Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(c-b)(c+b)$$
Which is at least a product of linear factors.
However I am now stuck as to how to proceed.
Also I would prefer the $(b^2-a^2)$ and the $(c^2-b^2)$ factors of my second line of working to be in the form $(a^2-b^2)$ and $(b^2-c^2)$, for 'neatness' , if possible
|
the resultat should be $$- \left( b-c \right) \left( a-c \right) \left( a-b \right) \left( a
+c+b \right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Finding $\sin^3\frac{x}{2}\cdot\cos^7\frac{x}{3}$ periodicity I need to find the period of $\sin^3\frac{x}{2}\cdot\cos^7\frac{x}{3}$ but couldn't find a nice method how? any ideas?
|
Notice that:
$$\sin\left(\frac{x}{2}\right) = \sin\left(\frac{x}{2} + 2k\pi\right) = \sin\left(\frac{x+4k\pi}{2}\right)$$
and
$$\cos\left(\frac{x}{3}\right) = \cos\left(\frac{x}{3} + 2h\pi\right) = \cos\left(\frac{x+6h\pi}{3}\right)$$
We need to find the minimum common period between $4k\pi$ and $6h\pi$ which is $12\pi$ (set $k=3$ and $h=2$).
Indeed:
$$\sin^3\left(\frac{x+12\pi}{2}\right)\cos^7\left(\frac{x+12\pi}{3}\right) = \sin^3\left(\frac{x}{2}+6\pi\right)\cos^7\left(\frac{x}{3}+4\pi\right) = \sin^3\left(\frac{x}{2}\right)\cos^7\left(\frac{x}{3}\right).$$
The period is $12\pi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1521127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the center of circle touching a line pair and a point. A circle passes through the point 3,$\sqrt{\frac{7}{2}}$ and touches the line pair $x^2-y^2-2x+1=0$. The co-ordinates of the centre of the circle are:-
My attempt:-
Using quadratic formula to separate the line pair into two lines.
$$x^2-y^2-2x+1=0$$
$$x=\frac{-(-2) \pm \sqrt{2^2-4(1-y^2)(1)}}{(2)(1)}$$
$$x=1\pm y$$
So we have two lines x-y-1=0 and x+y-1=0.I don't know that to do next?
|
So your two lines are $y=x-1$ and $y=-x+1$.
By symmetry we know the center of the circle must be on $x$-axis. Let the center be $(a,0)$ then the radius is $\large{a-1\over\sqrt{2}}$ because the angle between the two lines is 90 degrees and hence the two radius along with the two lines form a square where the $x$-axis is the diagonal.
Hence the circle is $(x-a)^2+y^2=\large{(a-1)^2\over2}$. Sub in your point, $(3-a)^2+{7\over2}=\large{(a-1)^2\over2}$.
$a^2-10a+24=0$ and you can find $a=4,6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1523541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
|
$$3^{3n+4}+2^{n+2}=3^{3n+1}\cdot 3^{3} +2^{n+1}\cdot 2\\
=(5k-2^{n+1})\cdot 3^{3}+2^{n+1}\cdot 2\\
=2^{n+1}(2-27)+5k\cdot 27=-2^{n+1}\cdot 25+5k\cdot 27\\
=5(-2^{n+1}\cdot 5+27k)$$ which is divisible by $5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1526409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
}
|
$x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}$. Then what is the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\}$?
If the equation $x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}\;,$ Where $x_{1}<x_{2}<x_{3}$.
Then the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\} = \;,$ Where $\{x\}$ Represent fractional part of $x.$
$\bf{My\; Try::}$ Let $f(x) = x^3-3x+1\;,$ Then Using Second Derivative Test, We get
$f'(x) = 3x^2-3$ and $f''(x) = 6x\;,$ Now for $\max$ and $\min\;,$ Put $f'(x) =0\;,$ So we get
$3x^2-3=0\Rightarrow x = \pm 1.$ Now $f''(1) = 6>0$
So $x=1$ is a Point of $\min$ and $f''(-1) = -6<0$
So $x=-1$ is a Point of $\max.$
And Rough graph of $f(x) = x^3-3x+1$ is Like
So we get $-2<x_{1}<-1$ and $0<x_{2}<1$ and $1<x_{3}<2$
So $\lfloor x_{1}\rfloor = -2$ and $\lfloor x_{2}\rfloor = 0$ and $\lfloor x_{3}\rfloor = 1$
Now I did not Understand How can I calculate value of fractional part
Help me, Thanks
|
$\left\{x_1\right\}+\left\{x_2\right\}+\left\{x_3\right\}=x_1+x_2+x_3-[x_1]-[x_2]-[x_3]=0-(-2+0+1)=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1527100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
|
We can clearly see that: $f(1)+1=0$ and $f(-1)-1=0$
We can write $f(x)-1=p(x)(x+1)^3$ and $f(x)+1=q(x)(x-1)^3$
By differentiation and double differentiation, you can see that
$f'(1)=0$ and $f''(1)=0$
AND
$f'(-1)=0$ and $f''(-1)=0$
You got six conditions and six unknowns!
[assume $f(x) = x^6+a_1x^5+a_2x^4+a_3x^3+a_4x^2+a_5x+a_6$]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1528699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
}
|
How to solve the equation $t = \sqrt{x^2 - 1} - x$ for $x$?
Let the equation $t = \sqrt {x^2 - 1} - x$. Find $x$.
So I've tried the following:
$$t^2 = x^2 - 1 -2\sqrt {x^2-1}x + x^2 = 2x^2 - 2\sqrt{x^2-1}x - 1 $$
What should I do next?
|
The following should be better :
$$\sqrt{x^2-1}=t+x\quad\Rightarrow\quad x^2-1=(t+x)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1529033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the following function:
$$\lim _{ x\rightarrow \infty }{ \frac { \sqrt { x+1 } }{ \sqrt { x } +1 } =1 } $$
Steps I took:
$$\left| \frac { \sqrt { x+1 } }{ \sqrt { x } +1 } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { \sqrt { x+1 } }{ \sqrt { x } +1 } -\frac { \sqrt { x } +1 }{ \sqrt { x } +1 } \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { \sqrt { x+1 } -\sqrt { x } -1 }{ \sqrt { x } +1 } \right| <\frac { 1 }{ 3 } $$
How can I manipulate the function inside of the absolute value in order to simplify this and find a lower bound $x$ (the M)?
|
First rationalization
i.e. $\left|\frac{\sqrt{x+1}-\sqrt{x}-1}{\sqrt{x}+1}\right|$
= $\left|\frac{(\sqrt{x+1}-\sqrt{x}-1)(\sqrt{x}-1)}{x-1}\right|$
=$\left|\frac{\sqrt{x(x+1)}-\sqrt{x+1}-x+1}{x-1}\right|$
Then, we could choose M to be a particular value to solve the problem(by taking maximum of M at last). In this case, take $M=2$
$\because x > M = 2$
$\frac{x}{2} > 1$
$\therefore$ $<\left|\frac{\sqrt{x+1}(\sqrt{x}-1)-x+1}{x-\frac{x}{2}}\right|$
=$\left|(\frac{2}{x})(\sqrt{x+1}(\sqrt{x}-1)-x+1)\right|$
<$\left|(\frac{2}{x})(\sqrt{x+x}(\sqrt{x}-1)-x+1)\right|,x>2$
=$\left|(\frac{2}{x})(2x-2\sqrt{x}-x+1)\right|$
=$\left|(\frac{2}{x})(1-\sqrt{x})^2\right|$
<$\left|(1-\sqrt{x})^2\right|$<$\frac{1}{3},x>2$
$\because (1-\sqrt{x})^2 < \frac{1}{3}$
$ x > (1-\sqrt\frac{1}{3})^2$
Therefore, take $M = max${${(1-\sqrt\frac{1}{3})^2},2$}$=2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1529490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Determine the minimal polynomial of $T$ with respect to $\begin{pmatrix} 0 &0& 1 \end{pmatrix}^t.$ Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(v)=\begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix}v.$ Determine the minimal polynomial of $T$ with respect to $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.$
$T(v)= \begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -4 \end{pmatrix}$
$T^2(v)= \begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix} \begin{pmatrix} 1 \\ -5 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ -2 \end{pmatrix}$
$T^3(v)= \begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix} \begin{pmatrix} 3 \\ -3 \\ -2 \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \\ -10 \end{pmatrix}$
The corresponding matrix is
$A = (v,T(v),T^2(v),T^3(v))= \begin{pmatrix} 0 & 1 & 3 & 7 \\ 0 & -5 & -3 & - 11 \\ 1 & -4 & -2 & -10 \end{pmatrix}$
After row reducing this matrix, we obtain $ \begin{pmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 2 \end{pmatrix}$, and so $\text{null}(A) = \text{span} \left\{\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right\}.$ From this we obtain polynomials $f(x) = x^2$ and $g(x) = x^3-x-2$ and since $\deg f(x) < \deg g(x),$ $x^2$ is the minimal polynomial.
The back of my book says the minimal polynomial is $x^2-3x+2.$ What am I doing wrong? I assume it is an arithmetical error, but I redid everything and got the same answer...
|
You got the nullspace of $A$ wrong. It is one-dimensional, since the rank of $A$ is $3$. The row-reduced form that you have indicates that the first three columns are linearly independent, and the fourth one is a certain linear combination of them. This yields
$$
\operatorname{null}(A) = \operatorname{span}( (2,-1,-2,1)^t)
$$
and the minimal polynomial of $v$ is
$$p(x) = x^3-2x^2-x+2$$
*
*Being of the same degree as the order of the matrix, this is also its minimal polynomial, and characteristic polynomial.
*I don't know where the book's answer comes from: it's easy to check that $T^2v-3Tv+2v\ne 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1529724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Sum of recursive series I am taking a course in algebra and this problem was on my problem set, and I had no idea how to solve it.
Suppose we have a sequence $s_n$ of real numbers such that $5s_{n+1}-s_{n}-3s_{n}s_{n+1}=1$ for $1 \leq n \leq 42$ and $s_1=s_{43}$.
What are the possible values of $s_1+s_2+ \ldots + s_{42}$?
|
Note that we have
$$s_{n+1} (5-3s_n) = 1 + s_n,$$
thus in particular $s_n \neq 5/3$ for all $n$ and so
$$\tag{1} s_{n+1} = \frac{1+s_n}{5-3s_n}.$$
The mapping
$$f(z) = \frac{z + 1}{-3z+5}$$
is a Mobius transform. One can check that
$$f^{(n)} (z) = \frac{az+ b}{cz+ d},$$
where
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -3 & 5\end{bmatrix}^n$$
Doing some linear algebra, we have
$$\begin{bmatrix} 1 & 1 \\ -3 & 5\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 3\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 4\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 3\end{bmatrix}^{-1}$$
When $n=42$, we have (direct computations)
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3\cdot 2^{42} -4^{42} & -2^{42} + 4^{42} \\ 3\cdot 2^{42} - 3\cdot 4^{42} & 3\cdot 4^{42} - 2^{42} \end{bmatrix}$$
So the fact that $s_1 = s_{43}$ is the same as
$$s_1 = \frac{(3\cdot 2^{42}-4^{42}) s_1 +(-2^{42} + 4^{42})}{(3\cdot 2^{42} - 3\cdot 4^{42})s_1 + 3\cdot 4^{42} - 2^{42}}$$
By some calculations,
$$3s_1^2 - 4s_1 + 1 = 0\Rightarrow s_1= 1 \text{ or } \frac 13.$$
Put into $(1)$, we get either $s_n = 1$ for all $n$ or $s_n = \frac 13$ for all $n$. So
$$s_1 + \cdots + s_{42} = 42 \text{ or } 14.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1531796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Calculate a limit $\lim_{x \to \pi/2} \frac{\sqrt[4]{ \sin x} - \sqrt[3]{ \sin x}}{\cos^2x}$ $$\lim_{x \to \pi/2} \frac{\sqrt[4]{ \sin x} - \sqrt[3]{ \sin x}}{\cos^2x}$$
I have an idea of replacing $\sin x$ to $n$ when $n \to 1$ but wolfram says that answer is $\frac{\pi}{48} $ so my suggestion is it's had to use trigonometry simplifications which I do not know so well.
Assuming that L'Hopital is forbidden but you can use asimptotical simplifications like big and small o notations and Taylor series.
|
Ian Miller's answer is the nicest and most efficient solution to the problem.
Just for your curiosity, I shall give you another one using Taylor series since you will use them a lot during your studies.
First, changing variable $x=\frac \pi 2+y$ $$\frac{\sqrt[4]{ \sin (x)} - \sqrt[3]{ \sin (x)}}{\cos^2(x)}=\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}$$ Now, using the generalized binomial expansion and Taylor series around $y=0$ $$cos^a(y)=1-\frac{a y^2}{2}+\left(\frac{a^2}{8}-\frac{a}{12}\right) y^4+O\left(y^6\right)$$ Using it, the numerator write $$\frac{y^2}{24}+\frac{y^4}{1152}+O\left(y^6\right)$$ and the denominator is $\approx y^2$. Then, the limit.
We could go further and use $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ So the denominator is $$y^2-\frac{y^4}{3}+O\left(y^5\right)$$ which makes $$\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}=\frac{\frac{y^2}{24}+\frac{y^4}{1152}+O\left(y^6\right) }{y^2-\frac{y^4}{3}+O\left(y^5\right) }$$ and the long division yields to $$\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}=\frac{1}{24}+\frac{17 }{1152}y^2+O\left(y^3\right)$$ which shows the limit and also how it is approached.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1534871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove $ 5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3) $ if $a^2+b^2+c^2=3$ Let $a,b,c\in\mathbb{R}$ such that $a^2+b^2+c^2=3$. Prove that:
$$
5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3)
$$
I tried to homogenize the inequality to get:
$$
5(a^4+b^4+c^4)+(a^2+b^2+c^2)^2≥\frac{8}{\sqrt3}(a^3+b^3+c^3)\sqrt{(a^2+b^2+c^2)}
$$
I hoped that one don't needs the condition anymore to prove this, but I couldn't get any further.
|
You can prove it as follows:
The inequality is equivalent to:
$$
\sum_{cyc}5a^4-8a^3+3a^2≥0\iff\sum_{cyc}5a^2\left(\left(a-\frac{4}{5}\right)^2-\frac{1}{25}\right)≥0\iff\sum_{cyc}a^2\left(\left(5a-4\right)^2-1\right)≥0\iff\\
\sum_{cyc}a^2(5a-4)^2≥3\iff\sqrt{\frac{1}{3}\sum_{cyc}a^2(5a-4)^2}≥1
$$
Now, by AM-QM:
$$
\sqrt{\frac{1}{3}\sum_{cyc}a^2(5a-4)^2}≥\frac{1}{3}\sum_{cyc}a(5a-4)=5-4\frac{a+b+c}{3}≥5-4\sqrt{\frac{a^2+b^2+c^2}{3}}=1
$$
And we're done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1537396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Prove $\liminf(M_n = -1) \subseteq ((\lim \frac{S_n}{n}) = -1)$
Let $M_1, M_2, \dots$ be iid RVs s.t. $P(M_n = n^2 - 1) = \frac{1}{n^2} = 1 - P(M_n = - 1)$. Define $S_n = \sum_{i = 1}^{n} M_i$. Prove that $P\{(\lim \frac{S_n}{n}) = -1\} = 1$.
$$\sum_n P(M_n = n^2 - 1) = \sum_n \frac{1}{n^2} < \infty$$
By BCL1, $$\to P\{\limsup(M_n = n^2 - 1)) = 0$$
$$\to P\{\liminf(M_n = -1)\} = 1$$
It seems intuitive, but how do we show rigorously that $$P\{(\lim\frac{S_n}{n}) = -1\} = 1?$$?
|
Let's denote $\liminf_n[M_n = -1]$ by $A$, by first Borel-Cantelli lemma, $P(A) = 1$. Therefore for each $\omega \in A$, there exists $N \in \mathbb{N}$, such that $M_n(\omega) = -1$ for all $n > N$. It then follows that
\begin{align}
& \left|\frac{S_n(\omega)}{n} - (-1)\right| \\
= & \left|\frac{M_1(\omega) + \cdots + M_N(\omega) - N \times (-1)}{n} \right|\\
\leq & \frac{1^2 - 1 + 2^2 - 1 + \cdots + N^2 - 1 + N}{n} \\
= & \frac{\frac{1}{6}N(N + 1)(2N + 1)}{n} \\
\to & 0
\end{align}
as $n \to \infty$. This shows that for each $\omega \in A$, $\lim\limits_{n \to \infty}\frac{S_n(\omega)}{n} = -1$. In other words, $\frac{S_n}{n} \to -1$ with probability $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1537706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Laplace Transforms Show that ${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\frac{(2n-1)!!}{2^n}\frac{1}{s^n} \sqrt{\frac{\pi}{s}}$
My Attempt:
${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\int_{0}^{\infty}\mathcal e^{-st}t^{n-\frac{1}{2}}dt$
Integration by parts gives:
${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\frac{(2n-1)}{2s}{\mathcal L} {\lbrace t^{n-\frac{3}{2}}\rbrace}$
${\mathcal L} {\lbrace t^{n-\frac{3}{2}}\rbrace}=\frac{(2n-3)}{2s}{\mathcal L} {\lbrace t^{n-\frac{5}{2}}\rbrace}$
${\mathcal L} {\lbrace t^{n-\frac{5}{2}}\rbrace}=\frac{(2n-5)}{2s}{\mathcal L} {\lbrace t^{n-\frac{7}{2}}\rbrace}$
I start to see the $\frac{(2n-1)!!}{2^n}\frac{1}{s^n}$, but not the $\sqrt{\frac{\pi}{s}}$
Thanks, in advance.
|
By the definition of the $\Gamma$ function,
$$\begin{eqnarray*}\mathcal{L}\left(t^{n-\frac{1}{2}}\right)&=&\int_{0}^{+\infty} t^{n-\frac{1}{2}}e^{-st}\,dt = \frac{1}{s^{n+\frac{1}{2}}}\int_{0}^{+\infty}u^{n-\frac{1}{2}}e^{-u}\,du\\&=&\frac{\Gamma\left(n+\frac{1}{2}\right)}{s^{n+\frac{1}{2}}}=\frac{\left(n-\frac{1}{2}\right)\Gamma\left(n-\frac{1}{2}\right)}{s^{n+\frac{1}{2}}}\\&=&\ldots = \frac{(2n-1)!!\cdot \Gamma\left(\frac{1}{2}\right)}{2^n s^{n+\frac{1}{2}}}\end{eqnarray*}$$
and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1541609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Alternate approaches to solve this Integral Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$
I have used parts taking first function as Integrand and second function as $1$ we get
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$
now $$\frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)=\frac{1}{2\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)} \times \left(\frac{\frac{1+\sqrt{x}}{-2\sqrt{x}}-\frac{1-\sqrt{x}}{2\sqrt{x}}}{(1+\sqrt{x})^2}\right)=\frac{-1}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})}$$ so
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\int\frac{xdx}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})} $$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int \frac{\sqrt{x}dx}{\sqrt{1-x}(1+\sqrt{x})}$$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}(1-\sqrt{x})dx}{(1-x)^{\frac{3}{2}}}$$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}-\frac{1}{2}\int\frac{xdx}{(1-x)^{\frac{3}{2}}}$$
Now $$\int\frac{xdx}{(1-x)^{\frac{3}{2}}}=\int\frac{(x-1+1)dx}{(1-x)^{\frac{3}{2}}}$$ and if we split is straight forward to compute.
Also $$\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}$$ can be evaluated using substitution $x=sin^2y$
I need any other better approaches to evaluate this integral.
|
$$
\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\times \dfrac{1-\sqrt{x}}{1-\sqrt{x}}}=\dfrac{1-\sqrt{x}}{\sqrt{1-x}}
$$
thus we get
$$
\int \dfrac{1}{\sqrt{1-x}}dx -\int \sqrt{\dfrac{x}{1-x}}dx
$$
So I got to your final integrals quickly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1542797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Finding the Maximum value.
