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Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable.
How does one find the exponential of a non-diagonalizable matrix?
My attempt:
Write
$\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix} = M + N$,
with $M = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ and $N = \begin{pmatrix}
0 & 1 & 2 \\
0 & 0 & -4 \\
0 & 0 & 0
\end{pmatrix}$.
We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus:
$$\begin{aligned}
e^{At}
&= e^{(M+N)t} = e^{Mt} e^{Nt} \\
&= \begin{pmatrix}
e^t & 0 & 0 \\
0 & e^t & 0 \\
0 & 0 & e^t
\end{pmatrix} \left(I + \begin{pmatrix}
0 & t & 2t \\
0 & 0 & -4t \\
0 & 0 & 0
\end{pmatrix}+\begin{pmatrix}
0 & 0 & -2t^2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\right) \\
&= e^t \begin{pmatrix}
1 & t & 2t \\
0 & 1 & -4t \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
e^t & te^t & 2t(1-t)e^t \\
0 & e^t & -4te^t \\
0 & 0 & e^t
\end{pmatrix}.
\end{aligned}$$
Is that the right answer?
| Hint:
Find the Jordan matrix, by which the exponential can always be found.
| {
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} |
prove that $(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$
For $a, b, c$ positive reals prove that $$(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$$
I used the Cauchy-Swartz inequality in the LHS so I was left to prove that $\dfrac{a}{\sqrt{a^2+2b^2}}+\dfrac{b}{\sqrt{b^2+2c^2}}+\dfrac{c}{\sqrt{c^2+2a^2}} \ge \sqrt3$ but apparently for $(a,b,c)=(1,2,1)$ it is not correct.
| Let $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, by Holder $(ab+ac+bc)\sum\limits_{cyc}\frac{c}{a(c^2+2a^2)}=(x+y+z)\sum\limits_{cyc}\frac{x^2}{y(x^2+2z^2)}=$
$=(x+y+z)\sum\limits_{cyc}\frac{x^3}{xy(x^2+2z^2)}\geq\frac{(x+y+z)^4}{3\sum\limits_{cyc}(x^3y+2x^2yz)}$.
Thus, it remains to prove that $(x+y+z)^4\geq9\sum\limits_{cyc}(x^3y+2x^2yz)$ or
$\sum\limits_{cyc}(x^4-5x^3y+4x^3z+6x^2y^2-6x^2yz)\geq0$ or
$\sum\limits_{cyc}(2x^4-x^3y-x^3z+12x^2y^2-12x^2yz)\geq9\sum\limits_{cyc}(x^3y-x^3z)$ or
$\sum\limits_{cyc}(x-y)^2(x^2+xy+y^2+6z^2)\geq9(x+y+z)(x-y)(x-z)(y-z)$.
Hence, it remains to prove the last inequality for $x\geq y\geq z$.
But $(x+y+z)(x-y)(x-z)(y-z)\leq(x+y)xy\Leftrightarrow z(x^2+y^2+xz-z^2)\geq0$, which is obvious.
Id est, it's enough to prove $(x+y+z)^4\geq9\sum\limits_{cyc}(x^3y+2x^2yz)$ for $z\rightarrow0$.
Let $x=ty$. Hence, we need to prove that $(t+1)^4\geq9t^3$, which is AM-GM:
$(t+1)^4=\left(3\cdot\frac{t}{3}+1\right)^4\geq\left(4\sqrt[4]{\frac{t^3}{27}}\right)^4>9t^3$.
Done!
| {
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How can I Show that, $\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$ Where $c^2=a^2+b^2$ is Pythagoras theorem. Sides a,b and c are of a right angle triangle.
Show that,
$$\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$$
How do I go about proving this identity? Can anybody help? I know of the arctan.
| *
*$\arcsin \dfrac bc -\arcsin \dfrac ac =2\arcsin \dfrac{b-a}{c \sqrt 2}$
*$a,b,c \gt 0$
*$a^2 + b^2 = c^2$
Let's rewrite this as
$\dfrac 12 \left(\arcsin \dfrac bc -\arcsin \dfrac ac \right)
=\arcsin \dfrac{b-a}{c \sqrt 2}$
Since you said that sides $a, b, $ and $c$ are sides of a right triangle, then we know that $a,b,c \gt 0$.
Let $\theta$ be the angle that corresponds to the point $(a,b)$ in the first quadrant.
Then $\sin \theta= \dfrac bc$ and $\cos \theta = \dfrac ac$
So $\dfrac 12 \left( \arcsin \dfrac bc -\arcsin \dfrac ac \right) =
\dfrac 12 \left( \theta - \left(\dfrac{\pi}{2} - \theta\right) \right) =
\theta - \dfrac{\pi}{4}$
If you can show that $\sin \left( \theta - \dfrac{\pi}{4} \right)
= \dfrac{b-a}{c \sqrt 2}$,
then you are done.
| {
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"source": "stackexchange",
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} |
Prove that $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divisible by $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}$
Prove that
$$(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$$
is divisible by
$$(a+b+c)^{3}-a^{3}-b^{3}-c^{3},$$
where $a,,b,c -$ integers, such that $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}\not =0$
My work so far:
$(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$
Let $333=3\cdot111=3t,$ where $t=111$.
$(a+b+c)^{333}-a^{333}-b^{333}-c^{333}=(a+b+c)^{3t}-a^{3t}-b^{3t}-c^{3t}$.
I need help here.
| Hint $1:$
Let $f(a)=(a+b+c)^{333}- a^{333} -b^{333} -c^{333}.$
$f(-b)=(-b+b+c)^{333} -(-b)^{333} -b^{333} -c^{333}=0$
So, by Factor theorem we get $(a+b)$ as a factor of the given expression.
Similarly, $(b+c)$ and $(c+a)$ are factors of the given expression.
Hint $2:$ Now, to prove the divisiblity of the expression by $3$, use Fermat's little theorem.
$a^2 \equiv 1 \pmod 3 \Rightarrow a^{333} \equiv a\pmod 3$
Similarly,
$b^{333} \equiv b \pmod 3$
$c^{333} \equiv c \pmod 3$
$(a+b+c)^{333} \equiv (a+b+c)\pmod 3$
So, the given expression is $\equiv 0 \pmod 3$
As the given expression is divisible by each of these four factors, it is divisible by $(a+b+c)^3 -a^3-b^3-c^3$.
| {
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"source": "stackexchange",
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} |
Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{n(1+x+\dots+x^n)} \\
&= \frac{1}{n}.
\end{align*}
Are my steps correct? Thanks.
| Without using L'Hospital,
$$\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}x$$
$$=-\lim_{x\to0}x^n\cdot\lim_{x\to0}\frac{\ln(1-x^{n+1})}{-x^{n+1}}+\lim_{x\to0}\dfrac{\ln(1-x)}{-x}$$
Use $\lim_{y\to0}\dfrac{\ln(1+u)}u=1$
| {
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"answer_id": 2
} |
Finding eigenvectors of a 3x3 matrix Struggling with this eigenvector problems. I've been using this SE article (Finding Eigenvectors of a 3x3 Matrix (7.12-15)) as a guide and it has been a very useful, but I'm stuck on my last case where $\lambda=4$.
Q: Find the eigenvalues $\lambda_1 < \lambda_2 < \lambda_3$ and corresponding eigenvectors of the matrix
$$A=\begin{bmatrix}-2 & 3 & 0 \\ 0 & -1 & -10 \\ 0 & 0 & 4\end{bmatrix}=\begin{bmatrix}-2-\lambda & 3 & 0 \\ 0 & -1-\lambda & -10 \\ 0 & 0 & 4-\lambda\end{bmatrix}$$
The eigenvalues for the $A$ matrix are $\lambda_1=-2$, $\lambda_2=-1$, $\lambda_3=4$ respectively.
Case $\lambda=-2$
$$A-2I_3=\begin{bmatrix}-2-(-2) & 3 & 0 \\ 0 & -1-(-2) & -10 \\ 0 & 0 & 4-(-2)\end{bmatrix}=\begin{bmatrix}0 & 3 & 0 \\ 0 & 1 & -10 \\ 0 & 0 & 6\end{bmatrix}$$
$$rref(A-2I_3)=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
$x_1$ is our free variable
$$\vec{v_1}=\begin{bmatrix}1\\0\\0\end{bmatrix}$$
Case where $\lambda=-1$.
$$A-I_3=\begin{bmatrix}-2-(-1) & 3 & 0 \\ 0 & -1-(-1) & -10 \\ 0 & 0 & 4-(-1)\end{bmatrix}=\begin{bmatrix}-1 & 3 & 0 \\ 0 & 0 & -10 \\ 0 & 0 & 5\end{bmatrix}$$
$$A^\prime=rref(A-I_3)=\begin{bmatrix}1 & 3 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
Finding the $ker(A-I_3)$ we find $x_2$ and $x_3$ as our free variables. Solving for each variable we get $x_1=3x_2$ and $x_2=1$.
$$\vec{v_2}=\begin{bmatrix}3 \\ 1 \\ 0\end{bmatrix}$$
Case $\lambda=4$
$$A+4I_3=\begin{bmatrix}-2-4 & 3 & 0 \\ 0 & -1-4 & -10 \\ 0 & 0 & 4-4\end{bmatrix}=\begin{bmatrix}-6 & 3 & 0 \\ 0 & -5 & -10 \\ 0 & 0 & 0\end{bmatrix}$$
$$rref(A-I_3)=\begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{bmatrix}$$
This is where I'm stuck. I checked wolframalpha and it says my last eigenvector should be $\vec{v_3}=\begin{bmatrix}-1 \\ -2 \\ 1\end{bmatrix}$ and I'm not sure how it can be that. Thanks in advance!
| The equations corresponding to that row-reduced form at the end are
$$
x - y/2 = 0 \\
y + 2z = 0
$$
Since $z$ is a free variable, you can pick $z = 1$ and back-substitute to get
$y = -2$, and then use this in the first equation to get $x = 1$. Of course, any nonzero multiple of this vector is also an eigenvector for $4$.
| {
"language": "en",
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"source": "stackexchange",
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} |
Largest root as exponent goes to $+\infty$ Let $a\geq 1$ and consider
$$
x^{a+2}-x^{a+1}-1.
$$
I am interested to see what is the largest root of this polynomial as $a\to +\infty$.
In order to find a root, we surely have to have
$$
x^{a+2}-x^{a+1}=x^{a+1}(x-1)=1.
$$
Hence, I guess we have to look for which $x$ we have that
$$
x^{a+1}(x-1)\to 1\text{ as }a\to+\infty.
$$
Intuitively, if $x$ tends to some value larger than $1$ for $a\to\infty$, the whole thing should diverge. On the other side, if $x$ tends to some value smaller than $1$, then the whole expression should converge to $0$. Hence I guess that $x\to 1$ as $a\to\infty$ in order to get a root.
| I will show,
by elementary means that
the root is beteween
$1+\dfrac1{\sqrt{a+1}}$
and
$1+\dfrac1{a+1}$.
Since the root of
$x^{a+2}-x^{a+1}-1
$
is close to $1$,
let
$x = 1+y$.
Then we want
$1
=(1+y)^{a+2}-(1+y)^{a+1}
=(1+y)^{a+1}(1+y-1)
=y(1+y)^{a+1}
$.
Since
$(1+y)^{a+1}
\gt 1+y(a+1)
$,
$1
\gt y(1+y(a+1))
\gt y^2(a+1)
$
so
$y \lt \dfrac1{\sqrt{a+1}}
$.
Since
$(1+1/(a+1))^{a+1}
< e
$,
if
$y < 1/(a+1)
$
then
$(1+y)^{a+1}
< e
$
so that
$y(1+y)^{a+1}
< ey
<\dfrac{e}{\sqrt{a+1}}
< 1
$
for
$a > 8$.
Therefore
$y > \dfrac1{a+1}$.
| {
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"source": "stackexchange",
"question_score": "8",
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Examine whether $\{a_n\}$ is a cauchy sequence where $a_n=\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}$
Examine whether $\{a_n\}$ is a cauchy sequence where $a_n=\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}$
Attempt: If $m>n$ where $m, n$ are natural numbers,
$|a_m-a_n|=|(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2m})-(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n})|=|\frac{1}{2(n+1)}+\frac{1}{2(n+2)}\cdots +\frac{1}{2m}|=\frac{1}{2(n+1)}+\frac{1}{2(n+2)}\cdots +\frac{1}{2m}>\frac{1}{2(n+1)}$
Is the steps correct? Then what to do. Please help.
| Note that
$$|a_m - a_n| = \sum_{k=n+1}^m \frac{1}{2k} > \frac{m-n}{2m} > \frac{1}{2}(1 - n/m).$$
For any $n \in \mathbb{N}$ we can choose $m > 2n$ and find
$$|a_m - a_n| > \frac1{4}.$$
Thus, it is not the case that for any $\epsilon > 0$ there exists $N$ such that $|a_m-a_n| < \epsilon$ for all $m > n > N$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782623",
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"source": "stackexchange",
"question_score": "1",
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} |
Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$
Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$
$\bf{My\; Try::}$ I have solved Using Direct formula::
$$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}$$
Now Put $n=7\;,$ We get
$$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}\cdot \sin \frac{4\pi}{7}\cdot \sin \frac{5\pi}{7}\cdot \sin \frac{6\pi}{7}=\frac{7}{2^{7-1}}$$
So $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}$$
Now my question is how can we solve it without using Direct Formula, Help me
Thanks
| Hint;
Assume
$$ \theta = (n\pi)/7$$
$$\implies 4\theta = n\pi -3\theta$$
Take the sine on both the sides
An apply the required manipulations
You should get a cubic in $\sin^2 \theta$
Since we have defined out $\theta$, we are quite aware what the roots are going to be
So the numerical part in the end divided by the coefficient of $x^3$ gives you
$$ {{[\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}}]}^2 = 7/64$$
That should be it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Points $P_i$ on an ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP_i^2}$$
I took $P_1$ as $(3,0)$. Let other points be $P_i(3\cos\theta_i,2\sin\theta_i)$.
$$\tan\frac{\pi}{n}=\frac{2}{3}\tan\theta_2$$
$$\tan\frac{2\pi}{n}=\frac{2}{3}\tan\theta_3$$
and so on.
$$OP_1^2=a^2$$
$$OP_2^2=9\cos^2\theta_2+4\sin^2\theta_2=9\cos^2\theta_2\left(1+\tan^2\frac{\pi}{n}\right)$$
$$OP_3^2=9\cos^2\theta_3+4\sin^2\theta_3=9\cos^2\theta_3\left(1+\tan^2\frac{2\pi}{n}\right)$$
and so on.
But I am not able to compute the required sum.
| If we set $P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$ we have
$$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$
from which:
$$ \tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) $$
and
$$ OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = 4+5\cos^2\theta_i=\frac{9+4 \tan^2\theta_i}{1+\tan^2\theta_i}$$
gives that the wanted sum is:
$$ \sum_{i=1}^{n}\frac{1+\tan^2\theta_i}{9+4\tan^2\theta_i}=\sum_{i=1}^{n}\frac{1+\frac{9}{4}\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) }{9+9\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) } $$
or:
$$ \frac{1}{36}\sum_{i=1}^{n}\left[4\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) +9\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) \right]$$
that can be computed by exploiting the duplication formulas for $\sin$/$\cos$, De Moivre's formula and geometric series, or simply by noticing that it does not really depend on $\vartheta_0$ by differentiation with respect to $\vartheta_0$:
$$\forall\, n,\vartheta_0,\qquad \sum_{i=1}^{n}\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=
\sum_{i=1}^{n}\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=\color{red}{\frac{n}{2}}\tag{1}$$
gives:
$$ \sum_{i=1}^{n}\frac{1}{OP_i^2} = \color{red}{\frac{13n}{72}}.\tag{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Estimation the integration of $\frac{x}{\sin (x)} $ How to prove that integral of $\frac{x}{\sin (x)} $ between $0$ to $\frac{\pi}{2}$ lies within the interval $\frac{\pi^2}{4}$ and $\frac{\pi}{2}$ ?
| $$f'(x)=\frac{\sin x-x\cos x}{\sin^2x}=\frac{\cos x(\tan x-x)}{\sin^2x}$$
The function is increasing in the given domain since $\tan x>x$ for $x\in\left(0,\frac{\pi}{2}\right)$.
When $x\to 0$, $\frac{x}{\sin x}=1$ and at $x=\frac{\pi}{2}$, $\frac{x}{\sin x}=\frac{\pi}{2}$.
The minimum area is the area of rectangle with sides $1,\frac{\pi}{2}$.
The maximum area is the area of a rectangle with sides $\frac{\pi}{2},\frac{\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786569",
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"source": "stackexchange",
"question_score": "3",
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If $a$ and $b$ are roots of $x^4+x^3-1=0$, $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I have to prove that:
If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.
I tried this :
$a$ and $b$ are root of $x^4+x^3-1=0$ means :
$\begin{cases}
a^4+a^3-1=0\\
b^4+b^3-1=0
\end{cases}$
which gives us :
$(ab)^4+(ab)^3=a^3+b^3+a^4+b^4+a^4b^3-a^3b^4-1$
can you help me carry on ? or propose another solution ? thanks in advance
| The companion matrix of $p(z) = z^4 + z^3 - 1$ is
$$ A = \pmatrix{0 & 0 & 0 & 1\cr
1 & 0 & 0 & 0\cr
0 & 1 & 0 & 0\cr
0 & 0 & 1 & -1\cr}$$
This is a matrix whose eigenvalues are the roots of $p(z)$. The Kronecker product $A \otimes A$ is a $16 \times 16$ matrix; all products $ab$ where $a$ and $b$ are roots of $p(z)$ are eigenvalues of $A \otimes A$. Moreover, if $a \ne b$, $ab$ will be an eigenvalue of multiplicity $\ge 2$, because
if $u$ and $v$ are eigenvectors of $A$ for eigenvalues $a$ and $b$,
both $u \otimes v$ and $v \otimes u$ are eigenvectors of $A \otimes A$ for eigenvalue $ab$.
Now the
characteristic polynomial of $A \otimes A$ turns out to be
$$\left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right) \left( {z}^{6}+{z}^{4}+{
z}^{3}-{z}^{2}-1 \right) ^{2}
$$
The products $ab$ with $a \ne b$ are roots of this characteristic polynomial
with multiplicity $\ge 2$, and therefore are roots of $ {z}^{6}+{z}^{4}+{
z}^{3}-{z}^{2}-1 $.
In the case $a=b$, $ab = a^2$ turns out to be a root of the other factor
$ {z}^{4}-{z}^{3}-2\,{z}^{2}+1 $.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
| Note that for $n\ge 2$ we have
$$\frac{n(n+1)}{n!}= \frac{1}{(n-2)!} + \frac{2}{(n-1)!}. $$
This lets us rewrite the sum as
$$
\frac{2}{0!} + \left(\frac{1}{0!} + \frac{2}{1!} \right)
+ \left(\frac{1}{1!} + \frac{2}{2!} \right)
+ \left(\frac{1}{2!} + \frac{2}{3!} \right)
+ \left(\frac{1}{3!} + \frac{2}{4!} \right)
+ \cdots
$$
Rearranging in the obvious way gives the sum as $3e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is there an easy way of showing $8\cos^3(12^o)-6\cos(12^o)=\phi$? Is there an easy way of showing (1)
(1)
$$8\cos^3(12^o)-6\cos(12^o)=\phi$$
with out substituting into the equation?
$8\left(\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\right)^3-6\cdot\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]=\phi$
Trigonometric constant expressed in real radicals-Wikipedia.
$\cos(12^o)=\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]$
$\phi=\frac{1+\sqrt5}{2}$
| Recall that $$\cos 3x=4\cos^3 x- 3\cos x$$
$$\implies 2(4\cos^3 12^{\circ}-3\cos 12^{\circ})=2\cos36^\circ$$
Using $$\cos 36^\circ=\frac{1+\sqrt{5}}4$$
We get the that
$$8\cos^3 12^{\circ}-6\cos 12^{\circ}=\frac{1+\sqrt{5}}2=\phi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794068",
"timestamp": "2023-03-29T00:00:00",
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Show that $(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6$ for $a^2 + b^2 + c^2 + d^2 = 1$. For $a, b, c, d \in \Bbb R$ such that $a^2 + b^2 + c^2 + d^2 = 1$, show that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6.$$
The answer uses the mysterious identity $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 + (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4 = 6(a^2 + b^2 + c^2 + d^2)^2.$$
But this just comes out of the blue. Are there any other solutions?
