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Proving a quadratic equation identity Continued question here
I encountered this question:
For quadratic polynomials such as $x^2\pm5x\pm6$ or $x^3\pm5x\mp6$, can be factorised over the integers. The main problem is to find a generator which can generate every polynomial that have this identity.
What have I tried:
For $x^2\pm mx+n=(x\pm p)(x\pm q)$ and $x^2\pm mx-n=(x\pm s)(x\mp t)$, solving a simultaneous equation for $q$ and $s$, getting
$$q=\frac {t(p+t)} {p-t}$$
$$s=\frac {p(p+t)} {p-t}$$
and applying into the general equation getting
$$x^2+(\frac {p^2+t^2} {p-t})x+\frac {pt(p+t)} {p-t}$$
but it doesn't suit for every integer $p$ and $t$, how can we get furthur than that?
|
Let $m$ and $n$ be two positive integers such that the two polynomials
$$x^2+mx+n\qquad\text{ and }\qquad x^2+mx-n,$$
both factor over the integers. Note that then also the two polynomials
$$x^2-mx+n\qquad\text{ and }\qquad x^2-mx-n,$$
factor over the integers. It means there exist integers $p$, $q$, $s$ and $t$ such that
\begin{eqnarray*}
x^2+mx+n&=&(x+p)(x+q)=x^2+(p+q)x+pq,\\
x^2+mx-n&=&(x+s)(x+t)=x^2+(s+t)x+st,\\
\end{eqnarray*}
and so in particular $m^2-4n$ and $m^2+4n$ are both perfect squares, because
\begin{eqnarray*}
m^2-4n&=&(p+q)^2-4pq=(p-q)^2,\\
m^2+4n&=&(s+t)^2-4st=(s-t)^2,\\
\end{eqnarray*}
and hence $(p-q)^2$, $m^2$ and $(s-t)^2$ are three perfect squares in arithmetic progression, with common difference $4n$. Conversely, if $a^2$, $b^2$ and $c^2$ are three perfect squares in arithmetic progression with common difference $4d$ then
\begin{eqnarray*}
p&=&-\frac{b-a}{2},\qquad\text{ and }\qquad q&=&-\frac{b+a}{2},\\
s&=&-\frac{b+c}{2},\qquad\text{ and }\qquad t&=&-\frac{b-c}{2},
\end{eqnarray*}
are integers satisfying
$$(x-p)(x-q)=x^2+bx+d,$$
$$(x-s)(x-t)=x^2+bx-d.$$
So your question is equivalent to characterizing all arithmetic progressions of perfect squares of length $3$. This was somewhat famously done by Fermat, see wikipedia, for example. The parametrization given there yields the parametrization
\begin{eqnarray*}
m&=&u^2+v^2,\\
n&=&uv(u^2-v^2),
\end{eqnarray*}
of the coefficients of such polynomials. Here $u$ and $v$ range over the integers.
|
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"url": "https://math.stackexchange.com/questions/4010511",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Evaluate the flux of $F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through surface $S$ ... $4y^2+z^2=4, \space x\in [-2,2]$ I am trying to find the flux of $\vec F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through the surface $S$ where $S$ consists of the elliptical cylinder defined by
$S$ ... $4y^2+z^2=4, \space x\in [-2,2]$
My first instinct is to parametarize the surface to get $\vec F \cdot \vec n$,
where
$$\vec r = \left\langle x,y,\pm \sqrt{1-y^2} \right\rangle$$
I get $$\vec n = \left\langle 0, \pm \frac{2y}{\sqrt{1-y^2}},1\right\rangle$$
so $$\vec F \cdot \vec n = \pm \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}.$$
I know that $x \in [-2,2]$ and $y \in \left[-\sqrt{1-\left(\frac{z}{2}\right)^2}, \sqrt{1-\left(\frac{z}{2}\right)^2} \right]$, and using a calculator I would be able to evaluate it, but with certain techniques the answer simplifies to
$$\frac{16\pi + 160 \sinh(2/5)}{3}.$$
I am not as familiar with integration involving hyperbolic functions and that is probably the reason why I am not seeing this.
Is there anyone that can confirm this fact?
|
While you can apply divergence theorem, this one is quite straightforward for direct surface integral as well. Given it is a cylinder, it is easier to set up the integral in cylindrical coordinates. An elliptic cylinder with form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be parametrized as $x = a \cos\theta, y = b \sin \theta, z$. Following the same parametrization but adjusting for the fact that axis of our cylinder is along $x-$axis instead of $z$.
We choose $r(t) = (x, \cos\theta, 2 \sin\theta)$, so you can find that $r'(t) = (0, 2\cos\theta, \sin\theta)$
So, $\vec{F} \cdot \vec{n} = (2x^2\cos^2\theta + 4\sin^3\theta)$. The integral becomes,
$\displaystyle \int_0^{2\pi} \int_{-2}^{2} (2x^2 \cos^2\theta + 4 \sin^3\theta e^{x/5}) \ dx \ d\theta$.
Please note that $\sin^3\theta$ is an odd function and over $(0,2\pi)$, its integral is zero. So all you are left with is to integrate $\displaystyle \int_0^{2\pi} \int_{-2}^{2} 2 x^2 \cos^2\theta \ dx \ d\theta$ which is a straightforward integral and the answer comes to $\frac{32\pi}{3}$.
Now coming to your working,
You should avoid parametrizing surface where you have $\pm$ sign and have to split the integral into two as we work with oriented surface and one will have to be very careful with orientation for both integral.
As far as the integral you have set up is not a hyperbolic integral. I am taking the plus sign part for example.
$\vec F \cdot \vec n = \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}$.
$x \in \mathbb{(-2,2)}$ is correct but $y \in \left[-\sqrt{1-\left(\frac{z}{2}\right)^2}, \sqrt{1-\left(\frac{z}{2}\right)^2} \right]$ is not correct as you have parametrized your surface in terms of $(x, y)$. So you have to just take limits of $y$ which is $y \in \mathbb{(-1,1)}$.
When we integrate wrt $x$, integral of $e^{x/5}$ is simply $5 e^{x/5}$ and between $-2, 2$, the integral comes to $5(e^{2/5} - e^{-2/5})$. Note that $\sinh({\frac{2}{5}}) = \frac{e^{2/5} - e^{-2/5}}{2}$ and hence the integral result can be written that way too.
$\int_{-1}^{1} \int_{-2}^{2} \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5} \ dx \ dy = \frac{16\pi}{3} + \frac{160}{3} \sinh(\frac{2}{5})$
Similarly for $\vec{F} \cdot \vec{n} = - \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}$, you get the integral of $\frac{16\pi}{3} - \frac{160}{3} \sinh(\frac{2}{5})$
Adding both, you get $\frac{32\pi}{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$ find the value:
$$\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$$
I try to integrate by parts:
$$-\int^\infty_0 \frac{1}{x}d\arctan(\cos x)$$
But it's not run,help me,thank you.
|
Observe that
\begin{align}
I=&\ \int^\infty_0 \frac{\sin x}{2x(1-\frac{1}{2}\sin^2x)}\ dx = \frac{1}{4}\int^\infty_{-\infty} \frac{\sin x}{x}\sum^\infty_{k=0}\frac{\sin^{2k}x}{2^k}\ dx\\
=&\ \frac{1}{4}\sum^\infty_{k=0}\frac{1}{2^k}\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx
\end{align}
If you believe that
\begin{align}
\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx = \binom{2k}{k}\frac{\pi}{4^k}
\end{align}
then it follows that
\begin{align}
I=\frac{\pi}{4}\sum^\infty_{k=0}\binom{2k}{k}\frac{1}{8^k} = \frac{\sqrt{2}\pi}{4} = \frac{\pi}{2\sqrt{2}}.
\end{align}
Note that the last line follows from the series expansion
\begin{align}
\frac{1}{\sqrt{1-x}}=\sum^\infty_{k=0}\binom{2k}{k}\frac{x^k}{4^k}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{x^2 + (y + h)^2}\right)\,dh$ I'm working on a diffraction simulator. In order to take into account for slit-width, I need to take this integral:
$$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{x^2 + (y + h)^2}\right)\,dh$$
Where $x$ and $y$ are the coordinates of the point for which the amplitude is computed and $w$ is the width of the slit. When developing and simplifying, I get:
$$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{h^2 + Ah + B}\right)\,dh$$
Where $A = 2y, B = x^2 + y^2$.
I tried u-substitution, with $u^2 = h^2 + Ah + B$, yielding $dh = \frac{2udu}{2h + A}$ and thus being rather useless.
Note: I can't approximate the integral by small steps for performance reasons.
Also, if there is no general solution (which seems pretty likely), is there any computationally efficient way of approximating it?
Thanks in advance
|
As you, I am more than skeptical about a closed form.
For an approximation, you could try this $1,400$ years old approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad \text{for} \qquad-\frac \pi 2 \leq t\leq\frac \pi 2$$
This would give
$$\int \cos \left(\sqrt{(h+y)^2+x^2}\right)\sim \frac{5 \pi ^2 }{\sqrt{x^2+\pi ^2}}\tan ^{-1}\left(\frac{h+y}{\sqrt{x^2+\pi
^2}}\right)-4 h$$
This would give for your integral
$$\frac{5 \pi ^2}{\sqrt{x^2+\pi ^2}}\tan ^{-1}\left(\frac{4 \omega \sqrt{x^2+\pi ^2} }{4 \left(x^2+y^2+\pi
^2\right)-\omega ^2}\right) -4 \omega$$
Edit
Concerning the approximation of $$\cos(x)\qquad \text{for} \qquad-\frac \pi 2 \leq t\leq\frac \pi 2$$ for easy integrations, we could use the $[2n,2]$ Padé approximants $(P_n)$. They write
$$\left(
\begin{array}{ccc}
n & \text{numerator} & \text{denominator} \\
1 & 12-5 x^2 & x^2+12 \\
2 & 3 x^4-56 x^2+120 & 4 x^2+120 \\
3 & -13 x^6+660 x^4-9720 x^2+20160 & 360 \left(x^2+56\right) \\
4 & 17 x^8-1680 x^6+65520 x^4-887040 x^2+1814400 & 20160 \left(x^2+90\right)
\end{array}
\right)$$
To check their quality, considering the norm
$$\Phi_n=\int_{-\frac \pi 2}^{+\frac \pi 2} \Big[\cos(x)-P_n\Big]^2\, dx$$
$$\left(
\begin{array}{cc}
n & \Phi_n \\
1 & 1.42 \times 10^{-4} \\
2 & 9.09 \times 10^{-8} \\
3 & 2.30 \times 10^{-11}\\
4 & 5.47 \times 10^{-15}
\end{array}
\right)$$ while
$$\int_{-\frac \pi 2}^{+\frac \pi 2} \Bigg[\cos(x)-\frac{\pi ^2-4x^2}{\pi ^2+x^2}\Bigg]^2\, dx \sim 2.98 \times 10^{-6}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$ Assume that all logarithms are natural.
Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$
This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$
My first inclination was to apply the well known inequality, $\log(z)\leq 2(\sqrt{z}-1)$ if $z>0$, to the left side by letting $z=\frac{x-y}{x+y}$, but this did not help me.
I also tried double induction on $(x,y)$ (since $x$ and $y$ are integers) but I could not finish it.
I appreciate any help, thank you.
|
The inequality
$$
\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right)
$$
holds for all real numbers $x, y$ satisfying $0 < y < x$.
With the substitution $t=y/x$ this is equivalent to
$$ \tag{*}
\log\left( \frac{1+t}{1-t}\right)>2\left(\sqrt{1+t}-\sqrt{1-t}\right)
$$
for $0 < t < 1$. This suggests to investigate the function
$$
f(t) = \log\left( \frac{1+t}{1-t}\right) -2\left(\sqrt{1+t}-\sqrt{1-t}\right) \, .
$$
Then $f(0) = 0$, and for $0 < t < 1$ is
$$
f'(t) = \frac{2}{1-t^2} - \left( \frac{1}{\sqrt{1+t}} + \frac{1}{\sqrt{1-t}} \right) \\
= \frac{2 - \sqrt{1-t^2}\left(\sqrt{1+t} + \sqrt{1-t}\right)}{1-t^2}
$$
and that is positive, because
$$
\sqrt{1-t^2} \cdot (\sqrt{1+t} + \sqrt{1-t}) < \sqrt{1+t} + \sqrt{1-t} \le 2
$$
as can be verified easily.
So $f$ is strictly increasing on $[0, 1)$, and that proves the inequality $(*)$.
|
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|
Compute the probability mass function of X There are two bags A and B. Bag A has 2 green marbles and 3 red marbles. Bag B has 2 green marbles and 3 blue marbles. Two marbles are drawn without replacement from bag A and put in bag B. Then two marbles are drawn without replacement from bag B and put in bag A. Let X be the number of green marbles in bag A after the above operations. Compute the probability mass function of X.
My attempt:
$X=\{0,1,2,3,4\}$
$X=0$: take all green from A and put non-greens back = $\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{7}\cdot\frac{2}{6}=\frac{17}{70}$
$X=1$: take all green from A and put one green back + take 1 green and put non-greens back = $\left(\frac{2}{5}\cdot\frac{1}{4}+\frac{4}{7}\cdot\frac{3}{6}\right)+\left(\frac{2}{5}\cdot\frac{3}{4}+\frac{4}{7}\cdot\frac{3}{7}\right)=\frac{228}{245}$
Then we continue on with $X=2,3,4$. My concern is that if we add up $\mathbb{P}(X=0)$ and $\mathbb{P}(X=1)$, we get a value that is greater than $1$, which should not be the case because everything should sum to $1$. Can someone tell me what I am doing wrong and possibly how I would approach this problem differently to get the right answer? Thank you.
|
You wrote:
take all green from A and put non-greens back
First sight first error found...
$$\mathbb{P}[X=0]=\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{70}$$
At present I did not check the rest
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum_{n \leq x} \tau (n) = 2(\sum_{n \leq \sqrt{x}} [\frac{x}{n}]) - [\sqrt{x}]^2$ Here $\tau(n)$ is the number of positive integers dividing $n$ and $[x]$ is the floor of $x$.
So I know if $n$ is not a perfect square, then half the positive integers dividing $n$ are less than $\sqrt{n}$, and the other half are greater than $\sqrt{n}$. Also, the amount of perfect squares that are less than or equal to $x$, is $[\sqrt{x}]^2$.
So let $d_n(k) = 1$ if $k|n$ and $d_n(k) = 0$ otherwise. Then
$\sum_{n \leq x} \tau (n) = \sum_{n \leq x} \sum_{k \leq n} d_n(k) = \sum_{n \leq x} 2(\sum_{k \leq \sqrt{n}} d_n(k) ) - [\sqrt{x}]^2 = 2(\sum_{n \leq x} \sum_{k \leq \sqrt{n}} d_n(k) ) - [\sqrt{x}]^2$.
So all I have to so is some how prove $\sum_{n \leq x} \sum_{k \leq \sqrt{n}} d_n(k) =\sum_{n \leq \sqrt{x}} [\frac{x}{n}]$
|
$$\begin{align}\sum_{n=1}^x\tau(n)&=\sum_{n=1}^x\#\{\,(a,b):ab=n\,\}\\
&=\#\{\,(a,b,n):ab=n\le x\,\}\\&
=\#\{\,(a,b):ab\le x\,\}\\&
=\#\{\,(a,b):ab\le x, a\le b\,\}+\#\{\,(a,b):ab\le x, a\ge b\,\}-\#\{\,(a,b):ab\le x, a= b\,\}\\&
=2\#\{\,(a,b):ab\le x, a\le b\,\}-\#\{\,(a,b):ab\le x, a= b\,\}\\&
=2\sum_{a\le\sqrt x}\#\{\,b:ab\le x, a\le b\,\}-\#\{\,a\mid a^2\le x\}\\&
=2\sum_{a\le\sqrt x}\left(\#\{\,b:ab\le x\,\}-\#\{\,b:ab\le x, a> b\,\}\right)-\#\{\,a\mid a^2\le x\}\\&
=2\sum_{a\le\sqrt x}\left(\left\lfloor\frac xa\right\rfloor-(a-1)\right)-\lfloor\sqrt x\rfloor\\&
=2\left(\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor-\frac{\lfloor \sqrt x\rfloor(\lfloor \sqrt x\rfloor-1)}2\right)-\left\lfloor\sqrt
x\right\rfloor\\
&=2\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor-\left\lfloor\sqrt
x\right\rfloor^2
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine if $x^3+y^3+z^3+t^3 = 10^{2021}$ has a solution I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions
PowersRepresentations[10^2021, 4, 3]
return
PowersRepresentations::ovfl: Overflow occurred in computation.
FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x < y < z < t}, {x,y,z,t}, Integers]
My computer runs too long. How can I reduce timing to solve this equation?
|
Two solutions may be found starting from:
$$10^{2021}=10^5\times10^{2016}=12500\times2^3\times10^{3\times672}$$
Since $12500=19^3+17^3+8^3+6^3=18^3+17^3+12^3+3^3$ we have:
$$10^{2021}=(38\times10^{672})^3+(34\times10^{672})^3+(16\times10^{672})^3+(12\times10^{672})^3$$
$$10^{2021}=(36\times10^{672})^3+(34\times10^{672})^3+(24\times10^{672})^3+(6\times10^{672})^3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4030057",
"timestamp": "2023-03-29T00:00:00",
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|
Point within the interior of a given angle The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.
We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^2+PM^2+OK^2+KM^2\\=x^2+b^2+y^2+a^2.$$ I can't approach the problem further. How can we use the given angle $\alpha$? Thank you in advance!
|
Let $d=OM$
$$ \arcsin(a/d)+ \arcsin(b/d) = \alpha$$
$$ a \sqrt{d^2-b^2} + b \sqrt{d^2-a^2} = d^2\sin \alpha$$
Let Mathematica solve fourth order equation for $d$
$$sal= \sin \alpha$$
{d -> 0.5` Sqrt[(4.` (1.` + 1.` b^2))/sal^2 +
2.` Sqrt[(4.` (1.` + 1.` b^2)^2)/sal^4 - (
16.` (0.25` - 0.5` b^2 + 0.25` b^4 + 1.` b^2 sal^2))/sal^4]]}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4032239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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|
Finding equilibrium points of system of equations I'm asked to solve system of ODE for finding equilibrium points of this
$$\frac{dx}{dt} = a_1x(1-x) + a_2x\frac{z}{1+z}+a_3xy$$
$$\frac{dy}{dt} = b_1y(1-y) + b_2y\frac{z}{1+z}+b_3xy$$
$$\frac{dz}{dt} = c_1z(1-z) + c_2z\frac{x}{1+x}+c_3z\frac{y}{1+y}$$
I checked some related links to solve this problem and i know that i need to extract variables from this system to matrix form: link:https://www.math24.net/linear-autonomous-systems-equilibrium-points/
but problem how to extract them
My question is how correctly do change of variable for further extraction to matrix form
|
Hint
An equilibrium point is a triple $\left(x , y , z\right)$ satisfying the non-linear system
\begin{equation}
\renewcommand{\arraystretch}{1.5} \left\{\begin{array}{rcl}{a}_{1} x \left(1-x\right)+{a}_{2} x \displaystyle \frac{z}{1+z}+{a}_{3} x y&=&0\\
{b}_{1} y \left(1-y\right)+{b}_{2} y \displaystyle \frac{z}{1+z}+{b}_{3} x y&=&0\\
{c}_{1} z \left(1-z\right)+{c}_{2} z \displaystyle \frac{x}{1+x}+{c}_{3} z \frac{y}{1+y}&=&0
\end{array}\right.
\end{equation}
The most interesting case is when $x \neq 0 , y \neq 0 , z \neq 0$. Then the system reduces to
\begin{equation}
\renewcommand{\arraystretch}{1.5} \left\{\begin{array}{rcl}{a}_{1} \left(1-x\right)+{a}_{2} \displaystyle \frac{z}{1+z}+{a}_{3} y&=&0\\
{b}_{1} \left(1-y\right)+{b}_{2} \displaystyle \frac{z}{1+z}+{b}_{3} x&=&0\\
{c}_{1} \left(1-z\right)+{c}_{2} \displaystyle \frac{x}{1+x}+{c}_{3} \frac{y}{1+y}&=&0
\end{array}\right.
\end{equation}
From the two first equations, one deduces in general a relation of the form
$x = {\alpha} y+{\beta}$. Substituting again in these two first
equations leads to
\begin{equation}
x = \frac{P \left(z\right)}{1+z} \qquad y = \frac{Q \left(z\right)}{1+z}
\end{equation}
where $P$ and $Q$ are polynomials of degree $1$ which coefficients
depend on ${a}_{1} , {a}_{2}, a_3 , {b}_{1} , {b}_{2}, b_3$.
Injecting these values of $x$ and $y$ in the third equation leads to
\begin{equation}
{c}_{1} \left(1-z\right)+{c}_{2} \displaystyle \frac{P \left(z\right)}{1+z+P \left(z\right)}+{c}_{3} \frac{Q \left(z\right)}{1+z+Q \left(z\right)} = 0
\end{equation}
This gives a cubic equation in $z$ with potentially three complex
roots, producing three equilibrium points.
With the help of a computer algebra system all the equilibrium points of this system can be computed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
An Elegant New Proof of Herons Formula? A triangle with side lengths $a, b, c$ an altitude($h$), where the height($h_a$) intercepts the hypotenuse($a$) such that it is the sum of two side lengths, $a = u +v$ and height($h_b$) intercepts hypotenuse($b$) such that it is also the sum of two side lengths $b = x + y$, we can find a simple proof of herons formula.
Image here:
(Note that $h = \sqrt{au}$ and $h = \sqrt{bx}$, giving $au = bx$)
First the equations we have are:
$u + v = a$
$x + y = b$
$au = bx$
$\sqrt{av} + \sqrt{by} = c$
Substituting $v$ and $y$ with $u$ and $x$ for the $4$th equation:
$\sqrt{a^2 - au} + \sqrt{b^2 - bx} = c$
Then using $au = bx$ we eliminate $x$ to find $u$:
$\sqrt{a^2-au}+\sqrt{b^2-b\left(\frac{au}{b}\right)}=c$
$u = \frac{1}{4ac^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)$
Finally, substituting this into $h = \sqrt{au}$:
$h = \sqrt{\frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)}$
Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:
$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $
$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $
$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $
$A = \sqrt{s(s - a)(s- b)(s- c)}$
Q.E.D.
Thoughts?
|
In my post, I had shown that
$$
r^{2}(x+y+z)=x y z \tag*{(result 1)}
$$
and
$$
K=r s \tag*{(result 3)}
$$
where $s=x+y+z$ is the semi-perimeter of the triangle ABC whose area is $K$ and sides $a=y+z, b=z+x $ and $c=x+y$.
From (1), $r^{2} s=x y z$ implies that
$$
K^{2}=r^2s^2 =sxy z=s(s-a)(s-b)(s-c)
$$
Taking square root on both sides yields the Heron’s Formula
$$
K=\sqrt{s(s-a)(s-b)(s-c)}
$$
|
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|
Find the function $f(x)$. Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?
I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.
|
You stated, that $f$ is a polynomial function, hence we can write $f$ as $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ plugging that into your given equation we get
$$f(x+2)-f(x)=a_n(x+2)^n+a_{n-1}(x+2)^{n-1}+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$
using the binomial formula, we can extend this to
$$a_n(x^n+2nx^{n-1}+...+2^{n-1}nx+2^n)+a_{n-1}(x^{n-1}+2(n-1)x^{n-2}+...+2^{n-2}(n-1)x+2^{n-1})+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$
We see, that the $a_nx^n$ cancel each other out; if we look at the coefficient of $x^{n-1}$ we have $2na_n+a_{n-1}-a_{n-1}=2na_n$. This is nonzero, as $a_n$ as the leading coefficient can't be zero.
Thus your polynomial is of degree $2$. That means $f(x)=a_2x^2+a_1x+a_0$. Since $f(0)=5$, we know that $a_0=5$. Plugging what we know so far back into our original equation we get:
$$f(x+2)-f(x)=a_2(x+2)^2+a_1(x+2)+5-a_2x^2-a_1x-5=4a_2x+4a_2+2a_1=8x-2$$
which yields $a_2=2$ and thus $a_1=-5$ giving us the complete polynomial $f(x)=2x^2-5x+5$.
We can plug that back in to verify our solution.
|
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|
How to remove roots from an equation? The question here is how (if it is even possible) to remove the square root terms and transform the following equation to a polynomial with one unknown $x$.
The coefficients $a$, $b$, $c$, and $d$ are known and also $r$.
$$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$$
|
Well the $r-x^2$ screams that they want a trig substitution as lone student's answer suggest.
But you can always remove roots but bringing terms over and squaring.
$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$
$\sqrt{x}(a + b\sqrt{r^2-x^2}) = - d - c\sqrt{r^2-x^2}$
$x(a^2 + b(r^2-x^2) + 2ab\sqrt{r^2-x^2})) = d^2 + c^2(r^2 - x^2) - 2cd\sqrt{r^2-x^2}$
$x(a^2 + b(r^2-x^2))- d^2 - c^2(r^2-x^2) = -(2cd+2abx)\sqrt{r^2 -x^2}$
$(x(a^2 + b(r^2-x^2))+ d^2 + c^2(r^2-x^2))^2 = (2cd+2abx)^2(r^2-x^2)$ so
$(x(a^2 + b(r^2-x^2))+ d^2 + c^2(r^2-x^2))^2 -(2cd+2abx)^2(r^2-x^2) =0$
Which is a $6$th degree polynomial
which 1) Answers exactly what you asked and 2) makes things much, much, much worse.
|
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|
How to evaluate from the number of solutions for $\sin 2x +\cos 3x - \sin 4x = 0$? The problem is as follows:
First find the number of solutions for the equation from below:
Assume $x \in [0,2\pi]$
$\sin 2x +\cos 3x - \sin 4x = 0$
Let $n$ be the number of solutions. Using this $n$ find the sum for:
$5\tan^2\left(\frac{n\pi}{18}\right)+1.5n^2$
I'm not sure how to solve this, what it came to my mind was this:
$\sin 2x +\cos 3x - \sin 4x = 0$
$\sin 4x - \sin 2x - \cos 3x = 0$
Using Prosthaphaeresis formulas then I'm getting to:
$\cos 3x \cdot (2\sin x - 1)=0$
From this it can be inferred that there will be six solutions for
$\cos 3x=0$
In the interval mentioned. Or at least I think so,
these will be:
$\frac{\pi}{6}$, $\frac{3\pi}{6}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$, $\frac{9\pi}{6}$, $\frac{11\pi}{6}$.
While for the other guy it will be this:
$\sin x = \frac{1}{2}$
There will be two solutions in that interval, namely:
$\frac{\pi}{6}$, $\frac{5\pi}{6}$
Thus the number of solutions between the two would be eight. But when replacing in what it is being asked it doesn't seem to yield something reasonable.
Therefore can someone help me here?. Which part went wrong?.
|
I Think $n=6$.
Algebraically:
$$\sin 2x +\cos 3x -\sin 4x = \\ 2\sin x \cos x +\cos 2x \cos x-2\sin x \cos x \sin x -4 \sin x
\cos x \cos 2x= \\ 2 \sin x \cos x +(1-2 \sin^2 x)\cos x - 2 \sin^2 x \cos x -2 \sin x \cos x(2-4 \sin^2 x) = \\ 2 \sin x \cos x (1-2+4 \sin^2 x) + \cos x (1- 2 \sin^2 x - 2 \sin^2 x)= \\ 2\sin x \cos x(-1+4\sin^2 x) + \cos x (1-4 \sin^2 x)= \\ 2\sin x \cos x(4\sin^2 x -1) - \cos x (4 \sin^2 x -1)= \\ (4 \sin^2 x -1)(\cos x)(2\sin x -1)=0 $$
so we have as solutions: $\frac{7\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6}$ for the first term, $\frac{\pi}{2},\frac{3\pi}{2}$ for the second, and $\frac{\pi}{6},\frac{5\pi}{6}$ for the third. Six solutions in total.
|
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|
Solution for approximating square roots? I just saw a post on instagram which said that
$$\sqrt{x}\approx \frac{x+y}{2\sqrt{y}}$$
I tried it out on a few values, and surprisingly, it came within 1 decimal point of the actual answer. Is there a reason for this or is it coincidental?
y is the closest perfect square to x
|
Google AM–GM inequality.
