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Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× 2\sin(x) \cos (x)×(1-2\sin^2(x)), \cos(x)>0$$
$$8\sin^3(x)-4\sin(x)+1=0$$
$$(2\sin(x)-1)(4\sin^2(x)+2\sin (x)-1)=0$$
$$4\sin^2(x)+2\sin (x)-1=0, \sin(x)≠\frac 12$$
$$4t^2+2t-1=0$$
$$t_{1,2}=\frac {-1±\sqrt 5}{4}$$
$$t=\frac {\sqrt5-1}{4} ,t>0$$
$$\sin 18°=\frac {\sqrt5-1}{4} .$$
Finally, $$\frac {1}{\sin 18°}=\frac {4}{\sqrt5-1}=\sqrt5+1$$
Is this way correct and is there a better/elegant way to do it? As always, it was an ugly solution.
Thank you!
| There are a couple of well-known approaches. The first parallels the answer of @egreg and goes back to Gauss. For a prime $p$ find a primitive root $g$ and then for and divisor $d$ of $p-1$ and integer $0\le j<d$ let
$$\sigma_{dj}=\sum_{k=0}^{\frac{p-1}d-1}\omega^{g^{j+kd}}$$
Where $\omega=e^{2\pi i/p}$. In our case $p=5$ and $g=2$ is a primitive root. Now, $\sigma_{10}=-1$ because the sum of the roots of $z^p-1=0$ is $\sigma_{10}+1=0$ because it's minus the coefficient of $z^{p-1}$. Then $\sigma_{20}+\sigma_{21}=\sigma_{10}=-1$ and we can work out the product $\sigma_{20}\sigma_{21}$ by constructing a table where each table entry is the product of its row and column headers:
$$\begin{array}{c|cc}
&\omega^1&\omega^4\\
\hline\omega^2&\omega^3&\omega^1\\
\omega^3&\omega^4&\omega^2\end{array}$$
And the product is the sum of the table entries, $\sigma_{20}\sigma_{21}=\sigma_{10}=-1$. Since we know the sum and product of $\sigma_{20}$ and $\sigma_{21}$ we can construct an equation they satisfy, $\sigma_{20}^2+\sigma_{20}-1=0$ with solution
$$2\cos\frac{2\pi}5=\omega+\omega^4=\sigma_{20}=\frac{-1\pm\sqrt5}2=\frac{-1+\sqrt5}2$$
because it's a first quadrant angle. So
$$\csc\frac{\pi}{10}=\frac1{\sin\frac{\pi}{10}}=\frac1{\cos\frac{2\pi}5}=\frac4{\sqrt5-1}=\sqrt5+1$$
It might seem like a lot of machinery to get this simple answer, but this approach is capable of getting to $\cos\frac{2\pi}{17}$ because just as in the above we can find an equation that $\sigma_{20}$ and $\sigma_{21}$ satisfy, then equations for $\sigma_{40}$ and $\sigma_{42}$ and for $\sigma_{41}$ and $\sigma_{43}$ and finally for $\sigma_{80}=2\cos\frac{2\pi}{17}$ and $\sigma_{84}$ and that's one construction they don't usually teach you in Euclidean geometry class.
The other approach is similar to that proposed in the original question but is perhaps a bit more systematic. We start with the identity
$$\cos(n+1)\theta=\cos n\theta\cos\theta-\sin n\theta\sin\theta=2\cos n\theta\cos\theta-\cos(n-1)\theta$$
Starting with $\cos(0\theta)=1$, $\cos(1\theta)=\cos\theta$ we progress to
$$\begin{align}\cos2\theta&=2\cos^2\theta-1\\
\cos3\theta&=4\cos^3\theta-3\cos\theta\\
\cos4\theta&=8\cos^4\theta-8\cos^2\theta+1\\
\cos5\theta&=16\cos^5\theta-20\cos^3\theta+5\cos\theta\end{align}$$
Since our target is $\theta=\frac{2\pi}5$ and we know that $\cos5\theta=1$ and $\cos0=1$ is a root and also that there are two double roots because $\cos\theta=\cos4\theta$ and $\cos2\theta=\cos3\theta$ we can factor
$$16\cos^5\theta-20\cos^3\theta+5\cos\theta-1=\left(\cos\theta-1\right)\left(4\cos^2\theta+2\cos\theta-1\right)^2=0$$
So we get
$$\cos\theta=\cos\frac{2\pi}5=\sin\frac{\pi}{10}=\frac{-2\pm\sqrt{20}}8=\frac{-1+\sqrt5}4$$
Because it's a first quadrant angle, as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is wrong. How can I solve this without using the binomial theorem?
| Let $u=x^2+1\implies x^2=u-1\implies du=2x\,dx$. So $$\begin{align}\int(x^2+1)^7x^3\,dx&=\int (x^2+1)^7\cdot\frac{x^2}2\cdot2x\,dx\\&=\int u^7\cdot\frac{u-1}2\,du=\frac12\int(u^8-u^7)\,du\\&=\frac12\left(\frac{u^9}9-\frac{u^8}8\right)+C\\&=\boxed{\frac1{144}(x^2+1)^8(8x^2-1)}+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root
set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begin{aligned}
\alpha^6&= \alpha^3 \alpha^3
\\&=(3\alpha-1) (3\alpha-1)
\\&= 3\alpha(3\alpha-1)-1(3\alpha-1)
\\&= 9\alpha^2-3\alpha-3\alpha+1
\\&=9 \alpha^2 -6 \alpha+1
\end{aligned} $$
Now looking at $\alpha^4$
$$
\begin{aligned}
\alpha^4= \alpha^3 \alpha^1
&=(3\alpha-1) \alpha
&=3 \alpha^2-\alpha
\end{aligned}
$$
Let us go back
$$ \begin{aligned}
&\alpha^6-9\alpha^4+26 \alpha^2 -24
\\
&=( 9 \alpha^2-6 \alpha+1 )
-9(3\alpha^2-\alpha)
+26 \alpha^2 -24
\\ &=9\alpha^2-6\alpha+1-27\alpha^2+9 \alpha+26 \alpha^2 -24
\\&=18 \alpha^2 + 3\alpha -23
\\& = \vdots?
\\&=0
\end{aligned}$$
| $(\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$
Um... that's not right.
$(\alpha^2-2)^3-3(\alpha^2-2)+1=$
$\alpha^6 +3(-2)\alpha^4 + 3(-2)^2\alpha^2+ (-2)^3 +$
$-3\alpha^2 + 6 + $
$1 =$
$\alpha^6- 6\alpha^4 + 9\alpha^2 -1$
And
$\alpha^6- 6\alpha^4 + 9\alpha^2 -1=$
$\alpha^6 - 3\alpha^4 + \alpha^3 -3\alpha^4 -\alpha^3 +9\alpha^2 -1=$
$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha^4 -\alpha^3 +9\alpha^2 -1=$
$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha^4 - \alpha^3 + 9\alpha^2 -1=$
$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha^4 + 9\alpha^2 -\alpha - \alpha^3+ \alpha -1=$
$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha(\alpha^3 - 3\alpha + 1) - \alpha^3+ 3\alpha -1=$
$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha(\alpha^3 - 3\alpha + 1) - (\alpha^3- 3\alpha +1)=$
$(\alpha^3 - 3\alpha - 1)(\alpha^3- 3\alpha +1)$
$(\alpha^3- 3\alpha +1)*0 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Common tangents of two ellipses I have two ellipses: $x^2/25+y^2/9=1$ and $x^2/16+y^2/25=1$. I have to find the equations of common tangents.
I understand that $xx_0/25+yy_0/9=1$ could be the equation of common tangents, ($x_0,y_0$) being the point of intersection. We will have 4 points of intersection thus 4 tangent equations.
I tried to find the points of intersection by writing
$x^2/25+y^2/9=x^2/16+y^2/25$
$9|x|=16|y|$
$x=16/9y$ and $x=-16/9y$
By inserting these into one of the ellipses equations, I got
$y=45/sqrt(481)$ and $y=45-/sqrt(481)$
I inserted ($x,y$) instead of ($x_0,y_0$) but this doesn't seem to be the right answer. What am I doing wrong?
| There exists a line $y = mx + b$
$\frac {x^2}{25} + \frac {(mx+b)^2}{9} = 1\\
(9+ 25m^2) x^2 + 50mbx +25b^2- 225 = 0$
Since the line is tangent
$x = \frac {-50mb \pm \sqrt {(50mb)^2 - 4(9+25m^2)(25b^2 + 225)}}{2(9+25m^2)}$
Since the line is tangent (and not intersecting) the discriminant is $0$
$(50mb)^2 - 4(9+25m^2)(25b^2 - 225) = 0$
Applying the line to the other equation
$(32mb)^2 - 4(25+16m^2)(16b^2 - 400) = 0$
Which simplifies to
$b^2 - 25m^2 - 9 = 0\\
b^2 - 16m^2 + 25$
Respectively
$m^2 = \frac {16}{9}\\
b^2 = \frac {481}{9}$
$y = \pm \frac {4}{3} x \pm \frac {\sqrt {481}}{3} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got ,
| Also, you can use Stolz: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem
$$\lim_{n\rightarrow+\infty}\frac{1^4+2^4+...+n^4}{n^5}=\lim_{n\rightarrow+\infty}\frac{n^4}{n^5-(n-1)^5}=\frac{1}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all the local extrema for $f(x,y)=x^3y^2+27xy+27y$ I have that $f'_x=3x^2y^2+27y$ and $f'_y=2x^3y+27x+27$. Setting $f'_x=0$ I get that
$$x^2=\frac{-27y}{3y^2}=-\frac{9}{y}\Leftrightarrow x=\pm\frac{3}{\sqrt{y}}i$$
Setting $x=\frac{3}{\sqrt{y}}i$ in the equation $f'_y=0$ gives
$$0=2\left(\frac{3}{\sqrt{y}}i\right)^3y+27\left(\frac{3}{\sqrt{y}}i\right)+27=\frac{27}{\sqrt{y}}i+27=0,$$
Which gives me the equivalent equation
$$i+\sqrt{y}=0\Leftrightarrow \sqrt{y}=-i \Leftrightarrow y=(-i)^2=-1.$$
This is a false root however, so for $x=\frac{3}{\sqrt{y}}i$ no roots for $y$ exists. Using $x=-\frac{3}{\sqrt{y}}i$ I get instead $i-\sqrt{y}=0$ and this equation as the root $y=-1$. This means that $x=-3.$ So, one stationary point is $(x,y)=(-3,-1).$ However the book says there is another stationary point, namely $(-1,0).$ However I've found all values for $y$ and the only working value that makes sense is $y=-1$, how can I find $y=0$? It doesn't make sense for $y$ to be $0$ since that would mean division by zero when I want to get my x-value.
| Here a way to solve the system $\;\begin{cases}3x^2y^2+27y=0,\\[1ex]2x^3y+27x+27=0.\end{cases}$
The first equation is equivalent to $\;y(x^2y+9)=0$. So:
*
*either $y=0$, and the second equation yields $\;x=-1$;
*or $x^2y=-9$. Plugging this relation into the second equation, multiplied by $xy$ (note that $x$ has to be $\ne 0$):
$$2x^4y^2+27x^2y+27xy=0 \Rightarrow \underbrace{2\cdot 81}_{6\mkern1mu\cdot 27}-27\cdot 9+27xy=0\Leftrightarrow 27xy= 3\cdot 27\Leftrightarrow xy=3.$$
Going back to the first equation, we get $\;x^2y=3x=-9$, whence $\;x=-3 $, then $\;y=-1$.
Thus, there are two candidates: $(-1,0)$ and $(-3,-1)$.
Now calculate the hessian:
$$H_f=\begin{vmatrix}f''_{x^2}&f''_{xy}\\f''_{yx}&f''_{yx^2}\end{vmatrix}=
\begin{vmatrix}6xy^2& 6x^2y+27\\ 6x^2y+27 & 2x^3\end{vmatrix}=\begin{cases}
{\scriptstyle\begin{vmatrix}0&27\\27&-2\end{vmatrix}}=-729,\\[1ex]
{\scriptstyle\begin{vmatrix}-18&-27\\-27&-2\cdot 27\end{vmatrix}}=9\cdot27\,{\scriptstyle\begin{vmatrix}2&1\\3&2\end{vmatrix}}=243.
\end{cases}$$
As the hessian is negative in the first case, we have a saddle point.
In the second case, it is positive, so we have a local extremum. Furthermore, the trace of the hessian is negative, which implies this extremum is a local maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Origin Triangle Tetrahedron Volume I have a problem that goes like this:
Triangle $ ABC$, with sides of length $ 5$, $ 6$, and $ 7$, has one vertex on the positive $ x$-axis, one on the positive $ y$-axis, and one on the positive $ z$-axis. Let $ O$ be the origin. What is the volume of tetrahedron $ OABC$?
I really can't wrap my head around this problem. I thought I could find the area of the triangle base by using Heron's Formula, and then multiply that by $1/3$ of the height, but I've got no way of finding the height.
The posted solution for this question was very confusing.
It says:
wlog use $A(a,0,0), B(0,b,0), C(0,0,c)$
(i) wlog set $ 5 = \sqrt{a^2+b^2} \qquad 6 = \sqrt{b^2+c^2} \qquad 7 = \sqrt{c^2+a^2}$
(ii) Square for $ 25 = a^2+b^2 \qquad 36 = b^2+c^2 \qquad 49 = c^2+a^2$
(iii) Adding and dividing by 2 gives $ a^2+b^2+c^2 = 55$
(iv) Subtracting each element of (ii) from (iii) gives $ a^2 = 19 \qquad b^2 = 6 \qquad c^2 = 30$
(v) Multiplying gives $ a^2b^2c^2 = 19 \cdot 6 \cdot 30 = 19 \cdot 5 \cdot 6^2$
(vi) Squareroot and divide by 6 to get $ V = \frac{1}{6} abc = \sqrt{95} \implies \boxed{\text{C}}$
What I don't understand about this solution is why they are using these $a, b$, and $c$ coordinates and taking the square roots of two coordinates added together and setting them equal to the triangle's side lengths.
Perhaps I just don't properly understand how tetrahedrons and $x$-$y$-$z$ coordinates work. I'd really appreciate some help wrapping my head around this problem.
Thanks :)
| Maybe this will help you to visualize
$OA^2 + OB^2 = 5^2\\
OB^2 + OC^2 = 6^2\\
OA^2 + OC^2 = 7^2$
And even though it looks quadratic, it is really a system of linear equations.
If it feels strange to treat variables with squared terms as linear equations, rewrite it as:
$x + y = 25\\
y + z = 36\\
x + z = 49$
And once you have $x, y, z$ finding the volume is $\frac 16 \sqrt {xyz}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos2x}dx}}$
$\implies \frac{1}{2\sqrt2}\log |\sec 2x + \tan 2x| + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos^2x-\sin^2x}dx} + C$
The answer that I'm supposed to get is:
$\frac{x}{\sqrt2}+C$
Please help, thanks!
|
$$\int { \frac { \cos ^{ 4 } x+\sin ^{ 4 } x }{ \sqrt { 1+\cos4x } } dx } =\frac { 1 }{ \sqrt { 2 } } \int { \frac { { \left( \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } \right) }^{ 2 }-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \sqrt { \cos ^{ 2 }{ 2x } } } dx } =\\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1-\frac { \sin ^{ 2 }{ 2x } }{ 2 } }{ \cos { 2x } } } dx=\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { 2-\sin ^{ 2 }{ 2x } }{ \cos { 2x } } } dx=\\ =\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { \cos ^{ 2 }{ 2x } +1 }{ \cos { 2x } } } dx=\frac { 1 }{ 2\sqrt { 2 } } \left[ \int { \cos { 2xdx } } +\int { \frac { dx }{ \cos { 2x } } } \right] =\\ =\frac { 1 }{ 4\sqrt { 2 } } \int { d\left( \sin { 2x } \right) } +\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { \cos { 2xdx } }{ \cos ^{ 2 }{ 2x } } } =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 4\sqrt { 2 } } \int { \frac { d\left( \sin { 2x } \right) }{ 1-\sin ^{ 2 }{ 2x } } } =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \left[ \int { \frac { d\left( \sin { 2x } \right) }{ 1-\sin { 2x } } } +\int { \frac { d\left( \sin { 2x } \right) }{ 1+\sin { 2x } } } \right] =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \left[ \int { \frac { d\left( 1+\sin { 2x } \right) }{ 1+\sin { 2x } } } -\int { \frac { d\left( 1-\sin { 2x } \right) }{ 1-\sin { 2x } } } \right] =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \ln { \left| \frac { 1+\sin { 2x } }{ 1-\sin { 2x } } \right| + } C\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta)-2\sin^3(\theta) $$ simply making the substitutions $$\sin(\theta)=\frac{y}{r},~\cos(\theta)=\frac{x}{r}$$ noting also that I can square both sides of the above to be substitutions we then can write down $$r=\dfrac{2yx}{r}\cdot \dfrac{x^2}{r^2}+\dfrac{y}{r}-\dfrac{2y^3}{r^3}$$ then multiplying through by $r^3$ we obtain $$r^4=2yx^3+yr^2-2y^3$$ then replacing $r^2$ with $x^2+y^2$ we get $$(x^2+y^2)^2=2xy^2+yx^2+y^3-2y^3 \\(x^2+y^2)^2=y(3x^2-y^2)$$
However I am unsure of where I have made a mistake as the true answer is$$(x^2+y^2)^2=4x^2y-(x^2+y^2)y$$ working backwards I've so far gotten to the point of asking how I would rearrange$$r=\sin (3\theta) \Rightarrow r=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ so that $$\sin(3\theta)=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ which I am afraid I'll have to ask help for the next steps. Thanks.
| $$r=sin(3\theta )$$
$$ r=3sin(\theta)-4(sin^3(\theta))$$
$$ r^2=3rsin(\theta)-4rsin(\theta)sin^2(\theta)$$
$$x^2+y^2=3y-4y(y^2/(x^2+y^2))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the taylor polynomial of $\log(1+x^2)$, which is $x^2-\frac{1}{2}x^4+$... to see if I could bound the absolute value of the fraction above with something like $|\frac{x^2-x^2}{x^2\sin^2x}|$ and apply the comparison lemma, but it looks like that does not work too...
| \begin{align*}
\lim_{x\rightarrow 0}\dfrac{x^{2}-\log(1+x^{2})}{x^{2}\sin^{2}x}&=\lim_{x\rightarrow 0}\dfrac{x^{2}-x^{2}+\dfrac{1}{2}x^{4}-\dfrac{1}{3}x^{6}\cdots}{x^{2}\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}-\dfrac{1}{3}x^{4}\cdots}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\left(\dfrac{1}{2}\cdot\dfrac{x^{2}}{\sin^{2}x}\right)\cdot\lim_{x\rightarrow 0}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)\\
&=\dfrac{1}{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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P.M.F and expected value/expected payout The question is:
Let a random variable X be the number of days that a certain patient needs to be in the hospital. Suppose that X has the p.m.f.
$$\displaystyle f(x) = \frac{5 - x}{10}, \quad x = 1, 2, 3, 4 $$
If the patient is to receive 166 dollars from an insurance company for each of the first two days in the hospital and $118 for each day after the first two days, what is the expected payment for the hospitalization?
The given answer is 312.8
I got: 284.
How I got this:
$$E(X) = 166(1(\frac{5-1}{10}) + 2(\frac{5-2}{10})) + 118(3(\frac{5-3}{10}) + 4(\frac{5-4}{10})$$
$$E(X) = 166(0.4 + 0.6) + 118(0.6 + 0.4) $$
$$E(X) = 166 + 118 = 284 $$
What am I doing wrong?
| There are a number of minor mistakes.
Part of your computation looks like the computation for "expected number of days."
$$1 \cdot \frac{5-1}{10} + 2 \cdot \frac{5-2}{10} + 3 \cdot \frac{5-3}{10} + 4 \cdot \frac{5-4}{10}.$$
You also seem to be computing "expected payment on the last day in the hospital."
$$166 \cdot \frac{5-1}{10} + 166 \cdot \frac{5-2}{10} + 188 \cdot \frac{5-3}{10} + 118 \cdot \frac{5-4}{10}.$$
But neither of these is what the question is asking.
$$E[X] = \sum_{n=1}^4 (\text{total cost of $n$ days}) \cdot P(\text{$n$ days in hospital})$$
For example, the cost of $3$ days is $166+166+118$.
| {
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"url": "https://math.stackexchange.com/questions/2642011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$: $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$
(3) Prove that this is true for n = k + 1:
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
$$ = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$= ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
| It is:
$$2^{5n+1}+5^{n+2}=2\cdot 32^n+25\cdot 5^n=$$
$$2\cdot (27+5)^n+25\cdot 5^n=2\cdot (27^n+\cdots +5^n)+25\cdot 5^n=$$
$$27m+2\cdot 5^n +25\cdot 5^n=27(m+5^n).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f:[a,b]\to\mathbb{R}$ is a function of first class, does it mean that $f$ is continuous everywhere except countably many points in $[a,b]?$ Let $a<b.$
We say that $f:[a,b]\to\mathbb{R}$ is a function of first class if $f$ is a pointwise limit of some sequence $(f_n)_{n=1}^\infty$ of real-valued continuous functions on $[a,b],$ that is, for each $x\in [a,b],$ we have
$$\lim_{n\to\infty}f_n(x)=f(x).$$
It can be shown that if $f$ is continuous everywhere except countably many points in $[a,b],$ then $f$ is a function of first class.