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$
The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$
$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$
$$
\left.
\begin{array}{l}
\text{}&2yx= 2y^2a^2\lambda\\
\text{}&
\end{array}
\right\}
*b^2x
$$
$$y^3a^2= 2yxa^2b^2 \lambda$$
$$2yx^2b^2 = 2y^2a^2b^2x\lambda$$
$\color{maroon}{\mathbf{Equalize}}$
$$2a^2b^2xy = 2a^2b^2xy^2$$
$$y=1$$
My main problem is which equations does one set each equal to. If anyone knows which one equals the other this will lead me to the right place in finding the solution.
|
just out of curiosity, we know that $x^2/a^2+y^2/b^2=1$, then by AM–GM inequality
$$1=\frac{x^2}{a^2}+\frac{y^2}{2b^2}+\frac{y^2}{2b^2}\geq 3\sqrt[3]{\frac{x^2}{a^2}\cdot\frac{y^2}{2b^2}\cdot\frac{y^2}{2b^2}}$$
the equality holds when
$$\frac{x^2}{a^2}=\frac{y^2}{2b^2}=\frac{y^2}{2b^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1543654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Partial sum of $\sum \frac {1} {k^2}$ It is pretty well-known that $\sum_{k = 1}^{\infty} \frac {1} {k^2} = \frac {\pi^2} {6}$. I am interested in evaluating the partial sum $\sum_{k = 1}^{N} \frac {1} {k^2}$. Here is what I have done so far. Since we have
$$\sum_{k = 1}^{N} \frac {1} {k^2} = 1 + \sum_{k = 2}^{N} \frac {1} {k^2}
< 1 + \sum_{k = 2}^{N} \frac {1} {k^2 - 1} =\\= 1 + \frac {1} {2} \left ( \sum_{k = 2}^{N} \frac {1} {k - 1} - \sum_{k = 2}^{N} \frac {1} {k + 1} \right )
= 1 + \frac {1} {2} \left ( \frac {3} {2} - \frac {2N + 1} {N^2 + N} \right )
= \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$
and
$$\sum_{k = 2}^{N} \frac {1} {k^2 - 1} - \sum_{k = 2}^{N} \frac {1} {k^2} < \int_{2}^{N} \left ( \frac {1} {x^2 - 1} - \frac {1} {x^2} \right ) \textrm {d}x = \int_{2}^{N} \frac {\textrm {d}x} {x^2 - 1} - \int_{2}^{N} \frac {\textrm {d}x} {x^2} < \frac {1} {N^2 - N} - \frac {1} {2} + \log \sqrt {3},$$
we have
$$\frac {7} {4} + \frac {1} {2} - \log \sqrt {3} - \frac {2 N^2 + N + 1} {2N^3 - 2N} < \sum_{k = 1}^{N} \frac {1} {k^2} < \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$
But how to obtain a more precise expression without using analytic methods such as Euler summation?
|
A quick fix on the upper bound is possible.
Just change everything from $\frac{1}{k^2-1}$ to $\frac{1}{k^2-\frac{1}{4}}$.
Then, we have $$ \sum_{k=1}^N \frac{1}{k^2} = 1+ \sum_{k=2}^N \frac{1}{k^2} < 1+\sum_{k=2}^N \frac{1}{k^2-\frac{1}{4}} = 1+\sum_{k=2}^N \frac{1}{k-\frac{1}{2}} - \sum_{k=2}^N \frac{1}{k+\frac{1}{2}} = 1+\frac{2}{3}-\frac{1}{N+\frac{1}{2}} = \frac{5}{3}-\frac{2}{2N+1}$$
Therefore, we have $$\sum_{k=1}^N \frac{1}{k^2} < \frac{5}{3}-\frac{2}{2N+1}$$
The upper bound should be more tight since $$\frac{1}{k^2}<\frac{1}{k^2-\frac{1}{4}}< \frac{1}{k^2-1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1544535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Expressing as a composite function to begin differentiating with the chain rule I'm trying to differentiate $y=(x+1)(x+2)^2(x+3)^3$ using the chain rule, but I am having trouble writing it as a composite function. Any help would be great as I can finish the problem on my own once I just get started in the right direction. Thanks.
|
When you have products such as $$y=(x+1)(x+2)^2(x+3)^3$$ logarithmic differentiation makes life simpler $$\log(y)=\log(x+1)+2\log(x+2)+3\log(x+3)$$ Differentiating $$\frac{y'}y=\frac 1{x+1}+\frac 2{x+2}+\frac 3{x+1}=\frac {6 x^2+22 x+18 }{(x+1) (x+2) (x+3)}$$ Multiply both sides by $y$ to get $$y'=\frac {6 x^2+22 x+18 }{(x+1) (x+2) (x+3)}(x+1)(x+2)^2(x+3)^3=(6 x^2+22 x+18)(x+2)(x+3)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1546525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find the limit of this trigonometric expression as x approaches 0 $$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})(\cot(x)+\frac{1}{x})$$
I have computed the limits
$$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})=0$$
$$\lim_{x\to 0^+} (\cot(x)+\frac{1}{x})=\infty$$
I can't multiply the limits together.
I rewrote the initial expression:
$$\lim_{x\to 0^+} (\frac{x\cos(x)-\sin(x)}{x\sin(x)})(\frac{x\cos(x)+\sin(x)}{x\sin(x)})$$
$$\lim_{x\to 0^+} \frac{x^{2}\cos^{2}(x)-\sin^{2}(x)}{x^{2}\sin^{2}(x)}$$
$$\lim_{x\to 0^+} \frac{\cos^{2}(x)-1}{\sin^{2}(x)}$$
Using Pythagorean Identity
$$\lim_{x\to 0^+} \frac{-\sin^{2}(x)}{\sin^{2}(x)}=-1$$
I also got the same result using L'Hospital's Rule.
I know the limit is $\frac{-2}{3}$
What did I do wrong?
|
It is possible to do not use Taylor series. First write
\begin{eqnarray*}
\cot ^{2}x-\frac{1}{x^{2}} &=&\left( \frac{x\cos x-\sin x}{x\sin x}\right)
\left( \frac{x\cos x+\sin x}{x\sin x}\right) \\
&=&\frac{x^{2}}{\sin ^{2}x}\left( \frac{x\cos x-\sin x}{x^{3}}\right) \left(
\frac{x\cos x+\sin x}{x}\right) \\
&=&\left( \frac{x}{\sin x}\right) ^{2}\left( \frac{x\left( \cos x-1\right)
+(x-\sin x)}{x^{3}}\right) \left( \cos x+\frac{\sin x}{x}\right) \\
&=&\left( \frac{x}{\sin x}\right) ^{2}\left( \frac{\cos x-1}{x^{2}}+\frac{%
x-\sin x}{x^{3}}\right) \left( \cos x+\frac{\sin x}{x}\right)
\end{eqnarray*}
Basic limits as
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x}{\sin x} &=&1 \\
\
lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}} &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6} \\
\lim_{x\rightarrow 0}\cos x &=&1
\end{eqnarray*}
imply that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \cot ^{2}x-\frac{1}{x^{2}}\right) &=&\left(
\lim_{x\rightarrow 0}\frac{x}{\sin x}\right) ^{2}\left( \lim_{x\rightarrow 0}%
\frac{\cos x-1}{x^{2}}+\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}\right)
\left( \lim_{x\rightarrow 0}\cos x+\lim_{x\rightarrow 0}\frac{\sin x}{x}%
\right) \\
&=&\left( 1\right) ^{2}\left( -\frac{1}{2}+\frac{1}{6}\right) \left(
1+1\right) =-\frac{2}{3}.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1549456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How do I find $\liminf$ and $\limsup$ if $a_{2n}=\frac {a_{2n-1}}2$ and $a_{2n+1}=\frac12+\frac {a_{2n}}2$? Its given that $a_1=a>0$ and that for any $n>1$ two things happen:
$$a_{2n}=\frac {a_{2n-1}}2$$
$$a_{2n+1}=\frac12+\frac {a_{2n}}2$$
How do I find $\lim\inf$ and $\lim\sup$
I am trying to look at
$a_{2n+1}$ and $a_{2n-1}$
$a_{2n}$ and $a_{2n-2}$
But I am unable to prove that they are bounded.
NOTE: Look at joeys answer for correct solution.
|
EDIT: This response was written for the recurrence equations
$$ a_{2n} = \frac{a_{2n-1}}{2}, a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2}. $$
The exact answer may differ depending on the exact edit to the question, but the techniques remain valid, assuming the coefficients are not drastically changed.
Since there are different rules for the even and odd terms, it makes sense to suspect that the even terms will behave differently from the odd terms. So let's investigate the two sets of terms separately.
For $n\in\mathbb{N}$, let $b_n = a_{2n-1}$ and $c_n = a_{2n}$. From the relations provided, we have
$$ b_{n+1} = a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2} = \frac{1}{2} + \frac{1}{2}\left(\frac{a_{2n-1}}{2}\right) = \frac{1}{2} + \frac{b_{n}}{4}$$
Similarly, we have
$$ c_{n+1} = a_{2n+2} = \frac{a_{2n+1}}{2} = \frac{1}{2}\left(\frac{1}{2} + \frac{a_{2n}}{2}\right) = \frac{1}{4} + \frac{c_n}{4}.$$
We now look at the behavior of $b_n$ as $n\rightarrow\infty$. Suppose it converged to some value, say $b$. What would $b$ have to be? If we take our equation
$$ b_{n+1} = \frac{1}{2} + \frac{b_n}{4} $$
and let $n\rightarrow\infty$, then both $b_{n+1}$ and $b_n$ would converge to $b$, and hence
$$ b = \frac{1}{2} + \frac{b}{4}\implies b=\frac{2}{3}.$$
So we have a hunch that $b_n$ converges to $\frac{2}{3}$. How do we prove it? Well, let $d_n = b_n - \frac{2}{3}$, and let's try to show that $d_n\rightarrow 0$. Now,
$$ b_{n+1} = \frac{1}{2} + \frac{b_n}{4}\implies d_n + \frac{2}{3} = \frac{1}{2} + \frac{d_n + \frac{2}{3}}{4} = \frac{2}{3} + \frac{d_n}{4}\implies d_{n+1} = \frac{d_n}{4}$$
which clearly shows that $d_n\rightarrow 0$. Hence, $b_n\rightarrow\frac{2}{3}$. Similar reasoning yields that $c_n\rightarrow c$, where
$$ c = \frac{1}{4} +\frac{c}{4}\implies c = \frac{1}{3}.$$
Hence, the odd terms tend to $\frac{2}{3}$, and the even terms tend to $\frac{1}{3}$. This shows that
$$ \limsup\limits_{n\rightarrow\infty}{a_n} = \frac{2}{3} $$
and
$$ \liminf\limits_{n\rightarrow\infty}{a_n} = \frac{1}{3}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1549581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
|
Let me add my own proof that I had in my mind.
$$Let\quad f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots$$
$$f(x)-\frac{1}{1-x}=\color{red}{\frac{-2x}{1-x^2}}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots\\
=\frac{3x^2}{1+x^3}+\color{red}{\frac{-4x^3}{1-x^4}}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots\\
=\cdots=\frac{3x^2}{1+x^3}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}
$$
$$f(x)-\frac{1}{1-x}-\frac{3x^2}{1-x^3}
=\frac{5x^4}{1+x^5}+\color{red}{\frac{-6x^5}{1-x^6}}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}\\
=\cdots=\frac{5x^4}{1+x^5}+\frac{7x^6}{1+x^7}+\cdots+\lim_{n\to\infty}\frac{-2^nx^{2^n-1}}{1-x^{2^n}}+\lim_{n\to\infty}\frac{-3\cdot2^nx^{3\cdot2^n-1}}{1-x^{3\cdot2^n}}$$
$$\cdots$$
$$f(x)-\sum_{k=0}^{\infty}\frac{(2k+1)x^{2k}}{1-x^{2k+1}}=\sum_{k=0}^{\infty}\lim_{n\to\infty}\frac{-(2k+1)\cdot2^nx^{(2k+1)\cdot2^n-1}}{1-x^{(2k+1)\cdot2^n}}=0$$
$$\therefore f(x)=\sum_{k=0}^{\infty}\frac{(2k+1)x^{2k}}{1-x^{2k+1}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1554777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
}
|
Simplify $\sqrt[3]{162x^6y^7}$ The answer is $3x^2 y^2 \sqrt[3]{6y}$
How does $\sqrt[3]{162x^6y^7}$ equal $3x^2 y^2\sqrt[3]{6y}$?
|
First note that 162 factors as $2\cdot 3^4$. So collecting our powers of 3 reveals
$$\sqrt[3]{162x^6y^7} = \sqrt[3]{3^3 \cdot (x^3)^2\cdot (y^3)^2 \cdot (2\cdot 3 \cdot y)} = 3x^2y^2\sqrt[3]{6y}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1556595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Integrate. $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How should i solve this integral?
i know that it is the same question like here
Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$
but I've tried solve it for more then 3 hours and i still have no idea ho to solve it. Thank for help.
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
I tried use $$2x+7 = \frac{\sqrt{61}}{\cos u}$$
$$\int\frac{\frac{\sqrt{61}}{\cos u}-\frac72+1}
{(\frac{(\frac{\sqrt{61}}{\cos u})^2-61}{4})^3}dx$$
Now i have
$$\int \frac{61\sin u - 5\sqrt{61}\sin u \cos u}{4\cos u
\frac{226981\tan^6u}{65}
}$$
|
HELP for the integrand $\frac{1}{(x^2-a)^3}$
Use the substitution $x = \sqrt{a}\sec(u)$ and $dx = \sqrt{a}\tan(u) \sec(u) du$ so then $(x^2 - a)^3 = (a\sec^2(u) - a)^3 = a^3\tan^6(u) du$.
Moreover you'll have to take in mind that $u = \text{arcsec}\left(\frac{x}{\sqrt{a}}\right)$
Thus you have
$$\sqrt{a}\int\frac{\cot^4(u)\csc(u)}{a^3} du = \frac{1}{a^{\frac{5}{2}}}\int\cot^4(u)\csc(u)$$
Now use the trigonometric identity
$$\csc^2(u) - \cot^2(u) = 1$$
and arranging the integrand you will obtain
$$
\frac{1}{a^{\frac{5}{2}}}\int \csc^5(u) - 2\csc^3(u) + \csc(u) du
$$
Then you can integrate term by term using the recurrence formula for the highest powers:
$$\int\csc^m(u) du = -\frac{\cos(u)\csc^{m-1}(u)}{m-1} + \frac{m-2}{m-1}\int \csc^{m-2}(u) du$$
where in your case $m = 5$ and next $m = 3$. Do the math and you will obtain:
$$
\frac{5\cot(u)\csc(u)}{8a^{5/2}} - \frac{\cot(u)\csc^3(u)}{4a^{5/2}} + \frac{3}{8a^{5/2}}\int\csc(u) du
$$
The integral of $\csc(u)$ is $-\log(\csc(u) + \csc(u))$
So you'll get in the end:
$$
\frac{5\cot(u)\csc(u)}{8a^{5/2}} - \frac{\cot(u)\csc^3(u)}{4a^{5/2}} + \frac{3}{8a^{5/2}}(-\log(\csc(u) + \csc(u))) du
$$
Now the tedious work: substitute back for $u$ to get the result in $x$: Do the math and get:
$$
-\frac{((\cot(\sec^{-1}(\frac{x}{\sqrt{a}}) \csc(\sec^{-1}(\frac{x}{\sqrt{a}}))^3)}{(4 a^{5/2}))} + \frac{(5 \cot(\sec^{-1}(\frac{x}{\sqrt{a}})) \csc(\sec^{-1}(\frac{x}{\sqrt{a}})))}{(8 a^{5/2})} -
\frac{3 \log(\cot(\sec^{-1}(\frac{x}{\sqrt{a}})) + \csc(\sec^{-1}(\frac{x}{\sqrt{a}}))))}{(8 a^{5/2})}
$$
Anyway, your integral is a pain.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1557054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
So $$(2x-3)^2-4(2x^2+2x+8-K)\geq 0$$
So $$-4x^2-20x-23+4K\geq0\Rightarrow 4x^2+20x+23-4K\leq0$$
Now How can I Solve after that, Help me
Thanks
|
since it has calculus tag im going to try solve it with calculus
heres my try:
we are going to find some extreme point and then check it if its minimum or maximum
let $Z = 2x^2 +y^2 +2xy + 2x -3y +8 $
so we got $\frac{dz}{dx} = 4x + 2y +2 $ and $\frac{dz}{dx} = 2y +2x -3 $
now we are going to find a possible extreme point from the equation above
$ 4x + 2y + 2 = 0 $ change this to $ 4x+2y = -2$
$2x + 2y - 3 = 0 $ change this to $ 2x + 2y = 3$
by eliminate 2 of the equation above we got x = $\frac{-5}{2}$ and y = $4$
and now we are going to check if this point is minimum or maximum .
by using second derivation $\frac{d^2z}{dx^2} = 4 $
now by subtitute $(x,y) = (\frac{-5}{2} ,4 )$ to the $\frac{d^2z}{dx^2} = 4 $
we got $\frac{d^2z}{dx^2} > 0 $, since $\frac{d^2z}{dx^2} > 0 $ this mean (x,y)
is a extreme minimum . this mean by subtituting $(x,y) = (\frac{-5}{2} ,4 )$ we
are going to get a minimum result
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1558291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Finding the number of permutations in$S_9$ of the form $(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$ How many permutations, ρ, are there in $S_9$(the group of permutations of order 9!) whose decomposition into disjoint
cycles consists of three 2-cycles (transpositions) and one 3-cycle? In other words, how
many permutations are there in $S_9$ whose decomposition into disjoint cycles is of the form
$(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$?
would I be able to assume that, $a_1 = 1$, so $a_2$ would have 8 possibile values, $a_3$ would have 7 and so on, for a total of $8!$ such permutations?
|
*
*You need to choose the support of the first transposition : $\begin{pmatrix}9\\2\end{pmatrix}$ choices.
*You need to choose the support of the second transposition in the remaining set : $\begin{pmatrix}7\\2\end{pmatrix}$ choices.
*You need to choose the support of the third transposition in the remaining set : $\begin{pmatrix}5\\2\end{pmatrix}$ choices.
*You need to divide this number by the number of ways to order the three transpositions (since they all give the same permutation) e.g. : $(1,2)(3,4)(5,6)=(3,4)(1,2)(5,6)$. Hence you need to divide by $3!$.
*Choose the support of the $3$-cycle in the remaining set $\begin{pmatrix}3\\3\end{pmatrix}=1$ and on a support of size $n$ you know that you have exactly $(n-1)!$ different $n$-cycles, here this gives $(3-1)!=2$ choices.
*Gather everything :
$$\frac{1}{3!} \begin{pmatrix}9\\2\end{pmatrix}\begin{pmatrix}7\\2\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}2=2\frac{9\times 8\times 7\times 6\times 5 \times 4}{8\times 6}=9\times 7\times 5\times 8$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1558934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
My question is, is this a double angle formula? And if so, how would I go about to solve it?
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
$$=2\sin{\pi \over 4}+\left(1-2\sin{\pi \over 4}\right)$$
Have I done this right so far? I feel I have not.
|
In general, you can write expressions of the form $a\sin \theta + b\cos\theta$ in the form $c\sin(\theta + \psi)$ for suitable choices of $c$ and $\psi$. To see this, look at the double-angle formula for sine; you obtain
$$a\sin\theta + b\cos\theta = (c\cos \psi)\sin\theta + (c\sin\psi)\cos\theta$$
Comparing coefficients gives $a=c\cos\psi$ and $b=c\sin\psi$.
Dividing the second equation by the first gives $\tan\psi = \frac{b}{a}$; and squaring both equations and adding them together gives
$$a^2+b^2 = c^2(\cos^2\psi+\sin^2\psi) = c^2$$
So you can work out the values of $c$ and $\psi$ from here.
Here, $a=b=1$, so we get
$$\tan\psi = 1 \quad \text{and} \quad c^2=1^2+1^2=2$$
so $\psi = \frac{\pi}{4}$ and $c = \sqrt{2}$ will work. Hence
$$\boxed{\sin \frac{\pi}{8} + \cos \frac{\pi}{8} = \sqrt{2}\sin\left(\frac{\pi}{8} + \frac{\pi}{4}\right)}$$
You can probably take it from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1559257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
}
|
Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$. Evaluate $$\int_0^1 f(x) dx$$ where
$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$
I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to
$$\sin^2y\cdot \ln (\sin y) dy$$
Then I used the property of definite integrals that
$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$
Then too it wasn't getting simplified.
I tried $e^z=\sin x$, but this gave no headway because after a while I reached a complete full-stop. How should I go about this?
|
$$I(\alpha)=\int_0^1\dfrac{x^\alpha\ln x}{\sqrt{1-x^2}}\ dx$$
Using Beta function we have
$$\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx=\dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)$$
then with Digamma function $\psi$
\begin{align}
I(\alpha)
= \dfrac{d}{d\alpha}\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx \\
= \dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)\left(\psi(\frac{\alpha+1}{2})-\psi(\frac{\alpha+2}{2})\right)
\end{align}
now let $\alpha=2$,
$$\psi(\frac{3}{2})-\psi(\frac{4}{2})=2\sum_{n=0}^\infty\left(\dfrac{1}{2n+4}-\dfrac{1}{2n+3}\right)=1-\ln4$$
therefore
$$I(2)=\color{blue}{\dfrac{\pi}{8}(1-\ln4)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1559380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
}
|
Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$
Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$ in $\{|z|>1\}$.
