Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How I find $\gcd(\frac{a^{2m+1}+1}{a+1}, a+1)$? $$\gcd\left(\frac{a^{2m+1}+1}{a+1}, a+1\right)$$
My answer until this moment is:
$$
For: a^{2m+1}+1^{2m+1} = (a + 1)^{m+1} - 2a = (a+1)^m(a+1)-2a
$$ where
\begin{align}
(a + 1)^{m} = a^{2m}+a^{2m-1}+\cdots+1
\end{align}
So,
\begin{align}
a^{2m+1}+1 = [a^{2m}+a^{2m-1}+\cdots+1](a+1)-2a
\end{align}
But, I don't know how to divide :
\begin{align}
\frac{[a^{2m}+a^{2m-1}+\cdots+1](a+1)-2a}{a+1}
\end{align}
| $$
\begin{align}
\frac{a^{2m+1}+1}{a+1}
&=\frac{((a+1)-1)^{2m+1}+1}{a+1}\tag1\\
&=\sum_{k=1}^{2m+1}(-1)^{k-1}\binom{2m+1}{k}(a+1)^{k-1}\tag2\\[6pt]
&\equiv2m+1\pmod{a+1}\tag3
\end{align}
$$
Explanation:
$(1)$: $a=(a+1)-1$
$(2)$: Binomial Theorem and algebra
$(3)$: the $k=1$ term is the only one without a factor of $a+1$
Therefore,
$$
\gcd\left(\frac{a^{2m+1}+1}{a+1},a+1\right)=\gcd(2m+1,a+1)\tag4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Second Method to Find the Volume of a Slice of a Cone
I was looking to formulate a general solution for a vertical slice of a cone. After failing to do this by integrating for the area of an ever shrinking chord segment, I eventually came up with a more geometric solution by subtracting the triangular part from the area of a sector which was easier to integrate to get the volume and appears to work fine. See my solution sketch. However, I would still like to formulate the integration area method of the chord segment.
From the information in the sketch, integrating along a $y$ axis, the area of the segment would be:
$$A_c = 2\int_0^{\sqrt(R^2-a^2)} \sqrt(R^2 - y^2) - a \ dy$$
How do I accomplish a second integration along a vertical $x$ axis to integrate these areas into a volume with $R$ decreasing in accordance with $$r = (R - Rx/h)\text{ from } 0 \text{ to } (h - ah/R)$$
I've tried a number of ways but now I'm going cross-eyed.
| A cone of radius $R$ and height $H$
would be:
$z = H - \frac {H}{R} \sqrt {x^2 + y^2}$
And we are going to slice it at some line $y = a$
$v = \int_a^R\int_{-\sqrt{R^2 - x^2}}^{\sqrt {R^2- x^2}} (H - \frac {H}{R} \sqrt {x^2 + y^2}) \ dx \ dy$
Convert to polar:
$x = r\cos \theta\\
y = r\sin \theta\\
z = z\\
dy\ dx = r \ dr\ d\theta$
$2\int_{\arcsin\frac {a}{R}}^{\frac {\pi}{2}}\int_{a\csc \theta}^{R} Hr -\frac{H}{R} r^2 \ dr d\theta\\
2\int_{\arcsin\frac {a}{R}}^{\frac {\pi}{2}} (\frac 16 HR^2 - \frac 12 Ha^2\csc^2\theta + \frac 13 \frac {H}{R}a^3\csc^3\theta \ d\theta\\
2(\frac 16 HR^2 \theta + \frac 12 Ha^2\cot\theta - \frac 16 \frac {H}{R}a^3\csc\theta\cot\theta + \frac 16\frac {H}{R}a^3(\ln|\csc\theta + \cot\theta|)|_{\arcsin\frac {a}{R}}^{\frac \pi 2}\\
2(\frac 16 HR^2 \arccos \frac {a}{R} - \frac 13 Ha \sqrt {R^2-a^2}+ \frac 16\frac {H}{R}a^3(\ln|\frac {R}{a} + \frac {\sqrt {R^2-a^2}}{a}|))\\$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating definite integral $\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$ Question: $$\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$ is equals to
(a) $-\frac\pi6$
(b) $-\frac{\pi}{12}$
(c) $\frac\pi{12}$
(d) $\frac\pi6$
My attempt: Denoting given integral by $I$ and letting $z=e^{iθ}$ then given integral becomes,
\begin{align*}
I&=\int_C\frac{1}{13-5(\frac{z-\bar{z}}{2i})}\frac{\mathrm dz}{iz}\\
&=\frac{1}{i}\int_C\frac{2i}{26iz-5z^2+5|z|^2}\mathrm dz\\
&=2\int_C\frac{\mathrm dz}{-5z^2+26iz+5}\hspace{0.5in}\text{As }C: |z|=1\\
&=2\int_C \frac{\mathrm dz}{(z-5i)(z-i/5)}\\
&=2\left(\frac{5}{24i}\int_C\frac{1}{z-5i}-\frac{5}{24i}\int_C \frac{1}{z-i/5}\right)
\end{align*}
Now as point $z=5i$ lies outside $C$ so it's integral evaluates to $0$ and by Cauchy integral formula, above becomes,
$$I=0-2\frac{5}{24i}2\pi i = -\frac{5\pi}{6}$$
But none of the given answer matches with mine. So is am i incorrect? Please help me..stuck on this from hours...
| COMMENT.-An unorthodox way. The function is continuous and its minimum and maximum are taken in $\dfrac{3\pi}{2}$ and $\dfrac{\pi}{2}$ respectively. Then we have (using areas of rectangles)
$$2\pi\cdot0.0555\approx\dfrac{\pi}{10}\lt\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta\lt 2\pi\cdot\frac 18=\frac{\pi}{4} $$
Taking into account that the answer is given to select between four possibilities, the solution is clearly now d) $\dfrac{\pi}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve the following trigonometrical equation? I have the following equations
\begin{align*}
R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\
R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}).
\end{align*}
From these equations how can I obtain the following solution
$$\omega T_2=\pm \arccos\left(\frac{Q_1^2+Q_2^2-R_1^2+R_2^2}{2R_2\sqrt{Q_1^2+Q_1^2}}\right)-\arctan\left(\frac{Q_2}{Q_1}\right)+\phi_2+2k\pi .$$
| Another way to tackle the same problem of solving $a \sin \theta + b \cos \theta + c = 0$ is to use the tan-half-angle substitutions
$$\begin{aligned}
t &= \tan \left( \frac{\theta}{2} \right)\, \rightarrow & \theta = 2 \arctan(t) \\
\sin(\theta) & = \frac{2 t}{1+t^2} \\
\cos(\theta) & = \frac{1-t^2}{1+t^2}
\end{aligned}$$
The problem is transformed into a polynomial equation
$$\left. a \frac{2 t}{1+t^2} + b \frac{1-t^2}{1+t^2} + c = 0 \right\}\;\;
t=\begin{cases} \frac{a+\sqrt{a^2+b^2-c^2}}{b-c} \\ \frac{a-\sqrt{a^2+b^2-c^2}}{b-c} \end{cases} $$
With the final and solution in terms of the angle
$$ \theta = \begin{cases} 2 \arctan\left( \frac{a+\sqrt{a^2+b^2-c^2}}{b-c} \right) \\ 2 \arctan\left( \frac{a-\sqrt{a^2+b^2-c^2}}{b-c} \right) \end{cases} $$
Example
$$ 5 \sin \theta + 2 \cos \theta - 3 = 0 $$
$$ t= \begin{cases}
\frac{5}{2-(-3)} + \frac{ \sqrt{5^2+2^2-(-3)^2}}{2-(-3)} = 1.89443 \\
\frac{5}{2-(-3)} - \frac{ \sqrt{5^2+2^2-(-3)^2}}{2-(-3)} = 0.10557
\end{cases} $$
$$ \theta= \begin{cases}
2 \arctan(1.89443) = 2.170\,{\rm rad} \\
2 \arctan(0.10557) = 0.2104\,{\rm rad}
\end{cases} $$
Check
$$ \begin{cases}
5 \sin(2.170)+2 \cos(2.170) -3 \approx 0 \\
5 \sin(0.2104) + 2\cos(0.2104) -3 \approx 0
\end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Convergence of integral with trigonometry
*
*$\displaystyle \int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}} dx$
Obviously, problem in $ -\frac{\pi}{4} $ as $ \lim_{x\to -\frac{\pi}{4}^+} = \infty $, integral is positive on whole segment, but I can't use any usual rules to prove it's convergence or simly divide it inside the root. I guess, i have to bound it with converging function, but can't find a good one aswell. Any thoughts?
| Recall that
*
*$\cos x - \sin x=\sqrt 2 \sin \left(\frac{\pi}4-x\right)$
*$\cos x + \sin x=\sqrt 2 \sin \left(\frac{\pi}4+x\right)$
then
$$ =\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)^{\frac{1}{3}} dx
=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi}4-x\right)}{\sin \left(\frac{\pi}4+x\right)}\right)^{\frac{1}{3}} dx$$
Assume $y=x+\frac{\pi}4$ then we have
$$\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi}4-x\right)}{\sin \left(\frac{\pi}4+x\right)}\right)^{\frac{1}{3}} dx
=\int_{0}^\frac{\pi}{2} \left(\frac{\sin \left(\frac{\pi}2-y\right)}{\sin y}\right)^{\frac{1}{3}} dy
=\int_{0}^\frac{\pi}{2} \left(\frac{1}{\tan y}\right)^{\frac{1}{3}} dy$$
and
$$\left(\frac{1}{\tan y}\right)^{\frac{1}{3}}\sim\frac1{\sqrt[3] y} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Try: Writting equation as $$4z^2-4(y-1)z+x^2+2y^2-2(x+y+xy)+5=0$$
Now if equation has real roots. Then $$z=\frac{(y-1)\pm \sqrt{2xy+2x-x^2-y^2-4}}{2}$$
So for integer roots $2xy+2x-x^2-y^2-4=k^2$ , Where $k\in\mathbb{Z}$.
So $$2x-(x+y)^2-4=k^2\Rightarrow (x+y)^2=2x-4-k^2$$
Now i did not understand how can i find non negative integer triplets $(x,y,z)$. Could some help me , Thanks
| The discriminant was:
$$\begin{align}
2xy+2x-x^2-y^2-4&=k^2\\
\text{let} \qquad y=x \pm &t \\
2x(x \pm t)+2x-x^2-(x \pm t)^2-4&=k^2\\
2x^2 \color{red}{\pm 2tx}+2x-x^2-x^2 \color{red}{\mp 2tx}-t^2-4&=k^2 \\
2x-t^2-4&=k^2 \\
x&=\frac{1}{2}(k^2+t^2+4) \\
y=x\pm t&=\frac{1}{2}(k^2+t^2 \pm 2t+4) \\
&=\frac{1}{2}\left[k^2+(t\pm 1)^2+3\right] \\
z=z=\frac{(y-1)\pm k}{2}&=\frac{\frac{1}{2}[k^2+(t\pm 1)^2+1] \pm k}{2} \\
&=\frac{(k\pm1)^2+(t\pm 1)^2}{4}
\end{align}$$
A solution set:
$$(x,y,z)=\frac{1}{2}\left(k^2+t^2+4,k^2+(t\pm 1)^2+3, \frac{(k\pm1)^2+(t\pm 1)^2}{2}\right)$$
It's apparent from $x$ and $y$ that both $k$ and $t$ should be the same parity, and it's apparent that from $z$ that that parity should be odd. If
$k=2m+1$ and $t=2n+1$ for $(m,n)\in \mathbb{Z}^2$:
$$\begin{align}
x&=\frac{1}{2}[(2m+1)^2+(2n+1)^2+4] \\
&=2m^2+2m+2n^2+2n+3 \\
\\
y&=\frac{1}{2}[(2m+1)^2+(2n+1\pm 1)^2+3] \implies \\
&\quad y_1=2m^2+2m+2(n+1)^2+2 \\
&\quad y_2=2m^2+2m+2n^2+2 \\
\\
z&=\frac{1}{4}[(2m+1\pm1)^2+(2n +1 \pm 1)^2] \implies \\
&\quad z_{(y_1,1)}=m^2+(n+1)^2 \\
&\quad z_{(y_1,2)}=(m+1)^2+(n+1)^2 \\
&\quad z_{(y_2,1)}=(m+1)^2+n^2 \\
&\quad z_{(y_2,2)}=m^2+n^2
\end{align}$$
To summarize:
$$\begin{align}
\text{for a given} \quad (m,n) &\to (x,y_1,z_{(y_1,1)}) \\
&\to (x,y_1,z_{(y_1,2)}) \\
&\to (x,y_2,z_{(y_2,1)}) \\
&\to (x,y_2,z_{(y_2,2)})
\end{align}$$
The bottom of Will Jagy's answer has the complete set of solutions, corresponding to these formulae
| {
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"url": "https://math.stackexchange.com/questions/2794625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
perpendicular distance from $ax-by = 1$ to origin I have a problem from a basic number theory book that asks for the perpendicular distance from the line $ax - by = 1$ to the origin.
My approach was to find the area of the triangle formed by the line and the axes, find the base of the triangle situated at the diagonal, then use those to find the height of the triangle, which would be the perpendicular distance.
I found that the $x$ intercept is $\frac1a$, and the y intercept is $-\frac1b$. So if you consider the base to be the diagonal, it is $\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}$. Then the area of the triangle is $\frac{1}{2ab}$. Thus the height is $\frac{1}{2ab\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$ = $\frac{1}{2\sqrt{(a^2 + b^2}}$, which I thought would be the answer.
However, the book states that the answer is $\frac{1}{\sqrt{a^2 + b^2}}$, so I'm off by $\frac12$. Did I make a mistake with the area of the triangle? If not, where did I go wrong?
Thank you!
| Minimum Distance
Since
$$
\begin{align}
1
&=\left|\,ax-by\,\right|\\
&=\left|\,(a,-b)\cdot(x,y)\,\right|\\
&\le\left|\,(a,-b)\,\right|\left|\,(x,y)\,\right|
\end{align}
$$
we have
$$
\begin{align}
\left|\,(x,y)\,\right|
&\ge\frac1{\left|\,(a,-b)\,\right|}\\
&=\frac1{\sqrt{a^2+b^2}}
\end{align}
$$
If $(x,y)=\frac{(a,-b)}{a^2+b^2}$, then $ax-by=1$ and $\left|\,(x,y)\,\right|=\frac1{\sqrt{a^2+b^2}}$
Thus, the minimum of $\left|\,(x,y)\,\right|$ is $\frac1{\sqrt{a^2+b^2}}$.
Perpendicular Distance
If $ax_1-by_1=1$ and $ax_2-by_2=1$, then
$$
(a,-b)\cdot(x_1-x_2,y_1-y_2)=1-1=0
$$
Thus, $(a,-b)$ is perpendicular to the line containing $(x_1,y_1)$ and $(x_2,y_2)$; i.e. the line $ax-by=1$.
This means the vector from the origin to $\frac{(a,-b)}{a^2+b^2}$ is perpendicular to the line $ax-by=1$ and the point $\frac{(a,-b)}{a^2+b^2}$ is on the line $ax-by=1$.
Thus, the perpendicular distance from the origin to the line is
$$
\left|\frac{(a,-b)}{a^2+b^2}\right|=\frac1{\sqrt{a^2+b^2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the number of integers $n$ such that the equation $xy^2 + y^2 - x - y= n$ has an infinite number of integer solutions $(x,y)$.
Find the number of integers $n$ such that the equation $xy^2 + y^2 - x - y= n$ has an infinite number of integer solutions $(x,y)$.
Firstly the equation can be rearranged into $(y-1)(y+x(y+1)) = n$. If $n$ is $0$, there's an infinite number of $(x,y)$ possible. However, I am struggling with the case when $n$ is a nonzero integer. I had a peek at the solution, and it says
There exists a divisor $k$ of $n$ such that $y-1 = k$ and $x(y+1) + y = n/k$ for infinitely many $x$. It forces $y+1 = 0$.
Yet I don't quite see how $y+1=0$... If anyone can help me understand that'd be great!
| If $x(y+1)+y=\frac{n}{k}$ for infinitely many values of $x$, then for any two such values $x_0$ and $x_1$ we have
$$(x_0-x_1)(y+1)=0.$$
where $x_0-x_1\neq0$, and so $y+1=0$.
The argument can be put more simply; given any integer $n$ and a solution $(x,y)$ to
$$n=xy^2+y^2-x-y=(y-1)(y+x(y+1)),$$
it follows immediately that $y-1$ divides $n$ and that
$$x=\frac{n+y-y^2}{y^2-1}.$$
So for there to be infinitely many solutions $n$ must have infinitely many divisors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If a,b,c are in harmonic progression value of $\frac{3a+2b}{2a-b} + \frac{3c+2b}{2c-b}$? If a, b and c form a harmonic progressions what is the value of $\frac{3a+2b}{2a-b} + \frac{3c+2b}{2c-b}$?
I tried to do this by substituting $a = \frac{1}{p-q}$, $b=\frac{1}{p}$ and $c=\frac{1}{p+q}$. I got the result $\frac{5p-2q}{p+q} + \frac{5p+2q}{p-q}$. According to my solution, I am supposed to get $10 + \frac{14q^2}{p^2-q^2}$. What should I do?
| Your result is correct.
Notice that $\frac{5p-2q}{p+q}+\frac{5p+2q}{p-q}$ can be written as
$$=\frac{5p^2-2pq-5pq+2q^2+5p^2+2pq+5pq+2q^2}{p^2-q^2}$$
$$=\frac{5p^2+5p^2+2q^2+2q^2-2pq+2pq-5pq+5pq}{p^2-q^2}$$
$$=\frac{10p^2+4q^2}{p^2-q^2}$$
which is nothing but $$10+\frac{14q^2}{p^2-q^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability of a $1000 \times 1000$ square matrix over $\mathbb{Z}_2$ having full rank There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus, only $16$ possible $2\times2$ matrices over $\mathbb{Z}_2$, and $6$ of them have full rank:
$$\begin{pmatrix}0&1\\
1&0\end{pmatrix} \quad
\begin{pmatrix}1&1\\
1&0\end{pmatrix} \quad
\begin{pmatrix}0&1\\
1&0\end{pmatrix} \quad
\begin{pmatrix}0&1\\
1&1\end{pmatrix} \quad
\begin{pmatrix}1&1\\
0&1\end{pmatrix} \quad
\begin{pmatrix}1&0\\
0&1\end{pmatrix}$$
Randomly generate a $n \times n$ matrix over $\mathbb{Z}_2$ (where $n$ is big, say, $1000$). What's the probability that the matrix has full rank?
| The general linear group $GL(n,q)$ is the group of invertible $n\times n$ matrices over a field with $q$ elements (note $q=p^k$ for some prime $p$).
The order of $$|GL(n,q)|=\prod_{k=0}^{n-1}(q^n-q^k)$$
So the probability is $$\dfrac{1}{q^{n^2}}\prod_{k=0}^{n-1}(q^n-q^k)$$
For $n=2$ $q=2$ you get $\frac{1}{2^4}(2^2-2)(2^2-1)=\frac{6}{16}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\int{\arctan x}\,\mathrm dx$ without substitution 2018-08-15: I'm still looking for an answer that does not rely on
$$\int{f\left[g(x)\right]g'(x)}\,\mathrm dx = F\left[g(x)\right]$$
I'm refreshing my old calculus skills, and the textbook (Kalkulus by Tom Lindstrøm, 3rd edition, a Norwegian book) asks me to find $\int{\arctan x}\,\mathrm dx$.
I start with integration by parts:
$$\int{\arctan x}\,\mathrm dx
= \int{1\cdot\arctan x}\,\mathrm dx
= x\cdot\arctan x - \int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$
Next, it would be natural to use substitution, and I can do that to get the answer:
$$x\cdot\arctan x - \frac{\ln(x^2+1)}{2}+c$$
Which is correct, but it's clear from context that the book wants me to do it without using substituation. I feel like I've tried everything, looking at the book's own examples, but I must be missing some essential trick.
Here are some of the approaches I've tried:
1. Integration by parts, again
1.1. $u = x$ , $v' = \frac{1}{x^2+1}$
$$x\cdot\arctan x - \left(x\cdot\arctan x - \int{1\cdot\arctan x}\,\mathrm dx\right)
= \int{\arctan x}\,\mathrm dx$$
Back where I started.
1.2. $u = \frac{1}{x^2+1}$ , $v' = x$
$$x\cdot\arctan x - \left(\frac{x^2}{2} \cdot \frac{1}{x^2+1} - \int{\frac{x^2}{2} \cdot \frac{-2x}{(x^2+1)^2}}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2(x^2+1)} - \int{\frac{x^3}{(x^2+1)^2}}\,\mathrm dx$$
I've tried hacking away at this, but it doesn't look like it's getting any easier.
2. Adding and subtracting $x^2$ in the numerator
$$x\cdot\arctan x - \int{x\cdot \frac{(x^2+1)-x^2}{x^2+1}}\,\mathrm dx$$
$$= x\cdot\arctan x - \int{x\cdot \left(1 - \frac{x^2}{x^2+1}\right)}\,\mathrm dx$$
$$= x\cdot\arctan x - \int{x - x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$
$$= x\cdot\arctan x - \frac{x^2}{2} + \int{x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$
The book uses a similar trick (adding and subtracting 1 in the numerator) to solve $\int{x\cdot\arctan x}\,\mathrm dx$. In the process it finds that $\int{\frac{x^2}{x^2+1}}\,\mathrm dx = x - \arctan x$. Perhaps I need to use this result:
Partial integration with $u = x$ , $v´ = \frac{x^2}{x^2+1}$
$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x\cdot(x-\arctan x) - \int{1\cdot (x-\arctan x)}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2} + x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx$$
$$= \int{\arctan x}\,\mathrm dx$$
Back where I started, again. I've been trying for days, different tricks and manipulations. The book doesn't even list it as a particularly tricky question, so I'm feeling a bit dumb.