Edit (for those who say this is lacking context): This comes from the beginning of a book named Algebraic Inequalities (not in English). After the preface to the first chapter, where the author introduced that many inequalities can be proved by using the non-negativity of $x^2$, this is the first example (without developing any other methods yet). Then the author notes: the reader is supposed to memorize identities like that.
| Note that $(a+b)^4+(a-b)^4=2a^4+12a^2b^2+2b^2$. Hence
$\begin{align}(a + b)^4 &+ (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4\\
&+ (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4\\
&=6(a^4+b^4+c^4+d^4)+12(a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2)\\
&= 6(a^2 + b^2 + c^2 + d^2)^2.\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794918",
"timestamp": "2023-03-29T00:00:00",
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Examine convergence of $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ Examine convergence of $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ for $a, b > 0$. There are 2 problems. $|\sin(x)|^b = 0$ for $x = k \pi$ and $x^a = 0$ for $x = 0$. We can write $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx = \int_0^{1} \frac{1}{x^a \cdot |\sin(x)| ^b}dx + \int_1^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ but what to do next?
| Note that $\lim_{x \to 0} x^{-1} \sin x = 1$ and on $[0,1]$ we have $\sin 1 \leqslant x^{-1} \sin x \leqslant 1.$
Hence, with $a,b > 0$,
$$(\sin 1)^b x^{a+b} \leqslant x^a |\sin x|^b = x^{a+b}|x^{-1} \sin x |^b \leqslant x^{a+b},$$
and
$$\frac{1}{x^{a+b}} \leqslant \frac{1}{x^a |\sin x|^b} \leqslant \frac{(\sin 1)^{-b}}{x^{a+b} }.$$
By the comparison test, the integral over $[0,1]$ converges if $a+b < 1$ and diverges if $a+b \geqslant 1.$
You can show the integral over $[1, \infty)$ diverges if $a \leqslant 1$ by comparing with a divergent $p$-series, using
$$\int_1^\infty \frac{1}{x^a |\sin x|^b} \, dx \geqslant \sum_{k=1}^\infty \int_{2k \pi + \pi/4}^{2k \pi + \pi/2} \frac{1}{x^a |\sin x|^b} \, dx. $$
Since $0 < 1/\sqrt{2}\leqslant \sin x \leqslant 1$ on $[2k\pi + \pi/4, 2k \pi + \pi/2],$ we have
$$\int_1^\infty \frac{1}{x^a |\sin x|^b} \, dx \geqslant \frac{\pi}{4}\sum_{k=1}^\infty \frac{1}{(2\pi)^a(k + 1/4)^a } \geqslant \frac{\pi}{4}\sum_{k=1}^\infty \frac{1}{(4\pi)^ak^a } . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796258",
"timestamp": "2023-03-29T00:00:00",
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How to solve the equation $(2x + 1)(2x + 3) = 143$ without using the Quadratic Formula? I have been a bit stuck on this question.
The product of two consecutive odd numbers is $143$. Find the next numbers.
I have made this into:
$$
(2x+1)(2x+3)=143.
$$
I got $x_1 = -7$ and $x_2 = -5$ for this, which doesn't seem right.
How can I fix this? (By the way we are only able to solve for $x$ by using factorising and not with the quadratic formula).
| \begin{align*}
(2x + 1)(2x + 3) & = 143\\
2x(2x + 3) + 1(2x + 3) & = 143 && \text{expand}\\
4x^2 + 6x + 2x + 3 & = 143 && \text{apply the distributive law}\\
4x^2 + 8x + 3 & = 143 && \text{combine like terms}\\
4x^2 + 8x - 140 & = 0 && \text{subtract $140$ from each side of the equation}\\
x^2 + 2x - 35 & = 0 && \text{divide each side of the equation by $4$}\\
(x + 7)(x - 5) & = 0 && \text{factor the quadratic}
\end{align*}
If a product is equal to zero, then one of the factors must be equal to zero. Hence,
\begin{align*}
x + 7 & = 0 & x - 5 & = 0\\
x & = -7 & x & = 5
\end{align*}
You can verify that these values are correct by substituting them into the equation $(2x + 1)(2x + 3) = 143$.
If $x = -7$, then the consecutive odd numbers are $2(-7) + 1 = -13$ and $2(-7) + 3 = -11$.
If $x = 5$, then the consecutive odd numbers are $2(5) + 1 = 11$ and $2(5) + 3 = 13$.
| {
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"url": "https://math.stackexchange.com/questions/1797659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A cube and a sphere have equal volume. What is the ratio of their surface areas? The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$
Since $$ \ a^3 = \frac{4}{3} \pi r^3 $$
I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$
and,
$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 \pi r^2 $$
But I am not really sure how to arrive at the desired result. I have tried to simplify it, but apparently I am missing some step in the process and come to a result that is far from the correct one.
Could you please help? Thank you.
| A sphere with equal volume to a cube of side $a$ must have radius $r$:
$$\frac{4\pi r^3}{3} = a^3$$
So
$$r = \sqrt[3]{\frac{3}{4 \pi}}a$$
Now just take
$$\frac{4\pi r^2}{6a^2} = \frac{4\pi\sqrt[3]{\frac{9}{16 \pi^2}}a^2}{6a^2} = \frac{2}{3}\pi\sqrt[3]{\frac{9}{16 \pi^2}} = \sqrt[3]{\frac{9\cdot 8 \pi^3}{27 \cdot 16 \pi^2}} = \sqrt[3]{\pi/6}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the problem. Any help is appreciated.
Find the value of $$\frac{a^2}{a^4+a^2+1}$$ if $$\frac{a}{a^2+a+1}=\frac{1}{6}$$
| Hint
$$\frac{a^2}{a^4+a^2+1}=\frac{a}{
(a^2+a+1)}\frac{a}{(a^2-a+1)}=\frac{1}{(\frac{a^2+1}{a}+1)}\frac{1}{(\frac{a^2+1}{a}-1)}=\frac{1}{((\frac{a^2+1}{a})^2-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Simplifying sum of powers of conjugate pairs The result of summing a conjugate pair of numbers each raised to the power $n$:
$$
(a + bi)^n + (a - bi)^n
$$
Produces a real number where $a + bi$ is a complex number.
Given the result is real, is there a simplified way to express the above expression in terms of $a$ and $b$ involving no imaginary number $i$ in the simplified expression?
| By De Moivre's Formula we have, for $a+bi=r(\cos \theta+i\sin \theta)$
$$(a+bi)^n+(a-bi)^n=\\r^n(\cos n\theta+i\sin n\theta)+r^n(\cos (-n\theta)+i\sin (-n\theta))=\\2r^n\cos n\theta$$
Since $\cos n\theta=\cos (-n\theta)$ and $\sin n\theta=-\sin(-n\theta)$
Now we use the fact that $$r=\sqrt{a^2+b^2}$$ and
$$L(a,b)=\cos n\theta=\begin{cases} \cos\theta\Big[1-\frac{(n^2-1^2)\sin^2\theta}{2!}+\frac{(n^2-1^2)(n^2-3^2)\sin^2\theta^4}{4!}-\cdots\Big] \text{n odd}\\1-\frac{n^2\sin^2\theta}{2!}+\frac{n^2(n^2-2^2)\sin^4\theta}{4!}-\cdots \text{n even} \end{cases}$$
where $$\cos\theta=\frac{a}{\sqrt{a^2+b^2}},\sin\theta=\frac{b}
{\sqrt{a^2+b^2}}$$
to obtain the expression $$S(a,b)=2(\sqrt{a^2+b^2})^nL(a,b)$$
which is not dependant on $i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Limit of $\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}$ when $n\to\infty$ I have to show the convergence of the series
$$\lim\limits_{n \to \infty}a_n=\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}.$$ I am quite sure that the limit is 1.5. I wanted to show this by estimation for another series smaller than $a_n$ and a further one that is bigger than $\lim\limits_{n \to \infty}a_n$.
I have been trying to find this estimation for hours without success. Can someone help me, please?
| break it up.
$a_n=\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} + \sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk}+\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk}.$
Since every element of each of those series is greater than zero,
the series for $a_n$ converges iff all three of the above converge, and diverges if any of the above diverge.
$\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} <\sum\limits_{k=1}^{n} \frac{1}{2n}$
$\sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk} <\sum\limits_{k=1}^{n} \frac{1}{2^{k}}$
$\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk} < \sum\limits_{k=1}^{n} \frac{1}{2^{k}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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All real values of $k$ in rational function
All real values of $k$ for which the range of function $\displaystyle f(x) = \frac{x-1}{k-x^2+1}$ does not contain the interval $\displaystyle \left[-1,-\frac{1}{3}\right].$
$\bf{}My\; Try::$ Let $\displaystyle y = \frac{x-1}{k-x^2+1}\Rightarrow ky-x^2y+y=x-1\Rightarrow x^2y+x-ky-y-1=0$
Now for real values of $y\;,$ We must have $x$ real. So Put $\bf{Discriminant\geq 0}$
So we get $1+4y(ky+y+1)\geq0\Rightarrow 4ky^2+4y^2+4y+1\geq0$
So We get $4(k+1)y^2+4y+1\geq0\;$
Now How can I solve after that, Help required, Thanks
| As you did, considering the discriminant, we get
$$(4k+4)y^2+4y+1\ge 0\tag1$$
Now, let us separate it into cases.
Case 1 : $k\lt -1$
$$(1)\iff (-4k-4)y^2-4y-1\le 0\iff \frac{1-\sqrt{-k}}{2(-k-1)}\le y\le \frac{1+\sqrt{-k}}{2(-k-1)}$$
So, we need
$$\frac{1+\sqrt{-k}}{2(-k-1)}\lt -1\quad\text{or}\quad \frac{1-\sqrt{-k}}{2(-k-1)}\gt -\frac 13$$
Solving these gives
$$k\lt -1\quad\text{or}\quad -1\lt k\lt -1/4$$
So, this case is sufficient.
Case 2 : $k=-1$
Since $(1)\iff y\ge -1/4\gt -1/3$, this is sufficient.
Case 3 : $-1\lt k\lt 0$
$$(1)\iff y\le\frac{-1-\sqrt{-k}}{2(k+1)}\quad\text{or}\quad y\ge \frac{-1+\sqrt{-k}}{2(k+1)}$$
So, we need
$$\frac{-1-\sqrt{-k}}{2(k+1)}\lt -1\quad\text{and}\quad \frac{-1+\sqrt{-k}}{2(k+1)}\gt -\frac 13$$
Solving these gives
$$-1\lt k\lt -\frac 14$$
So, in this case, we have $-1\lt k\lt -1/4$.
Case 4 : $k\ge 0$
$(1)$ holds for every $y$, so this case does not satisfy the condition.
Therefore, the answer is $\color{red}{k\lt -1/4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the last digit of $4n^5-5n^2+n$ is $0$
Prove that the last digit of $4n^5-5n^2+n$ is $0$ for all natural $n$
My attempt:
$$4n^5-5n^2+n\overset{?}\equiv 0\pmod {10}$$
using Fermat's little theorem
$$4n\cdot\pmod 5 -5n\pmod 2+n\overset ?\equiv 0 \pmod{10}$$
I am suck here
| You want to prove that $4n^5-5n^2+n\equiv 0\pmod {10}$.
So prove that $4n^5-5n^2+n\equiv 0\pmod {2}$, which is equivalent to $-n^2+n\equiv 0$, which is always true because $n^2\equiv n\pmod {2}$.
Then prove that $4n^5-5n^2+n\equiv 0\pmod {5}$, which is equivalent to $-n^5+n\equiv 0$, which is always true because $n^5\equiv n\pmod {5}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Definite integration of an algebraic expression
Evaluate $$\int_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}}$$
I think none of the properties of definite integral will be useful here so I think I will have to integrate. But I am unable to do so. Some hints on how to integrate. Thanks.
| Mathematica gives the indefinite integral:
$\int {1-x \over 1+x}{1 \over \sqrt{x+x^2+x^3}} dx =$
$-\frac{2 \sqrt[6]{-1} x^{3/2} \left(i \sqrt{1-\frac{(-1)^{2/3}}{x}}
\sqrt{\frac{x+\sqrt[3]{-1}}{x}} F\left(i \sinh
^{-1}\left(\frac{(-1)^{5/6}}{\sqrt{x}}\right)|(-1)^{2/3}\right)+\frac{2
\left(1+\sqrt[3]{-1}\right)
\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right) \left(\sqrt[3]{-1}
\sqrt{x}-1\right)}} \sqrt{\frac{(-1)^{2/3} \sqrt{x}-1}{\left(1+\sqrt[3]{-1}\right)
\left(\sqrt[3]{-1} \sqrt{x}-1\right)}} \sqrt{-\frac{(-1)^{2/3}
\sqrt{x}+1}{\sqrt{x}+\sqrt[3]{-1}-1}} \left(\sqrt[3]{-1} \sqrt{x}-1\right)^2
\left(\left(\sqrt[3]{-1}+i\right) F\left(\left.\sin
^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
\left(\sqrt[3]{-1} \sqrt{x}-1\right)}}\right)\right|-3\right)-2 \sqrt[3]{-1} \Pi
\left(\frac{-3 i+(1+2 i) \sqrt{3}}{(2+i)+\sqrt{3}};\left.\sin
^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
\left(\sqrt[3]{-1} \sqrt{x}-1\right)}}\right)\right|-3\right)\right)}{x}-\frac{2
\left(1+\sqrt[3]{-1}\right)
\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right) \left(\sqrt[3]{-1}
\sqrt{x}-1\right)}} \sqrt{\frac{(-1)^{2/3} \sqrt{x}-1}{\left(1+\sqrt[3]{-1}\right)
\left(\sqrt[3]{-1} \sqrt{x}-1\right)}} \sqrt{-\frac{(-1)^{2/3}
\sqrt{x}+1}{\sqrt{x}+\sqrt[3]{-1}-1}} \left(\sqrt[3]{-1} \sqrt{x}-1\right)^2
\left(\left(\sqrt[3]{-1}-i\right) F\left(\left.\sin
^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
\left(\sqrt[3]{-1} \sqrt{x}-1\right)}}\right)\right|-3\right)-2 \sqrt[3]{-1} \Pi
\left(-\frac{i \left(3+(2+i) \sqrt{3}\right)}{(-2+i)+\sqrt{3}};\left.\sin
^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
\left(\sqrt[3]{-1}
\sqrt{x}-1\right)}}\right)\right|-3\right)\right)}{x}\right)}{\sqrt{x
\left(x^2+x+1\right)}}$
which very strongly suggests that there's no simple method "by hand."
You can also do a Taylor series:
$\frac{1-x}{(x+1) \sqrt{x^3+x^2+x}} \approx \frac{1}{\sqrt{x}}-\frac{5 \sqrt{x}}{2}+\frac{23 x^{3/2}}{8}-\frac{37
x^{5/2}}{16}+\frac{203 x^{7/2}}{128}-\frac{355 x^{9/2}}{256}+O\left(x^{11/2}\right)$
and then integrate each term, but that will likely not give an adequate solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x.
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha gave me two incompatible answers
$$x=\pm\sqrt{n}, \ \ \ \ \ \ \ \ \sqrt{n}\not=0$$
$$x=\pm\frac{\sqrt{n}}{\sqrt{3}}, \ \ \ \ \ \ \ \ \sqrt{n}\not=0$$
The first answer is wrong the second is correct.
Can anyone explain why the first solution is wrong and how to derive the second?
| Let $z = \frac{1}{2}(n - x^2)$. Then the equation becomes $\Big(\frac{z}{x}\Big)^2 = z$ or better $z\Big(\frac{z}{x^2} - 1\Big) = 0$. Solutions are $z = 0$ and $z = x^2$. If $z = 0$, then $x = \pm \sqrt{n}$. If $z = x^2$, then $x = \pm \sqrt{n/3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you solve $x^2 - 4 \equiv 0 \mod 21$ There is an example in my textbook of how you solve:
$$ x^2 -4\equiv 0 \mod 21 \Leftrightarrow x^2-4\equiv 0 \mod 3 \times 7$$
and then 2 congruences can be formed out of this equation if:
$$x^2-4\equiv0 \mod 3 \\ x^2-4 \equiv 0 \mod 7$$
and from these 2 congruences result 2 more congruences, for each:
$$x - 2 \equiv 0 \mod 3 \Rightarrow x_1 = 2\\ x + 2 \equiv 0 \mod 3 \Rightarrow x_2 = 1\\ x - 2 \equiv 0 \mod 7 \Rightarrow x_3 = 2\\ x + 2\equiv 0 \mod 7 \Rightarrow x_4 = 5$$
and then 4 systems of linear congruences are formed:
$$\begin{cases} x \equiv 2 \mod 3 \\ x \equiv 2 \mod 7\end{cases}$$
$$\begin{cases} x \equiv 2 \mod 3 \\ x \equiv 5 \mod 7\end{cases}$$
$$\begin{cases} x \equiv 1 \mod 3 \\ x \equiv 2 \mod 7\end{cases}$$
$$\begin{cases} x \equiv 1 \mod 3 \\ x \equiv 5 \mod 7\end{cases}$$
What is the purpose of these systems? I already have the 4 solutions ($x_1, x_2, x_3, x_4$) of the congruence. Why do I need to form these systems?
| You need that $x^2\equiv 4\pmod{3}$ and $x^2\equiv 4\pmod{7}$. For instance, your $x_2=1$ doesn't satisfy $x_2^2\equiv 4\pmod{21}$.
You used or instead of and.
| {
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Number of solutions to $x+y+z = n$
If $\beta(n)$ is the number of triples $(x, y, z)$ such that $x + y + z = n$ and $0 \le z \le y \le x$, find $\beta(n)$.
Attempt:
I think there are many cases to look at to find $\beta(n)$. We know that the number of solutions without restriction is $\binom{n+2}{2}$, but we need one ordering. The number of solutions $0 \leq z \leq y \leq x$ is the same as the number of solutions $0 \leq y \leq x \leq z$, etc. Thus, we just need to worry about the case that $x=y=z$ since this is the only case they'll intersect. If $n$ is divisible by $3$, then $\beta(n) = \dfrac{\binom{n+2}{2}-1}{3}+1$ and otherwise $\dfrac{\binom{n+2}{2}}{3}$.
| Let $a=z$, $b=y-z$, $c=x-y$. We are looking for the number of solutions of $c+2b+3a=n$ with $a,b,c\geq 0$, so the answer is given by:
$$[x^n](1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)$$
i.e. by:
$$ [x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)} \tag{1}$$
that can be recovered through partial fraction decomposition. The meromorphic function $f(z)=\frac{1}{(1-z)(1-z^2)(1-z^3)}$ has a triple pole at $z=1$ and simple poles at $z\in\{-1,\omega,\omega^2\}$, so we know in advance that $(1)$ behaves like a $\color{green}{\text{second-degree polynomial in }n}$ plus $\color{blue}{\text{a small perturbation}}$ that depends on $n\pmod{6}$. Using the residue theorem we get:
$$ f(x) = \frac{17}{72}\cdot\frac{1}{(1-x)^3}+\frac{1}{8}\cdot\frac{1}{x+1}+\frac{1-i\sqrt{3}}{18}\cdot\frac{1}{x-\omega}+\frac{1+i\sqrt{3}}{18}\cdot\frac{1}{x+\omega}\tag{2} $$
hence:
$$ (1) = \frac{1}{72} \left(\color{green}{6n^2+36n+47}+\color{blue}{9 (-1)^n+4\cos\frac{2\pi n}{3}}\right)\tag{3}$$
and the answer is given by the closest integer to $\color{red}{\large\frac{(n+2)(n+4)}{12}}$.
| {
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Work done in a conservative vector field If my vector field is:
$F=(1-\frac{x}{x^2+y^2})i-(\frac{y}{x^2+y^2})j$
How would I go about finding the work done between A(3,2) and B(4, -3)?
