1{3^2}+\dfrac1{5^2}+\dfrac1{7^2}+\dfrac1{9^2}+\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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My approach to determine if the following series is convergent I've just started a course in real analysis, and I don't know if I'm doing this right. Here's the exercise:
Determine if the series $$\sum_{n=1}^{\infty}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}$$ converges or diverges.
Here's what I've done so far:
$$\sum_{n=1}^{\infty}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}\\\Downarrow\\\lim_{N\to\infty}\sum_{n=1}^{N}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}\\\Downarrow\\\underbrace{\lim_{N\to\infty}\sum_{n=1}^{N}\frac{3n}{(n^2-3n+20)^2}}_{\large\text{Part 1}}+\underbrace{\lim_{N\to\infty}\sum_{n=1}^{N}\frac{\sqrt{n^3+2}}{(n^2-3n+20)^2}}_{\large\text{Part 2}} $$
From here:
Part 1
We're going to use the limit comparison test. The term $\displaystyle\ a_n=\frac{3n}{(n^2-3n+20)^2}$ behaves like $\displaystyle\ b_n=\frac{3n}{n^4}=\frac{3}{n^3}$.
We know (using the results obtained in theory) that $$\sum_{n=1}^{\infty}\frac{3}{n^3}=3\sum_{n=1}^{\infty}\frac{1}{n^3}<\infty $$.
Furthermore:
$$\lim_{n\to\infty}\displaystyle\frac{\frac{3n}{(n^2-3n+20)^2}}{\frac{3n}{n^4}}=\lim_{n\to\infty}\displaystyle\frac{n^4}{(n^2-3n+20)^2}=1$$
Then, by the limit comparison test, the series $$\sum_{n=1}^{\infty}\frac{3n}{(n^2-3n+20)^2}$$
converges.
Part 2
We do the same. Using the limit comparison test, we compare $\displaystyle\ a_n=\frac{\sqrt{n^3+2}}{(n^2-3n+20)^2}$ with $\displaystyle\ b_n=\frac{\sqrt{n^3}}{n^4}.$ If we proceed as before, we will find that $$\lim_{n\to\infty}\frac{a_n}{b_n}=1.$$
So the series converges, and as both part converges, the series $$\sum_{n=1}^{\infty}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}$$ converges.
Is this right? Am I usin the limit comparison the right way?
Thank you.
| Note that ${3n+\sqrt{n^3+2} \over (n^2-3n+20)^2} = {n^{3 \over 2} \over n^4} { {3 \over \sqrt{n} }+ \sqrt{1+{2 \over n^3} } \over 1-{3 \over n} + {20 \over n^2} }$.
It is easy to see that ${3 \over \sqrt{n} }+ \sqrt{1+{2 \over n^3}} \le 5$ for all $n$, and for $n \ge 9$, we have
$1-{3 \over n} + {20 \over n^2} \ge {1 \over 3}$, so
${3n+\sqrt{n^3+2} \over (n^2-3n+20)^2} \le 15 { 1\over n^{5 \over 2}} $.
Since ${5 \over 2} > 1$ we know that the series $\sum_n { 1\over n^{5 \over 2}} $ is convergent, and so by the comparison test, so is the original series.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm. Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm.
I let two non-parallel sides and one of the parallel sides as $k$(shorter one).I know that area of the trapezium is $\frac{1}{2}\times $sum of parallel sides $\times$ height.But in this question,neither height is given nor longest side is given.How should i formulate the equation of the area?
|
The diagram shows the sides $k$ and included angles $\theta$ and $\pi-2\theta $ of the trapezium. Then its area $A$ can be expressed in terms of $k$ and $\theta$ now.
$$
\begin{aligned}
A &=k^{2} \sin \theta+\frac{k^{2}}{2} \sin (\pi-2 \theta) \\
&=\frac{k^{2}}{2}(2 \sin \theta+\sin 2 \theta)
\end{aligned}
$$
$$
\begin{aligned}
&\quad \frac{d A}{d \theta}= k^{2}(\cos \theta+\cos 2 \theta)=0 \\\Leftrightarrow & \quad \cos \theta+\cos 2 \theta=0 \\
\Leftrightarrow & \quad 2 \cos ^{2} \theta+\cos \theta-1=0 \\
\Leftrightarrow &\quad (2 \cos \theta-1) (\cos \theta+1)=0\\
\Leftrightarrow &\quad \cos \theta=\frac{1}{2} \text { or }-1 \textrm{ (rejected) }
\end{aligned}
$$
Hence
$$
\begin{aligned}
A_{\max } &=\frac{k^{2}}{2}\left(2 \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right) =\frac{3 \sqrt{3}}{4} k^{2} \quad cm^2
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove this inequality $\frac{a^2}{a+b}+\frac{b^2}{b+c}\ge\frac{3a+2b-c}{4}$ Let $a,b,c\ge 0$,show that
$$\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}\ge\dfrac{3a+2b-c}{4}$$
| By AM-GM:
$$\frac{a^2}{a+b}+\frac{a+b}{4}\ge a$$
and
$$\frac{b^2}{b+c}+\frac{b+c}{4}\ge b.$$
So
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}\ge a+b-\frac{a+b}4-\frac{b+c}4=\frac{3a+2b-c}4.$$
| {
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There is a smart way to solve this 2x2 system of linear equations? I was helping a high school student to study system of equations and he showed me this problem:
\begin{align}\frac{x-a}{b} + \frac{y-b}{a} &= x-a \\
\frac{x+y-b}{a}+\frac{y+b}{b} &= 3 .
\end{align}
I suggested to him to solve for one of the variables (e.g.. $x$) and then substitute, but the problem becomes painfully hard, because of all the literals, and fractions. So, there is a intelligent way to solve this system of equations?
| Given the system
$\frac{x-a}{b} + \frac{y-b}{a}=x-a\\
\frac{x+y-b}{a}+\frac{y+b}{b}=3$
it is worth rearranging prior to solving, as
$$ax-a^2-axb+yb-b^2+a^2b=0\\
bx+by-b^2+ya+ab-3ab=0$$
Separating the variables and constants,
$$a(1-b)x+yb=a^2+b^2-a^2b\tag 1$$
$$bx+(a+b)y=b^2+2ab\tag 2$$
The "slopes" of the two equations are generally unrelated, and except for multiplying by similar quantities across both equations, we can't isolate variables:
$$(1)\cdot (a+b)+(2)\cdot (-b)\\ \to a(a+b)(1-b)x+b(a+b)y-b^2x-b(a+b)y=(a+b)(a^2+b^2-a^2b)-b^3-2ab^2\\
(a(a+b)(1-b)-b^2)x=a^3+a^2b-ab^2-a^3b-a^2b^2\\
(a^2+ab-a^2b-ab^2-b^2)x=a(a^2+ab-a^2b-ab^2-b^2)\\
x=a\tag 3$$
With this solution in hand, we immediately see that $y=b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
| Let $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{4\sin^2 x+5\cos^2 x}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^2 x\;,$ We get
$$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{4\tan^2 x+5}dx\;,$$
Now Put $\displaystyle 2\tan x=t\;,$ Then $\displaystyle 2\sec^2 xdx =dt\Rightarrow \sec^2 xdx = \frac{1}{2}dt$
and Changing Limit , We get
$$\displaystyle I = \frac{1}{2}\int_{0}^{\infty}\frac{1}{t^2+5}dt = \frac{1}{2\sqrt{5}}\left[\tan^{-1}\left(\frac{t}{\sqrt{5}}\right)\right]_{0}^{\infty}=\frac{1}{2\sqrt{5}}\cdot \frac{\pi}{2}=\frac{\pi}{4\sqrt{5}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorizing $(11 - \sqrt{-14})$ in $\mathbb{Z}[\sqrt{-14}]$ as product of maximal ideals For an exercise I would like to:
Factorize $(11 - \sqrt{-14})$ in $\mathbb{Z}[\sqrt{-14}]$ as a product of maximal ideals
which is possible as it is a Dedekind domain (it is the ring of integers of $\mathbb{Q}(\sqrt{-14})$).
I do not know where to start however.
I thought that the norm may be useful, so I computed the norm of an element $a + b \sqrt{-14}$ as $a^2 + 14b^2$, which gives us that the norm of $11 - \sqrt{-14} = 135 = 3^3\cdot 5$. Is this also the norm of the ideal? (I know this is the case for the ideal $\alpha \mathcal{O}_K$)
Is there an easy way to find ideals with norm $3$ and $5$? According to my computation of the norm they must be of the form $(a,b)$ (non-principal).
EDIT:
In a linked question I found some more help, which gives me:
$\mathbb{Z}[\sqrt{-14}] = \mathbb{Z}[X]/(X^2 + 14)$ and modulo $3$ we get $(X^2 + 14) = (X + 1)(X+2)$. Does this imply that there are two prime ideals with norm $3$, namely $(3, -2 + \sqrt{-14})$ and $(3, 2 + \sqrt{-14})$?
I did the same modulo $5$, which gives us $(X^2 + 14) = (X+4)(X-4)$ which gives us the ideals $(5,-4 + \sqrt{-14})$ and $(5,4 + \sqrt{-14})$. So I assume these are again ideals of norm $5$ and that $(11-\sqrt{-14})$ must be some combination of these $4$ ideals.
| I think I have solved it myself.
We know that we must find $3$ prime ideals with norm $3$ and one with norm $5$.
We start with norm $5$, and see that $3\cdot 5 + 1\cdot(-4 + \sqrt{-14}) = 11- \sqrt{-14}$ which means that $(5,-4 + \sqrt{-14}) \supset (11- \sqrt{-14})$ and therefore $(5,-4 + \sqrt{-14}) | (11- \sqrt{-14})$ which covers the norm $5$ part.
Because $3\cdot 3 + 1\cdot (2 + \sqrt{-14}) = 11- \sqrt{-14}$ we can also say that $(3,2 + \sqrt{-14}) | (11- \sqrt{-14})$.
Now we shall show that $(3) \not \supset (11- \sqrt{-14})$ which implies that $(3) \not | (11- \sqrt{-14})$ and therefore $(3,-2 + \sqrt{-14}) \not| (11- \sqrt{-14})$.
Take $a + b\sqrt{-14} \in \mathbb{Z}[\sqrt{-14}]$ then all elements of $(11- \sqrt{-14})$ are of the form
$$(a + b\sqrt{-14})(11- \sqrt{-14}) = 11a + 14b + (11b-a)\sqrt{-14}$$
So certainly if $3$ was in this ideal, then $$11a + 14b = 3$$ $$11b - a = 0$$
But then $121b + 14b = 3$ for some $b \in \mathbb{Z}$, which is not possible.
Therefore $$(11- \sqrt{-14}) = (3,2 + \sqrt{-14})^3 (5,-4 + \sqrt{-14})$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ A question on submultiple angles states...
Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$
My efforts
I tried using the formula
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
and
$$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$
Then I tried simplifying it:
$$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$
I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.
Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.
Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.
| $$\sin^6\frac{\theta}{2}+\cos^6\frac{\theta}{2}=\cdots=1-3\left(1-\cos^2\frac{\theta}{2}\right)\cos^2\frac{\theta}{2}$$
is correct. From here, note that
$$\begin{align}\sin^6\frac{\theta}{2}+\cos^6\frac{\theta}{2}&=1-3\left(1-\color{red}{\cos^2\frac{\theta}{2}}\right)\color{red}{\cos^2\frac{\theta}{2}}\\&=1-3\left(1-\color{red}{\frac{\cos\theta+1}{2}}\right)\cdot\color{red}{\frac{\cos\theta+1}{2}}\\&=1-3\cdot\frac{1-\cos\theta}{2}\cdot\frac{1+\cos\theta}{2}\\&=\frac{4-3(1-\cos^2\theta)}{4}\\&=\frac{1+3\cos^2\theta}{4}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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square root of $\frac{2+\sqrt{3}}{4}$
Find the square root of $\frac{2+\sqrt{3}}{4}$.
Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?
| To add to the other answers, notice that:
$$\frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4},$$
where I've used zezanjee's hint
$$2 + \sqrt{3} = \frac{(\sqrt{3} + 1)^2}{2}$$
in the comments.
| {
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Generating functions for finding the coefficients I am new to the field of combinatorics and I recently came across a problem where it was asked to find the number of integer solutions to ${c_1 + c_2 + c_3 + c_4=20 }$ where ${c_i\ge 0}$ for all ${1\le i\le4}$ with ${c_2}$ and ${c_3}$ being even integers.
Using the generating function, we know that the solution would be the coefficient of ${x^{20}}$ in the expansion of ${(1 +x+x^2+x^3+...)^2*(1+x^2+x^4+x^6+...)^2}$ which is equivalent to: ${(1-x)^{-2}*(1-x^2)^{-2}}$
Now, we know that the first part could be solved by computing:${(-1)^{20}\dbinom{-2}{20} = \dbinom{2+20-1}{20}=21}$
Since we know that: ${(1+x^m)^n = \dbinom{n}{0} + \dbinom{n}{1}x^m + \dbinom{n}{2}x^{2m}+...+\dbinom{n}{n}x^{nm}}$
Would it be correct to assume that the coefficient of ${x^{20}}$ in ${{(1-x^2)^{-2}}}$ would be: ${\dbinom{-2}{10}}$
| Yes, the coefficient of $x^{20}$ in $(1-x^2)^{-2}$ is the coefficient of $y^{10}$ in $(1-y)^{-2}$ which is $(-1)^{10}\binom{-2}{10}=\binom{-2}{10}$.
As shown in this answer,
$$
\binom{-2}{10}=(-1)^{10}\binom{11}{10}=11
$$
Computation of the Coefficients
$$
\begin{align}
(1-x)^{-2}(1-x^2)^{-2}
&=\sum_{j=0}^\infty(-1)^j\binom{-2}{j}x^j\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{2k}\\
&=\sum_{j=0}^\infty\sum_{k=0}^{\lfloor j/2\rfloor}(-1)^{j-k}\binom{-2}{j-2k}\binom{-2}{k}x^j\\
&=\sum_{j=0}^\infty\sum_{k=0}^{\lfloor j/2\rfloor}\binom{j-2k+1}{j-2k}\binom{k+1}{k}x^j\\
&=\sum_{j=0}^\infty\sum_{k=0}^{\lfloor j/2\rfloor}(j-2k+1)(k+1)x^j\\
&=\sum_{j=0}^\infty\left[(j+1)\binom{\lfloor j/2\rfloor+2}{2}-4\binom{\lfloor j/2\rfloor+2}{3}\right]x^j\\
&=\sum_{j=0}^\infty\frac{j+3}{48}\left[\left(2j^2+12j+13\right)+(-1)^j3\right]x^j
\end{align}
$$
using the fact that $\lfloor j/2\rfloor=\frac{2j-1+(-1)^j}4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $z^5-32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5-32)=(z-2)(z^2-pz+4)(z^2-qz+4)$,then find $p^2+2p.$ If the expression $z^5-32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5-32)=(z-2)(z^2-pz+4)(z^2-qz+4)$,then find the value of $p^2+2p.$
I used $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$ to get $z^5-2^5=(z-2)(z^4+2z^3+4z^2+8z+16)$
Then i compared $z^4+2z^3+4z^2+8z+16$ with $(z^2-pz+4)(z^2-qz+4)$
to get $p+q=-2$ and $p^2q^2=2$
$pq=\pm\sqrt2$
But when i solve the two equations $pq=\pm\sqrt2$ and $p+q=-2$,i get the value of $p$ and $q$ whch are non-manageable difficult to simplify.And i am not able to calculate the final desired result $p^2+2p$.
Is my approach not correct?What is the simple and elegant method to do this?Please help me.
| Equating the coefficients of $z^2,$
$$4+pq+4=4\iff pq=-4$$
Equating the coefficients of $z^3,$
$$-2=p+q$$ So, $p,q$ are the roots of $$t^2+2t-4=0$$
| {
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I've formulated a proof, but also a counterexample? show that f(x) is injective.
$f(x)= \frac{x^2}{1+x^2}$
if $f(x)=f(y)$ then $\frac{x^2}{1+x^2}=\frac{y^2}{1+y^2}$
$(x^2)(1+y^2)=(y^2)(1+x^2)$
$x^2+x^2y^2=y^2+x^2y^2$
$x^2 = y^2$
$x = y$
but $f(1)=f(-1)$ due to the square roots. Where did I go wrong in the original proof?
| If $x^2=y^2$ then $x=y$ or $x=-y$.
You suppose only $x=y$.
There is the mistake.
| {
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Checking primality for $2 \uparrow \uparrow n + 3 \uparrow \uparrow n$ Is there a clever way to test primality for
$$2 \uparrow \uparrow n \quad + \quad 3 \uparrow \uparrow n$$
where $n \gt 3$? (Not surprisingly, I got stuck after that)
For $n \le 3$ we get:
$$
\begin{matrix}
2 \uparrow \uparrow 0 \quad + \quad 3 \uparrow \uparrow 0 & = & 1 + 1 & = & 2 & (\text{prime}) & (\text{...}) \\
2 \uparrow \uparrow 1 \quad + \quad 3 \uparrow \uparrow 1 & = & 2 + 3 & = & 5 & (\text{prime}) & (\text{yes}) \\
2 \uparrow \uparrow 2 \quad + \quad 3 \uparrow \uparrow 2 & = & 2^2 + 3^3 & = & 31 & (\text{prime}) & (\text{okay}) \\
2 \uparrow \uparrow 3 \quad + \quad 3 \uparrow \uparrow 3 & = & 2^{2^2} + 3^{3^3} & = & 7625597485003 & (\text{prime}) & (\text{hmm...}) \\
\end{matrix}
$$
Example:
One simple way to rule out primality is to check if the last digit of the sum is $5$. Now we know that the last digit of $2^k$ and $3^k$ follow a pattern such that for $2^k$ with $k \equiv 0 \pmod{4}$ the last digit is $6$ and for $3^k$ with $k \equiv 3 \pmod{4}$ the last digit is $7$. From this we can state that for $n \gt 2$ the last digit of $2 \uparrow \uparrow n$ will be $6$ and for $n \gt 1$ the last digit of $3 \uparrow \uparrow n$ will be $7$, which means we can't rule out primality this way.
| $$34,276,387$$ is a prime factor of $2\uparrow\uparrow 5+3\uparrow\uparrow 5$ ,found by Sheldon L
(See the link above).