3. Searching for the answer
I've looked at some other questions, including
*
*Integration of arctan(x) is itself?
*Indefinite integral of $\arctan(x)$, why consider $1\cdot dx$?
But they don't seem to answer my particular question.
| $\newcommand{\dx}{\mathrm dx\,}$The only other way I can see without using a u-substitution, which must I mention, is the easiest way to evaluate this integral, is using the infinite geometric sequence. Hopefully you remember that$$\sum\limits_{n\geq0}(-1)^n\, x^{2n}=\frac 1{1+x^2}$$
Therefore, calling the integral $\mathfrak{I}$, then
$$\begin{align*}\mathfrak{I} & =\int\dx\frac x{1+x^2}\\ & =\sum\limits_{n\geq0}(-1)^n\int\dx x^{2n+1}\\ & =\frac 12\sum\limits_{n\geq1}(-1)^{n-1}\frac {x^{2n}}n\\ & =\frac 12\log(1+x^2)+C\end{align*}$$
Where in the second to last line, we’ve made use of the taylor expansion series for $\log(1+x)$
$$\log(1+x)=\sum\limits_{n\geq1}(-1)^{n-1}\frac {x^n}n$$
| {
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"url": "https://math.stackexchange.com/questions/2799115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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cross product not associative, outer product associative The cross product is not associative. If $i=(1,0,0)$, $j=(0,1,0)$ and
$k=(0,0,1)$, then
\begin{eqnarray}
i \times (i \times j) = i \times k = -j \\
(i \times i) \times j = 0
\end{eqnarray}
However in Geometric Algebra, if $e_1=(1,0,0)$, $e_2=(0,1,0)$ and
$e_3=(0,0,1)$ then
\begin{eqnarray}
e_1 \wedge (e_1 \wedge e_2) = (e_1 \wedge e_1 ) \wedge e_2 = 0
\end{eqnarray}
According to D. Hestenes, New Foundations for Classical Mechanics
(equation 3.13)
\begin{eqnarray}
a \times b = -i a \wedge b \quad, i=\sqrt{-1}
\end{eqnarray}
So, where is the flaw here? Except for a complex scalar both definitions in $\mathbb{R}^3$ are the same. But.....one is associative and the other is not?
Thanks.
| The basic point is that a difference of associative bilinear products will still be bilinear but need not be associative at all. Still, it's instructive to get an explicit expression for the associator of the cross product.
Let $I := e_1 e_2 e_3$, so that
$$
\forall a,b \in \mathbb{R}^3, \quad a \times b = -I(ab - a \cdot b) = -\frac{1}{2}I(ab-ba).
$$
Then for any $a, b, c \in \mathbb{R}^3$,
\begin{align}
(a \times b) \times c &= -\frac{1}{2}I((a \times b)c - c(a \times b))\\ &= -\frac{1}{2}\left(\left(-\frac{1}{2}I(ab-ba)\right)c - c \left(-\frac{1}{2}I(ab-ba)\right) \right)\\ &= -\frac{1}{4}(abc-bac-cab+cba),
\end{align}
whilst
\begin{align}
a \times (b \times c) &= -\frac{1}{2}I(a(b \times c) - (b \times c)a)\\
&= -\frac{1}{2}I\left(a\left(-\frac{1}{2}I(bc-cb)\right) - \left(-\frac{1}{2}I(bc-cb)\right)a\right)\\
&= -\frac{1}{4}(abc-acb-bca+cba),
\end{align}
so that
\begin{align*}
(a \times b) \times c - a \times (b \times c) &= -\frac{1}{4}(abc-bac-cab+cba) + \frac{1}{4}(abc-acb-bca+cba)\\ &=\frac{1}{4}(bac+cab-acb-bca)\\
&= \frac{1}{2}(b(a \wedge c)-(a \wedge c)b)\\
&= b \wedge (a \wedge c)\\
&= -I(b \wedge (-I(a \wedge c)))\\
&= -b \times (a \times c),
\end{align*}
so that $(a \times b) \times c = a \times (b \times c)$ if and only if $b \times (a \times c) = 0$.
Anyhow, for example, $e_1 \times (e_1 \times e_2) = e_1 \times e_3 = -e_2 \neq 0$, so that $(e_1 \times e_1) \times e_2 \neq e_1 \times (e_1 \times e_2)$; more explicitly,
\begin{align*}
(e_1 \times e_1) \times e_2 = -\frac{1}{2}I(e_1e_1-e_1e_1) \times e_2 = 0 \times e_2 = 0,
\end{align*}
where, of course, $e_1 \times e_1 = -\frac{1}{2}I(e_1 \wedge e_1) = 0$, whilst
\begin{align*}
e_1 \times (e_1 \times e_2) &= -\tfrac{1}{2}I(e_1(-\tfrac{1}{2}I(e_1e_2-e_2e_1)) - (-\tfrac{1}{2}I(e_1e_2-e_2e_1))e_1)\\
&= -\frac{1}{4}(e_1e_1e_2-e_1e_2e_1-e_1e_2e_1+e_2e_1e_1)\\
&= -e_1^2e_2\\
&= -e_2 \neq 0.
\end{align*}
| {
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"source": "stackexchange",
"question_score": "4",
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Finding value of an algebraic expression. I am given $$2x = a-\frac{1}{a} \;\text{and} \; 2y=b-\frac{1}{b} .$$
I have to find the value of $ xy +\sqrt{(x²+1)(y²+1)}.$
One way to solve this is to simply putting the value of $x$ and $y$ and get the answer, but that's a very lengthy process though lucid. But I am looking for a short cut method or some kind of tricks to solve the problem.
If anyone can visualize it, I wil be grateful for sharing.
Thank you.
| I do not think there is any special trick in solving the problem.
$$x=\frac{a}{2}-\frac{1}{2a}$$
$$y=\frac{b}{2}-\frac{1}{2b}$$
$$xy=\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab}$$
$$x^2+1 = \frac{a^2}{4}-\frac{1}{2}+\frac{1}{4a^2}+1=\frac{a^2}{4}+\frac{1}{2}+\frac{1}{4a^2}=(\frac{a}{2}+\frac{1}{2a})^2$$
$$y^2+1 = \frac{b^2}{4}-\frac{1}{2}+\frac{1}{4b^2}+1=\frac{b^2}{4}+\frac{1}{2}+\frac{1}{4b^2}=(\frac{b}{2}+\frac{1}{2b})^2$$
Putting in the above values
$$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+\sqrt{(\frac{a}{2}+\frac{1}{2a})^2*(\frac{b}{2}+\frac{1}{2b})^2}$$
Assuming that $=\frac{a}{2}+\frac{1}{2a}>0$ and $\frac{b}{2}+\frac{1}{2b}>0$
$$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{a}{2}+\frac{1}{2a})*(\frac{b}{2}+\frac{1}{2b})$$
$$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{ab}{4}+\frac{a}{4b}+\frac{b}{4a}+\frac{1}{4ab})$$
$$=\frac{ab}{4}+\frac{1}{4ab}+\frac{ab}{4}+\frac{1}{4ab}$$
So,
$$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{2}+\frac{1}{2ab})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
Evaluate $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
I tried so many substitutions but none of them led me to the right answer:
$u=\frac 1{\sqrt{x+3}}$, $u=\frac 1{x+3}$, $u=\sqrt{x}$... I even got to something like $\int_0^1 \frac {u^2}{(u^2+3)^{\frac 32}}du$ or $\int_0^1 \frac {\sqrt{1-3u^2}}{u}du$... and I don't know how to solve these...
| If you change $u=\sqrt{x} \Rightarrow u^2=x \Rightarrow 2udu=dx$, then:
$$\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx=\int_0^1\frac {2u^2}{(u^2+3)^{3/2}}du=\int_0^1\frac {2u^2+6-6}{(u^2+3)^{3/2}}du=\\
2\int_0^1\frac {1}{(u^2+3)^{1/2}}du-6\int_0^1\frac {1}{(u^2+3)^{3/2}}du.$$
Both integrals you can evaluate by $u=\sqrt{3}\tan t$. See this and this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Orthonormal diagonalizable Let:
$$ A = \begin{pmatrix} 5 &2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5 \end{pmatrix} \in Mat_3(\mathbb{R})$$
1) Show that $0$ and $6$ are eigenvalues for $A$ and find the basis for the corresponding eigenspace.
2) Explain why $A$ is orthonormal diagonalizable and find an orthonormal basis for $\mathbb{R^3}$ consisting of eigenvectors for $A$
1) By solving the characteristic polynomial of $A$ it is possible to show that 0 and 6 are eigenvalues for A.
$det(A-\lambda \cdot I) = det( \begin{pmatrix} 5-t &2 & -1 \\ 2 & 2-t & 2 \\ -1 & 2 & 5-t \end{pmatrix} = (5-t)(t^2-7t+6)-2(-2t+12)-1(6-t)=-t^3+12t^2-36t = -t(t^2-12t+36) = -t((t-6)(t-6))=-t(t-6)^2$
To solve $-t(t-6)^2 = 0$ we either have $0$ or $6$, which means that 0 and 6 are eigenvalues for $A$.
The basis for the eigenspace can be found by calculating the null space and we get that: $E_A(0) = N(A) = span(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix})$ and $E_A(6) = N(A-6I) = span(\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix})$
2) We see that the geometric multiplicity and the algebraic multiplicity are equal to each other, which means that A is diagonalizable.
How do I go on from here? I know that I have an invertible matrix $P$ consisting of the eigenvectors $P=\begin{pmatrix} 1 & 2 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & -1 \end{pmatrix}$ such that $D = P^{-1}AP$. But this only means that it is diagonalizable and not orthogonal diagonalizable.
| Note that $A$ is symetric hence by the spectral theorem it's orthogonal diagonalizable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the initial value problem $ (x+1)^2 dx+(2xy+x^2-1)dy =0 , \ y(0)=1 \ $ Solve the initial value problem $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1 $$
Answer:
Consider the above equation with $ \ M(x,y)dx+N(x,y)dy=0 \ $ , then we get
$ M(x,y)=(x+1)^2, \ N(x,y)=2xy+x^2-1 \ $
Since $ \ \frac{\partial M}{\partial {y}}=0 \neq 2y+2x=\frac{\partial N}{\partial {x}} \ $ , the equation is not exact.
Let $ \ \mu =x^a y^b \ $ is an integrating factor.
Multiplying the differential equation by $ \ \mu=x^a y^b \ $ , we get
$$ (x^{a+2}y^b+2x^{a+1}y^b+x^ay^b)dx+(2x^{a+1} y^{b+1}+x^{a+2} y^b-x^ay^b)dy=0 \ $$ Consider it with $ M'dx+N'dy=0 \ $ , we gte
$ M'= x^{a+2}y^b+2x^{a+1}y^b+x^ay^b , \ N'=2x^{a+1} y^{b+1}+x^{a+2} y^b-x^ay^b \ $
Since the new equation is exact , we have
$ \frac{\partial M'}{\partial {y}}=\frac{\partial N'}{\partial {x}} \\ \Rightarrow bx^{a+2}y^{b-1}+2bx^{a+1} y^{b-1}+bx^a y^{b-1}=(a+1)2x^{a} y^{b+1}+(a+2)x^{a+1} y^{b}-ax^{a-1} y^{b} $
From this I have to find out the values $ \ a, \ b \ $.
But I can not find $ a,b \ $
Help me
| Hint:
$(x+1)^2~dx+(2xy+x^2-1)~dy=0$
$(2xy+x^2-1)\dfrac{dy}{dx}=-x^2-2x-1$
$\left(y+\dfrac{x}{2}-\dfrac{1}{2x}\right)\dfrac{dy}{dx}=-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$ with $y(0)=1$
This belongs to an Abel equation of the second kind.
Let $u=y+\dfrac{x}{2}-\dfrac{1}{2x}$ ,
Then $y=u-\dfrac{x}{2}+\dfrac{1}{2x}$
$\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{1}{2}-\dfrac{1}{2x^2}$
$\therefore u\left(\dfrac{du}{dx}-\dfrac{1}{2}-\dfrac{1}{2x^2}\right)=-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$
$u\dfrac{du}{dx}=\left(\dfrac{1}{2}+\dfrac{1}{2x^2}\right)u-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of trig identity using t-formulae Show that $$2\arctan x = \arccos\frac{1-x^2}{1+x^2}$$
if $ x > 0$, and
$$2\arctan x = -\arccos\frac{1-x^2}{1+x^2}$$ if $x <0.$
I have been able to equate the expressions by letting $x=\tan(y/2)$, but I do not know how to show the different signs of the equation for $x>0$ and $x<0$.
| The simplest way to go is probably to calculate $\;\cos(2\arctan x)$.
Set $\theta=\arctan x$ and use the duplication formula:
\begin{align}
\cos 2\theta&=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-x^2}{1+x^2},\\
\text{so }\qquad
2\arctan x&\equiv \pm\arccos\frac{1-x^2}{1+x^2}\mod 2\pi.
\end{align}
Now either $\;0\le \theta<\frac\pi2,\;$ so $\;0\le 2\theta<\pi$, which is (included in, strictly speaking) the range of arccos. In this case we conclude that
$$ 2\arctan x=\arccos\frac{1-x^2}{1+x^2},.$$
or $\;-\frac\pi 2<\theta<0$, so $\;-\pi<2\theta<0$, and for the same reason, we conclude in this case that
$$ 2\arctan x=-\arccos\frac{1-x^2}{1+x^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the matrix representation of a linear map Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix}
4 & 6\\
6 & 3
\end{pmatrix}$$
with respect to the basis $B=\bigg\{ \begin{pmatrix}
2 \\
2
\end{pmatrix} , \begin{pmatrix}
3\\
2
\end{pmatrix}\bigg\}$.
Find the matrix representation of $f$ with respect to the canonical basis.
MY ATTEMPT:
Let $b_1=\begin{pmatrix}
2 \\
2
\end{pmatrix}$, $b_2=\begin{pmatrix}
3\\
2
\end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis.
Then I have $f(b_1)=\begin{pmatrix}
4\\
6
\end{pmatrix} \Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$
$f(b_2)=\begin{pmatrix}
6\\
3
\end{pmatrix} \Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$
Solving this system I obtain that $f(e_1)=2e_1 -3e_2=\begin{pmatrix}
2\\
-3
\end{pmatrix}$ and $f(e_2)= 6e_2=\begin{pmatrix}
0\\
6
\end{pmatrix}$.
So the matrix rappresentation is $M_{E}^{E}=\begin{pmatrix}
2 & 0\\
-3 & 6
\end{pmatrix}$.
BUT my theacher says that the solution is $M_{E}^{E}=\begin{pmatrix}
13 & 10\\
0 & 0
\end{pmatrix}$. Where I am wrong?
| $$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$
This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$
The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is?
Let $x= \tan \theta$
$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$
I had to delete the rest of my attempt because it was totally wrong. What are the methods to solve this problem?
| Let
$$
f(x)=2\arcsin\frac{2x}{1+x^2}-\pi x^3.
$$
Then $f'(x)=-3\pi x^2-\frac{4}{1+x^2}<0$ if $|x|>1$, namely $f(x)$ is decreasing in $(-\infty,-1)$ and $(1,\infty)$. Also
$$ f'(x)=-3\pi x^2+\frac{4}{1+x^2}, |x|<1$$
and
$$ f''(x)=-\frac{2x[4+3\pi(x^2+1)^2]}{(x^2+1)^2}, |x|<1. $$
So $f''(x)>0$ if $x\in(-1,0)$ and $f''(x)<0$ if $(0,1)$ and hence $f(x)$ is strictly concave in $(-1,0)$ and strictly convex in $(0,1)$. Note that $f(-1)=f(0)=f(1)=0$. Thus $f(x)=0$ only has three roots $x=-1,0,1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The value of $\sum_{1\leq l< m How to solve this summation ? Also, I'm not sure what does $1\leq l< m <n$ supposed to imply in the development of summation form.
$$\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}$$
This one is from the Galois-Noether Contest in 2018:
Galois-Contest
| \begin{align*}\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\sum_{n=m+1}^\infty\frac{1}{2^n}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\cdot\frac{1}{2^m}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\cdot\frac{1}{6^l}\cdot\frac15\\&=\left(\frac{1}{1-\frac1{30}}-1\right)\cdot\frac15\\&=\frac{1}{29\cdot5}\end{align*}
or rather $$\frac{1}{(5\cdot3\cdot2-1)(3\cdot2-1)(2-1)}$$
| {
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Solving probability problem with complementation approach. Problem
If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt one pair (i.e. a,a,b,c,d where a,b,c, and d are all distinct)?
I was trying to solve this by complementation.
Let us find number of ways we can serve hand so that there is no pair. We can select five different denominations in $\binom{13}{5}$ ways. There are four cards of each domination. Thus total number of ways to serve pair-less hands is $=\binom{13}{5}\times 4^5$. To get number of ways to have at least one pair, we subtract it from total number of ways to serve a hand of five cards: $\binom{52}{5}$. So the final probability will be
$=\frac{\binom{52}{5}-\binom{13}{5}\times 4^5}{\binom{52}{5}} = 0.4929$
The solution given was as follows:
*
*Number of ways to choose one denomination out of 13: $\binom{13}{1}$
*
*Number of ways to select 2 cards of selected denomination: $\binom{4}{2}$
*Number of ways to choose 3 denominations out of remaining 12: $\binom{12}{3}$
*Number of ways to select 1 card each of selected denominations: $\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1}$
*Final probability $$=\frac{\binom{13}{1}\times\binom{4}{2}\times\binom{11}{3}\times\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1} }{\binom{52}{5}}\approx 0.4226$$
Whats going wrong in my solution?
Update
As stated by joriki in answer, I forgot to consider other arrangements. I am listing all below:
*
*all different denominations (i.e. a,b,c,d,e): $\binom{13}{5}4^5$
*two pairs (i.e. a,a,b,b,c): $\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}$
*three of a kind (i.e. a,a,a,b,c): $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}$
*four of a kind (i.e. a,a,a,a,b): $\binom{13}{1}\binom{12}{1}\binom{4}{1}$
So the desired probability becomes:
$$\frac{\binom{52}{5}-\binom{13}{5}4^5-\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}-\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}-\binom{13}{1}\binom{12}{1}\binom{4}{1}}{\binom{52}{5}}\approx 0.424$$
So this seems to closely match to the solution given. However, I am afraid that I might have missed something because the answer differs in fractions. So please check if I have mistaken somewhere.
| You ignored the part "where $a$, $b$, $c$, and $d$ are all distinct". You counted the hands that contain at least one pair, not the ones that contain exactly one pair.
| {
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Finding the pattern to a general matrix Consider the following $(n+1) \times (n+1)$ matrix:
$$\begin{bmatrix} n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n & n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n & n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & 6-n & n-2 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots &-3& 4-n & n-1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -2 & 2-n & n \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & -1 & -n \\\end{bmatrix}$$
Show the eigenvalue is $0$ with multiplicity $n+1$.
I have tried row reducing to no avail, it is doable for any particular instance but surely there is a simpler way to find the determinant with a more connected pattern across all integers $n$.
For clarity, the problem is equivalent to showing:
$\begin{bmatrix}x+ n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n &x+ n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n &x+ n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &x+n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots &x+ 6-n & n-2 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots &-3&x+ 4-n & n-1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -2 &x+ 2-n & n \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & -1 &x -n \\\end{bmatrix} = x^{n+1}$
| Add row 1 to row 2
and row 2 is
[0, n-1, 2, 0, ...].
Then add row 2 to row 3
and row 3 is
[0, 0, n-2, 3, 0, ...].
Then add row 3 to row 4
and row 4 is
[0, 0, 0, n-3, 4, 0, ...].
I think that,
with a proper specification
of each row,
this can reduce the matrix
to upper triangular
with the main diagonal being
[n, n-1, n-2, ...]
and the diagonal above it
being
[1, 2, 3, ...],
and the rest of the matrix being zero.
This might be enough
to show what is wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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find the length of the curve $y= \int_{-2}^x\sqrt{3t^4-1} \, dt$ I'm stuck for a while not sure how to continue, or if there's a mistake that I did that prevent me to continue.
$$y= \int_{-2}^x\sqrt{3t^4-1} \ dt \ , \ -2≤x≤-1$$
$$ y = F(x) - F(-2) $$
$$ y' = f(x) - f(-2) $$
$$ y' = \sqrt{3x^4-1} - \sqrt{47} $$
$$ (y')^2 = 3x^4-1 -2 \sqrt{(3x^4-1)\ 47} + 47$$
$$ 1+(y')^2 = 3x^4 -2 \sqrt{(3x^4-1)\ 47} + 47$$
Now how can I integrate this:
$$ \int_{-2}^{-1} \sqrt{3x^4 -2 \sqrt{(3x^4-1)\ 47} + 47} $$
And I cant get it to $\int_{-2}^{-1} \sqrt{ [\sqrt{3x^4-1} \ - \sqrt{47}]^2}$
| Notice by the FTC
$$ y' = \sqrt{3x^4-1}$$
and Thus,
$$ (y')^2 +1 = 3x^4 $$
it follows that
$$ \mathcal{L} = \int\limits_{-2}^{-1} \sqrt{3}x^2 dx = ... $$
| {
"language": "en",
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"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Showing that $P(x)$, where $P(x) = P(x+2)-x^2-2$ for all $x$, is a third-degree polynomial I want to show that $P(x)$ is a third degree polymomial:
$P(x) = P(x+2) -x^2-2$ for every real $x$
It is not so difficult, but I am not seeing the proof straight away.
| Let:
$$\begin{align}P(x+2)&=a_0(x+2)^n+a_1(x+2)^{n-1}+a_2(x+2)^{n-2}+\cdots+a_n\\
P(x)&=\color{red}{a_0}x^n+\color{blue}{a_1}x^{n-1}+a_2x^{n-2}+\cdots+a_n \\
x^2+2=P(x+2)-P(x)&=2na_0x^{n-1}+(2n(n-1)a_0+2(n-1)a_1)x^{n-2}+\cdots \Rightarrow \\
x^2&=2na_0x^{n-1} \Rightarrow \\
2&=n-1 \quad \text{and} \quad 2na_0=1\Rightarrow \\
n&=3 \quad \text{and} \quad \color{red}{a_0}=\frac16. \end{align}$$
Thus $n=3$. You can continue to find the next coefficient:
$$2n(n-1)a_0+2(n-1)a_1=0 \Rightarrow \color{blue}{a_1}=\frac{-2}{4}=-\frac12.$$
| {
"language": "en",
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"question_score": "5",
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Showing that $f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$ has no local extrema I have an exercise asking to find the local extrema of $$ f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$$if they exist... Wolfram Alpha tells me they don't, but I think I've spent too long trying to show it, so I'm wondering what I missed, resulting in a more time-consuming task. Below is my work:
$f$ is continuous on the whole plane except the origin.
