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Proving $CF = r(\frac{1-\cos x}{\sin x})$ in a geometry problem without using $\tan\frac{x}{2}$
This problem is from IMO $1985$ Problem $1$.
A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.
Let $\angle DAO$ be $\theta$. I want to show that $CG = r(\frac{1-\cos \theta}{\sin \theta})$. It is given that $$\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{1 - \cos \theta}{\sin \theta}$$ for $0 < x < 180^\circ$.
How do I use this fact to prove that $CG = r(\frac{1-\cos \theta}{\sin \theta})$? (without using the fact that $\frac{1-\cos \theta}{\sin \theta} = \tan \frac{\theta}{2}$.)
My observation:
*
*$\angle FOG = \theta$
*I also tried reverse engineering:
\begin{align}
CG = r\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \\
CG^2 = r^2\frac{1 - \cos \theta}{1 + \cos \theta} \\
CG^2 + CG^2\cos \theta = r^2 - r^2\cos \theta
\end{align}
but I can't find why this is true.
| Solution without trigonometry.
Let $P\in AB$ such that $AP=AD$.
We can assume also that $P$ is placed between $A$ and $O$ as on your picture.
Thus, $$\measuredangle APD=\frac{1}{2}(180^{\circ}-\measuredangle A)=x,$$ which says that $DPOC$ is cyclic, which gives:
$$\measuredangle CPO=\measuredangle CDO=y$$ and since $$\measuredangle B=180^{\circ}-2y,$$ we obtain $$BC=PB$$ and $$AD+BC=AP+PB=AB.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to show $\frac{\cos (3(x - \frac{\pi}{4}))}{\cos(x - \frac{\pi}{4})} = \frac{\sin 3x - \cos 3x}{\sin x + \cos x}$
$$\frac{\cos \left(3(x - \frac{\pi}{4})\right)}{\cos(x - \frac{\pi}{4})} = \frac{\sin 3x - \cos 3x}{\sin x + \cos x}$$
My attempt:
\begin{align}
LHS &=\frac{\cos \left(3(x - \frac{\pi}{4})\right)}{\cos(x - \frac{\pi}{4})} \\
&= 4\cos^2 (x - \frac{\pi}{4}) - 3 \\
&= 4(\cos x \cos\frac \pi 4 + \sin \frac \pi 4\sin x )^2 - 3 \\
&=-1 + 4\sin x \cos x \\
RHS &= \frac{\sin 3x - \cos 3x}{\sin x + \cos x} \\
&= \frac{2\sin x - 4\sin^3x - 4\cos^3x + 3\cos x}{\sin x + \cos x} \\
&= -1 + 4\sin x\cos x
\end{align}
But it seems that there is a way to show it more efficiently, but I am not sure how.
| Using the angle addition formula $$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$ we have
$$\frac{\cos(3(x-\frac{\pi}{4}))}{\cos(x-\frac{\pi}{4})}=\frac{\cos(3x)\cos(\frac{3\pi}{4})+\sin(3x)\sin(\frac{3\pi}{4})}{\cos(x)\cos(\frac{\pi}{4})+\sin(x)\sin(\frac{\pi}{4})}$$
$$=\frac{-\frac{\sqrt{2}}{2}\cos(3x)+\frac{\sqrt{2}}{2}\sin(3x)}{\frac{\sqrt{2}}{2}\cos(x)+\frac{\sqrt{2}}{2}\sin(x)}$$
$$=\frac{\sin(3x)-\cos(3x)}{\cos(x)+\sin(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $f(x,y) = x\sqrt{3 - x^2 - y^2}$ find and sketch the domain? I have been trying to do this but I am having a hard time. All I have is that $\sqrt{3-x^2-y^2}\geq0$
The next step I want to do is the following:
\begin{aligned}(\sqrt{3-x^2-y^2})^2\geq0^2\end{aligned}
\begin{aligned}3 -x^2-y^2\geq0\end{aligned}
\begin{aligned}(3 -x^2-y^2)-3\geq0-3\end{aligned}
\begin{aligned}-x^2-y^2\geq-3\end{aligned}
\begin{aligned} -1*(-x^2- y^2)\geq-3*-1\end{aligned}
\begin{aligned} x^2 + y^2 \leq3\end{aligned}
I would just assume that the domain is the circle inequality above and graph the circle in the $(x,y)$ plane including the radius. Maybe I am wrong. What do you guys think?
| The domain of the function is the set of points $(x,y)$ on which the function is well defined. Now, you know that you can only take the square root of a non-negative number
Hence,
$$ 3-x^2-y^2 \geq 0$$
$$\implies x^2+y^2 \leq 3$$
Only if $(x,y)$ satisfies this can you get a real value for $f(x,y)$
As for sketching, this would be a disc centered at origin of radius $\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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showing all gcd $(n^3-n, 2n^2-1)$ I've been stuck on this problem for a long time.
I must show all gcd of $(n^3-n, 2n^2-1)$. I know that $n^3-n=n(n+1)(n-1)$ and so I must find all gcd of $(n(n+1)(n-1), 2n^2-1)$. But I don't know how. Can somebody help me ?
| I'd do it the way I did below without factoring. But factor was a good thought and it will work:
$n^3-n = n(n+1)(n-1)$ so lets see what factors $n,n+1, n-1$ have with $2n^2 -1$.
First of all any (nontrivial) factor of $n$ will be a factor of $2n^2$ and will not be a factor of $2n^2 -1$. So $n$ and $2n^2 -1$ have no factors in common and $n$ doesn't "contribute" to the gcd. So $\gcd(n(n+1)(n-1), 2n^2-1)= \gcd((n+1)(n-1),1)$.
Now do $n+1$ and $2n^2 -1$ have any factors in common? Note: $2n^2 -1 = 2(n^2 -1) +1= 2(n+1)(n-1) + 1$. So any non trivial factor that $n+1$ will have will also be a factor of $2(n+1)(n-1)$ so it will not be a factor of $2(n+1)(n-1)-1$. $n+1$ and $2n^2 -1$ will have not factor in common.
And by the same argument neither will $n-1$ and $2(n+1)(n-1)+1 = 2n^2 -1$.
So $2n^2-1$ has no factors in common with any of the factors of the factorization of $n(n+1)(n-1)$ so the $\gcd(n^3 -n, 2n^2 -1) = 1$.
=====My way=====
use $\gcd(a,b) = \gcd(a \pm kb, b)$. And use if $\gcd(k,b) =1$ then $\gcd(ka,b)=\gcd(a,b)$
So $\gcd(n^2 -n, 2n^2-1)$; $2n^2-1$ is odd any common divisor is odd and multiplying $n^2-n$ by $2$ will not effect the $\gcd$. So:
$\gcd(n^3 -n, 2n^2-1)= \gcd(2n^3-2n,2n^2-1)$. Whatever common divisor they have divides $2n^2 -1$ and $2n^3-2n$ so it will divide $n(2n^2-1)$ and will divide $(2n^3-2n)-n(2n^2-1)$
So $ \gcd(2n^3-2n,2n^2-1)= \gcd([2n^3-2n]-n[2n^2 -1],2n^2-1) = \gcd(n,2n^2-1)$.
$\gcd(n,2n^2-1) = \gcd(n, [2n^2-1] -[(2n)n]) = \gcd(n, -1)$ and as the only thing that divide $-1$ are $1, -1$ and $1 > -1$ then $\gcd(n,-1) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| You have nice answers by expansion, so I present a solution without it.
$$
\sin^6x + \cos^6x
$$
$$
= \sin^6x + (1-\sin^2 x)^3
$$
$$
= \sin^6x + 1 - 3\sin^2 x+3\sin^4 x-\sin^6 x
$$
$$
= 1 - 3\sin^2 x + 3\sin^4 x
$$
$$
= 1 - 3\sin^2(1 - \sin^2 x)
$$
$$
= 1 - 3\sin^2 x\cos^2 x
$$
$$
= 1 - \frac{3}{4}\sin^22x
$$Hence Proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 1
} |
Calculate ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $ Let ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $, for n=1,2,3,.....Then
(A) ${S_n} < \frac{\pi }{{3\sqrt 3 }}$
(B) ${S_n} > \frac{\pi }{{3\sqrt 3 }}$
(C) ${T_n} < \frac{\pi }{{3\sqrt 3 }}$
(D) ${T_n} > \frac{\pi }{{3\sqrt 3 }}$
The official Answer is A and D
My approach is as follows
$\mathop {\lim }\limits_{n \to \infty } {S_n} = \sum\limits_{k = 1}^n {\frac{1}{{n\left( {1 + \frac{k}{n} + {{\left( {\frac{k}{n}} \right)}^2}} \right)}}} $
${S_n} < \int\limits_0^1 {\frac{{dx}}{{{x^2} + x + 1}}} = \int\limits_0^1 {\frac{{dx}}{{{x^2} + x + \frac{1}{4} + \frac{3}{4}}}} = \int\limits_0^1 {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{1}{{\frac{{\sqrt 3 }}{2}}}} \left( {\left. {{{\tan }^{ - 1}}\frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right|_0^1} \right) = \frac{2}{{\sqrt 3 }}\left( {{{\tan }^{ - 1}}\sqrt 3 - {{\tan }^{ - 1}}\frac{1}{{\sqrt 3 }}} \right) = \frac{2}{{\sqrt 3 }}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{3\sqrt 3 }}$
So A is correct
$\mathop {\lim }\limits_{n \to \infty } {T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{1}{{n\left( {1 + \frac{k}{n} + {{\left( {\frac{k}{n}} \right)}^2}} \right)}}} $
$t = k + 1\mathop {\lim }\limits_{n \to \infty } {T_n} = \sum\limits_{t = 1}^n {\frac{1}{{n\left( {1 + \frac{{t - 1}}{n} + {{\left( {\frac{{t - 1}}{n}} \right)}^2}} \right)}}} $
The answer is D but don't know to proceed.
| We have $$S_n \le I \le T_n$$ because $f(x)$ here is a decreasing function.
By squeez law, we have
$$I=\lim_{n \to \infty} T_n=\lim_{n \to \infty} S_n=\int_{0}^{1} f(x) dx $$
$$f(x)=\frac{1}{x^2+x+1}$$
So $$I=\int_{0}^{1} \frac{dx}{(x+1/2)^2+3/4}=\frac{2}{\sqrt{3}}\tan^{=1}\frac{(2x+1)}{\sqrt{3}}=\frac{2}{\sqrt{3}}[\pi/3-\pi/6]=\frac{\pi}{3\sqrt{3}}$$
So options (A) and (B) are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$
So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1+x+x^2+x^3+x^4)=0$$
Thus $$1+x+x^2+x^3+x^4=0$$ I tried using this condition to simplify the above expression but nothing interesting simplified. Please note i am looking for hints rather than complete solutions.
EDIT:I came to know its a duplicate ,but i feel that the answers given below are different from those in the original.
| Answer :
((x / (1+x² ))+((x³/ (1+x))=((x+x² +x³ +x⁵) / ((1+x)(1+x²) ))=((x+x² +x³ +x⁵) / (x+x² +x³ +1))=1
Because x^5=1
$((x² / (1+x⁴))+((x⁴/ (1+x³ )) = ((x² +x⁵+x⁴+x⁸) / (1+x³ +x⁴+x^7)) =((x² +1+x⁴+x³) / (1+x³+x⁴+x²)) =1$
Because $x^5 =1$
$\frac{x}{1+x^2 }+\frac{x^3 }{1+x} +\frac{x^2 }{1 +x^4}+\frac{x^4}{1+x^3 }=\frac{x+x^2 +x^3 +x^5}{(1+x)(1+x^2) }+\frac{x^2 +x^5+x^4+x^8}{(1+x^4)(1+x^3) }=\frac{x+x^2 +x^3 +x^5}{x+x^2 +x^3 +1}+\frac{x^2 +1+x^4+x^3 }{1+x^3 +x^4+x^2} =2$
Because $x^5 =1\Rightarrow x^7=x^2$ and $x ^8 = x^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.
My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.
| Hint:
$$x+y+z=xyz\iff x=\dfrac{y+z}{1-yz}$$
If $\dfrac{2x}{1-x^2}=a$ etc. assuming $x^2,y^2,z^2\ne1$, we need $$a+b+c=abc\iff\dfrac1{bc}+\dfrac1{ca}+\dfrac1{ab}=1$$
$$\implies\dfrac1a\left(\dfrac1b+\dfrac1c\right)$$
$$=\dfrac{(1-x^2)}{2x}\left(\dfrac{1-y^2}{2y}+\dfrac{1-z^2}{2z}\right)$$
$$=\dfrac{1-\left(\dfrac{y+z}{1-yz}\right)^2}{2\left(\dfrac{y+z}{1-yz}\right)}\cdot\dfrac{(1-yz)(y+z)}{2yz}$$
$$=\dfrac{(1-yz)(y+z)((1-yz)^2-(y+z)^2)}{4(y+z)(1-yz)yz}$$
$$=\dfrac{4yz-(1-y^2)(1-z^2)}{4yz}$$
$$=1-\left(\dfrac{1-y^2}{2y}\right)\left(\dfrac{1-z^2}{2z}\right)$$
$$=1-\dfrac1{bc}$$
So, we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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$A^{n}$ matrix problem without using $A=PDP^{-1}$ Eigendecomposition i have the following problem
Find $A^n$ for the following matrix
\begin{equation}
A=\begin{pmatrix}
0 & 0 & 1\\
0 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix}
\end{equation}
I have tried the following, calculating for $n=1,2,3,4,5,6$
\begin{equation}
A^1=\begin{pmatrix}
0 & 0 & 1\\
0 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix} \qquad \qquad
A^2=\begin{pmatrix}
1 & 1 & 1\\
1 & 2 & 2 \\
1 & 2 & 3
\end{pmatrix} \qquad \qquad
A^3=\begin{pmatrix}
1 & 2 & 3\\
2 & 4 & 5 \\
3 & 5 & 6
\end{pmatrix} \\
A^4=\begin{pmatrix}
3 & 5 & 6\\
5 & 9 & 11 \\
6 & 11 & 14
\end{pmatrix} \qquad \qquad
A^5=\begin{pmatrix}
6 & 11 & 14\\
11 & 20 & 25 \\
14 & 25 & 31
\end{pmatrix} \qquad \qquad
A^6=\begin{pmatrix}
14 & 25 & 31\\
25 & 45 & 56 \\
31 & 56 & 70
\end{pmatrix}
\end{equation}
That gives the following terms
\begin{equation}
A_{11} = 0,1,1,3,6,14,...\\
A_{12} = 0,1,2,5,11,25,...\\
A_{22} = 1,2,4,9,20,45,...
\end{equation}
But i can't figure out the succesion in terms of n.
| The following solution is using the eigenvalues in an implicit way.
$$A^3=2A^2+A-I_3$$
Which can be found by inspection or by calculating the characteristic polynomial.
Then, by long division
$$X^n=Q(X)(X^3-2X^2-X+1)+aX^2+bX+c$$
You can find $a,b,c$ by setting $X=\lambda_{1,2,3}$ the solutions to $X^3-2X^2-X+1$ in the above equation.
Then, setting $x=A$ in the equation we get:
$$A^n=aA^2+bA+C$$
Note The above implies that all entries of $A^n$ are of the form $c_1\lambda_1^n+c_2 \lambda_2^n+c_3 \lambda_3^n$ (which is consistent with diagonalisation). Since the roots
$\lambda_{1,2,3}$ of the equation
$$x^3-2x^2-x+1$$
are ugly, the formula is almost impossible to guess.
| {
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"url": "https://math.stackexchange.com/questions/3845140",
"timestamp": "2023-03-29T00:00:00",
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Stuck and confused with $\int \frac{u^{1/2}}{(u+1)^2} du$ I'm trying to find the expectation value $\langle x^2 \rangle_{\psi}$ for
$$\psi(x,0) = \sqrt{\frac{2}{\pi}}a \cdot \frac{1}{x^2+a^2}$$
and I know the result is $a^2$. I'm terribly stuck integrating this. Currently:
$$\langle x^2 \rangle_{\psi} = \int\limits_{-\infty}^{\infty}\, \psi^* x^2 \psi \, dx$$
$$= \left(\frac{2a^2}{\pi}\right) \int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^2} dx = \left(\frac{2a^2}{\pi}\right) \int_{-\infty}^{\infty} \frac{1}{a^2} \frac{(x/a)^2}{((x/a)^2+1)^2}dx = \left(\frac{4a}{\pi}\right) \int^\infty_0 \frac{u^2 }{(u^2+1)^2} du$$
where the change of variable $u=x/a$ is used.
With $v=u^2$, $u=\sqrt{v}$ and $du=\tfrac{dv}{2\sqrt{v}}$ and then,
$$\langle x^2 \rangle_{\psi} = \left(\frac{2a}{\pi}\right) \int^\infty_0 \frac{v^{1/2}}{(v+1)^2}dv = \frac{2a}{\pi} \int^\infty_0 \frac{v}{(v+1)^2}v^{1/2} \frac{dv}{v}$$
Then I checked the latter integral in Wolfram Mathematica but the results doesn't seem to match (not $\pi a$) and this is confusing me a lot. Any suggestions with this?
Also: is it possible to solve it analytically?
| The term $\langle x^2\rangle_\psi$ as defined by the OP can be obtained through standard method of integration of rational functions.
Here is one method:
$$
\int \frac{x^2}{(x^2+a^2)^2}\,dx =\int \frac{dx}{x^2+a^2} -a^2\int\frac{dx}{(x^2+a^2)^2}
$$
Setting $I_n=\int\frac{dx}{(x^2+a^2)^n}$, we get by integration by parts the recurrence formula
$$ I_{n+1}=\frac{1}{2na^2}\frac{x}{(x^2+a^2 )^n}+\frac{2n-1}{2na^2}I_n$$
For $n=1$ we obtain
$$ I_2=\frac{1}{2a^2}\frac{x}{x^2+a^2} +\frac{1}{2a^3}\arctan(x/a) + C $$
Putting things together, we het
$$
\frac{2a^2}{\pi}\int^\infty_{-\infty}\frac{x^2\,dx}{(x^2+a^2)^2}=\frac{2a^2}{\pi}\Big(\frac{1}{2a}\arctan(x/a)-\frac{1}{2}\frac{x}{x^2+a^2}\Big)\Big|^\infty_{-\infty}=a
$$
The OP, through a couple of change of variables, reduced the quantity of interest to calculating the integral $$\int^\infty_0 \frac{u^{1/2}}{(u+1)^2}\,du=\int^\infty_0 \frac{u}{(u+1)^2}u^{1/2}\,\frac{du}{u}$$
This integral can be obtained by calculus of residues. In this instance, the Mellin transform comes to mind (See Marsden's basic complex analysis, or Lang's complex analysis).
The Mellin transform of a function $f$ is defines as $M_f(s)=\int^\infty_0 f(x) x^s\,\frac{dx}{x}$, $s>0$. Under the following conditions
*
*$f$ is analytic on the complex plane, except for a finite number of poles none of which lies on the positive real axis.
*There are $0<c<s<b$ such that $f(z)=O(z^{-b})$ as $z\rightarrow\infty$, and $f(z)=O(z^{-c})$ as $z\rightarrow0$
we have that
$$\int^\infty_0 f(x) x^{s-1}\,dx = -\frac{\pi e^{-\pi i s}}{\sin \pi s}\sum^N_{j=1} \operatorname{Res}_{z_j}(f(z) z^{s-1})$$
where $z_1,\ldots, z_N$ are the poles of $f$ in $\mathbb{C}\setminus\{0\}$, and $z^{(s-1)}=e^{(s-1)\log z}$ with $\log$ is the logarithmic function with branch $0<\operatorname{arg}(z <2\pi$.
For the integral in the OP, it is easy to check that $f(z)=\frac{z}{(z+1)^2}$ satisfies condition (1) and (2) with $s=1/2$.
$g(z)=f(z) z^{1/2 -1}$ has a pole at $z=-1$, and
$$\operatorname{Res}_{-1}(g)=1\cdot e^{-\tfrac12\pi i} -(\tfrac12-1)e^{-\tfrac32 \pi i}=-\tfrac{i}{2}$$
Hence
$$
\int^\infty_0 \frac{u^{1/2}}{(u+1)^2}\,du =-\frac{\pi e^{-\tfrac12\pi i}}{\sin\tfrac12 \pi}(-\tfrac{i}{2})=\frac{\pi}{2}
$$
Thus the quantity $\langle x^2\rangle_{\psi}$, as defined by the OP, is
$\frac{2a}{\pi}\frac{\pi}{2}=a$.
Edit:
It seems that the OP is missing a factor $a$ in the normalization factor.
$$
\begin{align}
1=\int^\infty_{-\infty}\psi^*\psi&=c\int^\infty_{-\infty}\frac{dx}{(x^2+a^2)^2}\\
&=c\Big( \frac{1}{2a^2}\frac{x}{(x^2+a^2)^2}+\frac{1}{2a^3}\arctan(x/a)\Big)\Big|^\infty_{-\infty}\\
&=\frac{\pi}{2a^3}c
\end{align}
$$
implies that $c=\frac{2a^3}{\pi}$. The OP has $\frac{2a^2}{\pi}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 3,
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} |
Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$ How can we represent $2\cos(\frac{\pi}{11})$ and $2\cos(\frac{\pi}{13})$ as cyclic infinite nested square roots of 2
I have partially answered below for $2\cos(\frac{\pi}{11})$ which I derived it accidentally.
Currently I have figured out for $q$ as denominator with patterns of $2^{n\pm1}$here
Is there any way to figure out pattern of nested radicals for other rational number $p \over q$ in $2\cos(\frac{p}{q})\pi$.
