Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Square root of 9 from a technical perspective. When we say $$ \sqrt{9}= x $$
then $\;x = 3,\;$ right?
So why when we square both sides it becomes different:
$$ (\sqrt{9})^2 = x^2$$
$$9 = x^2$$
Here $\;x =\pm 3.$
So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$
| Just because $A$ is something that does $P$, does not mean that $A$ is the ONLY thing that does $P$.
It is true that $3^2 = 9$ but $3$ is not the only thing that when squared will equal $9$.
so $3^2 =9$ and $w^2 = 9$ does not in any way mean that $w = 3$.
This is no more valid or reasonable than saying $(1)^2- 3(1) +7 = 5$ and $(2^2 - 3(2) + 7 = 5$ and therefor $1 = 2$.
Or saying if Fred is a member of the chess club, and Betty is a member of the chess club, then Fred and Betty are the same person.
....
Your confusion is based on what is the definition of $\sqrt{9}$.
The definition of $\sqrt{k}$ has TWO parts. If $\sqrt{k} = m$ then TWO things must be true: 1) $m^2 = k$.... It's a common mistake to think this is the only thing that must be true but we must also have 2) $m \ge 0$.
So although $3^2 = 9$ and $(-3)^2 = 9$ so both have 1) true. $3>0$ but $-3< 0$ so only $3$ has 2) true and $-3$ has 2) false.
So $\sqrt 9 = 3$ and $-3\ne \sqrt 9$.
Now just because $3 = \sqrt 9$. ANd $(\sqrt 9)^2 = 9$, it does not follow that $\sqrt 9$ is the ONLY thing that when squared equals $9$. $\sqrt 9$ is only one thing that when squared equals $9$.
Just like Fred isn't the only member of the chess club and Betty can also be a member; $\sqrt 9$ is not the only thing that when squared is equal to $9$. $-3$ when squared is also equal to $9$ when squared.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the sum of the maximum of the trigonometric polynomial coefficient somedays ago,I have studying and discussing this Maximum Trigonometric polynomial coefficient
problem
Now I want consider a simple problem:
(The sum of the maximum of the trigonometric polynomial coefficient:) Let $n$ be give postive integers If for all real numbers $x$ we have $$f(x)=A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\ge -1$$
Find the maximum value of $A_{1}+A_{2}+\cdots+A_{n}$.
Here is what I tried:use this well kown :
$$\sum_{j=0}^{n-1}\xi^k_{j}=\begin{cases}
n&n|k\\
0&otherwise
\end{cases}$$
where $\xi_{i}$be the $n$th roots of unity
let $x_{k}=\dfrac{2k\pi}{n+1},k=0,1,\cdots,n$,so we have
$$\sum_{k=0}^{n}\cos{(mx_{k})}=0,m=1,2,\cdots,n$$
so $$\sum_{k=0}^{n}f(x_{k})=\sum_{m=1}^{n}\sum_{k=0}^{n}A_{m}\cos{(mx_{k})}=0$$
then we have
$$A_{1}+A_{2}+\cdots+A_{n}=f(0)=-\sum_{k=1}^{n}f(x_{k})\le n$$
I Conjecture the maximum is $n$.But I can't find example such reach this maximum
I have find example for $n=2$,then let $A_{1}=\dfrac{4}{3},A_{2}=\dfrac{2}{3},A_{1}+A_{2}=2$,then
$$f(x)=\dfrac{1}{3}(2\cos{x}+1)^2-1\ge -1$$
for all postive $n$,I can't find it.Thanks
| Solution due to @quasi:
Let $A_k = 2 - \frac{2k}{n+1}$. We have $A_1 + A_2 + \cdots + A_n = n$.
We have
\begin{align}
&\sum_{k=1}^n A_k \cos(kx) + 1\\
=\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\sum_{k=1}^n k \cos (kx) + 1\\
=\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\left(\sum_{k=1}^n \sin (kx)\right)' + 1\\
=\ & 2\left(\frac{-\sin \frac{x}{2} + \sin\frac{(2n+1)x}{2}}{2\sin \frac{x}{2}}\right)
- \frac{2}{n+1}\left(\frac{\cos \frac{x}{2} - \cos \frac{(2n+1)x}{2}}{2\sin \frac{x}{2}}\right)' + 1\\
=\ & \frac{\sin x \cos(nx) + \cos x \sin (nx) - \sin x + \sin(nx)}{\sin x}\\
&\quad - \frac{2}{n+1}\left(\frac{1 + \cos x - \cos(nx) - \cos x \cos (nx) + \sin x \sin (nx)}{2\sin x}\right)' + 1\\
=\ & \frac{1 - \cos((n+1)x)}{(n+1)(1-\cos x)}\\
\ge\ & 0.
\end{align}
Update: We can find $A_k = 2 - \frac{2k}{n+1}$ through the following steps.
Note that when $A_k$ is a polynomial in $k$, $\sum_{k=1}^n A_k \cos(kx)$ has a closed form.
We first try $A_k = a_n + b_n k$.
From $\sum_{k=1}^n A_k = a_n n + \frac{b_n n (n+1)}{2} = n$, we have $a_n + \frac{b_n (n+1)}{2} = 1$.
Thus, $b_n = \frac{2(1-a_n)}{n+1}$.
We need
\begin{align}
\sum_{k=1}^n \left(a_n + \frac{2(1-a_n)}{n+1}k\right) \cos(kx) + 1\ge 0, \ \forall x \in (0, 2\pi). \tag{1}
\end{align}
When $n=2$, (1) becomes
$$\frac{1 - \cos x}{3}(1 + 2\cos x)\Big(a_2 + \frac{4\cos x - 1}{1 - \cos x}\Big) \ge 0, \ \forall x \in (0, 2\pi). \tag{2}$$
It is easy to prove that (2) holds if and only if $a_2 = 2$.
When $n=3$, (1) becomes
$$2\cos x (1 - \cos^2 x)\Big(a_3 + \frac{3\cos x - 2}{1 - \cos x}\Big) \ge 0, \ \forall x \in (0, 2\pi). \tag{3}$$
It is easy to prove that (3) holds if and only if $a_3 = 2$.
We guess that when $a_n = 2$, (1) holds. Fortunately, it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got
$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$
$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?
What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?
update
When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:
$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$
$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$
$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.
2nd update
I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:
$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} + 1)^2=9 \implies
(\frac{y^2}{4} + 1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$
| A standard Cartesian graph deals only with real numbers "plotted against" real numbers, so in looking at the intersections of two curves in such a graph, one may not be "getting the whole story." Two distinct quadratic curves will have four intersections; since one of these (the circle) is bounded, all of the intersection points will be at finite distances from the origin.
If we use the curve equations with complex numbers $ \ x \ = \ a + bi \ $ and $ \ y \ = \ c + di \ \ , $ (with $ \ a \ , \ b \ , \ c \ , \ d \ $ real), they become
$$ (x-2)^2 \ + \ y^2 \ \ = \ \ 9 \ \ \rightarrow \ \ ( \ [a - 2] + bi \ )^2 \ + \ ( \ c \ + \ di \ )^2 \ \ = \ \ 9 $$
$$ \rightarrow \ \ [ \ (a^2 - 4a - b^2 + 4) \ + \ (2a - 4 )·b·i \ ] \ + \ [ \ (c^2 - d^2) \ + \ 2cd·i \ ] \ \ = \ \ 9 \ + \ 0·i \ \ ; $$
$$ y^2 \ \ = \ \ 4x \ - \ 4 \ \ \rightarrow \ \ (c + di)^2 \ \ = \ \ 4·(a + bi) \ - \ 4 \ \ = \ \ (4a - 4) \ + \ 4bi \ \ . $$
The calculation without the imaginary parts ($ \ b \ = \ d \ = \ 0 \ $) is the one you made (the one we learn to make in "standard" analytic geometry), giving $ \ \ a^2 \ = \ 9 \ \ , \ \ c^2 \ = \ 4·a - 4 \ \ . $ We take $ \ a \ = \ +3 \ \ , $ $ c^2 \ = \ 4·(+3) - 4 \ = \ 8 \ \ , $ and "discard" $ \ a \ = \ -3 \ $ , since it leads to $ \ c^2 \ = \ 4·(+3) - 4 \ = \ -16 \ \ $ , which is not permissible since $ \ c \ $ is taken to be a real number. Your graph then represents real part $ \ c \ $ for $ \ y \ $ as a function of real part $ \ a \ \ $ for $ \ x \ \ . $
But a "graph" using complex numbers needs to be made in $ \ \mathbb{C}^2 \ \ , $ which can be interpreted in $ \ \mathbb{R}^4 \ \ , $ making it challenging to visualize, as the "curves" are then treated as four-dimensional. The intersections we "rejected" occur in those extended parts of the curves. If we return to the curve equations and work with the two-dimensional "slice" that "graphs" the imaginary part $ \ d \ $ of $ \ y \ $ against the real part $ \ a \ $ of $ \ x \ \ $ (so $ \ b \ = \ c = \ 0 \ $ ) , we obtain
$$ ( \ [a - 2] + 0·i \ )^2 \ + \ ( \ 0 + di \ )^2 \ \ = \ \ 9 \ \ \rightarrow \ \ (a - 2)^2 \ - \ d^2 \ \ = \ \ 9 \ \ ; $$
$$ (0 + di)^2 \ \ = \ \ 4·(a + 0·i) \ - \ 4 \ \ \rightarrow \ \ -d^2 \ \ = \ 4a \ - \ 4 \ \ . $$
This produces $ \ a^2 \ = \ 9 \ \ , \ \ -d^2 \ = \ 4a - 4 \ \ , $ but using $ \ a \ = \ +3 \ $ will give us $ \ -d^2 \ = \ 8 \ \ , $ which we now "reject" because $ \ d \ $ is a real number; here, we take $ \ a \ = \ -3 \ \Rightarrow \ -d^2 \ = \ -16 \ \Rightarrow \ d \ = \ \pm 4 \ \ . $ As we see on the graph of $ \ d \ $ versus $ \ a \ $ below, we have found that a hyperbolic part of $ \ (x-2)^2 + y^2 \ = \ 9 \ $ intersects a parabolic part of $ \ y^2 \ \ = \ \ 4x \ - \ 4 \ $ at the points $ \ (-3 \ , \ \pm 4i) \ \ , $ as mentioned by Narasimham.
We have accounted then for the four intersections of the two curves, two of which are "invisible" on a "standard" graph in $ \ \mathbb{R}^2 \ \ . $ No other intersections appear on other "slices": for instance, the imaginary part $ \ d \ $ versus imaginary part $ \ b \ $ equations are
$$ ( \ [0 - 2] + b·i \ )^2 \ + \ ( \ 0 + di \ )^2 \ \ = \ \ 9 \ \ \rightarrow \ \ 4 \ - \ b^2 \ - \ d^2 \ \ = \ \ 9 \ \ \rightarrow \ \ b^2 \ + \ d^2 \ \ = \ \ -5 \ \ ; $$
$$ (0 + di)^2 \ \ = \ \ 4·(0 + b·i) \ - \ 4 \ \ \rightarrow \ \ -d^2 \ \ = \ 4bi \ - \ 4 \ \ , $$
which do not produce real solutions for $ \ b \ $ and $ \ d \ \ . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find what a Infinite product number approaches Find what the following number approaches:
$$\frac{6}{5}\times\frac{26}{25}\times\frac{626}{625}\times\frac{390626}{390625}\times...$$
What I found is that each fraction can be simplified as:$$1+\frac{1}{5^n}$$
Where $n$ is $1,2,4,8$ and so on (instead of $1,2,3,4,\ldots$)
What should I do next?
Note: Please use junior high school math if possible
| You have
$$\left(1+\frac{1}{5^1}\right)\left(1+\frac{1}{5^2}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$
$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$
$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}\right)\left(1+\frac{1}{5^8}\right)\cdots$$
$$=1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}+\frac{1}{5^8}+\cdots$$
$$= \frac{5}{5-1} \\= 1+\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
From the system of equations prove $a+b+c=0$
$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:
$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
{c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$
I added all the equations together and get:
$$a^3+b^3+c^3+(a+b+c)x+3y=0$$
It is similar to Euler identity (because we have $a^3+b^3+c^3$). if $a+b+c=0$ then from Euler identity we can conclude $a^3+b^3+c^3=3abc$. and equation change to:
$$3abc+3y=0$$
But it seems that doesn't work.
| Here's an alternative solution to Albus Dumbledore's. Note that the three equations imply that the nonzero vector $(1,x,y)$ is in the kernel of
$$\begin{pmatrix}a^3&a&1\\b^3&b&1\\c^3&c&1\end{pmatrix}$$
of determinant (after factoring)
$$(a-b)(a-c)(b-c)(a+b+c).$$
But also, the determinant must be 0 since it has nontrivial kernel. So $(a-b)(a-c)(b-c)(a+b+c)=0$ and by the assumption that $a,b,c$ are distinct, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding all natural $x$, $y$, $z$ satisfying $7^x+1=3^y+5^z$ The problem goes as follows:
Find all possible pairs of $x,y,z \in \mathbb{N}$ which satisfy the equation $7^x+1=3^y+5^z$
My first instinct was to continue by modding, but I don't think I can get anything out of it. The obvious solutions seem to be $x,y,z=1$ and $x,y,z=0$, but I am not really sure how to approach the problem.
Thanks in advance!
| The two solutions you found are indeed all of them. In fact, we can solve this with a straight mod bash, which was surprising to me.
If any of $x,y,z$ are $0$, then we easily get $x=y=z=0$. If $y=1$, then $7^x=5^z+2$, and we readily get that $z=1$, so $x=1$ too. So now assume $y>1$ and $x,z>0$. Let the mod bash begin.
Taking mod $3$ implies that $z$ is odd. Taking mod $4$ implies that $x$ and $y$ have the same parity. Taking mod $5$ then implies that $x\equiv y\equiv 1\pmod 4$.
Now taking mod $7$, we get that either (a) $y\equiv1\pmod{12}$ and $z\equiv1\pmod{6}$ or (b) $y\equiv5\pmod{12}$ and $z\equiv5\pmod6$.
Now taking mod $9$, in case (a) we have $x\equiv5\pmod{12}$ and in case (b) we have $x\equiv9\pmod{12}$.
Consider case (a). If $z\equiv1\pmod{12}$, then taking mod $13$ gives $11+1\equiv3+5\pmod{13}$; if $z\equiv7\pmod{12}$ we still get a contradiction $11+1\equiv8+5\pmod{13}$.
Similarly in case (b), we get a contradiction in mod $13$, considering both cases $z\equiv 5\pmod{12}$ and $z\equiv11\pmod{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
USA MO 1980's inequality Let $x;y;z$ be real numbers and $x;y;z \in [0;1]$
Prove: $\dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$
The official solution uses calculus, so I can't understand it.
I hope someone have a "simpler solution" (for middle-school student). Thank you.
| Assume $x+y+z>0$, else the inequality is easy. Do some rearranging:
\begin{align*}(1-x)(1-y)(1-z)\leq1-\sum\frac{x}{y+z+1}&=\sum\left(\frac{x}{x+y+z}-\frac{x}{y+z+1}\right) \\
&= \frac{1}{x+y+z}\sum\frac{x(1-x)}{y+z+1}
\end{align*}
Multiplying through by $x+y+z>0$, and moving everything to one side gives
$$\sum x(1-x)\left[\frac{1}{y+z+1}-(1-y)(1-z)\right]\geq0.$$
Now it's just AM-GM:
$$1=\frac{(1-y)+(1-z)+(y+z+1)}{3}\geq\sqrt[3]{(1-y)(1-z)(y+z+1)},$$
so $\frac{1}{y+z+1}\geq(1-y)(1-z)$, which implies the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
How to find all third roots of $-i$? How to find all third roots of $-i$?
I have no idea.
Could you give me some hint?
| If you know de Moivre's law, or representation of complex numbers as exponentials, this problem is simple.
If you don't, here is an algebraic approach.
There exists some complex number $a+bi$ such that $(a+bi)^3 = -i$
Multiplying it out and combining terms.
$a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3 = -i\\
(a^3 - 3ab^2) + (3a^2b - b^3)i = -i$
Set the real parts and the imaginary parts equal to one annother.
$a^3 -3ab^2 = 0\\
a^2b - b^3 = -1$
Working with the first equation.
$a(a^2-3b^2) = 0$
$a = 0$ or $a^2 = 3b^2$
If $a = 0$ substituting into the second equation.
$-b^3 = -1\\
b = -1\\
a+bi = -i$
This is one of our answers.
If $a^2 = 3b^2$
$9b^3 - b^3 = -1\\
8b^3 = -1\\
b^3 = -\frac {1}{8}\\
b = -\frac {1}{2}$
Substituting back into $a^2 = 3b^2$
$a^2 = \frac {3}{4}\\
a = \pm \frac {\sqrt 3}{2}\\
\frac {\sqrt 3}{2} - \frac 12 i, -\frac {\sqrt 3}{2} - \frac 12 i$
are our other 2 solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral involving ratios of polynomial and square roots Is there a simple way to compute the following integral
$$I(y,\beta,\gamma)=\int_1^y \frac{x^2}{\sqrt{\left(x^2-1\right)\left(x^2-\gamma ^2\right) \left(x^2+\beta ^2\right) }}\mathrm{d}x$$
with $y>1$, $0<\gamma<1$ and $\beta>0$?
|
Define the function $\mathcal{I}:\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)\rightarrow\mathbb{R}$ via the improper integral
$$\mathcal{I}{\left(p,q,y\right)}:=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}.$$
Suppose $\left(p,q,y\right)\in\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)$, and set $z:=y^{2}\land a:=1\land b:=p^{2}\land c:=0\land d:=-q^{2}$.
Note that $z>a>b>c>d$, and then also
$$0<\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}<1\land0<\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}<1.$$
The integral $\mathcal{I}$ can be transformed into an elliptic integral by means of a simple power substitution:
$$\begin{align}
\mathcal{I}{\left(p,q,y\right)}
&=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}\\
&=\int_{1}^{y^{2}}\mathrm{d}t\,\frac{t}{\sqrt{t}\sqrt{\left(t-1\right)\left(t-p^{2}\right)\left(t+q^{2}\right)}};~~~\small{\left[x=\sqrt{t}\right]}\\
&=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\sqrt{\left(t-a\right)\left(t-b\right)\left(t-c\right)\left(t-d\right)}}.\\
\end{align}$$
Next, consider the linear fractional transformation given implicitly by the relation
$$\left(t-b\right)\left(1-u\right)=\left(a-b\right).$$
Then,
$$t=\frac{a-bu}{1-u}\implies dt=du\,\frac{\left(a-b\right)}{\left(1-u\right)^{2}},$$
and
$$\frac{t-a}{t-b}=u\land\frac{t-c}{t-b}=\frac{a-c}{a-b}-\frac{b-c}{a-b}u\land\frac{t-d}{t-b}=\frac{a-d}{a-b}-\frac{b-d}{a-b}u.$$
Applying this LFT to $\mathcal{I}$, we obtain
$$\begin{align}
\mathcal{I}{\left(p,q,y\right)}
&=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\sqrt{\left(t-a\right)\left(t-b\right)\left(t-c\right)\left(t-d\right)}}\\
&=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\left(t-b\right)^{2}\sqrt{\left(\frac{t-a}{t-b}\right)\left(\frac{t-c}{t-b}\right)\left(\frac{t-d}{t-b}\right)}}\\
&=\int_{0}^{\frac{z-a}{z-b}}\mathrm{d}u\,\frac{\left(a-b\right)}{\left(1-u\right)^{2}}\cdot\frac{a-bu}{1-u}\cdot\frac{\left(1-u\right)^{2}}{\left(a-b\right)^{2}}\\
&~~~~~\times\frac{1}{\sqrt{u\left(\frac{a-c}{a-b}-\frac{b-c}{a-b}u\right)\left(\frac{a-d}{a-b}-\frac{b-d}{a-b}u\right)}};~~~\small{\left[t=\frac{a-bu}{1-u}\right]}\\
&=\int_{0}^{\frac{z-a}{z-b}}\mathrm{d}u\,\frac{a-bu}{1-u}\cdot\frac{1}{\sqrt{u\left[\left(a-d\right)-\left(b-d\right)u\right]\left[\left(a-c\right)-\left(b-c\right)u\right]}}\\
&=\int_{0}^{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\mathrm{d}v\,\frac{a-b\left(\frac{a-d}{b-d}\right)v}{1-\left(\frac{a-d}{b-d}\right)v}\\
&~~~~~\times\frac{1}{\sqrt{v\left(1-v\right)\left[\left(a-c\right)\left(b-d\right)-\left(b-c\right)\left(a-d\right)v\right]}};~~~\small{\left[u=\left(\frac{a-d}{b-d}\right)v\right]}\\
&=\frac{1}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\mathrm{d}v\,\left[\frac{a-b}{1-\left(\frac{a-d}{b-d}\right)v}+b\right]\frac{1}{\sqrt{v\left(1-v\right)\left[1-\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}v\right]}}\\
&=\frac{1}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\sin^{2}{\left(\varphi\right)}}\mathrm{d}v\,\left[\frac{a-b}{1-\eta\,v}+b\right]\frac{1}{\sqrt{v\left(1-v\right)\left(1-\kappa^{2}v\right)}}\\
&=\frac{2}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\sin{\left(\varphi\right)}}\mathrm{d}x\,\left[\frac{a-b}{1-\eta\,x^{2}}+b\right]\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-\kappa^{2}x^{2}\right)}};~~~\small{\left[v=x^{2}\right]}\\
&=\frac{2}{\sqrt{\left(a-c\right)\left(b-d\right)}}\left[\left(a-b\right)\Pi{\left(\varphi,\eta,\kappa\right)}+bF{\left(\varphi,\kappa\right)}\right],\\
\end{align}$$
where we've introduced the auxiliary parameters
$$\kappa:=\sqrt{\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}}\in\left(0,1\right),$$
$$\varphi:=\arcsin{\left(\sqrt{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\right)}\in\left(0,\frac{\pi}{2}\right),$$
$$\eta:=\frac{a-d}{b-d}\in\left(1,\infty\right).$$
Note: even though $\eta>1$, we can still define $\Pi$ in the usual way without worrying about Cauchy principal values since $0<1-\eta\sin^{2}{\left(\varphi\right)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation: $\lfloor a/2\rfloor+\lfloor a/3\rfloor+\lfloor a/5\rfloor=a$, where $a> 0$. Let $a\in\mathbb{Z}^+$, then $a$ we can write it as $a = 30k + i$, where $i\in\{0,1,2,\ldots, 29\}$ and $k\in\mathbb{Z}_0^+$. Thus substituting we have:
\begin{align*}
\lfloor(30k + i)/2\rfloor + \lfloor(30k + i)/3\rfloor + \lfloor(30k + i)/5\rfloor &= 30k + i\\
\lfloor 15k + (i/2)\rfloor + \lfloor 10k + (i/3)\rfloor + \lfloor 6k + (i/5)\rfloor &= 30k + i\\
15k + \lfloor i/2\rfloor + 10k + \lfloor i/3\rfloor + 6k + \lfloor i/5\rfloor &= 30k + i\\
31k + \lfloor i / 2 \rfloor + \lfloor i / 3 \rfloor + \lfloor i / 5 \rfloor & = 30k + i \\
k &= i - (\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5\rfloor)
\end{align*}
Thus we have for each $i$ we will have a value of $k$. Also note that if $i=0$,
\begin{align*}
k &= 0 -(\lfloor 0/2\rfloor + \lfloor 0/3\rfloor + \lfloor 0/5 \rfloor) \Rightarrow k=0.
\end{align*}
So $a=0$, but $a>0$. So $i=0$ is not taken, for all others we have a solution.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| Yes, this is an excellent solution.
It would perhaps be a good idea, after obtaining
\begin{align*}
k &= i -(\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5 \rfloor)
\end{align*}
to show that $k$ is non-negative. This is not totally trivial and, since$$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}>1,$$ it is only just true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
$\langle A,B \rangle$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$ Prove that the subgroup of $SL_2(F_3)$ generated by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$.
Proof: Let $G=SL_2(F_3)$. $|G|=24=2^3\cdot 3$. Let
$A =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
and
$B =\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
By the Sylow Theorems, let $n_2$ be the number of Sylow $2$-subgroup. $n_2\equiv 1 \mod 2$ and $n_2 | 3 \Rightarrow n_2=1,3$, and Sylow $2$-subgroup(s) have order $8$. We need to check $|\langle A,B \rangle | = 8$. Note $A^4=B^4=I$. We find
$$
\langle A,B \rangle
= \left\{ \pm I, \pm A, \pm B, \pm \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}\right\}
$$
What is left to show is $\langle A,B\rangle$ is the only one. We can do that by show $\langle A,B \rangle$ is normal.
i.e. $gng^{-1} \in \langle A,B \rangle$ $\forall g=\begin{pmatrix}
a & b \\ c & d \end{pmatrix}$ and $n\in N$.
So this is where I am stuck. I know I can do the computation for each element of $\langle A,B \rangle$ but that would suck. Is there a way to avoid that or not?
| Let $P = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$. Then $P^3 = I$ and $P^{-1} = P^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. Since $| \langle P \rangle | = 3$ and $| \langle A, B \rangle | = 8$ we have $\langle P \rangle \cdot \langle A, B \rangle = \langle P, A, B \rangle = SL_2(\mathbb{F}_3)$.
Calculating $PAP^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = -B \in \langle A, B \rangle$,
and
$PBP^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}= AB \in \langle A, B \rangle$.