Ok, so for $|z|>1 \iff |\frac{1}{z}|<1$ I can write
$\frac{1}{z-1}=\frac{1}{z}\cdot \frac{1}{1-\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{1}{z})^n=\frac{1}{z}\sum_{n<0}z^n$
Which is the Laureant Series for $\frac{1}{z-1}$.
But I don't know what to do with the $e^z$ multiplying. How should I proceed?
|
Looks like to me we can just do this,
\begin{align}
\frac{e^z}{z-1} &= \frac{1}{z-1}\sum_{n \geq 0} \frac{e(z-1)^n}{n!}.\\
&=\frac{e}{z-1}\left(1 + (z-1) + \frac{1}{2!}(z-1)^2 + \frac{1}{3!}(z-1)^2 +\dots \right) \\
&=e\left(\frac{1}{z-1} + 1 + \frac{1}{2!}(z-1) + \frac{1}{3!}(z-1)^2 +\dots \right)
\end{align}
So it suffices to find the sum of coefficient of $z$ in $(z-1)^n$, that is
\begin{align}
\sum_{n \geq 1} \frac{1}{(n+1)!} \binom{n}{n-1} (-1)^{n -1} &=\sum_{n \geq 1} \frac{n}{(n+1)!} (-1)^{n-1}\\
&=\sum_{n \geq 0} \frac{n+1}{(n+2)!} (-1)^{n}\\
&=\sum_{n \geq 0} \frac{1}{(n+2)n!} (-1)^{n}\\
&=\frac{e - 2}{e}.
\end{align}
Therefore the coefficient of $z$ is $e-2.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1560377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $1^{\infty}$.
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}}=\exp \left[{\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}-1\right)\times \frac{3x^2+8}{x^2}}\right]=e^{-8}$$
But i want to know,cant it be found using Taylor/Maclaurin series?What are the limitations of Taylor/Maclaurin series?
When can we not use them?
Is it true that we can use them whenever L Hospital rule is applicable(in $\frac{0}{0}$ and $\frac{\infty}{\infty}$ cases.)Can we use them in $0^0,\infty^0,1^\infty$ cases.
Please help me.Thanks.
|
Start by reducing to "known" Taylor series around $0$, and then compose them.
Here, you will rely on that of $\ln(1+x)$, $\frac{1}{1+x}$, and the fact that $a^b = e^{b\ln a}$:
$$
\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}} =
\left(\frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}}\right)^{\frac{3x^2+8}{x^2}}
=
e^{\frac{3x^2+8}{x^2}\ln\frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}}}
$$
Now,
$$
\frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}} = 1-x^2 +o(x^3)
$$
(using the known Taylor series expansion of $\frac{1}{1+x}$), and as $\ln(1+x)= x + o(x)$ around $0$ you get
$$
\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}}
=
e^{\frac{3x^2+8}{x^2}(-x^2 +o(x^2))} = e^{-8 + o(1)}
$$
To answer the more general question: in my experience, you can use Taylor series for limits around $0$, $\pm\infty$ (after some factorization to reduce to a limit around $0$), etc., whenever L'Hospital is applicable. (And I would argue in favor of the former over the latter almost every time, maybe out of personal taste.) Moreover, once you get to memorize the usual suspects (standard Taylor series of common functions), then it becomes quite fast (albeit sometimes cumbersome). The main advantage being: it is not always the most elegant method, but it is systematic -- if you don't see an other obvious way to compute the limit, try the Taylor expansions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1560950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Show that $\lim\limits_{x\rightarrow 0}f(x)=1$
Suppose a function $f:(-a,a)-\{0\}\rightarrow(0,\infty)$ satisfies $\lim\limits_{x\rightarrow 0}\left(f(x)+\frac{1}{f(x)}\right)=2$. Show that $$\lim\limits_{x\rightarrow 0}f(x)=1$$
Let $\epsilon>0$ , then there exists a $\delta>0$ such that $$\left(f(x)+\frac{1}{f(x)}\right)-2<\epsilon\;\;\;\text{and}\;\;\;|x|<\delta$$
Then
\begin{align}
&\left(f(x)+\frac{1}{f(x)}\right)-2\\
=&\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)<\epsilon\tag{1}
\end{align}
Squaring $(1)$ both sides gives $$\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2<\epsilon^2\tag{2}$$
Since $(f(x)-1)^2\leq\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2$, by using $(2)$, $(f(x)-1)^2<\epsilon^2\Rightarrow f(x)-1<\epsilon$; therefore, as $\epsilon$ is arbitrary, $\lim\limits_{x\rightarrow 0}f(x)=1$
Can someone give me a hint to do this question without using epsilon-delta definition? Thanks
|
Define $h(u) = \frac{u + \sqrt{u^2 - 4}}{2}$ (this is the inverse function of $x \mapsto x + \frac{1}{x}$). Note that $h(2) = 1$ and that $h$ is continuous at $u = 2$. Thus,
$$ 1 = \lim_{u \to 2} h(u) = \lim_{x \to 0} h \left( f(x) + \frac{1}{f(x)} \right) = \lim_{x \to 0} \frac{f(x) + \frac{1}{f(x)} + \sqrt{\left( f(x) + \frac{1}{f(x)} \right)^2 - 4}}{2} = \lim_{x \to 0} \frac{f(x) + \frac{1}{f(x)} + \sqrt{ \left( f(x) - \frac{1}{f(x)} \right)^2}}{2} = 1 + \lim_{x \to 0} \left| f(x) - \frac{1}{f(x)} \right| $$
which implies that
$$\lim_{x \to 0} \left( f(x) - \frac{1}{f(x)} \right) = 0. $$
Combining both results together we have
$$ 2 = 2 + 0 = \lim_{x \to 0} \left( f(x) + \frac{1}{f(x)} \right) + \lim_{x \to 0} \left( f(x) - \frac{1}{f(x)} \right) = \lim_{x \to 0} 2f(x) $$
which implies that $\lim_{x \to 0} f(x) = 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1561861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is .... Problem :
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is
(a) $2^{105}-2^{121}$
(b) $2^{121}-2^{105}$
(c) $2^{120}-2^{104}$
(d) $2^{110}-2^{108}$
My approach :
Tried to find pattern if any in the given expansion, so expanded first three terms :
$(1-x)(1-2x)(1-2^2x) = (1-3x+2x^2)(1-4x) = (1-3x+2x^2 -4x+12x^2 -8x^3)$
$= (1-7x+14x^2 -8x^3)$
But didn't find any pattern of this series please suggest how to proceed will be of great help thanks.
|
I'd rescale as $$2^0\cdot2^1\cdot2^2\cdots2^{15}\left[(1-x)\left(\frac12-x\right)\left(\frac14-x\right)\cdots\left(\frac1{2^{15}}-x\right)\right]$$ The coefficient of $x^{15}$ in the polynomial within the brackets is the negative of the sum of the roots (since there are $16$ linear factors). So the answer is $$-2^0\cdot2^1\cdot2^2\cdots2^{15}\left[1+\frac12+\frac14+\cdots+\frac1{2^{15}}\right]=-2^{120}\left[2-\frac{1}{2^{15}}\right]=2^{105}-2^{121}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1562138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
|
Let's try it straight from the binomial theorem:
$$
(1+x)^3\cdot(2+x^2)^{10} = \sum_{j=0}^3 {3 \choose j} x^{3-j} \cdot \sum_{k=0}^{10} {10 \choose k} 2^k (x^2)^{10-k}.
$$
Good so far? Now, note these terms can be arranged as
$$
\sum_{j=0}^3 \sum_{k=0}^{10} {3 \choose j} {10 \choose k}2^k x^{3-j} (x^2)^{10-k}.
$$
Note the $x$ terms combine as
$$
x^{23 - j - 2k}.
$$
We want the exponent of $x$ to be $3$, which means we want
$$
23 - j - 2k = 3 \iff j + 2k = 20,
$$
which is satisfied only when $j = 0$ and $k = 10$, and when $j = 2$ and $k = 9$. Thus the coefficient of $x^3$ is
$$
{3 \choose 0}{10 \choose 10}2^{10} + {3 \choose 2}{10 \choose 9}2^{9} = 16384.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1562811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Solve for $x$, correct to two significant figures, the equation: $4^{x}-2^{x+1}-3=0$ Solve for $x$, correct to two significant figures, the equation:
$$4^{x}-2^{x+1}-3=0$$
My answer: $x\log4=\log3+(x+1)\log2 \Rightarrow 0.602x-0.301x=0.477+0.301 \Rightarrow x = 2.6$ (Conflicting with book answer)
Answer in book: $x=1.6$
|
You made a mistake in your first step. Take a look at the following
$$\eqalign{
& {4^x} - {2^{x + 1}} - 3 = 0 \cr
& {\left( {{2^x}} \right)^2} - 2\left( {{2^x}} \right) - 3 = 0 \cr
& \left( {{2^x} - 3} \right)\left( {{2^x} + 1} \right) = 0 \cr
& \left\{ \matrix{
{2^x} = 3 \hfill \cr
{2^x} = - 1,\,\,\,\text{impossible} \hfill \cr} \right. \cr
& x\ln 2 = \ln 3 \cr
& x = {{\ln 3} \over {\ln 2}} = \log_{2}3 =1.584 \cr} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1563144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
Is the inverse of an invertible circulant matrix also circulant? The circulant matrices in $M_n(F)$ ($F$ field) form a subspace $\mathcal C_n$ spanned by $I,J,J^2,\cdots,J^{n-1}$ where
$$J=\begin{bmatrix} O & I_{n-1}\\1 & O\end{bmatrix}$$
This subspace $\mathcal C_n$ is also closed under multiplication, which makes it a subring of $M_n(F)$.
Now my question is, if $A\in \mathcal C_n$ and $A$ is non-singular, can we assert that $A^{-1}\in\mathcal C_n$ too?
I'm in particular encouraged to make this guess by observing these examples:
$$\begin{bmatrix}0 & 1 & 1 & \cdots & 1\\ 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 0 & \cdots & 1\\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 1 & 1 & \cdots & 0 \end{bmatrix}^{-1}=\frac{1}{n-1}\begin{bmatrix} 2-n & 1 & 1 & \cdots & 1\\ 1 & 2-n & 1 & \cdots & 1 \\ 1 & 1 & 2-n & \cdots & 1\\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 1 & 1 & \cdots & 2-n \end{bmatrix}$$
and
$$
\begin{bmatrix}
1 & 2 & 3 & \cdots & n-1 & n \\
n & 1 & 2 & \cdots & n-2 & n-1\\
n-1 & n & 1 & \cdots & n-3 & n-2 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
2 & 3 & 4 & \cdots & n & 1
\end{bmatrix}^{-1}=
\frac1{ns}
\begin{bmatrix}
1-s & 1+s & 1 & \cdots & 1 & 1\\
1 & 1-s & 1+s & \cdots & 1 & 1 \\
1 & 1 & 1-s & \cdots & 1 & 1\\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
1+s & 1 & 1 & \cdots & 1 & 1-s
\end{bmatrix}
$$
in which $s:=1+2+\cdots+n$.
|
By Cayley-Hamilton theorem, the inverse of every nonsingular matrix $A$ can be expressed as a polynomial in $A$ of degree $\le n-1$. Therefore, if $A$ is in the linear span of $I,J,\ldots,J^{n-1}$, then $A^{-1}$ is a polynomial in $J$ of degree $\le (n-1)^2$. Since $J^n=I$, it follows that $A^{-1}$ lies inside the linear span of $I,J,\ldots,J^{n-1}$. Hence it is circulant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1563907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must factorize the denominator:
$$x^3+2x^2 = (x+2)\cdot x^2$$
Great. So now we write three fractions:
$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$
Eventually we conclude that
$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$
So now we look at what happens when $x = -2$:
$$C = 12$$
When $x = 0$:
$$A = -1$$
And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:
$$B = -1$$
We replace the values here:
$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$
Which results in
$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$
But I fear the answer actually is
$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$
Can you tell me what did I do wrong, and what should I have done?
|
I can't resist to post this. I think this is the easiest way.
$$(B+C)x^2+(A+2B)x+2A=5x^2+3x-2$$
\begin{align}
2A&=-2&\rightarrow A&=-1\\
A+2B&=-1+2B=3&\rightarrow B&=2\\
B+C&=2+C=5&\rightarrow C&=3
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1565338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
}
|
Compute $\int \big( (4z)^5 + 4(4z)^2 \big ) \big( (4z)^3 +1 \big)^{12} dz$ by substitution The problem is $\int \big( (4z)^5 + 4(4z)^2 \big ) \big( (4z)^3 +1 \big)^{12} dz$. I know a substitution has to be used, but having the $4$ in front of the $4z$ is confusing me and I don't know how to get rid of it.
|
\begin{align}
&(4z \mapsto y)\\
&\int \left((4z)^5+4(4z)^2\right)\left((4z)^3+1\right)^{12}dz\\
&=\frac14\int y^2\left(y^3+4\right)\left(y^3+1\right)^{12}dy\\
&\\
&(y^3\mapsto x)\\
&=\frac1{12}\int (x+4)(x+1)^{12}dx\\
&=\frac1{12}\int (x+1)(x+1)^{12}dx+\frac1{12}\int 3(x+1)^{12}dx\\
&=\frac1{12\cdot14}(x+1)^{14}+\frac1{4\cdot13}(x+1)^{13}\\
&=\frac1{12\cdot14}(\sqrt[3]{4z}+1)^{14}+\frac1{4\cdot13}(\sqrt[3]{4z}+1)^{13}\\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1567542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Explain how to compute $\cos(2\pi/13)$ by solving quadratic and cubic equations only I know that we can express $2\pi/13$ as a root of unity on the unit circle taking $z^{13} = 1$ and $z=\cos(2\pi/13)+i\sin(2\pi/13)$ and that we should be able to find a polynomial for this over the rational numbers, and because its not constructible it will consist of irreducible polynomials of degrees 2 and 3. But I'm not sure of any method to go about doing this.
|
You can use the triple angle and the quadruple angle formula for cosine.
\begin{align}
&\cos x = \cos \frac{2\pi}{13}\\
&\cos 4x = 8 \cos^4 x - 8 \cos^2 x + 1\\
&\cos 3x = 4 \cos^3 x - 3 \cos x\\
&\cos 9x = 4 (4 \cos^3 x - 3 \cos x)^3 - 3 (4 \cos^3 x - 3 \cos x)\\
&\cos (4x+9x) = \cos 4x \cos 9x - \sin 4x \sin 9x \\
&= \cos 4x \cos 9x - \sqrt{1-\cos^2 4x} \sqrt{1-\cos^2 9x} = 1\\
&(\cos 4x \cos 9x -1)^2= (1-\cos^2 4x)(1-\cos^2 9x)\\
&\cos^2 4x + \cos^2 9x - 2\cos 4x \cos 9x = (\cos 4x -\cos 9x)^2=0\\
&\cos 4x = \cos 9x\
\end{align}
It was only after I reached the above line that I realized
$$4\times \frac{2\pi}{13}+9\times \frac{2\pi}{13}=2\pi$$
So we could have obtained $\cos 4x = \cos (2\pi-4x)= \cos 9x$ directly.
Anyway, let's keep going to find the equation
\begin{align}
&\cos x \mapsto y\\
&8y^4-8y^2+1=4(4y^3-3y)^3-3(4y^3-3y)\\
&(y-1)(4y^2+2y-1)(64y^6+32y^5-80y^4-32y^3+24y^2+6y-1)=0
\end{align}
$y-1=0$ is for the trivial solution of $x=0$.
And $4y^2+2y-1$ must be coming from one of those steps of squaring and tripling polynomials to get $\cos 3x$ and $\cos 4x$.
So the final equation you're looking for is
$$64y^6+32y^5-80y^4-32y^3+24y^2+6y-1=0$$
that has 6 real roots and one of them is the $\cos \frac{2\pi}{13}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1567755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
What is the sum of the digits of the sum of the digits? Problem
Let $(10^{2016}+5)^2=225N$. If $S$ is the sum of the digits of N, then find the sum of the digits of S
Attempt
Let's look at some smaller cases. We have $\dfrac{(10^{3}+5)^2}{225} = 4489$,$\dfrac{(10^{4}+5)^2}{225} = 444889$, and $\dfrac{(10^{5}+5)^2}{225} = 44448889$. Thus we see the pattern and $S = 4*2015+8*2014+9 = 24181$ and the sum of the digits of $S$ is $\boxed{16}$.
Question
Prove the result in the solution above by induction or some other method. That is, show that for $n > 1$
$$\dfrac{(10^{n}+5)^2}{225} =\underbrace{44\ldots4}_\text{n-1 4's}\underbrace{88\ldots8}_\text{n-2 8's}9.$$
|
$$
{1\over225}(10^{n+1}+5)^2 =
{1\over9}(2\cdot10^{n}+1)^2=
{1\over9}(4\cdot10^{2n}+4\cdot10^{n}+1)=\\
{1\over9}(4(10^{2n}-1)+4(\cdot10^{n}-1)+9)=
4{10^{2n}-1\over9}+4{10^{n}-1\over9}+1=\\
4{10^{2n}-1\over10-1}+4{10^{n}-1\over10-1}+1=
4\sum_{k=0}^{2n-1}10^k+4\sum_{k=0}^{n-1}10^k+1=\\
\sum_{k=n}^{2n-1}4\cdot10^k+\sum_{k=1}^{n-1}8\cdot10^k+9=\underbrace{44\ldots4}_\text{n}\underbrace{88\ldots8}_\text{n-1}9.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1569810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How can one show the inequality $\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$ How can one show the inequality
$$\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$$
Where $a,b,c$ are real and $ab+bc+ac$ is no equal to zero
This is what I did:
By Cauchy we have:
$$\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge\frac{(a+b+c)^2}{\sqrt{b^2-bc+c^2}+{\sqrt{a^2-ac+c^2}}+{\sqrt{a^2-ab+b^2}}}$$
So I need to prove that $${\sqrt{b^2-bc+c^2}+{\sqrt{a^2-ac+c^2}}+{\sqrt{a^2-ab+b^2}}}\ge(a+b+c)$$
We see that for example $b^2-bc+c^2=(b-c)^2+bc$
I still stuck here!
|
Note that:
$3(a-b)^{2}\ge0$
hence $4a^{2}-4ab+4b^{2}\ge a^{2}+b^{2}+2ab$.
And so:
$a^{2}-ab+b^{2}\ge\frac{a^{2}+b^{2}+2ab}{4}$
Since both sides are positive:
$\sqrt{a^{2}-ab+b^{2}}\ge\sqrt{\frac{a^{2}+b^{2}+2ab}{4}}$
We can do the same for each of $a,c$ and $b,c$ leading us to conclude that,
$\sqrt{b^{2}-bc+c^{2}}+\sqrt{a^{2}-ac+c^{2}}+\sqrt{a^{2}-ab+b^{2}}\ge\sqrt{\frac{(b^{2}+2bc+c^{2})}{4}}+\sqrt{\frac{(a^{2}+2ac+c^{2})}{4}}+\sqrt{\frac{(b^{2}+2ba+a^{2})}{4}}=\frac{a+b}{2}+\frac{a+c}{2}+\frac{b+c}{2}=a+b+c$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1570035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Mathematical Induction question: Prove divisibility by $4$ of $5^n + 9^n + 2$ Use mathematical induction to prove that $5^n + 9^n + 2$ is divisible by $4$, where $n$ is a positive integer.
|
With induction: $5^1+9^1+2=16$ is divisible by $4$. If $5^k+9^k+2$ is divisible by $4$ for some $k\ge 1$, then $5^{k+1}+9^{k+1}+2=\left(5^k+9^k+2\right)+4\left(5^k+2\cdot 9^k\right)$ is also divisible by $4$.
Without explicit induction: $$5^n+9^n+2=\left(5^n-1\right)+\left(9^n-1\right)+4$$
$$=(5-1)\left(5^{n-1}+5^{n-2}+\cdots+1\right)+(9-1)\left(9^{n-1}+9^{n-2}+\cdots+1\right)+4$$
$$=4\left(\left(5^{n-1}+5^{n-2}+\cdots+1\right)+2\left(9^{n-1}+9^{n-2}+\cdots+1\right)+1\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1570786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$
My questions:
$1)$ Is there an easy way to see that $x=-8$ is a root too?
