Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
| Your proof is correct, but could be made a little briefer, with a gain in clarity.
Here's a proof using the same ideas as yours, but I think a little easier to read . . .
\begin{align*}
&a+b=1\\[4pt]
\implies\;&(a+b)^2 = 1\\[4pt]
\implies\;&(a+b)^2 + (a-b)^2 \ge 1\\[4pt]
\implies\;&2a^2 + 2b^2 \ge 1\\[4pt]
\implies\;&a^2 + b^2 \ge \frac{1}{2}\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Convergence of the series $\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )$ Given $a, b \in \mathbb{R}$ with $b > 0$, is the series
\begin{equation}
\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )
\end{equation}
convergent or divergent?
If we drop out the $n=0$ term and fold the remaining sum, the question can be equivalently asked for the series
\begin{equation}
\sum_{n = 1}^{\infty} (\sqrt{(a+n)^2+b^2} + \sqrt{(a-n)^2+b^2} - 2n ).
\end{equation}
| $$
\begin{align}
\lim_{n \to \infty }\sqrt{a^2+2an+n^2+b^2}-n=\lim_{n \to \infty }\frac{\left ( a^2+2an+b^2 \right )}{\sqrt{a^2+2an+n^2+b^2}+n}=a
\end{align}
$$
evidently divergent for
$$
\begin{align}
a\neq 0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the Paramertric Coordinates of the Maximum and Minimum points of a Certain Curve I was studying for some quizzes when a wild question appears. It goes like this:
Find the paramertric coordinates of the maximum and minimum points of
the following curves. The curve is $x = 3 \cos \theta$ ;$\space$ $ y = 4 \sin \theta$
My work:
I need to get the rectangular equation described by the parametric equation.
I remember the relation $(\sin \theta)^2 + (\cos \theta)^2 = 1$
Getting the $\cos \theta:$ $\cos \theta = \frac{x}{3}$
Getting the $\sin \theta:$ $\sin \theta = \frac{y}{4}$
Substituting the newly-found variables to $(\sin \theta)^2 + (\cos \theta)^2 = 1$,
it becomes $$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1$$
By using the implicit differentiation, we got:
$$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1$$
$$\left( \frac{1}{9} \right)\left( 2x \right) + \left( \frac{1}{16} \right)\left( 2 y y' \right) = 0 $$
$$\left( \frac{2}{9} \right)\left( 2x \right) + \left( \frac{1}{8} \right)\left(y y' \right) = 0$$
$$\left( \frac{1}{8} \right)\left(y y' \right) = \left( \frac{-2}{9} \right)\left( 2x \right) $$
$$y' = \frac{-16x}{9y}$$
We got to make the derivative expressed as a single variable, so we need to get the $y$ in terms of $x$ in the equation
$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1.$
getting the $y$ from $\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1,$ it is:
$$y = \frac{4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
Then ,substituting it to the equation $y' = \frac{-16x}{9y},$ it becomes:
$$y' = \frac{-16x}{9\left( \frac{4}{3}\left((9 - x^2)^{\frac{1}{2}}\right) \right)}$$
$$ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
To get the minimum and maximum points of the rectangular equation, we set $\frac{dy}{dx} = y'$ to zero.
$$ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
$$ 0 = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
$$ 0 = (9 - x^2)^{\frac{1}{2}}$$
$$ 0 = (9 - x^2)$$
$$x^2 = 9$$
We get $x = \pm 3$
To determine whether $x = -3$ is a minimum or maximum point, we use the first derivative test.
Letting $x = -3$ as the critical/base point, we let $x = -99$ and $x = +99.$ If $x = -99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary.
If $x = +99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary too.
To determine whether $x = +3$ is a minimum or maximum point, we use the first derivative test.
Letting $x = -3$ as the critical/base point, we let $x = -99$ and $x = +99.$ If $x = -99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary.
If $x = +99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary too.
At this point, I'm stuck, because I can't determine if $x = \pm 3$ is a minimum or maximum point. I used large numbers
just to make sure I see the change of signs as I pass through the derivative (+, 0, -) or (-, 0, +)
How do you get the minimum and maximum points of the curve shown above?
| The max/min values occur at the values of $\theta$ for which
$$ \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=0 $$
which occurs when
$$ \frac{dy}{d\theta}=4\cos\theta=0 $$
So the solution is
$$\theta=\pm\frac{\pi}{2}$$
(or, more generally)
$$\theta=2\pi n\pm\frac{\pi}{2}$$
But $\theta=\pm\dfrac{\pi}{2}$ will be sufficient to find the max/min.
\begin{eqnarray}
\left(3\cos\frac{\pi}{2},4\sin\frac{\pi}{2}\right)&=&(0,4)\text{ max}\\
\left(3\cos\frac{-\pi}{2},4\sin\frac{-\pi}{2}\right)&=&(0,-4)\text{ min}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$. If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.
This equation becomes $x^4+3x^3-26x^2+3x+1=0$ which has four solutions. One of the solutions is $2+\sqrt3$ which has the form $a+\sqrt b$. So, $a+b=2+3=5$. How does this look?
| We can factor this equation as follows:
$$x^4+3x^3-26x^2+3x+1 = (x^2-4x+1) (x^2+7x+1).$$
One can check that $2\pm \sqrt3$ are solutions to the $(x^2-4x+1)$,
alse we can check that $\dfrac{-7\pm 3\sqrt5}{2}$ are solutions to the $(x^2+7x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What are some mathematically interesting computations involving matrices? I am helping designing a course module that teaches basic python programming to applied math undergraduates. As a result, I'm looking for examples of mathematically interesting computations involving matrices.
Preferably these examples would be easy to implement in a computer program.
For instance, suppose
$$\begin{eqnarray}
F_0&=&0\\
F_1&=&1\\
F_{n+1}&=&F_n+F_{n-1},
\end{eqnarray}$$
so that $F_n$ is the $n^{th}$ term in the Fibonacci sequence. If we set
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}$$
we see that
$$A^1=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix},$$
and it can be shown that
$$
A^n =
\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix}.$$
This example is "interesting" in that it provides a novel way to compute the Fibonacci sequence. It is also relatively easy to implement a simple program to verify the above.
Other examples like this will be much appreciated.
| If $(a,b,c)$ is a Pythagorean triple (i.e. positive integers such that $a^2+b^2=c^2$), then
$$\underset{:=A}{\underbrace{\begin{pmatrix}
1 & -2 & 2\\
2 & -1 & 2\\
2 & -2 & 3
\end{pmatrix}}}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}$$
is also a Pythagorean triple. In addition, if the initial triple is primitive (i.e. $a$, $b$ and $c$ share no common divisor), then so is the result of the multiplication.
The same is true if we replace $A$ by one of the following matrices:
$$B:=\begin{pmatrix}
1 & 2 & 2\\
2 & 1 & 2\\
2 & 2 & 3
\end{pmatrix}
\quad \text{or}\quad
C:=\begin{pmatrix}
-1 & 2 & 2\\
-2 & 1 & 2\\
-2 & 2 & 3
\end{pmatrix}.
$$
Taking $x=(3,4,5)$ as initial triple, we can use the matrices $A$, $B$ and $C$ to construct a tree with all primitive Pythagorean triples (without repetition) as follows:
$$x\left\{\begin{matrix}
Ax\left\{\begin{matrix}
AAx\cdots\\
BAx\cdots\\
CAx\cdots
\end{matrix}\right.\\ \\
Bx\left\{\begin{matrix}
ABx\cdots\\
BBx\cdots\\
CBx\cdots
\end{matrix}\right.\\ \\
Cx\left\{\begin{matrix}
ACx\cdots\\
BCx\cdots\\
CCx\cdots
\end{matrix}\right.
\end{matrix}\right.$$
Source: Wikipedia's page Tree of primitive Pythagorean triples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 26,
"answer_id": 4
} |
Line integral $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ Solve $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ =?
K: $x^2+y^2=1$ is oriented positively.
My attempt:
$\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy = |x=cost, y=sint,x'=- sint,y'=cost | $ = *
I put for $dx = -sint,dy=cost$
$= \int_0^{2\pi} \frac{sint}{cost^2+sint^2} -sint -\frac{cost}{cost^2+sint^2} cost dt= -\int_0^{2\pi} 1 dt = - 2 \pi $
We can't use Green's theorem because $P(x,y)$ and $Q(x,y)$ aren't continuous and their diveratives aren't continuous at point. $(0,0)$?
| Another approach is
$$\frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy=-\dfrac{x\,dy-y\,dx}{x^2+y^2}=-d\left(\arctan\dfrac{y}{x}\right)$$
so
$$\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy=-\int_K d\left(\arctan\dfrac{y}{x}\right)=-\int_0^{2\pi}dt=-2\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
use **similarity transformation** to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ . Consider the linear operator $ \ T(x,y)=(5x,x-y) \ \ on \ \ \mathbb{R}^2 $ . Find the matrix $ \ A \ $ of $ \ T \ $ relative to the standard basis on $ \ \mathbb{R}^2 \ $ . Then use similarity transformation to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ .
Note: The answer is given as $ B =\begin{pmatrix}-29 & -56 \\ 17 & 33 \end{pmatrix} $
Answer:
I have found $ \ A= \begin{pmatrix} 5 & 0 \\ 1 & -1 \end{pmatrix} $ .
Now $ \ B=P^{-1} AP \ $ ,
where $ \ P \ $ is the transition matrix from the basis $ \ \{(1,2) , (3,5) \} \ $ to the basis $ \ \{(1,0) , (0,1) \} \ $.
Now calculating I got ,
$ \ P=\begin{pmatrix}1 & 3 \\ 2 & 5 \end{pmatrix} $
Thus,
$ \ P^{-1} =\begin{pmatrix}-5 & 2 \\ 3 & -1 \end{pmatrix} $
Hence ,
$ B=P^{-1} AP = \begin{pmatrix}-5 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix}-27 & -79 \\ 16 & 47 \end{pmatrix} $
$
But the answer does not match with the given answer .
Was my calculation and appoach right ?
If not then how to solve the problem ?
Is there any help ?
| You made a "silly" minor mistake $$P=\left(
\begin{array}{ll}
1 & 2 \\
3 & 5 \\
\end{array}
\right)$$
If you use that then $$ P^{-1}AP=\left(
\begin{array}{cc}
-5 & 2 \\
3 & -1 \\
\end{array}
\right).\left(
\begin{array}{cc}
5 & 0 \\
1 & -1 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 2 \\
3 & 5 \\
\end{array}
\right)=\left(
\begin{array}{cc}
-29 & -56 \\
17 & 33 \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int\frac{x^b}{1+x^a}~dx$ for $a,b\in\Bbb N$ In this previous answer, MV showed that for $n\in\Bbb N$,
$$\int\frac1{1+x^n}~dx=C-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)$$
where
$$x_{kr}=\cos \left(\frac{(2k-1)\pi}{n}\right)$$
$$x_{ki}=\sin \left(\frac{(2k-1)\pi}{n}\right)$$
I am now interested in the case of $n=\frac ab\in\Bbb Q^+$. By substituting $x\mapsto x^b$, we get
$$\int\frac{bx^{b-1}}{1+x^a}~dx$$
Thus, the given integral in question is really
$$\int\frac{x^b}{1+x^a}~dx$$
By expanding with the geometric series and termwise integration, one can see that
$$\int_0^p\frac{x^b}{1+x^a}~dx=\sum_{k=0}^\infty\frac{(-1)^kp^{ak+b+1}}{ak+b+1}=\frac{p^{b+1}}a\Phi\left(-p^a,1,\frac{b+1}a\right)$$
where $\Phi$ is the Lerch transcendent.
A few particular cases that arise may be found:
\begin{align}\int\frac1{1+x^{1/n}}~dx&=C+(-1)^{n+1}n\left[\ln(1+x^{1/n})+\sum_{k=1}^{n-1}\frac{(-x^{1/n})^k}k\right],&a=1\\\int\frac1{1+x^{2/n}}~dx&=C+(-1)^nn\left[\arctan(x^{1/n})+\frac1{x^{1/n}}\sum_{k=1}^{(n-1)/2}\frac{(-x^{2/n})^k}{2k-1}\right],&a=2,n\ne2b\end{align}
Or, more generally, with $x=t^{an+1}$,
$$\int\frac1{1+x^{a/(an+1)}}~dx=(-1)^{n+a}(an+1)\left[\int\frac1{1+t^a}~dt+\frac1{x^{(a-1)/(an+1)}}\sum_{k=1}^{(n-1)/a}\frac{(-x^{a/(an+1)})^k}{a(k-1)+1}\right]$$
which reduces down to the previously solved problem.
But what of the cases when $n=a/b$ with $(b\bmod a)\ne0,1$?
For example,
$$\int\frac1{1+x^{3/2}}~dx=C+\frac16\left[\log(1-x^{1/2}+x)-2\log(1+x^{1/2})+2\sqrt3\arctan\left(\frac{2x^{1/2}-1}{\sqrt3}\right)\right]$$
| Let $\gamma=\exp(2\pi i/a)$.
The polynomial $Q(x)=x^a+1$ has $a$ simple roots $z_k=\gamma^{k+1/2}$,
$0\leq k<a$. Since $z_k^a=-1$, we have $Q'(z_k)=a z_k^{a-1}=-a z_k^{-1}$, so
$$ \frac{1}{Q(x)} =
\sum_{k=0}^{a-1} \frac{1}{Q'(z_k)} \frac{1}{x-z_k} =
-\frac{1}{a} \sum_{k=0}^{a-1} \frac{z_k}{x-z_k} $$
For the numerator we first make a reduction in the degree (when $b\geq a$).
Let $p= b \mod a \in \{0,1,...,a-1\}$ and $m=(b-p)/a$. Then
$$ \frac{x^b - (-1)^m x^p}{x^a + 1} =
\sum_{j=1}^m (-1)^j x^{b-ja}
$$
We deduce that
$$ \frac{x^b}{x^a+1} -
\sum_{j=1}^m (-1)^j x^{b-ja} = \frac{(-1)^mx^p}{x^a+1} =
- \frac{(-1)^m}{a} \sum_{k=0}^{a-1} z_k \frac{x^p}{x-z_k}=
- \frac{(-1)^m}{a} \sum_{k=0}^{a-1} \frac{z_k^{p+1}}{x-z_k} $$
The last equality follows from the fact that the difference is a polynomial
which must vanish since $p<a$. So a part from the trivial part on the LHS
(which I leave aside),
the problem is reduced to integrating
the RHS. We have
$$
- \int \frac{(-1)^m}{a} \sum_{k=0}^{a-1} \frac{z_k^{p+1}}{x-z_k}dx
=
- \frac{(-1)^m}{a} \sum_{k=0}^{a-1} z_k^{p+1}
\ln (x-z_k) $$
To avoid complex log and too long formulas,
let us write
$$ u_{k,p} =
\cos \left( \frac{2\pi (k+1/2)(p+1)}{a}\right) , \; \;
v_{k,p} =
\sin \left( \frac{2\pi (k+1/2)(p+1)}{a}\right) , \; \;
$$
Using
$\overline{z_{a-1-k}} = z_{k} = u_{k,0}+i v_{k,0}$ we obtain for $a$ even:
$$
- \frac{(-1)^m}{2a} \sum_{k=0}^{\lfloor a/2 \rfloor}
\left[ u_{k,p} \ln \left(x^2- 2 u_{k,0}
x+1\right) +
v_{k,p} \arctan
\frac{x-u_{k,0}}{v_{k,0}} \right]
$$
For $a$ odd you should add to this expression
the "middle term" (which has no arctan part)
$$
\frac{(-1)^{m+p}}{a}
\ln(x+1) $$
No guarantee for the above being free of errors ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms The quartic form
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $Q_1$, $Q_2$, $Q_3$ so that
$$(x^2 + y^2 + z^2)^2 - 3( x^3 y + y^3 z + z^3 x) = Q_1^2 + Q_2^2 + Q_3^2$$
I would like to find an explicit writing of the quartic forms, with rational quadratic forms $Q_i$. Maybe more than $3$ terms are necessary.
| $$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{6}\left ( \sum\limits_{cyc}\,(\,-\,3\,yz+ 3\,xy+ y^{\,2}+ z^{\,2}- 2\,x^{\,2}\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{2}\left ( \sum\limits_{cyc}\,(\,-\,yz- zx+ 2\,xy+ z^{\,2}- x^{\,2}\,)^{\,2} \right )$$
After that, using $ab+ bc+ ca= 0\!\therefore\!a\equiv (a- b)(b- c),b\equiv (b- c)(c- a),c\equiv (c- a)(a- b)$, so
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,x^{\,2}+ 2\,y^{\,2}- 3\,z^{\,2}- 4\,xy- yz+ 5\,zx\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,2\,x^{\,2}+ y^{\,2}- 3\,z^{\,2}- 5\,xy+ yz+ 4\,zx\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,3\,x^{\,2}- y^{\,2}- 2\,z^{\,2}- 5\,xy+ 4\,yz+ zx\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,3\,y^{\,2}- 2\,z^{\,2}- x^{\,2}- xy- 4\,yz+ 5\,zx\,)^{\,2} \right )$$
because we always have the following equality with $\sum\limits_{cyc}\left ( (\,a- b\,)(\,b- c\,) \right )\left ( (\,b- c\,)(\,c- a\,) \right )= 0$
$$\left ( \sum\limits_{cyc}\left ( (\,a- b\,)(\,b- c\,) \right )^{\,2} \right )(\,x^{\,2}+ y^{\,2}+ z^{\,2}\,)= \sum\limits_{cyc}\,\left ( \sum\limits_{cyc}\,\left ( x(\,a- b\,)(\,b- c\,) \right ) \right )^{\,2}$$
Furthermore, we have $\sum\limits_{cyc}\,(\,-\,3\,yz+ 3\,xy+ y^{\,2}+ z^{\,2}- 2\,x^{\,2}\,)= 0= \left ( \sum\limits_{cyc}(\,a- b\,) \right )$. To apply
$$\sum\limits_{cyc}\,\frac{(\!a- b\!)^{2}}{2}\!=\!\frac{(\!a+ b- 2c\!)^{2}\!+\!3(\!a- b\!)^{2}}{4}\!=\!\frac{(\!b+ c- 2a\!)^{2}\!+\!3(\!b- c\!)^{2}}{4}\!=\!\frac{(\!c+ a- 2b\!)^{2}\!+\!3(\!c- a\!)^{2}}{4}$$
$$(\!\sum\limits_{cyc}\,x^{\,2}\!)^{\,2}- 3(\!\sum\limits_{cyc}\,x^{\,3}y\!)= \frac{(\!2\,x^{\,2}- y^{\,2}- z^{\,2}- 3\,xy+ 3\,yz\!)^{\,2}+ 3(\!y^{\,2}- z^{\,2}- xy- yz+ 2\,zx\!)^{\,2}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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} |
Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}(3 \sin x+4 \cos x)+\frac{-1}{25}(3 \cos x-4 \sin x)$$
So
$$\cos 2x=\frac{7}{625}(3 \sin x+4 \cos x)^2+\frac{48}{625}\left((3 \sin x+4 \cos x)(3\cos x-4 \sin x)\right)-\frac{7}{625}(3 \cos x-4 \sin x)^2$$
So
$$I=\frac{7}{625}\int (3 \sin x+4 \cos x)dx+\frac{48}{625}\int (3 \cos x-4 \sin x)dx-J$$
where
$$J=\frac{7}{625}\int \frac{(3 \cos x-4 \sin x)^2dx}{3 \sin x+4 \cos x}$$
I got stuck up to integrate $J$.
| Note that $$3\sin x+4\cos x=5\Big( \dfrac{3}{5}\sin x+\dfrac{4}{5}\cos x \Big)$$ and $$\cos2x=1-2\sin^2x$$
also $\arcsin(4/5)=\arccos(3/5)=0.92$
combining all this, your integral becomes $\frac{1}{5}\int \frac{1 -2\sin^2x\: dx}{\sin(x+0.92)}$.
From here, writing $x=(x+0.92)-0.92$ in numerator and basic integration formula for $\csc x$, $\cot x$ will work.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Ordered pairs of positive integers $2^{2x} - y^2 = 60$ How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$?
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is always bigger than the latter. This means I only have to account for the positive factors of $60$, and so I had the following pairs $(60,1), (30,2), (20,3), (15,4), (12,5)$, and I plugged them in and solved the systems. Of the 5 possible cases, only $(30,2)$ worked and yielded $x=4, y=14$.
Am I right in having done so? They don't have an answer, and when I tried to graph this, it didn't exactly work out well (I didn't get an integer value for $x$).
| No you are not right.
You have neglected the case $\{ 2^x-y, 2^x+y \} = \{ 6, 10 \}$
; which gives you $(x,y)=(3,2)$.
Notice that:
*
*$(2^x+y)$ and $(2^x-y)$ have the same pairity;
*product of $(2^x+y)$ and $(2^x-y)$ is even; so at least one of them is even;
*by the above two remarks it follows that
both of $(2^x+y)$ and $(2^x-y)$ are even.
*
*So we can rewrite the equation as
$\dfrac{2^x+y}{2} \cdot \dfrac{2^x-y}{2} = 15$ ;
*the only possibilitis for $\{ \dfrac{2^x-y}{2}, \dfrac{2^x+y}{2} \}$
is $\{ 1, 15 \} $ and $\{ 3, 15 \} $ .
*So we can conlude that:
the only possibilitis for $\{ \ 2^x-y, \ 2^x+y \ \} $
is $\{ 2, 30 \} $ and $\{ 6, 10 \} $ .
The first possibility gives $(x,y)=(4,14)$
and the second gives you $(x,y)=(3,2)$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of this series? Show that the series $\displaystyle \sum_{n=0}^{\infty}\frac{x}{(1+x^2)^n} $ converges $\forall x\in \mathbb{R}$ and find the sum $\forall x\in \mathbb{R}$
So, I've shown that the series converges $\forall x\in \mathbb{R}$ and, through testing, suspect that the sum is equal to $\frac{x}{1+x^2}$. Heres what I did:
Let $s =\displaystyle \sum_{n=0}^{k}\frac{x}{(1+x^2)^n}$ = $\frac{x}{(1+x^2)}+ \frac{x}{(1+x^2)^2} + ... +\frac{x}{(1+x^2)^k} \Rightarrow ... \Rightarrow\frac{(x^2+1)^k}{x}s = (1+x^2)^{k-1}+(1+x^2)^{k-2} + ...+1 $.
Not quite sure what to do from here. Any tips ?
| Alternatively: If $x=0$, then $S=0$. If $x\ne 0$, then:
$$S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n}}\Rightarrow$$
$$(1+x^2)S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x(1+x^2)+\sum_{n=1}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x^2(1+x^2)+\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^n}=x(1+x^2)+S \Rightarrow$$
$$S=\frac{x(1+x^2)}{x^2}=\frac1x +x.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Are there any known methods to compute series $\sum_{k=0}^{\infty}2^k \big(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n2^{k+1}+(2k+1))^3}\big)$? I would like to ask if there are any known methods to compute series like this one ?
$$\sum_{k=0}^{\infty}2^k \bigg(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n2^{k+1}+(2k+1))^3}\bigg)$$
And their names so i can look for them if they exist.
I never studied double sums before that's why i am asking, thanks in advance.
| We have
$\displaystyle \sum\limits_{k=0}^\infty 2^k \sum\limits_{n=0}^\infty \frac{(-1)^n}{(n2^{k+1}+2k+1)^3} =\sum\limits_{k=0}^\infty \frac{2^k}{(2k+1)^3} - \sum\limits_{k=0}^\infty \left(\frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}\right)$
and
$\displaystyle\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+a)^3}=\frac{1}{8}\left(\zeta(3,\frac{1}{2}+\frac{a}{2})-\zeta(3,1+\frac{a}{2})\right)\,$ . $\hspace{1cm}$ (see: Hurwitz zeta function)
It follows
$\displaystyle\sum\limits_{k=0}^\infty \frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}=\sum\limits_{k=0}^\infty \frac{1}{2^{2k+6}}\left(\zeta(3,\frac{1}{2}+\frac{2k+1}{2^{k+2}})-\zeta(3,1+\frac{2k+1}{2^{k+2}})\right)$
$\hspace{5.5cm}\approx 0.05957499...$
$\displaystyle\sum\limits_{k=0}^\infty \frac{2^k}{(2k+1)^3}$ is divergent because $\displaystyle\left(\frac{2^k}{(2k+1)^3}\right)_k$ is not a null sequence and
$\displaystyle\sum\limits_{k=0}^\infty \left(\frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}\right)$ is convergent therefore the initial series is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2y^2=2\Longleftrightarrow a^2\cos^2{t}+2b^2\sin^2{t}=2.$$
Trig identities gives
$$a^2(1-\sin^2{t})+2b^2\sin^2{t}=a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
Solving for $\sin{t}$ gives
$$\sin{t}=\pm\sqrt{\frac{a^2-2}{a^2+2b^2}}=\pm f(a,b).$$
Since $$-1\leq\sin{t}\leq1\Longleftrightarrow -1\leq\pm f(a,b)\leq1.$$
Answer a) gives $f(2,1)=\frac{\sqrt{3}}{3} \in[-1,1]$. OK!
