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Help to use change of variables to solve the double integral I'm trying to evaluate:
$$\iint_D \left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy$$
where $D_{xy}$ is the disk $x^2+y^2\le a^2$.
The exercise is to use change of variables to solve this integral.
My solution
I chose $\varphi (r,\theta)=(ra\cos\theta,ra\sin\theta)$, where $0\le r\le 1$ and $0\le \theta\le 2\pi$ to be the change of variables.
The determinant of the Jacobian is $ra^2$ and
\begin{align*}
&\iint_{D_{xy}}\left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy \\&=\int_0^{2\pi}\int^1_0\left(\sqrt{a^2-r^2a^2}-ra\right)ra^2 drd\theta\\
&=2\pi a^3\int^1_0 \left(r\sqrt{1-r^2}-r^2 \right)dr\\
&=2\pi a^3\left(\int^1_0r\sqrt{1-r^2}dr-\int^1_0r^2dr\right)\\
&=2\pi a^3\left( \frac{1}{3}-\frac{1}{3} \right)\\
&=0
\end{align*}
I would like to know where I'm mistaken. The answer in the end of the book shows $\pi a^3/3$.
|
The answer is zero; the answer in the book is incorrect.
To verify, let's show that the following integrals (from your derivation above) are equal:
$$ A = \int_0^1 r\sqrt{1 - r^2} dr$$
$$ B = \int_0^1 r^2 dr $$
Let's change variable in $A$:
$$\quad t^2 = 1 - r^2 \Rightarrow 2tdt = -2rdr\Rightarrow rdr = -tdt.$$
Hence,
$$ A = \int_0^1 \sqrt{1 - r^2}\;rdr = - \int_1^0 t^2 dt = \int_0^1 t^2 dt$$
Thus, $A$ and $B$ are equal.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the systems of equations for real numbers $x$, $y$, $z$ Solve the system of equations in real solutions:
$x+2y+z$ $=$ $5(x+y)(y+z)$,
$y+2z+x$ $=$ $7(y+z)(z+x)$,
$z+2x+y$ $=$ $6(z+x)(x+y)$,
I substituded $a=x+y$, $b=y+z$ and $c=z+x$ and got:
$5ab-a-b$ $=$ $0$,
$7bc-b-c$ $=$ $0$,
$6ac-a-c$ $=$ $0$, and got to the conclusion where all equals $0$, but can't prove it
Thank you for your help!
|
Considering $$5ab-a-b=0\tag 1$$ $$7bc-b-c=0\tag 2$$ $$6ac-a-c=0\tag 3$$ extract $b=\frac{a}{5 a-1}$ from $(1)$ (assuming $a\neq \frac 15$), $c=\frac{a}{6 a-1}$ from $(3)$ (assuming $a\neq \frac 16$). Replace in $(2)$ and simplify to get $$\frac{2 (1-2 a) a}{30 a^2-11 a+1}=0$$
Lord Shark's solution is better
|
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|
A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence:
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many proofs for this one, but I think this is an elegant one. I'd like you to check it out, first because I consider it to be a nice and concise and elementary, and secondly because I'd like to make sure that there are no mistakes.
Let's get started: First of all, $a_n\geq 0 \ \forall \,n \in \mathbb{N}$. We write $a_n$ as
$$\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+2^k\left(\frac{\log\left(2\right)}{2}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}.$$
It is obvious that $\left(\frac{\log\left(n\right)}{n}\right)^k \to 0$. Let $\varepsilon >0$. Then $\exists\, n_0 \in \mathbb{N}$ such that $\forall \, n \geq n_0, \left(\frac{\log\left(n\right)}{n}\right)^k < \varepsilon$.
We have
\begin{align*}
a_n&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k+ n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+2^k+\dotsb +n^k} + \frac{n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k}\\ & \qquad \qquad+ \frac{\varepsilon\left(1^k+\dotsb +\left(n_0-1\right)^k\right)+n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k} + \frac{\varepsilon\left(1^k+2^k+\dotsb +n^k\right)}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}+\varepsilon.
\end{align*}
Now, by taking the limsup and the liminf as $n \to\infty$, and since
$$\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}\to 0,$$ we have
$$0 \leq \limsup\left(a_n\right) \leq \varepsilon \quad\text{and}\quad 0\leq \liminf\left(a_n\right) \leq \varepsilon.$$ But $\varepsilon$ was arbitrarily small, so
$$\liminf\left(a_n\right)=\limsup\left(a_n\right)=0=\lim\left(a_n\right).$$
This is more of a discussion and not so much of a question :)
|
Noting
\begin{eqnarray}
0&<&\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}=\frac{\frac{1}{n}\sum_{i=1}^n(\frac{\log i}{n})^k}{\frac1n\sum_{i=1}^n(\frac{i}{n})^k}\\
&\le& \frac{\frac{1}{n}\sum_{i=1}^n(\frac{\log n}{n})^k}{\frac1n\sum_{i=1}^n(\frac{i}{n})^k}=\frac{(\frac{\log n}{n})^k}{\frac1n\sum_{i=1}(\frac{i}{n})^k},
\end{eqnarray}
and
$$ \lim_{n\to\infty}\frac{\log n}{n}=0, \lim_{n\to\infty}\frac1n\sum_{i=1}^n(\frac{i}{n})^k=\int_0^1x^kdx=\frac1{k+1},$$
hence one has
$$ \lim_{n\to\infty}\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}=0. $$
|
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|
Finding the exact value of a radical How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
|
Wikipedia describes a simple method to write
$$
\sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e}
$$
with
$$
d=\frac{a + \sqrt {a^2-b^2c}}{2},
\qquad
e=\frac{a - \sqrt {a^2-b^2c}}{2}
$$
This works iff $a^2 - b^2c$ is a square.
For $\sqrt{97 +56\sqrt3}$ we have $a^2 - b^2c=1$ and so
$$
\sqrt{97 +56\sqrt3} = \sqrt{49}+\sqrt{48} = 7 + 4 \sqrt 3
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$ is a closed and bounded set.
Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.
The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\textbf{x}\in\mathbb R^n: f(\textbf{x})=0\}$ is closed. I would need to have a hint for the boundedness of $V$. Thanks!
|
Note that $(a-b)^2 \ge 0$, so $ab \le \frac{a^2+b^2}{2}$. Thus, since $x^2 - xy + y^2 - 1 = 0$ we have
$x^2 + y^2 = 1 + xy \le 1 + \frac{x^2+y^2}{2}$
and so
$\frac{1}{2} (x^2 + y^2) \le 1$
and finally
$x^2 + y^2 \le 2$.
|
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|
Find the condition that the diagonals of a parallelogram formed by $ax+by+c=0$. Find the condition that the.diagonals of a parallelogram formed by $ax+by+c=0$, $ax+by+c'=0$, $a'x+b'y+c=0$ and $a'x+b'y+c'=0$ are at right angles.
My Attempt:
The equation of diagonal passing through the point of intersection of $ax+by+c=0$ and $a'x+b'y+c=0$ is
$$(ax+by+c)+ K(a'x+b'y+c)=0$$
Where $K$ is any arbitrary constant.
Again, The equation of the diagonal passing through the point of intersection of $ax+by+c=0$ and $a'x+b'y+c'=0$ is
$$(ax+by+c)+L(a'x+b'y+c')=0$$
Where $L$ is any arbitrary constant.
How do I complete the rest?
|
$ax+by+c=0$ ...(1)
$a'x+b'y+c=0$ ...(2)
$ax+by+c'= 0$ ...(3)
$a'x+b'y+c'=0 $...(4)
Let P be the intersection of (1) and (2)
Equation of a line passing through P is given by
$ ax+by+c +K(a'x+b'y+c) = 0$ ...(5)
Let R be the intersection of (3) and (4) for which $ ax+by=-c'$ and $a'x+b'y=-c'$
Substituting in (5) $(c-c')+K(c-c') = 0$ or, $K = -1 $
Hence, from (5), $(a-a')x+(b-b')y=0$ ...(6).
This is the equation of the diagonal PR.
Let S be the intersection of (1) and (4)
Equation of a line passing through S is given by
$ax+by+c + L(a'x+b'y+c') = 0 $ …….(7)
Let Q be the intersection of (2) and (3) for which $ax+by = -c'$ and $a'x+b'y = -c$
Substituting these in (5) $c-c'+L(c'-c) = 0$ or, $L=1 $
Substituting in (7), $(a+a')x+(b+b')y+c+c'=0$ ...(8). This is the equation of the diagonal QS.
Now slope of PR is $$\frac{-(a-a')}{(b-b')}$$
slope of QS is $$\frac{-(a+a')}{(b+b')}$$
Since the diagonals are perpendicular,
Product of slopes=$$\frac{-(a-a')}{(b-b')}*\frac{-(a+a')}{(b+b')} = \frac{a^{2}-a'^{2}}{b^{2}-b'^{2}}=-1$$
Hence ${a^{2}+b^{2}}={a'^{2}+b'^{2}}$
|
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|
Using Ptolemy's Theorem to find length In $\triangle ABC$ we have $AB=7, AC=8, BC=9$. Point $D$ is the midpoint of the arc $BC$ of the circumcircle of $\triangle ABC$. Compute $\displaystyle\frac{AD}{BD}$, $BD$, and $CD$.
This is what I have so far but I am unsure if what I have is correct:
Since arc$BD$ = arc$CD$, $BD=CD=x$.
Using Ptolemy's Theorem in $ABDC$ we obtain $AB \cdot CD + AC \cdot BD = BC \cdot AD$. So $7 \cdot x + 8 \cdot x = 9 \cdot AD$ and $15x=9AD$.
Since $x=BD=CD$, $15BD = 9AD$ and $\displaystyle\frac{AD}{BD} = \frac{15}{9} = \frac{5}{3}$.
So this would mean that $BD=CD=3$.
|
$AD$ is the angle bisector of $\widehat{BAC}$, hence if we call $X$ the intersection of $AD$ and $BC$, we have $BX=\frac{7}{15}\cdot 9$ and $CX=\frac{8}{15}\cdot 9$ by the bisector theorem. By Stewart's theorem the length of $AX$ is given by
$$ AX^2 = \frac{7\cdot 8}{(7+8)^2}((7+8)^2-9^2) $$
hence $AX=\frac{8}{5}\sqrt{14}$ and since $AX\cdot XD=BX\cdot XC$ we have $XD=\frac{9}{10}\sqrt{14}$, from which $AD=\frac{5}{2}\sqrt{14}$. By Ptolemy's theorem it follows that $15\cdot BD = AD\cdot BC = \frac{45}{2}\sqrt{14}$, hence
$$ BD=CD=\frac{3}{2}\sqrt{14},\qquad \frac{AD}{BD}=\frac{5}{3}. $$
|
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|
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
|
$$-1<\frac{3x+1}{x-2}<1$$
Multiplying expression by $x-2$,
$-x+2<3x+1<x-2$
$-x+2<3x+1$ and $3x+1<x-2$
$1<4x$ and $2x<-3$
$\frac 14<x$ and $x<\frac{-3}{2}$
We have either $x<2$ or $x>2$.
|
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|
Simple inequality $\frac{4x+1}{2x-3}>2$
$$\frac{4x+1}{2x-3}>2$$
I have started with looking at positive/negative situations and there are 2 or that both expressions are positive or both are negative.
If both a positive we can solve for $$\frac{4x+1}{2x-3}>2$$
$${4x+1}>2(2x-3)$$
$${4x+1}>4x-6$$
$${0x}>-7$$
Or both are negative and then $${4x+1}<2(2x-3)$$
And again $$0x<-7$$
So there is not answer?
|
Let's use equivalent but simpler inequalities
$$
\frac{4x+1}{2x-3}>2 \quad \Leftrightarrow \quad
\frac{4x+1}{2x-3}-2>0 \quad \Leftrightarrow \quad
\frac{7}{2x-3}>0
$$
Therefore we must have
$$
2x-3>0 \quad \Leftrightarrow \quad x>{3\over2}.
$$
|
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|
Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime?
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
|
Hint: $(x+4)^4+4=0$ has roots $x = -4 \pm \sqrt{\pm 2i}=-4 \pm (1 \pm i)\,$, so the expression factors as: $$(x+4)^4+4=(x^2 + 6 x + 10) (x^2 + 10 x + 26)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
|
It is enough to prove the result for $x,y,z$ positive since
$$\frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3}$$
One can use the generalized AM-GM inequality: If
$$M_p = \left(\frac{x^p+y^p+z^p}{3}\right)^{\frac{1}{p}}$$ then for $p < q$, $M_p \leq M_q$ with equality holding if and only if $x=y=z$. Here, using $M_2 \geq M_1$, we get
$$\left(\frac{x^2+y^2+z^2}{3}\right)^{\frac{1}{2}} \geq \frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3} = \frac{1}{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\sin(\frac{\pi}{5})$ analytically How do I solve $\sin(\frac{\pi}{5})$ analytically? I looked up here and in the first step I have to
Show that: $\cos \left(\frac{\pi }{5}\right)-\sin
\left(\frac{\pi }{10}\right)=\frac{1}{2}$
My question is, why do I have to do that?
|
By repeated application of angle sum formulas we may get,
$$\sin (5x)=\sin^5 x+5 \cos^4 x\sin x-10 \sin^3 x \cos^2 x$$
Let $x=\frac{\pi}{5}$ and let $\sin (\frac{\pi}{5})=u$ then we have,
$$0=u^5+5(1-u^2)^2 u-10(1-u^2)u^3$$
It is safe to say $\frac{\sqrt{2}}{2}>u>0$. So that we may divide by $u$ to get.
$$0=u^4+5(1-u^2)^2-10(1-u^2)u^2$$
$$0=16u^4-20u^2+5$$
By solving this for $u^2$ first and then $u$ you get the only root in $(0, \frac{\sqrt{2}}{2})$ to be,
$$u=\frac{1}{2} \sqrt{\frac{5}{2}-\frac{\sqrt{5}}{2}}$$
|
{
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|
If $A= 2 \cdot \pi/7$ then show that $ \sec A+ \sec 2A+ \sec 4A=-4$ If $A= 2 \times \pi/7$ then how to show,
$$\sec A+ \sec 2A+ \sec 4A=-4$$ I have tried using formula for $\cos 2A$ but I failed.
|
$$\sec\frac{2\pi}{7}+\sec\frac{4\pi}{7}+\sec\frac{8\pi}{7}=\frac{1}{\cos\frac{2\pi}{7}}+\frac{1}{\cos\frac{4\pi}{7}}+\frac{1}{\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{8\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}}{2\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=\frac{2\sin\frac{\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)\cdot8\cos\frac{\pi}{7}}{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\left(\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}\right)\cdot8\cos\frac{\pi}{7}}{\sin\frac{16\pi}{7}}=$$
$$=\frac{-\sin\frac{\pi}{7}\cdot8\cos\frac{\pi}{7}}{\sin\frac{2\pi}{7}}=-4.$$
Done!
|
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|
Given: $\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$, find $f(x)$.
$$\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$$
On differentiating, I get,
$$f(x)\sin x\cos x = {f^\prime(x)\over f(x)}{1\over 2( b^2 - a^2)}$$
$$\sin 2x (b^2 - a^2) = {f^\prime( x)\over (f(x))^2} $$
On integrating,
$${-1\over f(x)} = {-(b^2 - a^2)\cos 2x\over 2} \implies f(x) = { 2\over(b^2 - a^2)\cos 2x}$$
The answer given is $\displaystyle f(x) = {1\over a^2 \sin^2 x + b^2 \cos^2 x}$.
I am unable to get the given result, the closest I got is, $$f(x) = {2\over b^2 \cos^2x -b^2\sin^2 x- a^2\cos^2x+ a^2\sin^2 x}$$.
How to simplify further to get the given answer ?
Related but not duplicate.
|
On differentiating you get indeed
$$
f(x)\sin x\cos x=\frac{f'(x)}{f(x)}\frac{1}{2(b^2-a^2)}
$$
so the differential equation
$$
\frac{f'(x)}{f(x)^2}=(b^2-a^2)\sin2x
$$
Integrating it you get
$$
-\frac{1}{f(x)}=-\frac{1}{2}(b^2-a^2)\cos2x+c
$$
hence
$$
f(x)=\frac{2}{(b^2-a^2)\cos2x-2c}
$$
You can expand $\cos2x=\cos^2x-\sin^2x$ and $2c=2c\cos^2x+2c\sin^2x$, so
$$
(b^2-a^2)\cos2x-2c=
(b^2-a^2-2c)\cos^2x+(a^2-b^2-2c)\sin^2x
$$
We can try for
$$
\begin{cases}
a^2-b^2-2c=2a^2 \\[4px]
b^2-a^2-2c=2b^2
\end{cases}
$$
which is valid for $-2c=a^2+b^2$, but I see no reason for choosing this particular solution. The only limitation is that
$$
(b^2-a^2)\cos2x-2c>0
$$
as far as I can see.
Is there any other condition in your problem?
|
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|
Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}-\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
But by symmetry $$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
so
$$2S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}$$ and by Cauchy Scwartz inequality
$$2S \le \sqrt{4-x+4-y+4-z}\times \sqrt{3}$$ so
$$2S \le \sqrt{24}$$
so
$$S \le \sqrt{6}$$
Is this approach correct?
|
No, this approach is not correct, since
$$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
does not necessarily hold. For example, consider $x=\frac12$, $y = \frac32$ and $z=2$, then $S=\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \right) \approx 2.420$, but $\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right) \approx 2.445$.
Also, if this were correct, you would need to provide $x, y$ and $z$ such that $S = \sqrt{6}$.
|
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|
A combinatoric problem
The question is to find which of the following option is correct regarding $$\left(\frac{\ \ 2^{10}}{11}\right)^{11}$$
$A)$ strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}$
$B)$ strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2$ but strictly smaller than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2 \binom{10}{5}$
$C)$ less than or equal to $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2$
$D)$ equal to $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}$
This question came in I.S.I. B.Stat 2013 Entrance exam in which calculators were not allowed.I tried using some combaratorial argument but failed.Any help shall be highly appreciated.Thanks.
|
Is this AM-GM? We have
$$\binom{10}1^2\binom{10}2^2\binom{10}3^2\binom{10}4^2\binom{10}5
=\prod_{k=0}^{10}\binom{10}k<\left(\frac1{11}\sum_{k=0}^{10}
\binom{10}k\right)^{11}=\left(\frac{2^{10}}{11}\right)^{11}.
$$
|
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|
Integral of exponential quadratic and linear Is there any closed form of the integral below?
$$
y = \int_0^\infty \exp\left[-\frac{(\mathbf{b}-\mathbf{Ax})^2}{2\sigma^2}-\lambda \sum_{i=0}^{N-1} x_i\right]\ \mathrm{d}\mathbf{x}
$$
where $\mathbf{x}$ is an $N\times1$ vector, $\mathbf{A}$ is an $M\times N$ matrix, and $\mathbf{b}$ is an $M\times 1$ vector. If $\mathbf{A}$ is a diagonal square matrix, I can solve it by separating the integral for each $x_i$. However, for general matrix $\mathbf{A}$, I have no idea how to solve it. Any ideas will be appreciated!
|
With a vector $\mathbf{1}$ with all components equal to 1, write is as
$$
y = \int_0^\infty \exp\left[-\frac{(\mathbf{b}-\mathbf{Ax})^2}{2\sigma^2}-\lambda \mathbf{x}\mathbf{1}\right]\ \mathrm{d}\mathbf{x}
$$
and further, for $M = N$,
$$
y = \int_0^\infty \exp\left[-\frac{1}{2\sigma^2}(\mathbf{b}^2-2 \mathbf{bAx} + \mathbf{xA^2x}+ 2 \sigma^2\lambda \mathbf{x}\mathbf{1})\right]\ \mathrm{d}\mathbf{x}\\
= \int_0^\infty \exp-\frac{1}{2\sigma^2}\left[(\mathbf{Ax} -(\mathbf{b} - \sigma^2\lambda \mathbf{A^{-1}}\mathbf{1}))^2 - (\mathbf{b} - \sigma^2\lambda \mathbf{A^{-1}}\mathbf{1}))^2 + \mathbf{b}^2\right]\ \mathrm{d}\mathbf{x}\\
= \exp-\frac{1}{2\sigma^2} (- (\mathbf{b} - \sigma^2\lambda \mathbf{A^{-1}}\mathbf{1}))^2 + \mathbf{b}^2) \int_{-\mathbf{A^{-1}}(\mathbf{b} - \sigma^2\lambda \mathbf{A^{-1}}\mathbf{1})}^\infty \exp-\frac{1}{2\sigma^2}(\mathbf{Ax} )^2 \mathrm{d}\mathbf{x}\\
$$
where the last integral is standard but however will leave you some work if $\mathbf{A}$ is not diagonal. It would be easier here to have the integrals $ \int_{-\infty}^\infty$.
