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In a cyclic $\square ABCD$, $BC, CD$ and $DA$ are three tangents of such a circle that its center is on the side $AB$. Proving that $AD + BC = AB$
In a cyclic quadrilateral $ABCD$, $BC, CD$ and $DA$ are three tangents of a circle. The center of the circle is located on the side $AB$. Prove that $$AD + BC = AB$$
Attempt:
First, I thought it to be very easy. So, I let the side $AB$ be the diameter of the large circle by which the quadrilateral $ABCD$ is circumscribed.
So, I made another quadrilateral congruent to $ABCD$ to the opposite side. So, I got a regular hexagon, the one side of which is denoted as $a$ and the radius of large circle = $R$ and the radius of small circle = $r$.
We know that area of regular polygon = $\frac{na^2}{4} \cot \frac{180}{n}$. $ABCD$ is the semi hexagon and so the area of $ABCD =\frac{1}{2}\cdot \frac{6a^2}{4} \cot (\frac{180}{6})^\circ = \frac{3a^2}{4} \cot 30^\circ = \frac{3a^2}{4}\cdot \sqrt3 = \frac {3\sqrt 3a^2}{4}........(i)$
Again $ABCD$ is a trapezium. So, $[ABCD] = \frac{1}{2}(2R + a)\cdot r..........(ii)$
Now from right angled triangle $DJA$: $\frac{r}{a} = \sin 60^\circ$ so $r = \frac{\sqrt 3a}{2}$
So, from equation $(ii)$ again, we get $[ABCD] = \frac{1}{2} (2R + a)\cdot\frac {\sqrt 3a}{2} = \frac{\sqrt 3a}{4} (2R + a)........(iii)$
Now from equation $(i)$ and $(iii)$ we get
$\frac{\sqrt 3a}{4} (2R + a) = \frac{3 \sqrt3 a^2}{4}$ and thus
$(2R + a) = 3a$ so $2R = 2a$.
Hence, $R = a$. And thus I proved that $2R = a + a \implies AD + BC = AB$.
But that wasn't a satisfactory solution for me in the case of letting $AB$ be the diameter of large circle and I reasonably made it specific. But I am very unaware of the fact that how could I solve that proof for any position of $AB$ such that other three sides of the quadrilateral $ABCD$ are tangents to the small circle?
Thanks in advance.
Source : IMO $1985$
|
Let $\mathcal{C}$ be a circle wichi is tangent to $BC,CD$ and $DA$ and let it center be $O$. Then $O$ must lie on angle bisector of angle $\angle ADC$ and angle $\angle DCB$. So these angle bisectors meet on $AB$.
Now let $E$ be on $AB$ so that $AE=AD$. We have to prove $BC=BE$ i.e. triangle $BCE$ is isosceles.
If $\angle ADE =\alpha$ then $\angle AED =\alpha$ and $\angle ADE =180^{\circ}-2\alpha$. So $\angle BCD =2\alpha$ and thus $\angle ECD =\alpha$. Since $\angle DEB =180^{\circ}-\alpha$ we have $CDEO$ is cyclic.
Let $\angle OEC =\beta$ then $\angle ODC =\beta$ and $\angle ADO = \beta $ so $\angle ABC =180^{\circ}-2\beta$ and thus $\angle BCE =\beta$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
\begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 \Rightarrow f^{\prime}(x) = 3x^2 + 18x + 36 \Rightarrow 3x^2 + 18x + 36 = 9 \Rightarrow\\\\
& 3x^2 + 18x = -27 \Rightarrow 3x ( x + 6 ) = -27 \Rightarrow 3x = -27 x + 6 = -27 \Rightarrow
x = -3 x = -33
\end{align*}
|
So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$.
First derive and set derivative to 9.
$ f'(x) = 3x^{2}+18x+36=9 $
$ \Rightarrow x^{2}+6x+12-3=0 $ (divide by 3 then subtract 3)
$ \Rightarrow (x+3)^{2}=0 $
$ \therefore x=-3$ ($x$-ordinate of the point on f(x))
To find the $y$-oordinate substitute $x$ back into $f(x)$:
$f(-3)=-44$
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
|
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"url": "https://math.stackexchange.com/questions/3141926",
"timestamp": "2023-03-29T00:00:00",
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|
Prove value of $\sin(15°)$ I'm doing the following exercise: prove that
$$
\sin(15°)=\frac{1}{2\sqrt{2+\sqrt{3}}}
$$
I'm using this formula:
$$
\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)
$$
to get:
$$
\sin(45-30)=\sin(45)\cos(30)-\cos(45)\sin(30) \\
=\frac{1}{\sqrt2}\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\frac{1}{2} \\
=\frac{\sqrt3-1}{2\sqrt2}
$$
However I'm stuck, can't seem to get to the desired result. I tried multiplying for $$\frac{\sqrt{3}+1}{\sqrt{3}+1}$$ but didn't get anything out of it. Any help will be really appreciated.
|
They are equivalent. Since $$(1 + \sqrt{3})^2 = 1 + 2 \sqrt{3} + 3 = 2(2 + \sqrt{3}),$$ it follows that $$\sqrt{2 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{2}}.$$ Then $$\frac{1}{2 \sqrt{2 + \sqrt{3}}} = \frac{\sqrt{2}}{2 (1 + \sqrt{3})} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}.$$
It is also worth noting that we can obtain the desired expression more directly via the half-angle identity for $0 \le \theta \le 90^\circ$: $$\sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} = \frac{\sqrt{1 - \cos^2 \theta}}{\sqrt{2 (1 + \cos \theta)}} = \frac{\sin \theta}{\sqrt{2 (1 + \cos \theta)}},$$ where upon selecting $\theta = 30^\circ$ immediately yields $$\sin 15^\circ = \frac{1}{2 \sqrt{2 + \sqrt{3}}}$$
|
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"url": "https://math.stackexchange.com/questions/3143138",
"timestamp": "2023-03-29T00:00:00",
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|
Finding expression for variable x in an equation (use Lambert function?) Let $a$, $b$ and $c$ be constants. How can one find an expression for variable $x$ in the following equation? $$\frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c})$$
From my research, I am being suggested to use the Lambert function, but I still feel quite helpless! It seems very complex.
|
$\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\e{\mathrm{e}}$
\begin{align}
\frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c})
\tag{1}\label{1}
\end{align}
Indeed, the solution to equation \eqref{1}
can be expressed in terms of the Lambert W function.
The usual approach in this case is to transform
original equation to the form
\begin{align}
t\cdot\exp(t)&=\dots
\tag{2}\label{2}
,
\end{align}
where the right-hand side of \eqref{2}
does not depend on $t$,
and then apply the Lambert W function
\begin{align}
\W(t\cdot\exp(t))&=\W(\dots)
\tag{3}\label{3}
\\
\text{to get }\quad
t&=\W(\dots)
\tag{4}\label{4}
.
\end{align}
First, simplifying \eqref{1} using substitution
\begin{align}
y&=1+\frac{ax}c
\tag{5}\label{5}
,\\
x&=\frac ca\cdot(y-1)
\tag{6}\label{6}
,
\end{align}
we get
\begin{align}
\ln(y)\cdot y&=\frac{ab}c-1+y
,\\
\ln(y)\cdot y-y&=\frac{ab}c-1
,\\
(\ln(y)-1)\cdot y&=\frac{ab}c-1
,\\
\ln(\tfrac y\e)\cdot y&=\frac{ab}c-1
,\\
\ln(\tfrac y\e)\cdot \tfrac y\e&=\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)
,\\
\ln(\tfrac y\e)\cdot
\exp\left(\ln(\tfrac y\e)\right)&=\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)
\tag{7}\label{7}
.
\end{align}
Now we have \eqref{7} in the form of \eqref{2}, hence
\begin{align}
\W\left[\ln(\tfrac y\e)
\cdot
\exp\left(\ln(\tfrac y\e)\right)
\right]
&=
\W\left[\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)\right]
,\\
\ln(\tfrac y\e)
&=
\W\left[\tfrac 1\e\cdot\Big(\frac{ab}c-1\Big)\right]
\tag{8}\label{8}
,
\end{align}
and from here $x$ can be trivially found.
$\endgroup$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
where $\rho$ is the "Golden Ratio" $\rho = \frac{1+\sqrt{5}}{2}$.
But I can't find a way to prove it.
Prove
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
|
Look at finite sums first (such that we do not subtract two diverging series):
$$
\sum_{n=1}^N\frac1{5n(5n-1)} =\\
\sum_{n=1}^N\frac{-1}{5n} + \frac{1}{5n-1} = \\
- \frac15 \sum_{n=1}^N(\frac{1}{n} - \frac{1}{n-1/5}) = \\
- \frac15 (\sum_{n=1}^N\frac{1}{n} - \sum_{n=0}^{N-1} \frac{1}{n+4/5} )= \\
- \frac15 (\sum_{n=1}^N\frac{1}{n} - \sum_{n=1}^N \frac{1}{n+4/5} - \frac54 + \frac{1}{N+4/5}) =\\
\frac{1}{4} - \frac{1}{5N+4} - \frac15 \sum_{n=1}^N(\frac{1}{n} - \frac{1}{n+4/5})
$$
The sum converges, so we can take the limit $N\to \infty$ to obtain
$$
\sum_{n=1}^\infty \frac1{5n(5n-1)} =\\
\frac{1}{4} - \frac15 \sum_{n=1}^\infty (\frac{1}{n} - \frac{1}{n+4/5})
$$
Now you can use a result (with proof by Achille Hui) from here which says
$$
\mathcal{S}_{k/p} \stackrel{def}{=}
\sum_{n=1}^\infty\left(\frac{1}{n} - \frac{1}{n+\frac{k}{p}}\right)
\\=
\frac{p}{k} - \log(2p) -\frac{\pi}{2}\cot\left(\frac{k\pi}{p}\right) +
\sum_{l=1}^{p-1}
\cos\left(\frac{2\pi k\ell}{p}\right) \log\sin\left(\frac{\ell\pi}{p}\right) \\
= \psi\left(\frac{k}{p}+1\right) + \gamma
$$
and conclude with $k=4$, $p=5$ that $\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac{1}{4} - \frac15 (\psi\left(\frac{9}{5}\right) + \gamma)$ or, without using the Digamma function,
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} =\\
\frac{1}{4}-\frac15 (\frac{5}{4} - \log(10) -\frac{\pi}{2}\cot\left(\frac{4\pi}{5}\right) +
\sum_{l=1}^{4}
\cos\left(\frac{2\pi 4\ell}{5}\right) \log\sin\left(\frac{\ell\pi}{5}\right)) =\\
\frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac15
\sum_{l=1}^{4}
\cos\left(\frac{2\pi 4\ell}{5}\right) \log\sin\left(\frac{\ell\pi}{5}\right) = \\
\frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac1{20} (\log(16/5) + \sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)))\\
\simeq 0.07756
$$
which is exactly as given in the question: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach :- Since rank is $2$ , So, the determinant of all the submatrices of order $3 \times 3$ must be zero.
So, $\begin{vmatrix} 1&1 &2 \\ 1&1 &1 \\ a&b &b \end{vmatrix}$ = $0$. After solving it, I am getting $a=b$
and $\begin{vmatrix} 1&2 &2 \\ 1&1 &3 \\ b&b &1 \end{vmatrix}$ = $0$. After Solving it , $b=\frac{1}{3}$
Now, Rank of a matrix is also defined as no. of non-zero rows in row echelon form of that matrix. So, If I convert it into Row Echelon form then
$\begin{bmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{bmatrix}$
Applying $R_{2} \leftarrow R_{2}-R_{1}$
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ a& b &b &1 \end{bmatrix}$
Now, Applying $R_{3}\leftarrow R_{3}-aR_{1}$
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&(b-a) &(b-2a) &(1-2a) \end{bmatrix}$
Now, to make rank of this matrix = $2$ means I should have 2 non-zero rows or all the elements of the last row must be zero simultaneously.
So, $(b-a) =0$ and $(b-2a) =0$ and $(1-2a) =0$
So, $b=a$ and $b=2a$ and $a=\frac{1}{2}$
Now, My doubt is how $b=a$ and $b=2a$ is possible here simultaneously and why this method is giving wrong result. I must be doing some mistake here but I am not getting what mistake I am doing. Please help.
|
If $b=a$, then the last row is $\begin{bmatrix}0&0&-a&1-2a\end{bmatrix}$. In which case you can use the second row to kill entries in this row and then get echelon form $$ \begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&0 &0 &(1-3a) \end{bmatrix}$$
Now for rank to be $2$, you need $a=1/3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_0^{\pi/2} \log \left| \sin^2 x - a \right|$ where $a\in [0,1]$.
How to evaluate
$$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
$$
where $a\in[0,1]$ ?
I think of this problem as a generalization of the following proposition
$$
\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2
$$
My try
Put
$$
I(a)=\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
$$
From the substitution $x \to \frac{\pi}{2}-x$ , we get
$$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
=
\displaystyle\int_0^{\pi/2} \log \left| \cos^2 x - a \right|\,dx
$$
Thus
$$
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx
=
\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - (1-a) \right|\,dx
$$
which means $$I(a)=I(1-a) \tag{1}$$
On the other hand,
\begin{align}
2I(a) &=
\displaystyle\int_0^{\pi/2} \log \left| (\sin^2 x - a)(\cos^2 x -a) \right|\,dx
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| a^2-a+\sin^2 x \cos^2 x \right|\,dx
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 (2x) \right|\,dx
-\pi \log 2
\\ &=
\frac{1}{2}\displaystyle\int_0^{\pi} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx
-\pi \log 2
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx
-\pi \log 2
\\ &=
\displaystyle\int_0^{\pi/2} \log \left| 1+4(a^2-a)-\sin^2 x \right|\,dx
-\pi \log 2
\\ &=
I((2a-1)^2)
-\pi \log 2
\end{align}
Thus
$$
2I(a)=I((2a-1)^2)-\pi \log 2 \tag{2}
$$
Let $a=0$ we get the proposition mentioned above $\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2.$
But how to move on ? Can we solve the problem only by $(1)$ and $(2)$? Or what other properties should we use to evaluate that?
Looking forward to your new solutions as well.
Thank you in advance!
Added:
As pointed out in the comments, it seems like that the integral is identical to $-\pi\log 2$.
From $(1)$ and $(2)$ we can also find many numbers such that $I(a)=-\pi\log 2$.
|
Using the symmetries and developing the $\sin^2$ term, we can express
\begin{align}
I&=\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \sin^2 x - a \right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1-2a}{2}-\frac{1}{2}\cos 2x\right|\,dx
%&=-\frac{\pi}{2}\ln 2+\frac{1}{4}\int_0^{2\pi} \log \left| \left( 2a-1 \right)+\cos 2x\right|\,dx
\end{align}
By denoting $2a-1=\cos 2\alpha$,
\begin{align}
I&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1}{2}\left( \cos 2\alpha+\cos 2x\right)\right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x+\alpha \right)\cos \left( x-\alpha \right)\right|\,dx\\
&=\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x+\alpha \right)\right|\,dx+\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x-\alpha \right)\right|\,dx
\end{align}
As the functions are periodic, the integration variables can be shifted, thus
\begin{equation}
I=\frac{1}{2}\int_0^{2\pi} \log \left| \cos \left( x \right)\right|\,dx
\end{equation}
Finally using the symmetries of the integrand,
\begin{align}
I&=2\int_0^{\pi/2} \log \left| \cos \left( x \right)\right|\,dx\\
&=2\int_0^{\pi/2} \log \left| \sin \left( x \right)\right|\,dx\\
&=-\pi\ln 2
\end{align}
from the quoted result.
|
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|
Help Solving Recurrence Relation: $a_n = n^3a_{n-1} + (n!)^3$ I'm trying to find an explicit formula for the following recurrence relation:
$a_0 = 1$, $\forall n \ge 1: a_n = n^3a_{n-1} + (n!)^3$
So far though, my attempts have been unsuccessful.
My Attempt
Let $A(x) = a_n\frac{x^n}{(n!)^3}$. Multiply by $\frac{x^n}{(n!)^3}$ and sum over $n$ to get:
$$\sum_{n=1}^{\infty}{a_n\frac{x^n}{(n!)^3}} = \sum_{n=1}^{\infty}{a_{n-1}\frac{x^n}{(n-1!)^3}} + \sum_{n=1}^{\infty}{x^n}$$
This can be simplified to:
$$
A(x) - a_0 = xA(x) + \frac{1}{1-x} \\
\implies A(x) = \frac{2-x}{(1-x)^2}$$
Using partial fraction decomposition you recover that $A(x) = \frac{1}{(1-x)} + \frac{1}{(1-x)^2}$, which means that $$A(x) = \sum_{n=0}^{\infty}{x^n} + \sum_{n=0}^{\infty}{(n+1)x^n} \\
\implies A(x) = \sum_{n=0}^{\infty}{(n+2)x^n}$$
Then, to put $A(x)$ back into the form that we defined it in in the beginning, we have:
$$A(x) = \sum_{n=0}^{\infty}{(n+2)(n!)^3\frac{x^n}{(n!)^3}} \\
\implies a_n = (n+2)(n!)^3$$
I'm wonderring if anyone can provide some insight into where I could have gone wrong. Is the original definition of $A(x)$ valid, or must I stick to either an ordinary / exponential generating function? If so, how could I clear the variable terms in the recurrence?
|
One simple way to find these sequences (so simple it is almost cheating!) is to generate the first few terms and search for the resulting sequence in the Online Encyclopedia of Integer Sequences (OEIS). In this case you have:
$$\begin{equation} \begin{aligned}
a_0 &= 1, \\[6pt]
a_1 &= 1^3 a_0 + (1!)^3 = 1^3 \cdot 1 + 1^3 = 2, \\[6pt]
a_2 &= 2^3 a_1 + (2!)^3 = 2^3 \cdot 2 + 2^3 = 24, \\[6pt]
a_3 &= 3^3 a_2 + (3!)^3 = 3^3 \cdot 24 + 6^3 = 864, \\[6pt]
a_4 &= 4^3 a_3 + (4!)^3 = 4^3 \cdot 864 + 24^3 = 69120, \\[6pt]
&\quad \vdots \\[6pt]
\end{aligned} \end{equation}$$
Searching the terms 1, 2, 24, 864, 69120 on OEIS leads to a single hit for the integer sequence A172492, which has the form:
$$a_n = (n!)^2 \cdot (n+1)! \quad \quad \quad \text{for all } n = 0,1,2,3,....$$
Inductive proof: It is now possible to prove that this form is correct using proof by weak induction. For the base case $n=0$ we have:
$$a_0 = (0!)^2 \cdot (0+1)! = 1^2 \cdot 1! = 1 \cdot 1 = 1.$$
For the inductive step we assume that $a_k = (k!)^2 \cdot (k+1)!$ and we then have:
$$\begin{equation} \begin{aligned}
a_{k+1}
&= (k+1)^3 a_k + (k+1)!^3 \\[6pt]
&= (k+1)^3 \cdot (k!)^2 \cdot (k+1)! + (k+1)!^3 \\[6pt]
&= (k+1) \cdot (k+1)!^2 \cdot (k+1)! + (k+1)!^3 \\[6pt]
&= (k+1) \cdot (k+1)!^3 + (k+1)!^3 \\[6pt]
&= [(k+1) + 1] \cdot (k+1)!^3 \\[6pt]
&= (k+2) \cdot (k+1)!^3 \\[6pt]
&= (k+1)!^2 \cdot (k+2)!. \\[6pt]
\end{aligned} \end{equation}$$
Since the base case is correct, and the inductive step is proved, this proves that the sequence you are dealing with is the integer sequence A172492, with explicit form given above.
|
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"url": "https://math.stackexchange.com/questions/3148902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Ways to simplify arctan() in integral results? Lately when I was computing $$\int\frac{\mathrm{d}x}{1+\sqrt{1-2x-x^2}},$$I got the result$$-2\arctan{\frac{\sqrt{1-2x-x^2}-1}{x}}-\ln\left(1-\frac{\sqrt{1-2x-x^2}-1}{x}\right)+\mathbf{C}.$$However, Wolfram|Alpha gives me its result in a different form, which may be written as$$\arcsin{\frac{x+1}{\sqrt{2}}}+\frac{1}{2}\ln{\frac{\sqrt{1-2x-x^2}-x+1}{\sqrt{1-2x-x^2}+x+3}}+\mathbf{C'},$$and I accordingly found that actually$$-2\arctan{\frac{\sqrt{1-2x-x^2}-1}{x}}=\arcsin{\frac{x+1}{\sqrt{2}}}+\frac{\pi}{4},$$a much neater expression! But apparently, without Wolfram|Alpha, I couldn't have found this neat relation that turns a complicated $2\arctan$ into a neat $\arcsin$ at all, so I've been wondering if there are any general way to simplify these $\arctan$s in integral expressions. Thanks!
|
Note that
$$
1-2x-x^2=2-(1+2x+x^2)=2-(1+x)^2
$$
so you can do the substitution $u\sqrt{2}=1+x$ and the integral becomes
$$
\int\frac{\sqrt{2}}{1+\sqrt{2}\sqrt{1-u^2}}\,du
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3149294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a - 2)(c^2 + a^2 - ca - c - a + 1)\\
&\ge (1 - b)\left[\dfrac{3}{4}(c + a)^2 - b + 4\right]\\
&=(1 - b)\left[\dfrac{3}{4}(3 - b)^2 - b + 4\right]\\
&= \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43)
\end{align*}
That means
\begin{align*}
(a - 1)^3 + (b - 1)^3 + (c - 1)^3 &\ge \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43) + (b - 1)^3\\
&= \dfrac{1}{4}(b - 1)(b^2 + 14b - 39)
\end{align*}
Since $c \ge a \ge b \implies b \le \dfrac{a + b + c}{3} = 1$.
And this is where I am stuck right now.
|
Hope it will help!
WLOG $ b\ge a\ge c\Rightarrow c\le 1\Rightarrow 1-c\ge 0$
We have: $$f\left(a,b,c\right)=\left(a-1\right)^3+\left(b-1\right)^3+\left(c-1\right)^3$$
And $$f\left(\frac{a+b}{2};\frac{a+b}{2};c\right)=2\left(\frac{a+b}{2}-1\right)^3+\left(c-1\right)^3$$
Note that:$$f\left(a,b,c\right)-f\left(\frac{a+b}{2};\frac{a+b}{2};c\right)=\frac{3}{4}\left(a-b\right)^2\left(a+b-2\right)=\frac{3}{4}\left(a-b\right)^2\left(1-c\right)\ge 0$$
So you need to prove this inequality in the case $a=b$ the condition gives $c=3-2a$
Thus we need to prove $$2\left(a-1\right)^3+\left(3-2a-1\right)^3\ge -\frac{3}{4}$$
Or $$-\frac{3}{4}\left(2a-3\right)\left(4a^2-6a+3\right)\ge 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3152853",
"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{cyc}\frac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$
If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$ \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$
Here's what I did.
