Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\lim_{x\to 1} (x^3+2x^2-2)=1$ using the definition to prove that $\lim_{x\to 1} (x^3+2x^2-2)=1$
let $\varepsilon>0$, and $x\in\mathbb{R}$
we must find $\delta>0$ such that if $|x-1|<\delta $ then $|x^3+2x^2-2-1|<\varepsilon$
i do :
$$
|x^3-1+2x^2-2|\leq |x^3-1|+2|x^2-1|=|x^3-1|+2 |(x-1)| |(x+1)|=
$$
$$
|x-1|(|x^2+x+1|+2|x+1|)\leq \delta (|x^2+x+1|+2|x+1|)
$$
how to continue?
| To make it easier
(I always prefer to have variables
go to zero),
let
$x = y+1$.
Then
$
f(x)=x^3+2x^2-2
=(y+1)^3+2(y+1)^2-2\\
=y^3+3y^2+3y+1
+2(y^2+2y+1)
-2\\
=y^3+5y^2+7y\\
=y(y^2+5y+7)
$
and it's easy to see
what happens as
$y \to 0$.
In particular,
if
$|y| < 1$
(i.e., $|x-1| < 1$),
then
$|f(x)|
\lt 13|y|
= 13|x-1|
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $1 - 3xy + x^3 + y^3 = 0$ I am trying to solve the equation
$$
1 - 3xy + x^3 + y^3 = 0
$$
in the real numbers. I can see several (infinitely many) solutions:
*
*$x=y = 1$
*$x+ y = -1$
I am trying to show that these are all of them.
For example, if $x = y$, then $x^2(-3+2x) = -1$ forcing $x = y = 1$
I can't seem to make much progress with the case where $x\neq y$ and showing that $x + y = 1$ in that case. (I am pretty sure that this must be true.)
| $$x=u+v$$
$$x^3=u^3+v^3+3uv(x)\iff x^3-3uv(x)-(u^3+v^3)=0$$
Comparing with $$x^3-3xy+y^3+1=0$$
$$uv=y$$
and $$ u^3+v^3=-(y^3+1)\iff u^3+\dfrac{y^3}{u^3}+y^3+1=0$$
$$0=u^6+(y^3+1)u^3+y^3=(u^3+1)(u^3+y^3)$$
If $u^3+1=0,u=-1,\implies v=-y$
$$\implies x=u+v=-1-y\iff x+y=-1$$
$$x^3+y^3-3xy+1=(x+y)^3-3xy(x+y+1)+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Bulgarian Mathematical Olympiad 1987 The sequence $x_1,x_2,\dots$ is defined by the equalities $x_1=x_2=1$ and $$x_{n+2}=14x_{n+1}-x_n-4, n\geq 1$$
Prove that each number of the given sequence is a perfect square.
I used the standard way to solve recurrence relations and arrived at
$$x_n=\frac{1}{6}\bigg( (2+\sqrt{3})^{n-\frac{3}{2}} + (2-\sqrt{3})^{n-\frac{3}{2}} \bigg)^2$$
which I think should be quite close. Any idea how to proceed?
| The equation is equivalent to
$$
x_{n+2}-14x_{n+1}+x_n=-4\tag1
$$
which is the same as
$$
\left(x_{n+2}-\tfrac13\right)-14\left(x_{n+1}-\tfrac13\right)+\left(x_n-\tfrac13\right)=0\tag2
$$
Using the standard methods for Linear Difference Equations, we note that
$$
x^2-14x+1=0\tag3
$$
has roots $7+4\sqrt3$ and $7-4\sqrt3$, so that $(2)$ has solutions
$$
x_n=\tfrac13+c_1\left(7+4\sqrt3\right)^{n-1}+c_2\left(7-4\sqrt3\right)^{n-1}\tag4
$$
Since $x_1=x_2=1$, we get
$$
\begin{align}
x_n
&=\tfrac13+\tfrac{2-\sqrt3}6\left(7+4\sqrt3\right)^{n-1}+\tfrac{2+\sqrt3}6\left(7-4\sqrt3\right)^{n-1}\\[3pt]
&=\tfrac13+\tfrac{2-\sqrt3}6\left(2+\sqrt3\right)^{2n-2}+\tfrac{2+\sqrt3}6\left(2-\sqrt3\right)^{2n-2}\\
&=\left(\tfrac{\sqrt3-1}{2\sqrt3}\left(2+\sqrt3\right)^{n-1}+\tfrac{\sqrt3+1}{2\sqrt3}\left(2-\sqrt3\right)^{n-1}\right)^2\\[3pt]
&=u_n^2\tag5
\end{align}
$$
Since $2+\sqrt3$ and $2-\sqrt3$ are roots of
$$
u^2-4u+1=0\tag6
$$
we have $u_1=u_2=1$ and
$$
u_n=4u_{n-1}-u_{n-2}\tag7
$$
Therefore, $u_n\in\mathbb{Z}$ and $x_n=u_n^2$.
For example, using $(7)$
$$
\{u_n\}=\{1,1,3,11,41,153,571,\dots\}\tag8
$$
and using $(1)$
$$
\{x_n\}=\{1,1,9,121,1681,23409,326041,\dots\}\tag9
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$\int\frac1{(x^2+1)^2}\ dx$ by partial fraction decomposition Is there any possible way to calculate the integral of $\frac{1}{(x^2+1)^2}$ by partial fraction decomposition? I do not know the formulas for the trigonometric method.
Thank you!
| When you let $x=\tan\theta$, and $dx=\dfrac{d\theta}{\cos^2\theta}$
$$\int\dfrac{1}{(1+x^2)^2}dx=\int\dfrac{1}{(1+\tan^2\theta)^2\cos^2\theta}d\theta=\int\dfrac{\cos^4\theta}{\cos^2\theta}d\theta=\int\cos^2\theta d\theta\\=\dfrac{\theta}{2}+\dfrac{\sin^2\theta}{4}+C=\dfrac{\arctan x}{2}+\dfrac{x}{2(1+x^2)}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Does the linear map represent a rotation, a shearing or a stretching? The unit square has the coordinates $\begin{pmatrix}0 \\ 0\end{pmatrix},\ \begin{pmatrix}1 \\ 0\end{pmatrix}, \ \begin{pmatrix}1 \\ 1\end{pmatrix}, \ \begin{pmatrix}0 \\ 1\end{pmatrix}$, and its graph is:
We consider the linear map $F_1:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $\vec{x}\mapsto A_i\vec{x}$ with $A_1:=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}$.
I want to find the image of the unit square under this map.
We have
\begin{align*}&F_1\begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix} \\
&F_1\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}2 \\ 0\end{pmatrix} \\
&F_1\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}2 \\ 1\end{pmatrix} \\
&F_1\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 1\end{pmatrix}\end{align*}
The image of the unt square under $F_1$ is:
Now I want to determine if this is a rotation, a shearing or a stretching.
I think that this is a stretching, but I am not sure because the result is not a square it is now an orthogonal parallelogram.
Could you please clarify if it indeed a stretching or something else?
| The transformation scales the first coordinate by a factor of two, so it's a stretch.
Shearing matrices have the form
$$
\begin{bmatrix}
1 & a \\
0 & 1
\end{bmatrix}
$$
or
$$
\begin{bmatrix}
1 & 0 \\
a & 1
\end{bmatrix},
$$
with $a \in \mathbb{R}$.
Rotation matrices are of the form
$$
\begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix},
$$
with $\theta \in [0,2\pi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\triangle$ is the area of triangle with side lengths $a,b,c$, then show that $\triangle \le\dfrac{1}{4}\cdot\sqrt{(a+b+c)abc}$ If $\triangle$ is the area of triangle with side lengths $a,b,c$, then show that $\triangle \le\dfrac{1}{4}\cdot\sqrt{(a+b+c)abc}$. Also show that equality occurs in the above inequality when $a=b=c$
My attempt is as follows:-
$$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\triangle=\dfrac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$$
$$b+c-a>0,a+c-b>0,a+b-c>0$$
So we can apply $A.M\ge G.M$
$$\dfrac{b+c-a+a+c-b+a+b-c}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}$$
$$\dfrac{a+b+c}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\tag{1}$$
As $a>0,b>0,c>0$, we can also say
$$\dfrac{a+b+c}{3}\ge(abc)^\frac{1}{3}\tag{2}$$
But we can't say from this that $(abc)^\frac{1}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}$
So I tried $G.M\ge H.M$ for $a,b,c$
$$(abc)^\frac{1}{3}\ge\dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{ab+bc+ca}{3}\ge(abc)^\frac{2}{3}\tag{3}$$
Seems like it didn't produce anything useful, so I tried $G.M\ge H.M$ for $b+c-a,a+c-b,a+b-c$
$$\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\ge\dfrac{3(b+c-a)(a+c-b)(a+b-c)}{((b+c-a)(a+c-b)+(a+c-b)(a+b-c)+(b+c-a)(a+b-c))}$$
$$\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\ge\dfrac{3(b+c-a)(a+c-b)(a+b-c)}{2ab+2bc+2ca-a^2-b^2-c^2}\tag{4}$$
But again it didn't produce anything useful, so I tried something else.
Applying $A.M\ge G.M$ for $a+b+c,a+b-c,b+c-a,a+c-b$
$$\dfrac{2(a+b+c)}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{4}$$
$$\dfrac{a+b+c}{2}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{4}$$
Squaring both sides
$$\dfrac{(a+b+c)^2}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{2}$$
By cachy's inequality:-
$$(1^2+1^2+1^2)(a^2+b^2+c^2)\ge(a+b+c)^2$$
$$3(a^2+b^2+c^2)\ge(a+b+c)^2$$
$$\dfrac{3(a^2+b^2+c^2)}{4}\ge\dfrac{(a+b+c)^2}{4}\tag{5}$$
Hence
$$\dfrac{3(a^2+b^2+c^2)}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{2}\tag{6}$$
Dividing by $4$
$$\dfrac{3(a^2+b^2+c^2)}{16}\ge\triangle$$
So I tried all these approaches, but unfortunately didn't arrive at the required result. What can we do here? Please help me in this.
| By your work we need to prove that
$$\sum_{cyc}(2a^2b^2-a^4)\leq\sum_{cyc}a^2bc$$ or
$$\sum_{cyc}(a^4-2a^2b^2+a^2bc)\geq0$$ or
$$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^3b+a^3c-2a^2b^2)\geq0$$ or
$$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}ab(a-b)^2\geq0,$$
which is true by Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{n\to\infty}{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}$ where $a>1$
$$\underset{n\rightarrow\infty}\lim{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}=?, \;\;a>1$$
In Shaum's Mathematical handbook of formulas and tables I've seen:
$$\;\;\;\;\;\;\;\;\;\;\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\;,x\in\langle-1,1]\;\;\;\;\;\;\;$$
$$\frac{1}{2}\ln{\Bigg(\frac{1+x}{1-x}\Bigg)}=1+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots\;\;\;,x\in\langle-1,1\rangle$$
The term in parentheses reminded me of the harmonic series. I thought of using the Taylor series. Is that a good idea?
It says $a>0$ so I probably can't use these two formulas.
On the other hand:
$$e^x=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\;\;\;\;\;\;,$$
but there are no factorials in the denominators.
Source in Croatian: 2.kolokvij, matematička analiza
| By Stoltz-Cesaro
$$\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+…+\frac{a^n}{n}\right)=\frac{\left(a+\frac{a^2}{2}+\frac{a^3}{3}+…+\frac{a^n}{n}\right)}{\frac{a^{n+1}}{n}}$$
we obtain
$$\frac{\frac{a^{n+1}}{n}}{\frac{a^{n+2}}{n+1}-\frac{a^{n+1}}{n}}=\frac1{\frac{na}{n+1}-1} \to \frac1{a-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| This is a direct application of harmonic vs. quadratic mean (HM-QM):
*
*$\frac{3}{\frac 1x + \frac 1y + \frac 1z} \stackrel{HM-QM}{\leq}\sqrt{\frac{x^2+y^2+z^2}{3}}$
Now, plugin $\frac 1x + \frac 1y + \frac 1z = 1$ and you get
$$3 \leq \sqrt{\frac{x^2+y^2+z^2}{3}} \Leftrightarrow 27 \leq x^2+y^2+z^2$$
Equality holds for $x=y=z = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
} |
Limit $\lim\limits_{x\to0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$ Without using L'Hopital's rule find:
$$\lim_{x\rightarrow 0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$$
I found the first derivative because I planned to use Taylor series:
\begin{align}
\frac{d}{dx}{\left(\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}\right)}
&=\frac{\left(\frac{1}{x+e}-e^x\right)\left(\cos^2{x} -e^x\right)-\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}
{(\cos^2{x} -e^x)^2}\\
&=\frac{\frac{1}{x+e}-e^x}{\cos^2{x}-e^x}-\frac{\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}{(\cos^2{x} -e^x)^2}
\end{align}
However, it seems I haven't gone so far. Should I start from the beginning and try a different method?
Source in Croatian: 2.kolokvij, matematička analiza
| We have that
$$\frac{\ln{(e+x)}-e^x}x=\frac{\ln{(e+x)}-1}x-\frac{e^x-1}x \to\frac1e-1$$
$$\frac{\cos^2{x} -e^x}x=\frac{\cos^2 x-1}{x}-\frac{e^x-1}x=$$
$$=x(\cos x+1)\frac{\cos x-1}{x^2}-\frac{e^x-1}x \to 0\cdot 2\cdot \left(-\frac12\right)-1=-1$$
then
$$\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}=\frac{\ln{(e+x)}-e^x}{x}\cdot\frac{x}{\cos^2{x} -e^x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A bound for $\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c$ in a triangle Assume that $ABC$ is a triangle with $a\geq b\geq c$, where the angle $A$ has a fixed value. We denote by $\Sigma$ the sum
$$\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c.$$
Then the only possible values of $A$ are $\pi/3\leq A<\pi$ and we have:
(i) The smallest possible value $\Sigma$ is
$$\frac{4\sin\frac A2+\sqrt{2\left( 1-\sin\frac A2\right)}}{\sqrt{2\sin\frac A2}}.$$
(ii) If $\pi/3\leq A<\pi/2$ then the largest possible value of $\Sigma$ is
$$\frac{4\cos A+\sqrt{2\left( 1-\cos A\right)}}{\sqrt{2\cos A}}.$$
(iii) If $\pi/2\leq A<\pi$ then there is no finite upper bound for $\Sigma$.
My question is how to prove (i), (ii), and (iii).
I firstly tried to square the $LHS$, but nothing. I also tried the Radulescu-Maftei theorem, but it didn't help. Hence, I am looking forward to seeing your ideas.
| From:
$$\begin{align}
\frac{b+c-a}a&=\frac{(b+c)^2-a^2}{a(a+b+c)}\\
&=\frac{2bc(1+\cos\alpha)}{a(a+b+c)}\\
&=\frac{4b^2c^2\cos^2\frac \alpha2}{abc(a+b+c)}\\
&=\frac{b^2c^2\sin^2\alpha}{abc(a+b+c)\sin^2\frac \alpha2}\\
&=\frac1{\sin^2\frac\alpha2}\frac{4A^2}{abc(a+b+c)}\\
&=\frac1{\sin^2\frac\alpha2}\frac{r}{2R},
\end{align}
$$
where $A,r,R$ are the area , incircle radius and circumcircle radius, respectively, and
recalling that
$$
\frac rR=4\sin\frac\alpha2\sin\frac\beta2\sin\frac\gamma2
$$
we can reformulate the problem as search of extrema of the function
$$
\left(\frac1{\sin x}+\frac1{\sin y}+\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}\tag1
$$
under restrictions
$$x+y+z=\frac\pi2,\tag2$$
$$z\le y\le x.\tag3$$
Here $x=\frac\alpha2$, $y=\frac\beta2$, $z=\frac\gamma2$. The latter restriction is due to the fact that $x$ is the largest angle in the triangle. We assumed without loss of generality $z\le y$.
Fixing the value of $x$ and applying the method of Lagrange multipliers one obtains that the minimum of the function is achieved at
$$
y=z=\frac\pi4-\frac x2\quad\text{ or }\quad\sin y=\sin z=\sqrt{\frac{1-\sin x}2}.\tag4
$$
Observe that both $y$ and $z$ satisfy the restriction $(3)$. Substituting the values into $(1)$ one obtains the expression (i).
As the point $(4)$ is the only critical point of the function $(1)$ its maximum lies on the boundary of the domain. In the case $\frac\pi6\le x<\frac\pi4$ the solution is:
$$
y=x,\; z=\frac\pi2-2x,\quad\text{ or }\quad \sin y=\sin x,\; \sin z=\cos 2x.
$$
Substituting the values into $(1)$ one obtains the expression (ii).
If $x\ge\frac\pi4$ there is no positive lower bound for the value of $z$. It is easy to see that the expression tends to infinity as $z\to0$. This is the statement (iii).
Appendix. Critical points.
To find the local extrema of the function $(1) $ subject to constraint $(2) $ we construct the function:
$$\left(\frac1{\sin x}+\frac1{\sin y}+\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}-\lambda \left(x+y+z-\frac\pi2\right),\tag {A1}
$$
where $x$ is assumed to be fixed. To find the extrema of the function (A1) we differentiate it wrt. $y$ and $z $ to obtain:
$$\begin{cases}
\frac12\frac{\cos y}{\sin y}\left(\frac1{\sin x}-\frac1{\sin y}+\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}=\lambda\\
\frac12\frac{\cos z}{\sin z}\left(\frac1{\sin x}+\frac1{\sin y}-\frac1{\sin z}\right)\sqrt{2\sin x\sin y\sin z}=\lambda
\end{cases}
$$
which amounts to:
$$
\frac{\cos y}{\sin y}\left(\frac1{\sin x}-\frac1{\sin y}+\frac1{\sin z}\right)
=\frac{\cos z}{\sin z}\left(\frac1{\sin x}+\frac1{\sin y}-\frac1{\sin z}\right).
$$
After straightforward algebraic transformation one obtains the equation:
$$\begin{align}
0&=\sin(y+z)\left(\frac1{\sin z}-\frac1{\sin y}\right)-\sin(y-z)\frac1{\sin x}\\
&=2\sin\frac{y-z}2\left(\frac{\sin(y+z)}{\sin y \sin z}\cos\frac{y+z}2
-\frac1{\sin x}\cos\frac{y-z}2\right)
\end{align}
$$
The fact that the equation holds for $y=z$ was of course obvious without any derivation. Somewhat harder is to prove that the expression inside the parentheses is strictly positive in
the whole domain of the variables $y,z$.
Setting $x=\frac\pi2-y-z$ and reducing to common denominator the claim boils down to
$$
\forall (y,z)\in{\cal D}_{yz}:
\sin(y+z)\cos(y+z)\cos\frac{y+z}2-\sin y \sin z \cos\frac{y-z}2>0\tag{A2}
$$
where ${\cal D}_{yz}$ is the triangle with vertex coordinates $(0,0)$, $\left(\frac\pi4,0\right)$, $\left(\frac\pi6,\frac\pi6\right)$.
The form of the inequality (A2) suggests the substitution $$u=\frac{y+z}2, v=\frac{y-z}2,\tag{A3}$$ so that it can be rewritten as:
$$
\sin4u\cos u +(\cos2u-\cos2v)\cos v>0.\tag{A4}
$$
Under the linear transform (A3) the domain will be mapped in the following way:
$$
{\cal D}_{yz}\mapsto {\cal D}_{uv}:
\quad (0,0)\mapsto(0,0),
\; \left(\frac\pi4,0\right)\mapsto\left(\frac\pi8,\frac\pi8\right),
\; \left(\frac\pi6,\frac\pi6\right)\mapsto\left(\frac\pi6,0\right).\tag{A5}
$$
Since $0<\cos v\le 1$ in the considered domain
$$\begin{align}
\sin4u\cos u +(\cos2u-\cos2v)\cos v
&\ge \sin4u\cos u +\cos2u-1:=\Phi(u).\tag{A4}
\end{align}
$$
Differentiating the expression (A4) twice wrt. to $u$ one obtains:
$$\begin{align}
\frac{\partial^2\Phi(u)}{\partial u^2}
&=-17\sin4u\cos u-8\cos4u\sin u-4\cos 2u\\
&=-9\sin4u\cos u-8\sin5u-4\cos 2u,
\end{align}
$$
which is negative in the whole domain:
$$
0<u\le\frac\pi6.
$$
This means that the function $\Phi(u)$ is concave in the domain. This in turn means that the values of the function in the domain lie above the line defined by the extreme points:
$$
(u,\Phi(u)):\quad (0,0),
\quad \left(\frac\pi6,\frac14\right),
$$
which confirms the claim (A2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
A homogeneous differential equation. Solving by substitution: $\frac{dy}{dx} = \frac{y-x}{y+x}$ I think i'm stuck on this differential equation. I have a differential equation:
$\frac{dy}{dx} = \frac{y-x}{y+x}$
I need to find the implicit general solution. This is what I've done so far:
$dy(y+x) = dx(y-x)$
divide both sides by $x$ until we have each $y$ divided by $x$
$dy(\frac{y}{x} + 1) = dx(\frac{y}{x} - 1)$
Let $y = ux$. Therefore $y' = u'x + u$
and $u = \frac{y}{x}$
I now have:
$dy(u + 1) = dx(u - 1)$
Therefore,
$\frac{dy}{dx} = \frac{u-1}{u+1}$
$u'x + u = \frac{u-1}{u+1}$
$\frac{du}{dx}x = \frac{u-1}{u+1} - u$
In order to make this seperable I decided to make the RHS into one fraction first:
$= \frac{(u-1) - u(u+1)}{u+1} = \frac{-u^2 - 1}{u+1} = -\frac{u^2 + 1}{u+1} = -\frac{1}{u+1}(u^2 + 1)$
$\frac{dx}{xdu} = -(u+1)(\frac{1}{u^2 + 1})$
$\frac{dx}{x} = -(u+1)(\frac{1}{u^2 + 1})du$
Now integrate both sides and get:
$ln(x) + c = ...$
Now For RHS: $-\int(u+1)(\frac{1}{u^2 + 1}) du$
Using integration by parts, let $a = u+1, b' = \frac{1}{u^2 + 1}$
Where: $\int ab' = ab - \int bda$
We have $b = arctan(u), a' = 1$
RHS = $uarctan(u) + arctan(u) - \int arctan(u)$
$=uarctan(u) + arctan(u) - (uarctan(u) - \frac{1}{2}ln|u^2 + 1|)$
RHS = $arctan(u) + \frac{1}{2}ln|u^2 + 1|$
Therefore the solution to our differential equation in terms of $u$ is:
$ln(x) + C = arctan(u) + \frac{1}{2}ln|u^2 + 1|$
Substituting back in, the implicit solution is:
$ln(x) + C = arctan(\frac{y}{x}) + \frac{1}{2}ln|\frac{y^2}{x^2} + 1|$
Yet in my answer book the answer is
$ln(x^2 + y^2) + 2arctan(\frac{y}{x}) = c$
Where did I go wrong?
| You forgot a negative sign on the right hand side.
Your RHS should have been $$RHS = -arctan(u) - \frac{1}{2}ln|u^2 + 1|$$
That will take care of the difference between your answer and the book's anwwer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{\sin{2\alpha}}+\frac{1}{\sin{2^2\alpha}}+\cdots+\frac{1}{\sin{2^n\alpha}}=\frac{1}{\tan{\alpha}}-\frac{1}{\tan{2^n\alpha}}$ Can someone help me with the problem?
Let $\alpha \in \mathbb{R} \backslash \{2^{-m}k\pi; k\in\mathbb{Z},m\in\mathbb{N} \}$.
Prove that the following equation is true for $\forall n\in N$:
$$\frac{1}{\sin{2\alpha}}+\frac{1}{\sin{2^2\alpha}}+\cdots+\frac{1}{\sin{2^n\alpha}}=\frac{1}{\tan{\alpha}}-\frac{1}{\tan{2^n\alpha}}$$
| This can be shown by Induction ( I will leave you to show the base case).
