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Proving that $a_n=\sqrt{n^2+n}-n$ converges, finding its limit and showing that its sequence ${(a_n)^{\infty}_{n=1}}$ is monotinic. As the title says, below I've proved the statement. Just posted here for verification and correction! And of course for other people to be inspired.
Let $a_n=\sqrt{n^2+n}-n$ with $n\in\mathbb{N}.$ Show that $a_n$ converges as $n\to\infty$ find its limit, and show that the sequence ${(a_n)^{\infty}_{n=1}}$ is monotinic.
Proof:
For convergence, we look for:
$\forall_{\epsilon\gt0}\exists_{n\in\mathbb{N}}\forall_{n\gt N}(|a_n-l|<\epsilon) \iff l-\epsilon<a_n<l+\epsilon.$ If this property holds for one and only one $l$, then $a_n$ convergence with its limit being unique.
$l-\epsilon<\sqrt{n^2+n}-n<l+\epsilon \iff (l+n-\epsilon)^2<n^2+n<(l+n+\epsilon)^2 \iff $
$n(n+l-\epsilon)+l(n+l-\epsilon)-\epsilon(n+l-\epsilon) <n^2+n< $
$n(n+l+\epsilon)+l(n+l+\epsilon)+\epsilon(n+l+\epsilon)$
$\epsilon$ can get randomly small, so it can be ommited for large $n$. Then;
$n(n+l)+l(n+l)<n^2+n< n(n+l)+l(n+l) \iff$
$n^2+2ln+l^2<n^2+n<n^2+2ln+l^2.$
As $n\to\infty$, all terms not containing $n$ can be omitted as they are unimpactfully small. Then;
$n^2+2ln<n^2+n<n^2+2ln \iff 2ln<n<2ln \iff n=2ln \iff l=\frac{1}{2}.$
This property only holds for one $l$, so $a_n$ is convergent, and with this, its limit $l=\frac{1}{2}$ has also been found.
For monotonity we look for:
$\forall_n(\sqrt{(n+1)^2+(n+1)}-(n+1)>\sqrt{n^2+n}-n)$.
For all $n$:
$1>0 \iff 4n^2+4n+1>4n^2+4n \iff (2n+1)^2>4(n^2+n) \iff$
$2n+1>2\sqrt{n^2+n} \iff n^2+2n+1>n^2+2\sqrt{n^2+n} \iff$
$(n+1)^2>n^2+2\sqrt{n^2+n} \iff (n+1)^2+n+1>n^2+n+2\sqrt{n^2+n}+1 \iff$
$\sqrt{(n+1)^2+n+1}>\sqrt{n^2+n}+1 \iff\sqrt{(n+1)^2+(n+1)}-(n+1)>\sqrt{n^2+n}-n$
So the series ${(a_n)^{\infty}_{n=1}}$ is monotonic.
Conclusion:
$a_n$ converges with its unique limit being $l=\frac{1}{2}$, and the series ${(a_n)^{\infty}_{n=1}}$ is monotinic. $\tag*{$\Box$}$
|
A simpler approach: $a_n>0$,
$$a_n^{-1}=\frac{1}{\sqrt{n^2+n}-n}=\frac{\sqrt{n^2+n}+n}{n}=\sqrt{1+n^{-1}}+1$$
is decreasing and $a_n^{-1}\to 2$. So $(a_n)$ is increasing and $a_n\to 1/2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I find the primitive of $(x^3+x^2)/(4-x^2)$? I’m trying to find the primitive of the below rational function. I’m stuck trying to separate the undetermined coefficients like:
$$\frac{x^3+x^2}{(2-x)(2+x)}=\frac{A}{2-x}+\frac{B}{2+x}\iff x^3+x^2=A(2+x)+B(2-x)$$
What am I doing wrong?
|
Hint :
$\dfrac{x^3+x^2}{(2-x)(2+x)}=-x-1+\dfrac{A}{2-x}+\dfrac{B}{2+x}$
|
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|
Finding the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?
Q. What is the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?
My approach : Let the matrix be denoted by $A$. Then $$A=
\left(\begin{matrix}
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
\end{matrix}\right)
$$
Some observations follow : $\text {Rank} A=2$, $\det A = 0$ and $\text {tr} A=0$.
Since $\det A=0$, there exists an eigenvalue which is $0$.
Since $\text {Rank} A=2$, number of non-zero eigenvalues can not exceed $2$.
Since $\text {tr} A=0$, number of non-zero eigenvalues are either $2$ or $0$.
Case 1. Number of non-zero eigenvalues is $2$: In this case they must be $\lambda$ and $-\lambda$ for trace should be zero. Thus the matrix is forced to be diagonalizable. This implies that the minimal polynomial is $P(x)=x(x-\lambda)(x+\lambda).$
Case 2. Number of non-zero eigenvalues is $0$: Here all diagonal entries of the Jordan form are $0$. We have either one $3\times 3$ jordan block or two $2\times 2$ Jordan blocks in order to preserve rank of $A$. Thus the minimal polynomial is $P(x)=x^3$ or $P(x)=x^2$ respectively.
However I checked that for even order matrices with order $\gt 2$ we have eigenvalues of the type we found in Case 1 above. That means for $\text {even $\times$ even}$ matrices, eigenvalues are always $0,\lambda,-\lambda$ for some $\lambda \neq 0$. How do I show this in order to discard the case 2 entirely?
|
Check that $(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)$ is an eigenvector with eigenvalue $10$.
|
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|
solution using synthetic geometry I managed to solve this problem only using complex numbers but I'd like to solve it using synthetic geometry and I can't.
Can someone help me to solve this problem using synthetic geometry?
Let $ABC$ an acute triangle with $AB > AC$ . Let $O$ its circumcenter and let $D$ the midpoint of $BC$. The circle of diameter $AD$ intersects again $AB$ and $AC$ in $E$ and in $F$, respectively. Let $M$ the midpoint of $EF$.
Prove that $MD$ is parallel to $AO$.
This is my solution. But, as I wrote above, I'd like to solve it using synthetic geometry and I can't.
Setting the origin of the plane in O and the points A, B and C on the circumference of unit radius, we have that
$ a \bar{a} = 1 \text{;} \ b \bar{b} = 1 \text{;} \ c \bar{c} = 1 $.
Because $ D $ is the midpoint of $ BC $, we can write
1) $ d = \dfrac{b + c}{2} $
and because $ AD $ is a diamtere of the new circle then said $ Q $ his midpoint we have
2) $ q = \dfrac{a + d}{2} = \dfrac{2a + b + c}{4}$
Said $ M_{1} $ the projection of $ Q $ on $ AB $, then
$ m_{1} = \frac{1}{2} \left[ \left( \dfrac{\bar{q} - \bar{a}}{\bar{b} - \bar{a}}\right) (b - a) + a + q \right] $
but
$\dfrac{1}{\bar{b} - \bar{a}} = \dfrac{1}{\dfrac{1}{b} - \dfrac{1}{a}} = \dfrac{ab}{a – b} $
and so
$ m_{1} = \frac{1}{2} \left[ \bar{q} ab (-1) - \dfrac{ab}{a} (- 1) + a + q \right] \Rightarrow $
$m_{1} = \frac{1}{2} \left( a + b + q - ab\bar{q} \right) $
In the same way, said $ M_{2} $ the projection of $ Q $ on $ AC $ we have
$ m_{2} = \frac{1}{2} \left[ \dfrac{\bar{q} - \bar{a}}{\bar{c} - \bar{a}} (c - a) + a + q \right] \Rightarrow $
$ m_{2} = \frac{1}{2} \left( a + c + q - ac\bar{q} \right) $
Because $ Q $ is the center of the new circle passing through $ A \text{,} D \text{,} E \text{,} F $ we have that $ M_{1}Q $ is axes of $ AE $ and that $ M_{2}Q $ in axes of $ AF $.
So we have that $ M_{1} $ is midpoint of $ AE $ and that $ M_{2} $ is midpoint of $ AF $.
So we can write
$ m_{1} = \dfrac{a + e}{2} \ \Rightarrow \ e = 2m_{1} – a $
and also
$ m_{2} = \dfrac{a + f}{2} \ \Rightarrow \ f = 2m_{2} - a$
The point $ M $ is defined as midpoint of $ EF $ so we have
$ m = \dfrac{e + f}{2} = \dfrac{2m_{1} + 2m_{2} - 2a}{2} = m_{1} + m_{2} - a$
therefore replacing $ m_{1} $ and $ m_{2} $ we have
$m = \frac{1}{2} a + \frac{1}{2} c + \frac{1}{2} q + \frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} q - \frac{ab\bar{q}}{2} - \frac{ac\bar{q}}{2} - a \ \Rightarrow $
$m = \dfrac{b + c}{2} + q - a\bar{q} \dfrac{b + c}{2}$
We know, furthemore, that $ \bar{q} $ is:
3) $\bar{q} = \dfrac{2\bar{a} + \bar{b} + \bar{c}}{4} = \frac{1}{4} \left( \frac{2}{a} + \frac{1}{b} + \frac{1}{c} \right) = \dfrac{2bc + ab + ac}{4abc}$
Now we write the equation of the parallel line through $D$ parallel to $AO$:
let $z$ be a generic point of this line it is possible to write
4) $ \dfrac{z - d}{a - 0} = \dfrac{\bar{z} - \bar{d}}{\bar{a} – 0} $
Replacing $d$ with the expression of 1) and taking into account that $ \bar {a} = \frac{1}{a} $, we get
5) $z - \dfrac{b + c}{2} = a^{2} \left( \bar{z} - \dfrac{\bar{b} + \bar{c}}{2} \right) = a^{2}\bar{z} - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}$
If the $ M $ point belongs to this line, $ m $ must satisfy equation 5). Substituting the value of $ m $ we have
$\dfrac{b + c}{2} + q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) - \dfrac{b + c}{2} = a^{2}\left( \dfrac{\bar{b} + \bar{c}}{2} + \bar{q} - \bar{a}q \cdot \dfrac{\bar{b} + \bar{c}}{2} \right) - \dfrac{a^{2} (\bar{b} + \bar{c})}{2} \Rightarrow $
$ q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) = a^{2}\bar{q} - \dfrac{aq( b + c)}{2bc} \ \Rightarrow $
$ q \left( 1 + \dfrac{ab + ac}{2bc} \right) = \bar{q}a \left( a + \dfrac{b + c}{2} \right)$
Substituting the values of $ q $ in 2) and the value of $ \bar{q} $ in 3) we have
$ \dfrac{(2a + b + c)(2bc + ab + ac)}{8bc} = \dfrac{a (2bc + ab + ac) (2a + b + c)}{a \cdot 8bc} $
which is, obviously, an identity, so $ MD $ is parallel to $ AO $, as we wanted to prove.
|
I was not able to solve it using synthetic geometry either. Is there anyone who can solve it and post a solution using synthetic geometry methods?
|
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|
Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
|
\begin{align}
x=|OA|&=|CO_1|=|A_1O_2|=|BO_3|
.
\end{align}
Let $|AD|=a$, $|DC|=2a$.
\begin{align}
|CO|&=2a\cos\theta
,\\
|AO|=|CO_1|&=
\sqrt{(3\,a)^2+(2a\,\cos\theta)^2-2\cdot3\,a\cdot(2\,a\,\cos\theta)
\cdot\cos\theta}
\\
&=a\,\sqrt{9-8\,\cos^2\theta}
,\\
|AO_1|&=\sqrt{|AC|^2-|CO_1|^2}
\\
&=\sqrt{(3a)^2-a^2\,(9-8\cos^2\theta)}
\\
&=2\sqrt2\,a\,\cos\theta
,\\
|OO_1|&=
\sqrt{|CO|^2-|CO_1|^2}
\\
&=\sqrt{12\,a^2\cos^2\theta-a^2\,(9-8\,\cos^2\theta)}
\\
&=a\,\sqrt{4\cos^2\theta-9}
,\\
[ACA_1B]=(3a)^2&=
2\,|CO_1|\,|AO_1|+|OO_1|^2
\\
&=
2\,a\,\sqrt{9-8\,\cos^2\theta}
\cdot 2\sqrt2\,a\,\cos\theta
+\left(a\,\sqrt{12\cos^2\theta-9}\right)^2
.
\end{align}
The last equation simplifies to
\begin{align}
-18&+4\sqrt2\sqrt{9-8\cos^2\theta}\,\cos\theta
+12\,\cos^2\theta=0
,\\
(18-12\,\cos^2\theta)^2
&=
32\,(9-8\cos^2\theta)\,\cos^2\theta
,\\
4\,(10\cos^2\theta-9)^2&=0
,\\
\cos^2\theta&=\tfrac9{10},
\\
\text{and }\quad |AO|&=
a\,\sqrt{9-8\,\cos^2\theta}
=
5\,\sqrt{9-8\cdot\tfrac9{10}}
=3\,\sqrt5
.
\end{align}
|
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|
Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$
My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+ab \leq210$$
Now what ?
|
$$a^4+b^4+a^2b^2=(a^2+b^2)^2-a^2b^2=60$$
$$(a^2+b^2-ab)(a^2+b^2+ab)=60$$
Solving equations $$a^2+b^2-ab=6$$$$a^2+b^2+ab=10$$ will give you the result.
|
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|
Finding: $\lim\limits_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$
I'm running into problems with this limit:
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$$
I've tried using l'Hospitals rule, however we will alway keep the $\cos(\pi x)$ expression, as well for $\sin(5\pi x)$. Also, terms do not cancel out with $\sin$ and $\pi$ since the product with $5$.
Can anyone give me a hint solving this?
Thanks in advance
Kind regards,
|
Easy using Squeeze theorem
$$ 6x^2-5\le6x^2+5\cos{\pi x}\le 6x^2+5$$
and
$$ 6x^4-1\le6x^2+\sin{5\pi x}\le 6x^4+1$$
then for $x<-1$ we have
$$\frac{6x^2-5}{\sqrt{x^4+1}}\le \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} \le \frac{6x^2+5}{\sqrt{x^4-1}}$$
By squeeze theorem we get
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} =6$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate the derivative using limit definition. This is the function $f(x)$$=\frac{1}{\sqrt{3x-2}}$ .
I wrote that $$\lim_{h\to 0}\frac{\frac{\sqrt{3x+3h-2}}{3x+3h-2}-\frac{\sqrt{3x-2}}{3x-2}}{h}.$$
I am not able to continue further.
|
HINT:
Note that
$$\begin{align}
\frac{1}{\sqrt{3(x+h)-2}}-\frac{1}{\sqrt{3x-2}}&=\frac{\frac{1}{3(x+h)-2}-\frac{1}{3x-2}}{\frac{1}{\sqrt{3(x+h)-2}}+\frac{1}{\sqrt{3x-2}}}\\\\
&=\frac{\frac{-3h}{(3(x+h)-2)(3x-2)}}{\frac{1}{\sqrt{3(x+h)-2}}+\frac{1}{\sqrt{3x-2}}}
\end{align}$$
Divide by $h$ and let $h\to 0$.
|
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|
If $x,y,z\in {\mathbb R}$, Solve this system equation: If $x,y,z\in {\mathbb R}$, Solve this system equation:
$$
\left\lbrace\begin{array}{ccccccl}
x^4 & + & y^2 & + & 4 & = & 5yz
\\[1mm]
y^{4} & + & z^{2} & + & 4 & = &5zx
\\[1mm]
z^{4} & + & x^{2} & + & 4 & = & 5xy
\end{array}\right.
$$
This is an olympiad question in Turkey (not international), which that, I could not solve it.
My idea:
$xy=a \\ yz=b \\ xz=c$
$$
\left\lbrace\begin{array}{ccccccl}
a^2c^3 & + & b^3a & + & 4b^2c & = & 5b^3c
\\[1mm]
a^{3}b^2 & + & bc^3 & + & 4c^2a & = &5c^3a
\\[1mm]
b^{3}c^2 & + & a^3c & + & 4a^2b & = & 5a^3b
\end{array}\right.
$$
Yes, I know, this is a stupid idea, because it did not work at all ( last system equation is more difficult).
|
I checked the Possible Solutions thoroughly. The following solution is possible:
$$x^4+y^2+4+y^4+z^2+4+z^4+x^2+4-5yz-5xz-5xy=0 \Rightarrow (x^4-4x^2+4)+(y^4-4y^2+4)+(z^4-4z^2+4)+\left(\frac {5x^2}{2}-5xy+\frac {5y^2}{2} \right)+\left(\frac {5x^2}{2}-5xz+\frac {5z^2}{2} \right)+\left(\frac {5y^2}{2}-5yz+\frac {5z^2}{2} \right)=0\Rightarrow (x^2-2)^2+(y^2-2)^2+(z^2-2)^2+\left(x \sqrt{\frac 52}-y \sqrt{\frac 52}\right)^2+
\left(x \sqrt{\frac 52}-z \sqrt{\frac 52}\right)^2+
\left(y \sqrt{\frac 52}-z \sqrt{\frac 52}\right)^2=0$$
You can continue from here.
Only solutions are $x=y=z=±\sqrt2$
|
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|
Can we find the limit $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^nx^2}{n^2+x^2}$ without evaluating the sum? How to find the limit $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}$ if we don't evaluate the sum?
I know the sum is actually an elementary function which we can find it using Fourier series or other methods, but I'm just curious about if there exists some alternative ways to find this limit.
I tried to write it as this form:
$$\displaystyle\lim_{x\rightarrow\infty}x\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right).$$
As we know, $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x}{\left(2n\right)^{2}+x^{2}}$ and $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x}{\left(2n-1\right)^{2}+x^{2}}$ must get a same value (we don't need to care about what the exact value is) , so this is in the form $``0\cdot\infty"$, which cannot be evaluated directly. This is where I get stucked.
After days of thinking, I'm getting closer to the answer.
We can use easy algebra to get that $$\left |\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right |\leq\frac{1}{n^2}\quad\forall n\in\mathbb{Z^+},x\in\mathbb{R}$$
Hence the series below converges uniformly on $\mathbb{R}$:$$\displaystyle\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)$$
Changing the order of sum and limit, we can get:$$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)=0$$
which is$$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)=\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$$
and we also know $$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)=\lim_{x\rightarrow\infty}\left(-\frac{x^2}{1+x^2}-\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)\right)$$
If the limit exists, there must be an equation for the limit $L=-1-L$ which solves $L=-1/2$.
So everything needed is to prove that the limit exists. This would require a bit of analysis.
I’m going to prove it via Cauchy’s rule ($\displaystyle\lim_{x\rightarrow+\infty}f\left(x\right)\ exists\Leftrightarrow\forall\epsilon>0\exists X>0 \forall x_1,x_2>X, \left|f(x_1)-f(x_2)\right|<\epsilon$).
|
Thanks for @Fimpellizieri , but I think I have got the answer.
We know $$\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}=\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}$$
Consider
$$\left|\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right|\leq\frac{1}{n^2},\ \forall x\in\mathbb{R},n\in\mathbb{Z^+}$$
By Weierstrass M test the series below converges uniformly on $\mathbb{R}$. We have $$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left[\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right]\\=\sum_{n=1}^{\infty}\lim_{x\rightarrow\infty}\left[\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right]=0.$$
So we just need to consider $\displaystyle\sum_{n=1}^{\infty}\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}=\frac{1}{x}\sum_{n=1}^{\infty}\frac{\frac{-4n}{x}}{\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$.
Then consider the "Riemann sum" of function $f\left(u\right)=\frac{-4u}{\left(4u^2+1\right)^2}$ on $\left[0,\infty\right)$ , we know the limit of the sum as $x\rightarrow\infty$ is actually the integral $$\int_{0}^{\infty}f\left(u\right)du=\frac{1}{8u^2+2}|_{0}^{\infty}=-\frac{1}{2}$$
which is the final answer.
Now I explain why the sum tends to the improper integral.
Notice that $f\left(u\right)$ is decreasing on $\left[0,\frac{1}{2}\right]$ and increasing on $\left[\frac{1}{2},+\infty\right)$, we split the sum into two parts $\displaystyle\sum_{0<2n\leq x}\frac{\frac{-4n}{x}}{x\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$ and $\displaystyle\sum_{2n>x}\frac{\frac{-4n}{x}}{x\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$ , noted respectively as $S_{1}\left(x\right)$ and $S_{2}\left(x\right)$.
$S_{1}\left(x\right)$ is obviously tending to the integral of $f\left(u\right)$ on $\left[0,\frac{1}{2}\right]$ , while $S_{2}\left(x\right)$ is bounded between $\displaystyle\int_{\frac{1}{x}\left(\left[\frac{x}{2}\right]+1\right)}^{\infty}f\left(u\right)du$ and $\displaystyle\int_{\frac{1}{x}\left(\left[\frac{x}{2}\right]+2\right)}^{\infty}f\left(u\right)du$ (due to monotonicity) , which both tend to $\displaystyle\int_{\frac{1}{2}}^{\infty}f\left(u\right)du$ .
|
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|
Transform complex equation into Cartesian coordinate Here is the problem
$$$$
Show the following equation is an ellipse
$$|z-1|=3-|z-2|$$
and I tried to solve it...
Square both sides,
$$(|z-1|)^2=(3-|z-2|)^2 \\
x^2+y^2-2x+1=9-6|z-2|+x^2+y^2-4x+4$$
Rearrange them...
$$6|z-2|=12-2x \\
3|z-2|=6-x$$
Square them again, and here is the problem i met...
$$(3|z-2|)^2=(6-x)^2 \\
9(x^2+y^2-4x+4)=36-12x+x^2 \\
$$
36 in both sides is going to be cancelled out, it seems like I must did something wrong. I have spent over 2 hours on this but still get the same result..please help!
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So you get $8x^2+9y^2-24x=0$, which is $8\Big(x^2-3x+\frac{9}{4}\Big)-18+9y^2=0$, which is
$$8\Big(x-\frac{3}{2}\Big)^2+9y^2 = 18$$
which is an ellipse.
|
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|
Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$
Suppose that $a,b,c,x,y,z$ are all positive real numbers. Show that
$$\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$$
Below are what I've done, which may be misleading.
*
*I've tried to analyze when the equality holds：
1.1 Under the condition that $a=b=c$, it reduces to
$$\frac{2}{y+z}+\frac{2}{z+x}+\frac{2}{x+y}\geq \frac{9}{x+y+z}$$
1.2 Under the condition that $x=y=z$, it reduces to
$$\frac{b+c}{2a}+\frac{c+a}{2b}+\frac{a+b}{2c}\geq 3$$
Both are easy to verify. However, $ax+by+cz$ is not easy to deal with. Is there any famous inequality that I can use here?
1.3 Under the condition that $(x,y,z)$ and $(a,b,c)$ are in proportion, i.e. $x=at, y=bt, z=ct$ for some $t>0$, the inequality reduces to
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3(a+b+c)}{a^2+b^2+c^2}$$
which can be proved by using Newton's inequality:
$$(ab+bc+ca)^2 \geq 3abc(a+b+c)$$
*I’ve also tried to construct a function
$$f(x,y,z)=\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}- \frac{3(a+b+c)}{ax+by+cz}$$
and analyze its global minimum. But the first-order condition is complicated
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{c+a}{ab(z+x)^2}+\frac{a+b}{ca(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ab(y+z)^2}+\frac{a+b}{bc(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ca(y+z)^2}+\frac{c+a}{bc(z+x)^2}$$
Maybe some convexity can be used here?
*Substitution has been considered. Let
$$u=\frac{a}{a+b+c},v=\frac{b}{a+b+c}, w=\frac{c}{a+b+c}$$
Then $u,v,w>0$ and $u+v+w=1$, which are just weights we assign to $x,y,z$. The inequality becomes
$$\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}\geq \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}$$
Define another function in variables $u,v,w$
$$g(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
and consider the constrained optimization problem:
$$\min g(u,v,w)\\ \text{s.t. } u>0\\ v>0 \\w>0\\ u+v+w=1$$
The corresponding Lagrangian function can be
$$L_{\lambda}(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}+\lambda(u+v+w-1)$$
And the first-order conditions give
$$\frac{1}{u^2(y+z)}-\frac{3x}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{u(y+z)}-\frac{3ux}{(ux+vy+wz)^2}=\lambda u\\ \frac{1}{v^2(z+x)}-\frac{3y}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{v(z+x)}-\frac{3vy}{(ux+vy+wz)^2}=\lambda v\\ \frac{1}{w^2(x+y)}-\frac{3z}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{w(x+y)}-\frac{3wz}{(ux+vy+wz)^2}=\lambda w\\ u+v+w=1$$
Summing up yields
$$\lambda=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
None of the methods I tried seems to work, and now I even doubt the truth of the inequality.
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Too long for a comment.