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{a(b^2 + c^2 + bc)}{(b^2c + c^2b + abc)^2} = \frac{1}{(a + b + c)^2}\sum_{cyc}\frac{ab^2 + ac^2 + 1}{(bc)^2} \le \frac{1}{9}\sum_{cyc}\left(\frac{a}{b^2} + \frac{a}{c^2} + a^2\right)$$
And this is where I gave up.
Thanks for reading this. Even more thanks if you can solve the problem.
(Oh wait, I did try again.)
$$\sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$= 3 - \sum_{cyc}\frac{b^4 + c^4}{a + b^4 + c^4} \le 3 - 4\sum_{cyc}\frac{a^4}{b + c + c^4 + 2a^4 + b^4} = 2\sum_{cyc}\frac{b + c + b^4 + c^4}{b + c + c^4 + 2a^4 + b^4} - 3$$
...and again...
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$= 3 - \sum_{cyc}\frac{b^4 + c^4}{a + b^4 + c^4} \le 3 - \frac{1}{2}\sum_{cyc}\frac{(b^2 + c^2)^2}{a + b^4 + c^4} = \sum_{cyc}\frac{2(a - bc) + (b^4 + c^4)}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{1}{2(a - bc) + (b^4 + c^4)}\frac{[2(a - bc) + (b^4 + c^4)]^2}{a + 2b^2c^2}$$
$$= \frac{1}{2}\sum_{cyc}\dfrac{(a - bc + b^4)^2 + (a - bc + c^4)^2}{[2(a - bc) + (b^4 + c^4)](a + 2b^2c^2)}$$
And I need help. I would be grateful if you could help with this problem.
|
Another way:
By C-S
$$\sum_{cyc}\frac{a}{a+b^4+c^4}-1=\sum_{cyc}\frac{a(a^3+2)}{(a+b^4+c^4)(a^3+1+1)}-1\leq$$
$$\leq\sum_{cyc}\frac{a^4+2a}{(a^2+b^2+c^2)^2}-1
=-\frac{2\sum\limits_{cyc}(a^2b^2-a^2bc)}{(a^2+b^2+c^2)^2}=-\frac{\sum\limits_{cyc}c^2(a-b)^2}{(a^2+b^2+c^2)^2}\leq0.$$
|
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|
inequality related to square the sum of any two sides of a triangle with respect to square of other side Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.
My approach
As we know that the sum of any two sides of triangle will always be greater then the third side i.e
$(a+b)>c $
$(b+c)>a$
$(a+c)>b$
Take square on both side then
$a^2+b^2+2ab>c^2$
$ c^2+b^2+2bc>a^2$
$a^2+c^2+2ac>b^2$
Add last three equations
$2a^2+2b^2+2c^2-a^2-b^2-c^2>-2(ab+bc+ac)$
$a^2+b^2+c^2>-2(ab+bc+ac)$
Edited :
Corrected equation
But the correct answer is this $a^2 + b^2 + c^2 < 2(ab + bc + ca)$
What am I doing wrong?Please guide.
|
The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing from David K's answer:
One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
which is the number of ways to put $3$ or fewer indistinguishable balls in $3$ numbered bins without that constraint, which is the number of ways of putting exactly $3$ indistinguishable balls in $4$ numbered bins.
Multiply by $8$ to take into account all the cases where $x < 0$ or $y < 0$ or $z < 0.$
$x,y,z >0 => x + y + z <= 3, n = Cr(3+4-1,3) * (8) = 160$
Now let $x = 0,$ $y > 0,$ and $z > 0.$
Count the number of ways to put up to $6$ balls in $2$ bins if each bin must contain at least one ball. Multiply by four to account for all the cases were $y < 0$ or $z < 0.$ Multiply that result by $3$ to account for the fact that we could have chosen $y= 0$ or $z=0$ instead of $x = 0.$
$x=0, x,y>0 => y+z <=6, n = Cr(4+3-1, 4) * 3 * 14 = 180$
Now let $x = y = 0$ and $z > 0.$ There are $6$ ways for that to happen. Multiply by $2$ to account for $z < 0,$ then by $3$ to account for the other choices of which variables are zeros.
$x,y = 0, z>0 => z<=6, n = 6 * 2 * 3 = 36$
$x,y,z=0, n = 1$
Sum of all of them is : 160 + 180 + 36 + 1 = 377
|
One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
which is the number of ways to put $3$ or fewer indistinguishable balls in $3$ numbered bins without that constraint, which is the number of ways of putting exactly $3$ indistinguishable balls in $4$ numbered bins.
Multiply by $8$ to take into account all the cases where $x < 0$ or $y < 0$ or $z < 0.$
Now let $x = 0,$ $y > 0,$ and $z > 0.$
Count the number of ways to put up to $6$ balls in $2$ bins if each bin must contain at least one ball. Multiply by four to account for all the cases were $y < 0$ or $z < 0.$ Multiply that result by $3$ to account for the fact that we could have chosen $y= 0$ or $z=0$ instead of $x = 0.$
Now let $x = y = 0$ and $z > 0.$ There are $6$ ways for that to happen. Multiply by $2$ to account for $z < 0,$ then by $3$ to account for the other choices of which variables are zeros.
Finally add $1$ for the case $x = y = z = 0,$ which was not covered by any of the other cases.
The total will be $377$ if you do all these calculations correctly.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Check if vectors are the the fundamental set of solutions $$\begin{cases} x_1+2x_2+x_3-x_4+x_5=0 \\ 2x_1+x_2-x_3+2x_4-x_5=0 \\ x_1+5x_2+4x_3-5x_4+4x_5=0 \\ 4x_1+5x_2+x_3+x_5=0 \end{cases}$$
$$\begin{cases} \vec{a}_1=(-5,4,3,3,-3) \\ \vec{a}_2=(-7,5,-3,6,6) \\ \vec{a}_3=(5,-4,6,-3,-6) \end{cases}$$
Solution:
\begin{bmatrix}
1&2&1&-1&1 \\
2&1&-1&2&-1 \\
1&5&4&-5&4 \\
4&5&1&0&1
\end{bmatrix}
Reducing to the row echelon form gives us
\begin{bmatrix}
1&2&1&-1&1 \\
0&3&-3&-4&-3
\end{bmatrix}
Let $x_3 = c_1$, $x_4=c_2$, $x_5=c_3$.
$$\begin{cases} x_1+2x_2+c_1-c_2+c_3=0 \\ 0x_1+3x_2-3c_1-4c_2-3c_3=0\end{cases}$$
$$x_2=c_1+\frac{4}{3}c_2+c_3$$
$$x_1=-2c_1-\frac{8}{3}c_2-2c_3-2c_1+c_2-c_3=-4c_1-\frac{5}{3}c_2-3c_2$$
However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?
|
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 &0 \\
-1 & 0 & 1 &0 \\
3 &-1 &-1 &0 \\
-5 & 0 & 1& 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1& 2 &1& -1 & 1\\
2& 1 &-1 & 2 &-1\\
1 &5 & 4 &-5 & 4\\
4 &5& 1 & 0& 1\\
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
1& 2 &1& -1 & 1\\
0& 3 &3 & -4 &3\\
0 &0 & 0 &0 & 0\\
0 &0& 0 & 0& 0\\
\end{array}
\right)
$$
|
{
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"url": "https://math.stackexchange.com/questions/3157512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the surface area of this function when rotated around the y axis. Stuck Say I have this equation:
$$y = \frac{1}{3} x^{\frac{3}{2}}$$ and the limits of the integration provided are $0 \leq x \leq 12$
So since we're rotating around the y axis, I'm going to change the limits of integration to use y.
$$0 \leq y \leq 8 \sqrt{3}$$
Also, arc length is:
$$\sqrt{1 + \frac{dy}{dx}^2}$$
and $\frac{dy}{dx} = \frac{1}{2} x^{\frac{1}{2}}$ and so $\frac{dy^2}{dx} = \frac{1}{4}x$
So I have this SA equation:
$$SA = 2 \pi \int_0^{8 \sqrt{3}} x \sqrt{1 + \frac{1}{4}x}dx$$
$$ = 2 \pi \int_0^{8 \sqrt{3}} x \sqrt{\frac{4 + x}{4}x}dx $$
taking out the $\sqrt{4}$ from the denominator:
$$ = \pi \int_0^{8 \sqrt{3}} x \sqrt{4 + x} $$
Then I get $u = x + 4$ and $ x = u - 4$
(this is odd... I guess this is another way to usub?)
new limits of integration are:
when $x = 0$, $u = 0$
when $x = 8 \sqrt{3}$, $u = 8 \sqrt{3} + 4$
$$ = \pi \int_0^{8 \sqrt{3} + 4} \sqrt{u} (u -4) du$$
$$ = \pi \int_0^{8 \sqrt{3} + 4} u^{\frac{3}{2}} - 4 \sqrt{u} du$$
$$ = \pi * \frac{2}{5}u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}} \biggr]_0^{8 \sqrt{3} + 4}$$
This comes out to be pretty ugly... did I make a mistake?
|
Yes, you did make a mistake in the very beginning, when you kinda decided to switch to integration with respect to $y$ … but you didn't actually switch. Note that your integral
$$2\pi\int_0^{8\sqrt{3}}x\sqrt{1+\frac{1}{4}x}\,dx$$
is with respect to $x$, but you're using the $y$-values from $0\le y\le8\sqrt{3}$ as your limits of integration. That doesn't make sense — effectively you're saying that $x$ ranges from $0$ to $8\sqrt{3}$, which isn't true.
Once you decided that
I'm going to change the limits of integration to use $y$,
you should actually do that. From the given equation you express $x$ as a function of $y$, $x=\ldots\text{something in terms of }y\ldots$, and then the arclength will be $ds=\sqrt{1+\left(\frac{dx}{dy}\right)^2}$, which you will use for setting up an integral with respect to $dy$.
|
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|
series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow \frac{\pi}{2}-x$
$\displaystyle I=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\ln(1-k\sin x)dx$
$\displaystyle I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\bigg[k\sin x-\frac{k^2\sin^2 x}{2}+\frac{k^3\sin^3 x}{3}-\cdots \bigg]dx$
$\displaystyle I =-2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{k^2\sin^2 x}{2}+\frac{k^4\sin^4 x}{4}+\cdots \bigg]dx$
$\displaystyle I =-\pi\bigg[\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots \bigg]$
How can I find sum of that series. Help me please.
|
METHODOLOGY $1$: Feynman's Trick
Let $I(a,k)$ be the integral given by
$$I(a,k)=\int_0^\pi \log(a+k\cos(x))\,dx\tag1$$
Differentiating $(1)$ with respect to $a$ reveals
$$\begin{align}
\frac{\partial I(a,k)}{\partial a}&=\int_0^\pi \frac{1}{a+k\cos(x)}\,dx\\\\
&=\frac{\pi}{\sqrt{a^2-k^2}}\tag2
\end{align}$$
Integrating $(2)$ with respect to $a$ and using $I(a,0)=\pi \log(a)$ yields
$$I(a,k)=\pi \log(a+\sqrt{a^2-k^2})-\pi\log(2)\tag3$$
Setting $a=1$ in $(3)$, we obtain the coveted result
$$\int_0^\pi \log(1+k\cos(x))\,dx=\pi\log\left(\frac{1+\sqrt{1-k^2}}{2}\right)$$
METHODOLOGY $2$: Series Evaluation
Enforcing the substitution $x\mapsto \pi/2-x$ in $(1)$, we have
$$\begin{align}
I(1,k)&=\int_{-\pi/2}^{\pi/2}\log(1+k\sin(x))\,dx\\\\
&=\int_{-\pi/2}^{\pi/2}\sum_{n=1}^\infty \frac{(-1)^{n-1}k^n\sin^n(x)}{n}\,dx\\\\
&=-\sum_{n=1}^\infty \frac{k^{2n}}{n}\int_0^{\pi/2}\sin^{2n}(x)\,dx\\\\
&=-\pi\int_0^k \sum_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}\,x^{2n-1}\,dx\\\\
&=-\pi\int_0^k \frac1x \sum_{n=1}^\infty \binom{-1/2}{n}(-x^2)^n\,dx\\\\
&=-\pi \int_0^k \frac1x\left(\frac1{\sqrt{1-x^2}}-1\right)\,dx\\\\
&=\pi \log\left(\frac{1+\sqrt{1-k^2}}{2}\right)
\end{align}$$
as expected!
|
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|
The generalised Catalan Numbers and Borel's Triangle I am currently reading "Counting with Borel’s Triangle" (https://arxiv.org/abs/1804.01597), and am very confused on a stated formula.
We know:
$C_{n,k}=\frac{n-k+1}{n+1}{n+k \choose n}$
$C(x) = \sum_{n=0}^{\infty}\frac{1}{n+1}{2n \choose n}x^{n}=\frac{1-\sqrt{1-4x}}{2x}$
$C(t,x)=\sum_{n,k}C_{n,k}t^kx^n =\frac{C(tx)}{1-xC(tx)}$
$C(2,i)=\sum_{k}C_{i-1,k}2^k$
These I can completely understand the derivation of.
However, I cannot under formula (7):
$\sum_{i\geq0}C(2,i)x^i=\frac{1+2xC(2x)}{1+x}=\frac{1}{1-xC(2x)}=\frac{4}{3+\sqrt{1-8x}}$
How are these equal? Clearly, the last equals holds, but how does the first or second?
To start, I assume we have:
$\sum_{i\geq0}C(2,i)x^i=\sum_{i\geq0}\sum_{k}C_{i-1,k}2^kx^i$
I then assume we reindex:
$\sum_{i\geq0}\sum_{k}C_{i-1,k}2^kx^i=\sum_{n+1\geq0}\sum_{k}C_{n,k}2^kx^{n+1}$
Then we can remove a factor of $x$:
$\sum_{n+1\geq0}\sum_{k}C_{n,k}2^kx^{n+1}=x\sum_{n+1\geq0}\sum_{k}C_{n,k}2^kx^{n}$
But now I am stuck. I feel like I've gone down a completely wrong path?
|
We show the validity of the equality chain
\begin{align*}
\sum_{i\geq0}C(2;i)x^i=\frac{1+2xC(2x)}{1+x}=\frac{1}{1-xC(2x)}=\frac{4}{3+\sqrt{1-8x}}
\end{align*}
with $$C(x)=\frac{1-\sqrt{1-4x}}{2x}$$ the generating function of the Catalan numbers and the generalized Catalan numbers $C(2;i)$ stored in OEIS as A064062.
First part: $\frac{1}{1-xC(2x)}=\frac{4}{3+\sqrt{1-8x}}$
We obtain
\begin{align*}
\color{blue}{\frac{1}{1-xC(2x)}}&=\frac{1}{1-x\left(\frac{1-\sqrt{1-8x}}{4x}\right)}\\
&=\frac{4}{4-(1-\sqrt{1-8x})}\\
&\,\,\color{blue}{=\frac{4}{3+\sqrt{1-8x}}}
\end{align*}
Second part: $\frac{1+2xC(2x)}{1+x}=\frac{4}{3+\sqrt{1-8x}}$
We obtain
\begin{align*}
\color{blue}{\frac{1+2xC(2x)}{1+x}}&=\frac{1}{1+x}\left(1+2x\left(\frac{1-\sqrt{1-8x}}{4x}\right)\right)\\
&=\frac{1}{1+x}\left(1+\frac{1-\sqrt{1-8x}}{2}\right)\\
&=\frac{1}{2(1+x)}\left(3-\sqrt{1-8x}\right)\\
&=\frac{1}{2(1+x)}\cdot\frac{9-(1-8x)}{3+\sqrt{1-8x}}\\
&\,\,\color{blue}{=\frac{4}{3+\sqrt{1-8x}}}
\end{align*}
For the next part it is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We start by inspecting $C(2;i)=\sum_{k=0}^{i-1}C_{i-1,k}2^k$ and looking at the relationship with the bivariate generating function
\begin{align*}
C(t;x)=\sum_{n=0}^\infty\sum_{k=0}^nC_{n,k}t^kx^k=\frac{C(tx)}{1-xC(tx)}.
\tag{1}
\end{align*}
Third part: $\sum_{i\geq0}C(2;i)x^i=\frac{1}{1-xC(2x)}$
We obtain for $i\geq 1$:
\begin{align*}
\color{blue}{C(2;i)}&=\sum_{k=0}^{i-1}C_{i-1,k}2^k\\
&=[x^{i-1}]\frac{C(2x)}{1-xC(2x)}\tag{2}\\
&=[x^i]\frac{xC(2x)}{1-xC(2x)}\\
&=[x^i]\left(\frac{1}{1-xC(2x)}-1\right)\\
&\,\,\color{blue}{=[x^i]\frac{1}{1-xC(2x)}}\tag{3}
\end{align*}
With $C(2;0)=1$ we finally obtain from (3)
\begin{align*}
\color{blue}{\sum_{i=0}^{\infty}C(2;i)x^i}&=\sum_{i=0}^\infty[u^i]\frac{1}{1-uC(2u)}x^i\\
&\,\,\color{blue}{=\frac{1}{1-xC(2x)}}\tag{4}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we evaluate (1) at $t=2$.
*In (4) we use the substitution rule
\begin{align*}
A(x)=\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty [u^n]A(u) x^n
\end{align*}
|
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|
Find all positive integers $n$ such that $\frac{3^n-1}{2^n}$ is an integer. I need to find all positive integers $n$ such that $\frac{3^n-1}{2^n}$ is an integer. So far I only found $n=1$, $n=2$ and $n=4$ and solutions to this problem? Is there a way to prove using modular arithmetic that there are no more solutions (or if there are more solutions), because I can't seem to find any more.
|
If $m$ is odd then the highest power of $2$ that divides $3^m-1$ is $1$ and dividing $3^m+1$ is $2$. If $n=2^k\cdot m $ where $m$ is odd then factorize the numerator as $ (3^m -1)(3^m +1)(3^{2m} +1)...(3^{2^{k-1}m} +1 )$ .From the third term onwards highest power of $2$ dividing each factor is $1$ and hence highest power of $ 2$ dividing the product is $3+k-1=k+2\geq 2^k\cdot m $ . This inequality gives you the desired solutions
|
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|
Prove that $\sum_{n} t_n x^n = \frac{x^k}{(1-x^2)^k(1-x)} $
Let $t_n$ be a number of sequences
$$ 1 \le a_1 < a_2 < ... < a_k \le n $$
such that $a_{2i}$ is even and $a_{2i+1}$ is odd.
Prove that $$\sum_{n} t_n x^n = \frac{x^k}{(1-x^2)^k(1-x)} $$
My attempt. I want to use there enumerators and I have 4 cases.
1. both $n$ and $k$ are even (other cases will be analogical)
Then enumerator will be $$(x+x^3+...+x^{n-1})^{k/2} (1+x^2+...+x^{n})^{k/2} = \left(x\cdot \frac{1-x^{\frac{n-1}{2}}}{1-x^2} \cdot \frac{1-x^{\frac{n}{2}}}{1-x^2} \right)^{k/2}$$
But how can I get from there my thesis?
|
We have that
$$f(x):=\frac{x^k}{(1-x^2)^k(1-x)}=\frac{1}{1-x}\left(\frac{x}{1-x^2}\right)^k
=\frac{1}{1-x}\left(\sum_{j=0}^{\infty}x^{2j+1}\right)^k.$$
Therefore the coefficient $t_n=[x^n]f(x)$ is the number of ways we can write a positive integer $\leq n$ as the sum of $k$ odd numbers.
Now note that $a_1,a_2-a_1,\dots,a_k-a_{k-1}$ are $k$ odd numbers whose sum is $a_k$ which is $\leq n$ and such $k$ odd numbers determine in unique way a sequence $ 1 \le a_1 < a_2 < ... < a_k \le n $ such that $a_{2i}$ is even and $a_{2i+1}$ is odd.
|
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|
Help verifying proof to proving that for all natural numbers $\sqrt{1} \leq$ than the sum Prove that for all natural numbers $n$,
$$\sqrt{n} \le 1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Solution: We must prove that $1 + \frac 1 {\sqrt{2}}, + \frac 1 {\sqrt{3}} +\,\cdots\, + \frac 1 {\sqrt{k+1}} \geq {\sqrt{k+1}}$
Add $\sqrt{k+1}-\sqrt{k}$ to both sides of first inequality, we get:
$1+\frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \sqrt{k+1}-\sqrt{k} \geq \sqrt{k+1}$
However,
$$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})}=\frac{k+1-k}{(\sqrt{k+1}+\sqrt{k})}=\frac{1}{(\sqrt{k+1}+\sqrt{k})} \leq \frac{1}{\sqrt{k+1}}$$
So:
$\sqrt{k+1} \leq 1 + \frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \sqrt{k+1}-\sqrt{k} \leq 1+\frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}}+\cdots + \frac{1}{\sqrt{k+1}}$
|
You are right. The inductive proof needs to establish the link between
$$\sqrt n\le1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}$$
and
$$\sqrt{n+1}-\frac1{\sqrt{n+1}}\le1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Obviously,
$$\sqrt{n+1}-\frac1{\sqrt{n+1}}=\frac n{\sqrt{n+1}}<\sqrt n$$ does the trick.
For a better estimate, use
$$\frac1{\sqrt{x+1}}<\frac1{\sqrt{\lfloor x\rfloor}}\le\frac1{\sqrt x},$$ then by integration
$$\int_1^{n+1}\frac{dx}{\sqrt{x+1}}<\sum_{k=1}^n\frac1{\sqrt k}\le\int_1^{n+1}\frac{dx}{\sqrt x},$$
$$2(\sqrt{n+2}-\sqrt2)<\sum_{k=1}^n\frac1{\sqrt k}\le2(\sqrt{n+1}-1).$$
|
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|
Simplify trigonometric expression using trigonometric identities I have the trigonometric expression: $$2\sin x +2\sin \left(\frac{\pi} {3} -x\right) $$
and it should simplified in: $$\sin x + \sqrt 3 \cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?
|
Using the formula for $\sin (\alpha - \beta)$ you obtain
\begin{align}
2&\sin x +2\sin \left(\frac{\pi} {3} -x\right)\\
= 2&\sin x +2\left[\sin \left(\frac{\pi} {3}\right) \cos x - \cos\left(\frac{\pi} {3}\right) \sin x\right]\\
= 2&\sin x + 2\left[{\sqrt 3 \over 2} \cos x - \frac 1 2 \sin x\right]\\[1ex]
= 2&\sin x + \sqrt 3 \cos x - \sin x\\[1em] =\ \, &\color{red}{\sin x + \sqrt 3 \cos x}
\end{align}
|
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|
A tricky integral involving hyperbolic functions Can anyone suggest a method for solving the integral below? I've tried numerous things but have had no luck yet. To be honest I'm not sure an analytical solution actually exists.
$$I=\int\cosh(2x)\sqrt{[\sinh(x)]^{-2/3}+[\cosh(x)]^{-2/3}}\,\textrm{d}x.$$
Thanks.