One can use 'polygonal approximation' technique to construct sequence of continuous functions $(f_n)_{n=1}^\infty.$
I would like to know whether the converse holds. In particular,
Question: If $f:[a,b]\to\mathbb{R}$ is a function of first class, does it mean that $f$ is continuous everywhere except countably many points in $[a,b]?$
One can consider the Cantor Middle Third set on $[0,1]$ to prove the statement above does not hold.
More precisely, let $C_1 = [0,1]$ and defined $f_1 = 1$ on $C_1.$
Let $C_2=[0,\frac{1}{3}]\cup[\frac{2}{3},1]$ and defined a continuous function $f_2$ on $[0,1]$ such that $f_2 = 1$ on $[0,\frac{1}{3}] \cup [\frac{2}{3},1] = C_2, f=0$ on $[\frac{1}{3}+\frac{1}{9},\frac{2}{3}-\frac{1}{9}]$ and $f$ is linear on $[\frac{1}{3},\frac{1}{3}+\frac{1}{9}]\cup [\frac{2}{3}-\frac{1}{9},\frac{2}{3}].$
Let $C_3 = [0,\frac{1}{9}]\cup [\frac{2}{9},\frac{3}{9}] \cup [\frac{6}{9},\frac{7}{9}]\cup [\frac{8}{9},1]$ and defined a continuous funtion $f_3$ on $[0,1]$ such that $f_3=1$ on $C_3, f=0$ on $[\frac{1}{9} + \frac{1}{27}, \frac{2}{9}-\frac{1}{27}]\cup [\frac{1}{3}+\frac{1}{27}, \frac{2}{3}-\frac{1}{27}]\cup [\frac{7}{9}+\frac{1}{27},\frac{8}{9}-\frac{1}{27}]$ and $f$ is linear on $[\frac{1}{9}, \frac{1}{9}+\frac{1}{27}] \cup [\frac{2}{9}-\frac{1}{27}, \frac{2}{9}] \cup [\frac{1}{3}, \frac{1}{3}+\frac{1}{27}]\cup[\frac{2}{3}-\frac{1}{27}, \frac{2}{3}]\cup [\frac{7}{9},\frac{7}{9}+\frac{1}{27}]\cup [\frac{8}{9}-\frac{1}{27},\frac{8}{9}].$
By defining inductively, one obtains a sequence of continuous functions $(f_n)_{n=1}^\infty$ on $[0,1]$ that converges to the characteristics function on Cantor set $\chi_C$ but $\chi_C$ has uncountably many discontinuities, namely the Cantor set.
Is the above function correct?
| Not a direct answer to your question, but this might be of interest: Let $C$ be the Cantor set, let $d(x,C)$ be the distance from $x$ to $C,$ and set $f_n(x) = (1-d(x,C))^n.$ Then each $f_n$ is continuous on $[0,1],$ and $f_n \to \chi_C$ pointwise on $[0,1].$
| {
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
| This is equivalent to multiplying $111111 \times 111111$. There is a principle in logic called "universal generalization". Since no property of the $10$ in $10^k$ the base of $11111$ is being used, because there are no carries, it can be generalized to $x^k$.
| {
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Is $\arccos\left (\frac{\sqrt 6 +1 }{2\sqrt3}\right)= \arctan\left(\frac{\sqrt3 - \sqrt 2}{1+\sqrt 6} \right)$?
Is $\arccos\left(\dfrac{\sqrt 6 +1}{2\sqrt3}\right)= \arctan\left(\dfrac{\sqrt3 - \sqrt 2}{1+\sqrt 6}\right)?$
They are equal according to the calculator but how?
I made a triangle with base $\sqrt 6+ 1$ and hypotenuse $2\sqrt 3$.
Then the height comes out to be $\sqrt{5- 2\sqrt 6}$ using Pythagoras theorem.
But it should come out to be $\sqrt 3 - \sqrt 2$
| All you did is fine. Now, notice that$$\left(\sqrt3-\sqrt2\right)^2=3+2-2\sqrt6=5-2\sqrt6.$$
| {
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Deriving the closed form of $M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$ I have the sequence, let $M_0=1$
$$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$$
Which I would like first to study the convergence and fine the closed form.
I failed to show that $M_n$ is bounded and monotone.
This could be easy if have the explicit expression of it.
Question:
Is there a closed a form of this sequence? does anyone has an idea?
FYI In the book it is mentioned that this sequence is used to approximate the area of the unit circle. may be some else has a more clever explanation to this connection
| Answer thanks to @Masacroso comment.
$$M_n = 2^n \tan\left(\frac{π}{2^{n+2}}\right)\to \frac{π}{4}$$
and
$$\lim_{n\to\infty}M_n = \lim_{n\to\infty} \frac{π}{4} \frac{\tan\left(\frac{π}{2^{n+2}}\right)}{\frac{π}{2^{n+2}}}=\frac{π}{4}\lim_{x\to 0} \frac{\tan\left(x\right)}{x} =\frac{π}{4}$$
One can show that $M_n\ge 0,$ let $X_n:=\frac{M_n}{2^n}$ then $$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)\implies X_{n+1}=\frac{1}{X_n}(\sqrt{1+X_n^2}-1)$$
There exists $a_n\in[0, \frac{π}{2}]$ such that, $$X_n=\tan(a_n)$$
hence we have,
$$\begin{align}\tan(a_{n+1})= X_{n+1}&=\frac{1}{X_n}\left(\sqrt{1+X_n^2}-1\right) \\&=\frac{1}{\tan(a_n)}\left(\sqrt{1+\tan^2(a_n)}-1\right)\\&= \frac{1}{\tan(a_n)}\left(\frac{1}{\cos(a_n)}-1\right) \\&= \frac{1-\cos(a_n)}{\sin(a_n)} = \frac{1-\cos^2(\frac{a_n}{2})+\sin^2(\frac{a_n}{2})}{2\cos(\frac{a_n}{2})\sin(\frac{a_n}{2})} \ \\&= \frac{1-\cos(a_n)}{\sin(a_n)} = \frac{2\sin^2(\frac{a_n}{2})}{2\cos(\frac{a_n}{2})\sin(\frac{a_n}{2})} \\&=\color{blue}{\tan(\frac{a_n}{2})}\end{align}$$
Finally we have
$$a_{n+1}= \frac{a_n}{2}\implies a_n=\frac{a_0}{2^{n}} $$
But we have $a_0 = π/4$ therefore we end up, with
$$M_n = 2^n \tan\left(\frac{π}{2^{n+2}}\right)\to \frac{π}{4}$$
| {
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Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
| Using Lagrange multipliers, write
$$F(x,y,z) = \frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3y^2+1}-\lambda(x+y+z)$$ Taking derivatives, we get $$\frac{6x}{(3x^2+1)^2}=\frac{6y}{(3y^2+1)^2}=\frac{6z}{(3z^2+1)^2}=-\lambda$$
Now it's straightforward to show that $f(t) = \frac{6t}{(3t^2+1)^2}$ is one-to-one on $t>1,$ so $x=y=z=2.$
| {
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Show that a polynomial is irreducible using Galois I need to show that the polynomial $X^4-3X^2+4$ is irreducible in $\mathbb{Q}[X]$. As Eisenstein's criterion fails and I hardly try to avoid something such as “Let's assume $X^4-3X^2+4$ is reducible, then $(X^2+aX+b)(X^2+cX+d) = X^4-3X^2+4$”.
My idea is that its roots $\pm\left(\sqrt{\frac{1}{2}(3\pm i \sqrt 7)}\right)$ and squared $\frac{1}{2}(3\pm i \sqrt 7)$ are obviously $\notin \mathbb{Q}$ (because if it was reducible, the degree $[\mathbb{Q}(\sqrt{\frac{1}{2}(3 + i \sqrt 7}): \mathbb{Q}]=2$, and no polynomial of degree $2$ then has $\sqrt{\frac{1}{2}(3 + i \sqrt 7)}$ as a root.
My question is simply if this argument is valid.
| We see that our polynomial has no rational roots.
Now, since coefficients before $x^3$ and before $x$ they are $0$, we have two cases only:
$$x^4-3x^2+4=(x^2+px+2)(x^2-px+2)$$ or
$$x^4-3x^2+4=(x^2+px-2)(x^2-px-2)$$ for $p\in\mathbb Z$ and easy to check that they are impossible even for a rational $p$.
| {
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More Questions from Mathematical Analysis by Apostol I was solving the exercise questions of the book "Mathematical Analysis - 2nd Edition" by Tom Apostol and I came across the questions mentioned below. While I was able to solve a few questions, the others I did not even get any hint of!
1.
(a) By equating imaginary parts in DeMoivre's Formula prove that
$$\sin {n\theta} = \sin^n\theta \left\lbrace \binom{n}{1} \cot^{n - 1}\theta - \binom{n}{3} \cot^{n - 3}\theta + \binom{n}{5} \cot^{n - 5}\theta - + \cdots \right\rbrace$$
(b) If $0 < \theta < \dfrac{\pi}{2}$, prove that
$$\sin{\left( 2m + 1 \right)\theta} = \sin^{2m+1}\theta . P_m\left( \cot^2 \theta \right)$$
where $P_m$ is the polynomial of degree $m$ given by
$$P_m(x) = \binom{2m + 1}{1} x^m - \binom{2m + 1}{3} x^{m - 1} + \binom{2m + 1}{5} x^{m - 2} - + \cdots$$
Use this to show that $P_m$ has zeros at $m$ distinct points $x_k = \cot^2 \left( \dfrac{k\pi}{2m + 1} \right)$ for $k = 1, 2, \dots, m$.
(c) Show that the sum of zeros of $P_m$ is given by
$$\sum\limits_{k = 1}^{m} \cot^2 \dfrac{k\pi}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$
and that the sum of there squares is given by
$$\sum\limits_{k = 1}^{m} \cot^4 \dfrac{k\pi}{2m + 1} = \dfrac{m\left( 2m - 1 \right) \left( 4m^2 + 10m - 9 \right)}{45}$$
*Prove that $z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right)$ for all complex $z$. Use this to derive the formula
$$\prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$
As far as the solutions are concerned, I am through with the 1st part of 1st question and even half of the second part. But, in the second question, proving the zeros and their sum (and the sum of their squares) is getting really difficult. I am not getting any sort of hint as to how to prove it further.
And for the second question, I could do the first half part since it was essentially finding the $n$ roots of unity. But for the second part, I have nearly proved everything but what was asked. Many times I came to the conclusion that $$\prod\limits_{k = 1}^{n} \sin \dfrac{k\pi}{n} = 0$$ which is obvious because at $k = n$, we have a term of $\sin \pi$ which is equal to $0$. I am not getting how to remove that last term from the product using the result we just proved above!
Help will be appreciated!
| With the 2nd question, 2nd part, you are asked to
$$z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right) \Rightarrow \prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$
Note that when $k=n$
$$e^{\frac{2ki\pi}{n}}=e^{2i\pi}=1$$
also
$$z^n-1=(z-1)(z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1)$$
altogether
$$\color{red}{(z-1)}(z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1)=\color{red}{(z-1)}\prod\limits_{k = 1}^{n-1} \left( z - e^{\frac{2ki\pi}{n}} \right)$$
which is
$$z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1=\prod\limits_{k = 1}^{n-1} \left( z - e^{\frac{2ki\pi}{n}} \right)$$
and substituting $z=1$
$$n=\prod\limits_{k = 1}^{n-1} \left( 1 - e^{\frac{2ki\pi}{n}} \right)=
\prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \left( e^{-\frac{ki\pi}{n}} - e^{\frac{ki\pi}{n}} \right)=\\
(2i)^{n-1} (-1)^{n-1} \cdot \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \left(\frac{ e^{\frac{ki\pi}{n}} - e^{-\frac{ki\pi}{n}}}{2i} \right)=\\
2^{n-1} (-i)^{n-1} \cdot \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \sin{\left(\frac{k\pi}{n}\right)}=\\
2^{n-1} (-i)^{n-1} e^{\sum\limits_{k=1}^{n-1}\frac{ki\pi}{n}} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=\\
2^{n-1} (-i)^{n-1} e^{\frac{i\pi}{n}\frac{n(n-1)}{2}} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=\\
2^{n-1} (-i)^{n-1} i^{n-1} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=2^{n-1} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$ but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind or in third kind.
Can someone explain me how to write in elliptical form
| Elliptic integral of the first kind is
$$E_1(\varphi, k) = \int\limits_0^\varphi \dfrac{d\theta}{\sqrt{1-k^2\sin^2\theta}} = \int\limits_0^{\sin\varphi}\dfrac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt.$$
Elliptic integral of the second kind is
$$E_2(\varphi, k) = \int\limits_0^\varphi \sqrt{1-k^2\sin^2\theta}\,d\theta = \int\limits_0^{\sin\varphi}\sqrt{\dfrac{1-k^2t^2}{1-t^2}}\,dt.$$
At first,
$$1 + x^2 + x^4 = \left(1 + \dfrac12 x^2\right)^2 - \left(i\dfrac{\sqrt3}{2}\right)^2x^4$$
$$ = \left(1 + \left(\cos\dfrac{\pi}3 - i\sin\dfrac{\pi}3\right)x^2\right)\left(1 + \left(\cos\dfrac{\pi}3 + i\sin\dfrac{\pi}3\right)x^2\right)$$
$$ = \left(1 + e{\large{^{-\frac{\pi}3i}}}x^2\right)\left(1 + e{\large{^{\frac{2\pi}3i}}}e{\large{^{-\frac{\pi}3i}}}x^2\right)
= \left(1 - e{\large{^{\frac{2\pi}3i}}}x^2\right)\left(1 - e{\large{^{\frac{2\pi}3i}}}e{\large{^{\frac{2\pi}3i}}}x^2\right)$$
$$ = \left(1 - \bigl(e{\large{^{\frac{\pi}3i}}}x\bigr)^2\right)\left(1 - \bigl(e{\large{^{\frac{\pi}3i}}}\bigr)^2\bigl(e{\large{^{\frac{\pi}3i}}}x\bigr)^2\right).$$
Let
$$k = e{\large{^{\frac{\pi}3i}}},\quad t = kx,$$
then
$$I = \int\dfrac{x^2 - 1}{\sqrt{x^4 + x^2 + 1}}dx = \frac1{k^3}\int\dfrac{t^2 - k^2}{\sqrt{(1-t^2)(1-k^2t^2)}}dt.$$
The ratio can be presented in the form of
$$\dfrac{t^2 - k^2}{\sqrt{(1-t^2)(1-k^2t^2)}} = \dfrac{A}{\sqrt{(1-t^2)(1-k^2t^2)}} + B\,\sqrt{\dfrac{1-k^2t^2}{1-t^2}},$$
where
$$t^2 - k^2 = A + B(1 - k^2t^2),\quad A + B = -k^2,\quad -k^2B = 1,$$
$$B = -\frac1{k^2},\quad A = \frac{1 - k^4}{k^2}.$$
Since $k^6 = 1,$ then
$$I = \left(k - \dfrac1k\right)\int\dfrac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}} - k\int\,\sqrt{\dfrac{1-k^2t^2}{1-t^2}}\,dt = {\left(k - \dfrac1k\right)E_1(\arcsin t, k) - kE_2(\arcsin t, k) + const},$$
$$\int\dfrac{x^2 - 1}{\sqrt{x^4 + x^2 + 1}}dx = \left(k - \dfrac1k\right)E_1\left(\arcsin \left(\frac{x}{k}\right), k\right) - kE_2\left(\arcsin\left(\frac{x}{k}\right), k\right) + const,\quad \text{ where } k = e^{\large{\frac\pi3i}}.$$
Differentiation shows the result is correct.
| {
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Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$
Why is the denominator under $C$ not simply $x+1$? It is $x$ times $(x+1)^2$ and not $(x+1)^3$
| Notice that the RHS after simplification must have an identical denominator as with the LHS.
The LHS denominator has a cubic term, hence the RHS must also be cubic.
So the choice for the term under $C$ has to be $(x+1)^2$, and not the ones you have suggested.
| {
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Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number of dices that are sixes after n rolls.
$$\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \end{pmatrix}$$
I want to figure out how many turns it takes for me on average to get all sixes.
I'm not even sure where to start with this problem. Is it a right approach to write a program where I calculate $P^n$ and see when the 6th column all equals to 1?
Any pointers would be greatly appreciated.
| If $E_n$ is the expected number required when $n$ out of $5$ are already a six then clearly $E_5=0$ and you can write (dropping your final column)
$$\begin{pmatrix}E_0 \\ E_1 \\ E_2 \\ E_3 \\ E_4 \end{pmatrix} =\begin{pmatrix}1 \\ 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + \begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} \\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} \\ 0& 0 & \frac{125}{6^3} &\frac{75}{6^3}& \frac{15}{6^3} \\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2} \\ 0 & 0 & 0 & 0 & \frac{5}{6} \end{pmatrix} \begin{pmatrix}E_0 \\ E_1 \\ E_2 \\ E_3 \\ E_4 \end{pmatrix}$$
which you can either solve by hand or, if $M$ is that square matrix, find the row sums of $(I-M)^{-1}$
I think it should give you $E_5 \approx 13.023662$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{n \rightarrow \infty}\int_0^1f_n(x)$ For $n=1, 2,...,$ let $f_n(x)=\frac{2nx^{n-1}}{x+1}, x\in [0, 1]$. Then $\lim_{n \rightarrow \infty}\int_0^1f_n(x)$
Function is unbounded at $1$, How do I solve?
| \begin{align*}
\int_{0}^{1}f_{n}(x)dx&=\int_{0}^{1}\dfrac{2nx^{n-1}}{x+1}dx\\
&=\dfrac{2}{x+1}\cdot x^{n}\bigg|_{x=0}^{x=1}-\int_{0}^{1}-\dfrac{2}{(x+1)^{2}}\cdot x^{n}dx\\
&=1+2\int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx,
\end{align*}
now
\begin{align*}
\dfrac{x^{n}}{(x+1)^{2}}\leq\dfrac{1}{(x+1)^{2}},~~~~x\in[0,1],
\end{align*}
and
\begin{align*}
\int_{0}^{1}\dfrac{1}{(x+1)^{2}}=-\dfrac{1}{x+1}\bigg|_{x=0}^{x=1}=\dfrac{1}{2}<\infty,
\end{align*}
and
\begin{align*}
\dfrac{x^{n}}{(x+1)^{2}}\rightarrow 0,~~~~x\in[0,1),
\end{align*}
so by Lebesgue Dominated Convergence Theorem,
\begin{align*}
\int_{0}^{1}f_{n}(x)dx\rightarrow 1.
\end{align*}
Another way:
\begin{align*}
\int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx\leq\int_{0}^{1}x^{n}dx=\dfrac{1}{n+1}\rightarrow 0.
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$
Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$.
Here is my solution I am looking to see if it is correct or if there is another way to do it, thanks!
The only way to form an $x^{25}$ term is to gather two $x^8$ and three $x^3$ . Since there are ${{10}\choose{2}} =45$ ways to choose two $x^8$ from the $10$ multiplicands and $8$ ways to choose three ${{8}\choose{3}}= 56$ ways to choose $x^3$ from the remaining $8$multiplicands, the answer is using the product rule $45×56 = 2520$.
| Expressions like above can be evaluated by the multinomial Theorem:
$(a+b+c)^n = \sum_{k,l,m;k+l+m=n} \frac{n!}{k!l!m!}a^kb^lc^m$.
This sum means you sum over all possible $k,l,m$ with the condition that $k+l+m=n$.
Now you set $n=10,a=1,b=x^3,c=x^8$ and you obtain now that the $x^{25}$ Terms are obtained when $3l+8m=25$ for integer numbers $l,m$. This equation is satisfied only for $m=2,l=3$. Thus you obtain $k = 10-2-3=5$. It is $a^kb^lc^m = x^{25}$ and the resulting binomial coefficient is $\frac{10!}{5!2!3!} = 2520$. With above formula you obtain that the factor is $2520$ and that is correct!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
Could some help me to solve it, Thanks in Advanced
| Durgesh: to complement Jack D'Aurizio fantastic answer above.
If you use Euler's identity
$$e^{ix} = cos x + i\cdot sin x$$
and then apply for $-x$
$$e^{-ix} = cos (-x) + i\cdot sin(-x) = cos x - i\cdot sinx$$
and sum both equations, you get the well known
$$cos x={e^{ix}+e^{-ix}\over2}$$
so
$$\frac12+cos x={1+e^{ix}+e^{-ix}\over2}$$
that is
$$\frac12+cos x={1+e^{ix}+e^{2ix}\over{2e^{ix}}}={1\over{2e^{ix}}}{{e^{3ix}-1}\over{e^{ix}-1}}$$
Substituting for $3x$, $9x$ and $27x$ and multiplying all 4 equations, you get Jack's magical leap :-)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to simplify the trigonometric term?
Simplify $$\frac{2\tan\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)-1\right)^2} + \frac{\tan^2\left(\frac{x}{2}\right)+1}{2\tan\left(\frac{x}{2}\right)}$$
It should be like: $$\frac{2\sin(x) - 2\sin^2(x) - 2}{2(\sin(x)-1)\sin(x)}$$
I can use any trigonometric identities, for example: $$\tan\left(\frac{x}{2}\right)=\frac{1-\cos(x)}{\sin(x)} $$
| Let $y=\dfrac x2$.