I do not know a prime factor of $2\uparrow\uparrow 4+3\uparrow\uparrow 4$, but the number is very probably composite, and if it is prime, even a probable-prime
test will be out of reach.
| {
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Calculate expression: $\cos\alpha-\cos2\alpha$ I attempt calculate this expression:
$$\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{5}$$
Please, help me, somebody. Thanks.
| $\cos\dfrac{\pi}{5}=\cos36^\circ$ and the next is $\cos\dfrac{2\pi}{5}=\cos72^\circ$. Now so by converting into $\sin$ we have to do
$$\sin54^\circ-\sin18^\circ,$$
which can be done by following.
Let $\alpha=18^\circ$, therefore $5\alpha=90^\circ$. So $\sin2\alpha=\sin(90^\circ-3\alpha)$, so $\sin2\alpha=\cos3\alpha$. Expanding both sides, and converting to $\sin$ we have:
$$4\sin^{2}\alpha+3\sin\alpha-1=0.$$ Hence, $\sin\alpha=\dfrac{\sqrt{5}-1}{4}$. Similarly find
$$\sin3\alpha=\dfrac{\sqrt{5}+1}{4}.$$
So answer is
$$\dfrac{\sqrt{5}+1}{4}-\dfrac{\sqrt{5}-1}{4}=\dfrac{1}{2}.$$
| {
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"url": "https://math.stackexchange.com/questions/1481369",
"timestamp": "2023-03-29T00:00:00",
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If $\int^1_0 \frac{\sin x}{1+x}dx =I $ then.... Problem :
If $\displaystyle \int^1_0 \dfrac{\sin x}{1+x}dx =I $ then $\displaystyle \int^{4\pi}_{4\pi-2} \dfrac{\sin\frac{x}{2}}{4\pi +2-x}dx $ =?
Options are :
(a) $I$ (b) $-I$ (c) $2I$ (d) $-2I$.
Please suggest how to proceed in such problem and how to make to integrals comparable to get the relation between them, will be of great help thanks.
| Given $$\displaystyle \int_{0}^{1}\frac{\sin x }{1+x} = I\;,$$ and $$\displaystyle \int_{4\pi-2}^{4\pi}\frac{\sin \frac{x}{2}}{4\pi+2-x}dx$$
Now Let $$\displaystyle J = \int_{4\pi-2}^{4\pi}\frac{\sin \frac{x}{2}}{4\pi+2-x}dx$$
Using the formula $$\displaystyle \bullet\; \int_{a}^{b}f(x)dx = (b-a)\int_{0}^{1}f\left[(b-a)x+a\right]dx$$
So we get $$\displaystyle J = 2\int_{4\pi-2}^{4\pi}\frac{\sin \left(\frac{2 x+4\pi-2}{2}\right)}{4\pi+2-\left(2x+4\pi-2\right)}dx = 2\int_{0}^{1}\frac{\sin \left(2\pi+x-1\right)}{2(2-x)}dx$$
So we get $$\displaystyle J = \int_{0}^{1}\frac{\sin (x-1)}{1+(1-x)}dx = -\int_{0}^{1}\frac{\sin (1-x)}{1+(1-x)}dx$$
Now Put $(1-x) = t$ and $dx = -dt$ and changing limit , we get
$$\displaystyle J = -\times -\int_{1}^{0}\frac{\sin t}{1+t} dt = -\int_{0}^{1}\frac{\sin t}{1+t}dt = -\int_{0}^{1}\frac{\sin x}{1+x}dx = -I$$
Above we have used the formulae
$$\displaystyle \bullet\ \int_{a}^{b} f(x)dx = -\int_{b}^{a}f(x)dx$$ and $$\displaystyle \bullet\; \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$
| {
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Find the maximum possible value of $\frac{A}{\gcd(A,B)} \text{ where } A={100\choose k} \text{ and } B={100\choose k+3}$ such that $30\leq k\leq70$:
For an integer $30\leq k\leq70$, let $M$ be the maximum possible value of
$$\frac{A}{\gcd(A,B)} \text{ where } A={100\choose k} \text{ and } B={100\choose k+3}$$
Find $M \mod 100$.
Okay, so directly, lets come to my attempt:
$$\gcd(a,b)=\frac{a.b}{[a,b]}$$ where $[.]$ denotes the LCM. And then we can apply differentiation to find out the min of the denominator. But what would be the LCM of the two?
| Slightly more generally, if $A = {n \choose k}$ and $B = {n \choose k+3}$, $0 \le k \le n-3$, we have
$$ \dfrac{A}{B} = \dfrac{(k+1)(k+2)(k+3)}{(n-k)(n-k-1)(n-k-2)}$$
If this is $s/t$ in lowest terms, then $\gcd(A,B) = A/s = B/t$.
You want to maximize (in the case $n=100$ with $30 \le k \le 70$) $A/\gcd(A,B) = s$, i.e. the numerator of $A/B$. Let $f(k) = (k+1)(k+2)(k+3)$ and
$g(k) = (100-k)(99-k)(98-k)$, so $$s = \dfrac{f(k)}{\gcd(f(k),g(k)}$$
$(k+1)(k+2)(k+3)$ increases pretty rapidly with $k$, so we should look at large $k$'s. On the other hand, we want to avoid common factors of $f(k)$ and $g(k)$ as much as possible. Some common factors are unavoidable: both $f(k)$ and $g(k)$ must be divisible by $6$, because at least one of $k+1,k+2,k+3$ is divisible by $2$ and at least one is divisible by $3$, and similarly for $100-k,99-k,98-k$. Can we make $\gcd(f(k), g(k)) = 6$?
To avoid having a common factor of $4$ we want to avoid $k \equiv 2$ or $3 (\text{mod } 4)$. To avoid having a common factor of $5$, we want to avoid $k \equiv 3$ or $4 (\text{mod } 5)$. This would rule out $k = 62, 63, 64, 66, 67, 68, 69, 70$. $k = 65$ would be a good candidate, but
$f(65)$ and $g(65)$ turn out to be divisible by $11$ and $17$. This leads
us to $61$, where indeed $f(61) = 2^7 \times 3^2 \times 7 \times 31$ and
$g(61) = 2 \times 3 \times 13 \times 19 \times 37$ have gcd $6$. Clearly
$$ \dfrac{f(k)}{\gcd(f(k),g(k))} \le \dfrac{f(61)}{6} \ \text{for } 0 < k \le 60 $$
while for $62 \le k \le 70$ we have $\gcd(f(k),g(k)) \ge 12$ and $f(k) \le f(70) < 2 f(61)$. Thus the maximum is
$$ \dfrac{f(61)}{6} = 41664$$
| {
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Verification of Green's Theorem homework help QUESTION: Verify Green's Theorem for $$\oint (x^2-2xy)dx+(x^2y+3)dy$$ around the curve $y^2=8x$ and $x=2$.
My attempt:
L.H.S. comes out to be $\frac{128}{5}$ which is correct acc. to the book.
For R.H.S. we have $$\int \int \left(\frac{\partial}{\partial x}(x^2y+3) -\frac{\partial}{\partial y}(x^2-2xy)\right)dx \,\ dy$$
$$=\int \int (2xy+2x)dx \,\ dy$$
$$=\int \int 2x(y+1)dx \,\ dy$$
$$=\int (y+1) \left(\int_0^{\frac{y^2}{8}} 2xdx\right) \,\ dy$$
$$=\int \frac{y^4}{64}(y+1) \,\ dy$$
$$=\frac{1}{64} \int_{-4}^4 (y^5+y^4) \,\ dy$$
$$=\frac{2}{64}\frac{4^5}{5}$$
$$=\frac{32}{5}$$
| It seems you made a mistake in identifying the integration limits. :)
In fact, you should have written
$$\eqalign{
& \int\!\!\!\int\limits_\Omega {\left( {{\partial \over {\partial x}}\left( {{x^2}y + 3} \right) - {\partial \over {\partial y}}\left( {{x^2} - 2xy} \right)} \right)da} \cr
& = \int\!\!\!\int\limits_\Omega {\left( {2xy + 2x} \right)da} = \int\!\!\!\int\limits_\Omega {2x\left( {y + 1} \right)da} \cr
& = \int_0^2 {\int_{ - \sqrt x }^{\sqrt x } {2x\left( {y + 1} \right)dydx} } = \int_0^2 {2x\left( {\int_{ - \sqrt {8x} }^{\sqrt {8x} } {\left( {y + 1} \right)dy} } \right)} dx \cr
& = ... \cr} $$
or this which is the solution you tried
$$\eqalign{
& \int\!\!\!\int\limits_\Omega {\left( {{\partial \over {\partial x}}\left( {{x^2}y + 3} \right) - {\partial \over {\partial y}}\left( {{x^2} - 2xy} \right)} \right)da} \cr
& = \int\!\!\!\int\limits_\Omega {\left( {2xy + 2x} \right)da} = \int\!\!\!\int\limits_\Omega {2x\left( {y + 1} \right)da} \cr
& = \int_{ - 4}^4 {\int_{{{{y^2}} \over 8}}^2 {2x\left( {y + 1} \right)dxdy} } = \int_{ - 4}^4 {\left( {y + 1} \right)\left( {\int_{{{{y^2}} \over 8}}^2 {2xdx} } \right)dy} \cr
& = ... \cr} $$
| {
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Inverse of half a polynomial I am given the assignment of finding the inverse of the following function:
$$f(x) = x^2 + 4x + 5$$
where $x > -2$.
I can get rid of the $5$ obviously, but then I get stuck. Hints are appreciated.
| You want to solve $y = x^2 + 4x + 5$ when $x > -2$, or $x^2 + 4x + (5 - y) = 0$. Solving for $x$ using the quadratic formula, we have
$$ x = \frac{-4 \pm \sqrt{16 - 4(5 - y)}}{2} = \frac{-4 \pm \sqrt{4(y - 1)}}{2} = -2 \pm \sqrt{y - 1}.$$
Since $x > -2$, we need to take the solution with the $+$ and so $x = -2 + \sqrt{y - 1}$.
| {
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
| $$\begin{align}z^6&=1\\
z^6-1&=0\\
(z-1)(\underbrace{z^5+z^4+z^3+z^2+z+1}_{f(z)})&=0
\end{align}$$
Hence the solutions to $f(z)$ are
$$\begin{align}z^6&=1=e^{i2\pi} &&(z\neq1)\\
z&=e^{i\frac {2n\pi}6}=e^{i\frac {n\pi}3}&& (n=1,2,3,4,5)\quad\blacksquare\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Remainder of $7^{7^{7}}$ divided by $32$ The problem is to find remainder of $7^{7^{7}}$ divided by $32$ the quickest way.
My attempt:
$32 = 8.4$
Mod 8: $7^{7^{7}}\equiv(-1)^{7^{7}}\equiv-1$
Mod 4: $7^{7^{7}}\equiv(-1)^{7^{7}}\equiv-1$
Hence in mod $32$: $7^{7^{7}}\equiv(-1)$ mod $8.(-1)$mod$ 4 (?) \equiv 7.3\equiv21 $
I am quite worried about the last step. I can't justify it.
Is there a theorem related to (?) or
Is there a better and correct way to do this ?
| You are correct that $7^{4}-1 = 48 \times 50 \equiv 0$ (mod $32$). Now $7^{7} \equiv (-1)^{7} \equiv -1$ (mod $4$). Hence $7^{7^{7}} \equiv 7^{-1}$ (mod $32$).
Now we have seen that $7^{4} \equiv 1$ (mod $32$), so that $7^{-1} \equiv 7^{3}$
(mod $32$). But $7^{2} = 49 \equiv 17$ (mod $32$), and $7^{3} \equiv 7 \times 17 \equiv 119 \equiv 23 \equiv -9$ (mod $32$).
| {
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How show this $\overrightarrow {AC}\cdot\overrightarrow {BD}=\frac{1}{2}[(b^2+d^2)-(a^2+c^2)]$ In Convex quadrilateral $ABCD$, such $|AB|=a,|BC|=b,|CD|=c,|DA|=d$, show that
$$\overrightarrow {AC}\cdot\overrightarrow {BD}=\dfrac{1}{2}[(b^2+d^2)-(a^2+c^2)]$$
I have one methods to solve this problem, following is my methods.
since
\begin{align*}&\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CD}+\overrightarrow {DA}=\overrightarrow {0}\Longrightarrow |AD|^2=(\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CD})^2\\
&=\overrightarrow {AB}^2+\overrightarrow {BC}^2+\overrightarrow {CD}^2+2(\overrightarrow {AB}\cdot\overrightarrow {BC}+\overrightarrow {BC}\cdot\overrightarrow {CD}+\overrightarrow {CD}\cdot\overrightarrow {AB})\\
&=a^2-b^2+c^2+2(\overrightarrow {BC}+\overrightarrow {CD})(\overrightarrow {AB}+\overrightarrow {BC})\\
&=a^2-b^2+c^2+2\overrightarrow {AC}\cdot\overrightarrow {BD}
\end{align*}
but I'm looking for other methods. It's not important to use what, only the time that it takes is important.
| Let $A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3}),D(x_{4},y_{4})$,so we have
$$\begin{cases}
(x_{1}-x_{2})^2+(y_{1}-y_{2})^2=a^2\\
(x_{2}-x_{3})^2+(y_{2}-y_{3})^2=b^2\\
(x_{3}-x_{4})^2+(y_{3}-y_{4})^2=c^2\\
(x_{4}-x_{1})^2+(y_{4}-y_{1})^2=d^2
\end{cases}$$
so we have
$$b^2+d^2-(a^2+c^2)=2(x_{1}x_{2}+x_{3}x_{4}-x_{2}x_{3}-x_{1}x_{4})+2(y_{1}y_{2}+y_{3}y_{4}-y_{4}y_{1}-y_{2}y_{3})$$
then
$$\overrightarrow{AC}\cdot\overrightarrow{BD}=x_{1}x_{2}+x_{3}x_{4}-x_{2}x_{3}-x_{1}x_{4})+y_{1}y_{2}+y_{3}y_{4}-y_{4}y_{1}-y_{2}y_{3}=\dfrac{b^2+d^2-a^2-c^2}{2}$$
| {
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Finding the value of $x+y$
$x^3+y^3=(x+y)(x^2-xy+y^2)=400$
and for
$x^2y+xy^2=(x+y)(x^2+y^2)=200$
I decided to multiply by $2$ on both sides: $(x+y)(x^2+y^2)=200$ or $2(x+y)(x^2+y^2)=2(200)$
So then I set the equations equal to each other:
$2(x+y)(x^2+y^2)=(x+y)(x^2-xy+y^2)=400$
I am having trouble solving though because there is two variables involved. Any ideas?
| Hint: $(x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy(x+y)$.
| {
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How do I prove that the inequality $\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $ holds for every $x,y>0$ Prove that for every $x,y>0$, the following inequality holds. Begin by proving the right side first and assume that $(\sqrt { x } -\sqrt { y } )^2\ge 0$
$$\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $$
What I did thus far:
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\sqrt { xy } \le \frac { x+y }{ 2 } $
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy }$
3) Then, $\frac { x+y }{ 2 } \ge \sqrt { xy } $
4) Therefore, $\sqrt { xy } \le \frac { x+y }{ 2 } $
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\frac { 2xy }{ x+y } \le \sqrt { xy } $
At this point, I get stuck. I don't know what do to prove the other side of the inequality. I also don't know if I am going about this proof the right way.
Hints are much more appreciated than the actual solution.
Edit: (added this after getting advice from answers below)
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy } $
3) By taking the reciprocal of both sides, we get: $\frac { 1 }{ x+y } \le \frac { 1 }{ 2\sqrt { xy } } $
4) Then, $(\sqrt { xy } )\frac { 1 }{ x+y } \le (\sqrt { xy } )\frac { 1 }{ 2\sqrt { xy } } \Rightarrow \frac { \sqrt { xy } }{ x+y } \le \frac { \sqrt { xy } }{ 2\sqrt { xy } } $
5) Then, $(2\sqrt { xy } )\frac { \sqrt { xy } }{ x+y } \le (2\sqrt { xy } )\frac { \sqrt { xy } }{ 2\sqrt { xy } } $
6) Therefore, $\frac { 2xy }{ x+y } \le \sqrt { xy } $
QED
| The left-hand side inequality (HGM inequality) is simply the right-hand side (AGM inequality), applied to $\dfrac1x$ and $\dfrac1y$, reversed.
| {
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Observation: $\big(\frac{\sin x }{x}\big) ^n \sim e^{-a_n x^2}$ I have observed that this
$$\big(\frac{\sin x }{x}\big) ^n \sim e^{-a_n x^2}$$
(It can be fitted better than this, but I did not want them to completely overlap)
I want to learn how I would find $a_n$, or if this relationship is technically asymptotic to begin with.
| This is a nice observation!