For $x,y\ne0$$$f(x,y)=\begin{cases}(x-y)\log\left(2x^2+y\right) &x,y>0 \\ (x+y)\log\left(2x^2-y\right) &x>0>y\\-(x+y)\log\left(2x^2+y\right) &x<0<y \\ -(x-y)\log\left(2x^2-y\right) &x,y<0\end{cases}$$hence $$f_x(x,y)=\begin{cases}\log\left(2x^2+y\right)+\dfrac{4x(x-y)}{2x^2+y} &x,y>0 \\ \log\left(2x^2-y\right)+\dfrac{4x(x+y)}{2x^2-y} &x>0>y \\ -\log\left(2x^2+y\right)-\dfrac{4x(x+y)}{2x^2+y} &x<0<y \\ -\log\left(2x^2-y\right)-\dfrac{4x(x-y)}{2x^2-y} &x,y<0 \end{cases} $$ and
$$f_y(x,y)=\begin{cases}-\log\left(2x^2+y\right)+\dfrac{x-y}{2x^2+y} &x,y>0 \\ \log\left(2x^2-y\right)-\dfrac{x+y}{2x^2-y} &x>0>y \\ -\log\left(2x^2+y\right)-\dfrac{x+y}{2x^2+y} &x<0<y \\ \log\left(2x^2-y\right)+\dfrac{x-y}{2x^2-y} &x,y<0 \end{cases}.$$ Setting both partial derivatives equal to $0$ I got $$\begin{cases}(x-y)(4x+1) &x,y>0 \\ (x-y)(4x+1) &x>0>y \\ (x+y)(4x-1) &x<0<y \\ (x-y)(4x-1) &x,y<0 \end{cases} \iff y=\pm x\ne0.$$ So for $x_0\ne0$ and small $h_1,h_2$ I examined \begin{align}\operatorname{sgn}\left\{f(x_o+h_1,\pm x_0+h_2)-f(x_0,\pm x_o)\right\}&:=\operatorname{sgn}(h_3\log(2x_0^2+|x_0|+h_4)) \\&= \operatorname{sgn}(h_3\log(2x_0^2+|x_0|)) \end{align}which obviously depends on how $h_3$ approaches $0$, so the $(x_0,\pm x_0)$ are saddle points.
EDIT: ok on second thought I forgot to show that $h_3$ doesn't depend on $h_1, h_2$ in a way that would make it approach $0$ only from the left/right, but let's pretend I did show it
Now I should study the behaviour around points $(x_0,0)$ and $(0,y_0)$ , and this is where I've been stuck. After a while I tried looking at things like $f(x_o,h)-f(x_0,0)$ but it didn't help... what am I missing?
| In the following I am assuming that $(x,y) \neq (0,0)$.
Since $f(x,y) = f(|x|,|y|)$, we need only examine $x\ge 0, y \ge 0$.
For $x>0,y>0$ we have
$f_x(x,y) = {1 \over 2x^2+y} (4x (x-y) + (2x^2+y) \log(2x^2+y))$,
$f_y(x,y) = {1 \over 2x^2+y} ((x-y) - (2x^2+y) \log(2x^2+y))$.
Suppose $f_x(x,y) = f_y(x,y) = 0$, then the above gives
$x = - { 1 \over 4}$, hence $f$ has no stationary points in $x>0, y>0$.
The axes need more care, as $f$ is not differentiable there.
Now consider the $x$ axis, we have $f_x(x,0) = \log(2 x^2)+2$, which has
a
zero at $x^*={1 \over \sqrt{2}e }$. A little work shows that $x^*$ is a strict local $\min$ (of $x \mapsto f(x,0)$).
If we consider the $y \ge 0$ side of the graph, we have $f_y(x^*,0) > 0$ (the one sided derivative) hence, with a little finesse, we can conclude that $f$ has a local $\min$ at $(x^*,0)$.
Now consider the $y$ axis, we have $f(0,y) = -y \log(y)$, which has a strict $\max$ at $y^* = {1 \over e}$. As above, by restricting to $x \ge 0$ and considering the one sided derivative, we see that
$f_x(0,y^*) = -1 <0$ from which we can conclude that $(0,y^*)$ is a strict
local $\max$.
To summarise, $f$ has strict local $\min$ at $(\pm {1 \over \sqrt{2} e}, 0)$ and
strict local $\max$ at $(0, \pm {1 \over e})$. There are no other
extrema.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.
PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
I LOVE YOU HONEY.
Anyone knows? Are we gonna get an integer number?
My attempt:
Does this help?
$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$
(long division)
\begin{align*}
I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\
&=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\
&=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2
\end{align*}
after the substitution
\begin{gather*}
(x-2)^{1/2}=u\\
du=\frac1{2(x-2)^{1/2}}dx\\
u^2=x-2\\
(x-1)^{1/2}=(u^2+1)^{1/2}
\end{gather*}
Update: This may help us proceed.
I tried to proceed:
$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$
after $u^2+1=t$ and $dt=2udu$
\begin{align*}
u^4-4u^2-4
&= (u^2+1)^2-(6u^2+5) \\
&= (u^2+1)^2-6(u^2+1)+1 \\
&= ((u^2+1)-3)^2-8
\end{align*}
I wonder whether this question can be solved from here?
Update:
This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.
Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.
| Looking at your previous deleted post, one answer suggested to use Euler subtitution
$$\sqrt{x^2-3x+2}=t+x\implies x=\frac{2-t^2}{2t+3}\implies dx=-\frac{2 (t+1) (t+2)}{(2 t+3)^2}\,dt$$ Replacing, we arrive to
$$\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}}=\frac{2 (t+1)^2 \left(3 t^4-4 t^3-2 t^2+56 t+60\right)}{(2 t+3)^4}$$ Now let $2t+3=u$ to make the integrand
$$\frac{3 u^2}{64}-\frac{25 u}{32}+\frac{317}{64}-\frac{135}{16 u}+\frac{317}{64
u^2}-\frac{25}{32 u^3}+\frac{3}{64
u^4}$$ and the antiderivative
$$\frac{u^3}{64}-\frac{25 u^2}{64}+\frac{317 u}{64}-\frac{135}{16} \log
\left({u}\right)-\frac{317}{64 u}+\frac{25}{64 u^2}-\frac{1}{64
u^3}$$ For $t$, the bounds were $(\sqrt 2,-1)$; so, for $u$, they are $(2\sqrt 2+3,1)$ giving as a result
$$ \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx=\frac{135}{16} \log \left(3+2 \sqrt{2}\right)-\frac{101}{4 \sqrt{2}}\approx -2.98127$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prime Numbers and Canonical Factorization. Find $n$ such that $2^{n} \mid 3^{1024}−1$, where $n$ is an integer. I factorized $3^{1024}−1$ as $a^{2}-b^{2}$ and showed $2 \mid 3^{1024}-1$. Help me to get $n$.
| $$2^n||3^{1024}-1$$
We know that $2^{10}=1024$ and $a^2-b^2=(a+b)(a-b)$
Now,
$$3^{2^{10}}=(3^{2^9}+1)(3^{2^9}1)$$
$$=(3^{2^9}+1)(3^{2^8}+1)(3^{2^8}-1)$$
$$=(3^{2^9}+1)(3^{2^8}+1)(3^{2^7}+1)(3^{2^7}-1)$$
$$.$$
$$.$$
$$=(3^{2^9}+1)(3^{2^8}+1)(3^{2^7}+1)....(3^{2^1}+1)(3^{2^0}+1)(3-1)$$
To find the largest $n$, $2^n||3^{1024}-1$, $2^n$ should divide $3^{1024}-1$
$$3^{2^{10}}=(3^{2^9}+1)(3^{2^8}+1)(3^{2^7}+1)(3^{2^6}+1)(3^{2^5}+1)(3^{2^4}+1)(3^{2^3}+1)(3^{2^2}+1)(3^{2^1}+1)(3^{2^0}+1)(3-1)$$
Note that each factor divides $2$ only once except $3^{2^0}+1$ which divides it by $2$ times
Therefore, there $11$ factors out of which $3^{2^0}+1$ divides $2$ times.
$$n=12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the PDE: $(xy-x^2)\frac{\partial u}{\partial x} + y^2 \frac{\partial u}{\partial y} + (e^{y/x}+yz)\frac{\partial u}{\partial z} = 0$ As stated in the title, I want to solve the following PDE:
$(xy-x^2)\frac{\partial u}{\partial x} + y^2 \frac{\partial u}{\partial y} + (e^{y/x}+yz)\frac{\partial u}{\partial z} = 0,$
with the condition that $u = \frac{y}{1+y}$ given $z=e^{y/x}$.
Finding two independent integrals $\psi_{1}$ and $\psi_{2}$ of this equation would give me a general solution $u=\Phi(\psi_{1}, \psi_{2})$, $\Phi \in C^{1}$.
The corresponding symmetrical system here is: $$ \frac{dx}{xy-x^2} = \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} = M. $$
By using $y^2dx = (xy-x^2)dy$, I was able to obtain the first integral of this equation as $$\psi_{1} = \frac{y}{x} + \ln{y}.$$
However, finding the second one eludes me. I've tried using $d(e^{y/x}) = -\frac{y}{x^2}e^{y/x}dx + \frac{1}{x}e^{y/x}dy$, to, for example, say $$\frac{-\frac{y}{x^2}e^{y/x}dx + (\frac{1}{x}e^{y/x}+z)dy-ydz}{-\frac{y^2}{x}e^{y/x} + ye^{y/x} +\frac{y^2}{x}e^{y/x} - ye^{y/x} -y^2z + y^2z} = M,$$ therefore $-\frac{y}{x^2}e^{y/x}dx + (\frac{1}{x}e^{y/x}+z)dy-ydz=0$, but I can't integrate this $1$-form to get the second integral.
Is there a good way to find the second integral in this equation?
| $$ \frac{dx}{xy-x^2} = \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} \quad \text{is OK.} $$
There is a sign mistake in your first integral. $\psi_{1} = \frac{y}{x} + \ln|y|$ is false and should be :
$$\psi_{1} = \frac{y}{x} -\ln|y|$$
Or, equivalently with $c_1=e^{\psi_1}$ :
$$c_1=\frac{1}{y}e^{y/x}$$
$ \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} = \frac{dz}{c_1y + yz}$
$$ \frac{dy}{y} = \frac{dz}{c_1 + z}$$
Integrating leads to $\frac{c_1 + z}{y}=c_2$
$$c_2=\frac{\frac{1}{y}e^{y/x} + z}{y}=\frac{1}{y^2}e^{y/x}+\frac{z}{y}$$
The general solution of the PDE is $\qquad u=\Phi(c_1,c_2)$.
Boundary condition: $\quad u=\frac{y}{1+y}\quad$ on $\quad z=e^{y/x}$
$c_1=\frac{1}{y}e^{y/x}=\frac{z}{y}\quad$ and $\quad c_2=\frac{1}{y^2}e^{y/x}+\frac{z}{y} = \frac{z}{y^2}+\frac{z}{y}=\frac{1+y}{y}\frac{z}{y}$
$\frac{y}{1+y}=\Phi\left(\frac{z}{y}\:,\:\frac{1+y}{y}\frac{z}{y}\right)$
An obvious solution is: $\quad u=\Phi(c_1,c_2)=\frac{c_1}{c_2} =\frac{\frac{1}{y}e^{y/x}}{\frac{1}{y^2}e^{y/x}+\frac{z}{y}}$
$$u=\frac{y}{1+y\,z\,e^{-y/x}}$$
| {
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"url": "https://math.stackexchange.com/questions/2815044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be the right.
| Let $\, X := e^{ia}, Y := e^{ib}, Z := e^{ic}. \,$ Now expand and factor
$$ W \!:=\! \sin^2 a \!+\! \sin^2 b \!+\! \sin^2 c \!-\! 2(1 \!-\! \cos a \cos b \cos c) \!=\!
\frac{(X Y \!-\! Z) (Y Z \!-\! X) (Z X \!-\! Y) (X Y Z \!-\! 1)}{(2 X Y Z)^2}.\,$$
This leads to
$$\, W = 4 \sin\frac{a+b-c}2 \sin\frac{a-b+c}2 \sin\frac{-a+b+c}2 \sin\frac{a+b+c}2 \,$$
which is zero iff at least one of
$\, a+b-c,\, a-b+c,\, -a+b+c,\, a+b+c \,$ are multiples of $\,2\pi.\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum_{n=1}^{\infty} 3^n \sin(\frac{1}{4^nx})$ I wish to prove the convergence of:
$$\sum_{n=1}^\infty 3^n \sin\left(\frac 1 {4^nx}\right)$$
for $1\le x \lt \infty$, using Cauchy's criterion.
Here is what I tried:
\begin{align}
|S_{n+p}-S_n| & = \left| 3^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right)
+ \cdots+3^{n+p} \sin\left(\frac 1 {4^{n+p}x}\right)\right| \\[10pt]
& \le \left|4^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right) \cdots 4^{n+p} \sin\left(\frac 1 {4^{n+p}x}\right) \sin\left(\frac 1 {4^{n+1}x}\right)(4^{n+1}+\cdots+4^{n+p}) \right|
\end{align}
I tried using geometric series sum from here but came empty handed.
how can I show that $|S_{n+p}-S_n|\lt \varepsilon$?
| If you absolutely want to use Cauchy's criterion, you can onserve that, if $n$ is large enough, $0<\frac1{4^n x}<\frac\pi2$, hence
$$0<\sin\frac1{4^n x}<\sin\frac1{4^{n+1}x}<\dotsm<\sin\frac1{4^{n+p}x},$$
and $\;\sin \dfrac1{4^{n+k}x}< \dfrac1{4^{n+k}x}<\dfrac1{4^{n+k}}$, so that, by the triangle inequality,
$$|S_{n+p}-S_n|\le 3^{n+1}\dfrac1{4^{n+1}}+\dots3^{n+p}\dfrac1{4^{n+p}}=\frac{\bigl(\frac 34\bigr)^{n+1}-\bigl(\frac 34\bigr)^{n+p+1}}{\frac14}<4\,\Bigl(\frac 34\Bigr)^{n+1}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$ where $a,b$ are finite natural numbers
Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers
I have spent about a day thinking over this problem. I tried integration by parts, differentiating under integral sign (Feynman's trick, with respect to $a, b$), using some trigonometric and logarithmic properties like changing $\cos^2\theta$ to $\cos2\theta$ and hereafter some logarithmic properties, etc., but failed miserably.
Also tried to use the property that $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ in-between, but still to no avail. I also tried to find similar questions on MSE but did not get a related one.
Can someone please help me to solve this integral?
Edit
My try (Feynman's trick) :
Let $$I(a)=\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) d\theta$$
Hence $$I'(a) =\frac 1a \int_0^{\pi/2} \frac {2a^2\cos^2\theta d\theta}{a^2\cos^2\theta +b^2\sin^2\theta}$$ $$=\frac 1a\left[ \frac {\pi}{2}+\int_0^{\pi/2} \frac {a^2\cos^2\theta -b^2\sin^2\theta}{a^2\cos^2\theta +b^2\sin^2\theta}\right]$$
Wherein between I broke $2a^2\cos^2\theta=a^2\cos^2\theta +b^2\sin^2\theta+a^2\cos^2\theta -b^2\sin^2\theta$
But now how do I continue further
| If we assume $a,b>0$ and set
$$ I(a,b)=\int_{0}^{\pi/2}\log(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta $$
we have $I(a,b)=I(b,a)$ from the substitution $\theta\mapsto\frac{\pi}{2}-\theta$. On the other hand $I(a,a)=\pi\log(a)$ is trivial, so $I(a,b)=\pi\log\left(\frac{a+b}{2}\right)$ is a very reasonable conjecture. Indeed, it can be proved by computing
$$ \frac{\partial I}{\partial a} = \int_{0}^{\pi/2}\frac{2a\cos^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta $$
as suggested in the comments, i.e. via $\theta\mapsto\arctan u$ and partial fraction decomposition.
An alternative approach is to notice that
$$\begin{eqnarray*} I(a,b) &=& 2\,\text{Re}\int_{0}^{\pi/2}\log\left(a\cos\theta+ib\sin\theta\right)\\&=&\pi\log\left(\frac{a+b}{2}\right)+\text{Re}\int_{0}^{\pi}\log\left(e^{i\theta}+\frac{a-b}{a+b}\right)\,d\theta\end{eqnarray*}$$
where the last integral is a purely imaginary number by the residue theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can't prove an equation using induction I have an equation
$$
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} = 2 - \frac{n + 2}{2^n}
$$
Below is what I have already done:
$$
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} + \frac{n + 1}{2^{n+1}} = 2 - \frac{n + 2}{2^n} + \frac{n + 1}{2^{n+1}}=
2 - \frac{3n + 5}{2^{n + 1}}
$$
I want the right part of the equation to be transformed to $$ 2-\frac{n + 1 + 2}{2^{n+1}}$$
Maybe do I do something wrong? I will be appreciated for any help.
| You forgot about the negative sign:
$$
-\frac{n+2}{2^{n}}+\frac{n+1}{2^{n+1}}=\frac{-(2n+4)+n+1}{2^{n+1}}=\frac{-n-3}{2^{n+1}}=-\frac{(n+1)+2}{2^{n+1}}.
$$
| {
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Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$.
I am rather stuck though. Is there a general method for going about this? I always find myself having to guess which is not so useful here. I notice in the first one that if $x=y=z$ then the polynomial is zero. I'm not sure how useful this is though in the case of three variables.
| And
$$
(x - a)^3 (b - c)^3 + (x - b)^3 (c - a)^3 + (x - c)^3 (a - b)^3 = 3 (a - b) (a - c) (b - c) (a - x) (b - x) (c - x)
$$
This is easy to conclude because $a,b,c$ are roots
| {
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Determining rational functions that simplify to a simple expression
Question: Is there a way to find, if possible, another way to write $(\sqrt[3]2-1)^2$ and $(\sqrt[3]4-1)^2$, in the form of $\frac {a+b}{c+d}$?
What I meant was that, let's take the second expression $(\sqrt[3]4-1)^2$ as an example. The expression is also equal to$$(\sqrt[3]4-1)^2=3\left(\frac {2-\sqrt[3]4}{2+\sqrt[3]4}\right)$$
You can check this with Wolfram Alpha. However, I'm not sure how to do that with $(\sqrt[3]2-1)^2$ and what the work is. I feel like this should be such a basic thing to do, but I'm struggling with where to begin.
| For $\,(\sqrt[3]4-1)^2\,$, using $(a-b)(a^2+ab+b^2)=a^3-b^3$ gives $\,\displaystyle
\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{4^2}+\sqrt[3]{4}+1\right) = 3
$.
Therefore $\,\displaystyle
\sqrt[3]{4}-1=\frac{3}{2 \sqrt[3]{2}+\sqrt[3]{4}+1} = \frac{3}{\left(\sqrt[3]{2}+1\right)^2}\,$.
Then, using $\,a^2-b^2=(a-b)(a+b)\,$ gives the following, which is equivalent to the one posted:
$$\require{cancel}
\left(\sqrt[3]4-1\right)^2=\frac {3\left(\sqrt[3]4-1\right)}{\left(\sqrt[3]{2}+1\right)^2} = \frac{3\left(\sqrt[3]2-1\right)\bcancel{\left(\sqrt[3]2+1\right)}}{\left(\sqrt[3]{2}+1\right)^\bcancel{2}} = 3 \cdot \frac{\sqrt[3]2-1}{\sqrt[3]2+1}
$$
For $\,\left(\sqrt[3]{2}-1\right)^2\,$, using similar identities:
$$\sqrt[3]4-2 \sqrt[3]{2}+1=-\left(\sqrt[3]{4^2}-\sqrt[3]{4}+1\right)+2 = \dfrac{-5}{\sqrt[3]{4}+1}+2=\dfrac{2\sqrt[3]{4}-3}{\sqrt[3]{4}+1}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\dfrac{4 - 3 \sqrt[3]{2}}{2+\sqrt[3]{2}}\,$$
| {
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Induction Problem from Putnam and Beyond I have copied the first step of the solution below in order to save time and space, the problem, with full solution, can be found in the book Putnam and Beyond.
