Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Roots in equation In the equation
$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} =0$
I got the roots of $x$ being $6$ and $7\sqrt{3}$.
Considering the graph shows only $6$ as being a valid solution, how should I go as figuring this out in the equation itself?
| Let $x=7\sqrt 3$ into $$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} $$ and you get $$ \sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} \approx -0.7869 \ne 0$$
Thus $x=7\sqrt 3$ is not a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and then solving for $x$ by using Cardano's method but I got the same equation at last that is
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Now I don't know how to solve this
| Let $a=\sqrt[3]{\sqrt{5}+2}$, $b=\sqrt[3]{\sqrt{5}-2}$ and $x=a-b$. Then,
$$x^3=\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab\left(a-b\right)\\=\sqrt{5}+2-\left(\sqrt{5}-2\right)-3x\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}=4-3x$$
Then the answer is the real root of the equation $x^3+3x-4=0$,
$$x^3-3x-4=0\\\left(x-1\right)\left(x^2+x+4\right)=0\\\because x^2+x+4>0\\ \therefore x=1$$ We get $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve Euler Project #9 only mathematically - Pythagorean triplet The "Euler Project" problem 9 (https://projecteuler.net/problem=9) asks to solve:
$a^2$ + $b^2$ = $c^2$
a + b + c = 1000
I find answers solving it with brute-force and programmatically, but is there a way to solve the problem ONLY mathematically? Can someone help, please?
Problem as explained in Project Euler website:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
| The triplets are all of the form
$a=u(n^2-m^2),
b=2umn,
c=u(n^2+m^2)
$
with $n > m$
so
$a+b+c
=u(2n^2+2mn)
=2un(n+m)
$.
We must have
$n > m$.
Therefore
$500
=un(n+m)
$.
If
$500 = rst
$
with
$s < t$
then
$u = r,
n = s,
n+m = t
$
so
$m = t-n
=t-s
$.
We must have
$n > m$
so
$s > t-s$
or
$s < t < 2s$.
Playing around a bit,
$500 = 1*20*25$,
so,
swapping $m$ and $n$,
$u = 1, m = 5,
n=20
$
and the sides are
$20^2-5^2 = 375 = 25\ 15,
2\ 20\ 5 = 200 = 25\ 8,
20^2+5^2 = 425 = 25\ 17
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Nesbitt by Nesbitt The title not says I'm Nesbitt but just only says it's a refinement of Nesbitt's inequality by itself so we have :
Let $a,b,c>0$ and $a\geq b \geq c$ then we have :
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{b}+\sqrt{a}}$$
I think (I have tested numerically ) we can use majorization like this (with $a,b,c>0$ and $a\geq b \geq c$):
First line of the majorization
$$\frac{a}{b+c}+\frac{b}{a+c}\geq \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}$$
Second line of the majorization
$$\Big(\frac{a}{b+c}+\frac{b}{a+c}\Big)\Big(\frac{a}{b+c}+\frac{c}{a+b}\Big)\geq \Big(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}\Big)\Big(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{b}+\sqrt{a}}\Big)$$
Third line of the majorization
$$\prod_{cyc}\Big(\frac{a}{b+c}+\frac{b}{a+c}\Big)\geq\prod_{cyc}\Big(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}\Big)$$
But this way's is too long so I forget this one .
I have tried also the Transformation of Ravi but without success .
So if you have a nice idea I take !
Thanks a lot for sharing your time and knowledge .
| We need to prove that: $$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq\sum_{cyc}\frac{a}{b+c},$$where a, b and c are positives.
Indeed, $$\sum_{cyc}\frac{a^2}{b^2+c^2}-\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{ab(a-b)-ac(c-a)}{(b^2+c^2)(b+c)}=$$
$$=\sum_{cyc}(a-b)\left(\frac{ab}{(b^2+c^2)(b+c)}-\frac{ab}{(c^2+a^2)(c+a)}\right)=$$
$$=\sum_{cyc}\tfrac{(a-b)^2ab(a^2+b^2+c^2+ab+ac+bc)}{(b^2+c^2)(b+c)(c^2+a^2)(c+a)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Does $\lim_{(x,y) \to (0,0)} \frac{x y \sin^2 y}{x^2 y}$ exists?
$\displaystyle\lim_{(x,y) \to (0,0)} \frac{x y \sin^2 y}{x^2 y}$
Along $x =y$, it becomes
\begin{equation}
\lim_{(x,y) \to (0,0)}\frac{y^2 \sin^2 y}{y^3} = 0.
\end{equation}
But along $x = y^3$,
\begin{equation}
\lim_{(x,y) \to (0,0)}\frac{y^4 \sin^2 y}{y^7} = \lim_{(x,y) \to (0,0)}\frac{ \sin^2 y}{y^3}.
\end{equation}
The limit is undefined.
This is what I thought, but using wolframalpha, it say the limit is $0$.
Where did I go wrong?
| You are right indeed
$$\frac{x y \sin^2 y}{x^2 y}=\frac{\sin^2y}{y^2 }\frac{y}{x}$$
and
*
*$\frac{\sin^2y}{y^2 }\to 1$
but
*
*$\frac{y}{x}$
has not limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question :
Solve for natural number the equation :
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try :
Let : $X=5^{x}$ and $Y=2^{y}$ so above equation
equivalent :
$2X^{2}+(Y-4)X-6Y^{2}-Y+2=0$
We solve this equation for $X$
$\Delta =(7Y)^{2}$ mean : $X_{1}=\frac{3}{2}Y+1$ and $X_{2}=1-2Y$
From here how I can find $X$ and $Y$ , this is
all my effort ?
Thanks!
| $5^{2x}-3\cdot 2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
$(5^{2x}-2\cdot 5^x + 1)+ (5^{x}2^{y-1}-2^{y-1})-3\cdot 2^{2y}=0$
$(5^x-1)^2 + 2^{y-1}(5^x - 1)-3\cdot 2^{2y}=0$
If we solve for $5^x-1$ we get
$5^x-1 =\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 12*2^{2y}}}{2}=$
$\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 48*2^{2y-2}}}{2}=\frac {-2^{y-1} \pm\sqrt{49*2^{2y-2}}}{2}=$
$\frac {-2^{y-1} \pm 7*2^{y-1}}{2}=3*2^{y-1}$
Or $5^x = 3*2^{y-1} + 1$.
$5^x = 2^y + 2^{y-1} + 1$
But $5^x = (4+1)^x = \sum_{j=0}^x C_{x,j} 4^j$
We have a solution for $x=2$ and $y=4$ ($5^2 = 3*2^3+1$) which seems to work as ${2\choose 1}4 + 4^2 = 2*2^2 + 2^4= 2^3+2^4$. But in general it doesn't seem like a likely occurrence.
For this to occur we need $\sum_{j=1}^x C_{x,j} 4^j = 2^{y-1} + 2^y$.
There's probably a really easy way to prove this is impossible if $x > 2$. If we divide the LHS by $4$ we get a remainder of $x\pmod 4$ and the RHS is $0\pmod 4$ if $y> 4$.... I think we have some chasing involved. $2^{y-2}|x$ but then $5^x$ is huge compared to $3*2^{y-1}+1$....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the value of $\frac{AN}{AD}$ and $\frac{ME}{MP}$.
$M, N$ and $P$ are respectively the midpoint of $BC$, the centroid of $\triangle ABC$ and a point on $CA$ such that $NP \parallel BC$. A line passing through point $B$ intersects $AM$ and $PM$ respectively at $D$ and $E$ such that $BD = 5DE$. Calculate the value of $\dfrac{AN}{AD}$ and $\dfrac{ME}{MP}$.
We have that $BD = 5DE \implies 5\overrightarrow{BE} = 6\overrightarrow{BD} \implies 5 \cdot \left(\overrightarrow{BM} - \overrightarrow{EM}\right) = 6 \cdot \left(\overrightarrow{BM} - \overrightarrow{DM}\right)$
$\implies \overrightarrow{BM} = 6\overrightarrow{DM} - 5\overrightarrow{EM} \implies - \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC} = 6\overrightarrow{DM} - 5\overrightarrow{EM}$
Let $\dfrac{\overrightarrow{DM}}{\overrightarrow{AM}} = x$ and $\dfrac{\overrightarrow{EM}}{\overrightarrow{PM}} = y \implies - \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC} = 6x\overrightarrow{AM} - 5y\overrightarrow{PM}$
Furthermore, we have that $\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC}$
and $\overrightarrow{PM} = \dfrac{1}{3}\cdot \left(\overrightarrow{PA} + \overrightarrow{AM}\right) + \dfrac{2}{3} \cdot \left(\overrightarrow{PC} + \overrightarrow{CM}\right)$
$ = - \dfrac{2}{9}\overrightarrow{AC} + \left(\dfrac{1}{6}\overrightarrow{AB} + \dfrac{1}{6}\overrightarrow{AC}\right) + \dfrac{2}{9}\overrightarrow{AC} + \left(\dfrac{2}{3}\overrightarrow{AB} - \dfrac{2}{3}\overrightarrow{AC}\right) = \dfrac{5}{6}\overrightarrow{AB} - \dfrac{1}{2}\overrightarrow{AC}$
So we have the solve the following equation $$- \dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC} = 6x \cdot \left(\dfrac{1}{2}\overrightarrow{AB} + \dfrac{1}{2}\overrightarrow{AC}\right) - 5y \cdot \left(\dfrac{5}{6}\overrightarrow{AB} - \dfrac{1}{2}\overrightarrow{AC}\right)$$
$\left\{ \begin{align} -\dfrac{1}{2} = 3x - \dfrac{25}{6}y\\ \dfrac{1}{2} = 3x + \dfrac{5}{2}y \end{align} \right. \implies (x, y) = \left(\dfrac{1}{24}, \dfrac{3}{20}\right) \implies \dfrac{AN}{AD} = \dfrac{16}{23}$, $\dfrac{ME}{MP} = \dfrac{3}{20}$
I wrote this solution almost at midnight and did all the calculations by hand so there might have been multiple mistakes. Please help me find them.
| I will go an other way, it is the simplest way to check.
We try to apply the theorems of Ceva and/or Menelaus. The situation is as follows.
Let $Q$ be the mid point of $AC$.
Let $Z$ be the intersection of the parallels
*
*to $BC$ through $A$,
*to $AB$ through $C$,
so that $ABCZ$ is a parallelogram. $Q$ is the mid point of the one diagonal $AC$, then also the mid point of the other diagonal, and the centroid $N$ is on the median $BQ$. So $B,N,Q,Z$ colinear.
The point $Z$ is also on $MP$, because of the match of the proportion
$$
\frac{NP}{AZ}=
\frac{NP}{BC}=
\frac{QN}{QB}=
\frac13=
\frac{MN}{MA}\ .
$$
$BDE$ with $AC$. Then using Menelaus for $\Delta EBZ$, intersected by the line $NDM$, we get:
$$
1 =
\frac{DB}{DE}\cdot
\frac{ME}{MZ}\cdot
\frac{NZ}{NB}
=
\frac{5}{1}\cdot
\frac{ME}{MZ}\cdot
\frac{2}{1}
\ .
$$
This leads to
$$
\frac{ME}{MP}=
\frac{ME}{MZ}\cdot
\frac{MZ}{MP}
=\frac 1{2\cdot 5}\cdot\frac31
=\color{blue}{\frac 3{10}}\ .
$$
We can use this or independently "do the same",
and apply Menelaus for $\Delta DBN$, intersected by the line $MEZ$, when we get:
$$
1 =
\frac{ED}{EB}\cdot
\frac{ZB}{ZN}\cdot
\frac{MN}{MD}
=
\frac{1}{6}\cdot
\frac{3}{2}\cdot
\frac{MN}{MD}
=
\frac{1}{4}\cdot
\frac{MN}{MD}
\ .
$$
So $\displaystyle MD=\frac 14 MN=\frac 14 \cdot \frac 13 AD=\frac 1{12}AD$.
This leads to
$$
\frac{AN}{AD}=
\frac{2/3}{11/12}=
\color{blue}{\frac{24}{33}}
\ .
$$
$\square$
I wrote this solution hours after mid night, it is 02:15 in Germany, hope things are all right, and easy to check in case errors came in. At any rate,
$MD:MN=1:4$, and $ME:MZ=1:10$. One last check is done by using the theorem of Menelaus in $\Delta MNZ$ w.r.t. the line $BDE$, and indeed,
$\displaystyle
\frac{DN}{DM}\cdot
\frac{EM}{EZ}\cdot
\frac{BZ}{BN}
=
\frac31\cdot
\frac19\cdot
\frac31
=1$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to find the basis (and cartesian equation(s)) of a sum of two vectorial subspaces? I have the following vectorial subspaces :
U = span $\left(\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix} \right)$ and V = span $\left(\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} \right)$
To calculate the basis of U + V should have use this method :
$\left(\begin{array}{cccc|c} 2 & 3 &0 & -1& 0 \\ 0 & 6 &2&1 & 0 \\ 1 & 9 &1&0&0 \\ -2 & -12&0&1 & 0 \end{array}\right)$ which generate this solution after reduction $\left(\begin{array}{cccc|c} 1 & 0 &0 & -\frac{1}{2}& 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$ $\sim$ $\begin{pmatrix} 1 & 0 & 0 & -\frac{1}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} \end{pmatrix}$
Now I don't understand how to have cartesian equation(s) of this hyperplane. Can someone help me understanding this?
| We have obtained that a basis is given by the following vectors
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}= a\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix} +b\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix}+ c\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}$$
to obtain the cartesian equation we need to eliminate $a$,$b$, and $c$ from the system
*
*$x_1=2a+3b$
*$x_2=6b+2c$
*$x_3=a+9b+c$
*$x_4=-2a-12b$
and determine the relation between $x_1$, $x_2$, $x_3$ and $x_4$.
We can also proceed by elimination using the augmented matrix
$$\left(\begin{array}{ccc|c} 2 & 3 &0 & x_1 \\ 0 & 6 &2 & x_2 \\ 1 & 9 &1&x_3 \\ -2 & -12&0 & x_4 \end{array}\right)$$
operating in the same way as you already did, you'll obtain the cartesian equation in the last row.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\sin^4(x)$ Consider the integral
$$\int \sin^4(x)dx$$
Now I could separate the $\sin^4(x)$ into two $\sin^2(x)$ terms and the use power reducing formula
$$\int \sin^2(x)\sin^2(x)dx $$
$$\int \frac{1-\cos(2x)}{2}*\frac{1-\cos(2x)}{2}dx $$
$$\int \frac{(1-\cos(2x))^2}{4} dx$$
$$\int\frac{1}{4}-\frac{2\cos(2x)}{4}+\frac{\cos^2(x)}{4}dx $$
$$\frac{1}{4} \int1-8\cos(2x)+\cos^2(2x)dx $$
Now would it be good to rewrite that $\cos^2(2x)$ as $1-\sin^2(2x)$ and then integrate or should I also separate the terms into their own integrands?
The answer I am looking for which is $\frac{3}{8}x -\frac{1}{4}\sin(2x)+\frac{1}{32}\sin(4x)+C$.
| First of all, I think you mistyped the last line of your integral, changing a 2 for an 8. Besides, you can use the linearity of the integral to write $$ \int(1-2\cos(2x)+\cos^{2}(2x))dx = \int 1 dx - 2\int \cos(2x)dx+\int \cos^{2}(2x)dx. $$ Now, note that $$\cos^{2}(2x) + \sin^{2}(2x) = 1$$ and $$ \cos^{2}(2x)-\sin^{2}(2x) = \cos(4x)$$ so that $$\cos^{2}(2x) = \frac{1+\cos(4x)}{2}.$$ Thus, we have $$ \int \cos^{2}(2x) = \int \frac{1+\cos(4x)}{2}dx = \frac{1}{2}x + \frac{\sin(4x)}{8}+C$$. Now, the previous integrals are given by $$\int 1 dx -2\int \cos(2x) = x -2\frac{\sin(2x)}{2}+C = x - \sin(2x) +C.$$
Finally, your integral is $$\frac{1}{4}[x-\sin(2x)+\frac{1}{2}x+\frac{\sin(4x)}{8}]+C = \frac{3}{8}x-\frac{1}{4}\sin(2x)+\frac{\sin(4x)}{32} + C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Calculate $\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$ without L'Hôpital's rule How can I calculate limit
$$\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$$
without using L'Hôpital's rule?
What I have tried so far:
I tried to use the fact that $\lim_{\alpha\to 0}(1 + \alpha)^{1/\alpha} = e$ and do the following:
$$\lim_{x\to \pi/4}\cot(x)^{\cot(4 \cdot x)} = \lim_{x\to \pi/4}(1 + (\cot(x) - 1))^{\cot(4 \cdot x)} = \lim_{x\to \pi/4}(1 + (\cot(x) - 1))^{\frac{1} {\cot(x) - 1} \cdot (\cot(x) - 1) \cdot \cot(4 \cdot x)} = \lim_{x\to \pi/4}e^{(\cot(x) - 1) \cdot \cot(4 \cdot x)} = e^{\lim_{x\to \pi/4}{(\cot(x) - 1) \cdot \cot(4 \cdot x)}} $$
But I have problems calculating limit
$$\lim_{x\to \pi/4}{(\cot(x) - 1) \cdot \cot(4 \cdot x)}$$
I tried to turn $\cot(x)$ into $\frac{\cos(x)}{\sin(x)}$ as well as turning it into $\tan(x)$, but I do not see any workaround afterwards.
I would appreciate any pieces of advice. Thank you!
| We substitute $c = \cot(x)$ and we get
$$
\lim\limits_{x\to\pi/4} \cot(x)^{\cot(4x)}
=\lim\limits_{c\to 1}\; c^{\frac{c^4-6c^2+1}{4c^3-4c}}
$$
because
$$
\cot(4x) = \frac{\cot^4x-6\cot^2x+1}{4\cot^3x-4\cot x}
$$
Then we have
$$
\lim\limits_{c\to 1} \;c^{\frac{c^4-6c^2+1}{4c^3-4c}}
=\lim\limits_{c\to 1} \;\exp\left(\ln(c) \cdot \frac{c^4-6c^2+1}{4c^3-4c}\right)
=\lim\limits_{c\to 1} \;\exp\left(\frac{\ln(c)}{c-1} \cdot \frac{c^4-6c^2+1}{4c^2+4c}\right) \\
=\exp\left(\lim\limits_{c\to 1} \;\frac{\ln(c)}{c-1} \cdot \lim\limits_{c\to 1} \;\frac{c^4-6c^2+1}{4c^2+4c}\right)
=\exp\left(1\cdot\frac{-4}{8} \right)= \frac{1}{\sqrt{e}}
$$
because $\lim\limits_{c\to 1} \;\frac{\ln(c)}{c-1}$ is the derivative of $\ln(x)$ evaluated at $x=1$, which is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Showing $\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=0$ $$\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=\lim_{x\to 0 }\frac{x^2\sin\frac{1}{x}}{x}=\lim_{x\to 0} x\sin\frac{1}{x}=0$$
Is this solution right?
Thank you very much!
| $$\lim_{x\to 0}|\frac{\sin(x^2\sin\frac{1}{x})}{x}|=$$
$$\lim_{x\to 0 }|\frac{\sin(x^2\sin\frac{1}{x})}{x^2} (x)|\le $$
$$\lim_{x\to 0 }|\frac{(x^2\sin\frac{1}{x})}{x^2} (x)| =$$
$$\lim_{x\to 0}| x\sin(\frac{1}{x})|=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Number of possible polynomials
Let $a,b,c,d$ be four integers (not necessarily distinct) in the set $\{1,2,3,4,5\}$. Find the number of polynomials of the form $x^4+ ax^3 + bx^2 + cx +d$ which is divisible by $x+1$.
My Try:
Let $f(x) = x^4+ ax^3 + bx^2 + cx +d$, then $f(-1) = 0$. Thus $1+ (b+d) = c+a$. On counting cases I got 80 permissible cases. Is there a way to solve the above equation $1+ (b+d) = c+a$?
| The counting can be a bit simplified using your intermediate result as follows:
*
*You have $a-b+c-d = 1 \Leftrightarrow (a-1) + (5-b) + (c-1) + (5-d) = 9$
So, your question is equivalent to counting the number of integer solutions of
$$a' + b' + c' + d' = 9 \mbox{ with } a',b',c',d' \in \{0,1,2,3,4\}$$
Now, let the following sink in first by considering the exponents:
This number is the same as the coefficient of $x^9$ in $(1+x+x^2+x^3+x^4)^4$.
Hence, using
*
*$1+x+x^2+x^3+x^4 = \frac{1-x^5}{1-x}$ and
*$\frac{1}{(1-x)^4} = \sum_{n=0}\binom{n+3}{3}x^n$ you get
\begin{eqnarray*}[x^9]\left(1+x+x^2+x^3+x^4\right)^4
& = & [x^9]\left(\frac{1-x^5}{1-x}\right)^4\\
& = & [x^9](1-4x^5)\sum_{n=0}\binom{n+3}{3}x^n\\
& = & \binom{9+3}{3} - 4\cdot \binom{4+3}{3}\\
& = & 80
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
quick way to count Suppose we have 10 sticks with length 1-10, respectively. Pick three from them, how many triangles can we form?
I counted one by one and got 50. Is there a quick way? Any help would be appreciated.
| Call the three sticks $a < b < c$. Pick $c$ first. The other two must sum to more than $c$. Pick $a$ second. So we have $$\sum_{c=1}^{10} \sum_{a=1}^{c-1} \sum_{b=a+1}^{c-1} [a+b > c]$$
Working from the inside out you should be able to tighten the bound on $b$ to remove that Iverson bracket and replace the inner sum with a $\max$ of $0$ and a linear expression in $a$ and $c$; then adjust the bounds on $a$ to remove the $\max$ and leave just the linear expression; then replace the sum over $a$ with a quadratic expression, and finally reduce the outer sum to a cubic expression in the largest stick length.