Of course $P$ commutes with $\pm I$ and since these elements generate $\langle A, B \rangle$ we have that conjugation by $P$ fixes $\langle A, B \rangle$. Since $\langle A, B \rangle$ is closed under conjugation by every element of itself, and by $P$, and since these elements generate $SL_2(\mathbb{F}_3)$, we have that $\langle A,B \rangle$ is closed under conjugation by any element, and thus is normal. Since all Sylow 2-subgroups are conjugate, there is only one, namely $\langle A, B \rangle$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$
I am aware of using the recursive method to find the limit: setting $x = \sqrt{2x}$ and getting $x = 2$, rejecting $x = 0$ due to the sequence being monotone and positive.
I now want to show that the limit of this sequence is $2$ using the epsilon-neighbourhood definition for a convergent series:
Let $(a_n)$ be a sequence that converges to a real number $a$. Then, for every number $\epsilon > 0$, there exists a number $N \in \mathbb{N}$ such that for all $n \geq N$, $|a_n - a| < \epsilon$.
I take the logarithm on both sides and arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \ln 2 - \ln 2 \right| < \epsilon. $$
Using triangle inequality, $|x - y| \geq |x| - |y|$, I arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| - |\ln 2| < \epsilon. $$
Add $\ln 2$ to both sides
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| < \epsilon + \ln 2. $$
The left hand side can be rewritten as a sum of $n$ terms in a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$,
$$ \frac{\frac{1}{2} \left( 1 - \left( \frac{1}{2} \right)^n \right)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} < \epsilon + \ln 2. $$
I am unsure of how to proceed beyond this point.
| Let's make an observation: for each natural number $n$, your sequence can be given explicitly by $$a_n = 2^{\sum_{i=1}^n (\frac{1}{2})^i}$$
For example, $a_2 = \sqrt{2\sqrt{2}} = (2(2^{1/2}))^{1/2} = 2^{1/2} \cdot 2^{1/4} = 2^{1/2 + 1/4}$
Furthermore, we know that $${\sum_{i=1}^n \left(\frac{1}{2}\right)^i} \to 1 \ \ \text{as}
\ \ n \to \infty$$
We also know that the function $f(x) = 2^x$ is a continuous function. In other words, if we pick $\epsilon > 0$, there is $\delta > 0$ such that $|x-1| < \delta$ implies $|f(x) - f(1)|< \epsilon$.
Thus, because $$\sum_{i=1}^n \left(\frac{1}{2}\right)^i \to 1$$there is $M > 0$ so that $n > M$ implies $$\left|{\sum_{i=1}^n \left(\frac{1}{2}\right)^i -1}\right| < \delta$$
Hence, letting $x = \sum_{i=1}^n \left(\frac{1}{2}\right)^i$ in the paragraph where I mentioned continuity, we obtain $$\left|f\left(\sum_{i=1}^n \left(\frac{1}{2}\right)^i\right) - 2\right| < \epsilon$$
This is precisely $|a_n -2 | < \epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$
In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$
(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)
Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.
I was referred by a friend to the identities
$$\begin{align}
a+b+c &= 2s \tag{1} \\[4pt]
ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt]
abc &= 4Rrs \tag{3}
\end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.
Any other idea?
| The area of the triangle can be expressed as $ \frac12r(a+b+c)= \frac{abc}{4R}$. Then
\begin{align}
\frac rR =\frac12 \frac{\frac{abc}{R^3}}{\frac{a+b+c}R}
=\frac{2\sin A \sin B \sin C}{\sin A +\sin B +\sin C}\tag1
\end{align}
Note
$$\sin A +\sin B +\sin C
= 2\cos\frac A2\sin\frac A2 + 2\sin\frac {B+C}2\cos\frac{B-C}2\\
= 2 \cos\frac A2 (\cos\frac {B+C}2+\cos\frac{B-C}2) = 4 \cos\frac A2 \cos\frac B2 \cos\frac C2
$$
Substitute into (1)
$$\frac rR = 4 \sin\frac A2 \sin\frac B2 \sin\frac C2\tag2
$$
Similarly
$$\cos A +\cos B +\cos C
= 1-2\sin^2\frac A2 +2\cos\frac {B+C}2\cos\frac{B-C}2\\
=1- 2 \sin\frac A2 (\cos\frac {B+C}2-\cos\frac{B-C}2)=1+ 4 \sin\frac A2 \sin\frac B2 \sin\frac C2
$$
Substitute into (2)
$$\cos A+\cos B+\cos C=1+\frac rR$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Writing a group as a product of its generators in MAGMA Let $G \subseteq S_8$ be generated by the elements $s=\begin{pmatrix} 1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 5 \end{pmatrix}\begin{pmatrix} 4 & 6 \end{pmatrix}\begin{pmatrix} 7 & 8 \end{pmatrix}$ and $t=\begin{pmatrix} 1 & 3 & 7 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5 & 8 & 6 \end{pmatrix}$.
Now I would like to see how to write the element
$$u = \begin{pmatrix} 1 & 8 \end{pmatrix}\begin{pmatrix} 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & 6 \end{pmatrix}\begin{pmatrix} 4 & 5 \end{pmatrix} \in S_8$$
can be written as a product of $s,s^{-1}$ and $t,t^{-1}$ in MAGMA (assuming that $u \in G$).
I wrote the following code
G<s,t> := PermutationGroup< 8 | (1, 2)(3, 5)(4, 6)(7, 8) , (1, 3, 7, 4)(2, 5, 8, 6) >;
u := G ! (1, 8)(2, 7)(3, 6)(4, 5);
Are there any commands to give me my desired result?
| The code below will do what you want - in fact the answer is rather easy! This method will work provided that the group is not too big - say maximum order about $10^7$ or maybe $10^8$. Although not really necessary for solving this problem, it computes a presentation of the group $G$.
> G<s,t> := PermutationGroup< 8 |
(1, 2)(3, 5)(4, 6)(7, 8) , (1, 3, 7, 4)(2, 5, 8, 6) >;
> u := G ! (1, 8)(2, 7)(3, 6)(4, 5);
> F<S,T>, phi := FPGroup(G);
> u @@ phi;
S * T^2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
In a course there's 10 boys and 10 girls and we need to divide them into 3 different groups.. We have $3$ groups $A,B,C$.
We have to divide all the boys and the girls into these three groups such that in group $C$, there's exactly $6$ or $7$ members, there's atleast one girl, and the number of the boys must be greater than the number of girls.
in how many ways we can do that?
My attempt: I'm trying to split this problem into cases:
Case 1: in group $C$ theres $5$ boys and $1$ girl, or $6$ boys and $1$ girl. for the rest of the boys and girls I would have to choose a group for each one from $A,B$ so we get $2^{14}+2^{13}$.
Case 2: in group $C$ theres $4$ boys and $2$ girls or $5$ boys and $2$ girls. so $2^{14}+2^{13}$.
Case 3: in group $C$ theres $4$ boys and $3$ girls: $2^{13}$.
Then I summed up all the results and got $57344$ ways and the answer was $50610$.
What am I missing? Appreciate any help, thanks in advance.
| It seems that the question is only about the possible combinations for group $C$.
This is where the answer 50610 comes from:
$$\binom{10}{5}\cdot\binom{10}{1} + \binom{10}{6}\cdot\binom{10}{1} + \binom{10}{4}\cdot\binom{10}{2} + \binom{10}{5}\cdot\binom{10}{2} + \binom{10}{4}\cdot\binom{10}{3} = 50610$$
For example: to choose 5 boys (out of 10) and 1 girl (out of 10) to put into group $C$ we have $\binom{10}{5}\cdot\binom{10}{1}$ possibilities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this using $x+2y=8$?
| The AM-GM inequality, as in the post tag, gives us a lower bound.
Indeed,
$$x+\frac{3}{x} + y +\frac{9}{2y}=2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}.$$
Since $x,y>0$ we can apply the AM-GM inequality to obtain
$$2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}\geq 2\sqrt{3}+2\sqrt{\frac{9}{2}}=2(\sqrt 3 + \sqrt\frac{9}{2})\sim 7.71.$$
However, this is just a lower bound.
To find the minimum, we have $$x+y+\frac{3}{x}+\frac{9}{2y}=x+2y-y + \frac{3}{x+2y-2y}+\frac{9}{2y}$$
which is equal to $$8-y +\frac{3}{8-2y}+\frac{9}{2y}.$$
Note that since $x=8-2y>0$, we want $0<y<4$.
A local minimum occurs at $y=3$.
One can see this by solving for when the derivative of $8-y +\frac{3}{8-2y}+\frac{9}{2y}$ is $0$.
At $y=3$, we have $x=2$ so that $2+3+\frac{3}{2}+\frac{9}{6}=8$ is the minimum desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solve $x^3-1=2y^2$ integer solution find the equation integer solution
$$x^3-1=2y^2$$
I have read this proof:proof, have simple methods? Because this problem is a Olympiad problem. So maybe have other methods to solve it
since $x>0.y>0$ so
$$(x-1)(x^2+x+1)=2y^2$$ since
$(x-1,x^2+x+1)=1$ or$3$.so
(1):if $x-1=2u^2,x^2+x+1=v^2,y=uv,u>0,v>0$,then wehave
$$(2v)^2=(2x+1)^2+3\Longrightarrow (2v-2x-1)(2v+2x+1)=3$$
then
$$\begin{cases}
2v+2x+1=3\\
2v-2x-1=1
\end{cases},~~\begin{cases}
2v+2x+1=1\\
2v-2x-1=3
\end{cases}$$
this two case has no solution $x>0$
(2): $$x-1=6u^2,x^2+x+1=3v^2,y=3uv,u,v>0$$
then $$(2v)^2-3(4u^2+1)^2=1
$$ it's pell equation, maybe this following I can't work. Thanks for your help!
| As suggested in the proof you link, it is easier to pass to $\Bbb{Z}[\sqrt{-2}]$ right from the start:
If $x$ and $y$ are integers satisfying $x^3-1=2y^2$, then
$$x^3=1+2y^2=(1+y\sqrt{-2})(1-y\sqrt{-2}),$$
where the two factors on the right hand side are coprime because their sum is $2$ and their product is odd. Then because $\Bbb{Z}[\sqrt{-2}]$ is a unique factorization domain with trivial unit group it follows that
$$1+y\sqrt{-2}=(a+b\sqrt{-2})^3,$$
for some integers $a$ and $b$. Expanding the right hand side and comparing coefficients then shows that
$$1=a^3-6ab^2=a(a^2-6b^2),$$
which clearly has no integral solutions apart from $(a,b)=(1,0)$, corresponding to $(x,y)=(1,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.
and please don't use any $\sum_{cyc}$ because I get confused with it.
| let $p=\frac{a^2}{a^2+bc},q=\frac{b^2}{b^2+ac},r=\frac{c^2}{c^2+ab}$ then easy to see $p,q,r\in (0,1)$ Now notice that $$\frac{(1-p)(1-q)(1-r)}{pqr}=1$$ so $$p+q+r=1+pq+qr+rp-2pqr$$ It remains to prove $$pq+qr+rp-2pqr\le 1$$ $$\iff \underbrace{p(1-q)(r-1)}_{\le 0}+\underbrace{q(1-r)(p-1)}_{\le 0}+\underbrace{(1-p)(q-1)}_{\le 0}\le 0$$ which is obvious as each term is $\le 0$ as shown
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$, $\frac{N^2+N}{2}$ has a remainder of 2
I'd like a correct solution as well. Thanks.
| $(2N+1)^{2} \equiv 1 $
$2N+1 \equiv 1$
This step is wrong, for example: $(2*3+1)^2 \equiv 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Close-form for triple integral $ \int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$ I am able to work out the double integral
$$\int_0^b \int_0^a \sqrt{x^2+y^2} dx dy $$
with brute-force (i.e. integrating $x$, then $y$) to arrive at the close-form result
$$\frac13ab\sqrt{a^2+b^2} +\frac16a^3\sinh^{-1}\frac ba +\frac16 b^3 \sinh^{-1}\frac ab$$
which has the expected parity between $a$ and $b$. However, it gets unwieldy to tackle the triple-integral extension $$\int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$$
this way and I am unable to slug it out. Does anyone know the corresponding close-form expression for the triple version?
| A start
Spherical coordinates. $(\rho,\theta,\phi)$. They are related to rectangular coordinates $(x,y,z)$ by:
\begin{align}
x &= \rho\sin\theta\cos\phi\\
y &= \rho\sin\theta\sin\phi\\
z &= \rho\cos\theta
\end{align}
and in reverse by
\begin{align}
\rho &= \sqrt{x^2+y^2+z^2}\\
\theta &=\arccos\frac{z}{\sqrt{x^2+y^2+z^2}}\\
\phi &= \arctan\frac{y}{x}
\end{align}
The box we want is
$$
0 \le x \le a,\\
0 \le y \le b,\\
0 \le z \le c.
$$
In spherical coorcinates:
$$
0 \le \rho\sin\theta\cos\phi \le a,\\
0 \le \rho\sin\theta\sin\phi \le b,\\
0 \le \rho\cos\theta \le c.
$$
If we think we will express our integrals using $\rho$ a function of $\theta,\phi$, do this as
$$
0 \le \rho \le a\csc\theta\sec\phi,\\
0 \le \rho \le b\csc\theta\csc\phi,\\
0 \le \rho \le c\sec\phi
$$
Our triple integral will have three terms, depending on which of the three
$a\csc\theta\sec\phi,b\csc\theta\csc\phi,c\sec\phi$ is smallest. That is,
dependingon which of the three faces the ray from the origin with angles $\theta,\phi$ intersects.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Proof of the Moreau Envelope of $l_1$ norm Given function $| \cdot | : \mathbb{R} \rightarrow \mathbb{R}_{+}$ and $\alpha > 0$, its Moreau envelope $e_{\alpha}|\cdot|: \mathbb{R} \rightarrow \mathbb{R}_{+}$, reads:
\begin{equation}
e_{\alpha}|\cdot|\left(x\right) = \inf_{y \in \mathbb{R}}{\bigg\{|y| + \frac{1}{2\alpha} | x - y|^2\bigg\}}
\end{equation}
How would you formally prove that the Moreau Envelope is given by?
\begin{equation}
e_{\alpha}|\cdot|\left(x\right) =
\begin{cases} \frac{1}{2\alpha}x^2 & |x| \leq \alpha \\ |x| - \frac{\alpha}{2} & |x| > \alpha \end{cases}
\end{equation}
| Suppose $y\ge 0$. Observe that
\begin{align}
|y|+\frac{1}{2\alpha}|x-y|^2 =&\ \frac{x^2+y^2}{2\alpha}+\frac{xy}{\alpha}+y = \frac{x^2-(x+\alpha)^2}{2\alpha} + \frac{1}{2\alpha}\left(y^2+2(x+\alpha)y+(x+a)^2 \right)\\
=&\ \frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha}.
\end{align}
If $x+\alpha>0$, then the minimum occurs when $y = 0$, i.e.
\begin{align}
\frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha} \ge \frac{x^2}{2\alpha}.
\end{align}
When $x+\alpha<0$ (i.e. $-x>\alpha>0$), then the minimum occurs when $y = -(x+\alpha)$ which means
\begin{align}
\frac{x^2-(x+\alpha)^2}{2\alpha} +\frac{(y+x+\alpha)^2}{2\alpha} \ge \frac{x^2-(x+\alpha)^2}{2\alpha} = -x-\frac{\alpha}{2}=|x|-\frac{\alpha}{2}.
\end{align}
For $y<0$. Observe that
\begin{align}
|y|+\frac{1}{2\alpha}|x-y|^2 =&\ \frac{x^2+y^2}{2\alpha}+\frac{xy}{\alpha}-y = \frac{x^2-(x-\alpha)^2}{2\alpha} + \frac{1}{2\alpha}\left(y^2+2(x-\alpha)y+(x-\alpha)^2 \right)\\
=&\ \frac{x^2-(x-\alpha)^2}{2\alpha} +\frac{(y+x-\alpha)^2}{2\alpha}.
\end{align}
I will leave you to complete the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$ I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.
| \begin{align}J&=\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x\\
&\overset{y=x^2}=2\int_0^1 \frac{\ln(1-y^2+y^4)}{1+y^2}dy\\
K&=\int_0^1 \frac{\ln\left(\frac{1-x^2+x^4}{x^2}\right)}{1+x^2}dx\\
&\overset{z=\frac{2}{x+\frac{1}{x}}}=\int_0^1 \frac{\ln\left(\frac{4-3z^2}{z^2}\right)}{2\sqrt{1-z^2}}dz\\
&=\frac{1}{2}\int_0^1 \frac{\ln(4-3z^2)}{\sqrt{1-z^2}}dz-\underbrace{\int_0^1 \frac{\ln z}{\sqrt{1-z^2}}dz}_{z=\sin t}\\
&=\frac{1}{2}\int_0^1 \left(\int_0^3 \frac{\partial}{\partial a}\frac{\ln(1+a(1-z^2))}{\sqrt{1-z^2}}da\right)dz-\int_0^{\frac{\pi}{2}}\ln(\sin z)dz\\
&=\frac{1}{2}\int_0^3 \left(\int_0^1\frac{\sqrt{1-z^2}}{1+a(1-x^2)}dz\right)da+\frac{1}{2}\pi\ln 2\\
&=\frac{1}{2}\int_0^3 \left[\frac{\arcsin x}{a}-\frac{\arctan\left(\frac{x}{\sqrt{1-x^2}\sqrt{1+a}}\right)}{a\sqrt{1+a}}\right]_0^1 da+\frac{1}{2}\pi\ln 2\\
&=\frac{\pi}{4}\int_0^3 \left(\frac{1}{a}-\frac{1}{a\sqrt{1+a}}\right)da+\frac{1}{2}\pi\ln 2\\
&=\frac{\pi}{2}\Big[\ln\left(1+\sqrt{1+a}\right)\Big]_0^3+\frac{1}{2}\pi\ln 2\\
&=\boxed{\frac{\pi}{2}\ln 3}
\end{align}
Therefore,
\begin{align}J&=2K+4\int_0^1\frac{\ln x}{1+x^2}dx\\
&=\boxed{\pi\ln 3-4\text{G}}
\end{align}
NB: i assume $\displaystyle \int_0^{\frac{\pi}{2}}\ln(\sin z)dz=-\frac{\pi}{2}\ln 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$? How to evaluate
$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$
The integral arises in the computation of
$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as
$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$
A similar Integral
$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$
Where $ C $ = Catalan Constant
Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and using Taylor Series as well as Mathematica.
Q = Is there a closed form for Integral $(1)$?
EDIT
$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$
Where $W$ is some value.
| By expanding the denominator and integrating term-by-term we obtain:
$$
-\sum _{n=0}^{\infty } \frac{(-1)^n }{4 n+3}H_{n+\frac{3}{4}},
$$
where $H_n$ is the $n$th harmonic number, in the formula above--the harmonic number of fractional order.
Not sure if it is useful, but the sum can be performed with the help of, e.g., Mathematica yielding a relatively simple expression:
$$
\tfrac{1}{3} \Gamma \big(\tfrac{7}{4}\big)\, {}_2F_1^{(0,0,1,0)}\!\left(\tfrac{3}{4},1,\tfrac{7}{4},-1\right)-\frac{\gamma}{4 \sqrt{2}}\left[\pi -2\log(\sqrt{2}-1)\right].
$$
The superscript means that derivative is taken with respect to the parameter $c$ of the regularized hypergeometric function ${}_2F_1(a,b,c,z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^4(g\circ f\circ g)(x)\mathrm{d}x$ where $f(x)=\sqrt[3]{x+\sqrt{x^2+1/27}}+\sqrt[3]{x-\sqrt{x^2+1/27}}$, $g(x)=x^3+x+1$
Let $$f(x)=\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}+\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}$$
and
$$g(x)=x^3+x+1$$
then, find $$\int_0^4(g\circ f\circ g)(x) \mathrm dx$$
My attempt:
Let $\displaystyle h(x)=\sqrt{x^2+\frac{1}{27}}$
$$(g\circ f)(x)=2x+3((2x)(x^2-[h(x)]^2))^{1/3}+(x+h(x))^{1/3}+(x-h(x))^{1/3}+1$$
Is finding $(g\circ f \circ g)(x)$ in term of $x$ necessary? Because it is quite a work to do.
| Note that $f^{-1}(x) = \dfrac{x^3+x}{2}\\$
$\therefore g(x) = 2f^{-1}(x)+1\\$
$\therefore g(f(g(x))) = 2f^{-1}(f(g(x)))+1\\$
and I'm sure you know $f^{-1}(f(x)) = x \\$
$\therefore g(f(g(x))) = 2g(x)+1\\$
$$\therefore \int_0^4g(f(g(x)))dx = \int_0^4(2g(x)+1)dx = \int_0^4(2x^3+2x+3)dx\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\[5pt]
&= \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{n+1}\right)\\[5pt]
&= \sum_{k=1}^n \frac{1}{2k-1} - \sum_{k=1}^{n+1} \frac{1}{2k} \\[5pt]
\end{align*}$$
How do I continue the problem from here? Is it even possible?
| It seems the expression $S$ needs to be revised somewhat. If we take a look at the first two summands
\begin{align*}
\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\cdots
\end{align*}
we see the left-hand factor in the denominator is odd and the right-hand factor is even. Taking $n$ summands of this kind we have
\begin{align*}
S_{1}=\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)2n}\tag{1}
\end{align*}
where the last summand has an odd and even factor of the given form in the denominator.
Here we calculate $S_1$. We obtain from (1)
\begin{align*}
\color{blue}{S_1=\sum_{k=1}^n\frac{1}{(2k-1)2k}}
&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\
&=\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\tag{2}\\
&=\left(\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{2k}\right)-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\tag{3}\\
&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\tag{4}\\
&\,\,\color{blue}{=H_{2n}-H_{n}}\tag{5}
\end{align*}
Comment:
*
*In (2) we split the sum into two sums and factor out the constant $\frac{1}{2}$ from the right-hand sum.
*In (3) we add and subtract summands with even denominator to the left-hand sum for convenient calculations in the further steps. This does not change the value of the sum since we are adding zero only.
*In (4) we simplify and collect the sums.
*In (5) we write the sums using the symbol for Harmonic numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving differential equation with boundary conditions $y=2$ at $x=0$ I need to find a solution to this differential equation $(1+x^2) $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = x-xy^2$, with boundary condition $y=2$ at $x=0$
I collect all terms on the RHS to get $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{1+x^2} (1-y^2)$$
I integrate then move the y terms to the left side and integrate both sides and simplify to get $$\frac{1}{\sqrt{1-y^2}} = k\sqrt{1+x^2}$$
I then isolate y in order to get it in a y= form and sub in the y and x value but I don't get the correct answer; I'm not sure where I'm going wrong
| $$\int\frac{dy}{1-y^2}=\int\frac{x\,dx}{1+x^2}$$
now you can do the left a few ways but we can use PFD:
$$\int\frac{dy}{1-y^2}=-\frac12\int\left(\frac{1}{y-1}-\frac{1}{y+1}\right)dy=\frac12\ln\left|\frac{y+1}{y-1}\right|$$
now do RHS:
$$\int\frac{x}{x^2+1}dx=\frac12\ln(x^2+1)+C_1$$
now equate:
$$\ln\left(\frac{y+1}{y-1}\right)=\ln(x^2+1)+C_2$$
exponentiate:
$$\frac{y+1}{y-1}=C_3(x^2+1)\tag{1}$$
now ideally get $y$ on its own, so you can use that:
$$\frac{y+1}{y-1}=\frac{y-1+2}{y-1}=1+\frac{2}{y-1}$$
bring it back into the equation:
$$1+\frac{2}{y-1}=C_3(x^2+1)$$
$$y-1=\frac{2}{C_3(x^2+1)-1}$$
$$y=1+\frac{2}{Cx^2+(C-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$
(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$
(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$
(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$
My Attempt:
(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.
But it's not helpful at all. :(
(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.
(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$
(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.
Plz give me some hints. Thx in advance.
| You can use matrix rank inequality:
$$r(AB) \le \min\{r(A),r(B)\}$$
For $D_1$ you wrote your matrix as $AB$, and both ranks of $A$ and $B$ are $\le 2$ so its rank is $\le 2$ by the above formula. Therefore if $n \ge 3$ then $D_1 = 0$. For $n=1, 2$ you can calculate manually. You get
$$|1+x_1y_1| = 1+x_1y_1, \quad \begin{vmatrix} 1+x_1y_1 & 1+x_1y_2 \\ 1 + x_2y_1 & 1+x_2y_2\end{vmatrix} = (x_1-y_1)(x_2-y_2).$$
Similarly for $D_2$ you can write your matrix as
$$\begin{bmatrix} \cos a_1 & \sin a_1 \\ \vdots & \vdots \\ \cos a_n & \sin a_1\end{bmatrix}\begin{bmatrix} \cos a_1 & \cdots & \cos a_n \\ \sin a_1 & \cdots & \sin a_n\end{bmatrix}$$
so again for $n \ge 3$ we get $D_2 = 0$, and for $n=1,2$ calculate manually.
For $D_3$ it's not hard to guess eigenvalues and eigenvectors. If
$$A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$$
then $$A\begin{bmatrix} -1 \\ 1 \\ 1 \\ 1\end{bmatrix} = -2\begin{bmatrix} -1 \\ 1 \\ 1 \\ 1\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0\end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0\end{bmatrix}=2\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0\end{bmatrix}$$
so $\det A = -2 \cdot 2 \cdot 2 \cdot 2 = -16$ and hence $D_3 = -16a^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$
Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.
So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.