The Taylor series for $$\operatorname{sinc} x := \frac{\sin x}{x}$$
at $x = 0$ is
$$\operatorname{sinc} x \sim 1 - \frac{1}{6} x^2 + \frac{1}{120} x^4 - \cdots + (-1)^{k / 2} \frac{1}{(2k + 1)!} + \cdots , $$
and so the series for $\operatorname{sinc}^n x$ is
$$\phantom{(\ast)} \qquad \operatorname{sinc}^n x \sim 1 - \frac{1}{6} n x^2 + \left( \frac{1}{72} n^2 - \frac{1}{180} n \right) x^4 - \cdots \qquad (\ast).$$
In principle it's not so hard to derive an expression for the coefficient of the $x^{2k}$ term, but its salient feature for our purposes is that its leading term (in $n$) is $$\frac{(-1)^{k / 2} n^k}{6^k k!}.$$
On the other hand, since $$\operatorname{sinc}^n 0 = \exp(-a_n \cdot 0^2) = 1 ,$$ we can naively attempt to align the two functions by choosing a parameter $a_n$ to match the Taylor series of the two functions to as high of an order as we can: The Taylor series of $\exp(-a_n x^2)$ is
$$\phantom{(\ast\ast)} \qquad \exp(-a_n x^2) \sim \sum_{k = 0}^{\infty} \frac{(-a_n x^2)^k}{k!} = 1 - a_n x^2 + \frac{1}{2} a_n^2 x^4 - \cdots + \frac{(-1)^{k / 2}}{k!} a_n^k x^{2k} + \cdots \qquad (\ast\ast),$$
so we can match the series $(\ast)$ and $(\ast\ast)$ to second order (in fact, third order) by taking $a_n$ so that their quadratic terms agree, that is, by setting $a_n = \frac{1}{6} n$. But then we see that
$$\exp\left(-\frac{n}{6} x^2\right) \sim 1 - \frac{1}{6} n x^2 + \frac{1}{72} n^2 x^4 - \cdots + \frac{(-1)^{k / 2} n^k}{6^k k!} x^{2k} + \cdots.$$
But the coefficient of this general term is precisely the same as the leading term (in $n$) of the coefficient of the $x^{2k}$ term for the series for $\operatorname{sinc}^n x$, which we identified above.
With this in hand, we can make precise in various ways the notion that the two functions are asymptotic as $n \to \infty$. One way to make this more tractable is to normalize the widths of the functions and compare
$$\operatorname{sinc}^n \left( \frac{x}{\sqrt{n}}\right) \sim 1 - \frac{1}{6} x^2 + \left(\frac{1}{72} - \frac{1}{180 n} \right) x^4 - \cdots$$
with
$$\exp\left[-\frac{n}{6} \left( \frac{x}{\sqrt{n}}\right)^2\right] = \exp\left(-\frac{1}{6} x^2\right) \sim 1 - \frac{1}{6} x^2 + \frac{1}{72} x^4 - \cdots;$$ these differ by a series $$- \frac{1}{180 n} x^4 + \cdots ,$$ which in particular is $O(n^{-1})$.
| {
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Show that $n^{n-3} \ge n!$ for n=9, 10,... Show that $n^{n-3} \ge n!$ for n=9, 10,...
I have tried to n=9
$9^{9-3} = 9^6 = 531411$
$9! = 362880$
So $9^6 \ge 9!$ is true
My question is how do I prove it by every for n=9, 10,... by induction? I have tried the following, but need some hints and or corrections.
Proof by induction
$ P(n): n^{n-3} \ge n!$
Base Case
$ n=9$
$9^6 \ge 9!$
Inductive Hypothesis
Assume P(k) is True
$k^{k-3} \ge k!$
Inductive Step
Show P(k+1) is True
$(k+1)^{(k+1)-3} \ge (k+1)!$
| You don't really need induction. Indeed
$$\frac{n^{n-3}}{n!}=\frac1{n^3}\cdot\frac n1\cdot\frac n2\dotsm\frac n6\dotsm\frac nn=\frac{n^3}{6!}\cdot\frac n7\cdot\frac n8\dotsm1 $$
Now if $n\ge 9$, $\dfrac{n^3}{6!}\ge\dfrac{9^3}{6!}=\dfrac{729}{720} $. Hence all factors are $\ge 1$, so that
$$\frac{n^{n-3}}{n!}\ge 1.$$
| {
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Find last digit of $7^{7^{7^7}}$ Find last digit of $7^{7^{7^7}}$
I know that the last digit of $7^x$ depends on the remainder $x$ leaves when divided by $4$: ($x = 7^{7^7}$)
$$7^{4k} \equiv 1 \bmod 10$$
$$7^{4k+1} \equiv 7 \bmod 10$$
$$7^{4k+2} \equiv 9 \bmod 10$$
$$7^{4k+3} \equiv 3 \bmod 10$$
And also that
$$7 \equiv -1 \bmod 4$$
So for all odd positive numbers, $7^{n} \equiv -1 \bmod 4$ and $7^{7}$ is odd. But which congruence relation should I use?
$4k+3$ and $4k+1$ are both odd
Please explain, thanks.
| (a)...$7^2\equiv 1$ mod $4\implies$ any odd power of $7$ is congruent to $7$ mod $ 4 \implies$ any odd power of $7$ is congruent to $3$ mod $4\implies 7^{7^7}\equiv 3$ mod $4$....(b)...$7^4\equiv 1$ mod $10\implies ( 7^n\equiv 7^3\equiv 3$ mod $10$ when $n\equiv 3$ mod $4)$...(c) Therefore $$7^{7^{7^7}}\equiv 3 \text { mod }10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).
How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
| Binomial Theorem and Beta Functions
$$
\begin{align}
\sum_{k=0}^n(-1)^k\frac1{2k+1}\binom{n}{k}
&=\int_0^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,\mathrm{d}x\tag1\\
&=\int_0^1\left(1-x^2\right)^n\,\mathrm{d}x\tag2\\
&=\frac12\int_0^1\left(1-x\right)^nx^{-1/2}\,\mathrm{d}x\tag3\\
&=\frac12\frac{\color{#C00}{\Gamma(n+1)}\,\color{#090}{\Gamma(1/2)}}{\color{#090}{\Gamma(n+3/2)}}\tag4\\
&=\frac12\color{#C00}{n!}\color{#090}{\frac{2^{n+1}}{(2n+1)!!}}\tag5\\
&=\frac{2^n\,n!}{(2n+1)!!}\tag6
\end{align}
$$
Explanation:
$(1)$: $\int_0^1x^{2k}\,\mathrm{d}x=\frac1{2k+1}$
$(2)$: Binomial Theorem
$(3)$: substitute $x\mapsto x^{1/2}$
$(4)$: apply the Beta Function integral
$(5)$: apply $\Gamma(x+1)=x\Gamma(x)$
$(6)$: cancel
Repeated Differences
Define a double factorial version of the binomial coefficients:
$$\newcommand{\dinom}[2]{\left\langle{#1\atop#2}\right\rangle}
\dinom{n}{k}=\frac{(2n+1)!!}{(2n-2k+1)!!(2k-2)!!}\tag7
$$
Note that $\dinom{n}{1}=2n+1$. Furthermore,
$$
\begin{align}
\frac1{\dinom{n}{k}}-\frac1{\dinom{n-1}{k}}
&=(2k-2)!!\left(\frac{(2n-2k+1)!!}{(2n+1)!!}-\frac{(2n-2k-1)!!}{(2n-1)!!}\right)\tag8\\
&=(2k-2)!!\frac{(2n-2k-1)!!}{(2n+1)!!}((2n-2k+1)-(2n+1))\tag9\\[3pt]
&=-(2k)!!\frac{(2n-2k-1)!!}{(2n+1)!!}\tag{10}\\[3pt]
&=-\frac1{\dinom{n}{k+1}}\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: apply $(7)$ while extracting a common factor
$\phantom{1}(9)$: extract more common factors
$(10)$: simplify the right hand term and incorporate into the left hand term
$(11)$: apply $(7)$
Apply $(11)$ $n$ times to get
$$
\begin{align}
\overbrace{\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{\dinom{n-j}{1}}}^\text{$n^\text{th}$ order backward difference}
&=\frac{(-1)^n}{\dinom{n}{n+1}}\tag{12}\\
&=\frac{(-1)^n(-1)!!(2n)!!}{(2n+1)!!}\tag{13}\\[3pt]
&=(-1)^n\frac{2^nn!}{(2n+1)!!}\tag{14}
\end{align}
$$
Explanation:
$(12)$: apply $(11)$ $n$ times
$(13)$: apply $(7)$
$(14)$: $(-1)!!=1$ and $(2n)!!=2^nn!$
Multiply $(14)$ by $(-1)^n$ and substitute $j\mapsto n-j$:
$$
\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{2j+1}=\frac{2^nn!}{(2n+1)!!}\tag{15}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Use mathematical induction to show that $ H_{2^n} \geq 1+ \frac{n}{2} $ An Inequality for Harmonic Numbers.
The harmonic numbers $H_j, j=1,2,3,...,$ are defined by $$H_j = 1 + \cfrac{1}{2}+\cfrac{1}{3}+...+\cfrac{1}{j}$$
Use mathematical induction to show that $$ H_{2^n} \geq 1+ \frac{n}{2} $$
whenever $n$ is a nonnegative integer.
BASIS STEP: $P(0)$ is true, because $H_{2^0}=H_1=1 \geq 1+\dfrac{0}{2}$
INDUCTIVE STEP: The inductive hypothesis is the statement that $P(k)$ is true, that is, $H_{2^k} \geq 1 +\dfrac{k}{2}$, where $k$ is an arbitrary nonnegative integer. We must show that if $P(k)$ is true, then $P(k+1)$, which states that $H_{2^{k+1}} \geq 1 +\dfrac{k+1}{2}$, is also true. So, assuming the inductive hypothesis, it follows that
$$H_{2^{k+1}} = 1+ \frac{1}{2} + \frac{1}{3} + ...+ \frac{1}{2^k} + \frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}}$$
$$=H_{2^k}+\frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}}$$
$$\geq (1+\frac{k}{2})+ \frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}} \qquad ...(?)$$
$$\geq (1+\frac{k}{2})+2^k \cdot \frac{1}{2^{k+1}}\qquad ...(??)$$
$$\geq (1+\frac{k}{2}) + \frac{1}{2}$$
$$=1+\frac{k+1}{2} $$
I don't understand what is going on at lines $(?)$ and $(??)$, why did it change from $=H_{2^k}$ to $\geq (1+\dfrac{k}{2})$ can somebody explain it to me?
| (base case, etc. omitted)
assume $H_{2^n}\ge 1+{n\over 2}$
$H_{2^n}+{1\over 2}\ge 1+{n\over 2}+{1\over 2}$
${1\over 2}={{2^n}\over{2^n+2^n}}={1\over {2^n+2^n}}_{(1)}+{1\over {2^n+2^n}}_{(2)}+...+{1\over {2^n+2^n}}_{(2^n)}$
$H_{2^{n+1}}=H_{2^n}+{1\over {2^n+1}}_{(1)}+{1\over {2^n+2}}_{(2)}+...+{1\over {2^n+2^n}}_{(2^n)}\ge H_{2^n}+{1\over {2^n+2^n}}_{(1)}+{1\over {2^n+2^n}}_{(2)}+...+{1\over {2^n+2^n}}_{(2^n)}=H_{2^n}+{1\over 2}\ge 1+{n\over 2}+{1\over 2}=1+{{n+1}\over 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$P$ is a prime number not equal to $2$ or $3$. Show that $p - 6$ is a quadratic residue $\pmod p$ whenever... $P$ is a prime number not equal to $2$ or $3$.
Show that $p - 6$ is a quadratic residue $\pmod p$ whenever $p \equiv 1, 5, 7, 11\pmod {24}$
And show that $p - 6$ is a quadratic non - residue $\pmod p$ whenever $p\equiv 13, 17, 19, 23\pmod {24}$
I know that $p - 6$ is a quadratic residue if there exists a $b$ such that $p - 6\equiv b^2\pmod p$
So for $p = 11$ we get $11 - 6\equiv b^2\pmod {11}$ and $b = 4$ satisfies this.
Similarly,
For $p = 1$ : $b = 0$
For $p = 5$ : $b = 2$
For $p = 7$ : $b = 1$
And for the quadratic non - residue:
Take $p = 13$ then we see that none of $1^2, 2^2, 3^2, \ldots, 12^2$ are congruent to $13 - 6 = 7 \pmod {13}$. So it is a quadratic non - residue. This is also the case for $17, 19, 23$.
What I do not understand is: Where does the number $24$ come from? And how do I use it in the explanation above, or is this not necessary?
| A short reason for the 24 is this: if $p \equiv 1,7 \pmod {24},$ then we always have an expression
$$ p = u^2 + 6 v^2 $$
in integers.
If $p \equiv 5,11 \pmod {24},$ then we always have an expression
$$ p = 2 r^2 + 3 s^2 $$
in integers.
When we have a quadratic form,
$$ f(x,y) = A x^2 + B xy + C y^2, $$
the discriminant, familiar from the quadratic formula, is
$$ \Delta = B^2 - 4 A C. $$
In the two earlier forms, $\Delta = -24.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$ How can I compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$?
| Recall that if $f$ is continuous, then
$$\lim_{x\rightarrow a}f(g(x))=f\left(\lim_{x\rightarrow a}g(x)\right).$$
Since the cube function $x\mapsto x^3$ is continuous, we have
\begin{align*}
\lim_{x\rightarrow 0}\frac{\sin^3 x}{(2x)^3} &= \left(\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin x}{x}\right)^3\\
&= \left(\frac{1}{2}\right)^3\\
&= \frac{1}{8}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve in integers the equation $2x+3y = 5$ If I have the following equation $$2x+3y = 5$$
I know all the integer solutions is $$x = 1+3n$$ $$y = 1-2n$$ $$n \in \Bbb Z$$
since I can just plug them in $$2(1+3n)+3(1-2n) = 5+6n-6n = 5$$
but I don't know how to derive the answer from the equation... also is there a name for algorithms to solve these integer function?
| First you find a particular solution. Say $(x, y) = (1, 1)$.
Then you suppose that $2x + 3y = 5$ for some $(x, y)$
So
\begin{align}
2x + 3y &= 2(1) + 3(1)\\
2(x-1) &= -3(y-1)\\
\end{align}
Hence $2 \mid -3(y-1)$.
Since $2$ and $-3$ are relatively prime, we must have $2 \mid y-1$. That is $y - 1 = 2t$ for some integer $t$.
So $y = 2t + 1$ for some integer $t$.
Substituting this back into $2(x-1) = -3(y-1)$, we find $x = -3t + 1$.
What we have shown is that, if $(x, y)$ is a solution to $2x + 3y = 5$, then
$(x,y) = (-3t+1, 2t+1)$ for some integer $t$.
It is easy to verify that if $(x,y) = (-3t+1, 2t+1)$ for some integer $t$, then
$(x, y)$ is a solution to $2x + 3y = 5$.
Hence $(x, y)$ is a solution to $2x + 3y = 5$, if and only if
$(x,y) = (-3t+1, 2t+1)$ for some integer $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Making a determinant easier to solve to find characteristic polynomial I have to find a characteristic polynomial for the follwoing matrix:
$$A=\left( \begin{matrix}0 & -1 & 1\\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{matrix}\right)$$
My goal to find the roots of the characteristic polynomial.
The characteristic polynomial is given by solving $det(\lambda I -A)$, and then we get:
$$det(\lambda I -A)=det\left( \begin{matrix}t & 1 & -1\\ 1 & t & -2 \\ -1 & -2 & t \end{matrix}\right)$$
The final polynomial I get is $$P_A=t^3-6t+4$$
While there is a way to solve this polynomial I believe there is a way to make this determinant more "comfortable" to solve.
Can I add the third column to the second column?
$$det(\lambda I -A)=det\left( \begin{matrix}t & 0 & -1\\ 1 & t-2 & -2 \\ -1 & t-2 & t \end{matrix}\right)$$
And Then, can I do the following (and why?)
$$ det(\lambda I -A)=(t-2)*det\left( \begin{matrix}t & 0 & -1\\ 1 & 1 & -2 \\ -1 & 1 & t \end{matrix}\right)$$
What kind of elementary operations, or other arithmetic ones I can do, to make the determinant easier to solve?
Thanks,
Alan
| Personally, I prefer this method
$$|A-\lambda I|=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
-1 & -\lambda & 2 \\
1 & 2 & -\lambda \\
\end{array}\right|$$
$$=-\lambda\left| \begin{array}{rr}
-\lambda & 2\\
2 & -\lambda \\
\end{array}\right|
+\left| \begin{array}{rr}
-1 & 2\\
1 & -\lambda \\
\end{array}\right|
+\left| \begin{array}{rr}
-1 & -\lambda\\
1 & 2 \\
\end{array}\right|$$
$$=-\lambda^3+6\lambda-4 =(2-\lambda)(\lambda^2+2\lambda-2)$$
$$=(2-\lambda)\left(\lambda-\sqrt3+1\right)\left(\lambda+\sqrt3+1\right)$$
You can also triangularize the matrix first. This makes the determinant trivial to compute
$$|A-\lambda I|=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
-1 & -\lambda & 2 \\
1 & 2 & -\lambda \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
0 & 2-\lambda & 2-\lambda \\
1 & 2 & -\lambda \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
0 & 2-\lambda & 2-\lambda \\
\lambda & 2\lambda & -\lambda^2 \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
0 & 2-\lambda & 2-\lambda \\
0 & 2\lambda-1 & 1-\lambda^2 \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
0 & 2-\lambda & 2-\lambda \\
0 & 0 & -\lambda^2-2\lambda+2\\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
1 & \lambda^{-1} & -\lambda^{-1}\\
0 & 2-\lambda & 2-\lambda \\
0 & 0 & \lambda^2+2\lambda-2\\
\end{array}\right|$$
$$=(2-\lambda)(\lambda^2+2\lambda-2)$$
$$=(2-\lambda)\left(\lambda-\sqrt3+1\right)\left(\lambda+\sqrt3+1\right)$$
In any case, I suggest you look up the properties of the determinant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Why is this sum what it is? I am trying to prove that
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = \frac{2n}{n+1}$$
I know the denominators have a closed form $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$ implying
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = 2\sum_{k=1}^{n}\frac{1}{k(k+1)}$$
Can't figure out where to go from here.
| Hint Notice that : $$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding minimal value of $\sqrt{5x^2 - 40x + 85} + \sqrt{5x^2 - 24x + 53}$ without using derivatives
Find minimal value of $f(x) = \sqrt{5x^2 - 40x + 85} + \sqrt{5x^2 - 24x + 53}$.