Ok, I was overly optimistic. The innermost sum is piecewise linear, and it gets more complicated from there.
$$\sum_{b=a+1}^{c-1} [a+b > c] = \sum_b [c-a < b][a < b][b < c] \\
= \sum_b [\max(c-a, a) < b < c] = \max(0, c - 1 - \max(c-a, a)) \\
= \begin{cases}
\max(0, a- 1) & \textrm{if } c-a > a \\
\max(0, c - a - 1) & \textrm{if } c-a \le a \\
\end{cases} \\
= \begin{cases}
a - 1 & \textrm{if } 2 < 2a < c \\
c - a - 1 & \textrm{if } c \le 2a < 2c-2 \\
0 & \textrm{otherwise}
\end{cases} \\
$$
Then $$\sum_{a=1}^{c-1} \sum_{b=a+1}^{c-1} [a+b > c] = \sum_{a=1}^{c-1} \begin{cases}
a - 1 & \textrm{if } 2 < 2a < c \\
c - a - 1 & \textrm{if } c \le 2a < 2c-2 \\
0 & \textrm{otherwise}
\end{cases} \\
= \left(\sum_{a=2}^{\frac {c-1}2} a-1\right) + \left(\sum_{a=\frac c2}^{c-2} c-a-1\right) \\
= \begin{cases}
\left(\sum_{a=2}^{\frac c2-1} a-1\right) + \left(\sum_{a=\frac c2}^{c-2} c-a-1\right) & \textrm{if } c \textrm{ is even} \\
\left(\sum_{a=2}^{\frac {c-1}2} a-1\right) + \left(\sum_{a=\frac {c+1}2}^{c-2} c-a-1\right) & \textrm{if } c \textrm{ is odd} \\
\end{cases} \\
% subst i:=a-1 / a=i+1, j:=c-a-1 / a=c-j-1
= \begin{cases}
\left(\sum_{i=1}^{\frac c2-2} i\right) + \left(\sum_{j=1}^{\frac c2-1} j\right) & \textrm{if } c \textrm{ is even} \\
\left(\sum_{i=1}^{\frac {c-3}2} i\right) + \left(\sum_{j=1}^{\frac {c-3}2} j\right) & \textrm{if } c \textrm{ is odd} \\
\end{cases} \\
= \begin{cases}
\frac{(c-2)^2}{4} & \textrm{if } c \textrm{ is even} \\
\frac {(c-1)(c-3)}4 & \textrm{if } c \textrm{ is odd} \\
\end{cases} \\
$$
Finally,
$$\sum_{c=1}^N \sum_{a=1}^{c-1} \sum_{b=a+1}^{c-1} [a+b > c]
= \sum_{c=1}^N \frac14 \begin{cases}
(c-2)^2 & \textrm{if } c \textrm{ is even} \\
(c-1)(c-3) & \textrm{if } c \textrm{ is odd} \\
\end{cases} \\
= \left(\sum_{k=1}^{\left\lfloor \frac N2\right\rfloor} \frac{(2k-2)^2}{4}\right) + \left(\sum_{m=0}^{\left\lfloor \frac{N-1}2\right\rfloor} \frac{(2m+1-1)(2m+1-3)}{4} \right) \\
% subst k'=k-1 / k=k'+1
= \left(\sum_{k'=0}^{\left\lfloor \frac{N-2}2\right\rfloor} k'^2\right) +
\left(\sum_{m=0}^{\left\lfloor \frac{N-1}2\right\rfloor} m(m-1) \right) \\
% PLAN A
= \left(\sum_{k=0}^{\left\lfloor \frac{N-2}2\right\rfloor} 2k^2 - k\right) + \left[\left\lfloor \frac{N-1}2\right\rfloor > \left\lfloor \frac{N-2}2\right\rfloor\right] \left\lfloor \frac{N-1}2\right\rfloor \left(\left\lfloor \frac{N-1}2\right\rfloor - 1\right)\\
= \frac{\left\lfloor \frac{N-2}2\right\rfloor \left(\left\lfloor \frac{N-2}2\right\rfloor + 1\right) \left(4\left\lfloor \frac{N-2}2\right\rfloor - 1\right)}{6}
+ [N\textrm{ is odd}] \frac{(N-1)(N-3)}4 \\
= \begin{cases}
\frac{(N-1)(N-3)(2N-7)}{24} + \frac{(N-1)(N-3)}4 & \textrm{if } N \textrm{ is odd} \\
\frac{N(N-2)(2N-5)}{24} & \textrm{if } N \textrm{ is even}
\end{cases} \\
= \begin{cases}
\frac{(N-1)(N-3)(2N-1)}{24} & \textrm{if } N \textrm{ is odd} \\
\frac{N(N-2)(2N-5)}{24} & \textrm{if } N \textrm{ is even}
\end{cases} \\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \le \frac{1}{216} \\
abc \le \frac{\sqrt{6}}{36}$$
However, I don't know where to go from here.
| $$a+b+c=\sqrt{\sum_{cyc}(a^2+2ab)}\leq\sqrt{\sum_{cyc}\left(a^2+\frac{1}{2}a^2+\frac{3}{2}ab\right)}\leq\sqrt{\frac{3}{2}}.$$
The equality occurs for $a=b=c$ and $\sum\limits_{cyc}(a^2+ab)=1,$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How to Solve This Exponential Limit without Derivate / L'Hôspital's Rule can someone teach me how can I solve this limit without using the L'Hopital's Rule?
$$\lim_{x\to 0} \left( \frac{2+x^{2}}{2-x^{2}} \right)^{\frac{1}{x^2}}$$
Thanks in advance.
| $$
\frac{2+x^2}{2-x^2}=1+\frac{2x^2}{2-x^2}=1+\frac{x^2}{1-\frac{x^2}{2}}
$$
and $y=\frac{x^2}{1-\frac{x^2}{2}}\to 0$, as $x\to 0$. Hence
$$
\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}=(1+y)^{1/y}\to e,
$$
as $x\to 0$.
Finally
$$
\left(\frac{2+x^2}{2-x^2}\right)^{1/x^2}=\left(\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}\right)^{\frac{1}{1-\frac{x^2}{2}}}\to e,
$$
since, if $f(x)\to e$ and $g(x)\to 1$, then $f(x)^{g(x)}\to e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum number of $k$-partitions of a set of size $n$ to enumerate all $n \choose k$ combinations Given a set $\mathcal{S}$ of size $n$, let a $k$-partition of $\mathcal{S}$ be a grouping into $k$ disjoint classes $$(S_1 ,S_2,...,S_k)$$
Where the $S_i$ do not necessarily contain the same number of elements.
Let $C(S_1, S_2, ... S_k)$ be the set of combinations which can be made by taking exactly one element of each class, (i.e. $C(S_1, S_2, ... S_k)$ is the Cartesian Product $S_1 \times S_2 \times ...\times S_k$). For a given $(n,k)$, what is the minimum number of k-partitions $N$ such that
$\bigcup\limits_{i=1}^{N} C(S_1^{(i)}, S_2^{(i)}, ...,S_3^{(i)}) = $ All Possible Combinations
Example:
for $n=4$ and $k=2$, $\mathcal{S} = \{1,2,3,4\}$ we can choose:
$$ S^{(1)} = \left\{\left\{ 1, 2\right\}, \left\{ 3, 4\right\}\right\}$$
$$ S^{(2)} = \left\{\left\{ 1, 3\right\}, \left\{ 2, 4\right\}\right\}$$
Then $$C_{1} = \left\{\left\{ 1, 3\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 2, 4\right\}\right\}$$
$$C_{2} = \left\{\left\{ 1, 2\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 3, 4\right\}\right\}$$
and $C_1 \cup C_2 =$ All possible combinations and $N=2$.
| Rather than $N$ I'm going to use the OEIS convention $T(n, k)$ (for triangle, I think).
Misha Lavrov's answer gives the bound $$T(n, k) \ge \frac{\log n - \log(k-1)}{\log k - \log(k-1)}$$
Another, easier, lower bound uses the auxiliary function A152072 which gives the largest possible size of the Cartesian product $S_1 \times S_2 \times \cdots \times S_k$: then clearly $$T(n, k) \ge \frac{\binom n k}{\textrm{A152072}(n, k)} = \frac{\binom n k}{ \left\lceil n/k \right\rceil^{n \bmod k} \left\lfloor n/k\right\rfloor^{k-n \bmod k}}$$
Neither of these bounds is always better than the other. In fact, we can go further. Note that when $k = n-1$ Misha's bound is $2$, whereas my bound is $\left\lceil\frac n2\right\rceil$ (and this is tight: consider $S_1^{(i)} = \{2i, (2i+1) \bmod n\}$, all other $S_j^{(i)}$ are singletons). On the other hand, when $k=2$ Misha's bound is $\left\lceil \log_2 n\right\rceil$, whereas my bound is $2$. Therefore neither bound is asymptotically tight.
Looking at actual figures, we have (starting at $n=1$, $k=1$)
$$\begin{matrix}1 \\
1& 1 \\
1& 2& 1 \\
1& 2& 2& 1 \\
1& 3& 3& 2& 1 \\
1& 3& 3& 3& 2& 1 \\
1& 3& 4& 3& 3& 2& 1 \\
1& 3& 4& 4& 4& 3& 2& 1\end{matrix} \textrm{ vs }\quad \begin{matrix}1 \\
1& 1 \\
1& 2& 1 \\
1& 2& 2& 1 \\
1& 2& 3& 3& 1 \\
1& 2& 3& 4& 3& 1 \\
1& 2& 3& 5& 6& 4& 1 \\
1& 2& 4& 5& 7& 7& 4& 1
\end{matrix}$$
for a combined lower bound of
$$\begin{matrix}1 \\
1& 1 \\
1& 2& 1 \\
1& 2& 2& 1 \\
1& 3& 3& 3& 1 \\
1& 3& 3& 4& 3& 1 \\
1& 3& 4& 5& 6& 4& 1 \\
1& 3& 4& 5& 7& 7& 4& 1\end{matrix}$$
By brute force, the actual start of the table should be
$$\begin{matrix}1 \\
1& 1 \\
1& 2& 1 \\
1& 2& 2& 1 \\
1& 3& 3& 3& 1 \\
1& 3& 3& 5& 3& 1 \\
1& 3& 4& 6& 6& 4& 1 \\
1& 3& 4& 6& 8& 8& 4& 1\end{matrix}$$
| {
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"url": "https://math.stackexchange.com/questions/3403318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Is $g(x_1, x_2) = (\alpha - x_1)^2 + (\max \{\alpha, x_1\} + \beta - x_2)^2$ convex?
Let $\alpha \geq 0$ and $\beta \geq 0$. Can we prove or disprove the following function is convex on $x_2 \geq x_1 \geq 0$?
$$
g(x_1, x_2) = (\alpha - x_1)^2 + (\max \{\alpha, x_1\} + \beta - x_2)^2
$$
My Approach: It is clear that the function $\max \{\alpha, x_1\} - x_2$ is convex. For $x \geq 0$, $x^2$ is convex and increasing, so composition of these two functions is convex. But this does not seem to work in general because of the region that $\max \{\alpha, x_1\} - x_2$ might go negative.
| As you already pointed out, $(\max \{\alpha, x_1\} + \beta - x_2)^2$ might descrease as $x_1$ increases when $x_1 + \beta - x_2 < 0$. So, to show a function is non-convex, one only needs to find a counter example.
Try the following three points (find a large enough $n > 2$ so that $\alpha - \frac{1}{n}\beta > 0$, and let $\lambda=\frac{1}{2}$).
*
*$g(\alpha - \frac{1}{n}\beta, \alpha + 2\beta) = \frac{\beta^2}{n^2} + \beta^2 = \frac{n^2+1}{n^2}\beta^2;$
*$g(\alpha, \alpha + 2\beta) = \beta^2;$
*$g(\alpha + \frac{1}{n}\beta, \alpha + 2\beta) = \frac{\beta^2}{n^2} + \frac{{(n-1)}^2\beta^2}{n^2} = \frac{{(n-1)}^2 + 1}{n^2}\beta^2.$
Now
$\frac{g(\alpha - \frac{1}{n}\beta, \alpha + 2\beta) + g(\alpha + \frac{1}{n}\beta, \alpha + 2\beta)}{2}=\frac{n^2+{(n-1)}^2+2}{2n^2}\beta^2<\beta^2=g(\alpha, \alpha + 2\beta)\\ =g(\frac{\left(\alpha - \frac{1}{n}\beta\right) + \left(\alpha + \frac{1}{n}\beta\right)}{2}, \frac{(\alpha + 2\beta) + (\alpha + 2\beta)}{2}).$
Hence the function is not convex.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the injectivity of function I have a subset $B = \{(x,y) : x^2 + y^2 \leq 1 \} \subseteq \mathbb{R}^2$.
I am trying to prove that function
$g(x,y) = (\frac{x^2 + y^2}{x^2 + y^2 + 1}x, \frac{x^2 + y^2}{x^2 + y^2 + 1}y)$
is an injective function $g: \mathbb{R}^2 \rightarrow B$.
I attempted to prove this by directly applying the definition of injection.
Suppose we have $(a,b),(c,d) \in \mathbb{R}^2$ such that
$g(a,b) = g(c,d)$.
If I can prove $g(a,b) = g(c,d) \implies (a,b) = (c,d)$,
then $g$ is an injective function.
$g(a,b) = (\frac{a^2 + b^2}{a^2 + b^2 + 1}a, \frac{a^2 + b^2}{a^2 + b^2 + 1}b)$, $g(c,d) = (\frac{c^2 + d^2}{c^2 + d^2 + 1}c, \frac{c^2 + d^2}{c^2 + d^2 + 1}d)$
Since $g(a,b) = g(c,d)$, $\frac{a^2 + b^2}{a^2 + b^2 + 1}a = \frac{c^2 + d^2}{c^2 + d^2 + 1}c$.
Furthermore, $0 \leq \frac{a^2 + b^2}{a^2 + b^2 + 1} < 1$ and $0 \leq \frac{c^2 + d^2}{c^2 + d^2 + 1} < 1$.
I want to have $\frac{a^2 + b^2}{a^2 + b^2 + 1} = \frac{c^2 + d^2}{c^2 + d^2 + 1}$,
and for that I need
$\frac{a^2 + b^2}{a^2 + b^2 + 1} \leq \frac{c^2 + d^2}{c^2 + d^2 + 1}$ and $\frac{c^2 + d^2}{c^2 + d^2 + 1} \leq \frac{a^2 + b^2}{a^2 + b^2 + 1}$.
which I am not so sure how to get.
Is this a valid approach to reaching the conclusion $a = c$?
Thank you for reading.
| We have $g(10,0)=\frac{1000}{101}(1,0)\not\in B$. So, $g$ does not map the plane to the disc $B$. Nevertheless, $g$ is an injective self-map of the plane.
To prove injectivity, try to pass to the polar coordinates $x=r\cos\theta$, $y=r\sin\theta$. Then $$g(x,y)=\frac{r^3}{r^2+1}(\cos\theta,\sin\theta).$$
Assume that $g(x_1,y_1)=g(x_2,y_2)$, i.e.
$$\frac{r_1^3}{r_1^2+1}(\cos\theta_1,\sin\theta_1)=\frac{r_2^3}{r_2^2+1}(\cos\theta_2,\sin\theta_2).$$
First, if $r_1\ne r_2$, then the above equality fails, because $r\mapsto\dfrac{r^3}{r^2+1}$ is an increasing function and then the euclidean norms of both sides are not equal. Then $r_1=r_2$. If $r_1=r_2=0$, then obviously $(x_1,y_1)=(x_2,y_2)=(0,0)$. So, we can now assume that $r_1=r_2>0$ and we arrive at
$$(\cos\theta_1,\sin\theta_1)=(\cos\theta_2,\sin\theta_2),$$
hence $\cos\theta_1=\cos\theta_2$ and $\sin\theta_1=\sin\theta_2$. Together with $r_1=r_2$ this means that $(x_1,y_1)=(x_2,y_2)$. This proves the injectivity of $g$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Following Fibonacci Prove that $ F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}} $?
I discovered this property from an attempt to solve the following problem:
Defining $ A_n = \sqrt{{F_n} ^ 2 + {F_{n + 2}} ^ 2}, $ the numbers $ A_n, $ $ A_ {n + 1} $ and $ A_{n + 2} $ are the length measures of the sides of a triangle whose area is $ \frac{1}{2} $ unit.
(I also put this in the Wikipedia article about the Fibonacci Sequence).
| To prove
$F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}}
$,
I would write it as
$F^2_{n + 2}(F_n ^ 2 + F_ {n + 1}^ 2)
= F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1
$
or
$(F_{n+1}+F_n)^2(F_n ^ 2 + F_ {n + 1}^ 2)
= F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1
$.
For $n=0$,
$F_n = 0, F_1 = 1$,
this becomes
$(1)^2(1)
= 1
$,
which is true.
For $n=1$,
$F_n = 1, F_{n+1} = 1$,
this becomes
$(2)^2(2)
= (3+4) + 1
$,
which is true.
For $n=2$,
$F_n = 1, F_{n+1} = 2$,
this becomes
$(2+1)^2(1 ^ 2 + 2^ 2)
= 2 ^ 2(3 +4\cdot 2) + 1
$,
or
$45
=45
$
which is true.
For $n=3$,
$F_n = 2, F_{n+1} = 3$,
this becomes
$(3+2)^2(2 ^ 2 + 3^ 2)
= 2\ 3 ^ 2(3\ 2+4\ 3) + 1
$.
or
$25\ 13
=18(18)+1
$
or
$325 = 325
which is true.
So, it looks good.
The next step would be
to expand the sides,
eliminate any common terms,
and try to prove it by induction.
I'll leave it at this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all rational triplets $(a,b,c)$ that are roots of the equation $x^3+ax^2+bx+c=0.$
Find all rational triplets $(a,b,c)$ that are roots of the equation $$x^3+ax^2+bx+c=0$$
My work so far:
By Vieta's formulas we have
$a+b+c=-a\;(1),$
$ab+ac+bc=b\;(2),$ and
$abc=-c\;(3).$
From $(3),$ either $c=0$ or $ab=-1$. If $c=0,$ then $(1)$ becomes $b=-2a,$ and $(2)$ becomes $b(1-a)=0$. Hence $a=b=0$ or $a=1\Rightarrow b=-2.$
So assume $c\neq 0$ and $ab=-1.$ $(1)$ becomes $c=-b-2a.$ Substituting in $(2)$ we have $-1-ab-2a^2-b^2-2ab=b\Rightarrow-1-(2a+b)(a+b)=b,$ so $-a^2-2a^4+3a^2-1=-a,$ or $2a^4-2a^2-a+1=0.$ So $a=1$ or $2a^3+2a^2-1=0.$ The first possibility gives $b=-1$ and $c=-1.$ Suppose $m/n$ is a root of $2a^3+2a^2-1=0$ with relatively prime integers. Then $2m^3/n^3+2m^2/n^2-1=0\Rightarrow 2m^3+2m^2n-n^3=0.$ So any prime factor of $n$ must divide $2$ and any prime factor of $m$ must divide $1.$ Hence the only possibilities are $-1/2,1/2,\pm 1$ and it is easy to check that they are not solutions. Thus, the only rational triplets are $\boxed{(0,0,0),(1,-2,0),(1,-1,-1)}.$
| Two years too late
$$(x^3+ax^2+bx+c)-(x-a)(x-b)(x-c)=$$ $$c(a b +1)- (c (a+b)+b(a-1))x+ (2 a+b+c)x^2=0\tag 1$$ So $c=-2a-b$ replaced in $(1)$ leads to
$$-(2 a+b) (a b+1)+[2 a (a+b)+b (b+1)]x=0 \tag 2$$
If $\color{red}{b=-2a}$, then $a(a-1)=0$, that is to say $a=0$ or $a=1$.
If $a=0$, then $b=0$ and $c=0$
If $a=1$, then $b=-2$ and $c=0$
If $\color{red}{b=-\frac 1a}$, we have $$\frac{\left(2 a^4-2 a^2-a+1\right) }{a^2}=\frac{(a-1) \left(2 a^3+2 a^2-1\right)}{a^2}=0$$
So, if $a=1$, $b=-1$ and $c=-1$
The last case is $(2a^3+2a^2-1)=0$, the real solution of which being
$$\frac{1}{3} \left(2 \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\frac{23}{4}\right)\right)-1\right)$$ which has very little chances to be rational.
Then, your solution and conclusions just using algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Two points on a curve that have a common tangent line I have a curve $y = x^3 - x^4$ with an oblique common tangent line across the top of the curve. What is the equation of the tangent?
Have spent hours on this but getting no-where.
| Generally, $y=x^3-x^4=ax+b$ have 4 roots for $x$.
$$x^3-x^4 = ax+b \Rightarrow (x-x_1)(x-x_2)(x-x_3)(x-x_4) = 0 $$
If $ax+b$ is a tangent line on 2 points of the curve, that means
$$x^3-x^4 = ax+b \Rightarrow (x-x_1)^2(x-x_2)^2 = 0 $$
Then we can have
$$
x^4 - x^3 +0\cdot x^2 + ax + b = 0
\Rightarrow
$$
$$
x^4 - 2(x_1+x_2)x^3 + (x_1^2+x_2^2+4x_1x_2)x^2 - 2x_1x_2(x_1+x_2)x + x_1^2x_2^2 = 0
$$
*
*$2x_1 + 2x_2 = 1$
*$x_1^2 + 4x_1x_2 + x_2^2 = 0$
*$-2x_1x_2(x_1+x_2) = a$
*$x_1^2x_2^2 = b$
not too hard to find $a=\frac{1}{8}$, $b=\frac{1}{64}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...
| $$\begin{align}
\left(1+{1\over\sqrt n}\right)^{2n}
&=\left(1+{2\over\sqrt n}+{1\over n}\right)^n\\
&\gt\left(1+{2\over\sqrt n} \right)^n\\
&\ge1+{n\choose1}{2\over\sqrt n}+{n\choose2}\left(2\over\sqrt n\right)^2\\
&\gt1+0+2(n-1)\\
&=2n-1\\
&\ge n
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D-4E+F)x^2+(4A-8+4B+4C+4D-8D+4E-4F)x+(-8A+4C+8D+4F)=1$$
So:
$$A+B+C+D+E=0$$
$$2A+4B+C-2D-4E+F=0$$
$$A+B+C-D+E=2$$
$$-8A+4C+8D+4F=1.$$
But how to find $A$, $B$, $C$, $D$, $E$, $F$?
I also tried substitution $$x=2\sec t$,$
but It caused some difficulty.
| Here is a compact approach with the substitution $x=2\cosh t$,
$$I= \int \frac{1}{(x^2-4)^2}dx$$
$$=\frac{1}{8} \int \text{csch}^3 tdt
= -\frac{1}{8} \int \text{csch} t \>d(\coth t)$$
$$= -\frac{1}{8} \left(\text{csch} t \coth t + \int \text{csch} t \coth^2 tdt \right)$$
$$= -\frac{1}{8} \left(\text{csch} t \coth t + \int \text{csch} t dt\right)-I$$
$$=-\frac{1}{16}\left( \text{csch} t \coth t + \int \text{csch} t dt\right)$$
$$=-\frac{1}{16} \left( \text{csch} t \coth t+ \ln \tanh\frac t2 \right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the minimum of :$P=a+b+c-ab-bc-ca$ soluLet $a,b,c$ be positive real numbers and $a+b+c+abc=4$.
We can rewrite the first equation as $a+b+c=4-abc.$ Then,
\begin{align*}
P&=a+b+c-ab-bc-ca\\&=(4-abc)-ab-bc-ca\\&=4-abc-ab-bc-ca-(a+b+c+1)+(a+b+c+1)\\&=4-(abc+ab+bc+ca+a+b+c+1)+a+b+c+1\\&=4-(a+1)(b+1)(c+1)+a+b+c+1\\&=5+a+b+c-(a+1)(b+1)(c+1)\\&=5+(4-abc)-(a+1)(b+1)(c+1)\\&=9-abc-(a+1)(b+1)(c+1).
\end{align*}It seems that both $abc$ and $(a+1)(b+1)(c+1)$ are maximized when $a=b=c,$ which would give $a=b=c=1$ and then $P=9-(1)(1)(1)-(2)(2)(2)=9-(1+8)=0.$ However, I'm not sure how to prove that statement.
| We may suppose that $a\le b\le c$.
Supposing that $a\gt 1$ gives $4=a+b+c+abc\gt 4$, so we have $a\le 1$.
Supposing that $c\lt 1$ gives $4=a+b+c+abc\lt 4$, so we have $c\ge 1$.