Observe that
\begin{eqnarray*}
\frac{1}{\sin (2^{n+1} \alpha)} &= & \frac{1}{ 2 \sin (2^{n} \alpha) \cos (2^{n} \alpha)} \\
&= & \frac{1}{ 2 \cos^2 (2^{n} \alpha) \tan (2^{n} \alpha)} \\
\end{eqnarray*}
Now add this term on
\begin{eqnarray*}
-\frac{1}{\tan (2^{n} \alpha)} +\frac{1}{\sin (2^{n+1} \alpha)} &= & -\frac{1}{\tan (2^{n} \alpha)} \left( \frac{1}{ 2 \cos^2 (2^{n} \alpha) } -1\right) \\
&= & \frac{1 -2 \cos^2 (2^{n} \alpha) }{ 2 \cos^2 (2^{n} \alpha) \sin (2^{n} \alpha)} \\
&= & -\frac{1 }{ \tan (2^{n+1} \alpha)}. \\
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a,b$ and $c$ are roots of $x^3-x-1=0$. Find $a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}$ $a,b$ and $c$ are the roots of
$$x^3-x-1=0$$
Find
$$a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}$$
To solve this question I called the quantity A and I calculated $A^3$. I have a feeling there is a better way.
Question From Jalil Hajimir
| There are one real and a pair of imaginary roots. Let $a$ be the real root available analytically and express $b=\frac1{a^{1/2}}e^{i \theta} $ and $c=\frac1{a^{1/2}}e^{-i \theta} $ based on $abc = 1$. Substitute $b$ and $c$ into $x^3-x-1=0$,
$$\frac1{a^{3/2}}e^{\pm i 3\theta} -\frac1{a^{1/2}}e^{\pm i \theta} -1=0$$
The difference of the two equations leads to
$$\frac1{a^{3/2}}\sin3\theta-\frac1{a^{1/2}}\sin\theta=0 \implies \cos2\theta = \frac{a-1}2$$
Then,
$$a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}
=a^{\frac{2}{3}} + a^{-\frac{1}{3}}(e^{ i \frac{2\theta}{3}}+e^{- i \frac{2\theta}{3}})=a^{\frac{2}{3}} + 2a^{-\frac{1}{3}}\cos\left(\frac{1}{3}\cos^{-1}\frac{a-1}2\right)$$
where the real root $a$ is given by,
$$a = \sqrt[3]{\frac12 -\frac{\sqrt{69}}{18}} + \sqrt[3]{\frac12 +\frac{\sqrt{69}}{18}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Induction of $\sum_{k=1}^n(2k-1)^2 = \frac{1}{3}n (2n-1)(2n+1), n \geq 1$ I want to understand the induction proof of
$$\sum_{k=1}^n(2k-1)^2 = \frac{1}{3}n (2n-1)(2n+1), n \geq 1$$
Here's the induction steps:
$$\sum_{k=1}^{n+1}(2k-1)^2 = \sum_{k=1}^n(2k-1)^2 + (2(n+1)-1)^2$$
$$= \frac{1}{3}n (2n-1)(2n+1)+(2n+2-1)^2$$
$$=\frac{1}{3}(2n+1)(n(2n-1)+3\cdot(2n+1))$$
$$=\frac{1}{3} (2n+1)(2n^2+5n+3)$$
$$=\frac{1}{3}(2n+1)(2n+3)(n+1)$$
What I don't understand is how you get from
$$= \frac{1}{3}n (2n-1)(2n+1)+(2n+2-1)^2$$
to
$$=\frac{1}{3}(2n+1)(n(2n-1)+3\cdot(2n+1))$$
| We have that
$$\ldots= \frac{1}{3}n (2n-1)(2n+1)+(2n+2-1)^2=\frac{1}{3}n (2n-1)(2n+1)+(2n+1)^2=$$
$$=\frac{1}{3}(2n+1)\cdot\left[n (2n-1)+3(2n+1)\right]=\ldots$$
which is obtained factoring out the $\frac{1}{3}(2n+1)$ term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $m \equiv 5\mod 10 $ prove that $1991 \mid 12^m + 9^m + 8^m + 6^m$ I tried to find the remainder of each one of $12$,$9$,$8$ and $6 \mod 5$ and then combine them but I didn’t get the answer
| Let $m=5+10n$ and let $x_n = 12^m + 9^m + 8^m + 6^m = 12^5 (12^{10})^n + 9^5 (9^{10})^n + 8^5 (8^{10})^n + 6^5 (6^{10})^n$.
Then $x_n$ satisfies a linear recurrence of the form $x_{n+4}=a_3 x_{n+3} + a_2 x_{n+2} + a_1 x_{n+1} + a_0 x_{n}$, where $t^4=a_{n+3}t^3 + a_2 t^2 + a_1 t^1 + a_0 $ is the equation having $2^{10}$, $9^{10}$, $8^{10}$, $6^{10}$ as roots. Hence, the $a_i$ are integers.
The claim that $1991$ divides $x_n$ for all $n$ follows by induction once you have verified it for $n=0,1,2,3$.
Not nice, but doable. WA can help. If you have to do it manually, it's probably easier to do it mod $11$ and $181$ separately.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $\theta$ be a root of $f(x) := x^3-3x +1$ . Then determine the other roots explicitly of terms of $\theta$.
Let $\theta$ be a root of $f(x) := x^3-3x +1$. Now , we can easily check that $\mathbb{Q}(\theta)$ is a splitting field of $f(x)$ and that the Galois group is cyclic of order $3$.
Then, how do I determine the other roots explicitly in terms of $\theta$ ? ( That is , $a \theta^2 + b \theta +c$ for some $a, b, c \in \mathbb{Q}$)
[My attempt]
(1) Let the Galois group be $<\sigma>$ , order $3$ cyclic group. Then, $\sigma(\theta)$ is also root of $f(x)$.
Since $[\mathbb{Q}(\theta) : \mathbb{Q}] = 3 $ (dimension) , so, $\sigma(\theta) = a \theta^2 + b \theta +c$ for some $a, b, c \in \mathbb{Q}$.
Using $\theta^3 = 3 \theta -1$, I tried to find $\sigma(\theta)^3 -3\sigma(\theta)+1 =0$ but this is so complicate.
(2) Let $\theta , \alpha , -\theta -\alpha$ be roots of $f(x)$. By relation of roots and coefficients of $f(x)$, we know that $$\theta^2 +\theta \alpha + \alpha^2 = 3 \;\;\; \text{and} \;\;\; \theta \alpha (\theta + \alpha) = 1 $$
First, using $\theta^2 +\theta \alpha + \alpha^2 = 3$ , $$ \alpha = \frac{-\theta \pm \sqrt{12-3\theta^2} }{2} $$
So, let $ \alpha = a \theta^2 + b \theta +c$ for some $a, b, c \in \mathbb{Q}$.
So $12 - 3 \theta^2 = \alpha^2 = (a \theta^2 + b \theta +c)^2$. This looks simple than (1) but also so complicate..
How can I solve this problem?
| There is a striking similarity between $f(2z)$ and $T_3(z)=4z^3-3z$, the third Chebyshev polynomial of the first kind. By setting $x=2z$ we have that the equation $f(x)=0$ is equivalent to
$$ T_3(z)=-\frac{1}{2} $$
or, by letting $z=\cos\theta$,
$$ \cos(3\theta)=-\frac{1}{2}.$$
Three distinct solutions are clearly given by $\theta\in\left\{\frac{2\pi}{9},\frac{4\pi}{9},\frac{8\pi}{9}\right\}$, so the roots of the original polynomial are
$$ \alpha=2\cos\frac{2\pi}{9},\quad\beta=2\cos\frac{4\pi}{9}=2T_2\left(\frac{\alpha}{2}\right),\quad \gamma=2\cos\frac{8\pi}{9}=2 T_4\left(\frac{\alpha}{2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Convergence of sequence $a(n+1) = \frac{3(1+a(n))}{3+a(n)}$ The sequence is defined by:
$a(1)=1, a(n+1) = \frac{3(1+a(n))}{3+a(n)}$
How do I show that the sequence is monotonic and bounded from above?
My approach to show its monotone was:
$a(n+1) - a(n) \geq 0$
$ \frac{3(1+a(n))}{3+a(n)} - a(n) = \frac{3+3a(n)}{3+a(n)} - \frac{3a(n)+ a(n)^2}{3+a(n)} = \frac{3(1+a(n)) - a(n)(3+a(n))}{3+a(n)} = \frac{3-a(n)^2}{3+a(n)} \geq ? \geq 0$
I know the sequence is bounded from above by $\sqrt{3}$ but I'm totally lost at how to show this.
| Notice that $$\dfrac{a_{n+1}-\sqrt{3}}{a_{n+1}+\sqrt{3}}=\dfrac{a_{n}-\sqrt{3}}{a_{n}+\sqrt{3}}\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$. We can calculate the accurate value of $a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which integers $n$ are there solutions to the equation $x^2-y^2=n$
For which integers $n$ are there solutions to the equation $x^2-y^2=n$
I believe $n$ must be divisible by $3$, since for any integers $x,y$ mod $3$. $x^2=3k+1$ and $y^2=3n+1$, thus $x^2-y^2=3(n-r)$.
Then I would have $n=0$ with $x,y=0$.
I also found that $n$ if it is even must be divisible by a power of $2$ greater then $1$. Since if $n$ is even then one of $(x+y),(x-y)$ is even, so both are even. Then $4\mid n$.
I'm not sure if there are more restrictions, or how to determine if there are no more. Assuming these are done correctly.
| As $x+y+x-y=2x,$
$x+y,x-y$ have the same parity
If both are odd, $(x+y)(x-y)=n$
we can choose $x+y=n, x-y=1$
If both are even $4|n,n=4m$ $$\dfrac{x+y}2=m,\dfrac{x-y}2=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$n$ be the smallest positive integer such that $1+ \frac{1}{2} + \frac{1}{3} + \frac {1}{4}+\cdots \geq 4$. $n$ be the smallest positive integer such that $1+ \frac{1}{2} + \frac{1}{3} +\frac14+\cdots \geq 4$.
My Attempt : I can show that $ \log(n+1) \leq 1+ \frac{1}{2} + \frac{1}{3} + \frac {1}{4} +\cdots+ \frac {1}{n}$. And by using this inequality I can say that the smallest number will be less than $54$. But I can not find the smallest number.
Can anyone please help me by giving some hints.
| We have
$$x-1<\lfloor x\rfloor\le x$$ and
$$\frac1x\le\frac1{\lfloor x\rfloor}<\frac1{x-1}.$$
Now integrating between $2$ and $n+1$, and adding $1$,
$$1+\log\frac{n+1}2\le\sum_{k=1}^n\frac 1k<1+\log n.$$
The lower and upper bounds equal $4$ for $n=41.17$ and $n=20.09$ and the desired solution will lie in that range.
We can narrow that range by computing more terms explicitly
$$1+\frac12+\frac13+\frac14+\frac15+\log\frac{n+1}6\le\sum_{k=1}^n\frac 1k<1+\frac12+\frac13+\frac14+\frac15+\log\frac n5,$$gives $32.40$ and $27.83$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $2^x≥x^2$ I could not understand this proof. How are we able to divide both sides of the inequality with different things and claim that the inequality stays the same? Plus, is this an induction proof?
If $x \in \mathbb{Z}$ and $x \ge 4$ then $2^x \ge x^2$ holds.
When $x = 4$ we have $2^4 = 16 \ge 16 = 4^2$.
We $x > 4$ we have $\frac{2^{x + 1}}{2^x} = 2$ and
$$
\frac{(x + 1)^2}{x^2}
= \left(1 + \frac{1}{x}\right)^2
\le \left(\frac{5x}{4}\right)^2
$$
since $2 \le 1.5625$
| If $$2^x\ge x^2$$ is known to hold, then
$$2^{x+1}\ge(x+1)^2$$ because
$$2^{x+1}=2\cdot2^x\ge 2\cdot x^2\ge\left(\dfrac{x+1}{x}\right)^2x^2=(x+1)^2$$ provided $$\frac{x+1}x\le\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find limit of $\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2}$ I have to show that $\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} = 0$. But I cannot figure out the trick you need to find an upper estimation which goes to $0$. Do you have any hints?
EDIT:
I think I got it:
$\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} = \lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2}{(x^4+2x^2 y^2 +y^4)} +\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|y^4}{(x^4+2x^2 y^2 +y^4)}\leq \lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2}{x^2 y^2}+\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|y^4}{y^4}=\lim_{{x\choose y}\to { 0 \choose 0}} |x|+ \lim_{{x\choose y}\to { 0 \choose 0}} |x|=0.$
What do you think?
| Your solution is fine indeed
$$\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} \le\frac{|x|^3 y^2}{x^2y^2}+\frac{|x|y^4}{y^4}=2|x| \to 0$$
as an alternative by polar coordinates we have
$$\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2}=r \cdot f(\theta) \to 0$$
since $f(\theta)$ is bounded.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$, for integers $a$ and $b$, then what is $\left(1-\sqrt2\right)^{2010}$ expressed using $a$ and $b$?
Being $ a $ and $ b $ integers such that $\left(1+\sqrt{2}\right)^{2011} =a+b\sqrt{2}, \left(1-\sqrt{2}\right)^{2010}$ equals:
a) $a+2b+(a-b)\sqrt{2}$
b) $a-2b+(a-b)\sqrt{2}$
c) $a+2b+(b-a)\sqrt{2}$
d) $2b-a+(b-a)\sqrt{2}$
e) $a+2b-(a+b)\sqrt{2}$
Solution:
Not going with any of the alternatives
| The key point is that $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$ implies
$\left(1-\sqrt2\right)^{2011}=a-b\sqrt{2}$.
Therefore,
$$
\left(1-\sqrt2\right)^{2010}
=\frac{\left(1-\sqrt2\right)^{2011}}{1-\sqrt2}
=\frac{\left(1-\sqrt2\right)^{2011}(1+\sqrt2)}{(1-\sqrt2)(1+\sqrt2)}
=-(a-b\sqrt2)(1+\sqrt2)
= \cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach:
First I calculated the antiderivative as follows:
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$
Using substitution we have:
$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$
$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$
Now we can calculate the definite integral as follows:
$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$
The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?
Here it shows that the correct answer is $\frac{4\pi}{3}$.
| Use $$\int_{0}^{2a} f(x) dx=2 \int_{0}^{a} f(x) dx,~ if~ f(2a-x)=f(x)$$ to get
$$I=\int_{0}^{4\pi} \frac{dx}{5+4\cos x}=4\int_{0}^{\pi} \frac{dx}{5+4 \cos x}~~~~(1)$$
Next use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$$ to get
$$I=4\int_{0}^{\pi} \frac{dx}{5-4 \cos x}~~~~(2)$$
Adding (1) and (2) we get
$$2I=40\int_{0}^{\pi} \frac{dx}{25-16 \cos^2 x} =40 \int_{0}^{\pi}\frac{\sec^2x dx}{25sec^2 x-16}=$$
$$40 \int_{0}^{\pi}\frac{\sec^2x dx}{25\tan^2 x-16}=\frac{8}{5} \int_{0}^{\infty}\frac{du}{9/25+u^2}=\left.\frac{8}{3} \tan^{-1}\frac{5u}{3}\right|_{0}^{\infty}=\frac{4 \pi}{3}.$$
| {
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Calculate the value of angle $ACB$ In an ABC triangle the angle $BAC$ is twice the angle $ACB.$ Consider a point $D$ in segment $AC$ so that the angle $DBA$ is twice the angle $BAD.$ Calculate the value of angle $ACB,$ knowing that the measurement of segment $CD$ is equal to the sum between twice the measurement of segment $BD$ and the length of segment $AD.$
Attemp:After using the law of sines on triangles ABD and BCD, I got the weird-looking equation attached.
I think my approach most likely is not correct.
$$4 \sin x \cos 2x= \sin(180 - 7x)$$
| An euclidean-trigonometric cocktail
Take on $DC$ a point $E$ such that $ED\cong BD$. By our hypotheses (that can be rewritten as $CD - BD \cong AD + BD$) we have that
$$CE\cong AE.$$
Produce $AB$ to $F$ so that $\triangle AEF$ is isososceles.
Call now $\angle CAB = x$, for simplicity. Then of course $\angle ABD = 2x$ and $\angle ACB = \frac{x}2$.
*
*$A$, $C$, and $F$ lie on the circle centered in $E$ with radius $\frac{\overline{AC}}2$, therefore $AF\perp CF$.
*Angle chasing yields $\angle BEF \cong\angle EBF=\frac{\pi}2-\frac{x}2$, therefore $BF\cong \frac{AC}2$.
*External angle theorem yields $\angle CBF = \frac{3x}2$.
We must have
$$\overline{AC} \sin x \cot \frac{3x}2 = \frac{\overline{AC}}2.$$
Trigonometric manipulation gives:
\begin{eqnarray}
\frac{\cos\frac{3x}2\sin x}{\sin\frac{3x}2} &=& \frac12\\
2\frac{\cos \frac{x}2\cos x - \sin\frac{x}2\sin x}{\sin \frac{x}2\cos x + \cos\frac{x}2\sin x}\sin\frac{x}2\cos\frac{x}2 &=& \frac12\\
2\frac{\cos\frac{x}2\left(2\cos^2\frac{x}2-1\right)-2\sin^2\frac{x}2\cos\frac{x}2}{\sin\frac{x}2\left(2\cos^2\frac{x}2-1\right)+2\cos^2\frac{x}2\sin\frac{x}2}\sin\frac{x}2\cos\frac{x}2 &=& \frac12\\
\frac{2\cos^2\frac{x}2\left(2\cos x-1\right)}{2\cos x +1}&=&\frac12\\
\frac{(\cos x+1)(2\cos x-1)}{2\cos x + 1}&=&\frac12,
\end{eqnarray}
which in turns yields
$$4\cos^2x - 3 = 0,$$
and thus $x = \frac{\pi}6$ as the only geometrically valid solution to the problem.
| {
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"url": "https://math.stackexchange.com/questions/3482790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Compute the $PV\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$ Problem :
Evaluate the closed form of : $PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$
Wolfram alpha give me :
$I=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx=\sqrt{3}-\coth^{-1}{\sqrt{3}}$
But i can't get by my try as following :
$\cos (4x)=8\cos^{4} x-8\cos^{2} x+1$
And
$\cos (3x)=4\cos^{3} x-3\cos x$
And I know that
$PV\displaystyle\int_0^{\frac{π}{3}}\frac{1}{\cos (3x)}dx=0$
So I need to find
$J=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos^{2} x}{\cos (3x)}dx=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\sin^{2} x}{\cos (3x)}dx$
Now take $y=\cos x$
$J=\displaystyle\int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{4x^{3}-3x}dx$
From here I don't know how I complete
| You already have that PV $\int_0^{\pi \over 3}{1 \over \cos 3x} = 0$, so your integral is equal to
$$PV \int_0^{\pi \over 3}{\cos 4x - 1 \over \cos 3x} \,dx$$
Using that ${1 + \cos 4x \over 2} = \cos^2 2x$, this is the
same as
$$PV \int_0^{\pi \over 3} {2\cos^2 2x - 2\over \cos 3x}\,dx $$
$$= PV \int_0^{\pi \over 3} -{2\sin^2 2x \over \cos 3x}\,dx $$
Now use identities $\cos 3x = 4\cos^3 x - 3\cos x$ and
$\sin^2 2x = 4\cos^2 x \sin^2 x$. Your integral then becomes
$$PV \int_0^{\pi \over 3} -{8\sin^2 x \cos x \over 4\cos^2 x - 3}\,dx $$
$$=PV \int_0^{\pi \over 3} -{8\sin^2 x \cos x \over 1 - 4\sin^2 x}\,dx $$
Now the subsitution $u = \sin x$ turns this into
$$PV \int_0^{\sqrt{3} \over 2} -{8u^2 \over 1 - 4u^2}\,du $$
$$= PV \int_0^{\sqrt{3} \over 2} 2 - {2 \over 1 - 4u^2}\,du $$
$$= \sqrt{3} - 2PV\int_0^{\sqrt{3} \over 2}{1 \over 1 - 4u^2}\,du$$
$$=\sqrt{3} - PV\int_0^{\sqrt{3} \over 2}{1 \over 1 - 2u} - \int_0^{\sqrt{3} \over 2}{1 \over 1 + 2u}\,du$$
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) - PV\int_0^{\sqrt{3} \over 2}{1 \over 1 - 2u}\,du$$
The principal value integral from $0$ to $1$ here is zero by symmetry of the kernel about $u = {1 \over 2}$, so the above is equal to
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) + \int_{\sqrt{3} \over 2}^1{1 \over 1 - 2u}\,du$$
$$=\sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) - \int_{\sqrt{3} \over 2}^1{1 \over 2u - 1}\,du$$
$$= \sqrt{3} - {1 \over 2}\ln(1 + \sqrt{3}) + {1 \over 2}\ln(\sqrt{3} - 1)$$
| {
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"url": "https://math.stackexchange.com/questions/3482853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
} |
Finding $\sum_{k=0}^{6}a_{3k} $ where $(x^2-x+1)^{10}=\sum_{k=0}^{20}a_k x^k$
If $(x^2-x+1)^{10}=\displaystyle\sum_{k=0}^{20}a_k x^k$ then find the value of the expression $\displaystyle\sum_{k=0}^{6}a_{3k}$.
My Attempt
Since the subscripts on terms we are supposed to find involve a difference of $3$, substituting in $x=\omega$ and $x=-\omega$, where $\omega$ is a cube root of unity seems likely.
$$\begin{align} x=\omega &\implies(\omega ^2-\omega +1)^{10}=2^{10}\omega=\sum_{k=0}^{6}a_{3k}+\omega \sum_{k=0}^{6}a_{3k+1}+\omega ^2\sum_{k=0}^{6}a_{3k+2}\tag1 \\ x=-\omega &\implies(\omega ^2+\omega+1)^{10}=0 \ \ \ \ \ =\sum_{k=0}^{6}a_{3k}-\omega\sum_{k=0}^{6}a_{3k+1}+\omega^2\sum_{k=0}^{6}a_{3k+2} \tag2\end{align}$$
Adding $(1)$ and $(2)$ gets us rid of the term with $\omega$ but $\omega^2$ term just adds up to give an expression containing an extra $\sum_{k=0}^{6}a_{3k+2}$.
How can I proceed? Any hints are appreciated. Thanks
| Use $x=1$, $x=\omega$ and $x=\omega^2$. You get
$$3\sum_{k=0}^6a_{3k}=f(1)+f(\omega)+f(\omega^2)$$
where $f(x)=(x^2-x+1)^{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a=\sin\theta+\cos\theta$ and $b=\cos\theta-\sin\theta$, express $\sum_k\sin^k\theta(\sin^k\theta-\cos^k\theta)$ in terms of $a$ and $b$
If $a=\sin\theta+\cos\theta$ and $b=\cos\theta-\sin\theta$, then express
$$\sin \theta(\sin \theta-\cos \theta)+\sin^2\theta(\sin^2\theta-\cos^2\theta)+\sin^3\theta(\sin^3\theta-\cos^3\theta) +\cdots$$
in terms of $a$ and $b$.