Therefore the probability of the triangle area being greater than 2 is $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$
However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!
| The area of the triangle is given by the function $\displaystyle A(X) = \frac{3}{2}(X+1)$, where X is uniformly distributed in $[0,3]$, so
\begin{equation}
\mathbb{P}(A(X) \geq 2) = \mathbb{P}\left(X\geq\frac{1}{3} \right) = \frac{1}{3} \left(3 - \frac{1}{3} \right) = \frac{8}{9}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$ Can someone prove this inequality for the real numbers a,b,c?
$$a^2+2b^2+8c^2\geq2a(b+2c)$$
I have tried simple manipulation of the terms to get quadratic expressions, but since one cannot factor the $4ac$ on the right side, I abandoned that approach. Then I tried turning it into an expression where I could apply AM-GM, but that did not work either.
Maybe something like Muirhead, though I do not know where to start with that?
Any and all help would be greatly appreciated!!
| Your inequation is equivalent to
$$(a-b-2c)^2+(b-2c)^2 \ge 0 \tag{1}$$
The advantage of expression (1) is that it allows to clearly obtain the limit cases where there is an equality sign in (1) instead of a $">"$ symbol, i.e., iff
$$a=2b=4c$$
Edit: One can wonder how I have found expression (1). If you happen to know the concept of matrix associated with a quadratic form, here is the explanation:
$$a^2+2b^2+8c^2-2a(b+2c)$$
$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix} 1& -1&-2\\
-1&2&0\\
-2&0&8 \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$
which can be transformed, using the so-called (incomplete) Cholesky factorization:
$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix} 1&0&0\\
-1&1& \ \ 0\\
-2&-2 & \ \ 0 \end{pmatrix}\begin{pmatrix} 1&-1&-2\\
\ \ 0&\ \ 1&-2\\
0&0 &0 \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$
$$=\begin{pmatrix}(a-b-2c)&(b-2c)&0\end{pmatrix}\begin{pmatrix}(a-b-2c)\\(b-2c)\\0\end{pmatrix}$$
$$=(a-b-2c)^2+(b-2c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
studying the function $f(x)= \sqrt{x^2+x}-x$ I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.
In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.
$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0 \Rightarrow 2x+1-2 \sqrt{x^2+x} >0 \Rightarrow 2x+1>2 \sqrt{x^2+x} \Rightarrow 1>0 $ . So it seems that the derivative is always positive in the domain.
But comparing with the result in wolfram alpha the function is decreasing in $(-\infty, -1]$
| $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0\iff 2x+1-2\sqrt{x^2+x}>0\iff 2x+1>2\sqrt{x^2+x}\iff \begin{cases}2x+1>0\\ x^2+x> 0\\ (2x+1)^2>4(x^2+x)\end{cases}\iff \begin{cases}x>\frac{-1}{2}\\x> 0 \vee x< -1\\ \forall x \end{cases}\iff x> 0 $$ Be careful to exclude $0$ in order to have the denominator not equal to $0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express the sum of a geometric series in the form $p + q\sqrt2$ I am working through a maths text book as a hobby and am stuck on the following problem:
An arithmetic series and a geometric series have r as a common difference and the common ratio respectively. The first term of the arithmetic series is 1 and the first term of the geometric series is 2. If the fourth term of the arithmetic series is equal to the sum of the third and fourth terms of the geometric series, find the three possible values of r.
When $|r| \lt 1$ find, in the form of $p + q\sqrt2$, (i) the sum to infinity of the geometric series (ii) the sum of the first ten terms of the arithmetic series.
This is my approach to the part (i), the geometric series:
The three roots are $1, -1 + \frac{1}{2}\sqrt 2$, $ -1 - \frac{1}{2}\sqrt 2$
Of these only $|-1 + \frac{1}{2}\sqrt 2| \lt 1$
$S_{\infty} = \frac{2}{1 -(-1 + \frac{\sqrt2}{2}}$
which I reduced to $\frac{1}{1 - \sqrt\frac{1}{8}}$
The book says the answer to this is $\frac{8}{7} + \frac{2}{7}\sqrt 2$
I cannot see how this is arrived at. If I apply the binomial expansion I get:
$1 + (-1)(-\frac{1}{\sqrt8}) + \frac{(-1)(-2)(-\frac{1}{\sqrt8})}{2!} +...$ so I am obviously on the wrong track.
| $$\frac{1}{1-\sqrt{1/8}} = \frac{\sqrt{8}}{\sqrt{8} - 1} = \frac{\sqrt{8}( \sqrt{8} + 1)}{(\sqrt{8} - 1)(\sqrt{8} + 1)} = \frac{8 + \sqrt{8}}{8 - 1} = \frac{8}{7} + \frac{2\sqrt{2}}{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show limit exists if and only if $m+n>2$. Let $m,n \in \mathbb{N}$.
Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.
Attempt:
$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\sqrt{x^2+y^2})^{n-2}(\sqrt{x^2+y^2})^m=(\sqrt{x^2+y^2})^{m+n-2}$.
So for $\epsilon>0$, choose $\delta=\epsilon^{\frac{1}{m+n-2}}$, then $0<\sqrt{x^2+y^2}<\delta \implies |\frac{x^ny^m}{x^2+y^2}|<\epsilon$.
$(\Rightarrow)$.If $n+m\leq 2$. $f(\frac{1}{k},\frac{1}{k})=(\frac{1}{2})(\frac{1}{k})^{n+m-2}$ which does not approach zero as $k \rightarrow 0$, while $f(\frac{1}{k},0)=0$ approaches $0$ as $k \rightarrow 0$. Since these limits are different, the limit does not exist in this case.
| Your proof seems to be correct! Another more intuitive way would be to use polar coordinates in this situation (which always is a good thing to consider when you have a $x^2+y^2$ term somewhere): $$\lim \limits_{(x,y) \to 0} \frac{x^ny^m}{x^2+y^2}=\lim \limits_{r\to 0}\frac{(r\cos(\theta))^n(r\sin(\theta))^m}{(r\cos(\theta))^2+(r\sin(\theta))^2}=\lim \limits_{r\to 0} r^{n+m-2}\cos(\theta)^n\sin(\theta)^m \rightarrow 0 \quad \text{when} \; n+m>2$$
(since $cos(\theta)^n\sin(\theta)^m$ is bounded)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A Euclidean algorithm problem Suppose that, after running the Euclidean algorithm on two integers $a$ and $b$, we find that $r_n=3$, where $r_n$ is the last remainder in the Euclidean algorithm. Furthermore, we find that $r_{n-1}=6$ and $r_{n-2}=21$. What can we say about $a$ and $b$?
Obviously, by the definition of the Euclidean algorithm, $\gcd(a, b)=3$. However, that isn't sufficient (isn't strong enough for an iff condition). I tried for a long time to think of more conditionns, but only came up with a bunch of false conjectures. Can someone help me?
| In the Euclidean algorithm, performed on the starting couple of positive integers $(n,m)$
the remainders follow the sequence
$$
\eqalign{
& n = r_{\, - 2} =
\left\lfloor {{n \over m}} \right\rfloor m + r_{\,0} \cr
& m = r_{\, - 1} =
\left\lfloor {{m \over {r_{\,0} }}} \right\rfloor r_{\,0} + r_{\,1} \cr
& \vdots \cr
& r_{\,k - 2} =
\left\lfloor {{{r_{\,k - 2} } \over {r_{\,k - 1} }}} \right\rfloor r_{\,k - 1} + r_{\,k} \cr
& \vdots \cr
& r_{\,h - 4} =
\left\lfloor {{{r_{\,h - 4} } \over {r_{\,h - 3} }}} \right\rfloor r_{\,h - 3} + r_{\,h - 2} \cr
& r_{\,h - 3} =
\left\lfloor {{{r_{\,h - 3} } \over {r_{\,h - 2} }}} \right\rfloor r_{\,h - 2} + r_{\,h - 1} \cr
& r_{\,h - 2} =
\left\lfloor {{{r_{\,h - 2} } \over {r_{\,h - 1} }}} \right\rfloor r_{\,h - 1} +
\left( {0 = r_{\,h} } \right)
\quad \left| {\;r_{\,h - 1} = \gcd (n,m) = g} \right. \cr}
$$
where the sequence shall be strictly decreasing down to $0$ at step $h$,
i.e.
$$
0 = r_{\,h} < \cdots < \,r_{\,k} \, < r_{\,k - 1} < \cdots < r_{\,0}
\quad \left\{ {\matrix{
{ \le \left| {\,n\,} \right|} & {\left| {\;0 = m} \right.} \cr
{ < \left| {\,m\,} \right|} & {\left| {\;0 \ne m} \right.} \cr
} } \right.
$$
That also implies that the $ r_k $ are all multiples of $g= \gcd(n,m)$
You are asking practically how the sequence can be reconstructed backwards.
Let's start from the last line
$$
\eqalign{
& s_{ - 2} = {{r_{\,h} } \over g} = 0 \cr
& s_{ - 1} = {{r_{\,h - 1} } \over g} = 1 \cr
& s_0 = {{r_{\,h - 2} } \over g} =
\left\lfloor {{{r_{\,h - 2} } \over {r_{\,h - 1} }}} \right\rfloor
= k_{\,0} = k_{\,0} s_{ - 1} + s_{ - 2} \cr
& s_1 = {{r_{\,h - 3} } \over g} =
\left\lfloor {{{r_{\,h - 3} } \over {r_{\,h - 2} }}} \right\rfloor {{r_{\,h - 2} } \over g}
+ {{r_{\,h - 1} } \over g} = k_{\,1} s_0 + s_{ - 1} \cr
& s_2 = {{r_{\,h - 4} } \over g} =
\left\lfloor {{{r_{\,h - 4} } \over {r_{\,h - 3} }}} \right\rfloor {{r_{\,h - 3} } \over g}
+ {{r_{\,h - 2} } \over g} = k_{\,2} s_1 + s_0 \cr
& \quad \quad \vdots \cr
& s_j = {{r_{\,h - j - 2} } \over g} =
\left\lfloor {{{r_{\,h - j - 2} } \over {r_{\,h - j - 1} }}} \right\rfloor {{r_{\,h - j - 1} } \over g}
+ {{r_{\,h - j} } \over g} = k_{\,j} s_{j - 1} + s_{j - 2} \cr}
$$
where the $ k_j$ are arbitrary integers greater than $1$.
In matrix form this recursion reads
$$
\left( {\matrix{ {s_{n - 1} } \cr {s_n } \cr } } \right) =
\left( {\matrix{ 0 & 1 \cr 1 & {k_n } \cr } } \right)
\left( {\matrix{ {s_{n - 2} } \cr {s_{n - 1} } \cr } } \right)
\quad \left| {\;\left( {\matrix{ {s_{ - 2} } \cr {s_{ - 1} } \cr } } \right)
= \left( {\matrix{0 \cr 1 \cr } } \right)} \right.
$$
In the example you give
$$
\eqalign{
& g = 3, \cr
& s_{ - 1} = {{r_{\,h - 1} } \over g} = {g \over g} = 1,\quad \cr
& s_0 = {{r_{\,h - 2} } \over g} = {6 \over g} = 2 = k_0 ,\; \cr
& s_1 = {{21} \over g} = \left\lfloor {{{r_{\,h - 3} } \over {r_{\,h - 2} }}}
\right\rfloor {{r_{\,h - 2} } \over g} + {{r_{\,h - 1} } \over g} = \cr
& = \left\lfloor {{{21} \over 6}} \right\rfloor {6 \over g} + {g \over g} =
3s_0 + s_{ - 1} = 7 \cr}
$$
then
$$
\eqalign{
& \left( {\matrix{ {s_{ - 1} } \cr {s_0 } \cr } } \right) =
\left( {\matrix{ 0 & 1 \cr 1 & {k_0 } \cr } } \right)\left( {\matrix{ 0 \cr 1 \cr } } \right) =
\left( {\matrix{ 0 & 1 \cr 1 & 2 \cr } } \right)\left( {\matrix{ 0 \cr 1 \cr } } \right) =
\left( {\matrix{ 1 \cr 2 \cr } } \right) \cr & \left( {\matrix{ {s_0 } \cr {s_1 } \cr } } \right) =
\left( {\matrix{ 0 & 1 \cr 1 & {k_1 } \cr } } \right)\left( {\matrix{ 1 \cr 2 \cr } } \right) =
\left( {\matrix{ 0 & 1 \cr 1 & 3 \cr } } \right)\left( {\matrix{ 1 \cr 2 \cr } } \right) =
\left( {\matrix{ 2 \cr 7 \cr } } \right) \cr}
$$
and from here we can proceed with multiplying by the matrix, with any selected constant, e.g.:
$$
\eqalign{
& \left( {\matrix{ 0 & 1 \cr 1 & 1 \cr } } \right)
\left( {\matrix{ 2 \cr 7 \cr } } \right) =
\left( {\matrix{ 7 \cr 9 \cr } } \right) \cr
& \left( {\matrix{ 0 & 1 \cr 1 & 2 \cr } } \right)
\left( {\matrix{ 2 \cr 7 \cr } } \right) =
\left( {\matrix{ 7 \cr {16} \cr } } \right) \cr
& \left( {\matrix{ 0 & 1 \cr 1 & 1 \cr } } \right)^{\,2}
\left( {\matrix{ 2 \cr 7 \cr } } \right) =
\left( {\matrix{ 9 \cr {16} \cr } } \right) \cr}
$$
which correspond to the couples $(21,27), \,(21,48) , \, (27,48)$ already indicated by @robjohn.
In conclusion we have the general solution to the recursion as
$$
\left( {\matrix{ {s_{n - 1} } \cr {s_n } \cr } } \right) =
\prod\limits_{j = 0}^n {\left( {\matrix{ 0 & 1 \cr 1 & {k_j } \cr } } \right)
\left( {\matrix{ 0 \cr 1 \cr } } \right)}
$$
so that when the $k$'s are unitary we get the Fibonacci numbers, which are known to have the longest
chain in the algorithm.
But for general values of the constants, there is not a simple characterization of the numbers
that can be obtained..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
3-variable 3-equations problem
For positive reals $x,y,z$, if $xy+yz+xz=\sqrt{\frac pq}$, given
$$x^2+xy+y^2=2$$
$$y^2+yz+z^2=1$$
$$z^2+zx+x^2=3$$
find $p-q$, where $p$ and $q$ are relatively prime integers.
I tried grouping the required sum by adding all the equations but it led me to nowhere. I tried applying basic AM-GM inequality to conclude that the sum is less than or equal to $2$, but this is of no use since $x,y,z$ are positive reals, not just positive integers.
I have exhausted all my ideas. Any hint/help is appreciated!
| If you like a more algebraic answer, consider the following linear combination of the three equations in the order written above $(x-y)(1)+(y-z)(2)+(z-x)(3)$. The LHS of the resultant equation vanishes and we get a linear relation:
$$ 2(x-y)+(y-z)+3(z-x)=0\iff z=\frac{x+y}{2}$$
Under this light we see that $\Pi=xy+xz+yz=6z^2-2$ because in fact equation $(1)$ can be rewritten in the form
$$x^2+xy+y^2-2=0\iff xy=4z^2-2$$
Thus we only need to compute $z^2$. In the following denote $\rho=\frac{x-y}{2}$. Then consider the combination $(3)-(2)$ we get
$$x^2-y^2+(x-y)z=2\iff z\rho=\frac{1}{3}$$
and finally we note that we can write
$$x^2+xy+y^2=(x+y)^2-xy=3z^2+\rho^2=2$$
The last two equations can be solved by substitution of the 1st into the 2nd, which yields the equation
$$3z^4-2z^2+\frac{1}{9}=0\iff 6z^2=2\pm\sqrt{\frac{8}{3}}$$
which since $\Pi>0$ yields immediately
$$\Pi=\sqrt{\frac{8}{3}}\Rightarrow p=8~,~q=3$$
and thus $p-q=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Diagonal element of the resolvent of bi-infinite tridiagonal Laurent operator For $\alpha, \beta, \gamma \in \mathbb{C}$ consider the bi-infinite tridiagonal Laurent operator $T$ with $\beta $ on the diagonal given by.
\begin{pmatrix}
\dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\
\dots & \alpha & \beta & \gamma & 0 & 0 & 0 & \dots \\
\dots & 0 & \alpha & \beta & \gamma & 0 & 0 & \dots\\
\dots & 0 & 0 & \alpha & \beta & \gamma & 0 & \dots\\
\dots & 0 & 0 & 0 & \alpha & \beta & \gamma & \dots \\
\dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\
\end{pmatrix}
General theory tells us that $T$ is invertible if and only the symbol curve (which is in this case is an ellipsis) given by
$
\{ z \in \mathbb{T} \mid \frac{\alpha}{z} + \beta + z \gamma \}
$
does not enclose $0$.
Suppose that this is the case and let $e_0$ denote the a unit vector in the standard basis. Then what is the value
$
\langle e_0 , T^{-1} e_0 \rangle ?
$
| As we explain in the Appendix here: https://arxiv.org/abs/2206.09879
this is a standard calculation:
"First, $\sigma(T)$ is the image of the symbol curve
$
a(z) = \alpha z^{-1} + \beta + \gamma z $
for $z \in \mathbb{T}$. Since $T$ is invertible it holds that $ a(z) \neq 0$ for all $z \in \mathbb{T}$ and therefore, the symbol curve of the inverse is given by
\begin{align*}
\frac{1}{a(z)} = \frac{1}{ \frac{\alpha}{ z} + \beta + \gamma z } = \frac{z}{ \alpha + \beta z + \gamma z^2} = \frac{z}{ \gamma( \frac{\alpha}{\gamma} + \frac{\beta}{\gamma} z + z^2) }.
\end{align*}
We can rewrite the denominator $ \gamma ( z - \lambda_+) (z- \lambda_-) $ with
\begin{align*}
\lambda_{\pm} = \frac{ - \beta}{2 \gamma} \pm \sqrt{ \left( \frac{\beta}{2 \gamma} \right)^2 - \frac{\alpha }{\gamma} }.
\end{align*}
Notice that
\begin{align*}
\lambda_+ \lambda_- = \frac{\alpha}{\gamma} \text{ , }
\lambda_+ + \lambda_- = - \frac{\beta}{\gamma},
\text{ and }
\lambda_+ - \lambda_- = 2 \sqrt{ \left( \frac{\beta}{2 \gamma} \right)^2 - \frac{\alpha }{\gamma} }.
\end{align*}
Now, assuming that $\vert \lambda_2 \vert < 1< \vert \lambda_1 \vert $ (where $\{\lambda_2, \lambda_1 \} = \{\lambda_+, \lambda_- \}$ has implications on how we write this up as a geometric series):
\begin{align*}
\frac{1}{a(z)} & = \frac{z}{ \gamma ( z - \lambda_+) (z- \lambda_-) } = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( \frac{1}{z-\lambda_1} - \frac{1}{z-\lambda_2} \right) \\
& = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( - \frac{1}{\lambda_1} \frac{1}{1- \frac{z}{\lambda} } - \frac{1}{z} \frac{1}{1-\frac{\lambda_2}{z} } \right) \\
& = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( - \frac{1}{\lambda_1} \sum_{n=0}^\infty (\frac{z}{\lambda_1})^n - \frac{1}{z} \sum_{n=0}^\infty (\frac{\lambda_2}{z})^n \right) \\
& = \frac{1}{ \gamma ( \lambda_2 - \lambda_1)} \left( \sum_{n=1}^\infty (\frac{z}{\lambda_1})^n + \sum_{n=0}^\infty (\frac{\lambda_2}{z})^n \right).
\end{align*}"
Then one can read of coefficients to obtain the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
By row reduction, I obtain:
$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$
Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.
Update: I have managed to solve this with the use of the determinant.
Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$
Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$
Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$
Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$
$(a - 5)(a + 1) = 0$
Thank you all for your help!
| The easiest thing is to use determinants: the system has a unique solution if and only if the matrix is invertible, i.e., if and only if the matrix has determinant not equal to $0$. So you can compute the determinant (either in your original or row-reduced matrix) in terms of $a$, set it equal to $0$, and solve for $a$. But if you do not know this yet, you can proceed as follows:
If $a=2$, then the system clearly has a unique solution, so you may assume $a\neq 2$. The reason to do this is so that we can continue the row reduction by multiplying by $a-2$. Multiply the second row by $9$ and the last row by $a-2$ to get:
$$\left(\begin{array}{ccc|c}
1 & 2 & 2 & 1\\
0 & 9(a-2) & 9 & 18\\
0 & 9(a-2) & (a-2)^2 & 2-a
\end{array}\right).$$
Subtracting the second row from the third row, we get
$$\left(\begin{array}{ccc|c}
1 & 2 & 2 & 1\\
0 & 9(a-2) & 9 & 18\\
0 & 0 & (a-2)^2-9 & -16-a
\end{array}\right).$$
So: you will have a unique solution if $(a-2)^2-9\neq 0$. If $(a-2)^2-9=0$, you either get infinitely many solutions (if $16+a=0$ as well), or no solutions (if $-16-a\neq 0$). So now we just need to figure out when that bottom right entry of the coefficient matrix is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\int_0^1 (1-x^2)^n dx$ In an exercise I'm asked the following:
a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.
b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$
I used the binomial theorem in $a$ and got:
$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$
and so in part (b) i got:
$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$
I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?
| My idea here is to use the Beta function:
$$ \int_0^1 t^{m}(1 - t)^n \;dt = \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}. $$
Where the Gamma function satisfies $\Gamma(n + 1) = n!$ if $n$ is an integer, and also $\Gamma$ satisfies the recurrence $\Gamma(n + 1) = n\Gamma(n)$.
So if we substitute $t = x^2$, $dt = 2x\; dx \iff \frac{1}{2\sqrt{t}}\;dt=dx$ then
\begin{align}
\int_0^1 (1 - x^2)^n \; dx &= \frac{1}{2} \int_0^1 t^{-1/2} (1 - t)^n \;dt \\
&= \frac{1}{2} \frac{\Gamma(\frac12)n!}{\Gamma(n+\frac32)}. \tag{1}
\end{align}
And now using the recurrence $\Gamma(n + 1) = n\Gamma(n)$, we find that
\begin{align}
\Gamma(n+\tfrac32) &= (n+\tfrac12)\Gamma(n + \tfrac12) \\
&= (n+\tfrac12)(n - \tfrac12) \Gamma(n - \tfrac12) \\
&\hspace{2.5mm}\vdots \\
&= (n+\tfrac12)(n - \tfrac12) \cdots \tfrac{5}{2} \cdot \tfrac{3}{2} \cdot \tfrac12 \cdot \Gamma(\tfrac12) \\
&= \frac{1}{2^n} (2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 \cdot \Gamma(\tfrac12) \tag{2}
\end{align}
The product $(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1$ is also known as the double factorial $(2n + 1)!!$. It is well-known that we can rewrite it in terms of $(2n + 1)!$ as
\begin{align}
(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 &= \frac{(2n + 1)(2n)(2n - 1)(2n - 2) \cdots 3 \cdot 2 \cdot 1}{(2n)(2n-2) \cdots 2} \\
&= \frac{(2n+1)!}{2^n \cdot n!}. \tag{3}
\end{align}
So now finally, if you combine $(1), (2), (3)$, you get the result you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 9,
"answer_id": 5
} |
Area of surface over $xy$-plane In my calculus book, it is claimed that the area $dS$ below, can be expressed as $$\frac{1}{\cos\gamma}dxdy.$$
Here $\vec{n}$ is perpendicular to the surface element, and $\vec{k}$ is given as $(0,0,1)$.
I understand how this can be concluded if either the area element's $x$- or $y$-component is parallell to the coordinate system. I do however have a hard time seeing how this can be concluded if the element is tilted in both the $x$ and $y$-direction.
Any tips on how this can be shown?
| WLOG we can translate things around so that the base of our rectangular prism has vertex $(0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0)$ and $(0, 0, 0)$ is the vertex with the smallest $z$-coordinate where the plane cut the rectangular prism.
Also, WLOG we can assume our normal vector $n = \begin{pmatrix} p \\ q \\ r \end{pmatrix}$ has length $|n| = 1$.
With this setup, all we need to do is to find the other 2 vertices $(a, 0, z_1), (0, b, z_2)$ where the plane cut the rectangular prism. Since the cross section is a parallelogram, the area is then simply the length of the cross product of the position vectors of the 2 vertices.
This is the hint part. You can stop reading now and try to solve it yourself.
Since $\gamma$ is the angle between the $k = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and $n$, by the property of dot product, $\cos \gamma = |k||n| \cos \gamma = k \cdot n = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} p \\ q \\ r \end{pmatrix} = r$
From now on, we can write $n = \begin{pmatrix} p \\ q \\ \cos \gamma \end{pmatrix}$
Observe $1 = |n| = \sqrt{p^2 + q^2 + \cos^2 \gamma}$
Therefore $p^2 + q^2 + \cos^2 \gamma = 1$ and hence $$p^2 + q^2 = 1 - \cos^2 \gamma = \sin^2 \gamma$$
Since the plane cut the rectangular prism at $(0, 0, 0)$, it passes throught the origin. Hence we can write its equation as $$\begin{pmatrix} p \\ q \\ \cos \gamma \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0$$ which is the same as $px + qy + (\cos \gamma)z = 0$.
As the plane cut through the rectangular prism at vertice $(a, 0, z_1)$, the vertice satisfies the equation of the plane. Therefore, $p a + q 0 + (\cos \gamma)z_1 = 0$ and hence $$z_1 = -\frac{pa}{\cos \gamma}$$
Similarly, at vertice $(0, b, z_2)$, we have $p 0 + q b + (\cos \gamma)z_2 = 0$ and hence $$z_2 = -\frac{qb}{\cos \gamma}$$
With these two $z$-coordinates, we now know the position vectors of the 2 vertices are $\begin{pmatrix} a \\ 0 \\ -\frac{pa}{\cos \gamma} \end{pmatrix}$ and $\begin{pmatrix} 0 \\ b \\ -\frac{qb}{\cos \gamma} \end{pmatrix}$.