Now using $$b=\frac{4-a-c}{1+ac}$$
we have
$$\begin{align}P&=a+b+c-ab-bc-ca
\\\\&=a+c-ac+(1-a-c)b
\\\\&=a+c-ac+(1-a-c)\cdot \frac{4-a-c}{1+ac}
\\\\&=\frac{(a+c-ac)(1+ac)+4-5(a+c)+(a+c)^2}{1+ac}
\\\\&=\frac{ac(a+c-ac)+(a+c)-ac+4-5(a+c)+(a+c)^2}{1+ac}
\\\\&=\frac{ac(a+c-ac-1)+4-4(a+c)+(a+c)^2}{1+ac}
\\\\&=\frac{ac(1-a)(c-1)+(a+c-2)^2}{1+ac}
\\\\&\ge 0\end{align}$$
$P$ equals $0$ when $a=b=c=1$, so the minimum of $P$ is $\color{red}0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$
Evaluate
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
I did this by $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\left(1+\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)^5$$ and get $0$
Does anyone have another idea?
Thanks
| Say $z = \sin\frac{\pi}{5}+i\cos\frac{\pi}{5}$.
$\big($We have $-iz = \cos\frac{\pi}{5}-i\sin\frac{\pi}{5} = \cos\frac{-\pi}{5}+i\sin\frac{-\pi}{5}$, so $\boxed{z^5 = -i}$ by De'Moivre formula.$\big) $
Notice that $\bar{z}={1\over z}$.
Then your expression is \begin{eqnarray}w &=& (1+z)^5+i(1+\bar{z})^5\\
&=& (1+z)^5+i(1+{1\over z})^5\\
&=& (1+z)^5+i{(z+1)^5\over z^5}\\
&=& (1+z)^5+i{(z+1)^5\over -i}\\
&=& 0\\
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the following limit probably by Riemann Sum: $\lim_{n\to\infty} \sum_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)$ I've tried solving the following limit and I think that it might be possible to transform the sequence into a Riemann Sum.
$$\lim_{n\to\infty} \sum_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right), \\ a \in \mathbb{R}$$
First, I've multiplied and divided by $n$ so I could use ${1\over n}$ as $\Delta x$ and turn it into an integral on $[0, 1]$. Since we have that $(2k - 1)$ in the limit, I thought it would be a good idea to use a midpoint Riemann Sum. So, $x_k^* = \frac{2k - 1}{2n}$
But I just have no idea what to do with the $n^2$ and the $n$ that I've multiplied.
Is this the way to go?
| Try using complex numbers:
$$\sum\limits_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)=\Im\left(\sum\limits_{k=1}^{n}z^{2k-1}\right)=
\Im\left(\frac{z (z^{2n} - 1)}{z^2 - 1}\right)= ...$$
where $z=e^{i\cdot \frac{a}{n^2}}$
$$...=\Im\left(\frac{\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)\left(\cos{\frac{2a}{n}}-1 + i\sin{\frac{2a}{n}}\right)}{\cos{\frac{2a}{n^2}}-1 + i\sin{\frac{2a}{n^2}}}\right)=\\
\Im\left(\frac{\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)
\left(\cos{\frac{2a}{n}}-1 + i\sin{\frac{2a}{n}}\right)
\color{red}{\left(\cos{\frac{2a}{n^2}}-1 - i\sin{\frac{2a}{n^2}}\right)}}{\left(\cos{\frac{2a}{n^2}}-1\right)^2 + \left(\sin{\frac{2a}{n^2}}\right)^2}\right)=\\
\Im\left(\frac{4\cdot\sin{\frac{a}{n^2}}\cdot\sin{\frac{a}{n}}\cdot\left(\cos{\frac{a}{n}} + i \sin{\frac{a}{n}}\right)}{2-2\cdot\cos{\frac{2a}{n^2}}}\right)=\\
\frac{2\cdot\sin{\frac{a}{n^2}}\cdot\sin{\frac{a}{n}}\cdot\sin{\frac{a}{n}}}{1-\cos{\frac{2a}{n^2}}}=
\frac{2\cdot\sin{\frac{a}{n^2}}\cdot\sin^2{\frac{a}{n}}}{2\cdot\sin^2{\frac{a}{n^2}}}=
\color{blue}{\frac{\sin^2{\frac{a}{n}}}{\sin{\frac{a}{n^2}}}}$$
Finally
$$\lim\limits_{n\to\infty}\frac{\sin^2{\frac{a}{n}}}{\sin{\frac{a}{n^2}}}=
\lim\limits_{n\to\infty}\frac{\sin^2{\frac{a}{n}}}{\left(\frac{a}{n}\right)^2} \cdot \frac{\frac{a}{n^2}}{\sin{\frac{a}{n^2}}}\cdot\frac{\left(\frac{a}{n}\right)^2}{\frac{a}{n^2}}=a$$
Some elaborations on the calculations above
$$\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)
\left(\cos{\frac{2a}{n}}-1 + i\sin{\frac{2a}{n}}\right)
\color{red}{\left(\cos{\frac{2a}{n^2}}-1 - i\sin{\frac{2a}{n^2}}\right)}=\\
e^{i\cdot \frac{a}{n^2}} \left(e^{i\cdot \frac{2a}{n}}-1\right) \color{red}{\left(e^{-i\cdot \frac{2a}{n^2}}-1\right)}=
\left(e^{i\cdot \frac{2a}{n}}-1\right) \color{red}{\left(e^{-i\cdot \frac{a}{n^2}}-e^{i\cdot \frac{a}{n^2}}\right)}=...$$
which from $\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}$ is
$$...=\left(e^{i\cdot \frac{2a}{n}}-1\right)\cdot\color{red}{(-2i)\cdot\sin{\frac{a}{n^2}}}=
e^{i\cdot \frac{a}{n}}
\color{blue}{\left(e^{i\cdot \frac{a}{n}}-e^{-i\cdot \frac{a}{n}}\right)}
\cdot\color{red}{(-2i)\cdot\sin{\frac{a}{n^2}}}=\\
\left(\cos{\frac{a}{n}}+i\sin{\frac{a}{n}}\right)\color{blue}{(2i)\cdot\sin{\frac{a}{n}}}\cdot\color{red}{(-2i)\cdot\sin{\frac{a}{n^2}}}=\\
4\cdot\left(\cos{\frac{a}{n}}+i\sin{\frac{a}{n}}\right)\cdot\color{blue}{\sin{\frac{a}{n}}}\cdot\color{red}{\sin{\frac{a}{n^2}}}$$
Other results used are
*
*$1-\cos{x}=2\sin^2{\frac{x}{2}}$ and
*$\lim\limits_{x\to 0}\frac{\sin{x}}{x}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Understand roots of $(z+i)^2=(\sqrt3+i)^3$ I'm trying to solve the equation
$$(z+i)^2=(\sqrt3+i)^3$$ but I don't know how to extract
the roots
$$(z+i)^2=(\sqrt3+i)^3 \rightarrow (z+i)^2=8i \rightarrow z^2+(2i)z-(8i+1)=0$$
$z_{1,2}=-i \pm \sqrt{8i}$.
According to my book the solutions are $2+i$ and $-2-3i$
and I think that they are another way to write them but I don't know how to have them.
Can someone help me to understand?
| You have $z_{1,2} = -i + K$ where $K^2 = 8i$.
So find $K$ where $K^2 = 8i$.....
Let $K = a+bi$ so $K^2 = (a^2 -b^2) + 2abi = 0 + 8i$ so $a^2-b^2 =0$ and $2ab=8$
So solve for $a$ and $b$.....
$a^2 - b^2 =0$ means $a=\pm b$ and $2ab = 8$ means $ab=4$ means $a$ and $b$ are both the same sign. So $a=b$ and $ab =a^2=4$ so $a =\pm 2$ and $b=\pm 2$.
So $K = \pm (2 + 2i)$ and $z_{1,2}=-i \pm(2+2i) = \begin{cases}2+i\\-2-3i\end{cases}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3419620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve a difficult system Problem: solve in $\mathbb R^{3}$ this system
$$
\begin{cases}y(1+x+x^2+x^3)=\dfrac{z}{16}\\ y^2x(1+x+2x^2+x^3+x^4)=\dfrac{2z+17}{16}\\
y^3x^3(1+x+x^2+x^3)=\dfrac{z}{16}\\
y^4x^6=1
\end{cases}
$$
Wolfram alpha gives $(x,y,z)=\left(4,\dfrac{1}{8},170\right)$.
I don't know how to solve it, but my attempt is as follows:
First equation:
$1+x+x^2+x^3=\dfrac{z}{16y}$
So by equation $2$ we find:
$y^{2}x\left(\dfrac{z}{16y}+x^2+x^4\right)=\dfrac{2z+17}{16}$
I don't know how to complete my work.
| The fourth equation is redundant. From the first and the third equation (by equating them and noting that $1+x+x^2+x^3=0$ iff $x=-1$, but $x=-1$ doesn't yield a real solution), you can already deduce that $y^2x^3=1$. If you set $x=t^2$ and $y=1/{t^3}$, then the first and the second equations are now $$\frac{z}{16}=t^3+\frac{1}{t^3}+t+\frac{1}{t}=s(s^2-2)$$ and $$\frac{2z+17}{16}=t^4+\frac{1}{t^4}+t^2+\frac{1}{t^2}+2=(s^2-2)(s^2-1),$$
where $s=t+\frac{1}{t}$. That is
$$(s^2-2)(s^2-1)-2s(s^2-2)=\left(\frac{2z+17}{16}\right)-2\left(\frac{z}{16}\right)=\frac{17}{16}.$$
Therefore
$$s^4-2s^3-3s^2+4s+\frac{15}{16}=0.$$
If $u=2s$, then
$$u^4-4u^3-12u^2+32u+15=0$$
or
$$(u-5)(u+3)(u^2-2u-1)=0.$$
That is,
$$u=5,-3,1\pm\sqrt{2}.$$
By AM-GM, $u=2s=2\left(t+\frac1t\right)$ satisfies $|u|\geq 4$. Thus, $u=5$ is the only solution. Therefore, $s=5/2$, so that $t=2$ or $t=1/2$.
In any case, this means
$$z=16s(s^2-2)=16\left(\frac52\right)\left(\frac{25}{4}-2\right)=170.$$
If $t=1/2$, then
$$(x,y,z)=(1/4,8,170).$$
If $t=2$, then
$$(x,y,z)=(4,1/8,170).$$
So wolframalpha missed a solution.
In fact, all solutions (real or complex) apart from the two solutions above are
$$(x,y,z)=\left(\left(\frac{-3\pm i\sqrt7}{4}\right)^2,\left(\frac{-3\mp i\sqrt7}{4}\right)^3,-6\right)$$
$$(x,y,z)=\left(\left(\frac{1+\sqrt{2}\pm i \sqrt{13-2\sqrt{2}}}{4}\right)^2,\left(\frac{1+\sqrt{2}\mp i \sqrt{13-2\sqrt{2}}}{4}\right)^3,-7-4\sqrt2\right)$$
$$(x,y,z)=\left(\left(\frac{1-\sqrt{2}\pm i \sqrt{13+2\sqrt{2}}}{4}\right)^2,\left(\frac{1-\sqrt{2}\mp i \sqrt{13+2\sqrt{2}}}{4}\right)^3,-7+4\sqrt2\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Writing polynomial using powers of (x-a) I have a question related to writing a polynomial using powers of binomial of form $(x-a).$
I found an example: polynomial $P(x) = x^4 + 2x^3-3x^2-4x+1$ can be written as
$ (x+1)^4-2(x+1)^3-3(x+1)^2+4(x+1)+1$ using powers of $(x+1)$ and Horner's Method. How do we obtain this representation of polynomial? How is Horner's Method used for that?
| This is similar to changing base of a number system.
So you can use long division.
$$P(x) = x^4 + 2x^3-3x^2-4x+1$$
To express it as powers of $(x+1)$, divide $P(x)$ by the new base $(x+1)$
$$\begin{align}
x^4 + 2x^3-3x^2-4x+1 &= (x+1)(x^3+x^2-4x) + \color{blue}{1} \\
x^3+x^2-4x &= (x+1)(x^2-4)+ \color{blue}{4}\\
x^2-4 &= (x+1)(x-1) \color{blue}{-3}\\
x-1 &= (x+1)(1) \color{blue}{- 2}\\
1 &= (x+1)0+ \color{blue}{1}\\
\end{align}$$
$$P(x) = \color{blue}{1}(x+1)^4 \color{blue}{-2}(x+1)^3\color{blue}{-3}(x+1)^2+\color{blue}{4}(x+1)+\color{blue}{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$
My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$
| $$ 8^{y+d} - 9^y = 431 $$
$$ 8^d = \frac{431 + 9^y}{8^y} $$
$$ 8^d = (\frac{9}{8})^y+\frac{431}{8^y} $$
$$ 8^d \sim (\frac{9}{8})^y $$
$$ 3d\cdot ln(2) \sim y \cdot(2ln(3)-3ln(2)) $$
$$ d \sim y \cdot(\frac{2\cdot ln(3)-3\cdot ln(2)}{3\cdot ln(2)}) $$
$$ d \sim y \cdot 0.0566416671474374543024926292985... $$
$$ \frac{3}{53} = 0,056603773... $$
First better aproximation, for $d=3$ and $y=53$:
$$ 8^{56} -9^{53} = -1,56579...\cdot10^{48} $$
Etc...
This problem is related to the approximation of $log_2 (3)$ to a rational number.
And as the different rational ones approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Heptadecagon Derivation I am currently very interested in the derivation of the constructability of the 17-gon by Carl Friedrich Gauß.
Has someone got an easy explanation for the solution of
$$x^{17} - 1=0?$$
That was the equation he solved with which he showed
\begin{align}\cos \frac{360^\circ}{17}&=\frac{1}{16}\left( -1 + \sqrt{17} + \sqrt{ 2\left(17 -\sqrt{17} \right)}+ 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right) \\&\approx 0.93247222940435580457311589182156.\end{align}
Can someone briefly explain his derivation, please?
| This is an elementary proof.
Let $\varphi=\frac\pi{17}$,
$$S=-\sum_{n=1}^8(-1)^n\cos(n\varphi)$$
Multiplication by $2\cos(\varphi/2)$ gives:
\begin{align}
2S\cos\left(\frac\varphi 2\right)
&=-\sum_{n=1}^8(-1)^n\left(\cos\left(\frac{2n-1}2\varphi\right)-\cos\left(\frac{2n+1}2\varphi\right)\right)\\
&=\cos\left(\frac 12\varphi\right)-\cos\left(\frac{17}2\varphi\right)\\
&=\cos\left(\frac\varphi2\right)
\end{align}
so that $S=\frac 12$.
Now let
\begin{align}
X&=\cos(3\varphi)+\cos(5\varphi)-\cos(6\varphi)+\cos(7\varphi)\\
Y&=-\cos(\varphi)+\cos(2\varphi)+\cos(4\varphi)+\cos(8\varphi)
\end{align}
so that $X-Y=\frac 12$.
Moreover, $XY=4S=2$, hence $XY=1$ which gives
\begin{align}
&X=\frac{\sqrt{17}+1}4&&Y=\frac{\sqrt{17}-1}4
\end{align}
Now let
\begin{align}
z&=\cos(3\varphi)+\cos(5\varphi)\\
x&=\cos(6\varphi)-\cos(7\varphi)
\end{align}
so that $X=z-w$.
Then $2zx=S=\frac 12$, so that we obtain
\begin{align}
z&=\frac{1+\sqrt{17}+\sqrt{34+2\sqrt{17}}}8\\
x&=\frac{-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}}8
\end{align}
Similarly, $y=\cos(\varphi)-\cos(4\varphi)$ and $v=\cos(2\varphi)+\cos(8\varphi)$ satisfy $Y=v-y$ and $yv=\frac 14$, thus giving
\begin{align}
y&=\frac{1-\sqrt{17}+\sqrt{34-2\sqrt{17}}}8\\
v&=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}8
\end{align}
Finally $\cos(2\varphi)+\cos(8\varphi)=v$ and $\cos(2\varphi)\cos(8\varphi)=\frac x2$ from which we get
$$\cos(2\varphi)=\frac 1{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the solution ?
| $n^2 = 1 + (n-1)!$
$n^2 -1 = (n-1)!$
$(n-1)(n+1) = (n-1)!$ (If we assume $n>1$)
$n+1 = \frac {(n-1)!}{n-1} = (n-2)!$.
$n-2 + 3 = (n-2)!$ (If we assume $n > 2$)
$1 + \frac {3}{n-2} = \frac {(n-2)!}{n-2} = (n-3)!$ is an integer.
So $n-2|3$. But $3$ is prime. So either $n-2=1$ or $n-2 = 3$. But we assume $n > 2$ so $n=5$.
.... But to be thourough we have to consider $n \le 2$ ....
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \equiv 2^6 \equiv 4\pmod{10} \\
(...)$$
I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?
| Since you found $572 \equiv 2\pmod{10} $ you have to check only what is $$2^{42} \equiv ?\pmod{10} $$
Since $2^5 \equiv 2$ we have $$2^{42} = 2^2\cdot (\color{red}{2^5})^8 \equiv 2^2\cdot \color{red}{2}^8 \equiv 2^{10} \equiv (\color{blue}{2^5})^2 \equiv \color{blue}{2}^2\equiv 4 \pmod {10}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If a variable chord of the hyperbola subtend a right angle at the centre, find the radius of the circle it is tangent to
If a variable line $x\cos\alpha+y\sin\alpha=p$ which is a chord of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $(b>a)$ subtend a right angle at the centre of hyperbola,then prove that it always touches a fixed circle whose radius is $\dfrac{ab}{\sqrt{b^2-a^2}}$
$x\cos\alpha+y\sin\alpha=p$ touches(tangent to) a fixed circle $x^2+y^2=p^2$ of radius $p$. So I need to find $p$.
Let $A$,$B$ be the points where the given lines intersects the hyperbola, ie. $\angle AOB=90^\circ$
How do I proceed further and find the radius of the required circle ?
Note: I really do not want to use some existing shortcut formula here
| The line equation seems a bit of a distraction. Here's how I'd solve the problem.
Suppose $H$ and $K$ determine a chord of the hyperbola that subtends a right angle at its center (the origin), $O$. We can write
$$H = (h \cos\phi, h \sin\phi) \qquad K = (k \sin\phi,-k \cos\phi)$$
where $h:=|OH|$, $k:=|OK|$, and $\phi$ is some arbitrary angle. Define $p := |OP|$, where $P$ is the foot of the altitude from $O$ to $\overleftrightarrow{HK}$. Calculating the area of $\triangle HOK$ in two ways, we have
$$\frac12|OH||OK| = \frac12|OP||HK|\quad\to\quad hk = p\sqrt{h^2+k^2}\quad\to\quad p = \frac{1}{\sqrt{\dfrac{1}{h^2}+\dfrac{1}{k^2}}} \tag{1}$$
Since $H$ and $K$ both lie on the hyperbola, we have
$$\begin{align}
\frac{h^2}{a^2}\cos^2\phi - \frac{h^2}{b^2}\sin^2\phi &= 1 \tag{2} \\[4pt]
\frac{k^2}{a^2}\sin^2\phi - \frac{k^2}{b^2}\cos^2\phi &= 1 \tag{3}
\end{align}$$
Therefore, $k^2 (2) + h^2 (3)$ becomes
$$\frac{h^2k^2}{a^2}(\cos^2\phi+\sin^2\phi)-\frac{h^2k^2}{b^2}(\sin^2\phi+\cos^2\phi) = h^2 + k^2
\quad\to\quad
\frac{1}{a^2}-\frac{1}{b^2} = \frac{1}{h^2} + \frac{1}{k^2} \tag{4}
$$
Thus, recalling $(1)$, we have
$$p = \frac{1}{\sqrt{\dfrac{1}{a^2}-\dfrac{1}{b^2}}} = \frac{ab}{\sqrt{b^2-a^2}} \tag{5}$$
Since $p$ is the distance from $\overleftrightarrow{HK}$ to the origin, we conclude that the line is always tangent to the circle with the radius given in $(5)$, as desired. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g(x)+5$ where $g(x)=(x+1)(x+2)(x+3)(x+4)$
Let's find the maximum value of g(x)
It can be clearly seen that from $x=-6$ to $x=6$, maximum value of $g(x)$ is at $x=6$.
$$g(6)=5040$$
Let's find the minimum value of $g(x)$
From the sign scheme one can see that negative value of $g(x)$ occurs in the interval $(-4,-3)$ and $(-2,-1)$
Hence intuitively it feels that the minimum value of g(x) would be at $x=-\dfrac{3}{2}$ or at $x=-\dfrac{7}{2}$ as in case of parabola also, minimum value is at average of both the roots.
So $g\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}=-\dfrac{15}{8}$
At $x=-\dfrac{7}{2}$, $g\left(-\dfrac{7}{2}\right)=-\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{8}$
So range of $g(x)$ would be $[\dfrac{-15}{8},5040]$
Hence range of $f(x)$=$\left[\dfrac{25}{8},5045\right]$
But it is given that the range is $[a,b]$ where $a,b\in N$
I am stuck here. I am also not able to prove mathematically that at $x=-\dfrac{3}{2}$ or $x=-\dfrac{7}{2}$, minimum value of $g(x)$ will occur.
Please help me in this.
| The function is symmetric about the line $x = -2.5$
We could write
$f(x) = ((x+2.5) + 1.5)((x+2.5)+0.5)((x+2.5)-0.5)((x+2.5) - 1.5) + 5$
Which equals
$f(x) = ((x+2.5)^2 - 1.5^2)((x+2.5)^2-0.5^2) + 5$
Now lets do the substituion $u = (x+2.5)^2$
$(u - 2.25)(u-0.25) + 5$
Which has its minimum when $u = \frac {5}{4}$
$(\frac {5}{4} - 2.25)(\frac {5}{4}-0.25) + 5 = 4$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need to find an $ N \in \mathbb{N}$ so that : $ \forall \epsilon > 0 n > N \implies \left| a_n - \frac{1}{2} \right| < \epsilon $
Next step I simplify $a_n$ : $ a_n = \frac{n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right) \cdot \sqrt{1+ \frac{1}{n}} + 1 } {\sqrt{1+ \frac{1}{n}} + 1} = \frac{1}{\sqrt{1+ \frac{1}{n}} + 1}$
And then I got stuck,what am I supposed to do with : $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|$
| You started well. You need to show that $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|<\epsilon$ for $n>N$ where $N$ is some number.
$$ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2}\right|=\left|\frac{\sqrt n}{\sqrt{n+1} +\sqrt n} - \frac{1}{2} \right|=\left|\frac{2\sqrt n -\sqrt{n+1} -\sqrt n}{\sqrt{n+1} +\sqrt n}\right|=\left|\frac{\sqrt n -\sqrt{n+1}}{\sqrt{n+1} +\sqrt n}\right|=\left|\frac{1}{(\sqrt{n+1} +\sqrt n)^2}\right|<\left|\frac{1}{(\sqrt{n} +\sqrt n)^2}\right|=\frac{1}{4n}$$
Now we want to find such $N$ that $\frac{1}{4N}<\epsilon \rightarrow N>\frac{1}{4\epsilon}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute $\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}$? How to prove
$$\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=\frac18\zeta(2)\zeta(3)-\frac{25}{32}\zeta(5)?$$
I came across this series while I was working on a nice integral $\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}dx$ and because I managed to calculate the integral in a different way, I got the closed form of the alternating series and solution will be posted soon.
