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Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to prove the solution.
I start by pointing out that it has the form
\begin{equation}
\dfrac{dy}{dx} + P(x)y = Q(x)y^3
\end{equation}
where
\begin{equation}
P(x) = \dfrac{1}{2x}, \qquad Q(x) = -\frac{1}{3x}
\end{equation}
Rewriting y(x) as
\begin{equation}
y(x) = (3x)^{\frac{1}{2}} (2x + 3c)^{-\frac{1}{2}}
\end{equation}
Getting rid of that square root, I'll need it later on to simplify things
\begin{equation}
[y(x)]^2 = 3x(2x + 3c)^{-1}
\end{equation}
Calculating dy/dx
\begin{align}
\dfrac{dy}{dx} &= \frac{1}{2}(3x)^{-\frac{1}{2}}(3)(2x + 3c)^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)(2x + 3c)^{-\frac{3}{2}}(2)(3x)^{\frac{1}{2}} \\
&= \frac{3}{2}(3x)^{-\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} - (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{3}{2}} \\
&= (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= y \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= \dfrac{y}{2x} - y(2x + 3c)^{-1} \\
&= \dfrac{y}{2x} - y\left(\dfrac{y^2}{3x}\right) \\
&= \dfrac{y}{2x} - \dfrac{y^3}{3x}
\end{align}
Finally
\begin{align}
\dfrac{dy}{dx} + P(x)y &= \dfrac{y}{2x} - \dfrac{y^3}{3x} + \dfrac{y}{2x} \\
&= \dfrac{y}{x} - \dfrac{y^3}{3x}
\end{align}
which obviously isn't the same as equation 2. I don't know where I screwed up.
| You could probably get an easier calculation by going over the logarithmic derivative,
$$
\log(y(x))=\frac12(\log(3x)-\log(2x+3c))
\\~\\
\implies
\frac{y'(x)}{y(x)}=\frac12\left(\frac1x-\frac2{2x+3c}\right)
=\frac1{2x}-\frac1{3x}y(x)^2
$$
The last step is obtained by doing the minimum to eliminate the constant $c$.
As you can see, this again confirms your result.
As to the original equation itself, it solves as
$$
(y^{-2})'=-2y^{-3}y'=\frac{y^{-2}}x+\frac2{3x}\\
\left(\frac{y^{-2}}x\right)'=\frac2{3x^2}\implies \frac{y^{-2}}x=-\frac2{3x}+c\\
y(x)=\pm\sqrt{\frac3{3cx-2}}
$$
which is quite different from your solution formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that the inequality $\left|\int_{0}^{1} f(x)\,dx\right| \leq \frac{1}{12}$ holds for certain initial conditions
Given that a function $f$ has a continuous second derivative on the interval $[0,1]$, $f(0)=f(1)=0$, and $|f''(x)|\leq 1$, show that $$\left|\int_{0}^{1}f(x)\,dx\right|\leq \frac{1}{12}\,.$$
My attempt: This looks to be a maximization/minimization problem. Since the largest value $f''(x)$ can take on is $1$, then the first case will be to assume $f''(x)=1$. This is because it is the maximum concavity and covers the most amount of area from $[0,1]$ while still maintaining the given conditions.
Edit: Because of the MVT and Rolle's Theorem, there exists extrema on the interval $[0,1]$ satisfying $f'(c)=0$ for some $c\in[0,1]$. These extrema could occur at endpoints.
Then $f'(x)=x+b$ and $f(x)=\frac{x^2}{2}+bx+c$. Since $f(0)=0$, then $c=0$ and $f(1)=0$, then $b=-\frac{1}{2}$. Remark: Any function with a continuous, constant second derivative will be of the form $ax^2+bx+c$ and in this case, $a=-b$ and $c=0$. Now, $$\begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(x^2-x)\,dx\\&=\frac{1}{2}\bigg[\frac{x^3}{3}-\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=-\frac{1}{12}\end{align*}$$
Next, we assume that $f''(x)=-1$ and repeating the process yields $$ \begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(-x^2+x)\,dx\\&=\frac{1}{2}\bigg[\frac{-x^3}{3}+\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=\frac{1}{12}\end{align*}$$ Thus we have shown that at the upper and lower bounds for $f''(x)$ that $\frac{-1}{12}\leq\int_{0}^{1}f(x)\,dx\leq \frac{1}{12}
\Longleftrightarrow \left|\int_{0}^{1}f(x)\,dx\right|\leq\frac{1}{12}$ because $f''(x)$ is continuous on $[0,1]$.
I was wondering if this was 'rigorous' enough to be considered a full proof and solution to the problem.
| Consider the following integral:
$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx. $$
By integrating by parts twice, you get
$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx = \underbrace{\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f'(x)\bigg|_0^1}_{0} - \int_0^1\bigg(x-\frac{1}{2}\bigg)f'(x)dx=$$$$= - \int_0^1\bigg(x-\frac{1}{2}\bigg)f'(x)dx= \underbrace{- \bigg(x-\frac{1}{2}\bigg)f(x)\bigg|_0^1}_{0} + \int_0^1f(x)dx$$
Therefore,
$$\boxed{\int_{0}^{1}f(x)\, dx = \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx}$$
Now use the following inequality:
$$\left|\int_{a}^{b}f(x)g(x)\,dx\right| \leq \int_{a}^{b}|f(x)||g(x)|\, dx$$
Since $g(x)=\frac{x^{2}}{2}-\frac{x}{2}$
is the expression you got, this should yield the desired result.
$$\left|\int_0^1 f(x)\,dx\right|=\left| \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx\right|\le\frac{1}{2}\int_{0}^{1}|x^2-x|\,dx=\frac{1}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding the area enclosed by three curves I need to find the area that is enclosed between three curves. I was trying to look for the smartest solution visually, but haven't gotten anywhere.
Curves:
$$y_1= 2x^2$$
$$y_2 = x^2$$
$$y_3 = \frac{1}{x}$$
Please help
| I'm not sure how to solve this visually, but we can solve it with integrals.
First, we need to find where the different equations intersect each other.
*
*$y_1=y_2\rightarrow2x^2=x^2\rightarrow x^2=0\rightarrow x=0$
*$y_1=y_3\rightarrow2x^2=\frac{1}{x}\rightarrow x^3=\frac{1}{2}\rightarrow x=\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}$
*$y_2=y_3\rightarrow x^2=\frac{1}{x}\rightarrow x^3=1\rightarrow x=1$
The area enclosed by the three curves is $\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}y_1-y_2dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1y_3-y_2dx=\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}(2x^2-x^2)dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1(\frac{1}{x}-x^2)dx=\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}x^2dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1\frac{1}{x}dx-\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1x^2dx=\int_0^\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}x^2dx+\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1\frac{1}{x}dx-\int_\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}^1x^2dx=\frac{1}{6}+\ln(\frac{1}{\frac{1}{\sqrt[\leftroot{-2}\uproot{2}3]{2}}})-\frac{1}{3}+\frac{1}{6}=\frac{1}{3}\ln(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\frac{\pi}{3}$, where $\{.\}$ is positive fractional part Here, $\{-3.4\}=0.6$.
The said integral can be solved using $\{z\}+\{-z\}=1$, if $z$ is a non-zero real number;
after using the property that $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [ f(x)+f(-x)] dx$$
So here $$I=\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\int_{0}^{1} \frac{[\{x^3\}+\{-x^3\}](x^4+1)}{(x^6+1)} dx =\int_{0}^{1} \frac{(x^4+1)}{(x^6+1)} dx.$$
$$\implies I= \int_{0}^{1} \frac{(1+x^2)^2-2x^2}{(x^6+1)}dx=\int_{0}^{1}\frac{(1+x^2) dx}{x^4-x^2+1}-\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=I_1-I_2.$$
In $I_1$, divide up and down by $x^2$ and use $x-1/x=u$, then
$$I_1=\int_{-\infty}^{0} \frac{du}{1+u^2}=\frac{\pi}{2}$$
Next, use $x^3=v$ $$I_2=\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=\frac{2}{3} \int_{0}^{1} \frac{dv}{1+v^2}=\frac{\pi}{6}$$
Finally $$I=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$$
It will be interesting to see other approaches/methods of proving this integral.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x} =
\int_{0}^{1}{1 + x^{4} - x^{6} - x^{10} \over 1 - x^{12}}\,\dd x
\\[5mm] = &\
{1 \over 12}\int_{0}^{1}{x^{-11/12} + x^{-7/12} - x^{-5/12} - x^{-1/12} \over 1 - x}\,\dd x
\\[5mm] = &\
{1 \over 12}\bracks{-\Psi\pars{1 \over 12} - \Psi\pars{5 \over 12} +
\Psi\pars{7 \over 12} + \Psi\pars{11 \over 12}}
\\[5mm] = &\
{1 \over 12}\braces{\bracks{%
\Psi\pars{7 \over 12} - \Psi\pars{5 \over 12}}
+
\bracks{\Psi\pars{11 \over 12} - \Psi\pars{1 \over 12}}}
\\[5mm] = &\
{1 \over 12}\bracks{\pi\cot\pars{5\pi \over 12} +
\pi\cot\pars{\pi \over 12}} =
{\pi \over 12}{\sin\pars{5\pi/12 + \pi/12} \over
\sin\pars{5\pi/12}\sin\pars{\pi/12}}
\\[5mm] = &\
{\pi \over 12}{1 \over
\bracks{\vphantom{\Large A}\cos\pars{5\pi/12 - \pi/12} - \cos\pars{5\pi/12 + \pi/12}}/\, 2}
\\[5mm] = &\
{\pi \over 6}{1 \over \cos\pars{\pi/3} - \cos\pars{\pi/2}} =
{\pi \over 6}{1 \over 1/2 - 0} =
\bbox[10px,#ffd,border:1px groove navy]{\large{\pi \over 3}} \\ &
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2 for which there exists $a,b,c$ such that $\left (2^m\right )^2=4^m=a^2+b^2+c^2$. Since $4^m$ is divisible by 4, $a^2+b^2+c^2$ has to be divisible by 4 too. This is only possible if $a^2\equiv b^2\equiv c^2\equiv 0\pmod 4$, so we can write $a=2a',b=2b',c=2c'$ with $a',b',c'\in \mathbb{N}$. But then we get $\left (2^{m-1}\right )^2=4^{m-1}=a'^2+b'^2+c'^2$, so $m=1$, otherwise $2^m$ wouldn't be the smallest power of two with this property. It is easy to check that $n=2$ doesn't work, so for $n=2^m$ the statement doesn't hold.
*I suspect (but can't prove) that for all other values the statement holds. It would be enough to prove that for all odd primes $p$ there exists $a,b,c$ such that $p^2=a^2+b^2+c^2$, since for all other values of $n$ there exist some $p,m$ such that $n=pm$. Then we get $n^2=(pm)^2=(ma)^2+(mb)^2+(mc)^2$.
| I think I found the solution.
Lebesgue's identity says $(k^2 + l^2 + m^2 + n^2)^2 = (2kn + 2lm)^2 + (2ln - 2km)^2 + (k^2 + l^2 - m^2 - n^2)^2$, so if every odd prime $p$ can be written as the sum of four primes such that non of the right-hand-side terms is equal to $p^2$ the question is solved.
Lagrange's four square theorem says that every integer can be written as the sum of four squares. We check that if $p=k^2+l^2+m^2+n^2$ then none of the right hand side terms is necessarily $p^2$.
*
*$(k^2 + l^2 + m^2 + n^2)^2 = (2km + 2ln)^2$ only if $m=k$ and $n=l$, but then $p$ should be even, so this is not possible.
*$(k^2 + l^2 + m^2 + n^2)^2 = (2ln - 2km)^2$ only if $n=l$ and $m=k=0$, which would give again that $p$ is even.
*$(k^2 + l^2 + m^2 + n^2)^2 = (k^2 + l^2 - m^2 - n^2)^2$ only if $m=n=0$ or $k=l=0$. Since $p$ can't be a perfect square (it's prime), at least 2 of $k,l,m,n$ aren't 0. So we can swap them in such a way that nor $k=l=0$ nor $m=n=0$.
Thus every square of an odd prime can be written as the sum of four squares. Because of the third observation in the question it follows that every number that isn't a power of 2 can be written as the sum of 3 squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$ Prove:
$$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$
First obvious sub is $t = 1 + x^{3}$:
$$
\frac{1}{3}\int_{1}^{9}{\left(\,{t - 1}\,\right)}^{-2/3}\, t^{-1/2}\, \mathrm{d}t
$$
From here I tried many things like $\frac{1}{t}$, $t-1$, and more. The trickiest part is the bounds! Reversing it from the answer the integral should be like
$$
\frac{1}{6}\int_{0}^{1}
x^{-2/3}\left(\,{1 - x}\,\right)^{-5/6}\,\mathrm{d}t
$$
I'm not sure where the $1/2$ comes from and the $0$ to $1$ bounds. Any idea or tip please ?.
| Consider
$$
y^2=4x^3+4,g_2=0,g_3=-4
$$
The real period of $\wp(z)$ is
$$
\omega_1=\int_{-1}^{\infty} \frac{1}{\sqrt{1+x^3}} \text{d}x
=\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{2\sqrt{\pi} }
$$
And
$$
\wp\left ( \frac{\omega_1}{2} \right ) =-1
$$
We have the addition formula:
$$
\begin{aligned}
&\wp(3z) = \frac{1}{4}\left [ \frac{\wp'(z)-\wp'(2z)}{\wp(z)-\wp(2z)} \right ]^2
-\wp(z)-\wp(2z)\\
&\wp(2z)=\frac{1}{4} \left [ \frac{\wp''(z)}{\wp'(z)} \right ] ^2-2\wp(z)\\
&\wp(u+v)
=\frac{1}{4} \left [ \frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)} \right ]^2
-\wp(u) -\wp(v)
\end{aligned}
$$
where $3z,2z,z$ is not at the period lattices.
Let $x=\wp\left ( \frac{\omega_1}{6} \right ) ,y=\wp\left ( \frac{\omega_1}{3} \right )$.And they have two relations:
$$
\begin{aligned}
&\frac{1}{4} \left [ \frac{\sqrt{4x^3+4}-\sqrt{4y^3+4} }{x-y} \right ]^2-x-y = -1\\
&y=\frac{9x^4}{4x^3+4} -2x
\end{aligned}
$$
Solve it and gives,
$$
\wp\left ( \frac{\omega_1}{6} \right )=2,
\wp\left ( \frac{\omega_1}{3} \right )=0
$$
So,
$$
\begin{aligned}
I:&= \int_{0}^{2} \frac{1}{\sqrt{1+x^3} } \text{d}x\\
&=2\left(\frac{\omega_1}{3} -\frac{\omega_1}{6} \right)\\
&=\frac{\omega_1}{3}
\end{aligned}
$$
We have $\wp\left ( \frac{\omega_1}{4} \right ) =\sqrt{3}-1$,hence
$$
\int_{0}^{\sqrt{3} -1} \frac{1}{\sqrt{1+x^3} }\text{d}x
=\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{12\sqrt{\pi} }
$$
We have
$t=\wp\left ( \frac{\omega_1}{8} \right ) =
-1+\sqrt{3} +\sqrt{3}\sqrt{2-\sqrt{3} }
+\sqrt{3} \sqrt{2-\sqrt{3}+2\sqrt{2-\sqrt{3} } }$,hence
$$
\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=\frac{5\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{24\sqrt{\pi} }
$$
We have
$t=\wp\left ( \frac{\omega_1}{12} \right )
=2+\sqrt{3}+\sqrt{3} \sqrt{3+2\sqrt{3} }$,hence
$$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{4\sqrt{\pi} }$$
We have
$
t=\wp\left ( \frac{\omega_1}{9} \right )
=2^{1/3}
\left ( 16+3\cdot2^{2/3}\sqrt[3]{37+i\sqrt{3} }+
6\cdot2^{-1/3}\sqrt[3]{37-i\sqrt{3} }
\right )^{1/3}\approx4.5707391614928...
$,hence
$$
\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=\frac{2\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{9\sqrt{\pi} }
$$
If $t\ge0$,and
$$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x
=R\cdot\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{\sqrt{\pi} }$$
where $R$ is rational with $0\le R\le\frac{1}{3}$.$t$ must be an algebraic number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
} |
Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $|\frac{1}{x-5}| \leq 1$ then condition $(1)$ and $(2)$ will not hold. So wouldn't the range be $x < 4$ and $5 \leq x$ ? I know this is not right since the answer should be $ 5\frac{1}{3} \leq x$ and $x < 4 \frac{2}{3}$ ... could someone provide a solution?
| What about the ratio test?
\begin{align*}
\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = \limsup_{n\to\infty}\frac{n3^{n}|x-5|^{n}}{(n+1)3^{n+1}|x-5|^{n+1}}\\\\
& = \limsup_{n\to\infty}\left(\frac{n}{n+1}\right)\frac{1}{3|x-5|} = \frac{1}{3|x-5|} < 1\\\\
& \Rightarrow |x - 5| > \frac{1}{3} \Rightarrow \left(x > \frac{16}{3}\right)\vee\left(x < \frac{14}{3}\right)
\end{align*}
Moreover, for $x = 16/3$, one has that
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}3^{-n}} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} < \infty
\end{align*}
due to the Leibniz test.
On the other hand, for $x = 14/3$, we have that
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}(-1)^{n}3^{-n}} = -\sum_{n=1}^{\infty}\frac{1}{n} = \infty
\end{align*}
since it is the harmonic series.
Hopefully this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \times 14$
*$4 \times 14$
I think we have
$$\det \left[\begin{array}{ll}6 & 2 \\ 2 & 6\end{array}\right] = 4 \times 8$$
$$\det \left[\begin{array}{lll}6 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 6\end{array}\right] = 4^{2} \times10$$
and for any $n$ we have $\det(A_n)= 4^{n-1} \times (6+2×(n-1))$ so "1" is true.
| According to the Matrix Determinant Lemma, $\det(A+uv^T)= \det(A)(1+v^T A^{-1}u)$, hence
$$\det(A) = \det(4I + 2\cdot\mathbf 1 \mathbf 1^T) = \det(4I)(1+2\cdot \mathbf 1^T(\tfrac{1}{4}I)\mathbf 1) = 4^n (1 +\tfrac{1}{2}n) $$
For $n=5$, that's $4^5\cdot \frac{7}{2} = 4^4 \cdot 14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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$p\equiv 1\pmod 4\Rightarrow p=a^2+b^2$ and $p\equiv 1\pmod 8\Rightarrow p=a^2+2b^2$, what about for $p\equiv 1\pmod {2^n}$ in general Primes $p$ with $p\equiv 1\pmod 4$ can be written as $p=a^2+b^2$ for some integers $a,b$. For $p\equiv 1\pmod 8$ we have $p=a^2+2b^2$. Can primes that satisfy $p\equiv 1\pmod{2^n}$ for $n>3$ be written in a similar form -- for example $p=a^2+4b^2$ for $n=4$?
| it gets harder, and we cannot just impose congruence conditions.
Added: one thing I've not seen in print is this: as soon as $p \equiv 1 \pmod 8,$ we find that $-1$ is a fourth power mod $p,$ and
$$ z^4 + 1 \equiv 0 \pmod p $$
has four distinct roots.
A prime can be expressed as $p = x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and
$$ z^4 - 2 z^2 + 2 \equiv 0 \pmod p $$
has four distinct roots.
A prime can be expressed as $p = x^2 + 64 y^2$ if and only if $p \equiv 1 \pmod 8$ and
$$ z^4 - 2 \equiv 0 \pmod p $$
has four distinct roots.
The 64 result can be found under "biquadratic reciprocity." Both results may be due to Gauss, but were not published until Jacobi and Eisenstein.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
$$ \Rightarrow 2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq (b+c-2a)^3 $$
But couldn't take it any further, but it looks as if Hölder's Inequality may help. And also I don't think that it is a "warm-up" problem.
Any help will be appreciated.
| My first SOS is same as Nguyen Huyen. Here is my second and third.
$$\displaystyle a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$
$$\displaystyle=\frac34 \left( b-c \right) ^{2} \left( b+c \right) +\frac34 \left( a-b
\right) ^{2}a+\frac14\, \left( a+b-2c \right) ^{2}a+\frac12\, \left(b+c-2a \right) ^{2}a$$
$$\displaystyle =\frac34\cdot {\frac { \left( 2\,{a}^{2}-2\,ac+{b}^{2}-{c}^{2} \right) ^{2}}{2
\,a+b+c}}+{\frac {3 a\left( a+b+c \right)\left( a-b \right) ^{2} }{
2\,a+b+c}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$ How to evaluate:
$$\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$$
Can somebody help me handle it?
I have no idea what to do.
