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Points on the hypotenuse of a right-angled triangle Points $K$ and $L$ are chosen on the hypotenuse $AB$ of triangle $ABC$ $(\measuredangle ACB=90^\circ)$ such that $AK=KL=LB$. Find the angles of $\triangle ABC$ if $CK=\sqrt2CL$.
As you can see on the drawing, $CL=x$ and $CK=\sqrt2x$.
I don't know how to approach the problem at all. Since $\measuredangle ACB=90^\circ$, it will be enough to find the measure of only one of the acute angles. If $\measuredangle ACK=\varphi_1$ and $\measuredangle BCL=\varphi_2$, I have tried to apply the law of sines in triangle $KCL$, but it seemed useless at the end. Thank you! I would be grateful if I could see a solution without using coordinate geometry.
|
Draw horizontal lines from points K and L of side $\overline{AB}$ to points $K_A$ and $L_A$ of side $\overline{BC}$. Because points $K$ and $L$ trisect side $\overline{AB}$, then points $K_A$ and $L_A$ trisect side $\overline{BC}$. So $\triangle{CK_AK}$ and $\triangle{CL_AL}$ are right triangles.
Draw vertical lines from points K and L of side $\overline{AB}$ to points $K_B$ and $L_B$ of side $\overline{AC}$. Because points $K$ and $L$ trisect side $\overline{AB}$, then points $K_B$ and $L_B$ trisect side $\overline{AC}$ So $\triangle{CK_BK}$ and $\triangle{CL_BL}$ are right triangles.
We can now argue that
\begin{align}
CL &= \sqrt 2 \; CK \\
\sqrt{\left(\dfrac 23b \right)^2 + \left(\dfrac 13a \right)^2}
&= \sqrt 2 \cdot \sqrt{\left(\dfrac 13b \right)^2
+ \left(\dfrac 23a \right)^2} \\
a^2 + 4b^2 &= 2(4a^2+b^2) \\
a^2 + 4b^2 &= 8a^2+2b^2 \\
2b^2 &= 7a^2 \\
b &= \sqrt{\dfrac{7}{2}} a
\end{align}
Since $\triangle{ACB}$ is a right triangle,
\begin{align}
c^2 &= a^2 + b^2 \\
c^2 &= a^2 + \dfrac 72 a^2 \\
c^2 &= \dfrac 92 a^2 \\
c &= \dfrac{3}{\sqrt 2} a
\end{align}
It follows that $a:b:c = \sqrt 2 : \sqrt 7 : 3$.
$m\angle A = \arcsin \dfrac{\sqrt 2}{3} \approx 28.13^\circ$.
|
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"url": "https://math.stackexchange.com/questions/4222311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$ If $f\colon\left[ { - 1,1} \right] \to \mathbb R$ be continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$, then $\mathop {\lim }\limits_{x \to 0} \frac{{f\left( {\cos x} \right)}}{{\sin x}}$ is _______.
My solution is as follow
$x = \cos \left( {\frac{\theta }{2}} \right)$
$f\left( {2{{\cos }^2}\left( {\frac{\theta }{2}} \right) - 1} \right) = \left( {{{\cos }^3}\left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right)} \right)f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
$\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} = \frac{{\left( {{{\cos }^3}\left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right)} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)\cos \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
$\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} = \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
How do I proceed from here
|
Notice that $f(1)=0$. Since the function is continuous, $\lim_{x\to1}f(x)=0$
Also, $4xf'(2x^2-1)=(x^3+x)f'(x)+f(x)(3x^2+1)\implies f'(1)=\lim_{\theta\to0}f'(\cos(\theta/2))=0$
Now
$$\lim_{\theta\to0}\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} =\lim_{\theta\to0} \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)=\frac00$$
since $\lim_{\theta\to0}f(\cos(\theta/2))=\lim_{x\to1}f(x)=0$
Therefore by L'Hopital's
$$\lim_{\theta\to0}\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} =\lim_{\theta\to0} \frac{{\left(\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)\right)'}}{{\left(2\sin \left( {\frac{\theta }{2}} \right)\right)'}}=\lim_{\theta\to0} \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)f'\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)}\frac{\sin(\theta/2)}{2}+f(\cos(\theta/2))\cos(\theta/2)\sin(\theta/2)}{{\cos \left( {\frac{\theta }{2}} \right)}}=??$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4226527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the sum of $5.5+55.55+555.555..$ up till n terms?
Find the sum of $5.5+55.55+555.555..$ up till n terms?
My attempt: $ 5.5+55.55+555.555 ... $
$ 5(1.1+11.11+111.111...) $
$ \dfrac{5}{9} \times 9(1.1+11.11+111.111..) $
$ \dfrac{5}{9} (9.9+99.99+999.999...) $
$ \dfrac{5}{9} (9+0.9+99+0.99+999+0.999...) $
$ \dfrac{5}{9} [(9+99+999 ...)+(0.9+0.99+0.999...)] $
$ \dfrac{5}{9} [(10-1+100-1+1000-1...)+(1-0.1+1-0.01+1-0.001...] $
So all the $1$ will cancel
$ \dfrac{5}{9} [(10+100+1000 ... n)+(n-(0.1+0.01+0.001...)] $
How to move forward? I can see two Geometric Progession in the 2 brackets but can't prove that? How do I continue? And is there any easier and less time taking method?
I have not as of yet learned summation.
|
You have to make a general formula so you may use an already wxisting formula for sum of $n$ terms of a $G.P.$
You arrived at $ \dfrac{5}{9} [(10+100+1000 ... n)+(n-(0.1+0.01+0.001...)] $
For a $G.P$ of type $a,ar,ar^2,\cdots ar^{n-1}$ The sum is given as
$S=a\displaystyle\frac{r^n-1}{r-1}$
In the first bracket $a=10$ and $r=10$ , for second bracket $ a=0.1$ and $r=0.1$ and hence we get
$\displaystyle S= \dfrac{5}{9} \bigg[(10*\frac{10^n-1}{10-1})+(n-\frac{1}{10}*\frac{1-\frac{1}{10})^n}{1-\frac{1}{10}})\bigg] $
$\displaystyle S= \dfrac{5}{9} \bigg[10*\frac{10^n-1}{9}+n-\frac{10^n-1}{9.10^n})\bigg] $
$S=\displaystyle\dfrac{50}{81}\bigg[10^n-1+\frac{9n}{10}-\frac{10^n-1}{10^{n-1}}\bigg]$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the value of $\sum_{n=1}^\infty \frac{1}{F_n F_{n+4}}$ where $F_n$ is nth Fibonacci number? I found this problem from my old note, with the memo 'The answer is very interesting."
But I forgot how to solve it!
I rummaged through all my bookshelf, but I couldn't find any clue.
The only thing I remember is
*
*Partial fraction decomposition is used.
*It would be easier to solve by using $\sum_{n=1}^\infty \frac{1}{F_n F_{n+2}}=\sum_{n=1}^\infty \left( \frac{1}{F_n F_{n+1}} - \frac{1}{F_{n+1} F_{n+2}} \right) = 1$.
*The solution is not that complicated.
Would you help me?
|
It can be reduced to $\sum \frac1{F_n F_{n+2}}$ as follows.
Multiply the numerator and denominator by $F_{n+2}$, and write this in terms of $F_n$ and $F_{n+4}$ by using that the sequences $(F_{2n})$ and $(F_{2n+1})$ are linear recurrences.
You will obtain something like
$$\frac1{F_{n}F_{n+4}} = \frac{F_{n+2}}{F_{n}F_{n+2}F_{n+4}} = \frac{a}{F_nF_{n+2}} + \frac b{F_{n+2} F_{n+4}}$$
so that we obtain roughly $(a+b) \sum \frac1{F_n F_{n+2}}$.
Details:
In fact, $F_{n+4} = c_1 F_{n+2} + c_2 F_n$ with $c_1$ and $c_2$ such that the roots of the polynomial $x^2 - c_1 x - c_2$ are $\varphi^2$ and $\overline{\varphi}^2 = \varphi^{-2}$. In particular:
*
*$c_1 = \varphi^2 + \varphi^{-2} = (\varphi - \varphi^{-1})^2 - 2 = (\sqrt 5 F_1)^2 - 2 = 3$;
*$c_2 = -1$.
We obtain
$$\begin{align*}
\sum_{n=1}^\infty \frac1{F_n F_{n+4}}
&= \sum_{n=1}^\infty \frac{F_{n+2}}{F_n F_{n+2} F_{n+4}} \\
&= \sum_{n=1}^\infty \frac{\frac1{c_1} F_{n+4} - \frac{c_2}{c_1} F_{n+1}}{F_n F_{n+2} F_{n+4}} \\
&= \frac13 \sum_{n=1}^\infty \left(\frac{1}{F_n F_{n+2}} + \frac{1}{F_{n+2} F_{n+4}} \right) \\
&= \frac23 \sum_{n=1}^\infty \frac{1}{F_n F_{n+2}} - \frac13 \frac1{F_1 F_3} - \frac13 \frac1{F_2 F_4} \\
&= \frac23 - \frac16 - \frac19 \\
&= \frac7{18} \,.
\end{align*}$$
|
{
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"url": "https://math.stackexchange.com/questions/4228802",
"timestamp": "2023-03-29T00:00:00",
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|
Proving $(x^2 - y^2) = (x + y)(x-y)$ I am trying to prove from the field axioms that $(x^2 - y^2) = (x+y)(x-y)$. I am going to take for granted the definition of subtraction as the addition of a negation and that $x(-y) = (-x)y = -(xy)$. Here is my attempt.
We have:
\begin{align*}
(x+y)(x-y) & = x(x-y) + y(x-y) & & \text{distributive law} \\
& = x(x + (-y)) + y(x + (-y)) & & \text{definition of subtraction} \\
& = (xx + x(-y)) + (yx + y(-y)) & & \text{distributive law} \\
& = (x^2 + (-(xy))) + (yx + (-y^2)) & & \text{uniqueness of additive inverse; definition of square} \\
& = (x^2 + (-(yx))) + (yx + (-y^2)) & & \text{commutativity of multiplication} \\
& = (x^2 + \left(-(yx) + yx \right) + (-y^2) & & \text{associativity of addition} \\
& = (x^2 + 0) + (-y^2) & & \text{additive inverse axiom} \\
& = x^2 + (-y^2) & & \text{additive identity axiom} \\
& = x^2 - y^2 & & \text{definition of subtraction}
\end{align*}
How does this look?
|
Looks good overall. I do have one nitpick:
At least for me, it is typically not given that the additive inverse is unique in fields, or that $x(-y) = -(xy)$ for sure. (Or the more general $(-1)(xy) = (-x)y = x(-y)$.)
I believe that technically needs to be shown if you want to go from the axioms.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$
Compute the limit:
$$\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$$
I tried applying the sandwich rule, constructing an upper and lower bound for the sequence, but I can't get the bounds to have the same limit.
Hints?
|
Let $S_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}$
The $k^{th}$ term = $t_k = \frac{kn}{n^3+k} \leq \frac{kn}{n^3} = \frac{k}{n^2}$, $\forall{k} \geq 1$
Then, $S_n=\sum\limits_{k=1}^{n}t_k \leq \sum\limits_{k=1}^{n}\frac{k}{n^2} = \frac{1}{n^2}\sum\limits_{k=1}^{n}k=\frac{1}{n^2}.\frac{n(n+1)}{2}=\frac{1}{2}+\frac{1}{2n}=U_n$ (let), $\forall{n} \in \mathcal{N}$
$\lim\limits_{n \to \infty} U_n = \frac{1}{2} \Rightarrow \lim\limits_{n \to \infty} S_n \leq \frac{1}{2}$.
Also, we have, $t_k = \frac{kn}{n^3+k} \geq \frac{kn}{n^3+n} = \frac{k}{n^2+1}$, $\forall{k} \leq n$
Again, $S_n=\sum\limits_{k=1}^{n}t_k \geq \sum\limits_{k=1}^{n}\frac{k}{n^2+1} = \frac{1}{n^2+1}\sum\limits_{k=1}^{n}k=\frac{1}{n^2+1}.\frac{n(n+1)}{2}=\frac{1}{2}.\frac{1+\frac{1}{n}}{1+\frac{1}{n^2}}=V_n$ (let), $\forall{n} \in \mathcal{N}$
$\lim\limits_{n \to \infty} V_n = \frac{1}{2} \Rightarrow \lim\limits_{n \to \infty} S_n \geq \frac{1}{2}$.
Hence, we have sandwich, $U_n \geq S_n \geq V_n$, $\forall{n} \in \mathcal{N}$ and $\lim\limits_{n \to \infty} U_n=\lim\limits_{n \to \infty} V_n=\frac{1}{2}$.
The next figure shows the convergence:
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $\frac{2}{3}$. But how to show?
|
I'll provide a proof that do not use integrals (and will probabily make you apreciate FTC). We can add an extra $\sqrt{n}$ without changing the limit. Will show now that the following limit exists.
\begin{gather*}
L=\lim_{n \to +\infty}\frac{\sum_{i=1}^n \sqrt{i}}{n\sqrt{n}}
\end{gather*}
To show the limit exist, notice that, notice that
\begin{align*}
\frac{i}{\sqrt{i+1} + \sqrt{i}} &\leq \frac{i}{2\sqrt{i}} \leq \frac{i}{\sqrt{i} + \sqrt{i-1}} \\
2i(\sqrt{i+1} - \sqrt{i}) &\leq \sqrt{i} \leq 2i(\sqrt{i} - \sqrt{i-1})
\end{align*}
Now
\begin{gather*}
\sum_{i=1}^n 2i(\sqrt{i+1} - \sqrt{i}) = 2\sum_{i=1}^n (i+1)\sqrt{i+1} - i\sqrt{i} - \sqrt{i+1} = \\
= 2 \left( (n+1)\sqrt{n+1} - 1 - \sum_{j=2}^{n+1}\sqrt{j} \right) = 2\left( n\sqrt{n+1} - \sum_{j=1}^{n}\sqrt{j} \right)
\end{gather*}
so
\begin{align*}
2n\sqrt{n+1} - 2\sum_{i=1}^{n}\sqrt{i} &\leq \sum_{i=1}^n \sqrt{i} \\
\frac{2}{3}n\sqrt{n+1} &\leq \sum_{i=1}^n \sqrt{i}
\end{align*}
Proceding similarly with the other inequality yields
\begin{gather*}
\sum_{i=1}^n 2i(\sqrt{i} - \sqrt{i-1}) = 2\sum_{i=1}^n i\sqrt{i} - (i-1)\sqrt{i-1} - \sqrt{i-1} = \\
= 2 \left( n\sqrt{n} - \sum_{j=0}^{n-1}\sqrt{j} \right) = 2\left( (n+1)\sqrt{n} - \sum_{j=1}^{n}\sqrt{j} \right)
\end{gather*}
so
\begin{align*}
\sum_{i=1}^n \sqrt{i} &\leq 2(n+1)\sqrt{n} - 2\sum_{i=1}^{n}\sqrt{i} \\
\sum_{i=1}^n \sqrt{i} &\leq \frac{2}{3}(n+1)\sqrt{n}
\end{align*}
Finally, we can apply squeeze theorem as
\begin{gather*}
\frac{2}{3} \sqrt{1 + \frac{1}{n}} \leq \frac{\sum_{i=1}^n \sqrt{i}}{n\sqrt{n}} \leq \frac{2}{3} (1 + \frac{1}{n})
\end{gather*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find min $ P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$ under $x,y,z >0$ and $xyz=1$ Let $x,y,z >0$ and $xyz=1$
Find min: $P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$
My trying but it false :<<
We have: $VT=\frac{1}{\frac{x+2}{x}}+\frac{1}{\frac{y+2}{y}}+\frac{1}{\frac{z+2}{z}}$
$=\frac{1}{1+\frac{2}{x}}+\frac{1}{1+\frac{2}{y}}+\frac{1}{1+\frac{2}{z}} $
-> $Bunhiacopxki$:
$⇒P \geq \frac{9}{3+2.(\frac{xy+yz+xz}{xyz})}=\frac{9}{3+2.(xy+yz+xz)}$
We have $\frac{(x+y+z)^2}{3} \geq xy+yz+xz $ and $x+y+z \geq 3\sqrt[]{xyz}=3$
(it flase, please don't talking about that :<<)
$⇒P \geq \frac{9}{3+2.(xy+yz+xz)} \geq \frac{9}{6}=\frac{3}{2}$
$=>P \geq \frac{3}{2}$
$x=y=z=1$
|
Let $ a = \ln x, b = \ln y, c = \ln z$
Want to minimize $ \sum \frac{ e^a}{e^a+2}$ subject to $ a + b + c = 0$.
Since the derivative is $ \frac{2e^a}{(e^a+2)^2} > 0$, hence we can apply Jensen's inequality to conclude that the minimum occurs at $ a = b = c = 0$, or when $ x = y = z = 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4232728",
"timestamp": "2023-03-29T00:00:00",
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|
Determinant of an interesting Toeplitz matrix Let $ab=1$. Find
$$\begin{vmatrix} c & a & a^2 & ... & a^{n-1} \\ b & c & a & \dots & a^{n-2} \\ b^2 & b & c& \dots &a^{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b^{n-1} & b^{n-2} & b^{n-3} & \dots & c \end{vmatrix}$$
I tried to decompose by line however it does not lead to a solution
|
Let $M$ be the matrix at hand. Its entries have the form
$$M_{ij} = (c-1)\delta_{ij} + b^{i-j} = (c-1)\delta_{ij} + b^i a^j$$
where $\delta_{ij} = \begin{cases}1, & i = j\\ 0,& i \ne j\end{cases}$ is the Kronecker delta. We can express $M$ as a rank-1 update of the diagonal matrix $(c-1)I_n$:
$$M = (c-1)I_n + uv^T
\quad\text{ where }\quad
\begin{cases}
u &= (b,b^2,\ldots,b^n)^T\\
v &= (a,a^2,\ldots,a^n)^T
\end{cases}
$$
When $c \ne 1$, by matrix determinant lemma, we have:
$$\det(M) = (c-1)^n \det\left(I_n + \frac{1}{c-1}uv^T\right)
= (c-1)^n\left( 1 + \frac{1}{c-1}v^Tu\right)$$
Since $v^T u = \sum\limits_{k=1}^n a^k b^k = \sum\limits_{k=1}^n 1 = n$, we obtain
$$\det(M) = (c-1)^n\left( 1 + \frac{n}{c-1}\right) = (c-1)^{n-1}(c + n - 1)\tag{*1}$$
When $c = 1$, $M = uv^T$ implies the columns of $M$ are linearly dependent. This forces $\det(M) = 0$ and formula $(*1)$ works at $c = 1$.
|
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"url": "https://math.stackexchange.com/questions/4233533",
"timestamp": "2023-03-29T00:00:00",
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|
Alternative and more direct proof that an integral is independent of a parameter I'm looking for alternative ways to calculate the integral
$$
\int\limits_0^\infty\frac{\tanh(\alpha x)}{\tanh(\pi x)}\sin(2\alpha x^2)\,dx=\frac{1}{4},\qquad \alpha>0.\tag {*}
$$
It was derived in a lengthy calculation using roundabout method from Fourier transform of a function of two variables. However, it seems like the integral (*) might have a simple proof? Note that when $\alpha=\pi$, the the integral reduces to Fresnel integral
$$
\int\limits_0^\infty\sin(2\pi x^2)\,dx=\frac{1}{4}.
$$
Thus if one could prove that it is independent of the parameter $\alpha$, then its value could be found.
|
Since no one has posted an answer, I'm going to use contour integration to show that the equation holds for the next "simplest" case, namely $\alpha = 2 \pi$.
$$\begin{align} \int_0^\infty \frac{\tanh(2 \pi x)}{\tanh(\pi x)} \sin (4 \pi x^2) \, \mathrm dx &= \int_{0}^{\infty} \frac{2 \cosh^2(\pi x)}{\cosh (2 \pi x)} \sin (4 \pi x^2) \, \mathrm dx\\ &= \int_0^\infty \left(1+ \frac{1}{\cosh (2 \pi x)} \right) \sin(4 \pi x^{2}) \, \mathrm dx \\ &= \frac{1}{2} \int_0^\infty \left(1+ \frac{1}{\cosh (\pi u)} \right) \sin (\pi u^2) \, \mathrm d u \\ &= \frac{1}{2}\left( \int_0^\infty \sin (\pi u^2) \, \mathrm du + \int_0^\infty \frac{\sin(\pi u^2)}{\cosh(\pi u)} \, \mathrm du \right) \\ &= \frac{1}{2} \left(\frac{\sqrt{2}}{4} + \int_0^\infty \frac{\sin(\pi u^2)}{\cosh(\pi u)} \, \mathrm du\right) \end{align}$$
To evaluate the integral $$\int_0^\infty \frac{\sin(\pi u^2)}{\cosh(\pi u)} \, \mathrm du $$ we can integrate the complex function $$f(z) = -\frac{e^{i \pi (z^{2}+1/4)}}{\sinh(2 \pi z)}$$ counterclockwise around an indented rectangular contour with vertices at $\pm R \pm \frac{i}{2} $.
As $R \to \infty$, the integral vanishes on the left and right sides of the rectangle since the hyperbolic sine function grows exponentially as $\Re(z) \to \pm \infty$.
And since $$f \left(x- \frac{i}{2} \right)- f \left(x+ \frac{i}{2} \right) = \frac{ e^{i \pi x^{2}}}{\cosh (\pi x)}, $$ we get
$$ \begin{align} \int_{-\infty}^\infty \frac{e^{i \pi x^2}}{\cosh(\pi x)} \, \mathrm dx &= \pi i \operatorname{Res}\left[f(z), -i/2\right] + 2 \pi i \operatorname{Res}[f(z), 0] + \pi i \operatorname{Res}[f(z), i/2] \\ &= \pi i \left(\frac{1}{2 \pi} \right) + 2 \pi i \left(- \frac{\sqrt{2}}{4 \pi} (1+i) \right) + \pi i \left(\frac{1}{2 \pi} \right) \\ &= \frac{\sqrt{2}}{2} + i \left( 1- \frac{\sqrt{2}}{2}\right). \end{align}$$
Therefore, $$\int_0^\infty \frac{\sin(\pi x^2)}{\cosh(\pi x)} \, \mathrm dx = \frac{1}{2}- \frac{\sqrt{2}}{4}, $$ and $$\int_0^\infty \frac{\tanh(2 \pi x)}{\tanh(\pi x)} \sin (4 \pi x^2) \, \mathrm dx = \frac{1}{2} \left(\frac{\sqrt{2}}{4} + \frac{1}{2} - \frac{\sqrt{2}}{4} \right) = \frac{1}{4}. $$
|
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|
Use Epsilon Definition of Limit to Prove Limit Exists We have the function
$$f(x,y)=
\begin{cases}
0 & \textrm{if } x=y=0 \\[2ex]
(x+y)\ln(x^2+y^2) & \textrm{otherwise}
\end{cases}$$
and need to prove $\lim\limits_{(x,y) \to (0,0)} f(x,y) = 0$.
We need to show that for any $\epsilon>0$ we can find a $\delta > 0$ such that for any $(x.y)$ with $0< \sqrt{x^2+y^2}<\delta$, we have $|f(x,y)| < \epsilon$
Some of the steps are:
$$\begin{align}|f(x,y)| &= \left|(x+y)\ln(x^2+y^2)\right| \\&= \left|\frac{2(x+y)}{\sqrt{x^2+y^2}}\sqrt{x^2+y^2}\ln(\sqrt{x^2+y^2})\right| \\&\le 2\sqrt{2}\left|\sqrt{x^2+y^2}\ln(\sqrt{x^2+y^2})\right|
\end{align}$$
but I do not understand how you get including expression including $\sqrt{x^2+y^2}$ after plugging in the value to the absolute value and how that relates to the inequality. There has to be a step I am missing.
|
You can actually skip the second equality and note that $ |x+y| \leq \sqrt {1^{2}+1^{2}} \sqrt {x^{2}+y^{2}}$ by Cauchy-Schwarz inequality.
|
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Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r = abc $
$\Rightarrow a^2+b^2+c^2 = p^2-2q $
$ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
or
$ p^2-2q +\sqrt{2} r + 2\sqrt{2} +3 \geq (2+\sqrt {2} )p $
or
$ (p - (1+\sqrt{2} ))^2 -2q +\sqrt{2} r \geq 0 $
or
The problem I am facing is exactly what I want to prove : $\sqrt{2} r \geq 2q$ is completely wrong .
I hope to get help from everyone. Thanks very much !
|
Proceeding along the OP's idea (pqr method):
First of all, we have
\begin{align*}
p^2 &\ge 3q, \tag{1}\\
p^3 - 4pq + 9r &\ge 0. \tag{2}
\end{align*}
Note: (2) is just Schur's inequality $a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\ge 0$.
If $q \le \frac{p^2}{4}$, it suffices to prove that
$$p^2 - 2 \cdot \frac{p^2}{4} + 2\sqrt2 + 3 \ge (2 + \sqrt2)p$$
or
$$\frac12(p - 2 - \sqrt2)^2 \ge 0$$
which is true.
If $q > \frac{p^2}{4}$, using (2), it suffices to prove that
$$p^2 - 2q + \sqrt2 \cdot \frac{4pq - p^3}{9}
+ 2\sqrt2 + 3 \ge (2 + \sqrt2)p$$
or
$$\left(\frac{4\sqrt2}{9}p - 2\right)q - \frac{\sqrt2}{9}p^3 + p^2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)p. $$
We split into two cases:
*
*$p \ge \frac{9\sqrt2}{4}$:
It suffices to prove that
$$\left(\frac{4\sqrt2}{9}p - 2\right)\cdot \frac{p^2}{4} - \frac{\sqrt2}{9}p^3 + p^2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)p$$
or
$$\frac12(p - 2 - \sqrt2)^2 \ge 0$$
which is true.
*$p < \frac{9\sqrt2}{4}$:
Using (1), it suffices to prove that
$$\left(\frac{4\sqrt2}{9}p - 2\right)\cdot \frac{p^2}{3} - \frac{\sqrt2}{9}p^3 + p^2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)p $$
or
$$\frac{\sqrt2}{54}(2p + 12 + 9\sqrt2)(p - 3)^2 \ge 0$$
which is true.
We are done.
|
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|
Integrating inverse of a polynomial using valid method gives wrong result The integral is:
$$\int \frac{1}{x^2+(x-1)^2}dx$$
One way to integrate inverses of polynomials is to find their roots and find the fractions which when added together, give the original fraction (with the roots being the denominators). For example:
$$\int \frac{1}{x^2-x-2}dx = \int \frac{1}{(x-2)(x+1)}dx\rightarrow \frac{a}{x-2}+\frac{b}{x+1}=\frac{a(x+1)}{(x-2)(x+1)}+\
\frac{b(x-2)}{(x-2)(x+1)}$$
$$\rightarrow ax+a+bx-2b=1 \rightarrow \left\{
\begin{array}{c}
ax+bx=0x \\
a-2b=1 \\
\end{array}
\right. \rightarrow a=\frac{1}{3}, b=-\frac{1}{3}$$
$$\rightarrow \int \frac{1}{x^2-x-2}dx=\int (\frac{1/3}{x-2}-\frac{1/3}{x+1})dx$$
And then solve.
I know this is known, I just wanted to show what I've been taught exactly, no ambiguation.
So about the question. We can open up the squares to get:
$$\int \frac{1}{x^2+(x-1)^2}dx = \int \frac{1}{2x^2-2x+1}dx=\int \frac{1}{2(x-0.5-0.5i)(x-0.5+0.5i)}dx$$
$$\rightarrow \frac{a}{x-0.5-0.5i}+\frac{b}{x-0.5+0.5i}=\frac{1/2}{(x-0.5-0.5i)(x-0.5+0.5i)}$$
I'll spare the details:
$$a=-\frac{i}{2},b=\frac{i}{2}$$
$$\rightarrow \int \frac{1}{x^2+(x-1)^2}dx=\int (\frac{i/2}{x-0.5+0.5i}-\frac{i/2}{x-0.5-0.5i})dx=i(\int \frac{1}{2x-1+i}dx-\int \frac{1}{2x-1-i}dx)$$
Once again sparing the details of some simple u substitution:
$$=\frac{i}{2}(ln\lvert 2x-1+i\rvert-ln\lvert 2x-1-i\rvert)$$
Absolute value (which's added from integrating something to the (-1)) always gives out a real number, and the entire "ln" expression is multiplied by an imaginary number, so the output is always imaginary.
The actual video I took this from contained a completely different solution. It's not even close to mine, which's always imaginary. Why? How did this happen?