Here is my question, is it possible to calculate the sum the same way @M.N.C.E calculated $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ or by series manipulations? All approaches appreciated though. Thank you
By the way, is this result known in the literature?
| Different approach
$$S=\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=-H_{1/2}+\sum_{n=2}^\infty\frac{(-1)^nH_{n/2}}{n^4},\quad H_{1/2}=2\ln2-2$$
use the fact that
$$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n)+\sum_{n=1}^\infty f(2n+1)$$
$$\Longrightarrow S=2-2\ln2+\frac1{16}\sum_{n=1}^\infty\frac{H_{n}}{n^4}-\sum_{n=1}^\infty\frac{H_{n+1/2}}{(2n+1)^4}$$
Lets compute the last sum,
notice that
$$H_{n+1/2}=2H_{2n+1}-H_n-2\ln2$$
$$\Longrightarrow \sum_{n=1}^\infty\frac{H_{n+1/2}}{n^4}=2\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^4}-\sum_{n=1}^\infty\frac{H_{n}}{(2n+1)^4}-2\ln2\underbrace{\sum_{n=1}^\infty\frac{1}{(2n+1)^4}}_{\frac{15}{16}\zeta(4)-1}$$
where
\begin{align}
2\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^4}&=2\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^4}-2\\
&=\sum_{n=0}^\infty\frac{H_{n+1}}{(n+1)^4}+\sum_{n=0}^\infty\frac{(-1)^nH_{n+1}}{(n+1)^4}-2\\
&=\sum_{n=1}^\infty\frac{H_{n}}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}-2
\end{align}
and
$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}8\ln2\zeta(4)-\frac{21}{16}\zeta(2)\zeta(3)$$
Which follows from using the generalization
$$\sum_{n=1}^\infty\frac{H_n}{(n+a)^2}=\left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2}$$
Combining the results gives
$$S=-\frac{15}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{31}{8}\zeta(5)-\frac{21}{16}\zeta(2)\zeta(3)$$
and by substituting the results of $\sum_{n=1}^\infty\frac{H_n}{n^4}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ we get the claimed closed form.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given a polynomial f(x)=x(x+1)(x+2)(x+3)+1 Given a polynomial $f(x)=x(x+1)(x+2)(x+3)+1$ and $p$ an odd prime, prove that it exists a number $n$ such that $p$ divides $f(n)$ if and only if it exists an integer $m$ such that $p$ divides $m^2-5$.
| You have that
$$f(x) = x(x + 1)(x + 2)(x + 3) + 1 \tag{1}\label{eq1A}$$
Since $p$ is an odd prime, then any $n$ with $p \mid f(n)$ is true iff only $p \mid 16f(n)$. Multiply $f(n)$ by $16$ and distribute the powers of $2$ to each factor with $n$ in it to get
$$16f(n) = (2n)(2n + 2)(2n + 4)(2n + 6) + 16 \tag{2}\label{eq2A}$$
Hint: Now, let
$$m = 2n + 3 \tag{3}\label{eq3A}$$
to get
$$\begin{equation}\begin{aligned}16f(n) = g(m) & = (m - 3)(m - 1)(m + 1)(m + 3) + 16 \\ & = (m - 3)(m + 3)(m - 1)(m + 1) + 16 \\ & = (m^2 - 9)(m^2 - 1) + 16 \\& = m^4 - 10m^2 + 9 + 16 \\& = m^4 - 10m^2 + 25 \\ & = (m^2 - 5)^2\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus $p \mid (m^2 - 5)^2$ and, since $p$ is a prime, this means $p \mid m^2 - 5$.
For the other direction, for any $m$ where $p \mid m^2 - 5$, if $m$ is even, then use a new $m$ with $p$ added to it so it's odd but where you still have $p \mid m^2 - 5$. Hint for remainder of the proof:
Then you can reverse the above steps in \eqref{eq4A} to get, from \eqref{eq3A}, that with the integer
$n = \frac{m - 3}{2}$, that $p \mid 16f(n) \implies p \mid f(n)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \right)^2+\dots$$
I have no idea how to start. Any hint would be really appreciated. THANKS!
| Notice, that:
$$\left(\frac{1}{1}\right)^2+\left(\frac{1}{2} + \frac{1}{3}\right)^2 + \dots<\left(\frac{1}{1}\right)^2+\left(\frac{2}{2}\right)^2+\left(\frac{3}{4}\right)^2+\left(\frac{4}{7}\right)^2=2+\sum_{n=2}^{\infty}\left(\frac{n}{\frac{n(n-1)}{2}+1}\right)^2<2+\sum_{n=2}^{\infty}\left(\frac{2}{n-1}\right)^2$$
Since it's bounded by converging series, it's convergent itself.
| {
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"url": "https://math.stackexchange.com/questions/3436804",
"timestamp": "2023-03-29T00:00:00",
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Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$
Solve over positive reals $$x+y^2+2xy=9$$
$$y+z^2+2yz=47$$
$$z+x^2+2xz=16$$
With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations,
$$(8-y-z)^2+y^2+2(8-y-z)y=9$$
$$y+z^2+2yz=47$$
Which becomes
$$-y^2-2yz+15y-z=1$$
$$y+z^2+2yz=47$$
Adding up gets
$$-y^2+16y+z^2-z=1$$
$$\implies z^2=z+1+y^2-16y$$
And this substitutes to
$$y+z+1+y^2-16y+2yz=47$$
$$\implies y+\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)+1+y^2-16y+2y\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)=47$$
However, this results into a nasty quartic which I don't know how to simplify. Thanks!
| $\begin{cases}
x+y^2+2xy=9\\
y+z^2+2yz=47\\
z+x^2+2xz=16
\end{cases} \implies
\begin{cases}
(2 x + 2 y)^2 - (2 x - 1)^2 = 35\\
(2 y + 2 z)^2 - (2 y - 1)^2 = 187\\
(2 z + 2 x)^2 - (2 z - 1)^2 = 63
\end{cases} \implies
\begin{cases}
x=1\\
y=2\\
z=5
\end{cases}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of a Cauchy Product I have the following series obtained via the Cauchy Product of the alternating harmonic series with itself
$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \right ) \cdot \left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \right ) = \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{(-1)^k}{k} \cdot \frac{(-1)^{n-k}}{n-k}
=
\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{(-1)^n}{k(n-k)}$
I wish to check the convergence of this Cauchy Product. I do not know how to handle double series. I do however know that I have to check if the following converges or diverges
$\sum_{n=1}^{\infty} c_n$
Where
$c_n = \sum_{k=1}^{n} \frac{(-1)^n}{k(n-k)}$
I just have no idea how to handle $c_n$. Can anyone please point out how to start?
P.S
The definition of the Cauchy Product I used is
$\left( \sum_{n=0}^{\infty} a_k \right) \cdot \left( \sum_{n=0}^{\infty} b_k \right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_k \cdot b_{n-k}$
| Since
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1},$$
we have, using the Cauchy product formula,
$$\begin{align}\left(\sum_{n=1}^\infty \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^\infty \frac{(-1)^n}{n}\right) &= \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1}\right)\left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1}\right)\\&= \sum_{n=0}^\infty \sum_{k=0}^n\frac{(-1)^{k+1}}{k+1}\frac{(-1)^{n+1-k}}{n+1-k}\\ &= \sum_{n=0}^\infty (-1)^n\sum_{k=0}^n\frac{1}{(k+1)(n+1-k)}\\ &= \sum_{n=0}^\infty (-1)^n c_n\end{align},$$
where
$$c_n = \sum_{k=0}^n\frac{1}{(k+1)(n+1-k)} = \sum_{k=0}^n\frac{1}{n+2} \left(\frac{1}{k+1} + \frac{1}{n+1-k} \right) \\= \frac{1}{n+2} \left(\sum_{k=0}^n \frac{1}{k+1} + \sum_{k=0}^n \frac{1}{n+1-k} \right) = \frac{2}{n+2}\sum_{k=0}^n\frac{1}{k+1} = \frac{2H_{n+1}}{n+2}$$
It is not difficult to show using the representation of $c_n$ in terms of the harmonic sum $H_{n+1}$ that $c_n$ is decreasing. We also have $c_n \to 0$ as $n \to \infty$. This follows from the fact that $H_{n+1} \sim \log(n+1)$. Alternatively, we can apply the Stolz-Cesaro theorem to get
$$\lim_{n \to \infty} \frac{H_{n+1}}{n+2}= \lim_{n \to \infty}\frac{H_{n+2} - H_{n+1}}{n+3 - (n+2)} =\lim_{n \to \infty} \frac{1}{n+2} = 0 $$
Therefore, the Cauchy product series converges by the alternating series test.
| {
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How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, however I can't find a way to prove it.
| For any $n \ge 8$, we have $$\dfrac{n-2}{\binom{n}{2}} = \dfrac{n-2}{\tfrac{n(n-1)}{2}} = \dfrac{2}{n} \cdot \dfrac{n-2}{n-1} \le \dfrac{2}{n} \le \dfrac{1}{4},$$
and thus, $$\sum_{k = 3}^{n-1}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n-1}{4} \implies \sum_{k = 3}^{n}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n}{4}.$$
So if you can prove that $\displaystyle\sum_{k = 3}^{n}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n}{4}$ for $3 \le n \le 7$, then you can use induction to finish the proof.
Remark: Obviously, Kevin Song's analysis gives a much better asymptotic bound. But if you only need to prove the upper bound $\dfrac{n}{4}$, this might be simpler.
| {
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Understanding derivation of Catalan Numbers I encountered a derivation for finding the nth Catalan Number $C_n$ using generating functions in the book Diskrete Strukturen, Band 1 (Steger 2007, p.178).
Given is the recursive function $C_0 := 1$, $C_n := \sum_{k=1}^n C_{k-1}C{n-k} \;\;\; (\forall n \geq 1)$.
The first part of the proof shows how to obtain the equality:
$$ C_n = -\frac{1}{2}\binom{1/2}{n+1}(-4)^{n+1}$$
I understand that part so no need to go into further detail on it. However, the following simplification steps are only sparsely commented in the book:
\begin{align*}
C_n &= -\frac{1}{2}\binom{1/2}{n+1}(-4)^{n+1}\\
&= -\frac{1}{2}\frac{\frac{1}{2} (\frac{1}{2} - 1) (\frac{1}{2} - 2) \ldots(\frac{1}{2} - n)}{ (n+1)!}(-4)^{n+1}\\
&= \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) \cdot 2^n}{ (n+1)!}\\
&= \frac{1}{n+1}\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{ n!} \frac{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}\\
&= \frac{1}{n+1}\binom{2n}{n}
\end{align*}
I can see how the first two equalities are justified, but I don't see how the the last three equalities came to be. I suspect that I am just missing some essential identities and that knowing these will make these steps rather easy. In any case, a more rigorous walkthrough would be much appreciated.
Thanks for the help & best wishes,
Rafael
| For the third equality,
\begin{align*}
-\frac{1}{2}\frac{\frac{1}{2} (\frac{1}{2} - 1) (\frac{1}{2} - 2) \ldots(\frac{1}{2} - n)}{ (n+1)!}(-4)^{n+1}&=
(-1)^n\cdot\frac{1}{2}\cdot\left(\frac{1}{2}\right)^{n+1}\frac{1(2-1)(4-1)\cdots(2n-1)}{(n+1)!}(-1)^n4^{2n+2}\\
&= 2^{-n-2}4^{2n+2}\frac{1\cdot 3\cdot 5\cdots 2n-1}{(n+1)!} \\
&= \frac{1\cdot 3\cdot 5\cdots (2n-1)\cdot 2^n}{(n+1)!}.
\end{align*}
The fourth equality results from multiplying top and bottom by $1\cdot 2\cdot 3 \cdots n$ and distributing the factor of $2^n$ across these terms in the numerator. Finally, this gives
\begin{align*}
\frac{1}{n+1}\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}{ n!} \frac{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}
&= \frac{1}{n+1}\frac{1\cdot 2\cdot 3\cdots 2n}{n!(1\cdot 2\cdot 3\cdots n)} \\
&= \frac{1}{n+1} \frac{(2n)!}{(n!)^2} \\
&= \frac{1}{n+1}\binom{2n}{n}.
\end{align*}
| {
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Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1}>=0 \text { and } \dfrac{x^2-1-x^2-1}{x^2+1}<=0$$
$$\dfrac{2x^2}{x^2+1}>=0 \text { and } \dfrac{-2}{1+x^2}<=0$$
$$x\in R$$
$$x^2-1\ne0$$
$$x\ne\pm1$$
$$\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$$
$$\pi-\cos^{-1}\dfrac{1-x^2}{1+x^2}-\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac{2\pi}{3}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}$$
Substituting $x$ by $\tan\theta$
$$x=\tan\theta$$
$$\tan^{-1}x=\theta$$
$$\theta\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)-\{-\dfrac{\pi}{4},\dfrac{\pi}{4}\}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-\tan^2\theta}{1+\tan^2\theta}+\tan^{-1}\dfrac{2\tan\theta}{1-\tan^2\theta}$$
$$\dfrac{\pi}{3}=\cos^{-1}(\cos2\theta)+\tan^{-1}(\tan2\theta)$$
$$2\theta\in(-\pi,\pi)-\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\}$$
Now we break the range of $2\theta$ into various parts:-
Case $1$: $2\theta\in\left(-\pi,-\dfrac{\pi}{2}\right)$,$\theta\in\left(-\dfrac{\pi}{2},-\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\pi+2\theta+\pi+2\theta$$
$$-\dfrac{8\pi}{3}=4\theta$$
$$-\dfrac{2\pi}{3}=\theta$$
But it is not the range of $\theta$ we assumed
Case $2$: $2\theta\in\left(-\dfrac{\pi}{2},0\right]$,$\theta\in\left(-\dfrac{\pi}{4},0\right]$
$$\dfrac{\pi}{3}=-2\theta+2\theta$$
$$\dfrac{\pi}{3}=0 \text { not possible }$$
Case $3$: $2\theta\in\left(0,\dfrac{\pi}{2}\right)$,$\theta\in\left(0,\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\theta+2\theta$$
$$\dfrac{\pi}{12}=\theta$$
It is coming in the range of $\theta$, so its a valid solution.
$$\tan^{-1}x=\dfrac{\pi}{12}$$
$$x=\tan\dfrac{\pi}{12}$$
$$x=2-\sqrt{3}$$
Case $4$: $2\theta\in\left(\dfrac{\pi}{2},\pi\right)$, $\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$
$$\dfrac{\pi}{3}=2\pi-2\theta+2\theta-\pi$$
$$\dfrac{\pi}{3}=\pi \text { not possible }$$
So only solution is $2-\sqrt{3}$, but actual answer is $2-\sqrt{3}, \sqrt{3}$
| Let $x=\tan\dfrac t2$ for $t\in(-\pi,\pi)$. The equation reads
$$\tan^{-1}(-\tan t)+\cos^{-1}(-\cos t)=\frac{2\pi}3$$
or
$$\tan^{-1}(\tan t)+\cos^{-1}(\cos t)=\frac{\pi}3.$$
By lazily checking on a plot, we see that the function has a null slope in the negatives, where there are no solutions.
In the positives, the slope is $2$ and there are two roots:
$$2t=\frac\pi3$$ and $$2t-\pi=\frac\pi3.$$
You deduce $x$ from $t$.
$\tan\dfrac\pi3=\sqrt3$ or $\tan\dfrac\pi{12}=2-\sqrt3$.
Note:
Despite the use of a plot, this resolution is rigorous, because we know that these functions are piecewise linear and we know the limits of the pieces, so we could check formally.
| {
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Prove that f(2000x)=2000f(x) Given function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+y+2xy)=f(x)+f(y)+2f(xy)$. Prove that $f(2000x)=2000f(x)$.
Letting $x=y=1$ yields $f(4)=4f(1)$, which means that $f(n)=nf(1)$. But I cannot prove this by induction.
And is it possible to prove that this function is monotonous and additive? If so, then the function is linear and the work must be much easier.
| We first prove by induction that we do in fact have that $f(n) = nf(1)$ for all integers $n$.
Taking $x = y = 0$ in the functional equation, we get that $f(0) = 4f(0)$, and so $f(0) = 0$.
Taking $y = -1$ in the functional equation gives us that
$$
f(-x - 1) = f(x) + f(-1) + 2f(-x) \label{a}\tag{1}
$$
for all real numbers $x$. In particular, taking $x = -1$ tells us that
$$
0 = f(0) = f(-(-1) - 1) = 2f(-1) + 2f(1)
$$
and so $f(-1) = -f(1)$.
Suppose that for some natural number $n$, we have that $f(n) = nf(1)$ and $f(-n) = -nf(1)$. We have already seen that this is true when $n \in \{0, 1\}$.
Taking $x = n$ in Equation $\ref{a}$ gives us that
$$
f(-(n + 1)) = f(n) + f(-1) + 2f(-n) = nf(1) - f(1) + 2(-n) f(1) = -(n + 1) f(1),
$$
and then taking $x = -(n + 1)$ in Equation $\ref{a}$ gives us
$$
nf(1) = f(n) = f(-(n + 1)) + f(-1) + 2f(n + 1) = -(n + 1) f(1) - f(1) + 2f(n + 1),
$$
and so $f(n + 1) = (n + 1) f(1)$.
We see by induction that $f(n) = n f(1)$ and $f(-n) = -n f(1)$ for all natural numbers $n$.
Let $y = -\frac{1}{2}$ in the original equation. We get that
$$
f\left(-\frac{1}{2} \right) = f(x) + f\left(-\frac{1}{2} \right) + 2f\left( -\frac{x}{2} \right)
$$
and so we have (by replacing $x$ with $-2x$) that
$$
f(-2x) = -2f(x) \label{c} \tag{2}
$$
for all $x$.
Returning to the original equation, taking $y = 1$ gives us that
$$
f(3x + 1) = 3f(x) + f(1) \label{b} \tag{3}.
$$
Taking $x \leftarrow -x - 1$ in Equation $\ref{b}$ gives us that
$$
f(-3x - 2) = 3f(-x - 1) + f(1),
$$
which upon making use of Equation $\ref{a}$ gives us that
$$
f(-3x - 2) = 3f(x) - 2f(1) + 6f(-x).
$$
Taking $y = -2$ in the original equation tells us that
$$
f(-3x - 2) = f(x) + f(-2) + 2f(-2x) = f(x) - 2f(1) + 2f(-2x)
$$
and so we have that
$$
2f(-2x) = 2f(x) + 6f(-x).
$$
Using Equation $\ref{c}$, this becomes
$$
-4f(x) = 2f(x) + 6f(-x)
$$
and so $f(-x) = -f(x)$ for all $x$.
Equation $\ref{c}$ now tells us that, in fact, $f(2x) = 2f(x)$ for all $x$.
Returning to Equation $\ref{a}$ and making use of the fact that $f(-x) = -f(x)$, we get
$$
-f(x + 1) = f(x) - f(1) - 2f(x)
$$
and so
$$
f(x + 1) = f(x) + f(1)
$$
for all $x$. By induction, this implies that $f(x + n) = f(x) + nf(1)$ for all natural numbers $n$.
Taking $y = 2$ in the original equation gives us that
$$
f(5x) + 4f(1) = f(5x + 4) = f(x) + f(4) + 2f(2x),
$$
and since $f(2x) = 2f(x)$ and $f(4) = 4f(1)$, we get that $f(5x) = 5f(x)$ for all $x$.
We thus have that
$$
f(2000x) = 2f(1000x) = \dots = 16 f(125x) = 80 f(25x) = 400 f(5x) = 2000f(x)
$$
for all $x$.
In fact, we can show by induction that $f(nx) = nf(x)$ for all natural numbers $n$ and all real numbers $x$.
We already know that $f(1x) = 1f(x)$. Suppose that $f(kx) = kf(x)$ for all $k \leq 2n + 1$. Then since $f(2x) = 2f(x)$, we have that
$$
f((2n + 2) x) = 2f((n + 1)x) = 2(n + 1) f(x)
$$
by the induction hypothesis.
Taking $y = n + 1$ in the original equation, we also have that
$$
f((2n + 3)x) + (n + 1) f(1) = f((2n + 3) x + (n + 1)) = f(x) + f(n + 1) + 2f((n + 1) x)
$$
and so
$$
f((2n + 3) x) = f(x) + 2f((n + 1) x) = (2n + 3) f(x)
$$
by the induction hypothesis.
We thus also have that $f(kx) = kf(x)$ for all $k \leq 2(n + 1) + 1$, and so we have that $f(kx) = kf(x)$ for all $k$ by induction.
This then also implies that $f(qx) = qf(x)$ for all rational numbers $q$.
| {
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An Analytic Inequality $$\frac 12\ln(AC-B^2)\leq\frac{1}{2\pi}\int_{0}^{2\pi}\ln(A\sin^2\theta+2B\sin\theta\cos\theta+C\cos^2\theta)d\theta(A,C>0)$$
I see this inequality from a book about Honeycomb structure.It can be proved by dividing $[0,2\pi]$.
$$|AC-B^2|^{\frac n2}\leq\prod_{l=1}^{n}\left[A\sin^2\frac{\pi(2l-1)}{n}-2B\sin\frac{\pi(2l-1)}{n}\cos\frac{\pi(2l-1)}{n}+C\cos^2\frac{\pi(2l-1)}{n}\right]$$
but I think there must be a direct way to prove this inequality.Any help will be thanked.
| The integral can be evaluated in closed form. The expression under the logarithm is $$A\frac{1-\cos 2\theta}{2}+B\sin 2\theta+C\frac{1+\cos 2\theta}{2}=\frac{A+C}{2}+\frac{C-A}{2}\cos 2\theta+B\sin 2\theta,$$ which is equal to $A'+B'\cos(2\theta-\phi)$ with $$A'=\frac{A+C}{2},\qquad B'=\sqrt{B^2+\Big(\frac{A-C}{2}\Big)^2},$$ and some value of $\phi$ we don't care about, since if $f$ is $2\pi$-periodic, we have $$\int_0^{2\pi}f(2\theta-\phi)~d\theta=\int_0^{2\pi}f(2\theta)~d\theta=\int_0^{2\pi}f(\theta)~d\theta.$$ Thus we're left with $\int_0^{2\pi}\ln(A'+B'\cos\theta)~d\theta$, which can be evaluated using $$\int_0^{2\pi}\ln(1+d\cos\theta)~d\theta=2\pi\ln\frac{1+\sqrt{1-d^2}}{2}\qquad(|d|<1)$$ where the principal values are taken (see this answer of mine); the result is $$2\pi\ln\color{lightgray}{\left[\color{black}{\frac{1}{2}\left(\frac{A+C}{2}+\sqrt{AC-B^2}\right)}\right]}.$$
| {
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Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$ Let $x,$ $a,$ $b,$ $c$ be nonnegative real numbers. Prove that
$$(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le \left( x + \frac{a + b + c}{3} \right)^3.$$
I know that this problem is a RHS-AM-GM-HM problem, but I am unsure how to solve it. I think that the leftmost part is AM, but that is as far as I was able to get. I also think that taking the cube root of everything might change it into something more manageable, but I am still unsure about that.