If we square any real number $z$, which can be expressed as a difference of two other real numbers $x$ and $y$, the result is always greater or equal $0$, this means $\color{blue}{z^2 \geq 0}$, assume $z = x-y$, it follows $$\color{blue}{z^2} = (x-y)^2 = x^2 \color{red}{-2xy} + y^2 = x^2 \color{red}{+ 2xy} + y^2 \color{red}{- 4xy} =(x+y)^2-4xy\color{blue}{\geq 0} $$
$$\Rightarrow (x+y)^2-4xy\geq 0 \Leftrightarrow(x+y)^2\geq 4xy \Rightarrow x+y \geq \sqrt{4xy} \Leftrightarrow \frac{x+y}{2\sqrt{y}} \geq \sqrt{x}$$
The result is:
$$\sqrt{x} \leq \frac{x+y}{2\sqrt{y}} $$
If $x = y$, then
$$\sqrt{x} =\frac{x+y}{2\sqrt{y}} $$
Example: $$\sqrt{4} =\frac{4+4}{2\sqrt{4}} = 2$$
|
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|
Find the last two digit of $5^{121}*3^{312}$ The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$
and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of $5^{143}*3^{312}$ are $25$.
Don't know why $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.
Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$
|
I have no idea what the book is trying to do.
I guess Id do it this way.
$5^2\cdot 3^4 = 25 *81 = 2025 \equiv 25\equiv 5^2 \pmod {100}$.
So $(5^2 \cdot 3^4)^{78} \equiv 25^{78} \pmod {100}$.
SO $5^{156}\cdot 3^{312} \equiv 5^{156} \pmod {100}$.
Now $5^2 \equiv 25 \pmod {100}$ so by induction if $5^k \equiv 25 \pmod {100}$ then $5^{k+1} \equiv 5\cdot 25 =125\equiv 25 \pmod {100}$.
So for any $k, m \ge 2$ you have $5^k \equiv 5^m \equiv 25 \pmod {100}$.
So $5^{156} \equiv 5^{121} \pmod {100}$ and
$5^{121}\cdot 3^{312} \equiv 5^{156}\cdot 3^{312} \equiv 5^{156}\equiv 25 \pmod{100}$.
So the last two digits are $25$.
......
Actually, that's me trying to guess what the book is trying to do.
I'd do:
So $5^{k} \equiv 25$ for all $k \ge 2$.
Then for all $k \ge 2$ then $5^k \cdot 3 \equiv 25 \cdot 3 \equiv 75 \pmod {100}$.
And $5^k 3^2 \equiv 75 \cdot 3 = 225 \equiv 25 \pmod {100}$.
So by induction $5^k 3^m \equiv \begin{cases} 75 & m\ odd\\25 & m\ even\end{cases}$.
So $5^{121} 3^{312} \equiv 25 \pmod {100}$.
|
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|
Unique solutions for $\frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}=1,\quad\text{and}\quad a+b+c=1.$
Let $A,B,C$ be strictly positive real numbers satisfying $A+B+C=1$ and let $a, b, c $ be real variables. Suppose $a, b, c$ satisfy the following system of equations:
$$\frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}=1,\quad\text{and}\quad a+b+c=1.$$
It is clear that $a=A$, $b=B$ and $c=C$ is one solution to the system. Can I conclude that the solution is unique and there exists no other solutions?
|
The solution is unique only if $a,b,c > 0$. This is because by Titu's lemma: $1 = \dfrac{A^2}{a}+\dfrac{B^2}{b}+\dfrac{C^2}{c} \ge \dfrac{(A+B+C)^2}{a+b+c}= \dfrac{1^2}{1} = 1$, and you have equality which means the $=$ must hold, and this occurs when $A = ak, B = bk, C = ck \implies k =1 \implies a = A, b = B, c = C$ as claimed. For if they are not required to be positive then there is another solution, namely: $a = \dfrac{A^2-B^2}{1-C^2}, b = \dfrac{B^2-A^2}{1-C^2},c = 1$.
|
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|
For two odd primes $pLet us fix two primes $p,q$ with $2<p<q$. How can we find positive integers $a,b$ which solve the equation $a^2+b^4=pq$ without brute force?
Interestingly there exist sometimes two solutions:
*
*$5\cdot13=7^2+2^4=8^2+1^4$
*$5\cdot821=3^2+8^4=53^2+6^4$
*$17\cdot113=25^2+6^4=36^2+5^4$
In more rare cases I even obtain three solutions, e.g.:
*
*$73\cdot89=49^2+8^4=64^2+7^4=79^2+4^4$
Is there a systematic way to obtain the solutions $a,b$ directly? Or at least, can we establish a relationship between the two primes and these natural solutions? I generated a CSV file containing some more cases, which maybe help in finding patterns or connections between $p,q$ and $a,b$.
Can we anticipate (state in advance) how many positive integer solutions we will get depending on $p,q$?
|
With kind assistance I have received a useful hint for a direction to investigate, wich I would like to share with you all:
In the case that $p\equiv3\pmod4$ or $q\equiv3\pmod4$ no solution exist.
Let us set $p=r^2+s^2$ and $q=u^2+v^2$. If $p\equiv1\pmod4$ and $q\equiv1\pmod4$ we exactly obtain one solution $r>s>0$ for $p$ and one solution $u>v>0$ for $q$.
If we now have these unique solutions $p=r^2+s^2$ and $q=u^2+v^2$, then the product of both primes is $pq=(r^2+s^2)(u^2+v^2)=(ru+sv)^2+(rv-su)^2=(ru-sv)^2+(rv+su)^2$.
Consider $b^2=c$, other integer solutions for $pq=a^2+c^2=a^2+b^4$ do not exist, unless one of the four integers $ru+sv$, $|rv-su|$, $|ru-sv|$ and $rv+su$ is a perfect square.
To retrace what happens in practice, I generated variuos cases, where three solutions exist (feel free to download the CSV):
*
*$p\cdot q=233\cdot12409=256^2+41^4=1616^2+23^4=1681^2+16^4$
*$p\cdot q=89\cdot233=1^2+12^4=129^2+8^4=144^2+1^4$
*$p\cdot q=17\cdot4241=81^2+16^4=256^2+9^4=264^2+7^4$
In the first case we have $p=233=8^2+13^2$ and $q=12409=72^2+85^2$.
In the second case we have $p=89=5^2+8^2$ and $q=233=8^2+13^2$.
In the third case we have $p=17=1^2+4^2$ and $q=4241=4^2+65^2$.
Hypothetically, there could be four solutions, although so far I have only found these cases with a maximum of three solutions?
|
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|
Find constant $c$ such that all intersection points of two spheres have perpendicular tangent planes I have come to a problem in a multivariable calculus book that I'm having trouble with.
The problem statement is :
"Find a constant $c$ such that for any point of intersection of the two spheres
$(x-c)^{2} + y^{2} + z^{2} = 3$ and $x^{2} + (y-1)^{2} + z^{2} = 1$, the corresponding
tangent planes will be perpendicular to each other."
So I define two functions :
\begin{align}
f(x,y,z) & = (x-c)^{2} + y^{2} + z^{2} - 3 \\
g(x,y,z) & = x^{2} + (y-1)^{2} + z^{2} - 1
\end{align}
We denote the partial derivative with respect to $x$ as $f_{x}$ and so on...
We see :
\begin{align}
f_{x}(x,y,z) & = 2(x-c) = 2x - 2c\\
f_{y}(x,y,z) & = 2y\\
f_{z}(x,y,z) & = 2z
\end{align}
and :
\begin{align}
g_{x}(x,y,z) & = 2x \\
g_{y}(x,y,z) & = 2(y-1) = 2y - 2 \\
g_{z}(x,y,z) & = 2z
\end{align}
I assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. So for every intersection point (x,y,z)
we have :
\begin{equation}
\bigtriangledown f(x,y,z) \cdot \bigtriangledown g(x,y,z) = 0
\end{equation}
So :
\begin{align}
\require{cancel}
(2x - 2c, 2y, 2z) \cdot (2x, 2y-2, 2z) & = 0\\
(2x-2c)2x + (2y)(2y) - 2(2y) + 4z^{2} & = 0\\
4x^{2} - 4xc + 4y^{2} - 4y + 4z^{2} & = 0\\
4(x^{2} + y^{2} + z^{2}) - 4(xc + y) & = 0\\
\cancel{4} \left[ (x^{2}+y^{2}+z^{2}) - (xc + y) \right] & = 0\\
(x^{2}+y^{2}+z^{2}) - (xc + y) & = 0 \\
x^{2} + y^{2} + z^{2} & = xc + y
\end{align}
We also see :
\begin{align}
\require{cancel}
f(x,y,z) & = x^{2} - 2xc + c^{2} + y^{2} + z^{2} - 3 \\
& = (x^{2} + y^{2} + z^{2}) + (c^{2} - 2xc - 3) \\
g(x,y,z) & = x^{2} + y^{2} - 2y + \cancel{1} + z^{2} - \cancel{1} \\
& = (x^{2}+y^{2}+z^{2}) - 2y
\end{align}
and :
\begin{align}
f(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = -(c^{2}-2xc-3) = -c^{2} + 2xc + 3\\
g(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = 2y
\end{align}
So we have :
\begin{equation}
xc + y = -c^{2} + 2xc + 3 = 2y
\end{equation}
It is here that I am stuck. It seems that the goal here would be to obtain an expression for $c$ that doesn't include $x$ or $y$, but I do not know how to obtain it.
Can someone help with this ?
|
Math is beautiful in that a good problem gives you everything that you need to solve it, and nothing more.
Taking this into consideration, let's see what we're given.
We have:
$$ (x-c)^2 + y^2 + z^2 = 3 $$
$$ x^2 + (y-1)^2 + z^2 = 1 $$
So, we can define two functions:
$$
f(x,y,z) = (x-c)^2 + y^2 + z^2 - 3
$$
$$
g(x,y,z) = x^2 + (y-1)^2 + z^2 - 1
$$
$$
$$
Taking the partial derivatives, we have:
$$ \frac{\partial f}{\partial x} = 2x-2c , \frac{\partial f}{\partial y} = 2y , \frac{\partial f}{\partial z} = 2z $$
$$ \frac{\partial g}{\partial x} = 2x , \frac{\partial g}{\partial y} = 2y-2 , \frac{\partial g}{\partial z} = 2z $$
Thus,
$$ \nabla f = (2x-2c,2y,2z) $$
$$ \nabla g = (2x,2y-2,2z) $$
You are correct to assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. Thus, we have:
$$ \nabla f(x,y,z) \cdot \nabla g(x,y,z) = 0 $$
$$ (2x-2c,2y,2z) \cdot (2x,2y-2,2z) = 0 $$
$$ 4x^2 - 4xc + 4y^2 - 4y + 4z^2 = 0 $$
Completing the square, we have:
$$ 4(x^2 - cx + \frac{1}{4} c^2) - c^2 + 4(y^2 - y + \frac{1}{4}) - 1 + 4z^2 = 0 $$
$$ 4(x - \frac{1}{2} c)^2 + 4(y - \frac{1}{2})^2 + 4z^2 = 1 + c^2 $$
Adding $ 1 + c^2 $ to both sides and simplifying, we have:
$$ 4x^2 - 4cx + c^2 + 4y^2 - 4y + 2 + 4z^2 = 2c^2 $$
$$
$$
Now we need to look back at what we're given. We know that
$$ (x-c)^2 + y^2 + z^2 = 3 $$
$$ x^2 + (y-1)^2 + z^2 = 1 $$
So, multiplying both equations by $ 2 $, we have:
$$ 2((x-c)^2 + y^2 + z^2) = 2(3) \implies 2x^2 - 4cx + 2c^2 + 2y^2 + 2z^2 = 6 $$
$$ 2(x^2 + (y-1)^2 + z^2) = 2(1) \implies 2x^2 + 2 - 4y + 2y^2 + 2z^2 = 2 $$
Substituting these equations into $ 4x^2 - 4cx + c^2 + 4y^2 - 4y + 2 + 4z^2 = 2c^2 $ , we can see that this simplifies to:
$$ 2c^2 + 2 = 8 $$
$$ 2c^2 = 6 $$
$$ c = \pm \sqrt3 $$
|
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|
Proof regarding convergence I am supposed to show that the sequence converges and to find the limit.
We've just started going over sequences and induction so I'm very limited in what I know/can use and I was wondering if the general pattern of the proof is okay and if it's correct.. Also it took me quiet a bit of time to prove it this way and I was wondering if there are any shortcuts I can do in the future?
$$ \{a_n\}=\left\{
\begin{array}{lcl}
a_1 = 3\\
a_{n+1} = 1+\sqrt{6+a_n}\\
\end{array}
\right.$$
First we show by induction $ \{a_n\} $ is increasing, that is, to prove that the following hold for every $n∈N$
$1+\sqrt{6+a_n}\ge a_n$
$\sqrt{6+a_n}\ge a_n-1$
$6+a_{n}\ge\left(a_{n}+1\right)^{2}$
$6+a_{n}\ge a_{n}^{2}-2a_{n}+1$
$0\ge a_{n}^{2}-3a_{n}-5$
Base case: $n=1$
$1+\sqrt{6+3}\ge 3$
$4\ge 3$
Inductive step: Assume $0\ge a_{n}^{2}-3a_{n}-5$, we show $0\ge a_{n+1}^{2}-3a_{n+1}-5$
$0\ge a_{n}^{2}-3a_{n}-5$
$6+a_{n}\ge a_{n}^{2}-2a_{n}+1$
$\sqrt{6+a_{n}}\ge a_{n}-1$
$0\ge a_{n}-1-\sqrt{6+a_{n}}$
$0\ge1+2\sqrt{6+a_{n}}+6+a_{n}-3-3\sqrt{6+a_{n}}-5$
$0\ge\left(1+\sqrt{6+a_{n}}\right)^{2}-3\left(1+\sqrt{6+a_{n}}\right)-5$
$0\ge a_{n+1}^{2}-3a_{n+1}-5$
Therefore by induction, $\{a_n\}$ is increasing for every $n∈N$.
Now we show by induction $\{a_n\}$ is bounded, that is, $a_n ≤ 4.2$ for every $n∈N$.
Base case: $n=1$, which is trivially true.
Inductive step: Assume $a_n ≤ 4.2$, we show $a_{n+1} ≤ 4.2$.
$a_n ≤ 4.2$
$6+a_{n}\le10.2$
$\sqrt{6+a_{n}}\le\sqrt{10.2}$
$\sqrt{6+a_{n}}\le3.2$
$1+\sqrt{6+a_{n}}\le4.2$
Therefore by induction $\{a_n\}$ is bounded from above.
As $\{a_n\}$ is both bounded and incresing for every $n∈N$, $\{a_n\}$ converges and $\lim _{n\to \infty }\left(a_n\right)=L$ for some $L∈R$
As $\lim _{n\to \infty }\left(a_n\right)=L$, then $\lim _{n\to \infty }\left(a_{n+1}\right)=L$, therefore
$L\ =\ 1+\sqrt{6+L}$
$L\ -1=\ \sqrt{6+L}$
$\left(L\ -1\right)^{2}=\ 6+L$
$L^{2}-2L+1=\ 6+L$
$L^{2}-3L-5=\ 0$
$L = \dfrac{ 3 \pm \sqrt{(-3)^2 - 4*(-5)}}{ 2 }$
$L_1 = 4.19258$
$L_2 = −1.19258$
As $a_n ≥ 0$ for every $n∈N$, $L = 4.19258$
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Your proof is honestly very tedious and should be replaced ! You must make use of the recursive property of the sequence directly. Here is how induction works for your problem. Base case: $a_2 = 4 \ge 3 = a_1$ is true. Assume $a_n \ge a_{n-1}$ for $n \ge 2$, then using the step $a_n \ge a_{n-1} \implies a_{n+1} = 1+\sqrt{6+a_n} \ge 1 +\sqrt{6+a_{n-1}} = a_n$. Hence by induction $a_n \ge a_{n-1}$ for all $n$ and the sequence is monotonically increasing. Next you can show too by induction again that the sequence is bounded above by $5$ and below by $4$. Here again, base case $a_1 = 3 < 5$, assume $a_n < 5$, then $a_{n+1} = 1 +\sqrt{6+a_n} < 1 +\sqrt{6+5} = 1+\sqrt{11} < 1+\sqrt{16} = 1+4 = 5$. Thus $a_n < 5$ for all $n$. Thus the $\{a_n\}$ has a limit $L$ which is the solution of the equation: $L = 1+\sqrt{6+L}$. It can be solve by squaring both sides and note that $4 \le L \le 5$. Can you solve for $L$ ?
|
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|
Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).
By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.
|
1875 book by Reuschle. The technique was initiated by Gauss, some 30 years before Galois theory.
Indeed, on page 249 in Cox, Galois Theory, we find
$$ \sum_{w=1}^{16} \; (w|17) \; \zeta_{17}^w \; = \sqrt{17}. $$
Letting $\eta_0$ refer to the quantity of interest, calling the sum of the others $\eta_1$ as in Reuschle below, we see $\eta_0 + \eta_1 +1=0, $ while $\eta_0 - \eta_1 = \sqrt{17}.$ So $2\eta_0 +1 = \sqrt{17}.$
For any prime $p \equiv 1 \pmod 4$ we have
$$ \sum_{w=1}^{p-1} \; (w|p) \; \zeta_{p}^w = \sqrt{p}. $$
while prime $q \equiv 3 \pmod 4$ gives
$$ \sum_{w=1}^{q-1} \; (w|q) \; \zeta_{q}^w = i\sqrt{q}. $$
Cox refers to Ireland and Rosen for quadratic Gauss sums, chapter 6. The bits just preceding are I+R Theorem 1, on page 75.
Anyway, double your displayed quantity ( that is, drop the $1/2$), call it $\eta.$ Write out $\eta^2$ in terms of powers of $a,$ taking every exponent from $-8$ to $8$ as you began. Add up $\eta^2 + \eta$ and reduce.
$$ \eta = a + a^2 + a^4 + a^8 + a^9 + a^{13} + a^{15} + a^{16} $$
If you write it with exponents from $0$ to $16,$ the result of $\eta^2 + \eta$ comes out to ( exponents $\pmod {17}$)
$$ \eta^2 + \eta = 4a^{16} + 4a^{15} + 4a^{14} + 4a^{13} + 4a^{12} + 4a^{11} + 4a^{10} + 4a^9 + 4a^8 + 4a^7 + 4a^6 + 4a^5 + 4a^4 + 4a^3 + 4a^2 + 4a + 8
$$
But this is ( we split the final $8$ into $4+4$)
$$ 4 \left(a^{16} + a^{15} + a^{14} + a^{13} + a^{12} + a^{11} + a^{10} + a^9 + a^8 + a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 \right) +4 = 4
$$
because
$$ a^{16} + a^{15} + a^{14} + a^{13} + a^{12} + a^{11} + a^{10} + a^9 + a^8 + a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 = 0$$
All together, as Reuschle tells us, $$ \eta^2 + \eta - 4 =0 $$
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|
Prove $\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$ for $a^{2}+b^{2}+c^{2}+abc=4$
Prove:
$$\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$$ for $a,b,c>0$ such that $$a^{2}+b^{2}+c^{2}+abc=4.$$
I tried to note $a=\cos A$,... because of a known identity, I used $s=\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} \geq \frac{a}{b} +\frac{b}{c} +\frac{c}{a}$ and $s\geq \frac{a}{c} +\frac{b}{a} +\frac{c}{b}$, and then I added them. Another suggestion? Please!
I also noted $s=a+b+c,p=ab+bc+ac, r=abc$. The condion is equivalent with $s^2-2p+r=4$. Using the 2 inequalities which I mentioned above $\implies s\ge(a+b)(b+c)(a+c)/(2abc)-1$, and then is enought to prove that $2sp+15r^2\ge33r$, but I don't know if the last one is correct inequality. You can try to prove that. I'm desperate.
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With the following link $a^2+b^2+c^2+abc=4 \Rightarrow abc \leq 1$ thus the maximum value is $1$ and we assume that $a=1,b=1,c=1$. $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{15abc}{4}$, we substitut $1$ we get exactly $\frac{27}{4}$, this is the maximum value that we could get. If $a,b,c$ is less that $1$ we could not reach $\frac{27}{4}$.So $\mathbf{a,b,c=1}$.
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Collatz Conjecture: Does it follow that since there is no $1$-cycle, if $ab > 1$, then $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be an integer? For me, one of the most interesting results of the Collatz Conjecture is that if an $m$-cycle exists, then $m > 68$. So, there are no non-trivial $1$-cycles, $2$-cycles, ..., up to $68$-cycles. The details behind this argument can be found in this paper by Simons & De Weger (2003).
Let:
*
*$\nu_2(x)$ be the 2-adic valuation of $x$.
*$u = 3^kt - 1$
*$b = \nu_2(u)$
*$a = \nu_2(\frac{u}{2^b}+1)$
If a $1$-cycle exists, then the maximum value of the cycle is equal to $3^kt - 1$ where $k,t$ are positive integer and the lowest value of the cycle is equal to $2^kt - 1$ characterized by the following:
$$u = \left(\frac{3}{2}\right)^a\left(\frac{u}{2^b} + 1\right)-1$$
So that:
$$\left(\frac{2}{3}\right)^a(u+1) = \frac{u}{2^b} + 1$$
And:
$$\frac{u(2^{a+b} - 3^a)}{3^a2^b} = \frac{3^a - 2^a}{3^a}$$
Which reduces to:
$$\frac{u}{2^b} = \frac{3^a - 2^a}{2^{a+b} - 3^a}$$
It seems to me that since no non-trivial $1$-cycle exists, it follows that if $a,b$ are positive integers where $ab > 1$, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not a positive integer.
Here's the argument:
(1) Assume that there exists positive integers $a,b$ with $ab > 1$ such that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is an integer.
(2) Let $x = \frac{3^a - 2^a}{2^{a+b} - 3^a}$
(3) Let $u = 2^bx$
(4) Let $k = a$, $t = \left(\frac{1}{2^a}\right)\left(\frac{u}{2^b}+1\right)$ which are both integers.
(5) So it follows that $u = 3^kt - 1$
(6) But then we have a contradiction since no $1$-cycle exists. Therefore, we can reject our assumption at step(1).
Is this argument valid? Does the result from Simons & De Weger (2003) prove that with the given assumptions, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not an integer?
Edit: I added the clarification that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be a positive integer.
@rukhin makes a great point. For positive integers $a=2$ and $b=1$, $\dfrac{3^a - 2^a}{2^{a+b} - 3^a} = -5$
This is correlated with the negative $1$-cycle with $k=1$ and $t=\left(\frac{1}{4}\right)\left(\frac{-10}{2} + 1\right) = -1$
The standard collatz sequence defined as:
*
*$T(x) = \begin{cases}
\frac{1}{2}(3x + 1), && \text{if }x\text{ is odd}\\
\frac{1}{2}x, && \text{if }x\text{ is even}\\
\end{cases}$
*$T_1(x) = T(x)$
*$T_{c+1}(x) = T(T_c(x))$
So that:
$$-1, -2, -1, \dots$$
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Say you start with $n_0=2^kt-1$, apply $a$ times the Collatz function $f(n)=\frac{3n+1}{2}$, than $b$ times the Collatz function $f(n)=\frac{n}{2}$, you can easily see that you end up with $$n=\frac{3^a}{2^{a+b}}n_0+\frac{3^a-2^a}{2^{a+b}}$$
or
$$n\cdot{2^{a+b}}-3^a\cdot n_0=3^a-2^a$$
and in the case of a cycle $n=n_0$
$$n=\frac{3^a-2^a}{2^{a+b}-3^a}$$
Note: $a=k$, the value $a = \nu_2(\frac{u}{2^b}+1)$ you gave is for the next iteration
Now in case of a cycle, it has been shown that (you can find it by noticing that $\frac{1}{2}<\frac{3^a}{2^{a+b}}<1$):
$$a+b=\lceil a\cdot \log_2(3)\rceil $$
So indeed, if there is such an integer $n=\frac{3^a-2^a}{2^{a+b}-3^a}$, it must be part of a 1-cycle, and you can conclude there is none (outside the trivial one), but it would only be valid within the above constraint on $a+b$.
Also notice that the question has already been raised.....by you
Is this argument valid for showing given integers $n>1$ and $2^m > 3^n$, $(2^m - 3^n) \nmid 3^n - 2^n$
Also note that a m-cycle does not mean m successive 1-cycles, so there is no such constraint on $a+b$ within the m-cycle (but in this case you don't have $n=n_0$ or the equation you mentioned)
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A limit of a given sequence Let $(a_n)_{n\ge1}$ a sequence of real positive numbers such that $a_n+\frac{n}{a_{n+1}^2} \le a_{n+1} \le \frac{a_{n-1}^2}{a_n}+\frac{n+1}{a_n^2}$ for any $n \ge 1$. Prove that $b_n = \frac{a_n}{\sqrt[3]{n^2}}$ is a convergent sequence and find its limit.
First of all, it's clear that $a_n$ is a increasing sequence from the first inequality given. So $a_n$ has a limit and it is either a finite number or infinity. If we suppose that $a_n$ is convergent and has a finite limit (let it be $l$), by using limit in the first inequality we obtain: $l+\infty \le l$, which is false and it means that $\lim_{n \to \infty} a_n = \infty$, so we may apply Cesaro-Stolz for finding the limit of $b_n$, but here I didn't know how to continue. Can you help me?
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We have $a_n$ increasing and positive, and
$$
a_{n} \leq a_{n+1} - \frac{n}{a_{n+1}^2},
$$
Thus $a_{n+1}^3 \geq n$, and
$$
\frac{a_{n-1}^2}{a_n} + \frac{n+1}{a_n^2} \leq a_n + \frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2},
$$
and using $a_{n}^3 \geq (n-1)$
$$
\frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2} \leq \frac{(n-1)^2}{a_n^5} \left(1-\frac{n-3}{(n-1)^2}a_n^3 \right) \leq 2 \frac{(n-1)}{a_n^5} < 2\frac{1}{(n-1)^{2/3}}
$$
so altogether
$$
a_n + \frac{n}{a_{n+1}^2}\leq a_{n+1} \leq a_n + \frac{2}{(n-1)^{2/3}} .
$$
So
$$
\sum_{k=2}^n a_{k+1} -a_{k} \leq 2\sum_{k=2}^n k^{-2/3} \leq 2 \int_{1}^{n-1} \frac{1}{t^{2/3}} dt = 6 n^{1/3}-6,
$$
So we have obtained that
$$
(n-1)^{1/3}\leq a_n \leq 6(n-1)^{1/3}+C,
$$
and the conclusion is clear.
|
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Two mysterious missing angles in the sine values of acute angle list? There is a well known list of 5 for trig values of special angles between 0$^o$ to $90^o$:
$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$
, $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$
, $\sin 60^o =\frac{\sqrt {\color{blue}{3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{blue}{4}}}{2}$
If wee add 15$^o$ and 75$^o$ , we can make the angles evenly spaced by 15$^o$ and expand the list of 5 to a list of 7:
$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$,$\sin 15^o =\frac{\sqrt {\color{red}{2-\sqrt3}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$
, $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$
, $\sin 60^o =\frac{\sqrt {\color{blue}{3}}}{2}$,$\sin 75^o =\frac{\sqrt {\color{red}{2+\sqrt 3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{blue}{4}}}{2}$
Reformat the list :
$\sin 0^o =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2}$,$\sin 15^o =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}$
, $\sin 45^o =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}$
, $\sin 60^o =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}$,$\sin 75^o =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}$
Now, we see a nice pattern of $\frac{\sqrt {{2\color{red}{\pm}\sqrt i}}}{2}$ with i=0,1,3,4.
But wait, why is 2 missing from this i list? Seems like we have two special angles lost. With them, we can expand this to list of 9. Where are these two mysterious angles? Can you find them and prove they fit into the pattern? Or, do you find ways to expand this list even further to a list of 11,13,...and find something amazing?
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The missing angles are:
$$\sin 22.5º = \frac{\sqrt{2 - \sqrt2}}{2}$$
$$\sin 67.5º = \frac{\sqrt{2 + \sqrt2}}{2}$$
There are other values listed here on Wikipedia, but the pattern is not obvious. For instance, $\sin 18º = \frac{\sqrt5 - 1}{4}$, but $\arcsin \frac{\sqrt5 + 1}{4}$ is not $90º - 72º = 36º$ but $54º$. In fact, $\sin 36º = \frac{\sqrt{10 - \sqrt{20}}}{4}$ and $\sin 72º = \frac{\sqrt{10 + \sqrt{20}}}{4}$.
The nice numbers for the sines of multiples of $18º$ come from the fact that $\sin (3\pi/10) = \cos (2 \pi/10)$, and solving $\sin(3x) = \sin(2x)$ using the triple-angle and double-angle formulas gives a nice quadratic in disguise after substituting $u = x^2$. The formulas involving $\sin(4x)$ and so on do not reduce into quadratics, and so if they can be reduced into radicals at all, the forms will be more complicated.