My observation :
$\underbrace{\sin \theta(\sin \theta-\cos \theta)}_1 +$$\underbrace{\sin^2\theta(\sin^2\theta-\cos^2\theta)}_2 +\underbrace{\sin^3\theta(\sin^3\theta-\cos^3\theta)}_3 +\cdots\\$
$i.)$$\;\sin \theta(\sin \theta-\cos \theta) =\sin \theta (-b) =\bbox[4px, yellow] {-b\sin \theta} $ $\\$
$ii.) \;\sin^2 \theta(\sin^2 \theta-\cos^2 \theta)=\sin^2 \theta(\sin\theta-\cos
\theta)(\sin \theta + \cos\theta)=\bbox[yellow, 4px]{-ab\sin^2\theta}
\\$
$\mathrm {Now} \\$
$iii.)$ $\sin^3\theta(\sin^3\theta-\cos^3\theta)=\sin^3\theta (\sin\theta-\cos\theta)(\sin^2\theta +\sin\theta \cos\theta+\cos^2\theta) \\$
$=\sin^3\theta(-b)\left(1+\boxed {\sin\theta\cos\theta} \rightarrow\left\{\frac{a^2-b^2} {4} \right\} \right) \\$
$=\underline{\bbox[yellow, 4px] {-b\sin^3\theta\left(1+\frac{a^2-b^2}{4}\right)}}$
Can anyone suggest a way out?
| $$\sin \theta(\sin \theta-\cos \theta)+\sin^2\theta(\sin^2\theta-\cos^2\theta)+\sin^3\theta(\sin^3\theta-\cos^3\theta) +\cdots$$
$$=-(\sin^2\theta)^0+\sum_{r=0}^\infty(\sin^2\theta)^r-\left((\sin\theta\cos\theta)^0+\sum_{r=0}^\infty(\sin\theta\cos\theta)^r\right)$$
$$=\dfrac1{1-\sin^2\theta}-\dfrac1{1-\sin\theta\cos\theta}$$
assuming $\sin^2\theta\ne1$ and $\sin\theta\cos\theta=\dfrac{\sin2\theta}2$ whose absolute value $\le\dfrac12$
Solve the two simultaneous equations to find $\sin\theta,\cos\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$n$-th power of a complex non-diagonalizable $2 \times 2$ matrix I have a complex matrix in the form $$ \begin{pmatrix} a-ib & ic \\ -ic & a+ib \end{pmatrix} $$ which is not diagonalizable according to my understanding. How to calculate the $n$-th power of this matrix? Can we use eigenvalue decomposition or Cayley-Hamilton theorem on this? Or is there a way to approximate the polynomials?
| The characteristic polynomial is $$X^2-2aX+a^2+b^2-c^2$$
To be nondiagonalizable, this would have to equal $(X-r)^2$ for some $r$. Examining the linear term, $r$ would have to be $a$. Examining the constant term, $b^2$ would have to equal $c^2$. So as noted in the comments, $c=\pm b$ is required for this to be not diagonalizable. And in that situation, $a$ is the one eigenvalue.
So now consider
$$M-aI=i\begin{pmatrix} -b & c \\ -c & b \end{pmatrix}$$
Its kernel is spanned by $\begin{pmatrix} c \\ b \end{pmatrix}$. And $\frac{-i}{2b}\begin{pmatrix} -c \\ b \end{pmatrix}\mapsto\begin{pmatrix} c \\ b \end{pmatrix}$.
It follows from the usual diagonalization/Jordan-normalization process that
$$M=\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}^{-1}$$
and then
$$M^n=\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}^n\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}^{-1}$$
$$M^n=\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}\begin{pmatrix} a^n & na^{n-1} \\ 0 & a^n \end{pmatrix}\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}^{-1}$$
| {
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"url": "https://math.stackexchange.com/questions/3486713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers.
It can probably be solved by a quick factoring trick, but I really can’t figure it out.
| Because $$a^2+b^2+c^2=a^2+b^2+\frac{a^2b^2}{(a+b)^2}=\frac{(a^2+ab+b^2)^2}{(a+b)^2}.$$
I used that:
$$(a^2+ab+b^2)^2=a^4+2a^3b+3a^2b^2+2ab^3+b^4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$ Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$.
I tried solving but got stuck.
Show that it is true for n=1
$$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$
Assume it true for $n = k$
$$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$
Show it is true for $n= k + 1$
$$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$
is a multiple of 24
$$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$
$$2\cdot7(24g + 5)3\cdot 5 - 5$$
I'm stuck and don't know how to proceed
| Notice that $5^2$ and $7^2$ are both one more than a multiple of $24.$ Prove the cases $k=1$ and $k=2$ for your basis step.
Then in the induction you have, for one term,
$$2(7^{k+2}) = 2(7^k)(48+1).$$
Do the same for the other term. Induct by twos instead of by ones.
| {
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"url": "https://math.stackexchange.com/questions/3488433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve this integral by using cover up method in partial fractions new user here. I am trying to decompose the following fraction so that it can be solved easily.
$$\frac{3x-2}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$$
General formula:
$$\frac{px+q}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$$
I am able to find the value of A by simply equating it to p (the coefficient of x) but I'm having trouble determining B. On substituting x=2 in either of the denominators, $$(x-2)$$ or $$(x-2)^2$$ They become 0 resulting in an indeterminant form of the fraction. Please help me out.
| $$\frac{3x-2}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$$
If and only if
$3x-2 = A(x-2) + B$
If you take $x = 2 \implies B = 4$
Then $3x-2 = A(x-2) + 4$
If you take $x = 0 \implies A = 3$
Then
$$\frac{3x-2}{(x-2)^2} = \frac{3}{x-2} + \frac{4}{(x-2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What's wrong in my calculation of $\int_0^{3 \pi/4} \frac{\cos x}{1 + \cos x}dx$? What's wrong in my calculation of $$\int_0^{3 \pi/4} \dfrac{\cos x}{1 + \cos x} dx\,?$$
I have to find:
$$\displaystyle\int_0^{3 \pi/4} \dfrac{\cos x}{1 + \cos x} dx$$
and I can't seem to get the right answer. This is what I did:
I decided to use the Weierstrass substitution with:
$$t = \tan \dfrac{x}{2}$$
$$\cos x = \dfrac{1 - t^2}{1 + t^2}$$
$$\sin x = \dfrac{2t}{1 + t^2}$$
$$dx = \dfrac{2}{1+ t^2}$$
I am pretty new to this type of substitution.
Anyway, the boundaries become:
$$t_1= \tan 0 = 0$$
$$t_2 = \tan \dfrac{3 \pi }{8}$$
We have
$$\displaystyle\int_0^{3 \pi/4} \dfrac{\cos x}{1 + \cos x} dx = $$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{\dfrac{1 - t^2}{1 + t^2}}{1 + \dfrac{1 - t^2}{1 + t^2}} \cdot \dfrac{2}{1 + t^2} $$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1-t^2}{1+t^2} dt$$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2}dt - \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{t^2}{1+t^2} dt $$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2} dt - \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{t^2 + 1 - 1}{1+t^2} dt$$
$$= \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2} dt - \displaystyle\int_0^{\tan 3 \pi/8} 1 dt + \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2} dt $$
$$= 2 \bigg [\arctan(t) \bigg ]_0^{\tan 3\pi/8} - \bigg [ t \bigg ]_0^{\tan 3\pi/8}$$
$$= 2\arctan \bigg ( \tan \dfrac{3 \pi}{8} \bigg ) - \tan \dfrac{3\pi}{8}$$
$$= 2 \dfrac{3 \pi}{8} - \tan \dfrac{3 \pi}{8}$$
$$=\dfrac{3 \pi}{4} - \tan \dfrac{3 \pi}{8}$$
So that's the answer that I got. However, my textbook claims that the correct answer is in fact $\dfrac{\pi}{4} + \tan \dfrac{3 \pi}{8} - 2$. So, what did I do wrong?
| It's easier to write $$\frac{\cos x}{1+\cos x}=1-\frac{1}{1+\cos x}\cdot\color{red}{\frac{1-\cos x}{1-\cos x}}=1-\frac{1-\cos x}{\sin^2x}=1-\csc^2 x+\frac{\cos x}{\sin^2x}$$
So
$$\int\frac{\cos x}{1+\cos x}dx=x+\cot x-\csc x+C$$
Added,
or we can use $$1+\cos x=2\cos^2(x/2)\Longrightarrow \frac1{1+\cos x}=\frac12\sec^2(x/2)$$
So $$\int\frac1{1+\cos x}dx=\tan(x/2)+C$$
| {
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"url": "https://math.stackexchange.com/questions/3489619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $\int \frac{1}{\cot \frac{x}{2}\cdot\cot\frac{x}{3}\cdot\cot\frac{x}{6}}\text{d}x$ $$\int \dfrac{1}{\cot \dfrac{x}{2}\cdot\cot\dfrac{x}{3}\cdot\cot\dfrac{x}{6}}dx$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\tan\dfrac{x}{6}dx$$
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\tan\left(\dfrac{x}{2}-\dfrac{x}{3}\right)dx$$
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\dfrac{\tan\dfrac{x}{2}-\dfrac{x}{3}}{1+\tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}}dx$$
$$\int \dfrac{\tan^2\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}}{\cos\dfrac{x}{6}}-\dfrac{\tan\dfrac{x}{2}\cdot\tan^2\dfrac{x}{3}\cdot\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}}{\cos\dfrac{x}{6}}dx$$
$$\int \dfrac{\sin^2\dfrac{x}{2}\cdot\sin\dfrac{x}{3}}{\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{6}}-\dfrac{\sin\dfrac{x}{2}\cdot\sin^2\dfrac{x}{3}}{\cos\dfrac{x}{3}\cos\dfrac{x}{6}}dx$$
$$2\left(\int \dfrac{\sin^2\dfrac{x}{2}\cdot\sin\dfrac{x}{6}}{\cos\dfrac{x}{2}}-\dfrac{2\sin\dfrac{x}{2}\cdot\sin^2\dfrac{x}{6}\cos\dfrac{x}{6}}{\cos\dfrac{x}{3}}\right)dx$$
This doesn't seem to be going anywhere.
Attempt $2$:
$$\int \dfrac{\sin\dfrac{x}{2}\cdot\sin\dfrac{x}{3}\cdot\sin\dfrac{x}{6}}{\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}\cdot\cos\dfrac{x}{6}}dx$$
$$\int \dfrac{\cos\dfrac{x}{6}\cdot\sin\dfrac{x}{6}-\cos\dfrac{5x}{6}\cdot\sin\dfrac{x}{6}}{\cos^2\dfrac{x}{6}+\cos\dfrac{x}{6}\cdot\cos\dfrac{5x}{6}}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{1+\cos\dfrac{x}{3}+\cos x+\cos \dfrac{2x}{3}}dx$$
As we know $\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+(n-1)\beta)=\dfrac{\sin\dfrac{n\beta}{2}}{\sin\dfrac{\beta}{2}}\cos\left(\dfrac{\alpha+\alpha+(n-1)\beta}{2}\right)$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{\cos(0)+\cos\left(0+\dfrac{x}{3}\right)+\cos\left(0+\dfrac{2x}{3}\right)+\cos\left(0+\dfrac{3x}{3}\right)}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}+\sin x+\sin\dfrac{2x}{3}-2\sin x}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \dfrac{\dfrac{\sin\left(\dfrac{3x}{6}\right)}{\sin\dfrac{x}{6}}\cdot\sin\dfrac{2x}{3}-2\sin x}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \tan\dfrac{x}{2}dx-\int \dfrac{4\sin\dfrac{x}{2}\sin\dfrac{x}{6}}{\sin\dfrac{2x}{3}}dx$$
$$\int \tan\dfrac{x}{2}dx-2\int \dfrac{\cos\dfrac{x}{3}-\cos\dfrac{2x}{3}}{\sin\dfrac{2x}{3}}dx$$
$$\int \tan\dfrac{x}{2}dx-\int \mathrm{cosec}\dfrac{x}{3}dx+2\int \cot\dfrac{2x}{3}dx$$
$$2\ln\left|\sec \dfrac{x}{2}\right|-3\ln\left|\mathrm{cosec}\dfrac{x}{3}-\cot\dfrac{x}{3}\right|+3\ln\left|\sin\dfrac{2x}{3}\right|+C$$
But this got too long,any short and better approach.
| If $A+B+C=m\pi,$ for some integer $m$
$\tan(B+C)=\tan(m\pi-A)=-\tan A$
$\implies\tan A+\tan B+\tan C=\tan A\tan B\tan C$
Set $A=\dfrac x3, B=\dfrac x6, C=-\dfrac x2\implies m=0$
$$\tan\dfrac x3+\tan\dfrac x6+\tan\left(-\dfrac x2\right)=\tan\dfrac x3\tan\dfrac x6\tan\left(-\dfrac x2\right)$$
As $\tan(-y)=-\tan y,$
$$\tan\dfrac x3+\tan\dfrac x6-\tan\dfrac x2=-\tan\dfrac x3\tan\dfrac x6\tan\dfrac x2$$
| {
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"url": "https://math.stackexchange.com/questions/3490871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots of complex number (2018 AMC 12A Problem 22)
The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$ -2018 AMC 12A Problem 22
The solution I found online was based on that the triangle formed by the origin and the two complex numbers in the first quadrant is $\frac{1}{4}$ area of the parallelogram, and that can be found since the $z^{2}=2+2\sqrt{3}i$ is easily convertible into polar coordinates and square-rooted with DeMoivre's, but it didn't find the coordinates of the $z^{2}=4+4\sqrt{15}i$.
In the solutions section of the AoPS problem page, however, one of the solutions is this:
The roots are $\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)$ (easily derivable by using DeMoivre and half-angle). From there, shoelace on $\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)$ and multiplying by $4$ gives the area of $6\sqrt{2}-2\sqrt{10}$, so the answer is $\boxed{20}$. (trumpeter)
This was what I tried to do at first, before realizing that I have no idea how $\sqrt{15}$ works into the trigonometric representation formulas. Apparently, though, it is "easily derivable by using DeMoivre and half-angle." Using this description of the solution process, how would one go about finding each of the roots of the two complex numbers given in the problem?
| Let $2\theta$ be an angle that satisfies $$\cos 2\theta = \frac{4}{\sqrt{4^2 + (4\sqrt{15})^2}} = \frac{1}{4}.$$ Then by the double-angle identity $$\cos 2\theta = 2 \cos^2 \theta - 1,$$ we readily obtain $$\cos \theta = \pm \sqrt{\frac{5}{8}}$$ hence $$\sin \theta = \pm \sqrt{\frac{3}{8}},$$ where the signs are chosen to be the same in each case because $z^2$ (hence $2\theta$) is in quadrant I, hence $z$ (and $\theta$) is in quadrant I or III. The rest is straightforward.
| {
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solve matrix equation $2AX-X^TA=B$ Given matrices $A=\begin{pmatrix}
2 & 4\\
3 & 6
\end{pmatrix},B=\begin{pmatrix}
3 & -6\\
7&-4
\end{pmatrix}$
Find all matrices $X\in M_2(\mathbb{R})$ such that
$$2AX-X^TA=B$$
I have no idea how to do it.
| Let $X = \begin{bmatrix}x_1&x_2\\x_3&x_4\end{bmatrix}$. Then we have
$\begin{bmatrix}4 & 8\\ 6&12\end{bmatrix}\cdot\begin{bmatrix}x_1&x_2\\x_3&x_4\end{bmatrix}-\begin{bmatrix}x_1&x_3\\x_2&x_4\end{bmatrix}\cdot\begin{bmatrix}2&4\\3&6\end{bmatrix}=\begin{bmatrix}3&-6\\7&-4\end{bmatrix}.$
Matching entries gives the system $(4x_1+8x_3)-(2x_1+3x_3)=3\tag{1}$
$(4x_2+8x_4)-(4x_1+6x_3)=-6\tag{2}$
$(6x_1+12x_3)-(2x_2+3x_4)=7\tag{3}$
$(6x_2+12x_4)-(4x_2+6x_4)=-4\tag{4}$.
We can represent this as the augmented matrix $\begin{bmatrix}2&0&5&0&3\\-4&4&-6&8&-6\\6&-2&12&-3&7\\0&2&0&6&-4\end{bmatrix}$
Rewriting this matrix in reduced row echelon form gives
$\begin{bmatrix}1&0&0&\dfrac{5}{2}&-\dfrac{7}{2}\\0&1&0&3&-2\\0&0&1&-1&2\\0&0&0&0&0\end{bmatrix}.$
Hence all desired matrices are of the form
$\begin{bmatrix}-\dfrac{5}{2}k-\dfrac{7}{2}&-3k-2\\k+2&k\end{bmatrix},$ where $k\in\mathbb{R}$.
| {
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on $p^n+q^n=(p+q)^k$ Evaluate the condition for integer solutions for $n$.
I found this question in an old book:
Find smallest number of form $2^n+3^n$ divisible by $625$. This solution is from that book; n must be odd and we may write:
$$2^n+3^n$$ $$=2^n+(-1)^n(2-5)^n$$ $$=2^n+(-1)^n\cdot 2^n-(-1)^n\cdot2^{n-1}\cdot5\cdot n+(-1)^n\cdot2^{n-2}\cdot5^2\frac{n(n-1)}{2}-(-1)^n\cdot2^{n-3}\cdot5^3\frac{n(n-1)(n-2)}{6}+625 N$$
$$2^n+3^n=5n\big[2^{n-1}-(n-1)2^{n-2}.5+\frac{(n-1)(n-2)}{3}2^{n-2}.5^2\big]+625 N;\ n\geq 4$$
The value inside box bracket is not divisible by $5$, so $n$ must be divisible by $125$ if $2^n+3^n$ must be divisible by $625$.
Similar reasoning can be used for any primes $p$ and $q$ such that:
$p^n+q^n≡0 \ mod (p+q)^k$
The condition is
$n=(p+q)^{k-1}$.
Now we try to apply Euler $\phi$ function:
$\phi(625)=625\big(1-\frac{1}{5}\big)=500$
$2^{500}≡1 \mod 625$
$3^{500}≡ 1 \ mod 625$
$3^{500}-2^{500}≡0 \mod 625$
$(3^{125}-2^{125})(3^{125}+2^{125})(3^{250}+2^{250})≡0 \mod 625$
Only $3^{125}+2^{125}$ can be divisible by 625. But $3^{125}+2^{125}$ can be factorized more and more.Suppose we can not use first method because p and q are too large, then how can we be sure the smaller factors are not divisible by $625$? is $n=125$ the smallest number?
| If $n$ is even, clearly, $$5\nmid 2^n+3^n.$$ For odd $n$, we can use Lifting The Exponent.
We have $$\nu_5\left(2^n+3^n\right)=\nu_5(2+3)+\nu_5(n),$$ so that $2^n+3^n$ is divisible by $625$ only when $n$ is divisible by $125$. That is, $\boxed{125}$ is the correct answer.
| {
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Proving that $x,y\in\mathbb{R}, x>0,$ and $x^2Suppose $x,y\in \mathbb{R}$ and $x>0$. We also know that ${{x}^{2}}<y$. I'm struggling to find a formal proof that there's always an $n\in \mathbb{N}$ such that ${{\left(x+\frac{1}{n} \right)}^{2}}<y$. A hint would be much appreciated.
Edit: Basically, any $n\in \mathbb{N}$ does the job as long as $n>\left\lfloor \frac{1}{\sqrt{y}-x} \right\rfloor $? That got me thinking... Would this still hold if both $x,y$ were rational numbers $x,y\in \mathbb{Q}$? I mean, we wouldn't be to use square roots.
| $\left(x+\frac{1}{n} \right)^{2}
=x^2+\frac{2x}{n}+\frac1{n^2}
$
so we want
$x^2+\frac{2x}{n}+\frac1{n^2}
\lt y
$
or
$\frac{2x}{n}+\frac1{n^2}
\lt y-x^2
$.
Let $d = y-x^2$.
We are done if
we can find an $n$
such that
$\frac{2x}{n}
\lt \frac{d}{2}
$
and
$\frac1{n^2}
\lt \frac{d}{2}
$.
The first is equivalent to
$n
\gt \frac{4x}{d}
$.
The second is equivalent to
$n^2 > \frac{2}{d}
$.
Since $n^2\ge n$,
this will be true if
$n > \frac{2}{d}
$.
Since $d > 0$
and
$x > 0$,
$ \frac{4x}{d}
\gt 0
$
and
$ \frac{2}{d}
\gt 0
$.
Therefore
if
$n
\ge \frac{4x}{d}+\frac{2}{d}
=\frac{4x+2}{d}
$,
then
${{\left(x+\frac{1}{n} \right)}^{2}}<y$.
| {
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Evaluate$\int\frac{x-1}{(x+1)\sqrt{x^4+x^3+x^2}}dx$
Evaluate the indefinite integral $$\int\frac{x-1}{(x+1)\sqrt{x^4+x^3+x^2}}dx$$
What I have tried
$$I = \int\frac{x^2-1}{(x^2+2x+1)\sqrt{x^4+x^3+x^2}}dx$$
$$I = \int\frac{x^2-1}{(x^2+2x+1)x\sqrt{x^2+\frac{1}{x}+1}}dx$$
How do I solve it? Help me please.
| Let $x=\frac1t$,
$$I=\int\frac{x-1}{(x+1)\sqrt{x^4+x^3+x^2}}dx=\int\frac{t-1}{(t+1)\sqrt{t^2+t+1}}dt$$
$$=\int\frac{dt}{\sqrt{t^2+t+1}}-2\int\frac{dt }{(t+1)\sqrt{t^2+t+1}} $$
where the first integral is straightforward,
$$\int\frac{dt}{\sqrt{t^2+t+1}}=\int\frac{dt}{\sqrt{(t+\frac12)^2+\frac34}}
=\sinh^{-1}\left(\frac{2t+1}{\sqrt3}\right)$$
and, with $u=\frac{2}{\sqrt3}(t+\frac12)=\tan y$, the second integral becomes
$$\int\frac{dt }{(t+1)\sqrt{t^2+t+1}}
=\int\frac{du }{(\frac{\sqrt3}2u+\frac12)\sqrt{u^2+1}}
= \int\frac{\sec y }{\frac{\sqrt3}2\tan y+\frac12}dy $$
$$\int\frac{dy }{\frac{\sqrt3}2\sin y+\frac12\cos y}dy
=\int\frac{dy }{\sin( y+\frac\pi6)}=\ln\left(\tan\left(\frac y2+\frac\pi{12}\right)\right) $$
Thus,
$$I=\sinh^{-1}\left(\frac{2t+1}{\sqrt3}\right) -2\ln\left(\tan\left(\frac y2+\frac\pi{12}\right)\right) +C$$
| {
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Sequence of functions inequality I need some help to prove that $f_n(x)\leq 1$ , where $f_n:[0,1]\to\mathbb{R}, f_n(x)=\frac{1+nx^2}{(1+x^2)^n}, n\in\mathbb{N}$.
I can write $(1+x^2)^n=\binom{0}{n}x^n+\binom{1}{n}x^{n-1}y+\binom{2}{n}x^{n-2}y^2+...+\binom{n-2}{n}x^2y^{n-2}+\binom{n-1}{n}xy^{n-1}+\binom{n}{n}y^n=\\x^n+nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+...+\frac{n(n-1)}{2!}x^2y^{n-2}+nxy^{n-1}+y^n$
$\\$
I unsuccessfully tried to find $1+nx^2$ in $(1+x^2)^n$ to show that $1+nx^2\leq(1+x^2)^n$.
| Differentiate the function $\frac{1+nx²}{(1+x²)^{n})}$ which comes out to be $\frac{2n(1-n)x^3(1+x²)^{n-1}}{(1+x²)^{2n}}$. Which obviously = 0 if n=1 and <0 if x >0 , n>1.We can conclude for any n maximum value of function in (0,1) occurs at 0 . At zero function is 1for any n. So maximum value of function is 1 for any value of n.
| {
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Find $T(a+bx+cx^2)$ Let the linear transformation $T: \mathbb P_2 \rightarrow M_{2\times2}$
Find $T(a+bx+cx^2)$
knowing that:
$$T(1+x)=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$$
$$T(x+x^2)=\begin{bmatrix} 0&1\\1&0\end{bmatrix}$$
$$T(1+x^2)=\begin{bmatrix} 0&0\\0&1\end{bmatrix}$$
I believe that it can be solved by combinations of the known transformations. But I've been stuck with this problem for a while.
| HINT
To start with, solve the following system in terms of $a,b$ and $c$.
\begin{align*}
a+bx+cx^{2} = p(1+x) + q(x+x^{2}) + r(1+x^{2}) \Longleftrightarrow
\begin{cases}
a = p + r\\
b = p + q\\
c = q + r
\end{cases}
\end{align*}
Once you know $p,q$ and $r$, you get
\begin{align*}
T(a + bx + cx^{2}) = pT(1+x) + qT(x+x^{2}) + rT(1+x^{2})
\end{align*}
Can you take it from here?
| {
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Maximum point of a function I am looking to find the local maximum value of the function
$$f(x,y,z) = 9+ \frac{(x-y)^2}{xy} + \frac{(y-z)^2}{yz}+ \frac{(z-x)^2}{zx}$$
at a point within the cube $[a,b] \times [a,b] \times [a,b],$ where $0<a<b$. I took the partial derivative of f with respect to x, y, and z. I set those to zero. I used the second-derivative test for multi-variable functions.
