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Expected value of dice game (with options) Setup
The question (modified from an old Jane Street interview):
(i) Two people throw a 6-sided dice. The person with a larger number wins $1. If the dice values are equal, neither person wins a reward. How much are you willing to pay to play this game?
(ii) A followup: say you now have the option to pay $0.25 to add 2 to the number you get. (You can choose this option after rolling your dice.) How much are you willing to pay to play this game now?
Part (i) seems easy: $E[game] = \frac{1}{6}(0 + \frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6}) = 15/36$. I was less sure about part (ii).
Question: what's the solution to part (ii)?
My initial hunch was that I should always exercise the option, even if I rolled a $6$ (since this would guarantee I win a dollar, as opposed to a $5/6$ chance). This would suggest $E[game]= \frac{1}{6}(\frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{6}{6}) - \frac{1}{4} = \frac{17}{36}$ is the game' value.
However, other answers I've seen online claim "when you roll 1, 2, 3, 4, 5, you should exercise the option; when you roll to 6, you shouldn't." This would suggest $E[game]= \frac{1}{6}(\frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{5}{6}) - \frac{5}{6} * \frac{1}{4} = \frac{35}{72}$. I'm unsure why/if this claim is true.
| If you have rolled a $6$ and do not pay to add $2$, your expectation is $\frac 56 \approx 0.83$ because you win $1$ unless your opponent has also rolled a $6$. By paying $0.25$ you increase this to $1$, so your expectation of rolling a $6$ and buying the $2$ is $1-0.25=0.75$. As this is less than $0.83$ you should not buy the $2$.
Both calculations of the game's value with the option are correct for the strategy employed. The fact that the second is greater than the first is another way to show that the second strategy is superior.
| {
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"url": "https://math.stackexchange.com/questions/4322580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
| Here $$ \frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} $$
you canceled $x^4$ in the numerator and denominator, so it should be $$\frac{-\frac{1}{3}+o\left(x^{-1}\right)}{-\frac{1}{2}+o\left(x\right)} $$
You'll need to consider more terms in the series to get more precision.
| {
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"timestamp": "2023-03-29T00:00:00",
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Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, when $a=b\ne 0$, $(a^2+ab+b^2)(a-b) = a^3-b^3 =0$.
| One way is to start from $$a^2+ab+b^2=b^2\left(\left(\frac{a}{b}\right)^2+\frac{a}{b}+1\right)=0$$ so letting $x=\frac{a}{b}$ we get the quadratic equation $x^2+x+1=0$ which has no real roots. This is similar to the idea hinted at in the comment by RiverX15.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327817",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the volume between the sphere $x ^2 + y^2 + z^2 = 4$ and the plane $z = 1$ Suppose $y \geq 3$. I want to compute the volume between the sphere $x ^2 + (y − 2)^2 + z^2 = 4$
and the plane $y = 3$.
So I move left the sphere and and the plan, and rotate it counterclockwise. I got the new sphere and the new plan:
Suppose $z \geq 1$. Then compute the volume between $x ^2 + y^2 + z^2 = 4$ and the plan $z = 1$.
Here is my attempt using spherical coordinates:
$$\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{6}} \int_1^2 \rho^2 \sin \phi dp d \phi d\theta = 2\pi (1- \frac{\sqrt{3}}{2})\frac{1}{3}$$ I am supposed to get $\frac{5 \pi}{3}$. Where am I wrong?
| There are two mistakes in your work. At the intersection of the sphere and the plane,
$z = 2 \cos\phi = 1 \implies \phi = \pi/3$
Also the lower bound of $\rho$ is defined by the plane
$z = \rho\cos\phi = 1 \implies \rho = \sec\phi$
So the integral should be,
$ \displaystyle \int_{0}^{2 \pi} \int_0^{\pi/3} \int_{\sec\phi}^2 \rho^2 \sin\phi ~ d\rho ~ d\phi ~d\theta = \frac{5\pi}{3}$
Also note that,
$x^2 + (y-2)^2 + z^2 = 4 \implies x^2 + y^2 + z^2 = 4 y$
So using $x = \rho \cos\theta \sin\phi, z = \rho\sin\theta\sin\phi, y = \rho\cos\phi$, we have
$\rho = 4 \cos\phi$
$y = \rho \cos\phi = 3 \implies \rho = 3 \sec\phi$
At intersection of the sphere and the plane,
$\rho = 4 \cos\phi = 3 \sec\phi \implies \cos\phi = \frac{\sqrt3}{2}$
So, $\phi = \pi/6$
So the integral can also be written as,
$ \displaystyle \int_{0}^{2 \pi} \int_0^{\pi/6} \int_{3\sec\phi}^{4\cos\phi} \rho^2 \sin\phi ~ d\rho ~ d\phi ~d\theta = \frac{5\pi}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $z^3 =-1$ I need to find the solution of $z^{3} =-1$.
Let$$z=[ r,\theta ]$$
then using de Moivre's theorm$$ z^{3} =\left[ r^{3} ,3\theta \right] $$
write the number $-1$ in modulus argument from,
$$[ 1,( 2n-1) \pi ]$$
the we can write$$\left[ r^{3} ,3\theta \right] =[ 1,( 2n-1) \pi ]$$
Therefore,$$ r^{3} =1 \space \text{and} \space \theta =\frac{( 2n-1) \pi }{3}$$
Then$$z_{n} =1\left( \cos\frac{( 2n-1) \pi }{3} +i\sin\frac{( 2n-1) \pi }{3}\right)$$
\begin{align}
&n=0 \quad z_{0} =\frac{1}{2} -i\frac{\sqrt{3}}{2}\\
&n=1 \quad z_{1} =\frac{1}{2} +i\frac{\sqrt{3}}{2}\\
&n=2 \quad z_{1} =-1\\
\end{align}
| $z=re^{i\theta}$
$z^3 = -1$
$r^3 (e^{3 i \theta})= 1\cdot e^{i\pi}$
Hence, $ r=1$ and $\theta= \frac{i(\pi + 2k\pi) }{3}$ for $k=0, 1,2$
$z= e^{\frac{i\pi}{3}} , e^{i\pi}, e^{\frac{5i\pi}{3}}$
| {
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"url": "https://math.stackexchange.com/questions/4332411",
"timestamp": "2023-03-29T00:00:00",
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Rankine-Hugoniot for Burgers equation with Lorentzian initial data We have the following equation with initial condition
\begin{align*}
u_y + uu_x &= 0\\
u(x,0)&=\frac{1}{1+x^2}\quad x\in\mathbb{R}
\end{align*}
let's say $g(x)=\dfrac{1}{1+x^2}$ and corresponding to the manifold $\Gamma$ in the $xyz$ space given by
\begin{equation}
x = s,\quad y = 0,\quad z = g (s)
\end{equation}
The characteristic differential equations
\begin{equation*}
\frac{dx}{dt}=z\quad \frac{dy}{dt}=1\quad \frac{dz}{dt}=0
\end{equation*}
Integrating each of the expressions we have
\begin{align*}
\int dx&=\int z dt\\
x&=zt
\end{align*}
also
\begin{align*}
\int dy&=\int 1 dt\\
y&=t
\end{align*}
and
\begin{align*}
\int dz&=\int 0 dt\\
z&=0
\end{align*}
combined with the initial condition for $t=0$ lead to the parametric representation
\begin{equation*}
x=s+zt,\quad y=t, \quad z=g(s)
\end{equation*}
now as
\begin{equation*}
s=x-zt\quad \text{and}\quad y=t \quad \text{and}\quad z=g(s)
\end{equation*}
For the solution $z=u(x,y)$ then yields the implicit equation
\begin{equation*}
u=g(x-uy)=\frac{1}{1+(x-uy)^2}
\end{equation*}
now the study is going to be done of Weak (or Integral) Solutions
as the characteristic is given by
\begin{equation*}
x=\frac{y}{1+\xi^2}+\xi
\end{equation*}
it is necessary that this characteristic presents shock to do the study according to the study made by the book (artial Differential Equations in Action From Modelling to Theory) page 199 to find $y_s$ (breaking time) and
the location $x_s$ the following is done where $q(u)=u^2 / 2$ then $q'(u)=u$ and $q''(u)=1$
\begin{equation*}
z(\xi)=-q''(g(\xi))g'(\xi)=-1\cdot \frac{-2x}{\left(1+x^2\right)^2}=\frac{2x}{\left(1+x^2\right)^2}
\end{equation*}
and the maximum point of $z(\xi)$ is $\left(\sqrt{\frac{1}{3}},\:\frac{3\sqrt{3}}{8}\right)$ then $\xi_M=\sqrt{\frac{1}{3}}$ and $z(\xi_M)=\frac{3\sqrt{3}}{8}$ then
\begin{equation*}
y_s=\frac{1}{z(\xi_M)}=\frac{1}{\frac{3\sqrt{3}}{8}}=\frac{8}{3\sqrt{3}}
\end{equation*}
and
\begin{equation*}
x_s=q'(g(\xi_M))y+\xi_M=\frac{1}{1 + (\sqrt{1/3})^2}\left(\frac{8}{3\sqrt{3}}\right)+\sqrt{\frac{1}{3}}=\frac{2}{\sqrt{3}}+\sqrt{\frac{1}{3}}
\end{equation*}
therefore $\displaystyle (x_s,y_s)=(\frac{2}{\sqrt{3}}+\sqrt{\frac{1}{3}},\frac{8}{3\sqrt{3}})$
1)My question is already having all these data, how can I do the study of Rankine-Hugoniot condition?
Your answers would be of great help. Thank you very much. I remain attentive to your answer.
| The characteristic curves satisfy
$$
x= x_0 + g(x_0) y, \qquad u = g(x_0)
$$
where $g(x) = (1+x^2)^{-1}$ is the initial data, i.e.
$
u = g(x-uy)
$
in implicit form. The shock time is given by
$$
y_s = \frac{-1}{\inf_{x_0} g'(x_0) } = \frac{8}{3\sqrt{3}} \simeq 1.54
$$
for $x_0 = 1/\sqrt{3}$. Therefore, the results in OP look correct, except the potential mistake in the expression of $x_s$. The Rankine-Hugoniot condition for the inviscid Burgers equation is still valid in the usual form
$$
x_s'(y) = \tfrac12 \left( u_\ell + u_r\right),$$
$$\text{with}\qquad x_s(y_s) = \frac{1}{\sqrt 3} + \frac{8}{3\sqrt{3}}\, g\left(\frac{1}{\sqrt 3}\right) = \sqrt 3 \simeq 1.73
$$
where $u_\ell$, $u_r$ represent the value of $u = g(x-uy)$ on the left and on the right of the discontinuity $(x_s(y), y)$ with $y \geq y_s$. See this post where a simpler case is tackled analytically.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of region ABC. For reference: In figure $T$ and $K$ are points of tangency,
$MT = a$ and $KN = b$; calculate area of region $ABC$.
(Answer:$2\sqrt{ab}(\sqrt a+\sqrt b)^2$)
My progress:
$$S_{ABC} = p \cdot r = \frac{r \cdot (AB+BC+AC)}{2}\\
AC +2R = AB+BC\\
S_{ABC} = AG \cdot GC \qquad \text{(property)} \\
S_{ABC} = (AC+R)R \qquad \text{(property)} \\
OTBQ:~\text{square} \implies TK = R\sqrt2 \\
\ldots ?$$
I'm not able to use segments a and b in the resolution
| I continue your process and we already have $\text{TK}=\sqrt{2} R$.
According to intersecting chords theorem, we have $$\text{AT} \cdot\text{TB}=\text{MT}\cdot \text{TN}$$$$\text{BK} \cdot\text{KC}=\text{KN} \cdot \text{MK}$$
So $$\text{AT}=\sqrt{2} a+\frac{\text{ab}}{R}$$$$\text{KC}=\sqrt{2} b+\frac{\text{ab}}{R}$$
Because $S_{ABC}$ is a rectangular triangle, we have $\text{AB}^2+\text{BC}^2=\text{AC}^2$
and $$\text{BC}=\text{BC}+\text{KC}=\text{KC}+R$$$$ \text{AB}=\text{AT}+\text{TB}=\text{AT}+R$$$$\text{AC}=\text{AG}+\text{GC}=\text{AT}+\text{KC}$$
So solving the equation we get $$R=\sqrt{\text{ab}}$$
Therefore $$S_{ABC} = \frac{\text{AB}\cdot \text{BC}}{2}$$$$=\frac{1}{2} \left(\sqrt{2} a+2 \sqrt{\text{ab}}\right) \left(\sqrt{2} b+2 \sqrt{\text{ab}}\right)$$$$=\boxed{\sqrt{2} \sqrt{\text{ab}} (a+b)+3 a b}$$
I think the answer you give might be wrong.
| {
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What is the Variance of $(1-X)/X$ and the Expectation of $(1-X)/X^{2}$? Knowing that
$X = \begin{cases}
0.05 \ \ \text{with probability} \ \ 0.7\\
0.2\ \ \text{with probability} \ \ 0.3\end{cases}$
What would be the Variance of $\frac{1-X}{X}$ and the Expectation of $\frac{1-X}{X^2}$ ?