EDIT:
Case $M \neq N$.
In this case there is no matrix $A^{-1}$.
The main task to do is to perform a completion of the square of the term already derived above,
$$
\mathbf{b}^2-2 \mathbf{bAx} + \mathbf{xA^2x}+ 2 \sigma^2\lambda \mathbf{x}\mathbf{1}
$$
When the inverse $\mathbf{A^{-1}}$ exists, the completion is possible by setting $\mathbf{y} = \mathbf{Ax}$ which inverts to $\mathbf{x} = \mathbf{A^{-1} y}$ and then continue as laid out above, where the term $ \mathbf{x}\mathbf{1}$ turns into $ \mathbf{1} \mathbf{A^{-1}} \mathbf{A}\mathbf{x} = \mathbf{1} \mathbf{A^{-1}}\mathbf{y}$.
Now, when the inverse $\mathbf{A^{-1}}$ does not exist, we can proceed along the lines of completions of the square, but in a modified way. We have the square term (with transposed matrices) $\mathbf{x^T A^T A x}$. Regardless of $M$, $\mathbf{ A^T A}$ is an $N \times N$ matrix which is symmetric. Therefore it can be decomposed into a product $\mathbf{ B^T B}$ where $\mathbf{B}$ will be a $N \times N$ matrix. Formally, write a matrix $U$ composed from eigenvectors and, if necessary, generalized eigenvectors of $\mathbf{ A^T A}$, which span the $N$-dimensional space. A diagonal matrix $\mathbf{\Lambda}$ will contain the eigenvalues and, if necessary, generalized eigenvalues of $\mathbf{ A^T A}$. Then $\mathbf{B} = \mathbf{\sqrt \Lambda U}$ where the square root indicates that the square root is taken of all elements on the diagonal. We have that $\mathbf{A^T A}$ = $\mathbf{B^T B}$.
With these considerations, $\mathbf{B}$ is invertible to $\mathbf{B^{-1}} = \mathbf{ U^{-1} \sqrt \Lambda^{-1}}$ and we can write $\mathbf{x} = \mathbf{B^{-1} y}$ and further, as above, $ \mathbf{1} \mathbf{x} = \mathbf{1} \mathbf{B^{-1}}\mathbf{y}$. The general result can then be taken directly from above,
$$
y
= \exp-\frac{1}{2\sigma^2} (- (\mathbf{b} - \sigma^2\lambda \mathbf{B^{-1}}\mathbf{1}))^2 + \mathbf{b}^2) \int_{-\mathbf{B^{-1}}(\mathbf{b} - \sigma^2\lambda \mathbf{B^{-1}}\mathbf{1})}^\infty \exp-\frac{1}{2\sigma^2}(\mathbf{Ax} )^2 \mathrm{d}\mathbf{x}\\
$$
where the square in the exponent can be written both ways, $\mathbf{xA^2x} = \mathbf{xB^2x}$.
P.S.: I believe that $\mathbf{B^{-1}}$ cannot be expressed as the Pseudoinverse.
|
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|
For which primes $p$ are there integers $a,b,q$ so that $pq^2=a^4+4b^4$? There is a right triangle with rational sides and are $p$ iff there exists a solution.
$$pq^2=a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$$
Without loss of generality, choose $a$ odd and $a,b$ coprime, then $a^2+2b^2+2ab$ and $a^2+2b^2-2ab$ are also coprime and $p\equiv 1 \pmod 4$. Divide into two cases:
*
*$a^2+2b^2+2ab=pk_1^2$, $a^2+2b^2-2ab=k_2^2$
*$a^2+2b^2+2ab=k_1^2$, $a^2+2b^2-2ab=pk_2^2$
These are the same under the transformation $a\to-a$, so I can ignore the second. Changing $a\mapsto a+b$ the system becomes:
$$a^2+b^2=pk_1^2$$
$$(a-2b)^2+b^2=k_2^2$$
There is no solution for $p=17$, and none for $p=29,61,73,89$ with $a,b<10^5$.
|
Unless I am getting really stupid in my old age, a rational triangle with area $p$ means that $p$ will be a Congruent Number. The first prime examples are $p=5$ and $p=7$. But a solution of $pq^2=a^4+4b^4$ must have $p \equiv 1 (\bmod 4)$, as the OP states. Thus the "iff" statement in the first sentence must be wrong.
|
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|
Find the radius of convergence of $\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$ Find the radius of convergence of the following:
$$\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$$
My attempt:
I used Ratio Test and managed to get until
$$\lim_{n\to\infty} \bigg|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \left(\frac{n}{n+1}\right)^{3} (x-2)\bigg|$$
I need to use L'Hopital Rule to get the answer.
|
L'Hospital's rule is not necessary here, one may just write, as $n \to \infty$,
$$
\left|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \cdot\left(\frac{n}{n+1}\right)^{3} (x-2)\right|=5\left|\frac{1 + \frac{(-1)^{n+1}}{5^{n+1}}}{1 + \frac{(-1)^{n}}{5^{n}}} \cdot\frac{1}{\left(1+\frac1n\right)^{3}} \right|\cdot|x-2| \to \color{red}{5|x-2|}.
$$
|
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|
Evaluate limit of sums Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + \sqrt{k^2+n}}$$
I saw this as a Riemann Sum and tried to rewrite as
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(\frac{k}{n})^2 + \frac{1}{n}}}$$
but I can't say this is equal to $$\int_0^1 \frac{1}{1 + \sqrt{x^2 + 1}}$$ because there would be a too big error.
The answer should be $\ln 2$.
|
A general approach for such "almost" Riemann sums is to evaluate as a double limit which, in this case, can be justified by uniform convergence of the inner limit:
$$\begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/n}} &= \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}}\\ &= \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + (k/n)} \\ &= \int_0^1 \frac{dx}{1 +x } \\ &= \log 2 \end{align}$$
The relevant theorem is if $x_{nm} \to y_n$ uniformly as $m \to \infty$ and $x_{nm}$ converges as $n \to \infty$, then the double limit exists and
$$\lim_{n \to \infty} x_{nn} = \lim_{n \to \infty} \lim_{m \to \infty}x_{nm} = \lim_{m \to \infty} \lim_{n \to \infty}x_{nm}$$
To show uniform convergence in this case, note that
$$\begin{align}\left|\frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}} - \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} \right| &\leqslant \frac{1}{n} \sum_{k=1}^n \frac{|1 + k/n - (1 + \sqrt{(k/n)^2 + 1/m})|}{|1+ k/n||1 + \sqrt{(k/n)^2 + 1/m}|} \\ &\leqslant \frac{1}{n} \sum_{k=1}^n (\sqrt{(k/n)^2 + 1/m}-k/n) \\ &=\frac{1}{n} \sum_{k=1}^n \frac{1/m}{\sqrt{(k/n)^2 + 1/m} + k/n}\\ &\leqslant \frac{1}{n} \sum_{k=1}^n \frac{1/m}{1/\sqrt{m}} \\ &= \frac{1}{\sqrt{m}}\end{align},$$
showing that
$$\lim_{m \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}} = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} $$
uniformly for all $n$.
Another approach would be to set bounds and apply the squeeze theorem.
|
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|
Minimal polynomial problems Let $V = \mathbb R^4, \phi \in$ End(V)
$$D(\phi)= \begin{pmatrix}
1 & -3 & 1 &-2 \\
2 & 1 & 1 & 1 \\
-2 & 4 & -2 & 3 \\
-2 & -2 & -1 & -2
\end{pmatrix}$$
The characteristic polynomial is $$h_\phi(x) = (x^2+1)(x+1)^2$$
After Cayley-Hamilton i know that the minimal polynomial divides the characteristic polynomial.
So let g(x) be the minimal polynomial.
Then it follows that $$g(x) = (x^2+1)(x+1)^2 \vee (x^2+1)(x+1) $$
I know that $g(\phi) = 0$ but neither of the suggested polynomials for equal $0$ for $\phi$.
Have i done something wrong?
|
I have made a mistake when calculating.
The minimal polynomial is indeed $g(x)=(x^2+1) \cdot (x+1)^2$
I still have another question.
Let $p_i^{e^i}$ be a primary factor of the characteristic polynomial.
Determining $V_i = \operatorname{Ker}(p_i^{e^i}(\phi))$, I get
$$V_1 = \left<\begin{pmatrix}1 \\ 0 \\ -1 \\ 0\end{pmatrix}
\begin{pmatrix}0 \\ 1 \\ 0 \\ -1\end{pmatrix} \right>,
V_2 = \left<\begin{pmatrix}1 \\ 0 \\ -2 \\ 0\end{pmatrix}
\begin{pmatrix}0 \\ 1 \\ 0 \\ -2\end{pmatrix}\right>
$$
If $T = V_1 \cup V_2$ then $T =
\left<\begin{pmatrix}1 \\ 0 \\ -1 \\ 0\end{pmatrix}
\begin{pmatrix}0 \\ 1 \\ 0 \\ -1\end{pmatrix}
\begin{pmatrix}1 \\ 0 \\ -2 \\ 0\end{pmatrix}
\begin{pmatrix}0 \\ 1 \\ 0 \\ -2\end{pmatrix}\right>$ since they are linearly independent.
Is the Transformation matrix $D_T(\phi) =
\begin{pmatrix}
1 &0 & 1 & 0 \\
0 & 1 & 0 &1 \\
-1 & 0 & -2 & 0 \\
0 & -1 & 0 & -2
\end{pmatrix}
$
correct ?
|
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|
Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my soliution:
$$S(x)=1+\frac{x^{4}}{4\cdot2^{4}}+\frac{x^{7}}{7\cdot2^{7}}+\frac{x^{10}}{10\cdot2^{10}}+\cdots=1+\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}=1+S_1(x)$$
$$(S_1(x))_x'=\left(\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}\right)_x'=\sum_{n=1}^\infty \frac{x^{3n}}{2^{3n+1}}=\frac{1}{x}\sum_{n=1}^\infty \left(\frac{x}{2}\right)^{3n+1}$$
Now let's take $\frac{x}{2}=y$, then
$$S_2(y)=\sum_{n=1}^\infty y^{3n+1}=y^4+y^7+y^{10}+\cdots=\frac{y^4}{1+y^3},|y|\le1$$
$$\left(S_1(y)\right)'=\frac{1}{2y}\cdot\frac{y^4}{1-y^3}=\frac{1}{2}\cdot\frac{y^3}{1-y^3}$$
$$S_1(y)=\frac{1}{2}\int\frac{y^3}{1-y^3}dy=\frac{1}{2}\int\left(-1+\frac{1}{1-y^3}\right)dy=-\frac{1}{2}y+\frac{\sqrt{3}}{6}\arctan\left(\frac{2\left(y+\frac{1}{2}\right)}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\lvert y-1\rvert+C$$
$$S_1(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|+C$$
$$S_1(0)=0, C=-\frac{\sqrt{3}\pi}{36}$$
$$S(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|-\frac{\sqrt{3}\pi}{36}+1$$
$$S(1)=\frac{3}{4}+\frac{\sqrt{3}}{3}\arctan\left(\frac{2\sqrt{3}}{3}\right)+\frac{1}{6}\ln(7)-\frac{\sqrt{3}\pi}{36}$$
By writing this for about 2 hours I deserve extra 50 points or at least good answers... Ha ha, thanks!
|
To apply a discrete Fourier transform to Taylor series of $-\log(1-x)$ is a good idea.
Since
$$ -\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n} $$
we have
$$ \sum_{n\geq 0}\frac{x^{3n+1}}{3n+1} = \int_{0}^{x}\frac{dt}{1-t^3}\\= -\frac{1}{3}\log(1-x)+\frac{1}{18} \left(-\sqrt{3} \pi +6 \sqrt{3} \arctan\left(\frac{1+2 x}{\sqrt{3}}\right)+3 \log\left(1+x+x^2\right)\right) $$
and by evaluating at $x=\frac{1}{2}$ it follows that
$$ \sum_{n\geq 0}\frac{1}{(3n+1)2^{3n+1}}=\frac{1}{2}\,\phantom{}_2 F_1\left(\frac{1}{3},1;\frac{4}{3};\frac{1}{8}\right)= \color{red}{\frac{1}{18} \left[-\pi\sqrt{3}+6 \sqrt{3} \arcsin\left(\frac{2}{\sqrt 7}\right)+3 \log(7)\right]}$$
|
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|
To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M G.M inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b^3+abc} \le \frac{1}{c(4ab-c^2)}$$ But since $a,b,c$ are positive reals
$$4abc-c^3 < 4abc$$ so
$$\frac{1}{4abc-c^3} \gt \frac{1}{4abc}$$ but i am unable to proceed here
|
The best proof is use $a^3+b^3 \geqslant ab(a+b).$ I have another proof
Because
$$(2c^2+ab)(a^3+abc+b^3)-abc\left[2(a^2+b^2+c^2)+ab+bc+ca\right]$$
$$=\frac{c(ab+bc-2ca)^2}2+\frac{c(3a^2+4ab+2ac+3c^2)(a-b)^2+b(a^2+5b^2)(c-a)^2}{4} \geqslant 0.$$
Thefore
$$\frac{abc}{a^3+abc+b^3} \leqslant \frac{2c^2+ab}{2(a^2+b^2+c^2)+ab+bc+ca},$$
so
$$\sum \frac{abc}{a^3+abc+b^3} \leqslant \sum \frac{2c^2+ab}{2(a^2+b^2+c^2)+ab+bc+ca}=1.$$
|
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|
Integral $ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$
$$ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$$
By taking $r-a=t^2$ we are getting
$$ \int_0^\infty \frac{2dt}{(a+t^2)^5}$$
Now I am stuck after this. Thanks.
|
Let $t = \sqrt{a}\tan(\theta), dt = \sqrt{a}\sec^2(\theta)$. Then your integral becomes
$$2\int \frac{\sqrt{a}\sec^2(\theta)}{(a+a\tan^2(\theta))^5} d \theta = \frac{2\sqrt a}{a^5} \int \cos^8(\theta)\, d\theta$$
$$ = \frac{2}{a^{9/2}} \int \left( \frac{1+\cos(2\theta)}{2}\right)^4 d\theta$$$$ = \frac{1}{8a^{9/2}}\int 1+4\cos(2\theta)+6\cos^2(2\theta)+4\cos^3(2\theta)+\cos^4(2\theta) d\theta$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2261708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $a,b,c$ are positive, prove that $\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \geq \frac{9}{a+b+c}$ If $a,b,c$ are positive real numbers, prove that
$$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$
|
Let $a+b+c = s$. Then we have to prove
$$\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c} \geq \dfrac{9}{2s},$$
or, equivalently,
$$\dfrac{3}{\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c}} \leq \dfrac{2s}{3}.$$
Note that the LHS is the harmonic mean of $s-a,s-b,s-c$ and the RHS is the arithmetic mean of the same numbers. This inequality is true by the AM-HM inequality.
|
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|
Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq y$, $A_{n \times n}$.
The determinant expression appears to be Dickson polynomial of second kind.
Let $D_n$ be the determinant of $A_n$. We can see that the appropriate recurrence relation is $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Base cases:
$$D_1=x+y=\frac{x^2-y^2}{x-y}$$
$$
D_2=(x+y)^2-xy=x^2+xy+y^2=\frac{x^3-y^3}{x-y}
$$
Suppose that $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Then we need to prove that $$D_{n+1}=(x+y)D_{n}-xyD_{n-1}$$
Which can be developed as:
$$
D_{n+1}=(x+y)((x+y)D_{n-1}-xyD_{n-2})-xyD_{n-1}=
$$
$$
=(x+y)^2D_{n-1}-xy(x+y)D_{n-2}-xyD_{n-1}=
$$
$$
=(x^2+xy+y^2)D_{n-1}-xy(x+y)D_{n-2}=
$$
$$
=\frac{x^3-y^3}{x-y}D_{n-1}-xy(x+y)D_{n-2}
$$
I tried doing this up to $D_{n-6}$ in order to get any insights into possible simplification but I'm pretty stuck.
|
You did the base cases $n=1,2$, then you do the inductive step using strong induction; suppose your claim is true for any $m \leq n$, then using the recursive formula, for $n \geq 3$:
\begin{align}
D_{n+1} &=(x+y)D_n-xyD_{n-1} =\\
&=(x+y)\frac{x^{n+1}-y^{n+1}}{x-y} -xy\frac{x^n-y^n}{x-y} = \\
&= \frac{1}{x-y}(x^{n+2}-xy^{n+1}+yx^{n+1}-y^{n+2}-x^{n+1}y+xy^{n+1})= \\
&= \frac{x^{n+2}-y^{n+2}}{x-y}
\end{align}
which proves the claim for $m=n+1$. By induction, the claim is true for any $n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given point $P=(x,y)$ and a line $l$, what is a general formula for the reflection of $P$ in $l$ I am doing Isometries i.e. Rotations, translations and reflections at the moment. General equations for 2 of them are "easy"
Translation: $f(x,y)=(x+a,y+b)$
Rotation about origin through the angle $\alpha$ is just $f(r,\theta)=(r,\theta+\alpha)$
However I struggle to find similiar function for a reflection in a line. For some simple cases like Reflection in line $y=x$ this is easy, $f(x,y)=(y,x)$ but I struggle to find the way for a general function $f(x)=mx+c$.
Thanks for any help
|
This is fun. If point P is given and a line $y=m x + c$, then the mirror point Q is
$$ \begin{pmatrix} x_Q \\ y_Q \end{pmatrix} =\begin{bmatrix} \frac{1-m^2}{1+m^2} & \frac{2 m}{1+m^2} \\ \frac{2 m}{1+m^2} & \frac{m^2-1}{1+m^2} \end{bmatrix} \begin{pmatrix} x_P \\ y_P \end{pmatrix} + \begin{pmatrix} -\frac{2 c m}{1+m^2} \\ \frac{2 c}{1+m^2} \end{pmatrix}$$
How
I decomposed the coordinates of point P into three vector parts
*
*From the origin to the line (perpendicular to the line)
$$ \begin{pmatrix} -\frac{c m}{1+m^2} \\ \frac{c}{1+m^2} \end{pmatrix}$$
*From there move along the line and closest to P
$$ \begin{pmatrix} \frac{m y_P+x_P}{1+m^2} \\ \frac{m ( m y_P+x_P)}{1+m^2} \end{pmatrix} $$
*From the closest point move perpendicular to P
$$ \begin{pmatrix} \frac{m (m x_P - y_P+c)}{1+m^2} \\ -\frac{m x_P-y_P+c}{1+m^2} \end{pmatrix} $$
You can confirm that the following is true
$$ \begin{pmatrix} x_P \\ y_P \end{pmatrix} = \begin{pmatrix} -\frac{c m}{1+m^2} \\ \frac{c}{1+m^2} \end{pmatrix} + \begin{pmatrix} \frac{m y_P+x_P}{1+m^2} \\ \frac{m ( m y_P+x_P)}{1+m^2} \end{pmatrix} + \begin{pmatrix} \frac{m (m x_P - y_P+c)}{1+m^2} \\ -\frac{m x_P-y_P+c}{1+m^2} \end{pmatrix} $$
To get to the mirror point Q flip the direction of the last part
$$ \begin{pmatrix} x_Q \\ y_Q \end{pmatrix} = \begin{pmatrix} -\frac{c m}{1+m^2} \\ \frac{c}{1+m^2} \end{pmatrix} + \begin{pmatrix} \frac{m y_P+x_P}{1+m^2} \\ \frac{m ( m y_P+x_P)}{1+m^2} \end{pmatrix} - \begin{pmatrix} \frac{m (m x_P - y_P+c)}{1+m^2} \\ -\frac{m x_P-y_P+c}{1+m^2} \end{pmatrix} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a, b, c$ are positive and $a+b+c=1$, prove that $8abc\le\ (1-a)(1-b)(1-c)\le\frac{8}{27}$ If $a, b, c$ are positive and $a+b+c=1$, prove that $$8abc\le\ (1-a)(1-b)(1-c)\le\dfrac{8}{27}$$
I have solved $8abc\le\ (1-a)(1-b)(1-c)$ (by expanding $(1-a)(1-b)(1-c)$)
but do not get how to show that $(1-a)(1-b)(1-c)\le\frac{8}{27}$
|
Hint: $a,b,c \ge 0$ and $a+b+c=1$ implies $a, b, c \le 1\,$ so $1-a, 1-b, 1-c \ge 0$, then by AM-GM:
$$\sqrt[3]{(1-a)(1-b)(1-c)} \le \cfrac{(1-a)+(1-b)+(1-c)}{3}$$
|
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"url": "https://math.stackexchange.com/questions/2264398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Continued Cosine Product. Is there a way to evaluate,
$$
\large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)}
$$
I asked this to one of my teachers and what he told is something like this,
Multiply and divide the last term of $(1)$ with $\boxed{\sin \frac{x}{2^{n-1}}}$
So,
$$
\large \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}} \\ \tag*{(2)}
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \sin \frac{x}{2^{n-1}}} \\
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \frac{\sin \frac{x}{2^{n-1}}}{\frac{x}{2^{n-1}}} \cdot \frac{x}{2^{n-1}}} \tag*{(3)}
$$
Now, as $n \to \infty$ , we have $x \to 0$,
Using this, $\lim$ we have,
$$
\boxed{ \lim_{x \to 0} \frac{\sin x}{x} = 1}
$$
Using this in $(3)$, we have,
$$
\large \boxed{\frac{\sin (2x)}{2x}} \tag*{(4)}
$$
All the steps sort of make sense. My doubts are,
*
*How do I do this for other trigonometric ratios?