Then using the identities $$\sec^2y=1+\tan^2y\tag{1}$$ $$\sin 2y=2\sin y\cos y\tag{2}$$ the expression becomes $$\begin{align}\frac{2\tan y}{(\tan y-1)^2}+\frac{\tan^2y+1}{2\tan y}&=\frac{2\tan y}{\tan^2y+1-2\tan y}+\frac{\sec^2y}{2\tan y}\tag{1}\\&=\frac{\frac{2\sin y}{\cos y}}{\frac1{\cos^2y}-\frac{2\sin y}{\cos y}}+\frac{\frac1{\cos^2y}}{\frac{2\sin y}{\cos y}}\tag{1}\\&=\frac{2\sin y\cos y}{1-2\sin y\cos y}+\frac1{2\sin y\cos y}\\&=\frac{\sin2y}{1-\sin2y}+\frac1{\sin2y}\tag{2}\\&=\frac{\sin^22y-\sin2y+1}{\sin2y(1-\sin2y)}\\&=\frac{\sin^2x-\sin x+1}{\sin x(1-\sin x)}\\&=\boxed{\frac{2\sin x - 2\sin^2 x - 2}{2(\sin x-1)\sin x}}\end{align}$$ as you have stated.
| {
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"url": "https://math.stackexchange.com/questions/2660667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n\right)}+\left(1-\frac{5}{n^4}\right)^4\right)=\lim _{n\to \infty }\left(\left(\left(1-\frac{5}{n^4}\right)^{n}\right)^{4*2018}+\left(1-\frac{5}{n^4}\right)^4\right)$$
so we get :
*
*$\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^4=1$
let:
$b_n=\frac{-5}{n^3}$
$\lim _{n\to \infty }\frac{-5}{n^3}=0$
*
*$\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^n\right)^{4\cdot 2018}=\lim \:_{n\to \:\infty \:}\left(\left(1+\frac{b_n}{n}\right)^n\right)^{4\cdot \:2018}=\left(e^0\right)^{4\cdot 2018}=1$
so the answer is :
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=1+1=2$$
Is this answer correct if not how can I find the limit ?
thanks.
| Let $m=2018n+1$ then $\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{m^4}$
=$\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{{m^4}\frac{(m-1)^4}{(m-1)^4}}$
=$\lim _{m\to \infty }\big(1-5\frac{2018^4}{(m-1)^4}\big)^{{(m-1)^4}{\frac{1}{(1-\frac{1}{m})^4}}}$
Known that $\lim _{m\to \infty }\big(1-\frac{x}{m}\big)^{m}=e^{-x}$
The limit is equal to $e^{-5*2018^4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4+y^4$$. I am stuck here. Is it wrong approach? is there an easier one?
| Firstly, $z=x+yi$, where $x$ and $y$ are reals.
Thus, it should be $$|x^2-y^2|=\frac{1}{2}$$ and $$x^2+y^2=1.$$
Finally, we obtain $$\left\{\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,\pm\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
& =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
& =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\
\end{align}
Therefor, I thought that:
$$\lim_{n \to \infty}\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}}$$
$$ = \lim_{n \to \infty}\frac{1}{\left(n^2-n^2\right)\sqrt{1}} = \infty$$
I also tried a different way where I got to the result of $-\dfrac{1}{2}$.
I am not going to show that method here but it starts with using the 3rd binomial formula. Then, having the squareroot at the top of the fraction, I factorized $n^2$ and it all worked.
Why does this method like shown above not work?
I am very happy for any help.
P.S. This is not the only example where this kind of getting to a solution does not work for me. Are there cases where I am not allowed to factorize something?
| write it as $$\frac{n^2+\sqrt{n^4+4n^2+n}}{n^4-n^4-4n^2-n}$$ and this is $$\frac{n^2(1+\sqrt{1+1/n^2+1/n^3})}{n^2(-4-1/n^2)}$$
| {
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"source": "stackexchange",
"question_score": "1",
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What is the value of $\sin^2 (\frac{\pi}{10}) \sin^2 (\frac{3\pi}{10})$?
PROBLEM
$$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$
My Try
$$
\left(1 + \cos \frac{\pi}{10}\right)
\left(1 + \cos \frac{9\pi}{10}\right)
\left(1 + \cos \frac{7\pi}{10}\right)
\left(1 + \cos \frac{3\pi}{10}\right)
= \sin^2 \left(\frac{\pi}{10}\right)
\sin^2 \left(\frac{3\pi}{10}\right)
$$
I am not able to proceed further.
Please help me.
| The roots of $\Phi_{10}(x)=1-x+x^2-x^3+x^4$ are the primitive tenth roots of unity, $\xi,\xi^3,\xi^7,\xi^9$ with $\xi=\exp\left(\frac{2\pi i}{10}\right)$. The roots of $\Phi_{20}(x)=\Phi_{10}(x^2)=1-x^2+x^4-x^6+x^8$ are the primitive $20$-th roots of unity, $\zeta,\zeta^3,\zeta^7,\zeta^9,\zeta^{11},\zeta^{13},\zeta^{17},\zeta^{19}$ with $\zeta=\exp\left(\frac{\pi i}{10}\right)$. By dividing $\Phi_{20}(x)$ by $x^4$ and by writing what we get as a polynomial in $\left(x+\frac{1}{x}\right)$ we get that
$$ x^4-\frac{5}{4}x^2+\frac{5}{16}=\prod_{k\in\{1,3,7,9\}}\left(x-\cos\frac{\pi k}{10}\right)$$
and by evaluating both sides at $x=-1$ we get:
$$ \prod_{k\in\{1,3,7,9\}}\left(1+\cos\frac{\pi k}{10}\right) = 1-\frac{5}{4}+\frac{5}{16} = \frac{1}{16}.$$
| {
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Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)\\
&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2}
\end{align}
| $$\lim_{x\rightarrow0}\frac{3^x+2^x-2}{4^x+2^x-2}=\lim_{x\rightarrow0}\frac{\frac{3^x-1}{x}+\frac{2^x-1}{x}}{\frac{4^x-1}{x}+\frac{2^x-1}{x}}=\frac{\ln3+\ln2}{\ln4+\ln2}=\log_86$$
I used $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\lim_{x\rightarrow0}\left(\frac{e^{x\ln{a}}-1}{x\ln{a}}\cdot\ln{a}\right)=\ln{a}$$ for all $a>0$.
| {
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"source": "stackexchange",
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decomposing and the summing a sum.
Prove:
$$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) \quad \text{ where } \{x\} = x-\lfloor x \rfloor $$
We have that $$\sum_{k\leq n} k ( \frac {n}{k} - \lfloor \frac {n}{k} \rfloor ) = \sum_{k\leq n} n -\sum_{k\leq n} k \lfloor \frac {n}{k} \rfloor $$
EDIT:
$$= n^2 - \sum_{k \leq n} \sigma (k)= n^2 - (\frac{\pi^2}{12}n^2 + O(n \log n)) $$
Hence $$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) $$
| We have that $$\sum_{k\leq n} k ( \frac {n}{k} - \lfloor \frac {n}{k} \rfloor ) = \sum_{k\leq n} n -\sum_{k\leq n} k \lfloor \frac {n}{k} \rfloor $$
$$= n^2 - \sum_{k \leq n} \sigma (k) $$
Since $ \sum_{k \leq n} \sigma (k)= (\frac{\pi^2}{12}n^2 + O(n \log n) )$
Hence we have that $$= n^2 - \sum_{k \leq n} \sigma (k)= n^2 - (\frac{\pi^2}{12}n^2 + O(n \log n)) $$
Hence $$\sum_{k\leq n} k \left\{ \frac {n}{k} \right\} = n^2\left(1 - \frac {\pi^2}{12}\right) + O (n \log n) $$
As desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\mathbf{s_n}$ term appearing in the denominator.
What I have so far:
Calculating a few terms: $s_1 = 1$, $s_2 = 2$, $s_3 = \frac{7}{4} < s_2$.
For this to be monotonic decreasing, we have to show $s_{k+2} < s_{k+1}$.
For the base case, we assume for $k$ we have $s_{k+1} < s_k.$
The inductive step: to show $s_{k+2} < s_{k+1}$, i.e. to show $$\color{grey}{s_{k+2} = }\frac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \frac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right) \color{grey}{= {s_{k+1}}_.}$$
In order to show $\dfrac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \dfrac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right)$, I am stuck:
Given, $s_{k+1} < s_k$, I cannot then say $s_{k+1} + \mathbf{\frac{3}{s_{k+1}}} < s_k + \mathbf{\frac{3}{s_{k}}}$, because the $s_k, s_{k+1}$ terms which appears in the denominator may reverse the inequality. Any pointers on how to proceed further?
Disclaimer: I am revising real analysis on my own from the Kenneth Ross book, this is not strictly homework.
| Assume $s>0$ and notice that for $s\ne\sqrt3$
$$\frac12\left(s+\frac3s\right)>\sqrt3$$ as can be established by computing the minium.
So for $n>1$, we have
$$s_n>\sqrt3\implies s_n^2>3\implies s_n>\frac3{s_n}\implies s_n>\frac12\left(s_n+\frac3{s_n}\right)=s_{n+1}.$$
At the same time this establishes that the sequence is decreasing and bounded below by $\sqrt3$, thus is convergent.
| {
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"url": "https://math.stackexchange.com/questions/2668041",
"timestamp": "2023-03-29T00:00:00",
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Prove that $ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+2 \cos2m\theta$ I am struggling with a part of my textbook can anyone help me? It states:
$$ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+ 2\cos2m\theta$$
Why is this obvious? Do I have to perform a Taylor series?
Any help appreciated
| You can use the standard formula:
$$1+\cos x+\cos 2x+\dots +\cos mx=\frac{\sin\dfrac{(m+1)x}2}{\sin\dfrac{x}2}\,\cos\dfrac{mx}2.$$
Here, with $x=2\theta$, one obtains
\begin{align}
1 +{} & 2 \cos2\theta + \dots+2 \cos2m\theta =2(1 + \cos2\theta + \dots+\cos2m\theta)-1 \cr
={} &\frac{2\sin(m+1)\theta}{\sin\theta}\,\cos m\theta -1 = \frac{\sin\bigl((m+1)+m\bigr)\theta+\sin\bigl((m+1)-m\bigr)\theta}{\sin\theta}-1 \cr
={}&\frac{\sin(2m+1)\theta+\sin\theta}{\sin\theta}-1 =\frac{\sin(2m+1)\theta}{\sin\theta}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is a proof using modular arithmetic in a question like this valid? It's been two years or so since I've finished my math undergrad (and I'm doing something non-math related now, unfortunately), so I apologize if what is to follow isn't a very good question!
Prove that for all Integers $n$, $n(n + 1)(2n + 1)$ will always be divisible by 6.
I can do that using induction, but I wanted to try a different way. Does it work to use modular arithmetic in the following way?
Let $f(n) = n(n+1)(2n+1) = 2n^3 + 3n^2 + n$. All we need to show is that $f(n)$ is divisible by both $2$ and $3$ for any choice of $n$.
Evaluate mod $2$.
$f(n) = n(n+1)(2n+1) = n(n+1)(0 + 1)$ mod $2 = n(n+1)$ mod $2$. Two consecutive numbers; one of them must be even, and so $f(n)$ is divisible by $2$.
Evaluate mod $3$
There are three possible residues for n modulo $3$: $0, 1,$ or $2$.
If the residue is $0$, then $f(n)$ is divisible by $3$.
If the residue is $1$, then $f(n) = n(n+1)(2n+1) = 1(1+1)(2*1+1) = 1(2)(3) = 0$ mod $3$.
If the residue is $2$, then $f(n) = 2(2+1)(2*2+1) = 2(3)(2) = 0$ mod $3$.
In any case, $f(n)$ is divisible by $3$.
Since $f(n)$ is divisible by $2$ and by $3$, it is divisible by $6$. The result follows.
Thank you!
| It's correct, but you can go further.
Fermat’s little theorem
If $p$ is a prime number, then, for every integer $n$,
$$n^p\equiv n \pmod{p}$$
There are several proofs. A simple one considers the fact that, if $n$ is coprime with $p$, then $n,2n,3n,\dots,(p-1)n$ are all distinct modulo $p$ and not congruent to $0$, so
$$
1\cdot2\cdot\dots\cdot(p-1)\equiv n\cdot2n\cdot\dots\cdot(p-1)n\pmod{p}
$$
which implies $n^{p-1}\equiv1\pmod{p}$. Therefore $n^p\equiv n$, which also holds when $n\equiv0\pmod{p}$.
Modulo $2$
$n(n+1)(2n+1)\equiv n(n+1)\equiv n^2+n\equiv n+n\equiv0\pmod{2}$ because $n^2\equiv n\pmod{2}$ (little Fermat).
Modulo $3$
$n(n+1)(2n+1)\equiv2n^3+3n^2+n\equiv2n^3+n\equiv2n+n\equiv0\pmod{3}$ because $n^3\equiv n\pmod{3}$ (little Fermat).
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sin x$ and $\cos x$ - linearly independent There is a lot of posts about that subject, but I'm not sure.
Task:
Check if
$$f(x)=\sin(x) $$
and
$$g(x)=\cos(x) $$
are linearly independent in the space of functions.
That's what I have done:
$a \sin(x) + b \cos(x) = 0$,
so $a=0$ and $b=0$.
for $x=0 $,
$a \sin(0) + b \cos(0) = 0$, then
$a\cdot 0 + b\cdot 1 = 0$.
for $x=\frac{\pi}{2}$
$a \sin(\frac{\pi}{2}) + b \cos(\frac{\pi}{2}) = 0$, then
$b \cdot 1 + a \cdot 0 = 0$
what am I supposed to write then? :)
| The Wronskian of $\sin$ and $\cos$ is not zero:
$$W(\sin, \cos)(x) = \begin{vmatrix} \sin x & \cos x \\ \cos x & -\sin x\end{vmatrix} = -\sin^2 x - \cos^2 x = -1 \ne 0$$
Therefore $\{\sin, \cos\}$ is linearly independent.
| {
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What is the mistake in this? What is wrong in this proof?
In $\triangle ABC$ with right angle at $B$, $BD$ is drawn as an altitude on $AC$. Let $AB=a$ and $BC=b$. So by similarity, $AD=\frac{a^2}{\sqrt{a^2+b^2}}$ and similarly $CD=\frac{b^2}{\sqrt{a^2+b^2}}$. Now, $\triangle ABD,BDC$ are also right angled at $D$. So applying the Pythagorean theorem,
$$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{a^2}{\sqrt{a^2+b^2}}$$
Solve these two equations as:
$$a^2-b^2=a^2-\frac{b^2}{\sqrt{a^2+b^2}}$$
So $\sqrt{a^2+b^2}=1$. That means for all values of $a$ and $b$, $a^2+b^2=1$.
But this is absurd.
| $$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{a^2}{\sqrt{a^2+b^2}}$$ is obviously wrong. Are you assuming a=b?
Also when using Pythagorean Theorem you forgot to square AD and CD.
| {
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
| So you have done some good work. Now the function is always positive for $x \not \in (0,1)$. Now lets analyse in this interval.
In $(0,1)$, $x^2+1$ lies in $(1,2)$ and $x(x-1)$ has a minimum of only $-1/4$ at $x=1/2$. Thats enough to conclude that $x(x-1)(x^2+1) > -1/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
| $$e^{ix} = i\sin{x}+\cos{x}$$
$$e^{4ix} = (i\sin{x}+\cos{x})^4$$
$$i\sin{4x}+\cos{4x} = \sin^4{x}-4i\sin^3x\cos{x}-6\sin^2x\cos^2{x}+4i\sin x\cos^3{x} + \cos^4x$$
As all coefficients are real, we can say the imaginary parts of each side are equal:
$$i\sin{4x} = -4i\sin^3x\cos{x}+4i\sin x\cos^3{x}$$
$$\sin{4x} = 4\sin x\cos^3{x}-4\sin^3x\cos{x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial Doubt Question :
$x^4 + px^3 + qx^2 + px + 1 =0 $ has real roots. Then what is the minimum value of $ p^2 +q^2 $ .
How I started ?
I started by dividing the whole equation by $x^2$ then we get
$ (x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0 $
Then put $(x + \frac{1}{x} ) = t$. Then discriminant should be greater than equal to zero. But now the problem arises that $t$ does not belong to $(-2,2)$ , so taking care of that part leads to solving inequality which I am unable to do .
Have I started the right way?
One more thing to notice is that the sum of roots of the equation is equal to the sum of reciprocal of the roots .
How to proceed further ?
| I don't know how much this is going to help, but if you use this you can find the roots
\begin{eqnarray}
x_1&=& -\frac{1}{4} \sqrt{p^2-4 q+8}-\frac{1}{2}
\sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4
q+8}}-q-2}-\frac{p}{4},\\
x_2&=& -\frac{1}{4} \sqrt{p^2-4
q+8}+\frac{1}{2} \sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2
\sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4},\\
x_3&=& \frac{1}{4}
\sqrt{p^2-4 q+8}-\frac{1}{2} \sqrt{\frac{p^2}{2}+\frac{-p^3+4 p q-8
p}{2 \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4}, \\
x_4&=&
\frac{1}{4} \sqrt{p^2-4 q+8}+\frac{1}{2}
\sqrt{\frac{p^2}{2}+\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4
q+8}}-q-2}-\frac{p}{4}
\end{eqnarray}
And from here the constraint
$$
p^2 - 4q+8 > 0 \tag{1}
$$
| {
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Solving three alike problems $a_{n+1} = \frac{1+a_n^2}{1+a_n}$ I have two problems alike:
*
*If $a_1=a$, $a>0$, $a_{n+1} = \dfrac{1+a_n^2}{1+a_n}$, study its convergence.
*If $a_0 >0$, $a_1>0$ and $a_{n+2} = \dfrac{2+a_{n+1}}{2+a_n}$, study this one's convergence.
I can solve this using two different situations: when $a$ is between $0$ and $1$ and when $a$ is between $1$ and $+\infty$ for the first problem and then say it is monotonic and it has an lower bond, therefore is convergent. But for me this solution does not help for the second, similar problem. Could you either show me how to solve the second one, not necessarily using my way, or show me a method that works for both? Thank you.
| Hint with the 2nd one, it is easy to show by induction that $a_n>0$, then
$$\color{red}{0<a_{n+2}}=\frac{2+a_{n+1}}{2+a_{n}}<
\frac{2+a_{n+1}}{2}=
1+\frac{a_{n+1}}{2}<
1+\frac{1+\frac{a_{n}}{2}}{2}=
1+\frac{1}{2}+\frac{a_n}{2^2}<\\
1+\frac{1}{2}+\frac{1}{2^2}+\frac{a_{n-1}}{2^3}<...<\\
1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n}}+\frac{a_{1}}{2^{n+1}}\color{red}{<
2+\frac{a_{1}}{2^{n+1}}}$$
so $(a_n)$ is bounded. More to come later ...
| {
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Powers for the sum of three squares Say $n$ can be represented as a sum of three non-zero squares. (i.e. $n = a^2 +b^2+ c^2$, for some $n,a,b,c \in \mathbb{N}$)
Is it possible that every natural power of $n$ is also a sum of three non-zero
squares? (i.e. $n^k = x^2+y^2+z^2$ for $x,y,z,k \in \mathbb{N}$)
| Yes, more generally, if,
$$N = x_1^2+x_2^2+\dots+x_m^2$$
then one can always find integer $y_i$ such that,
$$N^k = y_1^2+y_2^2+\dots+y_m^2$$
Proof: Use the expansion of
$$(a+i\sqrt{w})^k$$
To get,
$$a^2 + w = N,\\ (a^2 - w)^2 + (2 a)^2 w = (a^2 +
w)^2,\\ (a^3 - 3 a w)^2 + (3 a^2 - w)^2 w = (a^2 +
w)^3,\\ (a^4 - 6 a^2 w + w^2)^2 + (4 a^3 - 4 a w)^2 w = (a^2 + w)^4$$
and so on, then assume $w$ to be a sum of squares $w=b^2+c^2+\dots$
For example, for $k=3$,
$$\small\big(a^3 - 3 a (b^2+c^2+\dots)\big)^2 + \color{blue}{\big(3 a^2 - (b^2+c^2+\dots)\big)^2 \big(b^2+c^2+\dots\big)} = (a^2 + b^2+c^2+\dots)^3$$
then distribute the blue term. I trust the pattern for other $k$ is easy to discern.
| {
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$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction) I'm having some difficulty proving by induction the following statement.
$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) , but I am getting stuck on the induction step.
As far as I know I have to show $$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
implies
$$\sum_{i=0}^{n+1} \frac{1}{(i+3)(i+4)} = \frac{n+1}{4(n+5)}$$
To do this I think I should add the number $\frac{1}{(n+4)(n+5)}$ to $\frac{n}{4(n+4)}$ and see if it gives $\frac{n+1}{4(n+5)}$ , if I am not mistaken.
When trying to do that however I get stuck. I have:
$$\frac{n}{4(n+4)} +\frac{1}{(n+4)(n+5)} = \frac{n(n+4)(n+5)}{4(n+4)^2(n+5)} + \frac{4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+4)(n+5)+4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+5)+4}{4(n+4)(n+5)}$$
However beyond this point I don't know how to reach $\frac{n+1}{4(n+5)}$ I always just end up at the starting point of that calculation.