I have proved that the vector field is conservative:
$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{2xy}{(x^2+y^2)^2}-\frac{2xy}{(x^2+y^2)^2}=0$
Can Green's theorem be used in this case?
| I'll elaborate the idea of @RobBland .
Let us find $f$, so that $\nabla f=\mathbf{F}$. We will do that as follows:
$$
f(x,y)=\int P(x,y)dx=\int\left(1-\frac{x}{x^2+y^2}\right)dx=x-\frac{1}{2}\ln(x^2+y^2)+c(y).\tag 1
$$
We assume $y$ to be constant while performing integration $(1)$. $f(x,y)$ satisfies $\frac{\partial f}{\partial x}=P(x,y)$ condition if and only if it has the form $(1)$. Here $c(y)$ is some arbitrary (smooth enough) function; it depends only on $y$.
Now we should find $c(y)$ using $\frac{\partial f}{\partial y}=Q(x,y)$ condition. We will differentiate $(1)$ by $y$:
$$
\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\left(x-\frac{1}{2}\ln(x^2+y^2)+c(y)\right)=-\frac{y}{x^2+y^2}+c^\prime(y).\tag2
$$
But $\frac{\partial f}{\partial y}$ should equal $Q(x,y)=-\frac{y}{x^2+y^2}$:
$$
-\frac{y}{x^2+y^2}+c^\prime(y)=-\frac{y}{x^2+y^2}.
$$
Thus we obtain $c^\prime=0$, so $c=\mathrm{const}$. Now we got the potential:
$$
f(x,y)=x-\frac{1}{2}\ln(x^2+y^2)+c,
$$
here $c$ is an arbitrary constant, that may be omitted.
Now we obtain readily:
$$
\mathrm{Work}_{A\to B}=f(B)-f(A)=f(4,-3)-f(3,2)=\left[x-\frac{1}{2}\ln(x^2+y^2)\right]\Bigg|^{(4,-3)}_{(3,2)}=\\
=1-\frac{1}{2}\ln\frac{25}{13}
$$
REMARK
$f(x,y)$ may be found easier, if we recall, that $\left(-\frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2}\right)=-\frac{\mathbf{r}}{r^2}$ is the field of a charged straight wire (the wire is along $z$ axis). Its potential $f$ is $\ln r$ (up to some factor).
| {
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How does the Pearson correlation coefficient change under rotations I was reading on wikipedia about the pearson correlation coefficient. Assuming the data has zero mean it can be written as
$$
\rho = \frac{ \sum x_i y_i } {\sqrt{\sum x_i^2 \sum y_i^2}}
$$
The caption below this image says:
[...] Note that the correlation reflects the non-linearity and direction of a linear relationship (top row), but not the slope of that relationship (middle), nor many aspects of nonlinear relationships (bottom). [...]
(bold text emphasis added by me)
The middle row of the picture shows several distributions that are perfectly correlated ($\rho=1$) and illustrates, that in that case the correlation coefficient does not change when the slope changes (apart from the case if either $x$ or $y$ is constant).
However, I have my doubts whether the correlation coefficient is really independent of the slope, when the correlation is not perfect (ie $\rho<1$). In other words, how does the correlation coefficient change, when I apply a simple rotation
$$
x'_i = x_i \cos(\alpha) + y_i \sin(\alpha) \\
y'_i = -x_i \sin(\alpha) + y_i \cos(\alpha)
$$
to the data?
Note that the rotation does not change the mean values if $\sum x = \sum y = 0$, but even in the simple form as written above I didn't manage to derive an expression for
$$
\rho(\alpha) = ??
$$
yet. Or maybe I am just a bit confused and the correlation coefficient really does not change....
| I suppose that $\sum_i x_i = \sum_i = y_i = 0$.
Moreover, $P_x = \sum_i x_i^2$, $P_y = \sum_i y_i^2$ and $C_{xy} = \sum_i x_i y_i$.
Then, the sample Pearson coefficient $\rho$ based on data $x_i$ and $y_i$ produced by random variables $X$ and $Y$ is:
$$\rho = \frac{C_{xy}}{\sqrt{P_x P_y}}.$$
Notice that:
$$P_{x'} = \sum_i {x'}^2_i = \sum_i (x_i \cos \alpha + y_i \sin \alpha)^2 = \\
\cos^2 \alpha\sum_i x_i^2 + \sin^2 \alpha\sum_i y_i^2 + 2\sum_i x_i y_i \sin \alpha \cos \alpha = \\\cos^2 \alpha P_x + \sin^2 \alpha P_y + \sin(2\alpha) C_{xy},$$
$$P_{y'} = \sum_i {y'}^2_i = \sum_i (-x_i \sin \alpha + y_i \cos \alpha)^2 = \\
\sin^2 \alpha\sum_i x_i^2 + \cos^2 \alpha\sum_i y_i^2 - 2\sum_i x_i y_i \sin \alpha \cos \alpha = \\\sin^2 \alpha P_x + \cos^2 \alpha P_y - \sin(2\alpha) C_{xy},$$
and
$$C_{x'y'} = \sum_i x_i' y_i' = \sum_i (x_i\cos \alpha + y_i \sin \alpha)( -x_i \sin \alpha + y_i \cos \alpha) = \\
-\sum_i x_i^2\sin\alpha\cos \alpha + \sum_i x_i y_i (\cos^2 \alpha - \sin^2 \alpha) + \sum_i y_i^2\sin\alpha\cos \alpha = \\
\frac{1}{2}\sin(2\alpha)(P_y - P_x) + C_{xy} \cos(2\alpha).$$
Consider $\alpha = \frac{\pi}{2}$ and join all pieces togheter:
$$\rho' = \frac{C_{x'y'}}{\sqrt{P_{x'} P_{y'}}} = \frac{\frac{1}{2}\sin(2\frac{\pi}{2})(P_y - P_x) + C_{xy} \cos(2\frac{\pi}{2})}{\sqrt{(\cos^2 \frac{\pi}{2} P_x + \sin^2 \frac{\pi}{2} P_y + \sin(2\frac{\pi}{2}) C_{xy})(\sin^2 \frac{\pi}{2} P_x + \cos^2 \frac{\pi}{2} P_y - \sin(2\frac{\pi}{2}) C_{xy})}} = \\
= \frac{-C_{xy}}{\sqrt{P_yP_x}} = - \rho.
$$
Conclusion: rotation affects Peason coefficient.
Addition
In general, the new Pearson coefficient, as a function of $\alpha$, is
$$\rho' = \frac{C_{x'y'}}{\sqrt{P_{x'} P_{y'}}} = \frac{\frac{1}{2}\sin(2\alpha)(P_y - P_x) + C_{xy} \cos(2\alpha)}{\sqrt{(\cos^2 \alpha P_x + \sin^2 \alpha P_y + \sin(2\alpha) C_{xy})(\sin^2 \alpha P_x + \cos^2 \alpha P_y - \sin(2\alpha) C_{xy})}}.
$$
| {
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Graphing $\frac{(x-3)}{(x^2-3x)}$ and $1/x$ For graphing the first equation, in the solved example given in the textbook, we proceeded as follows:
$\frac{(x-3)}{x^2-3x} \implies \frac{(x-3)}{x(x-3)} \implies \frac{1}{x}$ for $x \neq 3$.
Now the graph of this equation is exactly the same as that of $\frac{1}{x}$ except that there's a hole in $f(3)$ because the equation is not defined for $x=3$. What I don't get is, if we have simplified the equation to $\frac{1}{x}$, why do we still need to consider for the values of $x$ where the value of the original equation was undefined? Both the equations are exactly the same and yet, have different graphs, why so?
| $f(x)=\frac{x-3}{x^2-3x}=\frac{(x-3)}{x(x-3)}$, only when $x \neq 3$, as division by $0$ is not possible.
*
*If $x \neq 3$, then we can divide both sides by $(x-3)$ to get $f(x)=\frac{1}{x}$. For example, for $x=2$, we have $f(2)=\frac{(2-3)}{2(2-3)}=\frac{1}{2}$(obtained by cancelling $(2-3) \neq 0$ )
*If $x =3$, then $f(3)=\frac{(3-3)}{3(3-3)} \neq \frac{1}{3}$, as we can't cancel $(3-3)=0$ on account of division by zero.
Now, as to why division by $0$ is not possible. Read this
| {
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Orthogonal diagonalisation of a $4\times 4$ matrix Can somebody help me to orthogonally diagonalise the matrix $\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$?
| The assiocated quadratic form is : $$q(a,b,c,d)=ad+bc+cb+da=2ad+2bc=\frac{1}{2}(a+d)^2-\frac{1}{2}(a-d)^2+\frac{1}{2}(b+c)^2-\frac{1}{2}(b-c)^2$$
So you can deduce that the eigen values are $1$ and $-1$, and by coming back to a matrix notation, you have :
$$\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}=P\begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}P^t$$
$$P=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 & 0 & 0\\0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & -1 & 0 & 0\end{bmatrix}$$
| {
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I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$
I've tried this far, and I'm stuck
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x (x+12)- \log_x 4^3 &= -1 \\
\log_x(x+12)- \log_x 64 &= -1 \end{align}$$
then I substituted $4^{y+3x} = 64$
$\log_x (x+12) - \log_x 4^{y+3x} = -1$
I don't know what should I do next. any ideas?
| $4^{y+3x} = 64$
$\log_4 ({4^{y+3x}}) = \log_4 (64)$
$(y+3x)\log_4 {4} = 3$
$y + 3x = 3$
$\log_x (x+12) - 3\log_x (4) = -1$
$\log_x (\frac{x+12}{4^3}) = -1$
$\frac{x+12}{64} = x^{-1}$
$x^2 + 12x - 64 = 0$
$(x+16)(x-4) = 0$
$x = 4$, as $x > 1$ for $\log_x (x+12) - 3\log_x (4) = -1$
(Why must the base of a logarithm be a positive real number not equal to 1?)
$y + 3(4) = 3$
$y = -9$
$4 + 2(-9) = -14$
| {
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How did we derive this general term for the series? We have this series of numbers:
$1, 3, 6, 10, 15$
The general term can be described wit: $\frac{r(r + 1)}{2}$
Apparently the following series:
$1, 4, 10, 20, 35$
Can be described with $\frac{r(r + 1)(r + 2)}{6}$ based on the first series.
But I am not clear how this is derived. Can someone please explain?
| $S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 1$
$S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 2$
Subtracting 1 from 2
$0 = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots + a_{n} - a_{n-1}) - a_n$
$a_n = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots + a_{n} - a_{n-1})$
$a_n = 1 + 3 + 6 + 10 + 15 + \cdots + (a_{n} - a_{n-1})$
$a_n = \sum_{r = 0}^n \frac{r(r + 1)}{2}$
$a_n = \frac{1}{2} * \sum_{r = 0}^n (r^2 + r)$
$a_n = \frac{1}{2} * (\sum_{r = 0}^n r^2 + \sum_{r = 0}^nr)$
$a_n = \frac{1}{2} * (\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2})$
$a_n = \frac{n(n+1)}{2} * (\frac{(2n+1)}{6} + \frac{1}{2})$
$a_n = \frac{n(n+1)}{2} * (\frac{(2n+1 +3)}{6})$
$a_n = \frac{n(n+1)}{2} * (\frac{(2n+4)}{6})$
$a_n = \frac{n(n+1)}{2} * (\frac{(n+2)}{3})$
$a_n = \frac{n(n+1)(n+2)}{6}$
This way is pretty easy.
| {
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Where did $\sqrt{x^2/x^2}$ come from in $\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} = \lim_{x \to -\infty}\frac{-1-1/x}{\sqrt{x^2/x^2}} = -1$? I'm reading a calculus book and I saw the following limit solution.
$$
\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} =
\lim_{x \to -\infty}
\left(\frac{x+1}{\sqrt{x^2}} \cdot \frac{-1/x}{-1/x}\right) =
\lim_{x \to -\infty}\frac{-1-1/x}{\sqrt{x^2/x^2}} = -1.
$$
I'm having some trouble to understand where did the $\sqrt{x^2/x^2}$ came from.
| I guess it's an exercise meant to show how things might go wrong without being careful doing algebraic manipulations.
Since they're evaluating the limit at $-\infty$, it's not restrictive to assume $x<0$. Divide numerator and denominator by $-x$:
$$
\frac{x+1}{\sqrt{x^2}}=
\frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{-x}}
$$
Since $x<0$, we have $-x>0$, so $-x=\sqrt{(-x)^2}=\sqrt{x^2}$ and so
$$
\frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{-x}}
=
\frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{\sqrt{x^2}}}
=
\frac{-1-\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}}}=
-1-\frac{1}{x}
$$
A very common error would be dividing numerator and denominator by $x$ and bringing $x$ in the square root:
$$
\frac{x+1}{\sqrt{x^2}}=
\frac{1+\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{x}}
\color{red}{=}
\frac{1+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}}}
$$
where the red equals sign is wrong.
On the other hand, recalling that $\sqrt{x^2}=|x|=-x$ (because we're in the interval $x<0$), the transformation is much simpler:
$$
\frac{x+1}{\sqrt{x^2}}=
\frac{x+1}{-x}=-1-\frac{1}{x}
$$
| {
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Prove $\forall n \geq 10, 2^n > n^3$ Prove $\forall n \geq 10, 2^n > n^3$
base case: $n = 10$
$2^{10} = 1024$
$10^3 = 1000$
$1024 > 1024$.
So $P(k)$ holds for $k = n$. We seek to show $P(k+1)$ holds:
We know $2^k > k^3$.
$2^k+3k^2+3k+1>k^3+3k^2+3k+1 = (k+1)^3$
$\iff 2^k+3k^2+3k+1>(k+1)^3$
Let us compare $k^3$ and $3k^2+3k+1$:
$k^3$ $\Box$ $3k^2+3k+1$
$k^3-3k^2-3k$ $\Box$ $1$
$k(k^2-3k-3)$ $\Box$ $1$.
$k^2-3k-3 > 0$, $\forall k \geq 10$,
$\implies k^3 > 3k^2+3k+1, \forall k \geq 10$.
Therefore $2^k + k^3 > 2^k + (3k^2+3k+1)$.
We know $2^k > k^3$,
so $2^{k+1} = 2^k+2^k > 2^k + k^3 > 2^k+3k^2+3k+1 > (k+1)^3$
$\implies 2^{k+1} > (k+1)^3$.
Therefore $P(k+1)$ holds and our hypothesis is proven.
I don't really know how else to segue into the argument about comparing $k^3$ to $3k^2+3k+1$, is this ok notation?
| I see your logic: you prove the chain
$$
2^{k+1}=2^k+2^k>k^3+k^3>k^3+3k^2+3k+1=(k+1)^3
$$
where the first inequality is due to the induction hypothesis and the second is because of the inequality $k^3>3k^2+3k+1$ ($k\geq 10$) for which you supply a separate proof. I think this is a good approach although you can write it more succinctly and clearly. You also can shorten the proof for $k^3>3k^2+3k+1$ as follows
$$
k\geq 10\implies k^3\geq10k^2>3k^2+3k^2+k^2>3k^2+3k+1.
$$
Alternatively, observe that
$$
2k^3>(k+1)^3\iff2>\left(1+\frac{1}{k}\right)^3\iff k>\frac{1}{\sqrt[3]{2}-1}\approx 3.84732
$$
so, of course, if $k\geq 10$ then you also have $2k^3>(k+1)^3$.
| {
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Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$
Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.
I've tried to use the cyclotomic polynomial as:
$$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$
So I have that my polynomial is
$$\frac{X^5-1}{X-1}$$ and now i have to prove that is irreducible.
The lineal change of variables are ok*(I don't know why) so I substitute $X$ by $X+1$ then I have:
$$\frac{(X+1)^5-1}{X}=\frac{X^5+5X^4+10X^3+10X^2+5X}{X}=X^4+5X^3+10X^2+10X+5$$
And now we can apply the Eisenstein criterion with p=5. So my polynomial is irreducible in $\mathbb{Q}$
Now let's prove that it has two different irreducible factors in $\mathbb{R}$
I've tryed this way: $X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D)$
and solve the system. But solve the system is quite difficult. Is there another way?
| A different route to the factorization over the reals (obviously the end result is same as in Egreg's post, but I give the factors explicitly).
Let $p(x)$ be your polynomial. By a direct calculation we see that
$$
(x^2+\frac x2+1)^2=x^4+x^3+\frac94x^2+x+1=p(x)+\frac54 x^2.
$$
This calculation is aided by palindromic symmetry of both $p(x)$ and this quadratic. Anyway, this gives us the factorization
$$
\begin{aligned}
p(x)&=(x^2+\frac x2+1)^2-(\frac{\sqrt5}2\,x)^2\\
&=(x^2+\frac{1-\sqrt5}2\, x+1)(x^2+\frac{1+\sqrt5}2\,x+1)
\end{aligned}
$$
by the usual
$$
a^2-b^2=(a-b)(a+b)
$$
formula.
So the Golden ratio (not surprisingly given that the zeros are vertices of a regular pentagon) makes an appearance.
| {
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Find a matrix $P$ such that $P^T H P$ is a diagonal matrix with non integer eigenvalues I have this problem in my textbook:
Find a matrix $P$ such that $P^T H P$ is a diagonal matrix, where $H := \begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}$
My attempt:
$$\det(A-\lambda I) = 6 - 5\lambda + \lambda ^2 -1$$
$$= 5 - 5 \lambda + \lambda ^2 $$
$$\therefore \lambda \in \left\{\frac{1}{2}(5-\sqrt{5}),\frac{1}{2}(5+\sqrt{5}) \right\}$$
Eigenvector for $\frac{1}{2}(5-\sqrt{5}):$
???
Eigenvector for $\frac{1}{2}(5+\sqrt{5}):$
???
$$\therefore \text{ let } P = \text{ ???} $$
Check: $P^tAP$
Something to show diagonal entries are the eigenvalues as expected.
I am having difficulty with the calculations with the non integer eigenvalues basically.
EDIT:
Thanks Thomas but what about finding $P$ I know there is a formula to find what goes in front of the matrix that comes from the eigenvectors (what do I put as $\frac{X}{Y}$ in $P=\frac{X}{Y}\begin{bmatrix}-1-\sqrt5 & 1+\sqrt5\\2 & 2 \end{bmatrix} $ for it to make sense
| Just do the normal procedure of finding eigenvectors for the complex numbers.
$$\left[\begin{array}{cc|c}2-\frac{5-\sqrt5}{2}&1&0\\1&3-\frac{5-\sqrt5}{2}&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}4-5+\sqrt5&2&0\\2&6-5+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}-1+\sqrt5&2&0\\2&1+\sqrt5&0\end{array}\right]$$Multiply row 1 by $-1-\sqrt5$,
$$\left[\begin{array}{cc|c}-4&-2-2\sqrt5&0\\2&1+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}-2&-1-\sqrt5&0\\2&1+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}2&1+\sqrt5&0\\0&0&0\end{array}\right]$$Thus $v_1=\begin{bmatrix}-1-\sqrt5\\2\end{bmatrix}$, correspondingly $v_2=\begin{bmatrix}-1+\sqrt5\\2\end{bmatrix}$
Now you can finish the rest of the problem.
| {
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Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$
$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roots of cubic equation, Then
$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$
So we get $\displaystyle t^3-4t^2-5t-c=0$
Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$
and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$
So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$
Now How can I solve it after that, Help required, Thanks
| Note that $$c=t((t-2)^2+1),\ t\in \mathbb{R}$$, so according to the calculations, $c\in [a,b]$ where $$a=t((t-2)^2+1)|_{t=5/3}=\frac{50}{27},\ b=t((t-2)^2+1)|_{t=1}=2$$
| {
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If $x^2 + 4 + 3\sin(ax+b)-2x=0$ has at least one real solution, ...
If the equation
$$x^2 + 4 + 3\sin(ax+b)-2x=0$$
has at least one real solution, where $a$ and $b$ belong to $(0,2\pi)$, then what can a possible value of $(a+b)$ be?
Can anyone say what exactly should be done in the question?