This is my attempt:
$$\cos24^\circ-\cos(60^\circ+24^\circ)-\cos12^\circ+\sin (12^\circ+30^\circ)$$
| $$\begin{split}
\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ &= \cos24^\circ-\cos84^\circ-\cos12^\circ+\cos48^\circ\\
&= \cos24^\circ +\cos48^\circ-\cos84^\circ-\cos12^\circ \\
&= 2\cos36^\circ\cos12^\circ-2\cos48^\circ\cos36^\circ\\
&=2\cos36^\circ(\cos12^\circ-\cos48^\circ)\\
&=2\cos36^\circ(2\sin18^\circ\sin30^\circ)\\
&=\frac{(2\sin18^\circ\cos18^\circ)2\cos36^\circ\sin30^\circ}{\cos18^\circ}\\
&= \frac{(2\sin36^\circ\cos36^\circ)\sin30^\circ}{\cos18^\circ}\\
&=\frac{\sin72^\circ\sin30^\circ}{\cos18^\circ}\\
&=\sin30^\circ=\frac{1}{2}
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
| Another way.
We need to prove that:
$$\sum_{cyc}a^{n+1}\geq\sum_{cyc}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}},$$ which is true by Muirhead because $$(n+1,0,0)\succ\left(n+\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
find the complex integral: $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$. Problem with integral formula.... Question I am trying to find the complex integral: $\displaystyle\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$.
My Attempt (and eventual question): $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz=\frac{1}{2}\int_\infty^\infty\frac{z^6}{(z^4+1)^2}dz$. Now, the singularities of $z^4+1$ are of the form $z_k=e^{\frac{i(\pi+2\pi k)}{4}}$, where $k=0,1,2,3$. Drawing the contour in the upper half plane, we see tht the only two singularities in our contour are $z_0=e^{\frac{i\pi}{4}}$ and $z_1=e^{\frac{i3\pi}{4}}$.
Let $f(z)=\frac{g(z)}{h(z)}$ where $g(z)=z^6$ and $h(z)=(z^4+1)^2$. Then, $h'(z)=8z^3(z^4+1)$. So, the value of the integral is $2\pi i\frac{1}{2}\sum_{k=0}^1\frac{z_k^6}{8z_k^3(z_k^4+1)}=\frac{\pi i}{8}\Big(\frac{e^{i\frac{3\pi}{2}}}{e^{i\frac{3\pi}{4}}(e^{i\pi}+1)}+\frac{e^{i\frac{\pi}{2}}}{e^{i\frac{\pi}{4}}(e^{i\pi}+1)}\Big)$. But, $e^{i\pi}+1=0$, so I must have done something wrong....? Any insight would be great! Thank you.
| Note
\begin{align}
\int_0^\infty \frac{z^6}{(z^4+1)^2}dz
&\overset{z\to\frac1z}
= \int_0^\infty \frac{dz }{(z^4+1)^2}
= \frac14\int_0^\infty \frac1{z^3}d \left(\frac{z^4}{z^4+1}\right)\\
&=\frac34 \int_0^\infty \frac{dz }{z^4+1}
\overset{z\to\frac1z} = \frac38\int_0^\infty \frac{1+z^2 }{z^4+1}dz\\
& = \frac38\int_0^\infty \frac{d(z-\frac1z)}{(z-\frac1z)^2+2}
= \frac{3\pi}{8\sqrt2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Why $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$ is equal to 1? $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$
By maths calculator it results 1.
I calculate and results $\sqrt{-\frac{1}{2}}$.
$\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$
$\sqrt{\frac{-{(3)}^{{\displaystyle\frac12}\times2}}{2^2}+\frac{1^2}{2^2}}=\sqrt{\frac{-3}4+\frac14}=\sqrt{\frac{-3+1}4}=\sqrt{\frac{-2}4}=\sqrt{-\frac12}$
Enlighten me what went wrong?
| The square of a negative number is positive.
$$\left(-\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that $8^{\frac{1}{3}}$ is $(8+i0)^{\frac{1}{3}}$
And, $r = 8$
And, $8\cos \theta = 8$ and $\theta = 0$.
So, $8^{\frac{1}{3}}\operatorname{cis} 0^\circ = 2\times (1+0)=2$
I just got only $2$. Where and how others $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ come from?
| Let $z^3=8$.
Thus, $$(z-2)(z^2+2z+4)=0,$$ which gives
$$\{2,-1+\sqrt3i,-1-\sqrt3i\}$$ or
$$\left\{2(\cos0+i\sin0),2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right), 2\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How were amplitudes of the $\cos$ and $\sin$ chosen? I don't understand why we use $\displaystyle\sqrt{1^2+\left(\frac{1}{2}\right)^2}$ in the below transformation. Can someone help to explain?
from
$$f(x)=\frac{3}{5}-\frac{3}{5}e^t\left(\cos(2t)+\frac{1}{2}\sin(2t)\right)$$
transform to
$$f(x)=\frac{3}{5}-\frac{3}{5}\sqrt{1^2+\left(\frac{1}{2}\right)^2}e^t\left(\frac{1}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}\cos(2t)+\frac{\frac{1}{2}}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}\sin(2t)\right)$$
let $\displaystyle\frac{1}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}=\cos\phi$ and $\displaystyle\frac{\frac{1}{2}}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}=\sin\phi$,
$$f(x)=\frac{3}{5}-\frac{3}{5}\sqrt{1^2+\left(\frac{1}{2}\right)^2}e^t(\cos\phi\cos(2t)+\sin\phi\sin(2t))$$
| Let’s concentrate on the important part, which is of the form
$$
f(x)=a\cos x+b\sin x
$$
which we want to express as
$$
f(x)=A(\cos\varphi\cos x+\sin\varphi\sin x)
$$
A necessary (and sufficient) condition is that
$$
A\cos\varphi=a,\qquad A\sin\varphi=b
$$
and therefore $a^2=A^2\cos^2\varphi$, $b^2=A^2\sin^2\varphi$. Hence
$$
A^2=a^2+b^2
$$
We want $A>0$ (not necessary, but convenient), so we get
$$
A=\sqrt{a^2+b^2},\quad \cos\varphi=\frac{a}{A},\quad \sin\varphi=\frac{b}{A}
$$
The last two requirements can be fulfilled, because $(a/A,b/A)$ is a point on the unit circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
| You should recall that $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$, since $(c-d)(c+d)=c^2-d^2$. This should help you simplify the expression with the square root.
Though I don't understand how you obtained what you wrote, from what I can see you should get:
$$ x+ \frac{2x^3}{3}-\frac{2(x^2+1)^{\frac{3}{2}}}{3}=x +\frac{2}{3}\Big( \frac{x^6-(x^2+1)^3}{x^3 +(x^2+1)^{ \frac{3}{2} } } \Big)=x+\frac{2}{3}\frac{-3x^4-3x^2-1}{x^3+(x^2+1)^{ \frac{3}{2} }}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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How to prove that $\sum_{n=0}^{\infty} (-1)^n \ln \frac{3n+2}{3n+1}=\frac{1}{2} \ln 3$ The sum $$\sum_{n=0}^{\infty} (-1)^n \ln \frac{3n+2}{3n+1}=\frac{1}{2} \ln 3$$ has been encountered in the post below:
How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$
I would like to know as to how this sum can be proved.
| We can obtain something analogous to the Wallis product using$$\prod_{n\ge1}(1-z^2/n^2)=\frac{\sin\pi z}{\pi z}\implies\prod_{n\ge1}(1-z^2/(2n+1)^2)=\frac{\frac{\sin\pi z}{\pi z}}{\frac{\sin\pi z/2}{\pi z/2}}=\cos\frac{\pi z}{2},$$so$$\prod_{n\ge0}\frac{(6n+2)(6n+4)}{(6n+1)(6n+5)}=\prod_{n\ge0}\frac{1-1/(6n+3)^2}{1-4/(6n+3)^2}=\frac{\cos\frac{\pi}{6}}{\cos\frac{\pi}{3}}=\sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that there are no integer solutions to $x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$ I have struggled on this problem for quite a bit of time now, asked some of my peers and teachers, and I am yet to find the solution. Here is the problem:
Prove that there are no integer solutions to the equation $$x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$$
Here is what I have tried:
*
*Expanding, moving things around, factoring (I wasn't able to factor it into something useful)
*Expanding, converting to a cubic equation (Too difficult to solve)
*Expanding, converting to a quadratic, using the quadratic formula (I was unable to simplify it enough)
It would be great if you guys could help!
| We have
$$xy^2-x = 2y+\frac{y}{x}$$
$$x^2y^2-x^2=2xy+y$$
$$x^2y^2-x^2-y-2xy = 0$$
Solve as a quadratic in $x$
$$(y^2-1)x^2-(2y)x-y = 0$$
Use quadratic formula
$$x = \frac{2y\pm \sqrt{4y^2+(4y^3-4y)}}{2(y^2-1)}$$
$$x = \frac{2y\pm \sqrt{4y^3+4y^2-4y}}{2y^2-2}$$
We can factor out a $2$ to get
$$x = \frac{y\pm \sqrt{y^3+y^2-y}}{y^2-1}$$
Take a look at the square root, the only rational root is $y = 0$ (by RRT), but testing this solution, $x = 0$, and the first expression has a $\frac{y}{x}$ in it, and obviously dividing by $0$ is illegal in this case.
Another way to see that $y = 0$ is the only rational root is to factor
$$y^3+y^2-y = y(y^2+y-1)$$
Then $y^2+y-1$ has no rational roots.
Therefore, there are no integer solutions.
| {
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"url": "https://math.stackexchange.com/questions/3796725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the stronger version of $9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0$ For $a,b,c \geqslant 0.$ Then $$9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0.$$
I use computer and found that the following stronger inequality holds for all reals of $a,b,c.$
$$\sum (a^2 -bc) \Big[9\, \left( a+b+c \right) ^{2} \left( ab+ac+bc \right) ^{2}+108\,{a}^{2 }{b}^{2}{c}^{2}-31\,abc \left( a+b+c \right) ^{3}\Big] \geqslant \frac{81}{4} \sum ab \prod \left( a-b \right) ^{2}$$
And also:
$$\sum (a^2 -bc) \Big[9\, \left( a+b+c \right) ^{2} \left( ab+ac+bc \right) ^{2}+108\,{a}^{2 }{b}^{2}{c}^{2}-31\,abc \left( a+b+c \right) ^{3}\Big] \geqslant {\frac {27}{4}}\, \left( a+b+c \right) ^{2} \prod \left( a-b \right) ^{2}$$
Any another inequality$?$
| There is the following stronger version.
Let $a$, $b$ and $c$ be non-negatives. Prove that:
$$9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3}\geq$$
$$\geq4(3\sqrt3-4)(a^3+b^3+c^3-3abc)abc.$$
The equality occurs for $a=b=c$ and for $(a,b,c)=t(6+4\sqrt3,1,1)$, where $t\geq0$ and for any cyclic permutations of the last.
Even if we'll replace $4(3\sqrt3-4)$ on $4$, the BW does not help here!
By the way, this inequality we can prove by $uvw$ immediately:
it's equivalent to $f(v^2)\geq0,$ where $f$ increases.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$729u^2v^4+108w^6-31\cdot27u^3w^3\geq4(3\sqrt3-4)(27u^3-27uv^2)w^3$$ or $f(v^2)\geq0,$ where
$$f(v^2)=27u^2v^4+4w^6-31u^3w^3-4(3\sqrt3-4)(u^3-uv^2)w^3.$$
But $$f'(v^2)=54u^2v^2+4(3\sqrt3-4)uw^3\geq0,$$
which says that $f$ increases and it's enough to prove our inequality for the minimal value of $v^2$, which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous and for $w^3=0$ it's obvious, it's e enough to assume $b=c=1$, which gives:
$$9(a+2)^2(2a+1)^2+108a^2-31(a+2)^3a\geq4(3\sqrt3-4)(a^3-3a+2)a$$ or $$(a-1)^2(a-6-4\sqrt3)^2\geq0$$ and we are done.
It seems that this inequality is true for any reals $a$, $b$ and $c$,
but it's another problem (the previous reasoning does not help because $v^2$ can be negative).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the set of all possible values of a function similar to Nesbitt's Inequality
Let $x,$ $y,$ $z$ be positive real numbers. Find the set of all possible values of
$$f(x,y,z) = \frac{x}{x + y} + \frac{y}{y + z} + \frac{z}{z + x}.$$
This seems extremely similar to Nesbitt's inequality, in which I did some reasearch on this problem to find. Nesbitt's states that for positive real $a, b, c,$ then $$\displaystyle\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$$ However I note that the function stated in the problem isn't in the same orientation to apply Nesbitt's, and just similar. I'm stumped on making any progression on this problem, as I've tried combining the denominators to form one big fraction as well as substituting variables to try clearing denominators. I would appreciate some help to start this problem.
| I think $1 < f(x,y,z) < 2.$ Indeed, because
$$\frac{x}{x+y} \geqslant \frac{x}{x+y+z}.$$
Equality occur when $x = 0$ or $z = 0.$
Therefore
$$f(x,y,z) \geqslant \frac{x+y+z}{x+y+z} = 1.$$
But $x,y,z$ are positive real numbers, so $f(x,y,z) > 1.$
Another
$$\frac{x}{x+y} < \frac{x+z}{x+y+z},$$
equivalent to
$$\frac{yz}{(x+y)(x+y+z)}\geqslant 0.$$
Equality occur when $yz=0.$ So
$$f(x,y,z) < \frac{2(x+y+z)}{x+y+z}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving limit using the definition of a limit Heres the question (provided for context) about proving a limit by using the definition of a limit:
$\mathop {\lim }\limits_{x \to 4} \left({x^2} + x - 11\right) = 9$
So, let’s get started. Let $\varepsilon > 0$ be any number then we need to find a number $\delta > 0$ so that the following will be true.
$\left| {\left( {{x^2} + x - 11} \right) - 9} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 4} \right| < \delta$
simplifying a bit
$\left| {\left( {{x^2} + x - 11} \right) - 9} \right| = \left| {{x^2}
+ x - 20} \right| = \left| {\left( {x + 5} \right)\left( {x - 4} \right)} \right| = \left| {x + 5} \right|\left| {x - 4} \right| < \varepsilon$
if, by some chance, we can show that $\left| {x + 5} \right| < K$ for some number $K$ then, we’ll have the following
$\left| {x + 5} \right|\left| {x - 4} \right| < K\left| {x - 4} \right|$
If we now assume that what we really want to show is $K\left| {x - 4} \right| < \varepsilon$ instead of $\left| {x + 5} \right|\left| {x - 4} \right| < \varepsilon$ we get the following,
$\left| {x - 4} \right| < \frac{\varepsilon }{K}$
All this is based on the assumption that we can show $\left| {x + 5} \right| < K$ for some $K$. To do this we assume that whatever $x$ is it must be close to $x=4$ since we are working with a limit. So lets assume that $x$ is within a distance of one of $x=4$. In terms of inequality we can assume
$\left| {x - 4} \right| < 1$
Starting with removing the absolute value bars we have
$- 1 < x - 4 < 1\hspace{0.5in} \Rightarrow \hspace{0.5in}3 < x < 5$
If we now add 5 to all parts of this inequality we get,
$8 < x + 5 < 10$
Now, since $x + 5 > 8 > 0$ (the positive part is important here) we can say that, provided $\left| {x - 4} \right| < 1$ we know that $x + 5 = \left| {x + 5} \right|$. Or, if take the double inequality above we have,
$8 < x + 5 = \left| {x + 5} \right| < 10$ $\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\left| {x + 5} \right| < 10\hspace{0.5in} \Rightarrow \hspace{0.5in}K = 10$
So, provided $\left| {x - 4} \right| < 1$ we can see that $\left| {x + 5} \right| < 10$ which in turn gives us,
$\left| {x - 4} \right| < \frac{\varepsilon }{K} = \frac{\varepsilon }{{10}}$
source
How did we go from this double inequality $8 < x + 5 < 10$ to this $8 < x + 5 = \left| {x + 5} \right| < 10$. From what I understand $|{x + 5}| < 10 $ can also be written as $-10<x + 5<10$, then the authors statement:
$8 < x + 5 = \left| {x + 5} \right| < 10$
should not be true since it is excluding a part of the interval that $|{x + 5}| < 10 $ includes (the interval which can be seen when the absolute value inequality is expanded into a double inequality i.e. $-10<x + 5<10$)
| You have
$$
|x+5|=|x-4+9|\leq|x-4|+9\leq\delta+9
$$
so
$$
|x-4||x+5|\leq\delta(\delta+9)<\varepsilon
$$
and you can see that you can choose
$$
0<\delta<\frac{9}{2} \left(\sqrt{1+\frac{4\varepsilon}{81}}-1\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$
Calculate the volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$.
My approach:
For $z = 0$:
\begin{equation*}
2x + y = 4 \implies y = -2x + 4
\end{equation*}
and for $y = 0$:
\begin{equation*}
2x = 4 \implies x = 2
\end{equation*}
So the integral for the volume are:
\begin{equation*}
\int_0^2 \int_0^{-2x +4} \left(4 - 2x - y\right)\,dy \,dx
= \frac{16}{3}
\end{equation*}
Is my approach correct? If not, can you provide the correct answer?
| Your approach is correct and you can calculate it also as
$$\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}\int\limits_{0}^{4-2x-y}\,dy\,dx\,dz=\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}(4-2x-y)\,dy\,dx =\int\limits_{0}^{2}\int\limits_{0}^{4-2x}(4-2x-y)\,dx\,dy$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the longest chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passing through $(0,-b)$? Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be an ellipse. How to find the longest chord in such ellipse which passes through the point $(0,-b)$ when
*
*$\ a> \sqrt{2}b>0$,
*$\ 0<a\leq \sqrt{2}b$
Frankly I do not even know what should I use. Is using the formula for the distance between two points and than maximization the expression a good idea? How the cases affect the solution?
| Let $(x_1,y_1)$ be the coordinates of the other end of the chord, with $x_1>0$ (the solution is symmetric with respect to $y$-axis, so this is not a limitation).
The condition to impose is: the tangent in $(x_1,y_1)$ should be orthogonal to the chord. The tangent is
$$
t:\ \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1
$$
A vector normal to the tangent is given from the coefficients of variables $x,y$ in the cartesian equation of the line,
$$
\mathbf{n}_t=\left(\frac{x_1}{a^2},\frac{y_1}{b^2}\right)
$$
while a directional vector of the line, orthogonal to $\mathbf{n},$ is
$$
\mathbf{u}_t=\left(-\frac{y_1}{b^2},\frac{x_1}{a^2}\right)
$$
The parametric equations of the line on which the chord lies is
$$
\begin{alignedat}{2}
x &= &&0+(x_1-0)t,\\
y &= -&&b+(y_1+b)t.
\end{alignedat}
$$
This is built with the formula for the parametric equations of the line through two points $A=(x_A,y_A)$ and $B=(x_B,y_B)$:
$$
c:\ \left\{
\begin{alignedat}{2}
x &= x_A+(x_B-x_A)t,\\
y &= y_A+(y_B-y_A)t.
\end{alignedat}
\right.
$$
Its directional vector is given by the coefficients of the parameter $t$
$$
\mathbf{u}_c=(x_1,y_1+b).
$$
The orthogonality condition
$$
\mathbf{u}_t\cdot\mathbf{u}_c=0
$$
gives
$$
-\frac{x_1y_1}{b^2}+\frac{x_1(y_1+b)}{a^2}=0
$$
and when $x_1\neq0$ we have
\begin{align}
y_1 &= \frac{b^3}{a^2-b^2}\\
x_1 &= a\sqrt{1-y_1^2/b^2}=\frac{a^2\sqrt{a^2-2b^2}}{a^2-b^2}
\end{align}
The square root requires $a\geq\sqrt{2}b>b.$
The length of the arc is
$$
l=\frac{a^2}{\sqrt{a^2-b^2}}.
$$
When $0<a\leq\sqrt{2}b$ the only solution is for $x_1=0$ and the chord is the vertical one.
This is an animation for fixed $b$ and $a$ decreasing
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da$
Evaluate
$$
\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da
$$
My try: rearranging original integral
$$
\int \frac{{1+\tan^{3/2}}(a)}{\sqrt{\sin^3(a)\sin(a+b)}}\,da
= \int \frac{{(1+\tan^{3/2}}(a))\sec^2(a)}{\sqrt{\tan^4(a)\cos(b)+\tan^3(a)\sin(b)}}\,da
$$
let $t=\tan a$ then it becomes
$$
\int \frac{{1+t^{3/2}}}{\sqrt{t^4\cos b+t^3\sin b}}\,dt
$$
now I am stuck.could anyone give me any hints to proceed further or any simpler method
| Hint:
$\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da$ = $\int \frac{1}{\sqrt{\sin^3(a)\sin(a+b)}}\,da + \int \frac{1}{\sqrt{\cos^3(a)\sin(a+b)}}\,da$
Take first part now.