Since $(0, 0, 0)$ is the vertex with the smallest $z$-coordinate where the plane cut the rectangular prism, these 2 position vectors are the sides of the parallelogram.
Now area $$\begin{align*} A
&= \left| \begin{pmatrix} a \\ 0 \\ -\frac{pa}{\cos \gamma} \end{pmatrix} \times \begin{pmatrix} 0 \\ b \\ -\frac{qb}{\cos \gamma} \end{pmatrix} \right| \\
&= \left| \det \begin{pmatrix}
i & j & k \\
a & 0 & -\frac{pa}{\cos \gamma} \\
0 & b & -\frac{qb}{\cos \gamma}
\end{pmatrix} \right| \\
&= \left| i \det \begin{pmatrix}
0 & -\frac{pa}{\cos \gamma} \\
b & -\frac{qb}{\cos \gamma}
\end{pmatrix} - a \det \begin{pmatrix}
j & k \\
b & -\frac{qb}{\cos \gamma}
\end{pmatrix}\right| \\
&= \left| \frac{pab}{\cos \gamma} i - a \left( -\frac{qb}{\cos \gamma} j - bk \right) \right| \\
&= \left| \frac{pab}{\cos \gamma} i + \frac{qab}{\cos \gamma} j + ab k \right| \\
&= \sqrt{\frac{p^2 a^2 b^2}{\cos^2 \gamma} + \frac{q^2 a^2 b^2}{\cos^2 \gamma} + a^2 b^2} \\
&= a b \sqrt{\frac{p^2}{\cos^2 \gamma} + \frac{q^2}{\cos^2 \gamma} + 1} \\
&= ab \sqrt{\frac{p^2 + q^2}{\cos^2 \gamma} + 1} \\
&= ab \sqrt{\frac{\sin^2 \gamma}{\cos^2 \gamma} + 1} \\
&= ab \sqrt{\tan^2 \gamma + 1} \\
&= ab \sqrt{\sec^2 \gamma} \\
&= ab \sec \gamma \\
&= \frac{ab}{\cos \gamma}
\end{align*}$$
which is exactly what we want to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Drawing chips from a bag A bag with contains $5$ chips that are either red or blue chips. The number of red chips is a random variable $X$ where $X$ can be either $2, 3$ or $4$ with equal probability. Given that when $5$ chips are drawn with replacement, the result is: $3$ of the draws are red and $2$ are blue, what is the conditional probability that $X = 4$?
My attempt:
Consider Bayes Theorem: $P(X = 4 | 3 red, 2 blue) = \frac{P(X = 4 \land 3 red, 2 blue)}{P(3 red, 2 blue)} = \frac{P(3 red, 2 blue | X = 4) \cdot P(X = 4)}{P(3 red, 2 blue)}$.
Assume $X = 4$ (then there are 4 reds and 1 blue) and let $Y$ be the number of red chips drawn. $Y$ is binomial with $n = 5$ and $p = \frac{4}{5}$. So $P(Y = 3) = 0.2048$ (by applying the binomial function). So $P(3 red, 2 blue | X = 4) = 0.2048)$.
Clearly $P(X = 4)$ = $\frac{1}{3}$.
All that remains in to determine $P(3 red, 2 blue)$. I tried this:
If $X = 2$ let $Z$ be the number of red chips drawn ($Z$ is binomial with $n = 5$ and $p = \frac{2}{5}$) and then $P(Z = 3) = 0.2304$
If $X = 3$ let $Z$ be the number of red chips drawn ($Z$ is binomial with $n = 5$ and $p = \frac{3}{5}$) and then $P(Z = 3) = 0.3456$
If $X = 4$, we already saw that $P(Z = 3) = 0.2048$.
So $P(3 red, 2 blue) = 0.2304 + 0.3456 + 0.2048 = 0.7808$
Plugging in the appropriate probabilities shows that $P(X = 4 | 3 red, 2 blue) = \frac{0.2048 \cdot \frac{1}{3}}{0.7808} \approx 0.0874$
Did I do this correctly? This is practice for an upcoming examination and no solutions are available. Whatever help I can get is appreciated.
| Your approach is correct but you made a mistake in calculation. Also I will suggest that your working should be using fractions instead of converting into decimals.
If $Y$ is the event of $3$ red and $2$ blue in $5$ draws with replacement,
$\small P(X = 4 | Y) = \displaystyle \frac{P(X = 4 \ \cap \ Y)}{P(Y)}$
$\small P(Y) = P(X=2 \cap Y) + P(X=3 \cap Y) + P(X=4 \cap Y)$
$ = \displaystyle \small \frac{1}{3} \cdot {5 \choose 3} \cdot \big[\big(\frac{2}{5}\big)^3 \big(\frac{3}{5}\big)^2 + \big(\frac{3}{5}\big)^3 \big(\frac{2}{5}\big)^2 + \big(\frac{4}{5}\big)^3 \big(\frac{1}{5}\big)^2\big]$
$\small P(X=4 \cap Y) = \displaystyle \frac{1}{3} \cdot {5 \choose 3} \cdot \big(\frac{4}{5}\big)^3 \big(\frac{1}{5}\big)^2$
$\small P(X = 4 | Y) = \displaystyle \frac{16}{61}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculation of a hard integral $\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx $ Calculate
$$\int^{1/2}_0\frac{1}{x^2-x+1} \cdot \ln\frac{1+x}{1-x}\,dx $$
I tried to subsitute $x=1/2-t$ so $dx=-dt$ but I just complicated more my problem.
| Mathematica gives:
$$\frac{\pi \log \left(\frac{9}{8}\right)-3 i \left(\text{Li}_2\left(-\frac{1}{2}
(-1)^{2/3}\right)+\text{Li}_2\left(\frac{1}{1-\sqrt[3]{-1}}\right)-\text{Li}_2\left(\frac{1}{1+\sqrt[3]{-1}}\right)+\text{Li}_2\left(\frac{1}{1-(-1)^{2/3}}\right)-\text{Li}
_2\left(\frac{1}{1+(-1)^{2/3}}\right)+\text{Li}_2\left(\frac{3}{4}-\frac{i
\sqrt{3}}{4}\right)-\text{Li}_2\left(\frac{1}{4}+\frac{i
\sqrt{3}}{4}\right)-\text{Li}_2\left(\frac{3}{4}+\frac{i \sqrt{3}}{4}\right)\right)}{3
\sqrt{3}}$$
which suggests you have a lot of work ahead of you if you want to do this by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The inequality $\,2+\sqrt{\frac p2}\leq\sum\limits_\text{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\,$ where $0\leq p\leq 2$ is: Probably true! Provably true? Let $p$ be a positive parameter in the range from $0$ to $2$.
Can one prove that
$$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}}
\,+\,\sqrt{\frac{b^2 +pca}{c^2+a^2}}\,+\,\sqrt{\frac{c^2 +pab}{a^2+b^2}}\quad?\tag{1}$$
Where $\,a,b,c\in\mathbb R^{\geqslant 0}\,$ and at most one variable equals zero.
The inequality $(1)$ is homogeneous of degree zero with regard to $a,b,c$.
Equality occurs if two variables coincide and the third one is zero.
To provide some plausibility to $(1)$ the two boundary cases $\,p=2\,$ and $\,p=0\,$ are
proved:
$p=2$
is the harder bit.
W.l.o.g. assume $\,a\geqslant b\geqslant c\,$ and $\,a,b>0$. Let
$u=\sqrt{\frac ab}\,+\,\sqrt{\frac ba}$, then $2\leqslant u$, and $u=2$ iff $a=b$.
I)$\:$ Let's show that
$$u\:\leqslant\:\sqrt{\frac{a^2 + bc}{b^2+c^2}} \,+\,\sqrt{\frac{b^2 +ac}{a^2+c^2}}\:.\tag{2}$$
The following expression is positive:
$$\begin{align}
& \frac{a^2+bc}{b^2+c^2} -\frac ab & +\quad &\frac{b^2+ac}{a^2+c^2} -\frac ba\\[2ex]
=\;\; & \frac{ab(a-b)+b^2c-ac^2}{b(b^2+c^2)} & +\quad &\frac{-ab(a-b)+a^2c-bc^2}{a(a^2+c^2)}\tag{3}\\[2ex]
\geqslant\;\; & \frac{ac(a-c)+bc(b-c)}{a(a^2+c^2)} \;\geqslant 0
\end{align}$$
The first summand in $(3)$ has been diminished by increasing the denominator, while
its numerator $\,ab(a-b) +b^2c -ac^2 =(a-b)(b-c)(a-c) + c(a-c)^2 + c^2(b-c)\,$ cannot get negative.
$(2)$ now follows from
$$\begin{split}
u^2\:=\:\frac ab + 2 +\frac ba \: & \leqslant\:\frac{a^2+bc}{b^2+c^2}
+ 2\,\underbrace{\sqrt{\frac{a^2+bc}{a^2+c^2}}}_{\geqslant 1}\;\underbrace{\sqrt{\frac{b^2+ac}{b^2+c^2}}}_{\geqslant 1}+ \frac{b^2+ac}{a^2+c^2}\\[2ex]
& =\:\left(\sqrt{\frac{a^2 + bc}{b^2+c^2}} +\sqrt{\frac{b^2 +ac}{a^2+c^2}}\:\right)^2
\end{split}$$
II)$\:$ The remaining square root summand in $(1)$ is also bounded below in terms of $u$ since one has
$$\frac 1{u^2-2} \:=\:\frac{ab}{a^2+b^2}\quad\implies\quad
\sqrt{\frac 2{u^2-2}} \:\leqslant\: \sqrt{\frac{c^2 +2ab}{a^2+b^2}}$$
III)$\:$ Applying $3$-AGM finally proves $(1)$:
$$\begin{split}\sum_\text{cyc}{\sqrt\frac{a^2 + 2bc}{b^2+c^2}} \;\geqslant\;
u+\sqrt{\frac 2{u^2-2}} &\:=\:\sqrt{\frac{u^2}4} +\sqrt{\frac{u^2}4} +\sqrt{\frac 2{u^2-2}}\\[2ex]
&\:\geqslant\:3\sqrt{\left(\frac{u^4}{8(u^2-2)}\right)^{1/3}} \:\geqslant\:3\end{split}$$
$p=0$
is more relaxing.
Only $2$-AGM in the form $\,a\sqrt{b^2+c^2}\leqslant\frac12\left(a^2+b^2+c^2\right)$ is needed:
$$\frac a{\sqrt{b^2+c^2}} + \frac b{\sqrt{c^2+a^2}}+ \frac c{\sqrt{a^2+b^2}}
\;=\;\sum_\text{cyc}\frac{a^2}{a\sqrt{b^2+c^2}}
\;\geqslant\;\sum_\text{cyc}\frac{2a^2}{a^2+b^2+c^2} \;=\;2$$
$0<p<2$
returns to the question.
With just some ideas how to catch the "remaining" $p$-values:
*
*The above method for $p=2$ may possibly be stretched down until the $p=1$ instance:
$$2 +\frac{\sqrt 2}{2} \;\leqslant\;\sum_\text{cyc}{\sqrt\frac{a^2+bc}{b^2+c^2}}$$
This has been detailed by mathlove in his answer.
*Interpolation with regard to $p$ (more a buzz word than substantial ...)
*A concavity argument as the two end points $p=0$ and $p=2$ are known: Could proving the second derivative with respect to $p$ being negative path a way towards a proof?
| This is a partial answer.
This answer proves that for $1\leqslant p\leqslant 2$, the inequality $(1)$ holds.
Proof :
(You have already proved that for $p=2$, $(1)$ holds. The idea used in the case $p=2$ works for the case $1\leqslant p\leqslant 2$.)
W.l.o.g. assume $\,a\geqslant b\geqslant c\,$ and $\,a,b>0$. Let
$u=\sqrt{\frac ab}\,+\,\sqrt{\frac ba}$, then $2\leqslant u$, and $u=2$ iff $a=b$.
You have already proved that
$$u\:\leqslant\:\sqrt{\frac{a^2 + bc}{b^2+c^2}} \,+\,\sqrt{\frac{b^2 +ac}{a^2+c^2}}\:.\tag{2}$$
Since
$$\sqrt{\frac{c^2 +pab}{a^2+b^2}}\geqslant\sqrt{\frac{pab}{a^2+b^2}}=\sqrt{\frac{p}{\frac ab+\frac ba}}=\sqrt{\frac{p}{u^2-2}}$$
one has, using $(2)$,
$$\sum_\text{cyc}{\sqrt\frac{a^2 + pbc}{b^2+c^2}} \geqslant
u+\sqrt{\frac p{u^2-2}}$$
Here, one can have
$$u+\sqrt{\frac p{u^2-2}}\geqslant 2+\sqrt{\frac p2}$$
since
$$\begin{align}&\bigg(u+\sqrt{\frac p{u^2-2}}\bigg)-\bigg(2+\sqrt{\frac p2}\bigg)
\\\\&=\frac{(u-2)\sqrt{2(u^2-2)}+\sqrt p(\sqrt 2-\sqrt{u^2-2})}{\sqrt{2(u^2-2)}}
\\\\&=\frac{(u-2)\sqrt{2(u^2-2)}+\sqrt p\cdot\dfrac{(2-u)(2+u)}{\sqrt 2+\sqrt{u^2-2}}}{\sqrt{2(u^2-2)}}
\\\\&=\frac{u-2}{\sqrt{2(u^2-2)}}\bigg(\sqrt{2(u^2-2)}-\dfrac{\sqrt p(2+u)}{\sqrt 2+\sqrt{u^2-2}}\bigg)
\\\\&=\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}\bigg(\frac{2\sqrt{u^2-2}+(u^2-2)\sqrt 2}{u+2}-\sqrt p\bigg)
\\\\&=\underbrace{\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}}_{\text{non-negative}}\bigg(\underbrace{\frac{2\sqrt{1-\frac{2}{u^2}}+(u-\frac 2u)\sqrt 2}{1+\frac 2u}}_{\text{increasing}}-\sqrt p\bigg)
\\\\&\geqslant\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}\bigg(\frac{2\sqrt{1-\frac{2}{2^2}}+(2-\frac 22)\sqrt 2}{1+\frac 22}-\sqrt p\bigg)
\\\\&=\frac{(u-2)(u+2)}{(\sqrt 2+\sqrt{u^2-2})\sqrt{2(u^2-2)}}\bigg(\sqrt 2-\sqrt p\bigg)
\\\\&\geqslant 0\:.\end{align}$$
Therefore, it follows that
$$\sum_\text{cyc}{\sqrt\frac{a^2 + pbc}{b^2+c^2}} \geqslant
u+\sqrt{\frac p{u^2-2}}\geqslant 2+\sqrt{\frac p2}\:.\quad\blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Problem with system of equations $(x,y,z)$ System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
$x^{2}+y^{2}+z^{2}=xy+yz+zx$
$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$
$(x+y+z)^{2}-2*3=3$
$(x+y+z)^{2}=9$
What should I do next?
| By the calculations you have made, your system is actually eqivalent to $$\begin{cases}x^2+y^2+z^2=3\\(x+y+z)^2=9\end{cases}$$Which is, at the same time, equivalent to the systems $$\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases} \quad \begin{cases}x^2+y^2+z^2=3\\x+y+z=-3\end{cases}$$But you can check using the distance formula between a point and a plane (or in many other ways) that the planes $x+y+z=3$ and $x+y+z=-3$ are both tangent to the sphere $x^2+y^2+z^2=3$, so the only solutions are $x=y=z=\pm1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the following limit? How can I calculate the following limit?
What do I miss here? What am I doing wrong?
$$\begin{align*}
\lim_{x\rightarrow0}\frac{6x\cdot\sin x-6\cdot\sin\left(x^{2}\right)+x^{4}}{x^{5}\left(e^{x}-1\right)}= & \lim_{x\rightarrow0}\frac{6x\cdot\sin x}{x^{5}\left(e^{x}-1\right)}-\lim_{x\rightarrow0}\frac{6\cdot\sin\left(x^{2}\right)}{x^{5}\left(e^{x}-1\right)}+\lim_{x\rightarrow0}\frac{x^{4}}{x^{5}\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin x}{x}}_{\rightarrow1}-\lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin\left(x^{2}\right)}{x^{2}}}_{\rightarrow1}+\lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\infty\neq\frac{21}{20}.
\end{align*}$$
According to WolframAlpha $21/20$ is the solution. What am I doing wrong?
| You replaced parts of the formula with their limits, so that some terms (not replaced with their limit) cancel. This may lead to wrong results, as you see.
I suggest finding equivalents for the numerator and the denominator.
*
*For the denominator, it is basid: it is known from Taylor-Young's formula that $\mathrm e^x-1\sim_0 x$, hence $\;x^5(\mathrm e^x-1)\sim_0x^6$
*For the numerator, apply Taylor-Young's formula to each term so as to obtain ultimately an expansion at order $6$:
\begin{align}
6x\sin x-6\sin x^{2}+x^4 &= 6x\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)-\Bigl(6x^2-\frac{6x^6}6+o(x^6)\Bigr)+x^4 \\
&=6x^2-x^4 +\frac{x^6}{20}+o(x^6)-\Bigl(6x^2-x^6+o(x^6)\Bigr)+x^4 \\
&=\frac{x^6}{20}+x^6+o(x^6)=\frac{21x^6}{20}+o(x^6)\sim_{0}\frac{21x^6}{20}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you
| There is a simpler method. Just use
$$\begin{align}a^3=b^3 \Longleftrightarrow a^3-b^3=0\Longleftrightarrow (a-b)(a^2+ab+b^2)=0\end{align}$$
To find the complex roots, just solve the quadratic:
$$a^2+ab+b^2=0.$$
In your case you can take $$a=2z+3, b=\dfrac 14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$\frac{1}{2\sqrt{2}}\int\frac{dz}{\sqrt{z^2-1}}=\frac{1}{2\sqrt{2}}\cosh^{-1}z$$ so my answer is
$$\frac{1}{2\sqrt{2}}\cosh^{-1}\left(\frac{1+x^2}{1-x^2}\right)^2$$
However the answer in the back is
$$\frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}$$
After much work (!!!) I have shown that the two forms are equal to a constant.
My question is: How can one solve the original integral to get the alternative answer directly. What substitution to use ?
| $$ I=-\int \dfrac{1}{\sqrt{x^4+1}}\dfrac{x^2+1}{x^2-1} dx=$$
$$=-\int \dfrac{1}{\sqrt{\left(x^2-1\right)^2+2x^2}}\dfrac{x^2+1}{x^2-1} dx$$
$$=-\int\frac{1}{(x^2-1) \sqrt{1+\frac{2x^2}{\left(x^2-1\right)^2}}}\dfrac{x^2+1}{x^2-1} dx dx$$
$$=-\int\frac{1}{\sqrt{1+\frac{2x^2}{\left(x^2-1\right)^2}}}\dfrac{x^2+1}{(x^2-1)^2}dx$$
Setting $u:=\sqrt{2} \dfrac{x}{x^2-1},$ as we have
$$\dfrac{du}{dx}=\sqrt{2}\dfrac{(x^2-1)-x(2x)}{(x^2-1)^2}=-\sqrt{2}\dfrac{x^2+1}{(x^2-1)^2}$$
We obtain finaly:
$$I=\dfrac{1}{\sqrt{2}}\int \frac{du}{\sqrt{1+u^2}}$$
Now, remember that $(\sinh^{-1})'(u)=\dfrac{1}{\sqrt{1+u^2}}$.
Remark: The initial inspiration came from this reference found by using (https://approach0.xyz/), a powerful search engine for formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trigonometric identity with sin, cos, and tan Very cool problem I was reading today!
$(1)$ $2\sin^2 A $ – $ 2\tan A \sin A + 2 $ – $ \sin^2 A $ – $ \cos^2 B = 0; $
$(2)$ $2\cos^2 B $ – $ 2\tan A \cos B + 2 $ – $ \sin^2 A $ - $ \cos^2 B = 0;$
If $\sin A ≠ \cos B$, find the value of $\boxed{\sin^2 A}$
$\\$
I realized there was an easy simplification in equation $(1)$; $2 \sin^2{A} - \sin^2{A} = \sin^2{A}$
There was another easy simplification in equation $(2); 2\cos^2{B} - \cos^2{B} = \cos^2{B}$
After simplifying $(1)$ and $(2)$, I got
$(3)$ $\sin^2 A $ – $ 2\tan A \sin A + 2 $ $ – \cos^2 B = 0$;
And $(4)$ $\cos^2 B $ – $ 2\tan A \cos B + 2 $ – $ \sin^2 A = 0$
Adding $(3)$ and $(4)$ gives
$(5)$ $- 2\tan A \sin A + 2\tan A \cos B + 4 = 0$ which can also be written as $\tan{A}(\sin{A} - \cos{B}) = 2$
Subtracting $(3)$ and $(4)$ gives
$(6)$ $2\sin^2{A} - 2\cos^2{B} - 2\tan{A}(\sin A - \cos{B}) = 0$
Substituting $(5)$ into $(6)$ gives
$(7)$ $2\sin^2{A} - 2\cos^2{B} - 4 = 0$
I am not really sure what to do after this. Could someone please explain what I should do? The $\sin{A}$ and $\cos{B}$ are very different and I can't seem to use any identities to solve the equation.
| You have a sign error in (5), it should be $-2\tan A \cos B$. Thus, $$ \tan A(\sin A+ \cos B)=2$$ and $$\sin^2 A-\cos^2 B=\tan A (\sin A - \cos B) \implies \sin A+ \cos B=\tan A$$ Can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
For $b$:
$$\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12} $$
$$\sqrt[3]{a x+b} - 2 = x\frac{5}{12} $$
$$\underset{x\to 0}{\text{lim}} \sqrt[3]{a x+b} - 2 = \underset{x\to 0}{\text{lim}} x\frac{5}{12} $$
$$\sqrt[3]{b} - 2 = 0 $$
$$\sqrt[3]{b} = 2 $$
$$ b = 8$$
For $a$:
$$a x+8 = (x\frac{5}{12} + 2)^3$$
$$a x+8 = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x+8$$
$$a x = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x$$
$$a = \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$\underset{x\to 0}{\text{lim}} a =\underset{x\to 0}{\text{lim}} \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$a = 5 $$
But the limit $$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{5x+8} - 2}{x}$$ doens't go to $\frac {5}{12}$.
May someone help.
| It appears from your working that you had the right idea (at least at the start) in mind and the issue was more related to the use of limit laws.
Let us fix your approach as follows. We have $$\lim_{x\to 0}(\sqrt[3]{ax+b}-2)=\lim_{x\to 0}x\cdot\frac {\sqrt [3]{ax+b} - 2}{x}=0\cdot\frac{5}{12}=0$$ which means that $$\sqrt[3]{b}-2=0$$ or $b=8$.
To find $a$ we need a bit more work. Observe that if $a=0$ then the limit in question becomes $0$. Hence $a\neq 0$ and let us put $t=ax+b$ so that $t\to 8$ as $x\to 0$. The limit in question can be written as $$\lim_{t\to 8}\frac{t^{1/3}-8^{1/3}}{t-8}\cdot\frac{t-8}{x}=\frac{5}{12}$$ ie $$\frac{1}{3}\cdot 8^{-2/3}\cdot a =\frac{5}{12}$$ or $a=5$.
To summarize $a=5,b=8$. We have use the standard limit $$\lim_{t\to c} \frac{t^n-c^n} {t-c} =nc^{n-1}$$ (with $c=8,n=1/3$) above.
You should also note that if limit of a function $f(x) $ is $L$ then it does not necessarily hold that $f(x) =L$. This is one crucial mistake which you made while finding $a$.
Food for thought: Why did I specifically eliminate the option $a=0$? The argument after that probably works without this assumption.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.
The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.
I know how to prove that it is a multiple of $60$
Lets write: $a_1+a_2+...+a_{10}=62$. Assuming $a_1<a_2<\cdots <a_{10}$ we search the maximal value of $a_{10}$. This is the difference of $62$ and the minimal value the sum $a_1+a_2+\cdots+a_9$ can take. As this value is given by $1+2+\cdots+9=45$, it follows that $a_{10}=62-45=17$. That means that for any $a_i>17$, there are no $9$ natural numbers to satisfy our initial sum.
So a solution would be $1,2,3,4,5,6,7,8,9,17$.
If $a_{10}=16$, the unique solution is $a_9=10$, the smaller solutions being unchanged as to preserve the distinctness.
If $a_{10}=15$, then $a_9=11$ or $a_8=10$.
If $a_{10}=14$, then $a_9=12$ or $a_8=11$ or $a_7=10$.
If $a_{10}=13$, then $a_8=12$ or $a_7=11$ or $a_6=10$.
If $a_{10}=12$, $a_9=13$ or $a_6=11$ or $a_5=10$. And from here on, the situation repeats.
We observe that the numbers $3,4,5$ or $6,10$ or $5,12$ repeat and as their products are $60$, then $60$ divides the product of $a_1,\cdots,a_{10}$. I don't know how to expand for $1440$. Any help, please?
| Experimental approach:
Sum of first 11 numbers is:
$S=\frac {11}2(11+1)=66$
$66-62=4$
So we delete 4 and our numbers are:
$1, 2, 3, 5, 6, 7, 8, 9, 10, 11$
which their product is a multiple of 1440.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For $z_1, z_2, \ldots, z_{2021}$ the roots of $z^{2021}+z-1$, evaluate $\sum_i\frac{z_i^3}{z_i+1}$
Let $z_1$, $z_2$, $\ldots$, $z_{2021}$ be the roots of the polynomial $z^{2021}+z-1$. Evaluate
$$\frac{z_1^3}{z_1+1} +\frac{z_2^3}{z_2+1} +\frac{z_3^3}{z_3+1} +\cdots +\frac{z_{2021}^3}{z_{2021}+1} $$
I'm not really sure where to go from here, I saw how the polynomial factors:
$$(z^2 - z + 1) (z^{n-2} + z^{n-3} - z^{n-5} - z^{n-6} + z^{n-8} - \cdots + z^2 - 1)$$
| This can be done using residues and the function
$$f(z) = \frac{z^3}{z+1} \frac{q(1-z)/z+1}{z^q+z-1}
= \frac{z^2}{z+1} \frac{q+(1-q)z}{z^q+z-1}.$$
where $q\ge 4,$ an integer.
