Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove: $\overline{(\frac{z_1}{z_2})}=\frac{\overline{z_1}}{\overline{z_2}}$
Prove: $$\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}$$
$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{z_1\cdot z_2^{-1}}=\overline{z_1}\cdot \overline{z_2^{-1}}=\overline{z_1}\cdot \overline{z_2}^{-1}=\frac{\overline{z_1}}{\overline{z_2}}$$
Just to be sure is it valid?
| Let's do this one in two parts.
If $z$ and $w$ are complex numbers, then $\overline{zw}=\overline{z}\,\overline{w}$.
Suppose $z=a+bi$ and $w=c+di$ with $a,b,c,d$ real numbers. Then
$$
\overline{zw} = \overline{(a+bi)(c+di)} = \overline{ac-bd + i(ad+bc)}
= ac-bd-i(ad+bc)
= (a-bi)(c-di)
= \overline{z}\,\overline{w}.
$$
If $z$ is a nonzero complex number, then $1/\overline{z}=\overline{(1/z)}$.
Suppose $z=a+bi$ with $a$ and $b$ real numbers such that $(a,b)\ne(0,0)$. Then
$$
\overline{\left(\frac{1}{z}\right)}
= \overline{\left(\frac{1}{a+bi}\right)}
= \overline{\left(\frac{a-bi}{a^2+b^2}\right)}
= \overline{\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i}
= \frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}i
= \frac{a+bi}{a^2+b^2}
= \frac{1}{a-bi}
= \frac{1}{\overline{z}}.
$$
Now
$$
\overline{\left(\frac{z}{w}\right)}
= \overline{\left(z\frac{1}{w}\right)} = \overline{z}\overline{\left(\frac{1}{w}\right)}
= \overline{z} \frac{1}{\overline{w}}
= \frac{\overline{z}}{\overline{w}}.
$$
This is essentially what you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Looking for a more efficient way to solve the Straight Line Equation problem.
If the straight line through the point $P(3,4)$ makes an angle $\dfrac{\pi}{6}$ with the $x$ axis and meets the line $12x +5y + 10=0$ at Q, find the length of PQ.
My method:
Equation of the given line is;
$y- \sqrt3x+3\sqrt3-4=0$ // using Point slope form.
Now, solving the simultaneous equations gives us the point of intersection $R$. And using the Distance formula, I can find the length $PR$. However, I feel this problem can be solved in a more simpler manner. What would be an efficient approach?
PS: The answer is: $\dfrac{132}{12\sqrt3+5}$
| The given line contains the point $P=(3,4)$ and has slope $m=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ so its equation is
$$ -\sqrt{3}x+3y-12+3\sqrt{3}=0\tag{1} $$
The second line has equation
$$ 12x+5y+10=0\tag{2} $$
The two equations intersect at a point $Q$ and we wish to find the distance from $P$ to $Q$.
The coordinates of $Q$ are quite messy. Sketching the graph might suggest an alternate way to approach the problem.
Notice that the horizontal line $y=4$ appears to cross the second line at the point $R=(-2.5,4)$ which we quickly verify by substitution into the second equation. Now we have a triangle $\triangle PQR$.
We know that $\angle QPR=\dfrac{\pi}{6}$ and that $PR=5.5$. If we can find $\angle QRP$ we can use the Law of Sines to find $PQ$.
The angle $\angle QRP$ is the same as the angle between the normal vectors of equation $(2)$ and equation $0\cdot x+1\cdot y=4$ which can be found using the vector version of the Law of Cosines.
The normal vector of $(2)$ is $\mathbf{u}=(12,5)$ and the normal vector of $0\cdot x+1\cdot y=4$ is $\mathbf{v}=(0,1)$. So we have
$$ \cos(\angle QRP)=\frac{(12,5)\cdot(0,1)}{\Vert(12,5)\Vert\,\Vert(0,1)\vert}=\frac{5}{13} $$
Therefore,
$$ \sin(\angle QRP)=\frac{12}{13} $$
Now we know that
$$\angle PQR=\pi-\left(\frac{\pi}{6}+\arcsin\left(\frac{12}{13}\right)\right)$$
So
\begin{eqnarray}
\sin\left(\angle PQR\right)
&=&\sin\left(\frac{5\pi}{6}\right)\cdot\frac{5}{13}-\cos\left(\frac{5\pi}{6}\right)\cdot\frac{12}{13}\\
&=&\frac{12\sqrt{3}+5}{26}
\end{eqnarray}
By the Law of Sines we have
$$ \frac{\vert PQ\vert}{12/13}=\frac{11/2}{\left(\frac{12\sqrt{13}+5}{26}\right)} $$
Working through the arithmetic gives
$$ \vert PQ\vert=\dfrac{132}{12\sqrt3+5}=\frac{12}{37}(12\sqrt{3}-5) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the absolute maximum and minimum of function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le2$
Find the absolute maximum and minimum of function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le2$.
I know how to find maxima/minima if the constraint was the boundary of the circle (by Lagrange's multiplier method). How do I find it for a region? Can I maximise/minimise $f(x,y)$ by finding its saddle points, and then using only those points that lie within the region?
| Rewrite constraint as: $\dfrac{x^2}{2} + y^2 \le 1\implies x =r \sqrt{2}\cos \theta, y = r\sin \theta, 0 \le r \le 1, 0 \le \theta \le 2\pi\implies f(x,y) = f(r,\theta) = 2r^2\cos^2 \theta+3r^2\sin^2 \theta- r\sin \theta\implies f_r = 0 = f_{\theta}\implies 4r\cos^2\theta+6r\sin^2\theta-\sin \theta = 0 = -4r^2\sin\theta\cos \theta+6r^2\sin \theta\cos \theta-r\cos \theta\implies 2r^2\sin \theta\cos \theta - r\cos \theta = 0\implies r\cos \theta(2r\sin \theta-1) = 0$. If $r = 0 \implies x = y = 0 \implies f(0,0) = 0$, if $\cos \theta = 0 \implies \sin \theta = \pm 1\implies r = 1/6\implies x = 0, y = 1/6\implies f(0,1/6) = -1/12$, if $2r\sin \theta - 1 = 0 \implies \sin \theta = \dfrac{1}{2r}\implies \sin^2 \theta = \dfrac{1}{4r^2}\implies \cos^2 \theta = \dfrac{4r^2-1}{4r^2}\implies \dfrac{4r^2-1}{r}+\dfrac{3}{2r}-\dfrac{1}{2r}=0\implies \text{no solution this case}$ for $r \ne 0$. Now calculate the values of $f(r,\theta)$ on the boundaries of the rectangle $A_{\theta, r}= \{(r,\theta): 0 \le r \le 1, 0 \le \theta \le 2\pi\}$:
a) $\theta = 0, 2\pi \implies x = r\sqrt{2}, y = 0\implies f(x,y) = 2r^2\le 2 = f(\sqrt{2},0)$
b) $r = 0\implies x= y = 0 \implies f(0,0) = 0$
c) $r = 1 \implies f(x,y) = f(\theta) = 2\cos^2 \theta + 3\sin^2\theta - \sin \theta = \sin^2\theta -\sin \theta + 2 \le \sin^2\theta + |\sin \theta| + 2 \le 1+1+2 = 4$, which attains a value of $4$ when $\theta = \dfrac{3\pi}{2}$.
Thus to sum it up, $f_{\text{min}} = -1/12, f_{\text{max}} = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Computing greatest common divisor and determining whether Diophantine equation has solutions Question:
(1) Compute gcd$(224, 12 + 2^{10})$
(2) Describe in words the set of all integers $ n$ for which the equation $ \ 4x + ny = 1$ has integer solutions.
(3)
Describe in words the set of all integers $ \ n$ for which the equation $ \ 3x + ny = 1$ has integer solutions.
My attempt:
For (1) the Euclidean algorithm will be difficult to use. Should I use the theorem that states gcd$ \ (a,b) = $ gcd($ a -bq, b)$ or should I prime factorize both sides? I am not quite sure.
For (2) we need need gcd$(4,n) = 1$ in order for the equation to have integer solutions. This means that $ \ n$ must belong to the set of all odd integers.
For (3) we need need gcd$(3,n) = 1$ in order for the equation to have integer solutions. This means that $ \ n$ must belong to the set of all integers that are not divisible by 3.
| $224= 4\cdot 7\cdot 8$ and $12+2^{10} = 4(3+2^8)=4\cdot 7\cdot 37$.
So $gcd(224,12+2^{10}) =28$.
The rest is fine.
Ok, ok, we can factor $$2^8+3 = 2^8-2^2+4+3= 2^2(2^6-1)+7 = 4(2^3-1)(2^3+1)+7 =4\cdot 7\cdot 9+7 = 7(36+1) =7\cdot 37$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
| Hint :divide into cases.
$$\forall x\geq 0 \to R.H.S=L.H.S \\2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \to \\2^{x+1}-2^x=2^x-1+1\\2(2^x)-2^x=2^x \checkmark\\ x\in[0,\infty)$$
Then $$for \space -\leq x\leq0 \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\
\to 2^{x + 1 } - 2^x = 1-2^x + 1 \implies x=-2\\2^x=2-2^x \to x=0 $$
$$for \space x\leq-1 \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\2^{-x-1}-2^x=1-2^x+1\\2^{-x-1}=2 \\\to -x-1=1 \to \\x=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?
I tried this by first taking total permutation as $\dfrac{8!}{2^4}$
Now $n_1$ as $22$ or $33$ or $44$ or $55$ occurs differently
$N_1 = \left(^7C_1\times \dfrac{7!}{8}\right)$
And $n_2 = \left(^4C_1 \times 4!\right)$
Using the inclusion-exclusion principle I got:
$\dfrac{8!}{16}-\left(^7C_1\times\dfrac{7!}{8}\right)+\left(^4C_1\times4!\right)$
But answer was wrong
Please help me solve the question
This question is from combinatorics and helpful for RMO
| Here is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
We encode the digits
\begin{align*}
2,3,4,5 \qquad\text{as}\qquad a,b,c,d
\end{align*}
and look for Smirnov words of length $8$ built from $a,b,c,d$ with each letter occurring exactly twice.
A generating function for the number of Smirnov words over a four letter alphabet $V=\{a,b,c,d\}$ is given by
\begin{align*}
\left(1-\frac{4z}{1+z}\right)^{-1}
\end{align*}
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series $A(z)$. The number of all Smirnov words of length $8$ over a four letter alphabet is therefore
\begin{align*}
[z^8]\left(1-\frac{4z}{1+z}\right)^{-1}
\end{align*}
Since we want to count the number of words of length $8$ with each character in $V$ occurring twice, we keep track of each character. We obtain with some help of Wolfram Alpha
\begin{align*}
[a^2b^2c^2d^2]\left(1-\frac{a}{1+a}-\frac{b}{1+b}-\frac{c}{1+c}-\frac{d}{1+d}\right)^{-1}=\color{blue}{864}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\sin^2(40^\circ)<3\cos^2(40^\circ)=3(1-\sin^2(40^\circ))$$
$$7\sin^2(40^\circ)<3$$
$$\sin(40^\circ)<\sqrt{\frac{3}7}$$
Is there another way to prove this inequality?
| The given inequality is equivalent to $\cos^2\left(\frac{2\pi}{9}\right)>\frac{4}{7}$, or to $\cos\left(\frac{4\pi}{9}\right)>\frac{1}{7}$.
The minimal polynomial of $\alpha=\cos\left(\frac{4\pi}{9}\right)$ over $\mathbb{Q}$ can be easily derived from $\Phi_9(x)=x^6+x^3+1$, and it is given by $p(x)=1-6x+8x^3$. Since $p'(x)=-6(1-4x^2)$, $p(x)$ is a decreasing function on $\left[0,\frac{1}{7}\right]$, hence the claim simply follows from checking that $p\left(\frac{1}{7}\right)>0$. A sharper inequality which can be proved through the same technique is
$$ \cos\left(\frac{4\pi}{9}\right)>\left(\frac{5}{12}\right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
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If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form $ z=\frac{1+it}{1-it}$ where $t\in \mathbb{R}$. If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form
$$ z=\frac{1+it}{1-it},$$
where $t\in \mathbb{R}$.
I don't know, how to begin. I started with the given conditions. Given that $|z|=1 \text{ and } z\neq-1\implies z=e^{it}, t\in[0,2\pi]/\{\pi\}.$
| Note that
$\dfrac{1 + it}{1 - it} = \dfrac{(1 + it)^2}{(1 + it)(1 - it)} = \dfrac{(1 + it)^2}{1 + t^2} = \dfrac{(1 - t^2) + 2it}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2} + \dfrac{2it}{1 + t^2}; \tag 1$
since
$(1 - t^2)^2 + (2t)^2 = 1 - 2t^2 + t^4 + 4t^2 = 1 + 2t^2 + t^4 = (1 + t^2)^2, \tag 2$
there is a unique $\theta \in (-\pi, \pi)$ such that
$\cos \theta = \dfrac{1 - t^2}{1 + t^2}, \tag 3$
and
$\sin \theta = \dfrac{2t}{1 + t^2}; \tag 4$
then we may take
$z = \cos \theta + i \sin \theta = e^{i\theta} \tag 5$
and thus
$\vert z \vert = 1. \tag 6$
We show every such $z \ne -1$ may be so represented. Observe that $z = -1$ is excluded since then $\sin \theta = 0$, implying $t = 0$, whence $\cos \theta = 1$ forcing $z = 1$; so $\theta \ne \pi, -\pi$. For all other $\theta \in (-\pi, \pi)$ the equation
$\cos \theta = \dfrac{1 - t^2}{1 + t^2} \tag 7$
may be written
$(1 + \cos \theta)t^2 = 1 - \cos \theta, \tag 8$
and we thus have
$t = \pm \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}, \tag 9$
yielding the two possible values for $\sin \theta$ via (4), showing every $z \ne -1$ with $\vert z \vert = 1$ may be so expressed.
Note: The correspondence $e^{i \theta} \longleftrightarrow (1 + it)/(1 -it)$ is important in the theory of operators on Hilbert space. See this wikipedia page on the Cayley Transform. End of Note.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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integral involving floor function $\int_0^1 \left \lfloor{ (a^{n}x) ^{ \frac{n}{2} }}\right \rfloor dx $ i need to find the result of this integral
$$\int_0^1 \left \lfloor{ (a^{n}x) ^{ \frac{n}{2} }}\right \rfloor dx $$
with $$a \in \mathbb{N}$$
i tried to transform it to a finite sum and i found this:
$$ \frac{1}{a^{n}} \sum_0^{ a^{n}-1 } k((k+1)^{ \frac{2}{n}}-(k)^{ \frac{2}{n} }) $$
but i couldn't calculate it and i don't know if it is right or false
| Let's put
$$
\eqalign{
& y = \left( {a^{\,n} x} \right)^{\,n/2} \cr
& x = {{y^{\,2/n} } \over {a^{\,n} }} \cr
& dx = {2 \over {na^{\,n} }}y^{\,2/n - 1} dy \cr
& u = y(1) = a^{\,n^{\,2} /2} \cr}
$$
then
$$
\eqalign{
& I(a,n) = \int_{x = 0}^1 {\left\lfloor {\left( {a^{\,n} x} \right)^{\,n/2} } \right\rfloor dx} = {2 \over {na^{\,n} }}\int_{y = 0}^u {\left\lfloor y \right\rfloor y^{\,2/n - 1} dy} = \cr
& = {2 \over {na^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {\int_{y = k}^{k + 1} {k\,y^{\,2/n - 1} dy} } \right)} + \int_{y = \left\lfloor u \right\rfloor }^u {\left\lfloor u \right\rfloor \,y^{\,2/n - 1} dy} } \right) = \cr
& = {2 \over {na^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {{{nk} \over 2}\left( {\left( {k + 1} \right)^{\,2/n} - k^{\,2/n} } \right)} \right)} + \left\lfloor u \right\rfloor {n \over 2}\left( {u^{\,2/n} - \left\lfloor u \right\rfloor ^{\,2/n} } \right)} \right) = \cr
& = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k\left( {\left( {k + 1} \right)^{\,2/n} - k^{\,2/n} } \right)} \right)} + \left\lfloor u \right\rfloor \left( {u^{\,2/n} - \left\lfloor u \right\rfloor ^{\,2/n} } \right)} \right) \cr}
$$
The last term is
$$
\eqalign{
& \left\lfloor u \right\rfloor \left( {u^{\,2/n} - \left\lfloor u \right\rfloor ^{\,2/n} } \right) = \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor \left( {a^{\,n} - \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor ^{\,2/n} } \right) = \cr
& = R(a,n) \cr}
$$
and is null for $a,n$ positive integers and $n$ even, but it is not if $n$ is odd. Let's call it $R(a,n)$.
The sum can be further simplified to give
$$
\eqalign{
& I(a,n) = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k\left( {\left( {k + 1} \right)^{\,2/n} - k^{\,2/n} } \right)} \right)} + R(a,n)} \right) = \cr
& = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {\left( {k + 1 - 1} \right)\left( {k + 1} \right)^{\,2/n} - k^{\,2/n + 1} } \right)} + R(a,n)} \right) = \cr
& = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n + 1} } - \sum\limits_{\left( {0\, \le } \right)\,1\, \le \,\,k\, \le \,\left\lfloor u \right\rfloor - 1} {k^{\,2/n + 1} } - \sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } + R(a,n)} \right) = \cr
& = {1 \over {a^{\,n} }}\left( {\left\lfloor u \right\rfloor ^{\,2/n + 1} + R(a,n) - \sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } } \right) = \cr
& = \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - {1 \over {a^{\,n} }}\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } \cr}
$$
So we can write
$$ \bbox[lightyellow] {
I(a,n) = \int_{x = 0}^1 {\left\lfloor {\left( {a^{\,n} x} \right)^{\,n/2} } \right\rfloor dx} = \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - {1 \over {a^{\,n} }}\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} }
} \tag{1}$$
Now, we have the sum of powers with constant exponent and variable basis.
This is related to Generalized Harmonic Numbers ( and to Bernoulli Polynomials, Digamma Function).
However, being the exponent fractional, the above are expressable through the Hurwitz zeta function, i.e.
$$ \bbox[lightyellow] {
\eqalign{
& \sum\limits_{0\, \le \,k\, \le \,\left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } = \cr
& = \sum\limits_{0\, \le \,k\,} {{1 \over {\left( {k + 1} \right)^{\, - \,2/n} }}} - \sum\limits_{0\, \le \,k\,} {{1 \over {\left( {k + \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor + 1} \right)^{\, - \,2/n} }}} = \cr
& = \zeta \left( { - 2/n,\;1} \right) - \zeta \left( { - 2/n,\;\left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor + 1} \right) \cr}
} \tag{2}$$
Example
$$
\eqalign{
& I(2,3) = 8.57135699... \cr
& I(2,4) = 84.84616346... \cr}
$$
checked with the original integral, eq. (1), and eq.(1) with substitution of (2).
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Generating function in combinatorics: combining ordinary and exponential generating functions I want to make a generating function for a combinatorics problem with $2$ different simultaneous constraints. In the one variable cases, one of the constraints would use an ordinary generating function, the other an exponential generating function. So I'm not sure how to combine them into a multivariate generating function.
Let's make up the following problem to illustrate:
Using the ten digits $0$-$9$, I can make a length $k$ string, eg. for $k=5$, the string $27485$.
Now say I want the number of $k$-strings (the number of all possible arrangements of length $k$ using the digits $1$-$9$), that satisfy not just one but $2$ conditions, for example:
*
*The $k$-string must have $1$ or $2$ occurrences of the digit "$4$".
*The sum of the digits in the $k$-string must be $N$.
Make things concrete: use $k=3$, and $N=10$:
If it was the just the first constraint, I think I could use the exponential generating function:
$$g(x) = \left[x + \frac{1}{2!}x^2\right]\cdot\left[1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3\right]^8.$$
The first term for the "$4$" digit which has to occur once or twice, the second term raised to the $8$ is for the other $8$ digits which can occur any number of times. Then I would look for the coefficient of $x^3$ and multiply by $3!$.
And if it were only the $2^{\text{nd}}$ constraint, I think I could do:
$$g(y) = \left[1 + y + y^2\right]\cdot \left[1 + y^2 + y^4\right]\cdot \left[1 + y^3 + y^6\right]\cdots\left[1+y^8+y^{16}\right]\cdot\left[1+y^9+y^{18}\right],$$
then look for the coefficient of $y^N$.
How do I combine these constraints? Do I just multiply them together, $g(x,y) = g(x)g(y)$ and then look for the coefficient of $(1/k!)x^ky^N$?
| If we let $x$ be the enumerator for the sum of digits and $y$ be the enumerator for the number of digits then we can form the generating function for occurrences of $4$
$$f(x,y)=(x^4)\frac{y}{1!}+(x^4)^2\frac{y^2}{2!}$$
Powers of $y$ count the number of $4$s in the string and powers of $x$ are multiples of $4$.
Similarly we can form the generating function for occurrences of each digit $0$-$9$ apart from $4$
$$\begin{align}g(x,y)&=\sum_{i\ge 0}(1+x+x^2+\underbrace{x^3+x^5}_{\text{no $x^4$ term}}+\cdots +x^9)^i\frac{y^i}{i!}\\[1ex] &=\sum_{i\ge 0}\left(\frac{1-x^{10}}{1-x}-x^4\right)^i\frac{y^i}{i!}\\[1ex] &=\exp\left(y\left(\frac{1-x^{10}}{1-x}-x^4\right)\right)\end{align}$$
Here powers of $y$ count occurrences of the other digits and $x$ counts the sum of those digits.