P.S. I am able to figure out that the rational number $p \over q$ must be as follows $1\over4$ < $p\over q$ < $ 1 \over 2$ as $1<$ $\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm...}}}$ $<2$
| Let $2\cos(\frac{\pi}{11})$ represented with the help of half angle cosine formula as follows
$2\cos(\frac{\pi}{11}) = \sqrt{2 + 2\cos(\frac{2\pi}{11})}=\sqrt{2 + \sqrt{2+ 2\cos(\frac{4\pi}{11})}}$
$\sqrt{2 + \sqrt{2+\sqrt{2+ 2\cos(\frac{8\pi}{11})}}}$
Now $\frac{8\pi}{11}$ is more than $\frac{\pi}{2}$ which makes $2\cos(\frac{8\pi}{11})$ to $-2\sin(\frac{5\pi}{22})$ and in turn this is $-2\cos(\frac{3\pi}{11})$
Next step plugging in leads to $\sqrt{2 + \sqrt{2+\sqrt{2- 2\cos(\frac{3\pi}{11})}}}$
Further expansion is $\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2+ 2\cos(\frac{6\pi}{11})}}}}$ and this is represented as $\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2- 2\sin(\frac{\pi}{22})}}}}$ which is equal to
$\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-2\cos(\frac{5\pi}{11})}}}}$
Final expansion is beautiful to observe
$\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+2\cos(\frac{10\pi}{11})}}}}}$ which is equal to
$\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2\cos(\frac{\pi}{11})}}}}}$
$\therefore 2\cos(\frac{\pi}{11}) = \sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2\cos(\frac{\pi}{11})}}}}}$
As these steps get repeated for ever we can conclude that
$$2\cos(\frac{\pi}{11})$$ can be represented as cyclic infinite nested square roots of 2 as follows and single cycle is
$$\sqrt{2 +\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-}}}}}...$$ which repeats infinitely.
This is represented simply as $$2\cos\frac{\pi}{11}=cin\sqrt2[2+3-]$$
[Refer here for reverse way to derive cosine angle from cyclic infinite nested square roots of 2][1]
| {
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Determinant by cofactor expansion Vs Row reduction For the following matrix
$$ \begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & -4 \\
-4 & 1 & 1 \\
\end{pmatrix} $$
The determinant by cofactor expansion is 25 . But by EROs reduction I'm always getting -25 ..here is one trial :
--> R3+4R2 & R1-R2
$$ \begin{pmatrix}
1 & 1 & 1 \\
0 & 0 & 5 \\
0 & 5 & -15 \\
\end{pmatrix} $$
And continuing by cofactor expansion
1(-25)=-25 !!
I aslo tried completing it by EROs , ECOs but the answer always is -25 ..WHY?
Also another example,
$$ \begin{pmatrix}
0 & 1 & 5\\
3 & -6& 9\\
2 & 6 & 1 \\
\end{pmatrix} $$
I tried R2-1.5R3
And got
$$ \begin{pmatrix}
0 & 1 & 5\\
3 & -6& 9\\
0 & -15 & 7.5 \\
\end{pmatrix} $$
Det()= -3(7.5+15×5) =-247.5
Where the real answer is -165!!
Please I need a justfication for these cases.
| You can only replace the row $R_i$ with $R_i + k R_j$ (not row $R_j$).
If you replaced row $R_j$ instead, the determinant is multiplied by a factor of $k$.
This is related to the elementary matrix multiplications that underlie the row reduction methods.
Hence for example $1$, under row operations $R_3 + 4R_2 \to R_3$ and $R_1-R_2 \to R_1$:
\begin{pmatrix}0 &0 &5\\1& 1& -4\\-4&1&1\end{pmatrix}
which by cofactor expansion evaluates to $25$.
The row operation $R_1-R_2 \to R_1$ can be represented as the matrix:
\begin{pmatrix}1 &-1 &0\\0& 1& 0\\0&0&1\end{pmatrix}
which has determinant $1$.
The row operation $R_1-R_2 \to R_2$ can be represented as the matrix:
\begin{pmatrix}1 &0 &0\\1& -1& 0\\0&0&1\end{pmatrix}
which has determinant $-1$.
Hence we must take note of which row we are replacing. The same applies for columns.
| {
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Proving an inequality is true (precalculus) How do I prove that
$(|x|+2)(|x^2+9|)-9|x^2-2| \ge 0$?
I tried using properties of absolute values such as triangle inequalities but so far I've got no luck.
The actual question was to prove that
$|\frac{x^2-2}{x^2+9}| \le \frac{|x|+2}{9}$. I tried using triangle inequality properties but only to the point where I get $|\frac{x^2-2}{x^2+9}| \le \frac{|x^2|+2}{9}$ which is different from what I wanted to prove.
| Using $f(x) = (|x|+2)(x^2+9)-9|x^2-2|$ as defined in the other answer and the observation that $f(x) = f(-x)$, we prove it when $x \ge 0$ using a slightly different way: completing squares.
For $0 \le x \le \sqrt 2$, we have:
$$f(x) = (x+2)(x^2+9)-9(2-x^2) = x^3+11x^2+9x\ge0 \text{ since }x \ge 0$$
For $x \ge \sqrt 2$ we have:
$$\begin{align}f(x) &= (x+2)(x^2+9)-9(x^2-2) \\&= x^3-7x^2+9x+36\\&= x^3-7x^2+\frac{49}4x-\frac{13}4x+36\\&=x(x-3.5)^2+\frac14(144-13x)\end{align}$$
which is positive when $\sqrt 2 \le x \le 10$.
We also have:
$$f(x) = x^3-7x^2+9x+36 > x^2(x-7)$$
which is positive when $x \ge 7$.
This proves the result.
| {
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Bias and MSE of $\hat{\theta} = \min(X_1, \ldots, X_n)$ Let $X_1, ... X_n$ iid with pdf given by $$p_{\theta, r} = r \theta^r x^{- (r+1)} \mathbb{1}\{x \geq \theta\}$$ for $\theta > 0$, and some $r > 2$ that is known. Then $\hat{\theta} = \min(X_1, \ldots, X_n) = X_{(1)}$.
I want to determine the bias and MSE of $\hat{\theta}$, so I need the pdf of $\hat{\theta}$.
If my calculations are correct, the pdf of $\hat{\theta}$ is given by:
$$f_{X_{(1)}} = n(r+1)r^n \theta^{rn} x^{-n(r+1) - 1} \mathbb{1}\{x \geq \theta\}.$$
Wondering if this pdf is correct, and how one would calculate the bias and MSE using this variance? I know that the bias is given by$E[\hat{\theta}] - \theta$, but I end up with a complicated expression, so I believe I am doing something wrong.
| $$
p_{\theta, r}(x) \, dx = r \left( \frac x \theta\right)^{- (r+1)}\left( \frac{dx} \theta \right) \mathbb{1}\{x \geq \theta\}.
$$
$$
\text{So for } x\ge\theta, \qquad \Pr(X>x) = \left( \frac x \theta \right)^{-r}
$$
$\min\{X_1,\ldots,X_n\} > x$ if and only if all of $X_1,\ldots, X_n$ are${}> x.$
That has probability $\big( \Pr(X_1>x)\big)^n = \left(\dfrac x \theta \right)^{-nr}.$
Therefore for $x>\theta,$
$$
p_{\min}(x) = \frac d {dx} \Pr(\min \le x) = \frac d {dx} \left( 1 - \left( \frac x \theta \right)^{-nr} \right) = nr\left( \frac x \theta \right)^{-(nr+1)}\cdot\frac 1 \theta.
$$
So
\begin{align}
& \operatorname E(\min) = \int_\theta^\infty x\cdot nr \left( \frac x \theta \right)^{-(nr+1)} \left( \frac{dx}\theta \right) \\[8pt]
= {} & nr\theta \int_\theta^\infty \frac x \theta \cdot \left( \frac x \theta \right)^{-(nr+1)} \, \left( \frac{dx} \theta \right) \\[8pt]
= {} & nr\theta \int_1^\infty u^{-nr} \, du = \theta \cdot \frac {nr} {nr-1} \\[8pt]
= {} & \theta \times \big(\text{something a bit more than 1} \big).
\end{align}
This is to be expected, since the minimum is always greater than $\theta;$ thus so is its expected value, and the fact that $\theta$ appears as a factor of this whole expression is necessary since $\theta$ is a scale parameter.
\begin{align}
& \operatorname E({\min}^2) = \int_\theta^\infty x^2\cdot nr \left( \frac x \theta \right)^{-(nr+1)} \left( \frac{dx}\theta \right) \\[8pt]
= {} & nr\theta^2 \int_\theta^\infty \left(\frac x \theta\right)^2 \cdot \left( \frac x \theta \right)^{-(nr+1)} \, \left( \frac{dx} \theta \right) \\[8pt]
= {} & nr\theta^2 \int_1^\infty u^{-nr+1} \, du = \theta^2 \cdot \frac {nr} {nr-2} \\[8pt]
= {} & \theta^2 \times \big(\text{something a bit more than 1} \big).
\end{align}
So
\begin{align}
& \operatorname{var}(\min) = \theta^2\left( \frac{nr}{nr-2} - \left( \frac{nr}{nr-1} \right)^2 \right) \\[8pt]
\text{and } & \frac{nr}{nr-2} - \left( \frac{nr}{nr-1} \right)^2 \\[8pt]
= {} & \frac{nr(nr-1)^2 - (nr)^2(nr-2)}{(nr-1)^2(nr-2)} \\[8pt]
= {} & \frac{nr}{(nr-1)^2(nr-2)}.
\end{align}
And then:
\begin{align}
& \text{m.s.e.} = \text{variance} + \big( \text{bias} \big)^2 \\[8pt]
= {} & \theta^2 \cdot \frac{nr}{(nr-1)^2(nr-2)} + \left( \theta\cdot \frac 1 {nr-1} \right)^2 \\[8pt]
= {} & \theta^2 \cdot \frac 2 {(nr-1)(nr-2)}.
\end{align}
| {
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Using integrating factors to solve difficult differential equations Consider the differential equatioN:
$$(2x^2 y -2y^4)dx+ ( 2x^3 + 3xy^3) dy = 0$$
This equation is of the form:
$$ Q dx + P dy=0$$
Now, it's easy to see that this differential is not exact by using the commutativity of second order partial condition. Multiply it by an integrating factor $\eta(x,y)$ such that it does:
$$ \eta_x P + P_x \eta = \eta_y Q + Q_y \eta$$
Or,
$$ \eta_x P - \eta_y Q + ( P_x - Q_y) \eta = 0$$
This becomes:(*)
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 -8y^3) \eta =0$$
Or,
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 - 11y^3) \eta =0$$
Now... I'm not quite sure what to do.. I don't think I can write $ \eta$ as function solely of $x$ or $y$.. did I miss something or... ?
As correctly observed by @Aleksas Domarkas. There is a mistake after the place where I put *. The current working is as follows:
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 +8y^3) \eta =0$$
Or,
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 + 11y^3) \eta =0$$
| I get ("+" in last term).
$$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 + 11y^3) \eta =0$$
Let $\eta=x^ay^b$. Then
$$\left( -2 b-3 a-11\right) \, {{x}^{a}}\, {{y}^{b+3}}+\left( 2 b-2 a-4\right) \, {{x}^{a+2}}\, {{y}^{b}}=0$$
Solve system $-2 b-3 a-11=0,\quad 2 b-2 a-4=0$.
Then $a=-3,\quad b=-1$. Integrating factor is
$$\eta=\frac{1}{x^3y}$$
We get general solution of differential equation
$$2\ln(xy)+\frac{x^3}{y^2}=C$$
| {
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| As written, no simplification is apparent, and this is due to the presence of the radicals. Now considering the identity
$$a-b=\frac{a^2-b^2}{a+b}$$ there is a hope of getting rid of them. In the case of your numerator,
$$\sqrt{6-x}-2=\frac{(\sqrt{6-x})^2-2^2}{\sqrt{6-x}+2}=\frac{2-x}{\sqrt{6-x}+2}.$$
Now the radical is gone from the numerator and has moved to the denominator, but it is important to notice that it does not cancel because the minus has turned to a plus.
Repeating this trick with the denominator of the original ratio, you will see a simplification.
| {
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How to increase speed of convergence of series? I'm trying to speed up convergence of series:
$$
a_{n} = \frac{n(n-0.3)}{(n^{2} + 0.3)^{2}}
$$
Without series speed up:
$$
R_{method} = \left|A - S_{N_{0}}\right| = \sum_{n = N_{0} + 1}^{\infty}\frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} \leq \sum_{n = N_{0} + 1}^{\infty}\frac{n^2}{n^{4}} < \int_{N_{0}}^{\infty}\frac{1}{x^2}dx = \frac{1}{N_{0}} < 0.25 \times 10^{7} \rightarrow N_{0} > 4 \times 10^{7}
$$
$$
S_{0} = \sum_{n = 1}^{4\times 10^{7}}\frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} \approx 0.959433..
$$
$$
R_{calc} = N_{0}\times A_{a_{n}} < 0.25 \times 10^{-7} \rightarrow A_{a_{n}} < 0.625 \times 10^{-15}
$$
First speed up:
$$
\alpha_{0} = \lim_{n \rightarrow \infty}\frac{a_{n}}{\frac{1}{n^{2}}} = \lim_{x \rightarrow \infty } = \lim_{n \rightarrow \infty} \frac{(n - 0.3)n^{3}}{(n^2 + 0.3)^{2}} = 1
$$
$$
b_{n} = \frac{n(n-0.3)}{(n^{2} + 0.3)^{2}} - \frac{1}{n^{2}} = \frac{-30n^{3}-60n^{2} -9 }{n^{2}(10n^2 + 3)^2}
$$
$$
B = \sum_{n = 1}^{\infty}b_{n}
$$
$$
R_{method} = \left|B - S_{N_{1}} \right| =\left| \sum_{n = N_{1} + 1}^{\infty} \frac{30n^3+60n^2+9}{n^2(10n^2+3)^2}\right| \leq \left| \sum_{n = N_{1} + 1}^{\infty}\frac{30n^3+60n^2+9}{10n^6}\right| <
$$
$$
< \int_{N_{1}}^{\infty}\frac{30n^3+60n^2+9}{10n^6} = \frac{25(3N_{1}+4)N_{1}^2+ 9}{25N_{1}^5} < 0.25 \times 10^{-7} \rightarrow N_{1} > 10995.1
$$
$$
S_{1} = \sum_{n = 1}^{10996}\frac{-30n^{3}-60n^{2} -9 }{n^{2}(10n^2 + 3)^2} + \frac{\pi^2}{6} \approx 18021
$$
There is a mistake but I cannot see it. Where did I go wrong and how should I have done it right?
| Your error seems to lie in this computation:
$$\int_{N_{1}}^{\infty}\frac{30n^3+60n^2+9}{10n^6}dn= \frac{25(3N_{1}+4)N_{1}^2+ 9}{25N_{1}^5}$$
Indeed a primitive function of the integrand is:
$$-\frac32 \frac{1}{n^2}-\frac12 \frac{1}{n^3}-\frac{9}{10} \frac{1}{n^5}=- \frac{75n^3+25n^2+9}{50 n^5}$$
| {
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What is the equation of the circle I've been experiencing a difficulties in answering this. I hope someone will help me in solving this
Find the equation of the circle through the points $(2,8),(7,3)$ and $(-2,0)$.
| HINT
Let start from the general equation
$$(x-x_C)^2+(y-y_C)^2=R^2$$
and plug in the values $(x,y)$ for the given points to obtain three equations in the three uknowns $x_C$, $y_C$ and $R$.
We obtain the system
*
*$2x_1x_C+2y_1y_C-x_1^2-y_1^2=x_C^2+y_C^2-R^2$
*$2x_2x_C+2y_2y_C-x_2^2-y_2^2=x_C^2+y_C^2-R^2$
*$2x_3x_C+2y_3y_C-x_3^2-y_3^2=x_C^2+y_C^2-R^2$
that is
*
*$2x_1x_C+2y_1y_C-x_1^2-y_1^2=2x_2x_C+2y_2y_C-x_2^2-y_2^2$
*$2x_2x_C+2y_2y_C-x_2^2-y_2^2=2x_3x_C+2y_3y_C-x_3^2-y_3^2$
from which we can find $x_C$ and $y_C$ and then $R$.
We can check directly the solution $(x-2)^2+(y-3)^2=25$ you have obtaind as follows
*
*$(2,8) \to (2-2)^2+(8-3)^2=25$
*$(7,3) \to (2-7)^2+(3-3)^2=25$
*$(-2,0) \to (-2-2)^2+(0-3)^2=25$
| {
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The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from Further Mathematics(9231) Paper 1, Question 1, 2009. I tried to solve it but was unable to figure it out, especially how to find the value of $\alpha^6+\beta^6+\gamma^6+\delta^6$. Could anyone try to solve this question and explain how they got the value?
| "Further mathematics" sounds like IB. So, they may suggest a solution using linear recursions.
The given polynomial
$$x^4-\color{green}{1}\cdot x^3 +\color{blue}{0}\cdot x^2 +\color{blue}{0}\cdot x - 1$$ belongs to the linear recursion
$$x_{n+4} = x_{n+3}+x_n$$
with (using Vieta)
*
*$x_0 = \alpha^0 + \beta^0 + \gamma^0 + \delta^0 = 4$
*$x_1 = \alpha + \beta + \gamma + \delta =\color{green}{1}$
*$x_2 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2=(\alpha + \beta + \gamma + \delta)^2 -2\cdot \color{blue}{0} = 1$
*$x_3 = \alpha^3 + \beta^3 + \gamma^3 + \delta^3 =(\alpha + \beta + \gamma + \delta)^3-3(\alpha + \beta + \gamma + \delta)\cdot\color{blue}{0}-3\cdot\color{blue}{0} =1$
Now, since
$$\alpha^6 + \beta^6 + \gamma^6 + \delta^6 = x_6$$
you just use the recursion
$$x_4= x_3+x_0 = 1+4 = 5$$
$$x_5= x_4+x_1 = 5+1 = 6$$
$$\boxed{x_6}= x_5+x_2 = 6+1 = \boxed{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861550",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Relation between prime roots and residues mod 7 in the prime root chapter from my book it was mentioned that the powers of the primitive root mod 7 equals the residues if you are calculating $\frac{1}{7}$.
It is easy to see that $3$ is a primitive root mod $7$ with
$$ 3^1 \equiv 3, 3^2 \equiv 2, 3^3 \equiv 6, 3^4\equiv 4, 3^5 \equiv 5, 3^6\equiv 1$$
So you get the sequence $3,2,6,4,5,1$.
Now let's calculate $1/7$ by hand. You get $10=1\cdot7+3, 30=4\cdot7+2, 20=2\cdot7+6 , \ldots.$
So the residues you get while calculating $1/7=\overline{142857}$ are identical to the powers of the primitive root $3$. If you calculate $2/7, 3/7, \ldots$, it just shift the sequence, but the pattern stays the same.
But this is not true for the other primitive root mod $7$ (namly $5$) and it is not true for any other primes but 7 i tested, like $5, 11,$ or $13$.
Now I really wonder, is this just a coincidence? And if not, what makes the number $7$ so special? Are there other numbers with a similar phenomenon?
| It does work for the other primitive root $5$, provided you work in base $12$ instead of $10$. The sequence of powers is
$$5, 4, 6, 2, 3, 1$$.
Now calculate $1/7$ in base $12$:
$$12 = 1\cdot7 + 5,\quad 5\cdot12 = 60 = 8\cdot 7 + 4,\quad 4\cdot12 = 48 = 6\cdot 7 + 6, \ldots$$
The sequence $5, 4, 6, \ldots$ turns up again. In general, if $r$ is a primitive root mod $p$, try calculating $1/p$ in base $p + r$.
| {
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Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$
I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked with WA
Is there an elgant solution using minimum calculus?. Substituiting $x+iy$ makes things very complex indeed.
Answer:
$[\sqrt{\frac{7}{2}},3\sqrt{\frac{7}{6}}]$
| Since $|z|=1$ we substitute $z=e^{i\theta}$, to get:
$$
f(\theta)=|1+e^{i\theta}|+|1-e^{i\theta}+e^{2i\theta}|\\
= \sqrt{ (1 + \cos \theta)^2 + \sin^2\theta} + \sqrt{(1 - \cos\theta+ \cos2\theta)^2 + (-\sin\theta + \sin2\theta )^2)}\\
=\sqrt{|2+2\cos\theta |}+|(1-2\cos\theta)|
$$
We can denote $u=\cos\theta$ and investigate $f(u)=\sqrt{|2+2u |}+|(1-2u)|$ for $u$ between $-1$ and $1$ by simple calculus. We get
$\max f=\frac{13}{4}$ for $u=-\frac{7}{8}$, and $\min f=\sqrt3$ for $u=\frac{1}{2}$. This does not agree with the answer given but I could not find any error.
| {
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Prove $(x^2 + xy + y^2)^2$ divides $(x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$. As stated in the title, I want to show the (bivariate) polynomial $g(x, y) = (x^2 + xy + y^2)^2$ divides the polynomial $f(x, y) = (x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$, where $n \geq 0$.
Naturally, I resorted induction. For $n = 1$, $f(x, y) = 7xy(x + y)g(x, y)$. From $n$ to $n + 1$,
I am only able to show that $x^2 + xy + y^2$ is a factor of $f(x, y)$. Is there any trick I am missing?
| Let $ y = k $ = constant and let $f(x)= (x+k)^{6n+1} - x^{6n+1}-k^{6n+1}$. We know that
$x^2 + xk + k^2= (x-kw)(x-kw^2)$ where $w$, $w^2$ are cube roots of unity. If we put $x=kw $ and $kw^2$ in $f(x)$, it gives the value $0$. Therefore $(x-kw)(x-kw^2)$ divides $f(x)$ . $\frac{df(x)}{dx}$ = $(6n+1)((x+k)^{6n} -x^{6n})$. Putting $x=kw $ and $kw^2$ in this expression we also get $0$. Therefore $(x-kw)(x-kw^2)$ divides both $f(x)$ and $f'(x)$. Therefore ${((x-kw)(x-kw^2))}^2$ divides $f(x)$ that is $(x^2 + xk + k^2)^2$ divides $f(x)$ . By varying $k$ we get that it is true for all values of $y$.
| {
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Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3)((k+1)+7) = (k+3)(k+4)(k+8)\\
= &(k+3)[(k+2)+2][(k+7)+1]\\
= &[(k+3)(k+2)+(2)(k+3)][(k+7)+1]\\
= &(k+2)(k+3)(k+7)+2(k+3)(k+7)+(k+2)(k+3)+2(k+3)\\
= &6P+2k^2+20k+42+k^2+5k+6+2k+6\\
= &6P+3k^2+27k+54\\
= &6p+3(k^2+9k+18)
\end{align*}
I'm not sure what to do, my proof turned out to be divisible by 3 instead of 6. Please let me know how I can move forward with this. Thank you!
| Let's look at 3 cases:
1) n is divisible by 3: (n+3) is divisible by 3 and one of (n+2) or (n+3) must be even and, therefore, divisible by 2. Being divisible by both 2 and 3, the expression is divisible by 6.
2) n is not divisible by 3 but (n-1) is: (n+2) is divisible by 3 and one of (n+2) or (n+3) must be even.