$\sqrt{\sin^3(a)\sin(a+b)} = \sqrt{\sin^4(a)(\cos b + \cot a \sin b)}$
say, $t = \cos b + \cot a \sin b$
$dt = - \dfrac {1}{\sin^2(a)} \sin b \,da$
So, first integral $\int \frac{1}{\sqrt{\sin^3(a)\sin(a+b)}}\,da = -\dfrac{1}{\sin b} \int \frac{1}{\sqrt t} \,dt$
You can do the second part similarly. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
subsets where consecutive difference is never $2$
Show that the number of subsets of $\{1,\cdots, n\}$ with size $k$ and where the difference between consecutive pairs of elements is never equal to $2$ is $\sum_{j=0}^{k-1} {k-1\choose j}{n-k-j+1\choose n-2j-k}.$
I got that the desired number is equal to $\sum_{j=0}^{k-1} {k-1\choose j} \sum_{i=j}^{2j} {j\choose i- j}(-1)^i {n-k-i+k\choose n-k-i}$ as follows, but this may be incorrect. Let $C_k$ be the set of ordered pairs $(A, n)$ so that $A$ is a $k$-subset of $\{1,\cdots, n\}$ where the difference between consecutive pairs of elements is never equal to $2$. Then each difference vector (e.g. the difference vector corresponding to the pair $(\{a_1,\cdots, a_k\},n)$ would be $(a_1,a_2-a_1,\cdots, a_k-a_{k-1}, n-a_k)$) would have a first element that's a positive integer, a nonnegative integer as the last argument, and $k-1$ differences in between, none of which are $2$. This gives the generating series $\dfrac{x}{1-x}\cdot (x+\frac{x^3}{1-x})^{k-1} (\frac{1}{1-x}) = x^k (1-x+x^2)^{k-1} (1-x)^{-(k+1)}.$ The coefficient of $x^n,$ denoted $[x^n]x^k (1-x+x^2)^{k-1} (1-x)^{-(k+1)}, $ is thus
\begin{align}
[x^{n-k}] (1-x+x^2)^{k-1} (1-x)^{-(k+1)} &= \sum_{i\geq 0} \big([x^i] (1-x+x^2)^{k-1}\big) \big([x^{n-k-i}](1-x)^{-(k+1)}\big)\\
&= \sum_{i\geq 0} \left([x^i] \sum_{j=0}^{k-1} {k-1\choose j}(x^2-x)^j \right) {n-i\choose k}\\
&= \sum_{i\geq 0} \left([x^i]\sum_{j=0}^{k-1} {k-1\choose j}(-x)^j \sum_{m=0}^j {j\choose m} (-x)^m\right)\binom{n-i}k\\
&= \sum_{i\geq 0} \left([x^i] \sum_{j=0}^{k-1} \sum_{m=0}^{k-1} {k-1\choose j}{j\choose m} (-1)^{j+m} x^{j+m}\right){n-i\choose k}\\
&=\sum_{i\geq 0} \left(\sum_{j=0}^{k-1} {k-1\choose j}{j\choose i-j}\right)(-1)^i {n-i\choose k}\\
&= \sum_{j=0}^{k-1} {k-1\choose j}\sum_{i=j}^{2j} {j\choose i-j} (-1)^i {n-i\choose k},
\end{align}
but apparently $\sum_{i=j}^{2j} {j\choose i-j} (-1)^i {n-i\choose k} \neq {n-k-j+1\choose n-2j-k},$ and despite checking this over many times, I am unable to figure out what went wrong. I used the binomial theorem and summation properties. Another possibility, though possibly even messier, would be to split the generating series up as $(1-\frac{x^2}{1-x})^k\cdot \dfrac{1}{(1-x-x^2)(1-x)}.$
| The generating function approach is sound. We show the following is valid for $M:=\min\{k-1,\lfloor(n-k)/2\rfloor\}$.
\begin{align*}
\color{blue}{[x^{n-k}]\frac{\left(1-x+x^2\right)^{k-1}}{(1-x)^{k+1}}=\sum_{j=0}^{M}\binom{k-1}{j}\binom{n-k-j+1}{n-2j-k}}\tag{1}
\end{align*}
In (1) we use the upper limit $M$ to assure that the lower index of $\binom{n-k-j+1}{n-2j-k}$ is non-negative.
We obtain
\begin{align*}
\color{blue}{[x^{n-k}]}&\color{blue}{\frac{\left((1-x)+x^2\right)^{k-1}}{(1-x)^{k+1}}}\\
&=[x^{n-k}]\sum_{j=0}^{k-1}\binom{k-1}{j}x^{2j}(1-x)^{-j-2}\tag{2.1}\\
&=\sum_{j=0}^{M}\binom{k-1}{j}[x^{n-k-2j}](1-x)^{-j-2}\tag{2.2}\\
&=\sum_{j=0}^{M}\binom{k-1}{j}\binom{-j-2}{n-k-2j}(-1)^{n-k-2j}\tag{2.3}\\
&\,\,\color{blue}{=\sum_{j=0}^M\binom{k-1}{j}\binom{n-k-j+1}{n-k-2j}}\tag{2.4}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (2.1) we apply the binomial theorem multiplying with the denominator $(1-x)^{-k-1}$.
*In (2.2) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and set the upper limit to $M$ to assure the power $n-k-2j$ of $x$ is non-negative.
*In (2.3) we select the coefficient of $x^{n-k-2j}$.
*In (2.4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
Note: We can derive a very similar binomial identity by using $1-x+x^2=(1-x)^2+x$.
We derive:
\begin{align*}
\color{blue}{[x^{n-k}]}&\color{blue}{\frac{\left((1-x)^2+x\right)^{k-1}}{(1-x)^{k+1}}}\\
&=[x^{n-k}]\sum_{j}\binom{k-1}{j}x^{j}(1-x)^{k-3-2j}\\
&=\sum_{j}\binom{k-1}{j}[x^{n-k-j}](1-x)^{k-3-2j}\\
&=\sum_{j}\binom{k-1}{j}\binom{k-3-2j}{n-k-j}(-1)^{n-k-j}\\
&=\sum_{j}\binom{k-1}{j}\binom{n-2k+j+2}{n-k-j}\tag{3.1}\\
&=\sum_{j}\binom{k-1}{j}\binom{n-k-j+1}{n-2k+j+1}\tag{3.2}\\
&\,\,\color{blue}{=\sum_{j}\binom{k-1}{j}\binom{n-k-j+1}{k-2j}}\tag{3.3}
\end{align*}
Comment:
*
*In (3.1) we use $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3.2) we change the order of summation $j\to k-1-j$.
*In (3.3) we use $\binom{p}{q}=\binom{p}{p-q}$.
From (2.4) and (3.3) we find the nice binomial identity
\begin{align*}
\color{blue}{\sum_{j}\binom{k-1}{j}\binom{n-k-j+1}{n-k-2j}=\sum_{j}\binom{k-1}{j}\binom{n-k-j+1}{k-2j}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is $\mathbb{E}\!\left(X^2 \mid X + 2Y \right)$ if $X,Y \sim \mathcal{U}\left(\left[-1,1\right]\right)$ independently? $X$ and $Y$ are independent random variables of the uniform distribution on $\left[-1,1\right]$, $X,Y \sim \mathcal{U}\!\left(\left[-1,1\right]\right)$.
What is $\mathbb{E}\!\left(X^2 \mid X + 2Y \right)$?
My attempt:
We need the joint pdf of $X$ and $X+2Y$ (as suggested). So we do a change of variables, let $W = X,~Z = X + 2Y$. This is a linear transformation, so the jacobian will be a constant. We need to find the image of the square $-1 \leq x, y \leq 1$ under this transformation. Since it is a linear transformation, the four edges of boundaries will be mapped to line segments. So we can find them by computing where the four corners of the square are mapped:
$$\begin{align}
\left(x,y\right)&|\left(w,z\right)\\
(-1,-1) &| (-1, -3)\\
(-1,+1) &| (-1,+1)\\
(+1,+1) &| (+1,+3)\\
(+1,-1) &| (+1,-1)
\end{align}$$
The image of the square is the parallelogram with vertices as above and in the picture below:
The joint density of $W$ and $Z$ will be uniform on this region, let A denote the interior of the parallelogram. Since it has area of 8, we can conclude
$f_{W,Z}\left(w,z\right) = \frac{1}{8}\unicode{x1D7D9}\left(\left(w,z\right) \in A\right) = \begin{cases}
\frac{1}{8}, & \text{if $-1\leq w \leq 1,~-2+w\leq z\leq2+w$} \\
0, & \text{otherwise}
\end{cases}.$
The computations for $\mathbb{E}\left(X|X+2Y\right)$ and $\mathbb{E}\left(X^2|X+2Y\right)$ are below.
(for $\mathbb{E}\left(X|X+2Y\right)$, $Z = X + 2Y$)
Now we can figure out what $f_{X|Z}\left(x|z\right)$ is. We must consider three cases:
*
*$-3 \leq z \leq -1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $z+2$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{-1+2+z}{2} = \frac{z}{2} + \frac{1}{2}$
*$-1 \leq z \leq 1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $1$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{-1+1}{2} = 0$
*$1 \leq z \leq 3$, given $Z = z$, $X$ is uniformly distributed between $1$ and $z-2$. So $\mathbb{E}\!\left(X|Z=z\right) = \frac{1-2+z}{2} = \frac{z}{2} - \frac{1}{2}$
Cases $1,2,3$ (with $0$ otherwise) gives us $\mathbb{E}\!\left(X|Z=z\right)$.
(for $\mathbb{E}\left(X^2|X+2Y\right)$, $Z = X + 2Y$)
We must consider three cases:
*
*$-3 < z \leq -1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $z+2$ (so probability density function $f_{X|Z}(x|z) = \frac{1}{(z+2) - (-1)} = \frac{1}{z+3}$). So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_{-1}^{z+2}x^2 \cdot \frac{1}{z+3}dx = \frac{1}{3} \cdot \frac{(z+2)^3+1}{z+3}$
*$-1 \leq z \leq 1$, given $Z = z$, $X$ is uniformly distributed between $-1$ and $1$ (probability density function $f_{X|Z}(x|z) = \frac{1}{2}$).
So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_{-1}^{1}x^2 \cdot \frac{1}{2}dx = \frac{1}{3}$
*$1 \leq z < 3$, given $Z = z$, $X$ is uniformly distributed between $1$ and $z-2$. So $\mathbb{E}\!\left(X^2|Z=z\right) = \int_1^{z-2}x^2\frac{1}{z-3}dx = \frac{1}{3}\cdot\frac{1}{z-3}\cdot\left((z-2)^3-1\right)$
Cases $1,2,3$ (with $0$ otherwise) gives us $\mathbb{E}\!\left(X^2|Z=z\right)$.
| Define $h:\mathbb{R} \to \mathbb{R}$ such that $h(x)=\frac 1 3 \chi_{[-3,3]}$.
The claim is that $\mathbb{E}\left(X^2 | X + 2Y \right)=h(X+2Y)$.
Suppose $B \in \mathscr{B}([-3,3])$.
Let us prove $\int \chi_{B}(x+2y)x^2dx dy=\int \chi_{B}(x+2y)h(x+2y)dxdy$.
If $g(x,y)=(x+2y,x)$ then $|\det g (x,y)|=2$ and thus we have to prove
$$\int_{[-3,3]\times [0,1]} \chi_{B}(z)x^2dz dx=\int_{[-3,3]\times [0,1]} \chi_{B}(z)h(z)dzdx$$
thanks to the change of variable.
Now $\int_{[-3,3]\times [0,1]} \chi_{B}(z)x^2dz dx=\mu(B)\frac 1 3$ and $\int_{[-3,3]\times [0,1]} \chi_{B}(z)h(z)dzdx=\int_{[-3,3]\times [0,1]} \chi_{B}(z)\frac 1 3 dzdx=\mu(B)\frac 1 3$ which conclude the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$ For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfrac{\sin^2\alpha}{\cos^2\alpha}+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}{\cos^2\alpha}.$
This doesn't seem to help much.
| Instead of dividing by $\cos^2\alpha$, you could do the following,
\begin{align}&\sin^2 \alpha + \tan^2 \alpha +\sin^2\alpha +\sin^2\alpha\cos^2\alpha + \cos^4\alpha\\ &= \sin^2 \alpha + \tan^2 \alpha +\sin^2\alpha +\cos^2\alpha(\sin^2\alpha + \cos^2\alpha) \\
&= \sin^2 \alpha + \tan^2 \alpha + \cos^2\alpha
\\ &=1+\tan^2\alpha
\\ &=\sec^2\alpha \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Number of digits of $xy$ is either $m+n$ or $m+n-1$ Given positive integers $x$ and $y$ with $m$ and $n$ digits respectively, must $xy$ have either $m+n$ or $m+n-1$ digits?
If $x$ has $m$ digits, then $10^{m-1} \le x < 10^m$. Likewise, if $y$ has $n$ digits, then $10^{n-1} \le y < 10^n$. Multiplying those two inequalities together (which is OK because all the numbers involved are positive), one gets $10^{m+n-2} \le xy < 10^{m+n}$.
So, from $10^{m+n-2} \le xy < 10^{m+n}$, one could conclude that $xy$ has either $m+n$ or $m+n-1$ digits, right?
| Yes, you are right.
From $10^{m-1} \le x$ and $10^{n-1} \le y$ it follows that $10^{m+n-2} \le xy$, so $xy$ has at least $m+n-1$ digits. Also, from $x<10^m$ and $y<10^n$ it follows that $xy < 10^{m+n}$. Since $10^{m+n}$ is the least integer having $m+n+1$ digits, $xy$ has less than $m+n+1$ digits and thus $xy$ has at most $m+n$ digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inverse of a matrix and matrix relation Let $A$ be the matrix $$ A= \begin{pmatrix}
2 & -1 & -1\\
0 & -2 & -1 \\
0 & 3 & 2 \\
\end{pmatrix} $$ I am trying to find $A^{-1}$ as a relation of $I_{3}, A$ and $A^{2}$ and also to prove that $A^{2006}-2A^{2005}=A^{2}-2A$. For the first one I noticed that $$A^{n}= \begin{pmatrix}
2^{n} & -(2^{n}-1) & -(2^{n}-1)\\
0 & -2 & -1 \\
0 & 3 & 2 \\ \end{pmatrix} , A^{k}= \begin{pmatrix}
2^{k} & -(2^{k}-1) & -(2^{k}-1)\\
0 & 1 & 0 \\
0 & 0 & 1\\ \end{pmatrix}$$ for $n$ odd and $k$ even but I don't know how to proceed from here. Also the the equality $A^{2006}-2A^{2005}=A^{2}-2A$ leads to $A^{2004}(A-2I_{3})=A-2I_{3}$ but again I don't know what to do next since $(A-2I_{3})$ is not invertible. Any help?
| In this case it is not necessary to calculate the characteristic polynomial of $\;A\;$ and apply the Cayley-Hamilton theorem.
Since $\;A=\begin{pmatrix}
2 & -1 & -1\\
0 & -2 & -1 \\
0 & 3 & 2 \\
\end{pmatrix}\;$ and $\;A^2=\begin{pmatrix}
4 & -3 & -3\\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}\;,\;$ the first column of the matrix $\;A-2I_3\;$ only contains zeros, moreover the second and the third row of the matrix $\;A^2-I_3\;$ only contain zeros, therefore
$\left(A-2I_3\right)\left(A^2-I_3\right)=0\;,$
$A^3-A-2A^2+2I_3=0\;,$
$I_3=\frac{1}{2}A+A^2-\frac{1}{2}A^3\;,$
and, by multiplying both sides of the previous equality by $\;A^{-1}$, we get that
$A^{-1}=\frac{1}{2}I_3+A-\frac{1}{2}A^2\;.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of ordered pairs $(a,b)$ of real numbers satisfying certain conditions with a polynomial How many ordered pairs $(a,b)$ of real numbers are there such that $$(2a^2+1)+(2a^2-1)i$$ is a solution to the equation $$x^2-10x+b^2+4b+20=0$$ There should be $6$.
Here is what I have so far:
$$\begin{align}f(x)&=x^2-10x+b^2+4b+20\\&=(x-5)^2+(b+5)(b-1)\\&=0\end{align}$$
and
$$\begin{align}f(x)&=(x-2a^2-1-(2a^2-1)i)(x-z)\\&= x^2-x\big[2a^2+1+(2a^2-1)i+z\big]+z\big[2a^2+1+(2a^2-1)i\big]\\&=0\end{align}$$
where $z=c+di$ and $z\in\mathbb{C}$.
Comparing coefficients,
$$2a^2+1+(2a^2-1)i+z=10\tag{*}$$
and
$$z\big[2a^2+1+(2a^2-1)i\big]=b^2+4b+20\tag{**}$$
This leads me to three cases: $\mathfrak{Re}(z)=0$ and $\mathfrak{Im}(z)\neq 0$, both $\mathfrak{Re}(z)$ and $\mathfrak{Im}(z)$ are non-zero, or $\mathfrak{Im}(z)=0$ and $\mathfrak{Re}(z)\neq 0$.
Case $1$: $\mathfrak{Re}(z)=0$ and $\mathfrak{Im}(z)\neq 0$
Now $z=c+di=0+di=di$ so by $(*)$, we have $a=\pm\frac{3}{\sqrt{2}}$ and $d=-8$.
Then from $(**)$, we have that $b^2+4b+20=64-80i$ and solving for $b$ yields non real solutions. So this case has no pairs$(a,b)$ of real solutions.
Case $2$: $\mathfrak{Re}(z)$ and $\mathfrak{Im}(z)$ are non-zero. Then, $z=c+di$ and from $(*)$, $c=9-2a^2$ and $d=1-2a^2$. Now from $(**)$, we get that $a=\pm\frac{1}{\sqrt{2}}$ or $a=\pm\sqrt{2}$.
If $a=\pm\sqrt{2}$, then the imaginary part of $(**)$ will not be zero and no real $b$ can be found.
If $a=\pm\frac{1}{\sqrt{2}}$, then this implies $c=8$ and $d=0$ so that $$16=b^2+4b+20\Longrightarrow b=2$$
This case gives the solutions $(\frac{1}{\sqrt{2}}, 2)$ and $(-\frac{1}{\sqrt{2}}, 2)$.
Case $3$: $\mathfrak{Im}(z)=0$ and $\mathfrak{Re}(z)\neq 0$. Now, $z=c\in\mathbb{R}$. This yields the same $2$ solutions as case $2$.
Some how I am missing solutions. Where am I going wrong with this? Am I missing something with symmetry, like if $(2a^2+1)+(2a^2-1)i$ is a solution, then the complex conjugate, $(2a^2+1)-(2a^2-1)i$ must also be a solution and $f(x)$ has a negative discriminant? Also, I was wondering what the fastest method for solving this problem is. Possibly something with group theory? Thanks in advance.
| Case 1. $2a^2-1=0$. Then $a=\pm 1/\sqrt{2}$. For each of these $a$, plug $2a^2+1=2$ into the equation and find $b=-2$ -we get $2$ solutions $(a,b)$: $(\frac{1}{\sqrt{2}},-2)$ and $(\frac{-1}{\sqrt{2}}, -2)$.
Case 2. $2a^2-1\ne 0$, then the number $(2a^2+1)+(2a^2-1)i$ is not real. So the number $(2a^2+1)-(2a^2-1)i$ is also a root of the equation. By Vieta, the sum of the roots is 10. So $4a^2+2=10$, $a=\pm \sqrt{2}$. So the root is $5+3i$. Plug this root into the equation $(5+3i)^2-10(5+3i)+b^2+4b+20=0.$
Get $b^2+4b-14=0$, get two solutions for $b$. So we get 4 pairs $(a,b)$ this way, and 6 pairs altogether.
| {
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Finding the domain of $f(x)=\sqrt\frac{x^2+6}{x-4}$
Given $f(x)=\sqrt{\dfrac{x^2+6}{x-4}}$, find the domain for the function $f(x)$.
My attempt failed:
So the first constraint is $\frac{x^2+6}{x-4}\geq0$ because of the radical.
Now, multiplying both sides by $x-4$ I get $x^2+6\geq0, x\neq4$.
And now $x^2\geq-6$, so $x\in\mathbb{R}-\{4\}$
But if I were to plug in say 3 for x I encounter a problem because in this case I get $\frac{x^2+6}{x-4}\lt0$
| $x^{2}+6 >0$ for all $x$ so $\frac {x^{2}+6} {x-4}$ exists and is $ \geq 0$ iff $x>4$.
| {
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Diophantine equation: $x^4+4=py^4$
find all primes p for which $x^4+4=py^4$ is solvable for integers.