Since the inequality is homogeneous, without loss of generality we may suppose $a+b+c=1$ and $x+y+z=1$. Then
$$(1-x)(1-y)(1-z)=1-x-y-z+xy+xz+yz-xyz=xy+xz+yz-xyz$$
Thus the left hand side of the inequality equals
$$\frac{1-a}{a(1-x)}+ \frac{1-b}{(1-y)}+ \frac{1-c}{c(1-z)}=$$
$$\frac{1}{a(1-x)}+ \frac{1}{b(1-y)}+ \frac{1}{c(1-z)}-\frac{1}{(1-x)}-\frac{1}{(1-y)}- \frac{1}{(1-z)}=$$
$$\frac{bc(1-y)(1-z)+ac(1-x)(1-z)+ab(1-x)(1-y)}{abc(1-x)(1-y)(1-z)}-
\frac{(1-y)(1-z)+(1-x)(1-z)+(1-x)(1-y)}{(1-x)(1-y)(1-z)}=$$
$$\frac{bc(1-y-z+yz)+ac(1-x-z+xz)+ab(1-x-y+xy)}{abc(xy+xz+yz-xyz)}-
\frac{(1-y-z+yz)+ (1-x-z+xz)+ (1-x-y+xy)}{xy+xz+yz-xyz }=$$
$$\frac{bc(x+yz)+ac(y+xz)+ab(z+xy)}{abc(xy+xz+yz-xyz)}-
\frac{(x+yz)+ (y+xz)+ (z+xy)}{xy+xz+yz-xyz}=$$
$$\frac{bc(x+yz)+ac(y+xz)+ab(z+xy)}{abc(xy+xz+yz-xyz)}-
\frac{1+ xy+yz+xz }{xy+xz+yz-xyz}=$$
$$\frac{\frac 1a(x+yz)+\frac 1b(y+xz)+\frac 1c(z+xy)-1-xy-yz-xz }{xy+xz+yz-xyz}.$$
So we have to show that
$$\left(\frac 1a(x+yz)+\frac 1b(y+xz)+\frac 1c(z+xy)-1-xy-yz-xz\right)(ax+by+cz)\ge 3(xy+xz+yz-xyz)$$
$$\left(\frac 1a(x+yz)+\frac 1b(y+xz)+\frac 1c(z+xy)\right)(ax+by+cz)\ge 3(xy+xz+yz-xyz)+(1+xy+yz+xz)(ax+by+cz)$$
$$x(x+yz)+ \frac bay(x+yz)+ \frac caz(x+yz)+
\frac abx(y+xz)+ y(y+xz)+ \frac cbz(y+xz)+
\frac acx(z+xy)+ \frac bcy(z+xy)+ z(z+xy)\ge $$
$$3(xy+xz+yz-xyz)+(ax+by+cz)+ax^2y+bxy^2+cxyz+axyz+by^2z+cyz^2+ax^2z+bxyz+cxz^2$$
$$x^2+\frac bay(x+yz)+ \frac caz(x+yz)+
\frac abx(y+xz)+ y^2+ \frac cbz(y+xz)+
\frac acx(z+xy)+ \frac bcy(z+xy)+ z^2+5xyz\ge $$
$$3(xy+xz+yz)+(ax+by+cz)+ax^2y+bxy^2+by^2z+cyz^2+ax^2z+cxz^2$$
or that (because $x^2+y^2+z^2+2xy+2xz+2yz=1$)
$$1+\frac bay(x+yz)+ \frac caz(x+yz)+
\frac abx(y+xz)+ \frac cbz(y+xz)+
\frac acx(z+xy)+ \frac bcy(z+xy)+ 5xyz\ge $$
$$5(xy+xz+yz)+(ax+by+cz)+ax^2y+bxy^2+by^2z+cyz^2+ax^2z+cxz^2$$
Finally, we have to show that
$$1+\left(\frac ba+\frac ab-5\right)xy+
\left(\frac ca+\frac ac-5\right)xz+
\left(\frac cb+\frac bc-5\right)yz+
\left(\frac ac-a\right)x^2y+\left (\frac bc-b\right)xy^2+
\left (\frac ab-a\right)x^2z+\left (\frac cb-c\right) xz^2+
\left (\frac ba-b\right)y^2z+\left(\frac ca-c\right)yz^2
- (ax+by+cz) + 5xyz \ge 0$$
|
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|
Solve the differential equation: $-x^2 y'' = \lambda y$ I need to solve the differential equation $-x^2 y'' = \lambda y$ by transforming the differential equation to a equation with constant coefficients. I need to do this by using $f(x) = y(e^x)$. If I do this, I become the equation:
$y''(e^x) + \frac{y'(e^x)}{e^x}+ \frac{\lambda y(e^x)}{(e^x)^2} = 0$
The I find a general solation of: $$y(e^x) = C \exp\left(\frac{-1+\sqrt{1- \frac{4\lambda}{x^2}}}{2(e^x)^2}\right) + C \exp\left(\frac{-1-\sqrt{1- \frac{4\lambda}{x^2}}}{2(e^x)^2}\right) $$
But Wolfram alpha gives the solution:
$$y(x) = c x^{\frac{1}{2} + \frac{1}{2}*\sqrt{1-4\lambda}} + c x^{\frac{1}{2} - \frac{1}{2}*\sqrt{1-4\lambda}} $$
Can someone help me figure this out?
|
It is true that using $f(X)=y(e^X)$ allows to transform the ODE into an ODE with constant coefficients, but with a little trick at first beginning :
$f(X)=y(e^X)$ is valid any symbol of variable, for example $f(t)=y(e^t)$
Let $e^t=x \quad\to\quad f(t)=y(x)$
$dx=e^tdt=xdt \quad\to\quad \frac{dt}{dx}=\frac{1}{x}$
$\frac{dy}{dx}=\frac{df}{dt}\frac{dt}{dx}=\frac{1}{x}\frac{df}{dt}$
$\frac{d^2y}{dx^2}=\frac{d\left(\frac{1}{x}\frac{df}{dt}\right)}{dx} =
-\frac{1}{x^2}\frac{df}{dt}+\frac{1}{x}\frac{d^2f}{dt^2}\frac{dt}{dx}=
-\frac{1}{x^2}\frac{df}{dt}+\frac{1}{x^2}\frac{d^2f}{dt^2}$
$-x^2\frac{d^2y}{dx^2}=\lambda y=x^2\left(-\frac{1}{x^2}\frac{df}{dt}+\frac{1}{x^2}\frac{d^2f}{dt^2}\right)= -\frac{df}{dt}+\frac{d^2f}{dt^2}$
$$\frac{d^2f}{dt^2}-\frac{df}{dt}-\lambda f(t)=0$$
This is a linear ODE with constant coefficients easy to solve.
$$f(t)= c_1e^{\frac{1+\sqrt{4\lambda+1}}{2}t}+c_2e^{\frac{1-\sqrt{4\lambda+1}}{2}t}=c_1(e^t)^{\frac{1+\sqrt{4\lambda+1}}{2}}+c_2(e^t)^{\frac{1-\sqrt{4\lambda+1}}{2}} $$
$$y(x)=c_1 x^{\frac{1+\sqrt{4\lambda+1}}{2}}+c_2x^{\frac{1-\sqrt{4\lambda+1}}{2}}$$
|
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|
Quadratic equation with different indices My maths teacher gave me a worksheet to work through as I was getting slightly bored in lessons. However, there was one question which I cannot do. The worksheet gives the answer, but you are supposed to show how you did it. Here is the question:
$729 + 3^{2x+1} = 4\times3^{x+2}$
The sheet said that the answer was $x=2$ or $x=3$
Here is my working - I 'solved' by completing the square, as you will see:
\begin{align}
3^{2x+1}-4\times3^{x+2}+729& = 0 \\
3(3^{2x})-4\times(3^2)(3^x)+729& = 0\\
3(3^x)^2-36(3^x)+729& = 0\\
y &= 3^x\\
3y^2-36y+729 & = 0\\
3(y^2-12y+243) &= 0\\
(y-6)^2-36+243 &= 0\\
(y-6)^2+207 &= 0\\
(y-6)^2&=-207\\
y-6 &= \sqrt{207}i\\
y &= 6+\sqrt{207}i\\
3^x &= 6+\sqrt{207}i\\
x &= \log_3(6+\sqrt{207}i)\\
\end{align}
Clearly, this is not the answer as stated on the sheet. Have I answered this question wrong, or is the answer on the sheet wrong? Thank you in advance.
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There was a typo in the question. It should have been
$$729+3^{2x+1}=4\times3^{x+3}$$
Note the $+3$ in the exponent instead of $+2$.
(You are correct that the question as stated does not have integer solutions, in fact it has no real solutions).
There's actually some pretty mathematical content here though. We can rewrite $729=3^6$ and $4=1+3$ to make this expression:
$$3^6+3^{2x+1}=3^{x+3}+3^{x+4}$$
Now the solutions $x=2$ and $x=3$ boil down to adding the same powers of $3$ on both sides. The completing the square method you outlined in your work will show that these are the only solutions.
|
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|
Integer solutions of $2x+3y=n$ What is the smallest $m$ such that for all $n\geq m$, the equation $2x+3y=n$ has solutions with $x,y \in \mathbb{Z}$ and $x,y\geq2$?
My approach. We can write the solutions in terms of the parameter $t$ as:
$$x(t)=-n-3t$$ and
$$y(t) = n+2t$$
Setting both of those greater than or equal to $2$, we have $-n-3t\geq2$ and $n+2t\geq2$. Solving for $t$, we find $\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}$. Solving the resulting inequality, we get
$$-2n-4\geq6-3n$$
So $n\geq10$. This seems to be right. For $n=10$, take $(2,2)$. But, suppose $n=11$. Then, $2x+3y=11$ clearly has no solution where both $x,y\geq2$. Is my solution incorrect? Can someone point me in the right direction?
EDIT: Thanks for the great answers!
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We consider two cases: $n$ is even and $n$ is odd.
If $n$ is even then $2x+3y = n \implies$ $y$ is an even number. The minimum possible value of even $n$ would be then $2(2)+3(2)=10$. For all even $n \ge 10$, we can set $x = n/2-3$ and $y=2$ to satisfy $2x+3y=n$.
If $n$ is odd then $2x+3y=n$ implies that $y$ is odd. Consequently, $y \ge 3$ and the minimum possible value of odd $n$ is $2(2)+3(3) = 13$. For odd $n \ge 13$, we set $y=3$ and $x= (n-9)/2$ to satisfy $2x+3y=n$.
For above, we conclude that for $n \ge 12$, we can find an integer solution to $2x+3y=n$ for $x\ge2$ and $y \ge 2$.
|
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|
Use of Taylor series Let $c>0$ a fixed parameter.
By using the Taylor's series I want to prove that there exists a constante $A>0$ which does not depend on $c$ such that $$\ln\left(1-\frac{2f(t)}{f(t)+g(t)}\right)\ge A(\frac{c+1}{4})t^3$$ for all $t\le \frac{1}{\sqrt c}$
where for all $t>0$:
\begin{align}
f(t) & =\frac{1-\cos(t\sqrt{c})}{c}-(\cosh(t)-1) \\[10pt]
g(t) & =\frac{\sin(t\sqrt{c})}{\sqrt{c}}+\sinh(t)
\end{align}
This is what i wrote:
for all $t\le \frac{1}{\sqrt{c}}$
$$sin(t\sqrt{c})=t\sqrt{c}-\frac{(t\sqrt{c})^3}{3!}+..+(-1)^n\frac{(t\sqrt{c})^{2n+1}}{(2n+1)!}+O((t\sqrt{c})^{2n+3})$$ for all $t\le \frac{1}{\sqrt{c}}$
for all $t\le \frac{1}{\sqrt{c}}$
$$cos(t\sqrt{c})=1-\frac{(t\sqrt{c})^2}{2!}+..+(-1)^n\frac{(t\sqrt{c})^{2n}}{(2n)!}+O((t\sqrt{c})^{2n+1})$$
$$\cosh(t)=1+\frac{t^2}{2!}+..+\frac{t^{2n}}{(2n)!}+O(t^{2n+1})$$
$$\sinh(t)=1+\frac{t^3}{3!}+..+\frac{t^{2n+1}}{(2n+1)!}+O(t^{2n+3})$$
But i don't know how to find the result.
|
I am not sure how much this could help you.
Using
$$f(t) =\frac{1-\cos(t\sqrt{c})}{c}-(\cosh(t)-1) $$
$$g(t) =\frac{\sin(t\sqrt{c})}{\sqrt{c}}+\sinh(t)$$ truncated Taylor series built around $t=0$ are
$$f(t)=-\frac{1}{24} (c+1) t^4+\frac{1}{720} \left(c^2-1\right)
t^6-\frac{\left(c^3+1\right)}{40320} t^8+O\left(t^{10}\right)$$
$$g(t)=2 t+\frac{1}{6} (1-c) t^3+\frac{1}{120} \left(c^2+1\right)
t^5+\frac{\left(1-c^3\right)}{5040} t^7+\frac{\left(c^4+1\right)
}{362880}t^9+O\left(t^{10}\right)$$ making
$$\frac{2f(t)}{f(t)+g(t)}=-\frac{1}{24} (c+1) t^3+\frac{1}{480} \left(1-c^2\right) t^5-\frac{(c+1)^2
}{1152}t^6-\frac{(c+1) (c^2-15c+1) }{40320}t^7-\frac{(c-1)
(c+1)^2 }{11520}t^8+O\left(t^9\right)$$
$$\log\left(1-\frac{2f(t)}{f(t)+g(t)}\right)=\frac{1}{24} (c+1) t^3+\frac{1}{480} \left(c^2-1\right) t^5+\frac{(c+1) (c^2-15c
+1) }{40320}t^7+O\left(t^9\right)$$
|
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|
Sequence : $a_{n+1}=2a_n-a_{n-1}+2$ Let $c$ be a positive integer. The sequence $a_1, a_2, \ldots$ is defined by $a_1=1, a_2=c$ and
$a_{n+1}=2a_n-a_{n-1}+2$ for all $n \geq 2$.
Prove that for each $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $a_na_{n+1} = a_k$.
My attempt :
Trying with small numbers, I see that $a_n=(n-1)c+(n-2)^2$ and will prove by induction.
$a_1 = 1 , a_2 = c+0 = c$ is true.
Suppose that $a_k=(k-1)c+(k-2)^2$ and $a_{k+1}=kc+(k-1)^2$ are true.
$a_{k+2} = 2a_{k+1}+a_k+2 = 2[kc+(k-1)^2]-[(k-1)c+(k-2)^2]+2=(k+1)c+k^2$
so $a_n=(n-1)c+(n-2)^2$ is true.
$a_na_{n+1} = [(n-1)c+(n-2)^2][nc+(n-1)^2] $
$= n(n-1)c^2 + (n-1)^3c + n(n-2)^2c + (n^2-3n+2)^2 = (k-1)c+(k-2)^2$
Please suggest how to solve this equation.
|
Since
$$
(a_{n+1}-a_n)-(a_n-a_{n-1})=2\tag1
$$
we have
$$
a_{n+1}-a_n=a_2-a_1+2n-2\tag2
$$
Thus,
$$
\begin{align}
a_{n+1}
&=a_1+(a_2-a_1)n+n^2-n\\
&=a_1+(a_2-a_1-1)n+n^2\tag3
\end{align}
$$
Using $(3)$, we can get by expansion
$$
a_na_{n+1}=a_{n^2+(a_2-a_1-2)n+2a_1-a_2+2}\tag4
$$
That is,
$$
\bbox[5px,border:2px solid #C0A000]{k=n^2+(a_2-a_1-2)n+2a_1-a_2+2}\tag5
$$
This does not require $a_1=1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding the maximum of $p^3 + q^3 +r^3 + 4pqr$
$p$,$q$,$r$ are $3$ non-negative real numbers less than or equal to $1.5$ such that $p+q+r = 3$, what will be the maximum of $p^3 + q^3 + r^3 + 4pqr$ ?
I tried AM-GM on $p,q,r$ to get the maximum of $pqr$ as $1$, but on doing it for $p^3 + q^3 + r^3 $, I get the minimum, so I won't be able to combine them.
I tried to assume symmetry $p=q=r=1$, and the maximum value to be 7, but putting $p=1.1$, $q=1.1$ and $r=0.8$, I get the value to be $7.046 (> 7) $
How do I solve this with concepts like AM-GM?
PS: I do not know multivariable calculus concepts, however if it's not possible to solve it without using it, please show how to solve it with these concepts.
|
A hint before the answer:
$p^3+q^3+r^3+4pqr=p^3+(q+r)^3+4pqr-3(q+r)(qr)$
Try to then eliminate both $q$ and $r$ to make this in terms of $p$.
Motivation
Firstly one might try $p=q=r=1$ and get $p^3+q^3+r^3+4pqr=7$
But the trying of more cases will reveal that this is not the maximum. One might then try to fix a term and see what is the maximum with the fixed term.
e.g. Set $p=3/2$. Then $p^3+q^3+r^3+4pqr=27/8+q^3+r^3+6qr$, $q+r=3/2$
The $q^3+r^3$ motivates writing $q^3+r^3+6qr$ in the form $(q+r)^3-something$ as $(q+r)^3=27/8$, dealing with the $q^3$ and $r^3$ terms.
i.e. $q^3+r^3+6qr=q^3+r^3+3(q+r)(qr)+(6-3q-3r)qr=(3/2)^3+(3/2)qr$
Then we are left to maximise $qr$, which is simple by AM-GM. (i.e. $q=r=3/4$) Trying to do this without fixing p at the start leads to the solution.
Solution
$$p^3+q^3+r^3+4pqr$$
$$=p^3+q^3+r^3+3(q+r)qr+(4p-3q-3r)qr$$
$$=p^3+(3-p)^3+(4p-(9-3p))qr$$
$$=p^3+(3-p)^3+(7p-9)qr$$
This is
less than or equal to $p^3+(3-p)^3+(7p-9)(\frac{3-p}{2})^2$, equality when $q=r$ if $p\geq 9/7$ and
less than or equal to $p^3+(3-p)^3+(7p-9)(1.5)(1.5-p)$, equality when one of $q, r$ equals $1.5$ if $p\leq 9/7$.
Note that among $p, q, r$, the smallest number has to be less than or equal to $1$. Without loss of generality $p\leq q \leq r$. Then $p\leq 9/7$. For $p^3+q^3+r^3+4pqr$ to be maximal, one of $q, r$ is 1.5. As $r$ is the largest, $r=1.5$.
Finally, note that $3/2>9/7$, so we can apply the other case, replacing $r$ with $p$ (which is fine since $p^3+q^3+r^3+4pqr$ is symmetric over the 3 terms), concluding that if $r=3/2$, $p^3+q^3+r^3+4pqr$ is maximal when $p=q=3/4$.
Therefore, the maximum achievable is when one of $p, q, r$ is $3/2$ and the other 2 are $3/4$, to get a value of $\frac{243}{32}$.
|
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|
Wrong Wolfram Alpha result for $\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}$?
I'm trying to solve this limit:
$$
\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}
$$
Here's my attempt:
$$0 \le |\frac{xy^4}{x^4+x^2+y^4} - 0| = \frac{|x|y^4}{x^4+x^2+y^4},$$ and since $x^4+x^2 \ge0$ then $\frac{y^4}{x^4+x^2+y^4} \le 1$ so
$$
\frac{|x|y^4}{x^4+x^2+y^4} \le |x|,$$ so
$$
0 \le \lim_{(x,y)\to(0,0)}|\frac{xy^4}{x^4+x^2+y^4} - 0| \le \lim_{(x,y)\to(0,0)} |x| = 0,
$$ and using the squeeze theorem the limit is $0$.
But if I input the limit in wolfram alpha, it says that the limit doesn't exist. Here is the link to the limit in Wolfram Alpha.
|
Both of
$$\frac{xy^{4}}{x^{4}+x^{2}+y^{4}},\,\,\frac{xy^{4}}{x^{4}+x^{2}+y^{2}}$$
have limit $0.$ For the first one, you gave a valid proof. The second one follows from this one since the denominator is at least as big, while the numerator is the same.
|
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|
Simplify $\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$ I know that the result of this expression is 16 but how do I get to that result?
$$\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$$
|
Use properties of logs first:
\begin{align*}
\dfrac{7^{\log_5 15} + 3^{2+\log_5 7}}{7^{\log_5 3}}
= \dfrac{7^{\log_5 5 + \log_5 3} + 3^{2}3^{\log_5 7}}{7^{\log_5 3}}
&= \dfrac{7^{1}7^{\log_5 3} + 3^{2}3^{\log_5 7}}{7^{\log_5 3}} \\
&= 7 \biggl( \dfrac{7^{\log_5 3}}{7^{\log_5 3}}\biggr) + 9 \biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr) \\
&= 7 + 9\biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr).
\end{align*}
Now we need to determine the quantity in the parentheses. Put $x=3^{\log_5 7}$ and $y = 7^{\log_5 3}$; then the quantity in the parentheses is $x/y$. We have $$\log_5 x = \log_5 (3^{\log_5 7}) = \log_5 7 \cdot \log_5 3 = \log_5 3 \cdot \log_5 7 = \log_5 (7^{\log_5 3}) = \log_5 y,$$ so $x=y$.
Therefore, $$\dfrac{7^{\log_5 15} + 3^{2+\log_5 7}}{7^{\log_5 3}} = 7 + 9\biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr) = 7 + 9\biggl(\dfrac{x}{y}\biggr) = 7+9(1) = 16.$$
|
{
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|
Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
|
($1-2x\geq 0\text{ and }2x+3>0$) or ($1-2x\leq0\text{ and }2x+3<0$). From the first condition we get $x\in (\frac{-3}{2},\frac{1}{2}]$. Second condition isn't possible.
|
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|
How to prove $\{n!e\} = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$ I have the definition $a_n = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$
I need to show that:
$a)$ $0 < a_n < \frac{1}{n}$
$b)$ $a_n = n!e - \lfloor{n!e}\rfloor$
So I know that
$\frac{1}{n} = \frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\dots$
$e = \sum_{n=0}^{+\infty}\frac{1}{n!}$
It's clear that $\{n!e\}=a_n<\frac{1}{n}$, but how to show it?
|
By comparison with the Geometric series we have $$0<a_n<\frac 1{n+1}+\frac 1{(n+1)^2}+\cdots=\frac 1{n+1}\times \frac 1{1-\frac 1{n+1}}=\frac 1n$$
Now assume $n>1$ (so $0<a_n<\frac 1n<1$). We write: $$n!e=n!+ \frac {n!}{2!}+\cdots \frac {n!}{n!}+a_n$$ and it is clear that the part preceding $a_n$ is an integer (each term in the sum is an integer) so we are done.
|
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|
How I do evaluate this polynomial problem? Let $f(x)=x^2+ax+b$.
Suppose $f(f(x))=0$ equation has $4$ different real solutions $x_1,x_2,x_3,x_4$
and that two of them sum up to $-1$ (i.e. $\exists\, i\neq j$ such that $x_i+x_j=-1$)
Prove that $b\lt-\frac14$
|
$f(x_k)$ are real roots of $f$, not all the same, so that $4b<-a^2$ is a necessary condition. If $z=f(x_2)=f(x_2)$ are the same root of $f$, then $x_1,x_2$ are the two solutions of $$f(x)=z\iff 0=x^2+ax+b-z$$ so that by Viete $a=-(x_1+x_2)=1$ and $b<-\frac14$ is necessary.
If $f(x_1)\ne f(x_2)$ then by Viete
\begin{align}
-a&=f(x_1)+f(x_2)=x_1^2+x_2^2-a+2b\\
\iff x_1^2+x_2^2&=-2b,\qquad 1+2b = 2x_1x_2\\
b&=f(x_1)f(x_2)\\
&=x_1^2x_2^2+ax_1x_2(x_1+x_2)+b(x_1^2+x_2^2)+ab(x_1+x_2)+b^2 \\
&=\tfrac14+b+b^2-a(\tfrac12+b)-2b^2-ab+b^2\\
\iff \tfrac14&=a(\tfrac12+2b)
\end{align}
so that in combination with the necessary condition
$$
\frac1{2a}-1=4b<-a^2
$$
If $b<-\frac14$ were not satisfied, $a$ would have to be positive. But then by the arithmetc-geometric mean inequality
$$
1>\frac1{2a}+a^2=3\,\frac{\frac1{4a}+\frac1{4a}+a^2}3\ge 3\,\sqrt[3\;]{\frac1{16}}=\sqrt[3\;]{\frac{27}{16}}>1
$$
which is impossible.
|
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|
How to approach this minimization problem? I see this problem somewhere (it is from an olympiad or something but Im not sure where it comes):
Minimize $$\frac1{\sqrt{y+\frac1x+\frac12}}+\frac1{\sqrt{z+\frac1y+\frac12}}+\frac1{\sqrt{x+\frac1z+\frac12}}\tag1$$ for $x,y,z>0$ and $xyz=1$.
My first thought was approach this problem trying to find the points where $\nabla f(x,y)=0$ and $\partial^2 f(x,y)$ is positive definite for $f$ being the function defined by $(1)$ and $xy=\frac1z$, that is, I wanted to find the local minima for such $f$ and after choose the absolute minimum.