3) n is not divisible by 3 but (n-2) is: (n+7) is divisible by 3 and one of (n+2) or (n+3) must be even.
That covers all possible scenarios for n. Thus, the expression is always divisible by 6.
| {
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"source": "stackexchange",
"question_score": "3",
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} |
The sum of the solutions of $\sin\left(2x\right)=\frac{\sqrt{3}}{2}$ over the interval$ [–π, d]$ is$–π.$ find the value of $d$ The sum of the solutions of $\sin\left(2x\right)=\frac{\sqrt{3}}{2}$ over the interval$ [–π, d]$ is$–π.$ find the value of $d$
I think the first thing I must know is the number of solutions. I know that $x=\:\frac{\pi }{6},\frac{\pi }{3}$, but how can I deduce how many more? How can I utilize $-\pi$ is the domain? What is the general approach to these kinds of questions?
Multiple choice options:
The options are
$a) o$,
$b)\frac{\pi }{6}$,
$c)\frac{3\pi }{4}$,
$d)\frac{7\pi }{6}$,
$e) \frac{3\pi }{2}$
| The solutions in
$$[-\pi,\infty)$$ are
$$-\frac{5\pi}6,-\frac{2\pi}3,\frac\pi6,\frac\pi3,\frac{7\pi}6,\frac{4\pi}3,\cdots$$
and if you accumulate them you obtain
$$-\frac{5\pi}6,-\frac{3\pi}2,-\frac{4\pi}3,-\pi,\frac\pi6,\frac{3\pi}2\cdots$$
| {
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I do not know how to solve for Induction Conclusion A sequence has $x_1=8$, $x_2=32$, $x_{n}= 2x_{n-1}+3x_{n-2}$ for $n \ge 3.$ Prove, for all $i$ of Naturals, $X_i = 2 (-1)^i + 10 \cdot 3^{i-1}$
I got bases covered, and I got the inductive step as $X_{i} = 2 * (-1)^k + 10 * 3^{k-1}$. I do not know how to follow up with k+1, as I get complete gibberish. Any suggestions
Could someone guide me through each step?
My work so far:—
Base Cases include $x = 1, 2.$
$X_{1} = 2(-1)^1 + 10 \cdot 3^0$
$8 = 8$
$X_2 = 2(-1)^2 + 10 \cdot 3^{1}$
$32 = 32$
Inductive Step: $i = k$
$X_k = 2(-1)^k + 10 \cdot 3^{k-1}$
Inductive Conclusion:
$X_{k+1} = X_k + X_{k-1}$
$X_{k+1} = 2(2(-1)^k + 10 \cdot 3^{k-1}) + 3(2(-1)^{k-1} + 10 \cdot 3^{k-2})$
$X_{k+1} = 4(-1)^k + 6(-1)^{k-1} + 20 \cdot 3^{k-1} + 10 \cdot 3^{k-1}$
At this point. I know I should factor, but the 2nd half of $3^{k-1}$ does not create $10\cdot3^{k+1}$ as needed.
| Let $A_n = x_n-x_{n-2}$ and $B_n = x_n+x_{n-1}$
We're given that $$x_n=2x_{n-1}+3x_{n-2}$$
$$\implies x_n-x_{n-2}=2(x_{n-1}+x_{n-2})$$
$$\implies A_n=2B_{n-1}$$
Also, using the definitions of $A_n$ and $B_n$:
$$B_n - A_n = B_{n-1}$$
Thus,
$$B_n = 3B_{n-1}$$
So, for general $B_n$:
$$B_n = 3^{n-2}\cdot B_{2}$$
$$\implies B_n = 3^{n-2}\cdot 40$$
Now we can use $B_n$ to calculate $x_n$.
$$\implies x_n+x_{n-1} = 3^{n-2}\cdot 40$$
And, $$ x_{n+1}+x_{n} = 3^{n-1}\cdot 40$$
So, subtracting the above equations:
$$x_{n+1}-x_{n-1} = 80 \cdot 3^{n-2}$$
And so,
$$x_{n+1} = x_1 + 80 \cdot (1+3^2+3^4+...+3^{n-2})$$
This is true if $n$ is even.
If $n$ is odd,
$$x_{n+1} = x_2 + 240 \cdot (1+3^2+...+3^{n-3})$$
Thus, for odd n (summing up the geometric progression):
$$x_{n+1}=10 \cdot 3^n +2$$
And for even n:
$$x_{n+1}=10 \cdot 3^n-2$$
Which implies that $x_n=10 \cdot 3^{n-1} + 2 \cdot (-1)^n$.
| {
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Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi i}{n}}$
So the roots are $2^\frac{1}{3}, -(2^\frac{1}{3}) ~and~ 2^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1).$
Is this correct?
Note: I don't like factoring method. I studied in schools time. So don't help with that method.
Thanks!
| If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi ik}{n}}$
So the roots are
$$x_1=(2)^\frac{1}{3} e^\frac{2\pi .1}{3}=(2)^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1),$$
$$x_3=(2)^\frac{1}{3} e^\frac{2\pi .2}{3}=(2)^\frac{1}{3},$$
$$x_2=(2)^\frac{1}{3} e^\frac{2\pi .2}{3}=
(2)^\frac{1}{3}(-\frac{1}{2})(\sqrt{3}i+1).$$
| {
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(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.
| $$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$
Let $f(k)=\lfloor\log_2k\rfloor$. Since $\log$ is increasing we know that
*
*$f(2^0)=f(1)=0$
*$f(2^1)=f(2)=f(3)=1$
*$f(2^2)=f(4)=f(5)=f(6)=f(7)=2$
*$f(2^3)=f(8)=f(9)=\cdots=f(15)=3$
*$f(2^4)=f(16)=f(17)=\cdots=f(31)=4$
*etc.
So there are 2 ones, 4 twos, 8 threes, 16 fours, 32 fives, 64 sixes, 128 sevens, 256 eights, etc. We can multiply these to form the sequence $(2,8,24,64,160,384,896,2048)$. Summing the first 7 terms of this sequence gives $1538$, so we're missing $1994-1538=456$. Since $456/8=57<2048$ then our $n$ should be $2+4+8+16+32+64+128+57=311$, but we must also account for the $f(1)=0$ term, so in fact $n=312$.
| {
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Why substitution in irrational equation doesn't give equivalent equation? I have two examples of irrational equations:
The first example: $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$
In solution, they take cube of both sides and do following:
\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x}) +6+x=27\\
&\iff&
3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27 \iff \sqrt[3]{(3-x)(6+x)} = 2\\
&\iff&
x^2+3x-10=0\\
&\iff& x=2\quad \text{ or }\quad x= -5
\end{eqnarray*}
They conclude that both values are solutions, they satisfy the original equation.
The second example: $\sqrt[3]{x+1} + \sqrt[3]{3x+1} = \sqrt[3]{x-1}$
Here they do the following:
\begin{eqnarray*}
&\sqrt[3]{x+1}& + \sqrt[3]{3x+1} = \sqrt[3]{x-1}\\
&\iff&
x+1 + 3 \sqrt[3]{(x+1)(3x+1)} (\sqrt[3]{x+1} + \sqrt[3]{3x+1} ) + 3x+1 = x-1\\
&\implies&
3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3\\
&\iff&
(x^2-1)(3x+1)= -(x+1)^3\\
&\iff&
x= 0 \quad\text{ or }\quad x = -1
\end{eqnarray*}
but the only solutions is $x= -1$, because $0$ doesn't satisfy the equation.
What is difference between these examples, why do we have all equivalence signs in the first, and implication in the secondnd example?
The explanation in solution is that we have substituted $\sqrt[3]{3-x} + \sqrt[3]{6+x}$ by a number in the 1st example, and in the 2nd example we substituted by another expression which depends on x. I don't understand that, when does substitution give equivalent equation? Do we always have to verify if solutions satisfy the original equation in the end (is that the answer)?
Please help with this explanation, I need to understand better the irrational equations. Thanks a lot in advance.
| These problems don't happen if you'll use the following identity.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Since, $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we see that $\sum\limits_{cyc}(a-b)^2=0$ for $a=b=c$ only and it can give a extraneous root of the equation.
Now, we can solve your equations by using this idea.
*
*$$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$$
Here $a=\sqrt[3]{3-x},$ $b=\sqrt[3]{6+x}$ and $c=-3.$
Since $$\sqrt[3]{3-x}=\sqrt[3]{6+x}=-3$$ is impossible, our equation is equivalent to:
$$3-x+6+x-27-3\sqrt[3]{(3-x)(6+x)}(-3)=0$$ or
$$\sqrt[3]{(3-x)(6+x)}=-2,$$ which gives the answer:
$$\{2,-5\}$$
2. $$\sqrt[3]{x+1} + \sqrt[3]{3x+1} = \sqrt[3]{x-1}.$$
Since $$x+1=3x+1=-(x+1)$$ is possible for $x=0$ and $0$ is not a root of our equation, we need to remove this root before we'll write the answer.
Id est, we obtain:
$$x+1+3x+1-(x-1)+3\sqrt[3]{(x+1)(3x+1)(x-1)}=0$$ or
$$\sqrt[3]{(x+1)(3x+1)(x-1)}=-x-1$$ or
$$x^2(x+1)=0,$$ which gives the answer:
$$\{-1\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$ One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$
I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * $\frac{1}{(1+x)^5}$ * $\frac{1}{(1-x)^6}$
Then got $f(x) = x^3 * \sum_{k=0}^\infty \binom{n+4}{4}(-1)^kx^k * \sum_{k=0}^\infty \binom{n+5}{5}x^k$
I dont know what to do from here, can you tell me what is the next step?
| Here is a variation using the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.
We start with $f(x)$ and obtain for $n\geq 3$:
\begin{align*}
\color{blue}{[x^n]f(x)}&=[x^n]x^3\sum_{k=0}^\infty\binom{k+4}{4}(-1)^kx^k\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{1}\\
&=[x^{n-3}]\sum_{k=0}^\infty\binom{k+4}{4}(-1)^kx^k\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{2}\\
&=\sum_{k=0}^{n-3}\binom{k+4}{4}(-1)^k[x^{n-3-k}]\sum_{l=0}^\infty\binom{l+5}{5}x^l\tag{3}\\
&=\sum_{k=0}^{n-3}\binom{k+4}{4}(-1)^k\sum_{l=0}^\infty\binom{l+5}{5}[x^{n-3-k}]x^l\\
&\,\,\color{blue}{=\sum_{k=0}^{n-3}\binom{k+4}{4}\binom{n-k+2}{5}(-1)^k}\tag{4}
\end{align*}
Comment:
*
*In (1) we write the sums using different indices $k$ and $l$ which helps to not mix them up.
*In (2) we absorb $x^3$ by applying the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ again. Note the upper index is set to $n-3$, since other indices do not contribute to the coefficient of $x^{n-3}$.
*In (4) we select the coefficient of $x^{n-3-k}$ which implies choosing $l=n-3-k$. Here we use $[x^p]x^q=\begin{cases}1&p=q\\0&p\ne q\end{cases}$.
| {
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Is it valid to apply L'hopital rule to evaluate the limit? $$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{x^2+x}\times\left(\sqrt{x^2+x+1}+1\right)$$
Then it used equivalence and wrote:
$$\lim_{x\to 0^+}\frac{2x\left(\frac12x^2\right)}{x^2+x}\times\left(\sqrt{x^2+x+1}\right)=\lim_{x\to 0^+}\frac{x^3}{x(x+1)}.\left(\sqrt{x^2+x+1}\right)=0$$
I wonder why we should do all these calculation? Is it possible to use L'hopital rule and get $\frac0{\tfrac12}=0$ ?
| Short answer:
$$\tan x\sim x$$
and
$$\sqrt{x^2+x+1}-1\sim\frac x2.$$
Hence the factor $1-\cos x$ makes the limit tend to zero.
| {
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Infinite product of $\sqrt{2}$ I am struggling to show the following relation
$$
\prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2.
$$
I have tried to compute the sum
$$
\sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right),
$$
by using the expansion for $\log(1+x)$, however, I was not able to evaluate the double sum. Furthermore, I tried to square and reorder (although it should not be possible), but haven't quite got the right track.
Could someone give me a hint for this problem?
| It follows from the Euler/Weierstrass factorisation for sine: for $x \in \mathbb{R}$
$$\frac{\sin{x}}{x} = \prod_{j \ge 1} \left(1-\frac{x^2}{\pi^2 j^2}\right)$$
Separating the odd even parts we can rewrite your product as:
$$\begin{aligned} \mathcal{P} & = \prod_{k \ge 1 } \left(1 - \frac{(-1)^k}{2k-1}\right) \\& = \frac{1}{2}\prod_{k \ge 0} \bigg(1-\frac{1}{4k-1}\bigg)\bigg(1+\frac{1}{4k+1}\bigg) \\& = \frac{1}{2} \prod_{k \ge 0} \frac{16k^2-4}{16k^2-1} \\& = 2 \prod_{k \ge 1} \frac{16k^2-4}{16k^2-1} \\& = 2 \prod_{k \ge 1}\frac{\bigg(1-(\dfrac{1}{2})^2/k^2\bigg)}{\bigg(1-(\dfrac{1}{4})^2/k^2\bigg)} \\& = 2 \cdot \frac{\sin{\frac{1}{2} \pi}}{\frac{1}{2}\pi} \bigg/\frac{\sin{\frac{1}{4} \pi}}{\frac{1}{4}\pi} \\& = 2 \cdot \frac{1}{\sqrt{2}} \\& = \sqrt{2}. \end{aligned}$$
| {
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Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $ Solve the ODE with the initial conditions:
$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$
I did the substitution:
$$
y’ = z
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$
Putting in the ODE:
$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = 1 \Rightarrow \frac{dz}{dy} = \frac{1}{y^2} \Rightarrow dz = \frac{1}{y^2}dy
$$
By integration we get:
$$
z = -\frac{1}{y} + C \Rightarrow y’ = - \frac{1}{y} + C
$$
Using the initial conditions we get:
$$
1 = -\frac{1}{1} + C \Rightarrow C = 2
$$
Therefore:
$$
y’ = -\frac{1}{y} + 2
$$
Now I am not sure how to continue.
| $$y’ = -\frac{1}{y} + 2$$
$$y’ = -\frac{1}{y} + \dfrac {2y}y$$
$$y’ = \frac{2y-1}{y}$$
$$\dfrac {y}{2y-1}dy = dx$$
Then integrate both sides.
$$\int \dfrac {y \, dy}{2y-1} = \int dx$$
Substitute $u=2y-1 \implies du=2dy$
and $y=\dfrac {u+1}2$
$$\dfrac 14 \int \dfrac {u+1}{u}du = \int dx$$
$$\dfrac 14 \int \left (1+\dfrac {1}{u} \right ) du = \int dx$$
| {
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I need to compute the integral $\int _0^s\:x^2$ using Riemann sums. I need to compute the integral $\int _0^s\:x^2$ using Riemann sums.
a) Consider the function $y=x^2$ on the interval [0, s] and divide it into n equal intervals. Each of these subintervals has length $\frac{s}{n}$.
b) Find the sum $U_n$ of all rectangles below the function $y=x^2$
$$U_n=\frac{s}{n}\left(0^3+\left(\frac{s}{n}\right)^3+\left(2\frac{s}{n}\right)^3+...+\left(n-1\right)\left(\frac{s}{n}\right)^3\right)= \frac{s^3}{n^3}\sum _{k=0}^{n-1}\:k^3$$
c) Now find the sum $O_n$ of all rectangles above the function $y=x^2$
$$O_n=\frac{s^3}{n^3}\left(1^3+2^3+3^3+...+n^3\right)=\frac{s^3}{n^3}\sum _{t=1}^nk^3$$
d) Take the limits to show that:
$$\lim _{n\to \infty }\left(U_n\right)=\lim \:_{n\to \:\infty \:}\left(O_n\right)$$
So My answer is:
$$U_n=\frac{s^3}{n^3}\sum _{k=0}^{n-1}\:k^3=\frac{s^3}{n^3}((\frac{n\left(n+1\right)}{6})-n^3$$
Therefore $U_n=\frac{s^3}{n^3}=\lim _{n\to \infty }\left(\frac{s^3}{3}+\frac{s^3}{2n}-\frac{s^3}{n}\right)=\frac{s^3}{3}$
Is it correct? I'd appreciate some guide and feedback if I am wrong.
| First a comment. If you assume that $x \mapsto x^2$ is Riemann integrable, you don't need to compute both upper and lower Riemann sums and verify that those are equal. This would be required to prove the Riemann integrability.
Now regarding the computation itself. You have
$$U_n=\frac{s}{n}\left(0^2+\left(\frac{s}{n}\right)^2+\left(2\frac{s}{n}\right)^2+...+\left(n-1\right)\left(\frac{s}{n}\right)^2\right)= \frac{s^3}{n^3}\sum_{k=0}^{n-1} k^2.$$
The power is $2$ and not $3$ as you're integrating $x \mapsto x^2$.
Using
$$\sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}$$ and taking the limit you get the right answer $$\int_0^s x^2 \ dx = \frac{s^3}{3}.$$
| {
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$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" way. So, I transform both complex numbers into trigonometric form and divide:
$$\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} = \frac{2[\cos(\frac{7}{4}\pi)+\sin(\frac{7}{4}\pi)i]}{2[\cos(\frac{5}{6}\pi)+\sin(\frac{5}{6}\pi)i]}= \cos \left(\frac{11}{12}\pi \right)+\sin \left( \frac{11}{12}\pi \right)i$$
Now, I see that this is $165^{\circ}$, and that module $= 1$. The problem is, that the only way I see to find the number of elements of the X set is to calculate degrees:
$$165^{\circ}, 330^{\circ}, 135^{\circ}, 300^{\circ}, 105^{\circ}, 270^{\circ}, 75^{\circ}, 240^{\circ}, 45^{\circ}, 210^{\circ}, 15^{\circ}, 180^{\circ}, 345^{\circ}, 150^{\circ}, 315,^{\circ} 120^{\circ}, 285^{\circ}, 90^{\circ}...$$
So I feel like there has to be some more effective way to solve that problem and here comes my question. Do you know any?
| A way is to find a polynomial satisfied by $z$. (For low degrees, this is likely to be a minimal polynomial. To show this requires computing elements of $X$ successively and searching for a linear relation among the powers of $z$ produced so far. The first such relation is the minimal polynomial. All we can actually promise is that we get a polynomial multiple of a minimal polynomial.) \begin{align*}
\frac{z}{1+\mathrm{i}\sqrt{3}}
&= \frac{\sqrt{2}(1+\mathrm{i})}{1+3} \\
4z &= \sqrt{2}(1+\mathrm{i})(1+\mathrm{i}\sqrt{3}) \\
16z^2 &= 2(1+\mathrm{i})^2(1+\mathrm{i}\sqrt{3})^2 \\
8z^2 &= 2\mathrm{i}(1+\mathrm{i}\sqrt{3})^2 \\
4z^2 &= \mathrm{i}(-2 + 2\mathrm{i}\sqrt{3}) \\
2z^2 &= \mathrm{i}(-1 + \mathrm{i}\sqrt{3}) \\
4z^4 &= \mathrm{i}^2(-1 + \mathrm{i}\sqrt{3})^2 \\
-4z^4 &= (-2 - 2\mathrm{i}\sqrt{3}) \\
-2z^4 &= (-1 - \mathrm{i}\sqrt{3}) \tag{$\star$} \\
4z^8 &= (-1 - \mathrm{i}\sqrt{3})^2 \\
4z^8 &= (-2 + 2\mathrm{i}\sqrt{3}) \\
2z^8 &= (-1 + \mathrm{i}\sqrt{3}) \\
2z^8 - 2z^4 &= -2 & & \text{using ($\star$)} \\
z^8 - z^4 + 1 &= 0 \\
z^8 &= z^4 - 1 \text{.}
\end{align*}
(Using other tools, we can show that this is the minimal polynomial of $z$, but this is not critical.)
So, $\require{enclose}$
$$ X = \{1,z, z^2, \dots, z^7, z^4-1, z^5-z, \dots, z^8 - z^4 = -1, -z, -z^2, \dots, -z^7, -z^4+1, -z^5+z, \dots, \enclose{horizontalstrike}{-z^8+z^4 = 1}\} \text{,} $$
a set of only $24$ elements.
We haven't explicitly shown that the norm of $z$ is $1$, it is implicit in the fact that the powers of $z$ are cyclic, equivalently, $X$ is finite.
| {
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"timestamp": "2023-03-29T00:00:00",
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how to prove transitive relation for this one? $R=\{(x,y) \mid 3 |(x-y^2) \}$ while R is relation on set A=$\mathbb{Z}$
I need to check if it's reflective , symmetric , transitive .
I figured already that it's not reflective or symmetric with (2,1) or (10,2),(2,10)
and I don't how to check if it's transitive or not.
I think we need to prove that, we have $(x,y), (y,c)$ thus $(x,c) $ then $3|(x-y^2) = 0$ and
3|$(y-c^2) = 0$ but $3|(x-c^2) \neq 0$
| If $3|x-y^2$ and $3|y-z^2$ does it follow that $3|x-z^2$?
$3|x-y^2\implies z\equiv y^2 \pmod 3$ and $3|y-z^2\implies y\equiv z^2 \pmod 3$.
So if $x\equiv y^2$ and $y\equiv z^2$ then $x \equiv z^4\equiv \pmod 3$.
Must it follow that $z^4 \equiv z^2 \pmod 3$?
If $z\equiv 0$ then $z^2 \equiv z^4 \pmod 3$ and, yes, $x\equiv y^2\pmod 3$ and $y\equiv 0^2\pmod 3\implies x\equiv 0\equiv z^2 \pmod 3$.
If $z \equiv \pm 1$ then $1\equiv z^2\equiv z^4 \pmod 3$ and yes $x\equiv y^2 \pmod 3$ and $y\equiv (\pm 1)^2 \equiv 1\pmod 3 \implies y^2 \equiv 1 \pmod 3$ and so $x\equiv y^2 \equiv 1\equiv z^2 \pmod 3$.
So yes, it is transitive.
......
Alternatively. If $3|x-y^2$ then we have $x-y^2 = 3k$ for some $k$. And $3|y-z^2$ then we have $y-z^2 = 3j$ for some $j$. So $y=3j-z^2$.