Answer b) gives $f(1,2)=\frac{i}{3}\in \mathbb{C}$. Disregard.
Answer c) gives $f(1,\sqrt{2}) = \frac{i\sqrt{5}}{5}\in \mathbb{C}$. Disregard.
So according to me, the correct answer should be a). But correct answer is d). Where did I go wrong?
| In order for $x^2+2y^2=2$ with $x=a\cos(t)$, and $y=b\sin(t)$ we have
$$(a\cos(t))^2+2(b\sin(t))^2=2\iff a^2\cos^2(t)+2b^2\sin^2(t)=2$$
Now we want $a^2=2b^2=2$ since then we can factor out the coefficients and use $\cos^2(t)+\sin^2(t)=1$. Hence, $a=\sqrt 2$ and $b=1$.
Also, just the first error I see in your solution is that we should have: $$a^2-a^2\sin^2(t)+2b^2\sin^2(t)=a^2-\sin^2(t)\left(a^2\color{red}{-}2b^2\right)$$
and not:
$$a^2-a^2\sin^2(t)+2b^2\sin^2(t)=a^2-\sin^2(t)\left(a^2\color{red}{+}2b^2\right)$$
| {
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"url": "https://math.stackexchange.com/questions/2421949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove by induction that $ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$ Question:
Prove by induction that $$ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$$
My attempt:
Base case is trivial.
Suppose that $ n \ge 2$ and $\ 2^{n+1} > n^{2} + 3$
WTS $ \ 2^{n+2} > (n+1)^{2} + 3$
$ 2^{n+2} = 2.2^{n+1}> 2(n^2 + 3) = 2n^2 + 6 = n^2 + n^2 + 2n -2n +6 = (n^2+2n+1) + 3 + (n^2 -2n+2) = (n+1)^2 + 3 + (n^2 -2n+2)$.
I am not sure what to do from here. How can I show $ (n^2 -2n+2) > 0?$
| $$n^2-2n+2=(n^2-2n+1)+1=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422155",
"timestamp": "2023-03-29T00:00:00",
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The License Plate Problem The 7-digit license plate of a car happens to be a palindrome. Its first digit on the far left is twice the digit next to it. The 3rd digit from the right is odd and 5 more than 2nd digit from the left. The middle digit is the average of the first three digits on the far left. What is this license plate number?
HINT: A palindrome is a number that reads the same forward and backward. For example, 123321 and 6776 are palindromes.
| The license plate is of the form $abcdcba$, where each $a,b,c,d$ is a digit.
From the information in the question, we know
$$
a=2b \quad\text{and}\quad c=5+b \quad\text{and}\quad c\ \text{is odd}
\quad\text{and}\quad d=(a+b+c)/3.
$$
Since $c$ is odd and $c=5+b$, we know $b$ is even. Thus $b$ is one of $0,2,4,6,8$. It can't be $6$ or $8$, because then $a$ would then not be a digit. This leaves that $b$ is $0$, $2$, or $4$.
If $b=0$, then $a=0$ and $c=5$ so that $d=(0+0+5)/3$ is not a digit.
If $b=2$, then $a=4$ and $c=7$ so that $d=(4+2+7)/3$ is not a digit.
If $b=4$, then $a=8$ and $c=9$ so that $d=(8+4+9)/3=7$.
Therefore the number is $8497948$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422952",
"timestamp": "2023-03-29T00:00:00",
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Definite integral $\int_0^1\sqrt{x^2+1}\, dx$. Use trig substitution? I have an integral:
$$\int_0^1\sqrt{x^2+1}\, dx$$
but I have gotten stuck. Here's the work I have done already:
I'm not sure where to go from here. Using a trig identity doesn't seem like it would work. Integration by parts doesn't work (at least I think so)
Any ideas on what I should do next?
Thank you!
| Hint:)
Since $\cosh^2u=\sinh^2u+1$ so let $x=\sinh u$.
It's better to do this
$$\int\sec^3t dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$
Edit:
\begin{align}
I
&= \int\sec^3t dt \\
&= \int\dfrac{1}{\cos^3t} dt \\
&= \int\dfrac{\cos t}{\cos^4t} dt \\
&= \int\dfrac{\cos t}{(1-\sin^2t)^2}dt
\end{align}
with substitution $\sin t=u$:
$$I=\int\dfrac{du}{(1-u^2)^2}$$
with fraction decomposition
\begin{align}
\dfrac{1}{(1-u^2)^2}
&= \dfrac{A}{1-u}+\dfrac{B}{1+u}+\dfrac{C}{(1-u)^2}+\dfrac{D}{(1+u)^2} \\
&= \dfrac{(A+B+C+D)+(A-B+2C-2D)u+(-A-B+C+D)u^2+(-A+B)u^3}{(1-u^2)^2}
\end{align}
equivalency of coefficients in numerator shows
\begin{cases}
A+B+C+D=1,\\
A-B+2C-2D=0,\\
-A-B+C+D=0,\\
-A+B=0.
\end{cases}
so $A=B=C=D=\dfrac14$ and thus with $u=\sin t$
\begin{align}
I
&= \dfrac14\int\left(\dfrac{1}{1-u}+\dfrac{1}{1+u}+\dfrac{1}{(1-u)^2}+\dfrac{1}{(1+u)^2}\right)du \\
&= \dfrac14\left(-\ln(1-u)+\ln(1+u)+\dfrac{1}{1-u}-\dfrac{1}{1+u}\right)+C \\
&= \dfrac14\left(\ln\dfrac{1+u}{1-u}+\dfrac{2u}{1-u^2}\right)+C \\
&= \dfrac14\left(\ln\dfrac{1+\sin t}{1-\sin t}+\dfrac{2\sin t}{1-\sin^2t}\right)+C \\
\end{align}
substitution $\sin^2t=\dfrac{x^2}{1+x^2}$ gives us
$$\color{blue}{\int\sqrt{x^2+1}\,dx=\dfrac12\left(\ln(x+\sqrt{x^2+1})+x\sqrt{x^2+1}\right)+C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Generator of an ideal in the polynomial ring $ F[x] $ for a field $ F $
Let $ F $ be a field and consider the polynomial ring $ F[x] $.
I know that $ F[x] $ is a principal ideal domain and so the ideal $\langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle $ is generated by a single element in $ F[x] $.
My guess is that
$$ \langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle = \langle 1 - x \rangle. $$
I have proven that $ \langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle \subset \langle 1 - x \rangle $, but I am struggling to prove the other direction.
My approach is to write an element $ f(x) = \sum_{i = 0} a_{i} x^{i} $ in $ \langle 1 - x \rangle $ in the form $ f_{1}(x^{n - 2} - x^{n - 1}) + f_{2}(x^{n - 3} - x^{n - 1}) + \dots + f_{n - 2}(x - x^{n - 1}) + f_{n - 1}(1 - x^{n - 1}) $ for $ f_{i} \in F[x] $.
If the degree of $ f $ does not exceed $ n - 2 $, then I can construct
$$ f(x) = a_{0}(1 - x^{n - 1}) + (a_{1} - a_{0})(x - x^{n - 1}) + (a_{2} - a_{1})(x^2 - x^{n - 1}) + \dots + (a_{n - 3} - a_{n - 4})(x^{n - 3} - x^{n - 1}) + (a_{n - 2} - a_{n - 3})(x^{n - 2} - x^{n - 1}).$$
This implies $ f(x) \in \langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle $, but this fails when $ \deg(f) > n - 2 $.
| We know that every element in the ideal $\langle 1-x\rangle$ is a multiple of $(1-x)$, so:
$$f(x) = (1-x) \sum_{i = 0}^m a_{i} x^{i}$$
Generate $1-x$ simply by subtracting the second-to-last element in the basis from the last:
$$(1-x^{n-1}) - (x-x^{n-1}) = 1-x$$
We're mostly done:
$$f(x) = [(1-x^{n-1}) - (x-x^{n-1})] \sum_{i = 0}^m a_{i} x^{i}$$
Just set the other coefficients to zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$.
From the assumption we get that $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}$.
So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?
| Taking squares it is equivalent to $(2k+1)^2(3k+4)\leq (3k+1)(2k+2)^2$, exapanding everything you get $12k^3+12k^2+3k+16k^2+16k+4\leq 12k^3+24k^2+12k+4k^2+8k+4$ which is equivalent to $19k\leq 20k$ which is trivially true since $k\geq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help finding the Marginal PDF of Y given a Density Function of Two Variables Problem:
The join pdf of $(X,Y)$ is given by
$
f(x,y) = \begin{cases}
\Big( \frac{5}{32} \Big) x^2(4 - y) & \text{for } x < y < 2x, 0 < x < 2 \\
0 & \text{otherwise} \\
\end{cases}
$
where $k = \frac{5}{32}$. Find the marginal pdf of $Y$.
Answer:
\begin{eqnarray*}
f_y(y) &=& \int_{-\infty}^{\infty} f(x,y) \,\, dx \\
f_y(y) &=& \int_{0}^{2} \Big( \frac{5}{32} \Big) x^2(4 - y) \,\, dx \\
f_y(y) &=& \Big( \frac{5}{32} \Big) \frac{x^3}{3}(4 - y) \Big|_{x = 0}^{x = 2} \\
f_y(y) &=& \Big( \frac{5}{32} \Big) \frac{2^3}{3}(4 - y) \\
f_y(y) &=& \Big( \frac{5}{12} \Big) (4 - y) \\
\end{eqnarray*}
My answer is:
\begin{eqnarray*}
f_y(y) &=& \begin{cases}
\Big( \frac{5}{12} \Big)(4x^3 - \frac{3x^4}{2}) & \text{for } 0 < y < 4 \\
0 & \text{otherwise} \\
\end{cases} \\
\end{eqnarray*}
but the books answer is:
\begin{eqnarray*}
f_y(y) &=& \begin{cases}
\Big( \frac{5}{12} \Big) (4 - y) & \text{for } 0 < y < 2 \\
\Big( \frac{5}{32} \Big)\Big( \frac{1}{3} \Big)(4-y)(8 - \frac{y^3}{8}) & \text{for } 2 < y < 4 \\
0 & \text{otherwise} \\
\end{cases} \\
\end{eqnarray*}
I would like to know what I did wrong.
Thanks
Bob
| It seems like you have a good idea of what to do but just made a tiny mistake, I'll clear that up and leave you the work of finishing up.
The problem is that the set of possible values of $y$ depends on the value of $x$. You are not taking this into account. To do so you can write the joint pdf in terms of characteristic functions.
So, rewrite the pdf as follows:
$$ f(x,y) = \left( \frac{5}{32} \right) x^2 (4-y) \mathbf{1}_{ [x,2x]} (y) \cdot \mathbf{1}_{[0,2]}(x) $$
Now aply the same method you did before, integrating with respect to all possible values of $x$.
Take into account that if $ y$ must be between $x$ and $2x$, this is equivalent to $x$ being smaller than $y$ and $2x$ being bigger than $y$. (Draw the region for easier integration)
Hope this helps!
| {
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How to Calculate the derivative of $\frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}$ I'm sorry to bother you with this easy problem. But I'm working alone and totally confused.
It is the 1378th Problem from "Problems in Mathematical Analysis" written by Demidovich. The standard answer is
$$\frac{60(1+x)^{99}(1+6x)}{(1-2x)^{41}(1+2x)^{61}}$$
But when I followed the rutine
$$\frac{d(P(x)/Q(x))}{dx}=\frac{\frac{dP(x)}{dx}Q(x)-P(x)\frac{dQ(x)}{dx}}{Q(x)^2}$$
I got
$$\frac{100(1+x)^{99}(1-2x)^{40}(1+2x)^{60}-(1+x)^{100}(-80(1-2x)^{39}+120(1+2x)^{59})}{(1-2x)^{80}(1+2x)^{120}}$$
and finally
$$\frac{60(1+x)^{99}P(x)}{(1-2x)^{41}(1+2x)^{61}}$$
$$P(x)=1-4x^2-\frac{3(1+x)}{(1-2x)^{39}}-\frac{2(1+x)}{(1+2x)^{59}}$$
Would you tell me what mistake I made. Best regards.
| You can just take the derivative of the logarithm:
$$y = \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}} \implies \ln y = 100\ln(1+x)-40\ln(1-2x)-60\ln(1+2x)$$
$$\frac{y'}{y} = (\ln y)' = \frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x}$$
\begin{align}y' &= y\left(\frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x} \right)\\
&= \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}\left(\frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x} \right)\\
&= \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}\frac{100(1-4x^2)+80(1+x)(1+2x)-120(1+x)(1-2x)}{(1+x)(1-2x)(1+2x)}\\
&= \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}\frac{60(1+6x)}{(1+x)(1-2x)(1+2x)}\\
&= \frac{60(1+x)^{99}(1+6x)}{(1-2x)^{41}(1+2x)^{61}}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_\limits{x\to 0} \frac{\sin x}{e^x -1 -\sin x}$ I have to find $$\lim_\limits{x\to 0} \dfrac{\sin x}{e^x -1 -\sin x}.$$
By using L'Hopital, $$\lim_\limits{x\to 0} \frac{\sin x}{e^x -1 -\sin x}= \lim_\limits{x\to 0} \frac{\cos x}{e^x -\cos x}.$$
Wolfram now states that $\nexists$ because $\lim_\limits{x\to 0^+} \frac{\sin x}{e^x -1 -\sin x}=\infty$ and $\lim_\limits{x\to 0^-} \frac{\sin x}{e^x -1 -\sin x}=-\infty$,
but I don't see why $\lim_\limits{x\to 0^-} \frac{\sin x}{e^x -1 -\sin x}=-\infty$
| Hint:
\begin{align}
& \lim_{x\to 0} \frac{\sin x}{e^x -1 -\sin x} \\[6pt]
= {} & \lim_{x\to 0} \frac{\sin x}{1+x+\frac{x^2}{2}+o(x^3) -1 -(x-\frac{x^3}{6}+o(x^5))} \\[6pt]
= {} & \lim_{x\to 0} \frac{\sin x}{\frac{x^2}{2}+o(x^3) } \\[6pt]
= {} & \lim_{x\to 0} \frac{x}{\frac{x^2}{2}+o(x^3) } \\[6pt]
= {} & \lim_{x\to 0} \frac{1}{\frac{x}{2}+o(x^2) }
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Evaluating this line integral Evaluate the following line integral
$\int 2x+y \,dx +xy\,dy$ from $(-1,2)$ to $(2,5)$ where $y=x^2 +1$
When I integrated it w.r.t $x$ I found the integral
$$\int 2x^4 +3x^2 +2x+1 \, dx=\frac{141}{5}$$
from $x=-1$ to $x=2$
But when I tried it w.r.t $y$ I got a confusion with evaluating the limits of $y$ since $y=x^2 +1 $ then $x=\sqrt{y-1}$ where $y$ belongs to $[1,5]$
and $x=-\sqrt{y-1}$ where $y$ belongs to $[1,2]$
| Your evaluation
$$\int_C (2x+y)dx +xydy=\int_{x=-1}^2 ((2x+x^2+1) +x(x^2+1)(2x))dx=\frac{141}{5}$$
is correct. Using $y$ as a parameter is more difficult, but you should obtain the same result.
\begin{align*}
\int_C (2x+y)dx +xydy&=\int_{y=2}^1 ((2f(y)+y)f'(y) +f(y)y)dy
+\int_{y=1}^5 ((2g(y)+y)g'(y) +g(y)y)dy\\
&=\int_{y=2}^1 \left(1-\frac{y}{2\sqrt{y-1}} -y\sqrt{y-1}\right)dy\\
&\qquad+\int_{y=1}^5 \left(1+\frac{y}{2\sqrt{y-1}} +y\sqrt{y-1}\right)dy
\end{align*}
where $f(y)=-\sqrt{y-1}$ and $g(y)=\sqrt{y-1}$.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$0.x3+0.3x=\frac{7}{9} \implies x=?$ An old book problem I was trying to solve some of problems in an old very old book , which not signed
$$0.3\times x ,\\0.\overline{3}\times x \\ 0.\overline{3x} $$ to solve below equation . I tried like below .
My question : How many way we can read this question ?
Am I right ?
$$0.x3+0.3x=\frac{7}{9}$$
$$\quad{0.\overline{x3}+0.\overline{x3}=\frac{7}{9}\\(\frac{\overline{x3}}{100}+\frac{\overline{x3}}{10000}+...)+
(\frac{\overline{3x}}{100}+\frac{\overline{3x}}{10000}+...)\\
(\dfrac{\dfrac{\overline{3x}}{100}}{1-\frac{1}{100}})+(\dfrac{\dfrac{\overline{x3}}{100}}{1-\frac{1}{100}})=\\\frac{\overline{x3}}{99}+\frac{\overline{3x}}{99}=\\
\frac{\overline{x3}+\overline{3x}}{99}=\\
\frac{10x+3+30+x}{99}=\\\frac{x+3}{9} \to \\\frac{x+3}{9}=\frac{7}{9} \\x=4}$$
| $$\quad{0.\overline{x3}+0.\overline{3x}=\frac{7}{9}\\0.\overline{x}\times 3+0.\overline{3x}=\frac{7}{9}\\
0.\overline{x3}+0.\overline{3}\times x=\frac{7}{9}
\\0.\overline{x}\times 3+0.\overline{3}\times x=\frac{7}{9}\\or \\0.3\times x+0.x\times 3=\frac{7}{9}\\0.(3x)+0.(x3)=\frac{7}{9}}$$ is there an other possible reading ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $ z^2 + \left\vert z \right\vert = 0 $, where $z$ is a complex number. I've tried solving this, but I'm stuck at one point.
Here's what I did:
Let $ z = x + yi $, where $x, y \in \mathbf R$
Then , $ (x + yi)^2 + \sqrt{x^2 + y^2} = 0 $
Or, $x^2 + {(yi)}^2 + 2xyi + \sqrt{x^2 + y^2} = 0 $
Or, $ x^2 - y^2 + 2xyi + \sqrt{x^2 + y^2} = 0 + 0i$
Thus, $ x^2 - y^2 + \sqrt{x^2 + y^2} = 0\qquad\qquad\qquad\qquad\qquad\qquad\ (i)$
and $2xy = 0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(ii)$
If $2xy = 0$, then either $x = 0 $ or $y = 0$
Now, if I take $ x = 0$, and subsitute in $(i)$, I get either $y = 0$ or $y = 1$.
So far, so good, but if I take $y = 0$, and substitute in $(ii)$:
We have $x^2 + \sqrt{x^2} = 0$
so $x^2 = -\sqrt{x^2} $
or $x^2 = -x$
or $\frac{x^2}{x} = -1 $
or $x = -1$
However, this solution doesn't satistfy the equation $x^2 + \sqrt{x^2}$ or the original equation.
What am I doing wrong here ?
| In $x^2 = -\sqrt{x^2}$, note that $x$ is real, so you have a non-negative on the left, and a non-positive on the right. Therefore they must both be $0$.
Alternative solution:
$$
z^2=-|z|\\
|z|^2=|z|\\
|z|=0\quad \text{or}\quad|z|=1
$$
then for $|z|=1$, solve $z^2+1=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into very complicated expressions, and are most likely wrong. Hints or solutions are welcomed. Sorry if this is a duplicate.
| Let $c = -(a+b)$. Consider following cubic polynomial having $a, b, c$ as roots:
$$t^3-At^2 + Bt - C \stackrel{def}{=} (t-a)(t-b)(t-c)
\quad\text{ where }\quad
\begin{cases}
A &= a + b + c = 0\\
B &= ab+bc+ca = -(a^2+b^2+ab)\\
C &= abc = -ab(a+b)
\end{cases}$$
When $t$ is one of $a, b, c$, we have $t^3 = C - Bt$. This implies
$$t^7 = (C-Bt)^2 t = C^2 t - 2CB t^2 + B^2t^3
= (C^2 - B^3)t + B^2C -2CB t^2$$
Substitute $t$ by $a,b,c$ and sum over them. Together with
$$\begin{align}a + b + c
&= A = 0\\
a^2 + b^2 +c^2 &= (a+b+c)^2 - 2(ab+bc+ca) = A^2 - 2B = -2B
\end{align}$$
we obtain
$$\begin{align}
&a^7 + b^7 + c^7 = (C^2-B^3)A + 3B^2C - 2CB(A^2 - 2B) = 7B^2C\\
\implies &
(a+b)^7 - a^7 - b^7 = -7B^2C = 7ab(a+b)(a^2+ab+b^2)^2
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $a, b, c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$ .
Let $a$, $b$ and $c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b).$$
This problem is from the Indian MO 2009, however, I do not know how to approach it, any help?
| We need to prove that
$$a^2+b^2-\frac{a^3+b^3}{c}>6(c-a)(c-b)$$ or
$$\frac{a^2(c-a)}{c}+\frac{b^2(c-b)}{c}>6(c-a)(c-b)$$ or
$$\frac{(c^3-b^3)(c-a)}{ac}+\frac{(c^3-a^3)(c-b)}{bc}>6(c-a)(c-b)$$ or
$$\frac{c^2+bc+b^2}{ac}+\frac{c^2+ac+c^2}{bc}>6.$$
Now, by AM-GM
$$\frac{c^2+bc+b^2}{ac}+\frac{c^2+ac+c^2}{bc}\geq2\sqrt{\frac{c^2+bc+b^2}{ac}\cdot\frac{c^2+ac+c^2}{bc}}\geq$$
$$\geq2\sqrt{\frac{3\sqrt[3]{b^3c^3}}{ac}\cdot\frac{3\sqrt[3]{a^3c^3}}{bc}}=6.$$
In our using AM-GM the equality occurs for $a=b=c$, which is impossible.
Thus, $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| Let $\frac{ax}{ax+by}=\alpha$ and $\frac{by}{ax+by}=\beta$.
Thus, $$\alpha+\beta=1,$$
$$\alpha x+\beta y=7,$$
$$\alpha x^2+\beta y^2=19$$ and
$$\alpha x^3+\beta y^3=58.$$
Hence, $$(x+y)(\alpha x+\beta y)=7(x+y)$$ or
$$19+xy(\alpha+\beta)=7(x+y)$$ or
$$19+xy=7(x+y).$$
In another hand
$$(x+y)(\alpha x^2+\beta y^2)=19(x+y)$$ or
$$58+xy(\alpha x+\beta y)=19(x+y)$$ or
$$58+7xy=19(x+y).$$
From here we obtain $x+y=2.5$ and $xy=-\frac{3}{2}$ and the rest is smooth.
I got $a+b=21$ and $$2014(x+y-xy)-100(a+b)=2014\cdot4-100\cdot21=5956.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| Let $x+2=a$ and $x=b$.
Thus, $a-b=2$ and we have
$$a^4+b^4=82$$ or
$$(a^2+b^2)^2-2a^2b^2=82$$ or
$$((a-b)^2+2ab)^2-2a^2b^2=82$$ or
$$(4+2ab)^2-2a^2b^2=82$$ or $$a^2b^2+8ab-33=0,$$ which gives
$ab=-11$, $x(x+2)+11=0,$ which is impossible or
$ab=-3$, $x(x+2)=3,$ which gives the answer:
$$\{1,-3\}.$$
We can end also your way.
We need to solve
$$2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$$ or
$$x^4+4x^3+12x^2+16x-33=0$$ or
$$x^4-x^3+5x^3-5x^2+17x^2-17x+33x-33=0$$ or
$$(x-1)(x^3+5x^2+17x+33)=0$$ or
$$(x-1)(x^3+3x^2+2x^2+6x+11x+33)=0$$ or
$$(x-1)(x+3)(x^2+2x+11)=0,$$
which gives the same answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the radical equation $ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$
Solve the following equation:
$$ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$$
I wanted to solve this equation. First I tried to change the equations under the roots to the complete square to simplify them out, but it just became more complicated.
Can someone help me with this equation, please?
| Assuming you want real roots, you can use this trick . . .