Thanks.
|
consider this instead:
$$\begin{align}
\phantom{=}&x^2+(x-1)^2\\
=&x^2+(x^2-2x+1)\\
=&2x^2-2x+1\\
=&2\left(x-\frac12\right)^2+\frac12\\
=&\frac12\left[4\left(x-\frac12\right)^2+1\right]
\end{align}$$
you can take this half out of the integral (so it becomes two) now first think of the substitution $u=x-\frac12$ so you get:
$$(2u)^2+1$$
now say $2u=\tan v$ and see what you get :)
|
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|
Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further expand the binomial coefficient and tried to write it as sum of integrals but a pesky "r" term ends up being multiplied
How can we evaluate this with and without using calculus?
|
The following approach is somewhat like taking a sledgehammer to crack a nut, but it can also conveniently used in more challenging situations. We use the coefficient of operator $[x^r]$ to denote the coefficient of $x^r$ of a series. This way we can write
\begin{align*}
\binom{2r}{r}&=[x^{r}](1+x)^{2r}\tag{1}\\
\end{align*}
We now use a clever change of variables stated as rule 5 in section 1.2 of Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev. Rule 5 adapted for this special case is:
\begin{align*}
[x^r]f^{r}(x)
&=[y^r]\left.\frac{f(x)}{f(x)-xf^{\prime}(x)}\right|_{x=g(y)}\tag{2}
\end{align*}
Here we have $f(x)=(1+x)^2$ and $g=g(y)$ is the inverse function of
\begin{align*}
\frac{x}{f(x)}=\frac{x}{(1+x)^2}=y
\end{align*}
We obtain
\begin{align*}
&yx^2+(2y-1)x+y=0\\
&x=\frac{1}{2y}\left(1-2y\pm\sqrt{1-4y}\right)\tag{3}\\
\end{align*}
We take from (3) the root $x$ with the minus sign, since this one represents a power series.
We obtain from (1),(2) and (3)
\begin{align*}
\color{blue}{\binom{2r}{r}}&=[x^{r}](1+x)^{2r}\\
&=[y^r]\left.\frac{(1+x)^2}{(1+x)^2-x\cdot 2(1+x)}\right|_{x=\frac{1}{2y}\left(1-2y-\sqrt{1-4y}\right)}\\
&=[y^r]\left.\frac{1+x}{1-x}\right|_{x=\frac{1}{2y}\left(1-2y-\sqrt{1-4y}\right)}\\
&\,\,\color{blue}{=[y^r]\frac{1}{\sqrt{1-4y}}}\tag{4}\\
\end{align*}
Since we can write a power series $A(x)=\sum_{r=0}^{\infty}a_r x^r$ as
\begin{align*}
A(x)=\sum_{r=0}^{\infty}a_r x^r=\sum_{r=0}^{\infty}\left([y^r]A(y)\right) x^r
\end{align*}
we obtain from (4)
\begin{align*}
\color{blue}{\sum_{r=0}^\infty { 2r \choose r} x^r} =\sum_{r=0}^\infty \left([y^r]\frac{1}{\sqrt{1-4y}}\right)x^r=\color{blue}{\frac{1}{\sqrt{1-4x}}}
\end{align*}
|
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|
Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?
|
There is an assumption in the derivation of the quadratic formula that $A \ne 0$. You can see it does not directly apply when this assumption is violated due to the $A$ in the denominator.
However, the solution of the linear case can nevertheless be obtained from the quadratic formula on the understanding one of the two solutions it yields is not valid.
First consider $A$ is very small, apply the binomial theorem to the square root
$$
\frac{-B\pm\sqrt{B^2-4AC}}{2A} \\
=\frac{-B\pm B\sqrt{1-4AC/B^2}}{2A} \\
\approx \frac{-B\pm B(1-2AC/B^2)}{2A} \\
= -B/2A\pm B/2A \mp C/B
$$
So far we still have 2 solutions. When $A=0$ we have a linear equation which has only has one solution, and one of the solutions from the quadratic formula is not valid.
Retaining the $+,-$ combination as the one valid solution,
$$
= -B/2A + B/2A - C/B \\
= -C/B.
$$
Note, there's also an assumption $B \ne 0$.
|
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|
Prove that $\sum{a^{2}bc}=\binom{n+3}{6}+\binom{n+2}{6}$ Question
We define $S$ as the set of positive integer triplets $(a,b,c)$ such that $a+b+c=n$. I want to prove that the following expression is correct:
$$
\sum_{S}{a^{2}bc}=\binom{n+3}{6}+\binom{n+2}{6}
$$
My Solution
Left Hand Side
Say that there is a line of $n$ balls. We color the first $a$ balls red, the next $b$ balls green, and the rest blue. We then choose two red balls randomly with replacement, choose one green ball randomly, and choose one blue ball randomly. The number of possibilities is given by $\sum_{S}{a^{2}bc}$
Right Hand Side
If the same red ball is chosen twice;
$$
\begin{align}
x_{1}&\text{ red balls lie on the left of the chosen red ball}\\
x_{2}&\text{ red balls lie on the right of the chosen red ball}\\
x_{3}&\text{ green balls lie on the left of the chosen green ball}\\
x_{4}&\text{ green balls lie on the right of the chosen green ball}\\
x_{5}&\text{ blue balls lie on the left of the chosen blue ball}\\
x_{6}&\text{ blue balls lie on the right of the chosen blue ball}
\end{align}
$$
Since $\sum_{i=1}^{6}{x_{i}=n-3}$, using stars and bars we get $\binom{n+2}{5}$ possibilities.
If different red balls are chosen, the approach is the same but we also need to count the number of red balls between the two chosen red ball and multiply by two because either red ball can be chosen first. This give us $2\binom{n+2}{6}$ possibilities.
Since I counted the same possibilities, then the expressions must be equal:
$$
\begin{align}
\sum_{S}{a^{2}bc}&=2\binom{n+2}{6}+\binom{n+2}{5}\\
\\
&=\binom{n+3}{6}+\binom{n+2}{6}
\end{align}
$$
I’d like to know if my solution is correct and if there is alternative solution or discussion.
|
Alternative solution: let us consider the following generating functions (power series with $\rho=1$):
$$ A(x)=\sum_{a\geq 1}a^2 x^a,\qquad B(x)=\sum_{b\geq 1} b x^b,\qquad C(x) = \sum_{c\geq 1} c x^c. $$
The sum $\sum a^2 bc$ over the triples of positive integers such that $a+b+c=n$ is exactly the coefficient of $x^n$ in $A(x)B(x)C(x)$. By stars and bars we have (formally, pointwise for any $x\in(-1,1)$ and uniformly over the compact subsets of $(-1,1)$)
$$ B(x)=C(x)=\frac{x}{(1-x)^2},\qquad A(x)=\frac{x(1+x)}{(1-x)^3} $$
so
$$ \sum_{a+b+c=n}a^2 bc = [x^n]\frac{x^3+x^4}{(1-x)^7}=[x^{n-3}]\frac{1}{(1-x)^7}+[x^{n-4}]\frac{1}{(1-x)^7} $$
where, as usual, $[x^n]f(x)$ stands for the coefficient of $x^n$ in the Maclaurin series of $f(x)$. Always by stars and bars we have
$$ \frac{1}{(1-x)^7}=\sum_{n\geq 0}\binom{n+6}{6}x^n $$
so
$$ \sum_{a+b+c=n}a^2 b c = \binom{n+2}{6}+\binom{n+3}{6} $$
as wanted.
|
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|
What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \frac{\sqrt3}{BD}\therefore BD = 2\sqrt3\\
cos 30 = \frac{BE}{BD}\rightarrow \frac{\sqrt3}{2}=\frac{BE}{2\sqrt3}\therefore BE = 3\implies BN = 3+\sqrt3\\
\triangle BNG:cos 30 = \frac{BN}{BG}\rightarrow \frac{\sqrt3}{2}=\frac{3+\sqrt3}{BG}\rightarrow BG = \frac{6+2\sqrt3}{\sqrt3}=2\sqrt3+2\\
BG = \frac{2BP}{3}\rightarrow BP = 3\sqrt3 + 3\\
\triangle BPC: tg30 = \frac{PC}{BP}\rightarrow \frac{\sqrt3}{3} = \frac{PC}{3\sqrt3+3} \implies \boxed{ PC = 3+\sqrt3}\\
\therefore \boxed{\color{red}AC = 2(3+\sqrt3) = 6+2\sqrt3}$
My question...only the equilateral triangle meets the conditions? Why if the quadrilateral is indescribable $\measuredangle MGN=120^o$
|
Since $$BM+GN=BN+MG,$$ in the standard notation we obtain:
$$\frac{c}{2}+\frac{1}{6}\sqrt{2b^2+2c^2-a^2}=\frac{a}{2}+\frac{1}{6}\sqrt{2a^2+2b^2-c^2}$$ or
$$3(a-c)+\sqrt{2a^2+2b^2-c^2}-\sqrt{2b^2+2c^2-a^2}=0$$ or
$$(a-c)\left(1+\frac{a+c}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2b^2+2c^2-a^2}}\right)=0,$$ which gives $a=c$ and our triangle is an equilateral triangle.
Thus, $$AC=2BN=2(EN+BE)=2(\sqrt3+3).$$
|
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|
Validity of binomial expansion for any power I know that
$$(1+x)^n=1+nx+\frac {n(n-1)}{2!}x^2+\frac {n(n-1)(n-2)}{3!}x^3+\frac {n(n-1)(n-2)(n-3)}{4!}x^2+\\
\dots+\frac {n(n-1)\dots(n-r+1)}{r!}x^r\text.$$
I know that this formula is valid for any real number n, provided that $|x|<1$. The formula extends infinitely for a negative value of $n$, and terminates for a positive value of $n$. Is there a formula for a case where $|x|>1$? Why is this formula applicable only for $|x|<1$?
|
The series you have written is valid for any real value of $x$ if $n$ is zero or a natural number. In this case your expansion is an identity and you get a polynomial of degree $n+1$ in $x$.
But if $n$ is negative integer or non integral, your expansion is valid only for |x|<1 and the series is infinite and convergent.
For example
$$(1+x)^4=1+4x+6x^2+4x^3+x^4.$$
$$(1+x)^{4.3}=1+4.3x+(4.3*3.3)x^2/2+(4.3*3.3*3.2)x^3/6+........$$
$$(1+x)^{-4.3}=1-4.3x+(4.3*5.3)x^2/2-(4.3*5.3*6.3)x^3/6+........$$
$$(1-x)^{-1}=1+x+x^2+x^3+......., \text{if}~ |x|<1$$
$$(1+1/3)^{1/2}=1+\frac{1}{2}(1/3)-\frac{1.1}{2.2.2!}(1/3)^2+\frac{1.1.3}{2.2.2.3!}(1/3)^3+.....$$
|
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|
Find $\int \frac{{\rm d}x}{1+e\cos x}$
Find $$\int \frac{{\rm d}x}{1+e\cos x}$$
*
*When $e$ lies between $0$ and $1$
*When $e$ is greater than $1$
I got $\int \frac{dx}{1+e\cos x}$ when $e$ lies between $0$ and $1$, by substituting $\tan\frac{x}{2} =t$.
$$\int \frac{dx}{1+e\cos x} = \frac{2}{\sqrt{1-e^2}} \tan^{-1}\left( \frac{\sqrt{1-e}}{\sqrt{1+e}}\tan \frac{x}{2} \right) +C,$$
which will not be valid for $e >1.$
What different attempt we can do for $e>1$?
|
Substituting $t=\tan(\frac{x}{2})$ we obtain for $e\neq1$ (and the case $e=1$ gives $\int \frac{dx}{1+\cos x}=\tan(\frac{x}{2})+c_1$)
$$\int \frac{dx}{1+e\cos x}=\int \frac{2}{1+t^2+e(1-t^2)}dt$$
$$=\int\frac{2}{t^2(1-e)+(1+e)}dt=\int\frac{2}{(1-e)\left(t^2+\frac{1+e}{1-e}\right)}dt=\frac{2}{1-e}\int\frac{1}{t^2+\frac{1+e}{1-e}}dt$$
$\bullet$ Then for $0<e<1$ you can use the $\tan$ substitution to obtain $\arctan$ function (as you did).
$\bullet$ For $e>1$, you can use partial fractions decomposition to obtain logarithms. To make the calculations simpler, you can let $c^2=-\frac{1+e}{1-e}$, then you have $\int\frac{1}{t^2-c^2}dt$
|
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|
Foci of Ellipse lies on Hyperbola and vice-versa Let the foci of the hyperbola $\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1$ , (A,B > 0) be vertices of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , (a, b > 0) and foci of ellipse be vertices of hyperbola. Let eccentricities of the ellipse and hyperbola be $e_1$ & $e_2$ , respectively $L_1$ and $L_2$ are length of Latus rectum of ellipse and hyperbola respectively. Then find the minimum value of $[e_1+e_2]$ (where [.] denotes greatest integer function)
My approach is as follow
Let $ {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} $
Foci of the ellipse are the vertices of the hyperbola and vice-versa then we get the following
$\frac{{{A^2} + {B^2}}}{{{a^2}}} = 1;\frac{{{a^2} - {b^2}}}{{{A^2}}} = 1 \Rightarrow {a^2} - {A^2} = {b^2}$ and ${B^2} = {a^2} - {A^2}$ hence $b=B$
${e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} \Rightarrow {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{b^2}}}{{{a^2} - {b^2}}}} $
Not able to proceed further
|
In the given notation the focus of ellipse is $S_1(ae_1,0)$ and vertex is $V_1(a,0)$.
For the hyperbola $S_2=(Ae_2,0)$ and $V_2=(A,0)$. Given that $S_1=V_2$ and S_2=V_1$.
we get $e_1=A/a$ and $e_2=a/A$ \implies $e_1 e_2=1$, so by AM-GM, we get
$$\frac{e_1+e_2}{2}\ge \sqrt{e_1 e_2} \implies [e_1+e_2]\ge 2.$$
|
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|
Given a prime number $p$ and integer numbers $x,y$. Prove that if a number $x(p-x)y(p-y)$ is a square of an integer then $x=y$.
Given a prime number $p>2$ and numbers $x,y \in \lbrace 1,2,...,\frac{p-1}{2}\rbrace$. Prove that if a number $x(p-x)y(p-y)$ is a square of an integer then $x=y$.
Firstly I noticed that $x$ and $p-x$ are relatively prime likewise $y$ and $p-y$. That means we have two cases, because if WLOG $x$ is a square then one of $y$ or $p-y$ is not a square due to $x$ and $p-x$ being coprime.
$$\begin{cases}
xy = k^2 \\
(p-x)(p-y)=l^2
\end{cases}
$$
and
$$\begin{cases}
x(p-y) = k^2 \\
y(p-x)=l^2
\end{cases}$$
The first system of equations gives us after subtraction $$l^2-k^2=(l+k)(l-k)=p(p-x-y)$$ we must have that $p|x+y$
also $2\le k+l \le \sqrt{\frac{p-1}{2}}+p$ implying $l+k=p$ and $l-k=p-x-y$ yielding $k=\frac{x+y}{2}$ and now using first equation in the first system of equations we get $xy=(\frac{x+y}{2})^2$ implying $(x-y)^2=0$ and giving $x=y$ i tried to solve the second system in similar fashion but I am unable to do so. How do I proceed? Any help appreciated.
|
WLOG, Let $x,y \leq \frac{p-1}{2}$. Clearly, $p-x,p-y<p$. So, $k^2,l^2<\frac{p-1}{2}\times p<\frac{p^2}{2}$. This implies that $k,l<\frac{p}{\sqrt{2}}$ or equivalently, $k+l<\sqrt{2}p$ and $k-l<\frac{p}{\sqrt{2}}$.
Substracting the two equations, we get that $(x-y)p=k^2-l^2$. From the inequality, we get that $p=k+l$ or $k-l=0$.
For the first:
From RMS-inequality, we have that $\frac{k+l}{2}\leq\sqrt{\frac{x(p-y)+y(p-x)}{2}}=\frac{\sqrt{(x+y)p-2xy}}{\sqrt{2}}$. This simplifies to $p^2/2\leq(x+y)p-2xy$. So, $x+y$ has to be atleast $p/2$.
Suppose $\frac{p+r}{2}= x+y$ for $0 < r < p/2$, then we have that one of $x,y$ has to be at least $r+1$, since the other is less than $\frac{p-1}{2}$. So, the product $2xy$ is at least $(p-1)(r+1)$. Therefore, $(x+y)p-2xy$ is at most $(\frac{p+r}{2})p-(p-1)(r+1)<p^2/2$, which is a contradiction for $r<p/2$. Since there is no possible $x+y$ for this inequality to hold, we can conlude that $k+l=p$ is not possible.
For the second: $k=l$. $(x-y)p=0$ and so, $x=y$.
$\mathbf{Edit:}$
The OP mentions that he has missed the case where $y$ is a square and $x(p-x)(p-y)$ is a square. This seems to be the hardest case to solve and took me quite a while.
It can be seen quite easily that if $x(p-x)(p-y)$ and $y$ are square, then we need to have $x=af^2, p-x=bg^2, p-y=abh^2, y=i^2$ for $a,b$ square free and some $f,g,h,i$. And we have the two relations: $$af^2+bg^2=p$$
$$abh^2+i^2=p$$
This reminded me a lot of the Brahmagupta identity and tried to establish something along those directions. Also, toying with some small values for such combinations gave me the feel that there was some identity involved that made any number which could be written in this form to have some factor. See this article on writing a number as sum of squares in two different ways.
So, I tried the following: $p^2=(af^2+bg^2)(abh^2+i^2)$ and applied a sort of Brahmagupta identity.
$$p^2=b((ahf)^2+(gi)^2)+a((fi)^2+(bgh)^2)$$
$$=b(ahf-gi)^2+a(fi+bgh)^2$$
$$=b(ahf+gi)^2+a(fi-bgh)^2$$
Now, all that is left is to see that this can't happen. Looking $mod(p)$, we can see that $(p-x)(p-y)$ is $xy(mod (p))$. So, $p$ divides $af^2i^2-ab^2g^2h^2$. So, $p$ divides $a(fi+bgh)(fi-bgh)$. $a<p$ and so, $p$ divides $fi+bgh$ or $fi-bgh$. Neither can be true because $p^2=b(ahf-gi)^2+a(fi+bgh)^2$ and $p^2=b(ahf+gi)^2+a(fi-bgh)^2$. So, one of them has to be zero. If not, then $a$ will be one, reducing it to an earlier case. This implies that $fi=bgh$. $b$ must divide one of the $x$ or $y$ which is not possible, so it must be that $b=1$. So, that means that $p-x=bg^2=g^2$ is a square and this reduces it to case to that was already done. This completes our proof.
$\mathbf{Remark:}$ As pointed out in a comment, it is entirely possible that none of $x,y,(p-x),(p-y)$ are squares. The general case can be done in the same way as follows. $x=gcd(a,p-y)gcd(a,y)f^2$, $p-x=gcd(b,p-y)gcd(b,y)g^2$, $p-y=gcd(a,p-y)gcd(b,p-y)h^2$ and $y=gcd(a,y)gcd(b,y)i^2$. And go into the same Brahmagupta type identity, to reduce it into the case where $a$ or $b$ is 1.
|
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|
Find $\int x^2\arcsin(2x)dx$ Find $\int x^2\arcsin(2x)dx$
My work.
$\frac{1}{3}\int \arcsin(2x)dx^3=\frac{1}{3}(x^3\arcsin(2x)dx-\int x^3d(\arcsin(2x))$
This yields to finding $\int \frac{2x^3}{\sqrt{(1-4x^2)}}dx$ with which I have problem finding.
$Edit$
$x=\frac{1}{2}sin\theta$
$\frac{2}{8}\int\frac{sin^3\theta cos\theta d\theta}{\sqrt{1-sin^2\theta}} = \frac{2}{8}\int\frac{sin^3\theta cos\theta d\theta}{|cos\theta|}$
Now what should I do with $|cosx|?$It is $cosx$ or $-cosx$.
|
I'd like to show you my way to calculate $\int x^2 \arcsin 2x$.
First, let $t=arcsin 2x$. We get $$\int x^2 \arcsin 2x=\int \frac18 \sin^2 t \cdot t \cdot \cos t dt$$.
Use partical integration twice, we get the following equation
$$\int \frac18 \sin^2 t \cdot t \cdot t\cos t dt=\frac{1}{24}t\sin^3t-\int\frac{1}{24}sin^3tdt\\
=\frac{1}{24}t\sin^3t+\frac{1}{24}\cos t\sin^2t-\int\frac{1}{12}\sin t\cos^2 tdt \\
=\frac{1}{24}t\sin^3t+\frac{1}{24}\cos t\sin^2t-\frac{1}{12}\int\sin t+\frac{1}{12}\int \sin^3 tdt$$
from the first equal sign and the last equal sign, we can find
$$-\frac{1}{24}\int \sin^3tdt=\frac{1}{24}\cos t\sin^2t-\frac{1}{12}\int\sin t+\frac{1}{12}\int \sin^3 tdt$$.
So you can get $$\int \sin^3 tdt=\frac13 \cos t\sin^2 t+\frac23\cos t$$
then you can get $\int x^2 \arcsin 2x dx$ by taking $t=\arcsin 2x$ into the equation.
|
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|
Proving $\left\lfloor \sqrt{x^2+3x+3}\right\rfloor = x+1$
Let $$f(x)=\sqrt{x^2+3x+3}$$
Claim: $$\lfloor f(x)\rfloor = x+1$$
I came across this expression in a question and made this claim based on observations
$$⌊f(1)⌋=\lfloor\sqrt{7}\rfloor=2$$
$$⌊f(2)⌋=\lfloor\sqrt{13}\rfloor=3$$
$$⌊f(3)⌋=\lfloor\sqrt{21}\rfloor=4$$
$$⌊f(4)⌋=\lfloor\sqrt{31}\rfloor=5$$
and so on ...
However, I can't think of a formal proof for my claim.
It would be appreciated if someone could help me with the proof or give some kind of hint.
Thank you.
|
The statement $[\sqrt{x^2 +3x + 3}] = x+1$ is true if and only if
*
*$x+1$ is an integer and
*$x+1 \le\sqrt{ x^2 + 3x + 3}^2 < x+2$.
The first is true if and only if $x$ is an integer. SO this not true if $x \not \in \mathbb Z$. But if $x \in\mathbb Z$ this might be true.
As for 2)
$(x+1)^2 = x^2 +2x + 1$ and $(x+2)^2 = x^2 + 4x + 4$.
At first blush it looks like we could say
As $x^2 + 2x + 1 < x^2 + 3x + 3 < x^2 + 4x + 4$ we always have $x+1 < \sqrt{x^2 + 3x + 3} < x+2$....
But there are two mistakes with that:
i) $x^2 + 2x + 1 < x^2 + 3x + 3 < x^2 + 4x + 4\iff x+2 > 0$ and $x+1 > 0$. If $x < -1$ this assumption isn't true and
ii) $M^2 < x^2 < N^2 \implies M < x < N$ only if all the terms are non-negative.
So if $x \in \mathbb Z$ and if $x > -1$ (or in other words if $x \in \mathbb N \cup \{0\}$ then that is a valid proof that $[\sqrt{x^2 +3x + 3}] = x+1$.
But if $x \not \in Z$ or if $x \le -1$ then that statement is not true.
.....
P.S.
If $x \le -2$ with $x\in \mathbb Z$ then we have
$(x+2)^2 = x^2 + 4x + 4 < x^2 +3x + 3 < x^2 + 2x + 1=(x+1)^2$ so
$-x - 2 = |x+2|< \sqrt{x^2 + 3x + 3} < |x+1| = -x-1$ so
$[ \sqrt{x^2 + 3x + 3}] = -x -2$.
If $x = -1$ then $[ \sqrt{x^2 + 3x + 3}]=[\sqrt{1-3+3}]=1= x+ 2=-x$
P.P.S
If $x\in \mathbb R$ and $x > -1$ we still have $x + 1 < \sqrt{x^2 + 3x + 3} < x+2$ so if $x \not \in \mathbb Z$ we have $[x]+1< x+1 < \sqrt{x^2 + 3x + 3} < x+ 2 < \lceil x\rceil + 2 \le [x] + 3$. So $[ \sqrt{x^2 + 3x + 3}] = \begin{cases} [x]+1\\ [x] + 2\end{cases}$ depending upon whether there is an integer between $[x]+1$ and $x + 1$ or not. (i.e. if $x \ge [x] +\frac 12$ or if $x < [x]+\frac 12$.
We can do a similar calculation for if $x < -2; x\not \in \mathbb Z$ but I don't see any great incentive to pin down that much detail.
|
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|
Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11, x>3,$ which doesn't make sense.
|
Alternative
Make sure you only multiply with positive number because otherwise the inequality sign will reverse
$$
\begin{align}
\left|\frac{7}{x+4}\right|\cdot\left|\frac{7}{x+4}\right|&<1\cdot\left|\frac{7}{x+4}\right|\\
\\
&<1\cdot 1\\
\\
\\
7^{2}&<(x+4)^{2}
\end{align}
$$
From here we get the solutions:
$$
x<-11\text{ or }x>3
$$
Notice that I only multiply with $\left|\frac{7}{x+4}\right|$ and $(x+4)^{2}$ which I know are positive numbers.
|
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|
Question about calculating a series involving zeta functions On this page it had shown that the sum of $\frac{1}{n^3(n+1)^3}=10-\pi^2$. I'm curious about, what is the value of $$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
For some positive integer $k$.
According to partial fraction expansion, we can show that $$\frac1{n^3(n+k)^3}= 6\bigg(\frac1{nk^5}-\frac1{(n+k)k^5}\bigg)-3\bigg(\frac1{k^4n^2}+\frac1{k^4(n+k)^2}\bigg)+\frac1{k^3n^3}-\frac1{k^3(n+k)^3}$$
It is obvious to show that the first part and the last part are telescoping series, and for the last part, we can see that $$\frac1{k^3n^3}-\frac1{k^3(n+k)^3}=\frac1{k^3}\bigg(\frac1{n^3}-\frac1{(n+k)^3}\bigg)=\frac1{k^3}\sum_{i=1}^{k}\frac1{i^3}=\zeta(6)+\sum_{i<j}\frac1{i^3j^3}=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
Which leads to the original question.
The particular values of the sum are
$k$
$$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
$1$
$10-\pi^2$
$2$
$\frac {21}{32}-\frac1{16}\pi^2$
$3$
$\frac {809}{5832}-\frac1{81}\pi^2$
We can easily know that the sum is in the form of $a+b\pi^2$ and $b=\frac1{k^4}$. So what about the value of $a$?
Edit: Some notes on $\zeta(3)$:
By squaring $\zeta(3)$,
$$(\zeta(3))^2=\zeta(6)+\sum_{i\ne j}\frac1{i^3j^3}$$.
Note that $i$ and $j$ are both integers and we can assume that $i$ is strictly larger than $j$, or we could say that $i=n$, $j=n+k$ for some positive integer $k$. Hence
$$(\zeta(3))^2=\zeta(6)+2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac1{n^3(n+k)^3}$$
Assume $\sum_{n=1}^\infty\frac1{n^3(n+k)^3} = a_k-\frac{\pi^2}{k^4}$.
Thus we can know
$$\begin{align}(\zeta(3))^2&=\zeta(6)+2\sum_{k=1}^{\infty}\bigg(a_k-\frac{\pi^2}{k^4}\bigg)\\&=\frac{\pi^6}{945}+2\sum_{k=1}^{\infty}a_k-2\pi^2\zeta(4)\\&=2\sum_{k=1}^{\infty}a_k-\frac{4\pi^6}{189}\end{align}$$
For $\sum_{k=1}^{10}a_k$, we can calculate that
$$\begin{align}(\zeta(3))^2&\approx 2\sum_{k=1}^{8}a_k-\frac{4\pi^6}{189}\\&\approx 1.42163941214...\end{align}$$
And $(\zeta(3))^2\approx1.44494079841...$
|
Using partial summation and asymptotics, we have $$a_k=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}+\frac {\pi^2}{k^4}$$
$$a_k=\frac{1}{2k ^5}\Big[k^2 (\psi ^{(2)}(k+1)+2 \zeta (3))+\pi ^2 k-6 k \psi ^{(1)}(k+1)+12 \psi^{(0)}(k+1)+12 \gamma \Big]$$ which generate the sequence
$$\left\{10,\frac{21}{32},\frac{809}{5832},\frac{2615}{55296},\frac{112831}{54000
00},\frac{168791}{15552000},\frac{17769701}{2823576000},\frac{22201623}{5619
712000},\dots\right\}$$
I did not find anything in $OEIS$ but the asymptotics is
$$a_k=\frac{\zeta (3)}{k^3}+\frac{\pi ^2}{2 k^4}+\frac{12 \log (k)+12 \gamma
-7}{2k^5}+\frac 5 {k^6}-\frac 5 {4k^7}+O\left(\frac{1}{k^9}\right)$$ which is not bad even for small values of $k$. Using this truncated series, we have
$$a_1=\zeta (3)+\frac{1}{4}+6 \gamma +\frac{\pi ^2}{2}=9.85015\cdots$$
$$a_2=\frac{64 \zeta (3)-21+96 \gamma +16 \pi ^2+96 \log (2)}{512} =0.65586\cdots$$
|
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|
Find a bijection between $[1,2)$ and $(1,2)$ I want to find a bijection between $[1,2)$ and $(1,2)$ and prove it.
