Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Combination and product question There are $5$ numbers, and each combination of $4$ numbers from those $5$ has a product of either $10, 20, 30, 40$, or $50$. What is the quotient of the sum of those $5$ numbers divided by the product of all $5$ numbers?
| The sum of these number divided by their product is:
$$
\frac{m_1+m_2+m_3+m_4+m_5}{m_1 m_2 m_3 m_4 m_5} = \frac{1}{m_2 m_3 m_4 m_5} + \frac{1}{m_1 m_3 m_4 m_5} + \frac{1}{m_1 m_2 m_4 m_5} + \frac{1}{m_1 m_2 m_3 m_5} + \frac{1}{m_1 m_2 m_3 m_4} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} + \frac{1}{40} + \frac{1}{50} = \frac{137}{600} \approx 0.228333
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Relationships between $\det(A+B)$ and $A+B$ When computing $\det(A+B)$ we notice that there is no relation between $\det A + \det B$.
However does the $\det(A+B)$ have any relation to the matrices $A+B$ as they stand?
| I can provide the following insight:
for every $A,B \in F^{n\times n}$
if we define $W=\left\{ M\in \left( \mathbb{Z} _{2}\right) ^{n\times n}| M \ is \ diagonal \right\} $ then the formal way of describing it is:
$$ det\left( A+B\right) =\sum _{M\in W}det\left( A\cdot M+B\cdot \left( I-M\right) \right) $$
in words, the sum of determinants above is the sum of all the determinants of the matrices that result from either removing the column from $B$ or from $A$ for each column in $A+B$ , the following is an example where $n = 2$ to better understand the idea:
$$
\left| A+B\right| =\begin{vmatrix}
a_{11}+b_{11} & a_{12}+b_{12} \\
a_{21}+b_{21} & a_{22}+b_{22}
\end{vmatrix}=\begin{vmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}
b_{11} & a_{12} \\
b_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}
a_{11} & b_{12} \\
a_{21} & b_{22}
\end{vmatrix}+\begin{vmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{vmatrix}
$$
we can notice that:
$$
\left| A+B\right| =det(A)+\begin{vmatrix}
\vert & \vert \\
B_{1} & A_{2} \\
\vert & \vert
\end{vmatrix}+\begin{vmatrix}
\vert & \vert \\
A_{1} & B_{2} \\
\vert & \vert
\end{vmatrix}+det(B)
$$
so in a way there is a relation between $det(A+B)$ and $det(A)+det(B)$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the solutions of this equation? How to find all the solutions of the following equation?
$1+\frac{x}{2!}+\frac{{{x}^{2}}}{4!}+\frac{{{x}^{3}}}{6!}+\frac{{{x}^{4}}}{8!}+...=0$
| $$ \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots $$
and
$$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots $$
This means that for $x > 0$, $$\cosh \sqrt x = 1 + \frac{x}{2!} + \frac{x^2}{4!} + \dots $$
which clearly has no roots. For $x < 0$, $$ \cos\sqrt{-x} = 1 + \frac{x}{2!} + \frac{x^2}{4!} + \dots $$
So you are after negative values of $x$ such that $\cos \sqrt{-x} = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Explain why there's no solution to the equation $2x-2x^2 = 1$ How can you tell there's no solution to the equation $2x - 2x^2 = 1$.
The supporting information goes like this:
The diagram above shows the graph of $y = 7 + 2x - 2x^2$.
I tried to do this:
$2 \cdot 1^2 = 2y$.
therefore $y = 2-2 = 0$;
| We want to solve $2x^2-2x+1=0$, presumably in the reals. Look at the upward-facing parabola $y=2x^2-2x+1$.
Complete the square. We have $x^2-x=(x-1/2)^2-1/4$. Thus our equation can be written as
$$y=2(x-1/2)^2+1/2.$$
The vertex of the parabola is at $x=1/2$. There, $y=1/2$, so the parabola is always above the $x$-axis. Thus it has no point in common with the $x$-axis, and therefore our equation has no real solutions.
The arithmetic is a little easier if we look at the equivalent equation $4x^2-4x+2=0$, which can be rewritten as $(2x-1)^2+1=0$. But $(2x-1)^2$ is always $\ge 0$ if $x$ is real. So we can't have $(2x-1)^2+1=0$ for real $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$ Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$
Im not sure should I use primitive root or quadratic residue.
For primitive root, $U_{33} = \{1,2,4,5,7,8,10,13,14,16,17,19,20,23,25,26,28,29,31,32\}$
Thus $\phi_{(3\cdot11)}=20=2^2\times{5}$
But it takes way too much time to test if each of them is primitive root or not...
But Im not quite familiar with the method of Quadratic Residue.
We have $x^8\equiv5\pmod3$ and $x^8\equiv 5 \pmod {11}$
so.... follows the rule of $x^2\equiv q \pmod n$,
we have $(x^4)^2 \equiv 5 \pmod 3$ and $(x^4)^2 \equiv 5 \pmod {11}$
and then how do I continue with it??
| I would think to work $\pmod 3$ and $\pmod {11}$, then combine the answers with the Chinese Remainder Theorem. The easy one is $\pmod 3$, where there are only two to check. We find $1^8 \equiv 1 \equiv 2^8 \not \equiv 5 \pmod 3$ so there are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Calculate: $\lim\limits_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$ How do I calculate the following limit without using l'Hôpital's rule?
$$\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$
| $$
\begin{aligned}
\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x
& = \lim _{x\:\to \infty }\left(e^{x\ln\left(\frac{x^2+2x+3}{x^2+x+1}\right)}\right)
\\& \approx \lim _{x\:\to \infty }\left(e^{x\left(\frac{x^2+2x+3}{x^2+x+1}-1\right)}\right)
\\& = \lim _{x\:\to \infty }\left(e^{\frac{x^2+2x}{x^2+x+1}}\right)
\\& = \color{red}{e}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Finding all the sets of three real numbers that satisfy specific equations I am having trouble finding a set of three numbers of real numbers $(x, y, z)$ satisfying $x + y + z = xy + xz + yz = 3$. I have tried factoring the equations around but I'm not having any luck and I don't think I'm going on the right path.
| Clearly $a=b=c=1$ is a solution.
Let $a=(1-\alpha)$, and $b=(1-\beta)$. Then $a+b+c=3$ implies
$c = (1 + \alpha + \beta)$
\begin{align}
ab + bc + ca &= 3 \\
(1-\alpha)(1-\beta) +
(1-\beta)(1 + \alpha + \beta) +
(1 + \alpha + \beta)(1-\alpha) &= 3 \\
-\alpha^2 - \alpha \beta - \beta^2 + 3 &= 3 \\
\alpha^2 + \alpha \beta + \beta^2 &= 0 \\
2\alpha^2 + 2\alpha \beta + 2\beta^2 &= 0 \\
\alpha^2 + \beta^2 + (\alpha^2+2\alpha \beta + \beta^2) &= 0 \\
\alpha^2 + \beta^2 + (\alpha + \beta)^2 &= 0
\end{align}
Which implies $\alpha = \beta = 0$. So $a=b=c=1$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Finding determinant for characteristic polynomial The question I'm currently working on has boiled down to
$\chi_A(t) = \det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix}$
I can see I want to use EROs (and taking note of the changes, if any, the EROs make to the determinant) to end up with a lower/upper triangular matrix, so then the determinant is just the product of the entries along the main diagonal, but I'm not sure how to get it.
Swapping the first column and the last column would multiply the determinant by $-1$,but would leave us with only a $t$ and $-1$ in the last column to get rid of, giving a lower triangular matrix - but I can't see how I'd do this.
Many thanks in advance!
EDIT:
Swapping the first and last column multiplies the determinant by $-1$, so
$\chi_A(t) = -\det \begin{bmatrix} -a_0 & 0 & 0 & \cdots & 0 & t \\ -a_1 & t & 0 & \cdots & 0 & -1 \\ -a_2 & -1 & t & \cdots & 0 & 0 \\ -a_3 & 0 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -a_{n-1} & 0 & 0 & \cdots & t & 0 \\ t-a_n & 0 & 0 & \cdots & -1 & 0 \end{bmatrix}$
Adding $t$ times row two to row one doesn't change the determinant and so
$\chi_A(t) = -\det \begin{bmatrix} -a_0 - ta_1 & t^2 & 0 & \cdots & 0 & 0 \\ -a_1 & t & 0 & \cdots & 0 & -1 \\ -a_2 & -1 & t & \cdots & 0 & 0 \\ -a_3 & 0 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -a_{n-1} & 0 & 0 & \cdots & t & 0 \\ t-a_n & 0 & 0 & \cdots & -1 & 0 \end{bmatrix}$
This doesn't seem to have helped...
| Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix, we see that:
$\det \begin{bmatrix} t & -a_0 \\ -1 & t-a_1 \end{bmatrix} = t(t-a_1) - a_0 = t^2 - a_1t - a_0$
and, by expansion along the first row:
$\det \begin{bmatrix} t & 0 & -a_0 \\ -1 & t & -a_1 \\ 0 & -1 & t-a_2 \end{bmatrix} = t \times\det \begin{bmatrix} t & -a_1 \\ -1 & t-a_2 \end{bmatrix} + (-a_0) \det\begin{bmatrix} -1 & t \\ 0 & -1 \end{bmatrix}$
$= t[t(t-a_2) - a_1] - a_0 = t^3 - a_2t^2 - a_1t - a_0 $
So it looks like:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$
Which we can prove by induction.
Assume that:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & t-a_{n-1} \end{bmatrix} = t^{n} - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - ... - a_2t^2 - a_1t - a_0$
Then, by expansion along the first row:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t \det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_1 \\ -1 & t & 0 & \cdots & 0 & -a_2 \\ 0 & -1 & t & \cdots & 0 & -a_3 \\ 0 & 0 & -1 & \cdots & 0 & -a_4 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} $
$+ (-1)^{n+1} \times (-a_0)(-1)^n $
$ = t[t^{n} - a_{n}t^{n-1} - a_{n-1}t^{n-2} - ... - a_3t^2 - a_2t - a_1] + (-1)^{2n+1} a_0$
$ = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$
Proof complete.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\sum^{7}_{k=1}\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}$ How do I evaluate this?
$$\sum^{7}_{k=1}\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}$$
I know I'm suppose to use substitutions but didn't work out so far. Help anyone?
| Indeed, make a substitution:
Let $\boxed{x=\sqrt[3]{k}}$ and $\boxed{y=\sqrt[3]{k+1}}$.
$$\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}=\frac{1}{x^2+xy+y^2}=\frac{y-x}{(y-x)(y^2+xy+x^2)}=\frac{y-x}{y^3-x^3}={y-x}{(k+1)-k}=y-x=\sqrt[3]{k+1}-\sqrt[3]{k}$$
Now, we know that the whole expression can be expressed as:
$$\therefore \sum^{7}_{k=1}\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}=(\sqrt[3]{2}-\sqrt[3]{1})+(\sqrt[3]{3}-\sqrt[3]{2})+\ldots +(\sqrt[3]{8}-\sqrt[3]{7})=\sqrt[3]{8}-1=\boxed{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Poincaré-Bendixon show periodic solutions.
Show that the system $x^{'}=x-y-x^{3}$,$y^{'}=x+y-y^{3}$ has a periodic solution.
I converted to polar:
$r r^{'}=x^{'}x+y^{'}y.$
Thus
$r r^{'}=x^{2}-x^{4}+y^{2}-y^{4}.$
Collecting squares:
$rr^{'}=x^{2}+y^{2}-(x^{2}+y^{2})^{2} +2x^{2}y^{2}.$
Substitute:
$r^{'}=r-r^{3}(1-(\sin^{2}(2\theta))/2)$
$r^{'}=r-r^{3}(1+\cos^{2}(2\theta))/2.$
Setting $r^{'}=0$ yields
$r^{2}=2/(1+\cos^{2}(2\theta).$
Thus $r_{max}= 2^{(1/2)}$ and $r_{min}=1.$
I suck at trig and this doesn't look right. Clearly $(0,0)$ is the only fixed point, so I have a bounded region. I just don't think I have found correct values for $r_{min}$ and $r_{max}$ to find my limit cycle.
| Dynamical system
Find the periodic solution for the dynamical system:
$$
%
\begin{align}
%
\dot{x} &= x - y - x^{3} \\
%
\dot{y} &= x + y - y^{3} \\
%
\end{align}
\tag{1}
%
$$
Fixed points
There is a lone fixed point at the origin
$$
\left[ \begin{array}{c}
\dot{x} \\ \dot{y}
\end{array} \right]_{(0,0)}
=
\left[ \begin{array}{c}
0 \\ 0
\end{array} \right]
$$
Methods
Compute $\dot{r}$.
Use the polar coordinate transform
$$
%
\begin{align}
%
x &= r \cos \theta \\
%
y &= r \sin \theta \\
%
\end{align}
%
$$
which implies
$$
r^{2} = x^{2} + y^{2}
\tag{2}
$$
Compute the derivative with respect to time for $(2)$ and use the definitions in $(1)$. This leads to
$$
\dot{r} = r - r^{3} \left( \cos^{4} \theta + \sin^{4} \theta \right)
= r
\left(
1 - \frac{1}{4} \left( 3 + \cos 4 \theta \right) r^{2}
\right)
\tag{3}
$$
Trapping region
To categorize the flow, identify regions where $r$ is expanding or shrinking. Consider the limiting values
$$1 \le \cos 4\theta \le 1.$$
*
*$\cos 4\theta = 1$
Equation $(3)$ reduces to
$$
\dot{r}
= r \left(
1 - r^{2}
\right)
$$
The conclusion is
$$
\begin{cases}
\color{blue}{\dot{r} > 0} & \color{blue}{r < 1} \\
\color{red}{\dot{r} < 0} & \color{red}{r > 1} \\
\end{cases}
$$
*$\cos 4\theta = -1$
Equation $(3)$ reduces to
$$
\dot{r}
= r \left(
1 - \frac{1}{2}r^{2}
\right)
$$
This implies
$$
\begin{cases}
\color{blue}{\dot{r} > 0} & \color{blue}{r < \sqrt{2}} \\
\color{red}{\dot{r} < 0} & \color{red}{r > \sqrt{2}} \\
\end{cases}
$$
The figures below depict the zone structure for the two limiting conditions. Combine these zone structures and identify where, in both figures, is $\dot{r}$ always positive, and always negative. There will be an annulus where the sign of $\dot{r}$ isn't specified. The third figure shows this trapping region in gray.
The trapping region is the set $ 1 \le r \le \sqrt{2}$
Results
The flow field is plotted next. The dashed, red lines are the nullclines. The gray annulus is the trapping region.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a_n = \{\left(1+\frac{1}{n}\right)^n\}$ is bounded sequence, $ n\in\mathbb{N}$ How to prove the following:
$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $ n\in\mathbb{N}$
| Let me denote $S_n=\{(1+\frac{1}{n})^n\}$
\begin{align*}
\bigg(1+\frac{1}{n}\bigg)^n &=1+ {_nC_1} \frac{1}{n}+ {_nC_2} \bigg(\frac{1}
{n}\bigg)^2+..........+{_nC_n} \bigg(\frac{1}{n}\bigg)^n\\
&=1+ n \frac{1}{n}+\frac{n(n-1)}{2!} \bigg(\frac{1}{n}\bigg)^2+ \frac{n(n-1)(n-2)}{2!} \bigg(\frac{1}{n}\bigg)^3+..........+\frac{n(n-1)(n-2)....(n-(n-1))}{n!}\bigg(\frac{1}{n}\bigg)^n\\
&=1+ 1+ \frac{(1-\frac{1}{n})}{2!} + \frac{(1-\frac{1}{n})(1-\frac{2}{n})} {3!}+..........+\frac{(1-\frac{1}{n})(1-\frac{2}{n}).....(n-\frac{n-1}{n})} {n!}\\
&<1+ 1+ \frac{1}{2!} + \frac{1}{3!}+..........+\frac{1}{n!}\\
&<1+ 1+ \frac{1}{2} + \frac{1} {2^2}+..........+\frac{1}{2^{n-1}}\\
&=1+\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}}\\
&=1+2\bigg(1-\frac{1}{2^n}\bigg)\\
&=3-\frac{1}{2^{n-1}}\\
&<3\\
\end{align*}
$S_n$ is bounded by 3.
Also note that $S_n>2$.
Here I'm not writing the reason in every step i hope u will understand.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Integral inequality: $\frac{1}{30}<\int_{2}^{\infty}\frac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\,ds<\frac{\sqrt{2}}{20}$ Prove that
$$\frac{1}{30}<\int_{2}^{\infty}\frac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\,ds<\frac{\sqrt{2}}{20}$$
I think this inequality can $s^5+s^2+1>( ) $
and $\sqrt{s^3-s^2+3}<()$?
| For right inequality, write
$$\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\mathrm{d}s < \int_{2}^{\infty}\dfrac{\sqrt{s^3}}{s^5}\mathrm{d}s = \int_{2}^{\infty}s^{-\frac{7}{2}} \mathrm{d}s=\frac{\sqrt{2}}{20}$$
For left one, you may try
$$\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\mathrm{d}s > \int_{2}^{\infty} \dfrac{\sqrt{7}}{s^5+s^3+s^2+1} \mathrm{d}s$$
Then using the fact that $s^5+s^3+s^2+1 = (s+1)(s^2+1)(s^2-s+1)$, you have
$$\frac{1}{s^5+s^3+s^2+1}=-\frac{2s-1}{3(s^2-s+1)}+\frac{s+1}{2(s^2+1)}+\frac{1}{6(s+1)}$$
Then you can compute the integral,
$$\int_{2}^{\infty} \dfrac{\sqrt{7}}{s^5+s^3+s^2+1} \mathrm{d}s = \sqrt{7} \left(-\frac{\log 5}{4}+\frac{\log 3}{6}+{1\over2}\arctan {1\over2}\right)$$
But it's a bit below $\frac{1}{30}$ ($\approx 0.997 \cdot \frac{1}{30}$) :-(
| {
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Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.
I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this.
Thank you!
Sorry, my approach:
$$\begin{align}
\frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\
\sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\
\theta + 45^\circ &= 45^\circ,\ 135^\circ \\
\theta &= 0^\circ, \ 90^\circ
\end{align}$$
| $\sin x+ \cos x=1 \implies \sin^2 x + \cos^2 x+ 2 \sin x \cos x = 1 \implies 2\sin x \cos x =0$, so either $\sin x=0$ or $\cos x = 0$, giving you the solutions $0$, $\dfrac {\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
How to solve the equation $\sqrt{x^{2}-x+2}+\sqrt{2x^{2}+2x+3}=\sqrt{2x^{2}-1}+\sqrt{x^{2}-3x-2}$
Solve in real numbers the equation $$\sqrt{x^{2}-x+2}+\sqrt{2x^{2}+2x+3}=\sqrt{2x^{2}-1}+\sqrt{x^{2}-3x-2}.$$
| Rewrite the equation as
$$\sqrt{x^{2}-x+2}-\sqrt{x^{2}-3x-2}=\sqrt{2x^{2}-1}-\sqrt{2x^{2}+2x+3}.\tag{$1$}$$
On each side, do the usual trick of multiplying "tops" and "bottoms" by the conjugates. A semi-miracle happens. We get
$$\frac{2x+4}{ \sqrt{x^{2}-x+2}+\sqrt{x^{2}-3x-2} }=\frac{-(2x+4)}{\sqrt{2x^{2}-1}+\sqrt{2x^{2}+2x+3}}.\tag{$2$}$$
After making note to check about $x=-2$ later, cancel the $2x+4$ terms on top of $(2)$, then flip both sides of $(2)$ over.
We get an equation that marries very nicely with $(1)$, and two of the square roots are gone. We get an equation of shape $\sqrt{A}=-\sqrt{B}$. So it comes down to checking $x=-2$ (works).
Remark: The trick could have been used directly with the equation as given. However, in more complicated cases, reduction of degree is useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\lim_{n \to \infty} \cos \frac {\pi}{2^2}\cos \frac {\pi}{2^3}\cos \frac {\pi}{2^4}......\cos \frac {\pi}{2^n}=\frac {2}{\pi}$ I came across the following problem that says:
prove that $$\lim_{n \to \infty} \cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}=\dfrac {2}{\pi}$$.
My Attempt: Let $$P=\lim_{n \to \infty} [\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}] \implies \log P=\lim_{n \to \infty} \sum_{r=2}^{n}\log (\cos \dfrac {\pi}{2^r})$$.