Value of $\sin (ax + b)$ could be found, but how can we know about the value of $(a+b)$?
| $x^2 + 4 + 3sin(ax+b)-2x=0$ can be written as $(x-1)^2+3= -3sin(ax+b)$ now you can observe that left hand side will always be greater than or equal to 3, while right hand side will always be less than or equal to 3.
hence for consistent system both sides will be equal to 3 .that implies $x=1$ further $sin(ax+b)=-1$ putting x in this will give $sin(a+b)=-1$
under given interval (0,2$\pi$) $a+b$ can be equal to ${\frac{3\pi}{2}}$ and ${\frac{7\pi}{2}}$ ;as $a+b$ will lie between $0$ and $4\pi$
| {
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Trigonometry- Triangles Let $A,B,C$ by the angles of a triangle.
Then how to prove that
$$a^3cos(B-C) + b^3cos(C-A) +c^3cos(A-B) = 3abc$$
I divided both sides by $abc$ and then tried to open the cosine function but nothing worked. I also took the cube of sine in the triple angle identity but that too just made the function more complicated and yielded no result.
| Let $2R=k$
$a=k\sin A, b=k\sin B,c=k\sin C$ (sine rule)
$$ a^3 \cos (B-C)= a^2\cdot a\cos (B-C)=$$
$$= a^2\cdot k\sin A\cos (B-C)= ka^2\sin (180-(B+C))\cos (B-C)=$$
$$= ka^2\sin (B+C)\cos (B-C)=\frac{ka^2}{2} (\sin 2B+\sin 2C) = $$
$$=\frac{ka^2}{2} (2\sin B\cos B+2\sin C\cos C) = ka^2(\sin B\cos B+\sin C\cos C) = $$
$$= a^2((k\sin B)\cos B+(k\sin C)\cos C)) =$$$$= a^2 (b\cos B+c\cos C)$$
The original expression now looks like this:
$$a^2(b\cos B+c\cos C)+b^2(c\cos C+a\cos A)+c^2(a\cos A+b\cos B) = $$
$$= a^2b\cos B + a^2c\cos C+b^2c\cos C+b^2a\cos A+c^2a\cos A+c^2b\cos B$$
$$=
a^2b\cos B +
b^2a\cos A +
b^2c\cos C +
c^2b\cos B +
a^2c\cos C +
c^2a\cos A
$$
$$=ab(a\cos B+b\cos A)+bc(b\cos C+c\cos B)+ca(a\cos C+c\cos A) =$$
$$= abc+bca+cab = 3abc$$
| {
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Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$ $$\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\frac{5!}{9!}+\frac{6!}{10!}+\cdots$$
This goes up to infinity. Trying finite cases may help.
My Attempt:It seems that it is going to be $\frac{1}{18}$. My calculations show that its going near $\frac{1}{18}$.
| Write $$\begin{align} {\frac{n!}{(n+4)!}}&=\frac{1}{(n+1)(n+2)(n+3)(n+4)}=\\&=\frac{A}{(n+1)(n+2)(n+3)}-\frac{B}{(n+2)(n+3)(n+4)}\end{align}$$
which gives $(n+4)A-B(n+1)=1 \implies A=B=1/3$
Then we get a telescoping series:
$$\begin{align}\sum\limits_{n=0}^{\infty }{\frac{n!}{(n+4)!}}=\frac{1}{3}\sum\limits_{n=0}^{\infty }\left({\frac{1}{(n+1)(n+2)(n+3)}}-\frac{1}{(n+2)(n+3)(n+4)}\right)=\frac{1}{18}\end{align}$$
| {
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Prove inequality for $a, b , c, d$ for $a,b, c, d\ge 0$ How does one prove the following inequality:
$$ \frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}+\frac{d}{d^2+1}\le\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{d^2+1}+\frac{d}{a^2+1}$$
without much computation? Is there a trick?
| Yes, there is.
You can consider the numbers $a,b,c,d$ on one side and the numbers $\frac1{a^2+1}$, $\frac1{b^2+1}$, $\frac1{c^2+1}$, $\frac1{d^2+1}$ on the other, and apply the Rearrangement Inequality.
| {
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Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q)^2]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P3| \frac{x\color{red}{q^{x+1}}+[-(x+1)]\color{red}{q^x}+1+[(x+1)(1-q)^2]\color{red}{q^x}}{(1-q)^2} = \frac{x\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2} | $$
Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' $\color{red}{q^x}$ are added next.
$$P4| \frac{xq^{x+1}+[-(x+1)+(x+1)(<1^2q^0+\binom{2}{1}1q-1^0q^2>)]q^x+1}{(1-q)^2}=\frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
$$P5 | \frac{xq^{x+1}+[2xq-xq^2+2q-q^2]q^x+1}{(1-q)^2} = \frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated.
Thanks in advance.
| Hint:
$1+2q+3q^2+\ldots+nq^{n-1}= (q+q^2+\ldots+q^{n})'$
| {
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In $\triangle ABC$ with $A = \frac{\pi}{4}$, what is the range of $\tan B\tan C$?
In a $\triangle ABC\;,$ If $\displaystyle A=\frac{\pi}{4}\;,$ and $\tan B\cdot \tan C = p\;,$ Then range of $p$
$\bf{My\; Try::}$ For a $\triangle ABC\;, A+B+C=\pi.$ So we get $\displaystyle A+B=\frac{3\pi}{4}$
So $$\tan(A+B)=-1\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B} = -1$$
So we get $$\tan A+\tan B=p-1$$
Now how can i solve after that help required, Thanks
| Like Triangular sides, if $\tan B=q,\tan C=\dfrac{q+1}{q-1}$
So, we have $p=\dfrac{q(q+1)}{q-1}$
$$\iff q^2+q(1-p)+p=0$$
As $q$ is real, the discriminant $$(1-p)^2-4p$$ must be $\ge0$
$$\implies p^2-6p+1\ge0$$
Now if $(x-a)(x-b)\ge0$ with $a\le b,$ we need $x\ge b$ or $x\le a$
| {
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Suppose that $a$ and $b$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.
Suppose that $a$ and $b \in \mathbb{Z}^+$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.
I have reduced the above formulation to these two cases. Assuming $b = a + k$. Proving that any of the below two implies that $a=b$ will be enough.
$$a^2b|(a+b)^3 - 3ab^2$$
$$a^2b|2a^3+3a(a+k)+k^3$$
I can't proceed from here. How should I proceed from here?
Thanks.
| Write $d=\gcd(a,b)$ and $a=dA$, $b=dB$, with $\gcd(A,B)=1$.
Then we can cancel $d^3$ on both sides of $a^2b \mid a^3+b^3$ and get $A^2B \mid A^3+B^3$.
This implies that $A^2 \mid B^3$ and $B \mid A^3$, and so $A=B=1$.
Indeed, $A^2 \mid B^3$ implies that every prime that divides $A$ divides $B$. Since $\gcd(A,B)=1$, we must have $A=1$. But then $B \mid A^3$ implies $B=1$.
| {
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Prove $\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$ $a,b,c \geqslant 0,$$ a+b+c=3$, and $(a+b)(b+c)(c+a) \neq 0$ , prove
$$\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$$
I try Bernouli's inequality but checking the case of $a>1$ and $0<a<1$ is truly complicated. I try to come up with some estimations (i.e., $x^x \geqslant \frac12 (1+x^2$)) but has not yet been successful.
| Yuri Negometyanov gave
$$\left(\frac{a+1}{a+b}\right)^a = \left(1 + \frac{b-1}{a+1}\right)^{-a} \ge 1 - a\cdot \frac{b-1}{a+1}
= 2 - b + \frac{b-1}{a+1}. \tag{1}$$
This follows from the Bernoulli inequality $(1+x)^r \ge 1 + rx$ for $x > -1$ and $r \le 0$.
Using (1), it suffices to prove that
$$\frac{b-1}{1+a}+ \frac{c-1}{1+b}+ \frac{a-1}{1+c}\ge 0. \tag{2}$$
We can prove (2) in another way as follows.
After clearing the denominators, it suffices to prove that
$$a^2b + b^2c + c^2a + a^2 + b^2 + c^2 - a - b - c - 3 \ge 0$$
or (since $a+b+c=3$)
$$a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 \ge 0.$$
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$a^2b + b^2c + c^2a \ge \frac{(ab + bc + ca)^2}{b + c + a} = \frac{(ab + bc + ca)^2}{3}.$$
Thus, we have
\begin{align}
a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 &\ge \frac{(ab + bc + ca)^2}{3} - 2(ab+bc+ca) + 3\\
&= \frac{1}{3}(ab + bc + ca - 3)^2\\
&\ge 0.
\end{align}
We are done.
| {
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How to obtain this factorization of $x^4+4$? $x^4 + 4 = (x^2 + 2x +2)(x^2 - 2x +2)$
I am curious how would one obtain this factorization?
Clearly, once the factorization is known it is routine to verify it, however the hard part is how to find the factorization in the first place?
Thanks!
I observed that $(A+B)(A-B)=A^2-B^2$ can be applicable here with $A=x^2+2$, $B=2x$.
Is there any other trick?
| You can try as follows (based on the $(a+b)^2=a^2+2ab+b^2)$):
\begin{align*}
x^4+4&=(x^2)^2+2^2\\
&=\big(x^2+2\big)^2-4x^2\\
&=\big(x^2+2\big)^2-(2x)^2\\
&=(x^2+2+2x)(x^2+2-2x)\\
&=(x^2+2x+2)(x^2-2x+2).
\end{align*}
| {
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How to find range of $\frac{\sqrt{1+2x^2}}{1+x^2}$? How to find range of $$\frac{\sqrt{1+2x^2}}{1+x^2}$$ ?
I tried put it equal to $y$ and squaring but I'm getting $4$th degree equation.
| Let $\sqrt{1+2x^2}=u\ge1,\implies1+x^2=\dfrac{1+u^2}2$
$$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2u}{1+u^2}=\dfrac2{u+\dfrac1u}$$
Now $u+\dfrac1u\ge2\sqrt{u\cdot\dfrac1u}=2$
Alternatively, let $\sqrt{1+2x^2}=\tan v$
Clearly, $\tan v\ge1+2\cdot0=1$ WLOG we can choose $\dfrac\pi4\le v<\dfrac\pi2\iff\dfrac\pi2\le2v<\pi$
Now $$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2\tan v}{1+\tan^2v}=\sin2v$$
| {
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An erroneous application of the Counting Theorem to a regular hexagon? I'm trying to count the unique orbits of a regular hexagon such that each vertex is either Black or White and each edge is either Red, Gree, or Blue. The group I've chosen to act on the hexagon is the dihedral group $D_7$, $$\{e,r,r^2,r^3,r^4,r^5,r^6,s,rs,r^2s,r^3s,r^4s,r^5s,r^6s\}$$ where $r$ is a rotation by $\frac{\pi}{3}$, and $s$ a reflection in the axis connecting two opposite vertices or the midpoints of two opposite edges. When chopped up, I get the following partition into conjugacy classes: $$\{e\} \hspace{0.5cm} \{r,r^6\}\hspace{0.5cm} \{r^2,r^5\}\hspace{0.5cm} \{r^3,r^4\}\hspace{0.5cm}\{s,r^2s,r^4s,r^6s\}\hspace{0.5cm} \{rs,r^3s,r^5s\}$$
Taking the first element of each conjugacy class as the representative, I then go about counting the permutations that are left fixed by that representative. Here's my count (note, $X^g$ denotes the set of all regular hexagons left fixed by group element $g$): $$\begin{align*}
|X^e|&=3^6\times 2^6 & |X^r|&=3\times 2 & |X^{r^2}|&=3^2\times 2^2\\ |X^{r^3}|&=3^3\times 2^3 & |X^s|&=3^4\times 2^3 + 3^3\times 2^4 & |X^{rs}|&=3^3\times 2^4 + 3^4\times 2^3 \end{align*}$$
Notice that the order of the last two sets, $X^s$ and $X^{rs}$, are sums: one addend counts the reflections through opposite vertices and the other through midpoints of opposite sides. When I apply the Counting Theorem (aka Burside's Lemma?) I obtain $$\frac{1}{14}[3^6\times 2^6 + 2(3\times 2) + 2(3^2 \times 2^2) + 2(3^3\times 2^3) + 4(3^4\times 2^3 + 3^3 \times 2^4) + 3(3^3\times 2^4 + 3^4 \times 2^3)]$$ and it is here I stumbled when I saw this product is not an integer.
| O negligence! Fit for a fool to fall by.
All credit to @ladisch for obviously making the obvious.
a) I should have used $D_6$ instead of $D_7$. So chop off the $r^6$ and $r^6s$ from the group and we get the dihedral group of order 12.
b) The conjugacy classes containing rotations only are adjusted acrodingly and $\{s,r^2s,r^4s,r^6s\}$ becomes $\{s,r^2s,r^4s\}$.
c) My counting of the elements of $X^g$, where $g$ is a representative of the conjugacy classes, remains the same...
d)... but the product $$\frac{1}{12}[3^6\times 2^6 + 2(3\times 2) + 3^2 \times 2^2 + 2(3^3\times 2^3) + 3(3^4\times 2^3) + 3(3^3\times 2^4)]=4183$$
| {
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Algebraic Manipulation Given that
$ a^2 - b^2 = 60 $
and
$a - b = 6 $
$a + b = 10$
Find value of $a\cdot b$
I tried
$(a-b)^2 = 6^2 \longrightarrow a^2 - 2ab + b^2 = 36$
| We can arrange the equations as a system of equations as follows
$$\begin{cases} a-b=6\\ a+b=10 \end{cases}$$
Now we can add the two equations
$$(a-b)+(a+b)=6+10$$
$$ 2a=16 \rightarrow a=8 $$
$$ a-b=6,\ 8-b=6\rightarrow b=2 $$
So the solutions are $a=8$ and $b=2$
| {
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Solve in integers the equation $\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$
Solve in integers the equation
$$\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$$
My attempt:
I used http://www.wolframalpha.com/:
$x=-2; y=\{3,4,5,6,7,..\}$ or $x=-1, y=\{-10,-9,...\}$.
1) Let $\lfloor\frac{x^2-y^3}{x+y^2} \rfloor=\frac{x^2-y^3}{x+y^2}-\{\frac{x^2-y^3}{x+y^2}\}$. Then
$$\frac{x^2-y^3}{x+y^2}-\{\frac{x^2-y^3}{x+y^2}\}=1+x-y$$
$$\frac{x^2-y^3}{x+y^2}-1-x+y=\{\frac{x^2-y^3}{x+y^2}\}$$
$$0\le\frac{x^2-y^3}{x+y^2}-1-x+y<1$$
2)
$$\lfloor\frac{x^2-y^3}{x+y^2} \rfloor=\lfloor\frac{x^2+2xy^2+y^4-2xy^2-y^4-y^3}{x+y^2} \rfloor=x+y^2-\lfloor\frac{2xy^2+y^4+y^3}{x+y^2} \rfloor$$
| If $x\ge0$ it is easy to show that the equation cannot hold. If $x=y=0$ that is obvious. So assume $x\ge0,y\ne0$. Then $x+y^2>0$, and we have $(x-y)(x+y^2)-(x^2-y^3)=xy(y-1)\ge0$, so dividing by $x+y^2$ we have $x-y\ge\frac{x^2-y^3}{x+y^2}$. The LHS is an integer, so we have $1+x-y>\lfloor\frac{x^2-y^3}{x+y^2}\rfloor$.
We now need to consider $x\le0$. For $|x|<y^2$, we see, looking carefully at the above, that it proves that $x-y\ge\frac{x^2-y^3}{x+y^2}$. Hence $1+x-y>\lfloor\frac{x^2-y^3}{x+y^2}\rfloor$ and there are no solutions.
For $|x|>y^2$ we need the slightly stronger $2+x-y\le\frac{x^2-y^3}{x+y^2}$ to conclude there are no solutions. We have $x+y^2<0$ so that is equivalent to $(x+y^2)(2+x-y)\ge x^2-y^3$ or $x(y^2-y+2)\ge-2y^2\ (*)$.
This case needs a little more care. If $x\le-3$ then LHS $\ge3(y^2-y+2)$ and so $(*)$ holds since $y^2-3y+6=(y-\frac{3}{2})^2+\frac{3}{2}>0$, so there are no solutions.
If $x=-1$, then the original equation becomes $\lfloor\frac{1-y^3}{y^2-1}\rfloor=-y$. That fails for $y=1,0,-1,-2$ (see Note at end). For $y>1$ we have $1-y^3<-y(y^2-1)$ and hence $\frac{1-y^3}{y^2-1}<-y$, so there are no solutions. For $y\le-3$ we have $y^2+y-2=(y+2)(y-1)>0$, so $1-y^3<-y^3+y^2+y-1$ and hence $\frac{1-y^3}{y^2-1}<-y+1$. But $\frac{1-y^3}{y^2-1}>-y$ and so $\lfloor\frac{1-y^3}{y^2-1}\rfloor=-y$. In other words all $(-1,y)$ are solutions for $y\le-3$.
If $x=-2$, then the original equation becomes $\lfloor\frac{4-y^3}{y^2-2}\rfloor=-y-1$. Again that fails for $y=-1,0,1,2$. For $y\le-2$, we have $y^2-2>0$ and hence $4-y^3>-y(y^2-2)$ implies $\frac{4-y^3}{y^2-2}>-y$ which implies $\lfloor\frac{4-y^3}{y^2-2}\rfloor>-y-1$ and so there are no solutions. For $y>2$ we have $4-y^3<-y^3+2y$ and so $\frac{4-y^3}{y^2-2}<-y$. On the other hand, $y^2-2y+1=(y-1)^2+1>0$, so $4-y^3>-y^3-y^2+2y+2=(y^2-2)(-y-1)$ and hence $\frac{4-y^3}{y^2-2}>-y-1$. Hence $\lfloor\frac{4-y^3}{y^2-2}\rfloor=-y-1$. So all of $(-1,y)$ are solutions for $y\ge3$.
[$\textbf{Note}$ If $x=-y^2$ and $x^2=y^3$, then the floor expression is indeterminate. That corresponds to $(x,y)=(-1,1)$. I do not regard that as a solution.]
| {
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"url": "https://math.stackexchange.com/questions/1837842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?
| The inequation is equivalent to
$$x-\frac1{x+2}=\frac{x^2+2x-1}{x+2}=\frac{(x-(-1-\sqrt2))(x-(-1+\sqrt2))}{x-(-2)}<0.$$
For the expression to be negative, you need an odd number of negative factors.
$$\begin{array}{l|ccccccc}
&&-1-\sqrt2&&-2&&-1+\sqrt2\\
\hline x-(-1-\sqrt2)&\color{green}-&0&+&+&\color{green}+&+&+\\
x-(-2)&\color{green}-&-&-&0&\color{green}+&+&+\\
x-(-1+\sqrt2)&\color{green}-&-&-&-&\color{green}-&0&+\\
\hline
&\color{green}-&0&+&|&\color{green}-&0&+\\
\end{array}$$
| {
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"url": "https://math.stackexchange.com/questions/1841125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Solve $4 \times2^x+3^x=5^x$ without any sort of calculator Is there a way i can solve the following equation only by using high school mathematics?
$$4 \times2^x+3^x=5^x$$
I tried writing $5$ as $2+3$ but didn't get any result.
After that i tried to divide by $5^x$ and see how the function goes, but, unfortunately didn't got me somewhere either.
| Assuming $x$ to be natural,
$$
5^x-3^x=(5-3)(5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^x\\
\implies (5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^{x-1}
$$
Each term in LHS looks like $3^{n}5^{x-1-n}>2^n2^{x-1-n}=2^{x-1}$.
There are $x$ number of terms in LHS. If there are $4$ or more terms in LHS, each of them would be greater than $2^{x-1}$ and the equation would not hold. So $1\le x\le 3$. Manually checking $x=1,2,3$ we see $x=2$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Minimum value of $\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$ Recently I was solving one question, in which I was solving for the smallest value of this expression
$$f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$$
My first attempt:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\theta \\
&=3(1-2\sin\theta \cos\theta)+2\sin^2\theta \\
&=3(\sin\theta-\cos\theta)^2 + 2\sin^2\theta
\end{align}$$
Hence the minimum value of $f(\theta)=2\sin^2\theta$ when $\theta=\pi/4$
hence minimum value of $f(\theta)=1$.