Then the desired count for strings length $k$ with sum $N$ is
$$\left[x^N\frac{y^k}{k!}\right]f(x,y)g(x,y)$$
So, for example inputting the following into sage
y=var('y')
f(x,y)=x^4*y+(x^4)^2*y^2/2
g(x,y)= e^(y*((1-x^10)/(1-x)-x^4))
taylor(f(x,y)*g(x,y),(x,0),(y,0),20).coefficient(x^10).coefficient(y^5)*factorial(5)
calculates the coefficient of $x^{10}y^5/5!$ and yields the output
360
So there are $360$ strings that use digits $0$-$9$, have either $1$ or $2$ occurrences of $4$, are length $5$ and have digit sum $10$.
Note that I have allowed for $10$ digits $0$-$9$ because the question mentions both cases of using $10$ digits and using $9$ digits.
This method is easily modified for the latter case: simply remove the leading $1$ from $1+x+x^2+x^3+x^5+\cdots +x^9$ in $g(x,y)$.
| {
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"url": "https://math.stackexchange.com/questions/2401793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Hard Integral $\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}$ I am trying to calculate the integral $$\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}$$
My idea is to use and try to calculate and prove that
$\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=-\int\limits_{0}^{1}{\frac{\ln \left( 1-x^{2} \right)}{1+x^{2}}\ln \frac{1+x^{2}}{2}dx}-\int\limits_{0}^{1}{\frac{2x\arctan x}{1+x^{2}}\ln \left( 1-x^{2} \right)dx}$
$\frac{2x\arctan x}{1+x^{2}}=\operatorname{Re}\left( \frac{\ln \left( 1+ix \right)}{1+ix} \right)-\operatorname{Re}\left( \frac{\ln \left( 1-ix \right)}{1+ix} \right)$
$\int\limits_{0}^{1}{\frac{\ln \left( 1-x^{2} \right)}{1+x^{2}}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{32}+\frac{\pi }{4}\ln ^{2}2-2C\ln 2$
$\int\limits_{0}^{1}{\frac{\ln \left( 1\pm ix \right)\ln \left( 1\pm x \right)}{1+ix}dx}$
| It seems the integral is designed to intentionally benefit from the following facts:
$$\log \frac{1+x^{2}}{2}=\log(\frac{1}{2}(x+i)(x-i))=\log(\frac{x+i}{\sqrt 2})+\log(\frac{x-i}{\sqrt 2})$$
and
$$\arctan (x)= \frac{1}{2}i\log(1-i x)-\frac{1}{2}i\log(1+i x)= \frac{1}{2i}(\log(\frac{x+i}{\sqrt 2})-\log(\frac{x-i}{\sqrt 2}))$$
Let's work on the anti derivative:
$$\int{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=-\frac{1}{2i}\int{\frac{1}{1+x}(\log^2(\frac{x+i}{\sqrt 2})-\log^2(\frac{x-i}{\sqrt 2}))dx}=\text{Im}{\int{\frac{1}{1+x}\log^2(\frac{x-i}{-i\sqrt 2})dx}}=\text{Im}{\int{\frac{1}{1+x}\log^2(\frac{1+ix}{\sqrt 2})dx}}$$
Encapsulating $y=\frac{1+ix}{\sqrt 2}$ (i.e. $1+x=1+(1-y\sqrt 2)i$ and $dx=-idy\sqrt{2}$), the anti derivative can be simplified to
$$\int{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\text{Im}{\int{-\frac{\log^2y}{1+(1-y\sqrt 2)i}i\sqrt{2}dy}}=\text{Im}{\int{\frac{\log^2y}{y+\frac{i-1}{\sqrt{2}}}dy}}$$
Using dilogarithm function and assuming $k=-\frac{i-1}{\sqrt{2}}$, yields
$$\int \frac{\log^2y}{y-k}dy=-2\text{Li}_3(ky)+2\text{Li}_2(ky)\log(y)+\log(1-ky)\log^2(y)+\text{constant}$$
Now, computing the finite integral should be doable. wolframalpha gives me $\pi^3/96$ as it is stated in the question. I won't be surprised if there are typos or errors - you're welcome to fix or improve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$
For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$
I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then inequality becomes
$$A=\frac{x^2}{1+xz}+\frac{y^2}{1+xy}+\frac{z^2}{1+zy} \ge \frac{3}{2}$$
By Cauchy-Schwarz
$$(3+xy+yz+zx)A \ge (x+y+z)^2 \ge 3(xy+yz+zx)$$
What then?
| If you let $B=xy+yz+zx$, by AM-GM, $B \ge 3$ and
$$A \ge \frac{3B}{B+3}=\frac{3}{1+3/B} \ge \frac32$$
| {
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Solving the differential equation $y''=y'^2+1$ I would like help solving the next differential equation.
$$y''=y'^2+1$$
I tried subsitiuting $p=y'$ and got to $\frac{1}{2}\ln|p^2+1|=\frac{1}{2}\ln|y|+\ln|c|$
but I don't know how to continue and maybe made a mistake.
any ideas?
| $$y''=y'^{ 2 }+1\\ { y }'=p\left( y \right) ,y''=pp'\\ \\ p{ p }'={ p }^{ 2 }+1\\ \int { \frac { pdp }{ { p }^{ 2 }+1 } } =\int { dy } \\ \frac { 1 }{ 2 } \int { \frac { d\left( { p }^{ 2 }+1 \right) }{ { p }^{ 2 }+1 } } =y+{ C }_{ 1 }\\ \ln { \left( { p }^{ 2 }+1 \right) =2\left( y+{ C }_{ 1 } \right) } \\ { p }^{ 2 }={ e }^{ 2\left( y+{ C }_{ 1 } \right) }-1\\ p=\sqrt { { e }^{ 2\left( y+{ C }_{ 1 } \right) }-1 } \\ \frac { dy }{ dx } =\sqrt { { e }^{ 2\left( y+{ C }_{ 1 } \right) }-1 } \\ \int { \frac { dy }{ \sqrt { { e }^{ 2\left( y+{ C }_{ 1 } \right) }-1 } } = } \int { dx } \\ \arctan { \left( \sqrt { { e }^{ 2\left( y+{ C }_{ 1 } \right) }-1 } \right) } =x+{ C }_{ 2 }\\ \\ \\ y=\frac { 1 }{ 2 } \ln { \left( \tan ^{ 2 }{ \left( x+{ C }_{ 2 } \right) } +1 \right) } +{ C }_{ 1 }$$ or
$$y=-\ln { \left( \cos { \left( x+{ C }_{ 2 } \right) } \right) + } { C }_{ 1 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?
| Here is an alternative proof that doesn't use induction - just for fun!
Let $a \in \mathbb{Z}$. Observe that $a^3 - a = a(a^2 - 1) = (a-1)a(a+1)$ is the product of three consecutive integers, hence divisible by three. Adding $3a$ to the initial expression does not alter divisibility by $3$.
In particular, $a^3 - a + 3a = a^3 + 2a = a(a^2 + 2)$ is divisible by $3$, which means that it is still an integer after dividing by $3$. QED
(A ridiculous version of the above: $a(a^2+2)$ is the sum of $a^3 - a$ and $3a$, where the latter addend is a multiple of three by observation, and the former is divisible by three by Fermat's Little Theorem; so, their sum is divisible by three, and its quotient by three is an integer. QED)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the value of $\frac{9}{5}(a+b)$ given that $a\sqrt{a}+b\sqrt{b}=183$ and $a\sqrt{b}+b\sqrt{a}=182$
Suppose $a,b$ are positive real numbers such that $a\sqrt{a}+b\sqrt{b}=183$, $a\sqrt{b}+b\sqrt{a}=182$. Find $\frac{9}{5}(a+b)$.
It is my equation. I subtracted the second equation from the first one and found $(a-b)(\sqrt{a}-\sqrt{b})=1$.or, $(a-b)^2=(\sqrt{a}+\sqrt{b})$.
Am I going to the right path? How to escape from this. Please give me hints. Thank you.
| Hint: let $\,u=\sqrt{a}\,$ and $\,v=\sqrt{b}\,$, then it's given that $\,u^3+v^3=183\,$ and $\,u^2v+uv^2=182\,$, and the problem asks for $\frac{9}{5}(u^2+v^2)$.
*
*$\;(u+v)^3=u^3 + v^3 + 3(u^2v+ uv^2) = 183 + 3 \cdot 182 = 729 \implies u+v = 9\,$
*$\;uv= (u^2v+uv^2)/(u+v) = 182 / 9 \,$
*$\;u^2+v^2 = (u+v)^2 - 2 uv = \cdots \,$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a solution, but hit and trial. Can this conclusion be derived using maximum/mininum concept?
| Let us consider the function $f(\theta)=\log\sin\theta$ on the interval $\theta\in\left(0,\frac{\pi}{2}\right)$. It is a concave function
since $f''(\theta)=-\frac{1}{\sin^2\theta}<0$, hence by Jensen's inequality
$$ f(a)+f(b)+f(c) \leq 3\,f\left(\frac{a+b+c}{3}\right) $$
holds for any $a,b,c\in\left(0,\frac{\pi}{2}\right)$. By considering $a=\frac{A}{2},b=\frac{B}{2},c=\frac{C}{2}$ and exponentiating both sides of the previous inequality turns into
$$ \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq \sin^3\left(\frac{A+B+C}{6}\right) = \frac{1}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why am i getting an incorrect result (0/0)? The given pair of equations are:
$$
3x-2y=0\\
kx+5y=0
$$
Clearly,here, $a_1=3$, $a_2=k$, $b_1=-2$, $b_2=5$, $c_1=0$ and $c_2=0$.
Now, here, I have to find a value of $k$ for which the given system of equations has infinite solutions.
Hence, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
Thus, $\frac{3}{k} = -\frac{2}{5} = \frac{0}{0}$
But, $\frac{0}{0}$ is nonsense!
Why am I getting such a wrong result?
Are $c_1$ and $c_2$ not equal to ZERO?
I will be thankful for help!
| First, if an equation has no unique solution,
$Δ = 0$,
$3×5 - (-2)k=0$
$2k=15$
Using Gaussian Elimination,
$ \left[\begin{array}{rr|r}3 & -2 & 0 \\k & 5 & 0 \\\end{array}\right]$~
$ \left[\begin{array}{rr|r}7.5 & -5 & 0 \\k & 5 & 0 \\\end{array}\right]$~$ \left[\begin{array}{rr|r}7.5 & -5 & 0 \\k + 7.5 & 0 & 0 \\\end{array}\right]$
Therefore, when there is infinite many solutions, $k+7.5=0$ must be true.
Then you have $k+7.5=0$ , i.e. $k=-7.5$
Let $x=t$, where $t$ is any real number, then the solutions of the equation are $(x,y)=$ $(t,\frac{3}{2}t)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine all pairs $(a,b)$ of positive integers such that $ab^2+b+7$ divides $a^2b+a+b$
Determine all pairs $(a,b)$ of positive integers such that $ab^2 + b + 7$ divides $a^2b + a +b$.
I guess the trivial solution is $a=b=7$.
Also any $a$ and $b$ which are divisible by $7$ will seem to suit.
we can also write
$b(\frac{a}{b}+1+\frac{7}{b}) \cdot x=a\cdot(ab+1+\frac{b}{a})$
From where all $a, b$ that satisfy: $\frac{b^2}{a}=7$ and $a>b$, are OK.
Is my solution full and complete? Are there any other solutions?
| $$%%ab^2 + b + 7 \ | \ a^2b + a + b$$
We know that:
$$
ab^2 + b + 7 \ | \ a(ab^2 + b + 7)
=
a^2b^2+ab+7a
;
$$
$$
ab^2 + b + 7 \ | \ b(a^2b + a + b)
=
a^2b^2+ab+b^2
;
$$
so we can conclude
$$
ab^2 + b + 7 \ | \
\Big[
a(ab^2 + b + 7) - b(a^2b + a + b) \Big]
\ \ \ \ \ \ \ \
\Longrightarrow
\\
ab^2 + b + 7 \ | \
\Big[
(a^2b^2+ab+7a) - (a^2b^2+ab+b^2) \Big]
\Longrightarrow
\\
ab^2 + b + 7 \ | \
7a - b^2
\ \ \ \ \
\color{Blue}{\star}
$$
First case:
$7a-b^2=0$; which implies that there exists an integer $n$; such that:
$$(a,b)=(7n^2,7n);$$
one can check that $(a,b)=(7n^2,7n)$ satisfies in;
the divisibility condition $ab^2 + b + 7 \ | \ a^2b + a + b$.
Second case:
$0 < 7a-b^2 $;
note that
$\color{Red}{0} < (b+\dfrac{1}{2})^2+\dfrac{27}{4} = \color{Red}{b^2+b+7}$,
so in this case we have:
$$
ab^2 + (b + 7) < |ab^2 + b + 7| < 7a-(b^2)
\Longrightarrow
\\
ab^2 + \color{Red}{0}
<
ab^2 + \color{Red}{(b^2 + b + 7)} < 7a
\Longrightarrow
ab^2 < 7a
\Longrightarrow
\\
b^2 < 7
\Longrightarrow
b=1
\ \
\text{or}
\ \
b=2
.
$$
*
*If $b=1$; then by replacing in $\color{Blue}{\star}$
we get:
$a + 8 \ | \ 7a - 1 $,
on the otherhand we have:
$a + 8 \ | \ 7(a + 8) $,
so we can conclude that
$a + 8 \ | \ 57 $,
which gives us the two posibilities $a=11$ and $a=49$;
one can check that each of the pairs $(a,b)=(11,1)$ and $(a,b)=(49,1)$ satisfies the divisibility condition.
*If $b=2$; then by replacing in $\color{Blue}{\star}$
we get:
$4a + 9 \ | \ 7a - 4 \ | \ 4(7a - 4) = 28a - 16 $,
on the otherhand we have:
$4a + 9 \ | \ 7(4a + 9) = 28a + 63 $,
so we can conclude that
$4a + 9 \ | \ 79 $,
which does not gives us any posibilities for $a$.
[Note that $79$ is a prime number.]
Third case:
$7a-b^2 < 0$;
note that in this case $0 < a$,
so we have:
$$
ab^2 + b + 7 < |ab^2 + b + 7| < b^2-7a
\Longrightarrow
\\
\color{Red}{0} <
(a-1)
\Bigg[ b + \dfrac{1}{2(a-1)} \Bigg]^2
+
(7a-\dfrac{1}{4(a-1)}+7)
=
(ab^2 + b + 7) +
(7a-b^2)
< \color{Red}{0}
;
$$
which is an obvious $\color{Red}{\text{contradiction}}$; so there is no solution in this case.
So the solutions are as follows:
$$
(11,1), \ (49,1)
\ \ \text{and} \ \
(7n^2,7n)
\ \ \text{for any} \ \
n \in \mathbb{N}
.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the value of $\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$. I am stuck with the following problem :
Find the value of $$\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$$
My try : Let $$p=\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}\implies \log p=\lim_{x \to 0}{\frac {1}{x^2}}\log\left(\frac {\sin x}{x}\right)$$...After applying l'hospitals rule few times I get
$$\log p= \lim_{x \to 0}\frac{1-\sec^2x}{4x \tan x+2x^2\sec^2x}$$.. we can again apply l'hospitals rule ,but the calculations get bigger and bigger..Is there any other easier way around or I am missing something?
| From the Taylor expansion of $\sin x$ around $0$, namely $\sin u = u -\frac{u^3}{6} + o(u^3)$; and that of $\ln(1+u)$, specifically $\ln(1+u)=u+o(u),$ we get
$$\begin{align}
\left( \frac{\sin x}{x}\right)^{1/x^2}
&= \left(1-\frac{x^2}{6} + o(x^2)\right)^{1/x^2}
= e^{\frac{1}{x^2}\ln\left(1-\frac{x^2}{6} + o(x^2)\right) }
= e^{\frac{1}{x^2}\left(-\frac{x^2}{6} + o(x^2)\right) }\\
&= e^{-\frac{1}{6} + o(1)}
\xrightarrow[x\to0]{}\boxed{e^{-\frac{1}{6}}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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A question regarding minimum value of... $a+b+c=2$
$a,b,c \in \mathbb{R}^+$
What is The min value of $1/a+4/b+9/c$?
I have tried to use geometric shapes and aritmetic harmonic geometric means but unfortunately ı am going nowhere?
What do you suggest?
| Lagrange multiplier
$$f(a,b,c,k)=\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+k (a+b+c-2)$$
Solve the system $f'_x=0,f'_y=0,f'_z=0,f'_k=0$
That is
$$k-\frac{1}{a^2}=0,k-\frac{4}{b^2}=0,k-\frac{9}{c^2}=0,a+b+c-2=0$$
there are $3$ solutions but only one leads to the minimum and respects constraints
$a= \frac{1}{3},b=\frac{2}{3},c= 1$ for which $\frac{1}{a}+\frac{4}{b}+\frac{9}{c}=18$
Hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
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An olympic mathematics problem regarding Cauchy-Schwarz This question was asked in Turkish National Maths Olympiad in 2008.
For all $xy=1$ we have $((x+y)^2+4)\cdot ((x+y)^2-2) \ge A\cdot(x-y)^2$.
What is the maximum value $A$ can get?
My efforts regarding this problem;
$(x+y)^2-8 \ge A\cdot(x-y)^2$
Using the property $xy=1$ ;
$x^2+y^2-6\ge A\cdot (x^2+y^2-2)$
$\sqrt{\dfrac{x^2+y^2}{2}} \ge\sqrt{\dfrac{A\cdot(x^2+y^2-2)}{2}}$
Therefore $\sqrt{\dfrac{A\cdot(x^2+y^2-2)}{2}}=\dfrac{x+y}{2}$
Although moving further that doesn't work I applied Cauchy-Schwarz by the way if I didn't mention it before.
How should I proceed?
What are you suggestions?
| Let $x^2+y^2=m$.
Thus, $$(m+6)m\geq A(m-2)$$ or
$$m^2+(6-A)m+2A\geq0,$$
for which we need $$(6-A)^2-8A\leq0,$$
which gives $2\leq A\leq18$.
For $A=18$ we get $m=6$ or $x^2+y^2=6$, which says that the equality occurs.
Thus, $18$ is our answer.
Done!
| {
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"url": "https://math.stackexchange.com/questions/2410952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Finding the Coefficient of $(x^3 + 2y^2)^n$ containing $x^{18} y^{12}$ I was helping somebody when I scratched my head because of this question. It goes like
this:
If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.
My work:
I let $u = x^3$ and $v = 2y^2.$ Then doing this: $$(u)^6 = (x^3)^6 \space and \space \space \left(\frac{v}{2} \right)^6 = (y^2)^6$$
we get $u^6 = x^{18}$ and $\left(\frac{v^6}{64} \right) = y^{12}$
We now conclude that the expression $(u + v)^n$ has a term $(u^6)\left(\frac{v^6}{64} \right)$ along its expansion when $u = x^3$ and $v = 2y^2$ We need to find its equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ when we go back to dealing with $(x^3 + 2y^2)^n.$
Everybody knows that in the binomial expansion of $(u + v)^n,$ in each term, the sum of the exponents of $u$ and $v$ is $n.$ and there are $n+1$ terms.
With that in mind, the sum of the particular term $(u^6)\left(\frac{v^6}{64} \right)$ is $n = 12$ and the number of terms in that particular expansion is $12+1 = 13.$ Since the problem asks for the coefficient of the middle term $C x^{18} y^{12},$ we need to find its middle term. Turns out, in the binomial expansion containing $13$ terms, the middle term would
be the $7$th term.
Now looking for for the expression of the $7$th term:
$$nth \space term = C(n, r-1) u^{n-r+1} v^{r-1}$$
$$expression \space of \space 7th \space term = C(12, 7-1) (x^3)^{12-7+1} (2y^2)^{7-1}$$
$$ = (924) (x^3)^{6} (2y^2)^{6}$$
$$ = (924) (x^{18}) (2)^{6}(y^2)^{6}$$
$$ = (924) (x^{18}) (64) (y^2)^{6}$$
$$ = (924) (x^{18}) (64) (y^{12})$$
$$ = 59136 x^{18} y^{12}$$
Therefore, we conclude that $C = 59136.$
Lastly, the equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ from $(u + v)^n$, when we go back to dealing with $(x^3 + 2y^2)^n$, is $59136x^{18}y^{12}$
I've done my best but I couldn't verify it. Is my solution correct?
|
If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.
You want the coefficient of the term $\binom{6+6}{6}(x^3)^{6}(2y^2)^6$, so ....$$C~=~ 2^6 \binom{6+6}{6} ~=~59\,136$$
Therefore: okay $\checkmark$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
| $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20(50)^n+18(12)^n\equiv-18(50)^n+18(12)^n\equiv-18(12)^n+18(12)^n\equiv0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
| Another way to solve $A \sin\theta + B \cos\theta = C$ is as follows.
Divide by $\sqrt{A^2 + B^2}$. Then, set $A / \sqrt{A^2 + B^2} = \sin\xi$
and $B / \sqrt{A^2 + B^2} = \cos\xi$.
This gives:
$$\sin\xi \sin\theta + \cos\xi \cos\theta = C / \sqrt{A^2 + B^2}.$$
So, the solution is given by
$$\cos(\theta - \xi) = C / \sqrt{A^2 + B^2},$$
where $\xi$ is the angle that the vector $(A,B)$ makes with the positive $x$ semi-axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \geqslant 5.$$
This is an assignment I got from a teacher. I've been baffled by the negative sign in the denominator.
Could you please provide insight as to how would one resolve this inequality please? (Do not solve yet, I've yet to turn in this assignment, I just want to get some tips)
| I think our inequality is true for all reals $x$, $y$ and $z$ such that $(x-y)(x-z)(y-z)\neq0$.