Let $f\colon \mathbb{R} \to \mathbb{R}$ be given by $f(t) = t\sqrt{t^2+5}$.
Then $f$ is an odd function.
Also, $f$ is strictly increasing, hence $f$ is one-to-one.
Then, letting $u=2x+1$,
\begin{align*}
&x\sqrt{x^2+5}+(2x+1)\sqrt{4x^2 + 4x + 6}=0
\qquad\qquad\qquad\qquad\;\;
\\[4pt]
\iff\;&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt]
\iff\;&f(x) + f(u)=0\\[4pt]
\iff\;&f(x) = -f(u)\\[4pt]
\iff\;&f(x) = f(-u)\qquad\text{[since $f$ is odd]}\\[4pt]
\iff\;&x = -u\qquad\qquad\;\;\text{[since $f$ is one-to-one]}\\[4pt]
\iff\;&x = -(2x+1)\\[4pt]
\iff\;&3x+1 = 0\\[4pt]
\iff\;&x = -{\small{\frac{1}{3}}}\\[4pt]
\end{align*}
As an alternative, using the same trick, you can get a factored form:
\begin{align*}
&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt]
\implies\;&x\sqrt{x^2+5}=-u\sqrt{u^2 + 5}\\[4pt]
\implies\;&x^2(x^2+5)=u^2(u^2 + 5)\\[4pt]
\implies\;&x^4+5x^2=u^4+5u^2\\[4pt]
\implies\;&(u^4-x^4)+5(u^2-x^2)=0\\[4pt]
\implies\;&(u^2-x^2)(u^2+x^2)+5(u^2-x^2)=0\\[4pt]
\implies\;&(u^2-x^2)(u^2+x^2+5)=0\\[4pt]
\implies\;&(u-x)(u+x)(u^2+x^2+5)=0\\[4pt]
\implies\;&(x+1)(3x+1)(5x^2+4x+6)=0\qquad\text{[replacing $u$ by $2x+1$]}
\end{align*}
The two candidate real roots, $x=-1,\;x=-{\large{\frac{1}{3}}}\;$need to be verified against the original equation since, when squaring both sides, extraneous real roots were potentially introduced. In this case, as it turns out, the candidate root $x=-{\large{\frac{1}{3}}}$ is ok, but the candidate root $x=-1$ fails, so is not an actual root.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $\sum\limits_{k=0}^{n} {2n+1 \choose k}$ I am stuck with this problem: $\sum_\limits{k=0}^{n} {2n+1 \choose k}$.
I used Wolframalpha for the answer and is $4^n$. So i searched for a similar way to expressed it which is:
$$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}{2n \choose k}$$
Is there a way to prove that $\sum_\limits{k=0}^{2n}{2n \choose k}=\sum_\limits{k=0}^{n} {2n+1 \choose k}$?
| Since $$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\left(\sum_{k = 0}^n \binom{2n + 1}{k} + \sum_{k = 0}^n \binom{2n + 1}{2n+1-k}\right) = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k}$$
Then we have:
$$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k} = \frac{1}{2} \cdot 2^{2n + 1} = 2^{2n} = 4^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Range of the function $f(x) = \frac{2x}{x^{2}+1}$ Plotting the function $f(x) = \frac{2x}{x^{2}+1}$ we can see the range is $[-1,1]$.
Now I was told to do $f(x)=y$ and isolate $x$ doing this I have, using the quadratic formula $x=\frac{1\pm \sqrt{1-y^2}}{y}$ then $$range f(x)= domain \frac{1\pm \sqrt{1-y^2}}{y}$$.
But this domain does not include $0$ and $0=f(0)$ so $0$ must be in the range, what am I missing?
| $y = \frac {2x}{x^2+1}\\
yx^2 - 2x + y = 0$
And then you use the quadratic formula to find values of $y,$ such that $x$ can be a real root of the polynomial.
All good.
What happens at $y = 0$?
$yx^2 - 2x + y = 0$ ceases to be a quadratic and the quadratic formula is not applicable.
| {
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For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer? For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer?
I managed to figure out:
$5p +49 = n^2$
$5p = (n+7)(n-7)$
But can't think of anything more than that, anyone able to solve this without bruteforcing?
For primenumbers under $10^7$, the solutions are $3$ and $19$.
| You're close.
You have $5p=(n+7)(n-7)$ for some integer $n$. Without loss of generality we may assume that $n$ is nonnegative, if $n$ were negative we could render $(n+7)(n-7)=(|n|-7)(|n|+7)$ where $|n|$ is nonnegative. Since $n+7$ is thereby positive we need $n-7$ positive too, so the nonnegative $n$ is at least $8$.
From the uniqueness of prime factorization over positive integers we are sure that the factors $n+7$ and $n-7$ must be $5$ and $p$ in some order, or else $1$ and $5p$ in some order. Obviously $n+7$ can't be $1$ or $5$ for nonnegative $n$, so $n-7$ must be $1$ or $5$ instead forcing $n\in\{8,12\}$. With $n-7=1, n=8$ we get $n+7=5p=15$ thus $p=3$. With $n-7=5,n=12$ we get $n+7=p=19$. The full solution set for the prime $p$ is then $\{3,19\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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I need some help with this bernoulli's equation that is given in my assignment Solve:
$$y′−\frac{6y}{x}=\frac{y^5}{x^{13}}$$
I have tried again and again but my answer is still wrong. Can someone show me the steps to solving this question?
| For Bernoulli equation of the form
$$y^\prime+p(x)y=q(x)y^\beta$$
we use substituation $u=y^{1-\beta}$. So with $\beta=5$ we have $u=y^{1-5}=\dfrac{1}{y^4}$ then $u'=-\dfrac{4y'}{y^5}$ then
\begin{eqnarray*}
-\frac{y^5u^\prime}{4}-6\frac{y}{x} &=& \frac{1}{x^{13}}y^5\\
-\frac{u^\prime}{4}-6\frac{1}{x}\frac{1}{y^4} &=& \frac{1}{x^{13}}\\
u^\prime+\dfrac{24}{x}u &=& -4\frac{1}{x^{13}}\\
x^{24}\left(u^\prime+\dfrac{24}{x}u \right) &=& -4 x^{24}\frac{1}{x^{13}}\\
x^{24}u'+24ux^{23}&=& -4 x^{11}\\
(x^{24}u)'&=& -4 x^{11}\\
x^{24}u&=& -\dfrac13 x^{12}+C\\
\frac{1}{y^4} &=& -\frac{1}{3x^{12}}+\frac{C}{x^{24}}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Sum of series $\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$
The problem is to find the sum of series $$\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$$
to infinite terms, if $|x|<1$
I tried taking $\frac{x}{1-x^2}$ outside but could not solve it.
How to proceed further?
| Notice $$\frac{y}{1-y^2} = \frac{(1+y)-1}{1-y^2} = \frac{1}{1-y} - \frac{1}{1-y^2}$$
The sum at hand is a telescoping one. The partial sums has the form:
$$\sum_{n=1}^p
\frac{x^{2^{n-1}}}{1-x^{2^n}} =
\sum_{n=1}^p
\left[\frac{1}{1-x^{2^{n-1}}} - \frac{1}{1-x^{2^n}}\right]
= \frac{1}{1-x} - \frac{1}{1-x^{2^p}}$$
When $|x| < 1$, $x^{2^p} \to 0$. This leads to
$$\sum_{n=1}^\infty
\frac{x^{2^{n-1}}}{1-x^{2^n}}
=
\lim_{p\to\infty}
\left[\frac{1}{1-x} - \frac{1}{1-x^{2^{p}}}\right]
= \frac{1}{1-x} - \frac{1}{1-0} = \frac{x}{1-x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the recurrence $T(n) = 2$ if $n=0$
$T(n) = 9T(n-1)-56n +63$ if $n>=1$
Repeated substitution
$k=1$
$T(n) = 9T(n-1)-56n+63 $
$k =2$
$T(n) = 81T(n-2) -560n + 1134$
$k =3$
$T(n) = 729T(n-3) -5096n + 15309$
I cant find the pattern for the n term and the integer
For now i just have
T(n) = $9^k(n-k)$
| Let's apply generating functions. It's not always the easiest method, but it solidly works. From
$$T_n=9T_{n-1}-56n+63$$
the generating function is
$$f(x)=\sum\limits_{n=0}T_nx^n \tag{1}$$
$$f(x)=T_0+\sum\limits_{n=1}T_nx^n=T_0+\sum\limits_{n=1}\left(9T_{n-1}-56n+63\right)x^n=\\
T_0+9\sum\limits_{n=1}T_{n-1}x^n-56\sum\limits_{n=1}nx^n+63\sum\limits_{n=1}x^n=\\
T_0+9x\sum\limits_{n=1}T_{n-1}x^{n-1}-56x\sum\limits_{n=1}nx^{n-1}+63\sum\limits_{n=1}x^n=\\
T_0+9xf(x)-\frac{56x}{(1-x)^2}+\frac{63x}{1-x}$$
Or
$$f(x)=T_0+9xf(x)-\frac{56x}{(1-x)^2}+\frac{63x}{1-x}$$
$$f(x)=\frac{T_0}{1-9x}-\frac{56x}{(1-9x)(1-x)^2}+\frac{63x}{(1-9x)(1-x)}$$
$$f(x)=\frac{T_0}{1-9x} +\frac{7}{(1 - x)^2} + \frac{7}{8(1 - x)} - \frac{63}{8(1 - 9x)} -\frac{63}{8(1 - x)} + \frac{63}{8(1 - 9x)}=\\
f(x)=\frac{T_0}{1-9x} +\frac{7}{(1-x)^2} -\frac{7}{1-x}$$
$$f(x)=T_0\sum\limits_{n=0}(9x)^n+7\sum\limits_{n=0}(n+1)x^n-7\sum\limits_{n=0}x^n=\sum\limits_{n=0}\left(T_0\cdot 9^n +7n \right)x^n \tag{2}$$
Or, comparing the terms of $(1)$ and $(2)$
$$T_n=T_0\cdot 9^n +7n$$
Some of the shortcuts are explained here.
The final step is to find $T_0$ ...
*
*from $\color{red}{2=T_0\cdot9+7}$, given $T(n)=2$ if $n=1$, this was the original question, as per the edit history. Or
*from $\color{red}{2=T_0}$, given $T(n)=2$ if $n=0$, this is from the updated question, as per the edit history.
This proves how flexible the generating functions are ...
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding an integral formula for intertwined recursive sequences I saw on Wikipedia that the arithmetic-geometric mean of two numbers is given by
$$\operatorname{agm}(x,y)=\frac{\pi}{2}\Bigg[\int_0^{\pi/2}\frac{d\theta}{\sqrt{x^2\cos^2\theta+y^2\sin^2\theta}}\Bigg]^{-1}$$
and I understand the proof of this little gem... however, how does one go about deriving something like this without knowing what to prove beforehand? For example, if I wanted to find an integral formula for the "geometric-harmonic mean" defined as
$$\operatorname{ghm}(x,y)=\lim_{n\to\infty} g_n=\lim_{n\to\infty} h_n$$
where
$$g_0=x$$
$$h_0=y$$
$$g_{n+1}=\sqrt{g_nh_n}$$
$$h_{n+1}=\frac{2}{\frac{1}{g_n}+\frac{1}{h_n}}$$
Then how would I go about finding an analogous integral formula?
| For any $a,b>0$, let $HG(a,b)=HG\left(\frac{2ab}{a+b},\sqrt{ab}\right)$ and for any $x>1$ let $f(x)=HG(1,x)$.
We have:
$$f(x) = HG(1,x) = HG\left(\frac{2x}{1+x},\sqrt{x}\right) = \frac{2x}{1+x}\,HG\left(1,\frac{1+x}{2\sqrt{x}}\right)\\= \frac{2x}{1+x}\,f\left(\frac{1+x}{2\sqrt{x}}\right)\tag{A} $$
and we may notice that the map $g:x\mapsto\frac{1+x}{2\sqrt{x}}$ sends the interval $(1,+\infty)$ into itself.
In particular the given problem is equivalent to finding an invariant measure $\mu$ such that
$$\forall x>1,\qquad \mu((1,x)) = \frac{2x}{1+x}\,\mu\left(\left(1,\frac{1+x}{2\sqrt{x}}\right)\right)$$
Additionally, the sequence $x,g(x),g(g(x)),g(g(g(x))),\ldots$ converges really fast to $1$ for any $x>1$.
In particular
$$ f(x) = \frac{4x}{(1+\sqrt{x})^2}\;f(g(g(x))) \tag{B}$$
leads to $f(x)\approx \frac{4x}{(1+\sqrt{x})^2}$ and
$$ HG(a,b) = a\cdot f\left(\tfrac{b}{a}\right) \approx \frac{4ab}{\left(\sqrt{a}+\sqrt{b}\right)^2}=H\left(\sqrt{a},\sqrt{b}\right)^2\tag{C}$$
where $H$ stands for the usual harmonic mean. We may also notice that
$$ h(x)=\text{AGM}(1,x)=\text{AGM}\left(\sqrt{x},\frac{1+x}{2}\right) = \sqrt{x}\,h\left(\frac{1+x}{2\sqrt{x}}\right) $$
hence
$$ \frac{f}{h}(x) = \frac{1}{g(x)}\cdot \frac{f}{h}(g(x))\tag{D} $$
and the problem of finding a closed form for $\text{HG}$ boils down to the problem of finding a closed form for the infinite product
$$ g(x)\cdot g(g(x))\cdot g(g(g(x)))\cdot\ldots $$
On the other hand:
$$ h(x) = \sqrt{x} h(g(x)) = \sqrt{x}\sqrt{g(x)} h(g(g(x))) = \ldots = \sqrt{x}\sqrt{g(x)g(g(x))\cdot\ldots} $$
hence $f(x)=\frac{x}{h(x)}$ and
$$\boxed{ HG(a,b) = \frac{ab}{\text{AGM}(a,b)} = \color{red}{\frac{2}{\pi}\int_{0}^{+\infty}\frac{dx}{\sqrt{\left(1+\frac{x^2}{a^2}\right)\left(1+\frac{x^2}{b^2}\right)}}}.}\tag{E}$$
Further proof of $(E)$: it is enough to check that $\frac{ab}{\text{AGM}(a,b)}$ is invariant with respect to the replacements $b\to\sqrt{ab}, a\to\frac{2ab}{a+b}$.
$$\begin{eqnarray*}\frac{\sqrt{ab}\frac{2ab}{a+b}}{\text{AGM}\left(\sqrt{ab},\frac{2ab}{a+b}\right)}&=&\frac{2ab\sqrt{ab}}{\text{AGM}\left((a+b)\sqrt{ab},2ab\right)}\\&=&\frac{2ab}{\text{AGM}\left(a+b,2\sqrt{ab}\right)}\\&=&\frac{ab}{\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)}=\frac{ab}{\text{AGM}(a,b)}\;\large\checkmark\end{eqnarray*} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find n'th derivative of polynomial and smallest possible degree Let $p(x)$ be a polynomial of degree strictly less than 100 and such that it does not have $x^3−x$ as a factor.If
$\frac{d^{100}}{d x^{100}} \frac{p(x)}{x^3-x} =\frac{f(x)}{g(x)}$
for some polynomials $f(x)$ and $g(x)$ then find the smallest possible degree of $f(x)$
| Using the division algorithm we have
$$
\frac{p(x)}{x^3 − x}= q(x) +\frac{r(x)}{x^3 − x}$$
As the degree of $q(x)$ is strictly less than that of $p(x)$ its $100-$th derivative is certainly zero. As $x^3 − x$ is not a factor of $p(x)$ one may assume (without loss of generality) that $x^2 − 1$ divides $r(x)$. In that case we have
$$
\frac{\mathrm d^{100}}{\mathrm dx^{100}}\frac{p(x)}{x^3 − x}=
\frac{\mathrm d^{100}}{\mathrm dx^{100}}\frac{k}{x}=\frac{100!k}{x^{100}}
$$
Hence the least possible degree of $f(x)$ is $0$.
If one assumes that $x^3 − x$ doesn't divide $p(x)$ then we have
$$\frac{r(x)}{x^3 − x}=\frac{A'}{x}+\frac{B'}{x-1}+\frac{C'}{x+1}$$
Consequently,
$$
\begin{align}
\frac{f(x)}{g(x)}&=\frac{A}{x^{101}}+\frac{B}{(x-1)^{101}}+\frac{C}{(x+1)^{101}}
\end{align}
$$
and then
$$
\begin{align}
f(x)&=A(x^2-1)^{101}+{B}{(x^2+x)^{101}}+{C}{(x^2-x)^{101}}\\
&=(A+B+C)x^{202}+101(B-C)x^{201}+\left[\binom{101}{2}B+\binom{101}{2}C-101A\right]x^{200}+\cdots
\end{align}
$$
Choosing $B = C$ and $A+B+C = 0$ we see that the coefficient of $x^{200}$ is $101\cdot 102 \ne0$. Hence the least possible degree of $f(x)$ in this case is $200$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple Probability Distribution Question Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent X $\sim$ Bin(n, 1/2) and Y $\sim$ Bin(n + 1, 1/2).
Can someone please explain to me, in simple english, why this statement is true?
$$P(X<Y)=P(n−X<n+1−Y)$$
The random variable Y is not inherently larger than X, so why is that statement true?
| Binomial coefficients have a symmetry that can be written
$$\binom{n}{k} = \binom{n}{n-k}$$
$Y$ is a random variable following a binomial distribution with parameters $n+1$ and $1/2$. When the second parameter is $1/2$, the symmetry in binomial coefficients extends to a symmetry in the probability mass function:
\begin{align}
P(Y=k) &= \binom{n+1}{k}\left(\frac{1}{2}\right)^k \left(1 - \frac{1}{2}\right)^{n+1-k} \\
&= \binom{n+1}{k} \left(\frac{1}{2}\right)^{n+1}\\\\
P(Y=n+1-k) &= \binom{n+1}{n+1- k }\left(\frac{1}{2}\right)^{n+1-k} \left(1 - \frac{1}{2}\right)^{n+1 - (n+1-k)}\\
&=\binom{n+1}{k} \left(\frac{1}{2}\right)^{n+1}\\\\
\end{align}
We have, for all $k$, the symmetric relationship
$$P(Y=k) = P(Y = n + 1 - k) = \binom{n+1}{k} \left(\frac{1}{2}\right)^{n+1}$$
Similarly, $X$ has parameters $n$ and $1/2$, so we have, for all $h$,
$$P(X = h) = P(X = n - h) =\binom{n}{h} \left(\frac{1}{2}\right)^{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Area enclosed by quarter circle arcs in a square
The problem I have here asks to find the area $A$ enclosed by the two quarter circle arcs in terms of $a$. The first circle has radius $a/2$, and the second has radius $a$. This is what I did.
Firstly I determined that $$A=\frac{a^2}{8}(2\theta_1-\sin(2\theta_1))+\frac{a^2}{2}(2\theta_2-\sin(2\theta_2))$$ $$=\frac{a^2}{4}(\theta_1+4\theta_2-\sin\theta_1\cos\theta_1-4\sin\theta_2\cos\theta_2)$$
Then $$h_1+h_2=a\sqrt2=\sqrt{\frac{a^2}{4}-x^2}+\sqrt{a^2-x^2}$$
$$(a\sqrt2-\sqrt{a^2-x^2})^2=\frac{a^2}{4}-x^2$$
$$12a^2-8a\sqrt{2a^2-2x^2}=a^2$$
$$11a=8\sqrt{2a^2-2x^2}$$
$$121a^2=128a^2-128x^2$$
$$x=a\sqrt{\frac{7}{128}}=\frac{a\sqrt{14}}{16}$$
So
$$h_1=\sqrt{\frac{a^2}{4}-\frac{14a^2}{256}}=\frac{5a\sqrt2}{16}$$
$$h_2=\sqrt{a^2-\frac{14a^2}{256}}=\frac{11a\sqrt2}{16}$$
Therefore
$$\sin\theta_1\cos\theta_1=\frac{\sqrt{14}}{8}\times\frac{5\sqrt2}{8}=\frac{5\sqrt7}{32}$$
$$4\sin\theta_2\cos\theta_2=4\times\frac{\sqrt{14}}{16}\times\frac{11\sqrt2}{16}=\frac{11\sqrt7}{32}$$
So
$$A=\frac{a^2}{4}(\theta_1+4\theta_2-\frac{5\sqrt7}{32}-\frac{11\sqrt7}{32})$$
$$=\frac{a^2}{4}(\theta_1+4\theta_2-\frac{\sqrt7}{2})$$
Now I can find $\theta_1$ and $\theta_2$ with my calculator, but there must be a way to express them in terms of surds but I don't know how. Also, is what I did actually correct and is there another shorter/nicer way to solve this?
|
Use the cosine rule
\begin{eqnarray*}
\cos ( A ) =\frac{b^2+c^2-a^2}{2bc}.
\end{eqnarray*}
\begin{eqnarray*}
\cos ( \theta_1 ) =\frac{(\sqrt{2}a)^2+(\frac{a}{2})^2-a^2}{2(\sqrt{2}a)(\frac{a}{2})}=\frac{5}{4 \sqrt{2}}
\end{eqnarray*}
\begin{eqnarray*}
\cos ( \theta_2 ) =\frac{(\sqrt{2}a)^2+a^2-(\frac{a}{2})^2}{2(\sqrt{2}a)a}=\frac{11}{8 \sqrt{2}}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$ Let $a,b,c$ be positive reals such that $ab+bc+ca+abc=4$.
Then prove
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c $
So high guys im a high schooler trying to solve this inequality. I did a few things such as try to set a> b>c and then try to generalize it but i couldn't make much progress at all. I would appreciate a clear solution to help me understand the problem in depth.
I did manage to generalize that
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq a+b+c $
By using the AM-GM inequality on each of a b and c however havent made much progress in proving that the LHS is less than 3 and the RHS is greater than 3.
| Both inequalities we can prove by the Contradiction method.
Indeed, for the right inequality let $a+b+c<3$, $a=kx$, $b=ky$ and $c=kz$,
where $k>0$ and $x+y+z=3$.
Thus, $k(x+y+z)<3$, which gives $k<1$.
Hence, $4=ab+ac+bc+abc=k^2(xy+xz+yz)+k^3xyz<xy+xz+yz+xyz$,
which is contradiction because we'll prove now that
$$4\geq xy+xz+yz+xyz$$ or
$$\frac{4(x+y+z)^3}{27}\geq\frac{(x+y+z)(xy+xz+yz)}{3}+xyz$$ or
$$\sum_{cyc}(4x^3+3x^2y+3x^2z-10xyz)\geq0,$$
which is true by Muirhead.
Also, we can use AM-GM.
$$\sum_{cyc}x^3=x^3+y^3+z^3\geq3\sqrt[3]{x^3y^3z^3}=3xyz;$$
$$\sum_{cyc}(x^2y+x^2z)=x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq6\sqrt[6]{x^6y^6z^6}=6xyz.$$
Thus, $$\sum_{cyc}(4x^3+3x^2y+3x^2z-10xyz)=$$
$$=4(x^3+y^3+z^3)+3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)-30xyz\geq$$
$$\geq4\cdot3xyz+3\cdot6xyz-30xyz=0.$$
The left inequality.
Let $\sqrt{ab}+\sqrt{ac}+\sqrt{bc}>3$, $a=kx$, $b=ky$ and $c=kz$,
where $k>0$ and $\sqrt{xy}+\sqrt{xz}+\sqrt{yz}=3$.
Thus, $k(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})>3$, which gives $k>1$ and
$$4=ab+ac+bc+abc=k^2(xy+xz+yz)+k^3xyz>xy+xz+yz+xyz,$$
which is contradiction because we'll prove now that
$$4\leq xy+xz+yz+xyz.$$
Indeed, let $\sqrt{xy}=r$, $\sqrt{xz}=q$ and $\sqrt{yz}=p$.
Thus, $p+q+r=3$ and we need to prove that
$$p^2+q^2+r^2+pqr\geq4$$ or
$$\frac{(p+q+r)(p^2+q^2+r^2)}{3}+pqr\geq\frac{4(p+q+r)^3}{27}$$ or
$$\sum_{cyc}(5p^3-3p^2q-3p^2r+pqr)\geq0$$ or
$$\sum_{cyc}(p^3-p^2q-p^2r+pqr)+2\sum_{cyc}(2p^3-p^2q-p^2r)\geq0,$$
which is true by Schur and Muirhead.