My try: I started out with $p=2$ . LHS must be even, thus $(x^4+4)\mod(16)=4$ similarly
$(2y^4)\mod (16)=${$0,2$}. Thus for $p=2$ there are no solutions.
Now I tried factoring i.e $x^4+4=(x^2+2x+2)(x^2-2x+2)=py^4.$ I dont know what to do next
Any ideas??
| It is easy to see that $x$ is odd (otherwise $4$ divides $py^4$ and $x, y$ are both even, which gives contradiction mod $16$). Then the $\gcd$ of $x^2 + 2x + 2$ and $x^2 - 2x + 2$ is $1$. Therefore we must have \begin{eqnarray}x^2 + 2x + 2 &=& u^4\\x^2 - 2x + 2 &=& pv^4\end{eqnarray} or \begin{eqnarray}x^2 + 2x + 2 &=& pu^4\\x^2 - 2x + 2 &=& v^4.\end{eqnarray} In the first case, we have $(x^2 + 1)^2 + 1 = (u^2)^2$ which is impossible. In the second case, we have $(x^2 - 1)^2 + 1 = (v^2)^2$ which leads to $x = 1$, and hence $y = 1$ and $p = 5$.
| {
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Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this problem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Similarly assuming $[x]=3,4$; we get correct roots as $\sqrt{12}$ and $\sqrt{18}$. But if er let $[x]=5$, it gives $x=\sqrt{24}$. which contradicts. So I get the sum of roots as $\sqrt{6}(1+\sqrt{2}+\sqrt{3}).$
The question is: Have I found all the real roots. In any case, what is(are) more appropriate method(s) of doing it.
| $x^2 = 6[x] - 6= 6([x]-1)$. Forget about the greatest aspect of GIF and concentrate on the integer aspect of GIF. $x^2$ is multiple of $6$ and $x = \pm\sqrt {6k}$ where $k=[x]-1$.
$x^2 \ge 0$ so $[x]-1\ge 0$ so $x\ge [x] \ge 1$ so $k \ge 1$.
And if $k \ge 6$ then $x =\sqrt{6k} \le \sqrt {k^2} = k$ so $[x]\le x \le k =[x]-1$ is impossible.
We can pick them off one by one. If $k=1$ and $x=\sqrt 6$ then $[x]=2$ and $6[x]-6=6=x^2$. So that's one solution. If $k=2$ and $x = \sqrt{12}$ then $[x]=3$ and $6[x]-6=12$ so that's another. If $k=4$ then $x=\sqrt{24}$ and $[x]=4=k = [x]-1$. Impossible. If $k=5$ then $x=\sqrt{30}$ and $[x]=5$ and $k\ne [x]-1$ so that's impossible.
$x = \sqrt 6$ or $\sqrt{12}$ and the sum of the roots is $\sqrt 6 + \sqrt {12} = \sqrt 3(2+\sqrt 2)$.
| {
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Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$ For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$
$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$
Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$
$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$
I wish to find a proof with $a:\neq {\rm mid}\left \{ a, b, c \right \},$ or another proof$?$
Actually$,$ I also found a proof true for all $a,b,c \in \Big[\dfrac{1}{3},3\Big],$ but very ugly.
After clearing the denominators$,$ need to prove$:$
$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$
but we have$:$
$$f=\dfrac{1}{32} \left( 3-a \right) \left( 3-b \right) \Big( c-\dfrac{1}{3} \Big) +
\left( 3-a \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \left(
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right) \left( 3-c
\right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right) \left( 3-c
\right) \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right) \left( c-\dfrac{1}{3}
\right) \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) + \left( 3-b \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
\left( 3-b \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
\left( 3-a \right) \left( c-\dfrac{1}{3} \right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
\right) \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
\left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right) \left(
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right) \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{2}$$
So we are done.
If you want to check my decomposition$,$ please see the text here.
| By AM-GM we have
$$
\frac{(a+b+c) + (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}.
$$
Note that by the assumption, we have
$$
3 + \frac{1}{3} \geq a + \frac{1}{a}
$$
and similarly for the other variables. Therefore
$$
3 \cdot \frac{10}{3} \cdot \frac{1}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)},
$$
as desired.
| {
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.
In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.
Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.
$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$
is what I have got so far, and from here I can’t continue.
Am I on the right direction? Any suggestions or hints will be much appreciated.
| Now, let $x^2+y^2=2uxy.$
Thus, since $xy\geq0$ and for $xy=0$ our inequality is true, we can assume that $xy>0$, which gives $u\geq1$ and we need to prove that:
$$8(x^2+xy+y^2)^3\geq6(-3xy(x+y))^2$$ or
$$2(2u+1)^3\geq27(u+1)$$ or
$$16u^3+24u^2-15u-25\geq0,$$ which is obvious for $u\geq1$.
| {
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If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}\leq\\
\frac{1}{4\sqrt[4]{a^3bcd}}+\frac{1}{4\sqrt[4]{ab^3cd}}+\frac{1}{4\sqrt[4]{abc^3d}}+\frac{1}{4\sqrt[4]{abcd^3}} = \\
\frac{1}{4\sqrt{a}}+\frac{1}{4\sqrt{b}}+\frac{1}{4\sqrt{c}}+\frac{1}{4\sqrt{d}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4\sqrt{abcd}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}
\end{equation}$$
This is as far as I got. Could you please help me finish off my thought pattern and finish the question the way I was attempting to solve it?
| Another way.
By C-S $$\sum_{cyc}\frac{1}{a^3+b+c+d}=\sum_{cyc}\frac{\frac{1}{a}+b+c+d}{(a^3+b+c+d)\left(\frac{1}{a}+b+c+d\right)}\leq$$
$$\leq\frac{\sum\limits_{cyc}\left(\frac{1}{a}+b+c+d\right)}{(a+b+c+d)^2}=\frac{\sum\limits_{cyc}\left(abc+3a\right)}{(a+b+c+d)^2}.$$
Thus, it's enough to prove that:
$$(a+b+c+d)^3\geq4\sum_{cyc}abc+12(a+b+c+d).$$
But by Muirhead (or Maclurin) $$(a+b+c+d)^3\geq16\sum_{cyc}abc$$
and by AM-GM $$\frac{3}{4}(a+b+c)^3\geq\frac{3}{4}\left(4\sqrt[4]{abcd}\right)^2(a+b+c+d)=12(a+b+c+d)$$ and we are done!
| {
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| We need to prove that $$c^2+ab\leq(a+b)c$$ or
$$a^2+b^2\leq(a+b)\sqrt{a^2-ab+b^2}$$ or
$$a^2+b^2\leq\sqrt{(a+b)(a^3+b^3)},$$ which is true by C-S.
| {
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Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$. This is Exercise 14 on page 110 of Analysis I by Amann and Escher.
The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.
If I attempt this, I get the following:
\begin{align*}
X^4 - 2X^3 - X^2 + 2X + 1 &= 0\\
\Rightarrow X^2 - 2X - 1 + \frac{2}{X} + \frac{1}{X^2} &= 0.
\end{align*}
My problem is that I don't understand how to make the suggested substitution. I'm wondering if there is something obvious I'm missing.
I appreciate any help.
| $\left(x-1/x\right)^2=x^2+1/x^2-2$. If you substitute $y=x-1/x$, your equation becomes $y^2+2-2y-1=0$, so $y^2-2y+1=0$, so $y=1$. Then you find $x$ from $x-1/x=1$: $x^2-x-1=0$, $x=(1\pm \sqrt{5})/2$. Each of these is a double root of the original equation.
| {
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Why is it enough to imply that $x^3+x+1|x^6+x^5+x^4+x^3+x^2+x+1$? I tried to understand a solution but I am stuck at how did the solution concluded that $x^3+x+1\mid x^6+x^5+x^4+x^3+x^2+x+1.$
Consider the field of element $8$ given by $\mathbb{F}=\frac{\mathbb{Z}_2[X]}{\langle X^3+X+1\rangle}.$ Now I can tell that each element $x$ in $\mathbb{F}$ satisfies the relation $x^8-x=0.$ Also I can see that $X^8-X=X(X-1)(X^6+X^5+X^4+X^3+X^2+X+1).$ The solution then says this implies that $X^3+X+1\mid X^6+X^5+X^4+X^3+X^2+X+1.$ However I do not understand how did they reach such conclusion?
I can see that $X^8-X$ can be factored into $\Pi _{i=1}^8X-q_i$ for every $q_i\in\mathbb{F}$ but I do not see how that would help to solve my problem.
| Well, the general result is that the elements of $\Bbb F_q$, $q=p^n$ with $p$ prime, satisfy $x^q-x=0$. Moreover, $x^q-x$ is the product of all irreducible polynomials of degree $d\geq 1$ over $\Bbb F_p$ with $d\mid n$.
In your case, $q=2^3$ and $x^8-x = x(x-1)(x^3+x+1)(x^3+x^2+1)$ over $\Bbb F_2$.
| {
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Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z+x) \geqslant 8xyz $$
Also, I got
$$ \frac{x+y+z+(x+y+z)}{4} \geqslant \bigg[ xyz(x+y+z) \bigg] ^{1/4} $$
$$ \therefore x+y+z \geqslant 2 $$
But, I am stuck here. Any hints ?
| Since $x+y+z \geqslant 3 \sqrt[3]{xyz}$, we have $xyz (x+y+z)\geqslant 3 (xyz)^{4/3}$.
Using the given condition, we have $1 \geqslant 3 (xyz)^{4/3}$. This is $xyz \leqslant \frac{1}{3^{3/4}} $
Also, we have $(x+y)(y+z)(z+x) = (x+y+z)(xy+ yz + zx) - xyz $
Now $ -xyz \geqslant - \frac{1}{3^{3/4}} $ and
$$ (xy+ yz + zx) \geqslant 3 (xyz)^{2/3} $$
$$ (x+y+z)(xy+ yz + zx) \geqslant 3 \frac{(x+y+z)xyz}{(xyz)^{1/3}} $$
$$ (x+y+z)(xy+ yz + zx) \geqslant \frac{3}{(xyz)^{1/3}} $$
But, we have
$$ xyz \leqslant \frac{1}{3^{3/4}} $$
$$ \therefore \frac{1}{xyz} \geqslant 3^{3/4} $$
$$ \therefore \frac{1}{(xyz)^{1/3}} \geqslant 3^{1/4} $$
So, we get
$$ (x+y+z)(xy+ yz + zx) \geqslant 3^{5/4} $$
So, we have
$$(x+y+z)(xy+ yz + zx) -xyz \geqslant 3^{5/4} - \frac{1}{3^{3/4}} $$
It follows that
$$ (x+y)(y+z)(z+x) \geqslant \frac{8}{3^{3/4}} $$
It can be seen that equality is achieved when $x=y=z$ in $xyz(x+y+z)=1$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the equation $z^4=(2+3i)^4$ without expanding the right side. So I tried :
$z={ \left| r \right| }e^{i \theta}$
$r^4e^{4 \theta i}=(\sqrt{13} e^{(2k\pi+\tan ^{-1}(\frac{3}{2}))i})^4$
$r=\sqrt{13}$
$4\theta=4 \times {(2k\pi+\tan ^{-1}(\frac{3}{2}))}$
$\theta=2k\pi+\tan ^{-1}(\frac{3}{2})$
But I calculated the value of $\theta$ wrongly. How can I fix it?
| Much simpler: $z^4 =(2+3i)^4= 1\cdot (2+3i)^4$
and $z = 1^{\frac 14} (2+3i)$, where $1^{\frac 14}$ is understood to mean the four complex fourth roots of $1$, namely $\pm 1, \pm i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
} |
How to solve system of equations using inverse matrix? System of equations is the following:
$$x + 4y + 2z = 10$$
$$4x - 3y+0z = 6$$
$$2x + 2y + 2z = 14$$
Here is my solution:
$$det(A) = 1 *(-3 * 2 - 0 * 2) -4 * (4 * 2 - 0 * 2) + 2 * (4 * 2 - (-3) * 2)$$
$$= -6 -32 + 28$$
$$= -10$$
$$
+\begin{pmatrix}
-3 & 0 \\
2 & 2 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
4 & 0 \\
2 & 2 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
4 & -3 \\
2 & 2 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
4 & 2 \\
2 & 2 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
1 & 2 \\
2 & 2 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
1 & 4 \\
2 & 2 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
4 & 2 \\
-3 & 0 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
1 & 2 \\
4 & 0 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
1 & 4 \\
4 & -3 \\
\end{pmatrix}
$$
Above equals to:
$$
\begin{pmatrix}
-6 & -8 & 14 \\
-4 & -2 & 6 \\
-6 & -8 & -15\\
\end{pmatrix}
$$
Then I multiply it by 1/-10 and the result is:
$$
\begin{pmatrix}
0,6 & 0,4 & 0,6 \\
0,8 & 0,2 & -0,8 \\
-1,4 & -0,6 & 1,5\\
\end{pmatrix}
$$
Then I multiply it by:
$$
\begin{pmatrix}
10 \\
4 \\
16\\
\end{pmatrix}
$$
Result is:
$$
\begin{pmatrix}
6 & 2,4 & 8,4 \\
8 & 1,2 & -11,2 \\
-14 & -3,6 & 21\\
\end{pmatrix}
$$
Which results in:
$$
\begin{pmatrix}
16,8 \\
-2\\
3,4\\
\end{pmatrix}
$$
So according to this logic:
$$x = 16,8$$
$$y = -2$$
$$z = 3,4$$
However when I test this solution it is incorrect, can anyone tell me what I'm doing wrong? Thanks.
| $$A=
\begin{matrix}
1 & 4 & 2 \\
4 & -3 & 0 \\
2 & 2 & 2 \\
\end{matrix}
$$
$$b=
\begin{matrix}
10 \\
6 \\
14 \\
\end{matrix}
$$
To find the inverse of A:
1.Find the Determinant: $\Delta A$ =-10
2.Find the matrix of cofactors:
$$\begin{matrix}
-6 & -8 & 14\\
-4 & -2 & 6\\
6 & 8 & -19\\
\end{matrix}$$
Cofactor for (i,j) the element =$(−1)^{i+j}\Delta M_{i,j}$ where $\Delta M_{i,j}$ is the minor for the (i,j) element
3.Take the transpose the matrix of cofactors:
$$\begin{matrix}
-6 & -4 & 6\\
-8 & -2 & 8\\
14 & 6 & -19\\
\end{matrix}$$
4. And in the end divide the matrix by the determine $\Delta A$ computed in step 1.
$$\begin{matrix}
\frac{3}{5} & \frac{2}{5} & \frac{-3}{5}\\
\frac{4}{5} & \frac{1}{5} & \frac{-4}{5}\\
\frac{-7}{5} & \frac{-3}{5} & \frac{19}{10}\\
\end{matrix}$$
And to find the solution multiply the inverse to the matrix b:
We get $x=0,y=-2, z=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. let's write some terms of the equation:
\begin{align}
x_0&=1 \\[2ex]
x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex]
x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex]
x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex]
\end{align}
So, we can write:
$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$
Here, I'm not sure how to prove it is bounded above.
| You have
$$
x_n = \prod_{k=1}^n \left(1+\frac{1}{2^k}\right)
= e^{\sum_{k=1}^n\ln \left(1+\frac{1}{2^k}\right)}
\leq e^{\sum_{k=1}^n\frac{1}{2^k}}
\leq e^{\sum_{k=1}^\infty\frac{1}{2^k}}
= e
$$
where for the first inequality we used that $\ln(1+x)\leq x$ for all $x>-1$. This shows the sequence is bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing the Change of Basis Matrix between two vector spaces For the two bases, $A$ and $B$ whose columns represent the basis vectors respectively, I was asked to find the transformation matrix from $A$ to $B$
$$ A=
\begin{Bmatrix}
1 & -1 & 1 \\
2 & 2 & 1 \\
1 & -1 & 3 \\
\end{Bmatrix} $$
$$
B= \begin{Bmatrix}
-3 & 1 & 5 \\
2 & -1 & 4 \\
-2 & -1 & 9 \\
\end{Bmatrix}
$$
I began by comparing the the basis vectors in $A$ to those in $B$ as per the definition of what the change of basis matrix is - it contains the coordinates of the new basis wrt to the old basis e.g.
$$
\left(\begin{matrix}
1 \\
2 \\
1 \\
\end{matrix}\right) = x_1\left(\begin{matrix}
-3 \\
2 \\
-3 \\
\end{matrix}\right) + x_2\left(\begin{matrix}
1 \\
-1 \\
-1 \\
\end{matrix}\right) + x_3\left(\begin{matrix}
5 \\
4 \\
9 \\
\end{matrix}\right) $$
And did so for all other columnns. This does seem like a tedious way to do things. Would I be right in thinking that I could identify the values of x, which are the coordinate representations of $A$ wrt $B$ by through the following relationship:
$A = XB$ and therefore, assuming $B$ is non-singular I could find the coordinates X by computing:
$AB^{-1} = X$ ?
If this is invalid could some explain why and perhaps suggest a quicker solution to such an equation?
| The equation $
\left(\begin{matrix}
1 \\
2 \\
1 \\
\end{matrix}\right) = x_{11}\left(\begin{matrix}
-3 \\
2 \\
-3 \\
\end{matrix}\right) + x_{21}\left(\begin{matrix}
1 \\
-1 \\
-1 \\
\end{matrix}\right) + x_{31}\left(\begin{matrix}
5 \\
4 \\
9 \\
\end{matrix}\right) $ can be written as $$
\left(\begin{matrix}
1 \\
2 \\
1 \\
\end{matrix}\right) = \left(\begin{matrix}
-3&1&5 \\
2&-1&4 \\
-3&-1&9 \\
\end{matrix}\right)\left(\begin{matrix}
x_{11} \\
x_{21} \\
x_{31} \\
\end{matrix}\right) $$ Do you see how you $X$ matrix should be on the right and not the left?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of sequence involving reciprocal of the sum of the first squares Let $(x_n)$ a sequence defined by $x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}.$ What is $\lim_{n\to \infty}x_n$?
My idea: $1^2+\dots+k^2=\frac{k(k+1)(2k+1)}{6}<\frac{k(k+1)(k+2)}{3}$. Thus
$$x_n>3\sum_{k=1}^n\frac{1}{k(k+1)(k+2)}=3\left(\frac 12 -\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{4}\right).$$
I couldn't continue any further. My assumption is that $x_n\to 2$ or something like that. Please help me solve it! Thank you!
| Hint: Note that
$$
\frac{1}{k(k+1)(2k+1)} = \frac 1{k+1} + \frac 1k - \frac 4{2k+1}.
$$
So for instance, with $n = 10$, we end up with the sum
$$
\left(1 + 2\cdot \frac12 + \cdots + 2\cdot \frac 1{10} + \frac 1{11} \right) - 4\left(\frac 13 + \frac 15 + \cdots + \frac 1{21} \right) = \\
1 + 2 \left( \frac 12 - \frac 13 + \cdots -\frac 19 + \frac 1{10}\right) + \frac 1{11} - 4\left(\frac 1{11} + \frac 1{13} + \cdots + \frac 1{21} \right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| Use the high-school formula
$$a^3+b^3=(a+b)(a^2-ab+b^2).$$
You'll obtain
\begin{align}
\cos^6x+\sin^6x&=(\cos^2+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\cr
&=\cos^4x-\cos^2x\sin^2x+\sin^4x
\end{align}
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 3
} |
$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
Find all pairs of integers (m, n) such that the integers $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
I assumed $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ to be $a^3,b^3$ and $c^3$ respectively and deduced that $6(2mn+4)=a^3+6b^3,$ $12(n-m)=a^3-6b^3+36$ but
I can't solve it further. Please help.
| Suppose $m$ and $n$ such that the expressions are perfect cubes. Let $a$, $b$ and $c$ be integers such that
\begin{eqnarray*}
a^3&=&6(m+1)(n-1)&=&6mn-6m+6n-6,\\
b^3&=&6+(m-1)(n+1)&=&mn+m-n-6,\\
c^3&=&(m-2)(n+2)&=&mn+2m-2n-4.
\end{eqnarray*}
Then comparing coefficients shows that
$$a^3-18b^3+12c^3=90,$$
so in particular $a$ is divisible by $6$, and so plugging in $a=6A$ then yields
$$36A^3-3b^3+2c^3=15,$$
which shows that $c$ is divisible by $3$, and so plugging in $c=3C$ then yields
$$12A^3-b^3+18C^3=5.$$
Now reducing mod $9$ shows that
$$3A^3-b^3\equiv5\pmod{9},$$
and so $A^3\equiv-1\pmod{9}$ and $b^3\equiv1\pmod{9}$, so $A\equiv2\pmod{3}$ and $b\equiv1\pmod{3}$. Then plugging in $A=3\alpha+2$ and $b=3B+1$ yields
$$81\alpha^3+54\alpha^2+36\alpha-B^3-3B^2-3B+18C^3=-2.$$
Reducing mod $3$ shows that $B^3\equiv2\pmod{3}$ and so plugging in $B=3\beta+2$ shows that
$$27\alpha^3+18\alpha^2+12\alpha-9\beta^3-27\beta^2-27\beta+6C^3=8.$$
The left hand side is divisible by $3$ but the right hand side is not, a contradiction. Hence no such integers $m$ and $n$ exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.
