Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Finding the radius of the smallest circle that can circumscribe an equilateral triangle Q:A puzzle board is in the form of an equilateral triangle that has an area of $7\sqrt{3}$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table. | Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7\sqrt{3}$, so we do:
$Area = \frac{1}{2}(base)(height) = \frac{1}{2}L\left(\frac{L\sqrt{3}}{2}\right) = 7\sqrt{3}$
Solving for L, we get $L = 2\sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = \left(\frac{L}{2}\right)^2 + h^2$
$h = \sqrt{L^2 - \frac{L^2}{4}} = \frac{L\sqrt{3}}{2}$
The radius of the circle is then:
$r = \frac{2}{3}h = \frac{2}{3}\frac{L\sqrt{3}}{2} = \frac{L\sqrt{3}}{3} = \frac{2\sqrt{7}\sqrt{3}}{3} = \frac{2\sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = \pi r^2 = \pi \left(\frac{2\sqrt{21}}{3}\right)^2 = \pi \frac{(4)(21)}{9} = \frac{28\pi}{3} \approx \frac{88}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $t = \tan (x/2)$, find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $3\sin x - 4\cos x = 2$.
If $$t = \tan \frac{x}{2},$$ find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $$3\sin x - 4\cos x = 2.$$
Attempt:
I have been solving a lot of trig questions lately but this is different. I don't know how to approach this. I'm thinking of getting $\sin x$ and $\cos x$ from $\tan x$ and replacing in the equation but not sure how because of the $t$. Help please.
| Indicated Solution
We can derive the Weierstrass Substitution:
Using the tangent double angle formula:
$$
\tan(x)=\frac{2t}{1-t^2}\tag{1}
$$
Then writing $\sec^2(x)$ in terms of $\tan^2(x)$
$$
\begin{align}
\sec^2(x)
&=1+\tan^2(x)\\
&=1+\frac{4t^2}{1-2t^2+t^4}\\
&=\frac{1+2t^2+t^4}{1-2t^2+t^4}\\
&=\left(\frac{1+t^2}{1-t^2}\right)^2\tag{2}
\end{align}
$$
Therefore, checking sign of $\cos(x)$ vs $\tan(x/2)$:
$$
\cos(x)=\frac{1-t^2}{1+t^2}\tag{3}
$$
Multiplying $(1)$ and $(3)$ gives
$$
\sin(x)=\frac{2t}{1+t^2}\tag{4}
$$
Then, as mentioned in comments, we simply need to solve for $t=\tan(x/2)$:
$$
3\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}-4\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)}=2\tag{5}
$$
which is simply a quadratic equation in $t$ giving
$$
\tan(x/2)=t=\frac{-3\pm\sqrt{21}}2\tag{6}
$$
Alternate Solution
Suppose $\theta$ is an angle so that $\sin(\theta)=\frac45$ and $\cos(\theta)=\frac35$; that is, $\theta=\sin^{-1}\!\left(\frac45\right)$.
Then,
$$
\begin{align}
\sin(x-\theta)
&=\cos(\theta)\sin(x)-\sin(\theta)\cos(x)\\[3pt]
&=\frac35\sin(x)-\frac45\cos(x)\\
&=\frac25
\end{align}
$$
which gives the two solutions
$$
x=\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right)
\implies\tan(x/2)=\frac{-3+\sqrt{21}}2
$$
and
$$
x=\pi-\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right)
\implies\tan(x/2)=\frac{-3-\sqrt{21}}2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Multivariable continuity in ball How do you show continuity for ball-based functions such as
$$f:B[(0,0),1)]\rightarrow\mathbb{R}, \space f(x,y) = \sqrt{(1-(x^2+y^2)}$$
| If you want an $\varepsilon$-argument:
Let $\varepsilon > 0$; let $(x',y') \in \mathbb{R}^{2}$ such that $x'^{2} + y'^{2} < 1$; then
$$
|\sqrt{1-x^{2}-y^{2}} - \sqrt{1- x'^{2}-y'^{2}}| \leq \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x^{2}-y^{2}} + \sqrt{1-x'^{2}-y'^{2}}} < \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x'^{2}-y'^{2}}}.
$$
If $|x-x'|, |y-y'| < 1$, then $2x'- 1 < x+x' < 2|x'| +1$ and $2y'- 1 < y+y'< 2|y'| + 1$, so $|x+x'| < \varepsilon_{1} := \max \{|2x'-1|, 2|x'| + 1\}$ and $ |y + y'| < \varepsilon_{2} := \max \{ |2y' - 1|, 2|y'| + 1 \}$, and hence
$$
\frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x'^{2}-y'^{2}}}
<
\frac{|x-x'|\varepsilon_{1} + |y-y'|\varepsilon_{2}}{\sqrt{1-x'^{2}-y'^{2}}},
$$
which is $< \varepsilon$ if in addition we have $|x-x'| < \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{1}}$ and $|y-y'| < \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{2}}$; we have proved that if $x,y$ are such that $x^{2}+y^{2}<1$, $|x-x'| < \min \{ 1, \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{1}} \}$, and $|y-y'| < \min \{ 1, \frac{\varepsilon\sqrt{1-x'^{2}-y'^{2}}}{2\varepsilon_{2}} \}$, then $|\sqrt{1-x^{2}-y^{2}} - \sqrt{1-x'^{2} - y'^{2}}| < \varepsilon.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer? $x,y,z \in \Bbb Z$, if $xy+yz+xz=1$ then prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer
$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+(xy)^2+(yz)^2+(xz)^2+(xyz)^2$
$(xy)^2+(yz)^2+(xz)^2+2xyz(x+y+z)=1$
| Use Brahmagupta–Fibonacci identity, $$(1+x^2)(1+y^2)=(1\pm xy)^2+(x\mp y)^2$$
Again, $$\{(1-xy)^2+(x+y)^2\}(1+z^2)=\{(1-xy)-z(x+y)\}^2+\{(1-xy)z+x+y\}^2$$
Do you see the destination?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to \infty}\left(x^2\ln\left(\cos \frac{4}{x}\right)\right)$ How do you evaluate $$\lim_{x\to\infty}\left(x^2\ln\left(\cos\frac{4}{x}\right)\right)$$
I know that you have to rewrite the expression as a quotient and use L'Hospital's Rule but I cant seem to figure it out.
| In this approach, we use some standard inequalities along with the squeeze theorem. First, recall that the log function satisfies the inequalities
$$\frac{x-1}{x}\le\log x\le x-1 \tag 1$$
for $x>0$. Then using $(1)$, it is straightforward to see that
$$x^2\frac{\left(\cos\left(\frac 4x\right)-1 \right)}{\cos\left(\frac 4x\right)}\le x^2\log \cos\left(\frac 4x\right)\le x^2\left(\cos\left(\frac 4x\right)-1 \right) \tag 2$$
Now, using the trigonometric identity $\cos x-1=-2\sin^2(x/2)$ in $(2)$ yields
$$-2x^2\left(\frac{\sin^2\left(\frac 2x\right)}{\cos\left(\frac 4x\right)}\right)\le x^2\log \cos\left(\frac 4x\right)\le -2x^2\sin^2\left(\frac 2x\right) \tag 3$$
Next, recall that the sine function satisfies the inequalities
$$x\cos x\le \sin x\le x \tag 4$$
for $x\ge 0$. So, using $(4)$ in $(3)$ yields
$$-8\left(\frac{\cos^2\left(\frac2x\right)}{\cos\left(\frac 4x\right)}\right)\le x^2\log \cos\left(\frac 4x\right)\le -8 \tag 5$$
Finally, we have from the squeeze theorem
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty} x^2\log \cos\left(\frac 4x\right)=-8}$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding a permutation given a composite of permutations Given the permutations $x = (1, 2)(3, 4)$ and $y = (5, 6)(1, 3)$ find a permutation $a$ such that $a^{−1}xa = y$.
I have an answer to this but I would like to know the process involved in getting the answer.
My thoughts thus far are that $a^{-1}(12)(34)a =(56)(13)$ which means that $a^{-1}((12)(34)a) =(56)(13)$. So $(a^{-1}(1(a)), a^{-1}(2(a))) (a^{-1}(3(a)), a^{-1}(4(a))) = (5,6)(1,3)$.
Any help would be appreciated.
| Let $\sigma=[i_1,\ldots, i_n]$ be a cycle. Use the fact that $\tau\sigma\tau^{-1}=[\tau(i_1),\ldots,\tau(i_n)]$. So, what permutation takes $1$ to $5$, $2$ to $6$, $3$ to $1$ and $4$ to $3$. The inverse of this permutation is the answer.
To find this, we can expess it as a matrix:
$$a^{-1}=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 6 & 1 & 3 & c & d
\end{pmatrix}$$
Here the image of a number in the first row, is the corresponding number in the second row.
We see that there are two possibilities, $c=2$, $d=4$ and $c=4$,$d=2$.
Let's take the case, $c=2$ and $d=4$.
So,
$$a^{-1}=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 6 & 1 & 3 & 2 & 4
\end{pmatrix}$$
and,
$$a=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 5 & 4 & 6 & 1 & 2
\end{pmatrix}$$
Writing these as cycles gives:
$$a=(134625)\quad\text{and}\quad a^{-1}=(152643).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric equation with parameter
Find $p$ for which $\cos^2(x) - \cos(x) + p + 1 = 0$ has EXACTLY two solutions for $0 \le x \le 2\pi$
I tried to substitute $t = \cos(x)$ and then I got two solutions, but I don't know what to do next.
$t_1 = \frac{1 + \sqrt{-3 - 4p}}{2}$
$t_2 = \frac{1 - \sqrt{-3 - 4p}}{2}$
| Since $\cos x=T$ has at most two real solutions in $0\le x\le 2\pi$ for a given real constant $T$, we have three cases.
(1) $\cos x=t_1$ has no solution and $\cos x=t_2$ has two solutions.
(2) $\cos x=t_1$ has exactly one solution and $\cos x=t_2$ has exactly one solution.
(3) $\cos x=t_1$ has two solutions and $\cos x=t_2$ has no solution.
Note here that
*
*$\cos x=T$ has no solution in $0\le x\le 2\pi$ if and only if $|T|\gt 1$.
*$\cos x=T$ has exactly one solution in $0\le x\le 2\pi$ if and only if $T=-1$.
*$\cos x=T$ has two solutions in $0\le x\le 2\pi$ if and only if $-1\lt T\le 1$.
So, noting that (2) does not occur, the answer is
$$\{p\mid -3-4p=0\}\cup \left\{p\mid -3-4p\gt 0,\frac{1+\sqrt{-3-4p}}{2}\gt 1,-1\lt\frac{1-\sqrt{-3-4p}}{2}\le 1\right\}\cup \left\{p\mid -3-4p\gt 0,-1\lt \frac{1+\sqrt{-3-4p}}{2}\le 1,\frac{1-\sqrt{-3-4p}}{2}\lt -1\right\},$$
i.e.
$$\color{red}{p=-\frac 34\quad\text{or}\quad -3\lt p\lt -1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series?
I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:
\begin{equation}
y(x) = \tan x
\end{equation}
\begin{equation}
y^{\prime} = \frac{1}{\cos^{2} x}
\end{equation}
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x}
\end{equation}
This can be rewritten as:
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime}
\end{equation}
Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$:
\begin{equation}
y^{\prime} = y^{2}
\end{equation}
And therefore
\begin{equation}
\frac{1}{\cos^{2} x} = \tan^{2} x
\end{equation}
Now, evalueting this equation at $x = \pi / 4$
\begin{equation}
\frac{1}{(\sqrt{2}/2)^{2}} = 1^{2}
\end{equation}
\begin{equation}
2 = 1
\end{equation}
| Another enjoyable "paradox": We first denote
$$S:=\sum_{n\in\mathbb N}\dfrac{(-1)^{n+1}}{n}$$
The fact that $0\neq S\in\mathbb R$ can be established using elementary tools.
We then write:
$S=\frac{1}{1}-\frac {1}{2}+\frac{1}{3}-...+...-...$
$2S = 2(\frac{1}{1}-\frac {1}{2}+\frac{1}{3}-...+...-...)=\frac{2}{1}-\frac {2}{2}+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-\frac{2}{6}+\frac{2}{7}-...=$
$=\color{red}{\frac{2}{1}}\color{red}{-\frac {2}{2}}\color{green}{+\frac{2}{3}}\color{blue}{-\frac{2}{4}}+\frac{2}{5}\color{green}{-\frac{2}{6}}+\frac{2}{7}-...=\color{red}{\frac{1}{1}}\color{blue}{-\frac{1}{2}}\color{green}{+\frac{1}{3}}-...=S$
And at last:
$$2S = S \Longrightarrow 2=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 1
} |
Regarding Leibniz formula for $\pi/4$ proof and its convergence \begin{align}
\frac{\pi}{4} & = \arctan(1)\;=\;\int_0^1 \frac 1{1+x^2} \, dx \\[8pt]
& = \int_0^1\left(\sum_{k=0}^n (-1)^k x^{2k}+\frac{(-1)^{n+1}\,x^{2n+2} }{1+x^2}\right) \, dx \\[8pt]
& = \sum_{k=0}^n \frac{(-1)^k}{2k+1}
+(-1)^{n+1}\int_0^1\frac{x^{2n+2}}{1+x^2} \, dx.
\end{align}
After this integration, $\frac{\pi}4\;=\;\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$
1.
I have trouble understanding why
$\int_0^1 \frac 1{1+x^2} dx = \int_0^1\left(\sum_{k=0}^n (-1)^k x^{2k}+\frac{(-1)^{n+1}\,x^{2n+2} }{1+x^2}\right) dx$
2.
How to compare the convergence of
$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$ and
$\int_0^1\left(\sum_{k=0}^n (-1)^k x^{2k}+\frac{(-1)^{n+1}\,x^{2n+2} }
{1+x^2}\right) dx$?
Which one converge faster and do not have the same range of convergence?
| Using long division, the integrand $\frac{1}{1+x^2}$ can be written
$$\begin{align}
\frac{1}{1+x^2}&=1-x^2+x^4-x^6+\cdots +(-1)^nx^{2n}+\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}\\\\
&=\sum_{k=0}^n (-1)^kx^{2k}+\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}
\end{align}$$
Thus, the integral of interest is
$$\begin{align}
\int_0^1\frac{1}{1+x^2}\,dx&=\int_0^1\left(\sum_{k=0}^n (-1)^kx^{2k}+\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}\right)\,dx\\\\
&=\sum_{k=0}^n (-1)^k\int_0^1x^{2k}\,dx+(-1)^{n+1}\int_0^1\frac{x^{2n+2}}{1+x^2}\,dx\\\\
&=\sum_{k=0}^n\frac{(-1)^k}{2k+1}+(-1)^{n+1}\int_0^1\frac{x^{2n+2}}{1+x^2}\,dx
\end{align}$$
Since $\left|\frac{x^{2n+2}}{1+x^2}\right| \le\frac{1}{1+x^2}$ and $\int_0^1\frac{1}{1+x^2}\,dx$ converges, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{n\to \infty}\int_0^1\frac{x^{2n+2}}{1+x^2}\,dx&=\int_0^1\lim_{n\to \infty}\left(\frac{x^{2n+2}}{1+x^2}\right)\,dx\\\\
&=0
\end{align}$$
Therefore, we have
$$\int_0^1\frac{1}{1+x^2}\,dx=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}=\pi/4$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Olympiad inequality problem with $a+b+c+abc=4$ If $a,b,c \in \mathbb R_{> 0}$ and $a+b+c+abc=4$, prove that
$$({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}})^2(ab+bc+ca) \ge {\frac 12}(4-abc)^3$$
This can be solved by AM-GM-HM or the Cauchy-Schwarz inequality. I'd tried for some hours but couldn't solve it. Can anyone help me? Thanks in advance:).
| Here is a possible approach without using Hölder's inequality. Assume without loss of generality that $a\ge b \ge c$. Then
$$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2\ge \frac{1}{2a}\left(a+b+c\right)^2$$
Now
$$ab+bc+ca=A=\frac{A+A}{2}=\frac{1}{2}(a(b+c)+b(c+a)+c(a+b))\ge c(a+b+c)$$
Since $a,b,c\in \mathbb{R}_+$ this implies
$$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2(ab+bc+ca) \ge \frac{c}{2a}(a+b+c)^3$$
However $0<\frac{c}{a}\le 1$ by assumption, so the desired inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction: $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
For $n=1$ equality is true.
For $n=m$
$m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{2}+...+\frac{1}{m}$
For $n=m+1$
$\left(\sum\limits_{k=1}^{m}(-1)^{k+1}{m\choose k}\frac{1}{k}\right)+(-1)^{m+2}\frac{1}{m+1}=1+\frac{1}{2}+...+\frac{1}{m+1}$
If $m$ is even, equality is true, but not if $m$ is odd.
Is this correct?
| We have:
$$ \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1} x^k = 1-(1-x)^n\tag{1}$$
hence:
$$ \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{1}{k} = \int_{0}^{1}\frac{1-(1-x)^n}{x}\,dx= \int_{0}^{1}\frac{1-x^n}{1-x}\,dx\tag{2}$$
and:
$$ \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{1}{k} = \int_{0}^{1}\left(1+x+\ldots+x^{n-1}\right)\,dx = 1+\frac{1}{2}+\ldots+\frac{1}{n}=H_n\tag{3}$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For f(x) = ((3x+7)^8)((4x-5)^3) , find f'(x) and use this answer to find the value(s) of x at which the graph of f(x) has a horizontal tangent line I know that $f'(x) = (24(3+7)^7)*(4x-5)^3+((3x+7)^8)*12(4x-5)^2$, but is there any easier way to find the horizontal tangent line without expanding the terms using pascal's triangle and solving for x by setting f'(x) equal to 0?
| $$f(x)=(3x+7)^8(4x-5)^3$$
$$f'(x)=8(3x+7)^7.3(4x-5)^3+(3x+7)^83.(4x-5)^2.4$$
$$f'(x)=(3x+7)^7(4x-5)^2[24(4x-5)+12(3x+7)]$$
$$f'(x)=(3x+7)^7(4x-5)^2[132x-36]$$
$$f'(x)=12(3x+7)^7(4x-5)^2[11x-3]$$
to find the tangent point
$$(3x+7)^7=0$$
$$(4x-5)^2=0$$
$$(11x-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving a recurrence relation with no real roots? $a_n - 2a_{n-1} -12a_{n-2} - 14a_{n-3} -5a_{n-4} = 0 $
I've tried a few method:
*
*setting the denominator to $1 -2x -12x^2 -14x^3 -5x^4$ and then finding the numerator by multiplying $(1 -2x -12x^2 -14x^3 -5x^4)*\sum a_nx^{n}$
*finding the roots of C(y) = $y^4 -4y^3 -12y^2 -14y -5$
What else can I do?
| A start: The given recurrence has characteristic equation $x^4-2x^3-12x^2-14x-5=0$. This has $x=-1$ as a solution.
Dividing our polynomial by $x+1$, we get $x^3-3x^2-9x-5$. Note that $-1$ is a zero of this polynomial. Continue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sequence of positive integers such that their reciprocals are in arithmetic progression Let $m_1 < m_2 < \ldots < m_k$ be $k$ distinct positive integers such that their reciprocals $\dfrac{1}{m_i}$ are in arithmetic progression.
*
*Show that $k < m_1 + 2$.
*Give an example of such a sequence of length $k$ for any positive integer $k$.
Any kind of help would be appreciated.
| Hint for the bound: Use the fact that the common difference $d$ of the AP is $\ge \frac{1}{m_1}-\frac{1}{m_1+1}$. Now if we take $(n-1)$ steps of length down from $a=\frac{1}{m_1}$, the result must be $\gt 0$.
Hint for the construction: The first interesting length is $k=3$. There we have the familiar AP $\frac{1}{2},\frac{1}{3},\frac{1}{6}$. Now the only challenge is constructing an AP of length $4$.
We look for an AP of the shape $\frac{1}{m},\frac{1}{2q},\frac{1}{3q},\frac{1}{6q}$.
So we want $\frac{1}{m}-\frac{1}{2q}=\frac{1}{6q}$, that is, $\frac{1}{m}=\frac{4}{6q}$. Take $q=4$.So our first term is $\frac{1}{6}$, and the others are $\frac{1}{8}$, $\frac{1}{12}$, and $\frac{1}{24}$.
Now on to $5$. In our example of length $4$, we had $q=4$, so common difference $\frac{1}{24}$ and first term $\frac{1}{6}$. Multiply the denominators of the length $4$ example by a new suitable $q$. We want to insert a new first term $\frac{1}{m}$, where $\frac{1}{m}-\frac{1}{6q}=\frac{1}{24q}$. Note that $q=5$ works. The new common difference is $\frac{1}{120}$ and the new first term is $\frac{1}{24}$.