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How to calculate the sum of a sequence of reciprocal of factorial I wonder how to calculate the sum of the below sequence
\begin{align}
S & = \frac{1}{2!} - \frac{2}{3!} + \frac{3}{4!} - \frac{4}{5!} + \frac{5}{6!} - ... \\
& = \frac{2 - 1}{2!} - \frac{3 - 1}{3!} + \frac{4 - 1}{4!} - \frac{5 - 1}{5!} + ...\\
& = 1 - \frac{1}{2!} - \frac{1}{2!} + \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} - \frac{1}{4!} + \frac{1}{5!}...\\
& = 1 - 2 * (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}...）\\
& = ？
\end{align}
There is a similar sequence, which is easy to solve
\begin{align}
S & = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \frac{5}{6!} + ... \\
& = \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \frac{4 - 1}{4!} + \frac{5 - 1}{5!} + ...\\
& = 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{4!} - \frac{1}{5!}...\\
& = 1 - \frac{1}{n!}
\end{align}
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$$\sum_{n!=2}^\infty (-1)^{n-1}\frac1{n!} = 1 - \frac1e.$$
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How to derive the sum-to-product formula for cos(a-b)-cos(a+b)? I know how to derive the sum-to-product formulas for cos+cos, sin+cos, and cos+sin. But I cannot seem to do it for cos-cos. My problem is the negative out front. I cannot product it from any of my proofs.
Here is my attempt:
I know that cos(a-b) = cosacosb + sinasinb
I know that cos(a+b) = cosacosb - sinasinb.
If I subtract these, I get cos(a-b)-cos(a+b) = 2sinasinb
However, my book tells me that cosa - cosb = -2sin(a+b / 2)sin(a-b /2)...
Where does the negative come from?
additional effort
My book also says I can make a + b = u and a - b = v. By adding and subtracting these from one another, this gets me u+v / 2 and u-v / 2. But none of those come with a negative that I can use to make sin(-x) = -sin(x).
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Develop the right side:
$$-2\sin\frac{a+b}2\,\sin\frac{a-b}2=-2\left[\sin\frac a2\cos\frac b2+\sin\frac b2\cos\frac a2\right]\left[\sin\frac a2\cos\frac b2-\sin\frac b2\cos\frac a2\right]=$$
$$\stackrel{\text{squares difference}}=-2\left[\sin^2\frac a2\,\cos^2\frac b2-\sin^2\frac b2\,\cos^2\frac a2\right]=$$
$$=-2\left[\left(1-\cos^2\frac a2\right)\cos^2\frac b2-\left(1-\cos^2\frac b2\right)\cos^2\frac a2\right]=$$
$$=-2\left[\cos^2\frac b2-\cos^2\frac a2\right]=-2\left[\cos^2\frac b2-\sin^2\frac b2+\sin^2\frac b2-\cos^2\frac a2+\sin^2\frac a2-\sin^2\frac a2\right]=$$
$$=-2\left[\cos b-\cos a+\sin^2\frac b2-\sin^2\frac a2\right]=-2\left[\cos b-\cos a-\left(\frac12-\sin^2\frac b2\right)+\frac12-\sin^2\frac a2\right]=$$
$$=-2\left[\cos b-\cos a-\frac12\cos b+\frac12\cos a\right]=\cos a-\cos b$$
We used several times in the above the super-identity
$$\cos2x=\begin{cases}\cos^2x-\sin^2 x\\{}\\2\cos^2 x-1\\{}\\1-2\sin^2x\end{cases}$$
Can you spot the places where we used this?
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Solving $|2x-5| + |7-2x| =2 $ $$|2x-5| + |7-2x| =2 $$
I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-
*
*1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.
*2nd case, where $2.5<x\le3.5$ :-
$$2x-5+2x-7 = 2 \Rightarrow x=3.5$$
We can say that one solution exists in this range.
*3rd case, where $x>3.5$ :-
$$2x-5+7-2x = 2 \Rightarrow 2=2$$
We can say that infinite solutions exist in this case.
So, as per above calculations, what I can conclude that for all $x\geq2.5 $ solution exists.
But the solution provided in the book says that $|2x-5| + |7-2x| =2 $ will only be valid when $2.5 \leq x \leq 3.5$.
Can someone please help me out on how to decide the solutions and solution ranges for these type of questions?
Thanks in advance !
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The short cut way: let $y=2x$, what are the values that $y$ can take such that the distance from $5$ and the distance from $7$ are exactly $2$. If $y$ is less than $5$, then the distance from $7$ is beyond $2$. SImilarly if $y$ is more than $7$, then the distance from $5$ is beyond $2$. We can easily check that any values of $y$ between $5$ and $7$ satisfies the condition and hence $x$ is in between $2.5$ and $3.5$.
Now, as a practice, let's work it out.
*
*If $x \le 2.5$, then $|2x-5|=5-2x$, $|7-2x|=7-2x$.
Hence $$5-2x+7-2x=2$$
$$10=4x$$
and hence the only solution is $x=2.5$.
*
*If $2.5 < x <3.5$, then $|2x-5|=2x-5$ and $|7-2x|=7-2x$.
$$2x-5+7-2x=2$$
which is always true.
*
*If $x \ge 3.5$, $|2x-5|=2x-5$ and $|7-2x|=2x-7$
$$4x-12=2$$
$$2x=7$$ the only solution is $3.5$.
Hence taking union: $\{2.5\} \cup (2.5, 3.5) \cup \{3.5\}=[2.5,3.5]$ is the desired result.
|
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Help with $\int \frac{1}{(1-t^2)t^2} \, dt$ $$\int \frac{1}{(1-t^2)t^2} \, dt$$
By using partial fractions I get:
$$\frac{1}{(1-t)(1+t)t^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+1} + \frac{D}{1-t}$$
$$1 = t^3 (-A-C+D) + t^2 (-B+C+D) + At+ B$$
$$-A-C+D = 0$$
$$-B+C+D=0$$
$$A=0$$
$$B=1$$
So, $A=0, B=1, C=\frac{1}{2}, D = \frac{1}{2}$
Then by replacing the variables with the results we get:
$$\int \frac{1}{(1-t^2)t^2} \, dt = \int \frac{1}{t^2} \, dt + \frac{1}{2} \int \frac{1}{t+1} \, dt + \frac{1}{2} \int \frac{1}{1-t} \, dt$$
$$- \frac{1}{t} + C + \frac{1}{2} \ln|t+1| + C + \frac{1}{2} \ln|1-t| + C$$
$$- \frac{1}{t} + \frac{1}{2} \ln|t+1| + \frac{1}{2} \ln|1-t| + C$$
Is this correct?
|
We may utilize $$\dfrac1{t^2(1-t^2)}=\dfrac{(1-t^2)+t^2}{t^2(1-t^2)}=\dfrac1{t^2}+\dfrac1{1-t^2}$$
Finally $$\dfrac1{1-t^2}=\dfrac{1+t+1-t}{2(1-t)(1+t)}=?$$
|
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|
Finding the orthogonal trajectories of $y^2=x^2(1-cx)$ Here is my initial attempt:
$c=\frac{x^2-y^2}{x^3}$
$2y\frac{dy}{dx}=2x-3cx^2$
$-2y\frac{dx}{dy}=2x-\frac{3x^2-3y^2}{x}$
$-2xydx=(2x^2-3x^2-3y^2)dy$
$-2xydx=(-x^2-3y^2)dy$
$2xydx=(x^2+3y^2)dy$
$\frac{dy}{dx}=\frac{2xy}{x^2+3y^2}$
Let $y=ux$; $\frac{dy}{dx}=u+x\frac{du}{dx}$
$u+x\frac{du}{dx}=\frac{2ux^2}{x^2+3u^2x^2}$
$u+x\frac{du}{dx}=\frac{2ux^2}{x^2(1+3u^2)}$
$u+x\frac{du}{dx}=\frac{2u}{1+3u^2}$
$x\frac{du}{dx}=\frac{u-3u^3}{3u^2+1}$
$\frac{dx}{x}=\frac{3u^2+1}{u-3u^3}du$
$\int\frac{dx}{x}=\int\frac{3u^2+1}{u-3u^3}du$
$\ln x=\ln u-\ln (3u^2-1) +\ln K$
$\ln(3u^2x-x)=\ln (uK)$
$3y^2-x^2=yK$
The textbook says that the correct answer is $x^2+3y^2=c_1y$. Can someone find what part of the solution I messed up.
|
I find always a little sad that such issues aren't associated with a graphical representation in order to have a geometrical feeling of what really happens.
In particular here, I wondered how a family of conical curves (a family of ellipses tangent to the $x$-axis, in red) could have an orthogonal family with third degree curves (with the origin as a double point, in blue and cyan).
This is how.
Please note that the origin is an exceptional point for both families.
This figure has been realized using Matlab but the dynamical way Desmos plots the curves is preferable.
|
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|
Solve definite integral $\int_{\frac{\sqrt2}2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$ by variable substitution I have this integral
$$\int_{\sqrt{2}/2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$$
I have to solve it by substituting x.
I believe that the easiest way would be to substitute with $t = \sqrt{x^2+1}$. The other ways I tried didn't help me much.
so.. let $t = \sqrt{x^2+1}$. Suppose $\sqrt{6}/2 \leq t \leq 2$. In this case, we can say that $x = \varphi(t) = \sqrt{t^2-1}$. $\varphi(\sqrt{6}/2) = \sqrt{2}/2$ and $\varphi(2) = \sqrt{3}$.
Now we can substitute.
$$\int_{\sqrt{6}/2}^{2} {\dfrac{1}{\varphi(t)\sqrt{\varphi^2(t) + 1}}\varphi'(t)dt} =
\int_{\sqrt{6}/2}^{2} {\dfrac{dt}{t\sqrt{t^2-1}}}$$
But I'm stuck here. Actually the indefinite integral of the integrand is equal to $-\ln{\sqrt{t^2-1}}$, but then by using the fundamental theorem of calculus I get a wrong result. What am I doing wrong?
The right answer is $\ln{\dfrac{3+\sqrt{6}}{3}}$. Thanks in advance.
|
Note that, with the substitution $t=\sqrt{x^2+1}$
$$ x = \sqrt{t^2-1}, \>\>\>\>\>dx = \frac t{\sqrt{t^2-1}}dt$$
Then, the integral becomes
$$\int_{\frac{\sqrt{2}}2}^{\sqrt{3}} \dfrac{1}{x\sqrt{x^2+1}}dx
= \int_{\sqrt{\frac32}}^{2} \dfrac{1}{t^2-1}dt
= \frac12 \ln\frac{t-1}{t+1}\bigg|_{\sqrt{\frac32}}^{2}\\
= \frac12\ln\left( \frac13\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\right)=\ln\frac{3+\sqrt6}3
$$
|
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|
Why is $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = 3$ This is a problem from SASMO Grade 8 (Secondary 2) Sample Questions.
Solve for $x$
$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 3$
Answer: $x=6$
I have tried this on a calculator: the more $x$ we add, the closer we are to $3$. But how can we prove it, and is there a way to figure this out by hand?
|
Let $x=\sqrt{6 + \sqrt{6 + \sqrt{6 + …}}}$ Next, square both sides: $x^2=6 + \sqrt{6 + \sqrt{6 + …}}$ Then, subtract 6 from each side: $x^2-6=\sqrt{6 + \sqrt{6 + …}}$ Wait a minute..... the right side of the equation is x! Substituting in, we get $x^2-6=x$, and $x^2-x-6=0$. This is a simple quadratic we can solve, yielding the solutions $x=3$ and $x=-2$. But which solution is the real one? Actually, when we squared both sides of the original equation, we introduced an extranneous solution. This extranneous solution is -2, and notice that when we substitute it back into the original equation we get $-2=\sqrt{6-2}$ -> $-2=2$. This is obviously wrong, but it makes sense that we got this solution from the quadratic because when we square both sides we get $4=4$, which is correct. Therefore $x=\sqrt{6 + \sqrt{6 + \sqrt{6 + …}}}=3$
EDIT
Notice that by solving for x using arithmetic we are assuming that x is a finite number and not positive or negative infinity. We can show this by letting $a_k$ represent a 'nest' of $k$ $\sqrt6$'s (e.g. $a_3=\sqrt{6 + \sqrt{6 + \sqrt{6}}}$) and proving via induction that if $a_n$ is between 0 and 3 then $a_{n+1}$ is also between 0 and 3: notice that the rule $a_{n+1}=\sqrt{6+a_{n}}$ applies to all integers $n>1$. We then apply the following logic:
$0<a_{n}<3$
$6<6+a_{n}<9$
$\sqrt6<\sqrt{6+a_{n}}<3$
$0<a_{n+1}<3$
Since $a_1=\sqrt6$ is obviously between 0 and 3, we know the sequence does not diverge.
However, we are not done here! Just because we know that $a_n$ is finite for all n does not tell us that the sequence converges, e.g. $1-1+1-1+1-...$ does not converge although all partial expressions $1, 1-1, 1-1+1, 1-1+1-1, ...$ are finite. We can show the aforementioned sequence does not behave like this by showing $a_{n+1}>a_n$ for all n. Both sides are greater than 0 so we can square both sides. Using the previously stated recursion $a_{n+1}=\sqrt{6+a_{n}}$ the inequality is equivalent to saying $6+a_{n}>a_{n}^2$, so $0>a_{n}^2-a_n-6$. If we solve the quadratic we see that the inequality is true for all $a_n$ such that $-2<a_n<3$, which is true. Therefore we know the series converges to a finite value.
Q.E.D.
tl; dr: THE SERIES CONVERGES, AND IT INDEED CONVERGES TO 3.
|
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In which of the intervals is $\sqrt{12}$ In which of the intervals is $\sqrt{12}:$
a) $(2.5;3);$
b) $(3;3.5);$
c) $(3.5;4);$
d)
$(4;4.5)$?
We can use a calculator and find that $\sqrt{12}\approx3.46$ so the correct answer is actually b. How can we think about the problem without a calculator (if on exam for example)? I was able to conclude that $$\sqrt{9}=3<\sqrt{12}<\sqrt{16}=4,\\3<\sqrt{12}<4,\\\sqrt{12}\in\left(3;4\right).$$ How can I further constrict the interval? Thank you in advance!
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$3^2=9.$
$3.5^2 = \frac{35}{10} \times \frac{35}{10} = \frac{35 \times 35}{100} = \frac{1225}{100} = 12.25.$
$9 < 12 < 12.25.$
$\sqrt{9} < \sqrt{12} < \sqrt{12.25}.$
$3< \sqrt{12} < 3.5.$
|
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Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
\cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3\cdot2}{2\sqrt2\sqrt5}= \frac{3}{2\sqrt2}\cdot\cos\beta$$
This is all I did till now. Could you go further with this to answer?
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Use trig identity: $\sin^2\theta+\cos^2\theta=1$ $$\implies\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}=\frac{1}{\sqrt{10}}\quad \forall \quad 0\le\alpha\le \frac{\pi}{2}$$
$$\implies \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}=\frac{1}{\sqrt{5}}\quad \forall \quad 0\le\beta\le \frac{\pi}{2}$$
Now, use trig identity
$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$
$$\cos(\alpha+\beta)=\frac{3}{\sqrt {10}}\frac{2}{\sqrt 5}-\frac{1}{\sqrt {10}}\frac{1}{\sqrt 5}
=\frac{1}{\sqrt2}$$
$$\implies \alpha+\beta=\cos^{-1}\frac{1}{\sqrt 2}=\frac{\pi}{4}
\quad \quad\left(\because \cos(\alpha+\beta)\in[-1,1]\right)$$
$$\therefore \cos^{-1}{\frac{3}{\sqrt {10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= \color{blue}{\frac{\pi}{4}}$$
Or alternatively use trig identity
$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$
$$\sin(\alpha+\beta)=\frac{1}{\sqrt {10}}\frac{2}{\sqrt 5}+\frac{3}{\sqrt {10}}\frac{1}{\sqrt 5}
=\frac{1}{\sqrt2}$$
$$\implies \alpha+\beta=\sin^{-1}\frac{1}{\sqrt 2}=\frac{\pi}{4}
\quad \quad\left(\because \sin(\alpha+\beta)\in[-1,1]\right)$$
$$\therefore \cos^{-1}{\frac{3}{\sqrt {10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}=\color{red}{\frac{\pi}{4}}$$
|
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False proof that $\frac{13}{6}=0$
At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, its length increased by $\frac{a_3}{3}$ (call the new length $a_2$) with respect to $a_1$. When $3$ hours have elapsed since start, its length increased by $\frac{a_3}{4}$ with respect to $a_2$. What is the value of $\frac{a_3}{a_1}$?
At $t=0$, the length is $a_0=0$. At $t=1$ hour, the length is $a_1=\frac{a_3}{2}$. At $t=2$ hours, the length is $a_2=\frac{a_3}{2}+\frac{a_3}{3}$. At $t=3$ hours, the length is $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$. So
$$\frac{a_3}{a_1}=\frac{\frac{13}{12}a_3}{\frac{a_3}{2}}=\frac{13}{6}$$
is the answer. But notice that the red equation enables solving for $a_3$ which gives $a_3=0$. Therefore $\frac{a_3}{a_1}=0$ and, by transitivity, we have:
Conclusion: $\frac{13}{6}=0$
Where is the mistake?
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If $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$ then $a_3$$=0$
So $a_1$ = $a_{3}/2$ $=0 .$ But you can’t divide by $0$ .
|
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|
Proving Proposition 2.4.3. in Artin I am trying to prove Proposition 2.4.3. in Artin. He leaves this fact unproved. It states:
Let $x$ be an element of finite order $n$ in a group, and let $k$ be an integer that is written as $k = nq + r$ where $q$ and $r$ are integers and $r$ is in the range $0 \leq r < n$. Then:
(a) $x^k = x^r$; (b) $x^k = 1$ if and only if $r = 0$; (c) Let $d$ be the greatest common divisor of $k$ and $n$. The order of $x^k$ is equal to $n/d$.
Here is my attempt. Part (a) is fairly straightforward.
We have $k = nq + r$. $x$ has order $n$, so $x^n = e$, and raising $x$ to any multiple of $n$ also gives $e$ by exponent rules, so we have
\begin{align*}
x^k = x^{nq + r} = x^{nq} x^r = (x^n)^q x^r = e^q x^r = ex^r = x^r.
\end{align*}
I'm only confident on half of part (b).
If $r = 0$, then $k = nq$, so
\begin{align*}
x^k = x^{nq} = (x^n)^q = e^q = e.
\end{align*}
Conversely, suppose $x^k = 1$. Then
\begin{align*}
e = x^k = x^{nq + r} = x^{nq} x^r = (x^n)^q x^r = x^r.
\end{align*}
So $n \mid r$, but $0 \leq r < n$, so this forces $r = 0$.
I am most uncertain on part (c ) and do not know how to start the proof.
|
You proofs of parts a) and b) are correct.
For part c), note first that $$\left(x^k\right)^{n/d}=\left(x^n\right)^{k/d}=e^{k/d}=e$$ so that the order of $x^k$ divides $\frac nd$.
On the other hand, let $m$ be a positive integer such that $\left(x^k\right)^m=e.$ Then $x^{km}=e$ so that $n\mid km$, and the $$\frac nd \mid\frac kd m\tag1$$ Since $d=\gcd(n,k)$, $\frac nd$ and $\frac kd$ are relatively prime, so $(1)$ implies $$\frac nd\mid m$$
EDIT
In response to OP's comment. $d$ is the greatest common divisor of $n$ and $k$, so there are positive integers $a$ and $b$ such that $n=ad$, $k=bd$. Let $c=gcd(a,b)$. There are positive integers $e,f$ such that $a=ce$, $b=cf$, which gives $n=ecd$, $k=fcd$, so $cd$ is a common divisor of $n$ and $k$. Since $d$ is the greatest common divisor, we must have $c=1$, which is to say that $\frac nd$ and $\frac kd$ are relatively prime.
|
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Number of possible values of $4x-z$ if $x+y+z=20$ Given that $x, y,z$ are non negative integers such that $x+y+z=20$. If $S$ is the set of all possible values of $4x-z$.Find $n(S)$.
My try:
By stars and bars number of non negative integer solutions of $x+y+z=20$ is $\binom{22}{2}=210$
Among these $210$ ordered triplets of $(x,y,z)$ we need to find number of possible values taken by $4x-z$.
Obviously the least value taken by $4x-z$ is $-20$ when $x=0,z=20$.
The next value taken by $4x-z$ is $-19$ when $x=0,z=19$ and so on.
So once we fix $x=0$ ,since $z$ varies from $[0,20]$ number of values taken by $4x-z$ is $21$.
But when we fix $x=1$, we get some overlaps. For example when $x=1,z=18$ we have $4x-z=-14$ and this value has already been counted in the previous case when $x=0,z=14$.
So any way to count total number of values taken by $4x-z$ excluding overlaps?
|
Well the very least $x$ can be is $0$ and if so, you can have $z$ be any value from $0$ to $20$ inclusive (by letting $y = 20-z$). SO all (integer) values from $-20$ to $0$ are possible but no smaller are possible.
Now we just have to find all possible positive values.
Every positive integer is a multiple of $4$ minus either $0, 1,2$ or $3$ so if we set $z= 0,1,2$ or $3$ we can get set $x$ to any value from $1$ to $17$ (by letting $y = 20 - x-z$) so get all values up to $4\cdot 17 - 0= 68$. So all values between $-20$ to $68$ inclusive are possible.
Now we just have to find all possible higher values.
And we can do those one by one. If $n = 4x -z > 68$ and $z \ge 0$ then $x > 17$ so as $0\le x \le 20$ we can have $x = 18,19, 20$.
If $x = 18$ then $z = 20 -18 -y = 2-z$ so $z$ can be $0,1$ or $2$. So that allows $6\cdot 18 - 0,1,2$ or $72,71,70$ but not $69$.
If $x = 19$ then $z =1-y$ and $z$ can be $0$ or $1$ so that allows $76,75$ (but not $74$ or $73$.
And $x = 20$ then $z=0$ and we have $80$ is possible but $79,78,77$ are not.
So the total number is $-20,....1$ and $0$ and $1.... 68$ and then $3$ and $2$ and $1$ more. For a total of $20 + 1 + 68 + 3+ 2 + 1= 95$ values.
=====
Or more systematic.
$x$ con range from $0$ to $20$ and $z$ from $0$ to $20$ so $4x-z$ can run from $4\cdot 0 -20=-20$ to $4\cdot 20 - 0=80$
The times $n = 4a - b$ are not possible is if $a + b>20$
As $4a - b= 4(a+1)- (b+4)$ so if we have $n$ not possible so $a + b > 20$ we must also have either $(a+1)+(b-4)>20$ or $a+1 > 20$ or $b-4 < 0$. So wolog we can assume that if $n=4a-b$ is impossible that we are assuming that $0\le b < 4$. And therefore $a+ b >20$ and $a > 20-b$.
For $a=20$ there are $3$ impossible values values for $b$. For $a = 19$ there are $2$ impossible values. For $a=18$ there is $1$. ANd for $a \le 17$ there are none.
So all values between $-20...,80$ are possible except for $1+2+3 = 6$ impossible values.
So there are $101-6 = 95$ possible values.
|
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|
How to find reduction formula for $I_n=\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$ I have to find the reduction formula for the following :
$$\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$$
I took this integral from wikipedia. By using parts this is what I got.
${2a^{-1}(px+q)^{n}\sqrt{ax+b}-2pna^{-1}\int\{px+q}^{n-1}\sqrt{ax+b}dx$
The answer is in this link in reduction formula table:
https://en.m.wikipedia.org/wiki/Integration_by_reduction_formulae
Any help is appreciated. Thanks in advance.
|
We first perform integration by parts on $\sqrt{ax+b}$.
\begin{aligned}
I_{n} &=\int \frac{(p x+q)^{n}}{\sqrt{a x+b}} d x \\
&=\frac{2}{a} \int(p x+q)^{n} d(\sqrt{a x+b}) \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int(p x+q)^{n-1} \sqrt{a x+b} d x
\end{aligned}
Rationalization gives back our integrals. $$
\begin{aligned}
I_{n} &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int \frac{(p x+q)^{n-1}(a x+b)}{\sqrt{a x+b}} d x \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int \frac{(p x+q)^{n-1}\left[\frac{a}{p}(p x+q)+b-\frac{a q}{p}\right]}{\sqrt{a x+b}} d x \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a}\left(\frac{a}{p} I_{n}+\frac{b p-a q}{p} I_{n-1}\right)
\end{aligned}
$$
Rearranging gives
$$
\begin{aligned}
\left(2n+1\right) I_{n} &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}+\frac{2 n(a q-bp)}{a } I_{n-1} \\
I_{n} &=\frac{2 p(p x+q)^{n}}{(2 n+1)\sqrt{a x+b}}+\frac{2 n(aq-b p)}{2 n+1} I_{n-1},
\end{aligned}
$$
which is the reduction formula of $I_n$.
|
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|
Is the matrix totally unimodular? Let $A=\begin{bmatrix}1&0&1&1&0&0&1&0&0\\0&1&0&1&0&0&0&0&0& \\0&0&1&1&1&0&0&0&0\\0&0&-1&-1&0&1&-1&0&0\\0&0&0&-1&0&0&0&1&0\\0&0&-1&-1&0&0&0&0&1\end{bmatrix}$, is the matrix totally unimodular?
I realised that if I change the order of columns in A I could get change the matrix I would get $A=( B I_{6x6})$, but I wasn't sure if I could use this for anything.
|
Successively applying column-swapping, $C_3 \leftrightarrow C_5$, $C_4 \leftrightarrow C_6$, $C_7 \leftrightarrow C_8$ and $C_8 \leftrightarrow C_9$, we get,
$A^{\prime}=\begin{bmatrix}1&0&0&0&0&0&1&1&1\\0&1&0&0&0&0&0&1&0& \\0&0&1&0&0&0&1&1&0\\0&0&0&1&0&0&-1&-1&-1\\0&0&0&0&1&0&0&-1&0\\0&0&0&0&0&1&-1&-1&0\end{bmatrix}=\left[I_6 \mid B\right]$, where $B=\begin{bmatrix}1&1&1\\0&1&0& \\1&1&0\\-1&-1&-1\\0&-1&0\\-1&-1&0\end{bmatrix}$
Now, from here, we have
*
*$A$ is unimodular iff $A^{\prime}$ is unimodular.
*$A^{\prime}$ is unimodular iff $B$ is unimodular.
Now, all remains to be shown is that $B$ is unimodular.
*
*All elements of $B \in \{-1,0,1\}$
*Any $2 \times 2$ submatrix of $B$ is of the form $\begin{bmatrix}x&x\\ x&x\end{bmatrix}$ or $\begin{bmatrix}x&x\\ -x&-x\end{bmatrix}$ or $\begin{bmatrix}y&0\\ x&x\end{bmatrix}$ or $\begin{bmatrix}0&y\\ x&x\end{bmatrix}$ or $\begin{bmatrix}x&x\\ y&0\end{bmatrix}$ or $\begin{bmatrix}x&x\\ 0&y\end{bmatrix}$ or $\begin{bmatrix}x&y\\ x&0\end{bmatrix}$ or $\begin{bmatrix}y&x\\ 0&x\end{bmatrix}$, or their transpose matrices, where $x,y \in \{-1,0,1\}$, clearly $det(B) = 0$ or $xy \in \{-1,0,1\}$
*Finally, show that all $3 \times 3$ submatrices (there are only ${6 \choose 3} = 20$ of them, each one having all cofactors $\in \{-1,0,1\}$) have determinant $\in \{-1,0,1\}$.
|
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|
$u_{n+1}=u_n^{u_n}$
*
*$u_0>1$
*$u_{n+1}=u_n^{u_n}$
We want to show that :
*
*the sequence $u_n$ diverges
*$ \sum \dfrac{1}{u_n}$ converges
*$ \exists N \in \mathbb{N} , \forall k : u_{N+k} \geq k+2$
*$\exists C >0, \forall n \geq N : \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k} \leq \dfrac{C}{u_{n+1}} $
*$\forall n \geq N : u_n \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k} \leq \dfrac{C} {u_n^{u_n-1} }$
My attempt :
$
\begin{align*}
\phi(x) &=x^x \\
\phi'(x)&= ( \ln x +1 ) x^x \\
\end{align*}
$
$\phi$ is decreasing on $]0, \dfrac{1} {e}]$ then increasing on $[ \dfrac{1} {e}, +\infty[$
1.
$
\begin{align*}
u_1&= u_0^{u_0} \\
u_2&=u_0^{ u_0 \times u_1} \\
u_n &> u_0^{u_0 ^n} \\
u_n &> \exp ( u_0^n \ln (u_0) ) \\
u_0 &>1 \\
u_n &\to \infty
\end{align*}
$
2.
