Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated!
Please suggest a simpler process.
| From sum and product, we have:$$a+b+ab+ab=0$$ $$a(b+1)=-b(a+1)$$$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\left(1-\dfrac{2a}{1+a}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1-\dfrac{2ab}{b(1+a)}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1+\dfrac{2b}{1+b}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3$$
Letting $x=\tfrac{2b}{b+1}$, we have $(1+x)^3+(1-x)^3=2+6x^2=2+6\left(\dfrac{2b}{1+b}\right)^2$.
As, $3b^2+2b+1=0$, so $(b+1)^2=-2b^2$, and finally $2+6\left(\dfrac{2b}{1+b}\right)^2=2+6\cdot(-2)=-10.$
| {
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"url": "https://math.stackexchange.com/questions/2344931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trig equation $\sin(x)+\sin(3x)=0$, the answers are given in a factored form $\sin(x)+\sin(3x)=0$
So to solve this I tried the following:
-First I transformed this expression from sum to product because it equals zero
$\sin(x)+\sin(3x)=0 => 2\sin(\frac{x+3x}{2})\cdot\cos(\frac{x-3x}{2})=0$
-Second part I got to
$2\sin(2x)\cos(x)=0$
-Third part:
I have separated the equation into two parts:
$2\sin(2x)=0$ and $\cos(x)=0$
From now on I don't know how to get to the correct answers, which option should i circle and why:
a) $x=(2k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$
b) $x=(k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$
c) $x=(3k+1)\frac{\pi}{3}$ or $x=\frac{3k\pi}{2}$
d) $x=(k-1)\frac{\pi}{2}$ or $x=\frac{3k\pi}{2}$
where $k$ is any integer.
| How I would have approached this problem,
$$\sin 3x = 3\sin x - 4\sin^3x$$
$$\sin x + \sin 3x = 4\sin x - 4 \sin^3x = 0 \implies \sin x(1-\sin^2 x) = 0 \implies \sin x \cos^2x = 0$$
$$\implies x = n\pi \text{ or } x = n\pi+\pi/2$$
$$\implies x = 2n\frac{\pi}{2} \text{ or } \frac{(2n+1)\pi}{2} \implies x = k\frac{\pi}{2} \text{ or } \frac{(k+1)\pi}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\begin{cases} z_1z_2=10\operatorname{cis}(\frac{4\pi}5)\\\frac{z_1}{\overline{z_2}^2}=\frac2{25}\operatorname{cis}(\frac{\pi}5)\end{cases}$
Solve
$$\begin{cases}z_1z_2=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\\frac{z_1}{\overline{z_2}^2}=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}$$
over $\mathbb C$. The answer should be in trigonometric form. $z=a+ib$.
Let:
$$
z_1=r_1(\cos\theta_1+i\sin\theta_1)\\
z_2=r_2(\cos\theta_2+i\sin\theta_2)\\
\frac{z_1}{\overline{z_2}^2}=\frac{z_1z_2^2}{\overline{z_2}^2z_2^2}=\frac{r_1r_2^2}{|z_2|^4}\stackrel{(\ast)}{=}\frac{r_1r_2^2}{r_2^4}=\frac{r_1}{r_2^2}
$$
then:
$$
\begin{cases}r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\[10pt]
\dfrac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2+i\sin2\theta_2)}\stackrel{(\ast\ast)}{=}\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}
$$
According to De Moivre's rule we know that $z^n=r^n \operatorname{cis}(n\theta)$ but in the transition $(\ast)$ why does $|z_2|^4=r_2^4$?
Lastly in the transition $(\ast\ast)$ is there some trig identity which leads to that?
Please explain the points $(\ast)$ and $(\ast\ast)$.
| The line where you have $(*)$ should better be
$$
\left|\frac{z_1}{\overline{z_2}^{\,2}}\right|=
\frac{|z_1|}{\bigl|\overline{z_2}^{\,2}\bigr|}=\frac{r_1}{r_2^2}
$$
There is no reason for the equality before the one you mark with $(*)$.
The equality you mark with $({*}{*})$ is indeed wrong and it should be
$$
\frac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2-i\sin2\theta_2)}=\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})
$$
because if $z=r(\cos\theta+i\sin\theta)$, then $\bar{z}=r(\cos\theta-i\sin\theta)$. Next apply the standard rules
\begin{gather}
\cos\theta-i\sin\theta=(\cos\theta+i\sin\theta)^{-1}\\[4px]
(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)=
\cos(\alpha+\beta)+i\sin(\alpha+\beta)
\end{gather}
Further notes.
The equality
$$
\frac{z_1z_2^2}{\overline{z_2}^{\,2}z_2^2}=\frac{r_1r_2^2}{|z_2|^4}
$$
is generally false, because there's no reason for $z_1z_2^2$ to be real.
When you expand $\dfrac{z_1}{\overline{z_2}^{\,2}}$, you should write
$$
\frac{r_1(\cos\theta_1+i\sin\theta_1)}
{\bigl(\overline{r_2(\cos\theta_2+i\sin\theta_2}\bigr)^2}
=
\frac{r_1(\cos\theta_1+i\sin\theta_1)}
{(r_2(\cos\theta_2-i\sin\theta_2)^2}
=
\frac{r_1(\cos\theta_1+i\sin\theta_1)}
{r_2^2(\cos2\theta_2-i\sin2\theta_2)}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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error in solutions? MIT single variable calculus - Simplifying formula for tangent line Question 1C-4 d) located here states:
Write an equation for the tangent line for the following questions:
d) $f(x) = \frac{1}{\sqrt x} $ at $x = a$.
By my workings:
$f'(a) = (\frac{-1}{2})a^\frac{-3}{2}$
Therefore, the equation of the tangent line at $x = a$ is:
$y = (\frac{-1}{2})a^\frac{-3}{2}[x-a] + \frac{1}{\sqrt a}$
My questions:
*
*The solutions pdf located here states that the formula for the tangent line at $x = a$ is
$y = (\frac{-1}{2})a^\frac{3}{2}[x-a] + \frac{1}{\sqrt a}$. The difference to mine is that I have a negative exponent to the first $a$. Is the solution correct?
*The solutions say that the equation for the tangent line at $x = a$ simplifies to $y = -a^\frac{-3}{2}x + (\frac{3}{2})a^\frac{-1}{2}$. Could you show me how to get to this simplified solution, explicitly stating each step and which rules have been applied to get to it?
Thank you.
| For the simplifcation:
$$ y = -\frac{1}{2}a^{-\frac{3}{2}}(x-a)+\frac{1}{\sqrt{a}} $$
$$ y = \frac{1}{2}a^{-\frac{3}{2}}\cdot a -\frac{1}{2}a^{-\frac{3}{2}} \cdot x + a^{-\frac{1}{2}} \\ \text{(multiplication, and definition of power and square root)} $$
$$ y = \frac{1}{2}a^{-\frac{1}{2}} + a^{-\frac{1}{2}} - \frac{1}{2}a^{-\frac{3}{2}}\cdot x \\ \text{(reordering)} $$
$$ y = \frac{3}{2}a^{-\frac{1}{2}} - a^{-\frac{3}{2}}x \\ \text{(summation)} $$
As was to be shown.
(I am not too familiar with the syntax, I hope it is clear though)
| {
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"timestamp": "2023-03-29T00:00:00",
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3abc$$
It would e helpful if someone could tell me how should I proceed.
| Hint:
$$a^2+b^2+c^2=ab+bc+ca \Leftrightarrow $$
$$\Leftrightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ca \Leftrightarrow$$
$$\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0 \Leftrightarrow$$
$$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0 \Leftrightarrow a=b=c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ cancel eachother out outside the integral, so we have:
$$-\int \frac{sec\ \theta\ tan\ \theta\ d\theta}{tan\ \theta}$$
The $tan\ \theta$ terms cancel, so we're left to integrate $sec\ \theta$ which is equal to:
$$- \ln\ (tan\ \theta + \ sec\ \theta) + C$$
Since $tan\ \theta = \sqrt{x^2-9}\ $ since it replaced it in the integral, and $sec\ \theta = \frac{x}{3}$, the answer is:
$$-\ ln\ (\sqrt{x^2-9}\ + \frac{x}{3})+C$$
However, using an online calculator, the answer turned out to be:
$$-\ ln\ (\sqrt{x^2-9}\ + x)+C$$
It seemed to come about due to their substitution of $u = \frac{x}{3}$ we led them to get the standard integral of $sec^{-1}x$, which would imply that
$$\sqrt{\frac{x^2}{9} - 9}$$
becomes $$\sqrt{u^2 - 1}$$
And I just don't see how. Can someone explain why what they did is valid and/or where my mistake was?
| $$I = -\int \frac{1}{\sqrt{x^2-9}}\,dx$$
$\sqrt{bx^2 - a} \implies x = \sqrt{\frac{a}{b}}\sec \alpha$
$$x = 3\sec\alpha \implies \frac{dx}{d\alpha}=3\sec\alpha\tan\alpha$$
$$ -\int \frac{1}{\sqrt{\left(3\sec \alpha \right)^2-9}}3\sec\alpha\tan\alpha\,\,d\alpha$$
$$ -\int \frac{\sec\alpha \tan\alpha}{\sqrt{\sec^2\alpha-1}}\,d\alpha$$
$\sec^2\alpha = 1 + \tan^2\alpha$
$$ -\int \frac{\sec\alpha \tan\alpha}{\sqrt{\tan^2\alpha}}\,d\alpha = -\int \frac{\sec\alpha \tan\alpha}{\tan\alpha}\,d\alpha = -\int\sec\alpha\,d\alpha$$
$$-\int\sec\alpha\,d\alpha = -\ln(\tan\alpha + \sec\alpha) +C$$
$$\alpha = arcsec\left( \frac{1}{3}x\right)$$
$$-\ln\left(\tan\left( arcsec\left( \frac{1}{3}x\right)\right)+ \sec\left( arcsec\left( \frac{1}{3}x\right)\right)\right) +C$$
$$-\ln\left(\sqrt{\frac{1}{9}x^2 -1} + \frac{1}{3}x\right)+C$$
$$-\ln\left(\frac{1}{3} \left(\sqrt{x^2 - 9} + x\right)\right)+C$$
$$-\ln\left(\sqrt{x^2 - 9} + x\right) - \ln(3)+C$$
$$I = -\ln\left(\sqrt{x^2 - 9} + x\right) +C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram. Problem
Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.
My Solution
B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogram.
A (Hypothesis): Let $A$, $B$, $C$, $D$ be four points such that they form a space quadrilateral.
B1: $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ where $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B}$ and $\dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ are congruent sides. The same can be said for the other two sides.
A1: $\mathbf{A} + \mathbf{B} = \mathbf{C} + \mathbf{D}$ by the definition of quadrilaterals.
$\implies \dfrac{1}{2} \left( \mathbf{A} + \mathbf{B} \right) = \dfrac{1}{2} \left( \mathbf{C} + \mathbf{D} \right)$
$\implies \dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$
$Q.E.D.$
I would greatly appreciate it if people could please review my proof for correctness.
| Hint:
since $\vec A+\vec B=\vec C+ \vec D$ we have:
$$
\frac{1}{2}(\vec A+\vec B)=\frac{1}{2}(\vec C+ \vec D)\iff \frac{1}{2}\vec A +\frac{1}{2}\vec B=\frac{1}{2}\vec C+\frac{1}{2}\vec D
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove $\sum_{cyc}\frac{1}{a(a+b)}\ge\frac{4}{ac+bd}$ Let $a,b,c,d$ be positives. Show that
$$\sum_{cyc}\dfrac{1}{a(a+b)}\ge\dfrac{4}{ac+bd}$$
Use Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{1}{a(a+b)} \cdot\sum_{cyc}a(a+b)\ge (1+1+1+1)^2$$
it suffient
$$\dfrac{16}{\sum\limits_{cyc}(a^2+ab)}\ge\dfrac{4}{ac+bd}$$
However this is wrong inequality
| $$\Leftrightarrow \left ( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right )^2\cdot \dfrac{1}{4+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}}\ge \dfrac{4}{ac+bd}$$
$$\left ( \left (\dfrac{1}{a}+\dfrac{1}{b} \right )+\left (\dfrac{1}{c}+\dfrac{1}{d}\right )\right )^2 \cdot (ac)\ge 4 \left( \dfrac{c}{b}+\dfrac{d}{a}\right)+\dfrac{2(a+b)(c+d)}{bd}$$
$$\left ( \left (\dfrac{1}{a}+\dfrac{1}{d} \right )+\left (\dfrac{1}{c}+\dfrac{1}{b}\right )\right )^2 \cdot (bd)\ge 4 \left( \dfrac{b}{a}+\dfrac{d}{c}\right)+\dfrac{2(d+a)(b+c)}{ac}$$
So:
$$ \left (\dfrac{1}{a}+\dfrac{1}{b} +\dfrac{1}{c}+\dfrac{1}{d}\right )^2\cdot(ac+bd)\ge 4\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}\right)+16$$
| {
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solving the inequality with three parameters Is there any way to show that the following inequality holds for the given function with constraints?
$\frac{(a x + y)^{y+1}}{a x (a x + y + 1)^y}\geq 1$ for $0.5 \leq a \leq 1$, $x >0,y \geq 0$. It can be easily checked that it saturates the inequality for $\lim_{x \rightarrow \infty}$.
Numerically, I've checked this for different range of $x,y$ and $a$. It always holds.. I'm not sure how I can prove this analytically!
Thanks!
| Assume WLOG that $a=1$ (otherwise replace $ax \mapsto x\,$).
$$
\begin{align}
\frac{(x + y)^{y+1}}{x (x + y + 1)^y}\geq 1 \;\;&\iff \;\; \frac{x+y+1}{x} \, \left(\frac{x+y}{x+y+1}\right)^{y+1} \ge 1 \\
&\iff\;\; \frac{x+y+1}{x}\,\left(1 - \frac{1}{x+y+1}\right)^{y+1} \ge 1
\end{align}
$$
The second factor on the LHS of the latter satisfies the conditions of Bernoulli's inequality:
for every real number $r \ge 1$ and real number $x \ge -1\,$: $\;(1+x)^r \ge 1+rx\,$.
Therefore
$$\require{cancel}
\begin{align}
\frac{x+y+1}{x}\,\left(1 - \frac{1}{x+y+1}\right)^{y+1} &\ge \frac{x+y+1}{x}\left(1-\frac{y+1}{x+y+1}\right) \\
&= \frac{\cancel{x+y+1}}{\bcancel{x}} \, \frac{\bcancel{x}}{\cancel{x+y+1}} \\
&= 1
\end{align}
$$
Note that the condition $0.5 \leq a \leq 1$ was not used and is not required. It is enough that $a \gt 0$.
| {
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| HINT:
If calculus is not mandatory,
Method$\#1:$
Set $x=\tan A$
Method$\#2:$
$$\implies x^2y-2x+y=0$$
As $x$ is real, the discriminant must be $\ge0$
Method$\#3:$
If $x>0,$ $$\dfrac{2x}{x^2+1}=\dfrac2{x+\dfrac1x}$$
Now using AM-GM inequality, $$\dfrac{x+\dfrac1x}2\ge\sqrt{x\cdot\dfrac1x}=?$$
If $x<0,$ set $x=-u$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does one expand $\sqrt{r^2+a^2}-r$ to $\frac{a^2}{2r}-\frac{a^4}{8r^3}+\frac{a^6}{16r^5}$? The problem and solution are attached as photos below.
I understand all parts of the solution except for the line above "By inspection..." where $V(r,\theta=0)$ is expanded. I tried a taylor expansion, taylor expansion + binomial, etc, to no avail, but maybe I don't know what I'm doing. Help is much appreciated!
Remember we assume r>~a in part b. The solution comes from a trusted source.
| Taylor expansion is indeed the key here, but you have to be mindful of what variable you are expanding! Here, depending on how far $(r)$ you are away from the disk, the smaller the disk is to you and less its extended structure matters so $a$ is 'small' for your purposes. We then should Taylor expand around $a=0$.
You don't want to Taylor expand around $r=0$ because that would be assuming that $r$ is small but it's given in the problem that $r>a$.
So we say
\begin{equation}
f(a) = \sqrt{r^2+a^2} = f(0) + f'(0)~a + \frac{1}{2}f''(0)~a^2+ \frac{1}{3!}f'''(0)~a^3 + \frac{1}{4!}f^{(4)}(0)~a^4 + \theta(a^5)
\end{equation}
Now the derivatives of $\sqrt{r^2+a^2}$ are
\begin{align}
\frac{\partial}{\partial a}\sqrt{a^2+r^2} = \frac{a}{\sqrt{r^2+a^2}}~,\\
\frac{\partial^2}{\partial a^2}\sqrt{a^2+r^2} = \frac{1}{\sqrt{r^2+a^2}}-\frac{a^2}{(r^2+a^2)^{3/2}}~,\\
\frac{\partial^3}{\partial a^3}\sqrt{a^2+r^2} = - \frac{3a r^2}{(r^2+a^2)^{5/2}}~,\\
\frac{\partial^4}{\partial a^4}\sqrt{a^2+r^4} = \frac{12 a^2 r^2}{(r^2+a^2)^{7/2}} - \frac{3r^4}{(r^2+a^2)^{7/2}}\\
...
\end{align}
so evaluating these at $a=0$ (only the 2nd and 4th derivatives are non-zero at $a=0$) and plugging into our Taylor expansion we get
$$
\sqrt{r^2+a^2} = \frac{a^2}{2r} - \frac{a^4}{8r^3} + \theta(a^5)~~.
$$
I'll leave it as an exercise to you to calculate the 5th and 6th derivatives with respect to $a$ to see if you can match the expression from your solution.
| {
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Finding all $ f : \mathbb R \to \mathbb R $ satisfying $ f \bigl ( f ( x ) f ( y ) \bigr ) + f ( x + y ) = f ( x y ) $ for all $ x , y \in \mathbb R $
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all reals $ x $ and $ y $:
$$ f \bigl ( f ( x ) f ( y ) \bigr ) + f ( x + y ) = f ( x y ) \text . $$
It was six hours ago in IMO 2017 (problem 2).
I tried the standard way: $ x = 0 $, $ x = y $, $ x = 1 $, etc. but without any success.
Let $ f ( 0 ) = a $. Hence, for all $ x \in \mathbb R $ we have $ f \bigl ( a f ( x ) \bigr ) + f ( x ) = a $. What is the rest?
Also we have
$$ ( x - 1 ) ( y - 1 ) - 1 + x + y - 1 = x y - 1 $$
and
$$ 1 - ( 1 - x ) ( 1 - y ) + 1 - x - y = 1 - x y \text . $$
| $$ f(f(x)f(y))+f(x+y)=f(xy). \ \ \ (1)$$
Let $f(0)=c$.
Choose $x=y=0$, we can get
$$f(c^2)=0.\ \ \ (2)$$
Choose $y=0$, we can get
$$f(cf(x))+f(x)=c.\ \ \ (3)$$
Choose $y=\frac{x}{x-1}(x\neq 1)$, we can get
$$f\left(f(x)f\left(\frac{x}{x-1}\right)\right)=0(x\neq 1).\ \ \ (4)$$
When $c=0$, from equation $(3)$ we get $f(x)=0.$
When $c\neq 0$, i.e. $f(0)\neq 0$. Form equation $(4)$ we know there esists $x_0\neq0$ such that $f(x_0)=0.$
We claim that :$x_0=1$.
Otherwise, choose $x=x_0$ in equation $(4)$, we get $f(0)=0$ which is a contradiction.
Combining equation $(2)$, we know $c=1$ or $-1$.
If $c=1$, that is to say $f(0)=1,$
choose $y=1$ in equqtion $(1)$, we get $f(x+1)=f(x)-1$.
So $f(n)=1-n$ and $f(x+n)=f(x)-n$ for all $n\in Z$.
By equation $(1)$ we get
$$f(f(x)f(y)+1)+f(x+y+n)=f(xy+n+1).\ \ \ (5)$$
In the following we prove that: $f$ is injective.