My attempt:
$[1,2) = \{x \in \mathbb R | 1 \leq x <2\}$
$(1,2) = \{x \in \mathbb R | 1 < x < 2\}$
$f(x) = x$ if $x \ne 1\frac{1}{n}$ for $n = 1,2,3,...$
and
$f(x) = 1\frac{1}{x+1}$, if $x =1\frac{1}{n}$ for $n = 1,2,3,...$
Proof - Injective - Prove $x_1 = x_2$ for $f(x) = x$
\begin{align*}
f(x_1) = f(x_2) &\implies x_1 = x_2.\\
\end{align*}
Proof - Injective - Prove $x_1 = x_2$ for $f(x) = 1\frac{1}{x+1}$
\begin{align*}
f(x_1) = f(x_2) &\implies 1\frac{1}{x_1+1} = 1\frac{1}{x_2+1}\\
&\implies \frac{1}{x_1+1} = \frac{1}{x_2+1}\\
&\implies x_2+1 = x_1+1\\
&\implies x_2 = x_1.\\
\end{align*}
Therefore $f$ is injective.
Any help will be appreciated.
Thanks.
|
In your question, you noted that card((1,2)) <= card([1,2))
Let $f: [1,2)\to(1,2)$
Fix $x \in (1,2)$. Let $\{A_i\}_{1}^{\infty}$ represent the non-terminating decimal expansion right of the decimal place (this is unique to each number).
Let $f(x) := 1 + \sum_{i=1}^{\infty}A_i*10^{-i-1}$. All values of f are of the form 1.0... Therefore, let $f(1):=1.1$. As each decimal expansion is unique the resulting function f is injective.
Therefore, there exists a bijection between (1,2) and [1,2)
|
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|
Sequence $(u_n)$: $\left\{\begin{array} .u_1=1\\ u_{n+1}=\frac{u_n}{2^nu_n+3} \quad , n=1,2,3,\dots \end{array}\right.$ The sequence $(u_n)$ is defined by the formula:
$$\left\{\begin{array}
.u_1=1\\
u_{n+1}=\frac{u_n}{2^nu_n+3} \quad , n=1,2,3,\dots
\end{array}\right.$$
Find $\lim\limits_{n\to \infty}\sqrt[n]{u_n}$.
The first, I see:
$$u_{n+1}=\frac{u_n}{2^nu_n+3}<u_n\Rightarrow\sqrt[n+1]{u_{n+1}}<\sqrt[n+1]{u_n}<\sqrt[n]{u_n}$$
But, I forgot that $0<u_n<1$ so $\sqrt[n+1]{u_n}>\sqrt[n]{u_n}$.
If I can prove that the $(\sqrt[n]{u_n})$ is a decrease/increase sequence, it's easy to see $\lim\limits_{n\to \infty}\sqrt[n]{u_n}=\frac{1}{2}$.
Please help me. Thank you.
|
If you don't insist on using a monotonicity argument, you can also derive an explicit form for $u_n$ and then calculate the limit:
Setting $v_n := \frac{1}{u_n}$ for all $n$, we get
\begin{alignedat}{2}
&&v_{n+1} &= 2^n + 3v_n\\
&\implies
&\frac{v_{n+1}}{3^{n+1}}-\frac{v_{n}}{3^{n}} &= \frac{2^n}{3^{n+1}}\\
&\implies
&\sum_{k=1}^{n-1} \left(\frac{v_{k+1}}{3^{k+1}}-\frac{v_{k}}{3^{k}}\right) &= \frac{1}{3}\sum_{k=1}^{n-1} \left(\frac{2}{3}\right)^k = \frac{1}{3}\frac{\frac{2}{3} -\left(\frac{2}{3}\right)^{n}}{1-\frac{2}{3}}\\
&\implies
& \frac{v_{n}}{3^{n}} - \frac{v_{1}}{3} &= \frac{2}{3} -\left(\frac{2}{3}\right)^{n}\\
&\implies
&v_{n} &= 3^n - 2^n\\
&\implies
&\sqrt[n]{u_n} &= \frac{1}{\sqrt[n]{3^n-2^n}} \xrightarrow{ n \to \infty } \frac{1}{3}.
\end{alignedat}
|
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|
How to show the following inequality $_2F_1\left(5.5, 1, 5;-x^2\right)>0$? Consider the function $_2F_1\left(5.5, 1, 5;-|x|^2\right)$ for $x\in \mathbb{R}^n.$ I want to show that this function is positive. I checked that it does not have any roots so can I conclude the inequality by using continuity in $x$ of the function $_2F_1\left(5.5, 1, 5;-|x|^2\right)$?
|
$$f(x)=\, _2F_1\left(\frac{11}{2},1;5;-x^2\right)=\, _2F_1\left(1,\frac{11}{2};5;-x^2\right)$$
As @Aaron Hendrickson already commented,
$$g(x)=\frac{315}{8} x^8 \left(x^2+1\right)^{3/2}\,f(x)=x^2 \sqrt{x^2+1} \left(35 x^6+5 x^4-6 x^2+8\right)-16 \left(\sqrt{x^2+1}-1\right)$$
$$g'(x)=315 x^7 \sqrt{x^2+1}$$ So, $g'(x)=0$ if $x=0$.
$$g''(x)=\frac{315 x^8}{\sqrt{x^2+1}}+2205 x^6\sqrt{x^2+1} \implies g''(0)=0$$ So $g(x) \geq 0 \quad \forall x$ and $f(x)$ too.
Now $g^{(7)}(0)=1587600$ so $x=0$ is the minimum.
|
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|
Prove or disprove that $\left(f'(x)+g'(x)\right)\geq 2$ It's a result found with geogebra .
Let us define :
$$f(x)=\left(\Gamma(1-x)\right)^{\frac{1}{1-x}}$$
And :
$$g(x)=\operatorname{W}\left(1-\frac{1}{x-1}\right)$$
Then it seems we have $2\leq x\leq \frac{215}{100}$ :
$$\left(f'(x)+g'(x)\right)\geq 2$$
My attempt :
One can show that :
$$\lim_{x\to 2^+}f''(x)=\infty$$
On the other hand we can show that the limit of $g''(x)$ is finite at $x=2$.We deduce the existence of a constant $\varepsilon$ such that the sum of the derivatives is increasing for $2<x\leq 2+\varepsilon$.
Second attempt :
I found a little trick :
If we show that the antiderivative of the sum of derivative is increasing we win because then the derivative is positive. And the good news we can split in two the problem as ($2\leq x \leq 2+k$) :
$$h(x)=f(x)-\alpha x$$
$$u(x)=g(x)-\beta x$$
Where $\alpha>\beta$ and $\alpha+\beta=2$ and $h(x)$ and $u(x)$ are increasing
As experiments I have tried $\alpha=1.1$ and $\beta=0.9$ and $k=0.03$
last edit :
I go a little bit further :
To show that the function $u(x)$ is increasing we can substitute :
$$y=\frac{xe^x-2}{xe^x-1}$$
Now find a substitution for $h(x)$ is delicate so I give you a link wich concern also my motivation :
https://mathoverflow.net/questions/12828/inverse-gamma-function
My Motivation:
The reason is simple .I have tried to find a "continuous extension" of the Gamma function with some nice properties around $x=2$.
How to (dis)prove this?
Thanks in advance for your effort.
|
Sketch of a proof:
Let
\begin{align*}
F(x) &= f(x) + \frac{27x^3 - 182x^2 + 394x - 316}{10} - 2x,\\
G(x) &= g(x) - \frac{27x^3 - 182x^2 + 394x - 316}{10}.
\end{align*}
We have
$f'(x) + g'(x) - 2 = F'(x) + G'(x)$.
It is not difficult to prove that $G'(x) \ge 0$
for all $x \in [2, 215/100]$.
It suffices to prove that
$F'(x) \ge 0$ for all $x \in [2, 215/100]$.
It suffices to prove that, for all $x \in [2, 215/100]$,
$$\Gamma(1 - x)^{\frac{1}{1 - x}}
\left(\frac{\ln \Gamma(1 - x)}{(1 - x)^2} - \frac{\psi(1 - x)}{1 - x}\right) + \frac{81}{10}x^2 - \frac{182}{5}x + \frac{197}{5} \ge 0$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
Fact 1: For all $x \in [2, 215/100]$,
$$\Gamma(1 - x)^{\frac{1}{1 - x}} \ge
(x - 2) + \frac{10}{11}(1 - \gamma)(x - 2)^2 - \frac{10}{11}(x - 2)^2\ln(x - 2) \ge 0$$
where $\gamma$ is the Euler-Mascheroni constant.
Fact 2: For all $x \in [2, 215/100]$,
$$(x - 2)\left(\frac{\ln \Gamma(1 - x)}{(1 - x)^2} - \frac{\psi(1 - x)}{1 - x}\right)
\ge \frac{1 + \frac67(2 - \gamma)(x - 2) - \frac67(x - 2)\ln(x - 2)}{(1 - x)^2} \ge 0.$$
Using Facts 1-2, it suffices to prove that, for all $x \in [2, 215/100]$,
\begin{align*}
&\left(1 + \frac{10}{11}(1 - \gamma)(x - 2) - \frac{10}{11}(x - 2)\ln(x - 2)\right)\\
&\quad \times \left(\frac{1 + \frac67(2 - \gamma)(x - 2) - \frac67(x - 2)\ln(x - 2)}{(1 - x)^2}\right)\\
&\qquad + \frac{81}{10}x^2 - \frac{182}{5}x + \frac{197}{5} \ge 0.
\end{align*}
|
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|
Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
|
Suppose that $ab = cd \neq 0$. Then we get $\frac{a}{c} = \frac{d}{b} = k$.
From the above, we get $a = kc$ and $d = bk$. Substituting in $a+b = c+d$, we get $kc + b = c + bk$. Thus $(k-1)c - (k-1)b = (k-1)(c-b) = 0$. Thus, we must have that $b = c$ or $k = 1$. In the first case, we would then get $a=d, b=c$. In the second case, we get $a = c, b = d$.
If $ab = 0$. The $a = 0$ or $b = 0$. WLOG $a = 0$. Similarly since $cd = 0$, $c = 0$ (in which case $a=c, b=d$) or $d=0$ (in which case $a=d, b=c$).
|
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|
Minimum of $(xy+a)^2+(bx-cy)^2$ Let $a,b,c>0$. Consider the function
$$f:\mathbb R^2 \to \mathbb R,\quad f(x,y)=(xy+a)^2+(bx-cy)^2.$$
I'm interested in the globel minimum of this function. I could show that for $a \le bc$ the minimum is attained at $(x,y)=(0,0)$ and hence
$$f(0,0)=a^2.$$
For $a>2bc$ however it seems that there exist $x<0<y$ such that $f$ attains the global minimum at $f(x,y)$ and $f(-x,-y)$.
Is this true? How do you prove it and what is $x,y$ in that case?
Of course I computed already the derivative of $f$. But I couldn't solve the equations you get. You get something like
$$
x(y^2+b^2)=y(bc-a),\\
y(x^2+c^2)=x(bc-a).
$$
Edit (one hour later): I could solve it. The minimum is given by $a^2-(a-2bc)$ in case $a \ge 2bc$. I can post an answer if anybody is interested. Edit: That's not quite right. Somewhere is still a mistake. I'll look tomorrow for it.
Edit (next day): I've found my mistake. I forgot a square, the minimum is $a^2-(a-2bc)^2=4bc(a-bc)$.
|
From completing the square we get
$$
f(x, y) = (y^2+b^2)x^2 -2y(bc-a) x + (a^2+c^2y^2) \\
= (y^2+b^2) \left[ \left( x - \frac{y(bc-a)}{y^2+b^2}\right)^2 + \frac{a^2+c^2y^2}{y^2+b^2} - \frac{y^2(bc-a)^2}{(y^2+b^2)^2}\right]
$$
and therefore
$$
f(x, y) \ge a^2+c^2y^2 - \frac{y^2(bc-a)^2}{y^2+b^2}
= \frac{(c y^2 +ab)^2}{y^2+b^2} \, ,
$$
with equality if $x= \frac{y(bc-a)}{y^2+b^2}$. Therefore it suffices to determining the minimum of
$$
g(u) = \frac{(cu+ab)^2}{u+b^2}
$$
for $u \ge 0$. From
$$
g'(u) = \frac{(cu+ab)(cu+2b^2c-ab)}{(u+b^2)^2}
$$
we see that:
*
*If $a \le 2bc$ then $g$ is increasing, and the minimum is $g(0) = a^2$.
*If $a > 2bc$ then $g$ has a minimum at $u^*=(a-2bc)b/c$, and the minimum is $g(u^*) = 4bc(a-bc)$.
|
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|
Finding the equation to tangent line to a function given certain info - Multiple answers?
The problem states: find the equation of the tangent line to the graph of continuous function $y = f(x)$ at $x=3$ given the following information.
$\lim_{x \to 3} {\dfrac{f(x)+7}{2x-6}}=8$ and $f(3) = -7$
How you are supposed to solve it
$\lim_{x \to 3} {{\dfrac{1}{2}}\dfrac{f(x)+7}{x-3}}=8$ Take out 1/2
${\dfrac{1}{2}}\lim_{x \to 3} {\dfrac{f(x)+7}{x-3}}=8$ Take 1/2 out of limit
${\dfrac{1}{2}}\lim_{x \to 3} {\dfrac{f(x)-f(3)}{x-3}}=8$ As we know $f(3)=-7$, plug that in
${\dfrac{1}{2}}f'(3)=8$ By definition of limit, change $\lim_{x \to 3} {\dfrac{f(x)-f(3)}{x-3}}$ to $f'(3)$
$f'(3)=16$
Therefore, the tangent line at $f(3)=-7$ is $y+7={16}(x-3)$
How I solved it
As $\lim_{x \to 3} {\dfrac{f(x)+7}{2x-6}}=8$ and this is a continuous function where it exists (as $f(x)$ is continuous and $\dfrac{f(x)+7}{2x-6}$ is a rational function so $\lim_{x \to a}{\dfrac{f(x)-f(a)}{x-a}}=f(a)$)
I know that the ratio between the numerator and the denominator of $ {\dfrac{f(x)+7}{2x-6}}$ approaches 8 as $\lim_{x \to 3}$.
Therefore, at $x=3$ (we know the function $f(x)$ exists at $f(3)$ as stated in the directions)
$8(f(x)+7)=2x-6$
$8(y+7)=2x-6$ Directions stated $f(x)=y$
$8y+56=2x-6$ Factoring in the 8
$8dy=2dx$ Via implicit differentiation
$dy={\dfrac{1}{4}}dx$
${\dfrac{dy}{dx}}={\dfrac{1}{4}}$
Therefore, the tangent line at $f(3)=-7$ is
$y+7={\dfrac{1}{4}}(x-3)$
Questions
Could someone find an example where either the 'correct' answer or 'my answer' gives a correct answer?
Or perhaps explain where I went wrong?
Or is the problem itself fundamentally flawed?
Thanks for the help.
|
As they have given the : $f(x=3) = -7$
And they have also given the value of $$\lim_{x\to3}\frac{f(x)+7}{2x-6} = 8$$
Now, this must be the undermined form($\frac{0}{0}$) $\implies$ you can use L'Hôpital's rule
Now,
$$\begin{align*} \frac{f'(x =3)}{2} =
8 \implies f'(x=3) = 8×2 = 16
\end{align*}$$
Which is the slope of the tangent at $x=3$
Equation of tangent at point: $P=(3, -7)$ As they have already given $f(3) = -7$
$$\begin{align*} \frac{y+7}{x-3} = 16
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What's the measure of the segment $MN$ in the question below? For reference:(exact copy of the question) In the triangle $ABC$, $H$ is the orthocenter, $M$ and $N$ are midpoints of $AC$ and $BH$ respectively.
Calculate $MN$, if $AH=14$ and $BC=48$ (answer: $25$)
My progress..my drawing according to the statement and the relationships I found
we have several similarities
$\triangle AKC \sim \triangle BKH\\
\triangle AHE \sim BHK \sim \triangle BCE\implies\\
\frac{14}{48} = \frac{HE}{CE}=\frac{AE}{BE}\\\frac{BH}{48} = \frac{KH}{CE}=\frac{BK}{BE}\\
\frac{14}{BH} = \frac{HE}{HK}=\frac{AE}{BK}\\
\triangle MKC \sim \triangle NKH\\
\triangle NBK \sim \triangle MAK$
|
$\triangle AFH \sim \triangle BFC$
$FC = \frac{48}{14} HF, BF = \frac{48}{14} AF$
$2 FN = BF+HF = \frac{24}{7} AF + HF \tag1$
$\triangle ABF \sim \triangle HCF$
$\frac{CF}{HF} = \frac{BF}{AF} \implies CF = \frac{48}{14} HF$
$2 FM = CF - AF = \frac{24}{7} HF - AF \tag2$
Squaring $1$ and $2$ and adding,
$4 (FN^2 + FM^2) = \left(\frac{25}{7}\right)^2 (AF^2 + HF^2) = \left(\frac{25}{7}\right)^2 \cdot 14^2$
$MN^2 = 25^2 \implies MN = 25$
|
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|
Find the sum of the series $3x+8x^2+15x^3 + ....$ I'm trying to express the following series $3x+8x^2+15x^3 + ....$ as a sum and hope to find its sum for $|x| < 1$
Here is what I have so far:
To me the series looks to be the derivative of the following form:
$1 + \frac{3}{2}x^2+\frac{8}{3}x^3+\frac{15}{4}x^4 ...$
Given that the denominator progresses as $\frac{1}{n}$ I had thought of taking it out like so:
$\frac{1}{n}(1+3x^2+8x^3+15x^4 ....)$
Now I'm unsure of the sequence that the constants follow but lets denote this as $a$ and take it out of the sequence of $x$ values, then I get:
$\frac{1}{n}(1+a(x^2+x^3+x^4+x^5 ...))$
We know that the sum of the series of $x$ follows the following geometric series: $\frac{x^2}{(1-x)}$, plugging this into the equation:
$\frac{1}{n}(1+\frac{ax^2}{1-x})$
Then taking the first derivative I get:
$\frac{ax(2+x)}{n(1-x)^2}$
I'm unsure as to whether the $n$ can still be introduced or whether it's removed - How is this answer optimally derived?
Following the hint below:
$$\sum_{n=2}^{\infty}(n^2-1)x^n = \sum_{n=1}^{\infty}(n^2-1)x^n =\sum_{n=1}^{\infty}n^2x^n - \sum_{n=1}^{\infty}x^n = \frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$$?
|
Hint $1$: First notice that the coefficients are $n^2-1$.
Hint $2$:
\begin{eqnarray*}
\sum_{n=1}^{\infty} n^2 x^n =\frac{x(1+x)}{(1-x)^3}.
\end{eqnarray*}
|
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|
Computing $\int_0^\infty \frac{dx}{(x^2 + 1)^2} $ with the residue theorem I am solving this integral and am stuck on proving an inequality, I believe I have the rest worked out.
My work:
First observe that the integrand is even, hence $$\int_0^\infty \frac{dx}{(x^2+1)^2} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2}$$
We now consider the integral on the r.h.s. above and integrate over the upper half circle with radius $R$ and the segment $[-R, R]$ on the real axis. We let $R$ tend towards infinity. Denoting this region by $D$, and the upper half circle by $C$, by Cauchy's Residue Theorem we have,
$$\int_{D} \frac{dz}{(z^2+1)^2} = 2\pi i \sum \text{Res} = \int_{-\infty}^\infty \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}$$
I first calculate the residues of the integrand in $D$. Notice there is a pole of order 2 at $z = i$. Hence,
$$\text{Res}_{z = i} = \lim_{z \rightarrow i} \Bigl( \frac{(z-i)^2}{(z^2+1)^2} \Bigr)' = \frac{1}{4i}$$
Thus,
$$\frac{\pi}{2} = \int_{-\infty}^\infty \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}$$
Now I would like to show that the complex integral on the r.h.s. tends to 0 as $R$ tends to infinity:
$$\int_C \frac{dz}{(z^2+1)^2} \leq \Biggl \vert \int_C \frac{dz}{(z^2+1)^2} \Biggr \vert \leq \sup f(X) \cdot L = \sup f(X) \cdot \pi R$$
This is as far as I have gotten. I know I have to play with the triangle inequality somehow to bound the integrand from above and determine $\sup f(x)$, but I cannot figure out how to do this.
One thought I had was to use the reverse triangle inequality since:
$$|(z^2+1)^2| = |z^4 +2z^2 + 1| \geq \bigl| |z^4| - |2z|^2 - 1\ \bigr|$$
so that,
$$\frac{1}{(z^2+1)^2} \leq \frac{1}{\bigl| |z^4| - |2z|^2 - 1\ \bigr|} = \frac{1}{\bigl| R^4 - 2R^2 - 1\ \bigr|}$$
This will wrap things up as
$$\sup f(x) \cdot R = \frac{\pi R}{\bigl| R^4 - 2R^2 - 1\ \bigr|}$$
which will tend to 0 as $R$ tends to infinity.
Putting it all together and diving by two,
$$\int_0^\infty \frac{dx}{(x^2 + 1)^2} = \frac{\pi}{4}$$
This is the correct numerical answer, but I am not sure if my use of the reverse triangle inequality is correct, can anyone confirm? Thanks!
|
The idea of your proof is correct, but there are some inaccuracies. First, it should be
$$
\frac{\pi}{2} = \int_R^R \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}
$$
for $R > 1$. You cannot replace the $R$ by $\infty$ in the first integral because the second integral depends on $R$. I would denote the half-circle with $C_R$ to emphasize the dependency on the radius.
Second,
$$
|(z^2+1)^2| = |z^4 +2z^2 + 1| \geq \bigl| |z^4| - |2z|^2 - 1 \bigr|
$$
is wrong as can be seen by substituting $z=i$. But
$$
|z^2+1| \ge |z^2| - 1 = R^2 - 1 > 0 \\
\implies |(z^2+1)^2| = |z^2+1|^2 \geq (R^2-1)^2
$$
for $R > 1$, so that
$$
\sup_{z \in C_R} |f(z)| \cdot L(C_R) \le \frac{\pi R}{R^4-2R^2+1}
$$
converges to zero for $R \to \infty$. (You wrote $\sup f(x) \cdot R = \ldots$, which is not correct because we have only an upper bound for the supremum, not the exact value.)
Then
$$
\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2} = \lim_{R \to \infty}\int_R^R \frac{dx}{(x^2+1)^2}
= \frac \pi 2 - \lim_{R \to \infty} \int_{C_R} \frac{dz}{(z^2+1)^2}
= \frac \pi 2 \, .
$$
|
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|
Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $ Find all complex numbers which make the following equations true:
$$ |z+1| =1 $$
$$ |z^2+1| =1 $$
Solution:
If $ |z+1| =1 $ holds true, then
$$z+1 = 1.e^{i2n\pi}$$
$$z = 1.e^{i2n\pi}-1$$
$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$
If $ |z^2+1| =1 $ holds true, then
$$z^2+1 = 1.e^{i2p\pi}$$
$$z^2 = 1.e^{i2p\pi}-1$$
$$z^2 = 1.e^{i2p\pi}-1e^{i2q\pi}$$
*
*How to proceed after this?
*Am I supposed to do in this manner or break complex number z into real part x and imaginary part y and get two equations and thus solve for x and y?
|
It is clear that $z=0$ is a solution. Hence we assume $z \ne 0$.
From $|z+1|^2=1$ we derive $ |z|^2= -(z+ \bar z) \quad (1)$.
From $|z^2+1|^2=1$ we derive $|z|^4= -(z^2+ \bar z^2) \quad (2)$.
Then $(1)$ gives: $|z|^4= z^2+ \bar z^2+2z \bar z=z ^2+ \bar z^2+2|z|^2$.
With $(2)$ we derive $|z|^4=-|z|^4+2|z|^2.$ Hence $|z|^4=|z|^2$. This gives $|z|=1.$
Use again $(1)$ to see that $Re(z)=-1/2.$
From $1=|z+1|^2=1/4 + (Im(z))^2$ we get $Im(z)= \pm \frac{\sqrt{3}}{2}.$
|
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|
Struggling with polynomial Long Division in $GF(256)$ I'm currently trying to build a QR Code generator, which has lead me to studying a bit of abstract algebra - more specifically, finite fields.
I've studied quite a bit on my own and I'm at the point where I'm trying to implement an algorithm for polynomial long division in the $GF(256)$, but my results don't match with the neither the QR Code specification examples nor with this tutorial that I have been following as well. So I thought I could perform a quick polynomial division here to see if I can figure out if I actually understand the whole process as a whole.
I'm going to divide $p(x) = 5x^4 + 32x^3 + 27x^2 + 7x + 3$ by $q(x) = x^2 + 5x + 1$.
I start out by multiplying $q(x)$ by $5x^2$, which gives me $5x^4 + 17x^3 + 5x^2$ (here I use the fact that each element can be represented as a power of the primitive. So $5 \cdot 5 = (\alpha^{50})^2 = 17$). Then, I add this result with $p(x)$, which gives me $49x^3 + 30x^2 + 7x + 3$ (also, here I use the fact that every element in this field can be represented as a polynomial with coefficients being $0$ or $1$. This means that an addition is equivalent to a XOR operation of bit the respective bit strings).
I then repeat this same process two more times:
*
*Multiply $q(x)$ by 49x, resulting in $49x^3 + 245x^2 + 49x$, to then add it to the remainder of the previous operation, giving me $235x^2 + 54x + 3$.
*Multiply $q(x)$ by 235, resulting in $235x^2 + 48x + 235$, to then add it to the remainder of the previous operation, giving me $6x + 232$, which is the final remainder.
Therefore, the quotient is $5x^2 + 49x + 235$ and the remainder is $6x + 232$.
Am I doing something wrong? If so, then what? I'd appreciate any kind of feedback.
Thanks in advance.
|
I made a crash course for Pari and used an online version and got
? c=ffgen(c^8+c^4+c^3+c^2+Mod(1,2))
% = c
? divrem((c^2+1)*x^4+(c^5)*x^3+(c^4+c^3+c+1)*x^2+(c^2+c+1)*x+(c+1),(x^2+(c^2+1)*x+1))
% = [(c^2 + 1)*x^2 + (c^5 + c^4 + 1)*x + (c^7 + c^6 + c^5 + c^3 + c + 1),
(c^6 + c^4 + c^2 + c)*x + (c^7 + c^6 + c^5 + c^3)]~
The symbols ? and % are the input and the output prompt.
I hope I did use this Pari calculations correctly.
Now I used Maxima (I used my Windows version) to check if I used the correct polynomials. I substituted $c$ by $2.$
(%i) (c^2+1)*x^4+(c^5)*x^3+(c^4+c^3+c+1)*x^2+(c^2+c+1)*x+(c+1), c=2;
(%o) 5*x^4+32*x^3+27*x^2+7*x+3
(%i) x^2+(c^2+1)*x+1, c=2;
(%o) x^2+5*x+1
(%i) (c^2 + 1)*x^2 + (c^5 + c^4 + 1)*x + (c^7 + c^6 + c^5 + c^3 + c + 1), c=2;
(%o) 5*x^2+49*x+235
(%i) (c^6 + c^4 + c^2 + c)*x + (c^7 + c^6 + c^5 + c^3), c=2;
(%o) 86*x+232
In Maxima %i is the input prompt and %o is the output prompt.
There is a Maxima online version here. Simply input
(c^2+1)*x^4+(c^5)*x^3+(c^4+c^3+c+1)*x^2+(c^2+c+1)*x+(c+1),c=2;
(x^2+(c^2+1)*x+1),c=2;
(c^2 + 1)*x^2 + (c^5 + c^4 + 1)*x + (c^7 + c^6 + c^5 + c^3 + c + 1),c=2;
(c^6 + c^4 + c^2 + c)*x + (c^7 + c^6 + c^5 + c^3),c=2;
and press Clic.
So in $GF(256)$ the quotient is $5x^2+49x+235$ and the remainder is $86x+232.$
These calculations were done very poorly, but I am not familiar with the tools. I only want to show you there are tools to make such calculation that are error prone if you do them by hand.
To calculate this by hand the coefficients should be seen as polynomials.
So $12$ is binary $1100$ is the polynomial $1 \cdot c^3+1\cdot c^2 +0 \cdot c + 0.$
If you ad two numbers you have to add the two polynomials which means for each power you add the coefficients. Because the calculations are modulo 2 you simpli can XOR the digits of the numbers.
To multiply the numbers you have to multiply the polynomials and if there highest exponent is 8 or higher you have to divide the result by your irreducible polynomial and take the remainder. A more efficient way is to use the Log-/Antilog-tables, e.g.
$$ 235×5=2^{235}×2^{50}=2^{235+50}=2^{285}=2^{30}=96 $$ The exponent must be reduced modulo 255.
|
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Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$
I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y = 5.$ (I found $5$ as a solution of the equation by hit and trial method).
Therefore, $x = 0$ or $3.$
I'm wondering if there's any another method to solve it as the repeated squaring step seems to be somewhat absurd.
|
We want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x $
So, we want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2x}}}}=x \quad \quad (E)$
*
*We see that if $\sqrt{x+2x} =x$, then, $x$ is a solution of $(E)$.
Therefore, $0$ and $3$ are two solutions of $(E)$.