My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.
| Define a new operation $p \otimes q = \frac{{p + q}}{{1 - pq}}$.
And it's easy to find that it's associative.
$p \otimes q \otimes r = \frac{{p + q + r - pqr}}{{1 - pq - pr - qr}}$.
It means $p + q + r = pqr \Leftrightarrow p \otimes q \otimes r = 0$.
And then $\begin{array}{l}
\frac{{2x}}{{1 - {x^2}}} \times \frac{{2y}}{{1 - {y^2}}} \times \frac{{2z}}{{1 - {z^2}}} - (\frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} + \frac{{2z}}{{1 - {z^2}}}) = 0\\
\Leftrightarrow \frac{{2x}}{{1 - {x^2}}} \otimes \frac{{2y}}{{1 - {y^2}}} \otimes \frac{{2z}}{{1 - {z^2}}} = 0\\
\Leftrightarrow (x \otimes x) \otimes (y \otimes y) \otimes (z \otimes z) = 0\\
\Leftrightarrow (x \otimes y \otimes z) \otimes (x \otimes y \otimes z) = 0\\
\Leftrightarrow 0 \otimes 0 = 0
\end{array}$
So the problem is solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 4
} |
Inverse formulas of rotations using the substitution method
Given the direct rotations
$$
\mathcal R \begin{cases}
x=X\cos\vartheta-Y\sin\vartheta\equiv x'\cos\vartheta-y'\sin\vartheta\\
y=X\sin\vartheta+Y\cos\vartheta\equiv x'\sin\vartheta+y'\cos\vartheta
\end{cases} \tag 1 $$
to find the inverse rotations,
$$\mathcal{R}^{-1}\begin{cases} x'\equiv X=x \cos\vartheta+y \sin\vartheta \\ y'\equiv Y=-x \sin\vartheta+y \cos\vartheta \end{cases} \tag 2$$
I can use Cramer's rule,
$$x'=X=\frac{ \left|\begin{matrix}
x & -\sin(\vartheta) \\
y & \cos(\vartheta)
\end{matrix}\right|}{\cos^2\vartheta+\sin^2\vartheta\equiv 1}=x \cos \vartheta +y\sin \vartheta$$
Same for $Y$, I will have:
$$y'=Y=-x \sin\vartheta+y \cos\vartheta$$
But with the substitution method to find the $(2)$ quickly, starting from the $(1)$, as with the Cramer's method (or thinking that $\cos(-\vartheta)=\cos\vartheta$ and $\sin(-\vartheta)=-\sin\vartheta$), I think that is more complicated. I have done some steps (see below),
$$
\begin{cases}
x=X\cos\vartheta-Y\sin\vartheta\\
y=X\sin\vartheta+Y\cos\vartheta
\end{cases}$$
If $\cos\vartheta\neq 0$
$$
\begin{cases}
X=\frac{x+Y\sin\vartheta}{\cos\theta}\\
Y\cos\vartheta=y-\left(\frac{x+Y\sin\vartheta}{\cos\theta}\right)\sin\vartheta
\end{cases}$$
$$
\begin{cases}
X=\frac{x+Y\sin\vartheta}{\cos\theta}\\
Y\cos\vartheta=y-\frac{x\sin\vartheta+Y\sin^2\vartheta}{\cos\theta}
\end{cases}$$
$$
\begin{cases}
X=\frac{x+Y\sin\vartheta}{\cos\theta}\\
Y=\frac{y}{\cos\vartheta}-\frac{x\sin\vartheta+Y\sin^2\vartheta}{\cos^2\vartheta}
\end{cases}$$
but, after, I have leaved all. Every solution for this method is welcome.
| Let $c=\cos\vartheta$ and $s=\sin\vartheta$. So, your system is$$\left\{\begin{array}{l}x=cX-sY\\y=sX+cY\end{array}\right.$$and we have\begin{align}\left\{\begin{array}{l}x=cX-sY\\y=sX+cY\end{array}\right.&\iff\left\{\begin{array}{l}X=\frac1c(x+sY)\\y=\frac sc(x+sY)+cY=\frac scx+\left(\frac{s^2}c+c\right)Y=\frac scx+\frac1cY\end{array}\right.\\&\iff\left\{\begin{array}{l}X=\frac1c(x+sY)\\Y=-sx+cy\end{array}\right.\\&\iff\left\{\begin{array}{l}X=\frac1c\bigl(x+s(-sx+cy)\bigr)=cx+sy\\Y=-sx+cy.\end{array}\right.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$ for positive reals $a+b=1$
Problem 1: Let $\lambda = \frac{\ln (7 + 3\sqrt{5})}{\ln 2} - 1 \approx 2.77697$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$$
with equality if and only if $a = b = \frac{1}{2}$
Problem 2(a weaker version of Problem 1): Let $\lambda_1 = \frac{25}{9}$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda_1 b} + b^{\lambda_1 a} + a^{\lambda_1 b^2} + b^{\lambda_1 a^2} \le 2.$$
Background Information: In Proposition 5.2 in [1], Vasile Cirtoaje gives the following result:
Problem 3: If $a, b$ are nonnegative real numbers satisfying $a + b = 1$, then $a^{2b} + b^{2a} \le 1$.
In Conjecture 5.1 in [1], Vasile Cirtoaje proposes the following conjecture:
Let $a, b$ be nonnegative real numbers satisfying $a + b = 1$. If $k \ge 1$, then $a^{(2b)^k} + b^{(2a)^k} \le 1$.
The case $k=2$ has been proved:
Problem 4: Let $a, b$ be positive real numbers with $a + b = 1$. Prove that $a^{4b^2} + b^{4a^2} \le 1$.
See If $a+b=1$ so $a^{4b^2}+b^{4a^2}\leq1$
I combine Proposition 5.2 and Problem 4 to come up with the problems.
I may use appropriate bounds to prove Problems 3 and 4 (something like
Inequality $a^{2b}+b^{2a}\leq \cos(ab)^{(a-b)^2}$). Now the problems are more difficult.
Reference
[1] Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938
| working the weaker case and using $b=1-a$, we have
$$f(a)=a^{\frac{25}{9} (1-a)}+(1-a)^{\frac{25}{9}a }+a^{\frac{25}{9} (1-a)^2}+(1-a)^{\frac{25 }{9}a^2}$$
The derivative cancels close to $a=\frac 14$. In fact, the first iterate of Newton method is extremely close to $\frac 6 {25}$
$$f\left(\frac{1}{4}\right)\sim 1.94050 \qquad \text{and} \qquad
f\left(\frac6{25}\right)\sim 1.94029$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve the equation $10x^3-6x^2-12x-8=0$ Solve the equation $10x^3-6x^2-12x-8=0$
I was doing this question but got stuck and hence looked at the solution, which went as follows $(x+2)^3=x^3+6x^2+12x+8$ and hence the equation is written as $11x^3=(x+2)^3$ from which we have that $\frac{x+2}{x}=\sqrt[3]{11}$ and hence $x=\frac{2}{\sqrt[3]{11}-1}$.
Could you please explain to me how they thought of using the fact that $(x+2)^3=x^3+6x^2+12x+8$ and if there is a more intuitive approach, could you please explain it to me?
| I imagine the first thing is to try to factor. That will fail as this has not rational roots but with $8 = 2^3$ as the last term of a third degree polynomial they think in terms of $(ax \pm 2)^3 = a^3 \pm 6a^2 + 12 a \pm 3$. That fails as there is no such possible $a$ but the coefficients of $6,12, 8$ are just too convenient to ignore.
$(x\pm 2)^3 = x^3 \pm 6x^2 + 12x \pm 8$ and $P(x) = 10x^3 - 6x^2 -12x -8$. So if we subtracted those then we'd wipe out every coefficient and something useful may come of it. We'd get, if nothing else, a shorter equation.
So $P(x) + (x+2)^3 = (10x^3 - 6x^2 -12x -8) + (x^3 + 6x + 12x + 8) = 11x^3$.
Well that's a clean expression $P(x) + (x+2)^3 = 11x^3$ but what can we do with it?
I'll be honest my first thought is not what the book did, but to set $P(x) = 11x^3 - (x+2)^3$ and factor knowing $m^3 - n^3 = (m-n)(m^2 + mn + n)$. But then I'd realize $11$ doesn't factor nicely. But... I suppose I must be honest. I didn't solve this before reading the books answer so if I hadn't read the solution .... I'd probably plow ahead.
$11x^3 - (x+2)^3 = (\sqrt[3]{11}x - (x+2))[(\sqrt[3]{11}^2x^2 + \sqrt[3]{11}x(x+2) +(x+2)^2]$ so either $\sqrt[3]{11}x - (x+2)=0$ and $x=\frac 2{\sqrt[3]{11}-1}$ (the book's solution)
or $(\sqrt[3]{11}^2x^2 + \sqrt[3]{11}x(x+2) +(x+2)^2=(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)x^2 +(2\sqrt[3]{11}+4)x + 4=0$
And by quadratic equation: $x = \frac{-(2\sqrt[3]{11}+4) \pm \sqrt{(2\sqrt[3]{11}+4)^2 - 16(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)}}{2(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)}$. But that has a negative determinate so no solution.
But the book made a simple observation that would have saved me a lot of time and headache
Well, we are attempting to solve $P(x) = 0$ so we need to solve $(x+2)^3 = 11x^3$ and now we are doing what the text did.
Maybe I'd do it differently $x+2 = \sqrt[3]{11} x$ so $x-\sqrt[3]{11}x = -2$ so $x = \frac {-2}{1-\sqrt[3]{11}} = \frac 2{\sqrt[3]{11}-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Please check my proof for $x^2+|x-2|>1$ Let $f(x)=x^2+|x-2|-1$.
If $x <2,$ then
$f(x)=x^2-x+1 \implies f'(x)=2x-1 \implies f_{min}=f(1/2)=\frac{3}{4} >1.$
If $x>2$, then $f(x)=x^2+x-3 \implies f'(x)=2x+1>0$. So the function is increasing for $x>2$
The function $f(x)$ has just one min, so $f(x)>f(1/3)=3/4 >0$, hence for all real values $x^2+|x-2| >1.$.
Is my proof correct. What can be other methods to prove this?
| Ehen $x>2$ the inequation is $$x*2+x-3 >0\implies x \in (-\infty, \frac{-1-\sqrt{13}}{2}) \cup (\frac{\sqrt{13}-1}{2}, \infty)$$
So the solution is $x \in (2, \infty).$
When $x \le 2$, the inequation becomes $x^2-x+1>0$ which is true as $B^2< 4AC$. Then the solutions are $(-\infty, 2]$.
Finally, the solutions of this inequation are all real numbers, hence proved.
| {
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Proving $ 1+2f'(x)+\frac{2}{x(1+x^2)}\left(\frac{3x}{2}+f(x) \right)\ge \frac{6x^2}{1+8x^2} $.
Put
\begin{align*}
f(x)=\left( -\frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}-\left( \frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}
\end{align*}
Prove that
$$
g(x):=1+2f'(x)+\frac{2}{x(1+x^2)}\left(\frac{3x}{2}+f(x) \right)\ge \frac{6x^2}{1+8x^2}
$$
My attempt
I put
\begin{align*}
A=\left( -\frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}\quad
B=\left( \frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}
\end{align*}
and then
\begin{align*}
f'(x)=-\frac{1}{3}\frac{1}{A^2-AB+B^2}
\end{align*}
where $AB=\frac13$. But I don't know how to continue. I know that $g(x)$ is an even function. Via mathematica I find that
$$\left[(1+x^2)g(x)\right]'\ge 0\quad \forall \,x>0$$
But I also can't prove this. Any hints? Thanks in advance!
| The hint.
Let $f(x)=y$.
Thus, $$y^3+y+x=0,$$ which gives $$3y^2y'+y'+1=0$$ or $$y'=-\frac{1}{1+3y^2}$$ and we need to prove a polynomial inequality of one variable $y$.
I got that finally we need to prove that:
$$y^2(6y^{14}+16y^{12}-10y^{10}+y^8+94y^6+94y^4+26y^2+1)\geq0,$$ which is obvious.
| {
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Equation of Conics The general equation of the conic section is : $ax^2+2hxy+by^2+2gx+2fy+c=0$,
where
$$\Delta=\begin{vmatrix}a&h&g\\h&b&f \\g&f &c\\
\end{vmatrix}$$
This equation can also be analysed to distinguish whether it is an equation of pair of straight lines, parabola, ellipse or hyperbola.
*
*If $\Delta=0$ and $h^2=ab$, it represents pair of straight lines
*If $\Delta \neq 0$ and $h^2=ab$, it represents a parabola
*If $\Delta \neq 0$ and $h^2<ab$, it represents an ellipse
*If $\Delta\neq 0$ and $h^2>ab$, it represents a hyperbola
So my question is
Can you represent a proof for why these are the relations for different shapes?
I would prefer a more analytic/geometric approach
| Let's go the other way:
Any pair of lines has equation $k(a_1 x+b_1 y+c_1)(a_2 x+b_2 y+c_2)=0,$ which corresponds to the matrix being $\begin{pmatrix}a_1 \\b_1\\c_1\end{pmatrix}\begin{pmatrix}a_2&b_2&c_2\end{pmatrix},$ which is of rank $1$ and hence has zero determinant. Also all the 2 by 2 minors are zero in particular $h^2-ab=0.$
Now any parabola can be written as a multiple of $\frac{(a_d x+b_d y+c_d)^2}{a_d^2+b_d^2}=(x-f_x)^2+(y-f_y)^2$ or $-k\frac{(b_d x-a_d y)^2+((-2b_d^2f_x)-2a_d^2f_x-2a_dc_d)x+((-2b_d^2f_y)-2a_d^2f_y-2b_dc_d)y+(b_d^2f_y^2+a_d^2f_y^2+b_d^2f_x^2+a_d^2f_x^2-c_d^2)}{a_d^2+b_d^2}=0.$ That is the first minor $\frac{k^2}{(a_d^2+b_d^2)^2}\begin{pmatrix}b_d^2&-a_db_d\\-a_db_d&a_d^2\end{pmatrix}$ vanishes or $h^2=ab.$
More generally any ellipse can be given as having equation proportional to $e^2\frac{(a_d x+b_d y+c_d)^2}{a_d^2+b_d^2}=(x-f_x)^2+(y-f_y)^2$ with the squared eccentricity $0<e^2<1$ and any hyperbola by the same equation with $e^2>1$. Making the first minor
$\frac{k^2}{(a_d^2+b_d^2)^2}\begin{pmatrix}a_d^2(e^2-1)-b_d^2&a_db_de^2\\a_db_de^2&b_d^2(e^2-1)-a_d^2\end{pmatrix},$ which has determinant proportional to $(e^2-1)(a_d^2+b_d^2)^2.$
| {
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Bisector of the acute angle formed between the lines $4x-3y+7=0$ and $3x-4y+14=0$
The bisector of the acute angle formed between the line $4x-3y+7=0$ and $3x-4y+14=0$ has the equation...
By calculating the intersection point, we get is as $(2,5)$.
But I couldn't proceed because I don't know how to find the equation on the acute side. Please help.
I would prefer a more geometrical approach
| Equation of angle bisector lines are given by normalized equations of the two lines.
So, $\frac{L1}{|L1|} \pm \frac{L2}{|L2|} = 0$
So, $\frac {4x-3y+7} {\sqrt{4^2+3^2}} \pm \frac {3x-4y+14} {\sqrt{4^2 + 3^2}} = 0$
That gives us both angle bisectors: $x - y + 3 = 0, \, x + y - 7 = 0$
Now the slopes of original lines are both positive: $\frac{4}{3}, \frac{3}{4}$.
So the angle bisector that we have to choose on the acute angle side should have a positive slope between $\frac{4}{3}$ and $\frac{3}{4}$.
So equation of desired angle bisector is $x - y + 3 = 0$
| {
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Verify if the following limit exists (by the formal definition): $\lim_{(x,y)\to(0,0)}\frac{x^3-y^3}{x^2+y^2}$ First, I tried approaching the point through a few curves: ($x=0,\,y=0,\, y=m\cdot x,\, y=x^2,\, x=y^2)$ and to all of those I got $0$ for my limit. Since this isn't enough to prove that the limit actually exists, I need to move to the formal definition.
The definition:
If $0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ then $\left|f(x,y)-L\right|<\epsilon$
But I'm having a lot of difficulty in using this definition of limits to prove if it exists or not.
| Theorem: If $\lim f(x) = 0$ and $g(x)$ is bounded, then $\lim f(x)g(x) = 0$.
$$\frac{x^3-y^3}{x^2+y^2} = \frac{(x-y)(x^2+xy+y^2)}{x^2+y^2} = \frac{(x-y)xy + (x-y)(x^2+y^2)}{x^2+y^2} = (x-y)\frac{xy}{x^2+y^2}+x-y$$
Because $\begin{aligned}\lim_{(x, y)\to(0, 0)} x-y = 0\end{aligned}$ we have $\begin{aligned}\lim_{(x, y)\to(0, 0)} \frac{x^3-y^3}{x^2+y^2} = \lim_{(x, y)\to(0, 0)} (x-y)\frac{xy}{x^2+y^2}\end{aligned}$.
By $MA\ge MG$ we have $\begin{aligned}|xy|\le \frac{x^2+y^2}{2} \text{ so } \left|\frac{xy}{x^2+y^2}\right|\le \frac{1}{2} \text{ i.e. } \frac{xy}{x^2+y^2}\end{aligned}$ is bounded, therefore, as $\begin{aligned}\lim_{(x, y)\to(0, 0)} x-y = 0\end{aligned}$, because of the Theorem we shall have
$$\lim_{(x, y)\to(0, 0)} (x-y)\frac{xy}{x^2+y^2} = 0$$
| {
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| If you replace with $t:= 2-x$, the question becomes
$$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1}$$
From wolframalpha, reading the diagram for $\sqrt{t+1}-1$, at the neighbour of $0$ it's something like $t$, or $$\sqrt{t+1}-1=t+O(t^2)$$; similarly, $$\sqrt{t+4}-2=\frac12 t+O(t^2)$$
, here we are using the big O notation.
Then
$$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1} = \lim _{t\to 0}\frac{\frac12 t+O(t^2)}{t+O(t^2)}=\frac12$$
Notice the last step is reduction of a fraction $t$.
In your first result,
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it
still gives the indetermination $\frac{0}{0}$
It's actually
$$\lim _{t\to 0} \left(\dfrac{(\sqrt{t+4}-2)(\sqrt{t+1}+1)}{t}\right) $$
, as the reduction hasn't been done, it's still $\frac{0}{0}$.
So the trick is to derive something as $t$ and reduce.
Sometimes, the problem is a bit tricky that need to do more than one round.
ps. in analysis, after limit, derivative would be explained, then its Taylor's expansion, at that time $\sqrt{t+1}-1=\frac12 t+O(t^2)$ will be more obvious.
| {
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Proving that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$ I have to show that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$.
What I know is that $\sum_{k=0}^{k=n} \binom{n}{k} \cdot k = 2^{n -1} \cdot n$.
How do I proceed from there?
| $$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{2n-1}+\dbinom{2n}{2n}=2^{2n}$$
$$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{n-1}+\dbinom{2n}{n}=2^{2n-1}$$
$$\dbinom{n}{k}=\dbinom{n}{n-k}$$
$$k\dbinom{n}{k}+(n-k)\dbinom{n}{n-k}=n\dbinom{n}{k}$$
$$0\dbinom{2n}{0}+1\dbinom{2n}{1}+2\dbinom{2n}{2}+...+(2n-1)\dbinom{2n}{2n-1}+2n\dbinom{2n}{2n}=2n\dbinom{2n}{0}+2n\dbinom{2n}{1}+2n\dbinom{2n}{2}+...+2n\dbinom{2n}{n-1}+2n\dbinom{2n}{n}=2n(\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{n-1}+\dbinom{2n}{n})=2n\times 2^{2n-1}$$
Hence
$$n\dbinom{2n}{0}+n\dbinom{2n}{1}+n\dbinom{2n}{2}+...+n\dbinom{2n}{n-1}+n\dbinom{2n}{n}=n\times2^{2n-1}$$
| {
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Calculating probability that $P(X\geq Y+1)$, where $X$ and $Y$ are independent random variables
Let $X$ and $Y$ be independent random variables with probability functions given by
$$
p_{X}(x)=\left\{\begin{array}{cc}
\frac{2^{x}}{15}, & x \in\{0,1,2,3\} \\
0, & \text { otherwise }
\end{array}\right. \text { e } p_{Y}(y)=\left\{\begin{array}{cc}
\frac{3^{y}}{40}, & y \in\{0,1,2,3\} \\
0, & \text { otherwise }
\end{array}\right.
$$
Determine $P(X\geq Y+1)$.