If $f(a)=f(b)$, choose integer $n$ such that $(b-n)^2>4(a-n-1)$,
such that there exists $x_0,y_0$ statisfying
$$x_0y_0+n+1=a,x_0+y_0+n=b.$$
From equation$(5)$ we get $f(x_0)f(y_0)+1=1$, so $f(x_0)=0$ or $f(y_0)=0$
,i.e. $x_0=1$ or $y_0=1$, and this implies $a=b$ which is the injectivity of function $f$.
From equation $(3)$, we know $f(f(x))=1-f(x)$.
On the one hand, $f(f(f(x)))=1-f(f(x))=1-(1-f(x))=f(x);$
on the other hand, $f(f(f(x)))=f(1-f(x))$.
Injectivity of $f$ implies $f(x)=1-x.$
If $c=-1$, we can get $f(x)=x-1$ in the same way as above!
In conclusion, all the solutions of the fucntioanl equation are the following:
$$f(x)=0; f(x)=1-x; f(x)=x-1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "26",
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Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists.
Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists.
I tried many different numbers and nothing is working but I don't know how to prove that it isn't possible.
| I assume you mean "difference of squares of positive integers".
Let $n$ be any positive integer, and $x,y$ positive integers such that $n = x^2 - y^2$. Then $n = (x - y)(x + y)$, so we get a factorization of $n$ into two integers. On the other hand, if we have a factorization $n = ab$, we can get $x$ and $y$ by $x = \frac{a + b}{2}$ and $y = \frac{b - a}{2}$. To get positive integers, we need $b > a$, and also for both $a$ and $b$ to have the same parity. (Both even or both odd). So we get:
The number of ways of writing $n = x^2 - y^2$ where $x$ and $y$ are positive integers is the same as the number of ways of writing $n = ab$, where $a$ and $b$ are positive integers with the same parity and $a < b$.
Now, if $n$ is even, we need both $a$ and $b$ to be even, meaning $n$ is divisible by $4$. Then factorizations $n = ab$ when $a$ and $b$ are both even are the same as factorizations $n/4 = a'b'$. So we get:
If $n$ is divisble by 4, the number of ways of writing $n = x^2 - y^2$ where $x$ and $y$ are positive integers is the same as the number of factorizations $n/4 = a'b'$, where $a' < b'$. (Explicitly, we can take $x = b' + a'$ and $y = b' - a'$)
There are two cases here: If $n/4$ is not a perfect square, the number of factorizations is half the number of factors. Thus we are looking at numbers with exactly $6$ factors. These are numbers of the form $p^5$, where $p$ is a prime, or $pq^2$, where $p$ and $q$ are distinct primes. If $n/4$ is a perfect square, the number of factorizations into distinct integers is half of one less than the number of factors, so we are looking at perfect squares with 7 factors. These are of the form $p^6$, where $p$ is prime. So there are three "families" of solutions:
$$n = 4pq^{2}; \qquad n = 4p^{5}; \qquad n = 4p^{6}$$
The smallest solution is therefore $4 \cdot 3 \cdot 2^{2} = 48$. The factorizations of $3 \cdot 2^{2}$ are $1 \cdot 12$, $2 \cdot 6$, $3 \cdot 4$. This yields the three solutions:
$$48 = 13^{2} - 11^{2} = 8^{2} - 4^{2} = 7^{2} - 1^{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\frac{-1+\sqrt{3}i}{2}$ is a cube root of 1? I know how to directly prove that complex number is one of the cube roots of 1. But the textbook solution also gives another solution as shown below,
However, I don't understand how it jumps to the conclusion that $\frac{-1+\sqrt{3}i}{2}$ is a root of the quadratic equation following the two equations at the top. Thank you!
| take $x=\frac{-1+\sqrt{3}i}{2}$ show that $x^3=1$
$$x=\frac12(-1+\sqrt{3}i)\\x^3=\frac{1}{8}((-1)^3+(\sqrt{3}i)^3+3(\sqrt{3}i)(-1)(-1+\sqrt{3}i))=\\
\frac18 (-1-3\sqrt3i+3\sqrt3i-9i^2)=\\\frac18(-1-9i^2)=\\\frac18(-1+9)=\\1$$
other method
$$x^3-1=(x-1)(x^2-x+1)=0\\x=1\\x^2-x+1=0\\\to (x-\frac12)^2-(\frac14)+1=0\\(x-\frac12)^2=-(\frac34)\\(x-\frac12)=\pm\frac {\sqrt{-3}}{2}\\
(x-\frac12)=\pm\frac {\sqrt{3}i}{2}\\
x=\frac{1}{2}\pm\frac {\sqrt{3}i}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Logical suite and inequalities This is a related problem (see here)
We have the following inequalities :
For $n=3$ with $a,b,c$ real numbers the following inequality holds.
$$\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(c-b)^2+(c-a)^2+(a-b)^2\geq \sqrt{54}$$
For $n=4$ with $a,b,c,d$ real numbers
$$\frac{1}{(a-d)^2}+\frac{1}{(d-b)^2}+\frac{1}{(d-c)^2}+\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(d-b)^2+(c-d)^2+(a-d)^2+(a-b)^2+(b-c)^2+(a-c)^2\geq \sqrt{288}$$
If we continue with $n=5,6,7\cdots$ there is a logical suite but I don't know how to prove this and what is the following numbers .
Thanks.
| To complete the answer of achille hui I will show that the lower bound
$$ \frac{n(n-1)}{2}\sqrt{2n}$$
is actually a minimum.
Let $z_j$ be the $n$ zeros of the Hermite polynomial of order n, then we have the so-called Stieltjes sums (see http://epubs.siam.org/doi/pdf/10.1137/0514028) :
$$ \sum_{j=1}^n \frac{1}{(z_k-z_j)^2} = \frac{1}{3}(2n-(z_k)^2-2)$$
(everytime the sums have to be understood without the singular terms $1/0$).
Now define
$$ x_j := \alpha\, z_j$$
for some $\alpha >0$. The above formula gives
$$ \sum_{j=1}^n \frac{1}{(x_k-x_j)^2} = \frac{1}{3\alpha^2}(2n-\alpha^{-2}(x_k)^2-2).$$
Notice that the $x_j$ are symetrics with respect to $x = 0$ (if $z$ is a zero of the Hermite polynomial of order $n$, then $-z$ too), we have
$$\sum_{ 1 \leq i < j \leq n} \frac{1}{(x_i-x_j)^2} = \frac{1}{2} \sum_{k=1}^n \sum_{j=1}^n \frac{1}{(x_k-x_j)^2}
= \frac{n(n-1)}{3\alpha^2} - \frac{1}{6\alpha^4} \sum_{k=1}^n (x_k)^2.$$
Now still using the symmetry of the $x_j$, it is not hard to show that
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 = n \sum_{k=1}^n (x_k)^2.$$
So one has
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 + \frac{1}{(x_i-x_j)^2} = \frac{n(n-1)}{3\alpha^2} + \left(n-\frac{1 }{6\alpha^4}\right) \sum_{k=1}^n (x_k)^2.$$
Now we use another "well-known"
formula (see Why the sum of the squares of the roots of the $n$th Hermite polynomial is equal to $n(n-1)/2$?)
$$\sum_{k=1}^n (z_k)^2 = \frac{n(n-1)}2 $$
so one has
$$ \sum_{k=1}^n (x_k)^2 = \alpha^2\frac{n(n-1)}2 $$
and
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 + \frac{1}{(x_i-x_j)^2} = \frac{n(n-1)}{4} \frac{2 n \alpha^4 +1}{\alpha^2}.$$
The above quantity is minimal for
$$ \alpha = (2n)^{-1/4}$$
and in this case we have
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 + \frac{1}{(x_i-x_j)^2} = \frac{n(n-1)}{2}\sqrt{2n} \quad \text{for $x_k = (2n)^{-1/4} z_k$}.$$
| {
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Which area is larger, the blue area, or the white area? In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
My Solution
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - $ area of the intersection of the two circles.
Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - t$.
The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $ = \frac{1}{4} * {\pi}r^2$.
Area of triangle $ = \frac{1}{2} * r^2$.
Area of segment $ = \frac{{\pi}r^2 - 2r^2}{4}$.
$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = {\pi}r^2 - \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = \frac{2{\pi}r^2 - {\pi}r^2 + 2r^2}{2}$.
White area $ = \frac{{\pi}r^2 + 2r^2}{2}$.
Blue area = $r^2\left(4 - \frac{{\pi} + 2}{2}\right)$.
Blue area = $r^2\left(\frac{8 - ({\pi} + 2)}{2}\right)$.
Blue area = $r^2\left(\frac{6 - {\pi}}{2}\right)$.
If White area $-$ Blue area $ \gt 0$, then the White area is larger.
$$r^2\left(\frac{{\pi}+2 - (6 - {\pi}}{2}\right)$$
$$r^2\left(\frac{2{\pi} - 4}{2}\right)$$
$$r^2(\pi - 2)$$
$\therefore$ the white area is larger.
My answer was wrong.
What is the error in my solution?
The provided solution:
| There was a mistake in my earlier solution. I correct that mistake here.
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - 2 {\times}$area of the intersection of the two circles.
Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - 2t$.
The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $ = \frac{1}{4} * {\pi}r^2$.
Area of triangle $ = \frac{1}{2} * r^2$.
Area of segment $ = \frac{{\pi}r^2 - 2r^2}{4}$.
$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = {\pi}r^2 - 2\left(\frac{{\pi}r^2 - 2r^2}{2}\right)$.
White area $ = \frac{2{\pi}r^2 - 2{\pi}r^2 + 4r^2}{2}$.
White area $ = \frac{4r^2}{2}$.
White area $ = 2r^2$.
Blue area = $r^2\left(4 - 2\right)$.
Blue area = $2r^2$.
The blue and white areas both have areas of $2r^2$, therefore the triangles have equal areas.
| {
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Probability that a length $8$ password is made by $2$ digits and $6$ characters
You have $10$ digits $(0,1,2,...,9)$ and $26$ characters (from the
American alphabet).
Calculate the probability to automatic generate a 8-length password
with $2$ digits and $6$ characters.
I know the answer is $ p=\frac{\binom{8}{2}\times10^2\times26^6}{36^8} $ and I know that $\binom{8}{2}$ are the picks where $2$ on $8$ elements are digits.
What I can't understand is why shouldn't I use $\binom{8}{6}$ to say that picks are made with $6$ characters on $8$ elements, instead of $\binom{8}{2}$?
The second question is: if I had to make a password with $1$ digits, $3$ special characters (chosen by $10$ special characters) and $4$ characters, the probability should change in: $ p=\frac{\binom{8}{1}\times\binom{8}{3}\times\binom{8}{4}\times10^1\times10^3\times26^4}{46^8} $?
| You could use $8 \choose 6$ instead of $8 \choose 2$ just fine. They are equal. In general ${n \choose r}={n \choose n-r}$ and your proof is a good one. For your second question, you should not use ${8 \choose 1}{8 \choose 3}{8 \choose 4}$. The first $8 \choose 1$ picks which character will be special. Having chosen that, you have only seven places left for the three digits, so the factor would be $7 \choose 3$. Having chosen the first two, you have only four places left for four letters, so the overall factor should be ${8 \choose 1}{7 \choose 3}{4 \choose 4}=\frac {8!}{1!3!4!}$
| {
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Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, can you guys help me out because I can't understand which formula to use, Do we have to use the Discriminant formula ($b^2-4ac$)?
| $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \frac{x}{2}$$
Using componendo dividendo
$$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}+({\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}})}{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}-(\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}})} = \frac{x+2}{x-2}$$
$$\frac{\sqrt{5+2\sqrt{6}} +({\sqrt{5+2\sqrt{6}} })}{\sqrt{5-2\sqrt{6}}+ \sqrt{5-2\sqrt{6}})} = \frac{x+2}{x-2}$$
$$\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}=\frac{x+2}{x-2}$$
$$\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=(\frac{x+2}{x-2})^2$$
Using it again
$$\frac{5+2\sqrt{6}+(5-2\sqrt{6})}{5+2\sqrt{6}-(5-2\sqrt{6})}=(\frac{(x+2)^2+(x-2)^2}{(x+2)^2-(x-2)^2})$$
$$\frac{10}{4\sqrt6}=\frac{2x^2+8}{8x}$$
$$\frac{10}{4\sqrt6}=\frac{x^2+4}{4x}$$
$$\frac{10}{\sqrt6}=\frac{x^2+4}{x}$$
$$\sqrt6(x^2+4)-10x=0$$
Just apply quadratic formula.
Here you go
| {
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To prove $n\binom{2n-1}{n-1}$ is divisible by $n(2n-1)$
Prove that for natural $n \ge 2$ $$n\binom{2n-1}{n-1}$$ is divisible by $$n(2n-1)$$
We have $$n\binom{2n-1}{n-1}=n \frac{(2n-1)!}{(n-1)! \:n!}=n(2n-1)\frac{(2n-2)!}{(n-1)! \:n!}$$
Now it suffices to prove $\frac{(2n-2)!}{(n-1)! \:n!}$ is an integer
Now
$$\frac{(2n-2)!}{(n-1)! \:n!}= \frac{1 \times 2 \times 3 \cdots \times (n-1) \times (n-2) \times (n-3) \cdots \times (2n-4) \times (2n-3) \times (2n-2)}{(n-1)! \: n!}$$
hence
$$\frac{(2n-2)!}{(n-1)! \:n!}=\frac{(n-2) \times (n-3) \cdots (2n-2)}{n!}$$
any clue to prove this is always an integer?
| Because $$\frac{\binom{2n-1}{n-1}}{2n-1}=\frac{\binom{2n-2}{n-1}}{n}$$ and
$gcd(2n-1,n)=1$.
| {
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Find the limit if it exists of $S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})$
Suppose that $S_0$ and A are positive numbers, let $$S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right)$$ with $n \geq 0 $.
(a)Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$
(b)Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists
(d) find s
(a) Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$
Given $$P_n: S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \geq \sqrt{A}$$
$$P_0: S_{1} = \frac{1}{2}\left(S_0 +\frac{A}{S_0}\right) \geq \sqrt{A} $$
We assume that $P_n$ is true
$$P_{n+1}: S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right)$$
by assumption
$$S_{n+2}= \frac{1}{2}\left(S_{n+1}\left(1 +\frac{A}{(S_{n+1})^2}\right)\right) \geq \frac{1}{2}\left(\sqrt{A}\left(1 +\frac{A}{(\sqrt{A})^2}\right)\right)$$
$$ S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right) \geq \sqrt{A} $$
It follows that $S_{n+1} \geq \sqrt{A} $
(b) Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
$$S_{n+1} \leq S_n$$
$$\frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \leq S_n $$
Dividing by $S_n$
$$\frac{1}{2}\left(1 +\frac{A}{S_n^2}\right) \leq 1 $$
$$\frac{A}{2S_n^2} \leq \frac{1}{2}$$
$$A \leq S_n^2$$
$$S_n \geq \sqrt{A}$$
As $S_{n+1} \leq S_n$ yields a true statement, it follows $S_{n+1} \leq S_n$ is true.
(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists
Since $S_{n+1} \leq S_n$, the sequence is non-increasing,
using the non-increasing theorem stating that
if $\{S_n\}$ is non-increasing then $$\lim\limits_{n \rightarrow
> \infty} S_n = \inf\{S_n\} $$
(d) find s
Is the argumentation in (a) and (b) appropriate? Also, I have to admit I m getting less confident in my argumentation (c) and (d). How to proceed in (c) and (d)?
Much appreciated for your input or help.
| Hint You can simplify a) by applying AM-GM, obviously showing that each $S_n>0$ which is easy to show
$$\frac{1}{2}\left(S_n+\frac{A}{S_n}\right)\geq \sqrt{S_n \cdot \frac{A}{S_n}}=\sqrt{A}$$
For c) more details here and you already proved the sequence is decreasing and bounded below in b) and a).
For d) as Eugen suggested, with a), b) and c) you are confident that the sequence is converging so:
$$\lim_{n \rightarrow \infty} S_{n+1}= \lim_{n \rightarrow \infty} \frac{1}{2}\left(S_n+\frac{A}{S_n}\right) \Rightarrow s=\frac{1}{2}\left(\lim_{n \rightarrow \infty}S_n+\frac{A}{\lim\limits_{n \rightarrow \infty} S_n}\right)\Rightarrow \\
s=\frac{1}{2}\left(s+\frac{A}{s}\right)$$
obviously $s \geq \sqrt{A}>0$, thus
$$2s^2=s^2+A \Rightarrow s=\sqrt{A}$$
| {
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Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
| I don't think there is a way to evaluate the limit without using the l'Hospital rule in some form. (Using Taylor series is basically the same, because they are obtained by calculating the derivatives.)
I would calculate the limit by applying l'Hospital once and then using known series expansions:
\begin{align}
&~~~~\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
\\
&=\lim_{x \to 0} \frac{ \sin(2x)- 2 \tan(x)}{4x^3}
\\\
&=\lim_{x \to 0} \frac{2x-\frac 8 6 x^3 + \mathcal O(x^5) - 2(x + \frac 1 3 x^3 +\mathcal O(x^5)) } {4 x^3}
\\
&= \lim_{x \to 0}\frac{-2 x^3 + \mathcal O(x^5)}{4x^3} = -\frac 1 2
\end{align}
| {
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"question_score": "3",
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On a determinant identity
Show that
$$\det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = (-1)^n(1+x-n)(1+x)^{n-1}$$
for all $n \in \mathbb{N}$.
I tried using induction but I didn't manage (using row/column swaps). Does anyone have any idea?
| Subtract the first row from all of the other rows ... we have
\begin{eqnarray*}
\left[\begin{array}{cccccc}
-x & 1 & 1 & 1 &\cdots &1\\
1+x & -(x+1) & 0 & 0 &\cdots &0\\
1+x & 0 &-(x+1) & 0 &\cdots &0\\
1+x & 0 &0 & -(x+1) & \cdots &0\\
\vdots & \vdots &\vdots &\vdots &\ddots &\vdots & \\
1+x & 0 &0 & 0 & \cdots &-(x+1)\\
\end{array}\right]
\end{eqnarray*}
The leading diagonal will give a contribution of $\color{blue}{-x(-1)^{n-1}(x+1)^{n-1}}$ and if we pick the $i^{th}$ $1$ in the first row then we must pick the $i^{th}$ $(1+x)$ in the first column and the rest of the terms on the leading diagonal, there are $\color{red}{(n-1)}$ of these & each will give a contribution of $\color{orange}{-(1+x)^{n-1}(-1)^{n-2}}$. Putting these together we have $(-1)^{n}(1+x-n)(1+x)^{n-1}$.
\begin{eqnarray*}
\color{blue}{-x(-1)^{n-1}(x+1)^{n-1}}+\color{red}{(n-1)} \color{orange}{(-(1+x)^{n-1}(-1)^{n-2})}=(-1)^{n}(1+x-n)(1+x)^{n-1}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is the determinant of the all one matrix minus the identity matrix n-1? Context (skippable)
I was asked (by a friend who is preparing for an exam) whether there was a special trick to compute the determinant of the following matrix. I didn't see anything beyond using the standard computations (like using "Gauss" to compute the value).
Then I asked another math student who, while quite bright, is a bit rusty in linear algebra and using sagemath we empirically found the below formula. Of course we were both confused as to a) whether it actually always holds and b) why it holds.
Actual question
Let $n\in\mathbb N$ be a positive integer. Let $I_n\in\mathbb R^{n\times n}$ be the identity matrix and let $1_n\in\mathbb R^{n\times n}$ be the all-one matrix, that is, the matrix for which every entry is $1$.