However this approach is messy, so I guess it must be a better way to approach this problem. So my question is, do you know a better way to approach this problem? I dont have a clue about how to handle it.
|
First, take $\displaystyle (x, y, z) = \left(t, t, \frac{1}{t^2}\right)$ and make $t \to +\infty$ to see that the infimum is no greater than $\sqrt{2}$. Next it will be proved that the infimum is $\sqrt{2}$.
Set $x = a^3$, $y = b^3$, $z = c^3$, then $abc = 1$ and\begin{align*}
(1) &= \frac{1}{\sqrt{\frac{b^3}{abc} + \frac{abc}{a^3} + \frac{1}{2}}} + \frac{1}{\sqrt{\frac{c^3}{abc} + \frac{abc}{b^3} + \frac{1}{2}}} + \frac{1}{\sqrt{\frac{a^3}{abc} + \frac{abc}{c^3} + \frac{1}{2}}}\\
&= \sqrt{\frac{ca^2}{ab^2 + bc^2 + \frac{1}{2} ca^2}} + \sqrt{\frac{ab^2}{bc^2 + ca^2 + \frac{1}{2} ab^2}} + \sqrt{\frac{bc^2}{ca^2 + ab^2 + \frac{1}{2} bc^2}}.
\end{align*}
Denote $u = bc^2$, $v = ca^2$, $w = ab^2$, then it suffices to prove$$
\sum_{\mathrm{cyc}} \sqrt{\frac{u}{\frac{u}{2} + v + w}} > \sqrt{2} \tag{2}
$$
for all $u, v, w > 0$, where $\sum\limits_{\mathrm{cyc}}$ means cyclic summation. Without loss of generality, assume that $u \geqslant v \geqslant w$. Since\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}} \frac{v}{\frac{v}{2} + w + u} + \frac{w}{\frac{w}{2} + u + v} > \frac{v}{\frac{v}{2} + u}\\
&\Longleftrightarrow \frac{w}{\frac{w}{2} + u + v} > \frac{v}{\frac{v}{2} + u} - \frac{v}{\frac{v}{2} + w + u} = \frac{vw}{\left(\frac{v}{2} + u\right) \left(\frac{v}{2} + u + w\right)}\\
&\Longleftrightarrow \left(\frac{v}{2} + u\right) \left(\frac{v}{2} + u + w\right) > v \left(\frac{w}{2} + u + v\right)\\
&\Longleftrightarrow u^2 - \frac{3}{4} v^2 + uw > 0,
\end{align*}
then\begin{align*}
\sum_{\mathrm{cyc}} \sqrt{\frac{u}{\frac{u}{2} + v + w}} &= \sqrt{\frac{u}{\frac{u}{2} + v + w}} + \sqrt{\frac{v}{\frac{v}{2} + w + u}} + \sqrt{\frac{w}{\frac{w}{2} + u + v}}\\
&> \sqrt{\frac{u}{\frac{u}{2} + v + w}} + \sqrt{\frac{v}{\frac{v}{2} + w + u} + \frac{w}{\frac{w}{2} + u + v}}\\
&> \sqrt{\frac{u}{\frac{u}{2} + v + w}} + \sqrt{\frac{v}{\frac{v}{2} + u}} \geqslant \sqrt{\frac{u}{\frac{u}{2} + 2v}} + \sqrt{\frac{v}{\frac{v}{2} + u}}\\
&= \sqrt{\frac{1}{\frac{1}{2} + 2t}} + \sqrt{\frac{t}{\frac{t}{2} + 1}} = \sqrt{\frac{2}{4t + 1}} + \sqrt{\frac{2t}{t + 2}},
\end{align*}
where $\displaystyle t = \frac{v}{u} \in (0, 1]$. Because\begin{align*}
\sqrt{\frac{2}{4t + 1}} + \sqrt{\frac{2t}{t + 2}} > \sqrt{2} &\Longleftrightarrow \sqrt{\frac{2}{4t + 1}} > \sqrt{2} - \sqrt{\frac{2t}{t + 2}}\\
&\Longleftrightarrow \frac{2}{4t + 1} > 2 \left(1 + \frac{t}{t + 2} - 2 \sqrt{\frac{t}{t + 2}}\right)\\
&\Longleftrightarrow 2 \sqrt{\frac{t}{t + 2}} > 1 + \frac{t}{t + 2} - \frac{1}{4t + 1} = \frac{8t (t + 1)}{(t + 2)(4t + 1)}\\
&\Longleftrightarrow 1 > \frac{4\sqrt{t} (t + 1)}{\sqrt{t + 2}(4t + 1)}\\
&\Longleftrightarrow (t + 2)(4t + 1)^2 > 16t(t + 1)^2\\
&\Longleftrightarrow 8t^2 + t + 2 > 0,
\end{align*}
then (2) holds. Therefore,$$
\inf_{\substack{x, y, z > 0\\xyz = 1}} \left\{\frac{1}{\sqrt{y + \frac{1}{x} + \frac{1}{2}}} + \frac{1}{\sqrt{z + \frac{1}{y} + \frac{1}{2}}} + \frac{1}{\sqrt{x + \frac{1}{z} + \frac{1}{2}}}\right\} = \sqrt{2}.
$$
|
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|
Why is my alternate method of calculating scalar products not working? The exercise is such: Given that $|\vec{a}| = 3$, $|\vec{b}| = 2$ and $\varphi = 60^{\circ}$ (the angle between vectors $\vec{a}$ and $\vec{b}$), calcluate scalar product $(\vec{a}+2\vec{b}) \cdot (2\vec{a} - \vec{b})$.
My initial thought was to solve the requested product by "sticking" separately calculated fragments together and then follow the definition of scalar product, which is
$$\vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos\varphi.$$
Via observation
$$\vec{a} \cdot \vec{a} = |\vec{a}| \cdot |\vec{a}| \cdot \cos0^{\circ} = |\vec{a}|^2 \Longrightarrow |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$$
we can get values of $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$ (mind the notation: $a^2$ in this case represents $a^2 = |\vec{a}|^2$, and $b^2 = |\vec{b}|^2$):
\begin{align*}
|\vec{a}+2\vec{b}| &= \sqrt{(\vec{a}+2\vec{b}) \cdot (\vec{a}+2\vec{b})} =\sqrt{a^2 + 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\
&= \sqrt{3^2 + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2}+2^2} = 5.
\end{align*}
Similarly,
\begin{align*}
|2\vec{a} - \vec{b}| &= \sqrt{(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b})} = \sqrt{4a^2 - 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\
&=\sqrt{4 \cdot 3^2 - 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} + 2^2} = 2\sqrt{7}.
\end{align*}
Now we plug both results in:
$$(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) = |\vec{a} + 2\vec{b}| \cdot |2\vec{a} - \vec{b}| \cdot \cos60^{\circ} = 5 \cdot 2\sqrt{7} \cdot \frac{1}{2} = 5\sqrt{7}.$$
However, this is not the right solution according to my textbook. The correct result is $19$. I thought about it for a bit and took a different route:
\begin{align*}
(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) &= \vec{a} \cdot 2\vec{a} - \vec{a} \cdot \vec{b} + 4 \vec{a} \cdot \vec{b} - 2\vec{b} \cdot \vec{b} = \\
&=2a^2 - |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + 4 |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} - 2b^2 = \\
&= 2 \cdot 3^2 - 3 \cdot 2 \cdot \frac{1}{2} + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} - 2 \cdot 2^2 = 19.
\end{align*}
The second method clearly worked, while the first one failed miserably. But my question is why did my first approach fail? Did I get the wrong results when calculating $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$? I have no idea. Please, help me understand my mistakes. Thank you in advance.
|
In the first method you assume at the end that the angle between $\vec{(a+2b)}$ and $\vec{(2a-b)}$ is $60^\circ$. If you were going to do this approach you would need the angle between them instead of $\cos 60^\circ$ for the last line.
|
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|
How to extend the matrix with determinant 1 to keep it Lets consider 2x2 integer matrix with determinant equal 1:
$$\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right)$$
I am working on the following:
How to extend this to 3x3 matrix in order to get another matrix with determinant 1:
$$\left(
\begin{array}{ccc}
a & b & e \\
c & d & f \\
i & h & g \\
\end{array}
\right)$$
And also is there any $a,b,c,d$ for which this extension is unique.
I even have no idea how to start solving this.
I have discovered the following so far on the web, but not sure how to use this:
Integer matrices with determinant equal to $1$
https://mathoverflow.net/questions/24131/is-the-semigroup-of-nonnegative-integer-matrices-with-determinant-1-finitely-gen
EDITED:
Actually I am looking for general algorithm, how to construct all 3x3 matricies from 2x2 matrix with determinant 1.
EDITED 2:
Some samples of such matricies:
$$\left(
\begin{array}{ccc}
1 & 1 & 1 \\
-1 & 0 & 1 \\
-1 & 0 & 2 \\
\end{array}
\right)
$$
$$\left(
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 3 \\
2 & 5 & 9 \\
\end{array}
\right)$$
$$\left(
\begin{array}{ccc}
1 & 1 & 1 \\
-6 & -5 & -4 \\
9 & 5 & 2 \\
\end{array}
\right)
$$
|
You can just set $g = 1$ and $e,f,i,h = 0$:
$$1 = \begin{vmatrix}
a & b \\
c & d \\
\end{vmatrix} = \begin{vmatrix}
a & b & 0\\
c & d & 0\\
0 & 0 & 1
\end{vmatrix}$$
Furthermore, the extension is never unique since:
$$1 = \begin{vmatrix}
a & b \\
c & d \\
\end{vmatrix} = \begin{vmatrix}
a & b & 0\\
c & d & 0\\
0 & 0 & 1
\end{vmatrix} = \begin{vmatrix}
a & b & 0\\
c & d & 0\\
a & b & 1
\end{vmatrix}$$
Note that $a$ and $b$ cannot both be $0$, so the two extensions are indeed different.
|
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|
Find $(a,b)$ such that $(a^2+b)(a+b^2)=2^n$. Find all pair of positive integer numbers $(a,b)$ such that $(a^2+b)(a+b^2)=2^n$ for some positive integer number $n$.
Attempt I guest that there is only $(a,b)=(1,1)$ satisfying the assumption.
Thank you for all solution.
|
We assume $a\geq b$. Let $a^2+b = 2^u, b+a^2 = 2^v$ with $u\geq v\geq 1$. Then
$$a^2+b \equiv a+b^2 \equiv 0 \pmod{2^v} \implies a^4+a \equiv b^4 + b \equiv 0 \pmod{2^v}$$
If $a,b$ are both odd, we have $$a^4 + a = a(a+1)(a^2-a+1) \equiv 0 \pmod{2^v} \implies a \equiv -1 \pmod{2^v}$$ similarly $b \equiv -1 \pmod{2^v}$. Let $$a = k_1 2^v - 1 \quad \quad b = k_2 2^v - 1$$ with $k_i \geq 1$. Then
$$2^v = b+a^2 = 2^v (k_1^2 2^v - 2 k_1 + k_2) \implies 2k_1 (k_1 2^{v-1} - 1) + k_2 = 1$$
Therefore $k_2 = k_1 = v = 1$, this gives $a=b=1, n=2$.
If $a,b$ are both even, then $$a^4 + a \equiv 0 \pmod{2^v} \implies a \equiv 0 \pmod{2^v}$$ similarly $b \equiv 0 \pmod{2^v}$.
Letting $a= k_1 2^v, b=k_2 2^v$ gives $$2^v = b+a^2 \implies k_2 + 2^v k_1^2 = 1$$ which is not solvable.
Thus the only solution of the equation is $a=b=1$.
|
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|
Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
|
Note that $\frac{x-1}{\sqrt{x}} = \sqrt{x} - \frac{1}{\sqrt{x}}$, for example, so the limit is equal to $$\lim_{x \to \infty} \sqrt{x} - \sqrt{x+1} - \frac{1}{\sqrt{x}} + \frac{2}{\sqrt{x+1}}$$
which by arithmetic of limits is $$\lim_{x \to \infty} \sqrt{x} - \sqrt{x+1}$$
which is well-known to be $0$.
|
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|
A Question About A Calculation Of A Determinant.
Calculate $\det(A)$.
$$A =\begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a\\
\end{bmatrix}.$$
This is an answer on a book:
$$A A^T = (a^2+b^2+c^2+d^2) I.$$
$$\det(A) = \det(A^T).$$
$$\det(A)^2 = (a^2+b^2+c^2+d^2)^4.$$
The coefficient of $a^4$ in $\det(A)$ is $1$.
Therefore,
$$\det(A) = (a^2+b^2+c^2+d^2)^2.$$
I cannot understand the above answer.
It is obvious that $\det(A)$ is a multivariable polynomial $p(a,b,c,d)$.
Maybe, the following is true:
$p(a,b,c,d) = (a^2+b^2+c^2+d^2)^2$ for $(a,b,c,d) \in A$,
$p(a,b,c,d) = -(a^2+b^2+c^2+d^2)^2$ for $(a,b,c,d) \in B$,
where, $A \cup B = \mathbb{R}^4, A \cap B = \emptyset$.
Please prove this is impossible.
Thank you very much, Mr. Kavi Rama Murthy.
Proof:
$p(a,b,c,d)$ is a continuous function.
$(a,b,c,d)=(0,0,0,0)$ is the only solution for $p(a,b,c,d)^2 = (a^2+b^2+c^2+d^2)^4 = 0$.
so,
$(a,b,c,d)=(0,0,0,0)$ is the only solution for $p(a,b,c,d) = 0$.
$p(1,1,0,0) = \begin{vmatrix}
1&1&0&0\\
-1&1&0&0\\
0&0&1&-1\\
0&0&1&1\\
\end{vmatrix} = \begin{vmatrix}
1&1&0&0\\
0&2&0&0\\
0&0&1&-1\\
0&0&0&2\\
\end{vmatrix}
= 4 > 0$.
Let $(w_0, x_0, y_0, z_0) \in \mathbb{R}^4 - (0,0,0,0)$.
We can assume $w_0 \ne 0$ without loss of generality.
$f(x) := p(x, 1, 0, 0)$ is a one variable continuous function.
By the intermediate-value theorem, $f(w_0) = p(w_0, 1, 0, 0) > 0$.
$g(x, y, z) := p(w_0, x,y,z)$ is a three variable continuous funtion.
Again, by the intermediate-value theorem, $g(x_0, y_0, z_0) = p(w_0,x_0,y_0,z_0) > 0$.
Hence, $p(w, x, y, z) \geq 0$ for all $(w,x,y,z) \in \mathbb{R}^4$.
So, $p(w, x, y, z) = (w^2+x^2+y^2+z^2)^2$ for all $(w,x,y,z) \in \mathbb{R}^4$.
|
why did I do like this? because above answer was formed by multiplying with transpose. if someone is interested in the direct solution, one can proceed as follows. It is too broad. I have learned about this technique today. I don't recommend this. Till $3\times 3$ this method is fine for this type of problems. For fun, I have solved like this.
Let $F$ be the field and let$D$ be an alternating $4-$linear function on $4\times 4$ matrices over the polynomial ring $F[x]$.
Let
$$A =\begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a\\
\end{bmatrix}.$$ If we denonte the rows of the identity matrix by $\epsilon_1$,$\epsilon_2$,$\epsilon_3$ and $\epsilon_4$. Then $$D(A)=D(a\epsilon_1-b\epsilon_2-c\epsilon_3-d\epsilon_4,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)$$
$$D(A)=aD(\epsilon_1,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)-bD(\epsilon_2,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)-cD(\epsilon_3,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)-dD(\epsilon_4,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)$$
By simplifying using the properties, we get $D(A)=a^4+b^4+c^4+d^4+2a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2=(a^2+b^2+c^2+d^2)^2.$
|
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|
If $(8+3\sqrt{7})^n = I + F $ where $ I $ is integer and $F $ is proper fraction, what is $I$? I am beginner to Binomial Theorem and I want to find out weather $I$ is even or odd in $$(8+3\sqrt{7})^n= I+F$$ if it can be expressed as a sum of an Integer $I$ and a proper fraction $F$
How could I find out ?
|
Notice that
$$\begin{align}
(8 +3 \sqrt{7})^n + (8 - 3 \sqrt{7})^n &= \sum_{i=0}^n \binom{n}{i} 8 ^i (3 \sqrt{7})^{n-i} + \sum_{i=0}^n \binom{n}{i} (-1)^i 8 ^i (3 \sqrt{7})^{n-i} \\
&= 2 \sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i} 8^{2i} 3^{2(n-i)} 7^{n-i} \qquad \text{(the odd numbered terms cancel out)}\\
&= 2 N
\end{align}$$
where $N$ is a positive integer. Since $8 - 3 \sqrt{7} \approx 0.06$, we have
$$(8 +3 \sqrt{7})^n + \epsilon = 2N $$
where $0 < \epsilon < 1$.
So the integer part of $(8 +3 \sqrt{7})^n$ is odd.
|
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|
Find the equations of the straight line through $(0,a)$ on which the perpendiculars dropped from the point $(2a,2a)$ are each of length $a$ unit. Find the equations of two straight lines drawn through the point $(0,a)$ on which the perpendiculars drawn from the point $(2a,2a)$ are each of length $a$.
My Attempt:
The equation of line passing through $(0,a)$ is
$y=mx+a$ where $m$ is the slope.
How do I get the value of $m$?
|
Solution,
Given,
Points: (0,a) and (2a, 2a)
Now,
y - y1 = m (x - x1)
or, y - a = m (x - 0)
or, mx - y + a = 0.........(i)
Since, perpendicular distance of (i) from (2a, 2a) is a.
a = +- 2am - 2a + a / (root under) m ^ 2 + (-1) ^ 2
or, a (root under) m ^ 2 + (-1) ^ 2 = +- a (2m - 1)
or, (root under) m ^ 2 + (-1) ^ 2 = +- (2m - 1)
Squaring on both sides,
m ^ 2 + 1 = (2m - 1) ^ 2
or, m ^ 2 + 1 = 4m ^ 2 - 4m + 1
or, m ^ 2 = 4m ^ 2 - 4m
or, 4m ^ 2 - m ^ 2 = 4m
or, 3m ^ 2 = 4m
or, 3m ^ 2 - 4m = 0
or, m (3m - 4) = 0
Either, m = 0 Or, 3m - 4 = 0
or, 3m = 4
or, m = 4 / 3
Putting m = 0 in equation (i),
0.x - y + a = 0
or, y = a
Putting m = 4 / 3 in equation (i),
4 / 3x - y + a = 0
or, 4x - 3y + 3a / 3 = 0
So, 3y = 4x + 3a
Therefore, the required equations are y = a and 3y = 4x + 3a.
|
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|
Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven? See here :
http://googology.wikia.com/wiki/Arrow_notation
for the definition of the up-arrow function.
Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 3$ ?
With Saibians theorem we get $$10\uparrow^n m<(2\uparrow^n 3)\uparrow ^n m<2\uparrow^n (m+3)$$ Also the claim is trivially true for $m=1$, but I did not manage to complete the induction step.
|
For $n=3$, we have:
For $m=1$, we have $10\uparrow^31=10$ and $2\uparrow^33=65536$.
Assume $10\uparrow^3m<2\uparrow^3(m+2)-3$ holds for some $m\ge1$. Then we have
\begin{align}10\uparrow^3(m+1)&=10\uparrow^210\uparrow^3m\\&<10\uparrow^2(2\uparrow^3(m+2)-3)\\&<(2\uparrow^23)\uparrow^2(2\uparrow^3(m+2)-3)-3\tag0\\&<2\uparrow^22\uparrow^3(m+2)-3\\&=2\uparrow^3(m+3)-3\end{align}
Hence $10\uparrow^3m<2\uparrow^3(m+2)-3<2\uparrow^3(m+2)$.
$(0)$ is due to the fact that up-arrows are strictly increasing in all arguments.
Assume it holds for some $n\ge3$ and all $m\ge1$. Then we get
\begin{align}10\uparrow^{n+1}m&=(10\uparrow^n)^{m-1}10\\&<(2\uparrow^n[2+)^{m-1}10]\tag1\\&<(2\uparrow^n)^{m-1}13\\&<(2\uparrow^n)^{m-1}2\uparrow^33\\&<(2\uparrow^n)^{m-1}2\uparrow^n2\uparrow^n2\\&=(2\uparrow^n)^{m+1}2\\&=2\uparrow^{n+1}(m+2)\end{align}
$(1)$ is the result of
\begin{align}2+2\uparrow^nm&<3+2\uparrow^nm\\&<2\uparrow^{n-1}2\uparrow^nm\\&=2\uparrow^n(m+1)\end{align}
|
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|
Product of length of segments in Ellipse.
If the normal at any point $P$ on the Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2} {b^2}=1$ meets the axis in $G$ and $g$ respectively, then find $PG\cdot Pg$, in terms of $a$ and $b$.
I tried considering the parametric point as $(a\cos\theta,b\sin\theta)$ on the ellipse, then constructed both tangent and normal at that point then found the coordinated where normal cut the axes, then found the length of the perpendicular from these coordinates on the tangent (at $(a\cos\theta,b\sin\theta)$), but I end up with following: $$PG\cdot Pg=a^2b^2(a^2\sin^2\theta+b^2\cos^2\theta)$$
please help.
|
Hint:
Using this, the equation of normal
$$\dfrac x{\dfrac{(a^2-b^2)\cos\theta}a}-\dfrac x{\dfrac{(a^2-b^2)\sin\theta}b}=1$$
$$|PG|=\sqrt{\left(\dfrac{(a^2-b^2)\cos\theta}a-a\cos\theta\right)^2+(0-b\sin\theta)^2}=?$$
|
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|
Newton binomial expansion Find the coefficient for ${x^{11}}$ in ${\sqrt{1+x}}$
The generic formula is
$$\sqrt{1+x}=(1+x)^\frac{1}{2}=1+\frac{1}{2}x+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2}x^2\dots$$
Solution of expansion of coeficient for ${x^{11}}$:
$${\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)(\frac{1}{2}-3)(\frac{1}{2}-4)(\frac{1}{2}-5)(\frac{1}{2}-6)(\frac{1}{2}-7)(\frac{1}{2}-8)(\frac{1}{2}-9)(\frac{1}{2}-10)}{11!}}$$
$$={\frac{\frac{654729075}{2^{11}}}{11!}}x^{11}$$
My question is: is this correct?
|
$(1+x)^a=1+ax+...+\dfrac{a(a-1)\cdots(a-(n-1))}{n!}x^n+o(x^n)$
Let show by induction $\sqrt{1+x}=\sum\limits_{k=0}^n a_kx^k+o(x^n)$ with $$a_n=\dfrac{(-1)^{n-1}}{(2n-1)}\dfrac{\binom{2n}{n}}{4^n}$$
$a_0=\dfrac{-1}{-1}=0\quad\checkmark$
$a_1=\dfrac{2}{4}=\frac 12\quad\checkmark$
$a_{n+1}=a_n\times\dfrac{(\frac 12-n)}{(n+1)}=\dfrac{(-1)^{n-1}(2n)!(-\frac 12)(2n-1)}{(2n-1)n!n!4^n(n+1)}=\dfrac{(-1)^n(2n+2)!}{(n+1)!n!(2n+1)(2n+2)4^n\times 2}=\dfrac{(-1)^n(2n+2)!}{4^{n+1}(n+1)!(n+1)!(2n+1)}\quad\checkmark$
So $a_{11}=\dfrac{\binom{22}{11}}{4^{11}\times 21}=\dfrac{4199}{524288}$, this is also the result you got!
|
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|
Simple exersice using CRT Calculate all solutions $0\leq x< 130$ for the following system of equations.
$x \equiv_2 1$
$x \equiv_5 2$
$x \equiv_{13} 3$
Solution:
$M=2\cdot 5\cdot 13=130$
So by CRT there exists a unique solution.
We get:
$M_1=M/2=65$
$M_2 = M/5 = 26$
$M_3=M/13=10$
Now:
$65\cdot N_1 \equiv_2 1 \ \ \Rightarrow \ \ N_1 = 1$
$26\cdot N_2 \equiv_5 2 \ \ \Rightarrow \ \ N_2=2$
$10\cdot N_3 \equiv_{13} 3 \ \ \Rightarrow \ \ N_3=12$
$x = R_{130}(1\cdot 65\cdot 1 + 2\cdot 26 \cdot 2 + 3 \cdot 10 \cdot 12)=79$
This is not correct, as one can easily see with the second equation.