I have computed it and would like to know if my development is correct :
\begin{aligned}
Var\left[\frac{1-X}{X}\right] &= E\left[\left(\frac{1-X}{X}\right)^2\right] - \left(E\left[\frac{1-X}{X}\right]\right)^2 \\
&= \left(\frac{0.95}{0.05}\right)^2\cdot 0.7 + \left(\frac{0.8}{0.2}\right)^2\cdot 0.3 - \left(\frac{0.95}{0.05}\cdot 0.7 + \frac{0.8}{0.2}\cdot 0.3\right)^2 = 47.25
\end{aligned}
and
$$E\left[\left(\frac{1-X}{X^2}\right)\right] = \left(\frac{0.95}{0.05^2}\right)\cdot 0.7 + \left(\frac{0.8}{0.2^2}\right)\cdot 0.3 = 272 $$
| Since
$$\operatorname{E}\left[\frac{1-X}{X}\right]=\sum_{i=1}^2 \frac{1-x_i}{x_i}p_i$$
$$=\frac{0.95}{0.05}\cdot 0.70+\frac{0.80}{0.20}\cdot 0.30=14.5$$
And
$$\operatorname{E}\left[\left(\frac{1-X}{X}\right)^2\right]=\sum_{i=1}^2 \left(\frac{1-x_i}{x_i}\right)^2 p_i$$
$$=\left(\frac{0.95}{0.05}\right)^2\cdot 0.70+\left(\frac{0.80}{0.20}\right)^2\cdot 0.30$$
$$=361\cdot 0.70+16\cdot 0.30=257.5$$
We have
$$\operatorname{Var}\left(\frac{1-X}{X}\right)=257.5-14.5^2=47.25$$
Also
$$\operatorname{E}\left[\frac{1-X}{X^2}\right]=\sum_{i=1}^2 \frac{1-x_i}{(x_i)^2}p_i$$
$$=\frac{0.95}{0.05^2}\cdot 0.70+\frac{0.80}{0.20^2}\cdot 0.30=272$$
So yes, as other users already mentioned, your methods and answers are indeed correct.
| {
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Minimum value of $a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$ Let $a,b,c$ be real numbers such that $a+b+c=3$, then find minimum value of
$$a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$$
I have solved many such question using A.M.-G.M. inequality but since $\cot (130^ \circ)$ is negative, hence it cannot be applied here. I also have a result in mind which is
$$\cot (60 ^\circ-\theta)\cot (\theta)\cot (60 ^\circ+\theta)=\cot (3\theta)$$ and in this question if we write $
\cot(130^\circ)=-\cot (50^\circ)$, then above mentioned result can be applied on $\cot (10^ \circ),\cot (70^ \circ),\cot (50^ \circ)$ but I am not able to put everything together and reach the final answer which is given as $\sqrt{27}$. Any help or hint would be appreciated.
| There's neither a maximum nor a minimum value. Take $a=1,b=x\in\mathbb{R},c=2-x$. Then
$$a^2\cot(10^\circ)+b^2\cot(70^\circ)+c^2\cot(130^\circ)\\
=(\cot(130^\circ)+\cot(70^\circ))x^2-4\cot(130^\circ)x+4\cot(130^\circ)+\cot(10^\circ)\xrightarrow{x\to\infty}-\infty $$
since the above is a quadratic function with negative leading coefficient: $\cot(130^\circ)+\cot(70^\circ)=\cot(70^\circ)-\cot(50^\circ)<0$. So the expression is unbounded below. Similarly, taking $a=2-x,b=x,c=1$ leads to
$$a^2\cot(10^\circ)+b^2\cot(70^\circ)+c^2\cot(130^\circ)\\
=(\cot(10^\circ)+\cot(70^\circ))x^2-4\cot(10^\circ)x+4\cot(10^\circ)+\cot(130^\circ)\xrightarrow{x\to\infty}\infty $$ since $\cot(10^\circ)+\cot(70^\circ)>0$. Thus, the expression is also unbounded above.
Where does the $\sqrt{27}$ you mentioned come from? If we set up Lagrange multipliers for this problem (the calculations here are relatively straightforward), we obtain the point $(a,b,c)=\left(-\cot(70^{\circ})\cot(130^{\circ}),-\cot(10^{\circ})\cot(130^{\circ}),-\cot(10^{\circ})\cot(70^{\circ})\right)$ and the expression is equal to $\sqrt{27}$ at this point. However, Lagrange multipliers do not always give us the extrema. In this particular case, we have a saddle point there.
| {
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Geometry problem to prove Euler line of a triangle $I$ is the incenter of $\triangle{ABC}$ and $D$ is the tangent point of the incircle with side $BC$. Points $E$ and $F$ are on side $BC$ so that $\angle{BAE}=\angle{ACB}, \angle{CAF}=\angle{ABC}$. $G$ and $H$ are incenters of $\triangle{ABE}$ and $\triangle{ACF}$ respectively. Prove that $AD$ is the Euler Line of $\triangle{GHI}$.
I tried to find related material to solve this problem but in vain. Any hints or helpful suggestion? Thanks!
With hint to use barycentric coordinates, here is what I have got:
Let $s=\dfrac{a+b+c}{2}, t=\sin A+\sin B+\sin C$, then
$A(1:0:0)\\
B(0:1:0)\\
C(0:0:1)\\
I(\dfrac{a}{2s}:\dfrac{b}{2s}:\dfrac{c}{2s})\\
D(0:\dfrac{s-c}{a}:\dfrac{s-b}{a})\\
E(0:1-\dfrac{c\sin C}{a\sin A}:\dfrac{c\sin C}{a\sin A})\\
F(0:\dfrac{b\sin B}{a\sin A}:1-\dfrac{b\sin B}{a\sin A})\\
G(\dfrac{\sin C}{t}:1-\dfrac{(a+c)\sin C}{at}:\dfrac{c\sin C}{at})\\
H(\dfrac{\sin B}{t}:\dfrac{b\sin B}{at}:1-\dfrac{(a+b)\sin B}{at})
$
| Oh, this is a nice exercise, i will remember it for didactic purposes. Below, all details will be given, and the computations are done explicitly.
As references i am using
bary-full and bary-short. Notations like $X(2)$ for the centroid are as in ETC.
Fix a (non-degenerated) triangle $\Delta ABC$ with sides $a,b,c$.
Below, $s=(a+b+c)/2$ is the semiperimeter.
Each point in plane is written uniquely as $P=xA+yB+zC$ with $x+y+z=1$.
This makes sense by replacing $P,A,B,C$ by their affixes in $\Bbb C$
or cartesian coordinates in $\Bbb R^2$. Also, it makes sense in vectorial notation after fixing some / any origin $O$, and plugging it in in the form
$OP=xOA+yOB+zOC$ (with vectors not explicitly decorated with arrows).
Then in (normalized) notation $P=(x,y,z)$. Sometimes, we use also $x,y,z$ with non-zero sum to describe a point. Then $P=[x:y:z]$ means that $P=(x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z))$. Abusively, i will also write
$\displaystyle P=\frac 1{x+y+z}(x,y,z)$ in such situations.
Note that $\Delta EBA\sim \Delta ABC$, so
$\displaystyle
\frac{BE}{AB}=\frac{AB}{BC}$, so
$\displaystyle BE=\frac{c^2}a$, and
$\displaystyle EC=a-\frac{c^2}a$, which gives
$\displaystyle
BE:EC=\frac{c^2}{a^2}:\left(1-\frac{c^2}{a^2}\right)=c^2:(a^2-c^2)$.
At this point, we already know the coordinates of $E$. Let us make the full list so far:
$$
\begin{aligned}
A &=(1,0,0)\\
B &=(0,1,0)\\
C &=(0,0,1)\\
I &= [a:b:c]=\frac 1{a+b+c}(a,b,c)\\
D &= [0:s-c:s-b]\\[3mm]
&\qquad\text{ so $AD$ has the equation }
0 = \begin{vmatrix}x&y&z\\1&0&0\\0&s-c&s-b\end{vmatrix}
\\
&\qquad\text{ which is simplified }
y(a-b+c) = z(a+b-c)
\\[3mm]
E &= [0:a^2-c^2:c^2]=\left(0\ :\ 1-\frac{c^2}{a^2}\ :\ \frac{c^2}{a^2}\right)\\
F &= [0:b^2:a^2-b^2]=\left(0\ :\ \frac{b^2}{a^2}\ :\ 1-\frac{c^2}{a^2}\right)\\
G &= \frac 1{a+b+c}(aE+bB+cA)
\\
&\qquad\qquad\text{ written parallel to }
I=\frac 1{a+b+c}(aA+bB+cC)
\text{ since }\Delta EBA\sim\Delta ABC
\\
&= \frac 1{a+b+c}\left(a\left(1-\frac{c^2}{a^2}\right)B +a\frac{c^2}{a^2}C+bB+cA\right)\qquad\text{ so}
\\
G &
=\left[c\ :\ a+b-\frac{c^2}a\ :\ \frac{c^2}a\right]
=\frac 1{a+b+c}\left(c\ ,\ a+b-\frac{c^2}a\ ,\ \frac{c^2}a\right)
\\
H &=\left[b\ :\ \frac{b^2}a\ :\ a+c-\frac{b^2}a\right]
=\frac 1{a+b+c}\left(b\ ,\ \frac{b^2}a\ ,\ a+c-\frac{b^2}a\right)
\\[3mm]
X(2)_{\Delta GHI}
&=\frac 13(G+H+I)\\
&=\frac 1{3(a+b+c)}
\left( a+b+c\ ,\ a+2b+\frac 1a(b^2-c^2)\ ,\ a+2c+\frac 1a(c^2-b^2)\right)
\\
&=\Big[a(a+b+c)\ :\ (a+b)^2-c^2\ :\ (a+c)^2-b^2\Big]
\\
&=[a\ :\ a+b-c\ :\ a-b+c]\ .
\end{aligned}
$$
Note that this centroid $X(2)$ of $\Delta GHI$ easily satisfies the equation of $AD$ computed above.
We need explicitly one more point on the Euler line of $\Delta GHI$, then chek again the equation of $AD$ for it, and we are done. I will compute its circumcenter $X(3)_{\Delta GHI}$ by intersecting its side bisectors. In this case, a picture is our friend:
Let $(x,y,z)$ be the coordinates of $X(3)_{\Delta GHI}$.
Then we build the displacement vectors for $IH$ and for the vector from the mid point $H'$ of $IH$ to $X(3)$. They are
$$
\begin{aligned}
\overrightarrow{IH}
&= H-I \\
&= \frac 1{a+b+c}\left(b-a\ ,\ \frac{b^2}a-b\ ,\ a-\frac{b^2}a\right)\\
&\sim (a(b-a)\ ,\ b(b-a)\ ,\ -(b-a)(b+a))\\
&\sim (a\ ,\ b\ ,\ -a-b)\\
\ ,
\\
\overrightarrow{H'X(3)}
&= X(3)-\frac 12(H+I) \\
&= (x,y,z)-
\frac 1{2(a+b+c)}\left(b+a\ ,\ \frac{b^2}a+b\ ,\ a+2c-\frac{b^2}a\right)
\\
&\sim (4asx,\ 4asy,\ 4asz) -(a^2+ab, \ b^2+ab\ ,\ a^2+2ac-b^2)
\ .
\end{aligned}
$$
We write the equation from EFFT (from bary-short) and get the point:
$$
X(3)
=\Big[ a(b^2+c^2-a^2)\ :\ bc(a+b-c)\ :\ bc(a-b+c)\Big]
\ .
$$
(If we already know the point, we may instead just verify it is indeed the circumcenter of $\Delta GHI$.)
It also easily satisfies the equation of the line $AD$. Job done.
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + 0 + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor \frac{1}{n+m} \rfloor = 1$
From here on I have no idea what to do!
Edit: It is easy to see that any value $1<x<2$ satisfies the equation, but can I find all the solutions?
| Hints:
*
*For $x>0$ and $x\neq 1$, one of the two values, $x$ or $\frac1x$, is between $0$ and $1$, so the floor of that value will be $0$.
*If $x$ is a solution, then so is $\frac1x$. This means you can first only focus on $x>1$, to get half of the solutions, then know that this is just half of the solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
(Dis)prove the inequality $2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)$ for $a,x>0$ Problem :
Let $a,x>0$ then define :
$$f\left(x\right)=x^{\frac{x}{x^{2}+1}}$$
And :
$$g\left(x\right)=\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}$$
Then prove or disprove that :
$$2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)\tag{E}$$
My attempt
From here New bound for Am-Gm of 2 variables we have :
$$h(x)=\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}\geq x^{\frac{x}{x^{2}+1}}$$
So we need to show :
$$\sqrt{2}\sqrt{h\left(xa\right)+h\left(\frac{1}{xa}\right)}\leq g(x)$$
Now using Binomial theorem (second order at $x=1$) for $1\leq x\leq 2$ and $0.5\leq a\leq 1$ we need to show :
$$\sqrt{2}\sqrt{r\left(xa\right)+r\left(\frac{1}{xa}\right)}\leq g(x)$$
Where :
$$r(x)=\left(1+\left(-\frac{1}{x}+\frac{1}{x^{2}}\right)x+\frac{1}{2}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right)^{2}\cdot x\cdot\left(x-1\right)\right)^{-\frac{1}{2}}$$
I have not tried but here show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ user RiverLi's provide some lower bound again I haven't checked but (perhaps?) it works with this inequality .If it doesn't work we need a higher order in the Pade approximation .