I can solve it using derivatives. Is there any other way to solve it? For example using some popular inequalities?
| HINT :
We can write
$$\frac{f(x)}{\sqrt 5}=\sqrt{(x-4)^2+(0-1)^2}+\sqrt{\left(x-\frac{12}{5}\right)^2+\left(0-\frac{11}{5}\right)^2}.$$
This represents the sum of the distance between $(x,0)$ and $(4,1)$ and the distance between $(x,0)$ and $\left(\frac{12}{5},\frac{11}{5}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that 17 divides 1111111111111111 (16 1's) and doesn't divide 11111111 I need to prove that $17$ divides $\underbrace{1111111111111111}_{\text{16 1's}}$ and doesn't divide $\underbrace{11111111}_{\text{8 1's}}$ by using congruence.
I know that $\underbrace{1111111111111111}_{\text{16 1's}}= \frac{10^{16}-1}{9}$ and that $10^{15}\equiv{1}\pmod {17}$. But how do I use these two facts to figure out if $17$ divides $1111111111111111$?
| $1111111111111111=10^{15} + ... + 1 = \large{(10^{16} -1)\over9}$
$10^2 \equiv -2\pmod {17}$
$10^{16} - 1 = (-2)^8 - 1 = 256 -1 \equiv 1 -1 = 0\pmod {17}$
$11111111= \large{(10^8 -1 )\over9} $
$10^2 \equiv -2\pmod {17}$
$10^8 - 1 \equiv (-2)^4 - 1 = 15 \pmod {17}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Maximum Value satisfying a variation of the Triangle Inequality The Triangle Inequality states that:
$$||\vec{x}+\vec{y}||\le ||\vec{x}||+||\vec{y}||$$
Now, suppose that we have a triangle with side lengths of $a,b,c$ such that $a+b+c=2$. Avoiding any assumptions of what kind of triangle we have, what the angle measures are, etc.,
Now suppose we have the following inequality:
$$k\le\dfrac{1-a}{b}+\dfrac{1-b}{c}+\dfrac{1-c}{a}$$
My question is that what can $k$ be. Meaning what is the maximum possible $k$ which satisfies the inequality I gave. I just don't want an answer, but a proof as to what the maximum value of $k$ will be.
| Suppose that $a\geq b\geq c$, so $1-a\leq 1-b\leq 1-c$ and $\frac{1}{a}\leq \frac{1}{b}\leq \frac{1}{c}$. By Tchebyshev inequality you have
$$\frac{1-a}{b}+\frac{1-b}{c}+\frac{1-c}{a}\geq 3\cdot \frac{1}{3}(1-a+1-b+1-c)\cdot \frac{1}{3}(1/a+1/b+1/c)=\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}).$$
By Cauchy-Schwarz you have $(a+b+c)(1/a+1/b+1/c)\geq 9$ so $(1/a+1/b+1/c)\geq 9/2$.
Then $k=\frac{3}{2}$ and its obtain for $a=b=c=2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Trace of Product of Powers of $A$ and $A^\ast$ Let $n$ be odd, $\displaystyle v=1,...,\frac{n-1}{2}$ and $\displaystyle \zeta=e^{2\pi i/n}$.
Define the following matrices:
$$A(0,v)=\left(\begin{array}{cc}1+\zeta^{-v} & \zeta^v+\zeta^{2v}\\ \zeta^{-v}+\zeta^{-2v}&1+\zeta^{v}\end{array}\right),$$
$$A(1,v)=\left(\begin{array}{cc}\zeta^{-1}+\zeta^{-v} & \zeta^{v}\\ \zeta^{-v}&\zeta^{-1}+\zeta^{v}\end{array}\right).$$
$$A(n-1,v)=\left(\begin{array}{cc}\zeta+\zeta^{-v} & \zeta^{2v}\\ \zeta^{-2v}&\zeta+\zeta^v\end{array}\right).$$
I am hoping to calculate for each of these $A$
$$\text{Tr}\left[\left(A^k\right)^*A^k\right]=\text{Tr}\left[\left(A^*\right)^kA^k\right].$$
All I have is that $A$ and $A^*$ in general do not commute so I can't simultaneously diagonalise them necessarily.
I do know that if we write $A=D+(A-D)$ (with $D$ diagonal), that
$$A^*=\overline{D}+(A-D).$$
I suppose anybody who knows anything about calculating $$\text{Tr}(A^kB^k)$$ can help.
Context: I need to calculate or rather bound these traces to calculate a distance to random for the convolution powers of a $\nu\in M_p(\mathbb{G}_n)$ for $\mathbb{G}_n$ a series of quantum groups of dimension $2n^2$ ($n$ odd). For $u=2,...,k-2$, $A(u,v)$ is diagonal so no problems there.
| Solution for $A_0 = A(0,v)$
\begin{align*}
A_0 &=
\begin{pmatrix}
1+\zeta^{-v} & \zeta^v + \zeta^{2v} \\
\zeta^{-v}+\zeta^{-2v} & 1+\zeta^v
\end{pmatrix}
=
(\zeta^{-\frac{1}{2}v}+\zeta^{+\frac{1}{2}v})
\begin{pmatrix}
\zeta^{-\frac{1}{2}v} & \zeta^{\frac{3}{2}v} \\
\zeta^{-\frac{3}{2}v} & \zeta^{\frac{1}{2}v}
\end{pmatrix}
\\
&=: \lambda \hat{A}_0
\\
\hat{A}^2_0
&=
\begin{pmatrix}
\zeta^{-\frac{1}{2}v} & \zeta^{\frac{3}{2}v} \\
\zeta^{-\frac{3}{2}v} & \zeta^{\frac{1}{2}v}
\end{pmatrix}^2
=
\begin{pmatrix}
1+\zeta^{-v} & \zeta^v + \zeta^{2v} \\
\zeta^{-v}+\zeta^{-2v} & 1+\zeta^v
\end{pmatrix}
\\
&= A_0
\\
\implies A_0^2 &= \lambda^2 \hat{A}_0^2 = \lambda^2 A_0
\\
\implies A_0^k &= \lambda^{2(k-1)} A_0
\end{align*}
Moreover note that $\lambda$ is real. In fact, $\lambda=2\cos(\frac{v}{n}\pi)$. Therefore:
\begin{equation*}
{\tt tr}((A_0^*)^kA_0^k)) = \lambda^{4(k-1)}{\tt tr}(A_0^* A_0) = 4\lambda^{4(k-1)}
\end{equation*}
Note that $\lambda^2 = (\zeta^{-\frac{1}{2}v}+\zeta^{+\frac{1}{2}v})^2 = (1+\zeta^v) (1+\zeta^{-v})$, hence the solution is equivalent to one by Start wearing purple.
Solution for $A_1 = A(1,v)$
\begin{align*}
A_1 &=
\begin{pmatrix}
\zeta^{-1}+\zeta^{-v}& \zeta^v \\
\zeta^{-v} & \zeta^{-1}+\zeta^v
\end{pmatrix}
=
\zeta^{-1}\begin{pmatrix} 1&0\\0&1 \end{pmatrix}
+
\begin{pmatrix}
\zeta^{-v} & \zeta^{v} \\
\zeta^{-v} & \zeta^{v}
\end{pmatrix}
\\
&=: \zeta^{-1} I + \hat{A}_1
\\
\hat{A}_1^2
&=
\begin{pmatrix}
\zeta^{-2v}+1 & \zeta^{2v}+1 \\ \zeta^{-2v}+1 & \zeta^{2v}+1
\end{pmatrix}
=
(\zeta^{-v} + \zeta^v)
\begin{pmatrix}
\zeta^{-v} & \zeta^v \\ \zeta^{-v} & \zeta^v
\end{pmatrix}
\\
&=:
\mu \hat{A}_1
\\
\implies \hat{A}_1^k
&=
\mu^{k-1} \hat{A}_1
\\
\implies A_1^k
&=
\big(\zeta^{-1}I + \hat{A}_1\big)^k
= \sum_{j=0}^{k}\binom{k}{j} \zeta^{j-k}\hat{A}_1^j
= \zeta^{-k}\bigg(I + \hat{A}_1 \sum_{j=1}^{k}\binom{k}{j} \zeta^{j}\mu^{j-1}\bigg)
\\&=
\zeta^{-k}\bigg(I + \frac{1}{\mu}\hat{A}_1\bigg[-1 + \sum_{j=0}^{k}\binom{k}{j} \zeta^{j}\mu^{j}\bigg] \bigg)
\\&=
\zeta^{-k}\bigg(I + \frac{(1-\zeta\mu)^k - 1}{\mu}\hat{A}_1\bigg)
\end{align*}
Here we again have that $\mu=2\cos(2\pi\frac{v}{n})$ is real; moreover we have ${\tt tr}(\hat{A}_1^*) = {\tt tr}(\hat{A}_1) = \mu$ and ${\tt tr}(\hat{A}_1^*\hat{A}_1)=4 $. Hence if we let $\alpha_k = (1-\zeta\mu)^k - 1$ the trace is
\begin{equation*}
{\tt tr}((A_1^*)^kA_1^k)) = 1 + \alpha_k + \bar\alpha_k + \frac{4}{\mu^2}\alpha_k\bar\alpha_k
\end{equation*}
Solution for $A_{n-1} = A(n-1,v)$
\begin{align*}
A_{n-1} &=
\begin{pmatrix}
\zeta+\zeta^{-v}& \zeta^{2v} \\
\zeta^{-2v} & \zeta + \zeta^{v}
\end{pmatrix}
= \zeta \begin{pmatrix} 1&0\\0&1 \end{pmatrix}
+
\begin{pmatrix}
\zeta^{-v}& \zeta^{2v} \\
\zeta^{-2v} & \zeta^{v}
\end{pmatrix}
\\&=:
\zeta I + \hat{A}_{n-1}
\\
\hat{A}_{n-1}^2
&=
\begin{pmatrix}
\zeta^{-2v} + 1 & \zeta^{v}+\zeta^{3v} \\
\zeta^{-3v} + \zeta^{-v} & \zeta^{2v}+1
\end{pmatrix}
=
(\zeta^{-v} + \zeta^{v})
\begin{pmatrix}
\zeta^{-v}& \zeta^{2v} \\
\zeta^{-2v} & \zeta^{v}
\end{pmatrix}
\\&=:
\mu \hat{A}_{n-1}
\\
\implies \hat{A}_{n-1}^k &= \mu^{k-1}\hat{A}_{n-1}
\\
\implies A_{n-1}^k &=
(\zeta I + \hat{A}_{n-1})^k = \ldots\\
&= \zeta^{k}\bigg(I + \frac{(1-\zeta^{-1}\mu)^k - 1}{\mu}\hat{A}_{n-1}\bigg)
\end{align*}
Here we immediately observe that this is remarkably similar to the previous solution. In fact, since ${\tt tr}(\hat{A}_{n-1}) = \mu$ and ${\tt tr}(\hat{A}_{n-1}^*\hat{A}_{n-1})=4$ aswell, the traces are the same!!
\begin{equation*}
{\tt tr}((A_{n-1}^*)^kA_{n-1}^k)) = 1 + \alpha_k + \bar\alpha_k + \frac{4}{\mu^2}\alpha_k\bar\alpha_k = {\tt tr}((A_1^*)^kA_1^k))
\end{equation*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate sum using generating function I'm trying to evaluate this sum:
$$ \sum\limits_{s = 0}^{500} (-1)^s \binom{3000 - 2s}{2000} \binom{2001}{s}$$
As I think we need to use an expansion of $(1 -x)^n (1+x)^k$, but I've tried several times, using different powers but stuck every time.
Could you help me, please?
| Note: Please note, this is essentially the same as Marko Riedels answer. Here we simply avoid the integral notation which might be somewhat more convenient.
In order to calculate OPs sum it's beneficial to consider the more general expression
\begin{align*}
S_n=\sum_{s= 0}^{\lfloor\frac{n}{2}\rfloor}(-1)^s\binom{3n-2s}{2n}\binom{2n+1}{s}
\end{align*}
We obtain OPs expression by setting $n=1000$. We need one additional small step before we start calculating $S_n$.
Since $\binom{m}{k}=\binom{m}{m-k}$ and $\binom{-m}{k}=\binom{m+k-1}{k}(-1)^k$ the following is valid
\begin{align*}
\binom{3n-2s}{2n}=\binom{3n-2s}{n-2s}=\binom{-(2n+1)}{n-2s}(-1)^n
\end{align*}
So, we can rewrite $S_n$ as
\begin{align*}
S_n=(-1)^n\sum_{s= 0}^{\lfloor\frac{n}{2}\rfloor}(-1)^s\binom{-(2n+1)}{n-2s}\binom{2n+1}{s}
\end{align*}
Observe the nice symmetry regarding $2n+1$. This will help simplifying $S_n$.
In the following we use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a polynomial or a series.
We obtain
\begin{align*}
S_n&=(-1)^n\sum_{s= 0}^{\lfloor\frac{n}{2}\rfloor}(-1)^s\binom{-(2n+1)}{n-2s}\binom{2n+1}{s}\\
&=(-1)^n\sum_{s= 0}^{\lfloor\frac{n}{2}\rfloor}[x^{n-2s}](1+x)^{-(2n+1)}[y^s](1-y)^{2n+1}\tag{1}\\
&=(-1)^n[x^n](1+x)^{-(2n+1)}\sum_{s= 0}^{\lfloor\frac{n}{2}\rfloor}x^{2s}[y^s](1-y)^{2n+1}\tag{2}\\
&=(-1)^n[x^n](1+x)^{-(2n+1)}(1-x^2)^{2n+1}\tag{3}\\
&=(-1)^n[x^n](1-x)^{2n+1}\tag{4}\\
&=\binom{2n+1}{n}
\end{align*}
Comment:
*
*In (1) we write the binomial coefficients as coefficients of polynomials (resp. series) $\binom{m}{k}=[x^k](1+x)^m$. Note, that the factor $(-1)^s$ is respected in $[y^s](1-y)^{2n+1}$.
*In (2) we see that the representation as coefficients of polynomials is convenient, since we can put parts of the expression, which do not depend on $s$ outside the sum. We also use the rule $$[x^{n+m}]p(x)=[x^n]x^{-m}p(x)$$
*In (3) we use the substitution rule (see also (1)):
\begin{align*}
(1-x^2)^{2n+1}&=\sum_{s\geq 0}\binom{2n+1}{s}(-1)^sx^{2s}\\
&=\sum_{s\geq 0}x^{2s}[y^s](1-y)^{2n+1}
\end{align*}
*In (4) we use the symmetry regarding $2n+1$ which helps cancelling a factor of $(1-x^2)^{2n+1}=(1-x)^{2n+1}(1+x)^{2n+1}$
Since OPs expression is $S_{1000}$ we finally conclude
\begin{align*}
S_{1000}&=\sum_{s=0}^{500}(-1)^s\binom{3000-2s}{2000}\binom{2001}{s}\\
&=[x^{1000}](1+x)^{2001}\\
&=\binom{2001}{1000}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$? Consider
$$x^4 + x^2 + y^4 + y^2 = 2$$
It is a smooth non-intersecting circle like curve in the plane.
A bit like a Hyperellipse.
See https://en.m.wikipedia.org/wiki/Superellipse
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$ ?
Wolfram|Alpha plot
| Easy estimation: $2\pi = 6.28\ldots \le s \le 8$.
$$
u = x^2 + 1/2 \Rightarrow u^2 = x^4 + x^2 + 1/4 \\
v = y^2 + 1/2 \Rightarrow v^2 = y^4 + y^2 + 1/4
$$
Then
$$
r^2 = (\sqrt{2+1/2})^2 = u^2 + v^2 = x^4 + x^2 + y^4 + y^2 + 1/2 \iff \\
2 = x^4 + x^2 + y^4 + y^2 \quad (*)
$$
So the transformed curve is a circle in the $u$-$v$-plane with radius $r = \sqrt{5/2}$ and has the known arc length $S = \pi \sqrt{10}$.
I looked for a way to transform this arc length into the wanted arc length $s$, but had no success.
For the upper half of the curve we get:
\begin{align}
v &= \sqrt{r^2 - u^2} \iff \\
y^2 + 1/2 &= \sqrt{9/4 - x^2 - x^4} \iff \\
y &= \sqrt{\sqrt{9/4 - x^2 -x^4} - 1/2} \quad (**) \\
\end{align}
This leads to the derivative
\begin{align}
y' &= \frac{1}{2\sqrt{\sqrt{9/4 - x^2 -x^4} - 1/2}}
\frac{-2x-4x^3}{2\sqrt{9/4 - x^2 -x^4}} \\
&= - \frac{\sqrt{2}\,x\,(1 + 2x^2)}{\sqrt{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}}
\end{align}
The arc length is
\begin{align}
s &= 4 \int\limits_0^1\sqrt{1 + (y')^2} \, dx \\
&= 4 \int\limits_0^1\sqrt{1 + \frac{2\,x^2\,(1 + 2x^2)^2}{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}} \, dx \\
&= 6.68188
\end{align}
where the rounded numerical value was obtained from WolframAlpha.