Question:
Let $a_1, a_2, \ldots, a_n, \ldots$ be a sequence of distinct positive integers. Prove that for any positive integer $n$,
$$a_1^2 + a_2^2 + \cdots + a_n^2 \ge \tfrac{2n + 1}{3}(a_1 + a_2 + \cdots + a_n).$$
Solution:
$$a_{n+1}^2 \ge \tfrac{2n + 3}{3}a_{n+1} + \tfrac{2}{3}(a_1 + a_2 + \cdots + a_n) \ldots$$
Note:
I understand how to make the jump from the question to the inductive step, namely replacing $n$ with $n+1$, but I don't see where the term $\tfrac{2}{3}(a_1 + a_2 + \cdots + a_n)$ came from in the solution. Any help would be appreciated.
| An inductive approach would use $$a_1{}^2 + \dots + a_n{}^2 \ge \frac{2n + 1}3\bigg(a_1 + \dots + a_n\bigg)$$ to establish $$a_1{}^2 + \dots + a_n{}^2 + a_{n+1}{}^2\ge \frac{2(n+1) + 1}3\bigg(a_1 + \dots + a_n + a_{n+1}\bigg)$$
by using transitivity of $x \ge y \text{ and } y \ge z \text{ implies } x \ge z$. So $x,y,z$ must be figured out. We can choose $x$ and $y$ as:
$$x = a_1{}^2 + \dots + a_n{}^2$$
$$y = \frac{2n+1}{3}\bigg(a_1 + \dots a_n\bigg)$$
but then to figure out what $z$ has to be, we have to choose $z$ to make the last equation $x \ge z$ :
$$z = \frac{2(n+1) + 1}3\bigg(a_1 + \dots + a_n + a_{n+1}\bigg) - a_{n+1}{}^2$$
So we are given $x \ge z$ by inductive hypothesis, we know the goal is $x \ge z$, we need to establish now $y \ge z$:
$$\frac{2n+1}{3}\bigg(a_1 + \dots a_n\bigg) \ge \frac{2(n+1) + 1}3\bigg(a_1 + \dots + a_n + a_{n+1}\bigg) - a_{n+1}{}^2$$
which is
$$a_{n+1}{}^2 \ge \bigg(\frac{2n + 1}3 + \frac{2}{3}\bigg)\bigg(a_1 + \dots + a_n + a_{n+1}\bigg) - \frac{2n+1}{3}\bigg(a_1 + \dots a_n\bigg)$$
which is
$$a_{n+1}{}^2 \ge \frac{2n + 1}3 a_{n+1} + \frac{2}{3}\bigg(a_1 + \dots + a_n + a_{n+1}\bigg) $$
| {
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Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$? Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ?
| Let $N = ab$, we count the number of integers $0\leq t \leq N$ that can be written as
$$
t = ma+nb
$$
such that $m,n\geq 0$. Let there be $R$ such numbers. Since the largest non-representable (positive)-integer is $ab-a-b$, all non-representable numbers are in $[0,N]$ and hence we know there are $N+1 - R$ of them.
If $n\geq a$, then we replace it with
$$
ma+nb = (m+b)a + (n-a)b = m'a+n'b
$$
So with repeated application we may assume that $0\leq n < a$. We require
$$
0 \leq ma+nb \leq N \implies 0 \leq ma \leq N-nb = (a-n)b
$$
Hence we get the bounds of $m$ as
$$
0 \leq m \leq \left \lfloor \frac{(a-n)b}{a}\right \rfloor
$$
This means the number of integers we can represent for a given $n$ is
$
\left \lfloor \frac{(a-n)b}{a} \right \rfloor + 1
$. Now summing this over all possibilities $n=0,1,\dots a-1$, the number of representable integers $R$ is
$$
\begin{align*}
R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1
&= \left\lfloor \frac{(a)b}{a} \right\rfloor + \left\lfloor \frac{(a-1)b}{a} \right\rfloor + \cdots + \left\lfloor \frac{(2)b}{a} \right\rfloor + \left\lfloor \frac{(1)b}{a} \right\rfloor + a\\
R &= a + b + \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\
\end{align*}
$$
Next we observe that if $a$ does not divide $kb$, then for some integer $t$
$$
t < \frac{kb}{a} < t+1 \Longleftrightarrow -t-1 < \frac{-kb}{a} < -t
$$
In particular this applies for $1\leq k\leq a-1$, since $\gcd(a,b)=1$ and $a>1$. We use the inequalities to obtain
$$
\left \lfloor \frac{-kb}{a}\right \rfloor = -t-1 = - \left \lfloor \frac{kb}{a}\right \rfloor -1,
$$
This lets us get another equation of $R$ as
$$
\begin{align*}
R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1
&= \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor + b + 1 \\
&= a(b+1) + \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor\\
&= a(b+1) + \sum_{k=1}^{a-1}\left\lfloor \frac{-kb}{a} \right\rfloor \\
&= a(b+1) + \sum_{k=1}^{a-1}-\left\lfloor \frac{kb}{a} \right\rfloor -1\\
R &= ab + 1 - \sum_{k=1}^{a-1}\left\lfloor \frac{kb}{a} \right\rfloor
\end{align*}
$$
Therefore adding both equations of $R$, we have
$$
2R = ab + a + b + 1 = (a+1)(b+1)
$$
giving us $R=(ab+a+b+1)/2$. Notice that $R$ is not an integer if and only if $a,b$ are both even, which is not possible since $\gcd(a,b)=1$.
Finally, the number of integers considered is $ab+1$, so the number of integers not representable is
$$
ab+1 - \frac{ab+a+b+1}{2} = \frac{ab -a - b+1}{2} = \frac{(a-1)(b-1)}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ is a UFD. Contradiction? Considering the ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$, I know that this ring is an euclidian domain and therefore a unique factorization domain.
Now, $23=(3+4\sqrt{2})(3-4\sqrt{2})=(11+7\sqrt{2})(11-7\sqrt{2})$.
$(3+4\sqrt{2}),(3-4\sqrt{2}),(11+7\sqrt{2}),(11-7\sqrt{2})$ are irreducibles in $\mathbb{Z}[\sqrt{2}]$, since for example: $$N(3+4\sqrt{2})=(3+4\sqrt{2})(3-4\sqrt{2})=-23$$Where $N$ is the norm $N(a+b\sqrt{d})=(a+b\sqrt{d})(a-b\sqrt{d})$. Since $23$ is prime, $(3+4\sqrt{2})$ is irreducible.
Therefore $23$ is not expressed by a unique product of irreducibles, doesn't this contradicts the fact that $\mathbb{Z}[\sqrt{2}]$ is UFD?
| There is just a calculation mistake, since
$$
(3+4\sqrt{2})(3-4\sqrt{2}) = 9 - 32 = -23
$$
| {
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Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t}+2}}}^{3 \left({\frac{1}{t}^2-4} \right)}$$
then I am left with:
$$\left ( e^{3} \right )^{3\left(\frac{1}{t^2}-4\right)}$$ which I get by using Euler number.
The answer is $e^9$, but clearly the answer I get is $(e^9)^{\text{expression}}$ which is not equal to the answer.
| To get the result notice that $3x-6 = 9 \cdot \frac{x+2}{3} - 12$ then you limit is
$$
\lim_{x \to +\infty} \left[ \left( 1 + \frac{1}{\frac{x+2}{3}} \right)^{\frac{x+2}{3}} \right]^9 \cdot \left( 1 + \frac{3}{x+2} \right)^{-12}.
$$
The second term goes to $1$, while the term inside the brackets goes to $e$. Therefore, you get that the limit is equal to $e^9$.
| {
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How to prove that $1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-1}(1+\frac1n))...)))=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+\frac1{n!}$? I'm trying to prove that
$$1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-1}(1+\frac1n))...)))=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+\frac1{n!}$$
Using induction, suppose that
$$1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-2}(1+\frac1{(n-1)}))...)))=1+\frac1{1!}+\frac1{2!}+...+\frac1{(n-1)!}$$
Then
$$1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-2}(1+\frac1{n-1}(1+\frac 1n))...)))\\
=1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-2}(1+\frac1{n-1}+\frac1{(n-1)n})...)))\\
=(1+\frac1{1!}+\frac1{2!}+...+\frac1{(n-1)!})+\frac1{n!}$$
But I couldn't completely justify the last equality. Could anyone explain this for me, please? Thanks!
| You can also start calculating the inner bracket first:
\begin{align}
&1+\frac11\left(1+\frac12\left(1+\cdots + \frac1{n-2}\left(1+\frac1{n-1}\left(1+\frac1n\right)\cdots\right)\cdots\right)\right) \\&= 1+\frac11\left(1+\frac12\left(1+\cdots + \frac1{n-2}\left(1+\frac{n+1}{n(n-1)}\right)\cdots\right)\right) \\&=
1+\frac11\left(1+\frac12\left(1+\cdots + \frac1{n-3}\left(1+\frac{n(n-1) + n+1}{n(n-1)(n-2)}\right)\cdots\right)\right) \\
&= 1+\frac11\left(1+\frac12\left(1+\cdots + \frac1{n-4}\left(1+\frac{n(n-1)(n-2) +n(n-1) + n+1}{n(n-1)(n-2)(n-3)}\right)\cdots\right)\right) \\
&= \cdots\\
&= \frac{n(n-1)(n-2) \cdots 2 \cdot 1 + n(n-1)(n-2) \cdots 3\cdot2 + \cdots n(n-1)(n-2) + n(n-1) + n+1}{n(n-1)(n-2) \cdots 2 \cdot 1} \\
&= 1 + \frac1{1!} + \frac1{2!} + \cdots + \frac1{(n-3)!} + \frac1{(n-2)!} + \frac1{(n-1)!} + \frac1{n!}
\end{align}
| {
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Evaluate $\lim_{n \to \infty} \int_{0}^{1} x^2 \left(1+\frac{x}{n}\right)^n dx$ Evaluate $$I=\lim_{n \to \infty} \int_{0}^{1} x^2 \left(1+\frac{x}{n}\right)^n dx$$
My try:
we have
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \int_{0}^{1} \left((x+n)^2-2n(x+n)+n^2\right)(x+n)^ndx$$ $\implies$
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \int_{0}^{1}(x+n)^{n+2}-2n(x+n)^{n+1}+n^2(x+n)^n dx$$ $\implies$
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \left(\frac{(n+1)^{n+3}}{n+3}-\frac{2n \times(n+1)^{n+2}}{n+2}+\frac{n^2 \times (n+1)^{n+1}}{n+1}\right)$$ $\implies$
$$I=\lim_{ n\to \infty} (n+1)\frac{(n+1)^{n}}{n^n} \frac{n^2+n+2}{(n+1)(n^2+5n+6)}=e$$
But the answer given is $e-2$
Any thing went wrong?
| I learned a thing or two about the DCT theorem, so let's give it a try. Observe that if $t \in [0,1] \implies \ln(1+t) \le t$ and substitute $t = \dfrac{x}{n}, 0 \le x \le 1, n \ge 1\implies \ln(1+\dfrac{x}{n}) \le \dfrac{x}{n}\implies 1+\dfrac{x}{n} \le e^{\frac{x}{n}}\implies (1+\dfrac{x}{n})^{n} \le e^x\implies |f_n(x)| = x^2(1+\dfrac{x}{n})^n\le x^2e^x=g(x), \text{g is integrable}\implies \displaystyle \lim_{n \to \infty} \displaystyle \int_{0}^1f_n(x)dx =\displaystyle \int_{0}^1\displaystyle \lim_{n\to \infty} f_n(x)=\displaystyle \int_{0}^1 g(x)dx= ...= e-2$ by applying the DCT .
| {
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Integrating $F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}$ along a circle
Let $F:\mathbb R^3 - \{0\}\to \mathbb R$ be given by $$F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}.$$ Compute $\int_C F\cdot ds$ where $C$ is the unit cirlce on the plane $x+y+z=3$ with the orientation from the point $$\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)$$ to $$\left(1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)$$ to $$\left(1 + \frac{2}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}, 1-\frac{1}{\sqrt{6}}\right)$$ and back to $$\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}, 1+\frac{2}{\sqrt{6}}\right)$$
What I've done: By Green's theorem, the integral is $$\iint_D \operatorname{curl}(F)\cdot k\ dA$$ I got that the curl is $$\frac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}}$$ so the integral is $$\int\int_D \frac{z}{(x^2+y^2+z^2)^{3/2}} \ dA$$ But I'm having difficulties with determining how to define the region $D$ to compute the last integral.
| Obviously $D$ is a disk of radius $1$ in the given plane. You just need to find it's origin. The format of the given points on the circle hints that the center is $(1,1,1)$, which you can verify.
I would do a change of variables, to some $X,Y,Z$ such that $X,Y$ are in the plane, then $Z=0$.
| {
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Triangles and Similarity. ADEN is a square. BMDF is a square such that F lies on AD and M lies on the extension of ED. C is the point of intersection of AD and BE. If the area of triangle CDE is 6 square units, what is the area of triangle ABC?
I tried solving this as :
tria.CDE is similar to tria.BME is similar to tria.CFB.
So, the squares of the ratios of the sides must be equal as :
(CD/BM) = (FC/BM) = (DE/ME) = (BF/ME)
But, this solution is just making me go around in circles with countless equations. Is there a smarter way of solving this problem? Please advise.
| Let DE = a, DM = b
*
*Area of $\triangle ABF = \frac{1}{2} \cdot b \cdot (a-b)$
*Area of $\triangle CBF = \frac{1}{2} \cdot b \cdot CF$
*Area of $\triangle CDE = \frac{1}{2} \cdot a \cdot CD$
*$CF + CD = b$ and $CF = \frac{b}{a} CD$. Thus, $CD = \frac{ab}{a+b}$ and $CF = \frac{b^2}{a+b}$
Now, from 1, 2 & 4: area of $\triangle ABC = \frac{1}{2} \cdot (b \cdot (a-b) + b \cdot \frac{b^2}{a+b}) = \frac{1}{2} \cdot b \cdot \frac{a^2}{a+b}$
From 3 & 4: Area of $\triangle CDE = \frac{1}{2} \cdot \frac{a^2}{a+b}$
Hence, Area of $\triangle ABC =$ Area of $\triangle CDE$ = 6
===
Edit:
A more intuitive way of doing this without much algebra is to notice that if you take away $\triangle CBF$ from $\triangle CDE$ you are left with a trapezoid. Area of trapezoid is the average of the two parallel sides multiplied by the height. Average of the two parallel sides (which are equal in length to CD and CF) is half the length of side DF. The height is the same as the difference in length of sides of the two squares. Thus, area of trapezoid is nothing but area of triangle ABF.
| {
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Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
*
*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
*Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.
| Put $x=\dfrac{b}{a}$
$x^2+x+1=\dfrac{b^2}{a^2}+\dfrac{b}{a}+1$
$=\dfrac{b^2+ab+a^2}{a^2}=\dfrac{\dfrac{1}{2}(a+b)^2+\dfrac{b^2}{2}}{a^2}+\dfrac{1}{2}>
0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 21,
"answer_id": 20
} |
Determine Minimum Value.
Find the minimum value of
$$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
for $x>0$.
When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks
I think differentiation would be really complicated
| Hint-$$\frac{(x+\frac1x)^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})} = \frac{(x+\frac{1}{x})^6-(x^3+\frac{1}{x^3})^2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}=(x+\frac{1}{x})^3-(x^3+\frac{1}{x^3})=3(x+1/x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Finding the $LU$ factorization of the matrix
Find the $LU$ factorization of the matrix:
$$\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}$$
I am aware that I need to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$ and $U=\begin{bmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{bmatrix}$
I did row transformations and got $U=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}$ but I couldn't understand how to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$
Can anyone explain how to find $L$
| Gaussian Elimination without Pivoting is as follows. The primary purpose of Gaussian elimination if you follow this is to find $\ell_{jk}$ which zeros out the row below. That is why it is the ratio of the two rows and then you subtract them. This continues on and on.
Suppose that
$$ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ A = LU $$
$$ U =A, L=I$$
$$ k=1,m=3,j=2$$
$$\ell_{21} = \frac{u_{21}}{u_{11}} = \frac{a_{21}}{a_{11}} = 3 $$
$$ u_{2,1:3} = u_{2,1:3} - 3 \cdot u_{1,1:3} $$
Then we're going to subtract 3 times the 1st row from the 2nd row
$$ \begin{bmatrix} 3 & 5 & 6 \end{bmatrix} - 3 \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\end{bmatrix} $$
Updating each of them
$$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
$$k=1,j=3,m=3 $$
$$\ell_{31} = \frac{u_{31}}{u_{11}} = \frac{-2}{1} = -2 $$
$$ u_{3,1:3} = u_{3,1:3} +2 \cdot u_{1,1:3} $$
Then we add two times the first row to the third row
$$ \begin{bmatrix} -2 & 2 & 7 \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 & 1& 1 \end{bmatrix} = \begin{bmatrix}0 & 4 & 9 \end{bmatrix} $$
Updating
$$ U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} $$
$$ k=2, j=3,m=3 $$
$$ \ell_{32} = \frac{u_{32}}{u_{22}} = \frac{4}{2} = 2$$
We're subtracting out little blocks
$$ u_{3,2:3} = u_{3,2:3} - 2 \cdot u_{2,2:3} $$
$$ \begin{bmatrix} 4 & 9 \end{bmatrix} - 2 \cdot\begin{bmatrix} 2& 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \end{bmatrix} $$
Updating
$$ U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix} $$
It now terminates
$$ A = LU $$
$$ \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}}_{L} \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}}_{U} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+b+c+d=x$. Then $x+e=8\implies e=8-x$. Also, $\frac{a^2+1}{2}+\frac{b^2+1}{2}+\frac{c^2+1}{2}+\frac{d^2+1}{2}\geq (a+b+c+d)=x$ by AM-GM inequality. Hence, $a^2+b^2+c^2+d^2\geq 2x-4$.
Now we have $2x-4+e^2\leq a^2+b^2+c^2+d^2+e^2=16$. Substituting $x=8-e$, we get $e^2-2e-4\leq 0$. We can easily calculate that the lowest value that $e$ can attain is $1 -\sqrt{5}$.
However, the answer given on the internet is $\frac{16}{5}$. Where am I going wrong?
EDIT $1$ -- Is this a case of how the value $1-\sqrt{5}$ can never be attained by $e$, although the inequality is true?
EDIT $2$ -- It seems that we need to find the maximum. By my method, I’ve found the maximum to be $1+\sqrt{5}$. This is greater than $\frac{16}{5}$. Have I found a sharper inequality?
| I don't understand, wouldn't the minimum value be $0$? if $$a=b=c=d=2$$ and $$e=0$$ both equations are satisfied. Clearly, zero would also be the smallest positive number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Triples $(x, y, z)$ that satisfy a set of equations
Suppose that $a$ is a fixed (but unknown) real number, with $a^2 \neq 1$. Determine all triples $(x, y, z)$ of real numbers that satisfy the system of equations:
$x + y + z = a$
$xy + yz + xz = -1$
$xyz = -a$
I've tried making substitutions but don't seem to be able to make much progress. Another thing I noticed is that:
$(x + y + z) = a \implies$
$(x + y + z)^2 = a^2 \implies$
$x^2 + y^2 + z^2 + 2(xy + yx + xz) = a^2 \implies$
$x^2 + y^2 + z^2 + 2(-1) = a^2 \implies$
$x^2 + y^2 + z^2 = a^2 + 2$
but I don't know if that actually is useful. Any suggestions on how to move forward?
| We have
\begin{align}
(u-x)(u-y)(u-z) &= u^3 - (x+y+z)u^2 + (xy+yz+xz)u - xyz\\
&= u^3-au^2-u+a\\
&= u^2(u-a)-(u-a)\\
&= (u^2-1)(u-a)\\
&= (u-1)(u+1)(u-a)
\end{align}
so $\{x,y,z\} = \{-1,1,a\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
general solution for a recurrence relation I have the following recurrence relation:
$$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\hspace{1cm}(*)$$
If we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that
$r=\frac{a+\sqrt{a^2-4}}{2}$ or $r=\frac{a-\sqrt{a^2-4}}{2}$.
By using the initial values $x_1=1, x_2=a$, I found that
$$x_n=\frac{1}{\sqrt{a^2-4}}\left(\frac{a+\sqrt{a^2-4}}{2}\right)^n-\frac{1}{\sqrt{a^2-4}}\left(\frac{a-\sqrt{a^2-4}}{2}\right)^n$$ is a solution for the recurrence relation (*).
Question: How do we know whether this is the only solution for the recurrence relation $(*)$?
Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions will have this form.
| A bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy
$$ x_n^2 - a \, x_n \, x_{n+1} + x_{n+1}^2 = 1 $$
Try consecutive values in
$$ 1, \; \; a, \; \; a^2 - 1, \; \; a^3 - 2a, \; \ldots $$
This comes from the matrix
$$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & a
\end{array}
\right)
$$
which has determinant $1$ and trace $a.$ It also gives an automorphism of the quadratic form $x^2 - a xy+y^2.$ An automorphism matrix $P$ for a quadratic form that has Hessian matrix $H$ satisfies $P^T HP = H$
In this case
$$
\left(
\begin{array}{cc}
0 & -1 \\
1 & a
\end{array}
\right)
\left(
\begin{array}{cc}
2 & -a \\
-a & 2
\end{array}
\right)
\left(
\begin{array}{cc}
0 & 1 \\
-1 & a
\end{array}
\right) =
\left(
\begin{array}{cc}
2 & -a \\
-a & 2
\end{array}
\right)
$$
The explicit relation with the sequence is
$$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & a
\end{array}
\right)
\left(
\begin{array}{c}
x_n \\
x_{n+1}
\end{array}
\right) =
\left(
\begin{array}{c}
x_{n+1} \\
x_{n+2}
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Complex Partial Fraction Decomposition The question I need help with is:
Prove that
$$\sum_{k=0}^{6}\frac{1-z^{2}}{1-2z\cos\left(\frac{2k\pi}{7}\right)+z^{2}}=\frac{7(z^{7}+1)}{1-z^{7}}$$
I have already tried brute forcing this by combining the LHS into a single fraction. While this worked, it is an extremely long proof.