By my calculations, we have
$$ \begin{aligned}
P(X \geq Y + 1) &= \sum_{i=0}^{2} P(X \geq i+1)P(Y=i) \\
&=P(X \geq 1)P(Y=0)+P(X \geq 2)P(Y =1) + P(X \geq 3)P(Y = 2) \\
&=(P(X=1) + P(X=2)+P(X=3)) \cdot \frac{1}{40} + (P(X=2)+P(X=3))\cdot \frac{3}{40}+P(X=3)\cdot \frac{9}{40} \\
&=\left(\frac{2}{15}+\frac{4}{15}+\frac{8}{15}\right)\cdot \frac{1}{40}+\frac{12}{15}\cdot \frac{3}{40} + \frac{8}{15} \cdot \frac{9}{40} \\
&=\frac{61}{300}
\end{aligned}
$$
Can someone check my work? I realise it's a pretty simple problem but I'd just like to know that I didn't make any mistakes.
| We have $$\{X\geqslant Y+1\} = \bigcup_{i=1}^3\bigcup_{j=0}^{i-1} \{X=i\}\cap\{Y=j\}, $$
and the union is of disjoint events. By independence, then
\begin{align}
\mathbb P(X\geqslant Y+1) &= \mathbb P\left(\bigcup_{i=1}^3\bigcup_{j=0}^{i-1} \{X=i\}\cap\{Y=j\}\right)\\
&= \sum_{i=1}^3\sum_{j=0}^{i-1}\mathbb P(X=i)\mathbb P(Y=j)\\
&= \frac1{15\cdot40}\sum_{i=1}^3\sum_{j=0}^{i-1} 2^i3^j\\
&= \frac{61}{300},
\end{align}
which agrees with your result. Indeed, a simple problem, just somewhat of a tedious computation.
| {
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Help with solving an elliptic integral I have the elliptic integral below, and currently I am stuck on how I can solve it, could someone please help me out:
$$\int^{\frac{L}{15}}_{0}2\pi\sqrt{\frac{a^{2}+b^{2}}{2}}dx$$
where: $a = W-\frac{0.608W^{2}}{L^{2}}x^{2}$ and $b = H-\frac{1.68H^{2}}{L^{2}}x^{2}$ ($H$, $W$ and $L$ are all positive constants).
Substituting into the initial equation:
$$\int^{\frac{L}{15}}_{0}2\pi\sqrt{\frac{(W-\frac{0.608W^{2}}{L^{2}}x^{2})^{2}+(H-\frac{1.68H^{2}}{L^{2}}x^{2})^{2}}{2}}dx$$
How can I solve the equation above?
| Using whole numbers
$$\frac{a^2+b^2}2=A-B x^2+C x^4$$
$$A=\frac{H^2+W^2}{2} \qquad B=\frac{2 \left(21 H^3+76 W\right)}{25 L^2} \qquad C=\frac{2 \left(441 H^4+5776\right)}{625 L^4}$$ and the antiderivative
$$I=\int \sqrt{\frac{a^2+b^2}2}\,dx$$ contains, as you expected , the elliptic integrals of the first and second kinds with very nasty arguments.
We can complete the square and write
$$A-B x^2+C x^4=C\left( \left(x^2-\frac{B}{2 C}\right)^2-\frac{B^2-4 A C}{4 C^2}\right)$$ Now, let
$$D=\frac{B}{2 C} \qquad \qquad E=\frac{4 A C-B^2}{4 C^2}$$ to make
$$I=\sqrt C \int \sqrt{(x^2- D)^2+E}\,dx$$
Have a look to the monster.
| {
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Solve the inequality $x^4-3x^2+5\ge0$
Solve $$\sqrt{x^4-3x^2+5}+\sqrt{x^4-3x^2+12}=7.$$
$D_x:\begin{cases}x^4-3x^2+5\ge0 \\x^4-3x^2+12\ge0\end{cases}.$ We can see that $x^4-3x^2+12=(x^4-3x^2+5)+7,$ so if $x^4-3x^2+5$ is non-negative, $x^4-3x^2+12$ is also non-negative (even positive). So I am trying to solve $$x^4-3x^2+5\ge0.$$Let $x^2=y,y\ge0.$ Now we have $$y^2-3y+5\ge0; D=9-4\times5<0,a=1>0$$ so the function $f(y)=y^2-3y+5>0$ for all $y$. I don't know what to do next. The solution of the inequality is indeed $x\in(-\infty;+\infty),$ but I have the restriction $y\ge0?$
| Hint :
First solve for $\sqrt{z} + \sqrt{z+7} = 7 $
Next substitute $z=x^4-3x^2+5$ and solve for $x$.
Actually easy to see $\sqrt{9} + \sqrt{9+7} = 7 $.
$9$ is only solution as monotonic function has unique root (when it does).
So only need to solve for $x^4-3x^2+5=9$.
| {
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Find all real values of $x, y$ and $z$ such that $x-\sqrt{yz}=42$, $y-\sqrt{xz}=6,z-\sqrt{xy}=-30$. (COMC) 1997 Part B Question 4 Find all real values of $x, y$ and $z$ such that $x-\sqrt{yz}=42$, $y-\sqrt{xz}=6,z-\sqrt{xy}=-30$.
I was preparing for my COMC Contest so I was doing past years exams. This question really stop me so I want to get some help from others.
I tried substitute $x$ to $a^2$, $y$ to $b^2$ and $z$ to $c^2$. And combining the equations so I can turn this function to 2 variables. But I am stuck because I can't find a way of turning this function to 2 variables. I add the three new equations with a, b and c in pairs, but I can't reduce the equations to 2 variables. I also tried to subtract the equations in pairs, too. I still can't figure it out. Can someone solve my question with the same method too?
Thank you very much!
This question is from Canadian Open Mathematic Challenge (COMC) 1997 Part B Question 4.
| Using your substitution and subtracting the equations pairwise, we end up with:
$72 = 42 - (-30) = a^2 - bc - (c^2 - ab ) = (a-c) ( a+b+c)$
$36 = 6 - (-30) = b^2 - ca - (c^2 - ab) = (b-c) (a+b+c )$
$ 36 = 42 - 6 = a^2 - bc - (b^2 - ca) = (a-b) ( a + b + c) $
Hence $ a-c = 2 (b-c) = 2(a-b)$, or that $ b = \frac{ a+c}{2}$.
Substituting this into the second equation gives us $(a-c)^2 = 24 $ so $|a-c| = 2 \sqrt{6} $.
Substituting this into the first equation gives us $84 = 2a^2 - ac - c^2 = (a-c)(2a + c) $, so $ |2a+c| = 7 \sqrt{6} $, where $a-c, 2a+c$ have the same sign.
Hence, the solutions are $(a, b, c) = \pm ( 3 \sqrt{6}, 2 \sqrt{6}, \sqrt{6} )$, which gives us $ (x,y,z) = (54, 24, 6 )$.
Notes:
*
*Given the symmetry in the equations, taking the pairwise difference and (attempting to) factoring out $a-b$ is a common approach. It is often harder to work with the pairwise sum.
| {
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Prove inequality $|a-b+c-d| \leqslant \frac{1}{16}$ Let $a,b,c,d$ be positive real numbers that fulfill two conditions:
$$a+b+c+d \leqslant 2$$$$ab+bc+cd+ad \geqslant 1$$
Prove that $|a-b+c-d|\leqslant \frac{1}{16}$
Let:
$a+c=x$ and $b+d=y$
Both $x$ and $y$ are positive.
$$x+y \leqslant 2$$$$xy \geqslant 1$$
$$-4xy \leqslant -4$$
$$(x+y)^2 \leqslant 4$$
$$(x-y)^2 \leqslant 0 \Rightarrow |a-b+c-d|=|x-y|=0\leqslant \frac{1}{16}$$
I do not really think that this is valid solution, but I can not find any mistakes.
| By AM-GM $$1\leq ab+bc+cd+da=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2\leq1,$$
which gives $$a+c=b+d$$ and $$a+b+c+d=2$$ or
$$a+c=b+d=1,$$ which gives $$|a-b+c-d|=0<\frac{1}{16}.$$
| {
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Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.
| Here is a different approach: Consider the change of variable $b_n=a_{n+4},$ so that the only initial condition is given by $b_0=0.$ Notice that the equation becomes
$$(n+5)a_{n+5}-(n+4)a_{n+4}+3a_{n+4}-12(n+5)H_{n+5}=0,$$
$$(n+5)b_{n+1}-(n+4)b_{n}+3b_{n}-12(n+5)H_{n+5}=0,$$
take $f(x)=(x+4)b_x$ so that $$\Delta (f)=f(x+1)-f(x)=(x+5)b_{x+1}-(x+4)b_x,$$
so we get that $$\Delta (f(x))=\frac{-3}{x+4}f(x)+12(x+5)H_{x+5}.$$
This looks like variation of parameters, check Remark 10.
We get then that the solution becomes
$$f(x)=\sum _{u=1}^{x-1}\left (\prod _{t=u+1}^{x-1} \left (1-\frac{3}{t+4}\right )\right )\cdot \left (12(u+5)H_{u+5}\right ),$$
notice that the product becomes
$$\left (\prod _{t=u+1}^{x-1} \left (1-\frac{3}{t+4}\right )\right )=\left (\prod _{t=u+1}^{x-1} \left (\frac{t+1}{t+4}\right )\right )=\frac{(u+2)(u+3)(u+4)}{(x+1)(x+2)(x+3)},$$
replacing we get
$$f(x)=12\sum _{u=0}^{x-1}\frac{\binom{u+5}{4}4!H_{u+5}}{(x+1)(x+2)(x+3)}=\frac{12\cdot 4!}{(x+1)(x+2)(x+3)}\left (\sum _{u=0}^{x+4}\binom{u}{4}H_{u}-\sum _{u=0}^4\binom{u}{4}H_{u}\right )$$
using your professors hint(which is a good exercise of integration by parts), we get that
$$f(x)=\frac{12\cdot 4!}{(x+1)(x+2)(x+3)}\left (\binom{x+5}{5}H_{x+5}-\frac{1}{5}\binom{x+5}{5}-\frac{25}{12}\right )$$
Taking back the change of variable, meaning plugging at $x-4,$ we get
$$na_n=f(n-4)=\frac{12\cdot 4!}{(n-3)(n-2)(n-1)}\left (\binom{n+1}{5}H_{n+1}-\frac{1}{5}\binom{n+1}{5}-\frac{25}{12}\right )$$
| {
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prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$
prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$
By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.
Hence it suffices to prove $$\sum {\left(z^{2/3} \right)}^{3}\ge 3^{2/3}$$ which is falsse.
I was able to get a weaker bound.since $xy+yz+zx=1$ we have $x^2+y^2\le 2$. By C-S and $x^2+y^2<2$
$$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge \frac{1}{2}\sum \frac{1}{x^2+y^2}>3/4$$
I am interested on a solution using standard inequalities (C-S AM-GM,chebyshov etc) rather than fully expanding and using uvw/pqr/schur.
| We need to prove that:
$$\sum_{cyc}\frac{1}{(x+y)^2}\geq\frac{9}{4(xy+xz+yz)}$$ or
$$\sum_{cyc}\left(\frac{1}{(x+y)^2}-\frac{3}{4(xy+xz+yz)}\right)\geq0$$ or
$$\sum_{cyc}\frac{4xz+4yz-2xy-3x^2-3y^2}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}\frac{(z-x)(3x+y)-(y-z)(3y+x)}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{3y+z}{(y+z)^2}-\frac{3x+z}{(x+z)^2}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq0.$$
Now, let $x\geq y\geq z$.
Thus, $$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq$$
$$\geq\sum_{cyc}(x-y)^2(xz+yz-z^2)(x+y)^2\geq$$
$$\geq(x-z)^2y(x+z-y)(x+z)^2+(y-z)^2x(y+z-x)(y+z)^2\geq$$
$$\geq(y-z)^2y(x-y)(x+z)^2+(y-z)^2x(y-x)(y+z)^2=$$
$$=(x-y)^2(y-z)^2(xy-z^2)\geq0.$$
| {
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Stronger than Nesbitt's inequality using convexity and functions Hi it's a refinement of Nesbitt's inequality and for that, we introduce the function :
$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$
With $a,b,x>0$
Due to homogeneity we assume $a+b=1$ and we introduce the function :
$$g(a)=\frac{a}{1-a+x}$$
Showing that $g(a)$ is convex on $(0,1)$ is not hard so we have :
$$\frac{b}{x+a}+\frac{a}{b+x}\geq 2\frac{a+b}{2(\frac{a+b}{2}+x)}=h(x)$$
So we have :
$$f(x)\geq h(x)+\frac{x}{a+b}$$
Now we put $u=\frac{x}{a+b}$ and we want to show :
$$h(x)+\frac{x}{a+b}=u+\frac{1}{0.5+u}\geq \frac{3}{2}$$
The last inequality is obvious.
My question :
It is correct?
Do you know other refinements?
Thanks in advance!
Ps: I add the tag reference request for the last question.
| Stronger than Nesbitts inequality for non-cyclic style
*
*If $a,\,b,\,c$ are positive real numbers, then
$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$
*If $a,\,b,\,c$ are positive real numbers, then
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{27(b-c)^{2}}{16(a+b+c)^{2}}.$$
*Let $a,\,b,\,c$ be nonnegative real numbers, no two of which are zero. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{(b-c)^{2}}{2(b+c)^{2}}.$$
*Let $a,b,c$ are non-negative real numbers such that $ab+bc+ca>0.$ Prove that
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geqslant \dfrac{3}{2}+\dfrac{7(b-c)^2}{16a(b+c)+7bc}.$$
| {
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How I can solve this limit only with algebra I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.
$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$
first all I use the logarithms properties and rewrite like this:
$$\lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$$
and tried to make to this expression goes to $$\lim_{x\to 0} \log e^\frac{1}{2}$$
and then limit will be 1/2
I can't reach to this because the limit goes to 0 instead of ∞
| Let $a = \lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$. Then:
$$\exp(a) = \lim_{x\to 0} \exp \left(\log({\frac{1 + x}{1 - x}})^{\frac{1}{2x}} \right) = \lim_{x\to 0} \left(\frac{1 + x}{1 - x} \right)^{\frac{1}{2x}} = \lim_{x\to 0} \left(\frac{1/x + 1}{1/x - 1} \right)^{\frac{1}{2x}}$$
Now let $u = \frac{1}{x}$. Then we have:
$$\lim_{u \to \infty} \left(\frac{u + 1}{u - 1} \right)^{\frac{u}{2}} =\lim_{u \to \infty} \left(\frac{u-1}{u-1} + \frac{2}{u - 1} \right)^{\frac{u}{2}} = \lim_{u \to \infty} \left(1 + \frac{2}{u - 1} \right)^{\frac{u-1}{2}} \left(1 + \frac{2}{u - 1} \right)^{\frac{1}{2}}$$
$$= e \cdot 1 = e$$
Therefore $\exp(a) = e \Rightarrow \boxed{a = 1}.$
| {
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Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$. I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?
| EDIT: The question was updated, so here's my new answer.
Notice the LHS is
and since $x>0$, it's clear that it is $<0$ for all $x$.
Old answer: This is false. Indeed, since $x>0$, we have
\begin{align*}
&x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 >0\\
\iff & (x+\sqrt x)^2\left(x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 \right) >0\\
\iff & x^3 + 2x^{5/2}-3x^2-4x^{3/2}+3x-3 >0\\
\iff & (\sqrt x+1)(x^{5/2}+x^2-4x^{3/2}+3x^{1/2}-3)>0,
\end{align*}
but this isn't true for all $x$, e.g. put $x=1$ and the quintic becomes $-2<0$.
| {
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indefinite integral of $\int \frac{dx}{\sqrt{(x)(x+1)(x+2)}}$ I tried integration by parts considering $dx=du$
$\frac{1}{\sqrt{(x)(x+1)(x+2)}}=v$ but im not getting the answer.
My attempt....
$uv=\sqrt{\frac{x}{(x+1)(x+2)}}$ and-$\int{dv u}$=$\frac{I}{2}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$
How do we calculate $\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$ and
$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$
How do we do it?
Please help!
| In general, an integral with square-root of a polynomial of degree $3$ or $4$ is an elliptic integral.
Maple evaluates this in terms of the elliptic integral of the first kind $F$:
$$
\int \!{\frac {1}{\sqrt {x \left( x+1 \right) \left( x+2 \right) }}}
\,{\rm d}x=2\;{ F} \left( \frac{\sqrt {x+2}}{\sqrt{2}},\sqrt {2} \right)
$$
| {
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Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.
Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.
Assume $y \ge 3$ and $x \ge 1$.
By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.
By $\mod 7$,we get $y \equiv 1 \mod 6$.
I also tried $\mod 2$ but it didn't work.
Please post hints ( not a solution). Thanks in advance!
| It's $3(3^a-1)=7(7^b-1)$ with $a=x-1$ and $b=y-1$.
Therefore $7\mid3^a-1$, so $a$ is a multiple of (what?).
Therefore, $3^a-1$ is a multiple of $13$.
Therefore, $7^b-1$ is a multiple of $13$.
Therefore, $b$ is a multiple of (what?).
Therefore, $7^b-1$ is a multiple of $9$.
Therefore, $3(3^a-1)$ is a multiple of $9$.
Therefore, $a$ is (what?).
Therefore, $x$ is (what?).
| {
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Proving Riordan's identity that $\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1$ In a book titled "Advances in Problem Solving," authored by Sailesh Shirali, Riordan's identity is mentioned which can also be written as
$$S=\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1~~~~(1)$$
Here, we prove it by the integral representation of $(j+1)!$ as:
$$S=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{(j+1)!}{n^{j+1}}=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{1}{n^{j+1}} \int_{0}^{\infty} x^{j+1} e^{-x} dx$$
$$\implies S=\frac{1}{n^{n}}\int_{0}^{\infty} x ~(n+x)^{n-1} e^{-x} dx=n^{-n}\int_{0}^{\infty} [(x+n)^n-n(x+n)^{n-1}]~ e^{-x} dx$$ $$S=-n^{-n}\left. (x+n)^n~ e^{-x}\right|_{0}^{\infty}=1.$$
It will be interesting to see other approaches for proving (1).
| We seek to show that
$$S= \sum_{k=0}^{n-1} {n-1\choose k} \frac{(k+1)!}{n^{k+1}} = 1.$$
The sum is
$$\frac{(n-1)!}{n}
\sum_{k=0}^{n-1} \frac{k+1}{(n-1-k)!} \frac{1}{n^k}.$$
We get without the factor in front
$$\sum_{k=0}^{n-1} \frac{-n+k+1}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{n}{(n-1-k)!} \frac{1}{n^k}
\\ = - \sum_{k=0}^{n-2} \frac{n-1-k}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = - \sum_{k=0}^{n-2} \frac{1}{(n-2-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = - \sum_{k=1}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = \frac{1}{(n-1)!} \frac{1}{n^{-1}}.$$
Restoring the factor in front we find
$$\frac{(n-1)!}{n} \times
\frac{1}{(n-1)!} \frac{1}{n^{-1}} = 1$$
as claimed.
| {
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Showing $\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$ for nonnegative integers $\alpha$ and $\beta$. I found this lemma in a paper I was reading and it was not proved. There doesn't seem to be any obvious factorized form so how would one go about proving the inequality holds?
$$\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$$
$$\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2-\alpha(\beta+1) \geq 0$$
$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geq 0$$
| Alternative proof inspired by Michael Hardy:
$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geqslant 0 \iff 20\alpha^2+4\beta^2-12\alpha
\beta-12\alpha \geqslant 0$$
$$20\alpha^2+4\beta^2-12\alpha \beta-12\alpha
= (2\beta-3\alpha )^2+11\alpha^2-12\alpha \\
=\begin{cases}
(2\beta-3\alpha )^2+\alpha (11\alpha-12)>0, & \text{if $\alpha>1$} \\
(2\beta-3)^2-1 \geqslant 0, & \text{if $\alpha=1$} \\
4\beta^2 \geqslant 0 & \text{if $\alpha=0$}
\end{cases}
$$
$$
"=" \iff \alpha=1, 2\beta-3=\pm 1 \text{ or } \alpha=\beta=0\\ \iff (\alpha, \beta)\in \{(0,0), (1,1), (1,2)\}
$$
| {
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Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$? If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?