Here is another attempt I have made:
Let $y=[\tanh(x)]^{2/3}$, and rewrite $I$ such that
$$I=\int\cosh(2x)[\sinh(x)]^{-1/3}\sqrt{1+[\tanh(x)]^{2/3}}\,\textrm{d}x.$$
Then $\textrm{d}x=(3/2)[\tanh(x)]^{1/3}[\cosh(x)]^{2}\,\textrm{d}y$. Therefore
\begin{align*}
I&=\frac{3}{2}\int\cosh(2x)[\sinh(x)]^{-1/3}[\tanh(x)]^{1/3}[\cosh(x)]^{2}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int\cosh(2x)[\cosh(x)]^{5/3}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int[\cosh(x)]^{11/3}\sqrt{1+y}\,\textrm{d}y
+\frac{3}{2}\int[\sinh(x)]^{2}[\cosh(x)]^{5/3}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int[\textrm{sech}(x)]^{-11/3}\sqrt{1+y}\,\textrm{d}y
+\frac{3}{2}\int[\tanh(x)]^{2}[\textrm{sech}(x)]^{-11/3}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int[\textrm{sech}(x)]^{-11/3}\sqrt{1+y}\,\textrm{d}y
+\frac{3}{2}\int[\textrm{sech}(x)]^{-11/3}y^3\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int\frac{(1+y)^{1/2}}{(1-y^3)^{11/6}}\,\textrm{d}y
+\frac{3}{2}\int y^{3}\frac{(1+y)^{1/2}}{(1-y^3)^{11/6}}\,\textrm{d}y
\\
&=\frac{3}{2}\int\frac{(1+y^3)(1+y)^{1/2}}{(1-y^3)^{11/6}}\,\textrm{d}y
\\
&=\frac{3}{2}\int\frac{(1-y^6)(1+y)^{1/2}}{(1-y^3)^{17/6}}\,\textrm{d}y
\end{align*}
|
Thanks to Yuri for his working. Very helpful. I believe I have arrived at the solution now
\begin{align*}
I&=\int\cosh(2x)\sqrt{[\sinh(x)]^{-2/3}+[\cosh(x)]^{-2/3}}\,\textrm{d}x
\\
&=\int\cosh(2x)\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1/6}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
\\
&=\frac{6}{5}\int\frac{\textrm{d}}{\textrm{d}x}\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}
[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
\\
&=\frac{6}{5}\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}
\\&\quad-\frac{6}{5}
\int\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}\frac{\textrm{d}}{\textrm{d}x}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x.
\end{align*}
Now
\begin{equation*}
\frac{\textrm{d}}{\textrm{d}x}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}
=\frac{1}{6}\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1}\frac{[\sqrt[3]{\tanh(x)}-\sqrt[3]{\coth(x)}]}{[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}}.
\end{equation*}
Therefore
\begin{align*}
I_{2}&=\frac{6}{5}
\int\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}\frac{\textrm{d}}{\textrm{d}x}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
\\&=
\frac{1}{5}\int\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1/6}
\frac{[\sqrt[3]{\tanh(x)}-\sqrt[3]{\coth(x)}]}{[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}}\,\textrm{d}x.
\end{align*}
Let $y=[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}$ then
\begin{equation*}
I_{2}=
\frac{6}{5}\int\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}\,\textrm{d}y=\frac{6}{5}\int[\coth(x)-\tanh(x)]^{-5/6}\,\textrm{d}y
.
\end{equation*}
After a bit of work it is possible to show that
\begin{equation*}
\coth(x)-\tanh(x)=(y^{4}-1)\sqrt{y^{4}-4}.
\end{equation*}
Therefore
$$I=\frac{6}{5}\biggl\{\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}y-\int(y^{4}-1)^{-5/6}(y^{4}-4)^{-5/12}\,\textrm{d}y\biggr\},$$
where $y$ is as above and the integral can be written in terms of the hypergeometric function.
A very interesting problem!
Thanks to all.
|
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|
Solve $\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$
Find $$\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$$
Looking at the numerator, combined with the surd, you can get $$\int\frac{e^x(1+x)\left(\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right)}{1-x^2}\mathrm dx$$Then this begins to look like the quotient rule, since the denominator is $\sqrt{1-x^2}^2$, and in the numerator, $1/\sqrt{1-x^2}$ is almost like the derivative of the square root. However, it isn't quite that - there are extra terms. $$\color{red}{e^x\left(\sqrt{1-x^2}+\frac{x}{\sqrt{1-x^2}}\right)}+\frac{e^x}{\sqrt{1-x^2}}+e^x x\sqrt{1-x^2}$$
The red terms are accounted for by quotient rule (giving an integral of $\frac{e^x}{\sqrt{1-x^2}}$). But then what do we do with the remaining terms?
|
Hint:
$$\dfrac{1+1-x^2}{(1-x)^{3/2}(1+x)^{1/2}}=\dfrac1{...}+f(x)$$
where $f(x)=\dfrac{\sqrt{1+x}}{\sqrt{1-x}},$
$f'(x)=?$
Recall $\dfrac{d(e^xf(x))}{dx}=?$
|
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|
Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$
The solution proposes that $x^2+y^2+1=k(xy)$ where $k$ is an integer.
It claims that there exists a minimum solution $(x,y)$ that has the minimum value of $x+y$.
So, they use $t$ to replace $x$ to show that $t^2-kty+y^2+1=0$
Then $t_1=x$ is one solution. By vieta formula,$t_1+t_2=ky$
Then $t_2=ky-x=\frac{x^2+y^2+1}{x}-x=\frac{y^2+1}{x}$
which implies $t_2\lt y$ then $t_1+t_2\lt x+y$.
So, the minimum condition only exists when $x=y$
I am ok as far, but after that it says, $x^2$ divided by $2x^2+1$, $x^2$divided by $1$.
So $k=3$.
But, why they can get $k=3$? $k=3$ only when $x$ and $y$ be the minimum solution. Why $k$ cannot be multiple of $3$?
|
Suppose there is a larger $k$. Then we can apply Viete's root jump and 'descend' the roots infinitely, which is impossible since there is no infinite descending chain in the naturals. This is the crux of the whole article you linked to.
|
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|
Upper bound estimate $ab^3\leq c(a^2+b^2)^2$. Find optimal, or good, constants $c$ I am trying to find a good upper bound estimate for the expression $ab^3$, where $a,b\in\mathbb R$, and it should be of the form $ab^3\leq c (a^2+b^2)^2, c\in\mathbb R.$
(The reason for the latter is, because in my application of this, $a^2+b^2$ is a physical quantity. So I would like to express the estimate in terms of it.)
By trial and error experiments with Maple, I found that $c=1/3$ yields quite a good estimate. But when I try to derive an estimate analytically, the best I got so far was $c=1/2$, since $ab\leq\frac{1}{2}(a^2+b^2)$ gives
$$b^2\leq a^2+b^2\implies ab^3=(ab)b^2\leq\frac{1}{2}(a^2+b^2)^2.$$
How can I either get the better $c=1/3$ estimate analytically, or prove it? (Or an even better one would do as well of course.)
|
By AM-GM
$$(a^2+b^2)^2=\left(a^2+3\cdot\frac{b^2}{3}\right)^2\geq\left(4\sqrt[4]{a^2\left(\frac{b^2}{3}\right)^3}\right)^2=\frac{16}{3\sqrt3}|ab^3|\geq\frac{16}{3\sqrt3}ab^3.$$
The equality occurs for $a=\frac{b}{\sqrt3},$ which says that $c=\frac{3\sqrt3}{16}.$
|
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|
Eigenvalues of two symmetric $4\times 4$ matrices: why is one negative of the other? Consider the following symmetric matrix:
$$
M_0 =
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & 4 & 3 \\
2 & 4 & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
and a very similar matrix:
$$
M_1 =
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & -4 & 3 \\
2 & -4 & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?
I also tried playing around with the values a little; for example, if the center block is $\begin{pmatrix}1 & \pm 4 \\ \pm 4 & 1\end{pmatrix}$ instead, then they do not share the same eigenvalues.
Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.
$$
M_2 =
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & e^{ix} & 3 \\
2 & e^{-ix} & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.
Thanks.
|
I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a =
\left[\begin{array}{rrrr}
0 & 1 & 2 & 0 \\
1 & 0 & a & 3 \\
2 & a & 0 & 1 \\
0 & 3 & 1 & 0
\end{array}\right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are
\begin{align*}
\chi_{M_a}(t)
&= t^{4} - \left(a^{2} + 15\right) t^{2} - 10 \, a t + 25 \\
\chi_{M_{-a}}(t)
&= t^{4} - \left(a^{2} + 15\right) t^{2} + 10 \, a t + 25
\end{align*}
Now, note that $\lambda$ is an eigenvalue of $M_a$ if and only if
\begin{align*}
0
&= \chi_{M_a}(t) \\
&= {\lambda}^{4} - {\left(a^{2} + 15\right)} {\lambda}^{2} - 10 \, a {\lambda} + 25\\
&= (-\lambda)^{4} - {\left(a^{2} + 15\right)} (-\lambda)^{2} + 10 \, a (-\lambda) + 25 \\
&= \chi_{M_{-a}}(-\lambda)
\end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=\left[\begin{array}{rrrr}
0 & 1 & 2 & 0 \\
1 & 0 & a + i \, b & 3 \\
2 & a - i \, b & 0 & 1 \\
0 & 3 & 1 & 0
\end{array}\right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
\begin{align*}
\chi_{M_{a+bi}}(t)
&= t^{4} + \left(-a^{2} - b^{2} - 15\right) t^{2} - 10 \, a t + 25 \\
\chi_{M_{-a+bi}}(t)
&= t^{4} + \left(-a^{2} - b^{2} - 15\right) t^{2} + 10 \, a t + 25
\end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
|
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|
Basic combinations logic doubt in probability
"If $3$ students are chosen at random from a class with $6$ girls and $4$ boys, what is the probability that all $3$ students chosen will be girls?"
$\left(\dfrac{6}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)$
So why can't we use that logic to answer this question?
"A bag holds $4$ red marbles, $5$ blue marbles, and $2$ green marbles. If $5$ marbles are selected one after another without replacement, what is the probability of drawing $2$ red marbles, $2$ blue marbles, and $1$ green marble?"
My answer: $\left(\dfrac{4}{11}\right)\left(\dfrac{3}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)\left(\dfrac{2}{7}\right)$
But the correct answer is $\dfrac{(_4C_2) \cdot (_5C_2) \cdot (_2C_1)}{_{11}C_5}$ (where $C$ is a combination).
Why doesn't the logic from the first problem work here?
The draws are without replacement in all cases.
|
The first question solved in the second method looks as follows:
$$\frac{{6\choose 3}}{{10\choose 3}}=\frac{6\cdot 5\cdot 4}{10\cdot 9\cdot 8}$$
Interpretation: There are ${6\choose 3}$ ways to choose $3$ girls out of $6$ and there are ${10\choose 3}$ ways to choose $3$ students out of $10$, hence the probability is the ratio of number of favorable outcomes to the total number of possible outcomes.
Now compare it with the answer of the second problem and try to interpret the selections.
|
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|
Understanding Mathematical Induction problems I have problem to understand this 3 formulas, I am new in this type of problems.
I have to solve this problems by induction.
\begin{align}
\sum_{j = 1}^{j = n} j^3 &= \left(\frac{n(n + 1)}{2}\right) ^ 2 &\text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j + 1) &= \frac{1}{3}n(n + 1)(n + 2) & \text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j!) &= (n + 1)! - 1
\end{align}
|
The general procedure for a proof by induction is to first show that the base case satisfies that proposition. Then you assume that your proposition holds for some $n$ and show that from their you can get $n+1$ to work.
I'll work the first the one and let you try the others using the same procedure.
We need to show that when $n=1$ the statement is true. clearly $1^3 = (\frac{1*2}{2})^2$.
Now we assume that the statement is true for some $n$. Then we compute:
$$\sum_{j=1}^{n+1}j^3 = \sum_{j=1}^{n}j^3 + (n+1)^3 = \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3 $$
$$ = \frac{n^2(n^2+2n+1)}{4} + (n^3+3n^2+3n+1)$$
$$ = \frac{n^4+2n^3+n^2}{4} + \frac{4(n^3+3n^2+3n+1)}{4}$$
$$ = \frac{n^4+6n^3+13n^2 + 12n+4}{4} = \left(\frac{(n+1)(n+2)}{2}\right)^2$$
Therefore the proposition holds for $n+1$ as well and so must hold for all $n$.
Conceptually think about it as a bunch of dominoes falling over one at a time. The base case guarantees that the first domino falls over, and the step of showing that you can get $n+1$ from the statement being true for $n$ is saying that any domino that falls will hit the next one.
|
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|
Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$
Find the degree of splitting field of $f=X^4+2X^2+2$ over $\mathbf{Q}$.
By Eisenstein, $f$ is irreducible. By setting $Y=X^2$, we can solve for the roots: $Y=-1\pm i \iff X=\sqrt[4]{2}e^{a\pi i/8}$, $a\in\{3,5,11,13\}$. Clearly $f$ splits in $\mathbf{Q}(\sqrt[4]{2},\zeta_{16})$. $\zeta_{16}$ is a zero of $X^8+1$, which is irreducible over $\mathbf{Q}$ by Eisenstein applied to $(X+1)^8+1$. I was not able to go further.
Could someone help me to proceed?
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The OP asked in a comment
is there some easy way to show that $−1+i$ is not a square in $\mathbb Q(\zeta_8)$?
Let $a,b,c,d \in \mathbb Q$ and $\zeta_8 = e^{\tfrac14\pi i}$.
Consider the number
$$\tag 1 \nu = a + b e^{\tfrac14\pi i}+ c e^{\tfrac12\pi i} + d e^{\tfrac34\pi i}$$
Then
$\quad \nu^2 = a^2 + c^2 e^{i \pi} + 2 b d e^{i \pi} +$
$\quad\quad\quad\quad 2 a b e^{(i/4) \pi} + 2 c d e^{(5 i/4) \pi} +$
$\quad\quad\quad\quad b^2 e^{(i/2) \pi} + 2 a c e^{(i/2) \pi} + d^2 e^{(3 i/2) \pi}+$
$\quad\quad\quad\quad 2 ad e^{(3 i/4) \pi} + 2 bc e^{(3 i/4) \pi} $
$\quad\quad\,=$
$\quad\quad\quad\quad a^2 - c^2- 2 b d +$
$\quad\quad\quad\quad (2 a b - 2 c d) e^{(1 i/4) \pi} +$
$\quad\quad\quad\quad (b^2 + 2 a c - d^2) e^{(1 i/2) \pi}+$
$\quad\quad\quad\quad (2 ad + 2bc) e^{(3 i/4) \pi} $
Assume that $\nu^2 = i -1$. Then $a b - c d = 0$ and $ad + bc = 0$. Multiplying the first equation by $c$ and the second equation by $a$ and then subtracting the first from the second and simplifying we get
$\quad (a^2 + c^2)d = 0$
Case 1: If both $a$ and $c$ are zero then $\nu^2 = -2bd + (b^2 -d^2)i$. So $b \ne 0$ and $d = \frac{1}{2b}$ and
$\quad 4b^4 -4b^2 - 1 = 0$
But that would imply that $\sqrt 2$ is a rational number.
Case 2: If $d = 0$ then $\nu^2 =
a^2 - c^2 + 2 abe^{πi/4} + (b^2 +2ac)i + 2bc e^{3πi/4}$. If we assume that $b \ne 0$ we must have both $a = 0$ and $c = 0$, which is impossible. But then $a^2 - c^2 = -1$ and $2ac = 1$. But then $a =\frac{1}{2c}$ and we wind up with,
$\quad 4c^4 -4c^2 - 1 = 0$
But that would imply that $\sqrt 2$ is a rational number.
It is interesting to note that $f(x) = 4x^4 -4x^2 - 1 $ has the same splitting field as the initial polynomial the OP was examining, and it has two real roots,
$$ x = \pm\sqrt{\frac{1+\sqrt2}2}$$
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Let $P (x )=x^4+ax^3+bx+c=0$ and have real coefficient and have all real roots . Prove that $ab \leq 0$ Let $P (x )=x^4+ax^3+bx+c=0$ and have real coefficient
and have all real roots . Prove that $ab \leq 0$
First Let the roots of this polynomial (call it P(x)) be $q,r,s,t$
By Vieta's, $a=-(q+r+s+t)$ and $b=-(qrs+rst+qrt+qst)$ So $ab=q^2(rs+st+rt)+r^2(qs+qt+st)+s^2(qr+qt+rt)+t^2(qr+qs+rs)+4qrst$
But how to find contradiction or use inequality to help ?
Thanks
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Let $p$, $q$, $r$ and $s$ be roots of our equation and
$$p+q+r+s=4u,$$
$$pq+pr+ps+qr+qs+rs=6v^2,$$ $$pqr+pqs+prs+qrs=4w^3$$ and $$pqrs=t^4,$$ where $v^2$ and $t^4$ can be negative.
Thus, $p$, $q$, $r$ and $s$ are roots of the equation:
$$x^4-4ux^4+6v^2x^2-4w^3x+t^4=0,$$ which says that by the Rolle's theorem the equation
$$(x^4-4ux^4+6v^2x^2-4w^3x+t^4)'=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ has three real roots.
Now, if $w^3=0$, so we are done because in this case $b=0$.
But for $w^3\neq0$ we see that the equation
$$w^3x^3-3v^2x^2+3ux-1=0$$ has three real roots, which says that the equation
$$(w^3x^3-3v^2x^2+3ux-1)'=0$$ or
$$w^3x^2-2v^2x+u=0$$ has two real roots, which says
$$v^4-uw^3\geq0.$$
But in our case $v^2=0$.
Id est, $$uw^3\leq0$$ or $$ab\leq0$$ and we are done!
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If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Is it necessarily the case that $m=1$ and that these two divisors are $2^k-1$ and $2^k+1$?
I've tested this up to $y\leq10^{10}$ but I haven't been able to make much progress with standard number theoretic techniques.
If $k=1$ then there are infinitely many solutions of the form $x=y-1$.
Let $(1)$ be the initial version of the problem and let $(2)$ be the supposedly equivalent version of the problem.
$(2)\implies(1)$: Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. We can write $y=m(2^k-1)+1$ for some $m\geq1$. Then
$$2^ky-1=2^k(m(2^k-1)+1)-1=(2^k-1)(2^km+1)$$
so $y-x$ and $y+x$ are two positive divisors of $(2^k-1)(2^km+1)$ which average to $y=m(2^k-1)+1$. By $(2)$, $y-x=2^k-1$ and $y+x=2^k+1$. Then $x=1$ and $y=2^k$.
$(1)\implies(2)$: Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Let $y=m(2^k-1)+1$. We can write the two divisors as $y-x$ and $y+x$ for some $0<x<y$. Thus,
\begin{align*}
y-x&\bigm|2^ky-1,\\
y+x&\bigm|2^ky-1,
\end{align*}
since $2^ky-1=(2^k-1)(2^km+1)$. Manipulating these divisibility relations shows that
\begin{align*}
y-x&\bigm|2^kx-1,\\
y+x&\bigm|2^kx+1,
\end{align*}
where $\gcd(2^kx-1,2^kx+1)=1$. Then $\gcd(y-x,y+x)=1$ so $y^2-x^2\bigm|2^ky-1$. We clearly have $2^k-1\bigm|y-1$. By $(1)$, $x=1$ and $y=2^k$. Then $m=1$ and the two positive divisors were $2^k-1$ and $2^k+1$.
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COMMENT.-We have $$2^ky-1=a(y^2-x^2)\\y-1=b(2^k-1)$$ where the given solution gives the identities $2^{2k}=2^{2k}$ and the equivalent $2^k=2^k$, not properly a system of two independent equations.
Suppose now a true (independent) system
The first equation gives a quadratic in $y$
$$ay^2+(-2^k)y+(-ax^2+1)=0$$ and the difference of the two equations gives another quadratic
$$ay^2+(2^k-1)y+(b-ax^2-b2^k)=0$$ Assuming these two quadratics have both roots $y$ equal we finish because the coefficients should be proportional and the first ones are equal ($a=a$) so the absurde with the seconds coefficients. Then there are not a true system.
Missing the case in which the two quadratics have only one common root. A known necessary condition of compatibility for this is
$$(ac'-a'c)^2=(ab'-a'b)(bc'-b'c)$$ when the quadratics are $$ax^2+bx+c=0\\a'x^2+b'x+c'=0$$
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Prove that $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+... +\frac{x_n}{n^b} $ is convergent Let $a\ge 0$ and $(x_n) _{n\ge 1}$ be a sequence of real numbers. Prove that if the sequence $\left(\frac{x_1+x_2+...+x_n}{n^a} \right)_{n\ge 1}$ is bounded, then the sequence $(y_n) _{n\ge 1}$, $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+... +\frac{x_n}{n^b} $ is convergent $\forall b>a$.
To me, $y_n$ is reminiscent of the p-Harmonic series, but I don't know if this is actually true. Anyway, I think that we may use the Stolz-Cesaro lemma on $\frac{x_1+x_2+...+x_n}{n^a}$, but I don't know if this is of any use.
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Let's do an Abel transformation. Denoting by $S_k = x_1 + ... + x_k$ for all $k \in \mathbb{N}$, you have
$$\sum_{n=1}^N \frac{x_n}{n^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}-S_n}{(n+1)^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}}{(n+1)^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} $$
$$= x_1 +\sum_{n=2}^{N} \frac{S_n}{n^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} =x_1 + \frac{S_N}{N^b} - \frac{x_1}{2^b} + \sum_{n=2}^{N-1} S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) $$
Now, by hypothesis, $S_N = O(N^a)$, so because $b > a$, you have $S_N = o(N^b)$. So you have only to study the convergence of the series
$$\sum_{n=2}^{N-1} S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) $$
But one has $S_n = O(n^a)$, so
$$S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) = O \left( \frac{n^a}{n^b}-\frac{n^a}{(n+1)^b}\right)$$
Moreover, $$\frac{n^a}{n^b}-\frac{n^a}{(n+1)^b} = \frac{n^a(n+1)^b - n^an^b}{n^b(n+1)^b} \sim \frac{n^a}{(n+1)^b} \left( \left(1+ \frac{1}{n} \right)^b-1\right)$$
$$ \sim \frac{n^a}{(n+1)^b} \left( \frac{b}{n} \right) \sim \frac{b}{n^{b-a+1}}$$
This last series converges, so you deduce that the series of general term
$$ S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right)$$
also converges, so you deduce that your original series converges.