I can't find my mistake :/ what's wrong?
|
If you set
$$26\cdot N_2 \equiv_5 1 \ \ \Rightarrow \ \ N_2=1 \\
10\cdot N_3 \equiv_{13} 1 \ \ \Rightarrow \ \ N_3=4
$$
Then you'll get the right number
$$x = R_{130}(1\cdot M_1 \cdot N_1 + 2\cdot M_2 \cdot N_2 + 3 \cdot M_3 \cdot N_3)\\=R_{130}(1 \cdot 65\cdot 1 + 2\cdot 26 \cdot 2 + 3 \cdot 10 \cdot 12)\\
=R_{130}(65+52+120)=R_{130}(107)
$$
This satisfies the question.
$$107 \equiv_2 1, 107 \equiv_5 2, 107 \equiv_{13} 3$$
|
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|
How to compute $\int_{0}^{1}\left (\frac{\arctan x}{1+(x+\frac{1}{x})\arctan x}\right )^2dx$ I want to calculate the following integral $$\int_{0}^{1}\left(\frac{\arctan x}{1+(x+\frac{1}{x})\arctan x}\right)^2 \, dx$$
But I have no way to do it, can someone help me, thank you.
|
Here is a way to arrive at the answer, and it is by no means obvious.
In finding the indefinite integral the reverse quotient rule will be used. Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
$$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$
it is immediate that
$$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$
Now the indefinite integral
$$I = \int \left (\frac{\arctan (x)}{1 + \left (x + \frac{1}{x} \right ) \arctan (x)} \right )^2 \, dx,$$
can be rewritten as
$$I = \int \frac{x^2 \arctan^2 (x)}{\left (x + (x^2 + 1) \arctan (x) \right )^2} \, dx.$$
Let $v = x + (x^2 + 1) \arctan (x)$. Then $v' = 2 + 2x \arctan (x)$. Now for the hard bit. We need to find a function $u(x)$ such that
$$u' v - v' u = u'[x + (x^2 + 1) \arctan (x)] - u [2 - 2x \arctan (x)] = x^2 \arctan^2 (x).$$
If
$$u = \frac{-x^2 + (x^2 + 1) \arctan^2 (x)}{2},$$
then
$$u' = x + x \arctan^2 (x) + \arctan (x),$$
and we find, miraculously, that
$$u' v - v' u = x^2 \arctan^2 (x).$$
Our indefinite integral can now be readily found as it can be rewritten in the form given by (1). The result is:
$$I = \int \left (\frac{-x^2 + (1 + x^2) \arctan^2 (x)}{2[x + (x^2 + 1) \arctan (x)]} \right )' \, dx = \frac{-x^2 + (1 + x^2) \arctan^2 (x)}{2[x + (x^2 + 1) \arctan (x)]} + C.$$
So for the definite integral on the interval $x \in [0,1]$ we have
$$\int_0^1 \left (\frac{\arctan (x)}{1 + \left (x + \frac{1}{x} \right ) \arctan (x)} \right )^2 \, dx = \frac{\pi^2 - 8}{8(2 + \pi)}.$$
|
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|
Find the ratio of segments using Ceva's theorem I am going through $IB$ further math geometry topic and just learned Ceva's theorem. Below is one of the questions in the exercise. I have thought for quite a while and cannot solve it. Can anyone give me some clues or hints, please? Thanks.
In the diagram, $BZ:ZC=2:1$ and $AR:RS:SZ=5:4:3$. Find the ratio in which $X$ divides $[AB]$.
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The proof of Ceva'sTheorem uses the fact that if $ A, B $ and $ C $ are three non-collinear points then any vector $ P $ can be expressed as $ P = xA + yB + zC $ where $ x+ y + z = 1 $. Also, a point on the line joining $ A $ and $ B $ is given by $ P = tA + (1-t)B $.
$ Z = \frac {2}{3} C + \frac {1}{3} B $
$ S = \frac {3}{4} Z + \frac {1}{4} A $
$ = \frac {1}{2} C +\frac {1}{4} B + \frac {1}{4} A $
Similarly, $ R = \frac {5}{18} C + \frac {5}{36} B + \frac {7}{12} A $
$ Y = tS + (1-t)B $
$ = \frac {t}{2} C + (\frac {t}{4} + 1-t)B + \frac {t}{4} A $
For $ Y $ to lie on $ AC $, $ t = \frac {4}{3} $ .
Then $ Y = \frac {2}{3} C + \frac {1}{3} A $
$ X = kR + (1-k)Y $
This gives $ X $ in terms of $ A, B $ and $ C $.
Find $ k $ for which $ X $ lies on $ AB $ etc.
|
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|
Calculate limit of $(1 + x^2 + y^2)^\frac{1}{x^2 + y^2 + xy^2}$ as $(x,y) \rightarrow (0,0)$ I'm trying to calculate the limit of
$(1 + x^2 + y^2)^\frac{1}{x^2 + y^2 + xy^2}$ as $(x,y) \rightarrow (0,0)$.
I know that the limit is supposed to be $e$, and I can arrive at this answer if I study the univariate limit by, for instance, setting $x=t, y=0$ and letting $t \rightarrow 0$ or vice versa.
I'm not sure how to calculate it for the multivariate case. I tried using polar coordinates $(x= r \cos{\theta}, y = r \sin{\theta})$ which gets me
$$ (1 + r^2)^{\frac{1}{r^2(1 + r \cos{\theta}\sin^2{\theta})}} $$, after symplifying the expression, but I don't know how to proceed from there. I assume that I am supposed to end up with an expression similar to $(1 + n)^\frac{1}{n} \rightarrow e$ as $n \rightarrow 0$.
|
I thought it might be instructive to present an approach that circumvents use of polar coordinates and relies instead on a straightforward application of the AM-GM inequality. To that end we proceed.
To begin, the AM-GM inequality guarantees that for $x>-1$
$$\begin{align}
\left|\frac{xy^2}{x^2+(1+x)y^2}\right|&\le \frac{|x|y^2}{2|x||y|\sqrt{1+x}}\\\\
&=\frac{|y|}{2\sqrt{1+x}}
\end{align}$$
Hence, applying the squeeze theorem reveals
$$\lim_{(x,y)\to(0,0)}\left(\frac{xy^2}{x^2+(1+x)y^2}\right)=0\tag1$$
Next, we can write
$$\frac{x^2+y^2}{x^2+y^2+xy^2}=1-\frac{xy^2}{x^2+(1+x)y^2}$$
Appealing to $(1)$, we have
$$\lim_{(x,y)\to(0,0)}\left(\frac{x^2+y^2}{x^2+y^2+xy^2}\right)=1\tag2$$
Finally, equipped with $(2)$, we find that
$$\begin{align}
\lim_{(x,y)\to(0,0)}\left(\left(\left(1+x^2+y^2\right)^{\frac1{x^2+y^2}}\right)^{\frac{x^2+y^2}{x^2+y^2+xy^2}}\right)&=\left(\lim_{(x,y)\to(0,0)}\left(\left(1+x^2+y^2\right)^{\frac1{x^2+y^2}}\right)^{\lim_{(x,y)\to(0,0)}\left(\frac{x^2+y^2}{x^2+y^2+xy^2}\right)}\right)\\\\
&=e^1\\\\
&=e
\end{align}$$
And we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2610870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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|
Question on complex numbers and cube root of unity
If $a$, $b$, $c$ are distinct integers and $\omega$ is a cube root of unity then minimum value of $|a + b\omega + c\omega^2| + |a+b\omega^2 + c\omega|$ is?
Now, I know the identity $a^3 + b^3 +c^3 -3abc=(a+b+c)(a + b\omega + c\omega^2)(a+b\omega^2 + c\omega)$.
I don't know how to apply this to the question. Or is there something else that I am missing?
Any help would be appreciated.
|
Let
$$A=|a + b\omega + c\omega^2|$$
$$B=|a+b\omega^2 + c\omega|$$
note that $B=\bar A=A\implies A+B=2A$
then
$$A+B=2A=2\left|a + b\left(-\frac12+i\frac{\sqrt3}{2} \right) + c\left(-\frac12-i\frac{\sqrt3}{2} \right)\right|=2\left|a -\frac12b-\frac12c+i\frac{\sqrt3}{2}\left(b-c\right)\right|=2\left(a^2+\frac14b^2+\frac14c^2-ab-ac+\frac12bc+\frac34b^2+\frac34c^2-\frac32bc\right)^{\frac12}=2\left(a^2+b^2+c^2-ab-bc-ca\right)^{\frac12}=2\left(\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\right)^{\frac12}$$
For $a=1,b=2,c=3$ we obtain $A+B=2\sqrt 3$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
\begin{align}
\left( \sum_{cyc} x \right) \left( \sum_{\text{cyc}} x^2
\right) &= \left( \sum_{\text{cyc}} x^3 \right) = 1 \\
\sum_{cyc} x^2y + \sum_{cyc} xy^2 &= 0
\end{align}
In the above equation, subtract $x^3+y^3+z^3$ from both sides.
\begin{align}
(x+y+z)^3 &= 1 \\
\sum_{cyc} x^3 + 3\sum_{cyc} x^2y + 3\sum_{cyc} xy^2 + 6xyz &= 1 \\
1 + 0 + 0 + 6xyz &= 1
\end{align}
Hence $xyz=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
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|
Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$
Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$?
For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down.
I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$
I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere.
Is there an algebraic proof/disproof of my above claim?
Thanks.
|
Assume $x\in\Bbb Z^+$.
If $x=m^2$ is a pefect square, then
$$m^2<x+\sqrt x=m^2+m<m^2+2m+1=(m+1)^2$$
an so $x+\sqrt x=R(x+\sqrt x)$ cannot be a perfect square.
Thus we need only consider the case that $x$ is not a perfect square, which makes $\lfloor x+\sqrt x\rfloor <\lceil x+\sqrt x\rceil$.
Let $n\in\Bbb Z^+$ be maximal with $n(n+1)<x$. Then $x=n^2+n+d$ with $1\le d\le (n+1)(n+2)-n(n+1)=2n+2$.
This makes
$$\lfloor x+\sqrt x\rfloor = n^2+n+d+n=(n+1)^2+d-1.$$
This is $\ge (n+1)^2$ and $\le n^2+4n+2<(n+2)^2 $.
Hence $ k^2=\lfloor x+\sqrt x\rfloor$
implies $k=n+1$, $x=n^2+n+1$. But $$(n+\tfrac12)^2=n^2+n+\tfrac14<x$$ implies that $x+\sqrt x$ should round up, not down.
Similarly, $k^2=\lceil x+\sqrt x\rceil = \lfloor x+\sqrt x\rfloor+1$ implies $k=n+2$, $d=(n+2)^2+1-(n+1)^2=2n+2$, $x=n^2+3n+2$.
But $$ (n+\tfrac32)^2=n^2+3n+\tfrac 94>x$$
implies that $x+\sqrt x$ should be rounded down, not up.
We conclude that $R(x+\sqrt x)$ is never a perfect square for $x\in\Bbb Z^+$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I find all integers that satisfy this condition? Suppose that $72x + 56y = 40$. Find all $x,y$ that satisfy this condition.
Here is what I did:
*
*I reduced the equation to give $9x + 7y = 5$
*But since 5 is relatively prime to 7 and 9, we know $x,y$ is divisible by 5. So we can reduce the equation to $9(5a) + 7(5b) = 5$.
*Simplifying, we get $9a + 7b= 1$. Would it be easier to continue along this direction, i.e. to find some $a,b$ such that their gcd is one? I cannot continue beyond this as I have no other insights left.
What kind of available theorems/techniques can I use to solve this?
|
$$9x+7y=5$$
First, you find a solution.
\begin{array}{rcll}
5-7(1) &= &-2 \\
5-7(2) &= &-9 \\
9(-1) + 7(2) &= &5
\end{array}
Because $\gcd(9,7)=1$, the general solution is
$$(x,y) = (-1+7t, 2-9t)$$
for all $t \in \mathbb Z$.
Now take the general case $9x+7y=5$ and subtract $9(-1)+7(2)=5$.
\begin{array}{cccc}
9x &+ &7y &= &5 \\
9(-1) &+ &7(2) &= &5 \\
\hline
9(x+1) &+ &7(y-2) &= &0 \\
&&9(x+1) &= &7(2-y)
\end{array}
Since $9$ and $7$ are relatively prime, then $9$ must divide $2-y$, let's say $ 2-y = 9t$ for some integer, $t$. Hence $y = -9t + 2$ for some integer, $t$. Substituting that back into $9(x+1) = 7(2-y)$, we get $9(x+1)=63t$. We can simplify that to $x = 7t - 1$.
So the most general solution is
$$9(7t-1) + 7(-9t+2) = 5$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sequence Convergence of $(s_n) = (\frac{1}{n^2})$ Part 1. Given $\varepsilon = 0.5$, find an $N \in \mathbb{N}$ s.t whenever $n \geq N$, $|s_n - 0| < \varepsilon$
&&
Part 2. Treat the epsilon in Part 1 with an arbitrary $\varepsilon > 0$
My work:
Let $s_n = \frac{1}{n^2}$, and suppose $s = 0$, implies $|\frac{1}{n^2} - 0| < \varepsilon$. Now is we assume $\varepsilon = 0.5$, then:
$|\frac{1}{n^2} - 0| < 0.2 \Rightarrow \frac{1}{n^2} < 0.5 \Rightarrow n^2 > \frac{1}{0.2} \Rightarrow n > \frac{1}{\sqrt{0.5}}$, we'll chose a $N$ s.t. $N > \frac{1}{\sqrt{0.5}}$.
So now by the Archimedes Prop of $\mathbb{R}$ let $N \in \mathbb{N}$ satisfying $N > \frac{1}{\sqrt{0.5}}$.
Then if
$n \geq N, |s_n - s| = |\frac{1}{n^2} - 0| = \frac{1}{n^2} \leq \frac{1}{N^2}$
$\frac{1}{(\frac{1}{\sqrt{0.5}})^2} = \frac{1}{\frac{1}{0.5}} = 0.5 = \varepsilon$
Part 2. Let $\varepsilon > 0$. By the Arch. Prop. of $\mathbb{R}$, let $N \in \mathbb{N}$ s.t. $N > \frac{1}{\sqrt{\varepsilon}}$. Now if $n \geq N$, then:
$|s_n - s| = |\frac{1}{n^2} - 0|$
$= \frac{1}{n^2}$ (since $n > 0$)
$\leq \frac{1}{N^2}$ (since $n \geq N)$
$< \frac{1}{(\frac{1}{\sqrt{\varepsilon}})^2}$ (since $N > \frac{1}{\sqrt{\varepsilon}})$
$= \varepsilon$
My issue is the fact that $N \in \mathbb{N}$, as that throws a problem in my part 1, where did go wrong?
|
Just use the ceiling function, we just have to choose an $N \in \mathbb{N}$ such that $N> \frac{1}{\sqrt{\epsilon}}$, you can choose it to be $1+\lceil \frac{1}{\sqrt{\epsilon}}\rceil.$
|
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|
help with sum of infinite series, stuck in problem the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:
$$\sum_{n=1}^{\infty} \frac{2^{n+1}+1}{3^n}$$
It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.
The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)
I read online after class, that what I should do is to find the limit of the sequence of partial sums.
This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.
It makes sense conceptually when you have a large value of k, for a finite series $S_{k}$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.
So that if I have partial sums in a sequence $S_{1}, S_{2}, S_{3}...S_{k} $
The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.
I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5
But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...
|
After 20 years, I still have to rederive the identity
$$ \sum_{n=1}^{\infty} \alpha^n = \frac{\alpha}{1-\alpha} \tag{1}$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.
Suppose that $|\alpha| < 1$, and let
$$ S_k := \sum_{n=1}^{k} \alpha^k = \alpha + \alpha^2 + \dotsb + \alpha^{k} $$
be the $k$-th partial sum. Then
$$ \alpha S_k = \sum_{n=1}^{k} \alpha^{k+1} = \alpha^2 + \alpha^3 + \dotsb + \alpha^{k+1}. $$
Subtracting, we obtain
\begin{align}
(1-\alpha) S_k
&= S_k - \alpha S_k \\
&= \left( \alpha + \alpha^2 + \dotsb + \alpha^{k} \right) - \left( \alpha^2 + \alpha^3 + \dotsb + \alpha^{k+1} \right) \\
&= \alpha + \left( \alpha^2 - \alpha^2\right) + \left( \alpha^3 - \alpha^3\right) + \dotsb + \left(\alpha^k - \alpha^k\right) - \alpha^{k+1} \\
&= \alpha - \alpha^{k+1}.
\end{align}
Solving for $S_k$, we get
$$ S_k = \frac{\alpha - \alpha^{k+1}}{1-\alpha}. $$
Since $|\alpha| < 1$, it follows that $\lim_{k\to\infty} \alpha^{k+1} = 0$, and so
$$
\sum_{n=1}^{\infty} \alpha^n
:= \lim_{k\to \infty} \sum_{n=1}^{k} \alpha^n
= \lim_{k\to\infty} \frac{\alpha - \alpha^{k+1}}{1-\alpha}
= \frac{\alpha}{1-\alpha}.
$$
But this is exactly the identity at (1). Huzzah!
Now, how do we use this here?
Observe that
$$ \frac{2^{n+1} + 1}{3^n}
= \frac{2^{n+1}}{3^n} + \frac{1}{3^n}
= 2 \left( \frac{2}{3} \right)^n + \left( \frac{1}{3} \right)^n.
\tag{2}$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $\sum a_n$ and $\sum b_n$ converge, and $C$ is any constant, then we have
$$ \sum_{n=1}^{\infty} \left[ a_n + b_n \right] = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n
\qquad\text{and}\qquad
\sum_{n=1}^{\infty} Ca_n = C \sum_{n=1}^{\infty} a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $\sum$ with $\int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
\begin{align}
\sum_{n=1}^{\infty} \frac{2^{n+1} + 1}{3^n}
&= \sum_{n=1}^{\infty} \left[ 2 \left( \frac{2}{3} \right)^n + \left( \frac{1}{3} \right)^n \right] && (\text{by application of (2)}) \\
&= 2\sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n + \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n && (\text{simplify the series}) \\
&= 2 \cdot \frac{\frac{2}{3}}{1-\frac{2}{3}} + \frac{\frac{1}{3}}{1- \frac{1}{3}} && (\text{by application of (1)}) \\
&= 2 \cdot \frac{\frac{2}{3}}{1-\frac{2}{3}}\cdot \frac{3}{3} + \frac{\frac{1}{3}}{1- \frac{1}{3}}\cdot \frac{3}{3} && (\text{silly, pedantic arithmetic}) \\
&= 2 \cdot \frac{2}{3-2} + \frac{1}{3-1} && (\text{more pedantic arithmetic}) \\
&= \frac{4}{1} + \frac{1}{2} && (\text{and some more}) \\
&= \frac{8+1}{2} && (\text{almost there}) \\
&= \frac{9}{2}. && (\text{done!})
\end{align}
|
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|
Computing the definite integral $\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$
Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$
This is what I did:
$u = x^2 + a^2 $
$du/dx = 2x$
$du = 2xdx$
$1/2 du = x dx$
$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$ from $0$ to $a$.
$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$ from $0$ to $a$.
I eventually got:
$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$
but this was incorrect.
The correct answer was:
$\frac{1}{3}\left(2\sqrt{2}-1\right)a^3$
Any help?
|
You can simply use the reverse chain rule for integration and you get:
$$\int_0^a x\sqrt{x^2+a^2}dx=\frac 12\int_0^a 2x\sqrt{x^2+a^2}dx=\left[\frac{(x^2+a^2)^\frac 32}3\right]_0^a=\\=\frac{(2a^2)^\frac 32}3-\frac{(a^2)^\frac 32}3=\color{red}{\frac{(2\sqrt 2-1)}3|a|^3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial Fraction problem solution deviates from the Rule Question:
Compute $\displaystyle \int\frac{x^2+1}{(x^2+2)(x+1)} \, dx$
My Approach:
As per my knowledge this integral can be divided in partial Fraction of form $\dfrac{Ax+B}{x^2+px+q}$ and then do the following as per to integrate it.
Solution:
Taking $\dfrac{x^2+1}{(x^2+2)(x+1)}=\dfrac{Ax^2+Bx+C}{x^2+2}+\dfrac{D}{x+1}$
Rule given: Denominator g(x) contains quadratic tractor (may not be factorisable).
To each non-repeated quadratic factor of the form $x^2+px+q$(or $x^2+q$, $q$ not equal to $0$), there
Should be a partial Fraction of the form $Ax+B/(x^2+px+q)$.
My problem:
I can't understand why the solution provided deviates from the rule that I have studied to solve these kind of problems.
Book:
ISC MATHEMATICS XII
Publishers:
Kalyani
|
From what I understand, the solution in your book points that the fraction can be writeen as
$$ \frac{x^{2}+1}{(x^{2}+2)(x+1)} = \frac{Ax^{2}+Bx+C}{x^{2}+2} + \frac{D}{x+1} $$
$$ = \frac{(Ax^{2}+Bx+C)(x+1) + D(x^{2}+2)}{(x^{2}+2)(x+1)} $$
Of course it can, you just need to find the appropriate constants :
$$ Ax^{3} + (A+B+ D)x^{2} + (B+C)x + (C+2D) = x^{2}+1 $$
$$ \implies A = 0, \:\: A+B+D = 1, \:\: B+C = 0, \:\: C+2D=1$$
Now, actually, from this step, we can already see that $A=0$is a necessity, so the partial fractions must be of the form
$$\frac{Bx+C}{x^{2}+2} + \frac{D}{x+1} $$
which is the rule that you already know.
Both the solution and the rule are correct, its just that using the rule will be more efficient.
|
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|
Subgroups of General Linear Group I have a question about the order of the subgroups of $GL_2(\mathbb{R})$ generated by the following matrix:
$\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}$
So I computed several powers of this matrix and realized that the order of the subgroup generated by this matrix is infinity. However, I do not know how to prove it in general that:
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^n\neq \begin{pmatrix}
1&0 \\
0&1
\end{pmatrix} \forall n\in\mathbb{Z}.$
Is there any way to prove this in general? I might have did not remember something from Linear Algebra to prove this. Any suggestion is really appreciated.
Here are what I got for my calculations:
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^2= \begin{pmatrix}
2&-1 \\
-1&1
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^3= \begin{pmatrix}
3&-2 \\
-2&1
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^4= \begin{pmatrix}
5&-3 \\
-3&2
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^5= \begin{pmatrix}
8&-5 \\
-5&3
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^6= \begin{pmatrix}
13&-8 \\
-8&5
\end{pmatrix} $
|
Hint: if $\lambda$ is an eigenvalue of a matrix $A$ (that is, there exists a nonzero column vector $v$ such that $Av = \lambda v$), then $\lambda^n$ is an eigenvalue of the matrix $A^n$.
Suppose that $A^n = I$ for some $n$. What are the eigenvalues of $I$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the following statements without using induction The first statement:
$$\tan(nx)=\frac{C^n_1\tan(x)-C^n_3\tan^3(x)+C^n_5\tan^5(x)-C^n_7\tan^7(x)+\dotsm}{1-C^n_2\tan^2(x)+C^n_4\tan^4(x)-C^n_6\tan^6(x)+\dotsm}$$
where
$$C^n_k=\frac{n!}{k!(n-k)!}$$
The second statement:
$$\sin(x)\sin\Bigl(x+\frac{\pi}{n}\Bigr)\sin\Bigl(x+\frac{2\pi}{n}\Bigr)\ ...\sin\Bigl(x+\frac{(n-1)\pi}{n}\Bigr)=2^{1-n}\sin(nx)$$
The third statement:
$$\DeclareMathOperator{\cotg}{cotg}\cotg(x)+\cotg\Bigl(x+\frac{\pi}{n}\Bigr)+\cotg\Bigl(x+\frac{2\pi}{n}\Bigr)+\dotsm+\cotg\Bigl(x+\frac{(n-1)\pi}{n}\Bigr)=n\cotg(nx)$$
|
Use Binomial & DeMoivre's Theorems
\begin{eqnarray*}
\cos(nx) &=& \operatorname{Re}(( \cos x+ i \sin x)^n) =\cos^n x - \binom{n}{2} \cos^{n-2} x \sin^2 x + \binom{n}{4} \cos^{n-4} x \sin^4 x - \cdots\\
\sin(nx) &=&\operatorname{Im}(( \cos x+ i \sin x)^n) =\binom{n}{1}\cos^{n-1} x \sin x- \binom{n}{3} \cos^{n-3} x \sin^3 x + \binom{n}{5} \cos^{n-5} x \sin^5 x - \cdots
\end{eqnarray*}
Now
\begin{eqnarray*}
\tan(nx) &=& \frac{\sin nx}{\cos{nx}} \\
&=& \frac{\binom{n}{1}\cos^{n-1} x \sin x- \binom{n}{3} \cos^{n-3} x \sin^3 x + \binom{n}{5} \cos^{n-5} x \sin^5 x - \cdots}{ \cos^n x - \binom{n}{2} \cos^{n-2} x \sin^2 x + \binom{n}{4} \cos^{n-4} x \sin^4 x - \cdots} \\
&=& \frac{\binom{n}{1} \tan x- \binom{n}{3} \tan^3 x + \binom{n}{5} \tan^5 x - \cdots}{ 1 - \binom{n}{2} \tan^2 x + \binom{n}{4} \tan^4 x - \cdots} \\
\end{eqnarray*}
|
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|
Beckenbach Introduction to Inequalities Chapter 2: Show $(a+b)/2 \le ( (a^2 + b^2 )/2)^{1/2}$ I'm having trouble understanding the following problem
Problem
Beckenbach, Chapter 2 Pg 24 Ex 1
$$
\text{Show the following for all a, b}\quad
\frac{(a+b)}{2} \le \left(\frac{a^2 + b^2}{2}\right)^\frac{1}{2}
$$
The book provides answers in the back, the answer is shown as
$$
\text{equivalent to}\quad(a - b)^2 \ge 0
$$
My attempt
Thanks to this answer: Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$ I was able to show the following
$$
\begin{align}
\frac{a+b}{2} &\le \sqrt{\frac{a^2 + b^2}{2}} \\
\frac{a+b}{2} &\le \sqrt{\frac{(a+b)^2 + (a-b)^2}{4}}\\
\frac{a+b}{2} &\le \sqrt{\left(\frac{a+b}{2}\right)^2\left(1 + \left(\frac{a-b}{a+b}\right)^2\right)}\\
\frac{a+b}{2} &\le \frac{a+b}{2}\sqrt{1 + \left(\frac{a-b}{a+b}\right)^2}\\
0 &\le \sqrt{1 + \left(\frac{a-b}{a+b}\right)^2} - 1 \\
\text{Since}\quad\left(\frac{a-b}{a+b}\right)^2 \ge 0\quad\text{we're done}
\end{align}
$$
This feels ugly and I don't think is the way Beckenbach intended us to solve, considering this is in Chapter 2.