The inequality $\sum\limits_{cyc}(2p^3-p^2q-p^2r)\geq0$ we can prove also without Muirhead:
$$\sum_{cyc}(2p^3-p^2q-p^2r)=\sum_{cyc}(p^3-p^2q-pq^2+q^3)=\sum_{cyc}(p-q)^2(p+q)\geq0.$$
Done!
Also, we can use the following trigonometric substitution.
Let $\sqrt{ab}=2\cos\gamma$, $\sqrt{ac}=2\cos\beta$ and $\sqrt{bc}=\cos\alpha$.
Thus, $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$
where $\alpha$, $\beta$ and $\gamma$ are measures of acute angles.
Hence, easy to see that $\alpha+\beta+\gamma=180^{\circ}$ and we need to prove that
$$\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}$$ and
$$\sum_{cyc}\frac{\cos\alpha\cos\beta}{\cos\gamma}\geq\frac{3}{2},$$
which is obvious.
Done again!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Napier analogy and algebra in triangle.
If in a $\triangle{ABC}$, we define $x=\tan\frac{B-C}{2}\tan\frac{A}{2}$, $y=\tan\frac{C-A}{2}\tan\frac{B}{2}$ and $z=\tan\frac{A-B}{2}\tan\frac{C}{2}$, then show that $x+y+z=-xyz$.
My attempts:
By Napier analogy,
$x=\frac{b-c}{b+c},\ y=\frac{c-a}{c+a},\ z=\frac{a-b}{a+b}$
Then one can simply put these values in LHS, but that is cumbersome, I need to use some beautiful algebra, please help.
I just need to solve that algebra stuff, if such problem exists somewhere on this site then please comment with that link, I'll delete this, then.
| $$\sum_{cyc}\frac{a-b}{a+b}=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}=$$
$$=\frac{\sum\limits_{cyc}c^2(a-b)}{\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}=-\prod_{cyc}\frac{a-b}{a+b}$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Limits of the strings What is the limit of this:
$a_n=\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}$, where n tends to infinity?
I would also like an intuitive explanation in addition to the logical one. Thanks in advance!
P.s.: Do you know where can I find the math API on this site?
| By induction prove that,
$$a_n\le \frac{1}{n}$$
*
*for $n=1$
$$a_1 = \frac{1}{2}\le \frac{1}{1}$$
Assume that
$$a_n=\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}\le \frac{1}{n} $$
$$a_{n+1}=\dfrac{(1!+2!+\dots+n!)+(n+1)!}{2(n+1)(2n+1)(2 \cdot n)!}\\=\frac{1}{2(n+1)(2n+1)}\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}+\frac{1}{2(n+1)}\dfrac{(n+1)n!}{(2n+1)(2 \cdot n)!}$$
But $$n!\le (2n)!~~~and~~~n+1\le 2n+1 \implies\dfrac{(n+1)n!}{(2n+1)(2 \cdot n)!}\le 1 $$
By assumption $$ \dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}\le\frac{1}{n}\le 1$$ we get
$$ a_{n+1}=\dfrac{(1!+2!+\dots+n!)+(n+1)!}{(2 n+2)!}\le \frac{1}{2(n+1)(2n+1)}+\frac{1}{2(n+1)}\le \frac{1}{n+1}$$
since $$\frac{1}{2n+1}\le 1$$
conclusion $$\forall~~n~~a_n\le \frac{1}{n}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the triple integral $\iiint_D \sqrt{x^2+y^2+z^2}\, dV$.
Calculate the triple integral of
$$\iiint_D \sqrt{x^2+y^2+z^2}\, dV$$ where $D$ is bounded by
(1) $x^2+y^2+z^2=2ay$ and (2) $y=\sqrt{x^2+z^2}$.
So far I was thinking that if I made $y$ turn to $z$ and $z$ turn to $y$ I could use cilyndrical coordinates
So (1)$a^2=r^2+(z-a)^2$ and (2) $z=r$ and both intersect at $z=a$
It is correct to propose the integral$$\int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-(z-a)^2}}r\sqrt{z^2+r^2} \,dr\,dz\,d\theta$$?
And if so I evaluate and get $a^3\pi$ but Im suspicious of this result
| Yes! Made $y$ turn to $z$ and $z$ turn to $y$, then $z=\sqrt{x^2+y^2}$ turns one time about $z-$axis. This gives us - with spherical coordinates - $0\leq\varphi\leq2\pi$, also $z=a$ shows $0\leq\theta\leq\dfrac{\pi}{4}$ and with $x^2+y^2+z^2=2az$ we have $\rho\leq2a\cos\theta$, hence
$$\iiint_D \sqrt{x^2+y^2+z^2} dV = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{2a\cos\theta} \rho^3\sin\theta \,d\rho \,d\theta \,d\varphi = \color{blue}{\dfrac{8-\sqrt{2}}{5}\pi a^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
minimum value of a two variable problem Suppose that $x$ and $y$ are in $(−2, 2)$ and $xy = −1$. The minimum value of
$\frac4{4-x^2}+\frac9{9-y^2}$ is ?
So I have tried to manipulate this equation but what I got is a complex equation with square roots in it...
| Let $x^2=\frac{2}{3}a$ and $y^2=\frac{3}{2}b$.
Hence, $ab=1$ and by C-S and AM-GM we obtain:
$$\frac{4}{4-x^2}+\frac{9}{9-y^2}=\frac{4}{4-\frac{2}{3}a}+\frac{9}{9-\frac{3}{2}b}=$$
$$=6\left(\frac{1}{6-a}+\frac{1}{6-b}\right)\geq\frac{6(1+1)^2}{6-a+6-b}=$$
$$=\frac{24}{12-a-b}\geq\frac{24}{12-2}=\frac{12}{5}.$$
The equality occurs for $a=b=1$, which says that $\frac{12}{5}$ is a minimal value.
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Summation of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$
Find the sum of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$ till n terms.
The $i^{th}$ term is given by $\dfrac{i}{1+i^2+i^4}$. I did a similar problem previously but there I was able to decompose the fraction into partial fractions. Here, I am unable to do so. What would be an efficeinent way to solve it then? (Just a hint would suffice)
| You can get the partial fraction decomposition by writing $x^4+x^2+1$ as the product of two quadratics $(Ax^2+Bx+C)(Dx^2+Ex+F)$, and doing some casework to get the variables. Since the coefficient of $x^4$ is $1$, and $1$'s only factor is $1$, then both $A$ and $D$ equal $1$. We then have:
$$(x^2+Bx+C)(x^2+Ex+F)$$
The coefficient of the constant term is $1$, so following the same logic, we get:
$$(x^2+Bx+1)(x^2+Ex+1)$$
Since there is no $x^3$ term, and we know that the only ways of getting a $x^3$ is from $Bx*x^2$ and $x^2*Ex$, then these terms must cancel each other. We can conclude $B = -E$, and we obtain:
$$(x^2+Bx+1)(x^2-Bx+1)$$
$$\Rightarrow (x^4-Bx^3+x^2)+(Bx^3-B^2x+Bx)+(x^2-Bx+1)$$
$$\Rightarrow x^4+(2-B^2)x^2+1$$
Then since $(2-B^2)x^2=x^2$, then $2-B^2=1$, or $B=1$. Then collecting all the variables into the same expression, we get that the factorisation of $x^4+x^2+1$ equals $(x^2+x+1)(x^2-x+1)$. You can multiply this out to verify my answer if you wish to.
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$ I am looking for a characterization of the solutions of
$$a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$$ in positive integers.
| In case you typed the thing correctly, it is integrally equivalent to $x^2 + 4yz.$ Setting this to zero is easy, setting it to a square not too bad.
Alright, this is exactly the comment by Joffan, with $x = a+b-c-d$ and $y=b$ and $z=c.$
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
-1 & 0 & 1 & 0 \\
-1 & 0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 1 & -1 & -1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{rrrr}
1 & 1 & -1 & -1 \\
1 & 1 & 1 & -1 \\
-1 & 1 & 1 & 1 \\
-1 & -1 & 1 & 1
\end{array}
\right)
$$
Remembered how to do this. Parametrization for $x^2 + 4yz = n^2$ by
$$ y = eg, $$
$$ z = fg, $$
$$ \frac{n+x}{2} = eh, $$
$$ \frac{n-x}{2} = fh. $$
To get something primitive, we need
$$ \gcd(e,f) = 1, $$
$$ \gcd(g,h) = 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove that $\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^1}=0$ I need prove that
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0$$
Can I use it?
if $\sqrt{x^2+y^2}< \delta$ then $|\frac{x^2y-xy^2}{x^2+y^2}|<\epsilon$
$$\left |\frac{x^2y-xy^2}{x^2+y^2} \right |=\frac{x^2 \left |y \right | -\left |x \right |y^2}{x^2+y^2}<\frac{x^2 \sqrt{x^2+y^2} -y^2\sqrt{x^2+y^2}}{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2-y^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2-y^2) }{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2) }{x^2+y^2}=\sqrt{x^2+y^2}<\delta=\epsilon$$
I use that $x^2<x^2+y^2$ twice, can I do it? or some identity to use when you have $x^2-y^2$
| The first inequality you're using has a flaw, as it is : $ |xy|\leqslant x^2+y^2$ and after that you can continue on.
In my answer,I assume that you have made a typo and in the denominator you have $y^2$, since I see you using it in the rest of your attempt.
Showing you some approaches here :
Re-writing the Limit using polar coordinates $x = r\sin\theta,y=r\cos\theta$, we get :
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0 \Rightarrow \lim_{r\to0} \frac{(r\cos\theta)^2(r\sin\theta)-(r\sin\theta)^2(r\cos\theta)}{r^2} \Rightarrow \lim_{r\to0} \frac{r^3\cos^2\theta\sin\theta-r^3\sin^2\theta\cos\theta}{r^2} \Rightarrow \lim_{r\to0} r(\cos^2(\theta)\sin(\theta)-\sin^2(\theta)\cos(\theta)) =0$$
In this case, the value of $\theta$ is insignificant.
In another approach, you have $\forall (x,y)\neq (0,0)$:
$$\left|\frac{xy}{x^2+y^2}\right|\leq 1 $$
So :
$$\left|\frac{xy(x-y)}{x^2+y^2}\right|\leq |x-y|\leq |x|+|y|\leq 2\sqrt{x^2+y^2}.$$
Now, for any $ε>0$ defining it as $δ=\frac{ε}{2}$, we have :
$$\sqrt{x^2+y^2}<\delta \Rightarrow \left|\frac{xy(x-y)}{x^2+y^2}\right|<\epsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form $\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx=12\log A-\frac{4}{3}\log 2$ Evaluate
$$\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx$$
I haven't been able to find references to the indefinite integral of the $\tanh$ term except for some similar forms that had solutions. see here and here.
Edit:
Following Random Variable's result we have the form
$$12\log A-\frac{4}{3}\log 2$$
| Let $$I(a,b) = \int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{e^{-ax}}{x} \, dx, $$ where $a, b >0$.
I will show that $$I(a,b) = 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)}\left(\frac{a}{4} \right)-a \log(a)+a + 2 a \log (2)+ \log(a+b)- 2 \log(4 \pi),$$ where $\psi^{(-2)}(x)$ is the polygamma function of order $-2$ defined by the integral $$ \psi^{(-2)}(x) = \int_{0}^{x} \log \Gamma (t) \, dt.$$
If we let $a \to 0^{+}$, we get $$\int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{dx}{x} = 8 \psi^{(-2)} \left(\frac{1}{2} \right)+ \log(b) - 2 \log(4\pi). $$
As explained in pisco's answer, $\psi^{(-2)} \left(\frac{1}{2} \right)$ can be expressed in terms of the Glaisher-Kinkelin constant.
Differentiating $I(a,b)$ under the integral sign with respect to $a$ (which is permissible for $a \ge c$, where $c$ is some positive value), we get $$ \frac{\partial}{\partial a}I(a,b) = -\int_{0}^{\infty}\left(\frac{\tanh (x)}{x}-e^{-bx} \right) e^{-ax} \, dx = -\int_{0}^{\infty} \frac{\tanh (x)}{x} \, e^{-ax} \, dx + \frac{1}{a+b}.$$
And using a property of the Laplace transform, we get
$$\begin{align}\int_{0}^{\infty} \frac{\tanh (x)}{x} \, e^{-ax} \, dx &= \int_{a}^{\infty} \int_{0}^{\infty} \tanh (x) e^{-px} \, dx \, dp \\ &=\int_{a}^{\infty} \int_{0}^{\infty} \frac{1-e^{-2x}}{1+e^{-2x}} \, e^{-px} \, dx \, dp \\ &= \int_{a}^{\infty} \int_{0}^{\infty} \left(2\sum_{n=0}^{\infty} (-1)^{n} e^{-2nx} -1\right) e^{-px} \, dx \, dp \\ &= \int_{a}^{\infty} \left( 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+p}- \frac{1}{p}\right) \, dp \\ &= \int_{a}^{\infty} \left(\frac{2}{p} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{2}{p}n +1} - \frac{1}{p}\right) \, dp \\ &= \int_{a}^{\infty} \left(\frac{1}{2} \left(\psi \left(\frac{p+2}{4} \right)- \psi\left(\frac{p}{4} \right)\right)- \frac{1}{p} \right) \, dp \tag{1} \\ &= -2 \log (2) -2\log \Gamma \left(\frac{a+2}{4} \right) +2 \log \Gamma \left(\frac{a}{4} \right) + \log (a). \tag{2} \end{align}$$
Therefore, $$I(a,b) =2a \log(2) + 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)} \left(\frac{a}{4} \right)- a \log(a) +a + \log(a+b) +C. $$
To determine the integration constant $C$, we can take the limit on both sides of the above equation as $a \to +\infty$.
Since $\psi^{(-2)}(x) $ can be expressed in terms of the Barnes G-function, we can use the asymptotic expansion of the Barnes G-function.
(Term-by-term integration of Stirling's formula for the log gamma function does lead to the same expansion but with an unknown constant.)
The integration constant $C$ turns out to be $-2 \log(4 \pi)$.
$(1)$ https://mathworld.wolfram.com/DigammaFunction.html (6)
$(2)$ Using Stirling's formula, we have
$$\log \Gamma\left(\frac{a+2}{4} \right) \sim \left(\frac{a+2}{4} - \frac{1}{2} \right) \log \left(\frac{a+2}{4} \right) - \frac{a+2}{4} + \frac{1}{2} \log(2 \pi) + \mathcal{O}\left(\frac{1}{a} \right), $$ where
$$\log \left(\frac{a+2}{4} \right)= \log \left(\frac{a}{4} \right) + \log \left(1+ \frac{2}{a} \right) \sim \log \left(\frac{a}{4} \right) + \frac{2}{a} + \mathcal{O}\left(\frac{1}{a^{2}} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forget the proof:
$$ (x^2 + y^2)^2 \geq 0 \text{ so that }x^4 + y^4 \geq 2 x^2 y^2 \text{ and }f(x,y,z) \geq - 4 x^2 y^2 \geq - 4 $$
I could try other rearrangements as well. This one gives me an upper bound of $3$.
$$ x^4 - 6x^2 y^2 + y^4 \leq x^4 + 6x^2 y^2 + y^4
= (x^2 + y^2)^2 + 4x^2 y^2 \leq 3\, \big( x^2 + y^2 \big)^2 \leq
3\, \big( x^2 + y^2 + z^2 \big)^2 = 3$$
If I use some real analysis we know that the sphere as a subset of Euclidean space is compact, so that:
$$ -\infty < -4 \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) < 3 < +\infty$$
I'm trying to avoid Lagrange multipliers unless the're really natural here. Observer also that:
$$ \left[ x^2 + y^2 + z^2 = 1 \right] \to \left[ x^2 + y^2 \leq 1\right] $$
as the original problem was defined on the unit sphere but the $z$ is extraneous.
They might not be extraneous we could set spherical coordinates:
$$ (x,y,z ) = \big(\cos \theta \, \cos \varphi, \;\cos \theta \sin \varphi, \;\cos \varphi\big)$$
and we could put into our inequality:
\begin{eqnarray*} x^4 - 6x^2 y^2 + y^4 &=& \cos^4 \theta \cos^4 \varphi - 6 \cos^4 \theta \sin^2 \varphi \cos^2 \varphi+ \cos^4 \theta \sin^4 \varphi \\ \\
&=& \cos^4 \theta \,\big( \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi \big) \\ \\
&\leq & \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
\end{eqnarray*}
This looks promising as I have reduced a three-dimensional problem to a problem with only an angle $\varphi$, but I may have lost something with the final "$\leq$" sign.
Just a tiny bit more:
$$
\cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
= (\cos^2 \varphi - \sin^2 \varphi)^2 - 4 \cos^2 \varphi \sin^2 \varphi
= \cos^2 2\varphi - \sin^2 2\varphi
$$
and if we use the double-angle identity.
$$ 1 \geq \cos^2 2\varphi - \sin^2 2\varphi
= \cos^2 2\varphi - (1 -\cos^2 2\varphi)
= 2\, \cos^2 2\varphi - 1 \geq - 1$$
This is very similar to what I obtained before.
| Another approach would be to use cylindrical coordinates. Then your expression is $r^4\cdot g(\theta)$ with $0\leq r\leq1$, so understanding $g$'s extrema, together with extremal values of $0$ and $1$ for $r^4$ would tell you the extreme values of $f$. Specifically,
$$\begin{align}
g(\theta)&=\cos^4(\theta)-6\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)\\
&=V^2-6V(1-V)+(1-V)^2\\
&=8V^2-8V+1\\
&=2(2V-1)^2-1
\end{align}$$
where $V=\cos^2(\theta)$ ranges from $0$ to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inductive proof of sum of falling factorials with Stirling number coefficients Concrete Mathematics (page 262 2nd ed.) demonstrates $x^n = \sum_k \left\{ \begin{matrix} n \\ k \\ \end{matrix} \right\} x^{\underline{k}}$ using a proof by induction:
$$
\begin{align}
x \sum_k \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} x^{\underline{k}} &=
\sum_k \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} x^{\underline{k+1}} +
\sum_k \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} kx^{\underline{k}}
\tag{1}
\\
&=
\sum_k \left\{ \begin{matrix} n-1 \\ k-1 \\ \end{matrix} \right\} x^{\underline{k}} +
\sum_k \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} kx^{\underline{k}}
\tag{2}
\\
&= \sum_k
\left(
k \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} + \left\{ \begin{matrix} n-1 \\ k-1 \\ \end{matrix} \right\}
\right) x^{\underline{k}}
= \sum_k \left\{ \begin{matrix} n \\ k \\ \end{matrix} \right\} x^{\underline{k}}
\tag{3}
\end{align}
$$
(1) is due to $ x \cdot x^{\underline{k}} = x^{\underline{k+1}} + kx^{\underline{k}}$.
(3) is an application of the recurrence relation for Stirling numbers of the second kind.
How do we get (2) from (1)?
| It is important to make the summation ranges explicit ( to be sure (to be sure))
\begin{eqnarray*}
x \sum_{k=1}^{n-1} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} x^{\underline{k}} &=&
\sum_{k=1}^{n-1} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} x^{\underline{k+1}} +
\sum_{k=1}^{n-1} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} kx^{\underline{k}}
\tag{1}
\\
&=&
\sum_{k'=2}^{n} \left\{ \begin{matrix} n-1 \\ k'-1 \\ \end{matrix} \right\} x^{\underline{k'}} +
\sum_{k=1}^{n-1} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} kx^{\underline{k}}
\tag{2a}
\\
&=&
\sum_{k=1}^{n} \left\{ \begin{matrix} n-1 \\ k-1 \\ \end{matrix} \right\} x^{\underline{k}} +
\sum_{k=1}^{n} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} kx^{\underline{k}}
\tag{2b}
\\
&=& \sum_{k=1}^{n}
\left(
k \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} + \left\{ \begin{matrix} n-1 \\ k-1 \\ \end{matrix} \right\}
\right) x^{\underline{k}}
= \sum_{k=1}^{n} \left\{ \begin{matrix} n \\ k \\ \end{matrix} \right\} x^{\underline{k}}
\tag{3}
\end{eqnarray*}
To go from $(1)$ to $(2a)$ a change of variable occur $k'=k+1$ in the first sum.
To go from $(2a)$ to $(2b)$ note that $\left\{ \begin{matrix} n-1 \\ 0 \\ \end{matrix} \right\}= 0 $ and $\left\{ \begin{matrix} n-1 \\ n \\ \end{matrix} \right\}= 0 $ so the summation ranges can be extended.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $a$, $b$ and $n$ are positive integers such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{n}$ then what is the total number of pair of $(a,b)$?
If $a$, $b$ and $n$ are positive integers such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{n}$ then what is the total number of pair of $(a,b)$? What if $a$, $b$ and $n$ are not necessarily positive integers?
My attempt:- If $n$ is small, like
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{5}$$
I just convert it to a suitable form like this
$$5a + 5b = ab$$
After this, I just try by hit and trial.
But I need a suitable method to solve these type of questions. I always get stuck on these for a long time.
| The answer is infinitely many pairs.
Let $a = b$. Then,
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{a} + \frac{1}{a} = \frac{2}{a}$$
From which we get $n = \frac{a}{2}$. This gives $a = b = 2n$ as sufficient to fulfill the equation. Since the last equation is itself fulfilled by infinitely many different values of $a$, it follows that there are infinitely many pairs of $(a,b)$, even if $a$, $b$ and $n$ are not necessarily positive, and we haven't even looked at the cases where $a \neq b$.
As for your other question, I can't really help as I came across this by chance. I haven't taken a course on this topic so it just looks like trickery to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Implicit differentiation question with a ladder and a house. A $13$ ft ladder is leaning against a house when its base starts to slide away from the house, at the instant when the base is $12$ ft from the house, the base is moving at a rate of $5$ ft/sec.
Question:
Let $ \theta $ be the angle between the ladder and the ground. at what rate is $\theta$ changing when the base is $12$ ft form the house?
I said $ \tan (\theta) = y x^{-1} $ then used implicit differentiation to take the derivative with respect to time.
\begin{align*}
\frac{d \theta}{dt} \sec^{2} (\theta) &= \frac {dy}{dt}\left( \frac {dx}{dt} -x^{-2} \right)\\
\implies\frac{d \theta}{dt} &= \cos^{2}(\theta) \left(\frac {dy}{dt} \left( \frac {dx}{dt} -x^{-2}\right) \right)
\end{align*}
Does this make sense or did I mess something up?
| With $\theta = \arctan \frac{y}{x}$ we get
$$d \theta =d \arctan \frac{y}{x} = \frac{ d (\frac{y}{x})}{1+ (\frac{y}{x})^2}= \frac{ x \cdot d y - y \cdot d x}{x^2 + y^2}$$
Now use $x^2 + y^2 = l^2= $ const. and get
$$ x dx + y d y = 0$$
Substituting from this $d y = -\frac{x}{y} dx$ into the first equality we get
$$ d \theta = \frac{ - \frac{x^2}{y} dx - y dx}{x^2 + y^2} = -\frac{d x}{y}$$
Note that this is the result we would get if we looked at the angle formed with the wall, approximated the angle with the tangent, and assumed that only the bottom part is moving.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a,b,c$ are rational roots for $x^3+a x^2+b x+c=0$, find the value of a, b, c
Given: $a,b,c$ are rational roots for the equation $x^3+a x^2+b x+c=0$
Find: all possible values of $a,b,c$.
Question used within the preparation for an entrance exam.
My attempt: from Vieta's formulas, we know that (1) $a+b+c=-a$, (2) $ab+bc+ac=b$ and (3) $abc=-c$. Developing this last equation, we get $c(ab+1)=0$. Let's assume that $c=0$. Replacing $c=0$ in (2) we get $ab=b$, or $a=1$. Replacing $a=1$, $c=0$ in (1) we get $b=-2$. Therefore $a=1,b=-2,c=0$ is a possible set of values. By inspection $a=b=c=0$ is another set.
Questions: are there other (rational) values for $a,b,c$ beyond these 2 ones I found? Are there other approaches to address this question?
| The way the question is phrased, it doesn't say that $a$, $b$, $c$ have to be all the roots of the equation (or in case some of them are equal, as in $(0,0,0)$, that they will list the roots with the correct multiplicities). In other words, Viet doesn't have to apply as $abc = -c$, it could be $aac = -c$ (if, for example, $a$ is a double root, while $b = c$ is a single root), or even $abd = -c$, where $d$ is another root (not one of $a$, $b$, $c$) while $a = c \neq b$ are two single roots.
Because of this, when starting with Viet, you might miss a solution. In fact you both (bluemaster and GAVD) have: it's $(-1, -1, 1)$. The roots are then $\{-1, 1\}$ but the double root is the $1$, not the $-1$. That's why the Viet approach misses it.