$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$
|
First advice: when you are looking for roots of a polynomial, factorize as much as you can. You have $(x-1)^2$ and $(x-1)^3$, so you can factorize $(x-1)^2$. You have $(x+2)$ and $(x+2)^2$, so you can factorize $(x+2)$. Thus you get:
$$(x+2)(x-1)^2 (2(x-1)-3(x+2))\,. $$ So:
*
*For question 1, the $-8$ root should be evident (develop the last term)
*For question 2: look at powers, $(x-1)^2$ means that $1$ is a root, twice. So your roots are $-8$, $-2$ and $1$ and $1$ again.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1572045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac {\pi} {4} = \lim_{N \to \infty} s_N.$$
By Euler-MacLaurin summation formula,
$$\sum_{k = 0}^{N} \frac {1} {4k + 1} = \int_{0}^{N} \frac {dx} {4x + 1} + \frac {1} {2} \left(1 + \frac {1} {4N + 1} \right) + o \left (\frac {1} {N^2} \right)$$
and
$$\sum_{k = 0}^{N} \frac {1} {4k + 3} = \int_{0}^{N} \frac {dx} {4x + 3} + \frac {1} {2} \left(\frac {1} {3} + \frac {1} {4N + 3} \right) + o \left (\frac {1} {N^2} \right).$$
We then have
$$s_N = \frac {1} {4} \log \left(3 - \frac {6} {4N + 3} \right) + \frac {1} {3} + \frac {1} {(4N + 1) (4N + 3)} + o \left (\frac {1} {N^2} \right)$$
and
$$\lim_{N \to \infty} s_N = \frac {\log 3} {4} + \frac {1} {3}.$$
But $\frac {\log 3} {4} + \frac {1} {3} \ne \frac {\pi} {4}$. How come? Where have I done the mistake?
|
The error surely lies in the $o\left({1\over N^2}\right)$ term(s). Look at it this way: Your use of Euler-Maclaurin would suggest
$$\sum_{k=1}^N{1\over k}=\int_1^N{dx\over x}+{1\over2}\left(1+{1\over N}\right)+o\left({1\over N^2}\right)$$
as well, which would suggest
$$\sum_{k=1}^N{1\over k}-\log N\to{1\over2}$$
instead of $\gamma\approx0.5772$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1575290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
}
|
How to find the Maclaurin series for $f(x) = \frac{1}{1 + \sin(x)}$? I have that $\frac{1}{1 + x} = 1 - x + x^2 - x^3 + ...$
So then $\frac{1}{1 + \sin(x)}$ should be $ 1 - \sin(x) + \sin^2(x) - \sin^3(x) + ...$ but clearly this is not the case.
So how does substitution into Maclaurin series work and why does this not work?
|
By substitution of the expansion of $\sin x$ at the required order in the expansion of $\dfrac1{1+u}$ at the same order.
Example for order 5:
$$\dfrac1{1+u}=1-u+u^2-u^3+u^4-u^5+o(u),\qquad \sin x=x-\frac{x^3}6+\frac{x^5}{120}+o(x5),$$
whence
\begin{align*}
\sin^2x&=x^2-\frac{x^4}3+o(x^5)&\sin^3x&=\Bigl(x^2-\frac{x^4}3+o(x^5)\Bigr)\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)=x^3-\frac{x^5}2+o(x^5)\\
\sin^4x&=x^4+o(x^5)& \sin^5x&=x^5,
\end{align*}
so we have
\begin{align*}
\dfrac1{1+\sin x}&=1-\Bigl(x-\frac{x^3}6+\frac{x^5}{120}\Bigr)+\Bigl(x^2-\frac{x^4}3\Bigr)-\Bigl(x^3-\frac{x^5}2\Bigr)+x^4-x^5+o(x^5)\\&=1-x+x^2-\frac{5x^3}6+\frac{2x^4}3-\frac{61x^5}{120}+o(x^5).
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1575784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Let $x^2+kx=0;k$ is a real number .The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set. Let $f(x)=x^2+kx;k$ is a real number.The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set.
The equation $x^2+kx=0$ has solutions $x=0,-k$.
So the solutions of the equation $f(f(x))=0$ should be $x=0,-k$.
$f(f(x))=(x^2+k x)^2+(x^2+k x)k=0$ gives
$x^4+k^2x^2+2kx^3+x^2k+x k^2=0$
$x^4+2kx^3+(k^2+k)x^2+xk^2=0$
$x(x^3+2kx^2+(k^2+k)x+k^2)=0$ has the solutions $x=0,x=-k$
Therefore $x^3+2kx^2+(k^2+k)x+k^2=0$ has the solution $x=-k$.
But putting $x=-k$ in the equation $x^3+2kx^2+(k^2+k)x+k^2=0$ gives me nothing.It just gives me $0=0$.
I am stuck here.Please help me.Thanks.
|
$f(f(x))$ has as solutions, the solutions of $f(x)$, plus the solution of the equation $x^2+kx+k=0$, which are $(1/2)[-k \pm \sqrt{k^2-4k}]$. So, to have $f(x)$ and $f(f(x))$ to have the same real solution set, these two solutions must be equal to the solutions of $f(x)$, which means $0$ and $-k$. So putting the previous solutions equal to $0$ and $-k$, lead to $k=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1576745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Differential Equation change of variable I'm investigating how you get from this:
$$ z\frac{d^2y}{dz^2}+(1-a)\frac{dy}{dz}+a^2z^{2a-1}y=0 $$
to this:
$$ \frac{d^2y}{dx^2}+y=0 $$
with a change of variable:
$$ z=x^{1/a}, (x\ge0) $$
I'm not quite 'getting' the substitution in the derivative...
In a normal situation where we have $ \frac{dy}{dx} $ and a function $ y(x) $, we basically perform $ \frac{d}{dx}y(x) $ to see how $ y $ changes with respect to $ x $.
Now, with the variable substitution are we now saying that $ \frac{dy}{dz} $ is changed to $ \frac{d}{dz}y(z) = \frac{d}{dx}y(x^{1/a}) $.
This leaves me with a derivative that can be solved using the Chain Rule?
So, $$ \frac{d}{dx}y(x^{1/a}) = \frac{dy}{dx}\frac{x^{\frac{1}{a}-1}}{a}$$
I would then just proceed to take the second derivative and substitute $ \frac{dy}{dx} $ , $ \frac{d^2y}{dx^2} $ and $ y $ into the equation original equation above.
Is my understanding correct on this please? Thankyou.
|
Note, that by the chain rule, we have
$$ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} $$
and, taking another derivative, by product and chain rules,
$$ \frac{d^2y}{dz^2} = \frac{d^2 y}{dx^2} \cdot \left(\frac{dx}{dz}\right)^2
+ \frac{dy}{dx} \cdot \frac{d^2 x}{dz^2} $$
As $x = z^a$, we have
$$ \frac{dx}{dz} = az^{a-1}, \quad \frac{d^2 x}{dz^2} = a(a-1)z^{a-2} $$
that is
\begin{align*}
\frac{dy}{dz} &= \frac{dy}{dx} \cdot az^{a-1}\\
\frac{d^2y}{dz^2} &= a^2z^{2(a-1)} \frac{d^2y}{dx^2} +
a(a-1)z^{a-2} \frac{dy}{dx}\\
\end{align*}
Giving
\begin{align*}
z\frac{d^2y}{dz^2} + (1-a)\frac{dy}{dz} + a^2z^{2a-1}y
&= a^2z^{2a-1}\frac{dy^2}{dx^2} + a(a-1)z^{a-1}\frac{dy}{dx}
- a(a-1)z^{a-1}\frac{dy}{dx} + a^2z^{2a-1}y\\
&= a^2z^{2a-1}\left(\frac{d^2y}{dx^2} + y\right)
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1576824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Integral $\int^\ell_{-\ell} r/\sqrt{L^2+r^2}^{\,3} \, dL$ We tried to solve a magnetics problem and ended up with
$$\int^\ell_{-\ell} \frac{r}{\sqrt{L^2+r^2}^{\,3}} \, dL.$$
How do I solve this integral?
|
$$\int_{-l}^{l}\frac{r}{\left(\sqrt{L^2+r^2}\right)^3}\space\text{d}L=$$
$$r\int_{-l}^{l}\frac{1}{\left(\sqrt{L^2+r^2}\right)^3}\space\text{d}L=$$
$$r\int_{-l}^{l}\frac{1}{\left(L^2+r^2\right)^{\frac{3}{2}}}\space\text{d}L=$$
For the integrand $\frac{1}{\left(L^2+r^2\right)^{\frac{3}{2}}}$, (assuming all variables are positive).
Substitute $L=r\tan(u)$ and $\text{d}L=r\sec^2(u)\space\text{d}u$.
Then $\left(L^2+r^2\right)^{\frac{3}{2}}=\left(r^2\tan^2(u)+r^2\right)^{\frac{3}{2}}=r^3\sec^3(u)$ and $u=\arctan\left(\frac{L}{r}\right)$.
This gives a new lower bound $u=-\arctan\left(\frac{l}{r}\right)$ and upper bound $u=\arctan\left(\frac{l}{r}\right)$:
$$r^2\int_{-\arctan\left(\frac{l}{r}\right)}^{\arctan\left(\frac{l}{r}\right)}\frac{\cos(u)}{r^3}\space\text{d}u=$$
$$\frac{1}{r}\int_{-\arctan\left(\frac{l}{r}\right)}^{\arctan\left(\frac{l}{r}\right)}\cos(u)\space\text{d}u=$$
$$\frac{1}{r}\left[\sin(u)\right]_{-\arctan\left(\frac{l}{r}\right)}^{\arctan\left(\frac{l}{r}\right)}=$$
$$\frac{1}{r}\left(\sin\left(\arctan\left(\frac{l}{r}\right)\right)-\sin\left(-\arctan\left(\frac{l}{r}\right)\right)\right)=$$
$$\frac{1}{r}\left(\frac{l}{\sqrt{l^2+r^2}}--\frac{l}{\sqrt{l^2+r^2}}\right)=$$
$$\frac{1}{r}\left(\frac{l}{\sqrt{l^2+r^2}}+\frac{l}{\sqrt{l^2+r^2}}\right)=$$
$$\frac{1}{r}\left(\frac{2l}{\sqrt{l^2+r^2}}\right)=$$
$$\frac{\frac{2l}{\sqrt{l^2+r^2}}}{r}=$$
$$\frac{2l}{r\sqrt{l^2+r^2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1577002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solving a simultaneous equation How can I solve the following simultaneous equations:
$$3x^4+3x^2y^2-6xy = 0\tag 1$$
$$-2x^3y+3x^2-y^2=0\tag 2$$
I have tried rearranging for $y$ in eq(1) and plugging it into eq(2), but the result did not give me the right answer.
|
Using the old Sylvester´s method of elimination, choosing $y$ (the lowest degree) to be eliminated, we have
$$\begin{vmatrix}
3x^2&-6x&3x^4&0\\
0&3x^2&-6x&3x^4\\
1&2x^3&-3x^2&0\\
0&1&2x^3&-3x^2
\end{vmatrix}=0$$ which gives $$36x^4(x^8+2x^4-3)=0$$ i.e.
$$x^4(x^4+3)(x^2+1(x+1)(x-1)=0$$
The $12$ values for $x$ are $$x=0\space \text{of order four}$$ $$x=\pm 1$$ $$x=\pm i$$ $$x= \pm \sqrt[4]{-3}$$ $$x=\pm i\sqrt[4]{-3}$$
(where $\sqrt[4]{-3}=\frac{(1+i)\sqrt[4]{-3}}{\sqrt 2}$)
To each of these values of $x$ it correspond correlative values of $y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1577952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
evaluate $\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$
$$\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$$
can I look at $\lim\limits_{x\to \infty} (3^{\frac{1}{3}}x^{\frac{2}{3}}-x+x)$?
|
HINT:
$$\left(1+z\right)^{1/3}=1+\frac13 z+O(z^2) \tag 1$$
SPOILER ALERT: Scroll over the highlighted area to reveal the full solution
$$\begin{align}x+\left(3x^2-x^3\right)^{1/3}&=x-x\left(1-\frac{3x^2}{x^3}\right)^{1/3}\\\\&=x-x\left(1-\frac3x\right)^{1/3}\\\\&=x-x\left(1-\frac1x+O\left(\frac1{x^2}\right)\right)\\\\&=1+O\left(\frac1x\right)\to 1\,\,\text{as}\,\,x\to \infty\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1578409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
How do you evaluate this sum of multiplied binomial coefficients: $\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} $? We have to find the value of x+y in:
$$\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} = \binom{x}{y} $$
My approach:
I figured that the required summation is nothing but the coefficient of $x^3$ is the following expression:
$$\sum_{r=1}^7 \frac{r(r+1)}{2}(1+x)^{10-r} $$
which looks like:
$$(1+x)^{10} + 3(1+x)^9 + 6(1+x)^8 + ..... + 36(1+x)^3$$
Hence we have a sequence in which the difference of the coefficients forms an AP.
So let the expression = $S$
So we have:
$$\frac {S}{1+x} = (1+x)^{9} + 3(1+x)^8 + 6(1+x)^7 + ..... + 36(1+x)^2$$
Subtracting $\frac{S}{1+x}$ from $S$, we get:
$$S(1-\frac{1}{1+x}) = (1+x)^{10} + 2(1+x)^9 + 3(1+x)^8 + ..... + 8(1+x)^3 - 36(1+x)^2$$
Here, all the terms except the last one are in AGP and hence we can find $S$ by applying the same method and isolating $S$ on one side. Then we can process to find the coefficient of $x^3$ but it got really lengthy. In the test that it was given, we have an average time of 3 minutes per question and not even the hardest question takes more than 7-8 minutes if you know how to do it.
Hence, I'm wondering if there is a better, shorter way to solve this question.
|
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Leftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{r = 2}^{9}{r \choose 2}{12 - r \choose 3}} =
\sum_{r = 0}^{\infty}{r + 2\choose 2}{10 - r \choose 3} =
\sum_{r = 0}^{\infty}{r + 2\choose r}{10 - r \choose 7 - r}
\\[3mm] = &\
\sum_{r = 0}^{\infty}{-r - 2 + r - 1 \choose r}\pars{-1}^{r}{-10 + r + 7 - r - 1 \choose 7 - r}\pars{-1}^{7 - r}
\\[5mm] = &\
-\sum_{r = 0}^{\infty}{-3 \choose r}{-4 \choose 7 - r} =
-\sum_{r = 0}^{\infty}{-3 \choose r}\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-4} \over z^{8 - r}}
\,{\dd z \over 2\pi\ic}}^{\ds{{-4 \choose 7 - r}}}
\\[5mm] = &\
-\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-4} \over z^{8}}\
\overbrace{\sum_{r = 0}^{\infty}{-3 \choose r}z^{r}}^{\ds{\pars{1 + z}^{-3}}}\
\,{\dd z \over 2\pi\ic} =
-\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-7} \over z^{8}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
-{-7 \choose 7} = -{-\pars{-7} + 7 - 1 \choose 7}\pars{-1}^{7} =
\color{#f00}{{13 \choose 7}} = 1716
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1579691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to use Mathematical Induction to prove $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$? $$\frac{1}{1 \cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$
What I have so far in the induction is:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
But I'm not sure where to go from there . . .
|
Get a common denominator. Your last expression simplifies to $\dfrac{k^2 + 2k + 1}{(k+1)(k+2)}$. Now factor it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1580303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Way to verify a least-squares solution without actually solving for $x$ and $y$? I just found the least squares solution of the system $\mathbf{x}A = \mathbf{b} = \begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 3 & 2 & 1 \\ 2 & 3 & 2\end{pmatrix} = \begin{pmatrix} 3 & 0 & 1\end{pmatrix}$ to be $\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} \frac{29}{21} \\ -\frac{2}{3}\end{pmatrix}$. I now am tasked with answering how this solution can be verified without solving for $x$ and $y$.
The system can be rewritten as follows: $\begin{pmatrix}3 & 2 \\ 2 & 3 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}$.
Since the least-squares solution is the vector $\mathbf{x}$ that makes $\Vert b - A\mathbf{x} \Vert_{2}$ a minimum, should I try graphing the equations $3x + 2y = 3$, $2x + 3y = 0$, and $x + 2y = 1$, then pointing out on the graph that $\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}\frac{29}{21} \\ -\frac{2}{3} \end{pmatrix}$ is close to the point where the three lines intersect?
I figure there must be a more formal way of verifying the solution, but it is escaping me. Could somebody please let me know how I should do it?
|
By construction, the least squares solution minimizes the sum of the squares of the residuals
$$
r^{2}(x,y) = \lVert A x - b \rVert_{2}^{2}.
$$
that is, the least squares solution can be defined as
$$
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]_{LS}
=
\left\{
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]
\in \mathbb{R}^{2} \colon \lVert A x - b \rVert_{2}^{2} \text{ is minimized}
\right\}
$$
Show the proposed solution is a minimum by establishing that $r^{2}(x,y)_{LS}$ is the smallest value in a $\delta-$neighborhood of $x_{LS}$.
Let
$$
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]
=
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]_{LS}
+
\left[
\begin{array}{c}
\delta_{1} \\ \delta_{2}
\end{array}
\right]
=
\left[
\begin{array}{r}
\frac{29}{21} + \delta_{1} \\ -\frac{2}{3} + \delta_{2}
\end{array}
\right]
.
$$
The norm of the residual error vector is
$$
\lVert A (x_{LS} + \delta) - b \rVert_{2}^{2} = \sqrt{\frac{32}{21} + 14 \delta_{1}^2 + 28 \delta_{1}\delta_{2} + 17\delta_{2}^2}.
$$
This establishes that the minimum value is attained when $\delta_{1} = \delta_{2} = 0$, that is when $x = x_{LS}$. Therefore, $x_{LS}$ is the solution of minimum norm.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1580713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Infinite sums of reciprocal power: $\sum\frac1{n^{2}}$ over odd integers The infinite series I need to solve is
$$\sum_{n=1,3,5...}^{\infty}\frac{1}{n^{2}}$$
and because the point of interest lies in the value of odd n,
the infinite series can be expressed as
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$
This came up in a quantum mechanics problem involving the expectation value of the Hamiltonian.
Is there a good idea to verify the solution is indeed $$\frac{\pi^{2}}{8}$$
or is this something with which I must refer to a math table?
Any good ideas would be helpful.
|
by using
$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(\frac{\pi^2}{6})$$
$$\frac{\pi^2}{6}-\frac{\pi^2}{24}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$
$$\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1580873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
}
|
Integrate $\int \frac{1}{x^4+4}dx$
Integrate $\displaystyle \int \dfrac{1}{x^4+4}dx$.
I could try breaking this up into two quadratic trinomials, but that seems like it would be a lot of work. If that is the best way here how do I do it most efficiently?
|
Notice
$$\frac{1}{x^4+4} = \frac{1}{(x^2+2)^2 - 4x^2}
= \frac{1}{(x^2+2x+2)(x^2-2x+2)}$$
Since
$$\begin{align}
\frac{1}{x^2-2x+2} - \frac{1}{x^2+2x+2} &= \frac{4x}{(x^2+2x+2)(x^2-2x+2)}\\
\frac{1}{x^2-2x+2} + \frac{1}{x^2+2x+2} &= \frac{2x^2+4}{(x^2+2x+2)(x^2-2x+2)}
\end{align}$$
We have
$$\frac{1}{(x^2+2x+2)(x^2-2x+2)}
= \frac14\left[\frac{1-\frac{x}{2}}{x^2-2x+2} + \frac{1+\frac{x}{2}}{x^2+2x+2}\right]\\
= \frac18\left[\frac{x+2}{x^2+2x+2} - \frac{x-2}{x^2-2x+2}\right]
= \frac18\left[\frac{(x+1)+1}{(x+1)^2+1} - \frac{(x-1)-1}{(x-1)^2+1}\right]
$$
Up to an integration constant, this give us
$$\begin{align}
\int\frac{dx}{x^4+1}
&= \frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right)
+ \frac18 \left[\tan^{-1}(x+1)+\tan^{-1}(x-1)\right]\\
&= \frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right)
+ \frac18 \left[\tan^{-1}(x+1)+\tan^{-1}(x-1)\right]\\
&=\frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right)
+ \frac18 \tan^{-1}\left(\frac{2x}{2-x^2}\right)
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1581136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Finding the vectors that form a basis of a span Let
$$\begin{pmatrix}
1 & 2\\
0 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 1\\ 3 & 0
\end{pmatrix},
\begin{pmatrix}
3 & 3\\ 4 & 0
\end{pmatrix},
\begin{pmatrix}
2 & 0\\ 1 & -1
\end{pmatrix}$$
be vectors in some vector space $U$. Which vectors form a basis of $Sp(U)$?