Now I am confused as to why the following (empirically found) statement holds (or does not):
$$\forall n\in\mathbb N:\det(1_n-I_n)=(-1)^{n-1}(n-1)$$
For illustration purposes, here is the matrix for $n=4$ (with the determinant being $-3$):
\begin{pmatrix}
0&1&1&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{pmatrix}
| Here is a matrix (here with $n=10$) with columns that are eigenvectors of your (symmetric) matrix. Indeed, given any constants $\alpha, \beta,$ this shows a basis of eigenvectors for
$\alpha I_n + \beta \, 1_n$
Note that $P$ is not orthogonal, although the columns are pairwise orthogonal.
$$
P =
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right).
$$
You get an evident basis of eigenvectors, so you can tell the eigenvalues.
The columns of $P$ are of varying lengths; dividing each by its length does give an orthogonal matrix.
For $n=4$
$$
P =
\left( \begin{array}{rrrr}
1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 \\
1 & 0 & 2 & -1 \\
1 & 0 & 0 & 3
\end{array}
\right).
$$
| {
"language": "en",
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} |
General solution of $\cos{4x}=\cos{2x}$ My attempt:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
\cos^2{2x}-\sin^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-1+\cos^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-\cos{2x}-1 &= 0 \\
(2\cos{2x}+1)(\cos{2x}-1) &= 0,
\end{align}
which leads to
$$\cos2x= -\frac{1}{2} \hspace{5mm} \text{or} \hspace{5mm} \cos2x=1,$$
$$2x=2n\pi\pm{2\pi\over3} \hspace{5mm} \text{or} \hspace{5mm} 2x=2n\pi$$
and
$$x=n\pi\pm{\pi\over3} \hspace{5mm} \text{ or } \hspace{5mm} x=n\pi.$$
Book Solution:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
-2\sin{3x}\sin{x} &=0 \\
\sin{3x}\sin{x} &=0.
\end{align}
$$\sin{3x}=0 \hspace{5mm} \text{or} \hspace{5mm} \sin{x}=0$$
$$3x=n\pi \hspace{5mm} \text{or} \hspace{5mm} x=n\pi$$
$$x=\frac{n\pi}{3} \hspace{5mm} \text{or} \hspace{5mm} x=n\pi$$
My question is: How can there be two general solutions of a same equation Or did I made a mistake somewhere?
| They are same answers.
I think it's better the following way.
$$4x=2x+2\pi k$$ or
$$4x=-2x+2\pi k,$$
where $k\in\mathbb Z$, which gives else the same answer:
$$\left\{\frac{\pi k}{3}|k\in\mathbb Z\right\}$$
| {
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Prove that $0.a_1b_1a_2b_2a_3b_3 \ldots = 0.21742615\ldots$ is irrational, where $(a_i,b_i)$ are the solutions of $x^2-3y^2=1$
Let $S$ be the set of all ordered pairs of positive integers $(x,y)$ such that $x^2-3y^2 = 1$. Let $(a_1,b_1),(a_2,b_2),\ldots$ be the elements of $S$ listed in ascending order, i.e., $a_i < a_{i+1}$ for all $i \geq 1$. Prove that $$k = 0.a_1b_1a_2b_2a_3b_3 \ldots = 0.21742615\ldots$$ is irrational.
The solutions to the equation $x^2-3y^2 = 1$ are the solutions to $x+y\sqrt{3} = (2+\sqrt{3})^m$. Therefore, $$x = \sum_{i=0}^{\left\lfloor\frac{m}{2}\right\rfloor} 2^{m-2i} \cdot 3^i \binom{m}{2i} \quad \text{and} \quad y = \sum_{i=0}^{\left\lfloor\frac{m-1}{2}\right\rfloor}2^{m-2i-1} \cdot 3^i\binom{m}{2i+1}.$$ How can we use this to find the value of $k$?
| Hints: If $k$ is rational then its decimal expansion repeats after some point. Let $p$ be the period, and let $N=10^p$. Then there is a finite set $F$ of pairs such that
$$
a_i,b_i\in\{(r+sN^n)/(N-1)\mid(r,s)\in F,\,n\geq0,\,s>0\}
$$
for $i$ sufficiently large. For some fixed $(r,s),(t,u)\in F$ with $s,u>0$, there are infinitely many $n$ for which
$$
\left(\frac{r+sN^n}{N-1}\right)^2-3\left(\frac{t+uN^n}{N-1}\right)^2=1.
$$
Now expand:
$$
(r^2-3t^2)+2(rs-3tu)N^n+(s^2-3u^2)N^{2n}=(N-1)^2
$$
and equate coefficients of powers of $N^n$:
$$
r^2-3t^2=(N-1)^2,\hspace{10mm}rs-3tu=s^2-3u^2=0.
$$
Then
$$
s^2(N-1)^2=s^2r^2-3s^2t^2=9t^2u^2-9t^2u^2=0,
$$
a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int{\frac{1}{x^2-x+1}dx}$ Original question was to integrate $$\int{\frac{(x+1)}{x^2-x+1}dx}$$
But I was able to break it into 2 parts: $$\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}+\frac{3}{2}\int{\frac{1}{x^2-x+1}dx} $$
The first part $\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}$ could be easily integrated by substituting $u={x^2-x+1}$ and thus, getting $\frac{\ln(x^2-x+1)}{2}$ as the answer. But I have no idea on how to integrate part 2 of the equation.
Any help will be appreciated.
| Hint : $x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}$.
| {
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How many natural numbers $n$ not greater than $10000$ are there such that $2^n - n^2$ is divisible by $7$? How many natural numbers $n$ not greater than $10000$ are there such that $2^n - n^2$ is divisible by $7$?
Please help me solve this question!
| Hint. Since $2^3\equiv 1 \pmod {7}$, consider $n=3q+r$ with the remainder $r=0,1,2$.
If $r=0$, then $1\leq q \leq 3333$ and
$$0\equiv 2^n - n^2=1-9q^2\equiv 1-2q^2\implies q\equiv 2, q\equiv 5\pmod{7}$$
Therefore the number of $n=6q$ is equal to
$$\left\lfloor\frac{3333-2}{7}\right\rfloor+1+\left\lfloor\frac{3333-5}{7}\right\rfloor+1=952.$$
Now consider the remaining similar cases. At the end you should find that for $r=1,2$ the number of $n$s are respectively $953$, $953$.
Therefore the answer should be $952+953+953=2858$.
| {
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Reorder the terms of the series This problem comes from a Brazilian book of Real Analysis.
Explicitly perform a reordering of the terms of the series $$1 -\frac{1}{2} + \frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots $$ so that its sum becomes zero.
It suggests to use the following reordering:
$$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{8} + \frac{1}{3} - \frac{1}{10} - \frac{1}{12} - \frac{1}{14} - \frac{1}{16} + \frac{1}{5} - \frac{1}{18} - \frac{1}{20} - \frac{1}{22}- \frac{1}{24} + \cdots$$
Where after each positive term is placed the first 4 negatives not used yet. But I don't know how to prove that this arrangement of the series converge to zero.
| Its not actually four terms. Perhaps, to make the hint more obvious:
$$\begin{align}0&<1\\0&>1-\frac12-\frac14-\frac16-\frac18\\0&<1-\frac12-\frac14-\frac16-\frac18+\frac13\\0&>1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac1{10}-\frac1{12}-\frac1{14}-\frac1{16}\end{align}$$
If you rearrange it so that the partial sums always go above zero and then below zero, and forever alternating in this manner, what should the limit be?
And can you prove that you can always pull the partial sums to the other side of zero? Hint: The harmonic series diverges.
| {
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Tartaglia triangle for real field? Let's say i need to find the coefficient of the 2nd term in a binomia, we can use pascal's triangle. This works only if the power is a natural number.
What if we want to know it for a real power binomial? We have something for that?
| We can derive a Pascal triangle as follows:
We expand
\begin{align*}
(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}
and obtain the coefficients
\begin{array}{lcccccccccccccccc}
(1+x)^0\qquad\qquad&&&&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&\ldots\\
(1+x)^1\qquad\qquad&&&&1&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&\ldots\\
(1+x)^2\qquad\qquad&&&1&&2&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&\ldots\\
(1+x)^3\qquad\qquad&&1&&3&&3&&1&&\color{lightgrey}{0}&\ldots\\
(1+x)^4\qquad\qquad&1&&4&&6&&4&&1&\ldots\\
\end{array}
In the series above we set the upper limit to $\infty$ and note that $\binom{n}{k}=0$ if $k>n$. This way we can better see the analogy with the generalised version which we consider now.
We can now derive a generalised Pascal triangle in a similar way:
We expand
\begin{align*}
(1+x)^{n+\frac{1}{2}}=\sum_{k=0}^\infty\binom{n+\frac{1}{2}}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}
and obtain the coefficients
\begin{array}{lcccccccccccccccc}
(1+x)^{\frac{1}{2}}\qquad&&&&&1&&\color{lightgrey}{\frac{1}{2}}&&\color{lightgrey}{-\frac{1}{8}}&&\color{lightgrey}{\frac{1}{16}}&&\color{lightgrey}{-\frac{5}{128}}&\ldots\\
(1+x)^{\frac{3}{2}}\qquad&&&&1&&\frac{3}{2}&&\color{lightgrey}{\frac{3}{8}}&&\color{lightgrey}{-\frac{1}{16}}&&\color{lightgrey}{\frac{3}{128}}&\ldots\\
(1+x)^{\frac{5}{2}}\qquad&&&1&&\frac{5}{2}&&\frac{15}{8}&&\color{lightgrey}{\frac{5}{16}}&&\color{lightgrey}{-\frac{5}{128}}&\ldots\\
(1+x)^{\frac{7}{2}}\qquad&&1&&\frac{7}{2}&&\color{blue}{\frac{35}{8}}&&\color{blue}{\frac{35}{16}}&&\color{lightgrey}{\frac{35}{128}}&\ldots\\
(1+x)^{\frac{9}{2}}\qquad&1&&\frac{9}{2}&&\frac{63}{8}&&\color{blue}{\frac{105}{16}}&&\frac{315}{128}&\ldots\\
\end{array}
Observe the sum of two adjacent entries in a row is the corresponding entry in the next row. They fulfill the binomial identity
\begin{align*}
\binom{\frac{9}{2}}{3}=\binom{\frac{7}{2}}{2}+\binom{\frac{7}{2}}{3}\qquad\qquad\qquad
\color{blue}{\frac{105}{16}=\frac{35}{8}+\frac{35}{16}}\\
\end{align*}
which is quite obvious when we consider
\begin{align*}
[x^k](1+x)^{\frac{9}{2}}
&=[x^k](1+x)(1+x)^{\frac{7}{2}}\\
&=[x^k](1+x)^{\frac{7}{2}}+[x^{k-1}](1+x)^{\frac{7}{2}}
\end{align*}
In the same way we can derive a generalised Pascal triangle for $\alpha\in\mathbb{C}$.
\begin{align*}
(1+x)^{n+\alpha}=\sum_{k=0}^\infty\binom{n+\alpha}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}
| {
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Evaluating $\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$ The question is to evaluate $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.
| Set $\cos 2x = \sin^2\theta$. Then
$$ \cos x = \sqrt{\frac{1+\sin^2\theta}{2}} \qquad \text{and} \qquad \left|\frac{dx}{d\theta}\right| = \frac{\sin\theta}{\sqrt{1+\sin^2\theta}}. $$
So we have
$$ \int_{0}^{\frac{\pi}{4}} \cos^{3/2}(2x) \cos x \, dx
= \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sin^4\theta \, d\theta = \frac{3\pi}{16\sqrt{2}}. $$
Of course, you may recognize that this is simply a shortened version of previous answers.
Alternative solution. Apply the substitution $x = \theta/2$ and exploit the symmetry to write
$$ I
= \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos\theta)^{3/2} e^{i\theta/2} \, d\theta
= \frac{1}{8\sqrt{2}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + e^{2i\theta})^{3/2} e^{-i\theta} \, d\theta. $$
Now we utilize the substitution $e^{i\theta} = iz$. If $\Gamma$ denotes the lower semicircular arc from $-1$ to $1$, the resulting integral is
$$ I = -\frac{1}{8\sqrt{2}} \int_{\Gamma} \frac{(1 - z^2)^{3/2}}{z^2} \, dz. $$
This integrand has the following antiderivative on the punctuated unit disc $\mathbb{D}\setminus\{0\}$:
$$ \frac{d}{dz} \left[ \left( \frac{1}{z} + \frac{z}{2} \right) \sqrt{1-z^2} + \frac{3}{2}\arcsin z \right] = - \frac{(1 - z^2)^{3/2}}{z^2}. $$
(This expression can be easily obtained by applying integration by parts twice. Of course, this back-of-the-envelope computation is then justified since the resulting function is holomoprhic on $\mathbb{D}\setminus\{0\}$ and has the desired derivative.) So $I$ can be computed as
$$ I = \frac{1}{8\sqrt{2}} \left[ \left( \frac{1}{z} + \frac{z}{2} \right) \sqrt{1-z^2} + \frac{3}{2}\arcsin z \right]_{z=-1}^{z=1} = \frac{3\pi}{16\sqrt{2}}.$$
| {
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How many positive integers less than $1000$ divisible by $3$ with sum of digits divisible by $7$?
How many positive integers less than $1000$ are divisible by $3$ with their sum of digits being divisible by $7$?
Well, I got Answer: $28$.
$a+b+c = 7k$ and $a,b,c$ are multiples of 3 so $a+b+c$ is a multiple of both $7$ and $3$.
Therefore $a+b+c=21$.
$(9-a)+(9-b)+(9-c)=6$
Then $^8C_2$. Can anyone explain why there is $9-a+9-b+9-c$?
If you have any other alternative method please explain?
| A number is divisible by $3$ iff the sum of its digits (in the decimal representation) is divisible by $3$. Since $3$ and $7$ are coprime, the problem boils down to finding the number of non-zero triples $(a,b,c)$ such that $a,b,c\in\{0,1,2,\ldots,9\}$ and $21\mid a+b+c$. $a+b+c$ is at most $27$, hence the problem is equivalent to finding the non-zero triples $(a,b,c)$ with $a,b,c\in\{0,1,2,\ldots,9\}$ and $a+b+c\color{blue}{=}21$. The answer is given by
$$ [x^{21}](1+x+x^2+\ldots+x^9)^3 = [x^{21}]\frac{(1-x^{10})^3}{(1-x)^3} $$
where $[x^k]\,f(x)$ stands for the coefficient of $x^k$ in the Taylor/Laurent series of $f(x)$ at the origin.
Since $\frac{1}{(1-x)^3}=\sum_{m\geq 0}\binom{m+2}{2}x^m$ and $(1-x^{10})^3 = 1-3x^{10}+3x^{20}-x^{30}$, the previous number equals
$$ \binom{21+2}{2}-3\binom{11+2}{2}+3\binom{1+2}{2}=28$$
so there are $\color{blue}{28}$ numbers in $\{1,2,\ldots,999\}$ whose digit sum is a multiple of both $3$ and $7$.
| {
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?
| If a is divisible by 3, then 3 divides a, and therefore a(a^2 + 2).
Otherwise, $a = 3k ± 1$, and 3 divides $(a^2 + 2)$ = $((3k ± 1)^2 + 2)$ = $((9k^2 ± 6k + 1) + 2)$ = $(9k^2 ± 6k + 3)$.
| {
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} |
How to show that $ \int_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} dt=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$ I've already known that
$$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$$
To get this answer, I let $\mathrm{d}u=\mathrm{d}\sqrt{t}$, then got
$$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=2\int \limits_0^\infty \frac{1}{\sqrt {u^4+21u^2+112}} \mathrm{d}u$$
$$2\int \limits_0^\infty \frac{1}{\sqrt {u^4+21u^2+112}} \mathrm{d}u=2\int \limits_0^\infty \frac{1}{\sqrt {u^2+\frac{21}{2}+\sqrt{\frac{7}{4}}i}{\sqrt {u^2+\frac{21}{2}-\sqrt{\frac{7}{4}}i}}} \mathrm{d}u$$
and it was pretty like Incomplete elliptic integral of the first kind.
But, how to carry on?
| As noted, this integral is reducible to an elliptic integral and the form of the proposed solution may suggest that its modulus could be a singular value. Indeed, from the form
\begin{align}
I&=2\int_0^\infty \frac{\mathrm{d}u}{\sqrt {u^4+21u^2+112}} \\
&=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {u^2+\frac{21}{2}+\sqrt{\frac{7}{4}}i}{\sqrt {u^2+\frac{21}{2}-\sqrt{\frac{7}{4}}i}}} \\
&=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {\left( u^2+u_+^2\right)\left( u^2+u_-^2 \right)}}
\end{align}
The opposite of the roots of the polynomial under the square root ($u_{\pm}^2=(21\pm\sqrt{7})/2$) are complex conjugate, as the coefficients of the polynomial are real. In the following, we first apply a Landen's transformation where these roots are replaced by their geometric and arithmetic means which are real.
Defining $p=\sqrt{u_+u_-}$ and $q=(u_++u_-)/2$, we find from the coefficients of the polynomial
\begin{align}
p^2&=\sqrt{112}=4\sqrt{7}\\
q^2&=\frac{1}{4}\left( u_+^2+u_-^2+2u_+u_- \right)\\
&=\frac{1}{4}\left(21+2\sqrt{112} \right)=\frac{1}{4}\left(21+8\sqrt{7} \right)
\end{align}
With $u=x+\sqrt{x^2+p^2}$, remarking that
\begin{equation}
\sqrt {\left( u^2+u_+^2\right)\left(u^2+u_-^2 \right)}=2u\sqrt{x^2+q^2}
\end{equation}
and that
\begin{equation}
du=\frac{u}{\sqrt{x^2+p^2}}
\end{equation}
The integral becomes
\begin{equation}
I=\int_{-\infty}^\infty\frac{dx}{\sqrt{\left( x^2+p^2 \right)\left( x^2+q^2 \right)}}
\end{equation}
Now, changing $x=q\tan\theta$ and taking into account the parity of the function, one obtains an integral representation of the complete elliptic integral of the first kind:
\begin{align}
I&=2\int_0^{\pi/2} \frac{d\theta}{\sqrt{p^2\cos^2\theta+q^2\sin^2\theta}}\\
&=\frac{2}{p}\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{p^2-q^2}{p^2}\sin^2\theta}}\\
&=\frac{2}{p}\mathbf{K}\left( \sqrt{1-\frac{q^2}{p^2}} \right)%\\
%&=\frac{2}{p}K'\left( \frac{q}{p} \right)
\end{align}
Now, $\sqrt{1-\frac{q^2}{p^2}} =\frac{\sqrt{2}}{8}\left( 3-\sqrt{7} \right)=\lambda^*(7)$, where $\lambda^*(r)$ gives a singular value of the elliptic modulus $k_r$ for which $\mathbf{K}\left( \sqrt{1-k_r^2} \right)=\sqrt{r}\mathbf{K}(k_r)$ see here, here and here. Then,
\begin{align}
I&=\frac{2}{p}\mathbf{K}\left( k_7 \right)\\
&=\frac{2}{2.7^{1/4}}\frac{\Gamma\left( \frac{1}{7} \right)\Gamma\left( \frac{2}{7} \right)\Gamma\left( \frac{4}{7} \right)}{4\pi 7^{1/4}}\\
&=\frac{\Gamma\left( \frac{1}{7} \right)\Gamma\left( \frac{2}{7} \right)\Gamma\left( \frac{4}{7} \right)}{4\pi \sqrt{7}}
\end{align}
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
Find $a_n$, given $a_1=1$ and $a_{n+1} =a_n/4+3/4$ The sequence is defined of rational numbers $a_1,a_2,a_3,...$. Let $a_1=1$ and $a_{n+1}=a_n/4+3/4$. What are the numbers $a_1,a_2,a_3,...$ ?
What I've done:
$a_1=1$
for $n=1$ we have $a_{n+1}=a_2=a_1/4+3/4=1$
for $n=2$ we have $a_{n+1}=a_3=a_2/4+3/4=1$
Therefore the sequence consists of the numbers $1,1,1,...$
I have two questions.
1) Is it correct or am I misunderstanding something?
2) How would I go about finding $a_n$?
| Claim: for every $n \in \mathbb{N}$; we have: $a_n=1$ .