So I think that either my approach must be wrong or I am missing some trick how to simplify $$\frac{(n(n+5)+4}{4(n+4)(n+5)}$$
I would be very grateful for any help, as this is a task on a preparation sheet for the next exam and I don't know anyone, that has a correct solution.
| $$\frac{n(n+5)+4}{4(n+4)(n+5)}=\frac{n^2+5n+4}{4(n+4)(n+5)}=\frac{(n+4)(n+1)}{4(n+4)(n+5)}=\frac{n+1}{4(n+5)}$$
| {
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Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2n-1)$
| $\displaystyle \sum_{k=1}^nk^2(2k-1)=2\sum_{k=1}^nk^3-\sum_{k=1}^nk^2=2\times\frac{1}{4}n^2(n+1)^2-\frac{1}{6}n(n+1)(2n+1)$
| {
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elimination part of a mathematical question During solving a math question I got stuck in the elimination part and I couldn't proceed! Can someone help me, please?
$$2(x+y)-3^\frac 1{2}(x-y)=0$$
$$4(x+y)+2(3)^{\frac 1{2}}(x-y)+9z=1$$
$$x+y+z=0$$
I want to get the values for $x$, $y$, and $z$.
| I would like to change the variables here.
let:
$u = x+y\\
v = \frac {\sqrt 3}{2} (x-y)$
then
$2u-2v = 0\\
4u + 4v + 9z = 1\\
u+z = 0$
$u=-1, v = -1, z = 1$
Now back to $x,y$
$x+y = u\\
x-y = \frac {2}{\sqrt 3}v$
$x+y = -1\\
x-y = -\frac {2}{\sqrt 3}$
$2x = -1-\frac {2}{\sqrt 3}\\
2y = -1 + \frac {2}{\sqrt 3}\\
x = -\frac {1}{2} - \frac {\sqrt 3}{3}\\
y = -\frac {1}{2} + \frac {\sqrt 3}{3}\\
z = 1$
| {
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Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| Since$$x^4+5x^2+3=(x^2+1)^2+3(x^2+1)-1,$$ $P(x)=x^2+3x-1$ and therefore $P(x^2-1)=\cdots$
| {
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Show that $\sum_{n \leq x} \frac{d(n)}{\log n} = x + 2E \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right).$
Show that $$\sum_{n \leq x} \frac{d(n)}{\log n} = x + 2E \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right),$$
where $d(n)$ is the number of divisors of $n$ and $E$ is Euler-Mascheroni constant.
I first showed a simpler statement since I had trouble with the denominator: $$\sum_{n \leq x} \frac{d (n)}{n} = \frac{1}{2} \log ^2 x + 2 E \log x + O(1).$$
\begin{align*}\sum_{n \leq x} \frac{d(n)}{n} & =
\sum_{d \leq x} \frac{1}{d} \sum_{q \leq x/d} \frac{1}{q} \\ & =
\sum_{d \leq x} \frac{1}{d} \left( \log \left (\frac{x}{d} \right) + E + O\left( \frac{d}{x} \right) \right) \\ & =
\sum_{d \leq x} \left( \frac{\log x + E}{d} - \frac{\log d}{d} + O\left( \frac{1}{x} \right) \right) \\ & =
(\log x + E) \sum_{d \leq x} \frac{1}{d} - \sum_{d \leq x} \frac{\log d}{d} + O(1) \\ & =
(\log x + E) \left( \log x + E + O\left( \frac{1}{x} \right) \right) - \left( \frac{1}{2} \log ^2 x + O(1) \right) + O(1) \\ & =
\log ^2 x + E \log x + O\left( \frac{\log x}{x} \right) + E \log x + E ^2 + O\left( \frac{1}{x} \right) - \frac{1}{2}\log ^2 x + O(1) \\ & =
\frac{1}{2} \log ^2 x + 2 E \log x + O(1).
\end{align*}
However, I am having trouble using this to show the result because of the $\log n$ in the denominator. How can we proceed and conclude the result at hand?
| If we write $D(x)=\sum_{n\leq x}d(n)$ then by the Dirichlet hyperbola method we have:
$$D(x)=x\text{log}(x)+x(2\gamma -1)+\mathcal{O}(x^{1/2})$$
Moreover applying Abel's summation formula to your partial sum gives us:
$$\sum_{2\leq n\leq x}\frac{d(n)}{\text{log}(n)}=\frac{D(x)}{\text{log}(x)}+\int_{2}^x\frac{D(t)}{t\text{log}(t)^2}dt+\mathcal{O}\left(\frac{1}{\log(x)}\right)$$
Now note that:
$$\frac{D(x)}{\text{log}(x)}=x+2\gamma\frac{x}{\text{log}(x)}-\frac{x}{\log(x)}+\mathcal{O}\left(\frac{x^{1/2}}{\text{log}(x)}\right)$$
$$\frac{D(t)}{t\text{log}(t)^2}=\frac{1}{\text{log}(t)}+\frac{(2\gamma-1)}{\text{log}(t)^2}+\mathcal{O}\left(\frac{t^{-1/2}}{\text{log}(t)^2}\right)\implies \int_{2}^x\frac{D(t)}{t\text{log}(t)^2}=\frac{x}{\log(x)}+\mathcal{O}\left(\frac{x}{\text{log}(x)^2}\right)$$
Therefore adding these together gives us:
$$\frac{D(x)}{\text{log}(x)}+\int_{2}^x\frac{D(t)}{t\text{log}(t)^2}dt=x+2\gamma\frac{x}{\text{log}(x)}+\mathcal{O}\left(\frac{x}{\text{log}(x)^2}\right)$$
Which by our first equality proves:
$$\sum_{2\leq n\leq x}\frac{d(n)}{\text{log}(n)}=x+2\gamma\frac{x}{\text{log}(x)}+\mathcal{O}\left(\frac{x}{\text{log}(x)^2}\right)$$
| {
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What is the integral of $\int\frac{dx}{\sqrt{x^3+a^3}}$?
What is the integral of $$\int\frac{dx}{\sqrt{x^3+a^3}}?$$
I came across this integration in a physics problem. I suspect role of complex numbers here.
'$a$' is a constant
| As the other users said, this integral is unlikely to have an elementary form, however, it can be expressed in terms of the well known Gauss hypergeometric function, which can be easily evaluated by most CAS or even Wolfram Alpha.
First, let's consider the case $|x|<|a|$, then we can substitute:
$$x=at, \qquad |t|<1$$
$$\int\frac{dx}{\sqrt{x^3+a^3}}=a^{-1/2}\int\frac{dt}{\sqrt{1+t^3}}=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \int t^{3k} dt=$$
$$=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \frac{t^{3k+1}}{3k+1}=a^{-1/2} \Gamma \left( \frac{1}{2} \right) \sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k+1}}{3k+1}=$$
$$=\frac{1}{3}\sqrt{\frac{\pi}{a}}~t~\sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k}}{k+1/3}$$
To find the hypergeometric form of the series above, we consider the ratio of the successive terms:
$$\frac{c_{k+1}}{c_k}=\frac{\left(-1/2-k \right)(k+1/3)}{ (k+4/3)} \frac{t^3}{k+1}=\frac{\left(k+1/2 \right)(k+1/3)}{ (k+4/3)} \frac{-t^3}{k+1}$$
$$c_0=\frac{3}{\sqrt{\pi}}$$
Which, by definition makes the series:
$$\sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k}}{k+1/3}=\frac{3}{\sqrt{\pi}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -t^3 \right)$$
Which makes the integral:
$$\int\frac{dx}{\sqrt{x^3+a^3}}=\frac{t}{\sqrt{a}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -t^3 \right)=\frac{x}{\sqrt{a^3}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -\frac{x^3}{a^3} \right)$$
This is a correct answer for $|x|<|a|$, as can be checked by numerical experiments.
Mathematica, or other advanced software, can directly evaluate and plot hypergeometric function, which makes this form more useful than the original integral.
For $|x|>|a|$ we can use the same method of binomial expansion to get the hypergeometric form.
$$x=at, \qquad |t|>1$$
$$\int\frac{dx}{\sqrt{x^3+a^3}}=a^{-1/2}\int\frac{t^{-3/2} dt}{\sqrt{1+1/t^3}}=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \int t^{-3k-3/2} dt=$$
$$=-a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \frac{t^{-3k-1/2}}{3k+1/2}$$
It's straightforward to continue in the same way and obtain another hypergeometric function.
| {
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Find the sum of the series $1+\frac12 z^2+\frac{1\cdot3}{2\cdot4}z^4+....$ Find the sum of the series $1+\frac12 z^2+\frac{1\cdot3}{2\cdot4}z^4+....$. We can rewrite n-th term as $a_n=\frac{(n-1)(n-3)\cdot...\cdot1}{n(n-2)\cdot..\cdot2}z^n=\frac{(2n)!}{2^{2n}\cdot(n!)^2}z^n$. But I don't know what should I do now. Consider the ratio $|\frac{a_n}{a_{n+1}}|$?
| Another approach (same origin as the other answer, but starting from a well-known formula): we have, for a real $z$ s. t. $|z|<1$,
$$\arcsin z=z+\frac12\frac{z^3}3+\frac{1\cdot 3}{2\cdot 4}\frac{z^5}5+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{z^7}7+\dotsm, $$
whence
$$(\arcsin z)'=\frac1{\sqrt{1-z^2\strut}}=1+\frac12z^2+\frac{1\cdot 3}{2\cdot 4}z^4+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}z^6+\dotsm. $$
The radius of convergence, of course, is $1$.
| {
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Approximate $\sqrt{(1.02)^3+(1.97)^3}$ using differential So $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$
So the differential will be $$\sqrt{1^3+2^3}+\frac{3(1)^2}{2\sqrt{1^3+2^3}}\Delta x+\frac{3(2)^2}{2\sqrt{1^3+2^3}}\Delta y=3+\frac{1}{2}\Delta x+2\Delta y$$
is $\Delta x=0.02 \text{ and } \Delta y=-0.03$ or $\Delta x=(0.02)^3 \text{ and } \Delta y=(-0.03)^3$?
| Note that $$\sqrt{1.02^3+1.97^3}=\sqrt{(x+\Delta x)^3+(y+\Delta y)^3}$$ where $\Delta x=0.02$ and $\Delta y= -0.03$ with $x=1$ and $y=2$.
| {
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A question about how to factorise a binomial. If $x=\frac{-1}{3}$
$(x+1)^3-{3(x+1)}^2+3(x+1)$ =?
Using the binomial theorem coefficients I'm going to add 1 at the end
[$(a-b)^3=a^3-3a^2b+3ab^2-b^3]
So we now have
$(x+1)^3-{3(x+1)}^2+3(x+1)+1$
How do I continue from here?
| Note that by
$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$
with
*
*$a=x+1$
*$b=-1$
we have
$$(x+1)^3-{3(x+1)}^2+3(x+1)=(x+1)^3-{3(x+1)}^2+3(x+1)-1+1=[(x+1)-1)]^3+1=x^3+1$$
| {
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What is the sum of $\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right) = $? $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=\sum_{n=1}^\infty\left(\frac{2n+1}{n^2}\frac1{{(n+1)}^2}\right)$$
I assume that I should get a telescoping sum in some way, but I'm couldn't find it yet.
| Since
\begin{align*}
\frac{2n+1}{n^4+2n^3+n^2}&=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\\
&=\frac 1{n^2}-\frac 1{(n+1)^2}
\end{align*}
then we have, for an integer $N>1$
$$\sum_{n=1}^N\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=1-\frac1{(N+1)^2}$$
| {
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How to find Jordan basis of a matrix Assume matrix
$$A=
\begin{bmatrix}
-1&0&0&0&0\\
-1&1&-2&0&1\\
-1&0&-1&0&1\\
0&1&-1&1&0\\
0&0&0&0&-1
\end{bmatrix}
$$
Its Jordan Canonical Form is
$$J=
\begin{bmatrix}
-1&1&0&0&0\\
0&-1&0&0&0\\
0&0&-1&0&0\\
0&0&0&1&1\\
0&0&0&0&1
\end{bmatrix}
$$
I am trying to find a nonsingular $P$, let $P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}$ s.t. $J=P^{-1}AP\Leftrightarrow AP=PJ$.
I came up with the Wikipedia article on JCF and I think I need to find the generalized eigenvectors so that
$AP=PJ=\begin{bmatrix}-\mathbf{p_1}&\mathbf{p_1}-\mathbf{p_2}&-\mathbf{p_3}&\mathbf{p_4}&\mathbf{p_4}+\mathbf{p_5}\end{bmatrix}$ yielding the systems
$$(A+I)\mathbf{p_1}=\mathbf{0}$$
$$(A+I)^2\mathbf{p_2}=\mathbf{0}$$
$$(A+I)\mathbf{p_3}=\mathbf{0}$$
$$(A-I)\mathbf{p_4}=\mathbf{0}$$
$$(A-I)^2\mathbf{p_5}=\mathbf{0}$$
I solved each of these systems making sure that the vectors $\mathbf{p_i}$ I chose are linearly independent. So I chose
$$P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}=\begin{bmatrix}1&2&-2&0&0\\1&1&2&0&1\\1&1&2&0&0\\0&0&0&1&1\\1&1&-2&0&0\end{bmatrix}$$
which even though is nonsingular I am not getting $AP=PJ$.
What am I doing wrong?
| I got the Jordan blocks in slightly different order.
What you seem to be missing is the consistency part: in my
$$ P =
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0
\end{array}
\right)
$$
we have a place where we do have $(A+I)^2 p_3 = 0,$ but we have consistency in that $p_2 =(A+I)p_3 .$ It follows automatically that $(A+I)p_2 = (A+I)^2 p_3 = 0.$
We also have $(A-I)^2 p_5 = 0,$ then $p_4 =(A-I)p_5.$ As a result $(A-I) p_4 = 0.$
$$
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{rrrrr}
-1 & 0 & 0 & 0 & 0 \\
-1 & 1 & -2 & 0 & 1 \\
-1 & 0 & -1 & 0 & 1 \\
0 & 1 & -1 & 1 & 0 \\
0 & 0 & 0 & 0 & -1
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
-1 & 0 & 0 & 0 & 0 \\
0 & -1 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1
\end{array}
\right)
$$
To get the Jordan form in the order they report, use my columns but permuted, $p_2p_3p_1p_4p_5$ and then correct $P^{-1}.$ We can correct $P^{-1}$ by permuting the rows to 23145.
| {
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Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
| Looking at the equation mod $7$ we see that $2^x = 1 \pmod 7$ which means $3 \mid x$ by Lagrange's theorem. Looking at the equation mod $3$ we see that $(-1)^x = 1 \pmod 3$ so $x$ is even.
Hence there is a positive integer $k$ such that $x = 6k$.
Rewrite the equation as $2^{6k} = 3^y7^z + 1$, which can also be written as $(2^k-1)(2^k+1)(2^{2k}+2^k+1)(2^{2k}-2^k+1)= 3^y7^z$
obtained by factoring the polynomial $X^6 - 1$.
If $k \gt 0$ then:
$\gcd(2^k - 1, 2^k + 1) = \gcd(2^k + 1, 2) = 1$,
$\gcd(2^k-1, 2^{2k}-2^k+1) = \gcd(2^k - 1, 2^{2k}) = 1$, and
$\gcd(2^k+1, 2^{2k}-2^k+1) = \gcd(2^k+1, 2^{k+1}-1) = \gcd(2^k + 1, 3) = 1$ or $3$.
Note that if all $3$ pairs were coprime then one of $2^k-1$, $2^k+1$ or $2^{2k}-2^k+1$ would need to be equal to $1$ since $3^y7^z$ has only two distinct prime factors. But for $k \gt 0$ the last two are always greater than $1$ so we have $2^k-1 = 1$ or $k = 1$.
This gives $x = 6$ and $3^y7^z = 3\cdot7\cdot3 = 3^27^1$ so we get $(x, y, z) = (6, 2, 1)$.
Otherwise we have $\gcd(2^k + 1, 2^{2k} - 2^k + 1) = \gcd(2^k + 1, 3) = 3$, in which case we can divide by $3^2$ to obtain $(2^k-1)\frac{2^k+1}{3}(2^{2k}+2^k+1)\frac{2^{2k}-2^k+1}{3} = 3^{y-2}7^z$ where now $2^k-1$, $\frac{2^k+1}{3}$ and $\frac{2^{2k}-2^k+1}{3}$ are pairwise coprime.
Proceeding as above, this means one of them must be one. We already saw the case $k = 1$ previously, and there are no other ways to make these equal to $1$.
| {
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"answer_count": 4,
"answer_id": 3
} |
Solving $x^4-15x^2-10x+ 24 = 0$ using Ferrari’s method
Ferrari’s method for solving a quartic equation
$$x^4-15x^2-10x+ 24 = 0$$
begins by writing:$$x^4= 15x^2+ 10x-24$$and then adding a term of
the form:$$-2bx^2+b^2$$to both sides.
(a) Explain why this is good idea and what it accomplishes.
(b) Use $b= 7$ to find the two quadratic equations that yield the
solutions.
$(a)$
I have
$$\begin{align*}x^4-2bx^2+b^2&=15x^2+ 10x-24-2bx^2+b^2\\\\
&=(15-2b)x^2+ 10x+(b^2-24)
\end{align*}$$
I then notice that $$x^4-2bx^2+b^2=(x^2-b)^2$$
so we get
$$(x^2-b)^2=(15-2b)x^2+ 10x+(b^2-24)$$
Pluggin in $b=7$ I get
$$(x^2-7)^2=x^2+ 10x+25$$
but I fail to see why this helps with regards to solving for $x$.
Any hints to lean me in the direction would be much appreciated.
| $$(x^2-7)^2=x^2+ 10x+25\implies (x^2-7)^2-(x+5)^2=0$$
$$(x^2-x-12)(x^2+x-2)=0$$
Now solve each quadratic equation for $ x$
| {
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Let a and b be such that $-2A^2 + 4A -3I_2 = aA + bI_2$. To find $a + b$. $A \in M_2$ with characteristic polynomial $p(x) = x^2 -3x - 5$. Let a and b be such that $-2A^2 + 4A -3I_2 = aA + bI_2$. To find $a + b$.
| A matrix satisfies its characteristic polynomial.
$$-2A^2+4A-3I=aA+bI$$
$$-2A^2+(4-a)A+(-3-b)I=0$$
$$-2\left(A^2+\frac{a-4}{2}A+\frac{3+b}{2}I\right)=0$$
$$A^2+\frac{a-4}{2}A+\frac{3+b}{2}I=0$$
From this we need:
$$\frac{a-4}{2}=-3$$
$$a-4=-6$$
$$a=-2$$
and
$$\frac{3+b}{2}=-5$$
$$3+b=-10$$
$$b=-13$$
So $a+b=-15$
| {
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} |
Quadratic equation formula help / simplification I have this quadratic equation,
$ x^{2} + \frac{10}{3}x -\frac{80}{3} = 0 $
I use the quadratic formula to solve and simplify
$-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ = $ \frac{-10 \pm \sqrt{1060}}{6}$
my book says it should simplify to
$ \frac{1}{3} ( -5 \pm \sqrt{73}) $
but i cant get this simplification can anyone show me if they can? Thank you.
| Your solution seems correct, indeed
$$x^{2} + \frac{10}{3}x -\frac{80}{3} = 0\iff3x^2+10x-80=0$$
$$ \frac{-10 \pm\sqrt{100-(4)(3)(-80)}}{6}=-\frac53\pm\frac{\sqrt{265}}{3}$$
| {
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"answer_id": 0
} |
Power Representation of A Series I am asked to find the power representation of the series $ f(x)=xln(2+3x)+1/(2-x)^2 $ about a = 0. I took the power series of $ xln(2+3x) $ and $ 1/(2-x)^2 $ separately and added them together to give me: $ xln(2)+\sum_1^\infty((-1)^{n-1}3^nx^{n+1}) /n2^n + 1/4\sum_1^\infty(n-1)(x/2)^{n-2} $ But im unsure how to make it all under one sum
| Hint: The power series expansion of $\frac{1}{(2-x)^2}$ at $x=0$ is
\begin{align*}
\frac{1}{(2-x)^2}&=\frac{1}{4}\cdot\frac{1}{\left(1-\frac{x}{2}\right)^2}\\
&=\frac{1}{4}\sum_{n=0}^\infty\binom{-2}{n}\left(-\frac{x}{2}\right)^n
=\frac{1}{4}\sum_{n=0}^\infty\binom{n+1}{n}\left(\frac{x}{2}\right)^n\\
&=\frac{1}{4}\sum_{n=0}^\infty\frac{n+1}{2^n}x^n
\end{align*}
Here we use the binomial series expansion and the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
Note: Valid representations of a power series expansion at $x=0$ are for instance
\begin{align*}
\sum_{n=0}^\infty a_n x^n\qquad\text{or}\qquad a_0+a_1x+\sum_{n=2}^\infty a_nx^n\tag{1}
\end{align*}
A representation putting everything in one sum is not always preferable. Sometimes it might be rather cumbersome to put everything in one sum and a representation as it is indicated in the right-hand side of (1) is more convenient. Nevertheless we usually collect all terms with equal power of $x$ into one term.