Now,I am stuck and not sure which way to go. Can someone point me in the right direction? Thanks in advance for your time.
| Take the term inside the limit:
$$\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}$$
Now multiply and divide by $2*\sin \dfrac {\pi}{2^n}$. You get:
$$\dfrac{\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......2*\sin \dfrac {\pi}{2^n}\cos \dfrac {\pi}{2^n}}{2*\sin \dfrac {\pi}{2^n}}$$
$2*\sin \dfrac {\pi}{2^n}\cos \dfrac {\pi}{2^n}$ = $\sin \dfrac {\pi}{2^{n-1}}$
Now multiply and divide by 2 and continue this process till you get :
$\dfrac{\sin \dfrac {\pi}{2}}{2^{n}*\sin \dfrac {\pi}{2^n}}$
Applying limit to this using the formula $$\lim_{x \to \infty}\dfrac{sin(x)}{x} = 1$$ will give you the answer as : $\dfrac{2}{\pi}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluation of a continued fraction Puzzle question... I know how to solve it, and will post my solution if needed; but those who wish may participate in the spirit of coming up with elegant solutions rather than trying to teach me how to solve it. [paraphrased from Lone Learner]
Prove (or disprove) the following equality:
$$1+\cfrac1{1+\cfrac2{1+\cfrac3{1+\ddots}}}=\frac1{\displaystyle e^{1/2}\sqrt{\frac{\pi}{2}}\;\mathrm{erfc}\left(\frac1{\sqrt2}\right)}\approx 1.525135276\cdots$$
(taken from Closed form for a pair of continued fractions)
| Let
\begin{align*}
p_n &= p_{n-1}+np_{n-2}, \quad p_{-1} = 1, \quad p_0=1,
\\
q_n &= q_{n-1}+nq_{n-2}, \quad q_{-1} = 0, \quad q_0=1,
\\
r_n &= \frac{p_n}{q_n}.
\end{align*}
So the $r_n$ are the convergents of the continued fraction:
$$
r_0 = 1,\qquad
r_1 = 1+\cfrac{1}{1},\qquad
r_2 = 1+\cfrac{1}{1+\cfrac{2}{1}},\qquad
r_3 = 1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1}}},
$$
and so on. Consider the exponential generating functions
$$
F(x) = \sum_{n=0}^\infty \frac{p_{n-1}}{n!}\,x^n,\qquad
G(x) = \sum_{n=0}^\infty \frac{q_{n-1}}{n!}\,x^n .
$$
They are solutions of the differential equations
\begin{align*}
&F'(x) + (-1-x)F(x) = 0,\quad F(0)=1,
\\
&G'(x) + (-1-x)G(x) = 1,\quad G(0)=0 .
\end{align*}
A solution for the homogeneous equation $y'+(-1-x)y=0$ is
$$
F_1(x) = e^{(1+x)^2/2}
$$
and a particular solution for the inhomogeneous equation
$y'+(-1-x)y=1$ is
$$
F_2(x) = \sqrt{\frac{\pi}{2}}\,e^{(1+x)^2/2}\,
\mathrm{erf}\,\frac{x+1}{\sqrt{2}}
$$
After all, there are formulas for the solution of first-order
linear differential equations.
Hayman's method (Wilf, generatingfunctionology, Theorem 4.5.1)
shows that the coefficients of $F_1$ are asymptotic to
$$
\frac{e^{n/2+\sqrt{n}+1/4}}{2n^{(n+1)/2}\sqrt{\pi}}
$$
as $n \to \infty$ and the coefficients of $F_2$ are asymptotic to
$$
\frac{e^{n/2+\sqrt{n}+1/4}}{2\sqrt{2}n^{(n+1)/2}}
$$
Applying the initial conditions, we conclude that
$$
F(x) = e^{-1/2}F_1(x),\qquad
G(x) = F_2(x)-\sqrt{\frac{\pi}{2}}\,\mathrm{erf}\,\frac{1}{\sqrt{2}}
\,F_1(x)
$$
So
\begin{align*}
\frac{p_{n-1}}{n!} &\sim \frac{e^{n/2+\sqrt{n}-1/4}}{2n^{(n+1)/2}\sqrt{\pi}};
\\
\frac{q_{n-1}}{n!} &\sim
\frac{e^{n/2+\sqrt{n}+1/4}}
{2\sqrt{2}n^{(n+1)/2}} -
\frac{e^{n/2+\sqrt{n}+1/4}\sqrt{\pi}\text{erf}(1/\sqrt{2})}
{2\sqrt{2}n^{(n+1)/2}\sqrt{\pi}} =
\frac{e^{n/2+\sqrt{n}-1/4}e^{1/2}\text{erfc}(1/\sqrt{2})}
{2\sqrt{2}n^{(n+1)/2}}
\end{align*}
with $\text{erfc}(x)=1-\text{erf}(x)$.
Finally,
$$
\frac{p_{n-1}}{q_{n-1}} \sim
\frac{1}{\displaystyle
e^{1/2}\;\sqrt{\frac{\pi}{2}}\;\text{erfc}\,\frac{1}{\sqrt{2}}}
\approx 1.525135276
$$
so that is the value of the continued fraction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Fast partial-fraction decomposition I'm studying Laplace transformations for my differential equations class and typically there's a partial fraction decomposition involved, which can be very long and demanding for calculations by hand, if done the standard way.
I am aware of some of the tricks used to speed-up this procedure (like the usage of limits at infinity, or multiplying with denominators and taking particular values of $s$), however I am not able to apply them at this example:
$$\frac {s} {(s^2+2s+5)(s^2+4)} = \frac {\alpha s + \beta} {s^2+2s+5} + \frac {\gamma s + \delta} {s^2+4}$$
For example, if I attempt to extract a relation for $\gamma$ and $\delta$, by multiplying with $s^2+4$ and taking $s=2i$, I get and equation that involves complex numbers; that makes me feel I have not gained much in terms of number of operations.
Are there any better tricks for this example?
| Well, the following is certainly a trick. Whether it's better or not remains to be decided....
First, I want to make the factors in the denominator symmetric with respect to the origin (right now they're of the form $s^2+4$ and $(s+1)^2+4$, hence symmetric with respect to $1/2$). So setting $s=t-1/2$, we have the equivalent problem of doing partial fractions on
$$
\frac{16 t-8}{(4 t^2-4t+17)(4 t^2+4t+17)}.
$$
Let's start by noting (using the symmetry) that
$$
\frac{1}{4 t^2-4t+17} - \frac{1}{4 t^2+4t+17} = \frac{8t}{(4 t^2-4t+17)(4 t^2+4t+17)}.
$$
How can we divide everything in sight by $t$? Notice that the inverse of $t$ modulo $4t^2-4t+17$ is $(-4t+4)/17$, while the inverse of $t$ modulo $4t^2+4t+17$ is $(-4t-4)/17$. So we get
\begin{align*}
\frac{8}{(4 t^2-4t+17)(4 t^2+4t+17)} &= \frac{1/t}{4 t^2-4t+17} - \frac{1/t}{4 t^2+4t+17} \\
&\equiv \frac{-4t+4}{17(4 t^2-4t+17)} + \frac{4t+4}{17(4 t^2+4t+17)} \pmod1.
\end{align*}
But this (mod 1)-equivalence must actually be an equality(!): by the theory of partial fractions, the left-hand side (whose numberator has larger degree than the denominator) has a representation in the form of the right-hand side.
Between the last two identities, we see that
$$
\frac{16 t-8}{(4 t^2-4t+17)(4 t^2+4t+17)} = \frac{42 t+30}{17 (4 t^2-4 t+17)}-\frac{4t+38}{17 (4 t^2+4 t+17)};
$$
now plugging $t=s+1/2$ back in yields
$$
\frac s{(s^2+2s+5)(s^2+4)} = \frac{-s-10}{17 (s^2+2 s+5)}+\frac{s+8}{17 (s^2+4)}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$? Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$?
The question is mentioned in my math olympiad. Please explain how to solve the problem. I have factorised the equation to $$-x^2+1, x-6x, -5x +1.$$
I am only in year 6.
| If $r,s,t$ be the roots of the equation $x^3 - 6x^2 + 5x + 1=0$, then $$x^3 - 6x^2 + 5x + 1 = (x - r)(x - s)(x - t)$$
Now, determine the value of $f(2)$ by evaluating $$x^3 - 6x^2 + 5x + 1\quad\text{at} \;x = 2\;\tag{$f(2)$}$$ Then set $$(2 - r)(2-s)(2 - t) = f(2)\tag{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions. The following problem was on a math competition that I participated in at my school about a month ago:
Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.
I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):
$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$
I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get
$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$
and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get
$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$
where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become
$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$
and
$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$
Then, by a short optimization argument, I showed that these last two equations have no real solutions.
First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?
| $$\cos(\sin x)=\sin(\cos x)=\cos\left(\frac\pi2-\cos x\right)$$
$$\text{So}, \sin x=2n\pi\pm \left(\frac\pi2-\cos x\right)$$
$$\text{Taking the '+' sign,} \sin x=2n\pi+ \left(\frac\pi2-\cos x\right)\implies \sin x+\cos x=\frac{(4n+1)\pi}2$$ which can be $\cdots,-\frac{3\pi}2,\frac{\pi}2,\frac{5\pi}2,\cdots$
$$\text{Taking the '-' sign,} \sin x=2n\pi- \left(\frac\pi2-\cos x\right)\implies \sin x-\cos x=\frac{(4n-1)\pi}2$$ which can be $\cdots,-\frac{5\pi}2,-\frac{\pi}2,\frac{3\pi}2,\cdots$
Now let, $1=r\cos\theta,1=r\sin\theta$ where $r>0$
So, $(r\cos\theta)^2+(r\sin\theta)^2=1+1=2\implies r^2=2\implies r=\sqrt2$
$\sin x\pm\cos x=r\cos\theta\sin x\pm r\sin\theta\cos x=\sqrt2\sin(x\pm \theta)$
So, $-\sqrt2\le \sin x\pm\cos x\le \sqrt 2 $
Now, $\sqrt 2<1.5<\frac\pi2$ as $3<\pi$
$\implies -\sqrt 2>-\frac\pi2$
$\implies -\frac\pi2<-\sqrt2\le \sin x\pm\cos x\le \sqrt 2<\frac\pi2 $
Hence, there is no real soultion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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How can I solve this system? Having this system:
$$\begin{cases}
2 x(x^2 + y - 11) + x + y^2 - 7 = 0,\\
x^2 + y - 11 + 2 y (x + y^2 - 7) = 0 .
\end{cases}$$
How can I solve it without using a computer?.
Thank you!
| Note first that a substitution of $x=0$ yields the system $$\begin{cases}y^2-7=0\\y-11+2y(y^2-7)=0,\end{cases}$$ which you can show has no solutions in $y$. Thus, we needn't worry about any problems that might arise from having $x=0$.
Now, multiplying the second equation in our original system by $2x$ gives us the equivalent system $$\begin{cases}2x(x^2+y-11)+x+y^2-7=0\\2x(x^2+y-11)+4xy(x+y^2-7)=0,\end{cases}$$ from which it follows that $$(4xy-1)(x+y^2-7)=0,$$ meaning $$y=\frac1{4x}\quad\text{or}\quad x=7-y^2.$$ Now, by $x=7-y^2$ and $x^2+y-11+2y(x+y^2-7)=0,$ we find $$0=(7-y^2)^2+y-11=y^4-14y^2+y+38.$$ Using the rational root test and a little trial and error, we find that $2$ is a root of that equation, and we can factor it as $$0=(y-2)(y^3+2y^2-10y-19).$$ Thus, $(3,2)$ is one solution to the system. We can explicitly calculate the roots of that cubic factor using Cardano's Method--be warned, they are real, even though they won't look like it, and they aren't very nice in any case--and from substituting those $y$-values back into $x=7-y^2$ find the corresponding $x$-values.
From $y=\frac1{4x}$ and $2x(x^2+y-11)+4xy(x+y^2-7)=0$ we find $$0=2x^3+\frac12-22x+x+\left(\frac1{4x}\right)^2-7=2x^3-21x-\frac{13}2+\frac1{16x^2},$$ whence multiplication by $16x^2$ yields $$0=32x^5-336x^3-104x^2+1.$$ This is messier, still. No rational roots exist, and quintics don't have a nice general method of solution. I wouldn't hold out much hope....
| {
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"timestamp": "2023-03-29T00:00:00",
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Pattern to last three digits of power of $3$? I'm wondering if there is a pattern to the last three digits of a a power of $3$? I need to find out the last three digits of $3^{27}$, without a calculator.
I've tried to find a pattern but can not see one? Am I missing something?
Thanks for your help in advance!
| Since $1000 = 2^3 5^3$, and those factors are relatively prime, you can determine $3^n \mod 1000$ by determining $3^n \pmod 8$ and $3^n \pmod {125}$,
$3^n \pmod 8$ is easy, as $3^2 = 9 \equiv 1 \pmod 8$.
$3^n \pmod {125}$ is tedious. Work with smaller powers of $5$ as exponent first:
$3^4 \equiv 1 \pmod 5$. The exponent must divide $\phi(5) = 4$.
$3^{20} \equiv 1 \pmod{25}$. The exponent must divide $\phi(25) = 20$ and be a multiple of $4$ from the above.
$3^{20} \equiv 26 \pmod {125}$. I cheated and used a calculator. But $3^{100} \equiv 1 \pmod {125}$. The exponent must divide $\phi(125) = 100$ and be a multiple of $20$.
So, $3^{27} \equiv 3 \pmod 8$ and $3^{27} \equiv 26\cdot3^7 \pmod {125}$.
$3^5 = 243 \equiv -7 \pmod {125}$. $3^7 \equiv -7 \cdot 9 \equiv -63 \equiv 62 \pmod {125}$. So $3^{27} \equiv 26 \cdot 62 \pmod {125}$.
Use the Chinese Remainder Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the formula of: $a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$? What is the formula of:
$$a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$$
Any ideas?
| If $a=1,a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}=1+1+\cdots$ up to $(n+1)$ terms hence $=n+1$
Else let $S=a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$
So, $a\cdot S=a^{1} + a^{2} + a^{3} + ... + a^{n} + a^{n+1}$
So, $S(a-1)=a^{n+1}-1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim_{n\to\infty}\frac{n+1}{n-2}=1$
By epsilon-delta definition, show that $\displaystyle\lim_{n\to\infty}\frac{n+1}{n-2}=1$.
I start with $\left|\frac{n+1}{n-2}-1\right|=\left|\frac 3 {n-2}\right|$, but I don't know how to carry on. Thank you.
| Write $\frac{n+1}{n-2} = 1+\frac{3}{n-2}$. Then $\left|\frac{n+1}{n-2} -1\right| = \left|\frac{3}{n-2}\right|$. Let $\epsilon>0$ and choose $N > 2+\frac{3}{\epsilon}$. Then if $n \ge N$, you have $\left|\frac{3}{n-2}\right|< \epsilon$.
To get a lower bound for $N$: We want $\left|\frac{3}{n-2}\right|< \epsilon$. This will be true if $3 < |n-2| \epsilon$, or equivalently $\frac{3}{\epsilon} < |n-2|$. Since we are interested in $n \to \infty$, we ask that $n$ satisfy $\frac{3}{\epsilon} < n-2$, or equivalently, $2+\frac{3}{\epsilon} < n$. So, if we pick $N > 2+\frac{3}{\epsilon}$, then if $n \ge N$, we will have the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Modular Algebra with Powers? Is it possible to calculate by modular algebra rules (with no calculators) numbers with powers $\textrm{mod}\ n$? Something like: $7^{127} \textrm{mod}\ 11$. If so, how it can be done?
Any explanation will be apriciated!
| First,
$$7^{127} \equiv 8 \cdot 7 \cdot 7^{127} \equiv 8 \cdot 7^{128} \equiv 8 \cdot 7^{2^7} \pmod{11}$$
Now, step by step:
$$7^2 \equiv 49 \equiv 5 \pmod{11}$$
$$5^2 \equiv 25 \equiv 3 \pmod{11}$$
$$3^2 \equiv 9 \equiv -2 \pmod{11}$$
$$(-2)^2 \equiv 4 \pmod{11}$$
$$4^2 \equiv 16 \equiv 5 \pmod{11}$$
You have still two squares to compute, but you can find them already computed above:
$$5^2 \equiv 25 \equiv 3 \pmod{11}$$
$$3^2 \equiv 9 \equiv -2 \pmod{11}$$
Now, we have squared $7$ times, so we have computed:
$$\left(\left(\left(\left(\left(\left(7^2\right)^2\right)^2\right)^2\right)^2\right)^2\right)^2 \equiv 7^{2^7} \equiv -2 \pmod{11}$$
And finally
$$7^{127} \equiv 8 \cdot 7^{128} \equiv 8 \cdot -2 \equiv -16 \equiv 6 \pmod{11}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$ How would I go about evaluating this integral?
$$\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx.$$
What I've tried so far: I tried a semicircular integral in the positive imaginary part of the complex plane, excluding the negative real axis, but had trouble calculating the residue at $z=i$ (perhaps there is a way of doing this that I don't know of). After that didn't work, I tried a rectangular box integral from $\epsilon$ to $R$, from $R$ to $R+i/2$, from $R+i/2$ to $-S+i/2$, from $-S+i/2$ to $-S+i\epsilon$, from $-S+i\epsilon$ to $-\epsilon+i\epsilon$ and finally a semicircle around the origin, radius $\epsilon$.
Any help would be appreciated.
| Here is a way by using contour integration and easy to understand:
Convention:
The branch of the logarithm is principal branch cut.
The argument is using under principal value.
$$\int_{0}^{\infty} \frac{\log(1+x^2)}{1+x^2}~dx = \\ \int_{0}^{\infty} \frac{\log(x^2+1)}{1+x^2}~dx = \\ \int_{0}^{\infty} \frac{\log((x+i)(x-i))}{1+x^2}~dx = \\\int_{0}^{\infty} \frac{\log(|x+i|) + Arg(x+i) + log(|x-i|)+ Arg(x-i) }{1+x^2}~dx = \\ \int_{0}^{\infty} \frac{\log(|x+i|) + log(|x-i|)+ Arg(2x) }{1+x^2}~dx = \\$$$$\int_{0}^{\infty} \frac{\log(|x+i|) + log(|x-i|)}{1+x^2}~dx \tag{1}
$$
Doing contour integration with respect to function: $f(x) = log(x+i)/(x^2+1)$
Define contour:
$\Gamma_1:= x \text{ from } 0 \text{ to } \infty$
$\Gamma_2:= x \text{ from } \infty \text{ to } -\infty \text{ along the upper semicircle}$
$\Gamma_3:= x \text{ from } -\infty \text{ to } 0$
Easy to see: $\int_{\Gamma_2} |f(x)| \leq 2 \pi \max_{\Gamma_2} {\frac{|\log(x+i)|}{|1+x^2|}} \leq \frac{log(\sqrt{x^2+1})}{x^2+1}+\frac{\pi^2}{x^2+1} = 0$
So $\int_{\Gamma_2} f(x) = 0$
The residue for $f(x)$ at $x = i$ is $\frac{\log(2)+ \frac{\pi}{2}}{2i}$
By residue theorem: $\int_{\Gamma_1} f(x) + \int_{\Gamma_3} f(x) = 2\pi Res(f,i) = \pi \log(2) + \frac{\pi^2 i}{2} $
$$\int_0^\infty \frac{log(x+i)}{(x^2+1)} + \int_{-\infty}^0 \frac{log(x+i)}{(x^2+1)} = \pi \log(2) + \frac{\pi^2 i}{2} \implies\\\int_0^\infty \frac{log|x+i|}{(x^2+1)} + \int_{-\infty}^0 \frac{log(|x+i|)+ \pi i}{(x^2+1)} = \pi \log(2) + \frac{\pi^2 i}{2} \implies\\\int_0^\infty \frac{log|x+i|}{(x^2+1)} + \int_{-\infty}^0 \frac{log(|x+i|)}{(x^2+1)}+ \int_{-\infty}^0 \frac{ \pi i}{(x^2+1)} = \pi \log(2) + \frac{\pi^2 i}{2} \\$$
Recall: $\int \frac{1}{x^2+1} = arctan(x)+c$
$$ \implies \int_0^\infty \frac{log|x+i|}{(x^2+1)} + \int_{0}^{\infty} \frac{log(|x-i|)}{(x^2+1)}+ \frac{i\pi^2}{2} = \pi \log(2) + \frac{i\pi^2}{2}\\\\\implies \int_0^\infty \frac{log|x+i| + log|x-i|}{(x^2+1)} = \pi \log(2)
$$
Substitute into $(1)$,$\int_{0}^{\infty} \frac{\log(|x+i|) + log(|x-i|)}{1+x^2}~dx = \pi \log(2)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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Mathematical Expressions; School Homework $y + 3$ is always $5$ more than $y – 2$
so $y + 3 – (y – 2) = 5$
$(y + 4 ) – (y – 3) = ~ ?$
$(y - 2) – (y - 3) = ~ ?$
How would you work it out? I know that $y + 3$ is 5 more than $y - 2$, so should I do trial method or is there a method that's easier and faster?
Oh and why type of mathematical equation is this? Please make it easy. Thanks!
| The distributive property says that $-\rm something = -1 \times \rm something$. Hence, $y + 3 - (y - 2) = y + 3 - 1(y - 2)$.
The original definition is that $a(b + c) = ab + ac$.
So $-1(y - 2) = (-1\times y) + (-1\times-2) = -y + 2 $.
You are left with
$y+3 - y + 2$
$= y - y + 3 +2$
$= 0 + 3+2$
$= 5$.