Can anyone help me with this or give me tips about how to solve it? Thank you.
| Using AM
$$\left( x + \frac{a + b + c}{3} \right)^3=
\left(\frac{(x+a) + (x+b) + (x+c)}{3} \right)^3\overset{AM}{\geq} \\
\left(\sqrt[3]{(x+a)(x+b)(x+c)}\right)^3=
(x+a)(x+b)(x+c)$$
Using HUYGEN’S INEQUALITY (which is easy to prove using AM-GM)
$$(x+a)(x+b)(x+c)=x^3\left(1+\frac{a}{x}\right)\left(1+\frac{b}{x}\right)\left(1+\frac{c}{x}\right)\geq\\
x^3\left(1+\sqrt[3]{\frac{a}{x}\frac{b}{x}\frac{c}{x}}\right)^3=
\left(x+\sqrt[3]{abc}\right)^3$$
| {
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Find $\lim_{x\to 0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$
If $$f(x)= \frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$$ is continuous at $x=0$ then find $f(0)$
$$
f(0)=\lim_{x\to0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}\\
=\lim_{x\to0}\big[\frac{a^x-1}{x}\big]^3.\frac{x\log a}{\sin(x\log a)}.\frac{x^2}{\log a.\log(1+x^2\log a^2)}\\
=(\log a)^3.1.\lim_{x\to0}\frac{x^2}{\log a.\log(1+x^2\log a^2)}\\
=(\log a)^2.1.\lim_{x\to0}\frac{x^2}{\log(1+x^2\log a^2)}=?
$$
Can I use the fact that $\log(1+x)=x-x2/2+x^3/3-....$ but is it not for $-1<x\leq1$ ?
| We can use that
$$\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log (a^2))}=\frac{(a^x-1)^3}{x^3}\frac{x^3}{\sin(x\log a)\log(1+x^2\log (a^2))}$$
and by standard limits
$$\frac{(a^x-1)^3}{x^3} \to \log^3 a$$
$$\frac{x^3}{\sin(x\log a)\log(1+x^2\log (a^2))}=\frac{x\log a}{\sin(x\log a)}\frac{x^2\log (a^2)}{\log(1+x^2\log (a^2))}\frac1{2\log^2 a} \to \frac1{2\log^2 a}$$
Edit
Yes we can also use the series expansion since $x^2\log^2 a \to 0$ and therefore eventually $0<x^2\log (a^2)<1$.
| {
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Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}$
and so $x=\frac{7\pi}{6},\frac{11\pi}{6}$
Solving for the other factor,
$\sin x-1=0 \Rightarrow \sin x=1$
And so $x=\frac{\pi}{2}$
Now we have found all our base solutions, and so ALL the solutions can be written as so:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$
| Yes your solution is very nice and correct, as a slightly different alternative
$$2\cos^2(x)+\sin (x)=1 \iff 2(1-\sin x)(1+\sin x)+\sin x-1=0 $$
$$\iff (\sin x-1)(-2-2\sin x)+(\sin x-1)=0 \iff (\sin x-1)(-1-2\sin x)=0$$
which indeed leads to the same solutions, or also from here by $t=\sin x$
$$2-2 \sin^2 x+\sin x=1 \iff 2t^2-t-1=0$$
$$t_{1,2}=\frac{1\pm \sqrt{9}}{4}=1, -\frac12$$
| {
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How do we prove this inequality? Suppose $a,b,c > 0$. Prove that
$$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$
I've tried multiplying everything by the denominator and then I tried to use the rearrangement inequality, but it didn't yield the result I was looking for. I couldn't really think of anything else to do.
| After clearing the denominators we have to prove that
$$a^4c^2+b^4a^2+b^2c^4\geq abc(a^2c+ab^2+bc^2)$$
Now use that $$x^2+y^2+z^2\geq xy+yz+zx$$
This is $$(a^2c)^2+(bc^2)^2+(ab^2)^2\geq a^2cab^2+a^2cbc^2+ab^2bc^2=abc(a^2b+ac^2+b^2c)$$
Since$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{a^2c+ab^2+a^2b}{abc}$$
| {
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Compute $\int\frac{du}{\sqrt{a^2-u^2}}$ Please, can you explain why isn't the $$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{du}{a\sqrt{1-(u/a)^2}}=\frac1a\sin^{-1}\bigg(\frac{u}{a}\bigg)+c$$ if $a > 0$ is a positive constant.
| Let $v = \dfrac ua \implies u = a\cdot v$, and $\mathrm du = a\cdot\mathrm dv$.
Substitute in original expression,
$$\require{cancel}\begin{align}\int\dfrac{\mathrm du}{\sqrt{a^2 - u^2}} &\equiv \int\dfrac{a\cdot\mathrm dv}{\sqrt{a^2 - a^2v^2}}\\ &= \int\dfrac{a\cdot\mathrm dv}{a\sqrt{1 - v^2}}\\ &= \int\dfrac{\cancel{a}\cdot\mathrm dv}{\cancel{a}\sqrt{1 - v^2}}\\ &= \int\dfrac{\mathrm dv}{\sqrt{1 - v^2}} = \sin^{-1}\left(v\right) + C\end{align}$$
Reverse substitution,
$$\sin^{-1}\left(v\right) + C \equiv \sin^{-1}\left(\dfrac ua\right) + C$$
What you are missing in your calculation is that
$$\dfrac {\mathrm d}{\mathrm du}\sin^{-1}\left(\dfrac ua\right) = \dfrac 1a\cdot\dfrac1{\sqrt{1 - \left(u/a\right)^2}}$$
and, therefore, the $1/a$ is consumed as part of the integral.
| {
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Prove that for all $n ≥ 1$, $\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$
Prove that for all $n ≥ 1 $
$$\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$$
My attempt:
By induction.
Base case: $n = 1$
$$1 ≤ 2 - 1 = 1$$
Induction step:
Suppose it is true for $n$, i.e
$$\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$$
Adding $\frac{1}{n^2+n}$ to both sides gives
$$\sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+n} ≤2 - \frac{1}{n} + \frac{1}{n^2+n} \implies $$
$$\sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+n} ≤2 - \frac{1}{n+1}$$
Since $\frac{1}{n^2 + 2n + 1} < \frac{1}{n^2 + n}$, we have
$$\begin{align}
2 - \frac{1}{n+1} & ≥ \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+n}\\
& ≥ \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+2n+1} \\
& = \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{(n+1)^2} \\
& = \sum_{i=1}^{n+1}\frac{1}{i^2}
\end{align}$$
$\Box$
Is it correct?
| Proving that
$$\sum_{k=1}^{n} \frac{1}{k^2} <2-\frac{1}{n}.$$
Notice that $$\frac{1}{k^2} <\frac{1}{k(k-1)}=\frac{1}{k-1}- \frac{1}{k},~~~ k \ge 2.$$Starting the telescopic summing from $k=2$ we get
$$\sum_{k=2}^n \frac{1}{k^2} \le 1-\frac{1}{n} \implies \sum_{k=1}^{n} \frac{1}{k^2} <2-\frac{1}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Smallest number of points required to estimate value of integral Find the smallest number of points required to estimate the value of $$\int_{0}^{1} sin(x^2\pi)dx$$ with error less than $10^{-8}$ using the Composite Simpson method. In the estimation n, use the absolute maximum of the corresponding derivative of $(1+x^2)^{-1}$
In this case I would find the maximum of the fourth derivative of $sin(x^2\pi)$ and then plug it into the following formula for Composite Simpson error: $ 10^{-8} ≥ $ $\dfrac {|f^4|(b-a)h^4}{180} $ and solve for h. Then, I would use $h= \dfrac{(b-a)}{n}$ where n is the number of points, and solve for n to get the number of points.
HOWEVER, the last sentence of the question confuses me because I do not know what it means to use the absolute maximum of the corresponding derivative for n.
| It might be easier than a direct computation of the sequence of derivative expressions to find the coefficient of $s^4$ in $f(x+s)=\sin(\pi(x+s)^2)$.
\begin{align}
\sin(\pi x^2 + \pi(2x+s)s)
&=\sin(\pi x^2)\cos(\pi(2x+s)s)+\cos(\pi x^2)\sin(\pi(2x+s)s)
\\
&=\sin(\pi x^2)\left[1-\frac12\pi^2(2x+s)^2s^2+\frac1{24}\pi^4(2x)^2s^4+O(s^5)\right]
\\&\qquad
+\cos(\pi x^2)\left[\pi(2x+s)s-\frac16\pi^3(2x+s)^3s^3+O(s^5)\right]
\end{align}
Thus
\begin{align}
\frac1{24}f^{(4)}(x)
&=\sin(\pi x^2)\left[-\frac12\pi^2+\frac1{24}\pi^4(2x)^2\right]
+\cos(\pi x^2)\left[-\frac12\pi^3(2x)^2\right]
\\
|f^{(4)}(x)|&\le4\pi^2\sqrt{(3-\pi^2x^2)^2+(12\pi x^2)^2}
\\
&\le4\pi^2\sqrt{9+138\pi^2+\pi^4}\le160\pi^2\le1600
\end{align}
giving $n\sim \sqrt[4]{10^8\cdot 1600/180}\le100\cdot\sqrt{3}\le 175$. An adaptive strategy finds a sequence of $147$ nodes with a minimal step of $0.00390625$ ($n=256$ if filled out) sufficient to get this accuracy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
$$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{4}+m_a^2a=\frac{ab^2}{2}+\frac{ac^2}{2}\implies m_a^2=\frac{2b^2+2c^2-a^2}{4},$$and likewise for the other two medians.
How to proceed?
| $$\begin{align}
m_a^4+m_b^4+m_c^4&=\frac1{16}\sum_\text{cyc.}(2b^2+2c^2-a^2)^2\\
&=\frac1{16}\sum_\text{cyc.}4b^4+4c^4+a^4+8b^2c^2-4a^2b^ 2-4c^2a^2\\
&=\frac1{16}(9a^4+9b^4+9c^4).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Use DeMoivre's Theorem to prove $ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$ I need to prove the following equalities using DMT:
$ \cos 5x = 16 \cos^5x$$ - 20\cos^3x$$+5\cos x$
and
$ \sin 5x = 16 \sin^5x$$ - 20\sin^3x$$+5\sin x$
Can someone help me with this question?
(Attempt: $\cos5x+i\sin5x=(\cos x+i\sin x)^5$)
| By DeMoivre's Theorem:
$$\begin{equation}\begin{aligned}
cos\,5x &= cos\,5x + i\,sin\,5x \\
&= (cos\,x + i\,sin\,x)^5 \\
&= cos^5\,x + 10\,cos^3\,sin^2\,x\,i + 5\,cos\,x\,sin^4\,x\,i^4\\
&= cos^5\,x - 10\,cos^3 x\,sin^2\,x + 5\,cos\,x\,sin^4\,x \\
&= cos^5\,x - 10\,cos^3\,x (1 - cos^2\,x) + 5\,cos\,x (1 - cos^2\,x)^2 \\
&= 16\,cos^5\,x - 20\,cos^3\,x + 5\,cos\,x \\
\end{aligned}\end{equation}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Knowing that $a,b,c\ge0$, prove that $\sum_\text{cyc}a^2 \cdot \left[\sum_\text{cyc}\frac{1}{(b - c)^2}\right] \ge\frac{11 + 5\sqrt5}{2}$.
Knowing that $a$, $b$ and $c$ are non-negatives, prove that
a/ $$(bc + ca + ab) \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge 4$$
b/ $$(a^2 + b^2 + c^2) \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge \frac{11 + 5 \sqrt 5}{2}$$
Well, it can be seen that $$[(a - b)^2 + (b - c)^2 + (c - a)^2] \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge 3 + 5 \sqrt 5$$
But nothing more, because typically, $$(x + y + z) \cdot \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9, \forall x, y, z > 0$$
Other than that, I'm not sure.
| One may consider the Buffalo Way because there are terms like $(a-b)$ which get simplified by the Buffalo substitution:
a) By cyclicity of the inequality we may assume $a\le b, c$. Hence, there are $x,y\geq 0$ such that $b=a+x$ and $c=a+y$. In fact, by well-definedness of your expression, we know that $x>0,y>0$ and $x\neq y$.
So your inequality is equivalent to $$\left(\frac{1}{x^2}+\frac{1}{(x-y)^2}+\frac{1}{y^2}\right) \big((a+x) (a+y)+a (a+x)+a (a+y)\big)-4\geq 0$$ which, by expansion and bringing everything on the same denominator, is the same as $$\frac{3 a^2 \left(x^2-x y+y^2\right)^2+2 a (x+y) \left(x^2-x y+y^2\right)^2+x y \left(x^2-3 x
y+y^2\right)^2}{x^2 y^2 (x-y)^2}\geq 0.$$
The last inequality is trivially true because $a,x,y\geq 0$.
b) This one is more tricky. Using the same procedure as above we get that this inequality is equivalent to $$\frac1{2 x^2 y^2 (x-y)^2}\cdot X\geq0,$$
where $$X=6 a^2 \left(x^2-x y+y^2\right)^2+4 a (x+y) \left(x^2-x y+y^2\right)^2+2 x^6-4 x^5 y-\left(3+5
\sqrt{5}\right) x^4 y^2+2 \left(7+5 \sqrt{5}\right) x^3 y^3-\left(3+5 \sqrt{5}\right) x^2 y^4-4 x y^5+2
y^6.$$
So the original inequality is equivalent to $X\geq 0$. Note that \begin{split}X&\geq 2 x^6-4 x^5 y-\left(3+5 \sqrt{5}\right) x^4 y^2+2 \left(7+5 \sqrt{5}\right) x^3 y^3-\left(3+5
\sqrt{5}\right) x^2 y^4-4 x y^5+2 y^6\\
&=\left(x^2+\sqrt{5} \left(1+\frac{1}{\sqrt{5}}\right) x y+y^2\right) \left(x^2-\frac{1}{2} \sqrt{5}
\left(1+\frac{3}{\sqrt{5}}\right) x y+y^2\right)^2 \\ &\geq0.\end{split}
As a conclusion: This is very far from a nice solution but it does the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the determinant of A find the determinant of A using only determinant properties
$$
A=\begin{bmatrix}
1 & 3 & -4 \\
-2 & 1 & 2 \\
-9 & 15 & 0 \\
\end{bmatrix}
$$
| Row reduction does not change the determinant:
$$
\begin{bmatrix} 1 & 3 & -4 \\ -2 & 1 & 2 \\ -9 & 15 & 0 \\ \end{bmatrix}
\to
\begin{bmatrix} 1 & 3 & -4 \\ 0 & 7 & -6 \\ 0 & 42 & -36 \\ \end{bmatrix}
\to
\begin{bmatrix} 1 & 3 & -4 \\ 0 & 7 & -6 \\ 0 & 0 & 0\\ \end{bmatrix}
$$
The determinant is zero because the last matrix has a row of zeros.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Mathematical Induction Involving The Floor Function I need to prove the identity $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$ for all natural numbers $n$.
I wanted to use mathematical induction. The identity is true for $n=1$. Then, I assume $$\lfloor \sqrt{k}+\sqrt{k+1}\rfloor=\lfloor\sqrt{4k+2}\rfloor$$
and need to show that it is also true for $n=k+1$
$$\lfloor \sqrt{k+1}+\sqrt{k+2}\rfloor=\lfloor\sqrt{4k+6}\rfloor.$$
I thought about the inequality $m\leq\sqrt{4k+6}<m+1$ for some integer $m$, and the same for the LHS, but this doesn't seem to help and I don't have other ideas.
| The identity is true for $n = 0$, so consider $n \ge 1$. Also, let
$$m = \sqrt{n} + \sqrt{n + 1} \tag{1}\label{eq1A}$$
With $m \gt 0 \; \to \; m = \sqrt{m^2}$, we get
$$\begin{equation}\begin{aligned}
m & = \sqrt{\left(\sqrt{n} + \sqrt{n + 1}\right)^2} \\
& = \sqrt{n + 2\sqrt{n(n+1)} + n + 1} \\
& = \sqrt{2n + 1 + 2\sqrt{n(n+1)}}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
We also have
$$\color{blue}{n} \lt \color{red}{\sqrt{n(n+1)}} \lt \color{green}{n + 1} \tag{3}\label{eq3A}$$
Thus, using this with \eqref{eq2A} results in
$$\begin{equation}\begin{aligned}
\sqrt{2n + 1 + 2(\color{blue}{n})} & \lt \sqrt{2n + 1 + 2\color{red}{\sqrt{n(n+1)}}} \lt \sqrt{2n + 1 + 2(\color{green}{n + 1})} \\
\sqrt{4n + 1} & \lt m \lt \sqrt{4n + 3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since natural numbers squared are congruent to $0$ modulo $4$ for even values and to $1$ modulo $4$ for odd values, neither $4n + 2$ or $4n + 3$ can be a perfect square. Thus, the largest perfect square less than or equal to values in \eqref{eq4A}, say it's $k^2$, must be less than or equal to $4n + 1$. This therefore gives
$$k \le \sqrt{4n + 1} \lt m \lt \sqrt{4n + 3} \lt k + 1 \tag{5}\label{eq5A}$$
Finally, we get that
$$k = \lfloor m \rfloor = \lfloor \sqrt{n} + \sqrt{n + 1} \rfloor = \lfloor \sqrt{4n + 2} \rfloor \tag{6}\label{eq6A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
matrix representations of complex, dual and split-complex numbers For complex, dual and split-complex numbers there are matrix representations:
$$a+b \cdot i \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\
\end{pmatrix} +b\begin{pmatrix}0 & -1\\1 & 0 \\
\end{pmatrix}, i^2=-1,$$
$$a+b \cdot \epsilon \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\
\end{pmatrix} +b\begin{pmatrix}0 & 1\\0 & 0 \\
\end{pmatrix}, \epsilon^2=0,$$
$$a+b \cdot j \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\
\end{pmatrix} +b\begin{pmatrix}0 & 1\\1 & 0 \\
\end{pmatrix}, j^2=1,$$
and I know (as a fact learned 20 years ago, not quite familiar with it) too, that these sets are isomorphic to $\Bbb R[X]/(X^2+1)$, $\Bbb R[X]/(X^2)$ and $\Bbb R[X]/(X^2-1)$. While the quotient ring contructions "looks quite complete", I wonder:
If one varies the "imaginary matrices", e.g. if one considers $\begin{pmatrix}0 & 0\\1 & 0 \\
\end{pmatrix}$ , are then the sets one gets in any way "special", "meaningful", well-known or named?
| In general, any matrix $B$ that you choose as the imaginary unit will have a minimal polynomial $p(x) $ such that $p(B) =0$. Any first or second degree minimal polynomial is possible. The resulting ring is $\mathbb R[x] /(p(x)) $. You can get rings that are not isomorphic to any of the three you mention this way, for example if $p(x) =(x-1)(x-2)$ and
$$B=\left(\begin{array} {cc} 1&0\\0&2\end{array}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to remove square roots from denominator in $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$? I have this math problem: Remove all square roots in the denominator of
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = ?$
The obvious solution is to multiply the fraction with $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$.
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} \cdot \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}\sqrt{x} - \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x}-\sqrt{y}\sqrt{y}}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x}-y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{\sqrt{x}\sqrt{x}-\sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{x}+\sqrt{y}\sqrt{y}}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x-\sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{x}+y}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x-2\sqrt{x}\sqrt{y}+y}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x} - \sqrt{y})^2}$
Then I tried to multiply the fraction by $\frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})^2}$.
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x} - \sqrt{y})^2} \frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x - 2\sqrt{x}\sqrt{y} + y} \frac{(\sqrt{x} - \sqrt{y})^2}{x - 2\sqrt{x}\sqrt{y} + y}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-x2\sqrt{x}\sqrt{y}+xy -2\sqrt{x}\sqrt{y}x + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y + yx - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-x2\sqrt{x}\sqrt{y} -2\sqrt{x}\sqrt{y}x + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-2x\sqrt{x}\sqrt{y} -2x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2y\sqrt{x}\sqrt{y} - 2y\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 4y\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + 2\cdot2\sqrt{x}\sqrt{x}\sqrt{y}\sqrt{y} - 4y\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + 4xy - 4y\sqrt{x}\sqrt{y}+y^2}$
This seems like a dead end. Where exactly did I make a mistake?
| Multiply denominator and numerator with $\sqrt x+\sqrt y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $
So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$
However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?
| If your limit exists, say it is equal to $L$. Then
$$
\begin{split}
\ln L &= \ln \left( \lim_{n \to \infty} \left(n^2+n\right)^{1/n} \right) \\
&= \lim_{n \to \infty} \ln \left( \left(n^2+n\right)^{1/n} \right) \\
&= \lim_{n \to \infty} \frac{\ln \left(n^2+n\right)}{n} \\
&= \lim_{n \to \infty} \left[ \frac{\ln n + \ln (n+1)}{n} \right] \\
&= 0,
\end{split}
$$
which implies that $L = e^0 = 1$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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How to find the overlapping area of a semicircle and a rectangle I am being asked to calculate the pink area that is overlapping between the semicircle and the rectangle. I was only given the radius of the circle (5), the equation of a circle $(x^2+y^2 = r^2).$ I have found the intersection points between the circle and the rectangle $(-4,3)$ and $(4,3)$ but I am lost afterwards, I no longer know what to do.
| There are several ways you could approach this problem, but I think this would be the most straight-forward:
*
*Split up the pink region into two: $0\le x\le 4,$ and $4\le x\le 5.$ For the first region, you have a rectangle of area $3\cdot 4=12.$
*For the other region, you can perform an integral. You have that $5^2=x^2+y^2$ on the circle, so that $y=\sqrt{25-x^2}$ above the $x$ axis. That is, you calculate
$$\int_4^5\sqrt{25-x^2}\,dx. $$
This integral screams "trig substitution", and since the $25$ and $x^2$ have opposite signs, you would substitute as follows:
\begin{align*}
\frac{\sqrt{25-x^2}}{5}&=\sin(\theta)\\
\frac{x}{5}&=\cos(\theta)\\
dx&=-5\sin(\theta)\,d\theta.