Indeed, if $z=0$ then
$$\sum_{cyc}\frac{(2x-y)^2}{(x-y)^2}=\frac{(2x-y)^2}{(x-y)^2}+5\geq5.$$
Let $xyz\neq0$, $\frac{2x-y}{x-y}=a$, $\frac{2y-z}{y-z}=b$ and $\frac{2z-x}{z-x}=c$.
Thus, $(2-a)x=(1-a)y$, $(2-b)y=(1-b)z$ and $(2-c)z=(1-c)x$, which gives
$$(2-a)(2-b)(2-c)xyz=(1-a)(1-b)(1-c)xyz$$ or
$$(2-a)(2-b)(2-c)=(1-a)(1-b)(1-c)$$ (now, try to prove the starting inequality by yourself without to see the rest) or
$$7-3(a+b+c)+ab+ac+bc=0$$ or
$$14-6(a+b+c)+2(ab+ac+bc)=0$$ or
$$5+9-6(a+b+c)+(a+b+c)^2=a^2+b^2+c^2$$ or
$$a^2+b^2+c^2=5+(a+b+c-3)^2,$$
which gives $$a^2+b^2+c^2\geq5$$ and we are done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
I have reached:
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$.
How do I continue from here? I am unable to find a suitable condition to proceed.
Edit:
vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$
and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$
but that ain't much helpful, is it?
| You have shown that $y$ is in the range of the function
$$y = \frac{x^2 + 3x + 1}{x + 1}$$
if $x$ is a real-valued root of the quadratic equation
$$x^2(y - 1) + x(y - 3) + (y - 1) = 0$$
For $x$ to be a real-valued root of the quadratic equation, its discriminant must be nonnegative. The discriminant is
\begin{align*}
\Delta & = b^2 - 4ac\\
& = (y - 3)^2 - 4(y - 1)(y - 1)\\
& = y^2 - 6y + 9 - 4(y^2 - 2y + 1)\\
& = -3y^2 + 2y + 5
\end{align*}
Hence, we require that
\begin{align*}
-3y^2 + 2y + 5 & \geq 0\\
3y^2 - 2y - 5 & \leq 0\\
3y^2 + 3y - 5y - 5 & \leq 0\\
3y(y + 1) - 5(y + 1) & \leq 0\\
(3y - 5)(y + 1) & \leq 0
\end{align*}
which holds if $y \in [-1,5/3]$. Therefore, the range of
$$y = \frac{x^2 + 3x + 1}{x^2 + x + 1}$$
is
$$\text{Ran} = \left[-1, \frac{5}{3}\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof verification: Determine whether $f(x)$ is one-to-one or not
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\
\text{Suppose for the sake of contradiction }f(x)=f(y)\text{ but }x\neq y.\\ \text{Therefore } 2x^3+3x-4 = 2y^3+3y-4\\
2x^3-2y^3+3x-3y=0\\
2(x-y)(x^2+xy+y^2)+3(x-y)=0\\
(x-y)\left [2(x^2+xy+y^2)+3)\right ]=0\\
\text{which implies that either } x=y\text{ or }2(x^2+xy+y^2)+3=0 \\
\text{which contradicts to the assumption that } x\neq y \Rightarrow \Leftarrow $$
Is there anything missing in my proof or there is a better solution? I got confused at the either or part because $x=y$ my or may not be true. My logic is given above, if there is something I misunderstand please give your feedback.
$\textbf{FINAL EDIT}$
Here is the continuation of the proof that it is never true that the right factor $x^2+xy+y^2=-\frac{3}{2}$.
$$\text{We know that } x^2\geq0 \text{ and } y^2\geq0\Rightarrow x^2+y^2\geq 0 \text{ and that there are two cases for }x,y.\\
\textbf{case 1: }\text{If nonzero x and y have the same sign then we are done since that implies that }\\
x^2+xy+y^2>0\\
\textbf{case 2: } \text{If nonzero x and y have differing signs }\\ xy<0
\text{ and also we know that } (x+y)^2>0\\
\Rightarrow x^2+2xy+y^2>0 \\
\Rightarrow x^2+xy+y^2>-xy\\
\text{ Since } xy<0 \Rightarrow -xy>0 \Rightarrow x^2+xy+y^2>0 \\
\text{Therefore it is never true that } x^2+xy+y^2=-\frac{3}{2}$$
Is my proof complete now?
| You have "$f(x) = f(y)$" implies that at least one of "$x=y$" or "$2(x^2 + x y+ y^2)+3 = 0$" is true. If you can show that the second is never possible for any $x,y \in \mathbb{R}$, you would be done.
Now $x^2 + xy + y^2$ has what minimum value? Can it ever be as small as $-3/2$?
EDIT
In your case 2, you go off the rails after "$x y < 0 \implies x^2 + xy + y^2 > x y$". You also have not addressed what happens when $x$ and $y$ don't have signs, that is, when $x=0$ or $y=0$. Something like ...
When $x = 0$, $2(x^2 + x y+ y^2)+3 = 2(y^2)+3 > 0$. Likewise, when $y=0$ by symmetry. Consequently, we need only consider nonzero $x$ and $y$.
Then, continuing on, ...
If nonzero $x$ and $y$ have the same sign, then $xy>0$ and, since $x^2 + y^2 >0$, $x^2 + x y+ y^2 > 0$.
If nonzero $x$ and $y$ have differing signs, then $xy < 0$ and, ... [what goes here?]
Further Edit
I would make small tweaks to your final edit.
If nonzero $x$ and $y$ have differing signs, then $xy < 0$, giving $-xy > 0$. Then, since $(x+y)^2$ is also positive,
$$ x^2 + xy + y^2 = (x+y)^2 - xy > 0 > -3/2 \text{,} $$
contradicting $2(x^2 + xy + y^2)+3 = 0$. Therefore, $x = y$ and we find $f$ is injective.
We immediately transform "$xy < 0$" into the form we will use, so that the reader knows what form of it they are looking for in the next sentence. We take this and another positivity result to produce the needed positivity result, separating the expression from $-3/2$, and explicit use of that constant reminds the reader what fact they were looking for. We observe that we have obtained our needed contradiction, then the conclusion forced by that contradiction, then the conclusion forced by that result, unrolling the nested conditional proofs that we are in.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\max(ab+bc+ac)$
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$
My attempt:
Squaring both the sides:
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly found the maximum.
How do I proceed?
| Lagrange multipliers will give the solution
\begin{eqnarray*}
L=ab+bc+ca+ \lambda (a+2b+c-4)
\end{eqnarray*}
Differentiating gives
\begin{eqnarray*}
b+c+ \lambda =0 \\
c+a +2\lambda =0 \\
a+b+ \lambda =0 \\
\end{eqnarray*}
Add these equation together and substitute to get $ \lambda = \frac{b-4}{2}$ now substitute this back into the equations above and we get $b=0, a=c=2$. So the maximum value is $\color{blue}{4}$.
Edit : Ooops ... it is a precalculus question ...
\begin{eqnarray*}
L=ab+bc+ca=ab+(a+b)(4-a-2b)=-a^2-2ab-2b^2+4(a+b) = \\ \underbrace{-\frac{1}{2}(2b+a-2)^2}_{2b+a=2}-\underbrace{\frac{1}{2}(a-2)^2}_{a=2}+\color{blue}{4}
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ended up with
$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
Which I'm not sure what to do with.
| I made the subsitution $y=\frac{1}{x}$. The limit as $y \rightarrow 0$ falls straight out as $\frac{a-b}{2}$. You'll need to use binomial expansion of $\sqrt{1+p} = 1 + \frac{p}{2}+...$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Show that $a_{n+1} =a_n + \frac{1}{3^n}$ defines a Cauchy sequence I just need help.
Let ${a_n}$ be the sequence defined by the relations $a_1 = 1$ and $a_{n+1} =a_n + \frac{1}{3^n}$ for $n = 1, 2,....$ Prove that ${a_n}$ is a Cauchy sequence using definition.
| It's easy to see that
$$a_n = \sum_{k=0}^{n}\frac{1}{3^k}$$
The RHS is the $n$'th partial sum of a convergent geometric series (with limit $3/2$), so $a_n$ is a convergent sequence, hence a Cauchy sequence.
Edit after the OP added "using definition":
Let $m > n$. Then
$$\begin{aligned}
a_m - a_n &= \sum_{k=n+1}^{m}\frac{1}{3^k} \\
&= \frac{1}{3^{n+1}}\sum_{k=0}^{m-n-1}\frac{1}{3^k} \\
&= \frac{1}{3^{n+1}}\frac{1 - 1/3^{m-n}}{1 - 1/3} \\
&= \frac{3}{2}\frac{1}{3^{n+1}}\left(1 - \frac{1}{3^{m-n}}\right) \\
\end{aligned}$$
Observe that
$$0 \leq 1 - \frac{1}{3^{m-n}} \leq 1$$
Therefore,
$$0 \leq a_m - a_n \leq \frac{3}{2} \frac{1}{3^{n+1}} = \frac{1}{2}\frac{1}{3^n}$$
Given $\epsilon > 0$, the RHS will be less than $\epsilon$ whenever
$$3^n > \frac{1}{2\epsilon}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Showing $ \mathbb{Q}[\sqrt[4]{2}]$ is a field Here i want to prove
\begin{align}
\mathbb{Q}[\sqrt[4]{2}] = \{ a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8} \mid a,b,c,d\in\mathbb{Q}\}
\end{align}
this is a field. To do that i want to prove
\begin{align}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} \in \mathbb{Q}[\sqrt[4]{2}]
\end{align}
Is there any ideas for showing this?
[edit] Many of you prove $\mathbb{Q}[\sqrt[4]{2}]$ is a field in indirect way. Is there any ways for showing this in direct way? For $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt[3]{2}]$, using some formulas i can check that indeed every nonzero elementn is unit. But this case does not seem that easy...
| Idea: use that $\sqrt[4]{2} = \sqrt{ \sqrt{2}}$, so first rationalize to a denominator in $\mathbb{Q}(\sqrt{2})$, and then finish off.
$$\frac{1}{a+b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}}= \frac{1}{a + c \sqrt{2} + (b + d \sqrt{2}) \sqrt[4]{2}}= \\
=\frac{a + c \sqrt{2} - (b + d \sqrt{2}) \sqrt[4]{2}}{(a + c \sqrt{2})^2 -\sqrt{2}(b + d \sqrt{2})^2}$$ and rationalize by amplifying the fraction by $(a - c \sqrt{2})^2 +\sqrt{2}(b - d \sqrt{2})^2$
| {
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$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0$, compute $ab+cd$
Consider that $\{a,b,c,d\}\subset \Bbb R$ and it is known that
$$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0.$$
Compute the value of $ab+cd$. (Edit: originally written with a typo as $ab+bd$),
This is from a list of training problems used within the preparation for the "Cone-Sul" math olympics. I tried to solve using newton identities, but without success. Any help will be appreciated. Sorry if this is a duplicate.
| Since
$$ac+bd=0\implies a=-\frac{bd}{c}; c\neq0$$
Then
$$1=a^2+b^2=b^2[\frac{d^2}{c^2}+1]=b^2\cdot\frac{c^2+d^2}{c^2}=\frac{b^2}{c^2}$$
$$\implies c=\pm b\implies d=\mp a$$
The solution is given by
$$(a,b,c,d)=(a,b,\pm b,\mp a)$$
So
$$ab+cd=ab+(\pm b)(\mp a)=ab-ab=0$$
(For $c=0$, the case is trivial.)
Another solution
Let $a=mb$, $c=nd$.
$$ac+bd=0\implies mbnd+bd=0\implies bd=0 \text{ or }mn=-1$$
For $mn=-1$
$$a^2+b^2=b^2(m^2+1)=1\implies b^2=\frac{1}{m^2+1}$$
Similarly, $\displaystyle d^2=\frac{1}{n^2+1}$
Then
$$ab+cd=mb^2+nd^2=\frac{m}{m^2+1}+\frac{n}{n^2+1}=\frac{(m+n)(mn+1)}{(m^2+1)(n^2+1)}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$ In the paper A Probabilistic Algorithm for k-SAT Based on Limited Local Search and Restart, by Uwe Schöning, I fail to understand an identity used in a proof:
$$\sum_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$$
for $z=q(1-q)$, with
$$B_2(z)=\sum_{i=0}^{\infty}\binom{2i+1}{i}\frac1{2i+1}z^i=\frac{1-\sqrt{1-4z}}{2z}$$
and
$$B_2(z)^r=\sum_{i=0}^{\infty}\binom{2i+r}{i}\frac{r}{2i+r}z^i$$
It would be great if someone could explain this to me.
|
We consider the generating functions
\begin{align*}
A(z)=\frac{1}{\sqrt{1-4z}}\quad\text{and}\qquad B_2(z)=\frac{1-\sqrt{1-4z}}{2z}
\end{align*}
and derive for non-negative integer $r$ the coefficients of
\begin{align*}
\left(B_2(z)\right)^r\qquad\text{ and }\qquad A(z)\left(B_2(z)\right)^r
\end{align*}
We apply the following
Change of variable formula: Let $f(z)$ be a Laurent series and $g(w)$ be a power series, $g(w)=g_1w+g_2w^2+\cdots$, where $g_1\ne 0$. Then
\begin{align*}
\color{blue}{[z^{-1}]f(z)=[w^{-1}]f(g(w))g^\prime(w)}\tag{1}
\end{align*}
See e.g. p.12 of this presentation by Ira Gessel.
Coefficients of $\left(B_2(z)\right)^r$: We use the transformation
\begin{align*}
z=\frac{w}{(1+w)^2}\qquad\qquad\frac{dz}{dw}=\frac{1-w}{(1+w)^3}\tag{2}
\end{align*}
and we get
\begin{align*}
B_2(w)=\frac{1-\sqrt{1-\frac{4w}{(1+w)^2}}}{\frac{2w}{(1+w)^2}}=1+w
\end{align*}
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ a generating function $A(z)$ and obtain with (1) and (2)
\begin{align*}
\color{blue}{[z^n]\left(B_2(z)\right)^r}&=[z^{-1}]z^{-n-1}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^r\\
&=[w^{-1}]\left(\frac{w}{(1+w)^2}\right)^{-n-1}(1+w)^r\frac{1-w}{(1+w)^3}\tag{3}\\
&=[w^n](1+w)^{2n+r-1}(1-w)\tag{4}\\
&=[w^n](1+w)^{2n+r-1}-[w^{n-1}](1+w)^{2n+r-1}\tag{5}\\
&=\binom{2n+r-1}{n}-\binom{2n+r-1}{n-1}\tag{6}\\
&=\binom{2n+r}{n}-2\binom{2n+r-1}{n-1}\tag{7}\\
&=\color{blue}{\binom{2n+r}{n}\frac{r}{2n+r}}\tag{8}
\end{align*}
and the claim follows. The coefficients $\binom{2n+1}{n}\frac{1}{2n+1}$ of $B_2(z)$ follow by setting $r=1$.
Comment:
*
*In (3) we use the substitution (2).
*In (4) we do some simplifications.
*In (5) we factor out and apply the rule $[z^p]z^{q}A(z)=[z^{p-q}]A(z)$.
*In (6) we select the coefficients accordingly.
*In (7) we use the binomial identity $\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}$.
*In (8) we do some final simplifications.
The same technique works for the other case.
Coefficients of $A(z)\left(B_2(z)\right)^r$:
We use the same transformation $z=\frac{w}{(1+w)^2}$ and we get
\begin{align*}
A(w)=\frac{1}{\sqrt{1-\frac{4w}{(1+w)^2}}}=\frac{1+w}{1-w}
\end{align*}
We obtain with (1) and (2)
\begin{align*}
\color{blue}{[z^n]\left(A(z)\left(B_2(z)\right)^r\right)}&=[z^{-1}]z^{-n-1}\frac{1}{\sqrt{1-4z}}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^r\\
&=[w^{-1}]\left(\frac{w}{(1+w)^2}\right)^{-n-1}\frac{1+w}{1-w}(1+w)^r\frac{1-w}{(1+w)^3}\\
&=[w^n](1+w)^{2n+r}\\
&=\color{blue}{\binom{2n+r}{n}}
\end{align*}
end the claim follows. The coefficients $\binom{2n}{n}$ of $A(z)$ follow by setting $r=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Prove that the ellipsoid $E = \{ (x,y,z) \in R^3 \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1 \}$ is convex I know what an ellipsoid is, but I do not know how to approach how to prove that one is convex. Any advice would be greatly appreciated.
| Let $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ be two points in the ellipsoid. We have to prove that $(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2,\lambda z_1+(1-\lambda)z_2)$ also lies in the ellipsoid, where $0 \le \lambda \le 1$.
Let $$A = \frac{\left(\lambda x_1+(1-\lambda)x_2\right)^2}{a^2}+\frac{\left(\lambda y_1+(1-\lambda)y_2\right)^2}{b^2}+\frac{\left(\lambda z_1+(1-\lambda)z_2\right)^2}{c^2}.$$
It is easy to observe that
$$A = \lambda^2\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} + \frac{z_1^2}{c^2}\right) + (1-\lambda)^2\left(\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} + \frac{z_2^2}{c^2}\right)+2\lambda(1-\lambda)\left(\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} + \frac{z_1z_2}{c^2}\right).$$
Consequently,
$$A \le \lambda^2 + (1-\lambda)^2+2\lambda(1-\lambda)\left(\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} + \frac{z_1z_2}{c^2}\right).$$
By Cauchy-Schwarz inequality,
$$\left(\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} + \frac{z_1z_2}{c^2}\right) \le \sqrt{\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} + \frac{z_1^2}{c^2}\right)\cdot \left(\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} + \frac{z_2^2}{c^2}\right)} \le 1.$$
Therefore,
$$A \le \lambda^2 + (1-\lambda)^2+2\lambda(1-\lambda)=1.$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
| Note: $(n+1)^2 = n^2 + 2n +1$.
So $(2n-1)^2 - (2n)^2 = 4n^2 -4n +1 - 4n^2 = -4n+1$.
So $\sum_{k=1}^{n=2m} (-1)^{k-1}k^2 = \sum_{j=1}^m [(2j-1)^2 - (2j)^2]=\sum_{j=1}^m [-4j +1]= -4(\sum j) + m = -4*\frac {m(m+1)}2 + m = -2m(m+1) +m=m(-2(m+1) + 1)=m(-2m-1)=-m(2m+1)= -\frac{n(n+1)}{2}$
..... if $n$ is even.
If $n=2m+1$ is odd then $\sum_{k=1}^{n=2m+1}(-1)^{k-1}k^2 = [\sum_{k=1}^{2m=2n-1}(-1)^{k-1}k^2] + n^2$
$= -\frac {(n-1)((n-1) + 1)}2 + n^2= n^2 - \frac {n(n-1)}2 = \frac {2n^2 -n(n-1)}2 = \frac {n(2n - (n-1)}2 = \frac {n(n+1)}2$
So either way ...
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
} |
Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gives
$$\cos{y}=1-\sin^2{\left(\arcsin{\frac{1}{\sqrt{10}}}\right)}=1-\frac{1}{10}=\frac{9}{10}$$
So I have reduced the problem to the following
$$\sin{x}\cdot \frac{9}{10}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Here I'm stuck. I cant just divide this expression by $(\sin{x}\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\sin(\arctan{2})$ and $\cos{\arctan2}.$ I also tried rewriting $\cos(\arctan{2})$ as $1-\sin^2(\arctan{2}),$ but to no avail.
Any suggestions on
*
*how to proceed from where I left;
*how tocompute this by means more effective;
*both of the above.
| $x=\arctan 2$ means $\tan x=2$ thus
$\sin x = \dfrac{\sqrt{\sin^2 x}}{\sqrt{1}}=\dfrac{\sqrt{\sin^2 x}}{\sqrt{\sin^2 x+\cos^2 x}}=\dfrac{\sqrt{\frac{\sin^2 x}{\cos^2 x}}}{\sqrt{\frac{\sin^2 x}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x}}}=\dfrac{\tan x}{\sqrt{\tan^2 x+1}}$
$\sin x=\dfrac{2}{\sqrt 5}$ and $\cos x =\sqrt{1-\sin^2 x}=\dfrac{1}{\sqrt 5}$
$y=\arcsin \dfrac{1}{\sqrt{10}}$ means $\sin y=\dfrac{1}{\sqrt{10}}$ and $\cos y=\sqrt{1-\sin^2 y}=\dfrac{3}{\sqrt{10}}$
Therefore $\sin{\left(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}}\right)}=\sin(x-y)=\sin x\cos y -\cos x\sin y=$
$=\dfrac{2}{\sqrt 5}\cdot \dfrac{3}{\sqrt{10}}-\dfrac{1}{\sqrt 5}\cdot \dfrac{1}{\sqrt{10}}=\dfrac{5}{\sqrt{50}}=\dfrac{5}{5\sqrt 2}=\dfrac{1}{\sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 5
} |
Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$1^2+3^2+...+(2n-1)^2+(2n+1)^2=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$n(2n-1)(2n+1)/3+3(2n+1)^2/3=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$4n(2n-1)(2n+1)/3=(n+1)(2n+3)/3$
What is wrong?
| For $n=1$ the formula holds,
$$S_n=1^2+3^2+5^2+\cdots+(2n-1)^2$$
$$S_{n+1}=1^2+3^2+\cdots+n^2+(2n+1)^2$$
$$\bbox[5px,border:2px solid red]{S_{n+1}-S_n=(2n+1)^2}$$
Now if,
$$S_n=\frac{n(2n-1)(2n+1)}{3}$$$$$$
$$S_{n+1}=\frac{(n+1)(2n+1)(2n+3)}{3}$$$$$$
$$S_{n+1}-S_n=\frac{(n+1)(2n+1)(2n+3)}{3}-\frac{n(2n-1)(2n+1)}{3}$$$$$$
$$=(2n+1)\frac{(n+1)(2n+3)-n(2n-1)}{3}$$$$$$
$$=(2n+1)\frac{(n)(2n+3)-n(2n-1)+(2n+3)}{3}$$$$$$
$$=(2n+1)\frac{6n+3}{3}$$$$$$
$$S_{n+1}-S_n=(2n+1)^2$$$$$$
Hence proved....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$\{a,b,c\}\subset \Bbb R$, $a^2-ab=1,b^2-bc=1, c^2-ca=1$. Find $abc(a+b+c)$.