In this case it is "easy to see" that $A_3 = A_1 + A_2 +A_4$, and $\sum_{i\in\{1,2,4\}} \alpha_{i}A_{i}=0$ has only the trivial solution. But what if I had $n>5$ matrices of order $k\times k$, $k>4$. How then can I found a basis and identify which matrices are a linear combination of the others?
Thank you.
|
Generally, a "basis" for a vector space must both span that space and be linearly independent. If the give spanning set is already independent, then it is a basis. If not you can delete one or more vectors until it is.
Here, for example, to see if these matrices are independent, look at the equation $a\begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ b\begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}+ c\begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}+ d\begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.
That gives us the four equations a+ 3c+ 2d= 0, 2a+ b+ 3c= 0, 3b+ 4c+ d= 0, and a- d= 0. An obvious, trivial, solution, to this is a= b= c= d= 0. If that is the only solution, then these vectors are independent so a basis. If we attempt to solve these equations, we can see that d= a from the last equation and then we hafe 3a+ 3c= 0, 2a+ b+ 3c= 0, and 4a+ 4c= 0. The first and third of those equations give c= -a. Putting that into the second equation, b- a= 0 so b= a. We have a= b= -c= d so we and write our equation as $a\begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ a\begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}- a\begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}+ a\begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.
We can divide through by a to get $\begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ \begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}- \begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}+ \begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.
The point now is that we can solve for any one of those vectors in terms of the other three- four example, we can write $\begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}= \begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ \begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}+ \begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$. Since that can be written as a linear combination if the other three, we can span the space with those three alone. If those three are independent, then they form a basis.
In general, if a spanning set is also independent, then they are a basis. If not then they can be written as $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ with at least some of the $a_i$ non-zero. If, say, $a_i$ is non-zero, we can solve for $a_iv_i= -(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)$ (where the sum on the right does NOT incude $a_1v_1$) and then, since $a_1\ne 0$, we can divide by it to get $v_i= -\frac{1}{a_i}(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)$ so that we can drop $v_i$ from the set and still span the space.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1582040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How can we find the area of the triangle which covers a finite point set in $\mathbb{R}^2$ by using the interior triangles with specified area? Suppose we have given a finite point set $X \subset \mathbb{R}^2$ in a way that any triangle made by vertices of $X$ has area at most 1. How can we prove that there is a triangle of area 4 which is covering the whole point set $X$?
And why we cannot take any other number less than 4 for the area of the covering triangle?
Hint: Suppose one of the triangles with vertices in $X$ and use the hyper planes which have intersections with mentioned triangle just in its vertices and find the bigger triangle.
|
Let $PQ$ be the segment with vertices in $X$ and let $R$ be another point of $X$ such that $PQR$ has the maximum area (of all possible triangles with $P$ and $Q$ being two of their vertices). Now let $l_R$ be the line passes through the point $R$ and parallel to the edge $PQ$, and let $H_R$ be the half plane through $l_R$ which does not contain the edge $PQ$. Since $PQR$ has the maximal area between the triangles with edge $PQ$, so any other point of $X$ does not belong to $H_R$, because if does? we will have a triangle with an area bigger than $PQR$; that is contradiction. So whole $X$ will stay in the other side of $H_R$. And if we repeat this process for other vertices $P$ and $Q$, we will find that $X$ will be inside the intersection of outers of the half plans $H_R$, $H_P$ and $H_Q$ which is the triangle $ABC$ with similarity ratio 2 as shown in the picture, which will give us:
$S_{ABC} = \frac{AC \times h_{ABC}}{2} = \frac{2 PR \times 2 h_{PQR}}{2} = 2 PR \times h_{PQR} = 4 S_{PQR}$.
It shows that area of $ABC$ is at most 4.
And for the next part of question, we need to prove that the triangle with maximum size inscribed in the circumference of a circle of radios $r$ is a regular (equilateral) triangle.
$\textbf{Solution:}$
$>$ suppose we have the following isosceles triangle inscribed in a circle with radios $r$ and want to maximize its area.
We first notice that the vertical line creates a right angle with the base of the triangle and bisects the
triangle since it is isosceles. Next, we write the equations
\begin{equation}
x^2 + y^2 = r^2
\end{equation}
from the Pythagorean theorem with the small triangles. Then, we write the equation
\begin{equation}
A = \frac{1}{2} (2x)(r + y)
\end{equation}
for the area of the triangle. Now, solve (0.0.1) for $x$, we obtain
$$x = \sqrt{r^2 - y^2}$$
by Substituting this into (0.0.2), we have
$$A(y) = \sqrt{r^2 - y^2} (y + r) $$
Taking the derivative by applying the product rule, we obtain
$$\frac{\partial A(y)}{\partial y} = \sqrt{r^2 - y^2} + (y + r) \frac{- 2 y}{2 \sqrt{r^2 - y^2}}$$
Simplifying this expression, we have
$$\frac{\partial A(y)}{\partial y} = \frac{r^2 - ry - 2 y^2}{\sqrt{r^2 - y^2}}$$
Now, to find the maximum, we set $\frac{\partial A(y)}{\partial y} = 0$ and solve:
$$\frac{r^2 - ry - 2 y^2}{\sqrt{r^2 - y^2}} = 0$$
Since $y = r$ and $y = −r$ give areas of $0$, we are not interested in the points where the derivative is not
defined, so we may assume that $r^2 - y^2 \neq 0$ . Therefore, we must solve
$$ r^2 - ry - 2 y^2 = 0 \Rightarrow y = \frac{r \pm \sqrt{9 r^2}}{-4}$$
Therefore, we have
$$y = −r or y = \frac{r}{2}$$
Clearly $y \neq −r$ since $A(−r) = 0$. Therefore, we must check to see that $y = \frac{r}{2}$ is a maximum. To do
this, we determine that
$$\frac{\partial A(y)}{\partial y} > 0 for y < \frac{r}{2} and \frac{\partial A(y)}{\partial y} < 0 for y > \frac{r}{2}$$
Therefore,$\frac{r}{2}$
is a maximum.
Now, we solve for the maximizing dimensions of the triangle. Using (0.0.1), we obtain
$$x^2 + (\frac{r}{2})^2= r^2 \Rightarrow x^2 = \frac{3}{4} r^2 \Rightarrow x = \frac{\sqrt{3}}{2} r$$
Now, writing the pythagorean theorem for the large triangles, we see that
$$x^2 + (r+ y)^2 = s^2 \Rightarrow \frac{3}{4} r^2 + (r + \frac{r}{2})^2 = \frac{12}{4} r^2 = s^2$$
Thus, we have $s = \frac{2\sqrt{3}}{2} r. $ Finally, we compute the area of the triangle and obtain
$$A(\frac{r}{2})= x(y + r) = \frac{\sqrt{3}}{2} r (\frac{r}{2} + r) = \frac{3\sqrt{3}}{4} r^2$$
Thus, the triangle with maximum area has dimensions
$$s = 2x = \sqrt{3} r, ~~~ h = r + y = \frac{3}{2}r.$$
Which shows that is a regular triangle.
Now we need to prove that the triangle with minimum area size circumscribed around the circumference of a circle of radios $r$ is a regular (equilateral) triangle.
$\textbf{Solution:}$
$>$ The area of the triangle is $A = \frac{1}{2} h b.$ By similar triangles (see figure)
$$\frac{\frac{b}{2}}{h} = \frac{r}{\sqrt{h^2 - 2rh}} \Rightarrow b = \frac{2rh}{\sqrt{h^2 - 2rh}}$$
So $A = \frac{rh^2}{\sqrt{h^2 - 2rh}}$ for $h > 2r,$ and then by taking the derivative, we have
$$\frac{\partial A(h)}{\partial h} = \frac{rh^2 (h - 3r)}{(h^2 - 2rh)^{\frac{3}{2}}}$$
Then $\frac{\partial A(h)}{\partial h} = 0$ for $h > 2r$ when $h = 3r$ and by the first derivative test $A$ is minimum when $h = 3r$.
If $h = 3r$ then $b = 2 \sqrt{3} r$, shows that the triangle is regular (equilateral).
Now the natural question which come to the mind is that " what is the value for the radios of the circle for to have inscribe triangles to have maximum area 1?"
$\textbf{Solution:}$
$>$ For to have this result we should have
$$x(y + r) \leq 1 \Rightarrow \frac{\sqrt{3} r}{2} \frac{3r}{2} \leq 1 \Rightarrow r \leq \frac{2}{3^{\frac{3}{4}}}$$
The another question will be the opposite one "what is the value for the radios of the circle for to have circumscribed triangle around the circumference circle to have minimum area 4?"
$\textbf{Solution:}$
$>$ the same as previous for to have the result we should have
$$x . 3 r \geq 4 \Rightarrow 3 \sqrt{3 } r^2 \geq 4 \Rightarrow r \geq \frac{2}{3^{\frac{3}{4}}}$$
And it will give the answer that $r$ should be equal to $ \frac{2}{3^{\frac{3}{4}}}$ .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1582436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Could you solve $x↑^n2=x↑^m2$? As my title asks, could you solve $x^2=2^x$?
But that's the worrisome part, as I noticed $x↑^n 2=x↑^m 2$ and $2↑^p x=2↑^q x$ will always have a solution at $x=2$. However, there is bound to be at least another solution.
For example:$$2x=x^2$$has a solution at $x=2,0$.
And $x^2=2^x$ has a solution at $x=2,4,-z$ where $z$ is some hard to determine value.
But I question $2^x=x^x$ having more solutions than just $x=2$.
$x^2=x^x$ appears to have solutions at $x=1,2$ and possibly a negative solution as well.
So what can one say about these solutions other than $x=2$ and some other number?
I've also noticed $2↑^n1=2$ as well.
Which makes my also wonder how odd $2$ is when it comes to higher towers.
And $1$, as that solution arises often.
Due to the confusion in what kind of answer I want, I will make a clarification.$$x↑^12=x\cdot x=x^2$$$$x↑^22=x^x=^2x$$$$x↑^32=^xx=f_2(x)$$$$x↑^42=f_x(x)=g_2(x)$$And so it goes...
|
As you have probably surmised, $x \uparrow^n 2 = x \uparrow^m 2$ and $2 \uparrow^p x = 2 \uparrow^q x$ have solutions at $x=1$ and $x=2$. If $m=n$ or $p=q$ then of course any $x$ will be a solution; if $m \neq n$ or $p \neq q$ then there are no other solutions.
First, I will prove the following useful theorems:
Theorem. For $a \ge 2$ and $b \ge 1$
a. $a \uparrow^b c > a\uparrow^b (c-1)$ for $c \ge 2$.
b. $a \uparrow^b c > c$ for $c \ge 1$.
Proof: By double induction on $b,c$. For $b=1$, we have $a^c = a * a^{c-1} > a^{c-1}$, and it is well known that $a^c \ge 2^c \ge c+1$. (Use the binomial theorem for example.)
For $c=1$, we have $a \uparrow^b 1 = a > 1$.
Assuming the theorem for $b$, and for $b+1$ and $c$, we will prove it for $b+1$ and $c+1$. Observe that
$$a\uparrow^{b+1}(c+1) = a\uparrow^b (a\uparrow^{b+1}c) > a\uparrow^{b+1} c$$
which proves part a. For part b, observe that $a\uparrow^{b+1}(c+1) > a\uparrow^{b+1} c > c$, so $a\uparrow^{b+1}(c+1) > c+1$, as desired.
Corollary. Given a fixed $a \ge 2$ and $b \ge 1$, $a \uparrow^b c$ is an increasing function of $c$.
The corollary follows directly from the theorem part a.
Theorem 2. Given $a,c \ge 2$ and $b \ge 1$, $a\uparrow^b c \ge c+2$.
Proof: By induction on $b$.
If $b=1$, then$a^c \ge 2^c \ge c+2$.
Assume the theorem for $b$. Then $$a\uparrow^{b+1} c = a\uparrow^b(a\uparrow^{b+1}(c-1)) \ge a\uparrow^b c \ge c+2$$
Now we can address the original equations. Observe that for $a \ge 2$, $b \ge 1$, $c \ge 3$,
$$a\uparrow^{b+1} c = a\uparrow^b(a\uparrow^{b+1}(c-1) \ge a\uparrow^b(c+1) > a\uparrow^b c$$
so that $a\uparrow^b c$ is an increasing function of $b$; in particular, $2\uparrow^p x =2\uparrow^q x$ has no solutions for $p \neq q$ and $x \ge 3$.
If $a \ge 3$, $b \ge 1$, $c = 2$,
$$a\uparrow^{b+1} 2 = a\uparrow^b a > a\uparrow^b 2$$
so in this case again $a\uparrow^b 2$ is an increasing function of $b$; in particular, $x \uparrow^n 2 = x \uparrow^m 2$ has no solutions for $m \neq n$ and $x \ge 3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1582600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather not $0$.
|
Rewrite each term into:
$$2\times\frac{4}{5}\times\frac{6}{7}\times\frac{8}{9}\times\frac{10}{11}\times\dots\times\frac{2n}{2n+1}\times\frac{1}{2n+3}<\frac{2}{2n+3}$$
Hence, the limit of the terms go to $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1583858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Solve the functional equation $f (2x)=f (x)\cos x$ Find all $f: \mathbb R\longrightarrow \mathbb R $
such that $f $ is a continuous function at $0$ and satisfies
$$\;\forall \:x \in \mathbb R,\; f\left(2x\right) = f\left(x\right)\cos x $$
My try: I just found the $f (x)$ is periodic, i.e.
$f (2\pi / 2)= f (\pi/2) \cos (\pi /2) $
And$ f (\pi)=f (3\pi)$ ... and so on,
Best I came up with is
$$f (2^n x) = f (x) \cos (x) \cos (2x) ... \cos (2^{n-1} x)$$
|
We have
$$f(x) = f\left(\dfrac{x}2\right)\cos\left(\dfrac{x}2\right) = f\left(\dfrac{x}4\right)\cos\left(\dfrac{x}4\right)\cos\left(\dfrac{x}2\right)$$
Hence, we have
$$f(x) = f\left(\dfrac{x}{2^{n}}\right) \prod_{k=1}^n \cos\left(\dfrac{x}{2^k}\right) = \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right)\dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n} \right)}$$
Hence, we have that
$$f(x) = \lim_{n \to \infty} \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right) \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \lim_{n \to \infty}f\left(\dfrac{x}{2^n}\right) \lim_{n \to \infty} \dfrac1{2^n} \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \dfrac{\sin(x)}{x} \lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right)$$
If we assume that $\lim_{x \to 0} f(x)$ exists at the origin, we then have that
$$\lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right) = c$$
from which we obtain that
$$f(x) = c\cdot \dfrac{\sin(x)}x$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1584120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x) = x \frac{d}{dx}[f_0(x)] = \frac{1}{(1-x)^2} = 0 + x + 2x^2 + 3x^3 +\cdots$$
$$f_2(x) = x \frac{d}{dx}[f_1(x)] = \frac{x^2+x}{(1-x)^3} = 0^2 + 1^2x + 2^2x^3 + 3^2x^3+\cdots,$$
and I assume I'm supposed to be able to do something similar in this case, but things get trickier when it's bounded by n and I keep getting stuck.
|
Given a generating function $f(x)$ for $a_n$, the generating function for $\sum_{i=0}^n a_i$ is $\sum_{n\geq 0} (\sum_{i=0}^n a_i)x^n = (\sum_{n\geq 0} x^n)(\sum_{n\geq 0} a_nx^n) = \frac{1}{1-x} f(x)$.
In this case, we have $f(x) = \frac{x(1+x)}{(1-x)^3}$ as you mention, so the desired generating function is $\frac{x(1+x)}{(1-x)^4}$. This has a partial fraction expansion:
$$\frac{x(1+x)}{(1-x)^4} = \frac{1}{(1-x)^2} - \frac{3}{(1-x)^3} + \frac{2}{(1-x)^4}$$
Since $1/(1-x)^{k+1}$ is the generating function for ${n+k}\choose{k}$ (we can see this by repeatedly differentiating $\frac{1}{1-x}$), this gives us:
$$1^2+2^2+\cdots+n^2 = {{n+1}\choose 1} - 3{{n+2}\choose 2} + 2{{n+3}\choose 3} = \frac{n(n+1)(2n+1)}{6}$$
By the way, we can avoid all this work with a good choice of basis. Since $n^2 = {n \choose 0} - 3{{n+1}\choose 1} + 2{{n+2}\choose 2}$, we have:
$$\sum_{i=0}^n i^2 = \sum_{i=0}^n \left({i \choose 0} - 3{{i+1}\choose 1} + 2{{i+2}\choose 2}\right) = {{n+1}\choose 1} - 3{{n+2}\choose 2} + 2{{n+3}\choose 3}$$
All we need here is the sum $\sum_{i=0}^n {{i+k}\choose k} = {{n+k+1}\choose {k+1}}$, which follows from the simple generating functions identity $\frac{1}{(1-x)^k}\cdot \frac{1}{1-x} = \frac{1}{(1-x)^{k+1}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1584854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
}
|
What's the method of integrating $\frac{1}{1-\ln(x)}$? How to do $\int\frac{1}{1-\ln(x)}dx$ manually? I mean I got its answer by running it through Mathematica, but is there anyway we can do it by hand?
|
$\frac{1}{1 - \ln x}$ has no elementary antiderivative. If you are so inclined you could express $1 - \ln x$ as a series.
$$\ln (1+x) = x - x^2/2 + x^3/3 - ...$$
$$\to \ln (1+x) = (-1+1+x) - (-1+1+x)^2/2 + (-1+1+x)^3/3 - ...$$
$$\to \ln (x) = (-1+x) - (-1+x)^2/2 + (-1+x)^3/3 - ...$$
$$\to -\ln (x) = -(-1+x) + (-1+x)^2/2 - (-1+x)^3/3 - ...$$
$$\to 1-\ln (x) = 1-(-1+x) + (-1+x)^2/2 - (-1+x)^3/3 - ...$$
$$\to 1-\ln (x) = 1+(1-x) + (-1+x)^2/2 - (-1+x)^3/3 - ...$$
$$\to 1-\ln (x) = 1+(x-1) + (x-1)^2/2 - (x-1)^3/3 - ...$$
Read more:
http://en.wikipedia.org/wiki/List_of_integrals_of_logarithmic_functions
http://en.wikipedia.org/wiki/Mercator_series
Alternatively:
Let $x=e^{1-z}, dx = -e^{1-z} dz$ where we can write:
$$e^{1-z} = \sum_{n=0}^{\infty} \frac{(1-z)^n}{n!}$$
$$\int\frac{dx}{1-\ln x}=\int\frac{-e^{1-z} }{z}dz = \int\frac{-1}{z}\sum_{n=0}^{\infty} \frac{(1-z)^n}{n!}dz = \sum_{n=0}^{\infty} \int\frac{-1}{z}\frac{(1-z)^n}{n!}dz$$
Since we can write
$$(1-z)^n = \sum_{k=0}^{n} \binom n k (-z)^k$$
we have
$$= \sum_{n=0}^{\infty} \int\frac{-1}{z}\frac{(1-z)^n}{n!}dz = -\sum_{n=0}^{\infty} \sum_{k=0}^{n} [\frac{1}{k!(n-k)!}(\frac{z^k}{k} + C)]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1587647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but had no luck. Thank you.
|
This is an incomplete answer, but I think the last inequality is (correct) and easier to prove.