Proof by induction: suppose that the assertion holds for $k=n$;
i.e. $a_n=1$ ; then one can see that
$$a_{n+1}=\dfrac{a_n}{4}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1;$$
so the assertion also holds for $k=n+1$.
Let $A \in \mathbb{R}$ be arbitrery, and let $a_1=A$;
Claim: for every $n \in \mathbb{N}$; we have: $a_n=\dfrac{A-1}{4^{n-1}}+1$ .
Proof by induction: suppose that the assertion holds for $k=n$;
i.e. $a_n=\dfrac{A-1}{4^{n-1}}+1$ ; then one can see that
$$a_{n+1}=\dfrac{a_n}{4}+\dfrac{3}{4}=\dfrac{\dfrac{A-1}{4^{n-1}}+1}{4}+\dfrac{3}{4}=
\dfrac{A-1}{4^{n}}+\dfrac{1}{4}+\dfrac{3}{4}
=
\dfrac{A-1}{4^{n}}+1
;
$$
so the assertion also holds for $k=n+1$.
Second proof:
Let's define $b_n=a_n-1$, so we get the second sequence as follows:
$$
b_0=A-1
\ \ \ \
\text{and}
\ \ \ \
b_n=\dfrac{b_n}{4}
;
$$
one can see easilly that $b_n=\dfrac{A-1}{4^{n-1}}$,
so we can conclude $a_n=\dfrac{A-1}{4^{n-1}}+1$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to solve the equation $\log_2(x-9)+\log_{(2x-18)}6=3$.
Solve the following equation. $$\log_2(x-9)+\log_{(2x-18)}6=3.$$
I tried this way,
\begin{align}
\log_2(x-9)+\log_{(2x-18)}6 & \ = \ 3\\
\log_2(x-9)+\log_{2(x-9)}6 & \ = \ 3\\
\log_2(x-9)+\frac{\log_{2}6}{\log_2(2(x-9))} & \ = \ 3\\
\log_2(x-9)+\frac{\log_{2}6}{1+\log_2(x-9)} & \ = \ 3\\
u+\frac{\log_{2}6}{1+u} & \ = \ 3&&\text{(when $ u=\log(x-9) $)}\\
u(1+u)+\log_{2}6 & \ = \ 3(1+u)\\
u^2-2u+(\log_{2}6-3) & \ = \ 0\\
u & \ = \ \frac{2\pm\sqrt{-4\log_26+16}}{2}\\
& \ = \ 1\pm\sqrt{-2\log_26+8}
\end{align}
| Now, $x=9+2^{1+\sqrt{4-\log_26}}$ or $x=9+2^{1-\sqrt{4-\log_26}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence of series $(\frac{1}{3})^{2}+(\frac{1.2}{3.5})^{2}+(\frac{1.2.3}{3.5.7})^{2}+...$ $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$
I am not able to find a general equation and that's creating problem for me as I can't proceed further without it.
| Yes it converges of course! Use $$\frac{n}{2n+1}<\frac{1}{2}$$
We obtain:
$$\left(\frac{1}{3}\right)^{2}+\left(\frac{1.2}{3.5}\right)^{2}+\left(\frac{1.2.3}{3.5.7}\right)^{2}+...<\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show $x=0$ and $x=1$ are the only integer solutions I'm trying to show that the only solutions to $x^2-x+1=y^2$ are when $x=0$ and $x=1$. All I can think of is completing the square gives $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}$, which is clearly not a perfect square. Does this suffice? Should I use induction?
| Multiply by $4$ to obtain $$(2x-1)^2+3=4y^2$$ or $$4y^2-(2x-1)^2=3$$ or $$(2y-2x+1)(2y+2x-1)=3$$
And the factors are in some order either $3,1$ or $-3, -1$
The sum of the two factors is $4y=4, -4$ respectively, so $y=\pm 1$
The difference between the two factors is $4x-2=\pm 2$ so $x=0,1$
All four combinations actually work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
| Without congruences, subtraction, or factoring a difference of common powers:
Inductive assumption:
For some $n$, there is a $k$ such that
$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1} = 19k $
i.e.
$20(50^n)+18(12^n) = 19k$
For this $n$, add $980(50^n) + 198(12^n)$ to both sides.
The left-hand side will turn into the desired $(n+1)^{th}$ expression. The right-hand side is now
$19k+11[20(50^n)+18(12^n)]+38[20(50^n)]$
Here just use the inductive assumption again, and these terms will have a common factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
A equation about determinant I read from the Wiki that:
When A = D and B = C, the blocks are square matrices of the same order and the following formula holds (even if A and B do not commute)
$$\det\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}=\det(A-B)\det(A+B).$$
How to prove this relation when $A$ and $B$ do not commute?
| Substitute the row $i$ by $R_i-R_{i+n}$ for each $i\in \{1,\dots,n\}$, where $n$ is the dimension of $A$ and $B$. We get
$$\det\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}=\det\begin{pmatrix}
A-B & B-A\\
B & A
\end{pmatrix}.$$
Now, consider the matrix $\begin{pmatrix}
A-B & B-A\\
B & A
\end{pmatrix}$ and substitute the column $R_{n+j}$ by $R_{n+j}+R_j$ for each $j\in \{1,\dots,n\}$ to get
$$
\det\begin{pmatrix}
A-B & B-A\\
B & A
\end{pmatrix}=\det\begin{pmatrix}
A-B & 0\\
B & A+B
\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the cosine of $\theta$ is $\frac{1 - t^2}{1 + t^2}$. Where am I going wrong? I've recently picked up the book Mathematics and it's History by John Stillwell due to a recent curiosity in the history of math.
I started doing one of the exercises in the book and got a little stumped (I'll admit, I'm quite rusty at math).
I'm asked to use this figure ...
... to show that $\cos \theta = \dfrac{1 - t^2}{1 + t^2}$.
I labeled the bottom side of the smaller triangle $a$, and the right side of the smaller triangle as $b$, which makes the bottom side of the bigger triangle $1 + a$.
In these terms, $\cos \theta = a$, since the hypotenuse of the smaller triangle is $1$.
I also came up with the following equations:
$$
a^2 + b^2 = 1
$$
$$
(a + 1)^2 + b^2 = t^2
$$
Solving the first equation for $b$ gives:
$$
b = \sqrt{1 - a^2}
$$
Substituting this equation into the second equation above gives:
$$
(a + 1)^2 + (\sqrt{1 - a^2})^2 = t^2
$$
$$
a^2 + 2a + 1 + 1 - a^2 = t^2
$$
$$
2a + 2 = t^2
$$
$$
a = \frac{t^2 - 2}{2}
$$
As you can tell, $\frac{t^2 - 2}{2}$ is not equal to the given $\frac{1 - t^2}{1 + t^2}$.
Where am I going wrong in my logic? Thanks!
| You never say what $t$ is. I suppose that it is the slope of the line. Since it is a line passing through $(-1,0)$ and $(\cos\theta,\sin\theta)$, you have $t=\frac{\sin\theta}{1+\cos\theta}$. So,$$t^2=\frac{1-\cos^2\theta}{(1+\cos\theta)^2}$$and therefore\begin{align}\frac{1-t^2}{1+t^2}&=\frac{(1+\cos\theta)^2-1+\cos^2\theta}{(1+\cos\theta)^2+1-\cos^2\theta}\\&=\frac{2\cos^2\theta+2\cos\theta}{2+2\cos\theta}\\&=\cos\theta.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{n \to \infty}\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right)=?$ What is the value of limit
$$\lim_{n \to \infty}\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right)$$
If we know that $a>0$?
I get stuck on this, it seems to be Riemann sum but I can't find relation. I am thankful if someone could guide me.
| Hint:
$$
\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right) \leq \left( \frac{\sqrt[n]a}{n}+\frac{\sqrt[n]{a^2}}{n}+\frac{\sqrt[n]{a^3}}{n}+\cdots+\frac{\sqrt[n]{a^n}}{n}\right)=\sqrt[n]{a}.\frac{a-1}{n\left(\sqrt[n]{a}-1\right)}
$$
and
$$
\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right) \geq \left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+1}+\frac{\sqrt[n]{a^3}}{n+1}+\cdots+\frac{\sqrt[n]{a^n}}{n+1}\right)=\sqrt[n]{a}.\frac{a-1}{(n+1)\left(\sqrt[n]{a}-1\right)}
$$
$\left(\text{ The answer would be }\dfrac{a-1}{\log a}.\right)$
| {
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"url": "https://math.stackexchange.com/questions/2420068",
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"source": "stackexchange",
"question_score": "5",
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} |
Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
I have reached:
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$.
How do I continue from here? I am unable to find a suitable condition to proceed.
Edit:
vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$
and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$
but that ain't much helpful, is it?
| $y = 1 +\frac{2x}{x^2 + x + 1} = 1 + \frac{2}{1+x+1/x}$, $x\ne 0.$
A) Let $x\gt 0:$
1) $x+1/x \ge 2$
AM-GM: $x+1/x \ge 2\sqrt{x×1/x}$
$y_{max} = 1 + \frac{2}{1+2} = 5/3$.
B) Let $x \lt 0:$
2) $x+1/x \le -2$
$y_{min} = 1 + \frac{2}{1- 2} = 1+ (-2) = -1.$
Since
$y = \frac{x^2 + 3x + 1}{x^2+x+1}$ is continuous on $\mathbb{R}$, we find:
Range$ = [-1,5/3].$
| {
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"url": "https://math.stackexchange.com/questions/2420197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ where $n$ is a positive integer.
We have \begin{align*}\lfloor(1+\sqrt{3})^{2n-1}\rfloor &= (\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}\\&=\dfrac{(4+2\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(4-2\sqrt{3})^n}{1-\sqrt{3}}\\&=2^n\left(\dfrac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(2-\sqrt{3})^n}{1-\sqrt{3}}\right).\end{align*} Thus we must show that the second term is an integer. How can we continue?
| Set $x_n=\frac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\frac{(2-\sqrt{3})^n}{1-\sqrt{3}}$. Show that $x_1, x_2$ are integer and $x_{n+2}=4x_{n+1}-x_{n}$ holds, thus by induction $x_n$ is an integer for all $n \in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (Click here.) $$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots\to~?$$
I tried like below
$$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots=\\\sum_{n=1}^{\infty}\frac{1}{1^2+2^2+3^2+\dots+n^2}=\\\sum_{n=1}^{\infty}\frac{1}{\frac{n(n+1)(2n+1)}{6}}=\\\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=\\
$$
Then I can use the fraction ,but $$\frac{1}{n(n+1)(2n+1)}=\frac{1}{n}+\frac{1}{n+1}+\frac{-4}{2n+1}$$ This is ugly to turn into telescopic series .
Can you help me to find :series converge to ?
Thanks in advance .
| Starting from
$$S=6 \sum _{n=1}^{\infty } \left(\frac{1}{n+1}-\frac{4}{2 n+1}+\frac{1}{n}\right)$$
I changed the index $n\to n+1$
$$S=6 \sum _{n=0}^{\infty } \left(\frac{1}{n+2}-\frac{2}{n+\frac{3}{2}}+\frac{1}{n+1}\right)$$
Then I applied an amazing property of
digamma function that can be found here
and I got
$$S=6 \left(-\psi (1)-\psi (2)+2 \psi \left(\frac{3}{2}\right)\right)$$
which gives ($\gamma $ is Euler-Mascheroni constant)
$$S=6 \left(2 \gamma -1+2 \psi \left(\frac{3}{2}\right)\right)=6 (3-2 \log 4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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solve equation with cos and powers
Solve the following equation.
$$3^{x^2-6x+11}=8+\cos^2\frac{\pi x}{3}.$$
Here is the problem that I am struggling with it
I tried to take a logarithm of both side but I kinda stuck
can someone help me with this?
Thanks
| Add 1 to each side of the equation to get
$$3^{x^2-6x+11} + 1 = 9 + \cos^2 \frac {\pi x} {3}$$
Take $\log_3$ of both sides to get
$$x^2-6x+11 = 2 + \log_3 \cos^2 \frac {\pi x} {3}$$
The maximum of $\cos^2 \frac {\pi x} {3} = 1$, thus $\log_3 1 = 0$ , so we have
$$x^2 - 6x + 11 = 2$$
Bringing the 2 to the right hand side we have
$$x^2 - 6x + 9 = 0$$
Factoring gives us
$$(x-3)^2 = 0$$
And thus we have $$x = 3$$
Check: $3^{3^2-18+11} = 3^{9-18+11} = 3^2 = 9; \cos^2 \pi = 0; \Rightarrow 9 + 0 = 9 \ \ \checkmark$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ended up with
$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
Which I'm not sure what to do with.
| $$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}=
\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} + \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}}=
\lim_{x\rightarrow \infty} \frac{ax-bx}{|x+\frac{a}{2}| + |x+\frac{b}{2}|}
$$
this is
$$\lim_{x\rightarrow +\infty} \frac{(a-b)x}{2x+\frac{a+b}{2}}=\color{blue}{\frac{a-b}{2}}$$
as $x\rightarrow +\infty$, and
$$\lim_{x\rightarrow -\infty} \frac{(a-b)x}{-2x-\frac{a+b}{2}}=\color{blue}{-\frac{a-b}{2}}$$
as $x\rightarrow -\infty$.
Of course I have done the first from you have been stucked, so you can do the main limit in this way
$$\lim_{x\rightarrow \infty} \sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} - \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}=\lim_{x\rightarrow \infty}|x+\frac{a}{2}| - |x+\frac{b}{2}|$$
and obtain the same results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
$66$ points in $100$ shots. I just received a probability problem from a friend via a text and by the time I took to read it, I was sent a solution as well - which is confusing.. The question goes like this..
A person shoots basketball 100 times. First time he scores a point and second time he misses it. For the following shots, the probability of him scoring a point is equal to number of points scored before this shot divided by number of shots taken before this shot, for ex: if he is into his 21st shot and he has scored 13 points in the first 20 shots then the probability of him scoring in 21st shot is 13/20. What is the probability of scoring 66 points in the 100 shots (including the first two)
The confusing solution:
$\dfrac1{(n-1)} \ge \dfrac1{99}$
Can someone please explain the logic behind this?
| Here is a slightly shorter version of Zubin Mukerjee's solution, with a (I believe) simpler argument for the edge cases $P(1,N)$ and $P(N-1,N)$. The arguments are effectively the same up to that point.
We denote the probability that we have a score of $x$ after $N$ shots as $P(x,N)$.
The problem states that
$$ P(1,2)=1,$$
hence
$$ P(0,2) = P(2,2) = 0.$$
We also have that the probability that we score on the $N$-th shot with a previous score $x$ is $\dfrac{x}{N-1}$.
We can write $P(x,N)$ as the sum of the chance of scoring on the $N$-th shot with a previous score of $x-1$, and the chance of not scoring the $N$-th shot with a previous score of $x$, or
$$
P(x,N) = P(x-1,N-1) \frac{x-1}{N-1}
+ P(x,N-1)\left(1 - \frac{x}{N-1}\right).
$$
We conjecture that, for $N>1$,
$$ P(x,N) =
\begin{cases}
\frac1{N-1} \quad & 1\leq x \leq N-1,\\
\,\,\, 0 & \text{otherwise}.
\end{cases}
$$
Base Case
Above, we see that $P(1,2) = 1$ and $P(0,2) = P(2,2) = 0$, which satisfies the conjecture.
Induction
Assuming the conjecture, we have
$$ P(x,N-1) =
\begin{cases}
\frac1{N-2} \quad & 1\leq x \leq N-2,\\
\,\,\, 0 & \text{otherwise}.
\end{cases}
$$
For $2\leq x \leq N-2$,
$$
\begin{aligned}
P(x,N) &= P(x-1,N-1) \frac{x-1}{N-1}
+ P(x,N-1)\frac{N-x-1}{N-1}\\
&= \frac1{N-2}\cdot\frac{x-1}{N-1}
+ \frac1{N-2}\cdot\frac{N-x-1}{N-1} \\
&= \frac1{N-1}.
\end{aligned}
$$
Edge cases (alternative argument) - for $x=1$,
$$
\begin{aligned}
P(1,N) &= P(0,N-1) \frac{1-1}{N-1}
+ P(1,N-1)\frac{N-1-1}{N-1}\\
&= 0 + \frac1{N-2}\cdot\frac{N-2}{N-1}\\
&= \frac1{N-1}.
\end{aligned}
$$
For $x=N-1$,
$$
\begin{aligned}
P(N-1,N) &= P(N-2,N-1) \frac{N-2}{N-1}
+ P(N-1,N-1)\frac{N-(N-1)-1}{N-1}\\
&= \frac1{N-2}\cdot \frac{N-2}{N-1} + 0\\
&= \frac1{N-1}.
\end{aligned}
$$
Finally, it should be clear that $P(0,N) = P(N,N) = 0$ (since for $N=2$ we have already scored one and missed one).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 5
} |
Showing $ \mathbb{Q}[\sqrt[4]{2}]$ is a field Here i want to prove
\begin{align}
\mathbb{Q}[\sqrt[4]{2}] = \{ a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8} \mid a,b,c,d\in\mathbb{Q}\}
\end{align}
this is a field. To do that i want to prove
\begin{align}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} \in \mathbb{Q}[\sqrt[4]{2}]
\end{align}
Is there any ideas for showing this?
[edit] Many of you prove $\mathbb{Q}[\sqrt[4]{2}]$ is a field in indirect way. Is there any ways for showing this in direct way? For $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt[3]{2}]$, using some formulas i can check that indeed every nonzero elementn is unit. But this case does not seem that easy...
| Through the magic of upper triangulation and backsubstitution...
\begin{align*}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} &= e + f \sqrt[4]{2} + g \sqrt[4]{4} + h \sqrt[4]{8} \\
\iff 1 &= (a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8})(e + f \sqrt[4]{2} + g \sqrt[4]{4} + h \sqrt[4]{8}) \\
&= (ae + 2bh + 2cg + 2df) \\ &\qquad + (af + be + 2ch + 2dg)\sqrt[4]{2} \\ &\qquad + ( ag + bf + ce + 2dh )\sqrt[4]{4} \\ &\qquad + ( ah + bg + cf + de )\sqrt[4]{8} \\
\iff \{ 1 &= ae + 2bh + 2cg + 2df, \\
0 &= af + be + 2ch + 2dg, \\
0 &= ag + bf + ce + 2dh, \\
0 &= ah + bg + cf + de \} \\
\implies \{ h &= \frac{-bg - \dots}{a}, \\
g &= \frac{-bf- \dots}{a}, \\
f &= \frac{-be- \dots}{a}, \\
e &= \frac{a^3-2 a \left(2 b d+c^2\right)+2 c \left(b^2+2
d^2\right)}{a^4-8 b d \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2 \left(2 a
c+b^2\right)+8 a b^2 c-2 b^4+4 c^4-8 d^4} \} \\
\implies f &= \frac{-b \left(a^2+2 c^2\right)+2 d \left(2 a
c+b^2\right)-4 d^3}{a^4-8 b d \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2
\left(2 a c+b^2\right)+8 a b^2 c-2 b^4+4 c^4-8 d^4} \\
\implies g &= \frac{-a^2 c+a b^2+2 a d^2-4 b c d+2 c^3}{a^4-8 b d
\left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2 \left(2 a c+b^2\right)+8 a b^2 c-2
b^4+4 c^4-8 d^4} \\
\implies h &= \frac{-d \left(a^2+2 c^2\right)+2 b \left(a
c+d^2\right)-b^3}{a^4-8 b d \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2
\left(2 a c+b^2\right)+8 a b^2 c-2 b^4+4 c^4-8 d^4}
\end{align*}
Maybe this isn't the best way to go...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction: $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ I want to prove that $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ using proof by induction.