*Suppose that $x$ is a solution of $(E)$ and $x\neq 0$. Then $x>0$ and $F(x)=1$ with
$F(x)=\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{ \dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}}}}}$
*$F$ is strictly decreasing on $(0,+\infty)$, so, the equation $F(x)=1$ has at most one solution.
We can conclude that $(E)$ has exactly two solutions, $0$ and $3$.
|
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|
Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be written as
$$\frac{dy}{dx} = \frac{3y}{2\{x+\sqrt{1+x²}\}}\{1+\frac{x}{\sqrt{1+x²}}\}\tag3$$
This is as far as I have got in terms of simplifying. The solution says that $(2)$ is equal to $(4)$ $$\frac{\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{3}{2}}{\sqrt{1+x²}}=\frac{3y}{2\sqrt{1+x^2}}\tag4$$
How do you get from $(2)$ to $(4)$? I have included $(3)$ in order to show my workings.
|
$$\frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(1+\frac{x}{\sqrt{1+x²}}\right) \\ = \frac{3}{2}\left(x+\sqrt{1+x²}\right)^\frac{1}{2}\left(\frac{\sqrt{1+x²}+x}{\sqrt{1+x²}}\right) \\ = \frac{3}{2}\frac{\left(x+\sqrt{1+x²}\right)^\frac{3}{2}}{\sqrt{1+x²}} \\ = \frac{3}{2}\frac{y}{\sqrt{1+x²}} \\ = \frac{3y}{2(y-x)}$$ though you did not ask for the final step
|
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|
The smallest value of the expression $4x^2y^2+x^2+y^2-2xy+x+y+1$ What is the smallest value that $4x^2y^2+x^2+y^2-2xy+x+y+1$ can take with real numbers $x$ and $y$?
I suspect the following transformation can be done: $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$.
|
$(2xy−\frac12)^2+(x+\frac12)^2+(y+\frac12)^2+\frac14 \geq \frac14$.
Because $(2xy−\frac12)^2\geq 0,(x+\frac12)^2 \geq 0,(y+\frac12)^2 \geq 0$.
So this minimum is attained in the original expression when
$(2xy−\frac12)^2=0,(x+\frac12)^2=0,(y+\frac12)^2=0 \iff x=y=-\frac12$.
|
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|
An uncommon continued fraction of $\frac{\pi}{2}$ I'm currently stuck with the following infinite continued fraction:
$$\frac{\pi}{2}=1+\dfrac{1}{1+\dfrac{1\cdot2}{1+\dfrac{2\cdot3}{1+\dfrac{3\cdot 4}{1+\cdots}}}}$$
There is an obscure clue on this: as one can derive the familiar Lord Brouncker’s fraction below
$$
\frac{4}{\pi}=1+\dfrac{1^{2}}{2+\dfrac{3^{2}}{2+\dfrac{5^{2}}{2+\dfrac{7^{2}}{2+\cdots}}}}
$$
from the Wallis' Formula:
$$
\dfrac{2}{\pi}=\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{3 \cdot 5}{4 \cdot 4} \cdot \frac{5 \cdot 7}{6 \cdot 6} \cdot \frac{7 \cdot 9}{8 \cdot 8} \cdots
$$
the first fraction can be proved in the same manner.
However, I'm not getting any close to it using the Wallis' Formula. Really appreciated if anyone could point me the right direction or explain further how to systematically derive those continued fractions from any given convergent cumulative product.
|
The result comes from a rather slight modification of Euler's continued fraction ,
$$a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+....=\cfrac{a_0}{1-\cfrac{a_1}{1+a_1-\cfrac{a_2}{1+a_2-\cfrac{a_3}{\cdots}}}}\label{1}\tag{1}$$
Note that for a product $a_0a_1a_2\cdots$; we can represent them as,
$$\begin{align}a_{1}a_{2}a_{3}...&=a_{1}+a_{1}\left(a_{2}-1\right)+a_{1}a_{2}\left(a_{3}-1\right)+\cdots\label{2}\tag{2}\end{align}$$
With \eqref{2} in \eqref{1},
$$a_{1}a_{2}a_{3}a_{4}\cdots=1+\cfrac{a_{1}-1}{1-\cfrac{a_{1}\left(a_{2}-1\right)}{a_{2}a_{1}-1-\cfrac{\left(a_{1}-1\right)a_{2}\left(a_{3}-1\right)}{a_{3}a_{2}-1-\cfrac{\left(a_{2}-1\right)a_{3}\left(a_{4}-1\right)}{\cdots}}}}\label{3}\tag{3}$$
Now apply the Wallis product, $$\frac{\pi}{2}=\frac{2\cdot2}{1\cdot3}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{6\cdot6}{5\cdot7}\cdots$$
in \eqref{3} with $a_1=2/1$, $a_2=2/3$ ... to get the first result
The second result is not obtained via the Wallis product, but from the Leibniz series for $\pi/4$ using \eqref{1}
|
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Decomposition of matrix occuring in problem of finding $n+1$ vectors in $\mathbb{R}^n$ with pairwise equal inner product I was toying with my intuition that there are always $n+1$ unit vectors in $\mathbb{R}^n$ such that every pair $v_i$ and $v_j$ ($i\neq j$) has the same angle between them. As those vectors are normalized this is equivalent to $$v_i\cdot v_j=\begin{cases} 1,& i=j \\m, & i\neq j \end{cases}$$ with $|m|<1$. So if $V$ is the $n\times (n+1)$ matrix whose columns are formed by the $v_i$ we have $$M:=V^TV=\begin{pmatrix}1 & m & \cdots &m \\m&1&\cdots&m\\ \vdots &&\ddots&\vdots \\ m&m&\cdots&1 \end{pmatrix}$$ $M$ is a $(n+1)\times (n+1)$ matrix and must be singular (as it is the product of two rank $n$ matrices). The sum of its rows is $(nm+1,nm+1,\ldots,nm+1)$. For $m=-\frac{1}{n}$ this is the null vector giving the required dot product $m$ as a function of $n$. E.g., for $n=2$, $m=-\frac{1}{2}$ corresponding to the expected angle of $120°=\arccos{-\frac{1}{2}}$. Now I'm asking is there a matrix decomposition so that I can get $V$ back from the product $M$? (I know that $V$ is not unique since I can freely rotate the vectors $v_i$ about an arbitrarily chosen fixed axis without changing their pairwise dot product.) I tried spectral decomposition of $M$ but that gives $n+1$ dimensional vectors and the $n$ nonzero eigenvalues of $M$ are all equal to $\frac{n+1}{n}$ which means that any linear combination of eigenvectors corresponding to that eigenvalue is also an eigenvector making the dot product between those eigenvectors rather arbitrary.
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Adding to the existing answer, you can also extract your vectors by orthogonally diagonalizing $M$. The matrix $M$ has rank $n$ with two eigenvalues: The eigenvalue $\lambda := \frac{n+1}{n}$ of multiplicity $n$ and the eigenvalue $0$ of multiplicity one. Let $U$ be an orthogonal matrix such that
$$ U \begin{pmatrix} \lambda I_{n \times n} & 0_{n \times 1} \\ 0_{1 \times n} & 0 \end{pmatrix} U^T = M. $$
The first $n$ columns of $U$ form an orthonormal basis for the eigenspace of $M$ associated to $\lambda$ which is
$$\{ (x_1, \dots, x_{n+1})^T \, | \, x_1 + \dots + x_{n+1} = 0 \}. $$
Write
$$ U = \begin{pmatrix} \hat{U}_{n \times n} & x_{n \times 1} \\ y_{1 \times n} & z \end{pmatrix}, V = \begin{pmatrix} \hat{V}_{n \times n} & v_{n+1} \end{pmatrix}. $$
The equation $V^T V = M$ then becomes
$$ \begin{pmatrix} \hat{V}^T \hat{V} & \hat{V}^T v_{n+1} \\ v_{n+1} V^T & v_{n+1}^T v_{n+1} \end{pmatrix} = \lambda \begin{pmatrix} \hat{U} \hat{U}^T & \hat{U} y^T \\ y \hat{U}^T & y \hat{y}^T \end{pmatrix}. $$
Hence if you take $\hat{V} = \sqrt{\lambda} \hat{U}^T, v_{n+1} = \sqrt{\lambda} y^T$ you get a solution. In other words, if you take the $n + 1$ rows of the eigenvector matrix $U$ (whose columns are the eigenvectors), forget the last coordinates and transpose, you get a solution. Since the eigenspace assosicated to the eigenvalue $\lambda$ is $n$-dimensional, the matrix
$$ \begin{pmatrix} \hat{U} \\ y \end{pmatrix} $$ is determined up to an orthogonal transformation of an $n$-dimensional subspace which corresponds to the freedom in rotating the columns of $V$ in an $n$-dimensional space.
Example: For $n = 2$, one possible decomposition is
$$ M = \begin{pmatrix} 1 & -1/2 & -1/2 \\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1 \end{pmatrix} = \underbrace{\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{pmatrix}}_{U} \begin{pmatrix} \frac{3}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 0 \\ 0 & 0 & 0 \end{pmatrix}
\underbrace{\begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{pmatrix}}_{U^T} $$
and the corresponding vectors are:
$$ v_1 = \sqrt{\frac{3}{2}} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}^T = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}, \\
v_2 = \sqrt{\frac{3}{2}} \begin{pmatrix} 0 & -\frac{2}{\sqrt{6}} \end{pmatrix}^T = \begin{pmatrix} 0 \\ -1 \end{pmatrix}, \\
v_3 = \sqrt{\frac{3}{2}} \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}^T = \begin{pmatrix} -\frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix} $$
|
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|
Small clarification Spivak Chapter 1, 18(b)
Suppose that ${b}^{2} - 4c < 0$. Show that there are no numbers $x$ satisfying ${x}^{2} + bx + c = 0$; in fact, ${x}^{2} + bx + c > 0$ for all $x$. Hint; Complete the square.
There is one answer here.
Solution (per Spivak)
We have
${x}^{2} + bx + c = \left (x + \frac{b}{2} \right)^{2} + \left(c - \frac{b^{2}}{4} \right) \ge c - \frac{b^{2}}{4} $
Analysis
$\left (x + \frac{b}{2} \right)^{2} $ is $\ge 0$ as any real number $a$, $a^{2} \ge 0$
$\left(c - \frac{b^{2}}{4} \right)$ See question below.
Spivak now states
but $c - \frac{b^{2}}{4} > 0$, so ${x}^{2} + bx + c > 0$ for all $x$
Question
Ignoring the conclusion (included for completeness), what allows him to assert $c - \frac{b^{2}}{4} > 0$?
|
Since the assumption (given condition) was,
$$b^2-4c<0\implies c-\frac{b^2}{4}>0.$$
|
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Where is the mistake? Finding an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
Find an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
We consider the foci in the coordinate system $XY$ such that $X=x-2$ and $Y=y-1-x$, the coordinates of the foci in this system are $(-1,0)$ and $(1,0)$, furthermore $2a=5$, the equation of the ellipse in $XY$ is
\begin{equation}
\left( \frac{X}{2.5}\right)^2 + \left( \frac{Y}{\sqrt{5.25}} \right)^2 = 1
\end{equation}
and this can be expressed in $xy$ as
$$\left( \frac{x-2}{2.5} \right) + \left( \frac{y-1-x}{\sqrt{5.25}}\right) = 1$$
I have made the graph of the last equation and it is not the case that foci are $(1,2)$ and $(3,4)$, so, can anyone help me to see the mistake please?
|
Let $r = [x, y]^T, F_1= [1, 2]^T , F_2 = [3, 4]^T $
Define a reference coordinate frame $O'uv$ with origin at $O' = \dfrac{1}{2}(F_1 + F_2) = [2,3]^T $, and let the $u$ axis extend along the vector $F_1 F_2$ while the $v$ vector is perpendicular to it and $+90^\circ$ away from it. If $r' =[u, v]^T $, then
$ r = O' + A r' $
where $A = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix} $
Now the equation describing the ellipse in the $O'uv$ frame is
$ \dfrac{u^2}{a^2} + \dfrac{v^2}{b^2} = 1 $
To determine $a,b$ we have that the two focii are at $(-c, 0)$ and $(c, 0)$ where $c = \dfrac{1}{2} \sqrt{ (3 - 1)^2 + (4 - 2)^2 } = \sqrt{2} $. In addition, we know that $2 a = 5$ , hence, $a = \dfrac{5}{2}$. Now $c^2 = a^2 - b^2$ from which it follows that $b^2 = a^2 - c^2 = \dfrac{25}{4} - 2 = \dfrac{17}{4} $
Therefore, the equation of the ellipse in the $O' uv $ plane is
$ r'^T Q r' = 1 $
where $Q = 4 \begin{bmatrix} \dfrac{1}{25} && 0 \\ 0 && \dfrac{1}{17} \end{bmatrix} $
Now since $ r = O' + A r' $ , then $ r' = A^{-1} (r - O') $
Substituting this in the equation of the ellipse results in
$ (r - O')^T A^{-T} Q A^{-1} (r - O') = 1 $
We now have: $A^{-1} = A^T = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && 1 \\ -1 && 1 \end{bmatrix} $
And $A^{-T} = A $. Therefore,
$A^{-T} Q A^{-1} = A Q A^T = \dfrac{1}{425} \begin{bmatrix} 84 && -16 \\ -16 && 84 \end{bmatrix}$
The resulting equation in the $xy$ plane is,
$ 84 (x - 2)^2 - 32 (x - 2)(y - 3) + 84 (y - 3)^2 = 425 $
|
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
|
$$\begin{align}
4x^2-2xy-4x+3y-3
&=(4x^2-4x-3)-(2x-3)y\\
&=(2x-3)(2x+1)-(2x-3)y\\
&=(2x-3)(2x+1-y)
\end{align}$$
If the quadratic $4x^2-4x-3$ hadn't had $2x-3$ as a factor, the polynomial in $x$ and $y$ would not have factored.
|
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When is the sum of the products of any three distinct numbers less than $n$, divisible by $n$? Define $A$ as the set of all integer triples $(a,b,c)$, such that $0<a<b<c<n$.
$$A=\left\{(a,b,c)\in \mathbb{Z^3}:1\le a<b<c\le n-1 \right\}$$
Define $S$ as the sum of the product $abc$ for each triplet in $A$.
$$S=\displaystyle \sum_{(a,b,c)\in A}a\cdot b\cdot c$$
For what values of $n$ is $S$ divisible by $n$? Find all possible remainders when $S$ is divided by $n$.
The smallest possible value of $n$ is $4$. If $n=4$, we have, $A=\{(1,2,3)\}$ and $S=6.$ Obviously, $4\nmid 6$.
Claim: $S\equiv 0\pmod n$ for all odd $n$.
Proof: If $(a,b,c)\in A$, then $(n-c,n-b,n-a)\in A$. Since,
$$abc+(n-c)(n-b)(n-a)\equiv abc+(-c)(-b)(-a)\equiv 0 \pmod n$$
and $(a,b,c)\not\equiv (n-c,n-b,n-a)$. Therefore, $n$ divides $S$.
If $n$ is even, the same proof does not work because $(a,b,c)$ and $(n-c,n-b,n-a)$ might be identical.
For example, if we set $a=1,b=\frac{n}{2}$ and $c=n-1$,we have, $(a,b,c)\equiv (n-c,n-b,n-a)$.
Also, I checked upto $n=100$ using a simple program:
L=[]
for n in range (4,101):
S=0
for a in range (1,n-2):
for b in range (a+1,n-1):
for c in range(b+1,n):
S=S+a*b*c
print('n:',n,' ','Sum:',S)
if S%n==0:
print(True)
else:
print(False)
L.append(n)
print()
print(L)
This is the output for the first few numbers
n: 4 Sum: 6
False
n: 5 Sum: 50
True
n: 6 Sum: 225
False
n: 7 Sum: 735
True
n: 8 Sum: 1960
True
n: 9 Sum: 4536
True
n: 10 Sum: 9450
True
n: 11 Sum: 18150
True
n: 12 Sum: 32670
False
n: 13 Sum: 55770
True
n: 14 Sum: 91091
False
This is the list of numbers $n $ for which $S$ is not divisible by $n$.
[4, 6, 12, 14, 20, 22, 28, 30, 36, 38, 44, 46, 52, 54, 60, 62, 68, 70, 76, 78, 84, 86, 92, 94, 100]
Why is $S\equiv 0\pmod n\;, \;\forall n\not\equiv 4 \;\text{or}\; 6 \pmod 8$? Also, when $n$ is even and does not divide $S$, is the below claim true ? $$S\equiv \frac{n}{2} \pmod n$$
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Suppose $n$ is even, then $(a,b,c)\in A\implies (n-c,n-b,n-a)\in A~~\forall (a,b,c)\ne (r,\frac n2,n-r)$,
here $(a,b,c)$ and $(n-c,n-b,n-a)$vare non identical.
So to calculate $S$ you only need to consider the cases where $(a,b,c)$ is of the form $(r,\frac n2,n-r)$
$$\begin{align*}S&\equiv \sum_{(a,b,c)=(r,\frac n2,n-r)}a\cdot b\cdot c \pmod n
\\&\equiv \sum_{r=1}^{ \frac{n-2}2}\frac{n^2r-r^2}{2}\pmod n
\\&\equiv \frac{12n^3(n-2)+n(n-2)(2n-2)}{96}\pmod n
\\&\equiv \frac{n(2n+1)(3n-1)(n-2)}{48}\pmod n\end{align*}$$
$n|S$ only if $48|n(2n+1)(3n-1)(n-2)$. It can be seen that $3|n(2n+1)(n-2)$ so the condition reduces to $16|n (n-2) $ which means that $n$ or $n-2$ must be a multiple of 8.
|
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|
How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x\\&=\displaystyle -\frac{1}{6} \int_{0}^{\infty} x d\left(\frac{1}{x^{6}+1}\right)\\&
=\displaystyle -\left[\frac{x}{6\left(x^{6}+1\right)}\right]_{0}^{\infty}+\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x \quad \textrm{ (Via Integration by Parts})\\&=\displaystyle \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x\end{aligned}$
$\textrm{Then I am planning to evaluate }\displaystyle I= \int_{0}^{\infty} \frac{1}{x^{6}+1}\text{ by resolving }\frac{1}{x^{6}+1} \text{ into partial fractions.}$
But after noticing that $$I=\int_{0}^{\infty} \frac{d x}{x^{6}+1}\stackrel{x\mapsto\frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{4}}{x^{6}+1} d x,$$
I changed my mind and started with $3I$ instead of $I$ as below:
$$
\begin{aligned}
3 I &=\int_{0}^{\infty} \frac{x^{4}+2}{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)} d x \\
&=\int_{0}^{\infty}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{4}-x^{2}+1}\right) d x \\
&=\left[\tan ^{-1} x\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&=\frac{\pi}{2}+\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\
&=\frac{\pi}{2}+\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\right] \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty}-0 \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\
&=\pi \\ \therefore I &=\frac{\pi}{3}
\end{aligned}
$$
Now I can conclude that
$$\boxed{\displaystyle \quad \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x=\frac{\pi}{18} }.$$
:|D Wish you enjoy the solution! Opinions and alternative methods are welcome.
|
First, evaluate
\begin{align}\int_{0}^{\infty} \frac{1}{x^{6}+1}dx\stackrel{x\to\frac{1}{x}}{=}&\ \frac12 \int_{0}^{\infty} \frac{x^{4}+1}{x^{6}+1} d x\\
=&\ \frac12 \bigg(\int_{0}^{\infty} \frac{1}{x^{2}+1} d x
+ \int_{0}^{\infty} \frac{x^2}{x^{6}+1}\overset{x^3\to x}{ d x}\bigg)\\
=&\ \frac12\cdot \frac43\int_{0}^{\infty} \frac{1}{x^2+1}dx
=\frac\pi3
\end{align}
Then,
\begin{aligned}\int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x \overset{ibp}=\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x =\frac\pi{18}\end{aligned}
|
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"url": "https://math.stackexchange.com/questions/4292775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
How to show $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$? I am trying to show that the function $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$, where $0.5 \leq L < H < 1$. Visual inspection in Mathematica suggests the conjecture is true. I tried showing this by differentiation, but am having trouble signing it. How do I show that $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$? Thanks!
|
You can write
$$
f(x) = \frac{(1-H)^x - (1-L)^x}{H^x - L^x} = \left(\frac{1-L}{H}\right)^x \cdot
\frac{1 - \left(\frac{1-H}{1-L}\right)^x}{1 - \left(\frac{L}{H}\right)^x}=
u(x)\cdot v(x).
$$
Then use the inequalities to obtain, $$(1-L)/H < 1 \quad\text{and}\quad
0 < \frac{1-H}{1-L}<\frac{L}{H}<1.$$
These inequalities imply both $u(x), v(x)$ are positive; moreover, from the first $u(x)$ is plainly decreasing and, with a little further investigation, the second means $v(x)$ is also decreasing. we can therefore conclude $f(x)$ is decreasing.
It remains to show $v(x)$ is decreasing.
We remark that $\log t / (t-1)$ is a decreasing function of $t$ for $ t > 0$, because its derivative is,
$$
\frac{\log(1/t) - 1/t +1 }{(t-1)^2}
$$
which is strictly negative when $t>0$.
Now to verify $v(x)$ is decreasing, introduce $a= (1-H)/(1-L), b = L/H$ so that $0 < a < b < 1$. Then, for $x >0$, the derivative,
\begin{align}
v'(x) &= \frac{-\log(a)a^x (1-b^x) + \log(b) b^x(1-a^x)}{(1-b^x)^2} \\
\end{align}
Focus on the numerator, and consider the ratio
\begin{align}
\frac{-\log(a) a^x (1-b^x)}{-\log(b) b^x (1-a^x)} &= \frac{x\log(1/a)(1/b^x - 1)}{x\log(1/b)(1/a^x-1)} \\
&=\frac{\log(1/a^x)}{1/a^x - 1} \cdot \frac{1/b^x - 1}{\log(1/b^x)}
\end{align}
As $x \to 0$, this expression has limit $1$ since left and right terms converge to the derivative of $\log(t) $ at $t=1$ and its reciprocal respectively. But we also have $1/b^x < 1/a^x$ when $x > 0$ and using our remark above,
\begin{align*}
\frac{\log(1/a^x)}{1/a^x - 1} < \frac{\log(1/b^x)}{1/b^x - 1}
\end{align*}
so that
\begin{align}
\frac{-\log(a) a^x (1-b^x)}{-\log(b) b^x (1-a^x)} < 1
\end{align}
from which it follows $v'(x) < 0$ when $x > 0$.
That leaves uncertainty when $x=0$ at which point $f$ and $v$ are not properly defined. However, using the power series for $e^z$, we can see
$$
v(x) = \frac{\log\left(\tfrac{1-H}{1-L}\right)x + \tfrac{1}{2!} \log\left(\tfrac{1-H}{1-L}\right)^2x^2 + O(x^3)}{\log\left(\tfrac{L}{H}\right)x + \tfrac{1}{2!} \log\left(\tfrac{L}{H}\right)^2 x^2 + O(x^3)}.$$
Because $L \neq H$ it is plain $v(x)$ thus extends to a continuously differentiable function at $x=0$, so that the derivative at $x=0$ would also be negative or zero.
Accordingly $v(x)$ is decreasing for $x \geqslant 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4293663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Simplifying $\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, $\cosh^2 x - \sinh^2 x$ using only the Taylor Series of $\cosh,\sinh$ I was trying to solve the following question:
\begin{align*}
\sinh x &= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\
\cosh x &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots
\end{align*}
Using only this information, calculate
$\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, and $\cosh^2 x - \sinh^2 x$.
Calculating $\cosh x + \sinh x$ was easy because it's just the Taylor Series of $e^x$, but dealing with squaring is where it gets difficult because the Taylor Series are infinite. How can I circumvent the infinite portion to get $\cosh^2 x$ and $\sinh^2 x$?
|
We have
$$
\begin{align}
\cosh^2(x)&=\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\right)\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\right)
\\&= \sum_{k=0}^\infty \sum_{l=0}^k \frac{x^{2l}x^{2(k-l)}}{(2l)!(2(k-l))!}
\\&= \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} \color{red}{\sum_{l=0}^k \binom{2k}{2l}}
\\&= 1 + \sum_{k=1}^\infty \frac{\color{red}{2^{2k-1}}x^{2k}}{(2k)!}
\\&= \frac 1 2 +\frac 1 2 \sum_{k=0}^\infty \frac{(2x)^{2k}}{(2k)!}
\\&= \frac 1 2 (\cosh{2x}+1)
\end{align}
$$
where the red is a well-known identity (when $2k \in \mathbb{Z^+}$). We similarly have
$$
\begin{align}
\sinh^2(x)&=\left(\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}\right)\left(\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}\right)
\\&= \sum_{k=0}^\infty \sum_{l=0}^k \frac{x^{2l+1}x^{2(k-l)+1}}{(2l+1)!(2(k-l)+1)!}
\\&= \sum_{k=0}^\infty \frac{x^{2(k+1)}}{(2(k+1))!} \color{red}{\sum_{l=0}^k \binom{2(k+1)}{2l+1}}
\\&= \sum_{k=0}^\infty \frac{\color{red}{2^{2k+1}}x^{2(k+1)}}{(2(k+1))!}
\\&= \frac 1 2 \sum_{k=0}^\infty \frac{(2x)^{2(k+1)}}{(2(k+1))!}
\\&= \frac 1 2 \sum_{k=1}^\infty \frac{(2x)^{2k}}{(2k)!}
\\&= \frac 1 2 (\cosh{2x} - 1)
\end{align}
$$
So
$$
\cosh^2(x)+\sinh^2(x)=\frac 1 2 (\cosh{2x}+1) + \frac 1 2 (\cosh{2x} - 1)=\cosh(2x)
$$
and
$$
\cosh^2(x)-\sinh^2(x)=\frac 1 2 (\cosh{2x}+1) - \frac 1 2 (\cosh{2x} - 1)=1
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4295089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
What is the series $ \dfrac{1}{\left( x-{y}^{2} \right) \left(1- yx \right) \left( {x}^{2}-y \right)} $ expansion? Consider the series expansion
$$
\frac{1}{\left( x-{y}^{2} \right) \left(1- yx \right) \left( {x}^{2}-y \right)}=P_0(y)+P_1(y)x+P_2(y)x^2+\cdots+P_n(y) x^n+\cdots.
$$
For small $n$ we have
\begin{gather*}
P_0(y)=\frac{1}{y^3},\\
P_1(y)={\frac {{y}^{3}+1}{{y}^{5}}},\\
P_2(y)={\frac {{y}^{6}+2\,{y}^{3}+1}{{y}^{7}}},\\
P_3(y)={\frac {{y}^{9}+2\,{y}^{6}+2\,{y}^{3}+1}{{y}^{9}}},\\
P_4(y)=\frac{{y}^{12}+2\,{y}^{9}+3\,{y}^{6}+2\,{y}^{3}+1}{y^{11}},\\
P_5(y)={\frac {{y}^{15}+2\,{y}^{12}+3\,{y}^{9}+3\,{y}^{6}+2\,{y}^{3}+1}{{y}^{
13}}}.
\end{gather*}
Question: What is a close expression for $P_n(y)$?
My attempt. The $P_n(y)$ has a form
$$
P_n(y)=\frac{1}{y^{2n+3}}(c_{n,1}+c_{n,2}y^3+c_{n,3}y^{6}+\cdots+с_{n,k} y^{3(k-1)}+\cdots+c_{n,n+1}y^{3n})
$$
for some unknown sequence $c_{n,k}.$
By empirical way I have found that
$$
c(n,k)= \left \{
\begin{array}{l}
\min(n{-}k+2,k) , 1 \leq k \leq n+1, \\
\\
0, \text{ otherwise, }
\end{array}
\right.
$$
but I can't find a rigorous proof.
Any help?
|
Your expression is a product of three geometric series (first two multiplied by $-1$ which does not affect the final product):
$$
\frac{1}{y-x^2}=\frac{1}{y}\frac{1}{1-\frac{x^2}{y}}=\frac{1}{y}+\frac{1}{y^2}x^2+\dots+\frac{1}{y^{n+1}}x^{2n}+\dots\\
\frac{1}{y^2-x}=\frac{1}{y^2}\frac{1}{1-\frac{x}{y^2}}=\frac{1}{y^2}+\frac{1}{y^4}x+\dots+\frac{1}{y^{2n+2}}x^{n}+\dots\\
\frac{1}{1-xy}=1+xy+x^2y^2+\dots+x^{n}y^n+\dots\\
$$
The product of the first two is
\begin{align}
\sum_{n=0}^{\infty}(\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{1}{y^{k+1}}\frac{1}{y^{2(n-2k)+2}})x^n
&=\sum_{n=0}^{\infty}(\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{1}{y^{2n-3k+3}})x^n
\end{align}
Multiplying with the third one we get
\begin{align}
\frac{1}{y-x^2}\frac{1}{y^2-x}\frac{1}{1-xy}
&=\sum_{n=0}^{\infty}\sum_{i=0}^{n}\sum_{k=0}^{\lfloor i/2 \rfloor}\frac{1}{y^{2i-3k+3}}y^{n-i} x^n\\
\end{align}
So the value you want is
\begin{align}
P_n(y)=\sum_{i=0}^{n}\sum_{k=0}^{\lfloor i/2 \rfloor}\frac{1}{y^{2i-3k+3}}y^{n-i}
&=y^{n-3}\sum_{i=0}^{n}\sum_{k=0}^{\lfloor i/2 \rfloor}\frac{1}{y^{3i-3k}}.