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If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align*}$$
By using this, we get value of $(3\csc3\theta - 4\sec3\theta) =9.95$ by using calculator.
I want know if there's any way to solve this problem without calculator.
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Hint: We get $$\theta=\frac{1}{9}\arctan(\frac{3}{4})$$ so we get
$$3\csc\left(\frac{1}{3}\arctan(\frac{3}{4})\right)-4\sec\left(\frac{1}{3}\arctan(\frac{3}{4})\right)=10$$
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Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$
Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$
So I started by combining the two fractions, which gave me:
$$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{2\sin\theta\cos\theta}{1-\cos^2\theta} = \frac{\sin2\theta}{1-\cos^2\theta}$$
I wasn't sure where to go from here considering I'm aiming for $2\cos\theta / \sin\theta$
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$$\frac{2\sin\theta\cos\theta}{1-\cos^2\theta} =\frac{2\sin\theta\cos\theta}{\sin^2\theta} = \frac{2\cos\theta}{\sin \theta} = 2\cot \theta$$
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Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$ Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of
$$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$
To find the minimum of $P$ I rewrite it as
$$P=2\left( ab+bc+ca+1\right)\left(2-abc\right)$$
Then I show that $P \geq 4$, equivalently show that
$$2\left( ab+bc+ca\right)-abc\left( ab+bc+ca\right) \geq abc$$
Indeed, we have
$$3abc\left( a+b+c\right) \leq \left( ab+bc+ca\right)^2 \Leftrightarrow 6abc \leq \left( ab+bc+ca\right)^2 $$
Thus we need to show
$$ \left( ab+bc+ca\right)^2 \leq 12\left( ab+bc+ca\right) -6abc\left( ab+bc+ca\right)$$
This is equivalent to
$$\left( ab+bc+ca\right)\left( ab+bc+ca+6abc-12\right) \leq 0$$
This implies from the inequalities that $ab+bc+ca \leq \left( a+b+c\right)^2/3=4/3$ and $abc \leq \left( a+b+c\right)^3/27=8/27$. The equality holds if $a=b=0$ and $c=1$
Is this right? And for the maximum value, I have no idea. I only guess it is $8$ attending at $a=0,b=c=1$. Please help me. Thank you.
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I would use that $$6=\frac{3}{2}(a+b+c)^2$$ and $$2=\frac{1}{4}(a+b+c)^3$$
Using this substitution we get the term
$$1/8\, \left( {a}^{2}+6\,ab+6\,ac+{b}^{2}+6\,bc+{c}^{2} \right)
\left( {a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}+2\,abc+3\,a{c}^{2}
+{b}^{3}+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3} \right)
$$
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Find the coefficient of $x^{32}$ in $(x^3 +x^4 +x^5 +x^6 +x^7)^7$ I don't understand the explanation in the book and why the final answer looks the way it does. I know I am supposed to factor and it will equal to $(x^3(1+x+\cdots+x^4))^7$. But after that, I am really confused about what happens. Can someone explain this problem step by step. Thank you!
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A strategy for solving problems like this one is to try to rewrite the expression as the quotient of a polynomial in $x$ and some power of $1-x$. You can then use either Newton’s generalized binomial theorem or the identity $${1\over(1-s)^{k+1}} = \sum_{n=0}^\infty \binom{n+k}k s^n$$ to expand the denominator as a power series and write the coefficient that you seek as a linear combination of a small number of binomial coefficients.
Before plunging into the calculations, a handy bit of notation: $[x^n]f(x)$ stands for “the coefficient of $x^n$ in the power series for $f(x)$.” The brackets “absorb” powers of $x$, that is, $[x^n]x^mf(x) = [x^{n-m}]f(x)$. Basically, pulling out a factor of $x^m$ lets you reduce the exponent in the bracket—the coefficients are shifted down $m$ places in what remains. Diving right in, then,
$$\begin{align} [x^{32}](x^3+x^4+x^5+x^6+x^7)^7 &= [x^{32}]\left(x^3(1+x+x^2+x^3+x^4)\right)^7 \\
&= [x^{32}]x^{21}(1+x+x^2+x^3+x^4)^7 \\
&= [x^{11}](1+x+x^2+x^3+x^4)^7 \\
&= [x^{11}]\left({1-x^5 \over 1-x}\right)^7 \\
&= [x^{11}]{\binom70-\binom71x^5+\binom72x^{10}-\dots \over (1-x)^7} \\
&= \binom70[x^{11}]{1\over(1-x)^7} - \binom71[x^6]{1\over(1-x)^7} + \binom72[x]{1\over(1-x)^7} \\
&= \binom70\binom{17}6 - \binom71\binom{12}6 + \binom72\binom{7}6 \\
&= 6055.\end{align}$$
The absorption rule is used in lines 3 and 6. When expanding the numerator using the binomial theorem in line five, we can ignore all terms that involve powers of $x$ greater than $11$ since they won’t contribute anything to the final result.
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Simplifying $\sin\frac{11\pi}{12}\sin\frac{29\pi}{12}-\cos\frac{13\pi}{12}\cos\frac{41\pi}{12}$. Why do I get the wrong answer? Can someone explain why I get wrong answer in simplifying this expression?
$$\sin\frac{11\pi}{12}\sin\frac{29\pi}{12}-\cos\frac{13\pi}{12}\cos\frac{41\pi}{12}$$
If we rewrite the expression with new angles,
$$\frac{29\pi}{12}\to\frac{5\pi}{12} \quad\text{and}\quad \frac{13\pi}{12}\to\frac{11\pi}{12}$$
we don't change the value of the expression. But, if we now use sine of sum of angles, we get $\frac{1}{2}$ instead of $0$.
Why does this happen? Do angles need to be in the same quadrant for the formula to work?
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No, you don't get $\frac12$. You get $0$ too, because\begin{align}\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)-\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{41\pi}{12}\right)&=\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)-\cos\left(\frac{11\pi}{12}\right)\cos\left(3\pi+\frac{5\pi}{12}\right)\\&=\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)+\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{5\pi}{12}\right)\\&=\cos\left(\frac{11\pi}{12}-\frac{5\pi}{12}\right)\\&=\cos\left(\frac\pi2\right)\\&=0.\end{align}
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Solve $a^3+b^3+3ab=1$ with $(a,b)\in \Bbb{Z}^2$
Solve the following equation for $(a,b)\in \Bbb{Z}^2$:
$$a^3+b^3+3ab=1$$
I tried all of the standard techniques I know. I tried modular arithmetic:
$$a^3+b^3+3ab\equiv 1 \pmod{3} $$
$$a^3+b^3\equiv 1 \pmod{3} $$
Now by Fermat's Little Theorem:
$$a^2 a+b^2 b\equiv 1 \pmod{3} $$
$$a+b\equiv 1 \pmod{3} $$
But I can't see the next move I have to do. I can't find any banal factorization of the first term(it would require solving a $3$ degree equation). I tried using classic scomposition such as the sum of $2$ cubes and the cube of a binomial. Thank you for your time :)
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The item worth memorizing is
$$ x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( x^2 + y^2 + z^2 - yz - zx - xy \right) $$
where the quadratic form is positive semidefinite because
$$ \left( x^2 + y^2 + z^2 - yz - zx - xy \right) = \frac{1}{2} \left( (y-z)^2 + (z-x)^2 + (x-y)^2 \right) $$
If you then take $z=-1$ you get
$$ x^3 + y^3 -1 + 3xy = (x+y-1)\left( x^2 + y^2 +1 + y + x - xy \right) $$
and the quadratic factor is
$$ \frac{1}{2} \left( (y+1)^2 + (-x-1)^2 + (x-y)^2 \right)= \frac{1}{2} \left( (y+1)^2 + (x+1)^2 + (x-y)^2 \right) $$
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If $f(x)f(y)+f(xy)\le -\frac{1}{4},\forall x,y\in[0,1)$, show that $f(x)=-\frac{1}{2}$ Let $f:[0,1) \to \mathbb{R}$ be a function such that
$$f(x)f(y)+f(xy)\le -\dfrac{1}{4} \quad \forall\, x,y\in[0,1).$$
Show that
$$f(x)=-\dfrac{1}{2} \quad \forall\, x \in[0,1).$$
I have proved that $f(0)=-\dfrac{1}{2}$: if $x=y=0$, we have
$$f^2(0)+f(0)\le-\dfrac{1}{4}\Longrightarrow \left( f(0)+\frac{1}{2} \right)^2\le 0\Longrightarrow f(0)=-\dfrac{1}{2}.$$
But I can't prove $f(x)$ be constant. Thanks.
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Plugging in $y=0$ gives $f(x) \ge -\frac{1}{2}$ for each $x$.
Let $y=x$ to get $f(x)^2+f(x^2) \le -\frac{1}{4}$. This implies $f(x^2) \le -\frac{1}{4}$ for each $x$, and so $f(x) \le -\frac{1}{4}$ for each $x$. But then $f(x)^2+f(x^2) \le -\frac{1}{4}$ implies $f(x^2) \le -\frac{1}{4}-(\frac{1}{4})^2 = -\frac{5}{16}$, and so $f(x) \le -\frac{5}{16}$ for each $x$. Doing this again gives $f(x) \le -\frac{1}{4}-(\frac{5}{16})^2 = -\frac{89}{256}$ for each $x$. If we keep doing this, we see that, for any $\epsilon > 0$, $f(x) \le -(\frac{1}{2}-\epsilon)$ for each $x$. Therefore, $f(x) \le -\frac{1}{2}$ for each $x$.
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|
Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force. I then found the next pair by using the equation:
$(x + \sqrt{7}y)^i(x-\sqrt{7}y)^i = 1$
$i = 1$ would give me the minimal solution $(8,3)$, so I went to $i = 2$.
$(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2 = 1$
$\left(x^2 + 7y^2 + 2xy\sqrt{7}\right) \left(x^2 + 7y^2 - 2xy\sqrt{7}\right) = 1$
$(x^2 + 7y^2)^2 - 7(2xy)^2 = 1$ (This is of the form $X^2 - 7Y^2 = 1$)
I then let $X = x^2 + 7y^2$ and $Y = 2xy$, so the equation becomes the desired
$X^2 - 7Y^2 = 1$
I then plugged in my minimal solution $(8,3)$ and found $X = (8^3+7(3^3)) = 127$ and $Y = 2(8)(3) = 48$, so my new pair is $(127,48)$. To find the next solution, I let $i = 3$.
$(x + \sqrt{7}y)^3(x-\sqrt{7}y)^3 = 1$
And this is where I got stuck. I tried a few methods to get this new equation into the form of $X^2 - 7Y^2 = 1$, but have been unsuccessful. I tried expanding out the equations completely getting:
$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1$
But I'm fairly sure that's not the right direction. The other method I tried was factoring out a $(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$, so I got:
$[(x + \sqrt{7}y)(x-\sqrt{7}y)](x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$
which simplifying I got
$(x^2-7y^2) \left( (x^2 + 7y^2)^2 - 7(2xy)^2 \right)$
I'm really not sure how to properly factor this cubic to get to the desired function form. Any help would be greatly appreciated!
|
Beginning with pairs $(1,0)$ and then $(8,3),$ all the other soltuions with positive variables satisfy
$$ x_{n+2} = 16 x_{n+1} - x_n \; , \; $$
$$ y_{n+2} = 16 y_{n+1} - y_n \; . \; $$
The $x_n$ begin
$$ 1, 8, 127, 2024, 32257, 514088, 8193151, $$
The $y_n$ begin
$$ 0, 3, 48, 765, 12192, 194307, 3096720, $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the area of the surface for the curve rotated about the x-axis
$y = \tan(x), 0\leq x \leq \frac{\pi}{3}$
I am struggling with constructing an integral for this formula.
Since the curve is rotated about the x-axis, I think this is the best way to setup the integral. Since the circumference is $2\pi r$, it makes sense to me that the radius is $y=f(x)$.
For the length of the curve,
$$ds = \sqrt{1 + \bigg(\frac{dy}{dx}\bigg)^2}dx = \sqrt{1 + \bigg(\frac{dx}{dy}\bigg)^2}dy $$
Since I can find $\frac{dy}{dx}$ relatively easily, $\frac{dy}{dx} = \sec^2x$, it would seem easiest for me to set up the integral as follows:
$$ S = \int 2\pi \cdot y \cdot ds = \int 2\pi \cdot y \cdot \sqrt{1+\frac{dy}{dx}^2}dx$$
Please let me know if my thinking is correct up to this point.
$$\begin{align}
S &= \int 2\pi y ds \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \tan x \sqrt{1+(\sec^2x)^2}dx \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \tan x \sqrt{1+\sec^4x}dx \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \tan x \sqrt{1+ \frac{1}{\cos^4x}}dx \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \frac{\sin x}{\cos x} \frac{\sqrt{1+cos^4x}}{\cos^2x}dx \\
&= 2\pi\int_{\frac{1}{2}}^{1} \frac{\sqrt{1+u^4}}{u^3}du
\end{align}$$
Substitution - $u = \cos x, -du = \sin x $
Edit: Here is where I am getting stuck
Edit: Edit: I am giving up on this problem. I just did a parts by substitution and then looks like I have to do another substitution + trigonometric substitution to solve that integral......My variables are a complete mess at this point.
|
Hint: Do a change of variable $\cos x=u$, then $du=-\sin x$, so $$S=2\pi\int_{1/2}^1\frac{\sqrt{1+x^2}}{x^2}du$$
Note: Since the derivative will need to be squared, $$S=2\pi\int_{1/2}^1\frac{\sqrt{1+u^4}}{u^3}du$$
Here are some hints how to solve the rest: Make substitution $v=u^2$, $dv=2udu$ and you get an integral$$\frac 12\int\frac{\sqrt{1+v^2}}{v^2}dv$$
Integrate by parts $f=\sqrt{1+v^2}$, $g'=\frac{1}{v^2}$, with $f'=\frac{v}{\sqrt{1+v^2}}$ and $g=-\frac 1v$. Then all you are left with is to calculate $$\int\frac 1{\sqrt{1+v^2}}dv$$
|
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|
Find $P(n+1)$ for a polynomial P
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.
My attempt:
I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce:
$$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$
I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
|
Your solution is fine.$$Q(-1) = \lambda (-1)(-2)\ldots (-(n+1))=\lambda (-1)^{n+1}(n+1)!=1$$
$$\lambda = \frac{(-1)^{n+1}}{(n+1)!}$$
$$Q(x)=\frac{(-1)^{n+1}x(x-1)\ldots (x-n)}{(n+1)!}=(x+1)P(x)-x$$
Let $x=n+1$
$$\frac{(-1)^{n+1}(n+1)n\ldots 1}{(n+1)!}=(n+2)P(n+1)-(n+1)$$
Hence $$P(n+1) = \frac{n+1 + (-1)^{n+1}}{n+2}$$
Let's check with small cases to see if $(-1)^{n+1}$ should be there.
Let $n=1$, then we have $P(0)=0$ and $P(1)=\frac12$. Then $P(x) = \frac{x}2$.
We have $P(2)= 1$.
Notice that $\frac{n+1}{n+2}<1$ and hence $P(n) = \frac{n+1}{n+2}$ is certainly wrong.
|
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"url": "https://math.stackexchange.com/questions/3203455",
"timestamp": "2023-03-29T00:00:00",
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|
Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$ Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$
I entered the equation in desmos.com and no roots were lying below or x=0 lines , hence all roots are imaginary.
Using Descartes rule for f(x) 4 sign change occurs , hence 4,2 or zero positive roots.
Using Descartes rule for f(-x) 0 sign change occurs , hence zero negative roots.
For here on how do I proceed
|
To expand on Dr. Sonnhard Graubner's answer, let $f(x) = 54x^4-36x^2+15x^2-6x+1$. Then:
$$54x^4-36x^3+18x^2-6x+1 > f(x)$$
$$\Rightarrow (3x-1)^2(6x^2+1)$$
by the rational root test.
The smaller function has only one root: $x = \frac{1}{3}$, and there $f''(x) = 648x^2-216x+30$ is positive. Therefore, for all real $x$, $f(x) > 0$ and therefore $54x^4-36x^3+18x^2-6x+1 > 0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Number of lattice points inside a circle
The number of lattice points inside the circle $x^2+y^2=a^2$ can not be
Options $(a)\; 202\;\;\; (b)\; 203\;\;\; (c)\; 204\;\;\; (d)\; 205$
Try: i have an idea of number of integer points on the circle $x^2+y^2=a^2$
Let $x,y\in\{4n,4n+1,4n+2,4n+3\}$
But no idea how to find number of integer points inside the circle.
Could some help me to solve it , Thanks
|
According to Gauss's circle problem, all choices cannot be ($r$ is radius, $N(r)$ is the number of lattice points):
$$\begin{array}{c|c|c}
r&0&1&2&3&4&5&6&7&8&9&10&11&12\\
\hline
N(r)&1&5&13&29&49&81&113&149&197&253&317&377&441
\end{array}$$
See the graph to verify the numbers $N(8)=197$ and $N(9)=253$:
$\hspace{5cm}$
$$\begin{align}N(8)&=1+(0+1+3+4+7+7+8+11+8)\cdot 4=197\\
N(9)&=1+(0+1+3+4+7+7+8+11+13+9)\cdot 4=253.\end{align}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$.
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \},$$ but I'm stuck proving that it's closed under inverse. If that is true, then the minimal polynomial should be of 3rd degree. True or not, I can't find any such polynomial.
Solution: As @José Carlos Santos mentioned below:
$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha,$$
from which it follows
$$(x - \sqrt[3]3-\sqrt[3]9)(x^2 + x(\sqrt[3]3+\sqrt[3]9) + (\sqrt[3]3+\sqrt[3]9)^2) - 9(x - \sqrt[3]3 - \sqrt[3]9)= x^3 - 12 - 9(\sqrt[3]3+\sqrt[3]9) -9x + 9(\sqrt[3]3+\sqrt[3]9) = x^3 - 9x - 12 = p(x);~~ p(\alpha) = 0.$$
|
Note that$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha.$$
|
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|
Integrating $\int_0^{\infty} \frac{\log x}{(x + a)^2 + b^2} \operatorname d\!x$ I'm trying to show that $\int_0^{\infty} \frac{\log x}{(x + a)^2 + b^2} \operatorname d\!x = \frac{1}{b}\arctan \sqrt{a^2 + b^2}.$ However, I am a bit confused applying the key hole "method." I consider $f(z) = \frac{\log^2(z)}{(z+a)^2+b^2}$ and can see that there are poles at $z_{1,2} = -a + bi, -a-bi.$ Then, I'm not sure what to do from there. I know that the integral on the outer circle vanishes as $R\to \infty$ and the integral of the inner circle also goes to zero as $\epsilon \to 0.$ I'm not sure what my integrands are for the contours $C_1$ going from $R$ to $\epsilon$ and $C_2$ going from $\epsilon$ to $R.$ I also don't know how to compute their residues.
|
without complex analysis$$I = \int^{\infty}_{0}\frac{\ln x}{(x+a)^2+b^2}dx$$
set $\displaystyle x = \frac{a^2+b^2}{y}$ and $\displaystyle dx=-\frac{a^2+b^2}{y^2}dy$
$$I = \int^{\infty}_{0}\frac{\ln(a^2+b^2)-\ln(y)}{(y+a)^2+b^2}dy$$
$$I = \ln(a^2+b^2)\int^{\infty}_{0}\frac{1}{(y+a)^2+b^2}dy-I$$
$$2I=\frac{\ln(a^2+b^2)}{b}\arctan\bigg(\frac{y+a}{b}\bigg)\bigg|^{\infty}_{0}$$
$$I=\frac{\ln(a^2+b^2)}{2b}\bigg[\frac{\pi}{2}-\arctan\bigg(\frac{a}{b}\bigg)\bigg]$$
$$I=\frac{\ln(a^2+b^2)}{2b}\cot^{-1}\bigg(\frac{a}{b}\bigg)=\frac{1}{b}\arctan\bigg(\frac{b}{a}\bigg)\ln\sqrt{a^2+b^2}$$
For $a,b>0$
|
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|
How to write $[(2+\sqrt{3})^n + (2-\sqrt{3})^n + (2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1}]/6$ to the form $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$). We know that $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ is an integer (See here).
However, we want to write the formula
\begin{align}
&\frac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \frac{3-\sqrt{3}}{6} (2-\sqrt{3})^n\\
&=\frac{1}{6} \left[(2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1} + (2+\sqrt{3})^n + (2-\sqrt{3})^n\right]
\end{align}
to the form
$$a^2 + 2\,b^2,\ (a, b \in \mathbb{N}).$$
How?
|
Call your formula $f(n)$. Since $2\pm\sqrt{3}$ are the roots of $x^2-4x+1$, it's easy to show that $f(n)$ is determined by $f(0)=1$, $f(1)=3$ and the recursive relation $f(n+1)=4f(n)-f(n-1)$.
After looking at @J.W.Tanner comment, I found out that we actually have
\begin{align*}
f(2n)&=f(n)^2+2\left(\sum_{0\leq k\leq n-1}f(k)\right)^2\\
f(2n+1)&=f(n)^2+2\left(\sum_{0\leq k\leq n}f(k)\right)^2
\end{align*}
Now, I don't really know how difficult is to prove this by induction, but it is an interesting pattern.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Power series differential check $(x-1)y'' + y' = 0$ I'm a bit stuck on how to solve this:
$$(x-1)y'' + y' = 0 $$
so assuming y is a solution in this form:
$$\sum_{n=0}^\infty C_nx^n$$
$$\sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
subbing in and distributing:
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty n(n-1)C_nx^{n-2} + \sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=1}^\infty (n+1)(n)C_{n+1}x^{n-1} \sum_{n=1}^\infty nC_nx^{n-1}$$
start at n=2
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty (n+1)(n)C_{n+1}x^{n-1} + \sum_{n=2}^\infty nC_nx^{n-1} - 2C_2 + C_1$$
so $-2C_2 + C_1 = 0$
and $$(n(n-1)C_n - (n+1)(n)C_{n + 1} + nC_n = 0$$
and $$(n^2C_n - (n+1)(n)C_{n + 1} = 0$$
and $c_{n+1} = \frac{nC_n}{n+1}$ for $n \ge 2$
so writing out the first few terms:
$c_0 = c_0$ and $c_1 = c_1$ and $c_2 = \frac{c_1}{2}$ and $c_3 = \frac{2c_2}{3}$ and $c_4 = \frac{3c_3}{4} = \frac{3 \cdot 2 \cdot c_1}{4!}$ and so $$c_n = \frac{c_1}{n}$$
so can I write that $$y = c_0 + c_1 + \sum_{n=2}^\infty \frac{x^n}{n}$$
|
Calling $Y = y' = \sum_{k=0}^{\infty}a_k x^k$ we have in $(x-1)Y'+Y = 0$
$$
(x-1)\sum_{k=1}^{\infty}k a_k x^{k-1}+\sum_{k=0}^{\infty}a_k x^k = a_0-a_1+\sum_{k=1}^{\infty}(k+1)a_k x^k-\sum_{k=1}^{\infty}(k+1)a_{k+1} x^k = 0
$$
hence $a_k - a_{k+1} = 0$ or $Y = a_0\sum_{k=0}^{\infty}x^k$ but $Y = y'$ so finally
$$
y = \int_0^x Y dx = a_0\sum_{k=1}^{\infty}\frac{x^k}{k}+ C_0
$$
NOTE
$$
\sum_{k=1}^{\infty}k a_k x^{k-1} = a_1+\sum_{k=1}^{\infty}(k+1) a_{k+1} x^k
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using De Moivre's law to compute $(-\sqrt3+i)^{2/3}$ Question:
If $z=-\sqrt{3}+i$, then $z^{2 / 3} = ?$
My work (which is wrong but I am not sure why):
We can write $z = r(\cos\theta + i\sin\theta)$
$$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$$
$$\theta = \tan^{-1} \left(\frac{-1}{\sqrt{3}}\right) = \frac{5\pi}{6} \quad (*)$$
Therefore
$$z = 2\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) \right) $$
so (applying De Moivre's)
$$ z^{\frac{2}{3}} = 2^{\frac{2}{3}} \left(\cos\left(\frac{5\pi}{9}\right) + i\sin\left(\frac{5\pi}{9}\right) \right)$$
However, the answer is apparently:
$$-2^{2 / 3} \sin \left(\frac{\pi}{18}\right)+i 2^{2 / 3} \cos \left(\frac{\pi}{18}\right)$$
*
*How does one arrive at the given answer?