First we need to show that the poles from the term in $q$ are all simple.
We write for a pole $\rho$ the Taylor series of $g(z) = z^q+z-1$
$$g(z) = g(\rho) + g'(\rho) (z-\rho) + \cdots$$
We have $g(\rho) = 0$ and
$$g'(\rho) = q \rho^{q-1} + 1 = q \frac{1-\rho}{\rho} + 1
= \frac{q}{\rho} - q + 1$$
Supposing this were zero we would have
$$\rho = \frac{q}{q-1}.$$
Keeping in mind that $g(\rho) = 0$ this yields
$$\frac{q^q}{(q-1)^q} + \frac{q}{q-1} - 1 = 0
\quad\text{or}\quad
\frac{q^q}{(q-1)^{q-1}} = -1.$$
This is clearly impossible with $q\ge 4$ a positive integer
and hence $g'(\rho) \ne 0$ and the poles are simple.
Note also that $-1$ is not a root of $z^q+z-1$ by inspection.
Observe that with $\rho$ the finite poles other than minus one we have
(note that there is in fact no pole at zero)
$$\sum_\rho \mathrm{Res}_{z=\rho} f(z)
= \sum_\rho \frac{\rho^3}{\rho+1}
(q(1-\rho)/\rho + 1)
\lim_{z\rightarrow \rho}
\frac{z-\rho}{z^q+z-1-(\rho^q+\rho-1)}
\\ = \sum_\rho \frac{\rho^3}{\rho+1}
(q(1-\rho)/\rho + 1) \frac{1}{q\rho^{q-1} + 1}
= \sum_\rho \frac{\rho^3}{\rho+1}.$$
Now residues sum to zero so our sum must be the residue at $-1$
with the sign flipped. The residue at infinity is zero by
inspection when $q\ge 4.$ We get
$$- \mathrm{Res}_{z=-1} f(z)
= - (-1)^3 \frac{1-2q}{(-1)^q-2}$$
for an end result of
$$\bbox[5px,border:2px solid #00A000]{
\frac{2q-1}{2-(-1)^q}.}$$
In particular with $q$ being the current year $2021$ we find
the value
$$1347.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double Integration in region $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
Calculate
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy\,,$$
where $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
I first found the intersection points that are $(1,0)$, $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{2}, \frac{1}{2}\right)$. And the determined the regions
\begin{split}
D & =\left\{(x,y)\in \mathbb{R}^{2}: x^{2}+y^{2}, y \le x\} = \{(r,\theta)\in \mathbb{R}^{2}: 0 \le r \le 1, 0 \le \theta \le \frac{\pi}{4} \right\}\\
B & =\left\{(x,y)\in \mathbb{R}^{2}: 0 \le x \le \frac{1}{\sqrt{2}}, 0 \le y \le x\right\}\\
C & =\left\{(x,y)\in \mathbb{R}^{2}: \frac{1}{\sqrt{2}} \le x \le 1, \frac{1}{\sqrt{2}} \le y \le 1-x \right\}
\end{split}
Then,
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy = \iint_{D} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{B} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{C} (x^{2}+y^{2})^{-3/2} \,dx\,dy$$
Is this right? Or is there any other easiest way?
| You are trying to integrate over the the shaded region in the diagram.
There are couple of ways to simplify this -
i) Use polar coordinates to avoid splitting the integral. In polar coordinates, $x = r \cos\theta, y = r \sin\theta, x^2+y^2 = r^2$
So line $x+y=1$ can be rewritten as $r ({\cos \theta + \sin\theta}) = 1$
leading to bounds $\frac{1}{\cos \theta + \sin\theta} \leq r \leq 1$
Also $y = x$ can be rewritten as $\tan\theta = 1, \theta = \frac{\pi}{4}$ leading to bounds $0 \leq \theta \leq \frac{\pi}{4}$
So the integral becomes,
$ \ \displaystyle \int_0^{\pi/4} \int_{1 / (\cos\theta + \sin\theta)}^1 \frac{1}{r^3} \cdot r \ dr \ d\theta = \int_0^{\pi/4} \int_{1 / (\cos\theta + \sin\theta)}^1 \frac{1}{r^3} \cdot r \ dr \ d\theta$
$\displaystyle = \int_0^{\pi/4} \big[-\frac{1}{r}\big]_{1 / (\cos\theta + \sin\theta)}^1 \ d\theta$
$\displaystyle = \int_0^{\pi/4} (\cos\theta + \sin\theta - 1) \ d\theta = 1 - \frac{\pi}{4} $
ii) Rotate the region by $\frac{\pi}{4}$ anti-clockwise (though not necessary here).
$x = r \cos (\theta - \frac{\pi}{4}), y = r \sin (\theta - \frac{\pi}{4}), x^2 + y^2 = r^2$
$x + y = 1$ becomes $\sqrt2 r \sin\theta = 1 \implies r = \frac{1}{\sqrt2 \sin\theta}$ and the integral becomes,
$\displaystyle \int_{\pi/4}^{\pi/2} \int_{\csc\theta / \sqrt2}^1 \frac{1}{r^2} \ dr \ d\theta $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Uncertain how to solve this trigonometric equation I am currently attempting to discover how to find the general solutions to $$\sqrt 3\tan^2x=2\tan x+\sqrt 3$$
The given solutions are $x= \frac{\pi}{3}+ \pi k$ , $\frac{5\pi}{6} + \pi k$
To solve this equation I removed the square root from $\sqrt 3\tan^2x$ leaving me with $\tan x$.
Then I subtracted $2\tan x$ from $3\tan x$ leaving me with $\tan x=\sqrt 3$.
Should I then not solve $\tan x = \sqrt 3$ for the generalized solutions?
This would give $x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$ which would then be generalized to $\frac{\pi}{3} + \pi k$ and $\frac{2\pi}{3} + \pi k$, which doesn't agree with the given answers.
| $$\text{Let }u=\tan x. $$
Divide through by $\sqrt{3}.$
$$ u^2 -\frac{2}{\sqrt{3}}u-1 = 0.$$
This is a quadratic equation, the solutions are:
$$u = \frac{1}{\sqrt{3}} \mp \frac{2}{\sqrt{3}}.$$
We thus have$$ \tan x = \sqrt{3} \quad \text{ or } \quad \tan x = - \frac{1}{\sqrt{3}}.$$
So $$x \in \left\{\frac{\pi}{3} + k\pi, \frac{5\pi}{6} + k\pi\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
iterated function $a(n) = \lfloor n\phi + 0.5\rfloor$ Let $a(n) = \lfloor n\phi + 0.5\rfloor$ for all $n \geqslant 1$, where $\phi$ is the golden ratio. Now let
$$a(n)^k = a(a(\ldots(a(n))))$$
where we have iterated $a(n)$, $k$ times in the RHS. I am looking for a nice closed form of $a(n)^k$. I am working on this problem for a long time but didn't get any satisfactory solution. Indeed, I was also able to find some sort of closed form but it again seems to be very painful to evaluate for some large $n$. I'm rather looking for a nice closed form.
I highhly appreciate your time and efforts. Thanks.
| Further to my comments, a proof of $a(a(n))=a(n)+n$.
Proposition 1. $\left\lfloor \frac{a(n)}{\phi} + \frac{1}{2}\right\rfloor=n$
Given $x-1< \lfloor x\rfloor \leq x$, we have
$$n\phi - \frac{1}{2} <a(n)\leq n\phi + \frac{1}{2} \iff\\
n-\frac{1}{2\phi} <\frac{a(n)}{\phi}\leq n+\frac{1}{2\phi} \iff\\
n+\frac{1}{2}-\frac{1}{2\phi} <\frac{a(n)}{\phi}+\frac{1}{2}\leq n+\frac{1}{2}+\frac{1}{2\phi}$$
Given $1+\frac{1}{\phi}=\phi$ and $\frac{1}{2}-\frac{1}{2\phi}>0$, then
$$n<n+\frac{1}{2}-\frac{1}{2\phi} <\frac{a(n)}{\phi}+\frac{1}{2}\leq n+\frac{\phi}{2}<n+1$$
or
$$n<\frac{a(n)}{\phi}+\frac{1}{2}<n+1 \Rightarrow
\left\lfloor \frac{a(n)}{\phi} + \frac{1}{2}\right\rfloor=n$$
Proposition 2. $a(a(n))=a(n)+n$.
Obviously $a(n)\in\mathbb{N}$, then
$$a(a(n))= \left\lfloor a(n)\phi + \frac{1}{2}\right\rfloor=
\left\lfloor a(n)\left(1+\frac{1}{\phi}\right) + \frac{1}{2}\right\rfloor=\\
a(n)+\left\lfloor \frac{a(n)}{\phi} + \frac{1}{2}\right\rfloor = ...$$
applying Proposition 1
$$...= a(n)+n$$
Now
$$a^{\circ 2}(n)=a(a(n))=F_2\cdot a(n) + F_1\cdot n$$
$$a^{\circ 3}(n)=a(a(\color{red}{a(n)}))=a(\color{red}{a(n)})+\color{red}{a(n)}=2\cdot a(n)+n=F_3\cdot a(n) + F_2\cdot n$$
$$a^{\circ 4}(n)=a(a(a(\color{blue}{a(n)})))=2\cdot a(\color{blue}{a(n)})+\color{blue}{a(n)}=3\cdot a(n)+2n=F_4\cdot a(n) + F_3\cdot n$$
By induction, this is
$$a^{\circ k}(n)=F_k\cdot a(n) + F_{k-1}\cdot n \tag{1}$$
because
$$a^{\circ (k+1)}(n)=a^{\circ k}(a(n))=F_k\cdot a(a(n)) + F_{k-1}\cdot a(n)=\\
(F_k+F_{k-1})\cdot a(n) + F_{k}\cdot n=F_{k+1}\cdot a(n) + F_{k}\cdot n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding value of a when the centroid of a region (p,0) is bounded by $y^2 = 4px$ and x = a The problem is basically the title, however I just got stuck when I did two solutions. Here is one of them:
$$y^2 = 4px\\
f(x) = \sqrt{4px}\\
g(x) = -\sqrt{4px}\\
\bar{x} = p = \frac{M_y}{M}\\
M_y = \int_0^ax(\sqrt{4px} + \sqrt{4px})dx\\
= \int_0^a2x\sqrt{4px}dx\\
=4\sqrt{p}\int^a_0x^{3/2}dx\\
\bf{M_y = \frac{8a^{5/2}\sqrt{p}}{5}}\\$$
\
$$M_x = \frac{1}{2}\int_0^af(x)^2-g(x)^2dx\\
=\frac{1}{2}\int_0^a4px-4px dx\\
=\frac{1}{2}\int_0^adx\\
\bf M_x = \frac{a}{2}$$
$$M = \int_0^af(x)-g(x)dx\\
= \int_0^a\sqrt{4px}+\sqrt{4px} dx\\
\bf M = 4\sqrt{pa}$$
\
$$\bar{y} = 0 = \frac{M_x}{M} =\frac{\frac{a}{2}}{4\sqrt{pa}}\\
= \bf\frac{1}{8\sqrt{pa}}$$
\
$$\bar{x} = p = \frac{M_y}{M} = \frac{\frac{8a^{5/2}\sqrt{p}}{5}}{4\sqrt{pa}}\\
= \bf \frac{2a^2}{5}$$
After all of this stuff, I get stuck. Trying different methods to find the value of using these values doesn't get me a concrete answer. Some of them even result in contradictions such as 1 = 0.
What did I do wrong in my solution? Is my solution even right in the beginning? Any answer would really help :D
| You have a few mistakes. It is easy to see by symmetry that $M_x = 0$.
$M = \displaystyle \int_0^a (\sqrt{4px}+\sqrt{4px}) \ dx$
$ = 4 \sqrt p\displaystyle \int_0^a \sqrt{x} \ dx = \frac{8 \sqrt p}{3} a^{3/2}$
As you calculated, $M_y = \displaystyle \frac{8 \sqrt p}{5} a^{5/2}$
So, $p = \overline {x} = \displaystyle \frac{3a}{5} \implies a = \frac{5p}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a closed-form solution to the following summation I am solving a summation that appears in a paper, it claims that
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{4z}{(1+\sqrt{1-4z})^2}\bigg)^k$$
I found this identity here in equation (66)
$$\sum_{n=0}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n+k}z^n =\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}$$
I didn't found the paper that has the proof of the identity above, but I notice that
$$\binom{2n}{n+k}+\sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}=\binom{2n+k}{n+k}$$
Therefore we obtain
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}-1-\sum_{n=1}^{\infty} \sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}z^n$$
Yet I have no idea for the last summation above, any help would be appreciated.
| The identity
$$\sum_{n\ge 0} {2n+k\choose n} z^n
= \left(\frac{2}{1+\sqrt{1-4z}} \right)^k
\frac{1}{\sqrt{1-4z}}
= Q_k(z)$$
can also be proved with the Cauchy Coefficient Formula, radius of
convergence is the distance to the nearest singularity which is at
$1/4.$
We obtain
$$[z^n] Q_k(z) =
\frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{1}{z^{n+1}}
\left(\frac{2}{1+\sqrt{1-4z}} \right)^k
\frac{1}{\sqrt{1-4z}} \; dz.$$
Now we put $w=\sqrt{1-4z}$ with the branch cut on $[1/4,\infty)$
(principal branch of the logarithm) so that we have
analyticity in a neighborhood of the origin and $dw = -2
\frac{1}{\sqrt{1-4z}} \; dz$. With $\sqrt{1-4z} = 1-2z-\cdots$ the image
of $|z|=\varepsilon$ makes one turn around $w=1$ plus lower order
fluctuations so that it may be deformed to a small circle and we get
with $z=(1-w^2)/4$
$$- \frac{1}{2} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{2}{1+w} \right)^k \; dw
\\ = - \frac{2^{2n+k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1-w)^{n+1}} \frac{1}{(1+w)^{n+1+k}} \; dw
\\ = \frac{(-1)^n\times 2^{n}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{n+1}} \frac{1}{(1+(w-1)/2)^{n+1+k}} \; dw$$
This yields by the Cauchy Residue Theorem
$$(-1)^n 2^n \times (-1)^n \frac{1}{2^n} {n+n+k\choose n}
= {2n+k\choose n}$$
as claimed.
Remark. This identity and the proof appeared at the following MSE link from which it may be obtained by a straightforward square root manipulation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\
v' &= (4n-1) x^{4n-2} \\
I_{n+1}&= \int_{0}^{ + \infty} \dfrac{-4n x^3 x^{4n-1} }{ 4 (1+x^4)^{n+1} (-n) }\\
I_{n+1}&= \left[ \dfrac{ x^{4n-1}}{-4n (1+x^4)^n } \right] + \int_{0}^{ + \infty} \dfrac{(4n-1) x^{4n-2}}{4n (1+x^4)^n } \\
I_{n+1} &= \dfrac{4n-1}{4n}I_n \\
I_2 &=\dfrac{4-1}{4 \times 1}I_1 \\
I_{n+1} &= \dfrac{\prod_{k=1}^{n} (4k-1) }{4^n n!} I_1 \\
I_1&= \int_{0}^{ + \infty} \dfrac{ x^{2}}{ (1+x^4) } dx \\
\end{align*}
$
Quanto's answer for $I_1$ :
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx = \frac12 \int_0^{\infty} \dfrac{ 1+\frac1{x^2}}{ x^2+\frac1{x^2} } dx \\
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
$I_{n+1} = \dfrac{ \prod_{k=1}^n (4k-1) }{ 4^n n!} I_1 =\prod_{k=1}^n \dfrac{4k-1}{4k} I_1=\prod_{k=1}^n (1- \dfrac{1}{4k}) I_1$
What is the limit ?
$\ln( 1 - u) \sim u$ donc $\ln (1 - \dfrac{1}{4k}) \sim - \dfrac{1}{4k}$
and $- \sum \dfrac{1}{k} = - \infty$ so $I_n \to 0$ ?
| Proceed as follows
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
find all triplets $(a,b,c)$ such that $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b$. Given that $a,b,c \in R$ find all triplets $(a,b,c)$ such that $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b$.
Attempt:
Case $I:$ Exactly one of $a,b,c$ is zero. WLOG, let $a=0$. Then we have
$$bc=1$$ and
$$c=b^2c=b$$
Solving we get
$$b=c=\pm 1$$
By the same symmetry we get $6$ triplets viz. $(1,1,0),(-1,-1,0),(1,0,1),(-1,0,-1),(0,1,1),(0,-1,-1)$
Case $II:$ None of $a,b,c $ is Zero.
Let us assume:
$$a^2b+c=b^2c+a=c^2a+b=p$$
Then we have:
$$a^2b^2+bc=bp$$
$$b^2c^2+ac=cp$$
$$c^2a^2+ab=ap$$
By adding all the above equations and
Using the fact that:
$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=1-2bac(a+b+c)$$
We get
$$2-2abc(a+b+c)=p(a+b+c)$$
$\textbf{ANY HINT PLZ}$
EDIT:
Case $III:$
If $a=b=c \ne 0$ Then we get two more triplets.
$$\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$$
| The case $a,b,c \neq 0 \implies c = \dfrac{a(1-ab)}{1-b^2}= \dfrac{ac(a+b)}{1-b^2}\implies a^2+ab=1-b^2\implies a^2+ab+b^2 = 1$. Thus it follows that: $b^2+bc+c^2 = 1 = c^2+ac+a^2\implies a^2+b^2+c^2 = 1$. But Cauchy-Schwarz inequality says: $1 = a^2+b^2+c^2 \ge ab + bc + ac = 1$. So $a = b = c = \pm \dfrac{1}{\sqrt{3}}$, and this is the only solutions for this case. The other cases are treated already.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let X and Y have a joint uniform distribution on the triangle with corners at $(0, 2)$, $(2, 0)$, and the origin. Find $E(Y | X = 1/2)$ so $f(x,y) = 1/2$ on the region of the triangle.
$f(x) = \int_0^{2-x}\cfrac{1}{2}dy = \cfrac{1}{2}(2-x)$
so $f(y|x) = \cfrac{1/2}{\cfrac{1}{2}(2-x)} = \cfrac{1}{2-x}$
So $E(Y|X=1/2) = \int_0^2y\cfrac{1}{3/2}dy = \int_0^2\cfrac{2}{3}ydy = [\cfrac{2}{6}y^2]^2_0 = 4/3$
But the answer is $3/4$
Where have I gone wrong?
| When $X = 1/2$, the support of the conditional density $$f_Y(y \mid x) = \frac{1}{2-x}$$ is not on $Y \in [0,2]$ as suggested by your integral, but rather, $Y \in [0, 2 - 1/2] = [0, 3/2]$. Therefore, the correct expression is $$\operatorname{E}[Y \mid X = 1/2] = \int_{y=0}^{3/2} y \cdot \frac{1}{3/2} \, dy = \frac{3}{4}.$$
You would not have been led astray had you used indicator functions:
$$f_Y(y \mid x) = \frac{1}{2-x} \mathbb 1 (0 \le y \le 2-x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I find the inverse of this problem u⊞v:=uv−2(u+v)+6 Let V=(2,∞). For u,v∈V and a∈R define vector addition by u⊞v:=uv−2(u+v)+6 and scalar multiplication by a⊡u:=(u−2)^a+2. It can be shown that (V,⊞,⊡) is a vector space over the scalar field R. Find the following:
the sum:
6⊞10=34
the scalar multiple:
−3⊡6=2.015625
the additive inverse of 6:
⊟6=
the zero vector:
0V=3
the additive inverse of x:
⊟x=
As you can see I found the sum and the scalar multiple as well as the zero vector.However, I tried doing the additive inverse of 6 and the additif inverse of x, but I have no idea on how to find the answers.
| Before you can find an additive inverse you must find the additive identity. That is $e\square x = x$ for all $x$.
so $ex -2(e+x) +6 = x$. Solver for $e$
so $e(x-2)=3x -6$ so $e=\frac {3x-6}{x-2} = 3$
The additive identity is $3$.
Verify: $k\square 3= 3k -2(k+3) + 6 = k -6 + 6 = k$ so that is indeed the indentity.
Now to find the additive inverse of $6$ we must solve: $x\square 6 = 3$.
So $6x - 2(x+6) + 6 = 3$
$4x =9$
$x = \frac 94$.
SO $x=\frac 94$ is the additive inverse of $6$.
Verify: $\frac 94 \square 6 = \frac 94\cdot 6 -2(6 +\frac 94) +6= \frac {27}2 - 12-\frac 93 + 6 = 9-12 + 6 = 3$. It checks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this question is given as
$$\lvert \sqrt{x-1} -1 \rvert + \lvert \sqrt{x-1}+1 \rvert$$
My efforts weren't great at all, I tried rationalizing them but the denominator becomes very similar to the question, which is totally not helpful for simplification.
| Just write
\begin{eqnarray*} \sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}
& = & \sqrt{x -1 - 2\sqrt{x -1} +1} +\sqrt{x -1 + 2\sqrt{x -1} +1} \\
& = & \sqrt{(\sqrt{x-1}-1)^2} + \sqrt{(\sqrt{x-1}+1)^2} \\
& = & \begin{cases} 2 & 1 \leq x\leq 2 \\ 2\sqrt{x-1} & x>2 \end{cases}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.
| Let $f$ be the conveniently translated function defined by :
$$f(t):=\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}$$
What I would like to underline is that a graphical representation provides a good insight for such questions. In particular, it can help to anticipate (or verify) a certain number of results, in particular the fact that the roots of $f(t)=0$ are all real, separated by integers $-2,-1,0,1,2$ and symmetric with respect to $0$ due to the fact that $f$ is an odd function.
Indeed, this function has a decreasing behavior (the derivative $f'$ is clearly negative) from $+\infty$ to $+\infty$ on each of the intervals: $[k,k+1], \ k=-2,-1,0,1$ delimitated by vertical asymptotes ; the two external branches do not provide any supplementary root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Prove that $\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n\geq 1$. Let $n\in\Bbb N$, then $\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n\geq 1.$
Notice that,
$$\begin{align*}\left (1+\dfrac 1 {2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n&=\sum_{k=0}^n \binom{n}{k} \left (\dfrac{1}{2n}\right )^k-\sum_{k=0}^n\binom{n}{k}\left (-\dfrac{1}{2n}\right )^k\\
&=\sum_{k=0}^n \binom{n}{k}\left (\dfrac{1}{2n}\right )^k[1-(-1)^k] \end{align*}$$
I don't know how to get from there, because if $ n $ falls into an even number, all that goes to 0, and it is not greater than or equal to 1, but if $ n $ is odd if it is true. How do I finish the problem?
| At least for large values of $n$
$$a_n=\left (1+\dfrac{1}{2n}\right )^n-\left (1-\dfrac{1}{2n}\right)^n$$
$$b=\left (1+\dfrac{1}{2n}\right )^n\implies \log(b)=n \log\left (1+\dfrac{1}{2n}\right )=\frac{1}{2}-\frac{1}{8 n}+\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
$$b=e^{\log(b)}=\sqrt e \left(1-\frac{1}{8 n}+\frac{19}{384 n^2}+O\left(\frac{1}{n^3}\right)\right)$$
$$c=\left (1-\dfrac{1}{2n}\right )^n\implies \log(c)=n \log\left (1-\dfrac{1}{2n}\right )=-\frac{1}{2}-\frac{1}{8 n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
$$c=e^{\log(c)}=\frac{1}{\sqrt{e}} \left(1-\frac{1}{8 n}-\frac{13}{384 n^2}+O\left(\frac{1}{n^3}\right) \right )$$
$$a_n=2 \sinh \left(\frac{1}{2}\right)-\frac{\sinh \left(\frac{1}{2}\right)}{4 n}+\frac{13+19
e}{384 \sqrt{e} n^2}+O\left(\frac{1}{n^3}\right)$$
$$2 \sinh \left(\frac{1}{2}\right)-\frac{\sinh \left(\frac{1}{2}\right)}{4 n}\gt 1 \implies n > \frac{1}{8-4 \text{csch}\left(\frac{1}{2}\right)}\sim 3.08774$$
Computed exactly
$$a_{10}=\frac{26372036801}{25600000000}=1.030158$$ while the truncated series gives
$$a_{10}=\frac{37939 e-37907}{38400 \sqrt{e}}\sim 1.030184$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$ I wrote out the integral for the convolution and it becomes a big exercise in partial fractions and integration of rational functions.
$$ \int(\frac{1}{a^2+t^2}) (\frac{1}{a^2+(x-t)^2}) dt=
\int\frac{x + 2 t}{x (4 a^2 + x^2) (a^2 + t^2)} + \frac{3 x - 2 t}{x (4 a^2 + x^2) (a^2 + x^2 - 2 xt + t^2)} dt =$$
$$=\frac{a (log(a^2 + t^2) - log(a^2 + (x - t)^2)) + x \cdot tan^{-1}(\frac{t}{a}) + x \cdot tan^{-1}(\frac{t - x}{a})}
{a x (4 a^2 + x^2)}
$$
Plugging in the bounds of plus/minus infinity makes the logs cancel out, and gives the 2$\pi x$ from the arctan terms.