\end{align}
For fixed $t$, how many ways we can choose $i,k$ to get $t=i-k$, subject to $0\leq i \leq n$, $0 \leq k \leq i/2$? Since $k=i-t$ we have $0 \leq i-t \leq i/2$ we get $i/2 \leq t \leq i \leq n$. This works in the other direction as well, for any $i$ with $i/2 \leq t \leq i$ we set $k=i-t$ and the original conditions are satisfied. So the question reduces to how many $i$ exist such that $i/2 \leq t \leq i \leq n$? We can see that if $t \geq n/2$, then any $t \leq i \leq n$ will work, hence there is $n-t+1$ of these. If on the other hand $t < n/2$, then any $i/2 \leq t \leq i$ will work, so $t\leq i \leq 2t$ and hence we have $t+1$ possibilities. We can combine these as $\min(t+1,n-t+1)=\min(t,n-t)+1$. So the value simplifies to
\begin{align}
\boxed{P_n(y)=\sum_{t=0}^{n}\frac{\min(t,n-t)+1}{y^{3t-n+3}}}
\end{align}
|
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"url": "https://math.stackexchange.com/questions/4295791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
$\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k\cdot \pi }{n}\right)-2\:+\:i\cdot \sin\left(\frac{2\cdot k\cdot \pi }{n}\right)\right)$ $\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k\cdot \pi }{n}\right)-2\:+\:i\cdot \sin\left(\frac{2\cdot k\cdot \pi }{n}\right)\right)$
Normally the general factor is $a(n)=\cos\left(\frac{2k\pi }{n}\right)-2+i\cdot \sin\left(\frac{2k\pi }{n}\right)$
All I was able to do to the sum above was rewrite it as:
$\sum _{k=1}^n\:\left(\cos\left(\frac{2k\pi }{n}\right)\right)\:+\:\sum _{k=1}^n\left(i\cdot \sin\left(\frac{2k\pi }{n}\right)\right)\: - 2n$
Expecting the sums to cancel each other out and a result of something around -2n but not sure how to do that.
|
$$S = \sum_{k =1}^n\left[\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right]$$
Consider vectors in complex plane:
*
*Analytical
$$\begin{align*}S
&= \sum_{k =1}^n\left[\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right] = \sum_{k =1}^ne^{i\frac {2k\pi}{n}}\\
& =\frac {e^{i(2\pi + \frac {2\pi}{n})}-e^{i(\frac {2\pi}{n})}}{e^\theta-1} =0 \text{ :Numerator = 0}\\
\end{align*}$$
$$\implies \sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot \:k\cdot \pi }{n}\right)-2\:+\:i\cdot \:\sin\left(\frac{2\cdot \:k\cdot \pi }{n}\right)\right) = -2n$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proof of identity related to sum of binomial coefficients and powers of $2$ Given two positive integers $a,b$, prove that $\left(\sum_{n=a}^{a+b-1}\binom{n-1}{a-1}2^{-n}\right)+\left(\sum_{n=b}^{a+b-1}\binom{n-1}{b-1}2^{-n}\right)=1$
The context comes from this MSE problem, where I indirectly showed that this is true with a combinatorial argument for the specific case of $a=3$ and $b=4$. This is because the first sum computes the probability of $A$ winning and the second sum computes the probability of $B$ winning. Since these are the only $2$ possible events, their sum must be $1$. This argument can be easily extended to a more general case.
While I understand the combinatorial argument, I would like to see an algebraic method. I'm not sure where to start for it.
Also, from the linked post, I (from the other answer) found that $\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}=2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}$. Proving this would readily prove the original identity, but I'm not sure how to prove this equation.
I'm looking for something like a generating function proof. I found that $\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}$ is the coefficient of $x^{b-1}$ in $\frac{1}{(2-x)^a(1-x)}$ and $2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}$ is the coefficient of $x^{b-1}$ in $\frac{2^{-a-b+1}(1+x)^{a+b-1}}{1-x}$
|
We prove that $$\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n} = 2^{-a-b+1} \sum_{n=a}^{a+b-1} \binom{a+b-1}{n}$$ for all positive integers $a$ and $b$ by inducting on $b$.
When $b = 1$, the LHS and RHS evaluate to $2^{-a}$, so the statement holds in this case.
We proceed with the induction step. Using the induction hypothesis,
\begin{align*}
\sum_{n=a}^{a+b} \binom{n-1}{a-1}2^{-n}
&= \left(\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}\right) + \binom{a+b-1}{a-1}2^{-a-b} \\
&= \left(2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}\right) + \binom{a+b-1}{a-1}2^{-a-b} \\
&= 2^{-a-b} \left[\left(2\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}\right) + \binom{a+b-1}{a-1} \right].
\end{align*}
Observe the terms in the brackets. Recall that $$\binom{m-1}{k-1} + \binom{m-1}{k} = \binom{m}{k}$$ for all integers $m \geq k \geq 1$. So, by cleverly rearranging the terms in the summation, we have
\begin{align*}
& \:\: \left(2\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}\right) + \binom{a+b-1}{a-1} \\
= & \:\: \binom{a+b-1}{a+b-1} + \sum_{n=a}^{a+b-1}\left(\binom{a+b-1}{n-1} + \binom{a+b-1}{n}\right) \\
= & \: \: 1 + \sum_{n=a}^{a+b-1} \binom{a+b}{n} \\
= & \: \: \sum_{n=a}^{a+b} \binom{a+b}{n}.
\end{align*}
Substituting this back, we get $$\sum_{n=a}^{a+b} \binom{n-1}{a-1}2^{-n} = 2^{-a-b} \sum_{n=a}^{a+b} \binom{a+b}{n}.$$ This completes the induction proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What's the measure of the segment $BC$ in the rhombus below? For reference: Given a rhombus $ABCD$, on $BC$
mark the point $P$ such that : $BP= 3PC$ and $AP^2+ 3DP^2 = 38$.
Calculate $BC$.(answer: $2\sqrt2$)
My progress:
$BP = 3CP\\
AP^2+3DP^2 = 38\\
AB=BC=CD=AD$
Th. Stewart:
$\triangle ABC:\\
AC^2.BP+AB^2.CP=AP^2BC+BC.CP.BP\\
AC^2. 3CP+AB^2,CP = BC(AP^2+3CP^2)\\
\boxed{CP(3AC^2+AB^2) = BC(AP^2+3CP^2)}(I)\\
\triangle DBC:\\
CD^2.BP+BD^2CP=DP^2.BC+BC.BP.CP\\
CD^2.3CP+BD^2.CP=BC(DP^2+3CP^2)\\
\boxed{CP(3CD^2+BD^2) = BC(DP^2+3CP^2}(II)$
(I)+(II):
$\boxed{CP(3(AC^2+CD^2)+AB^2+BD^2) = BC(AP^2+DP^2+6CP^2)(III)}$
...??
|
$ \small AB^2 \cdot CP + AC^2 \cdot 3 CP = 4CP \cdot(AP^2 + 3 CP^2)$
As $ \displaystyle \small AB = BC \text { and } BC = 4 CP$,
$ \displaystyle \small BC^2 + 3 AC^2 = 4 (AP^2 + \frac {3 BC^2}{16})$
$ \displaystyle \small BC^2 + 12 AC^2 = 16AP^2 \tag1$
Similarly,
$ \displaystyle \small BD^2 \cdot CP + CD^2 \cdot 3CP = 4 CP \cdot (DP^2 + 3CP^2)$
$ \displaystyle \small BD^2 + 3 BC^2 = 4 (DP^2 + \frac{3 BC^2}{16})$
$ \displaystyle \small 4BD^2 + 9BC^2 = 16 DP^2 \tag2$
By $(1) + 3 \cdot (2), ~$
$ \displaystyle \small 28BC^2 + 12 (AC^2 + BD^2) = 16 \cdot 38 $
Note that $ \displaystyle \small AC^2 + BD^2 = 4 BC^2$ given a rhombus.
So, $ \displaystyle \small 76 BC^2 = 16 \cdot 38 \implies BC^2 = 8$
$ \therefore BC = 2\sqrt2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A remarkable Continued Fraction for $\pi$ How to analyze the Continued Fraction $3+\cfrac{\color{red}1^2}{5+\cfrac{\color{blue}4^2}{7+\cfrac{\color{red}3^2}{9+\cfrac{\color{blue}6^2}{11+\cfrac{\color{red}5^2}{13+\cfrac{\color{blue}8^2}{15+\cfrac{\color{red}7^2}{17+\ddots}}}}}}}$
Numerically, it seems to evaluate to $\pi$. But why?
This remarkable Continued Fraction was sent to me by Paul Levrie.
The pattern in the nominators and denominators is obvious. Alternating $\color{red}3,\color{red}5,\color{red}7,\color{red}9,\color{red}{11}$ etc and $\color{blue}4^2,\color{blue}6^2,\color{blue}8^2,\color{blue}{10}^2$ etc.
Because of the appearance of squares, Euler's Differential Method can not be used. A clever trick is required here! Any idea's?
|
The difference between the even and odd squared coefficients is similar to that of the continued fraction of the ratio of two hypergeometric functions (DLMF):
\begin{equation}
\frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a,b+1;c+1;z\right)}=t_{0
}-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2}z}{t_{2}-\cfrac{u_{3}z}{t_{3}-\cdots}}}
\end{equation}
where
\begin{align}
t_n&=c+n\\
u_{2n+1}&=(a+n)(c-b+n)\\
u_{2n}&=(b+n)(c-a+n)
\end{align}
where $\mathbf{F}$ are regularized hypergeometric functions.
The continued fraction is adapted to this representation
\begin{align}
\mathcal J=&3+\cfrac{1^2}{5+\cfrac{4^2}{7+\cfrac{3^2}{9+\cfrac{6^2}{11+\cdots}}}}\\
\frac12 \mathcal J&=3/2+\cfrac{1^2/4}{5/2+\cfrac{4^2/4}{7/2+\cfrac{3^2/4}{9/2+\cfrac{6^2/4}{11/2+\cdots}}}}
\end{align}
It can be checked that $a=1/2,b=1,c=3/2,z=-1$ satisfies the recurrence relations, then
\begin{align}
\mathcal J&=\frac{2\,\mathbf{F}\left(1/2,1;3/2;-1\right)}{\mathbf{F}\left(1/2,2;5/2;-1\right)}\\
&=\frac{2\Gamma(5/2)}{\Gamma(3/2)}\,\frac{{}_2F_1\left(1/2,1;3/2;-1\right)}{{}_2F_1
\left(1/2,2;5/2;-1\right)}\\
&=3\,\frac{{}_2F_1\left(1/2,1;3/2;-1\right)}{{}_2F_1
\left(1/2,2;5/2;-1\right)}
\end{align}
From here and here
\begin{align}
{}_2F_1\left(1/2,1;3/2;-z\right)&=\frac{\tan^{-1}\sqrt{z}}{\sqrt{z}}\\
{}_2F_1\left(1/2,2;5/2;-z\right)&=\frac34\frac{(z-1)\tan^{-1}\sqrt{z}+\sqrt{z}}{z^{3/2}}
\end{align}
then
\begin{equation}
\mathcal J=\pi
\end{equation}
|
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|
Can we solve inequality without expanding both sides if each side has perfect square? The problem
$$\begin{align} (x+5)^2 &< (x-3)^2\tag{1} \end{align}$$
Normal people (most people) will solve the problem like this:
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
x^2 + 10x + 25 &< x^2-6x +9\\
10x+6x + 25-9 &< 0\\
16x &< -16\\
x &< -1
\end{align}$$
I want to know if I can solve $(1)$ like this:
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
(x+5)^2 - (x-3)^2 &<0\\
(2x+2)(7) &< 0 \tag{$a^2-b^2=(a+b)(a-b)$}\\
14x &< -14\\
x &< -1
\end{align}$$
Someone on facebook says like this:
You accidentally stumbled upon the right answer, but this process is
totally invalid and violates several properties of mathematics.
I want to know where the mistake is. Could you spot the mistakes please? Because I have tried on 3 different inequalities, and I still got the correct answers using this method. Thank you in advance!
P.S. edit version. I separate the edited version so that people can understand the problem and my mistakes.
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
(x+5)^2 - (x-3)^2 &<0\\
(2x+2)(8) &< 0 \tag{$a^2-b^2=(a+b)(a-b)$}\\
16x &< -16\\
x &< -1
\end{align}$$
|
I see no big mistake and you were told wrong. The process is totally valid and who made that comment probably didn't bother to look at it (you just have a minor computational mistake).
The inequality $(x+5)^2<(x+3)^2$ is equivalent to
$$
(x+5)^2-(x-3)^2<0
$$
and the left-hand side can be factored to give
$$
(x+5-x+3)(x+5+x-3)<0
$$
hence
$$
8(2x+2)<0
$$
(you mistakenly wrote $7$). Now expanding the multiplication is possible, but one can better note that this can be rewritten as
$$
16(x+1)<0
$$
and the factor $16$ can be removed to give
$$
x+1<0
$$
and so
$$
x<-1
$$
|
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|
Is it possible to find $\tan\theta$ by squaring both sides of $\sec\theta+\tan\theta=5$ right away? From $\sec\theta+\tan\theta=5$, in order to find $\tan\theta $ we can rewrite it as $\sec\theta=5-\tan\theta$ then square,
$$1+\tan^2\theta=\tan^2\theta-10\tan\theta+25\quad\Rightarrow\quad \tan\theta=\frac{24}{10}$$
But When I square the both sides of $\sec\theta+\tan\theta=5$ right away I get,
$$(1+\tan^2\theta)+\tan^2\theta+2\tan\theta\sec\theta=25$$
$$\tan^2\theta+\tan\theta\sec\theta=12$$
I'm wondering is it possible to find $\tan\theta$ from above equation?
|
The answer is no, not really.
In effect if $z=\tan \theta$ and $-\frac \pi 2 \lt \theta \lt \frac \pi 2$ then $\sec\theta+\tan\theta=5$ is equivalent to $\sqrt{1+z^2} +z=5$
If you just square both sides you then get $z^2 +z\sqrt{1+z^2}=12$ which does not take you forward; you still have the radical
while if you square both sides of $\sqrt{1+z^2} =5-z$ you get $1+z^2=z^2-10z +25$ and the solution $z=\frac{24}{10}$ as you found
|
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|
Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$. Prove the following $xyz\geq27,xy+yz+zx\geq27,x+y+z\geq9$ Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$.
Prove $$xyz\geq27,\\xy+yz+zx\geq27,\\x+y+z\geq9.$$
Here's what I tried:
$$\begin{align} \frac{x^3+y^3+z^3}{3}&\ge\sqrt[3]{x^3y^3z^3}\\
x^3+y^3+z^3&\ge3\sqrt[3]{x^3y^3z^3}\\
x^3+y^3+z^3&\ge3xyz \\ x^3+y^3+z^3&\ge 3(x^2+y^2+z^2)\end{align} $$
Not sure where to go from here, I think we have to use the arithmetic and geometrical means and their relation. ($A\ge G$)
|
By the AM-GM inequality, $$x^2 + y^2 + z^2 \ge 3 (x^2y^2z^2)^{1/3}$$
but $x^2 + y^2 + z^2 = xyz$, so $$xyz\ge 3(xyz)^{2/3} \implies (xyz)^{1/3} \ge 3 \implies xyz\ge 27$$
By two more applications of the AM-GM inequality and $xyz\ge 27$, you can show that $$x+y+z
\ge 3(xyz)^{1/3} \ge 9$$
and $$xy + yz + zx \ge 3(x^2y^2z^2)^{1/3} \ge 3\cdot 9 = 27$$
as required.
|
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|
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x+2}=0\\(x+1)(x+2)-2x\sqrt{3x+2}=0$$ What next? Thank you!
|
I'd recommend this method. You already got $(x+1)(x+2)-2x\sqrt{3x+2}=0$. And by this, $x\ge0$ is induced right away, because the left hand is $>0$.
Then...
$$
(x+1)(x+2)=2x\sqrt{3x+2}\\
(x+1)^2(x+2)^2=4x^2(3x+2)\\
x^4-6x^3+5x^2+12x+4=0
$$
This is sure a quatric, so seems hard to solve. But...
Let $x^2-3x=y$, then this quatric changes $y^2-4y+4=0$.So $y=2$, $x^2-3x-2=0$.Then we get two roots, $x=\frac{3\pm\sqrt{17}}2$. $\frac{3-\sqrt{17}}2$ is out of range.So the answer is ${x=\frac{3+\sqrt{17}}2}$.
|
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|
What's the rate of the stream's current? At his usual rate a man rows 15 miles downstream in five hours less than it takes him to return. If he doubles his usual rate, the time dowstream is only hour less than the time uqstream. What's the rate of the stream's current?
Let the man's rate in still water be $r$ and the rate of the current be $c$.
\begin{align*}
\frac{15}{r-c}-\frac{15}{r+c} &= 5 \tag{1}\\
\frac{15}{2r-c}-\frac{15}{2r+c} &= 1 \tag{2}
\end{align*}
Considering equation 1.
\begin{align*}
\frac{15(r+c)}{(r-c)(r+c)}-\frac{15(r-c)}{(r+c)(r-c)} &= 5 \\
\frac{15r+15c}{(r-c)(r+c)}-\frac{15r-15c}{(r+c)(r-c)} &= 5 \\
\frac{15r+15c}{r^2-c^2}-\frac{15r-15c}{r^2-c^2} &= 5 \\
\frac{30c}{r^2-c^2} &= 5 \\
5r^2-5c^2 &= 30c \tag{3}
\end{align*}
Considering equation 2.
\begin{align*}
\frac{15(2r+c)}{(2r-c)(2r+c)}-\frac{15(2r-c)}{(2r+c)(2r-c)} &= 1 \\
\frac{30r+15c}{(2r-c)(2r+c)}-\frac{30r-15c}{(2r+c)(2r-c)} &= 1 \\
\frac{30r+15c}{4r^2-c^2}-\frac{30r-15c}{4r^2-c^2} &= 1 \\
\frac{30c}{4r^2-c^2} &= 1 \\
4r^2-c^2 &= 30c \tag{4}
\end{align*}
We create a system of equations using 3 and 4.
\begin{align*}
5r^2-5c^2 &= 30c \tag{3}\\
4r^2-c^2 &= 30c \tag{4}
\end{align*}
\begin{align*}
5r^2 &= 30c +5c^2\\
\frac{5r^2}{5} &= \frac{30c +5c^2}{5}\\
\frac{5r^2}{5} &= \frac{30c +5c^2}{5}\\
r^2 &= 6c +c^2
\end{align*}
We subsitute into equation 4.
\begin{align*}
4r^2-c^2 &= 30c \\
4(6c +c^2)-c^2 &= 30c \\
24c +3c^2 &= 30c \\
\frac{3c^2}{3c} &= \frac{6c}{3c} \\
c &= 2
\end{align*}
I'm not feeling very confident of my answer.
|
Or eliminate $r^2$ by combining $5(4)-4(3)$:
$$ 5(4r^2-c^2)-4(5r^2-5c^2)=5\cdot 30c-4\cdot 30c$$
which simplifies to
$$ 15c^2=30c,$$
$$ 15c(c-2)=0,$$
a quadratic in $c$ with solutions
$$ c=2\quad \text{or}\quad c=0.$$
As $c=0$ makes no sense with the original equations, we must have $c=2$.
(You divided by $c$ somewhere, at which point you should mention a justification as to why we must have $c\ne0$).
|
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|
Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)=x^2+x(-2y-12)+6y^2+2y+41\ge 0, \forall x \in \Bbb{R}.$$
$$\implies B^2-4AC=(2y+12)^2-4(6y^2+2y+41)=-20(y-1)^2 \le 0,$$
which is true and the equality holds if $y=1$ this further means that $f(x)=(x-7)^2.$
So $f(x,y) \ge 0$, the equality holds if $x=7$ and $y=1$.
The question is: What could be other ways of proving this.
Edit: it will be interesting to note that this quadratic of $x$ and $y$ would represent just a point that is $(7,1)$. It is more clear by the solution of @Aqua given below.
|
$f(x,y)=x^2-2xy+6y^2-12x+2y+41$
$\frac{\partial f}{dx}=2x-2y-12$
$\frac{\partial f}{dy}=-2x+12y+2$
Stationary points are found when $\frac{\partial f}{dx}=0$ and $\frac{\partial f}{dy}=0$
$2x-2y-12=0$
$-2x+12y+2=0$
Adding the equations gives $10y-10=0\Rightarrow y=1$
Substitute to get $2x-2-12=0\Rightarrow 2x=14\Rightarrow x=7$
Stationary point is at $(7,1)$
Find $\frac{\partial^2 f}{dx^2}=2$ and $\frac{\partial^2 f}{dy^2}=12$ to determine that this is a minimum point (might otherwise have been a maximum or saddle point).
Work out the value $f(7,1)=49-14+6-84+2+41=0$
This is the minimum value of $f(x,y)$.
Therefore $f(x,y)\ge 0$
|
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|
System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z; \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Rightarrow 3y(x^2+1)=4x(y^2+1).$$ Now, from the second row, we have $$3y(x^2+xy+xz+yz)=4x(y^2+xy+xz+yz); \Rightarrow (x+y)(xy+4xz-3yz)=0.$$
Thus, $$x=-y; z=\frac {xy}{3y-4x}.$$
But if we replace $y$ with $-x$, we obtain a complex solution. The same happens when we replace $z$ with $\frac {xy}{3y-4x}$ in the second row because it takes us back to the first expression $x=-y$.
Finally, in case we try to deal with $$\frac{3(x^2+1)}x=\frac{5(z^2+1)}z, z=\frac {xy}{3y-4x},$$
then it complicates the problem with huge expressions.
Wolfram Alpha shows that the real solutions are $(x;y;z)=(\frac 13;\frac 12;1)\cup(-\frac 13;-\frac 12;-1)$, but I do not know how to get to them. I would appreciate any help.
Thank you in advance.
|
(Not a completely rigorous proof, but too long for a comment.)
Let $\,x=\tan(a)\,$ and similar, then $\,\frac{x^2+1}{x}=\frac{2}{\sin(2a)}\,$, so the equations can be written as:
$$
\frac{3}{\sin(2a)}=\frac{4}{\sin(2b)}=\frac{5}{\sin(2c)} \tag{*}
$$
Also, $\,xy+yz+zx=1 \implies a+b+c=(2k+1)\frac{\pi}{2}\,$, by proving a converse of if $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $, and therefore $\,2a+2b+2c=(2k+1)\pi\,$.
For positive $\,x,y,z\,$ the relation $\,(*)\,$ can be thought of as the law of sines in a right $3-4-5$ triangle with angles $\,2a,2b,2c = \frac{\pi}{2}\,$. It follows that the following are solutions:
*
*$\sin(2a)=\frac{3}{5} \implies \frac{x^2+1}{x} = 2 \cdot \frac{5}{3} \implies x = \frac{1}{3}$
*$\sin(2b)=\frac{4}{5} \implies \frac{y^2+1}{y} = 2 \cdot \frac{5}{4} \implies y = \frac{1}{2}$
*$\sin(2c)=1 \implies \frac{z^2+1}{z} = 2 \cdot 1 \implies z = 1$
This gives the positive solution $\,\left(\frac{1}{3}, \frac{1}{2}, 1\right)\,$, and
the given relations require $\,x,y,z\,$ to all have the same sign, so there is one symmetric solution in negative numbers $\,\left(-\frac{1}{3}, -\frac{1}{2}, -1\right)\,$.
|
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|
Proving the derivative of a function of another function at a point Is the following equality true?
$$ \left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0}
= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{x - x_0} $$
If it is true, how can I prove this?
It seems to work but I am unable to prove it. Take for example $ f(x) = x^2 $. Then $ f(ax + b) = (ax + b)^2 $.
\begin{align*} \text{LHS} &= \left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0} \\ \\
&= \left[ \frac{d}{dx} (ax + b)^2 \right]_{x = x_0} \\ \\
&= \left[ 2a(ax + b) \right]_{x = x_0} \\ \\
&= 2a(ax_0 + b). \\ \\
\text{RHS} &= \lim_{x \to x_0} \frac{(ax + b)^2 - (ax_0 + b)^2}{x - x_0} \\ \\
&= \lim_{x \to x_0} \frac{a^2x^2 + 2abx - a^2x_0^2 - abx_0}{x - x_0} \\ \\
&= \lim_{x \to x_0} \frac{a^2(x^2 - x_0^2) + 2ab(x - x_0)}{x - x_0} \\ \\
&= \lim_{x \to x_0} a^2(x + x_0) + 2ab \\ \\
&= 2a^2 x_0 + 2ab \\ \\
&= 2a(ax_0 + b).
\end{align*}
How do I prove this for any function $ f(x) $?
|
To do a formal proof, we can just use the chain rule.
\begin{align}
\left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0}
&= \left[f'(x) \frac d{dx}(ax+b) \right]_{x=ax_0+b} \\
&= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{a(x-x_0)} a \\
&= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{(x-x_0)}
\end{align}
The first line is the chain derivative, the second line is the definition of the derivative.
|
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|
Show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$ Let $\theta=2\pi/7$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$
I have found that $\cos(4\theta) = \cos(8\pi/7) = \cos(6\pi/7) = \cos(\theta)$ and,
similarly, $\cos(5\theta) = \cos(2\theta)$ and $\cos(6\theta) = \cos(\theta)$.Thus the expression
$$1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$$
can be written as
$$1+2\cos(\theta)+2\cos(2\theta)+2\cos(3\theta)= 0$$
Where do I get that $\cos(\theta)$ satisfies the polynomial $8x^3+4x^2-4x-1$.
But at the moment I don't see how it can help me with what I need. Is it the correct way? If not, could you give me a hint of where to go?
|
If you can use the property that $\cos(x) = Re(e^{i x})$, where $Re(z)$ is the real part of $z$, then:
\begin{align*}
\sum_{k=0}^6 \cos(k\theta)
&= Re\left(\sum_{k=0}^6 e^{ik\theta}\right) \\
&= Re\left(\frac{1-e^{i7\theta}}{1-e^{i\theta}}\right) \\
&= Re\left(\frac{1-e^{i2\pi}}{1-e^{i\frac{2\pi}{7}}}\right) \\
&= 0
\end{align*}
as $e^{i2\pi} = 1$.
|
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|
Coefficient extraction I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\alpha + k}}
\end{equation}
So far I got
\begin{equation*}
\frac{1}{(1 - z)^{\alpha + 1}} = (1-z)^{-(\alpha + 1)} = \sum_{n \geq 0}{\binom{-\alpha - 1}{k}(-1)^k z^k} = \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n}
\end{equation*}
where I used $\binom{-\alpha}{k} = (-1)^k \binom{\alpha + k - 1}{k}$and
\begin{equation*}
\log \frac{1}{1 - z} = - \log 1- z = \sum_{n \geq 1}\frac{z^n}{n} = z\sum_{n \geq 0}\frac{z^{n}}{n+1}
\end{equation*}
therefore
\begin{align*}
\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} &= z\left( \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n} \right) \left( \sum_{n \geq 0}\frac{z^{n}}{n+1} \right) \\
&= z \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n} \\
&= \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n + 1}.
\end{align*}
Now the $n$-th coefficient is
\begin{align*}
\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\frac{n!}{(n-k)! k!} \frac{1}{\binom{\alpha + n}{n - k}} \frac{1}{n - k + 1}} + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k} \frac{1}{\binom{\alpha + n}{n - k}} \frac{1}{n - k + 1}} + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k - 1} \frac{1}{\binom{\alpha + n}{n - k - 1}} \frac{k-1}{n - k -1} \frac{1}{\alpha + k + 1} } + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^{n+1}_{k = 2}{\binom{n}{k - 2} \frac{1}{\binom{\alpha + n}{n - k}} \frac{k- 2 }{n - k} \frac{1}{\alpha + k} } + \frac{1}{n+1}
\end{align*}
but from now on I can't see how to proceed.
I also know that
\begin{equation}
[z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n
\end{equation}
somehow I also tried to use some sort of transformation law for
coefficient extraction, but I am not aware of any kind of transformation law for coefficient extraction. Using $1-u = (1-z)^{\alpha + 1}$ or $z = 1 - (1-u)^{1/(\alpha + 1)}$gives
\begin{equation}
\frac{1}{\alpha + 1}\frac{1}{1-u} \log \frac{1}{1 - u}
\end{equation}
and for this I get the coefficient $\frac{1}{\alpha + 1} H_n$.