Question
*
*Is my attempt valid?
*Is there a more elegant way to show this, using $(a-b)^2 \ge 0$?
Thanks
|
If left side is not positive then inequality obviously hold.
So assume it is nonegative. Square it: $${a^2+2ab+b^2\over 4}\leq {a^2+b^2\over 2}$$
Get rid of denumerators and you get
$$ a^2+2ab+b^2\leq 2a^2+2b^2$$
or $(a-b)^2\geq 0$.
|
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|
How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$ Prove
$$
\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}
$$
and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.
My Attempt:
$$
\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}=2\tan\frac{x}{2}\\
\implies \tan^2\frac{x}{2}(\tan x)+2\tan\frac{x}{2}-\tan x=0\\
\implies \tan\frac{x}{2}=\frac{-2\pm2\sec x}{2\tan x}=\frac{-1\pm\sec x}{\tan x}
$$
Using this,
$$
\tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-1\pm\frac{4}{\sqrt{7}}}{\frac{3}{\sqrt{7}}}=\frac{-\sqrt{7}\pm4}{3}
$$
As $0\leq\frac{\sin^{-1}3/4}{2}\leq\frac{\pi}{4}\implies0\leq\tan(\frac{\sin^{-1}3/4}{2})\leq1$
$$
\implies \tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-\sqrt{7}+4}{3}
$$
Is my attempt correct and where is the value $\frac{4+\sqrt{7}}{3}$ to be excluded ?
|
Here's a nice (if I say so myself) geometric solution. Consider the right triangle $\triangle ABC$ with right $\angle ABC$, $|BC| = 3$ and $|AC| = 4$. $AD$ is the angle bisector of $\angle BAC$. Let $|BD| = x$ and hence $|CD| = 3-x$.
Now $\displaystyle |AB| = \sqrt{4^2 - 3^2} = \sqrt 7$ (Pythagoras). Note that $\displaystyle \theta = \frac 12 \arcsin \frac 34$ and $\tan \theta$ is the required ratio.
By the angle bisector theorem, $\displaystyle \frac x{3-x} = \frac{\sqrt 7}4$, giving $\displaystyle x = \frac{3\sqrt 7}{4 + \sqrt 7}$.
Hence the required ratio is $\displaystyle \tan \theta = \frac x{\sqrt 7} = \frac 3{4 + \sqrt 7} = \frac{4-\sqrt 7}3$ after rationalising the denominator.
|
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|
How to obtain the right-hand side of the following equation? How to obtain the right-hand side of the following equation\begin{align*}
&\mathrel{\phantom{=}}\sum_{m=1}^{w-2}\sum_{n=2}^{2w-2m-1} 2^{n-2} g(2m+n,2w-2m-n)\\
&=\frac{1}{6}\sum_{k=4}^{2w-1}[2^{k-1}-3+(-1)^k] g(k,2w-k)
\end{align*}
by substituting $k=2m+n$ into the left-hand side of the above equation?
I understand that $4 \le k \le 2w-1 $. But I don't know how to change the boundry of $n$ and obtain the right-hand side.
|
There might be an easier way, but this is what I could do. First of all, I will ignore the function $g$, since it is clear that it does not have to do with what we want to show, namely:
$$ \sum_{m=1}^{w-2}\sum_{n=2}^{2w-2m-1}2^{n-2} = \frac{1}{6} \sum_{k=4}^{2w-1} \left(2^{k-1}-3+(-1)^k \right)$$
We start from the LHS. First of all, with the substitution $k:=2m+n$ we obtain
$$ \sum_{m=1}^{w-2}\ \sum_{k=2m+2}^{2w-1} \ 2^{k-2m-2}$$
We want to exchange the summations, but this is not so easy because of the factor $2$ in $2m$. If you draw a grid with axes $m$ and $k$ and fix some $w$, and draw a point for all the values of $(m,k)$ that you want to consider, then you can see that the above sum is equal to
$$ \sum_{k=4}^{2w-1} \ \sum_{m=1}^{\lfloor\frac{k-2}{2}\rfloor} \ 2^{k-2m-2}$$
This is probably the hardest step. If it is not clear, try to exchange easier summations, like
$$ \sum_{k=1}^{w} \ \sum_{m=1}^{k} = \sum_{m=1}^{w} \ \sum_{k=m}^{w}$$
Anyway, once we have this sum, we are only left to show that
$$\sum_{m=1}^{\lfloor\frac{k-2}{2}\rfloor} \ 2^{k-2m-2} = \frac{1}{6}\left(2^{k-1}-3+(-1)^k\right)$$
But this is just a geometric series and the fact that $\lfloor\frac{k-2}{2}\rfloor = \frac{2k-5+(-1)^k}{4}$:
$$\sum_{m=1}^{\lfloor\frac{k-2}{2}\rfloor} \ 2^{k-2m-2} = \sum_{m=0}^{\lfloor\frac{k-2}{2}\rfloor-1} \ 2^{k-4}\left(2^{-2}\right)^{m}
= 2^{k-4} \frac{1-\left(2^{-2}\right)^{\lfloor\frac{k-2}{2}\rfloor}}{1-2^{-2}} = \frac{2^{k-1}}{6} \left( 1-2^\frac{-2k+5-(-1)^k}{2} \right) = \frac{1}{6}\left(2^{k-1}-3+(-1)^k\right).$$
|
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|
Find the no. of ways to color the matrix.. Given a $3 \times N$ matrix and three colors $a$, $b$ and $c$. In how many ways can all the cells be colored such that no row and no column contains all the cells of same color ?
I could not reach to any logical approach to the problem .
|
For $i=1,2,3$ let $R_i$ denote the set of colourings such that the cells in row $i$ get the same color.
For $j=1,\dots,N$ let $C_j$ denote the set of colourings such that the cells in column $j$ get the same color.
Then to be found is: $$3^{3N}-|R_1\cup R_2\cup R_3\cup C_1\cup\cdots\cup C_N|\tag1$$
For this we use inclusion/exclusion combined with symmetry. First an oversight of the terms that will show up in $(1)$
$\begin{array}{cccc}
\text{term in }\left(1\right) & & & \text{coefficient in (1)}\\
\left|C_{1}\cap\cdots\cap C_{k}\right|=3^{3N-2k} & \text{for }k=1,\dots,N & & \binom{N}{k}\binom{3}{0}\left(-1\right)^{k}\\
\left|R_{1}\cap C_{1}\cap\cdots\cap C_{k}\right|=3^{2N-2k+1} & \text{for }k=1,\dots,N & & \binom{N}{k}\binom{3}{1}\left(-1\right)^{k+1}\\
\left|R_{1}\cap R_{2}\cap C_{1}\cap\cdots\cap C_{k}\right|=3^{N-k+1} & \text{for }k=1,\dots,N & & \binom{N}{k}\binom{3}{2}\left(-1\right)^{k+2}\\
\left|R_{1}\cap R_{2}\cap R_{3}\cap C_{1}\cap\cdots\cap C_{k}\right|=3 & \text{for }k=1,\dots,N & & \binom{N}{k}\binom{3}{3}\left(-1\right)^{k+3}\\
\left|R_{1}\right|=3^{2N+1} & & & \binom{N}{0}\binom{3}{1}\left(-1\right)^{1}\\
\left|R_{1}\cap R_{2}\right|=3^{N+2} & & & \binom{N}{0}\binom{3}{2}\left(-1\right)^{2}\\
\left|R_{1}\cap R_{2}\cap R_{3}\right|=27 & & & \binom{N}{0}\binom{3}{3}\left(-1\right)^{3}
\end{array}$
Working this out we finally end up with:$$18\cdot3^N+24^N-9\cdot8^N+9\cdot2^N-24\tag2$$
possibilities. Note that $(2)$ gives $0$ for $N=1$ and gives $174$ for $N=2$.
|
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|
Mnemonics for typical integrations I am supposed to mug up these integrals for my upcoming exams:
$$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$$
$$\int\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|+C$$
$$\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+C$$
I have been trying all my past one year trying to mug them up, ever since this chapter of indefinite integration was introduced, yet I keep messing the terms/$\pm$ signs/etc. every single time.
I hope there could be a few mnemonics to help learn these integrals. I am not necessarily looking for a song or an animated gif, but perhaps, any thing that the mind can instantly hook up to, and connect to these equations, is appreciated.
|
As bames suggested above, the method of solving for these indefinite integrations is actually a helpful tool in quickly recalling them. Here, I will provide a simple derivation - for the three integrations I listed - that helped me recall them within thirty seconds on paper. This is a long post, because it captures the exact method that I currently use.
$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$
$$\begin{align}
I=\int\sqrt{a^2-x^2}~dx
&=\int\frac{a^2-x^2}{\sqrt{a^2-x^2}}dx\\
&=\int\frac{a^2}{\sqrt{a^2-x^2}}dx-\int\frac{x^2}{\sqrt{a^2-x^2}}dx\\
&=a^2\sin^{-1}\frac xa-\int x\cdot\frac{x}{\sqrt{a^2-x^2}}dx
\end{align}$$
Now, apply the Integration-By-Parts rule on the second integrand as follows:
$$
\begin{align}
\int x\cdot\frac{x}{\sqrt{a^2-x^2}}dx
&=x\int\frac{x}{\sqrt{a^2-x^2}}-\int\int\frac{x}{\sqrt{a^2-x^2}}dx\\
&=-x\sqrt{a^2-x^2}+\int \sqrt{a^2-x^2}dx\\
&=-x\sqrt{a^2-x^2}+I
\end{align}$$
Substituting it back, we get:
$$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$$
$\int\sqrt{x^2+a^2}~dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|+C$
$$
\begin{align}
I=\int\sqrt{x^2+a^2}~dx
&=\int\frac{x^2+a^2}{\sqrt{x^2+a^2}}dx\\
&=\int\frac{x^2}{\sqrt{x^2+a^2}}dx+\int\frac{a^2}{\sqrt{x^2+a^2}}dx\\
&=\int x\cdot\frac{x}{\sqrt{x^2+a^2}}dx+a^2\ln|\sqrt{x^2+a^2}+x|+C'
\end{align}
$$
Now, apply the Integration-By-Parts rule on the first integrand as follows:
$$
\begin{align}
\int x\cdot\frac{x}{\sqrt{x^2+a^2}}dx&=x\int\frac{x}{\sqrt{x^2+a^2}}dx-\int\int \frac x{\sqrt{x^2+a^2}}dx\\
&=x\sqrt{x^2+a^2}-\int\sqrt{x^2+a^2}\\
&=x\sqrt{x^2+a^2}-I
\end{align}$$
Substituting it back, we get: $$I=\frac x2\sqrt{x^2+a^2}+\frac{a^2}2\ln|\sqrt{x^2+a^2}+x|+C$$
$\int\sqrt{x^2-a^2}~dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+C$
$$
\begin{align}
I=\int\sqrt{x^2-a^2}~dx
&=\int\frac{x^2-a^2}{\sqrt{x^2-a^2}}dx\\
&=\int\frac{x^2}{\sqrt{x^2-a^2}}dx-\int\frac{a^2}{\sqrt{x^2-a^2}}dx\\
&=\int x\cdot\frac{x}{\sqrt{x^2-a^2}}dx-a^2\ln|\sqrt{x^2-a^2}+x|+C'
\end{align}
$$
Now, apply the Integration-By-Parts rule on the first integrand as follows:
$$
\begin{align}
\int x\cdot\frac{x}{\sqrt{x^2-a^2}}dx
&=x\int\frac{x}{\sqrt{x^2-a^2}}dx-\int\int \frac x{\sqrt{x^2-a^2}}dx\\
&=x\sqrt{x^2-a^2}-\int\sqrt{x^2-a^2}\\
&=x\sqrt{x^2-a^2}-I
\end{align}$$
Substituting it back, we get: $$I=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\ln|\sqrt{x^2-a^2}+x|+C$$
Quick references:
$$\begin{align}
\int\frac1{\sqrt{x^2+a^2}}dx&=\frac 1{a}\int\frac{a\sec^2\theta}{\sqrt{\tan^2\theta+1}}d\theta ~~(\because x=a\tan\theta)\\
&=\int\sec\theta d\theta\\
&=\ln|\sec\theta+\tan\theta|+C'\\
&=\ln|\sqrt{x^2+a^2}+x|+C
\end{align}$$
$$\begin{align}
\int\frac1{\sqrt{x^2-a^2}}dx&=\frac 1{a}\int\frac{a\sec\theta\tan\theta}{\sqrt{\sec^2\theta-1}}d\theta ~~(\because x=a\sec\theta)\\
&=\int\sec\theta d\theta\\
&=\ln|\sec\theta+\tan\theta|+C'\\
&=\ln|\sqrt{x^2-a^2}+x|+C
\end{align}$$
$$\begin{align}
\int\frac1{\sqrt{a^2-x^2}}dx&=
\frac 1{a}\int\frac{a\cos\theta}{\sqrt{1-\sin^2\theta}}d\theta ~~(\because x=a\sin\theta)\\
&=\int d\theta\\
&=\theta+C'\\
&=\sin^{-1}\frac xa+C
\end{align}$$
|
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Some implicit differentiation questions. Please check them [see desc.] I have a few implicit diffentiatial questions that I wanted to check. general question... how do you know that $y$ is a function of $x$? I assume the whole reason why these are called implicit differentiation questions is because $y$ is defined implicitly and it's hard or impossible to define $y$ as a function of $x$. Is that right?
1.
$$ \cos{xy} = 1+ \sin{y} $$
$$-\sin{xy} \cdot (\frac{dy}{dx} + y) = \cos{y} \cdot \frac{dy}{dx}$$
$$-\sin{xy} \cdot \frac{dy}{dx} - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx}$$
$$ - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx} + \sin{xy} \cdot \frac{dy}{dx}$$
$$ - y(\sin{xy}) = \frac{dy}{dx} ( \sin{xy} + \cos{y})$$
$$ \frac{- y(\sin{xy})}{( \sin{xy} + \cos{y})} = \frac{dy}{dx}$$
*$$x - y = x \cdot e^y$$
$$1 - \frac{dy}{dx} = x \cdot e^y \cdot \frac{dy}{dx} + e^y$$
$$ 1 - e^y = x \cdot e^y \frac{dy}{dx} + \frac{dy}{dx}$$
$$1 - e^y = \frac{dy}{dx}(x \cdot e^y + 1)$$
$$\frac{1-e^y}{x \cdot e^y + 1} = \frac{dy}{dx}$$
*$$y \cdot \cos{x} = x^2 + y^2$$
$$y(-\sin{x}) \cdot \cos{x} \cdot \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$$
$$ 2y \frac{dy}{dx} - \cos{x} \frac{dy}{dx} = y - \sin{x} - 2x$$
$$\frac{dy}{dx} (2y-cosx) = y - sinx - 2x$$
$$\frac{dy}{dx} = \frac{y - sinx - 2x}{2y - cosx}$$
Thank you.
|
1) As John pointed there is a mistake...
2) The second example seems correct to me
3) For the third differenciation...
$$y \cdot \cos{x} = x^2 + y^2$$
$$y'\cos(x)-y\sin(x)=2x+2yy'$$
$$y'\cos(x)-2yy'=y\sin(x)+2x$$
$$y'(\cos(x)-2y)=y\sin(x)+2x$$
$$\frac {dy}{dx}=\frac {y\sin(x)+2x}{\cos(x)-2y} $$
...
|
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How do I find the radius of convergence for this power series? I stumbled on this power series and was wondering how could I evaluate the interval of convergence:
$$\sum_{n=1}^{\infty }(-1)^nx^{3n}\frac{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}{n!\cdot 3^{n}}$$
I first tried using the ratio test to determine the $x$ interval for which it's convergent, but ran into problems when I couldn't express the numerator in terms of a factorial or some kind of closed expression:
$$\lim_{n\rightarrow \infty }\frac{x^{3(n+1)}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)!\cdot 3^{n+1}}}{x^{3n}\cdot \frac{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}{n!\cdot 3^{n}}}$$
$$=\lim_{n\rightarrow \infty }\frac{x^{3n}\cdot x^{3}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)\cdot n!\cdot 3^{n}\cdot 3}}{x^{3n}\cdot \frac{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}{n!\cdot 3^{n}}}$$
$$=\lim_{n\rightarrow \infty }\frac{ x^{3}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)\cdot 3}}{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}$$
$$=\lim_{n\rightarrow \infty }{ x^{3}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)\cdot 3}}\cdot\frac{1}{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}$$
The best idea I could come up with was trying to write the numerator of the first fraction as $\frac{(3n!)}{3^{n}\cdot{n!}\cdot{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}}$ but that didn't get rid of the problem, and I'm not even sure if it's correct. Any help would be appreciated :)
|
You misunderstood the notation when you wrote down the "$n+1$" term. The ratio should be
$$x^{3(n+1)}\frac{1\cdot4\cdot7\cdots(3n-2)\cdot(3(n+1)-2)}{(n+1)!\, 3^{n+1}}
\bigg/x^{3n} \frac{1\cdot 4\cdot 7\cdots (3n-2)}{n!\,3^{n}}$$
which simplifies to
$$\frac{(3n+1)x^3}{3(n+1)}\ ,$$
and I think you will find it easy enough from here.
|
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|
Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by
$$
u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.
$$
Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive.
Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$:
$$
u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\
u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\
u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\
u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.
$$
But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!
|
Notice that if the sequence $u_n$ is positive, then it must be decreasing. This can be proven by induction.
In this case, the sequence $u_n$ is monotonic and bounded, so it has a limit $L$ that has to be positive.
From the formula,
$$L=a-\frac{1}{L}$$
Thus
$$a=L+\frac{1}{L} \ge 2$$
So a necessary condition is $a \ge 2$.
To prove that the condition is suffecient, Take $L>0$ such that $a=L+\frac{1}{L}$ and prove by induction that $u_n$ is decreasing, positive and $u_n > L$.
|
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|
Want to understand how a fraction is simplified The fraction is used to determine the sum of a telescopic series.
$$\sum_{k=1}^\infty\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$
This is the solved fraction.
$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)^2k-k^2(k+1)} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k^3+2k^2+k-k^3-k^2} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)} = \frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1} = \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$
I want to understand what is being done on every step,mainly the last three.
Thank you.
|
Another way to look at it is first "declutter" the expression, since all those radicals obfuscate the simple structure. Let $a=\sqrt{k}\,$, $b=\sqrt{k+1}\,$, then:
$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} = \frac{1}{ab^2+a^2b}=\frac{1}{ab(a+b)}$$
Now consider that $\require{cancel}\,(b+a)(b-a)=b^2-a^2=(\cancel{k}+1)-\cancel{k}=1\,$, so $\,\dfrac{1}{a+b}=b-a\,$. Then:
$$\frac{1}{ab(a+b)}=\dfrac{b-a}{ab}=\dfrac{\cancel{b}}{a\cancel{b}}-\dfrac{\bcancel{a}}{\bcancel{a}b}=\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2640521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How would one show that the integral $\int_0^\infty \frac 1 {(1+x)\sqrt x}\,\mathrm d x$ either diverges or converges? I'm supposed to be showing that the integral
\begin{equation}
\int_0^\infty \frac 1 {(1+x)\sqrt x}\,\mathrm d x
\end{equation}
either converges or diverges, using the comparison test.
It can easily be verified that
$$
\frac 1 {(1+x)\sqrt x} \geq \frac 1 {(1+x)x} = \frac 1 x\frac 1 {(1+x)},
$$
so if I was able to show that the integral
$$
\int_0^\infty \frac 1 x\frac 1 {(1+x)}\,\mathrm d x
$$
diverges I would be in the clear.
I know this integral can be broken into two parts like so:
$$
\lim_{c\to0^+}\int_c^1 \frac 1 x\frac 1 {(1+x)}\,\mathrm d x
+
\lim_{c\to\infty}\int_1^c \frac 1 x\frac 1 {(1+x)}\,\mathrm d x\,.
$$
If either of these diverges, so does the sum.
Working with the left one using integration by parts:
\begin{align}
&&\lim_{c\to0^+} \int_c^1{ \frac 1 x \frac 1 {1+x} }{x}
&= \lim_{c\to 0^+}\left[ \frac 1 x \ln(1+x)\right]_c^1
- \int_c^1{ \frac {-1}{x^2} \ln({1+x}) }{x}\\
&& &= \lim_{c\to\infty}\left[ \frac 1 x \ln(1+x)\right]_c^1
+ \int_c^1{ \frac {1}{x^2} \ln(1+x) }{x}\\
&& &= \lim_{c\to\infty}\left[ \frac 1 x \ln(1+x)\right]_c^1
+ \left[ \frac {-1} x \ln(1+x)\right]_c^1
- \int_c^1{ \frac {-1} x \frac 1 {1+x} }{x}\\
\iff && 0&=0\,.
\end{align}
This doesn't exactly work. Is there an easier way to do this?
|
By letting $y=\sqrt{x}$, then
\begin{align*}
\int_0^\infty \frac 1 {(1+x)\sqrt{x}}\,dx=\int_0^\infty\frac{2}{1+y^2}\,dy,
\end{align*}
and
\begin{align*}
\int_0^\infty \frac{1}{1+y^2} \, dy = \lim_{M\rightarrow\infty} \tan^{-1} y \bigg|_{y=0}^{y=M} = \frac\pi 2<\infty.
\end{align*}
Maybe some sort of comparison:
\begin{align*}
\int_0^\infty \frac 1 {1+y^2}\,dy=\int_0^1 \frac 1 {1+y^2} \, dy + \int_1^\infty \frac 1 {1+y^2} \, dy,
\end{align*}
and
\begin{align*}
\int_1^\infty \frac 1 {1+y^2}\,dy \leq \int_1^\infty \frac 1 {y^2} \, dy = 1 < \infty.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
When is $(p^2-1)/8$ even? I am trying to find for what values of p $\frac{p^2-1}{8}$ will be even number $\frac{p^2-1}{8}=2m\implies (p-1)(p+1)=8(2m)$ then can I write $p\equiv 1\pmod8$ or $p\equiv -1\pmod8$ ?
Also trying to find for what values of p it will be an odd number
$\frac{p^2-1}{8}=(2m+1)\\\implies (p-1)(p+1)=8(2m+1)$$\text{ will it be } p\equiv 1\pmod8$ or $p\equiv -1\pmod8$ again? p is an odd prime
|
$\Box \ 4\mid p^2 - 1$ for all odd $p$.
Proof: For some $n\in\mathbb{Z}$, since $p$ is odd, then $p = 2n+1$. $$\begin{align} \therefore 4\mid (2n + 1)^2 - 1 &= (2n)^2 + 1^2 + 4n - 1\tag1 \\ &= 4n^2 + 4n \\ &= 4(2n^2 + 2n).\tag*{$\Box$}\end{align}$$ Notice though that $2n^2 + 2n = 2(n^2 + n)$. $$\therefore 4\times 2 = 8\mid (2n + 1)^2 - 1\tag*{$\because (1) = 8(n^2 + n)$}.$$ However, we want to show that $\dfrac{p^2 - 1}{8} = 2m$ for some $m\in\mathbb{Z}$, so this means that for all $n\in\mathbb{Z}$ we need to show that $n^2 + n = 2m$, and it is.