*How does the step 2 happen?
I need help looking into it more intuitionally.
Please provide necessary reading suggestions.
Regards.
|
By the sine duplication formula $\sin(2z)=2\sin(z)\cos(z)$ we have $\cos(z)=\frac{1}{2}\cdot\frac{\sin(2z)}{\sin(z)}$.
In particular
$$ \prod_{k=0}^{n-1}\cos\left(\frac{x}{2^k}\right)=\frac{1}{2^n}\prod_{k=0}^{n-1}\frac{\sin\frac{x}{2^{k-1}}}{\sin\frac{x}{2^k}}=\frac{\sin(2x)}{2^n\sin\frac{x}{2^{n-1}}} \tag{1} $$
is a simple telescopic product, that can be written in the form
$$ \frac{\sin(2x)}{2x}\cdot\frac{x}{2^{n-1}\sin\frac{x}{2^{n-1}}}\tag{2}$$
and since $\lim_{w\to 0}\frac{\sin w}{w}=\lim_{w\to 0}\frac{w}{\sin w}=1$ it follows that
$$ \prod_{k\geq 0}\cos\left(\frac{x}{2^k}\right)=\frac{\sin(2x)}{2x}.\tag{3}$$
In general, as soon as we have a differentiable function $f(x)$ such that $f(2x)=2 g(x)\,f(x)$ and $f(0)=0, f'(0)\neq 0$ we have the identity
$$ \prod_{k\geq 0}g\left(\frac{x}{2^k}\right) = \frac{f(2x)}{2x}\cdot \frac{1}{f'(0)}\tag{4} $$
by de l'Hospital theorem.
|
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|
null space of a matrix for a matrix with a 0 column I'm not a mathematician. Math is a hobby and currently I'm learning bits and pieces of linear algebra when I have time. Confused about a null space for the following matrix
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix}
How does one find a null space for this matrix? As far as I understand I need to a such values so that the total would equal to $0$. So what I'm getting is
$$
N = \alpha \begin{bmatrix} 0 \\ 0\end{bmatrix} + 0 \begin{bmatrix} 0 \\ 2\end{bmatrix}
$$
Wouldn't a coefficient of $0$ make it a trivial solution?
|
We find the null space by inspection. (Systematic methods are demonstrated in the linked pages below)
Given
$$
\mathbf{A} =
\left(
\begin{array}{cc}
0 & 0 \\
0 & 2 \\
\end{array}
\right)
$$
we note that
$$
\mathbf{A}
\left(
\begin{array}{cc}
1 \\
0 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 \\
0 \\
\end{array}
\right)
$$
Both $\color{red}{null}$ spaces are the same for your symmetric matrix:
$$
\color{red}{\mathcal{N}\left( \mathbf{A} \right)} = \color{red}{\mathcal{N}\left( \mathbf{A}^{*} \right)}
=\text{span }
%
\left\{ \,
\color{red}{\left(
\begin{array}{cc}
1 \\
0 \\
\end{array}
\right)}
\, \right\}
%
$$
Consider the linear system
$$
\begin{align}
\mathbf{A} x & = b \\
%
\left(
\begin{array}{cc}
0 & 0 \\
0 & 2 \\
\end{array}
\right)
%
\left(
\begin{array}{c}
x_{1} \\
x_{2} \\
\end{array}
\right)
%
&=
\left(
\begin{array}{c}
b_{1} \\
b_{2} \\
\end{array}
\right)
\end{align}
$$
and specify the data vector to classify the existence and uniqueness of solutions.
Existence and uniqueness
When the data vector has the form
$$
b = \left(
\begin{array}{c}
0 \\
b_{2} \\
\end{array}
\right)
$$
where $b_{2} \ne 0$, the solution exists and is unique. In fact, the solution is
$$
x=
\color{blue}{\left(
\begin{array}{c}
0 \\
\frac{1}{2} b_{2} \\
\end{array}
\right)}
$$
Existence without uniqueness
When the data vector has the form
$$
b = \left(
\begin{array}{c}
b_{1} \\
b_{2} \\
\end{array}
\right)
$$
where $b_{1} \ne 0$, and $b_{2} \ne 0$, the solution exists and is not unique.
The least squares solution is
$$
x_{LS} =
\color{blue}{\left(
\begin{array}{c}
0 \\
\frac{1}{2} b_{2} \\
\end{array}
\right)} +
\alpha
\color{red}{\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right)}, \qquad \alpha\in \mathbb{C}
$$
No existence
When the data vector has the form
$$
b = \color{red}{\left(
\begin{array}{c}
b_{1} \\
0 \\
\end{array}
\right)}
$$
where $b_{1} \ne 0$, the solution does not exist.
Read more on MSE
Formal steps for computing null spaces:
Deriving left nullspace of matrix from EA=R,
Find base and dimension of given subspace
|
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|
Finding the coefficient in an expansion? I'm reviewing for my combinatorics final and have completely forgotten how to find the coefficient in the expansion of a polynomial. Here's an example I'm struggling with
Find the coefficient of $x^{11}$ in the expansion of
$$(x+x^2+x^3+x^4+x^5)^7(1+x+x^2+x^3+\dots)^4$$
So first I know how to rewrite it as
$$x^7(1-x^5)^7\frac1{(1-x)^{11}}$$
But I have no idea how to find the coefficients from here.
|
Hint Pulling out the factor of $x^7$ we see that this is the same as finding the coefficient of $x^4$ in the expansion of $$(1 + x + x^2 + x^3 + x^4)^7 (1 + x + x^2 + x^3 + x^4 + \cdots)^4 .$$ Since no term of degree $\geq 5$ can contribute, this is the same as the coefficient of $x^4$ in
$$(1 + x + x^2 + x^3 + x^4)^7 (1 + x + x^2 + x^3 + x^4)^4 = (1 + x + x^2 + x^3 + x^4)^{11} .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof by induction $1\cdot 3 \cdot 5\cdots (2n-1) / (2 \cdot 4\cdot 6\cdots 2n) \leq 1/ \sqrt{n+1}, \forall n \geq 1, n\in \mathbb N$ Prove by induction for
$$ \frac{\left(1\right)\left(3\right)\left(5\right)...\left(2n-1\right)}{\left(2\right)\left(4\right)\left(6\right)...\left(2n\right)}\le\frac{1}{\sqrt{n+1}}\quad \text{for all integers}\quad n \ge 1$$
My current inductive step:
$$ \frac{(2(k+1))!}{2^{2\left(k+1\right)}((k+1)!)^2} = \frac{2k+1}{2(k+1)} \cdot\frac{(2k)!}{2^{2k}\left(k!\right)^2}$$
I'm not sure how to proceed to show
$$ \frac{2k+1}{2(k+1)} \cdot \frac{(2k)!}{2^{2k}\left(k!\right)^2} \le \frac{1}{\sqrt{k+2}}$$
|
There's no need to use factorials here, you can do it with basic arithmetic. The base case is
$$\frac{1}{2} \leq \frac{1}{\sqrt{2}},$$
which you can square both sides to find that it's true.
Then we note that if $f(n)$ is the fraction on the left of the inequality, then $f(n+1) = f(n)\frac{2n+1}{2n+2}$.
So we assume for induction that $f(n) \leq \frac{1}{\sqrt{n+1}}$, then:
$$f(n) \leq \frac{1}{\sqrt{n+1}}$$
$$f(n)\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{n+1}}\frac{2n+1}{2n+2}$$
$$f(n+1) \leq \frac{2n+1}{\sqrt{n+1}(2n+2)}$$
Now we want to know if
$$\frac{2n+1}{\sqrt{n+1}(2n+2)} \leq \frac{1}{\sqrt{n+2}}$$
$$(2n+1)\sqrt{n+2} \leq \sqrt{n+1}(2n+2)$$
$$(2n+1)^2(n+2) \leq (n+1)(2n+2)^2$$
$$4 n^3 + 12 n^2 + 9 n + 2 \leq 4 n^3 + 12 n^2 + 12 n + 4$$
$$9 n \leq 12 n + 2$$
Which is easy to see to be true for any positive $n$.
|
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|
Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^k \leq 5^k$ is true and I want to prove that implies $k+1$. Using the inductive hypothesis I multiply both sides by $4$ to get this:
$$4 \cdot 2^{k} + 4\cdot 4^{k} \leq 4 \cdot 5^{k}$$
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
I will use again the induction hypothesis, this time I'll multiply both side by $5$ to get:
$$5 \cdot (2^{k} + 4^{k}) \leq 5^{k+1}$$
I can say that $2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k})$ and $4 \cdot 5^{k} \leq 5^{k+1}$ so I concatenate them:
$$2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k} \leq 5^{k+1}$$
However this doesn't feel right. I'm assuming that $5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k}$ which there's no way I can be sure about. At this point I'm stuck since the whole reasoning seems wrong.
|
By Karamata for all $n \geq2$ we obtain:
$$2^n+4^n<5^n+1^n=5^n+1.$$
Thus, $2^n+4^n\leq5^n$.
Done!
|
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|
$a+b+c+d+e=79$ with constraints How many non-negative integer solutions are there to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$?
I get that for $a\ge7$ you do $79-7=72$, $\binom{72+5-1}{5-1}=\binom{76}4$. For $b\ge35$ I think it's $\binom{47}4$ and I'm not too sure what it is for $3\le c\le41$ and I also have no clue as to how to do them all at the same time.
|
Here is an answer based upon generating functions.
*
*$a\geq 7$ can be encoded as
\begin{align*}
z^7+z^8+z^9+\cdots=z^7\left(1+z+z^2+\cdots\right)=\frac{z^7}{1-z}\tag{1}
\end{align*}
*$b\leq 34$ can be encoded as
\begin{align*}
1+z+z^2+\cdots+z^{34}=\frac{1-z^{35}}{1-z}\tag{2}
\end{align*}
*$3\leq c\leq 41$ can be encoded as
\begin{align*}
z^3+z^4+\cdots+z^{41}=z^3\left(1+z+z^2+\cdots+z^{38}\right)=\frac{z^3\left(1-z^{39}\right)}{1-z}\tag{3}
\end{align*}
*$d,e\geq 0$ can be both encoded as
\begin{align*}
1+z+z^2+\cdots=\frac{1}{1-z}\tag{4}
\end{align*}
We want to find the number of non-negative integer solutions of
\begin{align*}
a+b+c+d+e=79
\end{align*}
with the constraints given above.
Denoting with $[z^n]$ the coefficient of $z^n$ we are looking for
\begin{align*}
[z^{79}]&\frac{z^7}{1-z}\cdot\frac{1-z^{35}}{1-z}\cdot \frac{z^3\left(1-z^{39}\right)}{1-z}\cdot \left(\frac{1}{1-z}\right)^2\tag{5}\\
&=[z^{79}]z^{10}\frac{(1-z^{35})(1-z^{39})}{(1-z)^5}\\
&=[z^{69}]\frac{(1-z^{35})(1-z^{39})}{(1-z)^5}\tag{6}\\
&=[z^{69}]\left(1-z^{35}-z^{39}\right)\sum_{k=0}^\infty\binom{-5}{k}(-z)^k\tag{7}\\
&=\left([z^{69}]-[z^{34}]-[z^{30}]\right)\sum_{k=0}^\infty\binom{k+4}{4}z^k\tag{8}\\
&=\binom{73}{4}-\binom{38}{4}-\binom{34}{4}\tag{9}\\
&=1088430-73815-46376\\
&=968239
\end{align*}
in accordance with the answer of @CYKwong.
Comment:
*
*In (5) we select the coefficient of $[z^{79}]$ of the product of the generating functions (1) to (4) which correspond to the valid ranges specified for $a$ to $e$.
*In (6) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (7) we multiply out the numerator and skip terms with powers greater than $69$ since they do not contribute to $[z^{69}]$. We also apply the binomial series expansion.
*In (8) we use the linearity of the coefficient of operator, apply the same rule as in (6) three times and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (9) we select the coefficients accordingly.
|
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|
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$.
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}{\sqrt[a+1]{b}}+\frac{c}{\sqrt[b+1]{c}}+\frac{a}{\sqrt[c+1]{a}}$$
but I do not yet know how to finish my proof from there (if this is helpful at all?!).
|
By AM-GM$$\sum_{cyc}\frac{1+ab}{1+a}\geq3\sqrt[3]{\prod_{cyc}\frac{1+ab}{1+a}}=3\sqrt[3]{\prod_{cyc}\frac{1+\frac{1}{c}}{1+a}}=$$
$$=3\sqrt[3]{\prod_{cyc}\frac{1+c}{1+a}}=3.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$
If $\cos^2\theta+\cos\theta = 1$, express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$.
We have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta &= \sin^6\theta(\sin^2\theta+1)+\sin^2\theta(\sin^2\theta+1)-2\\&=\sin^6\theta(2-\cos^2\theta)+\sin^2\theta(2-\cos^2\theta)-2,\end{align*} but I didn't see how to use that $\cos^2\theta+\cos\theta = 1$.
|
Note that $\cos^2\theta+\cos\theta =1$ gives $(1-\sin^2\theta)+\cos\theta = 1$, so $\cos\theta = \sin^2\theta$. Therefore, we have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2 &= \cos^4\theta+\cos^3\theta+\cos^2\theta+\cos\theta-2\\&=(1-\sin^2\theta)^2+\cos\theta(1-\sin^2\theta)+\sin^2\theta-2\\&=(1-\sin^2\theta)^2+\sin^2\theta(1-\sin^2\theta)+\sin^2\theta-2\\&=-\sin^2\theta.\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the number of non-negative integral solutions of $x+y+z≤ 20$
Find the number of non-negative integral solutions of $x+y+z≤ 20$.
Conventional method :
$ x+y+z=20$, then no. of ways= $^{22}C_2$ (using beggar's method)
$x+y+z=19$, then no. of ways= $^{21}C_2$
$x+y+z=18$, then no. of ways= $^{20}C_2$
...
...
...
...
$x+y+z=0$, then no. of ways= $^{2}C_2$
Therefore total number of ways= $^{22}C_2$ + $^{21}C_2$ +... $^{3}C_2$+ $^{2}C_2 = ^{23}C_2$
I was wondering if there is any other way to solve this question.
|
If $x+y+z \leq 20$ Then the number of solutions will be the same as solving for the number of solutions to $w + x + y + z = 20$.
One approach to compute this is to look at the coefficient of $x^{20}$ in the generating function $f(x) = (1 + x + x^2 + ... + x^{20})^{4}$.
Use the identity $\displaystyle \sum_{n=0}^{k}x^n = \frac{1-x^{k+1}}{1-x}$:
$$[x^{n}]f(x) = [x^{n}](1 + x + x^2 + ... + x^{20})^{4}$$
$$[x^{n}]f(x) = [x^{n}]\left(\dfrac{1-x^{21}}{1-x}\right)^{4}$$
$$[x^{n}]f(x) = [x^{n}]\dfrac{x^{84} - 4 x^{63} + 6 x^{42} - 4 x^{21} + 1}{(1-x)^4}$$
Use the identity $\displaystyle \frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty}\binom{n+m}{m}x^n$:
$$[x^{n}]f(x) = [x^{n}](x^{84} - 4 x^{63} + 6 x^{42} - 4 x^{21} + 1)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
$$[x^{n}]f(x) = [x^{n}]\sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+84} - 4 \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+63} + 6 \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+42} \\- 4 \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+21} + \sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
Shift indices:
$$[x^{n}]f(x) = [x^{n}]\sum_{n=84}^{\infty}\binom{n-81}{3}x^{n} - 4 \sum_{n=63}^{\infty}\binom{n-60}{3}x^{n} + 6 \sum_{n=42}^{\infty}\binom{n-39}{3}x^{n} \\- 4 \sum_{n=21}^{\infty}\binom{n-18}{3}x^{n} + \sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
Now we take the coefficient of $x^{20}$:
$$[x^{20}]f(x) = \binom{20+3}{3} = \binom{23}{3}$$
Note that most of the binomial sums dropped out because their starting indices were greater than $20$ (and therefore there were no $x^{20}$ terms present in any of them), but you could easily recompute the number of ways for any given boundary (not just $20$) and take the relevant binomial coefficients.
For example the number of ways for $w+x+y+z = 50$ where $0 \leq w, x, y, z \leq 20$ would be $[x^{50}]f(x) = 6\binom{50-39}{3}- 4 \binom{50-18}{3} + \binom{50+3}{3} = 4576$
|
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|
How can we tackle this integral $\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=-1?$ Something is wrong with this integral (in terms of splitting them out)
$$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$
My try:
Splitting the integral
$$\int_{0}^{1}{2x^2-2x\over x^3\sqrt{1-x^2}}\mathrm dx+\int_{0}^{1}{\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2\tag2$$
Note that $I_1$ and $I_2$ diverge, so how can we tackle it as a whole?
|
You have to split as
$$
I=\int_0^1\frac{x^2/2+x+\log(1-x)}{x^3\sqrt{1-x^2}}+\int_0^1\frac{3x^2/2-3x+3\log(1+x)}{x^3\sqrt{1-x^2}}=\color{red}{I_1}+\color{blue}{I_2}
$$
Now we can calculate
$$
\color{red}{I_1}=-\sum_{n=3}^{\infty}\frac{1}{n}\int_0^1\frac{x^{n-3}}{\sqrt{1-x^2}}=-\frac{\sqrt{\pi}}{2}\sum_{n=3}^{\infty}\frac{1}{n}\frac{\Gamma\left(\frac n2-1\right)}{\Gamma\left(\frac n2-\frac12\right)}\underbrace{=}_{(1)}\\
-\frac{1}{16}\sum_{n=3}^{\infty}\frac{2^n}{n}\frac{\Gamma^2\left(\frac n2-1\right)}{\Gamma\left(n-1\right)}=-\frac12\int_0^1dtt\sum_{k=1}^{\infty}(2t)^k\frac{ \Gamma^2 \left(\frac k2\right)}{k!}\underbrace{=}_{(2)}\\
\int_0^1dtt\arcsin(t)(\pi+\arcsin(t))=\color{red}{-\frac{1}{4}+\frac{3\pi^2}{16}}
$$
where we have used Legendre's duplication formula in $(1)$ and the results from this nice question in $(2)$ (the last integration is trivial after employing $t=\sin(q)$ so i spare it here).
Playing exactly the same game with $I_2$ we obtain
$$
\color{blue}{I_2}=-3\int_0^1dtt\arcsin(t)(\pi-\arcsin(t))=\color{blue}{-\frac{3}{4}-\frac{3\pi^2}{16}}
$$
or
$$
I=\color{red}{I_1}+\color{blue}{I_2}=-1
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n\left[\sqrt{4i/n}\right]$ $$
\lim_{n \to \infty}
\sum_{i = 1}^{n}{1 \over n}
\,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor\quad
\mbox{where}\ \left\lfloor\,x\,\right\rfloor\
\mbox{is the greatest integer function.}
$$
I approached the problem this way-
$$
\lim_{n \to \infty}
\sum_{i = 1}^{n}{1 \over n}
\,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor
=
\lim_{n \to \infty}\frac{1}{n}\int_{n/4}^{n}\mathrm{d}x =
\lim_{n \to \infty}\left(\frac{1}{n}\times\frac{3n}{4}\right) =
\frac{3}{4}
$$
I felt like a bad-ass doing this, only to find that the answer is actually $3$. Where did I go wrong? How do I make it right ?.
|
The Riemann integral is $$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + \frac{b-a}{n} k\right) = \int_{x=a}^b f(x) \, dx,$$ and with $a = 0$, $b = 1$,and $f(x) = \lfloor 2 \sqrt{x} \rfloor$, we obtain $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \lfloor 2 \sqrt{k/n} \rfloor = \int_{x=0}^1 \lfloor 2 \sqrt{x} \rfloor \, dx.$$ Since $2 \sqrt{x} = 1$ when $1/4 \le x < 1$, and $0$ if $0 \le x < 1/4$, this becomes $$\int_{x=1/4}^1 \, dx = \frac{3}{4}.$$ It is not possible for the given sum to equal $3$, since it is trivially bounded above by $$\frac{1}{n} \sum_{k=1}^n \left\lfloor 2\sqrt{\frac{k}{n}} \right\rfloor < \frac{1}{n} \sum_{k=1}^n 2 \sqrt{\frac{k}{n}} < \frac{1}{n} \sum_{k=1}^n 2 = \frac{2n}{n} = 2.$$
|
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|
If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $\tan{x}$'s and I don't know how to go to $t$, let alone express $\cos{x}$ in $t$.
|
In my opinion the simplest solution is as follows:
$\tan(\frac{x}{2})=t\implies\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}$;
this can be seen by constructing a triangle with opposite side of angle $\frac{x}{2}$ equal to $t$, and adjacent side equal to $1$.