I was wondering if there is a more elegant approach that uses partial fractions. I tried decomposing each term in the sum of the LHS into $$-1+\frac{B}{z-\omega^{k}}+\frac{C}{z-\omega^{-k}}$$but this gave me very complicated expressions for constants B and C so I gave up.
| Following the posts that were first to appear we introduce
$\zeta=\exp(2\pi i/n)$ and seek to evaluate
$$S = \sum_{k=0}^{n-1} \frac{1-z^2}{(z-\zeta^k)(z-1/\zeta^k)}.$$
where presumably $z$ is not a power of $\zeta$ and no singularity
appears. Introducing
$$f(v) = \frac{1-z^2}{(z-v)(z-1/v)} \frac{n/v}{v^n-1}
= \frac{1-z^2}{(z-v)(vz-1)} \frac{n}{v^n-1}
\\ = -\frac{1}{z} \frac{1-z^2}{(v-z)(v-1/z)} \frac{n}{v^n-1}$$
we have
$$S = \sum_{k=0}^{n-1} \mathrm{Res}_{v=\zeta^k} f(v).$$
The residue at infinity is zero by inspection and since residues
sum to zero we get
$$S = - \mathrm{Res}_{v=z} f(v) - \mathrm{Res}_{v=1/z} f(v).$$
This yields
$$\frac{1}{z} \frac{1-z^2}{z-1/z} \frac{n}{z^n-1}
+ \frac{1}{z} \frac{1-z^2}{1/z-z} \frac{n}{1/z^n-1}
\\ = \frac{1-z^2}{z^2-1} \frac{n}{z^n-1}
+ \frac{1-z^2}{1-z^2} \frac{z^n n}{1-z^n}
= \frac{n}{1-z^n}
+ \frac{z^n n}{1-z^n}.$$
We obtain
$$\bbox[5px,border:2px solid #00A000]{ S=n\frac{1+z^n}{1-z^n}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Series Convergence of Harmonic Means
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that
$$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$
| Note that
\begin{align}
& x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\
\iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\
\iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2}\\
\iff & a_n-a_{n-1} = -\frac{1}{2}\left(a_{n-1}-a_{n-2}\right)
\end{align}
where $a_i = \frac{1}{x_i}$. Consequently,
$$\sum_{n=3}^{k}(a_n-a_{n-1}) = a_k-a_2 = -\frac{1}{2}\sum_{n=3}^{k}(a_{n-1}-a_{n-2}) = -\frac{1}{2}(a_{k-1}-a_1)$$
Let $\alpha = \lim_{k\to \infty} a_k$, then the above equation implies
$$\alpha -a_2 =-\frac{1}{2}(\alpha-a_1) \implies \alpha = \frac{2a_2+a_1}{3}.$$
Therefore, $$\lim_{n\to\infty}x_n = \frac{1}{\alpha} = \frac{3x_1 x_2}{2x_1 + x_2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all the solutions of the polynomial equation $x^5+y^3 \equiv 1 \pmod9$. Find all the solutions of the polynomial equation $$x^5+y^3 \equiv 1\pmod9.$$ I have learned to find the solutions of the polynomials like $x^p \equiv n\pmod m$, but never been like two variables. Please help me to solve this. Thanks.
| Guide:
Let $y=3m+r, r=0,1,2$
$y^3=(3m+r)^3=(3m)^3+3(3m)^2r+3(3m)r^2+r^3\equiv r^3 \pmod{9}$
Hence we just have to consider $3$ cases,
Case $1$: $y \equiv 0 \pmod{3}$ and the problem reduces to $x^5 \equiv 1\pmod{9}$.
Case $2$: $y \equiv 1 \pmod{3}$ and the problem reduces to $x^5 \equiv 0\pmod{9}$.
Case $3$: $y \equiv -1 \pmod{3}$ and the problem reduces to $x^5 \equiv 2\pmod{9}$.
The problems are now of the form of $x^p \equiv n \pmod{m}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
$$\frac{1}{x+1}-\frac{1}{x+5}=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
| With factors following an arithmetic progression, you establish the rule
$$\frac1{a(a+b)}+\frac1{(a+b)(a+2b)}=\frac{2a+2b}{a(a+b)(a+2b)}=\frac2{a(a+2b)}.$$
Applying it three times, the sum reduces to
$$\frac4{(x+1)(x+5)},$$ giving the easy quadratic equation
$$(x+1)(x+5)=\frac4{0.8}$$
or $$x(x+6)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Trigonometry and quadratics : Possible mismatch? There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?
$2\sin^2{x} -5\cos{x} -4 =0 $
Here’s what I did:
$2\sin^2{x} -5\cos{x} -4 =0 $
$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$
$2 \cos^2{x} + 5 \cos{x} + 2 =0$
This is a quadratic function of $\cos x$, thus,
$\cos{x} = (-5 + 3)/4$ or $\cos{x}= (-5 - 3)/4$
But, the answer given is $\cos{x}=-\frac{1}{2} $ and WolframAlpha says the same but doesn’t show how to do it.
What did I do wrong?
Update: Thank you very much, everyone. Turns out that I wrote the squareroot of 9 as squareroot of 3. My bad
| Simple factorization..
$$2 \cos^2 x + 5 \cos x + 2 =0$$
$$(2 \cos^2 x + 4 \cos x) +(\cos x + 2) =0$$
$$2 \cos x(\cos x +2)+( \cos x + 2) =0$$
$$(\cos x +2)( 2\cos x + 1) =0$$
$$\implies ( 2\cos x + 1) =0$$
$$\implies \cos x =-\frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Geometry: Prove that two angles are not equal (This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $\triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $\alpha\neq\beta$.
Solution
$\triangle ABM$ and $\triangle AMC$ have the same area. They have a
common side $AM$.
Note that the area of either triangle is given by
$S=\frac12(AB)(AM)\sin\alpha=\frac{1}{2}(AC)(AM)\sin\beta$.
But $AB\neq AC$. So the equality holds only if $\alpha\neq\beta$.
| First assume that $\alpha$ = $\beta$ and that both would be equal to $\frac{\pi}{8}^\circ$ (since $\alpha = \beta$ and $\alpha + \beta = \frac{\pi}{4} \to 2\beta = \frac{\pi}{4} \to \beta = \frac{\pi}{8}$).
If we call side $\overline{AC}$, n, then side $\overline{MC}$ is $\frac{n}{2}$. Next, we solve for $\overline{AM}$ in terms of n to show that $\alpha = \beta$ for all values n.
$\overline{AM}^2 = n^2 + (\frac{n}{2})^2 = n^2 + \frac{n^2}{4} = \frac{4n^2 + n^2}{4} = \frac{5n^2}{4}$
$\overline{AM} = \sqrt{\frac{5n^2}{4}} = \frac{\sqrt{5n^2}}{2} = \frac{n\sqrt{5}}{2}$ by the pythagorean theorem
Now that we have $\overline{AM}$, we can find $cos{\beta}$.
$\cos{\beta}=\frac{n}{\overline{AM}} = \frac{n}{\frac{n\sqrt{5}}{2}} = \frac{2n}{n\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$
$\cos{\beta} = \frac{2\sqrt{5}}{5}$
We have that $\cos{\beta} = \frac{2\sqrt{5}}{5}$ and we already established that $\beta = \frac{\pi}{8}$.
This would mean that $\cos{\frac{\pi}{8}} = \frac{2\sqrt{5}}{5}$. We can find $\cos{\frac{\pi}{8}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$, which implies that $\frac{\sqrt{2 + \sqrt{2}}}{2} = \frac{2\sqrt{5}}{5}$.
If we square both sides, we get that $\frac{2+\sqrt{2}}{4} = \frac{20}{25} = \frac{4}{5}$. Then we can split the fraction, simplify, then subtract and simplify.
$\frac{2}{4} + \frac{\sqrt{2}}{4} = \frac{4}{5}$
$\frac{1}{2} + \frac{\sqrt{2}}{4} = \frac{4}{5}$
$\frac{\sqrt{2}}{4} = \frac{4}{5} - \frac{1}{2} $
$\frac{\sqrt{2}}{4} = \frac{3}{10}$
Next we can multiply both sides by 4 and conclude that...
$\sqrt{2} = \frac{12}{10} = \frac{6}{5}$
This is impossible since it has been proven that $\sqrt{2}$ is irrational and can't be written as a ratio of 2 integers. This would mean that $\sqrt{2} \neq \frac{6}{5}$ or $\frac{\sqrt{2 + \sqrt{2}}}{2} \neq \frac{2\sqrt{5}}{5}$ or $\cos{\frac{\pi}{8}}\neq \frac{2\sqrt{5}}{5}$
This means that $\beta \neq \frac{\pi}{8}$. Since $\alpha + \beta = \frac{\pi}{4}$, $\beta$ would have to be equal to $\frac{\pi}{8}$, but we proved that it cant be. This means that $\alpha \neq \beta$ since $\beta$ can no longer be multiplied by 2 to be $\frac{\pi}{4}$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
| {
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"url": "https://math.stackexchange.com/questions/2858658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 14,
"answer_id": 9
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Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$
Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.
My own solution
$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}~\frac{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}$$
$$\lim_{n\to\infty} \frac{5n^2+4~-~(5n^2+n)}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}~=~\lim_{n\to\infty} \frac{4-n}{n\left(\sqrt{5+\frac4{n^2}}~+~\sqrt{5+\frac1{n}}\right)}$$
$$=~\frac{-1}{2\sqrt{5}}~=~-\frac{\sqrt{5}}{10}$$
| Let $1/n=h$
As $n\to\infty,h\to0+,h>0$
$$\sqrt{5n^2+4}=\sqrt{\dfrac{5+4h^2}{h^2}}=\dfrac{\sqrt{5+4h^2}}{\sqrt{h^2}}$$
Now $\sqrt{h^2}=|h|=+h$ for $h>0$
So, we ahve
$$\lim_{h\to0^+}\dfrac{\sqrt{5+4h^2}-\sqrt{5+h}}h=\lim_{h\to0^+}\dfrac1{\sqrt{5+4h^2}+\sqrt{5+h}}\cdot\lim_{h\to0^+}\dfrac{5+4h^2-(5+h)}h=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Series convergence proof verification Show that the series $\sum_{k=0}^{\infty}(-1)^k\frac{1}{2k+1}$ converges to $\pi/4$.
Here is my draft of a proof.
Abel's theorem states that suppose for a nondegenerate $[a,b]$. If $f(x):=\sum_{k=0}^{\infty}a_x(x-x_0)^k$ converges on $[a,b]$, then $f(x)$ is continuous and converges uniformly on $[a,b]$. Define $f(x)$ as $\frac{1}{1-x}$ and additionally as the power series centered at $x_0=0$ as $1+x+x^2+x^3+\cdots$. In other words $$f(x)=\frac{1}{1-x}:=\sum_{k=0}^{\infty}1\cdot(x)^k.$$ The radius of convergence can be found by letting $$r=\lim_{k\to\infty}|\frac{1\cdot(x)^{k+1}}{1\cdot(x)^k}|=|x|\lim_{k\to\infty}|1|=|x|.$$ By the ratio test, $r$ converges if $0\leq r<1$, or $|x|<1$ expressed by the interval $(-1,1)$.
It is now shown that $f(X)$ converges on a nondegenerate interval $(-1,1)$ and $f(x):=\sum_{k=0}^{\infty}a_x(x-x_0)^k$ where $a_k=1$ and $x_0=0$, therefore $f(x)$ converges uniformly on (-1,1) by Abel's theorem.
Next let $g(x)=f(-x^2)$. $g$ will still converge uniformly on $(-1,1)$ as $|-x^2|\leq|x|$ when $0\leq x<1$. As a result, $$g(x)=\frac{1}{1-(-x)^2}=\frac{1}{1+x^2}:=1-x^2+x^4-x^6+\cdots=\sum_{k=0}^{\infty}(-1)^k(x^{2k}).$$ Because $g(x)$ is uniformly convergent, it follows that $g$ is integrable on $(-1,1)$ and one can perform term-by-term integration on the series by $$\int g(x)dx=\int (1-x^2+x^4-x^6+\cdots)dx$$ from theorem 7.14. Here, $$G(x)=\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Evaluating $G(x)$ at 1 gives $$G(1)=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots=\sum_{k=0}^{\infty}(-1)^k\frac{1}{2k+1}=\arctan(1)=\frac{\pi}{4}$$
| This is not entirely correct.
The series $\sum_{k=0}^\infty (-1)^kx^{2k}$ converges to $1/(1+x^2)$ on $(-1,1)$, but the convergence is not uniform. As a power series with convergence radius $1$, it does converge uniformly on any compact interval $[a,b] \subset (-1,1)$ -- a fact you cite but apply incorrectly. Consequently, you are premature in justifying termwise integration over $[0,1]$ on the basis of uniform convergence.
To see that the convergence is not uniform on the open interval, note that we have a geometric series where
$$\tag{*}\sum_{k=0}^n (-1)^kx^{2k} = \frac{1 - (-x^2)^{n+1}}{1+x^2},$$
and, as $n \to \infty$,
$$\sup_{x \in (-1,1)}\left|\sum_{k=0}^n (-1)^kx^{2k} - \frac{1 }{1+x^2}\right|= \sup_{x \in (-1,1)}\frac{(x^2)^{n+1}}{1+x^2} > \frac{((\sqrt{1-1/n})^2)^{n+1}}{1+(\sqrt{1-1/n})^2} \\ = \frac{(1-1/n)^{n+1}}{2-1/n} \to\frac{e^{-1}}{2} \neq 0$$
Solution 1
Using uniform convergence on the compact interval $[0,z] \subset (-1,1)$ we can integrate termwise to obtain
$$\sum_{k=0}^\infty \frac{(-1)^{k}z^{2k+1}}{2k+1} = \int_0^z \sum_{k=0}^\infty (-1)^{k}z^{2k} = \int_0^z \frac{dx}{1+x^2} = \arctan(z)$$
By Abel's limit theorem we have
$$\sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} = \lim_{z \to 1-}\sum_{k=0}^\infty \frac{(-1)^{k}z^{2k+1}}{2k+1} = \lim_{z \to 1-}\arctan(z) = \frac{\pi}{4},$$
since the series on the LHS converges by the alternating series test.
Solution 2
A more direct way is to use (*) and apply termwise integration to the partial sum, which is always permissible, to obtain
$$\tag{**}\sum_{k=0}^n \frac{(-1)^{k}}{2k+1} - \frac{\pi}{4}= \int_0^1 \sum_{k=0}^n (-1)^kx^{2k} \, dx - \int_0^1 \frac{dx}{1+x^2} = (-1)^{n+1} \int_0^1 \frac{x^{2n+2}}{1+x^2}\,dx$$
Note that the RHS of (**) converges to $0$ as $n \to \infty$, by virtue of the estimate
$$\left|(-1)^{n+1} \int_0^1 \frac{x^{2n+2}}{1+x^2}\,dx\right| \leqslant \int_0^1 x^{2n+2} \, dx = \frac{1}{2n+3}$$
Therefore,
$$\lim_{n \to \infty}\sum_{k=0}^n \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
| Let us use Roots of unity!
Here's a hint (by the way this question is from the ARML and I recommend you try it out because it is quite nice). We know that if $z\neq1$ is an $z$th root of unity, that $1+z+z^2+z^3+z^4+\cdots+z^{n-1}=0$. Also, note that the $n$th roots of unity form a regular $n$-gon with a vertex at $(1,0)$.
ANSWER (SPOILER):
After all this you will get an answer of $\frac{6\sqrt{2}+1}{4}$, so that $a+b+c+d=13$, so our answer is $\boxed{13}$
| {
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"timestamp": "2023-03-29T00:00:00",
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The sides of a triangle are in Arithmetic progression If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are $X$ and $Y$, then show that
$$4(1- \cos X)(1-\cos Y) = \cos X + \cos Y$$
I tried using sine rule but can't solve it.
| Let the sides be $a-d,a,a+d$ (with $a>d)$ be the three sides of the triangle, so $X$ corresponds to the side with length $a-d$ and $Y$ that to with length $a+d$. Using cosine formula
\begin{align*}
\cos X & = \frac{(a+d)^2+a^2-(a-d)^2)}{2a(a+d)}=\frac{a+4d}{2(a+d)}\\
\cos Y & = \frac{(a-d)^2+a^2-(a+d)^2)}{2a(a-d)}=\frac{a-4d}{2(a-d)}\\
\end{align*}
Then
$$\cos X +\cos Y=\frac{a^2-4d^2}{a^2-d^2}=4 \frac{(a-2d)}{2(a+d)}\frac{(a+2d)}{2(a-d)}=4(1-\cos X)(1-\cos Y).$$
| {
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Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$s_n=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)=\left(\frac{1}{6}+\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{3}\right)+...+\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$
I then grouped the terms by their position in each partial sum,
First terms: $\frac{1}{6},\frac{1}{8},\frac{1}{10},\frac{1}{12}...$
Second terms: $\require\cancel{\cancel{\frac{1}{2}}},\cancel{\frac{1}{4}},\cancel{\frac{1}{6}},\cancel{\frac{1}{8}}...$
Third terms: $\cancel{-\frac{1}{2}},-\frac{1}{3},\cancel{-\frac{1}{4}},-\frac{1}{5}...$
Cancelling leaves the series: $$\sum_{n=3}^{\infty}\frac{1}{2n}-\sum_{n=1}^{\infty}\frac{1}{2n+1}$$
However I'm stuck here since I see a divergent harmonic series summed with another harmonic series but I know the original series in question is convergent to $\frac{1}{4}$. What can I do? I suspect my error can be fixed somehow by adjusting the bounds of the sums...? Thanks in advance.
| $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$$=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{2n}-\frac{1}{2n}\right)$$
$$=1/2\sum_{n=2}^{\infty}\left(\frac{1}{(n+1)}-\frac{1}{n}+\frac{1}{(n-1)}-\frac{1}{n}\right)$$
Do you see two telescoping series?
| {
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How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| Note that $$\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{(49-x^2)-(25-x^2)}{\sqrt{49-x^2}-\sqrt{25-x^2}}=\frac{24}{3}=8\,.$$
Together with $\sqrt{49-x^2}-\sqrt{25-x^2}=3$, we conclude that
$$\sqrt{49-x^2}=\frac{3+8}{2}=\frac{11}{2}\,.$$
Therefore, $x^2=\dfrac{75}{4}$, or $x=\pm\dfrac{5\sqrt{3}}{2}$.
P.S.: Oopsie, nextpuzzle already made a comment, giving the same solution.
| {
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Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
| Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$\cos u=1-\frac{u^2}2+o(u^2),\qquad \sin u=u+o(u)$$
so
\begin{align}
\cos x-\cos 3x&=1-\frac{x^2}2+o(x^2)-\Bigl(1-\frac{9x^2}2+o(x^2)\Bigr)= 4x^2+o(x^2)\\
\sin 3x^2-\sin x^2&=3x^2+o(x^2)-\bigl(\sin x^2+o(x^2)\bigr)=2x^2+o(x^2).
\end{align}
Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence
$$\frac{\cos x-\cos 3x}{\sin 3x^2-\sin x^2}\sim_0\frac{4x^2}{2x^2}=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17
| Let $a=mc$ and $b=nc$ then
$$m+n=\dfrac{2}{c}-1~~~,~~~m^2+n^2=\dfrac{12}{c^2}-1$$
by Cauchy-Schwarz
$$(m+n)^2\leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20\geq0$ gives $c=-2,\dfrac{10}{3}$ leads us to difference $\dfrac{16}{3}$.
| {
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Application of Chebyschev inequality
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$\frac{a+b+c}{3}\cdot \frac{a+b+c}{3}\geq \frac{ca+b^2+ca}{3}$$
My problem
I would've wanted to get $\dfrac{ab+bc+ca}{3}$ on the right instead. Did I apply the inequality wrong, or does it follow that $\dfrac{ab+bc+ca}{3}$ is less than or equal to $\dfrac{a+b+c}{3}\cdot \dfrac{a+b+c}{3}$?
| Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(a\cdot a+b\cdot b+c\cdot c)\geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)\geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))\geq(a+b+c)^2+2(ab+ac+bc)$$ or
$$(a+b+c)^2\geq3(ab+ac+bc)$$ or
$$\frac{a+b+c}{3}\geq\sqrt{\frac{ab+ac+bc}{3}}.$$
By the same way we obtain:
$$3(ab\cdot ab+ac\cdot ac+bc\cdot bc)\geq (ab+ac+bc)(ab+ac+bc)$$ or
$$3(a^2b^2+a^2c^2+b^2c^2)\geq(ab+ac+bc)^2$$ or
$$(ab+ac+bc)^2\geq3abc(a+b+c).$$
Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain:
$$(ab+ac+bc)^3\geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)\geq$$
$$\geq3abc\cdot3(c\cdot ab+b\cdot ac+a\cdot bc)=27a^2b^2c^2,$$
which gives
$$\sqrt{\frac{ab+ac+bc}{3}}\geq\sqrt[3]{abc}.$$
| {
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Prove by definition $\lim_{n\to\infty} \sqrt{\frac{n}{n+1}}=1$
Prove by definition $\lim_{n\to\infty} \sqrt{\frac{n}{n+1}}=1$.