We obtain this way
\begin{align*}
\color{blue}{x}&\color{blue}{\ln(2+3x)+\frac{1}{(2-x)^2}}\\
&=(\ln 2)x+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{3}{2}\right)^nx^{n+1}
+\frac{1}{4}\sum_{n=0}^\infty \frac{n+1}{2^{n}}x^n\\
&=(\ln 2)x+\sum_{n=2}^\infty \frac{(-1)^{n}}{n-1}\left(\frac{3}{2}\right)^{n-1}x^{n}
+\sum_{n=0}^\infty \frac{n+1}{2^{n+2}}x^n\tag{2}\\
&\,\,\color{blue}{=\frac{1}{4}+\left(\frac{1}{4}+\ln 2\right)x
+\sum_{n=2}^\infty\left( \frac{(-1)^{n}}{n-1}\left(\frac{3}{2}\right)^{n-1}+\frac{n+1}{2^{n+2}}\right)x^n}\tag{3}
\end{align*}
Comment:
*
*In (2) we shift the index of the left series by one to obtain a repesentation with $x^n$ and we multiply the denominator of the coefficient of the right-hand series with $4$.
*In (3) we collect terms with equal power of $x$ keeping the constant and linear term out of the series for convenience only.
| {
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Inequality $\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$ Let $a,b,c>0$ and $\frac1{a+1}+\frac1{b+1}+\frac1{c+1}=1.\qquad $ Prove $$\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$$
The full solution (due to @Hanno) see at the very end in $\underline{\textbf{Conclusion}}\,\,$
My attemps:
*
*AM>HM gives $a+b+c\geqslant 6.$ Put $a=a_1t,b=b_1t,c=c_1t, \,t\geqslant 1, \, a_1+b_1+c_1=6$
As $\frac 1{x+1}$ is decreasing function we have $\sum\limits_{cyc}\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}\leqslant \sum\limits_{cyc}\frac{a_1}{\sqrt{b_1c_1}}\cdot\frac1{a_1+1}.$
So is it possible to prove the same inequality with new restriction $a+b+c=6?$
*$a=\tan^2\alpha, b=\tan^2\beta, c=\tan^2\gamma$ gives
$$\sum\limits_{cyc}\sin^2\alpha\cot\beta\cot\gamma\leqslant \sqrt2$$
if $0<\alpha,\beta,\gamma<\frac\pi2$ and $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.$
*After hint of Michael Rozenberg. Rewrite the substitution as linear system with determinant $d=\begin{vmatrix}-a&1&1\\
1&-b&1\\ 1&1&-c\\ \end{vmatrix}$
If $d\ne 0$ then $x=y=z=0$ and so the hint doesn't work
The equality $d=0$ gives editional restriction, but I am not sure that this case deserves attention.
In fact, I have an ugly solution with brute force, but I want something beautiful.
*Continue. Agree with $a=\frac{y+z}{x}$, $b$ and $c$ cyclically and $x+y+z=1.$ I'll try not to break the symmetry as in my ugly solution.
Substution gives
$$\sum\limits_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+z)(x+y)}}\leqslant\sqrt2$$
squaring gives
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+2\frac{(z+x)\sqrt{zx}}{\sqrt{(y+x)(y+z)}}\cdot\frac{(x+y)\sqrt{xy}}{\sqrt{(z+y)(z+x)}}\right)\leqslant2$$
or
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+
2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2$$
I do not see here where to use AM-GM, or did I miss the moment?
Thanks, @Michael Rozenberg, very cool. I did not miss, I did not get to the point, although I was already near. Why I stubbornly did not want to multiply by the common denominator? Although further my restriction $x+y+z=1$ would not help but would only hinders.
*Yesterday I did not check the last inequality but about this later.
We conside the case $d=0.$ In order for the linear system to have a unique solution, it is necessary to add a new constraint. I put $x+y+z=1$ and violated homogenity. To save it let $x+y+z=I$ where $I$ denotes the sum only.
Then from the system we have $x=\frac I{a+1}$ and $a=\frac Ix-1=\frac{y+z}{x}$ i.e. we simply forget that $I=1$ and save the homogenity.
we amend the previous one:
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+
2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2I^2$$
After AM-GM we have
$$\sum\limits_{cyc}\left({yz(y+z)^3}+
x(x+y)(x+z)(xy+xz+2yz)\right)\leqslant2I^2\prod\limits_{cyc}(x+y)$$
It remains to check the last inequality.
For this its sufficent to calculete the all summands in both sides. I do not want to apply brutal forse.
Avidently both sides have can have members only of type $x^\alpha y^\beta z^\gamma$ with $\alpha+\beta+\gamma=5$ and by symmetry we may assume $\alpha\leqslant\beta\leqslant\gamma.$ My quation is: Is there another way?
Start with memmber $yz^4$. It is contained in the LHS but not contained in the right. So the last inequality must contain negative term and the proof must be corrected if it is possibble yet. Or I'm wrong.
$\underline{\textbf{Conclusion}}\,\,$ The inequality turned out to be interesting and the proof gave @Hanno. Allow yourself to make some minor corrections.
The original restriction was
$$\sum_{\text{cyc}}{1\over a+1} =1 \qquad (1)$$
Now $\sum_{\text{cyc}}{a\over a+1} =\sum_{\text{cyc}}{a+1-1\over a+1} =3-1=2.$ So
$$\sum_{\text{cyc}}{a\over a+1} =2 \qquad (2)$$
Apply the Cauchy–Bunyakovsky–Schwarz inequality (CBS):
$$\sum_{\text{cyc}}{a\over\sqrt{bc}}\cdot\frac1{a+1}
\:= \:
\sum_{\text{cyc}}\sqrt{{a\over a+1}}\cdot\sqrt{\frac a{(a+1)bc}}\:\leqslant\:
{\sqrt{\sum_{\text{cyc}}{a\over a+1}}}\,\cdot\,
\sqrt{\sum_{\text{cyc}}{a\over (a+1)bc}}$$
Taking into account (2) it remains to proof inequality $\sum\limits_{\text{cyc}}{a\over (a+1)bc}\:{\leqslant}\:1$ which is equivalent to
$$\sum_{\text{cyc}}{a^2\over a+1}\:\leqslant\: abc $$
We have $\sum\limits_{\text{cyc}}{a^2\over a+1}=\sum\limits_{\text{cyc}}{a^2-1+1\over a+1}=\sum\limits_{\text{cyc}}(a-1)+\sum\limits_{\text{cyc}}{1\over a+1}=a+b+c-2<a+b+c+2=abc. $
The last equality $a+b+c+2=abc$ follows from (1) by bringing to a common denominator. DONE!
The inequality is exact. Take small $a$ and $b=c=1+\frac2a.$ Then (1) is true and the LHS of required inequality evaluates to ${\sqrt{a+2}\over a+1}\:+\:{a^2\over (a+1)(a+2)}$ which tends to $\sqrt2$ if $a\to0$.
| The presented proof relies alone on the Cauchy–Bunyakovsky–Schwarz inequality (CBS).
We exploit equivalent formulations of the given constraint
\begin{align*}
& \sum_{\text{cyc}}{1\over a+1} =1 \tag{1} \\[1ex]
\iff\quad & \sum_{\text{cyc}}{a\over a+1}=2 \tag{2} \\[1ex]
\iff\quad & abc = a+b+c+2 \tag{3}
\end{align*}
The latter one stems from clearing denominators and follow-up purge.
Upon application of (CBS) and using $(2)$ the upper limit of $\sqrt2$ materialises:
$$\sum_{\text{cyc}}{a\over\sqrt{bc}}\cdot\frac1{a+1}
\:= \:
\sum_{\text{cyc}}\sqrt{{a\over a+1}\cdot\frac a{(a+1)bc}}\:\leqslant\:
\underbrace{\sqrt{\sum_{\text{cyc}}{a\over a+1}}}_{\sqrt2}\,\cdot\,
\left(\sum_{\text{cyc}}{a\over (a+1)bc}\right)^{1/2}$$
Note the close similarity between the initial expression and the last factor. It remains to show that it is bounded above by $1$
$$\sum_{\text{cyc}}{a\over (a+1)bc}\:\stackrel{?}{\leqslant}\:1\,.$$
Taking advantage of $(3)$ we can write
$${a\over a+1}\cdot a+{b\over b+1}\cdot b+{c\over c+1}\cdot c
\;<\; a+b+c+2 \;=\; abc$$
and dividing by $\,abc\,$ gives the claim. This completes the proof.
Concluding remark
The upper bound of $\sqrt2$ on the RHS of the inequality is never attained, I'd guess. One gets close by via sending one variable to zero, say $a$, and keeping equal the remaining ones. Then $b=c=1+{2\over a}$ and the LHS evaluates to $${\sqrt{a+2}\over a+1}\:+\:{a^2\over (a+1)(a+2)}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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n-th element of recurrence relation I want to find formula for n-th element of recursively entered sequence using generating functions. It goes like this
$$a_{1} = 0, a_{2} = 1, a_{n+2} = a_{n+1} + a_{n} + 2$$
I rewrite this into generating functions like this:
$$a(x) = x(a(x)) + x^{2}(a(x)) + \frac{2}{1-x}-x-2$$
Then I end up with
$$ a(x) = -\frac{x}{-x^{2}-x+1}$$
I want to split this to partial fractions, but roots of denominator are
$$x_{1} = \frac{1+\sqrt{5}}{2}, x_{2} = \frac{1-\sqrt{5}}{2}$$
Problem is I cannot split it and if I could I cannnot say what sequence is generated by generating functions like
$$ \frac{A}{x + \frac{1+\sqrt{5}}{2}}, \frac{B}{x + \frac{1-\sqrt{5}}{2}} $$
I'd appriciate any help. Thanks in advance.
| As it almost always happens, the problem is in being very precise with the indexes. Start with
$$f(x)=\sum\limits_{n=1}a_nx^{n-1}=a_1+a_2x+\sum\limits_{n=3}a_nx^{n-1}=\\
x+\sum\limits_{n=3}(a_{n-1}+a_{n-2}+2)x^{n-1}=
x+\sum\limits_{n=3}a_{n-1}x^{n-1}+\sum\limits_{n=3}a_{n-2}x^{n-1}+2\sum\limits_{n=3}x^{n-1}=\\
x+\sum\limits_{n=2}a_{n}x^{n}+\sum\limits_{n=1}a_{n}x^{n+1}+2\sum\limits_{n=2}x^{n}=\\
x+x\sum\limits_{n=2}a_{n}x^{n-1}+x^2\sum\limits_{n=1}a_{n}x^{n-1}+2\sum\limits_{n=2}x^{n}=\\
x+x(f(x)-0)+x^2f(x)+2\left(\frac{1}{1-x}-x-1\right)$$
leading to
$$f(x)(1-x-x^2)=x+\frac{2x^2}{1-x}=\frac{x+x^2}{1-x}$$
or
$$\color{red}{f(x)=\frac{x+x^2}{(1-x)(1-x-x^2)}}=-\frac{x+x^2}{(1-x)\left(x+\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{1-\sqrt{5}}{2}\right)}=...$$
let's note by $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$
$$...=-\frac{1}{\sqrt{5}(1+\psi)(x+\psi)} + \frac{1}{\sqrt{5}(1+\varphi)(x+\varphi)} - \frac{2}{(1+\psi)(1+\varphi)(1-x)}=\\
-\frac{1}{\sqrt{5}\psi(1+\psi)\left(1+\frac{x}{\psi}\right)} + \frac{1}{\sqrt{5}\varphi(1+\varphi)\left(1+\frac{x}{\varphi}\right)} - \frac{2}{(1+\psi)(1+\varphi)(1-x)}=\\
-\frac{1}{\sqrt{5}\psi(1+\psi)}\sum\limits_{n=1}\left(-\frac{x}{\psi}\right)^{n-1}+\frac{1}{\sqrt{5}\varphi(1+\varphi)}\sum\limits_{n=1}\left(-\frac{x}{\varphi}\right)^{n-1}-\frac{2}{(1+\psi)(1+\varphi)}\sum\limits_{n=1}x^{n-1}=\\
\sum\limits_{n=1}\left(-\frac{1}{\sqrt{5}\psi(1+\psi)}\cdot \left(-\frac{1}{\psi}\right)^{n-1} +\frac{1}{\sqrt{5}\varphi(1+\varphi)}\cdot \left(-\frac{1}{\varphi}\right)^{n-1} -\frac{2}{(1+\psi)(1+\varphi)} \right)x^{n-1}$$
thus
$$\color{red}{a_n=}-\frac{1}{\sqrt{5}\psi(1+\psi)}\cdot \left(-\frac{1}{\psi}\right)^{n-1} +\frac{1}{\sqrt{5}\varphi(1+\varphi)}\cdot \left(-\frac{1}{\varphi}\right)^{n-1} -\frac{2}{(1+\psi)(1+\varphi)}=\\
\color{red}{\left(1+\frac{2}{\sqrt{5}}\right)\cdot \left(-\frac{1}{\psi}\right)^{n-1} +\left(1-\frac{2}{\sqrt{5}}\right)\cdot \left(-\frac{1}{\varphi}\right)^{n-1} -2}$$
and to confirm
$$a_1=\left(1+\frac{2}{\sqrt{5}}\right)\cdot 1 +\left(1-\frac{2}{\sqrt{5}}\right)\cdot 1 -2=0$$
$$a_2=\left(1+\frac{2}{\sqrt{5}}\right)\cdot \left(-\frac{1}{\psi}\right) +\left(1-\frac{2}{\sqrt{5}}\right)\cdot \left(-\frac{1}{\varphi}\right) -2=1$$
$$a_3=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arrange 6 adults and 12 children in 5 rooms, with at least 1 adult in each room Problem: How many ways can you arrange 6 adults and 12 children in 5 rooms of max 4 people such that there is at least 1 adult per room. Every person and room is distinguishable.
My take from the problem is the following:
*
*There are 2 distinct ways to place 18 people in room of 4: $\{4, 4, 4, 4, 2\}$ and $\{4,4,4,3,3\}$
*For the first sequence, I can start by fixing all the adults in every room in the following manner: $\binom{6}{2} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1} \times \binom{1}{1}$
*Now there are 2 cases:
*
*The room with 2 adults is the one with 2 people $\binom{12}{0} \times \binom{12}{3} \times \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}$
*The room where 2 adults is among one that contains 4 people $\binom{12}{2} \times \binom{10}{3} \times \binom{7}{3} \times \binom{4}{3} \times \binom{1}{1}$
*There are $5!$ ways to shuffle the rooms around since they are distinguishable
Adding the 2 cases in (3) and multiplying it with (2) and (4) should in my opinion give me the answer with the sequence $\{4, 4, 4, 4, 2\}$. Is there anything wrong with my reasoning?
| You could drive your consideration under 1. even further: you would have the following patterns (using a = adult, c = child):
$A = \{aacc, accc, accc, accc, ac\}$,
$B = \{accc, accc, accc, accc, aa\}$,
$C = \{aacc, accc, accc, acc, acc\}$,
$D = \{accc, accc, accc, aac, acc\}$.
Now let us consider first the rooms to be undistinguishable. Thus the filling pattern can be considered fixed in the above order.
Then we get groupings:
$gA = (\binom{6}{2}\cdot\binom{12}{2})\times(\binom{6-2}{1}\cdot\binom{12-2}{3})\times(\binom{6-3}{1}\cdot\binom{12-5}{3})\times(\binom{6-4}{1}\cdot\binom{12-8}{3})\times(\binom{6-5}{1}\cdot\binom{12-11}{1})$,
$gB = (\binom{6}{1}\cdot\binom{12}{3})\times(\binom{6-1}{1}\cdot\binom{12-3}{3})\times(\binom{6-2}{1}\cdot\binom{12-6}{3})\times(\binom{6-3}{1}\cdot\binom{12-9}{3})\times(\binom{6-4}{2}\cdot\binom{12-12}{0})$,
$gC = (\binom{6}{2}\cdot\binom{12}{2})\times(\binom{6-2}{1}\cdot\binom{12-2}{3})\times(\binom{6-3}{1}\cdot\binom{12-5}{3})\times(\binom{6-4}{1}\cdot\binom{12-8}{2})\times(\binom{6-5}{1}\cdot\binom{12-10}{2})$,
$gD = (\binom{6}{1}\cdot\binom{12}{3})\times(\binom{6-1}{1}\cdot\binom{12-3}{3})\times(\binom{6-2}{1}\cdot\binom{12-6}{3})\times(\binom{6-3}{2}\cdot\binom{12-9}{1})\times(\binom{6-5}{1}\cdot\binom{12-10}{2})$.
Finally you can mix up the room numbers as well, thus you get the sequence counts:
$sA = \frac{5!}{1! \cdot 3! \cdot 1!}$,
$sB = \frac{5!}{4! \cdot 1!}$,
$sC = \frac{5!}{1! \cdot 2! \cdot 2!}$,
$sD = \frac{5!}{3! \cdot 1! \cdot 1!}$.
Thus you get as individual possibilities:
$pA = gA \cdot sA$,
$pB = gB \cdot sB$,
$pC = gC \cdot sC$,
$pD = gD \cdot sD$.
Or finally as total possibility count:
$P = pA + pB + pC + pD$.
(The actual calculation is kept for you.)
---rk
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
| $$\sqrt {x^2-3}=x-3$$
Since, by definition, $\sqrt {x^2-3} \ge 0$, we must have $x-3\ge 0$. That is we must have $x \ge 3$.
If $x \ge 3$, then $x^2-3 \ge 6$. So there will be no complications caused by squaring both sides of the equation.
\begin{align}
x^2-3 &= x^2-6x + 9 \\
6x &= 12 \\
x &= 2
\end{align}
Since we need $x \ge 3$, there is no solution.
| {
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"url": "https://math.stackexchange.com/questions/2704837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of $\sqrt{x-\sqrt{x^2+1}}dx$ On Wolfram Alpha's website, I find
$$\int\sqrt{x-\sqrt{x^2+1}}dx=\frac23 \sqrt{x - \sqrt{1 + x^2}}( {2 x + \sqrt{1 + x^2}})+C$$
I am trying to get this answer by substitution by $x=\tan(\theta), y=\sqrt{x-\sqrt{x^2+1}},$ or $y={x-\sqrt{x^2+1}}$. But the integrated converted to more difficult form .
| Leonardo says this will not be easy to solve. With
$$
t=\sqrt{x-\sqrt{x^2+1}}
$$
I get
$$
\int \sqrt{x-\sqrt{x^2+1}} \;dx =
-2\int\sqrt{\frac{(t^4-1)^2}{4t^2}+1}\;dt
=-\int \left(t^2+\frac{1}{t^2}\right)\;dt
$$
and that integral is easy.
$$\begin{align*}
-\int \left(t^2+\frac{1}{t^2}\right)\;dt &= \frac{-t^3}{3}+\frac{1}{t}+C
\\
&=
\frac{-\left( x-\sqrt {{x}^{2}+1} \right) ^{3/2}}{3}+{\frac {1}{\sqrt {x-
\sqrt {{x}^{2}+1}}}}+C
\\
&=\frac23 \sqrt{x - \sqrt{1 + x^2}}\big( {2 x + \sqrt{1 + x^2}}\big)+C
\end{align*}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Showing a matrix inequality Can someone help me with showing the following
Let $X$, $B$, and $C$ all be positive definite matrices. Show that the following inequality is true:
$$ (X + C)^{-1} (X B X' + C ) ( X + C)^{-1} \ge (B^{-1} + C)^{-1}$$
for all $X$, where $A \ge B$ means $x' A x \ge x' B x$ for all column vector $x$.
I can show it for the 1-dimensional analogue, but I'm lost on what to do in the matrix space.
Thanks
| Put $D = B^{1/2}CB^{1/2}$ and $Y = B^{1/2}XB^{1/2}$, the inequality becomes
\begin{align}
(X+C)^{-1}(XBX+C)(X+C)^{-1} &\ge (B^{-1}+C)^{-1},\\
B^{1/2}(Y+D)^{-1}(Y^2+D)(Y+D)^{-1}B^{1/2} &\ge (B^{-1}+B^{-1/2}D^{-1}B^{-1/2})^{-1},\\
(Y+D)^{-1}(Y^2+D)(Y+D)^{-1} &\ge (I+D^{-1})^{-1},\\
(Y^2 + D) &\ge (Y+D)(I+D)^{-1}(Y+D).
\end{align}
Left and right multiply both sides by $D^{-1/2}$, the inequality is equivalent to
\begin{align}
D^{-1/2}(Y^2 + D)D^{-1/2}
&\ge D^{-1/2}(Y+D)(I+D)^{-1}(Y+D)D^{-1/2}\\
&= D^{-1/2}(Y+D)D^{-1/2}(D^{-1}+I)^{-1}D^{-1/2}(Y+D)D^{-1/2}.
\end{align}
Put $Z = D^{1/2}Y^{-1}D^{1/2}$, we get $Y = D^{1/2}Z^{-1}D^{1/2}$ and
\begin{align}
Z^{-1}DZ^{-1}+I
&\ge (Z^{-1}+I)(D^{-1}+I)^{-1}(Z^{-1}+I),\\
D+Z^2
&\ge (I+Z)(D^{-1}+I)^{-1}(I+Z),\\
&= (I+Z)^2 - (I+Z) (I+D)^{-1} (I+Z),\\
D + (I+Z) (I+D)^{-1} (I+Z) &\ge 2Z + I.