But then again I tried to do question differently by making substitutions in order to change the whole $f(\theta)$ in the form of $\cos x+\sin x$
Then $f(\theta)$ came out to be
$$f(\theta)= 4-(\cos(2\theta)+3\sin(2\theta))$$
The minimum value of this expression is surely $$(f(\theta))_{min}=4-\sqrt{10}$$
Can anybody explain me algebraically why my first method gave the wrong result?
| $f'(\theta)=3(sin\theta-cos\theta)^2 + 2sin^2\theta = 6 (\sin\theta -\cos\theta )(\cos\theta + \sin\theta) + 4 \sin \theta \cos\theta$
$
=6 (\sin^2 \theta - \cos^2 \theta) + 4 \sin \theta \cos\theta =-6\cos(2\theta) +2\sin(2\theta)\ge 0 \implies \frac{2}{6} \tan (2\theta)\ge 1
$
$
\implies \tan(2\theta)\ge 3 \implies 2\theta\ge \arctan(3) \implies \theta \ge \arctan(3)/2
$
What that tells us is that first derivative is bigger than $0$ when $\theta \ge \arctan(3)/2$ that is function is growing and after that its smaller than $0$ - meaning its decreasing so we know that the function has a minimum in point $\theta = \arctan(3)/2$
With that
$$
f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2=\frac{1+\cos(2\theta)}{2} - 3\sin(2\theta) +3\frac{1-\cos(2\theta)}{2} + 2=4-\cos(2\theta)-3\sin(2\theta) = 4-\frac{1}{\sqrt{10}} - 3\frac{3}{\sqrt{10}}=4 - \sqrt{10}
$$
Since we know that
$\cos^2 (2\theta)+\sin^2 (2\theta)=1 \implies 1+9=1/\cos^2 (2\theta) \implies cos^2(2\theta)=\frac{1}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the congruence $6x+15y \equiv 9 \pmod {18}$ Solve the congruence $6x+15y \equiv 9\pmod {18}$
Approach:
$(6,18)=6$, so $$15y \equiv 9\pmod 6$$
$$15y \equiv 3\pmod 6$$
So the equation will have $(15,6)$ solutions. Now we divide by 3
$$5y \equiv 1\pmod 2$$.
Solving the Diophantine equation we get $y \equiv1\pmod 2 $, so $y=1+2m$
$$6x \equiv 9-15y\pmod {18}$$
$$6x \equiv 9-15(1+2m)\pmod {18}$$
$$6x \equiv -6-30m\pmod {18}$$
Divide by 6
$$x \equiv -1-5m\pmod 3$$
The right solution is $x-m \equiv 2\pmod 3$. I have $x+5m \equiv 2\pmod 3$
$$x-m+6m \equiv 2\pmod 3$$ $$x-m \equiv 2\pmod 3$$
Wolfram alpha says $y=1+2t$ and $x=t+3d+2$
| Dividing everything by $3$, including the modulus, we get the equivalent congruence
$$2x+5y\equiv 3\pmod{6}.$$
It is convenient to rewrite this as $2x-y\equiv 3\pmod{6}$, or equivalently $y\equiv 2x-3\equiv 2x+3\pmod{6}$.
Now we have a parametric solution: $x\equiv t\pmod{6}$, $y\equiv 2t+3\pmod{6}$. To write it out at length, set $t=0,1,2,3,4,5$. The first two solutions are then $x\equiv 0\pmod{6}$, $y\equiv 3\pmod{6}$ and $x\equiv 1\pmod{6}$, $y\equiv 5\pmod{6}$. Four more to go.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \ x = 0 \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \ x = 0 ;\ x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \ x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x = \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi, \frac{\pi}{3} $$
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.
| In restricting $0 \le x < 2\pi$ you are were inadvertently also restricting $0 \le 2x < 2\pi$ which is not a stated restriction.
So $x = \pi$ is a solution as $\sin 2\pi$ does = $\sin \pi$ after all.
(Likewise $\sin 5\pi/3 = \sin 10\pi/3 = \sin 4\pi/3$).
So your restrictions are actually $0 \le 2x < 4\pi$ and for that matter $0 \le 3x/2 < 3\pi$.
So why did your answer not pick up $x = 5\pi/3$ as a potential answer.
As
Then as $\cos(3x/2) = 0$ is potential answer, $0 \le 3x/2 < 3\pi$, so $3x/2 = \pi/2, 3\pi/2, 5\pi/2$ so $x = \pi/3, \pi, 5\pi/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Prove that $\lim \limits_{x \to 5}\left(4x^2-7\right)=93$ So I first need to determine the limit and then prove it:
$\lim \limits_{x \to 5}\left(4x^2-7\right)$
So $L=93$
And thus $\left|f(x)-L\right|=\epsilon$ and $\left|x-c\right|=\delta$
Plugging in the values...
$\left|\left(4x^2-7\right)-93\right| \lt \epsilon$, which when factored gives $4(x+5)\left|x-5\right|\lt \epsilon$
So does this mean:
$\left|x-5\right|\lt 4(x+5)\delta$
| the next step.
Find $\delta$ such that when $|x-5|<\delta, |4(x + 5)(x-5)| < \epsilon$
let $\delta = \min (1, \frac {\epsilon}{44})$
Why 44?
Suppose $x-5 = \delta, |4(x + 5)(x-5)| = \delta \le 1 \implies |4(\delta + 10) \delta| = 4 \delta^2 + 40 \delta$
If $\delta \le 1$ then $|4(x + 5)(x-5)| < 44 \delta$
$\delta = \min (1, \frac {\epsilon}{44}) \implies |4(x + 5)(x-5)| < \epsilon.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to perform the following integration $\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$ $$\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$$
How to simplify the expression given? I tried using formulas for $\cos 2x$ and $\cos 3x$. I also tried the using $\cos x + \cos y = 2cos(\frac{x+y}{2})\cos(\frac{x+y}{2})$.
The options for answers are given are in form of half angles of trigonometric functions.
| If you write $$\cos w = \frac{1}{2}\left(e^{iw}+e^{-iw}\right),$$ you find the numerator is $$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)^5 = 16\cos^5(x)$$
The denominator is, likewise:
$$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)^6=32\cos^6(x)$$
So the integrand is $\frac{1}{2\cos x}$.
This all comes because $1,5,10,10,5,1$ is the fifth row of Pascal's triangle, and $1,6,15,20,15,6,1$ is the sixth row.
| {
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"url": "https://math.stackexchange.com/questions/1846580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
| A very long way... (Let $I$ equal the integral)
$$I=\frac{1}{\sqrt{2}}\int\frac{x^2-1}{x^3\sqrt{(x^2-1/2)^2+1/4}}dx$$
Let $(x^2-1/2)=u/2$ meaning $x=\sqrt{1/2(u+1)}$ and $dx=\frac{1}{4\sqrt{1/2(u+1)}}$ Hence
$$I=\frac{1}{\sqrt{2}}\int\frac{u-1}{(u+1)^2\sqrt{u^2+1}}du$$
Then let $u=\tan(v)$ and $du=\sec^2(v)$ giving
$$I=\frac{1}{\sqrt{2}}\int\frac{(\tan v -1)\sec v}{(\tan v+1)^2}dv$$
Using $\tan v=\frac{\sin v}{\cos v}$ and $\sec v=\frac{1}{\cos v}$ we have
$$I=\frac{1}{\sqrt{2}}\int\frac{\sin v-\cos v}{(\sin v +\cos v)^2}dv$$
Finally this can be evaluated using $m=\sin v + \cos v$ and $dm=\cos v - \sin v$ to get
$$I=-\frac{1}{\sqrt{2}}\int\frac{1}{m^2}dm=\frac{1}{\sqrt{2}m}$$
Doing substitutions back up to $x$ we have
$$I=\frac{1}{\sqrt{2}(\sin v + \cos v)}$$
$$=\frac{1}{\sqrt{2}(\sin(\arctan u)+\cos(\arctan u))}=\frac{\sqrt{u^2+1}}{\sqrt{2}\left(u + 1\right)}$$
$$=\frac{\sqrt{4 x^4-4 x^2+2}}{2
\sqrt{2}x^2}=\frac{\sqrt{2x^4-2x^2+1}}{2x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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surface area of $\left\{(x,y,z)\in R^3\,\mid\, x^2+y^2 =\frac{1}{z^2}, 1I want to calculate the surface area of the surface that bounds the solid $$K=\left\{(x,y,z)\in R^3\,\mid\, x^2+y^2 \leq\frac{1}{z^2}, 1<z<3\right\}$$
I'm stuck with the differential surface area that I shall consider so that I can solve $S=\iint dS$.
| Here is how I would do it:
The surface can be parametrized as follows
\begin{cases}
x=x\\
y=y\quad\quad\quad\quad\quad\quad\quad(x,y)\in D=\{(x,y)\;|\;\frac{1}{9}\le x^2+y^2\le 1\}\\
z=\frac{1}{\sqrt{x^2+y^2}}\\
\end{cases}
You can plot this surface and its domain $D$ with WolframAlpha:
Now, the surface area is given by
$$
A=\iint_D ||r_x\times r_y ||\; dA = \iint_{\{(x,y)\;|\;\frac{1}{9}\le x^2+y^2\le 1\}} \sqrt{ \frac{x^2+y^2+(x^2+y^2)^3}{(x^2+y^2)^3}}\; dA
$$
Switching to polar coordinates yields:
$$
\boxed{
A=\int_0^{2\pi}\int_{1/3}^1\sqrt{r^{-2}+r^2}\; drd\theta \approx 7.6030
}
$$
Alternatively you could proceed as follows:
\begin{cases}
x=\frac{\cos\theta}{z}\\
y=\frac{\sin\theta}{z}\quad\quad\quad\quad\quad\quad\quad0\le \theta\le 2\pi, \; 1\le z\le 3\\
z=z\\
\end{cases}
$$
A=\iint_{\{(\theta,z)|0\le\theta\le 2\pi, \; 1\le z\le 3\}} ||r_{\theta}\times r_z ||\; dA
$$
Computing the integral yields
$$
\boxed{A=\int_0^{2\pi}\int_1^3\sqrt{z^{-2}+z^{-6}}\;dzd\theta \approx 7.6030}
$$
Also note that using the change of variables $z=\frac{1}{r}$ (i.e., $dz=\frac{-dr}{r^2}=-z^2dr$):
$$
\int_{1/3}^1\sqrt{r^{-2}+r^2}\; dr = \int_1^3\sqrt{z^{-2}+z^{-6}}\;dz
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is conjugation of a positive semi-definite Hermitian matrix equal to conjugation by some rotation? Let $s \in GL_2(\Bbb R)$ be a symmetric positive definite matrix (is this roughly the stretching part of the polar decomposition of some other matrix $x=sk$?). Conjugate $s$ by the reflection $$\gamma=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$ Is there a $k_1 \in K = \left\{\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\right\}$, such that $\gamma s \gamma^{-1}=k_1 s k_1^{-1}$? How do I find it?
More generally, can we answer the question in the title?
| Notation: For square matrices $s$ and $t$ of the same order, let $t$-conjugation of $s$ be denoted by $s^t = tst^{-1}$. Then $(s^{t_1})^{t_2} = s^{t_2 t_1}$ [note the reversal of order — this could be remedied by defining conjugation differently, but this answer will be consistent with the definition used in the question].
Analysis of General Case
For a reflection $\gamma$ and a rotation $k$, $s^\gamma = s^k$ if and only if $(s^\gamma)^{k^{-1}} = s^{k^{-1}\gamma} = s$, or $s^\delta = s$, where $\delta = k^{-1}\gamma$, which is another reflection, being the composition of a rotation and a reflection. Now, $s^\delta = s$ is equivalent to $\delta s = s \delta$, so it suffices to find some reflection $\delta$ that commutes with $s$. Then $k = \gamma \delta^{-1}$.
Solution in Two Dimensions
In the two-dimensional special case given in the question, $\gamma = \begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}$, and $s$ is a symmetric positive definite matrix, so let $s = \begin{bmatrix}a & b\\ b & c\end{bmatrix}$ (where $a, c > 0$ and $ac > b^2$, but these will not be used). It is possible to start with a general reflection matrix $\delta$ and apply the commutativity condition obtained earlier to derive possible values of $\delta$. In this answer, I will define a particular $\delta$ and show that it satisfies the condition.
Let $$\cos \theta = \dfrac{a - c}{\sqrt{(a - c)^2 + 4b^2}}, \quad \sin \theta = \dfrac{2b}{\sqrt{(a - c)^2 + 4b^2}}$$
(noting that indeed, $\cos^2 \theta + \sin^2 \theta = 1$).
Let $$\delta = \begin{bmatrix} \cos \theta & \sin \theta\\ \sin \theta & -\cos \theta\end{bmatrix}$$
(noting that $\delta$ is orthogonal and $\det \delta = -1$, which makes it a reflection matrix).
Now
\begin{align*}
s \delta & = \begin{bmatrix}
a \cos \theta + b \sin \theta & -b \cos \theta + a \sin \theta\\
b \cos \theta + c \sin \theta & -c \cos \theta + b \sin \theta
\end{bmatrix}\\
\delta s & = \begin{bmatrix}
a \cos \theta + b \sin \theta & b \cos \theta + c \sin \theta\\
-b \cos \theta + a \sin \theta & -c \cos \theta + b \sin \theta
\end{bmatrix}.
\end{align*}
To verify that $s \delta = \delta s$, it is enough to check whether $-b \cos \theta + a \sin \theta = b \cos \theta + c \sin \theta$ (since both matrices have the same diagonal, and the $(1,2)$ elements are equal if and only if the $(2,1)$ elements are equal).
Rearranging this equation, we get $(a - c) \sin \theta = 2b \cos \theta$, which does hold since $\tan \theta = \dfrac{2b}{a - c}$ (by our definition of $\cos \theta$ and $\sin \theta$). Thus, $\delta$ is the required matrix, as claimed.
Now, $k = \gamma \delta^{-1}$, where $\delta^{-1} = \delta^T = \delta$, so we have
\begin{equation*}
k = \begin{bmatrix}
\cos \theta & \sin \theta\\
- \sin \theta & \cos \theta
\end{bmatrix}.
\end{equation*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Convexity of $\log f(e^s)$ where $f$ is a polynomial Let $f(t)$ be a monic real polynomial such that $f(t) > 0$ for all $t \ge 0$. Suppose that $\log f(e^x)$ is strictly convex on $\mathbb{R}$, i.e.
$f(s^2) \cdot f(t^2) > f(st)^2$
for all positive real numbers $s \neq t$.
Can one show that
$$\frac{d^2}{dx^2}(\log f(e^x)) > 0$$
for all $x \in \mathbb{R}$?
Remark: Since $\log f(e^x)$ is strictly convex (and hence convex), we always have $\frac{d^2}{dx^2}(\log f(e^x)) \ge 0$.
| This is false. Take $f(x) = x^4+bx^3+cx^2+bx+1$ where $-0.5<b<0.5$ and $c$ is the unique root of
$$216 b^2 + 108 b^4 - 324 b^2 c - 54 b^2 c^2 + 128 c^3 + b^2 c^3=0 \quad (\ast)$$
between $-1$ and $0$. The specific example I used to check this is $b=0.1$, $c\approx -0.2880429933200467$.
Then $\tfrac{d^2}{(dz)^2} \log f(e^z) = g(e^z)/f(e^z)^2$ where
$$g(x) = x(b + 4 c x + 9 b x^2 + b c x^2 + 16 x^3 + 4 b^2 x^3 + 9 b x^4 +
b c x^4 + 4 c x^5 + b x^6).$$
We rewrite this as $g(x) = x^4 h(x+1/x)$ where
$$h(y) = b y^3 + 4 c y^2 + b(c+6) y + 4 (b^2-2 c+4).$$
The discriminant of $h$ is
$$-4 (8 + b^2 - 4 c) (216 b^2 + 108 b^4 - 324 b^2 c - 54 b^2 c^2 +
128 c^3 + b^2 c^3)$$
so $(\ast)$ implies that $h$ has a double root. Some numerical verification shows that $h$ is positive on $[0,\infty)$ except at the double root, which occurs at some $y_0>2$. For example, with the values of $(b,c)$ above, the double root is at $y_0 \approx 7.42471$. Therefore,
$$\tfrac{d^2}{(dz)^2} \log f(e^z) = e^{4z} h(e^z+e^{-z})/f(e^z)^2$$
is also positive except for vanishing at the two solutions to $e^{z_0}+e^{-z_0} = y_0$. In our example, $z_0 = \pm 1.98616$.
Finally, it is also east to check that $f(x)$ has no real roots. (This is the part that fails if we let $b$ get bigger than $0.5$.)
In case you prefer exact solutions, take $b = \tfrac{8 \sqrt{2}}{17 \sqrt{17}}$ and $c = -7/17$. Then $g(x)$ has a double roots at $\tfrac{19 \sqrt{34} \pm \sqrt{11118}}{34} \approx 0.15724, 6.35971$ and is $\geq 0$ for all $x \geq 0$. Thus, $g(e^z)$ is $\geq 0$ for all real $z$, but has two double roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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From $ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} $ to $ \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}} $? Good evening to everyone. I have an expression that I don't know how to arrive at. $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} $$ to $$ \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}} $$
What I've tried: $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} = \frac{n!\left(n+\frac{1}{n!}\right)\left(n+1\right)}{n!\left(\left(n+1\right)^2+\frac{1}{n!}\right)} = \frac{\left(n+\frac{1}{n!}\right)\left(n+1\right)}{\left(\left(n+1\right)^2+\frac{1}{n!}\right)} $$ And from here I don't know what to do anymore. The second attempt: $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} = \frac{\left(n^2\cdot \:\:n!+n\cdot n!+n+1\right)}{n!\left(n+1\right)^2+1} = \frac{n!\left(n^2+n+\frac{n}{n!}+\frac{1}{n!}\right)}{n!\left(\left(n+1\right)^2+\frac{1}{n!}\right)} =\frac{\left(n^2+n+\frac{n}{n!}+\frac{1}{n!}\right)}{\left(\left(n+1\right)^2+\frac{1}{n!}\right)} $$
And again I don't know what to do anymore. Thanks for any response.
| The expression is not quite right. It is not true that
$$
\frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1}
= \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}}.
$$
For example, if you plug in $n = 1$ you get
$\frac{4}{5}$ and $\frac{2}{3}$, not equal.
What is true is that
$$
\frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1}
= \frac{n+\frac{1}{n!}}{n+1+\color{red}{\frac{1}{(n+1)!}}}.
$$
Your work gets you most of the way there. You want to factor out $(n+1)!$ from the top and bottom, instead of $n!$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\frac{G\ln 2}{2}$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ )
i need to show, in a simple way, that:
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\dfrac{G\ln 2}{2}$
$G$ being the Catalan constant.
The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:
$\displaystyle \int_0^1 \dfrac{\ln(1+x)\ln(1+x^2)}{1+x^2}dx$
$\displaystyle \int_0^1 \dfrac{\ln(1+x)^2}{1+x^2}dx$
$\displaystyle \int_0^{\tfrac{\pi}{4}}\big(\ln(\cos x)\big)^2 \dfrac{}{}dx$
and some constants.
| \begin{align}
I&:=\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x)\,dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)\,dx\\
&=\int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\cos\left(x+\frac {\pi}4\right)\right)\ln(\cos x)\,dx - \int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\sin\left(x+\frac {\pi}4\right)\right)\ln(\sin x)\,dx\\
&\ \quad\text{setting}\; x=\frac {\pi}4-y\;\ \text{in the first integral gives}\\
&=\int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\cos\left(\frac {\pi}2-y\right)\right)\ln\left(\cos\left(\frac {\pi}2-\frac {\pi}4-y\right)\right)\,dy - \int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\sin\left(x+\frac {\pi}4\right)\right)\ln(\sin x)\,dx\\
&=\int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\sin\left(y\right)\right)\ln\left(\sin\left(y+\frac {\pi}4\right)\right)\,dy - \int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\sin\left(x+\frac {\pi}4\right)\right)\ln(\sin x)\,dx\\
&=\int_0^{\tfrac{\pi}{4}}\left(\ln(\sqrt{2})+\ln\left(\sin x\right)\right)\ln\left(\sin\left(x+\frac {\pi}4\right)\right) - \left(\ln(\sqrt{2})+\ln\left(\sin\left(x+\frac {\pi}4\right)\right)\right)\ln(\sin x)\;dx\\
&=\ln(\sqrt{2})\int_0^{\tfrac{\pi}{4}}\ln\left(\sin\left(x+\frac {\pi}4\right)\right)-\ln(\sin x)\;dx\\
&=\dfrac{\ln 2}{2}G\\
\end{align}
It remains to prove that $\displaystyle \int_0^{\tfrac{\pi}{4}}\ln\left(\sin\left(x+\frac {\pi}4\right)\right)-\ln(\sin x)\;dx=G$.