The second-derivative test requires the computation of a 3 by 3 matrix. (Because the function is a three-variable function)
$D= \det \pmatrix{f_{xx} & f_{xy} &f_{xz} \\
f_{yx} & f_{yy} &f_{yz}
\\
f_{zx} & f_{zy} &f_{zz}}$
I found that the first dervatives equal to zero at $x=y=z$. On the other hand, Wolfram Alpha confirms that the max for the $[1,3] \times [1,3] \times [1,3]$ is for $y=z=\frac{x}{3}$ and $x=\frac{y}{3}=\frac{z}{3}$. I got confused
| We compute
\begin{align}
f_{xx} &= f_x\left(\frac1y-\frac y{x^2}\right) = \frac{2y}{x^3},\\\\
f_{xy} &= f_y \frac{(y+z) \left(x^2-y z\right)}{x^2 y z} = \frac{-(x^2+y^2)}{x^2y^2}\\
f_{xz} &= f_x \frac{(x+y) \left(z^2-x y\right)}{x y z^2} = \frac{-(x^2-z^2)}{x^2 z^2}\\
f_{yz} &= f_y \frac{(x+y) \left(z^2-x y\right)}{x y z^2} = \frac{-(y^2+z^2)}{y^2 z^2}.
\end{align}
Since the mixed partials are continuous, by Clairaut's theorem we need only compute
\begin{align}
f_{yy} &= f_y\frac{(x+z) \left(z^2-x z\right)}{x y^2 z} = \frac{2 (x+z)}{y^3}\\
f_{zz} &= f_z\frac{(x+y) \left(z^-2x y\right)}{x y z^2} = \frac{2 (x+y)}{z^3}.
\end{align}
Hence the Hessian matrix is
$$
H = \begin{pmatrix}
\frac{2y}{x^3} & \frac{-(x^2+y^2)}{x^2y^2} & \frac{-(x^2-z^2)}{x^2 z^2}\\
\frac{-(x^2+y^2)}{x^2y^2} & \frac{2 (x+z)}{y^3} & \frac{-(y^2+z^2)}{y^2 z^2}\\
\frac{-(x^2-z^2)}{x^2 z^2} & \frac{-(y^2+z^2)}{y^2 z^2} & \frac{2 (x+y)}{z^3}.
\end{pmatrix}
$$
We have $f_{xx} = \frac{2y}{x^3}>0$ on $[a,b]^3$ and
$$
\begin{vmatrix}
f_{xx} & f_{xy}\\\ f_{xy}& f_{xz}
\end{vmatrix} = \begin{vmatrix}
\frac{2y}{x^3} &\frac{-(x^2+y^2)}{x^2y^2}\\\
\frac{-(x^2+y^2)}{x^2y^2} & \frac{-(x^2-z^2)}{x^2 z^2}
\end{vmatrix} = \frac{\left(x^2+y^2\right) \left(-x \left(x^2+y^2\right)-\frac{2 y^5}{z^2}\right)}{x^5 y^4},
$$
which takes negative values. So $F$ admits no global minimum. As shown in @Michael Rozenberg's answer, $9$ is a local minimum.
| {
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Solve this equations system: $x^{5} + y^{5} + z^{5} = 1, x + y + z = 1$
Solve this equations system: \begin{cases} x^{5} + y^{5} + z^{5} =
1 \\ x + y + z = 1 \end{cases}
Sketch of solutions:
If x,y,z is solution of this equations system then $$(x+y+z)^{5} - (x^{5} + y^{5}+z^{5}) = 0$$ when $x+y= 0$, then equation system is true for any z.
So we have
$$(x+y+z)^5 - (x^{5}+y^{5}+z^{5})= A(x+y)(y+z)(x+y)(x^{2}+xy+xz+y^{2}+yz+z^{2})=$$
for $x=1, y=1, z=0$ we have
$$2^{5}-1-1 = 6A$$
$$ A = 5 $$
So we have
$5(x+y)(y+z)(x+z)(x^{2}+xy+xz+y^{2}+yz+z^{2})=0$
So the solutions:
1.
$$x+y= 0, z = 1 $$
2.
$$y+z= 0, x = 1 $$
3.
$$x+z= 0, y = 1 $$
4.
$$x^{2}+xy+xz+y^{2}+yz+z^{2} =$$
$$ \frac{1}{2}(x^{2}+2xy+y^{2}+x^{2}+2xy+z^{2}+y^{2}+2yz+z^{2}$$
$$ \frac{1}{2}((x+y)^{2}+(y+z)^{2}+(x+z)^{2})==0$$
this imply that (0,0,0) is solution. It's wrong because we have $0=1$
Do you see mistake?
Solutions from Mathematica:
$\left\{\{x\to 1,z\to -y\},\{y\to 1,z\to -x\},\{y\to -x,z\to 1\},\left\{y\to \frac{1}{2} \left(-\sqrt{-3 x^2+2 x-3}-x+1\right),z\to \frac{1}{2} \left(\sqrt{-3 x^2+2 x-3}-x+1\right)\right\},\left\{y\to
\frac{1}{2} \left(\sqrt{-3 x^2+2 x-3}-x+1\right),z\to \frac{1}{2} \left(-\sqrt{-3 x^2+2 x-3}-x+1\right)\right\}\right\}$
| Your answer of $0$ is not a mistake because it is a solution of $(x+y+z)^{5} - (x^{5} + y^{5}+z^{5}) = 0$ which is the equation you were solving.
It just happens not to be a solution which makes each of $(x+y+z)^{5}$ and $(x^{5} + y^{5}+z^{5})$ equal to $1$.
| {
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Prove that $\cos 2x + \cos 2y + \cos 2z = -4(\cos x \cos y \cos z) -1$ when $x+y+z=\pi$ I found on List of trigonometric identities on wikipedia that $\cos 2x + \cos 2y + \cos 2z = -4(\cos x \cos y \cos z) -1$ if $x+y+z=\pi = 180°$. I couldn't find a proof.
| We have $\cos2z=\cos2(\pi-x-y)=\cos2(x+y)=2\cos^2(x+y)-1$. Then use the sum-to-product identity twice:
$$\begin{align}
\cos2x+\cos2y+\cos2z&=2\cos(x+y)\cos(x-y)+2\cos^2(x+y)-1 \\
&=2\cos(x+y)\bigl(\cos(x-y)+\cos(x+y)\bigr)-1 \\
&=4\cos(x+y)\cos x\cos y-1 \\
&=4\cos(\pi-z)\cos x\cos y-1 \\
&=-4\cos z\cos x\cos y-1 \\
&=-4\cos x\cos y\cos z-1
\end{align}$$
| {
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Given $x \in \mathbb R$ such that $8x^3 - 4x^2 - 4x + 1 = 0$. Find all rationals $p, q, r$ such that $px + q(2x^2 - 1) + r(4x^3 - 3x) = -4$.
Given $x \in \mathbb R$ such that $8x^3 - 4x^2 - 4x + 1 = 0$. Find all rationals $p, q, r$ such that $$\large px + q(2x^2 - 1) + r(4x^3 - 3x) = -4$$
We have that $$8x^3 - 4x^2 - 4x + 1 = 0 \iff (-1) + 2x - 2 \cdot (2x^2 - 1) + 2 \cdot (4x^3 - 3x) = 0$$ and $$px + q(2x^2 - 1) + r(4x^3 - 3x) = -4 \iff (-4) \cdot (-1) + px + q(2x^2 - 1) + r(4x^3 - 3x) = 0$$
This implies that $(p + 8)x + (q - 8)(2x^2 - 1) + (r + 8)(4x^3 - 3x) = 0$.
One solution is that $p + 8 = q - 8 = r + 8 = 0 \implies (p, q, r) = (-8, 8, -8)$
I suspect to there be other solutions of $(p, q, r)$, for example, $$x = \frac{1}{6} \cdot \left[1 - \sqrt[3]{\frac{7}{2}(3\sqrt3i + 1)} + \sqrt[3]{\frac{7}{2}(3\sqrt3i - 1)}\right]$$, according to WolframAlpha, is a solution to the equation $8x^3 - 4x^2 - 4x + 1 = 0$.
Right... So WolframAlpha is untrustworthy, according to Michael Rozenberg, the solutions to $8x^3 - 4x^2 - 4x + 1 = 0$ are $\cos\dfrac{\pi}{7}, \cos\dfrac{3\pi}{7}, \cos\dfrac{5\pi}{7}$, which is correct when plugged into the original equation.
| Define $g(x) = 8x^3 - 4x^2 - 4x + 1$ and
$$\begin{align}h(x) &= px + q(2x^2 - 1) + r(4x^3 - 3x) +4\\
&= 4rx^3 + 2qx^2 + (p-3r)x + (4-q).\end{align}$$
In other terms, your question is: for what $p,q,r$ does $h(\alpha) = 0$, for every $\alpha: g(\alpha)=0$?
Well, $\mathbb C$ is algebraically closed, so $g(x)$ has $3$ (not necessarily distinct) roots; let $\alpha$ be one of them. Let's take a look at its derivative: $g'(x)= 24x^2-8x - 4$. If $\alpha$ has multiplicity greater than $1$, then we know that $g'(\alpha) = 0$. By polynomial long division, we have that $g(x) = d(x)g'(x) + (-\frac{28}{9}x+\frac79)$. So, if $\alpha$ has multiplicity greater than $1$,
$$\overbrace{g(\alpha)}^0 = d(\alpha)\underbrace{g'(\alpha)}_0 + \frac19(-28\alpha + 7) = \frac19(-28\alpha+ 7).$$
So $\alpha = \frac{7}{28} = \frac14$. But $g(\frac14) = -\frac{1}{8}$, so no root has multiplicity greater than one. Thus we have $3$ distinct roots $\alpha_1, \alpha_2$ and $\alpha_3$ to both $g$ and $h$. To prove that your result is the only possible, we'll use a trick to have a polynomial identity: define
$$\begin{align}
l(x) &= rg(x) - 2h(x)\\
&= (-4r-4q)x^2 + (2r-2p)x + (r-8+2q).\\
\end{align}$$
Note that $\operatorname{deg}l\leq 2$. Also, $\alpha_1, \alpha_2$ and $\alpha_3$ are roots. But the number of roots certainly exceeds the degree of this polynomial, so it must be identical to $0$, i.e., all coefficients are $0$. Finally,
$$
\begin{cases}
-4r -4q &= 0\\
2r -2p &= 0\\
r+2q-8 &= 0
\end{cases} \,\,\therefore \boxed{(p,q,r) = (-8,8,-8).}$$
You can check this linear system here, and it guarantees uniqueness.
| {
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"url": "https://math.stackexchange.com/questions/3503008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $\max(x^2y+y^2z+z^2x+xyz)(x^2z+y^2x+z^2y+xyz),$ where $x+2y+3z=4.$
Let $x$, $y$ and $z$ be non-negative numbers such that $x+2y+3z=4.$ Find:
$$\max(x^2y+y^2z+z^2x+xyz)(x^2z+y^2x+z^2y+xyz).$$
I took this problem here: https://dxdy.ru/topic18767-30.html a last post.
This problem is a similar to many contests problems.
On one of Canadians olimpiads was $x^2y+y^2z+z^2x\leq4$ for non-negatives $x$, $y$ and $z$ such that $x+y+z=3.$
Also, $x^2y+y^2z+z^2x+xyz\leq4$ with the same conditions.
My attempts:
For $(x,y,z)=(2,1,0)$ we get a value $8$, which looks as a maximal value.
I solved this problem for $x=\min\{x,y,z\}$ and for $y=\min\{x,y,z\}$.
But for $z=\min\{x,y,z\}$ we need to prove that
$$(x+2y+3z)^6\geq512(x^2y+y^2z+z^2x+xyz)(x^2z+y^2x+z^2y+xyz),$$ which after substitution $x=z+u$, $y=z+v$ gives something very hard:
$$38464z^6+64(473u+1202v)z^5+16(447u^2+3068uv+4092v^2)z^4+$$
$$+32(7u^3+234u^2v+1044uv^2+952v^3)z^3+$$
$$+4(7u^4-200u^3v+808u^2v^2+3040uv^3+2032v^4)z^2+$$
$$+4(9u^5-38y^4v-152u^3v^2+208u^2v^3+592uv^4+288v^5)z+$$
$$+(u-2v)^2(u^4+16u^3v+120u^2v^2+64uv^3+16v^4)\geq0.$$
Thank you!
| Solution by Tran Quoc Anh.
We need to prove
$$8\left(a^{2} b+b^{2} c+c^{2} a+a b c\right)\left(a b^{2}+b c^{2}+c a^{2}+a b c\right) \leqslant \left(\frac{a+2 b+3 c}{2}\right)^{6} .$$
By the AM-GM inequality we have
$$ \text{LHS} \leqslant \left[a^{2} b+b^{2} c+c^{2} a+a b c+2\left(a b^{2}+b c^{2}+c a^{2}+a b c\right)\right]^{2}.$$
and
$$(a+2 b)(b+2 c)(c+2 a)=2\left[\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c\right]+3 a b c$$
$$ \geqslant 2\left[\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c\right],$$
and
$$\begin{aligned}
(a+2 b)(b+2 c)(c+2 a) &=\frac{1}{4} \cdot(a+2 b) \cdot 4(b+2 c) \cdot(c+2 a) \\
& \leq \frac{1}{4}\left[\frac{(a+2 b)+4(b+2 c)+(c+2 a)}{3}\right]^{3} \\
&=\frac{1}{4}(a+2 b+3 c)^{3}.\end{aligned}$$
Therefore
$$\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c \leq \frac{1}{8}(a+2 b+3 c)^{3}.$$
The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to prove that $\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{\cdots}}}}<\sqrt[3] 2$ It's an estimation that I find interesting :
$$\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{1+\frac{1}{5}\sqrt{\cdots}}}}}<\sqrt[\leftroot{-0}\uproot{0}3]{2}$$
I think to evaluate this we need first to calculate the first ones terms .
In fact we have :
$$\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{3}\sqrt{1+\frac{1}{4}\sqrt{1+\frac{1}{5}\sqrt{1+\frac{1}{6}}}}}}\approx 1.259611$$
And :
$$\sqrt[\leftroot{-0}\uproot{0}3]{2}\approx 1.259921$$
But I can't prove that this first decimals of the infinite nested radical are fixed .
Maybe we can use series expansion to prove that with a sequence .
Thanks a lot for your help and your time .
| For $n\in\mathbb N$, let $$a_n = \frac1n\sqrt{1+\tfrac1{n+1}\sqrt{1+\tfrac1{n+2}\sqrt{\dots}}}$$
We are interested in $a_1$. Notice that $$a_1=\sqrt{1+a_2}=\sqrt{1+\tfrac12\sqrt{1+a_3}}=\dots$$
We have $$a_n<\frac1n\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\dots}}}}=\frac\phi n,$$ where $\phi=\frac{1+\sqrt 5}2$ is the golden ratio.
Hence, $$a_1<\sqrt{1+\frac{1}{2} \sqrt{1+\frac{1}{3} \sqrt{1+\frac{1}{4} \sqrt{1+\frac{\phi}{5}
}}}}\overset{\text{Def.}}=c$$
By numerical methods, we can guarantee that $$0.00016183\geq \sqrt[3]2-c\geq 0.00016182$$ and thus that $\sqrt[3]2> c>a_1$, which achieves the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find supremum and infimum for $C={\frac{x^2+1}{x^2+2}: x \in \mathbb{R}}.$ Find supremum and infimum for
$C={\frac{x^2+1}{x^2+2}: x \in \mathbb{R}}.$
We can easy see that $\frac{x^2+1}{x^2+2}$ is bounded from below with $m=\frac{1}{2}.$ This is also infimum. To show this I proceed like this.
Let us assume that $\frac{1}{2}$ is infinum (is not the largest bound for $\frac{x^2+1}{x^2+2}$), i.e there exists $m>\frac{1}{2}$ such that $\frac{x^2+1}{x^2+2}\geq m$. Let us take $x=0$. We have $\frac{0^2+1}{0^2+2}=\frac{1}{2}\geq m>\frac{1}{2}.$vSince we get $\frac{1}{2}>\frac{1}{2}$ which is not true, so we conclude that $\frac{1}{2}$ is infinum.
For supremum, we can see that $\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}$ is bounded from above with $M=1$ and I think this is supremum (the smallest bound). I do not know how to prove it.vI would be grateful for any help.
| $$\frac{1}{2}\leq\frac{x^2+1}{x^2+2}<\frac{x^2+2}{x^2+2}=1.$$
The equality in the right inequality occurs for $x=0$ and since $$\lim_{x\rightarrow+\infty}\frac{x^2+1}{x^2+2}=1,$$ we are done:
$$\inf_{x\in\mathbb R}\frac{x^2+1}{x^2+2}=\frac{1}{2}$$ and
$$\sup_{x\in\mathbb R}\frac{x^2+1}{x^2+2}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral with partial fractions I'm trying to calculate the following integral:
$$I=\int \frac{\arctan(x)}{x^4}dx$$
My steps so far are:
Per partes:
$$\frac{-\arctan(x)}{3x^3}+\int{\frac{1}{1+x^2} \frac{1}{3x^3}}dx
=\frac{-\arctan(x)}{3x^3}+\frac{1}{3}\int{\frac{1}{1+x^2} \frac{1}{x^3}}dx$$
and now I want to do partial fractions. However, with this integral, I fail to do partial fractions. Could you help me?
Thanks
| Without partial fractions
$$
\begin{aligned}
I=\int \frac{\arctan x}{x^4} d x & =-\frac{1}{3} \int \arctan x d\left(\frac{1}{x^3}\right) \\
& \stackrel{IBP}{=} -\frac{\arctan x}{3 x^3}+\frac{1}{3} \underbrace{\int \frac{1}{x^3\left(1+x^2\right)}}_J dx
\end{aligned}
$$
Letting $y=\frac{1}{x} $ yields
$$
\begin{aligned}
J & =\int \frac{1}{\frac{1}{y^3}\left(1+\frac{1}{y^2}\right)}\left(-\frac{d y}{y^2}\right) \\
& =-\int \frac{y^3}{y^2+1} d y \\
& =-\int \frac{y\left(y^2+1\right)-y}{y^2+1} d y \\
& =-\frac{y^2}{2}+\frac{1}{2} \ln \left(y^2+1\right)+C
\end{aligned}
$$
Hence $$
I=\frac{1}{6}\left[-\frac{2 \arctan x}{x^3}-\frac{1}{x^2}+\ln \left(1+x^2\right)-2 \ln |x|\right]+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Simplify trigonometric expression $\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}$
Simplify
$$\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}$$
How can I simplify this trigonometric expression?
Can you explain it if possible?
| Writing $A=2x,B=2y$ and using
http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$\dfrac{1-\cos2(x+y)}{\cos2x-\cos2y}=\dfrac{2\sin^2(x+y)}{-2\sin(x+y)\sin(x-y)}=\dfrac{\sin(x+y)}{-\sin(x-y)}$$
assuming $\sin(x+y)\ne0$
Use https://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 1
} |
Integral of $\frac{1}{(ax^2+bx+c)}$ from $-\infty$ to $\infty$ Can someone spot my error? I am trying to find a general solution for
$$\int_{-\infty}^\infty \frac {1}{ax^2 + bx + c} \, dx $$ using complex analysis.
I substitute a complex variable, $z$, for $x$ then find the roots and take only the root above the real axis
$$z = \frac{-b}{2a} + \frac{\sqrt{4ac - b^2}}{2a}i$$
This tells me that:
$$\int_{-\infty}^\infty \frac 1 {ax^2 + bx + c} \, dx = \int \frac {dz} {\left( z + \frac{b+\sqrt{4ac-b^2}i}{2a} \right) \left(z+\frac{b-\sqrt{4ac-b^2}i}{2a} \right)}$$
Finding the residue at $\ z = \displaystyle\frac {-b}{2a} + \frac{\sqrt {4ac - b^2}}{2a}i$
$$R\left(\frac {-b}{2a} + \frac{\sqrt{4ac - b^2}}{2a}i\right)
= \frac{1}{\left(\frac{-b+i\sqrt{4ac-b^2}}{2a}+\frac{-b+i\sqrt{4ac-b^2}}{2a}\right)} = \frac{1}{\frac{2i\sqrt{4ac-b^2}}{2a}} = \frac{a}{i\sqrt{4ac-b^2}}$$
$$\int_{-\infty}^\infty \frac {1}{(ax^2 + bx + c)}dx= {2\pi i} R = \frac{2a\pi}{\sqrt {4ac-b^2}}$$
but checking some example on wolfram alpha, I can see that the actual value is $\frac{2\pi}{\sqrt {4ac-b^2}}$. My question is where does that $a$ go? What mistake did I make?
Thanks!
| When you factored, you divided by $a$ in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Dividing a rope into three random pieces. Expected longest length A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
This question has been answered here:
Average length of the longest segment
Someone had a solution:
let the cuts be at $X, Y$, with $Y \gt X$:
Image of cut positions
Then each piece is equally likely to be the longest, and the expected length of the longest piece doesn't depend on which piece we choose. Then we can calculate $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$.
We have the three inequalities:
$$X \gt Y-X \implies Y < 2X$$
$$X \gt 1-Y \implies Y > 1-X$$
and, from our setup,
$$Y \gt X$$
These can be represented by the following diagram:
Diagram of inequalities
Then the area satisfying our inequalities is the two triangles A and B. So we wish to find the expected value of $X$ within this area.
The expected value of $X$ in A is $\bar{X}_A = \frac{1}{2}-\frac{1}{3}(\frac{1}{2}-\frac{1}{3}) = \frac{8}{18}$.
The expected value of $X$ in B is $\bar{X}_B = \frac{1}{2}+\frac{1}{3}(\frac{1}{2}) = \frac{4}{6} = \frac{12}{18}$
The area of A is $A_A = \frac{1}{2} \times \frac{1}{2}\times (\frac{1}{2}-\frac{1}{3}) = \frac{1}{24}$.
The area of B is $A_B = \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2} = \frac{1}{8} = \frac{3}{24} = 3 A_A$.
So $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} ) = \frac{\tfrac{8}{18} + 3\left(\tfrac{12}{18}\right)}{4} = \frac{11}{18}$
I understand everything but I got lost when he said calculate expected value of $X$ within triangles. How do you exactly find the expected value of $X$ over the triangle $A$ and $B$? How did he come up with $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$?
| In your problem, $\bar{X}_A$ is the $x$-coordinate of triangle A's center. To calculate it, use that the center of a triangle is situated at $1/3$ of any edge center.
| {
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"url": "https://math.stackexchange.com/questions/3508452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$ Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$
I have that $$(3 + \sqrt{13})^3 = 144 + 40 \sqrt{13} $$ and $$(3 - \sqrt{13})^3 = 144 - 40 \sqrt{13} $$
A cursory look into Bombelli's method led me to the following system of equations:
$$\sqrt[3]{18 + \sqrt{325}} = a + b^{1/2}$$
$$\sqrt[3]{18 - \sqrt{325}} = a - b^{1/2}$$
I am unsure how to solve this system of equations without making a mess of the radicals...I know however that the given cube roots on the LHS of the above system are solutions to the cubic $x^3 + 3x = 36 $
| Since $(3\pm\sqrt{13})^3=144\pm40\sqrt{13}=8(18\pm5\sqrt{13}),$
$\sqrt[3]{18\pm\sqrt{325}}=\sqrt[3]{18\pm5\sqrt{13}}=\dfrac{3\pm\sqrt{13}}2$.