We have:
$$\bigg (\frac{a}{b+c} \bigg)^2 + \bigg (\frac{b}{a+c} \bigg)^2 + \bigg (\frac{c}{b+a} \bigg)^2 = \frac{a^4}{a^2(b+c)^2} + \frac{b^4}{b^2(a+c)^2} + \frac{c^4}{c^2(b+a)^2} \geq \frac{(a^2+b^2+c^2)^2}{a^2(b+c)^2 + b^2(a+c)^2 + c^2(b+a)^2}$$
It suffices to show that:
$$\frac{(a^2+b^2+c^2)^2}{a^2(b+c)^2 + b^2(a+c)^2 + c^2(b+a)^2} \geq \frac{3}{4} \frac{a^2+b^2+c^2}{ab+bc+ac}$$
Or equivalently:
$$\frac{(a^2+b^2+c^2)(ab+bc+ac)}{a^2(b+c)^2 + b^2(a+c)^2 + c^2(b+a)^2} \geq \frac{3}{4} (*)$$
The denominator can be written as: $2(a^2b^2 + b^2c^2 + c^2a^2) +2abc(a+b+c) = 2(ab+bc+ac)^2 - 2abc(a+b+c)$
Thus $(*)$ becomes
$$\frac{\bigg ((a+b+c)^2-2(ab+bc+ac) \bigg)(ab+bc+ac)}{(ab+bc+ac)^2 - abc(a+b+c)} \geq \frac{3}{2} (**)$$
Let $S = a+b+c$, $P^3 = abc$, $X^2 = ab+bc+ac$, where $X^2 \leq \frac{S^2}{3}$ Thus we need to show that:
$$7X^4 - 2S^2X^2 - 3SP^3 \leq 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1588898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
particular solution of the second-order linear equation I'm trying to find particular solution of the second-order linear equation but I can't find $y_{1}$ and $y_{2}$ according to $y = c_{1}y_{1} + c_{2}y_{2}$
$$x^{2}y^{''}-2xy^{'}+2y=0, y(1) = 3, y'(1) = 1$$
If $r$ is used, $x^{2}r^{2}-2xr=-2$ then $xr(xr - 2) = -2$, I can't go on from there to find $y_{1}$ and $y_{2}$
|
$$y''(x)x^2+2y'(x)x+2y(x)=0\Longleftrightarrow$$
Assume a solution to this Euler-Cauchy equation will be proportional to $e^{\lambda}$ for some constant $\lambda$.
Substitute $y=x^{\lambda}$ into the differential equation:
$$x^2\frac{\text{d}^2}{\text{d}x^2}\left(x^{\lambda}\right)+2x\frac{\text{d}}{\text{d}x}\left(x^{\lambda}\right)+2x^{\lambda}=0\Longleftrightarrow$$
Substitute $\frac{\text{d}^2}{\text{d}x^2}\left(x^{\lambda}\right)=(\lambda-1)\lambda x^{\lambda-2}$ and $\frac{\text{d}}{\text{d}x}\left(x^{\lambda}\right)=\lambda x^{\lambda-1}$:
$$\lambda^2x^{\lambda}+\lambda x^{\lambda}+2x^{\lambda}=0\Longleftrightarrow$$
$$x^{\lambda}\left(\lambda^2+\lambda+2\right)=0\Longleftrightarrow$$
Assuming $x\ne0$, the zeros must come from the polynomial:
$$\lambda^2+\lambda+2=0\Longleftrightarrow$$
$$\lambda=-\frac{1}{2}\pm\frac{i\sqrt{7}}{2}$$
The roots $\lambda=-\frac{1}{2}\pm\frac{i\sqrt{7}}{2}$ give $y_1(x)=\text{C}_1x^{-\frac{1}{2}+\frac{i\sqrt{7}}{2}}$, $y_2(x)=\text{C}_2x^{-\frac{1}{2}-\frac{i\sqrt{7}}{2}}$ as solutions,
where $\text{C}_1$ and $\text{C}_2$ are arbitrary constants.
The general solution is the sum of the above solutions:
$$y(x)=y_1(x)+y_2(x)=\text{C}_1x^{-\frac{1}{2}+\frac{i\sqrt{7}}{2}}+\text{C}_2x^{-\frac{1}{2}-\frac{i\sqrt{7}}{2}}$$
Using $x^{\lambda}=e^{\lambda\ln(x)}$, apply Euler's identity $e^{a+bi}=e^a\cos(b)+ie^a\sin(b)$:
$$y(x)=\frac{\left(\text{C}_1+\text{C}_2\right)\cos\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}+\frac{i\left(\text{C}_1-\text{C}_2\right)\sin\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}\Longleftrightarrow$$
$$y(x)=\frac{\text{C}_1\cos\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}+\frac{\text{C}_2\sin\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}\Longleftrightarrow$$
Now using some algebra to find $\text{C}_1$ and $\text{C}_2$ with $y(1)=3$ and $y'(1)=1$:
$$y(x)=\frac{21\cos\left(\frac{1}{2}\sqrt{7}\ln(x)\right)+5\sqrt{7}\sin\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{7\sqrt{x}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1590376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Combinatorial Identity with Binomial Coefficients: $ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $ I got the following identity from commutative algebra.
I am curious to see elegant elementary methods.
$$ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $$
|
For variety's sake here is a slightly different approach.
Suppose we seek to evaluate
$$\sum_{k=0}^c {k+a-1\choose a-1} {b-1+c-k\choose b-1}.$$
Introduce
$${b-1+c-k\choose b-1} = {b-1+c-k\choose c-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c-k+1}} (1+z)^{b-1+c-k} \; dz.$$
Observe that this is zero when $k\gt c$ so we may extend $k$ to
infinity to get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{b-1+c}
\sum_{k\ge 0} {k+a-1\choose a-1} \frac{z^k}{(1+z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{b-1+c}
\frac{1}{(1-z/(1+z))^a}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{a+b-1+c}
\frac{1}{(1+z-z)^a}
\; dz
\\ = {a+b+c-1\choose c}.$$
We can take $\epsilon \lt 1/2$ for the series to converge.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1591041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Curious combinatorial summation Let $\gamma$ denote a grid walk from the upper left corner $(1,k)$ to the lower right corner $(\ell,1)$ of the $k\times\ell$ rectangle $\{1,..,k\}\times\{1,..,\ell\}$. There are $\binom{k+\ell-2}{k-1}$ such paths. Denote
$$
X_\gamma = \prod_{(i,j)\in\gamma} \frac{1}{i+j-1}\,.
$$
Claim:
$$\sum_\gamma X_\gamma = \frac{1}{(k+\ell-1)(k-1)!(\ell-1)!}\,.
$$
Equivalently, and more elegantly, for a random path $\gamma$, we have: $\ \Bbb E[X_\gamma] = 1/(k+\ell-1)!$
Example: $k=2$, $\ell=3$. There are $3=\binom{3}{1}$ paths $\,\gamma_1: (1,2) \to (1,1) \to (2,1) \to (3,1)$, $\,\gamma_2: (1,2) \to (2,2) \to (2,1) \to (3,1)$, $\,\gamma_3: (1,2) \to (2,2) \to (3,2) \to (3,1)$. Then:
$$
X_1 = \frac{1}{2\cdot 1\cdot 2\cdot 3} \ , \ X_2 = \frac{1}{2\cdot 3\cdot 2\cdot 3} \ , \ X_3 = \frac{1}{2\cdot 3\cdot 4\cdot 3} \ ,
$$
$$X_1+X_2+X_3 = \frac{1}{12}+\frac{1}{36}+\frac{1}{72} = \frac{1}{8} = \frac{1}{4\cdot 1!\cdot 2!}\,.
$$
Question: Is there a simple proof of this combinatorial summation? If it's known, does anyone have a reference?
P.S. I can in fact prove the claim but the proof is incredibly involved for such a simple looking result.
|
Suppose the start point is $(a,b)$ and the end point is $(c,d)$, where $a\geq c$ and $d\geq b$.
From the general form of $F((k,1),(1,l))$, and the fact that paths from $(k,1)$ go through $(k-1,1)$ or $(k,2)$, you can deduce the general form of $F((k,2),(1,l))$.
Then $F((k,3),(1,l))$ and so on.
I got this formula:
$$F((a,b),(c,d))=\frac{(a+d-b-c)!(b+c-2)!}{(a+d-1)!(a-c)!(d-b)!}$$
The base case of the induction proof is $F((a,b),(a,b))=1/(a+b-1)$ because there is one path of a single vertex.
The recursive equation is
$$F((a,b),(c,d))=\frac1{a+b-1}\left[F((a-1,b),(c,d))+F((a,b+1),(c,d))\right]$$
$$=\frac1{a+b-1}\left[\frac{(a+d-b-c-1)!(b+c-2)!}{(a+d-2)!(a-c-1)!(d-b)!}+\frac{(a+d-b-c-1)!(b+c-1)!}{(a+d-1)!(a-c)!(d-b-1)!}\right]\\
=\frac1{a+b-1}\frac{(a+d-b-c-1)!(b+c-2)!}{(a+d-1)!(a-c)!(d-b)!}\cdot\\
\left[(a+d-1)(a-c)+(b+c-1)(d-b)\right]$$
The final factor equals $(a+b-1)(a-b-c+d)$, so the final answer is $F((a,b),(c,d))$ given above, and we only assumed $F((a,b),(c,d))$ was correct for values with a lower value of $a-b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1591274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
}
|
How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5$$
where at least $2$ of the variables are $\geq 2$.
I found the total number of candidate solutions using the $\binom{B+N-1}{B} = 21$
However, only $9$ of them have two variables $\geq 2$.
\begin{align*}
2+0+3& =5\\
2+1+2& =5\\
3+0+2& =5\\
1+2+2& =5\\
3+2+0& =5\\
0+2+3& =5\\
0+3+2& =5\\
2+3+0& =5\\
2+2+1& =5
\end{align*}
I feel there is a connection to the Associated Stirling numbers of the second kind. But I can't place it :(
EDIT:
Here is my code for enumerating them all to count the number of ways of select B elements from a set of N (uniformly with replacement), such that you have at least C copies of K elements - also shows the output for this question I'm asking here as it's the core piece. Obviously can't be run for very large values of the parameters - that's why I'm here :) Code is here
Here is another example for B = 6, N = 3, C = 2 and K = 2 there are 16 solutions:
\begin{align*}
0+2+4& = 6\\
0+3+3& = 6\\
0+4+2& = 6\\
1+2+3& = 6\\
1+3+2& = 6\\
2+0+4& = 6\\
2+1+3& = 6\\
2+2+2& = 6\\
2+3+1& = 6\\
2+4+0& = 6\\
3+0+3& = 6\\
3+1+2& = 6\\
3+2+1& = 6\\
3+3+0& = 6\\
4+0+2& = 6\\
4+2+0& = 6\\
\end{align*}
There are a number of different and correct solutions below. I don't know which to accept.
|
A tailor-made approach by analytic combinatorics. The coefficient of $x^B$ in
$$ \frac{1}{(1-x)^N} = \left(1+x+x^2+x^3+\ldots\right)^N $$
obviously counts the number of ways of representing $B$ as a sum of $N$ non-negative integers. By stars and bars, or by the (negative) binomial theorem, such number is $\binom{N+B-1}{N-1}$. We may use an extra variable to mark the terms with exponent $\geq C$, and consider:
$$ g(z,x) = \left(1+x+x^2+\ldots+x^{C-1}+ z x^{C}+z x^{C+1}+ z x^{C+2}+\ldots\right)^N $$
that is:
$$ g(z,x) = \left(\frac{1-x^C}{1-x}+z\cdot\frac{x^C}{1-x}\right)^N = \frac{(1+(z-1) x^C)^N}{(1-x)^N}.$$
Now the coefficient of $x^B$ in $g(z,x)$ is a polynomial in the $z$ variable, $h_B(z)$, and we are interested in summing the monomials of $h_B(z)$ whose degree is $\geq K$. That sum evaluated at $z=1$ gives the answer to our problem. However, I suspect there is no nice closed formula for summarizing the process, since even the computation of $h_B(z)$ involves a convolution. Are you fine with an integral representation of the answer? That is not hard to achieve through Cauchy's integral formula.
If $B=5,N=3, C=2$ and $K=2$, we have $h_B(z)=12z+\color{red}{9}z^2$, hence the answer is $\color{red}{9}$ as you checked. The general form of the answer is given by:
$$\begin{eqnarray*} [x^B]\sum_{D\geq K}[z^D]\frac{\left((1-x^C)+z x^C\right)^N}{(1-x)^N}&=&[x^B]\frac{1}{(1-x)^N}\sum_{D\geq K}\binom{N}{D}x^{CD}(1-x^C)^{N-D}\\&=&\sum_{D\geq K}\binom{N}{D}[x^{B-CD}]\frac{(1-x^C)^{N-D}}{(1-x)^N}\\&=&\color{red}{\sum_{D\geq K}\binom{N}{D}\sum_{h=0}^{N-D}\binom{N-D}{h}(-1)^h\binom{N+B-CD-Ch-1}{N-1}}, \end{eqnarray*}$$
rather ugly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1591845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
}
|
Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1
\\
= &16k+4k+4 +1+4r+2+1
\\
=
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
|
Hint:
$$
5(5^n+2\cdot 3^{n-1}+1)=5^{n+1}+2\cdot 3^n+1+4(3^{n-1}+1)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1592108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
}
|
Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0$
This is not the starting inequality.
Is there something wrong in this method?
|
HINT:
Take $$f(x)=\ln x $$ and apply MVT in [$a,a+1$].
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1592906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
$\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$ I've got stuck at this problem:
Let $a$, $b$, $c$ be real numbers. Prove that
$$\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$$
Firstly, I've thought this :
$$a^2 - 2ab + b^2 \geq 0$$
$$4a^2 + ab + b^2 \geq 3(ab + a^2)$$
$$\frac{a(a+b)}{4a^2 + ab + b^2} \leq \frac{1}{3}$$
Similarly we obtain that
$$\frac{b(b+c)}{4b^2+bc+c^2} \leq \frac{1}{3}$$
And
$$\frac{c(c+a)}{4c^2+ca+a^2} \leq \frac{1}{3}$$
Summing all, we obtain the inequality.
Is this way correct?
(I have doubts about my solution because this problem was found in a math magazine which usually has difficult problems(at least for me)). Or is there a another way?
Thanks!
|
The inequality is equivalent to:
$$
\sum_{cyc}\frac{3a(a+b)}{4a^2+ab+b^2}≤3\iff \\
0≤\sum_{cyc}1-\frac{3a(a+b)}{4a^2+ab+b^2}=\sum_{cyc}\frac{a^2-2ab+b^2}{4a^2+ab+b^2}=\sum_{cyc}\frac{(a-b)^2}{(a-b)^2+3a(a+b)}\iff\\
0≤\sum_{cyc}\frac{(a-b)^2}{\left(\frac{7}{4}a-\frac{1}{4}b\right)^2+\frac{15}{16}(a+b)^2}
$$
Which is obvious.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1592972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Approximation of $\frac{x}{\sqrt{x^2+R^2}}$ How do you prove this statement?
If $x\gg R$ then
$$\frac{x}{\sqrt{x^2+R^2}}\cong 1-\frac{1}{2}\left(\frac{R}{x}\right)^2$$
I have no ideas even how to start.
|
Basically, this can be seen by a Taylor series expansion around $0$. Note that $x\gg R$ "means" $\frac{R}{x} \approx 0$: so you would need to make this quantity appear, first.
$$\begin{align}
\frac{x}{\sqrt{x^2+R^2}} &= \frac{x}{x\sqrt{1+\left(\frac{R}{x}\right)^2}}
= \frac{x}{x}\left(1-\frac{1}{2}\left(\frac{R}{x}\right)^2 + o\left(\left(\frac{R}{x}\right)^2\right)\right) \\
&= 1-\frac{1}{2}\left(\frac{R}{x}\right)^2 + o\left(\frac{R^2}{x^2}\right)
\end{align}$$
using the fact that $(1+u)^\alpha = 1+\alpha u + o(u)$ when $u\to0$ (for any fixed $\alpha \in\mathbb{R}$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1593752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$,
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
|
HINT:
the second integral $\int\frac{1}{2x^2+x+3}dx = \int\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{23}{8}}dx$
You should be able to use some substitutions to turn it into $C\int\frac{1}{u^2+1}du$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
}
|
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
|
HINT:
$$x^3+35-x^3+3x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3})=35+3\cdot30 $$
$$(x+\sqrt[3]{35-x^3})^3=5^3$$
Assuming $x$ to be real, $$x+\sqrt[3]{35-x^3}=5\iff\sqrt[3]{35-x^3}=5-x$$
Take cube in both sides
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1594109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement.
|
We have:
$$ \frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca} ≥ ab + bc + ca $$
Also note that
$$ \frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca} ≥ \frac{(a^2+b^2+c^2)^2}{ab+bc+ca}$$
(Because of the Cauchy Schwarz lemma or the Titu's Lemma)
So if we can prove that
$$\frac{(a^2+b^2+c^2)^2}{ab+bc+ca} ≥ ab + bc + ca$$
Then our work is done; so
$$ (a^2+b^2+c^2)^2 ≥ (ab+bc+ca)^2$$
$$ ⇒ a^2+b^2+c^2 ≥ ab+bc+ca \ \square$$
This can simply be proven using the rearrangement inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
}
|
how to solve the equation $\tan^2(2x)+2\tan(2x) \cdot \tan(3x)-1=0$ How to solve the equation $$\tan^2(2x)+2\tan(2x) \cdot \tan(3x)-1=0$$
Can anyone give some hints in this question ?
|
Let $t$ = tan $x$. Using trig, we can see that tan $2x = \frac{2t}{1-t^2},$ and tan $3x = t\frac{3-t^2}{1-3t^2}$.
Then the equation becomes:
$$\frac{4t^2}{(1-t^2)^2} + 2\frac{2t^2(3-t^2)}{(1-t^2)(1-3t^2)} - 1 = 0$$
Multiply by the GCD, which is $(1-t^2)^2(1-3t^2)$ to get
$$(4t^2)(1-3t^2) + 4t^2(3-t^2)(1-t^2) - (1-t^2)^2(1-3t^2) = 0$$
Expand, and solve for $t$, and then take the arctangent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1594713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Prove that $a\sqrt{a^2+bc}+b\sqrt{b^2+ac}+c\sqrt{c^2+ab}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}$
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a\sqrt{a^2+bc}+b\sqrt{b^2+ac}+c\sqrt{c^2+ab}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}.$$
I have a proof, but my proof is very ugly:
it's enough to prove a polynomial inequality of degree $15$.
I am looking for an easy proof or maybe a long, but a smooth proof.
|
$\sum\limits_{cyc}a\sqrt{a^2+bc}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\Leftrightarrow$
$\Leftrightarrow\sum\limits_{cyc}\left(a^4+a^2bc+2ab\sqrt{(a^2+bc)(b^2+ac)}\right)\geq\sum\limits_{cyc}(2a^3b+2a^3c+2a^2bc)\Leftrightarrow$
$\sum\limits_{cyc}(a^4-a^3b-a^3c+a^2bc)\geq\sum\limits_{cyc}\left(a^3b+a^3c+2a^2bc-2ab\sqrt{(a^2+bc)(b^2+ac)}\right)\Leftrightarrow$
$\Leftrightarrow\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+b-c)^2\geq\sum\limits_{cyc}ab\left(a^2+bc+b^2+ac-2\sqrt{(a^2+bc)(b^2+ac)}\right)\Leftrightarrow$
$\Leftrightarrow\sum\limits_{cyc}(a-b)^2(a+b-c)^2\geq2\sum\limits_{cyc}ab\left(\sqrt{a^2+bc}-\sqrt{b^2+ac}\right)^2\Leftrightarrow$
$\Leftrightarrow\sum\limits_{cyc}(a-b)^2(a+b-c)^2\left(1-\frac{2ab}{\left(\sqrt{a^2+bc}+\sqrt{b^2+ac}\right)^2}\right)\geq0$, which is obvious.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1595807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 1,
"answer_id": 0
}
|
Which integers can be represented as the most pair of difference of two squares? Let $f(x)$ be the number of $a,b,x\in \mathbb N$ where $a^2-b^2=x$.
For example, 1971 is not only $986^2-985^2$ but also $50^2-23^2$, $114^2-105^2$, $330^2-327^2$. So, $f(1971)=4$. Is there some sort of limit to $f(x)/x$ or $f(x)/ln(x)$ or similar? For which kind of $x$?
|
We're looking for the size of the set $S$ of ordered pairs $(a, b)$ with $a, b \geq 0$ and $a^2 - b^2 = (a + b)(a - b) = x$ for some fixed $x > 0$. (Removing the 'ordered' bit is trivial, since any such pair has $a > b$. Also, we can dispense with the case $x = a^2$ separately if we want to include the case $b = 0$.) Let $S'$ denote the set of ordered pairs $(n, m)$ with $n \geq m > 0, nm = x,$ and $n\equiv m\pmod{2}$. Then the map $f:S \to S'$ given by $f(a, b) = (a + b, a - b)$ is a bijection, with inverse $f^{-1}(n, m) = \frac{1}{2}(n + m, n - m)$.