Another approach I did was the following:
Since
\begin{equation*}
(1-z)^{-m} = \exp(-m \ln(1-z))
\end{equation*}
I get
\begin{equation*}
\frac{\partial }{\partial m} \exp(-m \ln(1-z)) = -\ln(1-z) \exp(- m \ln(1-z)) = - \ln(1-z) \frac{1}{(1-z)^m} = \frac{1}{(1-z)^m} \ln \frac{1}{1-z}
\end{equation*}
therefore
\begin{align*}
\frac{\partial }{\partial m} (1-z)^{-m} &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \binom{m + n - 1}{n} z^n} \\
&= \sum_{n \geq 0}{\frac{\partial }{\partial m} \frac{\Gamma(m + n)}{n!\Gamma(m)} z^n}
\end{align*}
But I don't know any identities for the derivative of the gamma function.
|
Perhaps similar to epi163sqrt's answer in that this uses Vandermonde's Identity. However, this avoids the use of a derivative.
Using the power series
$$
\log\left(\frac1{1-z}\right)=\sum_{k=1}^\infty\frac{z^k}{k}\tag1
$$
and the Generalized Binomial Theorem with negative binomial coefficents
$$
\begin{align}
(1-z)^{-\alpha-1}
&=\sum_{k=0}^\infty\binom{-\alpha-1}{k}(-1)^kz^k\tag{2a}\\
&=\sum_{k=0}^\infty\binom{k+\alpha}{k}z^k\tag{2b}
\end{align}
$$
we get
$$
\begin{align}
&\left[z^n\right] (1-z)^{-\alpha-1}\log\left(\frac1{1-z}\right)\tag{3a}\\
&=\sum_{k=0}^{n-1}\frac1{n-k}\binom{k+\alpha}{k}\tag{3b}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{\binom{k+\alpha}{k}}{\binom{n+\alpha}{n}}\tag{3c}\\
%&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{\frac{(k+\alpha)\dots(1+\alpha)}{k!}}{\frac{(n+\alpha)\dots(1+\alpha)}{n!}}\tag{3d}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{n\dots(k+1)}{(n+\alpha)\dots(k+1+\alpha)}\tag{3d}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{\binom{n}{n-k}(n-k)!}{(n+\alpha)\dots(k+1+\alpha)}\tag{3e}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac{\binom{n}{n-k}(n-k-1)!}{(n+\alpha)\dots(k+1+\alpha)}\tag{3f}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\sum_{j=k+1}^n\frac1{j+\alpha}\frac{\binom{n}{n-k}(n-k-1)!}{(n-j)\dots1(-1)(k+1-j)}\tag{3g}\\
&=\binom{n+\alpha}{n}\sum_{j=1}^n\sum_{k=0}^{j-1}\frac1{j+\alpha}(-1)^{j-k-1}\binom{n}{n-k}\binom{n-k-1}{j-k-1}\tag{3h}\\
&=\binom{n+\alpha}{n}\sum_{j=1}^n\sum_{k=0}^{j-1}\frac1{j+\alpha}\binom{n}{k}\binom{j-n-1}{j-k-1}\tag{3i}\\
&=\binom{n+\alpha}{n}\sum_{j=1}^n\frac1{j+\alpha}\tag{3j}\\
&=\binom{n+\alpha}{n}\left(H_{n+\alpha}-H_\alpha\right)\tag{3k}
\end{align}
$$
Explanation:
$\text{(3b)}$: $(1)$, $(2)$, and the Cauchy Product Formula
$\text{(3c)}$: factor $\binom{n+\alpha}{n}$ out front
$\text{(3d)}$: expand and cancel the binomial coefficients
$\text{(3e)}$: rewrite the product in the numerator
$\text{(3f)}$: cancel the $n-k$ in the numerator and denominator
$\text{(3g)}$: apply partial fractions
$\text{(3h)}$: collect the factors into $(-1)^{j-k-1}\binom{n-k-1}{j-k-1}$
$\text{(3i)}$: apply negative binomial coefficients
$\text{(3j)}$: apply Vandermonde's Identity
$\text{(3k)}$: rewrite the sum as a difference of Extended Harmonic Numbers
|
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|
Taking a limit inside an integral I would like to see a rigorous proof that:
$$\lim_{R\rightarrow \infty}\int_1^R \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x^2/R^2}}=\int_1^\infty \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$
Note that the second integral does converge (to $\log 2$). This probably follows from some basic theorems in integration theory but I am rusty in that area...I cannot see how to do it with the dominated convergence or monotone convergence theorems. Some reference to the relevant theorems would be great. Thanks for any help.
|
I finally was able to prove it, so I am posting the answer to my question. It was a case when doing it from scratch (the definition of limit) was easier (for me) than trying to find the right limit theorem that would apply. It is still possible it would follow right away from some limit theorem I don't know...
In my proof I replaced $R$ with $1/k$ and did the limit as $k\rightarrow 0^+$.
Let $$I(k)=\int_1^{1/k}\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{1}{\sqrt{1-k^2x^2}}dx$$
We show that
$$\lim_{k\rightarrow 0^+}I(k)=\int_1^\infty\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$
Let $\epsilon \gt 0$ be given and find $R\gt 0$ such that
$$\frac{1}{\sqrt{1-1/R^2}}-1 \lt \epsilon .$$
Then for $k\lt 1/R$,
$$\left|I(k)-\int_1^\infty\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx\right|\leq \left|\int_1^{1/k}\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\left(\frac{1}{\sqrt{1-k^2x^2}}-1\right)dx\right|+\int_{1/k}^\infty \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx$$
and the last integral on the right goes to zero as $k\rightarrow 0^+$. Using the identity
$$\frac{1}{\sqrt{1-k^2x^2}}-1=\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}$$
we can write the first integral on the right as $J_1(k)+J_2(k)$, where
$$J_1(k) =\int_1^R\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}dx\\
J_2(k)=\int_R^{1/k}\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}dx.$$
Note that for $1\leq x\leq R$,
$$\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}\leq \frac{k^2R^2}{\sqrt{1-k^2R^2}},$$
and for $R\leq x \leq 1/k$,
$$\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}\leq \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}}\\
=\left(\frac{1}{\sqrt{1-1/x^2}}-1\right)\frac{k^2x}{\sqrt{1-k^2x^2}} \leq \left(\frac{1}{\sqrt{1-1/R^2}}-1\right)\frac{k^2x}{\sqrt{1-k^2x^2}}. $$
Then
$$J_1(k)\leq \frac{k^2R^2}{\sqrt{1-k^2R^2}}\int_1^R\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx \rightarrow 0 \mbox{ as } k\rightarrow 0^+$$
and
$$J_2(k) \leq \left(\frac{1}{\sqrt{1-1/R^2}}-1\right)\int_R^{1/k}\frac{k^2x dx}{\sqrt{1-k^2x^2}}\leq \epsilon \int_0^{1-k^2R^2}\frac{du}{2\sqrt{u}}\leq \epsilon \int_0^1\frac{du}{2\sqrt{u}}=\epsilon$$
Since $\epsilon$ is arbitrary, we conclude that
$$I(k)\rightarrow \int_1^\infty\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$ as $k\rightarrow 0^+$.
|
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|
Proving an inflection point of a certain function
Show that the inflection points of the curve $y = \sin(x)/x$ lie on the curve $y^2(x^4+4)=4$
So, I know that to find the inflection points I have to set the second derivative of $f$ = $0$
I got $f''(x) = -((x^2-2)\sin(x)+2x\cos(x))/x^3 = 0$
Thus, $f''(x) = 2x\cos(x) = (x^2-2)\sin(x)$
What got me confused is that no matter how many times I tried, I could not get $y^2(x^4+4) = 4$ from that equation.
|
You know $2x\cos x = (x^2-2)\sin x$ and $y=\sin(x)/x$. Taking the square of the first implies $4x^2\cos(x)^2 = (x^2-2)^2\sin(x)^2$ and thus
$$ 4x^2(1-\sin(x)^2) = (x^2-2)^2\sin(x)^2$$
and thus
$$ \sin(x)^2 = \frac{4x^2}{(x^2-2)^2+4x^2} = \frac{4x^2}{x^4+4}$$
So $y^2(x^4+4) = \frac{\sin(x)^2}{x^2} (x^4+4) = \frac{4x^2}{x^2(x^4+4)}(x^4+4)=4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Check if $\sum_{n=1}^{\infty}\left|\frac{\ln(n+1)}{\ln(n^2+1)}-\frac{1}{2}\right|$ converges This is the series:
$$\sum_{n=1}^{\infty}\left|\frac{\ln(n+1)}{\ln(n^2+1)}-\frac{1}{2}\right|$$
We can drop the absolute value, because;
$$\frac{\ln(n+1)}{\ln(n^2+1)} > \frac{1}{2}$$
I've tried using inequalities below to form comparison tests, but to no avail.
$$\frac{n}{n+1} < \ln(n+1) < n$$
$$\frac{1}{n^2} < \frac{1}{\ln(n^2+1)} < \frac{n^2+1}{n^2}$$
I also tried d'Alembert's ratio test, but it resulted in limit of 1, which is inconclusive. I have no clues on how to approach this.
|
One has
\begin{align*}
\frac{\ln(n+1)}{\ln(n^2+1)} -\frac{1}{2} &= \frac{\ln(n)+\ln\left(1+\frac{1}{n}\right)}{\ln(n^2)+\ln\left(1+\frac{1}{n^2}\right)} -\frac{1}{2} \\
&= \frac{\ln(n)+ \frac{1}{n}+ o\left( \frac{1}{n}\right)}{2\ln(n)+ o\left( \frac{1}{n}\right)} -\frac{1}{2} \\
&= \frac{ \frac{1}{n} + o\left( \frac{1}{n}\right)}{2\ln(n) + o\left( \frac{1}{n}\right)} \sim \frac{1}{2n\ln(n)}
\end{align*}
This is the general term of a divergent series, hence by comparison, the original series diverges.
|
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|
How many ways can $720$ be decomposed into a product of two positive integers?
How many ways can $720$ be decomposed into a product of two positive integers?
My solution: There are $5 \cdot 3 \cdot 2 = 30$ ways to choose the exponents a, b, c, such that $2^a \cdot 3^b \cdot 5^c= 720$. Soon there are $30$ dividers. As there are $30$ dividers, there are $\frac{30}{2} = 15$ different products in the form $xy=720$
Question: We want $ab$ such that $a,b \in \{2^5 \cdot 3^3 \cdot 5^2\}$, that is, we want combinations of these possible values of $a$ and $b$, such that $ab=720$. How could I solve this problem following this line of reasoning? A different solution from mine (Using counting principles)
|
Since
\begin{align*}
720 & = 2 \cdot 360\\
& = 2 \cdot 2 \cdot 180\\
& = 2 \cdot 2 \cdot 2 \cdot 90\\
& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 45\\
& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 15\\
& = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5\\
& = 2^4 \cdot 3^2 \cdot 5
\end{align*}
each factor must be of the form $2^a3^b5^c$.
Suppose the factors are $2^{a_1}3^{b_1}5^{c_1}$ and $2^{a_2}3^{b_2}5^{c_2}$. Observe that
\begin{align*}
a_1 + a_2 & = 4 \tag{1}\\
b_1 + b_2 & = 2 \tag{2}\\
c_1 + c_2 & = 1 \tag{3}
\end{align*}
are equations in the nonnegative integers.
The equation
$$x_1 + x_2 = n$$
has $n + 1$ solutions in the nonnegative integers, namely $$(n, 0), (n - 1, 1), (n - 2, 2), \ldots, (2, n - 2), (1, n - 1), (0, n)$$
Hence, equation 1 has five solutions, equation 2 has three solutions, and equation 2 has two solutions. Hence, there are $5 \cdot 3 \cdot 2 = 30$ ordered pairs of factors with product $720$. Since $720$ is not a perfect square, each ordered pair consists of two distinct numbers. Since we do not care about the order of the factors, we must divide this result by two. Thus, the number of ways to decompose $720$ into two positive integer factors is
$$\frac{5 \cdot 3 \cdot 2}{2} = 15$$
|
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|
The monotony of an implicit sequence Context of the exercice:
Let us define $(f_n)_n$ for any $n\geq 3$, $\left\{\begin{matrix}
f_n:]1,2]&\rightarrow & \mathbb{R} \\
x &\mapsto & f_n(x):=x^n-x-n
\end{matrix}\right.$
It's easy to verify that $f_n$ is a bijection between $]1,2]$ and $]-n,f_n(2)]$ (since $f_n$ has a strictly positive derivative and $f_n(2)>0$).
Hence there exist a unique $u_n $ in $]1,2]$ such that $f_n(u_n)=0$.
Conclusion:
The sequence $$\left\{\begin{matrix}
u_n &\in & ]1,2]\\
u_n^n &=& u_n+n
\end{matrix}\right.$$
is well defined.
Now the goal is to study the monotony of $(u_n)_n$.
I'm almost sure that $(u_n)_n$ is a decreasing sequence, I tried to prove it but I could not.
The idea is to show that $\color{red}{^{*}}$: $\color{blue}{f_{n+1}(u_n)>0\text{(why!?)}}$ and use the fact that $0=f_{n+1}(u_{n+1})$ and that $f_{n+1}^{-1}$ is an increasing function to get $$u_n=f_{n+1}^{-1}\Big(f_{n+1}(u_{n})\Big)\geq f_{n+1}^{-1}\Big(f_{n+1}(u_{n+1})\Big)=u_{n+1}.$$
$$\color{red}{*}\color{blue}{: \begin{matrix}
f_{n+1}(u_n) &=& u_n^{n+1}-u_n-(n+1) \\
&=&u_n u_n^n-u_n-(n+1) \\
&=&u_n(u_n+n)-u_n-(n+1)\\
&=&u_n^2+nu_n-u_n-n-1
\end{matrix}}$$
|
First of all, note that $u_n>1+\frac{1}{n}$ since
$$
{f_n(1+\frac{1}{n})=(1+\frac{1}{n})^n-1-n-\frac{1}{n}<e-1-n<e-4<0
\\\text{and $f_n(x)$ is increasing}
}
$$
To prove why $u_n^2+(n-1)u_n-(n+1)$ is always positive for $n\ge 3$ and $u_n>1+\frac{1}{n}$, note that the roots of $x^2+(n-1)x-(n+1)$ are
$$
{r_1=\frac{1-n-\sqrt {n^2+2n+5}}{2}<0<u_n\\
r_2=\frac{1-n+\sqrt {n^2+2n+5}}{2}
\\=\frac{2n+2}{\sqrt {n^2+2n+5}+n-1}
\\<\frac{2n+2}{n+1+n-1}=1+\frac{1}{n}<u_n
}
$$
hence, both of the roots of $x^2+(n-1)x-(n+1)$ are less than $u_n$ and since the coefficient of $x^2$ is positive, then $u_n^2+(n-1)u_n-(n+1)>0$ and the proof is complete $\blacksquare$
|
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|
Find the roots of the polynomial $x^4+2x^3-x-1=0$. Problem
Find all $4$ roots of the polynomial: $$f(x)=x^4+2x^3-x-1.$$
My Attempt
Observe,
$$f(-2)=1,\; f(-1)=-1,\; f(0)=-1,\; f(1)=1.$$
Therefore, $f(x)$ has a root between $-2$ and $-1$, another root between $0$ and $1$.
Since $f(x)$ does not contain a second degree term, a clever substitution might change $f(x)$ into a polynomial that is easier to deal with. For example $(x\to x-1)$,
$$f(x-1)=x^4-2x^3+x-1=f(-x)$$
Also, I am avoiding computational or graphical assistance.
|
The "basic" tools for dealing with a polynomial with integer coefficients tell us that the zeroes will not be simple to locate. The Rational Zeroes Theorem only suggests the candidates $ \ \pm 1 \ \ , $ which you have already shown are not correct. The Rule of Signs indicates one positive real and three or one negative real zeroes. We may suspect that two of the zeroes form a "complex conjugate pair", since the function is not positive for values $ \ -1 \ < \ x \ < \ 0 \ \ , $ as $ \ x^4 - 1 \ $ is more negative in this interval than $ \ 2x^3 - x \ = \ x·(2x^2 - 1) \ $ is ever positive, so there are no further $ \ x-$intercepts.
We could naïvely factor the polynomial as $ \ x^4 + 2x^3 - x - 1 \ = \ (x^2 + ax + b)·(x^2 + cx + d) \ \ , $ one of these factors being irreducible over real numbers. Following on your observation and lhf's comment, it must be that each of these is the same under the transformations $ \ x \ \rightarrow \ -x \ $ and $ \ x \ \rightarrow \ x - 1 \ \ . $ [I am avoiding the implication of symmetry about $ \ x \ = \ -\frac12 \ , \ $ since Macavity and B. Goddard have already treated it.] This will mean, for instance, that
$$ x^2 \ - \ ax \ + \ b \ \ = \ \ (x-1)^2 \ - \ a·(x-1) \ + \ b \ \ = \ \ x^2 \ + \ (a-2)·x \ + \ (b - a + 1) \ \ $$
$$ \Rightarrow \ \ -a \ \ = \ \ a \ - \ 2 \ \ \Rightarrow \ \ a \ \ = \ \ 1 \ \ , $$
with $ \ b \ $ remaining unspecified. The product of the factors is then
$$ (x^2 + x + b)·(x^2 + x + d) \ \ = \ \ x^4 \ + \ 2x^3 \ + \ (b + d + 1)·x^2 \ + \ (b + d)·x \ + \ bd \ \ . $$
This presents us with the coefficient equations $ \ b + d + 1 \ = \ 0 \ $ and $ \ b + d \ = \ -1 \ \ , $ from the quadratic and linear terms, and $ \ bd \ = \ -1 \ \ $ as the constant term. Putting these together, we find
$$ ( b \ + \ d )^2 \ = \ 1 \ \ , \ \ 4bd \ = \ -4 \ \ \Rightarrow \ \ (b \ - \ d)^2 \ = \ 5 $$
$$ \Rightarrow \ \ b \ + \ d \ = \ -1 \ \ \text{[as already determined]} \ \ , \ \ b \ - \ d \ = \ \pm \sqrt5 $$
$$ \Rightarrow \ \ b \ = \ \frac12 · (-1 \ \mp \ \sqrt5) \ \ , \ \ d \ = \ \frac12 · (-1 \ \pm \ \sqrt5) \ \ . $$
This makes it clear why a straightforward application of the Viete relations, or even an attempt to solve directly for the coefficients in $ \ x^4 + 2x^3 - x - 1 \ = \ (x^2 + ax + b)·(x^2 + cx + d) \ , \ $ is not very enlightening. The factors may be written in either order, so, for instance,
$$ \left( \ x^2 + x - \frac12 · [1 \ + \ \sqrt5] \ \right) \ · \ \left( \ x^2 + x - \frac12 · [1 \ - \ \sqrt5] \ \right) \ \ = \ \ 0 \ \ , $$
the zeroes of the first factor being the real values $ \ -\frac12 \ \pm \ \frac{\sqrt{3 \ + \ 2·\sqrt5}}{2} \ \ \approx \ \ 0.8668 \ , \ -1.8668 \ $ and those of the second factor are the "conjugate pair" $ \ -\frac12 \ \pm \ i·\frac{\sqrt{2·\sqrt5 \ - \ 3}}{2} \ \ \approx \ \ -0.5 \ \pm \ i·0.6067 \ \ . $
|
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|
When is the projection of an ellipsoid a circle? Consider an ellipsoid in the three dimensional Euclidean space, say
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$
where $a$, $b$, $c$ are positive reals.
I'm counting the number of planes through the origin so that the image is a perfect circle. There may be divergent cases if we consider the case that some of $a$, $b$, $c$ are coincide. But at first, let us focus on the case that $a$, $b$, $c$ are all different, say $a>b>c$.
I guess the answer would be $4$. I have made many efforts but failed.
What I have observed is the that at least two such planes exists and the radius of the circle is $b$. Just consider rotating plane possesses $y$ axis and apply intermediate value theorem.
Causion! We are concerning projection, not intersection.
PS. Now I guess there are infinitely many...
PS2. According to one suggested answer, there are just two such planes for the non-degenerate case. I'm checking if it is correct.
PS3. Another opinion appeared that the selected answer may have fault. And it seems making sense. I think somewhat stronger analysis is required.
PS4. The above PS3 is about another answer which now have disappeared.
|
Considering an enveloping cylinder, namely
$$
\left(
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1
\right)
\left(
\frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2}
\right)=
\left(
\frac{l x}{a^2}+\frac{m y}{b^2}+\frac{nz}{c^2}
\right)^2$$
The eigenvalues $\lambda$ are given by
$$0=\det
\begin{pmatrix}
\frac{1}{a^2} \left( \frac{m^2}{b^2}+\frac{n^2}{c^2} \right)-\lambda &
-\frac{l m}{a^2 b^2} & -\frac{l n}{a^2 c^2} \\
-\frac{m l}{b^2 a^2} &
\frac{1}{b^2} \left( \frac{n^2}{c^2}+\frac{l^2}{a^2} \right)-\lambda &
-\frac{m n}{b^2 c^2} \\
-\frac{n l}{c^2 a^2} & -\frac{n m}{c^2 b^2} &
\frac{1}{c^2} \left( \frac{l^2}{a^2}+\frac{m^2}{b^2} \right)-\lambda \\
\end{pmatrix}$$
where $\lambda=0$ is one of the roots.
The discriminant (with last factor in cyclic permutations of $a$, $b$ and $c$) is given by
$$\Delta
=\frac{(l^2+m^2+n^2)^2}{a^8 b^8 c^8}
\left( \frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2} \right)
[(\alpha l^2-\gamma n^2)^2-
2\beta m^2 (\alpha l^2+\gamma n^2)+
\beta^2 m^4
]$$
where $\alpha=b^2-c^2$, $\beta=c^2-a^2$, $\gamma=a^2-b^2$.
Note that for $a>b>c>0$,
$$\fbox{$\alpha,-\beta,\gamma>0$}$$
For circular cylinder, there are two equal eigenvalues implying $\Delta=0$, hence
$$\fbox{$\frac{l^2}{n^2}=\frac{\gamma}{\alpha} \land m=0$}$$
The enveloping circular cylinder is given by
$$
\fbox{$\frac{b^2}{a^2 c^2}
\left(
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1
\right)=
\left(
\sqrt{\frac{a^2-b^2}{a^2-c^2}} \, \frac{x}{a^2} \pm
\sqrt{\frac{b^2-c^2}{a^2-c^2}} \, \frac{z}{c^2}
\right)^2
$}$$
Further points to be noticed
*
*The touching conic is not circular and lying on the plane
$$\color{green}{z=\pm \frac{c^2}{a^2} \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$
*
*The uniform cross section of the cylinder lies on the plane
$$\color{blue}{z=\pm \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$
*
*The circular section for the ellipsoid lies on the plane
$$\color{red}{z=\pm \frac{c}{a} \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$
|
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|
Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
|
i don't think you made a mistake. continuing from where you stopped,
if $(y+1)(x+1)>0$ , then $y+1=\frac{c_1}{x+1}$.
otherwise if $(y+1)(x+1)<0$ , then $y+1=-\frac{c_1}{x+1}$.
|
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|
Matrix written in terms of Kronecker delta What would be the matrix form of $P$ defined in equation (1.11) of this published work, where (I have chosen $n=2$, for simplicity)
$$P_{kl}=\begin{cases} \delta_{k,2l-1} ~~~ k \le 2\\\\ \delta_{2+k,2l} ~~~ l\le 2 \end{cases}$$
|
From the context of the paper (namely the text below Equation (1.2) that defines $\mathbf R$ and the text below Equation (1.4) defining $\mathbf S$), it seems that the permutation matrix $\mathbf P$ being discussed for which $\mathbf P \mathbf R = \mathbf S$ is the commutation matrix $\mathbf P = \mathbf K^{(2,n)}$.
Some examples:
$$n = 2: \mathbf P = \left[\begin{array}{cc|cc}1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\
\hline0 & 1 & 0 & 0\\0 & 0 & 0 & 1\end{array}\right]\\
n=3: \mathbf P = \left[\begin{array}{cc|cc|cc}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
\hline
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1\end{array}\right]\\
n=4: \mathbf P = \left[\begin{array}{cc|cc|cc|cc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
\hline
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\end{array}\right].
$$
With that, I suspect that the intended description of $\mathbf P$ was as follows.
$$
P_{kl} = \begin{cases}
\delta_{k,2l-1} & k \leq n,\\
\delta_{k-n,2l-2} & k > n.
\end{cases}
$$
|
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|
Computing a closed formula for a recurrent sequence using eigen -values and -vectors How would you use eigenvalues and eigenvectors to compute a closed formula for the following sequence: $$\{x_0=1, x_1=2, x_n=5x_{n-1} + 14x_{n-2}, n \ge 0 \}$$
I have come up with the following formula: $$ \begin{bmatrix}
0 & 1 \\
14 & 5
\end{bmatrix}
*
\begin{bmatrix}
x_{n-2}\\
x_{n-1}\\
\end{bmatrix}
=
\begin{bmatrix}
x_{n}\\
x_{n+1}\\
\end{bmatrix}$$
So to compute x_n I want to take the n'th power of the matrix A: $A=\begin{bmatrix}
0 & 1 \\
14 & 5
\end{bmatrix}$ and multiply it with the "starting state" vector: $\begin{bmatrix}
1\\
2\\
\end{bmatrix}$.
Therefore I used the theory that $A^n=PD^nP^{-1}$, where $D$ is the diagonal matrix holding eigenvalues for $A$ along its diagonal and $P$ is the matrix holding the corresponding eigenvectors for the eigenvalues in $A$.
I computed the eigenvalues and vectors for $A$ and got that $\begin{bmatrix}
1\\
\frac{-1}{7}
\end{bmatrix}$ is an eigenvector with eigenvalue $\lambda_1 = 7$. And $\begin{bmatrix}
1\\
\frac{1}{2}
\end{bmatrix}$ is an eigenvector with eigenvalue $\lambda_2 = -2$.
So filling out the formula $A^n=PD^nP^{-1}$:
$$
\begin{bmatrix}
0 & 1 \\
14 & 5
\end{bmatrix}^n=\begin{bmatrix}
1 & 1 \\
\frac{-1}{7} & \frac{1}{2}
\end{bmatrix} * \begin{bmatrix}
7^n & 0 \\
0 & -2^n
\end{bmatrix} * \frac{9}{14} \begin{bmatrix}
\frac{1}{2} & -1 \\
\frac{1}{7} & 1
\end{bmatrix}
$$
Which when multiplied gives: $$
\begin{bmatrix}
0 & 1 \\
14 & 5
\end{bmatrix}^n=\frac{9}{14} * \begin{bmatrix}
7^n*\frac{1}{2}+(-2)^n
*\frac{1}{7} & -9^n \\
-\frac{1}{7}*7^n*\frac{1}{2}+\frac{1}{2}*\frac{1}{7}*-2^n & \frac{1}{7}*-7^n+\frac{1}{2}*-2^n
\end{bmatrix}
$$
The problem is now that when I multiply the result onto the "starting vector" $\begin{bmatrix}
1\\
2\\
\end{bmatrix}$ I get the this formula for $x_n$: $\frac{9}{14}((7^n+\frac{1}{2}+(-2)^n*\frac{1}{7})+(2*(-9)^n))$
Which doesn't give the correct answer when checking for $n=2$ for example: $\frac{9}{14}((7^2+\frac{1}{2}+(-2)^2*\frac{1}{7})+(2*(-9)^2)) = \frac{26721}{196}$.
For $n=2$ I would expect $x_2=5*2 + 14*1=24$.
Can anyone help me figure out where I am doing it wrong?
|
Do you really need to use eigenvalues/eigenvectors? This is a linear difference equation for which you can easily identify a basis for the solution space. Looking at the roots of the characteristic polynomial $p(\lambda) = \lambda^2 -5 \lambda -14$, any solution is of the form
$$
x_n = c_1 \cdot (-2)^n + c_2 \cdot 7^n
$$
Now you can just compute $c_1, c_2$ from the initial condition $x_0 = 1$ and $x_1=2$.
|
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|
Determinant formula for coordinates of circumcenter and orthocenter of a triangle I've come across the following formulas for coordinates of circumcenter $O=(x_O,y_O)$ and orthocenter $H=(x_H,y_H)$ of a triangle, in a formula book, stated without derivation. For a triangle with vertices $(x_i,y_i),$ $i \in \{1,2,3\}$,
$$x_O=\frac{\begin{vmatrix} x_1^2+y_1^2 & y_1 & 1 \\ x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1 \end{vmatrix}}{2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \quad , \quad y_O=\frac{\begin{vmatrix} x_1 & x_1^2+y_1^2 & 1 \\ x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1 \end{vmatrix}}{2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \tag{A}$$
$$x_H=\frac{\begin{vmatrix} y_1 & x_2x_3+y_1^2 & 1 \\ y_2 & x_3x_1+y_2^2 & 1 \\ y_3 & x_1x_2+y_3^2 & 1 \end{vmatrix}}{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \quad , \quad y_H=\frac{\begin{vmatrix} x_1^2+y_2y_3 & x_1 & 1 \\ x_2^2+y_3y_1 & x_2 & 1 \\ x_3^2+y_1y_2 & x_3 & 1 \end{vmatrix}}{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \tag{B}$$
The circumcenter case is pretty clear to me. Comparing two different equations of circumcircle
$$\begin{vmatrix} x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0 = \lambda(x^2+y^2 -2x_{O}x -2y_{O}y+c)$$
and equating the coefficients of $x,y$, one obtains $(A)$.