An alternative query using the simpler equation $(**)$ is this. The result has to be multiplied by $4$ however.
Addendum
Alex Jordan pointed out that $y' \to -\infty$ for $x \to 1$, see WA plot. So we go only for the first 1/8 of the arc, were there is a point $(\xi,\xi)$:
$$
2 = \xi^4 + \xi^2 + \xi^4 + \xi^2 = 2\xi^2(\xi^2+1) \iff \\
1 = \xi^4 + \xi^2 = (\xi^2 + 1/2) - 1/4 \Rightarrow \\
\xi = \sqrt{(\sqrt{5} - 1)/2} = \sqrt{\varphi - 1} = 0.78615\ldots
$$
where $\varphi = 1.618\ldots$ is the golden ratio.
This leads to
\begin{align}
s &= 8 \int\limits_0^\xi\sqrt{1 + (y')^2} \, dx \\
&= 8 \int\limits_0^\xi\sqrt{1 + \frac{2\,x^2\,(1 + 2x^2)^2}{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}} \, dx \\
\end{align}
New query: link
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Computing the inverse a^-1 mod n using Euler or Fermats Theorem I'm trying to compute the inverse $a^{-1} \mod n$ with Fermat's Theorem (if applicable) or Euler's Theorem:
$a$ = 6, $n$ = 13
I know that since n is prime, we can use Fermat's Theorem to compute the inverse, using the following formula:
$a^{-1} = a^{p-2} \mod p$
So in our case, $p$ = 13.
So here's what I got, but it's obviously incorrect.
$6^{13-2} \mod 13 =$
$6^{11} \mod 13 =$
$362797056 \mod 13$
Can someone please explain where I went wrong?
| The remainder on dividing $362797056$ by $13$ is $11$. Showing this work in more detail and organizing it a little differently, and indicating the details of the reductions $\mathrm{mod} \ 13$...
$$\begin{align}
6 &\cong 6 \mod 13 \\
6^2 &\cong 36 \cong_{{}-2\cdot 13} 10 \mod 13 \\
6^3 &\cong 6 \cdot 6^2 \cong 6 \cdot 10 \cong 60 \cong_{{}-4 \cdot 13} 8 \mod 13 \\
6^9 &\cong (6^3)^3 \cong (8)^3 \cong 64 \cdot 8 \cong_{64 \rightarrow 64 - 4\cdot 13} 12 \cdot 8 \cong 96 \cong_{{}-7\cdot 13} 5 \mod 13 \\
6^{11} &\cong 6^9 \cdot 6^2 \cong 5 \cdot 10 \cong 50 \cong_{{}-3\cdot 13} 11 \mod 13
\end{align}$$
Then $a \cdot 11 \cong 6 \cdot 11 \cong 66 \cong_{{}-5\cdot 13} 1 \mod 13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1524981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Find generating function of a sequence Find generating function of a sequence $(1,1,2,2,2^2,2^2,2^3,2^3,...)$
Is it necessary to always look for a sub-sequence, e.g. $(1,2,2^2,2^3,...)$?
This is a geometric sequence which generating function is $\frac{1}{1-2x}$
$\frac{1}{1-2x}=\sum\limits_{k=0}^{+\infty}(2x)^n=1+2x+2^2x^2+2^3x^3+...$
How to find generating function for this sequence using geometric sub-sequence?
| Hint:
$$2^{0}+2^{0}x+2^{1}x^{2}+2^{1}x^{3}+2^{2}x^{4}+2^{2}x^{5}+2^3x^6+2^3x^7\cdots=$$$$\left[1+2x^{2}+\left(2x^{2}\right)^{2}+(2x^2)^3+\cdots\right]+x\left[1+2x^{2}+\left(2x^{2}\right)^{2}+(2x^2)^3+\cdots\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525536",
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"source": "stackexchange",
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Solution to a recurrence relation
Set $F_k(x) := \sum_{n\geq k} S(n,k)x^n$. Prove that $$F_1(x) = \frac{x}{1-x}, \space \space \space F_2(x) = \frac{x^2}{(1-x)(1-2x)} $$
Furthermore, show that the function $F_k(x)$ satisfy the recurrence relation $$F_k(x) = \frac{x}{1-kx}F_{k-1}(x)$$ and solve this recurrence.
Any ideas how to approach this problem?
| I prefer the notation $n\brace k$ for your $S(n,k)$. The first step is to observe that ${n\brace 1}=1$ for $n\ge 1$: there is only one partition of $[n]=\{1,\ldots,n\}$ into just one part. Thus,
$$F_1(x)=\sum_{n\ge 1}{n\brace 1}x^n=\sum_{n\ge 1}x^n=x\sum_{n\ge 1}x^{n-1}=x\sum_{n\ge 0}x^n=\frac{x}{1-x}\;.$$
To get $F_2(x)$, we need to figure out what $n\brace 2$ is. Suppose that we partition $[n]$ into two non-empty parts. One of the parts must contain the number $n$; call that part $A$ and the other part $B$. What can the rest of $A$ look like? It can’t be all of $[n-1]$, since then $B$ would be empty, but it can be any other subset of $[n-1]$. Since $[n-1]$ has $2^{n-1}$ subsets, there are $2^{n-1}-1$ possible choices for the rest of $A$ and hence for a $2$-part partition of $[n]$. Thus, ${n\brace 2}=2^{n-1}-1$ for $n\ge 2$, and
$$\begin{align*}
F_2(x)&=\sum_{n\ge 2}{n\brace 2}x^n\\
&=\sum_{n\ge 2}\left(2^{n-1}-1\right)x^n\\
&=\sum_{n\ge 2}2^{n-1}x^n-\sum_{n\ge 2}x^n\\
&=2x^2\sum_{n\ge 0}(2x)^n-x^2\sum_{n\ge 0}x^n\\
&=\frac{2x^2}{1-2x}-\frac{x^2}{1-x}\\
&=\frac{2x^2(1-x)-x^2(1-2x)}{(1-x)(1-2x)}\\
&=\frac{x^2}{(1-x)(1-2x)}\;.
\end{align*}$$
At this point we want the standard recurrence relation satisfied by the Stirling numbers of the second kind,
$${n\brace k}=k{{n-1}\brace k}+{{n-1}\brace{k-1}}\;,$$
substituting it into the definition of $F_k(x)$:
$$\begin{align*}
F_k(x)&=\sum_{n\ge k}{n\brace k}x^n\\
&=k\sum_{n\ge k}{{n-1}\brace k}x^n+\sum_{n\ge k}{{n-1}\brace{k-1}}x^n\\
&=kx\sum_{n\ge k-1}{n\brace k}x^n+x\sum_{n\ge k-1}{n\brace{k-1}}x^n\\
&=kxF_k(x)+xF_{k-1}(x)\;.
\end{align*}$$
Now just solve for $F_k(x)$ in terms of $F_{k-1}(x)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An inequality using convex functions: If $\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$ then $abc \le \frac1{2\sqrt 2}$ I see this in a Chinese convex analysis book.
Suppose that $a$,$b$,$c$ are positive real numbers satisfying
\begin{equation}
\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1.
\end{equation}
Show that $abc \le \dfrac1{2\sqrt 2}$.
Since it's from a convex analysis book, I tried proving this using Jensen's inequality. However, I can't think of a suitable convex function. Therefore, I tried AM–GM, but I can't get a product $abc$.
$$(abc)^2\le\frac18\iff8(abc)^2\le1$$
$$\iff\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right)^3 (abc)^2 \le 1$$
Finally, I used Lagrange multiplier to solve the problem, but I think there is some more elementary solution.
$$f(a,b,c)=abc$$
$$g(a,b,c)=\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}-2=0$$
$$\nabla f(a,b,c)=(bc,ca,ab)$$
$$\nabla g(a,b,c)=\left(-\frac{2a}{(1+a^2)^2},-\frac{2b}{(1+b^2)^2},-\frac{2c}{(1+c^2)^2}\right)$$
$$\because \nabla f = \lambda \nabla g$$
$$\therefore bc = -\frac{2a\lambda}{(1+a^2)^2} \iff abc = -\frac{2a^2\lambda}{(1+a^2)^2}$$
$$abc = -\frac{2a^2\lambda}{(1+a^2)^2} = -\frac{2b^2\lambda}{(1+b^2)^2} = -\frac{2c^2\lambda}{(1+c^2)^2}$$
$$\frac{a}{1+a^2}=\frac{b}{1+b^2}=\frac{c}{1+c^2}=\mu$$
$$a+\frac1a=b+\frac1b=c+\frac1c$$
$$\because \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}=2$$
$$\therefore \frac{\mu}{a}+\frac{\mu}{b}+\frac{\mu}{c}=2$$
$$\because \frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$$
$$\therefore a\mu + b\mu + c\mu = 1$$
$$\frac1a+\frac1b+\frac1c=2(a+b+c)$$
$$3(a+b+c)=a+\frac1a+b+\frac1b+c+\frac1c=3\left(a+\frac1a\right)$$
$$b+c=\frac1a$$
Similarly, $c+a=\dfrac1b$ and $a+b=\dfrac1c$. Substitute $a=\dfrac1{b+c}$ into the other two equations.
$$c+\frac1{b+c}=\frac1b$$
$$\frac1{b+c}+b=\frac1c$$
$$b(b+c)c+b=b+c$$
$$c+b(b+c)c=b+c$$
Subtracting one equation from another, we get $b=c$. Similarly, we have $a=b=c$. It remains to substitute it back to the original constraint and calculate the product.
Any alternative solution is appreciated.
| One possibility is to denote $A=\frac{a^2}{1+a^2}$, etc., so $a^2=\frac{A}{1-A}$, etc. We are given $A+B+C=1$ and want to prove that $8ABC\le(1-A)(1-B)(1-C)$, or, after opening the parentheses, replacing the sum by $1$, and putting all $ABC$ on the left,
$$
9ABC\le AB+BC+AC=(AB+BC+AC)(A+B+C)\,.
$$
Now just use AM-GM for each expression in parentheses on the right separately.
| {
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"url": "https://math.stackexchange.com/questions/1529750",
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"source": "stackexchange",
"question_score": "8",
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The equation of the plane which passes through the point of intersection of two space lines and at greatest distance from the point $(0,0,0)$ The equation of the plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and at greatest distance from the point $(0,0,0)$ is
$(A)4x+3y+5z=25\hspace{1cm}(B)4x+3y+5z=50\hspace{1cm}(C)3x+4y+5z=49\hspace{1cm}(D)x+7y-5z=2$
I found the point of intersection of two lines as $(4,3,5)$.Let the equation of the required plane is $ax+by+cz+d=0$.Squared distance of plane $ax+by+cz+d=0$ from $(0,0,0)$ is $\frac{d^2}{a^2+b^2+c^2}$
We need to maximize $\frac{d^2}{a^2+b^2+c^2}$ under the constraint $4a+3b+5c+d=0$.
I cannot solve it further.I am stuck here.Please help me.Thanks.
| You can use Cauchy–Schwarz inequality to continue your approach.
As you wrote, we want to maximize
$$\frac{d^2}{a^2+b^2+c^2}=\frac{1}{\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2}$$(This is because we may suppose that $d\not=0$.)
In other words, we want to minimize
$$\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2.$$
Since the plane $ax+by+cz+d=0$ passes through $(4,3,5)$, we have
$$4a+3b+5c+d=0,$$
i.e.
$$4\cdot\frac ad+3\cdot\frac bd+5\cdot\frac cd=-1\tag1$$
By Cauchy–Schwarz inequality,
$$\left(\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2\right)\left(4^2+3^2+5^2\right)\ge\left(4\cdot\frac ad+3\cdot\frac bd+5\cdot\frac cd\right)^2,$$
i.e.
$$\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2\ge\frac{\left(4\cdot\frac ad+3\cdot\frac bd+5\cdot\frac cd\right)^2}{4^2+3^2+5^2}=\frac{1}{50}.$$
The equality holds when
$$\frac{4d}{a}=\frac{3d}{b}=\frac{5d}{c}\tag2$$
From $(1)(2)$, the correct option is $(B)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To find the nth Derivative of $9\sqrt{x}$ Can you help me to proof that the nth derivative of $9\sqrt{x}$ is $$ (-1)^{(n-1)} \cdot \frac{9(2n-2)!}{(n-1)!} \cdot (4x)^{\frac{1-2n}{2}}$$
I've tried induction but didn't go very far.
Many thanks
| Firstly, the 9 is unimportant. Induction is the correct way to proceed, but perhaps we can disguise it in a computation:
$$\begin{align} \frac{d^n}{dx^n} \left( x^{\frac{1}{2}} \right) &= \frac{d^{n-1}}{dx^{n-1}} \left( \frac{1}{2} x^{\frac{1}{2} -1} \right) \\&= \frac{d^{n-2}}{dx^{n-2}}\left( \frac{1}{2} \left(\frac{1}{2} - 1 \right) x^{\frac{1}{2} -2} \right) \\&= \dots =\frac{d^{n-k}}{dx^{n-k}} \left( \frac{1}{2} \left( \frac{1}{2} -1\right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} -(k-1) \right) x^{\frac{1}{2} - k}\right) \\&= \frac{1}{2} \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} - (n-1) \right) x^{\frac{1}{2} -n }. \end{align} $$ Can you simplify this?
| {
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High school Math: confusion about the basic probability I am confused about the following two scenarios:
Out of a bag of 3 apples and 3 oranges, you pick 2 items.
1) What is the probability that you will have 2 apples?
2) What is the probability that you will have 1 apple and 1 orange?
My attempt:
1)
$$
\begin{aligned}
P(\mbox{2 apples})
&= P(\mbox{1st apple}) \times P(\mbox{2nd apple}) \\
&= \frac{3}{6} \times \frac{2}{5}.
\end{aligned}
$$
2)
$$
\begin{aligned}
P(\mbox{1 apple and 1 orange})
&= P(\mbox{1st apple}) \times P(\mbox{2nd orange}) \\
& + P(\mbox{1st orange}) \times P(\mbox{2nd apple}) \\
&= \frac{3}{6} \times \frac{2}{5} \times 2.
\end{aligned}
$$
My confusion is with case number 1: why you don't need to multiply the result by 2? Since your first pick could be apple #1, #2, #3.
| Consider all of the $6\times 5$ ways to pick two pieces of fruit. That's $30$:
$$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline
A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} \\
A_2 & \color{green}{A_2,A_1} & \times & \color{green}{A_2,A_3} & \color{blue}{A_2,O_1} & \color{blue}{A_2,O_2} & \color{blue}{A_2,O_3} \\
A_3 & \color{green}{A_3,A_1} & \color{green}{A_3,A_2} & \times & \color{blue}{A_3,O_1} & \color{blue}{A_3,O_2} & \color{blue}{A_3,O_3} \\ \hdashline
O_1 & \color{indigo}{O_1,A_1} & \color{indigo}{O_1,A_2} & \color{indigo}{O_1,A_3} & \times & \color{red}{O_1,O_2} & \color{red}{O_1,O_3} \\
O_2 & \color{indigo}{O_2,A_1} & \color{indigo}{O_2,A_2} & \color{indigo}{O_2,A_3} & \color{red}{O_2,O_1} & \times & \color{red}{O_2,O_3} \\
O_3 & \color{indigo}{O_3,A_1} & \color{indigo}{O_3,A_2} & \color{indigo}{O_3,A_3} & \color{red}{O_3,O_1} & \color{red}{O_3,O_2} & \times \\ \hline
\end{array}}$$
The ways to pick two apples are in the green quarter (upper left). There are $3\times 2$ of them; that is $6$ of $30$
The ways to pick an apple and an orange are in the blue quarter, but also in the indigo quarter (upper-right and lower-left). There are $3\times 3+3\times 3$ of them; that's $18$ of $30$
The final quarter are ways to pick two oranges. Again, just $3\times 2$ of these; that's $6$ of $30$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Solve modular arithmetic equation $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$ From another problem, I have reduced it to: This is the last step in solving:
Solve $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$
How should I begin, a major problem is the $1/24$
| $$\tfrac1{24}n(n+1)(n+2)(n+3)\equiv 1\mod 10\iff n(n+1)(n+2)(n+3)\equiv 24 \mod 240$$
Using the Chinese remainder theorem, this is equivalent to solving the system of congruences:
$$p(n)=n(n+1)(n+2)(n+3)\equiv\begin{cases}0\mod3 \\
4\mod5\\8\mod 16
\end{cases}$$
The only solution to the second congruence is $\;n\equiv 1\mod 5$, whence the solutions to the first two:
$$\color{red}{n_1\equiv 1,6,7\mod15}. $$
We'll determine the values of the factors and of $p(n)$, according to the value of $n\bmod 16$:
$$\begin{array}{r|cccc||r|cccc}
n\equiv&n+1&n+2&n+3&p(n)&n\equiv&n+1&n+2&n+3&p(n) \\
\hline
0&1&2&3&0&8&9&10&11&0\\
1&2&3&4&8&9&10&11&12&8\\
2&3&4&5&8&10&11&12&13&8\\
3&4&5&6&8&11&12&13&14&8\\
4&5&6&7&8&12&13&14&15&8\\
5&6&7&8&0&13&14&15&0&0\\
6&7&8&9&0&14&15&0&1&0\\
7&8&9&10&0&15&0&1&2&0
\end{array}$$
Thus the solutions of the second congruence are $\;\color{red}{n_2\equiv 1,2,3,4,9,10,11,12\mod16}$.