For the second example, you have
$y + 4 - (y - 3)$
$= y + 4 - 1(y - 3)$
$= y + 4 - y + 3$
$= y -y + 4 + 3$
$= 7$
Try doing the third yourself.
| {
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Find ${{{d^2}y} \over {d{x^2}}}$ when $x = {1 \over {t + 1}}$ and $y = {1 \over {t - 1}}$ The answer in the text book is:
${{{d^2}y} \over {d{x^2}}} = 4{\left( {{{t + 1} \over {t - 1}}} \right)^3}$
I've tried this:
$\eqalign{
& x = {1 \over {t + 1}},y = {1 \over {t - 1}} \cr
& x = {(t + 1)^{ - 1}},y = {(t - 1)^{ - 1}} \cr
& {{dx} \over {dt}} = - {(t + 1)^{ - 2}} = {{ - 1} \over {{{(t + 1)}^2}}} \cr
& {{dy} \over {dt}} = - {(t - 1)^{ - 2}} = {{ - 1} \over {{{(t - 1)}^2}}} \cr
& {{dy} \over {dx}} = {{dy} \over {dt}} \times \left( {{1 \over {{{dx} \over {dt}}}}} \right) \cr
& {{dy} \over {dx}} = {{ - 1} \over {{{(t - 1)}^2}}} \times {{{{(t + 1)}^2}} \over { - 1}} = {{{{(t + 1)}^2}} \over {{{(t - 1)}^2}}} \cr} $
My approach to ${{{d^2}y} \over {d{x^2}}}$ is:
$${{dy} \over {dx}} = {{{{(t + 1)}^2}} \over {{{(t - 1)}^2}}}$$
So using the quotient rule:
$\eqalign{
& u = {(t + 1)^2} \cr
& v = {(t - 1)^2} \cr
& {{du} \over {dt}} = 2(t + 1) \cr
& {{dv} \over {dt}} = 2(t - 1) \cr
& {{{d^2}y} \over {d{x^2}}} = {{2{{(t - 1)}^2}(t + 1) - 2{{(t + 1)}^2}(t - 1)} \over {{{(t - 1)}^4}}} \cr
& {{{d^2}y} \over {d{x^2}}} = {{2(t - 1)(t + 1)\left[ {(t - 1) - (t + 1)} \right]} \over {{{(t - 1)}^4}}} \cr
& {{{d^2}y} \over {d{x^2}}} = {{2(t - 1)(t + 1)( - 2)} \over {{{(t - 1)}^4}}} \cr
& {{{d^2}y} \over {d{x^2}}} = {{ - 4(t + 1)} \over {{{(t - 1)}^3}}} \cr} $
As the previous attempt yielded an incorrect answer I attempted to answer it this way:
$\eqalign{
& {{dx} \over {dt}} = - {(t + 1)^{ - 2}} \cr
& {{dy} \over {dt}} = - {(t - 1)^{ - 2}} \cr
& {{{d^2}x} \over {d{t^2}}} = 2{(t + 1)^{ - 3}} \cr
& {{{d^2}y} \over {d{t^2}}} = 2{(t - 1)^{ - 3}} \cr} $
$\eqalign{
& {{{d^2}y} \over {d{x^2}}} = {{{d^2}y} \over {d{t^2}}} \times \left( {{1 \over {{{{d^2}x} \over {d{t^2}}}}}} \right) \cr
& {{{d^2}y} \over {d{x^2}}} = {2 \over {{{(t - 1)}^3}}} \times {{{{(t + 1)}^3}} \over 2} \cr
& {{{d^2}y} \over {d{x^2}}} = {{{{(t + 1)}^3}} \over {{{(t - 1)}^3}}} \cr} $
I'd appreciate it greatly if someone could point out my misunderstanding(s) and show me the correct approach.
Thank you very much.
| One easy way is to use this: If $x(t),y(t)$ are differentiable functions of $t$ and $x'(t)\neq 0$, then
$$ \frac{dy}{dx}=\frac{y'(t)}{x'(t)}. $$
For this problem $x(t)=\frac{1}{t+1},y(t)=\frac{1}{t-1}$, then
\begin{eqnarray*}
\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{-\frac{1}{(t-1)^2}}{-\frac{1}{(t+1)^2}}=\frac{(t+1)^2}{(t-1)^2}=\left(\frac{t+1}{t-1}\right)^2
\end{eqnarray*}
and hence
\begin{eqnarray*}
\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}=\frac{\frac{-4(t+1)}{(t-1)^3}}{-\frac{1}{(t+1)^2}}=\frac{(t+1)^3}{(t-1)^3}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Complex proof (with sum) How can I prove $$\sum_{n=-\infty}^∞\frac{1}{(a+bn)^2}=\frac{π^2}{b^2} \csc^2 \frac{πa}{b}$$
When $\frac{a}{b}\notin$*Z*?
| You may do so with the residue theorem. Applied to infinite sums, it reads
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{Res}_{z=z_k} \pi \cot{\pi z}\, f(z)$$
where $z_k$ are the poles of $f$ in the complex plane outside of the integers. In your case, there is one pole of $f$, albeit a double pole, at $z=-a/b$. Then
$$\begin{align}\sum_{n=-\infty}^{\infty} \frac{1}{(a+b n)^2}&= \frac{1}{b^2} \sum_{n=-\infty}^{\infty} \frac{1}{(n + (a/b))^2}\\ &=-\frac{1}{b^2} \text{Res}_{z=-a/b} \frac{\pi \cot{\pi z}}{(z + (a/b))^2}\\ &= -\frac{1}{b^2} \lim_{z \rightarrow -a/b} \frac{d}{dz} \left [(z + (a/b))^2 \frac{\pi \cot{\pi z}}{(z + (a/b))^2} \right ] \\ &= -\frac{\pi}{b^2} \left[\frac{d}{dz} \cot{\pi z} \right ]_{z=-a/b}\\ &= \frac{\pi^2}{b^2} \csc^2{\pi \frac{a}{b}} \end{align}$$
as as to be shown.
| {
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Different ways of computing $\sum_{n=0}^{\infty}\binom{n+7}{n}\left(\frac{1}{3}\right)^{n}$ Calculate $\sum_{n=0}^{\infty}\binom{n+7}{n}\left(\frac{1}{3}\right)^{n}$ in various ways.$$$$
one more question: how about $\sum_{n=0}^{\infty}n\binom{n+7}{n}\left(\frac{1}{3}\right)^{n}$ ?
| $$\binom{n+7}{n}x^n=\frac{1}{7!}\frac{d^7}{dx^7}{x^{n+7}}$$
So you have $$\left[\frac{d^7}{dx^7}\sum_{n=0}^\infty\frac{1}{7!}x^{n+7}\right]_{x\to1/3}$$
which is $$\left[\frac{d^7}{dx^7}\frac{1}{7!}x^7\sum_{n=0}^\infty x^{n}\right]_{x\to1/3}$$
which is $$\left[\frac{d^7}{dx^7}\left(\frac{1}{7!}x^7\frac1{1-x}\right)\right]_{x\to1/3}$$
which is, according to the product rule for higher order derivatives, $$\left[\sum_{k=0}^7\binom{7}{k}\frac{d^k}{dx^k}\left(\frac{1}{7!}x^7\right)\frac{d^{7-k}}{dx^{7-k}}\left(\frac1{1-x}\right)\right]_{x\to1/3}$$
which is $$\left[\sum_{k=0}^7\binom{7}{k}\frac{1}{(7-k)!}x^{7-k}(7-k)!(1-x)^{-1-(7-k)}\right]_{x\to1/3}$$
which is $$\sum_{k=0}^7\binom{7}{k}(1/3)^{7-k}(2/3)^{k-8}$$
or just $$\frac{3}{2^8}\sum_{k=0}^7\binom{7}{k}2^k$$
where we see something familiar from the binomial theorem in the sum, and have at long last $$\frac{3}{2^8}(1+2)^7$$ aka $$\left(\frac{3}{2}\right)^8$$
For your second question,
$$\begin{align}n\binom{n+7}{n}x^n&=\frac{1}{7!}\frac{d^7}{dx^7}n{x^{n+7}}\\
&=\frac{1}{7!}\frac{d^7}{dx^7}\left(x^8\frac{d}{dx}x^n \right)\\
\end{align}$$
| {
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Prove that $\lvert\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\rvert<\frac{1}{n^2}$ Prove that $\lvert\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\rvert<\frac{1}{n^2}$ and deduce that $1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}$ is convergent.
Using algebra, the absolute value becomes $\lvert\frac{1}{2n}-\frac{1}{2n-2}\rvert$ which is $\lvert\frac{-(n+1)}{2n(n+1)}\rvert$.
Not entirely sure how to proceed with this proof..
Edit: Incorrect third term, changed to $\frac{1}{2n+2}$.
| I think you have it wrong there. The absolute value becomes $\dfrac{1}{2n^2-2n}$ which is $\dfrac{1}{2n^2-2n}=\dfrac{1}{n^2+n^2-2n}<\dfrac{1}{n^2}$ when $n^2>2n$.
| {
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Writing a series using Sigma notation How do I write $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}?$$
I have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.
| $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
Since everything else has a denominator, let's try putting a denominator for $2$. Since anything divided by one itself, we get:
$$\frac{2}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
Everything else has an $x$. Let's give $2$ an $x$ too! $x^0 = 1$, and anything multiplied by $1$ is itself, so:
$$\frac{2x^0}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
All the other $x$s have an exponent. Let's give one to that $x$ next to the $3$. $x^1 = x$, so we get:
$$\frac{2x^0}{1}+ \frac{3x^1}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
However, these denominators are just powers of two. Therefore, we have:
$$\frac{2x^0}{2^0}+ \frac{3x^1}{2^1} + \frac{4x^2}{2^2}+\frac{5x^3}{2^3}+\frac{6x^4}{2^4}$$
Now we see some patterns:
*
*$2, 3, 4, 5, 6$ (numerator of coefficient of $x$)
*$0, 1, 2, 3, 4$ (exponents of $x$)
*$1, 2, 4, 8, 16$ (denominator of coefficient of $x$. Powers of $2$, viz. $2^0, 2^1, 2^2, 2^3, 2^4$)
We'll make the sum range from $0$ to $4$, although in theory we could actually make it range over anything.
$$\sum_{k=0}^4 $$
We need exponents of $x$, so we'll put those in there.
$$\sum_{k=0}^4 x^k $$
We need those powers of two in the denominator, so add that too:
$$\sum_{k=0}^4 \frac{x^k}{2^k} $$
However, there are still those numerators we haven't checked off our list yet. We can't have them ranging from $0$ to $4$, we need them from $2$ to $6$. The solution is to add two!
$$\sum_{k=0}^4 \frac{(k+2)x^k}{2^k} $$
| {
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Radius of Convergence for $\sum \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}x^n$ I'm trying to find the radius of convergence for this series:
$$\sum \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}x^n$$
so I have,
$$R=\lim_{n\to\infty} \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}\frac{2^{2n+2}(2n+2)!}{[1\cdot 3 \cdots (2n+1)]^2}=\lim_{n\to\infty}\frac{4(2n+1)(2n+2)}{(2n)^2(2n+1)^2}=0$$
But the answer says $4$ and I'm not sure how to get that. Any help?
| $$\left(1 \cdot 3 \cdots (2n-1)\right)^2 = \left(\dfrac{(2n)!}{2^n \cdot (n!)} \right)^2$$
Hence, we have
$$\sum_{n} \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}x^n = \sum_{n} \frac{(2n)!^2}{2^{2n}2^{2n}(2n)! n! n!}x^n = \sum_n \dbinom{2n}n \left(\dfrac{x}{16} \right)^n$$
Can you now work it out?
| {
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Finding the limits related to $a_1=1$, $a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}$ Suppose $a_1=1, a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}, k \in \mathbb{N}$. Find the limits
$$i)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n\sqrt{n}}$$
$$ii)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n^2}$$
I'm puzzled with it. What to do?
| For $n > 1$, we have
$$a_{n+1}^2 = a_{n} + ( a_{n-1} + \ldots + a_{1} ) = a_{n} + a_{n}^2$$
This implies
$$\displaystyle a_{n+1} = \sqrt{a_n^2 + a_n} < \sqrt{a_n^2 + a_n +\frac14} = a_n + \frac12$$
Notice $a_2 = \sqrt{a_1} = 1$, this gives us an upper bound $a_m \le \frac{m}{2}$ for $m > 1$.
As pointed out by robjohn, there is a mistake in original derivation of a lower bound.
To obtain a correct lower bound, we will construct it in 3 stages.
*
*It is clear $a_k$ is an increasing sequence and since $a_1 = 1$, we have
$a_k \ge 1$ for all $k$.
*Since $\quad a_{n+1} = \sqrt{ a_n^2 + a_n } \ge \sqrt{ a_n^2 + \frac23 a_n + \frac13} > a_n + \frac13\quad$ and $a_1 = a_2 = 1 $,
we get $a_m > \frac{m}{3}$ for $m > 0$.
*Observe $\displaystyle\quad(a_n + \frac12 - \frac{1}{8a_n})^2 =
a_n^2 + a_n - \frac{1}{8a_n} ( 1 - \frac{1}{8a_n}) < a_{n+1}^2.$
The lower bound from $2^{nd}$ stage implies
$$\begin{align}
& a_{n+1} > a_n + \frac12 - \frac{1}{8a_n} > a_n + \frac12 - \frac{3}{8n}\\
\implies &
a_m \ge \frac{m}{2} - \frac{3}{8} \sum_{k=2}^{m-1}\frac{1}{k}\;\;\text{ for }m > 2
\end{align}$$
Since $\displaystyle \sum_{k=2}^{m-1}\frac{1}{k} \sim \log m + O(1)$ for large $m$, we find $\displaystyle\quad a_m = \frac{m}{2} + O(\log m)\quad$ as $m \to \infty$.
For the first limit, we have the estimate:
$$\frac{\sum_{k=1}^{n}a_k}{n\sqrt{n}} = \frac{a_{n+1}^2}{n\sqrt{n}} = \frac{\left(\frac{n}{2} + O(\log n)\right)^2}{n\sqrt{n}} \sim O(\sqrt{n})$$
This implies the first limit diverges. For the second limit, we have
$$\frac{\sum_{k=1}^{n}a_k}{n^2} = \frac{a_{n+1}^2}{n^2} = \frac{\left(\frac{n}{2} + O(\log n)\right)^2}{n^2} = \frac14 + O(\frac{\log n}{n})$$
Since $\lim_{n\to\infty} \frac{\log n}{n} = 0$, the second limit exists and equal to $\frac14$.
| {
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Find all solutions: $x^2 + 2y^2 = z^2$ I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem:
Find all solutions in positive integers to the following Diophantine equation
$x^2 + 2y^2 = z^2$
I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem
| Since the equation is homogeneous, we may WLOG assume $\gcd(x, y, z)=1$. (So that all solutions will be given by multiplying all the primitive solutions by any positive integer $k$)
Now if $x$ is even, then $z$ is even, so $y$ is also even, a contradiction. Thus $x$ is odd, so $z$ is odd, and so $x^2 \equiv z^2 \equiv 1 \pmod{4}$. Thus $4 \mid 2y^2$, so $y$ is even. Note that if $p \mid x, z$ for some prime $p$, then $p$ is odd and $p \mid 2y^2$, so $p \mid y$, a contradiction, so $\gcd(x, z)=1$.
Let $y=2y'$, so that $2y'^2=\frac{z^2-x^2}{4}=(\frac{z-x}{2})(\frac{z+x}{2})$. Now $\gcd((\frac{z-x}{2}),(\frac{z+x}{2}))=\gcd(x, z)=1$, so we have 2 cases:
Case 1: $4 \mid z-x$. Then we have $\frac{z-x}{2}=2a^2, \frac{z+x}{2}=b^2, y'=ab$ for some $a, b \in \mathbb{Z}^+$, so $z=b^2+2a^2, x=b^2-2a^2, y=2ab, \, b>a\sqrt{2}>0$. Checking, these are indeed solutions.
Case 2: $4 \mid z+x$. Then we have $\frac{z-x}{2}=b^2, \frac{z+x}{2}=2a^2, y'=ab$ for some $a, b \in \mathbb{Z}^+$, so $z=b^2+2a^2, x=2a^2-b^2, y=2ab, \, a\sqrt{2}>b>0$. Checking, these are indeed solutions.
Therefore all primitive solutions are given by $(x, y, z)=(|b^2-2a^2|, 2ab, b^2+2a^2), a, b \in \mathbb{Z}^+$.
Therefore all positive integer solutions are given by $$(x, y, z)=(k|b^2-2a^2|, k(2ab), k(b^2+2a^2)), a, b, k \in \mathbb{Z}^+$$
| {
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Finding the last digit of $3^{729}$ I'm practicing for my algebra exam but I stumbled on a question I don't know how to solve.
Let $N = 3^{729}$. What is the last digit of $N$?
The example answer says
Since $\gcd(3, 10) = 1$, check that
$3^4 = 81 = 1 \pmod {10}$:
Now, $729 = 182 \times 4 + 1,$ so we get we get (that might be a typo or missed a step)
$3^{729} = 3 \pmod {10}$.
Can anybody help me with this question? Thanks so much!
| $3^1 \equiv 3(\mod 10)$
$3^{2} \equiv -1 (\mod 10) $
$3^3 \equiv 7 (\mod 10)$
$3^{4} \equiv +1 (\mod 10) \implies 3^{4k} \equiv +1 (\mod 10)$
$3^{4k} \cdot 3 \equiv 3 (\mod 10)$
$3^{4k} \cdot 3^2 \equiv -1 (\mod 10)$
$3^{4k} \cdot 3^3 \equiv 7 (\mod 10)$
$729= 4k+1 \implies 3^{729} \equiv 3 (\mod 10)$
| {
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Need help solving $\int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx$ I have a complicated integral to solve. Can someone provide a better way to solve it than what i did - dividing by a inside the root, and then putting $ t = x / a $, and then putting $t^2 = \cos \theta $ and so many other substitutions.
$$
\int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx
$$
| Let $x^2=a^2\cos(y)$. We then have $2xdx = -a^2 \sin(y) dy$. The integral then becomes
\begin{align}
I & = \int x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}} dx = -\int \dfrac{a^2}2 \sin(y) \sqrt{\dfrac{1-\cos(y)}{1+\cos(y)}}dy = -\dfrac{a^2}2 \int \sin(y) \tan(y/2) dy\\
& = -a^2\int \sin^2(y/2)dy
\end{align}
I trust you can finish it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
limit of trigonometric function How to find the limit of this question
$$\lim_{x \rightarrow a} \left( \frac{\sin x}{\sin a} \right)^{\frac{1}{x-a}}$$ where $a \neq k\pi$ with k an integer.
We can write this as : $$\exp\left({\lim_{x\rightarrow a}\dfrac{1}{x-a}\left(\dfrac{\sin x}{\sin a} \right)} \right)$$
How to proceed further?
| If it is allowed, you can also use Taylor series around $a$:
$$\sin x = \sin a + (x - a) \sin' a + O((x - a)^2) = \sin a + (x - a) \cos a + O((x - a)^2).$$
Plugging this in, we get
$$\left(\frac{\sin x}{\sin a}\right)^{\frac{1}{x-a}} = \left(\frac{\sin a + (x - a) \cos a + O((x - a)^2)}{\sin a}\right)^{\frac{1}{x-a}} = \left(1 + (x - a) \cot a + O((x - a)^2)\right)^{\frac{1}{x-a}} = \left[\left(1 + (x - a) \cot a + O((x - a)^2)\right)^{\frac{1}{(x-a) \cot a}}\right]^{\cot a} \quad \stackrel{(x \to a)}{\longrightarrow} \quad e^{\cot a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:
*
*$a_0 = 1$,
*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.
It looks like this:
*
*$a_0 = 1$
*$a_1 = {1 \choose 2} + 3$
*$a_2 = {2 \choose 2} + 3{1 \choose 2} + 9$
*$a_3 = {3 \choose 2} + 3{2 \choose 2} + 9{1 \choose 2} + 27$
*$a_4 = {4 \choose 2} + 3{3 \choose 2} + 9{2 \choose 2} + 27 {1 \choose 2} + 81$
I know what the generating function of the sequence $3 ^n = (1, 3, 9, 27, 81, \dots)$ is, as well as what the generating functions for some sequences of combinatorial numbers are, but how do I split the sequence up into these pieces I know?
(The problem is those combinatorial numbers "move right" every time. If they were growing left-to-right along with their coefficients, it would be much easier. And there is no constant difference between $a_i$ and $a_{i + 1}$.)
| Let $A(x)=\sum_{n=0}^\infty a_nx^n$. Then
\begin{eqnarray}
A(x)&=&1+\sum_{n=1}^\infty a_{n}x^n\\
&=&1+\sum_{n=1}^\infty (3a_{n-1}+\frac{1}{2}n(n-1))x^n\\
&=&1+3xA(x)+\frac{1}{2}\sum_{n=1}^\infty n(n-1)x^n.