$
\begin{align*}
w_n&= u_0^{ u_0^n } \\
u_n &> w_n \\
\dfrac{1}{u_n} &< \dfrac{1}{w_n} \\
\dfrac{w_{n+1} }{w_n} &= u_0^{ u_0^n (1-u_0) } \\
\dfrac{w_{n+1} }{w_n} &< 1 \\
\sum \dfrac{1}{u_n} &< \infty \\
\end{align*}
$
3.
$
\begin{align*}
u_n &\underset{ n \to \infty} \to \infty\\
\exists N, u_N &> 2 \\
j >2 &\implies j^2 > j+1 \\
u_{N+1} &=u_N^{ u_N }\\
&> u_N^2 \\
&> u_N +1 \\
u_{N+k} &> k+1 \\
\end{align*}
$
4.
$\begin{align*}
\forall k \leq 0, u_{N+1+k}&=u_{k+N}^{ u_{k+N} } \\
&> u_{N+1}^{k+1} \\
\dfrac{1}{ u_{N+1+k}} &< \dfrac{1}{ u_{N+1} u_{N+1}^k} \\
\sum_{k=0}^{ \infty}\dfrac{1}{ u_{N+1+k}} &< \sum_{k=0}^{ \infty} \dfrac{1}{ u_{N+1} u_{N+1}^k} \\
&< \dfrac{1}{ u_{N+1}} \sum_{k=0}^{ \infty}\dfrac{1}{ u_{N+1}^k} \\
&< \dfrac{1}{ u_{N+1}} C \\
\end{align*}
$
5.
We multiply the relation given in (4) by $u_n$
|
We can show by induction that $u_n >u_0^{u_0^{n}}$ for all $n$. Since $u_0 >1$ this will certainly imply converegence of $\sum \frac 1 {u_n}$.
[$u_0^{n}=(1+(u_0-1))^{n}> (u_0-1)n$ by Binomial expansion so $u_0^{u_0^{n}}>u_0^{(u_0-1)n}$. Apply ratio test].
|
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|
Find integral $\int_{0}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx$ Find $$\int\limits_{0}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx$$
$$
\lim_{a\rightarrow 0^+}\int\limits_{a}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx=\lim_{a\rightarrow 0^+}-\int\limits_{0}^{1}\left ( 1-x \right ){}_3F_2\left ( 1,1,1;2,2;x^2-x \right )dx=$$
$$= -\sum\limits_{n=0}^{\infty }\frac{\Gamma \left ( n+1 \right )\Gamma \left ( n+1 \right )\Gamma \left ( n+1 \right )}{\Gamma \left ( n+2 \right )\Gamma \left ( n+2 \right )}\frac{\left ( -1 \right )^n}{n!}\int\limits_{0}^{1}x^n\left ( 1-x \right )^{n+1}dx=$$
$$\bigstar \; \int\limits_{0}^{1}x^n\left ( 1-x \right )^{n+1}dx=B\left ( n+1,n+2 \right )=\frac{1}{2}\frac{\left ( 1 \right )_n}{2^{2n}\left ( \frac{3}{2} \right )_n}$$
$$=-\frac{1}{2}\sum\limits_{n=0}^{\infty }\frac{\left ( 1 \right )_n\left ( 1 \right )_n\left ( 1 \right )_n\left ( 1 \right )_n}{\left ( \frac{3}{2} \right )_n\left ( 2 \right )_n\left ( 2 \right )_n}\frac{\left ( -1/4 \right )^n}{n!}=-\frac{1}{2}{}_4F_3\left ( 1,1,1,1;\frac{3}{2},2,2;-\frac{1}{4} \right )$$
$$\bigstar \; {}_4F_3\left ( 1,1,1,1;\frac{3}{2},2,2;-\frac{1}{4} \right )=\frac{4}{5}\zeta \left ( 3 \right )$$
$$\lim_{a\rightarrow 0^+}\int\limits_{a}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx=-\frac{2}{5}\zeta \left ( 3 \right )$$
I doubt I calculated the integral correctly ...
I have not found any other way but to solve through hypergeometric series ...
Did I make the right decision?
Can you solve it easier?
|
A simpler way to evaluate it is to use the following identity:
$$\boxed{\int _0^1\frac{x\ln ^n\left(t\right)}{1-xt}\:dt=\left(-1\right)^nn!\operatorname{Li}_{n+1}\left(x\right)}$$
Applying it to the desired integral yields:
$$\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=\int _0^1\ln \left(t\right)\int _0^1\frac{1-x}{1+x\left(1-x\right)t}\:dx\:dt$$
$$=-2\int _0^1\frac{\ln \left(t\right)\arctan \left(\frac{\sqrt{t}}{\sqrt{-4-t}}\right)}{\sqrt{-4-t}\sqrt{t}}dt=2\int _0^1\frac{\arctan ^2\left(\frac{\sqrt{t}}{\sqrt{-4-t}}\right)}{t}\:dt=-2\int _0^1\frac{\operatorname{arctanh} ^2\left(\sqrt{\frac{t}{4+t}}\right)}{t}\:dt$$
$$=-4\int _0^{\frac{1}{\sqrt{5}}}\frac{\operatorname{arctanh} ^2\left(x\right)}{x\left(1-x^2\right)}\:dx=-\frac{1}{2}\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{x}\:dx-\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{1-x}\:dx$$
$$=-\frac{4}{3}\ln ^3\left(\phi \right)-2\sum _{k=1}^{\infty }\frac{1}{k^3}+2\sum _{k=1}^{\infty }\frac{1}{k^3\:\phi ^{2k}}+4\ln \left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k^2\:\phi ^{2k}}+4\ln ^2\left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k\:\phi ^{2k}}$$
$$=\frac{8}{3}\ln ^3\left(\phi \right)-2\zeta \left(3\right)+2\operatorname{Li}_3\left(\frac{1}{\phi ^2}\right)+4\ln \left(\phi \right)\operatorname{Li}_2\left(\frac{1}{\phi ^2}\right)$$
By the use of standard functional equations we have:
$$=-\frac{4}{3}\ln ^3\left(\phi \right)-2\zeta \left(3\right)+\frac{8}{5}\zeta \left(3\right)+\frac{4}{3}\ln ^3\left(\phi \right)-\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)+\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)$$
And thus:
$$\boxed{\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=-\frac{2}{5}\zeta \left(3\right)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
In an acute angled triangle, prove that $\sin A + \sin B + \sin C > 2$, where $0Is there any way to prove this? When I check intuitively by taking $A=0$ and $B=C=\frac{\pi}{2}$, the value of the expression becomes $2$ and as I changed the angles the value kept increasing in $0$ to $\frac{\pi}{2}$. I tried using Jensen's inequality but we get $\sin A + \sin B + \sin C <\frac{3\sqrt{3}}{2}$, which is of no use.
|
$$\sin A + \sin B + \sin C=\sin A +\sin B + \sin(\pi-A-B)$$
$$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+\sin(A+B)$$
$$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+2\sin(\frac{A+B}{2})\cos(\frac{A+B}{2})$$
$$=2\sin(\frac{A+B}{2})\Bigl(\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})\Bigr)$$
$$=4\sin(\frac{A+B}{2})\Bigl(\cos\frac{A}{2}\cos\frac{B}{2}\Bigr)$$
$$=4\sin(\frac{\pi-C}{2})\Bigl(\cos\frac{A}{2}\cos\frac{B}{2}\Bigr)$$
$$=4\cos\frac{C}{2}\cos\frac{A}{2}\cos\frac{B}{2}$$
$$=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$
$$0<A<B<\frac{\pi}{2}$$
$$0<\frac{A}{2}<\frac{B}{2}<\frac{\pi}{4}$$
$$4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} >4\cos\frac{C}{2}$$
when $A=0, B=0$.
During this situation, $C=\frac{\pi}{2}$
So,
$$\sin A + \sin B + \sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} >4\cos\frac{\frac{\pi}{2}}{2}=2\sqrt{2}>2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respectively (note since this is a geometry question the lengths of a side cannot be negative so the negative solution sets do not matter)
|
Let
\begin{align}
a^2+9&=k
\tag{1}\label{1}
,\\
b^2+16&=k
\tag{2}\label{2}
,\\
1+(a+b)^2&=k
\tag{3}\label{3}
\end{align}
for some $k>0$.
Then \eqref{3}$-$\eqref{1}$-$\eqref{2} gives
\begin{align}
ab &= 12-\tfrac12\,k
\tag{4}\label{4}
,\\
a^2b^2 &= (12-\tfrac12\,k)^2
\tag{5}\label{5}
.
\end{align}
On the other hand, from \eqref{1},\eqref{2}
we have
\begin{align}
a^2b^2 &= (k-9)(k-16)
\tag{6}\label{6}
.
\end{align}
Combination of \eqref{5} and \eqref{6}
results in equation in $k$:
\begin{align}
k(3k-52)&=0
\tag{7}\label{7}
,
\end{align}
and since $k>0$, we have the only option
\begin{align}
k&=\frac{52}3
\tag{8}\label{8}
.
\end{align}
Then \eqref{1} and \eqref{2} gives possible values of $a$
and $b$:
\begin{align}
a&=\pm \tfrac53\sqrt3
\tag{9}\label{9}
,\\
b&=\pm \tfrac23\sqrt3
\tag{10}\label{10}
.
\end{align}
And since $a,b$ also must be positive, the solution is ready.
|
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|
How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$? My question is how can i expand
$$\sin^2 A + \sin^4 A = 1$$
into:
$$1 + \sin^2A = \tan^2A$$
I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometry. it is one of the extra test question from my textbook. I don't need it but cant control curiosity. so pls help me.
Thanks in advance!
EDIT:
found the solution, dropping it here,
\begin{align}
\sin^2 A + \sin^4 A & = 1 \\
\sin^4 A & = 1 - sin^2 A \\
\sin^2 A . \sin^2 A & = cos^2 A \\
\sin^2 A . (1 - \cos^2 A) & = cos^2 A \\
\sin^2 A - \sin^2 A.\cos^2 A & = cos^2 A \\
\sin^2 A & = cos^2 A + \sin^2 A.\cos^2 A \\
\sin^2 A & = \cos^2 A(1 + \sin^2 A) \\
1 + \sin^2 A & = \cfrac{\sin^2 A}{\cos^2 A} \\
1 + \sin^2 A & = \tan^2 A \\
\end{align}
|
Writing $\sin^2A=a,$ we have $a+a^2=1$
We need $$1+a=\dfrac a{1-a}$$
As $1-a\ne0,1+1^2\ne1$ $$\iff a=(1-a)(1+a)\iff a=1-a^2\iff a+a^2=1$$
Done!
|
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|
Volume inside paraboloid and sphere
Find the volume inside the paraboloid $az=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2a^2$
Since we have $x^2+y^2$ I thought about polar coordinates , but before that I showed the functions as $z_1=\frac {x^2+y^2}{a}$ for the paraboloid and $z_2=\sqrt{2a^2-x^2-y^2}$ and I am not sure if this is a right way to see it but if $x$ and $y$ are close to zero we can then see that the sphere is above.
$\iint z_2(x,y)-z_1(x,y)dA$ should be the integral that is the volume
$\iint(\sqrt{2a^2-x^2-y^2})-(\frac {x^2+y^2}{a})dA$ here it seems like the right choice to go with polar coordinates but I got stuck in this part , $\iint(\sqrt{2a^2-(x^2+y^2)})-(\frac {x^2+y^2}{a})dA$
we know that
\begin{cases}
x=rcos\theta\\
y=rsin\theta\\
r^2=x^2+y^2\\
tan\theta=\frac{y}{x}
\end{cases}
$\iint_{r\theta}(\sqrt{2a^2-r^2})-(\frac {r^2}{a})rdrd\theta$ according to the way I drew it I think that $\theta$ should be $0\leq \theta \leq \pi$ But I cannot find $r$.
Would appreciate any help and tips , other approaches are always great thank you!
|
The intersection between the sphere and the paraboloid takes place when $2a^2-z^2=az$, that is, when $z=a$. So, if $(x,y,z)$ belongs to the region whose volume we are trying to compute, then:
*
*if $z\leqslant a$, then $0\leqslant x^2+y^2\leqslant az$;
*if $z\geqslant a$, then $0\leqslant x^2+y^2\leqslant 2a^2-z^2$.
So, compute$$\int_0^{2\pi}\int_0^a\int_0^{\sqrt{az}}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta+\int_0^{2\pi}\int_a^{\sqrt2a}\int_0^{\sqrt{2a^2-z^2}}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta\left(=\frac{ \left(8 \sqrt{2}-7\right) \pi a^3}6\right).$$
|
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|
$4k^2+n$ is prime Find all $n$ positive integers such that $4k^2+n$ is prime for every nonnegative integer $k$ less than $n$
My progress $k=0$ we get $n$ is a prime number
let $k=n-l$ we get $n+4n^2+4l^2-8nl$ is a prime number
which means $l$ doesn't divide $n(4n+1)$ since $n$ is prime
$l$ doesn't divide $(4n+1)$ for every $l$ less than $n$
so suppose $4n+1=ab \leq n^2, a,b \gt 1, n(n-4) \leq 1$ this gives $4$ values still have not tried them
other than that we get $4n+1$ is prime
if $n \equiv 1 \pmod 4$ then $4(\frac{n-1}{4})^2+n = (\frac{n+1}{2})^2$
contradiction so whats left is $n=3 mod 4$
|
Note $n$ must be an odd integer so $n \equiv 1, 3 \pmod{4}$. As you already stated, for primes $n \equiv 1 \pmod{4} \implies n = 4j + 1, \; j \in \mathbb{N}$, then $k = j$ gives
$$4j^2 + 4j + 1 = (2j + 1)^2 \tag{1}\label{eq1A}$$
As $n \gt 1 \implies j \gt 0$, this is not a prime. Thus, this only leaves $n \equiv 3 \pmod{4}$. As Robert Shore's question comment stated, $n = 3$ works, and as I stated, $n = 7$ also works.
For primes $n \gt 7$, first consider $n \equiv 3 \pmod{8} \implies n = 8j + 3, \; j \in \mathbb{N}$. Then $k = j$ gives
$$4j^2 + 8j + 3 = (2j + 3)(2j + 1) \tag{2}\label{eq2A}$$
Since $j \gt 0$, this is not prime. This leaves checking $n \equiv 7 \pmod{8}$, with it being done in $2$ parts. First, $n \equiv 7 \pmod{16} \implies n = 16j + 7, \; j \in \mathbb{N}$. Here, $k = j$ gives
$$4j^2 + 16j + 7 = (2j + 1)(2j + 7) \tag{3}\label{eq3A}$$
As $j \gt 0$, this is not prime. Second, $n \equiv 15 \pmod{16} \implies n = 16j + 15, \; j \in \mathbb{N}$. Using $k = j$ gives
$$4j^2 + 16j + 15 = (2j + 3)(2j + 5) \tag{4}\label{eq4A}$$
with this also not being prime.
Since all of the remaining possibilities for odd prime integers $n$ have been covered, this shows $n = 3, 7$ are the only solutions.
|
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|
Find a lower bound for a Riemann integral. Let $a,b$ be strictly positive constants and let $p\in(1/4,3/4)$. Is there $c>0$ such that
$$\int_0^1(1-t^a)^{1/2+p}[2t-(2+b)t^{b+1}]dt>c \quad ?$$
*The constant $c>0$ may depend on $a,b$ or $p$.
I tried several values for $(a,b,p)$, and always observed a positive integral. So, I'm conjecturing that the integral is strictly positive.
Intuitively, $2t-(2+b)t^{b+1}$ is a concave function on $[0,1]$ which is weighted by a factor $(1-t^a)^{1/2+p}$. This factor is less than one, and decreases as $t\to 1$, and so it pushes the right tail of the function (which is negative) closer to zero. So it makes sense to think about a positive integral.
I wrote the integral in terms of Beta functions but didn't succeed.
|
Firstly, note that
\begin{align}
\int_0^1\lvert 1-t^a\rvert^{1/2+p}\lvert 2t-(2+b)t^{b+1}\rvert dt\leq\int_0^12t-(2+b)t^{b+1}=2<\infty.
\end{align}
Suppose that $1/4<p<1/2$. Now, using the Fubini's Theorem and the Binomial Theorem for the fractional exponent $q:=1/2+p$ ,
\begin{align}
&\int_0^1(1-t^a)^q[2t-(2+b)t^{b+1}]dt=\int_0^1\sum_{k=0}^\infty \binom{q}{k}(-t^a)^k[2t-(2+b)t^{b+1}]dt\\
=&\sum_{k=0}^\infty \binom{q}{k}(-1)^k \int_0^1\ t^{ak}[2t-(2+b)t^{b+1}]dt=\sum_{k=0}^\infty \binom{q}{k}(-1)^k\bigg[\frac{2}{ak+2}-\frac{2+b}{ak+2+b}\bigg]\\
=&\sum_{k=1}^\infty \binom{q}{k}(-1)^{k+1}\bigg[\frac{bak}{(ak+2)(ak+2+b)}\bigg]\\
=&\sum_{k\text{ odd}} \binom{q}{k}\bigg[\frac{bak}{(ak+2)(ak+2+b)}\bigg]-\binom{q}{k+1}\bigg[\frac{ba(k+1)}{(a(k+1)+2)(a(k+1)+2+b)}\bigg] \\
=&\sum_{k\text{ odd}} \binom{q}{k}ab\bigg[\frac{k}{(ak+2)(ak+2+b)}+\frac{(k-q)}{(ak+a+2)(ak+a+2+b)}\bigg] \ *obs. (k-q\geq k-1)\\
\geq & \sum_{k\text{ odd}} \binom{q}{k}ab\bigg[\frac{ak(2ak^2+ak+6k+2bk+a)+2a+4k+2bk}{(ak+2)(ak+2+b)(ak+a+2)(ak+a+2+b)}\bigg] \quad *obs. \binom{q}{k}\geq 0,\text{if } k \text{ odd}\\
>&\binom{q}{1}ab\bigg[\frac{a(2a+a+6+2b+a)+2a+4+2b}{(a+2)(a+2+b)(a+a+2)(a+a+2+b)}\bigg]>0.
\end{align}
where $\binom{q}{k}=q(q-1)\dotsc(q-k+1)/k!$.
When $1/2\leq p\leq 3/4$, similar argument can be applied rewriting the sum by means of even indices.
|
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|
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right) \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
|
This is a partial answer which leaves one integral unevaluated. I'm not sure if it really helps or if it is just a reformulation replacing an unevaluated sum by an unevaluated integral.
We are looking for the sum
$$s = \sum_{k=3}^{\infty} \frac{\log(k)}{k^2-4}$$
Inserting
$$\log(k) = \int_0^1 \frac{x^{k-1}-1}{\log (x)} \, dx\tag{1}$$
and interchanging the order of summation and integration the integrand is given by the sum
$$i(x):=\frac{1}{\log (x)}\sum _{k=3}^{\infty } \frac{x^{k-1}-1}{k^2-4}\tag{2}$$
for which Mathematica gives
$$\begin{align}i(x)=\frac{4 x+2 x^2 -7 x^3 +x^4+4 \left(1-x^4\right) \log (1-x)}{16 x^3 \log (x)}\end{align}\tag{3}$$
now integrating gives the integral representation
$$s = s_i := \int_{0}^{1} i(x)\,dx= s(x)|_{x\to 1}-s(x)|_{x \to 0}\tag{4}$$
where antiderivative is given by
$$s(x) = \int i(x)\,dx = s_1(x) + s_2(x)\tag{5}$$
where
$$\begin{align}s_1(x) & = \int \frac{x^4-7 x^3+2 x^2+4 x}{16 x^3 \log (x)} \, dx\\
= & \frac{\text{Ei}(-\log (x))}{4}+\frac{1}{16} \text{Ei}(2 \log (x))-\frac{7 \text{li}(x)}{16}\\
& +\frac{1}{8} \log (\log (x))\end{align}\tag{6}$$
and
$$s_2(x) = \frac{1}{4}\int \frac{ \left(1-x^4\right) \log (1-x)}{ x^3 \log (x)} \, dx\tag{7}$$
Here $\text{Ei}(z)$ is the exponential integral. The integral $s_2$ remains unevaluated. And I am stuck here.
That the integral $s_{i}$ is convergent can be seen from the expansions at the endpoints of the integration interval.
We have
$$\begin{align}i(x \to 0)& \simeq -\frac{25}{48 \log (x)}+\frac{x^2}{5 \log (x)}\\
& +\frac{x^3}{12 \log (x)}+O\left(x^4\right)\end{align}\tag{8}$$
and
$$\begin{align}i(x \to 1) & \simeq -\frac{9}{16}+\frac{19 (x-1)}{32}\\
& +(x-2) \log (1-x) + O((1-x)^2)\end{align}\tag{9}$$
|
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|
Discretization of second order ODE, apply Z transform and inverse I have the following ODE:
$\frac{\mathrm{d^2y(t)} }{\mathrm{d} t} + 2\frac{\mathrm{dy(t)} }{\mathrm{d} t}+4y(t)=e^{-2(t-2)}u(t-2)$
With $u(t)$ being the unit step function. I am than asked to discretize the ODE (with sampling T=1), which gave me:
$y(n+2) + 3y(n) = e^{-2(n-2)}u(n-2)$
Next, I'm told to apply the Z transform to the function and later apply the inverse Z transform to find y(n) (initial values y(0) = y(1) = 0):
$z^2Y(z) + 3Y(z) = \frac{1}{z(z-e^{-2})}$
$Y(z) = \frac{1}{z(z-e^{-2})(z^2+3)}$
The problem is, when I apply the inverse Z transform using Wolfram, I get an enormous y(n) which doesn't seem quite right (that is, I don't think the exercise actually expects this result):
$\frac{(e^{2 - 2 n} (i (-i)^n 3^{n/2 + 1/2} e^{2 n} - i i^n 3^{n/2 + 1/2} e^{2 n} + (-i)^n 3^{n/2 + 1} e^{2 n + 2} + i^n 3^{n/2 + 1} e^{2 n + 2} + 18 e^6) (1 - θ(1 - n)))}{(18 (1 + 3 e^4))}$
I imagine that I can simplify the result using some clever Z transform properties, but I couldn't get very far with that thought. Is there something I'm doing wrong? Is there, maybe, a nice property of the transform that I could apply here? I've been stuck in this exercise for the last few days now.
|
You can perform partial fractions and simplify first.
Let,
$${Y(z)} = \frac{1}{z(z-e^{-2})(z+\sqrt3i)(z-\sqrt3i)} =\frac{a}{z}+\frac{b}{z-e^{-2}}+\frac{c}{z+\sqrt3i}+\frac{c^*}{z-\sqrt3i}$$
$z = 0 \Rightarrow a = -\dfrac{e^2}{3}$
$z= e^{-2} \Rightarrow b = \dfrac{e^6}{1+3e^4}$
$z = -\sqrt3i \Rightarrow c =\dfrac{1}{-\sqrt3i(-\sqrt3i-e^{-2})(-2\sqrt3i)}= \dfrac{1}{6(\sqrt3i+e^{-2})} = \dfrac{e^{-2}-\sqrt3i}{6(3+e^{-4})} =\dfrac{e^2(1-e^2\sqrt3i)}{6(1+3e^4)}$
and $c^* = \dfrac{e^2(1+e^2\sqrt3i)}{6(1+3e^4)}$
So,
$\begin{align}\dfrac{c}{z+\sqrt3i}+\dfrac{c^*}{z-\sqrt3i} &= \dfrac{1}{6\sqrt3(1+e^{-4})(z^2+3)}\left[(z-\sqrt3i)(1+e^{-2}i)+(z+\sqrt3i)(1-e^{-2}i)\right]\\&=\dfrac{z+\sqrt3e^{-2}}{3\sqrt3(1+e^{-4})(z^2+3)}=\dfrac{1}{3\sqrt3(1+e^{-4})}\left[\frac{z}{z^2+3}\right]\end{align}$
The answer will depend on the Region of Convergence. Let us assume $|z| > \sqrt3 > e^{-2}$
$\begin{align}Y(z) &= \frac{a}{z}+\frac{b}{z-e^{-2}}+\frac{c}{z+\sqrt3i}+\frac{c^*}{z-\sqrt3i} \\&= \frac az + z^{-1}\left[b\frac{z}{z-e^{-2}}+c\frac{z}{z+\sqrt3i}+c^*\frac{z}{z-\sqrt3i}\right] \\
y(n) &= a\delta(n) +(be^{-2(n-1)}+c(-\sqrt3i)^{n-1}+c^*(\sqrt3i)^{n-1})u(n-1) \\
y(n)&=-\frac{e^2}{3}\delta(n)+\left[\frac{e^{8-2n}}{1+3e^4} + c(-\sqrt 3 i)^{n-1} + c^*(\sqrt 3 i)^{n-1}\right]u(n-1)\end{align} $
$$c(-\sqrt 3 i)^{n-1} + c^*(\sqrt 3 i)^{n-1} = \dfrac{e^23^{\frac n2-\frac12}\left[e^23^\frac 12 \left(i^{n} + (-i)^{n}\right) + i^{n-1}+(-i)^{n-1}\right]}{6(1+3e^4)}$$
So,
$$y(n) = -\frac{e^2}{3}\delta(n)+\dfrac{6e^{8-2n}+e^23^{\frac n2-\frac12}\left[e^23^\frac 12 \left(i^{n} + (-i)^{n}\right) + i^{n-1}+(-i)^{n-1}\right]}{6(1+3e^4)}u(n-1)$$
This can be further simplified using $\frac{i^n+(-i)^n}{2} = \frac{e^{i n \frac\pi2} + e^{-i n \frac\pi2}}{2} = \cos\left(\frac{n\pi}{2}\right)$
Finally (for the assumed ROC),
$$y(n) = -\frac{e^2}{3}\delta(n)+\dfrac{3e^{8-2n}+e^23^{\frac n2-\frac12}\left[e^23^\frac 12 \cos\left(\frac{n\pi}{2}\right) + \cos\left(\frac{(n-1)\pi}{2}\right)\right]}{3(1+3e^4)}u(n-1)$$
Similarly you can find for different ROC's
|
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|
Prove by induction that for $m, n ∈ N$, $m^{2n+1} − m$ is divisible by $6.$
Prove by induction that for $m, n ∈ N$, $m^{2n+1} − m$ is divisible by $6.$
What I have thus far:
Base case: $m=n=0$;
$0^{0+1}-0 = 0$, which is divisible by $6$.
Base case(2): $m=n=1$;
$1^{2+1}-1 = 0$, which is divisible by $6$.
Proposition: $m^{2n+1} − m = 6 \times k$ for $m,n ∈ N$.
Induction step: $m^{2n+1} − m$ => $m^{2(n+1)+1} -m$
And this is where I'm kinda stuck on.
I was also wondering if there maybe was an easier way to prove this by using $m^{2n+1} − m \mod 6 = 0$ as proposition.
|
Your induction step isn't right
$m^{2(n+1) + 1} - m = m^{2^n+1}m^2 - m=$
$([m^{2n+1} -m] + m)m^2 - m = $
$(6k + m)m^2 - m =$
$6km^2 + m^3- m$.
So we must prove $m^3 -m$ is always a multiple of $6$.
Hint: $m^3 - m = m(m^2-1) = m(m-1)(m+1)$
=====
In general you can do two proofs by induction, one where one variable is is fixed and the other is run through in the proof.
That is to say. Prove for a fixed $m$ that Base case $m^{2*0+1} - m$ is divisible by $6$; and induction step if $m^{2n+1} - m$ is divisible by $6$ then $m^{2(n+1) + 1} -m$ is divisible by $6$; ANd then to a second indcution proof for a fixed $n$ that Base case $0^{2n+1} - 0$ is divisble by $6$; and induction step if $m^{2n+1} - m$ is divisble by $6$ then $(m+1)^{2n+1} -(m+1)$ is divisible by $6$.
In this case it's not so straight forward. The induction on $n$ with $m$ fix will rely on us showing $m^3 - m$ is divisible but $6$. And the induction on $m$ with $n$ fix will rely on showing $\sum_{k=1}^{2n} (-1)^k{2n+1\choose k}m^k$ is divisible by $6$.
But I think the easiest thing is to first prove that $m^2 -m$ is divisible by $6$. This can be done by induction but it is easier to do it by modular arithmentic $\mod 2$ and $\mod 3$.
$m^3 -m = (m-1)m(m+1)$ so if $m\equiv 0,1,-1\pmod 3$ then $m,m-1,m+1 \equiv 0 \pmod 3$ so $3|(m-1)m(m+1)$ and if $m \equiv 0,1 \pmod 2$ then $m,m-1\equiv 0\pmod 2$ so $2|(m-1)m(m+1)$ and so $6|(m-1)m(m+1)$.
Then prove by induction for a fixed $m$ than induction on $n$ shows $6|m^{2m+1} - m$.
......
Or to use modular arithmetic from the start.
$m^{2m+1} - m = m(m^{2n} - 1)$.