Continue. At each stage we multiply the denominators of the length $l$ AP by a suitable $q$, and insert a new first term. Same calculation.
Now we can look back on the examples, and write down a simple explicit formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does $\lim\limits_{n \to \infty } ((1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$? Let $\left| x \right| < 1$.
Why does $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$
| $$(1+x)(1+x^2)(1+x^4)(1+x^8).....=1+x+x^2+x^3+x^4+x^5......$$
then you can use the fact of geometric series
$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5......$$
$$|x|<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Write the expression in Euler's formula
$a=2+2i$ , $b=5e^{i\frac{\pi}{3}}$
such that $$\frac{b^5}{a^3}=Re^{\theta i}$$
find: $R$ and $\theta$
$$R=\sqrt{2^2+2^2}=\sqrt{8}$$,
$$tan^{-1}=\frac{2}{2}\rightarrow \theta=\frac{\pi}{4}$$
$$a=\sqrt{8}e^{\frac{\pi}{4}i}, a^3=({\sqrt{8}e^{\frac{\pi}{4}i}})^3=(\sqrt{8})^3e^{\frac{3\pi}{4}i}$$
$$b^5=5^5e^{\frac{5\pi}{3} i}$$
$$\frac{b^5}{a^3}=\frac{5^5e^{\frac{5\pi}{3} i}}{(\sqrt{8})^3e^{\frac{3\pi}{4}i}}$$
$$\frac{b^5}{a^3}=\color{red}{5^5-(\sqrt{8})^3}*e^{(\frac{5\pi}{3}-\frac{3\pi}{4})i}=(5^5-(\sqrt{8})^3)*e^{\frac{11}{12}\pi i}$$
$$R=5^5-(\sqrt{8})^3, \theta=\frac{11}{12}\pi$$
I can not find my mistake
| Write $a = 2\sqrt 2 e^{\pi/4 i}$ Then
$$\frac {b^5}{a^3} = \frac{5^5e^{5i\pi/3}}{(\sqrt 8)^3 e^{3/4\pi i}} = \frac{5^5}{(\sqrt 8)^3} e^{ (5/3 - 3/4) i \pi} = \frac{5^5}{(\sqrt 8)^3} e^{ \frac {11}{12} i \pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Write on my own my first mathematical induction proof I am trying to understand how to write mathematical induction proofs. This is my first attempt.
Prove that the sum of cubic positive integers is equal to the formula
$$\frac{n^2 (n+1)^2}{4}.$$ I think this means that the sum of cubic positive integers is equal to an odd number. However, let's go on proving...
1) I start by proving the base case $n=1$ and I show that the formula holds.
2) I assume than any number $k$ other than $1$, which appartains at $N$, holds for the formula and I write the same formula but with $k$ which replaces $n$.
3) For mathematical induction, I assume that the formula holds also for $k+1$ = $n$
So, the left side of the equation should be:
$$\sum^{k+1}_{i=1} i^3 = 1^3 + 2^3 + 3^3 + ... + (k+1)^3$$
I am wondering about which one of these 2 forms (equivalents, I think) should have the right side :
this one, with $k+1$ in place of the $n$ of the original formula / or $k$ in the second version: $\frac{(k+1)^2[(k+1)+1]^2}{4}$ or this one: $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ ?
I think that, in order for the proof to be convincing, we should write an equivalent statement for the original form of the formula, namely $$\sum^{n}_{i=1} i^3= \frac{n^2(n+1)^2}{4}$$ and perhaps we do it by showing that after algebraic passages $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ is equal to $\frac{(k+1)^2[(k+1)+1]^2}{4}$ ?
Sorry for my soliloquy but it helps to understand and I would appreciate confirmation from you!
| You want to prove that $$\sum_{i=1}^{n}i^3 = \frac{n^2 (n+1)^2}{4}$$ using induction.
For $n=1$, $$\sum_{i=1}^{1}i^3 = 1^3=1=\frac{1^2(1+1)^2}4=1$$
So the formula does work for the base case $n=1$.
Now, assume the formula works for $n=k$ and show that this implies that the formula is correct for $n=k+1$ which will accomplish the prove by induction.
Thus,
\begin{align}
\sum_{i=1}^{k+1}i^3 & = \left(\sum_{i=1}^{k}i^3\right) +(k+1)^3 \\
& = \frac{k^2 (k+1)^2}{4}+(k+1)^3 \\
& = \frac{k^2 (k+1)^2+4(k+1)^3}{4} \\
& = \frac{ (k+1)^2(k^2+4(k+1))}{4} \\
& = \frac{ (k+1)^2(k+2)^2}{4} \\
& = \frac{ (k+1)^2((k+1)+1)^2}{4}
\end{align}
and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Hypergeometric series for $\sin(az)$ I am having trouble trying to show that: $$F\left(\frac{1}{2}+\frac{1}{2}a,\frac{1}{2}-\frac{1}{2}a;\frac{3}{2};\sin^2z\right) = \sin{az}.$$
I have found that the coefficients in the series are given by: $$\frac{\left(\frac{1}{2}+\frac{1}{2}a\right)_n \left(\frac{1}{2}-\frac{1}{2}a\right)_n}{\left(\frac{3}{2}\right)_n\, n!}=(-1)^n \frac{(a^2-1^2)(a^2-3^2)\ldots(a^2-(2n-1)^2)}{(2n+1)!},$$
but have no idea what to do with the $\sin^2z$ term?
Thanks!
| By comparing the Chebyshev differential equation with the hypergeometric differential equation you may check that:
$$ \phantom{}_{2}F_1\left(\frac{1+a}{2},\frac{1-a}{2};\frac{3}{2};z^2\right)=\frac{\sin(a\arcsin z)}{a z} \tag{1}$$
from which it follows that:
$$ \phantom{}_{2}F_1\left(\frac{1+a}{2},\frac{1-a}{2};\frac{3}{2};\sin^2 z\right)=\frac{\sin(a z)}{a\sin z}=\frac{1}{a}\,U_{a-1}(\cos z). \tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Reading cycle notation I need a little help understanding how we read cycle notation. I know it is a function so if we have something like
$\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\
2 & 5 & 4 & 3 & 1
\end{pmatrix} = (1 \quad 2 \quad 5)(3 \quad 4)=(3 \quad 4)(1 \quad 2 \quad 5)=(3 \quad 4)(5\quad 1 \quad 2) $
how is it read? Thanks!
| I think my original comment may have led to some confusion in regards to my intentions behind it--your function
\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\\tag{1}
2 & 5 & 4 & 3 & 1
\end{pmatrix}
may, perhaps, be more effectively communicated as
\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\
\downarrow & \downarrow & \downarrow & \downarrow & \downarrow\\\tag{2}
2 & 5 & 4 & 3 & 1
\end{pmatrix}
where each "$\downarrow$" in $(2)$ is really meant to be an upside down "$\mapsto$" (I'm not sure of the best way to rotate math symbols using MathJax). The point is that each number is being mapped to another. All of these mappings make up your function. Read aloud what happens in $(2)$:
*
*$1$ maps to $2$ which maps to $5$ which maps to $1$ which maps to $2$ which maps to $5$ ...
*$3$ maps to $4$ which maps to $3$ which maps to $4$ which maps to $3$ ...
The point is that both of the cases above lead to noticeable cycles (hence the name). By understanding how the different mappings in $(2)$ work, we can now understand how the cycles were created, yielding
$$
\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\
2 & 5 & 4 & 3 & 1
\end{pmatrix}
=\begin{pmatrix} 1 & 2 & 5\end{pmatrix}\begin{pmatrix} 3 & 4\end{pmatrix}
$$
Does that help?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1}$ and so on? Like where does the $n-3$ come from or the $n-1$.
| $$S = \sum_{n=5}^\infty \frac{8}{n^2 -1}=
\lim_{N\to\infty} \sum_{n=5}^N \frac{8}{n^2 -1}
= \lim_{N\to\infty} \sum_{n=5}^N \left\{ {4\over n-1} - {4\over n+1}\right\} = \lim_{N\to\infty} \sum_{n=5}^N {4\over n-1} - \sum_{n=5}^N {4\over n+1}$$
Now reindex to get
$$S = \lim_{N\to\infty} \sum_{n=4}^{N-1} {4\over n} - \sum_{n=6}^{N+1} {4\over n} = \lim_{N\to\infty} 1 + {4\over 5} - {4\over N}- {4\over N+1} = {9\over 5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to Completely Simplify the Derivative of $\sqrt{16-x^2}-4\cos^{-1}(x/4)$ I am trying to completely simplify the derivative of the following function:
So far, I have gotten the answer:
Apparently this is not simplified enough.
Does anyone know how to simplify this further, if it is at all possible?
All help is appreciated.
| \begin{align}
& \frac{d}{dx} \left(\sqrt{16-x^2}-4\cos^{-1}(\frac{x}{4})\right) \\[10pt]
= {} &\frac{d}{dx} \left(\sqrt{16-x^2}\right)-\frac{d}{dx} \left(4\cos^{-1} \left(\frac{x}{4}\right)\right) \\[10pt]
= {} & \frac{1}{2\sqrt{16-x^2}}\cdot(-2x)+\frac{4}{\sqrt{1-\frac{x^2}{16}}} \cdot \left(\frac{-1}{4}\right) \\[10pt]
= {} & \frac{-x}{\sqrt{16-x^2}}-\frac{4}{\sqrt{16-x^2}} \\[10pt]
= {} & \frac{-(4+x)}{\sqrt{16-x^2}}=\frac{-(4+x)}{\sqrt{(x+4)(x-4)}}=\frac{-1}{\sqrt{(x-4)}}.
\end{align}
I think this is right, double check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the last two digits of the number $7^{100}-8^{100}$. So I tried and found that $$7^{100} \equiv 1 \pmod{100}$$ but I got stuck with $8^{100}$. Help me out please.
| First, note that $7^2\equiv8^2\pmod{5}$. So $$7^{100}-8^{100}=(7^2-8^2)\left((7^2)^{49}+(7^2)^{48}(8^2)+\cdots+(8^2)^{49}\right)\equiv 0\pmod{25},$$
since the second term consists of $50$ summands of the same modulo $5$.
Second $7^{100}-8^{100}\equiv 7^{100}\pmod{4}\equiv 1\pmod{4}$.
Chinese remainder theorem then implies that the answer is $25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the differential equation $\frac{dy}{dx}=\frac{2x+y}{y}$ Solve
$$\frac{dy}{dx}=\frac{2x+y}{y}$$
Let
$$\frac{dx}{dt}=y \\ \frac{dy}{dt}=2x+y$$
Plugin $\frac{dx}{dt}=y$ into $\frac{dy}{dt}=2x+y$ I get $$2x+\frac{dx}{dt}=\frac{dy}{dt}$$
Using the fact that $\frac{d^2x}{dt^2}=\frac{dy}{dt}$ I get the 2nd order system:
$$\frac{d^2x}{dt^2}-\frac{dx}{dt}-2x=0$$
Solving this I get:
$(r-2)(r+1)=0$ , so $r=2,=1$
$$x(t)=Ae^{2t}+Be^{-1t}$$
How do I get my solution in terms of $y,x$? I can't seem to integrate $\frac{dy}{dx}$ by separating variables directly.
| I'm a bit rusty with differential equations, so I'll just assume that we have the required number of constants of integration...
$$\begin{align}
x & = Ae^{2t} + Be^{−t}\\
y & = \frac{dx}{dt}\\
y & = 2Ae^{2t} - Be^{−t}\\
x+y & = 3Ae^{2t}\\
2x-y & = 3Be^{−t}\\
e^{2t} = \frac{x+y}{3A} & = \left(\frac{3B}{2x-y}\right)^2\\
(2x-y)^2(x+y) & = 27AB^2\\
4x^3 - 3xy^2 + y^3 & = 27AB^2
\end{align}$$
Now to check that by differentiating:
$$\begin{align}
12x^2 - 3y^2 - 6xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} & = 0\\
\frac{dy}{dx}(3y^2 - 6xy) & = 3y^2 - 12x^2\\
\frac{dy}{dx} & = \frac{y^2 - 4x^2}{y^2 - 2xy}\\
& = \frac{(y - 2x)(y + 2x)}{y(y - 2x)}\\
& = \frac{y + 2x}{y}, \text{when } y \ne 2x\\
\end{align}$$
which agrees with the original equation.
Here's an alternative solution that doesn't use those parametric equations in $t$. Instead, we use a simple $y = vx$ substitution and separation of variables.
$$\begin{align}
\frac{dy}{dx}&=\frac{2x+y}{y}\\
ydy &= (2x+y)dx\\
\\
\text{Let } y &= vx\\
dy &= vdx + xdv\\
\\
\text{Substituting, }\\
vx(vdx + xdv) &= (2x+vx)dx\\
v^2dx + vxdv - (2+v)dx &= 0\\
(v^2-v-2)dx + vxdv &= 0\\
\\
\text{Separating variables, }\\
\frac{dx}{x} + \frac{vdv}{v^2-v-2} &= 0\\
\\
\text{Splitting the right term with partial fractions, }\\
\frac{dx}{x} + \frac{1}{3}\left(\frac{(v-2)+2(v+1)}{(v-2)(v+1)}\right)dv &= 0\\
\frac{dx}{x} + \frac{1}{3}\left(\frac{dv}{v+1} + 2 \frac{dv}{v-2}\right) &= 0\\
\\
\text{Integrating, }\\
\int\frac{dx}{x} + \frac{1}{3}\left(\int\frac{dv}{v+1} + 2 \int\frac{dv}{v-2}\right) &= k\\
\log(x) + \frac{1}{3}(\log(v+1) + 2 \log(v-2)) &= k\\
3\log(x) + \log(v+1) + 2 \log(v-2) &= 3k\\
x^3(v+1)(v-2)^2 &= e^{3k} = C\\
x^3(v+1)(v^2-4v+4) &= C\\
x^3(v^3-4v^2+4v + v^2-4v+4) &= C\\
x^3(v^3-3v^2+4) &= C\\
(xv)^3-3x(vx)^2+4x^3 &= C\\
\\
\text{Replacing $vx$ with $y$, }\\
y^3-3xy^2+4x^3 &= C\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
Second:
$$\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $$
Anyone able to provide the correct answer or method of solving these?
| I'll solve both in one shot here using the simplest methods I can think of:
For the first, we use the proof here
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
$$\Rightarrow \displaystyle \frac {\cos (4 x)- 4 \cos (2 x)+3 } 8 + \frac {\cos(4x)+4\cos(2x) + 3} 8 \geq \frac{5}{8}$$
$$\Rightarrow \displaystyle \frac {2\cos (4 x)+6 } 8 \geq \frac{5}{8}$$
$$\Rightarrow \displaystyle 2\cos (4 x)+6 \geq 5$$
$$\Rightarrow \cos(4 x) \geq -\frac{1}{2} $$
$$\Rightarrow 4x \leq \arccos(-1/2) $$
$$\Rightarrow x \leq \frac{1}{4}\bigg(\frac{2\pi}{3}\bigg) $$
$$\Rightarrow x \leq \frac{\pi}{6} $$
Because we used the inverse cosine the domain is restricted to non-negative numbers here, actually making this $0 \leq x \leq \frac{\pi}{6}$. However, since $\cos(x)$ is an even function, we also get $-\frac{\pi}{6} \leq x \leq 0$. Combining these, we get $\color{red}{-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}}$
For the second problem, $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52$, we set $u=\sqrt\frac{x^2+2x-2}{x^2+3x+2}$ and, as a result, $\frac{1}{u} = \sqrt\frac{x^2+3x+2}{x^2+2x-2}$. Plugging these into the original equation, we get
$$u + \frac{1}{u} = \frac{5}{2}$$
$$u^2 + 1 = \frac{5}{2}n$$
$$u^2 -\frac{5}{2}u + 1 = 0$$
Solving this, we get $u = 2, \frac{1}{2}$, both of which satisfy our equation in terms of $n$. Substituting, back,
$$\sqrt{\frac{x^2+2x-2}{x^2+3x+2}} = 2 \qquad \text{or} \qquad \sqrt{\frac{x^2+2x-2}{x^2+3x+2}} = \frac{1}{2}$$
$$\frac{x^2+2x-2}{x^2+3x+2} = 4 \qquad \text{or} \qquad \frac{x^2+2x-2}{x^2+3x+2} = \frac{1}{4}$$
The first equation has only complex solutions, but the latter has solutions $\color{red}{x = -\frac{5}{6} \pm \frac{\sqrt{145}}{6}}$, both of which satisfy our original equation in terms of $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the problem:
By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$
So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$
Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$.
Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$.
So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$.
I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.
| To solve this kind of problem systematically. If you want to solve $|f(x)|\le c$ or $|f(x)|\geq c$ (for some constant $c\geq 0$), you have to study the sign of $f(x)$. It can be longer for "simple" problem like yours, but it has the advantage to always give you the solution for more complex ones.
For your particular example, we have $f(x)=\frac{x}{x+2}$, which is:
$$\begin{array}{c|ccccc}
& &-2 & &0 & &\\
\hline
x & - &- & - &0 &+\\
x+2 & - & 0 & + & + & +\\
\hline
f(x) & + &\nexists &- & 0 &+
\end{array}$$
Which means that $f(x)\geq 0$ when $x\in(-\infty,-2)\cup[0,+\infty)$ and $f(x)\le 0$ when $x\in(-2,0]$.
1) For $x\in(-\infty,-2)\cup[0,+\infty)$, $f(x)\geq 0$, so that
$$\left\vert\frac{x}{x+2}\right\vert=\frac{x}{x+2}$$
and the inequation is
$$\frac{x}{x+2}\le 2\iff\frac{-x-4}{x+2}\le 0$$
which gives:
$$\begin{array}{c|ccccc}
& &-4 & &-2 & &\\
\hline
-x-4 & + &0 & - &- &-\\
x+2 & - & - & - & 0 & +\\
\hline
& - &0 &+ & \nexists &-
\end{array}$$
and the solutions are $$\left[(-\infty,-4]\cup(-2,+\infty)\right]\cap\left[(-\infty,-2)\cup[0,+\infty)\right]=(-\infty,-4]\cup[0,+\infty)$$
2) For $x\in]-2,0]$, $f(x)\le 0$, so that
$$\left\vert\frac{x}{x+2}\right\vert=-\frac{x}{x+2}$$
and the inequation is
$$\frac{x}{x+2}\le 2\iff\frac{-3x-4}{x+2}\le 0$$
which gives:
$$\begin{array}{c|ccccc}
& &-2 & &-4/3 & &\\
\hline
-3x-4 & + &+ & + &0 &-\\
x+2 & - & 0 & + & + & +\\
\hline
& - &\nexists &+ & 0 &-
\end{array}$$
and the solutions are $$\left[(-\infty,-2)\cup[-4/3,+\infty)\right]\cap(-2,0]=[-4/3,0]$$
And the final solution is given by $$\left\vert\frac{x}{x+2}\right\vert\le 2\iff x\in(-\infty,-4]\cup[0,+\infty)\cup[-4/3,0]=(-\infty,-4]\cup[-4/3,+\infty)$$
I know it seems long in this particular case but it is good to have a general method for this kind of problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $ \int_0^1 x^2 \psi(x) \, dx = \ln\left(\frac{A^2}{\sqrt{2\pi}} \right) $ Basically what the title says:
Prove that $ \displaystyle \int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right). $
where $A\approx 1.2824$ denotes the Glaisher–Kinkelin constant and $\psi(x) $ denote the digamma function.
I've tried reflection formula and some Feynmann method on the integrals in the wikipedia but nothing works.
(I've copied this from AoPS because there's no response there)
| Let $(A_n)$ and $(B_n)$ by
$$ A_n = \frac{1^1 2^2 \cdots n^n}{n^{n^2/2+n/2+1/12} e^{-n^2/4}}
\quad \text{and} \quad
B_n = \frac{n!}{e^{-n}n^{n+1/2}}. $$
In view of the Wikipedia page, we know that $A_n \to A$ and $B_n \to \sqrt{2\pi}$. We can also equivalently state these definitions as follows:
\begin{align*}
\sum_{k=1}^{n} k \log k
&= \log A_n + \left( \frac{n^2}{2} + \frac{n}{2} + \frac{1}{12} \right) \log n - \frac{n^2}{4}, \\
\sum_{k=1}^{n} \log k
&= \log B_n + \left(n + \frac{1}{2}\right) \log n - n.