Consider that $\{a,b,c\}\subset \Bbb R$ and it is known that
$$a^2-ab=1,\ b^2-bc=1,\ c^2-ca=1\ \ \text{(cond.1)}$$
Find $abc(a+b+c)$.
This question is from OBM 2007 (Brazilian Math Olympiad). Sorry if it is a duplicate. I will present my solution below but I think there might be easier approaches.
My attempt:
By multiplying the left and right sides from all equations from (cond. 1) we get:
$$abc(a-b)(b-c)(c-a)=1$$
Multiplying the 3 terms in parenthesis leads to
$$abc(abc-a²b-ac²+a²c-cb²+ab²+bc²-abc)
=1$$
$$abc(a²b-ac²+a²c-cb²+ab²+bc²)
=1$$
$$abc(-a(c²-ac)-c(b²-bc)-b(a²-ab))=1$$
Now by replacing the result from each equation in (cond. 1), and rearranging the expression, we get the desired answer
$$abc(a+b+c)=-1$$
Question: are there other ways to solve this problem? Please provide your own answer if it uses a different approach.
| My approach (not necessary simpler than yours):
Let $S=abc(a+b+c)$ then:
$$S=a^2bc+b^2ac+c^2ab=(1+ab)bc+(1+bc)ac+(1+ca)ab=ab+bc+ac+S$$
So: $ab+bc+ac=0$
Now add the three constraints
$$a^2+b^2+c^2-(ab+bc+ca)=3$$
So $a^2+b^2+c^2=3$
Then:
$$0=(ab+bc+ac)^2=a^2b^2+b^2c^2+c^2a^2+2S=a^2(1+bc)+b^2(1+ac)+c^2(1+ab)+2S=3+3S$$
So $S=-1$
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2.
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.
This is from a list of problems used for training a team for a math olympics. I tried to use known Newton identities and other symmetric polynomial results but without success (perhaps a wrong approach). Sorry if it is a duplicate. Hints and answers are always welcomed.
Edit: There is a problem with the original statement of the question in the original source. Under these assumptions it is impossible to have $a, b, c$ with value 2, as spotted in the comments and proved by the answers below
| There is a trouble with this question.
There are non-real solutions for example $a\approx 3.12128,\space b,c\approx -0.56064\pm 1.47835i$.
On the other hand, in $\mathbb R$ as in $\mathbb C$, if $a=2$ then the system
$$-1+(y-2)^3+(z-2)^3=0\\y^2+z^2=2\\y+z=0$$ is incompatible and if $b$ or $c$ are equal to $2$ then the system $$(x-3)^3+(y-2)^3=0\\x^2+y^2=2\\x+y=0$$ is incompatible too (i.e. none of the two systems has a solution).
Finally the conclusion is that there are no real numbers $a,b,c$ satisfying the three given equations
$$\begin{cases}(a-3)^3+(b-2)^3+(c-2)^3=0\\a+b+c=2\\a^2+b^2+c^2=6\end{cases}$$
Consequently there is no reason for this to imply that $a, b$ or $c$ must be equal to $2$.
(It could be argued that if the given system is incompatible then $a, b$ or $c$ must be equal to $2$ so that the resulting system is also incompatible.This is not true by doing $a = 4$. I stop here without make $b$ or $c$ distinc of $2$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Determine for what values of $x$ the given series converges The given series is
$\sum_{n=1}^∞
(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n}
)x^{n} $.
I tried it by using Cauchy Root Test as follows-
Let
$y=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )^{1/n}$,
then by taking logarithm both sides,we get
$\log(y)=\lim_{n\to\infty}\frac{1}{n}\log(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )$
Since, $\log(0)=-\infty$. So, $\log(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )=-\infty$.
Applying,L'Hôpital's rule,we get
$\log(y)=\lim_{n\to\infty}-\frac{(\frac{1}{n^2}+\frac{1}{(n-1)^2}+\frac{1}{(n-2)^2}+...+...+... )}{(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )}$.(Please Check this step!!)
Since,$\sum_{k=1}^\infty\frac{1}{k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$ & $\sum_{k=1}^\infty\frac{1}{k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}=\infty$.So,$\log(y)=0\implies y=1$.
Now let $a_n=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )x^{n} $. Then,$$(a_n)^{1/n}=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )^{1/n}x$$
$\implies \lim_{n\to\infty}\vert (a_n)^{1/n}\vert=1.x$.By ,Cauchy root test the given series converges if $\lim_{n\to\infty}(a_n)^{1/n}<1$.Hence, the given series converges if $\vert x\vert<1$.
I NEED TO KNOW WHETHER MY SOLUTION IS CORRECT OR NOT?
| Since both $\frac{1}{1-x}=\sum_{n\geq 0}x^n$ and $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$ have radius of convergence equal to one, the same holds for their (Cauchy) product
$$ \sum_{n\geq 1} H_n x^n = -\frac{\log(1-x)}{1-x}. $$
In particular the LHS is convergent for any $x\in(-1,1)$ and clearly is not convergent for any $x\in\mathbb{R}\setminus(-1,1)$ since in such a case the main term is not $o(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
determinant of the $7\times7$ matrix How to find determinant of the following $7\times7$ matrix
\begin{bmatrix}
a&1&0&0&0&0&1&\\
1&a&1&0&0&0&0&\\
0&1&a&1&0&0&0&\\
0&0&1&a&1&0&0&\\
0&0&0&1&a&1&0&\\
0&0&0&0&1&a&1&\\
1&0&0&0&0&1&a&
\end{bmatrix}
| Let $e_0, e_1, \ldots, e_6$ be the basis column vector with entries $1$ at $q^{th}$ row and zero otherwise. i.e
$$(e_q)_p = \begin{cases}1, & p = q\\0, & \text{ otherwise }\end{cases},\quad 0 \le p, q \le 6$$
Let $A$ be the matrix at hand. Let $L = (\ell_{pq})$ and $R = (r_{pq})$ be the $7 \times 7$ matrices given by
$$\ell_{qp} = r_{pq} = \begin{cases} 1, & p = q+1 \pmod 7\\0, & \text{ otherwise }\end{cases},\quad 0 \le p, q \le 6$$
$L$ and $R$ are matrices which shift the basis vector $e_q$ to $e_{q-1 \pmod 7}$
and $e_{q+1 \pmod 7}$ respectively.
In terms of $L$ and $R$, we have
$$A = a I_7 + L + R$$
It is easy to see $R = L^{T} = L^{-1}$ and $L$ has seven eigenvectors $u_0, u_1, \ldots, u_6$ of the form
$$u_p = \sum_{q=0}^6 e^{i\frac{2\pi pq}{7}} e_q\quad\iff\quad(u_p)_q = e^{i\frac{2\pi pq}{7}},\quad 0 \le q \le 6$$
with eigenvalue $e^{i\frac{2\pi p}{7}}$. This means $A$ has eigenvalues of the form $a + 2\cos\frac{2\pi p}{7}$ and
$$\det A = \prod_{p=0}^6 \left( a + 2\cos\frac{2\pi p}{7} \right)$$
Rewrite $a$ as $2\cosh\theta = e^\theta + e^{-\theta}$, we have
$$\begin{align}\det A
&= e^{-7\theta}
\prod_{p=0}^6 \left(e^{2\theta}+1 + 2e^\theta\cos\frac{2\pi p}{7}\right)
= e^{-7\theta}\prod_{p=0}^6 \left(e^\theta + e^{i\frac{2\pi p}{7}}\right)\left(e^\theta + e^{-i\frac{2\pi p}{7}}\right)\\
&= e^{-7\theta}(e^{7\theta} + 1)^2
= 2\cosh(7\theta) + 2 = 2T_7(\frac{a}{2}) +2\\
&= a^7-7a^5+14a^3-7a+2
\end{align}
$$
where
$T_n(x) = \cos(n\cos^{-1}(x)) = \cosh(n\cosh^{-1}(x))$ is the Chebyshev polynomial of first kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434577",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Polynomials with integration problem
Are there polynomials $P,Q$ with real coefficients satisfying the equalities
$$\int_0^{\ln n}\frac{P(x)}{Q(x)} \, dx
= 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$
for each integer $n\ge 2$?
I don't know how I show the existence of such polynomials or nonexistence. I took some example but I can't improve anything.
| Suppose there existed such $P,Q$. It is easy to see that $\deg P\le \deg Q$; otherwise the integral will blow up too quickly. This implies that $\frac{P(1/z)}{Q(1/z)}$ is a rational function which does not blow up at $0$, and hence has a Taylor expansion $\frac{P(1/z)}{Q(1/z)} = a_0+a_1z+a_2z^2+\dots$ valid for all small enough $z$. This implies that
$$ \frac{P(x)}{Q(x)} = a_0 + \frac{a_1}{x} + \frac{a_2}{x^2} + \dots $$
for all large enough $x$. I will now show that $a_0 = 1 $ and $a_k = 0$ for $k>0$, so that $\frac{P(x)}{Q(x)} = 1$, which is incompatible with the requirement that $\int\limits_{0}^{\ln n}{\frac{P(x)}{Q(x)}\,dx} = 1 + \frac{1}{2} + \dots + \frac{1}{n}$ for $n\ge 2$.
For $k\ge 0$, notice that
$$\frac{1}{(\ln(n+1))^k}\ln\left(\frac{n+1}{n}\right)\le\int_{\ln n}^{\ln(n+1)}{\frac{1}{x^k}\,dx}\le\frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right). $$
Furthermore, since
$$\frac{1}{(\ln n)^k}-\frac{1}{(\ln(n+1))^k} = \int_{\ln n}^{\ln(n+1)}{\frac{k}{x^{k+1}}\,dx}\le\frac{k}{(\ln n)^{k+1}}\ln\left(\frac{n+1}{n}\right) $$
it follows that
\begin{align} \left|\int_{\ln n}^{\ln(n+1)}{\frac{1}{x^k}\,dx} - \frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)\right|&\le\left(\frac{1}{(\ln n)^k}-\frac{1}{(\ln(n+1))^k}\right)\ln\left(\frac{n+1}{n}\right) \\
&\le\frac{k}{(\ln n)^{k+1}}\left(\ln\left(\frac{n+1}{n}\right)\right)^2
\end{align}
and hence $\int\limits_{\ln n}^{\ln(n+1)}{\frac{1}{x^k}\,dx} = \frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)(1+o(1))$.
Now, to show $a_0 = 1$, notice that
\begin{align} \int_{\ln n}^{\ln(n+1)}{\frac{P(x)}{Q(x)}\,dx} &= \int_{\ln n}^{\ln(n+1)}{a_0+O\left(\frac{1}{x}\right)\,dx} \\
&= a_0\ln\left(\frac{n+1}{n}\right) + O\left(\frac{1}{\ln n}\ln\left(\frac{n+1}{n}\right)\right) \\
&= \frac{a_0}{n} + o\left(\frac{1}{n}\right).
\end{align}
In order for $\int\limits_{\ln n}^{\ln(n+1)}{\frac{P(x)}{Q(x)}\,dx} = \frac{1}{n+1} = \frac{1}{n}(1+o(1))$ for all $n$, it follows that we must have $a_0 = 1$.
We now show $a_k = 0$ for $k>0$ by induction. Suppose $a_j = 0$ for all $0<j<k$ (note for the base case $k=1$ that this is vacuously true). Then
$$\frac{P(x)}{Q(x)} - 1 = \frac{a_k}{x^k} + \frac{a_{k+1}}{x^{k+1}} + \dots = \frac{a_k}{x^k} + O\left(\frac{1}{x^{k+1}}\right) $$
and hence
\begin{align} \int_{\ln n}^{\ln(n+1)}{\frac{P(x)}{Q(x)}-1\,dx} &= \int_{\ln n}^{\ln(n+1)}{\frac{a_k}{x^k}+O\left(\frac{1}{x^{k+1}}\right)\,dx} \\
&=\frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)(a_k+o(1)) + O\left(\frac{1}{(\ln n)^{k+1}}\ln\left(\frac{n+1}{n}\right)\right) \\
&=\frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)(a_k+o(1)).
\end{align}
By assumption, we have
$$\int_{\ln n}^{\ln(n+1)}{\frac{P(x)}{Q(x)}-1\,dx} = \frac{1}{n+1} - \ln\left(\frac{n+1}{n}\right) = O\left(\frac{1}{(n+1)^2}\right)$$
and $\frac{1}{(n+1)^2} = o\left(\frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)\right)$ for any $k>0$. It follows that
$$\frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)(a_k+o(1)) = \frac{1}{(\ln n)^k}\ln\left(\frac{n+1}{n}\right)o(1)$$
which yields $a_k = 0$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the limit to $\lim\limits_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$ My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the expression. My "toolbox" for solving problems like these is very limited... If polar form doesn't work, then I usually have no clue on how to continue.
Edit: I think this is the solution.
$$
\lim_{(x,y)\to(0,0)}\left|\frac{x^5+y^2}{x^4+|y|}\right| = \frac{|x^5+y^2|}{|x^4+|y||}
$$
Applying the triangle inequality gives
$$
\frac{|x^5+y^2|}{|x^4+|y||} \leq \left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right|
$$
Inspecting the denominators on the RHS gives:
$$
\left|\frac{x^5}{x^4+|y|}\right| \leq |x|, \quad\left|\frac{y^2}{x^4+|y|}\right| \leq |y|
$$
So
$$
\left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right| \leq |x| + |y|
$$
Since $|x| + |y| \to 0$ when $x,y\to 0$, the sandwich theorem states that $|\frac{x^5+y^2}{x^4+|y|}| \to 0$. And if $\lim |f(x)|=0$ then $\lim f(x)=0$ which solves the original problem.
| $0\le |\dfrac{x^5+y^2}{x^4 +|y|}| \le$
$\dfrac{|x^5| +|y^2|}{|x^4+|y||} \le$
$\dfrac{|x^5|}{x^4} + \dfrac{y^2}{|y|} =$
$|x| + |y|.$
And now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the product $\prod_{k=2}^n \left(1 - \frac{1}{k}\right)$ using Induction I'm trying to figure out how to prove by induction the following statement:
$$
\prod_{k=2}^n \left(1 - \frac{1}{k}\right) = \frac{1}{n}.
$$
| Inductive hypothesis:
Assume that
$$
\prod_{k = 1}^m \left( 1 - \frac{1}{k}\right) = \frac{1}{m}
$$
holds true for some $m$.
Then we compute
$$
\left(1 - \frac{1}{m+1} \right) \cdot \prod_{k = 1}^m \left( 1 - \frac{1}{k}\right) = \left(1 - \frac{1}{m+1} \right) \cdot \frac{1}{m} = \left(\frac{1}{m} - \frac{1}{(m + 1)m} \right) = \frac{(m+1) - 1}{m(m+1)} = \frac{1}{m+1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the area of a triangle using the sides lengths
The sides lengths of a triangle $a,$ $b$ and $c$ verify: $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=12b.$$ The task is to find the area of the triangle.
I'm trying to apply the heron's formula: $$\dfrac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}} {4}.$$
How do i get to heron's formula from what i know? Or it can be solved in another way?
| I think in you equation ,must be $24b$ $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=24b\\
\sqrt{a-24} + b^2 -24b+144+|c-12\sqrt{3} |=0\\
\sqrt{a-24} + (b-12)^2+|c-12\sqrt{3} |=0
\\\underbrace{\sqrt{a-24}}_{\geq 0} +\underbrace{(b-12)^2}_{\geq 0} + \underbrace{|c-12\sqrt{3} |}_{\geq 0} =0$$ some of positive expressions are zero , so all af them are zero at the same time
$$\begin{cases}\sqrt{a-24}=0 & a =24\\(b-12)^2=0 & b=12\\|c-12\sqrt{3} |=0 & c=12\sqrt3\end{cases}$$ now plug into Heron's formula
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
"answer_id": 1
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Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$?
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$
This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\sqrt{13}})^2 - (\sqrt[3]{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.
| $$\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)^3=$$
$$=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}} \sqrt[3]{18-5\sqrt{13}}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36+3\sqrt[3]{-1}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36-3\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right).$$
Now, let $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}=x.$
Thus, $$x^3=36-3x$$ or
$$x^3-3x^2+3x^2-9x+12x-36=0$$ or
$$(x-3)(x^2+3x+12)=0,$$ which gives $x=3$.
But I think the best way in this formulation it's the way by using
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Now, we need to prove that
$$18+5\sqrt{13}+18-5\sqrt{13}-27-3\cdot(-3)\sqrt[3]{18+5\sqrt{13}}\cdot\sqrt[3]{18-5\sqrt{13}}=0$$ or
$$36-27-3(-3)(-1)=0,$$
which is true.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Quadratic formula for complex numbers Consider the equation
$$
az^2+bz+c=0,
$$
where $a,b$ and $c$ are complex numbers and $a\ne 0$. Applying usual operations on $\mathbb{C}$ we have the following:
$$
az^2+bz+c=0\iff z^2+\frac{b}{a}z = -\frac{c}{a}\iff z^2+\frac{b}{a}z + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} -\frac{4ac}{4a^2} \\ \iff \left(z + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}.
$$
Now here's where my doubt comes. I know I can take the square root on both sides but wouldn't I have two results on both sides? Moreover, on the denominator of the right side (and on the whole left side) I have $z_0^2$; taking square root of this means I have only one solution ($z_0$)? Because I was told that the solution for the quadratic equation will be the same as for real numbers.
| Let's call $\sigma$ the square root of $z$ which means that $\sigma$ such as $\sigma^2=z$.
Thus, if $\rho = b^2 - 4ac$ and $\sigma^2 = \rho$ then
\begin{align*}
& \left(z + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\\
\iff & \left(z + \frac{b}{2a}\right)^2 = \frac{\rho}{4a^2}\\
\iff & z + \frac{b}{2a} = \frac{\pm\sigma}{2a}\\
\iff & z = \frac{\pm\sigma}{2a} - \frac{b}{2a}\\
\iff & z = \frac{-b\pm\sigma}{2a}
\end{align*}
Let's say we are looking for the square root $x + i y$ of $a + i b$ where $b$ is positive.
It is possible to find the square root of a complex number like this:
\begin{align*}
& (x + i y)^2 = a + i b \\
\iff & \begin{cases}
x^2-y^2 &= a\\
2xy &= b
\end{cases}
\iff \begin{cases}
x^2-y^2 &= a\\
y &= \frac{b}{2x}
\end{cases}
\iff \begin{cases}
x^2-\left(\frac{b}{2x}\right)^2 &= a\\
y^2 &= x^2 - a\\
2xy &= b
\end{cases}\\
\iff & \begin{cases}
x^2-\frac{b^2}{4x^2} &= a\\
y^2 &= x^2 - a\\
2xy &= b
\end{cases}
\iff \begin{cases}
\frac{4x^4-b^2}{4x^2} &= a\\
y^2 &= x^2 - a\\
2xy &= b
\end{cases}
\iff \begin{cases}
4x^4-b^2 &= 4ax^2\\
y^2 &= x^2 - a\\
2xy &= b
\end{cases}\\
\iff & \begin{cases}
4x^4-4ax^2-b^2 &= 0\\
y^2 &= x^2 - a\\
2xy &= b
\end{cases}\\
\end{align*}
We're looking for the $x$ such as $4x^4 - 4ax^2 - b^2 = 0$ or if we define $\lambda = x^2$ we have $4 \lambda^2 - 4a\lambda - b^2 = 0$.
$$\rho = (-4a)^2 - 4 (4) (-b^2) = 16 a^2 + 16 b^2 = 16 (a^2 + b^2)$$
$$\lambda = \frac{4a \pm 4\sqrt{a^2 + b^2}}{8} = \frac{a \pm \sqrt{a^2 + b^2}}{2}$$
Since $\lambda = x^2$, we have $\lambda \geq 0$ and thus $\lambda = \frac{a + \sqrt{a^2 + b^2}}{2}$.
It follows that $x = \pm\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}}$.
Knowing that we deduce the value of $y$: $y = \pm\sqrt{x^2-a} = \pm\sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}-a} = \pm\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}$.
Also we know that $2xy = b$ which explains why both $x = \pm \ldots$ and $y = \pm \ldots$ only produces 2 solutions instead of 4 (anyway we knew that being complex numbers there were only 2 solutions).
If $b > 0$, then $x$ and $y$ have the same sign and the solutions are $\left( \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} , \sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\,\right)$ and $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}},-\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\,\right)$.
If $b < 0$, then $x$ and $y$ have opposite signs and the solutions are $\left(\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} , -\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\,\right)$ and $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}},\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\,\right)$.
We may conclude that the equation $(x \pm i y)^2 = a \pm i b\mbox{ with } b \mbox{ positive}$ has the following solutions: $\left( \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} , \pm\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\,\right)$ and $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}},\mp\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}\,\right)$.
| {
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"url": "https://math.stackexchange.com/questions/2443067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Statistics Question involving Mode : Given the set of data $1,1,2,2,2,3,3,x,y$ .... I was studying for some quizzes when a wild question appears. It goes like this:
Given the set of data $1,1,2,2,2,3,3,x,y$, where $x$ and $y$ represent two different integers. If the mode is $2$, which of the following is true?
A. If $x = 1 \space or \space 2$, then $y = 3$
B. If $x = 1 \space or \space 3$, then $y = 2$
C. If $x = 1 \space or \space 2$, then $y = 2$
D. If $x = 1 \space or \space 3$, then $y = 3$
My work
The mode is the most common number occuring in a sequence of numbers. In this case, it's $2$.