\end{align*}
The integral becomes
$$\int_4^5\sqrt{25-x^2}\,dx=-25\int_{\arccos(4/5)}^{\arccos(1)}\sin^2(\theta)\,d\theta
=25\int_{0}^{\arccos(4/5)}\sin^2(\theta)\,d\theta. $$
You can integrate by parts to obtain
$$-6+\frac{25 \pi }{4}-\frac{25}{2} \arcsin\left(\frac{4}{5}\right), $$
for a final answer of
$$6+\frac{25 \pi }{4}-\frac{25}{2} \arcsin\left(\frac{4}{5}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$ The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different things, such as $1^2+1^2+1^2$, $(abc)^2+(abc)^2+(abc)^2$. But I still can’t solve it. Can someone help me?
| Just as what was done in Post 1, we can substitute;
$$a=\frac{1}{x}$$
$$b=\frac{1}{y}$$
$$c=\frac{1}{z}.$$
Then, we have that;
\begin{align*}
& \text{ } \text{ } \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ab+bc+ca \geq 1\\
&\implies \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\geq 1\\
&\implies \frac{xyz}{xy}+\frac{xyz}{yz}+\frac{xyz}{zx}\geq xyz\\
&\implies x+y+z\geq xyz.\hspace{7.7cm} \rightarrow 1
\end{align*}
We can now proceed as in this link Linked Question
We know by the QM-AM inequality that;
\begin{align*}
\sqrt{\frac{x^2+y^2+z^2}{3}}
&\geq \left(\frac{x+y+z}{3}\right)^2\\
&\geq \left(\frac{xyz}{3}\right)^2.\hspace{2cm} \left(\text{Note that }x+y+z\geq xyz\right)\\
\end{align*}
Squaring both sides and multiplying by 3 gives;
$$x^2+y^2+z^2\geq \frac{x^2y^2z^2}{3}$$
\begin{align*}
\implies\left(x^2+y^2+z^2\right)^\frac{1}{4}
&\geq \left(\frac{x^2y^2z^2}{3}\right)^\frac{1}{4}\\
&\geq \frac{\sqrt{xyz}}{3^\frac{1}{4}}.\hspace{6cm} \rightarrow 2\\
\end{align*}
By AM-GM, we know that;
$$\frac{x^2+y^2+z^2}{3}\geq \sqrt[3]{x^2y^2z^2}$$
$$\implies x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}$$
$$\implies \left(x^2+y^2+z^2\right)^3\geq 3^3\left(x^2y^2z^2\right)$$
\begin{align*}
&\implies \left(x^2+y^2+z^2\right)^\frac{3}{4}\geq 3^\frac{3}{4}\sqrt[4]{x^2y^2z^2}\\
&\implies \left(x^2+y^2+z^2\right)^\frac{3}{4}\geq 3^\frac{3}{4}\sqrt{xyz}.\hspace{5.05cm} \rightarrow 3
\end{align*}
We can now multiply Inequality 2 and 3 to have;
\begin{align*}
x^2+y^2+z^2
&=\left(x^2+y^2+z^2\right)^\frac{1}{4}\times \left(x^2+y^2+z^2\right)^\frac{3}{4}\\
&\geq \frac{\sqrt{xyz}}{3^\frac{1}{4}}\times 3^\frac{3}{4}\sqrt{xyz}\\
&\geq xyz\sqrt{3}.
\end{align*}
Substituting back the original variables yields the desired result;
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\sqrt{3}\frac{1}{abc}$$
$$\implies\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\frac{\sqrt{3}}{abc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$(x^2+1)(y^2+1)=z^2+1$
Prove that $$(x^2+1)(y^2+1)=z^2+1$$ has infinitely many solutions when $x,y,z \in \Bbb N$
I couldn't even find one solution for this equation.
Any ideas on how to prove it?
Thanks in advance.
| $$\boxed{\color{red}{(x^2+1)(4x^4+1)= (2x^3+x)^2+1}}$$
Some motivation for that "proof without words" $$z^2 = x^2+y^2+x^2y^2$$
Let $y=xt$ then we have $$z^2 = x^2(1+t^2+x^2t^2)$$
so if we put $t=2x$ we get $$z^2 = x^2(1+4x^2+4x^4)=x^2(2x^2+1)^2$$
So we have triples $(x,2x^2,2x^3+x)$ where $x$ is arbitrary integer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the analytic solution for $y' = -\frac{a}{y} + \frac{b}{y^2}$? For $y: \mathbb{R}_+ \to \mathbb{R}_+$, if $y'(x) = -\frac{a}{y} + \frac{b}{y^2}$, then $y(x) = ?$
Can anyone give an exact solution of this type?
Update: Yes, $a,b$ are some positive constants. By separating the variables and integrating over $x,y$ we have
\begin{align}
\frac{y^2}{-ay + b} dy &= dx \\
-\frac{2 b^2 \log(b-ay) + a^2 y^2 + 2 a b y}{2 a^3} &= x
\end{align}
For this can we get the explicit formula for $y(x)$?
| Suppose $y'=-\frac ay+\frac b{y^2}$. Then,
$$dx=\frac{y^2}{b-ay}\,dy$$
Now integrate and obtain an implicit equation for $y$.
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Question on drawing 4 tickets form 7 tickets A bag contains 7 tickets marked with the numbers 0, 1, 2, ..., 6 respectively. A ticket is drawn and replaced; find the chance that after 4 drawings the sum of the numbers drawn is 8.
| We can view $4$ drawings to be $4$ different boxes and the number on the ticket drawn at $i$-th drawing is the number of balls in the $i$-th box after the balls are thrown to the boxes where $1 \leq i \leq 4.$ Now since the numbers on the tickets are from $0,1, 2 , \cdots ,6$ so that means each box will contain at most $6$ balls.
So the number of ways to get $8$ as the sum of the tickets marked $0,1,2,.\cdots , 6$ in $4$ drawings is the number of ways $n$ to put $8$ indistinguishable balls in $4$ different boxes so that each box will contain at most $6$ balls.
Now let $n_7$ denote the number of ways to put $8$ indistinguishable balls in $4$ different boxes in which one of the boxes will contain $7$ balls and $n_8$ denote the number of ways to put $8$ indistinguishable balls in $4$ different boxes in which one of the boxes will contain $8$ balls. Then clearly $n_7 = 12$ and $n_8 = 4.$
Let $m$ be the number of ways to put $8$ indistinguishable balls in $4$ different boxes. Then $m = \binom {11} {3} = 165.$
Then $n = m - n_7 - n_8 = 165 - 12 - 4 = 149.$
So the required probability is $\frac {149} {7^4} = \frac {149} {2401}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+z^2}{3}+1\right)^3$.By equivalent transformations of $\left(\frac{x^2+y^2+z^2}{3}+1\right)^3\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$ I get $0\leq\frac{2}{3}\left(xy+yz+zx-x^2-y^2-z^2\right)$, which cannot be true. Where did I miss something? Is the AM-QM inequality too weak?
Thank you!
| Let $z = \max (x,y,z)$. Notice that
*
*$z^2 \geq zx \geq 1 $ so $ z \geq 1$.
*$ (x+y+z)^2 \geq 3 (xy+yz+zx) \geq 9$ so $(x+y+z) \geq 3$.
*Thus $z(x+y+z) \geq 3$.
*$(x+y)^2 \geq 4xy \geq 4 $.
*Thus $(x+y)(x+y+4z) = (x+y)^2 + 4z(x+y) \geq 4 + 8 = 12$.
Now, onto the problem.
First, as per Carl's hint, prove that if $ xy \geq 1$, then
$$(x^2 + 1) ( y^2 + 1) \leq \left( ( \frac{x+y}{2} ) ^ 2 + 1 \right)^2.$$
Second, since $ xy \geq 1$ and $ z \left( \frac{x+y+z}{3} \right) \geq 1$, then
$$(x^2 +1)(y^2 + 1) (z^2 + 1) \left( ( \frac{x+y+z}{3} ) ^ 2 + 1 \right) \leq \left( ( \frac{x+y}{2} ) ^ 2 + 1 \right)^2
\left( ( \frac{x+y+4z}{6} ) ^ 2 + 1 \right)^2 $$
Third, since $\left( \frac{x+y}{2} \right) \left( \frac{x+y+4z}{6} \right) \geq 1$, then
$$ \left( ( \frac{x+y}{2} ) ^ 2 + 1 \right)^2 \left( ( \frac{x+y+4z}{6} ) ^ 2 + 1 \right)^2 \leq \left( ( \frac{x+y+z}{3} ) ^ 2 + 1 \right)^4 $$
Hence, combining the last 2 inequalities, we get
$$(x^2 +1)(y^2 + 1) (z^2 + 1)\leq \left( ( \frac{x+y+z}{3} ) ^ 2 + 1 \right)^3.$$
Here's the general forward-backward induction approach:
If $ x_i x_j \geq 1$ for $ i \neq j$, then
$$ \prod (x_i^2 + 1 ) \leq \left( \frac{ \sum x_i } { n } \right)^2 + 1) ^ n$$
Base inequality: $n = 2$. Done.
Forward induction: $n = 2^m$.
We induct on $m$. We have
$$\prod_{i=1}^{2^{m+1}} (x_i^2 + 1 ) \leq \left( \frac{ \sum_{i=1}^{2^m} x_i } { 2^m } \right)^2 + 1) ^ {2^m} \left( \frac{ \sum_{i=2^m+1}^{2^{m+1}} x_i } { 2^m } \right)^2 + 1) ^ {2^m} $$
Now, $$ \left( \sum_{i=1}^{2^m} x_i \right) \left( \sum_{i=2^m+1}^{2^{m+1}} x_i \right) \geq 2^m \times 2^m$$
as the LHS consists of only cross terms, so
$$\left( \frac{ \sum_{i=1}^{2^m} x_i } { n } \right)^2 + 1) ^ {2^m} \left( \frac{ \sum_{i=2^m+1}^{2^{m+1}} x_i } { n } \right)^2 + 1) ^ {2^m} \leq
\left( \frac{ \sum_{i=1}^{2^{m+1}} x_i } { n } \right)^2 + 1) ^ {2^{m+1}}.$$
This establishes the induction hypothesis.
Backward induction: $ 2^m < n < 2^{m+1}$.
The issue, as Michael pointed out, is that we could have $x_1 < 1$ which could lead to $x_1 ( \frac{ \sum x_i }{n}) < 1$ which prevents us from applying the standard backward technique. We will use the above method to fix the gap.
Let $ S =\sum_{i=1}^n x_i$ and $ T = \sum_{i=1}^{2^m} x_i$, $K = 2^{m+1} - n$.
Standard inequality techniques show $ S \geq n $ and $ T \geq 2^m$, $S-T \geq n - 2^m$.
Define $x_{n+1} = x_{n+2} = \ldots x_{2^m} = \frac{ S} { n} \geq 1$.
Then, we have $ x_k \left( \sum_{i=2^m + 1} ^ {2^{m+1} } x_i \right) \geq 1 \times 2^m$ for $k \geq 2^m + 1$. (We use $x_1 < 1 $ here.)
So, we can apply the inequality for the $2^m$ case, which gives us
$$ \prod_{i=1}^{2^{m+1} } ( x_i ^2 + 1) \leq \prod (\left(\frac{T}{2^m}\right)^2 + 1)^{2^m} (\left( \frac{ S-T + K \frac{S}{n} }{2^m} \right)^2 +1)^2 $$
So, to apply the base inequality again, we need to show that the product of these 2 terms is $ \geq 1$. This follows from
$$ T ( S - T + K \frac{S}{n} ) \geq 2^m ( n-2^m + (2^{m+1}-n) \times 1 ) = 2^{m} \times 2^m$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the equation $x^3+2y^3+4z^3=9w^3$ has no solution $(x,y,z,w)\neq (0,0,0,0)$
Prove that the equation $x^3+2y^3+4z^3=9w^3$ has no solution $(x,y,z,w)\neq (0,0,0,0)$
So reduced $\mod 2$ I have $x^3\equiv w^3$, so if $x$ is odd, $w$ is odd and if $x$ is even, $w$ is even.
But I'm not really sure what else I should do.
I tried factoring the $2$ to get $x^3+2(y^3+2z^3)=9w^3$.
Reducing $\mod 3$ I get $x^3+2y^3+z^3=0$.
| Suppose there is an integer solution $(x, y, z, w)$ for your equation, then we have
$$ x^3+2y^2+4z^3\equiv 0 \pmod 9$$
However, $0^3\equiv 3^3\equiv 6^3\equiv 0 \pmod 9$, $1^3\equiv 4^3\equiv 7^3\equiv 1 \pmod 9$, $2^3\equiv 5^3\equiv 8^3\equiv -1 \pmod 9$. So this implies there exists $a, b, c \in \{-1,0,1\}$ such that $a+2b+4c\equiv 0 \pmod 9$. By enumerating all possible combinations of $a$, $b$, and $c$, we see that $a=b=c=0$ is necessary. This means $x$, $y$, $z$ are multiples of three. So is true for $w$, for if $x=3k$, $y=3l$ and $z=3m$, the original equation implies
$$ 27(k^3+2l^3+4m^3)=9w^3 $$
Hence $3$ divides $w^3$, and by the fact that $3$ is a prime number we have $3$ divides $w$.
If non-zero solutions exist, let $(x_0, y_0, z_0, w_0)$ be one of them such that $|x|+|y|+|z|+|w|$ is the smallest. Then we see that $(x_0/3, y_0/3, z_0/3, w_0/3)$ is another non-zero integer solution. But $|x_0/3|+|y_0/3|+|z_0/3| + |w_0/3|< |x_0|+|y_0|+|z_0| + |w_0|$ holds, a contradiction.
| {
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If $\tan2\theta=\frac{b}{a-c}$, then $\cos 2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}}$ and $\sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}}$ I am studying general equation of the second degree. While studying that chapter I came across
$$\tan2\theta=\frac{b}{a-c} \tag{1}$$
Now from (1), the author computed
$$\cos2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}} \quad\text{and}\quad \sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}} \tag{2}$$
I don't understand how did the author compute these terms?
If any member knows the correct answer may reply to this question?
| $$\tan(2\theta) = \frac{b}{a-c}$$
So we have a right triangle where the legs of the right triangle are of lengths $b$ and $a-c$, which means the hypotenuse should be $\sqrt{b^2 + (a-c)^2}$. We can, thus, compute the sine and cosine of the concerned angle using $\frac{\text{adj.}}{\text{hyp.}}$ and $\frac{\text{opp.}}{\text{hyp.}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that $[f]_{B_1}$ is congruent to $[f]_{B_2}$
Let $f$ be a bilinear from on $\mathbb{R}^2$ defined by $$f(u,v)=2u_1v_1-3u_1v_2+u_2v_2,\quad u=(u_1,u_2),v=(v_1,v_2)$$
$(i)$ Find the matrices $[f]_{B_1}$ and $[f]_{B_2}$ relative to the bases $B_1=\{(1,0),(1,1)\}$ and $B_2=\{(2,1),(1,-1)\}$, respectively.
$(ii)$ Hence show that $[f]_{B_1}$ is congruent to $[f]_{B_2}$.
I manage to do $(i)$. And find the matrices
$$[f]_{B_1}=\begin{pmatrix}2 &-1\\2 &0 \end{pmatrix} \qquad [f]_{B_2}=\begin{pmatrix}3 &9\\0 &6 \end{pmatrix}$$ But seems those matrices are not symmetric I don't know how to get any decomposition $[f]_{B_1}=P^T[f]_{B_2}P($in order to show congruent$)$.
Any help will be appreciated. Thanks in advance
Using @Trevor Gunn comment,
$$[f]_{B_2}=\underbrace{\begin{pmatrix}2&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\-1&1\end{pmatrix}}_{\begin{pmatrix}1&1\\ 2&-1\end{pmatrix}=P^T}[f]_{B_1}\underbrace{\begin{pmatrix}1&-1\\0&1\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix}}_{\begin{pmatrix}1&2\\1&-1\end{pmatrix}=P}$$ it seems lot of calculation needed to find those $P^T$ and $P$. Any easier approach do exist$?$
| The matrices are not symmetric because $f$ is not symmetric.
For (ii), it is perhaps helpful to think about $[f]_B$ where $B$ is the standard basis. We have
\begin{align}
\begin{pmatrix} 1 & 0 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= 2,&
\begin{pmatrix} 1 & 0 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 1 \end{pmatrix} &= -1 \\ \begin{pmatrix} 1 & 1 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= 2,&
\begin{pmatrix} 1 & 1 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 1 \end{pmatrix} &= 0.
\end{align}
So maybe you can see why
$$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^\top [f]_B \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = [f]_{B_1}. $$
If you can figure this out, you should be able to finish (ii).
| {
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Solve equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$ Find the value of $x$ given the equation,
$$(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$$
I think they are powers of $ \frac {1} {\Phi} $ or $ \Phi $. In case I think the first would be $ \Phi ^ 3 $ and the second $ (\frac {1} {\Phi}) ^ 4 $. Is it true?
How to solve the problem?
| Recognize
$$\left(\frac{7 - 3 \sqrt{5}}{2}\right)^{x/4}
= \left(\frac{3 - \sqrt{5}}{2}\right)^{x/2}
= \left(\frac{\sqrt{5}-1}{2}\right)^{x}$$
$$(2+ \sqrt{5})\left(\frac{3 - \sqrt{5}}{2}\right) = \frac2{\sqrt5-1}$$
and rewrite the given equation as
$$\left(\frac2{\sqrt5-1}\right)^{x/2}
+2 \cdot \left(\frac{\sqrt{5}-1}{2}\right)^{2x}=5\left(\frac{\sqrt{5}-1}{2}\right)^{x}$$
or
$$2a^5-5a^3+1 = 0, \>\>\>\text{where}\>\>\> a = \left(\frac{\sqrt{5}-1}{2}\right)^{x/2}$$
Then, factorize to get,
$$(a^2+a-1)(2a^3-2a^2-a-1)=0$$
The first factor $a^2+a-1=0$ yields,
$$a = \left(\frac{\sqrt{5}-1}{2}\right)^{x/2} = \frac{\sqrt{5}-1}{2}$$
which leads to the first solution $x=2$. On the other hand, $$2a^3-2a^2-a-1=0$$ has one real root given by
$$a = \left(\frac{\sqrt{5}-1}{2}\right)^{x/2}
= \frac13 + \frac13\sqrt[3]{10+\frac{15}4\sqrt6}+ \frac13\sqrt[3]{10-\frac{15}4\sqrt6}$$
which leads to the second solution,
$$x= \frac{2\ln\left( 1+ \sqrt[3]{10+\frac{15}4\sqrt6}+ \sqrt[3]{10-\frac{15}4\sqrt6}\right)-2\ln3}{\ln\frac{\sqrt{5}-1}{2} }$$
| {
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Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n+1}$
I couldn't come up with anything like that. I feel like I'm confused with something.
| As $\frac{1}{1-x}$ is the generating series for $1,1,1,\dots$ and $\frac{1}{1-x}$ is the generating series for $1,-1,1,-1,\dots$, the series for which you are calculating the generating function for is $2,0,2,0,\dots$.
But we can get the result you want by noticing that $2,1,2,1,\dots$ is actually $\frac{3}{2},\frac{3}{2},\frac{3}{2},\dots$ plus $\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2},\dots$.
We know the first series has generating function $\frac{3/2}{1-x}$ and the second has generating function $\frac{1/2}{1+x}$. Thus, the function you are looking for is$$\frac{\frac{3}{2}}{1-x}+\frac{\frac{1}{2}}{1+x}=\frac{\frac{3}{2}(1+x)+\frac{1}{2}(1-x)}{(1-x)(1+x)}=\frac{2+x}{1-x^2}$$
which is the result you were given.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it right?
| It is more or less correct, but how do you know that$$\lim_{n\to\infty}\left(1+\frac3{n^2+n^4}\right)^{n^4}=e^3?$$
It seems to me that it is more natural to do it as follows:\begin{align}\lim_{n\to\infty}\left(1+\frac3{n^2+n^4}\right)^{n^3}&=\lim_{n\to\infty}\left(\left(1+\frac3{n^2+n^4}\right)^{n^2+n^4}\right)^{\frac{n^3}{n^2+n^4}}\\&=(e^3)^0\\&=1.\end{align}
| {
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Simplify $\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$ Simplify $$\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$$
I have manage to arrive to $$\frac{x^3+2x+2x\sqrt{x^2+1}}{(x^2+2)\sqrt{x^2+1}+2x^2+2}$$
But Wolfram manage to simplify to $$\frac{x}{\sqrt{x^2+1}}$$
| Write $t:=\sqrt{x^2+1}$ so $\frac{x^2}{1+\sqrt{x^2+1}}=\frac{t^2-1}{1+t}=t-1$ has derivative $\frac{dt}{dx}=\frac{x}{\sqrt{x^2+1}}$.
| {
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Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$ I am trying to show the followng:
Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$.
I have shown it for a compact subset of $\mathbb{R}$ however, do not know how to extend it to the reals. Below is my proof for convergence on a compact subset.
I break up the sum into two parts and attack each individually
$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ =
$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1)}_{(1)}$ +
$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}[\sin(1 + \frac{x}{n})-\sin(1)]}_{(2)}$.
$\textbf{Equation (1)}$
$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1)$ converges because $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}$ converges by the alternating series test.
$\textbf{Equation (2)}$
we will first bound $\sin(1 + \frac{x}{n})-\sin(1)$ as follows:
\begin{align*}
|\sin(1 + \frac{x}{n})-\sin(1)| &= \Big|\int_{1}^{1+\frac{x}{n}} \cos(t)dt\Big| \\
&\leq \Big|\int_{1}^{1+\frac{x}{n}} |\cos(t)| dt\Big| \\
&\leq \Big|\int_{1}^{1+\frac{x}{n}} dt\Big| \\
&\leq \Big|1+\frac{x}{n} - 1 \Big| \\
&= \frac{|x|}{n} \\
\end{align*}
Therefore, on any compact interval $ x \in [-M, M]$
\begin{align*}
|\sin(1 + \frac{x}{n})-\sin(1)| \leq \frac{M}{n}
\end{align*}
It follows that
\begin{align*}
\Big|\frac{(-1)^{n}}{\sqrt{n}} \Big( \sin(1 + \frac{x}{n})-\sin(1)\Big)\Big| &= \Big|\frac{(-1)^{n}}{\sqrt{n}}\Big|\Big|\sin(1 + \frac{x}{n})-\sin(1)\Big| \\
&= \frac{1}{\sqrt{n}}\Big|\sin(1 + \frac{x}{n})-\sin(1)\Big| \\
&\leq \frac{1}{\sqrt{n}}\frac{M}{n} \\
&= \frac{M}{n^{\frac{3}{2}}}
\end{align*}
Hence,
\begin{align*}
\sum_{n}^{\infty}\frac{1}{n^{\frac{3}{2}}} < \infty &\implies \sum_{n}^{\infty}\frac{M}{n^{\frac{3}{2}}} < \infty \\
&\implies \sum_{n=1}^{\infty}\Big|\frac{(-1)^{n}}{\sqrt{n}} \Big[ \sin(1 + \frac{x}{n})-\sin(1)\Big]\Big| < \infty \\
&\implies \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}} \Big[ \sin(1 + \frac{x}{n})-\sin(1)\Big] < \infty
\end{align*}
Hence, we have that both $(1)$ and $(2)$ converge on a compact interval $[-M,M]$ which implies our original equation of interest does also.
I want to extend this proof to all $\mathbb{R}$ but I do not know how to. Could someone please help me with this. Anything is appreciated.
| The sum is not uniformly convergent on $\mathbb{R}$.