My question: Is there a way to simplify / make this any easier using the properties of the convolution? Or are these problems just 'plug and chug' like back in my engineering days?
| Here using Fourier transform may simplify things considerably:
Let $\phi(x)=\frac{1}{1+x^2}$, and $\phi_a(x)=\frac{1}{a}\phi(a^{-1}x)=\frac{a}{a^2+x^2}$.
Recall that $\pi e^{-2\pi|t|}=\int e^{-2\pi itx}\frac{1}{1+x^2}\,dx=\widehat{\phi_1}(t)$. Hence
$$\widehat{\phi_a}(t)=\frac{1}{a}\int e^{-2\pi itx}\phi(xa^{-1})\,dx=\frac{1}{a}\int e^{2\pi ia t(a^{-1}x)}\phi(a^{-1}x)\,dx=\widehat{\phi_1}(at)$$
Thus
$$\widehat{\phi_a*\phi_a}(t)=(\widehat{\phi_a}(t))^2=(\widehat{\phi_1}(at))^2=\pi^2e^{-4\pi a|t|}$$
That is,
$$
I_a=a^{-2}(\phi_a*\phi_a)(x)=a^{-2}\pi^2\mathcal{F}^{-1}(e^{-4\pi a|t|})(x)=\frac{2\pi}{a(a^2+x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are
*
*10 and 0
*-10 and 4
*10 and 4
*-10 and 0
This question is from the book, Mathematics, Class 9 (The IIT Foundation Series) , page number 1.25. The answer key present in the book says the first option as the correct answer. I need an explanation to solve this question. Till now I have tried to rationalize $\frac{1}{ \sqrt{k} + \sqrt{k+1}}$ found that $\frac{1}{ \sqrt{k} + \sqrt{k+1}} = \sqrt{k+1}-\sqrt{k}$. I am not sure what to do next.
Thanks!
| Your sum $$\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$$ converges to $10$ which means that $10 = a - \sqrt b$. After solving this equation we get $a = 10$ $and$ $b = 0$. Hence option(1) is correct. Your approach to rationalize it was correct. If we rationalize it we get $\sqrt(k+1) - \sqrt k$. If we substitute the value of 4 in the equation we get $\sqrt5 - \sqrt4$, we then move on to the next value of $k$ which is $5$ which becomes $\sqrt6 - \sqrt5$. Till now we got, $(\sqrt5 - \sqrt4) + (\sqrt6 - \sqrt5) + (\sqrt7 - \sqrt6) + ... + (\sqrt144 - \sqrt143)$. Now we can cancel out the $\sqrt5$ of the first bracket from the $\sqrt5$ of the second bracket and we will cancel out each and every term. After the cancellation we get $-\sqrt4 + \sqrt144$ and we get 10, which is your answer. Now we can equate it with $a - \sqrt b$ and we get $a = 10$ and $b = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Are these the only "intersections" between the following series, and why all of them are multiples of $10$ As a web developer that programs in PhP, I enjoy running some scripts to see some of math wonders, however, PhP is limited to large calculations.
I wanted to see if there are any intersections for the following $2$ series:
A) $1 + 2 + 3+ 4 + 5 + \ldots$
B) $(1) + (1 + 2) + (1 + 2 + 3) +(1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 + 5) + \ldots$
Out of the first $116410911$ positive integers, I have "found" the following "intersections" between A) and B):
At the number $10$:
$10 - (1+2+3+4)=0$
$10 - ((1)+(1+2)+(1+2+3))=0$
At the number $120$:
$120 - (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15) = 0$
$120 - ((1)+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+(1+2+3+4+5+6+7)+(1+2+3+4+5+6+7+8)) = 0$
And I am not going to list the other procedures, but also:
At the numbers $1540$ and $7140$.
Can anyone confirm if there are any more "intersections"? In other words, am I to expect a finite case or an infinite case? Also, is it a coincidence that all four of them are numbers that are multiples of $10$?
| These are the numbers that are both "triangular" ($n(n+1)/2$) and "tetrahedral" ($n(n+1)(n+2)/6$). They are tabulated at https://oeis.org/A027568. You have found all of them. There are several links to the literature at the OEIS site. It's unlikely that there is a wholly elementary proof, as elliptic curves are involved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\{a,b,c\}\subset\mathbb C$ such that $a^2+b^2+c^2=ab +bc +ca$ prove that they lie on the vertices of an equilateral triangle in the complex plane
Exercise 1.11 (from Friendly Approach to Complex Analysis).
If $a$, $b$, $c$ are real numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must be equal. Indeed, doubling both sides and rearranging gives $(a-b)^2+(b-c)^2+(c-a)^2=0$, and since each summand is nonnegative, it must be that case that each is $0$. On the other hand, now show that if $a$, $b$, $c$ are complex numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must lie on the vertices of an equilateral triangle in the complex plane. Explain the real case result in the light of this fact.
Hint: Calculate $\big((b-a)\omega+(b-c)\big)\big((b-a)\omega^2+(b-c)\big)$, where $\omega$ is a nonreal cube root of unity.
It is a solved exercise but I don't understand the solution (or the hint). More than that I don't get why the author shared this exercise.
I understand the cube roots of unity are $\{ (1,0), (-\frac{1}{2}, \frac{\sqrt{3}}{2}), (-\frac{1}{2}, -\frac{\sqrt{3}}{2}) \}$
So I have $a := (1,0)$ and $b := (-\frac{1}{2}, \frac{\sqrt{3}}{2}) $ and solved for $c$, and got two values. One of them was the last cube root of unity. However I have observed that the points (solutions for $c$ considered one at a time) lie on the vertices of an equilateral triangle.
Isn't this enough? What is the thought process behind the hint?
| Hint.
We have $(b-a)^2+(c-a)^2 = (b-a)(c-a)$ so
$$
c-a = (b-a)\left(\frac{1\pm\sqrt{3}i}{2}\right)
$$
then
$|b-a| = |c-a|$ and $\angle(b-a,c-a) = \pm\frac{\pi}{3}$
NOTE
The property $a^2+b^2+c^2 - a b- a c- b c = 0$ is translation invariant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How does this vandermonde identity proof works? Recently I have taken a look at this vandermonde identity proof Inductive Proof for Vandermonde's Identity?.
\begin{align*}
\binom{m + (n+1)}r &= \binom{m+n}r + \binom{m+n}{r-1}\\
&= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}\\
&= \binom mr + \sum_{k=0}^{r-1} \binom mk\biggl(\binom n{r-k} + \binom n{r-1-k}\biggr)\\
&= \binom mr\binom{n+1}0 + \sum_{k=0}^{r-1} \binom mk\binom{n+1}{r-k}\\
&= \sum_{k=0}^r \binom mk \binom{n+1}{r-k}
\end{align*}
I am confused mostly at this part
\begin{align*}
&= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}\\
&= \binom mr + \sum_{k=0}^{r-1} \binom mk\biggl(\binom n{r-k} + \binom n{r-1-k}\biggr)\\
\end{align*}
How did this
\begin{align*}
&= \sum_{k=0}^r \binom mk\binom n{r-k}
\end{align*}
turn into
\begin{align*}
&= \binom mr
\end{align*}
which identity did this person use? I am such that it could magically become C(m,r)?
I am also confused on how
\begin{align*}
\binom{n+1}0
\end{align*}
appears in here
\begin{align*}
&= \binom mr\binom{n+1}0 + \sum_{k=0}^{r-1} \binom mk\binom{n+1}{r-k}\\
\end{align*}
If someone could help that would be very helpful, I didn't comment on the post due to it being 8 years, posted already. If someone could help and point where I should study that would be very helpful. Thank you!
| Do you mean this step?
$\displaystyle\sum_{k=0}^r {m \choose k}{n \choose r-k} = {m \choose r} + \sum_{k=0}^{r-1}{m\choose k}{n\choose r-k}$
Also for your second question, as $0! \equiv 1$ we'll have ${n+1\choose 0} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$
Given a sequence $\{a_n\}_{n\in \mathbb N}$ of real numbers such that
$$\begin{align}
\lim_{n\to\infty}\frac{a_na_{n+1} - a_{n-1}a_{n+2}}{a_{n+1}^2 - a_na_{n+2}} &= p + q && (1)\\[1mm]
\lim_{n\to\infty} \frac{a_n^2 - a_{n-1}a_{n+1}}{a_{n+1}^2 - a_na_{n+2}} &= pq && (2)\end{align}$$
where $|p| < |q|$, prove that
$$\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$$
Attempts:
Idea #1: Let us denote $b_n = \dfrac{a_n}{a_{n+1}}$. If we divide both numerators and denominators of $(1)$ and $(2)$ by $a_na_{n+2}$ and $a_{n+1}a_n$, respectively, we have
$$
\begin{align}
\lim_{n\to\infty}\frac{\dfrac{a_{n+1}}{a_{n+2}} - \dfrac{a_{n-1}}{a_{n}}}{\dfrac{a_{n+1}}{a_n}\dfrac{a_{n+1}}{a_{n+2}} - 1} &= p+q && (1') \\[1mm]
\lim_{n\to\infty} \frac{\dfrac{a_n}{a_{a+1}} - \dfrac{a_{n-1}}{a_n}}{\dfrac{a_{n+1}}{a_n} - \dfrac{a_{n+2}}{a_{n+1}}} &= pq && (2')
\end{align}$$
which now can be written as
$$\begin{align}
\lim_{n\to\infty}\frac{b_{n+1}b_n - b_nb_{n-1}}{b_{n+1} - b_n} &= p+q && (1'') \\[1mm]
\lim_{n\to\infty} \frac{b_n - b_{n-1}}{\dfrac{1}{b_{n+1}} - \dfrac{1}{b_{n}}} &= -pq && (2'')
\end{align}$$
Now, every numerator and denominator contains the difference of consecutive terms of some sequence (reminds me of Cesaro-Stolz, but that cannot be applied here).
Idea #2: The given conditions can be written as
$$\begin{align}
\lim_{n\to\infty}\frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= p + q && (1')\\[2mm]
\lim_{n\to\infty} \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= pq && (2')\end{align}$$
Now, maybe a bit of Linear Algebra could be incorporated somehow.
If we set (B. Grossman)
$$\begin{align} x_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}}, \quad y_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} \end{align}$$
Then, we have
$$\pmatrix{a_{n+1} & a_{n+2}\\a_n & a_{n+1}} \pmatrix{x_n\\y_n} = \pmatrix{a_n\\a_{n-1}}$$
This gives
$$\frac{a_{n+1}x_n + a_{n+2}y_n}{a_nx_n + a_{n+1}y_n} = \frac{a_n}{a_{n-1}}$$
Dividing the numerator and the denominator by $a_{n+1}$, we get
$$\frac{x_n + \dfrac{a_{n+2}}{a_{n+1}}y_n}{\dfrac{a_n}{a_{n+1}}x_n + y_n} = \frac{a_n}{a_{n-1}}$$
Noting that $x_n \to p+q$, $y_n \to pq$ and assuming that $\dfrac{a_n}{a_{n-1}} \to A$, and sending $n$ to infinity, we get
$$\frac{p+q + Apq}{\dfrac{1}{A}(p+q) + pq} = A$$
which simplifies to
$$0=0$$
I think I did something wrong somewhere.
Any help is appreciated.
| HINT:
Call
$$
X_n=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n+1}^2-a_na_{n+2}}\\
Y_n=\frac{a_{n}^2-a_{n-1}a_{n+1}}{a_{n+1}^2-a_na_{n+2}}
$$
then
$$
\lim_nX_n=p+q\\
\lim_nY_n=pq
$$
thus
$$
\lim_n\frac{X_n}{Y_n}=\frac{p+q}{pq}
$$
but
$$
\frac{X_n}{Y_n}
=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n}^2-a_{n-1}a_{n+1}}\;.
$$
Then you get
$$
\frac{X_n}{Y_nY_{n-1}}=
\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n-1}^2-a_{n-2}a_{n}}
$$
and inductively
$$
\frac{X_n}{Y_nY_{n-1}\cdots Y_2}=
\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{1}^2-a_{0}a_{2}}
$$
from which
$$
\frac{X_n}{Y_nY_{n-1}\cdots Y_2}(a_{1}^2-a_{0}a_{2})+a_{n-1}a_{n+2}=a_na_{n+1}
$$
divide by $a_{n+1}^2$ and work on LHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$
I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$
| Here is another method to add to the list. Consider the summation
$$\sum\limits_{n=1}^{\infty }{\frac{1}{n\left( 2n+1 \right)}}$$
Note that
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}$$
for $n\ge 0$. So if we have a function $f$ that goes like $1/{{z}^{2}}$then taking $\left( \psi \left( -z \right)+\gamma \right)f\left( z \right)$over an infinitely large closed contour about $z=0$ just sums all the residues the total of which is zero. We have then
$$\sum\limits_{n=1}^{\infty }{f\left( n \right)}=-\sum\limits_{f\left( z \right)}^{{}}{res\left( \psi \left( -z \right)+\gamma \right)f\left( z \right)}$$
where the sum on the right is over all residues due to $f$. In this case its very simple: there are two simple poles at $z=-1/2,0$. The residue at $z=0$ is easy (use the expansion above). We have
$$\sum\limits_{n=1}^{\infty }{\frac{1}{n\left( 2n+1 \right)}}=2+\psi \left( \tfrac{1}{2} \right)+\gamma $$
At this point I’m just going to assume its well known that $\psi \left( \tfrac{1}{2} \right)=-\gamma -2\log \left( 2 \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Definite integral involving the exponential of an algebraic function We define the following integral for $\omega>0$ and $\lambda\neq0$ such that $\omega^2>\lambda$:
$$\begin{align}
f(\omega,\lambda)
&:= \int_\omega^\infty e^{-x\sqrt{x^2-\lambda}} \; \varphi(x) \, \text{d}x
\\[6pt]
\tag{1}
&=\frac{1}{\sqrt{2\pi}}\int_\omega^\infty e^{-\sqrt{x^4-\lambda x^2}-\frac{x^2} 2} \;\text{d}x
\end{align}$$
where $\varphi$ is the probability density function of a Gaussian random variable. I am wondering if it is possible to derive a closed-form expression, where by "closed-form" I do include expressions involving special functions such as the gamma function or elliptic integrals.
I am not sure how to proceed from expression $(1)$. I have scanned the book by Gradshteyn & Ryzhik to search for similarly-looking expressions, but have been unlucky $-$ or at least I haven't been able to find anything of the same vein. While it seems possible to solve this type of integral for exponents of the form $\sqrt{ax^2+bx+c}$, in my particular case the issues seem to be:
*
*The polynomial is of degree greater than $2$;
*The additional square term coming from the Gaussian pdf.
Any hints would be much appreciated. If a closed-form exists for a different domain of integration, please also do let me know.
| This is not a closed form solution.
As you already found it, I did not find anything for such integrands.
A possible solution would be a Taylor expansion around $\lambda=0$. This would give for the integrand
$$e^{-\sqrt{x^4-\lambda x^2}-\frac{x^2} 2}=e^{-\frac{3 }{2}x^2}\Bigg[1+\frac 12 \lambda+\sum_{n=2}^\infty\frac {P_{n}(x) }{2^n \,n!\, x^{2n-2}}\lambda^n\Bigg]$$ where the first polynomials are
$$\left(
\begin{array}{cc}
2 & x^2+1 \\
3 & x^4+3 x^2+3 \\
4 & x^6+6 x^4+15 x^2+15 \\
5 & x^8+10 x^6+45 x^4+105 x^2+105 \\
6 & x^{10}+15 x^8+105 x^6+420 x^4+945 x^2+945 \\
7 & x^{12}+21 x^{10}+210 x^8+1260 x^6+4725 x^4+10395 x^2+10395 \\
8 & x^{14}+28 x^{12}+378 x^{10}+3150 x^8+17325 x^6+62370 x^4+135135 x^2+135135
\end{array}
\right)$$ where some interesting patterns appear (o be explored with $OEIS$.
This means that we should face linear combinations of integrals $(k \geq 0)$
$$I_k=\int x^{-2k}\,e^{-\frac{3 }{2}x^2}\,dx=-\frac 12 \left(\frac{3}{2}\right)^{k-\frac{1}{2}}\,\Gamma \left(\frac{1}{2}-k,\frac{3 x^2}{2}\right)$$
$$J_k=\int_\omega ^\infty x^{-2k}\,e^{-\frac{3 }{2}x^2}\,dx=\frac 12 \left(\frac{3}{2}\right)^{k-\frac{1}{2}}\,\Gamma \left(\frac{1}{2}-k,\frac{3 \omega ^2}{2}\right)$$
Using $\lambda=\omega=\pi$ and only the terms given in the table, I obtained a value of $2.11383 \times 10^{-7}$ while numerical integration gives $2.11601 \times 10^{-7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that a Local minimum is Absolute minimum in $R^3$ Just trying to solve this question:
$f(x,y,z) = x^2 + y^2 +3z^2 -xy +2xz+ yz$.
Found the only critical point of the function and explain why she is an absolute minimum.
We learn at class how to found that a point is a critical point so i found her and its $(0,0,0)$.
I know how to say that its a Local minimum, but i don't know how i am supposed to explain why its an absolute minimum and i want to know how to approach it.
| I don't know if there's some immediate way to know that the local minimum is a global minimum, and I would like to see such answers if there are.
I would do some algebra to rewrite $f$ as:
$f(x,y,z) = 0.5(x-y)^2 + 0.5(y+z)^2 +(x+z)^2-0.5x^2+1.5z^2$
Then, the only potentially problematic term is $-0.5x^2$.
But minimizing $(x+z)^2-0.5x^2+1.5z^2$ over $z$, which occurs when $2(x+z)+3z=0$, i.e. $z=-0.4x$, we see that:
$(x+z)^2-0.5x^2+1.5z^2 \ge 0.36x^2 -0.5x^2 +0.24x^2 = 0.1x^2 \ge 0$
Therefore $f(x,y,z) \ge 0$
Note that, when I minimized $(x+z)^2-0.5x^2+1.5z^2$ over $z$ to show that $(x+z)^2-0.5x^2+1.5z^2 \ge 0.1x^2$, that was only to derive that inequality. I was not trying to minimize $f$ over only $z$, while ignoring $y$.
Another way to derive the same inequality would be to let $t = z + 0.4x$, with the choice inspired by the fact that we derived above that $z = -0.4x$ minimizes that expression.
Then we have $z = t - 0.4x$ and $z^2 = t^2 - 0.8tx + 0.16x^2$, and:
\begin{align}
(x+z)^2-0.5x^2+1.5z^2 &= x^2 + 2xz + z^2 - 0.5x^2+1.5z^2 \\
& = 0.5x^2 +2xz+2.5z^2 \\
& = 0.5x^2 +2x(t - 0.4x)+2.5(t^2 - 0.8tx + 0.16x^2) \\
& = 0.1x^2 +2.5t^2 \\
&\ge 0.1x^2
\end{align}
Using this inequality in $f$ gives:
\begin{align}
f(x,y,z) &= 0.5(x-y)^2 + 0.5(y+z)^2 +(x+z)^2-0.5x^2+1.5z^2 \\
&\ge 0.5(x-y)^2 + 0.5(y+z)^2 + 0.1x^2 \\
&\ge 0 \qquad \text{ since it is the sum of squares}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.
If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.
My try: $$F_Y(x) = P(X^2+1 < x) = \mathrm P(-\sqrt{x-1} < X < \sqrt{x-1}) = \mathrm P(X < \sqrt{x-1}) -\mathrm P(X < -\sqrt{x-1})$$
Now since $$F_X(x) = \begin{cases}\ 1-e^{-\lambda x}, & x \ge 0 \\ 0 , & x< 0\end{cases}$$
we have $\mathrm P(X < -\sqrt{x-1}) \equiv 0$. Hence
$$F_Y(x) = \mathrm P(X < \sqrt{x-1}) = F_X(\sqrt{x-1}) = \begin{cases} 1-e^{-\lambda \sqrt{x-1}}, & \sqrt{x-1} \ge 0 \\ 0 , & \sqrt{x-1}< 0\end{cases}$$
or
$$F_Y(x) = \begin{cases} 1- e^{-\lambda \sqrt{x-1}}, & x > 1 \\ 0 , & x \le 1\end{cases}$$
*
*I'm not sure putting the last intervals for $x$ though. Is there a mistake in this solution?
*Now, I'm going to differentiate $F_Y(x)$ to find the probability density function. Is there any shortcut/other approach to this?
| \begin{align}
F_Y(x) & = \Pr(X^2+1\le x) \\[8pt]
& = \begin{cases} \Pr(X\le \sqrt{x-1} ) & \text{if } x\ge1, \\
0 & \text{if } x<1
\end{cases} \\[8pt]
& = \begin{cases}
1 - e^{-\lambda\sqrt{x-1}} & \text{if } x\ge 1 \\
0 & \text{if } x<1 \quad \text{(not “if $\sqrt{x-1} < 0$”)}
\end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding which complex numbers satisfy $\left|z + 1\right| + \left|z - 1\right| = 4$ This problem is from the book Complex Analysis with Applications by Asmar. Currently I am stuck at, well, everything, as I don't know know how to manipulate the expression $\left|z + 1\right| + \left|z - 1\right| = 4$ into anything meaningful. Squaring and simplifying both sides yields $z\bar{z} + \left|z + 1\right|\left|z - 1\right| + 1 = 8$, and I don't see how that is useful.
The problem is asking us to find those complex numbers whose sum of distances from both $1$ and $-1$ is equal to $4$. I suppose the end result is a circle of some kind, as points $(2, 0)$ and $(-2, 0)$ satisfy the equality.
| Let $z = x + iy$, then $|z+1| = \sqrt{(x+1)^2 + y^2}$ and $|z-1| = \sqrt{(x-1)^2 + y^2}$.
$|z+1| = 4 - |z-1|$
$(x+1)^2 + y^2 = 16 + (x-1)^2 + y^2 - 8\sqrt{(x-1)^2 + y^2}$
$x - 4 = -2\sqrt{(x-1)^2 + y^2}$
$x^2 + 16 - 8x = 4(x-1)^2 + 4y^2$
$12 = 4y^2 + 3x^2$
$\frac{x^2}{4} + \frac{y^2}{3} = 1$ which is formula of ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following would be true If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following is true:
A) $\frac{\sin ^{-1} x}{1+\cos ^{-1} y}=\frac{\pi}{2}$
B) $\frac{x+y}{\sin 1}=-2$
C) $\sin ^{-1} x+\sin ^{-1} y=1$
D) $\frac{x+y}{\cos 1}=1$
My Approach: Well I couldn't do really much. I tried to consider approaches related to factorising the given expression but that doesn't work in this case. How should I try to approach this problem? Any hints/solutions are appreciated. Thanks!
| Consider the Gauss identity
\begin{align}
a^3+b^3+c^3-3abc &= (a+b+c)(a^2+b^2+c^2-ab-ac-bc) \\
& = (a+b+c) \frac{1}{2} ((a-b)^2+(a-c)^2+(b-c)^2).
\end{align}
If $a^3+b^3+c^3-3abc=0$, plugin $a=\sin^{-1}(x)$, $b=\sin^{-1}, c=-1$ we have the hypothesis and consequently two cases.
Case 1 $a+b+c=0$ then
$\sin^{-1}(x)+ \sin^{-1}(y)-1=0$ which is C).
Case 2 $(a-b)^2+(a-c)^2+(b-c)^2= 0$ which implies $a=b=c=0$ then $\sin^{-1}(x)=\sin^{-1}(y)=-1$ but this is not possible for all $x,y \in \mathbb{R}$ so the only posibility is C).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I wrote the series of $e^{a \sin^{-1}x}$ but it contained terms like $(\sin^{-1}x)^2$, $(\sin^{-1}x)^3$ and so on so I could not find the series. My idea was to find the series of the anti derivative of the function and then to derivate the obtained series. Any other way to do it?
| Hint
You face a problem of composition of series.
Start with
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^7\right)$$
$$a\sin ^{-1}(x)=ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right)$$ Now, use
$$e^{a \sin ^{-1}(x)}=\exp\Big[ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right) \Big]$$ when done, work the denominator and use long division.
Edit
Your idea of using the antiderivative is very good. You could even continue using the logarithm of it and then go backward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
synthetic divide polynomial $(2x^3+7x^2-13x-3) \div (2x-3)$ I need to figure out how synthetic division works. The problem is $$ (2x^3+7x^2-13X-3) \div (2x-3)$$ I can do the long division.
$$\require{enclose}
\begin{array}{r}
x^2+5x+1 \\
(2x-3) \enclose{longdiv}{2x^3+7x^2-13x-3}\\ \underline{-2x^3+3x^2} \phantom{100000000} \\ 10x-13x \phantom{1000} \\\underline{-10x+15x} \phantom{1000} \\ 2x-3 \\\underline{-2x+3} \\ 0
\end{array}$$
But when I do synthetic division my answer is slightly different;
$$ 3/2 \left[ \begin{array}{r}{2 \phantom{10}7 \phantom{2}-13\phantom{2}-3} \\ \underline{ \phantom{100}3 \phantom{100} 15 \phantom{1000} 3} \\ 2 \phantom{1}10\phantom{1000} 2 \phantom{1000} 0\end{array} \right] = 2x^2 +10x +2$$
I just realized that I could factor out the two from the polynomial after I synthetic divide. The answer then would be; $$2(x^2+5x+1)$$ The only problem I have is the two graphs would be different. The second graph would stretch along the y axis. If I put a 1 in for x then y =14 a
fter synthetic division. Y would = 7 if I put a 1 in for x after I use long division. I am not sure what to do next.
| When you do synthetic division, you're not dividing by $2x-3$, you're dividing by $x - 3/2$.