If I think about A, that would be false, because if I'm going to put $x = 1$ and $y = 3$, completing the sequence $1,1,2,2,2,3,3,1,3$.
The mode will be $2$ and $3$. The problem states that mode must be $2$. If I'm going to put $x = 2$ and $y = 3$, completing the sequence $1,1,2,2,2,3,3,2,3$,
the mode will be $2$, making A correct. In short, A is half-truth
If I think about B, that would be true, because if I'm going to put $x = 1$ and $y = 2$, completing the sequence $1,1,2,2,2,3,3,1,2$.
The mode will be $2$. The problem states that mode must be $2$. If I'm going to put $x = 3$ and $y = 2$, completing the sequence $1,1,2,2,2,3,3,3,2$,
the mode will be $2$, making B correct. In short, B is true
If I think about C, that would be true, because if I'm going to put $x = 1$ and $y = 2$, completing the sequence $1,1,2,2,2,3,3,1,2$.
The mode will be $2$. The problem states that mode must be $2$. If I'm going to put $x = 2$ and $y = 2$, completing the sequence $1,1,2,2,2,3,3,2,2$,
the mode will be $2$, making C true. In short, C is true
If I think about D, that would be false, because if I'm going to put $x = 1$ and $y = 3$, completing the sequence $1,1,2,2,2,3,3,1,3$.
The mode will be $2$ and $3$. The problem states that mode must be $2$. If I'm going to put $x = 3$ and $y = 3$, completing the sequence $1,1,2,2,2,3,3,3,3$,
the mode will be $3$, making D false. In short, D is false
The problem is, I got two choices that says they were TRUE, which is the choices B and C.
I don't know what would be the correct one, akin to choosing two equally-potential suitors.
What might be the correct choice in this problem?
| When you read something like:
If $x=a$ or $b$, then $y=c$
What this means is that for the values of $x$, $y$ must equal $c$ in order to satisfy the condition that the sequence has a mode of 2. And if at least one value of $x$ does not satisfy the condition for $y$, then the entire thing is false.
For example, $A$ is not true. It's false - let's see why.
As you correctly stated, we have sequences $$\begin{cases}1,1,2,2,2,3,3,1,y \\ 1,1,2,2,2,3,3,2,y\end{cases}$$
Now, for sequence 1, $y$ cannot be $3$, as this results in a mode other than $2$.
However, for sequence $2$, $y$ does not have to be $3$ - if $y=1,2$ or $3$ the mode will always be $2$. Thus, $y$ does not necessarily equal $3$.
Now, when $x=2$ in this case, we are not guaranteed that $y=2$, hence $A$ is false.
Notice that for $B$, we have sequences $$\begin{cases} 1,1,2,2,2,3,3,1,y \\ 1,1,2,2,2,3,3,3,y\end{cases}$$
For sequence 1 $y$ must be $2$, or else the mode of the sequence is not $2$.
For sequence 2, $y$ must be $2$ as well, or else the mode of the sequence does not equal $2$.
Thus, due to the property that the sequence has a mode of $2$, when $x$ equals $1$ or $3$ we are guaranteed that $y$ must equal $2$.
Hence $B$ is correct.
| {
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"url": "https://math.stackexchange.com/questions/2447214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that
T $\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
3 \\
1 \\
4 \\
\end{pmatrix}$, $T$ $\begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
-1 \\
-1\\
3\\
\end{pmatrix}$ , $T$ $\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix}$ = $\begin{pmatrix}
4 \\
-3 \\
-1\\
\end{pmatrix}$ then $T$ $\begin{pmatrix}
-5\\
4 \\
4 \\
\end{pmatrix}$ = $\begin{pmatrix}
? \\
? \\
? \\
\end{pmatrix}$
I do not know where to start with this question, I tried doing RREF of the transformation numbers but I feel that is wrong.
| Let $\{ e_1,e_2,e_3 \}$ be the canonical (standard) basis of $\mathbb{R}^3$. Note that for any $(x,y,z) \in \mathbb{R}^3$ we can write $(x,y,z)=ze_1+ye_2+ze_3$. Hence $T(x,y,z)=T(xe_1+ye_2+ze_3)$, and using lineality $$T(x,y,z)=T(xe_1)+T(ye_2)+T(ze_3)=xT(e_1)+yT(e_2)+zT(e_3)$$ and by hypotesis $$T(x,y,z)=x(3,1,4)+y(-1, -1, 3)+z(4,-3,-1)=(3x-y+4z, x-y-3z, 4x+3y-z).$$
Finally, $$T(-5,4,4)=(-15-4+16,-5-4-12, -20+12-4)=(-3, -21, -12).$$
| {
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In how many ways can $5$ balls of different colours be placed in $3$ boxes of different sizes if no box remains empty? 5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.
My attempt:-
First choose 3 balls to be placed in 3 boxes so that none of them remain empty in ${{5}\choose{3}}\cdot3! = 60$ ways.
Now remaining 2 balls can go into any of the 3 boxes in $3\cdot3 = 9$ ways.
Total number of ways $= 60\cdot9 = 540$.
Where am I going wrong ?
| There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are
$$\binom{3}{1}2^5$$
ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
by the Inclusion-Exclusion Principle.
Where am I going wrong?
You count each distribution in which one box receives three balls and the others receive one three times, once for each way you could place one of those three balls first.
You count each distribution in which two of the boxes receive two balls and the other box receives one four times, once for each way you could place one of the two balls in each of the two boxes with two balls first.
Three balls in one box and one ball in each of the others: There are three ways to choose which box receives three balls, $\binom{5}{3}$ ways to choose which three balls are placed in that box, and $2!$ ways to distribute the remaining balls. Hence, there are
$$\binom{3}{1}\binom{5}{3}2!$$
ways to distribute the balls so that three balls are placed in the same box.
Two boxes receives two balls and one box receives one ball: There are three ways to choose which box receives only one ball and five ways to choose the ball that is placed in that box. There are $\binom{4}{2}$ ways to choose which two of the remaining four balls are placed in the smaller of the two remaining boxes. The other two balls must be placed in the remaining box. Hence, there are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$
ways to distribute the balls so that two boxes receive two balls and one box receives one.
Observe that
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
Since you counted distributions in which one box receives three balls and the others receive one three times and distributions in which two boxes receive two balls and the other receives one four times, you obtained
$$3\binom{3}{1}\binom{5}{3}2! + 4\binom{3}{1}\binom{5}{1}\binom{4}{2} = \binom{5}{3} \cdot 3! \cdot 3^2$$
| {
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"source": "stackexchange",
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} |
How to find the closed form of $\int_{0}^{\infty}{(e^{-x}+x-1)^2\over x(e^{x\over n}-1)}\mathrm dx$ We partially guessed the closed form of $(1)$ to be
$$\int_{0}^{\infty}{(e^{-x}+x-1)^2\over x(e^{x\over n}-1)}\mathrm dx=f(n)+{\pi^2\over 6}n^2+\ln{{2n\choose n}}\tag1$$
where $n\ge1$
$f(1)=-2$
$f(2)=-6$
$f(3)=-11$
$f(4)=-{50\over 3}$
$f(5)=-{137\over 6}$
$f(n)=?$
How can we find the complete closed form for $(1)?$
| Hint by user reuns:
$$\begin{align}
\int_0^{\infty } \frac{(\exp (-x)+x-1)^2}{x \left(\exp \left(\frac{x}{n}\right)-1\right)} \, dx & =\int_0^{\infty } \frac{(\exp
(-x)+x-1)^2 \sum _{k=1}^{\infty } \exp \left(-\frac{k x}{n}\right)}{x} \, dx \\
&=\sum _{k=1}^{\infty } \int_0^{\infty }
\frac{(\exp (-x)+x-1)^2 \exp \left(-\frac{k x}{n}\right)}{x} \, dx \\
&=\sum _{k=1}^{\infty } \left(\frac{n^2 (-k+n)}{k^2 (k+n)}+2
\ln (k+n)-\ln (k (k+2 n))\right) \\
&=\sum _{k=1}^{\infty } \left(\frac{n^2 (-k+n)}{k^2 (k+n)}+\ln \left(\frac{(k+n)^2}{k (k+2
n)}\right)\right) \\
&=\sum _{k=1}^{\infty } \frac{n^2 (-k+n)}{k^2 (k+n)}+\sum _{k=1}^{\infty } \ln \left(\frac{(k+n)^2}{k (k+2
n)}\right) \\
&=\sum _{k=1}^{\infty } \frac{n^2 (-k+n)}{k^2 (k+n)}+\ln \left(\prod _{k=1}^{\infty } \frac{(k+n)^2}{k (k+2
n)}\right) \\
&=-2 \gamma n+\frac{n^2 \pi ^2}{6}-2 n \, \psi (1+n)+\ln \left(\frac{4^n \,\Gamma \left(\frac{1}{2}+n\right)}{\sqrt{\pi
} \,\Gamma (1+n)}\right) \\
&=-2 n H_n+\frac{n^2 \pi ^2}{6}+\ln \left(\binom{2 n}{n}\right)
\end{align}$$
where
*
*$\gamma$ is Euler’s constant $=0.577216...$
*$\psi (1+n) $ is the digamma function
*$H_n$ is the $n$th harmonic number
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.
| Any rational function can always be integrated by decomposing it into partial fractions. In this case this decomposition is particularly simple.
\begin{eqnarray}
\frac{1}{x^4-\sqrt{3} x^2+1} &=& \frac{1}{x^2-\theta_+} \cdot \frac{1}{\theta_+-\theta_-} + \frac{1}{x^2-\theta_-} \cdot \frac{1}{\theta_--\theta_+}\\
\frac{1}{x^4+\sqrt{3} x^2+1} &=& \frac{1}{x^2-\lambda_+} \cdot \frac{1}{\lambda_+-\lambda_-} + \frac{1}{x^2-\lambda_-} \cdot \frac{1}{\lambda_--\lambda_+}
\end{eqnarray}
where $\theta_\pm=(\sqrt{3}\pm \imath)/2$ and $\lambda_\pm=(-\sqrt{3}\pm \imath)/2$.
Therefore
\begin{eqnarray}
\int\limits_0^\infty \frac{dx}{x^4-\sqrt{3} x^2+1}&=& \frac{\pi}{\imath} \frac{1}{\sqrt{\frac{-\sqrt{3}-\imath}{2}}}\\
\int\limits_0^\infty \frac{dx}{x^4+\sqrt{3} x^2+1}&=& \frac{\pi}{\imath} \frac{1}{\sqrt{\frac{+\sqrt{3}-\imath}{2}}}
\end{eqnarray}
Now by using simple complex analysis we get the final result:
\begin{equation}
\int\limits_{-\infty}^\infty \frac{(x^4+1)^2}{x^{12}+1} dx = \sqrt{\frac{3}{2}}\pi
\end{equation}
| {
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Rationalize a fraction. Rationalize the denominator $\frac{1}{2^{\frac{1}{3}} + 3^{\frac{1}{3}} + 4^{\frac{1}{3}}}$. Is there a short solution for this task ? Thanks in advance.
| I found the minimal polynomial for $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}$
$x^9-18 x^7-27 x^6+108 x^5+162 x^4-459 x^3-972 x^2-81=0$
The I divided all terms by $81x$ and I got
$$\frac{1}{x}=\frac{x^8}{81}-\frac{2 x^6}{9}-\frac{x^5}{3}+\frac{4 x^4}{3}+2 x^3-\frac{17 x^2}{3}-12 x$$
Then I substituted $x$ with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}$ in the RHS getting after some trivial simplification
$$\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=\frac{1}{9} \left(6 \sqrt[3]{2}-3\sqrt[3]{4}-3\sqrt[3]{12}+\sqrt[3]{9}+2 \sqrt[3]{18} +3 \sqrt[3]{6}-2\sqrt[3]{36}\right)$$
Hope this helps
PS
Without a CAS it is unthinkable to simplify the fraction as I did
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of a triangle inscribed in the ellipse (Unicamp Mathematics Olympiad) Let $\zeta$ denote the Carteasian region given by $\zeta = \{(x,y): \frac{x^2}{4} + \frac{y^2}{9} = 1\}$. Let $A = (0,3), B = (x,y), C = (z,w)$ be points such that $A,B,C \in \zeta%$ and $\Delta ABC$ is equilateral. What is the area of $\Delta ABC$?
My approach: Since $B, C$ are homogeneous, I tried to do enough algebra so everything "cuts out". I used Gauss' formula for the area of a triangle given tree points, the fact that ${AB}^2 = {BC}^2 = {AC}^2$ and used the explicit equation of $\zeta$, but it ended up being too much algebra and I couldn't find an explicit result.
| The points of the ellipse are parameterised by $(2 \cos \theta, 3 \sin \theta)$ & by symmetry it is clear that the points $B$ and $C$ will have coordinates $(2 \cos \theta, 3 \sin \theta)$ & $(-2 \cos \theta, 3 \sin \theta)$. So the requirement of equilateralness gives
\begin{eqnarray*}
(4 \cos \theta)^2= (4 \cos \theta)^2+ (3 (1-\sin \theta))^2 \\
4 \cos^2 \theta= 3 (1-\sin \theta))^2 \\
4 (1+\sin \theta) = 3 (1-\sin \theta) \\
\end{eqnarray*}
The factor $(1-\sin \theta)$ can be discarded as it corresponds to the the points being concurrent. So we have $\sin \theta =- \frac{1}{7}$ and $\cos \theta =- \frac{4\sqrt{3}}{7}$. The length of the sides of the triangle are $\frac{16\sqrt{3}}{7}$ and the area of the triangle is $\color{blue}{\frac{192\sqrt{3}}{49}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e. $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}$$ and conversely $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
To my suspicions of which–either numerator or denominator–conjugate to multiply by I chose $$\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
This resulted in $$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{3-x-1}$$
Is it indeterminate? What is the reason for multiplying by a specific conjugate in a fraction (denominator or numerator) and the reason for the conjugate being either a) denominator b) numerator c) or both?
Am I simply practicing incorrect algebra by rationalizing the expression to:
$$\frac{(6-x)(3-x)-2\sqrt{3}+2x+2}{x-2}$$
Or am I failing to delve further and manipulate the expression out of the indeterminate form?
| By your idea
$$\lim_{x\rightarrow2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(6-x-4)}{(\sqrt{6-x}+2)(3-x-1)}$$
$$=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(2-x)}{(\sqrt{6-x}+2)(2-x)}=\lim_{x\rightarrow2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{2}{4}=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Problems involving LOCUS Okay so I'm really having a hard time finding the locus of curves or point of intersection of curves.
The problem : I am able to get the geometrical condition in most of the questions but then I have no idea what to do to get the locus.
The question :A variable straight line drawn through the point of intersection of the straight line $x+2y-4=0$ and $2x+y-4=0$ meet the coordinate axes in $A$ and $B$.Then what is the locus of midpoint of $AB$ ?
My try : I found the equation of the variable straight line as $3x+3y-8=0$,and the required mid-point in the question to be $(4/3,4/3)$.How do I proceed from here ?
| Let $M(a,b)$ be a point on our locus and $C\left(\frac{4}{3},\frac{4}{3}\right)$.
Thus, $$m_{MC}=\frac{b-\frac{4}{3}}{a-\frac{4}{3}}$$ and we have an equation of $AM$:
$$y-\frac{4}{3}=\frac{b-\frac{4}{3}}{a-\frac{4}{3}}\left(x-\frac{4}{3}\right),$$
which for $x=0$ gives
$$y=\frac{4}{3}-\frac{4}{3}\cdot\frac{b-\frac{4}{3}}{a-\frac{4}{3}}$$ and it gives an equation of the locus:
$$\frac{\frac{4}{3}-\frac{4}{3}\cdot\frac{b-\frac{4}{3}}{a-\frac{4}{3}}+0}{2}=b$$ or
$$b=\frac{2a}{3a-2}$$ or
$$y=\frac{2x}{3x-2}.$$
| {
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.
| It can be done by differentiation but it is much easier to use $x^2 \geq 0$.
Assuming $a,b,c $ are positive
\begin{eqnarray*}
a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0 \\
\end{eqnarray*}
Expand these out
\begin{eqnarray*}
ab(a+b)+bc(b+c)+ca(c+a) \geq 6abc \\
\end{eqnarray*}
Now divide by $abc$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve differential equation $yy'(yy'-2x)=x^2-2y^2$.
$$yy'(yy'-2x)=x^2-2y^2$$
I've tried to divide by $y^2$ and substitude $\frac{x}{y}=z$, but it led to:
$$y'^2-2zy'=z^2-2$$
$$(y'-z)^2=2(z^2-1)$$
$$|y'-z|=\sqrt{2}\sqrt{z^2-1}$$ hence
$y'=\sqrt{2}\sqrt{z^2-1}+z$ or $y'=-\sqrt{2}\sqrt{z^2-1}+z$ and I don't know any way to integrate it. Is there some hint for the very beginning maybe that I don't know?
| With $y^2=u$ then $2yy'=u'$ so
\begin{align}
2yy'(2yy'-4x)&=4x^2-8y^2 \\
u'(u'-4x)+4x^2&=8x^2-8u \\
(u'-2x)^2&=8(x^2-u) \\
-d(x^2-u)&=2\sqrt{2}\sqrt{x^2-u^2}dx \\
\sqrt{x^2-u}&=C-\sqrt{2}x \\
x^2-y^2=&(C-\sqrt{2}x)^2
\end{align}
| {
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Calculate $E(X)$ and the PMF of Number of Flips to Get $2$ Consecutive Heads or Tails with a Coin Flips Consider flipping a fair coin until either two consecutive heads appear or two consecutive tails appear, and let $X$ denote the total number of flips required to obtain the $2nd$ consecutive heads or the $2nd$ consecutive tails.
(a) Give the $pmf$ of $X$.
(b) Give the value of $E(X)$.
Attempted Solution:
(a) We need to have a sequence of tosses of length $n-1$ of the form HTHTHT,..., or THTHTH,..., and the probability that each sequence happens for $n - 1$ tosses is $\frac{1}{2^{n-1}}$. So the total probability of the two sequences is $\frac{2}{2^{n-1}}$. But then the next toss must be the same as the $(n-1)$th toss giving
$$\frac{2}{2^{n-1}}\frac{1}{2} = \frac{1}{2^{n-1}}.$$
I have then that $P(X=n)$ = $\frac{1}{2^{n-1}}$ where $n \geq 2$ so the $pmf$ would be
$$p_X(n) = \begin{cases} {\frac{1}{2^{n-1}}} & \text{$n \geq 2$} \\
{0} & \text{otherwise}\end{cases}$$
(b) Then would the $E(X)$ just be $\sum_2^\infty{\frac{n}{2^{n-1}}} = 3$?
| The Probability Mass Function (PMF)
Let $ X $ be the random variable that counts the number of tosses until we get two successive tosses of the same type.
We're basically dealing with Binary Chains which each chain has the probability of $ \frac{1}{ {2}^{n} } $ where $ n $ is the length of the chain.
The probability $ P \left( X = n \right) $ means having $ n - 1 $ chain of alternating tails and heads (HTHTHTH or THTHTHT for $ n = 7 $). Since there are 2 of those (One starts with H the other with T) their probability is $ \frac{2}{ {2}^{n} } $. The probability to have in the $ n $ -th toss H or T as required is $ \frac{1}{2} $ hence the Probability Mass Function is given by:
$$ {P}_{X} \left( n \right) = \begin{cases}
\frac{1}{ {2}^{n - 1} } & n \geq 2 \\
0 & n \leq 1
\end{cases} $$
The Expected Value
Solution 001
By definition the expected value is given by:
$$ \mathbb{E} \left[ X \right] = \sum_{n = 2}^{\infty} n {P}_{X} \left( n \right) = \sum_{n = 2}^{\infty} \frac{n}{ {2}^{n - 1}} = \sum_{n = 1}^{\infty} \frac{n + 1}{ {2}^{n} } = \sum_{n = 1}^{\infty} \frac{1}{ {2}^{n} } + \sum_{n = 1}^{\infty} \frac{n}{ {2}^{n} } $$
Now, the first term $ \sum_{n = 1}^{\infty} \frac{1}{ {2}^{n} } $ is a geometric series and it is known that $ \sum_{n = 1}^{\infty} \frac{1}{ {2}^{n} } = 1 $.
Looking at the second term:
$$ {S} = \sum_{n = 1}^{\infty} \frac{n}{ {2}^{n} } \Rightarrow \frac{S}{2} = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{n}{ {2}^{n} } = \sum_{n = 1}^{\infty} \frac{n}{ {2}^{n + 1} } = \sum_{n = 1}^{\infty} \frac{n - 1}{ {2}^{n} } $$
Now one could use another trick:
$$ \begin{align*}
\frac{S}{2} & = S - \frac{S}{2} \\
& = \sum_{n = 1}^{\infty} \frac{n}{ {2}^{n} } - \sum_{n = 1}^{\infty} \frac{n - 1}{ {2}^{n} } \\
& = \sum_{n = 1}^{\infty} \left( \frac{n}{ {2}^{n} } - \frac{n - 1}{ {2}^{n} } \right) \\
& = \sum_{n = 1}^{\infty} \frac{1}{ {2}^{n} } = 1 \\
& \Rightarrow S = 2
\end{align*} $$
Which implies $ \mathbb{E} \left[ X \right] = \sum_{n = 1}^{\infty} \frac{1}{ {2}^{n} } + \sum_{n = 1}^{\infty} \frac{n}{ {2}^{n} } = 1 + 2 = 3 $.