If $m\not \equiv 0 \pmod{2,3}$ its easy to show therefore than $m^2\equiv 1\pmod {2,3}$ and therefore $m^{2n} \equiv 1 \pmod{2,3}$ and if $2,3 \not \mid m$ then $2,3 \mid m^{2n} -1$ so either way $6|m(m^{2n} -1)$
i.e. either $m \equiv 0 \pmod 2$ or if $m\not \equiv 0\pmod 2$ then $m^2\equiv 1 \pmod 2$.
And either $m \equiv 0 \pmod 3$ or if $m\not \equiv 0 \pmod 3$ then $m^2 \equiv 1\pmod 3$.
======
In hindsight, I think the easiest answer should have been: It's often easiest to prove things are divisible by $2$ and by $3$ separately.
It's easy to convince ourselves that $m$ and $m^{2n}$ must be both even or both odd so $m^{2n} -m$ must be even. ($m\equiv 0, 1\pmod 2$ and $m^k\equiv 0^k, 1^k\equiv 0, 1\equiv m\pmod 2$ so $m^k-m \equiv m-m\equiv 0 \pmod 2$.) And it's not not much harder to convince ourselves that similarly that $m^2 \equiv \begin{cases} 0&m\equiv 0 \pmod 3\\1&m\equiv 1\pmod 3\\1&m\equiv 2\pmod 3\end{cases}\pmod 3$ so $m^{2k+1}=(m^2)^k m \equiv \begin{cases}0\cdot m =0\equiv m&m\equiv 0 \pmod 3\\1\cdot m \equiv m&\begin{cases}1\\2\end{cases}\pmod 3\end{cases}\equiv m \pmod 3$. So $m^{2n+1} - m\equiv m-m\equiv 0 \pmod 3$.
And that would be all.
|
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|
$\sum_{n=1}^{\infty}\frac{(2^n + 3^n)\sin(n)}{2^n + n^2\cdot3^n}$ Proving the convergence of a series by Weierstrass M-test.
$$\sum_{n=1}^{\infty}\frac{(2^n + 3^n)\sin(n)}{2^n + n^2\cdot3^n}.$$
$$\frac{(2^n + 3^n)\sin(n)}{2^n + n^2\cdot3^n} \leq \frac{2^n + 3^n}{2^n + n^2 \cdot 3^n} = \frac{1 + \left(\frac{3}{2}\right)^n}{1 + n^2 \left(\frac{3}{2}\right)^n}.$$
Further I got stuck how to evaluate and get convergence.
|
\begin{align*}
\left|\dfrac{(2^{n}+3^{n})\sin n}{2^{n}+n^{2}3^{n}}\right|\leq\dfrac{2^{n}+3^{n}}{2^{n}+n^{2}3^{n}}\leq\dfrac{3^{n}+3^{n}}{n^{2}3^{n}}=\dfrac{2}{n^{2}},
\end{align*}
now $\sum n^{-2}<\infty$.
|
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|
How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
|
General solution:
I will solve a specific quartic equation that is a specific case of a general quartic equation.
Let, $a≠0,~ b≠0$, then we have
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x^2+\frac e{ax^2}+\frac ba x+\frac d{ax}+\frac ca=0$$
$$\begin{align}&x^2+\frac ea \times \frac 1{x^2}+\frac ba \left(x+\frac {d}{bx}\right)+\frac ca=0&\end{align}$$
This quartic equation can be directly converted to the quadratic equation in the case below, avoiding the cubic equation.
$$\begin{align}&x+\frac {d}{bx}=t \\
\implies &t^2=x^2+\frac{d^2}{b^2}\times \frac {1}{x^2}+\frac{2d}{b}\end{align}$$
Then, if $$\frac ea =\frac{d^2}{b^2}$$
We have
$$t^2-\frac{2d}{b}+\frac ba t+\frac ca=0$$
$$t^2+\frac ba t+\left(\frac ca-\frac{2d}{b}\right)=0$$
The last equation is a quadratic equation.
After solving quadratic, you wil get
$$x+\frac {d}{bx}=t$$
$$bx^2-btx+d=0$$
The last equation is also a quadratic equation.
In your case we have
$$a=1,b=-2,c=-1, d=2, e=1$$
This means
$$\frac ea =\frac{d^2}{b^2}$$
holds.
|
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|
Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$
This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
Method 2:
$$S=\text{coefficient of $x^n$ in }:$$
$$2^n(1+x)^n\Bigg( 1+\frac{1+x}{2}+\left(\frac{1+x}{2}\right)^2+\cdots+\left(\frac{1+x}{2}\right)^n \Bigg)$$
$$=(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$
It looks like a heavy task to collect all the $x^n$ coefficients from this expression.
Im out of ideas. Any hint is appreciated.
From Putnam 2020, Q A2
|
Consider
$$\left( \begin{matrix}
n \\
k \\
\end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=R}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$
so
$$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix}
n+r \\
r \\
\end{matrix} \right)}=\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\sum\limits_{r=0}^{n}{{{2}^{n-r}}}\frac{{{\left( 1+z \right)}^{n+r}}}{{{z}^{r+1}}}dz}=\\\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\frac{{{2}^{n+1}}{{\left( 1+z \right)}^{n}}}{z-1}-\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( z-1 \right){{z}^{n+1}}}dz}\\=\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( 1-z \right){{z}^{n+1}}}dz}$$
Where for convenience we've chosen a circular contour enclosing the origin with radius less than $1$. We can do this because directly after the summation the only contribution comes from a residue at $z=0$ (note there is no residue at $z=1$ in the second line). The final line above reflects this. Now to get the residue at $z=0$...
$$res\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( 1-z \right){{z}^{n+1}}}=res\sum\limits_{m=0}^{\infty }{{}}\sum\limits_{k=0}^{\infty }{\left( \begin{matrix}
1+2n \\
k \\
\end{matrix} \right)\frac{1}{{{z}^{n+1-m-k}}}}\\=\sum\limits_{m=0}^{\infty }{\left( \begin{matrix}
1+2n \\
n-m \\
\end{matrix} \right)=}\sum\limits_{m=0}^{n}{\left( \begin{matrix}
1+2n \\
n-m \\
\end{matrix} \right)}=\sum\limits_{m=0}^{n}{\left( \begin{matrix}
1+2n \\
m \\
\end{matrix} \right)}={{4}^{n}}$$
hence
$$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix}
n+r \\
r \\
\end{matrix} \right)}={{4}^{n}}$$
|
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|
A system of equations with degree 2 Let $a,b,c,d \in \mathbb R$. Suppose the following holds
\begin{align}
a^2-c^2 &=1 \\
b^2-d^2 &=-1 \\
ad-bc &= \pm1 \\
ab-cd &=0
\end{align}
How can I find
$a,b,c$ and $d$. I'm trying to show something regarding differential geometry
and arrive with these equations. I tried to solved it but my algebra sucks.
Any help or hints on how to solved this system of equations would be much appreciated!
|
Just two by two matrices,
$$
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
1&0 \\
0&1 \\
\end{array}
\right)
$$
so that the right hand factor must be the inverse of $ M=
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)$
When $M$ has determinant $1,$ we are demanding
$$
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
d&-c \\
-b&a \\
\end{array}
\right) = M^{-1}
$$
so that $a=d$ and $b=c,$
or
$$
M=
\left(
\begin{array}{cc}
a&c \\
c&a \\
\end{array}
\right)
$$
When $M$ has determinant $-1,$ we are demanding
$$
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
-d&c \\
b&-a \\
\end{array}
\right) = M^{-1}
$$
so that $a=-d$ and $b=-c,$
or
$$
M=
\left(
\begin{array}{cc}
a&c \\
-c&-a \\
\end{array}
\right)
$$
One may fill in both versions using $\cosh t$ and $\sinh t$
|
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Set of all real values of $a$ for which the equation $(a-4)\sec^4x+(a-3)\sec^2x+1=0, (a\ne4)$ has real solutions
Find set of all real values of $a$ for which the equation $(a-4)\sec^4x+(a-3)\sec^2x+1=0, (a\ne4)$ has real solutions.
Let $\sec^2x=t$. So, the equation becomes $(a-4)t^2+(a-3)t+1=0$.
Since $\sec^2x\ge1\implies t\ge1\implies$ both the roots of the quadratic equation are $\ge1$.
Now, if we have $f(x)=ax^2+bx+c=0$ and its roots are $\ge k$ then (i) $af(k)\ge0$ and (ii) discriminant $\ge0$ and (iii) $\frac{-b}{2a}\gt k$.
(i) gives me $(a-4)(a-4+a-3+1)\ge 0\implies(a-3)(a-4)\ge0\implies a\in(-\infty,3]\cup[4,\infty)$.
(ii) gives me $(a-3)^2-4(a-4)\ge0\implies a^2+9-6a-4a+16\ge0\implies a^2-10a+25\ge0$. This is always true.
(iii) gives me $\dfrac{-a+3}{2(a-4)}\gt1\implies\dfrac{-a+3}{2(a-4)}-1\gt0\implies\dfrac{-a+3-2a+8}{2(a-4)}\gt0\implies\dfrac{3a-11}{2(a-4)}\lt0\implies a\in(\frac{11}3,4)$
On taking intersection, I do not get any common region but the answer given is $[3,4)$.
What is my mistake?
|
I think you overdid it. It is quite neat
$$\sec^2x=\frac{-(a-3)\pm\sqrt{(a-3)^2-4(a-4)}}{2(a-4)}=-1,\frac{1}{4-a}$$
So, obviously $\sec^2x\geq 1$
So $$\frac{1}{4-a}\geq 1$$
which gives $\frac{a-3}{a-4}\leq 0$
Thus $a\in[3,4)$
|
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Find the sum of $\sum\limits_{n=-\infty}^{\infty}\frac{2^n}{1+7^{2^n}}$ My attempt:
$\sum\limits_{n=-\infty}^{\infty}\frac{2^n}{1+7^{2^n}}=\sum\limits_{n=-\infty}^{1}\frac{2^n}{1+7^{2^n}}+\sum\limits_{1}^{\infty}\frac{2^n}{1+7^{2^n}}+\frac 18\\
=\sum\limits_{1}^{\infty}\frac1{2^n(1+7^{1/2^n})}+\sum\limits_{1}^{\infty}\frac{2^n}{1+7^{2^n}}+\frac 18\\
=\frac 18+\sum\limits_{1}^{\infty}\frac{2^n}{1+7^{2^n}}+\frac1{2^n(1+7^{1/2^n})}
\\$
I am stuck here, is my approach correct and leads to the desired result? Thanks in advance.
|
You can write this as a telescoping sum. Note that
$$
\frac{2^n}{7^{2^n}-1} - \frac{2^n}{7^{2^n}+1}=\frac{2^n(7^{2^n}+1)-2^n(7^{2^n}-1)}{7^{2^{n+1}}-1}=\frac{2^{n+1}}{7^{2^{n+1}}-1}.
$$
So
$$
\begin{eqnarray}
\sum_{n=-\infty}^{\infty}\frac{2^{n}}{7^{2^n}+1} &=& \sum_{n=-\infty}^{\infty}\left(\frac{2^n}{7^{2^n}-1} - \frac{2^{n+1}}{7^{2^{n+1}}-1}\right) \\ &=& \lim_{n\rightarrow-\infty}\frac{2^n}{7^{2^n}-1} - \lim_{n\rightarrow+\infty}\frac{2^n}{7^{2^n}-1}
\\ &=& \lim_{n\rightarrow-\infty}\frac{2^n}{{2^n}\log 7+o(2^n)} \\ &=&\frac{1}{\log 7}.
\end{eqnarray}
$$
|
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|
A Fibonacci conjecture: $\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}$ Given the Fibonacci sequence $F_n$, that is $$F_1=F_2=1,F_{n+2}=F_{n+1}+F_n.$$
I find that
$$
\log_21<\log_32<\log_53<\log_85<\cdots,
$$
i.e. we have $$
\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}, \quad \text{for} \; n\geqslant1\tag{1}
$$
If $n=2k$, $(1)$ is easy to get, but if $n=2k+1$, I have no idea to prove it.
What's more, if we let $F_1=a,F_2=b$ be two positive integers, then $(1)$ still exstis when $n$ is greater than some integer $m$.
I do not konw whether my conjecture is true.
|
Conjecture: $ \log_{F_{n+1} } F_n < \log_{F_{n+2} }F_{n+1}.$
This is equivalent to:
$$\dfrac{\log F_{n}}{\log F_{n+1}} < \dfrac{\log F_{n+1}}{\log F_{n+2}}$$
Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get $\log(F_n )\log (F_{n+2}) < (\log (F_{n+1}))^2$ ($\log$ here is the natural logarithm).
One can use the Cassini identity ($F_{n+1}^2 = F_nF_{n+2} \pm 1$) to prove your conjecture, but first let's establish an inequality equivalent to the above one,
$$\log(F_n)\log(F_{n+2})< (\log(F_{n+1}))^2 = (\frac{1}2\log(F_{n+1}^2))^2 = \frac{1}{4}(\log(F_{n+1})^2)^2$$
Therefore it suffices to prove that $\log(F_{n+1}^2)^2 > 4\log F_n \log F_{n+2}$
Case 1: when $F_{n+1}^2 = F_nF_{n+2} - 1$
$$(\log(F_{n+2}/F_{n}))^2 > (\log 2)^2 > \dfrac{2 \log F_n}{F_n} + \dfrac{2\log 3F_n}{F_n} - \dfrac{1}{F_n^2}$$
$$\geq \dfrac{2 \log F_n + 2\log (2F_n + F_{n-1})}{F_n}- \dfrac{1}{F_n^2} = \dfrac{2 \log F_n + 2\log (F_{n+2})}{F_n}- \dfrac{1}{F_n^2}$$
$$\implies (\log F_{n+2} - \log F_n)^2 > \dfrac{2 \log F_nF_{n+2}}{F_n} - \dfrac{1}{F_n^2}$$
for a sufficiently large $n$($n \geq 8$ for the Fibonacci sequence starting with $1$ and $2$). Now add $4\log F_n \log F_{n+2}$ to both sides:
$$(\log F_{n+2} + \log F_n)^2 > \dfrac{2 \log F_nF_{n+2}}{F_n} - \dfrac{1}{F_n^2} + 4\log F_n \log F_{n+2} $$
$$\implies (\log F_{n+2} + \log F_n)^2 - \dfrac{2 \log F_nF_{n+2}}{F_n} + \dfrac{1}{F_n^2} > 4\log F_n \log F_{n+2}$$
$$\implies (\log F_{n} + \log F_{n+2} - \dfrac{1}{F_n})^2 > 4\log F_n \log F_{n+2}$$
$$\implies (\log F_{n} + \log (F_{n+2} - \dfrac{1}{F_n}))^2 = (\log(F_n\cdot (F_{n+2} - \dfrac{1}{F_n})))^2 = \log(F_{n+1}^2)^2 > 4\log F_n \log F_{n+2} $$
Case 2: when $F_{n+1}^2 = F_nF_{n+2} + 1$
$$(\log F_{n+2} - \log F_n)^2 > 0 $$
$$\implies (\log F_{n+2} +\log F_n)^2 > 4\log F_n \log F_{n+2}$$
$$\implies (\log F_{n+1}^2)^2 = (\log F_{n+2}F_n + 1)^2 > (\log F_{n+2}F_n)^2 > 4\log F_n \log F_{n+2}$$
Verifying the inequality for $n < 8$ manually completes the proof using Cassini's identity.
|
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|
Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.
My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.
Edit:
From L.H.S -
$\dfrac{\frac{a^3}{b} + \frac{b^3}{a}}{2} \geqslant ab \implies \dfrac{a^3}{b} + \dfrac{b^3}{a} \geqslant 2ab$
|
Reduce to this:
$(a-b)(a^3-b^3)\ge 0$
Do you see why it is always satisfied?
|
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|
Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) \tag{1}
\end{equation}
This is an example problem and this was the solution:
\begin{gather}
\frac{x^{3} +x}{1+x^{3}} =\left( 1+\frac{1}{x^{2}}\right)\left( 1+\frac{1}{x^{3}}\right)^{-1} =\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{2}\\
=1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{3}
\end{gather}
And then
\begin{gather}
\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} =\left( 1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right)^{\frac{1}{7}} \tag{4}\\
=1+\frac{1}{7} .\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{5}
\end{gather}
\begin{equation}
\cos\frac{1}{x} =1-\frac{1}{2x^{2}} +\mathcal{O}\left(\frac{1}{x^{4}}\right) \tag{6}
\end{equation}
From above, we obtain:
\begin{equation}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{7}
\end{equation}
Hence the required limit is:
\begin{equation}
\lim _{x\rightarrow \infty }\left(\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right) =\frac{9}{14} \tag{8}
\end{equation}
I tried to fill in the details for the steps involved in the above steps.
I supplied the details for $\displaystyle ( 3)$ as below:
\begin{gather}
\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) =1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\left( x^{3} +x\right)\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{9}\\
=1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\frac{\left( x^{3} +x\right)}{x^{6}}\mathcal{O}( 1)\right) =1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{10}
\end{gather}
where in (9) and (10), I have used the standard result: $\displaystyle \frac{\mathcal{O}( f( x))}{g( x)} =\mathcal{O}\left(\frac{f( x)}{g( x)}\right)$, if $\displaystyle g( x) \neq 0$.
I tried to get (5) from (4) but failed.
Then I supplied details for $\displaystyle ( 7)$ as below:
\begin{gather*}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) -\mathcal{O}\left(\frac{1}{x^{4}}\right) =\frac{9}{14x^{2}} +\frac{1}{x^{3}}\left(\mathcal{O}( 1) -\mathcal{O}\left(\frac{1}{x}\right)\right)\\
=\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\\
\Longrightarrow \lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\lim _{x\rightarrow \infty }\left(\frac{9}{14} +\mathcal{O}\left(\frac{1}{x}\right)\right) =\frac{9}{14}
\end{gather*}
Any help in getting (5) from (4) is much appreciated. Thanks.
\begin{equation*}
\end{equation*}
\begin{equation*}
\end{equation*}
|
There is no need for so much work as you have done.
Just note that the expression under limit is of the form $$x^2(A^{1/7}-B)$$ where both $A, B$ tend to $1$. Then we can write the above expression as $$x^2\left(\frac{A^{1/7}-1}{A-1}\cdot(A-1)+1-B\right)$$ Now $x^2(1-B)\to 1/2$ and $$x^2(A-1)=\frac{x^2(x-1)}{x^3+1}\to 1$$ and the desired limit is $$\frac{1}{7}+\frac {1}{2}=\frac {9}{14}$$
The above can be converted into the kind of asymptotics you need. Thus for example you can write $$B=1-\frac{1}{2x^2}+o(x^{-2})$$ and $$A^{1/7}=1+\frac{A-1}{7}+o(x^{-2})$$ and you can get desired answer.
You should also observe that you don't need to multiply/divide/compose Taylor series here. For most typical limit problems that is usually the case.
Even after years of advertising (via my answers on mathse) the limit formula $$\lim_{x\to a} \frac {x^n-a^n} {x-a} =na^{n-1}$$ is not widely used. Perhaps I need to make more effort in this direction.
|
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|
Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$ Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$
This looks like a problem that would do well with a conversion to polar coordinates ($r^2 = x^2 + y^2$ and $x = r\cos(\theta)$ $y = r\sin(\theta)$). However, I am confused on how to change $y' = \frac{dy}{dx}$ to $\frac{dr}{d\theta}$, in order to do separation of variables. Could someone demonstrate a solution using this approach?
|
$x = r \cos\theta, y = r \sin\theta$
$dx = \cos\theta \ dr - r \sin\theta \ d\theta$
$dy = \sin\theta \ dr + r \cos\theta \ d\theta$
Given equation is $dy = \frac{\sqrt{x^2+y^2}-x}{y} dx$
So we get,
$r \sin\theta \ (\sin\theta \ dr + r \cos\theta \ d\theta) = r \ (1 - \cos\theta) \ (\cos\theta \ dr - r \sin\theta \ d\theta)$
On simplifying we get,
$\displaystyle \frac{1}{r} \ dr = - \frac{\sin\theta}{1-\cos\theta} \ d\theta$
To integrate, simply observe that $1 - \cos\theta = t \implies \sin\theta \ d\theta = dt$
So both sides are simple integration.
|
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|
Prove: $\cos^3 \frac x3 + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x$ I tried to solve the following identity:
$$\cos^3 \frac x3 + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x$$ I applied formulas, $\cos (a+b)$ and $\cos^3 x$ and I arrived at
$$\cos 3x + 2 \cos x - 3 \cos \frac{x}{3}=0 $$
After this I am stuck and any formula I applied, got complicated. Can someone give me an idea as to how I could prove this identity?
|
The formula you arrived at is wrong (put $x=\frac {\pi}{2}$ to see that). A correct method would be this:
$$\cos 3x=4\cos^3 x-3\cos x$$
So, $$\cos^3 x=\frac {\cos 3x+3\cos x}{4}$$
Now you just have to apply this formula to the given three angles.
|
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|
Prove that the number $1^k + 2^k + ... + 12^k$ is divisible by $13$ Prove that the number $1^k + 2^k + ... + 12^k$ is divisible by $13$ for $k = 1, 2, ..., 11$.
My attempt:
If k is odd number, then $1^k +...+6^k +7^k ... + 12^k \equiv 1^k + ... + 6^k - 6^k - 5^k - ... -1^k \equiv 0 (mod 13)$.
But if k is even number then $2(1^k + ... + 6^k)$
Please give a good hint or explain clearly how to solve. Thanks you
|
To be specific: $2$ is a primitive root modulo $13$. Hence any number from $1$ to $12$ can be written uniquely as $2^n$, for some $n$ between $1$ and $12$. Hence after a rearrangement your sum becomes:
$$2^k+ 2^{2k}+ \dots + 2^{12k} \pmod{13}.$$
If we denote your sum by $S$, then
$$ S \equiv \sum\limits_{n=1}^{12} 2^{nk} \pmod {13}.$$
Notice that the right hand sum is nothing but
$$ 2^k (2^{12k}-1) = 2^k (2^{12}-1)r.$$
Where $r$ is some integer. $2^{12}-1$ is divisible by $13$. Hence $S$ should be divisible by $13$.
|
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|
Prove that $\sqrt[n]{a^n+x}=a+\frac{x}{na^{n-1}}-r$ ,$a>0$,$x>0$ Prove that
$\sqrt[n]{a^n+x}=a+\frac{x}{na^{n-1}}-r$ ,$a>0$,$x>0$
given $0$$<$$r$$<$$\frac{n-1*x^2}{2n^2*a^{2n-1}}$.
What I did.
$\sqrt[n]{a^n+x}=a(1+\frac{x}{a^n})^\frac{1}{n}=a(1+\frac{1}{n}*\frac{x}{a^n}+\frac{1}{2}*\frac{1}{n}*\frac{1-n}{n}*(\frac{x}{a^n})^2)$ = $a+\frac{ax}{na^n}+\frac{a*(1-n)*x^2}{2n^2*a^{2n}}$ = $a+\frac{x}{n*a^{n-1}}+\frac{x^2(1-n)}{2n^2*a^{2n-1}}$
Now remainder is in Lagrange form.
$0$$<$$r$$=$$\frac{1-n}{2n^2*a^{2n-1}}x^2$ $<$ $\frac{(n-1)*x^2}{2n^2*a^{2n-1}}$
Because $n>1$,$r<\frac{(n-1)*x^2}{2n^2*a^{2n-1}}$ but I am getting $r<0$ can you please say where I made mistake.
|
The second term should be
$\frac{1}{2}\frac{1}{n}(\frac1{n}-1)(\frac{x}{a^n})^2
=-\frac{n-1}{2n^2}(\frac{x}{a^n})^2
$.
|
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|
System of two quadratics equation, $P(x)$ and $Q(x)$ If $P(x) = ax^2 + bx + c$ and $Q(x) = – ax^2 + dx + c$, $ac \ne 0$, then the equation $P(x) . Q(x) = 0$ has
(A) Exactly two real roots
(B) At least two real roots
(C) Exactly four real roots
(D) No real roots
My approach is as follow Let $T(x)=P(x).Q(x)$
$T\left( x \right) = - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$
Not able to approach from here
|
The parabolas represented by the quadratic functions $ \ P(x) \ = \ ax^2 + bx + c \ $ and $ \ Q(x) \ = \ – ax^2 + dx + c \ \ , \ \ ac \ne 0 \ \ , $ "open" in opposite "vertical" directions and have a common $ \ y-$intercept. The absolutely minimal number of $ \ x-$intercepts that could be arranged is one, by placing the vertices of both parabolas at the origin. But these are $ \ y \ = \ ax^2 \ $ and $ \ y \ = \ -ax^2 \ \ , $ which are not permitted by the condition $ \ ac \ne 0 \ \ ; \ $ a "vertical shift" in either direction necessarily produces two $ \ x-$intercepts, and so two real zeroes for $ \ (ax^2 + c)·(-ax^2 + c) \ \ . $
Thus, the unavoidable minimum number of real zeroes for $ \ P(x)·Q(x) \ $ is two. We can arrange for more than two by placing the two parabolas so that they are tangent at the origin at a point other than their vertices, with $ \ y \ = \ ax^2 + bx \ $ and $ \ y \ = \ -ax^2 + bx \ \ ( d = b ) \ \ . $ With their vertices then at $ \ x \ = \ \pm \frac{b}{2a} \ \ , $ there are three $ \ x-$intercepts, including one at the origin. Again, we may not have $ \ c = 0 \ \ , $ but a "vertical shift" in either direction, that is kept small enough that the vertices remain on opposite sides of the $ \ x-$axis, then leads to four $ \ x-$intercepts (the one at the origin "bifurcating" into two).
We demonstrate geometrically then that we must have at least two zeroes for $ \ P(x)·Q(x) \ $ and may easily obtain up to four [choice $ \ \mathbf{(B)} $ ] . If we generally "break" the symmetry of the vertices about the origin described in the previous paragraph, we can easily obtain more than two $ \ x-$intercepts for the parabolas.
We can also see this by putting the polynomials into "vertex form":
$$ a·\left(x \ + \ \frac{b}{2a} \right)^2 \ + \ \left(c \ - \ \frac{b^2}{4a} \right) \ \ , \ \ -a·\left(x \ - \ \frac{d}{2a} \right)^2 \ + \ \left(c \ + \ \frac{d^2}{4a} \right) $$
and finding their zeroes from
$$ \left(x \ + \ \frac{b}{2a} \right)^2 \ \ = \ \ -\frac{1}{a}·\left(c \ - \ \frac{b^2}{4a} \right) \ \ , \ \ \left(x \ - \ \frac{d}{2a} \right)^2 \ \ = \ \ \frac{1}{a}·\left(c \ + \ \frac{d^2}{4a} \right) \ \ . $$
Often, it will be the case that the right side of one of these equations is positive while the right side of the other is negative, giving us only two real zeroes. But it is also possible for both right sides to be positive when $ \ \frac{b^2}{4a^2} \ > \ \frac{c}{a} \ > \ -\frac{d^2}{4a^2} \ \ , $ producing four real zeroes. [This is equivalent to the condition for the polynomial discriminants.]
|
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|
Find all integers n such that $(\frac{n^3-1}{5})$ is prime
Find all integers n such that $(\frac{n^3-1}{5})$ is prime?
My Approach:
I wrote all the prime which i will get after dividing $(n^3-1)$ by $5$.
$n^3-1=10,15,25,35,55,...,215$
which lead me to $n^3=11,16,26,...,216$, then I obtained $n=6$
My doubt is that how to check more value of $n$ without using modular arithmetic because the book I'm referring has not introduced it yet.
Second approach: $\frac{(n^3-1)}{5}=\frac{(n-1)(n^2+n+1)}{5}$
But my second approach too does not lead me anywhere.
This problem is from the book Pathfinder for Olympiad Mathematics
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If we split all integer $n$ into $[5m+1,5m+2,5m+3,5m+4,5m+5]$ you can show that only numbers $(5m+1)^3-1$ are divisible by $5$ as $[2^3-1,3^3-1,4^3-1,5^3-1]$ are all not divisible by $5$
Now $$\frac{(5m+1)^3-1}{5}=3m+15m^2+25m^3=m(3+15m+25m^2)$$
and the product $m(3+15m+25m^2)$, can only be a candidate prime if $m=\pm1$
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Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$? Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$?
My thoughts: $\sqrt{2}$ is $\approx1,4$ and $\sqrt{3}$ is $\approx{1,7}$ so their sum must be of the interval $(3;4)$. Any more strict approaches? Thank you in advance!
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First we have $\sqrt{2} + \sqrt{3} < \sqrt{4} + \sqrt{4} = 4$. To show that $\sqrt{2} + \sqrt{3} > 3$, we square both sides, and see that it suffices to show that $5 + 2\sqrt{6} > 9$. But this follows from $\sqrt{6} > 2$.