Attempt
Prove true for $n = 1$
$2^{1+2} + 3^{2(1) +1} = 35$
35 is divisible by 7 so true for $n =1$
Induction step: Assume true for $n = k$ and prove true for $n = k+1$
$n = k$
$2^{k+2} + 3^{2k +1} = 2^k \cdot 2^2 + 3^{2k} \cdot 3^1 = (4) 2^k + (3) 3^{2k}$
$n = k+1$
$2^{k+3} +3^{2k+3} = (8) 2^k + (27) 3^{2k}$
Thoughts:
I know I need to use the expression for $n = k$ to prove that it is true for $n = k+1$ but I am not sure where to start. I think that using modular arithmetic would be too complex. Any ideas would be greatly appreciated!
Update
Thanks for all of the useful answers/suggestions. I understand the approach to take now!
| \begin{eqnarray*}
2^{k+3}+3^{2k+3} =(9-7)2^{k+2} +9 \times 3^{2k+1}= 9(\color{blue}{2^{k+2}+3^{2k+1}})-\color{blue}{7} \times2^{k+2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
How many committees of $2$ women and $3$ men can be formed if two of the men refuse to serve on the committe together? From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?
This is an example in the book that has already been solved in the text. I understand how to answer the first question. However, it goes on to say that
"Because a total of $\binom{2}{2}\binom{5}{1} = 5$ out of the $\binom{7}{3} = 35$ possible groups of 3 men contain both of the feuding men, it follows that there are $35-5=30$ groups that do not contain both of the feuding men. Because there are still $\binom{5}{2}=10$ ways to choose the 2 women, there are $30*10=300$ possible committees in this case"
I'm just not sure where they got the fact that $\binom{2}{2}\binom{5}{1} = 5$ out of the $\binom{7}{3} = 35$ possible groups of 3 men contain both of the feuding men.
Thanks!
| From a set of 5 women you always take two women.
From a set of two men you take one or none of them
Then you take 2 or 3 from a set of 5 men. So:
\begin{align}
P=\binom{5}{2}\binom{2}{1}\binom{5}{2} +\binom{5}{2}\binom{2}{0}\binom{5}{3} \\
P=\underbrace{\binom{5}{2}}_\text{Two women}(\underbrace{\binom{2}{1}\binom{5}{2} +\binom{2}{0}\binom{5}{3}}_\text{Three men})\\
P=\binom{5}{2}(\frac {2!}{1!1!}\frac {5!}{2!3!} +\frac {2!}{2!0!} \frac {5!}{3!2!})\\
P=\binom{5}{2}(2*10+1*10)\\
P=\frac {5!}{2!3!}(30)\\
P=10*30=300
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Explicit form of $\sin(x-y/2)$ I want to know if there exist and how to evaluate the explicit form of a curve given by $$y=\sin(x-y/2)$$
My intuition tells me there should exist one, because deivative of y w.r.t. x is always finite:
$$y'(x) = \frac{2\cos(x - y/2)}{2 + \cos(x - y/2)}$$
but I have no idea how tho evaluate this explicit form.
| First, solve for $x$ when $|x|\le\frac{\pi+1}2$, where the function is invert-able:
$$f(y)=\arcsin(y)+\frac12y=x$$
By Taylor expanding,
\begin{align}f(y)&=\frac12y+\sum_{k=1}^\infty\binom{2k}k\frac{2ky^{2k-1}}{4^k(2k-1)^2}\\&=\frac32y+\frac16y^3+\frac3{40}y^5+\dots\end{align}
Substitute $y=a_0x+a_1x^3+a_2x^5+\dots$ and apply series reversion to get
$$y=\frac23x-\frac2{27}x^3+\frac{239}{21870}x^5+\mathcal O(x^7),\quad|x|\le\frac{\pi+1}2$$
For $\frac{\pi+1}2\le x\le\frac{3\pi-1}2$, use
$$f(y)=-\arcsin(y)+\frac12y=x-\pi$$
From there, we apply a similar approach:
\begin{align}f(y)&=\frac12y-\sum_{k=1}^\infty\binom{2k}k\frac{2ky^{2k-1}}{4^k(2k-1)^2}\\&=-\frac12y-\frac16y^3-\frac3{40}y^5-\dots\end{align}
Substitute $y=a_0(x-\pi)+a_1(x-\pi)^3+a_2(x-\pi)^5+\dots$ and apply series reversion to get
$$y=-2(x-\pi)+\frac83(x-\pi)^3-\frac{88}{15}(x-\pi)^5+\mathcal O((x-\pi)^7),\quad \frac{\pi+1}2\le x\le\frac{3\pi-1}2$$
And finally,
$$y(x)=\begin{cases}\frac23x-\frac2{27}x^3+\frac{239}{21870}x^5+\mathcal O(x^7),&|x|\le\frac{\pi+1}2\\-2(x-\pi)+\frac83(x-\pi)^3-\frac{88}{15}(x-\pi)^5+\mathcal O((x-\pi)^7),&\frac{\pi+1}2\le x\le\frac{3\pi-1}2\\y(x\pm2\pi),&\text{else}\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$1^2+3^2+...+(2n-1)^2+(2n+1)^2=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$n(2n-1)(2n+1)/3+3(2n+1)^2/3=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$4n(2n-1)(2n+1)/3=(n+1)(2n+3)/3$
What is wrong?
| Here's a non-induction approach, provided for information only:
$$\begin{align}
&1^2+3^2+5^2+\cdots+(2n-1)^2\\
&=\sum_{r=1}^n(2r-1)^2\\
&=\sum_{r=1}^n\binom {2r-1}2+\binom {2r}2\\
&=\sum_{s=1}^{2n}\binom s2\\
&=\binom {2n+1}3
&&=\frac {(2n+1)(2n)(2n-1)}{1\cdot 2\cdot 3}\\
& &&=\frac 13n(2n-1)(2n+1)\\
& && =\frac 13 n(4n^2-1)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that the perpendicular from the origin upon the straight line Prove that the perpendicular drawn from the origin upon the straight line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ bisects the distance between them.
My Attempt:
Equation of the line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ is:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$
| Let $O$ be the origin.
Midpoint $M$ of $A$ and $B$:
$x_M =(1/2)c(\cos\alpha +\cos\beta)$;
$y_M =(1/2)c(\sin\alpha + \sin\beta)$;
Line $l_1$ joining points $O$(origin) and $M$ has slope:
$m: = \dfrac{y_M}{x_M}$.
Line $l_2$ joining $A$ and $B$ has slope:
$m':= \dfrac{\sin\beta - sin\alpha}{\cos\beta - \cos\alpha}$.
Remains to be shown:
$m'= -\dfrac{1}{m}$.
$m' = $
$m' × \dfrac{\cos\beta +\cos\alpha}{\cos\beta + cos\alpha}=$
$\dfrac{(\sin\beta -\sin\alpha)(\cos\beta + \cos\alpha)}{\cos^2 \beta - \cos^2\alpha} =$
$\dfrac{(\sin\beta -\sin\alpha)(\cos\beta +\cos\alpha)}{\sin^2\alpha - \sin^2\beta}=$
$- \dfrac{cos\beta +\cos\alpha}{\sin\beta +sin\alpha}=$
$-(1/m)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Arithmetic and Geometric progression in 3 numbers Suppose
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$
$$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally $$a=b=c=\frac{1}{2}.$$
Can there be any such triplet such that $a<b<c$ ?
| $2b=a+c,\;
a+b+c=\dfrac{3}{2}$
give $3b=\dfrac{3}{2}\to b=\dfrac{1}{2}$
and $a+c=1\to c=1-a$
Substitute in $b^4=a^2c^2$ and get
$a^2(1-a)^2=\dfrac{1}{16}$
$a(1-a)=\pm \dfrac{1}{4}$
$a (1-a)=\dfrac{1}{4}$ gives $a=b=c=\dfrac{1}{2}$ you have already found
$a(1-a)=-\dfrac{1}{4}$ gives
$\color{red}{a=\dfrac{1}{2} \left(1-\sqrt{2}\right),\;b=\dfrac{1}{2};\;c=\dfrac{1}{2} \left(1+\sqrt{2}\right)}$ which is the solution you was looking for $a<b<c$
the other one
$a=\dfrac{1}{2} \left(1+\sqrt{2}\right),b=\dfrac{1}{2},c=\dfrac{1}{2} \left(1-\sqrt{2}\right)$ does not satisfy the request
Hope this is useful
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
Question:
Find the values of the constants $a$,$b$ and $c$ such that the ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
What I've done so far:
Substitute $(-1,2)$ in $4x^2+y^2+ax+by+c=0$:
$$\begin{align}
&\implies 4+4-a+2b+c=0 \\
&\implies c=a-2b-8 \tag{1}
\end{align}$$
Rewriting the equation of the ellipse to get standard form:
$$\begin{align}
& 4x^2+y^2+ax+by=-c \\[6pt]
\implies& 4\left(x^2+\frac{a}{4}x+\frac{a^2}{64}\right)
+\left(y^2+by+\frac{b^2}{4}\right)=a-2b-8+\frac{a^2}{16}+\frac{b^2}{4} \\[6pt]
\implies& \frac{4\left(x+\frac{a}{8}\right)^2}{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}}+\frac{\left(y+\frac{b}{2}\right)^2}{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}}=1
\end{align}$$
Since the ellipse is tangent to the $x$-axis:
$$\begin{align}
&\quad\sqrt{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}} = \frac{b}{2} \\[6pt]
\implies&\quad a-2b-8+\frac{a^2}{16}+\frac{b^2}{4} = \frac{b^2}{4} \\
\implies&\quad 2b= a-\frac{a^2}{16}-8 \tag{2}
\end{align}$$
From $(1)$ and $(2)$:
$$\quad c= a-\left(a-\frac{a^2}{16}-8\right)- 8 \quad\implies\quad c= \frac{a^2}{16} \tag{3}$$
Therefore, the equation of the ellipse becomes
$$\frac{4\left(x+\frac{a}{8}\right)^2}{a-(a-\frac{a^2}{16}-8)-8+\frac{a^2}{16}+\frac{b^2}{4}}+\frac{(y+\frac{b}{2})^2}{a-(a-\frac{a^2}{16}-8)-8+\frac{a^2}{16}+\frac{b^2}{4}}=1$$
That is,
$$\frac{\left(x+\frac{a}{8}\right)^2}{b^2/16}+\frac{\left(y+\frac{b}{2}\right)^2}{b^2/4}=1 \tag{4}$$
I'm stuck at this point and don't know how to find $a$ (or what else to do in general).
All help is appreciated.
| Your equation $(1)$ is correct:
$$a - 2 b - c = 8 \tag{1}$$
For the tangency condition, you're thinking a little too hard. If the ellipse is tangent to the $x$-axis, then the equation should have a double-root for $x$ when $y=0$; that is, its discriminant must vanish:
$$4 x^2 + a x + c = 0 \quad\to\quad \text{discriminant} = a^2 - 4\cdot 4\cdot c = 0 \quad\to\quad a^2 = 16 c \tag{2}$$
... which is, of course, your $(3)$. Now, two independent equations aren't enough to specify three unknowns. There are lots of solutions here. Since $a^2 \geq 0$, we know $c$ is non-negative; take $c := (2d)^2$ for some $d$ (the factor of $2$ helps us avoid some fractions), and we have
$$a = \pm 8 d \qquad b = -2( d^2 \mp 2 d + 2 ) \qquad c = 4d^2 \tag{3}$$
For completeness, the equation of the ellipse is ...
$$4 \left(x \pm d \right)^2 + \left( y -(d^2\mp 2 d+2) \right)^2 = (d^2\mp 2 d + 2)^2 \tag{4}$$
Note that the vertical radius is $\left|d^2\mp 2 d + 2\right|$, which matches the vertical offset of the center, easily re-confirming the tangency property. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$? $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$
After evaluating the limit as $x \to 0$, I noticed that the problem comes up to be in an indeterminate form of $0/0$. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.
However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of $0/0$ indeterminants.
In my second attempt,
I have tried multiplying $\exp(x^2)$ in both the numerator and denominator with hopes to balance out the $\exp(x^{-2})$. However, an indeterminant is $0/0$ still resulting.
Any help would be appreciated, thank you all!
| Hint: Use the Taylor approximation for sine and the exponential function.
$$\sin(u)=u-u^3/6+O(u^5)$$
$$\exp(u)=1+u+u^2/2 + O(u^3).$$
| {
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Find the remainder when $f(x)$ divided by $(x^2 + x + 1)(x+1)$. When a polynomial $f(x)$ is divided by $x^2 + x + 1$ and $(x+1)^2$, the remainder are $x+5$ and $x-1$ respectively. Find the remainder when $f(x)$ is divided by $(x^2 + x + 1)(x+1)$.
First, I let the remainder be $Ax^2 + Bx +C$, then I try to find the values of $A$, $B$ and $C$. There is $3$ unknowns so we need three equations but I can only get two equations.
| We know that
$$f(x)=Q_1(x)(x^2 + x + 1)+x+5=Q_2(x)(x+1)^2+x-1$$
for some polynomials $Q_1$ and $Q_2$.
Moreover, for some polynomial $Q_3$,
$$f(x)=Q_3(x)(x^2 + x + 1)(x+1)+Ax^2+Bx+C.$$
Hence
$$f(-1)=-1-1=A(-1)^2+B(-1)+C.$$
Now consider the complex zeros $w_1$ and $w_2$ of $x^2 + x + 1$. Then
$$f(w_1)=w_1+5=Aw_1^2+Bw_1+C\quad,\quad f(w_2)=w_2+5=Aw_2^2+Bw_2+C.$$
By solving the linear system of the three equations we find that $A = -6$, $B = -5$, $C = -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of an expression in two variables $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}$$
$x$ and $y$ are positive real numbers, what is the minimum value?
I have tried finding the minimum value of the first expression and the third expression, with the first expression being $\frac{7}{\sqrt{2}}$ while the second expression being $5$, I do not know how to move on from this (note that the answer is an integer from $0$ to $999$).
| let $$f(x,y)=\sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2+y^2-\sqrt{2}xy}+\sqrt{50+y^2-10y}$$
then solve
$$f_x=1/2\,{\frac {2\,x-7\,\sqrt {2}}{\sqrt {{x}^{2}-7\,\sqrt {2}x+49}}}+1/2
\,{\frac {2\,x-\sqrt {2}y}{\sqrt {{x}^{2}+{y}^{2}-\sqrt {2}xy}}}
=0$$
and
$$f_y=1/2\,{\frac {2\,y-\sqrt {2}x}{\sqrt {{x}^{2}+{y}^{2}-\sqrt {2}xy}}}+1/
2\,{\frac {2\,y-10}{\sqrt {{y}^{2}-10\,y+50}}}
=0$$
| {
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Show that no GCD exists for $4$ and $2 \cdot (1 + \sqrt{-3})$ in $\{ x + y \sqrt{-3} | x, y \in \mathbb{Z}\}$ I need to show that there is no greatest common divisor for $4$ and $2 \cdot(1 + \sqrt{-3})$ in the ring $R := \{ x + y \cdot\sqrt{-3} \,\,|\,\, x, y \in \mathbb{Z}\}$.
I found that both $2$ and $(1 + \sqrt{-3})$ divide $4$ as well as $2 \cdot(1 + \sqrt{-3})$. Therefore if $x$ was a greatest common divisor of the two, it has to be divisible by both $2$ and $(1 + \sqrt{-3})$.
How can I proceed from here?
| Let the common divisor of $4$ and $2(1+\sqrt{-3})$ be $x+y\sqrt{-3}$. Then we have
\begin{align}
(x+y\sqrt{-3})(a+b\sqrt{-3})&=4,\\
(x+y\sqrt{-3})(c+d\sqrt{-3})&=2(1+\sqrt{-3}),
\end{align}
where $x,y,a,b,c,d\in\mathbb{Z}$. Multiplying $x-y\sqrt{-3}$ on both sides of the equations, one obtains
\begin{align}
a&=\frac{4x}{x^2+3y^2},\quad
b=-\frac{4y}{x^2+3y^2},\\
c&=\frac{2(x+3y)}{x^2+3y^2},\quad
d=\frac{2(x-y)}{x^2+3y^2}.
\end{align}
The denominators have higher degrees than the numerators, so the number of integer pairs of $(x,y)$ that makes $|a|\geq1$ or $|b|\geq 1$ or $|c|\geq 1$ or $|d|\geq 1$ is finite. One can therefore enumerate all such cases. After ruling out them all, the only possibility left is $a=b=c=d=0$, which cannot hold either (not allowed to have $x=y=0$ as they appear in the denominators), thus completing the proof.
| {
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How to solve system of equation, $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, $\sqrt{y-4}+\sqrt{z-4}=4\sqrt3$ and $ \sqrt{x-9}+\sqrt{z-9}=4\sqrt3$ . $$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ I tried somthing,like go to the power of two , and change of variables... but it became more complicated . Is there an idea to solve this system of equation ?
Thanks in advance
| So here are the 3 equations:
$$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$
As suggested by transcenmental,
$\sqrt{x-1}-\sqrt{y-1}=4\sqrt 3 - 2 \sqrt{y-1}$ and multiplying with the first eq. gives
$$
x - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})
$$
For the second eq., use
$$-\sqrt{y-4}+\sqrt{z-4}=4\sqrt 3 - 2 \sqrt{y-4}$$ and multiplying with the second eq.
$$
z - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})
$$
Plugging into the last one gives an eq. in y:
$$
\sqrt{y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})-9}+\sqrt{ y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})-9}=4\sqrt3
$$
This is pretty akward, but $y = 28/3$ is a solution (by computer). From here the others follow, namely
$$
x = 52/3$$
and
$$
z = 76/3$$
EDIT:
with a little bit of hindsight and a little bit of psychology, you could argue
as follows (with a twinkling of an eye):
suppose the person asking the question prefers a reasonably nicely looking solution (psychology 1). Then all variables should either be multiples of 3 or of $1/3$ to get rid of the $\sqrt 3$ on the RHS. Let's try $1/3$ (hindsight 1). So let $x = x' / 3$ etc. Now assume further that also the numerator of the variables should be nice, e.g. no roots etc. (psychology 2). Then we should have that $x'-3\cdot 1$ and $x'-3\cdot 9$ are "nice" squares (likewise with the other variables). If we want it even nicer, they should be squares of integers (hindsight 2).
So $x' = 3 + n^2$ and $x' = 27 + m^2$. Now start playing. "Nice" integers n and m will be reasonably small (psychology 3). $x' = 52$ does it nicely, with $n=7$ and $m=5$. A small number of trials, to match all three variables, will then give the solution....
| {
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Proving that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$ I need to show that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$. My attempted method was showing that as the distance between $(x,y)$ and $(0,0)$ approaches $0$, so does the function in the limit. I tried to show $\lvert{\frac{x^2y^3}{x^2+y^2}}\rvert \leq \sqrt{x^2+y^2}$ for $(x,y) \neq 0$. I expanded the terms to get $\frac{x^2\lvert y^3\rvert}{x^2+y^2} \leq \frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}$, or $(x^2+y^2)^\frac{3}{2}\geq x^2\lvert y^3\rvert$. I am unsure of how to prove this and whether this method is valid or would work in showing the existence of the above limit at all.
| Hint: Where defined, $\frac{1}{x^2 + y^2}\leq \frac{1}{y^2}$
| {
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Prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ How to prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ ? I have tried AM-GM but with no effect.
| By Cauchy inequality we have:
$$(a^2 + 1)(1+b^2) \ge (a + b)^2$$
$$(a^2 + 1)(1+c^2) \ge (a + c)^2$$
$$(c^2 + 1)(1+b^2) \ge (c + b)^2$$
Multiply these and take square root and thus the conclusion.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5$
Evaluate
$$5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5.$$
The options are $5^6$, $5^7$, $5^8$, $5^9$, $5^{10}$.