Therefore, $\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=\dfrac{3+\sqrt{13}}2+\dfrac{3-\sqrt{13}}2=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Finding the tangent planes to a torus parallel to $3x+4y-5z=20$. I've been tasked with finding all the points on the torus $S$ given by
$$
\left(6-\sqrt{x^2+y^2}\right)^2+z^2=2
$$
at which the tangent plane is parallel to the plane $P$ given by $3x+4y-5z=20$ (which is, itself, tangential to $S$ at $(3,4,1)$). The method that I have employed gives me another point, but since $S$ is a torus, and $P$ is not parallel to any of $x=0$, $y=0$ or $z=0$, I know there must be at least two more.
The method that I have employed is as follows. Let $f(x,y,z)=\left(6-\sqrt{x^2+y^2}\right)^2+z^2$, then we know that a vector normal to $S$ at $(x_0,y_0,z_0)$ is given by $\nabla f(x_0,y_0,z_0)$. We use this fact, in conjunction with the fact that for two planes to be parallel, their normal vectors must also be parallel, to arrive at the condition
$$
\nabla f(x_0,y_0,z_0)=\lambda\begin{pmatrix}3\\4\\-5\end{pmatrix}
$$
for some scalar, $\lambda\in\mathbb{R}$. We compare the components of these vectors to obtain expressions for $z$ and $x$ in terms of $y$, and then substitute these expressions into the equation for $S$ to find that $y=4$ or $y=28/5$. Since we already have a point at which $y=4$, we discard this value, and hence we have that another tangent to $S$ which is parallel to $P$ is given by $5x/5+8y/5-2z=16$. Graphically, these are the two tangent planes in question
However, as we can see, there is an entire other half of the torus, one which I'm confident has two more points at which the tangent planes are parallel to $P$. How come I have missed these points? Any help is appreciated.
| Note that the planes parallel to $3x+4y-5z=20$ are all of the form $3x+4y-5z=k$ for some $k$.
The gradient is
$$
(-\frac { 2(6-\sqrt {x^2+y^2})x}{\sqrt {x^2+y^2}},-\frac {2 (6-\sqrt {x^2+y^2})y}{\sqrt {x^2+y^2}},2z).
$$
You are looking for points in the torus where the gradient is a scalar multiple of $(3,4,-5)$. This gives you four equations (three from the gradient and one from the torus):
\begin{align}
z&=-5\lambda\\
(6-\sqrt {x^2+y^2})x&=-3\lambda\sqrt {x^2+y^2} \\
(6-\sqrt {x^2+y^2})y&=-4\lambda\sqrt {x^2+y^2}\\
(6-\sqrt {x^2+y^2})^2+z^2&=2 .
\end{align}
Dividing the second by the third you get
$$
\frac{x}{y}=\frac34,
$$
so $y=4x/3$. Then
$$
\sqrt{x^2+y^2}=\sqrt{x^2+\frac{16x^2}{9}}=\frac{5|x|}{3}.
$$
So now the system is
\begin{align}
z&=-5\lambda\\
(6-\frac{5|x|}{3})x&=-5\lambda{|x|} \\
(6-\frac{5|x|}{3})^2+z^2&=2 .
\end{align}
From the first and second equation you now get
$$
z=- {5\lambda} =5\,\frac{(6-\frac{5|x|}{3})x}{{5|x|}{}}=\frac{(6-\frac{5|x|}{3})x}{{|x|}{}}.
$$
Now the last equation is
$$
(6-\frac{5|x|}{3})^2+\left(\frac{(6-\frac{5|x|}{3})x}{{|x|}{}}\right)^2=2,
$$
so
$$
2(6-\frac{5|x|}{3})^2=2.
$$
This gives
$$
x=\pm\frac{18\pm3}5,\ \ \
y=\frac{4x}3=\pm\frac{24\pm4}5,\ \ \ z=\pm1.$$
So you get
*
*$(\tfrac{21}5,\tfrac{28}5,-1)$ and the plane is $3x+4y-5z=40$
*$(3,4,1)$ and the plane is $3x+4y-5z=20$
*$(-\tfrac{21}5,-\tfrac{28}5,1)$ and the plane is $3x+4y-5z=-40$
*$(-3,-4,-1)$ and the plane is $3x+4y-5z=-20$
| {
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"url": "https://math.stackexchange.com/questions/3512389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Possible To Do More Complex Trignometry Problems By Hand? Are there straightforward ways to do the following two problems by hand instead of using a calculator? Any shortcuts (i.e. the denominator of question 5?
Or is the expectation for these types of problems to just do them on the calculator ?
| $\cos \dfrac{23\pi}{4}
= \cos \left( 6\pi - \dfrac{\pi}{4} \right)
= \cos \left(- \dfrac{\pi}{4} \right)
= \dfrac{1}{\sqrt 2}$
$\sin \dfrac{15\pi}{4}
= \sin \left( 4\pi - \dfrac{\pi}{4} \right)
= \sin\left(- \dfrac{\pi}{4} \right)
= -\dfrac{1}{\sqrt 2}$
$$\cos \dfrac{23\pi}{4} - \sin \dfrac{15\pi}{4} = \sqrt 2$$
$$\dfrac{\cos 35^\circ}{\sin 20^\circ \cos 35^\circ + \cos 20^\circ \sin 35^\circ}
=\dfrac{\cos 35^\circ}{\sin 55^\circ}
= \dfrac{\cos 35^\circ}{\cos (90^\circ - 55^\circ)}
= 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{2n+1}\left ( \frac{1}{2} \right )^{n}$ Find sum of the series:
$\sum\limits_{n=0}^{\infty}\frac{1}{2n+1}\left ( \frac{1}{2} \right )^{n}$. I just thought that this series is the derivation of the original one, but it is not.
Then I know that: $\sum\limits_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$, so in my case $x=\frac{1}{2}$, but I really do not know, how I should get the fraction.
Can anyone please help me?
| \begin{align}
& \sum_{n=0}^\infty \frac{(1/2)^n}{2n+1} = \sum_{n=0}^\infty \frac{(1/\sqrt2\,)^{2n}}{2n+1} \\[8pt]
= {} & \sqrt 2 \sum_{n=0}^\infty \frac{(1/\sqrt2\,)^{2n+1}}{2n+1} \\[10pt]
& \frac d {dx} \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}
= \sum_{n=0}^\infty x^{2n} = \frac 1 {1-x^2}.
\end{align}
So you need an antiderivative of
$$
\frac 1 {1-x^2} = \frac 1 {(1-x)(1+x)} = \frac A {1-x} + \frac B {1-x}.
$$
You need to complete the partial-fractions problem. To find the value of the constant of integration, use the fact that when $x=0$ in the sum above, then the sum is $0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For $w\in\mathbb{C}$, prove that $w^3$ lies on the line passing through 1 and $w$ in the complex plane iff $\Re(w)=-\frac12$. For $w\in\mathbb{C}$, prove that $w^3$ lies on the line passing through 1 and $w$ in the complex plane if, and only if, $\Re(w)=-\frac12$.
I came across this fact while messing about with spirals of powers of a complex number. I've tried parametrising the straight line with a real number $t$, so, for example, $z(t)=(1-t)+tw$. But alas to no avail! Any insights are appreciated.
Here is a link to a Desmos graph showcasing the phenomenon:
https://www.desmos.com/calculator/gsugcdlkc9
| A different proof:
Let $\omega = x + iy\in\mathbb{C}, y \neq 0, \omega \neq 1$
$\omega^3 = (x^3 - 3xy^2) + i(3x^2y - y^3)$
The three points $\omega \equiv(x, y)$, $(1,0)$ and $\omega^3 \equiv (x^3 - 3xy^2, 3x^2y - y^3)$ are collinear, iff the following determinant is $0$
\begin{vmatrix}
x & y & 1 \\
1 & 0 & 1 \\
x^3 - 3xy^2 & 3x^2y - y^3 & 1
\end{vmatrix}
Simplifying,
$y(2x+1)[(x-1)^2 + y^2)] = 0$
Since $y \neq 0$, this is possible iff $ x = -\frac{1}{2}$
A shorter proof:
Let $\omega=x + iy, y \neq 0, \omega \neq 1$
$1,\omega,\omega^3$ are collinear iff $\frac{\omega^3-\omega}{\omega-1} = \omega^2 + \omega + 1$ is real.
Since $\omega^2 + \omega + 1 = (x^2 + x - y^2 + 1) + i y(2x + 1)$, we must have $y(2x + 1) = 0$
Now $y \neq 0 \implies x = -\frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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isn't right to prove that $\sqrt{2}$/4 is irrational number? Assume $\sqrt{2}$/4 is rational number.
rational number have p/q in the lowest term.
\begin{align}\sqrt{2}/4 = p/q\\
\sqrt{2}=4p/q\\
2=16p^2/q^2\\
2q^2=16p^2\\
2q^2=2(8p^2) \\
q^2=2(4p^2)\\
\end{align}
Then we know that q is even number according to if $q^2$ is even then $q$ is even.
Also,we have $2q^2$=$16p^2$ in the above,we know that $16p^2$ is a even number.
So we can construct the form in the below:
\begin{align}2(16a)=16p^2\\p^2=2a
\end{align}
Then we know that p is even number according to if $p^2$ is even then $p$ is even.
Since we assumed that $\sqrt{2}$/4 is in the lowest form of rational number,but the result showed p and q have the common factor 2.It causes to contradiction and means that our assumption is wrong.
Therefore,$\sqrt{2}$/4 is irrational number.
| Hint $:$ If $\frac {\sqrt 2} {4}$ is a rational number then so is $\sqrt 2 = \frac 1 2 \times \frac {4} {\sqrt 2}.$ Is $\sqrt 2$ a rational number?
| {
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Subspaces of the projection - matrix I have some problems with the following exercise:
Let $v_1 = [1,1,0]^T,v_2 = [1,0,1]^T, v_3 = [0,0,1]^T$. Find the dual basis $\bar{v}^1,\bar{v}^2,\bar{v}^3$ such that $\bar{v}^{i}(v_{j})= \delta^{i}_j$. Calculate $v_i \bar{v}^i$. Along which subspace are these projections and on which subspace are they projected into?
I have found the dual basis: $\bar{v}^1 = [\frac{1}{2},\frac{1}{2}, -\frac{1}{2}],\bar{v}^2 = [\frac{1}{2},-\frac{1}{2},\frac{1}{2}], \bar{v}^3 = [-\frac{1}{2},\frac{1}{2}, \frac{1}{2}]$ and the projection matrices: \begin{align*}
v_1 \bar{v}^1 = \left[
\begin{array}{ccc}
\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}\\
\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}\\
0& 0& 0
\end{array}
\right],\quad
v_2 \bar{v}^2 = \left[
\begin{array}{ccc}
\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\\
0& 0& 0\\
\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\\
\end{array}
\right],
\quad
v_3 \bar{v}^3 = \left[
\begin{array}{ccc}
0& 0& 0\\
-\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\
-\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\
\end{array}
\right].
\end{align*}
I don't know how from the projection operators one can determine along which subspace are these projections and on which subspace are they projected into?
| In general if you have a rank one matrix of the form $a\otimes b$, its annihilator is the subspace of vectors orthogonal to $b$ and its range is the subspace spanned by $a$. This is because $(a\otimes b)(x)=a(b\cdot x)$ by definition.
Another way: From your matrices you can tell that any vector orthogonal to $(1,1,-1)$ transforms to the origin, and the image of any vector is parallel to $(1,1,0)$ (being a linear combination of the columns).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Definite Integral of $(a^2-x^2)^\frac{3}{2}$
Prove the following:
$$\frac{4c}3\int\limits_0^a\left(a^2 - x^2\right)^\frac32\,\mathrm dx = \frac{\pi a^4c}4.$$
Taking $x = a\sin\theta$, how will the limit change?
| If you do $x=a\sin\theta$ then since $x$ goes from $0$ to $a$, $\theta$ goes from $0$ to $\frac\pi2$. So,\begin{align}\int_0^a(a^2-x^2)^{\frac32}\,\mathrm dx&=\int_0^{\frac\pi2}(a^2-a^2\sin^2\theta)^{\frac32}a\cos\theta\,\mathrm d\theta\\&=a^4\int_0^{\frac\pi2}\cos^4\theta\,\mathrm d\theta\\&=\frac{3a^4\pi}{16}.\end{align}The last equality comes from the fact that$$\cos^2\theta=\frac{\cos(2\theta)+1}2$$and therefore\begin{align}\cos^4\theta&=\frac{\cos^2(2\theta)+2\cos(2\theta)+1}4\\&=\frac{\frac{\cos(4\theta)+1}2+2\cos(2\theta)+1}4\\&=\frac{\cos(4\theta)+4\cos(2\theta)+3}8.\end{align}
| {
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There are $20$ books on Algebra & Calculus in our library. Find the condition of maximum selections each of which consist of $5$ books on each topic. There are $20$ books on Algebra & Calculus in our library. Prove that the greatest number of selections each of which consists of $5$ books on each topic is possible only when there are $10$ books on each topic in the library.
My attempt is as follows:-
Let there are $x$ books of Algebra and $20-x$ books of Calculus
So number of selections each of which consists of $5$ books on each topic= $\displaystyle{x\choose 5}{20-x\choose 5}$
$$y=\dfrac{x(x-1)(x-2)(x-3)(x-4)(20-x)(19-x)(18-x)(17-x)(16-x)}{5!\cdot5!}$$
Let's take pairs $x(20-x),(x-1)(19-x),(x-2)(18-x),(x-3)(17-x),(x-4)(16-x)$
Let's find the maximum value of $(x-1)(19-x)$
$$y=19x-x^2-19+x$$
$$y=-x^2+20x-19$$
$$\dfrac{dy}{dx}=-2x+20$$
$$2x=20,x=10$$
In the same way all the other pairs will get maximum value at $x=10$
So can we say that maximum value of the expression will be at $x=10$? Is there any better approach than this?
| Since $x$ will go from 5 to 15, ${x \choose 5}$ will increase while ${20-x \choose 5}$ will decrease as $x$ increases. You can intuitively arrive at the conclusion that the product will be maximum at $x=10$.
| {
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When $\frac {a^3-b^3}{a^2-b^2}$ is an integer? If $\frac {a^3-b^3}{a^2-b^2}$ is an integer, then supposing $a-b \ne 0$ we have that also$\frac {a^2+ab+b^2}{a+b}$ is an integer.
For which $a, b\in\mathbb Z$, the fraction $\frac {a^2+ab+b^2}{a+b}$ is an integer?
| Equation, $(a^2+ab+b^2)=p(a+b)$
where "p' is integral.
while solution given by "Jack D'Aurizio" is nice & since
'OP' needs $(a,b)$ to be integer's there is a
fraction $(k^2/d)$ to be taken care of in his solution.
If instead we take, $(a,b)=[d(k+1),dk(k+1)]$ then we get:
$p=d(k^2+k+1)$
For, $(d,k)=(5,2)$ we get:
$(a,b,p)=(15,30,35)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Angle Bisector Theorem Let $\triangle ABC$ be a triangle with angle bisector $CL$ $(L\in AB)$ and median $CM$ $(M\in AB)$. I should find $LM$ if $BC=a,AC=b,AB=c(a>b)$.
By the angle bisector theorem, $\dfrac{AL}{BL}=\dfrac{AC}{BC}$ or $AL=\dfrac{AC\cdot BL}{BC},BL=\dfrac{AL\cdot BC}{AC}$. We know $LM=AM-AL=BL-BM$ and I am not sure how to approach the problem further.
| Since $$\frac{AL}{LB}=\frac{b}{a},$$ we obtain:
$$AL=\frac{bc}{a+b}$$ and
$$LM=AM-AL=\frac{c}{2}-\frac{bc}{a+b}=\frac{c(a-b)}{2(a+b)}.$$
We can get $AL$ by the following way:
$$\frac{AL}{c-AL}=\frac{b}{a}$$ or
$$\frac{c-AL}{AL}=\frac{a}{b}$$ or
$$\frac{c}{AL}-1=\frac{a}{b}$$ or
$$\frac{c}{AL}=1+\frac{a}{b}$$ or
$$\frac{c}{AL}=\frac{a+b}{b}$$ or
$$\frac{AL}{c}=\frac{b}{a+b}$$ or
$$AL=\frac{bc}{a+b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528171",
"timestamp": "2023-03-29T00:00:00",
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How do I find the length of this curve? I have a curve $ y = \frac{3}{4}x^\frac{4}{3} - \frac{3}{8}x^\frac{2}{3} + 7$. I’m required to find out the length of this curve from $x=1$ to $x=8$.
Using the formula of length of curve,I am trying to evaluate the following integral —
$\int_1^8 \sqrt{1 + (x^\frac{1}{3} - \frac{1}{4x^\frac{1}{3}})^2} dx$ but I’m not sure how to proceed further. Please help.
| Hint: We note $x^{1/3}$ appearing twice in the integral. While not a sure-fire solution in general, here this works out partly: make the $u$-substitution $u=x^{1/3}$. Following this, what's under the radical will be a perfect square trinomial, and the result becomes easier to obtain.
Solution:
We wish to calculate
$$\mathcal I = \int_1^8 \sqrt{1 + \left(x^{1/3} - \frac{1}{4x^{1/3}} \right)^2} dx$$
Let $u=x^{1/3}$. Then $du = \frac 1 3 x^{-2/3}dx = \frac 1 3 u^{-2}dx$, giving us $dx=3u^2$. We change our bounds by noting that $u(1) = 1, u(8) = 2$. Thus,
$$\mathcal I = 3 \int_1^2 \sqrt{1 + \left(u - \frac{1}{4u} \right)^2} \cdot u^2du$$
Expand the square term in the radical:
$$\mathcal I = 3\int_1^2 \sqrt{u^2 - 2u\frac{1}{4u} + \frac{1}{16u^2}+1} \cdot u^2du$$
Simplify, and bring the $u^2$ term into the radical by noting $u^2 = \sqrt{u^4}$:
$$\mathcal I =3 \int_1^2 \sqrt{u^6 + \frac 1 2 u^4 + \frac{1}{16} u^2} \; du$$
Now, if we hope there is some stroke of luck, the radical will be a perfect square trinomial, so we test this theory. To see if a trinomial is a perfect square, note they take the form $a^2 + 2ab + b^2 = (a+b)^2$, so we work with the square roots of the first term ($u^3$) and the last ($u/4$) and test:
$$\left( u^3 + \frac u 4 \right)^2 = (u^3)^2 + 2u^3 \frac u 4 + \left( \frac u 4 \right)^2 = u^6 + \frac{u^4} 2 + \frac{u^2}{16}$$
Success!
$$\mathcal I = 3\int_1^2 u^3 + \frac u 4 \; du$$
Being a polynomial, evaluating it from this point will be easy for you.
| {
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How to prove that integral $\iint_{x\ge1,~y\ge1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy$ is divergent? How to prove that integral $$\iint\limits_{x\ge1,~y\ge1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy$$ is divergent?
Let $x=1+r\cos\varphi$, $y=1+r\sin\varphi$, then
$$\iint\limits_{x\ge1,~y\ge1}\dfrac{x^2-y^2}{(x^2+y^2)^2}\,dxdy=\iint\limits_{\substack{0\le\varphi\le\frac{\pi}2\\r\ge0}}\dfrac{r^3\cos2\varphi+2r^2(\cos\varphi-\sin\varphi)}{(r^2+2r(\cos\varphi+\sin\varphi)+2)^2}\,drd\varphi.$$
Because
$$\dfrac{|r^3\cos2\varphi+2r^2(\cos\varphi-\sin\varphi)|}{(r^2+2r(\cos\varphi+\sin\varphi)+2)^2}\ge\dfrac{|r^3\cos2\varphi-2r^2|}
{(r^2+4r+2)^2}\ge\dfrac{r^3|\cos2\varphi|-2r^2}{(r^2+4r+2)^2},$$
we get that the integral $$\iint\limits_{x\ge1,~y\ge1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy$$ will diverge if the integral diverges
\begin{multline*}
\iint\limits_{\substack{0\le\varphi\le\frac{\pi}2\\r\ge0}}\dfrac{r^3|\cos2\varphi|-2r^2}{(r^2+4r+2)^2}\,drd\varphi=\\
=\iint\limits_{\substack{0\le\varphi\le\frac{\pi}2\\r\ge0}}\dfrac{r^3|\cos2\varphi|}{(r^2+4r+2)^2}\,drd\varphi-\iint\limits_{\substack{0\le\varphi\le\frac{\pi}2\\r\ge0}}\dfrac{2r^2}{(r^2+4r+2)^2}\,drd\varphi=\\
=\int\limits_0^{\pi/2}|\cos2\varphi|\,d\varphi\int\limits_0^{+\infty}\dfrac{r^3\,dr}{(r^2+4r+2)^2}-\int\limits_0^{\pi/2}d\varphi\int\limits_0^{+\infty}\dfrac{2r^2\,dr}{(r^2+4r+2)^2}=\\
=\int\limits_0^{+\infty}\dfrac{r^3\,dr}{(r^2+4r+2)^2}-\frac{\pi}2\int\limits_0^{+\infty}\dfrac{2r^2\,dr}{(r^2+4r+2)^2}.
\end{multline*}
What follows is obvious.
| The integral does not exist (converge) in the sense of Lebesgue, since
$$\int_{[1,\infty)^2} \left|\frac{x^2-y^2}{(x^2+ y^2)^2} \right| = + \infty$$
The correct way to prove this when changing to polar coordinates is to consider the absolute value of the integrand. Since $\cos 2\varphi$ changes sign over the interval $[0,\pi/2]$, there is sufficient cancellation for the iterated improper integral to converge.
However, we can show that the iterated integrals converge directly without changing variables-- albeit to values with different signs depending on the order.
Note that
$$\frac {x^2-y^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right) = -\frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right), $$
and so
$$\begin{align}\int_1^\infty \left(\int_1^\infty \frac {x^2-y^2}{(x^2+y^2)^2}\, dx\right)\, dy &= \int_1^\infty \left(\int_1^\infty -\frac{\partial}{\partial x}\left(\frac {x}{x^2 + y^2}\right)\, dx\right)\, dy \\ &= \int_1^\infty \left.\frac{-x}{x^2+y^2}\right|_{x = 1}^{x = \infty} \, dy\\ &= \int_1^\infty \frac{dy}{1+y^2} \\ &= \frac{\pi}{4}\end{align}$$
By interchanging $x$ and $y$ it is easy to see that the iterated integral in reverse order takes the value $- \pi/4$.
Absolute divergence
We have
$$\int_{[1,\infty)^2}\left|\frac{x^2-y^2}{(x^2+ y^2)^2} \right| \geqslant \int_0^{\pi/2} \int_{\sqrt{2}}^R \frac{|\cos 2 \varphi|}{r} \, dr \, d \varphi = \int_0^{\pi/2}|\cos 2 \varphi| \, d \varphi \int_{\sqrt{2}}^R \frac{dr}{r}$$
and the RHS diverges as $R \to \infty$. Note that the iteration of integrals on the RHS is independent of order as the integrand is continuous and the region of integration is bounded.
| {
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Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$
Given that $\frac{1}{1-x}=\sum^{\infty}_{n=0}x^n$, what is the series representation of $\ln\left(\frac{1+x}{1-x}\right)$?