Byy the inverse isomorphism of the Chinese remainder theorem, from the Bézout's relation: $\;16-15=1$, we obtain the $24$ solutions:
$$\color{red}{n\equiv 16n_1-15n_2\mod 240}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Computing the integral I have the following integral:
$$\int ^{2\pi}_0 \frac{d\phi}{3+\sin\phi+\cos\phi}$$
I write $\cos\phi=\frac{1}{2}(z+\frac{1}{z})$, $\sin\phi=\frac{1}{2i}(z-\frac{1}{z}) \, d\phi=\frac{1}{i} \frac{dz}{z}$, Now I plug these in the equation and I get $$\frac{2 \, dz}{6iz+z^2(i+1)+(i-1)}$$
I have no idea to proceed from here on. A methodological answer would help a lot. Thanks in advance!
| Another approach that relies on real analysis only, is to use the Weierstrass Substitution $t=\tan (\phi/2)$. We first simplify the denominator. Note that
$$\sin \phi +\cos \phi =\sqrt{2}\cos (\phi -\pi/4)$$
Then, exploiting the periodicity and evenness of the cosine function we can write
$$\int_0^{2\pi} \frac{1}{3+\sin \phi +\cos \phi}\,d\phi=2\int_0^\pi\frac{1}{3+\sqrt{2}\cos \phi}\,d\phi$$
Now, we enforce the substitution $t=\tan \phi/2$ so that $\cos \phi =\frac{1-t^2}{1+t^2}$ and $d\phi=\frac{2}{1+t^2}\,dt$. Then,
$$\begin{align}
2\int_0^\pi\frac{1}{3+\sqrt{2}\cos \phi}\,d\phi&=4\int_0^\infty \frac{1}{3(1+t^2)+\sqrt{2}(1-t^2)}\,dt\\\\
&=4\int_0^\infty \frac{1}{(3-\sqrt{2})t^2+(3+\sqrt{2})}\,dt\\\\
&=\frac{4}{3-\sqrt 2}\int_0^\infty \frac{1}{t^2+\frac{3+\sqrt 2}{3-\sqrt 2}}\,dt\\\\
&=2\pi \sqrt{\frac{1}{(3+\sqrt 2)(3-\sqrt 2)}}\\\
&=\frac{2\pi}{\sqrt 7}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving Cubic when There are Known to be 3 Real Roots When solving for roots to a cubic equation, the sign of the $\Delta$ tells us when there will be 3 distinct real roots (as long as the first terms coefficient, $a$, is non-zero.) Namely when $\Delta$ is positive.
The equations to find the 3 roots are:
*
*$x_1 = -\frac{1}{3a}(b + C + \frac{\Delta_0}{C})$
*$x_2 = -\frac{1}{3a}(b + \frac{C(-1 + i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 + i\sqrt 3)})$
*$x_3 = -\frac{1}{3a}(b + \frac{C(-1 - i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 - i\sqrt 3)})$
Where: $$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}}$$
By the given equation, $\Delta_1^2 - 4\Delta_0^3 = -27a^2\Delta$, I know that when $\Delta$ is positive the square root in $C$ will produce an $i$. So when $\Delta$ is positive $C$ is effectively: $$C = \sqrt[3]{\frac{\Delta_1 + i\sqrt{4\Delta_0^3 - \Delta_1^2}}{2}}$$
So obviously the $i$ in $C$ cancels with the $i$s in the $x_2$ and $x_3$ roots, and we get 3 real roots. But for the life of me I cannot work out how. Can someone help me break those steps down?
| For example, let's try the equation $x^3 - 2 x^2 - x + 2 = 0$: $a= 1$, $b = -2$, $c = -1$, $d=2$, $\Delta_0 = 7$, $\Delta_1 = 20$.
$$C = \sqrt[3]{10 + 9 i \sqrt{3}} = \sqrt{7} e^{i\theta} = \sqrt{7}(\cos\theta + i \sin\theta)$$
where
$$ \theta = \dfrac{1}{3} \arctan(9 \sqrt{3}/10)$$
Then
$$ \Delta_0/C = \sqrt{7} e^{-i\theta} = \sqrt{7} (\cos \theta - i \sin \theta)$$
so
$$ \eqalign{x_1 &= -\dfrac{1}{3} \left( -2 + \sqrt{7} (\cos\theta + i \sin \theta) + \sqrt{7} (\cos\theta - i \sin\theta)\right)\cr
&= \dfrac{2}{3} - \dfrac{2\sqrt{7}}{3} \cos \theta}$$
$$ \eqalign{x_2 &= -\dfrac{1}{3} \left( - 2 + \sqrt{7} (\cos(\theta + 2 \pi/3) + i \sin(\theta + 2 \pi/3)) + \sqrt{7} (\cos(\theta - 2 \pi/3) + i \sin(\theta + 2 \pi/3))\right)\cr
&= \dfrac{2}{3} - \dfrac{2\sqrt{7}}{3} \cos(\theta + 2\pi/3)}$$
and similarly
$$ x_3 = \dfrac{2}{3} - \dfrac{2\sqrt{7}}{3} \cos(\theta - 2\pi/3)$$
It turns out that $x_1 = -1$, $x_2 = 2$, $x_3 = 1$ (which is no accident: I chose the polynomial to have those roots).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Divisibility by 30 How to show that if $30$ divides $a_1+a_2+ \ldots + a_n$, then $30$ divides $a_{1}^{5}+a_{2}^{5}+ \ldots + a_{n}^{5}$.
Since $30$ divides $a_1+a_2+ \ldots + a_n$, then $a_1+a_2+ \ldots + a_n \equiv 0 \pmod{30}$, thus $(a_1+a_2+ \ldots + a_n )^5 \equiv 0 \pmod{30}$ I don't know.
| Induction works. The proof for $n=2$ is identical to the one presented in the general case.
Say $30$ | $a_1 + \dots a_n$ implies $30$ | $a_1^5 + \dots a_n^5$ .
Now $a_1 + \dots + a_{n+1} = a_1 + \dots a_{n-1} + b$ if $b=a_n + a_{n+1}$ therefore $$30 \;|\; a_1^5 + \dots a_{n-1}^5 + b^5 $$ $$30 \; | \; a_1^5 + \dots a_{n-1}^5 + a_n^5 + a_{n+1}^5 + 5 a_n^4 a_{n+1}+10 a_n^3 a_{n+1}^2+10 a_n^2 a_{n+1}^3+5 a_n a_{n+1}^4$$
So to conclude we have to show that $30$ divides $5 a_n^4 a_{n+1}+10 a_n^3 a_{n+1}^2+10 a_n^2 a_{n+1}^3+5 a_n a_{n+1}^4$
Obviously, $5$ divides it. Also this is an even number (sum of even number of odd terms if $a_n$ and $a_{n+1}$ are both odd, if at least one of them is even all are since $a_n$ and $a_{n+1}$ are both present in all terms), so $2$ divides it.
Now, say $3$ doesn't divide both $a_n$ and $a_{n+1}$ (else we're done).
We have three cases mod $3$:
$a_n=a_{n+1}=1$ : $\quad$
Then $5+10+10+5 = 30 = 0$
$a_n=a_{n+1}=2$ : $\quad$
Then since $2^2 = 1$, we have $10+20+20+10 = 60 = 0$
$a_n=1, a_{n+1}=2$ or its symmetrical : $\quad$
Then $10+10+20+5 = 45 = 0$
So $2,3,5$ divide that, so $30$ does.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exponential Factorial vs Tetration I'm wondering whether there's a known way to compare the exponential factorial of n versus the tetration of a fixed number $($ e.g., $3$, since it appears in Graham's number $)$ with the same number of terms.
Here's an example: Find the smallest n such that $n^{!}$ is greater than $^{n-1}3$.
Both of these terms have the same number of exponents $($we ignore $1$ in the factorial exponential, thus $n-1$ terms$)$. A first computation shows that the tetration of $3$ grows much faster:
$4^{!} = 262144,~$ while $~^{3}3 = 3^{27} \approx 7.62 \cdot 10^{12}$
$5^{!} = 5^{262144} \approx 10^{183230},~$ while $~^{4}3 = 3^{3^{27}} \approx 10^{3.6 \cdot 10^{12}}$
Afterwards, numerical evaluation becomes more difficult, but what appears to be clear is that until the basis for the exponent in the exponential factorial becomes much, much larger, the tetration of $3$ will keep growing insanely faster, simply due to the higher last exponent.
Is this a known problem ?
| In fact, for no $n$ do we have $n^!$ greater than $^{n-1} 3$. For the following, we will use $\log$ to mean logarithm base 3.
We will prove by induction that $(n+2)^{n^!} < \ ^n 3$ for $n \ge 2$.
Base cases: $4^{2^1} = 16 < 27 = \ ^23$.
$5^{3^!} = 5^9 < 3^{27} = \ ^33$
$6^{4^!} = 6^{4^9} < 3^{2* 4^9} < 3^{3^{27}} = \ ^4 3$
Inductive step: Suppose that $(n+2)^{n^!} < \ ^n 3$ for some $n \ge 4$. Then
$\log\log ((n+3)^{(n+1)^!}) = \log((n+1)^! \log (n+3)) = \log((n+1)^!) + \log\log(n+3) = n^! \log(n+1) + \log \log (n+3)$.
Now, I claim that $n^! \log(n+1) + \log \log (n+3) < n^! \log(n+2)$. After rearranging, this is equivalent to $n^! > \frac {\log\log(n+3)}{\log(n+2) - \log(n+1)}$. But in fact
$\frac{\log\log(n+3)}{\log(n+2) - \log(n+1)} = \frac{\log\log(n+3)}{\log(1 + \frac{1}{n+1})} = \frac{\log\log(n+3)}{\frac{\ln(1+\frac{1}{n+1})}{\ln(3)}} < \frac{\log\log(n+3)}{\frac{1}{(n+2)\ln(3)}} = (n+2)\ln(3) \log\log(n+3) < n^9 \le n^!$.
(The last two inequalities follow because $n \ge 4$.)
So
$\log\log ((n+3)^{(n+1)^!}) < n^! \log(n+2) = \log((n+2)^{n^!}) < \log(^n 3) = \ ^{n-1} 3$
$(n+3)^{(n+1)^!} < \ ^{n+1} 3$, as desired.
So for $n \ge 3$, $n^! < (n+1)^{(n-1)^!} < \ ^{n-1} 3$, and of course one can verify this is true for $n = 2$ as well.
Going in the other direction, obviously $n^! > \ ^{n-2} 3$ for $n \ge 2$, since each term of the first tower is greater or equal to each term of the second, with equality in just one case. So for all $n \ge 2$ we have
$^{n-2} 3 < n^! < \ ^{n-1} 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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System of equations based question in matrix form If $\begin{bmatrix}1^2&2^2&3^2\\2^2&3^2&4^2\\3^2&4^2&5^2\\4^2&5^2&6^2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\4\\10\\\lambda\end{bmatrix}$,then $\lambda=$
$(A)17\hspace{1cm}(B)18\hspace{1cm}(C)19\hspace{1cm}(D)20$
I tried solving this question.
$1^2x+2^2y+3^2z=1$
$2^2x+3^2y+4^2z=4$
$3^2x+4^2y+5^2z=10$
I solved this system to get $x=\frac{25}{8},y=\frac{-5}{2},z=\frac{7}{8}$
$4^2x+5^2y+6^2z=\lambda$
I substituted $x,y,z$ in the above equation to get $\lambda=19$.
But this method is long and tedious.My book gives one method but that i cannot understand.Please help me understand this.Book's method is not self explanatory.
Book's solution:
$(n+3)^2=n^2-3(n+1)^2+3(n+2)^2$
$\lambda=1-3\times4+3\times 10=19$
OR if someone knows any better and elegant way to solve this question,please tell me,like using eigen values etc.Thanks.
| You can solve this by row reducing the augmented matrix into echelon form. The augmented matrix is given by:
$$ \left[
\begin{array}{ccc|c}
1^2&2^2&3^2&1\\
2^2&3^2&4^2&4\\
3^2&4^2&5^2&10\\
4^2&5^2&6^2&\lambda\\
\end{array}
\right] $$
Row reducing leads to:
$$ \left[
\begin{array}{ccc|c}
1^2&2^2&3^2&1\\
0&-7&-20&0\\
0&0&\frac{8}{7}&1\\
0&0&0&\lambda-19\\
\end{array}
\right] $$
This system is only consistent if $\lambda - 19 = 0$ , therefore $\lambda = 19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove a particular property of Laplacian operator I can't prove that Laplacian $\Delta(u(x))=\Delta(u(x_1,\ldots, x_n))=0$ also implies
$$
\Delta\left(|x|^{2-n}u\left(\frac{x}{|x|^2}\right)\right)=0
$$
for $\frac{x}{|x|^2}$ in the domain of definition of $u$. Help me!!
| Hint: Introduce $\mathbf{y}=\frac{\mathbf{x}}{|x|^{2}}$. Then
\begin{equation*}
(\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{\partial
\mathbf{y}})u(\mathbf{y})=0
\end{equation*}
Now express $\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{
\partial \mathbf{y}}$ in terms of $\mathbf{x}$ and $\frac{\partial }{
\partial \mathbf{x}}$.
Edit
It may help to use hyperspherical coordinates, see the contribution by
Kienzler in
What is the Laplace operator's representation in 3-sphere-coordinates?
Unless my calculation is incorrect it seems thar the desired property does
not follow.
Notation: $\mathbf{x}$ is a vector from $\mathbb{R}^{n}$, $x$ its absolute
value, $\Delta _{\mathbf{x}}=\frac{\partial }{\partial \mathbf{x}}\cdot
\frac{\partial }{\partial \mathbf{x}}=\partial _{\mathbf{x}}\cdot \partial _{
\mathbf{x}}=\nabla _{\mathbf{x}}\cdot \nabla _{\mathbf{x}}$. We note that
\begin{equation*}
\Delta _{\mathbf{x}}=x^{1-n}\partial _{x}(x^{d-1}\partial _{x})+\frac{1}{
x^{2}}L^{2}
\end{equation*}
where $L^{2}$ only contains derivatives wrt. angles.
The function $u(\mathbf{x})$ satisfies
\begin{equation*}
\partial _{\mathbf{x}}\cdot \partial _{\mathbf{x}}u(\mathbf{x})=0
\end{equation*}
Now the question is whether or not it is also true that
\begin{equation*}
\partial _{\mathbf{x}}\cdot \partial _{\mathbf{x}}\left( x^{2-n}u(\frac{%
\mathbf{x}}{x^{2}})\right) =0
\end{equation*}
Note that the part with angular derivatives gives no problems
\begin{equation*}
x^{-2}L^{2}\left( x^{2-n}u(\frac{\mathbf{x}}{x^{2}})\right)
=x^{2-n}x^{-2}L^{2}u(\frac{\mathbf{x}}{x^{2}})
\end{equation*}
We have
\begin{equation*}
\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{\partial \mathbf{%
y}}u(\mathbf{y})=0
\end{equation*}
so, setting
\begin{equation*}
\mathbf{y}=\frac{\mathbf{x}}{x^{2}}
\end{equation*}
we have to express $\frac{\partial }{\partial \mathbf{y}}\cdot \frac{
\partial }{\partial \mathbf{y}}$ in terms of the $\mathbf{x}$-derivatives
and we only need to consider the radial part
\begin{equation*}
y^{1-n}\partial _{y}y^{n-1}\partial _{y}=\partial _{y}^{2}+\frac{n-1}{y}%
\partial _{y}
\end{equation*}
Note that
\begin{equation*}
\mathbf{y}=\frac{\mathbf{x}}{x^{2}}\Rightarrow y=\frac{1}{x},\;x=\frac{1}{y}
,\;\mathbf{x}=\frac{\mathbf{y}}{y^{2}},\;\partial _{y}=\frac{\partial x}{
\partial y}\partial _{x}=-\frac{1}{y^{2}}\partial _{x}=-x^{2}\partial _{x}
\end{equation*}
so
\begin{equation*}
X=y^{1-n}\partial _{y}y^{n-1}\partial _{y}=x^{n-1}(-x^{2}\partial
_{x})x^{1-n}(-x^{2}\partial _{x})=x^{n+1}\partial _{x}x^{3-n}\partial
_{x}=x^{4}\partial _{x}^{2}+(3-n)x^{3}\partial _{x}
\end{equation*}
Now
\begin{eqnarray*}
Xx^{a} &=&x^{n+1}\partial _{x}x^{3-n}\partial _{x}x^{a}=x^{n+1}\partial
_{x}x^{a+3-n}x^{-a}\partial _{x}x^{a} \\
&=&x^{n+1}x^{a+3-n}x^{-(a+3-n)}\partial _{x}x^{a+3-n}(\partial _{x}+\frac{a}{%
x}) \\
&=&x^{a+4}\{\partial _{x}+(a+3-n)\frac{1}{x}\}(\partial _{x}+\frac{a}{x}) \\
&=&x^{a+4}\{\partial _{x}^{2}+(a+3-n)\frac{1}{x}\partial _{x}+\frac{a}{x}%
\partial _{x}-\frac{a}{x^{2}}+a(a+3-n)\frac{1}{x^{2}}\} \\
&=&x^{a+4}\{\partial _{x}^{2}+(2a+3-n)\frac{1}{x}\partial _{x}+a(a+2-n)\frac{
1}{x^{2}}\}
\end{eqnarray*}
This should be equal to $x^{a}X$ which requires
\begin{eqnarray*}
2a+3-n &=&3-n\Rightarrow a=0 \\
a(a+2-n) &=&0
\end{eqnarray*}
so $a=0$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $a\sqrt[3]{2} + b\sqrt[3]{4}$ irrational? I need to prove that $$ a\sqrt[3]{2} + b\sqrt[3]{4}$$ is irrational, while $a$,$b $ are non zero rationals.