I'm new to this site. I came across this question in an Olympiad foundation site. I have no idea how to solve it. Can I get the answer of this question. Thanks.
| $$a=5+4.5+4.5^2+4.5^3+4.5^4+4.5^5\to \times 5\\5a=25+4.5^2+4.5^3+4.5^4+4.5^5+4.5^6$$ now $5a-a= ?$
$$\quad{5a-a=(25-5)-(4.5)+(4.5^2-4.5^2)+(4.5^3-4.5^3)+(4.5^4-4.5^4)+(4.5^5-4.5^5)+4.5^6\\\to 5a-a=4.5^6 \\4a=4.5^6\\a=5^6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditions necessary for a function to contain multiple attractive fixed points Given a function $f(x)$, what are the necessary conditions for $f(x)$ to contain multiple attractive fixed points?
Another key condition is I want the construction of the fixed point $x^*$ to be repeated application of $f$: $\lim\limits_{x \rightarrow n} f^n(x) = x^*$
One approach I looked at was forcing $f$ to be a contraction mapping therefore having a Lipshitz constant be less than one. This will force the function to contain a fixed point (via Banach's fixed point theorem), but I don't think a contraction map is capable of having multiple attractive fixed points.
Any ideas on how to approach this construction?
| All attracting fixed points must be fixed points first of all. So, the function $f(x)$ must have multiple fixed points, or solutions for $f(x^*)=x^*$. If function $f(x)$ is differentiable in open neighbourhoods of those fixed points $\{x^*\}$ and $|f'(x^*)|<1$, then those points are also attracting. Let's use these statements to construct a function $f(x)$ with multiple contracting fixed points.
Consider $\left\{\frac{1}{4},\frac{2}{4},\frac{3}{4} \right\}\subset (0,1)$, the polynomial
$$P_{3}(x)=\left(x-\frac{1}{4}\right)\left(x-\frac{2}{4}\right)\left(x-\frac{3}{4}\right)$$
and
$$f(x)=x-P_{3}(x)$$
We have
$$f\left(\frac{1}{4}\right)=\frac{1}{4},f\left(\frac{2}{4}\right)=\frac{2}{4},f\left(\frac{3}{4}\right)=\frac{3}{4}$$
$$f'(x)=1-\left(x-\frac{2}{4}\right)\left(x-\frac{3}{4}\right)-\left(x-\frac{1}{4}\right)\left(x-\frac{3}{4}\right)-\left(x-\frac{1}{4}\right)\left(x-\frac{2}{4}\right)=\\
-3x^2+3x+\frac{5}{16}$$
And
$$\left|f'\left(\frac{1}{4}\right)\right|=\frac{7}{8}<1$$
$$\left|f'\left(\frac{2}{4}\right)\right|=\frac{17}{16}>1$$
$$\left|f'\left(\frac{3}{4}\right)\right|=\frac{7}{8}<1$$
So $f(x)$ has 2 attracting fixed points.
Now, try with $\left\{\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6} \right\}\subset (0,1)$ ... and continue with $\left\{\frac{1}{2n},\frac{2}{2n},...,\frac{2n-1}{2n} \right\}\subset (0,1)$ to obtain a function with $n$ attracting fixed points.
| {
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Show that $\overline{(2+\sqrt{3})}^{17} = \overline{2 - \sqrt{3}}$ in $\mathbb{Z}[\sqrt{3}]/(17)$ I am trying to solve the following problem:
Show that $\overline{(2+\sqrt{3})}^{17} = \overline{2 - \sqrt{3}}$ and $\overline{(2-\sqrt{3})}^{17} = \overline{2 + \sqrt{3}}$ in $\mathbb{Z}[\sqrt{3}]/(17)$, being $(17)$ the principal idela generated by $17$ in $\mathbb{Z}[\sqrt{3}]$ and $\overline{\alpha}$ the equivalence class of $\alpha$ in this quotient.
that is an intermediate step to calculating the remainder of $(2+\sqrt{3})^{17} + (2-\sqrt{3})^{17}$ in the division by 17.
My solution:
I am having some trouble on figuring out what is the set $\mathbb{Z}[\sqrt{3}]/(17)$. As I understand, given some $\alpha = a + b\sqrt{3} \in \mathbb{Z}[\sqrt{3}]$, then:
$$
\overline{\alpha} = \overline{a+b\sqrt{3}} = a + b\sqrt{3} + 17\mathbb{Z}[\sqrt{3}] = (a + 17\mathbb{Z}) + (b\sqrt{3} + 17\mathbb{Z}\sqrt{3}) = \overline{a} + \overline{b}\sqrt{3}
$$
where $\overline{a}$ is the equivalence class of $a$ in $\mathbb{Z}/17\mathbb{Z}$.
Question 1: is the above correct? If not, what is $\overline{\alpha} \in \mathbb{Z}[\sqrt{3}]/(17)$?
Then, assuming the above was right, i performed the following calculations:
$$
(\overline{2} + \overline{1}\sqrt{3})^2 = \overline{7} + \overline{2}\sqrt{3}
$$
$$
\therefore (\overline{2} + \overline{1}\sqrt{3})^3 = \overline{3}(\overline{1} - \overline{2}\sqrt{3})
$$
But then
$$
(\overline{2} + \overline{1}\sqrt{3})^{17} = (\overline{2} + \overline{1}\sqrt{3})^2(\overline{1} - \overline{2}\sqrt{3})^5\overline{3}^5 = \overline{11} - \overline{7}\sqrt{3}
$$
In this last step I omitted some computations that I believe are not important to show here, since I only wanted to show my idea for solving the problem.
Question 2: Is the above idea correct? If not, how to solve the problem?
Any hints and comments will be the most appreciated.
| Without answering your actual questions, here is another approach to the problem.
We have $\mathbb Z[\sqrt 3]/(17)\simeq \mathbb Z[x]/(x^2-3,17)\simeq \mathbb Z/(17)[x]/(x^2-3)$. We are looking to simplify $(2+x)^{17}$ in this ring. But because $17$ is prime, all the binomial coefficients $\binom{17}{i}$ with $0<i<17$ will be multiples of $17$ (if you have not come across this fact, try to prove it), and so in our ring, $(2+x)^{17}=2^{17}+x^{17}$ by the binomial theorem. By Fermat's little theorem, $2^{17}=2\pmod{17}$, and so $2^{17}+x^{17}=2+3^{8}x$, where we used $x^2=3$. Our calculation is finished by computing $3^{8}\equiv 9^{4}\equiv (-4)^2\equiv -1 \pmod{17}$, which gets you the answer you want.
Alternatively, we can compute $3^8\pmod{17}$ by quadratic reciprocity by noting that $$3^{8}\equiv 3^{(17-1)/2}\equiv\left(\frac{3}{17} \right)\equiv (-1)^{(3-1)(17-1)/4}\left(\frac{17}{3} \right)\equiv\left(\frac{2}{3} \right)\equiv -1\pmod{17}$$
N.B. The fact that $(a+b)^p=a^p+b^p$ in a ring of characteristic $p$ is a standard trick, and it implies that the map $x\mapsto x^p$ is a ring automorphism called the Frobenius automorphism. While it might look somewhat ad hoc here, it is a standard tool in the algebraist's toolbox.
| {
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Proving that a certain function approaches a certain limit at a certain point using epsilon-delta method (Proof Check) I have to prove that,
$\lim_{x\to 1}$ $\frac{x}{x+1}$ = $\frac{1}{2}$
using delta-epsilon method.
My answer: Let $\epsilon$ > $0$ and let $\delta$ = min( 1 , 2$\epsilon$)
Then, $\left|\frac{x}{x+1} - \frac{1}{2}\right|$ = $\left|x - 1\right|\frac{1}{2}\left|\frac{1}{x+1}\right|$
Since $\left|x-1\right| < 1$ implies that $0 < x < 2$
which, in turn implies that $\frac{1}{3} < \frac{1}{x+1} < 1$
So the result follows, $\left|x - 1\right|\frac{1}{2}\left|\frac{1}{x+1}\right| < \epsilon$
| Depending on where you are in your studies/class, more justification around some of these conclusions could be helpful for the sake of clarity. For example, writing:
$$\Big|\frac{x}{x+1} - \frac{1}{2}\Big| = \Big|\frac{2x}{2(x+1)} - \frac{x+1}{2(x+1)}\Big| = \Big|\frac{x-1}{2(x+1)}\Big| = \frac{1}{2}\Big|x-1\Big|\Big|\frac{1}{x+1}\Big|$$
Or, looking to the next line, that: $|x - 1| < 1$ means $-1 < x - 1 < 1$; so, we have $0<x<2$.
Similarly, this final inequality is equivalent to $1 < x+1 < 3$, whence
$$\frac{1}{3} < \frac{1}{x+1} < 1$$
as claimed. Acceptable proof writing is, to a nontrivial extent, a function of the reader.
| {
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Why is $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ rational? For this question,
Show the following irrational-looking expressions are actually rational numbers.
(a) $\sqrt{4+2\sqrt{3}}-\sqrt{3}$, and
(b) ...
I solved it as follows:
$$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\
x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\
(x+\sqrt{3})^{2} &= (\sqrt{4+2\sqrt{3}})^{2},\\
\end{align}\\
x^{2}+2\sqrt{3}x-(1+2\sqrt{3}) = 0,\\
(x-1)(x+(1+2\sqrt{3}))=0.$$
My question is that, there are two numbers satisfying $x = \sqrt{4+2\sqrt{3}}-\sqrt{3}$, but one of them is irrational. Then, how can we say $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational as a whole?
| $\sqrt{4+2\sqrt{3}}=\sqrt{1+2\sqrt{3}+3}=\sqrt{(1+\sqrt{3})^2}=1+\sqrt{3}$
| {
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Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e. $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}$$ and conversely $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
To my suspicions of which–either numerator or denominator–conjugate to multiply by I chose $$\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
This resulted in $$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{3-x-1}$$
Is it indeterminate? What is the reason for multiplying by a specific conjugate in a fraction (denominator or numerator) and the reason for the conjugate being either a) denominator b) numerator c) or both?
Am I simply practicing incorrect algebra by rationalizing the expression to:
$$\frac{(6-x)(3-x)-2\sqrt{3}+2x+2}{x-2}$$
Or am I failing to delve further and manipulate the expression out of the indeterminate form?
| Just another way:
Let $\sqrt{6-x}-2=h,\sqrt{3-x}-1=k\implies h\to0,k\to0$
and $$6-4-4h-h^2=x=3-1-2k-k^2\implies-h(4+h)=-k(k+2)\implies\dfrac hk=\dfrac{k+2}{h+4}$$
$$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{h\to0,k\to0}\dfrac hk=\lim_{h\to0,k\to0}\dfrac{k+2}{k+4}=?$$
| {
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p - Factorization modulo n I learned that ${n!} = p^{e}n_0$ where $e = \sum\limits_{k = 1}^\infty \lfloor\frac{n}{p^{k}}\rfloor$ where n is the number, p is the prime, $n_0$ is the product of factors of n! divided by $p^{e}$. Does it has something to do with this problem? Or Im referring to different thing?
Prove that: $\frac {n!}{p^{k}}\equiv(-1)^{k}k!(n - pk)!$mod p where $k = \lfloor\frac{n}{p}\rfloor$
| Case 1 : $k=\lfloor \frac{n}{p} \rfloor=0$ , so we have $\frac{n!}{p^0} = (-1)^0 * 0! *(n-0*p)! \mod p $ => $ n! = n! \mod p$ which is true.
Case 2 : $k=\lfloor \frac{n}{p} \rfloor=1$ , so we have $\frac{p!}{p} = (-1)^1 *1! *(0)! \mod p $ => $ (p-1)! = -1 \mod p$ which is true Wilson's Theorem
Assume its true for $n$ and we want to prove it for $n+1$,
There are two cases $ k=\lfloor \frac{n}{p} \rfloor = \lfloor \frac{n+1}{p} \rfloor$ or $ k=\lfloor \frac{n}{p} \rfloor , k+1 = \lfloor \frac{n+1}{p} \rfloor$
For the First case we need to prove that $\frac{(n+1)!}{p^k} = (-1)^k * k! *(n+1- p k)! \mod p$ given that $\frac{n!}{p^k} = (-1)^k * k! *(n-p k)! \mod p$
So we get that $\frac{(n+1) n!}{p^k} = (-1)^k *k!*(n-pk)! (n+1 -p k) \mod p $
From the assumption we get that $(n+1) (-1)^k *k! *(n-pk)! = (-1)^k *k!*(n-pk)! (n+1 -p k) \mod p$ which is just $(n+1-n-1+p k) (-1)^k *k! *(n-pk)! =0 \mod p$ which is true since $ pk =0 \mod p$.
For the second case we need to prove that $\frac{(n+1)!}{p^{k+1}} = (-1)^{k+1} (k+1)! (n+1-p (k+1))! \mod p$ given that $\frac{n!}{p^k} = (-1)^k * k! *(n-pk)! \mod p$
Obviously that $n+1 = (k+1)p$ since $\lfloor \frac{n}{p} \rfloor = k$ and $\lfloor \frac{n+1}{p} \rfloor = k+1$.
So $\frac{(n+1)!}{p^{k+1}} = (-1)^{k+1} (k+1)! (0)! \mod p$
Which is $\frac{(n+1) n!}{p^k p} = (-1)^{k+1} (k+1)! \mod p$
Which is $\frac{(k+1)p}{p} \frac{n!}{p^k} = (-1)^{k+1} (k+1)! \mod p$
By the assumption and the fact that $n+1=(k+1)p$ we get that its $(k+1) (-1)^k *k! *((k+1)p-1-p k)! = (-1)^{k+1} (k+1)! \mod p $ => $(k+1)! (-1)^k (p-1)! = (k+1)! (-1)^{k+1} \mod p $ by Wilson's Theorem we get that $(p-1)! =-1 \mod p$,
So we arrive at $(k+1)! (-1)^k *-1 = (k+1)! (-1)^{k+1} \mod p$
Which is true,thus concluding the proof.
Note : I did not use the Legendre formula in the proof which is $e =\sum \limits_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$, the main idea in my proof is Wilson's Theorem, yet there might be a proof using Legendre formula (i am not familiar with that).
| {
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"timestamp": "2023-03-29T00:00:00",
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Multiplicative group modulo $2^n$ I remember learning that the multiplicative group modulo $2^n$, namely the group $\mathbb{Z}_{2^n}^\times$of integers coprime with $2^n$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_{2^{n-2}}$, which is due to Gauss.
It boils down to proving that the multiplicative order of $3$ in the group is $2^{n-2}$. Can you give me a hint here?
| We want to show $$3^{2^{n-3}}\ne 1\mod 2^n$$ and $$3^{2^{n-2}}\equiv 1\mod 2^n$$
for $n\ge 3$
For $n=3$ , we have $3^{2^{n-3}}=3\ne 1\mod 8$ and $3^{2^{n-2}}=9\equiv 1\mod 8$
hence the claim is true for $n=3$
For $n=4$; we have $3^{2^{n-3}}=9\ne 1\mod 16$ and $3^{2^{n-2}}=81\equiv 1\mod 16$ , hence the claim is true for $n=4$
Assume the claim is true for $n$ with $n\ge 4$
Then, we have $$3^{2^{n-1}}-1=(3^{2^{n-2}}-1)(3^{2^{n-2}}+1)$$
By induction hypothesis, the first factor on the right side is divisible by $2^n$ and the second factor is even. Hence we have $2^{n+1}|3^{2^{n-1}}-1$ and therefore $$3^{2^{n-1}}\equiv1\mod 2^{n+1}$$
Now, we verify $$3^{2^{n-2}}\not\equiv 1\mod 2^{n+1}$$
We have $3^{2^{n-2}}-1=(3^{2^{n-3}}-1)(3^{2^{n-3}}+1)$
By induction hypothesis , the first factor on the right side is not divisible by $2^n$ and for the second, we have $3^{2^{n-3}}+1\equiv (-1)^{2^{n-3}}+1=2\mod 4$ hence the second factor is not divisble by $4$.
The exponents in the power of two in the prime factorizations are therefore a number smaller than $n$ in the first factor and $1$ in the second factor, so smaller than $n+1$ in total. Therefore, we have proven $3^{2^{n-2}}\ne 1\mod 2^{n+1}$ thus completing the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.
| By AM-GM for positive variables:
$$\sum_{cyc}\frac{b+c}{a}\geq2\sum_{cyc}\frac{\sqrt{bc}}{a}\geq6\sqrt[3]{\prod_{cyc}\frac{\sqrt{bc}}{a}}=6.$$
AM-GM it's the Jensen inequality for $f(x)=\ln{x}$ because $f''(x)=-\frac{1}{x^2}<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the following limit exist? (a)$ \lim_{(x,y) \to (0,0)} f(x,y) = \frac{x^2+y}{x^2+y^2}$? Does the following limit exist? If exist, evaluate (a)$ \lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}$ $(b)\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}$?
for (a)
approaching (0,0) using $y=0$ we have
$$\lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}\implies\lim_{x
\to 0} \frac{x^2}{x^2}= 1$$
approaching (0,0) using $y=-x^2$ we have
$$\lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}\implies\lim_{x
\to 0} \frac{x^2 -x^2}{x^2+x^4}= 0$$
Since the limits are different, $\lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}$ $D.N.E$
for (b) approaching using $y=0$
$$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}\implies\lim_{x \to 0} \frac{x(0)}{x^2+0} =0 $$
approaching (0,0) from $x=y^2$
$$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}\implies\lim_{y \to 0} \frac{y^4}{2y^4}=1/2 $$
Again limits are not equal so $\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}$ $D.N.E$
Is this correct ?
| we use $$x=\frac{1}{n},y=\frac{\sqrt{n}}{n}$$ and plug in this in your function
$$f(x,y)=\frac{x^2+y}{x^2+y^2}$$ we get
$${\frac {{n}^{2}+\sqrt {n}}{\sqrt {n} \left( n+1 \right) }}$$
and we obtain $$\lim_{n\to \infty}{\frac {{n}^{2}+\sqrt {n}}{\sqrt {n} \left( n+1 \right) }}=\infty$$ therefore the limit doesn't exist
| {
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Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation
| multiplying your given inequality by $$xyz>0$$ we get
$$x^2+y^2+z^2\geq xy+xz+zy$$
and this is equivalent to
$$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$ which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving systems of equation with unknown How do i solve this system?
$$
\left\{
\begin{array}{c}
ax+y+z=1 \\
x+ay+z=a \\
x+y+az=a^2
\end{array}
\right.
$$
Ive reduced to this form. How should i continue to get infinitely many solutions, no solution and unique?
$$ \left[
\begin{array}{ccc|c}
1&a&1&a\\
0&(a^2)-1&a-1&(a^2)-1\\
0&(a-1)(a-2)&0&a-1
\end{array}
\right] $$
| hint:$$\left\{
\begin{array}{c}
ax+y+z=1 \\
x+ay+z=a \\
x+y+az=a^2
\end{array}
\right.$$sum of the equations :
$$x(a+2)+y(a+2)+z(a+2)=1+a+a2\\x+y+z=\frac{1+a+a^2}{a+2}\\$$now $$ax+y+z=1-(x+y+z=\frac{1+a+a^2}{a+2})\\(a-1)x=1-\frac{1+a+a^2}{a+2}\\(a-1)=\frac{a+2-1-a-a^2}{a+2}\\x=\frac{1-a^2}{(a+2)(a-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Relationship between Pythagorean Theorem and leg-hypotenuse lengths With a standard right triangle, we have the legs $a$ and $b$ and the hypotenuse $c$, where typically $a$ is the shorter leg. I'm curious about the fact that
$$a + b > c$$
but
$$a^2 + b^2 = c^2$$
This discussion seems to be a simple proof -- assuming the Pythagorean Theorem is true. However, at an intuitive level, it seems odd that what wasn't equal is made equal by squaring each member. Is there any theoretical explanation from higher geometry?
| In $\triangle ABC$ with $\angle C$ denoting the interior angle at the vertex $ C$ and the sides' lengths being $a,b,c$ we have the Cosine Law: $$c^2=a^2+b^2-2ab\cdot \cos \angle C$$ which is very ancient. Therefore $$(a+b)^2=a^2+b^2+2ab=$$ $$=a^2+b^2-2ab \cdot \cos \angle C+2ab(1+\cos \angle C)=$$ $$=c^2+2ab (1+\cos \angle C).$$ Since $0<\angle C<\pi$ we have $1+\cos \angle C>0.$ Therefore $$(a+b)^2=c^2+2ab(1+\cos \angle C)>c^2$$ which implies $a+b>c.$
The fact that the interior angles of a triangle sum to $\pi\;$ ( implying $\angle C<\pi\;$ ) follows from the Parallel Postulate. The Cosine Law follows directly from the Theorem of Pythagoras, which follows from the Parallel Postulate, and the Theorem of Pythagoras is proved without needing to know that $a+b>c.$
The Theorem of Pythagoras can be considered a special case of the Cosine Law when $\angle C=\pi /2$ and $\cos \angle C=0.$ We can re-arrange the Cosine Law as $$\cos \angle C=\frac {a^2+b^2-c^2}{2ab}$$ from which we can see that if $\angle C<\pi /2$ then $a^2+b^2>c^2,$ and if $ \angle C>\pi /2$ then $a^2+b^2<c^2.$
| {
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If $a+b+c=0$ prove that $ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $$
I made this question as a more difficult (higher degree) version of this question. My idea was that algebraic brute force methods are easy to distinguish from more sophisticated ones if the degree of the terms in the question is higher.