\end{align*}
Now using the partial fractional decomposition of $\psi(x)$, we find that
\begin{align*}
\int_{0}^{1} x^2 \psi(x) \, dx
&= \lim_{n\to\infty} \int_{0}^{1} x^2 \left( -\gamma + H_{n+1} - \frac{1}{x} - \sum_{k=1}^{n} \frac{1}{k+x} \right) \, dx \\
&= \lim_{n\to\infty} \left( \frac{H_{n+1}-\gamma}{3} + \frac{n^2}{2} - \frac{1}{2} + \sum_{k=1}^{n} k^2 (\log k - \log(k+1)) \right) \\
&= \lim_{n\to\infty} \left( \frac{1}{3}\log n + \frac{n^2}{2} - \frac{1}{2} + \sum_{k=1}^{n} (2k-1)\log k - n^2 \log(n+1) \right) \\
&= \lim_{n\to\infty} \left(2\log A_n - \log B_n - n^2 \log\left(1+\frac{1}{n}\right) + n - \frac{1}{2} \right) \\
&= 2\log A - \log\sqrt{2\pi}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\int \sqrt{1+\cot x}dx$?
What is $$\int \sqrt{1+\cot x}dx$$
My friends and I tried using all possible trigonometric formula. We couldn't find a way to solve it
Please help me solve it.
| Put $(1+cotx)=t^2$ in the given integral and then do the reqd. Substitutions to get :
$$\int -\frac {2x^2dx}{x^4-2x^2+2}$$
I think we can avoid complex numbers.
Write it as:
$$\int -2(\frac {(x^2+ \sqrt2+x^2-\sqrt2)dx}{x^4-2x^2+2})$$
Split it into two integrals,
$$\int\frac {(x^2+\sqrt2)dx}{x^4-2x^2+2}$$
And
$$\int \frac {(x^2-\sqrt2)dx}{x^4-2x^2+2}$$
Simply divide by $x^2$ in both the integrals to get,
$$\int \frac {(1+\frac {\sqrt2}{x^2})dx}{x^2+2/x^2-2}$$
Now complete the square in the denominator as $(x-\sqrt2/x)^2 + 2(\sqrt2-1)$ . (for first integral)
Then substitute $x-\sqrt(2)/x = v$.
To get the integrals as
$$\int \frac{dv}{(v^2+ 2(\sqrt(2)-1))}$$
And $$\int \frac{dv}{(v^2-2(\sqrt(2)+1))}$$
Which are standard integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sign table in $2^k$ factorial experiment I have a factorial experiment with four factors $\{A,B,C,D\}$ ($k=4$) , just to ilustrate the case with $k=3$, the signal table is
\begin{matrix}
& I & A & B & C & AB & AC & BC & ABC\\
(1) & +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\
a & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
b & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
ab & +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\
c & +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\
ac & +1 & +1 & -1 & +1 & -1 & +1 & -1 & -1\\
bc & +1 & -1 & +1 & +1 & -1 & -1 & +1 & -1\\
abc & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\
\end{matrix}
I know this is just a signal table with produtcts of $+ $and $-$, but is there any quick way to assemble it without having to look at the signs?
I would like to know it, because with the table is easier and faster to calculate the contrasts, otherwise I would have to do
$$A=(a-1)(b+1)(c+1)(d+1)$$
$$B=(a+1)(b-1)(c+1)(d+1)$$
$$.....$$
$$ABCD=(a-1)(b-1)(c-1)(d-1)$$
Is there any easier way to assemble this table?
As I need to make the table a hand to study for the exam, I was looking at it and I saw a few things about the contrasts
$(A,B,C,D)$: $+1$ in which the letter appears
$(AB,AC,BC,AD,BD,CD)$: $+1$ in $(1)$, in the rows where the pair appears and in the rows where one or two of the letters does not belong to the pair appears
$(ABC,ABD,ACD,BCD)$:$+1$ in rows which appears each letter of triple alone, row in which the triple appears and rows where the letter does not belong to triple appears (only in pairs)
$ABCD$: $+1$ in all pairs and $ABCD$
| Let's start by isolating the basis:
$$\begin{array} {r|r|rrr|rrrr}
& I & A & B & C & AB & AC & BC & ABC\\
(1) & +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\
a & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
b & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
ab & +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\
c & +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\
ac & +1 & +1 & -1 & +1 & -1 & +1 & -1 & -1\\
bc & +1 & -1 & +1 & +1 & -1 & -1 & +1 & -1\\
abc & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\
\end{array}$$
Columns $A,B,C$ are just counting in binary, but if you reordered the rows then they're still simple: $A$ is $+1$ iff $a$ is in the row label; etc.
Column $AB$ is just column $A$ pointwise multiplied by column $B$; similarly column $ABC$ is just column $AB$ pointwise multiplied by column $C$; so no cell in the table requires more than one multiplication.
The result is that it's a question of parity. For even-length column labels ($I$ counts as even because it's really a substitute for the empty string) you get $+1$ in rows where an even number of the letters occur (e.g. $ABCD$ is even, so it will be $+1$ in rows $(1)$, $ab$, $ac$, $ad$, $bc$, $bd$, $cd$, $abcd$); for odd-length column labels you get $+1$ in rows where an odd number of the letters occur (e.g. $ABC$ is odd, so it will be $+1$ in rows $a$, $b$, $c$, $ad$, $bd$, $cd$, $abc$, $abcd$).
Really there's no reason to prepare a table like this by hand. Whatever software you're using for the analysis can probably do it for you, but as an online option this program will generate a CSV file of the table. Just change the input value and press Run.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine the smallest disc in which all the eigen values of a given matrix lie Let , $$A=\left[\begin{matrix}1&-2&3&-2\\1&1&0&3\\-1&1&1&-1\\0&-3&1&1\end{matrix}\right]$$Which of the following is the smallest disc in $\mathbb C$ which contains all eigen values of $A$.
*
*$|z-1|\le 7$
*$|z-1|\le 6$
*$|z-1|\le 4$.
Characteristic polynomial of $A$ is $x^4-4x^3+21x^2-48x+46$. From this computing eigen values is very difficult in hand. Without computing eigen values how we can detect the required interval ?
Does there any other process??
| (Too long for a comment.)
This looks like a trick question and I don't know the trick, but you may try the following approach. Let $B=A-I$. The characteristic polynomial of $B$ is $x^4 + 15x^2 - 14x + 16$. As $15x^2-14x+16>0$ over $\mathbb R$, $B$ has no real eigenvalues. Since $B$ has zero trace and nonreal eigenvalues of a real matrix must occur in conjugate pairs, it follows that
$$
x^4+15x^2-14x+16=(x^2-2ax+r^2)(x^2+2ax+R^2)
$$
where $a\ne0$ is the real part of one of the eigenvalues of $B$ and $r,R$ ( with $(0<|a|<r\le R$) are the moduli of the eigenvalues of $B$. By comparing coefficients of both sides, we get
\begin{cases}
R^2 + r^2 - 4a^2 = 15,\\
2a(R^2-r^2) = 14,\\
r^2R^2 = 16.
\end{cases}
Substitute the last two equations into the first, we get
$$
\left(\frac{16}{r^2} + r^2 - 15\right)\left(\frac{16}{r^2} - r^2\right)^2 = 196.
$$
If you can show that $r>1$, then we may conclude from $r^2R^2=16$ that $R<4$ and hence $(3)$ is the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the limit? How to calculate this limit ?
$$\lim _{x\to 1}\frac{\sqrt{x^2-1}}{\sqrt[3]{x-1}}$$
An image for clarification.
| We can rewrite the function:
$$\frac{\sqrt{x^{2}-1}}{\sqrt[3]{x-1}}=(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{2}}(x-1)^{-\frac{1}{3}}=(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{2}-\frac{1}{3}}$$
So that
$$\lim_{x\to 1}\frac{\sqrt{x^{2}-1}}{\sqrt[3]{x-1}}=\lim_{x\to 1}(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{6}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Modified Leibnitz integral: $\lim\limits_{a \to\infty}\frac1a\int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx=?$ $\lim\limits_{a \to \infty} \frac{1}{a} \int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx $ ,where $a$ is a parameter.
ATTEMPT:- Let $I(a)=\frac{1}{a} \int _0^\infty \frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx$
Now by Leibnitz theorem, $I'(a)= -\int _0^\infty\frac{(1+x^2)(arccot(x))}{(1+x^4)a^2}dx$
Substituting $x=cot\theta.$
$\implies$ $I'(a)= -\int _0^\frac{\pi}{2}\frac{(\theta)(cosec^4\theta)}{(1+cot^4\theta)a^2}d\theta$
Also $I'(a)= -\int _0^\frac{\pi}{2}\frac{(\frac{\pi}{2})(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta +\int _0^\frac{\pi}{2}\frac{(\theta)(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta$
$\implies $$2I'(a)= -\int _0^\frac{\pi}{2}\frac{(\frac{\pi}{2})(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta+\int _0^\frac{\pi}{2}\frac{(\theta)(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta -\int _0^\frac{\pi}{2}\frac{(\theta)(cosec^4\theta)}{(1+cot^4\theta)a^2}d\theta$
Note:The last two integrals add to ZERO.
$\implies$ $2I'(a)=-\int _0^\frac{\pi}{2}\frac{(\frac{\pi}{2})(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta$
This integral can be easily evaluated by putting $tan\theta =t$
However I am getting $I(a)=\frac{\pi^2}{4\sqrt{2}a}.$ which is not the correct answer.
| It is not necessary to calculate $I(a)$. Noting that $\lim_{a\to\infty}\frac{x^2+ax+1}{a(x^4+1)}=\frac{x}{x^4+1}$ and
$$ \frac{x^2+ax+1}{a(x^4+1)}-\frac{x}{x^4+1}=\frac{x^2+1}{a(x^4+1)} $$
and $\int_0^\infty\frac{(x^2+1)\arctan\frac1x}{x^4+1}dx$ converges, we have, for big $M>0$,
\begin{eqnarray}
&&\bigg|\int_0^\infty\frac{(x^2+ax+1)\arctan\frac{1}{x}}{a(x^4+1)}dx-\int_0^\infty\frac{x\arctan\frac{1}{x}}{x^4+1}dx\bigg|\\
&=&\frac1{a}\int_0^\infty\frac{(x^2+1)\arctan\frac{1}{x}}{x^4+1}dx\\
&\le&\frac{M}{a}\to 0\text{ as }a\to\infty,
\end{eqnarray}
and hence
\begin{eqnarray}
\lim_{a\to\infty}\frac{1}{a}\int_0^\infty\frac{(x^2+ax+1)\arctan\frac{1}{x}}{a(x^4+1)}dx=\int_0^\infty\frac{x\arctan\frac{1}{x}}{x^4+1}dx.
\end{eqnarray}
Let
$$ I\equiv\int_0^\infty\frac{x\arctan\frac{1}{x}}{x^4+1}dx, J=\int_0^\infty\frac{x\arctan x}{x^4+1}dx.$$
It is easy to see $I=J$ by using $x\to\frac1x$ and $I+J=\frac{\pi^2}{8}$ by usinng $\arctan x+\arctan\frac1x=\frac{\pi}2$. Hence $I=\frac{\pi^2}{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-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)$$
Which is at least a product of linear factors.
However I am now stuck as to how to proceed.
Also I would prefer the $(b^2-a^2)$ and the $(c^2-b^2)$ factors of my second line of working to be in the form $(a^2-b^2)$ and $(b^2-c^2)$, for 'neatness' , if possible
| The key is to look for symmetry, so write it as symmetrically as possible:$$a^3(c-b)+b^3(a-c) +c^3(b-a).$$This has cyclic symmetry; so, if it has linear factors, then they must be invariant under cyclic permutation of $a$, $b$, and $c$. The factors have to include negative signs, so try the simplest first: say $b-c$. Putting $b=c$ makes the first term zero, and it's easy to see that the remaining terms cancel too under this condition. So $b-c$ is a factor, and hence $c-a$ and $a-b$ must be too. The remaining factor has to be linear to make up the quartic; so it must be $\pm(a+b+c)$, which is the only cyclically symmetric linear factor possible (barring numerical multiples, because we don't have numbers other than $\pm1$). It's now easy to check whether the plus or minus sign should be taken.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove an inequality I want to show that following:
$$\left(\frac{n^2-1}{n^2}\right)^n\sqrt{\frac{n+1}{n-1}}\leq 1; ~~n\geq 2$$ and $n$ is an integer.
After some simplifications, I got left hand-side as
$$LHS:\left(1-\frac{1}{n}\right)^{n-\frac{1}{2}} \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$$
It is clear that the 1st term is less than 1, but I do not have any clue how I can show that multiplication is less than 1.
Can someone give me some hints?
| Consider on $(-1,1)$ the function (with motivation $x=\frac1n$)
$$
f(x)=\ln\bigl(1-x^2\bigr)+\frac x2\bigl(\ln(1+x)-\ln(1-x)\bigr)-\frac16\ln\bigl(1-x^4\bigr)
$$
Then
$$
f'(x)=-\frac{x}{1-x^2}+\frac12\bigl(\ln(1+x)-\ln(1-x)\bigr)+\frac16\frac{4x^3}{1-x^4}
$$
where we see that $f(0)=0=f'(0)$. The next derivative is
\begin{align}
f''(x)&=-\frac{2x^2}{(1-x^2)^2}+\frac{2x^2}{1-x^4}+\frac83\frac{x^6}{(1-x^4)^2}\\
&=-\frac{4x^4}{(1-x^4)(1-x^2)}+\frac83\frac{x^6}{(1-x^4)^2}\\
&=-\frac43·\frac{3x^4+x^6}{(1-x^4)^2}
\end{align}
which is always negative for $x\ne 0$. Then for $x>0$ the linear Taylor polynomial with quadratic remainder term gives
$$
f(x)=\frac12f''(\theta x)x^2<0
$$
and taking the exponential of $f(x)<0$ gives
$$
\bigl(1-x^2\bigr)·\left(\frac{1+x}{1-x}\right)^{\frac x2}<\bigl(1-x^4\bigr)^{\frac16}.
$$
Replacing $x=\frac1n$ and taking the $n$th power of the inequality results in the requested inequality,
$$
\left(1-\frac1{n^2}\right)^n·\sqrt{\frac{n+1}{n-1}}<\left(1-\frac1{n^4}\right)^{\frac n6}<1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Give bases for col(A) and null(A) I have
A= $\begin{bmatrix}1&1&-3\\0&2&1\\1&-1&-4\end{bmatrix}$
I row reduce it to
$\begin{bmatrix}1 &0& -3.5\\0&1&.5\\0&0&0\end{bmatrix}$
How do I find col(A) from the above info? Is it the pivot points correspond to the columns?
So col(A) would be $\begin{bmatrix}1\\0\\1\end{bmatrix}$and $\begin{bmatrix}1\\2\\-1\end{bmatrix}$
and for null(A) I got
$\begin{bmatrix}3.5\\-.5\\1\end{bmatrix}$
| To clear up confusion, work out the steps.
*
*Form the augmented matrix
$$
% A I
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}_{3} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rcr|ccc}
1 & 0 & 1 & 1 & 0 & 0 \\
1 & 2 & -1 & 0 & 1 & 0 \\
-3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right]
$$
*Clear column 1.
$$
% E
\left[
\begin{array}{rcc}
1 & 0 & 0 \\
-1 & 1 & 0 \\
3 & 0 & 1 \\
\end{array}
\right]
% in
\left[
\begin{array}{rcr|ccc}
1 & 0 & 1 & 1 & 0 & 0 \\
1 & 2 & -1 & 0 & 1 & 0 \\
-3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right]
=
% out
\left[
\begin{array}{ccr|rcc}
1 & 0 & 1 & 1 & 0 & 0 \\
0 & 2 & -2 & -1 & 1 & 0 \\
0 & 1 & -1 & 3 & 0 & 1 \\
\end{array}
\right]
%
$$
*Clear column 2.
$$
% E
\left[
\begin{array}{crc}
1 & 0 & 0 \\
0 & \frac{1}{2} & 0 \\
0 & -\frac{1}{2} & 1 \\
\end{array}
\right]
% in
\left[
\begin{array}{ccr|rcc}
1 & 0 & 1 & 1 & 0 & 0 \\
0 & 2 & -2 & -1 & 1 & 0 \\
0 & 1 & -1 & 3 & 0 & 1 \\
\end{array}
\right]
=
% out
\left[
\begin{array}{ccr|rrc}
1 & 0 & 1 & 1 & 0 & 0 \\
0 & 1 & -1 & -\frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & 0 & \frac{7}{2} & -\frac{1}{2} & 1 \\
\end{array}
\right]
%
$$
The
$$
\begin{align}
%
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}_{3} \\
\end{array}
\right]
&=
%
\left[
\begin{array}{rcr|ccc}
1 & 0 & 1 & 1 & 0 & 0 \\
1 & 2 & -1 & 0 & 1 & 0 \\
-3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right] \\
%
&\qquad \qquad \qquad
\Downarrow \\
%
\left[
\begin{array}{c|c}
\mathbf{E_{A}} & \mathbf{R} \\
\end{array}
\right]
&=
%
\left[
\begin{array}{ccr|rrc}
\boxed{1} & 0 & 1 & 1 & 0 & 0 \\
0 & \boxed{1} & -1 & -\frac{1}{2} & \frac{1}{2} & 0 \\\hline
0 & 0 & 0 & \color{red}{\frac{7}{2}} & \color{red}{-\frac{1}{2}} & \color{red}{1} \\
\end{array}
\right]
%
\end{align}
$$
The unit pivots (boxed) in the matrix $\mathbf{E_{A}}$ identifies the fundamental columns of the images. The red vector in $\mathbf{R}$ is the span of the null space:resolution
$$
\boxed{
\color{blue}{\text{col } \mathbf{A}} \oplus
\color{red} {\text{null } \mathbf{A}} =
\color{blue}{\mathcal{R}\left( \mathbf{A}\right)} \oplus
\color{red} {\mathcal{N}\left( \mathbf{A}^{*}\right)} =
%
\color{blue} {
\text{span } \left\{ \,
\left[
\begin{array}{r}
1 \\
1 \\
-3 \\
\end{array}
\right],
%
\left[
\begin{array}{r}
0 \\
2 \\
1 \\
\end{array}
\right]
\, \right\}}
% % %
\oplus
% % %
%
\color{red} {
\text{span } \left\{ \,
\left[
\begin{array}{r}
\frac{7}{2} \\
-\frac{1}{2} \\
1 \\
\end{array}
\right]
\, \right\}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Using hypergeometric functions to solve this integral After looking at calculations, I realized that the exponent needs to be 1/4 instead of -1/4
I have this equation and I am trying to solve the integral of it.
$$((R^2) - (y^2))^{1/4} dy$$
I tried to put it into wolfram alpha, and I got an answer, but I wanted to know how they arrived at the answer.
Any advice would be greatly appreciated. If you could please show me how to do this integral, I would be appreciate it very much.
I know you need to use hyper geometric functions; however, I am not sure how.
Thank-you very much
| I'm assuming the integral in question is
$$\int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy$$
as these types are quite common. If not, a very modest change can be made (just let the upper bound be $x$ instead of $R$ in the following analysis and leave the hypergeometric function in terms of $x$ and ignore the stuff concerning Gauss' theorem). Any time you come across an integral of something like $(x+y)^{\alpha}$, you should think about using the binomial theorem. The binomial theorem states that
$$ (x+y)^{\alpha} = \sum_{n=0}^{\infty} \frac{\Gamma(\alpha+1)}{\Gamma(\alpha-n+1)n!} x^{\alpha-n} y^n.$$
In our case, we get
$$ (R^2-y^2)^{-\frac{1}{4}} = \sum_{n=0}^{\infty} \frac{(-1)^n\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}-n\right) n!} R^{-\frac{1}{2}-2n}y^{2n}.$$
Therefore
$$\int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy = \sum_{n=0}^{\infty} \frac{(-1)^n\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}-n\right) n!}R^{-\frac{1}{2} -2n}\int_0^R y^{2n}\,dy.$$
We can interchange integral and sum via uniform convergence arguments (and considering the integral from $[0,R']$ where $R' < R$, then letting $R'\to R$). The integral gives nothing more than $\frac{1}{2n+1} R^{2n+1}$ so we are left with
$$ \int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy = R^{\frac{1}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}-n\right) n!(2n+1)}. $$
This series is more than a little daunting but it can be conveniently written as $_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};1\right).$ Strictly speaking, $_2F_1$ does not exist at $1$ as the series diverges, but by considering the limit from the left, we can assign it a value at $1$. (The value could be infinity.)