A more general approach is to use the theorem about rational roots to first narrow things down, then use the definition of what a root is to exhaust the possibilities. The theorem says that a rational root must be of the form $p/q$, where $p$ is a divisor of the constant term, while $q$ is a divisor oh the leading coefficient. In our case, this means $q = 1$ (so $a$, $b$, $c$ are all integers), and $p$, hence each of $a$, $b$, $c$, is a divisor of $c$.
You can now distinguish two cases:
*
*If $c=0$, after factoring out an $x$ we get a quadratic $x^2 + ax + b$ which must have both $a$ and $b$ as roots. Thus $2a^2 + b = 0$ and $b^2 + ab + b = 0$ which easily gives two integer solutions, $(0, 0)$ and $(1, -2)$.
*If $c \neq 0$, then since $c$ is a root, $c^3 + ac^2 + (b+1)c = 0$. The first two terms are divisible by $c^2$, so also must be the third one. This means $c$ is a divisor of $b+1$, and thus $b$ (which is a divisor of $c$) divides $b+1$. This is only possible if $b = 1$ or $b = -1$. Now consider two subcases, proceeding as in the previous case to make simultaneous equations for $a$ and $c$. You get the other two possibilities, $(1, -1, -1)$ and $(-1, -1, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Compute $\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$ NOTE 1: L'Hospitals and Taylor expansions are not allowed.
NOTE 2: I really appreciate if someone would correct my attempt, however any other easier method only involving single variable calculus (excluding the concepts in NOTE 1) are welcome.
PROBLEM: Compute $$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$$
I'll just manipulate without writing out the limit, for now. I have
\begin{array}{lcl}
\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n} & = & \left( 1+\frac{1}{n}\right)^{n^2}\cdot e^{-n} \\
& = & \exp\left( n^2\ln\left(1+\frac{1}{n}\right)-n\right) \\
& = & \exp((n\ln(1+\frac{1}{n})-1)n) \\
\end{array}
And proceeding:
\begin{array}{lcl}
\exp((n\ln(1+\frac{1}{n})-1)n) & = & e^{((n\ln(1+\frac{1}{n})-1)n} \\
& = & (e^{((n\ln(1+\frac{1}{n})-1)})^n \\
& = & \left(\frac{(e^{((n\ln(1+\frac{1}{n})-1)}-1+1)}{(n\ln(1+\frac{1}{n})-1}\cdot{((n\ln(1+\frac{1}{n})-1})\right)^n\\
\end{array}
It gets quite ugly very quickly as you can see. I'm trying to rewrite it so I can apply standard limits like $$\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.$$
| Let
$$a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \quad ;\quad b=\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}$$
We've;
\begin{align}
c= \frac ab
=\frac{\displaystyle\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2} {e^n} }
=\lim_{n \to \infty}\frac{\displaystyle \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle \left(1-\frac 1n\right)^{n^2} {e^n} }
&=\lim_{n \to \infty}\left(\frac {n+1}{n-1}\right)^{n^2} \frac{1}{e^{2n}}
\end{align}
Now, let $n=2m$. Since $n \to \infty$, $m \to \infty$ too.
\begin{align}
c&= \lim_{n \to \infty}\left(\frac {n+1}{n-1}\right)^{n^2} \frac{1}{e^{2n}}\\
&=\lim_{m \to \infty}\left(\frac {2m+1}{2m-1}\right)^{(2m)^2} \frac{1}{e^{2(2m)}}\\
&=\lim_{m \to \infty}\left(1+\frac {2}{2m-1}\right)^{4m^2} \frac{1}{e^{4m}}\\
\end{align}
Let $m-\frac 12=p$.
\begin{align}
c&=\lim_{m \to \infty}\left(1+\frac {2}{2m-1}\right)^{4m^2} \frac{1}{e^{4m}}\\
&=\lim_{p \to \infty}\left(1+\frac {1}{p}\right)^{4p^2+4p+1} \frac{1}{e^{4p+2}}\\
&=\lim_{p \to \infty}\left(1+\frac {1}{p}\right)^{4p^2} \left(1+\frac {1}{p}\right)^{4p}\left(1+\frac {1}{p}\right)\frac{1}{e^{4p} \cdot e^2}\\
&=\left(\left(1+\frac {1}{p}\right)^{p^2}\frac {1}{e^p}\right)^4 \left(1+\frac 1p\right)^{4p} \frac{1}{e^2}\\
&=a^4 \cdot e^4 \cdot \frac{1}{e^2}\\
&=a^4 e^2\\
\end{align}
Thus we have $\dfrac ab=a^4 e^2$ $$ \color{blue}{\implies a^3b=\frac{1}{e^2}} \tag 1$$
Now, we also have
\begin{align}
ab=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \cdot \lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}&=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n}\cdot \left(1-\frac 1n\right)^{n^2}{e^n}\\
&=\lim_{n \to \infty}\left(1-\frac {1}{n^2}\right)^{n^2}\\
&=\frac{1}{e}
\end{align}
Hence,
$$\color{red}{\implies ab=\frac{1}{e}} \tag 2$$
Using $(1)$ and $(2)$ we finally have
$$\bbox[5px,border:2px solid #6b2fed]{a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} =\frac{1}{\sqrt e}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.
| If
$a_n > 0,
s_n = \sum_{k=1}^n a_n,
r_n = s_{n+1}/s_n$,
then,
if $a_{n+1}/a_n \to 1$,
$r_n \to 1$.
Proof:
$r_n-1
=a_{n+1}/s_n
$.
Since
$a_{n+1}/a_n \to 1$,
then,
for any fixed $m$,
$a_{n+k}/a_n \to 1$
for $1 \le k \le m$.
Therefore,
for any fixed $m$,
$\dfrac{a_{n+1}}{s_n}
\lt \dfrac{a_{n+1}}{\sum_{k=0}^{m-1} a_{n-k}}
\to \dfrac1{m}
$
so that
$r_n-1
=\dfrac{a_{n+1}}{s_n}
\to 0
$.
Your case is
$a_n = n^4$,
and it is easy to show that
$a_{n+1}/a_n \to 0$.
This holds for
$a_n = n^p$
for any fixed $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Integrate $f(x)=\frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}$, for $x>-1$ Substituting $t^2=x+1$ I get $2t \ dt=dx.$ So therefore $$\int\frac{2t(1+t)}{1-t}dt=-2\int\frac{t^2+t}{t-1}dt.$$
Dividing the polynomials I get $$-2\int t+2+\frac{2}{t-1}dt=-2\int t \ dt-4\int1 \ dt-4\int\frac{1}{t-1}dt.$$
First integral equals $-t^2+C_1,$ second is $4t+C_2$ and the third is $4\ln{|t-1|+C_3},$ thus $$F(x)=-(x+1)-4\sqrt{x+1}-4\ln{|\sqrt{x+1}-1|+C}.$$
The answer is correct, except for the $-1$ in the beginning, it should not be there but I can't see why it is.
| Note that
$$
\frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}
=
\frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}\cdot \frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}
=
\frac{1+2\sqrt{x+1}+(x+1)}{1-(x+1)}
\\
=
-\frac{2+2\sqrt{x+1}+ x}{x}
=
-\frac{2}{x}
-\frac{2\sqrt{x+1}}{x}
-1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$ How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$
Can anyone help me,I really appreciate it.
| It is evident that
$$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$
and can be evaluated by following the pattern:
Consider the series
$$S(x) = \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)}$$
which upon differentiation leads to $S(0) = 0$, $S'(0) = 0$,
\begin{align}
S(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)} \\
S'(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+1}}{(2n+1)} \\
S''(x) &= \sum_{n=1}^{\infty} H_{n} \, x^{2n} = - \frac{\ln(1- x^2)}{1-x^2}.
\end{align}
Now,
$$ - 2 \, S''(x) = \frac{\ln(1-x)}{1-x} + \frac{\ln(1-x)}{1+x} + \frac{\ln(1+x)}{1-x} + \frac{\ln(1+x)}{1+x}$$
which, upon integration, leads to
\begin{align}
- 4 \, S'(x) &= \ln^{2}(1 + x) - \ln^{2}(1-x) + 2 \, Li_{2}\left(\frac{1-x}{2}\right) - 2 \, Li_{2}\left(\frac{1+x}{2}\right) + \ln4 \, \ln\left(\frac{1+x}{1-x}\right).
\end{align}
Integrating again leads to $S(x)$. The integrals
\begin{align}
\int_{0}^{x} \ln^{2}(1-t) \, dt &= (x-1) \, (\ln^{2}(1-x) - 2 \ln(1-x) + 2) + 2 \\
\int_{0}^{x} \ln^{2}(1+t) \, dt &= (x+1) \, (\ln^{2}(1+x) - 2 \ln(1+x) + 2) - 2 \\
\int_{0}^{x} \ln\left(\frac{1+t}{1-t}\right) \, dt &= x \, \ln\left(\frac{1+x}{1-x}\right) + \ln(1-x^2) \\
\int_{0}^{x} Li_{2}\left(\frac{1+t}{2}\right) \, dt &= (1+x) \, Li_{2}\left(\frac{1+x}{2}\right) + x \, \ln\left(\frac{1-x}{2}\right) - \ln(1-x) -x - Li_{2}\left(\frac{1}{2}\right) \\
\int_{0}^{x} Li_{2}\left(\frac{1-t}{2}\right) \, dt &= (x-1) \, Li_{2}\left(\frac{1-x}{2}\right) + (x+1) \, \ln\left(\frac{1+x}{2}\right) -x - Li_{2}\left(\frac{1}{2}\right) + \ln2
\end{align}
are needed for the evaluation. Once $S(x)$ is determined set $x=1$ to obtain
$$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
Evaluate $\lim_{n\rightarrow \infty} \Gamma(n+\frac{1}{2})/ \left( \sqrt{2n\pi} \Gamma(n) \right)$ using Stirling's formula. I am working on the limit
$$
\displaystyle\lim_{n\rightarrow \infty} \frac{\Gamma(n+\frac{1}{2})}{ \sqrt{2n\pi}\, \Gamma(n)}\,.
$$
I am thinking I may be able to use Stirling's formula, but they are slightly different, and I am having trouble relating the two. Any help is appreciated.
Stirling's formula says that the limit is 1 as $n$ approaches infinity of the following:
$$\Gamma(n) / ( \sqrt{2\pi} n^{n - \frac{1}{2}}e^{-n})$$
In particular, how do I relate $\Gamma(n)$ to $n^{n}$ and $e^{-n}$? Not sure how do deal with those two terms.
| By approaching this in a more general way, first we establish the following lemma.
Lemma: Let $p,q\in\mathbb{R}^{+}$. Then,
$$
\lim_{p\rightarrow \infty} \frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}}
= \frac{1}{2^{q/2}}\,.
$$
Proof: It is easy to verify using standard calculus that the following claim holds.
Claim: For $a,b,c\in\mathbb{R}$,
$$
\lim_{x\rightarrow \infty} \left(1+\frac{a}{x}\right)^{bx+c} = e^{ab}\,.
$$
Now, knowing that with Stirling's formula
$ \Gamma(n) \approx \sqrt{2\pi}(n-1)^{n-\frac{1}{2}}e^{-n}$, it follows that
\begin{align*}
\frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}}
&\approx
\frac{\sqrt{2\pi} \left(\frac{p+q}{2}-1\right)^{\frac{p+q}{2}-\frac{1}{2}}e^{-\frac{p+q}{2}}}
{\sqrt{2\pi} \left(\frac{p}{2}-1\right)^{\frac{p}{2}-\frac{1}{2}}e^{-\frac{p}{2}}\,p^{q/2}} \\
&= \frac{2^{-\frac{p+q}{2}+\frac{1}{2}}\, p^{\frac{p+q}{2}-\frac{1}{2}}
\left(1+\frac{q-2}{p}\right)^{\frac{p+q}{2}-\frac{1}{2}}e^{-\frac{p+q}{2}}}
{2^{-\frac{p}{2}+\frac{1}{2}}\, p^{\frac{p}{2}-\frac{1}{2}}
\left(1+\frac{-2}{p}\right)^{\frac{p}{2}-\frac{1}{2}}e^{-\frac{p}{2}}\,p^{q/2}}\\
&= \frac{1}{2^{q/2}}\cdot
\frac{\left(1+\frac{q-2}{p}\right)^{\frac{p+q}{2}-\frac{1}{2}}}{\left(1+\frac{-2}{p}\right)^{\frac{p}{2}-\frac{1}{2}}}\cdot
e^{-\frac{q}{2}}\,.
\end{align*}
Taking the limit both sides knowing that both converge to the same value, it easily follows with our claim that
\begin{align*}
\lim_{p\rightarrow \infty}
\frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}}
=& \lim_{p\rightarrow \infty} \frac{1}{2^{q/2}}\cdot
\frac{\left(1+\frac{q-2}{p}\right)^{\frac{1}{2}p + \frac{q}{2}-\frac{1}{2}}}{\left(1+\frac{-2}{p}\right)^{\frac{1}{2}p-\frac{1}{2}}}\cdot
e^{-\frac{q}{2}} \\
=&\, \frac{1}{2^{q/2}}\cdot \frac{e^{\frac{1}{2}(q-2)}}{e^{\frac{1}{2}(-2)}}\cdot e^{-\frac{q}{2}} \\
=&\, \frac{1}{2^{q/2}}
\end{align*}
This proves our lemma.
THE MOMENT YOU'VE BEEN WAITING FOR:
As a special case, when $p=2n$ and $q=1$
$$
\lim_{n\rightarrow \infty} \frac{\Gamma\left(n + \frac{1}{2}\right)}{\Gamma\left(n\right)\sqrt{2n}}
= \frac{1}{\sqrt{2}}\,.
$$
Therefore, it immediately follows from the limit of a constant times a function that
$$
\lim_{n\rightarrow \infty} \frac{\Gamma\left(n + \frac{1}{2}\right)}{\Gamma\left(n\right)\sqrt{2n\pi}}
= \frac{1}{\sqrt{2\pi}}\,.
$$
| {
"language": "en",
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"answer_count": 2,
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} |
Prove the following expression Prove that if $a,b,c$ are not equal:
$$\dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b=x$$
My Attempt:
Proving $ \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b=x$ is equivalent to proving
$$ \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x=0$$
i.e $f(x)=0$
where,
$$f(x)= \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x$$
..
| Let $f(x) = \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x = 0$
Note that $f(x)$ seems to be a quadratic equation in $x$ and can have no more than two roots. But $f(a) = f(b) = f(c) = 0$ (as Lord Shark stated in comment), meaning that $a,b,c$ are three roots. Since a quadratic can not have more than two roots, this is an identity in $x$, valid for all $x \in R$.
| {
"language": "en",
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"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the polynomial $x^8 -x^7+x^2-x+15$ has no real root I am not getting on how to approach this problem. Clearly, this polynomial can have atleast 2 real roots. And using Descartes's rule of signs, it can have a maximum of 4 positive real roots. But after that, how should I proceed ?
Any help would be highly appreciated...
| We have that $x^8-x^7$ and $x^2-x$ are bigger then zero for $x<0$ because $x^8>0$ and $-x^7>0$ similar logic for $x^2$ and $-x$ so $p(x)>15$ for $x<0$.
For $0<x<1$ we have that $x^8-x^7+x^2-x+15>-x^7-x+15>-1-1+15>13$ so $p(x)>13$
For $x\geq1$ we have that $x^8\geq x^7$ and $x^2\geq x$ so $x^8-x^7+x^2-x+15\geq 15$ so $p(x)\geq 15$.
Since $x=0$ is not a root we can conclude there aren't any roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
We play with $3$ dice. Which sum of pips on the dice is more probable, $10$ or $13$? (Question isn't as long as it looks.)
We play with $3$ dice. Which sum of pips on the dice is more
probable, $10$ or $13$?
Without doing any calculation, I would say that it is more probable to have $10$ as sum of pips because we only have numbers between $1$ and $6$ and because $10$ is lower than $13$, it might be easier to get it. I don't know if this makes sense but that's the very first thing that came to my mind. To not just make a wild guess, I tried to calculate it.
Did I do it correctly and is there a faster / better way of doing it? :)
Here are the combinations to get $10$ as sum ($25$ in total):
$$(1,4,5)$$ $$(1,5,4)$$ $$(1,6,3)$$ $$(2,2,6)$$ $$(2,3,5)$$ $$(2,4,4)$$ $$(2,5,3)$$ $$(2,6,2)$$ $$(3,1,6)$$ $$(3,2,5)$$ $$(3,3,4)$$ $$(3,4,3)$$ $$(3,5,2)$$ $$(3,6,1)$$ $$(4,1,5)$$ $$(4,2,4)$$ $$(4,3,3)$$ $$(4,4,2)$$ $$(4,5,1)$$ $$(5,1,4)$$ $$(5,2,3)$$ $$(5,3,2)$$ $$(6,1,3)$$ $$(6,2,2)$$ $$(6,3,1)$$
Combinations for sum $13$ (in total $21$):
$$(1,6,6)$$ $$(2,5,6)$$ $$(2,6,5)$$ $$(3,4,6)$$ $$(3,5,5)$$ $$(3,6,4)$$ $$(4,3,6)$$ $$(4,4,5)$$ $$(4,5,4)$$ $$(4,6,3)$$ $$(5,2,6)$$ $$(5,3,5)$$ $$(5,4,4)$$ $$(5,5,3)$$ $$(5,6,2)$$ $$(6,1,6)$$ $$(6,2,5)$$ $$(6,3,4)$$ $$(6,4,3)$$ $$(6,5,2)$$ $$(6,6,1)$$
So for sum $10$ we have probability of $p=\frac{25}{6^3} \approx 11.6$ %
For sum $13$ we have probability of $p=\frac{21}{6^3} \approx 9.72$ %
Thus, it is indeed more probable to get sum $10$.
| By the stars-and-bars formula, if $n,r$ are positive integers, the equation
$$x_1 + \cdots + x_n = r$$
has exactly ${\large{\binom{r-1}{n-1}}}$ solutions in positive integers $x_1,...,x_n$.
Let the values of the dice rolls be represented as an ordered triple $(d_1,d_2,d_3)$.
First count the number of triples $(d_1,d_2,d_3)$ for the case $d_1+d_2+d_3=10$ . . .
If we temporarily ignore the condition $d_1,d_2,d_3 \le 6$, the stars-and-bars formula for the equation
$$d_1+d_2+d_3=10$$
would yield ${\large{\binom{10-1}{3-1}}} = {\large{\binom{9}{2}}}$ positive integer triples $(d_1,d_2,d_3)$.
We need to subtract the count of the triples where at least one of $d_1,d_2,d_3$ exceeds $6$.
Given that $d_1+d_2+d_3=10$, at most one of $d_1,d_2,d_3$ exceeds $6$.
Consider the case $d_3 > 6$.
The number of positive integer triples $(d_1,d_2,d_3)$ such that $$d_1+d_2+d_3=10\;\;\text{and}\;\;d_3 > 6$$
is the same as the number of positive integer pairs $(d_1,d_2)$ such that $$d_1+d_2<4$$
which is the same, using $x_3$ as a dummy variable, as the number of positive integer triples $(d_1,d_2,x_3)$ such that
$$d_1+d_2+x_3=4$$
which, by the stars-and-bars formula, has ${\large{\binom{4-1}{3-1}}} = {\large{\binom{3}{2}}}$ solutions.
Hence the corrected count is
$${\small{\binom{9}{2}}} - {\small{\binom{3}{1}}}{\small{\binom{3}{2}}} = 27$$
where the factor ${\large{\binom{3}{1}}}$ accounts for the choice of which of $d_1,d_2,d_3$ exceeds $6$.
By analogous reasoning, the number of triples $(d_1,d_2,d_3)$ for the case $d_1+d_2+d_3=13$ is
$${\small{\binom{12}{2}}} - {\small{\binom{3}{1}}}{\small{\binom{6}{2}}} = 21$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Maximizing a polynomial expression involving the sidelengths of an inscribed pentagon
Given that $ABCDE$ is a convex pentagon and is inscribed in a circle of radius $1$ unit with $AE$ as diameter. If $AB=a,BC=b ,CD=c$ and $DE=d$, then maximum possible integral value of $a^2+b^2+c^2+d^2+abc+bcd$ is ?
What I tried
$AE =2$ , I joined points to form lines $BE,AC,CE$ and $AD$ and tried to use the fact that angle subtended by diameter would be a right angle.
However, I was unable to solve the problem.
I was looking for some hint or an alternate approach to the problem.
| We will use the fact: If a bunch of circular chords can be packed to "fit" a half-circle, so do any rearrangement of them.
Let $\mathcal{E} = a^2+b^2+c^2+d^2 + bc(a+d)$ be the expression at hand.
Choose points $P, Q, R$ on the half-circle such that
$AP = b$, $PQ = c$, $QR = a$, $RE = d$.
Let $u = AQ$, $v = QE$ and $\angle QEA = \theta$. When either $u$ or $v$ vanishes, $bc(a+d)$ also vanishes. It is easy to see $\mathcal{E}$ trivially reduces to $4$. This means we only need to consider the case $u, v \ne 0$.
Since $AE$ is a diameter of the half circle, $\angle AQE = \frac{\pi}{2}$. This leads to $u^2 + v^2 = 4$ and $v = 2\cos\theta$. The condition $v \ne 0$ implies $u \ne 2$.
Since $\angle APQ$ and $\angle QEA$ are opposite angles of cyclic quadrilateral $APQE$, these two angles sum to $\pi$. Apply cosine rule to triangle $APQ$, we find
$$\frac{b^2 + c^2 - u^2}{2bc} = \cos(\pi - \theta) = -\cos\theta = -\frac{v}{2}$$
This leads to
$$b^2 + c^2 + vbc = u^2\tag{*1a}$$
By a similar argument, we have
$$a^2 + d^2 + uad = v^2\tag{*1b}$$
Use these, we can rewrite the expression at hand as
$$\begin{align}
\mathcal{E} &= (a^2+ d^2) + (b^2+c^2) + bc(a+d)\\
&= (v^2 - uad) + (u^2-vbc) + bc(a+d)\\
&= 4 + bc(a+d-v) - uad\tag{*2}
\end{align}$$
To bound $\mathcal{E}$ from below and above, we look at two sub-problems.
We fix $a, d$ and then either minimize or maximize $\mathcal{E}$ by varying $b,c$ alone. When $a, d$ are fixed, so do $u, v$. Apply triangle inequality to triangle $QRE$, we have $a + d - v \ge 0$.
The sub-problem of minimizing $\mathcal{E}$ becomes easy, we just set either $b$ or $c$ to $0$ and get $$4 - uad \le \mathcal{E}$$
For non-degenerate configurations, $bc(a+d-v) > 0$ and above inequality becomes strict. It is easy to show $uad \le 1$ using the concavity of $\log\sin x$. In general, we have:
$$3 \le \mathcal{E}$$
and the inequality becomes strict for non-degenerate configurations.
For the sub-problem of maximizing $\mathcal{E}$, we need $bc$ as large as possible. Introduce variables $p, q$ such that $b+c = p$, $b-c = q$, condition $(*1a)$ becomes
$$\frac{p^2+q^2}{2} + v\frac{p^2-q^2}{4} = u^2
\quad\iff\quad (2+v)p^2 + (2-v)q^2 = 4u^2
$$
Notice
$$bc = \frac{p^2-q^2}{4}
= \frac{1}{4(2+v)}((2+v)p^2-(2+v)q^2) =
\frac{1}{2+v}(u^2 - q^2)$$
For this sub problem, $\mathcal{E}$ is maximized when $q = 0$.
When $q = 0$, $b = c$ and their common value satisfies:
$$b^2(2+v) = c^2(2+v) = u^2 \implies b^2 = c^2 = \frac{u^2}{2+v} = 2 - v$$
As a result, we find
$$\mathcal{E} \le 4 + (2-v)(a+d - v) - uad$$
Introduce variables $a + d = r, a - d = s$, condition $(*1b)$ becomes
$$\frac{r^2+s^2}{2} + u\frac{r^2-s^2}{4} = v^2
\quad\iff\quad (2+u)r^2 + (2-u)s^2 = 4v^2$$
We find
$$ad = \frac{r^2-s^2}{4} = \frac{1}{4(2-u)}((2-u)r^2 - (2-u)s^2)
= \frac{1}{2-u}(r^2-v^2)
$$
and the inequality becomes
$$\mathcal{E} \le 4 + (2-v)(r-v) - \frac{u}{2-u}(r^2-v^2)
= 4 + \frac{r-v}{2-u}\left[(2-u)(2-v) - u(r+v)\right]\\
= 4 - \frac{r-v}{2-u}\left[2(u+v-2) + ur\right]
$$
Apply triangle inequality to triangle $AQE$, we have $u + v - 2 \ge 0$.