Solution 002
By examining the term $ P \left( X \geq n \right) $:
$$ \begin{array} \\
P \left( X \geq 1 \right) & = & P \left( X = 1 \right) & + & P \left( X = 2 \right) & + & P \left( X = 3 \right) + \cdots \\
P \left( X \geq 2 \right) & = & & & P \left( X = 2 \right) & + & P \left( X = 3 \right) + \cdots \\
P \left( X \geq 3 \right) & = & & & & & P \left( X = 3 \right) + \cdots \\
\end{array} $$
One could see (Summing along the columns) that:
$$ \sum_{n = 1}^{\infty} P \left( X \geq n \right) = 1 P \left( X = 1 \right) + 2 P \left( X = 2 \right) + 3 P \left( X = 3 \right) + \cdots + n P \left( X = n \right) = \mathbb{E} \left[ X \right] $$
The term $ P \left( X \geq n \right) $ is the event of having $ n - 1 $ alternating chain as above. Namely:
$$ P \left( X \geq n \right) = \begin{cases}
\frac{1}{ {2}^{n - 2} } & n \geq 2 \\
1 & n \leq 1
\end{cases} $$
This implies:
$$ \mathbb{E} \left[ X \right] = \sum_{n = 1}^{\infty} P \left( X \geq n \right) = 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = 3 $$
| {
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Binomial Coefficient Question In the expression
$((. . . ((x − 2)^2 − 2)^2 − 2)^2 − · · · − 2)^2 − 2)^2$
,
there are $k$ pairs of parentheses, where k is a positive integer. In terms of $k$, find the coefficient of $x^2$
after expanding and collecting the terms
So far I have attempted to look for a recursion which would allow me to solve the problem. However, I cannot seem to find a relationship between the input of $k$ and the coefficient of the $x^2$ term. Can anybody help?
| Hint. Let $F_k(x)$ the the function at the $k$-th step, then
$$F_k(x)=a_kx^2+b_kx+4+o(x^2)$$
Then $F_1=x^2-4x+4$ implies that $a_1=1$ and $b_1=-4$.
Moreover
\begin{align*}a_{k+1}x^2+b_{k+1}x+4+o(x^2)&=F_{k+1}(x)=(F_k(x)-2)^2\\
&=(a_kx^2+b_kx+4+o(x^2)-2)^2\\
&=(4a_k+b_k^2)x^2+4b_kx+4+o(x^2).
\end{align*}
Therefore $b_{k+1}=4b_k=-4^{k+1}$ and
$$a_{k+1}=4a_k+b_k^2=4a_k+16^k.$$
It remains to solve the linear recurrence for $a_k$.
| {
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Finding the probability that a team roster contains Ted and Vanessa A school Science Fair team will consist of 2 male students, two female students and one alternate. The alternate can be of either sex. There are $4$ males (including Ted) and $5$ females (including Vanessa) trying out for the team. A team roster is a list of five students chosen without mention of which one of the five will be the alternate. If the five students are chosen randomly from among the nine trying out, find the probability that the team roster contains Ted and Vanessa.
I tried solving it by finding the situations where we pick 2 males out of 3 and 1 female out of 4 and 1 male out of 3 and 2 females out of 4 to find the possible outcomes and divided them with the total number of outcomes.
So i got $(3*4+3*6)/300$ which seems to be wrong as the answer is 3/10
| A team roster consists of three males and two females or two males and three females. Since there are four male and five female students available, the number of possible team rosters is
$$\binom{4}{3}\binom{5}{2} + \binom{4}{2}\binom{5}{3}$$
As you observed, if both Ted and Vanessa are selected, a team with three males and two females can be obtained by selecting two of the other three males and one of the other four females, and a team with two males and three females can be obtained by selecting one of the other three males and two of the other four females. Hence, the number of favorable cases is
$$\binom{3}{2}\binom{4}{1} + \binom{3}{1}\binom{4}{2}$$
Therefore, the desired probability is
$$\frac{\dbinom{3}{2}\dbinom{4}{1} + \dbinom{3}{1}\dbinom{4}{2}}{\dbinom{4}{3}\dbinom{5}{2} + \dbinom{4}{2}\dbinom{5}{3}}$$
| {
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.
| But it is not true, say $n=2$. :(
If $n$ is odd then $2^n+1 = (2+1)(2^{n-1}-...+1) =3k$, then is true.
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How to find $\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}$? How to find
$$
\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}?
$$
My try : $$(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=(x-1)$$
So we have :
$$\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1} \cdot\frac{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}=\lim_{x \to 1}\frac{(x+x^2+\sqrt{x}-3)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x-1}$$
Now what ?
| Setting $$t=\sqrt[6]{x}$$ then we have $$t^6=x$$ and we get
$$\frac{t^{12}+t^6+t^3-3}{t^2-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A fencing area question. A farmer creates a rectangular pen by using one side of a barn as one side of the pen and using fencing for the other three sides. The farmer has $80~\text{ft}$ of fencing, and the side of the barn is $40~\text{ft}$ long. If $x$ represents the length of the fenced side of the pen that is parallel to the barn, then the length of each of the two fences sides of the pen that are perpendicular to the barn is $40 -.5x~\text{ft}$. For what values of $x$ is the area of the pen at least $600~\text{ft}$?
I know that they got $40-.5x$ by $x+y+y=80$. Thus, $x+2y=80$ and $y=40-5.x$
Area is then equal to $lw$, where $A=(40-.5x)(x)$?
Would it be times $40-2x$ since fence cannot be longer than barn at that side? or $40-x$?
How do I proceed to find area at least $600~\text{ft}$?
| You correctly found that
$$A(x) = x\left(40 - \frac{1}{2}x\right)$$
Since we require that the area enclosed by the fence is at least $600~\text{ft}^2$ (you omitted the exponent),
\begin{align*}
x\left(40 - \frac{1}{2}x\right) & \geq 600\\
40x - \frac{1}{2}x^2 & \geq 600\\
x^2 - 80x & \leq -1200\\
x^2 - 80x + 1600 & \leq 400\\
(x - 40)^2 & \leq 400\\
|x - 40| & \leq 20
\end{align*}
Thus, $x - 40 \leq 20$ and $x - 40 \geq -20$.
\begin{align*}
x - 40 & \leq 20 & x - 40 & \geq -20\\
x & \leq 60 & x & \geq 20
\end{align*}
from which we conclude that $20 \leq x \leq 60$. That suggests that the side of the fence parallel to the barn should be between $20~\text{ft}$ and $60~\text{ft}$. However, if the side parallel to the barn exceeds $40~\text{ft}$, the pen cannot be closed, so the side parallel to the barn should be at least $20~\text{ft}$ and at most $40~\text{ft}$ long.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+z^4$ and the last equation is equivalent to the system of equations $$\begin{cases}
2x=4\lambda x^3 \\
2y=4\lambda y^3 \\
2z=4\lambda z^3
\end{cases}$$
After dividing into $2$ and multiplying to $x,y$ and $z$, respectively we get: $$\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}$$
Considerong the first equation we get two cases: $x=0$ or $1-2\lambda x^2=0$
After that I am stuck. How to rule out or consider each case?
Can anyone demonstrate it clearly?
Would be very thankful for help
| You have
\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}
Case 1: Assume $xyz\ne 0.$ Then $x^2=y^2=z^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{3}{4\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Case 2: Assume $z=0,xy\ne 0.$ Then $x^2=y^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{1}{2\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Case 3: Assume $y=z=0,x\ne 0.$ Then $x^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{1}{4\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Note that $x=y=z=0$ is not possible and that, because of symmetry, case $2$ cover all possibilities with one variable zero and case $3$ cover all possibilities with two variables zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac{(x+1)^2}{x-1}=x-1.$$ Which is far from the correct answer. I can't spot my mistake. I feel that the more I sit and study, the worse at math I become.
| $x-1$ doesn't divide $(x+1)^2$ exactly, so there must also be a remainder term of lower degree than $x-1$, i.e.
$$ \frac{(x+1)^2}{x-1} = ax+b + \frac{c}{x-1}. $$
Then you find
$$ x^2+2x+1 = ax^2+(b-a)x+(c-b), $$
giving three equations in three unknowns (you expect a unique solution anyway, so you should have the same number of equations as unknowns).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Proving properties of rings None of these are complete and I am looking for some guidance.
*
*In any ring if $ab=-ba$ then $(a+b)^2=(a-b)^2=a^2+b^2$.
So assuming $ab=-ba$ to be true, then squaring both sides gives $(ab)^2=(-ba)^2$, i.e. $abab=-ba-ba$. Not all rings are commutative under multiplication so I don't know if I can assume that $abab =aabb=a^2b^2$.
*In any integral domain, if $a^2=b^2$ then $a=\pm b$
I'm not positive on how the integral domain plays a part in this one? Something to do with not having zero divisors I'm sure, possibly that no two elements multiplied together will give zero unless one of the elements is zero? So we can just take the square root
*In any integral domain, only 1 and -1 are their own multiplicative inverses.
We must have a commutative ring with unity, thus we have a multiplicative inverse. Can we assume that 1 and -1 are the inverses?
*In any integral domain, if $a^n=0$ for some integer $n$ then $a=0$.
Just thinking, the only time anything to a power is zero is when $0^n$ where $n\neq 0$, thus $a$ must be 0.
| 1:
$(a + b)^2 = a^2 + ab + ba + b^2 = a^2 + ab - ab + b^2 = a^2 + b^2; \tag 1$
$(a - b)^2 = a^2 -ab - ba + b^2 = a^2 - ab + ab + b^2 = a^2 + b^2; \tag 2$
2:
$(a - b)(a + b) = a^2 - ab + ab - b^2 = a^2 - b^2 = 0, \tag 3$
which works since integral domains are commutative. Now if
$a \ne b, \tag 4$
then
$a - b \ne 0, \tag 5$
forcing
$a + b = 0 \tag 6$
or
$a = -b; \tag 7$
so
$a = \pm b. \tag 8$
3:
If
$a = a^{-1}, \tag 9$
then
$a^2 = 1, \tag{10}$
or
$(a - 1)(a + 1) = 0; \tag{11}$
now in an integral domain, if $a - 1 \ne 0$, (11) implies
$a = -1; \tag{12}$
so
$a = \pm 1. \tag{13}$
4:
Finally, if $a \ne 0$,
$a^n = 0 \Longrightarrow a(a^{n - 1}) = 0 \Longrightarrow a^{n - 1} = 0; \tag{14}$
a simple inductive argument now allows us to conclude that
$a = 0; \tag{15}$
i.e., if for some $k \in \Bbb N$,
$a^k = 0 \Longrightarrow a = 0, \tag{16}$
then
$a^{k + 1} = 0 \Longrightarrow a(a^k) = 0 \Longrightarrow a^k = 0 \Longrightarrow a = 0. \tag{17}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Combination of $n$ letter out of $3n$.
Show that the number of combinations of $n$ letters out of $3n$ letters of which $n$ are $a$s , $n$ are $b$s and the rest are unequal is $(n+2)2^{n-1}$.
My approach...
The number of ways of distributing $n$ distict object in $k$ distinct box is $\binom{n+k-1}{k-1}$ now when $k=2$ the logic stands as $\binom{n+2-1}{2-1}$=$\binom{n+1}{1}$.
Selection of $j$ distinct letter out of $n$ is $\binom{n}{j}$.
Hence my answer will be $\sum_{j=0}^{n} \binom{j+1}{1}\binom{n}{n-j}$.
Please help me with the answer
| Hint: The required number is coefficient of $x^n$ in $$[(1+x+x^2+\cdots x^n)^2(1+x)^n].$$
Do some simplification and find that the coefficient of $x^n$ is $$2^n\binom {n+2-1}{1} - \binom n1 2^{n-1} \binom{n+1-1}{0} = (n+2)2^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Polynomial sum: $x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$ Find sum of the expression,
$$x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$$
where $x,y$ are real numbers and $n$ is a natural number.
| It is:
$$\frac{x^{n+1}-y^{n+1}}{x-y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For which value of $k$ is the following matrix diagonalizable? \begin{pmatrix}7&k\\ \:0&7\end{pmatrix}
I could not figure out how to derive the eigenvalues and eigenvectors of the matrix above because of the letter $k$. How am I supposed to deal with a value in terms of $k$? And how would I be able to find out if the matrix is diagonalizable or not through that?
| When $k \neq 0:$
A column vector that is not sent to zero by
$$
\left(
\begin{array}{cc}
0 & k \\
0 & 0
\end{array}
\right)
$$
is
$$
\left(
\begin{array}{c}
0 \\
1
\end{array}
\right),
$$
and
$$
\left(
\begin{array}{cc}
0 & k \\
0 & 0
\end{array}
\right)
\left(
\begin{array}{c}
0 \\
1
\end{array}
\right) =
\left(
\begin{array}{c}
k \\
0
\end{array}
\right)
$$
Putting the columns in reverse order, we get
$$
P =
\left(
\begin{array}{cc}
k & 0 \\
0 & 1
\end{array}
\right)
$$
with
$$
P^{-1} =
\left(
\begin{array}{cc}
\frac{1}{k} & 0 \\
0 & 1
\end{array}
\right)
$$
after which
$$
\left(
\begin{array}{cc}
\frac{1}{k} & 0 \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{cc}
7 & k \\
0 & 7
\end{array}
\right)
\left(
\begin{array}{cc}
k & 0 \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{cc}
7 & 1 \\
0 & 7
\end{array}
\right)
$$
The point being that, as soon as $k \neq 0,$ the specific value of $k$ is not that important. The last matrix is the Jordan form of your original.
| {
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"url": "https://math.stackexchange.com/questions/2490001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\frac{6n^2+3n}{2n^2-5}$ converges to 3. Prove the following:
$$\lim_{n\to \infty}\frac{6n^2+3n}{2n^2-5}=3$$
Here's my solution:
Consider the function $\frac{6n^2+3n}{2n^2-5}$.
Then $$|\frac{6n^2+3n}{2n^2-5}-3|= |\frac{6n^2+3n}{2n^2-5}-\frac{6n^2+15}{2n^2-5}|=|\frac{3n-15}{2n^2-5}|$$
The upper bound for the numerator is $3n-15 \le 3n$ for all $n$ and the lower bound for the denominator is $2n^2-5 \le 4n^2$ if $n \le 2$.
It follows that
$|\frac{3n-15}{2n^2-5}| \le \frac{3n}{4n^2}$. Meaning $|\frac{3n-15}{2n^2-5}| \le \frac{3}{4n}$.
By the Archimedean Property, $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$
Therefore $(\frac{6n^2+3n}{2n^2-5})\rightarrow 3$.
$\blacksquare$
I feel like I'm missing a step after showing that $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$.
Can $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$ be simplified more so that it resembles $\frac{1}{n} \le \frac{1}{N} \lt \epsilon$ or is it fine how it is?
Please advise.
| To get an upper bound, you want to replace the numerator by something big and the denominator by something small. So instead of $2n^2-5\lt4n^2$, you should use $2n^2-5\gt2n^2$ (or even just $n^2$). Aside from that, everything is basically OK.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers. Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example. Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could reduce better to show there is a factor of $n+1$ left.
Examples show this is true for small n; for example
$$\binom{9}{6}=\binom{3\cdot 3}{2\cdot 3}=\frac{9\cdot8\cdot7}{3\cdot 2\cdot 1}=4(3\cdot 7)$$
$$\binom{16}{8}=\binom{4\cdot 4}{2\cdot 4}=\frac{16\cdot15\cdot...\cdot 10\cdot 9}{8\cdot 7\cdot...\cdot2\cdot 1}=5(2\cdot 3^2\cdot11\cdot 13)$$
| Here is a larger counterexample : $a=6,b=2,n=4$, then $n+1$ is not a multiple of $\binom{an}{bn}$.
Another one is given by $a=5,b=2,n=5$.
Also, note that $\binom{an}{bn} = \binom{an}{(a-b)n}$, therefore replacing $b$ above with $a-b$ would also work.
I am still thinking about $b > 2$(other than taking $b \to a-b$ obviously) though. I have not been able to find an example yet.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to find ord$[a]_m$ What are the steps to find minimum $k$ such that $\left([a]_m\right)^k=[1]_m$? In short, how to find ord$[a]_m?$
Specifically I need to find mod$\left( 173^{607},1147\right)$ but since $\varphi(1147)=1080>607$ I can't use Euler-Fermat's Th. on this one, and I don't feel like square-and-multiplying.
| Since $1147=31\cdot 37$, by Chinese Remainder theorem, you have a ring isomorphim
$$\psi:\mathbb Z/1147\mathbb Z\longrightarrow\mathbb Z/31\mathbb Z\times\mathbb Z/37\mathbb Z$$
Thus
$$\psi(173^{607}+1147\mathbb Z)=(173^{607}+31\mathbb Z,173^{607}+37\mathbb Z)$$
But
\begin{align*}
173^{607}
&\equiv 18^7\\
&=2^7\cdot 3^{14}\\
&\equiv 2^2\cdot (3^3)^4\cdot 3^2\\
&\equiv 2^2\cdot (-4)^4\cdot 3^2\\
&\equiv 2^{10}\cdot 3^2\\
&\equiv 9\pmod{31}
\end{align*}
because $2^5\equiv 1\pmod{31}$.
On the other hand:
\begin{align*}
173^{607}
&\equiv 25^{31}\\
&\equiv 3^{-31}\\
&\equiv 3^5\\
&\equiv 21\pmod{37}
\end{align*}
Consequently, $$173^{607}\equiv 9\cdot 37\cdot (-5)+21\cdot 31\cdot 6\pmod{1147}$$ because $-5\equiv 6^{-1}\equiv 37^{-1}\pmod{31}$ and $6\equiv (-6)^{-1}\equiv 31^{-1}\pmod{37}$.
Finally we get $173^{607}\equiv -45\cdot 37+126\cdot 31\equiv 1094\pmod{1147}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this identity of summation of ramanujan? In the first book of ramanujan him find this identity:
Let k , n $\in\mathbb{N}$* and define $A_k=3^k\left(n+\frac{1}{2}\right)-\frac{1}{2}$. the if r is a positive interger
$$\sum_{k=n+1}^{A_r}\frac{1}{k}=r+2\sum_{k=0}^{r-1}\left(r-k\right)\sum_{j=A_{k-1}+1}^{A_k}\frac{1}{\left(3j\right)^3-3j}$$
where $A_{-1}=0$
I want to know a demonstration for that identity, because he uses it to demonstrate his very popular affirmation:
$$\sum_{n=1}^{1000}\frac{1}{n}=7\frac{1}{2} "very\ nearly"$$
| First we need the result from entry $2$ of the reference. Lets start with the parial fractions
\begin{eqnarray*}
\frac{1}{(3j)^3-3j} =\frac{1/2}{3j-1}-\frac{1}{3j}+\frac{1/2}{3j+1}.
\end{eqnarray*}
So
\begin{eqnarray*}
1+2 \sum_{j=1}^{n} \frac{1}{(3j)^3-3j} =1+ \sum_{j=1}^{n} \left( \frac{1}{3j-1}\color{red}{+\frac{1}{3j}}+\frac{1}{3j+1} \color{red}{-\frac{3}{3j}} \right) \\
= \sum_{j=1}^{3n+1} \frac{1}{j} -\sum_{j=1}^{n} \frac{1}{j} = \sum_{j=n+1}^{3n+1} \frac{1}{j} .
\end{eqnarray*}
Now with $A_k=3^k n + \frac{1}{2} (3^k-1)$ then $A_{k+1}=3A_k+1$, the above result gives
\begin{eqnarray*}
\sum_{j=A_k +1}^{A_{k+1}=3A_k+1} \frac{1}{j} =1+ 2 \sum_{j=1}^{A_k} \frac{1}{(3j)^3-3j}.
\end{eqnarray*}
Now sum the above formula from $k=0,1,\cdots, r-1$ and we have
\begin{eqnarray*}
\sum_{j=n +1}^{A_{r}} \frac{1}{j} =r+ 2 \sum_{k=0}^{r-1} \sum_{j=1}^{A_k} \frac{1}{(3j)^3-3j}.
\end{eqnarray*}
Note that the inner sums on the RHS are contained within next sum, so this can be rearranged to the result
\begin{eqnarray*}
\sum_{j=n +1}^{A_{r}} \frac{1}{j} =r+ 2 \sum_{k=0}^{r-1} (r-k) \sum_{j=A_{k-1}+1}^{A_k} \frac{1}{(3j)^3-3j}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $ Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $
Show that$ a \equiv 1 \pmod{2^4 } \Rightarrow a^{2^{4-2}} \equiv 1 \pmod{2^4} $
Answer:
$ a \equiv 1 \pmod{2^3} \\ \Rightarrow a^2 \equiv 1 \pmod{2^3} \\ \Rightarrow a^{2^{3-2}}=a^{2^1} \equiv 1 \pmod{2^3} $
Am I right?
| That looks fine!
To generalise a bit, if $a \equiv1(\mathrm{mod}n)$, you see that $a^k \equiv 1(\mathrm{mod}n)$, and so $a^{k^l}\equiv 1(\mathrm{mod}n)$ whatever $k$ and $l$ are :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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let $f(x) = \left\lbrace\begin{array}{ll} x^2+3x, &x \geq1\\ x^2-3x+6, & x<1 \end{array}\right.$ then find the limit… Let
$f(x) =
\begin{cases}
x^2+3x, &x \geq1\\[2ex]
x^2-3x+6, & x<1
\end{cases}$
Then find the
$$\lim_{h \to0} \frac{f(1)-f(1-h^2)}{h^2}=?$$
My Try :
$$f(1)=4$$
$$f(1-h^2)=(1-h^2)^2-3(1-h^2)+6=h^4-h^2+4=h^2(h^2-1)+4$$
So we have :
$$\lim_{h \to0} \frac{h^2(h^2-1)}{h^2}=-1$$
it is right ?
| It should be observed that the limit in question is left hand derivative of $f$ at $1$ and hence it is equal to $2x-3$ evaluated at $x=1$. Thus the desired limit is $-1$. It should be also observed that the above works because $f$ is continuous at $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to differentiate $\sqrt[5]{1/x}$ Hey guys I could use some help with this derivative:
$$\sqrt[5]{1/x}$$
This is what I have so far:
$$=-\dfrac{1}{5}\left(\dfrac{1}{x}\right)^{-6/5}$$
Having trouble simplifying this to my given solution so I don't know if it is correct.