Edit 06/01/2022 :
Using the nice solution due to user RiverLi it seems we have for $0.7\leq a \leq 1$ and $1\leq x\leq 1.5$ :
$$\frac{1}{a}\cdot\frac{1+x+(x-1)a^{2}}{1+x-(x-1)a^{2}}+\frac{1}{x}\cdot\frac{1+a+(a-1)x^{2}}{1+a-(a-1)x^{2}}\geq \sqrt{2}\sqrt{r\left(a^{2}x^{2}\right)+r\left(\frac{1}{a^{2}x^{2}}\right)}$$
If true and proved it provides a partial solution .
Edit 07/01/2022:
Define :
$$t\left(x\right)=\left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-1}$$
As accurate inequality we have for $x\geq 1$ :
$$\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}\leq \left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-1}+h(1)-t(1)$$
Again it seems we have for $0<x\leq 1$ :
$$\left(\ln\left(\left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{\frac{x}{2}}+1\right)\right)^{-\frac{96}{100}}+h(1)-(t(1))^{\frac{96}{100}}\geq \left(-\frac{1}{x}+\frac{1}{x^{2}}+1\right)^{-\frac{x}{2}}$$
If true we can use the power series of $\ln(e^x+1)$ around zero and hope .
Last edit 08/01/2022:
I found a simpler one it seems we have firstly :
On $(0,1]$ :
$$\left(1+\frac{1}{x^{2}}-\frac{1}{x}\right)^{-\frac{x}{2}}-\frac{x^{2}}{x^{2}+1}-\left(1-\frac{0.5\cdot1.25x}{x+0.25}\right)\leq 0$$
And on $[1,8]$ :
$$\left(1+\frac{1}{x^{2}}-\frac{1}{x}\right)^{-\frac{x}{2}}-\frac{x^{2}}{x^{2}+1}-\frac{0.5\cdot1.25x}{x+0.25}\leq 0$$
Question :
How to show or disprove $(E)$ ?
Thanks really .
| Sketch of proof :
Define :
$$n\left(x\right)=\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{-\frac{x}{2}}-\sqrt{x^{1.1}}-3x\left(x-1\right)\left(1-\frac{x}{x+1}\right)^{3}+3\left(0.5-\frac{x}{x+1}\right)$$
Then we have on $(-\infty,\infty)$ :
$$\frac{d}{dx}\left(\frac{d}{dx}n\left(e^{x}\right)\right)<0$$
So we have by Jensen's inequality :
$$n(x)+n\left(\frac{1}{x}\right)\leq 2n(1)=0$$
Remains to show :
$$\sqrt{2}\sqrt{\sqrt{\left(ax\right)^{1.1}}+\frac{1}{\sqrt{\left(ax\right)^{1.1}}}}-\left(\sqrt{\frac{a^{x}}{x}}+\sqrt{\frac{x^{a}}{a}}\right)\leq 0$$
Wich need some constraint to be true .
A counter-part :
One can show on $(0,4]$ :
$$x^{\frac{x}{x^{2}+1}}-4x\left(x-1\right)\left(1-\frac{x}{x+1}\right)^{3}-1\geq 0$$
Wich conducts to :
$$2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\geq 2$$
To be continued .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Faster way to find the first four non-zero terms of the Maclaurin series for $\frac{1-x}{1+x}\cosh\sqrt{x}$ I want to find the first 4 non-zero terms for :
$$\frac{1-x}{1+x}\cosh\sqrt{x}$$
Before expanding, I rewrite this as $$(1-x)\left(\frac{1}{1+x}\right)\cosh\sqrt{x}$$
Then I expand to get $$(1-x)\left(1-x+x^2-x^3+\cdots\right)\left(1+\frac x {2!}+\frac {x^2} {4!}+\frac{x^3}{6!}+\cdots\right)$$
Now multiplying these brackets and simplifying takes a while (pretty long for my exam time), so is there a faster method to do this or to multiply these brackets?
| It's unnecessary to break up $1-x$ and $\frac{1}{1+x}$ into two separate series. You can just differentiate $\frac{1-x}{1+x}$ directly. As J.G. pointed out in his comment, it may be easier to rewrite the expression as $\frac{2}{1+x} - 1$ or $2(x + 1)^{-1} - 1$. This gives you $\frac{1-x}{1+x} = 1 - 2x + 2x^2 - 2x^3 + 2x^4 - 2x^5 + 2x^6 - 2x^7 + ...$.
Then you just have two series to multiply together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the sum (Includes binomial coefficients) I came across this question under summation:
Find the sum: $$\sum_{r=0}^n \left[\frac{r}{n} - \alpha \right]^2 {n \choose r}x^r(1-x)^{n-r} $$
To start with this, I wrote it as $\sum_{r=0}^n \left[\frac{r-n\alpha}{n} \right]^2 {n \choose r}\left(\frac{x}{1-x}\right)^r(1-x)^{n}$
= $(1-x)^{n}\sum_{r=0}^n \left[\frac{r^2+n^2\alpha^2-2n\alpha r}{n^2} \right] {n \choose r}\left(\frac{x}{1-x}\right)^r$
= $(1-x)^{n}\left[\sum_{r=0}^n \frac{r^2}{n^2} {n \choose r}\left(\frac{x}{1-x}\right)^r + \sum_{r=0}^n \alpha ^2 {n \choose r}\left(\frac{x}{1-x}\right)^r - \sum_{r=0}^n \frac{2\alpha r}{n} {n \choose r}\left(\frac{x}{1-x}\right)^r\right]$
Here, I found the values of individual sums by differentiating the binomial expansion of $\left(1+\frac{x}{1-x}\right)^n$.
I got :
$\sum_{r=0}^n \frac{r^2}{n^2} {n \choose r}\left(\frac{x}{1-x}\right)^r = \frac{nx+n(n-1)x^2}{(1-x)^nn^2}$
$\sum_{r=0}^n \alpha ^2 {n \choose r}\left(\frac{x}{1-x}\right)^r$ = $\alpha^2 \left(\frac{1}{1-x}\right)^n$
$\sum_{r=0}^n \frac{2\alpha r}{n} {n \choose r}\left(\frac{x}{1-x}\right)^r$ = $\frac{2\alpha}{n}\left(\frac{nx}{(1-x)^n}\right)$
Substituting the values, finally I got the sum to be $\frac{x(1-x)}{n}+(x-\alpha)^2$.
Can anyone please verify if the answer is correct?
| Correct, and here's an alternative approach that uses the binomial distribution. Let $X$ be a $\text{Bin}(n,x)$ random variable. Then $\mathbb{E}[X] = n x$ and $$\mathbb{E}[X^2] = \mathbb{V}[X] + \mathbb{E}[X]^2 = n x(1-x) + (nx)^2.$$
So
\begin{align}
\sum_{r=0}^n \left[\frac{r}{n} - \alpha \right]^2 {n \choose r}x^r(1-x)^{n-r}
&= \mathbb{E}\left[\frac{X}{n} - \alpha \right]^2 \\
&= \frac{\mathbb{E}[X]^2}{n^2} - \frac{2\mathbb{E}[X]}{n} + \alpha^2 \\
&= \frac{n x(1-x) + (nx)^2}{n^2} - \frac{2nx}{n} + \alpha^2 \\
&= \frac{x(1-x)}{n} + (x-\alpha)^2
\end{align}
As mentioned by @Essaidi, there is an implicit assumption that $x\in[0,1]$ when we use the binomial distribution. But both sides of the identity are polynomials in $x$ of degree at most $n$, so if it is true for $n+1$ points, it is true for all $x$. We have shown that it is true for $[0,1]$, so it is true for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_{-1}^{1} u^2(1-u^2)^{3/2}du$ The integral I'm having difficulty with solving is:
$I = \int_{-1}^{1} u^2(1-u^2)^{3/2}du$
I arrived at this integral in trying to solve
$\int_{0}^{\pi} sin^4(x)cos^{2}(x)dx$
by making the substitution $u=cos(x)$
I attempted integration by parts as follows:
$F=u, \: F'=1 \\G=-\frac{1}{5}(1-u^2)^{\frac{5}{2}}, \: G'=u(1-u^2)^{\frac{3}{2}},$
Such that
$I = FG|_{-1}^{1} - \int F' G = \frac{1}{5}\int_{-1}^{1}(1-u^2)^{\frac{5}{2}} $
After which I think the obvious thing to do is to again make a trig substitution $u=cos(x)$, which leads to:
$I= \frac{1}{5} \int_{\pi}^{0} sin^{6}(x) dx$
This seems like an arduous route to solving this integral. Can anyone come up with an easier one?
Thanks for your help in advance.
| The integral expression looks like differnetial binom
$$
I = \int{u^m(a + bu^n)^pdu}.
$$
Comparing that to the initial integral,
$$
m = 2, n = 2, p = \frac{3}{2}, a = 1, b = -1.
$$
$$
p + \frac{m+1}{n} = \frac{3}{2} + \frac{2+1}{2} = 3
$$
is an integer. Then the following substitution is used
$$
au^{-n} + b = x^s, \text{ where }s\text{ is a denominator of }p, \text{ i.e.,} s = 2.
$$
$$
\begin{aligned}
\int_{-1}^1{u^2(1 - u^2)^\frac{3}{2}du} &= 2\int_{0}^1{u^2(1 - u^2)^\frac{3}{2}du} = \\
&= \left|
\begin{aligned}
x^2 = u^{-2} - 1 &\Leftrightarrow u^2 = \frac{1}{x^2+1} \\
2udu = -\frac{2xdx}{(x^2+1)^2} &\Leftrightarrow du = -\frac{x}{(x^2+1)^2}\sqrt{x^2+1}dx \\
u = 0 &\Leftrightarrow x = +\infty \\
u = 1 &\Leftrightarrow x = 0 \\
\end{aligned}
\right| = \\
&= 2\int_0^{+\infty}\frac{1}{x^2+1}\left(\frac{x^2}{x^2+1}\right)^\frac{3}{2}\frac{x}{(x^2+1)^2}\sqrt{x^2+1}dx = \\
&= 2\int_0^{+\infty}\frac{x^4}{(x^2+1)^4}dx =
-\frac{1}{3}\int_0^{+\infty}x^3d\left\{\frac{1}{(x^2+1)^3}\right\} = |\text{integrating by parts}| = \\
&= -\frac{1}{3}\left(\underbrace{\left.x^3\frac{1}{(x^2+1)^3}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{(x^2+1)^3}3x^2dx\right) = \\
&= \int_0^{+\infty}x^2\frac{1}{(x^2+1)^3}dx = -\frac{1}{4}\int_0^{+\infty}xd\left\{\frac{1}{(x^2+1)^2}\right\} =|\text{intgrating by parts}| = \\
&= -\frac{1}{4}\left(\underbrace{\left.x\frac{1}{(x^2+1)^2}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{(x^2+1)^2}dx\right) = \\
&= \frac{1}{4}\int_0^{+\infty}\frac{1}{(x^2+1)^2}dx = \frac{1}{4}\int_0^{+\infty}\left(\frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\right)dx = \\
&= \frac{1}{4}\underbrace{\int_0^{+\infty}\frac{1}{x^2+1}dx}_{= \left.\arctan(x)\right|_0^{+\infty} = \frac{\pi}{2}}-\frac{1}{4}\int_0^{+\infty}\frac{x^2}{(x^2+1)^2}dx= \\
&= \frac{\pi}{8}+\frac{1}{8}\int_0^{+\infty}xd\left\{\frac{1}{x^2+1}\right\} = |\text{integrating by parts}| = \\
&=\frac{\pi}{8}+\frac{1}{8}\left(\underbrace{\left.x\frac{1}{x^2+1}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{x^2+1}dx\right) = \frac{\pi}{8} -\frac{1}{8}\underbrace{\int_0^{+\infty}\frac{1}{x^2+1}}_{=\frac{\pi}{2}} = \frac{\pi}{8}-\frac{\pi}{16} = \frac{\pi}{16}.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
"answer_id": 2
} |
Prove $\sum_{k=1}^n \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}$ by induction I'm not sure if my induction proof is correct. If anyone could kindly review it, I'd appreciate it.
Prove that $n \in \mathbb{N}$:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$
For $n=1$ the statement is true because $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
Let's assume that the statement is true for some natural number $k$, then:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{k(k+1)}=1-\frac{1}{k+1}$
Therefore,
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}
$$
$$=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
$$=1+\frac{1}{k+1}\cdot(\frac{-k-1}{k+2})$$
$$=1-(\frac{1}{k+1})\cdot\frac{k+1}{k+2}$$
$$=1-\frac{1}{(k+1)+1}$$
| There aren't a mistake in your third equalty. So you proof is right, clean and Nice, good work Enzo.
For $n=1$ the statement is true because $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
Here you are Right
Let's assume that the statement is true for some natural number $k$, then:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{k(k+1)}=1-\frac{1}{k+1}$
Your inductive hypotesis looks right.