Gauss' theorem states that for $_2F_1(a,b;c,z)$, the value at $1$ exists and is finite if $\Re c > \Re(a+b)$ and the value is
$$_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.$$
In our case, $a = \frac{1}{4}$, $b= \frac{1}{2}$ and $c=\frac{3}{2}$ so the conditions for Gauss' theorem are met and we get that the value of the series is
$$\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{2}-\frac{1}{4}-\frac{1}{2}\right)}{\Gamma\left(\frac{3}{2}-\frac{1}{4}\right)\Gamma\left(\frac{3}{2}-\frac{1}{2}\right)} = \frac{\frac{1}{2}\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}. = \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\frac{1}{4}\Gamma\left(\frac{1}{4}\right)} = \frac{2\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}.$$
Substituting this into $(1)$, we get
$$ \int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy = \frac{2\sqrt{\pi R}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Uniform convergence of $\sum_{n \ge 0} (-1)^n(1-x)x^n$ Let $f_n: [0,1] \to \mathbb{R}$, $f_n(x)=(-1)^n (1-x)x^n, \forall x\in [0,1], n \ge 0$. Prove that $\sum\limits_{n \ge 0} f_n(x)$ convergenes uniformly.
Let $S_n(x) = \sum\limits_{k = 0}^n f_k(x), \forall x \in [0,1], n \ge 0$.
Then,
$S_n(x)=1-x-x+x^2+x^2-x^3-x^3+x^4+...+(-1)^{n-1}x^{n-1}+(-1)^nx^n+(-1)^nx^n+(-1)^{n+1}x^{n+1}$
so
$S_n(x)=1-2x+2x^2-...+(-1)^n\cdot 2x^n+(-1)^{n+1}x^{n+1}, \forall x\in [0,1], n \ge 0$.
We know that $1+x^{2p+1}=(1+x)(x^{2p}-x^{2p-1}+x^{2p-2}-....+x^2-x+1)$, so, for $n=2p$, $p \ge 0$, we may write:
$$ S_{2p}= 2(1-x+x^2-...+x^{2p})-1-x^{2p+1}=2 \cdot \frac{1+x^{2p+1}}{1+x}-1-x^{2p+1}$$
Now, it's easy to see that $\lim\limits_{p \to \infty}S_{2p}(x)=\frac{1-x}{1+x}, \forall x\in [0,1]$.
For $n=2p+1$, $p \ge 0$, we have:
$$ S_{2p+1}(x)=2(1-x+x^2-...+x^{2p+2})-1-x^{2p+2}=2\cdot \frac{1+x^{2p+3}}{1+x} -1 -x^{2p+2} $$
In this case we have: $\lim\limits_{p \to \infty} S_{2p+1}(x) = \frac{1-x}{1+x}, \forall x\in [0,1]$.
So, $\lim\limits_{n \to \infty} S_n(x) =\frac{1-x}{1+x}, \forall x\in [0,1]$.
Now, it remains to prove that $\lim\limits_{n\to \infty} \sup\limits_{x \in [0,1]} |S_n(x)-\frac{1-x}{1+x}|=0$.
It's easy to see that $g_n(x)=|S_n(x)-\frac{1-x}{1+x}|=x^{n+1} \cdot \frac{1-x}{1+x}$.
How can I continue?
| Hints : with your notation, $g_n'(x) = \frac{ x^{n-1}(-nx^2+n-2x)}{(x+1)^2}$. The sign of $g_n'(x)$ on $[0,1]$ is the same as the sign of $-nx^2+n-2x$ and consequently is non-negative on $\left[0, \frac{\sqrt{n^2+1}-1}{n}\right]$ and non-positive on $\left[\frac{\sqrt{n^2+1}-1}{n},1\right]$. Consequently, the maximum of $g_n$ on $[0,1]$ is reached at $x= \frac{\sqrt{n^2+1}-1}{n}$. A study of the convergence of $g_n\left( \frac{\sqrt{n^2+1}-1}{n} \right)$ should lead to a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How do I complete the proof of proving the $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $ $$\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $$
Proof: Let $\epsilon > 0$
Then, $$ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -\frac { (x+1)^{ 2 } }{ (x+1)^{ 2 } } \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -\frac { x^{ 2 }+2x+1 }{ (x+1)^{ 2 } } \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -2x }{ (x+1)^{ 2 } } \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -2 }{ x^{ 3 }+2x^{ 2 }+x } \right| <\epsilon $$
$$\Longleftrightarrow \frac { 2 }{ \left| x^{ 3 }+2x^{ 2 }+x \right| } <\epsilon $$
Now, we will calculate the lower bound on $x$ to force $ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $ to hold true.
Without loss of generality, assume that $x>0$, then we get: $$\frac { 2 }{ x^{ 3 }+2x^{ 2 }+x } <\epsilon $$
$$\Longrightarrow 2<\epsilon (x^{ 3 }+2x^{ 2 }+x)$$
$$\Longrightarrow 2<\epsilon x^{ 3 }+\epsilon 2x^{ 2 }+\epsilon x$$
At this point I get stuck. How do I go from here? I can't find a way in which I could manipulate this to get it in the form $x>...$ A hint in the right direction would be beneficial.
| Given $\epsilon > 0$ we want to find an $M$ such that $x > M \Rightarrow | \cdots | < \epsilon$.
Remember you just need to find an $M$; make life as easy for yourself as you can and don't care about finding a low $M$.
Hence, I would simplify and write for all $x > 0$,
$$ \left| \frac{x^2+1}{(x+1)^2} -1 \right| = \left| \frac { -2x }{ (x+1)^{ 2 } } \right| \leq \frac{2x}{x^2} = \frac{2}{x}$$
Thus given $\epsilon > 0$, choose $M = 2/\epsilon$ (which is positive).
Then
$$x > M \ \Rightarrow \ x > \frac 2\epsilon \ \Rightarrow \ \frac 2x < \epsilon \ \Rightarrow \ \left| \frac{x^2+1}{(x+1)^2} -1 \right| < \frac 2x <\epsilon$$
Added (answer to question in comments):
We have for all positive $x$,
$$\left| \frac{x^2+1}{(x+1)^2} -1 \right| < \frac 2x .$$
Hence given $\epsilon > 0$ if we can find an $M > 0$ such that
$$x > M \quad \Rightarrow \quad \frac 2x < \epsilon$$
this implies
$$x > M \quad \Rightarrow \quad \left| \frac{x^2+1}{(x+1)^2} -1 \right| < \epsilon .$$
Now, bounding $\frac 2x$ by $\epsilon$ is straight forward:
$$\frac 2x < \epsilon \quad\text{ iff }\quad x > \frac 2\epsilon$$
Hence the choice of $M = 2/\epsilon$ does the trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the following function:
$$\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)?
| The usual rationalization:
$\begin{array}\\
\sqrt{x+1}-\sqrt{x}
&=(\sqrt{x+1}-\sqrt{x})\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\\
&=\frac{1}{\sqrt{x+1}+\sqrt{x}}\\
&< \frac1{2\sqrt{x}}
\end{array}
$
| {
"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": 0
} |
Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a_k = (-1)^k\sigma _k(x_1,x_2,x_3,x_4), k\in \{1,2,3,4\}$, where $\sigma _k$ is the $k$-th elementary symmetrical polynomial. Therefore:
$x_1+x_2+x_3+x_4 = 2$
$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = 0\ (*)$
$x_1x_2x_3 +x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = 1$
$x_1x_2x_3x_4 = 2/7$
Now how to determine the sum of the cubed roots?
$2^3 = 8= (x_1+x_2+x_3+x_4)(x_1+x_2+x_3+x_4)^2 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2 + 2(*))$
Here's where things go out of hand:
$(x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2) = (x_1^3 + x_2^3 + x_3^3+x_4^3) + x_1^2(x_2+x_3+x_4)+x_2^2(x_1+x_3+x_4)+x_3^2(x_1+x_2+x_4)+x_4^2(x_1+x_2+x_3) = 8$
What should I do here?
| Use Newton's relations between sums of powers $p_k=\sim_ix_i^k$ and the elementary symmetric functions:
\begin{align*}
p_1&=\sigma_1,\\
p_2&=\sigma_1p_1-2\sigma_2,\\
p_3&=\sigma_1p_2-\sigma_2p_1+3\sigma_3.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How should I integrate $\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ } $? How should I integrate $\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ } $ ?
I tried substituting $x=1/t$,but it is not working.Could someone help?Thanks.
| Let $x=\sin \theta$ and then
\begin{eqnarray}
\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ }&=&\int _{ 0 }^{ \arcsin\frac{1}{\sqrt{3}} } \frac {d\theta }{1+\sin^2\theta}\\
&=&\frac{1}{\sqrt 2}\arctan(\sqrt{2}\tan\theta)\bigg|_{ 0 }^{\arcsin\frac{1}{\sqrt{3}}}\\
&=&\frac{1}{\sqrt 2}\arctan(\sqrt{2}\cdot\frac{1}{\sqrt2})\\
&=&\frac{\pi}{4\sqrt 2}.
\end{eqnarray}
Here $\arctan(\arcsin\frac{1}{\sqrt{3}})=\frac{1}{\sqrt2}$ and
$$ \int\frac{1}{1+\sin^2\theta}d\theta=\int\frac{1}{1+2\tan^2\theta}d\tan\theta=\frac{1}{\sqrt2}\arctan(\sqrt2\tan\theta)+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing common volume of sphere and cylinder using triple integrals Find the volume of material cut from the solid sphere $x^2 + y^2 + z^2 \leq 9$ by the cylinder $r = 3\sin(\theta)$ using cylindrical coordinates.
| In cylindrical coordinates the equation of the sphere is: $r^2+z^2=9$ so, since $r\le 3 \sin \theta$ the volume is limited by the inequalities:
$$
0\le \theta \le 2\pi \qquad 0\le r \le 3 \sin \theta \qquad-\sqrt{9-r^2}\le z \le \sqrt{9-r^2}
$$
so, given the symmetry, the volume is:
$$
V=4 \int_0^{\frac{\pi}{2}}\int_0^{3\sin \theta}\int_0^{\sqrt{9-r^2}}rdzdrd\theta
$$
where $rdzdrd\theta$ is the volume element in cylindrical coordinates.
so we have:
$$
V=4\int_0^{\frac{\pi}{2}}\int_0^{3\sin \theta}r\sqrt{9-r^2}drd\theta=
4\int_0^{\frac{\pi}{2}}\sqrt{(9-9\sin^2 \theta)^3}drd\theta=-36\int_0^{\frac{\pi}{2}}\left(\cos^3 \theta-1 \right)d\theta
$$
can you do from this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this infinite product identity? How can I prove the following identity?
$$\large\prod_{k=1}^\infty\frac1{1-2^{1-2k}}=\sum_{m=0}^\infty\left(2^{-\frac{m^2+m}{2}}\prod_{n=1}^\infty\frac{1-2^{-m-n}}{1-2^{-n}}\right)$$
Numerically both sides evaluate to
$$2.38423102903137172414989928867839723877...$$
| The identity is in fact true for any $0 < x < 1$: $$\prod_{k=1}^{\infty}\frac{1}{1-x^{2k-1}}=\sum_{m=0}^{\infty}\left(x^{\frac{m^2+m}{2}}\prod_{n=1}^{\infty}\frac{1-x^{m+n}}{1-x^{n}}\right) = \sum_{m=0}^\infty\left(x^{\frac{m^2+m}{2}}\prod_{n=1}^{m}\frac{1}{1-x^{n}}\right)$$
We note the telescopic product: $$\displaystyle \prod_{k=1}^{\infty}\frac{1}{1-x^{2k-1}} = \prod_{k=1}^{\infty} \frac{1-x^{2k}}{1-x^k} = \prod_{k=1}^{\infty}(1+x^k)$$
On the other hand the expression:
\begin{align}\prod_{k=1}^{\infty}(1+x^k) &= \sum_{n=0}^{\infty} \left(\sum\limits_{1\le j_1 < j_2 < \cdots < j_n} x^{j_1+\cdots +j_n}\right)\tag{1}\\&= \sum_{n=0}^{\infty} \left(\sum\limits_{1\le k_1,k_2, \cdots ,k_n} x^{nk_1+(n-1)k_2\cdots +k_n}\right)\tag{2}\\&= \sum_{n=0}^{\infty} \left(\sum\limits_{k_1 \ge 1} x^{nk_1}\sum\limits_{k_2 \ge 1}x^{(n-1)k_2} \cdots \sum\limits_{k_n \ge 1}x^{k_n}\right)\tag{3}\\&= \sum\limits_{n=0}^{\infty} \left(\frac{x^n}{1-x^n}\cdot \frac{x^{n-1}}{1-x^{n-1}}\cdots\frac{x}{1-x}\right)\\&= \sum\limits_{n=0}^{\infty} x^{\frac{n^2+n}{2}}\prod\limits_{m=1}^{n}\frac{1}{1-x^m}\end{align}
Justifications:
$(1)$ Coefficient of $z^n:$ in the infinite product $\displaystyle \prod\limits_{k=1}^{\infty}(1+x^kz)$ being $\displaystyle \left(\sum\limits_{1\le j_1 < j_2 < \cdots < j_n} x^{j_1+\cdots +j_n}\right)$.
$(2)$ Made the change of variable $k_m = j_m - j_{m-1}$ for $m \ge 1$ where, $j_0 = 0$.
Then note that $j_1+\cdots +j_n = nk_1+(n-1)k_{2}+\cdots + k_n$
$(3)$ Used the formula for infinite geometric progression: $\displaystyle \sum\limits_{k\ge 1} x^{mk} = \frac{1}{1-x^m}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Distance between points - equation of a line I have worked on this particular example:
The distance between the point $M_1(3,2)$ and $A$ is $2\sqrt5$ and the distance between the point $M_2(-2, 2)$ and $B$ is $\sqrt5$. Come up with a equation for the line going through $A$ and $B$.
I have tried playing around with the equations for the circles with centers in $M_1$ and $M_2$ respectively (the radii being the distance between $M_1$ and $A$, that is the distance between $M_2$ and $B$ for the other circle), but I am stuck.
Would appreciate any help.
Thanks in advance.
P.S. I am probably wrong but, isn't there infinitely many solutions since with the data I'm given I am basically being asked to come up with a particular equation for a line connecting any two points on those circles ?
| Point on the circle with radius $2\sqrt{5}$ and center $(3,2)$ is at:
$$ A = (2\sqrt{5} \cos(a) + 3, 2\sqrt{5} \sin(a) + 2)$$
The other circle's point is at:
$$ B = (\sqrt{5} \cos(b) - 2, \sqrt{5} \sin(b) + 2) $$
Then slope of the line is:
$ m = \frac{B_y - A_y}{B_x-A_x} = \frac { \sqrt{5} \sin(b) + 2 - (2\sqrt{5} \sin(a) + 2) }{ \sqrt{5} \cos(b) - 2 - (2\sqrt{5} \cos(a) + 3)} = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5}$
Then the equation of the line is:
$$ y-A_y = m(x-A_x)$$
$$ y-(2\sqrt{5} \sin(a) + 2) = m(x-(2\sqrt{5} \cos(a) + 3)) $$
$$ y-(2\sqrt{5} \sin(a) + 2) = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5} (x-(2\sqrt{5} \cos(a)+3)) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Are these trigonometric expressions for the ceiling and floor functions correct? I believe that I have found a trigonometric expression for both the ceiling and floor function, and I seek confirmation that it is, indeed, correct.
Update.
$$\begin{align}
\lfloor x \rfloor &= x - \frac12+f(x) \\[4pt]
\lceil x \rceil &= x + \frac12+g(x)
\end{align}$$
where
$$\begin{align}
f(x) &= \begin{cases}
\frac12, & x\in\Bbb{Z} \\[4pt]
0, &x=\frac12n, n\in\Bbb{Z} \\[4pt]
\frac1\pi \tan^{-1}(\cot(\pi x)), &\text{otherwise}
\end{cases} \\[10pt]
g(x) &= \begin{cases}
-\frac12, & x\in\Bbb{Z} \\[4pt]
0, &x=\frac12n, n\in\Bbb{Z} \\[4pt]
\frac1\pi \tan^{-1}(\cot(\pi x)), &\text{otherwise}
\end{cases}
\end{align}$$
| If you're using the single-valued $\arctan$, i.e. $-\frac{\pi}{2} \leq \arctan(x) \leq \frac{\pi}{2}$, which is what all computer languages I've seen use, this works perfectly fine.
Proof:
Let $x = s(a + b)$, where $a \in \mathbb{N}$, $0 \leq b < 1$ and $s \in \{-1,1\}$. We can call $a$ the absolute integer part, $b$ the absolute fractional part, and $s$ is $x$'s sign.
Then:
$$
\begin{align*}
\frac{\arctan(\cot(\pi x))}{\pi}
&= \frac{\arctan(\cot(\pi s(a + b)))}{\pi}\\[1em]
&= \frac{\arctan(s\cdot\cot(\pi a + \pi b)))}{\pi}\\[1em]
&= \frac{\arctan(s\cdot\cot(\pi b))}{\pi}\\[1em]
&= \frac{\arctan(s\cdot\tan(\frac{\pi}{2} - \pi b))}{\pi}\\[1em]
&= \frac{\arctan(s\cdot\tan(\pi(\frac{1}{2} - b)))}{\pi}\\[1em]
&= \frac{\arctan(\tan(\pi s(\frac{1}{2} - b)))}{\pi}\\[1em]
&= \frac{\pi s(\frac{1}{2} - b)}{\pi}\\[1em]
&= s\bigg(\frac{1}{2} - b\bigg)\\[1em]
\end{align*}
$$
Note that we can assert that $\arctan(\tan(\pi s(\frac{1}{2} - b)) = \pi s(\frac{1}{2} - b)$ without issues only because
$$
\begin{alignat*}{5}
0 && \quad \leq && b \quad && < \quad && 1 &&\implies\\[0.5em]
0 && \quad \geq && -b \quad && > \quad && -1 &&\implies\\[0.5em]
\frac{1}{2} && \quad \geq && \quad \frac{1}{2} - b \quad && > \quad && -\frac{1}{2} &&\implies\\[0.5em]
\frac{\pi}{2} && \quad \geq &&\ \ \ \pi\Bigg(\frac{1}{2} - b\Bigg) \quad && > \quad && -\frac{\pi}{2} &&\\
\end{alignat*}
$$
$\arctan(\tan(x)) = x$ only holds for $x$ in that range. For example, $\arctan(\tan(x)) = x - \pi$ if $\frac{\pi}{2} < x < \pi$.
With that said...
$$
\begin{align*}
\lfloor x \rfloor &= x - \frac{1}{2} + s\bigg(\frac{1}{2} - b\bigg)\\[0.5em]
&= s(a + b) - \frac{1 - s}{2} - sb\\[0.5em]
&= s(a + b - b) - \frac{1 - s}{2}\\[0.5em]
&= sa - \frac{1 - s}{2}\\[0.5em]
\end{align*}
$$
If $s=1$, $\lfloor x\rfloor = a - \frac{1 - 1}{2} = a$.
If $s=-1$, $\lfloor x\rfloor = -a - \frac{1 + 1}{2} = -a - 1 = -(a + 1)$.
This works because $\lfloor \pm y.uwv \rfloor$ is y if the number is positive, and $-(y + 1)$ if the number is negative. e.g. $\lfloor 1.2\rfloor = 1$ but $\lfloor -1.2\rfloor = -2$.
Proving that your $ceil$ function works should go about the same way.
I realize I am three years late to this but I initially was going to post my own trigonometric floor:
$$
\lfloor x \rfloor = x - \text{frac}(x)\\[1.4em]
%
\text{frac}(x) = \text{sgn}\Big(\sin(2\pi x)\Big)\Bigg( \frac{\cos^{-1}(\cos(2\pi x))}{2\pi} - \frac{1}{2} \Bigg) + \frac{1}{2}\\[1.4em]
%
\text{sgn}(x) = \frac{1}{2}\Bigg(\frac{\cot^{-1}(x) - \cot^{-1}(-x)}{\big|\cot^{-1}(x)\big|}\Bigg)
$$
But yours seems a bit simpler.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Polynomial Equations Root Finding If one of the roots of the Equation :
$$2000x^6 + 100x^5 + 10x^3 + x - 2 = 0$$ ;
Is of the form $$(m + √n)/r$$. Where m is a non zero integer and n and r are relatively coprime.
We have to find $$(m+n+r)/100$$ .
I am not able to factorise this as I am not able to guess any roots.
I could only comment that there's a root between 0 and 1 which doesn't help
Using sum , products of roots seems too tough to even start...
Any hint would do ...