\end{eqnarray}
Note $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ for $|x|<1$. Differentiating this twice, you can give
$$ \sum_{n=2}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3}. $$
Thus
$$ A(x)=1+3xA(x)+\frac{x^2}{(1-x)^3} $$
from which you can get $A(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
definite Integration A integration is given,
$$M = \int_{- \infty}^{\infty} \left[\frac{1}{2} \left(\frac{d\phi}{dx}\right)^2 + \frac{\lambda}{4}(\phi^2-v^2)^2\right] dx,$$ where $$m=v\sqrt\lambda$$ and $$ \phi(x)= v\tanh\left[ \frac{m}{\sqrt 2} (x-x_0)\right]$$
How this integration gives =$\frac{2 \sqrt2}{3} \frac{m^3}{\lambda}?$
| Notice
$$\frac{d\phi}{dx} = \frac{vm}{\sqrt{2}} \frac{1}{\cosh(\frac{m}{\sqrt{2}}(x-x_0))^2}
= \frac{vm}{\sqrt{2}}\left(1 - \left(\frac{\phi}{v}\right)^2\right)
= \sqrt{\frac{\lambda}{2}}\left(v^2 - \phi^2\right)
$$
$M$ can be rewritten as:
$$M = \int_{-\infty}^{\infty}\left(\frac12\sqrt{\frac{\lambda}{2}}+\frac{\lambda}{4}\sqrt{\frac{2}{\lambda}}\right) (v^2 - \phi^2) \frac{d\phi}{dx} dx
= \sqrt{\frac{\lambda}{2}}\int_{-v}^{v} (v^2 - \phi^2) d\phi\\
= \sqrt{\frac{\lambda}{2}}\left(\frac43 v^3\right)
= \frac{2\sqrt{2\lambda}}{3}\left(\frac{m}{\sqrt{\lambda}}\right)^3 = \frac{2\sqrt{2}}{3}\frac{m^3}{\lambda}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
integral of normal distribution how to do this integral:
$$ \mathop{\int\int}_{y+2x>0} x y \frac1{2\pi\sigma_x\sigma_y}e^{ -\frac{(x-\mu_x)^2}{2\sigma_x^2}}\cdot e^{ -\frac{(y-\mu_y)^2}{2\sigma_y^2}} dx dy$$
Both x and y are normally distributed and mutually independent. I need to calculate the above integral.
| For simplicity, I let $\sigma_x=\sigma_y=1$. Then we have
\begin{align*}
&\mathop{\int\int}_{y+2x>0}\frac{xy}{2\pi}e^{ -\frac{(x-\mu_x)^2}{2}}\cdot e^{-\frac{(y-\mu_y)^2}{2}}dxdy\\
=&\int_{-\infty}^\infty dx\int_{-2x}^\infty\frac{xy}{2\pi}e^{-\frac{(x-\mu_x)^2}{2}}\cdot e^{-\frac{(y-\mu_y)^2}{2}}dy\\
=&\frac{1}{2\pi}\int_{-\infty}^\infty xe^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty ye^{-\frac{(y-\mu_y)^2}{2}}dy\\
=&\frac{1}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty ye^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x\mu_y}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy\\
=&\frac{1}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_y}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x\mu_y}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy
\end{align*}
You can have explicit expression for each term above except for the last one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/381758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Write a polynomial equation with the following roots. A quartic funciton with roots of $-3$,$-1$ and $4$? Write a polynomial equation with the following roots. A quartic funciton with roots of $-3,-1$ and $4$?
use each root as a factor
$$f(x) = a(x + 3)(x + 1)(x - 4)^2$$
$$f(x) = a(x^2 + 4x + 3)(x^2 - 8x + 16)$$
$$f(x) = a(x^4 - 4x^3 - 13x^2 + 40x + 48)$$
$$f(5) = 16$$
$$16 = a(5^4 - 4(5^3) - 13(5^2) + 40(5) + 48 = 48a \rightarrow a = 3$$
$$f(x) = 3x^4 - 12x^3 - 39x^2 + 120x + 144$$
Is that correct? Please show another way of doing it please with your steps.
| That's correct. You made $4$ a double root by including two $(x-4)$ factors. You could have equally well made $-3$ or $-1$ double roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/383621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| Showing that $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3} = 2$ is the same as showing that $2\pm\frac {10} 9\sqrt 3$ has a cube root of the form $1 \pm a\sqrt 3$ (with rational $a$).
$(1\pm a\sqrt 3)^3 = (1+9a^2)\pm(3a+3a^3)\sqrt 3$, and you quickly check that with $a = 1/3$ you simultaneously get $1+9a^2 = 2$ and $3a(1+a^2) = \frac{10}9$, hence $2 \pm \frac{10}9\sqrt 3 = (1\pm\frac 13\sqrt 3)^3$, and your original expression simplifies to $1+\frac 13\sqrt 3+1-\frac 13\sqrt 3=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 1
} |
Find Simple Convergent sum We have a sum
$\sum_{0}^{\infty} 2^{n+1}(x+1)^{3n+1}$
I am asked to find all values of x s.t this sum converges then compute the sum.
I have no idea what im doing so i did this. $\frac {2^{n}(x+1)^{3n}} {2^{n+1}(x+1)^{3n+1}}$
this yields $\frac {1}{2(x+1)^{3}}$ this looks alittle wierd but i just move it over and solved to get $x=-1 +2^{-1/3}$ this isnt right i also have no idea to find the sum if it was please help.
| First you should realize that just about the only series you've learned how to get an explicit sum for is a geometric series. So your first goal should be to make the given series looks like the standard form of a geometric series. One way to do this is as follows:
\begin{align}
\sum_{n = 0}^\infty 2^{n+1} (x+1)^{3n+1} & = 2(x+1) \sum_{n = 0}^\infty 2^{n} (x+1)^{3n} \\
& = 2(x+1) \sum_{n = 0}^\infty (2(x+1)^3)^n.
\end{align}
Now $\displaystyle \sum_{n = 0}^\infty (2(x+1)^3)^n$ is a geometric series with $a = 1$ and $r = 2(x+1)^3$. Hence
$$\sum_{n = 0}^\infty (2(x+1)^3)^n = \frac{1}{1 - 2(x+1)^3},$$
and from the work about it follows that
$$\sum_{n = 0}^\infty 2^{n+1} (x+1)^{3n+1} = 2(x+1) \sum_{n = 0}^\infty (2(x+1)^3)^n = \frac{2(x + 1)}{1 - 2(x+1)^3}.$$
A geometric series $\displaystyle \sum_{n=0}^\infty ar^n$ converges if and only if $|r| < 1$, so our given series converges if and only if
$$|2(x+1)^3| < 1 \iff -1 - 2^{-1/3} < x < -1 + 2^{-1/3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Least Squares solution on a plane in $\mathbb R^3$ Let $W$ be the subspace of $\mathbb R^3$ spanned by $(1,2,3)^T$ and $(1,1,1)^T$. Find the point in $W$ which lies closest to $(-4,1,2)^T$
I know that the least squares solution is $(A^T)A = (A^T)b$.
However I do not know how to set up my matrix of $A$.
The solution to the problem is : the point closest to $b$ which lies in $\mathrm{Ran}(A)$ is $Ax = (-10/3, -1/3, 8/3)$.
How is this solved?
| Let
$$
\mathbf{u}=\left[\begin{array}{c}
1\\
2\\
3
\end{array}\right]
$$
and
$$
\mathbf{v}=\left[\begin{array}{c}
1\\
1\\
1
\end{array}\right].
$$
We are trying to find $c$ and $d$ such that $c\mathbf{u}+d\mathbf{v}$
is closest (2-norm) to
$$
\mathbf{w}=\left[\begin{array}{c}
-4\\
1\\
2
\end{array}\right].
$$
That is,
$$
\min_{c,d}\left\Vert \left[\begin{array}{cc}
\mathbf{u} & \mathbf{v}\end{array}\right]\left[\begin{array}{c}
c\\
d
\end{array}\right]-\left[\mathbf{w}\right]\right\Vert=\min_{c,d}\left\Vert \left[\begin{array}{cc}
1 & 1\\
2 & 1\\
3 & 1
\end{array}\right]\left[\begin{array}{c}
c\\
d
\end{array}\right]-\left[\begin{array}{c}
-4\\
1\\
2
\end{array}\right]\right\Vert.
$$
As it turns out, this has the analytical solution (normal equations)
$$
\left[\begin{array}{cc}
1 & 1\\
2 & 1\\
3 & 1
\end{array}\right]^{T}\left[\begin{array}{cc}
1 & 1\\
2 & 1\\
3 & 1
\end{array}\right]\left[\begin{array}{c}
c\\
d
\end{array}\right]=\left[\begin{array}{cc}
1 & 1\\
2 & 1\\
3 & 1
\end{array}\right]^{T}\left[\begin{array}{c}
-4\\
1\\
2
\end{array}\right].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration by parts help Integration by parts
$$x\tan^{-1}x~dx=\frac12\left(x^2+1\right)\tan^{-1}x-\frac12x+c$$
and
$$x^2e^{-3x}dx =-\frac13e^{-3x}\left(x^2+\frac23x+\frac29\right)+c$$
I have
$$d(uv)=u~dv+v~du$$
Can anyone please help me?
| First one:
Recall that $d(\arctan(x)) = \dfrac{dx}{1+x^2}$.
$$\int x \arctan(x) dx = \underbrace{\int \arctan(x) d\left(\dfrac{x^2}2 \right)}_{\displaystyle \int v du} = \overbrace{\dfrac{x^2 \arctan(x)}2}^{\displaystyle uv} - \underbrace{\int \dfrac{x^2}{2(1+x^2)}dx}_{\displaystyle \int u dv}$$
Now
$$\int \dfrac{x^2}{2(1+x^2)}dx = \int \dfrac{1+x^2-1}{2(1+x^2)}dx = \int \dfrac{dx}2 - \int \dfrac{dx}{2(1+x^2)} = \dfrac{x}2 - \dfrac{\arctan(x)}2$$
Hence, we get that
$$\int x \arctan(x) dx = \dfrac{x^2 \arctan(x)}2 - \dfrac{x}2 + \dfrac{\arctan(x)}2 + c = \dfrac{(1+x^2) \arctan(x)}2 - \dfrac{x}2 + c$$
Second one:
$$\int x^2 e^{-3x} dx = \underbrace{\int x^2 d \left(\dfrac{e^{-3x}}{-3}\right)}_{\displaystyle \int v du} = \overbrace{\dfrac{e^{-3x}}{-3} x^2}^{\displaystyle uv} - \underbrace{\int \dfrac{e^{-3x}}{-3} (2x) dx}_{\displaystyle \int u dv} = \dfrac{e^{-3x}}{-3} x^2 + \dfrac23 \int xe^{-3x} dx$$
Now let us do integration by parts again to evaluate $\displaystyle \int xe^{-3x} dx$.
$$\int xe^{-3x} dx = \underbrace{\int x d \left(\dfrac{e^{-3x}}{-3}\right)}_{\displaystyle \int v du} = \overbrace{x \dfrac{e^{-3x}}{-3}}^{\displaystyle uv} - \underbrace{\int \dfrac{e^{-3x}}{-3} dx}_{\displaystyle \int u dv} = -\dfrac{x e^{-3x}}3 - \dfrac{e^{-3x}}{27}$$
Putting this back, we get that
$$\int x^2 e^{-3x} dx = -\dfrac13 e^{-3x}x^2 - \dfrac29 e^{-3x}x - \dfrac2{27} e^{-3x} + c = - \dfrac{e^{-3x}}3 \left(x^2 + \dfrac{2x}3 + \dfrac2{27}\right) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
if $\sin24^\circ = p$ what is $\cos24^\circ$? Let $p=\sin 24^\circ$
*
*Then what would $\cos (24^\circ)$ be in terms of $p$?
*What would $\sin (168^\circ) \cdot \sin(-78^\circ)$ be in terms of $p$?
I'm not sure how to approach these as we have only started this section.
Thanks.
| Part 2
Method 1:
$\sin168^\circ=\sin(180^\circ-168^\circ)$ ( as $\sin(180^\circ-\theta)=\sin\theta$)
So, $\sin168^\circ=\sin12^\circ$
and $\sin(-78^\circ)=-\sin78^\circ$ as $\sin(-\theta)=-\sin\theta$
So, $\sin(-78^\circ)=-\sin(90^\circ-12^\circ)=-\cos12^\circ$
$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\sin12^\circ\cdot(-\cos12^\circ)=-\frac{\sin (2\cdot12^\circ)}2=-\frac{\sin24^\circ}2$$
Method 2:
$\sin168^\circ=\sin(90^\circ+78^\circ)=\cos78^\circ$ as $\sin(90^\circ+\theta)=\cos\theta$
Again, $\cos78^\circ=\cos(-78^\circ)$ as $\cos(-\theta)=\cos\theta$
$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\cos(-78^\circ)\cdot\sin(-78^\circ)=\frac{\sin 2(-78^\circ)}2$$
$$=\frac{\sin(24^\circ-180^\circ)}2=-\frac{\sin24^\circ}2$$ as $\sin(\theta-180^\circ)=-\sin(180^\circ-\theta)=-\sin\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if a term $x+y$ comes out, but it looks hopeless at this point.
| $$
\begin{align}
x+\sqrt{x^2+3}
&=\frac3{y+\sqrt{y^2+3}}\\
&=\frac3{y+\sqrt{y^2+3}}\frac{y-\sqrt{y^2+3}}{y-\sqrt{y^2+3}}\\
&=\frac3{-3}\left(y-\sqrt{y^2+3}\right)\\
&=-y+\sqrt{y^2+3}\tag{1}
\end{align}
$$
Similarly
$$
y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\tag{2}
$$
Add $(1)$ and $(2)$ and cancel the radicals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
How to find $1+\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}$ Let $a_{i}>0,i=1,2,\cdots,n$,
If
$$1+\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\ge k\;\left(\dfrac{1}{\sqrt{1+a^2_{1}}}+\dfrac{1}{\sqrt{1+a^2_{1}+a^2_{2}}}+\cdots+\dfrac{1}{\sqrt{1+a^2_{1}+\cdots+a^2_{n}}}\right)$$
then find the maximum value for $k$
My idea was to use
$$\left(\sum_{i=1}^{n}\dfrac{1}{\sqrt{1+a^2_{1}+a^2_{2}+\cdots+a^2_{i}}}\right)^2\le\left[\sum_{i=1}^{n}\dfrac{a_{i}}{1+a^2_{1}+a^2_{2}+\cdots+a^2_{i}}\right]\left(\sum_{i=1}^{n}\dfrac{1}{a_{i}}\right)$$
and to use the approaches discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119163&sid=cd139cff36102a55e4f91a7c8dcd4f21#p119163
Alas! I failed to figure out how to do it.
| Substituting $a_i=\sqrt{2}^{i-1}$ gives
\begin{align}
1+\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}=1+\sum_{i=1}^{n}{(\frac{1}{\sqrt{2}})^{i-1}}=1+\sum_{i=1}^{n}{\frac{1}{a_i}} & \geq k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}}\right)\\
& =k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{2^{j-1}}}}}\right) \\
& =k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{2^i}}}\right) \\
& =k\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}
\end{align}
Thus
\begin{align}
k \leq \frac{1+\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}{\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}=\sqrt{2}+\frac{1}{\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}=\sqrt{2}+\frac{\sqrt{2}-1}{1-\frac{1}{\sqrt{2}^n}}
\end{align}
Since this must hold for all $n$, we have $k \leq 2\sqrt{2}-1$. We now show that $k=2\sqrt{2}-1$ works, so that $2 \sqrt{2}-1$ is the maximum value of $k$. In other words, we will show that
$$1+\sum_{i=1}^{n}{\frac{1}{a_i}} \geq (2\sqrt{2}-1)\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}}\right)$$
By Cauchy Schwarz inequality we have
$$\sqrt{2}^i\sqrt{1+\sum_{j=1}^{i}{a_j^2}}=\sqrt{1+\sum_{j=1}^{i}{a_j^2}}\sqrt{1+\sum_{j=1}^{i}{2^{j-1}}} \geq 1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}$$
By weighted AM-HM inequality we have
$$\frac{1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}}{2^i}=\frac{1+\sum_{j=1}^{i}{2^{j-1}\frac{a_j}{\sqrt{2}^{j-1}}}}{1+\sum_{j=1}^{i}{2^{j-1}}} \geq \frac{1+\sum_{j=1}^{i}{2^{j-1}}}{1+\sum_{j=1}^{i}{2^{j-1}\frac{\sqrt{2}^{j-1}}{a_j}}}=\frac{2^i}{1+\sum_{j=1}^{i}{\frac{(2\sqrt{2})^{j-1}}{a_j}}}$$
Thus
\begin{align}
\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}} \leq \sum_{i=1}^{n}{\frac{\sqrt{2}^i}{1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}}} & \leq \sum_{i=1}^{n}{\frac{1+\sum_{j=1}^{i}{\frac{(2\sqrt{2})^{j-1}}{a_j}}}{(2\sqrt{2})^i}} \\
&=\sum_{i=1}^{n}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{(2\sqrt{2})^{j-1}}{a_j}\sum_{i=j}^{n}{\frac{1}{(2\sqrt{2})^i}}} \\
&=\sum_{i=1}^{n}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{1}{a_j}\sum_{i=1}^{n-j+1}{\frac{1}{(2\sqrt{2})^i}}} \\
& \leq \sum_{i=1}^{\infty}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{1}{a_j}\sum_{i=1}^{\infty}{\frac{1}{(2\sqrt{2})^i}}} \\
&=\frac{1}{2\sqrt{2}-1}(1+\sum_{i=1}^{n}{\frac{1}{a_i}})
\end{align}
| {
"language": "en",
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Definite integration of a trigonometric function
How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$
my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$
I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.
| Often, this type of integral is workable using the substitution
$$
z=\tan(x/2)\quad\text{and}\quad\mathrm{d}x=\frac{\mathrm{2\,d}z}{1+z^2}\\
\sin(x)=\frac{2z}{1+z^2}\quad\text{and}\quad\cos(x)=\frac{1-z^2}{1+z^2}
$$
A couple of identities that simplify this integral are
$$
\sin^2(x)=\frac{1-\cos(2x)}{2}\quad\text{and}\quad\cos^2(x)=\frac{1+\cos(2x)}{2}
$$
Then, assuming $a,b\gt0$,
$$
\begin{align}
\int_0^{\pi/2}\frac{2a\sin^2(x)}{a^2\sin^2(x)+b^2\cos^2(x)}\,\mathrm{d}x
&=\int_0^{\pi/2}\frac{2a(1-\cos(2x))}{a^2(1-\cos(2x))+b^2(1+\cos(2x))}\,\mathrm{d}x\\
&=\int_0^\pi\frac{a(1-\cos(x))}{(b^2+a^2)+(b^2-a^2)\cos(x)}\,\mathrm{d}x\\
&=\int_0^\infty\frac{az^2}{b^2+a^2z^2}\frac{2\mathrm{d}z}{1+z^2}\\
&=\int_{-\infty}^\infty\frac{az^2}{b^2+a^2z^2}\frac{\mathrm{d}z}{1+z^2}\\
&=2\pi i\left(\frac1{2i}\frac{a}{a^2-b^2}-\frac1{2i}\frac{b}{a^2-b^2}\right)\\
&=\frac\pi{a+b}
\end{align}
$$
where we use the contour running along the real axis from $-\infty$ to $\infty$, circling back in the upper half-plane. The residue at $z=i$ is $\dfrac1{2i}\dfrac{a}{a^2-b^2}$ and at $z=i\dfrac{b}{a}$ is $-\dfrac1{2i}\dfrac{b}{a^2-b^2}$.
Removing the assumption on the signs of $a$ and $b$, we get
$$
\int_0^{\pi/2}\frac{2a\sin^2(x)}{a^2\sin^2(x)+b^2\cos^2(x)}\,\mathrm{d}x
=\mathrm{sgn}(a)\frac\pi{|a|+|b|}
$$
| {
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Integration $\int \left(x-\frac{1}{2x} \right)^2\,dx $ Evaluate $$\int\!\left(x-\frac{1}{2x} \right)^2\,dx. $$
Using integrating by substitution, I got $u=x-\frac{1}{2x},\quad \dfrac{du}{dx} =1+ \frac{1}{2x^2}$ , and $dx= 1+2x^2 du$. In the end, I came up with the answer to the integral as :
$$\left(\frac{1}{3}+\frac{2x^2}{3}\right)\left(x-\frac{1}{2x}\right)^3.$$
Any mistake ?
| Expanding the square bracket and integrating each term individually gives
\begin{equation*}
\int (x-\frac{1}{2x})^2dx=\int (x^2+\frac{1}{4x^2}-1)dx=\int x^2dx+\frac{1}{4}\int\frac{1}{x^2}dx-\int dx\\
=\frac{x^3}{3}-\frac{1}{4x}-x+C
\end{equation*}
for a constant $C$.
| {
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For what $n$ and $m$ is this number a perfect square? Find all positive integers $n$ and $m$ such that $$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3$$ is a perfect square.
Of course $n=m=1$ is a trivial solution and if you put $n=m+1$ (assume WLOG that $n \geq m$), the above expression is equal to $4n^4$ which is a perfect square - But I don't know if this is the only solution. I would really appreciate any help.
| Given the OP's form,
$$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3 = w^2\tag{1}$$
ccorn and Alyosha showed it is equivalent to the more aesthetic forms,
$$\tfrac{1}{4}\big((2m-1)^2+1\big)\big((2n-1)^2+1\big)-1 =w^2$$
$$(m+n-2mn-1)^2+(n-m)^2-1 =w^2$$
respectively. There are infinitely many positive integer solutions to (1) other than $n=m+1$ as we can use a Pell equation $x^2-dy^2=1$ to solve it. In fact, we can use infinitely many $d$. For any positive integer $m>1$, the solution is then,
$$\begin{aligned}
&n =(2m-3)x^2+4(m-1)^2xy-(m-2)\\
&w = 4(m-1)^2x^2+(2m-3)\big((2m-1)^2+1\big)xy-2(m-1)^2
\end{aligned}$$
where $x,y$ satisfy the Pell equation,
$$x^2-\big((2m-1)^2+1\big)y^2=1$$
For the simplest case when $m=1$, we have the alternative and simpler form,
$n = (x+1)/2,\;\; w=y,\;\; \text{where}\;\;x^2-2y^2=1.\tag{2}$
(Note: Since the $x$ of (2) is odd, then $n$ is an integer.)
| {
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Showing that a function has a certain absolute minimum. Suppose we have the function $$f(x) = \frac{x}{p} + \frac{b}{q} - x^{\frac{1}{p}}b^{\frac{1}{q}}$$ where $x,b \geq 0 \land p,q > 1 \land \frac{1}{p}+\frac{1}{q} = 1$
I am trying to show that $b$ is the absolute minimum of $f$.