I know that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational and I also know how to prove it, but I can't think of any reasonable implication that would state: $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational $\implies$ $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, so I could show it is a contradiction.
| Suppose $ a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational, ($=y$, say)
So $ a\sqrt[3]{2} + b\sqrt[3]{4} = y$ and
$ a\sqrt[3]{2} = y - b\sqrt[3]{4}$
Cubing both sides, we have
$$ (a\sqrt[3]{2})^3 = (y - b\sqrt[3]{4})^3$$
$$ 2a^3=y^3-4b^3 + 3y\cdot 4^{\frac{2}{3}} - 3y^2\cdot 4^{\frac{1}{3}}$$
Equating rational and irrational parts, we have $$3y\cdot 4^{\frac{2}{3}} - 3y^2\cdot 4^{\frac{1}{3}}=0$$ or,
$$3y \cdot 4^{\frac{1}{3}}\left[4^{\frac{1}{3}} - y\right] = 0$$
Now our first assumption that $y$ is rational holds iff $y=0$
But then $\frac{a}{b}=-\frac{\sqrt[3]{4}}{\sqrt[3]{2}}$ implies $a,b$ irrational.
CONTRADICTION.
| {
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Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| $$A+\sqrt{B}=\frac{A+C}{2}+\frac{A-C}{2}+\sqrt{A^2-(A^2-B)}$$
$$={\sqrt{\frac{A+C}{2}}}^2+2\sqrt{\frac{(A+\sqrt{A^2-B})(A-\sqrt{A^2-B})}{4}}+{\sqrt{\frac{A-C}{2}}}^2$$
$$={\sqrt{\frac{A+C}{2}}}^2+2\sqrt{\frac{(A+C)(A-C)}{4}}+{\sqrt{\frac{A-C}{2}}}^2$$
$$=\left(\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
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Does $a_n = \sin (\tfrac{\pi}{2}n + \tfrac{\pi}{3})\sqrt{n}$ converge? The question sounds like:
Let $a_n = \sin (\frac{\pi}{2}n + \frac{\pi}{3})n^{0.5}$ for $n = 1, 2, 3,...$
Check if $\{a_n\}$ from $n = 1$ to infinity is convergent or divergent.
So far, I have the following:
$$\{a_n\} = \sin \left(n\frac{\pi}{2} + \frac{\pi}{3}\right)\sqrt{n}$$
$$= \lim\limits_{n\to \infty} \sin \left(n\frac{\pi}{2}\right)\cdot\lim\limits_ {n\to \infty} \left(\frac{\pi}{3}\right) \sqrt{n}$$
I know that the limit of $\sin$ is bounded between $-1$ and $1$. Am I on the right path?
| I assume your sequence is $\{a_n\}$ where $a_n = \sin(\frac{\pi n}{2} + \frac{\pi}{3}) \cdot \sqrt{n}$, for $n = 1, 2, 3, ...$.
Using the angle sum formula, we see that
$\sin(\frac{\pi n}{2} + \frac{\pi}{3}) = \cos(\frac{\pi n}{2}) \; \sin(\frac{\pi}{3}) + \sin(\frac{\pi n}{2})\; \cos(\frac{\pi}{3}) = $
$\frac{\sqrt{3}}{2}$ for $n = 4k$,
$ \frac{-\sqrt{3}}{2}$ for $n = 4k + 2$,
$ \frac{1}{2}$ for $n = 4k + 1$, and
$\frac{-1}{2}$ for $n = 4k + 3$.
We have
$\lim_{k\to \infty} a_{4k} = \lim_{k \to \infty} a_{4k + 1} = \frac{\sqrt{3}}{2} \cdot(\lim_{k \to \infty} \sqrt{4k}) = \frac{1}{2} \cdot(\lim_{k \to \infty} \sqrt{4k+1}) = \infty$,
and
$\lim_{k \to \infty} a_{4k+2} = \lim_{k \to \infty} a_{4k + 3} = -\frac{\sqrt{3}}{2} \cdot(\lim_{k \to \infty} \sqrt{4k+2}) = -\frac{1}{2} \cdot (\lim_{k \to \infty} \sqrt{4k + 3}) = - \infty$.
So the limits of different index sequences are different. I believe we say that the limit does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$ then what is $f(x)$? assume that:
$$2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$$
Then what is $f(x)$?
| Since $$(x^2+x+1)^2=x^4+x^2+1+2x(x^2+x+1)$$
we have
$$x^4+x^2+1=(x^2+x+1)^2-2x(x^2+x+1).$$
So,
$$\frac{x^2}{2(x^4+x^2+1)}=\frac{x^2}{2((x^2+x+1)^2-2x(x^2+x+1))}=\dfrac{\left(\dfrac{x}{x^2+x+1}\right)^2}{2-4\cdot\dfrac{x}{x^2+x+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
| We can use the basic limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \
\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}}=\dfrac{1%
}{24}
\end{equation*}
as follows
\begin{equation*}
\lim\limits_{x\rightarrow 0}\frac{x^{2}+2\cos x-2}{x\sin ^{3}x}%
=\lim\limits_{x\rightarrow 0}\left( \frac{x}{\sin x}\right)
^{3}\lim\limits_{x\rightarrow 0}\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}%
}\right) \left( 2\right) =1^{3}\cdot \frac{1}{24}\cdot 2=\frac{1}{12}.
\end{equation*}
The basic limits can be computed by repeated use of L'Hospital's rule.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving by induction that $2^n \le 2^{n+1}-2^{n-1} - 1$ . Does my proof make sense? I'm not sure if this proof is valid or not and could use some feedback.
Let P be the supposition that $2^n \le 2^{n+1}-2^{n-1} - 1$ for all positive integers n.
Base Case:
Let n=1
$2^1 \le 2^2-2^0-1$
$2\le2$
Suppose that P holds true for some positive integer n=k
$$2^{k+1}=2*2^k$$
By inductive hypothesis:
$$2*2^k \le 2(2^{k+1}-2^{k-1}-1)$$
$$2^{k+1} \le 2^{k+2}-2^{k}-2$$
Since we know $2^{k+1} \le 2^{k+2}-2^{k}-2$, we also know $2^{k+1} \le 2^{k+2}-2^{k}-1$ because the former is always smaller.
This proves that:
$$2^k \le 2^{k+1}-2^{k-1} - 1\Rightarrow 2^{k+1} \le2^{k+2}-2^{k}-1 $$
| It looks fine!
There's really no particular need for induction, though, if you know that $x\mapsto 2^x$ is an increasing function. Instead, show that $$2^{n+1}=2^n+2^n=2^n+2^{n-1}+2^{n-1},$$ so since $n-1\ge0$ for all positive integers $n,$ we have $2^{n-1}\ge 2^0=1,$ so that $$2^{n+1}\ge 2^n+2^{n-1}+1,$$ from which the result follows.
More generally, we can use this approach to show that, for real numbers $x,$ we have $2^x\le2^{x+1}-2^{x-1}-1$ if and only if $x\ge1.$
| {
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IMO 1997 problem 6 For each positive integer $n$ , let $f (n)$ denote the number of ways of representing $n$ as a sum of powers of $2$ with non-negative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. Prove that, for any integer $n \geqslant 3$, $$2^{n^2/4} < f (2^n) < 2^{n^2/2}.$$
| Observe that $k f (2k) < f (4k) \leqslant 2k f (2k)$ for $k \geqslant 4$ a power of $2$. We use induction on $n$ to prove the inequality. If $n = 3$, then it is obvious that $$2^{3^2/4} < f(8) = 10 < 2^{3^2/2}.$$ Suppose the inequality holds for $n = m$, we now prove it for $n = m + 1$. Since $f (2^m) < 2^{m^2/2}$ and using $f (4k) \leqslant 2k f( 2k)$ with $k = 2^{m - 1}$, we have $$f (2^{m + 1}) \leqslant 2^m f (2^m) < 2^m 2^{m^2/2} = 2^{(m^2 + 2m)/2} < 2^ {(m + 1)^2/2},$$ as desired. Now, by $f (2^m) > 2^{m^2/4}$ and $f (4k) > k f (2^k)$ with $k = 2^{m - 1}$, we have $$f (2^{m + 1}) > 2^{m - 1} f (2^m) > 2^{m - 1} 2^{m^2/4} = 2 ^ {(m^2 + 4m - 4)/4} > 2^{(m + 1)^2/4}$$ for $m > 2$. Proof is complete. $\Box$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Verification of my solution to $\frac{dy}{dx}=\frac{2xy^2+x}{x^2y+y}$ I'm not certain that the following solution is entirely correct:
$$\frac{dy}{dx}=\frac{2xy^2+x}{x^2y+y}=\frac{1}{y}\frac{x(2y^2+1)}{1+x^2}$$
$$\frac{y\frac{dy}{dx}}{2y^2+1}=\frac{x}{1+x^2}$$
$$\frac{y}{2y^2+1}dy=\frac{x}{1+x^2}dx$$
$$\int\frac{y}{y^2+1}dy=2\int\frac{x}{1+x^2}dx$$
$$\frac{1}{2}\ln\left|y^2+1\right|=\ln\left|x^2+1\right|+\mathrm{c}$$
$$y=\sqrt{2e^c(x^2+1)-1}$$
Any comments are greatly appreciated.
| You get a wrong turn in your 3th to your 4th equation, it should be:
$$y'(x)=\frac{2xy(x)^2+x}{x^2y(x)+y(x)}\Longleftrightarrow$$
$$\frac{\text{d}y(x)}{\text{d}x}=\frac{2xy(x)^2+x}{x^2y(x)+y(x)}\Longleftrightarrow$$
$$\frac{\text{d}y(x)}{\text{d}x}=\frac{x\left(\frac{1}{y(x)}+2y(x)\right)}{x^2+1}\Longleftrightarrow$$
$$\frac{\frac{\text{d}y(x)}{\text{d}x}}{\frac{1}{y(x)}+2y(x)}=\frac{x}{x^2+1}\Longleftrightarrow$$
$$\int\frac{\frac{\text{d}y(x)}{\text{d}x}}{\frac{1}{y(x)}+2y(x)}\space\text{d}x=\int\frac{x}{x^2+1}\space\text{d}x\Longleftrightarrow$$
$$\frac{1}{4}\ln(2y(x)^2+1)=\frac{1}{2}\ln(x^2+1)+\text{C}\Longleftrightarrow$$
$$y(x)=\pm\frac{\sqrt{\text{C}(x^2+1)^2-1}}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving inequality with constraint $abc=1$ For positive reals $a,b,c$, with $abc=1$, prove that $$\frac{a^3+1}{b^2+1}+ \frac{b^3+1}{c^2+1}+\frac{c^3+1}{a^2+1} \geq 3$$ I tried the substitution $x/y,y/z,z/x$, but it didn't give me anything. What else to do? Thanks.
| By re-arrangement inequality, we have
$$\sum_{cyc} \frac{a^3+1}{b^2+1} \ge \sum_{cyc} \frac{a^3+1}{a^2+1}$$
So it is enough to show that for positive $x$, the function
$$f(x) = \frac{x^3+1}{x^2+1}-1-\frac12 \log x \ge 0$$
as the inequality is equivalent to $f(a)+f(b)+f(c) \ge 0$.
From $\displaystyle f'(x) = (x-1)\frac{2x^4 + x^3 + 7x^2 + x + 1}{2x (x^2+1)^2}$, it is clear that
$f$ is decreasing for $x \in (0, 1)$ and increasing for $x > 1$.
Now as $f(1) = 0$, we must have $f(x) \ge 0$ for all $x > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$. Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$.
I need to solve this with triple integrals, however I'm having a lot of problems. Anyone knows how to solve?
| The hard part of such problems is to imagine the volume enclosed by the surfaces and describing the points inside the volume in a mathematical language so that you can determine the limits of integration.
It seems that the two surfaces are some parabolic surfaces facing toward each other. You may imagine some picture in your mind like the figure below
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
So, first we find the intersection of the two surfaces as it bounds the variation of $x$ and $y$ coordinates of the points inside this volume. Hence, consider the following
$$\eqalign{
& {x^2} + 3{y^2} = 8 - {x^2} - {y^2} \cr
& 2{x^2} + 4{y^2} = 8 \cr
& {{{x^2}} \over 4} + {{{y^2}} \over 2} = 1 \cr
& {\left( {{x \over 2}} \right)^2} + {\left( {{y \over {\sqrt 2 }}} \right)^2} = 1 \cr} $$
and hence all points $(x,y,z)$ which lie on both surfaces satisfy
$$\matrix{
{\left\{ \matrix{
{\left( {{x \over 2}} \right)^2} + {\left( {{y \over {\sqrt 2 }}} \right)^2} = 1 \hfill \cr
z = {x^2} + 3{y^2} \hfill \cr} \right.\,\,\,\,\,or\,\,} & {\left\{ \matrix{
{\left( {{x \over 2}} \right)^2} + {\left( {{y \over {\sqrt 2 }}} \right)^2} = 1 \hfill \cr
z = 8 - {x^2} - {y^2} \hfill \cr} \right.} \cr
} $$
which are two different equations of the same intersection curve. Now, we are ready to describe the coordinates of the points inside the volume in some mathematical language
$$\Omega = \left\{ {\left( {x,y,z} \right)| - 2 \le x \le 2, - \sqrt {2 - {1 \over 2}{x^2}} \le y \le \sqrt {2 - {1 \over 2}{x^2}} ,{x^2} + 3{y^2} \le z \le 8 - {x^2} - {y^2}} \right\}$$
and hence your triple integral will be
$$\eqalign{
& \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt \Omega }
{dV} = \int\limits_{x = - 2}^2 {dx\int\limits_{y = - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {dy\int\limits_{z = {x^2} + 3{y^2}}^{8 - {x^2} - {y^2}} {dz} } } = \int\limits_{x = - 2}^2 {\int\limits_{y = - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {\int\limits_{z = {x^2} + 3{y^2}}^{8 - {x^2} - {y^2}} {dzdydx} } } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_{x = - 2}^2 {\int\limits_{y = - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {\left[ {\left( {8 - {x^2} - {y^2}} \right) - \left( {{x^2} + 3{y^2}} \right)} \right]} } dydx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_{x = - 2}^2 {\int\limits_{y = - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {\left( {8 - 2{x^2} - 4{y^2}} \right)dydx} } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\int\limits_{x = - 2}^2 {\int\limits_{y = 0}^{\sqrt {2 - 0.5{x^2}} } {\left( {4 - {x^2} - 2{y^2}} \right)dydx} } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\int\limits_{x = - 2}^2 {\left[ {\left( {4 - {x^2}} \right)y - {2 \over 3}{y^3}} \right]} _0^{\sqrt {2 - 0.5{x^2}} }dx = 4\int\limits_{x = - 2}^2 {\left[ {\left( {\left( {4 - {x^2}} \right) - {2 \over 3}{y^2}} \right)y} \right]} _0^{\sqrt {2 - 0.5{x^2}} }dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\int\limits_{x = - 2}^2 {\left[ {\left( {{8 \over 3} - {2 \over 3}{x^2}} \right)\sqrt {2 - 0.5{x^2}} } \right]dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{32} \over 3}\int\limits_{x = 0}^2 {\left[ {\left( {2 - 0.5{x^2}} \right)\sqrt {2 - 0.5{x^2}} } \right]dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{32} \over 3}\left( {{{3\sqrt 2 \pi } \over 4}} \right) = 8\sqrt 2 \pi \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I simplify $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ )$? $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ ) =$ ?
The answer is $3$.
My progress so far:
\begin{align}
\sqrt{3}(\cot 70^\circ+4\cos 70^\circ)&= \sqrt{3}(\tan 20^\circ+4\sin 20^\circ) \\
&= \sqrt{3}(\sin 20^\circ)\left(\frac{1}{\cos 20^\circ}+4\right)\\
&= \sqrt{3}\frac{(\sin 20^\circ)(1+4\cos 20^\circ)}{\cos 20^\circ}\\
&= \sqrt{3}\frac{(\sin 20^\circ+2\sin 40^\circ)}{\cos 20^\circ}\\
\end{align}
| This is a repeating use of the sum-to-product identities.
\begin{align}
\frac{\sin 20^\circ + 2\sin 40^\circ}{\cos 20^\circ} &= \frac{2\sin 30^\circ \cos 10^\circ + \sin 40^\circ}{\cos 20^\circ} \\
&= \frac{\cos 10^\circ + \cos 50^\circ}{\cos 20^\circ} \\
&= \frac{2\cos 30^\circ \cos 20^\circ}{\cos 20^\circ} \\
&= \sqrt{3}
\end{align}
so the answer of the original expression is $3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find out the number of solutions to the equation- $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ under certain constraints How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ where each $x_i$ is a non-negative integer, $3 \leq x_1 \leq 10, \ 2 \leq x_2 \leq 7$ and $x_3 \geq 5$
I have been able to do all my counting problems but this one. I can not find an equation for max number per variety.
I know that I will subtract the min for each $x$ from the total, but I can not do $\begin{pmatrix} 15 \\ 6\end{pmatrix}$ cause that would not factor in the max per variety.