Thanks in advance.
| Given $$2000x^6+100x^5+10x^3+x-2=0\Rightarrow (2000x^6-2)+(100x^5+10x^3+x)=0$$
Now Divide it by $x^3\;,$ Where $x\neq 0\;,$ We get
$$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\left[(10x)^2+1+\frac{1}{x^2}\right]=0$$
So $$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\frac{(10x)^3-\frac{1}{x^3}}{\left[10x-\frac{1}{x}\right]}=0$$
So $$\left[(10x)^3-\frac{1}{x^3}\right]\cdot \left[2+\frac{1}{\left(10x-\frac{1}{x}\right)}\right]=0$$
So $$\left[(10x)^2+1+\frac{1}{x^2}\right]\cdot \left(20x^2+x-2\right)=0$$
So we get $100x^4+10x^2+1=0$ or $20x^2+x-2=0$
Now $100x^4+10x^2+1>0\;\forall x\in \mathbb{R}$. So here $$\displaystyle 20x^2+x-2=0\Rightarrow x = \frac{-1\pm \sqrt{1+160}}{2\times 20} = \left(\frac{-1\pm \sqrt{161}}{40}\right)$$
So after camparing with $\displaystyle \frac{m+\sqrt{n}}{r}\;,$ We get $m+n+r=-1+61+40=200$
So $$\displaystyle \frac{m+n+r}{100} = \frac{200}{2} =2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\frac{n}{2}\rfloor\\\lfloor\frac{n}{2}\rfloor\end{pmatrix} = F_{(n+1)}$$
Example: ${6\choose0} + {5\choose1} + {4\choose2} + {3\choose3} = 13 = F_{7}$
| Another way is using generating functions. The g.f. of $\displaystyle a_n = \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k} $ is
$$ \eqalign{G(x) & =\sum_{n=0}^\infty \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k} x^n\cr &= \sum_{k=0}^\infty \sum_{n=2k}^\infty {n-k \choose k} x^n\cr
&= \sum_{k=0}^\infty x^{2k} \sum_{m=0}^\infty {m+k \choose k} x^m\cr
&= \sum_{k=0}^\infty \dfrac{x^{2k}}{(1-x)^{k+1}} \cr
&= \dfrac{1}{1-x}\; \dfrac{1}{1 - x^2/(1-x)}\cr& = \dfrac{1}{1-x-x^2}}$$
The g.f. of the Fibonacci numbers $F_n$ is
$$ F(x) = \sum_{n=0}^\infty F_n x^n = \dfrac{x}{1+x-x^2} $$
with $F_0 = 0$, so the g.f. of $F_{n+1}$ is $F(x)/x = 1/(1+x-x^2) = G(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Inverse Trigonometry Question (Stuck in Algebraic Simplification) The original question:
If x and y are of same sign, then find the value of $$\cfrac{x^3}{2} \csc^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right) + \cfrac{y^3}{2} \sec^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right) $$
This is my attempt to the question:
$$\cfrac{x^3}{2\sin^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right)} + \cfrac{y^3}{2\cos^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right)} $$
$$
=\cfrac{x^3}{1-\cos(\tan^{-1} (x/y))} + \cfrac{y^3}{1+\cos(\tan^{-1}(y/x))} $$
$$
=\cfrac{x^3}{1 - \cfrac{|y|}{\sqrt{x^2+y^2}}} + \cfrac{y^3}{1+\cfrac{|x|}{\sqrt{x^2+y^2}}} $$
I'm not sure how to simplify this further. I've been trying to simplify this further for a long time but couldn't get any desirable answer.
Options are:
(A) $(x-y)(x^2+y^2)$
(B) $(x+y)(x^2-y^2)$
(C) $(x+y)(x^2+y^2)$
Also, any other better/alternate way to approach the question is welcomed.
| i will take $x,y$ to have the same sign. using double angle formula, we have $$\tan^{-1}(y/x) = 2t, \tan^{-1}(x/y) = \pi/2 - 2t, r = \sqrt{x^2 + y^2}.$$ then
$$\begin{align} \cfrac{x^3}{2} \csc^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right) + \cfrac{y^3}{2} \sec^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right) &= \frac{x^3}{2\sin^2(\pi/4 - t)} + \frac{y^3}{2\cos^2 t}\\
&=\frac{x^3}{1 - \cos (\pi/2 - 2t)} + \frac{y^3}{1 + \cos 2t }\\
&=\frac{x^3}{1 - \sin 2t} + \frac{y^3}{1 + \cos 2t }\\
&=\frac{x^3}{1 - \frac y r} + \frac{y^3}{1 + \frac xr }\\
&= \frac{r\left(x^3(r+x)+y^3(r-y)\right)}{(r-y)(r+x)}\\
&=\frac{r\left( r(x^3+y^3)+(x^4 - y^4)\right)}{(r-y)(r+x)} \\
&=\frac{r(x+y)\left( r(x^2-xy+y^2)+(x - y)(x^2+y^2)\right)}{(r-y)(r+x)}\\
&=\frac{r^2(x+y)\left( r^2-xy +r(x - y)\right)}{(r-y)(r+x)}\\
&=r^2(x+y)=(x^2+y^2)(x+y) \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$ Find the value of the expression $\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$
I identified that $\frac{\pi}{16}+\frac{13\pi}{16}=\frac{5\pi}{16}+\frac{9\pi}{16}=\frac{14\pi}{16}$
$\tan(\frac{\pi}{16}+\frac{13\pi}{16})=\tan(\frac{5\pi}{16}+\frac{9\pi}{16})$
$\frac{\tan\frac{\pi}{16}+\tan\frac{13\pi}{16}}{1-\tan\frac{\pi}{16}\tan\frac{13\pi}{16}}=\frac{\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}}{1-\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}}$
But i am stuck here.Please help me.Thanks.
| As $\tan4\left(\dfrac\pi4+x\right)=\tan(\pi+4x)=\tan4x,$
If $\tan4x=\tan4A,4x=n\pi+4A\implies x=\dfrac{n\pi}4+A$ where $n=0,1,2,3$
as $$\tan4x=\dfrac{4\tan x-\binom41\tan^3x}{1-\binom42\tan^2x+\tan^4x}$$
$$\dfrac{4\tan x-\binom41\tan^3x}{1-\binom42\tan^2x+\tan^4x}=\tan4A$$
$$\iff\tan4A\tan^4x+4\tan^3x-6\tan4A\tan^2x-4\tan x+\tan4A=0$$
$$\implies\sum_{r=0}^1\tan\left(r\dfrac\pi4+x\right)=-\dfrac4{\tan4A}$$
Here $4A=\dfrac\pi4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more complicated then.
Is there a way to solve this using only basic formulas?
| There is nothing simple about this. My take would be to write
$$
\frac{x+1}{x^2-x+8}=\frac{x+1}{((x-1/2)^2+31/4)^3}
=\frac12\,\frac{2(x-1/2)+3}{((x-1/2)^2+31/4)^3}\\
=\frac12\,\frac{2(x-1/2)}{((x-1/2)^2+31/4)^3}
+
\frac12\,\frac{3}{((x-1/2)^2+31/4)^3}
$$
Now the first term is a straightforward substitution. The second one would require a couple rounds of integration by parts.
According to Wolfram Alpha, an antiderivative is given by
$$
\int \frac{x+1}{(x^2-x+8)^3}\, dx =
\frac{\frac{(31 (18 x^3-27 x^2+246 x-599)}{(x^2-x+8)^2}+36 \sqrt{31} \arctan\left(\frac{2 x-1}{\sqrt{31}}\right)}{59582}+c.
$$
So you cannot expect anything too straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\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\\$$
| Check that for $|x| < r < 1$,
$$\frac{nx^{n-1}}{1+x^n}, \frac{nx^{n-1}}{1-x^n}, \frac{2nx^{2n-1}}{1-x^n}$$ are all $O(nr^n)$, which is summable.
Then, you can switch the order of summation in the following infinite sum :
$$0 = \sum_{n \ge 1} (\frac{ nx^{n-1}}{1+x^n} - \frac {nx^{n-1}}{1-x^n} + \frac {2nx^{2n-1}}{1-x^{2n}})
= (\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n}) - \sum_{n \ge 1}(\frac {nx^{n-1}}{1-x^n} - \frac {2nx^{2n-1}}{1-x^{2n}})$$
The second sum is a telescoping sum $\sum (a_n - a_{2n})$ so reordering again, it is $$(\sum a_n) - (\sum a_{2n}) = \sum_{n \text{ odd}} a_n + \sum_{n \text{ even}} (a_n - a_n) = \sum_{n\text{ odd}} a_n$$
Hence $$\sum_{n \ge 1} \frac{ nx^{n-1}}{1+x^n} - \sum_{n \text{ odd}}\frac {nx^{n-1}}{1-x^n}=0$$
| {
"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": 1
} |
What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| Here is a slightly different way.
We start with
$$
\frac{x - 1}{(x+3)(x^2+1)} =
\frac{A}{x+3} + \frac{B}{x+i} + \frac{B^*}{x-i},
\qquad (1)
$$
where $B^*$ is the complex conjugate of $B$. The coefficient for $(x-i)^{-1}$ must be $B^*$: since the left-hand side is real, the right-hand side is real too, and hence is invariant under complex conjugation.
Multiplying both sides by $x+3$ and taking the limit of $x \rightarrow -3$, we get
$$
A = \frac{-3-1}{(-3)^2 + 1} = -\frac{2}{5}.
\qquad (2)
$$
Similarly, multiplying both sides by $x + i$ and taking the limit of $x \rightarrow -i$ yields
$$
B = \frac{-i - 1}{(-i + 3)(-i - i)} = \frac{1}{5} - \frac{i}{10}.
\qquad (2)
$$
Now the right-hand side of (1) is easy to integrate
$$
\begin{aligned}
\int \frac{x - 1}{(x+3)(x^2+1)} dx
&=
A\log|x+3| + B\log(x+i) + B^*\log(x-i) + C\\
&=
A \log|x+3| + (\Re B) \log(x+i)(x-i) + (\Im B) \, i\log\frac{x+i}{x-i} + C \\
&=
A \log|x+3| + (\Re B) \log(x^2+1) -2 (\Im B) \arctan \frac{1}{x} + C \\
&=
-\frac{2}{5} \log|x+3| + \frac{1}{5} \log(x^2+1)
+\frac{1}{5} \arctan \frac 1 x + C.
\end{aligned}
$$
The third line needs some explanation. Since
$$
\begin{aligned}
\log \frac{x + i}{\sqrt{x^2+1}} &= i \arctan\frac{1}{x}, \\
\log \frac{x - i}{\sqrt{x^2+1}} &= -i \arctan\frac{1}{x},
\end{aligned}
$$
the difference
$$
\log \frac{x + i}{x - i} = 2 i \arctan\frac{1}{x}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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 $3x^2 y^2\sqrt[3]{6y}$?
| \begin{align}
\sqrt[3]{162x^6y^7}&=\sqrt[3]{27x^6y^6*6y}\\
&=\sqrt[3]{27x^6y^6}\times\sqrt[3]{6y}\\
&=3x^2y^2\sqrt[3]{6y}
\end{align}
| {
"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": 0
} |
How to prove $\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m$? Question: How to prove the following identity?
$$
\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m.
$$
I'm also looking for the generalization of this identity like
$$
\sum_{s=k}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=?
$$
Proofs, hints, or references are all welcome.
| Suppose we seek to verify that
$$\sum_{s=0}^m {2s\choose s} {s\choose m-s}
\frac{(-1)^s}{s+1} = (-1)^m$$
without using the generating function of the Catalan numbers.
Re-write the sum as follows:
$$\sum_{s=0}^m {2s\choose m} {m\choose s}
\frac{(-1)^s}{s+1}$$
which is
$$\frac{1}{m+1}
\sum_{s=0}^m {2s\choose m} {m+1\choose s+1} (-1)^s$$
which turns into
$$- \frac{1}{m+1}
\sum_{s=1}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}
\\ = \frac{1}{m+1} {-2\choose m}
- \frac{1}{m+1}
\sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}
\\ = \frac{1}{m+1} (-1)^m \frac{(m+1)!}{m!}
- \frac{1}{m+1}
\sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}
\\ = (-1)^m
- \frac{1}{m+1}
\sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}.$$
Now introduce
$${2s-2\choose m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{2s-2} \; dz.$$
We get for the sum
$$- \frac{1}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2}
\sum_{s=0}^{m+1} {m+1\choose s} (-1)^{s} (1+z)^{2s} \; dz
\\ = -\frac{1}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2}
(1-(1+z)^2)^{m+1}\; dz
\\ = \frac{(-1)^m}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2}
(z^2+2z)^{m+1} \; dz
\\ = \frac{(-1)^m}{m+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1+z)^2}
(z+2)^{m+1} \; dz
= 0.$$
This concludes the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Order of $5$ in $\Bbb{Z}_{2^k}$ Is it true that the order of $5$ in $\Bbb Z_{2^k}$ is $2^{k-2}$? I was unable solve the congruence $5^n\equiv 1\pmod {2^{k}}$ nor see why $5^{2^{k-2}}\equiv 1\pmod {2^{k}}$. I'm not sure if this is the correct approach to this problem.
| Lemma 1: If $a$ is odd, $k\ge 3$, then $a^{2^{k-2}}\equiv 1\pmod{2^k}$.
Proof: Use induction. Base case: $a^2\equiv 1\pmod{8}$ for all odd $a$,
because if $a=2k+1$ for some $k\in\mathbb Z$, then $a^2-1=4k(k+1)$,
where $k(k+1)$ is a product of two consecutive integers, so $2\mid k(k+1)$.
If $a^{2^{m-2}}=2^mt+1$ for some $m\ge 3$, then
$a^{2^{m-1}}\equiv \left(2^mt+1\right)^2\equiv 2^{2m}t^2+2^{m+1}t+1\equiv 1\pmod{2^{m+1}}$.
Lemma 2: If $a\not\equiv 1\pmod{8}$, $k\ge 3$, then $a^{2^{k-3}}\not\equiv 1\pmod{2^k}$.
Proof: Use induction. Base case: $a\not\equiv 1\pmod{8}$.
If $a^{2^{m-3}}=2^mt+r$, $2\le r<2^m$ for some $m\ge 3$, then
$a^{2^{m-2}}\equiv \left(2^mt+r\right)^2\equiv 2^{2m}t^2+2^{m+1}tr+r^2\equiv r^2\pmod{2^{m+1}}$.
But $4\le r^2<2^{m+1}$, so $a^{2^{m-2}}\not\equiv 1\pmod{2^{m+1}}$.
Corollary: If $a$ is odd, $a\not\equiv 1\pmod{8}$, $k\ge 3$, then $\text{ord}_{2^k}(a)=2^{k-2}$.
In your case, it also turns out that $\text{ord}_{2^2}(5)=2^0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determinants: divisibility by 6 without remainder and $n\times n$ matrix
*
*Compute the determinant of \begin{pmatrix}
1 & 2 & 3 & ...& n\\
-1 & 0 & 3 & ...& n\\
-1 & -2 & 0 & ...& n\\
...& ...& ...& ...& \\
-1 & -2 & -3 & ...& n
\end{pmatrix}
after some elementary raw operation, one can reach:
\begin{pmatrix} 1 & 2 & 3 & ...& n\\
-2 & -2 & 0 & ...& 0\\
0& -2& -3& ...&0& \\
...& ...& ...& ...& \\
0 & 0 & 0 & 1-n & n
\end{pmatrix}
but I don't sure how to proceed.
*Why the det. of the following matrix is divisible by 6 without remainder?
\begin{pmatrix} 2^0 & 2^1 & 2^2 \\ 4^0 & 4^1 & 4^2\\ 5^0 & 5^1 & 5^2 \end{pmatrix}
So I know that I have to show that its det. is divisible by $2$ and $3$, or equivalently that the sum of its digits divisible by $3$ and last digit is even. But I don't sure how to start the process.
Thank you.
| Regarding (2), you can reduce the matrix $A$ modulo $2$ to obtain
$$ A' = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix}. $$
Clearly, $\det(A') = 0$ which shows that $\det(A) \mod 2 = \det(A') = 0$. Similarly, reducing $A$ modulo 3, we obtain
$$ A'' = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 1 \end{pmatrix} $$
and this implies that $\det(A) \mod 3 = \det(A'') = 0$ which shows that $\det(A)$ is divible by both two and three.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is my $\epsilon$-$\delta$ proof incorrect? Prove that $f(x)= \sqrt{4+x^2}$ is continuous at $x_0$ using the $\epsilon -\delta$ definition of continuity.
Textbook proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\underbrace{\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}}_{K(x,x_0)}|x-x_0|$
$K(x,x_0)=\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq \frac{|x|+|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}+\frac{|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}$
$\leq\frac{|x|}{\sqrt{4+x^2}}+\frac{|x_0|}{\sqrt{4+{x_0}^2}}$
$\frac{|x|}{\sqrt{4+x^2}}\leq 1$ and $\frac{|x_0|}{\sqrt{4+{x_0}^2}}\leq 1\Rightarrow K(x,x_0)\leq 2\Rightarrow |f(x)-f(x_0)|\leq2|x-x_0|<2\delta=\epsilon$
So, for all $\epsilon$ we can choose a $\delta$ such that $\delta=\frac{\epsilon}{2}$.
My proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq |x^2-{x_0}^2|$
$=|x-x_0||x-x_0+2x_0|<\delta^2+2|x_0|\delta$
Case $\delta<1$:
$\delta^2+2|x_0|\delta<\delta(1+2|x_0|)=\epsilon$
$\Rightarrow$ If $1+2|x_0|>\epsilon$ we have $\delta=\frac{\epsilon}{1+2|x_0|}$
Case $\delta=1$:
$\Rightarrow$ If $1+2|x_0|\leq\epsilon$ we have $\delta = 1$
Why do I get a different result for delta? What did I do wrong?
| As others have pointed out the things that matter, I provide an argument for your reference; it is more succinct:
Let $x_{0} \in \Bbb{R}$. If $x \in \Bbb{R}$, then
$$
|\sqrt{4+x^{2}} - \sqrt{4+x_{0}^{2}}| = \frac{|x-x_{0}||x+x_{0}|}{\sqrt{4+x^{2}} + \sqrt{4 + x_{0}^{2}}} < |x-x_{0}||x+x_{0}|;
$$
if in addition $|x-x_{0}| < 1$, then $|x| - |x_{0}| \leq |x-x_{0}| < 1$, implying $|x+x_{0}| \leq |x|+|x_{0}| < 1 + 2|x_{0}| =: M_{0}$, implying
$|x-x_{0}||x+x_{0}| < |x-x_{0}|M_{0}$; given any $\varepsilon > 0$, we have
$|x - x_{0}|M_{0} < \varepsilon$ if in addition $|x-x_{0}| < \varepsilon/M_{0}$. All in all, for every $x_{0} \in \Bbb{R}$ and every $\varepsilon > 0$, it holds that $|x-x_{0}| < \min \{ 1, \varepsilon/M_{0} \}$ implies $|f(x) - f(x_{0})| < \varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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?
| You got $A$ correct.
You have incorrectly evaluated $C$. When you sub in $x=-2$ you should get:
$$5(-2)^2+3(-2)-2=(-2)^2C$$
$$12=4C$$
$$C=3$$
Then to evaluate $B$ you can pick an arbitrary value for $x$ then solve for $B$. E.g. you selected $x=1$ this gives:
$$5\cdot1^2+3\cdot1-2=A\cdot(1+2)+B\cdot(1+2)\cdot1+C\cdot1^2$$
Replace $A$ and $C$ with the values you already found to give:
$$5+3-2=-3+3B+3$$
$$6=3B$$
$$B=2$$
| {
"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": 1
} |
If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations:
$$
7=3S+2T\\
18=7S+3T\\
47=18S+7T
$$
Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?
| From the recurrence relation we have
$$
a_{n+1} a_{n-1} - a_n^2 = 5
=
a_{n+2} a_{n} - a_{n+1}^2.
$$
So
$$
a_{n+1} \, (a_{n-1} + a_{n+1}) = a_n \, (a_n + a_{n+2}).
$$
or
$$
\frac{a_{n-1} + a_{n+1}}{a_n} = \frac{a_n + a_{n+2}}{a_{n+1}},
$$
This means
$$
c_n \equiv \frac{a_{n-1} + a_{n+1}}{a_n}.
$$
satisfies $c_n = c_{n+1}$, and thus is a constant.
Let $c_n = S$.
By $a_0 = 2$, $a_1 = 3$, $a_2 = (a_1^2 +5)/a_0 = 7$, we determine $S = 3$.