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Logarithmic integral $ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\ \mathrm{d}x $ I found this integral weeks ago.
$$ \int_0^1 \dfrac{x\ln(x)\ln(1+x)}{1+x^2}\ \mathrm{d}x $$
I tried to solve this integral using various series representation and ended up with a complicated double series which I have asked here. How can I solve this integral?
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Here is a magical solution where no advanced results are used.
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$
$$=-\underbrace{\int _0^{\infty }\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx}_{J}+\underbrace{\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx}_{I}+\int _0^1\frac{x\ln ^2\left(x\right)}{1+x^2}\:dx.$$
Let's try differentiation under the integral sign for $I$.
$$I=\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx,\:I\left(a\right)=\int _0^1\frac{\ln \left(x\right)\ln \left(1+ax\right)}{x\left(1+x^2\right)}\:dx$$
$$I'\left(a\right)=-G\frac{1}{1+a^2}+\frac{a\operatorname{Li}_2\left(-a\right)}{1+a^2}+\frac{1}{8}\zeta \left(2\right)\frac{a}{1+a^2}$$
$$I=-G\int _0^1\frac{1}{1+a^2}\:da+\underbrace{\int _0^1\frac{a\operatorname{Li}_2\left(-a\right)}{1+a^2}\:da}_{\mathcal{A}}+\frac{1}{8}\zeta \left(2\right)\int _0^1\frac{a}{1+a^2}\:da$$
$$=-\frac{\pi }{4}G+\int _0^1\frac{\operatorname{Li}_2\left(-a\right)}{a}\:da-\frac{\pi }{4}\int _0^1\frac{\ln \left(t\right)}{1+t^2}\:dt+\frac{1}{2}\ln \left(2\right)\int _0^1\frac{t\ln \left(t\right)}{1+t^2}\:dt$$
$$-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t\right)}{1+t^2}\:dt+\frac{1}{16}\ln \left(2\right)\zeta \left(2\right)$$
$$=-\frac{3}{4}\zeta \left(3\right)-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t\right)}{1+t^2}\:dt.$$
Note that for $\mathcal{A}$ I made use of the well-known representation $\int _0^1\frac{a\ln \left(t\right)}{1+at}\:dt=\operatorname{Li}_2\left(-a\right)$ and interchanged the order of integration.
With the same technique and parameter we can prove that $J$ is:
$$J=\int _0^{\infty }\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx=-\frac{3}{4}\zeta \left(2\right)\int _0^1\frac{a}{1+a^2}\:da-\frac{1}{2}\int _0^1\frac{a\ln ^2\left(a\right)}{1+a^2}\:da$$
$$=-\frac{3}{32}\zeta \left(3\right)-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right).$$
Plugging these results into the original expression we obtain:
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx=\frac{3}{32}\zeta \left(3\right)+\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)-\frac{3}{4}\zeta \left(3\right)-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t\right)}{1+t^2}\:dt$$
$$+\int _0^1\frac{x\ln ^2\left(x\right)}{1+x^2}\:dx.$$
Therefore:
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx=-\frac{15}{64}\zeta \left(3\right)+\frac{3}{16}\ln \left(2\right)\zeta \left(2\right).$$
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Solve system of equations $3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$, $xy+yz+zx = 1$ Find all $x,y,z>0$ such that
$$3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$$ $$xy+yz+zx = 1$$
The only solution should be $x=\frac{1}{3}$, $y = \frac{1}{2}$, $z=1$.
There is a way to do it with $x = \tan \alpha$, etc., but I would like to find an even more elementary way.
So far, I have written
$$x^2 - \frac{k}{3}x + 1 = y^2 - \frac{k}{4}y + 1 = z^2 - \frac{k}{5}z + 1 = 0$$
and noticed that at least two of $x$, $y$, $z$ must be smaller than $1$. (So in the bad cases I can express uniquely $x,y,z$.) But then?
Any help appreciated!
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Rewrite the second equation $xy+yz+zx = 1$ as $z=\frac{1-xy}{x+y}$ and evaluate
$$z+\frac1z= \frac{(x^2+1)(y^2+1)}{(x+y)(1-xy)}$$
Substitute above $z+\frac1z = \frac{z^2+1}z$ into the first equation
$$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y = \frac{5(z^2+1)}z \tag1$$
to get
\begin{align}
&(x^2+1)\left(\frac3{5x}- \frac{y^2+1}{(x+y)(1-xy)}\right)=0\\
&(y^2+1)\left(\frac4{5y}- \frac{x^2+1}{(x+y)(1-xy)}\right)=0\\
\end{align}
which reduce to
\begin{align}
&3x^2y+8xy^2+2x-3y=0\tag2\\
&9x^2y+4xy^2-4x+y=0\tag3\\
\end{align}
Note that 3$\times$(2) -(3) and 2$\times$(3) -(2) simplify the equations to
\begin{align}
&2xy^2+x -y=0\tag4\\
&3x^2y-2x+y=0\tag5
\end{align}
and, furthermore, with 3$x\times$(4) -2$y\times$(5)
$$3x^2+xy-2y^2=(3x-2y)(x+y)=0$$
Substitute the resulting $y=\frac32 x$ and $y=-x$ into (1) to obtain the real solutions
$$(x,y,z)=\pm \left(\frac13,\frac12,1\right)$$
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Let $c$ be a positive real number for which the equation $x^4-x^3+x^2-(c+1)x-(c^2+c)=0$ has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$ Let $c$ be a positive real number for which the equation
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$
I tried to to solve using relation between roots and coefficients but unable to progress much. Please help. Thanks in advance.
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Suppose that $\alpha$ is a root to the equation:
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
Okay, then we have:
$\alpha^4-\alpha^3+\alpha^2-(c+1)\alpha-(c^2+c)=0$
$\begin{align}
\text{You can} & \text{put it in standard form as:} \\
& \\
& c^2+(1+\alpha)c^1 + (-\alpha^4+\alpha^3-\alpha^2 + \alpha)c^0=0 \\
\end{align}$
$\begin{align}
\text{Use the quadratic formula} & \\
& \\
\forall b,c, x \in \mathbb{R},&\\
& x^2 + bx^1+cx^{0} = 0 \text{ if and only if } x = \dfrac{-b\pm\sqrt{b^2-4c}}{2} \\
\end{align}$
$c=\dfrac{-(1+\alpha)\pm\sqrt{(1+\alpha)^2-4(-\alpha^4+\alpha^3-\alpha^2 + \alpha)}}{2}$
$c=\dfrac{-(1+\alpha)\pm\sqrt{(1+2\alpha+\alpha^2)+(4\alpha^4-4\alpha^3+4\alpha^2 -4 \alpha)}}{2}$
$c=\dfrac{-(1+\alpha)\pm\sqrt{4\alpha^4-4\alpha^3+5\alpha^2 -2 \alpha + 1}}{2}$
Maybe, after some additional work, We have that that $c=\alpha ^2 - \alpha$
I don't know.
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Find a linear transformation $T\begin{pmatrix}-1\\-2\end{pmatrix}$ $T:\mathbb{R}^2 \Rightarrow\mathbb{R}^3, T\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\4\\3\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}4\\-1\\1\end{pmatrix}$. Find $T\begin{pmatrix}-1\\-2\end{pmatrix}$ and $T\begin{pmatrix}-3\\-1\end{pmatrix}$.
For $T\begin{pmatrix}-1\\-2\end{pmatrix}=-1T\begin{pmatrix}1\\0\end{pmatrix}-2T\begin{pmatrix}0\\1\end{pmatrix}=-\begin{pmatrix}-1\\4\\3\end{pmatrix}-2\begin{pmatrix}4\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-4\\-3\end{pmatrix}+\begin{pmatrix}-8\\2\\-2\end{pmatrix}=\begin{pmatrix}-7\\-2\\-5\end{pmatrix}$.
Is this correct? I would like to know if my procedure is correct by finding T.
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It's easy to see that the matrix representation of $T$ is $A=\pmatrix{-1 & 4 \\ 4 &-1 \\ 3 &1}$
so applying this transformation to a vector $(x,y)$ you get $$T_A(x,y)=\pmatrix{-1 & 4 \\ 4 &-1 \\ 3 &1} \pmatrix{x \\ y}=\pmatrix{-x+4y \\ 4x-y \\ 3x+y }$$
so now just plug the values you want $$T_A(-1,-2)=\pmatrix{-7 \\-2 \\-5}$$
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Let $a, b$ and $c$ be the $7$th, $11$th and $13$th terms respectively of an AP. If these are three consecutive terms of a GP, then $\cfrac{a}{c}$ is
Let $a, b$ and $c$ be the $7$th, $11$th and $13$th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\cfrac{a}{c}$ is equal to:
APPROACH:
$a = A+6d = a\ \ \ $ ; $ \ \ \ b = A+10d = ar \ \ \ $ ; $\ \ \ c = A+12d = ar^2$
Where: $A$ is the first term of the AP, $ \ \ d$ is the common difference of the AP, and $ \ \ r$ is the common ratio of the GP.
$\cfrac {c}{a}= \dfrac {1}{r^2}$
Also,
$$(A+6d)r = (A+10d) \implies Ar+6rd = A+10d $$
$$\implies \cfrac{A}{d} = \cfrac{6r-10}{r-1}$$
Now, we know that:
$$(A+10d)^2 = (A+6d)(A+12d)$$
$$[k(6r-10)+ 10k(r-1)]^2 = [k(6r-10)+6k(r-1)] \ \cdot \ [k(6r-10)+12k(r-1)]$$
where $k$ is the proportionality constant.
$k$ gets cancelled leaving us with the following:
$$(16r - 20)^2 = (12r-16)(18r-22)$$
on solving further, I got:
$$ 5r^2 - 11r +6 = 0$$
Which gives $r = \cfrac{6}{5}$ or $5$.
so, $\cfrac{1}{r^2} = \cfrac{25}{36}$ or $\cfrac{1}{25}$.
Unfortunately, this isn't the answer. I do not see any flaw in my reasoning. Please help me with my mistake and suggestions of better methods would be most appreciated.
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It is unfortunate that the complete question was not posted, though one might suspect this was a multiple-choice question; it would be helpful, but not essential (as we'll see), to know whether we are to arrive at a numerical value or just an expression in terms of the arithmetic difference and/or the geometrical factor.
In any case, it seems likely that we won't actually care what the values of $ \ a \ $ and $ \ c \ $ are. So saying, we will try to work as far as possible only with the ratio $ \ \frac{a}{c} \ \ . $ It will be helpful to express the specified terms as $ \ \ c-6d \ , \ c - 2d \ , \ c \ \ . $ We can then describe the geometric ratio $ \ r \ $ by
$$ \frac{a}{c} \ = \ \frac{1}{r^2} \ \ , \ \ \frac{a}{c-2d} \ = \ \frac{1}{r} $$ $$ \Rightarrow \ \ \frac{a}{c} \ = \ \left( \ \frac{a}{c-2d} \ \right)^2 \ = \ \left(\frac{a}{c} \right)^2 · \frac{1}{ \left[ \ 1 \ - \ 2· \left(\frac{d}{c} \right) \ \right]^2} \ \ \Rightarrow \ \ \frac{a}{c} \ = \ \left[ \ 2\left(\frac{d}{c} \right) \ - \ 1 \ \right]^2 \ \ . $$
It remains to find a value or expression for $ \left(\frac{d}{c} \right) \ . $ Returning to the ratio between the indicated terms, we have
$$ r \ = \ \frac{c-2d}{c-6d} \ = \ \frac{c}{c-2d} \ \ \Rightarrow \ \ c^2 - 6cd \ = \ c^2 - 4cd + 4d^2 $$ $$ \Rightarrow \ \ 4d \ = \ -2c \ \ \Rightarrow \ \ \frac{d}{c} \ = \ -\frac12 \ \ \Rightarrow \ \ \frac{a}{c} \ = \ \left[ \ 2\left(-\frac12 \right) \ - \ 1 \ \right]^2 \ = \ \mathbf{4} \ \ . $$
We see from this that the arithmetic series is decreasing , with the selected terms being $$ c-6d \ = \ c \ - \ 6·\left(-\frac12 c \right) \ = \ 4c \ \ , \ \ c-2d \ = \ c \ - \ 2·\left(-\frac12 c \right) \ = \ 2c \ \ , \ \ c \ \ , $$
confirming the geometric progression among these.
[It might be remarked that the issue of a "constant" series was avoided in this development. We can see, however, that setting $ \ d = 0 \ $ does imply that $ \ \frac{a}{c} \ = \ \left[ \ 2\left(\frac{0}{c} \right) \ - \ 1 \ \right]^2 \ = \ 1 \ \ , $ a separate indication that this expression "behaves correctly". ]
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$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$ to function I want to calculate $3-\frac{5}{2}+\frac{7}{24}-\frac{9}{720}+...$, while using $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$.
Here is my attempt:
Firs I proved that the radius of converges for this series is $R= \infty$.
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n} =~_{k=n-1} ~ = \sum_{k=0}^\infty \frac{(-1)^{k+2}}{(2k)!}(2k+3)x^{2k+2} = \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k)!}(x^{2k+3})'$$
$$= (\sum_{k=0}^\infty \frac{(-1)^{k}}{(2k)!}x^{2k+3})'= (x^3 \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k)!}x^{2k})'=(x^3 \cos(x))'=3x^2 \cos(x)-x^3 \sin(x)$$
So $f(x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}=3x^2 \cos(x)-x^3 \sin(x)$.
If I did everything correct, than I should find a value for $x$ which will help me to calculate the initial sum.
Did I do it ok?
Thanks a lot!
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$$S=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$$
Let $k=n-1$, then
$$S=\sum_{k=0}^{\infty} (-1)^k (2k+3)\frac{x^{2k+2}}{(2k)!}$$
We konw that $$x^3\cos x=\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+3}}{(2k)!}$$
D.w.r.t. $x$, then
$$-x^3\sin x+3x^2 \cos x=S$$
OP's summation is verified. The sum of the numerical series is nothing but
$$3\cos 1-\sin 1$$ which is when $x=1$.
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Finding the coordinate of a point after multiple transformations
Find the new coordinates of $(2,3)$ if following transformations take place:
(i) Origin is shifted to $(1,1)$
(ii) Axes are rotated by an angle $45^{\circ}$ in anticlockwise sense
(iii) Origin is shifted to $(1,1)$ and then axes are rotated by angle $45^{\circ}$ in clockwise sense
(iv) Coordinate axes become $x+y+1=0$ and $x-y+2=0$
Here is what I have tried:
At first step, coordinates become $(1,2)$, then using $X=x\cos\theta+y\sin\theta, Y=y\cos\theta-x\sin\theta$, we get at the second step $(3/\sqrt{2}, 1/\sqrt{2})$. Now again translating origin to $(1,1)$, and applying rotation, this time $45^{\circ}$ clockwise we get $(1,2-\sqrt{2})$. Finally, changing axes to given lines, we get using the following formula, the point as $(2\sqrt{2}-1,1+1\sqrt{2})$
$$X=\frac{lx+my+n}{\sqrt{l^2+m^2}}\\ Y=\frac{mx-ly+n^{'}}{\sqrt{l^2+m^2}}$$
Turns out the actual answer given is $(1/\sqrt{2}, 6/\sqrt{2})$. Can anybody spot the mistake? Thanks.
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(i) Origin is shifted to $(1, 1)$, so coordinates become $(1, 2)$.
(ii) If the axes are rotated by $45^\circ$ anticlockwise then the relation between
the original coordinates $(x,y)$ and the new coordinates $(x', y')$ is
$ x = \cos 45^\circ x' - \sin 45^\circ y' $
$ y = \sin 45^\circ x' + \cos 45^\circ y' $
which implies,
$ x' = \cos 45^\circ x + \sin 45^\circ y $
$ y' = -\sin 45^\circ x + \cos 45^\circ y $
Hence, the new coordinates are
$(x', y') = (\dfrac{5}{\sqrt{2}} , \dfrac{1}{\sqrt{2}})$
(iii) In this case, we have a combination of shifting and rotation, so
$(x, y) = (1,1) + R (x', y') $
solving,
$(x', y') =( \cos (-45^\circ) (x-1) + \sin (-45)^\circ (y-1) , -\sin (-45^\circ) (x-1) + \cos (-45^\circ) (y-1)) $
And this evaluates to,
$(x', y') = (-\dfrac{1}{\sqrt{2}}, \dfrac{3}{\sqrt{2}})$
(iv) The point of intersection of the axes is $(-\dfrac{3}{2} , \dfrac{1}{2} )$
The axes orientation is rotated by $45^\circ$ clockwise with respect $xy$ axes.
Hence,
$(x', y') = \cos (-45^\circ) (x+\dfrac{3}{2}) + (\sin -45^\circ) (y-\dfrac{1}{2}) , -\sin( -45^\circ ) (x+\dfrac{3}{2}) + \cos (-45^\circ )(y-\dfrac{1}{2}) ) $
And this evaluates to,
$(x', y') = ( \dfrac{1}{\sqrt{2}} , \dfrac{6}{\sqrt{2}} ) $
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$\int_0^\infty \frac{x^a}{(x^2+1)^2}dx$ The following exercise is from the book of Churchill of complex analysis. Solve the integral
$$\int_0^\infty \frac{x^a}{(x^2+1)^2}dx$$
where $-1<a<3$ the book of complex analysis of Churchill suggests the indented countour:
And also the book gives you the answer $\frac{\pi(1-a)}{4\cos(a\pi/2)}$.
Progress: I did $\int_\Gamma \frac{z^a}{(z^2+1)}dz$ where $\Gamma$ is the region above. I already have
$$\int_\Gamma \frac{z^a}{(z^2+1)^2}dz=\frac{i\pi e^{ai\pi/2}(1-a)}{2},$$
$$\int_\Gamma \frac{z^a}{(z^2+1)^2}dz=\int_{-R}^{-r} \frac{z^a}{(z^2+1)^2}dz+\int_r^R \frac{z^a}{(z^2+1)^2}dz-\int_{C_r} \frac{z^a}{(z^2+1)^2}dz+\int_{C_R} \frac{z^a}{(z^2+1)^2}dz.$$
$$\lim_{R\rightarrow\infty}\int_{C_R} \frac{z^a}{(z^2+1)^2}dz=0.$$
and
$$\lim_{R\rightarrow\infty,r\rightarrow 0}\left(\int_{-R}^{-r} \frac{z^a}{(z^2+1)^2}dz+\int_r^R \frac{z^a}{(z^2+1)^2}dz\right)=(1+(-1)^a)\int_0^\infty\frac{z^a}{(z^2+1)^2}dz=(1+e^{i\pi a})\int_0^\infty\frac{z^a}{(z^2+1)^2}dz$$
So I get
$$\int_0^\infty\frac{z^a}{(z^2+1)^2}dz=\frac{\pi(1-a)}{4\cos(a\pi/2)}+\lim_{r\rightarrow 0}\int_{C_r} \frac{z^a}{(z^2+1)^2}dz.$$
Where $C_R$ is the upper semicircle of radius $R$ with positive orientation and $C_r$ is the upper semicircle of radius $r$ with positive orientation.
So that tells me it must happen that
$$\lim_{r\rightarrow 0}\int_{C_r} \frac{z^a}{(z^2+1)^2}dz=0,$$
but I haven't been able to prove it, I have tried to use Jordan's lemma doing a change of variable, integrating and using L'Hopital and I don't get that this integral converges to $0$. Does somebody knows how to achieve it?
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$$ \left | \int_{C_r}\frac{z^a}{(z^2+1)^2} dz \right |\le \int_{C_r} \left | \frac{z^a}{(z^2+1)^2} dz \right | \le \int_{C_r}\frac{r^a}{(1-r^2)^2} |dz|= \frac{\pi r^{a+1}}{(1-r^2)^2} \to 0$$
as $a+1 \gt 0$.
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Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $b^4+b^3+b^2+b+1=0$)
$uv=(b+b^4)(b^2+b^3)$
$=(b^3+b^4+b^6+b^7)$
$=b^3(1+b+b3+b4)$
$=b^3(-b^2)$
$=-b^5$
$=-1^5$
$=-1$ (using $z^5=1$)
For ii)
$u-v=(b+b^4)-(b^2+b3)>0$
$(b+b^4)-(-b-b^4-1)>0$
$2b+2b^4+1>0$
$b+b^4>\frac{-1}{2}$
Although it's a dead-end after that.
What's the general idea for going about solving part ii)?
|
Just compute
$$
(u-v)^2=(b+b^4-b^2-b^3)^2=-(b^4+b^3+b^2+b+1)+5=5.
$$
We have used $b^5=1$ in the second step.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4172762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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|
Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\iff (ab+bc+ca)(a+b+c)=abc \iff (a+b)(b+c)(c+a)=0 $
I'm curious to prove this some other ways. I tried to use function and inequality but still no progress.
|
Another approach, as you requested.
I write your question as (L)$\iff $(R). Since (L) is clearly
valid if $a+b=0,$ it suffices to prove (L) implies (R).
I now assume that (L) is valid. By scaling, we can assume that $c=1.$
From
$$\frac{1}{a} +\frac{1}{b} +1=\frac{1}{a+b+1},$$
we get
$$\frac{a+b}{ab}=\frac{1}{a+b+1}-1=-\frac{a+b}{a+b+1}\tag1$$
We're done if $a+b=0$ so cancel $a+b$ in (1) to obtain
$-ab=a+b+1.$ Thus, $(a+1)(1+b)=0,$ proving (R).
(Scaling uses that the right side of (R) is zero.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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|
If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neither of these known angles worked. But by doing it I realized that for $\theta=\frac{\pi}4+k\pi\quad$, $\tan^2\theta+4\tan\theta=5\quad$ and $2\sin\theta+\cos\theta\neq\sqrt3$ Hence the fourth choice is wrong.
Also tried to expanding,
$$\tan^2\theta+4\tan\theta=\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{4\sin\theta}{\cos\theta}=\dfrac{\sin^2\theta+4\sin\theta\cos\theta}{\cos^2\theta}$$But can't continue even writing $4\sin\theta\cos\theta=2\sin2\theta$ doesn't help.
|
$(2\sin\theta+\cos\theta)^2 = 3$
$4\sin^2\theta + \cos^2\theta + 4 \sin\theta \cos\theta = 3 \sin^2\theta + 3\cos^2\theta$
$\sin^2\theta + 4\sin\theta \cos\theta = 2\cos^2\theta$
Now dividing both sides by $\cos^2\theta$,
$\tan^2\theta + 4\tan\theta = 2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4177501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.
|
Let the polynomial be $P(x)$. It is given that for some polynomials $Q(x),Q_1(x)$
$$P(x)=(x^2+x+1)Q(x)+1-x$$
$$P(x)=(x^2-x+1)Q_1(x)+3x+5$$
Now it is well known that $x^2+x+1$ has the zeros as $\omega,\omega^2$ and $x^2-x+1$ has the zeroes $-\omega,-\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$
So We get
$$P(\omega)=1-\omega$$
$$P(\omega^2)=1-\omega^2$$
$$P(-\omega)=3\omega+5$$
$$P(-\omega^2)=-3\omega^2+5$$
Now let us assume for some polynomial $Q_2(x)$ we have
$$P(x)=(x^4+x^2+1)Q_2(x)+Ax^3+Bx^2+Cx+D$$
Using the fact that $x^4+x^2+1$ has the zeros $\omega,\omega^2,-\omega,-\omega^2$ we get four linear equations as:
$$\left[\begin{array}{cccc}
1 & \omega^{2} & \omega & 1 \\
1 & \omega & \omega^{2} & 1 \\
-1 & \omega^{2} & -\omega & 1 \\
-1 & \omega & -\omega^{2} & 1
\end{array}\right]\left[\begin{array}{c}
A \\
B \\
C \\
D
\end{array}\right]=\left[\begin{array}{c}
1-\omega \\
1-\omega^{2} \\
5-3 \omega \\
5-3 \omega^{2}
\end{array}\right]$$
Now let us solve by Cramer's rule. The determinant of the matrix is $\Delta =12$,
$\Delta_1=-24$, $\Delta_2=24$, $\Delta_3=12$ and $\Delta_4=60$.
Thus the values of $A,B,C,D$ are
$$\begin{aligned}
&A=\frac{\Delta_{1}}{\Delta}=-2 \\
&B=\frac{\Delta_{2}}{\Delta}=2 \\
&C=\frac{\Delta_{3}}{\Delta}=1 \\
&D=\frac{\Delta_{4}}{\Delta}=5
\end{aligned}$$
So the required remainder is
$$-2x^3+2x^2+x+5$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4179472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How Can I calculate $\int_{0}^{1}\frac{x^{2n}}{x^2+3}dx$ While simplifying a big integral I had to calculate the following two integrals
$$\int_{0}^{1}\frac{x\space \tan^{-1}(x)}{x^2+3}, \int_{0}^{1}\frac{x\space \tan^{-1}(x)}{3x^2+1}$$
$$I=\int_{0}^{1}\frac{x\space \tan^{-1}(x)}{x^2+3}$$
$$\implies I=\int_{0}^{1}\frac{x}{x^2+3}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1} x^{2n-1} dx$$
$$\implies I=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{x^2+3} dx$$
Doing same thing with the second integral, we get
$$J=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{3x^2+1} dx$$
But I am stuck at those 2 integrals
$$\int_{0}^{1}\frac{x^{2n}}{x^2+3} dx,
\int_{0}^{1}\frac{x^{2n}}{3x^2+1} dx$$
Since they are almost the same so I think that doing one is enough to evaluate the other integral( Or Correct me If I'm wrong).
Thank you for your help!
|
I think we can get a closed form. For that we simply solve
$$\Delta I_n +4 I_n = \frac{1}{2n+1},\ I_0 = \frac{\pi}{3\sqrt{3}}$$ (credit of @Martund)
with which we obtain
$$I_n = (-3)^n\left(C-\frac{1}{3}\sum\left(-\frac{1}{3}\right)^{n}\frac{1}{2n+1}\right)=(-3)^n\left(C-\frac{1}{3}\int^{-\frac{1}{3}} \frac{t^{2n}}{t^2-1}dt\right)$$ and so
$$C=\frac{\pi}{3\sqrt{3}}+\frac{\ln\left(\frac{4}{3}\right)-\ln\left(\frac{2}{3}\right)}{6}=\frac{\pi}{3\sqrt{3}}+\frac{\ln(2)}{6}$$ thus
$$I_n =(-3)^n\left(\frac{\pi}{3\sqrt{3}}+\frac{\ln(2)}{6}+\left(-\frac{1}{3}\right)^{2n+2}\frac{{_2}F_1\left(1,n+\frac{1}{2},n+\frac{3}{2};\frac{1}{9}\right)}{2n+1} \right)$$
The hypergeometric series above should always have a closed form:
|
{
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"url": "https://math.stackexchange.com/questions/4186035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
$(\frac{x}{x+1})^x$ decreasing I want to show that $f(x) = \left(\dfrac{x}{x+1}\right)^x$ is decreasing for $x > 0$; this is clear from plotting its graph. Taking its derivative,
$$f'(x) = \frac{x^x}{(x+1)^{x+1}}\left(1+(x+1)\log\left(\frac{x}{x+1}\right)\right).$$
So we need to show that $1+(x+1)\log\left(\frac{x}{x+1}\right) < 0$, which seems to require taking another derivative. Is there an easier way to show that $f(x)$ is decreasing?
|
By taking a derivative, you can easily show that
$$\log(1+y) \le y$$
for all $y > -1$, with equality iff $y=0$ (this is a standard inequality). In particular, since $-1<-\frac{1}{x+1}<0$,
$$\log\left(\frac{x}{x+1}\right) = \log\left(1 - \frac{1}{x+1}\right) < - \frac{1}{x+1}.$$
Therefore
$$1+(x+1)\log\left(\frac{x}{x+1}\right) < 1 + (x+1)\cdot\frac{-1}{x+1} = 0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4186184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Minimum of $\cos^2(x) (\csc^2(\frac{\pi}{2n} - \frac{x}{n}) + \csc^2(\frac{\pi}{2n} + \frac{x}{n}))$ for $0 \leq x < \pi/2$ and $n \geq 3$ I am trying to proove that $x=0$ is the minimum point of the function
$$f(x) = \cos^2(x)\left( \csc^2(x_+) + \csc^2(x_-) \right)$$
in the interval $0 \leq x < \pi/2$ where $x_\pm = \frac\pi{2n} \pm \frac x n$,
and $n \in \mathbb{N}$ such that $n \geq 3$.
I have unsuccessfully tried the following approaches.