Together with $r - v = a + d - v \ge 0$, this leads to
$$\frac{r-v}{2-u}\left[2(u+v-2)+ur\right] \ge 0 \quad\implies\quad \mathcal{E} \le 4
$$
If one look at last inequality carefully, we find it is strict unless
$$(r - v = 0) \lor ( u+v-2 = 0 \land ( u = 0 \lor r = 0 ))$$
The case $r = 0$ can be ruled out because it contradict with our assumption $v \ne 0 $. We can simplify last condition to
$$a + d = v \lor u = 0$$
In both cases, the configuration is degenerate (i.e at least one of $a,b,c,d$ vanishes).
Combine these two bounds, we can conclude
$$3 \le \mathcal{E} \le 4$$
For non-degenerate configurations, we can strengthen this to $3 < \mathcal{E} < 4$.
So the answer to the original question "What is the maximum possible integral value of $\mathcal{E}$?" is $4$ for generate configurations and doesn't exist for non-degenerate configurations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How to efficiently compute $\,25^{37}\pmod{55}$? Most efficient way to do this question?
$25^{37}\cong (mod 55)$
Attempt:
$25^2 \cong 20 (mod55)$
Then I don't know the best way to continue... I know how to brute force it but I'm just curious what the best way is?
| First, $55 = 5 \cdot 11$ and $(5,11) = 1$, so we will use the Chinese remainder theorem to assemble the result after working modulo $5$ and $11$, separately.
Modulo $5$, $25 \cong 0$, so $25^{37} \cong 0 \pmod{5}$.
Modulo $11$, $25 \cong 3$ and (using Fermat's little theorem) $37 \cong 7 \pmod{10}$, so we compute
$$ 25^{37} \cong 3^{37} \cong 3^7 \pmod{11} \text{.} $$
Then \begin{align*}
3^{7} &\cong 3^8 \cdot 3^{-1} \pmod{11} \\
&\cong ((3^2)^2)^2 \cdot 3^{-1} \pmod{11} \\
&\cong (9^2)^2 \cdot 3^{-1} \pmod{11} \\
&\cong 4^2 \cdot 3^{-1} \pmod{11} \\
&\cong 5 \cdot 3^{-1} \pmod{11} \\
&\cong 5 \cdot 4 \pmod{11} \\
&\cong 9 \pmod{11} \text{.}
\end{align*}
Then, using the CRT, we want a multiple of $5$ that is congruent to $9$ modulo $11$. Check sequentially: $5$, $10$, $15$, $20$. We find that $20$ works. (There are direct methods, but it's easy to work through the (at most) eleven multiples of $5$.)
Our answer is $25^{37} \cong 20 \pmod{55}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Induction proof: $2^n + 3^n ≡ 5^n (mod 6)$ I'm trying to prove that $2^n + 3^n ≡ 5^n\ (mod\ 6)$ using induction.
$n=1$:
$2+3≡5\ (mod\ 6)$
$n=k$:
$2^k + 3^k ≡ 5^k\ (mod\ 6)$
$n=k+1$:
$2^{k+1} + 3^{k+1} ≡ 5^{k+1}\ (mod\ 6)$
$2*2^k + 3*3^k ≡ 5*5^k\ (mod\ 6)$
$6\ |\ 5*5^k - 2*2^k - 3*3^k$
$6\ |\ (2+3)*5^k - 2*2^k - 3*3^k$
$6\ |\ 2*5^k + 3*5^k - 2*2^k - 3*3^k$
$6\ |\ 2*(5^k - 2^k) + 3*(5^k - 3^k)$
Not quite sure if I'm going in the right direction or where to go from here...
| Without induction:
$$5^n=(2+3)^n=2^n+3^n+6k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.
Plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
and assuming the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.
I've not been able to find the value for $d$, putting the three points in the equation of sphere I got $3$ equations with $4$ variables.
| First, let's work out the plane of the circle.
The three triangle vertices are
$$\vec{a} = \left [ \begin{array}{c} a \\ 0 \\ 0 \end{array} \right ], \qquad \vec{b} = \left [ \begin{array}{c} 0 \\ b \\ 0 \end{array} \right ], \qquad \vec{c} = \left [ \begin{array}{c} 0 \\ 0 \\ c \end{array} \right ] \tag{1}\label{na1}$$
The equation for the plane is
$$\vec{p} \cdot \vec{n} - d = 0$$
where the plane normal $\vec{n}$ is
$$\vec{n} = ( \vec{b} - \vec{a} ) \times ( \vec{c} - \vec{a} ) = \left [ \begin{array}{c} -a \\ b \\ 0 \end{array} \right ] \times \left [ \begin{array}{c} -a \\ 0 \\ c \end{array} \right ] = \left [ \begin{array}{c} b c \\ a c \\ a b \end{array} \right ]$$
and the distance from plane to origin $d$ is the same at all points on the plane, including the vertices,
$$\vec{a} \cdot \vec{n} = \vec{b} \cdot \vec{n} = \vec{c} \cdot \vec{n} = a b c$$
i.e., the full equation of the plane is
$$\vec{p} \cdot \vec{n} - a b c = 0, \qquad \vec{n} = \left [ \begin{array}{c} b c \\ a c \\ a b \end{array} \right ] \tag{2}\label{na2}$$
Now, one way to treat the circle is to treat it as a great circle of a sphere that passes through the points. The key point is that because it is a great circle, the center of the sphere must be on the same plane as the vertices.
This also means that we can use two coordinates, say $u$ and $v$, to describe the center of the circle (and sphere):
$$\vec{p} = \vec{a} + u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) \tag{3}\label{na3}$$
If the radius of the circle (and therefore the radius of the sphere) is $r$, then we have a system of equations,
$$\begin{cases}
\left \lVert \vec{p} - \vec{a} \right \rVert^2 = r^2 \\
\left \lVert \vec{p} - \vec{b} \right \rVert^2 = r^2 \\
\left \lVert \vec{p} - \vec{c} \right \rVert^2 = r^2
\end{cases}$$
i.e.
$$\begin{cases}
\left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) \right \rVert^2 - r^2 = 0 \\
\left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) + \vec{a} - \vec{b} \right \rVert^2 - r^2 = 0 \\
\left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) + \vec{a} - \vec{c} \right \rVert^2 - r^2 = 0
\end{cases} \tag{4}\label{na4}$$
This system of equations has two solutions, one of which has a negative radius $r$; thus only one solution that makes sense:
$$\begin{cases}
u = \frac{a^2 b^2 + b^2 c^2}{2 (a^2 b^2 + a^2 c^2 + b^2 c^2)} \\
v = \frac{a^2 c^2 + b^2 c^2}{2 (a^2 b^2 + a^2 c^2 + b^2 c^2)} \\
r = \sqrt{\frac{(a^2 + b^2)(a^2 + c^2)(b^2 + c^2)}{4 (a^2 b^2 + a^2 c^2 + b^2 c^2 )}} \tag{5}\label{na5}\end{cases}$$
To find the center $\vec{p}$ in Cartesian coordinates, insert $u$ and $v$ and $\eqref{na1}$ to $\eqref{na3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How to see the order relation for $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$ as $n$ very large? I have two sequences $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$. How do I check whether these two sequence have a fixed order relation when $n$ getting larger? Say, for all $n$ that is large than $N$, $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}\leq\frac{n^4}{n^7}$.
I know that $\lim_{n\to\infty}\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}=0$, but this doesn't help to see the relation for these two sequence(only helpful to see that the nominator$<$denominator). Also, even I know $$\lim_{n\to\infty}\frac{\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}}{\frac{n^4}{n^7}}=1$$
it appears that we can't deduce which is necessarily bigger than the other for large $n$.
Need help.
| Let $n$ be positive and large. Then
$$
\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2} > \frac{n^4}{n^7}
\\
\Longleftrightarrow
\\
(n^4+99999n^3-99n^2+3)n^7 > n^4(n^7+9999n^5-9999n^2)
\\
\Longleftrightarrow
\\
n^{11}+99999n^{10}-99n^{9}+3n^7 > n^{11}+9999n^{9}-9999n^{6}
\\
\Longleftrightarrow
\\
99999n^{10}-99n^{9}+3n^7 > 9999n^{9}-9999n^{6}
$$
which is true for large $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Change of basis of linear transformation $ℝ^2 \to ℝ^3$ Linear transformation $T:ℝ^2\to ℝ^3$ in bases
$\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right\}$ and $\left\{ \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\right\}$ has matrix:
$\begin{bmatrix}
2 & 5 \\
1 & 1 \\
8 & 1 \\
\end{bmatrix}$. What is the standard matrix?
I thought this problem will be easy, but it got me horribly confused. I don't know where to start.
| let $x_1, x_2$ be the canonical basis in \mathbb R^2.
and
$y_1, y_2, y_3$ be the canonical basis in \mathbb R^3.
$T(x_1 + x_2) = T(x_1) + T(x_2) = $$(2\cdot 2 + 1\cdot 1 + 8\cdot 1) y_1 + (2\cdot 1 + 1\cdot 0 + 8\cdot -1) y_2 + (2+1+8) y_3\\
13 y_1 - 6y_2 + 11y_3$
$T(x_1) + 3T(x_2) = 12y_1 + 4y_2 + 7y_3\\
2T(x_2) = -1y_1 +10y_2-4 y_3\\
T(x_2) = \begin{bmatrix}{- \frac 12\\5\\-2}\end{bmatrix}\\
T(x_1) = \begin{bmatrix}{13\\-6\\11}\end{bmatrix}-\begin{bmatrix}{- \frac 12\\5\\-2}\end{bmatrix} = \begin{bmatrix}{13.5\\-11\\13}\end{bmatrix}$
Alternatively
$\begin{bmatrix} 2&1&1\\1&0&-1\\1&1&1\end{bmatrix}\begin{bmatrix}2&5\\1&1\\8&1\end{bmatrix}\begin{bmatrix}1&1\\1&3\end{bmatrix}^{-1} = \begin{bmatrix} 13.5 &-0.5\\-11&5\\13&-2\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Inequality estimate involving series Given $b_{j}>0$ and $b_{j}=b_{j+1}$ for $j=1,\ldots,n$, with $n\ge 2$ an integer and for some $\Psi>1$, does the following estimate or inequality hold?
\begin{equation}
\sum_{j=1}^n j(j-1)b_{j}|\Psi|^{j-2}~~\le~~\frac{ \left(\sum_{j=1}^n jb_j|\Psi|^{j-1}\right)^2}{\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)}\nonumber
\end{equation}
Or are there constants $C>0$ and $D\ge 0$ such that the following estimate always holds for all $\Psi>1$?
\begin{equation}
\Psi^4(\Psi-1)\left(\frac{\sum_{j=1}^n j(j-1)b_j|\Psi|^{j-2}}{\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)}~-~\frac{ \left(\sum_{j=1}^n j~b_j|\Psi|^{j-1}\right)^2}{\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)^2}\right)
~~\le~~C\ln\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)+D\nonumber
\end{equation}
Is there a neat way to prove/disprove this inequality? Note that the inequalities need only hold for some $\Psi > 1$ or $\Psi \gg 1$.
| Substituting $x=|\Psi|$, consider a polynomial $p(x)=1+\sum_{j=1}^n b_j x^j$. Then the first inequality your asking is $$p’’(x)p(x)\le p’(x)^2$$ and the second is $$x^4(x-1)\left(\frac{p’’(x)}{p(x)}-\frac{p’(x)^2}{p(x)^2}\right)\le C\ln p(x)+D.$$
Moreover, since all $b_j=b$ are equal and $x\ne 1$, we have $p(x)=1-b +b\cdot \frac {x^{n+1}-1}{x-1}$. Therefore $p'(x)= b\cdot\frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$ and $p''(x)= b\cdot\frac {(n-1)nx^{n+1}-2(n-1)(n+1)x^n+n(n+1)x^{n-1}-2}{(x-1)^3}$. Hence
$$p’(x)^2-p’’(x)p(x)=$$ $$\frac b{(x-1)^4}\left(bn x^{2n+2}+(-2bn-2b)x^{2n+1}+(bn+b)x^{2n}+((b-1)n^2+(b-1)n)x^{n+2}+((3-2b)n^2+(2b-1)n+4b-2)x^{n+1}+((b-3)n^2+(-b-1)n-2b+2) x^n+(n^2+n)x^{n-1}+(-2b+2)x+b-2)\right.$$
Since the highest coefficient of the long polynomial at the right hand side is positive, $p’’(x)p(x)-p’(x)^2$ is positive for sufficiently large $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$ $P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$
No idea how to prove this. Would really appreciate any help, hints...
Thanks in advance :)
| Since
$P(z)= z^n - 1, \tag 1$
we have
$P(z) = (z - 1)(z^{n - 1} + z^{n - 2} + \ldots + z + 1) = (z - 1)Q(z), \tag 2$
where
$Q(z) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i \tag 3$
it then follows that if $1 \ne b \in \Bbb C$ satisfies
$P(b) = (b - 1)(b^{n - 1} + b^{n - 2} + \ldots + b + 1) = (b - 1)Q(b) = 0, \tag 4$
we must have
$b^{n - 1} + b^{n - 2} + \ldots + b + 1 = \sum_0^{n - 1} b^i = Q(b) = 0; \tag 5$
thus any root $b$ of $P(z)$ other than $1$ must satisfy $Q(z)$; now $\deg Q(z) = n - 1$, so it has at most $n - 1$ zeroes, so if we assume the list of roots of $P(z)$,
$1, a_1, a_2, \ldots, a_{n - 1}, \tag 6$
is irredundant, that is, does not repeat itself (as seems to be the intention of our OP kjhg), then the zeroes of $Q(z)$ are the $a_i$, $1 \le i \le n - 1$. Thus we may write
$Q(z) = \displaystyle \prod_1^{n - 1} (z - a_i); \tag 7$
comparing (3) and (7), we see that
$\displaystyle \prod_1^{n - 1} (z - a_i) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i;\tag 8$
taking $z = 1$ in (8), we reach
$\displaystyle \prod_1^{n - 1} (1 - a_i) = 1^{n - 1} + 1^{n - 2} + \ldots + 1^1 + 1 = n, \tag 8$
the desired result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a system, please?
| Setting $a=bt$ then we get
$$b^3+t^3+39b^3t-18=0$$
$$3b^3t^2+13b^3-5=0$$ eliminating $b^3$ then we get
$$5t^3-54t^2+195t-234=0$$
one solution is $t=3$ then you will get $a=3b$
plugging this in the given equation we get $$a=\frac{3}{2},b=\frac{1}{2}$$
| {
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Computing the determinant of a large matrix? How would I go about computing the determinant of large matrices, such as $6 \times 6$. I believe that I need to use multilinear maps, but I am not sure how I can go about computing the determinant in a nice and efficient way.
Can anyone show me how I can determine the determinant of the matrix below in a simple and efficient way?
\begin{pmatrix}
0 & 0& 1& 1& 1 & 1\\
1 & 0 & 0 & 0& 0& 1\\
1 & 0 & 1& 1 & 1 &1 \\
0 & 1 & 1 & 1 & 0 &1 \\
0 & 1 & 0& 1 & 0& 0\\
0 & 0& 1& 0 & 0& 0
\end{pmatrix}
| You can use Laplace's expansion of the determinant.
More precisely, begin expanding by the last row, then some row manipulation and expanding by convenient rows/columns ends up in a $2\times2$ determinant:
\begin{align}
\begin{vmatrix}
0&0&1&1&1&1 \\ 1&0&0&0&0&1 \\ 1&0&1&1&1&1 \\
0&1&1&1&0&1 \\ 0&1&0&1&0&0 \\ 0&0&\color{red}1&0&0&0\end{vmatrix}&=
-\begin{vmatrix}
0&0&1&1&1 \\ 1&0&0&0&1 \\ 1&0&1&1&1 \\
0&1&1&0&1 \\ 0&1&1&0&0\end{vmatrix}=
-\begin{vmatrix}
0&0&1&1&1 \\ 1&0&0&0&1 \\ 1&0&1&1&1 \\
0&0&0&0&\color{red}1 \\ 0&1&1&0&0\end{vmatrix}=
+\begin{vmatrix}
0&0&1&1 \\ \color{red}1&0&0&0 \\ 1&0&1&1 \\ 0&1&1&0 \end{vmatrix}\\
&=-\begin{vmatrix}
0&1&1 \\ 0&1&1 \\ \color{red}1&1&0 \end{vmatrix}=-\begin{vmatrix}
1&1 \\ 1&1 \end{vmatrix}=0.
\end{align}
| {
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} |
How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.
| We have
\begin{eqnarray*}
1^4+\cdots + n^4 &=& \frac{1}{5} n^5+\frac{1}{2}n^4+\frac{1}{3}n^3+\frac{1}{30}n) \\
&=& \frac{1}{30}(6 n^5+15n^4+10n^3-n). \\
\end{eqnarray*}
So
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 n^5+15n^4+10n^3-n +30n^4+120n^3+180n^2+120n+30). \\
\end{eqnarray*}
Now check this really does equal the expression below
\begin{eqnarray*}
\frac{1}{30}(6 (n^5 +5n^4+10n^3+10n^2+5n+1) \\
+15(n^4+4n^3+6n^2+4n+1) \\
+10(n^3+3n^2+3n+1) \\
-(n+1) ). \\
\end{eqnarray*}
So
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 (n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)). \\
\end{eqnarray*}
| {
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Prove that, if $a,b,c$ are positive real numbers $\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$
Prove that, if $a,b,c$ are positive real numbers:
$$\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$$
Can the proof be written without the need for high mathematics?
| The inequality is homogeneous, i.e. $(a,b,c)$ satisfies the inequality if and only if $(ta,tb,tc)$ $\forall t>0$ satisfies the inequality. So we can assume WLOG $a+b+c=1$.
$$\sum_{\text{cyc}}\frac{a}{2a+b+c}=\sum_{\text{cyc}}\frac{a}{a+1}=$$
$$=\sum_{\text{cyc}}\left(1-\frac{1}{a+1}\right)=3-\sum_{\text{cyc}}\frac{1}{a+1}$$
You can use Cauchy Schwarz (CS) inequality to prove that for real $a_i$, positive $b_i$:
$$\sum_{i=1}^n\frac{a_i^2}{b_i}\ge \frac{(\sum_{i=1}^n a_i)^2}{\sum_{i=1}^n b_i}$$
To prove it, multiply both sides by $\sum_{i=1}^n b_i$ and use CS.
$$3-\sum_{\text{cyc}}\frac{1}{a+1}\le 3-\frac{9}{4}=\frac{3}{4}<\frac{19}{25}$$
| {
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$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$ $$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$
I can't find $$? \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}} \leq ?$$ I remeber there was a sharp bound , Am i right ?
I am thankful if someone help me to find the bound , or give another idea
Last question : Is it possible to find $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}$ by integral ?
| Since $$\frac1{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n})=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac1{\sqrt{n}},$$
we have $$2(\sqrt{n+1}-\sqrt{2})<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{1}),$$ i.e.
$$2\sqrt{n+1}-2\sqrt{2}+1<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n}-1.$$
For $n=1024$, the bounds are $2\sqrt{1025}-2\sqrt{2}+1\approx62.2$ and $63$, so the result would be $62$.
| {
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} |
Residue theorem for a rational function
Evaluate
$$\int_{-\infty}^{\infty}\frac{x^4}{1+x^8}\mathrm{d} x.$$
My concern: One of the consequences of the residue theorem states that, given a polynomial of the form $P/Q$ such that the degree of $Q$ exceeds $P$ by at least two, the integral can then be expressed as $$\int fdz=2\pi i\sum_{U}{\mathrm{Res}\left ( f;z_{i} \right )}.$$
The zeros of $Q$ in the Upper half plane is here given by $z=e^{\frac{i \pi}{8}\left ( 2n+1 \right )}$ for $n\in \left \{0,1,2,3\right \}$. The residue at $z_{n}$ is now given by $\frac{P(z_{n})}{Q´(z_{n})}$.
Should I derive the polynomial or exponential function?
| You can use the following reasoning.
$$\frac{x^4}{x^8+1}=\frac{x^4}{(x^4+\sqrt2x^2+1)(x^4-\sqrt2x^2+1)}=$$
$$=\frac{1}{2\sqrt2}\left(\frac{x^2}{x^4-\sqrt2x^2+1}-\frac{x^2}{x^4+\sqrt2x^2+1}\right)=$$
$$=\frac{1}{2\sqrt2}\left(\tfrac{x^2}{\left(x^2+\sqrt{2-\sqrt2}x+1\right)\left(x^2-\sqrt{2-\sqrt2}x+1\right)}-\tfrac{x^2}{\left(x^2+\sqrt{2+\sqrt2}x+1\right)\left(x^2-\sqrt{2+\sqrt2}x+1\right)}\right)=...$$
| {
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"question_score": "8",
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} |
For each $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix}1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^n$
If $a \in \mathbb{R}$, evaluate $$ \lim_{n \to \infty}\left(\begin{matrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{matrix}\right)^{n}$$
My attempt: Let $$A = \left(\begin{matrix} 0&a\\-a&0\end{matrix}\right) = -a\left(\begin{matrix} \cos(\frac{\pi}{2})&-\sin(\frac{\pi}{2})\\\sin(\frac{\pi}{2})&\cos(\frac{\pi}{2})\end{matrix}\right)$$ so that $$A^k = (-a)^k \left(\begin{matrix} \cos(\frac{k\pi}{2})&-\sin(\frac{k\pi}{2})\\\sin(\frac{k\pi}{2})&\cos(\frac{k\pi}{2})\end{matrix}\right)$$
Thus,
\begin{align}\displaystyle \lim_{n \to \infty}\left(\begin{matrix} 1&\dfrac{a}{n}\\\dfrac{-a}{n}&1\end{matrix}\right)^{n} &=\displaystyle \lim_{n \to \infty} \left(I+\dfrac{A}{n}\right)^n =e^A=\displaystyle \sum_{k=0}^{\infty}\dfrac{A^k}{k!}\\&= \sum_{k=0}^{\infty} \left(\begin{matrix} \dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}&-\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}\\\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}&\dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}\end{matrix}\right) \end{align}
and since $\displaystyle \sum_{k=0}^{\infty}\dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}=1+0-\dfrac{a^2}{2!}+0+\dfrac{a^4}{4!}+\cdots= \cos a$ and
$\displaystyle \sum_{k=0}^{\infty}\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}=0-a+0+\dfrac{a^3}{3!}+0-\dfrac{a^5}{5!}+\cdots= -\sin a$ therefore the required answer is
$\left(\begin{matrix} \cos a&\sin a\\-\sin a&\cos a\end{matrix}\right).$
However the above answer does not match the choices provided which are $I, 0$ and none of the above. So my question is: Is my answer correct?
| Your answer is correct as well. but there is a short way to reach it. Merely by putting ourselves in the complex plan where we identify
$$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{matrix}\right).$$
Thus, $$\begin{align}\displaystyle \lim_{n \to \infty}\left(\begin{matrix} 1&\dfrac{a}{n}\\\dfrac{-a}{n}&1\end{matrix}\right)^{n} &= \displaystyle \lim_{n \to \infty}\color{blue}{\left(1+\dfrac{ai}{n}\right)^n}
\\&=\color{red}{e^{ai} = \cos a+i\sin a} \\&= \left(\begin{matrix} \cos a&\sin a\\-\sin a&\cos a\end{matrix}\right).\end{align}$$
Given that, for any $z\in \Bbb C$ we have, $\lim\limits_{n \to \infty}\left(1+\dfrac{z}{n}\right)^n =e^{z} $
See here
also here,
| {
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"answer_count": 2,
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Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$
Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$
1)$\dfrac{\cos(x)}{1+\sin(x)}$
2)$\dfrac{\cos(x)}{1-\sin(x)}$
3)$\dfrac{1-\sin(x)}{\cos(x)}$
4)$\dfrac{1+\sin(x)}{\cos(x)}$
Of course we can find the correct choice by testing each of them,but I'm looking for an analytical solution assuming we don't know the final expression.