This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
\begin{align}
\int_0^{\tfrac{\pi}{4}}\ln\left(\sin\left(x+\frac {\pi}4\right)\right)-\ln(\sin x)\,dx&=\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ln\left(\sin x\right)\,dx-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\
&=\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\
&=-\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx\\
&=-\int_0^1\frac{\ln t}{1+t^2}\,dt,\quad\text{integrated by parts}\\
&=-\left.\ln(t)\;\arctan(t)\right|_0^1+\int_0^1\frac{\arctan(t)}{t}\,dt\\
&=\int_0^1\sum_{n=0}^\infty (-1)^n\frac{t^{2n}}{2n+1}\,dt\\
&=\left.\sum_{n=0}^\infty (-1)^n\frac{t^{2n+1}}{(2n+1)^2}\right|_0^1\\
&=G\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is $e^{A}$ where A is an anti-diagonal matrix I am trying to get a closed form for the matrix produced by the following operation: $$e^A$$ where $A$ is an anti diagonal matrix, say, of size $2\times 2$:
$$A=\begin{pmatrix}
0 &b \\
c &0
\end{pmatrix}$$
Using Mathematica MatrixExp I got
$$ \left(
\begin{array}{cc}
\frac{1}{2} e^{-\sqrt{b} \sqrt{c}}+\frac{1}{2} e^{\sqrt{b} \sqrt{c}} & \frac{\sqrt{b} e^{\sqrt{b} \sqrt{c}}}{2 \sqrt{c}}-\frac{\sqrt{b} e^{-\sqrt{b} \sqrt{c}}}{2 \sqrt{c}} \\
\frac{\sqrt{c} e^{\sqrt{b} \sqrt{c}}}{2 \sqrt{b}}-\frac{\sqrt{c} e^{-\sqrt{b} \sqrt{c}}}{2 \sqrt{b}} & \frac{1}{2} e^{-\sqrt{b} \sqrt{c}}+\frac{1}{2} e^{\sqrt{b} \sqrt{c}} \\
\end{array}
\right)
$$
But here one can find a formula for the computation on a general $2\times 2$ matrix, at the bottom of the page. Using that formula I got a different result.
What is the correct answer?
| Recall that: $$e^A=\sum_{k=0}^{+\infty}\frac{A^k}{k!}.$$
Moreover, notice that in your case, one has: $$A^2=\begin{pmatrix}bc&0\\0&bc\end{pmatrix}.$$
Therefore, you can compute $e^A$ summing on even and odd integers. Indeed, one has: $$A^{2k}=\begin{pmatrix}b^kc^k&0\\0&b^kc^k\end{pmatrix},A^{2k+1}=\begin{pmatrix}0&b^{k+1}c^k\\b^kc^{k+1}&0\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Is $77!$ divisible by $77^7$?
Can $77!$ be divided by $77^7$?
Attempt:
Yes, because $77=11\times 7$ and $77^7=11^7\times 7^7$ so all I need is that the prime factorization of $77!$ contains $\color{green}{11^7}\times\color{blue} {7^7}$ and it does.
$$77!=77\times...\times66\times...\times55\times...\times44\times...\times33\times...\times22\times...\times11\times...$$
and all this $\uparrow$ numbers are multiples of $11$ and there are at least $7$ so $77!$ contains for sure $\color{green}{11^7}$
And $77!$ also contains $\color{blue} {7^7}:$
$$...\times7\times...\times14\times...\times21\times...\times28\times...\times35\times...42\times...49\times...=77!$$
I have a feeling that my professor is looking for other solution.
| Your solution is right on!
You might be careful how you present it to your professor. I think it is fine, but here is other words saying the same thing. (I usually try to avoid too many $\dots$ in number theory proofs.)
$$
11 = 1\cdot 11 \\
22 = 2\cdot 11 \\
33 = 3\cdot 11 \\
44 = 4\cdot 11 \\
55 = 5\cdot 11 \\
66 = 6\cdot 11 \\
77 = 7\cdot 11
$$
are all distinct factors in $77!$, so $11^7$ divides $77!$.
Likewise
$$
7 = 1\cdot 7 \\
14 = 2\cdot 7 \\
21 = 3\cdot 7 \\
28 = 4\cdot 7 \\
35 = 5\cdot 7 \\
42 = 6\cdot 7 \\
49 = 7\cdot 7.
$$
are factors of $77!$, so $7^7$ divide $77!$
Now since $7$ and $11$ are coprime then also are $7^7$ and $11^7$, we conclude that $7^711^7$ divide $77!$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 3
} |
Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off and rearranging the problem to $1/3(k+2)(k+3)$ is where I always get stuck.
| Here is an easy way of going about the inductive step:
\begin{align}
\sum_{i=1}^{k+1}i(i+1)&= \sum_{i=1}^ki(i+1)+(k+1)(k+2)\\[1em]
&= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\\[1em]
&= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\\[1em]
&= \frac{(k+1)(k+2)(k+3)}{3}.
\end{align}
See if you can determine how each line was obtained (where the inductive hypothesis was applied, etc.).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find range of $f(x)=3^x+5^x-8^x$ Find range of $f(x)=3^x+5^x-8^x$.
My attempt:
On observation one sees that $f(1)=0$.
On taking $g(x)=\left(\frac{3}{8}\right)^x+\left(\frac{5}{8}\right)^x-1$ and then observing that
$g'(x)=\left(\frac{3}{8}\right)^x \ln\left(\frac{3}{8}\right)+\left(\frac{5}{8}\right)^x \ln\left(\frac{5}{8}\right)<0$
we can conclude that $f(x)$ is negative for $x\in (1,\infty)$.
Also $\lim_{x\rightarrow-\infty}f(x)=0$.
Only thing to be done is to find maximum value of $f(x)$
| Considering the function $$f(x)=3^x+5^x-8^x$$ by inspection we have $$f(-\frac 12)=-\frac{1}{2 \sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\approx 0.67101$$ $$f(0)=1$$ $$f(\frac 12)=-2 \sqrt{2}+\sqrt{3}+\sqrt{5}\approx 1.13969$$ $$f(\frac 34)=-4 \sqrt[4]{2}+3^{3/4}+5^{3/4}\approx 0.86638$$ So, as already answered by Ahmed S. Attaalla, we can safely start Newton method for the zero of $f'(x)$ using $x_0=\frac 12$.
The iterates will then be
$$\left(
\begin{array}{cc}
n & x_n \\
1 & 0.412626 \\
2 & 0.397630 \\
3 & 0.397236 \\
4 & 0.397235
\end{array}
\right)$$ and, for the converged solution, $f(x)\approx 1.15811$ is the maximum.
From a numerical point of view, instead of looking for the zero of $$f'(x)=\ln 3 (3^x)+\ln 5 (5^x)-\ln 8 (8^x)$$ it would be more efficient to look at at zero of $$g(x)=x \log (8)-\log \left(\frac{3^x \log (3)+5^x \log (5)}{\log (8)}\right)$$ which is almost a straight line in the area of interest. Using the same $x_0=\frac 12$, the iterates would then be instead
$$\left(
\begin{array}{cc}
n & x_n \\
1 & 0.396751 \\
2 & 0.397235
\end{array}
\right)$$
As you can see, the first iterate is "almost" the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the equation of this hyperbola? What is the equation of the hyperbola that satisfies these conditions:
Asymptotes $y=2x$ and $y=-2x$, centre $(0,0)$, and the point $(1,1)$ lies on the curve.
This isn't a homework question; I study maths for the fun of it. The question is taken from the book "Delta Mathematics" by David Barton and Anna Cox, 2013 edition, page 22 question 11 b.
The answer in the back of the book is $$\frac{4x^2}{3} - \frac{y^2}{3} = 1$$
But how do I work out the answer from the given information?
I started by sketching a graph to see where the point $(1,1)$ falls.
I don't know how to continue at this point.
| The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $y = \pm \frac{bx}{a}$.
Since $(1, 1)$ lies on your curve then $\frac{1}{a^2} - \frac{1}{b^2} = 1 \iff b^2 -a^2 = a^2b^2$.
But you know that $\frac{b}{a} = 2\iff b = 2a$. Plug this into the above equation to get ($a\neq 0$) $$3a^2 = 4a^4 \Rightarrow a^2 = \frac{3}{4}$$ and $b^2 = 3$.
So the hyperbola is $$\frac{x^2}{\frac{3}{4}} - \frac{y^2}{3} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using row operations to compute the following 3x3 determinant
Use row operations to compute the following determinant $\begin{bmatrix}3&3&-3\\3&4&-4\\2&-3&-5\end{bmatrix}$
I know how to easily compute the determinant using $i - j + k$ method... The problem is I have put the matrix in LTF (Lower Triangular Form) and then used the product of the diagonal. So I did:
$R_2 \leftarrow R_1 + (-1)R_2$
$R_3 \leftarrow 2R_1 + (-3)R_3$
$R_3 \leftarrow 15R_2 + R_3$
to get the matrix in LTF $\begin{bmatrix}3&3&-3\\0&-1&1\\0&0&23\end{bmatrix} \implies \det(A) = (3)(-1)(23) = -69$ I know that since I used these operations I changed the determinant, how exactly can I fix it back? I also know that the $\det(A)$ should be $= -24$ .
| The final matrix should be
$$
\begin{bmatrix}
3 & 3 & -3 \\
0 & -1 & 1 \\
0 & 0 & 24
\end{bmatrix}
$$
However, you have multiplied the determinant by $-1$ with the first operation and by $-3$ with the second one, so you get
$$
\frac{3\cdot(-1)\cdot24}{(-1)\cdot(-3)}=-24
$$
I use a different method, reducing the pivots to $1$:
\begin{align}
\begin{bmatrix}
3 & 3 & -3 \\
3 & 4 & -4 \\
2 & -3 & -5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & -1 \\
3 & 4 & -4 \\
2 & -3 & -5
\end{bmatrix} && R_1\gets\color{red}{\frac{1}{3}}R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
2 & -3 & -5
\end{bmatrix} && R_2\gets R_2-R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & -5 & -3
\end{bmatrix} && R_3\gets R_3-2R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & -8
\end{bmatrix} && R_3\gets R_3+5R_2
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{bmatrix} && R_3\gets \color{red}{-\frac{1}{8}}R_3
\end{align}
The red numbers tell that the determinant has been multiplied by
$$
-\frac{1}{24}
$$
so it is $-24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Embeddings of pure cubic field in complex field I know that the complex embeddings (purely real included) for quadratic field $\mathbb{Q}[\sqrt{m}]$ where $m$ is square free integer, are
*
*$a+b\sqrt{m} \mapsto a+b\sqrt{m}$
*$a+b\sqrt{m} \mapsto a-b\sqrt{m}$
So, norm of $a+b\sqrt{m}$ is $a^2-mb^2$.
Motivated by this , I want to calculate norm of $a+\sqrt[3]{n}$ in $\mathbb{Q}[\sqrt[3]{n}]$ where $n$ is positive cubefree integer. I am able to calculate the norm to be $a^3+n$ using the fact that it's equal to the negative of constant term of the minimal polynomial. But I don't get same answer when I assume embeddings to be
*
*$a+\sqrt[3]{n} + 0\sqrt[3]{n^2} \mapsto a+\sqrt[3]{n} + 0\sqrt[3]{n^2}$
*$a+\sqrt[3]{n} + 0\sqrt[3]{n^2} \mapsto a-\sqrt[3]{n} + 0\sqrt[3]{n^2}$
*$a+\sqrt[3]{n} + 0\sqrt[3]{n^2} \mapsto a+\sqrt[3]{n} - 0\sqrt[3]{n^2}$
So what are the correct conjugation maps for pure cubic field case?
| The conjugates of $\sqrt[3]{n}$ are $\omega\sqrt[3]{n}$ and $\omega^2\sqrt[3]{n}$, where $\omega$ is a primitive cube root of unity. Therefore the norm of $a+\sqrt[3]{n}$ is
$$ (a+\sqrt[3]{n})(a+\omega\sqrt[3]{n})(a+\omega^2\sqrt[3]{n})=a^3+n$$
using the fact that $1+\omega+\omega^2=0$.
Edit: the three complex embeddings of $\mathbb{Q}(\sqrt[3]{n})$ are
$$ a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\mapsto a+b\sqrt[3]{n}+c\sqrt[3]{n^2} $$
$$ a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\mapsto a+b\omega \sqrt[3]{n}+c\omega^2\sqrt[3]{n^2} $$
$$ a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\mapsto a+b\omega^2\sqrt[3]{n}+\omega \sqrt[3]{n^2}$$
Here $a,b,c\in\mathbb{Q}$ and $\sqrt[3]{n}$ is the real cube root of $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0$
Show that for any real number $x$:
$$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0.$$
$\bf{My\; Try::}$ Using $a\sin x+b\cos x\geq -\sqrt{a^2+b^2}$
So $$x^2\sin x+x\cos x\geq -\sqrt{x^4+x^2}=-x\sqrt{1+x^2}$$
and $$4x^4+4x^2+1>4x^4+4x^2\Rightarrow (2x^2+1)^2>4x^2(x^2+1)$$
So $$(2x^2+1)>2x\sqrt{x^2+1}\Rightarrow x^2+\frac{1}{2}>x\sqrt{x^2+1}$$
So $$x^2\sin x+x\cos x+x^2+\frac{1}{2}>-x\sqrt{x^2+1}+x\sqrt{x^2+1}=0$$
So $$x^2\sin x+x\cos x+x^2+\frac{1}{2}\gt 0\;\forall x\; \in \mathbb{R}$$
Is my solution is right, If not Then how can we solve it, Help required, Thanks in Advance
| If we consider $a=(1+\sin x),\, b=\cos(x),\, c=\frac{1}{2}$ we have that
$$ b^2-4ac = \cos^2(x)-2-2\sin(x)=-4\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)^4 \leq 0$$
hence $ax^2+bx+c$ is never negative, since $a\geq 0$ and $\Delta=b^2-4ac\leq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Locus of intersection point of perpendicular tangents Here is the question which I am referring to:
A tangent is drawn to the circle $(x-a)^2+y^2=b^2$ and a perpendicular tangent to the circle $(x+a)^2+y^2=c^2$, find locus of their point of intersection.
What I did:
First I supposed the intersection of the perpendicular tangents to be $(h,k)$ and then from that point I found the equation of tangents to respective circles and after that I found slopes of each tangent using the condition that distance from center is equal to radius for a tangent and in the end. I multiplied the slopes of each tangent received from respective circles and set it equal to $-1$ because product of slopes of perpendicular lines is $-1$. I've found the locus but it doesn't seem as the answer. Can you tell me what mistake I made or is there any other way to approach this question?
Below are the images of my work:
Part 1 of work
Part 2 of work
The answer:
| There're two pairs of tangents for the two circles are perpendicular. Let the contact points on the circle be
$\begin{pmatrix} a+b\cos t \\ b\sin t \end{pmatrix}$,
$\begin{pmatrix} -a-c\sin t \\ c\cos t \end{pmatrix}$,
$\begin{pmatrix} a-b\cos t \\ -b\sin t \end{pmatrix}$ and
$\begin{pmatrix} -a+c\sin t \\ -c\cos t \end{pmatrix}$.
The tangent from the first point:
$$(a+b\cos t)x-a(x+a+b\cos t)+a^2+by\sin t=b^2$$
The tangent from the second point:
$$(-a-c\sin t)x+a(x-a-c\sin t)+a^2+cy\cos t=c^2$$
On solving,
$$\begin{pmatrix} x \\ y \end{pmatrix}
= \begin{pmatrix}
a\cos 2t+b\cos t-c\sin t \\
a\sin 2t+b\sin t+c\cos t \end{pmatrix}$$
Similarly, we have another branch
$$\begin{pmatrix} x \\ y \end{pmatrix}
= \begin{pmatrix}
a\cos 2t+b\cos t+c\sin t \\
a\sin 2t+b\sin t-c\cos t \end{pmatrix}$$
Each branch correspond to one diagonal of the rectangle.
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x',y')$ is given by
$$ax'x+h(y'x+x'y)+by'y+g(x+x')+f(y+y')+c=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ Evaluate the following limit:
$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$
Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$
I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$
Could some tell me how to proceed further?
| Indeed, the expansion of $(1+x)^{1/x}$ about $x=0$ is $e - \frac{ex}{2} + \frac{11e}{24} x^2 + O(x^3)$, so the limit is then $\dfrac{11e}{24}$.
--
To get this expansion, start with
$$(1+x)^{1/x} = \exp \left({1 - x/2 + x^2 / 3 - \dots}\right)$$
which you already found and put this into the power series of $\exp$ to get
$$\sum_{n=0}^{\infty} \frac{(1 - x/2 + x^2/3 - \dots)^n}{n!}$$
The coefficient of $x^0$ is
$$\sum_{n=0}^{\infty} \frac{1}{n!} = e$$
The coefficient of $x^1$ in $\dfrac{(1-x/2+x^2/3 - \dots)^n}{n!}$ is $-\frac{n/2}{n!}$, hence the coefficient in total is
$$\sum_{n=1}^{\infty} -\frac{1}{2 (n-1)!} = -\frac{e}{2}$$
I'll leave the coefficient of $x^2$ to you. (It's the hardest, but not too bad).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.
| Since $$\sin { x } =2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } $$
so
$$\cos \left( x \right) +\sin \left( x \right) =\frac { 7 }{ 5 } \\ 5\left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 } } } \right) +10\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } =7\left( \cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 } } } \right) \\ 12\sin ^{ 2 }{ \frac { x }{ 2 } - } 10\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } +2\cos ^{ 2 }{ \frac { x }{ 2 } =0 } \\ 6\tan ^{ 2 }{ \frac { x }{ 2 } -5\tan { \frac { x }{ 2 } +1 } =0 } \\ \tan { \frac { x }{ 2 } =\frac { 5\pm 1 }{ 12 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P
If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P<Q)$. Find the value of $4Q-aP$?
How will I begin with the solution just a hint would be enough.
| $$a^3 +b^3 +(-9)^3-3ab(-9)=(a+b-9)(a^2+b^2-ab+9a+9b+81)=0$$
therefore
\begin{cases}
a+b-9=0\\
\qquad\operatorname{or}\\
a=b=-9
\end{cases}
since $a>b>0$ thus
$$a+b-9=0$$
Set $f(x)=ax^2+bx-9$. We have $f(1)=a+b-9=0$, thus
$Q=1$ and $P=\frac{-9}{a}$
finally
$$4Q-aP=4+9=13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $a^{4n+1}-a$ is divisible by 30 for any a and $n\ge1$ It is valid for n=1, and if I assume that $a^{4n+1}-a=30k$ for some n and continue from there with $a^{4n+5}-a=30k=>a^4a^{4n+1}-a$ then I try to write this in the form of $a^4(a^{4n+1}-a)-X$ so I could use my assumption but I can't find any $X$ that would set the two expressions equal.
Then I tried factoring $a^{4n+1}-a=30k$ as $a(a^n-1)(a^n+1)(a^{2n}+1)=30k$ and using that as my assumption. Then I also factor as $a^{4n+5}-a=a(a^{n+1}-1)(a^{n+1}+1)(a^{2n+2}+1)$ and again I'm stuck not being able to use my assumption.
Please note that I strictly need to use induction in this problem.
| Don't factor entirely. Just factor enough to realize
$a^{4n+1} - a = a(a^{4n} - 1)=a(a - 1)(a^{4n-1} + a^{4n-2} + .... + a + 1)$.