The question was specifically made using the method from my answer to the linked question.
| Let's define
$$S_n=a^n+b^n+c^n$$
and consider the generating function
$$F(t)=\sum_{n=0}^\infty S_nt^n=\frac1{1-at}+\frac1{1-bt}+\frac1{1-ct}.$$
Using $a+b+c=0$ gives
$$(1-at)(1-bt)(1-ct)=1+pt^2+qt^3$$
for some $p$ and $q$, and
$$F(t)=\frac{3+pt^2}{1+pt^2+qt^3}.$$
From this we expand as a power series
$$F(t)=(3+pt^2)(1-(pt^2+qt^3)+(pt^2+qt^3)^2-\cdots)
=3-2pt^2-3qt^3+2p^2t^4+5pqt^5+\cdots$$
and now you can read off this, and several other identities...
| {
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If $-1$is a root for $ax^2+bx−3$, find $a^2+b^2$
Given: -1 is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$.
Find: the numeric value for $a^2+b^2$.
Background: question asked in an entrance exam (Colégio Militar 2005).
My attempt: the other root is $3/a$ and by substitution we can easily find that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2ab=9.$$
I got stuck at this point... how to get the value for $ab$? Hints please.
| Starting from $a-b=3$, which is odd, this implies the positive primes cannot be both odd. So one of them is $2$, the other is an odd prime. As $a-b>0$, it's $b$ which is equal to $2$, so $a=3+b=5$, and $a^2+b^2=29$.
| {
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Solving the cubic $x^3-x^2-2x+1 = 0$ Solving the cubic $x^3-x^2-2x+1 = 0$.
Using the Cubic Formula I get the following three solutions.
$x_1 = \frac{1}{3} - \frac{1}{3}\left(\frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3} - \frac{7}{3}\frac{1}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong -1.2469796037174670610+2.10^{-20}i$
$x_2 = \frac{1}{3} + \frac{1}{6}\left( \frac{7}{2}+\frac{21}{2}i\sqrt{3} \right)^{1/3}(1+i\sqrt{3}) + \frac{7}{6}\frac{1-i\sqrt{3}}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong .44504186791262880859 - 3.10^{-20}i$
$x_3 = \frac{1}{3} + \frac{1}{6}\left( \frac{7}{2}+\frac{21}{2}i\sqrt{3} \right)^{1/3}(1-i\sqrt{3}) + \frac{7}{6}\frac{1+i\sqrt{3}}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong 1.8019377358048382525 + 3.10^{-20}i$
It is clear that the three solutions are all real solutions, but is there a way I can remove the complex components algebraically? My ultimate goal is to describe what the Galois group from this polynomial would like.
| I'm sorry, but I'm not entirely sure if there's a way make a transformation to get the appropriate trigonometric roots. However, there is another way to derive its roots using Euler's formula.
Starting with your cubic$$y=x^3-x^2-2x+1$$First, make the substitution $x=t+t^{-1}$. Expanding and collecting in $t$, we find that$$\begin{align*}\left(t+\tfrac 1t\right)^3-\left(t+\tfrac 1t\right)^2-2\left(t+\tfrac 1t\right)+1 & =\frac {t^6-t^5+t^4-t^3+t^2-t+1}{t^3}\\ & =\frac {t^7+1}{t^3(t+1)}\end{align*}$$Therefore, we have that$$t^7+1=0\implies t=e^{\tfrac {\pi i(2k+1)}7}$$Hence, the solutions are given by$$t+t^{-1}=e^{\tfrac {\pi(2k+1)}7i}+e^{-\tfrac {\pi(2k+1)}7i}=2\cos\left(\frac {2k\pi+\pi}7\right)$$for $k=0,1,2$. And it follows immediately that your three solutions are$$\begin{align*} & x_1=2\cos\left(\frac {\pi}7\right)\\ & x_2=2\cos\left(\frac {3\pi}7\right)\\ & x_3=2\cos\left(\frac {5\pi}7\right)\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the ratio of the areas
I'm having a trouble in solving this problem. How am I supposed to approach the problem?
| Here's a "coordinate geometry" method: Set up a Cartesian coordinate system so that A is (0, 0), B is (1, 0) and R is (0, -1). Then Q is (1, -1). Each interior angle of a regular hexagon is 120 degrees so angles FAR and BAS are both 120- 90= 30 degrees. Then F is $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ and S is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
From that, the two lines, through A and S and through F and P, have slope $-\frac{1}{\sqrt{3}}= -\frac{\sqrt{3}}{3}$. In particular, the line through F and P can be written as $y+ \frac{\sqrt{3}}{2}= -\frac{1}{\sqrt{3}}\left(x- \frac{1}{2}\right)$$= -\frac{1}{\sqrt{3}}x+ \frac{1}{2\sqrt{3}}$ or $y= -\frac{1}{\sqrt{3}}x- \frac{1}{\sqrt{3}}$.
The lines through A and F and through S and P have slope $\sqrt{3}$. In particular, the line through S and P can be written $y+ \frac{1}{2}= \sqrt{3}\left(x- \frac{\sqrt{3}}{2}\right)= x\sqrt{3}- \frac{3}{2}$.
P is the intersection of those two lines: $y= -\frac{1}{\sqrt{3}}x- \frac{1}{\sqrt{3}}= \sqrt{3}x- 2$ so $y= \frac{2\sqrt{3}- 1}{4}$, $x= \frac{2\sqrt{3}+ 5}{4\sqrt{3}}$.
Now that you know the coordinates of the vertices of each triangle, you can find the lengths of the sides and calculate the area using "Heron's formula": The area of a triangle with side lengths, a, b, and c, is given by $\sqrt{s(s- a)(s- b)(s- c)}$ where s is the "half perimeter, $s= \frac{a+ b+ c}{2}$.
| {
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.
| This is only true for odd integers $n=2k+1$:
$$2^{2k+1}+1 = 2^{2k+1}-(-1)^{2k+1} = (2+1) (2^{2k}-2^{2k-1}\pm\cdots+1)$$
| {
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How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$? How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$
My Try :
$$x^2+x-2=(x-1)(x+2)$$
$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\sqrt{x(x+3)}-6}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)}=\\\lim_{x \to 1} \frac{2(x-3+\sqrt{x(x+3)})}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)} $$
Now what ?
| HINT: write
$$\frac{2(x-3+\sqrt{x(x+3)})(x-3-\sqrt{x(x+3)})}{(x-1)(x+2)(\sqrt{x}+\sqrt{x+3}+3)(x-3-\sqrt{x(x+3)}}$$
and this is equal
$$\frac{-9(x-1)}{(x-1)(x+2)(\sqrt{x}+\sqrt{x+3}+3)(x-3-\sqrt{x(x+3)})}$$
| {
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Solve the differential equations: $$y' = {y \over {\sin x}}+\tan{x \over 2}$$
I was trying to do this by substitution $u=y/x $ and it did not work and also with
$$y' - {y \over {\sin x}} = 0$$
$${dy \over dx} = {y \over {\sin x}} $$
$${\ln(y) = \ln\left|\tan\frac{x}{2}\right|+c } $$
$${y = c\cdot\tan\frac{x}{2} } $$
but then when im trying to calculate $y'$ I have a problem and I have too many equations. Is there some easier way or am I making some mistakes.
| I don't think there "too many equations"
$${ y=C\left( x \right) \tan \frac { x }{ 2 } }\\ y'=C'\left( x \right) \tan { \frac { x }{ 2 } } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } }{ 2 } \\ C'\left( x \right) \tan { \frac { x }{ 2 } } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } }{ 2 } =\frac { C\left( x \right) \tan \frac { x }{ 2 } }{ \sin { x } } +\tan { \frac { x }{ 2 } } \\ \\ C'\left( x \right) \tan { \frac { x }{ 2 } } =\tan { \frac { x }{ 2 } } \\ C\left( x \right) =x+{ C }_{ 1 }\\ { y=C\left( x \right) \tan \frac { x }{ 2 } }={ \tan \frac { x }{ 2 } }\left( x+{ C }_{ 1 } \right) \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of sixth power of roots of $x^3-x-1=0$ Question:
Find the sum of sixth power of roots of the equation $x^3-x-1=0$
My First approach:
Let $S_i$ denote the sum of $i^{th}$ power of roots of the given equation.
Now, multiplying given equation by $x^3$ , putting each of the three roots and adding the three formed equations
we get, $S_6=S_4+S_3$
repeating same above procedure to obtain,
$S_4=S_2+S_1$ , $S_3=S_1+1$
hence, $S_6=S_2+2S_1+1=2+0+1=3$
My Second Approach:
Let $a,b,c$ be the roots of $f(x)=x^3-x-1$
Clearly, $f^/(x)/f(x)=1/(x-a)+1/(x-b)
+1/(x-c)$
=$\sum(1/x+a/x^2+a^2/x^3+a^3/x^4+\cdots)=3/x+S_1/x^2+S_2/x^3+\cdots$
hence we get, $S_6=5$
$\rule{200px}{0.5px}$
I am getting wrong answer through First Approach, Please point out my mistake or post a better solution.
Thank You
| Squaring $x^3=x+1$, we have
$$
x^6=x^2+2x+1
$$
Furthermore, we know that the coefficient of $x^2$ in $x^3-x-1$ is the negative of the sum of the roots, and the coefficient of $x$ is the sum of the products of pairs of distinct roots.
Therefore, we need to find the sum of the squares of the roots, which is
$$
\left(\sum_{k=1}^3x_k\right)^2-2\sum_{j\lt k}x_jx_k=0^2-2(-1)=2
$$
The sum of the roots is $0$.
The sum of $1$ is $3$.
Thus, the sum of the sixth power of the roots is $2+0+3=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac{(x+1)^2}{x-1}=x-1.$$ Which is far from the correct answer. I can't spot my mistake. I feel that the more I sit and study, the worse at math I become.
| $$(x+1)^2=x^2+2x+1=x^2-x+3x -3+4=(x-1)(\underbrace{x+3}_\text{quotient})+\underset{\strut{\rlap{\text{remainder}}\qquad}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Given $\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0$, $\{a,b\}\subset \mathbb R^+$, show the roots are real and within defined intervals
Given: $\{a,b\}\subset \mathbb R^+$, $P(x)=\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0.$ (correction: last term is $\frac{1}{x{\color{red}+}b})$
Show: all roots of $P(x)$ are real, one in the interval $(-b,0)$ and the other in the interval $(0,a)$.
Question asked in an entrance exam.
My attempt: developing $P(x)$, we get to
$$x^2-\frac{2}{3}(a+b)x+\frac{ab}{3}=0$$
with discriminant defined by
$$\Delta=\frac{4}{9}(a+b)^2-\frac{4}{3}ab=\frac{4}{9}((a+b)^2-3ab)=\frac{4}{9}(a^2+b^2-ab)$$
As it is easy to show that $a^2+b^2-ab\ge 0$ it follows that $\Delta\ge 0$. This solves the first part of the question.
I tried, unsuccessfully, to solve the second part, which asked to show that the roots are within two specific intervals. Hints and full answers are welcome.
Edit 1. The term $\frac{1}{x-b}$ is, most likely, $\frac{1}{x+b}$. See comments below for an explanation.
| I am using calculus. Please tell me if it is not allowed in the exam, then I can post an alternate solution.
Case $1$ : $P(x) = \frac 1{x} + \frac 1{x-a} + \frac 1{x-b} = 0$ (This was done before the edit, but is included for completeness)
$P(x)$ is a continuous function on $\mathbb R \backslash\{0,a,b\}$. Without loss of generality, let us assume that $0 < a < b$. Then, in the interval $(0,a)$, note that as $x \to 0^+$, $P(x)$ eventually becomes positive, and as $x \to a^-$, $P(x)$ eventually becomes negative, and therefore, by the intermediate value theorem, $P(x)$ has a root in $(0,a)$.
Similarly, in the interval $(a,b)$, note that as $x \to a^+$, $P(x)$ eventually becomes positive, and as $x \to b^-$, $P(x)$ eventually becomes negative, and therefore, by the intermediate value theorem, $P(x)$ has a root in $(a,b)$.
Finally, we note that for $x < 0$, all three quantities $\frac{1}{x},\frac{1}{x-a},\frac{1}{x-b}$ are strictly negative, hence their sum is strictly negative. Similarly, for $x>b$, the sum is strictly positive. Hence, no roots lie in these regions.
Furthermore, $P(x)$ has no complex roots. To see this, note that by taking the common denominator, $$\frac 1{x} + \frac 1{x-a} + \frac 1{x-b} = \frac{3x^2+x(-2a-2b)+ab}{x(x-a)(x-b)}$$
Therefore, $P(x)=0$ precisely when $3x^2 + x(-2a-2b) + ab = 0$. This can have at most two roots in the entire complex plane because it has degree $2$. However, we have already located those roots above, showing that there are no other roots available.
Therefore , there are precisely two roots of this equation, both real.
Case $2$ : $P(x) = \frac 1x + \frac 1{x-a} + \frac 1{x+b} = 0$.
We will use calculus, yet again. Now, we have the following regions to consider :
*
*$x<-b$ : Each of $\frac 1{x},\frac{1}{x-a},\frac 1{x+b}<0$, therefore there is no root in this interval.
*$-b<x<0$ : As $x \to -b^+$, $P(x) \to +\infty$, while as $x \to 0^-$, $P(x) \to -\infty$. Therefore, by the intermediate value theorem, there is a root in $(-b,0)$.
*$0<x<a$ : As $x \to 0^+$, $P(x) \to +\infty$, while as $x \to a^-$, $P(x) \to -\infty$. Therefore, by the intermediate value theorem, there is a root in $(0,a)$.
*$x>a$ : Each of $\frac 1x , \frac 1{x-a},\frac{1}{x+b}$ is positive in this interval, hence there is no root.
However, similar to the argument in Case $1$, $P(x)$ has only two roots in total : and we found both above. Hence, these are the roots of the equation. Furthermore, if one needs to find them, then one can use the quadratic formula on the equation derived as the numerator of $P(x)$ after one takes the common denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$ Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
I already tried to use the Sqeeze theorem on it, but I just was not able to find some reasonable upper series for it, only lower:
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
$$\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right) \leq n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
Is there anyone who can give me a hint as to how to solve it?
| $$\lim_{n \to \infty} (\sqrt[3]{n^3+1}-\sqrt{n^2+1})=$$
$$=\lim_{n\to \infty}((\sqrt[3]{n^3+1}-n)-(\sqrt{n^2+1}-n))=$$
Use the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1}), k\ge 2, k\in\mathbb Z, a,b\in\mathbb R$$
with, e.g., $a=\sqrt[3]{n^3+1}$, $b=n$, $k=3$, etc.
$$=\lim_{n\to \infty}\left(\frac{(\sqrt[3]{n^3+1})^3-n^3}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{(\sqrt{n^2+1})^2-n^2}{\sqrt{n^2+1}+n}\right)=$$
$$=\lim_{n\to \infty}\left(\frac{1}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{1}{\sqrt{n^2+1}+n}\right)=0-0=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to compute the following limit$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$ I am trying to find
$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$
where $p>0$. I have tried to factor out as
$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
but still was not able to make progress. Any other approach to this is welcome.
| We can use the expansion
$$(1+y)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} \cdot y^n.$$
Taking $y = \frac{1}{x^p}, \ \alpha = \frac{1}{p}$, we get
$$\left( 1 + \frac{1}{x^p} \right)^{\frac{1}{p}} = 1 + \frac{1}{p x^p} + o \left( \frac{1}{x^{p}} \right).$$
Similarly
$$\left(1+\frac{1}{x^{p+1}} \right)^{\frac{1}{p+1}} = 1+\frac{1}{(p+1)x^{p+1}} + \mathcal{o} \left( \frac{1}{x^{p+1}} \right) = 1 + o \left( \frac{1}{x^p} \right).$$
Hence
$$\begin{align*}
\left( 1+x^{p+1} \right)^{\frac{1}{p+1}} - \left(1+x^p\right)^{\frac{1}{p}} & = x \left[ \left( 1 + \frac{1}{x^p} \cdot o(1) \right) - \left( 1 + \frac{1}{x^p} \left( \frac{1}{p} + o(1) \right) \right) \right] \\[1ex]
& = -\frac{x}{x^p} \left( \frac{1}{p} + o(1) \right).
\end{align*}$$
*
*If $p > 1$, the limit is $0$.
*If $p = 1$, the limit is $-1$.
*If $p < 1$, the limit is $-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Nonlinear first order differential equation (exact)
Find an implicit solution to the initial value problem $$(x+y)^2+(2xy+x^2-1)\dfrac{dy}{dx} = 0 \\ y(1) = 1$$
Attempted solution
The equation can be written as $M(x,y)+N(x,y)y'=0$. We try to find find the potential function $\psi(x,y)$: $$\psi_x = M(x,y) = (x+y)^2 \\ \psi_y = N(x,y) = 2xy+x^2-1 $$
Integrating last eq. wrt $y$ yields $\psi(x,y) = xy^2+x^2y-y+f(x)$, so that $M(x,y)=\psi_x \Leftrightarrow y^2+2xy+x^2= y^2+2xy+f'(x) \Rightarrow f'(x) = x^2 \Rightarrow f(x) = \dfrac{x^3}{3}+C $
Okay so now we have the potential $\psi(x,y) = xy^2+x^2y-y + \dfrac{x^3}{3}+C$.
The problem
Plugging in $y(1)=1$ yields $\psi(1,1) = \dfrac{4}{3}+C$. What should I set this equal to?
| Conventionally an implicit solution is of the form $\psi(x,y) =0$. To this end, you should set $$C=-\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $ \lim_{n \to \infty} \frac{5^n}{5^n + 3^n} $ How can I compute the next limit of sequences:
$$ \lim_{n \to \infty} \frac{5^n}{5^n + 3^n} $$
It is $\frac{\infty}{\infty}$ indeterminate when evaluating in $n$, but using L'Hopital never eliminates the $5^n, 3^n$ terms. But because the denominator is greater than the numerator, it seems that the limit should be 1 or zero.
| $$\lim_{n \to \infty} \frac{5^n}{5^n + 3^n} = \lim_{n \to \infty} \frac{5^n}{5^n\left(1 + \left(\frac{3}{5}\right)^n\right)} = \lim_{n \to \infty} \frac{1}{1 + \left(\frac{3}{5}\right)^n} = 1,$$
since $\left(\frac{3}{5}\right)^n$ goes to $0$ as $n$ goes to $+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry Problems Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac{1}{2}$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
My student left me with a list of questions just now (he also said that he wanted to be helped at school :P)
| \begin{align}
{\sin2x\over\sin2y}+{\cos2x\over\cos2y}=
{}& {2\sin x\cos x\over2\sin y\cos y}+{\cos^2x-\sin^2x\over\cos^2y-\sin^2y} \\[10pt]
= {} &
{3\sin y\cos y\over2\sin y\cos y}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y} \\[10pt]
= {} &
{3\over2}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y}.