I tried the following techniques,all failed: multiplying by $\frac{\sin(x)}{\sin(x)}$ , multiplying by $\frac{\cos(x)}{\cos(x)}$ , using $\sin^2(x)+\cos^2(x)=1$
|
First Method: Using the identities
$$\cos a+ \cos b =2\cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right)\tag{I}$$
and
$$\cos a + \sin a =\sqrt2\cos\left(a-\frac{\pi}4\right)\tag{II}$$
we have, \begin{align}\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}&= \frac{1+\sqrt2\cos\left(x+\frac{\pi}4\right)}{1+\sqrt2\cos\left(x-\frac{\pi}4\right)}
~~~~\text{using (II)}
\\&= \frac{\cos(\frac\pi4)+\cos\left(x+\frac{\pi}4\right)}{\cos(\frac\pi4)+\cos\left(x-\frac{\pi}4\right)}
\\&=\frac{\cos\left(\frac{x}2+\frac{\pi}4\right)\cos\left(\frac{x}2\right)}{\cos\left(\frac{x}2-\frac{\pi}4\right)\cos\left(\frac{x}2\right)}~~~~\text{using (II)}
\\&=
\frac{2\cos\left(\frac{x}2+\frac{\pi}4\right)\cos\left(\frac{x}2-\frac{\pi}4\right)}{2\cos^2\left(\frac{x}2-\frac{\pi}4\right)}
\\&=\frac{\cos\left(x\right)+\cos\left(\frac{\pi}2\right)}{1+\cos\left(\frac{\pi}2-x\right)}~~~~\text{using (I) and $\cos2x = 2\cos^2 x -1$ }
\\&\color{red}{=\frac{\cos\left(x\right)}{1+\sin\left(x\right)}}~~~~\text{since}~~~~\cos( \frac{\pi}2-x) = \sin x
\\&\color{red}{=\frac{\cos\left(x\right)}{1+\sin\left(x\right)} \frac{1-\sin\left(x\right)}{1-\sin\left(x\right)} = \frac{1-\sin\left(x\right)}{\cos\left(x\right)}}\end{align}
second Method:
$$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}\times\frac { \cos { x } }{ 1-\sin { x} } = \frac{\cos (x)+\cos^2(x)-\sin(x)\cos (x)}{(1+\sin(x))(1-\sin(x))\cos(x)(1-\sin(x))} \\= \frac{\cos (x)+\cos^2(x)-\sin(x)\cos (x)}{\cos(x)+\color{blue}{(1-\sin^2(x)}-\cos(x)\sin(x)} =\frac{\cos (x)+\cos^2(x)-\sin(x)\cos (x)}{\cos(x)+\color{blue}{\cos^2(x)}-\cos(x)\sin(x)} =\color{blue}{1} $$
| {
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Bounding $\|f(a)-f(b)\|$ for $a,b$ in the unit ball Let $f:\Bbb{R}^3 \to \Bbb{R}$ be given by
$$f(x,y,z) = x^2 +\cos(xyz) -z^2$$
I want to show that for any $a,b \in B(0,1)$ (the unit open ball around the origin) $$\|f(a)-f(b)\|\le 2\sqrt{5}$$
In a addition I wonder if this is the best bound I could get?
My attempt:
I know $f(a+h) \approx f(a) + \nabla f(a) \cdot h$. If i put $h = b-a$ then I get
$$\|f(b)-f(a)\| \approx \|\nabla f(a) \cdot h\| \le \|\nabla f(a)\|\|b-a\|$$
by Cauchy Schwarz. The distance between $b$ and $a$ is less than $2$ so can I show that the magnitude of the gradient is bounded above by $\sqrt 5$? But I think there is a better way to find the bound because when I use "$\approx$" I am not really sure if it is a lower or upper bound, and furthermore I know the approximation is only good when $h$ (or $b-a$) is pretty small.
| Your approach is fine.
By the mean value theorem exists $ a \in B(0,1) $ such that
$$ \|f(b)-f(a)\| \le \|\nabla f(a)\|\|b-a\| < 2 \|\nabla f(a)\| $$
Having in mind that $ x^2 + y^2 + z^2 < 1 $ we can bound $ \|\nabla f(a)\| $ as follows
$$ \begin{align}
\|\nabla f(a)\| &= \sqrt{\left(x^2 \left(y^2+z^2\right)+y^2 z^2\right) \sin(x y z)^2 +4 \left(x^2+z^2\right)} \\
&< \sqrt{ \left(x^2 \left( 1-x^2 \right) + y^2 z^2\right) \sin(x y z)^2 + 4 \left( 1 - y^2 \right)} \\
\end{align} $$
Now since
$$ x^2 \left( 1-x^2 \right) \le \frac{1}{4} $$
$$ y^2 z^2 \le \frac{1}{4} $$
$$ \sin(x y z)^2 < 1 $$
and
$$ 1 - y^2 \le 1 $$
then
$$ \begin{align}
\|\nabla f(a)\| &< \sqrt{ \left( \frac{1}{4} + \frac{1}{4} \right) \cdot 1 + 4 \cdot 1} \\
&< \sqrt{5}
\end{align} $$
As you wanted to show.
Furthermore I have a strong intuition that we can lower the upper bound to $\require{enclose}\enclose{horizontalstrike}{\|f(a)-f(b)\| \le 1} $ $ \|f(a)-f(b)\| \le 2 $ and that it occurs in the boundary of the unit ball. But for that we would need Lagrange multipliers.
| {
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Derivating with respect the parameter calcule $\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$ Derivating with respect the parameter calcule $$\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$$
| To complement the post by @MarkViola, let us show
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}= \frac{\pi}{2(a+1)}
\end{align}
assuming $a>0$.
Observe
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x} = \int^{\pi/2}_0 \frac{\sec^2 x- \tan^2x}{1+a^2\tan^2x}\ dx
\end{align}
which means
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}+ \frac{1}{a^2}\int^{\pi/2}_0 \frac{a^2\tan^2 x}{1+a^2\tan^2x}\ dx = \int^{\pi/2}_0 \frac{\sec^2 x}{1+a^2\tan^2x}\ dx.
\end{align}
Next, notice the left hand side of the above identity is equal to
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}+ \frac{1}{a^2}\int^{\pi/2}_0 1- \frac{1}{1+a^2\tan^2 x}\ dx = \left(1-\frac{1}{a^2} \right)\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x}+ \frac{\pi}{2a^2}.
\end{align}
and the right hand side by $u$-sub equals
\begin{align}
\int^{\pi/2}_0 \frac{\sec^2 x}{1+a^2\tan^2 x}\ dx = \frac{\pi}{2a}.
\end{align}
Hence it follows
\begin{align}
\left(1-\frac{1}{a^2} \right)\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x} = \frac{\pi(a-1)}{2 a^2} \ \ \ \implies \ \ \ \int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x} = \frac{\pi}{2(a+1)}.
\end{align}
Remark: If we drop the condition $a>0$, then we see that
\begin{align}
\int^{\pi/2}_0 \frac{\sec^2 x}{1+a^2\tan^2 x}\ dx = \frac{\pi}{2|a|}
\end{align}
which leads to the conclusion
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x} = \frac{\pi}{2(|a|+1)}.
\end{align}
| {
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Find $6^{273} + 8^{273}\pmod{49}$
The number $6^{273} + 8^{273}$ divided by $49$ has a remainder, what is its value?
I used the totient function to compute for modulo 49.
$6^{42}$ and $8^{42}$ are $-1$ and $1$ mod $49$ respectively, $273/49$ is equal to $5$ with a remainder of $21$.
We would then look for the remainder of $-6^{21}$ + $8^{21}$ which I do not know how to solve.
I am aware of other solutions such as factoring odd exponents, but I wanted to know if we can use this kind of approach.
| Observe that
\begin{align*}
6^{273}+8^{273} & \equiv (7-1)^{273}+(7+1)^{273} \pmod{7^2}\\
& \equiv \sum_{r=0}^{136}\binom{273}{2r}7^{273-2r}\pmod{7^2}\\
& \equiv \binom{273}{272}7\pmod{7^2}\\
& \equiv 7(273) \pmod{7^2}\\
& \equiv 0 \pmod{7^2}.
\end{align*}
| {
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$\sum_{k=\frac{n}{2}}^{n}\binom{k}{n-k}=F_n$?
Is it true:
$$\sum_{k=\frac{n}{2}}^{n}\binom{k}{n-k}=F_n? $$ where $F_n$ is a
Fibonacci number.
Wolfram alpha says that the answer must be $F_{n+1}$.
I've seen many posts with this question but none of them was helpful to me. And this time the sum is different. Can anybody give me a breath explanation how should I solve this equation?
Edit: $n \ge 0, \quad F_1=F_2=1$
| It depends. If you assume that the Fibonacci numbers start with $F_1=1$ and $F_2=1$, then Wolfram alpha is correct, i.e.
$$
\sum_{k=\frac{n}{2}}^n \binom{k}{n-k}=F_{n+1}.
$$
If you assume that the Fibonacci numbers start with $F_1=0$ and $F_2=1$, then we have
$$
\sum_{k=\frac{n}{2}}^n \binom{k}{n-k}=F_{n}.
$$
Actually, with $F_1=F_2=1$, we have
$$
\sum_{k=\lceil\frac{n}{2}\rceil}^n \binom{k}{n-k}=F_{n+1}.
$$
Let
$$
y_n = \sum_{k=\lceil\frac{n}{2}\rceil}^n \binom{k}{n-k}.
$$
What I want to proof is that $y_n = y_{n-2}+y_{n-1}$. I will use the fact that $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$. First, I proof this for $n$ is even, such that I can use the fact that $\lceil \frac{n+1}{2} \rceil = \lceil \frac{n}{2} \rceil + 1$:
\begin{align}
y_n &= 1 + \sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k}{n-k} + 1 \\
&=\sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k-1}{n-k-1}+1 + \sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k-1}{n-k}+1 \\
&=\sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k-1}{n-2-(k-1)}+1 + \sum_{k=\lceil\frac{n+1}{2}\rceil}^{n-1} \binom{k-1}{n-1-(k-1)}+1 \\
&= \sum_{k=\lceil\frac{n-2}{2}\rceil+1}^{n-2} \binom{k}{n-2-k}+1 + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-2} \binom{k}{n-1-k} + 1 \\
&= \sum_{k=\lceil\frac{n-2}{2}\rceil}^{n-2} \binom{k}{n-2-k} + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-1} \binom{k}{n-1-k} \\
&=y_{n-2} + y_{n-1}.
\end{align}
Now we can do the same for $n$ being odd. I will assume that $\lceil\frac{n}{2}\rceil-1 = \lceil\frac{n-1}{2}\rceil$:
\begin{align}
y_n &= \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k}{n-k} + 1 \\
&= \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-k-1} + \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-k} + 1 \\
&= \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-2-(k-1)} + \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-1-(k-1)} + 1 \\
&= \sum_{k=\lceil\frac{n-2}{2}\rceil}^{n-2} \binom{k}{n-2-k} + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-2} \binom{k}{n-1-k}+1 \\
&= y_{n-2} + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-1} \binom{k}{n-1-k} \\
&= y_{n-2} + y_{n-1}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\left(t+\frac{\pi}{6}\right)}\le 2+\dfrac{5}{4}\sqrt{3}$$
My Question:this function have other methods to find this maximum? such as AM-GM,Cauchy-Schwarz inequality and so on?
| There's always the classic, but perhaps computationally icky derivative approach.$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
$$f'(x)= \frac{\sqrt3}2 x + \frac{20 x - 4 x^3}{
8 \sqrt{(9 - x^2) (-1 + x^2)}} = \frac 12 x \, \Big(\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }}\Big)$$
Let $f'(x) = 0$ and solve to find the local extrema:
Clearly $x=0$ is a solution, but doesn't exist in the domain of $f$, which is $[−3,−1]∪[1,3]$.
So we find solutions to $$\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }} = 0$$ or equivalently $$ \sqrt3 \sqrt{-x^4+10 x^2-9} = 5 - x^2 $$ which comes to
$$3 (-x^4+10 x^2-9)=(5-x^2)^2$$
$$-3 x^4+30 x^2-27=x^4-10 x^2+25$$
$$-4 x^4+40 x^2-52=0$$
Complete the square to get
$$(x^2-5)^2=12$$ so
$$x = \pm \sqrt{5 \pm 2 \sqrt 3}$$
Checking for extraneous solutions reveals that the only values of $x$ satisfying $f'(x) = 0$ are $$x = \pm \sqrt{5 +2 \sqrt 3}$$
Now plug in these values to get
$$f_{\text{max}} = f(\pm \sqrt{5 +2 \sqrt 3}) = \frac{8 + 5 \sqrt3}4$$ which is in line with MyGlasses's observation that $f_{\text{max}}$ must be no more than $\dfrac{\sqrt{3}}{4}9+\dfrac{8}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Does there exist a closed form for the sinc function series $\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}}$? Here I want to get the closed form solution of the following summation
$$
\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}} \qquad(1)
$$
Or the more general form ($x$ be an arbitrary real number, and $a\geq0$ is a constant):
$$
f_a(x) = \sum_{n=1}^\infty \frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(2)
$$
I tried the numeircal simulations before I post the question. I truncated the first $1,000,000$ terms of equation (1) and it turned $0.781233190560320$.
Anyone can help me?
In fact, I made it the reduced case when $a=0$. It can be proved by Fourier series:
$$
f_0(x)=\sum_{n=1}^\infty \frac{\sin nx}{n} = \left\{ \matrix{\dfrac{\pi-x}{2}, 0<x<2\pi\\0, x=0,2\pi} \right.
$$
And the function is periodical:
$$
f_0(x) = f_0(x+2\pi)
$$
Edit: How about this one?
$$
g_a(x) = \sum_{n=1}^\infty \frac{\cos\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(3)
$$
We get the "diverging wave solution" in physics when we combine the equation (2) and (3):
$$
h_a(x)=g_a(x)+\text{i}f_a(x)=\sum_{n=1}^\infty \frac{\exp\left(\text{i}x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(4)
$$
Edit #2:
I tested the solution solved by Random Variable (see the most ranked answer and thousands thanks to it!) compared with the truncating results:
$$
\sum_{n=1}^N\frac{\sin \left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}, N = 1,000,000, a=1
$$
Here is the solution by @Random Variable:
the solution of equation (2) as the following:
$$
\sum_{n={1}}^\infty \frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}
=
\frac{\pi}{2} J_0(ax) -\frac{\sin(ax)}{2a}, a>0, 0<x<2\pi\qquad(2*)
$$
where $J_0(ax)$ is the Bessel function of the first kind of order zero.
Here is the comparison:
It can be found that the both agree well when $0 <x<2\pi$,but differ in other domain. So, how about the solution beyond $(0,2\pi)$?
Edit #3:
Inspired by Random Variable's answer, I found the solution of equation (3) as the following:
$$
\sum_{n={1}}^\infty \frac{\cos\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}
=
-\frac{\pi}{2} Y_0(ax) -\frac{\cos(ax)}{2a}, a>0, 0<x<2\pi\qquad(3*)
$$
where $Y_0(ax)$ is the Bessel function of the second kind of order zero.
Here is the comparison:
Note that equation (3) is divergent when $x=0$.
Possible relating QUESTIONS:
Does there exist a closed form for the non-integer shifted sinc-function series: $\frac{\sin(n+a)x}{(n+a)x}$?
| To go beyond the limitation $-2\pi<x<2\pi$ for the sine series, we can use the representation G&R (6.677.6) (or Ederlyi TI p.57 1.13.47)
\begin{equation}
\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}=\int_0^x J_0\left( n\sqrt{x^2-t^2} \right)\cos at\,dt\tag{1}
\end{equation}
valid for $x>0$. (For $x<0$, we will use the fact that the series is an odd function of $x$, as remarked by @RandomVariable).
The summation can be computed using the Schlömilch series (G&R 8.521.1):
\begin{equation}
\sum_{n=1}^\infty J_0(nz)=-\frac{1}{2}+\frac{1}{z}+2\sum_{m=1}^p\frac{1}{\sqrt{z^2-4\pi^2m^2\pi^2}}
\end{equation}
for $2p\pi<z<2(p+1)\pi$, which defines $p=\lfloor \frac{z}{2\pi}\rfloor$. Choosing $z=\sqrt{x^2-t^2}$,
\begin{align}
S(x)&=\sum_{n=1}^\infty\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}\\
&=\int_0^x\sum_{n=1}^\infty J_0\left( n\sqrt{x^2-t^2} \right)\cos at\,dt\\
&=\int_0^x\left[-\frac{1}{2}+\frac{1}{\sqrt{x^2-t^2}}+2\sum_{m=1}^{\lfloor \frac{\sqrt{x^2-t^2}}{2\pi}\rfloor}\frac{1}{\sqrt{x^2-t^2-4\pi^2m^2\pi^2}}\right] \cos at\,dt\\
&=\int_0^x\left[-\frac{1}{2}+\frac{1}{\sqrt{x^2-t^2}}\right]\cos at\,dt+2\sum_{1\le m< \lfloor \frac{x}{2\pi}\rfloor}\int_0^{\sqrt{x^2-4\pi^2m^2}}\frac{\cos at\,dt}{\sqrt{x^2-4\pi^2m^2-t^2}}
\end{align}
(The summation does not exist if $\lfloor \frac{x}{2\pi}\rfloor=0$). With the classic integral representation
\begin{equation}
\frac{\pi}{2}J_0(s)=\int_0^1\frac{\cos sx}{\sqrt{1-x^2}}\,dx
\end{equation}
the result can be written as
\begin{equation}
\sum_{n=1}^\infty\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}=\frac{\pi}{2}J_0(ax)-\frac{\sin ax}{2a}+\pi\sum_{1\le m< \lfloor \frac{x}{2\pi}\rfloor}J_0\left( a\sqrt{x^2-4\pi^2m^2-t^2}\right)
\end{equation}
which seems to be numerically correct. Unfortunately, no corresponding form of (1) for the cosines seems to exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
} |
Solve $h_n = 3h_{n-1} -4n$ using generating functions Solve $h_n = 3h_{n-1} -4n$, where $h_0 = 2$ using generating functions.
I am struggling to figure out how to solve this using generating functions. I know the answer should be $h_n = -3^n +2n + 3$.
Here is the method that both my textbook and professor used.
Let $g(x) =\sum_{n=0}^{\infty} h_nx^n $. Then let $h_n = 0$ if $n>0$.
Then $3x g(x)= \sum_{n=0}^{\infty} 3h_nx^{n+1} = \sum_{n=0}^{\infty} 3h_{n-1} x^n $
Subtracting these two equations we get
$(1-3x)g(x) = \sum_{n=0}^{\infty}(h_n - 3h_{n-1})x^n $
$= h_0 + (h_1-3h_0)x + ... + (h_n -3h_{n-1})x^n + ... $
$= 2-4x-8x^2 -...-4nx^n -...$
From here we are suppose to find some sort of pattern to simplify, but all the example have coefficients that can be written as a number to the power n, but here we have -4n instead. How do I compensate for this?
| The best approach to get the generating function of a recurrence
is to include directly in it the initial conditions.
In your case, we have
$$
\left\{ \matrix{
h_{\,n} = 0\quad \left| {\;n < 0} \right. \hfill \cr
h_{\,n} = 3h_{\,n - 1} - 4n + 2\left[ {n = 0} \right] \hfill \cr} \right.
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
Then, multiplying by $x^n$ and summing, you get
$$
\eqalign{
& g(x) = \sum\limits_{0\, \le \,n} {h_{\,n} x^{\,n} } = \cr
& = 3\sum\limits_{0\, \le \,n} {h_{\,n - 1} x^{\,n} } - 4\sum\limits_{0\, \le \,n} {nx^{\,n} } + 2\sum\limits_{0\, \le \,n} {\left[ {n = 0} \right]x^{\,n} } = \cr
& = 3x\sum\limits_{0\, \le \,n} {h_{\,n - 1} x^{\,n - 1} } - 4x\sum\limits_{0\, \le \,n} {nx^{\,n - 1} } + 2 = \cr
& = 3x\left( {h_{\, - 1} x^{\, - 1} + \sum\limits_{1\, \le \,n} {h_{\,n - 1} x^{\,n - 1} } } \right) - 4x{d \over {dx}}\sum\limits_{0\, \le \,n} {x^{\,n} } + 2 = \cr
& = 3x\sum\limits_{0\, \le \,n} {h_{\,n} x^{\,n} } - 4x{d \over {dx}}\left( {{1 \over {1 - x}}} \right) + 2 = \cr
& = 3xg(x) - 4x\left( {{1 \over {\left( {1 - x} \right)^{\,2} }}} \right) + 2 \cr}
$$
that is
$$
\eqalign{
& \left( {3x - 1} \right)g(x) = {{4x} \over {\left( {1 - x} \right)^{\,2} }} - 2 = - 2\,{{x^{\,2} - 4x + 1} \over {\left( {1 - x} \right)^{\,2} }} \cr
& g(x) = 2\,{{x^{\,2} - 4x + 1} \over {\left( {1 - x} \right)^{\,2} \left( {1 - 3x} \right)}} \cr}
$$
At this point you can decompose $g(x)$ into partial fractions
$$
g(x) = 2\,{{x^{\,2} - 4x + 1} \over {\left( {1 - x} \right)^{\,2} \left( {1 - 3x} \right)}} = {2 \over {\left( {1 - x} \right)^{\,2} }} + {1 \over {\left( {1 - x} \right)}} - {1 \over {\left( {1 - 3x} \right)}}
$$
and finally get
$$
h_{\,n} = 2\left( \matrix{
n + 1 \cr
1 \cr} \right) + 1 - 3^{\,n} = 2n + 3 - 3^{\,n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find value of $x^2+y^2$ using given expression
Consider two real numbers $x , y$ such that $$\left(x^2+1\right)\left(y^2+1\right)+9=6\left(x+y\right)$$ Hence find the value of $x^2+y^2$.
At first I tried to factorise the condition but the $(xy)^2$ created much problems. I also tried to create a complete square to get a simpler expression but to no avail. And just out of curiosity, can this question be interpreted geometrically so as to find the required value?
| Hint. You can rewrite this as $$(y^2 + 1) x^2 - 6x + (y^2 - 6y + 10) = 0.$$
Taken as a quadratic function of $x$ this has discriminant $$36 - 4(y^2 + 1)(y^2 - 6y+10) = -4 (y^2 - 3y + 1)^2;$$ in particular, the only way to get a real solution is with $y^2 - 3y + 1 = 0$, i.e. $y = \frac{3 \pm \sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to factorize polynomials to the 5th degree? I have the polynomial:
$$2x^5-x^4+10x^3-5x^2+8x-4$$
and I know that the final result is:
$$(2x-1)(x^4+5x^2+4) = (2x-1)(x^2+1)(x^2+4)$$
But how would you do it step by step? I've seen a couple of videos and blogs about it, but they mostly use examples, where their is a common factor between the expressions but in this case their are none.
| Laboring under the assumption that such a polynomial is not irreducible and can be factored (which is clear in this case since you already have a factorization), we would generally start by looking for rational roots.
The rational root theorem asserts that if $z = \frac{a}{b}$ is a rational root of the polynomial (in lowest terms), then $b$ must divide the leading coefficient, and $a$ must divide the constant term. This gives us several potential rational roots:
$$ \left\{ \pm 1, \pm 2, \pm 4, \pm \frac{1}{2} \right\}. $$
By trial and error, it is possible to determine that the only one of these that is, in fact, a root is $\frac{1}{2}$. From this, we know that
$$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right) q(x), $$
where $q$ is a polynomial of degree 4. Via polynomial long division, Horner's algorithm / synthetic division, or some other technique, we can deduce that $q(x) = 2x^4 + 10x^2 + 8$, thus
$$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right)\left( 2x^4 + 10x^2 + 8 \right). $$
The last term is quadratic in $x^2$. Via the quadratic formula, completing the square, or some other technique, we can determine that it, too, factors. For example, we can "factor by grouping:"
$$ 2x^4 + 10x^2 + 8
= 2x^4 + 8x^2 + 2x^2 + 8
= 2x^2(x^2 + 4) + 2(x^2+4)
= (2x^2 + 2)(x^2 + 4).
$$
Note that both of these factors are irreducible over $\mathbb{R}$ (they have complex roots, but no purely real roots), hence we can stop here. Putting all of the pieces back together, we get
$$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4
= \left( x - \frac{1}{2} \right) (2x^2 + 2)(x^2 + 4),
$$
which is equivalent to the factorization you gave.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $ First, the original problem follows,
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}$$
where $$H_{k}=\sum_{j=1}^k \frac{1}{j}$$ is the $k$-th partial sum of harmonic series.