Thus,
$$
a_{n+1} = 3 \, a_n - a_{n-1} = S \, a_n + T \, a_{n-1},
$$
where $S = 3$, $T = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Sequence corresponding to the generating function
Find the sequence corresponding to the generating function $$G(x) = \frac{2x^4}{2x^3-x^2-2x+1}$$
First of all, I wrote this equation like that;
$\sum\limits_{n=0}^\infty(a_n)x^n = \frac{2x^4}{2x^3-x^2-2x+1}$
Then, I think right hand side should be;
$\frac{2x^4}{2x^3-x^2-2x+1} = \frac{A}{2x-1} + \frac{B}{x-1} + \frac{C}{x+1}$
I found
$A = \frac{-2x^2}{3}$ , $B = x^2$ , $C = \frac{x^2}{3}$
However, I cannot continue after that point.
| Assuming the partial fraction decomposition is correct (I haven't checked it, but I suppose you have), remember that $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$. Hence $$\frac{1}{2x-1}=-\frac{1}{1-2x}=-\sum_{n=0}^\infty (2x)^n=-\sum_{n=0}^\infty 2^nx^n$$
and $$\frac{1}{x-1}=-\frac{1}{1-x}=-\sum_{n=0}^\infty x^n$$
and $$\frac{1}{x+1}=\frac{1}{1-(-x)}=\sum_{n=0}^\infty (-1)^nx^n$$
Remember also that multiplying a power series by $x^n$ shifts the coefficients $n$ steps upwards. Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt{2} + \sqrt[3]{3}$ is irrational $\sqrt{2} + \sqrt[3]{3}$ is irrational ?
These are my steps -
$\sqrt{2} + \sqrt[3]{3} = a$
$3 = (a-\sqrt{2})^{3}$
$3 = a^{3} -3a^{2}\sqrt{2} + 6a -2\sqrt{2}$
$3a^{2}\sqrt{2}+2\sqrt{2} = a^{3}+6a-3$
$\sqrt{2}(3a^{2}+2) = a^{3}+6a-3$
Then, $\sqrt{2}$ in the left side is irrational , and mulitply irratinal with rational is irrational.
The right side is rational.
So, $irrational \neq rational$.
This is a good proof ?
| Taking powers of $\alpha=\sqrt2+\sqrt[\large3]{3}$ and putting them into matrix form, we get
$$
\begin{bmatrix}
\alpha^0\\\alpha^1\\\alpha^2\\\alpha^3\\\alpha^4\\\alpha^5\\\alpha^6
\end{bmatrix}
=
\begin{bmatrix}
1&0&0&0&0&0\\
0&1&1&0&0&0\\
2&0&0&2&1&0\\
3&2&6&0&0&3\\
4&12&3&8&12&0\\
60&4&20&15&3&20\\
17&120&90&24&60&18
\end{bmatrix}
\begin{bmatrix}
1\\2^{1/2}\\3^{1/3}\\2^{1/2}3^{1/3}\\3^{2/3}\\2^{1/2}3^{2/3}
\end{bmatrix}\tag{1}
$$
We can use the method from this answer to get a vector perpendicular to all the columns in the matrix above:
$$
\begin{bmatrix}
1\\-36\\12\\-6\\-6\\0\\1
\end{bmatrix}^{\large T}
\begin{bmatrix}
1&0&0&0&0&0\\
0&1&1&0&0&0\\
2&0&0&2&1&0\\
3&2&6&0&0&3\\
4&12&3&8&12&0\\
60&4&20&15&3&20\\
17&120&90&24&60&18
\end{bmatrix}=0\tag{2}
$$
$(1)$ and $(2)$ imply that
$$
\alpha^6-6\alpha^4-6\alpha^3+12\alpha^2-36\alpha+1=0\tag{3}
$$
$(3)$ says that $\alpha$ is an algebraic integer. A rational algebraic integer must be an integer. However, $1\lt\sqrt2\lt\frac32$ and $1\lt\sqrt[\large3]3\lt\frac32$, thus $2\lt\alpha\lt3$. Therefore, $\alpha$ must be irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Circle of radius of Intersection of Plane and Sphere The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius?
I tried putting value of y from plane in sphere but then I get a $zx$ term. How to proceed?
| $x+2y-4=z$ and $x^2+y^2-x=-(z^2+z-2)$
$$x^2+y^2-x=(z+2)(1-z)$$
Substituting, $$x^2+y^2-x=(x+2y-2)(1-(x+2y-4))$$
$$x^2+y^2-x=-(x+2y-2)(x+2y-5)$$
And this way you get an equation in $x(s)$ and $y$(s).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The roots of the equation $z^n=(z+3)^n$
Prove that the roots of the equation $z^n=(z+3)^n$ are collinear
Taking modulus on both sides, $$\vert z-0\vert =\vert z-3\vert$$
This represents the perpendicular bisector of line joining $x=0$ and $x=3$
That was easy. But I tried to solve it using algebra:
$$\frac{z+3}{z}=1^{\frac{1}{n}}=\cos{\frac{2k\pi}{n}}+i\sin{\frac{2k\pi}{n}}$$
After simplifying, I got $$z=\frac{3i ( \cos{\frac{k\pi}{n} }+i\sin{ \frac{k\pi}{n}})}{2\sin{ \frac{k\pi}{n}}}$$
What should I do next?
| $\left(\dfrac{z+3}{z}\right)=(1)^n= e^{\dfrac{2\pi i k}{n}}$ where k=0,1,2.....(n-1)
$\left(1+\dfrac{3}{z}\right)=(1)^n= \alpha^k$ where $\alpha=e^{\dfrac{2\pi i}{n}}$
$\implies z=\dfrac{3}{\alpha^k-1} ;k\neq0$
roots of your equation will be $z=\dfrac{3}{\alpha-1},\dfrac{3}{\alpha^2-1},.......\dfrac{3}{\alpha^{n-1}-1}$
but we know, for $n^{th}$ roots unity$\implies$ $\alpha^{n-1}=\dfrac{\alpha^n}{\alpha}=\dfrac{1}{\alpha};\alpha^{n-2}=\dfrac{1}{\alpha^2} $and so on
thus roots of your equation will be $z=\dfrac{3}{\alpha-1},\dfrac{3}{\alpha^2-1},.......\dfrac{3\alpha^2}{1-\alpha^2},\dfrac{3\alpha}{1-\alpha}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"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.
| Apart the obvious solution $(x=0,y=0)$ :
$$x^3+xy^2-2y=0$$
$$3x^2-y^2-2x^3y=0$$
From eq.(2) : $y^2=3x^2-2x^3y$ that is plugg into eq.(1)
$x^3+x(3x^2-2x^3y)-2y=0$
$$y=\frac{2x^3}{x^4+1}$$
Plugging it into eq.(1) leads to :
$$x^8+2x^4-3=0$$
$$x^4=-1\pm 2$$
$$x=(-1\pm 2)^{1/4}$$
$$y=\frac{2x^3}{x^4+1}=(-1\pm 2)^{3/4}$$
So, eight solutions (real and complex).
The real solutions are $(x=1,y=1)$ and $(x=-1,y=-1)$
| {
"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": 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.
| $$\begin{align}\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3}
&=\sum_{r=2}^9\binom r{r-2}\binom {12-r}{9-r}\\
&=\sum_{r=2}^9(-1)^{r-2}\binom{-3}{r-2}(-1)^{9-r}\binom {-4}{9-r}&&\text{(upper negation)}\\\
&=-\sum_{r=2}^9\binom {-3}{r-2}\binom{-4}{9-r}\\
&=-\binom {-7}7&&\text{(Vandermonde)}\\
&=-(-1)^7\binom {13}7&&\text{(upper negation)}\\
&=\binom {13}7=\binom{13}6\quad\blacksquare
\end{align}$$
| {
"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": 0
} |
Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$
Let $x$, $y$, $a$, $b$ be real numbers such that $a^2+b^2 \leq 1$ and $x^2+y^2 \leq 1$. Show that $$(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$$
I am unable to find a solution to this problem. My initial thoughts were to have a trigonometric substitution of variables, but that didn't lead me further. Please help.
Thank you.
| This is a quadratic inequality in $x,y$, so all you have to do
is complete the squares.
Formally, if $\Delta=(ax+by-1)^2 -(x^2+y^2-1)(a^2+b^2-1)$ then you
have
$$
(1-a^2)\Delta=(1-a)^2(1+ay-bx)^2+(1-(a^2+b^2))(x-a)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Finding the value of an expression given three equations. $$x+y+2z=22$$
$$3x-2y+z=6$$
$$7x+3y-5z=1$$
above are three equations and we have to find the value of $$x^2+y^2+z^2=?$$
| $$
\begin{cases}
x+y+2z=22\\
3x-2y+z=6\\
7x+3y-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
3x-2y+z=6\\
7x+3y-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
3x-2(22+(-x-2z))+z=6\\
7x+3(22+(-x-2z))-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
-44+5x+5z=6\\
66+4x-11z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
66+4x-11z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
66+4(10-z)-11z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
106-15z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-7\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=3\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-3-2\cdot7)\\
x=3\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=5\\
x=3\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=3\\
y=5\\
z=7
\end{cases}
$$
So:
$$x^2+y^2+z^2=3^2+5^2+7^2=9+25+49=83$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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.
| Brute force:
$$f(x) \triangleq \sum_{r=0}^{n+1}x^r=\frac{x^{n+2}-1}{x-1} $$
$$f'(x) = \sum_{r=0}^{n+1}rx^{r-1} = \sum_{r=0}^{n}(r+1)x^r$$
$$f''(x) = \sum_{r=0}^{n}r(r+1)x^{r-1} \therefore xf''(x)=\sum_{r=0}^{n}r(r+1)x^{r} = \sum_{r=0}^n\left(r^2+r\right)x^r$$
$$f'(x)-(f(x)-x^{n+1}) = \sum_{r=0}^{n}rx^r$$
So $$\sum_{r=1}^nr^2x^r =xf''(x) - (f'(x)-(f(x) - x^{n+1}))$$
Finally take the limit as $x \to 1$, which might get ugly, and Bob's your uncle.
[edit]
Changed $$\sum_{r=1}^nr^2x^r =f''(x) - (f'(x)-(f(x) - x^{n+1}))$$ to $$\sum_{r=1}^nr^2x^r =xf''(x) - (f'(x)-(f(x) - x^{n+1}))$$
| {
"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": 0
} |
find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$ If the equation of the tangent to the graph of the function $y=f(x)$ at $x=2$ is $4x-y-3=0$ and at this point tangent cuts the graph also,then find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$$
As the denominator is tending to zero,so numerator must br tending to zero.
So I applied L Hospital rule two times to get
$$\lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)}$$ for the first time and $$\lim_{x\to 2}\frac{4x^2f''(x^2-2)+2f'(x^2-2)-f'(f(x)-3).f''(x)-f''(f(x)-3).(f'(x))^2}{2}$$
As the equation of tangent at the point $x=2$ is $y=4x-3$
So slope of tangent at $x=2$ is $4$ i.e. $f'(2)=4$ and as the tangent cuts the graph at $x=2$,so put $x=2$ in $y=4x-3$ to get $y=5$, so $f(2)=5$ but
I don't know how to find $f''(2)$.
I am stuck here, please help me. Thanks.
| The numerator is approaching 0 as you can see
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2^2-2)-f(f(2)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2)-f(5-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2)-f(2)}{(x-2)^2}$$
By applying L'Hôpital's rule you have $$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)} = \lim_{x\to 2}\frac{f'(2^2-2) \cdot 2x-f'(f(2)-3) \cdot f'(2)}{2(x-2)} = \lim_{x\to 2}\frac{f'(2) \cdot 2x-f'(5-3) \cdot f'(2)}{2(x-2)} = \lim_{x\to 2}\frac{4 \cdot 2x-4 \cdot 4}{2(x-2)} = 4 \lim_{x\to 2}\frac{x-2}{x-2} = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Approximation of $\pi$ using Brahmagupta's Identity Brahmagupta, an ancient Indian Mathematician, gave an pretty efficient algorithm for finding integer solutions to the famous Pell's Equation, far before Fermat propounded this before the European mathematicians' community.
Brahmagupta's Identity:
If $(x_1,y_1)$ is a solution to $Dx^2+m=y^2$ and $(x_2,y_2)$ is a solution to $Dx^2+n=y^2$, then $(x_1y_2\pm x_2y_1,y_1y_2\pm D x_1x_2)$ is a solution to the equation $Dx^2+mn=y^2$.
Famous mathematician André Weil denoted this more efficiently by $$(x_1,y_1;m)\oplus (x_2,y_2;n)=(x_1y_2\pm x_2y_1,y_1y_2\pm D x_1x_2;mn)$$
One can easily prove this by writing $m=y_1^2-Dx_1^2$ and $n=y_2^2-Dx_2^2$, and then multiplying them $$mn=(y_1^2-Dx_1^2)(y_2^2-Dx_2^2)=(y_1y_2\pm D x_1x_2)^2-D(x_1y_2\pm x_2y_1)^2$$ and also notice that it is a group. A $600 AD$ mathematician is solving problems using Group Theory!
I am giving an example, integer solutions to $83x^2+1=y^2$.
We know that, $$83\times 1^2-2=9^2. $$
So, we here get $$(1,9;-2)\oplus (1,9;-2)=(18,81+83;4)=(18,164;4).$$
So, we get the equation, $$\begin{align}
&83(18)^2+4=(164)^2\\
\implies &83\left(\frac {18}2\right)^2+1=\left(\frac {164}2\right)^2\\
\implies &83\times 9^2+1=82^2
\end{align}$$
Now, we have, if $Da^2+1=b^2$, then, $$\frac ba-\sqrt D=\frac {b-\sqrt D a}a=\frac {b^2-Da^2}{a(b+\sqrt D a)}=\frac 1{a(b+\sqrt D a)} $$
So, for sufficiently large $(a,b)$, $\frac ba$ is a good approximation for $\sqrt D$.
One, can verify by finding solutions to $2a^2+1=b^2$, i.e. $(2,3),(12,17),\dots$. So $$\color{red}{\sqrt 2\approx \frac 32,\frac {17}{12},\frac {577}{408}}.$$
So, I was trying to approximate $\sqrt \pi$ or $e$ or $\pi$ using this identity, but could not came to result. I am trying my way, but you folks please help me sharing your idea.
As, $\pi$ is irrational, we can't have $\pi =D$, so, I used $3\lt \pi\lt 4$, so $\sqrt 3\lt \sqrt \pi\lt 2$, so, $1\lt \sqrt \pi \lt 2$,
But feeling some difficulty.
| The procedure for calculating the square root of the number can be used to calculate the number $\pi$ to arbitrary precision. You can use the ratio $$\tan\dfrac x2 = \dfrac{\tan x} {\sqrt{\tan^2 x+1} + 1}$$ with initial data $\tan{\dfrac{\pi}4} = 1$.
After $n$ iterations you will have $\tan\dfrac{\pi}{2^{n+2}}$ and then can use the formula $$\pi=\lim\limits_{n\to\infty} 2^{n+2}\tan\dfrac{\pi}{2^{n+2}}$$ or Maclaurin series for the arctangent function:
$$\pi=2^{n+2}\left(\tan\dfrac{\pi}{2^{n+2}}-\dfrac13\left(\tan\dfrac{\pi}{2^{n+2}}\right)^3+\dots\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why are the factors of some solutions to a Pell equation also a solution? I came across this observation while trying to answer this post using the Pell equation $x^2-2y^2=1$. Define,
$$P(m) = \frac{ (3+2\sqrt{2})^m+(3-2\sqrt{2})^m}{2}$$
$$Q(m) = \frac{ (3+2\sqrt{2})^m-(3-2\sqrt{2})^m}{2\sqrt{2}}$$
such that $P(m)^2-2Q(m)^2=1$. We have,
$$P(2^k) = \color{brown}{3, 17, 577, 665857, 886731088897,\dots}$$
$$Q(2^k) = \color{blue}{2, 12, 408, 470832, 627013566048,\dots}$$
Notice that,
$$\color{blue}{12} = 2^2\cdot\color{brown}3$$
$$\color{blue}{408} = 2^3\cdot\color{brown}{3\cdot17}$$
$$\color{blue}{470832} = 2^4\color{brown}{\cdot3\cdot17\cdot577}$$
and so on. In general, given the roots of $ax^2+bx+c=0$,
$$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
and let $D=b^2-4ac$.
Question: Is it true that,
$$\frac{2^{n}}{a}\prod_{k=0}^n \frac{x_1^{2^k}+x_2^{2^k}}{2}=\frac{x_1^{2^{n+1}}-x_2^{2^{n+1}}}{2\sqrt{D}}$$
| Note that $x^2-y^2=(x-y)(x+y)$ and by induction $$x^{2^{n+1}}-y^{2^{n+1}}=(x^{2^n}-y^{2^n})(x^{2^n}+y^{2^n})=(x-y)\prod_{k=0}^n(x^{2^k}+y^{2^k}).$$ For roots $x_1, x_2$ of a quadratic equation and $x_1\leq x_2$ we have $x_2-x_1 = \sqrt{D}/a$. This proves your identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sum of factors of a huge number. I recently appeared in a math olympiad and it had this one question which had me stumped. This was a few weeks back and I have been looking for a way to find its answer ever since, but with no success. Searched the internet for the solution, but couldn't find any on it too! Anyway, here's how the question goes:
The value of $2^{96} - 3^{16}$ has two factors between 60 and 70. What is the sum of these two factors?
BTW, I should add that I did use wolframalpha to actually find the answer so I am more interested in knowing how to work it out manually than just knowing the answer. Any feedback would be appreciated.
Thanks!
| Using $a^2-b^2 = (a+b)(a-b)$
\begin{align*}
2^{96}-3^{16}
={} &
(2^{48}+3^8)(2^{48}-3^8)
\\
={} &
(2^{48}+3^8)(2^{24}+3^4)(2^{24}-3^4)
\\
={} &
(2^{48}+3^8)(2^{24}+3^4)(2^{12}+3^2)(2^{12}-3^2)
\\
={} &
(2^{48}+3^8)(2^{24}+3^4)(2^{12}+3^2)(2^6+3)(2^6-3)
\end{align*}
and $2^6+3=67$, $2^6-3=61$, ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Existence of Solutions to a $2-$Equation System of Congruences Do there exist $a, b> 1$, such that
$$ a^4 \equiv 1 \pmod{b^2}$$ and $$ b^4 \equiv 1 \pmod{a^2}.$$
| If the given $2$ relations hold true, then
$$b^2\big|a^4-1 \Rightarrow b^2\big|(a^2+1)(a+1)(a-1)$$
and
$$a^2\big|b^4-1 \Rightarrow a^2\big|(b^2+1)(b+1)(b-1)$$
Also we have the relations below for suitable positive integers $k,l$ that
$$a^4=kb^2+1$$
$$b^4=la^2+1$$
Hence $$a^4-b^4=kb^2-la^2$$ or $$a^2(a^2+l)=b^2(b^2+k)$$
For this to hold, $a^2$ divides both L.H.S. and R.H.S.
But $a^2$ and $b^2$ do not divide each other, as per the given relations.
So $$a^2\big|b^2+k$$ and similarly $$b^2\big|a^2+l$$
Since $k,l$ are positive, $$a^2\big|b^2+k \Rightarrow b^2>a^2$$ and similarly $$b^2\big|a^2+l \Rightarrow a^2>b^2$$
Both of these cannot hold true together.
Hence such $a,b$ do not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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.
| You can actually do it with Cauchy:
$$
\left(\sum_{cyc}\left(\frac{a}{b+c}\right)^2\right)\left(\sum_{cyc}\left(a(b+c)\right)^2\right)≥\left(a^2+b^2+c^2\right)^2
$$
Thus it is enough to prove:
$$
\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)≥\frac{3}{4}\sum_{cyc}\left(a(b+c)\right)^2=\frac{3}{2}\sum_{cyc}a^2b^2+\frac{3}{2}\sum_{cyc}a^2bc
$$
After little simplification, this reduces to:
$$
\sum_{sym}a^3b≥\frac{3}{2}\sum_{cyc}a^2b^2+\frac{1}{2}\sum_{cyc}a^2bc
$$
Now we can see that:
$$
a^3b+b^3a≥2a^2b^2
$$
And
$$
a^3b+a^3c+ab^3+ac^3≥4a^2bc
$$
By AM-GM. By symmetrically adding these inequalities, we obtain:
$$
2\sum_{sym}a^3b≥4\sum_{cyc}a^2b^2\\
4\sum_{sym}a^3b≥8\sum_{cyc}a^2bc
$$
Which yields directly the inequality remaining to prove.
| {
"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": 0
} |
If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem.
If x, y are nonzero natural
numbers with $x>y$ such that
$$x-y = 5y^2 - 4x^2,$$
prove that $x - y$ is perfect square.
What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) + y^2$$
$$x - y + 4(x-y)(x+y) = y^2$$
$$(x-y)(4x+4y+1) = y^2$$
So $4x+4y+1$ is a divisor of $y^2$.
I also take into consideration that $y^2$ modulo $4$ is $0$ or $1$ (I don't know if this can help.)