Proof: Set $n = 2k$ for some $k\in\mathbb{Z}$ then, $$\begin{align}(2k)^2 + 2k &= 4k^2 + 2k \\ &= 2(2k^2 + k)\tag*{$\begin{align} \therefore \,\,&m = k^2 + k \\ \Leftrightarrow \,\,&m\in\mathbb{Z}\end{align}$}.\end{align}$$ Set $n = 2k + 1$ then, $$\begin{align}(2k + 1)^2 + (2k + 1) &= (2k)^2 + 1^2 + 4k + 2k + 1 \\ &= 4k^2 + 6k + 2 \\ &= 2(2k^2 + 3k + 1).\tag*{$\begin{align} \therefore\,\, &m = 2k^2 + 3k + 1 \\ \Leftrightarrow\,\, &m\in\mathbb{Z}\end{align}$}\end{align}$$ Therefore, the thesis remains true. $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\Box$
We arrive at the conclusion that $p$ is simply just an odd number, and since the set of prime numbers is contained within the set of odd numbers, then it is allowable for $p$ to be an odd prime. There is, however, the exception of $2$ but $2^2 - 1 = 3 < 4$ so we do not include this as a part of our set since $3/4\notin\mathbb{Z}$. And besides, it is an even number.
Also, if $p^2 - 1 = 8$ then $2m = 1\Leftrightarrow m\notin \mathbb{Z}$ so here we have to be careful! The only value for $p$ that satisfies this equation is $3$, so we have to exclude that from our set as well. In general, $p$ can be any prime of the form $6t\pm 1$ for some $t\in\mathbb{Z^+}$ since of course $p > 0$, and so furthermore, we clarify that $m\in\mathbb{Z^+}$.
In the first part, it is be wrong writing it in modulo $8$ because we know that there is a bigger number than $8$ which divides into $p^2 - 1$, namely $16$ as @Lubin mentioned.
In the second part, expand $8(2m + 1)$ as $16m + 8$ so you obtain $$\frac{p^2 - 1}{8} - 8 = \frac{p^2 - 65}{8} = 16m.$$ Now we have that $p^2\equiv 65 \pmod{128 = 2^7}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2645398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculate the determinant $\left|\begin{smallmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{smallmatrix}\right|$ Question: Calculate the following determinant
$$A=\begin{vmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$
Progress: So I apply $R1'=R1+R2+R3+R4$ and get
$$A=(a+b+c+d)\begin{vmatrix} 1&1&1&1\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$
Then, I apply $C2'=C2-C1,\, C3'=C3-C1$, etc to get
$$A=(a+b+c+d)\begin{vmatrix}1&0&0&0\\b&a-b&d-b&c-b\\c&d-c&a-c&b-c\\d&c-d&b-d&a-d \end{vmatrix}$$
Thus, $$A=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix}$$
But now I'm stuck here. I don't really want to expand the whole thing, because it is really messy. Is there any approach that don't require much calculation?
|
\begin{align}A&=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix} \\&=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\0 &a+b-c-d&a+b-c-d \end{vmatrix} \\
&= (a+b+c+d)(a+b-c-d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\0 &1&1\end{vmatrix} \\
&= (a+b+c+d)(a+b-c-d)\begin{vmatrix}a-b&d-b&c-b\\a+d-b-c&a+d-b-c& 0\\0 &1&1\end{vmatrix} \\
&=(a+b+c+d)(a+b-c-d)(a+d-b-c)\begin{vmatrix}a-b&d-b&c-b\\1&1& 0\\0 &1&1\end{vmatrix} \\
&=(a+b+c+d)(a+b-c-d)(a+d-b-c)\begin{vmatrix}a-b&d-c&0\\1&1& 0\\0 &1&1\end{vmatrix} \\
&=(a+b+c+d)(a+b-c-d)(a+d-b-c)(a-b+c-d)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2647193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$. Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$.
I tried this:
$$
\mathbb{Z}[x]/(2x^2+1,2x-3)=:R \\
(2x^2+1,2x-3)=:I \\
\because (2x^2+1)-x(2x-3)=3x+1, \\
2(3x+1)-3(2x-3)=11 \\
\therefore I=(2x^2+1,2x-3,11) \\
\therefore R \cong \mathbb{Z}_{11}[x]/(2x^2+1,2x-3) \\
\cong \mathbb{Z}_{11}[x]/(x^2+6,x-7) \\
\cong \mathbb{Z}_{11}
$$
but I'm totally unsure. Thank you for your help :)
|
$2x-3=0 \implies x=\frac 3 2$. Substituting $\frac {11} 2=0$, thus of course $11=0$. Our ring becomes $\mathbb Z_{11}[\frac 3 2]$. Now we have to check that we already have an element in $\mathbb Z_{11}$ such that multiplying it by $2$ it becomes $3$: $$2y \equiv 3 (mod \hspace{0,2cm}11)$$ we multply by $6$ and we get $$y \equiv 18 \equiv 7(mod \hspace{0,2cm}11)$$ and we already have $7 \in \mathbb Z_{11}$, thus $\mathbb Z_{11}[\frac 3 2] \simeq \mathbb Z_{11}$.
In the same way you could add the two equations and get $x^2 + x - 1=0$ and get $\mathbb Z_{11}[\frac{-1+\sqrt5}{2} ]$. The inverse of $2$ is already in $\mathbb Z_{11}$, it is $6$, and the square roots of $5$ are $(\pm 4)^2 \equiv 5$, so again $\mathbb Z_{11}[\frac{-1+\sqrt5}{2} ] \simeq \mathbb Z_{11}$.
|
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"url": "https://math.stackexchange.com/questions/2648387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
VERIFICATION: Apply Step-deviation method to find the average mean of the following frequency distribution Q: Apply Step-deviation method to find the average mean of the following frequency distribution
\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|} \hline
Variate (x_i) & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 \\ \hline
Frequency(f_i) & 20 & 43 & 75 & 67 & 72 & 45 & 39 & 9 & 8 & 6 \\ \hline
\end{array}
My Solution
\begin{array}{|l|l|l|l|l|l|}
\hline
\textbf{h=10} & \textbf{$f_i$} & \textbf{$xi$} & \textbf{$di=xi-a$ ; a=30} & \textbf{$ui=\frac{di}{a}$} & \textbf{$f_iu_i$} \\
5-15 & 63 & 10 & -20 & -2 & -126 \\ \hline
15-25 & 142 & 20 & -10 & -1 & -142 \\ \hline
25-35 & 117 & 30 & 0 & 0 & 0 \\ \hline
35-45 & 48 & 40 & 10 & 1 & 48 \\ \hline
45-55 & 14 & 50 & 20 & 2 & 28 \\ \hline
& \Sigma f_i=384 & & & & \Sigma f_iu_i=-192 \\ \hline
\end{array}
So, $$\bar x = a + \left(\frac{\Sigma f_iu_i}{\Sigma f_i}\right)(h)$$
$$ \bar x= 30 + \left(\frac{-192}{384}\right)(10) $$
$$\bar x = 30 -5$$
$$ \therefore \bar x = 25$$
but the correct answer turns out to be 22.214.
Where am i worng?
|
The source of error is the wrong combination of the $2i-1$ and $2i$-th data into one class of wrong width $h=10$. You're in fact calculating
$$\sum_{i=1}^5 x_{2i}(f_{2i-1}+f_{2i}) \tag{overestimated}$$
instead of
$$\bar x = \sum_{i=1}^5 (x_{2i-1}f_{2i-1} + x_{2i}f_{2i}) = \sum_{i=1}^{10} x_i f_i.$$
Since the common difference of successive terms in $x_i$ is $5$, we should choose $h=5$.
\begin{array}{|r|r|r|r|r|} \hline
\text{Marks } x_i & \text{freq. } f_i & {\text{Deviation } \\ d_i=x_i-a \\ a = 30} & {u_i=\frac{d_i}{h} \\ \text{Here } h = 5} & f_iu_i \\ \hline
5 & 20 & -25 & -5 & -100 \\ \hline
10 & 43 & -20 & -4 & -172 \\ \hline
15 & 75 & -15 & -3 & -225 \\ \hline
20 & 67 & -10 & -2 & -134 \\ \hline
25 & 72 & -5 & -1 & -72 \\ \hline
30 & 45 & 0 & 0 & 0 \\ \hline
35 & 39 & 5 & 1 & 39 \\ \hline
40 & 9 & 10 & 2 & 18 \\ \hline
45 & 8 & 15 & 3 & 24 \\ \hline
50 & 6 & 20 & 4 & 24 \\ \hline
& \sum f_i=384 & & & \sum f_i u_i=-598 \\ \hline
\end{array}
\begin{align}
\bar x &= a + \left(\frac{\sum f_iu_i}{\sum f_i}\right)(h) \\
&= 30 + \left(\frac{-598}{384}\right)(5) \\
&= 30 - \frac{1495}{192} \\
&= \frac{4265}{192} \\
&= 22.2135 \tag{cor. to 4 d.p.}
\end{align}
Reference: https://gradestack.com/CBSE-Class-10th-Course/Statistic/Step-deviation-Method/15052-3000-5351-study-wtw
|
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|
Trig Subs issues Problem:
Use a trigonometric substitution to find
$$I = \int_\sqrt{3}^2{\frac{\sqrt{x^2-3}}{x}}\;dx$$
Give both an exact answer (involving $\pi$ and a square root) and a decimal estimate to 3 significant digits.
Here is the work I have so far.
$$I = \int_\sqrt{3}^2{\frac{\sqrt{x^2-3}}{x}}\;dx$$
$$1 + \tan^2\Theta = \sec^2\Theta$$
$$\sec^2\Theta - 1 =\ tan^2\Theta$$
$$x = \sqrt{3}\sec\Theta$$
$$\sqrt{(\sqrt{3}\sec\Theta)^2 - 3}$$
$$\sqrt{3(1+\tan^2\Theta)-3}$$
$$\sqrt{3\tan^2\Theta}$$
$$\sqrt{3} \ tan\Theta$$
$$I = \int_\sqrt{3}^2{\frac{\sqrt{3}\tan\Theta}{\sqrt{3}\sec\Theta}}\;dx$$
$$\frac{dx}{d\Theta} = (\sqrt{3}\sec\Theta)'$$
$$dx = \sqrt{3}\tan\Theta\sec \Theta d\Theta$$
I feel like around here I may have messed up and I'm not sure if I should switch the limits of integration to $\Theta$ numbers of $0$ to $\frac{\pi}{6}$ yet or not. Any help is appreciated =) Thanks.
|
You did well.
Swith the limit when you replace $dx$ by $d\Theta$.
\begin{align}
I&= \int_0^{\frac{\pi}6} \frac{\tan \Theta}{\sec \Theta} \sqrt{3} \tan \Theta \sec \Theta \, d\Theta \\
&= \sqrt{3} \int_0^{\frac{\pi}{6}} \tan^2\Theta\, d\Theta\\
&= \sqrt{3} \int_0^{\frac{\pi}6} \sec^2 \Theta -1 \,d\Theta\\
&= \sqrt{3} \left( \tan \Theta - \Theta\right) \mid_0^\frac{\pi}{6} \\
&= \sqrt{3} \left( \frac1{\sqrt3}-\frac{\pi}6\right)\\
&= 1 - \frac{\pi\sqrt{3}}{6}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $(x,y) \in \Bbb R^+$ Prove that $\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$ Let $(x,y) \in \Bbb R^+$ Prove that $$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
My try
Well, i didn't see a way to factorize this, so i put it in WolframAlpha and i got that we can rewrite it like this, which is obviously true for $(x,y) \in \Bbb R^+$:
$$\frac{(y-x)^2(x+y+1)}{xy(x+y)^2}\ge 0$$
But i don't see the way to get there manually, im stuck here:
$$\frac{(x+y)(y+1)}{xy}\ge \frac{(x+y+2)^2}{(x+y)^2}$$
|
It's just Jensen for $f(x)=\ln\left(1+\frac{1}{x}\right)$.
Indeed, $f''(x)=\frac{(2x+1)}{x^2(x+1)^2}>0$ and we get
$$\frac{\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{x}\right)}{2}\geq\ln\left(1+\frac{1}{\frac{x+y}{2}}\right),$$
which is your inequality.
Also, it's
$$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\geq1+\frac{4}{x+y}+\frac{4}{(x+y)^2},$$ which is true by C-S
$$\frac{1}{x}+\frac{1}{y}\geq\frac{(1+1)^2}{x+y}=\frac{4}{x+y}$$ and AM-GM
$$\frac{1}{xy}\geq\frac{1}{\left(\frac{x+y}{2}\right)^2}=\frac{4}{(x+y)^2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sangaku Circle Geometry Problem I'm having difficulties with this Sangaku problem and was hoping for some help!
Five circles (1 of radius c, 2 of radius b, and 2 of radius a) are inscribed in a segment of a larger circle (note: this segment does not have to be a semi-circle). Given a and b, find c. For example, if a = 72 and b = 32, find c.
Spoiler for answer to a = 72 and b = 32:
c = 25
Spoiler for general formula:
$c = \frac{a(\sqrt{a} + \sqrt{b})^2}{4(3b + \sqrt{ab} )}$
I was able to solve the problem by using a computer algebra system to solve a system of nonlinear equations, but the result I got was not as pretty as the above solution, and it wasn't clear how to reduce it to the above formula. I also tried circle inversion using the tangent of circles A and B as the point of inversion but the equations quickly grew unwieldy. Is there a good way to tackle this problem that arrives at the above formula?
|
Got the answer... Brutally... Took about 1 hour work and a page of letter size by hand. It at least seemed to me that using inversion did not make things simpler... But maybe someone can work this out?
Anyway this is what I did. Suppose that $a < 9b$. (Need to check what should be corrected if $a > 9b$).
Step 1: Compute $l$.
See there is a right triangle. By the Pythagorean theorem,
$$
l^2 = (a+b)^2 - (a-b)^2 \quad \Rightarrow \quad l = 2\sqrt{ab}.
$$
Step 2: Compute $d$. This is the bloody part.
If the radius of the giant circle is $R$, the two segments connecting the centers should be of length $R-a$ and $R-b$, respectively. Use two Pythagorean theorem to calculate their length gives the equation
$$
a - b = \sqrt{(d+b)^2 + (a+2\sqrt{ab})^2} - \sqrt{(d+a)^2 + a^2}.
$$
Typically this involves squaring the equation twice. First, move the terms:
$$
\sqrt{(d+a)^2 + a^2} + a - b = \sqrt{(d+b)^2 + (a+2\sqrt{ab})^2}.
$$
Take square on both sides. Simplify it, YOLO!
$$
-a(\sqrt a - 3\sqrt b) - (\sqrt a - \sqrt b)d = (\sqrt a - \sqrt b)\left(\sqrt{(d+a)^2 + a^2}\right)
$$
Take square another time. Observe that $d^2$ term cancelled and it is indeed a linear equation. Thank god...
$$
\begin{aligned}
2(\sqrt a - \sqrt b)d(2\sqrt b) &= a(-a - 2\sqrt{ab} + 7b)\\
d &= \frac{a(-a - 2\sqrt{ab} + 7b)}{4\sqrt b (\sqrt a - \sqrt b)}.
\end{aligned}
$$
Step 3: Compute $R$.
Well there is nothing interesting... $R = \sqrt{(d+a)^2 + a^2} + a$. One thing worth mentioning is that the formula $(m^2 - n^2)^2 + (2mn)^2 = (m^2+n^2)^2$. Here
$$
d+a = \frac{-a(\sqrt a - 3\sqrt b)(\sqrt a + \sqrt b)}{4\sqrt b(\sqrt a - \sqrt b)}.
$$
Using $m = \sqrt a - \sqrt b$, $n = 2\sqrt b$, we get
$$
(d+a)^2 + a^2 = \left(\frac{a}{4\sqrt b(\sqrt a - \sqrt b)}\right)^2(a - 2\sqrt{ab} + 5b)^2.
$$
Therefore,
$$
R = \sqrt{(d+a)^2 + a^2} + a = \frac{a(\sqrt a + \sqrt b)^2}{4\sqrt b(\sqrt a - \sqrt b)}.
$$
Step 4: Compute $c$.
Using the Pythagorean theorem on the right triangle shown above, we have
$$
(a+c)^2 = a^2 + (R-a-d-c)^2.
$$
As
$$
R-a-d = \frac{a(\sqrt a + \sqrt b)}{2\sqrt b},
$$
we get the equation
$$
2ac + c^2 = \left(\frac{a(\sqrt a + \sqrt b)}{2\sqrt b} - c\right)^2.
$$
Finally,
$$
c = \frac{a(\sqrt a + \sqrt b)^2}{4\sqrt b(\sqrt a + 3\sqrt b)}.
$$
|
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|
How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
$$\bigg(\frac{x}{8\sqrt{x^2+8}}\bigg)'=\frac{1}{8}\cdot\frac{\sqrt{x^2+8}-\frac{x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{\frac{x^2+8-x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{8}{\sqrt{(x^2+8)^3}}=\frac{1}{\sqrt{(x^2+8)^3}}$$
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?
|
As mentioned in another answer, the solution to the given integral can be found by making a substitution with a trigonometric function, then integrating, calculating, substituting back and simplifying.
For a general integral of the form
$I =\displaystyle \int \frac{1}{\left(a x^n+b\right)^p} \, dx$ ,
the solution can be obtained with the help of Mathematica in terms of a special function:
$\displaystyle I = x \left(a x^n+b\right)^{-p} \left(\frac{a x^n}{b}+1\right)^p \, _2F_1\left(\frac{1}{n},p;1+\frac{1}{n};-\frac{a x^n}{b}\right)+C$,
where $\, _2F_1(a,b;c;z)$ is the Gaussian hypergeometric function.
The integral in this question is the special case where
$a =1, b = 8, n = 2,p =3/2$
A function can be defined with Mathematica:
solh[a_, b_, n_,
p_] = (x*(1 + (a*x^n)/b)^p*
Hypergeometric2F1[1/n, p, 1 + 1/n, -((a*x^n)/b)])/
(b + a*x^n)^p;
Then taking the special values for the integral in this question:
FullSimplify[solh[1, 8, 2, 3/2]]
It can be verified that the solution is
$\displaystyle \frac{x}{8\sqrt{x^2+8}} + constant$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I find the singular solution of the differential equation $y' = \frac{y^2 + 1}{xy + y}$? I start out with the separable differential equation,
$$y' =\frac{dy}{dx} = \frac{y^2 + 1}{xy + y} = \frac{y^2 + 1}{y(x+1)}$$
Thus, $\frac{1 }{x+1}dx = \frac{y }{y^2 + 1}dy$.
Then integrating both sides of the equation, I get
$$\ln(x+1) = \frac{1}{2}\ln(y^2 +1) + C$$
Now, $e^{\ln(x+1)}$ = $e^{\frac{1}{2}\ln(y^2 +1) + C}$. So...
$$(x+1) = e^C(y^2 + 1)^{\frac{1}{2}}$$
I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.
|
I believe that $x + 1 = \pm e^C\sqrt{y^2 + 1}$ is the correct general solution. You can rewrite it as: $x + 1 = C\sqrt{y^2 + 1}$, where $C = \pm e^c \neq 0$.
As for your concern about the singular solution, I don't think the original equation has $x = -1$ as the singular solution because $x = -1$ makes the original equation undefined.
If only the original equation is initially given in the form:
\begin{align}y'(x + xy) = y^2 + 1\end{align}
then the solution $x = -1$ does not make the original equation undefined and therefore is singular.
Apologize for bad Engish.
|
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|
Maximum value of $5\sin x - 12 \cos x + 1$ Given that $5\sin x - 12\cos x = 13\sin (x-67.4)$
Find the maximum value of $5\sin x - 12 \cos x + 1 $ and the corresponding value of x from 0 to 360.
Maximum value = $13+1=14$
Corresponding value of $x$
$13\sin (x-67.4) + 1 = 14$
$\sin(x-67.4) = 1 $
$x = 157.4 , 337.4 $
I found the value of $x$ and there’s 2 values. However , the answer is only$157.4$ why is that the case ?
|
Note that we have the maximum value for $x-67.4°=90°$ indeed for $x-67.4°=270°$ we have that $\sin=-1$ is minimum.
|
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"url": "https://math.stackexchange.com/questions/2657921",
"timestamp": "2023-03-29T00:00:00",
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|
How to extract $(x+y+z)$ or $xyz$ from the determinant Prove
$$\color{blue}{
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3}
$$
using elementary operations and the properties of the determinants without expanding.
My Attempt
$$
\Delta\stackrel{C_1\rightarrow C_1+C_2+C_3}{=}\begin{vmatrix}
xy+y^2+yz+zx+yz+z^2&xy&zx\\
x^2+xy+xz+xz+yz+z^2&(x+z)^2&yz\\
x^2+xy+xz+xy+y^2+yz&yz&(x+y)^2\\
\end{vmatrix}\\
=\begin{vmatrix}
y(x+y+z)+z(x+y+z)&xy&zx\\
x(x+y+z)+z(x+y+z)&(x+z)^2&yz\\
x(x+y+z)+y(x+y+z)&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)\begin{vmatrix}
y+z&xy&zx\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
\stackrel{R_1\rightarrow R_1+R_2+R_3}{=} \\(x+y+z)\begin{vmatrix}
2(x+y+z)&x^2+xy+xz+xz+yz+z^2&x^2+xy+zx+xy+y^2+yz\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)^2\begin{vmatrix}
2&x+z&x+y\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2
\end{vmatrix}
\stackrel{R_2\rightarrow R_2-(x+z)R_1|R_2\rightarrow R_3-(x+y)R_1}{=}
2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
$$
How do I extract $x+y+z$ or $xyz$ from the determinant to find the solution ?
Note: In a similar problem How to solve this determinant there seems to be solutions talking about factor theorem and polynomials. I am basically looking for extracting the terms $(x+y+z)$ or $xyz$ from the given determinant to solve it only using the basic properties of determinants.
|
Let $s=x+y+z$, then
$$
\begin{align}
&\det\begin{bmatrix}
(y+z)^2&xy&zx\\
xy&(z+x)^2&yz\\
zx&yz&(x+y)^2
\end{bmatrix}\\[9pt]
&=\det\begin{bmatrix}
s(y+z)&xy&zx\\
s(z+x)&(z+x)^2&yz\\
s(x+y)&yz&(x+y)^2
\end{bmatrix}\tag1\\[9pt]
&=\det\begin{bmatrix}
2s^2&s(z+x)&s(x+y)\\
s(z+x)&(z+x)^2&yz\\
s(x+y)&yz&(x+y)^2
\end{bmatrix}\tag2\\[9pt]
&=s^2\det\begin{bmatrix}
2&z+x&x+y\\
z+x&(z+x)^2&yz\\
x+y&yz&(x+y)^2
\end{bmatrix}\tag3\\[9pt]
&=s^2\det\begin{bmatrix}
2&-(z+x)&-(x+y)\\
z+x&0&yz-(x+y)(z+x)\\
x+y&yz-(x+y)(z+x)&0
\end{bmatrix}\tag4\\[9pt]
&=s^2\det\begin{bmatrix}
2&-(z+x)&-(x+y)\\
z+x&0&-sx\\
x+y&-sx&0
\end{bmatrix}\tag5\\[9pt]
&=s^3\det\begin{bmatrix}
2s&z+x&x+y\\
z+x&0&x\\
x+y&x&0
\end{bmatrix}\tag6\\[9pt]
&=s^3\det\begin{bmatrix}
y+z&z+x&x+y\\
z&0&x\\
y&x&0
\end{bmatrix}\tag7\\[9pt]
&=s^3\det\begin{bmatrix}
0&z&y\\
z&0&x\\
y&x&0
\end{bmatrix}\tag8\\[18pt]
&=2xyz(x+y+z)^3\tag9
\end{align}
$$
Explanation:
$(1)$: add columns $2$ and $3$ to column $1$
$(2)$: add rows $2$ and $3$ to row $1$
$(3)$: factor $s$ out of row $1$ and then out of column $1$
$(4)$: subtract $z+x$ times column $1$ from column $2$
$\phantom{(4)\text{:}}$ subtract $x+y$ times column $1$ from column $3$
$(5)$: $yz-(x+y)(z+x)=-sx$
$(6)$: factor $-s$ out of columns $2$ and $3$
$\phantom{(6)\text{:}}$ distribute one factor of $s$ over row $1$
$(7)$: subtract columns $2$ and $3$ from column $1$
$(8)$: subtract rows $2$ and $3$ from row $1$
$(9)$: the determinant is now simple to compute
|
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|
Probability of at least 3 identical numbers from an array I have an array {1, 2, 3, 4, 5} and I extract 4 random numbers. I want to find out the probability that at least 3 numbers are identical.
I concluded that it's the probability of having exactly 3 identical numbers added to the probability of having exactly 4 identical numbers.
The probability of having exactly 3 numbers is: $$\binom{4}{3}*p^{3}*(1-p)^{4-3}$$
Now, my issue is, who is p? As far as I understand, p is: $$\frac{1}{5}$$
Since it represents the probability of extracting any random number, but I'm not sure.
|
You calculated the probability of getting at least three of a $particular$ number. However, we can get at least three of $any$ of the five numbers.