Now, it is known $\cos(x)=2\cos^2(\frac{x}{2})-1$.
$\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}\implies\cos^2(\frac{x}{2})=\frac{1}{{t^2+1}}$.
Thus $\cos(x)=\frac{2}{{t^2+1}}-1=\frac{1-t^2}{1+t^2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Matrices and Divisibility Let $p$ be an odd prime number and $T_p$ be the following set of $2$ x $2$ matrices
$$T_p=\{A=\left(\begin{array}{cc} a & b\\ c & a \end{array}\right); a,b,c \in\{0,1,2,3...,p-1\}\}$$
Q.1) The no. of $A$ in $T_p$ such that $det(A)$ is not divisible by p.
(A) $2p^2$
(B) $p^3-5p$
(C) $p^3-3p$
(D) $p^3 - p^2$
Q.2) The no. of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but $det(A)$ is divisible by $p$
(A) $(p-1)(p^2-p+1)$
(B) $p^3-(p-1)^2$
(C) $(p-1)^2$
(D) $(p-1)(p^2-2)$
Q.3) The no. of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both and $det(A)$ is divisible by $p$
(A) $(p-1)^2$
(B) $2(p-1)$
(C) $(p-1)^2 +1$
(D) $2p-1$
I wa able to solve Q.3 only.
Approach:-
Considering the values of $a,b,c,$ $A$ can never be skew-symmetric.
Now $det(A) = a^2-bc$
For symmetric matrix, $b=c$
So, $det(A)=a^2-b^2=(a+b)(a-b)$
Case $I$: $a=b$
There are $p$ ways of selecting $a$ or $b$ (Select any no. in $\{1,2,3...,p-1\}$
Case $II$: $a \neq b$
$a+b$ must be a multiple of $p$ since $a-b$ will always given a no. less than $p$ according to the given set of $a,b,c$ and also $p$ is a prime no.
So there are $p-1$ ways to select $a$ and $b$.
Possible ordered pairs of $(a,b)$ $(1,p-1), (2,p-2),...(p-1,1)$
Total ways: $2p-1$
Need help for Q.1 and Q.2
|
The answers are
Q1: $ \ p^3-p^2$,
Q2: $ \ (p-1)^2$.
To obtain these answers, let us solve an auxiliary problem first:
How many combinations of $a,b,c\in[0,p-1]$ result in $\det A$ divisible by $p$?
Example $1$: $ \ p=3. \ $ We have only $p^2=9$ combinations with $\det A$ divisible by $p$, namely:
$$
a=0, \quad b=0, \quad c=0 \\
a=0, \quad b=0, \quad c=1 \\
a=0, \quad b=0, \quad c=2 \\
a=0, \quad b=1, \quad c=0 \\
a=0, \quad b=2, \quad c=0 \\
a=1, \quad b=1, \quad c=1 \\
a=1, \quad b=2, \quad c=2 \\
a=2, \quad b=1, \quad c=1 \\
a=2, \quad b=2, \quad c=2 \\
$$
Of these, $(p-1)^2=4$ combinations have $a\ne0$,
and $2p-1=5$ combinations have $a=0$. Indeed,
$$
(p-1)^2 + (2p-1) = p^2.
$$
Example $2$: $ \ p=5. \ $ We have $p^2=25$ combinations with $\det A$ divisible by $p$. Similar to the previous example, $(p-1)^2=16$ combinations have $a\ne0$,
and $2p-1=9$ combinations have $a=0$.
Here are all the combinations with $p\,|\,\det A\,$ for $p=5$:
$$
a=0, \quad b=0, \quad c=0 \\
a=0, \quad b=0, \quad c=1 \\
a=0, \quad b=0, \quad c=2 \\
a=0, \quad b=0, \quad c=3 \\
a=0, \quad b=0, \quad c=4 \\
a=0, \quad b=1, \quad c=0 \\
a=0, \quad b=2, \quad c=0 \\
a=0, \quad b=3, \quad c=0 \\
a=0, \quad b=4, \quad c=0 \\
a=1, \quad b=1, \quad c=1 \\
a=1, \quad b=2, \quad c=3 \\
a=1, \quad b=3, \quad c=2 \\
a=1, \quad b=4, \quad c=4 \\
a=2, \quad b=1, \quad c=4 \\
a=2, \quad b=2, \quad c=2 \\
a=2, \quad b=3, \quad c=3 \\
a=2, \quad b=4, \quad c=1 \\
a=3, \quad b=1, \quad c=4 \\
a=3, \quad b=2, \quad c=2 \\
a=3, \quad b=3, \quad c=3 \\
a=3, \quad b=4, \quad c=1 \\
a=4, \quad b=1, \quad c=1 \\
a=4, \quad b=2, \quad c=3 \\
a=4, \quad b=3, \quad c=2 \\
a=4, \quad b=4, \quad c=4. \\
$$
This pattern holds in the general case; there are $p^2$ combinations with
$p\,|\,\det A$, namely:
*
*Each $a\in[1,p-1]$ corresponds to
$(p-1)$ combinations with $\det A$ divisible by $p$. This gives us $(p-1)^2$
combinations with $p\,|\,\det A$ and $a\ne0$.
*In addition, there are also $2p-1$ different combinations where $p\,|\,\det A$ and $a=0$.
Now it is easy to answer Questions 1 and 2
Question 1: There are $p^3$ combinations altogether. Of these, $p^2$ combinations correspond to $\det A$ divisible by $p$. Hence there are $p^3-p^2$ combinations with $\det A$ not divisible by $p$.
Question 2: We easily see that the trace of $A$ (which is $2a$) is divisible by $p$ if and only if $a=0$. So out of $p^2$ combinations where $p\,|\,\det A$ we need to exclude the $(2p-1)$ combinations where $a=0$. This leaves us with
$$
p^2 - (2p-1) = (p-1)^2
$$
combinations where $a\ne0$ and $p\not|\,\,{\rm tr}\, A$ while $p\,|\,\det A$.
|
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|
showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$, where $f$ is a periodic arithmetical function of period $4$, with the values $f(1)=f(3)=f(4)=1$ and $f(2)=-3$. Since $\sum_{1 \le i \le 4} f(i)=0$, it is assured that this series is convergent.
|
Your series is
$$\begin{align}
S &=\sum_{k=0}^\infty \left( \frac{1}{4k+1}-\frac{3}{4k+2} + \frac{1}{4k+3} + \frac{1}{4k+4}\right)\\
&=\sum_{k=0}^\infty \int_0^1 (x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3})dx\\
&=\int_0^1 \left( \sum_{k=0}^\infty \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right) \right)dx\\
&=\int_0^1 \frac{1-3x+x^2+x^3}{1-x^4} dx\\
&=\int_0^1 \left(\frac{1}{1+x}-\frac{2x}{1+x^2}\right)dx\\
&=\log(2)-\log(2)\\
&=0\\
\end{align}
$$
and this zero relation was the motivation for OEIS sequence http://oeis.org/A176563
It can also be constructed from $2\log(2)-\log(4)$ as follows
$$\begin{align}
2\log(2)
&=2\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\
&=2\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right)\\
&=\sum_{k=0}^\infty \left(\frac{2}{4k+1}-\frac{2}{4k+2}+\frac{2}{4k+3}-\frac{2}
{4k+4}\right)\\
\log(4)&=\sum_{k=0}^\infty \left(\frac{1}{4k+1}+\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{3}{4k+4}\right)\\
2\log(2)-\log(4)&=\sum_{k=0}^\infty \left(\frac{2-1}{4k+1}-\frac{2+1}{4k+2}+\frac{2-1}{4k+3}-\frac{2-3}{4k+4}\right)\\
&=\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)\\
\end{align}$$
For the formula for $\log(4)$, see Do these series converge to logarithms?
|
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|
If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \frac{2x}{1-x^2} \right) $ where $|x| < \frac {1}{\sqrt {3}}$, then what is $y$? If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right)$ where $|x| < \dfrac {1}{\sqrt {3}}$ then find the value of $y$.
. . .
Let $\tan^{-1} y= A$
$$y=\tan A$$
$$\tan^{-1} x=B$$
$$x=\tan B$$
$$
\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right) =C
$$
$$\dfrac {2x}{1-x^2}=\tan C$$.
|
We use the identity $$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}.$$
Letting $a=\tan^{-1}(x)$ and $b=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, we have $$y=\tan(a+b)=\frac{x+\frac{2x}{1-x^2}}{1-\frac{2x^2}{1-x^2}}=\frac{3x-x^3}{1-3x^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
...
Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
|
Let $y=\frac{1}{\sqrt{1+x^2}}$ and let $\tan^{-1}y=w$
Then, $$\begin{align}
\ \cos{\tan^{-1}y} & =\cos{w} \\
\ \text{Now since $1 \ge y \gt 0$ we have $\cos{w} \gt 0$}\\
\ &= \frac{1}{\sec{w}}\\
\ &= \frac{1}{\sqrt{1+\tan^2{w}}}\\
\ &= \frac{1}{\sqrt{1+y^2}}\
\end{align}$$
And then substitute $y$.
|
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|
Prove: $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}$ Let $a,b,c$ be the lengths of the sides of triangle $ABC$ opposite $A,B,C$, respectively, and let $m_a,m_b,m_c$ be the lengths of the corresponding angle medians. Prove:
$$\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}.$$
Source: I thought about it while solving the American Mathematical Monthly's problem 11945, which asked to prove $\frac{a}{w_a}+\frac{b}{w_b}+\frac{c}{w_c}\ge 2\sqrt{3}$, where $w_a,w_b,w_c$ are respective angle bisectors.
I think it can be solved by making up (inequality-) constrained optimization problem, but I am interested in solution within elementary geometry.
|
By AM-GM $$\sum_{cyc}\frac{a}{m_a}=\frac{2a}{\sqrt{2b^2+2c^2-a^2}}=\sum_{cyc}\frac{4\sqrt3a^2}{2\sqrt{3a^2(2b^2+2c^2-a^2)}}\geq$$
$$\geq\sum_{cyc}\frac{4\sqrt3a^2}{3a^2+2b^2+2c^2-a^2}=2\sqrt3\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}=2\sqrt3.$$
Done!
|
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|
Function defined by its own integral at a given point. If $y=f(x)$ is a differentiable function such that
$$f(x)=x+ \int_0^1x^2zf(z)dz + \int_0^1xz^2f(z)dz$$ Then what is the value of $f\left(\frac{-9}{4}\right)$
The options were
(A)$\frac{-4}{9}$ (B)$\frac{4}{9}$
(C)$-1$ (D)$0$
First I tried differentiating with respect to $x$. I believe we can take out the x terms from definite integral. I could not deduce anything meaningful from it.
Hence I assumed $x=z$ , but solving required the values of $f(0)$ and $f(1)$, which is not given in the question.
How can this question be solved?
NOTE:$z$ is NOT a complex variable.
|
Let $\displaystyle a=\int_0^1zf(z)dz$ and $\displaystyle b=\int_0^1z^2f(z)dz$. Then
$$f(x)=ax^2+(b+1)x$$
We have
\begin{align*}
\int_0^1zf(z)dz&=\int_0^1[az^3+(b+1)z^2]dz\\
a&=\left[\frac{az^4}{4}+\frac{(b+1)z^3}{3}\right]_0^1\\
a&=\frac{a}{4}+\frac{b+1}{3}\\
\frac{3a}{4}&=\frac{b+1}{3}\\
b+1&=\frac{9a}{4}
\end{align*}
So
$$f(x)=ax^2+\frac{9a}{4}x$$
$$f\left(\frac{-9}{4}\right)=a\left(\frac{-9}{4}\right)^2+\frac{9a}{4}\left(\frac{-9}{4}\right)=0$$
|
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|
Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$ that is
$$\sqrt{57-x}-25=\sqrt{x+40}-10 \sqrt[4]{x+40}$$ again squaring both sides we get
$$682-x-50\sqrt{57-x}=x+40+100\sqrt{x+40}-20(x+40)^{\frac{3}{4}}$$
i got messed up here any better way and just a hint please
|
Since you found two roots you may try to argue that there are no more roots.
One way to do this is study the function $f(x)=(57-x)^{1/4}+(x+40)^{1/4}$.
Now,
we can calculate $$f''(x)= \dfrac{3}{16} \left (-\dfrac{1}{(57 - x)^{7/4}} - \dfrac{1}{(40 + x)^{7/4}} \right )<0.$$
Therefore $f$ cannot have three roots in $(-40,57)$, since that would give two roots for $f'$ contradicting the fact that $f'$ is decreasing.
|
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|
How can I find maximum and minimum modul of this complex? Let be given the complex $z$ such that $$\left |z + \dfrac{1}{z}\right |=3.$$ Find maximum and minimum modul of complex $z$.
I tried. Put $z=x+i y$. From $$\left |z + \dfrac{1}{z}\right |=3.$$ We have $z^2 + 3z|$ = 3|z|.$ Threrefore
$$1 + 2 x^2 + x^4 - 2 y^2 + 2 x^2 y^2 + y^4 - 9 (x^2 + y^2)=0.$$
From here, I cann't find max and min of the $\sqrt{x^2 + y^2}.$
|
Hint. By using samjoe's comment $z\bar{z}=|z|^2$, we have that the given equation is equivalent to
$$|z|^2+2\cos(\theta)+\frac{1}{|z|^2}=\left |z + \dfrac{1}{z}\right |^2=3^2=9, \;\;|z|^2+\frac{1}{|z|^2}=9-2\cos(\theta)$$
where $z=|z|(\cos(\theta)+i\sin(\theta))$. Note that
$$7=9-2\cos(0)\leq|z|^2+\frac{1}{|z|^2}\leq 9-2\cos(\pi)= 9+2=11.$$
Now try to show that the maximum of $|z|$ is attained when
$$|z|^2+\frac{1}{|z|^2}=11,$$
that is for $|z|=(11+3\sqrt{13})/2$.
What about the minimum?
|
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|
Evaluate the following triple summation
Evaluate $$\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \sum_{i=0}^{\infty} \frac{1}{3^i3^j3^k}$$ for $i\neq j\neq k$
My Attempt
Evaluate the summation with no restriction on $i,j,k$. Let this be $a_0$
When $i= j=k$. Let this be $a_1$.
When $i\neq j=k$, let this be $a_2$.
The required answer would be $a_0-a_1-3a_2$.
I got \begin{align}a_0&=\frac{27}{8}\\\\
a_1&=\frac{27}{26}\\ \\
a_2&=\frac{135}{208}\end{align}
And therefore, my answer is
\begin{align}a_0-a_1-3a_2&=\frac{27}{8}-\frac{27}{26}-3\times\frac{135}{208}\\\\
&=\frac{81}{208}\end{align}
Is everything correct? The method and the calculation? Also if anyone has a better method, please tell.
|
Let's rearrange the sum:
$$ S := \sum_{i,j,k \ge 0, \ i \neq j \neq k} 3^{-i} 3^{-j} 3^{-k} = \sum_{j \ge 0} \left( 3^{-j} \sum_{i \ge 0, \ i \neq j} 3^{-i} \sum_{k \ge 0, \ k \neq j} 3^{-k} \right). $$
The two inner sums are equal
$$ \sum_{i \ge 0, \ i \neq j} 3^{-i} = \left( \sum_{i \ge 0} 3^{-i} \right) - 3^j = \frac{3}{2} - 3^{-j}, $$
so we're left with
\begin{align*}
S & = \sum_{j \ge 0} \left( 3^{-j} \left(\frac{3}{2} - 3^{-j}\right)^2 \right) \\
& = \left( \frac 32 \right)^2 \sum_{j \ge 0} 3^{-j}
- 2 \cdot \frac 32 \cdot \sum_{j \ge 0} 3^{-2j}
+ \sum_{j \ge 0} 3^{-3j} \\
& = \frac 94 \cdot \frac 32 - 3 \cdot \frac 98 + \frac{27}{26} \\
& = \frac{27}{26}.
\end{align*}
If I understand your notation correctly, the sum should be equal to
$$ S = a_0 - a_1 - 2 a_2, $$
but substituting the values of $a_0,a_1,a_2$ you calculated gives a negative result. Since you haven't included your reasoning, I cannot say more.
If you're interested in $i,j,k$ pairwise distinct, this is a different problem. By the inclusion-exclusion principle, we get
\begin{align*}
\sum_{i,j,k \text{ distinct}} 3^{-i-j-k}
& = \left( \sum_{i,j,k} 3^{-i-j-k} \right)
- 3 \left( \sum_{i=j, \ k} 3^{-i-j-k} \right) +
2 \left( \sum_{i=j=k} 3^{-i-j-k} \right) \\
& = \left( \frac 32 \right)^3 - 3 \cdot \frac 32 \cdot \frac 98 + 2 \cdot \frac{27}{26} \\
& = \frac{81}{208},
\end{align*}
which happens to coincide with your result.
|
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|
Simplifying $\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1}$ I was attempting to go through a worked solution from my lecturer but I can't really understand this step:
$\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1} = \lim_{n\to \infty} \frac{1}{\sqrt[n]{4}+\sqrt[n]{2}+1}$
What was actually even done here? And why?
Any help is appreciated.
|
Hint:
$$\frac{\sqrt[n]{2} - 1}{\sqrt[n]{8} - 1}=\frac{\sqrt[n]{2} - 1}{\sqrt[n]{8} - 1}\cdot\frac{\sqrt[n]{4} + \sqrt[n]{2} + 1}{\sqrt[n]{4} + \sqrt[n]{2} + 1}=\frac{\sqrt[n]{8} - 1}{\bigg(\sqrt[n]{8} - 1\bigg)\bigg(\sqrt[n]{4} + \sqrt[n]{2} + 1\bigg)}$$
Can you take it from here?
|
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|
Functions and Sequences Problem
The function $F(k)$ is defined for positive integers as $F(1) = 1$,
$F(2) = 1$, $F(3) = -1$ and $F(2k) = F(k)$, $F(2k + 1) = F(k)$ for $k \geq
2$. Then $$F(1) + F(2) + \dotsb + F(63)$$ equals
$\begin{array}{lr}
(\text{A}) & 1 \\
(\text{B}) & -1 \\
(\text{C}) & -32 \\
(\text{D}) & 32 \\
\end{array}$
My approach: $F(4)=1$, $F(5)=1$, $F(6)=-1$, $F(7)=-1$ (i.e., all values of $F$ are either $1$ or $-1$). I tried to find a pattern for which $F(x)$ repeats after a certain integer but tried till $F(30)$ and cannot find a solution. Where am I going wrong?
|
We have $F(1)=1, F(2)=1, F(3)=-1$ and $F(2k)=F(k)$ and $F(2k+1)=F(k)$.