\end{align}
Put $E = I+D$ and $W=Z+I$, the inequality further reduces to
\begin{align}
E + WE^{-1}W &\ge 2W,\\
E^{1/2}(I-E^{-1/2}WE^{-1/2})^2E^{1/2} &\ge 0,
\end{align}
which is true because both $E^{1/2}$ and $I-E^{-1/2}WE^{-1/2}$ are Hermitian. Note that in the course of proof, we do not need the positive definiteness of $X$. The inequality still holds if $X$ is only Hermitian and $X+C$ is invertible.
| {
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"source": "stackexchange",
"question_score": "3",
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Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is
Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So we have
$OA=a,OB=b,OC=c$ and using cosine formula we have $AB=3,BC=7,CA=\sqrt{52}$.
Now using Area of triangle is $\displaystyle ab+bc+ca=\frac{42}{\sqrt{3}}$.
Could some help me to solve it , Thanks
| $a^2+ab+b^2=9 \implies a^2 = 9 - ab - b^2 $
$b^2+bc+c^2=52 \implies b^2 = 52 -bc - c^2$
$c^2+ac+a^2=49 \implies c^2 = 49 - ac - a^2$
Substituting $a^2$ in $(iii)$
$(c-b)(a+b+c) = 40 --(iv)$
Similarly,
$(b-a)(a+b+c) = 3 -- (v)$
$(c-a)(a+b+c)= 43 -- (vi)$
Now adding $(i),(ii),(iii)$
$2(a^2 + b^2 + c^2) + ab + bc + ca = 110$
$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca + a^2 + b^2 + c^2- ab - bc - ca = 110$
$(a+b+c)^2 + \cfrac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2} = 110$
Now, substitute $(a-b),(b-c),(c-a)$ from equation $(iv),(v),(vi)$,
$(a+b+c)^2 + \cfrac{3^2+40^2+ 43^2}{2(a+b+c)^2}= 110$
Take,$(a+b+c)^2 = t$ and solving quadratic,
$t = 91,19$
$(a+b+c) = \sqrt{91}$ or $\sqrt{19}$
clearly, $a+b+c$ can't be $\sqrt{19}$
Now, substitue $a+b+c$ in $(iv),(v),(vi)$
we get, the values of (c-b),(b-a),(c-a)
put the values of b,c in $a+b+c = \sqrt{91} $
We will get the value of a,b,c.
| {
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| Given that $b^2-4c<0$,
$$x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\ge-\frac{b^2}{4}+c=-\frac{1}{4}\left(b^2-4c\right)>0$$
| {
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"source": "stackexchange",
"question_score": "7",
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Expectation and variance of 4th generation Let $z_1,z_2,...,z_n$ describe a branching process in which each parent has j offspring with probability $p_j$, for
$ p_0 = \frac{1}{12}, p_1 = \frac{1}{6}, p_2 = \frac{3}{4}$.
Find the expectation and variance of the $4$th generation.
$H(s) = \frac{1}{12} + \frac{1}{6} s + \frac{3}{4}s^2$
I start by finding $ H_2(s), H'_2 (s), H''_2 (s)$.
$H_2(s) = H(\frac{1}{12} + \frac{1}{6} s + \frac{3}{4}s^2)\\
H_2(s) = \frac{59}{576} + \frac{7s}{144}+ \frac{41s^2}{288}+\frac{3s^3}{16}+\frac{27s^4}{16} \\
H'_2(s) = \frac{7}{144}+\frac{41s}{144}+\frac{9s^2}{16}+\frac{27s^3}{16} \ \rightarrow \mathbb{E}(z_2) = H'_2(1) = \frac{31}{12}\\$
Similarly,
$H''_2(s) = \frac{41}{144} + \frac{9s}{8} + \frac{81s^2}{16} \ \rightarrow H''_2(1) = \frac{895}{144}$
Now, I am presuming it works like this:
$H_4(s)= H_2[H_2(s)] \\
H'_4(s) =H'_2[H_2(s)]H'_2(s) \\
\ \ \ \ \ \ \ \ \ = \frac{31}{12} \times H'_2[H_2(s)] \\
\ \ \ \ \ \ \ \ \ = \frac{31}{12} \times H''_2[H_2(s)].H'_2(s) \\
\ \ \ \ \ \ \ \ \ = \left[\frac{31}{12}\right]^2 \times H_2[H''_2(s)] \\
H'_4(1) = H_2 \left[\frac{895}{144}\right] \left[\frac{31}{12}\right]^2 \rightarrow \mathbb{E}(z_4)= 4541.16 $
Is this logic right. Is it the right direction?
| Let $$P(s) = \sum_{k=0}^\infty \mathbb P(Z_1=k)s^k=\frac1{12}+\frac16s+\frac34s^2$$ be the generating function of the offspring distribution. Then
$$
m:=\mathbb E[Z_1] = \lim_{s\uparrow 1}P'(s) = \frac53
$$
and
$$
\mathbb E[Z_1(Z_1-1)] =\lim_{s\uparrow 1}P''(s)=\frac32,
$$
so
$$
\sigma^2:= \mathsf{Var}(Z_1) = P''(1) + P'(1) - P'(1)^2 = \frac7{18}.
$$
Let $P_n(s)$ be the generating function of $Z_n$. Since $Z_n = \sum_{i=1}^{Z_{n-1}} Z_{n,i}$ where the $Z_{n,i}$ are independent with the same distribution as $Z_1$, we have $P_n(s)=P_{n-1}(P(s))$. To compute $m_n:=\mathbb E[Z_n]$, note that
$$
P_n'(s) = P_{n-1}'(P(s))P'(s),
$$
and letting $s\uparrow 1$ yields $m_n=m_{n-1}m$. This recursion implies that $m_n=m^n$. From $P_n(s) = P(P_n(s))$ we find by a rather tedious computation that
$$
\mathsf{Var}(Z_n) = \sigma^2m^{n-1}\left(\frac{1-m^n}{1-m} \right).
$$
Substituting $m=\frac53$ and $\sigma^2=\frac7{18}$, we have
\begin{align}
m_n &= \left(\frac53\right)^n\\
\mathsf{Var}(Z_n) &= \frac7{20}\left(\left(\frac{25}9\right)^n - \left(\frac53\right)^n\right).
\end{align}
For $n=4$, this gives
\begin{align}
\mathbb E[Z_4] &= \left(\frac53\right)^4 = \frac{625}{81}\approx 7.71605\\
\mathsf{Var}(Z_n) &= \frac7{20}\left(\left(\frac{25}9\right)^4 -\left(\frac53\right)^4\right) = \frac{119000}{6561}\approx 18.1375.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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GCD of the elements of a set
Let $S$ be the set of numbers of the form
$n(n + 1)(n + 2)(n + 3)(n + 4),$
where $n$ is any positive integer. The first few terms of $S$ are
\begin{align*}
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\
3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2520,
\end{align*}
and so on. What is the GCD of the elements of $S$?
Every other number should be even, so $n(n + 1)(n + 2)(n + 3)(n + 4)$ has at least a factor of $2^2=4$ (or $2^3=8$ when $n, n+2$ and $n+4$ are all even). Every third number is divisible by $3$, so $n(n + 1)(n + 2)(n + 3)(n + 4)$ is at least divisible by $3$. Every 5 number is divisible by $5$, so $n(n + 1)(n + 2)(n + 3)(n + 4)$ is divisible by $5$ as well. So I put $3\times 4\times 5 = 60$ as the answer, which is wrong. But I don't think numbers above $5$ are guaranteed to be a factor of $n(n + 1)(n + 2)(n + 3)(n + 4)$...
| For $n\ge 1$, the number $$\frac{n(n+1)(n+2)(n+3)(n+4)}{120}$$ is just the binomial coefficient $\binom{n+4}{5}$ which is always an integer. Hence all the numbers are divisible by $120$.
Because $120$ is the smallest , the $\gcd$ of all the numbers is $120$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use the method of "changing variables" to solve the following recurrence: $T(n) = 2\cdot T(n-2) + n$
I tried doing this, but I don't know how to continue from here. I think it doesn't work and is not correct.
Please help, thanks in advance!
| Let $f(x)=T_0+T_1x+T_2x^2+....+T_nx^n+...$. Thus, $x^2f=T_0x^2+T_1x^3+T_2x^4+...+T_{n-2}x^n+....$. Also, note that $\frac{x}{(1-x)^2}=x+2x^2+3x^3+...nx^n+...$
Thus,
$$f-2x^2f-\frac{x}{(1-x)^2}=T_0+(T_1-1)x+(T_2-2T_0-2)x^2+...+(T_n-2T_{n-2}-n)x^n+...$$
$$=T_0+(T_1-1)x$$
Thus,
$$f=\frac{T_0}{1-2x^2}+\frac{x}{(1-2x^2)(1-x)^2}+\frac{(T_1-1)x}{1-2x^2}$$
Now, $\frac{x}{(1-2x^2)(1-x)^2}= \frac{3}{\left(x-1 \right)}+{\frac {-6\,x-4}{2\,{x}^{2}-1}}- \frac{1}{\left(
x-1 \right) ^{2}}$
Thus atlast we have:
$$f=\frac{T_0+4}{1-2x^2}+\frac{(T_1+5)x}{1-2x^2}+\frac{3}{x-1}-\frac{1}{(x-1)^2}$$
All the above functions are closed forms of well knows geometric series, from which we can find the coefficient of $x^n$ which gives us $T_n$(From the definition of $f$ in the beginning of the answer)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why $\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$? Why does an integral $$\int \frac{dx}{1-x^2}$$ with the limitless (undefined) interval equal to $$\frac 12\ln\frac{1+x}{1-x},$$ yet an integral $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}$$ with an interval from $\sin\frac{-5π}{12}$ to $\sin\frac{5π}{12}$ has $$\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$$ without one half attached to ln?
| $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\left.\frac12\log\frac{1+x}{1-x}\right|_{\sin\left(-\frac{5\pi}{12}\right)}^{\sin\left(\frac{5\pi}{12}\right)}=\frac12\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}-\frac12\log\frac{1+\sin\left(-\frac{5\pi}{12}\right)}{1-\sin\left(-\frac{5\pi}{12}\right)}=$$$${}$$
$$=\frac12\log\left[\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\cdot\frac{1-\sin\left(-\frac{5\pi}{12}\right)}{1+\sin\left(-\frac{5\pi}{12}\right)}\right]\;(**)$$
But $\;\sin(-x)=-\sin x\;$ , so...
$$(**)=\frac12\log\left(\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\right)^2=\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed by using the letters of the word DIFFERENTIATION
Find the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed by using the letters of the word DIFFERENTIATION.
Vowels are IIIEEAO. Consonants are DFFNNTTR.Vowels can be selected in $\binom{4}{3}$ ways and consonants can be selected in $\binom{5}{4}$ ways and total ways are $\binom{4}{3}\binom{5}{4}7!=100800$ ways, but the answer is 532770. Where did I go wrong?
| Another approach worth mentioning is that of exponential generating functions (egfs). Let $x$ enumerate vowels, $y$ enumerate consonants and $z$ enumerate total letters. Then the egf factors for vowels break down as follows
III
$$1+xz+\frac{1}{2!}(xz)^2+\frac{1}{3!}(xz)^3$$
EE
$$1+xz+\frac{1}{2!}(xz)^2$$
A, O
$$1+xz$$
and consonants:
FF, NN, TT
$$1+yz+\frac{1}{2!}(yz)^2$$
D, R
$$1+yz$$
The product of these is the egf for words
$$\begin{multline}f(x,y,z)=\left(1+xz+\frac{1}{2!}(xz)^2+\frac{1}{3!}(xz)^3\right)\times\\\left(1+xz+\frac{1}{2!}(xz)^2\right)\left(1+\vphantom{\frac{1}{1}}xz\right)^2\left(1+yz+\frac{1}{2!}(yz)^2\right)^3\times\\\left(1+\vphantom{\frac{1}{1}}yz\right)^2\end{multline}$$
and we want the $x^3y^4z^7/7!$ coefficient of $f(x,y,z)$. Using the sage input
y,z=var('y','z')
show(factorial(7)*expand((1+z*x+(z*x)^2/2+(z*x)^3/6)*(1+z*x+(z*x)^2/2)*(1+z*x)^2*(1+z*y+(z*y)^2/2)^3*(1+z*y)^2).coefficient(z^7).coefficient(x^3).coefficient(y^4))
gives our answer:
$$532\,770\tag{Answer}$$
| {
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"url": "https://math.stackexchange.com/questions/2714503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4x(2/3(900 -x^2)^1/2 = (8/3)x(900 -x^2)^1/2$
Taking the derivative
$A'(x) = 8/3(900 -x^2)^{1/2} + (8/3)x(1/2)(900 -x^2)^{-1/2}(-2x)
= (2400 - (16/3)x^2)/(\sqrt{900-x^2})$
Set the $A' = 0$
$2400 - (16/3)x^2 = 0$
$(16/3)x^2 = 2400$
$(16/3)x = \sqrt{2400} = (20\sqrt{6}) / 3$
$x = (5\sqrt{6})/12$
Then i put the value of x in the equation and get
$A = (8/3)((5 \sqrt 2)/12)(900 - ((5\sqrt 2)/12)^2)^{1/2} = 81.6...$
Is this right or?
| You seem to be going in the right direction. There is a mistake somewhere, but your algebra is a bit convoluted, so I cannot find it exactly. Here is what I got following your procedure:
Solve $4x^2+9y^2=3600$ for $y$ to get $$y=\frac23\sqrt{(30-x)(x+30)}.$$ Then, we want to maximize $$A=4xy=\frac{8x}{3}\sqrt{(30-x)(x+30)}.$$ Taking the first order condition we get $$\frac{d}{dx}\left(\frac{8x}{3}\sqrt{(30-x)(x+30)}\right)=0\;\Rightarrow\;-\frac{16}{3}\frac{x^2-450}{\sqrt{900-x²}}=0\;\Rightarrow\;x=15\sqrt{2}.$$
Plugging this back into the formula for $A$ we get
$$A=\frac{8*15\sqrt{2}}{3}\sqrt{(30-15\sqrt{2})(15\sqrt{2}+30)}=1200.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+3$
What to do next?
Tried it using trigonometry by replacing $z^2$ by $tan^2\theta$ but could not get promising results.
Is there any trick to such genre of problems?
|
Given $\;x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}\,$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+\color{red}{3}$
That's actually a good approach, and it does in fact work out nicely, but:
*
*in order to reverse the fractions, you have to assume $\,x,y,z \ne 0\,$; however $\,x=y=z=0\,$ is a solution, which you lose if you don't state that assumption upfront;
*you made a mistake in the calculations, the second line should rather be:
$$
\begin{align}
&\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2} \cdot \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\color{red}{\frac{3}{2}} \\[5px]
\iff\quad &\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} - 2 \cdot \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) + 3 = 0 \\[5px]
\iff\quad &\left(\frac{1}{x}-1\right)^2 + \left(\frac{1}{y}-1\right)^2 + \left(\frac{1}{z}-1\right)^2 = 0 \\[5px]
\iff\quad &x=y=z=1
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| \begin{align*}
\sum_{k=1}^\infty\frac{1}{k^3}&<\sum_{k=2}^\infty\frac{1}{(k-1)(k)(k+1)}+1\\
&=\frac{1}{2}\sum_{k=2}^\infty\left[\frac{1}{k-1}-\frac{2}{k}+\frac{1}{k+1}\right]+1\\
&=\frac{1}{2}\left(1-\frac{1}{2}\right)+1\\
&=\frac{5}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The difference between the radii of the largest and smallest circles having centres on the circumference of $x^2+2x+y^2+4y=4$ Owing to restriction of 150 characters in the title section I include the latter part of the problem here below in bold and italics
Also given that both the circles(largest and smallest) pass through a point $(a,b)$ lying outside the given circle, $x^2+2x+y^2+4y=4$
My attempt:
$x^2+2x+y^2+4y=4$
$\implies\space (x+1)^2+(y+2)^2=(3)^2$
I have drawn a circle with center $(-1,-2)$
& Also the largest circle with a center at the circumference of $x^2+2x+y^2+4y=4$ is the circle with $radius=6$
Also this circle passes through $(a,b)$
Now the smallest circle with center on the circumference of $x^2+2x+y^2+4y=4$ and passing through $(a,b)$, let us assume, has its center $(\alpha,\beta)$ and radius $=r$.
Thus $(\alpha-a)^2+(\beta-b)^2=r^2\cdot\cdot\cdot(1)$
Again if the largest circle has radius $R=6$
Then it has its center(by observation) at $(2,-2)$
Thus Equation of the largest circle is $(x-2)^2+(y+2)^2=36$
Again this largest circle passes through $(a,b)$
$\therefore$ $(a-2)^2+(b+2)^2=36 \cdot\cdot\cdot(2)$
I am stuck here. Please throw some light.
| Any point on the circle $$(x+1)^2+(y+2)^2=3^2$$ can be $P(3\cos t-1,3\sin t -2)$
So, if $r$ is the radius of the new circle $$r^2=(a+1-3\cos t)^2+(b+2-3\sin t)^2$$
$$=(a+1)^2+(b+2)^2+9-6\{(b+2)\sin t+(a+1)\cos t\}$$
$$=(a+1)^2+(b+2)^2+9-6\sqrt{(b+2)^2+(a+1)^2}\sin\left(t+\arctan\dfrac{a+1}{b+2}\right)$$
$$r\ge|\sqrt{(b+2)^2+(a+1)^2}-3|\text { for } \sin\left(t+\arctan\dfrac{a+1}{b+2}\right)=1$$
If $(a,b)$ resides outside $$(x+1)^2+(y+2)^2=3^2,(a+1)^2+(b+2)^2>3^2,$$
$$\implies|\sqrt{(b+2)^2+(a+1)^2}-3|=+(\sqrt{(b+2)^2+(a+1)^2}-3)$$
Similarly, $$r\le\sqrt{(b+2)^2+(a+1)^2}+3\text { for } \sin\left(t+\arctan\dfrac{a+1}{b+2}\right)=-1$$
| {
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"url": "https://math.stackexchange.com/questions/2725315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is there a closed form of this sequence? The sequence is $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{2}{1}$ , $\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ , $\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ , ...
and I need to find n th..
Here's my approach..
I bound them with who has same sum of denominator and numerator.
So $\frac{1}{1}$ is set_1 .
$\frac{1}{2}$ , $\frac{2}{1}$ are set_2..
$\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ are set_3 .
$\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ are set_4 and so on..
If I suppose that n is in set_K , I can conclude n th number is $$ \frac{1+( (n-1)-\sum_{i=1}^K (i-1) ) }{ K - ( (n-1)-\sum_{i=1}^K (i-1) ) } $$
( because set_K starts with $\frac{1}{K}$ and n is ( n - $\sum_{i=1}^K (i-1)$ ) th member of set_K.. )
As a result my conclusion is $$\frac{2n-(K-1)K}{K(K+1) - 2(n-1) }$$ but I don't know the relationship between n and K..
Is there anyone to help me? I really want to know.. :(
| The $n$-th term can be expressed as ${\large{\frac{a}{b}}}$, where
\begin{align*}
a&=n-\frac{k(k+1)}{2}\\[4pt]
b&=k+2-a\\[4pt]
k&=
\left\lfloor
{
\frac
{
\left\lfloor
\sqrt{8n-7}
\right\rfloor
-1
}
{2}
}
\right\rfloor\\[4pt]
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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sum (difference) of polynomials to the power n Is there a way to simplify the expression:
$D = (f_1(\omega)+f_2(\omega))^n-(f_1(\omega)-f_2(\omega))^n$
where $n$ is a positive integer.
In this particular problem:
$f_1(\omega)=-\omega^2+2$
$f_2(\omega)=\omega \sqrt{\omega^2-4}$
Expanding $D$ for some values of $n$:
$n=1$: $\sqrt{{\omega}^{2}-4}(2 \omega)$
$n=2$: $\sqrt{{\omega}^{2}-4}(-4\omega^3 + 8\omega)$
$n=3$: $\sqrt{{\omega}^{2}-4}\left( 8\,\omega^{5}-32\,\omega^{3}+24\,\omega\right) $
$n=4$: $\sqrt{{\omega}^{2}-4}\left(-16\omega^7+96\omega^5-160\omega^3+64\omega\right)$
$n=5$: $\sqrt{{\omega}^{2}-4}\left(32\omega^9-256\omega^7+672\omega^5-640\omega^3+160\omega\right)$
| By the binomial theorem
$$
(f_1+f_2)^n -(f_1-f_2)^n = \sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}f_2^k -\sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}(-f_2)^k.
$$
Notice that since we have $(-f_2)^k$ the second sum is alternating, so every other term cancels, leaving
$$
2\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2j+1}f_1^{n-(2j+1)}f_2^{2j+1}=2\left( \binom{n}{1}f_1^{n-1}f_2 + \binom{n}{3}f_1^{n-3}f_2^3 + \dots \right)
$$
where $\lfloor\ \rfloor$ is the floor function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int x \, \tan^{-1} (x) \, dx$ Integrate $\int x \, \tan^{-1} (x) \, dx$
My Attempt:
\begin{align}
\int x \, \tan^{-1} (x) \, dx &= \int x \, \tan^{-1} (x) \, dx \\
&= \dfrac {x^2}{2} \tan^{-1} (x) - \int \dfrac {x^2}{x^2+1} \, dx.