I tried this problem for a while but made no progress. I don't know how $\frac{a}{b}+\frac{c}{d}$ can take only certain values. The answer was given to be $17$. Can someone help me with this?
| Hint: Note that
$$
\frac ac + \frac ca + \frac bd + \frac db = 20 \implies\\
\frac ab \cdot \frac bc + \frac cd \cdot \frac da + \frac bc \cdot \frac cd + \frac da \cdot \frac ab = 20.
$$
With that in mind, compare the expanded sums
$$
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^2, \quad
\left(\frac{a}{b}-\frac{b}{c}+\frac{c}{d}-\frac{d}{a}\right)^2.
$$
| {
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Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.
| Hints:
Put Let $a=\frac{x}{3\sqrt2},$ $b=\frac{y}{3\sqrt2}$ and $c=\frac{z}{3\sqrt2}.$
And $x=\tan(u)$,$y=\tan(v)$,$z=\tan(w)$ with $u+v+w=\pi$
Then the function $f(u)=\sqrt{3+\frac{18}{\tan(u)^2}}$ is convex on $(0,\pi)$ see here
Remains to apply Jensen's inequality .
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Proof: not a perfect square Let $y$ be an integer.
Prove that
$$(2y-1)^2 -4$$
is not a perfect square.
I Found this question in a discrete math book and tried solving it by dividing the question into two parts:
$$y = 2k , y = 2k + 1$$
But that got me nowhere.
| $(2y-1)^2-4=4(y^2-y)-3$ If it were a perfect square it would be $=c^2$, where c is an integer. Solve for $y$ in $4(y^2-y)-3-c^2=0$ and get $y=\frac{4\pm \sqrt{16+16(3+c^2)}}{8}=\frac{1\pm \sqrt{4+c^2}}{2}$.
However $c^2+4$ cannot be a square, unless $c=0$ (where $y$ is not an integer). Assume $c^2+4=b^2$ so $b=c+a$ with $(c+a)^2=c^2+2ac+a^2$. $2ac+a^2=4$ has no possible integer solutions. ($a=1$ LHS is odd, $a\gt 1$ LHS $\gt 4$).
Therefore no possible integer $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone out there will have a super smart elegant solution to this!
Base step: test when n = 1 ...
Assume true for $n = 1$
ie . $12^k - 4^k - 3^k +1 = 6M$, where $m$ is an integer
RTP: also true for $n = k+1$
ie. $12^{k+1} - 4^{k+1} - 3^{k+1} +1 = 6N$ where $N$ is an integer
LHS: $12^{k+1} - 4^{k+1} - 3^{k+1} +1$
$= 12( 4^k + 3^k - 1 + 6M) - 4^{k}(4) - 3^{k}(3) +1$ (from assumption)
$= 6(12M) + 12(4^k) + 12(3^k) -12 -4^{k}(4) - 3^{k}(3) +1$
Here is where I break down and go around in circles.
| We are going to replace $4^n=12^n-3^n+1-6M$
$\begin{align}12^{n+1}-4^{n+1}-3^{n+1}+1
&=12.12^n-4.4^n-3.3^n+1\\
&=12.12^n-4(12^n-3^n+1-6M)-3.3^n+1\\
&=8.12^n+3^n-3+24M
\end{align}$
$12^n$ and $24M$ are obviously divisible by $6$
Notice $3^n-3=3\times\underbrace{(3^{n-1}-1)}_{\text{even}}$ is also divisible by $6$ so you can finish your induction step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| For me it's just Vieta's formulas (or comparing coefficients as if $x^2+5x$ were a single variable):
$(x^2 + 5x + \color{red}4)(x^2 + 5x + \color{red}6) \color{red}{-48} = (x^2 + 5x + \color{red}{12})(x^2 + 5x \color{red}{-2})$?
$4+6=12+(-2)$, and $4\cdot 6 - 48 = 12 \cdot (-2)$, so yes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Cross section of cylinder $5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$ Prove that
$$5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$$
represents a cylinder whose cross-section is an ellipse of eccentricity $\frac{1}{\sqrt 2}$.
I know how to find the plane of a cross section if I know the direction ratios of at least one generator of the cylinder, but I don't know how to find that out from just the equation of the cylinder alone.
| The equation can be written $5(x+1)^2+5(y-1)^2+8z^2-2(x+1)(y-1)+8(y-1)z+8(x+1)z-6=0,$ and we can translate.
From @WillJagy's comment look at the plane $x+y-z=0.$ We're interested in the intersection $\langle 5x^2+5y^2+8z^2-2xy+8yz+8xz-6,x+y-z\rangle.$
Now rotate so that the normal vector $[\frac1{\sqrt{3}},\frac1{\sqrt{3}},-\frac1{\sqrt{3}}]$ goes to $[0,0,1],$ by way of the rotation matrix $\begin{pmatrix}\frac{1
}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\ -
\frac{1}{\sqrt{2}\sqrt{3}}&-\frac{1}{\sqrt{2}
\sqrt{3}}&-\frac{\sqrt{2}}{\sqrt{3}}\\ \frac{1
}{\sqrt{3}}&\frac{1}{\sqrt{3}}&-
\frac{1}{\sqrt{3}}\end{pmatrix}.$
This produces an equation in the new $XY$-plane: $6(X^2+2Y^2-1)=0,$ which has eccentricity $\frac1{\sqrt{2}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3920923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.
Any help would be appreciated.
| Start with $x-x^3/6<\sin(x)<x$; there are many nice proofs here.
Then we have
$$
\sum_{k=1}^{n}(k/n^2)-(k/n^2)^3/6 <\sum_{k=1}^{n}\sin(k/n^2) < \sum_{k=1}^{n}k/n^2
$$Using the fact that $\sum_{k=1}^{n}k^p = \frac{n^{p+1}}{p+1}+\text{ lower order terms}$, we have
$$
\frac{n^2+\cdots}{2 n^2}- \frac{n^4+\cdots}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2) < \frac{n^2+\cdots}{2 n^2},
$$and the result follows by squeezing. We can actually do a bit better with the sums using closed-form identites:
$$
\frac{n^2+n}{2 n^2}- \frac{n^4+2n^3+n^2}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2) < \frac{n^2+n}{2 n^2}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$? What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
I know that $289 = 17 \times 17$
so would it be $2^8\equiv 256\bmod17 =1$
and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this
| $256 \equiv 1 \pmod {17}$ but $256\not \equiv 1 \pmod {289}$ which we need.
But not $289 = 17\times 17$ so $\phi (289) = 17\cdot16$ so $2^{17\cdot 16}\equiv 1\pmod {289}$ by Eulers theorem.
But the order might be something smaller that divides $17\cdot 16$.
We can figure that $2^8 = 17*15 + 1 \equiv 17*(-2) + 1\pmod{17^2}$ so
$2^{16} \equiv 17^2 *4 + 2*(-2)*17 + 1 \equiv -67 \pmod {289}$.
So the order of $2$ is not $16$ and thus not anything that divides $16$. So the order of $2$ will be a multiple of $17$. be a multiple of $17$ that divides $16*17$.
And $2^{17} \equiv -8*17+2$
$2^{2*17} \equiv (-8*17+2)^2 \equiv -32*17+ 4\equiv 2*17+4 \equiv 38\pmod{289}$.
$2^{4*17} \equiv 4^2*17^2 + 16*17 + 4^2 \equiv 16*17 +16\equiv 18*16\equiv 1*(-1)\equiv -1 \pmod {289}$.
And so $2^{8*17}\equiv (-1)^2 \equiv 1 \pmod {289}$.
So the order of $2$ is $8*17= 136$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to find the sum of cosines I have been trying to calculate this sum, could someone confirm if all my working is correct please :) :
$$S_k(x)=\sum_{n=0}^k\cos(nx)=\Re\sum_{n=0}^ke^{inx}=\Re\sum_{n=0}^k\left(e^{ix}\right)^n=\Re\left(\frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)=\Re\left(\frac{1-\cos((k+1)x)-i\sin((k+1)x)}{1-\cos(x)-i\sin(x)}\right)\tag{1}$$
$$=\Re\left(\frac{\left[1-\cos((k+1)x)-i\sin((k+1)x)\right]\left[(1-\cos x)+i\sin x\right]}{(1-\cos x)^2+\sin^2x}\right)=\frac{(1-\cos x)(1-\cos((k+1)x))+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$
$$=\frac{1-\cos((k+1)x)-\cos (x)+\cos((k+1)x)\cos (x)+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$
$$\sin((k+1)x)\sin(x)=\frac{\cos(kx)-\cos((k+2)x)}{2}$$
$$\cos((k+1)x)\cos(x)=\frac{\cos(kx)+\cos((k+2)x)}{2}$$
$$\therefore\sin((k+1)x)\sin(x)+\cos((k+1)x)\cos(x)=\cos(kx)$$
$$S_k(x)=\frac{1-\cos((k+1)x)-\cos(x)+\cos(kx)}{1-2\cos(x)}\tag{2}$$
$$\cos(kx)-\cos((k+1)x)=-2\sin\left(\frac{(2k+1)x}{2}\right)\sin\left(\frac{-x}{2}\right)=2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)$$
$$S_k(x)=\frac{1-\cos(x)+2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)}{1-2\cos(x)}$$
| You could easily use $$\cos(nx)=\frac{e^{inx}+e^{-inx}}{2}$$ to make calculations bit easier. Let me propose you an alternative less messy solution, which uses telescoping method. Note that $$\sin((n+1)x)-\sin((n-1)x)=2\cos(nx)\sin(x).$$ Now telescoping gives $$\sin((n+1)x)+\sin(nx)-\sin(x)=2\sin(x)\sum_{k=1}^n\cos(kx)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$ for natural numbers $a,b,c$ Find natural numbers $a<b<c$ such that:
$$\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$$
I found infinitely many solutions when $a=1$, where $b$ and $c$ can be found in the following way:
Calculate, in the simplest form, the continued fractions of the series: $1+\frac{1}{2}$,$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}$, where each term contains an odd number of $2$'s.
Then $b$ is $\frac{1}{2}D$ and $c$ is $\frac{N-1}{2}$ where $D$ and $N$ are the numerator and the denominator of any term of the previous series, respectively.
For example,$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}$, So $b=\frac{1}{2}(12)=6$, and $c=\frac{17-1}{2}=8$.
So, $a=1,b=6,c=8$ is satisfied.
Calculating continued fractions by hand is easy, but tedious!, So is there a general way to calculate $b$ and $c$ given any natural number $a$?
Your help would be appreciated. Thanks!
| Not an answer, but rather a long comment.
Hint. Use the well-known formula $$\sum_{i=1}^n i=\frac{n\cdot (n+1)}{2}$$ in order to deduce $$\sum_{k=a}^{b-1} k=\sum_{k=1}^{b-1}k-\sum_{k=1}^{a-1}k=\ldots=\frac{(b-a)(a+b-1)}2$$ Similarly $$\sum_{k=b+1}^c k=\frac{(c-b)(c+b+1)}2$$Thus, your problem comes down to solving the following Diophantine equation $$(b-a)(a+b-1)=(c-b)(c+b+1)\iff2b^2=a^2-a+c^2+c$$ Given some $a\in\mathbb N$, observe that the latter equation is equivalent to $$\begin{align*}8b^2+2=(2a-1)^2+(2c+1)^2&\iff(2a-1)^2-2=8b^2-(2c+1)^2\\&\iff ... \end{align*}$$
I've been playing with Wolfram Alpha for a while and can't find any pattern for the solutions of the respective $a$ values, neither can the OEIS.
| {
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"timestamp": "2023-03-29T00:00:00",
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understanding the solution to a particular differential equation? so I ran across this problem:
$(3x^2+1)y' -2xy = 6x$
I solved it by integrating factor $ e^{\int\frac{-2x}{3x^2+1} dx}$
.
.
.
$\int D[(3x^2+1)^{\frac{-1}{3}}y] = \int 6x(3x^2+1)^{\frac{-4}{3}} dx$
.
.
.
$(3x^2+1)^{\frac{-1}{3}}y = -3(3x^2+1)^{\frac{-1}{3}} + C$
$y = -3 + C(3x^2+1)^{\frac{1}{3}}$
my question is interpreting the solution $y=-3 + C(3x^2+1)^{\frac{1}{3}}$. So I understand why the integrating factor method works in solving the problem. Why did the answer come out so simple after doing so much work? Could I have solved it 'faster?'
| You could transform the equation simply as
$$
(3x^2+1)y'=2x(y+3)
\\~\\
y'=\frac{2x}{3x^2+1}(y+3)
$$
This is a separable equation and the case of the constant solution $y=-3$ appears directly as the one to exclude for performing the separation of variables. Then continue
$$
\int\frac{dy}{y+3}=\int\frac{2x\,dx}{3x^2+1}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that if $3x$ is even then $3x+5$ is odd I'm learning the absolute basics of how to do proofs, and am really struggling.
If 3x is even then 3x+5 is odd.
This is the solution:
I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but nothing I found is this basic.
Work:
-Assumptions-
3x = 2n
3x+5 = 2k+1
-Trying to make sense of 3x-
3x+5 = 2k+1
3x = 2k-4
-Plugging in 2k-4 for 3x-
2k-4 = 2n
2k = 2n+4
k = n+2
-Plugging in n+2 for k-
3x+5 = 2(n+2)+1
...This is where I gave up. I don't know where I'm going with this anymore.
| Your problem is that $3x + 5 = 2k +1$ is not an assumption. It is the conclusion you need to prove.
Your one and only assumption is that $3x = 2n$ for some integer $n$.
so you start with
$3x = 2n$.
.... then you do a bunch of steps ....
.... steps .....
.... and get in the end ........
Conclusion: $3x + 5 = 2(????????) + 1$ where $??????$ is some integer you come up with in you steps.
Let's see what happens when we try. Let's take it nice and slow:
=======
$3x = 2n$.
$3x +5 = 2n + 5$
....hmmm, we want $2(??????) + \color{red}1$ in the end so let's pull out the $+\color{red}1$ first.....
$3x + 5 = 2n+5 = 2n + (4 + \color{red}1)=(2n+4) +\color{red}1$
.... hmmm, okay that's the $+1$ now we want $2(\color{red}{??????}) + 1$. To get the So we need to factor then $2$ out of $2n+4$ and see what we have left.... that will bee the $\color{red}{??????}$
$3x + 5 = (\color{red}{2n+4}) + 1$
$3x + 5= 2(\color{red}{n + 2}) + 1$
.... and that's it......
Conclusion: $3x+5 = 2(\color{red}{n + 2})+1$.
The $??????$ we wanted turns out to be $\color{red}{n+2}$ an we have
$3x + 5 = (3x+4) + 1 = (2n+4) + 1 = 2(\color{red}{n+2}) + 1$.
And because $\color{red}{n+2}$ is an integer if we let $k = n+2$ be that integer $3x+5 = 2k + 1$ and so... $3x + 5$ is odd.
=======
Although if you want to work backwards
Conclusion: $3x + 5 = 2k +1$ ..... and we want to solve for $k$ to show it is possible...
$3x + 5 -1 =2k + 1-1$
$3x +4 = 2k$
$k = \frac {3x + 4}2 = \frac {3x}2 + 2$.
.... but is $\frac {3x}2 + 2$ an integer?????
Well, $3x$ is even. So there is an integer $n$ so that $3x = 2n$ so
$k = \frac {3x}2 +2 = \frac {2n}2 + 2 = n+2$.
So $k=n+2$ is the integer we want to conclude $3x+5 =2k +1$.
If we did it this way our proof would go:
$3x$ is even so there is an integer $n$ so that $3x = 2n$. Let $k = n+2$; that is an integer.
$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$.
So $3x+5 = 2k +1$ and that is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Exponents of Generating Function $A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $ I have the following generating function:
$$A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $$
Among multiple of $2$, we will get a multiple of 6 whenever we take a multiple that is a multiple of 3: possible multiples: $3m, 3m+1, 3m+2$, where $3m$ cancels with the numerator.
So, we will have $1$ left in the numerator. What is left are factors in which the exponent is either $2\times(3m+1) $, $2\times(3m+2)$ or a multiple of $3$ ($3\times(2m+1), 3\times(3m+2)$).
Questions:
*
*Why $3m$ cancels with the numerator?
*Why what is left are factors in which the exponent is either $2\times(3m+1) $, $2\times(3m+2)$ or a multiple of $3$: $3\times(2m+1), 3\times(3m+2)$.
| We obtain
\begin{align*}
\color{blue}{A(x)}&=\prod_{k=1}^\infty \frac{1-x^{6k}}{\left(1-x^{2k}\right)\left(1-x^{3k}\right)}\\
&=\prod_{k=1}^\infty \frac{1+x^{3k}}{1-x^{2k}}\tag{1}\\
&=\prod_{k=1}^\infty \frac{1-x^k+x^{2k}}{1-x^k}\tag{2}\\
&\,\,\color{blue}{=\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\tag{3}
\end{align*}
Comment:
*
*In (1) we cancel $1-x^{3k}$.
*In (2) we cancel $1+x^k$.
Using the coefficient of operator $[x^N]$ to denote the coefficient of $x^N$ of a series, we derive from (3)
\begin{align*}
\color{blue}{[x^N]A(x)}&\color{blue}{=[x^N]\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\\
&=[x^N]\left(1+x^2+x^3+x^4+\cdots\right)\\
&\qquad\qquad\cdot\left(1+x^4+x^6+x^8+\cdots\right)\\
&\qquad\qquad\cdot\left(1+x^6+x^9+x^{12}+\cdots\right)\\
&\qquad\qquad\cdot\ \cdots
\end{align*}
which is the number of partitions of $N$ where each part of $N$ occurs at least twice.
| {
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What is the $n$th term of the sequence $1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1 ...$ So there exist a lot of similar questions which ask the $n$th term of
$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider each number as a single term. I tried making a pyramid of terms and got this
$1^2\\
1^2~2^2\\
1^2~2^2~3^2\\
1^2~2^2~3^2~4^2\\
...$
I got the $n$th term as$$T_n= (n\mod(m(m-1)/2)^2$$
Where $m$ is the row number.
But now I am stuck, is there a way to eliminate $m$ and write the general term only in terms of $n$?
| OEIS has an entry for this sequence. And they give a formula:
a(n) = A000290(m+1), where m = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2)
The sequence named A000290 is just the perfect squares, so A000290(m+1) means $(m+1)^2$. And $m$ is given as $n-\frac{t(t+1)}2$, where $t=\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor$. Nesting out all of this, the finished formula becomes
$$
\left(n-\frac{\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor\left(\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor+1\right)}{2}+1\right)^2
$$
where the brackets indicate the floor function.
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem in limits to show that $\lim_{n\to\infty}$ $p_n$/$q_n$ = $π/4$ Define two sequences of real numbers $p_n$ and $q_n$ with $n = −1, 0, 1, 2, \ldots$ as follows:
$$p_n = 2p_{n−1} + (2n−1)^2p_{n−2}$$ and
$$q_n = 2q_{n−1} + (2n−1)^2q_{n−2}$$
for every $n \geq 1$, and starting with $p_{−1}= 0, q_{−1}= 1, p_0 = q_0 = 1$.
Show that
$$\lim_{n \to \infty}\frac{p_n}{q_n} = \frac{\pi}{4}.$$
I did:
Suppose $a_n$ is a sequence that satisfies $a_n = 2a_{n−1} + (2n − 1)^2a_{n−2}$ for all
$n\geqslant1$. Subtracting $(2n + 1)a_{n−1}$ from both sides we derive a simple recursion for a new
sequence $b_n = a_n − (2n + 1)a_{n−1}$, namely $b_n = −(2n − 1)b_{n−1}$.
What to do after this... Is it correct?