Let $X$ denote the number of identical numbers. First we compute $P(X=3)$.
We must get three of one number and one of another. First choose one of the five numbers to be obtained three times. Then denote $p=\frac{1}{5}$. Thus we have
$$\begin{align*}
P(X=3)
&= {{5 \choose 1} \cdot {4 \choose 3} \cdot\frac{1}{5}^3 \cdot\frac{4}{5}}\\\\
&=0.128
\end{align*}$$
Next compute $P(X=4)$. Choose one number of the five to be selected all four times. We have
$$\begin{align*}
P(X=4)
&= {5 \choose 1} \cdot {4 \choose 4} \cdot\frac{1}{5}^4\\\\
&=0.008
\end{align*}$$
Finally,
$$P(X=3)+P(X=4)=0.128+0.008=0.136$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int-\sin x\mathrm e^{-\sin x}$ I want to integrate the function $$-\sin x\mathrm{e}^{-\sin x}.$$
I tried integration by parts with $u=-\sin x$ but I got another difficult integration which is $$\int (\cos x)^{2}\mathrm{e}^{-\sin x} \,\mathrm{d}x.$$
Could someone help me out?
|
$-\int\mathrm e^{-\sin x}\sin x~dx=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx$
For $n$ is any non-negative integer,
$\int\sin^{2n+2}x~dx=\dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}-\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$
This result can be done by successive integration by parts.
$\int\sin^{2n+1}x~dx$
$=-\int\sin^{2n}x~d(\cos x)$
$=-\int(1-\cos^2x)^n~d(\cos x)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$
$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n+2)!x}{4^{n+1}(2n+1)!((n+1)!)^2}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}(2n+1)!((n+1)!)^2(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{2(n+1)x}{4^{n+1}((n+1)!)^2}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2(n+1)(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x}{2^{2n+1}n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\sin^{2k+1}x\cos x}{2^{2n-2k+1}n!(n+1)!(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}+C$
|
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|
Verify the Limit: $\lim_{x\rightarrow 9} (\sqrt{x} +9) = 12$ I am using the definition of the limit of a function. So far I have:
$$|f(x)-L|<\epsilon \iff |(\sqrt{x}+9)-12|<\epsilon \iff |\sqrt{x}-3|<\epsilon.$$
This is where I get stuck. Any help is appreciated.
|
Keep in mind that in deriving these proofs we are always working backwards. We start with $\epsilon$ and try to find $\delta$ that will make it true.
Let $\epsilon > 1$. Since $\epsilon $ may be as small as we like we may as well assume $1 > \epsilon > 0$. I'm saying this now, because later we will have a reason for this.
So we want to find a $\delta$ (expressed in terms of $\epsilon$) so that whenever $|x - 9|< \delta$ we will have to have $|\sqrt{x} - 3|< \epsilon$.
We can find this $\delta$ one of two ways:
Forward:
$|x - 9|<\delta \implies$
$-\delta < x - 9 < \delta \implies $
$9 -\delta < x < 9+\delta \implies $
$\sqrt{9-\delta} < x <\sqrt{9+\delta}\implies$
Now if we can somehow find some way that $3 - \epsilon \le \sqrt{9-\delta}$ and then $\sqrt{9+\delta} \le 3+ \epsilon$ we'll be on our way.
So we want $3-\epsilon \le \sqrt{9 -\delta}$ so $9 - 6\epsilon + \epsilon^2 \le 9- \delta$. So we need $\delta \le 6\epsilon - \epsilon^2$.
And we want $\sqrt {9-\delta} \le 3 + \epsilon$ so $9-\delta \le 9 + 6\epsilon +\epsilon^2$. So we need $-6\epsilon -\epsilon^2 \le \delta$. Well, that's true for any positive $\delta$.
How can we find a $\delta$ where both will be true. Well, if we assume $0 < \epsilon < 1$, then $\epsilon^2 < \epsilon$ so if we can define $\delta$ so that $\delta \le 5\epsilon < 6\epsilon - \epsilon^2$ we'd be done.
So define $\delta = 5\epsilon$. If $|x - 9| < \delta$ then $-5\epsilon < x-9 < 5\epsilon$ and $9-6\epsilon + \epsilon^2 < 9-5\epsilon < x < 9+5\delta < 9 + 6\delta + \delta^2$. So $ 3-\epsilon < \sqrt {9-5\epsilon} < \sqrt x< \sqrt{9+5\delta} < 3 + \epsilon$. So $-\epsilon < \sqrt x < \epsilon$ and $|\sqrt x - 3| < \epsilon$.
The proof is done.
Or backwards:
$|\sqrt{x} - 3| <\epsilon $ is implied by
$-\epsilon < \sqrt{x} - 3 <\epsilon \Leftarrow$
$3-\epsilon < \sqrt{x} < 3 + \epsilon\Leftarrow$
$9-6\epsilon +\epsilon^2 < x < 9 + 6\epsilon + \epsilon^2 \Leftarrow$
$-6\epsilon +\epsilon^2 < x -9 < 6\epsilon + \epsilon^2 \Leftarrow$
$-5\epsilon < -6\epsilon +\epsilon^2 < x-9 < 5\epsilon < 6\epsilon + \epsilon^2$.
Note: $\Leftarrow$ means is implied by. We are going in the opposite direction then we are used to.
Normally we say $x -9 < 5\epsilon \implies x-9 < 5\epsilon < 6\epsilon + \epsilon^2\implies x< 6\epsilon + \epsilon^2$ because we are saying: if A then B.
But $\Leftarrow$ means: A can come about from B. Or: if B then A. $x-9 < 5\epsilon \Leftarrow x-9 < 6\epsilon + \epsilon^2$.
$\Leftarrow |x-9|< 5\epsilon$.
S if we let $\delta = 5\epsilon$ we are done.
So define $\delta = 5\epsilon$. If $|x - 9| < \delta$ then $-5\epsilon < x-9 < 5\epsilon$ and $9-6\epsilon + \epsilon^2 < 9-5\epsilon < x < 9+5\delta < 9 + 6\delta + \delta^2$. So $ 3-\epsilon < \sqrt {9-5\epsilon} < \sqrt x< \sqrt{9+5\delta} < 3 + \epsilon$. So $-\epsilon < \sqrt x < \epsilon$ and $|\sqrt x - 3| < \epsilon$.
The proof is done.
|
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|
Calculating the summation $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$ I need to find explicitly the following summation
$$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}, \quad
H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
From Mathematica, I checked that the answer is $2$. The same result is returned by WolframAlpha.
A thought from afar:
$$
\begin{align}
\frac{H_{n+1}}{n(n+1)}
&=H_{n+1}\left(\frac1n-\frac1{n+1}\right)\tag1\\
&=\color{#C00}{H_{n+1}}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag2\\
&=\color{#C00}{\frac1{n+1}}\color{#090}{\frac1n}+\color{#C00}{H_n}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag3\\
&=\frac1n-\frac1{n+1}+\frac{H_n}{n}-\frac{H_{n+1}}{n+1}\tag4\\
&=\frac{H_n+1}{n}-\frac{H_{n+1}+1}{n+1}\tag5
\end{align}
$$
What was done:
$(1)$: partial fractions
$(2)$: used $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(3)$: used $\color{#C00}{H_{n+1}=\frac1{n+1}+H_n}$ and $\color{#090}{\frac1n=H_n-H_{n-1}}$
$(4)$: partial fractions and $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(5)$: combined terms
Now this looks like Clement C's hint.
|
Noting that
$$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1},$$
the sum may be rewritten as
$$\sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} = \sum_{n = 1}^\infty \left (\frac{H_{n + 1}}{n} - \frac{H_{n + 1}}{n + 1} \right ).\tag1 \label1$$
Now as the harmonic numbers satisfy the recurrence relation
$$H_{n + 1} = H_n + \frac{1}{n + 1},$$
we have
\begin{align}
\sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} &= \sum_{n = 1}^\infty \left [\frac{H_n}{n(n+1)} + \frac{1}{n(n +1)} - \frac{1}{(n + 1)^2} \right ]\\
&= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n + 1)} - \underbrace{\sum_{n = 1}^\infty \frac{1}{(n + 1)^2}}_{n \, \mapsto n - 1}\\
&= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n+1)} - \sum_{n = 1}^\infty \frac{1}{n^2} + 1\tag2\\
&= \zeta (2) + 1 - \zeta (2) + 1\\
&= 2
\end{align}
In (2) an evaluation for the first sum can be found here while the second sum telescopes and has a sum equal to one.
|
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|
real value of $k$ inirrational equation Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.
solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$
$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$
$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$
$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$
$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$
Help me how i solve it after that point
|
Since $z\geq3$, we obtain:
$$\sqrt{3z+3}=\sqrt{2z+k}+\sqrt{3z-9}=\sqrt{5z+k-9+2\sqrt{(2z+k)(3z-9)}}=$$
$$=\sqrt{3z+3+k-6+2(z-3)+2\sqrt{(2z+k)(3z-9)}}\geq\sqrt{3z+3+k-6},$$
which gives that for $k>6$ our equation has no roots.
|
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|
Creating 6 Digit Numbers Whose Adjacent Digits are Relatively Prime Six-digit integers will be written using each of the digits $1$ through $6$ exactly once per six-digit integer. How many different positive integers can be written such that all pairs of consecutive digits of each integer are relatively prime? (Note: $1$ is relatively prime to all integers.)
The pairs of numbers that are relatively prime from $1-6$ are: $1:2,3,4,5,6$ and
$2:3,5$ and $3,4,5$ and $4,5$ and $5,6$. Now, what should I do from here.
|
If $6$ is an interior entry then it has to appear in one of the subwords $165$ or $561$. If all three of $\{2,3,4\}$ are on the same side of these the digit $3$ has to be in the middle of the three (gives $4$ ways). If one of $\{2,3,4\}$ stands alone it has to be one of $2$ and $4$ (gives another $8$ ways). In all there are $24$ admissible words of this kind.
If $6$ is the first or last entry the entry next to $6$ has to be $1$ or $5$. The remaining four digits could be arranged in $4!=24$ ways, but we have to exclude the $2\cdot 3!=12$ ways containing $24$ or $42$ as a subword. It follows that there are $2\cdot 2\cdot(24-12)=48$ admissible words of this kind.
The total number of admissible words therefore is $72$.
|
{
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Proving that if $x>0$ then $\frac{x-(x^2+1)\arctan(x)}{x^2(x^2+1)}$ is less than $0$ How do I show that for $x>0$:
$$\frac{x-(x^2+1)\arctan(x)}{x^2(x^2+1)} < 0$$
I tried to do it somehow using the fact that $$\frac{\arctan(x)}{x} < 1$$ but still didn't figure it out...
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We have $$\begin{align}\frac{x-(x^2+1)\arctan x}{x^2(x^2+1)} < 0&\iff x<(x^2+1)\arctan x\\&\iff (1+x^2)\arctan x-x>0\end{align}$$ This is true iff $$\frac d{dx}\left[(1+x^2)\arctan x-x\right]=2x\arctan x+\frac{1+x^2}{1+x^2}-1=2x\arctan x>0$$ which is true since both $x$ and $\arctan x$ are greater than $0$ when $x>0$. Hence result.
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Roots of $f(x)=x^3-x^2-2x+1$ We can prove using a monotony study that the function $f(x)=x^3-x^2-2x+1$ has three real roots. However, when I solve the equation $f(x)=0$ using Mathematica, I get
$$x_1=\frac{1}{3}+\frac{7^{2/3}}{3 \left(\frac{1}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}}+\frac{1}{3} \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$
$$x_2=\frac{1}{3}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i \sqrt{3}\right)}{3 \left(-1+3 i \sqrt{3}\right)^{1/3}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$
$$x_3=\frac{1}{3}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1-i \sqrt{3}\right)}{3 \left(-1+3 i \sqrt{3}\right)^{1/3}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$
Since those roots are real, then maybe there's a way to write them without using the complex number $i$?
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This is Example 9.2.2 in the first edition of Galois Theory by David A. Cox. He gives the polynomial, on page 239, as $y^3 + y^2 - 2 y -1.$
Easy enough to confirm the next bit with trig identities for $\cos 2t$ and $\cos 3t$ in terms of $\cos t \; .$ We have $\cos 2t = 2 \cos^2 t - 1$ and $\cos 3t = 4 \cos^3 t - 3 \cos t.$ The easiest proof is what Gauss intended, take a nontrivial seventh root of unity $\omega,$ so that $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0.$ Then take $x = \omega + \frac{1}{\omega},$ calculate $x^3 + x^2 - 2x - 1,$ and confirm that $(x^3 + x^2 - 2x - 1) \omega^3 = \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1,$ which is then zero.
The roots of $x^3 + x^2 - 2 x -1$ are $$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; 2 \cos \left( \frac{4 \pi}{7} \right) \; , \; 2 \cos \left( \frac{6 \pi}{7} \right) \; . \; $$ Yours are negative of these, therefore
$$ 2 \cos \left( \frac{ \pi}{7} \right) \; , \; 2 \cos \left( \frac{3 \pi}{7} \right) \; , \; 2 \cos \left( \frac{5 \pi}{7} \right) \; . \; $$
from Cox:
From Reuschle (1875)
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Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$
I have tried, but I don't get anything. Can you please give me a hint?
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This method of infinite ascent is probably wrong.. but...
Suppose none of $a = -b, b = -c, c = -a$ is true.
Then multiplying both sides by $abc$ and inverting we get:
$\frac{a+b+c}{abc} = \frac{1}{ab + bc + ca} \implies \frac{1}{ab + bc + ca}= \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$
We can now write $x = ab, y = bc, z = ca$ and do this all over again which is probably absurd (needs better logic here).
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Calculating a multivariable limit. Find if it exists the limit : $$\lim_{(x,y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$$
I've tried the following :
1st attempt : Using polar coordinates:
Set $x= r\cos\theta$ and $y=r\sin\theta$
$$\frac{x^4y^4}{(x^2+y^4)^3}=\frac{r^8\cos^4\theta \sin^4\theta}{(r^2\cos^2\theta+r^4\sin^4\theta)^3}=\frac{r^2\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$
$$\lim_{r\to0}\frac{r^2\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}=\frac{0}{\cos^6\theta}$$
Now the limit mentioned above would be equal to zero if and only if the denominator is different than zero.
Hence we need to calculate the limit in the case where $\theta = \frac{\pi}{2}+k\pi$ and compare it with the precalculated limit.
However I was not able to get rid of the indeterminate form.
2nd attempt : Choosing a specific path $y = ax$.
$$\frac{x^4y^4}{(x^2+y^4)^3}=\frac{a^4x^2}{(1+a^4x^2)^3}$$
$$\lim_{x\to0}\frac{a^4x^2}{(1+a^4x^2)^3}=0.$$No conclusion about the limit.
If anyone could give me hints or point me in the right direction I would be grateful.
Thanks in advance.
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Approaching along the path $x = y^2$ (motivated by trying to simplify $x^2 + y^4$, we have
$$\lim_{y \to 0} \frac{(y^2)^4 y^4}{((y^2)^2 + y^4)^3} = \lim_{y \to 0} \frac{y^{12}}{(2y^4)^3} = \frac 1 8 \ne 0.$$
Hence, the limit does not exist.
Alternative interpretation in polar coordinates: $\cos \theta \to 0$ in such a way that the denominator vanishes at the same order that the numerator does, and so the limit is non-zero.
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|
Angle between diagonals of an irregular pentagon
In pentagon $ABCDE$, $DC$ is parallel to $BE$ and $BC$ is parallel to $AE$. $△DCE$ and $△BCE$ are both isoscles with $$∠CBE= ∠DEC = α,\ ∠EAB= α+\frac{π}{2}.$$ Suppose $AD$ meets $BE$ at $G$. Find $∠BGA$.
I have tried producing $AD$ to meet $BC$ at $F$ and showing that $∠DFC$ must be $α$, but I haven't been able to show definitively a value for $∠BGA$, which I think is $2α$. I have also tried producing lines from $D$ to $BC$ and from $A$ to $BC$ both inclined at an angle $α$ to $BC$, but I haven't been able to show that they are the same line.
Any hints about how to proceed would be much appreciated!
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Let $DC=ED=a$.
Hence, $$EC=CB=2a\cos\alpha$$ and
$$EB=2EC\cos\alpha=4a\cos^2\alpha.$$
Thus, by law of sines for $\Delta EAB$ we obtain:
$$\frac{EA}{\sin\left(90^{\circ}-2\alpha\right)}=\frac{4a\cos^2\alpha}{\sin\left(90^{\circ}+\alpha\right)},$$
which gives $$EA=4a\cos\alpha\cos2\alpha.$$
Now, since $\alpha<60^{\circ}$, by law of cosines for $\Delta DEA$ we obtain:
$$DA=\sqrt{a^2+16a^2\cos^2\alpha\cos^22\alpha-8a^2\cos\alpha\cos2\alpha\cos3\alpha}=$$
$$=a\sqrt{1+8\cos\alpha\cos2\alpha(2\cos\alpha\cos2\alpha-\cos3\alpha)}=a\sqrt{1+8\cos^2\alpha\cos2\alpha}=$$
$$=a\sqrt{1+4(1+\cos2\alpha)\cos2\alpha}=a\sqrt{(1+2\cos2\alpha)^2}=a(1+2\cos2\alpha).$$
Thus, by law of cosines again we obtain:
$$\cos\measuredangle DAE=\frac{AE^2+AD^2-DE^2}{2AE\cdot AD}=\frac{16a^2\cos^2\alpha\cos^22\alpha+a^2(1+2\cos2\alpha)^2-a^2}{2\cdot4a\cos\alpha\cos2\alpha\cdot a(1+2\cos2\alpha)}=$$
$$=\frac{16\cos^2\alpha\cos^22\alpha+4\cos^22\alpha+4\cos2\alpha}{8\cos\alpha\cos2\alpha(1+2\cos2\alpha)}=\frac{4\cos^2\alpha\cos2\alpha+\cos2\alpha+1}{2\cos\alpha(1+2\cos2\alpha)}=$$
$$=\frac{4\cos^2\alpha\cos2\alpha+2\cos^2\alpha}{2\cos\alpha(1+2\cos2\alpha)}=\cos\alpha,$$
which says that indeed, $$\measuredangle BGA=2\alpha.$$
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Surd inside Surd equality I was trying a problem where I got the following surd as my answer:
$$ {\sqrt{6 - 2 \sqrt{5} }\over 4} \approx 0.309016.... $$
The answer listed was:
$$ {\sqrt{5} - 1 \over 4} \approx 0.309016.... $$
Is there a simple way that you could get to the second expression from the first?
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Suppose you assume that you can write $\sqrt{6-2\sqrt5} = \sqrt a - \sqrt b$ for some rational numbers $a$ and $b$. Squaring both sides will give you $6-2\sqrt5 = a+b-2\sqrt{ab}$, and so
\begin{align*}
6 &= a+b \\
\sqrt{5} &= \sqrt{ab}
\end{align*}
So we want $ab=5$ and $a+b=6$. An obvious solution is, indeed, $a=5$ and $b=1$. (Taking $a=1$ and $b=5$ gives a negative result for $\sqrt a-\sqrt b$.)
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Greatest Common Divisor of two Polynomials. Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$. Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$
I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$
Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$. Now to find $a(x)$ and $b(x)$ I have no clue how to do that. I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!
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You have the complete extended Euclid algorithm (which IMHO is the best way) given already in Will Jaggy's post.
Given that you already extracted $x-3$ as GCD, here is a perhaps quicker way (which may not generalise much). We simplify a bit by noting we seek linear $a(x), b(x)$, s.t.:
$a(x)\color{blue}{(x^2-3x+5)} + b(x) \color{blue}{(x-3)(x-2)}=1$. Setting $x=2, 3$ we get $a(x) = \frac1{15}(-2x+9)$.
Thus we are left with
$\tfrac1{15}(-2x+9)\color{blue}{(x^2-3x+5)} + (px+q)\color{blue}{ (x-3)(x-2)}=1$
Comparing coeffs of the cubic term and the constant, $\implies p=\frac2{15},\; 3+6q = 1\implies q = -\frac13$.
So
$$\tfrac1{15}(-2x+9)\color{blue}{(x^2-3x+5)} + \tfrac1{15}(2x-5)\color{blue}{(x-3)(x-2)}=1$$
Multiply this by $\color{blue}{x-3}$ if you want to see the equation in form $a(x) p_1(x) + b(x) p_2(x) = \gcd(p_1, p_2)$.
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Finding $\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ I am having some diffuculty finding the limit for this expression and would appreciate if anyone could give a hint, as to how to continue. I know the limit must be $e^{14}$ (trough an engine) and I can show it for
$(1+\frac{2}{x}+\frac{1}{x^2})^{7x}$ like
$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{1}{x^2})^{7x} = ((1+\frac{1}{x})^2)^{7x} = (1+\frac{1}{x})^{x \cdot 7 \cdot 2} = e^{7 \cdot 2} = e^{14}$
However the $\frac{3}{x^2}$ in $(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ is causing problems for me I have:
$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x} = (1+\frac{2}{x}+\frac{1}{x^2} +\frac{2}{x^2})^{7x} = ((1+\frac{1}{x})^2+\frac{2}{x^2})^{7x} =...$
but I'm not sure how to continue (how to get rid of $\frac{2}{x^2}$) I was thinking that using the binomial theorem might somehow reduce the$\frac{2}{x^2}$ for $\lim_{x\to \infty}$, but I'm not sure how to do that for an x in the exponent, or if it is useful, or if it is even allowed in this case.
If anyone could point me in the right direction, as to how to continue I would be very grateful.
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It can be convenient to use Taylor series when doing problems like this:
$$\lim_{x \to \infty} \left(1 + \frac{2}{x} + \frac{3}{x^2}\right)^{7x} = \lim_{x \to \infty} e^{7x\ln(1+\frac{2}{x}+\frac{3}{x^2})} = \lim_{x\to \infty} e^{7x\left( \frac{2}{x} + \frac{3}{x^2} + \mathcal{O}\left(\left((\frac{2}{x} + \frac{3}{x^2}\right)^2\right) \right)} = \lim_{x\to \infty} e^{14 + \frac{21}{x}} = e^{14}$$
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If $\sum\limits_{k=1}^n\frac1{x_k+y_k}\leqslant\frac n2$, prove that $\sum\limits_{k=1}^n(\frac{x_k}{1+y_k}+\frac{y_k}{1+x_k})\geqslant n$
Let $n \in \mathbb{N}, n\geqslant 2$ and $x_1,x_2,\cdots,x_n,y_1,y_2,\cdots,y_n>0$ with $$\sum_{k=1}^{n}{ 1 \over{x_k+y_k} } \leqslant {n\over 2}.$$
Show that $$\sum_{k=1}^{n}\left({ x_k \over{1+y_k} }+{y_k\over{ 1+x_k}}\right) \geqslant n.$$
I have tried a lot of ways, but I cannot find a good solution. Does anyone have an idea?
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For each $k$,\begin{align*}
\frac{x_k}{1 + y_k} + \frac{y_k}{1 + x_k} &= \frac{(x_k + y_k + 1)(x_k + y_k + 2)}{(1 + x_k)(1 + y_k)} - 2\\
&\geqslant \frac{(x_k + y_k + 1)(x_k + y_k + 2)}{\dfrac{1}{4}((1 + x_k) + (1 + y_k))^2} - 2 = \frac{2(x_k + y_k)}{x_k + y_k + 2},
\end{align*}
thus$$
\sum_{k = 1}^n \left( \frac{x_k}{1 + y_k} + \frac{y_k}{1 + x_k} \right) \geqslant \sum_{k = 1}^n \frac{2(x_k + y_k)}{x_k + y_k + 2}.
$$
By Cauchy's inequality,\begin{align*}
&\mathrel{\phantom{=}} \left( \sum_{k = 1}^n \left( 1 + \frac{2}{x_k + y_k} \right) \right) \left( \sum_{k = 1}^n \frac{2(x_k + y_k)}{x_k + y_k + 2} \right)\\
&= \left( \sum_{k = 1}^n \left( 1 + \frac{2}{x_k + y_k} \right) \right) \left( \sum_{k = 1}^n \frac{2}{1 + \frac{2}{x_k + y_k}} \right) \geqslant \left( \sum_{k = 1}^n \sqrt{2} \right)^2 = 2n^2,
\end{align*}
and by condition,$$
\sum_{k = 1}^n \left( 1 + \frac{2}{x_k + y_k} \right) = n + \sum_{k = 1}^n \frac{2}{x_k + y_k} \leqslant 2n,
$$
thus$$
\sum_{k = 1}^n \left( \frac{x_k}{1 + y_k} + \frac{y_k}{1 + x_k} \right) \geqslant \sum_{k = 1}^n \frac{2(x_k + y_k)}{x_k + y_k + 2} \geqslant n.
$$
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What are the maximum and minimum values of $4x + y^2$ subject to $2x^2 + y^2 = 4$? $$ 2x^2 + y^2 = 4 $$
$$ Y = \sqrt{4-2x^2} $$
$$4x + y = 2x^2 + \sqrt{4-2x^2}$$
Find the derivative of $$ 2x^2 + \sqrt{4-2x^2} $$ set as = 0
$$X^2 = 64/33$$
$$ F(64/33) = 34\sqrt{33}/33 $$
How to solve it the right way?