I proceeded as follows:
$$\frac{df(x)}{dx} = \frac{1}{p} - \frac{x^{\frac{1}{p}-1}}{p} b^{\frac{1}{q}} = \frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px}$$
Now I will look for critical points by searching for the zeros of this function.
$$\frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px} = 0 \iff x - x^{\frac{1}{p}} b^{\frac{1}{q}} = 0 \iff x = x^{\frac{1}{p}} b^{\frac{1}{q}}$$.
Now I can see that $b$ is a critical point.
How ever when I continue my calculations to check whether there are any other critical points
$$x = x^{\frac{1}{p}} b^{\frac{1}{q}} \implies x^p = b^{\frac{p}{q}}x \implies x^{p-1} = b^{\frac{p}{q}} \implies x = b^{\frac{p}{(p-1)q}}$$
But this could not be equal to $b$, where did I go wrong?
| Let $\dfrac{1}{p}=a$ and $\dfrac{1}{q}=k$
$ax+bk-x^a \cdot b^k=f(x) $
$\dfrac{x + \dots x_{ath}+ b+ \dots b_{kth}}{a+k} \ge (x^ab^k)^{1/(a+k)}$ (By AM-GM inequality)
$x+ \dots x (a$ times)$=ax$ and $b+ \dots b (k$ times)$=bk$
$a+k=1 \implies ax+bk \ge x^a\cdot b^k \implies ax+bk-x^ab^k \ge 0$
Now you have $f(x) \ge 0$, the minimum is achieved when $x=b$ and $a=k$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$ $\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$
*
*$x=2\tan^{-1}t$
*$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$
*$dx=\dfrac{2}{1+t^2}\,dt$
*$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$
Is this the right idea? If so what do I do next?
$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$
$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$
$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$
| Hint:
When $\tan \dfrac{x}{2}=t$ you have $ \cos x= \dfrac{1-t^2}{1+t^2}$
| {
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Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.
I tried many method, such as AM-GM, but all of them failed.
Thank you.
| Geometrically the given equation represents a sphere and $x+y+z=t$ represents a plane which can be shifting around. So the min will be achieved on $x,y$ or $z$-axes intersecting with the sphere. Hence the possible points are $(\frac{\sqrt{14}-1}{2},0,0),(0,\frac{\sqrt{17}}{2}-1,0),(0,0,\frac{\sqrt{22}-3}{2})$, and the min is $\frac{\sqrt{22}-3}{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side
Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
I got $$s + 13 = 4 + 4\sqrt{7-s}+ 7 – s$$
I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$
I’m stuck at this point. Does anyone have an idea how to solve this equation?
| From where you left off, squaring both sides again yields
$(2s + 2)^2 = (4 √(7-s))^2$ Which becomes
$4s^2 +8s +4 = 16(7-s)$
And you have a quadratic to solve, take all terms to one side and factorise or use the quadratic formula to find solutions to $s$, there will be two solutions of course. And then plug values of $s$ found back into your original equation to make sure you dont have a negative value under the square root.
| {
"language": "en",
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Simplifying Differentiation I'm working off my textbook and I've followed the steps easily enough until it gets to this
$ \dfrac {dy}{dx} = \dfrac{(x^2 + 1)^3}{2 \sqrt{x - 1}} + \sqrt{x-1}~(6x)~(x^2 + 1)^2$
$= \dfrac{(x^2 + 1)^2}{2\sqrt{x - 1})}[(x^2 + 1) + 12x(x - 1)]$
$= \dfrac{(x^2 + 1)^2(13x^2 - 12x + 1)}{2\sqrt{x - 1}}$
How do they go from the first line onwards?
Thanks in advance (sorry about formatting)
| I will assume that we are looking at
$$\frac{(x^2+1)^3}{2\sqrt{x-1}} +\sqrt{x-1}(6x)(x^2+1)^2.$$
Take out the common factor $(x^2+1)^2$, and multiply and divide by $2\sqrt{x-1}$. We get
$$\frac{(x^2+1)^2}{2\sqrt{x-1}}\left((x^2+1)+(2)(x+1)(6x) \right).$$
Now simplify the expression on the right.
That seems to be the way it was thought of. I would prefer to first bring the expression to the common denominator $2\sqrt{x-1}$. We then get
$$\frac{(x^2+1)^3 +2(x-1)(6x)(x^2+1)^2}{2\sqrt{x-1}}.$$
Now take out the common factor
$\dfrac{(x^2+1)^2}{2\sqrt{x-1}}$. Perhaps that was the way it was done, with a step skipped.
Remark: If you have seen the (natural) logarithm already, the following approach, called logarithmic differentiation, may appeal to you. It is of at most marginal utility in this problem, but could be useful when differentiating a longer product.
We were differentiating $(x^2+1)^3 \sqrt{x-1}$. Let
$$y=(x^2+1)^3\sqrt{x-1}.$$
Take the natural logarithm of both sides. Using "laws of logarithms," we get
$$\ln y=3\ln(x^2+1)+\frac{1}{2}\ln(x-1).$$
Differentiate, using the Chain Rule. We get
$$\frac{1}{y}\frac{dy}{dx}=\frac{6x}{x^2+1} +\frac{1}{2(x-1)}.$$
Finally, multiply by $y$ to get $\frac{dy}{dx}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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trigonometric representation of a complex number. Let $z=e^{it}+1$ where $0\leq t\leq \pi$, Find the trigonometric representation of $z^2+z+1$.
(The trigonometric representation should be in the form of : $r(\cos \theta +i \sin \theta)$, where $r,\theta \in \mathbb{R}$ and $r>0$.
What I have done : $$z^2+z+1=e^{2it}+3z=(\cos (t)+i \sin (t)) (-2 i \sin (t)+4 \cos (t)+3).$$
The problem is that the expression in the right bracket is not real always.
| The complex number $a+bi$ can be written in trigonometric form as
$$a+bi=\sqrt{a^{2}+b^{2}}(\cos \theta +i\sin \theta ),\qquad\text{with } \tan \theta =\frac{b}{a}.$$
For $z=e^{it}+1=(\cos t+1)+i\sin t$ we have
\begin{eqnarray*}
z^{2}+z+1 &=&\left( e^{it}+1\right) ^{2}+(e^{it}+1)+1= (e^{i2t}+2e^{it}+1)+(e^{it}+1)+1 \\
&=&e^{2it}+3e^{it}+3 \\
&=&((\cos 2t+1)+i\sin 2t)+3\left( (\cos t+1)+i\sin t\right) +3 \\
&=&(\cos 2t+3\cos t+3)+i\left( \sin 2t+3\sin t\right) \\
&=&u+iv,\qquad u=\cos 2t+3\cos t+3,v=\sin 2t+3\sin t \\
&=&r\left( \cos \theta +i\sin \theta \right) ,
\end{eqnarray*}
where
\begin{eqnarray*}
r &=&\sqrt{u^{2}+v^{2}}=\sqrt{(\cos 2t+3\cos t+3)^{2}+\left( \sin 2t+3\sin
t\right) ^{2}} \\
\tan \theta &=&\frac{v}{u}.
\end{eqnarray*}
For $0\leq t\leq \pi ,v=\sin 2t+3\sin t\geq 0$ and $u=\cos 2t+3\cos t+3\geq 0
$. So
$$
\begin{equation*}
\theta =\arctan \frac{v}{u} =\arctan \left( \frac{\sin
2t+3\sin t}{\cos 2t+3\cos t+3}\right) \geq 0,
\end{equation*}
$$
because $w=z^2+z+1=u+iv=\operatorname{Re}(w)+i\operatorname{Im}(w)$ is in the first quadrant.
Plots of $u,v$:
$$u=\cos 2t+3\cos t+3 \text{ (blue) },\quad v=\sin 2t+3\sin t \text{ (red) },\quad 0\leq t\leq \pi.$$
| {
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Is there a simpler closed form for $\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}$ I have the following infinite sum that can be expressed in terms of the generalized hypergeometric function:
$$\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}=\frac{31}8-4\times{_4F_3}\left(-\frac12,\frac12,1,1;\ 2,2,2;\ 1\right)\\\ \\\approx0.008749644047541935203478962326551903908774780849356243615274...$$
I wonder if it can be expressed in terms of simpler functions and well-known mathematical constants.
| Ookay... Consider just the hypergeometric function and its closed form.
Take 16.5.2 from DLMF (with $a_0=1/2$, $b_0=2$) and write (I used Mathematica to substitute the special form for the hypergeometric function in the integrand; I don't really know how to do it by hand):
$$ F(-1/2,1/2,1,1;2,2,2;1) = \int_0^1 \frac{8 \sqrt{1-t} \left(4-4 \sqrt{1-t}+t\sqrt{1-t}-\log8+3 \log(1+\sqrt{1-t})\right)}{9 \pi t^{3/2}}\,dt. $$
Mathematica can then do the integral in closed form to give
$$\frac{1}{9 \pi }8 \left(\frac{70}{3}+\frac{11 i \pi ^2}{4}+\pi \left(-7+\log 512-6\log\left(1+\frac{1+i}{\sqrt{2}}\right)\right)+24 i \mathrm{Li}_2\left(-\frac{1+i}{\sqrt{2}}\right)-24 i \mathrm{Li}_2\left(1-\frac{1+i}{\sqrt{2}}\right)\right),$$
where $\mathrm{Li}_2$ is the polylogarithm.
Of this we take the real part only and do some more FunctionExpand:
$$-\frac{56}{9}+\frac{560}{27 \pi }-\frac{16 C}{3 \pi }+\frac{64 \Im\left(\mathrm{Li}_2(1-(-1)^{1/4})\right)}{3 \pi }+\frac{8 \log512}{9}-\frac{8}{3} \log\left(\frac{1}{2}+\left(1+\frac{1}{\sqrt{2}}\right)^2\right)+\frac{\psi_1\left(\frac{1}{8}\right)}{3 \sqrt{2} \pi }+\frac{\psi_1\left(\frac{3}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{5}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{7}{8}\right)}{3 \sqrt{2} \pi }, $$
where $\psi_1$ is the polygamma function and $C$ is the Catalan constant.
Now, the polylogarithm term there can be simplified using the identity (DLMF 25.12.6)
$$ \mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\log x\log(1-x), $$
because $\mathrm{Li}_2((-1)^{1/4})$ is simpler.
After a further FunctionExpand (which gets rid of the polygamma functions also), ComplexExpand to get the real part, and FullSimplify to simplify the expression, the answer is
$$ -\frac{56}{9}+\frac{560}{27 \pi }-\frac{32C}{3 \pi }+\frac{16}{3} \log2$$
| {
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How to evaluate the following limit? How to evaluate this limit?
$$\underset{n\to \infty }{\mathop{\lim }}\,\left( {{n}^{-2}}\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{{{n}^{2}}}{\frac{1}{\sqrt{{{n}^{2}}+ni+j}}}} \right)$$
Thanks!
| The idea here is to rearrange in terms of a Riemann sum that leads to a double integral. Here, a little manipulation produces
$$\frac{1}{n} \sum_{i=1}^n \frac{1}{n^2} \sum_{j=1}^{n^2} \frac{1}{\sqrt{1+\frac{i}{n}+\frac{j}{n^2}}}$$
Then as $n \to \infty$, the double sum becomes
$$\begin{align}\int_0^1 dx \, \int_0^1 dy \frac{1}{\sqrt{1+x+y}} &= 2 \int_0^1 dx \left [ \sqrt{2+x}-\sqrt{1+x}\right ]\\ &= \frac{4}{3} \left [\left (3^{3/2}-2^{3/2} \right ) - \left (2^{3/2}-1 \right )\right ]\\&=4 \sqrt{3} - \frac{16}{3} \sqrt{2}+1\end{align}$$
| {
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Is it possible to isolate p in $"a = 2bpq + 2apq + a(p^2) + 2aqq + bqq"?$ Is it possible to simplify this equation so that $p$ is isolated on one side instead of $a$? I tried factoring out $p$ on the right side, but I get stuck with the $ap^2$.
$$a = 2bpq + 2apq + ap^2 + 2aq^2 + bq^2$$
| By solving p in this equation you get:
$p = \frac{\sqrt{a^2(-q)^2+a^2+abq^2+b^2q^2} -aq - bq}{a}$
or
$p = \frac{-\sqrt{a^2(-q)^2+a^2+abq^2+b^2q^2} -aq - bq}{a}$
EDIT:
Steps as requested:
$ap^2+2apq+2bpq+2aq^2+bq^2=a$
$p^2 + \frac{p(2aq+2bq)}{a }+ \frac{2aq^2+bq^2}{a} = 1$
$p^2 + \frac{p(2aq+2bq)}{a} = 1 - \frac{2aq^2+bq^2}{a}$
$p^2 + \frac{p(2aq+2bq)}{a} + \frac{(2aq+2bq)^2}{4a^2} = 1 + \frac{(2aq+2bq)^2}{4a^2} - \frac{2aq^2+bq^2}{a}$
$(p + \frac{(2aq+2bq)}{2a})^2 = 1 + \frac{(2aq+2bq)^2}{4a^2} - \frac{2aq^2+bq^2}{a}$
Can you take it from here?
| {
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Can you tell which are correct terms of the sum of solution of the integral $\int (x^2+1)^n dx $? According to WolframAlpha,
$$\int (x^2+1)^n dx = x \cdot _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)$$
and
$$_2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{n=0}^{\infty} \frac{1}{3}(-n)\frac{(-x^2)^n}{n!}.$$
Can you tell which are correct terms of the sum of solution of the integral $\int (x^2+1)^n dx $?
these
$$(n=0): x, (n=1): 1/3x(x^2+3), (n=2): 1/3 x(x^4+x^2), \dots$$
or (by using $_2F_1(a,b;c;z)= \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$(1)), these which I got $$(n=0): 0, (n=1): 1/3x(x^2), (n=2): 1/3 x(-x^4+x^2), \dots$$
If mine are incorrect, can you compute some terms of the sum (1)?
| You are getting the wrong answer. The answer should be
$$ x\,{\mbox{$_2$F$_1$}\left(\frac{1}{2},-n;\,\frac{3}{2};\,-{x}^{2}\right)}. $$
Now, you won't have problems. For $n=0,1,2,3$, we get the corresponding answers
$$ x,\,
\frac{x}{3} \left( {x}^{2}+3 \right),
\, \frac{x}{15} \left( 3\,{x}^{4}+10\,{x}^{2}+15 \right),\, \,
\frac{x}{35} \left( 5\,{x}^{6}+21\,{x}^{4}+35\,{x}^{2}+35 \right). $$
Added: You should use different index for the sum as
$$ _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{k=0}^{\infty} \frac{(1/2)_k(-n)_k}{(3/2)_k}\frac{(-x^2)^k}{k!}. $$
Now, you need this identity
$$ (a)_m = \frac{\Gamma(m+a)}{\Gamma(a)}= \frac{(m+a-1)!}{(a-1)!}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determinant of $4\times4$ Matrix I tried to solve for a $4 \times 4$ matrix, but I'm unsure if I did this properly, can anyone tell me if I did this correct? Or if there were any mistakes where at? Also, I know this is an inefficient method for finding the determinant, however I want to get practice with solving like so:
$$A=
\begin{bmatrix}
2 & 4 & 0 & 1 \\
0 & 8 & 0 & 2 \\
0 & 3 & 0 & 5 \\
1 & 2 & 1 & 1
\end{bmatrix}
$$
$$\begin{align*}
\det(A)&=2
\begin{vmatrix}
8 & 0 & 2 \\
3 & 0 & 5 \\
2 & 1 & 1
\end{vmatrix}
-4
\begin{vmatrix}
0 & 0 & 2 \\
0 & 0 & 5 \\
1 & 1 & 1
\end{vmatrix}
+0
-1
\begin{vmatrix}
0 & 8 & 0 \\
0 & 3 & 0 \\
1 & 2 & 1
\end{vmatrix}\\[0.3in]
&=2\left(8
\begin{vmatrix}
0 & 5 \\
1 & 1
\end{vmatrix}-0+2
\begin{vmatrix}
3 & 0\\
2 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}-4\left(0-0+2
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1\left(0-8
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}+0\right)\\[0.3in]
&=2(8(0-5)-0+2(3-0))\\[0.1in]
&\quad{}-4(0-0+2(0))\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1(0-8(0)+0)\\[0.3in]
&= 2(8(-5)-0+2(3))\\[0.3in]
&=2(-45+6)\\[0.3in]
&=2(-39)\\[0.3in]
&=-78
\end{align*}$$
Sorry for the long post, I tried to make the readability easy for everyone.
| I know you said that you know that this is not the most efficient way of finding the determinant, but may I point out a quick way? In fact, you use the same technique, but apply it to the third column.
$$\det(A) = (-1)^{3+4}
\left| \begin{array}{lll}
2 & 4 & 1 \\
0 & 8 & 2 \\
0 & 3 & 5
\end{array}\right| = -2 \left| \begin{array}{lll}
8 & 2 \\
3 & 5
\end{array}\right| = -68
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Solving a algebraic equation I am to solve the following equation third order equation
$x^3+x^2+x+1=0$
What I've tried so far is writing the equation as
$ x \cdot (x^2+x+1)+1=0$
but that didn't lead anywhere. How do I solve this without using a computer?
| Two approaches to finding a root: (We can confirm a real root exists, by the rational root theorem).
$$x^3+x^2+x+1=0$$
*
*We can recognize that $x = -1$ is a zero by using a quick check at $x = \pm 1$ (again, by the rational root theorem, those are possible zeros). Then we know that since $x = -1$ is a zero, $(x - (-1)) = (x + 1)$ is a factor, and we can just use polynomial long division to obtain the remaining factor of $(x^2 + 1)$.
*We can "group" the equation as follows: $$x^3+x^2+x+1=0 \iff x^2(x + 1) + (x + 1) = (x+1)(x^2 + 1) = 0$$
Finally, we know that the remaining factor in $(x+1)(x^2 + 1)=0$, namely $(x^2 + 1),\;$ has no real root (and hence, cannot be factored) by evaluating the discriminant $$\Delta = b^2 - 4ac = 0 - 4 = -4 < 0$$
(Recall the discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by the $\color{blue}{\bf radicand}$ of the quadratic formula $$\frac{-b \pm \sqrt{\color{blue}{\bf b^2 - 4ac}}}{2a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Closed form of $\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor^2$? Does $\displaystyle\sum_{i=1}^n\left\lfloor\dfrac{n}{i}\right\rfloor^2$ admit a closed form expression?
| To see how the first term in the asymptotic expansion is obtained, put $$a(n) = \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor^2$$ and note that
$$a(n+1)-a(n) = 1 + \sum_{k=1}^n
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor^2 - \bigg\lfloor \frac{n}{k} \bigg\rfloor^2\right) \\=
1 + \sum_{d|n+1 \atop d<n+1} \left(\left(\frac{n+1}{d}\right)^2 - \left(\frac{n+1}{d}-1\right)^2\right)
= \sum_{d|n+1} \left(2\left(\frac{n+1}{d}\right)-1\right) \\=
2\sigma(n+1)-\tau(n+1).$$
It now follows that $$a(n) = 2\sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k) =
\sum_{k=1}^n \left(2\sigma(k)-\tau(k)\right).$$
We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series
$$L(s) = \sum_{n\ge 1} \frac{2\sigma(n)-\tau(n)}{n^s} =
2\zeta(s-1)\zeta(s)-\zeta(s)^2.$$
We have $$\operatorname{Res}(L(s); s=2) = \frac{\pi^2}{3},$$
so that by the aforementioned theorem,
$$a(n) \sim \frac{\pi^2/3}{2} n^2 = \frac{\pi^2}{6} n^2.$$
In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting
$$a(n) = \left(\sigma(n)-\frac{1}{2}\tau(n)\right) +
\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} L(s) n^s \frac{ds}{s}$$
which yields
$$a(n) \sim \left(\sigma(n)-\frac{1}{2}\tau(n)\right)
+\frac{\pi^2}{6} n^2 - (\log n + 2 \gamma)n - \frac{1}{6}.$$
This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
$\mathbb C[X]/(X^2)$ is isomorphic to $\mathbb R[Y]/((Y^2+1)^2)$ This question led me to the following:
Prove that $\mathbb C[X]/(X^2)$ is isomorphic to $\mathbb R[Y]/((Y^2+1)^2)$.
| We want to construct an isomorphism of $\mathbb{R}$-algebras $\mathbb{R}[T,X]/(T^2+1,X^2) \cong \mathbb{R}[Y]/((Y^2+1)^2)$. By the universal properties of polynomial and quotient algebras, as well as the Yoneda Lemma, this is equivalent to a natural bijection
$\alpha : \{(a,b) \in A^2 : a^2=-1, b^2 = 0\} \cong \{c \in A : (c^2+1)^2=0\}$,
where $A$ runs through all $\mathbb{R}$-algebras. In my opinion, this reformulation catches the real content of the isomorphism: It is a really elementary statement about solutions of polynomial equations. By the way, $\mathbb{R}$ can be replaced by any ring in which $2$ is invertible (but not by an arbitrary ring, consider $A=\mathbb{Z}/4$).