Here is the critical step
Therefore,
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}
$$
$$=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
well at this part
$$=1+\frac{1}{k+1}\cdot(\frac{-k-1}{k+2})$$
Here there aren't any mistake (Thanks to @Nem49 for notice my fault) so you approach is correct, clean and valid (I don't look that you simplify the expression so well and basically you do a nice work in your proof. Good work Enzo and continue working hard in your questions as now you show you big effort and welcome to Math SE. I'm apologize with you and the community ).
The way that you chose Is right and done, but other way is proceed as follow (which is more work that the work that you already do done , but is other approach for get the same result ).
You can also proceed as follow:
You know by the hypotesis
$$=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{(k+1)(k+2)}{(k+1)(k+2)}-\frac{(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{k^2+3k+2-k-2+1}{(k+1)(k+2)}$$
$$=\frac{k^2+2k+1}{(k+1)(k+2)}$$
$$=\frac{(k+1)^2}{(k+1)(k+2)}$$
$$=\frac{k+1}{k+2}$$
$$=\frac{k+2-1}{k+2}$$
$$=\frac{k+2}{k+2}-\frac{1}{k+2}$$
$$=1-\frac{1}{(k+1)+1}$$
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
| Just take the difference
of the expression and its
proposed limit
and conjugate that.
Works out very nicely.
$\begin{array}\\
d(n)
&=\sqrt{(n+a)(n+b)}-(n+\frac{a+b}{2})\\
&=(\sqrt{(n+a)(n+b)}-(n+\frac{a+b}{2}))\dfrac{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{(n+a)(n+b)-(n+\frac{a+b}{2})^2}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{n^2+(a+b)n+ab-(n^2+n(a+b)+\frac{(a+b)^2}{4})}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{ab-\frac{(a+b)^2}{4}}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{-\frac{(a-b)^2}{4}}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{-(a-b)^2}{4\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2}))}\\
\text{so}\\
|d(n)|
&=|\dfrac{(a-b)^2}{4(\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2}))}|\\
&\lt|\dfrac{(a-b)^2}{4(2n)}|\\
&=|\dfrac{(a-b)^2}{8n}|\\
&\to 0\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/4364410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\sum_{i=1}^ni\sin{ix}=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{(n+1)\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}$ by induction Prove by induction that:
$$\sum_{i=1}^ni\sin{ix}=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{(n+1)\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}$$
I tried to prove it for the basis $n=1$:
For the left side, it's $\sin{x}$
The right side is:
$$A=\frac{\sin(2x)}{4\sin^2{\frac{x}{2}}}-\frac{2\cos{\frac{3x}{2}}}{2\sin{\frac{x}{2}}} = \frac{2\sin{x}\cos{x}-4\sin{x}\cos{\frac{3x}{2}}}{4\sin{\frac{x}{2}}}$$
$$= 2\sin{x}(\frac{\cos{x}-2\cos{\frac{3x}{2}}}{4\sin^2{\frac{x}{2}}})=2\sin{x}(\frac{\cos^2{\frac{x}{2}}}{4\sin^2{\frac{x}{2}}}-\frac{1}{4}-\frac{\cos{\frac{3x}{2}}}{2\sin^2{\frac{x}{2}}})$$
Then I showed that
$$\frac{\cos^2{\frac{x}{2}}}{4\sin^2{\frac{x}{2}}}=\frac{1-sin^2{\frac{x}{2}}}{4\sin^2{\frac{x}{2}}}=\frac{1}{4\sin^2{\frac{x}{2}}}-\frac{1}{4}$$
So:
$$A=2\sin{x}(\frac{1-2\cos{\frac{3x}{2}}}{4\sin^2{\frac{x}{2}}}-\frac{1}{2})$$
Then I tried to prove that the expression inside the parentheses is equal to $\frac{1}{2}$ i.e. the first fraction is equal to 1 i.e. the numerator is equal to the denominator. But I couldn't.
| Okay, as you say prove by induction...
Base case $n=1$, we need to prove:
$$\sin x=\frac{\sin{(2x)}}{4\sin^2{\frac{x}{2}}}-\frac{2\cos{(\frac{3x}{2})}}{2\sin{\frac{x}{2}}}.$$
We have
$$\frac{2\cos{(\frac{3x}{2})}}{2\sin{\frac{x}{2}}}=\frac{2\cos{(\frac{3x}{2})}\sin \frac x2}{2\sin^2{\frac{x}{2}}}=\frac{\sin 2x-\sin x}{2\sin^2{\frac{x}{2}}}.$$
So we have
$$RHS=\frac{\sin{(2x)}}{4\sin^2{\frac{x}{2}}}-\frac{\sin 2x-\sin x}{2\sin^2{\frac{x}{2}}}=\frac{2\sin x-\sin 2x}{2(1-\cos x)}=\frac{2\sin x-2\sin x\cos x}{2(1-\cos x)}=\sin x=LHS.$$
Induction step, we need to prove:
\begin{align}
n\sin{nx}&=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{(n+1)\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}-\frac{\sin{((n)x)}}{4\sin^2{\frac{x}{2}}}+\frac{(n)\cos{((2n-1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}
\\
&=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{\sin{((n)x)}}{4\sin^2{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\\
&+n\left(\frac{\cos{((2n-1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\right).
\end{align}
We know that
\begin{align}
&\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{\sin{((n)x)}}{4\sin^2{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\\
=&\frac{\sin{((n+1)x)}-\sin(nx)}{4\sin^2{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\\
=&\frac{2\sin\frac x2 \cos\frac{2n+1}2x}{4\sin^2{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\\
=&\frac{ \cos\frac{2n+1}2x}{2\sin{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}=0.
\end{align}
Also we have
\begin{align}
&\frac{\cos{((2n-1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\\
=&\frac{\cos{((2n-1)\frac{x}{2})}-\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}
=\frac{2\sin nx\sin \frac x2}{2\sin{\frac{x}{2}}}=\sin nx.
\end{align}
Therefore, we have
$$RHS=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{\sin{((n)x)}}{4\sin^2{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\\
+n\left(\frac{\cos{((2n-1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}-\frac{\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}\right)=0+n\sin nx=n\sin nx=LHS.$$
So, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and $a$ is positive.
First of all, let’s start with the ‘simple’ case, $$
I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}.
$$
However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula
$$
\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
which I can’t prove by elementary method yet. Please tell me if you have any.
Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]}
=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$
Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right)
\cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}}
\end{aligned}
$$
Rearranging and simplifying yields
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)}
$$
Putting $a=1$ gives our formula
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)}
$$
For verification, $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26
=\frac{1118260 \pi}{4782969 \sqrt{3}},
\end{aligned}
$$
which is checked by Wolframalpha
$$\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\
&=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\
&\doteq 0.845906950943631,
\end{aligned}
$$
which is checked by Wolframalpha
My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$
| You could make it faster using
$$I=\int \frac{d x}{\left(x^{n}+a\right)^{m}}=x\, a^{-m} \, _2F_1\left(m,\frac{1}{n};1+\frac{1}{n};-\frac{x^n}{a}\right)$$ which gives
$$J=\int_0^\infty \frac{d x}{\left(x^{n}+a\right)^{m}}=a^{\frac{1}{n}-m}\,\frac{\Gamma \left(1+\frac{1}{n}\right) \Gamma
\left(m-\frac{1}{n}\right)}{\Gamma (m)}$$ provided $\Re(m n)>1\land \Re(n)>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Geometrical meaning of $x^2+y^2+z^2-xy-xz-yz$ I am looking for a geometrical interpretation of the symmetrical expression
$$f=x^2+y^2+z^2-xy-xz-yz\tag{1}$$
with $x,y,z \in \mathbb{R}$. I could for example $f$ interprete as dot products of a vector with its permuted vector
$$f=\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}-\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}y\\z\\x\end{pmatrix}\tag{2}$$
however I think there are more symmetrical ways to represent $f$ geometrically. Maybe $x,y,z$ can be thought of sides of a triangle, etc. Geometrical interpretations in $\mathbb{R},\mathbb{R^2}, \mathbb{R^3}$ are of interest.
| Render $(x-y)^2=x^2-2xy+y^2$ and similarly for the $x,z$ and $y,z$ pairs. When you add up the squares you get
$2(x^2+y^2+z^2-xy-xz-yz).$
So $x^2+y^2+z^2-xy-xz-yz$ is half the sum of squared distances between three points on a line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve $x^2+2^{17}=y^3$ over the integers I am asked to solve the equation $x^2+2^{17}=y^3$ over the integers. The technique I was taught for such equations uses algebraic numbers and a little algebraic number theory/ring theory depending on how you look at it. However, I got stuck at some point because I have never had such big numbers as $2^{17}$ in this type of equations and I really don't know what to do.
So, I began by rewriting the equation as $(x+256\sqrt{-2})(x-256\sqrt{-2})=y^3$. This is good because $\mathbb{Z}[\sqrt{-2}]$ is an Euclidean domain, so it is an UFD in particular.
Now the next step I was taught is to look at $d=\operatorname{gcd}(x+256\sqrt{-2}, x-256\sqrt{-2})$. It is easy to see that $d | 512\sqrt{-2}$. Now $2$ and $(\sqrt{-2})^2$ are associates (I will denote this by $2\sim (\sqrt{-2})^2$, I don't know if this is standard), so $d\sim (\sqrt{-2})^{19}$.
Thus, $d\sim (\sqrt{-2})^{j}$ for $j\in \{0, 1, 2, ..., 19\}$. I could even eliminate the associate sign and just write $d=\pm (\sqrt{-2})^j$, but it doesn't really make much of a difference.
Now I suppose that I am to see which $j$ work and which don't. The case $j=0$ is the easiest one, so I tried assuming that $j\ge 1$ to see what happens. Then $\sqrt{-2}|d|x\pm 256\sqrt{-2}|y^3$. This easily implies that $x$ and $y$ are both even, so $x=2x_1$ and $y=2y_1$ for some integers $x_1$ and $y_1$.
Substituting into the equation this gives $x_1^2+2^{15}=2y_1^3$. Thus, $x_1^2$ is even, so $x_1$ is even. This implies that $4|x_1^2$, so $2|y_1^3$. From here we get that $2|y_1$, so $8|y_1^3$. After all of this, $16|x_1^2$, so $4|x_1$. Now we write $x_1=4x_2$ and $y_1=2y_2$ for some integers and we get that $$x_2^2+2^{11}=y_2^3.$$ Here is where I am stuck. This equation looks just like the one at the beginning. I could write $d=\operatorname{gcd}(8(x+32\sqrt{-2}), 8(x-32\sqrt{-2}))\sim 8\operatorname{gcd}(x+32\sqrt{-2}, x-32\sqrt{-2})$ and this basically tells me that if $j\ge 1$ then $j\ge 6$ in fact, but that's all I can do.
| Note, your approach is about powers of $2$ that divide $x.$ How many factors of $\sqrt{-2}$ divide $x+512\sqrt{-2}$ is entirely about powers of two that divide $x.$
Review
Let $\nu_2(m)$ be the largest integer $n$ such that $2^n\mid m.$
A few key results:
$$\nu_2(m^k)=\nu_2(m)^k$$
and,
If $\nu_2(m_1)\ne \nu_2(m_2)$ then: $$\nu_2(m_1+m_2)=\min(\nu_2(m_1),\nu_2(m_2))$$
In your equation, $\nu_2(y^3)$ is a multiple of $3$ and, since $\nu_2(x^2)$ is even, it can't equal $\nu_2(2^{17})=17.$ So $$\nu_2(x^2+2^{17})=\min(2\nu_2(x),17).$$
So $\nu_2(x)\leq 8$ and must be divisible by $3.$
Likewise, $\nu_2(y)$ must be even.
So either:
*
*$x$ is odd
*$x=8x_1,$ $x_1$ odd.
*$x=64x_2,$ $x_2$ odd.
So either:
*
*$x$ is odd and $x+512\sqrt{-2}$ is a perfect cube in $\mathbb Z[\sqrt{-2}].$
*$x_1$ is odd and $x_1+32\sqrt{-2}$ is a perfect cube
*$x_2$ is odd and $x_2+4\sqrt{-2}$ is a perfect cube.
Solve each of these independently.
Expanding $(a+b\sqrt{-2})^3,$ you get the equation:
$$b(3a^2-2b^2)=2^{3k+2}$$ for some $a,b$ and $k=0,1,2.$
So $b=\pm 2^i$ for some $i\leq 3k+2.$ Plugging in and re-arranging, you get:
$$3a^2=2^{2i+1}\pm 2^{3k+2-i}$$ Since $\nu_2(3a^2)$ is even, and $3k+2-i\neq 2i+1$ you must have $2i+1>3k+2-i$ or $i>k,$ and $3k+2-i$ must be even. Modulo $3,$ $2^{2i+1}\equiv 2$ and $2^{3k+2-i}\equiv 1,$ so the sign in the expression is $+.$
So: $$3a^2=2^{3k+2-i}\left(2^{3(i-k)+1}+1\right)$$
And you need $k<i\leq 3k+2$ and $k\equiv i\pmod 2$ and $$2^{3(i-k)+1}+1=3a_0^2.$$
If $i=k+2j,$ this becomes $0<i\leq k+1$ and: $$2^{6j+1}+1=3a_0^2.$$
But none of $2^{7}+1, 2^{13}+1, 2^{19}+1$ is three times a square. In fact, all three are of the form $3p,$ where $p$ is prime.