\end{align}
| Write the integrand as $\dfrac{x^2+1-1}{x^2+1}$ and rewrite it as $1-\dfrac{1}{x^2+1}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$ I have encountered this integral and I am stuck evaluating it:$I=\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$
My try was to expand the numerator into power series, indeed: $$x\cos^2x=\frac{x}{2}(1+\cos(2x)) =\frac{x}{2} +\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1} x^{2n+1}}{(2n)!}$$ And using $\zeta{(z)} \Gamma{(z)} =\int_0^{\infty} \frac{x^{z-1}}{e^x-1}dx$ gives: $$I=\frac{1}{2}\zeta{(2)}+ \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1}}{(2n)!} \zeta{(2n+2)} \Gamma{(2n+2)}=\frac{1}{2}\zeta{(2)}+ 2 \sum_{n=0}^{\infty} (-1)^n \zeta{(2n+2)}$$ Is there a way to simplify this? Or maybe another approach to this integral?
Edit: According to the answer in the comment, would this show that $\sum_{n=1}^{\infty} (-1)^{n-1} \zeta{(2n)}=\frac{\pi^2}{6}(2-3\text{ csch}^2(2\pi))+\frac{1}{16} $ ?
| Hint. Another way is to use the residue theorem. Since, for a suitable function, we have $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ we get $$\sum_{n\geq1}\frac{1}{2n^{2}}+\sum_{n\geq1}\left(\frac{1}{2\left(4+n^{2}\right)}-\frac{4}{\left(4+n^{2}\right)^{2}}\right)$$ $$=\frac{\pi^{2}}{12}+\frac{1}{16}+\frac{1}{2}\sum_{n\in\mathbb{Z}}\left(\frac{1}{2\left(4+n^{2}\right)}-\frac{4}{\left(4+n^{2}\right)^{2}}\right).$$ Now you have just to check that $f\left(z\right)=\frac{1}{2\left(4+z^{2}\right)}-\frac{4}{\left(4+z^{2}\right)^{2}}$ is an admissible function and $$\mathrm{Res}_{z=2i}\pi\cot\left(\pi z\right)f\left(z\right)=\mathrm{Res}_{z=-2i}\pi\cot\left(\pi z\right)f\left(z\right)=\frac{1}{4}\pi^{2}\mathrm{csch}^{2}\left(2\pi\right).$$
| {
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"url": "https://math.stackexchange.com/questions/2730723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\sin x \cos x -2x}{2x \sin^2 x+ 2x^2\sin x \cos x})$
=$\lim_{x\to 0}(\frac{\sin 2x -2}{2x \sin^2 x+ x^2\sin 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+2x\sin x \cos x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ 2(x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x+x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+x\sin 2x+ x\sin 2x+x^2\cos 2x})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+2x\sin 2x+x^2\cos 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-2\sin 2x}{2\sin x \cos x+2(\sin 2x+ 2x\cos 2x)+(2x\cos 2x-2x^2\sin 2x})$
=$\lim_{x\to 0}(\frac{-2\sin 2x}{3\sin 2x+6x\cos 2x-2x^2\sin 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-4\cos 2x}{6\cos 2x+6(\cos 2x-2x\sin 2x)-2(2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{-4\cos 2x}{12\cos 2x-12x\sin 2x-4x\sin 2x-4x^2\cos 2x)})$
=$\frac{-4\cos 0}{12 \cos 0-0-0-0}$
=$\frac{-4}{12}$
=$\frac{-1}{3}$
My problem: This method is very lengthy and involves a lot of calculations. Is there is any other (better and efficient) method to evaluate this limit?
| $$\frac{1}{x^2}-\frac{1}{\sin^2(x)}=\frac{\sin^2(x)-x^2}{x^2\sin^2(x)}$$
You have that $\sin(x)=x-\frac{x^3}{3!}+o(x^3)$,
and thus $$\sin^2(x)=x^2-\frac{x^4}{3}+o(x^4).$$
Finally,
$$\frac{1}{x^2}-\frac{1}{\sin^2(x)}=\frac{-x^4+o(x^4)}{3x^2\sin^2(x)}\underset{x\to 0}{\longrightarrow }-\frac{1}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 3,
"answer_id": 0
} |
$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$
Given the limit:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \alpha$$
Find the value of $\alpha$
I know the series does not converge (it is equivalent to the harmonic series. Correct me, please, if I am wrong).
Does it let me apply the following:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \lim_{n\to+\infty}\frac{2n}{n\log(n)(n^2-2n-3)}$$
Or not?
Thank you in advance.
| Note that
$$\frac{2k}{k^2-2k-3}=\frac{2k}{(k+1)(k-3)}= \frac{1}{2}\left(\frac{1}{k+1} +\frac{3}{k-3} \right)$$
So, with $H_n = \sum_{k=1}^{n}\frac{1}{n}$
$$\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \frac{1}{2}\sum_{k=4}^{n}\left(\frac{1}{k+1} +\frac{3}{k-3} \right)\leq \frac{1}{2}H_{n+1} + \frac{3}{2}H_{n+1} = 2H_{n+1}\leq 2(\ln{(n+1)}+1)$$
If follows:
$$0\leq\frac{1}{n\ln(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}\leq2\frac{\ln{(n+1)}+1}{n\ln(n)}\stackrel{n\rightarrow\infty}{\longrightarrow}0$$
| {
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} |
Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$ Find Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$
My try:
First of all $x$ belongs to First quadrant
Also $$\sqrt{\sin x}+\sqrt{\cos x}-1 \ge 0$$ Squaring both sides we get
$$\sin x+\cos x+2\sqrt{\sin x\cos x} \ge 1$$
Any clue here?
| For
$f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$,
you first of all have to
restrict $x$ to the reals.
Then,
since $\sin$ and $\cos$
are periodic
with period $2\pi$,
you can restrict $x \in [0, 2 \pi)$.
Any restrictions there
are replicated in
$[2k\pi, 2(k+1)\pi)$.
Since you need
$\sin(x) \ge 0$,
this gives
$x \in [0, \pi]$.
Since you need
$\cos(x) \ge 0$,
this gives
$x \in [0, \pi/2]\cup [3\pi/2, 2\pi)$.
Combining these gives
$x \in [0, \pi/2]$.
Finally,
you want
$\sqrt{\sin x}+\sqrt{\cos x}-1
\ge 0$.
If $f(x)
= \sqrt{\sin x}+\sqrt{\cos x}-1
$,
then
$f(x) = f(\pi/2-x)$,
so we only need to look at
$x \in [0, \pi/4]$.
$f(0) = 0$,
$f(\pi/4)
=\sqrt{2}-1
\gt 0$.
$\begin{array}\\
f'(x)
&=\frac12 \cos(x) \sin^{-1/2}(x)-\frac12 \sin(x) \cos^{-1/2}(x)\\
&=\frac12 \dfrac{\cos^{3/2}(x)- \sin^{3/2}(x)}{\sin^{1/2}(x) \cos^{1/2}(x)}\\
\end{array}
$
and since
$\cos(x) > \sin(x)$
for
$x \in [0, \pi/4)$,
$f'(x) > 0$ for
$0 \le x \lt \pi/4$
so $f(x)
\ge 0$
for
$0 \le x \le \pi/2$
with equality only at
$x=0$
and
$x = \pi/2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with real analysis methods. Partial summation of the series gets nasty pretty quickly since it involves harmonic numbers. Another interesting fact to note about this series is that this part
$$\sum_{n=1}^{\infty} \left ( \frac{1}{2n-1} + \frac{1}{2n+1} \right )$$
diverges. This came as a complete surprise to me since I expected this to telescope. What remains now to prove the series I want is by using generating function.
\begin{align*}
\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )x^n &=\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{2n-1} - \sum_{n=1}^{\infty} \frac{x^n}{2n+1} \\
&= -\log \left ( 1-x \right ) - \sqrt{x} \;\mathrm{arctanh} \sqrt{x} - \frac{{\mathrm {arctanh} }\sqrt{x} - \sqrt{x}}{\sqrt{x}}
\end{align*}
I cannot , however, evaluate the limit as $x \rightarrow 1^-$ of the last expression. Can someone finish this up?
Of course alternatives are welcome.
| $$\sum_{n\ge 1}\left(\frac1n - \frac1{2n-1} - \frac1{2n+1}\right) = \sum_{n\ge 1}\left(\frac1{2n} - \frac1{2n-1}\right) + \sum_{n\ge 1}\left(\frac1{2n} - \frac1{2n+1}\right) = \sum_{n \ge 1} \frac{(-1)^n}{n} + \sum_{n\ge 2}\frac{(-1)^n}{n} = 2\sum_{n\ge 1} \frac{(-1)^n}{n} + 1 = -2\log(1-(-1)) + 1 = 1 - 2\log2$$
| {
"language": "en",
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} |
How to properly find a critical point for $xy^2$? Let $f(x,y) = xy^2$ and the domain $D = \lbrace (x,y)| x,y\geq0, x^2 + y^2 \leq 3 \rbrace$
$f_x(x,y) = y^2$ and $f_y(x,y) = 2xy$
Therefore, the critical points should be $\lbrace (x,y)| y = 0, \sqrt{3} \geq x \geq 0 \rbrace$.
The determinant of the Hessian is
$$\det(Hf(x,y))=
\begin{vmatrix}
0 & 2y \\
2y & 2x
\end{vmatrix} = 0-4y^2$$
But this doesn't make sense to me because, if $y = 0$, then $f(x,0) = 0$. But on this interval, suppose I chose $(x,y) = (1,\sqrt{2})$, this would be larger and would be the maximum value of the system.
Why didn't finding the first partials and the Hessian allow me to compute the critical point?
| On the circle of radius $r$, we have $x^2+y^2=r^2$. Therefore,
$$
xy^2=r^2x-x^3\tag1
$$
This implies that the interior critical points are at
$$
(x,y)=\frac r{\sqrt3}\left(\pm1,\pm\sqrt2\right)\tag2
$$
with the corresponding values of
$$
xy^2=\pm r^3\frac2{3\sqrt3}\tag3
$$
At the endponts of $x=\pm r$, we get the values of $xy^2=0$. Therefore,
$$
-r^3\frac2{3\sqrt3}\le xy^2\le r^3\frac2{3\sqrt3}\tag4
$$
Thus, the maxima and minima are on the bounding circle.
If $x^2+y^2=3$, then $r=\sqrt3$ and so $(4)$ says that
$$
-2\le xy^2\le2\tag5
$$
| {
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"source": "stackexchange",
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"answer_count": 2,
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} |
How many $m$ such that : $\sum\limits_{k=1}^m \left\lfloor\frac{m}{k}\right\rfloor$ be even? Find how many $m \le 1000$ such that :
$$\sum\limits_{k=1}^m \left\lfloor \frac{m}{k}\right\rfloor$$
be even ( $\lfloor x\rfloor$ is the largest integer smaller than $x$.)
I think that one case is $\sum\limits_{k=1}^m \left\lfloor \frac{m}{k}\right\rfloor$ is even if for every integer $k$ ( $ 1 \le k \le m $) $\left\lfloor\frac{m}{k}\right\rfloor$ be even or the number of odd number be even in $\sum\limits_{k=1}^m \left\lfloor\frac{m}{k}\right\rfloor$. $\big($ $
\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor=\sum\limits_{i=1}^n d(i)
$
where $d(\ )$ is the divisor function $\big)$
| Hint, further to the comments
$$\sum\limits_{k=1}^{m} \left \lfloor \frac{m}{k} \right \rfloor=\sum\limits_{k=1}^{m}d(k) \tag{1}$$
*
*$m=1 \Rightarrow \sum\limits_{k=1}^{1} \left \lfloor \frac{1}{k} \right \rfloor=\sum\limits_{k=1}^{1}d(k)=\color{blue}{1}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\color{blue}{\left \lfloor \sqrt{1} \right \rfloor=1}$ in total.
*$m=2 \Rightarrow \sum\limits_{k=1}^{2} \left \lfloor \frac{2}{k} \right \rfloor=\sum\limits_{k=1}^{2}d(k)=1+2=\color{blue}{3}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\color{blue}{\left \lfloor \sqrt{2} \right \rfloor=1}$ in total.
*$m=3 \Rightarrow \sum\limits_{k=1}^{3} \left \lfloor \frac{3}{k} \right \rfloor=\sum\limits_{k=1}^{3}d(k)=1+2+2=\color{blue}{5}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\color{blue}{\left \lfloor \sqrt{3} \right \rfloor=1}$ in total.
*$m=4 \Rightarrow \sum\limits_{k=1}^{4} \left \lfloor \frac{4}{k} \right \rfloor=\sum\limits_{k=1}^{4}d(k)=1+2+2+3=\color{blue}{8}$, i.e. $\color{red}{\{1, 4\}}$ are the only perfect squares or $\color{blue}{\left \lfloor \sqrt{4} \right \rfloor=2}$ in total.
*$m=5 \Rightarrow \sum\limits_{k=1}^{5} \left \lfloor \frac{5}{k} \right \rfloor=\sum\limits_{k=1}^{5}d(k)=1+2+2+3+2=\color{blue}{10}$, i.e. $\color{red}{\{1, 4\}}$ are the only perfect squares or $\color{blue}{\left \lfloor \sqrt{5} \right \rfloor=2}$ in total.
Is the pattern visible now? It should be easy to show it by induction.
| {
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"source": "stackexchange",
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} |
Inequality with Sum of Binomial Coefficients My friend noticed that for $n>12$, we have the following pattern in the binomial coefficients.
$$\sum_{i=0}^{\lfloor n/3 \rfloor}\binom{n}{i} < \binom{n}{\lfloor n/3 \rfloor + 1}$$
$$\sum_{i=0}^{\lfloor n/3 \rfloor + 1}\binom{n}{i} > \binom{n}{\lfloor n/3 \rfloor + 2}$$
We've checked these inequalities up to $n=200$ with a computer, but have not been able to come up with a proof.
Attempt: We've attempted a asymptotic approach using Stirling approximation
$$\binom{n}{k} \approx \sqrt{\frac{n}{2\pi k (n-k)}} \frac{n^n}{k^k (n-k)^{n-k}}$$
but approximating with an integral doesn't seem to help very much, as it seems quite hard to compare
$$\int_0^{n} \sqrt{\frac{3n}{2\pi k (3n-k)}} \cdot \frac{(3n)^n}{k^k (3n-k)^{3n-k}} dx \quad \text{and} \quad \sqrt{\frac{3}{4\pi}} \frac{3^{3n}}{2^{2n}}$$
One thing I have realized which seems important is that
$$2 \binom{3n}{n} \approx \binom{3n}{n+1} $$
However, I have not been able to work this into a proof. Any ideas?
2nd Attempt:
Using B. Mehta's linked post, in particular this inequality,
$$\sum_{i=0}^k \binom{n}{i} \leq \binom{n}{k} \frac{n-(k-1)}{n-(2k-1)}$$
subbing in $k=\lfloor\frac{n}{3}\rfloor$, we can almost get the inequality as follows
$$\sum_{i=0}^{\lfloor n/3 \rfloor} \binom{N}{i} \leq \binom{n}{\lfloor n/3 \rfloor} \frac{n-\lfloor n/3 \rfloor + 1}{n - 2 \lfloor n/3 \rfloor + 1}$$
Now we can use that $\binom{n}{k+1} = \binom{n}{k} \frac{n-k}{k+1}$ to get
$$\sum_{i=0}^{\lfloor n/3 \rfloor} \binom{N}{i} \leq \binom{n}{\lfloor n/3 \rfloor + 1}\frac{\lfloor n/3 \rfloor+1}{n-\lfloor n/3 \rfloor}\frac{n-\lfloor n/3 \rfloor + 1}{n - 2 \lfloor n/3 \rfloor + 1}$$
But unfortunately, the inequality
$$\frac{\lfloor n/3 \rfloor+1}{n-\lfloor n/3 \rfloor}\frac{n-\lfloor n/3 \rfloor + 1}{n - 2 \lfloor n/3 \rfloor + 1} \leq 1$$ does not hold for any multiples of $3$. Any other ideas?
| This is partially a comment that is slightly too long. In the Math Overflow article, we want to bound
$$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots \over {N \choose k}}
= {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$
The author chooses to use a geometric series as an upper bound starting with the first term. However, we can slightly delay the geometric series to get a smaller upper bound as follows:
$$ 1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots = \\1 + \frac{k}{N-k+1} \left(1 + \frac{k-1}{N-k+2} + \frac{(k-1)(k-2)}{N-k+2(N-k+3)} \right) + \cdots$$
Using a geometric series for an upper bound in the inner parenthesis gives us an upper bound of
$$ \dbinom{N}{k} \left(1 + \frac{k}{N-k+1} \cdot \frac{N-k+2}{N-2k+3} \right). $$
We can easily check that for large enough $N$,
$$1 + \frac{k}{N-k+1} \cdot \frac{N-k+2}{N-2k+3} < \frac{N-(k-1)}{N-(2k-1)}. $$
Finally we can check that the steps used by the OP in the 2nd attempt portion of the question does go through if $N = 3k$ (which was an issue last time) for large enough $N$. This is just some algebraic manipulations so I won't post it here. Hopefully I haven't made an error.
| {
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"url": "https://math.stackexchange.com/questions/2738804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Analytical solution of Laplace's equation with robin/third boundary condition I'm new to PDE solution, and feel difficult to solve the Laplace's equation with robin boundary conditions.
The equation $\Delta u=0$ with boundary conditions shown in the picture:
picture of problem.
Does this problem has an analytical solution? if so, what is it?
| Using separation of variables $u = X(x)Y(y)$, we obtain the ODEs
\begin{align}
X'' &= -\lambda X \\
Y'' &= \lambda Y
\end{align}
where we used $-\lambda$ for our separation constant, with associated boundary conditions
\begin{align}
u(L,y) &= 0 \implies X(L) = 0 \\
u_{x}(0,y) &= 0 \implies X'(0) = 0 \\
u_{y}(x,0) - hu(x,0) &= 0 \implies Y'(0) - hY(0) = 0
\end{align}
Solving the ODE in $X$, we find non-trivial solutions only if $\lambda > 0$ which yields
$$X = A \cos \sqrt{\lambda} x + B \sin \sqrt{\lambda} x$$
Now
\begin{align}
X'(0) &= \sqrt{\lambda} B \\
&= 0 \\
\implies B &= 0 \quad \text{(why?)} \\
\therefore X(L) &= A \cos \sqrt{\lambda} L \\
&= 0 \\
\implies \sqrt{\lambda} L &= \frac{(2n + 1) \pi}{2}, \quad n \ge 0 \quad \text{($A \ne 0$ for non-trivial solutions)} \\
\implies \lambda &= \frac{(2n + 1)^{2} \pi^{2}}{4L^{2}}, \quad n \ge 0 \quad(*)
\end{align}
and hence
$$X_{n} = A \cos \left( \frac{(2n + 1) \pi x}{2L} \right)$$
Using the eigenvalue $(*)$ and solving the ODE in $Y$ yields
$$Y = \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi y}{2L} \right) + \sinh \left(\frac{(2n + 1) \pi y}{2L} \right)$$
and so the general solution is given by
$$u(x,y) = \sum_{n \ge 0} A_{n} \cos \left( \frac{(2n + 1) \pi x}{2L} \right) \left[ \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi y}{2L} \right) + \sinh \left(\frac{(2n + 1) \pi y}{2L} \right)\right]$$
Applying the inhomogeneous condition, we find
\begin{align}
u(x,L) &= u_{0} \\
&= \sum_{n \ge 0} A_{n} \cos \left( \frac{(2n + 1) \pi x}{2L} \right) \left[ \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi}{2} \right) + \sinh \left(\frac{(2n + 1) \pi}{2} \right)\right] \\
&= \sum_{n \ge 0} C_{n} \cos \left( \frac{(2n + 1) \pi x}{2L} \right)
\end{align}
where
$$C_{n} = A_{n} \left[ \frac{1}{h} \cdot \frac{(2n + 1) \pi }{2L} \cosh \left(\frac{(2n + 1) \pi}{2} \right) + \sinh \left(\frac{(2n + 1) \pi}{2} \right)\right]$$
You can now solve for the coefficients $C_{n}$ using orthogonality relations. Note that when doing the integrals, a change of variable might help.
| {
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"question_score": "1",
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Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in advanced.
| First we simplify the numerator of $$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
$$(5+6)(5^2+6^2)=(6-5)(5+6)(5^2+6^2)=(6^4-5^4)$$
$$(5+6)(5^2+6^2)(5^4+6^4)=(6^4-5^4)((5^4+6^4)=(6^8-5^8)
\\.\\.\\ (5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})=(6^{2048}-5^{2048})$$
$$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}=\frac { 6^{2048} }{ 3^{1024}} = 12^{1024} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Prove that $6$ is a divisor of $n^3 - n$ for all natural numbers. How would you approach such a problem? Induction perhaps? I have been studying proof by induction, but so far I have only solved problems of this nature:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$
| You can decompose $n^3-n = (n-1)n(n+1)$. So $n^3-n$ is a product of three consecutive integers, and in such a sequence you always have an even number and a multiple of $3$. Hence $n^3-n$ is a multiple of $6$.
By induction: the base case is $n = 0$, which is trivial. Now suppose $6$ divides $n^3-n$. Claim: $6$ divides $(n+1)^3-(n+1)$.