Now here is a direct proof: We define $\alpha(a,b) = a+b$. Then $\alpha$ is well-defined, since $c:=a+b$ satisfies $c^2=a^2+2ab=2ab-1 \Rightarrow (c^2+1)^2=0$. Clearly $\alpha$ is natural.
In order to find or motivate the definition the inverse map, let us solve $\alpha(a,b)=c$ for $a$: We have $0=b^2=(c-a)^2=c^2-2ac+a^2=c^2-2ac-1$, hence $a=(c^2-1)/(2c)$. Here, $c$ is invertible with $1/c = -(c^3+2c)$ since $0=c^4+2c^2+1$. It follows $-2a=(c^2-1)(c^3+2c)=c^5+c^3-2c=c (-1-2c^2)+c^3-2c=-c^3-3c$, hence $a=c(c^2+3)/2$, and therefore $b=c-a=-c(c^2+1)/2$.
So let us define $\alpha^{-1}(c):=(c(c^2+3)/2,-c(c^2+1)/2)$. By construction, this is a map inverse to $\alpha$, as soon as we have shown that it is well-defined. Writing $\alpha^{-1}(c)=(a,b)$, we have $b^2=c^2 (c^2+1)^2/4=0$, and $4a^2+4=c^2(c^2+3)^2+4=(c^2+3)^2 (c^2+1)^2=0$, i.e. $a^2=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 4,
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Prove $BA - A^2B^2 = I_n$. I have a problem with this. Actually, still don't have the right way to start :/
Problem :
Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$.
Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.
Thanks for any help.
| The two given conditions $AB-B^2A^2=I$ and $A^3+B^3=0$ can be rewritten as
$$
\pmatrix{A&B^2\\ -B^2&A} \pmatrix{B&A^2\\ -A^2&B} = \pmatrix{I&0\\ 0&I}.
$$
Since $XY=I$ implies that $YX=I$ for any two square matrices $X$ and $Y$, we have
$$
\pmatrix{B&A^2\\ -A^2&B} \pmatrix{A&B^2\\ -B^2&A} = \pmatrix{I&0\\ 0&I}
$$
and the assertion follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 4
} |
Determinant of a matrix Having some problems with a determinant of a 4x4 matrix M.
$
M =
\left( {\begin{array}{cc}
1 & 2 & 3 &-1 \\
0 & 1 & 2 & 2 \\
1 &1 &0 &0 \\
3&1&2&0
\end{array} } \right)
$
Went along and developed it according to the 4th column. So I end up with two matrixes A and B.
$
A = -1 \cdot det
\left( {\begin{array}{cc}
0 & 1 & 2 \\
1 & 1 & 0 \\
3 &1 &2 \\
\end{array} } \right)
$
$
B = 2 \cdot det
\left( {\begin{array}{cc}
1 & 2 & 3 \\
1 & 1 & 0 \\
3 &1 &2 \\
\end{array} } \right)
$
I get $A= (-1) \cdot((0 \cdot1\cdot2)+(1\cdot0\cdot3)+(2\cdot1\cdot1)-(3\cdot1\cdot2)-(1\cdot2\cdot2)-(1\cdot1\cdot0)) \\$
$A=(-1) \cdot(-6)=6$
$B= 2 \cdot((1\cdot1\cdot2)+(2\cdot0\cdot3)+(3\cdot1\cdot1)-(3\cdot1\cdot3)-(1\cdot2\cdot2)-(1\cdot1\cdot0))
\\$
$B = 2\cdot8=16$
$A+B=22$
which is wrong. Where is my mistake?
The correct answer should be $-22$ but I don't get why my solution keeps being positive.
Edit: im such a moron: A = 1* det and B = -2 * det. Everythings clearing up while in bed. Hehe!
| Your mistake was that you forgot the $(-1)^{i+j}$ for the cofactors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $m+n=5$ and $mn=3$, find $\sqrt{\frac{n+1}{m+1}} + \sqrt{\frac{m+1}{n+1}}$? It is known that $m+n=5$ and $mn=3$. So what is the value of:
$$
\sqrt{\dfrac{n+1}{m+1}} + \sqrt{\dfrac{m+1}{n+1}}
$$
I think we're suppose to solve for the system of equations first, but I'm not getting any results that's useful.
| \begin{align}
\sqrt{\frac{n+1}{m+1}}+\sqrt{\frac{m+1}{n+1}}&=\frac{(n+1)+(m+1)}{\sqrt{(m+1)(n+1)}}
\\ \\
&=\frac{m+n+2}{\sqrt{mn+m+n+1}}
\\ \\
&=\frac{5+2}{\sqrt{3+5+1}}
\\ \\
&=\frac{7}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\displaystyle \lim_{x \to{+}\infty}{ax^{3}+bx^{2}+cx+d} = {+}\infty$, $a>0$. I need some help proving that:
$\displaystyle\lim_{x \to{+}\infty}{ax^{3}+bx^{2}+cx+d} = {+}\infty$, with $a,b,c,d$ real numbers, $a>0$.
Its the first time i work this kind of limits and a I feel a little lost!
| We have, for non-zero $x$,
$$ax^3+bx^2+cx+d=x^3\left(a+\frac{b}{x}+\frac{c}{x^2}+\frac{d}{x^3}\right).$$
For large enough $x$, we have $\left|\frac{b}{x}\right|\lt \frac{a}{4}$, and $\left|\frac{c}{x^2}\right|\lt \frac{a}{4}$, and $\left|\frac{d}{x^3}\right|\lt \frac{a}{4}$.
Thus by the Triangle Inequality, $a + \frac{b}{x}+\frac{c}{x^2}+\frac{d}{x^3}\gt \frac{a}{4}$.
It follows that for large enough $x$, we have
$$ax^3+bx^2+cx+d \gt \frac{ax^3}{4}.$$
Remark: The intuition is that when $x$ is large, the leading term $ax^3$ is "dominant," and the others are (comparatively) small. The answer formalizes this intuition.
To be even more formal, there is an $N_1$ such that if $x\gt N_1$, then $\left|\frac{b}{x}\right|\lt \frac{a}{4}$. Similarly, there are $N_2$ and $N_3$ for the other two desired inequalities. If $N=\max(N_1,N_2,N_3)$ then all of the inequalities hold. So if $x\gt N$ then $ax^3+bx^2+cx+d\gt \frac{ax^3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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An integral about trigonometric function. Recently I met an integral which is $\int_0^\infty \left(\frac{\sin x}{x}\right)^3 \; dx$. I get the result is $3\pi/8$ by using Mathematica,
but I cannot derive it independently.
So I hope someone can help me.
It is my first time to ask questions on math.stackexchange.
Please tell me if I have done something wrong, thanks.
At last, thanks all the people for your precious time.
| Integrating by parts:
$$
\int_0^\infty \frac{\sin^3(x)}{x^3} \mathrm{d}x = \int_0^\infty \sin^3(x) \mathrm{d} \left(-\frac{1}{2 x^2} \right) = \frac{3}{2} \int_0^\infty \cos(x) \frac{\sin^2(x)}{x^2} \mathrm{d}x = \ldots
$$
Integrating by parts again:
$$
\ldots = \frac{3}{2} \int_0^\infty \frac{2 \cos^2(x) \sin(x) - \sin^2(x)}{x} \mathrm{d}x = \ldots
$$
Using
$$
2 \cos^2(x) \sin(x) - \sin^2(x) = \frac{1}{4} \left(3 \sin(3x)-\sin(x)\right)
$$
we continue:
$$
\ldots = \frac{3}{8} \left( \underbrace{\int_0^\infty \frac{3 \sin(3x)}{x} \mathrm{d}x }_{3 \frac{\pi}{2}}- \underbrace{\int_0^\infty \frac{\sin(x)}{x} \mathrm{d}x}_{\frac{\pi}{2}} \right) = \frac{3}{8} \pi
$$
where we used the value of the Dirichlet integral $\int_0^\infty \sin(x) \frac{\mathrm{d}x}{x} = \frac{\pi}{2}$ and a change of variable
$$
\int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x \stackrel{3x = y}{=} \int_0^\infty \frac{\sin(y)}{y} \mathrm{d}y = \frac{\pi}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$ $$ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$$
I know that it should be the chain rule and product rule used together to get the answer
$$ y = \frac{dx}{dy}((2x - 3)^4) \cdot (x^2 + x + 1)^5 + \frac{dx}{dy}(x^2 + x + 1)^5 \cdot (2x - 3)^4 $$
this gives me something ridiculous like this
$$8(2x-3)^3 \cdot (x^2 + x + 1)^5 + (x^2 + x + 1)^4 \cdot (2x+1) (2x-3)^4$$
This is wrong and I keep getting it, I don't know how to simplify it without expanding everything.
The book Houdini's out $(2x -3)^3 (x^2 + x + 1)^4 (28x^2 - 12x - 7)$
| We have $$8(2x-3)^3 * (x^2 + x + 1)^5 + 5(x^2 + x + 1)^4 (2x+1) (2x-3)^4\\=
(2x-3)^3(x^2+x+1)^4(8(x^2+x+1)+5(2x+1)(2x-3))
\\=(2x-3)^3(x^2+x+1)^4(8x^2+8x+8+20x^2-20x-15)\\=
(2x-3)^3(x^2+x+1)^4(28x^2-12x-7)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there an algorithm to find the number of digits in 2^n, where n is a positive integer? Is there an algorithm to find the number of digits in 2^2030 ?
$2^1=2$
$2^2=4$
$2^3=8$
$2^4=16$
$2^5=32$
$2^6=64$
$2^7=128$
...
$2^{10}=1024$
| Just for fun, let's suppose you don't have access to floating-point computations, but you can compute with rational numbers.
We have the series (for $|x|<1$)
$$ \ln \left( \frac{1+x}{1-x}\right) = 2 x + \frac{2}{3} x^3 + \frac{2}{5} x^5 + \frac{2}{7} x^7 + \ldots$$
and in particular, if $0 < x < 1$,
$$ 2 x + \frac{2}{3} x^3 + \frac{2}{5} x^5 + \frac{2}{7} x^7
< \ln\left( \frac{1+x}{1-x}\right) < 2 x + \frac{2}{3} x^3 + \frac{2}{5} x^5 +
\frac{2 x^7}{7(1 - x^2)}$$
For $x = 1/3$ this gives us
$$ \dfrac{53056}{76545} < \ln(2) < \dfrac{23581}{34020}$$
while for $x = 1/9$ we have
$$ {\frac {37355104}{167403915}} < \ln \left( 5/4 \right) < {\frac {
3689393}{16533720}}
$$
Then $2030 \log_{10}(2) = \dfrac{2030 \ln 2}{\ln(5/4) + 3 \ln(2)}$ is between
$\dfrac{23263994880}{38070491}$ and $\dfrac{94221399069}{154182208}$, both of which are between $611$ and $612$.
So, $612$ digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $xyz=1$ , find $\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}$? I think I solved this problem, but I don't feel $100$ percent sure of my solution. We have:
$xy=\large {\frac 1z}$
$xz=\large \frac 1y$
$yz=\large \frac 1x$
So let's substitute these into our sum:
$\large \frac{1}{1+x+\frac 1z}+\frac{1}{1+y+\frac 1x}+\frac{1}{1+z+\frac 1y}$
If we rewrite with a common denominator we get
$\large \frac{z}{1+z+xz}+\frac {x}{1+x+xy}+\frac{y}{1+y+zy}$
If $\large \frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac {x}{1+x+xy} +\frac{y}{1+y+zy}+\frac{z}{1+z+xz}$ , $;$ then $(x, y, z)=1$ and we can compute that the sum is equal to $1$.
The problem I have with this is that I found what $(x, y, z)$ $had$ to be, but I but I only had one weak restriction on the variables. Secondly, what are other way of doing this that I can learn from? Thanks.
| We only need to continue a little further on the path you took. The first term is
$$\frac{1}{1+x+xy}.\tag{1}$$
Consider the second term
$$\frac{1}{1+y+yz}.$$
Using your substitution, we (you) find that this is
$$\frac{x}{1+x+xy}.\tag{2}$$
Consider the third term $\dfrac{1}{1+z+zx}$ and replace $z$ by $\frac{1}{xy}$, and $zx$ by $\frac{1}{y}$. We end up with
$$\frac{xy}{1+x+xy}.\tag{3}$$
Add up the first term, the second, and the third, or equivalently (1), (2), and (3). We get
$$\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+x+xy},$$
which is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Show that $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b} \ge 3$
If $a,b,c$ are positive numbers then show that
$$\dfrac{a^2+1}{b+c}+\dfrac{b^2+1}{a+c}+\dfrac{c^2+1}{a+b} \ge 3$$
I am stuck at the first stage. Please give me some hints so that I can solve the problem. Thanks in advance.
| Direct Question of Titu's Lemma.
$\dfrac{1}{b+c}+ \dfrac{1}{c+a}+\dfrac{1}{a+b} \ge \dfrac{(1+1+1)^2}{2(a+b+c)}$
Similarly,
$\dfrac{a^2}{b+c}+ \dfrac{b^2}{c+a}+\dfrac{c^2}{a+b} \ge \dfrac{(a+b+c)^2}{2(a+b+c)}$
Let, $a+b+c=k$
$\dfrac{3^2}{2k}+\dfrac{k}{2} \ge 2\sqrt{3^2 \cdot \dfrac{k}{2} \dfrac{1}{2k}} = 3$, we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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} |
An inequality involving complex numbers I would like to prove the following inequality:
$$\large{
\left| {1 + re^{ - \frac{2}{3}\theta i} } \right|^2 \left| {1 + re^{\frac{2}{3}\left( {\pi - \theta } \right)i} } \right|^2 \ge \cos ^2 \theta ,
}$$
for any $r>0$ and $-\frac{\pi}{2}<\theta<\frac{3\pi}{2}$. I expanded the left-hand side as a polynomial in $r$ but it seems hopeless. Any idea? Thank you very much in advance!
| Unfortunately I haven't found a shorter proof so far.
We have
\begin{align}
f(r,\theta)&=\left|1+re^{-2i\theta/3}\right|^2\left|1+re^{2i(\pi-\theta)/3}\right|^2=\\&=\left(1+r^2+2r\cos\frac{2\theta}{3}\right)\left(1+r^2+2r\cos\frac{2(\pi-\theta)}{3}\right)=\\
&=\frac{3}{4}\left(1-r^2\right)^2+4r^2\left(\frac{1+r^2}{4r}+\cos\frac{\pi-2\theta}{3}\right)^2.
\end{align}
On the other hand,
$$\cos^2\theta=\frac{1+\cos2\theta}{2}=\frac{1-\cos\left(3\times\frac{\pi-2\theta}{3}\right)}{2}=\frac12\left(1-4\cos^3\frac{\pi-2\theta}{3}+3\cos\frac{\pi-2\theta}{3}\right).$$
Therefore, if we denote $q=\cos\frac{\pi-2\theta}{3}$, all we need to show is that for $r>0$ and $q\in(-\frac12,1]$ one has
$$\frac{3}{4}\left(1-r^2\right)^2+4r^2\left(\frac{1+r^2}{4r}+q\right)^2
\geq \frac12\left(1-4q^3+3q\right)\tag{1}$$
or, equivalently,
$$g(q,r)=4q^3+8r^2q^2+(4r^3+4r-3)q+(2r^4-2r^2+1)\geq0.\tag{2}$$
Now since it can be easily shown that
\begin{align}
&g(-1/2,r)=2(r-1)^2(r^2+r+1)\geq0,\\
&g(1,r)=2(r^2+r+1)^2>0,
\end{align}
the only possibility for (2) to be violated is to have $\displaystyle\frac{\partial g}{\partial q}=0$ at some point $q_{min}\in(-\frac12,1)$. Differentiating $g$ and solving the resulting quadratic equation for $q$, one can show that such point (+ giving a minimum) can exist only if $r\in(0,\frac34]\cup[(\frac34)^{1/3},\infty)$ and is explicitly given by
$$q_{min}(r)=\frac{1}{6}\left(\sqrt{(4r^3-3)(4r-3)}-4r^2\right).$$
Finally, studying the function
$$h(r)=g(q_{min}(r),r)=-\left(\frac{\sqrt{(4r^3-3)(4r-3)}}{3}\right)^3+\frac{64r^6-72r^5+54r^4-72r^3+27}{27},$$
it can be shown by standard single-variable methods that $h(r)\geq0$ for $r\in(0,\frac34]\cup[(\frac34)^{1/3},\infty)$, and the minimal value $0$ is attained only for $r=0$ and $r=1$. $\blacksquare$
| {
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Need further help and verification regarding a dynamic model problem We have the model $y= ax(1-x)$ and we want to find the period 2 solutions such that $X_{n} = X_{n+2} $ and $X_n \neq X_{n+1}$. My teacher told us to do this problem with the quadratic formula. This is my attempt:
$y = ax (1-x)$, so for period 2 we'd want $x = a(ax(1-x))(1-ax(1-x))$. Factor out the parentheses and you get $$x = -a^3x^4 + 2a^3x^3-a^3x^2 -a^2x^2 + a^2x$$
Here I am confused because I am supposed to use a quadratic formula for a quartic function.
| The two fixed points, $z_0 = 0$ and $z_1 = (a-1)/a$ are also solutions of $X_{n+2} = X_n$, hence you can divide those two out of the equation
$$a^3x^4 - 2a^3x^3 +a^2(a+1)x^2 - (a^2-1)x = 0$$
(you have miscalculated the coefficient of $x^2$), to obtain the quadratic equation
$$a^2x^2 - a(a+1)x + (a+1) = 0$$
for the two points of the 2-cycle.
First, for the fixed points, we have the equation $x = ax(1-x)$, which, expanded and rearranged becomes
$$ ax^2 - (a-1)x = 0$$
and we can read off the two solutions $z_0 = 0$ and $z_1 = (a-1)/a$ immediately.
Then, for the points with $X_{n+2} = X_n$, we have the equation $x = f(f(x))$, which, substituting $f(w) = aw(1-w)$ expands to
$$x = af(x)\bigl(1-f(x)\bigr) = a^2x(1-x)\bigl(1-ax(1-x)\bigr) = a^2x\bigl(1-x -ax(1-x) + ax^2(1-x)\bigr)$$
and then
$$x = a^2x(1 -(a+1)x +2ax^2 - ax^3) = x(-a^3x^3 + 2a^3x^2 - (a+1)a^2x + a^2).$$
Rearranging yields
$$x\bigl(a^3x^3 - 2a^3x^2 + (a+1)a^2x - (a^2-1)\bigr) = 0.$$
We already know that $z_0$ and $z_1$ are solutions (I have already factored out $x - z_0$), and thus we can write the polynomial expression on the left hand side as $(x - z_0)(x - z_1)Q(x)$ with a quadratic polynomial $Q$. Now we split off the factor $x - z_1$, or, better $a\cdot(x - z_1) = ax - (a-1)$ from the cubic factor.
$$\bigl(a^3x^3 - 2a^3x^2 + (a+1)a^2x - (a^2-1)\bigr) = \bigl(ax - (a-1)\bigr)\cdot Q(x)$$
To determine $Q$, we start with the highest-order term, and it's clear that must be $a^2x^2$ to obtain the highest-order term $a^3x^3$ of the cubic polynomial. $\bigl(ax - (a-1)\bigr)\cdot a^2x^2 = a^3x^3 - a^2(a-1)x^2$, so
$$\bigl(a^3x^3 - 2a^3x^2 + (a+1)a^2x - (a^2-1)\bigr) = \bigl(ax-(a-1)\bigr)\cdot a^2x^2 + R(x)$$
with $R(x) = -a^2(a+1)x^2 + a^2(a+1)x - (a^2-1)$. Splitting off $ax - (a-1)$ from $R(x)$ yields the next term of $Q(x)$, namely $-a(a+1)x$.
$$\bigl(ax - (a-1)\bigr)\cdot a(a+1)x = a^2(a+1)x^2 -a(a+1)(a-1)x$$
then gives us
$$\bigl(a^3x^3 - 2a^3x^2 + (a+1)a^2x - (a^2-1)\bigr) = \bigl(ax - (a-1)\bigr)\cdot\bigl(a^2x^2 - a(a+1)x\bigr) + S(x)$$
with $S(x) = a(a+1)x - (a^2-1) = (a+1)\bigl(ax - (a-1)\bigr)$, so wrapping up, we obtain the factorisation
$$x\bigl(ax - (a-1)\bigr)\cdot\bigl(a^2x^2 - a(a+1)x + (a+1)\bigr) = 0$$
and the two zeros of the last factor are the points with $X_{n+2} = X_n \neq X_{n+1}$.
| {
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Prove that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^2\ge \frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{12}{a^2+b^2+c^2}+\frac{18}{a^2+b^2+c^2+d^2}$
Let $a,b,c,d$ be positive numbers. Show that
$$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{12}{a^2+b^2+c^2}+\dfrac{18}{a^2+b^2+c^2+d^2}$$
I have seen this Similar Problem
$$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{9}{a^2+b^2+c^2}+\dfrac{16}{a^2+b^2+c^2+d^2}$$
This problem pf:
\begin{align}
LHS &=\sum\dfrac{1}{a^2}+\sum\dfrac{2}{ab}\\
&=\dfrac{1}{a^2}+\dfrac{2}{ab}+(\dfrac{1}{b^2}+\dfrac{2}{ac}+(\dfrac{2}{ad}+\dfrac{2}{bc})++(\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{2}{ab}+\dfrac{2}{bd}+\dfrac{2}{cd})\\
&\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+(\dfrac{1}{b^2}+\dfrac{4}{a^2+c^2})+(\dfrac{4}{a^2+d^2}+\dfrac{4}{b^2+c^2})+(\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{2}{ab}+\dfrac{2}{bd}+\dfrac{2}{cd})\\
&\ge\dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{9}{a^2+b^2+c^2}+\dfrac{16}{a^2+b^2+c^2+d^2}
\end{align}
| There is the solution in the document you provided indeed - the proof is exactly the same. Note that on the top of p.25, there are four inequalities. Replace the third one by
$$\frac{1}{b^2} + \frac{1}{c^2} \ge \frac{4}{b^2+c^2} \ge \frac{4}{a^2+b^2+c^2}$$
The fourth one by
$$\frac{18}{ad+bd+cd} \ge \frac{18}{a^2+b^2+c^2+d^2}$$
and you get the result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\sqrt{2\sqrt{2\sqrt{2\cdots}}} = 2$? Why is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}}}$ equal to 2? Does this work for other numbers?
| Let $x=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots2}}}}}}$ then
$x^{2}=2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots2}}}}}}=2x$
so $x(x-2)=0$ since $x\neq 0$
$x=2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $abc=1$ and $\prod_{\text{cyc}}(a^2+b^2)=8$, then $(a+b+c)^3\prod_{\text{cyc}}(a^2+b^2)\ge\left (\sum_{\operatorname{cyc}} (a^2b+b^2a) \right) ^3$.