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you will get $$4x+y^2=4x+4-2x^2$$
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A first course in Complex Analysis with Applications Assume for the moment that $\sqrt{1+i}$. makes sense in the complex number system.
How would you then demonstrate the validity of the equality
$$\sqrt{1+i} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i\sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}$$?
Source: Dennis G. Zill & Patrick D. Shanahan
I'm Confused that what to do to solve this one or what the question asks for?
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In order to prove o$$\sqrt{1+i} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i\sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}$$
we square the right hand side and show that it is the same as $1+i.$
Note that $$\bigg (\sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i\sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}\bigg )^2 =$$
$$ \bigg( \frac{1}{2} +\frac{1}{2} \sqrt{2}\bigg) -\bigg( \frac{-1}{2} +\frac{1}{2} \sqrt{2}\bigg) + 2i \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} \sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}=$$
$$ 1 +2i(1/2)=1+i$$
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Simplifying $\frac{y}{y^2+b^2}$ where $y=b \cot\theta$ Could appreciate some help with this question.
I want to simplify the following trigonometric equation.
$$\frac{y}{y^2+b^2}$$
where $y=b \cot\theta$.
The solution I got was
$$\frac{1}b \cos\theta$$
Can someone verify and try and guide me through the solution?
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Hint:
$$b^2+b^2\cot^2(x)=b^2(1+\cot^2(x))=b^2\left(1+ \frac{\cos^2(x)}{\sin^2(x)}\right)=b^2\left(\frac{\sin^2(x)}{\sin^2(x)}+ \frac{\cos^2(x)}{\sin^2(x)}\right)$$
|
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$6$ digit numbers formed from the first six positive integers such that they are divisible by both $4$ and $3$. The digits $1$, $2$, $3$, $4$, $5$ and $6$ are written down in some order to form a six digit number. Then (a) how many such six digits number are even? and (b) how many such six digits number are divisible by $12$?
My attempt: "In some order" means that the six digits number we get do not have repeated digits. Then there are $6!$ possible digits, that is we can make $720$ six digits number. Then the number of even numbers must be end with one of $2$ or $4$ or $6$. Then the number of even numbers can be formed by the given digits are $3\times 5!=360$. Then it is almost done. Now for (b) we have to find the possible numbers of six digits which is divisible by $12$, that is the number must be divisible by both $3$ and $4$. I know that a number is divisible by $3$ if the sum of the numbers are divisible by $3$ and a number is divisible by $4$ if its last two digit is divisible by $4$. But how can I find that how many of such common numbers are there that are divisible by both three and four? Please help me to solve this.
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Since digits are not repeated, all $6$ digits must be used.
Sum of all digits is $21$ $(1+2+3+4+5+6)$. So this number is divisible by $3$.
Now we only have to check its divisibility by $4$.
Number should end with any of $8$ combination $(12,16,24,32,36,52,56,64)$.
Starting $4$ digits can be in any order.
So total count $= 8*4! = 192$
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Calculate $ \arctan(x_{1}) \cdot \arctan(x_{2}) $
Let $ x_{1} $ and $ x_{2} $ be the roots of the equation : $ x^2-2\sqrt{2}x+1=0 $
Calculate $ \arctan(x_{1}) \cdot \arctan(x_{2}) $
The answer should be $ \dfrac{3\pi^2}{64} $.
How does the fact that $ x_{1} $ = $ 1 + \sqrt2 $ = $ \dfrac{1}{x_{2}} $ help?
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We have $$x^2-2\sqrt{2}x+1=(x-(1+\sqrt2))(x-(-1+\sqrt2))=0$$ so let $x_1=1+\sqrt2$ and $x_2=-1+\sqrt2$.
Let $a=\tan^{-1}(1+\sqrt2)$. Using the double angle tangent formula, $$\begin{align}\tan2a=\frac{2\tan a}{1-\tan^2a}&\implies\tan2a=\frac{2(1+\sqrt2)}{1-(1+\sqrt2)^2}=-1\end{align}$$ and hence taking the principal angle gives $$\tan^{-1}(1+\sqrt2)=\frac38\pi$$
Let $b=\tan^{-1}(-1+\sqrt2)$. Using the double angle tangent formula, $$\begin{align}\tan2b=\frac{2\tan b}{1-\tan^2b}&\implies\tan2b=\frac{2(-1+\sqrt2)}{1-(-1+\sqrt2)^2}=1\end{align}$$ and hence taking the principal angle gives $$\tan^{-1}(-1+\sqrt2)=\frac18\pi$$ Therefore $$\boxed{\tan^{-1}(x_1)\cdot\tan^{-1}(x_2)=\frac38\pi\cdot\frac18\pi=\frac3{64}\pi^2}$$ as desired.
|
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|
Is $x^2 \equiv 1 \pmod{p^k} \iff x \equiv \pm 1 \pmod {p^k}$ for odd prime $p$? Problem: As the title says, is $x^2 \equiv 1 \pmod {p^k} \iff x \equiv \pm 1 \pmod {p^k}$ for odd prime $p$? If not, is it at least true that there are two solutions to the congruence equation?
Attempt: I am a beginner in number theory, and I am not sure if the statement in question is true. The following is my attempt. Assume $x^2 \equiv 1 \pmod {p^k}$. Then $p^k | (x-1)(x+1) \Rightarrow p^k|(x-1)$ or $p^k|(x+1)$, because if $p^k$ can divide both, then $p^k|2$ which is false. This shows the forward direction.
The converse is easy.
For any $y^2 \equiv 1 \pmod {p^k}$, $y^2 \equiv x^2 \pmod {p^k}$ which yields the conclusion before, so $\pm 1$ are the only solutions $\pmod {p^k}$.
|
Assume $x^2 \equiv 1 \pmod {p^k}$. Then $x^2-1 \equiv 0 \pmod {p^k}$, meaning $p^k | x^2-1=(x-1)(x+1)$. Since $p$ is prime, we can conclude that $p^k\vert(x-1)$ or $p^k\vert(x+1)$. Yet, $p^k\vert(x-1) \Rightarrow x-1 \equiv 0 \pmod {p^k} \Rightarrow x \equiv +1 \pmod {p^k}$ and $p^k\vert(x+1) \Rightarrow x+1 \equiv 0 \pmod {p^k} \Rightarrow x \equiv -1 \pmod {p^k}$.
Thus, there are two solutions: If $x^2 \equiv 1 \pmod {p^k}$, then, $x \equiv -1 \pmod {p^k}$ and $x \equiv +1 \pmod {p^k}$.
|
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|
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$ My question:
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$.
My work so far:
$a^2+160=b^2+5\Rightarrow (b-a)(a+b)=155=31\times 5$
$a^2+320=c^2+5\Rightarrow (c-a)(c+a)=315=5\times3^2\times 7$
And now, I'm stuck.
($a,b,c$ are a members of $\mathbb Z$ and are positive)
|
Hint:
So, $a+b> a-b$ and $a+b=\dfrac{31\cdot5}{a-b}$
Either $a+b=155, a-b=?$ or $a+b=31,a-b=?$
Check which values of $a$ keep $c$ integer
|
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|
Solve $\frac{x^2}{\left(a+\sqrt{a^2+x^2}\right)^2}+\frac{x^2(1+x^2)}{\left(a+\sqrt{a^2+x^2(1+x^2)} \right)^2}=1$ I have been trying to solve the following equation for $x$:
\begin{align}
\frac{x^2}{\left(a+\sqrt{a^2+x^2}\right)^2}+\frac{x^2(1+x^2)}{\left(a+\sqrt{a^2+x^2(1+x^2)} \right)^2}=1,
\end{align}
for some fixed $a \in (0,1/2]$.
This equation came from an analysis of an electrical circuit. The solution is the current.
However, after trying to solve it by hand using the software it doesn't appear that there is a good way of finding a solution.
My question: Can we at least give a good estimate of the position of the positive zero? For example, can show that the zero belong to the specific interval?
Here is an equaivalent polynomial:
\begin{align}
x^4 - 4a^4 - 4a^3(a^2 + x^2)^{1/2} + x^6 - 4a^3(x^2(x^2 + 1) + a^2)^{1/2} - 4a^2(x^2(x^2 + 1) + a^2)^{1/2}(a^2 + x^2)^{1/2}=0.
\end{align}
|
Hints:
Let $x = a \tan (2 A)$ and $x\sqrt{1 +x^2} = a \tan (2 B)$, then, due to the $\tan$- half-angle theorem, the equation in question transforms into
$$
\tan^2 (A) + \tan^2(B) = 1
$$
This can be used for numerical analysis, e.g. (with fixed $a$) a Banach type fixed point iteration $x^{(n)} \to B \to A \to x^{(n+1)} \to \dots $
Re-inserting $A$ and $B$ gives
$$
\tan^2 \left(\frac12 \arctan \frac{x}{a}\right) + \tan^2\left(\frac12 \arctan \frac{x\sqrt{1 +x^2}}{a}\right) = 1
$$
where one might check for further simplification.
EDIT (March 18,2018)
Matlab is able to plot the solutions by implicit plot:
A rough (underestimated) linearization is $x = 2.3 a$; near $a=0$ we have the pretty exact linearization $x \simeq 2.8 a$.
|
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|
Are there any "nonstandard" special angles for which trig functions yield radical expressions? Everyone learns about the two "special" right triangles at some point in their math education—the $45-45-90$ and $30-60-90$ triangles—for which we can calculate exact trig function outputs. But are there others?
To be specific, are there any values of $y$ and $x$ such that:
*
*$y=\sin(x)$;
*$x$ (in degrees) is not an integer multiple of $30$ or $45$;
*$x$ and $y$ can both be written as radical expressions? By radical expression, I mean any finite formula involving only integers, addition/subtraction, multiplication/division, and $n$th roots. [Note that I require $x$ also be a radical expression so that we can't simply say "$\arcsin(1/3)$" or something like that as a possible value of $x$, which would make the question trivial.]
If yes, are they all known and is there a straightforward way to generate them?
If no, what's the proof?
|
Yes, a $15-75-90$ triangle may be the one you want.
Assume we have a right $\Delta ABC$ with $\widehat{BAC}=15^0;\widehat{ABC}=90^0;\widehat{ACB}=75^0$.
Put an extra point $D$ like above so that $B,C,D$ are collinear and $AC$ is the angle bisector of $\widehat{DAB}$, this means $\widehat{DAB}=30^0;\widehat{BDA}=60^0$.
Let $DB=a$. Then the special right triangle $\Delta ABD$ will have $AD=2a$ and $AB=\sqrt{3}a$.
Because $AC$ is the angle bisector of $\widehat{DAB}$, we have $\frac{CB}{CD}=\frac{AB}{AD}=\frac{\sqrt{3}}{2}$.
We have this set of equations: ${\begin{cases}DB=CB+CD=a\\\frac{CB}{CD}=\frac{\sqrt{3}}{2}\end{cases}} \Rightarrow {\begin{cases}CD=\left(4-2\sqrt{3}\right)a\\CB=\left(-3+2\sqrt{3}\right)a\end{cases}}$
Apply the Pythagorean theorem: $CA=\sqrt{AB^2+BC^2}=\sqrt{(\sqrt{3a})^2+((-3+2\sqrt{3})a)^2}=\sqrt{(24-12\sqrt{3})a^2}=\sqrt{24-12\sqrt{3}}a$
We conclude that $sin(15)=sin\widehat{BAC}=\frac{BC}{CA}=\frac{-3+2\sqrt{3}}{\sqrt{{24-12\sqrt{3}}}}=\frac{-3+2\sqrt{3}}{3\sqrt{2}-\sqrt{6}}=\frac{\sqrt{6}-\sqrt{2}}{4}$.
|
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|
$\left|m^2 -\frac{n\cdot(n+1)}{2}\right|=1$ and A001110 The pairs of squares and triangular numbers differing by $1$ are so far: $(0,1)$, $(3,4)$, $(9,10)$, $(15,16)$, $(120,121)$, $(528,529)$. Do you think there are more terms than are found in A001110? If so, could there be a reason for this?
|
Let's start with an existing solutuion $(m,n)$:
$$\left|m^2 -\frac{n\cdot(n+1)}{2}\right|=1 \tag{1}$$
Then
$$\left|(m+1)^2 -\frac{n\cdot(n+1)}{2}-(2m+1)\right|=1$$
$$\left|(m+2)^2 -\frac{n\cdot(n+1)}{2}-(2m+1)-(2m+3)\right|=1$$
$$\left|(m+3)^2 -\frac{n\cdot(n+1)}{2}-(2m+1)-(2m+3)-(2m+5)\right|=1$$
$$... \text{by induction}$$
$$\left|(m+k)^2 -\frac{n\cdot(n+1)}{2}-(2km+k^2)\right|=1$$
which is
$$\left|(m+k)^2 -(1+2+...+n+2km+k^2)\right|=1$$
but we want it to be
$$\left|(m+k)^2 -(1+2+...+n+(n+1)+...+(n+p))\right|=1 \tag{2}$$
leading to
$$2km+k^2=k(2m+k)=\frac{p(2n+p+1)}{2} \iff 2k(2m+k)=p(2n+p+1)$$
or, we could ask for
$$\left\{\begin{matrix}
2m+k=p\\
2k=2n+p+1
\end{matrix}\right. \Rightarrow 2k=2n+2m+k+1$$
leading to
$$\left\{\begin{matrix}
k=2m+2n+1\\
p=4m+2n+1
\end{matrix}\right. \tag{3}$$
Jumping back to $(2)$ which is
$$\left|(m+k)^2 -\frac{(n+p)(n+p+1)}{2}\right|=1 \tag{4}$$
and using $(3)$ we have
$$\left|(3m+2n+1)^2 -\frac{(4m+3n+1)(4m+3n+2)}{2}\right|=1 \tag{5}$$
which is easy to expand to check that it reduces to $(1)$.
To conclude:
The mapping/recurrence $$(m,n) \mapsto (3m+2n+1, 4m+3n+1)$$ leads to infinity of pairs satisfying $(1)$
Example:
$$(0,1)\mapsto (3,4)\mapsto (18,25)\mapsto (105,148)\mapsto ...$$
or, in style of the question, using $m \mapsto m^2$ and $n \mapsto \frac{n(n+1)}{2}$:
$$(0,1)\mapsto (9,10)\mapsto (324,325)\mapsto (11025,11026)\mapsto ...$$
|
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|
Find the rational solution of the equation Find the rational solution of the equation:
$\frac{2x - 1}{2016} + \frac{2x - 3}{2014} + \frac{2x - 5}{2012}+ ...+ \frac{2x - 2011}{6} +\frac{2x - 2013}{4} + \frac{2x - 2015}{2} =\\ \frac{2x - 2016}{1} + \frac{2x - 2014}{3} + \frac{2x - 2012}{5}+ ...+ \frac{2x - 6}{2011} + \frac{2x - 4}{2013} + \frac{2x - 2}{2015}$
The problem is from a competition for seven graders. I tried various algebraic manipulations for no avail. Any hint will be highly appreciated.
|
Alpha gives $x=\frac {2017}2$. With that value all the fractions become $1$ so each side sums to $1008$
|
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|
Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose function $f$ and $h.$
|
A better upper bound: by the convexity of the integrand in $[0,1]$:
$$\forall x\in[0,1]:\qquad\frac1{1 + x + x^2}\le 1 - \frac{2x}3,$$
and this implies
$$\int_0^1\frac{1}{1+x+x^2}dx\le\int_0^1(1 - 2x/3)\,dx = \frac23.$$
|
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|
What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.
I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?
|
Notice that your matrix $$\begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}$$
has a special block decomposition as
$$ P=\begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}$$
Where A,B,C,D are non zero submatrices.
To find $P^2$ we use the new form to get
$$ P^2 = \begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}\begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}=\begin {pmatrix} AB&0 &AC\\0&BA+CD&0\\DB&0&DC\end{pmatrix} $$
As you know we can find powers of P by multiplying the new form of P as many times as we wish.
The alternating repeating form of powers of P is due to the block decomposition form of P.
|
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|
Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$ Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ , let lines $PB$ and $CA$ intersect at $E$, and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF $ is twice the area of $\triangle ABC$.
I had used my method and it required that $\triangle DEF $ have to be isosceles triangle which means $ P $ should be the mid point of $\widehat {BC}$ or $\widehat{AC}$ or $\widehat {BC}$. But the problem does not said so. So, how to solve it correctly?
|
Proof (for the case when $P$ lies on arc $BC$):
Let $AB=BC=CA=1$ and $\angle PAC=\theta$.
$$\frac{PD}{\sin\angle PBD}=\frac{PB}{\sin \angle PDB} \implies \frac{PD}{\sin\theta}=\frac{PB}{\sin (120^\circ-\theta)} $$
$$\frac{PB}{\sin\angle PAB}=\frac{AB}{\sin \angle APB}\implies \frac{PB}{\sin(60^\circ-\theta)}=\frac{AB}{\sin 60^\circ}$$
So, $\displaystyle PD=\frac{\sin\theta\sin(60^\circ-\theta)}{\sin60^\circ\sin(120^\circ-\theta)}=\frac{\sin\theta\sin(60^\circ-\theta)}{\sin60^\circ\sin(60^\circ+\theta)}$.
$$\frac{PE}{\sin\angle PAC}=\frac{AE}{\sin \angle APE} \implies \frac{PE}{\sin\theta}=\frac{AE}{\sin 120^\circ} $$
$$ \frac{AE}{\sin\angle ABE}=\frac{AB}{\sin\angle AEB} \implies \frac{AE}{\sin (60^\circ+\theta)}=\frac{AB}{\sin(60^\circ-\theta)}$$
So, $\displaystyle PE=\frac{\sin\theta\sin(60^\circ+\theta)}{\sin120^\circ\sin(60^\circ-\theta)}=\frac{\sin\theta\sin(60^\circ+\theta)}{\sin60^\circ\sin(60^\circ-\theta)}$.
$$\frac{PF}{\sin\angle PBF}=\frac{PB}{\sin \angle PFB} \implies \frac{PF}{\sin(120^\circ-\theta)}=\frac{PB}{\sin \theta}$$
So, $\displaystyle PF=\frac{\sin(120^\circ-\theta)\sin(60^\circ-\theta)}{\sin60^\circ\sin\theta}=\frac{\sin(60^\circ+\theta)\sin(60^\circ-\theta)}{\sin60^\circ\sin\theta}$.
The area of $\triangle DEF$ is
\begin{align*}
&\;\frac{1}{2}(PD\cdot PE+ PE\cdot PF+PF\cdot PD)\sin 120^\circ\\
=&\;\frac{\sqrt{3}}{4}\left[\frac{\sin^2\theta+\sin^2(60^\circ+\theta)++\sin^2(60^\circ-\theta)}{\sin^260^\circ}\right]\\
=&\;\frac{1}{\sqrt{3}}\left[\sin^2\theta+(\sin60^\circ\cos\theta+\cos60^\circ\sin\theta)^2+(\sin60^\circ\cos\theta-\cos60^\circ\sin\theta)^2\right]\\
=&\;\frac{1}{\sqrt{3}}\left[\sin^2\theta+2\sin^260^\circ\cos^2\theta+\cos^260^\circ\sin^2\theta\right]\\
=&\;\frac{1}{\sqrt{3}}\left[\frac{3}{2}\sin^2\theta+\frac{3}{2}\cos^2\theta\right]\\
=&\;\frac{\sqrt{3}}{2}
\end{align*}
The area of triangle $\triangle ABC$ is
$$\frac{1}{2}(1)^2\sin60^\circ=\frac{\sqrt{3}}{4}$$
|
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|
Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent? Here is Prob. 10, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Establish the convergence or the divergence of the sequence $\left( y_n \right)$, where
$$ y_n \colon= \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \ \mbox{ for } n \in \mathbb{N}. $$
My Attempt:
For each $n \in \mathbb{N}$, we have
$$
\begin{align}
y_{n+1} - y_n &= \left( \frac{1}{ (n + 1) +1} + \frac{1}{ (n+1) +2 } + \cdots + \frac{1}{2(n+1)} \right) - \left( \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right) \\
&= \left( \frac{1}{ n + 2} + \cdots + \frac{1}{2n+ 2 } \right) - \left( \frac{1}{ n+1} + \cdots + \frac{1}{2n} \right) \\
&= \frac{ 1 }{2n+1} + \frac{ 1}{2n+2} - \frac{1}{n+1} \\
&= \frac{1}{2n+1} - \frac{1}{2n+2} \\
&= \frac{1}{ (2n+1) (2n+2) } > 0,
\end{align}
$$
which shows that our sequence is monotonically increasing.
Also, for each $n \in \mathbb{N}$, we have
$$ y_n = \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} < \underbrace{\frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n} }_{ \mbox{ $n$ times } } = 1. $$
Thus our sequence, being monotonically increasing and bounded, is convergent.
Is what I've done so far all correct?
How to determine the limit of this sequence?
|
Your proof is correct.
It is too early (in the book) to obtain the limit. Some time later it will be easy to see an integral sum of
$$
\int_{0}^{1}\frac{1}{1+x}\,dx.
$$
|
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|
How do I calculate this kind of expressions $(-1)^{-\frac{2}{5}}$? I used a tool and found that it is equal to $0.31 - 0.95i$
Can someone please tell me the way I reach this result step by step?
|
That's only one possible value.
$(-1)^{-\frac 25}$ will be a number $z$ so that $z^5 = (-1)^{-2} = 1$. $z = 1$ is one such number. $z = 1$ is the only real such number. There are four other such complex numbers.
If we view a non-zero complex number, $a + bi$ as a point $(a,b)$ in a plane the point has a distinct distance $r = \sqrt{a^2 + b^2} > 0$ from $(0,0)$. And the angle between the real axis, the origin, and the point is a distcint angle $\theta$ were $\cos \theta = \frac {a}r$ and $\sin \theta = \frac {b}r$.
So the complex number can be written as $r(\cos \theta + i\sin \theta)$.
And here's the astonishing thing about it: If $z = r(\cos \theta + i\sin \theta)$ and $w = s(\cos \eta + i\sin \eta)$ then $z\cdot w = rs(\cos(\theta + \eta) + i \sin (\theta + \eta)$.
[Just do the trig: $z\cdot w = rs(\cos \theta + i\sin \theta)(\cos \eta + i\sin \eta)$ and $(\cos \theta + i\sin \theta)(\cos \eta + i\sin \eta) = (\cos \theta\cos \theta - \sin \theta\sin \eta)+ i(\sin\theta \cos \eta + \cos \theta \sin \eta)$ and $\cos (\theta + \eta) + i\sin(\theta + \eta) = (\cos \theta\cos \theta - \sin \theta\sin \eta)+ i(\sin\theta \cos \eta + \cos \theta \sin \eta)$]
So that means $z^k = r^k(\cos k\theta + i\sin k \theta)= r^k(\cos k(\theta+ 2m\pi) + i\sin k (\theta+ 2m\pi))$.
And that means $z^{\frac 1k} = r^{\frac 1k}(\cos (\frac {\theta}k + m*\frac {2\pi}k) + i\sin (\frac {\theta}k + m*\frac {2\pi}k))$.
So if $z^5 = 1= \cos 0 + i \sin 0$ then $z = \cos {m \frac {2\pi}5} + i \sin{m\frac {2\pi}5}$.
$0.31 - 0.95i\approx \cos \frac {8\pi}5 + i\sin \frac {8\pi}5$. i.e. the solution when $m = 4$. The other four are $\cos \frac {2\pi}5 + i\sin \frac {2\pi}5\approx 0.31 + 0.95i$ and $\cos \frac {4\pi}5 + i\sin \frac {4\pi}5 \approx -0.81 + 0.59i$ and $\cos \frac {6\pi}5 + i\sin \frac {6\pi}5 \approx -0.81 - 0.59i$ and $\cos 0 + i\sin 0 =1$
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When we get comfortable with these ideas we define
$e^{\theta i} :=\cos \theta + i\sin \theta $ (because that satisfies the property that $\frac {d e^z}{d z} = e^z$).
From there we can classify all complex numbers as $r*e^{i\theta}$. This allows us to derive the famous Euler's identity $e^{\pi i} = \cos \pi + i \sin \pi = -1$.
As $z^5 = 1$ has five solutions which do we declare to be the fifth root. Well.... the real root is $1$ of course but as the expressing was $(-1)^{-\frac 25}$ and fractiona powers of negative numbers implies complex roots and for one reason or another the concept of "principal root" and "principal" Logarithms leads to Markus Shauer's answer.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2706275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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