*Forgive my remembrance of high school trigonometry. In $(*)$ I understand that we can take the negative out of $\tan^{-1}$ resulting in
$$\theta=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6} = \frac{11\pi}{6}$$
which results in a different $\theta$ than I had. My intuition for why $\theta =\frac{5 \pi}{6}$ is because $\frac{-1}{2} \div \frac{\sqrt{3}}{2}$ as $\sin \frac{5 \pi}{6} = -\frac{1}{2}$ and $\cos \frac{5 \pi}{6} = \frac{\sqrt{3}}{2}$. But I am not sure if this is the correct reasoning
|
You forgot $i$ when writing $z = r(\cos \theta + i \sin \theta)$. Arctangent was computed correctly, so
$$
z = 2\left(\cos(\frac{5}{6}\pi) + i \sin(\frac{5}{6}\pi)\right),
$$
$$
z^{2/3} = 2^{2/3}\left(\cos(\frac{5}{9}\pi) + i \sin(\frac{5}{9}\pi)\right).
$$
Now some trigonometry:
$$
\cos(\frac{5}{9}\pi) = \cos(\frac{1}{2}\pi + \frac{1}{18}\pi) = -\sin(\frac{1}{18}\pi).
$$
Similarly,
$$
\sin(\frac{5}{9}\pi) = \cos(\frac{1}{18}\pi),
$$
and you arrive to the answer.
|
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|
limit $\sqrt{-n^4+4n^2+4}-in^2 = -2i$? What is the limit of the following complex sequence? I get a different result than Wolfram Alpha. My approach was:
$$\sqrt{-n^4+4n^2+4}-in^2 = \sqrt{i^2n^4+4n^2+4}-in^2 = in^2\sqrt{1-\frac{1}{4n^2}-\frac{4}{n^4}}-in^2 \xrightarrow{n\xrightarrow{}\infty} 0$$
Wolfram Alpha says the limit is $-2i$. I'm confused. I must've made a stupid mistake, but cannot find it. Thanks in advance!
|
Let's forget about those pesky $i$s and consider the limit of
$$n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}-n^2$$
as $n\to\infty$. This equals
$$\frac{n^4\left(1-\frac4{n^2}-\frac4{n^4}\right)-n^4}{n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}+n^2}
=\frac{-4n^2-4}{n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}+n^2}
=\frac{-4-4/n^2}{\sqrt{1-\frac4{n^2}-\frac4{n^4}}+1}
$$
which tends to $-2$ as $n\to\infty$.
|
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|
using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}
but i can't try more
|
$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$
$\iff c^4+s^2=c^2+s^4=P$(say) where $c=\cos x,s=\sin x$
$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-\dfrac{\sin^22x}2=2-\dfrac{1-\cos4x}4=?$
using $\sin2x=2\sin x\cos x,\cos2y=2-2\sin^2y$
We need $$\dfrac1P+\dfrac2P=?$$
|
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|
Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
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For positive $n$, let $a=\sqrt[3]{n^3+3n^2}$, and let $b=\sqrt{n^2+2n}$.
Note that $a < n+1$ and
\begin{align*}
3n +1 &= (n+1)^3-a^3\\[4pt]
&= \bigl((n+1)-a\bigr)\bigl((n+1)^2+a(n+1)+a^2)\\[4pt]
&> \bigl((n+1)-a\bigr)(n+1)^2\\[4pt]
\end{align*}
hence $0 < n+1-a < {\large{\frac{3n+1}{(n+1)^2}}}$, which implies $\displaystyle{\lim_{n\to \infty}}\;(n+1)-a = 0$.
Also, $b < n+1$ and
\begin{align*}
1&=(n+1)^2-b^2\\[4pt]
&= \bigl((n+1)-b\bigr)\bigl((n+1)+b\bigr)\\[4pt]
&> \bigl((n+1)-b\bigr)(n+1)\\[4pt]
\end{align*}
hence $0 < (n+1)-b < {\large{\frac{1}{n+1}}}$, which implies $\displaystyle{\lim_{n\to \infty}}\;(n+1)-b = 0$.
Then we get
$$\lim_{n\to \infty}\;a-b = \lim_{n\to \infty}\;\bigl(a - (n+1)\bigr) + \bigl((n+1)-b\bigr) = 0 + 0 = 0$$
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Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$. If $f(x), g(x)$ are defined as following on $[0 , \infty)$,
$$\tag 1 f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$$
$$\tag 2 g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}.$$
. Then how to show that $f,g$ are bounded function on $[0 , \infty)$?
I found this problem on this
Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$?.
In solution's last steps, i don't know why this is correct answer.
Could you explain for me to elaborate?
Thank you.
|
A simple proof:
Clearly:
$$
n^3+x^3\ge \frac{1}{4}(n+x)^3 \tag{1}
$$
and hence
$$
0\le \frac{nx}{n^3+x^3} \le \frac{4nx}{(n+x)^3}\le\frac{4(n+x)x}{(n+x)^3}=\frac{4x}{(n+x)^2}.
$$
Hence
$$
\sum_{n=1}^\infty\frac{nx}{n^3+x^3}\le
\sum_{n=1}^\infty\frac{4x}{(n+x)^2}=\frac{4x}{(1+x)^2}+\sum_{n=2}^\infty\frac{4x}{(n+x)^2}\le \frac{4x}{(1+x)^2}+\int_1^\infty \frac{4x\,ds}{(s+x)^2} \tag{2}\\=
\frac{4x}{(1+x)^2}+\frac{4x}{(1+x)}.
$$
Meanwhile
$$
\frac{x^4n}{(x^3+n^3)^2}=\frac{nx}{n^3+x^3}\cdot\frac{x^3}{x^3+n^3}\le
\frac{nx}{n^3+x^3}.
$$
Notes. Inequality $(1)$ holds since
$$
4(n^3+x^3)=4(n+x)(n^2-nx+x^2)=(n+x)(4n^2-4nx+4x^2)
=(n+x)(n^2+2nx+x^2+3n^2-6nx+3x^2)=(n+x)^3+3(n+x)(n-x)^2\ge(n+x)^3.
$$
The last inequality in $(2)$ holds since if $f: [1,\infty)\to (0,\infty)$ is decreasing, then $\sum_{n=2}^\infty f(n)\le \int_1^\infty f(x)\,dx.$
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|
Proving $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$ The inequality:
$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$$
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
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Since squaring is monotonic, this is equivalent to
$$
a^2 + b^2 + c^2 + d^2 + 2\sqrt{a^2+c^2}\sqrt{b^2+d^2} \ge a^2+b^2+c^2+d^2 + 2(ab + cd)
$$
which is in turn equivalent to
$$
ab + cd \le \sqrt{a^2+c^2}\sqrt{b^2+d^2},
$$
the Cauchy-Schwarz inequality.
Putting it in vector form is even more elegant. For $\mathbf{a} = (a,c)$ and $\mathbf{b} = (b,d)$, this is
$$
|\mathbf{a}| + |\mathbf{b}| \ge |\mathbf{a}+\mathbf{b}|,
$$
the triangle inequality.
|
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|
In diophantine $3b^2=a^2$ where $a$ and $b$ are coprime, does $3|a$? Integers $a$ and $b$ are co-prime and $3\cdot b^2=a^2$.
$3\cdot b^2=a^2$, implies $a^2$ is divisible by 3 since, $3b^2$ is divisible by 3.
Is $a$ divisible by 3?
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Yep; in general, if $p|ab$ then $p|a$ or $p|b$ ($p$ is a prime). In your case, $3|a^2$ so $3|a$ or $3|a$; thus $a$ is divisible by $3$
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|
Volume of cylinder cut by two planes I try to find volume of cylinder $x^2 + y^2=9$ cut by inclined plane $z = x + y - 4$ and plane $z=1$
I should solve it in Cartesian, so my idea is to rotate the plane $z=x + y - 4$, and to split the cylinder in two symmetrical portions, which allows using only the portion over $x$ axis. Resulting in two new functions: $j: z=x-2\sqrt{2}$ and $i: y = \sqrt{9-x^2}$.
My guess is that I can calculate the volume of the cylinder minus the little portion between $z = 0$ and $j$
$V = 2\pi(\text {radius})(\text {height}) - \text {little Area}.$
Can you help me solve this?
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$V = \int^3_{x =-3}\int^{\sqrt{9-x^2}}_{y = -\sqrt{9-x^2}}\int^{1}_{z = x+y-4}dzdydx=\int^3_{x =-3}\int^{\sqrt{9-x^2}}_{y = -\sqrt{9-x^2}} \bigg( 1 - x - y +4\bigg)dydx $
$V = \int^3_{x =-3}\bigg(5y-yx-\frac{y^2}{2}\bigg)^{\sqrt{9-x^2}}_{-\sqrt{9-x^2}}dx =\int^3_{x =-3} \big[(5-x)(2\sqrt{9-x^2}) - \frac{1}{2}(9-x^2 - 9 +x^2)\big]dx $
$V = \int^3_{-3}10\sqrt{9-x^2}dx -\int^3_{-3}2x\sqrt{9-x^2}dx $
$V = 2\cdot 10\bigg[\frac{x}{2}\sqrt{x^2-9}+\frac{9}{2}\sin^{-1}(\frac{x}{3})\bigg]^3_0 + 0 $
(As, $2x\sqrt{9-x^2}$ is an odd function of $x$)
$V = 20[0+\frac{9}{2}\frac{\pi}{2}] +0$
$$V = 45\pi $$
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|
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
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Note: $1-x^2\ge 0 \Rightarrow -1\le x\le 1$.
Square both sides and denote $1-x^2=t^2, -1\le t\le 1$:
$$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} \Rightarrow \\
(1+\sqrt{1-x^2})(2+6x^2+2\sqrt{(1-x^2)^3}=5-x^2+4\sqrt{1-x^2} \Rightarrow \\
(1+t)(2+6(1-t^2)+2t^3)=4+t^2+4t \Rightarrow \\
2t^4-4t^3-7t^2+4t+4=0 \Rightarrow \\
t\approx 0.93, x\approx \pm 0.38.$$
Note: Other roots are rejected as they are outside of domain.
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|
Sec. 6 in G.F. Simmons' INTRO TO TOPOLOGY & MODERN ANALYSIS: A formula for the general terms of these two sequences please? The set $\mathbb{Q}^+$ of all the positive rational numbers is given by
$$ \mathbf{Q}^+ = \left\{ \ \frac{p}{q} \ \colon \ p \in \mathbf{N}, q \in \mathbf{N} \ \right\}.$$
This set is countably infinite; so we can list its elements as a sequence (or an infinite sequence). One such sequence is as follows:
$$ 1, \ \frac{1}{2}, \ \frac{2}{1} = 2, \ \frac{1}{3}, \ \frac{3}{1} = 3, \ \frac{1}{4}, \ \frac{2}{3}, \ \frac{3}{2}, \ \frac{4}{1} = 4, \ \frac{1}{5}, \ \frac{5}{1} = 5, \ \frac{1}{6}, \ \frac{2}{5}, \ \frac{3}{4}, \ \frac{4}{3}, \ \frac{5}{2}, \ \frac{6}{1} = 6, \ \frac{1}{7}, \ \frac{3}{5}, \ \frac{5}{3}, \ \frac{7}{1} = 7, \ \frac{1}{8}, \ \frac{2}{7}, \ \frac{4}{5}, \ \frac{5}{4}, \ \frac{7}{2}, \ \frac{8}{1} = 8, \ \ldots.$$
Here we have ordered the positive rationals by the size of the sum of the numerator and denominator, making sure to avoid repetition. Is this sequence correct as far as I've given it?
Now my question is, can we find a formula for the general term of this sequence? If so, then how to? I'd appreciate if the entire process is explained in detail.
Similarly, the set $\mathbf{Q}$ of all rational numbers is given by
$$ \mathbf{Q} = \left\{ \ \frac{p}{q} \ \colon \ p \in \mathbf{Z}, q \in \mathbf{Z}, q \neq 0 \ \right\}.$$
And, since $\mathbf{Q}$ is also countably infinite, we can list its elements as an infinite sequence. One such sequence is as follows:
$$ 0, 1, -1, \frac{1}{2}, -\frac{1}{2}, 2, -2, \frac{1}{3}, -\frac{1}{3}, 3, -3, \frac{1}{4}, -\frac{1}{4}, \frac{2}{3}, -\frac{2}{3}, \frac{3}{2}, -\frac{3}{2}, 4, -4, \frac{1}{5}, -\frac{1}{5}, 5, -5, \frac{1}{6}, -\frac{1}{6}, \frac{2}{5}, -\frac{2}{5}, \frac{3}{4}, -\frac{3}{4}, \frac{4}{3}, -\frac{4}{3}, \frac{5}{2}, -\frac{5}{2}, 6, -6, \frac{1}{7}, -\frac{1}{7}, \frac{3}{5}, -\frac{3}{5}, \frac{5}{3}, -\frac{5}{3}, 7, -7, \frac{1}{8}, -\frac{1}{8}, \frac{2}{7}, -\frac{2}{7}, \frac{4}{5}, -\frac{4}{5}, \frac{5}{4}, -\frac{5}{4}, \frac{7}{2}, -\frac{7}{2}, 8, -8, \ldots. $$
This sequence is based on the scheme of the earlier sequence, except that (1) we begin with $0$ and (2) we follow each positive rational number as we come to it by its negative.
My second question is, how to find out a formula for the general term of this particular sequence of the rational numbers? I'd again really appreciate if the whole process is explained in detail.
Last but not the least, any references for reading up on this?
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Such formulae necessarily involve the Euler's totient function, because the number of positive fractions $p/q$ (in their lower terms, of course) such that $p+q=n$ is precisely $\varphi(n)$. Then, for the first sequence, the $k$-th term can be obtained this way:
Let $n=\max\{t\in\Bbb N:\sum_{j=1}^t\varphi(j)\le k\}$. In other words, sum $\varphi(1)+\varphi(2)+\cdots$ until you exceed $k$. Then, the terms of the fraction sum $n$. Now, let
$$r=k-\sum_{j=1}^{n-1}\varphi(j)$$
Then, the numerator of the fraction is the $r$-th number coprime with $n$.
Example:
To find the $100$-th term first we see that $\sum_{j=1}^{19}\varphi(j)=98$ and $\sum_{j=1}^{20}\varphi(j)=106$, so the sum of the terms of the fraction is $19$. Since $100-98=2$ the numerator if the fraction is the second number coprime with $19$, namely $2$, so the $100$th term is $2/17$.
The formula (or, rather, the algorithm) for the second sequence is very similar. You have to count the first term, $0$, and sum $2\varphi(j)$ instead of $\varphi(j)$.
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Find the amplitude of a complex number
If
$$z = (1-\cos{\theta}) + i \sin{\theta},$$
then the amplitude of $z$ is:
$$a) \frac{\pi}{4}-\frac{\theta}{2}$$ $$b)
\frac{\pi}{2}-\frac{\theta}{2}$$ $$c) \frac{\pi}{2}-\theta$$ $$d)
\frac{\pi}{2}-\frac{\theta}{4}$$ where 0<$\theta$ <$\pi$.
I tried :
$x= 1-\cos{\theta}, y= \sin{\theta}$ then
$\tan{\phi}=\frac{\sin{\theta}}{1-\cos{\theta}}$.
However I could not solve the last equation.
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$\sin\theta = 2\sin\frac{\theta}{2}\cdot \cos\frac{\theta}{2}$
$1-\cos\theta = 2\sin^2\frac{\theta}{2}$
So, $\tan\phi = \frac{\sin\theta}{1-\cos\theta} = \frac{2\sin\frac{\theta}{2}\cdot \cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$
$\tan\phi = \cot\frac{\theta}{2} = \tan(\frac{\pi}{2} - \frac{\theta}{2})$
$\implies\phi =\frac{\pi}{2} - \frac{\theta}{2} $
For half-angle formulas: https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php
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How do I solve this log equation for x? $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$ $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$
I only managed to solve up to this step: $\left(2+x\right)^{\frac{3}{2}}\sqrt{2-x}=\sqrt{4-x^2}\cdot \:125$
However, I'm not too sure about squaring both sides as I would have to handle ridiculously large numbers to find x, so I'm not too sure that I'm doing it correctly.
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You are almost there.
$$\left(2+x\right)^{\frac{3}{2}}\sqrt{2-x}=\sqrt{4-x^2}\cdot125$$
$$\implies (2+x)\sqrt{2+x}\cdot\sqrt{2-x} = 125\sqrt{(2+x)(2-x)}$$
Here, notice that $x = 2$ and $x = -2$ are not the solutions for the original equation because $\log_50$ is not defined. Therefore, we can divide both sides by $\sqrt{2+x}\sqrt{2-x}$. Then, we have
$$2+x = 125 \implies x = 123$$
But this solution makes $\sqrt{2-x}$ a non-real number. Therefore this equation has no real solution.
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Simultaneous equation of trig functions If I have the two following equations:
$$\begin{align}
A &= F\sin(x)+ G\cos(x)\\
B &= G\sin(x)+ F\cos(x)
\end{align}$$
and $A$,$B$,$F$,$G$ and are all constants, what is the easiest method of calculating $x$?
I've had a look at Cramers rule and the $R\cos(x-\alpha)$ rule. I need to find the most effective method for a program I am writing!
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If $F\ne G$, then subtracting the second equation from the first gives
\begin{eqnarray}
(F-G)\sin x-(F-G)\cos x&=&A-B\\\
\sin x-\cos x&=&\frac{A-B}{F-G}\\\
\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\right)&=&\frac{A-B}{F-G}\\\
\sqrt{2}\left(\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}\right)&=&\frac{A-B}{F-G}\\\
\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)&=&\frac{A-B}{F-G}\\\
\sin\left(x-\frac{\pi}{4}\right)&=&\frac{\sqrt{2}}{2}\cdot\frac{A-B}{F-G}\tag{1}
\end{eqnarray}
In the case $F=G$ it follows that $A=B$ and the equation reduces to the form
$$\sin x+\cos x=C $$
which can be reformed into
$$ \cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}C $$
ADDENDUM Due to the comment by @GCab
If $F\ne-G$ and the first equation is added to the second a similar process gives the result
$$ \cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\cdot\frac{A+B}{F+G} \tag{2}$$
These equations can be solved for $x$ over whichever range is required.
In the case that $G=-G$ it follows that $A=-B$ and the equation will be of the form
$$\sin x-\cos x=C $$
which has solution
$$ \sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}C $$
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|
Find $ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$ How to compute the limit
$$ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$$
Does exist an explicit formula for finding the numerator?
|
The required limit is $1/(n+2)$. There probably does not exist a closed formula for the numerator. However, the numerator can be estimated by use of an integral, as follows.
Note that
$$\frac{k^{n+2}}{n+2}=\int_0^{k}x^{n+1}\,\mathrm dx < \sum_{x=0}^kx^{n+1}<\int_1^{k+1}x^{n+1}\,\mathrm dx=\frac{(k+1)^{n+2}}{n+2}-\frac{1}{n+2}$$
Therefore,
$$\frac{k}{(n+2)(k+1)}<\frac{\sum_{x=0}^kx^{n+1}}{(k+1)k^{n+1}}<(1+\frac{1}{k})^{n+1}\frac{1}{n+2}-\frac{1}{(k+1)k^{n+1}(n+2)}$$
and both the r.h.s and the l.h.s tend to $\frac{1}{n+2}$ as $k\to\infty$.
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Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expression can be reduced as sum of sine functions as:
$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$
Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.
Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,
$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$
For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!
So what's the problem here? Can it be solved?
Thanks :)
Edit : I'd like to thank @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.
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We can use a trick to express the function as something easier to deal with. We claim that we can write the function as $(d \cos \theta + e \sin \theta)^2 + f$ or $-(d \cos \theta - e \sin \theta)^2 + f$ for some constants $d, e, f$. From here, converting $d \cos \theta + e \sin \theta$ into a single trigonometric function is simple, and we can find the bounds exactly.
Note that for our expression to work, we must have $de = \frac{b}{2}$. So we expand $(d \cos \theta + \frac{b}{2d} \sin \theta)^2$ to get $d^2 \cos^2 \theta + b \sin \theta \cos \theta + \frac{b^2}{4d^2} \sin^2 \theta$. If we can find a $d$ satisfying $d^2 - \frac{b}{4d^2} = a^2 - c^2$, then the expression will be exactly $a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta - f(\sin^2 \theta + \cos^2 \theta)$ for some $f$. But solving for $d$ is now a quadratic in $d^2$. A similar approach can be taken with $-(d \cos \theta - e \sin \theta)^2$; one of the two will always work, and yield the exact bounds required.
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Why is: $1+2+2^2+\cdots+2^{n-2} = 2^{n-1}-1$? I know that this might be elementary, but I don't know how to prove this equality, some hint?