Let $a,b,c>0 $ where $abc=1$ and $(a^2+b^2)(b^2+c^2)(c^2+a^2)=8$. Prove that $$ (a+b+c)^3(a^2+b^2)(b^2+c^2)(c^2+a^2) \ge \big((a^2b+b^2a)(b^2c+c^2b)(c^2a+a^2c)\big) ^3.$$
Thanks
| New proof
I would like to start by looking at the second limitation, $(a^2+b^2)(b^2+c^2)(c^2+a^2)=8$. Multiplying the parentheses and noting that $abc = 1$ gives $a^4b^2 = \frac{a^4b^2c^2}{c^2} = \frac{a^2}{c^2}$, we get
\begin{align}
8 &= (a^2+b^2)(b^2+c^2)(c^2+a^2)\\\\
8 &= a^4 b^2+a^4 c^2+a^2 b^4+a^2 c^4+b^4 c^2+b^2 c^4+2a^2b^2c^2\\\\
9 &= \frac{b^2 + c^2}{a^2}+\frac{a^2 + c^2}{b^2}+\frac{a^2 + b^2}{c^2} + 3\\\\
9 &= \frac{b^2 + c^2}{a^2}+\frac{a^2 + c^2}{b^2}+\frac{a^2 + b^2}{c^2} +\frac{a^2}{a^2}+\frac{b^2}{b^2}+\frac{c^2}{c^2}\\\\
9 &= (a^2 + b^2 + c^2) \bigg( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)
\end{align}
By the Cauchy-Schwarz inequality this means that the vectors $(a, b, c)$ and $(\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$ are linearly dependant, which again means that $a^2 = b^2 = c^2$, and thus $a=b=c = 1$. Now the original inequality shouldn't be too hard to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
| Here is a very calculus way to do the problem. (Your "taking" $n$ out is useful.) Write $h$ for $1/n$, and rewrite our limit as
$$\lim_{h\to 0}\frac{(\sqrt{1+h} -1)-(\sqrt[3]{1+h}-1)}{h}, \tag{1}$$
because
$$\frac{\sqrt{1+h} -1}{h} \quad\text{and}\qquad \frac{\sqrt[3]{1+h}-1}{h}$$
are familiar expressions whose limits we can compute.
We recognize
$$\lim_{h\to 0}\frac{\sqrt{1+h} -1}{h}$$
as the derivative of $\sqrt{1+x}$ at $x=0$, and
$$\lim_{h\to 0}\frac{\sqrt[3]{1+h} -1}{h}$$
as the derivative of $\sqrt[3]{1+x}$ at $x=0$.
These two derivatives, evaluated at $0$, are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. It follows that the limit (1) is equal to
$\dfrac{1}{2}-\dfrac{1}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =-\frac{\pi^2 \sqrt{2}}{16}$ I could prove it using the residues but I'm interested to have it in a different way (for example using Gamma/Beta or any other functions) to show that
$$
\int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x
=-\frac{\,\pi^{2}\,\sqrt{\,2\,}\,}{16}.
$$
Thanks in advance.
| Another approach, we can split the denominator part as follows
$$
\frac{1}{x^4+1}=\frac{1}{2i}\left(\frac{1}{x^2-i}-\frac{1}{x^2+i}\right).
$$
Consequently, the integral becomes
$$
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{2i}\int_{0}^{\infty }\left(\frac{\ln x}{x^2-i}-\frac{\ln x}{x^2+i}\right)\ dx.
$$
Using formula from here,
$$
\int_0^{\infty}\frac{\ln x}{x^2+a^2}\ dx=\frac {\pi \ln a}{2a},
$$
we obtain
$$
\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx&=\frac{1}{2i}\left(\frac {\pi \ln \sqrt{-i}}{2\sqrt{-i}}-\frac {\pi \ln \sqrt{i}}{2\sqrt{i}}\right)\\
&=\frac{\pi}{4i}\left(\frac {\ln i^{\frac{3}{2}}}{i^{\frac{3}{2}}}-\frac {\ln i^{\frac{1}{2}}}{i^{\frac{1}{2}}}\right).
\end{align}
$$
Taking $0\le\theta\le2\pi$, from Euler's formula we have
$$
e^\frac{i\pi}{2}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i.
$$
Thus
$$
\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx
&=\frac{\pi}{4i}\left(\frac {\ln e^\frac{3i\pi}{4}}{e^\frac{3i\pi}{4}}-\frac {\ln e^\frac{i\pi}{4}}{e^\frac{i\pi}{4}}\right)\\
&=\frac{\pi}{4i}\left(-\frac{i\pi}{2\sqrt{2}}\right)\\
&=\boxed{\color{blue}{-\Large\frac{\pi^2}{16}\sqrt{2}}}
\end{align}
$$
$$\\$$
$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
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How to check if the lines are coplanar? I have two lines:
$\frac{x-1}{3}=\frac{y+2}{-2}=\frac{z}{1} = L_1$
$ \frac{x+1}{4} = \frac{y-3}{1}=\frac{z}{\alpha} =L_2$
How can I find the value of $\alpha$ for which these two lines lie on the same plane?
Just notice that this is not a homework question. I took it from an exam held a few years ago, and got no idea about it.
Thanks!
| $$\begin{align*}(L_1)&\;\;z=\frac{x-1}3\;,\;\;\;\;y=\frac{-2x+2}3-2=\frac{-2x-4}3\\
(L_2)&\;\;z=\frac{\alpha x+\alpha}4\;,\;\;y=\frac{x+13}4\end{align*}$$
Suppose there exists a plane $\,ax+by+cz+d=0\;$ which contains both lines, then we have
$$ax+b\left(\frac{-2x-4}3\right)+c\left(\frac{x-1}3\right)=ax+b\left(\frac{x+13}4\right)+c\left(\frac{\alpha x+\alpha}4\right)\iff$$
$$\left(\frac{11}{12}b+\frac{3\alpha-4}{12}c\right)x+\left(\frac{55}{12}b+\frac{3\alpha+4}{12}c\right)=0$$
and from here we get
$$\begin{align*}11b&=(-3\alpha+4)c\\
55b&=(-3\alpha-4)c\end{align*}$$
Divide second eq. by first one above and get your value of $\,\alpha=2\,$ ...
| {
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Evaluating $\int \frac{\mathrm dz}{z^3 \sqrt{z^2 - 4}}$ $$\int \frac{dz}{z^3 \sqrt{z^2 - 4}}$$
$z = 4\sec\theta$
$dz = 4\sec\theta \tan d\theta$
$$\int \frac{\sec\theta \tan\theta}{4^3 \sec^3 \theta \tan \theta}$$
$$ \frac{1}{4^3} \int \frac{d \theta}{\sec^2 \theta}$$
I am again stuck, I have no idea how to proceed.
| Hint:
$$\frac{1}{\sec^2(\theta)}=\cos^2(\theta)=\frac{1}{2}(1+\cos(2\theta))$$
$$z=4\sec(\theta)$$
$$2\theta=2\sec^{-1}(\frac{z}{4})$$
$$\sin(2\theta)=\sin(2\sec^{-1}(\frac{z}{4}))=\frac{8\sqrt{z^2-16}}{z^2}$$
| {
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How prove $\frac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\frac{1}{t^2}
Let $k,n\in \mathbb{N},n\ge k$, prove that
$$\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\dfrac{1}{t^2}<e.$$
I got the impression that this inequality is very sharp.
My idea:
$$\sum_{t=k+1}^{n}\dfrac{1}{t^2}\le\sum_{t=k+1}^{n}\dfrac{1}{t(t-1)}=\dfrac{1}{k}-\dfrac{1}{n},$$
$$\Longleftrightarrow \dfrac{(k+1)^{k+1}}{k^k}\left(\dfrac{1}{k}-\dfrac{1}{n}\right)<\dfrac{(k+1)^{k+1}}{k^k}\dfrac{1}{k}<e$$
$$\Longleftrightarrow \left(1+\dfrac{1}{k}\right)^{k+1}<e.$$
It is well konwn that
$$(1+1/x)^{x+1}>e,x>0.$$
so I failed at this direction. Thank you everyone help.
Idea 2: use the well-known
$$\left(1+\dfrac{1}{x}\right)^x<e\left(1-\dfrac{1}{2(1+x)}\right)$$
$$\Longleftrightarrow \dfrac{n-k}{k+1}\left(1-\dfrac{1}{2(k+1)}\right)<1,$$
but this can not prove the inequality either.
| A sketch. First note that:
$$\sum_{r=m+1}^{\infty} \frac{1}{r^2}=\int_0^1 \frac{x^m \ln x}{x-1}\,dx$$
Now:
$$\begin{aligned}\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^n \frac{k}{t^2}&<\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^{\infty} \frac{k}{t^2}\\&=\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 \frac{kx^k \ln x}{x-1}\,dx\\&<\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 kx^{k-\frac{1}{2}} \,dx\\&=\left(1+\frac{1}{k}\right)^{k+1}\left(\frac{2k}{2k+1}\right)\\&=\left(1+\frac{1}{k}\right)^k\frac{2k+2}{2k+1}\\&<e\end{aligned}$$
| {
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Show that $\frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Given $a,b,c>0$ and $(a+b)(b+c)(c+a)=8$. Show that $\displaystyle \frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Obviously, AM-GM seems to be suitable for LHS.
For RHS, $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)=(a+b+c)^3-24$, then I don't know what to do.
Can someone please teach me? Thank you.
p.s. That $\sqrt [27]{}$ is really terrible...
| Let $s = a+b+c$. Then, as you noted, the inequality is equivalent to:
$$ \frac{s}{3} \geq \sqrt[27]{\frac{s^3-24}{3}} \qquad (\ast)$$
Because you know that $(a+b)(b+c)(c+a) = 8$, or equivalently $\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{c+a}{2} = 1$, it follows that $s = \frac{a+b}{2} + \frac{b+c}{2} + \frac{c+a}{2} \geq 3$. In fact, this is all you can say about $s$, so our initial problem is equivalent to showing that $(\ast)$ holds for all $s \geq 3$. You can rewrite $(\ast)$ as:
$$ \left( \frac{s}{3} \right)^2 - \frac{s^3-24}{3} \geq 0 \qquad (\ast\ast)$$
For $s = 3$ there is equality in $(\ast\ast)$, and we can in fact divide out $(s-3)^2$ from the left side. The resulting function:
$$ \frac{\left( \frac{s}{3} \right)^2 - \frac{s^3-24}{3}}{(s-3)^2} $$
is a polynomial which takes strictly positive values for $s \geq 3$ (in fact, for $s \geq -3$). You can directly verify that by a variety of brute-force methods (possibly with help of some softwere like Mathematica)
A more satisfactory idea: Put $t = (s/3)^3 \geq 1$ for ease of notation. We can easily see that $(\ast\ast)$ is equivalent to $$t^{9} - 9t + 8 \geq 0 $$
But this follows directly by applying AM-GM to $t^9,1,1,1,1,1,1,1,1$!.
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial fraction $\int \frac{x^2 + 11x dx}{(x-1)(x+1)^2}$ I have been using the cover up method from this
video lecture
$$\int \frac{x^2 + 11x}{(x-1)(x+1)^2} dx$$
$$\frac{x^2 + 11x }{(x-1)(x+1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{x-1}$$
What do I do with the double $x-1$ denominator? With the cover up method it seems useless. I can get that $A$ is $3$, $B$ might be 5 but then so is $C$? How do I use the cover up method to solve this?
| You need to decompose as follows
$${x^2 + 11 x\over (x-1)(x+1)^2} = {A\over x + 1} + {B\over x - 1} + {C\over (x+1)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence relation: $x_{n+1} = \frac 12 x_n + \frac 1 {x_n},$ $$x_{n+1} = \frac 12 x_n + \frac 1 {x_n}, x_0 \neq 0$$
$$ a = \frac a2 + \frac 1a \Rightarrow a = \frac {a^2 + 2} {2a} \Rightarrow 2a^2 = a^2 + 2 \Rightarrow a^2 = 2 \Rightarrow a = \pm \sqrt 2$$
If $x_0 > 0$, then any subsequent terms will be positive too. If $x_0 < 0$ , then - negative. Hence, if $x_0 > 0$, then $\lim_{n \rightarrow \infty} x_n = \sqrt 2$. If $x_0 < 0$, then $\lim_{n \rightarrow \infty} x_n = - \sqrt 2$
Is it complete solution? I know that it's correct answer, but may be I missed something in proof?
| You can show that the sequence indeed converges by the following simple way:
From the recursion relation we get, $$\left(\frac{x_{n+1}+\sqrt{2}}{x_{n+1}-\sqrt{2}}\right)=\left(\frac{x_{n}+\sqrt{2}}{x_{n}-\sqrt{2}}\right)^2, \forall n\geq1 $$ So iteratively, we get, $$\left(\frac{x_{n+1}+\sqrt{2}}{x_{n+1}-\sqrt{2}}\right)=\left(\frac{x_{0}+\sqrt{2}}{x_{0}-\sqrt{2}}\right)^{2^{n+1}}$$ Rearranging, we get, $$x_{n+1}=\sqrt{2}\frac{(x_0+\sqrt{2})^{2^{n+1}}+(x_0-\sqrt{2})^{2^{n+1}}}{(x_0+\sqrt{2})^{2^{n+1}}-(x_0-\sqrt{2})^{2^{n+1}}}$$ Now if $x_0\ne \sqrt{2},\ x_0>0$, then $|\frac{x_0-\sqrt{2}}{x_0+\sqrt{2}}|<1$. Hence $$\lim_{n\rightarrow \infty}x_{n+1}=\sqrt{2}$$
If if $x_0\ne -\sqrt{2},\ x_0\leq0$, then $|\frac{x_0+\sqrt{2}}{x_0-\sqrt{2}}|<1$. Hence $$\lim_{n\rightarrow \infty}x_{n+1}=-\sqrt{2}$$ Clearly, if $x_0={}_{-}^+\sqrt{2}$ then
$$\lim_{n\rightarrow \infty}x_{n+1}={}_{-}^+\sqrt{2}$$ respec.
| {
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The value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$? How to find value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$ ?
I've calculated it by MATLAB for some finite terms and I've got : $0.3001 - 0.4201i$, but I don't know how to find the value analytically! Would you mind helping me find it? Thanks
| First of all let's assume the series is convergent. Looking for fixed points we have:
$$x=\sqrt{1-\sqrt{1+x}}$$
Now we will try to solve this equation. First squaring both sides:
$$1-x^2=\sqrt{1+x} \\
\left(\left( 1-x\right)\left( 1+x\right) \right)^2=1+x$$
Note that $x$ must be nonnegative, therefore:
$$(1-x)^2(1+x)-1=0 \\
\Rightarrow x^3-x^2-x=0$$
So $x=0$ is a solution. The other solutions are:
$$x^2-x-1=0 \Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$$
where only $x=\frac{1 + \sqrt{5}}{2}$ is greater than or equal to zero and may look valid. But as people pointed out, one has to check if the answers actually fit into the initial equation. In this case $\frac{1 + \sqrt{5}}{2}$ doesn't, therefore the only fixed point we have found is $x=0$.
But $x=0$ cannot be the convergence limit(it doesn't converge smoothly). Assume we deflect $x=0$ with the tiny amount of $\epsilon$(or rather starting with a tiny $x_1=\epsilon$). Putting it back into our initial equations and getting the next $x$:
$$x_2=\sqrt{1-\sqrt{1+\epsilon}}\approx \sqrt{1-\left( 1+\frac{\epsilon}{2}\right)} \approx \frac{i\sqrt{\epsilon}}{\sqrt{2}}$$
Now for $\epsilon < \frac{1}{2}$, $|x_2|>|x_1|$; ergo $x=0$ cannot be the convergence limit. We have proved that this infinite radicals doesn't have a single limit.
| {
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Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove more general identity:
$$
\begin{array}{|c|}
\hline \\
\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2 \\
\hline
\end{array}
, \qquad \mbox{ where } \ n \ \mbox{ is odd.}$$
| Since $n$ is odd, the numbers $u_k=\cos k\pi/n$ are the same as the numbers $\cos 2k\pi/n$, i.e. the distinct angles $\theta$ satisfying $n\theta=0$ (mod $2\pi$). We think of them as the roots of the equation $\cos n\theta=1$. Writing
$$\cos n\theta = \cos^n\theta - \binom{n}{2}\cos^{n-2}\theta\sin^2\theta \cdots \pm n\cos\theta \sin^{n-1}\theta$$
and using $\sin^2\theta=1-\cos^2\theta$, we see that the $u_k$'s are the roots of the polynomial
$$p(u)=u^n - \binom{n}{2}u^{n-2}(1-u^2) \pm n u(1-u^2)^{(n-1)/2} + 1.$$
Note that all powers of $u$ which occur are odd (except for the constant term).
The reciprocals $1/u_k$ are the roots of the "reverse polynomial"
$$r(u)=u^n p(1/u) = u^n + a_{n-1} u^{n-1} + \cdots,$$
where $a_{n-1}=\pm n$ and $a_{n-2}=0$.
The sum in question is the sum of the squares of the roots of $r(u)$, i.e. $a_{n-1}^2 - 2a_{n-2} = n^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation : $2013x+\sqrt[4]{(1-x )^7}=\sqrt[4]{(1+x )^7}$. Solve the equation : $2013x+\sqrt[4]{(1-x )^7}=\sqrt[4]{(1+x )^7}$.
Show that it has percisely one root: $x=0$.
| $x=0$ is an obvious root and we need to restrict $x$ to $|x|\le 1$ for the roots to make (real) sense in the first place.
First proceed with standard techniques, i.e. rearrange:
$$ 2013 x=\sqrt[4]{(1+x)^7}-\sqrt[4]{(1-x)^7}$$
Multiply by the conjugate:
$$ 2013 x\left(\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\right)=\sqrt[2]{(1+x)^7}-\sqrt[2]{(1-x)^7}$$
Multiply by the conjugate again:
$$ 2013 x\left(\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\right)\left(\sqrt{(1+x)^7}+\sqrt{(1-x)^7}\right)={(1+x)^7}-{(1-x)^7}.$$
Since $|x|\le 1$, twe can estimate
$$ \begin{align}\left|{(1+x)^7}-{(1-x)^7}\right|&=2\left|{7\choose 1}x+{7\choose 3}x^3+{7\choose 5}x^5+{7\choose 7}x^7\right|\\
&=2|x|\cdot \left|7+35x^2+21x^4+x^6\right|\\&\le128|x|. \end{align}$$
On the other hand, at least one of $1+x$, $1-x$ is $\ge 1$, hence $\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\ge1$ and $\sqrt{(1+x)^7}+\sqrt{(1-x)^7}\ge1$ so that
$$ \left|2013 x\left(\sqrt[4]{(1+x)^7}+\sqrt[4]{(1-x)^7}\right)\left(\sqrt{(1+x)^7}+\sqrt{(1-x)^7}\right)\right|\ge 2013|x|.$$
In combniation, this gives us $2013|x|\le 128|x|$, i.e. $x=0$.
| {
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How to show that $f(x,y)$ is continuous. How to show that $f(x,y)$ is continuous.
$$f(x,y)=\frac{4y^3(x^2+y^2)-(x^4+y^4)2x\alpha}{(x^2+y^2)^{\alpha +1}}$$ for $\alpha <3/2$.