So there are no solutions.
| {
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"url": "https://math.stackexchange.com/questions/4371783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving Math Limits Questions with Square Roots I seek your help on this question.
$$\lim_{x\to\infty}\left(\sqrt{9x^6+7x^3+5x}-\sqrt{9x^6+6x^2+3x}\right)$$
This is what I have done so far.
I tried to divide both numerator and denominator with the largest power but I am not too sure if I am doing it right. Please help! Thank you!
| We know that, by asymptotic relations:
$$(1+f(x))^{\alpha} \;\; \sim \;\; \alpha \cdot f(x) \;\; \text{when}\;\; f(x) \to 0$$
One can write:
$$\lim_{x\to\infty}\left(\sqrt{9x^6+7x^3+5x}-\sqrt{9x^6+6x^2+3x}\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{\frac{9x^6+7x^3+5x}{9x^6+6x^2+3x}}-1\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{\frac{9x^6+6x^2+3x+(7x^3-6x^2)}{9x^6+6x^2+3x}}-1\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{1+\frac{7x^3-6x^2}{9x^6+6x^2+3x}}-1\right)\;\; \sim\;\;\lim_{x\to\infty}3x^3\cdot\left(\sqrt{1+\frac{7x^3}{9x^6}}-1\right)\;\; \lim_{x\to\infty}3x^3\cdot\frac{1}{2}\cdot\frac{7}{9x^3}=\frac{7}{6}$$
So, the limit is $\frac{7}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit: $ \lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}$
Calculate the limit:
$$\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}
$$
I tried to change to polar coordinates like that:
\begin{align*}
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}} &=
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2-2x^2y^2}}\\
&=\lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}\\
\end{align*}
and I'm not sure how to continue from this point.
Thank you!
| Hint: $x^{4}+y^{4} \geq \frac 1 2 ({x^{2}+y^{2}})^{2}$. Now use polar coordinates to see that the limit is $0$.
This allows you to get rid of $\theta$. You will also need the fact that $t^{4}e^{-t} \to 0$ as $ t \to \infty$ (Take $t=\frac 1r$). This fact is proved using L'Hopital's Rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$.
Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$. Then $z+\frac{1}{y}=\frac{m}{n}$, where $m$ and $n$ are coprime. Find $m+n+1$.
I tried this, $$y=29-\frac{1}{x}$$ $$y=\frac{29x-1}{x}$$
And, $$x+\frac{1}{z}=5$$ $$z=\frac{1}{5-x}$$
What can I do next?
| Just substitute the values of $y$ and $z$ in $xyz=1$.
$$\frac{x(29x-1)}{x(5-x)}=1$$
$$29x-1=5-x$$
$$x=\frac{1}{5}$$
$$y=\frac{29/5-1}{1/5}=24$$
$$z=\frac{1}{5-1/5}=\frac{5}{24}$$
$$z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$$
Thus, $$m+n+1=1+4+1=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve derivative/limit of $f(x)=x\sqrt{4-x^2}$ I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.
So it would be something like
$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$
I was trying to solve and I just can end up with something like
$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
Now if I group on h, I will have some tricky 3 instead of 2.
The idea is I should have something like $h(2x^2+4)$ that would cancel up.
I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".
The final result shall be
$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$
| As Spivak points out in his Calculus, the proofs of limit theorems are strategies for implementing the definition of a limit.
Here, guided by the trick
\begin{align*}
\frac{f(x + h)g(x + h) - f(x)g(x)}{h}
&= \frac{f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x)}{h} \\
&= f(x + h)\frac{g(x + h) - g(x)}{h} + \frac{f(x + h) - f(x)}{h}g(x)
\end{align*}
in the proof of the power rule, we can add and subtract from the numerator either of
$$
f(x + h)g(x) = (x + h)\sqrt{4 - x^{2}},\qquad
f(x)g(x + h) = x\sqrt{4 - (x^{2} + 2xh + h^{2})}.
$$
(Alex's (+1) answer is another approach, but the one here applies more generally.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Rectangle in a circle of radius a that maximizes x^n+ y^n
Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $$x^2 + y^2 = a^2,$$ and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $$z = x^n + y^n.$$ If $n = 2$, it is clear that $z$ has the constant value $a^2$ for all rectangles.
*
*If $n < 2$, show that the square maximizes $z$; and
*if $n > 2$, show that $z$ is maximized by a degenerate rectangle in which $x$ or $y$ is $0$.
My approach.
$$x^2 + y^2 = a^2$$
$$\therefore\, y = (a^2 - x^2)^{1\over 2}$$
\begin{align}
z &= x^n + y^n \\
{d\over dx}z & = nx^{n-1} - nx(a^2-x^2)^{n-2\over 2}
\end{align}
Equating with $0$ gives:
$$nx^{n-1} = nx(a^2-x^2)^{n-2\over 2}$$
$$x^{n-2} = (a^2 - x^2)^{n-2\over 2}$$
Squaring on both sides gives:
$$(x^2)^{n-2} = (a^2 - x^2)^{n-2}$$
Taking $\ln$ on both sides gives
$$(n-2) \ln(x^2) = (n-2)\ln(a^2 - x^2)$$33
Equating $x^2 = a^2 - x^2$ , we get $x = {a\over \sqrt{2}}$ and $y = {a\over\sqrt{2}}$. In particular $x=y$ implies that the optimal rectange is a square.
My computation appears to be independent of $n$, so I am puzzled about the statement in the question for $n>2$ and $n<2$.
| Alternative way, without calculus.
*
*If $\,n \le 2\,$ then by the generalized means inequality $\,\left(\frac{x^n+y^n}{2}\right)^{1/n} \le \left(\frac{x^2+y^2}{2}\right)^{1/2}=\frac{a}{\sqrt{2}}\,$. The maximum is attained when the equality is achieved, for $\,x=y\,$ i.e. the square.
*If $\,n \gt 2\,$ then $\,x^n+y^n=(x^2)^{n/2}+(y^2)^{n/2} \le (x^2+y^2)^{n/2} = a^n\,$ with equality for $\,x=0\,$ or $\,y=0\,$, since $\,x^2/a^2, \,y^2/a^2 \le 1\,$ so $\,(x^2/a^2)^{n/2}+(y^2/a^2)^{n/2} \le x^2/a^2+y^2/a^2 = 1\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the points whose distance from the center of the square = distance from the boundary.
Let's find the locus of the boundary of the shaded region
From the second figure
${\sqrt {h^2+k^2} } = {\sqrt {(h-h)^2 + (k-{a\over 2})^2 }}$
This simplifies to be:
$k = {a^2 - 4h^2\over 4a}$
$y = {a^2 - 4x^2\over 4a}$
Also the curve intersects the line( the hypotenuse of the triangle) y= x
For point of intersection :
$ {a^2 - 4x^2\over 4a} = x $
$4x^2 +4ax - a^2 = 0$
$x = {-4a ± \sqrt{16a^2 - 4×4×(-a^2)}\over 8}$
$x = a{(\sqrt{2} ± 1)\over 2}$
Solution 1: $x = a{(\sqrt{2} + 1)\over 2}$
Solution 2: $x = a{(\sqrt{2} - 1)\over 2}$
Solution 1 can be discarded as $ x = a{(\sqrt{2} + 1)\over 2} ≈ 1.207106 a > {a\over 2} $
Solution 2: $ a{(\sqrt{2} - 1)\over 2} ≈ 0.2071 a < {a\over 2} $
from 0 to $a{(\sqrt{2} - 1)\over 2}$ :
The curve (boundary of shaded region) lies above the line $ y=x$
so $dA = ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
$dA = ({a\over 4} - {x^2\over a} - x)dx $
$\int_{0}^{A} dA = \int_{0}^{a{(\sqrt{2} - 1)\over 2}} ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
A = $({a\over 4}x - {x^3\over 3a} - {x^2\over 2})]_{0}^{{a{(\sqrt{2} - 1)\over 2}}}$
$ A = {a^2\over 8}(\sqrt{2}-1) - {a^2\over 24}(\sqrt{2}-1)^3 - {a^2\over 8}
(\sqrt{2}-1)^2 $
$ = {a^2\over 8}(\sqrt{2}-1) \Biggl( 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)\Biggr)$
This simplifies to be equal to
${a^2\over 8}(\sqrt{2}-1)({3+5\sqrt{2}\over 3})$
$= {(7- 2\sqrt{2})\over 8×3}a^2$
Area of shaded figure = 8A
$A_{total}$ = ${(7- 2\sqrt{2})\over 3}a^2$
But the answer is :
${4\sqrt{2}-5\over 3}a^2 $
I don't know where I got it wrong, and also I have re calculated this and the result is same.
Did I miss something important or calculated wrongly
Any help of hint or suggestion or worked out solution would be appreciated.
| Note that
$$ 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)= \frac{3-\sqrt2}3
$$
not ${3+5\sqrt{2}\over 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Alternative way to solve a limit problem $$
\lim _{n \rightarrow \infty} \frac{1}{1+n^{2}}+\frac{2}{2+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
I want to find the limit of this infinite series which I found in a book. The answer is $1/2$.
The solution to this limit was given by Sandwich/Squeeze Theorem, which was basically that the above function lies between:
$$
\frac{1}{n+n^{2}}+\frac{2}{n+n^{2}}+\frac{3}{n+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
And,
$$
\frac{1}{1+n^{2}}+\frac{2}{1+n^{2}}+\cdots+\frac{n}{1+n^{2}}
$$
series and the limit of both of these series tend to $1/2$ as $n \to \infty$.
I fully understood the solution, but I find that this isn't something that naturally/intuitively comes to your mind. I mean we need to find two different series by trial and error, both of which need to converge to a single number.
Is there any different solution to this limit problem, like dividing by powers of n, or maybe telescoping sums?
| This answer uses the same idea as the answers of TurlocTheRed and Koro, but shows that one doesn't need infinite series or big-$O$ notation. Each denominator is extremely close to $n^2$, so let's compare explicitly:
\begin{align*}
\sum_{k=1}^n \frac{k}{k+n^2} &= \sum_{k=1}^n \frac{k}{n^2} - \sum_{k=1}^n \bigg( \frac{k}{n^2} - \frac{k}{k+n^2} \bigg) \\
&= \frac1{n^2} \sum_{k=1}^n k - \sum_{k=1}^n \frac{k^2}{n^2(k+n^2)}.
\end{align*}
By standard formulas, the first sum is exactly $\frac12(1+\frac1n)$; while the second sum is positive and bounded above by the sum
$$
0 < \sum_{k=1}^n \frac{k^2}{n^2(k+n^2)} < \sum_{k=1}^n \frac{k^2}{n^2(0+n^2)} = \frac1{n^4} \sum_{k=1}^n k^2 = \frac1{6n}\bigg(1+\frac1n\bigg)\bigg(2+\frac1n\bigg).
$$
The desired limit now follows from the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
Solving recursive formula dependent on a periodic function Say we have some initial values $a_1, a_2, \dots, a_k$ and a recursive formula $a_n = f(n) + c_{1}a_{n-1} + c_{2}a_{n-2} + \dots + c_{k}a_{n-k}$ Where $f(n)$ is periodic and $a_n,f(n),c_i \in \mathbb{N}$.
For example let $a_1 = 1$ and $a_n = f(n) + 2a_{n-1}$ where $f(n) = 1,2,1,2 \dots$
How can we generally solve this kind of formula?
I was initially thinking to solve the linear recurrence relation for each value of $f(n)$ and then merge the subsequences, but not sure how this should be done. The main issue is how to describe the extra shifting done each time we get a higher value for $f(n)$
In the example we can solve $a_{n,1} = 1 + 2a_{n-1,1}$ and get $a_{n,1} = 2^n - 1$
And solve $a_{n,2} = 2 + 2a_{n-1,2}$ and get $a_{n,2} = 3\cdot2^{n-1} - 2$
Though I am not sure how to merge them to get $a_n$ which first values are $a_n= 1, 3, 8, 17, 36,..$ given by multiplying by $2$ and adding $1$ or $2$ alternately.
| Suppose $\,f(n)\,$ has period $\,K\,$, then eliminating $\,f(n+K)=f(n)\,$ between the two recurrence relations for $\,a_n+K\,$ and $\,a_n\,$ gives a linear homogeneous recurrence that no longer includes $\,f(n)\,$, which can be solved with the usual methods.
$$
\begin{cases}
\begin{align}
a_{n+K} &= f(n+K) + c_{1}a_{n+K-1} + c_{2}a_{n+K-2} + \dots + c_{k}a_{n+K-k}
\\ a_n &= f(n) + c_{1}a_{n-1} + c_{2}a_{n-2} + \dots + c_{k}a_{n-k}
\end{align}
\end{cases}
$$
$$
\begin{align}
\implies\quad\quad a_{n+K} = a_n &+ (c_{1}a_{n+K-1} + c_{2}a_{n+K-2} + \dots + c_{k}a_{n+K-k}) \quad\quad\quad\quad\quad\quad
\\ &- (c_{1}a_{n-1} + c_{2}a_{n-2} + \dots + c_{k}a_{n-k})
\end{align}
$$
It can be shown that if the characteristic polynomial of the original recurrence is $\,p(t)\,$ then the characteristic polynomial of the derived recurrence is $\,(t^K-1)\,p(t)\,$.