You have
$$(n+1)^3-(n+1) =n^3+3n^2+3n+1-n-1=n^3-n+3n(n+1).$$
Now observe that since $6$ divides $n^3-n$ you just need to show that $3n(n+1)$ is a multiple of $6$. But one between $n$ and $n+1$ is even, so $3n(n+1)$ is a multiple of $6$. Hence $6$ divides $(n+1)^3-(n+1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find variance of $X$ given its cumulative distribution function A rv $X$ has the cumulative distribution function
$$ F(x) = \begin{cases} 0 \; \; \; \; x<1 \\ \frac{x^2-2x+2}{2} \; \; \; \; 1 \leq x < 2 \\ 1 \; \; \; \; x \geq 2 \end{cases} $$
Calculate the variance of $X$
attempt
First since $F'(x) = f(x)$, then
$$ f(x) = x-1 $$
on $[1,2)$ and $0$ otherwise. Now,
$$ Var(X) = E(X^2) - E(X)^2 = \int\limits_1^2 (x^3 - x^2) dx - \left( \int\limits_1^2 (x^2-x )\right)^2 $$
After solving this easy integral I get $0.73$ whereas my answer key says the answer is $\boxed{0.139}$. What is my mistake here? Am I applying the formulas wrong?
| It doesn't have a density. Note that $F(1)=\frac12$. If you assume that there is a density and compute according to your formula, you will notice that $E(X) <1$, which imply that something went wrong.
Let $y \in [1,4)$,$$P(X^2 \le y)=P(X \le \sqrt{y})=\frac{y-2\sqrt{y}+2}{2}$$
$$E[X^2]=\int_0^1 \, dx+\int_1^4 \frac{2\sqrt{x}-x}{2}\, dx=\left. \frac23x^\frac32-\frac{x^2}4\right|_1^4+1=1+\frac{14}{3}-\frac{15}4=1+\frac{56-45}{12}=\frac{23}{12}$$
$$E[X]=\int_0^1 \, dx + \int_1^2\frac{2x-x^2}{2}\, dx = \left. \frac{x^2}2-\frac{x^3}{6} \right|_1^2+1=1+\frac32-\frac76=\frac{4}{3}$$
Hence the variance is
$$\frac{23}{12}-\frac{16}{9} \approx 0.139$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $P(X + 10/X > 7 )$ where $X$ is uniform
Let $X$ be a continuous r.v. with a continuous uniform distribution on
$[0,10]$. What is $P \left( X + \frac{10}{X} > 7 \right)$?
Attempt
We notice that $X+ \frac{10}{X} > 7$ can be rewritten as $X^2 + 10 - 7X > 0$ which is equivalent to $(X-5)(X-2) > 0$. Thus,
$$ P \left( X + \frac{10}{X} > 7 \right) = P[(X-5)(X-2)>0] = P[ \{X>2 \} \cup \{ X < 5 \} ] $$
Since we have two cases: either $X>5$ and $X>2$ OR $X<5$ and $X<2$. now, using inclusion-exclusion we get
$$ P(X>2) + P(X<5) - P(2<X<5) = 1 - F(2) + F(5) - F(5) + F(2) = 1$$
What is wrong with my solution? According to my solution key, I should get $\frac{7}{10}$. What am I missing here?
| Note: $\mathsf P(X+\tfrac {10}X>7) ~{=~ \mathsf P(X^2+10>7X)\\ =~ \mathsf P((X-5)(X-2)>0) \\=~ \mathsf P(X<2~\cup~5<X)}$
Plot $~y=(x-5)(x-2)~$ and look at where it is above the x-axis.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integrate $\int \ln(x^2 +1)\ dx$ $$\int \ln(x^2 +1)\ dx$$
I done it using integration by parts where
$\int u\ dv = uv - \int v\ du$
Let $u$ = $\ln(x^2 +1)$
$du = \frac{2x}{x^2+1} dx $
Let $dv = dx$ so $v=x$
$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1} $
I integrate $2\int \frac{x^2}{x^2+1} $ separately using substitution.
Let $u$ = $x^2 + 1$ -> $dx= \frac{1}{2x} du $
substitute $u$ back into it I get a simplified $\int \frac{x}{x^2 + 1} dx$
I used substitution again and got $ \frac{1}{2} \int \frac{1}{u} du$
When I sub it back into the original question, this is my final answer,
$x\ln(x^2 +1) + \frac{1}{2} \ln(x^2 +1) + C$
However, this answer is wrong,
The answer is, $x\ln(x^2 +1) - 2x + 2\tan^{-1} (x) + C $
I believe my integration of $2\int \frac{x^2}{x^2+1} $ is wrong. Where did I went wrong ?
| Your integration by parts in the beginning is correct. Truly :
$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x =x\ln\left(x^2+1\right)-{\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x$$
Now, handling the second integral, first of all let's factor out the constant and write $x^2$ as $x^2 + 1 - 1$ to split it up :
$${\displaystyle\int}\dfrac{2x^2}{x^2+1}\,\mathrm{d}x =\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{x^2}{x^2+1}\,\mathrm{d}x ={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-4}{x^2+1}}}{x^2+1}-\dfrac{\class{steps-node}{\cssId{steps-node-5}{1}}}{x^2+1}\right)\mathrm{d}x $$
$$=$$
$$={\displaystyle\int}\left(1-\dfrac{1}{x^2+1}\right)\mathrm{d}x ={\displaystyle\int}1\,\mathrm{d}x-{\displaystyle\int}\dfrac{1}{x^2+1}\,\mathrm{d}x =2x-2\arctan\left(x\right)$$
The integral is now solved and we yield the result :
$${\displaystyle\int}\ln\left(x^2+1\right)\,\mathrm{d}x=x\ln\left(x^2+1\right)+2\arctan\left(x\right)-2x + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the range correct? What is the range, using interval notation, of:
$$
f(x) = \frac{2x-4}{x^2+x-2}
$$
The answer I have is
$$
(-\infty, 2/9] \cup [2, \infty)
$$
but i am not sure if it is correct.
| First, you decompose your function w.r.t. the poles
$$
f(x)=\frac{2x-4}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
$$
you get $A+B=2, 2A-B=-4$ hence $3A=-2, A=-\frac{2}{3}, B=\frac{8}{3}$. The derivative of $f$ is
$$
\frac{-2 x^2 + 8 x}{x^4 + 2 x^3 - 3 x^2 - 4 x + 4}
$$
Now, as the denominator is $((x-1)(x+2))^2$ the derivative has the sign of its numerator $-2 x^2 + 8 x=-2x(x-4)$ and we get the variation table
$$
\begin{array}{c|c|c|c|c|c|c|c|c}
x & (-\infty,-2)& -2 &(-2, 0) & 0 & (0,1) &1 & (1,4) & 4 & (4,+\infty)
\\
sign(f') & - & ND & - & 0 & + & ND & + & 0 & -
\\
f & 0\searrow -\infty & ND &+\infty\searrow 2 & 2 & 2\nearrow +\infty & ND & -\infty\nearrow \frac{2}{9}&\frac{2}{9} &\frac{2}{9} \searrow -\infty
\end{array}
$$
ND means ``Non Defined''.
Hope it helps
| {
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"timestamp": "2023-03-29T00:00:00",
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The smallest distance between any point on a curve and the parabola $y=x^2$ is 1. What is the equation of the curve? Find the equation of the function $f(x)$, where:
*
*the optimal distance between any one point $P$ on the curve $y=f(x)$ and the parabola $y=x^2$ is always equal to $1$
*$f(x)>x^2$ for all $x$ (looking for the solution curve above the parabola, not below)
Here is a visual approximation of the construction on Desmos: https://www.desmos.com/calculator/pyomsrazo7
This problem arose when attempting to find the equation of motion of a ball with radius $1$ as it rolls along the concave side of the parabola $y=x^2$. Point $P$ is the center of the ball whose path of motion is the curve $y=f(x)$.
| Let a point on the curve be $(a,f(a)$ and the tangent point on the parabola be $(b,b^2)$. Two conditions must be true for all $a \in \mathbb R$
First, the distance is $1$
$$ (b-a)^2 + (b^2-f(a))^2 = 1 $$
Second, the slope between the two points must be normal to the tangent slope, which is $2b$
$$ \frac{b^2-f(a)}{b-a} = -\frac{1}{2b} $$
Substituting $b^2-f(a) = -\frac{b-a}{2b}$, we get
$$ (b-a)^2 = \frac{4b^2}{4b^2+1} $$
Since $(a,f(a))$ is always on the interior of the parabola, we have $b-a > 0$ if $b > 0$, and $b-a < 0$ if $b < 0$. Therefore $b-a$ should have the same sign as $b$ and we can simplify
$$ b-a = \frac{2b}{\sqrt{4b^2+1}} $$
and
$$ b^2 - f(a) = -\frac{1}{\sqrt{4b^2+1}} $$
From here, you have the solution in parametric form
\begin{align} x &= b - \frac{2b}{\sqrt{4b^2+1}} \\
y &= b^2 + \frac{1}{\sqrt{4b^2+1}} \end{align}
Here is a visualization. Two "bends" occur when $b = \pm \frac12$
Also note that this does not represent one continuous motion in your original proposed problem, as the ball is too large and would get stuck at the peak, where the tangent points are $b = \pm \frac{\sqrt{3}}{2}$
EDIT: A natural question you might ask is, what radius does the ball need to be to not get stuck at the peak? To find out, redo the problem with a radius of $r>0$ to get the parametric solution
\begin{align} x &= b - \frac{2br}{\sqrt{4b^2+1}} \\
y &= b^2 + \frac{r}{\sqrt{4b^2+1}} \end{align}
A simple fix is to restrict $r$ so we have only one tangent point at $x=0$. Solving $x=0$ gives
$$ (4b^2 + 1 - 4r^2)b^2 = 0 $$
The only solution needs to be $b=0$, which means
$$ 1-4r^2 \ge 0 $$
Therefore $r \le \frac12$
Here's a solution curve with $r=0.4$
Actually, $r=\frac12$ does not result in a smooth curve, so you may not want to include it. But $\forall r < \frac12$ is valid
EDIT 2: You can also obtain an "outer" solution by switching the signs
\begin{align} x &= b + \frac{2br}{\sqrt{4b^2+1}} \\
y &= b^2 - \frac{r}{\sqrt{4b^2+1}} \end{align}
This solution has no restriction on $r$ (the curve is always smooth no matter how large the radius)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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throwing a two-dice 7 with in total 6 dice We have been stuck on the following questions regarding the game Qwixx:
What is the probability that there is a combination of two dice which add up to exactly 7 when throwing with a total of 6 dice once?
Thank you in advance!
| For $k=1,2,3$ let's say that we are in status $S_k$ if exactly $k$ distinct faces have shown up and no pair exists that adds up to $7$.
Now throw a die.
It does not matter what outcome it has and we have landed in status $S_1$.
From here $5$ steps are made giving a total of $6^5=7776$ possible routes.
Note that:
*
*there is $1$ possible step $S_1\to S_1$ and $4$ possible steps $S_1\to S_2$.
*there are $2$ possible steps $S_2\to S_2$ and $2$ possible steps $S_2\to S_3$
*there are $3$ possible steps $S_3\to S_3$.
This results in the following scheme:
$$\begin{array}{ccccc}
S1\\
1\uparrow\\
S1 & \stackrel{4}{\to} & S2\\
1\uparrow & & 2\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3\\
1\uparrow & & 2\uparrow & & 3\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3\\
1\uparrow & & 2\uparrow & & 3\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3\\
1\uparrow & & 2\uparrow & & 3\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3
\end{array}$$
That gives:
*
*$1^5=1$ route that ends in $S_1$
*$4\times[1^42^0+1^32^1+1^22^2+1^12^3+1^02^4]=124$ routes that end in $S_2$
*$4\times2\times[1^3+1^22^1+1^23^1+1^12^2+1^12^13^1+1^13^2+2^3+2^23^1+2^13^2+3^3]=720$ routes ending in $S_3$
That gives probability $\frac{845}{7776}$ that we end up in $S_1$, $S_2$ or $S_3$ hence probability $$\frac{6931}{7776}$$ that we do not end in $S_1$, $S_2$ or $S_3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Indefinite integral involving the Lambert W or product log function. How to calculate the integral
$\displaystyle \int \frac{1-x}{x W\left(\frac{1-x}{x}\right)} \, dx$ ?
I tried making a substitution but it doesn't seem to work.
Does this integral have a symbolic solution or a series expansion?
| If you change variable $$\frac{1-x}x=t\implies x=\frac{1}{t+1}\implies dx=-\frac{dt}{(t+1)^2}$$ you end with
$$I=\int \frac{1-x}{x W\left(\frac{1-x}{x}\right)} \, dx=-\int \frac {t}{(1+t^2) \, W(t)} \,dt$$ You can expand the integrand around $t=0$ using composition of Taylor series. This would give
$$ \frac {t}{(1+t^2) \, W(t)}=1-t+\frac{1}{2}t^2+\frac{2 }{3}t^3-\frac{71 }{24}t^4+\frac{443 }{60}t^5-\frac{11627
}{720}t^6+\frac{86111 }{2520}t^7+O\left(t^8\right)$$
Edit
Thinking more about it, using Taylor around $x=1$, you have
$$\frac{1-x}{x W\left(\frac{1-x}{x}\right)} =1-(x-1)+\frac{1}{2} (x-1)^2-\frac{2}{3} (x-1)^3+\frac{3}{8} (x-1)^4-\frac{19}{30}
(x-1)^5+\frac{35}{144} (x-1)^6-\frac{601}{840} (x-1)^7+O\left((x-1)^8\right)$$
Integrating between $\frac 12$ and $\frac 32$,the above series would lead to $\frac{48241}{46080}\approx 1.0468967$ while the numerical integration would give $1.0467994$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Distribution of similar objects of one type and distinct object of other type
$4$ identical oranges and $6$ distinct apples(each of different variety) are to be distributed into $5$ distinct boxes. Then probability that each box get a total of $2$ objects is
Try: Number of ways in which $4$ identical object of one type and $6$ Distinct objects of other type are to be distributed into $5$ persons such that each have $2$ objects is
$$=\binom{5}{2}\times 1 \times \binom{6}{2}\times \binom{4}{2}\times \binom{2}{2}$$
Now i did not understand how to solve it, Help me to solve it , thanks
| Number of distributions of four identical oranges and six distinct apples to five distinct boxes: Since the oranges are identical, what matters is how many are placed in each box. Let $x_k$ be the number of oranges that are placed in the $k$th box. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 = 4$$
is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of four addition signs in a row of four ones. For instance,
$$+ 1 1 + 1 + + 1$$
corresponds to the solution $x_1 = 0$, $x_2 = 2$, $x_3 = 1$, $x_4 = 0$, $x_5 = 1$. The number of such solutions is
$$\binom{8}{4}$$
since we must select which four of the eight positions required for four ones and four addition signs will be filled with addition signs.
Since the apples are distinct, what matters is which apple is placed in which box. There are five distinct boxes in which each of the six distinct apples could be placed. Thus, there are $5^6$ ways to distribute the apples.
Thus, there are
$$\binom{8}{4}5^6$$
ways to distribute four identical oranges and six distinct apples to five distinct boxes.
Number of distributions of four identical oranges and six apples to five distinct boxes in which two pieces of fruit are placed in each box: We place the oranges, then distribute the apples.
Two oranges are placed in each of two boxes and two apples are placed in each of the other three boxes: The two boxes that each receive two oranges can be selected in $\binom{5}{2}$ ways. Of the remaining three boxes, there are $\binom{6}{2}$ ways to select which two apples are placed in the lowest numbered of those boxes and $\binom{4}{2}$ ways to select which two of the remaining four apples are placed in the next lowest numbered of those boxes. The remaining two apples must be placed in the other box. There are
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
such distributions.
Two oranges are placed in one box, one orange and one apple are placed in each of two other boxes, and the other two boxes each receive two apples: There are five ways to select the box that receives two oranges and $\binom{4}{2}$ ways to select which two of the four remaining boxes receive one orange each. There are six ways to place an apple in the lower numbered of the two boxes with one orange and five ways to place an apple in the higher numbered of those boxes. There are $\binom{4}{2}$ ways to place two of the four remaining apples in the lower numbered of the two remaining boxes. The other box must receive the other two apples. There are
$$\binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
such distributions.
One orange and one apple are placed in each of four boxes and the remaining two apples are placed in the remaining box: There are $\binom{5}{4}$ ways to select which boxes receive an orange. There are $\binom{6}{2}$ ways to select which two apples are placed in the empty box. The remaining four apples can be distributed to the four boxes that contain an orange so that one apple is placed in each such box in $4!$ ways. Hence, there are
$$\binom{5}{4}\binom{6}{2}4!$$
such distributions.
Total: The number of ways of distributing four identical oranges and six distinct apples to five distinct boxes in such a way that each box receives two pieces of fruit is
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{6}{2}4!$$
Probability that four identical oranges and six distinct apples are distributed to five distinct boxes in such a way that two pieces of fruit are placed in each box: Divide the number of ways of distributing the pieces of fruit in such a way that two pieces of fruit are placed in each box by the number of ways of distributing the fruit.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logarithm simultaneous equation
Let $(x_1, y_1,z_1)$ and $(x_2, y_2, z_2)$ - where $x_1\ge y_1\ge z_1$ and $x_2\le y_2\le z_2$ - be two triplets satisfying the following simultaneous equations:
$$
\begin{align}
\log_{10}(2xy)&=\log_{10}x\cdot\log_{10}y\\
\log_{10}(xy)&=\log_{10}z\cdot\log_{10}y\\
\log_{10}(2xz)&=\log_{10}x\cdot\log_{10}z
\end{align}
$$
Find $(x_1+y_1+z_1)^{x_2y_2z_2}$. (answer is an integer)
My attempt:
Put $a=\log_{10}(x),b=\log_{10}(y),c=\log_{10}(z),k=\log2$. We get:
$$
\begin{align}
k+a+b&=ab\tag{1}\\
a+b&=bc\tag{2}\\
k+a+c&=ac\tag{3}
\end{align}
$$
Subtracting (1) and (3), we get: $b-c=a(b-c)$. This implies $b=c$ or $a=1$.
If $a=1$, we get $k=-1$ from (1) which is invalid. If $b=c=d$ (let), we get: $a=d^2-d$ (from (2)) and from (1), we have: $k=ad-d-a=(d^2-d)d-d-(d^2-d)=d^3-2d^2$.
But now we have a cubic in $d$ which I've no clue how to solve. What is the correct way to solve this problem?
Update: I'll be happy to offer any clarification on the question. Also, there may be a typo or two (my books are not 100% perfect), so if you're getting an integer answer by making a small adjustment, please feel free to share that. Thanks!
| Too long for a comment.
I think that you did a good job and that the solution $b=c=\color{red}{x}$ is the one to consider (I changed notation on purpose).
You ended with the cubic equation $x^3-2x^2-k=0$ which you do not know how to solve.
Let us follow the steps given here for $a=1$, $b=-2$, $c=0$, $d=-k=$. So, we have $$\Delta=-k\,(27 k+32) <0$$ "then the equation has one real root and two non-real complex conjugate roots".
We also have $p=-\frac{4}{3}$ and $q=-k-\frac{16}{27} <0$
Using the hyperbolic method for one real root, we have
$$t_0=\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{27
}{16}k\right)\right)$$ and $$x=t-\frac b {3a}=\frac 23+\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{27
}{16}k\right)\right)$$
So, back to your notations we have $b=c=x$ and $a=x^2-x$
Now, $?\cdots ?\cdots ?\cdots ?$
| {
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"url": "https://math.stackexchange.com/questions/2762912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find a line equation that divides a triangle in half Let $(0,0), (2,2), (4,0)$ be the vertices of a triangle. Then find the line $l$ that passes through $(1,-1)$ and divides the triangle in half (equal area).
I think can solve this problem by
Letting $P$ and $Q$ be the intersection of the line and the triangle and $m$ be the slope of $l$.
$$
\frac{y-1}{x+1}=m\\
$$
Then the height of the $\triangle APQ $ is
$$
\frac{|mx-y+m+1|}{\sqrt{m^2+1}}=\frac{|5m+1|}{\sqrt{m^2+1}}
$$
and then finding the distance between $P$ and $Q$... Finding the half area in terms of $m$...But is there more clean and easy way to do this? Am I missing something obvious?
| Just an alternative solution. Not very nice.
Let $A=(0,0)$, $B=(4,0)$, $C=(2,2)$, $P=(1,-1)$ and $D=(2,0)$.
Suppose that $l$ meets $AB$ at $H$, $CD$ at $K$ and $BC$ at $L$. Let$ DK=a$.
Then $m=a+1$, $K=(0,a)$, $\displaystyle H=\left(2-\frac{a}{a+1},0\right)$ and $L=(4-b,b)$, where $\displaystyle \frac{b-a}{2-b}=a+1$. Hence $\displaystyle L=\left(\frac{a+6}{a+2},\frac{3a+2}{a+2}\right)$.
Note that $\triangle HKD$ and $\triangle KLC$ have the same area.
\begin{align*}
\frac{1}{2}\left(\frac{a}{a+1}\right)(a)&=\frac{1}{2}(2-a)\left(\frac{a+6}{a+2}-2\right)\\
a^3+2a^2&=a^3-3a^2+4\\
a&=\frac{2}{\sqrt{5}}
\end{align*}
So, $\displaystyle m=1+\frac{2}{\sqrt{5}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist? My solution is the following:
approaching by the y-axis:
$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$
approaching by $y=x$
$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$
So I think,that this limit exists.
is it correct in this form?
| Hint:
Use polar coordinates: $(x,y)\to (0,0)\iff r\to 0$. In polar coordinates, thexpression becomes
$$\frac{x^3+y^3}{x^2+y^2}=r(\cos^3\theta+\sin^3\theta).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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