Note that $2(x-3/2) = 2x-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?
| You do not have to use implicit differentiation. $x$ and $y$ are explicitly given in terms of parameter $t$.
$\displaystyle \frac{dy}{dt} = 12t^3 + 12t^2, \frac{dx}{dt} = 2t + 2$.
From the above, and chain rule, you get: $\displaystyle \frac{dy}{dx} = \frac{dy}{dy} \cdot \frac{dt}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 6t^2$, after simplification. Note that there is the assumption $t \neq -1$ made here to permit cancellation.
Now, $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$, again by chain rule.
Utilising results from before, $\displaystyle \frac{d^2y}{dx^2} = \frac{12t}{2t + 2} = \frac{6t}{t+1}$.
To find the correct value of $t$, solve for $x = 8 \implies t^2 + 2t = 8$, giving $t = 2$ or $t = -4$. By putting this into the expression for $y$, you note that only $t = 2$ gives the expected value of $y = 80$, so you need $t = 2$.
Putting that into the expression for the second derivative, you get $4$ as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$.
Let $f(x)=a(x-2)(x-b)$, where $a,b\in R$ and $a\ne0$. Also, $f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$. Find the minima and maxima of $f(x)$.
$f(f(x))$ is a quartic equation. So, it would have $4$ roots. Its real root should occur in pair because complex roots occur in pair. So, I failed to understand the meaning of 'exactly one real zero' in the question. Does that mean its real roots are equal and they are equal to $5$?
Since $f(x)$ has a zero at $2$, $f(f(2))=f(0)=2ab$. But even RHS of $f(f(x))$ is coming out to be $2ab$ at $x=2$.
Putting $x=5$ in $f(f(x))$ gives an equation in $a^2$ and $b^2$. Don't know what to do with it.
Also, $f(x)=a(x^2-(2+b)x+2b)$
$f'(x)=a(2x-2-b)=0\implies x=\frac{2+b}2$ is the minima or maxima.
How to proceed next? Do the critical points of $f(x)$ tell us anything about $f(f(x))$?
| If $a > 0$, then there are two solutions to $a(x-2)(x-b) = 2$, and thus $f(f(x))$ has more than one zero. So we know $a < 0$.
Since $f(f(x))$ is a downward-opening quartic with only one zero, it takes on no positive values. Thus, the maximum of $f$ must be no larger than its smallest zero. Additionally, since $f(f(x))$ actually achieves the value $0$, that means that its maximum value is equal to its smaller zero. Since have $f(f(x_{max})) = 0$ and we want $f(f(5)) = 0$, we must have $x_{max} = 5$. It's not hard to see $x_{max} = 1 + b/2$, so we must have $b= 8$. Since that's larger than $2$, we must have $f(x_{max}) = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
On $\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x $ The Sine Integral, $\operatorname{Si}(x)$, is defined as $$\operatorname{Si}(x)=\int_0^x \frac{\sin t}{t} \ \mathrm{d}t $$
It follows that $$\int_0^{\pi} \operatorname{Si}(x) \ \mathrm{d}x = \bigg[x \operatorname{Si}(x) \bigg]_0^{\pi}-\int_0^{\pi} \sin x \ \mathrm{d}x = \pi \operatorname{Si} (\pi)-2$$ and further $$\int_0^{\pi} \operatorname{Si}^2 (x) \ \mathrm{d}x = \bigg[ x \operatorname{Si}^2(x) \bigg]_0^{\pi} -2\int_0^{\pi} \operatorname{Si} (x) \sin x \ \mathrm{d}x \\ = \pi \operatorname{Si}^2(\pi) -2 \left( \big[ -\cos x \operatorname{Si} (x) \big]_0^{\pi}+\int_0^{\pi} \frac{\sin x \cos x}{x} \ \mathrm{d}x \right)\\ = \pi \operatorname{Si}^2(\pi)-2 \operatorname{Si} (\pi)-\int_{2x=0}^{2\pi}\frac{\sin 2x}{2x} \ \mathrm{d}(2x)\\ = \pi\operatorname{Si}^2(\pi)-2 \operatorname{Si} (\pi)-\operatorname{Si} (2\pi)$$
Following a similar procedure on $\int_0^{\pi} \operatorname{Si}^3(x) \ \mathrm{d}x$ doesn’t quite work, because it requires evaluating the integral $\int_0^{\pi} \operatorname{Si}^2(x) \sin x \ \mathrm{d}x$, or equivalently, $\int_0^{\pi} \frac{\operatorname{Si}(x) \sin 2x}{x} \ \mathrm{d}x$.
This leads me to ask, can (and how can) $\int_0^{\pi} \operatorname{Si}^3(x) \ \mathrm{d}x$ be expressed in terms of elementary or special functions? How about the general case, i.e. $$\color{purple}{\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x} \; ? $$
| Here is an observation that seems to hint a negative answer to OP's question.
Considering that $\pi$ does not play a particular role in the special values for $\operatorname{Si}(\cdot)$, the question would be almost equivalent to investigating the antiderivatives of $\operatorname{Si}(x)^k$'s. Now let us write
$$F(x) = \int_{x}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t = \frac{\pi}{2} - \operatorname{Si}(x)$$
for the complementary version of $\operatorname{Si}(\cdot)$. In light of the binomial theorem, it is clear that $\int\operatorname{Si}(x)^n\,\mathrm{d}x$ can be computed if all of $\int F(x)^k\,\mathrm{d}x$ for $k=0,\dots,n$ can be computed. So we will shift our focus to finding the antiderivatives of $F(x)^k$'s.
Now employing OP's technique, we find that
\begin{align*}
\int F(x) \, \mathrm{d}x
&= x F(x) + \int \sin x \, \mathrm{d}x \\
&= x F(x) - \cos x + C
\end{align*}
and
\begin{align*}
\int F(x)^2 \, \mathrm{d}x
&= x F(x)^2 + 2\int F(x) \sin x \, \mathrm{d}x \\
&= x F(x)^2 - 2F(x)\cos x - \int \frac{2\sin x\cos x}{x} \, \mathrm{d}x \\
&= x F(x)^2 - 2F(x)\cos x + F(2x) + C.
\end{align*}
These computations, together with the asymptotic formula for $F(x)$, reveal the closed-forms for the following integrals:
$$ I_1 = \int_{0}^{\infty} F(x) \, \mathrm{d}x = 1 \qquad\text{and}\qquad I_2 = \int_{0}^{\infty} F(x)^2 \, \mathrm{d}x = \frac{\pi}{2}. $$
Similarly, if $\int F(x)^3 \, \mathrm{d}x$ admits a closed-form expression involving $F(\cdot)$, then we can expect that the integral
$$ I_3 = \int_{0}^{\infty} F(x)^3 \, \mathrm{d}x $$
will admit an elementary closed-form (or at least assume a simpler formula than $\int_{0}^{\pi}\operatorname{Si}(x)^3\,\mathrm{d}x$). However, we can prove that
$$I_3 = \frac{\pi^2}{4} - \frac{3}{2}\log^2 2 - \frac{3}{4}\operatorname{Li}_2\left(\frac{1}{4}\right) = 1.545982100082988\dots $$
holds, where $\operatorname{Li}_2(\cdot)$ is the dilogarithm. Since this expression seems non-elementary, I am skeptical about the ideal that $\operatorname{Si}(x)^3$ has an elementary antiderivative.
Addendum: Proof of the formula for $I_3$. We find that
\begin{align*}
\int F(x)^3 \, \mathrm{d}x
&= x F(x)^3 + 3\int F(x)^2 \sin x \, \mathrm{d}x \\
&= x F(x)^3 - 3F(x)^2\cos x - 6\int \frac{F(x)\sin x\cos x}{x} \, \mathrm{d}x \\
&= x F(x)^3 - 3F(x)^2\cos x - 3\int \frac{F(x)\sin(2x)}{x} \, \mathrm{d}x.
\end{align*}
From this, we get
$$ I_3 = \frac{3\pi^2}{4} - 3\int_{0}^{\infty} \frac{F(x)\sin(2x)}{x} \, \mathrm{d}x. $$
To compute this, we adopt the Feynman's trick. More precisely, we consider the parametrized integral
$$ I_3(s) = \frac{3\pi^2}{4} - 3\int_{0}^{\infty} \frac{F(x)\sin(sx)}{x} \, \mathrm{d}x. $$
Then
\begin{align*}
I_3'(s)
&= -3\int_{0}^{\infty} F(x)\cos(sx) \, \mathrm{d}x \\
&= -\frac{3}{s}\int_{0}^{\infty} \frac{\sin x \sin(sx)}{x} \, \mathrm{d}x \\
&= -\frac{3}{2s}\int_{0}^{\infty} \frac{\cos((s-1)x) - \cos((1+s)x)}{x} \, \mathrm{d}x
\end{align*}
The last integral can be computed using the Frullani's integral, yielding
\begin{align*}
I_3'(s)
&= -3\int_{0}^{\infty} F(x)\cos(sx) \, \mathrm{d}x \\
&= -\frac{3}{s}\int_{0}^{\infty} \frac{\sin x \sin(sx)}{x} \, \mathrm{d}x \\
&= \frac{3}{2s}(\log\left|s-1\right| - \log(s+1)).
\end{align*}
So it follows that
\begin{align*}
I_3 = I_3(2)
&= I_3(0) + \int_{0}^{2} I_3'(s) \, \mathrm{d}s \\
&= \frac{3\pi^2}{4}+ \int_{0}^{2} \frac{3}{2s}(\log\left|s-1\right| - \log(s+1)) \, \mathrm{d}s.
\end{align*}
By computing the last integral using the dilogarithm, we can complete the proof of the formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2}$ the problem of $k = 0$. Am I doing it right? I would like to calculate:
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
We have that:
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n \frac{\sqrt{2n^2+kn-k^2}}{n} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n \sqrt{2+\frac{k}{n}-(\frac{k}{n})^2} $$
Now, since we have: $\lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^n f(\frac{k}{n})$ I would like to use the fact that:
$$
\lim_{n \to \infty} \frac{b-a}{n} \sum^n_{k=1}f(a + k \frac{b-a}{n})= \int^b_a f(x) \ dx,
$$
But here I have $k = 0$ and not $k = 1$. So do I have to put $l = k+1 \to k = l-1$?
In that case we have:
$$\lim_{n \to \infty} \frac{1}{n}\sum_{l=1}^n \sqrt{2+\frac{l+1}{n}-(\frac{l+1}{n})^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{l=1}^n \sqrt{2 - \frac{l}{n}-\frac{l^2}{n}} $$
Now I can use the equation:
$$\lim_{n \to \infty} \frac{1}{n}\sum_{l=1}^n \sqrt{2 - \frac{l}{n}-\frac{l^2}{n}} = \int^1_0 \sqrt{2 - \frac{l}{n}-\frac{l^2}{n}} \ dx$$
Is that correct up to this point? Am I allowed to do change $k$ for $l$ like that?
| No.
$$\sum_{k=0}^n\frac{\sqrt{2n^2+kn-k^2}}{n^2}=$$
$$\frac{\sqrt{2}}{n}+\sum_{k=1}^n\frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
the first term goes to zero and the second to
$$\int_0^1\sqrt{2+x-x^2}dx=$$
$$\int_0^1\sqrt{\frac 94-(x-\frac 12)^2}dx$$
now put
$$x-\frac 12=\frac 32\sin(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
*
*A) $x^2-2x+2=0$
*B) $x^2-5x+5=0$
*C) $x^2-7x+7=0$
*D) $x^2-9x+9=0$
Method $1$:$$\sec^2\theta+\csc^2\theta=\frac1{\cos^2\theta}+\frac1{\sin^2\theta}\\=\frac1{\sin^2\theta\cos^2\theta}\\=\frac4{\sin^22\theta}\ge4$$
Also, $\sec^2\theta\csc^2\theta=\dfrac1{\sin^2\theta\cos^2\theta}$
So, options $B),C),D)$ are correct.
Method $2$: Let the quadratic equation be $x^2-px+q=0$
So, $\sec^2\theta+\csc^2\theta=p, \sec^2\theta\csc^2\theta=q\implies \csc^2\theta=\dfrac{q}{\sec^2\theta}$
Putting that in the sum of roots, we get $$\sec^2\theta+\frac{q}{\sec^2\theta}=p\\\implies\sec^4\theta-p\sec^2\theta+q=0\\\implies\sec^2\theta=\frac{p\pm\sqrt{p^2-4q}}2\ge1\\\implies p\pm\sqrt{p^2-4q}\ge2\\\implies\pm\sqrt{p^2-4q}\ge2-p\\\implies p^2-4q\ge4+p^2-4p\\\implies p-q\ge1$$
What's wrong in this method?
| It seems that what we are asked to notice is that
$$ \sec^2 x \ + \ \csc^2 x \ \ = \ \ \frac{1}{\cos^2 x} \ + \ \frac{1}{\sin^2 x} \ \ = \ \ \frac{\sin^2 x \ + \ \cos^2 x}{\cos^2 x·\sin^2 x} \ \ = \ \ \sec^2 x · \csc^2 x \ \ , $$
a perhaps not overly-familiar trigonometry identity. This is why all of the polynomials in the question choices have the form $ \ x^2 \ - \ m·x \ + \ m \ \ . $ Since we are apparently to assume that $ \ \theta \ $ is real, we must have of course $ \ \sec^2 \theta \ge 1 \ , \ \csc^2 \theta \ge 1 \ \ . $ But for the equation $ \ x^2 - m x + m \ = \ 0 \ \ , $ this is not possible unless the quadratic polynomial is at least a "binomial-square" $ \ (x - p)^2 \ = \ x^2 - 2px + p^2 \ \ , $ which requires $ \ p^2 \ = \ 2p \ \Rightarrow \ p \ = \ 2 \ \Rightarrow \ m \ = \ 4 \ \ . $ Since choice $ \ \mathbf{(A)} \ $ thus does not even have real roots, it may be eliminated. The balance of the choices indicate that $ \ \sec^2 x \ + \ \csc^2 x \ = \ \sec^2 x · \csc^2 x \ = \ m \ > \ 4 \ \ , $ they all seem to be acceptable.
We ought to check that the other given values for $ \ m \ $ can be attained. Writing $$ \sec^2 \theta \ + \ \csc^2 \theta \ \ = \ \ \frac{1}{\cos^2 \theta} \ + \ \frac{1}{(1 - \cos^2 \theta)} \ \ = \ \ \frac{1}{\cos^2 \theta · (1 - \cos^2 \theta)} \ \ = \ \ m $$
$$ \Rightarrow \ \ \cos^4 \theta \ - \ \cos^2 \theta \ + \ \frac{1}{m} \ \ = \ \ 0 \ \ , $$
we have the (presumed) solutions $ \ \cos^2 \theta \ = \ \frac12 \ \pm \ \frac{\sqrt{1 - \frac{4}{m}}}{2} \ \ . $ The solutions are real for $ \ m \ \ge \ 4 \ \ $ and, since the term $ \ \frac{\sqrt{1 - \frac{4}{m}}}{2} \ $ only asymptotically approaches $ \ \frac12 \ $ as $ \ m \ \rightarrow \ \infty \ \ , $ we have valid results for $ \ \cos^2 \theta \ \ . $ All of the other equation choices have $ \ m \ > \ 4 \ \ , $ so they all have admissible real roots with $ \ \sec^2 \theta \ge 1 \ , \ \csc^2 \theta \ge 1 \ \ . $
The problem in your second method then for the choice $ \ \sec^2\theta + \csc^2\theta \ = \ p \ , \ \sec^2\theta·\csc^2\theta \ = \ q \ $ is that $ \ p = q \ \ , $ which implies that we also have
$$ \sec^2 \theta \ + \ \frac{\mathbf{p}}{\sec^2\theta} \ \ = \ \ p \ \ \Rightarrow \ \ \sec^2 \theta \ \ \overbrace{=}^{?!} \ \ p · (1 \ - \ \cos^2 \theta) \ \ , $$ which is certainly non-negative, but not always greater than or equal to $ \ 1 \ \ . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the solution of equation $z^2 - 2 \bar{z} = 3$? my question is how to solve the equation $z^2 - 2 \bar{z} = 3$, where $z \in \mathbb{C}$?
I know that polar form of any complex number is $z = r \cos(\varphi) + i \cdot r \sin (\varphi)$. And similarly $\bar{z} = r \cos(\varphi) - i \cdot r \sin(\varphi)$. But now, when I plug in into equation, I'm getting a little confused.
So, $(r \cos(\varphi) + i \cdot r \sin(\varphi))^{2} - 2(r \cos(\varphi) - i \cdot r \sin(\varphi)) = 3$, but now I don't know how to solve this equation.
| In a problem like this, I prefer Cartesian coordinates, where $z = x + iy.$
$z^2 - 2 \bar{z} = 3$
Therefore $(x + iy)^2 - 2(x - iy) = 3 \implies $
$(x^2 - y^2 - 2x - 3) + i(2xy + 2y) = 0 + i(0)$.
Therefore, $(2y)(x + 1) = 0$.
Therefore, either $y = 0$ or $x = -1$.
$y = 0 \implies (x^2 - 2x - 3) = 0 \implies (x - 3)(x + 1) = 0 \implies x \in \{3, -1\}.$
$x = -1 \implies 3 - y^2 - 3 = 0 \implies y = 0.$
Therefore, the possible answers are $(-1 + i[0], 3 + i[0]).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $a$ and $b$ such that $x^3-4ax^2-8bx-3+a+b$ has a triple root $$P(x)=x^3-4ax^2-8bx-3+a+b$$
NOTE: a & b are both real and the solution is not necessarily unique
So far i've done this:
If $P(x)$ is a $3$rd degree polynomial and has a triple root, then it has the form:
$$
P(x)=(x-n)^3=x^3-3x^2 n+3xn^2-n^3.
$$
Then
$$
x^3-3x^2 n+3xn^2-n^3=x^3-4ax^2-8bx+(-3+a+b).
$$
And from this, I got three equations:
$$-3n=-4a$$
$$3n^2=-8b$$
$$-n^3=(-3+a+b)$$
I don't know what to do from here. Is my reasoning correct?
| Let $(x−k)^3=x3−4ax^2+(−8b)x−(3−a−b)$
$⟹x^3−(3k)x^2+(3k^2)x−k^3=x^3−4ax^2+(−8b)x−(3−a−b)$
$⟹x^3−(3k)x^2+(3k^2)x−k^3=x^3−4ax^2+(−8b)x−(3−a−b)$
Comparing like terms
$3k=4a$
$⟹a=3k/4 $
$3k^2=−8b$
$⟹b=−3k^2/8$
$k^3=3−a−b$
$⟹k^3=3−3k/4+3k^2/8$
$⟹8k^3−3k^2+6k−24=0 $
Solving this gives $k≈1.38935$
$a≈1.04201$
$b≈−0.723859$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How do I prove this inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$? If $a,b,c\gt 0$ and $abc=1$, how do I prove the following inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$?
My version:
\begin{gathered}
\sqrt{\frac{a^{2}+1}{a+1}} \leq \sqrt{1+\frac{a^{2}-a}{a+1}} \leq 1+\frac{1}{2} \frac{a^{2}-a}{a+1}=\frac{a}{2}+\frac{1}{a+1} \\
\Leftrightarrow 4\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \leq 3+(a+b+c), a b c=1 \\
a+b+c=3 t^{2}=p \geq 3, a b+b c+c a=q \geq \sqrt{3 p r}, a b c=r=1 \\
4\left(1+\frac{1+p}{2+p+q}\right) \leq 4\left(1+\frac{1+p}{2+p+\sqrt{3 p}}\right) \leq 3+p \\
(t-1)\left(3 t^{3}+6 t^{2}+3 t+2\right) \geq 0
\end{gathered}
| Remarks: Here is a trick for the inequality
of the form $f(a) + f(b) + f(c) \ge 0$ under the constraints $a, b, c > 0$ and $abc = 1$.
Let $F(x) = f(x) + m \ln x$. If we can find
an appropriate $m$ such that $F(x)\ge 0$ for all $x > 0$, then we have $F(a) + F(b) + F(c)
= f(a) + f(b) + f(c) + m\ln (abc) \ge 0$
and we are done.
We apply the trick for our problem.
Let $F(x) = x + 1 - \frac{4}{1 + x} - 2\ln x$.
We have $F'(x) = \frac{(x - 1)(x^2 + x + 2)}{x(x + 1)^2}$.
Thus, $F'(1) = 0$, and $F'(x) < 0$ on $(0, 1)$, and $F'(x) > 0$ on $(1, \infty)$.
Thus, $F(x) \ge F(1) = 0$ for all $x > 0$.
Thus, we have $F(a) + F(b) + F(c) \ge 0$
which results in $4\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \leq 3+(a+b+c)$.
We are done.
How to determine the coefficient $m$:
Let $F(x) = x + 1 - \frac{4}{1 + x} + m\ln x$.
We have $F'(x) = 1 + \frac{4}{(1 + x)^2} + \frac{m}{x}$. Let $F'(1) = 0$ and we have $m = -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $a_n=\sqrt{1+\sqrt{2+\cdots\sqrt{n}}}$ and $\lim\limits_{n\to\infty}a_n=\ell$, prove $\lim_{n\to\infty}[(\ell-a_n)^{1/n}n^{1/2}]=\frac{\sqrt e}2$ $$a_n=\sqrt{1+\sqrt{2+\cdots\sqrt{n}}}$$
We can prove that $\{a_n\}$ is convergent
(using mathematical induction, $\sqrt {k+\sqrt{k+1+\cdots\sqrt{n}}}\leq k-1, for \ k\geq3$).
If
$$
\lim\limits_{n\to\infty} a_n=\ell,
$$
prove:
$$\lim\limits_{n\to\infty} \left[\,(\ell-a_n)^{1/n}\cdot n^{1/2}\,\right]=\frac{\sqrt e}{2}$$
| Let's denote for $x_i\ge 0$:
$$ [x_1, x_2,\cdots ,x_n] = \sqrt{x_1 + \sqrt{x_2 + \cdots \sqrt{x_n}}}$$
Let
$$c_{k,n} = [k, k+1, \cdots, n] \text{, for } n \ge k$$
from your inequality proved by induction we deduce:
$$c_{k,n} = [k + c_{k+1,n}] \le \sqrt{2k} \text{, for } n \ge k \ge 3$$
And injecting this enquality again, we deduce
$$\sqrt{k} \le c_{k,n} \le \sqrt{k + \sqrt{2(k+1)} }$$
We have in particular $\ell = c_{1,\infty}$
From
$$c_{k,\infty} - c_{k,n} = \frac{c_{k+1, \infty} - c{k+1, n}}{c_{k,\infty} + c_{k,n}}$$
we deduce
$$\ell -a_n = \frac{c_{n+1,\infty}}{\prod_{k=1}^{n}\left(c_{k,\infty} + c_{k, n}\right)}$$
Let be $\varepsilon > 0$, there exists $N_0$ large enough, such that
$$\sqrt{k} < \sqrt{k+\sqrt{2(k+1)}} < (1 + \varepsilon)\sqrt{k}$$
Let be $K_{N_0} = \frac{\sqrt{N_0!}}{\prod_{k=1}^{N_0}c_{k,\infty}}$
So for all $n\ge N_0$, we have
$$
K_{N_0} \frac{ \sqrt{n+1} }{ (1+\varepsilon)^{n - N_0} 2^n \sqrt{n!} } \le K_{N_0}\frac{ c_{n+1,\infty} }{ (1+\varepsilon)^{n - N_0} 2^n \sqrt{n!}} \le
\ell - a_n \le
\frac{ c_{n+1,\infty} }{ 2^n \sqrt{n!} } \le
\frac{ \sqrt{2(n+1)} }{ 2^n\sqrt{n!} }$$
Hence by taking the nth square root, we deduce
$$ K_{N_0}^{\frac1{n}} \frac{\left(n+1\right)^{\frac1{2n}}}{2\left(1+\varepsilon\right)^{1 - n/N_0}\left(n!\right)^{\frac1{2n}}} \le
\left(\ell - a_n\right)^{\frac1{n}} \le
\frac{(2(n+1))^{\frac1{2n}}}{2\left(n!\right)^{\frac1{2n}}}$$
We have then by Sterling formula
There exists $N_1 \ge N_0$ s.t. for all $n \ge N_1$, we have
$$ (1 - 2\varepsilon)\frac{\sqrt{e}}{2} \le \sqrt{n}\left(\ell - a_n\right)^{\frac1{n}} \le (1 + 2\varepsilon)\frac{\sqrt{e}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}, a \neq b \neq c$; What is $a^3+b^3+c^3$?
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}\\a \neq b \neq c\\a^3+b^3+c^3=$
Since $\frac{1-a^3}{a}=\frac{1-b^3}{b}$, $a^3=b^2a-\frac{a}{b}-1$.
$\therefore a^3+b^3+c^3=ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1$
Since $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$,
$ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
(stuck here)
| $a, b, c$ are roots of $$\frac{1-x^3}{x}=m$$
$$x^3+mx-1=0$$
Hence, we know that $a+b+c=0$ and $abc=1$ from Vieta's formula. Now you can use the identity that you stated to solve for $a^3+b^3+c^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Finding $a$ and $b$ such that the graph of $y=-2\sin3x+1$ passes through $(-\frac\pi{12},a)$ and $(b,1)$ Please look at the attached image through the link. I already figured out $b$, which I'm pretty sure is $2\pi/3$. However, I'm confused on figuring out $a$. This is what I tried:
\begin{align*}
y & = -2\sin(3x)+1\\
y & = -2\sin\left(3 \cdot -\frac{\pi}{12}\right)+1\\
y & = -2\sin\left(-\frac{\pi}{4}\right)+1
\end{align*}
I used the unit circle to find $\sin\left(-\frac{\pi}{4}\right)$.
\begin{align*}
y & = -2\left(-\frac{\sqrt{2}}{2}\right)+1\\
y & = \sqrt{2} + 1\\
y & = 2.41421356
\end{align*}
My answer isn't correct, so could anyone please explain how I should solve for $b$? Thanks!
| You have the correct value for $a$. Perhaps it was marked incorrect if you entered your approximation rather than the exact value of $\sqrt{2}+1$.