We have to evaluate $S=\sum_{i=1}^{63}F(i)$
Notice that $S=F(1)+F(2)+F(3) + \sum_{i=2}^{31}[F(2k) + F(2k+1)]$
$\Rightarrow S=F(1)+F(2)+F(3) + \sum_{i=2}^{31}[2F(k)]$
$\Rightarrow S=1+2\sum_{i=2}^{31}F(k)$
$\Rightarrow S=1+2[F(2)+F(3)+\sum_{i=2}^{15}[F(2k)+F(2k+1)]]$
$\Rightarrow S=1+2\sum_{i=2}^{15}[F(2k)+F(2k+1)]$
$\Rightarrow S=1+4\sum_{i=2}^{15}F(k)$
$\Rightarrow S= 1+ 4[F(2)+F(3) + \sum_{i=2}^{7}[F(2k)+ F(2k+1)]]$
$\Rightarrow S = 1+ 8\sum_{i=2}^7 F(k)$
Similarly proceeding, we get
$\Rightarrow S = 1+ 16\sum_{i=2}^3F(k)$
$\Rightarrow S =1$
|
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|
Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$ For $n \ge 5, \frac{9n^2 - 9n + 2}{4n^2-2n} > 2$
$9n^2 -9n +2 -8n^2+4n = n^2 -5n +2 \ge 5^2-25+2$
Here's the basis step:
$${{9} \choose {6}} = 84 > \frac{6^3}{3} = 72$$
Here's the inductive step:
$${{3n} \choose {2n}} = {{3(n-1)}\choose {2(n-1)}}\left(\frac{3n}{n}\right)\left(\frac{3n-1}{2n-1}\right)\left(\frac{3n-2}{2n}\right) > \left(\frac{6^{n-1}}{n-1}\right)\left(\frac{3n}{n}\right)\left(\frac{9n^2 -9n +2}{4n^2-2n}\right) > \left(\frac{6^{n-1}}{n}\right)(6) = \frac{6^n}{n}$$
|
I will show that
$0.850
\lt \dfrac{\binom{3n}{2n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
More precise bounds,
of the form
$1+O(\frac{c}{n})$,
can be easily gotten
by the method below.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{an}{bn}
&=\dfrac{(an)!}{(bn)!((a-b)n!}\\
&\sim \dfrac{\sqrt{2\pi an}(an/e)^{an}}
{(\sqrt{2\pi bn}(bn/e)^{bn})(\sqrt{2\pi n(a-b)}(((a-b)n)/e)^{(a-b)n})}\\
&= \sqrt{\dfrac{2\pi an}{2\pi bn2\pi n(a-b)}}\left(\dfrac{(an)^ae^be^{a-b}}{e^a(bn)^b((a-b)n)^{a-b}}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^an^a}{b^b(a-b)^{a-b}n^a}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n\\
&=b(n, a, b)\\
\end{array}
$
If $a=3, b=2$,
$\dfrac{a}{b(a-b)}
=\dfrac{3}{2}
$
and
$\dfrac{a^a}{b^b(a-b)^{a-b}}
=\dfrac{3^3}{2^2}
=\dfrac{27}{4}
=6\frac34
$
so
$\binom{3n}{2n}
\sim \sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n
$.
Actually
(https://en.wikipedia.org/wiki/Stirling%27s_approximation),
if
$f(n)
=\dfrac{n!}{\sqrt{2\pi n}(n/e)^n}$,
then the following inequality holds:
$$1
\lt f(n)
\lt e^{1/(12n)}
\lt \frac{e}{\sqrt{2\pi}}
= 1.0844...
$$
So,
if $u < f(n) < v$,
then
$\frac{u}{v^2}
\lt \dfrac{\binom{an}{bn}}{b(n, a, b)}
\lt \frac{v}{u^2}
$.
From the simple bounds above,
we can take
$u = 1$
and
$v = \frac{e}{\sqrt{2\pi}} $,
so the bounds are
$\dfrac{2\pi}{e^2}
=0.850...
$
and
$\dfrac{e}{\sqrt{2\pi}}
=1.0844...
$.
Therefore
$0.850
\lt \dfrac{\binom{3n}{2n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
|
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|
How to find the inverse Laplace transform of the following fraction? Please help me to find the inverse Laplace transform of the following:
$$\frac{s^2+6s+9}{(s-1)(s-2)(s-3)}$$
Do I have to divide the fractions and use partial fractions decomposition?
Like this:
$$\frac{s^2}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$
$$\frac{6s}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$
$$\frac{9}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3}$$
|
We perform partial fraction decomposition on the expression first:
\begin{align*}
\frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} &= \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3} \\
&= \frac{A(s-2)(s-3) + B(s-1)(s-3) + C(s-1)(s-2)}{(s-1)(s-2)(s-3)} \\
s^2 + 6s+9 &= A(s^2-5s+6) + B(s^2-4s+3) + C(s^2-3s+2) \\
&= (A+B+C)s^2 + (-5A-4B-3C)s + (6A+3B+2C)
\end{align*}
So we have the following system of equations, which can be solved by row reduction (the simpler way) or matrix inversion (the harder way):
\begin{align*}
\begin{pmatrix} 1 & 1 & 1 \\ -5 & -4 & -3 \\ 6 & 3 & 2 \end{pmatrix}
\begin{pmatrix} A \\ B \\ C\end{pmatrix} &= \begin{pmatrix} 1 \\ 6 \\ 9\end{pmatrix} \\
\begin{pmatrix} A \\ B \\ C\end{pmatrix} &= \begin{pmatrix} 8 \\ -25 \\ 18\end{pmatrix}
\end{align*}
So we have:
\begin{align*}
\frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} &= \frac{8}{s-1} - \frac{25}{s-2} + \frac{18}{s-3} \\
\mathcal{L}^{-1} \left\{ \frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} \right\} &= \mathcal{L}^{-1} \left\{ \frac{8}{s-1} - \frac{25}{s-2} + \frac{18}{s-3} \right\} \\
&=8 \cdot \mathcal{L}^{-1} \left\{ \frac{1}{s-1} \right\} - 25 \cdot \mathcal{L}^{-1} \left\{ \frac{1}{s-2} \right\} + 18 \cdot\mathcal{L}^{-1}\left\{ \frac{1}{s-3} \right\} \\
&= 8e^t - 25e^{2t}+18e^{3t}
\end{align*}
|
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|
Evaluate $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$ How do you evaluate $$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$$ using the identity $$\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)?$$
I assume I'll have to express $1+\frac{1}{n^2}+\frac{1}{n^4}$ as $\left(1+\frac{a}{n^2}\right)\left(1+\frac{b}{n^2}\right)$ for some $a,b\in\mathbb{C}$ so that I could use the identity given, but I can't seem to factor the above appropriately.
|
Observing that
$$
\left(1+\frac1{n^2}+\frac1{n^4}\right)
=\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right)
$$
where
$$
a^2+b^2=-1\\
a^2b^2=1
$$
find $a$ and $b$ by a direct computation.
We get
\begin{align*}
\prod_{n\ge1}\left(1+\frac1{n^2}+\frac1{n^4}\right)
=&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right)\\
=&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right)
\prod_{n\ge1}\left(1-\frac{b^2}{n^2}\right)\\
=&\frac{\sin(\pi a)}{\pi a}\frac{\sin(\pi b)}{\pi b}
\end{align*}
|
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|
Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$
Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$
My Try: By Cauchy Schwarz Inequality we have
$$\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)\le \sqrt{3}\sqrt{S}$$
$\implies$
$$\sqrt{3S} \ge 12+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
Now by $AM \ge HM $ inequality we have
$$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ $\implies$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{4}$$ hence
$$\sqrt{3S} \ge 12+\frac{3}{4}=\frac{51}{4}$$
hence
$$3S \ge \frac{2601}{16}$$
so $$S \ge \frac{867}{16}$$
is this approach correct and any better approach please share.
|
I think it's correct.
By C-S we obtain:
$$\sum_{cyc}\left(a+\frac{1}{b}\right)^2=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}\left(a+\frac{1}{b}\right)^2\geq$$
$$\geq\frac{1}{3}\left(\sum_{cyc}\left(a+\frac{1}{b}\right)\right)^2=\frac{1}{3}\left(12+\frac{1}{12}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)^2\geq$$
$$=\frac{1}{3}\left(12+\frac{1}{12}\cdot9\right)^2=\frac{867}{16}.$$
The equality occurs for $a=b=c=4$, which says that $\frac{867}{16}$ is a minimal value.
Also we can use Jensen.
Let $f(x)=x^2+\frac{1}{x^2}$.
Hence, $f''(x)=2+\frac{6}{x^4}$, which says that $f$ is a convex function.
Thus, by Jensen and by AM-GM we obtain:
$$\sum_{cyc}\left(a+\frac{1}{b}\right)^2=\sum_{cyc}\left(a^2+\frac{1}{a^2}\right)+2\sum_{cyc}\frac{a}{b}\geq$$
$$\geq3\left(\left(\frac{a+b+c}{3}\right)^2+\frac{1}{\left(\frac{a+b+c}{3}\right)^2}\right)+2\cdot3\sqrt[3]{\prod_{cyc}\frac{a}{b}}=$$
$$=3\left(16+\frac{1}{16}\right)+6=\frac{867}{16}.$$
|
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|
equation involving absolute value function has $4$ solutions
If the equation $|x^2-5|x|+k|-\lambda x+7 \lambda = 0$ has exactly $4$ solution, Then $(\lambda,k)$ is
Attempt:$x^2-5|x|+k = \lambda x-7 \lambda$
$\star$ For $x>0,$ then $x^2-5x+k=\lambda x- 7 \lambda.$
$\star$ For $x\leq 0,$ then $x^2+5x+k=\lambda x- 7 \lambda.$
could some help me how to go further,thanks
|
First, notice that if $\lambda=k=0$ then the only solutions are $0,\,-5$ and $5$.
Suppose $\lambda=0$ and $0<k<\dfrac{25}{4}$.
Then
\begin{eqnarray}
{\Big\vert}\, |x|^2-5|x|+k\, {\Big\vert}&=&0\\
{\Huge\vert}\left(|x|-\frac{5}{2}\right)^2+k-\frac{25}{4}{\Huge\vert}&=&0\\
\left(|x|-\frac{5}{2}\right)^2&=&\frac{25}{4}-k\\
|x|&=&\frac{5}{2}\pm\frac{5}{2}\sqrt{1-\frac{4k}{25}}>0
\end{eqnarray}
Thus there are only four solutions
$$ x=\pm\frac{5}{2}\left(1\pm\sqrt{1-\frac{4k}{25}}\,\right) $$
provided that $(\lambda,k)=(0,k)$ and $0<k<\dfrac{25}{4}$.
This does not exhaust the solutions. Graphing techniques indicate that if $k=0$ then there is an open interval containing $\lambda=-1$ for which there are only four real solutions. But perhaps we are including complex solutions?
Was the object to characterize all possible solutions?
|
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|
Find $xyz$, given that the value of $x^2+y^2+z^2$, $x+y+z=x^3+y^3+z^3=7$
Given that $$x^2+y^2+z^2=49$$ $$x+y+z=x^3+y^3+z^3=7$$
Find $xyz$.
My attempt,
I've used a old school way to try to solve it, but I guess it doesn't work.
I expanded $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+yz^2)+6xyz$
Since I know substitute the given information into the equation and it becomes $112=x^2y+xy^2+xz^2+yz^2+2xyz$
In another hand, I also expanded $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$7^2=49+2(xy+xz+yz)$,
So from here, I know that $xy+xz+yz=0$.
It seems that I stuck here and don't know how to proceed anymore.
How to continue from my steps? And is there another trick to solve this question? Thanks a lot.
|
Ok, so you've got $$xy+yz+zx=0$$
Now, we know that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\
\implies7-3xyz=7(49-0)\\
\implies xyz=-\dfrac{7\times48}{3}=-112$$
|
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|
solve the set of simultaneous congruences using Chinese Remainder Theorem 2x= 1 (mod 5), 3x= 9 (mod 6), 4x = 1 (mod 7), 5x = 9 (mod 11) Solve the set of simultaneous congruences using Chinese Remainder Theorem
$\begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 9 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \\
\end{cases}$
This is what I got so far:
I simplify:
$2x \equiv 1 \pmod 5$ becomes $x \equiv 3 \pmod 5$
$3x \equiv 9 \pmod 6$ has $3$ solutions since $\gcd(3,6)=3$ they are $x = 17$, $19$, or $21 \pmod 6$.
$4x \equiv 1 \pmod 7$ becomes $x \equiv 2 \pmod 7$
$5x \equiv 9 \pmod {11}$ becomes $x \equiv 4 \pmod {11}$
By CRT, and substituting the $3$ different solutions, I got:
$x \equiv 653 \pmod{2310}$, $x \equiv 1423 \pmod{2310}$ and $x \equiv 2193 \pmod {2310}$.
But, the key answer from the book is $x \equiv 653 \pmod{770}$.
I am struggling to figure out how to get $x = 653 \pmod{770}$. I need help. Thank you.
|
$$ \begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 3 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \\
\end{cases} \implies \begin{cases}
x \equiv 3 \pmod{5} \\
x \equiv 1 \pmod{2} \\
x \equiv 2 \pmod{7} \\
x \equiv 4 \pmod{11} \\
\end{cases} \implies x \equiv 653 \pmod{770} $$
|
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|
Given that $a\cos\theta+b\sin\theta+A\cos 2\theta+B\sin 2\theta \leq 1$ for all $\theta$, prove that $a^2+b^2\leq 2$ and $A^2+B^2 \leq 1$ I tried to solve this question but things got too much complicated and hence my efforts were completely futile.
Let $a,A,b,B \in \mathbb{R}$ and $$F(\theta)= 1- a\cos \theta - b\sin \theta- A\cos 2\theta- B\sin 2\theta$$
It is given that $$F(\theta) \ge 0 \;\forall\; \theta $$
and we have to prove that $\color{red}{a^2+b^2 \le 2}$ and $\color{green}{A^2+B^2 \le 1}$.
MY ATTEMPT
We need to prove that $$a\cos \theta + b\sin \theta+A\cos 2\theta+ B\sin 2\theta \le 1$$
$$\begin{align}
& = a\cos \theta + b\sin \theta+A( \cos^2 \theta- \sin^2 \theta)+ B \sin \theta \cdot \cos \theta + B \sin \theta \cdot \cos \theta \le 1 \\
& =\ cos \theta (a+A \cos \theta+ B \sin \theta)+\sin \theta(b-A \sin \theta+B \ \cos \theta) \le1 \\
\end{align}$$
We know that $-\sqrt{x^2 + y^2} \le x \cos \theta + y \sin \theta \le \sqrt{x^2 + y^2}$
$$\Rightarrow (a+A \cos \theta+ B \sin \theta)^2 + (b-A \sin \theta+B \ \cos \theta)^2 \le 1$$
After solving this equation we get
$$a^2 + b^2 + 2(A^2 +B^2)+ \cos \theta (2aA+2bB) + \sin \theta (2aB-2bA) \le1$$
Now If I again apply the same property, certainly the things are going to become more complicated and hence I think my approach is not at all right. Kindly Help me with this question.
|
We have
$$0 \leq F(\theta) + F(\theta + \pi) = 2 - 2(A \cos 2\theta + B \sin 2\theta).$$
Let $u$ be the vector $(A,B)$. Pick $\theta$ so that $v = (\cos 2\theta, \sin 2\theta)$ is a unit vector in the same direction as $u$. Then
$$\sqrt{A^2 + B^2} = |u| = u \cdot v = A \cos 2\theta + B \sin 2\theta \leq 1.$$
For the second part, write
$$0 \leq F(\theta) + F(\theta + \pi/2) = 2 - (b + a)\cos \theta - (b- a)\sin \theta.$$
Now pick $\theta$ so that $v = (\cos \theta,\sin \theta)$ is a unit vector in the same direction as $u = (b + a,b - a)$. Then
$$\sqrt{2}\sqrt{a^2 + b^2} = \sqrt{(b + a)^2 + (b - a)^2} = |u| = u \cdot v \leq 2.$$
|
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|
Evaluate $\int_{|z|=1} \frac{1}{z^2 -\frac{3}{2}z + 1} dz$ Evaluate :
$$\int_{|z|=1} \frac{1}{z^2 -\frac{3}{2}z + 1} dz$$
Using residue method :
$$z=\frac{3}{4} \pm i \frac{\sqrt{7}}{4}$$
The problem is the two roots on the boundary $|z|=1$
|
As stated in the comments, the integral $\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz$ is not defined.
However, we can define the Cauchy Principal Value of the integral as the limit
$$\begin{align}
\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)&=\lim_{\epsilon \to 0}\int_{C_\epsilon} \frac{1}{z^2-\frac32 z+1}\,dz\tag 1
\end{align}$$
where $C_\epsilon$ is the contour $|z|=1$, $\arg(z)\notin (\pm \arctan(\sqrt{7}/3)-\epsilon,\pm \arctan(\sqrt{7}/3)+\epsilon)$. That is to say that $C_\epsilon$ excludes the poles of the integrand.
We can evaluate the limit in $(1)$ by using Cauchy's Integral Theorem. We close the contour $C_\epsilon$ with contours that are circular arcs around the poles such that neither pole is enclosed.
As $\epsilon\to 0^+$ of the integration of the arc around the pole at $z=\frac34+i\frac{\sqrt 7}{4}$ can be easily evaluated using the parametric description $z=\frac34+i\frac{\sqrt 7}{4}+2\sin(\epsilon/2)e^{i\phi}$, $\pi/2 +\epsilon<\phi<3\pi/2-\epsilon$. Proceeding we have
$$\lim_{\epsilon\to 0^+}\int_{\pi/2 +\epsilon}^{3\pi/2-\epsilon}\frac{i}{2i\frac{\sqrt 7}{4}+2\sin(\epsilon/2)e^{i\phi}}\,d\phi=\frac{2\pi}{\sqrt 7}$$
Similarly, for the integration of the arc around the pole at $z=\frac34-i\frac{\sqrt 7}{4}$ we find
$$\lim_{\epsilon\to 0^+}\int_{\pi/2 +\epsilon}^{3\pi/2-\epsilon}\frac{i}{-2i\frac{\sqrt 7}{4}+2\sin(\epsilon/2)e^{i\phi}}\,d\phi=-\frac{2\pi}{\sqrt 7}$$
Obviously, since the sum of the residues is zero, then we find
$$\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)=0 \tag 2$$
Another interpretation is to take the average of $\oint_{|z|=1+\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz$ and $\oint_{|z|=1-\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz$ and then take the limit as $\epsilon\to 0^+$.
Since the contour $|z|=1-\epsilon$ encloses neither pole, Cauchy's Integral Theorem guarantees that
$$\oint_{|z|=1-\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz=0 \tag 3$$
And since the contour $|z|=1+\epsilon$ encloses both poles we find from the reside theorem that
$$\begin{align}
\oint_{|z|=1\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz&=2\pi i \text{Res}\left(\frac{1}{z^2-\frac32 z+1}, z=\frac34\pm i \frac{\sqrt 7}{4}\right)\\\\
&=2\pi i \left(\frac{1}{2i\frac{\sqrt 7}{4}}+\frac{1}{-2i\frac{\sqrt 7}{4}}\right)\\\\
&=0\tag 4
\end{align}$$
The average of $(3)$ and $(4)$ is also $0$ and we find
$$\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)=0 \tag 5$$
And we see that $(5)$ agrees with $(2)$, as expected!
|
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|
The circumference of the circle is $C$, what is the area of circle in terms of $C$? The circumference of the circle is $C$, what is the area of circle in terms of $C$?
a). $\dfrac {C^2}{4\pi }$
b). $2\pi C$
c). $\dfrac {4}{3} \pi C^2$
d). $2\pi C^2$
My Attempt:
$$\textrm {Circumference}=2\pi r$$
$$C=2\pi r$$
$$\dfrac {C}{2\pi }= r$$
Now,
Area$=\pi r^2$
$$=\pi \dfrac {C^2}{4\pi^2}$$
|
Your attempt is correct, you should just cancel out the $\pi$ in the expression:
$$\require{cancel} \dfrac {\pi C^2}{4\pi^2} = \dfrac {\bcancel{\pi} C^2}{4 \bcancel{\pi} \pi} = \dfrac {C^2}{4\pi}$$
|
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|
Intersection of two parabolas Given $a>0$ and $b>0$, I want to find the points of intersection of the two parabolas
\begin{align}
y&=1-ax^2 \\x&=1-by^2
\end{align}
Clearly I can just eliminate one of the variables, and I'll get a quartic equation, whose general solutions will be an enormous mess (according to Mathematica, anyway).
I also tried using this approach, but again got stuck in a quagmire of algebra.
Or, I could just use numerical methods, but that's what I'm trying to avoid.
The general problem of intersecting two conic section curves is well understood, and can only be solved by the techniques I described above (as far as I know). But my problem is not the general one, it's a very specific special case, and I'm wondering if someone can see some clever shortcut.
According to this question, the intersection points all lie on a circle, but I don't know if that helps.
|
Re-writing the second equation to be in terms of $y$, we have the two equations $y = 1 − a x^2$ and $y = \sqrt{\frac{1 − x}{b}}$.
Set them equal to each other, giving $1 − a x^2 = \sqrt{\frac{1 − x}{b}}$.
Square both sides and expand to get $a^2 x^4 − 2 a x^2 + 1 = \frac{1}{b} − \frac{1}{b} x$.