*
*Attempted to show that $f(x) - f(0) \geq 0$, i.e.
$$\cos^2(x)\left(
\csc^2\left(\frac\pi{2n} - \frac x n\right) +
\csc^2\left(\frac\pi{2n} + \frac x n\right)
\right) - 2\csc^2\left(\frac\pi{2n}\right) \geq 0;$$
*Computed $f'(x) = \frac\partial{\partial x} f(x)$ and tried to show that $f'(x) \geq 0$ in the interval, where
\begin{align}
\frac\partial{\partial x} f(x) = 2\cos^2(x) \left(
\frac{1}{n} \cot(x_-)\csc^2(x_-) -
\frac{1}{n} \cot(x_+)\csc^2(x_+) -
\tan(x)\left(\csc^2(x_-) + \csc^2(x_+)\right)
\right);
\end{align}
*Used Taylor Series Expansion in $f(x)$ attempting to find a tight lower bound for $f(x)$.
Any help or hint would be appreciated.
|
Trying to proove that $x=0$ corresponds to $\large \color{red}{a}$ minimum point is not so difficult if you make a Taylor expansion of
$$f_n(x)=\cos ^2(x) \left(\csc ^2\left(\frac{\pi -2 x}{2 n}\right)+\csc ^2\left(\frac{\pi+2
x }{2 n}\right)\right)$$
$$f_n(x)=2 \csc ^2\left(\frac{\pi }{2 n}\right)\Bigg[1+ \left(\frac{1+3 \cot ^2\left(\frac{\pi }{2
n}\right)}{n^2}-1\right)x^2+O\left(x^4\right) \Bigg]$$ So, we have
$$f_n(0)=2 \csc ^2\left(\frac{\pi }{2 n}\right)\qquad f'_n(0)=0\qquad f''_n(0)=\frac{2 \left(\left(n^2+2\right) \cos \left(\frac{\pi }{n}\right)+4-n^2\right)
\csc ^4\left(\frac{\pi }{2 n}\right)}{n^2}$$ and
$$f''_n(0) > 0 \qquad \text{ } \qquad \forall n \quad \geq 3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4186322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do you prove the Cauchy product (multiplication) of two infinite power series (generating functions) which have different exponents/indices? I'm trying to multiply 2 generating functions ( (1/(1-x) ) and ( 1/(1-x^5) ) which have different denominations so that I can find the coefficient. When browsing math.stackexchange I found the equation:
\begin{align*}
\frac{1}{1-x}\cdot\frac{1}{1-x^r}=\sum_{n\ge 0} \left(1+\left\lfloor\frac{n}{r}\right\rfloor\right)x^n
\end{align*}
Could someone go through all of the steps of the equation and explain how and why the product of these two power series are equivalent to the summation shown on the right?
|
\begin{align*}
\frac{1}{1-x} \frac{1}{1-x^r}
&= (1+x+x^2+\cdots)(1+x^r+x^{2r}+\cdots) \\
&= 1+x+x^2+\cdots \\
&+ x^r+x^{r+1}+x^{r+2}+\cdots \\
&+ x^{2r} + x^{2r+1} + x^{2r+2} + \cdots \\
&+ \cdots
\end{align*}
Every exponent from $x^0, \ldots, x^{r-1}$ is included once, every exponent from $x^r, \ldots, x^{2r-1}$ is included twice, etc. So the coefficient of $x^n$ is $1 + \lfloor n/r\rfloor$.
|
{
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"url": "https://math.stackexchange.com/questions/4189058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Computing the series $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}$ I want to know how to compute the closed form for the series
$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}, \quad |A| < 1,$$
with or without special functions. When I plugged this into Mathematica, it gave me
$$\frac{1}{\sqrt{1 - |A|^2}}.$$
I also would really, really like to know if there are exercises/lessons all in one place (it's a bit hard to search the SE for something like this) or texts out there that can help me learn methods of doing these computations. I know I won't be able to always find closed forms of expressions like this, but I'm always amazed at what I see folks are able to compute here on Stack Exchange.
I thought to use a modified Bessel function $I_\alpha(z)$ with $\alpha = 0$,
$$I_0(z) = \sum_{k=0}^\infty \frac{1}{(k!)^2} \left(\frac{z}{2}\right)^{2k}$$
and perhaps differentiate and then evaluate at $|A|$, but I haven't figured out how to manipulate that. Any suggestions, general and specific?
Note: For some context, I am generally trying to compute norms for squeezed coherent states (quantum mechanics application) which have the form $\psi(z) = e^{Az^2/2} e^{-|z|^2/2}$, $z \in \mathbb{C}$, and I wind up with series like this all the time.
|
Just for reference, I guess, I finally did go back and take the suggestion of @samario28 and use the generalized binomial theorem,
$$ (1 + z)^\alpha = \sum_{n=0}^\infty \frac{(\alpha)_n}{n!} z^n = \sum_{n=0}^\infty \binom{\alpha}{n} z^n,$$
where $(\alpha)_n$ denotes the falling factorial. Just for good measure, a ratio test of $a_n = (2n)!/(n!)^2 z^{2n}$ will show that radius of convergence is for $|z| < 1/2$, which corresponds to our values for $|A|$. We have
\begin{align*} \binom{2n}{n} &= \frac{(2n)!}{n!n!} \\
&= \frac{2n (2n-1)(2n-2) \cdots 3 \cdot 2 \cdot 1}{n!n!} \\
&= \frac{2^{2n} n (n-\tfrac{1}{2}) (n-1) (n- \frac{3}{2}) \cdots \frac{3}{2} \cdot 1 \cdot \frac{1}{2} \cdot 1}{n!n!} \\
&= \frac{2^{2n} (n-\tfrac{1}{2}) (n- \frac{3}{2}) \cdots \frac{3}{2} \cdot \frac{1}{2}}{n!} \\
&= \frac{2^{2n} (-1)^n (\tfrac{1}{2}-n) (\frac{3}{2}-n) \cdots (-\frac{3}{2}) \cdot (-\frac{1}{2})}{n!} \\
&= \frac{2^{2n} (-1)^n (-\frac{1}{2})(-\frac{1}{2}-1) \cdots (-\frac{1}{2}-n+2)(-\frac{1}{2} - n+1)}{n!} \\
&= \frac{2^{2n} (-1)^n (-\frac{1}{2})_{n}}{n!}.
\end{align*}
Thus, $\alpha = -1/2$, and so
\begin{align*} \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \left(\frac{|A|}{2}\right)^{2n} &= \sum_{n=0}^\infty \frac{\big(-\!\frac{1}{2}\big)_n}{n!} \cdot 2^{2n} (-1)^n \left(\frac{|A|}{2}\right)^{2n} \\
&= \sum_{n=0}^\infty \frac{\big(-\!\frac{1}{2}\big)_n}{n!} \cdot (-|A|^2)^n \\
&= \frac{1}{\sqrt{1 - |A|^2}}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4192000",
"timestamp": "2023-03-29T00:00:00",
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|
Derivative of an integral with change of variable. I have this problem.
I want to compute the following integral in terms of $z$, where $x = az$
$$
\frac{d}{da}\int a^2x^2dx
$$
Computing the integral first, the derivative as the last passage, I obtain what I believe to be the correct solution, i.e.,
$$
\frac{d}{da}\int a^2x^2dx = \frac{d}{da}\left(a^2 \frac{x^3}{3}\right) = \frac{d}{da}\left(a^2 \frac{a^3z^3}{3} + c\right) = 5a^4\frac{z^3}{3}
$$
and the same by substituting first in the integral, keeping in mind that $dx = a dz$,
$$
\frac{d}{da}\int a^2x^2dx = \frac{d}{da}\int a^2(az)^2 adz \frac{d}{da}\int a^5z^2 dz = \frac{d}{da}\left(a^5 \frac{z^3}{3}\right) = 5a^4\frac{z^3}{3}.
$$
Things start to get messy when I take the derivative inside the integral, i.e.,
$$
\frac{d}{da}\int a^2x^2dx = \int 2ax^2dx = \int 2a a^2z^2 adz = 2a^4\frac{z^3}{3}.
$$
One might argue that I need to keep in consideration also of $d/da x$, since $x$ depends on $a$, after all. However, I still obtain a wrong result
$$
\frac{d}{da}\int a^2x^2dx = \int a^2 \frac{d}{da}\left(x^2\right) + 2a x^2 dx = \int \left( a^2 \frac{d}{da}\left(a^2z^2\right) + 2a a^2z^2 \right)adz \\
= \int 2a^4z^2 + 2a^4z^2dz = 4a^4\frac{z^3}{3}.
$$
Since I am obtaining three different results, I must be making at least two different mistakes; I wonder where these mistakes are. (sorry for being lazy and not including the constant $c$ from the integrals)
|
We substitute $x=az$ right from the beginning and obtain
\begin{align*}
\color{blue}{\frac{d}{da}\int a^2x^2dx}&=\frac{d}{da}\int a^2(az)^2d(az)\\
&=\frac{d}{da}\left(a^5\int z^2 dz\right)\tag{1}\\
&=\left(\int z^2 dz\right)\left(\frac{d}{da} a^5\right)\tag{2}\\
&\,\,\color{blue}{=\left(\frac{1}{3}z^3+C\right)\left(5a^4\right)}\tag{3}\\
\end{align*}
Comment:
*
*In (1) we use $d(az)=adz$ and factor out $a$.
*In (2) we factor out the constant part.
*In (3) we integrate and differentiate.
Next we look at OPs final approach somewhat more detailed. Since OP wants to compute the integral with respect to $z$ we consider $x=x(z)=az$ as function of $z$. We obtain
\begin{align*}
\frac{d}{da}\int a^2x^2\,dx&=\frac{d}{da}\int a^2x^2(z)\,d\left(x(z)\right)\\
&=\frac{d}{da}\int a^2 x^2(z) x^{\prime}(z)\,dz\tag{4}
\end{align*}
We are now ready to do the differentiation with respect to $a$ before the integration. To better see what's going on we write
$x=x(z,a)$ and $x_z(z,a)$ instead of $x^{\prime}(z)$ to indicate that we differentiate with respect to $z$.
We obtain from (4) by successively applying the multiplication rule $(u\,v)^{\prime}=u^\prime v+uv^\prime$
\begin{align*}
\color{blue}{\frac{d}{da}}&\color{blue}{\int a^2 x^2(z,a) x_z(z,a)dz}\\
&=\int \frac{d}{da} \left(a^2 \cdot \left(x^2(z,a) x_z(z,a)\right)\right)dz\\
&=\int 2a \left(x^2(z,a) x_z(z,a)\right)dz\\
&\qquad +\int a^2\frac{d}{da}\left(x^2(z,a) x_z(z,a)\right)dz\\
&=\int 2a \left(x^2(z,a) x_z(z,a)\right)dz\\
&\qquad +\int a^2 \left(2x(z,a) \left(\frac{d}{da}x(z,a)\right) x_z(z,a) +x^2(z,a)\frac{d}{da} x_z(z,a)\right)dz\\
&=\int 2a \left((az)^2 a\right)dz\\
&\qquad +\int a^2 \left(2(az)\left(z\right) a + (az)^2\cdot 1\right)dz\\
&=2a^4\int z^2\,dz+2a^4\int z^2\,dz + a^4 \int z^2dz\\
&=5a^4\int z^2dz\\
&\,\,\color{blue}{=5a^4\left(\frac{1}{3}z^3+C\right)}
\end{align*}
in accordance with (3).
|
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"url": "https://math.stackexchange.com/questions/4192281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
What is the origin and meaing of the term "Telescoping Series"? I looked into Carl. B. Boyer and Morris Kline books of math history, some calculus books like Apostol and Swokowski, many pages on the internet and even the Tractatus de Seriebus Infinitis of Jacobi Bernoulli with no sucess to find out the origin and meaning of the term "Telescoping Series".
May someone help?
|
Picture an old telescope (a.k.a. spyglass) that is retractable.
When the terms of a series contain differences, internal terms can be canceled, much like the segments of the telescope overlapping as it is contracted. For example, using the identity
$$
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1},
$$
\begin{alignat*}{8}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4}
+ \cdots + \frac{1}{(n-1) \cdot n}
&= \frac{1}{1} &{}- \frac{1}{2} & & & & & \\
& & {}+ \frac{1}{2} &{}- \frac{1}{3} & & & & \\
& & & {}+ \frac{1}{3} & &{}- \frac{1}{4} & & \\
& & & & & & \ddots & \\
& & & & & & & {}+ \frac{1}{n-1} - \frac{1}{n} \\
&= 1 \rlap{{}- \frac{1}{n}} & & & & & &
\end{alignat*}
|
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|
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$. The following question is taken from the practice set of JEE exam.
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$.
$$\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x\\\implies {^6}C_1\alpha^0+{^6}C_2\alpha^1+{^6}C_3\alpha^2+{^6}C_4\alpha^3+{^6}C_5\alpha^4+{^6}C_6\alpha^5=x\\\implies {^6}C_1\alpha^1+{^6}C_2\alpha^2+{^6}C_3\alpha^3+{^6}C_4\alpha^4+{^6}C_5\alpha^5+{^6}C_6\alpha^6=\alpha x\\\implies {^6}C_0\alpha^0+{^6}C_1\alpha^1+{^6}C_2\alpha^2+{^6}C_3\alpha^3+{^6}C_4\alpha^4+{^6}C_5\alpha^5+{^6}C_6\alpha^6=\alpha x+{^6}C_0\\\implies (1+\alpha)^6=1+\alpha x$$
How shall I proceed next? I understand $\alpha$ is $6^{th}$ root of unity. How to use that here? Maybe $\alpha^6=1$ can be written as $$\alpha^6-1=0\\\implies (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1)=0$$
Also, $\alpha\ne1\implies \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$. Is this of some help here? Thanks.
|
$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$
$\implies\alpha^3(\alpha^2+\alpha+1)+(\alpha^2+\alpha+1)=0$
$\implies(\alpha^3+1)(\alpha^2+\alpha+1)=0$
$\implies\alpha^3=-1, \alpha^2+\alpha+1=0$
Now from your equation, $(1+\alpha)^6=1+x\alpha$
$(1+\alpha)^6-1=x\alpha$
$\alpha(\alpha+2)(\alpha^2+3\alpha+3)(\alpha^2+\alpha+1)=x\alpha$
$0=x\alpha$
$\therefore|x|=0$
Here I didn't talked about the equation $\alpha^3=-1$. So this is just a special case.
|
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|
How to evaluate $\lim\limits_{x\to y}\frac{\tan x-\tan y}{x-y}$?
Evaluate $$\lim_{x\to y}\frac{\tan x-\tan y}{x-y}.$$
I used elementary methods to evaluate the limit. Here is my solution:
Let $d=x-y$. Then as $x\to y$, we have $d\to 0$. Now,
$\lim\limits_{x\to y}\frac{\tan x-\tan y}{x-y}\\ =\lim\limits_{d\to 0}\frac{\tan x-\tan (x-d)}d\\ =\lim\limits_{d\to 0}\frac{\tan x-\frac{\tan x-\tan d}{1+\tan x\tan d}}{d}\\ =\lim\limits_{d\to 0}\frac{\tan x(1+\tan x\tan d)-\tan x+\tan d}{d(1+\tan x\tan d)}\\ =\lim\limits_{d\to 0}\frac{\tan x(1+\tan x\tan d-1)}{d(1+\tan x\tan d)}+\frac{\tan d}d \cdot \frac1{1+\tan x\tan d}\\ =\lim\limits_{d\to 0}\frac{\tan^2x}{1+\tan x\tan d}\cdot \frac{\tan d}d+\frac{\tan d}d \cdot \frac1{1+\tan x\tan d}\\ =\lim\limits_{d\to 0}\frac{\tan^2x}{1+\tan x\tan 0}\cdot 1+1\cdot\frac1{1+\tan x\tan 0}\\ =\tan^2x+1$
I want to know whether my solution is correct or not. And can this limit be evaluated using L'Hôpital's rule? (I've just learnt L'Hôpital's rule and I don't know how to use this to evaluate a limit which contains two variables). And some other methods to solve the problem are also welcome.
|
You can use the definition of the derivative to evaluate such a limit :
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$$
But here's an approach if you want to exercise more, by the end of this answer I'll mention the important properties you have to know :
\begin{align}
\lim_{x\to y} \frac{\tan x -\tan y}{x-y}&=\lim_{x\to y} \frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{x-y}\\
&=\lim_{x\to y} \frac{\sin x \cos y-\sin y\cos x}{\cos y\cos a}\frac{1}{x-y} \\
&=\lim_{x\to y} \frac{\frac{1}{2}\left(\sin(x+y)+\sin(x-y) -\sin(y+x)-\sin(x-y)\right)}{\frac{1}{2}\left(\cos(x+y)+\cos(x-y)\right)}\frac{1}{x-y} \\
&=\lim_{x\to y} \frac{-2\sin(y-x)}{\cos(2y)+1}\frac{-1}{y-x}\\
&=\lim_{x\to y} \frac{\sin (y-x)}{y-x} \frac{1}{\cos^2 y}\\
&= \frac{1}{\cos^2 y}
\end{align}
Recall :
$$\sin \alpha \cos \beta =\frac{1}{2} \left(\sin (\alpha -\beta)+\sin(\alpha +\beta)\right)$$
$$\cos \alpha \cos \beta =\frac{1}{2} \left(\cos(\alpha -\beta)+\cos(\alpha+\beta)\right)$$
And this usual limit :
$$\lim_{\alpha\to 0}\frac{\sin(\alpha)}{\alpha}=1$$
Note that :
$$\frac{1}{\cos^2 \alpha}=\frac{\cos^2 \alpha +\sin^2 \alpha}{\cos^2 \alpha}
=1+\frac{\sin^2 \alpha}{\cos^2\alpha}
=1+\tan^2 \alpha $$
Good luck !
|
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|
Given $3$ points on a unit circle, figure out something about them. Question: Given three points $(a, b), (c, d)$ and $(x, y)$ on the unit circle in a rectangular coordinate plane, find the maximum possible value of the expression $(ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy + a)^2 + (dx - cy - b)^2. $
Answer: We will prove that the only value for this expression is $4$.
Without loss of generality, assume that the unit circle is the graph $x^2 + y^2 = 1$. This means that $b^2 = 1 - a^2, d^2 = 1 - c^2,$ and $y^2 = 1 - x^2$.
Expanding the expression, we get
$a^2x^2 + b^2y^2 + c^2 + 2axby - 2acx - 2byc + b^2x^2 + a^2y^2 + d^2 - 2bxay - 2ayd + 2bxd + c^2x^2 + d^2y^2 + a^2 + 2cxdy + 2ady + 2acx + d^2x^2 + c^2y^2 + b^2 - 2dxcy + 2bcy - 2dxb$.
This was a long expression! Fortunately, we see that most of the terms cancel out and we are left with $a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 + c^2x^2 + c^2y^2 + d^2x^2 + d^2y^2 + a^2 + b^2 + c^2 + d^2$.
This can be factored to $(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + 1)$.
Using the fact from above that $b^2 = 1 - a^2, d^2 = 1 - c^2,$ and $y^2 = 1 - x^2$, this expression simplifies to $(a^2 + 1 - a^2 + c^2 + 1 - c^2)(x^2 + 1 - x^2 + 1) = 2(2) = 4$.
So, the answer to our original question is $\boxed{4}$.
Please verify my proof to see if there are any flaws or mistakes with the proof. Thanks in advance!
|
It is correct. Here is a way to see it by complex numbers and geometry. By abuse of notations, using complex numbers to represent points, one writes $$A=(a,b)=a+bi,B=(c,d)=c+di,C=(x,y)=x+iy,$$ three complex numbers on the unit circle, so $\overline{A}A=1,$ etc. Then it is easy to see using multiplication of complex numbers that the given expression equals $$|A\overline{C}-\overline{B}|^2+|B\overline{C}+\overline{A}|^2$$
$$=|A-\overline{B}C|^2+|A+\overline{B}C|^2$$
$$=|A-D|^2+|A+D|^2,$$ where $D=\overline{B}C.$ Then by the known theorem for parallelogram, the square sum of diagonals equals the square sum of the sides, which is $4$. QED
|
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|
Calculate surface of a triangle It's quite a simple question solvable with many methods but the $2$ of them I picked don't agree.
First of I have chosen the triangle with corners: $(3,0,0) \quad (0,3,0) \quad (0,0,6)$
Thus, this one here:
First Method is by evaluating the surface Integral: $A = \int_{S}1\,\mathrm{dS} = \int_{(u,v)}1\,|n|\,\mathrm{du\,dv}$
As parametrization I came up with: $\psi(u,v) = \left(\begin{array}{cc}3\,u\\ 3\,v \\ 6-6\,u-6\,v\end{array}\right) \quad u \in[0,1] \quad v \in [0,1] \quad u+v<1$
Hence $n = \partial_u\psi \times \partial_v\psi = \left(\begin{array}{cc}18\\ 18 \\ 9\end{array}\right)\quad$ $\Rightarrow \quad \displaystyle{A = \int_0^1 \int_0^{1-u}} \sqrt{2\cdot18^2+9^2}\,\mathrm{dv\,du} = 13,5$
Next Method is by using straight geometry: $A= \frac{1}{2}\,g\,h$ (base times hight)
It should be: $A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\sqrt{3^2-\frac{1}{2}\sqrt{18}}\right)^2} \approx 13,9$
Now again, where does this difference come from? Probably one could check by calculation $\textbf{det}$
|
You must have:
$$A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\sqrt{3^2-\left(\frac{1}{2}\sqrt{18}\right)^\color{red}{2}}\right)^2} =13.5.$$
Or you can note that the triangles at the base are isosceles:
So $AC=BC=\frac{\sqrt{18}}{2}$ is what you need:
$$A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\frac{\sqrt{18}}{2}\right)^2} =13.5.$$
Alternatively, you can use Heron's formula:
$$a=3\sqrt{2},b=c=3\sqrt5,\\
S=\sqrt{p(p-a)(p-b)(p-c)}=\\
\sqrt{\frac{3\sqrt{2}+6\sqrt5}{2}\cdot \frac{6\sqrt{5}-3\sqrt2}{2}\cdot \frac{3\sqrt{2}}{2}\cdot \frac{3\sqrt{2}}{2}}=\\
\sqrt{\frac{162\cdot 18}{16}}=\frac{27}{2}=13.5.$$
|
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|
Selecting identical and distinguishable balls when the order matters I made up a question such that
There are $3$ identical blue balls ,$2$ identical red balls , $5$ distinguishable yellow balls ,$4$ distinguishable green balls. We want to make a mixture consisting of $8$ balls by using these balls. How many ways are there
a-) If selection order does not matter
b-) If selection order matters
MY WORK =
a-) This part was easy by using generating functions such that
Blue balls = $(1+x +x^2 +x^3 ) $
Red balls = $(1+x +x^2 )$
Yellow balls = $(1+5x +10x^2 +10x^3 +5x^4 +x^5) $
Green balls = $(1+4x +6x^2 +4x^3 +x^4)$
We should find the coefficient of $[x^{8}]$ in the expansion of them .
b-) This part is where i hesitated my solution. I used exponential generating functions , but i am not sure about the exponential generating function of distinguishable balls.
Blue balls = $$\bigg(1+ \frac{x}{1!} +\frac{x^2}{2!} + \frac{x^3}{3!} \bigg ) $$
Red balls = $$\bigg(1+ \frac{x}{1!} +\frac{x^2}{2!} \bigg)$$
Yellow balls = $$ \bigg (1 + \frac{P(5,1) \times x}{1!} +\frac{P(5,2) \times x^2}{2!} + \frac{P(5,3) \times x^3}{3!} + \frac{P(5,4) \times x^4}{4!} + \frac{P(5,5) \times x^5}{5!} \bigg)$$
Green balls = $$ \bigg (1 + \frac{P(4,1) \times x}{1!} +\frac{P(4,2) \times x^2}{2!} + \frac{P(4,3) \times x^3}{3!} + \frac{P(4,4) \times x^4}{4!} \bigg)$$
So , i should find the coefficient of $\frac{x^{8}}{8!}$ or find the coefficient of $x^{8}$ and multiply it by $8!$.
I am not sure about part $b$ .I want you to check my solution.
|
Your work is all correct.
Here is a way to simplify your work. Give each of the yellow and green balls its own generating functions, since $5$ distinguishable yellow balls is the same as 5 different colors each with one ball. The g.f. for each of these colors with a single ball is $1+x$, for both the ordered and unordered cases. Therefore, the unordered case is
$$
[x^8](1+x+x^2+x^3)(1+x+x^2)(1+x)^{5+4}
$$
and the ordered case is
$$
8!\cdot [x^8](1+x+x^2/2!+x^3/3!)(1+x+x^2/2!)(1+x)^{5+4}
$$
|
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For any $ a \in \mathbb{N}$ ; $f(x)=x^4+(4a+2)x^2+1$ is irreducible in $\mathbb{Z}[X]$ & reducible mod $p$ for all $p$. A.1 First we show $f(x)$ is reducible mod $p$ for all $p$.
Taking $y=x^2$ ; $f(x)$ becomes $y^2+(4a+2)y+1$. Solving the equation $y^2+(4a+2)y+1=0$ we get
$$
y=-(2a+1)\pm2\sqrt{a(a+1)}
$$
Now replacing $y$ by $x^2$ we have
$$
x=\pm \sqrt{-(2a+1)+2\sqrt{a(a+1)}} = \pm \ (i\sqrt{a} + i\sqrt{a+1})
$$
&
$$
x=\pm \sqrt{-(2a+1)-2\sqrt{a(a+1)}} = \pm \ (i\sqrt{a} - i\sqrt{a+1})
$$
Hence we have obtained all the roots of the polynomial $f(x)$ say $x_1, x_2, x_3, x_4$ where
$$
x_1= (i\sqrt{a} + i\sqrt{a+1}) \ ; \ x_3= -(i\sqrt{a} + i\sqrt{a+1})
$$
$$
x_2= (i\sqrt{a} - i\sqrt{a+1}) \ ; \ x_4= -(i\sqrt{a} - i\sqrt{a+1})
$$
A.2 The three ways of expanding this quartic are as follows:
*
*$\{(x-x_1)(x-x_2)\}\{(x-x_3)(x-x_4)\}=(x^2+1-2\sqrt{a}ix)(x^2+1+2\sqrt{a}ix)\\
= (x^2+1)^2-(2\sqrt{a}ix)^2 \\
= (x^2+1)^2-2^2\mathbf{(-a)}x^2$
*$\{(x-x_1)(x-x_4)\}\{(x-x_2)(x-x_3)\}=(x^2-1-2\sqrt{a+1} \ ix)(x^2-1+2\sqrt{a+1} \ ix)\\
= (x^2-1)^2-(2\sqrt{a+1} \ ix)^2\\
=(x^2-1)^2-2^2 \ \mathbf{(-(a+1))} \ x^2$
*$\{(x-x_1)(x-x_3)\}\{(x-x_2)(x-x_4)\}\\
=(x^2+(2a+1)+2\sqrt{a(a+1)})(x^2+(2a+1)-2\sqrt{a(a+1)})\\
=(x^2+(2a+1))^2-(2\sqrt{a(a+1)})^2\\
=(x^2+(2a+1))^2-2^2\mathbf{(a(a+1))}$
A.3 Now the final piece.
$f(x)$ can be factored in any of the above 3 ways. Consider any prime $p$.
*
*If $\mathbf{-a}$ is a square element in $\mathbb{F}_p$; there exist $b \in \mathbb{F}_p$ such that $-a=b^2$. So factoring $f(x)$ as in form A.2.1. Hence considering $f(x)$ in $\mathbb{F}_p[x]$ we have
$$
f(x)=(x^2+1)^2-2^2(-a)x^2=(x^2+1)^2-(2bx)^2=(x^2+1-2bx)(x^2+1+2bx)
$$
Hence $f(x)$ is reducible modulo $p$ if $-a$ is a square element.
*If $\mathbf{-(a+1)}$ is square element in $\mathbb{F}_p$, there exist $c \in \mathbb{F}_p$ such that $-(a+1)=c^2$. So we factor $f(x)$ as in form A.2.2. Now considering $f(x)$ in $\mathbb{F}_p[x]$ we have
$$
f(x) =(x^2-1)^2-2^2 \ \mathbf{(-(a+1))} \ x^2=(x^2-1)^2-(2cx)^2=(x^2-1+2cx)(x^2-1+2cx)
$$
Hence we get $f(x)$ is reducible modulo $p$ if $2$ is a square element.
*Now if both $\mathbf{-a}$ and $\mathbf{-(a+1)}$ are non square elements in $\mathbb{F}_p$ then $\mathbf{a(a+1)}$ is a square element in $\mathbb{F}_p$. Hence factoring $f(x)$ as in the form A.2.3 we get that $f(x)$ is reducible modulo $p$.
Hence we have $f(x)$ is reducible modulo $p$ for all $p$.