Assume for $n = k$ that
$a^{4k + 1}-a= a(a^{4n} - 1)=a(a - 1)(a^{4k-1} + a^{4k-2} + .... + a + 1)= 30M$
then $a^{4(k+1) + 1} - a = a(a-1)(a^{4k + 3} + ...)=$
$a(a-1)(a^{4k + 3} + a^{4k + 2} + a^{4k+1} + a^{4k}) +a(a - 1)(a^{4k-1} + a^{4k-2} + .... + a + 1)$
$= a(a-1)(a^{4k + 3} + a^{4k + 2} + a^{4k+1} + a^{4k}) + 30M$.
$= a(a-1)(a^3 + a^2 + a + 1)a^{4k} + 30M$
$= (a^{4*1 + 1} - a)a^{4k} + 30M$.
But $30|(a^{4*1 + 1} - a) $ By our initial step.
So $30|(a^{4*1 + 1} - a)a^{4k} + 30M$.
====
In hind sight it's easy to see that $a^{4(n+1) + 1} - a = a^{4(n+1) + 1} - a^{4n+1} + a^{4n + 1} - a = (a^{4*1 + 1} - a)a^{4n} - (a^{4n+1} - a)$.
So $a^{4*1 + 1} -a|a^{4*m + 1} - a;m \ge 1$ via induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
find $f\in L^2([0,\pi])$ such that its $L^2$ distance to $\sin(x)$ and $\cos(x)$ are both bounded by specific constants I want to find all $f\in L^2([0,\pi])$ such that
$$
\begin{align}
\int_0^\pi\lvert f(x)-\sin(x)\rvert^2\,dx &\le \frac{4\pi}{9}\\
\int_0^\pi\lvert f(x)-\cos(x)\rvert^2\,dx &\le \frac{\pi}{9}\\
\end{align}
$$
I started by taking the Fourier transform of $\sin(x)$ and $\cos(x)$, and then used Parseval's theorem. This led to
$$
\begin{align}
\frac{4\pi}{9} \ge \lvert\lvert \,f(x) - \sin(x) \rvert\rvert_{L^2} &= \left|\left|\, \hat{f}(\kappa) - \frac{1+e^{-2\pi i \kappa}}{(1-4\kappa^2)\sqrt{\pi}} \right|\right|_{\ell^2}\\
\frac{\pi}{9} \ge \lvert\lvert \,f(x) - \cos(x) \rvert\rvert_{L^2} &= \left|\left|\, \hat{f}(\kappa) - \frac{2i\kappa(1+e^{-2\pi i \kappa})}{(1-4\kappa^2)\sqrt{\pi}} \right|\right|_{\ell^2}
\end{align}
$$
Maybe that was going down the wrong path, but that's where I'm at now. I don't see what to do next, or have any other ideas.
Any thoughts? Thanks.
| First, note that
\begin{align}
\|\sin-\cos\|^2&=\|\sin(x)\|^2+2\langle\sin,\cos\rangle+\|\cos\|^2 \\
& =\|\sin\|^2+\|\cos\|^2 \\
& = \|1\|^2=\pi.
\end{align}
You want
$$
\|f-\sin\|+\|f-\cos\| \le \frac{2}{3}\sqrt{\pi}+\frac{1}{3}\sqrt{\pi}=\sqrt{\pi}
$$
If you have the above, then
$$
\sqrt{\pi}=\|\sin-\cos\| \le \|\sin-f\|+\|f-\cos\| \le \sqrt{\pi}.
$$
The only way that happens is if $f$ lines on the segment connecting $\sin$ and $\cos$. That means there is a real scalar $\alpha$ such that $0 \le \alpha \le 1$ and
$$
f = \alpha\sin +(1-\alpha)\cos.
$$
The $\alpha$ is unique because
\begin{align}
\|f-\sin\|=\frac{2}{3}\sqrt{\pi},&\;\;\; \|f-\cos\|=\frac{1}{3}\sqrt{\pi} \\
\|(\alpha-1)\sin+(1-\alpha)\cos\|=\frac{2}{3}\sqrt{\pi},
&\;\;\; \|\alpha(\sin-\cos)\|=\frac{1}{3}\sqrt{\pi} \\
(1-\alpha)\|\sin-\cos\|=\frac{2}{3}\sqrt{\pi}, &\;\;\;
\alpha\|\sin-\cos\|=\frac{1}{3}\sqrt{\pi} \\
(1-\alpha)=\frac{2}{3},&\;\;\; \alpha=\frac{1}{3}.
\end{align}
So that works out to give the solution
$$
f = \frac{1}{3}\{\sin+2\cos\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trigonometry Olympiad problem: Evaluate $1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$
Find the value of
$$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$
My attempt
I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!
I also tried using double angle formula for the angles which are even.
| Use $\sin(\theta)=\sin(180^\circ-\theta)$. Let the summation as $S$. Then
\begin{align}
2S&=\sum_{k=1}^{90}(k\sin(2k^\circ)+(90-k)\sin((180-2k)^\circ)) \\
&=90\sum_{k=1}^{90}\sin(2k^{\circ})
\end{align}
Now use $-2\sin(2k^\circ)\sin(1^\circ)=\cos((2k+1)^\circ)-\cos((2k-1)^\circ)$, then
\begin{align}
S=45\sum_{k=1}^{90}\sin(2k^{\circ})=-\frac{45}{2\sin(1^\circ)}\sum_{k=1}^{90}(\cos((2k+1)^\circ)-\cos((2k-1)^\circ)=\frac{45\cos(1^\circ)}{\sin(1^\circ)}=45\cot(1^\circ)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Show $\ln\left(\frac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\frac{a-b}{n}+\mathcal{O}\left(\frac{1}{n^2} \right) $
I would like to prove the following:
$$\ln\left(\dfrac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\dfrac{a-b}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) $$
My attempt
i tried this way but i didn't get what i want :
\begin{align*}
\ln\left(\dfrac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)&=\ln\left(\dfrac{1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}}{1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}} \right) \\
&= \ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}\right)^{-1} \right) \\
&= \ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1-\dfrac{(-1)^{n}}{\sqrt{n}}-\dfrac{b}{n}+\mathcal{O}\left( \dfrac{1}{n}\right)\right) \right) \\
&=\ln\left( 1+\dfrac{a-b}{n}-\dfrac{(-1)^{n}}{\sqrt{n}}-\dfrac{1}{n}+\mathcal{O}\left( \dfrac{1}{n}\right) \right)
\end{align*}
| One way could be $$\ln { \left( \frac { n+{ \left( -1 \right) }^{ n }\sqrt { n } +a }{ n+{ \left( -1 \right) }^{ n }\sqrt { n } +b } \right) } =\ln { \left( n+{ \left( -1 \right) }^{ n }\sqrt { n } +a \right) -\ln { \left( n+{ \left( -1 \right) }^{ n }\sqrt { n } +b \right) } } =\\ =\ln { \left( n\left( 1+\frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { a }{ n } \right) \right) } -\ln { \left( n\left( 1+\frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { b }{ n } \right) \right) } =\\ =\ln { \left( n \right) +\ln { \left( 1+\frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { a }{ n } \right) } } -\ln { \left( n \right) -\ln { \left( 1+\frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { b }{ n } \right) } } =\\ =\frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { a }{ n } -\frac { 1 }{ 2 } { \left( \frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { a }{ n } \right) }^{ 2 }+O\left( \frac { 1 }{ { n }^{ 2 } } \right) -\left( \frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { b }{ n } -\frac { 1 }{ 2 } { \left( \frac { { \left( -1 \right) }^{ n } }{ \sqrt { n } } +\frac { b }{ n } \right) }^{ 2 }+O\left( \frac { 1 }{ { n }^{ 2 } } \right) \right) =\\ =\frac { a-b }{ n } +\frac { { \left( -1 \right) }^{ n }\left( a-b \right) }{ n\sqrt { n } } +O\left( \frac { 1 }{ { n }^{ 2 } } \right) \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$
Find $$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}.$$
My attempt:
$$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{3}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{1}{\frac{n-1}{3}}\right)^{2n+3}$$
Now we should do something to change the power to $\frac{n-1}{3}$ because:
$\lim_{x\to \infty}(1+\frac{1}{x})^x=e$
But I cannot get the answer(the answer is $e^2$). Please give small hints not full answers.
Here is a picture from my answer:
Note that in persian $2=۲$ and $3=۳$.
edit:The answer is mistaked and take $n+1$ instad of $n+1$.
|
$$\lim _{ n\to \infty } \left( \frac { n+2 }{ n-1 } \right) ^{ 2n+3 }=\lim _{ n\to \infty } \left( 1+\frac { 3 }{ n-1 } \right) ^{ 2n+3 }=\\ =\lim _{ n\to \infty }{ \left[ \left( 1+\frac { 1 }{ \frac { n-1 }{ 3 } } \right) ^{ \frac { n-1 }{ 3 } } \right] } ^{ \frac { 3 }{ n-1 } \left( 2n+3 \right) }=\lim _{ n\rightarrow \infty }{ { e }^{ \frac { 6n+9 }{ n-1 } } } ={ e }^{ 6 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't belong simultaneously to the interval $(1,2)$ I have to solve the following problem but I don't know how to :
Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't belong simultaneously to the interval $(1,2).$
Here's what I've tried:
I worked on cases
$$\begin{cases} a \ge b \ge c \vee a \ge c \ge b \rightarrow \frac{a+b}{b+c-a}<0 \\ b \ge a \ge c \vee b \ge c \ge a \rightarrow \frac{c+a}{c+a-b}<0 \\ c \ge a \ge b \vee c \ge b \ge a \rightarrow \frac{a+b}{a+b-c}<0 \end{cases} $$
Therefore $a,b,c$ can't be simultaneously in the interval $(1,2)$
| Assume $a\leq b\leq c$
Notice $\frac{a+b}{a+b-c}\in(1,2)\implies \frac{a+b-c}{a+b}\in(\frac{1}{2},1)\implies \frac{c}{a+b}\in(0,\frac{1}{2})$, but $\frac{c}{a+b}\geq\frac{c}{2c}=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find the determinant of this $n \times n$ matrix in a clever way? Is there any clever and short way to find out the determinant of the following matrix?
\begin{bmatrix}
b_1 & b_2 & b_3 & \cdots & b_{n-1} & 0 \\
a_1 & 0 & 0 & \cdots & 0 & b_1 \\
0 & a_2 & 0 & \cdots & 0 & b_2\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
\end{bmatrix}
Any help will be appreciated.
| $b_1 \det\begin{pmatrix}
0 & 0 & \cdots & 0 & b_1 \\
a_2 & 0 & \cdots & 0 & b_2\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
\end{pmatrix}
- a_1 \det\begin{pmatrix}
b_2 & b_3 & \cdots & b_{n-1} & 0 \\
a_2 & 0 & \cdots & 0 & b_2\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
\end{pmatrix}$
$(-1)^nb_1\cdot a_2\cdot a_3\cdots a_{n-1}\cdot b_1-\\
a_1\cdot b_2 \det\begin{pmatrix}
0 & 0 & \cdots & 0 & b_2 \\
a_3 & 0 & \cdots & 0 & b_3\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
\end{pmatrix}+\\
a_1\cdot a_2 \det\begin{pmatrix}
b_3 & b_4 & \cdots & b_{n-1} & 0 \\
a_3 & 0 & \cdots & 0 & b_3\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
\end{pmatrix}$
I think we have a pattern.
$b_1^2\cdot a_2\cdots a_{n-1} + a_1\cdot b_2^2\cdot a_3\cdots a_{n-1}+a_1\cdot a_2\cdot b_3^2\cdot a_4\cdots a_{n-1}+\cdots$
$(-1)^n\sum_\limits{j=1}^{n-1} b_j^2 \frac{\prod_\limits{k=1}^{n-1} a_k}{a_j}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove using induction the following equation is true. If $$(1-x^2)\frac{dy}{dx} - xy - 1 = 0$$
Using induction prove the following for any positive integer n$$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$$
I know Leibtniz can be used to solve it easier but I need the proof to use induction.
| It obviously holds for $n=0$; now assume it holds for a general $n$. That is, we have $$(1-x^2)\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (2n+3)x\frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} - (n+1)^2\frac{\mathrm{d}^ny}{\mathrm{d}x^n} = 0$$
Let us differentiate (all differentiation in this answer is with respect to $x$) each term separately for clarity. The first term differentiates to (via use of the product rule) $$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}} - 2x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}}$$
The second term, again (via use of the product rule) differentiates to $$-(2n+3)\left(x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} + \frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}}\right)$$
and the third term differentiates to $$-(n+1)^2 \frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}}$$
The RHS of the equality differentiates trivially to $0$. So, putting all this together gives us
$$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}} - 2x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (2n+3)x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - 2n\frac{\mathrm{d}^{n+1}y}{\mathrm{d} x^{n+1}} -3\frac{\mathrm{d}^{n+1}y}{\mathrm{d} x^{n+1}} - (n+1)^2\frac{\mathrm{d}^{n+1}y}{\mathrm{d} x^{n+1}} $$
(all of that equal to $0$) Cleaning up and grouping terms gives
$$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}} - (2(n+1) +3)x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (n^2 + 4n + 4)\frac{\mathrm{d}y^{n+1}}{\mathrm{d}x^{n+1}} = 0$$
And hence, since $n^2 + 4n+4 = (n+2)^2$ we have $$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}} - (2(n+1) +3)x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (n +2)^2\frac{\mathrm{d}y^{n+1}}{\mathrm{d}x^{n+1}} = 0$$
as required, rather uneventfully.
| {
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Proving an expression is an integer
Prove that $$A^2 \cdot \dfrac{\sum_{m=1}^{n}(a^{3m}-b^{3m})}{\sum_{m=1}^n(a^m-b^m)}-3A^2$$ is an integer if $a=\frac{k+\sqrt{k^2-4}}{2}, b=\frac{k-\sqrt{k^2-4}}{2}$, where $k>2$ is a positive integer, and $A = \dfrac{1}{\sqrt{k^2-4}}$ for all positive integers $n$.
I thought about expanding the numerator and denominator. For the denominator we get $\sum_{m=1}^n(a^m-b^m) = \left(a \cdot \dfrac{a^n-1}{a-1}-b \cdot \dfrac{b^n-1}{b-1}\right)$ and we can similarly do the same thing with the numerator. How do we use all this to solve the question?
| Partial answer:
Observe that $a,b$ are the roots of the quadratic equation $x^2 - kx + 1 = 0$, and hence $ab = 1$.
This implies that
$$ \frac{\sum (a^{3m}- b^{3m})}{\sum (a^m - b^m)} - 3 = \frac{\sum (a^m - b^m)^3}{\sum (a^m - b^m)} $$
Next, we also have the recurrence relation
$$ k(a^m - b^m) - (a^{m-1} - b^{m-1}) = a^{m+1} - b^{m+1} $$
so defining $c_m$ by $c_1 = 1$, $c_2 = k$ and
$$ c_{m+2} = k c_{m+1} - c_m $$
we have that it suffices to show that
$$ \frac{\sum_{m = 1}^n (c_m)^3}{\sum_{m = 1}^n c_m} $$
is an integer. This last step however I don't know how to treat.
| {
"language": "en",
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Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:
$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$
I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also am unable to think substitution.
| Hint: $$\int\frac{1+x^{2}}{1+x^{4}}dx=\frac{1}{2}\int\left(\frac{1}{x^{2}+\sqrt{2}x+1}+\frac{1}{x^{2}-\sqrt{2}x+1}\right)dx$$ $$=\frac{1}{2}\int\left(\frac{1}{\left(x+\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}+\frac{1}{\left(x-\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}\right)dx.$$
| {
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Solve the second order equation $$\frac{d^2u}{dt^2} = \begin{bmatrix} -5 & -1 \\ -1 & -5 \end{bmatrix}u
$$
with
$$
u(0)=\begin{bmatrix}1 \\ 0 \end{bmatrix}
$$
and
$$
u'(0) = \begin{bmatrix}0 \\ 0 \end{bmatrix}
$$
Okay I tried to create a $u$ vector but couldn't. And believe I can't get $\frac{du}{dt}$ by finding the eigenvalues and eigenvectors of the matrix $A$ right? Since matrix $A$ multiplied by $u$ gives the second derivate, not matrix
$A$ multiplied by the first derivative.
What to do here?
| An equivalent method without (explicitly) using eigenvalues, as the matrix is symmetric:
Writing $u=(u_1,u_2)^T$ the d.e. is equivalent to:
\begin{align}
\frac{d^2 u_1}{d t^2}&=-5u_1-u_2\\
\frac{d^2 u_2}{d t^2}&=-u_1-5u_2\\
\end{align}
Consider instead $u_1+u_2$, and $u_1-u_2$,
\begin{align}
\frac{d^2 }{d t^2}(u_1+u_2)&=-6(u_1+u_2)\\
\frac{d^2 }{d t^2}(u_1-u_2)&=-4(u_1-u_2)\\
\end{align}
Thus $$u_1+u_2=A\cos(t\sqrt{6})+B\sin(t\sqrt{6})$$
and $$u_1-u_2=C\cos(2t)+D\sin(2t)$$
Thus
\begin{align}
u_1&=\frac{1}{2}(A\cos(t\sqrt{6})+B\sin(t\sqrt{6})+C\cos(2t)+D\sin(2t))\\
u_2&=\frac{1}{2}(A\cos(t\sqrt{6})+B\sin(t\sqrt{6})-C\cos(2t)-D\sin(2t))\\
\end{align}
$u(0)=(1,0)^T \implies \frac{1}{2}(A+C)=1$ and $\frac{1}{2}(A-C)=0$, so $A=C=1$.
$u'(0)=(0, 0)^T \implies \frac{1}{2}(\sqrt{6}B+2D)=0$ and $\frac{1}{2}(\sqrt{6}B-2D)=0$, so $B=D=0$.
Thus the solution is:
$$\begin{pmatrix} u_1\\u_2 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} \cos{t\sqrt{6}}+\cos{2t}\\\cos{t\sqrt{6}}-\cos{2t} \end{pmatrix}$$
Alternatively we can follow the clever answer here by user: Start wearing purple.