Move everything to the left to get $a^2 x^4 − 2 a x^2 + \frac{1}{b} x + 1 − \frac{1}{b} = 0$.
Comparing that to the standard form of a quartic equation $A x^4 + B x^3 + C x^2 + D x + E = 0$, we see that:
*
*$A = a²$,
*$B = 0$,
*$C = −2 a$,
*$D = \frac{1}{b}$, and
*$E = 1 − \frac{1}{b}$.
The next steps use the equations for the quartic formulae found on this webpage, simplified due to the fact that $B = 0$ and thus any terms containing $B$ vanish.
Define the new variables:
*
*$p = \left(128 − \frac{144}{b}\right) a^3 + 27 \frac{a^2}{b^2}$,
*$q = \left(16 − \frac{12}{b}\right) a^2$, and
*$s = \frac{1}{3 a^2} \left(\frac{q \sqrt[3]{2}}{\sqrt[3]{p + \sqrt{p^2 − 4 q^3}}} + \frac{\sqrt[3]{p + \sqrt{p^2 − 4 q^3}}}{\sqrt[3]{2}} + 4 a\right)$
Then the $x$-values for the four potential intersections are:
*
*$x_1 = \frac{1}{2} \sqrt{s} + \frac{1}{2} \sqrt{\frac{4}{a} − \frac{2}{a² b \sqrt{s}} − s}$
*$x_2 = \frac{1}{2} \sqrt{s} − \frac{1}{2} \sqrt{\frac{4}{a} − \frac{2}{a² b \sqrt{s}} − s}$
*$x_3 = −\frac{1}{2} \sqrt{s} + \frac{1}{2} \sqrt{\frac{4}{a} + \frac{2}{a² b \sqrt{s}} − s}$
*$x_4 = −\frac{1}{2} \sqrt{s} − \frac{1}{2} \sqrt{\frac{4}{a} + \frac{2}{a² b \sqrt{s}} − s}$
These can be plugged into the first of the original pair of equations, $y = 1 − a x^2$, to get the corresponding $y$-values.
Depending on the values of $a$ and $b$, anywhere from none to all-four intersections may be complex-valued.
Notably, all of the real-valued intersections lie on a circle with radius $r = \sqrt{\frac{a^2 + 4 a^2 b + 4 a b^2 + b^2}{4 a^2 b^2}}$ and center $\left(\frac{−1}{2 b}, \frac{−1}{2 a}\right)$, including for cases with negative values of $a$ and/or $b$.
|
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|
Solutions of $\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$
:Solutions of:
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$$
My try :
$$\lfloor 2x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac12 \rfloor $$
$$\lfloor 3x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac13 \rfloor +\lfloor x+\dfrac23 \rfloor $$
$$\lfloor 4x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor $$
Now :
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0 $$
$$(\lfloor x \rfloor)^2+(\lfloor x \rfloor)(\lfloor x+\dfrac13 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac23 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac12 \rfloor)+(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac13 \rfloor)\\(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac23 \rfloor) +\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor=0$$
Now what ?
|
Just do it in 6 cases:
$x = [x]+\{x\}$
Case 1: $\{x\} < 1/4$
Then $$[2x]\times[3x] + [4x] = 2[x]3[x] + 4[x] = 0$$ so $[x] = 0$ of $[x] = -\frac 32$. So $[x]$ is an integer $[x] = 0$ and $0 \le x < \frac 14$.
Case 2: $1/4 \le \{x\} < 1/3$
Then $$[2x]\times[3x] + [4x] = 2[x]3[x] + 4[x]+ 1 = 6[x]^2 + 4[x] + 1 = 0.$$
$[x]$ must be negative but then $6[x]^2 +1 > 4|[x]| $ so that is impossible.
Case 3: $1/3 \le \{x\} < 1/2$
then $$[2x]\times[3x] + [4x] =2[x](3[x] + 1) + 4[x] + 1 = 6[x]^2 + 6[x] + 1 = 0.$$ Then $[x]^2 + [x] = -\frac 16$ which is not an integer so impossible.
Case 4: $\frac 12 \le \{x\} < 2/3$
then $$[2x]\times[3x] + [4x] = (2[x] + 1)(3[x] + 1) + 4[x] + 2 = 6[x]^2 + 9[x] + 3 = 0$$ so $2[x]^3 + 3[x] + 1 = 0$ so $(2[x] + 1)([x] + 1) = 0$ so $[x] = -1$ or $[x ] = -\frac 12$. So $[x] = -1$ and $- \frac 12 \le x < -1/3$.
Case 5: $\frac 23 \le \{x\} < 3/4$
then $$[2x]\times[3x] + [4x] = (2[x] + 1)(3[x] + 2) + 4[x] + 2= 6[x]^2 + 11[x] + 4 = 0$$
$$[x] = \frac {-11 \pm \sqrt{121 - 4\times4\times6}}{12} = \frac {-11 \pm 6}{12} \not \in \mathbb Z.$$
Case 6: $\frac 34 \le \{x\} < 1$
then $$[2x]\times[3x] + [4x] = (2[x] + 1)(3[x] + 2) + 4[x] + 3= 6[x]^2 + 11[x] + 5 = 0$$
$$[x] = \frac {-11 \pm \sqrt{121 - 4\times4\times5}}{10} = \frac {-11 \pm \sqrt{41}}{12} \not \in \mathbb Z.$$
So $- \frac 12 \le x < -\frac 13$ or $0 \le x < \frac 14$
|
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|
Constructing Series : Alternating between Rational Number and Irrational Number In the range of $[0,1] \in \Bbb R$, I'd like to construct a series which is strictly increasing and rational number and irrational numbers are alternatively appearing, like $0< \sqrt{1\over 3} < 2/3 < \sqrt{1 \over 2} < ... $
I want to make the range being partitioned in this way, with "infinitely many terms".
Any recommendation?
|
You may use numbers of the form $\frac{n}{n+1}$ such as $\frac{2}{3}$ for rational numbers and their geometric mean $\sqrt{\frac{n}{n+1}\frac{n+1}{n+2}}=\sqrt{\frac{n}{n+2}}$ in between.
This way, your sequence would start
$$\frac{1}{2}<\sqrt{\frac{1}{3}}<\frac{2}{3}<\sqrt{\frac{1}{2}}<\frac{3}{4}<\sqrt{\frac{3}{5}}<\frac{4}{5}<\sqrt{\frac{2}{3}}<...$$
Starting at $n=0$, with rational $a(2n)$ and irrational $a(2n+1)$, the closed form is
$$a(n)=\frac{1+(-1)^n}{2}\frac{n+2}{n+4}+\frac{1-(-1)^n}{2}\sqrt{\frac{n+1}{n+5}}$$
Equivalently,
$$\begin{align}
a(n)&=\sqrt{\frac{(2n+3+(-1)^n)(2n+5-(-1)^n)}{(2n+7+(-1)^n)(2n+9-(-1)^n)}}\\
&=\sqrt{\frac{2n^2+8n+7+(-1)^n}{2n^2+16n+31+(-1)^n}}\\
\end{align}$$
|
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|
Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\frac{b}{3x-2}\right) dx=k$$
then
$$\lim_{N \to \infty} \left[\frac{2}{3}ln(x)+(a-\frac{2}{3})ln (x+3)-\frac{b}{3}ln(3x-2)\right]_3^N =k$$
By logarithm properties:
$$\lim_{N \to \infty} \left[ln\frac{(x)^\frac{2}{3}(x+3)^\left(a-\frac{2}{3}\right)}{(3x-2)\frac{b}{3}}\right]_3^N =k$$
Evaluating:
$$\lim_{N \to \infty} \left[ln\frac{(N)^\frac{2}{3}(N+3)^\left(a-\frac{2}{3}\right)}{(3N-2)^\frac{b}{3}}\right]-\left[ln\frac{(3)^\frac{2}{3}(6)^\left(a-\frac{2}{3}\right)}{(7)^\frac{b}{3}}\right] =k$$
But, I do not know what to do after this step. Any recommendation will be appreciated
|
Since $\frac{ax+2}{x^2+3x}=\frac{a}x+O\!\left(\frac1{x^2}\right)$ and $\frac{b}{3x-2}=\frac{b}{3x}+O\!\left(\frac1{x^2}\right)$, it should be that $a=\frac{b}3$.
$$
\begin{align}
&\int_3^\infty\left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right)\,\mathrm{d}x\\
&=\int_3^\infty\left(\frac{\frac23}{x}+\frac{a-\frac23}{x+3}-\frac{b}{3x-2}\right)\,\mathrm{d}x\\
&=\left[\frac23\log(x)+\left(a-\frac23\right)\log(x+3)-\frac{b}3\log(3x-2)\right]_3^\infty\\[6pt]
&=\left[\frac23\log\left(\frac{x}{x+3}\right)+a\log\left(1+\frac3x\right)-\frac{b}3\log\left(3-\frac2{x}\right)+\left(a-\frac{b}3\right)\log(x)\right]_3^\infty
\end{align}
$$
So we see that indeed, the convergence is when $a=\frac{b}3$.
|
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|
Integer solutions of equation $x^3+10x-1=y^3+6y$? I want to find all integer solution of the following equation in integers$$x^3+10x-1=y^3+6y$$ I tried to factor it somehow, but failed. Also $x$ and $y$ are have different parity. How can one solve this?
|
Comparing terms $6y$ and $10x$ tells us that $x$ can not be greater than $y$. We'll find out that, let $y-x=z$ and equation becomes
$$x^3+10x-1=(x+z)^3+6(x+z)$$ $$-3zx^2-(3z^2-4)x-(z^3+6z+1)=0 $$
Discriminant of the obtained quadratic in $y$ must be non-negative, that is
$$(3z^2-4)^2-4(3z)(z^3+6z+1)\ge0,$$which gives $z=0$ only. Thus $y=x$, which gives $4x=1$. No integer solutions!
|
{
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|
Find Volume V bounded by surface $\sqrt[3]{ x^2}+\sqrt[3]{ y^2}+\sqrt[3]{ z^2}=1$
I try this substitution $x={u}^3$,$y={v}^3$ and $z={k}^3$ ,and I get sphere with radius 1
|
Let $x=r\sin^3\alpha\cos^3\beta$, $y=r\sin^3\alpha\sin^3\beta$
and $z=r\cos^3\alpha$.
Thus, $J=9r^2\sin^5\alpha\cos^2\alpha\sin^2\beta\cos^2\beta$ and the volume it's
$$9\int_{0}^1r^2dr\int_{0}^{\frac{\pi}{2}}\sin^5\alpha\cos^2\alpha d\alpha\int_{0}^{2\pi}\sin^2\beta\cos^2\beta d\beta=\frac{4\pi}{35}$$
About Jacobian see here: https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
|
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Solving $Т(n)=\frac{1}{\frac{1}{T(n-1)}+n^2}$ and $T(1) = 1$ $$T(n) = \frac1{\frac1{T(n-1)}+n^2}$$
I did find $T(2) = \frac{1}{1+n^2}$, but I don't know how to proceed.
How do I go on from here? How can I find the solution? Thanks. :)
|
$$\frac{1}{t_{n}}=\frac{1}{t_{n-1}}+n^2$$ and the rest is smooth I think:
$$\frac{1}{t_2}=\frac{1}{t_1}+2^2$$
$$\frac{1}{t_3}=\frac{1}{t_2}+3^2...$$
$$\frac{1}{t_n}=\frac{1}{t_{n-1}}+n^2.$$
A summing of these equalities gives:
$$\frac{1}{t_n}-\frac{1}{t_1}=2^2+3^2+...+n^2$$ or
$$\frac{1}{t_n}=1^2+2^2+3^2+...+n^2$$ or
$$\frac{1}{t_n}=\frac{n(n+1)(2n+1)}{6}.$$
The answer is:
$$t_n=\frac{6}{n(n+1)(2n+1)}$$
|
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|
Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing
Since there are 20 presents to be distributed among 4 children,
$$C_1 + C_2 + C_3 + C_4 = 20$$
By bars and stars,
$$\binom{ 20 + 4-1 }{ 4-1 } = \binom{ 23 }{ 3 } = 1771$$
$C_2 + C_3 + C_4 = 14 $ (after giving 6 presents to $C_1$)
Assuming $C_1$ gets 6 present = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_2$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_3$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_4$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
No one gets exactly 6 presents = Total - Everyone gets 6 presents
= $\binom{ 23 }{ 3 } - \binom{ 4 }{ 1 }\binom{ 16 }{ 2 } = 1771 - 480 = 1291$
I am new to combinatorics and not sure if I'm on the right path any help would be appreciated.
|
For future reference here is a verification of the result by
N.F. Taussig, which I upvoted. This is the case where it is possible
for a child not to receive any presents. The combinatorial species
here is
$$\mathfrak{S}_{=4}(\mathfrak{P}_{\ne 6}(\mathcal{Z})).$$
We get the generating function
$$G(z) = \left(\exp(z)-\frac{z^6}{6!}\right)^4
\\ = \exp(4z) - 4\frac{\exp(3z)z^6}{6!}
+ 6\frac{\exp(2z)z^{12}}{6!^2}
- 4\frac{\exp(z)z^{18}}{6!^3} + \frac{z^{24}}{6!^4}.$$
Extracting coefficients we find
$$20! [z^{20}] G(z) =
20! \left(\frac{4^{20}}{20!} - 4 \frac{3^{14}}{14! \times 6!}
+ 6 \frac{2^8}{8!\times 6!^2} - 4\frac{1^2}{2!\times 6!^3}\right).$$
This evaluates to
$$\bbox[5px,border:2px solid #00A000]{523708416736.}$$
|
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|
How to parametrise intersection of two cylinders I have a question that asks me to verify Stokes Theorem on
$S:=$portion of the cylinder $ y^2+z^2=4$ with $ z>0$ and $x^2+y^2\leq1$
How would I parametrise the curve and surface so that I can apply Stokes Theorem. I have plotted this surface and can imagine what it looks like but I am finding it difficult to parametrise it. Any help will be appreciated. I can complete the question once I have found a suitable parametrisation.
The parametrisation I came up with is $r(\theta,x)=(x,2cos\theta,2sin\theta)$ with $0<\theta<\pi$, $-\sqrt{1-(2cos\theta)^2}\leq x \leq\sqrt{1-(2cos\theta)^2}$ If this is incorrect what is wrong with this?
Or would a better parametrisation be the usual cylinderical coordinates with $0\leq\theta\leq2\pi$ and $0<z<\sqrt{4-sin^2\theta}$ If either of these are correct, why is the other one incorrect?
|
Based on the fact that $z = \sqrt{4-y^2}$ :
$$
\sigma(r,\theta) = (x,y,z) = (r\cos\theta, r\sin\theta, \sqrt{4-(r\sin\theta)^2}) \\
r \in [0,1],\ \theta \in [0,2\pi[
$$
According to WA, (And my understanding of the surface), this seems right.
The cross product is painful, I agree...
$$
\sigma_r = (\cos\theta, \sin\theta, -\frac{r\sin^2\theta}{\sqrt{4-r^2\sin^2\theta}}) \\
\sigma_\theta = (-r\sin\theta, r\cos\theta, -\frac{r^2\sin\theta\cos\theta}{\sqrt{4-r^2\sin^2\theta}}) \\
\sigma_r \times \sigma_\theta = (0,\frac{r^2(\cos^2\theta\sin\theta + \sin^3\theta)}{\sqrt{4-r^2\sin^2\theta}},r)
$$
Regarding your parametrization it doesn't work for $\theta < \pi/3, \theta > 2\pi/3$, the square roots in the bounds give complex numbers but I see what you meant to do. The problem is that the bound for $x$ doesn't define a circle of radius $1$ as it should be. Here you are mixing the cylinder of radius $2$ (which gives you the $y=2\cos\theta$) and the equations $x = \pm \sqrt{1-y^2}$ and to make this equation work $y$ needs to be between $-1$ and $1$ but by its nature its range is $[-2,2]$.
Checking Stokes theorem with this surface:
$$
A= \iint_S rot \vec{F} \cdot d\vec{n} = \iint_S (0,x,1) \cdot d\vec{n} =
\iint_S\left( r\cos\theta \cdot \frac{r^2(\cos^2\theta\sin\theta + \sin^3\theta)}{\sqrt{4-r^2\sin^2\theta}} + r \right)dS = \\
\underbrace{\int_0^1 dr \int_0^{2\pi} d\theta \frac{r^3(\cos^3\theta\sin\theta + \sin^3\theta\cos\theta)}{\sqrt{4-r^2\sin^2\theta}}}_{=I_1} + \int_0^1 dr \int_0^{2\pi} d\theta\ r = 0 + \pi
$$
Then
$$
\partial S(\theta) = (\cos\theta,\sin\theta, \sqrt{4-\sin^2\theta}), \ \partial S'(\theta) = (-\sin\theta,\cos\theta, -\frac{\sin\theta \cos\theta}{\sqrt{4-\sin^2\theta}}),\ \theta\in [0,2\pi[ \\
B = \oint_{\partial S} \vec{F}\cdot d\vec{l} = \oint_{\partial S} (zx-y,0,0) \cdot d \vec{l} = \int_0^{2\pi} (\cos\theta \sqrt{4-\sin^2\theta} -\sin\theta) (-\sin\theta)\; d\theta = \pi
$$
Therefore $A = B$. However I agree that the integrals are not be that simple but clearly doable which may indicate that the parametrization I chose was maybe not the best, but at least it worked !
$$
I_1 = \int_0^1 dr\ r^3 \int_0^{2\pi}d\theta \cos\theta \frac{(1-\sin^2\theta)\sin\theta + \sin^3\theta}{\sqrt{4-r^2\sin^2\theta}} \overset{w = \sin\theta, dw = \cos\theta d\theta}{=} \int_0^1 dr\ r^3\int_0^0 dw \frac{w}{\sqrt{4-r^2 w^2}} = 0
$$
|
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|
Integrating an expression by parts and by substitution giving two different solutions? So I'm given this example in my notes:
$$\int x(x+3)^5dx$$
And when I integrate by parts and by substitution I end up getting two different answers, but I don't know why.
Integrating by parts:
If I let $u = x$ and $dv = (x + 3)^5$, then:
$$
\begin{align*}
\int x(x+3)^5dx &= {x \over 6 }(x+3)^6 - \int {1 \over 6 }(x+3)^6dx\\
&= {x \over 6 }(x+3)^6 - {1 \over 6 }({1 \over 7 }(x+3)^7) + C\\
&= {x \over 6 }(x+3)^6 - {1 \over 42 }(x+3)^7 + C
\end{align*}
$$
Integration by u-substitution:
If I let $u = x + 3$, then ${du\over dx } = 1$ and $du = dx$:
$$
\begin{align*}
\int x(x+3)^5dx &= \int (u-3)u^5 \ du\\
&= \int u^6 - 3u^5\ du\\
&= {(x+3)^7 \over 7 } - {(x+3)^6 \over 2 } + C
\end{align*}
$$
Where am I going wrong?
|
In fact, the two solutions are equivalent (if you put the forgotten $C$ on the first integral):
\begin{align*}
\frac{x}{6}(x+3)^6 - \frac{1}{42}(x+3)^7 + C & = \frac{(x+3)-3}{6}(x+3)^6 - \frac{1}{42}(x+3)^7 + C \\
& = \frac{1}{6} (x+3)^7 - \frac{1}{2} (x+3)^6 - \frac{1}{42} (x+3)^7 + C \\
& = \frac{1}{7} (x+3)^7 - \frac{1}{2} (x+3)^6 + C.