For example let $a_1 = 1$ and $a_n = f(n) + 2a_{n-1}$ where $f(n) = 1,2,1,2 \dots$
This is the case $\,k=1\,$, $\,p(t)=t-2\,$, $\,K=2\,$.
$$
\begin{cases}
\begin{align}
a_{n+2} &= f(n+2) + 2a_{n+1}
\\ a_n &= f(n) + 2a_{n-1}
\end{align}
\end{cases}
$$
$$
\implies\quad\quad a_{n+2} = 2a_{n+1} + a_n - 2a_{n-1} \quad\quad\;\;
$$
The characteristic polynomial is $\,t^3-2t^2-t+2 = (t^2-1)(t-2)\,$, with roots $\,\pm 1, 2\,$, so $\,a_n = C_1 + C_2 \,(-1)^n + C_3 \,2^n\,$ where $\,C_1,C_2,C_3\,$ are determined from the initial conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
How to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$? I want to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$.
My attempt:
We know that $\sqrt{ab}≤\frac{a+b}{2}$ because $(\sqrt{a}-\sqrt{b})^2≥0$ and that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$ by expanding $(1+x)^n$ and substituting $x=1$.
Next, I re-wrote $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2$ as $\binom{n}{1}+\binom{n}{2}+...+\binom{n}{1}+\sum_{0<i<j}\sqrt{\binom{n}{i}\binom{n}{j}}$
$≤2^n-1+\sum_{0<i<j}\frac{\binom{n}{i}+\binom{n}{j}}{2}$ using the two term $A_m G_m$ inequality.
$=2^n-1+\frac{1}{2}\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{1}+\binom{n}{3}+...+\binom{n}{1}+\binom{n}{n}+\binom{n}{2}+\binom{n}{3}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{n-1}+\binom{n}{n}\right)$
$=2^n-1+\frac{n}{2}\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n}\right)$
$=2^n-1+\frac{n}{2}(2^n-1)$
$=\frac{n+2}{2}(2^n-1)$, which is clearly not the result.
Could someone point out the mistake in my working and help me finish the proof?
| Hint: Use the CS inequality: $(1\cdot x_1+1\cdot x_2+\cdots+1\cdot x_n)^2 \le (1+1+\cdots+1)(x_1^2+x_2^2+\cdots+x_n^2)$, with $x_k = \sqrt{\binom{n}{k}}$
| {
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"url": "https://math.stackexchange.com/questions/4388240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How many ways you arrange $12$ marbles where two marbles of the same color are considered the same? Problem:
Suppose that we have $6$ red marbles, $4$ blue marbles, $2$ green marbles and
$2$ black marbles. How many different ways are there to arrange them?
Note: Two marbles of the same color are considered identical.
Answer:
Let $c$ be the count we seek.
\begin{align*}
c &= \dfrac{ 12! }{6!4!2!2!} \\
c &= \dfrac{ 12(11)(10)(9)(8)(7)}{ 4(3)(2)(2)(2)} \\
c &= \dfrac{ 11(10)(9)(8)(7)}{ 2(2)(2)} = 11(10)(9)(7) \\
c &= 6930
\end{align*}
Is my solution correct?
| Your solution is correct.
We can choose six of the twelve positions for the red marbles in $\binom{12}{6}$ ways, four of the remaining positions for the blue marbles in $\binom{6}{4}$ ways, and two of the remaining four positions for the green marbles in $\binom{4}{2}$ ways. Both of the remaining positions must be occupied by the two black marbles. Hence, the number of distinguishable arrangements of $6$ red, $4$ blue, $2$ green, and $2$ black marbles is
$$\binom{12}{6}\binom{6}{4}\binom{4}{2} = \frac{12!}{6!6!} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} = \frac{12!}{6!4!2!2!} = 6930$$
as you found. The factors in the denominator represent the number of ways marbles of the same color could be permuted among themselves in a particular permutation without producing a permutation which is distinguishable from that permutation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate $\sum^{\infty}_{r=1} \frac{8}{(2r-1)\sqrt{(2r+3)(2r+5)}+\sqrt{(2r+1)(2r-1)}(2r+3)}$ If $\sum^{\infty}_{r=1} \frac{8}{(2r-1)\sqrt{(2r+3)(2r+5)}+\sqrt{(2r+1)(2r-1)}(2r+3)}=\sqrt{a}+\sqrt{\frac{5}{3}}-\sqrt{c}$, where $a$ and $c$ are coprime numbers, then the value of $a+\frac{c}{4}$ is
(1) 3
(2) 5
(3) 8
(4) 4
I'm completely lost here, some pointers would be nice.
I tried putting it into Wolfram Alpha to see if it could get a partial sum formula, but that didn't seem to work.
And putting in the entire series itself just gave me a decimal approximation (unless I'm missing something).
| Hint: Use telescoping series. Multiply numerator and denominator by $\sqrt{\frac{2r+5}{2r+3}}-\sqrt{\frac{2r+1}{2r-1}}$, to get final simplified expression for $t_{r}$ as under.
$$\begin{aligned}t_{r}&=\frac{8}{(2r-1)(2r+3)\left[\sqrt{\frac{2r+5}{2r+3}}+\sqrt{\frac{2r+1}{2r-1}}\right]} \\ &= \sqrt{\frac{2r+1}{2r-1}}-\sqrt{\frac{2r+5}{2r+3}}\end{aligned}$$
Now define $a_{r}=\sqrt{\frac{2r+1}{2r-1}}$, so that $t_{r}=a_{r}-a_{r+2}$. The sum would be $\sum_{r=1}^{\infty}(a_{r}-a_{r+2})=a_{1}+a_{2}-2a_{\infty}$. Since $a_{\infty}\to 1$, $a_{1}=\sqrt{3}$ and $a_{2}=\sqrt{5/3}$. So the answer seems to be $\sqrt{3}+\sqrt{\frac{5}{3}}-\sqrt{4}\approx 1.01808$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiability of $f(x,y)= \frac{xy^3}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $0$ for $(x,y)=(0,0)$.
Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ with\begin{align*} f:=\begin{cases} \frac{xy^3}{x^2+y^2},&(x,y)\neq (0,0)\\0, & (x,y)=(0,0). \end{cases} \end{align*}
Show that $f$ is differentiable at $(0,0)$.
My approach:
We prove that ${0\choose 0}$ is the (total) derivative at point $(0,0)$.
If $|x|\geq|y|>0$ then
$$
\Big|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\Big|\leq\frac{|x||y|y^2}{(2y^2)|y|\sqrt{2}}=\frac{|x|}{4}.
$$
If $|y|\geq|x|>0$ then
$$
\Big|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\Big|=\frac{|y|}{(\frac{x^2}{y^2}+1)\sqrt{1+\frac{y^2}{x^2}}}\leq |y|.
$$
Keeping those bounds in mind we look at the limit:
\begin{align*} \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{f(x,y)-f(0,0)-{0\choose 0}\left({x\choose y}-{0\choose 0}\right)}{\Vert{x\choose y}-{0\choose 0}\Vert_2 }=\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{\frac{xy^3}{x^2+y^2}}{\Vert{x\choose y}\Vert_2 }=\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}},\text{ where } (x,y)\neq (0,0).
\end{align*}
If $x=0$ or $y=0$ then the above limit is $0$. If $|y|\geq|x|>0$ or $|x|\geq|y|>0$ then we use the above upper bounds and see that the limit is again $0$. Hence, $f$ is (total) differentiable at $(0,0)$.
Is this correct? Our tutor told us that when it comes to total differentiability we should rather prove that the partials are continuous. Maybe in this case it was just luck that one could easily see the matrix/derivative.
| Noting
$$ \frac{|x|}{\sqrt{x^2+y^2}}\le 1, \frac{y^2}{x^2+y^2}\le1, $$
one has
$$ \bigg|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\bigg|=|y|\frac{|x|}{\sqrt{x^2+y^2}}\cdot\frac{y^2}{x^2+y^2}\le|y|$$
which gives
$$ \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}=0. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Limits Question $$\lim _{x\to 0}\frac{\left(\cos \left(\frac{\pi }{2\cos \left(x\right)}\right)\right)}{\sin \left(\sin \left(x\right)^2\right)}=l$$
then find the value of {l}, where {l} denotes fractional part.
I tried multiplying and dividing the denominator with $\sin \left(x\right)^2$ and $x^2$ to simply the denominator, transformed the cos function to sin, repeated the multiplying and dividing once again then I got stuck after that. I reached here, I am not sure how to proceed.
$$\lim _{x\to 0}\frac{\left(\sin \left(\frac{\pi }{2}-\frac{\pi }{2\cos x}\right)\right)}{\frac{\left(\sin \:\left(\sin \:\left(x\right)^2\right)\sin \:\left(x\right)^2x^2\right)}{\sin \left(x\right)^2x^2}}$$
$$=\lim _{x\to 0}\frac{\left(\frac{\sin \left(\frac{\pi }{2}-\frac{\pi }{2\cos x}\right)}{\frac{\pi }{2}-\frac{\pi }{2\cos \left(x\right)}}\right)\left(\frac{\pi \:}{2}-\frac{\pi \:}{2\cos \left(x\right)}\right)}{x^2}$$
$$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}$$
$$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{x^2}$$
Could anyone please explain how to proceed with this question and correct any mistakes up to what I solved if there are any?
|
$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{x^2}$
This step is incorrect .It should have been:
$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{1}$
as $\lim _{x\to 0} \frac{\ cosx-1}{x^2}=-\frac{1}{2} $ and $\lim _{x\to 0} {\cos x=1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving PDE using characteristic method without polar coordinate. I'm trying to solve this PDE using method of characteristic without coordinate transformation :
$xu_x+yu_y = \sqrt{x^2+y^2}$ for $x^2+y^2>1$, $u(x,y)=x$ on $x^2+y^2 = 1$.
I only know how to use the method of characteristic for this kind of PDE such that initial condition is constant in one variable e.g., $u(0,y) = $ or $u(x,1) = $. I tried to mimick the technique to solve the above pde : Introducing new variable $(\xi,\eta)$ such that $\xi = x,\eta = y$ on $x^2+y^2 = 1$ and using chain rule,
$${\partial u\over\partial \eta} = {\partial u\over\partial x}{\partial x\over\partial\eta}+{\partial u\over\partial y}{\partial y\over\partial\eta} = \sqrt{x^2+y^2}$$
so ${\partial x\over\partial\eta}= x,{\partial y\over\partial \eta} = y$. So $x = \phi_1(\xi)e^{\eta}$ and $y = \phi_2(\xi)e^{\eta}$ ..? I'm stuck here and I'm not sure I'm in the right direction. Please help.
| $$xu_x+yu_y = \sqrt{x^2+y^2}$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{\sqrt{x^2+y^2}}$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ :
$$y=c_1x$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{\sqrt{x^2+(c_1x)^2}}$
$$u-x\sqrt{1+c_1^2}=c_2$$
General solution of the PDE on the form of implicit equation $c_2=F(c_1)$ :
$$u-x\sqrt{1+\left(\frac{y}{x}\right)^2}=F\left(\frac{y}{x}\right)$$
$F$ is an arbitrary function.
$$\boxed{u(x,y)=\sqrt{x^2+y^2}+F\left(\frac{y}{x}\right)}$$
Condition : $u(x,y)=x$ on $x^2+y^2 = 1$
$$\sqrt{x^2+(1-x^2)}+F\left(\frac{\sqrt{1-x^2}}{x}\right)=x$$
$$F\left(\frac{\sqrt{1-x^2}}{x}\right)=x-1$$
Let $X=\frac{\sqrt{1-x^2}}{x}\quad\implies\quad x=\frac{1}{1+X^2}$
$$F(X)=\frac{1}{\sqrt{1+X^2}}-1$$
Now the function $F(X)$ is determined. We put it into the above general solution where $X=\frac{y}{x}$
$$u(x,y)=\sqrt{x^2+y^2}+\frac{1}{\sqrt{1+\left(\frac{y}{x}\right)^2}}-1$$
$$\boxed{u(x,y)=\frac{x^2+y^2+x}{\sqrt{x^2+y^2}}-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What rationale am I missing with this simple boundary value problem? Consider $$u_{xy} = xy,\ u(0,\ y) = 0,\ u_x(x,\ 0) = 0$$
Antidifferentiating the PDE with respect to $y$ yields $u_x = \frac{xy^2}{2} + a(x)$, and antidifferentiating this equation with respect to $x$ yields $u = \frac{x^2 y^2}{4} + A(x) + b(y)$, where $A(x)$ is the antiderivative of $a(x)$ with respect to $x$.