But I have not been able to make progress towards derivation of $(B)$. Can somebody provide a proof of it?
Please avoid an algebra-tedious proof. Thank you!
|
Consider instead triangle $A_1B_1C_1$ whose midtriangle is precisely triangle $ABC$.
A classical result says that the orthocenter $H$ of $ABC$ is the circumcenter of $A_1B_1C_1$.
It is easy to express the coordinates of points $A_1,B_1,C_1$ as functions of the coordinates of $A,B,C$ using relationships
$$\vec{AB}+\vec{AC}=\vec{AA_1} \implies A_1=B+C-A, \ \text{etc.} $$
Last step: one applies the first set of formulas to triangle with vertices $A_1,B_1,C_1$ .
Please note that my first explanation using barycentric coordinates was a much too indirect way.
|
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|
Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$
Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$ such that
$$\left\{\begin{array}{l} a_{1}+a_{2}+\ldots+a_{n}>0 \\a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}<0 \\a_{1}^{5}+a_{2}^{5}+\ldots+a_{n}^{5}>0 \end{array}\right.$$
I don't have a solution, but I suspect $n_{\min}=5$ is optimal. I think so mainly because I could not find anything by brute force for either $4$ or $3$. There are quite a few solutions for $n=5$, for example:
$$a_{1}=1, a_{2}=-0.88, a_{3}=-0.88, a_{4}=0.5, a_{5}=0.5$$
$$a_{1}=-8, a_{2}=-7, a_{3}=3, a_{4}=4, a_{5}=9$$
It is also not difficult to prove by contradiction that $n_{\min}>2$.
|
n = 3 seems to be fine. Let $c$ be a parameter to be determined later.
Put $a_1 =-1-6c^3$, $a_2=-1+6c^3$, and $a_3=6c^2$.
Clearly $a_1 + a_2 + a_3 = -1 +6c^2$ which is positive for $c$ large enough.
Now, $a_1^3 + a_2^3 +a_3^3 = (-1-6c^3)^3 + (-1 + 6c^3)^3 + (6c^2)^3 = -2 < 0 $
Also, $a_1^5 + a_2^5 + a_3^3 = (-1 - 6c^3)^5 + (-1 + 6c^3)^5 + (6c^2)^5$ has the leading coefficient $2*6^4*c^{12}$ that is it, it is positive for $c$ large enough.
|
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|
Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a multiple of 11?
Dont know where to progress after this. All help appreciated.
|
Let $x=a-11,y=b-11$ in $xy+11x+11y=0$ to get
$$ab=121$$
Since the change of variables is completely reversible in the integers, every divisor of $121$ corresponds one-to-one to a solution: $a=\pm1,\pm11,\pm121$, correspinding to
$$(x,y)=(0,0),(-22,-22),(-10,110),(110,-10),(-12,-132),(-132,-12)$$
|
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|
$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.
I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second function, but it did not work.
Need some help to compute it.
|
A complex analysis approach is to integrate the function $$f(z) = \frac{\left(\log(z+1) - \log(z-1)\right)^{\color{red}{2}}}{1-az}, \quad 0< a< 1, $$ where $0 \le \arg(z+1), \arg(z-1) < 2 \pi$, around a dog bone contour that goes around branch cut on $[-1, 1]$.
(The nicest picture I can find of the contour in this answer.)
Since $f(z) \sim \mathcal{O}\left(\frac{1}{z^{2}} \right)$ as $|z| \to \infty$, the residue of $f(z)$ at $\infty$ is zero.
Integrating clockwise around the contour, we get $$ \begin{align} \oint f(z) \, \mathrm dz &=\small \int_{-1}^{1} \frac{\left(\log(1+x)-\log(1-x) - \pi i \right)^{2}}{1-ax} \, \mathrm dx + \int_{1}^{-1}\frac{\left(\log(1+x)+ 2 \pi i - \log(1-x) - \pi i\right)^{2}}{1-ax} \mathrm dx\\ &= -4 \pi i \int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right) }{1-ax} \, \mathrm dx \\ &= 2 \pi i \operatorname{Res}\left[f(z), \frac{1}{a}\right] \\ & = -2 \pi i \, \frac{\log^{2} \left(\frac{\frac{1}{a}+1}{\frac{1}{a}-1} \right)}{a} \\&=-2 \pi i \, \frac{\log^{2} \left(\frac{1+a}{1-a} \right)}{a} . \end{align}$$
Therefore, $$\int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right) }{1-ax} \, \mathrm dx = \frac{\log^{2} \left(\frac{1+a}{1-a} \right)}{2a} =\frac{2 \operatorname{artanh}^{2}(a)}{a}. $$
|
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|
Find all primes $m$ and $l$ such that $2m-1, 2l-1, 2ml-1$ are perfect squares Find all primes $m$ and $l$ such that the integers $2m-1$, $2l-1$ and $2ml-1$ are perfect squares.
I have been trying to solve this problem from the 2022 Moroccan Maths Olympiad training program, but I wasn’t able to find anything.
By some few calculations, I found that (5,5) is the only couple between 1-100 satisfying the conditions, so I tried proving that it is the only solution using congruence but nothing worked. Help me please.
|
I don't know of any way to solve this using congruences. However, it can be done using the uniqueness, plus the format & number of representations, of sums of squares. First, the stated conditions mean there is a positive integer $a$ such that
$$\begin{equation}\begin{aligned}
2m - 1 & = (2a + 1)^2 \\
2m - 1 & = 4a^2 + 4a + 1 \\
m & = 2a^2 + 2a + 1 \\
m & = a^2 + (a^2 + 2a + 1) \\
m & = a^2 + (a + 1)^2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Similarly, there are positive integers $b$ and $c$ with
$$l = b^2 + (b + 1)^2 \tag{2}\label{eq2A}$$
$$lm = (a^2 + (a + 1)^2)(b^2 + (b + 1)^2) = c^2 + (c + 1)^2 \tag{3}\label{eq3A}$$
With \eqref{eq1A} and \eqref{eq2A}, note Fermat's Two Squares Theorem states a "prime number $p$ can be represented as a sum of two nonzero squares if and only if $p = 2$ or $p \equiv 1 \pmod{4}$; and that this representation is unique". Since neither $m$ or $l$ are $2$, this means $m \equiv l \equiv 1 \pmod{4}$.
Next, Wikipedia's "Sum of squares function" article's $k = 2$ Formulae section shows, if $l \neq m$, there are $4(1 + 1)(1 + 1) = 16$ ways to write $lm$ as a sum of $2$ squares (if $l = m$, then there are $4(2 + 1) = 12$ ways, with this reduction due to $a - b = 0$ not having distinct positive & negative values). Note the Brahmagupta–Fibonacci identity shows how to write the product of two sums of $2$ squares as a sum of two squares in two different ways. The order of the terms in each way can be switched to give a factor of $2$. Also, the values being squared in each term in each way can be made negative, giving another factor of $2 \times 2 = 4$. Thus, altogether this gives all of the $2 \times 2 \times 4 = 16$ ways to write $lm$ as a sum of $2$ squares. Note that switching the orders of the factors of $l$ or $m$, or using negatives for $a$, $a + 1$, $b$ and/or $b + 1$, will not add any additional ways. This is also indicated in Sum of two squares theorem which states
The product of any two representable numbers is another representable number. Its representation can be derived from representations of its two factors, using the Brahmagupta–Fibonacci identity.
Using the ordering in \eqref{eq3A} with the first Brahmagupta–Fibonacci identity option gives
$$\begin{equation}\begin{aligned}
(a^2 + (a + 1)^2)(b^2 + (b + 1)^2) & = ((ab) - (a + 1)(b + 1))^2 + (a(b + 1) + (a + 1)b)^2 \\
& = (a + b + 1)^2 + (a + b + 2ab)^2
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This can match the RHS of \eqref{eq3A} only if $c = a + b + 1$ and $a + b + 2ab = c + 1$, which means $2ab = 2 \; \to \; a = b = 1$. This gives $l = m = 5$, which is the solution you found.
Using the second identity option gives
$$\begin{equation}\begin{aligned}
(a^2 + (a + 1)^2)(b^2 + (b + 1)^2) & = ((ab) + (a + 1)(b + 1))^2 + (a(b + 1) - (a + 1)b)^2 \\
& = (a + b + 2ab + 1)^2 + (a - b)^2
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
There are no positive integer values of $a$ and $b$ for which the terms being squared can be consecutive integers. As stated earlier, this comprises all of the ways to write the sum of squares.
Nonetheless, to help confirm there are no different results, continuing by switching the $a^2$ and $(a + 1)^2$ terms around, then switching $b^2$ and $(b + 1)^2$, and finally switching both at the same time, gives these results:
$$\begin{equation}\begin{aligned}
& (b - a)^2 + (a + b + 2ab + 1)^2 \\
& (a + b + 2ab)^2 + (a + b + 1)^2 \\
& (a - b)^2 + (a + b + 2ab + 1)^2 \\
& (a + b + 2ab)^2 + (- a - b - 1)^2 \\
& (a + b + 1)^2 + (a + b + 2ab)^2 \\
& (a + b + 2ab + 1)^2 + (b - a)^2
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
As expected, these just repeat the possibilities expressed earlier in \eqref{eq4A} and \eqref{eq5A}. Thus, the only primes which work are $l = m = 5$.
|
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|
Compute the sum $ \sum_1^{\infty} \ln{ \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} }$ I need to prove that the sum $ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) $ converges and equals $ \ln(\frac{4}{3})$. I tried expanding the fraction into four terms, which cancel each other out, but I'm left with $\ln(4)$. Here's my attempt:
$
\sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right)
= \sum_{1}^{\infty} \ln{(n+1)} - \sum_{1}^{\infty} \ln{n} + \sum_{1}^{\infty} \ln{(3n+1)} -\sum_{1}^{\infty} \ln{(3n+4)} \\
= \sum_{1}^{\infty} \ln{(n+1)} - \sum_{2}^{\infty} \ln{n} + \ln(1) + \sum_2^{\infty} \ln{(3n+1)} + \ln(4) -\sum_{1}^{\infty} \ln{(3n+4)} = \ln(4)$
The last step is simply about playing correctly with the indices and the lower bounds of the series. These telescopic series almost completely cancel each other out, though I can't find the right answer.
Please tell me what I missed.
Thanks
|
The issue with your calculation is that the sums are not individually convergent. Therefore, you cannot write
$$\sum_{n=1}^\infty \log \frac{3n+1}{3n+4} = \sum_{n=1}^\infty \log (3n+1) - \sum_{n=1}^\infty \log(3n+4),$$ because each of those sums on the RHS are infinite.
Instead, you need to look at the limiting behavior for a corresponding finite partial sum. Define
$$f(m) = \sum_{n=1}^m \log \frac{3n+1}{3n+4}.$$ Then $f(m)$ is finite for any positive integer $m$, and now we have no problem computing the telescoping sum:
$$\begin{align}
f(m) &= \sum_{n=1}^m \log (3n+1) - \sum_{n=1}^m \log (3(n+1) + 1) \\
&= \sum_{n=1}^m \log (3n+1) - \sum_{n=2}^{m+1} \log (3n+1) \\
&= \log 4 - \log (3(m+1)+1) \\
&= \log \frac{4}{3m+4}.
\end{align}$$
Similarly,
$$g(m) = \sum_{n=1}^m \log \frac{n+1}{n} = \log (m+1) - \log 1 = \log (m+1).$$
Consequently,
$$f(m) + g(m) = \log \frac{4(m+1)}{3m+4} = \log \frac{4 + 4/m}{3 + 4/m},$$ for all positive integers $m$,
and now taking the limit as $m \to \infty$, we get
$$\sum_{n=1}^\infty \log \frac{(n+1)(3n+1)}{n(3n+4)} = \lim_{m \to \infty} f(m) + g(m) = \lim_{m \to \infty} \log \frac{4 + 4/m}{3 + 4/m} = \log \frac{4}{3}.$$
|
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|
Laurent series of $\frac{z^2 + iz - 1}{(z - 2)(z + 3)}$ I want to expand the function following into Laurent series:
$$\frac{z^2 + iz - 1}{(z - 2)(z + 3)}$$
for $2 < |z| < 3$
My solution
First thing to apply is partial function decomposition:
$$\frac{1}{(z - 2)(z +3)} = \frac{\frac 1 5}{z - 2} + \frac{-\frac 1 5}{z + 3}$$
From which we have that:
$$\frac{z^2 + iz - 1}{(z - 2)(z +3)} = \frac{\frac 1 5 (z^2 + iz - 1)}{z - 2} + \frac{-\frac 1 5 (z^2 + iz - 1)}{z + 3}$$
Now we expand those two series:
$$\frac{1}{z - 2} = \frac{1}{z}\frac{1}{1 - \frac 2 z} = \sum_{n = 0}^\infty \frac{2^n}{z^{n + 1}}$$
$$\frac{1}{z + 3} = \frac{1}{3(\frac z 3 + 1)} = \frac 1 3 \frac{1}{1 - ( - \frac z 3)} = \sum_{n = 0}^\infty (-1)^n \frac{z^n}{3^{n + 1}}$$
So final result is:
$$\frac{z^2 + iz - 1}{(z - 2)(z + 3)} = \sum_{n = 0}^\infty \frac{z^2 + iz - 1}{5}[\frac{2^n}{z^{n + 1}} - \frac{(-1)^nz^n}{3^{n + 1}}]$$
Does it make any sense to you?
|
An idea for you: using Taylor's polynomial, or whatever, write the numerator as power of $\;z-2\;$ (first...), so you get:
$$f(z)=z^2+iz-1=f(2)+f'(2)(z-2)+\frac{f''(2)(z-2)^2}{2!}$$
and now multiply termwise the above by the development around $\;z=2\;$ to get an actual (Laurent) power series:
$$\left(f(2)+f'(2)(z-2)+\frac{f''(2)(z-2)^2}{2!}\right)\sum_{n=0}^\infty\frac{2^n}{z^{n+1}}=\frac{f''(2)(z-2)}{2}+f'(2)+\ldots$$
(you can write the above more detailed), and then do the same with $\;z=3\;$ ...
|
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|
Implicit differentiation: why I cannot multiple both sides by an expression? I was solving a problem on implicit differentiation and I was wondering why my answer does not match the given answer. Apparently I multiplied both parts of an equation by an expression to get rid of some fractions before differentiating. Here is my illustration:Differentiating $$\frac{x}{\sqrt y}+\frac{y}{\sqrt x}=xy$$ we get $$\frac{\sqrt y-\frac{xy'}{2\sqrt y}}{y}+\frac{y'\sqrt x-\frac{y}{2\sqrt x}}{x}=y+xy'$$ or $$y'=\frac{y-\frac{1}{\sqrt y}+\frac{y}{2x\sqrt x}}{\frac{1}{\sqrt x}-x-\frac{x}{2y\sqrt y}}=\frac{2xy^2\sqrt{xy}-2xy\sqrt x+y^2\sqrt y}{2xy\sqrt y-2x^2y\sqrt {xy} -x^2\sqrt x}$$However if I multiply both parts by $\sqrt{ xy}$ then differentiate $x^{\frac{3}{2}}+y^{\frac{3}{2}}=x^{\frac{3}{2}}y^{\frac{3}{2}}$, the result will be: $$\frac{3}{2}x^{\frac{1}{2}}+\frac{3}{2}y^{\frac{1}{2}}y'=\frac{3}{2}x^{\frac{1}{2}}y^{\frac{3}{2}}+\frac{3}{2}x^{\frac{3}{2}}y^{\frac{1}{2}}y'$$ or
$$y'=\frac{y\sqrt{ xy}-\sqrt x}{\sqrt y-x\sqrt{ xy}}$$
Graphing both equations produces the same graph so the slope of a tangent line should be no different. Why do I get two different expressions for the slope?
|
Consider a simpler example.
$$
f(x,y) = y-x=0,\qquad\text{implies}\qquad g(x,y)=y^2-xy = 0,
$$
so from $f$ we find $y'(x)=1$; from $g$ we find $y'(x)=\frac{y}{2y-x}$, which gives the same answer if $y=x$ but is a different function of $x,y$ in general.
The equation $\frac{d}{dx}f(x,y(x))=0$ allows you to find $y'(x)$ not only on curve $f(x,y)=0$ but on the whole family of curves $f(x,y)=c$ (the same is true for $g$). When $c=0$ both of the families produce the same curve (*). However, in general, they produce different curves, so the slopes of those curves will be different as well.
(*) in the appropriate region
|
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|
Find general form of eigenvalues of a circulant matrix I have an $n \times n$ matrix in a general form:
$$ A = \begin{pmatrix}
\alpha^{n-1} & 1 & \alpha & \alpha^2 & \dots & \alpha^{n-2} \\
\alpha^{n-2} & \alpha^{n-1} & 1 & \alpha & \dots & \alpha^{n-3} \\
\alpha^{n-3} & \alpha^{n-2} & \alpha^{n-1} & 1 & \dots & \alpha^{n-4} \\
\alpha^{n-4} & \alpha^{n-3} & \alpha^{n-2} & \alpha^{n-1} & \dots & \alpha^{n-5} \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha & \alpha^2 & \alpha^3 & \dots & \alpha^{n-1}
\end{pmatrix},$$
where $\alpha \in (0,1)$ is some parameter. Is there a chance to have eigenvalues of this matrix in some general form (as an expression in $n$ and $\alpha$)? Or, at least, is that possible to say something about them?
I tried to numerically generate some of them but I don't really see any pattern.
|
Notably, $A$ is a circulant matrix. As such, its eigenvalues will be of the form
$$
\lambda_j = \alpha^{n-2}\omega^{(n-1)j} + \cdots + \alpha \omega^{2j} + \omega^j +\alpha^{n-1},
$$
where $\omega = e^{2 \pi i/n}$ and $j = 0,1,\dots,n-1$. We can simplify $\lambda_j$ to
$$
\lambda_j = \alpha^{-1}((\alpha \omega^j)^{n-1} + \cdots + (\alpha \omega^j)^2 + (\alpha \omega^j) + 1) + \alpha^{n-1} - 1
\\ = \alpha^{-1}\frac{1-(\alpha \omega^j)^n}{1-\alpha\omega^j} + \alpha^{n-1} - 1\\
= \frac{1-\alpha^n \omega^{jn}}{\alpha - \alpha^2 \omega^j} + \alpha^{n-1} - 1.
$$
For $j = 0$, this yields the real eigenvalue
$$
\lambda_0 = 1 + \alpha + \cdots + \alpha^{n-1} = \frac{1 - \alpha^n}{1 - \alpha}.
$$
|
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|
How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful trick to evaluate the integral.
Noting that
$$I(1):= \int_{0}^{\infty} \frac{d x}{x^{4}-x^{2}+1} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{2}}{x^{4}-x^{2}+1}
$$
Combining them yields
\begin{aligned}
I(1)&=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\\&= \frac{1}{2}\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&= \frac{1}{2}\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
&= \frac{1}{2}\tan ^{-1}\left(x-\frac{1}{x}\right)_{0}^{\infty} \\
&= \frac{\pi}{2}
\end{aligned}
Later, I started to investigate the integrands with higher powers.
Similarly,
$$
I(2):= \int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{2}} \stackrel{x \mapsto \frac{1}{x}}{=}\int_{0}^{\infty} \frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}} d x
$$
By division, we decomposed $x^6$ and obtain $$
\frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}}=\frac{x^{2}+1}{x^{4}-x^{2}+1}-\frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}
$$$$
I(2)=\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x-\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}dx
$$
We can now conclude that $$I(2)=I(1)=\frac{\pi}{2} $$
My Question:
How about the integral $$\displaystyle I_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}$$ for any integer $n\geq 3$?
|
Differentiation sometimes is easier than integration!
Inverse substitution rewrites the integral
$\displaystyle \int_{0}^{\infty} \frac{x^{r}}{x^{m}+a} d x \stackrel{x\mapsto\frac{1}{\sqrt[m]{a}}}{=}\frac{1}{a^{1-\frac{r+1}{m}}} \int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x \tag*{} $
where $0<a<2.$
Using the formula proved in my post
$ \displaystyle \int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m } \csc \frac{(r+1) \pi}{m},\tag*{} $
we get
$ \displaystyle \int_{0}^{\infty} \frac{x^r}{x^{m}+a} d x=a^{-\left(1-\frac{r+1}{m}\right)} \frac{\pi}{m} \csc \frac{(r+1) \pi}{m}\tag*{} $
Differentiating the integral w.r.t. $ a$ by$n-1 $ times yields
$ \displaystyle \int_{0}^{\infty} \frac{(-1)^{r-1}(n-1) ! x^{r}}{\left(x^{m}+a\right)^{n}}=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m}(-1)^{n-1}\left(1-\frac{r+1}{m}\right)\left(2-\frac{r+1}{m}\right)\cdots\left(n-1-\frac{r+1}{m}\right)a^{-\left(n-\frac{r+1}{m}\right)} \tag*{}$
Now we can conclude that
$$\boxed{ \displaystyle \int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+a\right)^{n}}=\frac{\pi}{m(n-1) !} \csc \frac{(r+1) \pi}{m} \prod_{j=1}^{n-1}\left(j-\frac{r+1}{m}\right) a^{-\left(n-\frac{r+1}{m}\right)}}$$
Come back to our integral and use Binomial Theorem.
\begin{aligned}\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}} d x =& \int_{0}^{\infty}\left(\frac{x^{2}+1}{x^{6}+1}\right)^{n} d x =\sum_{k=0}^{n}\left(\begin{array}{l}n \\k\end{array}\right) \int_{0}^{\infty} \frac{x^{2 k}}{\left(x^{6}+1\right)^{n}} d x\end{aligned}
Putting $ m=6, r=2k $ and $ a=1 $ yields
$\displaystyle \boxed{\begin{aligned}\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}} d x &=\sum_{k=0}^{n}\left(\begin{array}{c}n \\k\end{array}\right) \frac{\pi}{6(n-1) !} \csc \frac{(2k+1) \pi}{6} \prod_{j=1}^{n-1}\left(j-\frac{2k+1}{6}\right)\\&=\sum_{k=0}^{n}\left[\left(\begin{array}{l}n \\k\end{array}\right) \frac{\pi}{6} \csc \frac{(2 k+1) \pi}{6} \prod_{j=1}^{n-1}\left(1-\frac{2 k+1}{6 j}\right)\right] \end{aligned} }\tag*{} $
|
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|
Rhombus, find ratio of angles $KLMN$ is a rhombus with $\measuredangle MNK=120^\circ$. On the side $MN$ point $C$ is taken such that $CN=2CM$. Find the ratio $\dfrac{\cos\measuredangle CKN}{\cos\measuredangle CLM}$.
The Cosine Rule in triangle $CNK$ gives $$CK^2=KN^2+CN^2-2KN\cdot CN\cos120^\circ=\\=a^2+\dfrac49a^2+\dfrac23a^2=\dfrac{19}{9}a^2,CK=\dfrac{\sqrt{19}}{3}a$$ and in triangle $CML$ $$CL^2=CM^2+LM^2-2CM\cdot LM\cos60^\circ=\\=\dfrac19a^2+a^2-\dfrac13a^2=\dfrac79a^2,CL=\dfrac{\sqrt7}{3}a$$ Now $$\cos\measuredangle CKN=\dfrac{KN^2+CK^2-CN^2}{2KN\cdot CK}=\dfrac{a^2+\frac{19}{9}a^2-\frac{4}{9}a^2}{2a\frac{\sqrt{19}}{3}a}=\dfrac{4}{\sqrt{19}}$$ and $$\cos\measuredangle CLM=\dfrac{LM^2+CL^2-CM^2}{2LM\cdot CL}=\dfrac{a^2+\frac79a^2-\frac19a^2}{2a\cdot\frac{\sqrt7}{3}a}=\dfrac{5}{2\sqrt7}$$ Now the ratio that we are supposed to find is $$\dfrac{\cos\measuredangle CKN}{\cos\measuredangle CLM}=\dfrac{4\cdot5}{\sqrt{19}\cdot2\sqrt7}=\dfrac{10}{\sqrt{133}}=\dfrac{10\sqrt{133}}{133}$$
The given answer in my book for the ratio is $$\dfrac85\sqrt{\dfrac{7}{19}}$$ Am I wrong?
|
you have multiplied rather than divided the two ratios at the last step. If you divide them you get the right answer
|
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|
Why $(\sec^2(x)+1)^{\frac{3}{2}}=\sec^3(x), x>0$? I was trying to find the intermediaries steps to this equality, considering $x$ is a positive number, but I couldn't. Using the identities $\sec^2(x)=1+\tan^2(x)$ and $\sin^2(x)+\cos^2(x)=1$, here some of my work:
First try:
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\left(\dfrac{1}{\cos^2(x)}+1\right)^{\frac{3}{2}}=\left(\dfrac{\cos^2(x)+1}{\cos^2(x)}\right)^{\frac{3}{2}}=\left(\dfrac{(1-\sin^2(x))+1}{\cos^2(x)}\right)^{\frac{3}{2}}$$
Second try:
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\left((\tan^2(x)+1)+1\right)^{\frac{3}{2}}=\left(\tan^2(x)+2\right)^{\frac{3}{2}}$$
Third try (using the first try):
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\left((\cos^2(x)+1)\sec^2(x)\right)^{\frac{3}{2}}=\sec^3(x)\sqrt{(\cos^2(x)+1)^3}$$
In the last and third try, I need that $\sqrt{(\cos^2(x)+1)^3}=1$, then I think the problem is solved...
|
It is not true in general. Indeed, one has:
\begin{align*}
\left(\sec^{2}\left(\dfrac{\pi}{3}\right)+ 1\right)^{3/2}= \dfrac{5\sqrt{5}}{8} \neq \frac{1}{8} = \sec^{3}\left(\dfrac{\pi}{3}\right)
\end{align*}
EDIT
As @JetfiRex has mentioned, you may have switched $\tan^{2}(x)$ by $\sec^{2}(x)$.
If this is the case, the corresponding identity holds:
\begin{align*}
\tan^{2}(x) + 1 & = \frac{\sin^{2}(x)}{\cos^{2}(x)} + 1 = \frac{\sin^{2}(x) + \cos^{2}(x)}{\cos^{2}(x)} = \sec^{2}(x)
\end{align*}
And consequently we also get
\begin{align*}
(\tan^{2}(x) + 1)^{3/2} = (\sec^{2}(x))^{3/2} = |\sec^{3}(x)|
\end{align*}
Now it remains to determine for each values of $x$ one has that $\sec(x) \geq 0$.
Can you take it from here?
|
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|
Given $a,b,c>0$, $a+b+c=ab+bc+ca$, prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$ Given $a,b,c>0$, $$a+b+c=ab+bc+ca$$, prove $$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$$
I tried with derivatives but haven't solved it yet.
Is there a more natural and elementary proof?
My progress with derivatives:
First $a+b+c=ab+bc+ca\le \frac{(a+b+c)^2}{3}$, so $a+b+c\ge 3$.
Then $a+b+c=ab+bc+ca \Leftrightarrow \frac1a+\frac1b+\frac1c=\frac1{ab}+\frac1{bc}+\frac1{ca}$, so similarly $\frac1a+\frac1b+\frac1c\ge 3$. Hence $ab+bc+ca\ge 3abc$.
WOLOG, we assume $a>b>c$. Obviously $a>1>c$.
We have $b+c > 1$, otherwise $ab+bc+ca=(b+c)a+bc<a+bc<a+b+c$: contradiction. Now we have $(\sqrt b +\sqrt c)^2 > b+c >1$, so $\sqrt b+ \sqrt c>1$.
With notation $s:=\sqrt a + \sqrt b + \sqrt c$ and $t:=\sqrt{abc}$, we have $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2 \Leftrightarrow (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2 \ge (2+\sqrt t)^2 $, which is equivalent to $$ab+bc+ca + 2s \sqrt{t} \ge 4 + 4\sqrt t + t$$, call it (*), now we discuss by case a) $abc \ge 1$ and case b) $abc<1$.
Case a) $abc \ge 1$. Inequation (*) holds if $$3t+6\sqrt t \sqrt[6] t\ge 4 + 4\sqrt t + t$$ holds, which again is equivalent to $$-2t+4\sqrt t -6t^{\frac23}+4\le 0$$ , but this can be proved by computing the derivative.
Case b) $abc<1$. I haven't solved this yet.
|
Remark: @Calvin Lin's comment reminds me that, in 2019, I proved some similar inequalities under the condition $a + b + c = ab + bc + ca$.
WLOG, assume that $c = \min(a, b, c)$.
Using AM-GM, we have
\begin{align*}
&\sqrt{ab} + \sqrt{bc} + \sqrt{ca} - \sqrt{abc} - 2\\
\ge\,& \sqrt{ab}
+ 2\sqrt{\sqrt{bc}\sqrt{ca}} - \sqrt{abc} - 2\\
=\,& (1 - \sqrt{c})\sqrt{ab} + 2\sqrt{c}\sqrt[4]{ab} - 2. \tag{1}
\end{align*}
From $ab + bc + ca = a + b + c$, we have
$(1 - c)(a + b) = ab - c$. Thus, $c \le 1$ (easy).