\end{align*}
|
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|
Evaluate $\int_0^{2\pi} \frac{\cos 2x}{1-2a\cos x+a^2} $ with $a^2<1$ How do I compute the integration for $a^2<1$,
$$\int_0^{2\pi} \dfrac{\cos 2x}{1-2a\cos x+a^2}dx=? $$
I think that:
$$\cos2x =\dfrac{e^{i2x}+e^{-2ix}}{2},
\qquad\cos x =\dfrac{e^{ix}+e^{-ix}}{2}$$
But I cannot. Please help me.
|
One can factorise the denominator:
$$ 1-2a\cos{x}+a^2 = (e^{ix}-a)(e^{-ix}-a). $$
Then using partial fractions gives
$$ \frac{1}{2}\frac{e^{2ix}+e^{-2ix}}{(e^{ix}-a)(e^{-ix}-a)} = -\frac{a^2+1}{2 a^2} + \frac{a^4+1}{2 a (a^2-1)}\left( -\frac{1}{e^{i x}-a}-\frac{1}{1-a e^{i x}} \right)-\frac{e^{i x}+e^{-i x}}{2 a} $$
Now, since $a^2<1$, the remaining fractions can be expanded in a power series such as $\sum_{k=0}^{\infty} (ae^{ix})^n $. But
$$ \int_0^{2\pi} e^{inx} \, dx = \begin{cases} 2\pi & n=0 \\ 0 & n \neq 0 \end{cases}, $$
so only the constant terms contribute, and hence the integral is
$$ 2\pi \left( -\frac{a^2+1}{2 a^2} + \frac{a^4+1}{2 a (a^2-1)}\left( \frac{1}{a}-1 \right) \right) = \frac{2\pi a^2}{1-a^2}. $$
It may be easier to use the formula
$$ \frac{1-a^2}{1-2a\cos{x}+a^2} = 1+2\sum_{k=0}^{\infty} a^k\cos{kx} $$
for $\lvert a \rvert <1$.
|
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|
Prove that $\sum_{cyc}\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}\ge \frac{\sqrt{181}}{5}$ For $a,b,c>0$ satisfy $ab+bc+ca\ge \frac{4}{3}$. Prove that $$\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}+\sqrt{b^2+\frac{1}{\left(c+1\right)^2}}+\sqrt{c^2+\frac{1}{\left(a+1\right)^2}}\ge \frac{\sqrt{181}}{5}$$
My try: By Minkowski:
$LHS\ge \sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)^2}$
We have: $(a+b+c)^2\ge 3(ab+bc+ca)=3.\frac{4}{3}=4$
And by C-S: $\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)^2\ge \left(\frac{9}{a+b+c+3}\right)^2$
I can't continue, help me
|
Let $a=\frac{2x}{3}$, $b=\frac{2y}{3}$ and $c=\frac{2z}{3}$.
Hence, the condition gives $xy+xz+yz\geq3$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{(2y+3)^2}}\geq\frac{\sqrt{181}}{5}.$$
Now, by C-S
$$\sum_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{(2y+3)^2}}=\frac{15}{\sqrt{181}}\sum_{cyc}\sqrt{\left(\frac{4}{9}+\frac{9}{25}\right)\left(\frac{4x^2}{9}+\frac{9}{(2y+3)^2}\right)}\geq$$
$$\geq\frac{15}{\sqrt{181}}\sum_{cyc}\left(\frac{4x}{9}+\frac{9}{5(2y+3)}\right).$$
Thus, it remains to prove that
$$\frac{15}{\sqrt{181}}\sum_{cyc}\left(\frac{4x}{9}+\frac{9}{5(2y+3)}\right)\geq\frac{\sqrt{181}}{5}$$ or
$$20(x+y+z)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)\geq\frac{543}{5}.$$
Now, let $x+y+z=3u$ and $xy+xz+yz=3v^2$, where $v>0$.
Hence, $u\geq v\geq1$ and by C-S we obtain:
$$20(x+y+z)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)-\frac{543}{5}\geq$$
$$\geq20(x+y+z)+81\cdot\frac{(1+1+1)^2}{\sum\limits_{cyc}(2x+3)}-\frac{543}{5}=60u+\frac{729}{6u+9}-\frac{543}{5}=$$
$$=3\left(20u+\frac{81}{2u+3}-\frac{181}{5}\right)=\frac{6(u-1)(100u+69)}{5(2u+3)}\geq0.$$
Done!
|
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|
Recursive relationship for incomplete beta function Consider the incomplete Beta function $I_x(a, b)$
$$ I_x(a, b) =
\dfrac{B(x; a, b)}{B(a, b)} =
\dfrac{\int_0^x t^{a-1} \left( 1-t \right)^{b-1} dt}{\int_0^1 t^{a-1} \left( 1-t \right)^{b-1} dt}.
$$
How can I prove the recursive property given in Wikipedia
$$
I_x(a+1, b) = I_x(a, b) - \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)}
$$
|
Using the Beta function property $B(a+1,b)=aB(a.b)/(a+b)$ compute the derivative
$$I_x'(a+1, b) = \frac{x^a (1-x)^{b-1}}{B(a+1,b)} =
\frac{x^a (1-x)^{b-1}}{B(a,b)a/(b+a)}=
(a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a}$$
Now let $$
f(a,b,x) = I_x(a+1, b) - I_x(a, b) + \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)}
$$
Then $f(a,b,0)=0$ and
$$f'(a,b,x) = I_x'(a+1, b) - I_x'(a, b) + \frac{d}{dx}\Big(\frac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)}\Big)$$
$$f'(a,b,x)= (a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a}
- \frac{x^{a-1} (1-x)^{b-1}}{B(a,b)}
+ \frac{x^{a-1}(1-x)^b}{B(a,b)}
- \frac{b x^a(1-x)^{b-1}}{aB(a,b)}$$
$$f'(a,b,x)=
\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)}
\Big((a+b)x/a - 1 + 1 -x -bx/a\Big) = 0
$$
therefore $f \equiv 0$ and the recursive property is proven.
|
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Inequality "A la Rozenberg" Hello I want to solve this
The inequality is equivalent to this :
$$\frac{a^2}{b^2}cos(arctan(\sqrt{\frac{b}{a}}))^2+\cos(\arctan(\sqrt{\frac{c}{b}}))^2+\frac{c^2}{b^2}\cos(\arctan(\sqrt{\frac{a}{c}}))^2\geq \frac{3}{2b^2}$$
We put :
$\sqrt{\frac{b}{a}}=\frac{x+y}{1-xy}$
$\sqrt{\frac{c}{b}}=\frac{z+y}{1-yz}$
$\sqrt{\frac{a}{c}}=\frac{x+z}{1-xz}$
Where $x$,$y$,$z$ are positive real numbers with the condition
$1>xy$,$1>xz$,$1>zy$.
We get :
$$(\frac{1-xy}{x+y})^4\cos(\arctan(\frac{x+y}{1-xy}))^2+\cos(\arctan(\frac{z+y}{1-yz}))^2+(\frac{z+y}{1-yz})^4\cos(\arctan(\frac{x+z}{1-xz}))^2\geq \frac{3^{\frac{1}{3}}}{2}(1+(\frac{1-xy}{x+y})^6+(\frac{z+y}{1-yz})^6)^{\frac{2}{3}}$$
Wich is equivalent to :
$$(\frac{1-xy}{x+y})^4(\frac{1}{(\frac{x+y}{1-xy})^2+1})+\frac{1}{(\frac{z+y}{1-yz})^2+1}+(\frac{z+y}{1-yz})^4\frac{1}{(\frac{x+z}{1-xz})^2+1}\geq \frac{3^{\frac{1}{3}}}{2}(1+(\frac{1-xy}{x+y})^6+(\frac{z+y}{1-yz})^6)^{\frac{2}{3}}$$
And if we make a substitution like this :
$A=\frac{1-xy}{x+y}$
$B=\frac{z+y}{1-yz}$
$C=\frac{x+z}{1-xz}$
We get :
$$A^4(\frac{1}{(\frac{1}{A})^2+1})+\frac{1}{(B)^2+1}+B^4\frac{1}{(C)^2+1}\geq \frac{3^{\frac{1}{3}}}{2}(1+(A)^6+(B)^6)^{\frac{2}{3}}$$
With the condition :
$(\frac{\frac{1}{A^6}+1+B^6}{3})^{-1}+(\frac{C^6+1+\frac{1}{B^6}}{3})^{-1}+(\frac{\frac{1}{C^6}+1+A^6}{3})^{-1}=3$
Wich is equivalent to :
$(\frac{3A^6}{A^6B^6+1+A^6})+(\frac{3B^6}{C^6B^6+1+B^6})+(\frac{3C^6}{A^6C^6+1+C^6})=3$
After that I can't continue.
Thanks.
|
To start we study the following function :
$$f(x)=\frac{x^3}{x^2+1}+\frac{1}{h^2+1}+\frac{h}{y^2+1}-\frac{3^{\frac{2}{3}}}{2}(x^3+h^3+1)^{\frac{1}{3}}$$
The derivative of function is :
$$f'(x)=\frac{1}{2}x^2[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]$$
So we study this part of the derivative:
$$[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]$$
We remark this :
$$[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]\geq [\frac{-3^{\frac{2}{3}}}{(x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]>0$$
So the derivative is positive and the function is increasing :
So we have this :
$$f(x)\geq f(0)$$
Wich is equivalent to :
$$f(x)\geq \frac{1}{h^2+1}+\frac{h}{y^2+1}-\frac{3^{\frac{2}{3}}}{2}(h^3+1)^{\frac{1}{3}}$$
Now we combine this with the initial condition :
$$\frac{3h}{hy+1+y}+\frac{3y}{1+y}=3$$
We get :
$$\frac{1}{h^2+1}+\frac{h}{(h-1)^2+1}-\frac{3^{\frac{2}{3}}}{2}(h^3+1)^{\frac{1}{3}}\geq 0$$
Done!
|
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|
Prove that for any $n \neq 4$, if for any $k < n$ we have that $a_k$ is relatively prime to $a_n$, then $a_n$ is prime
Consider the sequence $a_n = |n(n+1)-19|$ for $n = 0,1,2,\ldots$. Prove that for any $n \neq 4$, if for any $k < n$ we have that $a_k$ is relatively prime to $a_n$, then $a_n$ is prime.
The first few terms are $19,17,13,7,1,11,23,37,53,71,91,\ldots$ and we can verify the property for each of these integers. Suppose that $a_n$ where $n \neq 4$ is relatively prime to $a_k$ for any $k < n$. Then $$\gcd(a_n,a_{n-1}) = \gcd(n(n+1)-19,(n-1)n-19) = 1.$$ Then $\gcd(n^2+n-19,n^2-n-19) = \gcd(2n,n^2-n-19) = \gcd(n,n^2-n-19) = \gcd(n,19)$ so $19 \nmid n$. Similarly $$\gcd(a_{n-1},a_{n-2}) = \gcd((n-1)n-19,(n-2)(n-1)-19).$$ Then $\gcd(n^2-n-19,n^2-3n-17) = \gcd(2n+2,n^2-3n-17) = \gcd(n+1,n^2-3n-17)$, but I didn't see how to generalize it.
|
If $a_n$ is not prime and $n > 4$, then $a_n>1$ and there exists a prime natural number $p\leq n$ such that $p\mid a_n$. Then, $p$ divides both $a_n$ and $a_{n-p}$. (In fact, one can show that $p=n$ implies $n=19$.)
|
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|
Finding the sum of digits of $9 \cdot 99 \cdot 9999 \cdot ... \cdot (10^{2^n}-1)$ Find the sum of the digits (in terms of $n$) of $$9 \cdot 99 \cdot 9999 \cdot 99999999 \cdot ...
\cdot (10^{2^n} - 1)$$
Where every term has two times more digits than the previous term.
I tried to factorise as $$(10-1)(10^2-1)(10^4-1)...(10^{2^n}-1)$$
and they're all differences of two squares (except for the first term)
$$(10-1)(10-1)(10+1)(10^2-1)(10^2+1)(10^4-1)(10^4+1)...(10^{2^n-1}-1)(10^{2^n-1}+1)$$
but this doesn't seems to help that much. Also i tried with base 10 but it didn't work at all. Any ideas?
|
Let us assume that $N$ is a number with $2^k-1$ decimal digits, whose last digit is $\geq 1$.
Let $S(N)$ be the sum of digits of $N$. Let us study the sum of digits of
$$ N\cdot(10^{2^k}-1) = N\cdot 10^{2^k}- N = N\cdot 10^{2^k} - 10^{2^k} + (10^{2^k}-1-N)+1. $$
We have:
$$ S(N\cdot(10^{2^k}-1)) = S(N)-1+\left(9\cdot(2^k-1)-S(N)+9\right)+1 $$
and it is very interesting to notice that such sum does not depend on $S(N)$, but simply is $9\cdot 2^k$. The number
$$ N = 9 \cdot 99 \cdot 9999 \cdots (10^{2^{k-1}}-1) $$
has $2^k-1$ decimal digits, the last of them being $1$ or $9$. By induction it follows that
$$ S\left(9\cdot 99\cdots (10^{2^k}-1)\right) = \color{red}{9\cdot 2^{k}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2319051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Is this an ellipse?
Is this parameterisation an ellipse:
\begin{align}x(t) &= \frac{2 \cos(t)}{1 + a \sin(t)}\\
y(t) &= \frac{2 \sin(t)}{1 + a \sin(t)}\end{align}
where $a$ is a real positive parameter.
I tried to do it the naive way but couldn't find a definitive answer.
Plotting our curve with the help of Geogebra gives the following very ellipse like graph:
Any help would be appreciated.
|
$$\begin{align}
x&=\frac {2\cos t}{1+a\sin t}\tag{1}\\
y&=\frac {2\sin t}{1+a\sin t}\tag{2}\\
(2)/(1):\hspace{3cm}\\
\frac yx&=\tan t\tag{3}\\
(1)^2+(2)^2:\hspace{3cm}\\
x^2+y^2&=\frac 4{(1+a\sin t)^2}\\
&=\frac {4(x^2+y^2)}{\big(\sqrt{x^2+y^2}+ay\big)^2}
&&\scriptsize \bigg(\sin t=\frac y{\sqrt{x^2+y^2}}\bigg)\\
(x^2+y^2)\big[\big(\sqrt{x^2+y^2}+ay\big)^2-4\big]&=0\\
\because{x^2+y^2}\neq 0\therefore \qquad
\big(\sqrt{x^2+y^2}+ay\big)^2-4&=0\\
x^2+y^2&=(\pm2-ay)^2\\
\color{red}{x^2+(1-a^2)y^2\pm 4ay-4}&\color{red}{=0}
\end{align}$$
which is an ellipse if $a^2<1$, per criteria outlined here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2319777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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|
Smith Normal Form I would like to put this matrix below into Smith Normal Form over $\mathbb{Q}[x]: $
$$\left(
\begin{array}{ccc}
7 & x & 0 & -x \\
0 & x-3 & 0 & 3\\
0 & 0 & x-4 & 0 \\
x-6 & -1 & 0 & x+1
\end{array}
\right)$$
but I am stuck here:
$$\left(
\begin{array}{ccc}
7 & 0 & 0 & -x \\
0 & x & 0 & 3\\
0 & 0 & x-4 & 0 \\
8x+22 & x & 0 & -x^2-3x+1
\end{array}
\right)$$
I'm not sure how to proceed. Any help is appreciated.
|
Here are the first few steps as per my suggestion in the comments:
\begin{align*}
\left(\begin{array}{rrrr}
x & 7 & 0 & -x \\
x - 3 & 0 & 0 & 3 \\
0 & 0 & x - 4 & 0 \\
-1 & x - 6 & 0 & x + 2
\end{array}\right) &\leadsto
\left(\begin{array}{rrrr}
-1 & x - 6 & 0 & x + 2 \\
x - 3 & 0 & 0 & 3 \\
0 & 0 & x - 4 & 0 \\
x & 7 & 0 & -x
\end{array}\right)\\
\left(\begin{array}{rrrr}
1 & -x + 6 & 0 & -x - 2 \\
x - 3 & 0 & 0 & 3 \\
0 & 0 & x - 4 & 0 \\
x & 7 & 0 & -x
\end{array}\right) &\leadsto
\left(\begin{array}{rrrr}
1 & -x + 6 & 0 & -x - 2 \\
0 & x^{2} - 9 x + 18 & 0 & x^{2} - x - 3 \\
0 & 0 & x - 4 & 0 \\
0 & x^{2} - 6 x + 7 & 0 & x^{2} + x
\end{array}\right)\\
\left(\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & x^{2} - 9 x + 18 & 0 & x^{2} - x - 3 \\
0 & 0 & x - 4 & 0 \\
0 & x^{2} - 6 x + 7 & 0 & x^{2} + x
\end{array}\right)
&\leadsto
\left(\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & x - 4 & 0 & 0 \\
0 & 0 & x^{2} - 9 x + 18 & x^{2} - x - 3 \\
0 & 0 & x^{2} - 6 x + 7 & x^{2} + x
\end{array}\right)
\end{align*}
Can you take it from here? In the end, I get
$$
\left(\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & x^{4} - 3 x^{3} - 11 x^{2} + 7 x + 84
\end{array}\right)
$$
for the Smith normal form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2320137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do you integrate $\frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)}$?
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)} \,dx$$
I tried by dividing the terms in both the numerator and denominator by $\cos^5x$ but still cant find my way.
|
$$I =\int_{0}^{\frac{\pi}{2}} \frac{\cos^5x}{\sin^5x + \cos^5x} dx\tag{1}$$
Let $u =\frac{\pi}{2}-x$
$$I = -\int_{\frac{\pi}{2}}^{0} \frac{\sin^{5}u}{\sin^5u+\cos^5u}du\tag{2} =\int_{0}^{\frac{\pi}{2}} \frac{\sin^5x}{\sin^5x + \cos^5x} dx$$
Then $$(1)+(2)\implies 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^5x+\cos^5x}{\sin^5x+\cos^5x}dx=\int_{0}^{\frac{\pi}{2}}1dx=\frac{\pi}{2}\implies I=\frac{\pi}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2321809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
$e^i.e_j=\delta_{ij}$ not true in a two dimensional coordinate system with dual. Suppose vector space $\mathbb{R}^2$ with standard basis $E=\{e_1,e_2\}$, vector space $V$ with basis $E^\prime=\{e^\prime_1,e^\prime_2\}$, $V^*$ (dual of vector space $V$) with basis $E^*=\{e^1,e^2\}$ and a linear transformation $A^\prime_X:\mathbb{R}^2\rightarrow V$.
suppose we have vector $u$ with components $(x,y)$, $(x^\prime,y^\prime)$ and $(x^1,y^1)$ for $\mathbb{R}^2$, $V$ and $V^*$ respectively, and :
$$A^\prime_X:\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$
In this figure, the red lines and blue lines represent coordinate system for $V$ and $V^*$ respectively.
I found :
$$\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$
$$\left[\begin{array}{} x^1 \\ y^1 \end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \cos\beta & \sin\beta\\ -\cos\gamma &-\sin\gamma\end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$
$$\left[\begin{array}{} x^1 \\ y^1 \end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} 1 & \cos\theta\\ -\cos\theta &-1\end{array} \right]\left[\begin{array}{} x^\prime\\ y^\prime\end{array} \right]$$
$$\left[\begin{array}{} e^\prime_1 \\ e^\prime_2\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\sin\beta\\ -\cos\gamma & \cos\beta \\ \end{array} \right]\left[\begin{array}{} e_1\\ e_2\end{array} \right]$$
$$\left[\begin{array}{} e^1 \\ e^2\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \cos\beta & -\cos\gamma \\ \sin\beta & -\sin\gamma \\ \end{array} \right]\left[\begin{array}{} e_1\\ e_2\end{array} \right] .$$
My problem: I know $e^i(e^\prime_j)=\delta_{ij}$ but it is not true in this example, I do not know why. where and what are my mistakes?
Figure source
|
I get
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]=
\left[ \begin{matrix}
\cos\beta & \sin\beta \\
\cos\gamma & \sin\gamma
\end{matrix} \right]
\left[\begin{matrix} e_1\\ e_2\end{matrix} \right]
$$
At least that makes
$$
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\begin{matrix} x' \\ y' \end{matrix} \right]
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[ \begin{matrix}
\cos\beta & \cos\gamma \\
\sin\beta & \sin\gamma
\end{matrix} \right]
\frac{1}{\sin\theta}
\left[ \begin{matrix}
\sin\gamma & -\cos\gamma \\
-\sin\beta & \cos\beta
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
\\
=
\frac{1}{\sin\theta}
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[ \begin{matrix}
\cos\beta\sin\gamma - \cos\gamma \sin\beta & -\cos\beta \cos\gamma + \cos\beta \cos\gamma \\
\sin\beta \sin\gamma - \sin\beta \sin\gamma & -\sin\beta \cos\gamma + \sin\gamma \cos\beta
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
\\
=
\frac{1}{\sin\theta}
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[ \begin{matrix}
\sin(\gamma-\beta) & 0 \\
0 & \sin(\gamma-\beta)
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
\\
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
$$
since $\gamma-\beta = \theta$.
How did I find the equation at the top of my post?
We wanted $\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\begin{matrix} x' \\ y' \end{matrix} \right]
= \left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]$.