Now the boundary conditions produce the system $\begin{cases}0 = A(0) + b(y) \\ 0 = a(x)\end{cases}$. Antidifferentiating the second equation, $A(x) = C$, where $C$ is constant with respect to both $x$ and $y$ (because $A(x) = C(y)$ makes no sense, except in a trivial way when $C$ is constant). Plugging this into the the first equation, $0 = C + b(y)$, or lumping the minus sign, $b(y) = C$. This suggests the solution $u = \frac{x^2 y^2}{4} + C$, which might be thought to require another boundary condition to pin down $C$.
However, only $C = 0$ seems to work when checking the solution. Where is the mistake in my reasoning which leads to the conclusion that any $C$ should solve the problem?
| $$u(x,y) = \frac{x^2 y^2}{4} + A(x) + b(y) \quad\text{is OK.}$$
Condition $u(0,y) = 0$ :
$$u(0,y) = A(0) + b(y) \quad\implies\quad b(y)=-A(0)=\text{constant}$$
$$u(x,y) = \frac{x^2 y^2}{4} + A(x) -A(0)$$
Condition $u_x(x, 0) = 0$ :
$u_x(x,y)=\frac{x y^2}{2} + A'(x)$
$$u_x(x,0)= A'(x)=0$$
$$A(x)=\text{constant}=C\quad\implies\quad A(0)=C$$
$$u(x,y) = \frac{x^2 y^2}{4} + C -C$$
$$\boxed{u(x,y) = \frac{x^2 y^2}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can we prove that the equation $2^n+1=5^m$ has no integral solution other than $(2, 1)$? $\textrm{Assume, on the contrary, that }n\geq 4,\text{then }$
\begin{aligned} 2^{n} &=5^{m}-1 \\ &=(5-1)\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ &=4\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ 2^{n-2} &=5^{m-1}+5^{m-2}+\cdots+1 \\ 2^{n-2} & \equiv \underbrace{1+1+\cdots+1}_{m \text { ‘1’s }}(\bmod 4) \\ 0 & \equiv m \quad(\bmod 4) \\ \therefore \quad m &=4 k \text { for some integer } k . \end{aligned}
$\text {Again, }$
$\begin{aligned}2^{n} &=5^{4 k}-1 \\ \displaystyle \quad &=\left(5^{4}\right)^{k}-1 \\ \displaystyle \quad &=\left(5^{4}-1\right)\left(625^{k-1}+625^{k-2}+\cdots+1\right) \\ \displaystyle \quad &=624\left(625^{k- 1}+625^{k-2}+\cdots+1\right) \\&\Rightarrow 624 | 2^{n},\text {which leads to a contradiction. } \\\therefore \quad n=0,1,2 \text{ or }3.\end{aligned}$
After checking one by one, we can conclude that $(n, m)=(2,1) $
is the unique integer solution.
Your opinions and alternative proofs are highly appreciated.
| Another way to do this. Assume that $n \ge 3$. Then for the equation $5^m \pmod {2^n} = 1$ to hold for positive integral $m$, the equation $5^m \pmod 8 = 1$ must hold, which gives $m$ even, and $n \ge 3$ gives $m$ positive as well.
So it now remains to show that there are no positive integral [in $n$ and $m$] solutions to the equation
$$2^n = 5^m-1 = (5^k-1)(5^k+1),$$
where $n$ is an integer at least $3$, and $m$ is an even and positive integer, and thus $k = \frac{m}{2}$ is a positive integer i.e., difference of two squares.
If there were such solutions, both $5^k-1$ and $5^k+1$ would have to be powers of $2$, so let $b$ and $a$ be integers such that $5^k+1=2^b$ and $5^k-1=2^a$. Note however, that $(5^k+1)-(5^k-1)=2$, and that $2^b-2^a= 2^a(2^{b-a}-1)$, is $2$, only if $a=1$ and $b=2$. However, there is no integral $k$ that gives $5^k+1=4$ and $5^k-1=2$. So indeed, there are no solutions to the equation
$$2^n = (5^k-1)(5^k+1),$$
where $n$ is an integer at least $3$, and $k$ is a positive integer. Thus from the top paragraph, there are no nontrivial positive integral solutions in $m,n$ to $2^n+1=5^m$ with $n \ge 3$ that is.
The trick of establishing that [whenever the problem allows it that is], all solutions must satisfy an equation of the form $$(A-c)(A+c) = k^n$$ for some integral $k \ge 2$ and some small $c$, is used often. If something like this can be established, then one can conclude from this that the possible values that $n$ may take, falls into a [VERY] *bounded set, which leaves us only a few possible choices to go through. Above we established above, for example, that both $5^k-1$ and $5^k+1$, which differ by only $2$, must both be powers of $2$. That ended up leaving few choices for $k$ and $n$. Indeed, for two integers $A=5^k+1$ and $A-2=5^k-1$ that differ by only $2$, to both be powers of $2$, both $A$ and $A-2$ must be quite small, and thus in turn, so must $A=5^k+2$ and $A-2=5^k-2$, and thus in turn, so must their product $2^n$. Then as $n$ must be small, this reduces finding all the solutions to the Diophantine equation, to checking only the choices inside a very small set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do I find the centralizer for this matrix?
Let $B=\begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix} \in GL_2(\mathbb{R})$. Determine the centralizer for $B$.
Here's what I've done so far:
Let $D=\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \in GL_2(\mathbb{R})$.
$\begin{align}
BD&=DB\\
\begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix} \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} &= \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix}\\
\begin{pmatrix}
2(a+c) & 2(b+d) \\
c& d \\
\end{pmatrix}&= \begin{pmatrix}
2a & 2a+b \\
2c& 2c+d \\
\end{pmatrix}
\end{align} $
From the above, we have:
$\begin{aligned}
2a+2c&=2a\\
2b+2d&=2a+b\\
c&=2c\\
d&=2c+d
\end{aligned}
$
From which we derive:
$\begin{aligned}
c&=0\\
b+2d-2a&=0\\
d&=d \in \mathbb{R} \backslash \{0\}
\end{aligned}$
As you can see, I've managed to determine $c$ and $d$, but I can't seem to figure out $a$ and $b$. What am I missing?
| $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\R}{\mathbb{R}}$Allow me to sketch a slightly more conceptual approach.
$B$ has clearly the eigenvalues $2$ and $1$, hence it can be diagonalised, and one sees immediately that
$$
T^{-1} B T = \begin{bmatrix}2 & 0\\0 & 1\end{bmatrix},
\quad
\text{for }
T
=
\begin{bmatrix}1 & -2\\0 & 1\end{bmatrix}.
$$
Now the centraliser of $T^{-1} B T$ is immediately seen to be
$$
\mathbf{C}
=
\Set{
\begin{bmatrix}a & 0\\0 & d\end{bmatrix}
:
a, d \in \R \setminus \Set{0}}.
$$
Hence the centraliser of $B$ is
$$
T \mathbf{C}
T^{-1}
=
\Set{
\begin{bmatrix}a & 2 a - 2 d\\0 & d\end{bmatrix}
:
a, d \in \R \setminus \Set{0}}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving by Induction that $x_{2n+1} = 1+1/2+1/2^3+\cdots+1/(2^{2n-1})$ Let $x_1 = 1$, $x_2 = 2$ and $x_n = \frac{x_{n-1}+x_{n-2}}{2}$. The sequence $(x_n)$ is Cauchy, which I can easily prove. To find its limit I must first show (by Induction) the assertation
$$
P(n): x_{2n+1} = 1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2n-1}},\quad\forall n\in\mathbb{N}.
$$
Base case.
For $n=1$, $x_{2n+1} = x_{3} = \frac{1+2}{2} = 1+ \frac{1}{2},$ thus $P(1)$ holds.
Inductive step. Let $P(k)$ be true. Then for $P(k+1)$ we have
$$
x_{2(k+1)+1} = \frac{1}{2}(x_{2(k+1)}+x_{2(k+1)-1}) = \frac{1}{2}(x_{2k+2}+x_{2k+1}).
$$
But then I'm stuck. I am missing something here but, at present, I cannot think of a different approach.
| After re-checking the algebra I finally found the solution I was looking for. The rest of the proof is thus as follows.
By definition $x_{2k+2} = \frac{x_{2k}+x_{2k+1}}{2}$ and $x_{2k+1} = \frac{x_{2k}+x_{2k-1}}{2}$. Thus $x_{2k} = 2x_{2k+1}-x_{2k-1}$; noting that $x_{2k-1} = x_{2(k-1)+1}$ and replacing $x_{2k+2}$ in the equation of the Inductive step gives
\begin{align*}
x_{2(k+1)+1} & = \frac{5}{4}x_{2k+1}-\frac{1}{4}x_{2(k-1)+1}\\
& = \frac{5}{4}(1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2k-1}})-\frac{1}{4}(1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2k-3}})\\
&= 1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2(k+1)-1}},
\end{align*}
which is the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
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The palindrome counting function Let $b \geq 2$ be a positive integer and consider the function $f_b : \mathbb{N}^+ \to \mathbb{N}^+$ given by $$f_b (n) = |\{ k \in \mathbb{N}^+ : k \leq n \mbox{ and } k \mbox{ is palindromic in base } b\}|.$$
A simple counting argument shows that:
$$f_b (b^k) = \begin{cases} (1+b) b^{\frac{k-1}{2}} - 2& \mbox{if } k \mbox{ is odd } \\ 2 \cdot b^{\frac{k}{2}} - 2 & \mbox{if } k \mbox{ is even} \end{cases}$$
Using the monotonicity of $f_b$ and the previous fact, together with the inequality $b^{\lfloor \log_b n \rfloor} \leq n \leq b^{1 + \lfloor \log_b n \rfloor}$ we get the following
$$\frac{2\sqrt{b}}{b} \sqrt{n} - 2\leq 2 b^{\frac{\lfloor \log_b n \rfloor}{2}} - 2 \leq f_b(n) \leq (1+b)b^{\frac{\lfloor \log_b n \rfloor}{2}} \leq (1+b) \sqrt{n} - 2$$
From this it follows that
$$ \frac{2\sqrt{b}}{b} - 2\frac{\sqrt{n}}{n} \leq \frac{f_b(n)}{\sqrt{n}} \leq (1+b) - 2\frac{\sqrt{n}}{n}$$
Consider the function $g_b: \mathbb{N}^+ \to \mathbb{N}^+$ given by
$$g_b (n) = \frac{f_b(n)}{\sqrt{n}} \hspace{3mm} \forall n \in \mathbb{N}^+$$
The previous facts imply that $g_b$ is a bounded positive function, and that
$$\frac{2\sqrt{b}}{b} \leq \liminf_{n \to \infty} g_b (n) \leq \limsup_{n \to \infty} g_b(n) \leq 1 + b$$ Computational evidence suggests that $$\liminf_{n \to \infty} g_b(n) = 2 \hspace{3mm} \mbox{ and } \hspace{3mm} \limsup_{n \to \infty} g_b(n) = \frac{(1+b)\sqrt{b}}{b}$$
How can i prove it? I think i need tighter bounds, and therefore a better and non trivial way to count palindromes.
| The proof is by me, not from any source, sorry.
Let $n$ be $k$-digit number in base $b$. Let $n=a_{k-1}a_{k-2}\dots a_0$. We split the interval $[0,n]$ to $[0,b^{k-1}), [b^{k-1},a_{k-1}a_{k-2}\dots a_{\lfloor k/2\rfloor}0\dots 0),[a_{k-1}a_{k-2}\dots a_{\lfloor k/2\rfloor}0\dots 0,n]$.
We count the number of palindrome numbers in each interval.
(1) $k$ is odd
The first interval $[0,a_{k-1}000\dots 0)$ counts all the palindromes with $\le k-1$ digits and all the palindromes with $k$ digits and first digit is $1,\dots,a_{k-1}-1$. The number of the palindromes with $\le k-1$ digits is $2b^{(k-1)/2}-2$.
The last interval has first $\frac{k+1}2$ numbers fixed, so it has at most one palindrome numbers. Let $r$ be the number of palindromes in this interval, and $r\in\{0,1\}$
The second interval has $a_{k-1}a_{k-2}\dots a_{(k-1)/2}-b^{(k-1)/2}$ palindromes, since we can map any palindrome, by taking its first $(k+1)/2$ digits, to a number from $b^{(k-1)/2}$ to $a_{k-1}a_{k-2}\dots a_{(k-1)/2}-1$. Also, for each number in $b^{(k-1)/2}$ to $a_{k-1}a_{k-2}\dots a_{(k-1)/2}-1$, we can write its first $(k-1)/2$ digits reversely to its end, and it is in the second interval.
So the number of palindrome is
$$2b^{(k-1)/2}-2+a_{k-1}a_{k-2}\dots a_{(k-1)/2}-b^{(k-1)/2}+r\\=b^{(k-1)/2}+a_{k-1}a_{k-2}\dots a_{(k-1)/2}+O(1)=b^{(k-1)/2}+\frac{n}{b^{(k-1)/2}}+O(1)$$
So
$$g_b(n)=\frac{b^{(k-1)/2}+\frac{n}{b^{(k-1)/2}}}{\sqrt{n}}+O(\frac{1}{\sqrt n})=\frac{b^{(k-1)/2}}{\sqrt{n}}+\frac{\sqrt{n}}{b^{(k-1)/2}}+O(\frac{1}{\sqrt n})$$
Notice that $n$ is from $[b^{k-1},b^k]$, so $g_b(n)$ takes value from $[2+O(\frac{1}{\sqrt n}),\frac{1+b}{\sqrt{b}}+O(\frac{1}{\sqrt n})]$
(2) $k$ is even
Everything is similar. The first interval $[0,a_{k-1}000\dots 0)$ counts all the palindromes with $\le k-1$ digits and all the palindromes with $k$ digits and first digit is $1,\dots,a_{k-1}-1$. The number of the palindromes with $\le k-1$ digits is $(b+1)b^{(k-2)/2}-2$.