So how do I prove that $4x+4y+1$ is a perfect square (this would involve $x-y$ - a perfect square)? While taking examples, I couldn't find any perfect square with a divisor that is $M_4 + 1$ and is not perfect square.
If there are any mistakes or another way, please tell me.
Some help would be apreciated. Thanks!
| Solve,
$$x - y=5y^2 - 4x^2\tag1$$
hence,
$$x = \frac{-1\pm\sqrt{1+16y+80y^2}}{8}$$
Let,
$$1+16y+80y^2 = \big(\tfrac{2p}{q}y-1\big)^2$$
Expand and factor to get,
$$y = \frac{pq+4q^2}{p^2-20q^2}$$
with the relevant $x$ as,
$$x = \frac{pq+5q^2}{p^2-20q^2}$$
and $x,y$ will be integers if $p,q$ satisfy the Pell equation,
$$p^2-20q^2 = 1$$
One can then see that $x>y$ and,
$$x-y = \frac{q^2}{p^2-20q^2}=q^2$$
The first few $p,q$ are,
$$9, 2 \\161, 36 \\ 2889, 646$$
and these yield $x,y$,
$$38, 34 \\ 12276, 10980 \\ 3952874, 3535558$$
and so on, consistent with the numerical search made in the comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
How to solve the equation $\sin^3x+\sin^3(\frac{2\pi}{3}+x)+\sin^3(\frac{4\pi}{3}+x)+\frac{3}{4}\cos2x=0$. Solve the equation $$\sin^3x+\sin^3\left(\frac{2\pi}{3}+x\right)+\sin^3\left(\frac{4\pi}{3}+x\right)+\frac{3}{4}\cos2x=0$$
May I have a hint on how to solve this equation?
| Using the identity $\sin^3 \theta = \dfrac{3}{4}\sin\theta-\dfrac{1}{4}\sin 3\theta$, we get:
$\sin^3 x + \sin^3(x+\tfrac{2\pi}{3}) + \sin^3(x+\tfrac{4\pi}{3})$
$= \dfrac{3}{4}\left[\sin x + \sin(x+\tfrac{2\pi}{3})+\sin(x+\tfrac{4\pi}{3})\right] - \dfrac{1}{4}\left[\sin 3x + \sin(3x+2\pi) + \sin(3x+4\pi)\right]$
$= \dfrac{3}{4} \cdot 0 - \dfrac{1}{4} \cdot 3\sin3x$
$= -\dfrac{3}{4}\sin 3x$.
Therefore, your equation is equivalant to $-\dfrac{3}{4}\sin 3x + \dfrac{3}{4}\cos 2x = 0$.
Can you take it from here? (The next thing you should do is use one of these sum-to-product identities.)
Using the identities $\sin\theta = \cos(\tfrac{\pi}{2}-\theta)$ and $\cos A - \cos B = -2\sin\tfrac{A+B}{2}\sin\tfrac{A-B}{2}$, we have:
$\cos 2x - \sin 3x= 0$
$\cos 2x -\cos(\tfrac{\pi}{2}-3x) = 0$
$2\sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right) = 0$
$\sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right) = 0$ OR $\sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right) = 0$
$\dfrac{x}{2}-\dfrac{\pi}{4} = k\pi$ OR $\dfrac{5x}{2}-\dfrac{\pi}{4} = k\pi$ for some $k \in \mathbb{Z}$
$x = \dfrac{(4k+1)\pi}{2}$ OR $x = \dfrac{(4k+1)\pi}{10}$ for some $k \in \mathbb{Z}$
It's not hard to see that $x = \dfrac{(4k+1)\pi}{10}$ covers all the values which satisfy $x = \dfrac{(4k+1)\pi}{2}$. So the solution is simply $x = \dfrac{(4k+1)\pi}{10}$ for some $k \in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int{\sin^{13}x}dx$ Evaluate $\int \sin^{13}x\,dx$.
My solution so far:
$$\int\sin^{13}x\,dx=\int\sin x\sin^{12}x\,dx=\int\sin x(\sin^{2}x)^6\,dx$$
Let $t=\cos x$, then $dt=-\sin x\,dx$.
Now we have: $$\int\sin^{13}x\,dx=-\int(1-t^2)^6,\ dt$$.
How do I proceed from here? I need detailed answer.
| \begin{align}
\int(1-t^2)^6 &= \int \sum_{n=0}^6 {6 \choose n} (-t^2)^n \\
&= \sum_{n=0}^6 {6 \choose n} (-1)^n \int t^{2n} \\
&= \sum_{n=0}^6 {6 \choose n} (-1)^n \left( \frac{t^{2n+1}}{2n+1}\right) + C \\
&= \frac{t^{13}}{13} - \frac{6 t^{11}}{11} + \frac{5t^9}{3} - \frac{20t^7}{7} + 3 t^5 - 2t^3 + t + C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$.
I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
| We can easily verify that $n=1$ is a solution.
Assume for some $k$ that,
$3^{k+1}|2^{3^{k}}+1$, hence there exists an integer $m$, such that,
$m(3^{k+1})=2^{3^{k}}+1$.
Then we have,
$2^{3^{k+1}}+1=(2^{3^{k}})^{3}+1=(2^{3^{k}}+1)[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]$. (*)
Now, $3^{k}$ is clearly odd for all integers $k$, hence,
$(2^{3^{k}})^{2}-2^{3^{k}}+1 \equiv ((-1)^{3^{k}})^{2}-(-1)^{3^{k}}+1 \equiv 1+1+1 \equiv 0\mod 3$.
Thus let,
$(2^{3^{k}})^{2}-2^{3^{k}}+1=3n$ where $n$ is an integer.
Continuing from (*), we are left with,
$m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[3n]=m(3^{k+2})(n)$
And so,
$3^{k+2}|2^{3^{k+1}}+1$.
Thus our inductive proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Solve this equation $xy-\frac{(x+y)^2}{n}=n-4$ Let $n>4$ be a given positive integer. Find all pairs of positive integers $(x,y)$ such that
$$xy-\dfrac{(x+y)^2}{n}=n-4$$
What I tried is to use
$$nxy-(x+y)^2=n^2-4n\Longrightarrow (n-2)^2+(x+y)^2=nxy+4$$
| If you frame the problem as a quadratic equation in $y$, the condition that $y$ is an integer implies that
\begin{equation*}
n(n-4)(x^2-4)=\Box
\end{equation*}
By parameterising this quadric and searching for integer pairs, we get what looks like an infinite number of pairs $(x,y)$ for each $n$.
Investigating these we find $2$ simple parametric solutions
\begin{equation*}
x=n^2-4n+2 \hspace{2cm} y=n-2 \hspace{1cm} \mbox{or} \hspace{1cm} y=(n-2)(n^2-4n+1)
\end{equation*}
and
\begin{equation*}
x=n^4-8n^3+20n^2-16n+2 \hspace{1cm} y=(n-2)(n^2-4n+1) \hspace{0.5cm} \mbox{or} \hspace{0.5cm} y=(n-2)(n^4-8n^3+19n^2-12n+1)
\end{equation*}
There are almost certainly an infinite number of such parametric solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What will be the remainder when $2^{31}$ is divided by $5$? The question is given in the title:
Find the remainder when $2^{31}$ is divided by $5$.
My friend explained me this way:
$2^2$ gives $-1$ remainder.
So, any power of $2^2$ will give $-1$ remainder.
So, $2^{30}$ gives $-1$ remainder.
So, $2^{30}\times 2$ or $2^{31}$ gives $3$ remainder.
Now, I cannot understand how he said the last line. So, please explain this line.
Also, how can I do this using modular congruency?
| From Fermat's Little Theorem-
$a^{p-1}\equiv1\pmod p$
Putting $a=2$ and $p=5$ we have,
$2^4\equiv 1\pmod 5$
$\implies 2^{28}\equiv 1\pmod 5$
Now,$2^{31}=2^{28}\times 2^{3}$.So,it is equivalent to finding the remainder when $2^3$ is divided by $5$..which is $3$.So,the remainder when $2^{31}$ is divided by $5$ is $3$..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Find all $(x,y)$ satisfying $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$ Find all pairs $(x,y)$ of real numbers that satisfy the equation $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$
I supposed $a=\sin^2x$ and $b=\cos^2x$
So the equation becomes $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=12+\frac{1}{2}\sin y$
As $a+\frac{1}{a}\geq 2$ and $b+\frac{1}{b}\geq 2$
$12+\frac{1}{2}\sin y\geq 8$
$\sin y\geq -8$
I am stuck here.I could not solve further.Please help me.Thanks.
| As JMoravitz notes in the comments, it's probably the case that the minimum value of the LHS is 12.5. Let's prove this.
Using the inequality $a^2+b^2\ge\frac{(a+b)^2}{2}$, note that
\begin{align*}\left(\sin^2x + \frac{1}{\sin^2x}\right)^2 + \left(\cos^2x + \frac{1}{\cos^2x}\right)^2&\ge \frac{1}{2}\left(\sin^2x + \frac{1}{\sin^2x} + \cos^2x + \frac{1}{\cos^2x}\right)^2\\
&=\frac{1}{2}\left(1 + \frac{\cos^2x+\sin^2x}{\sin^2x\cos^2x}\right)^2 \\
&=\frac{1}{2}\left(1 + \frac{4}{(2\sin x\cos x)^2}\right)^2 \\
&=\frac{1}{2}\left(1 + \frac{4}{\sin^2{2x}}\right)^2\\
&\ge\frac{1}{2}\left(1 + \frac{4}{1}\right)^2 = \frac{25}{2}.
\end{align*}
I leave it to you to determine when this equality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$C$ be curve of intersection of hemisphere $x^2+y^2+z^2=2ax$ and cylinder $x^2+y^2=2bx$ ; to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$ $C$ be the curve of intersection of the hemisphere $x^2+y^2+z^2=2ax$ and the cylinder $x^2+y^2=2bx$ , where $0<b<a$ ; how to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$ using Stoke's theorem ? I can't parametrize the curve $C$ nor able to get the surface $S$ . Please help . Thanks in advance
| There are at least to ways you can parametrize curve $C$.
Method 1. Since the curve is at the intersection of the sphere and the cylinder, it belongs to the cylinder, which has radius $b$ and is shifted $b$ units along the $x$ axis, so you can write:
$$
x=b\cos\theta+b, \quad y =b\sin\theta, \quad 0\le \theta \le 2\pi,
$$
but since it also belongs to the sphere, you can write
$$
z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(b\sin\theta)^2-(b\cos\theta+b-a)^2},
$$
which gives you a first parametrization.
Method 2. The cylinder has polar equation $r(\theta)=2b\cos\theta$, so
$$
x=r(\theta)\cos\theta=2b\cos^2\theta, \quad y =2b\cos\theta\sin\theta, \quad -\pi/2\le \theta \le \pi/2,
$$
and again
$$
z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(2b\cos\theta\sin\theta)^2-(2b\cos^2\theta-a)^2}.
$$
Note the difference of variations of $\theta$ in the two parametrizations.
This being said, Stokes theorem is probably more appropriate for this question, so you are going to need to parametrize the surface. You could do it in spherical coordinates, but I think cartesian coordinates are the way to go here. Take $x$ and $y$ as parameters and you have:
$$
x=x, \quad y=y, \quad z=\sqrt{a^2-y^2-(x-a)^2},
$$
where $x$ and $y$ belong to the projection of the cylinder in the $xy$ plane, that is,
$$
(x,y)\in D:=\{(r,\theta)\;|\; -\pi/2\le \theta \le \pi/2,0\le r\le 2b\cos\theta \}
$$
It follows by the Stokes theorem that your integral equals
$$
\iint_D \nabla\times\vec{F}(x,y)\cdot \vec{r_x}\times \vec{r_y}\; dA,
$$
where $\nabla\times\vec{F}(x,y)$ is the curl of your field expressed as a field depending only on $x$ and $y$. I am sure you can takes computations from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.
| your inequalitiy is equivalent to
$$a^4c+b^4a+c^4b-a^2b^2c-a^2bc^2-ab^2c^2\geq 0$$
this is equivalent to
$$(a^2-b^2)(a^2c-bc^2)+(b^2-c^2)(ab^2-bc^2)\geq 0$$ we assume that $$a\geq b\geq c$$ thus $$a^2\geq b^2$$ and $$a^2\geq bc$$ and $$b^2\geq c^2$$ and $$ab\geq c^2$$
in the case $$a\geq c\geq b$$ we have
$$(a^2-c^2)(a^2c-c^2b)+(a^2c-b^2a)(c^2-b^2)\geq 0$$
| {
"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": 4
} |
How to solve the functional equation $ f(f(x))=ax^2+bx+c $ Find all real numbers $a,b,c\in\mathbb{R}$ for which there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that:
$$
f(f(x))=ax^2+bx+c
$$
for all $x\in\mathbb{R}$.
The only thing I could deduce is:
$$
f(ax^2+bx+c)=af(x)^2+bf(x)+c
$$
Which doesn't help much. How to tackle the problem?
| Here are some reflections for continuous and differentiable $f$. It turns out that there is a unique solution if some conditions are satisfied (you can look at the end for the final answer).
*
*$$
f(f(x))=ax^2+bx+c\implies f(ax^2+bx+c)=af(x)^2+bf(x)+c
$$
*If $f(x)=f(y)$, $x\neq y$ then:
$$
a(x+y)=b.
$$
If $a(x+y)=b$ then in any case $ff(x)=ff(y)$ which means that either $f(x)=f(y)$ or $a(f(x)+f(y))=b$. In any case there are two non-equal numbers $x'$ and $y'$ such that $f(x')=f(y')$ and they lie on two different sides of $\frac{-b}{2a}$.
*If $f(x)=x$ then: $f(x)=ax^2+bx+c$ hence:
$$
ax^2+(b-1)x+c=0\implies (b-1)^2\geq 4ac; x=\frac{-b+1\pm\sqrt{(b-1)^2-4ac} }{2a}.
$$
*$f'(ax^2+bx+c)(2ax+b)=f'(x)(2af(x)+b)$ hence:
$$
x=\frac{-b}{2a}\implies f'(\frac{-b}{2a})=0\text{ or } f(\frac{-b}{2a})=\frac{-b}{2a}$$
On the other hand:
$$
f'(f(x)).f'(x)=2ax+b.
$$
So:
If $f'(x)=0$ then $x=\frac{-b}{2a}$.
Note that from large enough $x$ and $y$ satisfying $a(x+y)=b$ and $f(x)=f(y)$
*
*If $f(\frac{-b}{2a})=\frac{-b}{2a}$, then $f\circ f(\frac{-b}{2a})=\frac{-b}{2a}$ which means:
$$
c-\frac{b^2}{4a}=-\frac b{2a}\implies b^2-4ac=2b.
$$
moreover:
$$
f'(f(\frac{-b}{2a})).f'(\frac{-b}{2a})=\implies f'(\frac{-b}{2a})=0.
$$
If $b^2-4ac\neq 2b$, $f'(\frac{-b}{2a})=0$; so in any case:
$$
f'(\frac{-b}{2a})=0.
$$
This means that the function is strictly increasing on the one side of $\frac{-b}{2a}$ and strictly decreasing on the other side.
Moreover if $f'(f(x))=0$ then $x=\frac{-b}{2a}$; so if there is $x$ such that $f(x)=\frac{-b}{2a}$ then $f'(f(x))=0$ which means that $x=\frac{-b}{2a}$. So:
$$f(\frac{-b}{2a})=\frac{-b}{2a}\implies b^2-4ac=2b$$
Without this condition there will be no answer.
*
*But if $a(x+y)=b$ and $f(x)\neq f(y)$, then $a(f(x)+f(y))=b$; but this means that both of them cannot be in the same time bigger (or smaller) than $\frac{-b}{2a}$; which is a contradiction since $f'(\frac{-b}{2a})=0$ and $\frac{-b}{2a}$ is an extremum point of $f$. Hence:
$$ f(x)=f(y), x\neq y\iff a(x+y)=b$$
Therefore it is enough to find the function for $x>\frac{-b}{2a}$.
*
*See that after using the information derived above:
$$
f(ax^2+bx+\frac{b^2-2b}{4a})=af(x)^2+bf(x)+\frac{b^2-2b}{4a}\implies\\
f\left(a(x+\frac{b}{2a})^2-\frac{b}{2a}\right)=a\left(f(x)+\frac{b}{2a}\right)^2-\frac{b}{2a}.
$$.
Note that if $a>0$, then $x>-\frac{b}{2a}\implies f(x)>-\frac{b}{2a}$; w.l.o.g. we assume $a>0$ and $x>-\frac b{2a}$.
$$
af\left(a(x+\frac{b}{2a})^2-\frac{b}{2a}\right)+\frac{b}{2}=\left(af(x)+\frac{b}{2}\right)^2
$$
See that defining $g(x)=af(x)+\frac b2$, we get:
$$
g(a(x+\frac{b}{2a})^2-\frac{b}{2a})=g(x)^2
$$
Define the function:
$$
h(x)=\frac{\log g(x)}{\log(ax+\frac b2)}.
$$
Then we get:
$$
h(a(x+\frac{b}{2a})^2-\frac{b}{2a})=h(x)
$$
This function can be shown to be a constant by constructing a decreasing sequence $u_n$ from an arbitrary $u$ and showing that $h(u)$ is equal to the limit of $u_n$ which is a constant value independent of $u$. Hence:
$$
\frac{\log g(x)}{\log(ax+\frac b2)}=C\implies g(x)=(ax+\frac b2)^C\\
\implies f(x)=\frac 1a\left((ax+\frac b2)^C-\frac b2\right).
$$
By plugging in to the original question, it turns out that $C=\sqrt 2$. Extending it for all $x$'s, we get:
$$
f(x)=\frac 1a\left(|ax+\frac b2|^{\sqrt 2}-\frac b2\right) \text{ if } b^2-4ac=2b.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Ratios Of Multiple Angles If $2\tan A=3\tan B$ then prove that $$\tan(A-B)=\frac{\sin2B}{5-\cos 2B}.$$
I found that $\tan A=\frac{3}{2} \tan B$ and after that used the formula of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ but could not reach to the required answer.
| Notice, here is another method: $$RHS=\frac{\sin 2B}{5-\cos 2B}$$
$$=\frac{\frac{2\tan B}{1+\tan^2B}}{5-\frac{1-\tan^2 B}{1+\tan^2A}}$$
$$=\frac{2\tan B}{4+6\tan^2B}$$
$$=\frac{\tan B}{2+3\tan^2B}$$
$$=\frac{(3\tan B)-2\tan B}{2+(3\tan B)\tan B}$$
setting $3\tan B=2\tan A$,
$$=\frac{2\tan A-2\tan B}{2+2\tan A\tan B}$$
$$=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
$$=\tan(A-B)=LHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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 hope and wish that you will get a simpler solution. For the time being, here is mine.
I started writing $$\sqrt{7+2\sqrt{7-2\sqrt{7-2y} } } =x$$ which I solved for $y$; this gives $$y=\frac{1}{128} \left(-x^8+28 x^6-238 x^4+588 x^2+7\right)$$ Replacing $y$ by $x$ and expanding leads to $$x^8-28 x^6+238 x^4-588 x^2+128 x-7=0$$ I suspected that this could be factorized and, after some attempts, a CAS found that this reduces to $$(x^2+2 x-7)(x^3-x^2-9 x+1)^2=0$$ As alex.jordan answered, because of the acceptable range, the solution corresponds to one of the roots of $x^3-x^2-9 x+1=0$. Using Cardano method, we know that there are three real roots. Using the trigonometric method for solving the cubic equation when three real roots are present, we arrive for the only acceptable root to the result $$x=\frac{1}{3}+\frac{4\sqrt{7}}{3} \cos \left(\frac{1}{3} \tan ^{-1}\left(3
\sqrt{3}\right)\right)\approx 3.49396$$ which is almost equal to $\frac 72$ and to $\sqrt{7+2 \sqrt{7}}\approx 3.50592$ !!
| {
"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": 1
} |
Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?
| You want calculate $\sum_{k=1}^\infty\frac{k(k+1)}{2(k!)}=\frac{1}{2}\sum_{k=1}^\infty\frac{k+1}{(k-1)!}=\frac{1}{2}\sum_{k=0}^\infty\frac{k+2}{k!}$...(*)
But, as $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, then $x^2e^x=x^2+x^3+\frac{x^4}{2!}+\frac{x^5}{3!}+\cdots$. Deriving both sides give us $x^2e^x+2xe^x=2x+3x^2+\frac{4x^3}{2!}+\frac{5x^4}{3!}+...$. Thus, evaluating at $x=1$: $e+2e=\sum_{k=0}^\infty\frac{k+2}{k!}$.