Inserting $\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]=
\frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$ gives
$$\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\begin{matrix} x' \\ y' \end{matrix} \right]
=
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\frac{1}{\sin\theta}
\left[ \begin{matrix}
\sin\gamma & -\cos\gamma \\
-\sin\beta & \cos\beta
\end{matrix} \right]
\left[\begin{matrix} x \\ y \end{matrix} \right]
$$
so we must have
$$\left[\begin{matrix} e_1 & e_2\end{matrix} \right] =
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\frac{1}{\sin\theta}
\left[ \begin{matrix}
\sin\gamma & -\cos\gamma \\
-\sin\beta & \cos\beta
\end{matrix} \right]
$$
Solving for $\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]$ gives
$$
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\frac{\sin\theta}{\sin\gamma \cos\beta - \sin\beta \cos\gamma}
\left[\begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix}\right]
=
\left[\begin{matrix} e_1 & e_2\end{matrix} \right]
\left[\begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix}\right]
$$
Finally, taking the transpose results in
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]
=
\left[\begin{matrix} \cos\beta & \sin\beta \\ \cos\gamma & \sin\gamma \end{matrix}\right]
\left[\begin{matrix} e_1 \\ e_2\end{matrix} \right]
$$
One can also see from the picture that we must have
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]
=
\left[\begin{matrix} a \cos\beta & a \sin\beta \\ b \cos\gamma & b \sin\gamma \end{matrix}\right]
\left[\begin{matrix} e_1 \\ e_2\end{matrix} \right]
$$
for some constants $a$, $b$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2322398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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|
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.
LHS $= 2\cos^{2}3x-1$
$=2{(4\cos^{3}x-3\cos{x})}^2-1$
$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$
$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$
Still thank you for the answers!
|
Another way:
$$\cos(3\cdot2x)=4\cos^3(2x)-3\cos2x$$
Now use $\cos2x=2\cos^2x-1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2324685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\frac{c}{a}$
$α+δ+β+δ=\frac{-2B}{A}$
⇒ $α+β+2δ =\frac{-2B}{A}$
$(α+δ)(β+δ)=\frac{C}{A}$
⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$
⇒$\frac{c}{a}+αδ+βδ+δ^2=\frac{C}{A}$ ⇒ $αδ+βδ+δ^2=\frac{Ca-cA}{Aa}$
$(α+β)^2=\frac{4b^2}{a^2}$
⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$
$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$
⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)
$(α+β+2δ)^2 =\frac{4B^2}{A^2}$
⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$
⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$
⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)
From (1) and (2) we get,
$\frac{4b^2-2ac}{a^2}=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$
My problem: I tried simplifying it further but could not reach the required result. A continuation of my method would be more appreciated compared to other methods.
|
By the formula for the roots of a quadratic equation, the squared difference between them is $$\left(\frac{\pm\sqrt{b^2-ac}}{a}\right)^2=\frac{b^2-ac}{a^2}$$ (factor $4$ omitted) and is invariant by translation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Prove by this inequality to with $x^2+y^2+z^2=1$ Let $x,y,z>0,x^2+y^2+z^2=1$ show that
$$\sum_{cyc}\sqrt{\dfrac{x^2+y}{2}}\ge\sqrt{2}$$
Here's what I have done. The expression $x^2+y\ge x^2+y^2$. it is is equivalent to
$$\sum_{cyc}\sqrt{1-x^2}\ge 2$$ shows that the function is neither concave or convex. So I don't think Jensen useful here.
|
We need to prove that
$$\sum_{cyc}\sqrt{\frac{x^2+y^2}{2}}\geq\sqrt2$$ or
$$\sum_{cyc}\sqrt{x^2+y^2}\geq2,$$
which is true because
$$\sum_{cyc}\sqrt{x^2+y^2}=\sqrt{\sum_{cyc}(x^2+y^2+2\sqrt{(x^2+y^2)(x^2+z^2)})}\geq$$
$$\geq\sqrt{\sum_{cyc}(x^2+y^2+2x^2)}=2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2329911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{(x+4)}{\sqrt{6}})+C$
However the result is $\frac{arctan(\frac{(x+4)}{\sqrt{6}})}{\sqrt{6}}+C$
Why is my result wrong? I can't see any mistake.
|
Start U-substitution:
$u$ = $x$+4 and $dx$= 1$dx$ $$\\$$
Making the integral:
$$\int \frac{1}{6+(u)^2} du$$ $$\\$$
Now do another U-substitution:
$u$ = $\sqrt6v$ and $du$= $\sqrt6 dv$ $$\\$$
Making the integral:
$$\int \frac{1}{\sqrt6+(v^2+1)} dv$$ $$\\$$
Take out the constant:
$$\frac{1}{\sqrt6}\int \frac{1}{(v^2+1)} dv$$
The integrand is a common integral.
The integral of $\frac{1}{(v^2+1)} dv$ is $arctan(v)$.
So now you have:
$\frac{1}{\sqrt6} arctan(v)$
Don't forget to substitute back in $v$ and $u$.
$v$= $\frac{u}{\sqrt6}$ and $u$= $x$+4. $$\\$$
Your final answer is:
$$\frac{1}{\sqrt6} arctan(\frac{x+4}{\sqrt6}) + C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
convergence of $\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $ For which value $\alpha$ does the following integral converges:
$\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $
I tried to use: $ \sqrt{x^2 + 2x + 2} - 1 = \frac{x^2 + 2x + 1}{\sqrt{x^2 + 2x + 2} + 1} $
I wanted to simplify it more but that didn't lead to anything.
Thank you for your help.
|
$$\sqrt{x^2+2x+2}-x-1=\frac{x^2+2x+2-(x^2+2x+1)}{\sqrt{x^2+2x+2}+x+1}=\frac{1}{\sqrt{(x+1)^2+1}+x+1}\sim \frac{1}{2(x+1)}$$
So our integral converges if
$$\int_0^\infty\frac{dx}{(2(x+1))^\alpha}$$
converges, do you know when this integral converges? You can compute the anti derivative to assure yourself.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
|
HINT:
You could further find $$z+\frac{1}{z}=3$$
and note that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$
|
{
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"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
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|
Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square? Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square?
I haven't been able to figure out which properties of perfect squares disallows such a number from being a perfect square.
|
Let $x$ be a positive integer with $2n$ digits when written in base $b>1$, where the $n+1$ first digits are 1's, and where we will assume that $b-1=d^2$ is a perfect square. Note that this holds in the decimal system ($b=10$), as $10-1=9=3^2$. It also holds in the binary system ($b=2$), and with respect to several other bases.
Assume that $x$ is a perfect square, $x=c^2$. Then
$$b^{2n-1}+\dots+b^{n-1}\leq c^2\leq b^{2n-1}+\dots +b^{n-1}+(b-1)\Bigl(b^{n-2}+\dots+1\Bigr)$$
Summing the geometric series, we get
$$b^{n-1}{b^{n+1}-1\over b-1}\leq c^2\leq b^{n-1}{b^{n+1}-1\over b-1}+(b^{n-1}-1)$$
Multiply by $b-1=d^2$ and subtract $b^{2n}$. This gives
$$-b^{n-1}\leq c^2d^2-b^{2n}\leq -b^{n-1}+(b-1)(b^{n-1}-1)$$
It follows that
$$-b^{n-1}\leq (cd+b^n)(cd-b^n)<b^{n}$$
Now $d$ divides $b-1$, and hence $d$ cannot divide $b^n$. So $cd-b^n$ is a nonzero integer, but then one of the inequalities above must be violated.
It follows that the assumption that $x$ was a perfect square cannot hold.
Edit: An example to show that the result is not true in all bases. Let $b=n=3$, and observe that $(111101)_3=1\cdot 3^5+1\cdot 3^4+1\cdot 3^3+1\cdot 3^2+1\cdot 3^0=361=19^2$ is a perfect square with $n+1=4$ initial 1's. But then $b-1=2$ is not a perfect square, as was required in the argument above.
|
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|
$a-b \nmid a^2+b^2$ if $a>b+2$ and $\gcd(a,b)=1$ Any two co-prime number $a,b$ with $a>b+2$ we have $a^2+b^2$ is not divisible by $a-b$, $a,b \in \mathbb{N}$.
But how to prove this?
|
If $a-b \mid a^2+b^2$ then $\gcd(a^2+b^2,a-b)=a-b \ge 3$. Now
$$
\gcd(a^2+b^2,a-b)=\gcd(a^2+b^2-(a-b)^2,a-b)=\gcd(2ab,a-b).
$$
Since $\gcd(a,b)=1$ then every prime dividing $a$ or $b$ cannot divide $a-b$. Hence
$$
\gcd(a^2+b^2,a-b) \in \{1,2\}.
$$
In particular, your conjecture is true.
|
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|
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
positive point of maximum lies in the interval
$(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$
I found $f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$ and Let the point of maximum is $x_1>0$
So $f''(x)=6x+6(a-7)<0$
$\implies 6x_1+6(a-7)<0$
But i am stuck here.The answer given in my book is $a_2+11a_3+70 a_4=320$.From this answer i cannot know what is the interval $(a_1,a_2)\cup(a_3,a_4)?$
|
The point of maximum is the least root of $f'(x)=0$. In order for it to be positive, both roots must be. So we need, after writing the equation as
$$
x^2+2(a-7)x+(a^2-9)=0,
$$
that the following conditions are satisfied:
\begin{cases}
-2(a-7)>0 & \text{sum of roots is positive} \\
a^2-9>0 & \text{product of roots is positive} \\
4(a-7)^2-4(a^2-9)>0 & \text{discriminant is positive}
\end{cases}
|
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|
If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this equation are $a,b,c$. If I were to find the equation having the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$ then I can use the root co-efficient relations and use the rule of creating equations from roots.
This gives me the result : $$4x^3 + 2x^2 -3x -1 =0$$
A much much simpler way to solve this math is: if we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of $x$ are $a,b$ and $c$ then $f\left(\frac{1}{2x}\right)$ will denote an equation (if we just write $f\left(\frac{1}{2x}\right)=0$ ) which has the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$. And this way the result also matches the result of my previous work.
But now I face an equation
$$x^3+3x+1=0$$ and if the roots are $a,b,c$ then I have to find the equation with the roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
If I use the root-coefficient relations then I have the result
$$x^3+6x^2+9x+5=0$$ (which is correct I think).
But here $f\left(\frac{1-x}{x}\right)= 3x^3-6x^2+3x-1$. So writing $f\left(\frac{1-x}{x}\right)=0$ won't give me the actual equation.
Where am I mistaking actually?
|
Hint: let $y=(1-x)/x \iff x=1 / (y+1)\,$, then:
$$
P(x)=P\left(\frac{1}{y+1}\right)=\frac{1}{(y+1)^3}\big(1+3(y+1)^2+(y+1)^3\big)
$$
The polynomial in $y$ is therefore:
$$
1+3(y+1)^2+(y+1)^3=y^3 + 6 y^2 + 9 y + 5
$$
This can be verified in WA by calculating the resultant[ x^3+3x+1, 1-x-xy, x ].
|
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|
Simplify $5(3 \sin(x) + \sqrt3 \cos(x))$ So Wolfram tells me I can reach $10 \sqrt3 \sin(x + π/6)$ from this expression, but I cant grasp how to do it. Any help is appreciated.
|
We know
$$\sin (x + a) = \sin x \cos a + \sin a \cos x$$
While we have
$$15 \sin x + 5 \sqrt 3 \cos x \tag {*}$$
Therefore if we can find an numbers $a, C$ such that $15 = C \cos a$ and $5\sqrt3 = C\sin a$, we can rewrite $(*)$ as
$$C \cos a \sin x + C \sin a \cos x = C\sin (x+a)$$
To find appropriate $C$ and $a$, notice that we have a system of two equations
$\left\{
\begin{array}{c}
C \cos a = 15 \\
C \sin a = 5 \sqrt 3 \\
\end{array}
\right.$
Square both sides of both equations and add the resulting equations. We get:
$C^2 \sin ^2 a + C^2 \cos ^2 a = 15^2 + 25 \cdot 3$
$C^2(\sin^2 a + \cos^2 a) = 300$
$C^2 = 300$
$C = 10 \sqrt 3$
Now subsitute in one of the equations
$10 \sqrt 3 \sin a = 5 \sqrt 3$
$\sin a = \dfrac 12$
$a = \dfrac {\pi}{6}$
Another possible value for $a$ is $\dfrac {5 \pi}{6}$, but only $\dfrac {\pi}{6}$ satisfies both equations.
|
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|
Definite integral of exponential of square root Does anyone have a clue as to how integrals of the kind
$$
I(a) = \int_{-\infty}^{\infty} dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
$$
or
$$
I_2(a) = \int_{-\infty}^{\infty} dt e^{-\sqrt{t^2 + a^2}}
$$
can be solved? I realize there is a problem as the square root function has branch cuts, but I do not have a clear idea how to deal with them in this case.
These integrals often arise when calculating matrix elements between electronic wavefunctions in cylindrical coordinates.
|
For this one $$
I(a) = \int_{-\infty}^{\infty} dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
$$
Note that $(\ln |t+ \sqrt {t^2 + a^2} |)' = \frac {1}{\sqrt {t^2 +a^2}}$
Next notice that $(e^{-\sqrt{t^2 + a^2}})' = -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2}$
so $(\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}})' = \frac {e^{-\sqrt{t^2 + a^2}}}{\sqrt {t^2 +a^2}} + (\ln |t+ \sqrt {t^2 + a^2} |) * -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2}$ hence the $ \int dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
= (\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}}) -\int (\ln |t+ \sqrt {t^2 + a^2} | * -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2})$ using integration by parts let $dv= (-e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2})$ so $v =e^{-\sqrt{t^2 + a^2}} $.
wanted to post this as a comment but it wouldnt fit if its nonsense lemme know and ill delete it.
$ \int dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
= (\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}}) - (\ln |t+ \sqrt {t^2 + a^2} | * e^{-\sqrt{t^2 + a^2}}) - \int e^{-\sqrt{t^2 + a^2}}$
which jack already solved $\int e^{-\sqrt{t^2 + a^2}}$ above.
|
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|
Simplify $x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$ to $x^2 +45x-8=0$ Simplify ($x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$) to ($x^2 +45x-8=0$)
These two equations have the same solutions.
If given the first equation how would you go about simplifying such that you end up with the quadratic form ?
I tried cubing the first expression but it got really messy.
|
Cubing is not a mess. Look here:
$$x^{2/3}+3 x^{1/3}=2$$
cubing both sides
$$\left(x^{2/3}+3 x^{1/3}\right)^3=8$$
Expand:
$$27 x^{4/3}+9 x^{5/3}+x^2+27 x=8$$
Collect $x$ between (among?) the first two terms:
$$9 \left(x^{2/3}+3 x^{1/3}\right) x+x^2+27 x=8$$
Inside the parenthesis there is the left side of the given equation, which is 2
$$9 \cdot 2 \cdot x+x^2+27 x=8$$
Rearrange and get the second equation
$$x^2+45 x-8=0$$
The two equations have the same real solutions. Usually cubing an equation creates fake solutions, as in the following example:
$$x=1$$
has the solution $x=1$, while
$$x^3=1$$
has three solutions, the cubic roots of 1
$$1,\frac{1}{2} \left(1+i\sqrt{3} \right),\frac{1}{2} \left(-1+i\sqrt{3} \right)$$
Two of them are fake solutions.
|
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|
Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? And why?
\begin{pmatrix}5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}
|
It is easy to see, that $v=(1,1,1,1,1,1,1)^T$ is an eigenvector of that matrix. By calculation the corresponding eigenvalue is $35$ (just calculate $Av$).
since the rank of the matrix is $1$ and it has the eigenvalues with their multiplicities as zeros it has to be of the form $p(t) = a t^6 (t-35)$ with $a\neq 0$
|
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|
what is $\sqrt{(i^4)}$ We know that,
$(a^m)^n = a^{mn} = (a^{n})^m$
So ,
*
*$\sqrt {i^4} = (i^{1/2})^4 = \left(\pm\frac{1+ i}{\sqrt2} \right)^4 = -1$
*$\sqrt {i^4} = (i^{2}) = -1$
*$\sqrt {i^4} = \sqrt1 = 1$
I think only no $3$ is right. I must have violated some rule in $1$ & $2$. Please let me know.
Maybe $(a^m)^n = a^{mn} = (a^{n})^m$ is true only for Real Numbers.
|
$$\sqrt{i^4}=i^2 = -1\tag1$$
Or
$$\sqrt{i^4}= \sqrt{(\sqrt{-1})^4} = (\sqrt{-1})^2 = -1 = i^2 \tag2$$
Or
$$\sqrt{i^4} = \sqrt{i^2\cdot i^2} = \sqrt{i^2}\cdot \sqrt{i^2}= i\cdot i = i^2 = -1\tag3$$
|
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|
In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
Alright, so if d is the midpoint then my two triangles are ADB and ADC.
Using the law of cosines I get:
$$3^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos \angle ADB$$
for the first triangle,
$$4^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos\angle ADC$$
for the second one. Then I'm stuck. Any help?
|
By Stewart's theorem the squared length of the median from $A$ is given by $\frac{2b^2+2c^2-a^2}{4}$.
In our case we know that $c=3, b=4$ and $2b^2+2c^2-a^2 = 4a^2$, hence
$$ a^2 = \frac{2}{5}(b^2+c^2) = 10 $$
and $a=\color{red}{\sqrt{10}}$.
|
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|
second derivative of a function equals a power of said function Is it possible to find an expression for $f(x)$ so that:
$$f''(x) = (f(x))^{-2}$$
and
$$f(0) = 1$$
I suspect that no such expression exists but i might be wrong.
Thanks in advance.
|
One can take the hard path and let
$$f(x) = 1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots,$$
since $f(0) = 1 = a_{0}$, such that
\begin{align}
1 &= f^2 \, f'' = (1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots)^2 \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\
&= (1 + 2 a_{1} x + (2a_{2} + a_{1}^2) x^2 + \cdots) \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\
&= 2 a_{2} + 2(2 a_{1} a_{2} + 3 a_{3} ) x + 2 (6 a_{4} + 6 a_{1} a_{2} + a_{1}^2 a_{2} + a_{2}^3) x^2 + \cdots.
\end{align}
Equating the equations for the coefficients yields
\begin{align}
f(x) &= 1 + a_{1} x + \frac{x^2}{2!} - 2 a_{1} \, \frac{x^3}{3!} + \left( 6 a_{1}^2 - \frac{1}{2} \right) \, \frac{x^4}{4!} + a_{1} \, (13 - 24 a_{1}^2) \, \frac{x^5}{5!} \\
& \hspace{10mm} + ( 30 \, a_{1}^{2} \, (4 a_{1}^{2} - 5) + 1) \, \frac{x^6}{6!} + \cdots.
\end{align}
Further terms may be obtained by considering more terms. Here $a_{1}$ is arbitrary unless another condition is applied. It may be of interest that if $a_{1} = 0$, or $f'(0) = 0$, the the function $f(x)$ is an even function given by
$$f(x) = 1 + \frac{x^{2}}{2!} - \frac{1}{2} \, \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \mathcal{O}\left(\frac{x^{8}}{8!}\right).$$
|
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|
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
|
If you just want the value of expression $(a+b)^2 + (b+c)^2 + (a+c)^2$than set the vale of $a=0$, $b=1$ and $c=1$ (these satisfy the required conditions ) and you are going to get the answer.
$$(1)^2 + (2)^2 + (1)^2 = 6$$
|
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|
Where was my mistake (integration by trig-substitution problem)? I am attempting to solve the problem
$$\int \frac{dx}{x^2+x+1}$$
First, I complete the square, then factor out a $\frac {3}{4}$:
$$\int \frac{dx}{\frac{3}{4}(\frac{4}{3}(x+\frac{1}{2})^2+1)}$$
Let $u = \sqrt{\frac{4}{3}}(x+\frac{1}{2})$
$$\frac{du}{dx} = \frac{2}{\sqrt{3}}$$
$$dx = \frac{\sqrt{3}}{2} du$$
Thus, we now have the integral:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \int \frac{du}{u^2+1}$$
Let $u = \tan \theta$
$$du = \sec^2\theta \ d\theta$$
What follows is obvious now, and the solution should be:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \theta + C$$
$$\theta = \tan^{-1}(u)$$
Thus, the final solution is:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) + C$$
However, according to online calculator integral-calculator, the answer is:
$$\frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right)+C$$
Any indication as to where my mistake falls would be very beneficial.
|
Your only mistake appears to be a failure to notice that $\displaystyle \frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) $ is exactly the same thing as $ \displaystyle \frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right).$
First you have
$$
\frac 4 3 \cdot \frac{\sqrt 3} 2 = \frac{4\sqrt 3}{\sqrt 3\sqrt 3 \cdot 2} = \frac 2 {\sqrt 3}.
$$
And then
$$
\sqrt{\frac 4 3} \left( x + \frac 1 2 \right) = \frac 2 {\sqrt 3} \left( x + \frac 1 2 \right) = \frac 1 {\sqrt 3} (2x+1).
$$
|
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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.