Using the following identity,
$$H_{k+1} = \sum_{j=1}^\infty (\frac{1}{j} - \frac{1}{k+j+1}).$$
I was able to get this one.
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}=\sum_{j=1}^\infty [\frac{1}{j} - \frac{1}{j+1} + \frac{1}{j+1}\sum_{k=1}^\infty \frac{1}{(k+1)(k+1+j)}] $$
So, If I get the sum of the double series,
$$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}.$$
I can also find the original problem.
What method can I use at this problem?
| We show here how to solve the original problem.
Partial fraction decomposition gives
$$\frac{1}{k (k+1)}=\frac{1}{k}-\frac{1}{k+1}\tag{1}$$
Hence the sum can be written as
$$s = \sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right) H_{k+1}$$
$$=\sum _{k=1}^{\infty } \left(\frac{H_{k+1}}{k}-\frac{H_{k+1}}{k+1}\right)$$
Now we have
$$H_{k+1}=H_{k}+\frac{1}{k+1}$$
so that $s=s_{1} + s_{2}$ where
$$
\begin{align}
s_{1}&=\sum _{k=1}^{\infty } \left(\frac{H_k}{k}-\frac{H_{k+1}}{k+1}\right)\\
&=( \frac{H_1}{1}-\frac{H_2}{2})+(\frac{H_2}{2}-\frac{H_3}{3}) +\text{...} =\frac{H_1}{1}=1\\
s_{2}&=\sum _{k=1}^{\infty } \frac{1}{k (k+1)}=\sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right)\\
& =( \frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3}) +\text{...} =\frac{1}{1}=1
\end{align}$$
As we can see both series telescope so that we get $s_{1}=1$ and $s_{2}=1$
so that finally $s=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Transformation matrices in $\mathbb{R}^2 \to \mathbb{R}^2$ Let $v_1=(1,1)$ and $v_2=(-1,1)$ vectors in $\mathbb{R}^2$. They are clearly linearly independent since each is not an scalar multiple of the other. The following information about a linear transformation $f: \mathbb{R}^2 \to \mathbb{R}^2$ is given: $$f(v_1)=10 \cdot v_1 \text{ and } f(v_2)=4 \cdot v_2$$
*
*Give the transformation matrix $_vF_v$ with respect to ordered basis $\mathcal{B}=(v_1,v_2)$
*Give the transformation matrix $_eF_e$ with respect to the ordered standard basis $e=(e_1,e_2)$ of $\mathbb{R}^2$
Recall that
$$ \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} $$
We need a matrix $_eF_e$ such that:
$$_eF_e\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 10 & 0 \\ 0 & 4 \end{bmatrix}$$
then
$$_eF_e=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}^{-1}
=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}=\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix}$$
Okay so I'm pretty sure that $$_eF_e=_eF_v \cdot _vF_v \cdot _vF_e$$
And i figured I could find $_eF_e=\begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix}$ in the following equation $$\begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix} \text{ } \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}= \begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix} \\ \Rightarrow
\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix} \text{ } \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}= \begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} \\ \Rightarrow {}_eF_e=\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} $$
Now, how can I find ${}_v{F}_v$? I got a feeling that I'm making it more difficult than necessary
| Recall that $v_1$ and $v_2$ are linearly independent eigenvectors of $f$ since:
$$f(v_1)=10\cdot v_1$$
$$f(v_2)=4\cdot v_2$$
The vectors $v_1$ and $v_2$ form a basis of $\mathbb{R}^2$ consisting of eigenvectors of $f$. So $_vF_v$ is similar to a diagonal matrix whose elements in the diagonal are the eigenvalues $\lambda_1=10$ and $\lambda_2=4$ of $f$.
So your matrix $_vF_v$ that represents $f$ w.r.t. the ordered basis $\mathcal{B}=(v_1,v_2)$ should be the diagonal form bellow:
$$_vF_v=\left[\begin{array}{cc} 10&0\\ 0&4\end{array}\right]$$
Also, if you let $P$ to be the matrix whose columns are $v_1$ and $v_2$:
$$P=\left[\begin{array}{c} \\ \ v_1\\ \\ \end{array} \middle\vert \begin{array}{c} \\ \ v_2\\ \\ \end{array}\right]=\left[\begin{array}{cc} 1&-1\\ &\\1& \ 1\end{array}\right]$$
then
$$_{e}F_e = P \cdot{}_vF_v\cdot P^{-1}$$
or you can do this:
$$(x,y)=\alpha\cdot v_{1}+\beta\cdot v_{2}
=\alpha\cdot(1,1)+\beta\cdot(-1,1)=(\alpha-\beta,\alpha+\beta)$$
so
$$\alpha=\frac{x+y}{2}$$
$$\beta=\frac{y-x}{2}$$
then:
$$ (x,y)= \frac{(x+y)}{2}\cdot v_1 + \frac{(y-x)}{2}\cdot v_2$$
Apply $f$ in both sides to get:
$$f(x,y) = \frac{(x+y)}{2}\cdot f(v_1) + \frac{(y-x)}{2}\cdot f(v_2) $$
$$f(x,y) = \frac{(x+y)}{2}\cdot 10\cdot v_1 + \frac{(y-x)}{2}\cdot 4\cdot v_2$$
$$ = 5(x+y)\cdot v_1+2(y-x)\cdot v_2$$
$$ = (5x+5y,5x+5y)+(2x-2y,2y-2x)=(7x+3y,7y+3x)=(7x+3y,3x+7y)$$
So
$$f(x,y)=(7x+3y,3x+7y)$$
then
$$f(1,0)=(7,3)$$
$$f(0,1)=(3,7)$$
this implies that:
$$_{e}F_e = \left[\begin{array}{cc} 7&3\\ &\\3& 7\end{array}\right] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit
I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
| Here's a more elementary argument. Note that mod $10$ we have
$$a^{4n+1}-a = a(a^{4n}-1) = a(a^{2n}-1)(a^{2n}+1) \equiv a(a^{2n}-1)(a^{2n}-9) =a(a^n-1)(a^n+1)(a^n-3)(a^n+3) \pmod{10}.$$
If $\gcd(a,10)=1$ or $10$, then $a^n \equiv 0, \pm1 \mbox{ or }\pm3 \pmod{10}$, making the above product congruent to $0$ mod $10$.
If $\gcd(a,10) = 5$, then $a$ is odd and each of the other factors in the product is even, so again the product is congruent to $0$ mod $10$.
If $\gcd(a,10) = 2$, then $a^n \equiv 2, 4, 8, \mbox{ or } 6 \pmod{10}$ Each of these is $\pm 1$ or $\pm 3$ from $5$, so $5$ is a factor of the product and again the product is congruent to $0$ mod $10$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Checking for convergence of $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ I have the following series $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ which I am checking for convergence.
My working:
$\lim_{n\to\infty}|a_n|=0$ where $a_n=\frac{(-1)^nn}{n^2+1}$.
Doesn't that mean $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ is absolutely convergent?
My book says it's conditionally convergent. What am I missing?
| The sum of each
even-odd pair is
$\begin{array}\\
\dfrac{2n}{(2n)^2+1}-\dfrac{2n+1}{(2n+1)^2+1}
&=\dfrac{2n((2n+1)^2+1)-(2n+1)(4n^2+1)}{((2n)^2+1)((2n+1)^2+1)}\\
&=\dfrac{4n^2+2n-1}{(4n^2+1)(4n^2+4n+2)}\\
&\lt \dfrac1{8n^2}\\
\end{array}
$
and the sum of these converges.
Since
$\dfrac{2n}{(2n)^2+1}
\to 0$,
the sum of the whole series
converges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is the coefficient of $x^{9}$ in the expansion of $(1+x)^{12}(1+x)^{4}$? Answer in the textbook: $\sum_{k=0}^{4} \binom{12}{9-k}\binom{4}{k}$
Can I just multiply the two terms together to get $(1+x)^{16}$ then apply the binomial coefficient theorem?
I don't know how they got the sum at the end, I was only taught how to find the coefficient of a term in one binomial expansion using the binomial coefficient theorem...
| The factorization $(1+x)^{12}(1+x)^4$ is not accidental here. Let's expand by using Binomial Theorem:
$$\sum_{k=0}^{12}\binom{12}{k}x^k\cdot\sum_{m=0}^{4}\binom{4}{m}x^m.$$
What about $x^9$ in this expansion? We get this power when $k=9,m=0$, $k=8,m=1$, $k=7,m=2$, $k=6,m=3$ and finally $k=5,m=4$. Then our coefficient is $$\binom{12}{9}\cdot\binom{4}{0}+\binom{12}{8}\cdot\binom{4}{1}+\binom{12}{7}\cdot\binom{4}{2}+\binom{12}{6}\cdot\binom{4}{3}+\binom{12}{5}\cdot\binom{4}{4}.$$ It's that!!!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the proof I am using, sufficient/ correct for the system of equation? I want to ask MSE to confirm the correctness of the alternate solution and its mistake.
I know possible solution: https://math.stackexchange.com/a/2557094/456510
If $x,y,z\in {\mathbb R}$, Solve the system equation:
$$
\left\lbrace\begin{array}{ccccccl}
x^4 & + & y^2 & + & 4 & = & 5yz
\\[1mm]
y^{4} & + & z^{2} & + & 4 & = &5zx
\\[1mm]
z^{4} & + & x^{2} & + & 4 & = & 5xy
\end{array}\right.
$$
I wrote a solution myself (after more work).
My attempts / solution:
It is obvious that, if $x>0,y>0,z>0$ are solutions, $x<0,y<0,z<0$ are also solutions and it is obvious $x≠0,y≠0,z≠0$.
If the equations have a solution, then $ x = y = z $ should be.
Proof:
I will accept $x,y,z\in {\mathbb R^+}$
a-1)
Let $x≥z>y$
We can write :
$z^4>y^4 \\ x^2≥z^2 \\ z^4+x^2+4>y^4+z^2+4 \\ 5xy > 5zx \\ y>z$
We get the contradiction : $y>z$
Because, it must be $z>y$
a-2)
Let $x>z≥y$
We can write:
$z^4≥y^4 \\ x^2>z^2 \\ z^4+x^2+4>y^4+z^2+4 \\ 5xy > 5zx \\ y>z$
We get the same contradiction : $y>z$
Because, it must be $z≥y$
b)
$y≥x>z$
We can write:
$x^4>z^4 \\ y^2≥x^2 \\ x^4+y^2+4>z^4+x^2+4 \\ 5yz > 5xy \\ z>x$
But, this is contradiction, because it must be $z<x$.
We get the same contradiction for : $y>x≥z$
c)
$y>z≥x$
We can write:
$y^4>z^4 \\ z^2≥x^2 \\ y^4+z^2+4>z^4+x^2+4 \\ 5zx > 5xy \\ z>y$
But, this is contradiction, because it must be $z<y$.
We get the same contradiction for : $y≥z>x$
d)
$z>x≥y$
We can write:
$z^4>x^4 \\ x^2≥y^2 \\ z^4+x^2+4>x^4+y^2+4 \\ 5xy > 5yz \\ x>z$
But, this is contradiction, because it must be $z>x$.
We get the same contradiction for : $z≥x>y$
e)
$z≥y>x$
We can write:
$y^4>x^4 \\ z^2≥y^2 \\ y^4+z^2+4>x^4+y^2+4 \\ 5zx > 5yz \\ x>y$
But, this is contradiction, because it must be $x<y$.
We get the same contradiction for : $z>y≥x$
f)
$x>y≥z$
We can write:
$x^4>y^4 \\ y^2≥z^2 \\ x^4+y^2+4>y^4+z^2+4 \\ 5yz > 5zx \\ y>x$
But, this is contradiction, because it must be $x>y$.
We get the same contradiction for : $x≥y>z$
Then, solution must be $x=y=z$ (if there is a solution).
The proof is completed.
Finally,
$$x^4+x^2+4-5x^2=0 \Rightarrow x^4-4x^2+4=0 \Rightarrow (x^2-2)^2=0 \Rightarrow x=±\sqrt2\Rightarrow x=y=z=±\sqrt2 .$$
Is my proof/ solution correct?
Thanks.
| Good job. Just too verbose.
You are correct into saying you can assume $x$, $y$ and $z$ all positive (there will be a corresponding solution with their negatives). The case when two are positive and one negative cannot appear, nor can the case of two negative and one positive, because the positivity of the left-hand sides forces positivity of the right-hand sides, so all three numbers must share the sign.
However, there is another simplification, namely, you can also assume $x$ is the maximum solution, because the equations are cyclic. Thus
$$
x\ge y\ge z \qquad\text{or}\qquad x\ge z>y
$$
You have already excluded the second case, so we can concentrate on the first.
In order to show that for a solution you need $x=y=z$, you just have to exclude $x>y$ and $y>z$.
In the case $x>y\ge z$, we have, according to your method,
$$
x^4>y^4
\qquad
y^2\ge z^2
$$
Then
$$
5yz=x^4+y^2+4>y^4+z^2+4=5zx
$$
which implies $y>x$: a contradiction.
In the case $x\ge y>z$ we have
$$
y^2>z^2
\qquad
x^4\ge y^4
$$
which implies
$$
5yz=x^4+y^2+4>y^4+z^2+4=5zx
$$
implying $y>x$, again a contradiction.
We saw that assuming either $x>y$ or $y>z$ leads to a contradiction. Since $x\ge y\ge z$ by assumption and we cannot have neither $x>y$ nor $y>z$, we deduce that $x=y$ and $y=z$.
Now, finding what's the common value is easy: we have
$$
x^4-4x^2+4=0
$$
so $x^2=2$ and $x=\pm\sqrt{2}$. The problem has exactly two solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What can we learn as we reduce the fraction at the end of the quadratic formula process? In the final throes of the quadratic formula, you reduce a fraction. Consider the following two examples.
*
*$y = 6x^2 + 11x + 3$; the quadratic formula reveals the roots $x = -4/12$ or $x = -18/12$. The parts of the first fraction (4 and 12) have a common factor of 4, so the fraction reduces to -1/3. The parts of the second fraction (18 and 12) have a common factor of 6, so the fraction reduces to -3/2. The common factor of all three parts (4, 12 and 18) is 2.
*$y = 42x^2 + 77x + 21$; the quadratic formula reveals the roots $x = -28/84 $or $x = -126/84$. The parts of the first fraction (28 and 84) have a common factor of 28, so the fraction reduces to -1/3. The parts of the second fraction (126 and 84) have a common factor of 42, so the fraction reduces to -3/2. The common factor of all three parts (28, 84 and 126) is 14.
Now. Compare the two solutions.
Equation 2 is the product of Equation 1 times 7. Apparently as a result, the common factors of the equations were all multiplied by 7, and after reducing the fractions we discover the same roots.
My question is whether that step, reducing the fraction, reveals anything about the numerical factors of the equations.
It seems to me that any common numerical factor of a, b and c (like 7 in Equation 2) will also be a common factor of the parts of the fractions, and cancel when we reduce the fraction. But the parts of the fraction can have other common factors (like 2 in both Equations).
So when we come to that step, of reducing the fraction, does the common factor reveal anything about the numerical common factors of a, b and c?
| Suppose that we have some quadratic $q(x) = Ax^2 + Bx + C$, where $A, B, C$ are integers with a common factor $k$. Since they have a common factor $k$, there exist integers $a, b, c$ such that $A = ak$, $B = bk$, $C = ck$. Applying the quadratic formula to $q(x)$ gives
$$ \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-bk \pm \sqrt{k^2b^2 - 4k^2ac}}{2ka} = \frac{-bk \pm k\sqrt{b^2 - 4ac}}{2ka}$$
and so we can see that the common factor $k$ will definitely show up in both the numerator and denominator of the roots given by the quadratic formula.
However, you may get "false positives" this way. For example, consider $q(x) = x^2 + 2x + 1$. Then, the quadratic formula gives the roots as
$$\frac{-2 \pm \sqrt{4 - 4}}{2} = -\frac{2}{2}$$
but $2$ is not a common divisor of $a, b, c$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of possible matrices with the following property? I have a $8\times 8$ matrix, where every element of the matrix is either $0$, $1$ or $2$, so I define $A$ as a $8\times 8$ ternary matrix.
The matrix needs to have these properties:
*
*In the rows $1,2$ and $3$ the number $1$ must appear exactly once and no number $2$ can appear (per row).
*In the rows $4,5$ and $6$ the number $2$ must appear exactly once and no number $1$ can appear.
*In the rows $7$ and $8$ the numbers $1$ and $2$ must appear exactly once per row.
*All properties of the rows are held in the columns too.
An example would be
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 & 0 & 0 & 2 & 1
\end{pmatrix}
How many matrices can we construct having these properties? What I find is that every configuration of an element is strongly connected to all the others, so I think the number cannot be very big.
| Hint:
The top-left $3\times3$ minor is the identity or a permutation of the identity.
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor series for $\sqrt{1+x^2}$ I want to expand
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\frac{1}{2}x^2+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{x^4}{2!}+\frac{1}{2} \left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\frac{x^6}{3!}$$
Then
$$\sqrt{1+x^2}=\sqrt{1+(x-2)^2-4+4x}=\sqrt{(x-2)^2-3+4x}=\sqrt{(x-2)^2-3+4(x-2)+8}$$
I could not complete , what can we do then ?
(I know that we can differentiate the function and substitute in the Taylor formula , but I want a shorter way)
for example :
if we want to expand $$g(x)=\frac{1}{1-x}$$ around $x=2$ we can start by
\begin{align}
g(x) & =\frac{1}{1-x}=\frac{1}{1-(x-2)-2}=\frac{-1}{1+(x+2)} \\[10pt]
& =-[1-(x-2)+(x-2)^2-(x-2)^3+\cdots]
\end{align}
So I want to get convert $f(x)$ to a form that we can write its expansion without getting derivatives, like I did with $g(x)$ above
| The Generalized binomial formula for $a\in \Bbb R\setminus \{\Bbb N\}$ says
$$(1+x)^a= \sum_{k=1}^{\infty}{a\choose k} x^k$$
where: $$ \color{blue}{{a\choose k} =\frac{a(a-1)\cdots(a-k+1)}{k!}}$$
see here https://en.wikipedia.org/wiki/Binomial_series
just take $a=1/2$ and replace $x$ by $x^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Proof of Wilsons theorem How to understand the grouping of integers from $2$ to $p-2$ in the proof af Wilsons theorem?
I don't understand how you can group the integers $2$ to $p−2$ into $(p-3)/2$ and next how you become the $2*3*...=1 (mod p)$
| I think the best way to illustrate what's happening is to apply the proof to a specific value of $p$. Take, for example, $p = 11$.
We can solve $2x \equiv 1 \pmod {11}$ with $x = 6$. We can solve $3x \equiv 1 \pmod {11}$ with $x = 4$. We can solve $5x = 1 \pmod {11}$ with $x \equiv 9$. We can solve $7x \equiv 1 \pmod{11}$ with $x = 8$.
Note that in the above, we have accounted for all numbers from $2$ to $(11-2) = 9$. We can now rearrange the product
$$
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 =\\
(2 \cdot 6) \cdot (3 \cdot 4) \cdot (5 \cdot 9) \cdot (7 \cdot 8) \equiv\\
1 \cdot 1 \cdot 1 \cdot 1 \equiv 1 \pmod{11}
$$
That is, by grouping the integers from $2$ to $11 - 2$ into $\frac{11 - 3}{2} = 4$ pairs of multiplicative inverses, we have shown that
$$
2 \cdot 3 \cdots (11 - 2) \equiv 1 \pmod{11}
$$
which allows us to proceed with the proof.
| {
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"url": "https://math.stackexchange.com/questions/2565184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratic equation estimating changes in b and c If $x^2 + bx + c=0$ is an equation the roots are
$$
\begin{aligned}
x&=\frac{-b + \sqrt{b^2 - 4c}}{2}\\
x&=\frac{-b - \sqrt{b^2 - 4c}}{2}
\end {aligned}$$
What is the effect of small changes in $b$ or $c$?
For example if I have $\sqrt{a}$ then to estimate a small change in this equation I would write
$$\sqrt{a + \Delta a}$$
and set $v = \sqrt{a}$ and
$$v + \Delta v =\sqrt{a + \Delta a}$$
Therefore $a + \Delta a = (v + \Delta v)^2$ which is roughly
$v^2 + 2v \times \Delta v$ (ignoring $\Delta v^2$ as it's too small) and thus
$$\Delta a = 2\sqrt{a} \times \Delta v$$
so with some rearranging $\Delta v$ is roughly $\frac{1}{2\sqrt{a}} \times\Delta a$
(to the first order) and this lets me estimate square roots.
How can I use a method like this to estimate small changes in $b$ and $c$?
| Let us consider $$x_1=\frac{-b + \sqrt{b^2 - 4c}}{2} \qquad \text{and} \qquad x_2=\frac{-b - \sqrt{b^2 - 4c}}{2}$$ and use Taylor expansions around, say, $b_0$ or $c=c_0$.
For a change in $b$, what we would obtain is
$$x_1=\frac{1}{2} \left(\sqrt{b_0^2-4 c}-b_0\right)+(b-b_0) \left(\frac{b_0}{2 \sqrt{b_0^2-4
c}}-\frac{1}{2}\right)+O\left((b-b_0)^2\right)$$
$$x_2=-\frac{1}{2}\left( \sqrt{b_0^2-4 c}+{b_0}\right)+(b-b_0) \left(-\frac{b_0}{2
\sqrt{b_0^2-4 c}}-\frac{1}{2}\right)+O\left((b-b_0)^2\right)$$ that is to say
$$\Delta x_{1,2}=\left(\frac{b_0}{2 \sqrt{b_0^2-4
c}}-\frac{1}{2}\right)\Delta b$$
For a change in $c$, what we would obtain is
$$x_1=\frac{1}{2} \left(\sqrt{b^2-4 c_0}-b\right)-\frac{c-c_0}{\sqrt{b^2-4
c_0}}+O\left((c-c_0)^2\right)$$
$$x_2=-\frac{1}{2} \left(\sqrt{b^2-4 c_0}+{b}\right)+\frac{c-c_0}{\sqrt{b^2-4
c_0}}+O\left((c-c_0)^2\right)$$ that is to say
$$\Delta x_{1,2}=\pm {\sqrt{b^2-4
c_0}}\,\Delta c$$
Edit
I did focus too much on the quadratic equation. So, let us do it for a general equation $$F(x,a_1,a_2,\cdots,a_n)=0$$ (considered as an implicit function) and let compute the partial derivatives. We have
$$\frac{\partial x}{\partial a_i}=-\frac{\frac{\partial F(x,a_1,a_2,\cdots,a_n)}{\partial a_i}} {\frac{\partial F(x,a_1,a_2,\cdots,a_n)}{\partial x}}$$ which makes $$\Delta x=\frac{\partial x}{\partial a_i}\Delta a_i$$ which can be applies to any root (analytical or not) and any parameter.
Applied to $F(x,b,c)=x^2+bx+c=0$, this would lead to $$\Delta x=-\frac{x}{b+2 x} \Delta b \qquad \text{and}\qquad \Delta x=-\frac{1}{b+2 x}\Delta c$$ Now, replace $x$ by $x_1$ and $x_2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x) &= (x^2c) \cdot \int \frac{x^3}{x^2} dx \\
&= (x^2c) \cdot \frac{x^2}{2}
\end{align}
but this solution is wrong. Where is the mistake?
| If $y'-p(x)y=0$ is a homogeneous equation then its general solution is given by $y=Ce^{\int p(x)dx}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then have: $$\forall \epsilon > 0 \exists N\in\mathbb{N}_0 \forall n \geq N: \left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon$$
and after doing a bit of math we have $$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon \Leftrightarrow \left|\frac{1}{n}+2-2\right| < \epsilon \Leftrightarrow \frac{1}{n} < \epsilon \Leftrightarrow \frac{1}{\epsilon} < n $$ which means that we need a $N\in\mathbb{N}$ with $N > \frac{1}{\epsilon}$ and we have $\forall n\geq N$:$$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| = \dots=\frac{1}{n}\leq\frac{1}{N}<\epsilon\,.$$
And hence our sequence $a_n \longrightarrow a_\infty$ with $a_\infty = 2$$._{\,\,\square}$
Did I do this right?
| Note that $$a_n = \dfrac{(n+1)^2-n^2}{n} = \dfrac{(n^2+2n+1)-n^2}{n} = 2 + \dfrac{1}{n}\,.$$ Therefore, $$\lim a_n = \lim {}\biggl( 2 + \dfrac{1}{n} \biggr) = \lim 2 + \lim \dfrac{1}{n} = 2 + 0 = 2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.