I got to $\left|\sqrt{\frac{n}{n+1}}-1\right|< ε$, but I don't know how to proceed.
| Let me explain Kenny Lau's answer a bit. The first step, he does
$$\left|\sqrt{\frac{n}{n+1}}-1\right|=\left|\frac{\sqrt{n}}{\sqrt{n+1}}-1\right|=\left|\frac{\sqrt{n}}{\sqrt{n+1}}-\frac{\sqrt{n+1}}{\sqrt{n+1}}\right|=\left|\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}}\right|$$
Now he proceeds to multiply both sides of the fraction with $\sqrt{n}+\sqrt{n+1}$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes
$$(\sqrt{n}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1})=\sqrt{n}^2-\sqrt{n+1}^2=n-(n+1)=-1$$
so that we get
$$\left|\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}}\right|=\left|\frac{-1}{{\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}}\right|$$
He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $\sqrt{n}<\sqrt{n+1}$ so that
$$\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})>\sqrt{n}(\sqrt{n}+\sqrt{n})=2n$$
hence,
$$\frac{1}{\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}<\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n})}=\frac{1}{2n}$$
so in the end
$$\left|\sqrt{\frac{n}{n+1}}-1\right|<\left|\frac{1}{2n}\right|$$
the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $\epsilon$, because you can just make $\frac1{2n}$ smaller and then
$$\left|\sqrt{\frac{n}{n+1}}-1\right|$$
must be smaller too.
| {
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$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$? On a problem book solution I was faced with the following step:
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$
I identified $\frac{1}{n+1}+\frac{1}{(n+1)^2}...$ as a geometric series so the sum would be $\frac{1}{1-r}$ so that $\frac{1}{1-\frac{1}{n+1}}=1+\frac{1}{n}$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
| How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + \cdots = \dfrac{1}{1-x}\text{.} \tag{*}$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $\dfrac{1}{n+1}$ that
$$\dfrac{1}{n+1} + \dfrac{1}{(n+1)^2} + \cdots = \dfrac{1}{n+1}\left(1+\dfrac{1}{n+1}+\dfrac{1}{(n+1)^2}+\cdots \right)\text{.}\tag{A}$$
The sum
$$1+\dfrac{1}{n+1}+\dfrac{1}{(n+1)^2} + \cdots = \dfrac{1}{1-\frac{1}{n+1}} = \dfrac{1}{(n+1-1)/(n+1)} = \dfrac{n+1}{n}\tag{B}$$
as long as $$\left|\dfrac{1}{n+1} \right| < 1 \implies |n+1| > 1 \implies n+1 > 1 \text{ and } -(n+1) < -1 \implies n > 0\text{.}$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$\dfrac{1}{n+1} \cdot \dfrac{n+1}{n} = \dfrac{1}{n}$$
as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
| {
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Computing the determinant of a $4\times4$ matrix
Compute the determinant of \begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}
My try:
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}_{R_2\rightarrow2R_2-R_1\\R_3\rightarrow R_3-R_1\\R_4\rightarrow2R_4+R_1}$$
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 0 & 5 & -1 & -1 \\ 0 & 2 & 0 & -4 \\ 0 & 1 & 5 & -1 \end{vmatrix}$$
Now I took $$\begin{vmatrix} 5 & -1 & -1 \\ 2 & 0 & -4 \\ 1 & 5 & -1 \end{vmatrix}=92$$
But the answer is $46$. Where did I go wrong?
| Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $\frac{92*2}{4}$=46.
| {
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What is the domain of the function $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
What is the domain of the function
$f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
I started using the fact $-\frac{\pi}{2}\le f(x) \le \frac{\pi}{2}\implies-1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$.Now,on dissecting it into two cases.
$CASE (1): -1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}$
$CASE (2): \frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
| Examine that,
$$\lim_{x\to-\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
$$\lim_{x\to\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=1$$
$$8(3)^{x-2}=1-3^{2(x-1)}$$
$$8(3)^{x-2}+3^{2(x-1)}=1$$
$$3^{x-1}(\frac{8}{3}+3^{x-1})=1$$
Let $3^{x-1}=t$
$$t(8+3t)=3$$
$$t=\frac{1}{3} \implies x=0 $$
$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=-1$$
$$8(3)^{x-2}=-1+3^{2(x-1)}$$
$$8(3)^{x-2}-3^{2(x-1)}=-1$$
$$3^{x-1}(\frac{8}{3}-3^{x-1})=-1$$
Let $3^{x-1}=t$
$$t(8-3t)=-3$$
$$t=3 \implies x=2 $$
Therefore, Domain is, $$(-\infty,0]\cup[2,\infty)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
| We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)\rightarrow a+b+4c+7=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| I assume that by "dish" you mean "quotient"; that's not terminology I'm familiar with, but it seems to make sense in context.
When you're dividing polynomials, remember that you are not done until the remainder is of lower degree than the denominator. So you're correct that the first step of dividing $n^2 + 1$ into $n^6 - 7$ gives you a remainder of $-n^4 - 7$ (with $n^4$ in the quotient); but since $-n^4 - 7$ is of higher degree than $n^2 + 1$, we have to keep going.
To go into $-n^4 - 7$, we need to multiply by $-n^2$; so now our quotient so far is $n^4 - n^2$, and our remainder is $(-n^4 - 7) - (-n^2 (n^2 + 1)) = n^2 - 7$.
Since $n^2 - 7$ is still at least as high in degree as $n^2 + 1$, we have to divide one more time; this gives us a quotient of $n^4 - n^2 + 1$ and a remainder of $(n^2 - 7) - (n^2 + 1) = -8$.
Since the remainder now has degree $0$, which is less than the degree of our divisor, we can stop now; so the quotient is $n^4 - n^2 + 1$ and the remainder is $-8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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what is the multiplicative order of the following elements in $F_{16}$ For $f(x) = x^4 + x + 1 \in F_2[x]$ . Let $a \in F_{16}$ be a root of $f$.
How can I find the multiplicative order of $a$ and $a^2+a+1$
My attempt:
$f$ clearly has not root in $F_2$ so if it is reducible it must be reduced to polynomials of degree 2. Then by considering the only few cases we see that $f$ is irreducible. Now since $a\in F_{16}$ and the multiplicative group of $ F_{16}$ has only 15 elements we deduce that the order of $a$ is one of $3,5,15$ and it cannot be 3 since the minimal polynomial is of degree 4 so it can only be 5 and 15, from there I cannot continue and for $a^2+a+1$ I have no idea
Any help would be much appreciated
Thanks
| Since $x^4+x+1$ is irreducible over $\mathbb{F}_2$ the roots of $x^4+x+1$ in $\mathbb{F}_{16}\simeq \mathbb{F}_2[x]/(x^4+x+1)$ are of the form $a,a^2,a^4,a^8$. Since $a$ is a root of $x^4+x+1$ we have $a^4=-a-1=a+1$. We cannot have $a^3=1$ or $a^5=1$ since $x^3-1$ and $x^5-1$ share no common factor with $x^4+x+1$ over $\mathbb{F}_2$. It follows that the multiplicative order of $a$ is $15$. Similarly
$$(a^2+a+1)^3 \equiv 6a^3+2a^2-6a-5 \pmod{a^4+a+1} $$
implies that the multiplicative order of $a^2+a+1$ is three.
As a brute-force alternative, since $16$ elements are not that much, we may just list the elements of $\mathbb{F}_{16}^*$ as the group generated by $a$:
$$ 1,a,a^2,a^3,a^4,a+1,a^2+a,a^3+a^2,a^3+a+1,a^2+1,a^3+a,\color{red}{a^2+a+1},a^3+a^2+a,a^3+a^2+a+1,a^3+a^2+1,a^3+1$$
we get $a^2+a+1 = a^{10}$ and the same conclusion as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$? Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic
$f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that $f(4)=0$. Therefore, by uniqueness of the real root, one must have
$$
\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4\tag{1}
$$
I had the observation above when I solved the cubic equation $x^3-6x-40=0$. My question is as follows:
without referring to the unique real root of the cubic, can we show (1) directly?
[An attempt.]
When taking the cube on both sides of (1) and simplifying further, I ended up with (1) again.
| Well, $(2\pm\sqrt2)^3=20\pm14\sqrt2$, that is $\sqrt[3]{20\pm14\sqrt2}
=2\pm\sqrt2$. Therefore
$$\sqrt[3]{2+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction for $f^{n+1}(x)=\frac{x}{\sqrt{1+(n+1)x^2}}$ Look at the following function f: $\mathbb{R} \to \mathbb{R}: x \mapsto \frac{x}{\sqrt{1+x^2}}.$
Show with the complete induction that the recursive ( given by $f^1:=f$ and $f^{n+1}:=f\circ f^n$) composition $f^n$ from $f$ has the following explicit illustration:
$f^n(x)=\frac{x}{\sqrt{1+nx^2}}$
My solution:
Base case: For $n=1$
$f^1(x)=f(x)=\frac{x}{\sqrt{1+x^2}}=\frac{x}{\sqrt{1+1\cdot x^2}}$
Inductive step:
$f^{n+1}(x)=(f\circ f^n)(x)=f(f^n(x))=\frac{f^n(x)}{\sqrt{1+(f^n(x))^2}}$
So now I would have to plug in $f^n(x)$ for $x$ I think.
$\frac{x}{\sqrt{1+nx^2}}$ $\frac{\frac{x}{\sqrt{1+nx^2}}}{\sqrt{{1+(\frac{x}{\sqrt{1+nx^2}}})^2}}$ But how do I get from there to $\frac{x}{\sqrt{1+(n+1)x^2}}$ = $\frac{x}{\sqrt{1+nx^2+x^2}}$
I have encountered this problem in an old math exam from 2016 and it is of interest to me because I am currently practicing a lot of exercises related to convergences and mathematical induction.
Any hints guiding me to the right direction I much appreciate.
| \begin{align}
(f\circ f^n)(x)&= \dfrac{ \dfrac{x}{\sqrt{1+nx^2}} }{\sqrt{1+\dfrac{x^2}{1+nx^2}}}\\
& = \dfrac{ \dfrac{x}{\sqrt{1+nx^2}} }{\sqrt{\dfrac{1+nx^2+x^2}{1+nx^2}}}\\
& =\dfrac{x}{ \sqrt{1+nx^2+x^2}}\\
&=\dfrac{x}{ \sqrt{1+(n+1)x^2}}\\
&= f^{n+1}(x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it
Find the inverse Laplace transform of
$$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$
I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,
After factoring the denominator, I got a case of non repeating linear factors
$\dfrac{2s+12}{ (s+2)(s+3)(s+1)} = \dfrac{A}{s+2} + \dfrac{B}{s+3} + \dfrac{C}{s+1} $
$2s + 12 = A (s+3)(s+1) + B (s+2)(s+1) + C (s+2)(s+3) $
$2s + 12 = (As^2 + Bs^2 + Cs^2) + (4As + 3Bs + 5Cs) + (3A + 2B+6C) $
So...
$A + B + C = 0$
$ 4A + 3B + 5C = 2$
$3A + 2B + 6C = 12$
how do I solve this complicated equations ? This is where i got stuck and cannot continue.
| You can use the methods associated with the "Partial fraction decomposition" (see https://en.wikipedia.org/wiki/Partial_fraction_decomposition). After writing :
$$\frac{2s+12}{ (s+2)(s+3)(s+1)} = \frac{A}{s+2} + \frac{B}{s+3} + \frac{C}{s+1} $$
You multiply and the right and left hand side by $(s+2)$, which gives :
$$\frac{2s+12}{(s+3)(s+1)} =A + \frac{B(s+2)}{s+3} + \frac{C(s+2)}{s+1} $$
Taking $s = -2$, you find :
$$ \frac{2(-2)+12}{(-2+3)(-2+1)} = \frac{8}{-1} = -8 = A $$
You can find $B$ and $C$ using the same method (which is usually easier than solving the system).
| {
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"url": "https://math.stackexchange.com/questions/2885818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots of unity and large expression
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} \\
&= \dfrac{\omega(1+\omega^3) + \omega^4(1+\omega^2)}{(1+\omega^2)(1+\omega^3)} + \dfrac{\omega^2(1+\omega) + \omega^3(1+\omega^4)}{(1+\omega^4)(1+\omega)} \\
&= \dfrac{\omega + 2\omega^4 +\omega^6}{1+\omega^2 + \omega^3 + \omega^5} + \dfrac{\omega^2 + 2\omega^3 + \omega^7}{1+\omega + \omega^4 + \omega^5} \\
&= \dfrac{2\omega + 2\omega^4}{2+\omega^2 + \omega^3} +
\dfrac{2\omega^2 + 2\omega^3}{2+\omega+\omega^4}
\end{align*}
OR
\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} \\
&= \dfrac{\omega(1+\omega) + \omega^3(1+\omega^2)}{(1+\omega)(1+\omega^2)} + \dfrac{\omega^2(1+\omega^3) + \omega^4(1+\omega^4)}{(1+\omega^3)(1+\omega^4)} \\
&= \dfrac{\omega + \omega^2 + \omega^3 + \omega^5}{1+\omega + \omega^2 + \omega^3} + \dfrac{\omega^2 + \omega^4 + \omega^5 + \omega^8}{1 + \omega^3 + \omega^4 + \omega^7} \\
&= \dfrac{2\omega+\omega^2+\omega^3}{1+\omega+\omega^2+\omega^4} + \dfrac{1+\omega+\omega^2+\omega^4}{1+2\omega^3+\omega^4}
\end{align*}
| Alt. hint: let $\,z=\omega+\dfrac{1}{\omega}\,$ so that $\,z^2=\omega^2+\dfrac{1}{\omega^2}+2\,$, then use that $\,\omega^4=\bar\omega\,$ and $\,\omega^3=\bar\omega^2\,$ so the sum is:
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\bar\omega^2}{1 + \bar\omega^4} + \frac{\bar\omega}{1 + \bar\omega^2} = 2 \operatorname{Re}\left(\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} \right) = 2 \operatorname{Re}\left(\frac{1}{z} + \frac{1}{z^2-2}\right)$$
But $\,0=\omega^5-1=(\omega-1)\left(\omega^4+\omega^3+\omega^2+\omega+1\right)=\omega^2(\omega-1)\left(z^2 + z - 1\right)\,$, so $\,z^2+z-1=0\,$ and:
$$\require{cancel}
\frac{1}{z} + \frac{1}{z^2-2} = \frac{1}{z}+\frac{1}{-z-1} = \frac{\cancel{-z}-1+\cancel{z}}{-z^2-z} = \frac{-1}{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Write as the sum of a series The question asks to write
$\dfrac{1}{1-i-z}$ as the sum of a series such that $\left|-z-i\right|<1$
but I genuinely have no idea how to do it, or even where to start.
| $$
\frac 1 {1-i-z}
$$
You want the series to converge if $\left| -z-i \right|<1.$ That is the same as $\left|z+i\right|<1,$ and so the same as $\left|z-(-i)\right|<1.$ We therefore want to write this as a sum of powers of $z-(-i),$ i.e. of $z+i.$ The function is
$$
\frac 1 {1 - (z+i)}.
$$
Recall that
$$
\frac 1 {1-r} = 1+r+r^2+r^3+r^4+\cdots. \tag 1
$$
Thus we need
$$
1 + (z+i) + (z+i)^2 + (z+i)^3 + (z+i)^4 + \cdots.
$$
The answer would be different if, for example, we had wanted it to converge if $|z-5|<\text{something}.$ In that case we would want a sum of powers of $z-5.$ We would have
$$
|1-i-z| = |z- 1+i| = |(z-5)-(i-4)|
$$
and then we would apply line $(1)$ above to write
\begin{align}
& \frac 1 {1-i-z} = \frac 1 {(z-5) - (i-4)} = \frac 1 {4-i} \cdot \frac 1 {1 - \left(\dfrac{z-5}{i-4}\right)} \\[12pt]
= {} & \frac 1 {4-i} \left( 1 + \frac{z-5}{i-4} + \left( \frac{z-5}{i-4} \right)^2 + \left( \frac{z-5}{i-4} \right)^3 + \left( \frac{z-5}{i-4} \right)^4 + \cdots \right).
\end{align}
This converges if $\left|\dfrac {z-5}{i-4} \right| < 1,$ i.e. if $\left|z-5\right| < |i -4|,$ i.e. if $|z-5|< \sqrt{17}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the Fourier series of the function: $f(x)=\begin{cases}x+1 & -1\leq x< 0,\\1-x &0\leq x< 1\end{cases},\;\;f(x+2)=f(x)$ I want to find the Fourier series of the function. For now, I am clueless on how to handle the function, int that it has $f(x+2)=f(x).$
$$f(x)=\begin{cases}x+1 & -1\leq x< 0,\\1-x &0\leq x< 1\end{cases},\;\;f(x+2)=f(x)$$
Question:
Can anyone explain what this term ($f(x+2)=f(x)$) means in relation to the function? Solution or references will be highly appreciated.
| BACKGROUND
First $f(x+2) = f(x)$ is just a way of saying that you got to repeat your function, i.e. it is periodic of period $2$.
This is needed so that we could derive the fourier series,
\begin{equation}
a_0 + \sum\limits_{k=1}^{\infty}
a_k\cos(k x) + b_k \sin(k x)
\end{equation}
where
\begin{align}
a_0 &= \frac{1}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2} }
f(x) \ dx \\
a_k &= \frac{2}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2}}
f(x) \cos(kx) \ dx \\
b_k &= \frac{2}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2} }
f(x) \sin(kx) \ dx \\
\end{align}
with $T$ being the period of our function, which in our case is $2$, right?
COMPUTING $a_0$
Let's compute $a_0$
\begin{equation}
a_0 = \frac{1}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2} }
f(x) \ dx
=
\frac{1}{2} \int\limits_{-1 }^{1 }
f(x) \ dx
=
\frac{1}{2} \int\limits_{-1 }^{0}
(x+1) \ dx
+\frac{1}{2} \int\limits_{1}^{0}
(1-x) \ dx
=
\frac{1}{4}
+
\frac{1}{4}
=
\frac{1}{2}
\end{equation}
COMPUTING $a_k$
Now, to compute $a_k$, we use the formula
\begin{equation}
a_k = \frac{2}{2} \int\limits_{-1 }^{1}
f(x) \cos(kx) \ dx
=
\int\limits_{-1 }^{0}
(x+1) \cos (kx) \ dx
+
\int\limits_{0 }^{1}
(1-x) \cos (kx) \ dx
\end{equation}
that is after starightforward math
\begin{equation}
a_k = \frac{2}{2} \int\limits_{-1 }^{1}
f(x) \cos(kx) \ dx
=
-2\dfrac{\cos\left(k\right)-1}{k^2}
\end{equation}
COMPUTING $b_k$
Now, to compute $b_k$, we use the formula
\begin{equation}
b_k = \frac{2}{2} \int\limits_{-1 }^{1}
f(x) \sin(kx) \ dx
=
\int\limits_{-1 }^{0}
(x+1) \sin (kx) \ dx
+
\int\limits_{0 }^{1}
(1-x) \sin (kx) \ dx
\end{equation}
that is after staright-forward math
\begin{equation}
b_k
=
\dfrac{\sin\left(k\right)-k}{k^2}
-\dfrac{\sin\left(k\right)-k}{k^2}
=
0
\end{equation}
NOTE on $b_k$
Since $f(x) = f(-x)$ (easy to prove), then $f(x)$ is even, then $b_k = 0$ for all $k$. That's a property btw.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 x$
Solving and re-arranging we get
$-64\sin^6 x+80\sin^4 x-20\sin^2 x+1=0$
Using the substitution $\sin^2 x=t$
$-64t^3+80t^2-20t+1=0$
I am not able to solve it from hence forth
| First thing first, if you make the substitution $t=\sin^2x$ the polynomial you get is $$-64t^3+80t^2-20t+1=0$$ Now, to decompose it, you could use Ruffini's rule: first we find a zero of the polynomial that divides the constant term (in our case $\pm 1$). Let us call the polynomial $P(t)$, then $$P(1) =-64+80-20+1 \neq 0 \\P(-1) = 64+80+20+1 \neq 0$$ clearly we have no integer solutions! So what we can do is substitute $z=\frac{1}{t}$ and what we get is the following polynomial in $z$ $$Q(z) = z^3-20z^2+80z-64$$ and now we apply the same rule: let us find a zero of $Q(z)$ in the divisors of the constant term $64$ which are $\pm1,\pm2,\pm4,\cdots$. You can easily see that $$Q(4) = 0$$ then we can go on with the simplification and get $$(z-4)(z^2-16z+16)=0$$ which is definetly easier to solve. We find $$z_1=4\implies t_1={1\over 4}\\ z_2=4(2-\sqrt{3})\implies t_2 = {1\over {4(2-\sqrt{3})}}\\ z_3 = 2(2+\sqrt{3})\implies t_3 = {1\over {4(2+\sqrt{3})}}$$ from which you can find the values of $\sin^2x$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that the highest and lowest order terms will be $z^{1 + 2 + \cdots + m} = z^{\frac 1 2m(m+1)}$ and its reciprocal, respectively.
Here's the situation for low values of $m$:
$$\begin{split}
f_1(z) = z &+ \frac 1 z \\
f_2(z) = z^3 + z &+ \frac 1 z + \frac 1 {z^3} \\
f_3(z) = z^6+z^4+z^2+\ &\color{red}2 + \frac 1 {z^2}+ \frac 1 {z^4} + \frac 1 {z^6} \\
f_4(z) = z^{10} + z^8 + z^6 + \color{red}2z^4 + \color{red}2z^2 +\ &\color{red}2 + \frac {\color{red}{2}} {z^2} + \frac {\color{red}2} {z^4} + \frac 1 {z^6} + \frac 1 {z^8} + \frac{1}{z^{10}} \\
f_5(z)=z^{15} + z^{13}+ z^{11} + \color{red}2z^{9} + \color{red}2z^7 + \color{orange}3z^5 + \color{orange}3z^3 + \color{orange}3z &+ \frac{\color{orange}3} z + \frac {\color{orange}3}{z^3} + \frac {\color{orange}3} {z^5} + \frac {\color{red}2} {z^7} + \frac {\color{red}2} {z^9} + \frac 1 {z^{11}} + \frac 1 {z^{13}} + \frac 1 {z^{15}}
\end{split}$$
It seems that
*
*In each $f_m$ the exponent of $z$ falls from $\tau_m = \frac 1 2 m(m+1)$ down to $-\tau_m$, skipping every other value;
*For all $m$ the coefficient $a_{m,n}$ of the $n$-th term in the analytic part is equal to the coefficient $a_{m,-n}$ of the $n$-th term in the singular part (which makes sense intuitively).