Suppose for simplicity that $x$ is not a perfect square. Now note that
\begin{align*}
\#S' = \frac{1}{2}\#\left\{n > 0:\, n|x\text{ and } n \equiv x/n\!\!\!\!\pmod{2}\right\}
\end{align*}
If $x$ is odd, then the equivalence above holds for any $n$, and $\#S = \#S' = \frac{1}{2} d(n)$, where $d$ denotes the number of divisors of $n$. If instead $x = 2^p x'$ with $p > 0$ and $x'$ odd, then the equivalence holds iff $n = 2^r n'$ with $n' | x'$ odd and $r\not = 0, p$. Thus
\begin{align*}
\#S = \#S' = \begin{cases}
0 & \text{ if $p = 1$;} \\
\frac{1}{2}(p - 1) d(x') & \text{ if $p \not = 1$} \\
\end{cases}
\end{align*}
Since $d$ is multiplicative, we have $d(2^p x') = d(2^p) d(x') = (p + 1)d(x')$. Thus we can combine the previous results to give
\begin{align*}
\#S &= \frac{1}{2}\left(\frac{p-1}{p+1}\right) d(x)
\end{align*}
where $2^p$ is the highest power of $2$ dividing $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1596024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
solution of nested radical $\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$ This question is from my friend. he think that there is trigonometry involved to this equation.
$\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$
that is from $\ \ \ x = \sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt...} } }} } }$
I try to compare between this nested radical and Ramanujan's nested radicals form ,but solution is for all positive terms.
Thank you for all comments , thank you so much
|
I notice that $ x^3-x^2-9 x+1=0 $ has three roots:
$ x_1=1-4 \cos \left(\frac{\pi }{7}\right)=-2.60388... $
$ x_2=1+4 \cos \left(\frac{2 \pi }{7}\right)=3.49396... $
$ x_3=1-4 \cos \left(\frac{3 \pi }{7}\right)=0.109916... $
Therefore, the above result $ \sqrt{7+2 \sqrt{7-2 \sqrt{7-2 \left(1+4 \cos \left(\frac{2 \pi }{7}\right)\right)}}}=1+4 \cos \left(\frac{2 \pi }{7}\right) $ can be extended to:
$ \sqrt{7-2 \sqrt{7-2 \sqrt{7+2 \left(-1+4 \cos \left(\frac{\pi }{7}\right)\right)}}}=-1+4 \cos \left(\frac{\pi }{7}\right) = -x_1 $
$ \sqrt{7+2 \sqrt{7-2 \sqrt{7-2 \left(1+4 \cos \left(\frac{2 \pi }{7}\right)\right)}}}=1+4 \cos \left(\frac{2 \pi }{7}\right) =x_2 $
$ \sqrt{7-2 \sqrt{7+2 \sqrt{7-2 \left(1-4 \cos \left(\frac{3 \pi }{7}\right)\right)}}}=1-4 \cos \left(\frac{3 \pi }{7}\right) =x_3 $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1596824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.
If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
|
I will use a method similar to yours. Letting $m,n$ respectively equal $2k+1$ and $2p+1$ and subsequently expanding the expression, we get $2k^2p^2+ 2k^2 p+ \frac{k^2}{2} + 2kp^2 +2kp + \frac{k}{2} +\frac{p^2}{2} + \frac{p}{2}$. Of course, $k$ and $p$ are both integers, so the only terms we have to worry about are the fractions.
Note $\frac{p}{2} + \frac{p^2}{2} = \frac{p(p+1)}{2}$ which is clearly integer since one of $p$ and $p+1$ must be even. Hence $\frac{p}{2} + \frac{p^2}{2}$ is integer; with similar reasoning, $\frac{k}{2} + \frac{k^2}{2}$ is integer.
This implies the expression is integer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1597394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 7
}
|
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
|
I think the simplest way is by expanding
to get terms with degrees below 4 we have to choose which combination of terms of original cos(x) series must be chosen and sum up the posibble combination
here is the answer:
$$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
$$=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)\times\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)\times...\times\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)$$
so there is this combinations:
1-$$choosing \; all \; 1's \; from \; 6 \; parentheses \;that\;can \;be \;done \;in \;one \;uniqe \;way\; so \; term \;1 \;is \;produced :\;1$$
2-$$choosing \; 1's \; from \; 5 \; parentheses \; and- \frac{1}{2}x^2 \; or \; \frac{1}{24}x^4 from \;other \;parentheses \;which \;can \;be \;done \;in\; 6 \;defferent \;ways\; so \; terms \; - \frac{6}{2}x^2 and \frac{6}{24}x^4\;are \;produced :\;- 3\times x^2 + \frac{1}{4}x^4$$
3-$$choosing \; 1's \; from \; 4 \; parentheses \; and- \frac{1}{2}x^2 \; from\; 4\;other \;parentheses \;which \;can \;be \;done \;in\; \binom{6}{2} \;defferent \;ways\; so \; term \; \frac{15}{4}x^4 is \;produced :\;+ \frac{15}{4}x^4$$
4-there is no other term possible with degree below 6. summing up above terms results in the truncated series as below:
$$1 + 3\times x^2 + 4\times x^4 + O(x^6)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
}
|
How can you prove the inequality $2x^3 (x^3 + 8y^3) + 2y^3 (y^3 + 8z^3) + 2z^3 (z^3 + 8x^3) ≥ 9x^4 (y^2 + z^2) + 9y^4 (z^2 + x^2) + 9z^4 (x^2 + y^2)$ I changed the RHS as $\displaystyle \sum_{cyc} 9x^4(S-x^2) $ for $S = x^2 + y^2 + z^2$
Then I thought I could apply Jensen's inequality for $f(x) = 9x^4(S-x^2) = -9x^6 + 9Sx^4$ in the RHS as follows (i think $f(x)$ is concave):
$f(\frac{x +y +z}{3}) \leq \frac{f(x) + f(y) + f(z)}{3}$
$\Rightarrow 3(-9(\frac{x +y +z}{3})^6+9S(\frac{x +y +z}{3})^4) \leq RHS$
$\Rightarrow -\frac{(x +y +z)^6}{27}+S(\frac{(x +y +z)^4}{3}) \leq RHS$
So I think what it remains to be proved is that
$LHS \geq -\frac{(x +y +z)^6}{27}+S(\frac{(x +y +z)^4}{3})$
But I have no clue how to go on :P
|
In fact, we have $$ LHS-RHS=\sum\limits_{cyc}{\left(x-y\right)^4\left(x^2+4xy+y^2\right)} \ge 0 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1598570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Difficult Integral I have a problem with solving this integral:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx$$
I tried to use substitution but I got stuck. Can anyone help me?
|
The easiest way is using undetermined coefficients:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$
You must differentiate and you have:
$$\frac{2x^2+3x+1}{\sqrt{x^2+1}}=A\sqrt{x^2+1}+(Ax+B)\frac{2x}{2\sqrt{x^2+1}}+\frac{\lambda}{\sqrt{x^2+1}}$$
So you have an equation:
$$2x^2+3x+1=A(x^2+1)+(Ax+B)x+\lambda$$
$$2x^2+3x+1=Ax^2+A+Ax^2+Bx+\lambda$$
from which you can find out an simultaneous equation:
\begin{cases}
2=2A\\
3=B\\
1=A+\lambda
\end{cases}
After solving it you have:
\begin{cases}
A=1\\B=3\\\lambda=0
\end{cases}
To remind, your integral grom the beginning is:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$
So
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (x+3)\sqrt{x^2+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1599403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
}
|
Walk me through some basic modular arithmetic? The problem is as follows:
Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?
The solution is as follows:
First we factor $abc + ab + a$ as $a(bc + b + 1)$, so in order for the number to be divisible by 3, either $a$ is divisible by $3$, or $bc + b + 1$ is divisible by $3$.
We see that $a$ is divisible by $3$ with probability $\frac{1}{3}$. We only need to calculate the probability that $bc + b + 1$ is divisible by $3$.
We need $bc + b + 1 \equiv 0\pmod 3$ or $b(c + 1) \equiv 2\pmod 3$. (I get it up to this point because of my lacking modular arithmetic knowledge). Using some modular arithmetic, $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$. The both cases happen with probability $\frac{1}{3} * \frac{1}{3} = \frac{1}{9}$ so the total probability is $\frac{2}{9}$.
Then the answer is $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$
Is there a rule relating that two things being multiplied (b and (c+1)), if you mod them and add the result, would equal the other side? As in, I don't know why $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$. Trying to find a pattern (2+0 = 2 and 1+1 = 2) lead me to that question. Also, where did the 2/3 come from?
I would appreciate someone walking me through the solution as I am not experienced with mod (and my brain is having a hard time functioning without food).
|
You can find the solutions just by plugging the all the options ($b = 0, 1, 2$, $c = 0, 1, 2$) and seeing which satisfy $b(c+1) \equiv 2 \pmod{3}$.
It often helps me, when working with mod, to forget the definition in terms of divisibility entirely and just imagine I am working with objects called $0, 1, 2$ such that $2 + 1 = 0$, $2 + 2 = 1$, etc.
(The $\frac{2}{3}$ comes of splitting the original situation into three cases. $\frac{1}{3}$ of the time, $a$ is divisible by $3$. The other $\frac{2}{3}$ of the time, it is not. It took me a while of staring at the solution to understand that as well! It would have helped if they noted it specifically.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1599875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solve $\sin 2x=-\cos x$ I'm working on solving the problem $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\pi }{2}-x +2\pi k \Leftrightarrow 3x=\frac{\pi }{2} +2\pi k\Leftrightarrow x_1=\frac{\pi }{6}+\frac{2\pi}{3} k\ $
$2x =\pi -(-\frac{\pi }{2}-x) +2\pi k \Leftrightarrow x_2=\frac{3\pi}{2}+2\pi k$
None of these answers are correct.
The correct answer is $x=\frac{\pi}{2}+k\pi$ and $x=-\frac{\pi}{2}\pm\frac{\pi}{3}+2 k\pi$
I don't know what I'm doing wrong in my attempt.
|
Begin by using the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$ to make the equation $2\sin(x)\cos(x) = -\cos(x)$, or $2\sin(x)\cos(x) + \cos(x) = 0$. Now we can factor to obtain $\cos(x)(2\sin(x) + 1) = 0$. This means that either $\cos(x) = 0$ or $\sin(x) = -\frac{1}{2}$. Thus, the solutions are $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6},$ and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1603427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Interval for area bounded by $r = 1 + 3 \sin \theta$ I'm trying to calculate the area of the region bounded by one loop of the graph for the equation
$$
r = 1 + 3 \sin \theta
$$
I first plot the graph as a limaçon with a maximum outer loop at $(4, \frac{\pi}{2})$ and a minimum inner loop at $(-2, -\frac{3 \pi}{2})$. I then note the graph is symmetric with respect to the $\frac{\pi}{2}$ axis and the zero for the right half is at $\theta = \arcsin(-\frac{1}{3})$.
So, I chose the interval $[\arcsin(-\frac{1}{3}),\frac{\pi}{2}]$ to calculate the area which can then be multiplied by $2$ for the other half. The problem is that the answer in the book seems to use $\arcsin(\frac{1}{3})$ instead, note the change of sign.
Just to make sure I'm not misunderstanding where I went wrong, I get the answer
$$
\frac{11 \pi}{4} - \frac{11}{2} \arcsin(-\frac{1}{3}) + 3 \sqrt 2
$$
Whereas the book gets
$$
\frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2
$$
It's a subtle change of sign but I'd really like to understand where I went wrong.
|
Notice how $\arcsin(-\frac{1}{3}) = - \arcsin(\frac{1}{3})$, so your answer now looks like
$$
\frac{11 \pi}{4} + \frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2 \\
$$
That means your area is greater than the answer in your book by:
$$
2 \left(\frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2\right)
$$
This might indicate you are calculating the area of the outer loop whereas your book is calculating the inner loop. If you choose the interval $[\frac{3 \pi}{2}, 2 \pi - \arcsin(\frac{1}{3})]$ to calculate the half as you did before, you get:
$$
\begin{eqnarray}
A &=& 2 \times \frac{1}{2} \int_{\frac{3 \pi}{2}}^{2 \pi - \arcsin \frac{1}{3}} (1 + 3 \sin \theta)^2 \, \textrm{d}\theta \\
&=& \left[\frac{11 \theta}{2} - 6 \cos \theta - \frac{9 \sin(2 \theta)}{4} \right]_{\frac{3 \pi}{2}}^{2 \pi - \arcsin \frac{1}{3}} \\
&=& \frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2 \\
\end{eqnarray}
$$
This seems to agree with the answer in your book.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1604676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Where is the mistake in my solution? Trigonometry proof
To prove: $$1 + 2 \sin 70^\circ = \frac{1}{2\sin 20^\circ}$$
My attempt:
$$\begin{align}
1 + 2 \sin 70^\circ &= 1 + \frac{\sin 140^\circ}{\cos 70^\circ} \\[6pt]
&= 1 + \frac{\sin 40^\circ}{\sin 20^\circ} \\[6pt]
&= \frac{\sin 20^\circ + \sin 40^\circ}{\sin 20^\circ} \\[6pt]
&= \frac{2\sin 30^\circ \cos 10^\circ}{\sin 20^\circ} \\[6pt]
&= \frac{\cos 10^\circ}{2\sin 10^\circ \cos 10^\circ} \\[6pt]
&= \frac{1}{2\sin 10^\circ} \\
\end{align}$$
Can anyone explain where my mistake is?
(original solution image)
|
As noted in the comments:
*
*The original identity is wrong (LHS $\approx 2.88$, RHS $\approx 1.46$).
*Your derivation of $1 + 2 \sin 70^\circ = \dfrac{1}{2\sin 10^\circ}$is correct.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1604939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$?
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$ ?
MY ATTEMPT:
$A\cdot A=A$
$B\cdot B=B$
$(A+I)\cdot (A+I)=A+I$ or, $A\cdot A+A\cdot I+I\cdot A+I\cdot I=A+I$ which implies $A\cdot I+I\cdot A=0$ (using above equations)
$(A+B)\cdot (A+B)=A+B$ or,$A\cdot A+A\cdot B+B\cdot A+B\cdot B=A+B$ which implies $A\cdot B+B\cdot A=0$
What's next?
|
Well, $AI=IA=A$, so your equation $AI+IA=0$ actually says $2A=0$. Over a field of characteristic $\neq 2$, this implies $A=0$, so $AB=BA=0$ trivially. Over a field of characteristic $2$, your equation $AB+BA=0$ gives $AB=-BA$, and $-BA=BA$ since the characteristic is $2$.
(Incidentally, if $A$, $B$, and $A+B$ are all idempotent this already implies that $AB=BA$, and that $AB=BA=0$ if the characteristic is not $2$. As you derived, you get $AB+BA=0$ and hence $AB=-BA$. But then $$AB=(A^2)B=A(AB)=-A(BA)=-(AB)A=(BA)A=B(A^2)=BA.$$ If the characteristic is not $2$, $AB=BA=-BA$ implies $AB=BA=0$.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1605229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Double Integral With a Change of Variables I need to calculate:
$$
\iint _D \frac{2y^2+x^2}{xy}~\mathrm dx~\mathrm dy
$$
over the set $D$ which is:
$$
y\leq x^2 \leq 2y , \quad 1\leq x^2 +y^2 \leq 2 , \quad x\geq 0
$$
can someone help me understand what possible change of variables can I do here?
Thanks a lot in advance .
|
Use this change of variables
$$\begin{align}
u&=x^2+y^2\\
v&=\frac{x^2}{y}
\end{align}$$
So we may compute the Jacobian first
$$\frac{\partial(u,v)}{\partial(x,y)}=
\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix}
=
\begin{vmatrix}
2x & 2y \\
2\frac{x}{y} & -\frac{x^2}{y^2}
\end{vmatrix}
=-2\frac{x^3}{y^2}-4xy = -2(\frac{x^3+2xy^3}{y^2})=-2\frac{x}{y^2}(x^2+2y^2)
$$
We also know that
$$\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{\partial(x,y)}}=\frac{y^2}{-2x(x^2+2y^2)}$$
Consequently, your integral becomes
$$\begin{align}
I&=\int_{1}^{2}\int_{1}^{2}\frac{x^2+2y^2}{xy} \cdot \frac{y^2}{-2x(x^2+2y^2)}dudv \\
&=-\frac{1}{2}\int_{1}^{2}\int_{1}^{2} \frac{y}{x^2} dudv \\
&=-\frac{1}{2}\int_{1}^{2}\int_{1}^{2} \frac{1}{v} dudv \\
&=-\frac{1}{2}\left(\int_{1}^{2}\frac{1}{v}dv\right)\left(\int_{1}^{2}du\right) \\
&=-\frac{1}{2}(\ln2)(1) \\
&=\boxed{-\ln\sqrt{2}}
\end{align}$$
The domain of integration is showed in the following figure.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1605718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integer solution of $ y-x = \sqrt{y+x} $ I have a problem with how to find integer solution of $\ \ \ y-x = \sqrt{y+x} \ \ \ which \ \ \ \ y>x$
$ y^2 -2xy + x^2 -y -x = 0 $
$ y^2 - (2x+1)y + (x^2 - x) = 0 $
$ y = \frac{(2x+1) \pm \sqrt{8x+1}}{2} $ and then ? I cannot solve this anymore.
because If it had a solution , I will bring it to apply for $\ \ (x^b \ mod \ \ y )= x $
It's so cool. Thank you so much for every comments.
|
Set $a=\sqrt{y+x}$. Then $a^2=y+x$ and $a=y-x$ by the condition. Since $y>x$ in fact $a$ is a positive integer. Adding and subtracting, respectively, we have $2y=a^2+a$ and $2x=a^2-a$. Whether $a$ is even or odd, $a^2+a$ and $a^2-a$ are even, so we get integer $x,y$. Hence the solution set is given by the following (where $a$ is any positive integer):
$$(x,y)=\left(\frac{a^2-a}{2},\frac{a^2+a}{2}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1607146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Cantor set as an intersection I've seen that the Cantor set $C$ can be expressed as a countable intersection of $C_n$'s where $C_n=C_{n-1}-\bigcup_{k=0}^{3^{n-1}-1}(\frac{3k+1}{3^n},\frac{3k+2}{3^n})$.
Taking $C_1,C_2$ into consideration, this is clear, however, when trying to come up with this specific equation on my own, I cannot. I mean, how can we be sure that inductively, this formulation gives the correct $C_n$'s as we know.
|
What might help is to see that the formula in the question is a piece of the inductive proof of a closed form description of $C_n$, expressed using trinary expansions of integers.
For each $n$, each integer $k \in \{0,...,3^n-1\}$ can be written as a trinary number $a_{n-1}a_{n-2}...a_0$ with entries in the set of trigits $\{0,1,2\}$, such that
$$k = a_{n-1}3^{n-1} + ... + a_1 3^1 + a_0 3^0
$$
Let $T_n$ be the set of such integers whose trinary expansion contains no $1$'s. For example, $T_1 = \{0,2\}$. Also, $T_2 = \{0,2,7,9\}$ which in trinary is $T_2=\{00,02,20,22\}$.
Let me sketch the proof that
$$(*) \qquad\qquad C_n = \bigcup_{k\in T_n} \biggl[ \frac{3k}{3^{n+1}}, \frac{3k+3}{3^{n+1}} \biggr]
$$
You can check easily enough that $C_0=[0,1]$, and $C_1=[0,\frac{1}{3}] \cup [\frac{2}{3},1]$, and
$$C_2=\biggl[0,\frac{1}{9}\biggr] \cup \biggl[\frac{2}{9},\frac{1}{3}\biggr] \cup \biggl[\frac{2}{3},\frac{7}{9}\biggr] \cup \biggl[\frac{8}{9},1\biggr]
$$
and that this agrees with $(*)$.
So, let's assume as an induction hypothesis that
$$C_{n-1} = \bigcup_{k=T_{n-1}} \biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]
$$
Next, we use that $C_n$ equals what you get by removing the open middle third of each of the component intervals of $C_{n-1}$. Thus, for each $k \in T_{n-1}$, we want to remove the open middle third of the interval
$$\biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]
$$
which is open interval
$$\biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
leaving
$$\biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] - \biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
and taking their union gives us $C_n$ and the formula
$$C_n = \bigcup_{k \in T_{n-1}} \biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] - \biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
Collecting all the first terms of each summand we get the formula in your question:
$$C_n = C_{n-1} - \bigcup_{k=1}^{3^{n-1}-1}\biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
Thus, your formula is the exact expression of the statement that $C_n$ is equal to $C_{n-1}$ with the open middle thirds removed from each component interval of $C_{n-1}$.
To complete the induction, we have to do a little bit of rewriting, expressing
$$\biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] - \biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr) = \biggl[ \frac{3k}{3^{n}}, \frac{3k+1}{3^{n}} \biggr] \bigcup \biggl[\frac{3k+2}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]
$$
and taking the union we get
$$C_n = \bigcup_{k\in T_{n-1}} \biggl[ \frac{3k}{3^{n}}, \frac{3k+1}{3^{n}} \biggr] \bigcup \biggl[\frac{3k+2}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]
$$
Now with a bit more rewriting of the union, you can finish the induction and get the formula for $C_n$ at the beginning of this answer, using the trinary arithmetic fact that
$$T_n = \bigcup_{k \in T_{n-1}} \{3k,3k+2\}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1608473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Using Cauchy's integral formula to evaluate $\int_0^{2\pi} \frac{1}{a^2 \cdot {\cos}^2(t) + b^2 \cdot {\sin}^2(t)} dt$ I have to solve this integral using Cauchy's integral formula.