So $x-y^2 = x-(9j^2 - 6jz^2 + z^4)= 3k$
So $x-z^2 = 3k + 9j^2 - (6j+1)z^2 + z^4=$
$3(k + 3j^2 -2jz^2) +z^4-z^2$.
so $3|x-z^2$ if and only if $z^4-z^2 = z^2(z^2 - 1)=z^2(z+1)(z-1)$ is divisible by $3$.
But either $z, z+1$ or $z-1$ must be divisible by $3$ as they are three consectutive integers and one of them must be divisible by $3$.
So $3|z^2(z+1)(z-1)$.
| {
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Guessing pattern of Picard's iteration of ODE Given the following initial value problem $x'(t) = x(t) + e^t$ with $x(0) = 0$. I've calculated its picard's iteration, but having trouble guessing the pattern: $$\begin{cases}
x_1 = e^t - 1, \\[3pt]
x_2 = 2e^t - t - 2, \\[3pt]
x_3 = 3e^t - 2t - 3 - \frac{t^2}{2}, \\[3pt]
x_4 = 4e^t - 3t - 4 - t^2 - \frac{t^3}{6}, \\[3pt]
x_5 = 5e^t - 4t - 5 - \frac{3t^2}{4} - \frac{t^3}{3} - \frac{t^4}{24}, \\[3pt]
x_6 = 6e^t - 5t - 6 - 2t^2 - \frac{t^3}{2} - \frac{t^4}{12} - \frac{t^5}{120},...
\end{cases} $$
So maybe we have something like $$x_n(t) = n(e^t - 1) - (n-1)t - \sum_{k = 2}^{n-1} \frac{t^k}{??}$$
Since the solution is actually $x(t) = t e^t$, I also don't see how the $n$ in the first two elements should disappear as $n \to \infty$, because otherwise, it diverges.
Any hint/help would be appreciated. Thanks.
| If you expand with Maclaurin $$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+O\left(t^6\right)$$
and plug this in the, let's say, $6$th step, you get
$$6 \left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}\right)-\frac{t^5}{120}-\frac{t^4}{12}-\frac{t^3}{2}-2 t^2-5 t-6+O\left(t^6\right)=\\=t+t^2+\frac{t^3}{2}+\frac{t^4}{6}+\frac{t^5}{24}+O\left(t^6\right)$$
the last one is the expansion of $te^t$
| {
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Probably $x_1 \leq x_3 \leq x_2$ From set ${1,2...n}$ we choose three random different numbers, $x_1, x_2,x_3$. If $x_1 \le x_2$, find probability $x_1 \le x_3 \le x_2$.
I set A is event $x_1 \leq x_3 \leq x_2$, and B $x_1 \leq x_2$, and find P(AB)=$\frac{1}{6}$, P(B)=$\frac{1}{2}$. Is it okay?
| The problem has two different solutions, depending on whether you take them with replacement or without it (your question is quite ambiguous in that sense: the word 'different' suggests that they are taken without replacement, and the sign '$\leq$' - that they are taken with replacement):
Solution for numbers taken without replacement:
$$P(x_1 < x_2) = \frac{1}{2}$$
$$P(x_1 < x_3 < x_2) = \frac{1}{6}$$
$$P(x_1 < x_3 < x_2| x_1 < x_2) = \frac{1}{3}$$
Solution for numbers taken with replacement:
$$P(x_1 \leq x_2) = P(x_1 = x_2) + P(x_1 < x_2) = P(x_1 = x_2) + \frac{1 - P(x_1 = x_2)}{2} = \frac{1}{n} + \frac{n-1}{2n} = \frac{n+1}{2n}$$
$$P(x_1 \leq x_3 \leq x_2) = P(x_1 < x_3 < x_2) + P(x_1 = x_3 < x_2) + P(x_1 < x_3 = x_2) - P(x_1 = x_3 = x_2) = \frac{1 - P(x_1 = x_3 < x_2) - P(x_1 < x_3 = x_2) + P(x_1 = x_3 = x_2) }{6} + P(x_1 = x_3 < x_2) + P(x_1 < x_3 = x_2) - P(x_1 = x_3 = x_2) = \frac{1}{6} + \frac{5}{6}P(x_1 = x_3 < x_2) + \frac{5}{6}P(x_1 < x_3 = x_2) - \frac{5}{6}P(x_1 = x_3 = x_2) = \frac{1}{6} + \frac{5(n+1)}{12n^2} + \frac{5(n+1)}{12n^2} - \frac{5}{6n^2} = \frac{n+5}{6n}$$
$$P(x_1 \leq x_3 \leq x_2| x_1 \leq x_2) = \frac{n+5}{6n}\frac{2n}{n+1} = \frac{n+5}{3(n+1)}$$
| {
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Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried to subtract one equation from the other, but got nothing. No idea what to do.
| We have following divisibility rules :
*
*$\overline{abcdef} \equiv \overline{def} - \overline{abc} \pmod 7$
*$\overline{abcdef} \equiv a-b+c-d+e-f \pmod {11}$
*$\overline{abcdef} \equiv \overline{def} - \overline{abc} \pmod {13}$
So one obtains
$$ \overline{xyz} \equiv 5 \pmod 7$$
$$ x+y+z \equiv 6 \pmod {11}$$
$$ \overline{xyz} \equiv 1 \pmod {13}$$
Can you take it from here?
| {
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Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.
| let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$
$$xy+yz+zx=18 \tag{given}$$
$$\color{red}{x+y+z\ge 3\sqrt{6}} \tag1$$
notice that $${(x-\sqrt{6})}^2\ge 0\Rightarrow \sqrt{3+x^2}\ge \frac{\sqrt{6}}{3}x+1$$
Thus $$\sum \sqrt{3+x^2}\ge \sum \frac{\sqrt{6}}{3}x+1\ge\frac{\sqrt{6}}{3}\color{red}{3\sqrt{6}}+3=9$$
Note the inequality marked $(1)$ is left as an exercise
| {
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Consider a stick, cut twice, probability the smallest is 1/5 Consider a stick of unit length, you take two points (uniform independent), and then break the stick on these two points. You now got 3 segment. What is the probability the smallest segment is less than 1/5?
$X, Y$ be the two cuts.
$P(X < 1/5 \text{ and } Y-X < 1/5 \text{ and } (1-Y-X)<1/5). $
| Let $P(S)$ be the probability that the smallest segment is less than $\frac{1}{5}$; let $X$ be the location of the first cut, and $Y$ be the location of the second cut.
If $X < \frac{1}{5}$ or $X > \frac{4}{5}$, then you already know that one of the segments has length less than $\frac{1}{5}$, so the conditional probability $P\big(S \vert X \in [0,\frac{1}{5}) \cup (\frac{4}{5},1]\big)=1$.
If $\frac{2}{5} < X < \frac{3}{5}$, then the possible locations where the second cut could facilitate $S$ are the regions of length $\frac{1}{5}$ on either side of $X$, and at either end of the stick, all of which are non-overlapping.
Therefore, the conditional probability $P(S \vert X \in (\frac{2}{5},\frac{3}{5}]) = \frac{4}{5}$.
Moving $X$ left of $\frac{2}{5}$, you linearly lose area in which $Y$ can land to facilitate $S$, until right before $X=\frac{1}{5}$, the good region is only length $\frac{3}{5}$. The same argument follows on the other side.
Combining all of these results, I end up with a piecewise function for the conditional probability of $P(S|X)$:
$$
P(S \vert X) = \begin{cases}
1 & 0 \leq X < \frac{1}{5} \\
\frac{2}{5} + X & \frac{1}{5} \leq X < \frac{2}{5} \\
\frac{4}{5} & \frac{2}{5} \leq X < \frac{3}{5} \\
\frac{7}{5} - X & \frac{3}{5} \leq X < \frac{4}{5} \\
1 & \frac{3}{5} \leq X \leq 1
\end{cases}
$$
and by integrating that function over $X$, I find that $P(S)=\frac{21}{25}=84\%$.
| {
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How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone out there will have a super smart elegant solution to this!
Base step: test when n = 1 ...
Assume true for $n = 1$
ie . $12^k - 4^k - 3^k +1 = 6M$, where $m$ is an integer
RTP: also true for $n = k+1$
ie. $12^{k+1} - 4^{k+1} - 3^{k+1} +1 = 6N$ where $N$ is an integer
LHS: $12^{k+1} - 4^{k+1} - 3^{k+1} +1$
$= 12( 4^k + 3^k - 1 + 6M) - 4^{k}(4) - 3^{k}(3) +1$ (from assumption)
$= 6(12M) + 12(4^k) + 12(3^k) -12 -4^{k}(4) - 3^{k}(3) +1$
Here is where I break down and go around in circles.
| Work mod 3, and put aside induction for now. Note that $4 \equiv_3 1$, and so for each integer $n$:
$$4^n \equiv_3 1^n = 1.$$
Thus $12^n -4^n -3^n +1 \equiv_3 (0 -1 -0 -1) = 0$.
So this quantity is divible by 3 for each positive integer $n$.
Lets now get back to mod 6 though. So $12^n -4^n -3^n +1$ is divisible by 3. But it is also even so it is divisible by 6 as well.
| {
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Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| $$(x^2 + 5x + 5-1)(x^2 + 5x + 5+1) - 48 = (x^2 + 5x + 5)^2-7^2$$
| {
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the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$
Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$
Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$
Now I can use the squeeze theorem and get:
*
*$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$
*$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$
And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?
Unfortunately, I can not use integrals while doing that exercise.
| If you are familiar with generalized harmonic numbers.
The expression you are considering is
$$a_n=\frac{1}{\sqrt{n}}\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}$$ and, for large $p$, we have
$${H_p^{\left(-\frac{1}{2}\right)}}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta
\left(-\frac{1}{2}\right)+\frac{1}{24 \sqrt{p}}+O\left(\frac{1}{
p^{5/2}}\right)$$
$$\frac{p}{H_p^{\left(-\frac{1}{2}\right)}}=\frac {3}{2\sqrt{p}}-\frac {9}{8p\sqrt{p}}+\cdots$$
$$\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}=3 \sqrt{n}+\left(\frac{3 \zeta \left(\frac{1}{2}\right)}{2}-\frac{9 \zeta
\left(\frac{3}{2}\right)}{8}\right)+O\left(\frac{1}{n^{1/2}}\right)$$
$$a_n=3+\left(\frac{3 \zeta \left(\frac{1}{2}\right)}{2}-\frac{9 \zeta
\left(\frac{3}{2}\right)}{8}\right)\frac 1 {\sqrt n}+O\left(\frac{1}{n^{3/2}}\right)$$ which shows the limit and how it is approached.
| {
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"url": "https://math.stackexchange.com/questions/3918907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx} = \_\_\_\_\_\_\_$.
My approach is as follow
$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
y=h
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$
How will I proceed from here
| Continuing from your method $$f'(x)=\lim_{h\to 0} \frac{f(h)+1 -(e^{-x}-1)(e^{-h}-1))}{h}$$ for derivative to exist $f(0)=-1$ hence by L-Hospitals rule $$f'(x)=\lim_{h\to 0}\frac{f(h)+1}{h}-\lim_{h\to 0}(e^{-x}-1)\frac{(e^{-h}-1)}{h}$$
$$f'(x)=f'(0)+e^{-x}-1=e^{-x}+C\Rightarrow f(x)=-e^{-x}+Cx$$
Now for the given limit to exist $f'(1)+f(0)=e^{-1}$...hence find $C$...
| {
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"timestamp": "2023-03-29T00:00:00",
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We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
What I Tried: Here is a picture :-
Let $AC = CD = x$. As $\angle ACD = 90^\circ$, we have $AD$ the diameter of the circle, so $AD = 2r$.
From here, using Pythagorean Theorem :- $$2x^2 = 4r^2$$
$$\rightarrow x = r\sqrt{2}$$
We have the green angles to be $45^\circ$, now as $ADCB$ is cyclic, $\angle ABC = 135^\circ$ and each of the brown angles are $22.5^\circ$. So we can use :-
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\rightarrow a \div \frac{\sqrt{2 - \sqrt{2}}}{2} = r\sqrt{2} * \frac{2}{\sqrt{2}}$$
$$\rightarrow \frac{2a}{\sqrt{2 - \sqrt{2}}} = 2r$$
$$BC = a = r\sqrt{2 - \sqrt{2}}$$
However, the answer to my question is given as $r\sqrt{\sqrt{2}}$ , so where did I go wrong?
| The given answer is wrong and you are correct. $ABC$ happens to be two consecutive sides of a regular octagon.
| {
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"source": "stackexchange",
"question_score": "3",
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Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.
Any help would be appreciated.
| Another method, because it wasn't mentioned, is using Lagrange's trigonometric identities (which should not be difficult to prove using complex numbers, like here)
$$\sum _{k=1}^{n}\sin(k\theta )=
{\frac {1}{2}}\cot {\frac {\theta }{2}}-{\frac {\cos \left(\left(n+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac {\theta }{2}}\right)}}=...$$
with $\theta=\frac{1}{n^2}$. Given $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$$...={\frac {1}{2}}\cot {\frac {\theta }{2}}-\frac{1}{2}\left(\cos\left(n\theta\right)\cot\frac{\theta}{2}-\sin(n\theta)\right)=\\
\frac{1}{2}\cdot\left(\sin(n\theta)+\cot\frac{\theta}{2}\cdot\left(1-\cos\left(n\theta\right)\right)\right)=\\
\frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cot\frac{1}{2n^2}\cdot\left(1-\cos\frac{1}{n}\right)\right)=...$$
considering that $1-\cos{a}=2\sin^2\frac{a}{2}$
$$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cot\frac{1}{2n^2}\cdot \sin^2\frac{1}{2n}\right)=...$$
also considering $\color{red}{\lim\limits_{x\to0}\frac{\sin x}{x}=1}$
$$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cos\frac{1}{2n^2}\cdot \frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\cdot \frac{\sin^2\frac{1}{2n}}{\frac{1}{4n^2}}\cdot\frac{\frac{1}{4n^2}}{\frac{1}{2n^2}}\right)=\\
\frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cos\frac{1}{2n^2}\cdot \color{red}{\left(\frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\right)}\cdot \color{red}{\left(\frac{\sin\frac{1}{2n}}{\frac{1}{2n}}\right)^2}\right)\to \frac{1}{2}, n\to\infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Generalisation of IMO 1990/P3:For which $b $ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?
For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?
It was from my LTE/ Zsigmondy handout.
By taking examples, it looks like for $b= 2^k-1 , 2$ it's not true .
Here's my progress:
*
*I got $b=4,5,6,8,9$ works ( $2,3,7$ doesn't )
*$n$ is odd : If not then $4\mid b^n-1$ , but $b^n \equiv 0,1 \mod 4$
*If $b+1$ is a power of $2$ and $n^2\mid b^n+1$, then $n=1$ : Let $p$ smallest prime dividing $n$ ( note that it can't be $2$) . Then $n^2\mid b^{2n}-1 \implies p\mid b^{2n}-1 \implies p\mid b^{\gcd(2n,p-1)}-1 \implies p\mid b^2-1\implies p|b-1 \text{(since $b+1$ is power of $2$)}\implies p\mid b^n+1 \implies p|2.$ Contradiction.
Any hints? Thanks in advance!
| We can construct b, we want:
$b^n+1\equiv 0 \mod (n^2)$
If $b=t.n^2-1$ and $n =2k+1$ is odd, then we have:
$(t.n^2-1)^n+1=M(n^2)-1+1=M(n^2)\equiv 0 \ mod (n^2)$
Where $M(n)$ is a multiple of $n^2$.
The condition is that n must be odd.
| {
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"source": "stackexchange",
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Perfect Square Number, prime I want to show that $p^{p+1}+(p+1)^p$ is not perfect square number. ($p$ is prime number)
This is what I do
Assume $p^{p+1}+(p+1)^p=k^2$.
$(p+1)^p=(k+p^{\frac {p+1} 2})(k-p^{\frac {p+1} 2})$.
Let $2ab=p+1, gcd(a,b)=1$.
$k+p^{\frac {p+1} 2}=2^{p-1}a^{p}, k-p^{\frac {p+1} 2}=2b^p$.
Therefore $p^{\frac {p+1} 2}=|2^{p-2}a^p-b^p|$.
By Fermat's little theorem, $a\equiv 2b \pmod{p}$, and $2ab=p+1$, so $a=2b$. In this case I can show contradiction.
but, if $a=1$ I don't know how show contradiction.
| Your proof is quite faulty (mathematically speaking, you omitted 2-3 cases i think), so I will reweite it entirely.
$$p^{p+1}+(p+1)^p=k^2\Leftrightarrow (p+1)^p=\big(k-p^{\frac{p+1}{2}}\big)\big(k+p^{\frac{p+1}{2}}\big)$$
Notice that $\text{gcd}(p+1;p)=1$ and $\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)|2\cdotp^{\frac{p+1}{2}}$, thus leading to $$\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)|2$$
$(p+1)^p$ is even so at least one of $k-p^{\frac{p+1}{2}}$ and $k+p^{\frac{p+1}{2}}$ is even, thus $$\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)=2$$
Note that case $p=2$ does not work, so from now on we will only work with $p\geq 3$
We can now safely assume that we have $2$ cases:
$1)$ $k-p^{\frac{p+1}{2}}=2\cdot a^p$ and $k+p^{\frac{p+1}{2}}=2^{p-1}\cdot b^p$ with $2ab=p+1$
$2)$ $k-p^{\frac{p+1}{2}}=2^{p-1}\cdot a^p$ and $k+p^{\frac{p+1}{2}}=2\cdot b^p$ with $2ab=p+1$
(notice that $a$ abd $b$ are not divisible by $p$)
Case $1)$: Using Fermat's little theorem, we get $k\equiv2a\pmod{p}$ and $k\equiv b\pmod{p}$ so $2a\equiv b\pmod{p}$ but $2ab\equiv 1\pmod{p}$ so either $b=1$ and $a=\frac{p+1}{2}$ or $b=p-1$ and $a=\frac{p+1}{2(p-1)}$.
Clearly when $a=\frac{p+1}{2(p-1)}$ we get a contradiction (unless $p=3$, which is a special case, we can easily see that $4^3+3^4$ is not a perfect square)
So $b=1$ and $a=\frac{p+1}{2}$ But $k=2^{p-1}b^p-p^{\frac{p+1}{2}}$ so if $b=1$ then $2^{p-1}> p^{\frac{p+1}{2}}$ which is clearly false for all primes $p$.
So we only get contradictions in this case.
Case $2):$ Again, using the same method we get $2b\equiv a\pmod{p}$ so with the same reasoning $a=1$ and $b=\frac{p+1}{2}$
We know $k=2\cdot a^p+p^{\frac{p+1}{2}}=2^{p-1}\cdot b^p-p^{\frac{p+1}{2}}$ so we get $$2+p^{\frac{p+1}{2}}=2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p-p^{\frac{p+1}{2}}$$ which can we rewritten as $$2(1+p^{\frac{p+1}{2}})=2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p$$
We assumed $p\geq 3$ so $p\equiv 1$ or $3\pmod{4}$
If $p\equiv 1\pmod{4}$, $1+p^{\frac{p+1}{2}}\equiv 2\pmod{4}$ so $$v_2(2(1+p^{\frac{p+1}{2}})=2$$ but $$v_2(2(1+p^{\frac{p+1}{2}})=v_2(2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p)\geq p-1$$ giving us $$3\geq p$$ so contradiction; we assumed $p\equiv 1\pmod{4}$
If $p\equiv 3\pmod{4}$ then $\frac{p+1}{2}$ is even, so $p^\frac{p+1}{2}\equiv 1\pmod{4}$ (any odd perfect square is $1\pmod{4}$) so we have again that $1+p^{\frac{p+1}{2}}\equiv 2\pmod{4}$ so $$v_2(2(1+p^{\frac{p+1}{2}})=2$$ giving us the same result, $3\geq p$, which would lead to $p=3$, which we established is a contradiction.
Thus, there is no prime $p$ for which $p^{p+1}+(p+1)^p$ is a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
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Drawing numbers from a set, quantile We have a set containing $20$ numbers from $1$ to $20$. Each time we draw only one number, and repeat it $15$ times (without replacement). Let's denote $X-$ the largest drawn number. Find the smallest ${{16} \choose {15}}/{20 \choose 15}$-fractile of a random variable $X$.
So, we look for such $x_p$ that $F(x_p)\geq {{16} \choose {15}}/{20 \choose 15}.$ If $X$ is maximum of all drawn numbers, then $F_X(t)=\Bbb P(X\leq t)=\Bbb P(X_1\leq t, ...,X_m\leq t)$ and, if all of $X_1,...X_m$ are iid then we have $\Bbb P(X_1\leq t)\cdot...\cdot\Bbb P(X_m\leq t)$. But... how to proceed?
| There are $\binom {20} {15}$ ways to choose 15 numbers from the set.
$\binom{16}{15}$ is the number of ways to choose 15 numbers only from $1$ to $16$ inclusive. The maximum of such subset is at most $16$. So the cumulative probability up to and including $t=16$ is the $\binom {16} {15} / \binom {20} {15}$.
Alternatively, consider breaking down to individual probabilitites $\Pr(X=t)$, where $15\le t \le 20$.
To choose a subset with a maximum of $t$, the subset will have 14 elements between $1$ and $t-1$. The number of subsets with a maximum $t$ is $\binom {t-1}{14}$.
Verify that the total number of subset is $\binom {20}{15}$: (the hockey-stick identity)
$$\sum_{t=15}^{20}\binom{t-1}{14} = \sum_{t'=14}^{19}\binom{t'}{14} = \binom {20}{15}$$
And similarly the $\binom {16}{15}$ is the sum
$$\sum_{t=15}^{16}\binom{t-1}{14} = \sum_{t'=14}^{15}\binom{t'}{14} = \binom {16}{15}$$
And so
$$\begin{align*}
\Pr(X \le 16) &= \sum_{t=15}^{16} \Pr(X = t)\\
&= \sum_{t=15}^{16} \frac{\binom{t-1}{14}}{\binom {20}{15}}\\
&= \frac{\binom{16}{15}}{\binom {20}{15}}
\end{align*}$$
| {
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Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
I'm having trouble solving this equation. I've tried squaring both sides and got this
$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$
after simplifying
$$19 - 6\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} = a^2$$
and that's where I got stuck.
| Hint:
$11=(\sqrt7)^2+2^2$ and $\sqrt7-2>0$
$8=(\sqrt7)^2+1^2$ and $\sqrt7-1>0$
Finally
$$\sqrt{a^2+b^2-2ab}=|a-b|=a-b\text{ if } a-b\ge0$$
| {
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Closed form for continued fraction I am working with a recursive algorithm and I have realized that in each step it computes something equivalent to:
$A_{0}=\frac{a}{b}$
$A_{1}=a/(b-\frac{a^{2}}{b})$
$A_{2}=a/(b-\frac{a^{2}}{b-\frac{a^{2}}{b}})$
$\vdots$
$A_{n+1}=a/(b-aA_{n})$
And so on. I was wondering if there could exist some closed form for $A_{n}$ to avoid unnecessary iterations
| The closed form does exist, I found it with Mathematica and it's amazing, to use in a program I mean.