How to show $f(x)$ is irreducible in $\mathbb{Z}[X]$?
|
$f$ has no real roots, so the only way to factor it over $\mathbb Z[x]$ is as two second degree polynomials of the form $(x-r)(x-\bar r)$, where $r$ is a complex root. You already found the roots, so the two factors would have to be
$$
(x-x_1)(x-x_3) = x^2+2a+1+2\sqrt{a(a+1)}
$$
and
$$
(x-x_2)(x-x_4) = x^2+2a+1-2\sqrt{a(a+1)}.
$$
But $a(a+1)$ can't be a square number (for example since $\gcd(a,a+1)=1$, which means both $a$ and $a+1$ would have to be square). So the polynomials above are not in $\mathbb Z[x]$.
|
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|
Inequality: $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$ I am asking this question on behalf of another student, who sought my help through a school mentoring scheme, and claims that the question is similar to that found in Question $1$ of the British Maths Olympiad.
Prove that for all positive real numbers $a, b, c$:
a) $(a+b)^2 \geq 4ab$
b) $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$
The person mentioned solved a) easily but struggled with b). To solve b), the person had attempted the following:
"...putting the right hand side as a single fraction and then cross multiplying as well as cancelling terms. But it turns out to be a repeated process which makes the original question even more complicated."
|
Looking at part a), I think the intention of part b) is
\begin{eqnarray*}\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}
& \stackrel{a)}{\leq} & \frac{(a+b)^2}{4(a+b)} +\frac{(b+c)^2}{4(b+c)} + \frac{(a+c)^2}{4(a+c)} \\
& = & \frac 14(a+b + b+c + a+c) \\
& = & \frac{a+b+c}{2}
\end{eqnarray*}
|
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|
How can I establish the convergence (or divergence) of $ \int_{0}^{\infty} \frac{x^2 \ln(\sin^2(x))}{(1+x^2)^2} \mathrm{d}x $ I want to find out wether the following integral converges
$ \displaystyle \int_{0}^{\infty} \frac{x^2 \ln(\sin^2(x))}{(1+x^2)^2} \mathrm{d}x \tag{1} $
The integral above converges iff the following integral converges.
$ \displaystyle \int_{0}^{\infty} \frac{x^2 \ln \left( \frac{\sin^2(x)}{x^2} \right) }{(1+x^2)^2} \mathrm{d}x \tag{2} $
Since we have $ \frac{\sin^2(x)}{x^2} <1 $ for all $ x \in \mathbb{R} $, we can say that
$ \displaystyle \ln \left( \frac{\sin^2(x)}{x^2} \right) =- \sum_{k=1}^{\infty} \frac{1}{k} \left(1- \frac{\sin^2(x)}{x^2} \right)^k \tag*{} $
At this point, if we can somehow prove that
$\displaystyle \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} \left(1- \frac{\sin^2(x)}{x^2} \right)^k < \frac{1}{k^{\epsilon}} \tag*{} $
For some $ \epsilon >0 $, Then the claim of convergence of the original integral will follow, but I can't seem to find a way to do so.
P.S. Wolframalpha has given up on providing a numerical estimate for the integral in (1), but says that the integral in (2) diverges, which is kind of puzzling.
|
Note that $x\to \ln(\sin^2(x))\leq 0$ and it is periodic of period $\pi$. Notice that
$$\int_0^{\pi}\ln(\sin^2(x))\,dx=-2\pi\ln(2).$$
(see for example Complex integration with trigonometric and logarithm? ).
Moreover $x\to\frac{x^2 }{(1+x^2)^2}\geq 0$, it is increasing in $[0,1]$ and decreasing in $[1,+\infty)$. Hence
$$\begin{align}\int_{0}^{\infty} &\frac{x^2 |\ln(\sin^2(x))|}{(1+x^2)^2}\, dx=
\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{x^2 |\ln(\sin^2(x))|}{(1+x^2)^2}\, dx\\
&\leq \frac{1}{4}\int_{0}^{\pi}|\ln(\sin^2(x))| dx+\sum_{k=1}^{\infty}\frac{(k\pi)^2 }{(1+(k\pi)^2)^2}\int_{k\pi}^{(k+1)\pi}|\ln(\sin^2(x))|\, dx\\
&\leq \frac{\pi\ln(2)}{2}+2\pi\ln(2)\sum_{k=1}^{\infty}\frac{(k\pi)^2 }{(k\pi)^4}\\
&= \frac{\pi\ln(2)}{2}+\frac{2\ln(2)}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi\ln(2)}{2}+\frac{\pi\ln(2)}{3}=\frac{5\pi\ln(2)}{6}
<+\infty\end{align}$$
Hence the integral is convergent.
|
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What is the order of magnitude of $\sum\limits_{n=1}^kn\binom{k}{n}\frac{(2k-2n-1)!!}{(2k-1)!!}$ as $k\to\infty$? What is the order of magnitude of the function $f(k)$ below in the limit as $k \rightarrow \infty$? Does it diverge, converge to a positive limit, or converge to zero?
$$
f(k) = \sum\limits_{n=1}^{k} n \binom{k}{n} \frac{(2k - 2n - 1)!! }{(2k-1)!! }
$$
Here $!!$ represents the double factorial. Does anyone have an hint?
|
We are interested in the asymptotics of
$$g(n) = \sum_{k=1}^{n-1} k {n\choose k}
\frac{(2n-2k-1)!!}{(2n-1)!!}
= n \sum_{k=1}^{n-1} {n-1\choose k-1}
\frac{(2n-2k-1)!!}{(2n-1)!!}.$$
Now we have
$$(2n-1)!! = \frac{(2n-1)!}{2^{n-1} \times (n-1)!}$$
so we get for our sum
$$\frac{n! \times 2^{n-1}}{(2n-1)!}
\sum_{k=1}^{n-1} {n-1\choose k-1}
\frac{(2n-2k-1)!}{2^{n-k-1} \times (n-k-1)!}
\\ = \frac{n! \times 2^{n-1}}{(2n-1)!}
\sum_{k=0}^{n-2} {n-1\choose k}
\frac{(2n-2k-3)!}{2^{n-k-2} \times (n-k-2)!}
\\ = \frac{n! \times 2^{n-1}}{(2n-1)!}
\sum_{k=0}^{n-2} {n-1\choose n-2-k}
\frac{(2k+1)!}{2^k \times k!}
\\ = 2^{n-1} {2n-1\choose n}^{-1}
\sum_{k=0}^{n-2} \frac{1}{(n-2-k)!} \frac{1}{2^k}
{2k+1\choose k}
\\ = 2^{n-1} {2n-1\choose n}^{-1} [z^{n-2}] \exp(z)
\sum_{k=0}^{n-2} z^k \frac{1}{2^k}
{2k+1\choose k}.$$
Here the coefficient extractor enforces the upper limit of the sum and
we obtain
$$2^{2n-2} {2n\choose n}^{-1} [z^{n-2}] \exp(z/2)
\sum_{k\ge 0} z^k \frac{1}{2^{2k}} {2k+1\choose k}$$
The sum is
$$-\frac{2}{z} + \frac{2}{z} \frac{1}{\sqrt{1-z}}.$$
We get from the first piece
$$-2^{2n-1} {2n\choose n}^{-1} [z^{n-1}] \exp(z/2)
= -2^{n} {2n\choose n}^{-1} \frac{1}{(n-1)!}$$
Now from the asymptotic ${2n\choose n}^{-1} \sim \sqrt{\pi n}/2^{2n}$ we
get for the modulus $\sqrt{\pi n}/2^n/(n-1)!$ so this vanishes quite
rapidly. Continuing with the second piece we obtain
$$2^{2n-1} {2n\choose n}^{-1}
[z^{n-1}] \frac{\exp(z/2)}{\sqrt{1-z}}.$$
We apply the Darboux method here as documented on page 180 section 5.3 of
Wilf's generatingfunctionology where we expand $\exp(z/2)$ about $1$
and take the first term, extracting the corresponding factor from the
singular term. This yields
$$\exp(1/2) \times 2^{2n-1} {2n\choose n}^{-1}
[z^{n-1}] \frac{1}{\sqrt{1-z}}
\\ = \exp(1/2) \times 2^{2n-1} {2n\choose n}^{-1}
{n-3/2\choose n-1}
\\ = \exp(1/2) \times 2^{2n-1} {2n\choose n}^{-1}
\frac{n}{n-1/2} {n-1/2\choose n}.$$
Using the Gamma function approximation of the second binomial
coefficient from the Wilf text we get
$$\exp(1/2) \times 2^{2n-1} {2n\choose n}^{-1}
\frac{n}{n-1/2} \frac{1}{\sqrt{n} \times \Gamma(1/2)}
\\ \sim \exp(1/2) \times 2^{2n-1}
\times \frac{\sqrt{\pi n}}{2^{2n}}
\frac{n}{n-1/2} \frac{1}{\sqrt{\pi n}}
\sim \frac{1}{2} \exp(1/2).$$
We have obtained
$$\frac{\sqrt{e}}{2}$$
the same as in the contributions that were first to appear.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4212878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Show that $\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$
Show that $$\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$$
My book wrote that
$$=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{4})} \cdot \frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{2})}{2\Gamma(3/4)}$$
My work says that there wasn't $\Gamma(5/4)$ in the denominator of the left one.
$$\beta(\frac{\frac{1}{2}+1}{2},\frac{1}{2})\beta(\frac{-\frac{1}{2}+1}{2},\frac{1}{2})$$
$$\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{3}{4}+1)}\cdot \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{1}{4}+\frac{1}{2})}$$
$$\frac{\Gamma(\frac{3}{4})\pi}{2\Gamma(\frac{7}{4})} \cdot \frac{\Gamma(\frac{1}{4})}{2\Gamma(\frac{3}{4})}$$
For that reason, my work doesn't match with their even answer.
|
Your method was correct, you just made a small mistake in the simplification. It is true that
\begin{align}
\int_0^{\dfrac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\dfrac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta
\end{align}
equals
\begin{align}
& \beta(\frac{3}{4}, \frac{1}{2})\; \beta(\frac{1}{4}, \frac{1}{2}) \\
\\
&=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{3}{4}+\color{red}{\frac{1}{2}})}\cdot \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{1}{4}+\frac{1}{2})} \\
\\
&=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{4})} \cdot \frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{2})}{2\Gamma(3/4)}
\end{align}
which is the step from your book. Then using $\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}) = \Gamma(\frac{1}{2})\Gamma(1-\frac{1}{2}) = \frac{\pi}{sin(\pi / 2)} = \pi$ is correct, and will lead to the answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4214088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Angle chasing problem: Find $\angle Q_{2024}Q_{2025}P_{2025}$ in the quadrilateral.
$ABCD$ is a convex quadrilateral where $BC = CD$, $AC = AD$, $\angle BCD = 96^\circ$ and $\angle ACD = 69^\circ$. Set $P_0 = A, Q_0 = B$ respectively. We inductively define $P_{n+1}$ to be the center of the incircle of $\triangle CDP_n$, and $Q_{n+1}$ to be the center of the incircle of $\triangle CDQ_n$. If $\angle Q_{2024}Q_{2025}P_{2025} - 90^\circ = \frac{2k-1}{2^n}$, compute $k+n$.
This is a question from a national Olympiad. I've been struggling to solve this. Here is my attempt to solve the problem:
In quadrilateral $ABCD$, $\angle CAD=\angle CBD=42^\circ$. So, $A,B,C,D$ are concyclic. So, it can be shown that $C,D,P_n,Q_n$ are concyclic for all $n\geq0$ by induction. So, we have $\angle CQ_nP_n=180^\circ -\angle CDP_n=180^\circ-\frac{69}{2^n}$.
Again, because $\triangle CQ_nQ_{n-1}\cong \triangle CDQ_n$ we have $\angle CQ_nQ_{n-1}=\angle CQ_nD=180^\circ-\angle CDQ_n-\angle DCQ_n$.
Now, $\angle Q_{n-1}Q_{n}P_n - 90^\circ\\=360^\circ -\angle CQ_nQ_{n-1}-\angle CQ_nP_n-90^\circ\\ =270^\circ-(180^\circ-\angle CDQ_n-\angle DCQ_n)-(180^\circ-\angle CDP_n)\\ =\angle CDQ_n +\angle DCQ_n+\angle CDP_n-90^\circ\\ =\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}-90^\circ.$
So, I am getting a negative number as answer which is not possible. And I am quite sure that there is nothing wrong in the question. What am I missing and what is the correct solution?
|
Let us start with the triangle $P_{n-1}CD$ with angles $2x$, $2y$, $2z$ in $P_{n-1},C,D$ respectively. Let $I=P_n$ be its incenter. Then:
$$
\hat P_n=
\hat I
:=
\widehat{CID}=180^\circ-y-z=90^\circ+\frac12(180^\circ-2y-2z)=90^\circ+\frac 12\cdot 2x=90^\circ+\frac 12\hat P_{n-1}\ .
$$
The same relation applies for $\hat Q_n$. This gives
$$
\hat P_n=\hat Q_n=90^\circ\cdot\frac{1-\frac 1{2^n}}{1-\frac 12}+42^\circ\cdot\frac 1{2^n}\ .
$$
Then we have:
$$
\begin{aligned}
\widehat{Q_{n-1}Q_nP_n}
&=
360^\circ
-\widehat{CQ_nQ_{n-1}}
-\widehat{Q_{n-1}Q_nP_n}
\\
&=
360^\circ
-\widehat{CQ_nD}
-\left(180^\circ-\widehat{CDP_n}\right)
\\
&=
180^\circ
-\widehat{CQ_nD}
+\frac 1{2^n}\widehat{CDA}
\\
&=
180^\circ
-90^\circ\cdot\frac{1-\frac 1{2^n}}{1-\frac 12}-42^\circ\cdot\frac 1{2^n}
+69^\circ\cdot\frac 1{2^n}
\\
&=
180^\circ
-180^\circ\cdot\left(1-\frac 1{2^n}\right)
+27^\circ\cdot\frac 1{2^n}
\\
&=
\left(
180^\circ
+27^\circ
\right)
\cdot\frac 1{2^n}
\ .
\end{aligned}
$$
Where is the error in the following lines?
$$
\begin{aligned}
\angle Q_{n-1}Q_{n}P_n\color{blue}{ - 90^\circ}
&=360^\circ -\angle CQ_nQ_{n-1}-\angle CQ_nP_n\color{blue}{ - 90^\circ}
\\
&=270^\circ-(180^\circ-\angle CDQ_n-\angle DCQ_n)-(180^\circ-\angle CDP_n)\\ &=\angle CDQ_n +\angle DCQ_n+\angle CDP_n\color{blue}{ - 90^\circ}\\
&=\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}\color{blue}{ - 90^\circ}
\end{aligned}
$$
No error, $\angle Q_{n-1}Q_{n}P_n \color{blue}{ - 90^\circ}=
\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}\color{blue}{ - 90^\circ}$
gives the same
$\angle Q_{n-1}Q_{n}P_n =
\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}$.
The angle is so small, since the arcs $CP_{n-1}Q_{n-1}D$ and $CP_nQ_nD$ are hardly distinguishable in some computer picture, they are plotted slightly over the segment $CD$, so we expect an angle which is either almost $0^\circ$ or $180^\circ$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4214264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Integral of exponential within a region Are there methods to compute the following integral for $a \leq b$? Here $x\in\mathbb{R}$
$$
\int\limits_{a \leq -\frac{x^2}{2} \leq b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx
$$
Substitution
The error function is
$$
\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt
$$
Using the substitution $t^2=\frac{x^2}{2}$ we have $2dt = dx$ and $\sqrt{-b}\leq t \leq \sqrt{-a}$
$$
\frac{1}{\sqrt{2}}\cdot \frac{2}{\sqrt{\pi}}\int_{\sqrt{-b}}^{\sqrt{-a}} e^{-t^2}dt = \frac{1}{\sqrt{2}}\cdot \left[\frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{-a}} e^{-t^2}dt + \frac{2}{\sqrt{\pi}}\int^0_{\sqrt{-b}} e^{-t^2} dt\right]
$$
Using the definition of error function and substituting $t' = -t$
$$
\frac{1}{\sqrt{2}}\left[\text{erf}(\sqrt{-a}) - \frac{2}{\sqrt{\pi}}\int_0^{-\sqrt{-b}} e^{-t^2} dt\right] = \frac{1}{\sqrt{2}}\left[\text{erf}(\sqrt{-a}) - \text{erf}(-\sqrt{-b})\right]
$$
|
You can not compute it directly. But you can use normal distribution in other to have a very accurate estimate. In fact by supposing that $b\leq0$, the region $a\leq -\frac{x^2}{2}\leq b$ is the same as $\{\sqrt{-2b}\leq x\leq \sqrt{-2a}\}\cup\{-\sqrt{-2a}\leq x\leq -\sqrt{-2b}\}$.
So, $\begin{align*}
\int_{a\leq -\frac{x^2}{2}\leq b} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx &= \int_{\sqrt{-2b}}^{\sqrt{-2a}} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx + \int_{-\sqrt{-2a}}^{-\sqrt{-2b}} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\
&= \left[\Phi(\sqrt{-2a}) - \Phi(\sqrt{-2b})\right] + \left[\Phi(-\sqrt{-2b}) - \Phi(-\sqrt{-2a})\right]\\
&= 2\left[1+\Phi(\sqrt{-2a}) - \Phi(\sqrt{-2b})\right]
\end{align*}$
where $\Phi(.)$ is the normal cumulating distribution function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4215085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Calculate the integral with divergence theorem
Given $F=(xz,y,z^2)$ and $E=\{(x,y,z)\;|\;y\geq0,x^2+y^2+z^2\leq4\}$ calculate the integral $\int_{\partial E} \langle F,N\rangle$ in two ways using the divergence theorem.
First $\text{div}(F)=3z+1$ .
Second $\int_E \text{div}(F)=\int_{\partial E} \langle F,N\rangle$ , the left part I calculated and got $16\pi$ ,for right part.
$$\int_{\partial E} \langle F,N\rangle=\int_{y=0}\langle F,N_1\rangle+\int_{x^2+y^2+z^2=4}\langle F,N_2\rangle$$
with $N_1=(0,-y,0),N_2=\frac{1}{2}(x,y,z)$. For the integral with $N_1$ wasn't sure how to proceed from here.
|
As already pointed out by Math Lover, the left part is
$$\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}(3z+1)\, dV=\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}1\, dV=\frac{1}{2}\left(\frac{4 \pi}{3}\cdot 2^3\right)=\frac{16 \pi}{3}$$
where the integral of $3z$ is zero by symmetry.
For the right part, the integral over the disc is
$$\iint_{y=0,x^2+z^2\leq 4}(xz,y,z^2)\cdot(0,-1,0)\,dxdz=-\iint_{y=0,x^2+z^2\leq 4}y\,dxdz=0.$$
As regards the integral over the half surface of the sphere, by the using spherical coordinates
$$x=2\sin(\phi)\cos(\theta),\quad y=2\sin(\phi) \sin(\theta),\quad z= 2\cos(\phi),\quad dS=2\sin(\phi)d\theta d\phi,$$ we get
$$\begin{align}\iint_{y\geq 0,x^2+y^2+z^2= 4}(xz,y,z^2)\cdot\frac{(x,y,z)}{2}\,dS
&=\iint_{y\geq 0,x^2+y^2+z^2= 4}\frac{x^2z+y^2+z^3}{2}\,dS\\
&=\iint_{y\geq 0,x^2+y^2+z^2= 4}\frac{y^2}{2}\,dS\\
&=\int_{\phi = 0}^{\pi} \int_{\theta = 0}^{\pi}4\sin^2(\phi) \sin^2(\theta)\,\sin(\phi)d\theta d\phi\\
&=4\int_{0}^{\pi} \sin^3(\phi)\, d\phi \cdot\int_{0}^{\pi}\sin^2(\theta)\,d\theta \\
&=4\,\frac{4}{3}\,\frac{\pi}{2}=\frac{16\pi}{3}.
\end{align}$$
Note that the integral of $\frac{x^2z+z^3}{2}$ is zero by symmetry.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4216393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Find Value of $(3\tan^2\alpha \;+ 4\tan^2\beta)$ Let $\vec V_1$ and $\vec V_2$ are two vectors such that $\vec V_1= 2(\sin\alpha+\cos\alpha) \hat i+\hat j$ and $\vec V_2=\sin\beta \; \hat i +\cos\beta \; \hat j$, where $\alpha$ and $\beta$ satisfy the relation $2 (\sin \alpha \; + \cos\alpha)\sin\beta=3-\cos\beta, $ Find Value of $(3\tan^2\alpha \;+ 4\tan^2\beta)$
My Approach:
I took Dot Product of $\vec V_1$ and $\vec V_2$ $\;$ and I obtained $\vec V_1.\vec V_2=3 $ using the given Relation.
Solving Given Relation too did not lead me anywhere
I am stuck now.
How to processed further?
|
Trick question! Doesn't really have anything to do with vectors.
Note that the maximum value of $a \sin \theta+b \cos \theta$ is $\sqrt {a^2+b^2}$, which occurs when $\tan \theta =\frac ab$.
Here, we've been given that:
$$2(\sin \alpha+ \cos \alpha)\sin \beta+\cos \beta=3$$
Now, maximum value of LHS is $$\sqrt{4(\sin \alpha+\cos \alpha)^2+1}=\sqrt {5+4\sin 2\alpha}$$
Notice that $\sqrt {5+4\sin 2\alpha}\leq 3$, reaching maximum value at $\sin 2\alpha=1$.
Thus, maximum value of LHS, is $3$, which is reached when $\sin 2\alpha=1$ and $\tan \beta=2(\sin \alpha+\cos \alpha)$.
Now, $\sin 2\alpha=1 \implies \tan^2 \alpha=1$, and $\tan^2 \beta=4(1+\sin 2\alpha)=8$.
Thus, $3\tan^2 \alpha +4\tan^2 \beta=3\cdot 1+4\cdot 8=35$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4216542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$ The question is
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$.
My attempt:
This reduces to finding primes $p,q$ such that p divides $q^3+1=(q+1)(q^2-q+1)$ and q divides $p^3+1=(p+1)(p^2-p+1)$.
Now we have 4 cases: If p divides $q+1$ and q divides $p+1$, then $\{p,q\}=\{2,3\}$.
I don't know how to deal with the other cases. For example, what happens if p divides $q+1$ and q divides $p^2-p+1$?
|
$Solution\ :$
\begin{gather}
So,\ given \notag\\
p^{3} +q^{3} +1\ \equiv 0\bmod pq \notag\\
And\ we\ know\ that, \notag\\
( a+b+c)^{3} -a^{3} -b^{3} -c^{3} =3\cdotp ( a+b)( b+c)( c+a) \notag\\
\notag\\
So,\ a^{3} +b^{3} +c^{3} =( a+b+c)^{3} -3\ ( a+b) \ ( b+c) \ ( c+a) \notag\\
\notag\\
\therefore p^{3} +q^{3} +1\ \equiv 0\bmod pq \notag\\
\Longrightarrow ( p+q+1)^{3} -3\ ( p+1)( q+1)( p+q) \equiv 0\bmod pq\ ....(1)\\
\notag\\
So,\ p^{3} +q^{3} \ can\ be\ written\ as\ ( p+q)\left( p^{2} -pq+q^{2}\right) \notag\\
In\ the\ modulo\ form,\ ( p+q)\left( p^{2} -pq+q^{2}\right) +1\equiv 0\bmod pq \notag\\
And\ ( p+q)\left( p^{2} -pq+q^{2}\right) \equiv -1\bmod pq\ ....(2)\\
\notag\\
So\ we\ have\ four\ cases,\ \notag\\
Case\ I\ :\ ( p+q) \equiv -1\bmod pq \notag\\
Case\ II\ :\ \left( p^{2} -pq+q^{2}\right) \equiv -1\bmod pq \notag\\
\\
Taking\ both\ the\ cases\ separately. \notag
\end{gather}
$ \begin{array}{l}
Case\ I:\\
\\
( p+q) \equiv -1\bmod pq
\end{array}$
\begin{gather*}
We\ see\ that\ p+q+1\equiv 0\bmod pq\\
So\ we\ have\ ( p+q) \equiv -1\bmod pq,\ Substituting\ this\ in\ ( 1)\\
We\ get,\ ( 0)^{3} -3\ ( p+1)( q+1)( -1) \equiv 0\bmod pq\\
\Longrightarrow 3\ ( p+1)( q+1) \equiv 0\bmod pq\ \\
( If\ we\ again\ substitute\ \ p+q\equiv -1\bmod pq,\ we\ get\ 3( -p)( -q) \equiv 0\bmod pq)\\
So\ even\ after\ substituting\ p+q\equiv -1\bmod pq,\\
We\ get\ the\ remaining\ product\ again\ as\equiv 0\bmod pq.\\
So\ we\ can\ write\ 3( p+1)( q+1) =pq\cdotp k\\
So\ again\ we\ have\ two\ cases,\\
\ i) \ 3\mid pq,\ \&\ ii) \ When\ 3\mid k\
\end{gather*}
$ \begin{array}{l}
Case\ ( i)\\
\\
When\ 3\mid pq,\
\end{array}$
\begin{gather*}
So,\ If\ 3\mid pq,\ then\ either\ p=3\ or\ q=3,\ since\ both\ are\ prime.\\
So\ let\ p=3.\ Then,4( q+1) =qk.\\
Since\ q+1\nmid q\Longrightarrow q+1\mid k.\ Also,\ we\ know\ 4\mid qk.\\
If\ 4\mid k,\ then\ 4( q+1) \mid k\Longrightarrow y=1,\ which\ leads\ to\ a\ contradiction\ as\ 1\ is\ not\ a\ prime.\\
Thus,\ we\ must\ have\ 2\mid q\ and\ 2\mid k.\\
This\ leads\ to\ the\ solutions\ ( p,q) \equiv \ ( 3,2) \ and\ ( 2,3) \ ( by\ symmetry) .
\end{gather*}
$ \begin{array}{l}
Case\ ( ii)\\
\\
When\ 3\mid k,\
\end{array}$
\begin{gather}
So,\ if\ 3\mid k\ \Longrightarrow 3m=k, \notag\\
\therefore 3\ ( p+1)( q+1) =pq\cdotp k\equiv ( p+1)( q+1) =pq\cdotp m \notag\\
And\ we\ also\ know\ p+1\nmid p\Longrightarrow p+1\mid qm,\ and\ we\ can\ write\ it\ as,( p+1) \ n=qm. \notag\\
\Longrightarrow q+1=p\cdotp n\ \equiv p\mid ( q+1) \ and\ by\ symmetry\ we\ can\ say\ q\mid ( p+1) . \notag\\
\notag\\
Now\ we\ can\ see\ that\ as\ both\ p\ and\ q\ are\ primes,\ and\ q\leqslant p+1\ and\ p\leqslant q+1.\ ....(3)\\
From\ ( 3) \ we\ get\ that\ p\ and\ q\ should\ be\ consecutive\ to\ hold\ the\ inequality. \notag\\
( OR) \notag\\
And\ if\ q\ is\ an\ odd\ prime\ then, \notag\\
p\ should\ be\ 2\ from\ p\mid ( q+1) \ and\ q\ should\ be\ 3\ from\ q\mid ( p+1) . \notag\\
And\ if\ q=2\ then\ p=3. \notag\\
This\ leads\ to\ the\ solutions\ ( p,q) \equiv \ ( 3,2) \ and\ ( 2,3) . \notag
\end{gather}
$And\ even\ ( 2,3) \ satisfies\ \left( p^{2} -pq+q^{2}\right) \equiv 1\bmod pq$
$ \begin{array}{l}
Case\ II:\\
\\
\left( p^{2} -pq+q^{2}\right) \equiv -1\bmod pq
\end{array}$
\begin{gather}
We\ have\ \left( p^{2} -pq+q^{2}\right) \equiv -1\bmod pq,\ and\ adding\ 3pq \notag\\
we\ get,\ p^{2} +2pq+q^{2} \equiv -1\bmod pq\Longrightarrow ( p+q)^{2} \equiv -1\bmod pq\\
So,\ we\ get\ ( p+q)^{2} \equiv -1\bmod pq,\ ....(4) \notag\\
now\ looking\ eq\ ( 2) \ we\ see\ that\ if\ \left( p^{2} -pq+q^{2}\right) \equiv -1\bmod pq, \notag\\
then\ this\ must\ be\ ( p+q) \equiv 1\bmod pq,\ squaring\ both\ sides\ we\ get, \notag\\
( p+q)^{2} \equiv 1\bmod pq\ ....(5)\\
from\ eq\ ( 3) \ and\ ( 4) \ we\ get \notag\\
\left( p^{2} -pq+q^{2}\right) \equiv -1\bmod pq,\ have\ No\ solution. \notag
\end{gather}
Conclusion: Only $(2,3)$ is the solution
\begin{equation*}
\end{equation*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4219435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.