Let $$A=\begin{pmatrix} -5 & -1\\-1 & -5\end{pmatrix}$$
A is symmetric so we can diagonalize it with an orthogonal matrix $P$ whose columns are eigenvectors. The eigenvalues of $A$ are found by solving $\det(\lambda I-A)=0$, which gives $(\lambda+5)^2-1=0$, i.e. $\lambda=-4$ or $\lambda=-6$. Plugging back in lambda and solving for $(A-\lambda I)\vec{x}=0$, gives two eigenvectors $(1, 1)^T$ and $(1, -1)^T$. So take $$P=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}$$
As you can check $P$ is orthogonal, $P^TP=PP^T=I$. Then $$PAP^T=\begin{pmatrix}-6 & 0 \\ 0 & -4\end{pmatrix}=D$$
So if we act on our differential equation by $P^T$ then we have that
$$\frac{d^2}{dt^2}P^T \vec{u}=PA{u}=DP^T\vec{u}$$
Let $\vec{v}=P^T \vec{u}$, then we have the new differential equation:
$$\frac{d^2}{dt^2}\vec{v}=D\vec{v}$$
with new initial conditions $\vec{v}(0)=P^T \vec{u}(0)$ and $\vec{v}'(0)=P^T \vec{u}'(0)$. This decouples into:
\begin{align}
\frac{d^2 v_1}{d t^2}&=-6v_1\\
\frac{d^2 v_2}{d t^2}&=-4v_2\\
\end{align}
We can solve the way we did before $\textit{or}$ to keep everything in terms of matrices and using that $f(\operatorname{diag}(\lambda_1,\lambda_2))=\operatorname{diag}(f(\lambda_1),f(\lambda_2))$ we can say that:
$$\vec{v}=\cos\left(t\sqrt{\strut-D} \right)\vec{v}(0)+\left(\sqrt{\strut-D}\right)^{-1}\sin\left(t\sqrt{\strut-D} \right)\vec{v}'(0)$$
Using that $P\vec{v}=\vec{u}$ we find that
\begin{align}
\vec{u}&=P\cos\left(t\sqrt{\strut-D} \right)P^T\vec{u}(0)+P\left(\sqrt{\strut-D}\right)^{-1}\sin\left(t\sqrt{\strut-D} \right)P^T\vec{u}'(0)\\
&=\cos\left(t\sqrt{\strut-A} \right)\vec{u}(0)+\left(\sqrt{\strut-A}\right)^{-1}\sin\left(t\sqrt{\strut-A} \right)\vec{u}'(0)\\
&=\cos\left(t\sqrt{\strut-A} \right)\vec{u}(0)
\end{align}
using that $P^T=P^{-1}$, and $f(PDP^{-1})=f(D)$. Of course it will be easier to evaluate $P\cos\left(t\sqrt{\strut-D} \right)P^T\vec{u}(0)$ which is
\begin{align}
&=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}\cdot\begin{pmatrix}\cos(t\sqrt{6}) & 0 \\ 0 & \cos{2t}\end{pmatrix} \cdot\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}\\
&=\frac{1}{2}\begin{pmatrix} \cos{t\sqrt{6}}+\cos{2t}\\\cos{t\sqrt{6}}-\cos{2t} \end{pmatrix}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?
| WLOG, $a=b=1$. Then factoring the denominator, we decompose in simple fractions
$$\frac y{y^3+1}=A\frac1{y+1}+B\frac{2y-1}{y^2-y+1}+C\frac1{y^2-y+1}=\frac{(A+2B)y^2+(-A+B+C)y+(A-B+C)}{y^3+1}.$$
Identifying,
$$\frac y{y^3+1}=\frac13\frac1{y+1}-\frac16\frac{2y-1}{y^2-y+1}+\frac12\frac1{y^2-y+1}.$$
The first two terms have an obvious antiderivative. The third one also becomes easy by completing the square
$$\frac1{y^2-y+1}=\frac1{\left(y-\frac12\right)^2+\frac34}=\frac43\frac1{\left(\sqrt{\frac43}(y-\frac12)\right)^2+1}$$ which yields an arctangent.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Constructing the 11-gon by splitting an angle in five In "Angle Trisection, the Heptagon, and the Triskaidecagon", published in the American Mathematical Monthly in March 1988, Andrew Gleason discusses what regular polygons can be constructed with compass, straightedge and angle trisector. At the end of that article he notes that the angle p-sectors required for a regular n-gon are the odd primes p dividing $\varphi(n)$.
For the heptagon, which only requires an angle trisector, he gives the minimal polynomial of $2\cos(2\pi/7)$
$$x^3+x^2-2x-1$$
and transforms it into the Chebyshev polynomial expression
$$7\sqrt{28}(4\cos^3\theta-3\cos\theta)=7$$
leading to the final identity
$$\sqrt{28}\cos\left(\frac{\cos^{-1}(1/\sqrt{28})}{3}\right)=1+6\cos(2\pi/7).$$
I am interested in the hendecagon (11 sides), which requires an angle quinsector (that splits an angle into five equal parts).
Is there a similar transformation between the minimal polynomial for $2\cos(2\pi/11)$
$$x^5+x^4-4x^3-3x^2+3x+1$$
and the relevant Chebyshev polynomial
$$\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$
and how do I find it? If I had such a transformation, I could construct an exact hendecagon with the quinsector.
I have tried Tschirnhaus transforms on the former polynomial, without success.
| The question essentially asks about transforming solvable equations from one form to another.
I. Cubic
Using just a linear transformation, the general cubic $P(x)=0$ can be transformed to the form,
$$y^3+3ay+b = 0\tag1$$
with solution,
$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{3}+\tfrac{1}{3}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^3}}\big)\right)\tag2$$
for $k=0,1,2$. Undoing the transformation establishes a relation between the roots $x,y$.
II. Quintic
Similarly, an appropriate Tschirnhausen transformation can transform a solvable quintic $P(x)=0$ to the Demoivre form (essentially the Chebyshev polynomial mentioned by the OP),
$$y^5+5ay^3+5a^2y+b = 0\tag3$$
with analogous solution,
$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{5}+\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big)\right)\tag4$$
for all five roots $y_k$. A cubic Tschirnhausen gives us three degrees of freedom to transform a solvable quintic to Demoivre form.
III. Transformations
For $p=7$:
$$x=2\cos\big(\tfrac{2\pi}{7}\big)\tag5$$
$$\color{blue}{y=3x+1} = 2\sqrt{7}\cos\left(\tfrac{1}{3}\,\cos^{-1}\big(\tfrac{1}{2\sqrt{7}}\big)\right)=4.7409\dots$$
then $x,y$ solves,
$$x^3+x^2-2x-1=0$$
$$y^3-21y-7=0$$
For $p=11$:
Let $\phi=\tfrac{1+\sqrt{5}}{2}$ be the golden ratio.
$$x=2\cos\big(\tfrac{2\pi}{11}\big)\tag6$$
$$\color{blue}{y=x^3-\phi\,x^2-\tfrac{7+\sqrt{5}}{2}x+\tfrac{5+4\sqrt{5}}{5}} = 2\,\phi\sqrt{\tfrac{11}{5}}\cos\left(-\tfrac{6\pi}{5}+\tfrac{1}{5}\,\cos^{-1}\big(\tfrac{-89-25\sqrt{5}}{44\sqrt{11}}\big)\right)=-4.7985\dots$$
then $x,y$ solves,
$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1=0$$
$$y^5-5ay^3+5a^2y+b=0$$
where $a=\tfrac{11}{5}\phi^2,\;\;b=\tfrac{11(125+89\sqrt{5})}{250}\phi^5$.
Thus as you can see, the transformation (in blue) which relates the quintic roots $x,y$ is more complicated than the cubic version, but is nonetheless doable in radicals.
| {
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"timestamp": "2023-03-29T00:00:00",
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Permutations and Combinations, Picking discs A bag contains 6 red, 5 blue and 4 green discs. If 3 discs are chosen at random, find the probability that there is at least one blue disc given that at least one red disc is chosen.
I thought that on the denominator is 6*14C2 as you pick one red and from the rest you pick 2. On the numerator you do 6*5*13, as you pick one red, one blue and one of the 13 rest.
This would give 5/7.
But the answer is 255/371.
| A red disc is selected unless only blue and green discs are selected. Therefore, the number of selections in which at least one red disc is selected is
$$\binom{6 + 5 + 4}{3} - \binom{5 + 4}{3} = \binom{15}{3} - \binom{9}{3}$$
Selections that include both a blue disc and a red disc either contain one disc of each color, two red discs and a blue disc, or a red disc and two blue discs. The number of such selections is
$$\binom{6}{1}\binom{5}{1}\binom{4}{1} + \binom{6}{2}\binom{5}{1} + \binom{6}{1}\binom{5}{2}$$
Hence, the probability that a blue disc is selected given that at least one red disc is selected is
$$\frac{\dbinom{6}{1}\dbinom{5}{1}\dbinom{4}{1} + \dbinom{6}{2}\dbinom{5}{1} + \dbinom{6}{1}\dbinom{5}{2}}{\dbinom{15}{3} - \dbinom{9}{3}}$$
| {
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How to solve the limit $\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$? Hi I got an examination at the school which was so arduous that I'm stumped.
This problem is the toughest for me :
$$\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$$
I could not any approaching in order to solve problem . I need help!
| We have that
$$\begin{align*}
n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{\left(n+1\right)}{n}\right)
&=n^2\left(\exp\left(n\ln\left(1+\frac{1}{n\left(n+2\right)}\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\exp\left(n\left(\frac{1}{n\left(n+2\right)}+o\left(\frac{1}{n^3}\right)\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\exp\left(\frac{1}{n+2}+o\left(\frac{1}{n^2}\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(1+\frac{1}{n+2}+\frac{1}{2(n+2)^2}+o\left(\frac{1}{n^2}\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\frac{1/n}{1+2/n}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)-\frac{1}{n}\right)\\
&=n^2\left(\frac{1}{n}\left(1-\frac{2}{n}+o\left(\frac{1}{n}\right)\right)+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)-\frac{1}{n}\right)\\
&=n^2\left(-\frac{2}{n^2}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\\
\end{align*}=-\frac{3}{2}+o\left(1\right).$$
Hence your limit as $n\to+\infty$ is $-\frac{3}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^n+n^n+\cdots+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n+1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{1}{n}$$
But is still does not help to conclude about converges/diverges
| You can make your approach work if you do the following:
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^{\color{red}{n-1}}+n^{\color{red}{n-1}}+\cdots+n^{\color{red}{n-1}}+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n-1}(n-1)+n^n}{n^{n+2}}$$
$$=\sum_{n=1}^\infty \frac{n^{n}+n^n-n^{n-1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{2}{n^2}-\sum_{n=1}^\infty \frac{1}{n^3}$$
| {
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Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
\begin{align}\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}&=3\sum_{k=2}^{\infty}\frac{1}{5^{k-1}}\\&=3\sum_{k=2}^{\infty}\frac{1}{5^{k}}\cdot\frac{1}{5^{-1}}\\&=3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5\\&= 15\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )\\&=15\left( \frac{1}{1-\frac{1}{5}}-2 \right )\\&=-\frac{45}{4}\end{align}
Did I do it correctly?
| This step is wrong: $3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5= 15\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )$
It should be: $3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5= 15\left( \sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k} \right)=15(\sum_{k=0}^\infty\frac{1}{5}^k-\frac{6}{5})=15(\frac{1}{1-\frac{1}{5}}-\frac{6}{5})=15(\frac{5}{4}-\frac{6}{5})=15\frac{1}{20}=\frac{3}{4}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Basic Joint probability distribution calculation example? (X, Y) has the Joint probability distribution:
\begin{array}{|c|c|c|}
\hline
X / Y & 1 & 2 \\ \hline
1& \frac{1}{4}& \frac{1}{3} \\ \hline
2& \frac{1}{6}& a \\ \hline
\end{array}
So $P(X=1\mid Y=1)=\frac{3}{5}$ and $P(X=2\mid Y=2)=\frac{3}{7}$.
My challenge is via calculation, how $\frac{3}{5}$ and $\frac{3}{7}$
was calculated in this example?
| $$
P(X=1|Y=1)=\frac{P(X=1, Y=1)}{P(Y=1)}= \frac{1/4}{1/6+1/4}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is my solution of $e^{\sin x } - e^{- \sin x} - 4 = 0$ wrong? $$e^{\sin x } - e^{- \sin x} - 4 = 0$$
Substitute $e^{\sin x} = y$:
$$y - \frac{1}{y} - 4 = 0 \implies y^2 - 4y - 1 = 0$$
Solve for $y$:
$$y = 2 \pm \sqrt{5}$$
$e^{\sin x}$ can't be negative:
$$\therefore y = 2 + \sqrt{5} \implies e^{\sin x} = 2 + \sqrt{5}$$
Differentiating both sides with respect to $x$:
$$\frac{d e^{\sin x}}{d x} = \frac{d (2 + \sqrt{5})}{d x} \implies \frac{d e^{\sin x}}{d \sin x} \times \frac{d \sin x}{d x} = 0 \implies e^{\sin x} \times \cos x = 0$$
If $e^{\sin x} = 0$ then $x$ has no solution. However, when $\cos x = 0$ then $x = 90^{\circ}$
The actual answer of the question states that there is no real solution of the equation. Then why am I getting the solution as $90^{\circ}$ too?
| Let $e^{\sin x} = 2 + \sqrt{5}$. Then, $\sin{x}=\ln{(2+\sqrt{5})}>\ln{3}>1$, which isn't possible (since $\sin{x}\in[-1,1]$ for every $x$). Therefore, there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$
Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$
My work: $d|n^3+5n,n^3-5n^2+6n\implies\ d|5n^2-n,5n^2+25\implies d|n+25 $
Now I can't go further!!
| Let $\,f(n) = n^2+5.\,$ Then $\,d\mid \color{#c00}{f(n),n\!+\!25}\,\Rightarrow\, d\mid \overbrace{{f(n)\ \rm mod}\ (n\!+\!25) = f(-25)}^{\textstyle\!\! \color{#c00}{f(n)} - (\color{#c00}{n\!+\!25})q(n) = r(n)} = 630\, $ where we used the Remainder theorem $\ f(n)\equiv f(a)\pmod{n-a}$
Remark $\ $ It is simpler to work modulo the common divisor $\,d,\,$ namely
$ n^2\!+5\equiv 0\,\Rightarrow\, \color{#c00}{n^2\equiv -5}\,\Rightarrow\, 0\equiv n(\color{#c00}{n^2})\!-5\color{#c00}{n^2}\!+6n\equiv (\color{#c00}{-5})n-5(\color{#c00}{-5})+6n\equiv \color{#0a0}{n+25} $
thus $\ \color{#0a0}{n\equiv -25}\,\Rightarrow\, 0\equiv \color{#0a0}n^2+5\equiv (\color{#0a0}{-25})^2+5\equiv 630,\ $ therefore $\ d\mid 630$.
Alternatively $\, 25\!+\!n\equiv 0\,\Rightarrow\ 0 \equiv (25 + n)\,(25 - n)\equiv 25^2\!-n^2\equiv 625-(-5)\equiv 630$
essentially $\,\ 25\!+\!\sqrt5\equiv 0\,\Rightarrow\, 0\equiv (25\!+\sqrt5)(25\!-\!\sqrt5)\equiv 625,\, $ like taking a norm in $\,\Bbb Z[\sqrt 5]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $a^{2013} + b^{2013} + c^{2013}$ Problem Statement
Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$
Suppose that the following conditions hold:
*
*$f(1)=0$
*the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$
$$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$
My attempt
Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$
Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$
$$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$
And so we have $$a = -(p+q+r) = -p^2q^2r^2$$
$$b = pq + pr + qr = p^2 + q^2 + r^2$$
$$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$
Then we get the following relations
$$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$
From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$
However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?
Edit
For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$
| The line
$$b = pq + pr + qr = p^2 + q^2 + r^2$$
is wrong.
$$b = pq + pr + qr =-( p^2 + q^2 + r^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Optimizing continued fraction of square root From this question, I learned that the square root of a number $n$ can be written as a continued fraction of the form:
$$\sqrt n=a+\frac{n-a^2}{a+\sqrt n}$$
where $a$ can have any value. By jumping to conclusions and testing, I believe that the optimal value for $a$ for a rapid convergence is the largest integer such that $a^2 < n$, but I haven't even been able to start trying to prove this.
Any insight would be helpful. Thanks!
| \begin{align*}
\frac{p_{n}}{q_{n}} &= \sqrt{n}+e_{n} \\
&= a+\frac{n-a^2}{a+\frac{p_{n-1}}{q_{n-1}}} \\
&= \frac{a\left(a+\frac{p_{n-1}}{q_{n-1}} \right)+n-a^2}
{a+\frac{p_{n-1}}{q_{n-1}}} \\
&= \frac{a \frac{p_{n-1}}{q_{n-1}}+n}{\frac{p_{n-1}}{q_{n-1}}+a} \\
&= \frac{a (\sqrt{n}+e_{n-1})+n}{\sqrt{n}+e_{n-1}+a} \\
&= \frac{\sqrt{n}(\sqrt{n} \color{red}{+e_{n-1}}+a)+
(a \color{red}{-\sqrt{n}})e_{n-1}}
{\sqrt{n}+e_{n-1}+a} \\
&= \sqrt{n}+\frac{(a-\sqrt{n})e_{n-1}}{a+\sqrt{n}+e_{n-1}} \\
\frac{p_{n}}{q_{n}}-\sqrt{n} &=
\frac{(a-\sqrt{n})e_{n-1}}{a+\sqrt{n}+e_{n-1}} \\
e_{n} &=
\frac{(a-\sqrt{n})e_{n-1}}{a+\sqrt{n}+e_{n-1}} \\
\frac{e_{n}}{e_{n-1}} & \approx \frac{a-\sqrt{n}}{a+\sqrt{n}} \\
\end{align*}
In usual practice, we take $a=\lfloor \sqrt{n} \rfloor \implies a-\sqrt{n}<0 $,
so that
$$\frac{p_n}{q_n} \lessgtr \sqrt{n} \lessgtr \frac{p_{n+1}}{q_{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving a non-homogeneous linear recurrence relation So i have this non-homogeneous linear recurrence relation to solve:
$$a_{n}=2a_{n-1}-a_{n-2}+2^n+2$$
$a_{1}=7$ and $a_{2}=19$
I know that the non-homogeneous part is $2^n$ and i know how to solve homogeneous linear recurrence relations, but when i get a non-homogeneous one i have no idea how to approach it.
I have this as well: $b_{n}=Aq^n$ when the non-homogeneous part is an exponential function look-alike. So yeah i know it should be of help, but i don't know how to apply it in solution. I need to make a characteristic polynomial as well(if that's what it is called) but i am really stuck at using all that in practical example.
So i am not sure how to solve it, any hints or help would really be appreciated.
| Define:
$g_n := a_{n}-a_{n-1}$
Then $a_{n}=2a_{n-1}-a_{n-2}+2^n+2 \implies a_{n}-a_{n-1}=a_{n-1}-a_{n-2}+2^n+2 \\\ \implies g_n = g_{n-1} + 2^n +2 \implies g_n- g_{n-1} = 2^n +2$
$$g_n- g_{n-1} = 2^n +2 $$
$$g_{n-1}- g_{n-2} = 2^{n-1} +2 $$
$$.$$
$$.$$
$$.$$
$$g_3- g_{2} = 2^3 +2 $$
Since $g_2 = 19-7=12$, then $g_n = \sum_{i=3}^n g_i-g_{i-1} +g_2 = 2^3\times \frac{1-2^{n-3+1}}{1-2} + 2\times (n-3+1) + 12= 2^{n+1}+2n$ Therefore $a_n-a_{n-1} = 2^{n+1}+2n$.For all $n \geq 3$.
$$a_n-a_{n-1} = 2^{n+1}+2n$$
$$a_{n-1}-a_{n-2} = 2^{n}+2(n-1)$$
$$.$$
$$.$$
$$.$$
$$a_3-a_{2} = 2^4+2 \times 3$$
$a_n = \sum_{i = 3}^{n} a_i - a_{i-1} + a_{2} = 2^4 \times \frac{1-2^{n-3+1}}{1-2}+ (3+n) \times (n-3+1) + 19 = n^2+n +2^{n+2}-3$. For all $n \geq 3$.
EDIT:
For generalization:
$$a_n = Aa_{n-1}+Ba_{n-2} + P(n)$$
Assert: $$a_n - C a_{n-1} = D (a_{n-1} - Ca_{n-2}) + P(n) \implies
$$ Where $C,D \in \mathbb C$
\begin{cases}C+D = A \\ CD = -B\end{cases}
so $C,D$ is roots of $x^2-Ax-B=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
hypergeometric 2F2 function close form for specific case I am trying to find the closed form for the following
\begin{equation}
{_2F_2}(2,4;3,4+a;a)
\end{equation}
any help would be greatly appreciated. thank you
I find the closed form for
\begin{equation}
{_2F_2}(1,3;2,3+a;a)=1+\frac{a}{2}
\end{equation}
But don't success to solve the first sum
| $_2F_2(2,4;3,4+a;a)$
$=\sum\limits_{n=0}^\infty\dfrac{(2)_n(4)_na^n}{(3)_n(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{(n+3)(2)_na^n}{3(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{(n+2)(2)_na^n}{3(a+4)_nn!}+\sum\limits_{n=0}^\infty\dfrac{(2)_na^n}{3(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{(2)_{n+1}a^n}{3(a+4)_nn!}+\sum\limits_{n=0}^\infty\dfrac{(2)_na^n}{3(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{2(3)_{n}a^n}{3(a+4)_nn!}+\sum\limits_{n=0}^\infty\dfrac{(2)_na^n}{3(a+4)_nn!}$
$=\dfrac{2}{3}{}_1F_1(3;a+4;a)+\dfrac{1}{3}{}_1F_1(2;a+4;a)$
$=\dfrac{(a+3)(a+2)(a+1)}{3}\int_0^1x^2(1-x)^ae^{ax}~dx+\dfrac{(a+3)(a+2)}{3}\int_0^1x(1-x)^{a+1}e^{ax}~dx$
Which is expected to related to incomplete gamma function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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