\end{align}
On the other hand:
$$
1=\cos^2x+\sin^2x={1\over4}\cos^2y+9(1-\cos^2y),
\quad\hbox{whence:}\quad
\cos^2y={32\over35},\quad\sin^2y={3\over35}.
$$
We finally have
$$
{\sin2x\over\sin2y}+{\cos2x\over\cos2y}=
{3\over2}+{{1\over4}{32\over35}-9{3\over35}\over{32\over35}-{3\over35}}={49\over58}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
| Use the fact that
$$
a^3-b^3=(a-b)(a^2+ab+b^2)
$$
to get that
\begin{align}
\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}
&=\lim_{x\to 3}\left(\frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}\times
\frac{(x+5)^{2/3}+2(x+5)^{1/3}+4}{(x+5)^{2/3}+2(x+5)^{1/3}+4}
\right)\\
&=\lim_{x\to 3}\left(\frac{\cos (\pi x)+1}{x-3}\times [(x+5)^{2/3}+2(x+5)^{1/3}+4]\right)\\
&=0\times 12=0.
\end{align}
since
$$
\lim_{x\to 3}\left(\frac{\cos (\pi x)-(-1)}{x-3}\right)
$$
is the definition of $f'(3)$ for $f(x)=\cos(\pi x)$. But
$$
f'(3)=-\pi\sin(3\pi)=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prime counting function formulas Are there any elementary (including floor, ceiling, mod) representations of the prime counting function. Or one without an integral.
| I made this elementary formula, although it really has no practical importance:
for $n > 3$
$\pi (n) = 2 + \sum \limits_{m=4}^{n} \left\lfloor \frac{\displaystyle 1}{\displaystyle 1+ \displaystyle\sum_{p=2}^{\left\lfloor \sqrt{m} \right\rfloor}{\left( \left\lfloor\frac{m}{p}\right\rfloor- \left\lfloor\frac{m-1}{p}\right\rfloor \right)}} \right\rfloor$
if we neglect the multiples of 2 and 3, it can be transformed into the following formula:
$\pi (n) = 2 + \sum \limits_{k=1}^{\left\lfloor{\frac{n+1}{6}}\right\rfloor} {\left\lfloor \frac{\displaystyle 1}{ \displaystyle 1 + \displaystyle\sum_{a=1}^{\left\lfloor\frac{1+\sqrt{-1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{-1+6\cdot k}{-1+6\cdot a}\right\rfloor - \left\lfloor\frac{-2+6\cdot k}{-1+6\cdot a}\right\rfloor \right)} + \displaystyle\sum_{a=1}^{\left\lfloor\frac{-1+\sqrt{-1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{-1+6\cdot k}{1+6\cdot a}\right\rfloor - \left\lfloor\frac{-2+6\cdot k}{1+6\cdot a}\right\rfloor \right)} }\right\rfloor}+\sum \limits_{k=1}^{\left\lfloor{\frac{n-1}{6}}\right\rfloor} {\left\lfloor \frac{\displaystyle 1}{\displaystyle 1 + \displaystyle\sum_{a=1}^{\left\lfloor\frac{1+\sqrt{1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{1+6\cdot k}{-1+6\cdot a}\right\rfloor - \left\lfloor\frac{6\cdot k}{-1+6\cdot a}\right\rfloor \right)} + \displaystyle\sum_{a=1}^{\left\lfloor\frac{-1+\sqrt{1+6\cdot k}}{6}\right\rfloor} {\left( \left\lfloor\frac{1+6\cdot k}{1+6\cdot a}\right\rfloor - \left\lfloor\frac{6\cdot k}{1+6\cdot a}\right\rfloor \right)} }\right\rfloor}$
This is the Matlab code to verify the formula:
n=10000;
Pin=2;
for k=1:floor((n+1)/6)
Sk=1;
for a=1:floor((1+sqrt(-1+6*k))/6)
Sk = Sk + floor((-1+6*k)/(-1+6*a)) - floor((-2+6*k)/(-1+6*a));
end
for a=1:floor((-1+sqrt(-1+6*k))/6)
Sk = Sk + floor((-1+6*k)/(1+6*a)) - floor((-2+6*k)/(1+6*a));
end
Pin=Pin+floor(1/Sk);
end
for k=1:floor((n-1)/6)
Sk =1;
for a=1:floor((1+sqrt((1+6*k)))/6)
Sk = Sk + floor((1+6*k)/(-1+6*a)) - floor((6*k)/(-1+6*a));
end
for a=1:floor((-1+sqrt((1+6*k)))/6)
Sk = Sk + floor((1+6*k)/(1+6*a)) - floor((6*k)/(1+6*a)) ;
end
Pin=Pin+floor(1/Sk);
end
Pin
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Concerning this integral $\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}$ Motivation from this question $(1)$
$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{1\over 2}\right){x^s\over 1-x}\mathrm dx=F(s)\tag1$$
setting $s=0$ then $F(0)=-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over 2}\gamma$
by a slight variation of $(1)$, we have
$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}={1\over 2}H_s-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over 2}\gamma\tag2$$
$H_s$ is the harmonic number, $H_0=0$
How do we go about to prove $(2)?$
| Use
\begin{eqnarray*}
H_v= \int_0^1 \frac{1-x^v}{1-x} dx.
\end{eqnarray*}
We have
\begin{eqnarray*}
\int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{x^s}{2}\right)\frac{ dx}{1-x} &=& \int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{1}{2} +\frac{1-x^s}{2}\right)\frac{ dx}{1-x} \\
&=& F(0)+ \int_0^1 \frac{1-x^s}{2(1-x)}dx \\
&=& -\frac{1}{2}+ \frac{1}{2} \ln( 2 \pi) -\frac{1}{2} \gamma + \frac{1}{2} H_s.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program. I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.
| Here's a way
to get bounds on $x$
for a generalized version of this.
We want to solve
$x
=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor
$.
Upper bound:
If
$a = \max(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\le\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\le n \sqrt{x+a} \\
\text{so}\\
x^2
&\le n^2(x+a)\\
\end{array}
$
Therefore
$x^2-n^2x+n^4/4
\le n^2(a+n^2/4)
$
or
$|x-n^2/2|
\le n\sqrt{a+n^2/4}
= (n/2)\sqrt{4a+n^2}
$
so
$x
\le (n/2)(n+\sqrt{4a+n^2})
$.
For this case,
$n=3, a=2$,
so
$x
\le (3/2)(3+\sqrt{17})
\lt 10.7
$
so
$x \le 10$.
Lower bound:
If
$a = \min(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\ge\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\gt n \sqrt{x+a}-1 \\
\text{so}\\
(x+1)^2
&\gt n^2(x+a)\\
&= n^2(x+1+a-1)\\
\end{array}
$
Therefore
$(x+1)^2-n^2(x+1)+n^4/4
\gt n^2(a-1+n^2/4)
$
or
$|x+1-n^2/2|
\gt n\sqrt{a-1+n^2/4}
= (n/2)\sqrt{4a-4+n^2}
$
so
$x+1-n^2/2
\gt (n/2)\sqrt{4a-4+n^2}
$
or
$x+1-n^2/2
\lt -(n/2)\sqrt{4a-4+n^2}
$.
Therefore
$x
\gt n^2/2-1+(n/2)\sqrt{4a-4+n^2}
$
or
$x
\lt n^2/2-1-(n/2)\sqrt{4a-4+n^2}
$.
For this case,
$n=3, a=0$,
so
$(n/2)\sqrt{4a-4+n^2}
=\frac32\sqrt{-4+9}
=\frac32\sqrt{5}
\approx 3.35
$
and
$n^2/2-1
=3.5
$
so
$x > 6.85$
or
$x < 0$.
Therefore
$x \ge 7$,
so that,
combining the bounds,
$7 \le x \le 10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? More generally, why does a negative exponent turn positive when moved to the denominator? Would appreciate kindergarten language ;-D
| Notice that $7^6\cdot 7^{-6}=7^{6-6}=7^0=1$
Notice also that $7^6\cdot\frac{1}{7^6}=\frac{7^6}{7^6}=1$
So, we learned that $7^6\cdot 7^{-6}=7^6\cdot\frac{1}{7^6}$.
Remembering that $x\cdot a=x\cdot b$ for nonzero $x$ implies that $a=b$ by cancelling this tells us that $7^{-6}=\frac{1}{7^6}$
In general the following properties are useful to know:
*
*$x = \frac{x}{1}=x^1$
*$x^n = \frac{1}{x^{-n}}$
*$x^{-n}=\frac{1}{x^n}$
Another useful identity is $x^0 = 1$ which is true for all nonzero $x$
Tangentially, depending on context it can also be correct to say that $x^0=1$ for $x=0$ as well, for example in the field of combinatorics. There are some other situations though where we leave $0^0$ undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 1
} |
Exponential Generating Functions With Coefficients Given the exponential generating function, $B(x)=\sum_{n\geq 0}b_n\frac{x^n}{n!}=b_0+b_1\frac{x}{1!}+b_2\frac{x^2}{2!}+\cdots$, I want to find a formula for the generating function
\begin{equation*}
\tilde{B}(x)=b_1+b_0\frac{x}{1!}+b_3\frac{x^2}{2!}+b_2\frac{x^3}{3!}+b_5\frac{x^4}{4!}+b_4\frac{x^5}{5!}+\cdots
\end{equation*} in terms of $B(x)$. Now to know if the formula is correct then if \begin{equation*}B(x)=e^x=\tilde{B}(x).
\end{equation*} Likewise if \begin{equation*} B(x)=xe^x=\sum_{n\geq 0}n\frac{x^n}{n!}\end{equation*}
then \begin{equation*} \tilde{B}(x)=xe^x+e^{-x}=1+0x+3\frac{x^2}{2!}+2\frac{x^3}{3!}+5\frac{x^4}{4!}+\cdots.\end{equation*}
How would i go about swapping coefficients, would the formula involve a negative exponent like \begin{equation*}\frac{x^{-n}}{n!}?\end{equation*}
| You can bisect a generating function into its even and odd parts with the formulae
$$ \sum_{n \text{ even}} b_n \frac{x^n}{n!} = \frac{B(x) + B(-x)}{2} $$
and
$$ \sum_{n \text{ odd}} b_n \frac{x^n}{n!} = \frac{B(x) - B(-x)}{2}. $$
You can shift the terms in an exponential generating function by integrating or differentiating.
If you put this together you get the formula
$$ \bar B(x)= \frac{C(x) - C(-x)}{2} + \frac{D(x) + D(-x)}{2} $$
where $C = \int B$ and $D = \frac{d}{dx} B$.
For $B(x) = xe^x$ we have $C(x) = (x - 1)e^x + 1$ and $D(x) = e^x + xe^x$. Therefore
$$ \bar B(x) = \frac{(x - 1)e^x - (-x - 1)e^{-x}}{2} + \frac{e^x + xe^x + e^{-x} - xe^{-x}}{2} = xe^x - e^{-x}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)$.
Thus, $$z^{1/3} = (1351)^{2/3} \left(\cos \left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3}+ \frac{2k\pi}{3} \right] + i\sin\left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3} + \frac{2k\pi} {3}\right]\right)$$ for $k = 0,1,2.$
This does not seem like I am heading in the right direction. Any clues or hints would be greatly appreciated. Thank you!
| $$\sqrt[3]{26+15\sqrt3}=\sqrt[3]{8+3\cdot4\cdot\sqrt3+3\cdot2\cdot3+3\sqrt3}=\sqrt[3]{(2+\sqrt3)^3}=2+\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate: $\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}$ Evaluate: $\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}$
My Attempt:
$$=\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}$$
$$=\lim_{y\to x} \dfrac {y\cos x-x\cos y}{\cos x\cos y \cdot (y-x)}$$
| $\lim_\limits{y\to x} \frac {y\sec y - x\sec x}{y-x}$ is the definition of $\frac {d}{dx} x\sec x$
But since you requested a solution to this limit that was algebraic, here goes:
$\frac {y\cos x - x\cos y}{(y-x)\cos x\cos y}\\
\frac {y\cos x - y\cos y + y \cos y- x\cos y}{(y-x)\cos x\cos y}\\
\frac {y(\cos x - \cos y) + (y-x)\cos y}{(y-x)\cos x\cos y}\\
\frac {-2y \sin(\frac {y+x}{2})\sin (\frac {x-y}{2})} {(y-x)\cos x\cos y}+\frac {1}{\cos x}\\
\frac {y\sin(\frac {y+x}{2})}{\cos x\cos y}\frac {(-2 \sin (\frac {x-y}{2}))} {(y-x)}+\frac {1}{\cos x}\\
\lim_\limits{y\to x}\frac {y\sin(\frac {y+x}{2})}{\cos x\cos y}\lim_\limits{y\to x}\frac {(2 \sin (\frac {y-x}{2}))} {(y-x)} + \lim_\limits{y\to x}\frac {1}{\cos x} = \frac {x\sin x}{\cos^2 x} + \frac{1}{\cos x}\\
x\sec x\tan x + \sec x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximum and minimum values of function
For any number $c$ we let $f_c(x)$ be the smaller of the two numbers
$(x-c)^2$ and $(x-c-2)^2$. Then we define $g(c)=\int_0^1f_c(x)dx$.
Find the maximum and minimum values of $g(c)$ if $-2\le c \le 2$.
So the maximum/minimum values are either $g(\pm2)$ (the endpoints) or $g(c)$ where $g'(c)=0$.
$$g(2)=\int_0^1f_2(x)dx=\int_0^1\min((x-2)^2,(x-4)^2)dx\\=\int_0^1(x-2)^2dx=\frac{(1-2)^3}{3}-\frac{(0-2)^3}{3}=\frac{7}{3}$$
The last step I was thinking was because on the interval $[0,1]$, $|x-2|$ is always smaller than $|x-4|$.
In the same way, $$g(-2)=\int_0^1f_{-2}(x)dx=\int_0^1\min((x+2)^2,x^2)dx\\=\int_0^1x^2dx=\frac{1}{3}$$
So $7$ thirds and $1$ third for the endpoints. When the derivative is zero, I am having trouble. I am struggling to understand how the derivative and antiderivative of $g$ and $f_c$ would operate. Thanks.
Problem Source: James Stewart, Calculus, Integrals Chapter
| Hint $f_c(x)$ depends only on $x-c$, in fact $f_c(x) = f_0(x-c)$, hence
$$g(c)= \int_0^1 f_0(x - c) d x = \int_{-c}^{1-c}f_0(t) d t$$
Edit: As ${f}_{0}$ is continuous, it follows that $g$ has the
derivative
$${g'} \left(c\right) =-{f}_{0} \left(1-c\right)+{f}_{0} \left({-c}\right) = \min \left({d}^{2} \left(c , {-2}\right) , {d}^{2} \left(c , 0\right)\right)-\min \left({d}^{2} \left(c , {-1}\right) , {d}^{2} \left(c , 1\right)\right)$$
where $d \left(a , b\right) = \left|a-b\right|$ is the usual distance in $\mathbb{R}$.
It follows easily that
$$\renewcommand{\arraystretch}{1.5} {g'} \left(c\right) = \left\{\begin{array}{l}{d}^{2} \left(c , {-2}\right)-{d}^{2} \left(c , {-1}\right) \quad \text{ if } c \in \left[{-2} , {-1}\right]\\
{d}^{2} \left(c , 0\right)-{d}^{2} \left(c , {-1}\right) \quad \text{ if } c \in \left[{-1} , 0\right]\\
{d}^{2} \left(c , 0\right)-{d}^{2} \left(c , 1\right) \quad \text{ if } c \in \left[0 , 1\right]
\end{array}\right.$$
Now we use that given $a , b \in \mathbb{R}$, the antiderivative of
${\varphi} \left(x\right) = {d}^{2} \left(x , a\right)-{d}^{2} \left(x , b\right) = 2 \left(b-a\right) \left(x-\frac{a+b}{2}\right)$
is ${\phi} \left(x\right) = \left(b-a\right) {d}^{2} \left(x , \frac{a+b}{2}\right)+K$ where $K$ is a constant.
Note that $K = {\phi} \left(\frac{a+b}{2}\right)$ and that
${\phi} \left(a\right) = {\phi} \left(b\right) = \frac{{\left(b-a\right)}^{3}}{4}+{\phi} \left(\frac{a+b}{2}\right)$. Also note that
${\phi}$ has a unique extremum at $\frac{a+b}{2}$. It follows that
$$\renewcommand{\arraystretch}{2} g \left(c\right) = \left\{\begin{array}{l}g \left({-\frac{3}{2}}\right)+{d}^{2} \left(x , {-\frac{3}{2}}\right) \quad \text{ if } c \in \left[{-2} , {-1}\right]\\
g \left({-\frac{1}{2}}\right)-{d}^{2} \left(x , {-\frac{1}{2}}\right) \quad \text{ if } c \in \left[{-1} , 0\right]\\
g \left(\frac{1}{2}\right)+{d}^{2} \left(x , \frac{1}{2}\right) \quad \text{ if } c \in \left[0 , 1\right]
\end{array}\right.$$
Hence $g$ has a local minimum at $x =-\frac{3}{2}$ and $g \left({-\frac{3}{2}}\right) = g \left({-2}\right)-\frac{1}{4} = \frac{1}{12}$, a local maximum at $x =-\frac{1}{2}$ and $g \left({-\frac{1}{2}}\right) = g \left({-1}\right)+\frac{1}{4} = \frac{7}{12}$, and a local minimum at $x = \frac{1}{2}$ and $g \left(\frac{1}{2}\right) = g \left(0\right)-\frac{1}{4} = \frac{1}{12}$
Thus, the maximum of $g$ is $g \left(2\right) = \frac{7}{3}$, the minimum is
$g \left(\frac{1}{2}\right) = g \left({-\frac{3}{2}}\right) = \frac{1}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $\int_0^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx$ converges I have to show that the above improper integral is convergent and I'm stuck. But this is my work so far:
*
*Split it: $$\int_0^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx = \int_0^\frac{\pi}{2} \frac{e^{\sin x}-1}{\sqrt{x}x}dx + \int_\frac{\pi}{2}^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx$$
The split point $\frac{\pi}{2}$ is for obvious reasons.
*Then I work on the right term: $$\frac{e^{\sin x}-1}{\sqrt{x}x} \leq \frac{e}{\sqrt{x}x}$$
*$$\lim_{B \to \infty}\int_\frac{\pi}{2}^B \frac{e}{\sqrt{x}x}dx = \sqrt{\frac{2}{\pi}}2e$$ So that part of the integral is convergent.
But the left term does not yield as easily. Since $\sin x \approx x$ when $x \to 0$, I'm doing the following comparison:
$$\frac{e^{\sin x}-1}{\sqrt{x}x} \leq \frac{e^x-1}{\sqrt{x}x}$$
So now I "only" have to show that
$$\int_0^\frac{\pi}{2} \frac{e^x-1}{\sqrt{x}x}dx$$
is convergent. But I have worked for a long while on it and I'm getting nowhere so please help me. :)
| \begin{align*}
\frac{e^{x}-1}{x^{3/2}}=\frac{1}{x^{1/2}}+\sum_{k=2}\frac{1}{k!}x^{k-3/2}
\end{align*}
we know that
\begin{align*}
\int_{0}^{\pi/2}\frac{1}{x^{1/2}}dx<\infty
\end{align*}
and
\begin{align*}
\sum_{k=2}\frac{1}{k!}\int_{0}^{\pi/2}x^{k-3/2}dx=\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}x^{k-(1/2)}\bigg|_{x=0}^{x=\pi/2}=\left(\frac{2}{\pi}\right)^{1/2}\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}\left(\frac{\pi}{2}\right)^{k}
\end{align*}
but
\begin{align*}
\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}\left(\frac{\pi}{2}\right)^{k}\leq\sum_{k=2}\frac{1}{k!}\left(\frac{\pi}{2}\right)^{k}<\infty.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.