Differentiating $\ln\left(\frac{1+x}{1-x}\right)$ results in: $\frac2{(1-x)^2}$. This means that $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$
Can I now say that $\ln\left(\frac{1+x}{1-x}\right)=2\left(\sum^{\infty}_{n=0}\int x^n\right)^2$ ? This would get: $2\sum^{\infty}_{n=0}\frac{x^{2(n+1)}}{(n+1)^2}$.
I'm assuming the answer is wrong because the answer key did not agree with me. Did I mess up somewhere in this problem?
The answer key only doubles the $x$ in this step: $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$ so instead of $2\int\frac1{(1-x)^2}\ dx$ they get just $2\int\frac1{1-x^2}\ dx$. Why is this correct? (or is it incorrect?)
| Differentiate:
$$f'(x)=\left ( \ln \frac {1+x}{1-x} \right)'=\frac {1-x}{1+x} \frac {2}{(1-x)^2} $$
$$f'(x)=\frac 2 {1-x^2}$$
$$f'(x)= 2\sum_{n=0}^\infty x^{2n}$$
Integrate:
$$\boxed {f(x)=2\sum_{n=0}^\infty \dfrac { x^{2n+1}}{2n+1}}$$
| {
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Evaluate limit of the form $0^{\infty}$ $$\lim_{n\to\infty}\left(\frac{n^4-3\cdot n^3-n^2+2\cdot n-1}{n^5+n^4-n^3-3\cdot n^2-3\cdot n+1}\right)^{\frac{6\cdot n^5-2\cdot n^4-2\cdot n^3+n^2-2\cdot n}{9\cdot n^4-2\cdot n^3+n^2+3\cdot n}}$$
It is $$\lim_{n\to \infty}\left(\frac{1}{n}+o\left(\frac{1}{n^2}\right)\right)^{(n+o(1))}$$
How can I bring it to a form that I can compute?
| There is no hesitation with $0^\infty$, which can be read as $\dfrac1{\infty^\infty}$.
| {
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Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line Section 2.5 #14
Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line.
Okay, so having a horizontal tangent line at a point on the graph means that the slope of that tangent line is zero. The derivative of a function is another function that tells us the slope of the tangent line at any given point on the graph of the original function.
Thus, to find where the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line, we need to take the derivative, set it equal to zero, and solve for $x$. This will give us the $x$-coordinate of where the graph of $f(x)$ has a horizontal tangent line. To find the corresponding $y$ value, we plug the $x$ value that we found into the original equation.
In this problem, when we plug the $x$ value we find into the original equation, we get an imaginary number, which means that no point on the graph of $f(x)$ has a horizontal tangent line, and thus our answer is DNE, does not exist. Let's go through the motions!!!
$f(x) = \sqrt{8x^2+x-3}=(8x^2+x-3)^{1/2}$
$f'(x) = \frac{d}{dx}(8x^2+x-3)^{1/2}$
Time do the chain rule!!!
$$\begin{align}
f'(x) &= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$
Alright, we have our derivative. We want to find horizontal tangent lines, so we set this equal to zero and solve for $x$
$$0 = \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}$$
multiplying both sides of the equation by $2(8x^2+x-3)^{1/2}$ we get
$0 = (16x+1)$
And thus $x = \frac{-1}{16}$
Now, we plug this value into the original equation to get the corresponding $y$ value, because remember, we are looking for a point on the graph where the horizontal line is tangent, so our answer will be in $(x,y)$ format, is it exists, (which in this case, it won't)..
$f(\frac{-1}{16}) = \sqrt{8(\frac{-1}{16})^2+\frac{-1}{16}-3}$
But $8(\frac{-1}{16})^2+\frac{-1}{16}-3<0$, so taking its square root will give us an imaginary number. Thus the answer is DNE
| We know the answer by calculus. Here an alternative method is explored, based on analytic geometry.
Recognizing the graph as (part of) a conic curve, we square both sides of the equation and bring the result into standard form:
$8x^2-y^2+x-3=0$
The quadratic terms factor as $(2\sqrt2x+y)(2\sqrt2x-y)$ so the conic is, of course, a hyperbola, and the full equation of the hyperbola will have the product form
$(2\sqrt2x+y+a)(2\sqrt2x-y+b)=c$
Expanding the left side and matching like terms to the standard form equation rendered above leads to
$(2\sqrt2)(a+b)=1$ matching linear terms in $x$
$b-a=0$ matching linear terms in $y$
$ab-c=-3$ matching constant terms
The two linear equations give $a=b=1/(4\sqrt2)$, then the last equation gives $c=97/32>0$. Thus
$(2\sqrt2x+y+(1/(4\sqrt2)))(2\sqrt2x-y+(1/(4\sqrt2)))=97/32$
We now draw the asymptote given by each factor on the left being set to zero; then since their product is actually positive the hyperbola must lie in the regions where the factors have identical signs. This corresponds to the shaded regions in the sketch below (not drawn to scale):
We see that the horizontal line through the center cuts through the allowed quadrants and thus through the actual hyperbola, and then any other horizontal line similarly cuts through. There cannot be any horizontal tangents because the hyperbola ends up in the wrong pair of "quadrants" defined by its asymptotes.
To have the hyperbola in the right quadrants and get a horizontal tangent we would have needed $c$ to be negative. This would have been equivalent to identifying a nonzero real value of $y$ at the zero derivative point rendered by calculus. Instead of that happy ending, the actual positive value of $c$ corresponds to an imaginary value of $y$ instead and thus no (real) horizontal tangent.
| {
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How would you approach this question on definite integral?
$$\int_{\frac \pi 4}^{\frac \pi 2}\left(2\cos\left(\frac x 2\right) - e^{-x}\right)^2dx$$
I have tried expanding it and then do the integral but it leads to $4 cos(\frac{x}{2})e^{-x}$ in the middle term of the expansion.
Am i missing something? Is there a "trick" to this? If so how would you approach the question? What are the things that i should keep in mind solving this kind of question? PS. the answer to this is $1.41$
| $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(2\cos\left(\frac{x}{2}\right)-e^{-x}\right)^{2}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(4\cos^{2}\left(\frac{x}{2}\right)+e^{-2x}-4\cos\left(\frac{x}{2}\right)e^{-x}\right)dx$$$$=4\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos^{2}\left(\frac{x}{2}\right)dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-2x}dx-4\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos\left(\frac{x}{2}\right)e^{-x}dx$$$$=2\color{green}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1+\cos\left(x\right)dx}+\color{blue}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-2x}dx}-4\color{red}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx}$$
For the red part (where you have a problem with) using integration by parts we have:
$$\text{I}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx$$$$=-e^{-x}\cos \left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}-\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\sin\left(\frac{x}{2}\right)dx$$$$=-e^{-x}\cos\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}-\frac{1}{2}\left[-e^{-x}\sin\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}+\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx\right]$$$$=-e^{-x}\cos\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}+\frac{1}{2}e^{-x}\sin\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}-\frac{1}{4}\underbrace{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx}_\textrm{I}$$
Hence
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos\left(\frac{x}{2}\right)e^{-x}dx=\frac{4}{5}\left(-e^{-x}\cos\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}+\frac{1}{2}e^{-x}\sin\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}\right)\simeq \color{red}{{0.208396}
}$$
For the blue part setting $-2x \mapsto u$ follows:
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-2x}dx=\frac{1}{2}\int_{-\pi}^{-\frac{\pi}{2}}e^{u}du=\frac{1}{2}\left(e^{\left(-\frac{\pi}{2}\right)}-e^{\left(-\pi\right)}\right)\simeq \color{blue}{0.0823328290435}$$
For the green part using some elementary integrals follows:
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1+\cos\left(x\right)dx=\frac{\pi}{2}-\frac{\pi}{4}+\sin\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{4}\right)\simeq \color{green}{1.07829138221}$$
Summing the results gives the final answer:
$$2\left(1.07829138221\right)+\left(0.0823328290435\right)-4\left({0.208396}
\right)=1.4053315934635
$$
Which is the the answer you mentioned.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the minimum of $abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$ when $ab+bc+cd+da+ac+bd=6$ If $a,b,c,d\geq0$ are such that $ab+bc+cd+da+ac+bd=6$, then what is the minimum value of $$f(a,b,c,d)=abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$$ ?
My attempts: I think the minimum $5$ is obtained at $a=b=c=d=1$. I tried the Hölder inequality: $$f(a,b,c,d)\geq abcd + (1+\sqrt{a b c d})^2,$$ but the last term is $1<5$ for $d=0$.
Also, direct AM-GM on the two terms of $f$ did not work either. What to do?
Remark: This problem is from AoPS. There, it is also asked what happens for $ab+bc+cd+da+ac+bd=7$, which seems even harder.
| Preliminary remark. I just noticed that we can use the square root algorithm for polynomials to show in general that $$(a^2 + 1) (b^2 + 1) (c^2 + 1) (d^2 + 1) - (a b + b c + c d + d a +
a c + b d - a b c d - 1)^2=(a + b + c +d- a b c - a b d - a c d - b c d)^2\geq0$$
which immediately implies that $f(a,b,c,d)+1\geq a b + b c + c d + d a +
a c + b d$ for all $a,b,c,d\geq0$ with equality if and only if $$d=\frac{a+b+c-abc}{a b+a c+b c-1}.$$
But I noticed this too late, so here is some tedious manual work for you, which basically executes the above idea for the special cases $a b + b c + c d + d a +
a c + b d=6$ and $7$.
Tedious calculations:
Indeed the minimum is achieved at $5$. We need to prove $$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}\geq 5-abcd.$$ By direct AM-GM, we get $6=ab+bc+cd+da+ac+bd\geq 6\sqrt{abcd}$ so that $abcd\le1<5$. So we can square both sides of the inequality to prove and we now have to prove $$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\geq(5-abcd)^2.$$ Since $6=ab+bc+cd+da+ac+bd$, we get $d=\frac{6-a b-a c-b c}{a+b+c}$ and so we have to prove $$\left(a^2+1\right) \left(b^2+1\right) \left(c^2+1\right) \left(\frac{(-a b-a c-b
c+6)^2}{(a+b+c)^2}+1\right)-\left(5-\frac{a b c (-a b-a c-b
c+6)}{a+b+c}\right)^2\geq0$$
which is true because the left-hand side (amazingly!) equals $$\frac{\left(6 + a^2 - 5 a b + b^2 + a^2 b^2 - 5 a c - 5 b c + a^2 b c +
a b^2 c + c^2 + a^2 c^2 + a b c^2 + b^2 c^2\right)^2}{(a+b+c)^2}.$$
If we replace the condition by $7=ab+bc+cd+da+ac+bd$, a very similar approach works: Indeed I claim that the minimum of $f(a,b,c,d)$ is $6$ with that side constraint. Again by direct AM-GM, $a b c d \le \left(\frac{ab+bc+cd+da+ac+bd}6\right)^2=\frac{49}{36}<6$ so we can subtract $abcd$, square both sides, and need to prove $$(1 + a^2) (1 + b^2) (1 + c^2) (1 + d^2)-(6 - a b c d)^2\geq 0.$$
By the side constraint we have $d=\frac{7-a b-a c-b c}{a+b+c}$. Substituting this we need to prove $$\left(a^2+1\right) \left(b^2+1\right) \left(c^2+1\right) \left(\frac{(-a b-a c-b
c+7)^2}{(a+b+c)^2}+1\right)-\left(6-\frac{a b c (-a b-a c-b
c+7)}{a+b+c}\right)^2\geq 0.$$
But very nicely, the left-hand side can again be written as $$\frac{\left(7 + a^2 - 6 a b + b^2 + a^2 b^2 - 6 a c - 6 b c + a^2 b c +
a b^2 c + c^2 + a^2 c^2 + a b c^2 + b^2 c^2\right)^2}{(a+b+c)^2}.$$
We have equality if and only if the side constraint is fulfilled and $$7 + a^2 - 6 a b + b^2 + a^2 b^2 - 6 a c - 6 b c + a^2 b c +
a b^2 c + c^2 + a^2 c^2 + a b c^2 + b^2 c^2=0.$$
Note that this is a quadratic polynomial in each variable if the other variables are fixed. So for given $a$ and given $b$, we can solve for $c$. As an example, $a=1,b=1,c=\frac12$ leads to $d=2$ and indeed equality as $$f\left(1,1,\frac12,2\right)=6.$$
| {
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"url": "https://math.stackexchange.com/questions/3540491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$ Prove $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$
Just want to see if my "reasoning is sound.
1) Showing $1 < \frac{1 + \sqrt{5}}{2}$
Consider $\frac{1}{2}$:
$$\frac{1}{2} < \frac{1}{2} + \frac{1}{2} < \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2} $$
2) Showing $\frac{1 + \sqrt{5}}{2} < 2$
We know $$\sqrt{5} < 3 \\ \Rightarrow 1 + \sqrt{5} < 4 \\ \Rightarrow \frac{1 + \sqrt{5}}{2} < 2$$
Therefore: $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$
| Another way
As $\varphi>0,$
$\varphi=1+\dfrac1{\varphi}>1$
$\implies1+\dfrac1{\varphi}<1+1$
| {
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For real numbers $x,y$ and $z$ if $9x^2+16y^2+4z^2=108$ and $6xy+4yz+3zx=54$. Find value of $x^2+y^2+z^2$. Question
For real numbers $x,y$ and $z$ if $9x^2+16y^2+4z^2=108$ and $6xy+4yz+3zx=54$. Find value of $x^2+y^2+z^2$.
My calculations
I found
$$(3x+4y+2z)^2=108+4×54=324$$
I have no clue how to proceed further .
But for check of validity of question I let wolfram solve it . But results left me more puzzled .
Now how can I get to this result .
There is one more picture
Now why I don't get same result $x^2+y^2+z^2$
Feel free to edit , comment and advise . Thank You
| Let $a=3x,b=4y,c=2z$. Then we have $a^2+b^2+c^2=108$ and $\frac{ab+bc+ac}2=54$ i.e. $$a^2+b^2+c^2=ab+bc+ac=108,$$ which implies $$a^2+b^2+c^2-ab-bc-ac=\frac{(a-b)^2+(b-c)^2+(c-a)^2}2=0$$ and thus $a=b=c$ and since $a^2+b^2+c^2=108$, we have $a^2=b^2=c^2=\frac{108}3=36$ so that $$x^2+y^2+z^2=a^2\cdot\left(\frac19+\frac1{16}+\frac14\right)=\frac{61}4.$$
| {
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"url": "https://math.stackexchange.com/questions/3541765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum value of $x$ s.t. $\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$ Let $x,y\in \mathbb{R}$ such that $$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$. Find the minimize value of $x$.
[Edit by Michael Rozenberg] I tried to use AM-GM to retire the radical but failed:
$$27=\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\geq2\sqrt[4]{\left(\frac{x+y}{2}\right)^3\left(\frac{x-y}{2}\right)^3}=$$
$$=2\sqrt[4]{\frac{(x^2-y^2)^3}{64}}=\sqrt[4]{\frac{(x^2-y^2)^3}{4}}.$$
Help me
| For any two real numbers $a$ and $b$, we have
$$(a^2+b^2)^3 \geq (a^3+b^3)^2$$
because
$$(a^2+b^2)^3 - (a^3+b^3)^2 = a^2b^2[2a^2+2b^2+(a-b)^2]\geq 0$$
Setting $a=\sqrt{\frac{x+y}{2}}$ and $b=\sqrt{\frac{x-y}{2}}$, we get:
$$\left[\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\right]^2 \leq \left(\frac{x+y}{2}+\frac{x-y}{2}\right)^3=x^3$$
Therefore $x^3 \geq 27^2\Rightarrow x\geq 9$. This minimum is attained when $x=y=9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$? How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$?
\begin{align*}
A_1 & = \left\lfloor{\frac{200}{3}} \right\rfloor = 66 && \text{(divisible by $3$)}\\
A_2 & = \left\lfloor{\frac{200}{5}} \right\rfloor = 40 && \text{(divisible by $5$)}\\
A_3 & = \left\lfloor{\frac{200}{2}} \right\rfloor = 100 && \text{(odd)}\\
| A_1 \cap A_2 | & = \left\lfloor{\frac{200}{3 \cdot 5}}\right\rfloor = 13\\
| A_1 \cap A_3 | & = \left\lfloor{\frac{200}{3 \cdot 2}}\right\rfloor = 33\\
| A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2}} \right\rfloor= 20\\
| A_1 \cap A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2 \cdot 3}}\right\rfloor = 6
\end{align*}
Therefore, by the principle exclusion inclusion theorem
$= 66 + 40 + 100- (13 + 33 + 20) + 6 = 146$
Is this logically right?
| Your answer is incorrect. While it is true that there are positive odd integers less than or equal to $200$, what you have actually calculated with the expression
$$\left\lfloor \frac{200}{2} \right\rfloor$$
is the number of even integers less than or equal to $200$. It just so happens that $200 - 100 = 100$. Similarly, there are
$$\left\lfloor \frac{200}{2 \cdot 3} \right\rfloor = 33$$
positive integers less than or equal to $200$ that are divisible by both $2$ and $3$,
$$\left\lfloor \frac{200}{2 \cdot 5} \right\rfloor = 20$$
positive integers less than or equal to $200$ that are divisible by both $2$ and $5$, and
$$\left\lfloor \frac{200}{2 \cdot 3 \cdot 5} \right\rfloor = 6$$
positive integers less than or equal to $200$ that are divisible by $2$, $3$, and $5$.
However, we can work with these numbers.
Let $A$ be the set of positive odd integers less than or equal to $200$ which are odd; let $B$ be the set of positive integers less than or equal to $200$ which are multiples of $3$; let $C$ be the set of positive integers less than or equal to $200$ which are multiples of $5$. Then
\begin{align*}
|A| & = 200 - \left\lfloor \frac{200}{2} \right\rfloor = 100\\
|B| & = \left\lfloor \frac{200}{3} \right\rfloor = 66\\
|C| & = \left\lfloor \frac{200}{5} \right\rfloor = 40\\
|A \cap B| & = 66 - \left\lfloor \frac{200}{2 \cdot 3} \right\rfloor = 66 - 33 = 33\\
|A \cap C| & = 40 - \left\lfloor \frac{200}{2 \cdot 5} \right\rfloor = 40 - 20 = 20\\
|B \cap C| & = \left\lfloor \frac{200}{3 \cdot 5} \right\rfloor = 13\\
|A \cap B \cap C| & = 13 - \left\lfloor \frac{200}{2 \cdot 3 \cdot 5} \right\rfloor = 7
\end{align*}
where we obtain $|A \cap B|$ by subtracting the number of even multiples of $3$ less than or equal to $200$ from the number of positive integer multiples of $3$ which are at most $200$, $|A \cap C|$ by subtracting the number of even multiples of $5$ less than or equal to $200$ from the number of positive integer multiples of $5$ which are at most $200$, and $|A \cap B \cap C|$ by subtracting the number of even multiples of $3$ and $5$ less than or equal to $200$ from the number of positive integer multiples of $3$ and $5$ less than or equal to $200$.
Hence, by the Inclusion-Exclusion Principle, the number of positive integers less than or equal to $200$ which are odd or divisible by $3$ or divisible by $5$ is
$$100 + 66 + 40 - 33 - 20 - 13 + 7 = 147$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed-form expression for $F(x,y) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 (t+y)} \mathrm{d}t$? I am considering the following function
$$F(x,y) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 (t+y)} \mathrm{d}t,$$
which is well-defined for any $x > 0$ and $y \geq 0$.
Is there a hope to obtain a closed form formula with respect to $x$ and $y$?
For instance, according to Mathematical, we have that
$$F(x,0) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 t} \mathrm{d}t = \frac{\pi}{x \sqrt{x (x+2)}}.$$
Remark: To give a bit of context, the function $F$ appears when I consider the quadratic optimization problem of the form $\min_{\mathbf{x} \in \mathrm{R}^N} \lVert \mathbf{A} \mathbf{x} - \mathbf{y} \rVert_2^2 + \lambda \lVert \mathbf{x} \rVert_2^2$ and I try to understand the behavior of $\lVert \widehat{\mathbf{x}} - \mathbf{x}_0 \rVert_2^2$ with $\widehat{\mathbf{x}}$ the unique optimizer and $\mathbf{x}_0$ the vector we aim at recovering, with $\mathbf{y} = \mathbf{A} \mathbf{x}_0 + \mathbf{n} \in \mathbb{R}^M$ and $\mathbf{n}$ an i.i.d. Gaussian vector. The values $x$ and $y$ above appear as functions of $\lambda$ and $\gamma = \lim M/N$ when $N\rightarrow \infty$ when the matrix $\mathbf{A}$ is i.i.d. Gaussian and its spectrum behaves according to the Marchenko-Pastur law.
| Let's change notation to $\displaystyle F(a,b) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+a)^2 (t+b)}\, \mathrm{d}t$ to emphasise the constant variables.
Let $\displaystyle t = \frac{u-b}{u+a} \implies \mathrm{d}t = \frac{a+b}{(u+a)^2}\,\mathrm{d}u$; moreover, this maps $[0, 1]$ to $[b, \infty)$. We've
$$\displaystyle F(a,b) = (a+b)\sqrt{a+b}\int_b^{\infty} \frac{\sqrt{u-b}}{(a b + b u - b + u) (a^2 + a u - b + u)^2}\,\mathrm{d}u $$
Now, letting $u = v^2+b$ we obtain integral of a rational function
$$\displaystyle F(a,b) ={2(a+b)\sqrt{a+b}}\int_0^{\infty} \frac{v^2}{(a^2 + a b + v^2 + a v^2)^2 (a b + b^2 + v^2 + b v^2)}\mathrm{d}v $$
So using partial fractions on the integrand we find
$$\displaystyle \frac{F(a,b)}{2(a+b)\sqrt{a+b}} = \frac{a}{a-b}I+\frac{(a+1)b}{(a-b)^2(a+b)}J -\frac{b(b+1)}{(a-b)^2(a+b)}K ~~~~~~~~~ (1)$$
where $I, J, K$ are the integrals $$\displaystyle I= \int_0^\infty \frac{1}{(a^2+ab+av^2+v^2)^2} \,\mathrm{d}v = \frac{\pi}{2} \cdot\frac{\sqrt{a (a + b)}}{2 a^2 \sqrt{1 + a} (a + b)^2}$$
$$J = \int_0^\infty \frac{1}{(a^2+ab+av^2+v^2)} \,\mathrm{d}v = \frac{\pi}{2} \cdot \frac{1}{\sqrt{a (1 + a) (a + b)}}.$$
$$K = \int_0^\infty \frac{1}{(ab+b^2+bv^2+v^2)} \, \mathrm{d}v = \frac{\pi}{2} \cdot \frac{1}{\sqrt{b (1 + b) (a + b)}}. $$
Which routinely fall-out as standard arctangent integrals. Putting these values in $(1)$ we get:
$$ F(a,b) = \frac{\pi}{2}\cdot \frac{a + b + 2 a b - 2 \sqrt{ab (1 + a) (1 + b)}}{2 \sqrt{a (1 + a)} (a - b)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $ \int\limits_{-1}^{1} \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx$ Is there a nice way to substitute something in this double integral:
$$ \int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx$$
Can calculate it easier?
Wolfram Alpha gives me $\sqrt{2}\pi$ as solution.
| $$\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (1-2x^2-y^2-2y) \,dy\,dx\\
\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2-y^2-2y-1) \,dy\,dx\\
\int\limits_{-1}^1 \int\limits_{-1-\sqrt{2-2x^2}}^{-1+\sqrt{2-2x^2}} (2-2x^2)- (y+1)^2 \,dy\,dx$$
This now integrates more elegantly
$$\int\limits_{-1}^1 (2-2x^2)- \frac 13 (2-2x^2)^{\frac 32} \,dx$$
We should probably break this up into two integrals at this point, but each is pretty straightforward.
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum value of $ac+bd$
If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is
what i try
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$
$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$
but answer is $3$
How do i get right answer help me please
| You are correct that the maximum value is $2\sqrt 2<3$. As another answer (now deleted; it applied AM-GM separately to $a^2+c^2$ and $b^2+d^2$) points out, you can easily get an upper bound of $3$ using AM-GM, but this bound can't actually be attained, so isn't the correct maximum.