I found it with this function
RSolve[{A[n + 1] == a/(b - a A[n]), A[0] == a/b}, A[n], n]
Set:
$$
\begin{cases}
D=\sqrt{b^2-4 a^2}\\
x=\frac{b-D}{a}\\
y=\frac{b+D}{a}\\
\end{cases}
$$
Then
$$A_n=\frac{a (b+D ) y^n-a (b-D ) x^n}{x^n \left(2 a^2-b (b-D )\right)+y^n \left(b (b+D )-2 a^2\right)}$$
You will surely find a way to optimize better this formula in order to use it in a program.
Edit
This is the formula as output by Mathematica
$$A_n=\frac{a \left(b \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+(-b) \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n\right)}{-2 a^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n-b^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+2 a^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n}$$
| {
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Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
And I'm trying to find a invertible matrix $Q$ such that $AQ$ = $B$, but I'm stuck.
Can you help me?
I have not seen determinats yet
| From linear independence of the rows we see that $A$ is invertible and so $Q = A^{-1}B$. You can use adjugate matrix and determinant to compute $A^{-1}$
$$
A^{-1} = \frac{{\rm adj} A}{\det A} = \frac{\begin{pmatrix}
1/3 & 1/3 & 0 \\
-1/3 & 1/3 & 1/3 \\
0 & -1 & 0
\end{pmatrix}}{\frac{1}{3}} =
\begin{pmatrix}
1 & 1 & 0 \\
-1 & 1 & 1 \\
0 & -3 & 0
\end{pmatrix}.
$$
Alternatively, you can find $A^{-1}$ using Gaussian elimination. Finally, we get
$$
Q = A^{-1} B = \begin{pmatrix}
1 & 0 & 1 \\
0 & 2 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$
| {
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USAJMO 2017 P4: triples $(a,b,c)$ such that $(a-2)(b-2)(c-2)+12$ is a prime number that divides the positive number $a^2+b^2+c^2+abc-2017$?
Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2)+12$ is a prime number that properly divides the positive number $a^2+b^2+c^2+abc-2017$?
My progress: Let $x=a-2$, $y=b-2$, $z=c-2$, then we get $xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 \implies xyz+12 \mid (x+y+z-41)(x+y+z+49)$. So $xyz+12 \mid (x+y+z-41)$ or $xyz+12|(x+y+z+49)$ .
Also $xyz+12 $ can't divide both or else then $xyz+12 |90$ ( which doesn't work ).
WLOG $x\ge y\ge z$, then clearly $x\ge 9$ , since $11^2\cdot 3+ 11^3\le 2017$ ( not satisfying properly divide property).
Also I got $x,y,z$ odd or not divisible by $1 \mod 3$, or else $(a-2)(b-2)(c-2)+12$ is not prime.
| If we continue from here: $\;\;\;xyz+12 \mid x+y+z-41\;\;\;$ or $\;\;\;xyz+12|x+y+z+49$.
We can assume $x\leq y\leq z$. We see that $x,y,z\notin \{0,2,3,4,6,8,9\}$ since $xyz+12$ is prime.
*
*If $x>1$ then $x,y\geq 5$ and so $xyz+12\geq 25z+12$ so in
*
*first case: $$25z+1\leq |x+y+z-41|\leq 3z+41\implies z\le 1$$ and thus no solution.
*second case: $$25z+1\leq x+y+z+49\leq 3z+49\implies z\leq 2$$ so no solution again.
*If $x=1$ then $\;\;\;yz+12 \mid y+z-40\;\;\;$ or $\;\;\;yz+12|y+z+50$.
*
*If $y\neq 1$ then $y\geq 5$. In first case we have $5z+12\leq 2z+40\implies z\leq 9$ so $z=7$ (and $y=5$ or $y=7$) or $z=5$ (and $y=5$). None works. In second case we have $5z+12\leq 2z+50\implies z\leq 12$. So $z\in\{11,7,5\}$ but none works.
*If $y=1$ then $\;\;\;z+12 \mid z-39\;\;\;$ then $z+12\mid 51 \implies z=5$ or $\;\;\;z+12|z+51$, then $z+12\mid 39$ so $z=1$.
*If $x=-1$ then we proccede similary like in a previous case...
| {
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"url": "https://math.stackexchange.com/questions/3942754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
I need to solve $12x-30y+24z=18$ equation as a diophantine equation where $x,y,z$ is whole number I know i can divide with $6$ so I get $2x-5y+4z=3$, I replaced $2x-5y$ with $u$ so I got $u+4z=3$. I see that $u=3$ and $z=0$ is a solution but I got stuck. How can I continue? Is there a way to solve this using extended euclidean algorithm?
Thank you for your help.
| Experiments in a spreadsheet show that $x$ $y$ $z$ combinations only add up to a given value if we increment $y$ by $4$ and $z$ by $5$ for a given value of $x$. Further experimentation shows offsets of $1$ and $2$, respectively to make the result zero.
There are infinite solutions starting with $x\in\mathbb{Z}$ and
$$y = 4 n + 2 x + 1,\qquad z = 5 n + 2 x + 2\qquad n \in Z$$
Here are samples with $x\in\{-2,-1,0,1,2,3\}$
For $n=-1$
\begin{equation}
\cdots\quad
(-2,-7,-7)\quad
(-1,-5,-5)\quad
(0,-3,-3)\quad
(1,-1,-1)\quad
(2,1,1)\quad
(3,3,3)\quad\cdots
\end{equation}
For $n=0$
\begin{equation}
\cdots\quad
(-2,-3,-2)\quad
(-1,-1,0)\quad
(0,1,2)\quad
(1,3,4)\quad
(2,5,6)\quad
(3,7,8)\quad\cdots
\end{equation}
For $n=1$
\begin{equation}
\cdots\quad
(-2,1,3)\quad
(-1,3,5)\quad
(0,5,7)\quad
(1,7,9)\quad
(2,9,11)\quad
(3,11,13)\quad\cdots
\end{equation}
for $n=2$
\begin{equation}
\cdots\quad
(-2,5,8)\quad
(-1,7,10)\quad
(0,9,12)\quad
(1,11,14)\quad
(2,13,16)\quad
(3,15,18)\quad\cdots
\end{equation}
For $n=3$
\begin{equation}
\cdots\quad
(-2,9,13)\quad
(-1,11,15)\quad
(0,13,17)\quad
(1,15,19)\quad
(2,17,21)\quad
(3,19,23)\quad\cdots
\end{equation}
and so on.
| {
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"question_score": "1",
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Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$ I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4} }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4} } |<ϵ.$$
I have this:
$\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4}} \cdot\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}} $
= $\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac{\sqrt[3]{2}}{6} $
=$ \dfrac{6\cdot \sqrt[3]{2+h}-6\cdot \sqrt [3]{2}-\sqrt[3]{2}\cdot h}{6h} $
=$ \left |\dfrac{1}{(\sqrt[3]{2+h})^2 + \sqrt[3]{2+h}\sqrt[3]{2}+(\sqrt[3]{2})^2 } - \dfrac {\sqrt[3]{2}} {6}\right | $
=$ $
I don't know how continue for use the fact $|h|<\delta$
The original exercise says: $f(x)=\sqrt[3]{x}$. Found $\displaystyle\lim_{h→0} \dfrac {f(2+h)-f(2)}{h}$
| Have you tried multiplying top and bottom of $\frac{\sqrt[3]{2+h}-\sqrt[3]{2}}{h}$ by $(2+h)^{\frac{2}{3}}+\sqrt[3]{2}\cdot\sqrt[3]{2+h}+2^{\frac{2}{3}}$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from there to apply the Euler's theorem
| Induction:
\begin{align*}2^{2^{6(k+1)}\cdot 4}=2^{2^{6k+6}\cdot 4}=\left(2^{2^{6k}\cdot 4}\right)^{64}&\equiv (2^4) ^{64}\mod 19\\&\equiv 2^{256}\mod19\end{align*}
By Fermat's little theorem, $2^{18}\equiv 1\mod 19$. Since $256=18*14+4$, then
\begin{align*}2^{2^{6(k+1)}\cdot 4}&\equiv (2^{18})^{14}2^4\mod19\\&\equiv 1^{14}2^4\mod 19\\&\equiv 2^4\mod19\end{align*}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$ The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$
My attempt
I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.
| The denominator of $1+e\cos\theta$ occurs in inverse $r^2$ force problems and when $e\gt1$ as here, corresponds to a hyperbolic orbit, like Rutherford scattering or ʻOumuamua. We can zap this denominator depending on whether the orbit is elliptical, parabolic, or hyperbolic. In the latter case we may let
$$\sin y=\frac{\sqrt{e^2-1}\sinh\theta}{e\cosh\theta-1}=\frac{\sqrt3\sinh\theta}{2\cosh\theta-1}$$
$$\cos y=\frac{e-\cosh\theta}{e\cosh\theta-1}=\frac{2-\cosh\theta}{2\cosh\theta-1}$$
The reader may check that $\cos^2y+\sin^2y=1$. Taking differentials we have
$$\cos y\,dy=\frac{e-\cosh\theta}{e\cosh\theta-1}dy=\frac{\sqrt{e^2-1}(e-\cosh\theta)}{(e\cosh\theta-1)^2}$$
So
$$dy=\frac{\sqrt{e^2-1}}{e\cosh\theta-1}d\theta=\frac{\sqrt3}{2\cosh\theta-1}d\theta$$
And then
$$1+e\cos y=1+2\cos y=\frac{e^2-1}{e\cosh\theta-1}=\frac3{2\cosh\theta-1}$$
$$\cosh\theta=\frac{\cos y+e}{1+e\cos y}=\frac{\cos y+2}{1+2\cos y}$$
$$\sinh\theta=\frac{\sqrt{e^2-1}\sin y}{1+e\cos y}=\frac{\sqrt3\sin y}{1+2\cos y}$$
So we can just clean up the denominator a little by letting $x=y+\frac{\pi}6$ so that $\sin\left(x+\frac{\pi}3\right)=\sin\left(y+\frac{\pi}2\right)=\cos y$ and $\cos x=\cos\left(y+\frac{\pi}6\right)=\frac{\sqrt3}2\cos y-\frac12\sin y$ so we have
$$\begin{align}I&=\int\frac{\cos x+\sqrt3}{\left(1+2\sin\left(x+\frac{\pi}3\right)\right)^2}dy=\int\frac{\frac{\sqrt3}2\cos y-\frac12\sin y+\sqrt3}{(1+2\cos y)^2}dy\\
&=\int\frac{(2\cosh\theta-1)^2}9\left[\frac{\sqrt3}2\frac{(2-\cosh\theta)}{(2\cosh\theta-1)}-\right.\\
&\quad\left.\frac{\sqrt3}2\frac{\sinh\theta}{(2\cosh\theta-1)}+\sqrt3\frac{(2\cosh\theta-1)}{(2\cosh\theta-1)}\right]\frac{\sqrt3\,d\theta}{(2\cosh\theta-1)}\\
&=\int\left(\frac12\cosh\theta-\frac16\sinh\theta\right)d\theta=\frac12\sinh\theta-\frac16\cosh\theta+C_1\\
&=\frac{\frac{\sqrt3}2\sin y-\frac16\cos y-\frac13}{1+2\cos y}+C_1=\frac{\frac{\sqrt3}2\sin y-\frac14}{1+2\cos y}+C\\
&=\frac{\frac34\sin x-\frac{\sqrt3}4\cos x-\frac14}{1+2\cos\left(x-\frac{\pi}6\right)}+C=\frac{\frac34\sin x-\frac{\sqrt3}4\cos x-\frac14}{1+2\sin\left(x+\frac{\pi}3\right)}+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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How do I find a plane that is tangent to two given spheres and passes through a given point? My problem is the following:
Given two spheres: $$(x-6)^2+(y-1)^2+(z+1)^2=1$$ and $$x^2+(y-5)^2+(z-4)^2=9$$ find a plane that is tangent to both of those spheres and passes through the point $$(5;2;0)$$
I tried plugging all the information in formulas but then I get a scary-looking system of equation which I don't know how to solve.
My problem book says that the correct answer is $$x+2y+2z-9=0$$ or $$y-2=0$$
Whatever I did I couldn't get the answer.
I would really appreciate any help.
Thanks in advance!
| $C_1(6,1,-1);\;r_1=1$
$C_2(0,5,4);\;r_2=3$
A generic plane $\pi$ has equation
$$\pi:ax+by+cz+d=0$$
The distance from the plane to the center of the spheres must be equal to the respective radius. Furthermore the plane must pass through $P(5,2,0)$
$$\begin{cases}
\frac{|6a+b-c+d|}{\sqrt{a^2+b^2+c^2}}=1\\
\frac{|5b+4c+d|}{\sqrt{a^2+b^2+c^2}}=3\\
5a+2b+d=0&\\
\end{cases}$$
$$\begin{cases}
(6 a+b-c+d)^2=\frac{1}{9} (5 b+4 c+d)^2\\
(6 a+b-c+d)^2=a^2+b^2+c^2\\
d=-5a-2b\\
\end{cases}$$
Two solutions $a=0,b=-d/2,c=0;\;a=-d/9,b=-2 d/9,c=-2 d/9$
Two planes
$-d/2 y + d = 0\to y-2=0$
$-d/9 x -2d/9 y -2 d/9 z + d=0 \to x+2 y+2 z-9=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
2D random walk - first hits A 2D symmetric random walk $(X_k,Y_k)$ $k\ge 0$ (Markov chain) where $(X_{k+1},Y_{k+1})$ takes one value of the next ones: $(X_k,Y_k+1),(X_k,Y_k-1),(X_k+1,Y_k),(X_k-1,Y_k)$, all have the same probability and the initial values of X and Y are zero.
a) Find the value of $P(X_T=3,Y_T=0)$ and $E[T]$ if $T=inf(k\ge 0:\left| X_k \right|+\left| Y_k \right|=3)$
b) Find the value of $P(X_T=-2,Y_T=0)$ and $E[T]$ if $T=inf(k\ge 0:max(-X_k,\left| Y_k \right|=2))$
Any help is greatly appreciated!
| I am not sure my answer for part a) is right, just for your reference.
a) First, it is important to observe that the point $(X_{T-2}, Y_{T-2})$ must be one of the four:
\begin{align}
(1,0),\,(-1,0),\,(0,1),\,(0,-1).\tag{1}
\end{align}
Now, if $(X_T, Y_T)=(3,0)$, then $(X_{T-2}, Y_{T-2})=(1,0)$. Consider (1), and by symmetry,
\begin{align}
P((X_{t-2}, Y_{T-2})=(1,0))=\frac{1}{4}.
\end{align}
On the other hand, if we start the random walk at $(X_{T-2}, Y_{T-2})=(1,0)$, then the point $(X_T,Y_T)$ must be one of the following five:
\begin{align}
\begin{array}{ll}
(1,2)&\mbox{with $1$ path},\\
(2,1)&\mbox{with $2$ paths},\\
(3,0)&\mbox{with $\color{red}{1}$ path},\\
(2,-1)&\mbox{with $2$ paths},\\
(1,-2)&\mbox{with $1$ path}.
\end{array}\tag{2}
\end{align}
Since each path has equal probability, we get that
\begin{align}
P(X_T=3, Y_T=0)=P((X_{t-2}, Y_{T-2})=(1,0))\cdot\frac{\color{red}{1}}{1+2+1+2+1}=\frac{1}{28}.
\end{align}
To calculate $E[T]$, we omit the detail, just note that $T\ge 3$, $T$ must be odd, and moreover, $(X_3,Y_3), (X_5,Y_5),\ldots,(X_{T-2},Y_{T-2})$ are one of the following four:
\begin{align}
(1,0),\,(-1,0),\,(0,1),\,(0,-1).\tag{3}
\end{align}
From the above observations, we may obtain
\begin{align}
P(T=3)&=\frac{28}{4^3}=\frac{7}{16},\\
P(T=5)&=\frac{9}{16}\cdot\frac{7}{16},\\
P(T=7)&=\left(\frac{9}{16}\right)^2\cdot\frac{7}{16},\\
&\vdots
\end{align}
and therefore,
\begin{align}
E[T]&=\sum_{n=3,5,7,\ldots} nP(T=n)\\
&=\frac{7}{16}\left(3+5\cdot\frac{9}{16}+7\cdot\left(\frac{9}{16}\right)^2+\cdots\right)\\
&=\frac{39}{7}.
\end{align}
| {
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Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$
Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$
Attempt. Using ratio test for $a_n:=\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}>0$, we have:
$$\frac{a_{n+1}}{a_n}=\frac{\frac{2}{3}+n+1}{\frac{3}{2}+n+1}\to 1,$$
so we do not conclude. Also $\sqrt[n]{(1+a)\ldots(n+a)}\to +\infty$ for $a>0$, so the root test also is inconclusive.
Thanks for the help.
| Let us use Gauss's test.
$$\frac{a_n}{a_{n+1}} = \frac{n+\frac52}{n+\frac53} = 1 + \frac{\frac56}{n(1+\frac{5}{3n})} = 1 + \frac{\alpha}{n} + \underline{O}(n^{-2})$$
with $\alpha = \frac{5}{6} < 1$. Hence $\sum_n a_n$ diverges.
| {
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What is $\left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$? I did
\begin{align}
& \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t(t^2+2)k}{(t^2+2)^2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t^3k+4tk}{(t^2+2)^2} \\[8pt]
= {} & \frac{-4tj+4tk}{(t^2+2)^2}
\end{align}
Is this true?
| $(\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$
$\frac {1}{(t^2 + 2)^2} ((-2(t^2+2)+4t^2) \mathbf i + (4t^2+t^2(t^2+2))\mathbf j + (-2t^3-4t) \mathbf k)$
$\frac {1}{(t^2 + 2)^2} ((2t^2 - 4) \mathbf i + (t^4+6t^2) \mathbf j -(2t^3+4t) \mathbf k)$
$\frac {2t^2 - 4}{(t^2 +2)^2} \mathbf i + \frac {t^4+6t^2}{(t^2 + 2)^2} \mathbf j - \frac {2t^3+4t}{(t^2 + 2)^2} \mathbf k$
| {
"language": "en",
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How to show that $\frac{a + b}{\gcd(a,b)^2}$ is a Fibonacci number when $ \frac{a+1}{b}+\frac{b+1}{a}$ is an integer?
Let $a, b$ be positive integers such that the number $ \dfrac{a + 1}{b} + \dfrac{b + 1}{a}$ is also integer. Then, show that $\dfrac{a + b}{\gcd{(a, b)^{2}}}$ is a Fibonacci number.
I prove that : $ \displaystyle{\forall a, b \in \mathbb{N}} $ such that $ \displaystyle{ k = \frac{a + 1}{b} + \frac{b + 1}{a} \in \mathbb{N} \Rightarrow k = 3 }$ or $ \displaystyle{ k = 4 } $. Also, we have that (I use the method Vieta Jumping) :
$ \displaystyle{ \mathbb{S} = \Big \{ \big( a, b \big) \in \mathbb{N} \times \mathbb{N} : \frac{a + 1}{b} + \frac{b + 1}{a} = k \in \mathbb{N} \Big \} = \Big \{ \big( a, b \big) \in \mathbb{N} \times \mathbb{N} : \frac{a + 1}{b} + \frac{b + 1}{a} = 3 \text{ } or \text{ } \frac{a + 1}{b} + \frac{b + 1}{a} = 4 \Big \} } $
Fibonacci sequence : $\displaystyle{f_{0} = 0, f_{1} = 1}$ and $\displaystyle{f_{n+2} = f_{n+1} + f_{n}, \forall n \in \mathbb{N} \cup \{ 0 \} = \{ 0, 1, 2, 3, . . . \}}$
$ \displaystyle{ k = 3 } $ : If $ \displaystyle{ \big( a, b \big) \in \mathbb{S} } $ with $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 3 } $, then $ \displaystyle{ \big( 3 \cdot a - 1 - b, a \big) \in \mathbb{S} } $ so we have $ \displaystyle{ \big( 2, 2 \big), \big( 3 = f_{3} + 1, 2 \big), \big( 6 = f_{5} + 1, 3 \big), \big( 14 = f_{7} + 1, 6 \big), \big( 35 = f_{9} + 1, 14 \big), . . . \in \mathbb{S} } $ (I checked some couples and I found that the request is valid) but I can not porve that $ \displaystyle{ \frac{a + b}{\gcd{\big(a, b \big)^{2}}} } $ is a Fibonacci number.
$ \displaystyle{ k = 4 } $ : If $ \displaystyle{ \big( a, b \big) \in \mathbb{S} } $ with $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 4 } $, then $ \displaystyle{ \big( 4 \cdot a - 1 - b, a \big) \in \mathbb{S} } $ so we have $ \displaystyle{ \big( 1, 1 \big), \big( 2, 1 \big), \big( 6, 2 \big), \big( 21, 6 \big), \big( 77, 21 \big), . . . \in \mathbb{S} } $ (I checked some couples and I found that the request is valid) but also I can not porve that $ \displaystyle{ \frac{a + b}{\gcd{\big(a, b \big)^{2}}} } $ is a Fibonacci number.
I need some help, because I can not solve the problem. So my questions are :
if $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 3 } $ where $ \displaystyle{ a = f_{2 \cdot n - 1} + 1, b = f_{2 \cdot n + 1} + 1 \in \mathbb{N}, \forall n \geqslant 2, \forall n \in \mathbb{N} } $, how I can show that $ \displaystyle{ \frac{a + b}{\gcd{\big( a, b \big)^{2}}} } $ is a Fibonacci number ?
and
if $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 4} $ where $ \displaystyle{ a, b \in \mathbb{N} } $, how I can show that $ \displaystyle{ \frac{a + b}{\gcd{\big( a, b \big)^{2}}} } $ is a Fibonacci number ?
| First of all we assume that $gcd(a,b)=d $ so we may assume that there exist coprime integers $x,y$ such that $a=dx , b=dy$ then from the hypothesis of the problem: $ab|a^2+b^2+a+b
\rightarrow ab|(a+b)(a+b+1)
\rightarrow dxy|(x+y)(dx+dy+1)$
because $d$ is coprime to $dx+dy+1$ , $d|x+y$ (first result) and because $x,y$ are coprime it is easy to conclude that $xy$ is coprime to $x+y$ and therefore $xy|dx+dy+1$ (second result) now instead of the first result, assume that there is a positive integer $t$ such that $y=td-x$ then the second result will be $x(td-x)|td^2+1 $.