Given solution:
$$-\frac{1}{5x\sqrt[5]{x}}$$
| Here's my take on it with each step spelled out as clearly as possible:
$$
\left(\sqrt[5]{\frac{1}{x}}\right)'=
\left(x^{-\frac{1}{5}}\right)'=
-\frac{1}{5}x^{\left(-\frac{1}{5}-1\right)}=
-\frac{1}{5}x^{\left(-\frac{1}{5}-\frac{5}{5}\right)}=
-\frac{1}{5}x^{-\frac{6}{5}}=\\
-\frac{1}{5\sqrt[5]{x^6}}=
-\frac{1}{5\sqrt[5]{x^{5+1}}}=-\frac{1}{5\sqrt[5]{x^5\cdot x}}=
-\frac{1}{5\sqrt[5]{x^5}\cdot\sqrt[5]{x}}=
-\frac{1}{5x\sqrt[5]{x}}
$$
As you can see, it matches the provided solution on the nose.
The tricky algebra below is probably what threw you off:
$$
a^{-n}=\frac{1}{a^n}\\
a^{\frac{n+m}{n}}=\sqrt[n]{a^{n+m}}=\sqrt[n]{a^n\cdot a^m}=
\sqrt[n]{a^n}\cdot\sqrt[n]{a^m}=a\sqrt[n]{a^m}
$$
| {
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Proof by Contradiction: let a,b $\in$Z . Prove that if 3 $\nmid$ a and 3 $\nmid$ b then $3 | a^2 - b^2$ Just wondering if this works for this question, book had a different answer and I couldn't find another answer for the question.
Assume, to the contrary, that 3 | a and 3 | b, then a = 3k, and b = 3x for x,k $\in$ Z, then $a^2 - b^2$ = $(3k)^2 - (3x)^2$ which is equal to $3(3k^2 - 3x^2)$, since $3k^2 - 3x^2$ is an integer, 3 | $a^2 - b^2$ which is a contradiction.
Is this correct?
| Proof by contradiction:
Suppose 3 does not divide $a^2-b^2$ the we must have:
$ 3 ł (a-b)(a+b)$
Suppose:
$a+b=3k_1 + r_1; r_1<3 ⇒ r_1 =1 , 2$
$a-b=3k_2 +r_2; r_2<3 ⇒ r_2=1, 2$
$2a=3(k_1 +k_2) + r_1 +r_2$
$2b=3(k_1-k_2) +r_1 -r_2$
1): $r_1 =1 $ and $r_2 =1$ ⇒ $2b=3(k_1-k_2) +1 - 1=3(k_1-k_2)$ ; cntradicts 3łb
2): $r_1 =1 $ and $r_2 =2$ ⇒ $2a=3(k_1 +k_2) + 1 +2=3(k_1 +k_2+1)$; contradicts 3ła
3): $r_1 =2 $ and $r_2 =2$ ⇒ $2b=3(k_1 -k_2) + 2 -2=3(k_1 -k_2)$; contradicts 3łb
4): $r_1 =2 $ and $r_1 =1$ ⇒ $2a=3(k_1 +k_2) + 2 +1=3(k_1 +k_2+1)$; contradicts 3ła
That is 3 must divide $a^2 - b^2$.
| {
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Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program. I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.
| Side comment: $\frac{\sqrt{m-1} + \sqrt{m+1}}2 \ge \sqrt{\sqrt{m^2 - 1}}>\sqrt m$ by $AM-GM$ theorem.
So $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} \approx 3\sqrt{x+1}$ but slightly larger.
$x \le \sqrt{x} + \sqrt{x+1} + \sqrt{x+2}< 3\sqrt{x+1} < x+1$ implies $x^2 < 9x + 9 < x^2 + 2x +1$ (the first inequality is definite, the second is aproximate).
$x^2 <9x+9$ (which is definitely true) implies $x^2 - 9x -9 = (x -\frac {9+\sqrt{81 + 4*9}}2)(x -\frac {9-\sqrt{81 + 4*9}}2) < 0$ which implies $x < \frac {9+\sqrt{81 + 4*9}}2= \frac {9+3\sqrt{13}}2\approx 9.9$ but is *definitely less than $10$. $9$ is a definite upper limit of $x$
$9x + 9 < x^2 + 2x + 1$ (which is only approximate) implies $x^2 - 7x -8 = (x-8)(x+1) > 0$ implies $x > 8$ (or $x < 1$ which would not be possible.)
So $9$ is a solution and $8$ is possible if $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} < 9 \le 3\sqrt{x+1}$ ... which, ... actually is the case for $x= 8$. (As $3\sqrt{8+1} = 9$ exactly.)
So solutions are $8$ and $9$.
Hmmm.... I haven't actually shown it fails for $x < 8$ but for that to hold the difference between $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$ and $3\sqrt{x+1}$ must be larger than $1$ and that's .... well, clearly not true for $x \approx 8$. We can just check that $7$ fails.
I'm also assuming the rate of which $x$ increases will "outstrip" $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$. Calculus verifies this.
| {
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Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such that $\binom{k}{0}=1^2$ etc. but that didn't work. Does anybody have a push in the right direction? Thanks!
| (Not using the combinatorial identity but without induction)
You can start saying that $$(k+1)^{3}-k^3=3k^2+3k+1$$
then you take the sum in both sides to get:
$$\sum_{k=1}^{n}\bigg((k+1)^{3}-k^3\bigg)=\sum_{k=1}^{n}\big(3k^2+3k+1\big)$$
The LHS is a telescopic sum and RHS will give you the sum you are looking for:
$$ (n+1)^3-1^3=\sum_{k=1}^{n}\bigg((k+1)^{3}-k^3\bigg)=3\sum_{k=1}^{n}k^2+3\sum_{k=1}^{n}k+\sum_{k=1}^{n}1
=3\sum_{k=1}^{n}k^2+3\frac{n(n+1)}{2}+n$$
and this implies:
$$ \frac{(n+1)^3-1-n-3\frac{n(n+1)}{2}}{3}=\sum_{k=1}^{n}k^2 $$
You can compute $\sum_{k=1}^{n}k$ without induction by saying that $$(k+1)^{2}-k^{2}=2k+1$$ and repeat the last argument.
| {
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Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$) Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$).
I know by Fermatt's Little Theorem, $x^{58} \equiv1$ (mod 59), and I know that $x^{58} = (x^{11})^5(x^3)$, but I dont know where to go from here. What do I do next?
| As you figured out
$$(x^{11})^5 \equiv (21)^5 \equiv 3 \pmod{59} \Rightarrow (x^{11})^5x^3 \equiv 3x^3 \pmod{59} \Rightarrow \\
1 \equiv 3x^3 \pmod{59} \Rightarrow 1 \equiv 3^3x^9 \pmod{59}\Rightarrow \\
35 \equiv (35\cdot 27) \cdot x^9 \equiv x^9 \pmod{59}$$
since
$$35\cdot 27 \equiv 1 \pmod{59}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determining the limit of a function containing $(x-1)^5$ without l'hopital`s rule. Find the limit of $$\begin{equation*}
\lim_{x \rightarrow 0}
\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}
\end{equation*}$$
Shall I use the binomial theorem?
Any hint will be appreciated!
| Ok, now that OP's conditions are better known, let's try again.
Numerator is
$$(x-1)^5+(1+5x)\equiv x^5-5x^4+10x^3-10x^2+5x-1+(1+5x)$$$$\equiv x^5-5x^4+10x^3-10x^2+10x$$
so the given expression is
$$\frac{x^5-5x^4+10x^3-10x^2+10x}{x^5+x^2}\equiv\frac{x^4-5x^3+10x^2-10x+10}{x^4+x}.$$
Now, numerator does not approach $0$ as $x\to 0$ but denominator does, so the limit does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing: $\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx$ I would like to compute the exact value of the integral below.
$$\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx$$
I have proved the convergence already. but failed to the residues theorem in other to get the exact value. It will be great if somebody could provide with some hint.
| In fact
\begin{eqnarray}
&&\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx\\
&=&\int_0^1\sin(x^2)dx+\int_0^1\frac{\sin(\frac{1}{x^2})}{x^2}dx.
\end{eqnarray}
For the second integral, under $x\to\frac1x$, one has
\begin{eqnarray}
&&\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx\\
&=&\int_0^1\sin(x^2)dx+\int_0^1\frac{\sin(\frac{1}{x^2})}{x^2}dx\\
&=&\int_0^1\sin(x^2)dx+\int_1^\infty\sin(x^2)dx\\
&=&\int_0^\infty\sin(x^2)dx\\
&=&\frac{\sqrt{2\pi}}{4}.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate?
| May be, you could make life a bit simpler using $x=y+a$ which makes
$$\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}=\frac{\sqrt{a+y}-\sqrt{a}+\sqrt{y}}{\sqrt{y (2 a+y)}}$$ Now, use the generalized binomial theorem or Taylor series around $y=0$ to get
$$\sqrt{a+y}=\sqrt{a}+\frac{y}{2 \sqrt{a}}+O\left(y^2\right)$$
$${\sqrt{y (2 a+y)}}= \sqrt{2a} \sqrt{y}+O\left(y^{3/2}\right)$$ which make finally
$$\frac{\sqrt{a+y}-\sqrt{a}+\sqrt{y}}{\sqrt{y (2 a+y)}}=\frac{1}{\sqrt{2a} }+\frac{\sqrt{y}}{2 \sqrt{2}\,a}+O\left(y^1\right)$$ which shows the limit and also how it is approached when $y\to 0$.
| {
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Show that $(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx=G+\sum_{k=1}^{n}{(-1)^k\over (2k-1)^2}$ How do we show that $(1)$
$$(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx=G+\sum_{k=1}^{n}{(-1)^k\over (2k-1)^2}?\tag1$$
$n\ge0$ and G is the Catalan's constant
$u=\tan{x}$ then $\mathrm \cos^2{x}du=\mathrm dx$ [0,1]
$$-\int_{0}^{1}(u)^{2n}\ln(u)\cos^2{x}\mathrm du\tag2$$
$$-\int_{0}^{1}u^{2n}\ln(u){\mathrm du\over 1+u^2}\tag3$$
$$-\int_{0}^{1}\ln(u){\mathrm du\over 1+u^2}=G\tag4$$
| This is a partial solution. I can take the integral as far as expressing it in terms of polygamma functions of order one (the trigamma function) but are currently unable to simplify it further into the form the OP gives in terms of Catalan's constant $G$ and a finite sum term.
Let
$$I_n = \int^{\pi/4}_0 \tan^{2n} x \ln (\cot x) \, dx,$$
and setting $u = \tan x$ the integral can be rewritten as
$$I_n = -\int^1_0 \frac{u^{2n}}{1 + u^2} \ln u \, du.$$
Writing the term $1/(1+u^2)$ that appears in the integrand as a geometric series, namely
$$\frac{1}{1+u^2} = \sum^\infty_{k = 0} (−1)^k u^{2k},\quad |u|<1,$$
after interchanging the summation with the integration (by Tonelli's theorem) we have
$$I_n = -\sum^\infty_{k = 0} (-1)^k \int^1_0 u^{2k + 2n} \ln u \, du.$$
After integrating by parts, we find
$$I_n = \sum^\infty_{k = 0} \frac{(-1)^k}{(2k + 2n + 1)^2}.$$
To evaluate the sum we write
\begin{align*}
I_n &= \sum^\infty_{k = 0} \frac{(-1)^k}{(2k + 2n + 1)^2} = \frac{1}{4} \sum^\infty_{k = 0} \frac{(-1)^k}{(k + n + \frac{1}{2})}\\
&= \frac{1}{4} \sum^\infty_{k = 0, k \in \text{even}} \frac{1}{(k + n + \frac{1}{2})^2} - \frac{1}{4} \sum^\infty_{k = 0, k \in \text{odd}} \frac{1}{(k + n + \frac{1}{2})^2}.
\end{align*}
Reindexing, in the first sum let $k \mapsto 2k$ while in the second sum let $k \mapsto 2k + 1$. Thus
\begin{align*}
I_n &= \frac{1}{4} \sum^\infty_{k = 0} \frac{1}{(2k + n + \frac{1}{2})^2} - \frac{1}{4} \sum^\infty_{k = 0} \frac{1}{(2k + n + \frac{3}{2})^2}\\
&= \frac{1}{16} \sum^\infty_{k = 0} \frac{1}{(k + \frac{n}{2} + \frac{1}{4})^2} - \frac{1}{16} \sum^\infty_{k = 0} \frac{1}{(k + \frac{n}{2} + \frac{3}{4})^2}\\
&= \frac{1}{16} \left [\zeta \left (2, \frac{2n + 1}{4} \right ) - \zeta \left (2, \frac{2n + 3}{4} \right ) \right ],
\end{align*}
where $\zeta (s,z)$ is the Hurwitz zeta function.
Now the polygamma function $\psi^{(m)}(z)$ is related to the Hurwitz zeta function by
$$\zeta (1 + m, z) = \frac{(-1)^{m + 1}}{m!} \psi^{(m)}(z),$$
so in terms of polygamma (trigamma) functions our integral can be expressed as
$$I_n = \frac{1}{16} \left [\psi^{(1)} \left (\frac{2n + 1}{4} \right ) - \psi^{(1)} \left (\frac{2n + 3}{4} \right ) \right ].$$
It now remains to prove that
$$\psi^{(1)} \left (\frac{2n + 1}{4} \right ) - \psi^{(1)} \left (\frac{2n + 3}{4} \right ) = (-1)^n 16 \left [G + \sum^n_{k = 1} \frac{(-1)^k}{(2k - 1)^2} \right ].$$
| {
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Proof by induction that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$ Through Induction I tried to prove that:
$$\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$$
I first let $n=1$, so that on the left hand side and the right hand side you get 1.
Then I tried to prove that this also works when $$n=n+1$$
So the equation then becomes:
$$\sum_{k=1}^{n+1} (-1)^{k-1}k^2 = \sum_{k=1}^n (-1)^{k-1}k^2 +(-1)^{n-1}\frac{(n)(n+1)}{2}$$
As we know that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$, we can replace the first term after the $=$ with this. The equation becomes:
$$=(-1)^{n-1}\frac{(n)(n+1)}{2} + (-1)^{n-1}\frac{(n)(n+1)}{2}$$
The I subsitute $n+1$ for $n$, so that:
$$=(-1)^{n+1-1}\frac{(n+1)(n+1+1)}{2} + (-1)^{n+1-1}\frac{(n+1)(n+1+1)}{2}$$
$$=(-1)^{n}\frac{(n+1)(n+2)}{2} + (-1)^{n}\frac{(n+1)(n+2)}{2}$$
$$=(-1)^{n}[\frac{(n+1)(n+2)}{2}+\frac{(n+1)(n+2)}{2}]$$
$$=(-1)^{n}[\frac{(2n^2+6n+4)}{2}]$$
$$=(-1)^{n}[\frac{(2n^2+2n+4n+4)}{2}]$$
$$=(-1)^{n}[\frac{((n+1)(2n+4))}{2}]$$
But how do I proceed from here?
| i think it must be
$$\sum_{k=1}^{n+1}(-1)^{k-1}k^2=\sum_{k=1}^n(-1)^{k-1}k^2+(-1)^n(n+1)^2$$
and we get $$(-1)^{n-1} \frac {n(n+1)}{2}+(-1)^n(n+1)^2=(-1)^n\frac{(n+1)(n+2)}{2}$$
| {
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By using Wallis’s product, prove $\zeta(2)=\frac{\pi^2}{6}$ $\sin x = x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\cdots$
This equal is used in both the solution of Basel’s problem and the proof of Wallis’s product.
So I wonder if I can prove one by using the other.
Help me solve this problem.
Or if you are sure this assumption is wrong, tell me the reason.
I’m sorry if I put wrong tugs.
| You get the $\,$ Wallis product $\,$, if you set $\,\displaystyle x=\frac{\pi}{2}\,$ .
Basel problem :
The coefficient for $\,z\,$ of the infinite product $\,(1+a_1 z)(1+a_2 z)...\,$ is $\,a_1+a_2+...\,$ .
Therefore with $z:=-x^2$ and using the series of sine we get $\,\displaystyle\frac{1}{3!}=\frac{1}{(1\cdot \pi)^2}+\frac{1}{(2\cdot \pi)^2}+...\,$ .
Both calculations are based on partial fraction expansion or equivalent on the sine product :
$($ Wallis product $\displaystyle )^2 \,\,= \left(\prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)^{-1}|_{x=\frac{1}{2}}\right)^2=\left(\frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}}\right)^2=\frac{\pi^2}{4}$
$\displaystyle \frac{\pi^2}{6}=\frac{1-\pi x \cot(\pi x)}{2 x^2}|_{x\to 0}=\sum\limits_{k=0}^\infty \frac{1}{k^2-x^2}|_{x=0}=\zeta(2) $
Both calculations can be put together to an equation chain. The core thing is:
$\displaystyle \frac{2}{3}\prod\limits_{k=1}^\infty \left(1-\left(\frac{x+\frac{1}{2}}{k}\right)^2\right)^{-2} \bigg|_{x=0} =\frac{\pi^2}{6} = -\frac{1}{2}\frac{d^2}{dx^2}\ln \prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)\bigg|_{x=0} $
| {
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Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
| May be somehow tricky but I love this kind ...
$$\quad{\frac{2x}{(x+2)^2} = \\\frac{2(x)}{(x+2)^2} = \\
\frac{2(x+2)-4}{(x+2)^2} = \\
\frac{2(x+2)}{(x+2)^2} +\frac{-4}{(x+2)^2} = \\
\frac{2}{(x+2)} +\frac{-4}{(x+2)^2} \\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of series $\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}$ I was trying to solve this series and I have an exam in a week. I can't understand how to find its sum although I managed to rework it by transforming $\frac{1}{m^2 - n^2}$ into $\frac{1}{2m}(\frac{1}{m+n} - \frac{1}{n-m})$.
I'm sure there is a way, as my book explicitly confirms this as being convergent and having a sum.
Any help is appreciated.
| We have the series were $n<m$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1}- (\frac {1}{1-m} + \frac {1}{2-m} +\frac {1}{3-m}+\cdots+ \frac {1}{-1}))$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{1} + \frac {1}{2} +\frac {1}{3}+\cdots+ \frac {1}{m-1}))$
Then we skip $n = m$
And we go from $m<n<2m$
$\frac 1{2m}(\frac {1}{2m + 1} + \frac {1}{2m + 2} +\frac {1}{2m + 3}+\cdots+ \frac {1}{3m - 1}- (\frac {1}{1} + \frac {1}{2} +\frac {1}{3}+\cdots+ \frac {1}{m-1}))$
and we have some stuff that cancels with the first partial sum
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{2m + 1} + \frac {1}{2m + 2} +\frac {1}{2m + 3}+\cdots+ \frac {1}{3m - 1})$
Note that we have skipped $\frac {1}{2m}$
$n\ge 2m$
$\frac 1{2m}(\frac {1}{3m} - \frac {1}{m} + \frac {1}{3m+1} - \frac {1}{m+1}\cdots)$
Now we are subtracting every term from $\frac {1}{m+1}$ on that we added since the beginning.
$\frac {1}{2m}(-\frac {1}{m}-\frac {1}{2m}) = -\frac {3}{4m^2}$
| {
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} |
If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression:
$$\min \{a(1-c)+b(1-a)+c(1-b)\}$$
The given expression can be rewritten as:
$$a+b+c-ab-ac-bc = S_1-S_2 \tag{1}$$
Then from the given condition:
$$\begin{align}
abc &= 1-(a+b+c)+ab+ac+bc-abc \\
S_3 &= 1-S_1+S_2-S_3\\
S_1-S_2 &= 1-2S_3 \tag{2}
\end{align}$$
Using $AM \ge GM$
$$\frac{1-S_3 -S_3}{3} \ge\sqrt[3]{S_3^2}$$
for extreme value I set them equal to get:
$$\begin{align}
\frac{(1-2S_3)^3}{27}-S^2_3 &= 0 \\
8S_3^3 +15S_3^2 +6S_3-1 &=0 \\
(S_3 +1)(8S_3^2+7S_3-1) &= 0\\
(S_3+1)^2\left(S_3-\frac{1}{8}\right) &= 0
\end{align}$$
From here we get for minimum value, $S_3 = \frac{1}{8}$. We ignore $S_3 = -1$ because all $a,b,c \ge 0$.
So we can say:
$$\min \{1-2S_3\} = \frac{3}{4}$$
Q. Is there error in any argument (highly probable)
Q. Please provide a better method for inequality, if possible!
NOTE :
I hope you understand my terminology, cannot put it in a better way:
$$S_i = \sum_{cyc} a_1 a_2 a_3 ... ('i'\text{ factors})$$
| We'll prove that $\frac{3}{4}$ is a minimal value.
Let $c=0$.
Hence, $(1-a)(1-b)=0$ and $\sum_{cyc}a(1-c)=a+b-ab=1>\frac{3}{4}.$
Now, let $abc\neq0$, $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z$.