Thus, we have $(1 - c)\cdot 2\sqrt{ab} \le ab - c$
or $(\sqrt{ab} - 1 + c)^2 \ge c^2 - c + 1$ which results
in
$$\sqrt{ab} \ge \sqrt{c^2 - c + 1} + 1 - c. \tag{2}$$
Note: $\sqrt{ab} - 1 + c \le -\sqrt{c^2 - c + 1}$ is impossible, since $1 - c - \sqrt{c^2 - c + 1} < 0$ for all $c > 0$.
We split into two cases:
*
*$c = 1$:
From (2), we have $\sqrt{ab} \ge 1$.
From (1), the desired result follows.
*$0< c < 1$:
From (1), it suffices to prove that
$$(1 - \sqrt{c})\sqrt{ab} + 2\sqrt{c}\sqrt[4]{ab} - 2 \ge 0$$
which is written as
$$(1 - \sqrt c)\left(\sqrt[4]{ab} + \frac{\sqrt c}{1 - \sqrt c}\right)^2 \ge \frac{2 + c - 2\sqrt c}{1 - \sqrt c}.$$
It suffices to prove that
$$\sqrt[4]{ab} + \frac{\sqrt c}{1 - \sqrt c} \ge \frac{\sqrt{2 + c - 2\sqrt c}}{1 - \sqrt c}$$
or
$$\sqrt[4]{ab}
\ge \frac{\sqrt{2 + c - 2\sqrt c} - \sqrt c}{1 - \sqrt c}
= \frac{2}{\sqrt{2 + c - 2\sqrt c} + \sqrt c}$$
or
$$\sqrt{ab} \ge \frac{4}{(\sqrt{2 + c - 2\sqrt c} + \sqrt c)^2} = \frac{4}{2 + 2c - 2\sqrt c + 2\sqrt c \sqrt{1 + (1 - \sqrt c)^2}}.$$
It suffices to prove that
$$\sqrt{ab} \ge \frac{4}{2 + 2c - 2\sqrt c + 2\sqrt c } = \frac{2}{1 + c}.$$
Using (2), it suffices to prove that
$$\sqrt{c^2 - c + 1} + 1 - c \ge \frac{2}{1 + c}$$
or
$$\sqrt{c^2 - c + 1} \ge \frac{1 + c^2}{1 + c}$$
or
$$c^2 - c + 1 \ge \left(\frac{1 + c^2}{1 + c}\right)^2$$
or
$$\frac{c(1 - c)^2}{(1 + c)^2}\ge 0.$$
We are done.
|
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|
$x^2\frac{dy}{dx}+xy+1=0$ Solve the following differential equation by the form of homogeneous equation. Letting $y=vx$
The equation:
$x^2\frac{dy}{dx}+xy+1=0$
I can’t separate variables
My solution steps are:
$x^2(v+x \frac{dv}{dx})+x^2v+1=0$
$(v+x \frac{dv}{dx})+v=\frac{-1}{x^2}$
$2v+x \frac{dv}{dx}=\frac{-1}{x^2}$
How to separate variables x and v?
|
Divide the proposed DE by $x$ in order to obtain
\begin{align*}
x^{2}y' + xy + 1 = 0 & \Longleftrightarrow xy' + y = -\frac{1}{x}\\\\
& \Longleftrightarrow (xy)' = -\frac{1}{x}\\\\
& \Longleftrightarrow xy = -\ln|x| + k\\\\
& \Longleftrightarrow y(x) = -\frac{\ln|x|}{x} + \frac{k}{x}
\end{align*}
and we are done.
Hopefully this helps !
|
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|
Given $a,b,c$ are real numbers such that $a-b=8$ and $ab + c^2 +16 =0$. The numerical value of $a^{2021} + b^{2021} +c^{2021}$ is? I saw this question and was intrigued by it. I wanted to see if what I have so far made sense.
So from the second equation $ab + c^2 +16 =0$ we get $c^2 = -(16+ab)$. Which implies $16+ab\leq0$ or $ab\leq-16$. Then using the first equation $a-b=8$ and solving for $a$ we get $a=8+b$. Then substituting in $a$ in the inequality we get:
$\begin{align}
ab&\leq-16\\
(8+b)b&\leq-16\\
8b+b^2&\leq-16\\
b^2+8b+16&\leq0\\
(b+4)^2&\leq 0\\
\end{align}$
So we must have $b=-4$ for this inequality to be true. Then $a=4$, and $c^2 = 0 \implies c = 0$. Hence
$\begin{align}
a^{2021} + b^{2021} +c^{2021} = 4^{2021} + (-4)^{2021}= 4^{2021} -4^{2021} = 0.
\end{align}$
If anyone sees anything wrong let me know or if there is another solution I'm curious!
|
Yes, you are correct. The numerical value is $0$, with $a=4,b=-4$ and $c=0$.
How I did it [before opening this post]: First note that for there to be a real solution to the equation $ab+16+c^2=0$, the product $ab$ must satisfy $ab \le -16$. The only $(a,b)$ that satisfies $ab \le -16$ with $a-b=8$ is $a=4,b=-4$, and so $c=0$.
So same idea as yours.
If however the "$16$" were replaced with any number larger, there would be no such $a,b,c$ that simultaneously satisfy both of the resulting constraints, and if the "$16"$ were replaced with anything smaller, there would not only be an infinite number of $a,b,c$ that simultaneously satisfy both of the resulting constraints, but there would also be an infinite number of values that $a^{2021}+b^{2021}+c^{2021}$ could take for such $a,b,c$.
|
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|
To prove: $\bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2 $ How could one show that
$$ \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2
$$
I have tried the below approach:
\begin{align*}
LHS & = \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr)\\
& = 1-\sin(A-B)+\sin(A+B)-\sin(A+B)\sin(A-B)\\
& = 1 - \sin{A}\cos{B} + \sin{B}\cos{A} + \sin{A}\cos{B} + \sin{B}\cos{A}\\
& \qquad - (\sin{A})^2 + (\sin{B})^2\\
& = 1 + 2\sin{B}\cos{A} - 1 + (\cos{A})^2 + (\sin{B})^2\\
& = 2\sin{B}\cos{A} + (\cos{A})^2 + (\sin{B})^2\\
& = (\cos{A} + \sin{B})^2
\end{align*}
|
$$(1+\sin A\cos B+\cos A\sin B)(1-\sin A\cos B+\cos A\sin B)$$
$$=(1+\cos A\sin B)^2-(\sin A\cos B)^2$$
$$=1+2\cos A\sin B+\cos^2A\sin^2B-(1-\cos^2A)(1-\sin^2B)$$
$$=\cdots$$
$$=(1+\cos A\sin B)^2$$
|
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"url": "https://math.stackexchange.com/questions/4383653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.
First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $2$ that isn't the multiple of $4$. Since the problem is asking for a minimal case that satisfies these conditions, $n+1$ should be the multiple of $4$, $n+2$ should be the multiple of $3$, and $n+3$ should be the multiple of $2$, thus giving $4 \cdot 3^2 \cdot 2^3 = 288.$
However, I am unsure as to whether or not this answer is correct, as testing cases has consistently yielded higher GCDs than this.
|
First write $f(n)=(n+1)(n+1)^2(n+2)^3(n+4)^4$ The gcd of the $f(n)$s; $n \in \mathbb{N}$, is of the form $2^b3^a$, for some positive integers $a$ and $b$. Indeed, no prime $p \ge 7$ divides $f(1)$, and $5$ does not divide $f(5)$. Thus indeed, the common prime divisors are restricted to $2$ and $3$, and thus the gcd of the $f(n)$s is indeed of the form $2^b3^a$.
We now find $b$. Observe that the largest power of $2$ that divides $f(3)$ is $5$. Meanwhile, the largest power of $3$ that divides $f(1)$ is $2$. So $a \le 2$ and $b \le 5$. As you have shown $a \ge 2$ and $b \ge 5$, it follows that the equations $a=2$ and $b=5$ hold. This leaves us with the gcd of the $f(n)$s; $n \in \mathbb{N}$, is indeed $2^53^2$.
|
{
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"url": "https://math.stackexchange.com/questions/4385677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show this recursively defined integer sequence $a_{n+1} = (2b_1 - a_1)a_n + (a_1b_1 - b_1^2 - 2a_1^2)a_{n-1}$ is never $0$. Let $a_1, b_1 \in \mathbb{Z}$ be arbitary integers with $a_1 \neq 0$, and define sequences $\{a_n \}_{n \geq 1}$ and $\{b_n \}_{n \geq 1}$ to be solutions to the following linear system of recursions:
\begin{cases}
a_{n+1} = (b_1 - a_1)a_n + (a_1)b_n \\
b_{n+1} = (-2a_1)a_n + (b_1)b_n
\end{cases}
I have done the following manipulations which give a recursive formula for $a_n$.
First, observe the first equation gives us $b_{n} = \dfrac{a_{n+1} - (b_1 - a_1)a_n}{a_1}$.
Substituting this into the second equation, we see $$b_{n+1} = (-2a_1)a_n + b_1 \Big(\dfrac{a_{n+1} - (b_1 - a_1)a_n}{a_1}\Big).$$
Reindexing $n + 1 \rightarrow n$, we achieve $$b_n = (-2a_1)a_{n-1} + b_1 \Big(\dfrac{a_{n} - (b_1 - a_1)a_{n-1}}{a_1}\Big).$$
Finally, substituting this back into the first equation, we get:
$$ a_{n+1} = (b_1 - a_1)a_n + a_1 \Big[(-2a_1)a_{n-1} + b_1 \Big(\dfrac{a_{n} - (b_1 - a_1)a_{n-1}}{a_1}\Big) \Big].$$
Simplifying, we find $$a_{n+1} = (2b_1 - a_1)a_n + (a_1b_1 - b_1^2 - 2a_1^2)a_{n-1}.$$
I suspect, and would like to prove, that $a_n \neq 0$ for all $n \geq 1$ and for all choices $a_1, b_1 \in \mathbb{Z}$ where $a_1 \neq 0$.
As corrected in the comments, $\lim\limits_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$ does not appear to be $-3$, so my claim of well-behavedness asymptotically isn't even true. Any suggestions/hints for forward movement are appreciated.
|
Set $a_0=0$ and $b_0=1$. Then the original recurrence works for $n\ge0$. Write it in matrix form:
$$
\begin{pmatrix}
a_{n+1} \\ b_{n+1}
\end{pmatrix}
=
\begin{pmatrix}
b_1 - a_1 & a_1 \\ -2a_1 & b_1
\end{pmatrix}
\begin{pmatrix}
a_{n} \\ b_{n}
\end{pmatrix}
=
A
\begin{pmatrix}
a_{n} \\ b_{n}
\end{pmatrix}
$$
Then
$$
v_{n}=
\begin{pmatrix}
a_{n} \\ b_{n}
\end{pmatrix}
=
A^n
\begin{pmatrix}
a_{0} \\ b_{0}
\end{pmatrix}
=
A^n
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
= A^n v_0
$$
Therefore
$$
A^n
=
\begin{pmatrix}
* & a_n \\ * & b_n
\end{pmatrix}
$$
If $a_n=0$, then $v_0$ is an eigenvector of $A^n$ with eigenvalue $b_n$. The eigenvectors of $A^n$ are the same as the eigenvectors of $A$, and these have nonzero first coordinates, unlike $v_0$.
The calculations were made with WA.
|
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"timestamp": "2023-03-29T00:00:00",
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|
show $k<2^m$ if $2^{2^m}+1=\binom{n}{k}$ let $m$ be give positive integer,and $n,k$ be positive integer ,such $n\ge 2k\ge 2$,and
$$2^{2^m}+1=\binom{n}{k}$$
show that: $k<2^m$
Some of my ideas and attempts:
$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}=\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}=\left(1+\dfrac{n-k}{k}\right)\left(1+\dfrac{n-k}{k-1}\right)\cdots\left(1+\dfrac{n-k}{1}\right)\ge (1+\dfrac{n-k}{\sqrt[k]{k!}})^k\ge\left(1+\dfrac{k}{\sqrt[k]{k!}}\right)^k\ge 1+\dfrac{k^2}{\sqrt[k]{k!}}$$
at last two step use Holder inequality and Bernoulli’s inequality
so we have
$$2^{2^m}\ge \dfrac{k^2}{\sqrt[k]{k!}}$$Now, I can’t prove it $k<2^m$
|
If $k = 1$, clearly, $k < 2^m$.
In the following, assume that $k \ge 2$.
We have, for all $q > k \ge 0$,
$$\binom{q + 1}{k} =
\frac{(q + 1)!}{k! (q + 1 -k)!} = \frac{q!}{k! (q - k)!} \frac{q + 1}{q + 1 - k}
\ge \binom{q}{k}.$$
Thus, we have
$$\binom{n}{k} \ge \binom{2k}{k}.$$
Using Mathematical Induction, it is easy to prove that, for all $k \ge 2$,
$$\binom{2k}{k} > 2^k + 1.$$
(Note: Use $\binom{2k+2}{k+1} = \binom{2k}{k}\frac{(2k+1)(2k+2)}{(k+1)^2} > 2\binom{2k}{k}$.)
Thus, we have
$$\binom{n}{k} > 2^k + 1.$$
Thus, we have
$$2^{2^m} + 1 > 2^k + 1$$
which results in
$k < 2^m$.
We are done.
|
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"url": "https://math.stackexchange.com/questions/4395940",
"timestamp": "2023-03-29T00:00:00",
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|
Prove that $x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$
Let $x,y,z\ge0$ satisfy $\max\left \{ x,y,z \right \}\ge 1$. Prove that $$x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$$
My attempts:
From the condition we can deduce $x+y+z\ge 1$
The inequality can be written as $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+(x+y+z)^2-2(x+y+z)\ge0$$ or $$x^2+y^2+z^2-xy-yz-zx+x+y+z-2\ge0$$
Let $x+y+z=p\ge1;xy+yz+zx=q$, the problem is: $$p^2-3q+p-2\ge0$$
I don't know how to find the relation between $p$ and $q$, because this inequality is not symmetric (The inequality hold iff $(x;y;z)=(1;0;0);(0;1;0);(0;0;1)$)
Please give me a hint in the comments, no need to give a full answer
|
Continuing from OP's work. WLOG let $x \geq 1$, we have
$$q = xy+yz+zx = x(y+z) + yz \leq x(p-x) + (\frac{p-x}{2} ) ^2,$$
with equality iff $ y = z = \frac{ p-x}{2}$.
We WTS in the domain $ p \geq x \geq 1$,
$$p^2 + p - 2 - \frac{1}{4} ( p-x)(p+3x) \geq 0.$$
Conditioning on each value of $p$, we have a much simpler quadratic in $x$:
$$ \frac{3}{4} ( x -p)(x+ \frac{p}{3}) + p^2 + p - 2 \geq 0.$$
By considering the quadratic on the restricted domain $ 1 \leq x \leq p$, the minimum occurs at $ \frac{ \frac{p}{3} + p} {2} = \frac{p}{3} $ if that's within the domain, else check the endpoints, giving us :
*
*Case 1 if $p \geq 3$: Minimum of $\frac{2p^2}{3} + p - 2 $ occurring at $x = \frac{p}{3}. $
*Case 2 if $3 \geq p \geq 1$: Minimum of $ \frac{1}{4} (p-1)(3p+5)$ occurring at $ x = 1$.
Thus, for all $ p \geq 1$, the minimum at least 0.
Equality holds iff $ p = x = 1$, which gives us $ (x,y,z) = (1, 0, 0)$ (and cyclic forms).
|
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"timestamp": "2023-03-29T00:00:00",
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|
What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m$, determine the length of $AC$.(Answer$:\sqrt[3]{2})$
My progress:
$ AF = h\\
a=m+1\\
h^2 =m.FC = m.1 = m\\
\triangle AFC: AC^2 =b= h^2+FC^2 \implies b = \sqrt{m+1}\\
\triangle ABC: a^2 = b^2+ c^2 \implies (m+1)^2=c^2+(\sqrt{m+1})^2\\
\therefore c = \sqrt{m^2+m}\\
\triangle BAD: BD^2=AD^2+c^2\implies 1 = (\sqrt{m+1}-1)^2+m^2+m\\
\therefore: m^2+2m-2\sqrt{m+1}+1=0$
...????
|
Following on from $m^2+2m-2\sqrt{m+1}+1=0$, you already have $b=\sqrt{m+1}$, so substitute this in to get $b^4-2b=0 \implies b=\sqrt[3]2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4407067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solutions of Triangles inequalities - Duplicate A,B,C are angles of a triangle we are supposed to prove that $sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$ $\leq$ $\frac{1}{8}$. I used trigonometric ratios of half angles which would give $\frac{(s-a)(s-b)(s-c)}{abc}$. How can I proceed after this step?
Any help would be appreciated ( I need the solution where angles are made in terms of sides and then the inequality is simplified)
|
Here is an alternative trigonometric proof of the inequality for fun. Denote $$
x = \frac{A}{2}, \qquad y = \frac{B}{2}, \qquad \{x, y, x + y\} \subset [0, 90^\circ].
$$
Then, using the product-to-sum formula
$$
\sin x \sin y = \frac{\cos (x - y) - \cos (x + y)}{2}
$$
as well as the AM-GM inequality,
we have
$$ \begin{split}
\sin \frac{A}{2} \, \sin \frac{B}{2} \, \sin \frac{C}{2}
&= \sin x \sin y \sin (90^\circ - x - y) \\
&= \sin x \sin y \cos (x + y) \\
&= \frac{\cos (x - y) - \cos (x + y)}{2} \cos(x + y) \\
&\le \frac{1 - \cos(x + y)}{2} \cos(x + y) \\
&\le \frac{1}{2} \Bigl( \frac{1}{2} \Bigr)^2 = \frac{1}{8},
\end{split} $$
where equality holds for the first $\le$ sign if and only if $x = y$, and equality holds for the second $\le$ sign if and only if $\cos(x + y) = \frac{1}{2}$. Therefore, we have
$$
\sin \frac{A}{2} \, \sin \frac{B}{2} \, \sin \frac{C}{2} \le \frac{1}{8}
$$
where equality holds if and only if $A = B = C = 60^\circ$.
|
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"url": "https://math.stackexchange.com/questions/4408491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\int_{0}^{1} \frac{ \ln^n x}{1+x^{m}} d x$ =? Inspired by the post, I try to generalize the result to
$$
I(m, n)=\int_{0}^{1} \frac{ \ln ^nx}{1+x^{m}} d x
$$
Instead, I first investigate its partner integral $$
\begin{aligned}
I(a): &=\int_{0}^{1} \frac{x^{a}}{1+x^{m}} d x . \\
&=\sum_{k=0}^{\infty}(-1)^{k} \int_{0}^{1} x^{a} \cdot x^{m k} d x \\
&=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{a+m k+1}
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ by $n$ times at $a=0$ yields $$
\begin{aligned}
I(m, n)&=\left.\frac{{\partial}^{n}}{{\partial }a^{n}} I(a)\right|_{a=0}\\ &=\left.(-1)^{n} n ! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(a+m k+1)^{n+1}}\right|_{a=0} \\
&=(-1)^{n} n ! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(m k+1)^{n+1}} \\
&= (-1)^{n} n !\frac{(-1)^{n} n !}{m^{n+1}} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{\left(k+\frac{1}{m}\right)^{n+1}} \\ &= \sum_{k=0}^{\infty} \frac{1}{\left(2 k+\frac{1}{m}\right)^{n+1}}-\sum_{k=0}^{\infty} \frac{1}{\left(2 k+1+\frac{1}{m}\right)^{n+1}}\\& = (-1)^{n} n !\frac{1}{2^{n+1}}\left[\sum_{k=0}^{\infty} \frac{1}{\left(k+\frac{1}{2 m}\right)^{n+1}}-\sum_{k=0}^{\infty} \frac{1}{\left(k+\frac{1}{2}\left(1+\frac{1}{m}\right)\right)^{n+1}}\right]\\
&=\frac{(-1)^{n} n !}{m^{n+1} 2^{n+1}}\left[ \zeta \left(n+1, \frac{1}{2 m}\right)-\zeta \left( n + 1 , \frac { 1 } { 2 } ( 1 + \frac { 1 } { m } ) \right)\right]
\end{aligned}
$$
My Question
I don’t know how to simplify the last answer. Would you help?
|
Take the $n$th derivative of both sides of
$$\int_0^1\frac{y^{s}}{1+y}\mathrm{d}y=\frac12\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\quad \mathfrak{R}(s)>-1$$
w.r.t $s$, we get
$$\int_0^1\frac{y^{s}\ln^n(y)}{1+y}\mathrm{d}y=2^{-n-1}\left[\psi^{(n)}\left(\frac{s+2}{2}\right)-\psi^{(n)}\left(\frac{s+1}{2}\right)\right]$$
let $y=x^m$
$$\int_0^1\frac{x^{sm+m-1}\ln^n(x)}{1+x^m}\mathrm{d}x=(2m)^{-n-1}\left[\psi^{(n)}\left(\frac{s+2}{2}\right)-\psi^{(n)}\left(\frac{s+1}{2}\right)\right]$$
Finally set $s=\frac{1-m}{m}$
$$\int_0^1\frac{\ln^n(x)}{1+x^m}\mathrm{d}x=(2m)^{-n-1}\left[\psi^{(n)}\left(\frac{1+m}{2m}\right)-\psi^{(n)}\left(\frac{1}{2m}\right)\right],\quad n\in \mathbb{Z}^{+},\quad m>0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$p+7=2\cdot x^2$ and $p^2+7=2\cdot y^2$. Let $p$ a prime number such that there exists $x,y$ positive integers such that $p+7=2\cdot x^2$ and $p^2+7=2\cdot y^2$.
Find $p$.
I found that $p=11$ . I noticed that $p|(x+y)$. I tried with quadratic residues but I am stuck.
|
Subtracting the second problem equation minus the first one gives
$$p^2 - p = 2y^2 - 2x^2 \; \; \to \; \; p(p - 1) = 2(y - x)(y + x) \tag{1}\label{eq1A}$$
Note $p$ being an odd prime means $p \mid y - x$ or $p \mid y + x$. The first case gives $y - x \ge p$ and $y + x \gt p$, so $p^2 - p = 2(y - x)(y + x) \gt 2p^2$, which is not possible. Thus, as you already deduced, this means
$$p \mid x + y \tag{2}\label{eq2A}$$
If $y \ge p \; \to \; 2y^2 \ge 2p^2$, then the second equation gives $p^2 + 7 \ge 2p^2 \; \to \; p^2 \le 7$, but $p \ge 3$, which means this is not possible. Thus, $y \lt p$, and since $p + 7 \lt p^2 + 7$, then $x \lt y \; \to \; x \lt p$, so
$$x + y \lt 2p \; \; \to \; \; x + y = p \; \; \to \; \; y - x = p - 2x \tag{3}\label{eq3A}$$
Substitute this into the RHS of \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
p(p - 1) & = 2(p - 2x)p \\
p - 1 & = 2p - 4x \\
p & = 4x - 1
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Using this in the problem's first equation gives
$$\begin{equation}\begin{aligned}
4x + 6 & = 2x^2 \\
x^2 - 2x - 3 & = 0 \\
(x - 3)(x + 1) & = 0
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
With $x$ being a positive integer, then $x = 3$, so using \eqref{eq4A} gives your one result of $p = 11$.
|
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|
How to integrate $\frac{\cos2x}{ \sin x + \sin 3x}$? I thought of using trig identities to get rid of $\cos2x$ and $\sin3x$, then use Weierstrass substitution, but I got myself into big trouble as the expression got too complicated.
Is there a simpler way to tackle this problem?
|
The method using the Weierstass substitution isn't so bad.
$$t=\tan\left(\frac x2\right) \implies \begin{cases}\sin(x) = \frac{2t}{1+t^2} \\ \cos(2x) = 2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1 \\ \sin(3x) = \frac{6t(1-t^2)^2-8t^3}{(1+t^2)^3}\end{cases}$$
where we use the identities
$$\begin{cases}\sin(x) = 2\sin\left(\frac x2\right)\cos\left(\frac x2\right) \\ \cos(x) = \cos^2\left(\frac x2\right) - \sin^2\left(\frac x2\right) \\ \cos(2x) = 2\cos^2(x) - 1 \\ \sin(3x) = 3\cos^2(x)\sin(x) - \sin^3(x)\end{cases}$$
Now $dt = \frac12\sec^2\left(\frac x2\right)\,dx \iff dx = \frac{2\,dt}{1+t^2}$, so
$$\begin{align}
\int \frac{\cos(2x)}{\sin(x) + \sin(3x)} \, dx &= \int \frac{2\left(\frac{1-t^2}{1+t^2}\right)^2-1}{\frac{2t}{1+t^2} + \frac{6t(1-t^2)^2-8t^3}{(1+t^2)^3}} \frac2{1+t^2} \, dt \\[1ex]
&= 2 \int \frac{2(1-t^2)^2-(1+t^2)^2}{2t(1+t^2)^2 + 6t(1-t^2)^2-8t^3} \, dt \\[1ex]
&= \frac14 \int \frac{t^4-6t^2+1}{t^5-2t^3+t} \, dt \\[1ex]
&= \frac14 \int \left(\frac1t + \frac1{(t+1)^2} - \frac1{(t-1)^2}\right) \, dt \\[1ex]
\end{align}$$
|
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|
Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. Any other method?
When I calculate an integral $I_n=\int_0^\frac{\pi}{4}\frac{\cos nx}{\cos^nx}dx$, we find $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$. So we need to find the $J_n$, as the problem states.
The proof of $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$ is as follows.
\begin{align}
I_n/2^{n-1}-I_{n-1}/2^{n-1}
&=\frac{1}{2^{n-1}} \int_0^\frac{\pi}{4}\left(\frac{\cos nx}{\cos^nx}-\frac{2\cos(n-1)x}{\cos^{n-1}x}\right)dx\\
&=\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos[(n-1)x+x]-2\cos(n-1)x\cos x}{\cos^nx}dx\\
&=-\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos(n-2)x}{\cos^nx}dx
=-1/2^{n-1}J_n
\end{align}
|
I'll do the change of variable $n-2 \to n$ to simplify the derivation. You can recover your original integral by reversing this substitution.
Using the multiple-angle formulas shown here we know that for a positive integer $n$
\begin{align}
\sin(nx) &= \sum_{k=0}^{\lfloor{\frac{n-1}{2}\rfloor}}(-1)^k \binom{n}{2k+1}\sin^{2k+1}(x) \cos^{n-2k-1}(x) \tag{1}\\
\cos(nx) &= \sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}}(-1)^k \binom{n}{2k}\sin^{2k}(x) \cos^{n-2k}(x) \tag{2}
\end{align}
Also recall that
\begin{equation}
\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = 2^{-\frac{1}{2}} \tag{3}
\end{equation}
With this we get that
$\require{cancel}$
\begin{align*}
\int_0^\frac\pi4 \frac{\cos(nx)}{\cos^{\color{green}{n+2}}(x)}\mathrm{d}x & \overset{\color{blue}{(2)}}{=} \sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}}(-1)^k \binom{n}{2k}\int_0^\frac\pi4\sin^{2k}(x) \cos^{\cancel{n}-2k \color{green}{-} \cancel{\color{green}{n}}\color{green}{-2}}(x)\mathrm{d}x\\
& = \sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}}(-1)^k \binom{n}{2k}\int_0^\frac\pi4 \underbrace{\frac{\sin^{2k}(x)}{\cos^{2k}(x)}}_{\color{purple}{\tan^{2k}(x)}} \sec^2(x)\mathrm{d}x\\
& \overset{\color{blue}{u = \tan(x)}}{=} \sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}}(-1)^k \binom{n}{2k} \int_0^1 \color{blue}{u}^{2k}\, \mathrm{d}\color{blue}{u}\\
& = \sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}}(-1)^k \binom{n}{2k} \frac{1}{2k+1} \left(\color{purple}{\frac{n+1}{n+1}}\right)\left(\color{green}{ \frac{ 2^{\frac{n+1}{2}} }{ 2^{\frac{n+1}{2}} } }\right)\\
& = \frac{\color{green}{2^{\frac{n+1}{2}}}}{\color{purple}{n+1}}\sum_{k=0}^{\lfloor{\frac{n}{2}\rfloor}}(-1)^k \frac{(\color{purple}{n+1})n!}{(2k+1)(2k)! (n-2k)!} \color{green}{\left(2^{-\frac{1}{2}}\right)^{2k+1}\left(2^{-\frac{1}{2}}\right)^{(n+1) -2k-1}}\\
& \overset{\color{blue}{(3)}}{=} \frac{2^{\frac{n+1}{2}}}{n+1} \sum_{k=0}^{\lfloor{\frac{(n+1)-1}{2}\rfloor}}(-1)^k \binom{n+1}{2k+1}\sin^{2k+1}\left(\frac{\pi}{4}\right)\cos^{(n+1) -2k-1}\left(\frac{\pi}{4}\right)\\
& \overset{\color{blue}{(1)}}{=}\frac{2^{\frac{n+1}{2}}}{n+1}\sin\left((n+1)\frac{\pi}{4}\right)
\end{align*}
valid for $n \in \mathbb{N}$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4413657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 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.