Now assume a positive integer $k$ such that $td^2+1=ktdx-kx^2$ which is the equation $kx^2-ktdx+(td^2+1)=0$ and by multiplying both sides by $4k$, $4k^2x^2-4k^2tdx+(4ktd^2+4k)=0$ and this is actually $(2kx-ktd)^2=(k^2t^2d^2-4k-4ktd^2) $.
From this result, the existance of such $x$ is clearly equivalent to these two:
i) $k^2t^2d^2-4k-4ktd^2$(we'll call this $r$) is a perfect square
ii) $2k|\sqrt{r}+ktd$. Now just let's say that we choose $k$ to be odd. Then this condition will be changed to the following two conditions:
*
*$2|\sqrt{r}+ktd$ which is clear because $\sqrt{r}$ has the same parity with $r$ and $r$ has the same parity with $k^2t^2d^2$ so actually $\sqrt{r}$ has the same parity with $k^2t^2d^2$ and thus, $\sqrt{r}+ktd$ is even because $ktd+k^2t^2d^2$ is even, so the first condition is satisfied.
*$k|\sqrt{r}+ktd$ iff $k|\sqrt{r}$ iff $k^2|r$ iff $k^2|4k+4ktd^2$ iff $k|4+4td^2$ but because we've chosen $k$ to be odd, that's equivalent to $k|td^2+1$
So, summing up; if we choose $s$ to be a number that satisfies $td^2+1=ks$, it's enough to prove that $r=4k^2s-k^2t^2d^2$ is a perfect square, which is again equivalent to $4s-t^2d^2$ being a perfect square.
So, for avoiding any confusion here I restate the new problem:
We have positive integers that satisfy $td^2+1=ks$ and $4s-t^2d^2$ is a perfect square and want to prove that $t$ is a fibbonachi number.
Now $k$ doesn't have a role anymore so instead of $td^2+1=ks$, we say that $s|td^2+1$.
Now again assume $u$ to be a number that $4s-t^2d^2=u^2$.
Mixing the two hypotheses, we'll have $u^2+t^2d^2=4s|4td^2+4$.Now because RHS shall be bigger than or equal to the LHS, $4td^2+4\leqslant u^2+t^2d^2$.
Now considering $t$ as our variable,
$t^2d^2-4td^2+(u^2-4)$ is less than or equal to zero.
On the other hand that is actually $4+4d^2-u^2\leqslant(t-2)^2d^2$.
Now two important notes here:
*$d\not=1$ because then the problem's original statement leads to a contradiction so I assume it meant $d\not=1$.
*$u\leqslant2$:
If $u=0$ , then back to the $u^2+t^2d^2|4td^2+4$, $t|t^2d^2|4td^2+4$ and thus $t|4$ but if $t=4$ then $16d^2|16d^2+4$ a contradiction. Otherwise $t=1or2$ which are both fibbonachi numbers.
If $u=1$, back to the definition of $u$, $4s=t^2d^2+1$ which is a contradiction $mod4$. So $u\leqslant2$.
Now back to our inequality, $4+4d^2-u^2\leqslant(t-2)^2d^2$. This is equivalent to $sqrt{\frac{4+4d^2-u^2}{d^2}}+2\leqslant t$
Now by the second note we just mentioned, the LHS is maximized when u=2 , and so clearly $2\leqslant t$ and so $t$ is a fibbonachi number.
In the same way you can consider that $k$ is even and get to the numbers {3,5} and so all of the possible values of $t$ are {1,2,3,5} which are all fibbonachi numbers.
| {
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Proof by induction for a sequence, given $a_1$ and $a_2$ Question: prove by induction that this statement holds.
Define the sequence $(a_{k}), k∈\mathbb N ,$
by $a_1 = 2, a_2 = 5 $
and
$a_{k+2} = 5a_{k+1} − 6a_{k} $ for all $k≥1$. Then $a_{k} =2^{n−1} +3^{n-1}$ for all $n≥1 $.
What I did was find the base case -- $ P(1)=2 $ -- holds
Then made an assumption that $P(k)$ holds. Thus we can assume $P(k+1)$ and I got that
$= 5×2^{k−1} +5×3^{k−1} −3×2^{k−1} −2×3^{k−1}.$
Where do I go from here?
| Say $P(n)$ is $a_n=2^{n-1}+3^{n-1}$.
Base: $P(1): a_1=2=2^0+3^0$, $P(2): a_2=5=2^1+3^1$.
Now assuming $P(k)$ and $P(k+1)$, we have
$P(k+2): a_{k+2}=5a_{k+1}-6a_k=5(2^k+3^k)-6(2^{k-1}+3^{k-1})$
$=(5-3)2^{k}+(5-2)3^{k}=2^{k+1}+3^{k+1}.$
| {
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Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$
What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?
All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am stuck and would like to know how we would proceed with this.
Also how would we approach this problem if repetition was allowed ?
| We'll find sums at each place (units, tens, etc.) and add them.
At thousands place each of $1,2,3$ occurs $6$ times. Sum is $6(1+2+3)\cdot 1000=36000$.
For hundreds place, we first look at numbers with thousands digit $1$. These are six numbers in which each of $2,3,0$ occurs twice. Hence in the $18$ nos, each of $1,2,3$ occurs $4$ times while $0$ occurs $18-3\cdot 4=6$ times as expected. Sum at hundreds place is $4(1+2+3)\cdot 100=2400$.
Sum at tens and units places are similar - $240$ and $24$ respectively.
Desired sum is $36000+2400+240+24=38664$.
The repetition case is exactly similar and probably easier.
We have $3\cdot 4^3$ nos in all, with $4^3$ starting with $1,2,3$ each. Sum at thousands place is $4^3(1+2+3)\cdot 1000$
For hundreds place, lets fix any digit as $$\square \, 0 \, \square \, \square$$ There'll be $3\cdot 4^2=48$ occurences of the fixed digit. Thus sum at hundreds place is $48(0+1+2+3)\cdot 100$.
Similarly at tens and units place - $48(0+1+2+3)\cdot 10$ and $48(0+1+2+3)\cdot 1$ resp.
Desired sum is $$(1+2+3)(64000 + 48\cdot 111) = 415968$$
| {
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Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.
What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?
| $x^2 - 1 = (x+1)(x-1)$.
So you have $\sqrt{(x+1)(x-1)} =(x+5) \sqrt\frac {x+1}{x-1}$.
Now my first thoughts are dividing both sides by $\sqrt{x+1}$ and multiplying both sides by $\sqrt{x-1}$ to get $x-1 = x+5$.
BUT BE CAREFUL!
First off. When we divide both sides by $\sqrt{x+1}$ we are assuming that $\sqrt{x+1}\ne 0$. We must consider the case where $\sqrt{x+1} = 0$ so $x = -1$ as a potential solution.
Indeed, if $x = -1$ we have $\sqrt{x^2 -1} =0$ and $(x+5) \sqrt{\frac {x+1}{x-1}} = 0$. So $x=-1$ is a solution.
And we could divide both sides by $\sqrt{x+1}$.... except....
$\frac {\sqrt{(x+1)(x-1)}}{\sqrt{x+1}} \ne \sqrt{x+1}$ and $\sqrt{\frac {x+1}{x-1}}{\sqrt{x+1}}\ne \frac 1{\sqrt{x-1}}$ unless both $x+1$ and $x-1$ are positive. If they are both negative the terms we are refering to simply are not designed.
So we could two cases where we consider both $x+1, x-1$ are positive and a case were they are both negative, in which case we divide both sides by $\sqrt{-(x+1)}$ and get $\sqrt{1-x} = (x+5)\frac 1{\sqrt {1-x}}$.
But more sophisticatedly we shouldn't divide and multiply in two steps. We should simply multiply both sides by $\sqrt{\frac{x-1}{x+1}}$.
In this case we still must make the case where $x+1 = 0$ for a potential solution.
So if $x = -1$ that is a solution. We did that above.
If $x \ne -1$ multiply bot sides by $\sqrt{\frac{x-1}{x+1}}$ to get:
$\sqrt{(x^2-1)\frac {x-1}{x+1}}=(x+5)\sqrt 1$ or
$\sqrt{(x-1)^2} = x+5$ or
$|x-1| = x+ 5$.
(Note my initial thought of $x-1$ being positive was erroneous and we get $|x-1| = x+5$ not $x-1=x+5$)
If $x -1 \ge 0$ we get $x-1=x+5$ which has no solutions.
If $x -1<0$ we get $1-x = x+5$ or $x=-2$.
So our two solutions are $x=-1$ and $x=-2$ both of which fit in the domain that you calculated.
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Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!
| From the equation solve for $\sqrt{x^2-1}$ and get
$$\sqrt{x^2-1} = \frac{x}{\frac{35}{12}-x}$$
now raise to the square and get an equation of degree $4$. Only $2$ of its roots will be roots of the initial equation, since they also include the roots of $x-\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$.
Worth plotting the graph of the function $x + \frac{x}{\sqrt{x^2-1}}$ on $(1, \infty)$. It it convex, with a minimum at $\sqrt{2}$, which is in between the roots $\frac{5}{4}$ and $\frac{5}{3}$ of our equation.
| {
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Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$ The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \cdot 3 \cdot x^4}{2^2 \cdot 2! } + \frac{1 \cdot 3 \cdot 5 \cdot x^6}{2^3 \cdot 3!} +... \right)$$
Here I run into a problem. How come all the terms are negative while $1$ isn't? This would mean I can't add it to the infinite sum. So continuing:
$$\sqrt{1-x^2} = 1- \sum_{n=0}^\infty \frac{(2n+1)!!}{2^n n!}x^{2n} = 1-\sum_{n=0}^\infty \frac{(2n+1)!!}{(2n)!!}x^{2n} $$
When I plot this on desmos my approximation is a) shifted by $1$ unit down b) inaccurate. On the image below I have used 50 terms so I doubt the error is due to a lack of terms.
Where have I made a mistake. Thank you
| For $n\ge2$,$$[x^{2n}]\sqrt{1-x^2}=(-1)^n\binom{1/2}{n}=(-1)^n\frac{1}{n!}\frac12\prod_{k=1}^{n-1}(\tfrac12-k)=-\frac{(2n-3)!!}{n!2^n},$$so$$\sqrt{1-x^2}=1-\tfrac12x^2-\sum_{n\ge2}\frac{(2n-3)!!}{n!2^n}x^{2n}.$$See how that looks in Desmos.
| {
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Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$
This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$
I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.
| Wouldn't be integration by parts much easier? Start with
$$\frac12\int x^2\cdot2x\sqrt{x^2-9}\,dx.$$
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All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$ Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.
I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$
I think that a geometric solution might be possible, please help me to proceed.
Thanks.
| Let $f(x) = \sqrt{1-x^{2}}-p x-q$, then the condition becomes $$|f(x)| \leq \frac{\sqrt{2}-1}{2}$$
Thus, the extrema of $f$ should lie in $\left[ -\frac{\sqrt{2}-1}{2}, \frac{\sqrt{2}-1}{2}\right]$. Now, let's see how its extrema look like:
$$f'(x) = \frac{-x}{\sqrt{1-x^2}} - p = 0 $$ After bringing to a common denominator, numerator should be equal to zero: $$ -x - p\sqrt{1-x^2} = 0 \\[2mm] x = r\sqrt{1-x^2}, \ \ (r=-p) \\[2mm] x^2 = r^2(1-x^2) \\[2mm] (1+r^2)x^2 = r^2 \\[2mm] x_0 = \pm \frac{r}{1+r^2} = \mp\frac{p}{\sqrt{1+p^2}}$$
However, since $x \ge 0$, we get $$x_0 = -\frac{p}{\sqrt{1+p^2}} $$
Therefore, $$\begin{align}f(x_0) &= \sqrt{1-\frac{p^2}{1+p^2}} + \frac{p^2}{\sqrt{1+p^2}}-q \\[1mm]&= \frac{1 + p^2}{\sqrt{1+p^2}} -q \\[1mm]&= \sqrt{1+p^2}-q \end{align}$$
Hence, the condition seems to be satisfied for all $p$ and $q$ such that
$$\left| \sqrt{1+p^2} -q \right| \le \frac{\sqrt2 - 1}{2} \tag{1}$$
Also, we need to see the edge cases when $x = 0$ and $x = 1$, which give
$$|f(0)| = \left|1 - q\right| \le \frac{\sqrt{2}-1}{2} \tag{2}$$ and $$|f(1)| = \left|-p - q\right| \le \frac{\sqrt{2}-1}{2} \tag{3}$$ Finally, combining $(1)$, $(2)$ and $(3)$ we get the final answer.
I didn't solve the last three inequalities putting all together, but after graphing I noticed (but still needs to be justified rigorously by solving the inequalities) that only one pair satisfies the statement, namely, $$(p,q) = \left(-1, \frac{\sqrt 2 + 1}{2}\right)$$
Below is the graph with those $p$ and $q$:
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Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ This problem $5b$ from Chapter $14$ of Spivak's Calculus:
Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ (Note $g$ is not assumed to be continuous at $0$).
I think this question is impossible? First of all there's no restrictions on $x$, so letting $x = 1$ and $x = -1$, we get $0 = 2$.
So maybe we have to restrict to $x \geq 0$? In that case, we can take the square root:
Find $g$ such that $\forall x \geq 0 : \int_0^{x}tg(t)dt = x + \sqrt{x}$
Let $f(t) = tg(t)$, so we need $\int_0^{x}f(t)dt = x + \sqrt{x}$
Now $\frac{d}{dx} (x + \sqrt{x}) = 1 + \frac{1}{2\sqrt{x}}$. But $f(t) = 1 + \frac{1}{2\sqrt{t}}$ won't work because $1 + \frac{1}{2\sqrt{t}}$ isn't bounded as $t \rightarrow 0$, so the integral won't be defined.
Am I missing something?
| The function $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3}$$ seems to satisfy the condition. Below is how to find it:
Let $G(x) = \int xg(x)dx$. Then, we have
$$\int_0^{x^2}tg(t)dt \overset{\text{(FTC)}}= G(x^2) - G(0) \overset{*}= x + x^2
$$
Now, differentiate both sides of $(*)$ to get
$$\begin{align} 2x[x^2g(x^2)] &= 1 + 2x \\[1mm] 2x^3g(x^2) &= 1+ 2x \\ g(x^2) &= \frac{1+2x}{2x^3}.\end{align}$$ Therefore, we get $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3} $$
Indeed,
$$\int_0^{x^2} t \frac{1 + 2\sqrt t}{2(\sqrt t)^3} dt = \int_0^{x^2}\left( 1 + \frac{1}{2\sqrt t}\right)dt = \left[\sqrt t + t\right]_0^{x^2} = x + x^2$$ as desired.
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How many rectangles can be found in this shape?
Question: How many rectangles can be found in the following shape?
My first solution:
Let $a_n$ be the number of rectangles in such a shape with side length $n$. Then, the number of the newly added $1 \times k$ rectangles while expanding the shape with side length $n-1$ to the shape with side length $n$ is, $n^2$. Similarly, the number of the newly added $2 \times k$ rectangles while expanding is, $(n-2)^2$, the number of the $3 \times k$ ones is $(n-4)^2$, and so on. So, we have, $$a_n = a_{n-1} + 1^2+3^2+\ldots +n^2$$ or, $$a_n = a_{n-1} + 2^2+4^2+\ldots +n^2$$ depending on the parity of $n$. In either case, we have $$a_n=a_{n-2}+1^2+2^2+ \ldots n^2$$ So, $a_{n+1} = a_{n-1} + 1^2+2^2+ \ldots (n+1)^2$. Subtracting both sides, we have, $a_{n+1}-a_n = a_{n-1} - a_{n-2} + (n+1)^2$. Repeating the same method, we obtain $$a_{n+1}-4a_n+5a_{n-1} = 5a_{n-3} - 4a_{n-4}+a_{n-5}$$ This recurrence relation has characteristic equation $r^6-4r^5+5r^4-5r^2+4r-1=0 \iff (r-1)^5 \cdot (r+1) = 0$. So, using the usual ways to solve recurrence relations, we can obtain $a_n=\frac{n^4}{24} + \frac{n^3}{4} + \frac{11n^2}{24} + \frac{n}{4} = \binom{n+3}{4}$.
However, since the answer is so simple, I tried to solve it with a method which can give the result in a more 'direct' way:
Let us find these rectangles by choosing 2 rows and 2 columns and looking at their intersection. If we choose the right most column, we have $n$ choices for the other column and only $1$ choice for the rows. If we choose column which is the second from right, we have $n-1$ choices for the other column (in order not to over count), and, $\binom{2+2-1}{2}$ (number of ways to choose 2 rows out of 2 rows including repetitions) choices for the row. Using this argument, we can obtain, $$a_n = \sum_{k=0}^{n-1} (n-k) \cdot \binom{(k+1)+2-1}{2} = \sum_{k=0}^{n-1} (n-k) \cdot \binom{k+2}{2}=\sum_{k=0}^{n-1} \frac{(n-k) \cdot (k+2) \cdot (k+1)}{2}$$ After some computation, we can see from here that $a_n=\binom{n+3}{4}$.
However, I think this solution is also overcomplicated and I think that there is a really elegant and/or 'direct' solution for this problem. So, my question is, how can this problem be solved with a 'direct' method (for example, a combinatorial method which immediately gives the number as $\binom{n+3}{4}$)? It would be nice to see your 'nice' solutions too, thank you.
| Searching ${n+3 \choose 2}$ at OEIS you can find in A000332 the reference to A004320 and there the document Counting the lattice rectangles inside Aztec diamonds and square biscuits by Teofil Bogdan and Mircea Dan Rus and the solution to problem 3, on page 3. Basically, for each rectangle you have to choose four coordinates for the vertices $(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2)$ with $x_1 \lt x_2 \lt y_1 \lt y_2$ among $n+3$ values.
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How many N-digit numbers have the sum of odd places digits equal the sum of even places digits? We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.
My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.
Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.
We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.
After you find a nice way, please provide some example calculations for N=4 and N=6.
| This answer is intended to be simple and self-contained but please ask if it has not been expressed sufficiently clearly.
The expansion of $(1+x+x^2+...+x^9)^n$ as $\sum a_ix^i$
Consider multiplying out $(1+x+x^2+...+x^9)(1+x+x^2+...+x^9)...(1+x+x^2+...+x^9).$
For a given integer $M$, a term $a_Mx^M$ in the product is given by the sum of terms $$x^ux^v...$$
where $x^u$ is from the first bracket, $x^v$ is from the second bracket and so on. Thus $a_M$ is precisely the number of ways of expressing the number $M$ as the sum of $n$ digits.
The sum of squares
We require $\sum a_M^2$. This is the constant coefficient in the product $$\sum a_Mx^M\sum a_Mx^{-M}$$ or, equivalently, the coefficient of $x^{9n}$ in $(1+x+x^2+...+x^9)^{2n}$
The calculation for $N=4$ i.e. n=2
We require the coefficient of $x^{18}$ in $(1+x+x^2+...+x^9)^{4}=(1-x^{10})^4(1-x)^{-4}$.
Expanding, we obtain
$$(1-4x^{10}+...)(1+4x+...+\begin{pmatrix}11\\8\\\end{pmatrix}x^{8}+...+\begin{pmatrix}21\\18\\\end{pmatrix}x^{18}+...)$$
The required number is $\begin{pmatrix}21\\18\\\end{pmatrix}-4\begin{pmatrix}11\\8\\\end{pmatrix}=670.$
The calculation for $N=6$ i.e. n=3
We now require the coefficient of $x^{27}$ in $(1-x^{10})^6(1-x)^{-6}$. This is
$$(1-6x^{10}+15x^{20}-...)(1...+\begin{pmatrix}12\\7\\\end{pmatrix}x^{7}+...+\begin{pmatrix}22\\17\\\end{pmatrix}x^{17}+...\begin{pmatrix}32\\27\\\end{pmatrix}x^{27}+...)$$
The required number is $\begin{pmatrix}32\\27\\\end{pmatrix}-6\begin{pmatrix}22\\17\\\end{pmatrix}+15\begin{pmatrix}12\\7\\\end{pmatrix}=55252.$
A general formula
Let $K=9N/2$. Then the calculation we require is just $$\begin{pmatrix}K+N-1\\K\\\end{pmatrix}-\begin{pmatrix}N\\1\\\end{pmatrix}\begin{pmatrix}K+N-11\\K-10\\\end{pmatrix}+\begin{pmatrix}N\\2\\\end{pmatrix}\begin{pmatrix}K+N-21\\K-20\\\end{pmatrix}-....$$
| {
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If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$ If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$
I attempted to solve this question using binomial coefficients, saying:
$(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}=\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a$
and from here I was thinking that I had to do something with $\sqrt{m}^{n-a}$ with $\sqrt{k}$ and do something else with $\sqrt{k+1}$ and $\sqrt{m+1}^a$. Unfortunately I couldn't think of what to do with this circumstance.
I then thought of maybe taking $(\sqrt{k}+\sqrt{k+1})^2=(\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a)^2$, but this immediately over-complicated the question.
Could you please explain to me how to solve this question and how to solve similar questions in the future?
| This is a USAMTS problem. The official solution can be found here. You are actually on the right track.
Below is my solution. Notice that I used the same notation as in the original problem.
My proof:
We prove by induction that: If $m$ is odd then
$$x^m = a_m\sqrt n + b_m \sqrt{n+1}, na_m^2+1 = (n+1) b_m^2 $$
If $m$ is even then
$$x^m = c_m + d_m \sqrt{n(n+1)}, c_m^2 = n(n+1) d_m^2 +1 $$
When $m=1$ it's true.
If it's true for $m$, we prove it's true for $m+1$.