I tried to substitute it with several different attempts but without a solution. Can anyone help?
$$\int_0^{2\pi} \frac{1}{a^2 \cdot {\cos}^2(t) + b^2 \cdot {\sin}^2(t)} dt$$
|
First, note that $\cos(t)=\frac{e^{it}+e^{-it}}{2}$ and $\sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Substituting this into the integral yields: $$\int_o^{2\pi}\frac{1}{a^2\cdot(\frac{e^{it}+e^{-it}}{2})^2+b^2\cdot(\frac{e^{it}-e^{-it}}{2i})^2}dt$$
Simplifying yields: $$=\int_0^{2\pi}\frac{4e^{2it}}{(a^2-b^2)e^{4it}+2(a^2+b^2)e^{2it}+(a^2-b^2)}dt$$
Note that this is the same as integrating over the unit circle twice, i.e. we have: $$=\int_{\partial D(0,1)}\frac{2}{i}\cdot\frac{1}{(a^2-b^2)z^2+2(a^2+b^2)z+(a^2-b^2)}dz$$
The poles of the integrand occur at: $\frac{a^2+b^2\pm\sqrt{(a^2+b^2)^2-(a^2-b^2)^2}}{a^2-b^2}=\frac{(a\pm b)^2}{a^2-b^2}$. Here you should apply any assumptions you have on $a$ and $b$ to determine whether either pole occurs in the unit disc, and neither pole occurs on its boundary. For the sake of example, suppose only the pole at $\frac{(a- b)^2}{a^2-b^2}$ is inside of the unit disc. Then using the calculus of residues, we have: $$=2\pi i\cdot\frac{2}{i}\cdot\frac{-(a^2-b^2)}{4ab}\cdot 2=\frac{2\pi(b^2-a^2)}{ab}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1610996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find all natural numbers $x,y$ such that $3^x=2y^2+1$. Find all natural numbers $x,y$ such that
$$3^x=2y^2+1$$
solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$.
|
There is a "standard" way to attack this using Thue's theorem, by writing the left hand side as $z^3$, $3 z^3$, or $9 z^3$, depending on the assumed residue $x \pmod 3$. I can't imagine doing a search up to any of the effective versions of Thue's bound on the size of the solutions during an Olympiad.
Some of the comments are close to suggesting that we attack this as a Pell equation, so I have started that below. I don't know how to finish either case. Perhaps one of the commenters would be willing to edit in a completion to either case below.
If $x$ is even, $x = 2k$, this is the Pell equation $(3^k)^2 - 2 y^2 = 1$. The usual technique for finding the minimal solution, constructing convergents of the continued fraction for $\sqrt{2}$, gives the integer solutions $$\begin{align}
3^k &= \frac{1}{2} \left( (3-2\sqrt{2})^n + (3+2\sqrt{2})^n \right) \\
y &= \frac{-1}{2\sqrt{2}} \left( (3-2\sqrt{2})^n - (3+2\sqrt{2})^n \right),
\end{align}$$ for any integer $n$ (and not yet applying the restriction that $k$ be a positive integer). Note that $y>0$ requires $n \geq 0$. In fact, $n=0$ gives $3^k = 1$, so $k$ and $x$ are both zero (and $x = 0$ is not allowed). Hence, $n > 0$.
If $x$ is odd, $x = 2k+1$, we multiply through by $3$ to get $3^2 (3^k)^2 - 6 y^2 = 3$, another Pell equation, whose integer solutions are $$\begin{align}
3 \cdot 3^k &= \frac{1}{2}\left( (3+\sqrt{6}) (5-2\sqrt{6})^n + (3 - \sqrt{6})(5+2\sqrt{6})^n \right) \\
y &= \frac{-1}{4} \left( (2+\sqrt{6})(5-2\sqrt{6})^n + (2 - \sqrt{6})(5+2\sqrt{6})^n\right),
\end{align}$$ for any integer $n$ (and not yet applying the restriction that $k$ be a positive integer). Note that $y>0$ requires $n>0$.
It's not clear to me that the $k \in \Bbb{N}$ restriction is straightforward to apply to either of these. But I would be happy for someone(s) to edit in a completion to either of the above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1612956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
}
|
Integrating $\int \frac{u \,du}{(a^2+u^2)^{3/2}}$ How does one integrate $$\int \frac{u \,du}{(a^2+u^2)^{3/2}} ?$$
Looking at it, the substitution rule seems like method of choice. What is the strategy here for choosing a substitution?
|
You can take the solution as I've presented here:
\begin{equation}
\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n} \right)\right]
\end{equation}
Here $a$ is $a^2$, $k = 1$, $m = \frac{3}{2}$, thus:
\begin{align}
\int_0^x \frac{t}{\left(t^{2} + a^{2}\right)^{\frac{3}{2}}}\:dt &= \frac{1}{2}\left(a^2\right)^{\frac{1 + 1}{2} - \frac{3}{2}} \left[B\left(\frac{3}{2} - \frac{1 + 1}{2}, \frac{1 + 1}{2}\right) - B\left(\frac{3}{2} - \frac{1 + 1}{2}, \frac{1 + 1}{2}, \frac{1}{1 + a^2x^2} \right)\right] \\
&= \frac{1}{\left|a\right|}\left[B\left(\frac{1}{2}, 1\right) - B\left(\frac{1}{2},1, \frac{1}{1 + a^2x^2} \right) \right]
\end{align}
Using the relationship between the Beta and the Gamma function we find that:
\begin{equation}
B\left(\frac{1}{2},1 \right) = \frac{\Gamma\left(\frac{1}{2} \right)\Gamma(1)}{\Gamma\left(1 + \frac{3}{2}\right)} = \frac{\Gamma\left(\frac{1}{2} \right)\Gamma(1)}{\frac{1}{2}\Gamma\left(\frac{1}{2}\right)} = 2\Gamma(1) = 2
\end{equation}
Thus,
\begin{equation}
\int_0^x \frac{t}{\left(t^{2} + a^{2}\right)^{\frac{3}{2}}}\:dt = \frac{1}{\left|a\right|}\left[2 - B\left(\frac{1}{2},1, \frac{1}{1 + a^2x^2} \right) \right]
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1613689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}{x+4} & \quad -4 < x < 0 \\
\frac{x}{x+4} & \quad x < −4
\end{array}
\right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
|
Case $\#1:$ For $x+4\ne0$
As for real $y, a>0$
$$|y|<a\iff -a<y<a$$
$$\implies\left|\dfrac x{x+4}\right|<4\iff-4<\dfrac x{x+4}<4$$
Now $-4<\dfrac x{x+4}\iff0<\dfrac x{x+4}+4=\dfrac{5x+16}{x+4}=5\cdot\dfrac{(x+4)\left(x+\dfrac{16}5\right)}{(x+4)^2}$
$\iff(x+4)\left(x+\dfrac{16}5\right)>0$
Now if $(z-a)(z-b)>0$ with $a<b$ for real $z,$ either $z<a$ or $z>b$ (Prove this)
Similarly deal with $$\dfrac x{x+4}<4$$
Case $\#2:$ Check for $x+4=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
Reduction of a quadratic form to a canonical form I'm supposed to reduce following polynomial to its canonical form. But my result differs from the one given in my book, so I'm not sure if it's correct too.
$$
q = u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0
$$
So, the characteristic quadratic polynomial is
$$
x^2 - xy -2y^2
$$
Here I'm using Lagrange's reduction method for quadratic polynomial:
\begin{align}
x^2 - xy -2y^2 &= (x^2 - xy + \frac{1}{4}y^2) - \frac{1}{4}y^2 - 2y^2\\
&= (x-\frac{1}{2}y)^2 - \frac{9}{4}y^2\\
&= \xi^2 - \frac{9}{4}\eta^2\\\\
&\xi = x-\frac{1}{2}y\\
&\eta = y
\end{align}
Now I can define the function $u$ like this:
$$
u(x,y) = u(\xi(x,y), \eta(x,y))
$$
Now I need to deduce $u_x$ and $u_y$ too:
\begin{align}
u_x &= u_\xi \cdot \xi_x + u_\eta \cdot \eta_x = u_\xi\\
u_y &= u_\xi \cdot \xi_y + u_\eta \cdot \eta_y = -\frac{1}{2}u_\xi + u_\eta
\end{align}
So, my canonical form looks like this:
\begin{align}
q &= u_{\xi\xi} - \frac{9}{4}u_{\eta\eta} + u_\xi - \frac{1}{2}u_\xi + u_\eta\\
&= \underline{\underline{u_{\xi\xi} - \frac{9}{4}u_{\eta\eta} + \frac{1}{2}u_\xi + u_\eta}}
\end{align}
Is my solution correct or is there a mistake?
|
We reduce second order PDE with constant coefficients to canonic form
$ u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0$
Quadratic form:
$Q=x^2 - xy -2y^2=\left(x-\frac{y}{2}\right)^2-\left(\frac{3y}{2}\right)^2$
We get transform
$\xi=x,\quad\eta=\frac{x}{3}+\frac{2y}{3}$
and canonical form of PDE:
$u_{\xi\xi}-u_{\eta\eta}+u_\xi+u_\eta=0$
see:
V.S.Vladimirov, A Collection of Problems on the Equations
of Mathematical Physics, Springer, 1986
In our case
$$B=\begin{pmatrix}1 & \frac{1}{3}\\
0 & \frac{2}{3}\end{pmatrix}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1614549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find the volume of the solid bounded by $(x^2+y^2+z^2)^2=x^2+y^2-z^2$ As Title.
Find the volume of the solid bounded by $(x^2+y^2+z^2)^2=x^2+y^2-z^2$
--
By Spherical Coordinates, I got $\rho^4=\rho^2(\sin^2\phi-\cos^2\phi)$
Then $\rho^2=-\cos2\phi$
How to find the integral bounded?
|
As $\rho^2>0$
It must be $-1 \leq \cos{2\phi} \leq 0$, then $\frac{\pi}{4} \leq \phi \leq \frac{3\pi}{4}$
--
$\displaystyle V=\int_{0}^{2\pi} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_{0}^{\sqrt{-\cos2\phi}} \rho^2 \sin\phi \, d\rho d\phi d\theta = \frac{-2\pi}{3} \int_{\frac{\pi}{3}}^{\frac{4\pi}{3}} (-\cos{2\phi})^{3/2} \sin{\phi} \, d\phi \\ \displaystyle = \frac{2\pi}{3} \int_{\frac{\pi}{3}}^{\frac{4\pi}{3}} (1-2\cos^2{\phi})^{3/2} \, d(\cos{\phi}) = \frac{2\pi}{3} \times \frac{3\pi}{8\sqrt{2}}=\frac{\pi^2}{4\sqrt{2}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1615254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate this integral using cylindrical coordinates Find the volume of the solid bounded above by the paraboloid of revolution
$z^{2}=x^{2}+y^{2}$
And below by the $xy$ plane, and on the sides by the cylinder $x^{2}+y^{2}=2ax$
We take $a>0$.
I'm struggling to understand what this would look like graphically, I understand how to find limits of integration for $x$ and $y$ but struggling to find them for $z$. So far i have equated the two terms, but i have got no where with that.
Thanks
|
This is what I have right now, so correct me if I'm wrong:
I think you meant $z=x^2+y^2$ if you are trying to indicate a paraboloid of revolution. The graph of $z^2=x^2+y^2$ would be two cones.
The cylinder $x^2 + y^2 = 2ax$ can be represented as follows:
$\begin{align} x^2 + y^2 &= 2ax \\
x^2 - 2ax + y^2 &= 0 \\
\left(x^2 - 2ax + a^2\right) + y^2 &= a^2 \\
\left(x - a\right)^2 + y^2 &= a^2 \end{align}$
Since integrating over this circular region would be very complicated, let's translate both objects to the left by $a$ units and move the centre of the circle to the origin. The cylinder would become $x^2 + y^2 = a^2$ and your paraboloid of revolution would be $z = \left(x + a\right)^2 + y^2$.
Now we evaluate the integral:
$\begin{align} \iint\limits_{R}{f\left(x, y\right)\,dA} &= \iint\limits_{R}{r\,f\left(r\cos\theta, r\sin\theta\right)\,dr\,d\theta} \\
&= \int^{2\pi}_{0}{\int^{a}_{0}{r^3 - 2ar^2\cos\theta + a^2r\,dr\,d\theta}} \\
\, &= \int^{2\pi}_{0}{a^{4}\left(\frac{3}{4} - \frac{2\cos{\theta}}{3}\right)d\theta} \\
&= a^{4}\int^{2\pi}_{0}{\frac{9 - 8\cos{\theta}}{12}\,d\theta} \\
&= \boxed{\boxed{\frac{3\pi a^4}{2}.}} \end{align}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1615585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(x)=f\left( \frac{x+1}{2} \right)+f\left( \frac{x-1}{2} \right)$.
But how to find all functions?
|
Let's investigate the special case that $f$ is continuous.
Let $k=f(1)$ and
$$S=\{\,t\in\Bbb R\mid f(t)=kt\,\}. $$
Trying $x=y=0$ we find $f(0)=0$. Then with $y=-x$ we find that $f(x)=-f(-x)$, i.e., $f$ is odd. Then also $S=-S$. So far we have $\{-1,0,1\}\subseteq S$.
With $y=0$, we have
$$\tag1f(x^2)=xf(x)$$
hence
$$\begin{align}f(x^2-y^2)&=(x-y)f(x)+(x-y)f(y)\\
&=\left(1-\frac yx\right)xf(x)+\left(\frac xy-1\right)yf(y)\\
&=\left(1-\frac yx\right)f(x^2)+\left(\frac xy-1\right)f(y^2).\end{align} $$
Now if $x^2,y^2\in S$ then
$$f(x^2-y^2)=\left(1-\frac yx\right)kx^2+\left(\frac xy-1\right)ky^2=kx^2-ky^2$$
and so also $x^2-y^2\in S$. More generally, if two of the numbers $x^2,y^2,x^2-y^2$ are $\in S$ then all three are $\in S$.
We conclude that for two nonnegative numbers $\in S$, also their sum and difference is $\in S$, which makes $S$ a subgroup of $\Bbb R$. Then at least $\Bbb Z\subseteq S$.
From $(1)$, we see that $S$ is closed under taking square roots of positive elements (i.e., $x^2\in S$ implies $x\in S$). As a subgroup, $S$ must then be a dense subset of $\Bbb R$.
By continuity of $f$, the set $S$ must be closed, hence all of $\Bbb R$, as was to be shown.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1617496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
To obtain the condition for vanishing of the given determinant If $a,b,c$ are distinct real numbers obtain the condition under which the following determinant vanishes.
$$\left|
\begin{array}{cc}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{array}
\right|$$
My answer: After a little calculation I was able to show that $D=0$ reduces to $abc=-1$.
Is there a simpler one line answer to this? especially since this matrix looks suspiciously similar to the Vandermonde matrix?
|
There is a pretty easy way to obtain the determinant:
First, use the linearity of the determinant to split up the matrix: $$\det (A)=
\begin{vmatrix}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{vmatrix}
=\begin{vmatrix}
a & a^2 & 1\\
b & b^2 & 1\\
c & c^2 & 1\\
\end{vmatrix}+\begin{vmatrix}
a & a^2 & a^3\\
b & b^2 & b^3\\
c & c^2 & c^3\\
\end{vmatrix}$$
Then, use $\det(A \cdot B) = \det(A) \cdot \det(B)$ to split up the right matrix; also, switch the colums in the left matrix. What you get are two identical Vandermonde matrices and a diagonal matrix:
$$\det (A)=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}+\begin{vmatrix}
a & 0 & 0\\
0 & b & 0\\
0 & 0 & c\\
\end{vmatrix}\begin{vmatrix}
1 & a & a^2\\
1 & b & b^2\\
1 & c & c^2\\
\end{vmatrix}$$
Now, you can get the determinants with the Vandermonde formula and the formula for diagonal matrices:
$$\det (A)=(b-a)(c-a)(c-b)+abc((b-a)(c-a)(c-b))$$
$$=(b-a)(c-a)(c-b)(abc+1)$$
Now you can easily see when $\det (A)$ vanishes:
$$\det (A) = 0 \iff (b = a) \lor (c = a) \lor (c = b) \lor (abc = 1) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1617875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
|
Since $2 \equiv -1 \pmod{3}$, therefore $2^{k} \equiv (-1)^k \pmod{3}$. When $k$ is odd this becomes $2^k \equiv -1 \pmod{3}$. Thus $2^k+1 \equiv 0 \pmod{3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 7
}
|
Find the value of $ \int_0^{\infty} \frac{2x^2-1}{4x^4+1}\,{dx} $
Find $ \displaystyle \int_0^{\infty} \frac{2x^2-1}{4x^4+1}\,{dx}
$
I can find the primitive, but the fact that the answer is $0$ made me suspect there might be a way of getting answer without finding the primitive. I tried letting $x = 1/y$ but it didn't get me anywhere.
|
Instead of substituting $x = \dfrac{1}{y}$, try substituting $x = \dfrac{1}{2y}$.
This gives you:
\begin{align*}I &= \displaystyle\int_{0}^{\infty}\dfrac{2x^2-1}{4x^4+1}\,dx \\ &= \displaystyle\int_{\infty}^{0}\dfrac{\tfrac{1}{2y^2}-1}{\tfrac{1}{4y^4}+1} \cdot -\dfrac{1}{2y^2}\,dy \\ &= \displaystyle\int_{0}^{\infty}\dfrac{\tfrac{1}{2y^2}-1}{\tfrac{1}{4y^4}+1} \cdot \dfrac{2y^2}{4y^4}\,dy \\ &= \displaystyle\int_{0}^{\infty}\dfrac{1-2y^2}{4y^4+1}\,dy \\ &= -\displaystyle\int_{0}^{\infty}\dfrac{2y^2-1}{4y^4+1}\,dy \\ &= -I.\end{align*}
Since $I = -I$, we have $I = 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1618930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n=2$,
$$
x^2+y^3=z^4
$$
I'm already pretty lost.
After this it gets even more complicated, has anyone encountered this problem before?
I want all the solutions, or at least infinitely many for every equation.
|
If $n+2$ is composite, say $n+2=mk$ with $m,k>1$, take any $z$, $x_m = z^k$, all other $x_j = 0$.
EDIT:
Suppose we have a solution $(a_1, \ldots, a_{n+1})$ in positive integers with
$a_1^{d_1} + \ldots + a_n^{d_n} = a_{n+1}^{d_{n+1}}$, and positive integer $k$ so that $d_i+k$ are pairwise coprime. By the Chinese Remainder Theorem there are positive integers $m_1, \ldots, m_{n+1}$ such that $m_i \equiv 0 \mod d_j + k$ for $j \ne i$ while $m_i \equiv k \mod d_i + k$. Let $b = a_1^{m_1}\ldots a_{n+1}^{m_{n+1}}$ and $t_i = (b/a_i^k)^{1/(d_i+k)}$, which is a positive integer.
Then
$$(t_1 a_1)^{d_1+k} + (t_2 a_2)^{d_2+k} + \ldots + (t_n a_n)^{d_n+k}
= b (a_1^{d_1} + \ldots + a_n^{d_n}) = b a_{n+1}^{d_{n+1}} = (t_{n+1} a_{n+1})^{d_{n+1}+k}$$
For example, starting from
$$ 28^2 + 8^3 = 6^4$$ ($a_1 = 28$, $a_2 = 8$, $a_3 = 6$), for $k=1$
where $3,4,5$ are pairwise coprime (it will work for any odd $k$) we take $m_1,m_2,m_3 = 40, 45, 36$.
Then $$\eqalign{b &= 28^{2} 8^3 6^4 = 2^{251} 3^{36} 7^{40} \cr
t_1 &= (b/a_1)^{1/3} = 2^{83} 3^{12} 7^{13}\cr
t_2 &= (b/a_2)^{1/4} = 2^{62} 3^9 7^{10}\cr
t_3 &= (b/a_3)^{1/5} = 2^{50} 3^7 7^8\cr
(t_1 a_1)^{3} &+ (t_2 a_2)^4 = (t_3 a_3)^5 \cr
0^2 + (2^{85} 3^{12} 7^{14})^3 &+ (2^{65} 3^9 7^{10})^4 = (2^{51} 3^8 7^8)^5 \cr}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1619089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.