Thus, $xyz=1$, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y}$, $c=\frac{z}{1+z}$ and we need to prove that
$$\sum_{cyc}\frac{x}{1+x}\left(1-\frac{z}{1+z}\right)\geq\frac{3}{4}$$ or
$$\sum_{cyc}\frac{x}{(1+x)(1+z)}\geq\frac{3}{4}$$ or
$$4\sum_{cyc}x(1+y)\geq3\prod_{cyc}(1+x)$$ or
$$4(x+y+z)+4(xy+xz+yz)\geq3+3(x+y+z)+3(xy+xz+yz)+3xyz$$ or
$$x+y+z+xy+xz+yz\geq6,$$ which is true by AM-GM.
The equality occurs for $x=y=z=1$, which says that $\frac{3}{4}$ is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Calculating Laurent series expansion I have to calculate the Laurent series expansion of $$f(z) = \frac {2z−2}{(z+1)(z−2)}$$ in $1 < |z| < 2$ and $|z| > 3$.
For first annulus, I know I must manipulate the given expression to contain terms $1/z$ and $z/2$ so that some expansion is valid for $|\frac1{z}|<1$ and $|\frac z{2}|<1$, so I try doing that.
Decomposing into partial fractions,
$$f(z) = \frac 4{3} (\frac 1{z+1}) + \frac 2{3}(\frac 1{z-2})$$
$$= \frac4{3z} \frac1{1-\frac {(-1)}{z}} - \frac1{3} \frac 1{1 - \frac z{2}}$$
So can I now expand the $2$ terms by a G.P. for $|\frac1{z}|<1$ and $|\frac z{2}|<1$ to get the Laurent series?
For second case of $|\frac 3{z}|<1 $, how should I proceed?
| The function
\begin{align*}
f(z)&=\frac{2z-2}{(z+1)(z+2)}\\
&= \frac{4}{3}\left( \frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\
\end{align*}
is to expand around the center $z=0$ in $1<|z|<2$ and $|z|>3$.
Since there are simple poles at $z=-1$ and $z=2$ we have to distinguish three regions of convergence
\begin{align*}
D_1:&\quad 0\leq |z|<1\\
D_2:&\quad 1<|z|<2\\
D_3:&\quad |z|>2
\end{align*}
*
*The first region $D_1$ is a disc with center $z=0$, radius $1$ and the pole at $z=-1$ at the boundary of the disc. It admits for both fractions a representation as power series.
*The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It admits for the fraction with pole at $z=-1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=2$ a power series.
*The region $D_3$ contains all points outside the disc with center $z=0$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Since we are interested in an expansion for $1<|z|<2$ we consider the expansion in $D_2$.
Expansion in $D_2$:
\begin{align*}
f(z)&=\frac{4}{3}\left(\frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\
&=\frac{4}{3z}\left(\frac{1}{1+\frac{1}{z}}\right)-\frac{1}{3}\left(\frac{1}{1-\frac{1}{2}z}\right)\\
&=\frac{4}{3}\sum_{n=0}^\infty\frac{(-1)^{n}}{z^{n+1}}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{2^n}z^n\\
&=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{2^n}z^n\\
\end{align*}
Since we are interested in an expansion for $|z|>3$ we consider the expansion in $D_3$.
Expansion in $D_3$:
\begin{align*}
f(z)&=\frac{4}{3}\left(\frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\
&=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{2}{3z}\left(\frac{1}{1-\frac{2}{z}}\right)\\
&=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=0}^\infty\frac{2^{n+1}}{z^{n+1}}\\
&=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=1}^\infty\frac{2^n}{z^{n}}\\
&=\frac{1}{3}\sum_{n=1}^\infty\left(2^n-4(-1)^{n}\right)\frac{1}{z^{n}}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Probability that the minimum is $2$ and the maximum is $5$ after rolling a die $4$ times
Suppose that a fair $6$-sided die is rolled $4$ times. Letting $N$ be
the minimum number of spots observed and $X$ be the maximum number of
spots observed, give the value of $P(N= 2, X= 5)$
For this to happen, we need all $4$ die to be greater than or equal to $2$ but we need at least one to be $2$. Similarly, we need all $4$ die to be less than or equal to $5$ but we need at least one to be $5$. So I need to find the probability of getting a $2$, a $5$, and a $2,3,4$, or $5$, twice. I have
$${4\choose{1}} \cdot{3\choose{1}} \cdot{2\choose{2}}\cdot {1\over{6}} \cdot{1\over{6}}\cdot {4\over{6}}^2 =.148$$
since we are choosing one spot of $4$ to be a $2$, one spot of the remaining $3$ to be a $5$, and $2$ spots of the remaining $2$ to be a $2,3,4,$ or $5$.
I do not think this is right because I did an excel simulation and got a probability of a little over $.09$.
| Your enumeration counts some rolls repeatedly. For instance a 2 2 4 5 roll is counted twice, on account of the two different choices for 2.
I prefer an inclusion-exclusion-style argument for this, which yields a probability of
$$
\frac{4^4-2*3^4+2^4}{6^4} = \frac{55}{648} \approx 0.0849
$$
Here the $4^4$ is all rolls with all values from 2 to 5, the two copies of $3^4$ are the rolls with all values from 2 to 4 and all rolls from 3 to 5, and the $2^4$ are the rolls with all values either 3 or 4. You want to exclude values without a 5 and without a 2, but after subtracting the $2*3^4$ representing both of these possibilities, you need to add in the intersection of these two situations since those got subtracted twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519286",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is the polynomial $\frac{1}{64}x^6 + 3x^4 - \frac{1}{4}x^3 - x + 6$ irreducible over $\Bbb Q$? How do I approach proving irreducibility/reducibility in a polynomial with rational coefficients? Can I apply Eisenstein in some way?
Is the polynomial $\frac{1}{64}x^6 + 3x^4 - \frac{1}{4}x^3 - x + 6$ irreducible over $\Bbb Q$?
| Replace $x$ by $2y$ gives
$$\frac{1}{64}x^6 + 3x^4 - \frac{1}{4}x^3 - x + 6 = y^6+48y^4-2y^3-2y+6$$
Now apply Eisenstein criterion with $p=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to solve a quartic equation using trigonometric power identitiy? According to this post, we can solve a cubic equation
$$t^3+pt+q=0$$
by the trigonometric identity
$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$
So I've tried to solve the quartic equation
$$t^4+pt^2+qt+r=0$$
using the identity
$$8\cos^4\theta-4\cos2\theta-\cos4\theta-3=0$$
$$8\cos^4\theta-4(2\cos^2\theta-1)-\cos3\theta\cos\theta+\sin3\theta\sin\theta-3=0$$
$$8\cos^4\theta-8\cos^2\theta-\cos3\theta\cos\theta+1+\sin3\theta\sin\theta=0$$
If we let $t:=A\cos\theta$, then we have
$$A^4\cos^4\theta+A^2p\cos^2\theta+Aq\cos\theta+r=0$$
or
$$8\cos^4\theta+\frac{8p}{A^2}\cos^2\theta+\frac{8q}{A^3}\cos\theta+\frac{8r}{A^4}=0$$
now $\frac{8p}{A^2}=-8$ implies that $A=\sqrt{-p}$ and we need to find $\theta$ in the following system
$$\begin{cases}\cos3\theta=-\frac{8q}{A^3}\\1+\sin3\theta\sin\theta=\frac{8r}{A^4}\end{cases}$$
What would you think on solving this system? Is it possible to solve this system? I couldn't find anything on the web about the solving a quartic equation by trigonometric power identity, so I don't know if this way works ...
| it`s possible to solve this system. Just rewrite
$$\cos 3\phi = \cos\phi(2\cos2\phi-1)$$
and
$$\sin3\phi \sin\phi=\sin^2(\phi)(2\cos 2\phi+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Trig integral evaluations with Complex analysis I am currently trying to evaluate this integral with residues but I'm not quite there. The questions asks me to show
$$\int_0^{2\pi} \frac{\sin^2 \theta}{5 + 3\cos\theta}\ d\theta = \frac{2\pi}{9} $$
So I let $z = e^{i\theta}$ and so $\frac {dz}{iz} = d\theta$
I let $\sin^2\theta = \frac {z^2 + z^{-2} -2}{-4}$ but I can't seem to get the answer! Are there any other tips to this?
| $\int_0^{2\pi} \frac{\sin^2 \theta}{5+3\cos\theta} \ d\theta$
$z = e^{i\theta}\\
dz = ie^{i\theta}\ d\theta\\
d\theta = \frac 1{i z} \ dz$
$\sin \theta = \frac {1}{2i}(z - z^{-1})\\
\sin^2 \theta = \frac {1}{4}(2-z^2 - z^{-2})\\
\cos\theta = \frac 1{2}(z + z^{-1})$
$\oint_{|z| = 1} \frac{2-z^2 - z^{-2}}{(4iz)(5 + \frac 12 (3z + 3z^{-1}))} \ dz\\
\oint_{|z| = 1} \frac{-z^4 +2z^2 - 1 }{(2i)(z^2)(3z + 1)(z+3)} \ dz\\
\frac 1{2i}\oint_{|z| = 1} -\frac 13 - \frac 1{3z^2} + \frac {10}{9z} - \frac {8}{9(z+\frac 13)} + \frac 8{9(z+3)} \ dz$
$z = 0, z =-\frac 13$ are the only poles inside the countour
$\pi (\frac {10}{9} - \frac {8}{9}) = \frac {2\pi}{9}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find angle between $2$ points in $3D$ space? I think I am not able to convey what I need so here is the edited question:
As you can see in the image, a point on the left is at $(-3,-3,0)$ and the point on the right is at $(3,3,0)$.
And when I draw the cylinder it spawns at $(0,0,0)$, the center point of the cylinder is at $(0,0,0)$ and size of the cylinder is the distance between $2$ points. Now my goal is to rotate the cylinder in all $x$, $y$ and $z$-axis so that it connects both the points.
Here this is a $2$D image so I know that I will need to rotate this only in z-axis but I want a generalized for the points anywhere in the coordinate system.
| Angle between $(a,b,c)$ and $(x,y,z)$ is given by,
$$\cos\theta =\frac{ax+by+cz}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}$$
The unit vector from $(a,b,c)$ to $(x,y,z)$ will be$$\frac{a-x}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}i+\frac{b-y}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}j+\frac{c-z}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}k$$
Here the $i^{th}$ term is Cosine of angle with x-axis,
$j^{th}$ term is Cosine of angle with y-axis ,
$k^{th}$ term is Cosine of angle with z-axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question related to the simplification of a nested radical $\sqrt[n]{\frac{\sqrt[n]{a}}{a^2}\dots}$
Given that $a\in \mathbb R^+\setminus\{0\}$ and $n\in \mathbb N$, $n>2$, find the denominator of the irreducible fraction that represents the exponent for $a$ resulting from the simplification of the following nested radical expression:
$$
\sqrt[\leftroot{+4}\uproot{-35}\scriptstyle n]{
\frac{
\sqrt[\leftroot{+4}\uproot{-25}\scriptstyle n]{
\frac{
\sqrt[\leftroot{+2}\uproot{-15}\scriptstyle n]{
\frac{
\begin{array}{c}
\vdots\\
\sqrt[n]{\frac{\sqrt[n]{\frac{\sqrt[n]{a}}{a^2}}}{a^3}}\\
\vdots\\
\end{array}
}{a^{n-2}}
}
}{a^{n-1}}
}
}{a^{n}}
}
$$
This is a question from a math contest. Correct answer unknown. I have a problem at the end in the definition of the denominator of the "irreducible" fraction.
My attempt:
First, by analyzing the intimidating expression, starting from $n=2$ and greater, we can find that it is equivalent to
$$a^{S(n)},~~S(n)=\frac{1-V(n)}{n^n},~~V(n)=2n+3n^2+4n^3+\ldots+n\times n^{n-1}.$$
A simpler expression for $V(n)$ can be found noticing that
$$V(n)-nV(n)=2n+Q(n)-n^{n+1},~~Q(n)=n^2+n^3+\ldots+n^{n-1}.$$
It is easy to see that $Q(n)=\frac{n^2-n^n}{1-n}$ following that
$$V(n)=\frac{2n+\frac{n^2-n^n}{1-n}-n^{n+1}}{1-n}=\frac{(1-n)(2n-n^{n+1})+n^2-n^n}{(1-n)^2}.$$
Finally, we can we get
$$a^{S(n)},~~S(n)=\frac{1-V(n)}{n^n}=\frac{(1-n)^2-(1-n)(2n-n^{n+1})-n^2+n^n}{n^n(1-n)^2}$$
Another possible way to express the exponent $S(n)$ is
$$S(n)=\frac{n^{-n}(1-4n+2n^2)+1+n-n^2}{(1-n)^2}$$
Questions: (a) this development is correct up to the definition of $S(n)$? (b) are there ways to simplify it? (c) how to uniquely define the irreducible fraction representing the exponent in this case, so that the denominator is uniquely defined? Note: some numerical exploration suggests that the asked denominator is $n^n$.
| Your $S(n)$ is correct.
For $n=3$, it is easy to see that the asked denominator is $3^3$.
For $n\ge 4$, we have
$$\begin{align}&\frac{1-4n+2n^2+n^n(1+n-n^2)}{(1-n)^2}\\\\&=\frac{-n^n(n^2-2n+1+n-2)+2n^2-4n+2-1}{(n-1)^2}\\\\&=-n^n+2+\frac{-n^{n-1}(n^2-2n+1-1)-1}{(n-1)^2}\\\\&=-n^n+2-n^{n-1}+\frac{n^{n-1}-1}{(n-1)^2}\\\\&=-n^n-n^{n-1}+2+\frac{n^{n-2}+n^{n-3}+\cdots +n+1}{n-1}\\\\&=-n^n-n^{n-1}+n+1+\sum_{j=1}^{n-3}(n-2-j)n^{j}\end{align}$$
So, we get
$$S(n)=\frac{-n^n-n^{n-1}+n+1+\displaystyle\sum_{j=1}^{n-3}(n-2-j)n^{j}}{n^n}$$
It is easy to see that the numerator is coprime to the denominator.
Therefore, the answer is $\color{red}{n^n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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To evaluate $\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$ Evaluate $$\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$$
I tried taking log on both sides and then using L'Hospital rule but its giving complex results.Are there any simpler methods to approach this?
| From this answer, we can see that the Taylor coefficients of $\tan(x)$ expanded around $0$ will be positive, which implies that every truncation of the Taylor series acts as a lower bound for $\tan(x)$ on $[0, \pi/2)$. Therefore, $$\tan(x)\geq x+\frac{x^3}{3}$$ for $x\in [0, \pi/2)$. As $\tan(x)/x$ is an even function, this implies $$\frac{\tan(x)}{x}\geq 1+\frac{x^2}{3}$$ for $x\in (-\pi/2, 0)\cup (0, \pi/2)$. Therefore, as $1/x^3$ is negative for $x < 0$, $$\left(\frac{\tan(x)}{x}\right)^{1/x^3}\leq \left(1+\frac{x^2}{3}\right)^{1/x^3}$$ By Bernoulli's inequality, $$\left(1+\frac{x^2}{3}\right)^{-1/x^3}\geq 1-\frac{1}{x^3}\cdot \frac{x^2}{3} = 1-\frac{1}{3x}$$ for $-1\leq x < 0$, so $$\left(1+\frac{x^2}{3}\right)^{1/x^3}\leq \left(1-\frac{1}{3x}\right)^{-1} = \frac{3x}{3x-1}$$ Therefore, $$0\leq \lim_{x\to 0^-} \left(\frac{\tan(x)}{x}\right)^{1/x^3}\leq \lim_{x\to 0^-} \left(1+\frac{x^2}{3}\right)^{1/x^3}\leq \lim_{x\to 0^-} \frac{3x}{3x-1} = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the real roots for $\displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.$
Find the real roots for $\displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.$
Question from a Math Olympiad (ES, 2005). Answer: $(3\pm \sqrt{2})^4$.
My attempt: I will make my attempt below, but I think the approach might be too complicated... are there simpler approaches?
I started making the substitution
$$u=\sqrt[4]{386-x},~~v=\sqrt[4]{x}$$
so that $u^4+v^4=386$.
Therefore we can define a system of 2 equations, in order to hopefully find $u$ and $v$:
$$\left\{
\begin{array}{l}
u+v=6 \\
u^4+v^4=386\\
\end{array}
\right.
$$
If we use $p_k=u^k+v^k$, $e_1=(u+v)$ and $e_2=uv$, by Newton-Girard identities we get
$$
\left\{
\begin{array}{l}
p_4=e_1p_3-e_2p_2\\
p_3=e_1p_2-e_2p_1\\
p_2=e_1p_1-e_2\times 2\\
\end{array}
\right.
$$
But as we know that $p_1=e_1=u+v=6$, we can make the substitutions backwards,
so that
$$
\left\{
\begin{array}{l}
p_2=36-2 uv\\
p_3=6(36-2uv)-6uv=216-18uv\\
p_4=6(216-18uv)-uv(36-2 uv)\\
\end{array}
\right.
$$
But, as we know that $p^4=u^4+v^4=386$, so making the substitution $z=uv$ and some simplifications the last equation is equivalent to
$$-z^2+72z-455=0.$$
Solving this last equation we find the roots $7$ and $65$, therefore $uv$ will potentially have these two values. Now as we know that $u+v=6$ we can set a system to solve for $u$ and $v$:
$$
\left\{
\begin{array}{l}
u+v=6\\
uv=7~~\text{or}~~65\\
\end{array}
\right.
$$
That can be solved noticing the $u$ and $v$ are the roots of $P(z)=z^2-(u+v)z+uv$. When $uv=65$, $P(z)$ has only complex roots. When $uv=7$, $P(z)$ has 2 real roots.
When using $uv=7$ the roots $(u,v)$ for $P(z)$ are, by the symmetry, either
$$(3-\sqrt{2},3+\sqrt{2})~~\text{or}~~(3+\sqrt{2},3-\sqrt{2}).$$
But as $x=v^4$ the final solution would be
$$x=(3\pm \sqrt{2})^4=193\pm 132\sqrt{2}.$$
Both solutions comply with the initial restritions for the argument of the radicals: $0\le x\le 386$.
Questions: (a) is the development correct? at least the final answer matches Wolphram Alpha; (b) are there other approaches? other solutions are welcomed.
Sorry if this is a duplicate.
| I would simply let $x=u^4$ and hope that $386-u^4=(6-u)^4$ is easy to factor. Expanding and collecting first to
$$2u^4-24u^3+216u^2-864u+910=0$$
we can hope for a factorization
$$u^4-12u^3+108u^2-432u+455=(u^2-au+b)(u^2-cu+d)$$
from which it would be easy to identify the real roots. (Hoping for an integer root would leave a cubic with a real root to solve for, so it's better to hope for a factorization into a pair of quadratics.)
The factorization $u^4+1\equiv(u^2+1)^2$ mod $2$ tells us $a$ and $c$ are both even, and the factorization $u^4\equiv (u^2-1)(u^2+1)$ mod $3$ tells us $a$ and $c$ are multiples of $3$. This leaves two possibilities:
$$u^4-12u^3+108u^2-432u+455=(u^2+b)(u^2-12u+d)$$
and
$$u^4-12u^3+108u^2-432u+455=(u^2-6u+b)(u^2-6u+d)$$
The second of these somehow looks more likely. It's satisfied if $108=b+d+36$, $432=6(b+d)$, and $bd=455$. We're in luck: the requirements $108=b+d+36$ and $432=6(b+d)$ are in accord; both say $b+d=72$. So $b$ and $d$ are roots of the quadratic
$$v^2-72v+455=(v-7)(v-65)$$
We thus have
$$u^4-12u^3+108u^2-432u+455=(u^2-6u+7)(u^2-6u+65)$$
and we're just about done. The quadratic $u^2-6u+65$ has complex roots whose real and imaginary parts are both non-zero (so that their fourth powers are not non-negative real numbers), while the roots of $u^2-6u+7$ are $u=3\pm\sqrt2$. Thus $x=(3+\sqrt2)^4$ and $x=(3-\sqrt2)^4$ are the real roots of the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum of the value $f=(1+\sin^2{x})(1+\sin^2{y})$ Let $\tan{x}\tan{y}=-\frac{1}{2}$,find the minumin of the value $$f=(1+\sin^2{x})(1+\sin^2{y})$$
My ugly solution:
$$f=\dfrac{1+\sin^2{x}}{\sin^2{x}+\cos^2{x}}\cdot\dfrac{1+\sin^2{y}}{\sin^2{y}+\cos^2{y}}=\dfrac{2m^2+1}{m^2+1}\cdot\dfrac{2n^2+1}{n^2+1}$$where $m=\tan{x},n=\tan{y},mn=-\frac{1}{2}$,so
$$f=2\cdot\dfrac{4m^4+4m^2+1}{4m^4+5m^2+1}=2\left(1-\dfrac{m^2}{4m^4+1+5m^2}\right)\ge 2\left(1-\dfrac{m^2}{4m^2+5m^2}\right)=\dfrac{16}{9}$$
some other simple methods?such as AM-GM and Cauchy-Schwarz inequality kill it?
| Let $\tan^2x=a$ and $\tan^2y=b$.
Thus, $ab=\frac{1}{4}$ and by AM-GM we obtain:
$$(1+\sin^2x)(1+\sin^2y)=(2-\cos^2x)(2-\cos^2y)=$$
$$=\left(2-\frac{1}{1+a}\right)\left(2-\frac{1}{1+b}\right)=\frac{(2a+1)(2b+1)}{(1+a)(1+b)}=$$
$$=\frac{2+2(a+b)}{\frac{5}{4}+a+b}=2-\frac{\frac{1}{2}}{\frac{5}{4}+a+b}\geq$$
$$\geq2-\frac{\frac{1}{2}}{\frac{5}{4}+2\sqrt{ab}}=\frac{16}{9}.$$
The equality occurs for example for $\tan{x}=\frac{1}{\sqrt2}$ and $\tan{y}=-\frac{1}{\sqrt2}$,
which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.