I tried to extract the common two:
$$ 1 + 2^1+2^2 + \cdots + 2^{n-2} = 1 + 2(1+2+\cdots+2^{n-3}) $$
But this didn't help
|
The sum $1+2+2^2+\cdots+2^{n-2} $ doesn't change if we multiply it by $1$ i.e. $2-1$:
\begin{eqnarray}1+2+2^2+\cdots+2^{n-2} &=& (1+2+2^2+\cdots+2^{n-2})(2-1)\\
&=& (\color{red}{2+2^2+2^3+\cdots+2^{n-2}}+2^{n-1})-(1+\color{red}{2+2^2+2^3+\cdots+2^{n-2}})\\
&=&2^{n-1}-1
\end{eqnarray}
|
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|
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1}$$
upon expanding them we get:
$$f_1(x)=1 + x + x^2$$
$$f_2(x)=1 + 2 x + 3 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6$$
$$f_3(x)=1 + 3 x + 6 x^2 + 10 x^3 + 15 x^4 + 18 x^5 + 19 x^6 + 18 x^7 +
15 x^8 + 10 x^9 + 6 x^{10} + 3 x^{11} + x^{12}$$
$$f_4(x)=1 + 4 x + 10 x^2 + 20 x^3 + 35 x^4 + 56 x^5 + 80 x^6 + 104 x^7 +
125 x^8 + 140 x^9 + 146 x^{10} + 140 x^{11} + 125 x^{12} + 104 x^{13} +
80 x^{14} + 56 x^{15} + 35 x^{16} + 20 x^{17} + 10 x^{18} + 4 x^{19} + x^{20}$$
$$\vdots$$
$$f_{n-1}(x)=1 + ?x + ?x^2 + ?x^3 +?x^4+ ?x^5+\cdots+?x^{n(n-1)}$$
I'm wondering how to determine the coefficients for the n-th order? I can observe that the coefficients are symmetric.
|
$\sum_{i=0}^{k\cdot m}c_ix^i=(1+x^1+x^2+...+x^m)^{k}$ is the generating function of the number of weak integer compositions (integer compositions with repetitions of $0$) of integer $i$ with $k$ parts where all parts are lower equal to $m$.
Unfortunately, it is not yet in OEIS.
$k,m>0$
Their coefficients are:
$$c_i=\sum_{j=0}^{\frac{i+k-1}{m+1}}(-1)^{j}\binom{k}{j}\binom{i+k-j(m+1)-1}{k-1}.$$
[Stanley 1999], Mistake in the closed form formula for the number of restricted compositions?
with $n>0$, $k=n$, $m=n+1$:
$$c_i=\sum_{j=0}^{\frac{i+n-1}{n+2}}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-1}{n-1}$$
$\ $
[Stanley 1999] Stanley, R. P.: Enumerative Combinatorics Vol. I. Cambridge University Press, 1999
|
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|
When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$ When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$, gives $\sum\limits_{n=1}^{\infty} \left(\frac{nx^{n-1}}{2^n}\right)$. so, $n =0$ , becomes $n = 1$. Then, if we were to differentiate $\sum\limits_{n=1}^{\infty} \left(\frac{x}{2}\right)^n$, does it become $\sum\limits_{n=2}^{\infty} \left(\frac{nx^{n-1}}{2^n}\right)$? (from $n= 1$, to $n= 2$?)
If this is how it works,
is the above equation false? and how could we solve $\sum\limits_{n=1}^{\infty} \left(\frac{n^2x^n}{2^n}\right)$?
|
No, because the reason the sum lower bound goes from 0 to 1 is because the derivative kills the first term in the sum. When we differentiate the sum starting from $n=1$, the first term is not killed by the derivative and hence you shouldn't drop it from the sum.
To deal with the sum
$$
F(x)=\sum_{n=1}^\infty n^2 x^n 2^{-n},
$$
Do the following:
Define
$$
S(x)=\frac2{2-x}=\sum_{n=0}^\infty\left(\frac x2\right)^n.
$$
We have shown
$$
S'(x)=\sum_{n=1}^\infty n\left(\frac x2\right)^n=\frac2{(2-x)^2}.
$$
Similarly, we can show
$$
S''(x)=\sum_{n=2}^\infty n(n-1)\left(\frac x2\right)^n=\frac4{(2-x)^3}.
$$
Finally, note that
\begin{equation}\begin{aligned}
F(x)&=\frac x2+\left(\frac x2\right)^2\sum_{n=2}^\infty n(n-1)\left(\frac{x}2\right)^{n-2}+\frac x 2\sum_{n=2}^\infty n\left(\frac x2\right)^{n-1}\\
&=\frac x2+\left(\frac x2\right)^2S''(x)+\frac x2S(x)-\frac x2\\
&=\left(\frac x2\right)^2\frac4{(2-x)^3}+\frac x2\frac2{(2-x)^2}=\frac{4x}{(2-x)^2}.
\end{aligned}\end{equation}
|
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|
$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$.
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$
Before you complain, this problem is adapted from a recent competition. I have put my solution down below, there might be more practical and correct answers. In that case, please post them.
|
$$\frac{xy}{z^2(x+y)}+\frac{yz}{x^2(z+y)}+\frac{xz}{y^2(x+z)}$$
$$=\frac{2x^2y^2}{z(x+y)}+\frac{2y^2z^2}{x(z+y)}+\frac{2x^2z^2}{y(x+z)}$$
$$\geq 2\frac{(xy+yz+xz)^2}{2(xy+yz+zx)}=xy+yz+xz$$
Using Titu's Lemma which is a variant of the Cauchy-Schwarz inequality.
|
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|
Polar decomposition of a general matrix How can I calculate the polar decomposition for a general matrix? For example for this simple one:
$$
\begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix}
$$
I know how to calculate it for a matrix with numbers, via eigenvalues, eigenvectors. I have been searching for the answer on the internet for a while but I don't fully understand it.
|
So first let $A = \begin{pmatrix}a&-b\\b&a\end{pmatrix}=QS$.
Then, we have $A^{T}A=\begin{pmatrix}a&b\\-b&a\end{pmatrix}\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}a^2+b^2&0\\0&a^2+b^2\end{pmatrix}$.
The Eigenvalue(s) of this matrix is $a^2+b^2$, sometimes they're different but not too hard to handle.
The Eigenvectors are $\begin{pmatrix}1\\0 \end{pmatrix}$ and $\begin{pmatrix}0\\1 \end{pmatrix}$.
So then we can write $A^TA=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}a^2+b^2&0\\0&a^2+b^2\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}^{-1}$.
The first factor is just the Eigenvectors, the second is a diagonal matrix consisting of each of the corresponding eigenvalues, and the third is the inverse of the first.
We now define a second matrix $S=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}^{-1}$.
The only difference here is that $S$ is constructed using the singular values, which are basically the square roots of the eigenvalues.
We have that $S=\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}$.
We also have that $Q=AS^{-1}= \begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}^{-1}=\begin{pmatrix}\frac{a}{\sqrt{a^2+b^2}}&-\frac{b}{\sqrt{a^2+b^2}}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{pmatrix}$.
So finally, we have the polar decomposition: $\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}\frac{a}{\sqrt{a^2+b^2}}&-\frac{b}{\sqrt{a^2+b^2}}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{pmatrix}\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}$
|
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|
How can i get the least $n $ such that $17^n \equiv 1 \mod(100$)? When I solve the problem:
$17^{2018}\equiv r \pmod{100} $
used Euler theorem since $\gcd (17,100)=1$ and so
$\phi(100)=40$
and so
$17^{40}\equiv 1 \pmod{100}$
But i also found that: $17^{20}\equiv 1 \pmod{100}$
How can i get the least n such that $17^{n}=1\pmod{100} $?
Is there any theorm or generalization to this problem?
Thanks for your help
|
Euler's phi function, $\varphi(n)$, is a multiplicative function for which, if $a$ and $N$ are relatively prime, then $a^{\varphi(N)} \equiv 1 \pmod N$. In particular
$$\varphi(100) = \varphi(4)\varphi(25) = (4-2)(25-5)=40$$
So $17^{40} \equiv 1 \pmod{100}$. The smallest $n$ such that $17^n \equiv 1 \pmod{100}$ must therefore be a divisor of $40$.
The divisors of $40$ are $$1, 2, 4, 5, 8, 10, 20, 40$$
\begin{align}
17^1 &\equiv 17 \pmod{100} \\
17^2 &\equiv 17\times17 \equiv 89 \pmod{100} \\
17^4 &\equiv 89\times 89 \equiv 21 \pmod{100} \\
17^5 &\equiv 17\times 21 \equiv 57 \pmod{100} \\
17^8 &\equiv 21\times 21 \equiv 41 \pmod{100} \\
17^{10} &\equiv 57\times 57 \equiv 49 \pmod{100} \\
17^{20} &\equiv 49\times 49 \equiv 1 \pmod{100}
\end{align}
Added 6/8/2019
In response to the comment of @labbhattacharjee, $\lambda(n)$, the Carmichael function of n, is defined by
For any power of a prime number, $n = p^\alpha$,
$$\lambda(n) = \begin{cases}
\frac 12\varphi(n) & \text{If n is a power of $2$ that is $\ge 8$ }\\
\varphi(n) & \text{Otherwise}
\end{cases}$$
For any prouduct of powers of unique prime numbers,
$n=p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$,
$$\lambda(n) = \operatorname{LCM}
\{\lambda(p_1^{\alpha_1}),
\lambda(p_2^{\alpha_2}), \dots,
\lambda(p_k^{\alpha_k})\}$$
So
\begin{align}
\lambda(100)
&= \lambda(2^2 \cdot 5^2) \\
&= \operatorname{LCM}\{\lambda(2^2),\lambda(5^2)\} \\
&= \operatorname{LCM}\{\varphi(2^2),\varphi(5^2)\} \\
&= \operatorname{LCM}\{2,20 \} \\
&= 20
\end{align}
|
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|
Solve for integer $m,n$: $2^m = 3^n + 5$
Solve for integer $m$ and $n:$ $$2^m = 3^n + 5$$
My Attempt:
Easy to guess two solutions namely $(3,1)$ and $(5,3)$.
Also easy to see that a solution will exist iff $m > 0$ and $n > 0$.
Rewriting it as $2^m - 2 = 3^n + 3$ we get $2^m = 2 \mod 3 \Rightarrow m = 1 \mod 2$ and $3^n = 1 \mod 2 \Rightarrow n = 1 \mod 2$, hence $m$ and $n$ are both odd.
Beyond this, I could not figure out what approach to use.
Source: Past IMO shortlisted problem.
|
Rename $m\to x$ and $n\to y$
We see $x\geq 3$, $y\geq 1$.
Modulu 3 implies $x$ is odd. For $x\leq 5$ we get only $(3,1)$, $(5,3)$.
Say $x\geq 6$, then
$$3^y\equiv -5\;({\rm mod}\; 64)$$ It is not difficult to see
$$3^{11}\equiv -5\;({\rm mod}\; 64)$$
so $3^{y-11}\equiv 1\;({\rm mod}\; 64)$. Let
$r=ord_{64}(3)$, then since $\phi(64)=32$,
we have (Euler)
$$3^{32}\equiv 1\;({\rm mod}\; 64)$$
We know $r\;|\;32$. Since
$$3^{32} -1 = (3^{16}+1)\underbrace{(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)}_{(3^{16}-1)}$$
we get $r=16$ so $16\;|\;y-11$ and thus $y=16k+11$ for some integer $k$.
Look at modulo 17 now. By Fermat little theorem:
$$2^x\equiv 3^{16k+11}+5\equiv (3^{16})^k \cdot 3^{11}+5\equiv 1^k\cdot 7+5\equiv 12\;({\rm mod}\; 17)$$
Since $x$ is odd we have
\begin{eqnarray*}
2^1&\equiv & 2\;({\rm mod}\; 17)\\
2^3&\equiv & 8\;({\rm mod}\; 17)\\
2^5&\equiv & -2\;({\rm mod}\; 17)\\
2^7&\equiv & -8\;({\rm mod}\; 17)\\
2^9&\equiv & 2\;({\rm mod}\; 17)
\end{eqnarray*}
so upper congurence is never fulfiled, so no solution for $x\geq 6$.
|
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|
Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$ Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$.
I tried Euler and hyperbolic substitutions, but both lead to complicated calculations and yet WolframAlpha is able to generate quite simple form of integral.
|
Your integral is equivalent to
$$ \frac{1}{2} \underbrace{\int \frac{2x+1}{\sqrt{x^2 + x + 1}} \mathrm dx}_{=:I_1} - \frac{1}{2} \underbrace{\int \frac{\mathrm d x}{\sqrt{x^2 + x + 1}}}_{=: I_2}.$$
Note that $I_1$ can be easily solved with the substitution $u = x^2 + x + 1$.
For $I_2$ write
$$I_2 = \int \frac{ \mathrm d x}{\sqrt{(x + \frac{1}{2})^2 + \frac{3}{4}}}.$$
Now the substitution $u = \frac{2x+1}{\sqrt{3}}$ will lead you to the integral
$$\int \frac{\mathrm d u}{\sqrt{u^2 + 1}},$$
which is a standard integral - an antiderivative is $\ln (\sqrt{u^2 + 1} + u)$.
|
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|
Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$
Solve system of congruences
$$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$
Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? When we have this table it is very fast job but making it is... time-consuming
\begin{array}{|c|c|c|c|}
\hline
i^2 \mod 17 & i^3 \mod 17 \\ \hline
0 & 0 \\ \hline
1 & 1 \\ \hline
4 & 8 \\ \hline
9 & 10 \\ \hline
16 & 13 \\ \hline
8 & 6 \\ \hline
2 & 12 \\ \hline
15 & 3 \\ \hline
13 & 2 \\ \hline
13 & 15 \\ \hline
15 & 14 \\ \hline
2 & 5 \\ \hline
8 & 11 \\ \hline
16 & 4 \\ \hline
9 & 7 \\ \hline
4 & 9 \\ \hline
1 & 16 \\ \hline
0 & 0 \\ \hline
1 & 1 \\ \hline
4 & 8 \\ \hline
\end{array}
|
$\left.\begin{align}\rm Squaring &\ \ k^{\large 3}\equiv -l^{\large 3}\,\Rightarrow\, k^{\large 6}\,\equiv\, l^{\large 6}\\
\rm\&\ cubing &\ \ k^{\large 2}\equiv -l^{\large 2}\,\Rightarrow\, k^{\large 6}\!\equiv -l^{\large 6}\end{align}\right\} $ $\Rightarrow\, 2l^{\large 6}\equiv 0\,\Rightarrow\, l\equiv 0\ $ (in any domain with $2\not\equiv 0)$
|
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|
Find K such that the line is tangent to $y=-\frac{3}{4}x+4$ The function is $f(x)= \frac{k}{x}$ and the line is $y =-\frac{3}{4}x+3$
So the derivative of $f(x)$ should equal $-\frac{3}{4}$
If I take $k$ to be a constant I can remove it from the fraction like so
$$f'(x)=k \frac{d}{dx} \frac{1}{x}$$
to get $-\frac{k}{x^2}$ but then I try to solve for $k$
$$-\frac{k}{x^2}=-\frac{3}{4}$$
and this is where I get stuck. So should I not even remove $k$ from the differentiation or have I gone in a completely wrong direction?
|
From the given equation we obtain
$$-\frac{3}{4}x+3=\frac{k}{x}$$ so we get
$$-\frac{3}{4}x^2+3x-k=0$$ solving for $x$ we get
$$x=2\pm \frac{2}{3}\sqrt{9-3k}$$ since the line is a tangent line, we get
$$9-3k=0$$
|
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|
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2\pi}{7}=\sqrt{7}$$
$$\tan\frac{\pi}{11}+4\sin\frac{3\pi}{11}=\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}=\sqrt{11}$$
$$\tan\frac{6\pi}{13}-4\sin\frac{5\pi}{13}=\tan\frac{2\pi}{13}+4\sin\frac{6\pi}{13}=\sqrt{13+2\sqrt{13}}$$
Does anyone know of any analogous identities for larger primes? I have not been able to find anything similar for $p=17$ or $p=19$.
(I am not asking for proofs of the above equations.)
|
$$\tan\frac{2\pi}{37}+4\left(-\sin\frac{4\pi}{37}+\sin\frac{14\pi}{37}+\sin\frac{16\pi}{37}+\sin\frac{24\pi}{37}-\sin\frac{28\pi}{37}+\sin\frac{32\pi}{37}-\sin\frac{36\pi}{37}\right)=\sqrt{37+6\sqrt{37}} $$
|
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|
How would the most general $2 \times 2$ normal matrix look like?
How would the most general $2 \times 2$ normal matrix look like?
The normal matrix satisfy equation: $A^*A=AA^*$ where $A^*$ denotes
conjugate transpose.
I was thinking about the matrix:
$$
\begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix}
$$ because its columns are orthogonal to each other and it satisfies the given equation:
$$
\begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix}
\begin{pmatrix}
a & b \\
-b & a \\
\end{pmatrix} =
\begin{pmatrix}
a^2 + b^2 & 0 \\
0 & b^2 + a^2 \\
\end{pmatrix}
$$
$$
$$
$$
\begin{pmatrix}
a & b \\
-b & a \\
\end{pmatrix}
\begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix} =
\begin{pmatrix}
a^2 + b^2 & 0 \\
0 & b^2 + a^2 \\
\end{pmatrix}
$$
$$
$$
It is true for real matrices, and I suppose for the complex one too. But is this the most general case, or is there something else?
|
Let $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$ For $A$ to be normal we require that $$\begin{bmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{bmatrix},$$ or equivalently $$\begin{bmatrix} |a|^2 + |c|^2 & \overline{a}b + \overline{c}d \\ \overline{b}a + \overline{d}c & |b|^2 + |d|^2 \end{bmatrix} = \begin{bmatrix} |a|^2 + |b|^2 & a\overline{c} + b\overline{d} \\ c\overline{a} + d\overline{b} & |c|^2 + |d|^2 \end{bmatrix}.$$
Firstly notice that by equality of the diagonal entries we must have $$|b|^2 = |c|^2.$$ And furthermore comparing the off-diagonal entries we must have
$$ \overline{a}b + \overline{c}d = a\overline{c} + b\overline{d}, $$
$$ \overline{b}a + \overline{d}c = c\overline{a} + d\overline{b}. $$
In the real case the first requirement reduces to $c=|b|$, but for complex matrices we could also have for example $b=1$ and $c=i$. For now let's just consider the real case: the most general case is to fix $b$ and let $c=|b|$. Then we have two cases:
*
*If $b$ is nonnegative $a$ and $d$ could be any real values.
*If $b$ is negative $a=d$.
The second case above corresponds to your example - I wouldn't consider this to be more general than the first case - for real numbers 1 and 2 together is most general.
So now for complex numbers there is a lot more work to be done: you could explore this further by for example setting $a = a_1 + ia_2, b=b_1 + ib_2, \ldots$, fixing some and then working out by the constraints what values the other variables could assume...
|
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|
Summation closed form I have the following sum,
$$\sum_{j=0}^{\lfloor\frac{i+n-1}{n+2}\rfloor}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-1}{n-1}+\sum_{j=0}^{\lfloor\frac{i+n-2}{n+2}\rfloor}2(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-2}{n-1}+\sum_{j=0}^{\lfloor\frac{i+n-3}{n+2}\rfloor}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-3}{n-1}$$
Notice that the upper bound of each sum is slightly different in each term. I wonder if there is a closed form for this? If we don't have the last binomial in each summation then there would be a closed form but at this level I am not sure how to get the closed form.
Edit: $n$ is non-negative integer, and the constraint on $i$ is as follow: $0\leqslant i\leqslant n(n+1)+2$.
|
Note that we can better write each term as
$$
\eqalign{
& S(k,n)=\sum\limits_{j = 0}^{\left\lfloor {{{i + n - k} \over {n + 2}}} \right\rfloor } {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
n - 1 \cr} \right)} = \cr
& = \sum\limits_{0\, \le j} {\left[ {0 \le i + n - k - j\left( {n + 2} \right)} \right]\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
n - 1 \cr} \right)} = \quad \quad (1) \cr
& = \sum\limits_{0\, \le j} {\left[ {0 \le i + n - k - j\left( {n + 2} \right)} \right]\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} = \quad \quad (2) \cr
& = \left[ {1 \le n} \right]\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} \quad \quad (3) \cr}
$$
where $[P]$ denotes the Iverson bracket
and where:
- (1) we replace the upper bound with Iverson bracket;
- (2) we can apply reflection, since the upper term of the binomial is non-negative;
- (3) we can omit the Iverson bracket, because for $1 \le n$ it is implicit in the binomial.
Note that $S(k,n)$ is equivalent to
$$
S(k,n) = N_{\,b} (i + 1 - k,\,n + 1,n)
$$
where
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{k}
\binom
{ s + m - 1 - k\left( {r + 1} \right) }
{ s - k\left( {r + 1} \right)}\ }
$$
is the number discussed in this related post, and is also called the "r-nomial coefficient" since
$$
F_b (x,r,m) = \sum\limits_{0\,\, \leqslant \,\,s\,\,\left( { \leqslant \,\,r\,m} \right)} {N_b (s,r,m)\;x^{\,s} }
= \left( {1 + x + \cdots + x^{\,r} } \right)^m = \left( {\frac{{1 - x^{\,r + 1} }}{{1 - x}}} \right)^m
$$
Now, the cumulative sum of $N_b$ has a similar expression
$$
M_{\,b} (t,\,r,m) = \sum\limits_{0\, \le s\; \le \,\,t} {N_{\,b} (s,\,r,m)}
= \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{t \over {r + 1}}\, \le \,m} \right)} {\left( { - 1} \right)^k \binom {m}{k}
\binom{ t + m - k\left( {r + 1} \right)} {t - k\left( {r + 1} \right) } }
$$
but this does not help to simplify much your sum, since it will always leave three terms.
The attempt to directly perform the sum will give:
$$
\eqalign{
& S(n) = \left[ {1 \le n} \right]\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( \matrix{
2 \cr
k \cr} \right)\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
2 \cr
k \cr} \right)\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,k + j} \left( \matrix{
k - 3 \cr
k \cr} \right)\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - \left( {k + j} \right) - j\left( {n + 1} \right) \cr
i + 1 - \left( {k + j} \right) - j\left( {n + 1} \right) \cr} \right)} } = \cr
& = \sum\limits_{0\, \le l} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,l} \left( \matrix{
l - j - 3 \cr
l - j \cr} \right)\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - l - j\left( {n + 1} \right) \cr
i + 1 - l - j\left( {n + 1} \right) \cr} \right)} } \cr}
$$
and we cannot proceed to simplify by applying double correlation, due to the presence of $\left( { - 1} \right)^{\,l}$
We could instead simplify the related sum
$$
\eqalign{
& S_{\, - } (n) = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
2 \cr
k \cr} \right)S(k,n)} = \cr
& = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
2 \cr
k \cr} \right)\sum\limits_{} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( \matrix{
k - 3 \cr
k \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - 2 - j\left( {n + 2} \right) \cr
i + 1 - j\left( {n + 2} \right) \cr} \right)} \cr}
$$
so that the actual sum may be reduced to two terms.