So:
*
*Is there a closed-form expression for $a_{m,n}$?
*How can I prove fact 1. above?
| A proof of 1. could go as follows.
We obtain for integral $m\geq 1$
\begin{align*}
\color{blue}{f_m(z)}&=\prod_{k=1}^m\left(z^k+\frac{1}{z^k}\right)\\
&=\prod_{k=1}^m\left(z^{-k}\left(1+z^{2k}\right)\right)\\
&=z^{-\sum_{k=1}^mz^k}\prod_{k=1}^m\left(1+z^{2k}\right)\tag{1}\\
&=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\sum_{k=1}^m k}a_jz^{2j}\tag{2}\\
&\,\,\color{blue}{=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\frac{1}{2}m(m+1)}a_jz^{2j}}\\
\end{align*}
In the last line we observe the smallest power of the expression is $-\frac{1}{2}m(m+1)$ while the greatest power is
$-\frac{1}{2}m(m+1)+m(m+1)=\frac{1}{2}m(m+1)$.
Comment:
*
*In (1) we have a finite product where each term is either $1$ or $z^{2k}$. We so obtain a polynomial with smallest power $0$ and greatest power of $z^2$ equal to $\sum_{k=1}^m k$. This polynomial is written more conveniently in (2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Convergence and limit of recursive sequence Consider the sequence $(a_n)_{n \in \mathbb{N}}$ which has startvalue $a_0 > -1$ and recursive relation:
$$a_{n+1} = \frac{a_n}{2} + \frac{1}{1+ a_n}$$
How to prove the convergence and find the limit?
_
I think you need to show the convergence with a Cauchy sequence. It is also possible to show that the sequence is decreasing and bounded, but I found that if $-1 < a_0 < 0$ that the sequence first increases and after $a_n > 1$ it starts to decrease. So that method seems more complicated to me.
For the limit I found the values $1$ and $-2$ but don't know how to show which is the right one.
| $$a_0>-1$$
$$a_{n+1}=\dfrac{a_n}{2}+\dfrac{1}{1+a_n}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}$$
We know that $x^2+x+2>0, \forall x\in \mathbb{R}$.
$$a_1=\dfrac{a_0^2+a_0+2}{2(1+a_0)}>0\rightarrow a_{n+1}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}>0\rightarrow \boxed{a_n>0, \forall n\geq 1}$$
We are going to prove by induction that $\boxed{a_n<1,\forall n\geq 1}$
If $x<1\rightarrow \dfrac{x^2+x+2}{2(1+x)}<1$
Then: $a_{n+1}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}<1, \blacksquare .$
$$a_{n+1}-a_n=\dfrac{a_n^2+a_n+2}{2(1+a_n)}-a_n=\dfrac{a_n^2+a_n+2-2a_n-2a_n^2}{2(1+a_n)}=-\dfrac{a_n^2+a_n-2}{2(1+a_n)}=$$
$$=-\dfrac{(a_n+2)(a_n-1)}{2(a_n+1)}>0,\forall n\geq 1\rightarrow \{ a_n\} \, \text{is increasing and bounded.}$$
$$L=\dfrac{L}{2}+\dfrac{1}{1+L}\rightarrow L=\dfrac{L^2+L+2}{2(1+L)}\rightarrow 2L^2+2L=L^2+L+2\rightarrow$$
$$L^2+L-2=0\rightarrow L=-2\quad \text{or}\quad \boxed{L=1}$$
| {
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} |
Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
| Observe that
$$a-\sqrt a=b-\sqrt b\implies a-b=\sqrt a-\sqrt b\iff\frac{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)}{\sqrt a-\sqrt b}=1\iff$$
$$\iff \sqrt a+\sqrt b=1\;(\text{ assuming $\,a\neq b\,$)}\implies b=(1-\sqrt a)^2$$
So you need the maximum of $\;f(a):=a+b=a+(1-\sqrt a)^2=2a-2\sqrt a+1\;$ ...can you take it from here?
| {
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"answer_count": 7,
"answer_id": 1
} |
Disjoint sets in a combinatoral sum (continued) Let $f_S(m)$ be defined as in my previous question:
Let $S = \{1/n^2 : n \in \mathbb{N} \}$. Let $f_S(m)$ be the sum of the products of all $m$-tuples chosen from $S$. That is
$$f_S(m) = \sum_{X \in {S \choose m}} \prod_{x \in X} x $$
I have used inclusion-exclusion to derive a formula for $f_S(m)$.
For example: Let $a,b,c$ be the $3$ multiplicands when expanding
$$\zeta(2)^3 = \left( 1 + \frac 1 4 + \frac 1 9 + \cdots \right)^3$$
Then $f_S(3)$ can be solved by excluding all products that satisfy at least one of these rules: $a=b, \ b=c, \ a=c$.
*
*The values that satisfy $a=b$ let $c$ be any value, so the sum of their products is $\left(1 + \left(\frac 1 4\right)^2 + \left(\frac 1 9\right)^2 + \cdots \right) \left( 1 + \frac 1 4 + \frac 1 9 + \cdots \right)
=\zeta(4)\zeta(2)$. The same applies for $b=c$ and $a=c$.
*The values that satisfy $a=b \wedge b=c$ have $a=b=c$, so the sum of the products is $\zeta(6)$. The same applies for $a=b \wedge a=c$ and $a=c \wedge b=c$.
*For $a=b \wedge b=c \wedge a=c$ the result is again $\zeta(6)$.
By inclusion-exclusion and symmetry ordering $a,b,c$
$$f_S(3) = \frac{1}{3!}(\zeta(2)^3 - 3\zeta(4)\zeta(2) + 3\zeta(6) - \zeta(6))
= \frac 1 6 (\zeta(2)^3 - 3\zeta(4)\zeta(2) + 2\zeta(6)) $$
In fact, for larger values of $m$ satisfying rules like $a=b$ and $c=d$ turns out to be the disjoint union problem. If the rules we satisfy are $a=b$ and $c=d$, then we have the disjoint sets $(a,b), (c,d)$ and the result will be $\zeta(4)^2$. If the rules are like $a=b$ and $b=c$ then the disjoint sets are $(a,b,c), (d)$ and the result is $\zeta(6)\zeta(2)$.
I wrote a Mathematica program to calculate $f_S(m)$. Unfortunately it's quite slow, from evaluating all the inclusion-exclusion subsets.
altTotal[x_] := Total@x[[;; ;; 2]] - Total@x[[2 ;; ;; 2]];
inex[n_, zeta_] :=
Module[{productPairs, summands},
productPairs = Subsets[Range[n], {2}];
summands = Table[
conComp = ConnectedComponents /@
Map[Join[#, Table[{i, i}, {i, 1, n}]] &,
Subsets[productPairs, {depth}]];
Total[Times @@@ Map[zeta[2 Length[#]] &, conComp, {2}]],
{depth, 1, Length[productPairs]}];
Return[(zeta[2]^n - altTotal[summands])/n!]
]
I have determined the following:
\begin{align}
f_S(3) &= \frac 1 6 (\zeta(2)^3 - 3\zeta(4)\zeta(2) + 2\zeta(6)) \\
f_S(4) &= \frac{1}{24} \left(\zeta(2)^4-6 \zeta(4) \zeta(2)^2+8 \zeta(6) \zeta(2)+3 \zeta(4)^2-6 \zeta(8)\right) \\
f_S(5) &= \frac{1}{120} \left(\zeta (2)^5-10 \zeta (4) \zeta (2)^3+20 \zeta (6) \zeta (2)^2+15 \zeta (4)^2 \zeta (2)-30 \zeta (8) \zeta (2)-20 \zeta (4) \zeta (6)+24 \zeta (10)\right) \\
f_S(6) &= \frac{1}{720} \left(\zeta (2)^6-15 \zeta (4) \zeta (2)^4+40 \zeta (6) \zeta (2)^3+45 \zeta (4)^2 \zeta (2)^2-90 \zeta (8) \zeta (2)^2-120 \zeta (4) \zeta (6) \zeta (2)+144 \zeta (10) \zeta (2)-15 \zeta (4)^3+40 \zeta (6)^2+90 \zeta (4) \zeta (8)-120 \zeta (12)\right)
\end{align}
My question is how can I derive these expressions? (my linked answer has the closed-form already) The coefficients match A181897. Wikiversity: Permutations by cycle type looks insightful; unfortunately I don't know any group theory.
What is the connection between the cycle index of the symmetric group and this inclusion-exclusion process for $f$?
(Program that calculates much more efficiently)
f[m_, zeta_] := ((-1)^m) CycleIndexPolynomial[SymmetricGroup[m], -zeta /@ Range[2, 2 m, 2]]
| These may be computed with the unlabeled set operator
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}$$
applied to the Riemann Zeta function
$$\zeta(s) = \sum_{n\ge 1} \frac{1}{n^s}.$$
We use the recurrence by Lovasz for the cycle index $Z(P_n)$ of the
set operator $\textsc{SET}_{=n}$ on $n$ slots, which is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$
This recurrence lets us calculate the cycle index $Z(P_n)$ very
easily. We are interested in
$$f(s, m) = Z(P_m; \zeta(s))$$
and hence the recurrence becomes
$$f(s,m) = \frac{1}{m} \sum_{l=1}^m (-1)^{l-1} \zeta(ls) f(s, m-l)
\quad\text{where}\quad
f(s,0) = 1.$$
This yields e.g. for $f(2,4)$
$$1/24\, \left( \zeta \left( 2 \right) \right) ^{4}-1/4
\,\zeta \left( 4 \right) \left( \zeta \left( 2
\right) \right) ^{2}+1/3\,\zeta \left( 6 \right)
\zeta \left( 2 \right) +1/8\, \left( \zeta \left( 4
\right) \right) ^{2}-1/4\,\zeta \left( 8 \right)$$
and for $f(2,5)$
$${\frac { \left( \zeta \left( 2 \right) \right) ^{5}}{
120}}-1/12\,\zeta \left( 4 \right) \left( \zeta
\left( 2 \right) \right) ^{3}+1/6\,\zeta \left( 6
\right) \left( \zeta \left( 2 \right) \right) ^{2}+1
/8\,\zeta \left( 2 \right) \left( \zeta \left( 4
\right) \right) ^{2}\\-1/4\,\zeta \left( 8 \right)
\zeta \left( 2 \right) -1/6\,\zeta \left( 4 \right)
\zeta \left( 6 \right) +1/5\,\zeta \left( 10 \right)$$
We also have for $f(3,7)$
$$-1/12\, \left( \zeta \left( 3 \right) \right) ^{2}
\zeta \left( 6 \right) \zeta \left( 9 \right) +1/8\,
\zeta \left( 3 \right) \zeta \left( 6 \right) \zeta
\left( 12 \right) -1/6\,\zeta \left( 18 \right) \zeta
\left( 3 \right)\\ -{\frac {\zeta \left( 6 \right)
\left( \zeta \left( 3 \right) \right) ^{5}}{240}}+{
\frac {\zeta \left( 9 \right) \left( \zeta \left( 3
\right) \right) ^{4}}{72}}+1/48\, \left( \zeta
\left( 3 \right) \right) ^{3} \left( \zeta \left( 6
\right) \right) ^{2}-1/24\,\zeta \left( 12 \right)
\left( \zeta \left( 3 \right) \right) ^{3}\\+1/10\,
\zeta \left( 15 \right) \left( \zeta \left( 3 \right)
\right) ^{2}-1/48\,\zeta \left( 3 \right) \left(
\zeta \left( 6 \right) \right) ^{3}\\+1/18\,\zeta
\left( 3 \right) \left( \zeta \left( 9 \right)
\right) ^{2}+1/24\, \left( \zeta \left( 6 \right)
\right) ^{2}\zeta \left( 9 \right) -1/10\,\zeta
\left( 6 \right) \zeta \left( 15 \right) \\-1/12\,\zeta
\left( 9 \right) \zeta \left( 12 \right) +1/7\,\zeta
\left( 21 \right) +{\frac { \left( \zeta \left( 3
\right) \right) ^{7}}{5040}}.$$
(Sorted and formatted by Maple.)
| {
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"answer_count": 2,
"answer_id": 0
} |
Critical points and extremum of $f(x,y)=y^2-x^2+x^3+x^2y+\frac{y^3}{3}\;\;\forall\;(x,y)\in\Bbb{R}^2$ Let $f:\Bbb{R}^2\to \Bbb{R}$ be a function defined by
\begin{align}f(x,y)=y^2-x^2+x^3+x^2y+\frac{y^3}{3}\;\;\forall\;(x,y)\in\Bbb{R}^2\end{align}
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?
My work:
\begin{align}\frac{\partial f}{\partial x}=-2x+3x^2+2xy\qquad (1)\end{align}
\begin{align}\frac{\partial f}{\partial y}=2y+x^2+y^2\qquad (2)\end{align}
At \begin{align}\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\end{align}
we have
\begin{align}x=\frac{1}{2}(3x^2+2xy)\end{align}
Substituting into $2$, we get
\begin{align}0=y(8+12x^3)+y^2(4x^2+4)+9x^4\end{align}
\begin{align}y=\frac{-2-3 x^3\pm\sqrt{4+12 x^3-9 x^4}}{2 \left(1+x^2\right)}\end{align}
I don't know where to go from here. Can someone please, help?
| \begin{align}\frac{\partial f}{\partial x}=-2x+3x^2+2xy\qquad (1)\end{align}
\begin{align}\frac{\partial f}{\partial y}=2y+x^2+y^2\qquad (2)\end{align}
At \begin{align}\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\end{align}
we have (1) implies
\begin{align}x(3x+2y-2)=0\end{align}
So there are two cases $x=0$ or ... continue from there.
| {
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} |
Solve for $x$ in the equation containing ${\lfloor{x}\rfloor}$ and $\{x\}$
Calculate all possible values of $x$ satisfying,
$$\frac{\lfloor{x}\rfloor}{\lfloor{x-2}\rfloor}-\frac{\lfloor{x-2}\rfloor}{\lfloor{x}\rfloor}=\frac{8\{x\}+12}{\lfloor{x}\rfloor
\lfloor{x-2}\rfloor}$$
where $\{x\}$ stands for fractional part of $x$.
My Attempt:
$${\lfloor{x}\rfloor}^2-{\lfloor{x-2}\rfloor}^2={8\{x\}+12}$$
${\lfloor{x}\rfloor}^2-{\lfloor{x-2}\rfloor}^2$ is an integer, so ${8\{x\}+12}$ must also be an integer, i.e. $8\{x\}$ must be an integer.
As, $0\leq{\{x\}}\lt{1}$, Therefore the only integer values of $8\{x\}$ will be,
$$\{x\}=0 \implies 8\{x\}=0 \implies 8\{x\}+12=12$$
$$\{x\}=\frac{1}{2} \implies 8\{x\}=4 \implies 8\{x\}+12=16$$
$$\{x\}=\frac{1}{4} \implies 8\{x\}=2 \implies 8\{x\}+12=14$$
$$\{x\}=\frac{1}{8} \implies 8\{x\}=1 \implies 8\{x\}+12=13$$
On calculating , we get that $12$,$16$ can be expressed as,
$$16=5^2-3^2 \implies 5\leq{x}\lt 6$$
but $\{x\}=\frac{1}{2}, \therefore x=5+\frac{1}{2}=\frac{11}{2}$
$$12=4^2-2^2 \implies 4\leq{x}\lt 5 $$
but $\{x\}=0, \therefore x=4+0=4$
Therefore the only solution set is,
$$\bigg\{4,\frac{11}{2}\bigg\}$$
Is this the correct approach? I am afraid that I am missing some values.
| Decompose $x=n+f$. Then
$$\frac n{n-2}-\frac{n-2}n=\frac{8f+12}{n(n-2)}$$ or
$$n^2-(n-2)^2=8f+12$$
or
$$f=\frac{n-4}2.$$
The possible values for $n$ are $4$ and $5$ (so that $0\le f<1$).
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is $f $ decreasing? Let $f:(0,\infty)\rightarrow (0,\infty) $, $f (x)+f (y)\geq 2f (x+y), \forall x, y>0$.
Is $f $ decreasing?
I need to show that $f (x)+f (y)+f (z)\geq 3f (x+y+z) $ and it's enough to show that $f$ is decreasing.
| We have:
$$f(x)+f(y)+f(z) \ge 2f(x+y)+f(z) = f(x+y) +f(z) + f(x+y)+f(z)-f(z) \ge 2f(x+y+z)+2f(x+y+z)-f(z)=4f(x+y+z)-f(z)$$
In the same manner:
$$f(x)+f(y)+f(z) \ge 2f(x+z)+f(y) = f(x+z) +f(y) + f(x+z)+f(y)-f(y) \ge 2f(x+y+z)+2f(x+y+z)-f(y)=4f(x+y+z)-f(y)$$
And:
$$f(x)+f(y)+f(z) \ge 2f(y+z)+f(x) = f(y+z) +f(x) + f(y+z)+f(x)-f(x) \ge 2f(x+y+z)+2f(x+y+z)-f(x)=4f(x+y+z)-f(x) $$
In the end we have:
$$ f(x)+f(y)+f(z) \ge 4f(x+y+z)-f(z)$$
$$ f(x)+f(y)+f(z) \ge 4f(x+y+z)-f(y)$$
$$ f(x)+f(y)+f(z) \ge 4f(x+y+z)-f(x)$$
Summing the we get:
$$3(f(x)+f(y)+f(z)) \ge 12f(x+y+z)-f(x)-f(y)-f(z) \Leftrightarrow 4(f(x)+f(y)+f(z)) \ge 12f(x+y+z) \Leftrightarrow f(x)+f(y)+f(z) \ge 3f(x+y+z)$$
I wish I helped!
| {
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"answer_count": 2,
"answer_id": 1
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If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?
I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it.
Another thought: If $(a^2+b^2+c^2)^2 \geq 1$, then this statement will be true. But how can I prove it?
| By Rearrangement inequality we have that
$$(a^2+b^2+c^2)^2\ge a^4+b^4+c^4\ge a^2bc+b^2ac+c^2ab=abc(a+b+c)=a+b+c$$
| {
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"answer_count": 4,
"answer_id": 3
} |
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now
| Another way to look:
\begin{eqnarray*}
x^{4}-2 x^{2} \cos^{2}\left(\frac{\pi x}{2}\right) +1 &=& x^{4}-2x^2+1 + 2 x^{2} \cos^{2}\left(\frac{\pi x}{2}\right) \\
&=& \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} + \cos^{2}\left(\frac{\pi x}{2}\right)
\end{eqnarray*}
Now $ \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} + \cos^{2}\left(\frac{\pi x}{2}\right)=0$ implies,
\begin{equation}
0 \ge - \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} = \cos^{2}\left(\frac{\pi x}{2}\right) \ge 0
\end{equation}
Bounding from both side means, it has to be equality.
\begin{equation}
0 = - \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} = \cos^{2}\left(\frac{\pi x}{2}\right) = 0
\end{equation}
This is satisfied only with $x=\pm 1$.
| {
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"answer_count": 6,
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Factorization $\cos\left(\tfrac{\pi z}{4}\right)-\sin\left(\tfrac{\pi z}{4}\right)$
Prove that
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right)$$
My try:
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\cos\left(\frac{\pi z}{4}\right)-\cos\left(\frac{\pi}{2}-\frac{\pi z}{4}\right)=-\sqrt{2}\sin\left(\frac{\pi z-\pi}{4}\right) $$
Now, substitute the expression of the $\sin(\pi z)$ and get
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=-\sqrt{2}\pi\left(\frac{z-1}{4}\right)\prod_{n=1}^\infty\left(1-\frac{(z-1)^2}{16n^2}\right).$$
How should I proceed now?
| $\displaystyle \prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right) =\prod_{n=0}^\infty\left(1-\frac{z}{4n+1}\right)\prod_{n=1}^\infty\left(1+\frac{z}{4n-1}\right)=$
$\displaystyle =(1-z)\prod_{n=1}^\infty\left(\frac{4n(1-\frac{z-1}{4n})}{4n(1+\frac{1}{4n})}\right)\prod_{n=1}^\infty\left(\frac{4n(1+\frac{z-1}{4n})}{4n(1-\frac{1}{4n})}\right)= (1-z)\frac{\prod_{n=1}^\infty\left(1-(\frac{z-1}{4n})^2\right)}{\prod_{n=1}^\infty\left(1-(\frac{1}{4n})^2\right)}$
$\displaystyle =(1-z)\frac{\sin(\frac{z-1}{4}\pi)}{\frac{z-1}{4}\pi}\frac{\frac{1}{4}\pi}{\sin(\frac{1}{4}\pi)}=-\sqrt{2}\sin\left(\frac{\pi z-\pi}{4}\right)= -2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi z-\pi}{4}\right)$
$\displaystyle = \cos\left(\frac{\pi z-\pi}{4}+\frac{\pi}{4}\right) - \cos\left(\frac{\pi z-\pi}{4} -\frac{\pi}{4}\right) = \cos\left(\frac{\pi z}{4}\right) - \sin\left(\frac{\pi z}{4}\right)$
because of $\enspace\displaystyle \sin x\sin y=-\frac{1}{2}\left(\cos(x+y)-\cos(x-y)\right)$
Note:
The decisive step was $\enspace\displaystyle \frac{1}{\sin(\frac{1}{4}\pi)}=2\sin\left(\frac{\pi}{4}\right)\,$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$
and therefore $\ \frac{z-1}{z+1}$ is imaginary
now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..