Take $\epsilon \in (0,\frac{1}{10})$. Suppose $N$ is large such that $|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})| \le \epsilon$ for all $x \in \mathbb{R}$. Take $X = 2\pi\prod_{N \le n \le 2N} n$. Independent of the specific choice of $X$, we of course have $$\frac{1}{X}\int_0^{X} \left|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})\right|^2dx \le \epsilon^2.$$ But due to the specific choice, we have $$\frac{1}{X}\int_0^{X} \left|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})\right|^2dx$$ $$= \frac{1}{X}\int_0^X \sum_{N \le n,m \le 2N} \frac{(-1)^{n+m}}{\sqrt{nm}}\sin(1+\frac{x}{n})\sin(1+\frac{x}{m})dx$$ $$ = \frac{1}{X}\sum_{N \le n \le 2N} \frac{1}{n}\left(\frac{X}{2}-\frac{1}{4}n\sin(\frac{2(n+X)}{n})+\frac{1}{4}n\sin(2)\right),$$ where we used $\int \sin(1+\frac{x}{n})\sin(1+\frac{x}{m})dx = \frac{1}{2}mn\left(\frac{\sin(x(\frac{1}{n}-\frac{1}{m}))}{m-n}-\frac{\sin(\frac{x}{m}+\frac{x}{n}+2)}{m+n}\right)$ for $m \not = n$ (which takes the same value at $0$ and $X$ by our choice of $X$) and $\int \sin^2(1+\frac{x}{n})dx = \frac{x}{2}-\frac{1}{4}n\sin(\frac{2(n+x)}{n})$. And once again by our choice of $X$, we end up with $$\frac{1}{X}\sum_{N \le n \le 2N} \frac{X}{2n},$$ which is around $\frac{1}{2}\ln(2)$, much greater than $\frac{1}{100}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos^23x+\cos x\cos3x\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\dfrac{1+\cos6x}{2}+\dfrac{1}{2}\left(\cos4x+\cos2x\right)\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{4}\left(1+\cos6x+\cos4x+\cos2x\right)}{x^2}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-6\sin6x-4\sin4x-2\sin2x\right)}{2x}$$
Again applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-36\cos6x-16\cos4x-4\cos2x\right)}{2}=\dfrac{36+16+4}{8}=\dfrac{56}{8}=7$$
But actual answer is $3$, what am I messing up here?
| My preferred way is using Taylor expansion of cosine
$$\cos{x}=1-{x^2\over 2}+o(x^3)$$
This yields
$$\begin{align}
\cos{x}\cos{2x}\cos{3x}=&\left(1-{x^2\over 2}+o(x^3)\right)\cdot\\
& \left(1-{4x^2\over 2}+o(x^3)\right)\cdot\\
&\left(1-{9x^2\over 2}+o(x^3)\right)=1-7x^2+o(x^2)
\end{align}$$
And so
$${1-\cos{x}\cos{2x}\cos{3x}\over x^2}=7+o(x)$$
This method generalises to
$$F_n(x)={1-\cos{x}\cdot\cos{2x}\cdots\cos{nx}\over x^2}$$
And we can prove that
$$\lim_{x\to 0}F_n(x)={n(n+1)(2n+1)\over 12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Sum of coefficients of $x^i$ (Multinomial theorem application)
A polynomial in $x$ is defined by
$$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$
Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is
$$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$
I don't know how to proceed. I know the Multinomial theorem, however, I have problems in applying it. Any help will be appreciated as it will help me understand the theorem well.
Thanks!
| Write $(x+2x^2+\cdots+nx^n)^2=(x+2x^2+\cdots+nx^n)(x+2x^2+\cdots+nx^n)$ and for each coefficient in the first factor find the sum of the coefficients in the second factor with which it will enter into the desired sum: The coefficient $k$ in the first factor is paired with $n+1-k$ through $n$ in the second factor, so it contributes
\begin{eqnarray*}
k\sum_{j=n+1-k}^nj
&=&
k\left(\sum_{j=1}^nj-\sum_{j=1}^{n-k}j\right)
\\
&=&
k\left(\frac{n(n+1)}2-\frac{(n-k)(n-k+1)}2\right)
\\&=&
\left(n+\frac12\right)k^2-\frac12k^3\;.
\end{eqnarray*}
Then summing over $k$ yields
\begin{eqnarray*}
\sum_{k=1}^n\left(\left(n+\frac12\right)k^2-\frac12k^3\right)
&=&\left(n+\frac12\right)\sum_{k=1}^nk^2-\frac12\sum_{k=1}^nk^3
\\
&=&
\left(n+\frac12\right)\frac{n(n+1)(2n+1)}6-\frac12\left(\frac{n(n+1)}2\right)^2
\\
&=&
\frac{n(n+1)(5n^2+5n+2)}{24}\;.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have $x^2-y^2=5$, therefore $xy=6$. What to do next?
| Rewrite $\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}$ as $6x^2-5xy-6y^2=0$ and then factorize,
$$(2x-3y)(3x+2y)=0$$
to have $x=\frac32y$ and $x=-\frac23 y$. Plug them into $x^2-y^2=5$ to obtain the real solutions $(3,2)$ and $(-3,-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Point at infinity on quartic elliptic curve Elliptic curve defined by
$$E_1: y^2=7 x^4+x^3+x^2+x+3, P_1=(-1,3)$$
can be transformed to
$$E_2: v^2=u^3-\frac{250 u}{3}-\frac{1249}{27}$$
Substitutions used are:
$$\left(x\to \frac{15 u-9 v+217}{39 u+9 v+209},y\to \frac{9 \left(54 u^3+639 u^2-27 v^2+592 v-16501\right)}{(39 u+9 v+209)^2}\right)$$
$$\left(u\to \frac{2 \left(20 x^2+x+9 y+8\right)}{3 (x+1)^2},v\to -\frac{81 x^3+3 x^2+26 x y-3 x-10 y-33}{(x+1)^3}\right)$$
Then we can check, as an example, that point $P_2=(\frac{6}{7},-3)$ on $E_1$ correcponds with point $Q_2=(-\frac{2}{3},3)$ on $E_2$.
Questions:
*
*Point $P_{\infty}=(0,1,0)$ on $E_1$ corresponds with what point on $E_2$?
*Point $Q_{\infty}=(0,1,0)$ on $E_2$ corresponds with what point on $E_1$?
*Point $P_1=(-1,3)$ on $E_1$ corresponds with what point on $E_2$?
*Point $Q_1=(-\frac{71}{9},\frac{296}{27})$ on $E_2$ corresponds with what point on $E_1$?
EDIT:
Maybe it was not clear, but for points $P_2=(\frac{6}{7},-3)$ and $Q_2=(-\frac{2}{3},3)$ I used the substitutions to verify they correspond to each other. For points in my question the same method did not work for me because of singularities (division by zero).
|
$$E_1(-1,3)\to E_2(0,1,0)$$
$$E_1(-1,-3)\to E_2(-\frac{71}{9},-\frac{296}{27})$$
$$E_1(-\frac{5413}{16069},-\frac{434267883}{258212761})\to E_2(-\frac{71}{9},\frac{296}{27})$$
$$E_1(1,-\sqrt{7},0)\to E_2(\frac{40}{3}-6 \sqrt{7},-81+26 \sqrt{7})$$
$$E_1(1,\sqrt{7},0)\to E_2(\frac{40}{3}+6 \sqrt{7},-81-26 \sqrt{7})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$. If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
Here's what I have got,
By Vieta's rule
$p+q+r=1\text{. ...........}(1)$
$pq+qr+pr=1\text{. ...........}(2)$
$pqr=2\text{. ...........}(3)$
Squaring $(1)$,
$p^2+q^2+r^2+2(pq+qr+pr)=1\text{. ...........}(4)$
From $(2)$,
$p^2+q^2+r^2=-1\text{. ...........}(5)$
Putting the roots and adding these equations,
$p^3-p^2+p-2=0$
$q^3-q^2+q-2=0$
$r^3-r^2+r-2=0$
We get,
$(p^3+q^3+r^3)-(p^2+q^2+r^2)+(p+q+r)-6=0$
Putting the values,
$(p^3+q^3+r^3)-(-1)+1-6=0$
$(p^3+q^3+r^3)=4$
Am I doing something wrong in my solution?
Because the answer given is -5.
Any help would be appreciated.
| $$(x^3-2)^3=(x^2-x)^3$$
$$(x^3)^3-8-3(x^3)^22+3(x^3)2^2=(x^3)^2-(x^3)-3x^2\cdot x(x^3-2)$$
Replace $x^3=y$ to find
$$y^3-8-3y^2\cdot2+3y\cdot2^2=y^2-y-3y(y-2)$$
$$\iff y^3-y^2(6+1-3)+\cdots=0$$ whose roots are $ p^3,q^3,r^3$
$$\implies p^3+q^3+r^3=\dfrac{6+1-3}1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Birationally transforming general curve of genus 1 to Weierstrass form
What are general rules to birationally transform general curve of any
degree of genus 1 to Weierstrass form, provided we have one rational
point?
Example of curve of degree 12:
$$x^9 y^3+9 x^9 y^2+27 x^9 y+27 x^9+9 x^8 y^3+81 x^8 y^2+243 x^8 y+243 x^8+35 x^7 y^3+318 x^7 y^2+963 x^7 y+972 x^7+74 x^6 y^3+687 x^6 y^2+2124 x^6 y+2187 x^6+90 x^5 y^3+871 x^5 y^2+2799 x^5 y+2988 x^5+67 x^4 y^3+692 x^4 y^2+2358 x^4 y+2655 x^4+39 x^3 y^3+415 x^3 y^2+1466 x^3 y+1717 x^3+21 x^2 y^3+211 x^2 y^2+723 x^2 y+840 x^2+4 x y^3+47 x y^2+180 x y+228 x-3 y^3-20 y^2-40 y-20=0$$
This curve has (geometric) genus 1. It has rational point $(-3, -\frac{17}{5})$.
I am familiar with transforming $y^2=a x^4+b x^3+c x^2+d x+e$ to Weierstrass form but have never seen similar process for curves of higher degree than $4$.
| $$C: x^9 y^3+9 x^9 y^2+27 x^9 y+27 x^9+9 x^8 y^3+81 x^8 y^2+243 x^8 y+243 x^8+35 x^7 y^3+318 x^7 y^2+963 x^7 y+972 x^7+74 x^6 y^3+687 x^6 y^2+2124 x^6 y+2187 x^6+90 x^5 y^3+871 x^5 y^2+2799 x^5 y+2988 x^5+67 x^4 y^3+692 x^4 y^2+2358 x^4 y+2655 x^4+39 x^3 y^3+415 x^3 y^2+1466 x^3 y+1717 x^3+21 x^2 y^3+211 x^2 y^2+723 x^2 y+840 x^2+4 x y^3+47 x y^2+180 x y+228 x-3 y^3-20 y^2-40 y-20=0$$
$$E: v^2=u^3+2 u^2+3 u+7$$
$$C\to E: \{x,y\}=\left\{\frac{2-u-v}{u+v},\frac{(3 u-3 v+2) (u+v)^2+24}{(v-u) (u+v)^2-8}\right\}$$
$$E\to C: \{u,v\}=\left\{\frac{-x^3 y-3 x^3-3 x^2 y-9 x^2-3 x y-10 x-1}{(x+1) (y+3)},\frac{x^3 y+3 x^3+3 x^2 y+9 x^2+3 x y+10 x+2 y+7}{(x+1) (y+3)}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to prove this formula for the determinant of a $4 \times 4$ tridiagonal matrix? This following is a problem from B. S. Grewal's Higher Engineering Mathematics.
Show
$$\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1
& 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta)
\end{vmatrix} = \frac{\sin(5\theta)}{\sin(\theta)}.$$
If I take the $3 \times 3$ matrix after deleting the first row and first column, the value is $\frac{\sin(4\theta)}{\sin(\theta)}$, but I am unable to solve this $4\times 4$ matrix. I tried solving the RHS but I am still unable to solve.
| By an expansion on the first column one gets
\begin{align}\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1 & 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta) \end{vmatrix} &= 2 \cos(\theta) \begin{vmatrix} 2 \cos(\theta) & 1 & 0 \\ 1 & 2 \cos(\theta) & 1 \\ 0 & 1 & 2 \cos(\theta) \end{vmatrix}\\
&- \begin{vmatrix} 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1 \\ 0 & 1 & 2 \cos(\theta) \end{vmatrix} \\
&=2\cos(\theta) \bigg ( 2 \cos(\theta) \begin{vmatrix} 2\cos(\theta) & 1 \\ 1 & 2 \cos(\theta) \end{vmatrix} - \begin{vmatrix} 1 & 0 \\ 1 & 2 \cos(\theta) \end{vmatrix} \bigg)\\
&= 4 \cos^2(\theta) (4\cos^2(\theta) - 1) - 4 \cos^2(\theta) - (4 \cos^2\theta) - 1) \\
&= 16 \cos^4(\theta) - 12 \cos^2(\theta) + 1.\end{align}
Now we use the trigonometric Pythagoras $\sin^2(\theta) + \cos^2(\theta) = 1$. With this formula, the above expression can be rewritten as
\begin{align} 16 (1 - \sin^2(\theta))^2 - 12 (1 - \sin^2(\theta)) + 1 &= 16 - 32 \sin^2(\theta) + \sin^4(\theta)-12 + 12 \sin^2(\theta) + 1 \\
&= \sin^4(\theta) - 20 \sin^2(\theta) + 5.\end{align}
The last formula is valid for every value of $\theta$. The result which you have to show is true if and only if $\theta \neq k\pi$, where $k \in \mathbb{Z}$. For those $\theta$, the last term is equivalent to
$$\frac{\sin^5(\theta) - 20 \sin^3(\theta) + 5\sin(\theta)}{\sin(\theta)}.$$
Now it is a basic trigonometric addition formula that $\sin^5(\theta) - 20 \sin^3(\theta) + 5\sin(\theta) = \sin(5\theta)$. This shows that the last term is equal to
$$\frac{\sin(5\theta)}{\sin(\theta)},$$
which was to be proven.
| {
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"url": "https://math.stackexchange.com/questions/3489325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)
| $$x^4=-(x^2+1)\Rightarrow x^6=-x^4-x^2=x^2+1-x^2=1$$
So $$\sum x^6=\sum 1=1+1+1+1=4$$
Where $\displaystyle \sum x^6=\alpha^6+\beta^6+\gamma^6+\delta^6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $n>3$ is not prime, show that we can find positive integers $a,b,c$ such that $n=ab+bc+ca+1$.
If $n>3$ is not prime, show that we can find positive integers $a,b,c$ such that $n=ab+bc+ca+1$.
When $n=4$, pick $a=b=c=1$. When $n=6,$ pick $a=2$ and $b=c=1$. When $n=8$, pick $a=b=c=2$. I don't get the pattern here. I know that if $n>3$ is not prime, then $n=xy,$ where $1<x<n$ and $1<y<n$. But how do I proceed from here?
| Notice that $xy = (x-1)(y-1)+(y-1) + (x-1)+1.$ Choose $a=x-1,b=y-1, $ and $c=1$. All are positive as $x,y>1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\int\limits_0^x \exp\left(-\frac{t^2}2\right)dt = \frac{f(x)}{g(x)}$
Let $f(x) = x+\dfrac{x^3}{1\cdot 3} + \dfrac{x^5}{1\cdot 3 \cdot 5} + \dfrac{x^7}{1\cdot 3\cdot 5\cdot 7}+\dots$ and let $g(x) = 1+\dfrac{x^2}{2} + \dfrac{x^4}{2\cdot 4} + \dfrac{x^6}{2\cdot 4\cdot 6} + \dots$ Show that $\displaystyle\int_0^x \exp\left(-\frac{t^2}2\right)dt = \frac{f(x)}{g(x)}.$
I was thinking of finding a closed form for $f(x)$ and $g(x)$ but I couldn't do that (I did get that $\sinh(x) = \displaystyle\sum_{i=0}^\infty\dfrac{x^{2i+1}}{(2i+1)!}$ and $\cosh(x) = \displaystyle\sum_{i=0}^\infty \dfrac{x^{2i}}{(2i)!}$).
Edit: Apparently I need to differentiate both sides and show that they vanish at $0$.
| Since $f^\prime+xf=1$, integration factors tell us a constant $c$ exists for which $f=\exp-\frac{x^2}{2}\cdot\int_c^x\exp\frac{t^2}{2}dt$. Since $f(0)=0$, $c=0$. The desired claim then follows from $g=\exp\frac{x^2}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Complex power series in Stein's book Let $$ F(z)=\sum\limits_{n=1}^{\infty}d(n)z^{n}$$for $ \left|z\right|<1 $,where $d(n)$ denotes the number of divisors of $ n $.Obeserve that the radius of convergence of this series is 1.Verify the identity $$\sum\limits_{n=1}^{\infty}d(n)z^{n}=\sum\limits_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}} $$Using the identity,show that if $ z=r $ with $ 0<r<1 $,then $$\left|F(r)\right|\geqslant c\frac{1}{1-r}log(\frac{1}{1-r})$$as $ r\longrightarrow1 $.Similarly,if $ \theta=2\pi p/q $ where $ p $ and $ q $ are positive integers and $ z=re^{i\theta} $,then$$ \left|F(re^{i\theta})\right|\geqslant c_{p/q}\frac{1}{1-r}log(\frac{1}{1-r}) $$as $ r\longrightarrow1 $.Conclude that $ F $ cannot be continued analytically past the unit disc. I solve the problem when $ z=r $ is real,my method is here: $$\left|F(r)\right|=\left|\sum\limits_{n=1}^{\infty}\frac{r^{n}}{1-r^{n}}\right|=\left|\frac{1}{1-r}\right|\left|\sum\limits_{n=1}^{\infty}\frac{r^{n}}{1+r+\cdots+r^{n-1}}\right|\geqslant\frac{1}{1-r}\sum\limits_{n=1}^{\infty}\frac{r^{n}}{n}\geqslant c\frac{1}{1-r}log(\frac{1}{1-r})$$But when $ z=re^{i\theta} $ i found my method didn't work.So how can I solve the problems?
| Let be $
z=z\left( r \right) = re^{2\pi i\frac{p}
{q}}
$
We will prove that
$$
\mathop {\lim }\limits_{r \to 1^ - } \left| {\left( {1 - r} \right)F\left( {z\left( r \right)} \right)} \right| = + \infty
$$
As first we write
$$
F(z) = \sum\limits_{n = 1}^{ + \infty } {\frac{{z^n }}
{{1 - z^n }} = \sum\limits_A {\frac{{z^n }}
{{1 - z^n }} + } } \sum\limits_B {\frac{{z^n }}
{{1 - z^n }}}
$$
where
$$
A = \left\{ {n \in \mathbb{N}:n \equiv 0\,\,(\bmod \,\,q)} \right\}
$$
and
$$
B = \left\{ {n \in \mathbb{N}:n\not \equiv 0\,\,(\bmod \,\,q)} \right\}
$$
We will prove that
$$
\mathop {\lim }\limits_{r \to 1^ - } \left| {\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}} } \right| = + \infty
$$
and that
$$
\left| {\left( {1 - r} \right)\sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} } \right| \leqslant M
$$
where $M$ is a suitable constant.
We note that if $
n \equiv 0\,(\bmod \,\,q)
$ then we can put $n=kq$ with $k$ integer and greater than $0$. Therefore
$$
z=\left( {z\left( r \right)} \right)^n = \left( {re^{2\pi i\frac{p}
{q}} } \right)^{kq} = r^{kq}
$$
thus
$$
\begin{gathered}
\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}} = \left( {1 - r} \right)\sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} }}
{{1 - r^{kq} }}} = \hfill \\
\hfill \\
= \frac{{\left( {1 - r} \right)}}
{{1 - r^q }}\sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} \left( {1 - r^q } \right)}}
{{1 - r^{kq} }}} = \hfill \\
\hfill \\
= \frac{1}
{{1 + r + \cdots r^{q - 1} }} \cdot \sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} \left( {1 - r^q } \right)}}
{{\left( {1 - r^q } \right)\left( {1 + r^q + \cdots r^{q\left( {k - 1} \right)} } \right)}}} = \hfill \\
\hfill \\
= \frac{1}
{{1 + r + \cdots r^{q - 1} }} \cdot \sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} }}
{{\left( {1 + r^q + \cdots r^{q\left( {k - 1} \right)} } \right)}}} \hfill \\
\end{gathered}
$$
But $1+r+...r^{q-1} \leq q$ as well as $1+r^q+...r^{q(k-1)}\leq k$. Thus
$$
\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}} \geqslant \frac{1}
{q}\sum\limits_{k = 1}^{ + \infty } {\frac{{\left( {r^q } \right)^k }}
{k}} = \frac{1}
{q}\log \left( {\frac{1}
{{1 - r^q }}} \right)
$$
so that
$$
\mathop {\lim }\limits_{r \to 1^ - } \left[ {\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}} } \right] \geqslant \mathop {\lim }\limits_{r \to 1^ - } \left[ {\frac{1}
{q}\log \left( {\frac{1}
{{1 - r^q }}} \right)} \right] = + \infty
$$
Now we consider
$$
\left| {\left( {1 - r} \right)\sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} } \right|
$$
We have that
$$
\begin{gathered}
\left| {1 - z^n } \right|^2 = \left| {1 - z^n } \right|\left| {1 - \bar z^n } \right| = \hfill \\
\hfill \\
= 1 - 2r^n \cos \frac{{2\pi pn}}
{q} + r^{2n} = \hfill \\
\hfill \\
= \left( {1 - r^n } \right)^2 + 2r^n \left( {1 - \cos \frac{{2\pi pn}}
{q}} \right) \geqslant \hfill \\
\hfill \\
\geqslant 2r^n \left( {2\sin ^2 \frac{{\pi pn}}
{q}} \right) = 4r^n \sin ^2 \frac{{\pi pn}}
{q} \hfill \\
\end{gathered}
$$
Now, let be
$$
np = aq + b\,\,\,\,\,0 < b < q
$$
It is
$$
\frac{{\pi pn}}
{q} = a\pi + \frac{{\pi b}}
{q}
$$
so
$$
\sin ^2 \frac{{\pi pn}}
{q} = \sin ^2 \frac{{\pi b}}
{q} \geqslant \sin ^2 \frac{\pi }
{q}
$$
Hence,
$$
\left| {1 - z^n } \right|^2 \geqslant 4 r^n \sin ^2 \frac{\pi }
{q}
$$
and so
$$
\left| {1 - z^n } \right| \geqslant 2 r^{\frac{n}
{2}} \sin \frac{\pi }
{q}
$$
It follows that
$$
\begin{gathered}
\left| {\left( {1 - r} \right)\sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} } \right| \leqslant \left( {1 - r} \right)\sum\limits_{n = 0}^{ + \infty } {\frac{{\left| z \right|^n }}
{{\left| {1 - z^n } \right|}}} \leqslant \hfill \\
\hfill \\
\leqslant \frac{{\left( {1 - r} \right)}}
{{2\sin \frac{\pi }
{q}}}\sum\limits_{n = 0}^{ + \infty } {\frac{{r^n }}
{{r^{\frac{n}
{2}} }}} = \frac{{\left( {1 - r} \right)}}
{{2\sin \frac{\pi }
{q}}}\sum\limits_{n = 0}^{ + \infty } {r^{\frac{n}
{2}} } = \hfill \\
\hfill \\
= \frac{{\left( {1 - r} \right)}}
{{2\sin \frac{\pi }
{q}}}\frac{1}
{{1 - r^{\frac{1}
{2}} }} = \frac{{\left( {1 + r^{\frac{1}
{2}} } \right)}}
{{2\sin \frac{\pi }
{q}}} \leqslant \frac{2}
{{2\sin \frac{\pi }
{q}}} = M \hfill \\
\end{gathered}
$$
From this it follows that
$$
\mathop {\lim }\limits_{r \to 1^ - } \left| {\left( {1 - r} \right)F(z\left( r \right))} \right| = + \infty
$$
which is even more of what you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) =
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.
| The key insight is to recognize roots of unity in the expression.