Besides these, and other related manipulations, I do not see better ways.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Area of a Polygon in a Polygon I'm dealing with a regular polygon with 7 corners. In this polygon is another polygon defined by connecting one point with the two opposite points of the same polygon.
I made a small sketch of the polygons
I need to calculate the ratio of the area of the big and of small polygon. I know how to calculate the area of the big polygon but I don't know how to get area of the small polygon.
|
Easy to see that $P_1Q_2P_3P_2$ and $P_1P_7Q_1P_2$ are parallelograms, which says that
$$P_1P_7=Q_1P_2=P_1P_2=P_1Q_2=P_2P_3.$$
Now, let $P_1P_4\cap P_6P_2=\{A\}$.
Thus, from $\Delta Q_1Q_2A$ we obtain:
$$Q_1Q_2=2AQ_2\sin\frac{\measuredangle Q_1AQ_2}{2}=2AQ_2\sin\frac{3\pi}{14}.$$
Also, from $\Delta P_1P_2A$ we obtain:
$$P_1P_2=2AP_1\sin\sin\frac{\measuredangle P_1AP_2}{2}=2AP_1\sin\frac{3\pi}{14}.$$
Since $$AQ_2+AP_1=P_1Q_2=P_2P_3=P_1P_2,$$ we obtain:
$$\frac{Q_1Q_2}{2\sin\frac{3\pi}{14}}+\frac{P_1P_2}{2\sin\frac{3\pi}{14}}=P_1P_2,$$ which gives
$$\frac{P_1P_2}{Q_1Q_2}=\frac{1}{2\sin\frac{3\pi}{14}-1},$$ which gives the answer:
$$\frac{1}{\left(2\sin\frac{3\pi}{14}-1\right)^2}.$$
Another way.
Let $R$ be a radius of the circumcircle of $P_1P_2...P_7.$
Thus, $$Q_1Q_2=P_7P_3-2P_1P_2=2R\sin\frac{3\pi}{7}-4R\sin\frac{\pi}{7}$$
and $$P_1P_2=2R\sin\frac{\pi}{7}.$$
Id est, $$\frac{P_1P_2}{Q_1Q_2}=\frac{\sin\frac{\pi}{7}}{\sin\frac{3\pi}{7}-2\sin\frac{\pi}{7}}=\frac{1}{3-4\sin^2\frac{\pi}{7}-2}=$$
$$=\frac{1}{1-2\left(1-\cos\frac{2\pi}{7}\right)}=\frac{1}{2\cos\frac{2\pi}{7}-1},$$ which gives a ratio of areas again.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3260761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove that $3 \le \sum_{cyc}a\sqrt{b^3 + 1} \le \sum_{cyc}ab^2 + 3$ where $a, b, c \ge 0$ and $a + b + c = 3$.
$a$, $b$ and $c$ are non-negatives such that $a + b + c = 3$. Prove that $$\large 3 \le a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \le \frac{ab^2 + bc^2 + ca^2}{2} + 3$$
This problem is adapted from a competition... that happened two years ago. Yup, I've changed things up.
I have provided a solution below if you want to check out. (I am typing this at midnight so my consciousness is drifting away and I can't focus.)
|
We have that $a, b, c \ge 0 \iff \sqrt{a^3 + 1} - 1, \sqrt{b^3 + 1} - 1, \sqrt{c^3 + 1} - 1 \ge 0$
$$\implies c\left(\sqrt{a^3 + 1} - 1\right) + a\left(\sqrt{b^3 + 1} - 1\right) + b\left(\sqrt{c^3 + 1} - 1\right) \ge 0$$
$$\iff a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \ge a + b + c = 3$$
The equality sign occurs when $\left[ \begin{align} a = b = 0 &\text{ and } c = 3\\ b = c = 0 &\text{ and } a = 3\\ c = a = 0 &\text{ and } b = 3 \end{align} \right.$.
Furthermore, we have that $$\sqrt{x^3 + 1} = \sqrt{(x^2 - x + 1)(x + 1)} \le \dfrac{(x^2 - x + 1) + (x + 1)}{2} = \dfrac{x^2 + 2}{2}$$
$$\implies a\sqrt{b^3 + 1} + b\sqrt{c^3 + 1} + c\sqrt{a^3 + 1} \le \frac{a(b^2 + 2) + b(c^2 + 2) + c(a^2 + 2)}{2}$$
$$= \frac{ab^2 + bc^2 + ca^2}{2} + (a + b + c) = \frac{ab^2 + bc^2 + ca^2}{2} + 3$$
The equality sign happens when $\left[ \begin{align} a = 0, b = 1 \text{ and } c = 2\\ a = 0, b = 2 \text{ and } c = 1\\ a = 1, b = 2 \text{ and } c = 0\\ a = 1, b = 0 \text{ and } c = 2\\ a = 2, b = 0 \text{ and } c = 1\\ a = 2, b = 1 \text{ and } c = 0\\ \end{align} \right.$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3261281",
"timestamp": "2023-03-29T00:00:00",
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|
If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
I have provided a (dumbfounding) solution down below if you want to check out. There should be simpler solutions, I believe so.
|
We have that $\left\{ \begin{align} 5 &\mid x^2 - 2xy - y\\ 5 &\mid xy - 2y^2 - x \end{align} \right.$$\iff \left\{ \begin{align} 5 &\mid (x^2 - 2xy - y) + (xy - 2y^2 - x)\\ 5 &\mid (x^2 - 2xy - y) - (xy - 2y^2 - x) \end{align} \right.$
$\iff \left\{ \begin{align} 5 &\mid (x + y)(x - 2y - 1)\\ 5 &\mid (x - y)(x - 2y + 1) \end{align} \right.$$\iff \left\{ \begin{align} 5 \mid x + y &\text{ or } 5 \mid x - 2y - 1\\ 5 \mid x - y &\text{ or } 5 \mid x - 2y + 1 \end{align} \right.$
$\implies \left[ \begin{align} 5 \mid x + y &\text{ and } 5 \mid x - y\\ 5 \mid x + y &\text{ and } 5 \mid x - 2y + 1\\ 5 \mid x - y &\text{ and } 5 \mid x - 2y - 1 \end{align} \right.$$\iff \left[ \begin{align} x + y &\equiv x - y \equiv 0 \text{ (mod 5)}\\ x \equiv -y &\text{ (mod 5) and } x \equiv 2y - 1 \text{ (mod 5)}\\ x \equiv y &\text{ (mod 5) and } x \equiv 2y + 1 \text{ (mod 5)} \end{align} \right.$
$\iff \left[ \begin{align} x &\equiv y \equiv 0 \text{ (mod 5)}\\ x \equiv -y &\text{ (mod 5) and } -y \equiv 2y - 1 \text{ (mod 5)}\\ x \equiv y &\text{ (mod 5) and } y \equiv 2y + 1 \text{ (mod 5)} \end{align} \right.$
$\iff \left[ \begin{align} x &\equiv y \equiv 0 \text{ (mod 5)}\\ x \equiv -y &\text{ (mod 5) and } y \equiv 2 \text{ (mod 5)}\\ x \equiv y &\text{ (mod 5) and } y \equiv -1 \text{ (mod 5)} \end{align} \right.$$\iff \left[ \begin{align} x &\equiv y \equiv 0 \text { (mod 5)}\\ x \equiv -2 &\text{ (mod 5) and } y \equiv 2 \text{ (mod 5)}\\ x \equiv -1 &\text{ (mod 5) and } y \equiv -1 \text{ (mod 5)} \end{align} \right.$
$\implies \left[ \begin{align} 2x^2 + y^2 + 2x + y \equiv 2 \cdot 0^2 + 0^2 + 2 \cdot 0 + 0 &\equiv 0 \text{ (mod 5)}\\ 2x^2 + y^2 + 2x + y \equiv 2 \cdot (-2)^2 + 2^2 + 2 \cdot (-2) + 2 &\equiv 0 \text{ (mod 5)}\\ 2x^2 + y^2 + 2x + y \equiv 2 \cdot (-1)^2 + (-1)^2 + 2 \cdot (-1) + (-1) &\equiv 0 \text{ (mod 5)} \end{align} \right.$
$\iff 5 \mid 2x^2 + y^2 + 2x + y$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
|
It's clear that the inequality
$$
\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\ge 0
$$holds for a triple $(a,b,c)$ if and only if it holds for the triple $(ta,tb,tc)$, for any $t > 0$.
Since $a,b,c > 0$, we can assume, by an appropriate scaling, that$\;a+b+c=1$.
With that assumption, it suffices to prove the inequality
$$\frac{3a-1}{2(1-a)^2}+\frac{3b-1}{2(1-b)^2}+\frac{3c-1}{2(1-c)^2}\ge 0$$
for $a,b,c > 0\;$such that$\;a+b+c=1$.
Letting $f{\,:\,}(0,1)\to\mathbb{R}$ be given by
$$f(x)=\frac{3x-1}{2(1-x)^2}$$
we get
$$f''(x)=\frac{3(1+x)}{(1-x)^4}$$
hence, since $f''(x) > 0$ for all $x\in (0,1)$, it follows that $f\;$is convex.
Then for $a,b,c > 0\;$such that$\;a+b+c=1$, we get
\begin{align*}
&\frac{3a-1}{2(1-a)^2}+\frac{3b-1}{2(1-b)^2}+\frac{3c-1}{2(1-c)^2}\\[4pt]
=\;&f(a)+f(b)+f(c)\\[4pt]
\ge\;&3f\left(\frac{a+b+c}{3}\right)\\[4pt]
=\;&3f\left({\small{\frac{1}{3}}}\right)\\[4pt]
=\;&0
\end{align*}
|
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|
Find $n \in \mathbb{Z}$ that maximizes $ (n+10)/(n^3+10)$ $(n+10)/(n^3+10)=(n+10)/(n^3+1000-990)$
I have thought about something like this. But am not sure about how to proceed with this. So I would like some help.
|
You have that for $n=1$ that $11/11=1$ so you want to find $$\frac{n+10}{n^3+10}> 1$$
Otherwise the maximum value is $1$.
If $n^3+10>0$ we can multiply both sides of inequality to get
$$n+10> n^3+10\\n> n^3\\0> n(n-1)(n+1)$$
This happens when $n< -1$ or $0< n< 1$ The only integer satisfying $n^3+10>0$ and $n< -1$ is $-2$ and there are no integers satisfying $0<n<1$. For $n=-2$ we get $8/2=4$ so it's greater.
Now if $n^3+10<0$ then when we multiply we must also flip signs so
$$n+10<n^3+10\\n(n-1)(n+1)>0 $$
This happens when $-1<n<0$ or $n>1$ none of those satisfy $n^3+10<0$ hence for $n=-2$ is the maximum.
|
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|
Evaluate the definite integral $\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx$ Writing:
Integrate[ArcTan[(a Cos[x] + b Sin[x])^2], {x, 0, 2 Pi}, Assumptions -> a^2 + b^2 > 0]
$$\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx,$$
where $a $ and $b $ are real numbers.
I get:
2 Pi ArcTan[Sqrt[1/2 (-1 + Sqrt[1 + (a^2 + b^2)^2])]]
$$2\pi\arctan\sqrt{\frac{\sqrt{1 + (a^2 + b^2)^2}-1}2} $$
How to derive this result on paper?
|
*
*We have $a\cos x+b\sin x=r\cos(x-\phi)$, where $$r=\sqrt{a^2+b^2},\quad\cos\phi=a/r,\quad\sin\phi=b/r,$$ and we can simply replace $x-\phi$ by $x$ in the integrand (because of its $2\pi$-periodicity). Denoting $c=(a^2+b^2)/2$, we see that the given integral is equal to $$\int_{0}^{2\pi}\arctan(2c\cos^2 x)\,dx=\int_{0}^{2\pi}\arctan\big(c(1+\cos x)\big)\,dx.$$
*Recall that, for $d\in\mathbb{C}$ such that $|d|<1$, $$\int_{0}^{2\pi}\ln(1-2d\cos x+d^2)\,dx=0$$ (assuming the principal branch taken). This can be seen, after $$1-2d\cos x+d^2=(1-de^{ix})(1-de^{-ix}),$$ as an application of the Cauchy integral theorem. (Alternatively, one can use the above and the power series for $\ln(1+z)$, or even just split $\int_{0}^{2\pi}=\int_{0}^{\pi}+\int_{\pi}^{2\pi}$ and substitute $x=y+\pi$ in the second integral, to get $I(d)=I(d^2)/2$ from which $I(d)=0$ follows easily.) This implies $$\int_{0}^{2\pi}\ln(1+d\cos x)\,dx=2\pi\ln\frac{1+\sqrt{1-d^2}}{2}.$$
*Write $$\arctan\big(c(1+\cos x)\big)=\frac{1}{2i}\ln\frac{1+ic}{1-ic}\frac{1+d\cos x}{1+\bar{d}\cos x},\qquad d=\frac{ic}{1+ic}$$ (where $\bar{d}$ is complex conjugate to $d$); the integral then equals $$2\pi\arg(1+ic+\sqrt{1+2ic})=2\pi\arctan\frac{c+v}{1+u}=2\pi\arctan v,$$ where $\sqrt{1+2ic}=u+iv$, and we use $u=c/v$. This is the answer.
|
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|
What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, because I went and drew this graph. The correct result is $x\in (-\infty,\frac{4}{3}]$.
I'm totally confused on how to solve irrational inequalities now, because the official solving in my textbook looks like this:
$\sqrt{x^2+1}> 2x-1$. This inequality is fulfilled , if the right side is negative, therefore $x<\frac{1}{2}$. If $x\geq\frac{1}{2}$, the right side is positive or equals to $0$ and we get $0> 3x^2-4x$ which is true for $x\in (0,\frac{4}{3}]$. Now with previous condition $x\geq\frac{1}{2}$ we get the solution $[\frac{1}{2},\frac{4}{3})$.The complete solution set is $x\in (-\infty,\frac{4}{3}]$.
I never solved this type of inequality in this way, because it's messy- Why would I look at the conditions $x<\frac{1}{2}$ and $x\geq\frac{1}{2}$ when I can immediately tell where irrational part is defined and where not?
In the end, this textbook solution process only confused me. Could anyone please explain it why the correct solution is $(-\infty,\frac{4}{3}]$ or more concretely: Where did I miss the part of the solution $(-\infty,0]$?
|
First of all, we should check for $x \ge \frac{1}{2}$ and $x < \frac{1}{2}$ because in this step:
$$\sqrt{x^2+1}> 2x-1$$
we are squaring both sides. So sign of $2x-1$ may change the direction of inequality (precisely when $|2x-1| > \sqrt{x^2+1}$).
Now, for the case where $x < \frac{1}{2}$, we also have a solution for
$$x^2+1 <4x^2-4x+1 \implies 3x^2-4x > 0 \implies x < 0 \lor x > \frac{4}{3}$$
but $x > \frac{4}{3}$ doesn't satisfy our first assumption $x < \frac{1}{2}$. Therefore, we get a solution $x < 0$ here.
|
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|
Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$ The sum of series is
$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$
I used to solve this problem as its $n$th term
$${1\over (2n+1)(2n+3)}$$
Now how can I proceed??
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This is a telescoping series:
$$
\frac{1}{(2n+1)(2n+3)}=\frac{1}{2}\frac{(2n+3)-(2n+1)}{(2n+1)(2n+3)}=\frac{1}{2}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)
$$
So:
$$
\sum_{n=1}^N\frac{1}{(2n+1)(2n+3)}=\frac{1}{2}\sum_{n=0}^N\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)=\frac{1}{2}\left(\frac{1}{2\cdot 1+1}-\frac{1}{2N+3}\right)
$$
In the limit, we get:
$$
\sum_{n=1}^\infty\frac{1}{(2n+1)(2n+3)}=\frac{1}{6}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\
&=\frac{n+2}{(n+1)^2} \\
&=\frac{n+2}{n^2 + 2n + 1} \\
&<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\
&=\frac{n+2}{n(n+2)}\\
&=\frac{1}{n} \end{align} $$
I am just wondering if there is a simpler way of doing this.
|
Instead of comparing $\frac{1}{(n+1)^2} + \frac{1}{n+1}$ and $\frac{1}{n}$, we can compare $\frac{1}{(n+1)^2}$ and $\frac{1}{n}-\frac{1}{n+1}$. Then, what we have is
$$\frac{1}{(n+1)^2} = \frac{1}{(n+1)(n+1)} < \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \implies \frac{1}{(n+1)^2} < \frac{1}{n}-\frac{1}{n+1}$$
$$\implies \frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$$
Alternatively, by using the same way, we could try to prove
$$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)} > 0$$
For this one, we have
$$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)} = \frac{(n+1)^2-n-n(n+1)}{n(n+1)^2} = \frac{1}{n(n+1)^2} > 0$$
since $n \ge 1$. Therefore, the result follows.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Help calculating a sum with exponential function So I have a sum to calculate:
$$\sum_{n=1}^{\infty} \frac{n-2}{n2^{n-1}+2^{n+1}}$$
So I first did this:
$$\sum_{n=1}^{\infty} \frac{n-2}{2^{n-1}(n+4)}$$
So first i thought to make this into a function of some sort and use derivation and integration to make it into something easier to calculate, but am having trouble doin so.
Is this a long shot?
$$ f(x)= \sum_{n=1}^{\infty} \frac{n-x}{x^{n-1}(n+x^2)}$$
Since $f(2)$ equals the original sum.
Is this the correct way to define and use it?
If not any help would be appreciated, thank you in advance.
|
Let's see if we can split this up:
$$\dfrac{n-2}{2^{n-1}(n+4)} = \dfrac{(n+4)-6}{2^{n-1}(n+4)} = \dfrac{1}{2^{n-1}}-\dfrac{6}{2^{n-1}(n+4)}$$
So, this sum becomes:
$$\sum_{n\ge 1} \dfrac{n-2}{2^{n-1}(n+4)} = 2-\sum_{n\ge 1} \dfrac{6}{2^{n-1}(n+4)}$$
Now, let's work with this summation:
$$\sum_{n\ge 1} \dfrac{6}{2^{n-1}(n+4)} = 192\sum_{n\ge 5} \dfrac{1}{n2^n} = 192\left(\sum_{n\ge 1} \dfrac{1}{n2^n}-\sum_{n=1}^4 \dfrac{1}{n2^n}\right)$$
This final sum is easy to calculate:
$$\sum_{n\ge 1} \dfrac{1}{n2^n} = \log 2$$
So, putting it all together, we have:
$$2-192\left(\log 2 - \dfrac{1}{2}-\dfrac{1}{8}-\dfrac{1}{24}-\dfrac{1}{64}\right) = 133-192\log 2$$
|
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|
unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?
The solution as far as I figured out is this:
first, from the division of $(x-1)^2$, I got that $f(1) = 3$
in the same way from division of $x^2$, I got $f(0) = 4.$
I can write the polynomial as follows:
$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$
$ax^2 + bx + c$ is the remainder. And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$
This leaves $a + b = -1,$ and I can't figure out how to continue; please help.
Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$
|
Adapting the Extended Euclidean Algorithm, as implemented in this answer, to polynomial division and rotating to go down the page rather than across:
$$
\begin{array}{c|cc|c}
\color{#C00}{x^2}&\color{#C00}{1}&0\\
\color{#090}{x^2-2x+1}&0&\color{#090}{1}\\
2x-1&1&-1&1\\
\color{#C90}{\frac14}&\color{#C00}{-\frac12x+\frac34}&\color{#090}{\frac12x+\frac14}&\frac12x-\frac34\\
0&4x^2-8x+4&-4x^2&8x-4
\end{array}\tag1
$$
which says
$$
\overbrace{(2x+1)}^{4\left(\frac12x+\frac14\right)}(x-1)^2+\overbrace{(-2x+3)}^{4\left(-\frac12x+\frac34\right)}x^2=\overbrace{\ \quad1\ \quad}^{4\cdot\frac14}\tag2
$$
Therefore
$$
(2x+1)(x-1)^2\equiv\left\{\begin{align}1&\pmod{x^2}\\0&\pmod{(x-1)^2}\end{align}\right.\tag3
$$
and
$$
(-2x+3)x^2\equiv\left\{\begin{align}0&\pmod{x^2}\\1&\pmod{(x-1)^2}\end{align}\right.\tag4
$$
Thus, mod $x^2(x-1)^2$, the polynomial is
$$
\begin{align}
&(2x+4)\overbrace{(2x+1)(x-1)^2}^{(3)}+(x+3)\overbrace{(-2x+3)x^2}^{(4)}\\
&=2x^4-x^3-3x^2+2x+4\\
&\equiv3x^3-5x^2+2x+4&&\bmod x^2(x-1)^2\\
&\equiv-2x^2+2x+4&&\bmod x^2(x-1)\tag5
\end{align}
$$
which we can say since $\left.x^2(x-1)\,\middle|\,x^2(x-1)^2\right.$.
|
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|
Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation I'm trying to calculate the improper Fresnel integral $\int\limits_0^\infty\sin(x^2)dx$ calculation.
It uses several substitutions. There's one substitution that is not clear for me.
I could not understand how to get the right side from the left one. What subtitution is done here?
$$\int\limits_0^\infty\frac{v^2}{1+v^4} dv = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du.$$
Fresnel integral calculation:
In the beginning put $x^2=t$ and then: $$\int\limits_0^\infty\sin(x^2) dx = \frac{1}{2}\int\limits_0^\infty\frac{\sin t}{\sqrt{t}}dt$$
Then changing variable in Euler-Poisson integral we have: $$\frac{2}{\sqrt\pi}\int_0^\infty e^{-tu^2}du =\frac{1}{\sqrt{t} }$$
The next step is to put this integral instead of $\frac{1}{\sqrt{t}}$.
$$\int\limits_0^\infty\sin(x^2)dx = \frac{1}{\sqrt\pi}\int\limits_0^\infty\sin(t)\int_0^\infty\ e^{-tu^2}dudt = \frac{1}{\sqrt\pi}\int\limits_0^\infty\int\limits_0^\infty \sin (t) e^{-tu^2}dtdu$$
And the inner integral $\int\limits_0^\infty \sin (t) e^{-tu^2}dt$ is equal to $\frac{1}{1+u^4}$.
The next calculation: $$\int\limits_0^\infty \frac{du}{1+u^4} = \int\limits_0^\infty \frac{v^2dv}{1+v^4} = \frac{1}{2}\int\limits_0^\infty\frac{1+u^2}{1+u^4} du = \frac{1}{2} \int\limits_0^\infty\frac{d(u-\frac{1}{u})}{u^2+\frac{1}{u^2}} $$
$$= \frac{1}{2} \int\limits_{-\infty}^{\infty}\frac{ds}{2+s^2}=\frac{1}{\sqrt2}\arctan\frac{s}{\sqrt2} \Big|_{-\infty}^\infty = \frac{\pi}{2\sqrt2} $$
In this calculation the Dirichle's test is needed to check the integral $\int_0^\infty\frac{\sin t}{\sqrt{t}}dt$ convergence. It's needed also to substantiate the reversing the order of integration ($dudt = dtdu$). All these integrals exist in a Lebesgue sense, and Tonelli theorem justifies reversing the order of integration.