If not kindly provide a detailed answer in the answer section.. As it helps a new beginner like me to understand who has just started to solve advanced problems.
| Let $r_n := p_n/q_n$. We can compute the first few times to observe that: $r_0 = 1$, $r_1 = \frac{2}{3} = 1-\frac{1}{3}$, $r_2 = \frac{13}{15} =1-\frac{1}{3} + \frac{1}{5}$, $r_3 = \frac{76}{105} = 1-\frac{1}{3} + \frac{1}{5} -\frac{1}{7}$. So this gives us enough motivation to conjecture that: $$\frac{p_n}{q_n} = 1 -\frac{1}{3} + \frac{1}{5} - \cdots + (-1)^n \frac{1}{2n+1} $$
for all $n \ge 0$. The first case is done, as stated above. Suppose the hypothesis for $n-1$ is true. Then \begin{align*}
\frac{p_n}{q_n} & = \frac{2 p_{n-1}}{q_n} + \frac{(2n-1)^2 p_{n-2}}{q_n} \\
& = 2\bigg(1-\frac{1}{3} + \cdots (-1)^{n-1}\frac{1}{2n-1}\bigg)\frac{q_{n-1}}{q_n} +(2n-1)^2 \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3}\bigg) \frac{q_{n-2}}{q_n} \\
&= \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3}\bigg) + 2\frac{q_{n-1}}{q_n} (-1)^{n-1}\frac{1}{2n-1} \\
&= \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3} +(-1)^{n-1}\frac{1}{2n-1}\bigg) + (2n-1)\frac{q_{n-2}}{q_n} (-1)^n \quad -\star
\end{align*}
Where I applied IH to both terms $p_{n-1}$ and $p_{n-2}$ in the second line. I now claim that $\frac{q_{n-2}}{q_n} =\frac{1}{(2n-1)(2n+1)}$ for all $n$. As usual, the base case is easily verified. Now suppose the hypothesis for $n-1$ is true. Then \begin{align*}
q_{n-2} &= 2q_{n-3} + (2n-5)^2 q_{n-4} \\
& = \frac{2q_{n-1}}{(2n-3)(2n-1)} + (2n-5)\frac{q_{n-2}}{2n-3} \quad \text{Apply IH to both terms} \\
& =\frac{1}{(2n-3)(2n-1)}\bigg(2 q_{n-1} + (2n-5)(2n-1) q_{n-2} \bigg) \\
&= \frac{1}{(2n-3)(2n-1)} \bigg(q_n - 4(2n-1)q_{n-2}\bigg)
\end{align*}
You can check that this easily simplifies to $\displaystyle \frac{q_{n-2}}{q_n}=\frac{1}{(2n-1)(2n+1)}$, which is what we want! Now substituting this back into the $(\star)$ we have proven by induction that:
$$\frac{p_n}{q_n} = 1 -\frac{1}{3} + \frac{1}{5} - \cdots + (-1)^n \frac{1}{2n+1} $$
So if we let $n\rightarrow \infty$, we see that this term is just equal to the series $$1-\frac{1}{3}+\frac{1}{5} - \frac{1}{7} + \cdots $$ This is precisely equal to the $\arctan 1$! (Search up the Taylor series expansion for arctan for this.) So this value is just $\displaystyle \frac{\pi}{4}$, which is what we desired.
| {
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Question related to the following function $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$ Let $f:\left( {0,1} \right) \to R$ be defined by $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
, where 'b' is a constant such that $0 < b < 1$. Then
(A) $f$ is not invertible on (0,1)
(B) $f\ne f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$
(C) $f= f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$
(D) $f^{-1}$ is differentiable on (0,1)
My approach is as follow
$f:\left( {0,1} \right) \to R$, $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$,$0 < b < 1$
$f'\left( x \right) = \frac{{\left( {1 - bx} \right)\frac{d}{{dx}}\left( {b - x} \right) - \left( {b - x} \right)\frac{d}{{dx}}\left( {1 - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}}$
$f'\left( x \right) = \frac{{\left( {1 - bx} \right) \times - 1 - \left( {b - x} \right) \times \left( { - b} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( {1 - bx} \right) \times - 1 + \left( {b - x} \right) \times b}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( { - 1 + bx} \right) + \left( {{b^2} - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{{b^2} - 1}}{{{{\left( {1 - bx} \right)}^2}}}$
$x\in(0,1)$ and $b\in(0,1)$, then $bx\in(0,1)$, hence $f'(x)<0$, hence my assumption is that $f$ is invertible as it is decreasing function when $0<x<1$ but official answer is (A) that is $f$ is not invertible on (0,1)
| You also may show that $f(f(x))=x$.
| {
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How to factor $x^5-1$ for binary codes Currently, I have that I can do:
$$ x^5-1 = (x-1)(x^4+x^3+x^2+x+1) $$
Then I thought I could do:
$$ x^5-1 = (x-1)(x(x+1)(x^2+1)+1) $$
However I want to use the factorization to find all cyclic codes in $[5, k]$. For that, I need to find all irreducible factors of $ x^5-1 $. I am not seeing if I can actually reduce $x^4+x^3+x^2+x+1$ any further
| Over GF(2)$=\Bbb F_2$ we have $1=-1$ and $2=0$. The factorization (using prime factors in $\Bbb F_2[x]$) of the given polynomial is already
$$
x^5+1 =(x+1)(x^4+x^3+x^2+x+1)\ .
$$
To see that $f=(x^4+x^3+x^2+x+1)$ is irreducible (= prime in $\Bbb F_2[x]$), it is enough to check there is no factor of degree one or two. The irreducible factors in degrees one and two are $x$, $(x+1)$, and $(x^2+x+1)$. The factors in degree one are easily excluded. (Since $f(0)=f(1)=1$.) And the rest by division with rest of $f$ by $(x^2+x+1)$ is $x+1$, so the factor in degree two is also excluded.
A computer algebra system is useful in such cases, for instance sage delivers the factorization:
sage: R.<x> = PolynomialRing(GF(2))
sage: factor(x^5+1)
(x + 1) * (x^4 + x^3 + x^2 + x + 1)
sage: # note that x^2 + 1 is reducible...
sage: factor(x^2+1)
(x + 1)^2
| {
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Compute $\iint _S (y^2+x^2)\, dA$ by change of variables.
Let $S$ be the region in the first quadrant bounded by the curves $xy=1$, $xy=3$, $x^2-y^2=1$, and $x^2-y^2=4$. Compute $$\iint _S (y^2+x^2)\, dA.$$
I have tried to use the change of variables $u=xy$ and $v=x+y$. Then $$\begin{align}x^2+y^2=v^2-2u \\ x^2-y^2=v^2-2u \end{align}$$
Since $v^2=x^2-y^2+2u$, $$0 \le v^2 \le 4-1=3$$, so $$0 \le v \le \sqrt{3}$$
The integral becomes
$$\begin{align} \iint _S (y^2+x^2)\, dA= \int_0^\sqrt{3} \int_1^3 (v^2-2u) \left|J \right| \,du\,dv \end{align}$$
but I can't solve $x,y$ in terms of $u,v$.
I have tried another one, $u=x^2+y^2$ and $v=x^2-y^2$.
I get $x=\sqrt{\frac{u+v}{2}}$ and $y=\sqrt{\frac{u-v}{2}}$
Then $1\le xy=\frac{\sqrt{u^2-v^2} }{2} \le 3$, and $$4\le u^2-v^2 \le 36$$
The integral becomes $$\begin{align} \iint _S (y^2+x^2)\, dA= \int_1^4 \int_{\sqrt{4+v^2}}^{\sqrt{36+v^2}} u \left|J \right| \,du\,dv \end{align}$$
Hope someone could give me some suggestion how to do it.
| EDIT
The substitution $u=xy$, $v=x^2-y^2$ does the job. The Jacobian with respect to $x$ and $y$ is
$$
J=\left(\begin{matrix}
y & x\\
2x & -2y
\end{matrix}\right)
$$
so $|\det(J)|=2(x^2+y^2)$, which means $dudv=2(x^2+y^2)dxdy$ and your integral becomes
$$\begin{align} \iint _S (y^2+x^2) dxdy= \int _1^3 \int_1^4 2 \;dudv \end{align}$$
As Ian said, this is no more than a special case of the "inverse method", used also in order to calculate one-dimensional integrals. E.g., if you have
$$
\int \tan(x)^4 dx
$$
and you apply the substitution $t=\tan(x)$, then it is easy to obtain $dt=(1+\tan(x)^2)dx=(1+t^2)dx$, and then $dx=\frac{dt}{1+t^2}$ (of course, in this case it is equally easy to start from $x=\arctan(t)$ and so on, but sometimes you can avoid some annoying calculation).
| {
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Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$ Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable
Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$
| HINT
\begin{align*}
y'' + y = \frac{1}{y^{3}} & \Longleftrightarrow y''y' + yy' = \frac{y'}{y^{3}}\\\\
& \Longleftrightarrow (y')^{2} + y^{2} = -\frac{1}{y^{2}} + c\\\\
& \Longleftrightarrow (y')^{2} = \frac{cy^{2} - y^{4} - 1}{y^{2}}\\\\
& \Longleftrightarrow y' = \pm\sqrt{\frac{cy^{2} - y^{4} - 1}{y^{2}}}
\end{align*}
| {
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General term of a sequence and its limit. Having a sequence $\;\sqrt[3]5\;,\;\sqrt[3]{5\sqrt[3]5}\;,\;\sqrt[3]{5\sqrt[3]{5\sqrt[3]5}}\;,\;\ldots\;,\;$ what could be the general term and its limit ?
Note that the second term is cube root of $5$ and inside that there is cube root of $5$ again and the third term is cube root of $5$ and two cube root of $5$ inside the outer cube root of $5$.
I tried to define it recursively as
$c_{n+1}=\sqrt[3]{5c_n}\;$ where $\;c_1=\sqrt[3]5\;.$
| Observe the pattern that, if the $n$th term of the sequence is denoted $c_n$, then
$$c_n = 5^{1/3 + (1/3)^2 + (1/3)^3 + \cdots + (1/3)^n}$$
To see this, notice that, for instance:
\begin{align*}
c_2 &= \sqrt[3]{ 5 \sqrt[3] 5} = \left( 5 \cdot 5^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} = 5^{1/3 + 1/3^2} \\
c_3 &= \sqrt[3]{ 5 \sqrt[3]{5\sqrt[3]{5}}} = \left( 5 \cdot \left(5 \cdot 5^{1/3}\right)^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} \cdot 5^{1/3^3} = 5^{1/3 + 1/3^2 + 1/3^3} \\
\end{align*}
and so on and so forth. The exponent of $c_n$ is a geometric series with ratio $1/3$ (just starting at $1/3$ and not $1$), and thus
\begin{align*}
\sum_{k=1}^n \left( \frac 1 3 \right)^k
&= \frac{(1/3)^{n+1} - 1}{1/3 - 1} - 1\\
&= \left( -\frac 3 2 \right)\cdot \frac{1}{3^{n+1}} - \left( -\frac 3 2 \right)\cdot1 - 1\\
&= \frac 1 2 \left( 1 - \frac{1}{3^n} \right)
\end{align*}
and so
$$c_n = 5^{(1 - 3^{-n})/2}$$
Taking the limit $n \to \infty$ is trivial enough, and gives the limit $c_n \to 5^{1/2} = \sqrt 5$.
| {
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Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$ $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$
How can I think this for k=3 ?
Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}$
then
$\frac{3}{14}\sum _{n=1}^{3}\frac{{\left(-4\right)}^{n+2}}{{\left(n-2\right)}^{2}}=\frac{3}{14}.\left(\frac{{\left(-4\right)}^{3}}{{\left(1-2\right)}^{2}}+\frac{{\left(-4\right)}^{5}}{{\left(3-2\right)}^{2}}\right)=-233,1428571$
Is my idea correct?
| *Note: * For $k=1$ to $3$ we have to evaulate the sum for $k\in\{1,2,3\}$:
\begin{align*}
\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}\tag{1}
\end{align*}
Hint: Evaluation at $n=2$ is not feasible, since in that case we have division by zero, which is not admissible. In order to overcome this problem we can exclude the value which causes the indeterminate expressions;.
We consider the valid expression:
\begin{align*}
\sum _{{n=1}\atop{n\ne 2}}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}
\prod _{i=n}^{2k}\frac{i}{i+1}\tag{2}
\end{align*}
which can be simplified, since
\begin{align*}
\prod _{i=n}^{2k}\frac{i}{i+1}&=\frac{n}{n+1}\cdot\frac{n+1}{n+2}\cdots\frac{2k}{2k+1 }\\
&=\frac{n}{2k+2}
\end{align*}
| {
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Does $\int \tan^3x\sec^2x \space dx$ have 2 solutions? So normally, you evaluate $\int\tan^3x\sec^2x \space dx$ by substituting $u = \tan x$ and $du = \sec^2x\space dx$ right? So,
$$\begin{equation}\begin{aligned}
\int\tan^3x\sec^2x \space dx &= \int u^3 \space du \\
&= \frac{u^4}{4} + C\\
&= \frac{tan^4x}{4} + C \\
\end{aligned}\end{equation}$$
But, I tried to solve it this way instead,
$$\begin{equation}\begin{aligned}
\int\tan^3x\sec^2x \space dx &= \int\tan^2x\tan x\sec x\sec x\space dx \\
&= \int(\sec^2x-1)\tan x\sec x\sec x\space dx \\
&= \int\sec^3x\tan x\sec x-\sec x\tan x\sec x\space dx \\
&= \int\sec^3x\tan x\sec x-\sec x\tan x\sec x\space dx \\
&= \int\sec^3x\tan x\sec x\space dx - \int\sec x\tan x\sec x\space dx \\
\end{aligned}\end{equation}$$
and then substituting $u = \sec x$, thus $du = \tan x\sec x\space dx$. So,
$$\begin{equation}\begin{aligned}
\int\sec^3x\tan x\sec x\space dx - \int\sec x\tan x\sec x\space dx &= \int u^3 du - \int u\space du\\
&= \frac{u^4}{4} - \frac{u^2}{2} + C \\
&= \frac{\sec^4x}{4} - \frac{\sec^2x}{2} + C \\
\end{aligned}\end{equation}$$
What?! Can someone explain to me where I made the mistake?
| There's no mistake. Both methods resulted in the same solution, up to a constant of integration. Indeed, notice that:
\begin{align*}
\frac{1}{4}\tan^4 x + C
&= \frac{1}{4}(\tan^2 x)^2 + C \\
&= \frac{1}{4}(\sec^2 x - 1)^2 + C \\
&= \frac{1}{4}(\sec^4 x - 2\sec^2 x + 1) + C \\
&= \frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 x + \frac{1}{4} + C \\
&= \frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 x + D \\
\end{align*}
where $D = \frac{1}{4} + C$ is an arbitrary constant of integration.
| {
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What's wrong with this proof of $3=0$ starting from $x^2+x+1=0$? What am I missing?
$$ x^2 + x + 1 = 0$$
Then
$$ x^2 + x = -1$$
$$ x(x+1) = -1$$
$$ x(-x^2) = -1$$
$$x^3 = 1 $$
$$x = 1 $$
but
$$(1)^2 + (1) + 1 = 3 $$
So
$$ 3 = 0$$
| Nothing wrong: it is true that if $x^2+x+1=0$ and $x\in\Bbb R$, then $x=1$. Keeping all the pieces together $$x^2+x+1=0\Leftrightarrow\begin{cases}x^2+x=-1\\ x+1=-x^2\end{cases}\Leftrightarrow \begin{cases}-x^3=-1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ 2=-1\end{cases}$$
So the equation has no solutions in $\Bbb R$.
Of course, we may argue that there is some degree of circularity in proving that $x^2+x+1$ has no real roots by using the fact that $x^3-1$ has exactly one root in $\Bbb R$...
| {
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Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."
If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?
| Since every term but the last is divisible by $a$, so we only need to deal with $(2x)^n \bmod \left(x^2-x+1\right)$. Let's start from small ones, and try to find patterns:
$$
\begin{aligned}
((2 x)^1 &\bmod \left(x^2-x+1\right))=\color{gray}{2x} \\
((2 x)^2&\bmod \left(x^2-x+1\right)) =\color{red}{-4+4 x} \\
((2 x)^3 &\bmod \left(x^2-x+1\right)) = \color{blue}{-8}\\
((2 x)^4 &\bmod \left(x^2-x+1\right)) = \color{green}{-16x}\\
((2 x)^5 &\bmod \left(x^2-x+1\right)) = 32-32x\\
((2 x)^6 &\bmod \left(x^2-x+1\right)) = 64\\
((2 x)^7 &\bmod \left(x^2-x+1\right)) =\color{gray}{128x}\\
((2 x)^8 &\bmod \left(x^2-x+1\right)) = \color{red}{-256+256x}\\
((2 x)^9 &\bmod \left(x^2-x+1\right)) = \color{blue}{-512} \\
((2 x)^{10} &\bmod \left(x^2-x+1\right)) =\color{green}{-1024 x}
\end{aligned}
$$
What do you find? Hint: Divide into groups by $n \bmod 6$. Prove by induction separately.
| {
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I can't understand the result of this limit $$ \lim_{n\to\infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
$$
I think the answer is $-\infty$ because I take the highest degree coefficients $$n^4 (numerator)$$ and $$-n ^ 3(denominator)$$
and I was thinking $+/- = -$, but the actual answer is $+\infty$, this is probably a dumb question, but I didn't understand why is $+\infty$ instead of $-\infty$.
| \begin{align} \lim_{n\to\infty} \left(\frac{n^4-n^3+1}{-n^3+n^2+\sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \frac{n^3-n^2+\frac1n}{-n^3 + n^2 + \sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \left(-\ \frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}} \ .\ \frac{ \frac{1}{n^3} }{ \frac{1}{n^3} }\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{1-\frac{1}{n}+\frac{1}{n^4}}{1-\frac{1}{n} -\frac{\sqrt{n}}{n^3}} \ \right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n\right) \ .\ 1 \\
\\
=-\infty.
\end{align}
| {
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Mistake when substituting constraint $4x^2+y^2=1$ into a function $f(x,y)=x^2+y^2$ in extrema problem Consider the function $f(x,y)=x^2+y^2$ under the constraint $4x^2+y^2=1$. The extrema of $f$ under that constraint can be easily found with Lagrange multipliers, and they are attained for $(0,1)$, $(0,-1)$ for maximum and $(1/2,0)$, $(-1/2,0)$ for minimum; however, if we isolate $y^2=1-4x^2$ and we substitute it in the function, we get a wrong result (in particular, the one variable function $g(x)=1-3x^2$ obtained has only a maximum for $x=0$ and Weierstrass theorem assures that the are both maximum and minimum for $f$). Can someone explain me why this fails?
| Solving
$$
\min_{x,y}(\max_{x,y})(x^2+y^2)\ \ \text{s. t.}\ \ \ 4x^2+y^2=1
$$
using the Lagrange multipliers method, reduces to determine the stationary points for
$$
\nabla(x^2+y^2)+\lambda\nabla(4x^2+y^2-1)=0
$$
by solving for $x,y,\lambda$
$$
\cases{
2x+8x\lambda=0\\
2y+2y\lambda=0\\
4x^2+y^2-1=0
}
$$
giving as solutions the four tangency points between the level curves for $z=x^2+y^2$ and the ellipse $4x^2+y^2-1=0$. Those points can be depicted in the attached plot.
Now by making the substitution $y = 1-4x^2$ into $z = x^2+y^2$ giving $z=1-3x^2$ we are searching extrema along the $x$ axis at $x^*=0$ with value $1$ and the corresponding $y^*$ we obtain from the restriction $4(x^*)+(y^*)^2=1$. Also with the substitution $x^2=\frac{1-y^2}{4}$ we are searching extrema along the $y$ axis with value $\frac 14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| Rewriting the inequality $$2\sin^2(\frac{x}{2})\ge\frac{x^2}3 \iff (\sqrt2 \sin(\frac{x}{2}) - \frac{x}{\sqrt3})(\sqrt2 \sin(\frac{x}{2}) + \frac{x}{\sqrt3}) \ge 0 \tag{1}$$
Assume $0\le x \le 1$. Obviously we have $$\sqrt2 \sin(\frac{x}{2}) + \frac{x}{\sqrt3}\ge 0$$ So we should analyze $$f(x,a) = \sin(\frac{x}{2}) - \frac{x}{a}$$It's interesting to see this visually:
When $a=2$, the line $y_1 = \frac{x}{2}$ touches $y_2 = \sin(\frac{x}{2})$ only at $x = 0$(remember that $|\sin(\frac{x}{2})|\le |\frac{x}{2}|$). Increasing $a$ further, leads to another intersection point(we can actually see the concavity of $\sin(\frac{x}{2})$ here).
Note that $$f(x,a) = \int_{0}^{x}(\frac{1}{2}\cos (\frac{t}{2}) - \frac{1}{a})dt$$Let $a = \sqrt6$. We can show that the integrand is nonnegative when $0\le t \le 1$. So when $0\le x \le 1$ we have $f(x,\sqrt6)\ge 0 $ which implies $$\sqrt2 \sin(\frac{x}{2}) - \frac{x}{\sqrt3}\ge 0$$In order to show nonnegativity, consider $$0\le t \le 1 \implies 0\le \frac{t}{2} \le \frac{1}{2} \lt \frac{\pi}{2} \implies 1 \ge \cos(\frac{t}{2}) \ge \cos(\frac{1}{2}) \implies \frac{1}{2} \ge \frac{1}{2}\cos(\frac{t}{2}) \ge \frac{1}{2}\cos(\frac{1}{2}) \approx 0.439 \gt \frac{1}{\sqrt6} \approx 0.408$$ Now that we have shown $(1)$ holds for $0 \le x \le 1$, let $x = -t$ and conclude it holds for $-1 \le t \le 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
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