Substituying the last in (*) give us finally $\frac{3e}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Prove that for positive real numbers $a,b,c$ we have $\frac{a}{b+c}+ \frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}.$
Prove that for positive real numbers $a,b,c$ we have $$\dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}.$$
Attempt
I tried using AM-GM and got $ \dfrac{a}{2\sqrt{bc}}+\dfrac{b}{2\sqrt{ac}}+\dfrac{c}{2\sqrt{ab}} \geq \dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b}$ but that doesn't seem to help since that gives an upper not lower bound.
| $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = \frac{a^2}{ab+ac} + \frac{b^2}{ab+bc} + \frac{c^2}{ac+bc} \geq \frac{(a+b+c)^2}{2(ab+bc+ac)}$$
Recall Buniakowsky inequality:
$$ (a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2$$
by expanding and regrouping the terms of:
$$(ay-bx)^2 + (az-cx)^2 + (bz - cy)^2 \geq 0$$
So now:
$$ \bigg (\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \bigg )(x+y+z) \geq (a+b+c)^2, \forall x,y,z > 0$$
or
$$ \frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \geq \frac{(a+b+c)^2}{x+y+z}, \forall x,y,z > 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can one show that : $|u_{n+1}-\sqrt{2}|\le\frac{1}{4}|u_n-\sqrt{2}|$ $U_n$ numerical sequence such that :
( For all natural numbers $n$ ) $U_{n+1}=1+\dfrac{1}{1+U_n}$ and $U_0=1$
How can one show that : $|U_{n+1}-\sqrt{2}|\le\frac{1}{4}|U_n-\sqrt{2}|$
I arrived to show that
$$|U_{n+1}-U_n|\le\frac{1}{4}|U_n-U_{n-1}|$$
Using the fact that function $f(x)=1+\frac{1}{1+x}$ is decreasing in interval $[1,\frac{3}{2}]$
| Let $x=\sqrt{2}$ .
Note that this isn't an ordinary number , it's the root of $t=1+\frac{1}{1+t}$ which looks exactly as the recurrence :
$$U_{n+1}-x=1+\frac{1}{1+U_n}-\left (1+\frac{1}{1+x} \right )=\frac{x-U_n}{(1+x)(1+U_n)}$$
Now we need only to prove that :
$$(1+x)(1+U_n) \geq 4$$
But $U_n \geq 1$ for every $n$ and $x>1$ so :
$$(1+x)(1+U_n) \geq 2 \cdot 2 =4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find Complex Roots
Question:
Find the complex roots of $$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
What I have attempted:
$$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
$$ {(z^{6} -1)(z^{6} +1)\over (z^2-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^{3} -1)(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)} = 0 $$
$$ {(z+1)(z^2-z+1)(z^6+1)\over (z+1)(z-1)(z^2+1)} = 0 $$
$$ {(z^2-z+1)(z^6+1)\over (z-1)(z^2+1)} = 0 $$
How should I continue?
| Only $$z^{12}-1$$ must equal 0.
Also, find the roots of $$z^{4}-1$$ and $$z^{3}-1$$ and remove them from your final solution, since the denominator ≠ 0
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I integrate this? (for calculate value of L-function ) I want to calculate the definite integral:
$$
\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx.
$$
Indeed, I already know that $\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx=L(\chi,1)=\frac{\pi}{\sqrt{5}}$ where $\chi$ is the Dirichlet character for $\mathbb{Z}[\sqrt{-5}]$.
I have some trouble with this actual calculation.
My calculation is as following:
\begin{eqnarray}
&&\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx \cr
&=& \int_{0}^{1} \frac{1+x^{6}}{1-x^{2}+x^{4}-x^{6}+x^{8}} dx\cr
&=& \int_{0}^{1} \frac{x^{2}(1+x^{2})(\frac{1}{x^{2}}-1+x^{2})}{x^{4}(\frac{1}{x^{4}}-\frac{1}{x^{2}}+1-x^{2}+x^{4})}dx
\end{eqnarray}
Substitute $x-\frac{1}{x}=t$. Then, $(1+\frac{1}{x^{2}})dx= dt$.
So The integral is
$$
\int_{-\infty}^{0} \frac{t^{2}+1}{t^{4}+3t^{2}+1}dt
$$
However, it has different value with the value which is calculated by the site
http://www.emathhelp.net/calculators/calculus-2/definite-integral-calculator/?f=%28t%5E%7B2%7D%2B1%29%2F%28t%5E%7B4%7D%2B3t%5E%7B2%7D%2B1%29&var=&a=-inf&b=0&steps=on
So I've just given up to try more.
Is there any fine solution?
| Noting that
$$ t^4+3t^2+1=(t^2+\frac{3}{2})^2-(\frac{\sqrt5}{2})^2,$$
and
$$ \frac{t^2+1}{t^4+3t^2+1}=\frac{5-\sqrt{5}}{10 \left(t^2-\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}-\frac{-5-\sqrt{5}}{10\left(t^2+\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}$$
we have
\begin{eqnarray}
&&\int_{-\infty}^0\frac{t^2+1}{t^4+3t^2+1}dt\\
&=&\frac{1}{10}\int_{-\infty}^0\left(\frac{5-\sqrt{5}}{t^2+\frac{3}{2}-\frac{\sqrt5}{2}}+\frac{5+\sqrt5}{t^2+\frac{3}{2}+\frac{\sqrt5}{2}})\right)dt\\
&=&\frac{1}{10}\left(\frac{5-\sqrt{5}}{\sqrt{\frac{3}{2}-\frac{\sqrt5}{2}}}\arctan(\frac{x}{\sqrt{\frac{3}{2}-\frac{\sqrt5}{2}}})+\frac{5+\sqrt{5}}{\sqrt{\frac{3}{2}+\frac{\sqrt5}{2}}}\arctan(\frac{x}{\sqrt{\frac{3}{2}+\frac{\sqrt5}{2}}})\right)\bigg|_{-\infty}^0\\
&=&\frac{1}{10}\left(\frac{5-\sqrt{5}}{\sqrt{\frac{3}{2}-\frac{\sqrt5}{2}}}+\frac{5+\sqrt{5}}{\sqrt{\frac{3}{2}+\frac{\sqrt5}{2}}}\right)\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt5}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| As its an equation and open end it will have infinite solutions so Assume $x=\sin(2\theta)$, $y=\cos(2\theta)$ as we all know that sum of their squares is $1$ now by double angle $\sin(2\theta)=\frac{2t}{1+t^2}$, $\cos(2\theta)=\frac{1-t^2}{1+t^2}$ and now you can prove it trigonometrically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0}\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$
I tried L'hopital, but the denominator gets more and more complicated.
How does one calculate this limit?
| Notice, you can find this limit by using 'Hôpital's rule (do it twice):
$$\lim_{x\to0}\frac{\cos(2x^2)-1}{x^2\sin(x^2)}=\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(\cos(2x^2)-1\right)}{\frac{\text{d}}{\text{d}x}\left(x^2\sin(x^2)\right)}=$$
$$\lim_{x\to0}\frac{-4x\sin(2x^2)}{2x^3\cos(x^2)+2x\sin(x^2)}=-2\left[\lim_{x\to0}\frac{\sin(2x^2)}{x^2\cos(x^2)+\sin(x^2)}\right]=$$
$$-2\left[\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(\sin(2x^2)\right)}{\frac{\text{d}}{\text{d}x}\left(x^2\cos(x^2)+\sin(x^2)\right)}\right]=-2\left[\lim_{x\to0}\frac{4x\cos(2x^2)}{4x\cos(x^2)-2x^3\sin(x^2)}\right]=$$
$$-2\left[\lim_{x\to0}-\frac{2\cos(2x^2)}{x^2\sin(x^2)-2\cos(x^2)}\right]=-2\left[-\frac{2\cos(2\cdot0^2)}{0^2\sin(0^2)-2\cos(0^2)}\right]=$$
$$-2\left[-\frac{2\cdot1}{0-2\cdot1}\right]=-2\left[-\frac{2}{-2}\right]=-2\left[--\frac{2}{2}\right]=-2\left[\frac{2}{2}\right]=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
De Moivre Theorem Express the following function in the form of $a+ib$, $z$ being written for $\cos \theta+i \sin \theta$
$$\frac{1}{1+z^2}$$
My attempt,
$$\frac{1}{1+z^2}=\frac{1}{1+(\cos \theta +i \sin \theta)^2}
=\frac{1}{1+\cos 2\theta+i \sin 2\theta}$$
How to proceed? The given answer is $\frac{1}{2}(1-i \tan \theta)$
| Continuing from where you stopped:
$$\begin{align}
\frac 1 {1+\cos 2\theta + {\rm i} \sin 2\theta} = \\
\frac {1+\cos 2\theta - {\rm i} \sin 2\theta} {(1+\cos 2\theta + {\rm i} \sin 2\theta) (1+\cos 2\theta - {\rm i} \sin 2\theta)} = \\
\frac {1+\cos 2\theta - {\rm i} \sin 2\theta} {(1+\cos 2\theta)^2 + \sin^2 2\theta} = \\
\frac {1+\cos 2\theta - {\rm i} \sin 2\theta} {1 + 2 \cos 2 \theta + \cos^2 2 \theta + \sin^2 2 \theta} = \\
\frac {1+\cos 2\theta - {\rm i} \sin 2\theta} {2 + 2 \cos 2 \theta} = \\
\frac {1+\cos 2\theta} {2 + 2 \cos 2 \theta} - {\rm i} \frac {\sin 2\theta} {2 + 2 \cos 2 \theta} = \\
\frac 1 2 - {\rm i} \frac 1 2 \frac {\sin 2\theta} {1 + \cos 2 \theta} = \\
\frac 1 2 \left( 1 - {\rm i} \frac {\sin 2\theta} {1 + \cos 2 \theta} \right) = \\
\frac 1 2 \left( 1 - {\rm i} \frac {2 \sin \theta \cos \theta} {1 + 2 \cos^2 \theta - 1} \right) = \\
\frac 1 2 \left( 1 - {\rm i} \tan \theta \right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the difference between the largest and smallest possible values of any of these numbers?
If $x,y,z$ are real numbers such that $x+y+z = 5$ and $xy+yz+zx = 3$, what is the difference between the largest and smallest possible values of any of these numbers?
What is wrong with this approach?
We have that $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz) = 25 \implies x^2+y^2+z^2 = 19$. Thus the points $(x,y,z)$ lie on a sphere with radius $\sqrt{19}$. Therefore the maximum value of any of these is $\sqrt{19}$ and the smallest is $-\sqrt{19}$ and the difference is $2\sqrt{19}$.
| $(x,y,z)$ is not $\bf{three}$ points. It is a single point in three dimensional space, where $x,y,z$ are the coordinates.
Now think of a sphere of radius $\sqrt{19}$. If the $x$ coordinate is $\sqrt{19}$, then is it possible for any other coordinate to be $-\sqrt{19}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
long division with no numbers Im stumped by the following brain teaser:
Solve the following long division problem. Each letter represents a unique digit ($0$ - $9$)
$$
\require{enclose}
\begin{array}{r}
B \\[-3pt]
XZD \enclose{longdiv}{BUMG} \\[-3pt]
\underline{APZK} \\[-3pt]
ABU \\[-3pt]
\end{array}
$$
im just looking for a first step hint, not the entire solution, or a similar simpler example.
| You can separate it into cases on $B$ (it is easy to see that $B\not=0,1,5,9$), and find the values in the following order : $B\rightarrow A\rightarrow X\rightarrow Z,D,K\rightarrow P\rightarrow U,M,G$.
For $B=2$, we have $A=0,1$. But if $A=0$, then
$$(BU)=(2UMG)-(PZK)\ge 2000-999=1001$$
So, $A=1$. Also, if $X\le 8$, then
$$(2UMG)=2\times (XZD)+(12U)\le 2\times 899+129=1927$$
So, $X=9$.
Here, let $[n]$ be the right-most two digits of $n$. Then, since we have
$$[2\times (ZD)]=[ZK]$$
we have $Z=0$. But $(2UMG)=2\times (90D)+(12U)\le 2\times 909+129=1947$.
Similarly, you can do for $B=3,4,6,7,8$. In the following, I'll write the outlines.
For $B=3$, we have $A=2,X=9$. Now $Z=4,5$.
Case 1 : If $Z=4$, then $D=7,K=1,P=8,U=0,M=7\quad\Rightarrow\quad D=M$.
Case 2 : If $Z=5$, then $D=0,P=8,K=0\quad\Rightarrow\quad D=K$.
For $B=4$, we have $A=3,X=9$. Now $Z=0,6$.
Case 1 : If $Z=0$, then $D=2,P=6,K=8$. So, $$(4UMG)=4\times 902+(34U)\le 4\times 902+349=3957.$$
Case 2 : If $Z=6$, then $D=5,P=8,K=0,U=2,M=0\quad \Rightarrow\quad K=M$.
For $B=6$, we have $A=5,X=9$. Now $(Z,D,K)=(2k,1,6),(2k-1,9,4)\quad\Rightarrow\quad B=K\quad\text{or}\quad X=D$.
For $B=7$, we have $A=6, X=9$. Now $(Z,D,K)=(3,4,8),(8,3,1)$.
Case 1 : If $(Z,D,K)=(3,4,8)$, then $P=5,U=2,M=1,G=0$. This is sufficient.
Case 2 : If $(Z,D,K)=(8,3,1)$, then $P=8\quad\Rightarrow \quad Z=P$.
For $B=8$, we have $A=7,X=9, Z=1,D=4,K=2,P=3,M=9\quad\Rightarrow\quad M=X$.
Hence, the answer is
$$A=6,\quad B=7,\quad D=4,\quad G=0,\quad K=8,$$$$\quad M=1,\quad P=5,\quad U=2,\quad X=9,\quad Z=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$
Let $x=\cos\theta$ and $y=\sin\theta$,because $\sin^2\theta+\cos^2\theta=1$.
Then we need to find the minimum and maximum value of the expression $\frac{4-\sin\theta}{7-\cos\theta}$.
I differentiated it and equated it to zero to find the critical points or points of extrema.
They are $\theta_1=\arcsin(\frac{1}{\sqrt{65}})-\arctan(\frac{7}{4})$ and $\theta_2=\arccos(\frac{1}{\sqrt{65}})+\arctan(\frac{4}{7})$
I found $\frac{4-\sin\theta_1}{7-\cos\theta_1}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}$.
$\frac{4-\sin\theta_1}{7-\cos\theta_1}=\frac{3}{4}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}=\frac{5}{12}$
This method is full of lengthy calculations.I want to know is there an elegant solution possible for this problem which is short and easy.
| Let $\displaystyle k=\frac{4-y}{7-x}\Rightarrow 7k-kx=4-y\Rightarrow kx-y = 7k-4$ and given $x^2+y^2=1$
Now using the Cauchy-Schwarz inequality, we get $$[k^2+(-1)^2](x^2+y^2)\geq (kx-y)^2$$
So $$k^2+1\geq (7k-4)^2\Rightarrow 49k^2+16-56k\leq k^2+1$$
So $$48k^2-56k+15\leq 0\Rightarrow (4k-3)(12k-5)\leq 0$$
So we get $\displaystyle \frac{5}{12}\leq k\leq \frac{3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$
I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But this does not seem to work here.I am stuck.What should i do?
| First write $$x - 1 = x(2x - 1) - (2x^2 - 2x + 1).$$ Then our integrand becomes: $$\frac{x-1}{x^2 \sqrt{2x^2 - 2x + 1}} = \frac{2x-1}{x\sqrt{2x^2 - 2x + 1}} - \frac{\sqrt{2x^2 - 2x + 1}}{x^2}.$$ Now with the choice $$u = \sqrt{2x^2 - 2x + 1}, \quad v = \frac{1}{x},$$ we have $$u' = \frac{2x-1}{\sqrt{2x^2-2x+1}}, \quad v' = -\frac{1}{x^2},$$ hence the integrand is of the form $$u'v + uv' = (uv)'.$$ It immediately follows that an antiderivative is $$\int \frac{x-1}{x^2\sqrt{2x^2-2x+1}} \, dx = uv + C = \frac{\sqrt{2x^2 - 2x + 1}}{x} + C.$$ The rest is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$
I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equation gives RHS dependent on $y$.
| We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\sqrt{6}xy \leq A^2x^2+B^2y^2$, and $4yz \leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \dfrac{\sqrt{6}}{2}, CD = 2 = AC$. Thus $A^4 = A^2\cdot A^2 = A^2(B^2+C^2) = (AB)^2 + (AC)^2 = \left(\dfrac{\sqrt{6}}{2}\right)^2 + 2^2 = \dfrac{11}{2}=\dfrac{22}{4}$. So $A^2 = \dfrac{\sqrt{22}}{2}$, and clearly this value of $A^2 = \dfrac{\sqrt{22}}{2}$ is the maximum we are looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Rational solutions to $e^x - \frac{1}{2} = \sqrt{ x^3 + 1/4}$ Show that the only rational solution to the title curve is $x=0$.
My attempt: Squaring both sides we have $e^{2x} - e^x + \frac{1}{4} = x^3 + \frac{1}{4}$, which yields $e^{2x} - e^x = x^3$. I suspect that the left hand side is irrational for any rational $x\neq 0$, and if so, we reach a contradiction and conclude that the only possibility is $x=0$.
| We can write equation as $$e^x(e^x-1)=x^3$$ therefore $e^x-1=\frac{x^3}{e^x}$ now we know that $e$ is irrational(2.781...) so any value of $x$ else than $0$ makes the the equation irrational so $0$ is the only rational solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate this definite integral? How can I calculate this
$$\int_1^{+\infty} \frac {dx}{x^3\sqrt{x^2+x}}$$
I have no idea what to do with it, integration by parts or by substitution doesn't work for me.
| First of all, I've checked my answer with Mathematice and I'm right.
$$\int_{1}^{\infty}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=$$
$$\lim_{n\to\infty}\int_{1}^{n}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=$$
$$\lim_{n\to\infty}\int_{1}^{n}\frac{1}{x^3\sqrt{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}}}\space\text{d}x=$$
Substitute $u=x+\frac{1}{2}$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=1+\frac{1}{2}=\frac{3}{2}$ and upper bound $u=n+\frac{1}{2}$:
$$\lim_{n\to\infty}\int_{\frac{3}{2}}^{n+\frac{1}{2}}\frac{1}{\left(u-\frac{1}{2}\right)^3\sqrt{u^2-\frac{1}{4}}}\space\text{d}u=$$
Substitute $s=\text{arcsec}(2u)$ and $\text{d}u=\frac{\tan(s)\sec(s)}{2}\space\text{d}s$.
This gives a new lower bound $s=\text{arcsec}\left(2\cdot\frac{3}{2}\right)=\text{arcsec}(3)$ and
upper bound $s=\text{arcsec}\left(2\cdot\left(n+\frac{1}{2}\right)\right)=\text{arcsec}\left(2n+1\right)$:
$$\frac{1}{2}\lim_{n\to\infty}\int_{\text{arcsec}(3)}^{\text{arcsec}\left(2n+1\right)}\frac{2\sec(s)}{\left(\frac{\sec(s)}{2}-\frac{1}{2}\right)^3}\space\text{d}s=$$
$$\lim_{n\to\infty}\int_{\text{arcsec}(3)}^{\text{arcsec}\left(2n+1\right)}\frac{\sec(s)}{\left(\frac{\sec(s)}{2}-\frac{1}{2}\right)^3}\space\text{d}s=$$
Substitute $p=\tan\left(\frac{s}{2}\right)$ and $\text{d}p=\frac{\sec^2\left(\frac{s}{2}\right)}{2}\space\text{d}s$.
This gives a new lower bound $p=\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)$ and upper bound $p=\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)$:
$$\lim_{n\to\infty}\int_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}\frac{(p^2-1)^2}{p^6}\space\text{d}p=$$
$$2\lim_{n\to\infty}\int_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}\left[\frac{1}{p^6}-\frac{2}{p^4}+\frac{1}{p^2}\right]\space\text{d}p=$$
$$2\lim_{n\to\infty}\left[\left[-\frac{1}{5p^5}\right]_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}-2\left[-\frac{1}{3p^3}\right]_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}+\left[-\frac{1}{p}\right]_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}\right]=$$
$$\frac{14\sqrt{2}}{15}-\lim_{x\to\infty}\space\frac{2\sqrt{\frac{1+n}{1+2n}}\sqrt{1+2n}\left(3-4n+8n^2\right)}{15\sqrt{n^5}}=$$
*
*Apply L'Hôpitals rule;
*Use the product rule;
*Use the power rule;
$$\frac{14\sqrt{2}}{15}-\frac{16}{15}$$
So:
$$\int_{1}^{\infty}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=\frac{14\sqrt{2}}{15}-\frac{16}{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.