For equality in the AM-GM bound, you would need $a=c$ and $b=d$, but this is clearly impossible since $a^2+b^2\neq c^2+d^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex number inequality $\left|\sum_{k=1}^n\Im(z^k)\right| < \sqrt{\frac{1}{2}\Re\left(n-\sum_{k=1}^n\frac{z^{2k}}{|z|^{2k}}\right)}$
If $z\in \mathbb{C}^*$ with $|z|< \frac{\sqrt{2}}{2}$ and $n$ is a
positive integer, prove that:
$$\left|\sum_{k=1}^n\Im(z^k)\right| < \sqrt{\frac{1}{2}\Re\left(n-\sum_{k=1}^n\frac{z^{2k}}{|z|^{2k}}\right)}$$
I coould not prove this inequality. It is from an old preparation test at my school (2001). I tried with polar form $z=|z|(\cos a+i\sin a)$ and I arrive at:
$$\left(\sum_{k=1}^n|z|^k\sin(ka)\right)^2 < \frac{1}{2}\left(n-\sum_{k=1}^n\cos(2ka)\right)$$
but no further success.
| Proof: We have
\begin{align}
\Big(\sum_{k=1}^n |z|^k \sin k a\Big)^2
&\le \Big(\sum_{k=1}^n |z|^k |\sin k a|\Big)^2\\
&\le \sum_{k=1}^n |z|^{2k} \cdot \sum_{k=1}^n |\sin ka|^2 \tag{1}\\
&\le \sum_{k=1}^n \frac{1}{2^k} \cdot \sum_{k=1}^n \sin^2 ka\tag{2}\\
&< \sum_{k=1}^n \sin^2 ka \tag{3}.
\end{align}
Explanation: in (1), we have used Cauchy-Bunyakovsky-Schwarz inequality;
in (2), we have used $|z| < \frac{1}{\sqrt{2}}$; in (3), we have used $\sum_{k=1}^n \frac{1}{2^k}
= 1 - \frac{1}{2^n} < 1$.
On the other hand, we have
\begin{align}
\frac{1}{2}\Big(n - \sum_{k=1}^n \cos 2k a\Big)
&= \frac{1}{2}\Big(n - \sum_{k=1}^n (1 - 2\sin^2 ka)\Big)\\
&= \sum_{k=1}^n \sin^2 ka.
\end{align}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Roots of the equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ The equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ has
(A) all its solution real but not all positive
(B) only two of its solution real
(C) two of its solution positive and two negative
(D) none of its solution real
My approach is as follow $(x^2+3x+4)^2+3(x^2+3x+4)+4-x=y$
$f(x)=y=x^4+6x^3+20x^2+32x+32$
$f(-x)=y=x^4-6x^3+20x^2-32x+32$
Using Descartes rule no positive roots but the possible ways we can have either 4,2,0 negative roots.
$f'(x)=4x^3+18x^2+40x+32$
$f''(x)=12x^2+36x+40$ which is imaginary
From here I am not able to approach.
| Let $t=x^2+2x+4$, then we can write the equation as:
$$(t+x)^2+3(t+x)+4-x=0$$
or
$$t^2+2tx+x^2+3t+2x+4=0$$
or
$$t^2+2tx+3t+t=0$$
or
$$t(t+2x+4)=0$$
so
$$(x^2+2x+4)(x^2+4x+8)=0$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3553739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of $f(a,b) = a^2+ab+b^2-3a-6b+11$
Minimum value of
$f(a,b) = a^2+ab+b^2-3a-6b+11$ for all $a,b\in \mathbb{R}$
what i try
Let $$k=a^2+ab+b^2-3a-6b+11$$
$$k=a^2+(b-3)a+b^2-6b+11$$
$\displaystyle k=\bigg[a^2+(b-3)a+\frac{(b-3)^2}{4}+b^2-6b+11-\frac{(b-3)^2}{4}\bigg]$
$$\displaystyle k = \bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3b^2-18b+35\bigg]$$
$$k=\bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3(b-3)^2+8\bigg]\geq 8$$
but answer is $20$
Help me please
| Your way works very well. The problem is here:
$$\displaystyle k=\bigg[a^2+(b-3)a+\frac{(b-3)^2}{4}+b^2-6b+11-\frac{(b-3)^2}{4}\bigg]$$
when you group the remaining terms in $b$, you lose the $\frac{1}{4}$ factor:
$$\displaystyle k = \bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+\color{red}{\frac{1}{4}}\left(3b^2-18b+35\right)\bigg]$$
With this you arrive at the right result which you probably mean as $2$, not $20$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\min \{a^7,b^3,c^2,1\} \le abc$ If $a,b,c$ are three non-negative real numbers, prove that:
$$\min \{a^7,b^3,c^2,1\} \le abc$$
I tried several ideas with the geometric mean:
$$\min \{a^7,b^3,c^2,1\} \leq \sqrt[4]{a^7b^3c^2}$$
but $\sqrt[4]{a^7b^3c^2} \leq abc$ is not true. I tried with other means but nothing significant. I think these ideas fail because the inequality is not homognenous.
| For the sake of contradiction, assume that:
$$\min\{a^7,b^3,c^2,1\} > abc$$
Then:
$$a^7 > abc \Rightarrow a > (abc)^{\frac{1}{7}}$$
$$b^3 > abc \Rightarrow b > (abc)^{\frac{1}{3}}$$
$$c^2 > abc \Rightarrow c > (abc)^{\frac{1}{2}}$$
Also, because $\frac{1}{2}+\frac{1}{3}+\frac{1}{7} < 1$, we have:
$$1 > abc\Rightarrow 1 > abc^{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}}$$
Multiplying these four inequalities, it follows that:
$$abc > (abc)^{\frac{1}{7}} \cdot (abc)^{\frac{1}{3}}\cdot (abc)^{\frac{1}{2}} \cdot abc^{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}} = abc$$
a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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An indeterminate limit form of infinity/infinity I am trying to solve the limit:
$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)$$
I was trying to find a way to bring it into a fraction form to apply L'Hospital's rule, and I tried using
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$
But it made it even more complex and after applying L'Hospital's rule I got stuck with all the terms. Is there a smarter way to evaluate it?
| $$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)= \lim_{x\to\infty}x^2\left(\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-1\right)=\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-1}{\frac{1}{x^2}}$$
Now using L'Hopital we get
$$\lim_{x\to\infty}\frac{\frac{1}{3}\left(1+\frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^\frac{-2}{3}\left(\frac{-1}{x^2}\right)\left( \sin\left(\frac{1}{x}\right) +\frac{1}{x}\cos\left(\frac{1}{x}\right) \right)}{\frac{-2}{x^3}}=\frac{1}{6}\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right) +\frac{1}{x}\cos\left(\frac{1}{x}\right)}{\frac{1}{x}}=\\ \frac{1}{6}\lim_{x\to\infty}\frac{\frac{-2}{x^2}\cos\left(\frac{1}{x}\right)-\frac{-1}{x^3}\sin\left(\frac{1}{x}\right)}{\frac{-1}{x^2}}=\frac{1}{6}\times2=\frac{1}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exponential Equation with strange terms Solve for all $x\in\mathbb{R}$
$(x^4-x^3-x+1)5^{x^2-1}+(x^3-x^2-x+1)5^{x^3-1}+(x^5-x^3-x^2+1)5^{x-1}=x^5+x^4-x^3-2x^2-2x+3$
I rewrote is as $(x^3-1)(x-1)5^{x^2-1}+(x^2-1)(x-1)5^{x^3-1}+(x^2-1)(x^3-1)5^{x-1}=(x^3-1)(x-1)+(x^2-1)(x-1)+(x^3-1)(x^2-1)$
This is equivalent with $(x^3-1)(x-1)(x^2-1)({5^{x^3-1}\over x^3-1}-{1\over x^3-1}+{5^{x^2-1}\over x^2-1}-{1\over x^2-1}+{5^{x-1}\over x-1}-{1\over x-1})=0$
I tried considerind the function $f(x)={5^{x^3-1}\over x^3-1}-{1\over x^3-1}+{5^{x^2-1}\over x^2-1}-{1\over x^2-1}+{5^{x-1}\over x-1}-{1\over x-1}$ which is increasing on $(1,\infty)$(and the function has always positive values) and so x=1 is the only positive solution.
But i don't think this is correct and also im not sure what to do on $(-\infty,1)$ cause -1 is also a solution .
Please help me out it looks kind of ugly and i don't know what to do.
| We obtain:
$$(x-1)^2(x^2+x+1)5^{x^2-1}+(x-1)^2(x+1)5^{x^3-1}+(x-1)^2(x+1)(x^2+x+1)5^{x-1}=$$
$$=(x-1)^2(x^3+3x^2+4x+3),$$
which gives $x=1$ or
$$(x^2+x+1)5^{x^2-1}+(x+1)5^{x^3-1}+(x+1)(x^2+x+1)5^{x-1}=x^3+3x^2+4x+3$$ or
$$(x^2+x+1)\left(5^{x^2-1}-1\right)+(x+1)\left(5^{x^3-1}-1\right)+(x+1)(x^2+x+1)\left(5^{x-1}-1\right)=0.$$
Easy to see that $-1$ is a root.
Let $x\neq-1.$
Thus, we need to solve $f(x)=0,$ where $$f(x)=\frac{5^{x^2-1}-1}{x+1}+\frac{5^{x^3-1}-1}{x^2+x+1}+5^{x-1}-1.$$
Now, prove that $f(1)=0,$ $f(x)>0$ for $x>1$ and $f(x)<0$ for $x<1,$ $x\neq-1$.
| {
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"url": "https://math.stackexchange.com/questions/3560111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving $\int_0^1 \exp \left(x-\frac{1}{x}\right) \, dx=\frac{1}{2} (1+\pi \pmb{H}_{-1}(2)+\pi Y_1(2))$ and generalize How to prove
$$\int_0^1 \exp \left(x-\frac{1}{x}\right) \, dx=\frac{1}{2} (1+\pi \pmb{H}_{-1}(2)+\pi Y_1(2))$$
Where $\pmb{H}, Y$ denotes Struve and Bessel function respectively? Any help will be appreciated.
| Let $u=x-\frac{1}{x} \implies dx = \frac{1}{2}\left(1+\frac{u}{\sqrt{u^2+4}}\right)du$. We get the integral
$$\frac{1}{2}\int_{-\infty}^0 e^u\left(1+\frac{u}{\sqrt{u^2+4}}\right)du = \frac{1}{2} - \int_0^\infty \frac{t}{\sqrt{t^2+1}}e^{-2t}\:dt$$
by the further substitution $u=-2t$. Now using integration by parts on the remaining integral
$$-\sqrt{t^2+1}e^{-2t}\Bigr|_0^\infty - 2\int_0^\infty \sqrt{t^2+1}e^{-2t}\:dt = 1-2\int_0^\infty(1+t^2)^{1-\frac{1}{2}}e^{-2t}\:dt$$
we also have that
$$\int_0^\infty (1+t^2)^{\alpha-\frac{1}{2}}e^{-xt}\:dt = \frac{\sqrt{\pi}\Gamma\left(\alpha+\frac{1}{2}\right)}{2\left(\frac{x}{2}\right)^\alpha}K_\alpha(x)$$
from the integral representation of $K_\alpha(x)$, the Struve function of the second kind. Plugging in, the integral becomes
$$1-2\left(\frac{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}{2}K_1(2)\right)=1-\frac{\pi}{2}K_1(2)$$
leaving the final answer as
$$\frac{3}{2}-\frac{\pi}{2}K_1(2) \equiv \frac{1}{2}(3 - \pi H_1(2) + \pi Y_1(2)) \approx 0.31594$$
which is numerically equivalent to your form of the answer.
| {
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if $x$ is odd, show that $x^3+x$ has a remainder 2 when divided by 4 I did part of this question but am stuck and don't know how to continue
I let $x= 2k +1$
Also noticed that $x^3+x = x(x^2+1)$
therefore
$4m+2 = 2k+1((2k+1)^2+1)$
I simplified this and ended up with
$4m+2 = 8k^3+12k^2+8k+2$
I don't know how to continue from and prove that $x^3+x$ has remainder 2 when divided by 4
| Let $x=2n+1, then
f(x)=x^3+x=(2n+1)^3+(2n+1)=2+8n^3+12n^2+8n \implies \frac{f(x)}{4}=\frac{2}{4}+2n^3+3n^2+2n.$ So the remaner is 2.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Search for maximum and minimum functions on the surface in 3D Find all global and local minimums and maximums of $f(x,y,z)=x-y+2z$ on the set $M=${$x^2+2y^2+2z^2=1$}
I have an idea to parameterize the surface with spherical coordinates:
$x=cos\phi cos \theta$
$y=\frac{1}{\sqrt2} sin\phi cos \theta$
$z=\frac{1}{\sqrt2}sin \theta$
Thus, the function is converted to
$f(\phi,\theta)=cos\phi cos \theta-\frac{1}{\sqrt2} sin\phi cos \theta+\sqrt2sin \theta$
where $phi$ from $0$ to $2\pi$, $\theta$ from $0$ to $\pi$
Then I calculate derivative of $\phi$ and $\theta$, and first gives
$cos\theta=0$
or $tg\phi=\frac{1}{\sqrt2}$
but what should I do with the theta derivative?
| $x^2+2y^2+2z^2=0$ gives $x=y=z=0$ and from here: $$x-y+2z=0.$$
If $x^2+2y^2+2z^2=k>0$, so by C-S
$$\sqrt{k}=\sqrt{x^2+2y^2+2z^2}=\sqrt{\frac{2}{7}}\sqrt{\left(1+\frac{1}{2}+2\right)(x^2+2(-y)^2+2z^2)}\geq$$
$$\geq \sqrt{\frac{2}{7}}\sqrt{(x-y+2z)^2}=\sqrt{\frac{2}{7}}|x-y+2z|.$$
The equality occurs for $\left(1,\frac{1}{\sqrt2},\sqrt2\right)||\left(x,-\sqrt2y,\sqrt2z\right),$
which gives $$\max_{x^2+2y^2+2z^2=k}(x-y+2z)=\sqrt{3.5k}$$ and
$$\min_{x^2+2y^2+2z^2=k}(x-y+2z)=-\sqrt{3.5k}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the matrix A that satisfies the equation Find the matrix A that satisfies the equation
$$
\left[
2\left[
\begin{array}{}
1&1\\
-2&3
\end{array}
\right] -5A^{-1}
\right]
^T=
\left(4A^T
\right)^{-1}$$
I'm not sure how to start solving this question. where/how do i start and how do i think?
| Let the given matrix $$\begin{pmatrix} 1 & 1 \\ -2 & 3 \end{pmatrix}=B$$ and call $(A^T)^{-1}=C=(A^{-1})^T$, then we have $$(2B^T-5(A^{-1})^T)=(4A^T)^{-1} \implies 2B^T-5C= \frac{1}{4}C\implies B^T=\frac{21}{8} C \implies (B^T)^T=\frac{21}{8} C^T=\frac{21}{8} A^{-1} \implies B=\frac{21}{8} A^{-1} \implies A= \frac{21}{8}B^{-1}$$
$$\implies A=\frac{21}{40}\begin{pmatrix} 3 &-1 \\ 2 & 1 \end{pmatrix}$$
Here we have used the interchangeability of transpose and inverse.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$ Given that
$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$
find the value of rationals $a,b,c.$
Solution I tried: I tried to rationalize it, but I'm not getting the answer:
$$\frac{1}{(1+\sqrt[3]{2})}\times \frac{(1-\sqrt[3]{2})}{(1-\sqrt[3]{2})}$$
$$\frac{(1-\sqrt[3]{2})}{1-2^{\frac{2}{3}}}$$
so doing so not getting answer; also, I tried to expand it, but we have condition that $(1+x)^n$ where $n$ is in fraction can be expandable only when $x < 1$, but here the cube root of $2$ is not less than $1.$
Thank you
| Apply $a^3+1 = (a+1)(a^2-a+1)$, or
$$\frac1{a+1}=\frac{a^2-a+1}{a^3+1}$$ to $a = \sqrt[3]{2}$ to obtain,
$$\frac1{1+\sqrt[3]{2}}= \frac{(\sqrt[3]{2})^2-\sqrt[3]{2}+1}{(\sqrt[3]{2})^3+1 }=\frac13(\sqrt[3]{2^2}-\sqrt[3]{2}+1) $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What do elements of this set look like? I was under the impression that elements of quotient rings such as $\mathbb F[x]/(f)$ were of the form $h(x)+(f)$ where $h(x)\in\mathbb F[x]$. Is this correct? If so could somebody explain why elements of $\mathbb F[x]/(x^3+x+1)$ have degree 2 or less? Why isn’t $2x+x^3+x+1$ an element? Is my understanding of what $\mathbb F[x]/(f)$ means wrong? I have looked at several answers and still don’t understand
| In fact, in the quotient ring $R/I$, some elements are eaten by the ideal $I$.
For example, in the case of $\mathbb{F}[x]/(x^3+x+1)$, since
$$x^3+x+1\in(x^3+x+1)\text{ therefore }\\ x^3+(x^3+x+1)=-x-1+(x^3+x+1)$$
and thus
$$ax^3+(x^3+x+1)=-ax-a+(x^3+x+1)$$
also we have
$$x^4+(x^3+x+1)=x(-x-1)+(x^3+x+1)=-x^2-x+(x^3+x+1)$$
and $$x^5+(x^3+x+1)=x(-x^2-x)+(x^3+x+1)\\=-x^3-x+(x^3+x+1)\\=x+1-x+(x^3+x+1)=1+(x^3+x+1)$$
and so on.
As you see, any monomial of degree greater than or equal to $3$, reduces to a polynomial of degree at most $2$.
In fact the ring $\mathbb{F}[x]/(x^3+x+1)$ is
$$\{a+bx+cx^2:a,b,c\in\mathbb{F}, x^3=-x-1\}$$
For example, the product of $x+1$ and $x^2+x$ is
$$(x^2+1)(x^2+x-1)=x^4+x^3-x^2+x^2+x-1\\
=x(x^3)+x^3+x^2+x-1\\
=x(-x-1)+(-x-1)+x^2+x-1\\
=-x^2-x-x-1+x^2+x-1\\=-x-2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$
Solve over reals:
$$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$
I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:
$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$
Here, there are two cases: all three variables are greater than $1$ or only one.
If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.
How can I prove there are no solutions when only variable, say $x$, is greater than $1$?
| If $xyz=0$ so $(x,y,z)=(0,0,0).$
Let $xyz=r\neq0.$
Thus, by the WE Tutorial School's work we obtain:
$$(x+1)(y+1)(z+1)=1.$$ or
$$x+y+z+xy+xz+yz=-r.$$
In another hand, by your work we obtain:
$$(x-1)(y-1)(z-1)=x^2y^2z^2,$$ which gives
$$x+y+z-xy-xz-yz=r^2-r+1,$$ which gives
$$x+y+z=\frac{(r-1)^2}{2}$$ and
$$xy+xz+yz=-\frac{r^2+1}{2}.$$
But after summing of our equations we obtain:
$$\sum_{cyc}(x^3-x^2+x)=0$$ and since
$$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz$$ and $$x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz),$$ we have:
$$\frac{(r-1)^6}{8}+\frac{3(r-1)^2(r^2+1)}{4}+3r-\frac{(r-1)^4}{4}-r^2-1+\frac{(r-1)^2}{2}=0$$ or
$$r^6-6r^5+19r^4-24r^3+11r^2+6r+1=0$$ or
$$(r^3-3r^2+3r+1)^2+4r^2(r^2-2r+2)=0,$$ which is impossible for real value of $r$.
Id est, $(0,0,0)$ is an unique solution of the system.
| {
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"timestamp": "2023-03-29T00:00:00",
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At which values of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$?
Problem:
At which values of the parameter $k$, there is no solution to the inequality $$(k+1)x^2-2kx+2k+2<0.$$
The solution in my textbook is as follows:
$a=k+1, D=4\left[k^2-2(k+1)^2\right]$
$\begin{cases} k+1>0 \\k^2-2(k+1)^2<0 \end{cases} \Longrightarrow \begin{cases} k>-1 \\ -k^2-2k-2 <0 \end{cases} \Longrightarrow -1<k<+\infty.$
Answer: That is, there is no solution to the inequality of $ (k + 1) x ^ 2-2kx + 2k + 2 <0$ for the $ k $ `s that satisfy the $ -1 <k <+ \infty $ condition.
Firstly, I understand the question as follows:
At which values of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$ , for all $x\in\mathbb{R}.$
The last sentence is logically equivalent to :
At which values of the parameter $k$, the inequality $(k+1)x^2-2kx+2k+2\geq 0$ holds on for all $x\in\mathbb{R}.$
If I understand the question correctly, here is my solution:
It is obvious that, for $k=-1$ is not a solution.
$\color{black}{\large\text{Case} \thinspace 1:}$ $k+1>0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0, ∀ x\in\mathbb {R}$$
Then, applying $x=\dfrac{k}{k+1}$ we get, $\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$. We have,
$$\begin{cases} k+1>0 \\ \dfrac{k^2+4k+2}{(k+1)^2}\geq 0 \end{cases} \Longrightarrow k\geq -2+\sqrt2.$$
$\color{black}{\large\text{Case} \thinspace 2:}$ $k+1<0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\leq 0, ∀ x\in\mathbb {R}$$
For sufficiently large $ x $ `s, we have $\left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$, which gives a contradiction.
Finally we deduce that, for all $x\in\mathbb{R}$ satisfying the condition $k\in\mathbb[\sqrt 2-2; +\infty)$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0.$
It does not match the solution in my book. Probably, maybe I misunderstand the question or my solution is wrong. Or the book says it wrong.
Did I get the question right? If so, is my solution correct?
Thank you very much.
| Well, in your textbook as can be seen in your post there are two inconsistencies.
k+1>0 is not k>-2 and
more importantly
k^2−2(k+1)^2<0 isn't equal to −k^2−2k−2<0 but it's similar to condition which you got in your solution i.e. k^2+4k+2.
So, you have done it correctly. However, your textbook method if corrected is shorter.
| {
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"source": "stackexchange",
"question_score": "1",
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How to find the number of different m-colorings of vertices?
These are the textbook answers for d) and e), but I'm not exactly sure how they get it. Can someone show me with diagram like a picture perhaps. Thank you. You can just show me one part, don't need to be both parts.
| We will do the second problem. This question looks like it is asking for
the Burnside lemma. The term floating suggests rotational symmetries
only but there are twelve terms in the proposed closed form which means
we probably have dihedral symmetry. Recall the cycle index of the
cyclic group:
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
as well as the cycle index of the dihedral group
$$Z(D_n) =
\frac{1}{2} Z(C_n) +
\begin{cases}
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$
Therefore we get for the cycle index of the hexagon without the central
node
$$Z(D_6) = \frac{1}{12} (a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6)
+ \frac{1}{4} (a_1^2 a_2^2 + a_2^3).$$
The central node is always fixed so we get for our cycle index of the
wheel
$$Z(W_6) = \frac{a_1}{12} (a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6)
+ \frac{a_1}{4} (a_1^2 a_2^2 + a_2^3).$$
By Burnside we must be constant on the cycles and hence we have $m$
choices for every cycle:
$$\frac{m}{12} (m^6 + m^3 + 2 m^2 + 2 m)
+ \frac{m}{4} (m^4 + m^3)
\\ = \frac{1}{12} (m^7 + m^4 + 2 m^3 + 2 m^2 + 3 m^5 + 3 m^4)
\\ = \frac{1}{12} (m^7 + 3 m^5 + 4 m^4 + 2 m^3 + 2 m^2).$$
This is the claim.
| {
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Solve $x(2^x+2^{-x})=\frac{17}{2}$ analytically Is it possible to solve $x(2^x+2^{-x})=\frac{17}{2}$ analytically? I was able to rearrange to get $x\cosh(x\ln(2))=\frac{17}{4}$ but can't get any further. WA gives 2 as the solution, but no steps provided.
| First we simplify the question
$$2^x + \frac{1}{2^x} = \frac{17}{2x}$$
$$\frac{4^x + 1}{2^x} = \frac{17}{2x}$$
Assigning $2^x$ as $t$
$$\frac{t^2 + 1}{t} = \frac{17}{2x}$$
Comparing numerators and denominators
$$t = 2x$$
$$t^2 + 1 = 17$$
$$t^2 = 16$$
$$x^2 = 4$$
$$x = +2, -2$$
But, on substituting the obtained values in the main equation, we get only $x= 2$ as solution.
| {
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If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$
Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$.