What is the equation I would use for this?
| The answer will be the coefficient of $x^{25}$ from the below expression-
$$(x^3+x^4+ \dots+x^{10})(x^2+x^3+ \dots+x^7)(x^5+x^6+ \dots+x^{20})(x^0+x^1+ \dots+x^{15})(x^0+x^1+ \dots+x^{15})(x^0+x^1+ \dots+x^{15}).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$n^{th}$ determinant Find determinant $D_n$ of matrix $$
\begin{bmatrix}
1 & 1 & \cdots & 1 & -n \\
1 & 1 & \cdots & -n & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & -n & \cdots & 1 & 1 \\
-n & 1 & \cdots & 1 & 1
\end{bmatrix}
$$
After multiplying first row by $-1$ and adding to $n-1$ rows:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & 0 & \cdots & -n-1 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & -n-1 & \cdots & 0 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Second and $n-1$ row are changing place:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Multiplying first row by $n$ and adding it to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & n+1 & \cdots & n+1 & 1-n^2
\end{vmatrix}
$$
Adding second to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & n+1 & (1+n)(2-n)
\end{vmatrix}
$$
Adding $n-1$ to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & 0 & (1+n)(3-n)
\end{vmatrix}
$$
Product on main diagonal gives $$D_n=(n+1)(3-n)(-n-1)^{n-2}$$
This is not correct. What is wrong with this upper triangular transformation?
| You have only dealt with the second and $n-1$th row. There is also the third and $n-2$nd row; 4th and $n-3$rd row, and so on.
| {
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How to solve this inequality involving trig functions: $\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$? Source - Larsen 1.3.12
Problem:
Prove using $y=\sin(x)$ graph, for $A,B,C$ as angles of triangle, that:
$$\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$$
Attempt:
(without graph)
Using sum-to-product formula, $\sin B+\sin C=2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$
So,
$$\frac{\sin B + \sin C}{2}=\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$$
$$=\sin(\frac{180-A}{2})\cos(\frac{B-C}{2})$$
$$=\cos(\frac{A}{2})\cos(\frac{B-C}{2})$$
$$=\frac{\cos(\frac{A+B-C}{2})+\cos(\frac{A-B+C}{2})}{2}$$
$$=\frac{\cos(\frac{180-2C}{2})+\cos(\frac{180-2B}{2})}{2}$$
$$=\frac{\sin C+\sin B}{2}$$
And I think I arrived where I had begun.
Question:
How to solve this inequality? (without graph)
| Based on Ekaveera's hint, right at the first step we have:
$$\frac{\sin B + \sin C}{2}=\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$$
Now, since $-1<\cos A<1$ for any $0<A<180$, we have:
$$\frac{\sin B + \sin C}{2} \leq \sin(\frac{B+C}{2})$$
| {
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Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$... Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$.
So far I have the base case completed, and believe I am close to completing the proof itself.
Base case:$(n=1)$
$3^1 + 7^1 - 2 = 8/8 = 1 $
Inductive Hypothesis: Assume that $3^n +7^n −2$ is divisible by 8 for all positive integers n.
Induction step $(n+1)$ case:
$$ 3^{n+1} + 7^{n+1} - 2 $$
$$3(3^{n}) + 7(7^{n}) - 2$$
$$3^n + 7^n = 8x $$
-It seems to me that this could be the end of the proof because whatever the answer is would be a multiple of 8: but I am unsure, any help is appreciated.
| $\begin{array}\\
3^{n+1}+7^{n+1}-2
&=3^{n+1}-1+7^{n+1}-1\\
&=(3-1)\sum_{k=0}^n 3^k+(7-1)\sum_{k=0}^n 7^k\\
&=\sum_{k=0}^n (2\ 3^k+6\ 7^k)\\
&=\sum_{k=0}^n (6\ 3^k+6\ 7^k-4\ 3^k)\\
&=\sum_{k=0}^n (6(3^k+7^k-2)+12-4\ 3^k)\\
&=\sum_{k=0}^n (6(3^k+7^k-2)+4(3-3^k))\\
\end{array}
$
Since
$3-3^k$
is even for $k \ge 0$
(being the difference of
two odd numbers),
$4(3-3^k)$
is divisible by $8$.
Therefore,
if
$3^k+7^k-2$
is divisible by $8$
for
$0 \le k \le n$,
then
$3^{n+1}+7^{n+1}-2$
is also divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do I rationalize the following denominator $$\frac{-2}{3\sqrt\frac{5}{12u}}$$
What I did:
turned denominator and numerator into square roots
$\frac{\sqrt5}{\sqrt{12u}}$
simplified denominator to
$2\sqrt{3u}$ and $2$ is multiplied by $-2/3$ to make $-4/3 \sqrt 5/\sqrt{12u}$
I then multiplied denominator and numerator by denominator to get
$\frac{4\sqrt{15u}}{9u}$
correct answer: $\frac{-\sqrt{15u}}{9u}$
what did i do wrong thanks?
| Your 'correct' answer and the other answer are wrong. Please have a look at this:
$$
\begin{align}
\frac{-2}{3\sqrt\frac{5}{12u}}&=-\frac{2}{3}\times\frac{\sqrt{12u}}{\sqrt{5}}\\\\
&=-\frac{2}{3}\times\frac{\sqrt{5}\times \sqrt{12u}}{\sqrt{5}\times \sqrt{5}}\\\\
&=-\frac{2}{15}\times \sqrt{5}\times \sqrt{12u}\\\\
&=-\frac{2}{15}\times \sqrt{5}\times \sqrt{4\times 3 \times u}\\\\
&=-\frac{4}{15}\sqrt{15\: u}.
\end{align}
$$ Hoping it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the condition $a+b=-c$ are such that $a^2=bc+1$ and $b^2=ca+1$. Also if $a-b=b-a,$ it means $a=b.$ Then both $b-a$ and $a-b$ are $0.$ Hence $c$ can thus be any integer. Hence, there are two groups of integers satisfying the required condition $-a=b$ with any $c$ and $a+b=-c.$
Please let me know whether my solution is correct or not.
| I tried solving came out with following results:
$(i)-(ii)$
gives
$a+b+c=0$
which implies,
$a^3+b^3+c^3=3abc$
Multiplying equation $(i)$ with $a$ and multiplying equation $(ii)$ with $b$
and then subtracting $(i)-(ii)$
we get
$a^2+b^2+ab=1$
which can be written as
$(a+b)^2-ab=1$ , which can be further written as $c^2-ab=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that all level sets of the Hamiltonian are bounded Consider the system
$$
\dot{x}=y,~~~~~~~~~~\dot{y}=x^2-x^3.
$$
Its Hamiltonian is
$$
H(x,y)=\frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}.
$$
Show that all level sets of the Hamiltonian are bounded.
As far as I see, it is meant to show that the set
$$
\left\{(x,y)\in\mathbb{R}^2: H(x,y)=c\right\}
$$
is a bounded set.
My computation gives that $H(x,y)=c\Leftrightarrow y=\sqrt{2c+\frac{2}{3}x^3-\frac{1}{2}x^4}$, i.e.
$$
\left\{(x,y)\in\mathbb{R}^2: H(x,y)=c\right\}=\left\{(x,y): y=\sqrt{2c+\frac{2}{3}x^3-\frac{1}{2}x^4}\right\}
$$
How can I see that this is a bounded set?
| We could think about different ways to show this, and after a while we might think that it would be nice to show that any level set of $H$ is contained within some ellipse. That would definitely mean that any level set is bounded.
In formulas, this would mean showing that any $x,y$-pair satisfying
$$
\frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} = c
$$
satisfies
$$
ay^2 + bx^2 \leq d
$$
for some $a,b,d > 0$ which depend on $c$. The first equation already has a $y^2$ term, so we might be able to get away with $a=1/2$. For now let's also take the simplest choice for $b$, namely $b=1$, and hope this doesn't complicate finding $d$.
So now we want to show that if $(x,y)$ satisfies
$$
\frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} = c \tag{1}
$$
then
$$
\frac{y^2}{2} + x^2 \leq d \tag{2}
$$
for some $d$ depending on $c$.
With our target $(2)$ in mind, let's rearrange $(1)$ like
$$
\frac{y^2}{2} + x^2 = -\frac{x^4}{4} + \frac{x^3}{3} + x^2 + c = p(x) + c. \tag{3}
$$
The polynomial $p(x) = - x^4/4 + x^3/3 + x^2$ is bounded above, and to find this bound we first calculate
$$
p'(x) = -x^3 + x^2 + 2x = -x(x+1)(x-2),
$$
and then check that $p(0) = 0$, $p(-1) = 5/12$, and $p(2) = 8/3$ to conclude that $p(x) \leq 8/3$, and hence, from $(3)$, we conclude that if
$$
\frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} = c
$$
then
$$
\frac{y^2}{2} + x^2 \leq c + \frac{8}{3},
$$
which completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Number Theory: Show that $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$ Show that for all $n\in\mathbb{N}$, $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$.
I think I've proved this problem, but I was unsure if my proof was correct:
Proof
Let $n=1$. Then, $10^3=1000\equiv1\pmod{3^3}$ since $3^2=9\mid999$ and $3\mid111\implies 3^3\mid999$; but, $3^4\not\mid999$.
Now, assume $10^{3^n}\equiv1\pmod{3^{n+2}},3^{n+3}\not\mid10^{3^n}-1$.
We want to prove that this is true for $n+1$:
$10^{3^{n+1}}\equiv1\pmod{3^{n+3}}$
$10^{3^{n+1}}-1\equiv0\pmod{3^{n+3}}$
$(10^{3^n})^3-1\equiv0\pmod{3^{n+3}}$
$(10^{3^n}-1)((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod{3^{n+3}}$.
By our assumption, $3^{n+2}\mid(10^{3^n}-1)$ but $3^{n+3}\not\mid10^{3^n}-1$, so $\gcd(3^{n+3},10^{3^n}-1)=3^{n+2}$.
So,
$((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod3$.
But $((10^{3^n})^2+10^{3^n}+1)$ is of the form $100\dots00100\dots001$, so the sum of its digits is $3$, so it can be divided by $3\implies ((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod3$. (Just added this:) However, it cannot be divided by $3^2=9$ since the sum of these digits is not divisible by $9$.
Thus, $10^{3^{n+1}}\equiv1\pmod{3^{n+3}}$ and $3^{n+4}\not\mid 10^{3^{n+1}}-1$.
So, by induction, $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$ for all $n\in\mathbb{N}$.
| The "divides" part is fine. You have enough machinery set up to prove that $3^{n+3}$ does not divide $10^{3^n}-1$, but you have not proved it explicitly. Hint: You showed that a certain expression is divisible by $3$. Show it is not divisible by any higher power of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$
How to solve the differential equation
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$$
Hints please.
| Let's pose $y = \arcsin(xz)$. Then:
$$\begin{cases}
\displaystyle\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = \frac{x}{\sqrt{1-x^2z^2}}\frac{dz}{dx}\\
\tan(y) = \tan(\arcsin(xz)) = \displaystyle\frac{xz}{\sqrt{1-x^2z^2}}\\
\sin(y) = \sin(\arcsin(xz)) = xz
\end{cases}$$
Then:
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y) \Rightarrow\\
\frac{x}{\sqrt{1-x^2z^2}}\frac{dz}{dx} + \frac{1}{x} \frac{xz}{\sqrt{1-x^2z^2}}= \frac{1}{x^2} \frac{xz}{\sqrt{1-x^2z^2}}xz \Rightarrow\\
\frac{dz}{dx} + \frac{1}{x}z = \frac{1}{x}z^2.$$
This is a Bernoulli differential equation with $Q(x) = P(x) = \frac{1}{x}$. The solution is $$z(x) = \frac{1}{ax + 1}.$$
Finally:
$$y(x) = \arcsin\left(\frac{x}{ax + 1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this sum $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$ Show that
$$\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$$
| Suppose we seek to verify that
$$\sum_{k=0}^n (-1)^k 2^{2n-2k} {2n-k+1\choose k}
= n+1.$$
Introduce
$${2n-k+1\choose k} = {2n-k+1\choose 2n-2k+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-k+1}}{z^{2n-2k+2}} \; dz.$$
Observe that this is zero when $k\gt n$ so we may extend $k$ to
infinity.
We thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+1}}{z^{2n+2}}
\sum_{k\ge 0} (-1)^k 2^{2n-2k} \frac{z^{2k}}{(1+z)^{k}} \; dz
\\ = \frac{2^{2n}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+1}}{z^{2n+2}}
\frac{1}{1+z^2/(1+z)/4}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+2}}{z^{2n+2}}
\frac{1}{4(1+z)+z^2}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+2}}{z^{2n+2}}
\frac{1}{(z+2)^2}
\; dz.$$
Evaluate the integral by computing the negative of the residue
at $z=-2$. This requires
$$\left(\frac{(1+z)^{2n+2}}{z^{2n+2}}\right)'
= \frac{(2n+2)(1+z)^{2n+1}}{z^{2n+2}}
- (2n+2)\frac{(1+z)^{2n+2}}{z^{2n+3}}.$$
Setting $z=-2$ we obtain
$$2^{2n+2} (2n+2) \frac{(-1)^{2n+1}}{(-2)^{2n+2}}
- 2^{2n+2} (2n+2) \frac{(-1)^{2n+2}}{(-2)^{2n+3}}
\\ = -(2n+2) + (n+1) = - (n+1).$$
Therefore the integral is $$n+1$$ as claimed.
Addendum. We also need to verify that the residue at infinity is
zero. We obtain
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{2n+2}
\frac{(1+z)^{2n+2}}{z^{2n+2}} \frac{1}{(2+1/z)^2}
\\ = -\mathrm{Res}_{z=0} (1+z)^{2n+2} \frac{1}{(2z+1)^2} = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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When do we expand the numerator of the Bayes' Theorem I am trying to understand why the proposed solution below to the following question is wrong:-
A box contains three cards: a card that is black on both sides, one that is white on both sides and a card that is black on one side and white on the other. You withdraw a card at random and see that the visible face is black. What is the probability that the other side of this card is also black?
Proposed solution:
Let A = {other side is Black} and B = {observed side is Black}, then
$$ P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ where,
$$ P(B|A) = \frac{1}{2}$$
$$ P(A) = \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot 0 = \frac{1}{2}$$
$$P(B) = P(B|A)P(A) + P(B|\tilde{A}) P(\tilde{A})$$
$$ = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot \frac{1}{2}= \frac{1}{2}$$
where $\tilde{A}$ ={other side is not Black}. Therefore,
$$ P(A|B) = \frac{\frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}} = \frac{1}{2} \textbf{ (wrong!)}$$
The proposed solution is incorrect. The correct solution is as follows:
$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
where,
$$ P(A\cap B) = \frac{1}{3}$$
$$ P(B) = P(B|C_{B}) P(C_{B}) + P(B|C_{W}) P(C_{W}) + P(B|C_{M}) P(C_{M})$$
$$ =\frac{1}{3}\cdot1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot\frac{1}{2} = \frac{1}{2}$$
where $C_{B}$ = {Card with both sides Black}, $C_{W}$ = {Card with both sides White} and $C_{M}$ = {Card with Mixed sides}. This gives,
$$P(A|B) = \frac{2}{3} \textbf{ (correct!)}$$
It seems that expanding the numerator $P(A \cap B)$ as $P(B|A) P(A)$ gives the wrong solution. Where is the proposed solution going wrong?
| Your error is in asserting that : $\mathsf P(B\mid A) = \tfrac 1 2$ It does not.
The probability that the observed side is black given that the other side is black is $2/3$. Of the six (equally probable) sides that the unobserved side could be, three of them are black. Of these three black sides, only two of them have a black side on the flip-side of their card.
Then $\qquad\mathsf P(A\mid B) = \dfrac{\mathsf P(B\mid A)\mathsf P(A)}{\mathsf P(B)} = \dfrac{\tfrac 2 3\cdot\tfrac 1 2}{\tfrac 1 2} = \dfrac 2 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the difference between the two real numbers that satisfy this equation?
What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form
I tried guessing solutions but seeing how there are no common zeroes to both the left- and right-hand sides I don't know what to do.
| $(x^2-1)(x-2)-[(x^2-9)(x+2)]=0$ thus $x^3-x-2x^2+2-x^3+9x-2x^2+18=0$ so $-4x^2+8x+20=0$ thus Solving we get $x=1\pm \sqrt{6}$ thus difference =$2\sqrt{6}\approx 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
What is the number of real solutions of the equation $ | x - 3 | ^ { 3x^2 -10x + 3 } = 1 $? I did solve, I got four solutions, but the book says there are only 3.
I considered the cases $| x - 3 | = 1$ or $3x^2 -10x + 3 = 0$.
I got for $x\leq 0$: $~2 , 3 , \frac13$
I got for $x > 0$: $~4$
Am I wrong? Is $0^0 = 1$ or NOT?
Considering the fact that : $ 2^2 = 2 \cdot2\cdot 1 $
$2^1 = 2\cdot 1$
$2^0 = 1$
$0^0$ should be $1$ right?
| First of all notice that $0^0\ne 1$
$$|x-3|^{3x^2-10x+3}=1\Longleftrightarrow$$
$$\ln\left(|x-3|^{3x^2-10x+3}\right)=\ln(1)\Longleftrightarrow$$
$$\ln\left(|x-3|\right)\left(3x^2-10x+3\right)=0\Longleftrightarrow$$
$$\ln\left(|x-3|\right)\left(x-3\right)\left(3x-1\right)=0$$
Split $\ln\left(|x-3|\right)\left(x-3\right)\left(3x-1\right)$ into seperate parts with additional assumptions:
So we got:
*
*$$\ln\left(|x-3|\right)=0\Longleftrightarrow$$
$$|x-3|=1\Longleftrightarrow$$
$$x-3=\pm 1\Longleftrightarrow$$
$$x=\pm 1+3$$
*$$x-3=0\to\text{not a valid solution}$$
*$$3x-1=0\Longleftrightarrow$$
$$3x=1\Longleftrightarrow$$
$$x=\frac{1}{3}$$
So at the end we got 3 solutions:
$$x_1=\frac{1}{3}$$
$$x_2=2$$
$$x_3=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.