For $b$, you can simply let $y=1$ and solve:
$$y=-2\sin(3x)+1$$
$$1=-2\sin(3b)+1$$
$$-2\sin(3b)=0$$
$$\sin(3b)=0$$
In your diagram, it looks like $\frac{\pi}{4}<b<\frac{\pi}{2}$. Use your unit circle to determine which values of $3b$ produce $\sin(3b)=0$, and find a value of $b$ that satisfies both $\sin (3b) = 0$ and $\frac{\pi}{4}<b<\frac{\pi}{2}$. Your answer of $b=\frac{2\pi}{3}$ satisfies the first criterion, but not the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing $\int_{0}^{2\pi}\cos(x)\cos(2x)\cos(3x)\,dx = \frac\pi2$ An integral from MIT Integration Bee:
Show that
$$I = \int_{0}^{2\pi}\cos(x)\cos(2x)
\cos(3x)\,dx = \frac\pi2$$
This integral appeared in the 2019 paper. Below is my own solution:
$$\begin{align}
I &= \int_{0}^{2\pi}\cos(x)\cos(2x)(\cos x\cos2x-\sin x\sin2x)\,dx \\[6pt]
&= \int_{0}^{2\pi}\cos^2(x)\cos^2(2x) \,dx -\int_{0}^{2\pi}\cos(x)\cos(2x)\sin(x)\sin(2x)\, dx
\end{align}$$
Replacing $\cos^2(x)= \frac{1+\cos(2x)}{2}$ for the first integral and $\sin(x)\cos(x)= \sin(2x)/2 $ for the second, we get
$$\begin{align}
&\int_{0}^{2\pi}\frac{1+\cos(2x)}{2}\cos^2(2x) dx-\frac{1}{2}\int_{0}^{2\pi}\cos(2x)\sin^2(2x) dx \\[6pt]
=\; &\frac{1}{2}\left(\int_{0}^{2\pi}\cos^2(2x)dx \, + \int_{0}^{2\pi}\cos(2x)\cos(4x)dx \right) \\[6pt]
=\; &\frac{1}{2} \left( \pi + 0\right) \qquad \text{$\because$ the orthogonality of $\cos(mx)$} \\[6pt]
=\; &\frac\pi2
\end{align}$$
This solution is rather awkward, and I'm sure there's a better and faster approach to this integral. Could anyone provide a more elegant solution(or a sketch of it)? Thanks.
| Answer inspired by Aaron's answer.
Required formulas from handbook:
$\cos(2x) = 2\cos^2(x) - 1.$
$\cos(3x) = 4\cos^3(x) - 3\cos(x).$
$\cos^2(x) = \frac{1}{2}[\cos(2x) + 1].$
$\cos^4(x) = \frac{1}{8}[\cos(4x) + 4\cos(2x) + 3).$
$\cos^6(x) = \frac{1}{32}[\cos(6x) + 6\cos(4x) + 15\cos(2x) + 10].$
The first thing to notice is that when integrating $\cos^2(x), \cos^4(x),$ or $\cos^6(x)$, the indefinite integral will look like $A\sin(6x) + B\sin(4x) + C\sin(2x) + D.$
Since the integral is being taken from $(0)$ to $(2\pi)$, all but the constant term can be ignored. That is, for example, at $(x = 2\pi), \sin(6x) = 0.$
Therefore
$\cos(x) \cos(2x) \cos(3x) =
[2\cos^3(x) - \cos(x)] [4\cos^3(x) - 3\cos(x)]$
$= 8\cos^6(x) - 10\cos^4(x) + 3\cos^2(x).$
Therefore
$\displaystyle I = \int_0^{2\pi} \left[\frac{8}{32}(10)
- \frac{10}{8}(3)
+ \frac{3}{2}(1)\right] dx.
$
This equals
$\displaystyle (2\pi) \times \left[\frac{10}{4} - \frac{15}{4} + \frac{6}{4}\right] = (2\pi) \times \frac{1}{4} = \frac{\pi}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
} |
Given $\triangle ABC$ with $m\angle A = 90^{\circ}$. $D$ is midpoint of $BC$ and $F$ is midpoint of $AB$. Given $\triangle ABC$ with $\angle A = 90^{\circ}$. $D$ is midpoint of $BC$ and $F$ is midpoint of $AB$. Points $E$ and $G$ lie on the line $AB$ such that $AE:EF = 1:1$ and $FG:GB = 1:1$ as shown on the picture. What is the ratio of $PQ:PR$?
What I have done so far.
Using menelaus theorem, I get
$\frac{AE}{EF} \cdot \frac{FC}{CQ} \cdot \frac{QP}{PA} = 1$ or $\frac{PQ}{PA} = \frac{2}{3}$
and $\frac{AE}{EG} \cdot \frac{GC}{CR} \cdot \frac{PR}{PA} = 1$ or $\frac{GC}{CR} \cdot \frac{PR}{PA} = \frac{2}{1}$
I don't have any clue to find $\frac{GC}{CR}$ in order to find $\frac{PQ}{PR}$ after that.
| Construct $GX\parallel AD$ ($X\in BC$). We can now use the intercept theorem and see that $AG:GB=DX:XB=4:1$. Let $XB=x$ and so $DX=4x$. We therefore have $DX+XB=5x=DB=CD$ and from here we observe $CD:DX=5:4$. However, $CD:DX=CR:RG$ by the intercept theorem and thus $CG:CR=9:5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove or disprove $\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$ let $x,y>0$.prove or disprove
$$\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$$
By this inequality it seem Cauchy-Schwarz inequality
$$(1+x^2)(1+y^2)\ge (1+xy)^2$$
But this is Log function.so How to prove it? Thanks
| It seems to me that the inequality is not correct.
Putting $y=\frac 1x$ we get,
$$\ln{\left(1+x^2\right)}\times\ln{\left(1+\frac {1}{x^2}\right)}\ge \ln^2{2}$$
But,
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$
Evaluation of the limit:
Let, $1+x^2=e^u$. This implies, $1+\frac{1}{x^2}=1+\frac{1}{e^u-1}.$
Then, we have
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right)
&=\lim_{u\to\infty}\left(u\times \ln {\left(1+\frac {1}{e^u-1}\right)}\right)\\
&=\lim_{u\to\infty }\frac{\ln {\left(1+\frac {1}{e^u-1}\right)}}{\frac 1u}\\
&=\lim_{u\to\infty}\frac{-\frac{1}{e^u-1}}{-\frac {1}{u^2}}\\
&=\lim_{u\to\infty}\frac{u^2}{e^u-1}\\
&=2\lim_{u\to\infty}\frac{u}{e^u}\\
&=0.\end{align}$$
But, we can calculate the limit faster.
Using the inequality,
$$\ln (1+u)<u, ~u>0$$
We have,
$$\begin{align}0&\le\ln\left(1+x^2\right)\ln\left(1+\frac{1}{x^2}\right)\\
&\le\ln\left(\left(1+x\right)^2\right)\ln\left(1+\frac{1}{x^2}\right)\\
&=2\ln\left(1+x\right)\ln\left(1+\frac{1}{x^2}\right)\\
&\le \frac{2x}{x^2}\\
&=\frac{2}{x}\end{align}$$
and $$\lim_{x\to \infty}\frac 2x=0.$$
Then, the Squeeze theorem tells us that,
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac{1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$
This effective solution using the Squeeze theorem was made by @BarryCipra.
Since I found this method nicer and more useful, I added this to my answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\
x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\
x+1&>&-3x+6\\
4x&>&5\\
x&>&\frac{5}{4}
\end{eqnarray}
$$
which cannot be true, as $x$ can equal $2$ with these restrictions which is obviously not allowed. Doing the left side the correct way:
$$
\begin{eqnarray}
\frac{x+1}{x-2} &>& -3\\
\frac{x+1}{x-2} + 3 &>& 0\\
\frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\
\frac{4x-5}{x-2} &>& 0\\
\Rightarrow x < \frac{5}{4}\lor x>2\\
\end{eqnarray}
$$
leads to the solution set $(-\infty,\frac{5}{4})\cup(2,\infty)$. My question is, what happened with the first method where I failed to come up with a solution for $(2,\infty)$, (and why the signage for $x>\frac{5}{4}$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.
| In the first method, when you go from $\dfrac{x+1}{x-2}\gt -3$ to $x+1\gt -3(x-2)$, what you're doing is multiplying both sides of the inequality by $x-2$ and you assume that $x-2\gt 0$ when doing that because otherwise the inequality sign flips to $\le$
So, assuming $x\gt 2$, you get from that inequality that $x\gt 5/4$ and the second inequality $\dfrac{x+1}{x-2}\lt 3$ gives $x\gt 7/2$
So, assuming $x\gt 2$, you get $x\gt 5/4$ and $x\gt 7/2$; together they imply $x\gt\max\{2,5/4,7/2\}=7/2$
Similarly, assume $x-2\lt 0$, ie, $x\lt 2$ and solve $x+1\lt -3(x-2)$ and $x+1\gt 3(x-2)$ simultaneously to obtain the other solution set. You get $x\lt\min\{2,5/4,7/2\}=5/4$
The complete solution set is thus $(-\infty,5/4)\cup (7/2,\infty)$
The answer you arrived at is wrong. Note that $x=3\in (2,\infty)$ doesn't satisfy the original equation.
In the second method, you're only solving one of the two inequalities that are supposed to hold. If you solve the second inequality $\dfrac{x+1}{x-2}\lt 3$ in a similar way, you get $$\frac{x+1}{x-2}-3\lt 0\iff\frac{7-2x}{x-2}\lt 0\iff x\in (-\infty, 2)\cup (7/2,\infty)$$
and the solution set to your original problem is the intersection of this with $(-\infty,5/4)\cup (2,\infty)$ so that
$$[(-\infty,2)\cup (7/2,\infty)]\cap[(-\infty,5/4)\cup (2,\infty)]=(-\infty,5/4)\cup (7/2,\infty)$$
which is the solution set to the original problem $\left|\dfrac{x+1}{x-2}\right|\lt 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit of sequence with greatest integer function Calculate $\lim \limits_{x \to 0} \ x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big)$, where [.] represents greatest integer function.
Since we know that $x-1\lt [x]\leq x,$ I tried squeezing the sequence between $\ x^2\Big(1+2+3+....+\ \frac{1}{\vert x \vert} -1\Big)$ and $\ x^2\Big(1+2+3+....+\ \frac{1}{\vert x \vert}\Big)$ which gives me the limit as $0$. But the correct answer is given as $\frac{1}{2}$ .
| The sum of the first $n$ integers is $\frac {n(n+1)}2$. Thus, with $n=\Big[\frac{1}{\vert x \vert}\Big]$,
$$x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big) = x^2\frac {n(n+1)}2$$
Since, by definition of $n$, we have $n\leq \frac 1 x < n+1$
$$ x^2\frac{\left(\frac 1 x - 1\right) \frac 1 x}2< x^2\Big(1+2+3+....+\Big[\frac{1}{\vert x \vert}\Big]\Big) \leq x^2\frac{\frac 1 x \left(\frac 1 x +1\right)}2$$
Taking the limit (squeezing) gives that the limit is $\frac 1 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the maximum value of $x\cos^{-1}\left(x\right)$? I wanted to find the maximum value of $x\cos^{-1}\left(x\right)$ through differentiation, but upon on differentiation I get $$f'(x) = \arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}}$$ to find the maximum I equated the function to zero: $$\arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}} = 0$$
But I am unable to find the roots of this equation. Could someone show how to find the zeros for this equation?
| \begin{align*}
f(x) &=xcos^{-1}(x) \\
f'(x) &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
0 &=cos^{-1}(x) - \frac{x}{\sqrt{1-x^{2}}} \\
\frac{x}{\sqrt{1-x^{2}}} &=cos^{-1}(x) \\
cos(\frac{x}{\sqrt{1-x^{2}}}) &= x \\
\end{align*}
Now substitute $x = cos\alpha$:
\begin{align*}
cos(\frac{cos\alpha}{\sqrt{1-(cos\alpha)^{2}}}) &= cos\alpha \\
cos(\frac{cos\alpha}{sin\alpha}) &= cos\alpha \\
\frac{cos\alpha}{sin\alpha} &= \alpha \\
cot\alpha &= \alpha \\
\end{align*}
I will admit that I do not know how to analytically find the $\alpha$ that makes this true (I looked online and the only answers I found were using Newtons method or other ways of approximating the answer). Either way, the cosine of this $\alpha$ is the maximum point of $f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$
I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}-\frac{(7x)^6}{6!}+\dots\}}{3x^2}\\ =\lim\limits_{x\to 0}\frac{x^2(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1}{3}(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)\\ =\frac {49}{6}$
Can this be solved without using the Taylor series?
| Multiplying top and bottom by $1 + \cos 7x$:
$$ \lim_{x\to 0}\frac{1-\cos 7x}{3x^2} = \lim_{x\to 0}\frac{(\sin 7x)^2}{3x^2} \frac{1}{1 + \cos 7x} = \lim_{x\to 0}\frac{(\sin 7x)^2}{(7x)^2} \frac{49/3}{1 + \cos 7x} = 1 \cdot \frac{49/3}{2} = \boxed{\frac{49}{6}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
$
| The sides of a right triangle with integer sides are of the form
$a(m^2-n^2), 2amn, a(m^2+n^2)$
for integer $a, m, n$
with $m > n$.
We therefore want
$p
=2a(m^2+mn)
=2am(m+n)$
where $p=30$ here,
or
$p/2
=am(m+n)$.
For p=30,
possible values for $a$ are
1, 3, 5, 15.
For these,
$m(m+n)
=15, 5, 3, 1
$.
If
$m(m+n)=15$
with $m > n$,
$m=3, n=2,$
sides $=5, 12, 13$.
If
$m(m+n)=5, 3, 1$,
no solutions.
Here are bounds on
$m$ and $n$
with $m > n$
and
$m(m+n)=r$
in terms of $r$.
If
$m(m+n)=r$,
$n
=\dfrac{r}{m}-m
$
so,
since $m > n$,
$\dfrac{r}{m}-m
\lt m$
or
$r < 2m^2$
or
$m > \sqrt{r/2}
$
and $m | r$.
Also,
if
$m(m+n)=r$,
$m^2+mn-r=0$,
$m
=\dfrac{-n+\sqrt{n^2+4r}}{2}
$.
Since $m > n$,
$-n+\sqrt{n^2+4r}
\gt 2n$
or
$n^2+4r > 4n^2
$
or
$n < \sqrt{4r/3}
$.
Similarly,
$r \ge m(m+1)$
or
$4r+1 \ge 4m(m+1)+1
=(2m+1)^2
$
so
$m \le \sqrt{4r+1}/2
=\sqrt{r+\frac14}
\lt \sqrt{r}+\frac12
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calc $P=\int_C \frac{e^{-x^2y}}{\sqrt{( 1+x^2y^4 )^3}}( xy[ 2( 1+x^2y^4 )-y^3 ]dx+x^2[ e^{-x^2y}\sqrt{(1+x^2y^4)^3}+1+x^2y^4-2y^3]dy)$ Calc the integral:
$$P=\int \limits_C \dfrac{e^{-x^2y}}{\sqrt{\left( 1+x^2y^4 \right)^3}}\Bigg( xy\left[
2\left( 1+x^2y^4 \right)-y^3 \right]dx+x^2\left[ e^{-x^2y}\sqrt{\left(1+x^2y^4\right)^3}+1+x^2y^4-2y^3\right]dy\Bigg),$$
C is half circle $x^2+y^2=1, y\ge0$, direction from $A(1,0)$ to $B(-1,0).$
If set $x=\cos t$ and $y=\sin t$ $\left( \text{or} \,y=\sqrt{1-x^2} \right)$, can calc P by single integral. But, it's complex and may be not calc.
I also thought about Green's theorem. But it's easy to go wrong when calc derivative.
I wanna find a beautiful solutions. I hope some hints from you. Thanks.
| If you parametrize the half circle as,
$r(t) = (t, \sqrt{1-t^2}), t \in (1,-1)$
$r'(t) = \left(1, - \frac{t}{\sqrt{1-t^2}}\right)$.
So the line integral is, $ \ P = \displaystyle - \int_{-1}^1 \vec F(r(t)) \cdot r'(t) \ dt$
$\vec F(r(t)) \cdot r'(t) = t \ g(t^2)$
The integrand is an odd function and its integral over $t \in (-1, 1)$ will be zero due to symmetry.
where, $ \ g(t^2) = \dfrac{e^{-t^2 \sqrt{1-t^2}}}{\sqrt{\left(1+t^2(1-t^2)^2 \right)^3}} \cdot $
$\Bigg( \sqrt{1-t^2}\left[
2\left( 1+ t^2 (1-t^2)^2 \right) - (1-t^2)^{3/2} \right]$
$- \frac {t^2}{\sqrt{1-t^2}}\left[e^{-2t^2 \sqrt{1-t^2}}\sqrt{\left(1+t^2(1-t^2)^2\right)^3}+1 + t^2 (1-t^2)^2 - 2 (1-t^2)^{3/2}\right]\Bigg)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
| A more general inequality:
$$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 3(1+r)}{a^2+b^2+c^2+r(a+b+c)}$$
for $0\le r\le\frac{5}{2}$ is proved in Vasile Cirtoaje's "Algebraic Inequalities - Old and New Methods" using Equal-Value Theorem (Section 5.2, Problem 8, page 217).
Although for this case ($r=2$) Vo Quoc Ba Can's solution is much neater.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- Nothing special really came in my mind. I did not find a way to use Triangle Inequality. What I did was, by AM-GM :-
$$a^2 + b^2 + c^2 \geq \frac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg) > \frac{72}{35}(\sqrt{sabc}).$$
But I couldn't proceed from this.
Another Idea I had was :- $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$.
$\rightarrow a^2 + b^2 + c^2 = 4s^2 - 2(ab + bc + ca).$
But I did not know how to use this here, and would make the calculations a bit messy, especially of the $\dfrac{36}{35}$ part present there.
Can Anyone Help me? Thank You.
| After expanding, the expression is equivalent to $$13(a^3+b^3+c^3)+4(a^2b+b^2c+c^2a)+4(ab^2+bc^2+ca^2)-63abc\geq 0$$
$$\Leftrightarrow 13[a^3+b^3+c^3-3abc]+4[(a^2b+b^2c+c^2a)-3abc]+4[(ab^2+bc^2+ca^2)-3abc]\geq 0,$$ which is clear, as follows from AM-GM.
[EDIT] With more thoughts and less work, it suffices by AM-GM to prove a stronger result: $$a^2+b^2+c^2\geq \frac{36}{35}\left(s^2+\frac{(2s/3)^3}s\right),\qquad (1)$$ where one replaces $abc$ by $((a+b+c)/3)^3=(2s/3)^3\geq abc.$
Clearly (1) is equivalent to $$3(a^2+b^2+c^2)\geq (a+b+c)^2,$$ which is true by C-S.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$ Question: Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
My Attempt:
On rationalizing both fractions separately and then adding (since denominator becomes $\sqrt{3}$) I got $$-2\sqrt{6}+2\sqrt{2+\sqrt{3}}+2\sqrt{2-\sqrt{3}}$$
However the given answer is $$\frac{\sqrt{6}}{3}$$
Even more confusing is that on inputting the problem into wolframalpha the solution is given as $$\sqrt{2}$$
I have broken my head over this for a couple of hours and I just can't find a solution. Hope someone can help.
| You are right, the correct result for the sum of the two given ratios is $\sqrt{2}$. It is $\frac{\sqrt{6}}{3}$ when you take the difference. Presumably there is a typo in your book.
Revise your work and notice that
$$\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2
=2+\sqrt{3}+2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}_1+2-\sqrt{3}=6,$$
and
$$\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2
=2+\sqrt{3}-2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}_1+2-\sqrt{3}=2.$$
P.S. As a complement you may take a look at nested radicals. The above computations imply that
$$2\sqrt{2\pm\sqrt{3}}=\sqrt{6}\pm\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is my method correct to answering this question? Is there a quicker method to solve this question?
What is the smallest number that can be written as the sum of three, four and five consecutive numbers?
I encountered this question while doing my Math summer homework. I have tried to make progress on this question.
Sum of three consecutive numbers = $x + x+1 + x+2 = 3x+3$
Sum of four consecutive integers = $x + x+1 + x+2 + x+3 = 4x+6$
Sum of five consecutive integers = $x + x+1 + x+2 + x+3 + x+4 = 5x+10$
The number must be the lowest common multiple of $3x+3$ , $4x+6$ and $5x+10$, which is $60x + 30$.
Substituting $x = 0$ gives us the lowest positive, non-zero and whole number, which is $30$.
$$30 = 9 + 10 + 11\\
30 = 6 + 7 + 8 + 9\\
30 = 4 + 5 + 6 + 7 + 8$$
Is my answer correct? If not, where in my method have I produced an error? Is there an ever quicker method to solve this question?
| If we are talking about a non-negative integer, I have another method to propose.
Sum of three consecutive numbers = $++1++2=3+3$
Sum of four consecutive integers = $++1++2++3=4+6$
Sum of five consecutive integers = $++1++2++3++4=5+10$
Another way of formulating a solution from this step can be as follows:
$y = 3x_1 + 3 = 4x_2+6 = 5x_3+10$
$y= 3x_1+3\equiv 0 \pmod{3}$
$y= 4x_2+6\equiv 2 \pmod{4}$
$y= 5x_3+10\equiv 0 \pmod{5}$
$y$ is divisible by $5$ and $3$ which gives us $y=15k$
The remainder left from dividing $y$ by $4$ is $2$. $y=15$ doesn't satisfy this condition, but $y=30$ does.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqrt2}{2})$
My progress:
Point P is on the circumcircle of ABC because the angle $\measuredangle ABP = 135^o$
$ Where~ AC = 2R\\
\triangle CBP \rightarrow
PC^2 = BP^2 + BC^2 - \sqrt2BCBP\\
but~PC = PA = R\sqrt{2} \text{(since P is in the bisector AC)}\\
2R^2 = BP^2 +BC^2 - \sqrt{2}BCBP\\
0= BP^2 - \sqrt{2}BCBP + BC^2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2}2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2-4R^2}2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2\\
$
I can't find the relationship between BC, AB and a...
If anyone finds another way to solve by geometry I would be grateful
| $BP \parallel AF \implies ∠PBF= ∠BFA=45° \text{ (alternate interior angles)}$
$ \implies ∠BAF=∠BFA \text{ (sum of angles in triangle)} $
$\implies AB=FB \text{ (sides opposite to equal angles in a triangle)}$
$\therefore BC= CF+FB=a+AB \implies AB=BC-a$
Using this, in the equation you got
$(BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2 =0 \implies (BP - \frac1{ \sqrt{2}}BC)^2 - (\frac{BC-a}{\sqrt{2}})^2=0$
$\implies \{(BP - \frac1{ \sqrt{2}}BC)+(\frac{BC-a}{\sqrt{2}})\}\{(BP - \frac1{ \sqrt{2}}BC)-(\frac{BC-a}{\sqrt{2}})\}=0$
$\implies (BP - \frac{a}{ \sqrt{2}})(BP - \frac2{ \sqrt{2}}BC+\frac{a}{ \sqrt{2}})=0$
$\implies BP=\frac{a}{ \sqrt{2}} \text{ or }BP=\sqrt{2} BC-\frac{a}{ \sqrt{2}} \tag{i} \label{i}$
Let us assume $BP=\sqrt{2} BC-\frac{a}{ \sqrt{2}} \tag{ii}$
$$\text{In }\triangle ABF, AB+BF>AF \implies 2(BC-a)>AF (\because AB=BF=BC-a$$
$$\implies 2BC-2a+a>AF+a>AC (\because AF+a>AC \text{ in } \triangle AFC$$
$$\implies \sqrt{2} BC-\frac{a}{ \sqrt{2}}>\frac{AC}{\sqrt{2}} \text{
(divided both sides by }\sqrt{2})$$
$$\implies BP>\frac{AC}{\sqrt{2}}=\frac{2R}{\sqrt{2}}=\sqrt{2}R\text{ (using (ii))}$$
$\implies BP>AP \implies \overset\frown{BP}>\overset\frown{AP}$ which is a contradiction. Therefore our assumption (ii) is wrong.
$\therefore BP\neq \sqrt{2} BC-\frac{a}{ \sqrt{2}}$ and (i) $\implies BP =\frac{a}{ \sqrt{2}} $
$\textbf{Method 2: Using similarity of triangles}$
$\begin{array}{l}
\text{In } \triangle ABP \text{ and } \triangle AFC\\
\angle ABP= \angle AFC =135° (\because \angle AFC=180°-\angle BFA)\\
\angle BAP=\angle FAC (\because \angle BAP=\angle BAC - \angle PAC =\angle BAC - 45°=\angle BAC - \angle BAF=\angle FAC)
\end{array}$
$\therefore \triangle ABP \sim \triangle AFC$ by AAA similarity criterion.
$\implies \frac{BP}{FC}=\frac{AP}{AC} \implies \frac{BP}{a}=\frac{\sqrt{2} R}{2R}=\frac1{\sqrt{2}}$
$\therefore BP=\frac{a}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.