The final result is $$\frac{1}{\sqrt\pi}\frac{\pi}{2\sqrt2}=\frac{1}{2}\sqrt\frac{\pi}{2}$$
|
There's a neat trick to evaluate the integral
$$S_n(t)=\int_0^\infty \sin(tx^n)dx.$$
First, take the Laplace transform:
$$\begin{align}
\mathcal{L}\{S_n(t)\}(s)&=\int_0^\infty e^{-st}S_n(t)dt\\
&=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dxdt\\
&=\int_0^\infty\int_0^\infty \sin(tx^n)e^{-st}dtdx\\
&=\int_0^\infty \frac{x^n}{x^{2n}+s^2}dx\tag1\\
&=s^{1/n-1}\int_0^\infty \frac{x^n}{x^{2n}+1}dx\\
&=\frac{s^{1/n-1}}{n}\int_0^\infty \frac{x^{1/n}}{x^2+1}dx\\
&=\frac{s^{1/n-1}}{n}\int_0^{\pi/2} \tan(x)^{1/n}dx\\
&=\frac{s^{1/n-1}}{n}\int_0^{\pi/2} \sin(x)^{1/n}\cos(x)^{-1/n}dx\\
&=\frac{s^{1/n-1}}{2n}\Gamma\left(\frac12(1+1/n)\right)\Gamma\left(\frac12(1-1/n)\right)\tag2\\
&=\frac{s^{1/n-1}\pi}{2n\cos\frac{\pi}{2n}}\tag3\\
&=\frac{\pi \sec\frac{\pi}{2n}}{2n\Gamma(1-1/n)}\mathcal{L}\{t^{-1/n}\}(s).
\end{align}$$
Thus, taking the inverse Laplace transform on both sides,
$$S_n(t)=\frac{\pi \sec\frac{\pi}{2n}}{2nt^{1/n}\Gamma(1-1/n)}.$$
Choose $n=2$ and $t=1$ to get your integral:
$$S_2(1)=\frac12\sqrt{\frac\pi2}\ .$$
Explanation:
$(1)$: for real $q$ and $s$,
$$\begin{align}
\int_0^\infty \sin(qt)e^{-st}dt&= \text{Im}\int_0^\infty e^{iqt}e^{-st}dt\\
&=\text{Im}\int_0^\infty e^{-(s-iq)t}dt\\
&=\text{Im}\left[\frac{1}{s-iq}\right]\\
&=\frac{1}{s^2+q^2}\text{Im}\left[s+iq\right]\\
&=\frac{q}{s^2+q^2}
\end{align}$$
$(2)$: See here.
$(3)$: See here.
|
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|
Compute $\lim_{(x,y) \to (0,0)} \frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2}$
Does the limit $$\lim_{(x,y) \to (0,0)} \left( \frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2} \right) $$ exist?
I think it does and it's equal to $1$, but I don't know how to prove it. I tried to use Taylor expansion of $\cos(x)$ and $\cos(y)$, but it doesn't help me to compute the limit.
|
Note that, since $\cos(t)=1-\frac{t^2}{2}+O(t^4)$ as $t\to 0$, we have that
$$\frac{x^2 - 2\cos(y) + 2}{y^2 - 2\cos(x) + 2}=\frac{x^2 - 2+y^2+O(y^4) + 2}{y^2 - 2+x^2+O(x^4) + 2}=\frac{r^2+O(y^4)}{r^2+O(x^4)}=\frac{1+\frac{O(y^4)}{r^2}}{1+\frac{O(x^4)}{r^2}}$$
where $r^2=x^2+y^2$. Can you take it from here?
|
{
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"url": "https://math.stackexchange.com/questions/3278934",
"timestamp": "2023-03-29T00:00:00",
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|
Compute $ \int_0^1\frac{\ln^2(1+x)}{1+x^2}\, dx$ How to prove
$$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$
Where $ \operatorname{Li}_3(x)$ is the the trilogarithm function and $G$ is the Catalan constant.
Variant approaches are appreciated.
|
I proved here in Eq $(1)$:
\begin{align}
\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx&=\Im\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}+\overset{\text{IBP}}{\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}\ dx}\\
&=\Im\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}-\int_0^1\frac{\ln x\tan^{-1}x}{1+x}\ dx-\int_0^1\frac{\ln(1+x)\tan^{-1}x}{x}\ dx\\
\end{align}
FDP beautifully calculated here the first integral: $\displaystyle\int_0^1\frac{\ln x\tan^{-1}x}{1+x}\ dx=\frac12G\ln2-\frac{\pi^3}{64}$
and I managed here to find the second integral: $$\displaystyle\int_0^1 \frac{\ln(1+x)\tan^{-1}x}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2-3\text{Im}\operatorname{Li}_3(1+i)$$
Plugging the results of the two integrals, we get the closed form of the original integral.
|
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|
Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$ I have to simplify the following expression:
$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$
Answer: $\sqrt{a+b}-\sqrt{a-b}$
I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)
|
You can write $$\frac{(a+b)^2-(a^2-b^2)}{\sqrt{a+b}(a+b+\sqrt{a^2+b^2})}$$
|
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|
Solve $\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx$ using $\sinh x$
How to solve definite Integral
$$\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$
with hyperbolic function $\sinh(x)$?
Below is using $\tan x$ to do it. I thought it is same with this because $\sinh^2(x)+1=\cosh^2(x)$ is similar with $\tan^2(x)+1=\sec^2(x)$ ,but It isn't...
Substituting $$x=\tan t, dt=\sec^2{t} \ dx ,(0\leq t\leq\pi/2)$$
So, we get
$$
\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx =\int_{0}^{\frac{\pi}{2}}\frac{\sec{t}}{1+\tan{t}+\tan^2{t}}dt \\=\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}}{1+\sin{t}\cos{t}}dt\\=I
$$
Note that $\cos{\left(\frac{\pi}{2}-t\right)}=\sin{t}$, hence substituting $t=\frac{\pi}{2}-\theta$, we have
$\cos{\left(\frac{\pi}{2}-t\right)}=\sin{t}$, hence substituting $t=\frac{\pi}{2}-\theta$, we have
$$
\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}}{1+\sin{t}\cos{t}}dt=\int_{0}^{\frac{\pi}{2}}\frac{\sin{\theta}}{1+\sin{\theta}\cos{\theta}}d\theta=I
$$
Therefore, by substitution $ \sin{t}-\cos{t}=\xi, (\sin{t}+\cos{t})dt=d\xi $
$$
2I=\int_{0}^{\frac{\pi}{2}}\frac{\cos{t}+\sin{t}}{1+\sin{t}\cos{t}}dt\\=\int_{-1}^{1}\frac{1}{1+\frac{1-\xi^2}{2}}d\xi \\=\int_{-1}^{1}\frac{1}{3-\xi^2}d\xi \\=\frac{1}{\sqrt3}\int_{-1}^{1}\frac{1}{\sqrt3-\xi}+\frac{1}{\sqrt3+\xi}d\xi \\=\frac{1}{ \sqrt3 }\left(\ln(\sqrt3-\xi)-\ln(\sqrt3+\xi) \bigg\rvert_{-1}^{1}\right)\\=\frac{2\ln(2+\sqrt{3})}{\sqrt3}
$$
Because of the result, given integral is
$$I=\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$
And,what is actually different with this and using $\sinh(x)$?
|
\begin{align}
&\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}\overset{x\to 1/x}{dx}\\
=&\ \frac12\int_0^\infty \frac{1+x}{\sqrt{x^2+1}(x^2+x+1)} \overset{x=\sinh t}{dx}\\
=& \ \frac12\int_0^\infty \frac{1+\sinh t}{\cosh^2 t+\sinh t}dt
= \frac12 \int_0^\infty
\frac{\text{sech}\,t(\text{sech}\,t +\tanh t)}{1+\text{sech}\,t\tanh t}dt\\
=& \int_0^\infty \frac{d(\tanh t- \text{sech}\,t )}{3-(\tanh t- \text{sech}\,t)^2}= \frac{\ln(2+\sqrt{3})}{\sqrt{3}}
\end{align}
|
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|
What is the vector $x ∈ \mathbb{R^3}$ that achieves $max||x||_1$ subject to $||x||_2 = 1$? I'm trying to answer the questions "What is the vector $x ∈ \mathbb{R^3}$
that achieves $max||x||_1$ subject to $||x||_2 = 1$?" and "What is the vector x ∈ $R^3$ that achieves $max||x||_∞$ subject to $||x||_2 = 1$?
I think the first question is asking me to find a vector with three components that will have the maximum $||x||_1$ norm value where $\sqrt{x_1^2 + x_3^2 + x_2^2} = 1$, so $x_1^2 + x_3^2 + x_2^2 = 1$. I know the The L1 norm is just the sum of the absolute values of the vector's components. After trial and error I came up with $x = [\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}]$ , but also $[-\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}]$, and $[-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}]$, etc.
For my the second question, I think I need to find the vector in $\mathbb{R^3}$ that will give me the maximum value of the absolute value of the vector's components given $x_1^2 + x_3^2 + x_2^2 = 1$. I came up with $[1, 0, 0]$, $[0, 1, 0]$ , $[0, 0, 1]$, $[-1, 0, 0]$, $[0, -1, 0]$ , and $[0, 0, -1]$.
Am I correct? Is there a more formal way to figure this out and write my solution?
|
For the solution of your first question, you should also add the follwoing points: $\left(\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right), \left(\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right), \left(\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right), \left(-\sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right)$ and $\left(-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{3}}\right)$.
To intuitively guess them, you need to think that both L1 and L2 are symmetric metrics. So you should consider points $(x_1, x_2, x_3)$ where $|x_1| = |x_2| = |x_3|$. The constraint of $x_1^2 + x_2^2 + x_3^2 = 1$ would give you $|x_1| = |x_2| = |x_3| = \sqrt{\frac{1}{3}}$.
For a geomteric solution, let us first consider the case in 1st octant i.e. where $x_1 \geq 0, x_2 \geq 0, $ and $x_3 \geq 0$. Thus $L_1 = |x_1| + |x_2| + |x_3|$ is equivalent to $L_1 = x_1 + x_2 + x_3$. Now geometrically, one needs to find the point where the plane $x_1 + x_2 + x_3 = constant$ touches the sphere $x_1^2 + x_2^2 + x_3^2 = 1$ in the first octant.
If you want to formally and algebraically compute the values in the 1st oactant, then you need to write the Lagrangian function $L(x_1, x_2, x_3, \lambda) = x_1 + x_2 + x_3 + \lambda(x_1^2 + x_2^2 + x_3^2 - 1)$. Now equate all the partial derivatives to 0. This will get you $\lambda = -\frac{1}{2x_1} = -\frac{1}{2x_2} = -\frac{1}{2x_3}$. This leads to $x_1 = x_2 = x_3 = \frac{1}{\sqrt{3}}$. In the other octant (where $x_1 \le 0, x_2 \geq 0, $ and $x_3 \geq 0$) the Lagrangian function will change sign: $L(x_1, x_2, x_3, \lambda) = -x_1 + x_2 + x_3 + \lambda(x_1^2 + x_2^2 + x_3^2 - 1)$. This would yield an answer of $-x_1 = x_2 = x_3 = \frac{1}{\sqrt{3}}$.
For the second problem, let us take the first case where the $L_\infty$ norm $= Max(x_1, x_2, x_3) = x_1$. Then the Lagrangian function $L(x_1, x_2, x_3, \lambda) = x_1 + \lambda(x_1^2 + x_2^2 + x_3^2 - 1)$.
|
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|
On the Sum of Some of the Factors of $2310$
Given 3 natural numbers a,b,c such that $a\times b\times c=2310$. Find the sum of $\sum a+b+c$
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
So now since $a,b,c$ are symmetric , their sum should be the same , and I should get , $\sum a+b+c= 3 \sum a$
But computing this is becoming tedious. As when $a$ is $1$ ,then $b,c$ will have $2^5 $combinations. When $a = 2$ or $3$ or $5$ or $7$ or $11$ ,then $b,c$ will have $2^4$ combination. So, the sum. Will become $$3 \sum a = 3 \{2^5 + (2+3+5+7+11) 2 ^4 + (2×3 +2×5... +3×5 +....) 2^3 + (2×3×5 +... ) 2^2 ..... +( 2×3×5×7×11)2^0 \} $$
What next ?? , Because this sum really starts to become too big. I'm sure there must be a shorter way as it is an exam type problem supposed to be solved in a limited time. Or is there any better approach ?? Please help :)
|
Consider $f(x)=2^5(x+\dfrac{2}{2})(x+\dfrac{3}{2})(x+\dfrac{5}{2})(x+\dfrac{7}{2})(x+\dfrac{11}{2})$ Your sum equals $3f(1)=3[2^5(1+\dfrac{2}{2})(1+\dfrac{3}{2})(1+\dfrac{5}{2})(1+\dfrac{7}{2})(1+\dfrac{11}{2})]=3(2+2)(2+3)(2+5)(2+7)(2+11)$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Taylor series at infinity I'm required to make a Taylor series expansion of a function $f(x) = \arctan(x)$ at $x = +\infty$. In order to do this I introduce new variable $z = \frac{1}{x}$, so that $x \to +\infty$ is the same as $z \to +0$. Thus I can expand $f(z)$ at $z = 0$: $$f(z) = z - \frac{z^3}{3}+\frac{z^5}{5}-...$$
Then I try to reverse the substitution but this is either incomplete or incorrect: $$f\bigg(\frac{1}{x}\bigg) = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - ...$$
|
Your expansion is not correct. You should find
$$f(z) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^2} - \frac{1}{5 x^5} + \mathcal{O}\left(\left(\frac{1}{x}\right)^7\right),$$
which is a perfectly good expansion in the small parameter $\frac{1}{x}$ for $x \rightarrow \infty$. See also Taylor expansion at infinity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3292694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Commutator subgroup of Heisenberg group. Dears,
Let $H$ be Heisenberg group, a group of $3\times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $\Bbb R$ above the main diagonal.
Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.
My question is - is it also a commutator subgroup of that group?
The quotient group $H/Z(H)$ is abelian (this group is nilpotent of class two), so commutator subgroup must be inside the center. I can't imagine other subgroups being CS, but I want someone smarter to let me know.
Have a nice day.
|
$$\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&-x&xz-y\\ 0&1&-z\\ 0&0&1\end{pmatrix}$$
Therefore \begin{align}ABA^{-1}B^{-1}&=\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&s&t\\ 0&1&u\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-x&xz-y\\ 0&1&-z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-s&su-t\\ 0&1&-u\\ 0&0&1\end{pmatrix}=\\&=\begin{pmatrix}1&s+x&t+ux+y\\ 0&1&u+z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-x-s&ux+xz-y+su-t\\ 0&1&-z-u\\ 0&0&1\end{pmatrix}=\\&=\begin{pmatrix}1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\\ 0&1&0\\ 0&0&1\end{pmatrix}=\\&=\begin{pmatrix}1&0&ux-zs\\ 0&1&0\\ 0&0&1\end{pmatrix}\end{align}
It is therefore apparent that commutators are exactly the elements in the form $\begin{pmatrix}1&0&\alpha\\0&1&0\\ 0&0&1\end{pmatrix}$, which incidentally form a subgroup.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3293033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $\cot(A-B)=3$ and $\cot(A+B)=2$, then evaluate $\tan A + \frac{1}{\cot^2A}+\tan^3A+\frac{1}{\cot^4A}+\cdots$
Given:
$$\cot (A-B) = 3 \qquad \cot (A+B) = 2$$
Evaluate:
$$\tan A + \frac{1}{\cot^2 A} + \tan^3 A + \frac{1}{\cot^4 A} + \cdots $$
Attempts:
I think that the given information is used to get the value of $A$ so I can input $A$ to the problems and using the formula of infinite sum of geometric sequence. I haven't touched that step yet. I am still trying to get the value of $A$.
From
$$\cot (A-B) = \frac{\cot A \cot B-1}{\cot A + \cot B}$$
I get to $\cot A - 5\cot B = 2$.
Now, I am stuck. Where should I do next?
|
Let $\tan{A}=x$ and $\tan{B}=y$.
Thus, we obtain the following system.
$$\frac{x-y}{1+xy}=\frac{1}{3}$$ and
$$\frac{x+y}{1-xy}=\frac{1}{2}$$ or
$$3(x-y)=1+xy$$ and
$$2(x+y)=1-xy,$$ which gives
$$5x-y=2$$ or $$y=5x-2.$$
Now, since $$2(x+y)=1-xy,$$ we obtain:
$$2(x+5x-2)=1-x(5x-2)$$ or
$$12x-4=1-5x^2+2x$$ or
$$5x^2+10x-5$$ or
$$x^2+2x-1=0.$$
Can you end it now?
Actually.
For $\tan{A}=-1-\sqrt2$ your series diverges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3293938",
"timestamp": "2023-03-29T00:00:00",
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|
Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$ Prove the following for all real $x$
i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$
ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$
For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant, simpler method.
For ($i$) I tried cases like in part ($ii$) but since there are 2 variables that would lead to 4 cases.
An elegant and easy solution will be much appreciated
|
Let's denote with $\{ x \}$ the fractional part:
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}
$$
Then I would suggest that you first master the addition
$$
\eqalign{
& \left\lfloor {x + y} \right\rfloor = \left\lfloor {\left\lfloor x \right\rfloor + \left\{ x \right\} + \left\lfloor y \right\rfloor + \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \cr}
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
Thereafter you simply have
i)
$$
\left\lfloor {2x} \right\rfloor = \left\lfloor {x + x} \right\rfloor = 2\left\lfloor x \right\rfloor + \left[ {1/2 \le \left\{ x \right\}} \right]
$$
so
$$
\left\{ \matrix{
\left\lfloor {2x} \right\rfloor + \left\lfloor {2y} \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1/2 \le \left\{ x \right\}} \right] + \left[ {1/2 \le \left\{ y \right\}} \right] \hfill \cr
\left\lfloor {x + y} \right\rfloor + \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \cr} \right.
$$
and clearly
$$
\left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \le \left[ {1/2 \le \left\{ x \right\}} \right] + \left[ {1/2 \le \left\{ y \right\}} \right]
$$
ii)
$$
\eqalign{
& \left\lfloor x \right\rfloor = \left\lfloor {x/2 + x/2} \right\rfloor = \cr
& = 2\left\lfloor {x/2} \right\rfloor + \left[ {1/2 \le \left\{ {x/2} \right\}} \right] \cr}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3294982",
"timestamp": "2023-03-29T00:00:00",
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|
Compute the integral $\int\limits_0^1 \frac{3x}{\sqrt{4-3x^2}} dx $? I am struggling to compute the following equation.
\begin{equation}
\displaystyle\int_0^1 \dfrac{3x}{\sqrt{4-3x^2}} dx
\end{equation}
We are expected to use u-substitution, but I'm stuck and don't know how to proceed.
I don't know how to compute the integral of an equation in the form of:
\begin{equation}
\frac{1}{\sqrt{1-ax^{2}}}
\end{equation}
|
In general, we would like to simplify the equation by use of u-substitution.
That being said, it would be smart to choose a value for $u$ that simplifies the equation to one that we know how to solve.
There are many options for $u$ in this case.
@Allawonder pointed out that $u=4-3x^{2}$ might be a good place to start, so let's use that.
\begin{align}
\int_0^1 \frac{3x}{\sqrt{4-3x^2}} dx \\
3\int_0^1 \frac{x}{\sqrt{4-3x^2}} dx \\
u = 4-3x^2 \tag{declare} \\
du = -6x dx \\
-\frac{1}{6}du = x dx
\end{align}
Now, let's find the new values for $a$ and $b$ for this u-substitution.
\begin{align}
a = 4 \\
b = 1 \\
\end{align}
Now substitute the computed values:
\begin{align}
&=3\int_4^1 -\frac{1}{6\sqrt{u}} du \\
&=-\frac{1}{2}\int_4^1 \frac{1}{\sqrt{u}} du \\
&=\frac{1}{2}\int_1^4 \frac{1}{\sqrt{u}} du \\
\end{align}
Now we have something that looks more familiar.
\begin{align}
&=\frac{1}{2}\left[ 2\sqrt{u}\right]^{1}_{4} \\
&=\left[ \sqrt{u} \right]^{1}_{4} \\
&=\left[ \sqrt{(4)} - \sqrt{(1)}\right]
&=1
\end{align}
And there we go!
Hope this helped
|
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|
Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t X^t
\end{matrix}$$
Let $B$ be the standard basis for $\mathbb{R}^{2 \times 2}$ :
$$B =\left\{ \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
1 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
0 & 1\\
\end{pmatrix} \right\}$$
Calculate $\textsf{M}_B(\varphi)$ we come to $$\textsf{M}_B(\varphi) = \begin{pmatrix}
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
We calculate a basis for the kernel like this:
If
$$X:= \begin{pmatrix}
a & b\\
c & d\\
\end{pmatrix}$$
then $$\varphi(X) = \begin{pmatrix}
a-b & -a+b\\
c-d & -c+d\\
\end{pmatrix}+\begin{pmatrix}
a-b & c-d\\
-a+b & -c+d\\
\end{pmatrix} = \begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
Now we have to look, for what values $$\begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
by definition, the kernel of a linear transformation is $\varphi(X) = 0$, therefore our basis for $\ker(\varphi)$ should be
$$\left\{ \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \right\}$$
Now here comes the part where I'm confused. How do I calculate a basis for $\operatorname{im}(\varphi)$ ?
$\textsf{M}_B(\varphi)$ is the transformation matrix. I've read that you'd just transpose the matrix $\textsf{M}_B(\varphi)$ and row reduce to calculate a basis. I just don't get it.
The solution according to the solution it should be the basis $$ \left\{\begin{pmatrix}
0 & 1 \\
1 & -2 \end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & -1 \end{pmatrix} \right\}$$
Also little side questions. I do know about
$$\dim(A) = \dim(\operatorname{im} A) + \dim(\ker A)$$
but how exactly do you know the dimension of the Kernel/Image?
|
I think if you take your linear map $\varphi$ and write it as a matrix in terms of the basis you provided, then it becomes
$$\begin{pmatrix}2 & -2&0 &0\\ -1 &1&1&-1\\-1&1&1&-1\\ 0&0&-2&2\end{pmatrix}$$
The column space of this linear map is spanned by two vectors, namely $(2,-1,-1,0)$ and $(0,1,1,-2)$. Rewriting these as matrices will give you your answer.
This also gives you an easier method for finding the basis of the kernel, by row reducing this matrix and studying the resulting linear equations.
|
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"url": "https://math.stackexchange.com/questions/3296084",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.