We have
$$
z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m} = \frac{(2\omega)^n}{(2\omega^5)^m} = 2^{n-5m}\omega^{n-5m}
$$
where $\omega^6=-1$. Therefore, we need $n-5m \equiv 0 \bmod 6$, or $n+m \equiv 0 \bmod 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Convergence of $\sum_{n=1}^{\infty} \frac{x^n}{n+1}$ Would anyone happen to know what $\sum_{n=1}^{\infty} \frac{x^n}{n+1}$ converges to? My method goes as
$$
f(x)
= \sum_{n=1}^{\infty} \frac{x^n}{n+1}
= \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}
= \frac{1}{x} \sum_{n=1}^{\infty} \int_0^x t^ndt
= \frac{1}{x} \int_0^x \sum_{n=1}^{\infty} t^ndt
= \frac{1}{x} \int_0^x \frac{t}{1-t}dt
= \frac{-x - \ln (1-x)}{x}
= \frac{1}{x}\ln\big(\frac{1}{1-x}\big) - 1 ~.
$$
Can anyone confirm?
| $$\sum_{n=1} \frac{x^n}{n+1} = \frac{1}{x}\sum_{n=2} \frac{x^n}{n} = \frac{1}{x}\left(\sum_{n=1} \frac{x^n}{n} - x\right) = -1 - \frac{\log(1-x)}{x}$$
for $0<|x| < 1$ (the limit to $0$ exists, and equals $0$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Numbers 'poetics' 1 and 2019
One number is called 'poetic' when it can be represented of only one form how $2^a+2^b+2^c$ such that $a\ge b \ge c\ge 0.$ How many poetic numbers are between $1$ and $2019?$
Attempt: I separate in cases, I find that every $2^k$ with $k\ge2 , 2^k + 1$ with $k\ge1$ and every case with $a>b>c$
My final answer was $184$. Check that please .
$a\ge b\ge c\ge 0$
Numbers of form $2^k, k\ge 2$ and $2^k+1,k\ge 1$ are also unique re presentable
$2^k=2^{k-1}+2^{k-2}+2^{k-2}$
$2^{k}+1=2^{k-1}+2^{k-1}+2^0$
| As $2^{11}>2047>2020>2^{10}+2^9$, there are no poetic numbers from $2020$ to $2047$.
All poetic numbers from $1$ to $2047$ actually lie between $1$ and $2019$.
Integers in this range can be written as binary numbers of no more than $11$ digits.
A poetic number should have at most three $1$'s in the digits when it is expressed as a binary number.
A binary number with exactly three $1$'s in the digits when it is expressed as a binary number is clearly a poetic number.
A binary number with exactly two $1$'s in the digits when it is expressed as a binary number is a poetic number if and only if the last digit is $1$. [Note: $2^a+2^b=2^{a-1}+2^{a-1}+2^b=2^a+2^{b-1}+2^{b-1}$.]
A binary number with exactly one $1$ in the digits when it is expressed as a binary number is a poetic number if and only if the last two digits are $0$'s. [Note: $2^a=2^{a-1}+2^{a-2}+2^{a-2}$.]
So the number of poetic numbers from $1$ to $2019$ is $\displaystyle \binom{11}{3}+\binom{10}{1}+\binom{9}{1}=184$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please?
| $$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)} = \lim_{x \to 1} \frac{(2x+1)\cdot(x^3+x-1)}{(3x^2+1)\cdot(x^2+x-1)} = \frac{3}{4}$$ (using L'Hospital)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Help with inequality problem
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(ab + bc + ca)} $$
And I am stuck here.
| By C-S $$\sum_{cyc}\frac{a^2}{(a+b)(a+c)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b)(a+c)}\geq\frac{3}{4},$$ where the last inequality it's
$$4\sum_{cyc}(a^2+2ab)\geq3\sum_{cyc}\left(a^2+3ab\right)$$ or
$$\sum_{cyc}(a^2-ab)\geq0$$ or
$$\sum_{cyc}(2a^2-2ab)\geq0$$ or
$$\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or $$\sum_{cyc}(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find density function for a variable $Y$, if $Y=X^2$ and $f_X=30x^2(1-x)^2, 0 I have to find $f_Y$ if $f_X$ is given as $f_X=30x^2(1-x)^2, 0 <x < 1$ and $Y=X^2$. How do I do this?
Ok, then I get: $F_Y=P(x \le y^{\frac{1}{2}}) \Rightarrow F_X(y^{\frac{1}{2}})$
$F(y)=\int^{\sqrt{y}}_0 30x^2(1-x)^2dx=\left[ u'=30x^2 \Rightarrow u=10x^3 \\ v=(1-x)^2 \Rightarrow v'=-2(1-x) \right]=\left. 10x^3 (1-x^2) + 60\int x^3(1-x^2)dx=10x^3-10x^2+15x^4-\frac{60}{7}x^7 \right|^{\sqrt{y}}_0=10y^{\frac{3}{2}}-10y+15y^2-\frac{60}{7}y^3\sqrt{y}$
IS at least $F(y) $ correctly calculated?
| $$F_Y(y)=P(X^2\le y)=P(-\sqrt y\le X\le\sqrt y)=P(X\le\sqrt y)=F_X(\sqrt y)$$
Since
$$f_X(x)=\begin{cases}30x^2(1-x)^2&\text{for }0<x<1\\0&\text{otherwise}\end{cases}$$
the CDF for $X$ is
$$F_X(x)=\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<0\\10 x^3 - 15 x^4 + 6 x^5&\text{for }0\le x<1\\1&\text{for }x\ge1\end{cases}$$
so that the CDF for $Y$ is
$$F_Y(y)=F_X(\sqrt y)=\begin{cases}0&\text{for }y<0\\10y^{3/2} - 15y^2 + 6 y^{5/2}&\text{for }0\le y<1\\1&\text{for }y\ge1\end{cases}$$
and hence its PDF is
$$f_Y(y)=\frac{\mathrm dF_Y}{\mathrm dy}=\begin{cases}15\sqrt y-30y+15y^{3/2}&\text{for }0<y<1\\0&\text{otherwise}\end{cases}$$
You made a mistake while integrating $f_X(x)$. Integrating by parts gives
$$\int_0^{\sqrt y}30x^2(1-x)^2\,\mathrm dx=vu\bigg|_0^{\sqrt y}-\int_0^{\sqrt y}u\,\mathrm dv$$
where
$$\begin{matrix}v=(1-x)^2&\mathrm dv=-2(1-x)\,\mathrm dx\\\mathrm du=30x^2\,\mathrm dx&u=10x^3\end{matrix}$$
$$\begin{align*}
\int_0^{\sqrt y}30x^2(1-x)^2\,\mathrm dx&=10x^3(1-x)^2\bigg|_0^{\sqrt y}+\int_0^{\sqrt y}20x^3(1-x)\,\mathrm dx\\[1ex]
&=10y^{3/2}(1-\sqrt y)^2+(5x^4-4x^5)\bigg|_0^{\sqrt y}\\[1ex]
&=10y^{3/2}(1-2\sqrt y+y)+5y^2-4y^{5/2}\\[1ex]
&=10y^{3/2}-15y^2+6y^{5/2}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integral representation for $\sum_{k=0}^{x}k^{p}$ How the following integral representation can be derived?
$$\sum_{k=0}^{x}k^{p}=\int_{0}^{x+1}B_{p}\left(t\right)dt=\frac{B_{p+1}\left(x+1\right)-B_{p+1}}{p+1}$$
I know Faulhaber's formula which is as follows:
$$\sum_{k=0}^{x}k^{p}=\frac{1}{p+1}\sum_{j=0}^{p}B_{j}{{p+1}\choose{j}}\left(N\right)^{\left(p+1-j\right)}$$
where $N=x+1$
or another formula:
$$\sum_{p=1}^{k}p^{n}=\sum_{m=1}^{p}{{k+1}\choose{m+1}}{n\brace m}m!$$
but I don't know if they are useful or not.
| We use the generating function
\begin{align*}
\frac{te^{tx}}{e^t-1}=\sum_{n\geq 0}B_n(x)\frac{t^n}{n!}\tag{1}
\end{align*}
of Bernoulli polynomials to derive the integral representation.
From (1) we can show for $n\geq 1$
\begin{align*}
B_{n}(x+1)-B_n(x)&=nx^{n-1}\tag{2}
\end{align*}
as follows:
We obtain
\begin{align*}
\sum_{n\geq 0}\left(B_n(x+1)-B_n(x)\right)\frac{t^n}{n!}
&=\frac{te^{t(x+1)}}{e^t-1}-\frac{te^{tx}}{e^t-1}\tag{3}\\
&=\frac{te^{tx}e^t}{e^t-1}-\frac{te^{tx}}{e^t-1}\\
&=te^{tx}\\
&=t\sum_{n\geq 0}x^n\frac{t^n}{n!}\\
&=\sum_{n\geq 1}nx^{n-1}\frac{t^n}{n!}\tag{4}
\end{align*}
Comparison of the coefficient of $t^p$ in (3) and (4) gives for $p\geq 1$:
\begin{align*}
B_{p}(x+1)-B_p(x)=px^{p-1}
\end{align*}
and the claim (2) follows. Division by $p$ and shifting $p$ by one gives
\begin{align*}
x^p=\frac{B_{p+1}(x+1)-B_{p+1}(x)}{p+1}\tag{5}
\end{align*}
We obtain from (5) for $N\geq 0, p\geq 0$:
\begin{align*}
\color{blue}{\sum_{k=0}^Nk^p}&=\sum_{k=0}^N\frac{B_{p+1}(k+1)-B_{p+1}(k)}{p+1}\\
&=\frac{B_{p+1}(N+1)-B_{p+1}(0)}{p+1}\tag{6}\\
&\,\,\color{blue}{=\frac{B_{p+1}(N+1)-B_{p+1}}{p+1}}\tag{7}\\
\end{align*}
In (6) and (7) we use the telescoping property and $B_p(0)=B_p, p\geq 0$.
We obtain from (1)
\begin{align*}
\sum_{n\geq 0}B_n^{\prime}(x)\frac{t^n}{n!}&=\frac{d}{dx}\left(\frac{te^{tx}}{e^t-1}\right)\tag{8}\\
&=\frac{t^2e^{tx}}{e^t-1}\\
&=\sum_{n\geq 0}^{\infty}B_n(x)\frac{t^{n+1}}{n!}\tag{9}
\end{align*}
In the following we use the coefficient of operator $[t^p]$ to denote the coeffficient of $t^p$ in a series. Coefficient comparison of $t^p$ in (8) and (9) gives for $p\geq 1$:
\begin{align*}
p![t^p]\sum_{n\geq 0}B_n^{\prime}(x)\frac{t^n}{n!}&=B_p^{\prime}(x)\tag{10}\\
p![t^p]\sum_{n\geq 0}B_n(x)\frac{t^{n+1}}{n!}&=p!B_{p-1}(x)\frac{1}{(p-1)!}\\
&=pB_{p-1}(x)\tag{11}
\end{align*}
We obtain from (10) and (11)
\begin{align*}
\int_{0}^{x+1}B_{p+1}^{\prime}(u)du&=B_{p+1}(x+1)-B_{p+1}(0)\\
&=(p+1)\int_{0}^{x+1}B_{p}(u)du\\
\end{align*}
from which
\begin{align*}
\color{blue}{\int_{0}^{x+1}B_{p}(u)du=\frac{B_{p+1}(x+1)-B_{p+1}}{p+1}}
\end{align*}
follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx$
Evaluate $$
\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx
$$
$$
\int\frac{x}{(1+x^2)^2}dx=\frac{-1}{2(1+x^2)}\\
\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx=
\bigg[\log x\frac{-1}{2(1+x^2)}\bigg]_0^\infty-\int_0^\infty\frac{1}{x}.\frac{-1}{2(1+x^2)}dx\\
=0+\int_0^\infty\frac{1}{2x(1+x^2)}dx=\frac{1}{2}\int\bigg[\frac{1}{x}-\frac{x}{1+x^2}\bigg]dx\\
=\frac{1}{2}[\log x]^\infty_0-\frac{1}{4}[\log|1+x^2|]_0^\infty=\frac{1}{4}\log\frac{x^2}{1+x^2}\\
=\bigg[\frac{1}{4}\log\frac{1}{\dfrac{1}{x^2}+1}\bigg]_0^\infty
$$
Is there a better substitution that I can make to evaluate thi definite integral ?
Note: The solution given in my reference is $0$
Similar question is asked @Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ but that is about the convergence of the given integral, here I am looking for a better and obvious substitution that will make the above definite integral easier to solve. But, hint about the solution is found there, Thanks @lab bhattacharjee
| Note that, with $x=\frac1t$,
$$\int_1^\infty\frac{x\log x}{(1+x^2)^2}dx
=- \int_0^1\frac{t\log t}{(1+t^2)^2}dt$$
Thus, the integral is evaluated to zero.
| {
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"url": "https://math.stackexchange.com/questions/3506717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find area of ellipse $5x^2 -6xy +5y^2=8$
Find the area of ellipse whose equation in the $xy$- plane is given by $5x^2 -6xy +5y^2=8$
My attempt : I know that area of ellipse $ = \pi a b$ ,where $a$ is semi-major axis and $b$ is semi minor axis
Now if we make matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmatrix}$ Here eigenvalue $\lambda_1= 8 ,\lambda_2=2$
That is area of ellipse$ = \pi \frac{1}{\sqrt\lambda_2} \frac{1}{\sqrt\lambda_1}= \pi \frac{1}{2\sqrt 2}\frac{1}{\sqrt2}$
Is its correct ?
| Center of ellipses is $(0,0)$ and any point on the chord passing through center is $(r\cos{\theta} , r\sin{\theta} )$ . Let chord intersect ellipses at points $A$ and $B$ . Put the point in equation of ellipses we get
$$(5\cos^2{\theta})r^2-(6\cos{\theta} \sin{\theta} ) r^2 + (5\sin^2{\theta} )r^2 =8$$
$$(5-3\sin^2{\theta})r^2=8$$
$$r^2=\frac{8}{5-3\sin^2{\theta} }$$
Maximum $r = a $ and minimum $r=b$ and maximum $r$ we get when $\sin^2{\theta} = 1$ and minimum $r$ we get when $\sin^2{\theta}=-1$ .
Hence $$a^2=\frac{8}{2} \,\,and \,\,b^2=\frac{8}{8}$$
$$\therefore a=2\, \,and \,\,b=1$$
Hence area is $\pi\times 2\times 1=2\pi$ Square units.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solution verification:$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$
Evaluate without L'Hospital:$$\lim_{x\to
2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$$
My attempt:
I used: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$$
$$
\begin{split}
L &= \lim_{x\to 2} \frac{\ln(x-1)}{3^{x-2}-5^{-x+2}} \\
&= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}\cdot(x-2)}
{(x-2)\cdot\dfrac{3^{x-2}-1+1-5^{-x+2}}{x-2}} \\
&= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}}
{\dfrac{3^{x-2}-1}{x-2}+\dfrac{5^{2-x}-1}{2-x}} \\
&=\frac{1}{\ln3+\ln5} \\
&=\frac{1}{\ln(15)}
\end{split}
$$
Is this correct?
| This is fine. Here is an alternative approach using taylor series:
First substitute $x-2=y$ to simplify it. Let the required limit be $l$. Then
$$l = \lim_{y\to0}\left(\dfrac{\ln(1+y)}{3^y-5^{-y}}\right)$$
$$ = \lim_{y\to0}\left(\dfrac{y-\dfrac{y^2}{2}+\cdots}{(1+y\ln3+\cdots)-(1-y\ln5+\cdots)}\right)$$
$$=\dfrac{1}{\ln3+\ln5}=\dfrac{1}{\ln15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1]$$
$$x_1 = -1/2,~~~x_2 = -1$$
therefore:
$$2x^2 + 3x + 1 = (x + 1/2)(x+1)$$
Now I check it:
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x \cdot x + x \cdot 1 + \frac{1}{2}\cdot x + \frac{1}{2}\cdot 1$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + \frac{3}{2}x + \frac{1}{2}$$
but:
$$2x^2 + 3x + 1 \ne x^2 + \frac{3}{2}x + \frac{1}{2}$$
Why does't quadratic formula work when $a \ne 1$?
however, I can pull out the 1/2.
$$2x^2 + 3x + 1 \ne \frac{1}{2}[2x^2 + 3x + 1]$$
I feel that this is something I must have missed in grade school...
Does this mean you need to multiple by "a" if $a \ne 1$?**
| You are not "solving" $$2x^2+3x+1$$
but $$
2x^2+3x+1 \mathbf{=0}.
$$
For the specific values of $x$ you found, it is indeed true that
$$
2x^2+3x+1 = x^2+\frac{3}{2}x+\frac{1}{2}=0 \quad \text{ where }x\in \left\{- \frac{1}{2}, -1\right\}
$$
Since we solved for when this expression is equal to zero, we are free to multiply everything by any factor to obtain a different true equation:
$$
2x^2+3x+1 = A\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=0
$$
for any real number $A$ when $x$ is one of $-1/2$ or $-1$.
It is not true that in general
$$
2x^2+3x+1 = x^2+\frac{3}{2}x+\frac{1}{2}
$$ for any $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that if $x$ and $y$ are both not $0$ Prove that if $x$ and $y$ are both not $0,$ then
$$x^4+x^3y+x^2y^2+xy^3+y^4>0$$
I know this seems fairly easy but I'm fairly new to calculus and need some help proving that this is true. Appreciate the help!
| Another way:
Note that $0\leq x^2(x+y)^2=x^4+2x^3y+x^2y^2$, and similarly $y^4+2xy^3+x^2y^2\geq 0$. Adding these and dividing by $2$, we get $\frac12x^4+x^3y+x^2y^2+xy^3+\frac12y^4\geq 0$. Then add $\frac{x^4+y^4}{2}$, which is positive since $x$ and $y$ are not both $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$.
Prove that:
For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$
After many hours, I found a general formula satisfying the statement for all $A$ and $B$. $$A^2=(3A+B)^2+(9A+2B)^2-(5A+B)^2-(8A+2B)^2$$ This was derived by noticing the following pattern, to which I used the first equation below as a seed and multiplied through $A^2$. I was then able to find a second paramater $B$ due to the arithmetic progression of the pattern, in order to prove the infinitude of $\{a_n\}_{n=1}^{4}$. $$\begin{align}1^2+5^2+8^2 &= 3^2+9^2 \\ 1^2+6^2+10^2 &= 4^2+11^2 \\ 1^2+7^2+12^2 &= 5^2+13^2 \\ &\vdots\end{align}$$
My question is, is there a way one could prove the statement without the involvement of such curious patterns surrounding square numbers? Apologies if this question is somewhat vague.
Edit:
Fun fact, it appears there is also a general formula for the equation $$A=a_1^2+a_2^2+a_3^2-a_4^2+a_5^2$$ That is, $$A^2=(A+B)^2+(A+3B)^2+(A+8B)^2-(A+5B)^2-(A+7B)^2$$
Edit 2: It appears that the first general equation at which I arrived in this question is actually part of an even more general equation $$(pq +s)^2=\big\{p(3q+r)+3s\big\}^2+\big\{p(9q+r)+3s\big\}^2-\big\{p(5q+r)+s\big\}^2-\big\{2p(4q+r)+4s\big\}^2+4pqs$$ where $(p,q,r,s)=(1,A,B,0)$. Note that an interesting fact implies when $p$, $q$ and $s$ are square numbers.
| Note that
$$a_1^2+a_2^2-a_3^2-a_4^2=\underbrace{(a_1+a_3)(a_1-a_3)}_{=:M}+\underbrace{(a_2+a_4)(a_2-a_4)}_{=:N} $$
where $M$ and $N$ can be any integer that is either odd or a multiple of $4$. In particular, negatives are allowed so that for each number ($A^2$ or otherwise), we find infinitely many such $M,N$.
Concretely, let $C=A^2$ (which in fact need not really be a perfect square).
Pick $R,S$ with $R>S>\max\{2,C\}$ and $R\equiv S\not\equiv C\pmod 2$. Let $N=RS-C$. Note that $N$ is odd and $>2R$.
Now let
$$a_1=\frac{R+S}2,\quad a_3=\frac{R-S}2, a_2=\frac{N-1}2, a_4=\frac{N+1}2. $$
Then
$$\begin{align} a_1^2+a_2^2-a_3^2-a_4^2&=(a_1+a_3)(a_1-a_3)-(a_2+a_4)(a_4-a_2)\\&=RS-N=C,\end{align}$$
as desired.
Also,
$$a_3<a_1<R\le a_2<a_4,$$
i.e., the numbers are distinct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Evaluating $\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;)$ $$\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;)$$
I know that due to continuity of sine I can split it in:
$$\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;) = \lim_{n \to \infty}\sin\sqrt{n+1} -\lim_{n \to \infty}\sin\sqrt{n}$$
I also know that sine has its values in $[-1;1]$, but I don't know how to combine these pieces of information.
| Your line,
$$ \lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;) \\
\qquad \qquad = \lim_{n \to \infty}\sin\sqrt{n+1} -\lim_{n \to \infty}\sin\sqrt{n} \text{,} $$
is only valid if the two limits on the right exist. Neither limit on the right exists, so this is an invalid operation.
Generally, you can make progress on this type of expression using the sum-to-product trigonometric identiites.
$$ \sin \sqrt{n+1} - \sin \sqrt{n} \\
\qquad = 2 \sin\left(\frac{\sqrt{n+1} - \sqrt{n}}{2}\right) \cos \left( \frac{\sqrt{n+1} + \sqrt{n}}{2} \right) \text{.} $$
In this particular case, we need a little more to see the limit. We use conjugates to rewrite the arguments to be decreasing towards $0$. \begin{align*}
&2 \sin\left(\frac{\sqrt{n+1} - \sqrt{n}}{2}\right) \cos \left( \frac{\sqrt{n+1} + \sqrt{n}}{2} \right) \\
&\quad{}= 2 \sin\left(\frac{\sqrt{n+1} - \sqrt{n}}{2} \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \right) \cos \left( \frac{\sqrt{n+1} + \sqrt{n}}{2} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} \right) \\
&\quad{}= 2 \sin\left(\frac{(n+1) - (n)}{2(\sqrt{n+1} + \sqrt{n})} \right) \cos \left( \frac{(n+1) + (n)}{2(\sqrt{n+1} - \sqrt{n})} \right) \\
&\quad{}= 2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})} \right) \cos \left( \frac{2n+1}{2(\sqrt{n+1} - \sqrt{n})} \right) \text{.}
\end{align*}
Now we can use the squeeze theorem. $-1 \leq \cos(\text{anything}) \leq 1$, so $$ -2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})} \right)
\\{}\leq 2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})} \right) \cos \left( \frac{2n+1}{2(\sqrt{n+1} - \sqrt{n})} \right)
\\{}\leq 2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})} \right) \text{.} $$
Now we study \begin{align*}
& \lim_{n \rightarrow \infty} 2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})} \right) \\
&= 2 \sin\left( \lim_{n \rightarrow \infty} \frac{1}{2(\sqrt{n+1} + \sqrt{n})} \right) \\
&= 2 \sin 0 \\
&= 0 \text{.}
\end{align*}
Therefore, the desired limit is squeezed between $-0$ and $0$ as $n \rightarrow \infty$. That is,
$$ \lim_{n \rightarrow \infty} \left( \sin\sqrt{n+1} - \sin\sqrt{n} \right) = 0 \text{.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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