Case 1: If $m$ is odd, then
$$x^{m+1} = (a_m\sqrt n + b_m \sqrt{n+1}) (\sqrt n + \sqrt{n+1})$$
$$=na_m + (n+1) b_m + (a_m+b_m) \sqrt{n(n+1)}$$
So $c_{m+1} = na_m+(n+1)b_m, d_{m+1} = (a_m+b_m)$ and
$$c_{m+1}^2-n(n+1)d_{m+1}^2-1$$
$$= n^2 a_m^2 + 2n(n+1)a_mb_m + (n+1)^2b_m^2 - n(n+1)a_m^2 - 2n(n+1)a_mb_m - n(n+1)b_m^2 - 1$$
$$
= - n a_m^2+(n+1)b_m^2 - 1 = 0
$$
Therefore $x^{m+1}$ is groovy.
Case 2: If $m$ is even then
$$x^{m+1} = (c_m+d_m\sqrt{n(n+1)})(\sqrt n + \sqrt{n+1}) $$
$$=(c_m+(n+1)d_m)\sqrt n + (c_m+nd_m)\sqrt{n+1}$$
So $a_{m+1} = c_m+(n+1)d_m, b_{m+1} = c_m + nd_m$ and
$$na_{m+1}^2+1 - (n+1) b_{m+1}^2$$
$$=nc_m^2+2n(n+1)c_md_m+n(n+1)^2 d_m + 1 - (n+1)c_m^2 - 2n(n+1) c_md_m - (n+1) n^2 d_m^2$$
$$=-c_m^2 + (n+1) d_m^2 + 1=0$$
Again $x^{m+1}$ is groovy.
| {
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Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$
Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?
| By the definition of cosine as the x-coordinate of a circle, cosine is negative in the interval $(90^{\circ},180^{\circ})$
$\cos 2x =2\cos^2x-1=1-2\sin^2x$
$\cos 2x =(\sqrt{2} \cos x-1)(\sqrt{2} \cos x+1)=-(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1)$
Using the fact that $\cos 2x$ is -ve in the interval $x \in (45^{\circ},90^{\circ})$
Prove the fact that $\cos x < \frac{1}{\sqrt{2}}< \sin x$ in the interval $x \in (45^{\circ},90^{\circ})$
Proof 2:
There is 1 more way people define $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$
Let the angles be $x,90-x,90$
Use the fact that the side opposite to the greater angle is greater.
Therefore, adjacent < opposite (for $x \in (45^{\circ},90^{\circ})$)
Therefore, $\cos x < \sin x$ for $x \in (45^{\circ},90^{\circ})$
| {
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Why does the Minus-1 trick to get particular and general solutions work? Minus-1 trick: A practical trick where the reduced row-echelon form of a system of equations is augmented with -1 rows. It is used to get particular and general solutions.
While it is a handy trick, I was wondering why does it work.
The following is an example of the minus-1 trick from the book Mathematics for Machine Learning Book
| In the example, set the non-pivot variables to $x_2=-\lambda_1$, $x_5=-\lambda_2$ and solve the system for other variables. You will get the exact same result.
The minus-1 trick is just a schematic way to achieve the same with less writing.
Another way to look at this: The columns $1$, $3$ and $4$ of $A$ are the standard basis of the column space. Using these you can uniquely express the remaining two columns:
\begin{align*}
\color{blue}{\begin{pmatrix} 3\\0\\0\end{pmatrix}} &= 3 \begin{pmatrix}1\\0\\0\end{pmatrix} + 0 \begin{pmatrix}0\\1\\0\end{pmatrix} + 0 \begin{pmatrix}0\\0\\1\end{pmatrix},
\\
\color{blue}{\begin{pmatrix} 3\\9\\-4\end{pmatrix}} &= 3 \begin{pmatrix}1\\0\\0\end{pmatrix} + 9 \begin{pmatrix}0\\1\\0\end{pmatrix} -4 \begin{pmatrix}0\\0\\1\end{pmatrix}.
\end{align*}
We can rewrite these two equations as
\begin{align*}
3 \begin{pmatrix}1\\0\\0\end{pmatrix} \color{blue}{{}- 1 \begin{pmatrix} 3\\0\\0\end{pmatrix}} + 0 \begin{pmatrix}0\\1\\0\end{pmatrix} + 0 \begin{pmatrix}0\\0\\1\end{pmatrix} + \color{red}{0 \begin{pmatrix} 3\\9\\-4\end{pmatrix}} &= \begin{pmatrix}0\\0\\0\end{pmatrix},
\\
3 \begin{pmatrix}1\\0\\0\end{pmatrix} + \color{red}{0 \begin{pmatrix} 3\\0\\0\end{pmatrix}} + 9 \begin{pmatrix}0\\1\\0\end{pmatrix} - 4 \begin{pmatrix}0\\0\\1\end{pmatrix} \color{blue}{{}- 1 \begin{pmatrix} 3\\9\\-4\end{pmatrix}} &= \begin{pmatrix}0\\0\\0\end{pmatrix},
\end{align*}
Hence, we have two (linearly independent) solutions to $Ax=0$ given by
$$
\begin{pmatrix} 3\\\color{blue}{-1}\\0\\0\\\color{red}{0} \end{pmatrix},
\begin{pmatrix} 3\\\color{red}{0}\\9\\-4\\\color{blue}{-1} \end{pmatrix}.
$$
Again, the minus-1 trick is just a schematic way to achieve the same result mechanically.
| {
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difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 & 4 \\ \hline
1 & 6 & 6 & 0 \\ \hline
2 & 26 & 30 & -4 \\ \hline
3 & 94 & 102 & -8 \\ \hline
4 & 306 & 318 & -12 \\ \hline
\end{array}
We can deduce that $f(n) - g(n) = 4 - 4n$
But now we have to prove it.
Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$
We can simply check this by induction but I will skip it, so the question won't be so long.
Now lets put it together:
$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$
As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.
How to prove that:
$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$
All I could do is this:
\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}
And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$
What to do next? Did I go the wrong direction anywhere?
Thank you for your fast responses.
| There is no need for induction, use a straight proof.
The equation is
$$f(n+1)-g(n+1)=3(f(n)-g(n))+8n-12$$ with $$f(0)-g(0)=4$$ and it does verify the solution
$$f(n)-g(n)=4-4n$$ as is shown by substitution,
$$4-4(n+1)=3(4-4n)+8n-12,$$ equivalent to $$-4n=-4n.$$
Furthermore, $$4=4-4\cdot 0.$$
| {
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Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$ Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$
where $a$ and $b$ are real numbers and not equal. Find the value of $\frac{a+b}{a\cdot b}$.
At first I applied integration by parts but then it got even more complicated and I couldn't solve it. How can I solve this problem?
| Note that the log function in the integrand can be decomposed as a sum of odd and even ones
\begin{align}
\ln\frac{1+ax}{1+bx}& =\frac12\ln\frac{1+ax}{1-ax}-\frac12\ln\frac{1+bx }{1-bx} + \frac12 \ln\frac{1-a^2x^2}{1-b^2x^2} \\
& =\tanh^{-1}(ax)-\tanh^{-1}(bx) + \frac12 \ln\frac{1-a^2x^2}{1-b^2x^2}
\end{align}
where the odd functions $\tanh^{-1}()$ do not contribute to the integral and the even function has to vanish for the integral to vanish, i.e.
$$\ln\frac{1-a^2x^2}{1-b^2x^2} =0$$
which leads to $a=-b$, hence $\frac{a+b}{ab }=0$.
| {
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How to get the value of a function dependant on $\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$? The problem is as follows:
Find the value of:
$$R=\frac{\sec^2B-\cot A}{4+\csc^2 A}$$
Where $A$ is given by:
$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot
> \left(\frac{A}{2}\right)}$
Also assume this follows the relationship seen in the figure from
below:
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\frac{1}{2}\\
2.&1\\
3.&\frac{1}{4}\\
4.&\frac{1}{3}\\
\end{array}$
My main issue with this problem is the half angle which is presented in the right side of this equation:
$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$
I was able to spot that the figure might intend to say that $A$ and $B$ are complementary thus:
$A+B=90^{\circ}$
Therefore this makes the previous expression into:
$\tan A = \frac{2+\tan \left(\frac{A}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$
This helps a bit but again the half angle would produce a square root. I tried to follow such path. But it didn't ended very well.
After squaring and using other identities I ended up having a 8th degree equation. And certainly this wasn't intended.
So what's the trick here?. So far I'm still stuck and all sorts of manipulations don't give me results. Can someone guide me in the right path?. I'm also confused on how to better use the figure. Those other letters don't seem to help much.
It would really help me alot a method explained step by step so I can understand.
| Alternative approach
Using the diagram and half-angle formulas:
$$\cot(A/2) = \frac{1 + \cos(A)}{\sin(A)}
~~\text{and}~~ \cot(B/2) = \frac{1 + \cos(B)}{\sin(B)} = \frac{1 + \sin(A)}{\cos(A)} \implies $$
$$\frac{\sin(A)}{\cos(A)} = \frac{2 + \frac{1 + \sin(A)}{\cos(A)}}{4 + \frac{1 + \cos(A)}{\sin(A)}} =
\frac{2\cos(A) + 1 + \sin(A)}{4\sin(A) + 1 + \cos(A)} \times
\frac{\sin(A)}{\cos(A)} \implies$$
$$2\cos(A) + 1 + \sin(A) = 4\sin(A) + 1 + \cos(A) \implies$$
$$\cos(A) = 3\sin(A)\implies$$
*
*$\displaystyle \cot(A) = 3$
*$\displaystyle 10\sin^2(A) = 1.$
Therefore,
$$R = \frac{\frac{1}{\cos^2(B)} - 3}{4 + \frac{1}{\sin^2(A)}}
= \frac{\frac{1}{\sin^2(A)} - 3}{4 + \frac{1}{\sin^2(A)}}
= \frac{10-3}{4 + 10} = \frac{1}{2}.$$
| {
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Find real solution $x$ in order to $x + \sqrt{2020}$ and $\dfrac{5}{x} -\sqrt{2020}$ are integers $x + \sqrt{2020}$ and $\dfrac{5}{x} - \sqrt{2020}$ are integers
$\Rightarrow x + \dfrac{5}{x}$ is an integer
$\Rightarrow \dfrac{x^2 + 5}{x}$ is an integer
$\Rightarrow x^2 + 5\ \vdots\ x$
$\Rightarrow x^2 + 5 - x^2\ \vdots\ x$
$\Rightarrow 5\ \vdots\ x$
Is it the right way?
Any idea? Or any better way to solve this problem?
Please help !!!
| As Edward H's question comment hint suggests, let $a$ be the integer where
$$a = x + \sqrt{2020} \implies x = a - \sqrt{2020} \tag{1}\label{eq1A}$$
Then let $b$ be the other integer, and use \eqref{eq1A}, to get
$$\begin{equation}\begin{aligned}
b & = \frac{5}{a - \sqrt{2020}} - \sqrt{2020} \\
b + \sqrt{2020} & = \frac{5}{a - \sqrt{2020}} \\
(b + \sqrt{2020})(a - \sqrt{2020}) & = 5 \\
ab - b\sqrt{2020} + a\sqrt{2020} - 2020 & = 5 \\
ab + \sqrt{2020}(-b + a) & = 2025 \\
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $ab$ and $2025$ are integers, this means $\sqrt{2020}(-b + a)$ must also be an integer. However, due to $\sqrt{2020}$ being irrational and $-b + a$ being an integer, this is only possible if it's $0$, i.e.,
$$b = a \tag{3}\label{eq3A}$$
Using this in \eqref{eq2A} gives
$$a^2 = 2025 \implies a = \pm 45 \tag{4}\label{eq4A}$$
Finally, \eqref{eq1A} gives that
$$x = a - \sqrt{2020} \implies x = \pm 45 - \sqrt{2020} \tag{5}\label{eq5A}$$
| {
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Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
\left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$
I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.
Please, advice.
| Just another solution (similar to Albus Dumbledore's).
$$x^4+4x^3+4x^2-4x+3=x^2(x+2)^2-4x+3 = x^4+4x^3+(2x-1)^2 +2$$
For $x<0$ we have $-4x+3>0$ and therefore $x^2(x+2)^2-4x+3>0$.
For $x>0$ it follows from $x^3>0$ that $x^4+4x^3+(2x-1)^2 +2>0$.
| {
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Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$ Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$
Attempt:
We can write $x^5+x+1 = (x^2+x+1)(x^3-x^2+1)$ which could be a hint to be reducible, if we look clously we see that $(x^2+x+1),(x^3-x^2+1)$ are both irreducible in $Q$, if we use rational root theorem, we see there is no root in $Q$, so its irreducible over $Q$?
| Let $x=w,w^2$, where $w$ cube root of unity. The $f(x)=x^5+x+1 \implies f(w)=w^2+w+1=0, f(w^2)=w+w^2+1=0$. So $g(x)=(x-w)(x-w^2)=x^2+x+1$ is a factor of $f(x)$ Further $\frac{f(x)}{g(x)}=x^3-x^2+1.$
So $f(x)$ is reducible.
| {
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Find the general form of the $n$th derivative of $\ln^2(x+1)$ $f(x)=\ln^2(x+1)$.
I tried to write $f(x)=\ln (x+1)\cdot\ln (x+1)$ and then apply Leibniz but i think maybe i did something wrong.
Could you explain me the right method?
| Ok we talked a bit about it in the comments: we can start with the chain rule to find $f'$. Then we calculate a few more derivatives, and eventually we find a nice form that looks like a good guess. That guess can be proven by induction. I will leave the guessing and the proof by induction for you to do. Alright, here we go:
$$f(x) = \ln^2(x+1) $$
$$f'(x) = 2 \times\frac{\ln(x+1)}{x+1} $$
Now we can use $(\frac{u}{v})'=\frac{u'v-v'u}{v^2}$ and the fact that $(\ln(x+1))'=(x+1)^{-1}$ :
$$f''(x) = 2 \times \frac{(x+1)^{-1+1}-\ln(x+1)}{(x+1)^2}=2 \times \frac{1-\ln(x+1)}{(x+1)^2}$$
$$f'''(x) = 2 \times \frac{-(x+1)^{-1+2}-2(x+1)(1-\ln(x+1))}{(x+1)^4}=2 \times \frac{-1-2+2\ln(x+1)}{(x+1)^3}$$
$$f^{(4)}(x)=2\times\frac{2(x+1)^{-1+3}-3(x+1)^2(-1-2+2\ln(x+1))}{(x+1)^6}$$
$$ = 2 \times \frac{2+3\times1+3\times2-3\times2\ln(x+1)}{(x+1)^4}$$
Almost feeling like there's a pattern, right? Let's keep going:
$$f^{(5)}(x) = 2 \times \frac{-3\times2(x+1)^{-1+4}-4(x+1)^3(2+3\times1+3\times2-3\times2\ln(x+1))}{(x+1)^8} $$
$$= 2 \times \frac{-3\times2-4\times2-4\times3\times1-4 \times 3 \times 2 + 4\times3\times2 \ln(x+1)}{(x+1)^5} $$
Curious...let's organize:
$$f^{(5)}(x) = 2 \times \frac{-(4\times3\times2 + 4\times3\times1 + 4\times2\times1+3\times2\times1)+4\times3\times2\times1\times\ln(x+1)}{(x+1)^5} $$
$$= 2 \times \frac{-4!(\frac11+\frac12+\frac13+\frac14)+4!\times\ln(x+1)}{(x+1)^5} $$
Now the general form is evident. I still recommend that you do $f^{(6)}(x)$ yourself before moving on to writing the general form and proving by induction. If you find difficulties with induction, let us know and we will help you. Good luck!
| {
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theory of equation let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
( c+d)^{-1}+ (c+d)^{-1}$$
my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck
| Notice that
$$\frac{1}{a+b} + \frac{1}{c+d} = \frac{a+b+c+d}{(a+b)(c+d)} = \frac{3}{(a+b)(c+d)}.$$
Repeat for two more sets of two fractions. Then add the three results.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
Show that $(\mathbf{a}·\mathbf{a})(\mathbf{b}·\mathbf{b})-(\mathbf{a}·\mathbf{b})^2=(\mathbf{a}\times\mathbf{b})·(\mathbf{a}\times\mathbf{b})$? I need to prove that, for two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$,
$$(\mathbf{a} · \mathbf{a}) (\mathbf{b} · \mathbf{b}) - (\mathbf{a} · \mathbf{b})(\mathbf{a} · \mathbf{b}) = (\mathbf{a} \times \mathbf{b})·(\mathbf{a} \times \mathbf{b})$$
Could someone give me a demonstration of this identity? Or a hint to prove it?
| The left hand side is the determinant of the Gramian matrix of $\mathbf a$ and $\mathbf b$:
$$
\det\begin{pmatrix}\mathbf a\cdot\mathbf a & \mathbf a\cdot\mathbf b \\
\mathbf b\cdot\mathbf a & \mathbf b\cdot\mathbf b\end{pmatrix}.
$$
In general, the determinant of a Gramian matrix is the square of the $n$-dimensional volume of a parallelotope, in this case the square of the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$.
The right hand side is $\|\mathbf a\times\mathbf b\|^2$, where $\|\mathbf a\times\mathbf b\|$ also is the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$, hence both sides are equal.
| {
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"answer_count": 3,
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Prove by induction: for every integer $n > 0$, $a_n = 2 \cdot 3^n + n^2-6$ I am working on what should be a simple induction proof exercise. However, I'm not sure if I am failing to simplify an expression after substitution correctly, or whether a prior step is incorrect.
The exercise:
Let $a_1 = 1$ and for every integer $n > 0$ let $$a_{n+1} = 3a_n - 2n^2 + 2n + 13$$
Use induction to prove that for every integer $n > 0$ $$a_n = 2 \cdot 3^n + n^2 - 6$$
My steps:
*
*Let $P(n)$ be the statement that for every integer $n > 0$, $a_n = 2 \cdot 3^n + n^2 - 6$
*For $n = 1$, $a_1 = 2 \cdot 3^1 + 1^2 - 6 = 1$. Thus, $P(1)$ is true.
*Suppose that $P(k)$ is true. That is, $a_k = 2 \cdot 3^k + k^2 - 6$.
*Then (using substitution to replace $a_n$ in the given definition for $a_{n+1}$), $$a_{k+1} = 3(2 \cdot 3^k + k^2 - 6) - 2k^2 + 2k + 13 = (2 \cdot 3^{k+1}) + k^2 + 2k -5$$
This is not the needed result.
Is my error in the simplification of the expression in step 4? Or is a previous step (also) incorrect?
| It is the needed result if you notice $$(2\cdot 3^{k+1}) + k^2 + 2k - 5 = (2\cdot 3^{k+1}) + k^2 + 2k +1 - 6 = (2\cdot 3^{k+1}) + (k+1)^2 - 6$$
| {
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$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)
| Hint:
Let $\sqrt{x^2+12y}-\sqrt{y^2+12x}=1+2a$
$$\implies\sqrt{x^2+12y}=\dfrac{23+1+2a}2,\sqrt{y^2+12x}=11-a$$
$\implies11\ge a\ge-12\ \ \ \ (1)$
$$(12+a)^2-(11-a)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$
Replace the value of $x+y$ to find $x-y$
Consequently we find $x,y$ in terms of $a$
But we have $$x^2+12y=(12+a)^2$$
Replace the values of $x,y$ in terms of $a$ to from a quadratic equation $a$
Solve and remember $(1)$
| {
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Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$ Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$.
I drew the picture and found two surfaces: $S_1$ and $S_2$,
where $S_1$ is $\{ 0 \leq y \leq 1 \land -\sqrt{(\frac{1}{2})^2 - (y-\frac{1}{2})^2} \leq x \leq \sqrt{(\frac{1}{2})^2 - (y-\frac{1}{2})^2} \} \cap y + z = 1 \} $
and
$S_2$ is $\{ x^2 + y^2 + z^2 \leq 1 \cap x^2 + y^2 \leq y \} $ .
I wonder if this is correct:
$P(S_1) = \begin{gather*}
\iint_D \sqrt{1+ 0^2 +(-1)^2}\,dx\,dy = ... = \frac{\sqrt{2}}{4}\pi
\end{gather*}
$
and
$P(S_2) = \begin{gather*}
\iint_D \sqrt{1+ \frac{x^2}{1-x^2 - y^2} + \frac{y^2}{1-x^2 - y^2}} \,dx\,dy = \cdots =\pi
\end{gather*}
$
In both cases $D$ is $\{ -\frac{1}{2} \leq x \leq \frac{1}{2} $ and $ \frac{1}{2} - \sqrt{(\frac{1}{2})^2 - x^2} \leq y \leq \frac{1}{2} + \sqrt{(\frac{1}{2})^2 - x^2}\}$
Could someone please say if the idea is good, and if it isn't, what other approach should I take?
I am most familiar with this one.
Thanks in advance.
**EDIT **
There was a mistake in text above, I've corrected it.
| In cylindrical coordinates, the cylinder $x^2 + y^2 = y \ $ is
$r^2 = r \sin \theta \implies r = \sin \theta, \ 0 \leq \theta \leq \pi$.
This is a cylinder of radius $\frac{1}{2}$ with center at $(0,\frac{1}{2})$.
Now parametrization of cylinder is
$r(\theta, z) = (r\cos\theta, r\sin\theta, z) = (\frac{1}{2} \sin 2\theta, \sin^2\theta, z)$
$|r'_\theta \times r'_z| = 1$
Also note that at the intersection of the cylinder and the plane $y +z = 1$,
$\sin^2\theta + z = 1 \implies z = \cos^2\theta$
Also the intersection of the cylinder and the sphere $ \ \sin^2\theta + z^2 = 1 \implies z = \pm \cos\theta$
So the integral to find surface area of the cylindrical surface satisfying the given condition is
$\displaystyle 2 \int_{0}^{\pi/2} \int_{\cos^2\theta}^{\cos\theta} dz \ d\theta$
You may have to find for other surfaces similarly. The integral for other surfaces based on their parametrization are given by,
For spherical surface: $\displaystyle \int_0^{\pi} \int_0^{\sin\theta} \frac{r}{\sqrt{1-r^2}} \ dr \ d\theta$
For planar bottom: $\displaystyle \int_0^{\pi} \int_0^{\sin\theta} \sqrt2 \ r \ dr \ d\theta$
| {
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Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.
| $K$ has $2^4$ elements:
$$ax^3+bx^2+cx+d, \ (a,b,c,d) \in {\mathbb{Z}_2}^4$$
We have $$g(0)^2=g(1)^2=1$$
So the only possible candidates for $g$ are
$$1\\
x^3+x^2+1\\
x^3+x+1\\
x^2+x+1
$$
First we calculate
$$x^4\equiv x^2+x$$
$$x^6\equiv (x^4+x^2+x)x^2 +x^4+x^3\equiv x^3+x^2+x$$
We can skip the constant polynomial $1$ immediately. We use $$(a+b)^2=a^2+b^2 \bmod(2)$$
to check the remaining 3 candidates.
$$(x^3+x^2+1)^2=x^6+x^4+1=(x^3+x^2+x)+(x^2+x)+1=x^3+1$$
$$(x^3+x+1)^2=x^6+x^2+1=(x^3+x^2+x)+x^2+1=x^3+x+1$$
$$(x^2+x+1)^2=x^4+x^2+1=(x^2+x)+x^2+1=x+1$$
| {
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"source": "stackexchange",
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Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3 - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3 - x^2 + x + 1)$.
Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.
| Note
\begin{align}
\frac{(-a^3 + a^2)^6 - 1 }{(-a^3 + a^2)^3 - 1 }&
=(-a^3 + a^2)^3 +1 = (-a^3 + a^2 +1 )[ (a^3 -a^2)^2 + (a^3 - a^2)+ 1]\tag1\\
\end{align}
where
\begin{align}
& (a^3 -a^2)^2 +(a^3 - a^2)+ 1 \\
= & (a^3 -a^2+a-a)^2 +(a^3 - a^2+a -a)+ 1 \\
=& (a^3 -a^2+a)^2 -2a (a^3 -a^2+a) +(a^3 - a^2+a ) +a^2-a+ 1 \\
=& (a^2 -a+1)(a^4-a^3-a^2+a+1)=0
\end{align}
Plug into (1) to obtain
$$(-a^3 + a^2)^6 =1$$
| {
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If $p$ is an odd prime (different than $5$) prove that $10 \mid p^{2}-1$ or $10 \mid p^{2}+1$ I was reading some properties on prime numbers in a book and I’ve found this problem.
Using Modular Arithmetic isn’t allowed
Using the division algorithm I have found:
$p \in \{5m+1, 5m+2, 5m+3, 5m+4\}$
I’ve tried to plug those values of $p$ to :$\frac{p^{2}\pm1}{10}$, but I didn’t get an integer.
Thank you for your help
| Why on earth would modular arithmetic not be allowed. As I am of the theory all arithmetic is modular arithmetic that may be hard.
But as $p$ is an odd prime $p^2 -1$ and $p^2 +1$ are both even and $2|p^2 \pm 1$ so it is sufficient to show that $5|p^2 -1$ or $5|p^2 + 1$. .... Oh, wait, that was modular arithmetic! Well, too bad.
Now $p$ is odd so $p=2k-1$ for some integer $k$ so $p^2\pm 1 = 4k^2 - 4k+1 \pm 1 =\begin{cases}4k^2 - 4k = &5k^2 - 5k -k^2 + k = &5(k^2-k) -k(k-1)\\4k^2+4k+2 = &5k^2 - 5k - k^2 + k +2=&5(k^2-k) -(k+1)(k-2)\end{cases}$.
Now $k-2,k-1,k,k+1,k+2$ are five consecutive numbers so one of them is divisible by $5$. (Oh wait... that was modular arithmetic!... Oh, well.) If the one divisible by $5$ is $k$ or $k-1$ then $p^2 -1 = 4k^2 -4k = 5(k^2-k) -k(k-1)$ is divisible by $5$
If $k+1$ or $k-2$ is divisible by $5$ then $p^2+1 = 4k^2 -4k + 2 = 5(k^2-k) -(k+1)(k-2)$ is divisible by $5$.
And if $k+2$ is divisible by $5$, then there is an integer $m$ so that $5m = k+2$ so $p = 2k -1=2(5m-2)-1=10m-5 = 5(2m-1)$. But that means $p$ is not a prime other than $5$. So this is impossible. $k+2$ is not divisible by $5$ and one of the other four cases must be true.
.....
But that used modular arithmetic.
| {
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Proving combinatorial formula by induction and combinatorially I need to prove this equation by induction and by combinatorial explanation. $$\sum_{k=0}^{n} \left( \begin{array}{c} 4n \\ 4k \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1}$$
My induction proof:
Base case: p(1): $$\left( \begin{array}{c} 4 \\ 0 \end{array} \right) + \left( \begin{array}{c} 4 \\ 4 \end{array} \right) = 2^{4*1-2} + (-1)^1 \cdot 2^{2*1-1}\;\;\;\;\;\;\;\checkmark$$
Inductive step: $p(k) \Longrightarrow p(k+1)$: $$\sum_{k=0}^{n} \left( \begin{array}{c} 4n \\ 4k \end{array} \right) + \left( \begin{array}{c} 4(n+1) \\ 4k \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1} + \left( \begin{array}{c} 4(n+1) \\ 4k \end{array} \right) \\ \sum_{k=0}^{n} \left( \begin{array}{c} 4(n+1) \\ 4(k+1) \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1} + \left( \begin{array}{c} 4(n+1) \\ 4k \end{array} \right)$$
I got stuck in this part and I can't find a combinatorial explanation to prove this.
| The problem is that the $n+1$ case isn't actually obtained from the $n$ case by adding an extra term to the end, the way you're thinking of it.
Here are the first few cases of this formula:
*
*For $n=1$, $\binom 40 + \binom 44 = 2^2 - 2^1$.
*For $n=2$, $\binom 80 + \binom 84 + \binom 88 = 2^6 + 2^3$.
*For $n=3$, $\binom{12}0 + \binom{12}{4} + \binom{12}{8} + \binom{12}{12} = 2^{10} - 2^5$.
*For $n=4$, $\binom{16}{0} + \binom{16}{4} + \binom{16}{8} + \binom{16}{12} + \binom{16}{16} = 2^{14} + 2^7$.
To prove this by induction, you might use Pascal's formula: $\binom nk = \binom{n-1}{k-1} + \binom{n-1}{k}$. But then you have to look at some intermediate cases as well: sums where the tops and/or the bottoms of the binomial coefficients aren't multiples of $4$. It's possible to get there, but the binomial theorem argument suggested in the comments is easier.
| {
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"question_score": "4",
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The matrix of $T(x_1,x_2)=(x_1+x_2,x_1-x_2)$ with respect to a basis Consider the linear map $T:\mathbb{R}^2\to \mathbb{R}^2, T(x_1, x_2)=(x_1+x_2,x_1-x_2)$. Let $B_1$ be the canonical base of $\mathbb{R}^2$ and consider another basis $B_2=\{f_1,f_2\}$, where $f_1=(1,1)$ and $f_2=(1,2)$.
So, according to my computations, the matrix of $T$ with respect to $B_1$ is $\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}$ and the matrix of $T$ with respect to $B_2$ is $\begin{pmatrix}
2 & 3 \\
0 & -1
\end{pmatrix}$. However, I tried computing the transition matrices. I got that the transition matrix from $B_1$ to $B_2$ is $\begin{pmatrix}
1 & 1\\
1 & 2
\end{pmatrix}$ and the transition matrix from $B_2$ to $B_1$ is $\begin{pmatrix}
2 & -1\\
-1 & 1
\end{pmatrix}$. I should have that $\begin{pmatrix}
2 & 3 \\
0 & -1
\end{pmatrix}= \begin{pmatrix}
2 & -1\\
-1 & 1
\end{pmatrix}\cdot \begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}\cdot \begin{pmatrix}
1 & 1\\
1 & 2
\end{pmatrix}$, but this is not true. Where did I go wrong?
| The second column of your matrix of $T$ with respect to $B_2$ is wrong.
Indeed, $T(f_2) = T(1,2) = (3,-1) \neq 3f_1 - f_2 = (2,1).$
| {
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Factorize polynomial of degree 4 given statements If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?
Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$
Can I solve $a,b,c,d$ from there? How can I do so?
| Let $$f(x) = x^4-x^3-13x^2+26x-8$$
Observe that $f(2) = 0$ and $f(-4) = 0$
Hence $(x-2)$ and $(x+4)$ are two factors of $f(x)$
Also $f(x)$ is divisible by $(x-2)(x+4) = x^2 + 2x - 8$
Now divide $f(x)$ by $x^2 + 2x - 8$ to obtain the other quadratic factor $x^2 - 3x +1$
If you solve $x^2 - 3x +1 = 0$, you'll find $x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$
Finally
$$f(x) = \left(x - \left(\frac{3}{2} + \frac{\sqrt{5}}{2}\right) \right) \left(x - \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right) \right) (x + 4)(x-2)$$
| {
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Solve the last part of a diferential equation I'm trying to solve this problem but I've some doubts about the final steps.
Solve the initial value problem using matrix functions:
$$
\frac {dy}{dx} =
\begin{pmatrix}
1 & a & 1 \\
0 & 1 & a \\
0 & 0 & 1 \\
\end{pmatrix}
y
\text{ ; }
y(0) =
\begin{pmatrix}
0 \\
0 \\
-1 \\
\end{pmatrix}
$$
First of all, I'm not sure what do they mean with "matrix functions" but I think I've to do the exponential development.
The way I've tried:
$$
\frac {dy}{dx} = Ay
\text{ }\to\text{ }
\frac {dy}{y} = Adx
\text{ }\to\text{ }
ln(y) = Ax+K
\text{ }\to\text{ }
y(x) = Ce^{Ax}
$$
$$
e^{Ax} = \sum_{n=0}^{\infty} \frac{(Ax)^n}{n!}
$$
I try to get a recurring rule for the powers of the matrix A:
$$
A^2 =
\begin{pmatrix}
1 & 2a & 2+a^2 \\
0 & 1 & 2a \\
0 & 0 & 1 \\
\end{pmatrix}
\text{ }\text{ }
A^3 =
\begin{pmatrix}
1 & 3a & 3+3a^2 \\
0 & 1 & 3a \\
0 & 0 & 1 \\
\end{pmatrix}
\text{ }\text{ }
A^4 =
\begin{pmatrix}
1 & 4a & 4+6a^2 \\
0 & 1 & 4a \\
0 & 0 & 1 \\
\end{pmatrix}
$$
So that, it looks the n-power of A is:
$$
A^n =
\begin{pmatrix}
1 & na & n+\frac{n(n-1)}{2}a^2 \\
0 & 1 & na \\
0 & 0 & 1 \\
\end{pmatrix}
=
I+
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{pmatrix}
na+
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
(n + \frac {n(n-1)}{2}a^2)
$$
I just rename:
$$
\alpha =
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{pmatrix}
\text{ }\text{ }
\beta =
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
$$
So that $A^n$ becomes:
$$
A^n = I + na\alpha + (n + \frac {n(n-1)}{2}a^2)\beta
$$
$$
A^n =
I + na\alpha + n\beta + \frac{n^2}{2}a^2 \beta - \frac{n}{2}a^2 \beta =
I + (a\alpha + \beta -\frac{1}{2}a^2 \beta)n + \frac{a^2 \beta}{2}n^2
$$
Once this is done, I go back to the exponential development:
$$
e^{Ax} =
\sum_{n=0}^{\infty} \frac{(Ax)^n}{n!} =
\sum_{n=0}^{\infty} (I + (a\alpha + \beta -\frac{1}{2}a^2 \beta)n + \frac{a^2 \beta}{2}n^2) \frac{x^n}{n!}
$$
$$
e^{Ax} =
Ie^{x} + (a\alpha + \beta -\frac{1}{2}a^2 \beta) \sum_{n=0}^{\infty} \frac{nx^n}{n!}
+ \frac{a^2 \beta}{2} \sum_{n=0}^{\infty} \frac{n^2x^n}{n!}
$$
From here I've tried to solve both sums:
$$
\sum_{n=0}^{\infty} \frac{nx^n}{n!} =
0 + \sum_{i=1}^{\infty} \frac{nx^n}{n!} =
\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}x =
\sum_{m=0}^{\infty} \frac{x^m}{m!}x = xe^x
$$
$$
\sum_{n=0}^{\infty} \frac{n^2x^n}{n!} =
0 + \sum_{i=1}^{\infty} \frac{n^2x^n}{n!} =
\sum_{n=1}^{\infty} \frac{nx^{n-1}}{(n-1)!}x =
\sum_{m=0}^{\infty} \frac{(m+1)x^m}{m!}x =
\sum_{m=0}^{\infty} \frac{mx^m}{m!}x +
\sum_{m=0}^{\infty} \frac{x^m}{m!}x =
x^2e^x+xe^x
$$
So that, the solution should be:
$$
e^{Ax} =
Ie^{x} + (a\alpha + \beta -\frac{1}{2}a^2 \beta) xe^x
+ \frac{a^2 \beta}{2} (x^2e^x+xe^x)
$$
$$
e^{Ax} =
Ie^{x} + (a\alpha + \beta) xe^x + \frac{a^2 \beta}{2} x^2e^x
$$
On a matrix representation:
$$
e^{Ax} =
\begin{pmatrix}
1 & ax & x + \frac{a^2}{2}x^2 \\
0 & 1 & ax \\
0 & 0 & 1 \\
\end{pmatrix}
e^x
$$
On the last part, I'm not sure how to get the solution because the constant C should be a row vector instead of a column vector. Is this right? How to get the final solution?
Thanks in advance.
| A simple solution:
$$
\frac {dy}{dx} =
\begin{pmatrix}
1 & a & 1 \\
0 & 1 & a \\
0 & 0 & 1 \\
\end{pmatrix}
y
\text{ ; }
y(0) =
\begin{pmatrix}
0 \\
0 \\
-1 \\
\end{pmatrix}
$$
Let $y=\begin{pmatrix} y_1 \\ y_2 \\y_3 \end{pmatrix}$
Then you have three couples ODEs as
$\frac{dy_1}{dx}=y_1+ay_2+y_3~~~(1), \frac{dy_2}{dx}=y_2+ay_3~~~(2), \frac{dy_3}{dx}=y_3~~~~(3)$
Fist solve (3) to get $y_3=C e^x \implies y_3=-e^{x}$. Put this in (2) and solve the ODE:
$$\frac{dy_2}{dx}=y_2-ae^x \implies y_2=e^{x}\int -ae^x e^{-x}dx+ Be^{x}=-axe^{x}+Be^{x} \implies y_2=-axe^x$$
Putting these solutions in (1), we get
$$\frac{dy_1}{dx}=y_1-a^2 xe^x-e^{x} \implies y_1=e^{x}\int (-a^2xe^x e^{-x}-1) dx+Ae^x$$ $$ \implies y_1= -\frac{a^2x^2}{2} e^{x}-xe^x+Ae^x, y_1(0)=0 \implies A=0.$$ So the solutions of the Matrix ODE are
$$y(x)=\begin{pmatrix} -\frac{a^2x^2}{2}e^x-xe^x \\ -axe^x \\ -e^x \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse of fourth root function I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$
but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!
| The equation is easier to work into the quadratic equation if we first get rid of radicals and avoid trying to "simplify" by using the $y^4/a^4$ rational. Expansion gives us the best view of a,b,c.
$$y=a\sqrt[^4]{b^2-(x-c)^2}\implies y^4=a^4\big(b^2-(c^2 - 2 c x + x^2)^2\big)\\
\implies 0=a^4 b^2 - a^4 c^2 + 2 a^4 c x - a^4 x^2-y^4\\
\implies (a^4)x^2 - (2 a^4 c)x + (y^4 - a^4 b^2 + a^4 c^2) = 0\\
\implies x= \frac{(2 a^4 c) \pm\sqrt{(2 a^4 c)^2-4(a^4)(y^4 - a^4 b^2 + a^4 c^2)}}{2a^4}\\
= \frac{(2 a^4 c) \pm\sqrt{4 a^8 b^2 - 4 a^4 y^4}}{2a^4} \\
\implies x = c \pm \frac{\sqrt{(a^2 b - y^2) (a^2 b + y^2)}}{a^2 }
\quad a,b,c,y\in\mathbb{R}\land a\ne 0\\
x\in\mathbb{R}\iff y\le \sqrt{a^2b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$ Calculate
$\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$.
The only idea that I have is to substitute $t=x+1$, but I do not think it is a good one.
| As @Tito Eliatron commented, integrating by parts
$$I=\int\frac{\tan ^{-1}(x)}{x^2-4 x-1}dx$$
$$I=\tan ^{-1}(x)\frac{\log \left(-x+\sqrt{5}+2\right)-\log \left(x+\sqrt{5}-2\right)}{2 \sqrt{5}}+\frac J{2 \sqrt{5}}$$
$$J=\int\frac{\log \left(-x+\sqrt{5}+2\right)-\log \left(x+\sqrt{5}-2\right)}{x^2+1}dx$$
$$\frac 1{x^2+1}=\frac 1{(x+i)(x-i)}=\frac{i}{2 (x+i)}-\frac{i}{2 (x-i)}$$All of these make that we face now four integrals
$$I=\int \frac{\log (a x+b)}{x+c}dx=\text{Li}_2\left(\frac{a x+b}{b-a c}\right)+\log (a x+b) \log \left(1-\frac{a
x+b}{b-a c}\right)$$ Integrated between the given bounds, this leads to a quite complicated expression which is hard to simplify because of the bunch of polylogarithms of complex arguments.
After a pretty tedious work
$$\int_0^2\frac{\tan ^{-1}(x)}{x^2-4 x-1}dx=-\frac{\tan ^{-1}(2) \sinh ^{-1}(2)}{2 \sqrt{5}}\sim -0.3573950303$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:
Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.
I first came across the problem in a school book targeted towards students just exposed to factorisation and it intrigued me how one could prove this without Sturm's Theorem. Maybe one could try to factorise into squares, but I can get no result from that. Thanks in advance.
| I came up with a weird proof, but the idea behind is kind of simple: the fact is that I expect that for $x$ outside an interval the terms $x^8+x^7+x^6$ will be greater than the negative terms, and in this range the remaining terms will be positive. On the other side, for small $x$ the -5 will paly his role and make this negative.
I don't know if this actually happens, but this led me to factorize in this way:
$$ (x^3-5)(x^5+x^4+x^3) + 7x^2+7x+6 = 0$$
Now, continuing with algebraic manipulation:
$$ (x^3-5)x^3(x^2+x+1) + 7(x^2+x+1) = 1 $$
$$ (x^6 - 5x^3 + 7)(x^2+x+1) = 1 $$
The nice thing here is that we have two big terms and they have to multiplicate to 1. Weird. Let us calculate the minima of the two terms. Imposing derivatives equal to zero, we see that the minima of the first is reached in $ x^3 =(15/6)$, which yields
$$ \frac{225}{36} - \frac{450}{36} + 7 = - \frac{225}{36} + \frac{252}{36} = \frac{27}{36} = \frac{3}{4} $$
The second reaches its minimus at $x= -1/2$, which yields $ 1/4 -1/2+1 = 3/4$.
The strategy now is the following: it is enough to show that when one term is $\le 1$, the other is $> 4/3$: in this way the product will be always $>1$.
Case 1. $x^6-5x^3+7 \le 1$. Substituting $t=x^3$, this is equivalent to
$$ (t-3)(t-2) \le 0$$
that is $2 \le t \le 3$. Substituing back $t=x^3$ we get $\sqrt[3]{2} \le x \le \sqrt[3]{3}$. Since the derivative of $x^2+x+1$ is positive for $x \ge 1$, in the above interval it will yield at least
$$ \sqrt[3]{4} +\sqrt[3]{2} +1 \ge 3 > 4/3 $$
Case 2. $x^2+x+1 \le 1$. This is equivalent to $x(x+1) \le 0$, that is $-1 \le x \le 0$. Using again $t=x^3$, we see that $(t-3)(t-2)+1$ is decreasing in the above interval. Thus the minima is attained at $t=0$, for which it is $7 > 4/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.
There are $2$ cases to consider:
Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:
\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) + abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}
Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?
| The standard substitution is $x=b+c-a , y=c+a-b , z=a+b-c $ ($x,y,z \geq 0$ by the triangle inequality.) These invert to give
\begin{eqnarray*}
2a&=&y+z \\ 2b&=&z+x \\ 2c&=&x+y.
\end{eqnarray*}
Subbing these gives
\begin{eqnarray*}
x^3z+y^3x+z^3x-xyz(x+y+z) \geq 0
\end{eqnarray*}
which you could $\color{red}{\text{not}}$ say is true by Muirhead's inequality ... but can be shown to be true because
\begin{eqnarray*}
xy(y-z)^2+yz(z-x)^2+zx(x-y)^2 \geq 0.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$? If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?
I need to solve it using combinatorics.
I've started with calculating the number of possible solutions without restrictions and I am stuck.
| The straightforward way to get the result is to compute the coefficient of $X^{12}$ in $(1+X+X^2+X^4+X^5+X^6)^4$. Though computing that coefficient basically amounts to counting solutions explicitly in a systematic manner, it is not so hard to do even by hand: $$(1+X+X^2+X^4+X^5+X^6)^2=\\1+2X+3X^2+2X^3+3X^4+4X^5+6X^6+4X^7+3X^8+2X^9+3X^{10}+2X^{11}+X^{12},$$
and from there one easily sees the required coefficient is $2\times(1^2+2^2+3^2+2^2+3^2+4^2)+6^2=122$.
I can also propose a more ad hoc method that might be slightly easier (because fewer and smaller numbers are involved) if one were not allowed to use pen and paper. Most of the solutions will involve two of $a,b,c,d$ being${}<3$ and the other two being${}>3$, the exceptions being the permutations of $(0,4,4,4)$ and of $(2,2,2,6)$, which are $8$ solutions. For the remaining solutions that do have a $2$-$2$ split, there are $\binom42=6$ ways to choose the larger positions, and after we have attributed the minimum of $4$ units to those two positions, we must still distribute the remaining $12-8=4$ units over the $4$ positions, while never putting more than $2$ of those units together. By an argument similar to the one used above, that amounts to finding the coefficient of $X^4$ in $(1+X+X^2)^4$, and from $(1+X+X^2)^2=1+2X+3X^2+2X^3+X^4$ one finds that the number is $1^2+2^2+3^2+2^2+1^2=19$. All in all $6\times19+8=122$ solutions are possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $ If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $
Prove $ abc \geq 8 $
I have tried AM-GM and substituting the 1, without results.
| Let
$$x = \frac{1}{a+1}, \quad y = \frac{1}{b+1}, \quad z = \frac{1}{c+1},$$
then $x+y+z=1$ and
$$a = \frac{1-x}{x} = \frac{y+z}{x}.$$
The inequality become
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant 8.$$
Which is true because
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant \frac{2\sqrt{xy} \cdot 2\sqrt{yz} \cdot 2\sqrt{zx}}{xyz} = 8.$$
The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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