Please show me Thanks :)
| Notice that $|x|,|y|\leq (x^2+y^2)^{1/2}=||(x,y)||$ so we have
\begin{align}|f(x,y)|&=\frac{|4y^3(x^2+y^2)-(x^4+y^4)2x\alpha|}{(x^2+y^2)^{\alpha +1}}\leq \frac{4|y|^3(x^2+y^2)+(x^4+y^4)2|x||\alpha|}{(x^2+y^2)^{\alpha +1}}\\
&\leq \frac{4||(x,y)||^3||(x,y)||^2+(||(x,y)||^4+||(x,y)||^4)2||(x,y)|||\alpha|}{||(x,y)||^{2\alpha +2}}\\
&= \frac{4(|\alpha|+1)||(x,y)||^5}{||(x,y)||^{2\alpha +2}}=4(|\alpha|+1)||(x,y)||^{3-2\alpha}\rightarrow 0\end{align}
So if $f(0,0)=0$ then $f$ is continuous at $(0,0)$ and hence everywhere.
| {
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Prove limit of function I need to prove this limit:
Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)>0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.
| Write $c(x) = f(x) + \frac{1}{f(x)}$. Solve a quadratic equation to see that
$f(x)$ is either $(c(x)+\sqrt{c(x)^2-4}\;)/2$ or $(c(x)-\sqrt{c(x)^2-4}\;)/2$ . So, for all $x$, we have
$$
\frac{c(x)-\sqrt{c(x)^2-4}}{2} \le f(x) \le \frac{c(x)+\sqrt{c(x)^2-4}}{2}
$$
But we are told that $\lim c(x) = 2$, so that
$$
\lim \frac{c(x)-\sqrt{c(x)^2-4}}{2} = 1\quad\text{and}\quad \frac{c(x)+\sqrt{c(x)^2-4}}{2} = 1.
$$
Our function $f(x)$ is between these, so $\lim f(x) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{3}{2}\le\frac{1}{a+ab}+\frac{a}{1+ab}+\frac{ab}{1+a}\le\frac{19}{10}$ for $a, b \in [1/2, 2]$
Let
$a,b\in [\frac{1}{2},2]$. Prove that
$$\dfrac{3}{2}\le\dfrac{1}{a+ab}+\dfrac{a}{1+ab}+\dfrac{ab}{1+a}\le\dfrac{19}{10}.$$
my idea:
$$\dfrac{1}{a+ab}+\dfrac{a}{1+ab}+\dfrac{ab}{1+a}-\dfrac{3}{2}\ge 0?$$
and this problem is from《Mathematics Studying》(2012.7).
See: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487281&p=2730329#p2730329
Thank you everyone.
I have see this same problem
| Remarks: Here is a proof without using calculus.
@Calvin Lin gave a nice proof of the left inequality.
Here we only prove the right inequality.
Let $c = 1/a \in [1/2, 2]$. We have
$$
\frac{1}{a + ab} + \frac{a}{1 + ab} + \frac{ab}{1 + a} =
\frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c}.
$$
It suffices to prove that, for all $b, c \in [1/2, 2]$,
$$\frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c} \le \frac{19}{10}.$$
(1) If $b + c \le 5/2$, letting $x = b + c \in [1, 5/2]$, we have
\begin{align*}
\frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c}
&= \frac{(b + c)^2 + b + c - 2bc}{1 + b + c + bc} + \frac{1}{b+c}\\[5pt]
&\le \frac{(b + c)^2 + b + c - 2\cdot \frac12(b + c - \frac12)}{1 + b + c + \frac12(b + c - \frac12)} + \frac{1}{b+c} \tag{1}\\[5pt]
&= \frac{4x^2 + 2}{3 + 6x} + \frac{1}{x}\\[5pt]
&\le \frac{19}{10}
\end{align*}
where in (1) we have used $bc - \frac12(b + c - \frac12) = \frac14(2b - 1)(2c - 1) \ge 0$, and the last inequality follows from
$$\frac{19}{10} - \frac{4x^2 + 2}{3 + 6x} - \frac{1}{x}
= \frac{(5x+2)(5-2x)(4x-3)}{30x(1+2x)} \ge 0.$$
(2) If $b + c > 5/2$, letting $x = b + c \in (5/2, 4]$, we have
\begin{align*}
\frac{c}{1+b} + \frac{1}{b+c} + \frac{b}{1+c}
&= \frac{(b + c)^2 + b + c - 2bc}{1 + b + c + bc} + \frac{1}{b+c}\\[5pt]
&\le \frac{(b + c)^2 + b + c - 2\cdot 2(b + c - 2)}{1 + b + c + 2(b + c - 2)} + \frac{1}{b+c} \tag{2}\\[5pt]
&= \frac{x^2 - 3x + 8}{3x - 3} + \frac{1}{x}\\[5pt]
&\le \frac{19}{10}
\end{align*}
where in (2) we have used $bc - 2(b + c - 2) = (2 - b)(2 - c) \ge 0$, and the last inequality follows from
$$\frac{19}{10} - \frac{x^2 - 3x + 8}{3x - 3} - \frac{1}{x} = \frac{(6-x)(2x-5)(5x-1)}{30x(x-1)} \ge 0.$$
We are done.
Also, the original inequality can be proved using some substitutions.
Note that
$$a \in [1/2, 2)
\iff a = \frac12 \cdot \frac{1}{1+s} + 2\cdot \frac{s}{1 + s}, \quad s \ge 0$$
and
$$b \in [1/2, 2)
\iff b = \frac12 \cdot \frac{1}{1+t} + 2\cdot \frac{t}{1 + t}, \quad t \ge 0.$$
If $a \in [1/2, 2)$ and $b \in [1/2,2)$,
we have
$$\frac{1}{a + ab} + \frac{a}{1 + ab} + \frac{ab}{1 + a}
= \frac{f(s, t)}{g(s, t)}$$
where $f(s, t)$ and $g(s, t)$
are both polynomials with non-negative coefficients.
The desired result follows.
It remains to prove the case
$a = 2$ or $b = 2$. It is easy.
We are done.
| {
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How to show the monotonicity of this function? Thanks for your help. I want to show $f(x)=\displaystyle \frac{1-x^{n+1}(n+2)+(n+1)x^{n+2}}{(1-x^{n+1})(1-x)}$ (strictly) increases w.r.t. $x$ for $x>0$. Here, $n=1,2,3,\dots.$. So far, I can only show $f(x)$ is increasing when $x>1$. Can anyone give me a hint to complete the proof? Thanks a lot.
| Here's a solution that doesn't use calculus. As user7530 points out, we may simplify the expression to $(n+1) + \frac{1}{1-x} - \frac{n+1}{1-x^{n+1} } $.
Now, for $ a> b$, we want to show that (the constant term $(n+1)$ cancels out)
$$ \frac{1}{1-a} - \frac{n+1}{1-a^{n+1} }> \frac{1}{1-b} - \frac{n+1}{1-b^{n+1} }$$
This is equivalent to
$$\frac{ (a-b) } { (1-a)(1-b) } > \frac{ (n+1) (a^{n+1}-b^{n+1} )} { (1-a^{n+1})(1-b^{n+1})} $$
which is equivalent to
$$ \frac{ (1-a^{n+1}) } {(1-a)} \times \frac{ (1-b^{n+1}) } { (1-b) } > (n+1) \frac{ a^{n+1} - b^{n+1} } { a-b}$$
which is equivalent to
$$( 1 + a + a^2 + \ldots + a^n) ( b^n + b^{n-1} + \ldots + 1) > (n+1) (a^n + a^{n-1}b + \ldots + ab^{n-1} + b^n) $$
which is true by the rearrangement inequality.
In hindsight, the last statement is similar to 7530's AM-GM.
| {
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Why can't I solve this homogenous second order differential equation? I've been banging my head on the wall for quite some time trying to come up with a solution to the following:
$$\frac {\partial^2 y(x)} {\partial x^2} + (A-B*V(x)) y(x) = 0 $$
$$V(x) = (36 + (2 - x)^2)^{-1/2}$$
With A and B constants, and $y$ solely a function of $x$.
If it helps, in my area of concern $0 \leq x \leq 4$, you can treat V as:
$$V(x) = (-1/432)*(x - 2)^2 + 1/6$$
With no real loss of accuracy (that I care about). Generally, I know this is equvalent to the form:
$$y''(x) + p(x)y' + q(x)y = 0$$
With $p(x) = 0$. I can find plenty of examples of constant coefficients and solutions for the form of $q(x)=0$.
Can anybody recommend an anzatz/method/approximation solution that might help me solve this?
| Suppose $B\neq0$ :
$\dfrac{d^2y}{dx^2}+\left(A-B(36+(2-x)^2)^{-\frac{1}{2}}\right)y=0$
$\dfrac{d^2y}{dx^2}+\biggl(A-\dfrac{B}{\sqrt{(x-2)^2+36}}\biggr)y=0$
$\sqrt{(x-2)^2+36}\dfrac{d^2y}{dx^2}+\left(A\sqrt{(x-2)^2+36}-B\right)y=0$
Let $r=x-2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=\dfrac{dy}{dr}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}=\dfrac{d^2y}{dr^2}$
$\therefore\sqrt{r^2+36}\dfrac{d^2y}{dr^2}+\left(A\sqrt{r^2+36}-B\right)y=0$
Let $s=\sqrt{r^2+36}$ ,
Then $\dfrac{dy}{dr}=\dfrac{dy}{ds}\dfrac{ds}{dr}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{dy}{ds}$
$\dfrac{d^2y}{dr^2}=\dfrac{d}{dr}\biggl(\dfrac{r}{\sqrt{r^2+36}}\dfrac{dy}{ds}\biggr)=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d}{dr}\biggl(\dfrac{dy}{ds}\biggr)+\dfrac{36}{(r^2+36)^\frac{3}{2}}\dfrac{dy}{ds}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d}{ds}\biggl(\dfrac{dy}{ds}\biggr)\dfrac{ds}{dr}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{r}{\sqrt{r^2+36}}\dfrac{d^2y}{ds^2}\dfrac{r}{\sqrt{r^2+36}}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{r^2}{r^2+36}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}=\dfrac{s^2-36}{s^2}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}$
$\therefore s\biggl(\dfrac{s^2-36}{s^2}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^3}\dfrac{dy}{ds}\biggr)+(As-B)y=0$
$\dfrac{(s+6)(s-6)}{s}\dfrac{d^2y}{ds^2}+\dfrac{36}{s^2}\dfrac{dy}{ds}+(As-B)y=0$
$\dfrac{d^2y}{ds^2}+\dfrac{36}{s(s+6)(s-6)}\dfrac{dy}{ds}+\dfrac{s(As-B)}{(s+6)(s-6)}y=0$
$\dfrac{d^2y}{ds^2}+\left(\dfrac{1}{2(s-6)}+\dfrac{1}{2(s+6)}-\dfrac{1}{s}\right)\dfrac{dy}{ds}+\left(\dfrac{6A-B}{2(s-6)}-\dfrac{6A+B}{2(s+6)}+A\right)y=0$
| {
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Suppose $xyz=8$, try to prove that $\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2$ Who can help with the following inequality? I can prove it but using some rather ugly approach (e.g. by leveraging the derivative of $\frac{1}{\sqrt{t+1}}+\frac{1}{2}\sqrt{1-\frac{8}{t^2+8}}$ to show this is always less than 1 for $t>0$.
I'm just wondering if we can have some elegant simple prove. I guess we should use Jensen's inequality. Thanks.
Suppose $x,y,z\in R^+$ and $xyz=8$, try to prove that $$\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2.$$
Please note that the usual AM-GM inequality may not do its trick here as the equality is rather hold on the boundary.
Thanks.
| Let $f(x,y,z)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}$ and WOLOG assume that $x\geq y\geq z$ so we have $x\geq 2$. Consider the following two cases:
*
*$yz\geq 1$ : In this case we have $z\geq1/y$ and thus
$$f(x,y,z)\leq f(x,y,1/y)\\ \leq\sqrt{3\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{y}{1+y}\right)}\\
= \sqrt{3\left(\frac{1}{1+x}+1\right)}\\
\leq \sqrt{3\left(\frac{1}{3}+1\right)}=2.$$ Equation in this case cannot be attained as we cannot have $z=1/y$ and $x=2$ simultaneously.
*$yz\leq 1$: Fix the product $yz=p^2$. Since we assumed $y\geq z$ we must have $y\geq p$. Furthermore, we have $p\leq 1$ for which the function $\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}$ would be decreasing with respect to $y$. Therefore, the mentioned function is minimized at $y=z=p$, i.e., $$\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}\leq\frac{2}{\sqrt{1+\sqrt{yz}}}\\=\frac{2}{\sqrt{1+\sqrt{8/x}}}.$$ Thus $f(x,y,z)\leq \frac{1}{\sqrt{1+x}}+\frac{2}{\sqrt{1+\sqrt{8/x}}}.$ The RHS is increasing in $x$ so the upper bound as $x\to \infty$ would be $f(x,y,z)<2$.
There might be neater way to show the second case.
| {
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"url": "https://math.stackexchange.com/questions/446863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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How to find the number of intersection for $ \rho =\frac{\theta} {2\pi+1} $ and $\rho =\frac {1} {2-\cos\theta} $ How to Find the number of intersection for curve $ \rho =\frac{\theta} {2\pi+1} $ and curve $\rho =\frac {1} {2-\cos\theta} $ .
| $\cos{\theta}=2-\dfrac{2\pi+1}{\theta}, 1 \ge \cos{\theta} \ge -1 \to 1 \ge 2-\dfrac{2\pi+1}{\theta} \ge -1 \iff 3 \ge \dfrac{2\pi+1}{\theta} \ge 1 \iff 2\pi+1 \ge \theta \ge \dfrac{2\pi+1}{3}$
$\dfrac{\pi}{4} < 1 < \dfrac{\pi}{3} \to 2\pi + \dfrac{\pi}{3} > \theta > \dfrac{2\pi}{3}+\dfrac{\pi}{12}=\dfrac{3\pi}{4}$, now divide the domain into 4 segments:
$[\dfrac{3\pi}{4},\pi] ;[\pi,\dfrac{3\pi}{2}],[\dfrac{3\pi}{2},2\pi],[2\pi,2\pi + \dfrac{\pi}{3}]$
let $g(\theta)=2-\dfrac{2\pi+1}{\theta}$
in $[\dfrac{3\pi}{4},\pi], \cos{\theta} $ from $-\dfrac{\sqrt{2}}{2}$ to $-1,g(\theta)$ from $-1$ to $-\dfrac{1}{\pi}$ , they will corss each.(explain yourself)
in $[\pi,\dfrac{3\pi}{2}], \cos{\theta} $ from $-1$ to $0,g(\theta) $ from $-\dfrac{1}{\pi}$ to $\dfrac{2(\pi-1)}{3\pi}$,what happen to $ \cos{\theta},g(\theta)$.
I think you can do rest now.
Edit: I add two graphics to show the result:
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ Please help to prove this inequality.
Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ and each of $x,y,z$ is greater than 1.
Thanks.
| Let $x^2-1=\frac{8bc}{a^2}$, $y^2-1=\frac{8ac}{b^2}$ and $z^2-1=\frac{8ab}{c^2}$, where $a$, $b$ and $c$ are positives.
Hence, by Holder we obtain:
$$\sum_{cyc}\frac{1}{x}=\sum_{cyc}\frac{1}{\sqrt{1+\frac{8bc}{a^2}}}=\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}=$$
$$=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\sum\limits_{cyc}a(a^2+8bc)}{\sum\limits_{cyc}a(a^2+8bc)}}\geq\sqrt{\frac{(a+b+c)^3}{\sum\limits_{cyc}a(a^2+8bc)}}=$$
$$=\sqrt{\frac{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}{\sum\limits_{cyc}(a^3+8abc)}}=\sqrt{\frac{\sum\limits_{cyc}(a^3+8abc)+3\sum\limits_{cyc}c(a-b)^2}{\sum\limits_{cyc}(a^3+8abc)}}\geq1.$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/447495",
"timestamp": "2023-03-29T00:00:00",
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show that $\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}(1-\frac{e^{-ab}(ab+2)}{2})$ show that
$$\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}\left(1-\frac{e^{-ab}(ab+2)}{2}\right)$$
for $a,b> 0$
I would like someone solve it using contour but also I would to see different solution using different way to solve it
is there any help
thanks for all
| $$\int_0^{\infty} \frac{\sin ax\,dx}{x(x^2+b^2)^2}=\frac{1}{b^4}\left(\int_0^{\infty}\frac{\sin ax}{x}\,dx-\int_0^{\infty}\frac{x\sin ax\,dx}{x^2+b^2}\right)-\frac{1}{b^2}\int_0^{\infty}\frac{x\sin ax\,dx}{(x^2+b^2)^2}$$
The first integral is well known. For any $a>0:$
$$\int_0^{\infty} \frac{\sin ax}{x}\,dx=\frac{\pi}{2}$$
The second, consider:
$$\begin{aligned}f(t)=\int_0^{ \infty} \frac{x\sin axt\,dx}{x^2+b^2}\,dx \Rightarrow \mathcal{L} \{ f(t)\} &=\int_0^{ \infty}e^{-st}\int_0^{ \infty}\frac{x\sin axt\,dx}{x^2+ b^2}\,dx\,dt\\&=\int_0^{ \infty}\frac{x}{x^2+ b^2}\int_0^{\infty}e^{-st}\sin axt\,dt\,dx\\&=\int_0^{\infty} \frac{ax^2}{(x^2+ b^2)(a^2x^2+s^2)} \,dx\\&= \frac{\pi}{2(s+ab)}\end{aligned}$$
$$\frac{\pi}{2}\cdot\mathcal{L}^{-1}\left\{ \frac{1}{s+ab}\right\}\Bigg|_{t=1}= \frac{\pi}{2e^{ab}}$$
The third, using the same parameter (call the function $g(t)$ now) one obtains:
$$\begin{aligned}\mathcal{L} \{ g(t)\} &=\int_0^{ \infty}\frac{x}{(x^2+ b^2)^2}\int_0^{\infty}e^{-st}\sin axt\,dt\,dx\\&=\int_0^{\infty} \frac{ax^2}{(x^2+ b^2)^2(a^2x^2+s^2)} \,dx\\&= \frac{a\pi}{4b(s+ab)^2}\end{aligned}$$
$$\frac{\pi a}{4b}\cdot\mathcal{L}^{-1}\left\{ \frac{1}{(s+ab)^2}\right\}\Bigg|_{t=1}= \frac{a\pi}{4be^{ab}}$$
Therefore:
$$\int_0^{\infty} \frac{\sin ax\,dx}{x(x^2+b^2)^2}=\frac{\pi}{2b^4}\left(1-\frac{2+ab}{2e^{ab}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\sqrt 5$ is irrational I have to prove that $\sqrt 5$ is irrational.
Proceeding as in the proof of $\sqrt 2$, let us assume that $\sqrt 5$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1,
$$\frac{p}{q} = \sqrt5$$
$$\Rightarrow \frac{p^2}{q^2} = 5$$
$$\Rightarrow p^2 = 5 q^2$$
This means that 5 divides $p^2$. This means that 5 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 5 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 5, which is a contradiction.
Is this proof correct?
| Let us assume $√5$ is rational.
$$√5=\frac{x}{y}$$
Square both sides of the equation above
$$5 =\frac{x^2}{y^2}$$
Multiply both sides by $y^2$
$$5 y^2 =\frac{x^2 }{y^2}$$
We get $5 y^2 = x^2$
Another important concept before we finish our proof: Prime factorization.
Key question: is the number of prime factors for a number raised to the second power an even or odd number?
For example, $6^2$, $12^2$, and $15^2$
$6^2 = 6 × 6 = 2 × 3 × 2 × 3$ ($4$ prime factors, so even number)
$12^2 = 12 × 12 = 4 × 3 × 4 × 3 = 2 × 2 × 3 × 2 × 2 × 3$ ($6$ prime factors, so even number)
$15^2 = 15 × 15 = 3 × 5 × 3 × 5$ ($4$ prime factors, so even number)
There is a solid pattern here to conclude that any number squared will have an even number of prime factors
In order words, $x^2$ has an even number of prime factors.
Let's finish the proof then!
$5 y^2 = x^2$
Since $5 y^2$ is equal to $x^2$, $5 y^2$ and $x^2$ must have the same number of prime factors.
We just showed that
$x^2$ has an even number of prime factors, $y^2$ has also an even number of prime factors.
$5 y^2$ will then have an odd number of prime factors.
The number $5$ counts as $1$ prime factor, so $1$ + an even number of prime factors is an odd number of prime factors.
$5 y^2$ is the same number as $x^2$. However, $5 y^2$ gives an odd number of prime factor while $x^2$ gives an even number of prime factors.
This is a contradiction since a number cannot have an odd number of prime factors and an even number of prime factors at the same time.
The assumption that square root of $5$ is rational is wrong. Therefore, square of $5$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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