Thanks
| Note that\begin{align}\frac{z-1}{z+1}&=\frac{(z-1)\left(\overline z+1\right)}{(z+1)\left(\overline z+1\right)}\\&=\frac{|z|^2+z-\overline z-1}{|z+1|^2}\\&=\frac{|z|^2-1+2i\operatorname{Im}(z)}{|z+1|^2}\end{align}and that therefore $\frac{z-1}{z+1}$ is purely imaginary if and only $|z|^2-1=0$, which is the same thing as asserting that $|z|=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Linear dependence and independence solution. How was it found? I am reading this text:
I feel like my text skipped a step to find the coefficients/solutions. Did they write a system of linear equations and use Gaussian elimination? How did they do this?
| In order to see whether these vectors are dependent examine the system:
$a(1,3,1)+b(0,1,2)+c(1,0,-5)=(0,0,0)$
which is equivalent with
$$\begin{pmatrix}
1 & 0 & 1\\
3 & 1 & 0\\
1 & 2 & -5
\end{pmatrix}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$$
then you can use elimination to find $a,b,c$
$$\begin{pmatrix}
1 & 0 & 1\\
3 & 1 & 0\\
1 & 2 & -5
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
1 & 2 & -5
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
0 & 2 & -6
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
0 & 0 & 0
\end{pmatrix}$$
So the above system is equivalent to the following
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}\Leftrightarrow \begin{cases} a+c=0 \\ b-3c=0 \end{cases}\Leftrightarrow \begin{cases} a=-c \\ b=3c \end{cases}\Leftrightarrow \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
-c\\
3c\\
c
\end{pmatrix}=c\begin{pmatrix}
-1\\
3\\
1
\end{pmatrix} $$
And we can choose $c=1$ so $(a,b,c)=(-1,3,1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get the equation where a circle goes through three points
If I have the equation $ax^2+ay^2+bx+cy+d=0$ how do I get the equation where the circumference goes through the points P = (1,1), Q = (−1,−1) and
R = (−1,1)
I have it in mind to solve it with a matrix, but the instructions seem confusing to me, can I get some help?
| If you need to use matrices/linear algebra, consider this approach: If you move $d$ to the other side of the equation $ax^2 + ay^2 + bx + cy + d = 0$, you get $ax^2 + ay^2 + bx + cy = -d$. We can view the left hand side of this equation as the product of two matrices:
\begin{eqnarray}
\left[ax^2 + ay^2 + bx + cy\right] \;=\begin{bmatrix} x^2 & y^2 & x & y \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix}
\end{eqnarray}
Since $ax^2 + ay^2 + bx + cy = -d$, we get the matrix equation
\begin{eqnarray}
\begin{bmatrix} x^2 & y^2 & x & y \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right].
\end{eqnarray}
Now we have 3 points that satisfy this matrix equation. Taking the first point, $P=(1,1)$, and plugging it into this equation, we obtain
\begin{eqnarray}
\begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right].
\end{eqnarray}
Similary, we plug in $Q=(-1, -1)$ to get
\begin{eqnarray}
\begin{bmatrix} 1 & 1 & -1 & -1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right]
\end{eqnarray}
and $R=(-1,1)$ to get
\begin{eqnarray}
\begin{bmatrix} 1 & 1 & -1 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right].
\end{eqnarray}
Now we have 3 $simultaneous$ $linear$ $equations$, which can be combined into the matrix equation
\begin{eqnarray}
\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & -1 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \begin{bmatrix}-d \\ -d \\ -d \end{bmatrix}.
\end{eqnarray}
From here, one can use standard Gaussian Elimination to obtain
\begin{eqnarray}
\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \begin{bmatrix}-d \\ 0 \\ 0 \end{bmatrix}.
\end{eqnarray}
This tells us that $b=c=0$, and $2a = -d$. Coming back to our original equation we insert these values to obtain
\begin{equation}
ax^2 + ay^2 + 0 x + 0 y = 2a
\end{equation}
We can divide by $a$ since $a$ must be nonzero (otherwise we don't have a circle!) to obtain the final equation
\begin{equation}
x^2 + y^2 = 2.\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all triples $(p,x,y)$ such that $ p^x=y^4+4$ Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x = (y^2+2y+2)(y^2-2y+2)$
Then: $p^k = y^2+2y+2$ and $p^j= y^2-2y+2$
The solutions I saw were different from here and they just wanted to prove that $(5,1,1)$ was the only solution. I´ll just ask what I didn't understand:
1)
Therefore, $p^k = (y+1)^2+1$ and $p^j= (y-1)^2+1$
And then: $(y+1)^2 \equiv (y-1)^2 \equiv -1 \pmod p$
So here $-1$ is a quadratic residue, and therefore $p= 4n+1$. I understood his solution until this part.
*
*I know that $p= 4n+1$ because I saw that was one of the properties of quadratic residues, but would like to see a proof of it if possible.
*He then says that by$\mod 8$ , $ k$ and $j$ had different parity, so $x$ was odd.
Then he assumes that $p, k >1$ and says:
*
*If $k=2m$, then: $(y+1)^2= (p^m+1)(p^m -1)$ or if $j= 2m$: then $(y-1)^2= (p^m+1)(p^m -1)$ and affirms those equations have no solution, that might be a property but want to know about it anyway.
Then he says there are no solution for $x>1$, then $p = (y^2+2y+2)(y^2-2y+2)$, so $1=y^2-2y+2$ and that was the triple $(5,1,1)$, therefore that is the only solution.
2)
He started off by saying: if $y$ is even then $ p^k \equiv p^j \equiv 2 \pmod 4$ then $a = b =1$ and that is not possible, therefore $y$ is odd.
*
*Since $ 4y = p^{b} - p^{a}$, therefore $ p^{b-a}=5$, $ p^{a}=y$
*$ 2y^{2}+4=p^{b}+p^{a}$, if $ a\neq 0$, then $ y^2 \equiv 3\mod 5$, which is impossible.
I don´t know how you can deduce those things. After that he affirmed that $(5,1,1)$ was the only solution.
3) He showed that $y$ was odd and said:
*
*$gcd (y^2+2y+2, y^2-2y+2) = gcd (y^2+2y+2, 4y) = 1$ (I think this is also a property but I don´t even know it).
And since $y^2+2y+2 > y^2-2y+2$, then $y^2+2y+2=p^x$ and $y^2-2y+2 =1$, so $y=1$ and $(5,1,1)$ was the only solution.
Sorry if I asked many things, I just thought those things may be useful for other problems. I pointed out the things I didn´t understood so that it was easy to see them. Thanks in advance.
| Let $p$ be a prime number, and $x$ and $y$ natural numbers, such that $p^x=y^4+4$. Then as you note
$$p^x=(y^2+2y+2)(y^2-2y+2),$$
and hence both factors are powers of $p$, say
\begin{eqnarray*}
y^2+2y+2&=&p^u,\\
y^2-2y+2&=&p^v.
\end{eqnarray*}
First note that if $v=0$ then $y=1$, and we find that $p=5$ and $x=1$, the solution you already found.
Now suppose $v>0$. Because $y>0$ we have $p^u>p^v$ and so $u>v$, so $p^{u-v}-1$ is an integer. Then from
$$4y=(y^2+2y+2)-(y^2-2y+2)=p^u-p^v=p^v(p^{u-v}-1).$$
we see that $p$ divides $4y$ because $v>0$. So either $p$ divides $4$ or $p$ divides $y$. If $p$ divides $y$ then from
$$p^x=y^4+4,$$
it follows that $p$ also divides $4$, so either way we find that $p=2$. Then clearly $x>2$ because $y>0$, and reducing mod $2$ shows that $y$ is even, say $y=2z$. Plugging this in shows that
$$2^x=p^x=y^4+4=(2z)^4+4=2^4z^4+4,$$
and dividing by $4$ shows that
$$2^{x-2}=4z^4+1.$$
In particular $2^{x-2}$ is odd, so $x=3$. But then $4z^4+1=1$ and so $z=0$, which implies $y=0$, a contradiction.
This shows that the unique solution is indeed $(p,x,y)=(5,1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$?
Question: If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if
$q=3.$
I am certain the answer to the question is yes, and I believe I have worked out a solution - although I am not certain it is correct; whence the solution-verification tag. I feel it is ok to also ask if there are solutions that are "deeper" - maybe using better applications of modular arithmetic or making use of group theory ?
Solution To Question: Since $q\equiv 3 \pmod 4$ then there exist a positive number $m\in\mathbb{N}$ such that $q=4m+3$. Then after substitution, ${q+1 \choose 2}={4m+4 \choose 2}=\frac{(4m+4)(4m+3)}{2}=(2m+2)(4m+3);$ which has even parity. In particular if $p$ is a prime number I can write $(2m+2)(4m+3)=2p.$ Division by two yields $\left(\frac{2m+2}{2}\right)\left(4m+3\right)=p.$ Recall a prime number has as divisors the numbers $1$ and itself. Note that $4m+3\neq 1$ for positive $m$ and so $\frac{2m+2}{2}=1;$ which implies $m=0$ in which case $4m+3=3.$
On the other hand if ${q+1 \choose 2}=6$ then explicitly $\frac{q(q+1)}{2}=6$ and so $q^2+q-12=0.$ Equivalently $(q-3)(q+4)=12;$ which has roots $q=3$ or $q=-4.$ We require the positive root $q=3.$ This completes the solution to the question. $\blacksquare$
| If $q \equiv 3 \mod 4$, $q+1$ is divisible by $4$. Say $q+1=4k$ Then
$$ {q+1 \choose 2} = \frac{(q+1)q}{2} = 2 k (4k-1)$$
This is divisible by $2$, and by any primes that divide $k$ or $4k-1$. That makes at least three (not necessarily distinct) primes, unless either $k = 1$ or $4k-1 = 1$ (the latter being impossible).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\Im(z) > 0$then $\Im (-1/(z+1)) >0$ My work so far:
Letting $z = x+iy$, where $x,y$ are real,
then $\Im(z) = y > 0$
$y+1 > 0$
$1/(y+1) > 0$
But I am not sure how to proceed. I also attempted to multiply $-1/(z+1)$ by its conjugate, but that led to $(-x-1+iy)/((x+1)^2 + y^2)$, which I feel is the wrong way to approach the problem.
Thank you for your help.
| $z = x + iy, \; x, y \in \Bbb R, \; y > 0; \tag 1$
$z + 1 = (1 + x) + iy; \tag 2$
$\dfrac{1}{z + 1} = \dfrac{1}{(1 + x) + iy}$
$= \dfrac{(1 + x) - iy}{(1 + x)^2 + y^2} = \dfrac{1 + x}{(1 + x)^2 + y^2} -i\dfrac{y}{(1 + x)^2 + y^2}; \tag 3$
$-\dfrac{1}{z + 1} = -\dfrac{1 + x}{(1 + x)^2 + y^2} +i\dfrac{y}{(1 + x)^2 + y^2}; \tag 4$
we see that
$\Im \left ( -\dfrac{1}{1 + z} \right ) = \dfrac{y}{(1 + x)^2 + y^2}; \tag 5$
we are given that $y > 0$; manifestly $(1 + x)^2 + y^2 > 0$, so . . .
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $n\in\mathbb{N}$ such that the nonzero roots of $(z+1)^n-z^n-1$ are all on the unit circle. I try to use Mathematica to find such $n$, and I believe that only $n=2,3,4,5,6,7$ meet the requirement but I don't know how to prove that for $n\ge 8$, there is always some root lying outside the unit ball by using the basic complex analysis knowledge.
In fact, I do have a dirty way to deal with this problem, which only involves the property of polynomial. But I don't think it is an efficient and elegant solution.
| It is clear that the result is true for $n=2$. Now we consider $n\ge 3$.
Note that $(z+1)^n-z^n-1=\sum_{k=0}^n\binom{n}{k} z^k-z^n-1 = \sum_{k=1}^{n-1}\binom{n}{k} z^k= nz\sum_{k=1}^{n-1}\frac{\binom{n}{k}}{n}z^{k-1}$. Suppose that $\sum_{k=1}^{n-1}\frac{\binom{n}{k}}{n}z^{k-1}=\prod_{k=1}^{n-2}(z-z_k)$. Then we have $\sum_{k=1}^{n-2}z_k=-\frac{\binom{n}{n-2}}{n}=-\frac{n-2}{2}$ and $\sum_{j\ne l}z_jz_l=\frac{\binom{n}{n-3}}{n}=\frac{(n-1)(n-2)}{6}$. Then
$$\sum_{k=1}^{n-2}z_k^2 = \left(\sum_{k=1}^{n-2}z_k\right)^2-2\sum_{j\ne l}z_jz_l = \frac{(n-1)^2}{4}-\frac{(n-1)(n-2)}{3}= -\frac{n^2-6n+5}{12}.$$
Write $z_k=x_k+iy_k$, where $x_k,y_k\in\mathbb{R}$ for $k=1,\cdots, n-1$. Note that for each root $z_k$, $\overline{z_k}$ is also a root and $z_k^2+\overline{z_k}^2 = x_k^2-y_k^2+2ix_ky_k+x_k^2-y_k^2-2ix_ky_k=2(x_k^2-y_k^2)$, so
$$\sum_{k=1}^{n-2}z_k^2=\sum_{k=1}^{n-2}(x_k^2-y_k^2)=-\frac{n^2-6n+5}{12}.$$
Suppose that all the nonzero roots lying on the unit circle, then
$$\sum_{k=1}^{n-2}|z_k|^2 = n-2=\sum_{k=1}^{n-2}(x_k^2+y_k^2),$$
with the above equation, we have
$$2\sum_{k=1}^{n-2}x_k^2 = n-2-\frac{n^2-6n+5}{12}=-\frac{n^2-18n+29}{12}\ge 0,$$
which follows that $n\le 9+2\sqrt{13}\approx 16.2$.
By using the Wolfram Mathematica, for $n=2,3,4,5,6,7$, the nonzero roots of $(z+1)^n-z^n-1=0$ lying on the unit circle and for $8\le n\le 16$, there is some root not lying on the unit circle.
| {
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"url": "https://math.stackexchange.com/questions/2899959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $
I understand every step until here
$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$
How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
| $\sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! }
$
For region of convergence,
just replace $n!$ with
$(n/e)^n$
and then use the
n-th root test.
The additional
$\sqrt{n}$
only affects the boundaries.
In this case
(why the annoying spaces?),
$\dfrac{n!n^{n}}{(2n)!}
$
becomes
$\dfrac{(n/e)^nn^{n}}{(2n/e)^{2n}}
=\left(\dfrac{(n/e)n}{(2n/e)^{2}}\right)^n
=\left(\dfrac{n^2e^2}{4n^2e}\right)^n
=\left(\dfrac{e}{4}\right)^n
$
Since
$\dfrac{e}{4} < 1$,
the sum converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$, show that $3^{16}-33$ and $3^{15}+5$ are divisible by 4.
My proof:
$$p^{2n}-2nq-1=(1+q)^{2n}-2nq-1$$
$$=[1+2nq+\frac{(2n)(2n-1)}{2!} q^2+\frac{2n(2n-1)(2n-2)}{3!}q^3+...]-2nq-1$$
$$=n(2n-1)q^2+\frac{2}{3} n(2n-1)(n-1)q^3+...$$
$$=q^2[n(2n-1)+\frac{2}{3} n(2n-1)(n-1)q+...]$$
Hence the expansion has a common factor $q^2$
Taking $n=8$ and $p=3$, and given that $p=q+1, q=2$, by substitution,
$$3^{16} -33=4[120+1120+...]$$
By factoring a 3:
$$3(3^{15}-11)=4[120+1120+...]$$
Dividing both sides by 3 and adding 15 to both sides:
$$3^{15} +5=4[\frac{1}{3}(120+1120+...)+4]$$
Then it is proven that it is also divisible by 4.
The only problem I have with the proof is that how do I know that each term in the brackets $(120+1120+...)$ are divisible by 3?
| It's better if you use
$$
(1+q)^{2n}=1+2nq+\sum_{k=2}^{2n}\binom{2n}{k}q^k
$$
so
$$
p^{2n}-2nq-1=q^2\sum_{k=2}^{2n}\binom{2n}{k}q^{k-2}
$$
is divisible by $q^2$.
With $q=2$ and $n=8$ you have $p=3$ and $p^{2n}-2nq-1=3^{16}-33$.
For the second case, consider
$$
3^{15}+5=3(3^{14}-29)+3\cdot29+5
$$
and set $q=2$, $n=7$, noting that $3\cdot 29+5=92$, which is divisible by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\ \left({x \over x^2+x}+{x \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$
Can we do this without squeeze theorem?
| We could do it using harmonic numbers
$$S_x=\sum_{k=1}^x \frac x {x^2+k}=x\sum_{k=1}^x \frac 1 {x^2+k}=x \left(H_{x^2+x}-H_{x^2}\right)$$
Now, using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^4}\right)$$ we should get
$$S_x=1-\frac{1}{2 x}-\frac{1}{6 x^2}+\frac{1}{4
x^3}+O\left(\frac{1}{x^4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Whats wrong in this approach to evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin x\cos xdx}{\sin x+\cos x}$ Whats wrong in this approach to evaluate $$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x\cos xdx}{\sin x+\cos x}$$
Since $f\left(\frac{\pi}{2}-x\right)=f(x)$ we have
$$I=2\int_{0}^{\frac{\pi}{4}} \frac{\sin x\cos xdx}{\sin x+\cos x}$$
Now applying $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get
$$I=\sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{(\sin x-\cos x)(\sin x+\cos x)dx }{\sin x-\cos x+\cos x+\sin x}$$ $\implies$
$$I=\sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{2\sin^2 x-1}{\sin x}$$
But second integral viz $$\int_{0}^{\frac{\pi}{4}} \csc xdx $$ is not defined when we substitute lower limit?
| We have that
*
*$\sin(\pi/4-x)=\frac{\sqrt 2}2(\cos x-\sin x)$
*$\cos(\pi/4-x)=\frac{\sqrt 2}2(\cos x+\sin x)$
therefore
$$I=2\int_{0}^{\frac{\pi}{4}} \frac{\sin x\cos xdx}{\sin x+\cos x}=\sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{(\cos x-\sin x)(\cos x+ \sin x)dx }{\cos x-\sin x+\cos x+ \sin x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
| You reduced the inequality to $$\frac{4(x+1)(x-2)}{(x^2+2)(x+4)} \ge 0$$
which is equivalent to $$\frac{(x+1)(x-2)}{x+4} \ge 0$$
This is true if and only if an even number of terms $(x+1), (x-2), (x+4)$ are $\le 0$. There are four options:
*
*$x+1 \ge 0$, $x-2 \ge 0$, $x+4 > 0$ which gives $x \in [2, +\infty\rangle$
*$x+1 \le 0$, $x-2 \le 0$, $x+4 > 0$ which gives $x \in \langle -4, -1]$
*$x+1 \le 0$, $x-2 \ge 0$, $x+4 < 0$ which gives no solutions
*$x+1 \ge 0$, $x-2 \le 0$, $x+4 < 0$ which gives no solutions
Therefore the solutions are $x \in \langle -4, -1] \cup [2, +\infty\rangle$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solve $a^2 - 2b^2 - 3 c^2 + 6 d^2 =1 $ over integers $a,b,c,d \in \mathbb{Z}$ Are we able to completely solve this variant of Pell equation?
$$ x_1^2 - 2x_2^2 - 3x_3^2 + 6x_4^2 = 1 $$
This has an interpretation as is related to the fundamental unit equation of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$ as well as various irreducible quartics. Is this the same as solving three separate Pell equations?
\begin{eqnarray*}
x^2 - 2y^2 &=& 1 \\
x^2 - 3y^2 &=& 1 \\
x^2 + 6y^2 &=& 1 \tag{$\ast$}
\end{eqnarray*}
Our instinct suggests there should be three degrees of freedom here, and setting different variables to zero we could find three two of these (the third equation has no solutions over $\mathbb{R}$). Does that generate all the solutions?
Perhaps I should remark this quadratic form is also a determinant
$$ a^2 - 2b^2 - 3c^2 + 6d^2 = \det \left[
\begin{array}{cc}
a + b \sqrt{2} & c - d \sqrt{2}\\
3(c + d \sqrt{2}) & a - b \sqrt{2}
\end{array}
\right]$$
This might not even contain Oscar's solution. Extending Keith's solution We could have:
$$
\left[
\begin{array}{cc}
a + b \sqrt{2} & c - d \sqrt{2}\\
3(c + d \sqrt{2}) & a - b \sqrt{2}
\end{array}
\right] =
\left[
\begin{array}{cc}
3 + 1 \sqrt{2} & 2 - 1 \sqrt{2}\\
3(2 + 1 \sqrt{2}) & 3 - 1 \sqrt{2}
\end{array}
\right]^n
$$
| Equation given above is shown below:
$a^2-2b^2-3c^2+6d^2=1$
As shown by "Individ" , what "OP" can do is to take (a,b,c,d) as shown below:
$a=p(6w^2+4w+3)$
$b=p(2w^2+6w+1)$
$c=p(4w^2+4w+2)$
$d=p(2w^2+4w+1)$
Where $(p)= [1/(2w^2-1)]$
For suitable value's of 'w' we get the numerical solutions below:
w=(1), (a,b,c,d)=(13, 9, 10, 7)
w=(3/4), (a,b,c,d)= (75, 53, 58, 41)
w=(5/7), (a,b,c,d)= (437, 309, 338, 239)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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