My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies \left(\frac{\sin^4x}{a}+\frac{\cos^4x}{b}\right)^2=\frac{1}{(a+b)^2} \\ \implies \frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}-\sin^2x\cos^2x\left(\frac{\sin^2x}{a}-\frac{\cos^2x}{b}\right)^2=\frac{1}{(a+b)^2}.$$
Therefore, if we can prove that $$\frac{\sin^2x}{a}-\frac{\cos^2x}{b}=0,$$ then we are done. But, I am not able to prove the same.
| We know, $~\sin^2x + \cos^2x = 1 ⇒\cos^2x = 1 - \sin^2x~$ and hence $~\cos^4x = (1 - \sin^2x)^2 = 1 + \sin^4x - 2\sin^2x~$.
Now
$~\dfrac{\sin^4x}{a}+\dfrac{\cos^4x}{b}=\dfrac{1}{a+b}~$
$~\implies\dfrac{\sin^4x}{a} + \dfrac 1b(1 + \sin^4x - 2\sin^2x) = \dfrac{1}{a + b}~$
$~\implies \dfrac{\sin^4x}{a} + \dfrac 1b + \dfrac{\sin^4x}{b} - \dfrac{2}{b}\sin^2x = \dfrac{1}{a + b}~$
$~\implies \sin^4x\left(\dfrac 1a + \dfrac 1b\right) - \dfrac{2}{b}\sin^2x = \dfrac{1}{a + b} - \dfrac 1b~$
$~\implies\dfrac{a+b}{ab}\sin^4x- 2a\dfrac{\sin^2x}{ab} = \dfrac{b - a - b}{(a + b)b}~$
$~\implies(a + b)^2(\sin^2x)^2 - 2a(a + b) \sin^2x = -a²~$
$~\implies \{(a + b)\sin^2x\}^2 -2.a.(a + b)\sin^2x + a^2 = 0~$
$~\implies \{(a + b)\sin^2x - a\}^2 = 0~$
$~\implies \sin^2x = \dfrac{a}{a + b}~$
$~⇒1 - \sin^2x = \cos^2x = 1 - \dfrac{a}{a + b} = \dfrac{b}{a + b}~$
Therefore, $~\sin^2x = \dfrac{a}{a + b}~$ and $~\cos^2x = \dfrac{b}{a + b}~$.
Now
$~\dfrac{\sin^6x}{a^2}+\dfrac{\cos^6x}{b^2}~$
$~=\dfrac 1{a^2}\left(\dfrac{a}{a + b}\right)^3+\dfrac 1{b^2}\left(\dfrac{b}{a + b}\right)^3~$
$~=\dfrac{a}{(a + b)^3}+\dfrac{b}{(a + b)^3}~$
$~=\dfrac{1}{(a + b)^2}~$
| {
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The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ and product of two roots is unity, then:
$Q-1$: Find $x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4$
$Q-2$: Find $x_2^3+x_4^3$
My attempt is as follows:-
$A-1$ : First I tried to find any trivial root, but was not able to find any. After that I tried following:-
$$x_1\cdot x_2+x_1\cdot x_3+x_1\cdot x_4+x_2\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2\cdot x_3\cdot x_4=-1$$
$$x_1\cdot x_4=\dfrac{-1}{x_2\cdot x_3}$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_2\cdot x_3+\dfrac{1}{x_2\cdot x_3}$$
But from here I was not able to proceed as I was not able to calculate value of $x_2\cdot x_3$
$A-2$ : $(x_2+x_4)(x_2^2+x_4^2-x_2\cdot x_4)$
Now here I was not getting any idea for how to proceed.
Please help me in this.
| Also, we can use the following way.
For any value of $k$ we obtain: $$x^4-2x^3-3x^2+4x-1=(x^2-x+k)^2-x^2-k^2+2kx-2kx^2-3x^2+4x-1=$$
$$=(x^2-x+k)^2-((2k+4)x^2-(2k+4)x+k^2+1),$$ which for $k=0$ gives:
$$x^4-2x^3-3x^2+4x-1=(x^2-x)^2-(2x-1)^2=(x^2-3x+1)(x^2+x-1).$$
Can you end it now?
| {
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When finding relative minimum/maximum, what is the point of using the second derivative test? Why use the second derivative test over the first derivative test when finding maxima and minima if it's uncertain what $f''(x) = 0$ is? Why not just always use the first derivative since we need to take the first derivate either way?
| Let $f(x) = \sin(\mathrm{e}^{16-x^2})$ and determine whether $x = 0$ is a critical point, and if so, whether $f$ has a local minimum, local maximum, or neither at $x = 0$.
(The small voids near the upper and lower bounds are numerical artifacts. A more accurate plot would look like a solid rectangle.) I immediately notice a difficulty with the first derivative attack on the problem: the function oscillates vigorously near $x = 0$.
Then $f'(x) = -2x\mathrm{e}^{16-x^2}\cos(\mathrm{e}^{16-x^2})$. We can immediately see $x = 0$ is a critical point. The first derivative has a lot of zeroes near $x = 0$.
It's "easy enough" to solve for the zeroes of the derivative: $x = 0$, $x = \pm \sqrt{16 + \ln(2/\pi)}$, $x = \pm \sqrt{16+\ln\left( \frac{-2}{\pi - 4 \pi k} \right)}$, or $x = \pm \sqrt{16+\ln\left( \frac{2}{\pi + 4 \pi k} \right)}$, for any integer $k$ with $1 \leq k \leq 1\,414\,268$. The fourth case with $k = 1\,414\,268$ gives the critical points closest to (and distinct from) $x=0$: $x = \pm 0.00034073{\dots}$. (How much work would one show to justify the claims in the previous sentence?) But all of this analysis has been a stupendous waste of time.
Instead,
$$ f''(x) = (4x^2 - 2)\mathrm{e}^{16-x^2}\cos(\mathrm{e}^{16-x^2}) \\ - 4 x^2 \mathrm{e}^{32-2x^2}\sin(\mathrm{e}^{16-x^2}) $$ and $f''(0) = -2 \mathrm{e}^{16}\cos(\mathrm{e}^{16}) = 1.5251 \times 10^{7}$, so $f$ has a local minimum at $x = 0$.
We don't only run into problems with the "first derivative strategy" only when we use transcendental functions. Consider
$$ f(x) = \frac{1}{6} x^6 - \frac{6}{5} x^5 - \frac{\mathrm{e}^2 + \pi^2 - 14}{4} x^4 + \frac{4}{3}(\mathrm{e}^2 + \pi ^2 - 4) x^3 + \frac{1}{2}(9 - 5\mathrm{e}^2 - 5\pi ^2 +\mathrm{e}^2\pi^2)x^2 - 2 (\mathrm{e}^2 -1)(\pi ^2 - 1) x + 7 \text{.} $$
It's just a polynomial. All its derivatives are polynomials, how bad could it be?
$$ f'(x) = x^5 - 6x^4 - (\mathrm{e}^2 + \pi^2 - 14)x^3 + 4(\mathrm{e}^2 + \pi^2 - 4)x^2 + (9 - 5\mathrm{e}^2 - 5 \pi^2 + \mathrm{e^2}\pi^2) x - 2 (\mathrm{e}^2 - 1)(\pi^2 - 1) \text{.} $$
If we abuse the rational root test, applying it even though the coefficients are not integers, we can find that $x = 2$ is a root of $f'$.
$$ f'(x) = (x-2)(x^4 - 4x^3 - (\mathrm{e}^2 + \pi^2 - 6)x^2 + (2\mathrm{e}^2 + 2\pi^2 - 4)x + \mathrm{e}^2\pi^2-\mathrm{e}^2 - \pi^2 + 1) \text{.} $$
Good luck factoring that. (It's $f'(x) = (x-2)((x-1)^2 - \pi^2)((x-1)^2-\mathrm{e}^2)$.)
However, we can characterize the critical point at $x = 2$ immediately.
$$ f''(2) = (\pi^2 - 1)(\mathrm{e}^2 - 1) = 56.668{\dots} \text{.} $$
| {
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"answer_id": 5
} |
Calculate the limit (verifying my answer). I have to calculate
$$\lim_{x\rightarrow 1^+} \frac{\sin(x^3-1)\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
Making the substitution $x-1=y$ and doing the math, I get that,
$$=\lim_{y\rightarrow 0^+} \frac{\sin(y(y^2+3y+3))}{y(y^2+3y+3)}\cdot\cos\Big(\dfrac{1}{y}\Big)\cdot\sqrt{y}(y^2+3y+3)$$
Since the first fraction goes to $0$. I have to worry with the $\cos(1/y)$, but I realized that $\cos(x)$ is bounded above and below, then, this kind of function times something that goes to $1$ results in $0$ (I studied this theorem). Since $\sqrt{y}$ goes to $0$. Then, the asked limit is $0$. Is that correct?
| First of all, since $ \left(\forall x\in\left]1,+\infty\right[\right),\ \left|\cos{\left(\frac{1}{x-1}\right)}\right|\leq 1 $, we get that $ \left(\forall x\in\left]1,+\infty\right[\right),\left|\sqrt{x-1}\cos{\left(\frac{1}{x-1}\right)}\right|\leq\sqrt{x-1} $, meaning : $ \lim\limits_{x\to 1^{+}}{\sqrt{x-1}\cos{\left(\frac{1}{x-1}\right)}}=0 \cdot $
Thus,
\begin{aligned} \lim_{x\to 1^{+}}{\frac{\sin{\left(x^{3}-1\right)}\cos{\left(\frac{1}{x-1}\right)}}{\sqrt{x-1}}}&=\lim_{x\to 1^{+}}{\frac{\sin{\left(x^{3}-1\right)}}{x^{3}-1}\left(x^{2}+x+1\right)\sqrt{x-1}\cos{\left(\frac{1}{x-1}\right)}}\\ &=1\times 3\times 0 \\ \lim_{x\to 1^{+}}{\frac{\sin{\left(x^{3}-1\right)}\cos{\left(\frac{1}{x-1}\right)}}{\sqrt{x-1}}}&=0\end{aligned}
| {
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Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution. Section 5.2
Can somebody verify this solution for me?
Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution.
Let $u=x^2+2x+8$. Then $\frac{du}{dx}=2x+2$ and so $\frac{du}{2x+2}=dx$. Thus we have:
$\int \frac{16x+16}{(x^2+2x+8)^9}dx$
$= \int \frac{16x+16}{u^9}\frac{du}{2x+2}$
$= \int \frac{8(2x+2)}{u^9}\frac{du}{2x+2}$
$= \int \frac{8}{u^9}du$
$= 8 \int u^{-9} du$
$= 8 \frac{u^{-8}}{-8}+C$
$= 8 \frac{(x^2+2x+8)^{-8}}{-8}+C$
$= -(x^2+2x+8)^{-8}+C$
| You don't need online calculators to check your answers in indefinite integrals, the best thing you can learn is to derive again your result and hoping it will give you back the integrand function (of course when the calculations aren't too much brutal).
Indeed
$$\frac{\text{d}}{\text{d}x} \left(-(x^2+2x+8)^{-8}+C\right)=-(-8)(x^2+2x+8)^{-9}(2x+2)=$$
$$=8(x^2+2x+8)^{-9}(2x+2)=\frac{16x+16}{(x^2+2x+8)^9}$$
| {
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Floor function repeated addition I've come across this problem: $$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$ $$\text{Find} \
\lfloor 100r\rfloor. \textit{(Source: AIME)}$$
Here is my work:
$\lfloor r + \frac{19}{100}\rfloor = r + \frac{19}{100} - \{r + \frac{19}{100}\}$, so the figure can be restated as $\{r + \frac{19}{100}\} = r + \frac{19}{100} + a - 546$, where $a = \lfloor r + \frac{20}{100}\rfloor + \lfloor r + \frac{21}{100}\rfloor + \dots + \lfloor r + \frac{91}{100}\rfloor$.
Because $\{r + \frac{19}{100}\}$ is the fractional part, $0 \le r + \frac{19}{100} + a - 546 < 1$, so after some more maniuplation, $545 + \frac{81}{100} \le r + a < 546 + \frac{81}{100}$. $a$ is an integer, so the fractional part of $r$ must be $\frac{81}{100}$.
$r = \lfloor r\rfloor + \{r\}$, so $\lfloor r + \frac{19}{100}\rfloor = \lfloor \lfloor r\rfloor + \frac{81}{100} + \frac{19}{100}\rfloor$ = $\lfloor \lfloor r\rfloor + 1\rfloor$. Because both of the terms inside that floor function are integers, it must equal $\lfloor r\rfloor + 1$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $\lfloor r\rfloor + 1$.
Therefore, $73 \lfloor r\rfloor + 73 = 546$, so $73\lfloor r\rfloor = 473$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $\lfloor r\rfloor$ is not an integer.
If you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred.
| $$ \left\lfloor r + \frac{19}{100}\right\rfloor
+ \left\lfloor r + \frac{20}{100} \right\rfloor
+ \left\lfloor r + \frac{21}{100}\right\rfloor
+ \dots
+ \left\lfloor r + \frac{91}{100}\right\rfloor
= 546$$
First, let $r = s - \dfrac{19}{100}$. Then the sum becomes
$$ \left\lfloor s + \frac{0}{100}\right\rfloor
+ \left\lfloor s + \frac{1}{100} \right\rfloor
+ \left\lfloor s + \frac{2}{100}\right\rfloor
+ \dots
+ \left\lfloor s + \frac{72}{100}\right\rfloor
= 546$$
If $s$ were an integer, then you would get $73s = 546$, which has solution $s = 7 \dfrac{35}{73}$.
Noting that $7 \cdot 73 = 511$ and $8 \cdot 73 = 584$, we see that $7 < s < 8$. Now $546-511 = 35$. So where are you going to pick up that extra $35$?
| {
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Exponential generating function to find recurrence formula $a_n$ Given $a_0 = 1$ and $a_n = (n + 1)a_{n - 1} + 3^n$ for $n \geq 1$, find formula for $a_n$.
First remind the definition of power series $F(x) = \sum_{n \geq 0}f_n \frac{x^n}{n!}$
and exponential generating function $G(x) = \sum_{n \geq 0}\frac{x^n}{n!} = e^x$
Note we have to use exponential generating function to solve this question. So let $f(x) = \sum_{n \geq 0}a_n\frac{x^n}{n!}$, multiplying both sides of $a_n = (n + 1)a_{n - 1} + 3^n$ by $\frac{x^n}{n!}$ and sum over all $n \geq 1$. We get, $$\sum_{n \geq 1}a_n\frac{x^n}{n!} = \sum_{n \geq 1}(n + 1)a_{n - 1}\frac{x^n}{n!} + \sum_{n \geq 1}3^n\frac{x^n}{n!}$$
$$\sum_{n \geq 1}a_n\frac{x^n}{n!} = \sum_{n \geq 1}na_{n - 1}\frac{x^n}{n!} + \sum_{n \geq 1}a_{n - 1}\frac{x^n}{n!} + \sum_{n \geq 0}\frac{(3x)^{n}}{n!} - 1$$
$$\underbrace{\sum_{n \geq 0}a_n\frac{x^n}{n!}}_{f(x)} - 1 = x\underbrace{\sum_{n \geq 1}a_{n - 1}\frac{x^{n-1}}{(n-1)!}}_{f(x)} + \underbrace{\sum_{n \geq 1}a_{n - 1}\frac{x^{n}}{(n)!}}_{\int{f(x)} = F(x) + C} + e^{3x} - 1$$
$$F'(x) - 1 = xF'(x) + F(x) + e^{3x} - 1 + C$$
$$F'(x)(1 - x) - F(x) = e^{3x} + C$$
I think I'm essentially stuck here, I'm trying my best to manipulate each term so that we can express each term in terms of $f(x)$. But I can't go further from here because it doesn't like look this form can lead us to the solution.
Any feedback or suggestion?
| Given:
$$a_{n+1} = (n+2) \, a_{n} + 3^{n+1} \quad \text{with} \quad a_{0}=1$$
and
$$f(t) = \sum_{n=0}^{\infty} a_{n} \, \frac{t^n}{n!}$$
then the exponential generating functions is obtained as follows.
\begin{align}
\sum_{n=0}^{\infty} a_{n+1} \, \frac{t^n}{n!} &= \sum_{n=0}^{\infty} (n+2) \, a_{n} \, \frac{t^n}{n!} + 3 \, \sum_{n=0}^{\infty} \frac{(3 t)^n}{n!} \\
\sum_{n=0}^{\infty} (n+1) \, a_{n+1} \, \frac{t^n}{(n+1)!} &= \sum_{n=0}^{\infty} (n+1) \, a_{n} \, \frac{t^n}{n!} + f(t) + 3 \, e^{3 t} \\
\frac{d}{dt} \, \sum_{n=1}^{\infty} a_{n} \, \frac{t^n}{n!} &= \frac{d}{dt} \, \sum_{n=0}^{\infty} a_{n} \, \frac{t^{n+1}}{n!} + f(t) + 3 \, e^{3 t} \\
\frac{d}{dt} (f(t) - 1) &= \frac{d}{dt} \, (t \, f(t)) + f(t) + 3 \, e^{3 t} \\
(1 - t) \, f' - 2 \, f &= 3 \, e^{3 t} \\
(1 - t)^2 \, f' - 2 \, (1-t) \, f &= 3 \, (1-t) \, e^{3 t} \\
\frac{d}{dt} \, ( (1-t)^2 \, f) &= 3 \, (1-t) \, e^{3 t} \\
(1-t)^2 \, f(t) &= 3 \, \int (1-u) \, e^{3 u} \, du + c_{1}
\end{align}
or
$$f(t) = \frac{1}{(1-t)^2} \, \left( 3 \, \int (1-u) \, e^{3 u} \, du + c_{1} \right).$$
This leads to
$$f(t) = \frac{c_{0} + (4 - 3 t) \, e^{3 t}}{3 \, (1-t)^2}$$
and using $f(0) = 1$ yields
$$f(t) = \frac{(4 - 3 t) \, e^{3 t} - 1}{3 \, (1-t)^2}.$$
Edit
Coefficients and/or solution:
The first few terms of $a_{n}$ are $a_{n} \in \{ 1, 5, 24, 123, 696, \cdots \}_{n \geq 0}$ and, with some work, lead to the form
$$ a_{n} = \frac{(n+1)!}{3} \, \sum_{k=1}^{n+1} \frac{3^k}{k!}$$
or
$$ a_{n} = \frac{(n+1)!}{3} \, ( e_{n+1}(3) - 1 ),$$
where the finite exponential function is defined by
$$ e_{n}(x) = \sum_{k=0}^{n} \frac{x^k}{k!}.$$
In general the difference equation
$$ a_{n+1} = (n+2) \, a_{n} + b^{n+1} \quad a_{0} = 1$$
has the solution
$$ a_{n} = \frac{(n+1)!}{b} \, ( e_{n+1}(b) - 1 ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that: $\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2=\frac{1}{\phi^3}$ How to show that?
$$S=\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2=\frac{1}{\phi^3}$$
Where $F_n$ Fibonacci number
$F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}$
$$F_n^2=\frac{\phi^{2n}-2\phi^n(-\phi)^{-n}+(-\phi)^{-2n}}{5}$$
$$S=\sum_{n=1}^{\infty}(-1)^{n+1}2\left(\frac{\phi^n}{F_{n+1}F_{n+2}}\right)^2+\sum_{n=1}^{\infty}\left(\frac{1}{F_{n+1}F_{n+2}}\right)^2$$
| Lemma: Let $F_n$ be Fibonacci Numbers and $L_n$ be Lucas Numbers, then
$$
F_nF_{n+1}-F_{n-k}F_{n+k+1}=\frac{(-1)^{n+1}+(-1)^{n-k}L_{2k+1}}5\tag1
$$
Proof: Using $F_n=\frac{\phi^n-(-1/\phi^n)}{\sqrt5}$ and $L_n=\phi^n+(-1/\phi)^n$, we get
$$
\begin{align}
&5(F_nF_{n+1}-F_{n-k}F_{n+k+1})\\
&=\left(\phi^n-(-1/\phi)^n\right)\left(\phi^{n+1}-(-1/\phi)^{n+1}\right)
-\left(\phi^{n-k}-(-1/\phi)^{n-k}\right)\left(\phi^{n+k+1}-(-1/\phi)^{n+k+1}\right)\\
&=(-1)^{n+1}(\phi-1/\phi)+(-1)^{n-k}\left(\phi^{2k+1}-1/\phi^{2k+1}\right)\\
&=(-1)^{n+1}+(-1)^{n-k}L_{2k+1}
\end{align}
$$
$\large\square$
Therefore,
$$
\begin{align}
\sum_{k=0}^\infty\frac1{F_{2k+1}F_{2k+3}}
&=\sum_{k=0}^\infty\frac{F_{2k+1}F_{2k+2}-F_{2k}F_{2k+3}}{F_{2k+1}F_{2k+3}}\tag{2a}\\
&=\sum_{k=0}^\infty\left(\frac{F_{2k+2}}{F_{2k+3}}-\frac{F_{2k}}{F_{2k+1}}\right)\tag{2b}\\
&=\frac1\phi\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply the Lemma substituting $(n,k)\mapsto(2k+1,1)$
$\text{(2b)}$: algebra
$\text{(2c)}$: the partial sum to $n$ telescopes to $\frac{F_{2n+2}}{F_{2n+3}}$, which limits to $\frac1\phi$
$$
\begin{align}
\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2
&=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{F_{n+2}-F_{n+1}}{F_{n+1}F_{n+2}}\right)^2\tag{3a}\\
&=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac1{F_{n+1}^2}+\frac1{F_{n+2}^2}-\frac2{F_{n+1}F_{n+2}}\right)\tag{3b}\\
&=\sum_{n=0}^\infty(-1)^{n+1}\frac1{F_{n+1}^2}-\sum_{n=1}^\infty(-1)^{n+1}\frac1{F_{n+1}^2}\\
&+2\sum_{n=0}^\infty\left(\frac1{F_{2n+1}F_{2n+2}}-\frac1{F_{2n+2}F_{2n+3}}\right)\tag{3c}\\
&=-1+2\sum_{n=0}^\infty\frac1{F_{2n+1}F_{2n+3}}\tag{3d}\\[3pt]
&=-1+\frac2\phi\tag{3e}\\[6pt]
&=\frac1{\phi^3}\tag{3f}
\end{align}
$$
Explanation:
$\text{(3a)}$: $F_n=F_{n+2}-F_{n+1}$
$\text{(3b)}$: algebra
$\text{(3c)}$: break the sum into three pieces
$\phantom{\text{(3c):}}$ substitute $n\mapsto n-1$ in the second piece
$\phantom{\text{(3c):}}$ break the third piece into two to remove $(-1)^{n+1}$
$\text{(3d)}$: cancel identical terms in the first two sums
$\phantom{\text{(3d):}}$ combine terms in the third sum using $F_{2k+3}-F_{2k+1}=F_{2k+2}$
$\phantom{\text{(3d):}}$ and cancel $F_{2k+2}$ in the numerator and denominator
$\text{(3e)}$: apply $(2)$
$\text{(3f)}$: $2-\phi=\frac1{\phi^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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