The last interval has first $\frac{k}2$ numbers fixed, so it has at most one palindrome numbers. Let $r$ be the number of palindromes in this interval, and $r\in\{0,1\}$
The second interval has $a_{k-1}a_{k-2}\dots a_{k/2}-b^{(k-2)/2}$ palindromes, since we can map any palindrome, by taking its first $k/2$ digits, to a number from $b^{(k-2)/2}$ to $a_{k-1}a_{k-2}\dots a_{(k-1)/2}-1$. Also, for each number from $b^{(k-2)/2}$ to $a_{k-1}a_{k-2}\dots a_{(k-1)/2}-1$, we can write all its digits reversely to its end, and it is in the second interval.
So the number of palindrome is
$$(b+1)b^{(k-2)/2}-2+a_{k-1}a_{k-2}\dots a_{k/2}-b^{(k-2)/2}+r\\=b^{k/2}+a_{k-1}a_{k-2}\dots a_{(k-1)/2}+O(1)=b^{k/2}+\frac{n}{b^{(k-1)/2}}+O(1)$$
So
$$g_b(n)=\frac{b^{k/2}+\frac{n}{b^{k/2}}}{\sqrt{n}}+O(\frac{1}{\sqrt n})=\frac{b^{k/2}}{\sqrt{n}}+\frac{\sqrt{n}}{b^{k/2}}+O(\frac{1}{\sqrt n})$$
Notice that $n$ is from $[b^{k-1},b^k]$, so $g_b(n)$ takes value from $[2+O(\frac{1}{\sqrt n}),\frac{1+b}{\sqrt{b}}+O(\frac{1}{\sqrt n})]$
Therefore, we can say that $\limsup g_b(n)\le \frac{1+b}{\sqrt{b}}$ and $\liminf g_b(n)\ge 2$. The "minimum" case achieved when $n=b^{2k}$, the "maximum" case achieved when $n=b^{2k+1}$. Thus, the problem is solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding roots of the equation $8x^3+4x^2-4x-1=0$ which are in form of cosine of angles. So, I start this by reducing it to a cubic with 0 coefficient of the 2nd degree term which gives me :
$$\left(x + \frac{1}{6}\right)^3 - \frac{7}{12}\left(x + \frac{1}{6}\right)-\frac{7}{216}=0$$
Replacing $\left(x + \frac{1}{6}\right)$ by $y$:
$$y^3-\frac{7}{12}y -\frac{7}{216}=0$$
Then replacing $y$ by $r \cos(\alpha)$ and comparing the equation with :
$$\cos^3 (\alpha) - \frac{3}{4}\cos(\alpha)-\frac{1}{4}\cos(3\alpha)=0$$
I get:
$$r = \sqrt{\frac{7}{3}}, \alpha = \frac{1}{3}\cos^{-1}\left(\frac{1}{2\sqrt{7}}\right)$$
which finally give:
$$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{9}\cos^{-1} \left(\frac{1}{2\sqrt{7}}\right)-\frac{1}{6}\right)$$
which is approximately 1.329 on putting it in the calculator.
But if I consider the common problem:
$8x^3-4x^2-4x+1=0$ having roots $\cos(\pi/7), \cos(3\pi/7), \cos(5\pi/7)$.
If I replace $x$ by $-x$ in this equation, we get:
$$8x^3+4x^2-4x-1=0$$
which is same the expression in the actual problem. So its roots must be $-\cos(\pi/7), -\cos(3\pi/7), -\cos(5\pi/7)$. This clearly doesn't matches with the answer I got.
Where I go wrong? Also can someone confirm while comparing coefficients in $cos(\alpha)$ variable cubic equation, we treat $\cos(3\alpha)$ as a constant. My book give a special section in which it shows this method of comparing in this way. So I guess its probably right? But I am not sure how is it right, isn't $\cos(3\alpha)$ also dependent on the main variable $\cos(\alpha)$.
| if we take $t=2x$ we have $t^3 + t^2 - 2t - 1$
next, take complex number $w$ with $w \neq 1$ but $w^7 = 1.$ Factoring out a $w-1$ tells us $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0.$ Now take $t = w + w^6$ which is the same as $w + \frac{1}{w}.$ We find $ t^2 = w^2 + 2 + w^5,$ while $t^3 = w^3 + 3 w + \frac{3}{w} + \frac{1}{w^3} = w^3 + 3 w + 3 w^6 + w^4 $
Adding,
$$ t^3 + t^2 = w^3 + 3 w + 3 w^6 + w^4 + w^2 + 2 + w^5 ,$$
in order
$$ t^3 + t^2 = 3 w^6 + w^5 + w^4 + w^3 + w^2 + 3w + 2$$ so that
$$ t^3 + t^2 -2t = w^6 + w^5 + w^4 + w^3 + w^2 + w + 2$$ and
$$ t^3 + t^2 -2t - 1 = w^6 + w^5 + w^4 + w^3 + w^2 + w + 1$$
finally
$$ t^3 + t^2 -2t - 1 = 0$$
Taking $w = \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7},$ so that
$ 1/w = \cos \frac{2\pi}{7} - i \sin \frac{2\pi}{7},$ we see
$$ t = 2\cos \frac{2\pi}{7} $$
is one of the roots $t.$ Next we may take $w = \cos \frac{4\pi}{7} + i \sin \frac{4\pi}{7},$ so that $ t = 2\cos \frac{4\pi}{7} $ is another root. The last one can be $w = \cos \frac{8\pi}{7} + i \sin \frac{8\pi}{7},$so that $ t = 2\cos \frac{8\pi}{7} $ is the third root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
(a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
(a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are
*
*Are there any other simple nontrivial examples? I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.
*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream? Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?
| I don't know if this may interest you or answer your question but if you play around with field different from $\mathbb{R}$, special things happens. For example working with field $\mathbb{F}$ with characteristic equal to 2 it's always true that
$$(a+b)^2 = a^2 + b^2, \space \forall a,b \in \mathbb{F}$$
And if we take another finite sum $(a + b+ c+ d)^2$ in order to obtain the claim what we want is
$$2(ab + ac + ad + ...) = 0$$
and since that term is even we can choose a field with proper even characteristic and we obtain a series of examples that are non trivial.
| {
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"url": "https://math.stackexchange.com/questions/4411628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Sum of "alternating-like" series I tried to prove that the following series converges.
$$1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}+ \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} +\frac{1}{9} - ...$$
I found a expression for this sum in sigma form by summing pair of numbers with the same sign:
$$1 + \sum_{n=1}^{\infty} \frac{8n +1}{16n^{2}+4n} - \frac{8n -3}{16n^{2} - 12n +2} = 1 + \sum_{n=1}^{\infty} \frac{(8n +1)(16n^{2}+4n) - (8n -3)(16n^{2}+4n)}{(16n^{2}+4n) (16n^{2} - 12n +2)}$$
But I'm getting a very long expression that can't find a way to prove a convergence of it , I feel that there is more elegant way to prove convergences of this series.
I will thank for your help.
| $$S_p=1 + \sum_{n=1}^{p} \frac{8n +1}{16n^{2}+4n} - \sum_{n=1}^{p}\frac{8n -3}{16n^{2} - 12n +2} $$
Notice that
$$ \frac{8n +1}{16n^{2}+4n}=\frac{1}{4 n+1}+\frac{1}{4 n}$$
$$\frac{8n -3}{16n^{2} - 12n +2}=\frac{1}{4 n-1}+\frac{1}{4n-2}$$ Computing the four summations
$$\sum_{n=1}^{p}\frac{1}{4 n+1}=\frac{1}{4} \left(\psi \left(p+\frac{5}{4}\right)-\psi \left(\frac{5}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4 n}=\frac{1}{4} H_p$$
$$\sum_{n=1}^{p}\frac{1}{4 n-1}=\frac{1}{4} \left(\psi \left(p+\frac{3}{4}\right)-\psi \left(\frac{3}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4n-2}=\frac{1}{4} \left(\psi\left(p+\frac{1}{2}\right)-\psi\left(\frac{1}{2}\right)\right)$$ Summing and using harmonic numbers in place of digamma's
$$S_p=\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{1}{4} \left(-H_{p-\frac{1}{2}}-H_{p-\frac{1}{4}}+H_p+H_{p+\frac{1}{4}}\right)$$ Using the asymptotics
$$S_p=\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{1}{4 p}-\frac{3}{32 p^2}+O\left(\frac{1}{p^3}\right)$$gives the limit, shows how the limit is approached and provides a shortcut method for the evaluation of the partial sums.
Using $p=10$
$$S_{10}=\frac{101405260644075743}{219060189739591200}=0.462910\cdots$$ while the truncated expansion gives
$$S_{10}\sim\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{77}{3200}=0.462887\cdots$$ to which corresponds a relative error of $0.005$%.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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Investigate whether the sequence $\frac{nx^n + 3\sin{2n\pi x}}{n}$ is uniformly convergent Let $(f_n)$ be a sequence of functions defined by
\begin{equation*}
f_n(x) = \frac{nx^n + 3\sin{(2n\pi x)}}{n}
\end{equation*}
for all $x \in [0,1]$ and $n \in \Bbb N$.
Show that $f_n$ is converges pointwise to a function $f:[0,1]\to \Bbb R$ on $[0,1]$.
Investigate whether $f_n$ is uniformly convergent to $f$ on $[0,1]$ or not.
Attempt:
*
*For $x=0$: We have $f_n(0)=0 \to 0$.
*For $x=1$: We have $f_n(1)=1 \to 1$.
*For $x \in (0,1)$: I claimed that $f_n \to 0$. The proof is goes as follows:
Let $\varepsilon > 0$ and $x \in (0,1)$ be arbitrary. Choose $k \in \Bbb n$ such that
$k > \max\left\{\frac{2x}{\varepsilon - \varepsilon x}, \frac{6}{\varepsilon} \right\}$. Then, for any $n \in \Bbb n$ with $n \ge k$, we have
\begin{align*}
\left|f_n(x)-f(x) \right| &= \left|\frac{nx^n + 3\sin{(2n\pi x)}}{n}-0 \right| \\
&\le x^n + \frac{3}{n} \\
&= \frac{1}{\left(1+\left(\frac{1}{x}-1\right)\right)^n} + \frac{3}{n} \\
&\le \frac{1}{1+n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\
&< \frac{1}{n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\
&\le \frac{1}{k\left(\frac{1}{x} -1 \right)} + \frac{3}{k} \\
&< \frac{\frac{\varepsilon}{2}\left(\frac{1}{x} -1 \right)}{\frac{1}{x}-1} + \frac{\varepsilon}{2} \\
&= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align*}
Hence, $f_n \to f$ where
\begin{equation*}
f(x) =
\begin{cases}
0, \qquad \text{if} \ 0 \le x < 1 \\
1, \qquad \text{if} \ x = 1
\end{cases}.
\end{equation*}
Now, my intuition is said that $f_n$ is not uniformly convergent to $f$ on $[0,1]$. But, I didn't be able yet to find the subsequence such that the sequence did not uniformly convergent. On the other hand, I tried using the uniform norm
\begin{align*}
||f_n - f||_{[0,1]} =
\sup \left\{\begin{array}{lr}
\frac{nx^n + 3\sin{(2n\pi x)}}{n}, & \text{for} \ 0 \le x < 1\\
0, & \text{for} \ x=1
\end{array}\right\} = \frac{13}{4} \ne 0.
\end{align*}
But, is this correct? Any ideas? Thanks in advanced.
| There is simple way of showing that the convergence is not uniform.
Note that $f_n(x)=x^{n}+g_n(x)$ where $g_n(x)=\frac {3\sin (2\pi nx)} n$. Now $g_n(x) \to 0$ uniformly so $f_n$ converges uniformly if and only if $x^{n}$ does. Can you show that $x^{n}$ does not converge uniformly?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$
I tried to full expand but the rest seems complicated for me.
Can anyone help me full my idea? Every thinking is welcomed, thanks!
| Futhermore, I tried C-S inequality and saw something "tricky":
It is obvious to get equivalent inequality: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{ca+abc}}{b+\sqrt{ac+abc}}\ge4$$
By C-S: $$2(a+b+c)=\sqrt{\left(a(b+c)+bc+abc\right)\left((b+c-a)^2+4a(b+c)\right)}\ge 2a(b+c)+\sqrt{bc+abc}(b+c-a)$$
Or: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge \frac{(a+b+c)\sqrt{bc+abc}+2a(b+c)+\sqrt{bc+abc}(b+c-a)}{(a+b+c)(a+\sqrt{bc+abc})}=\frac{2(b+c)}{a+b+c}$$
The rest is similar to above solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Find the value of $x^2+y^2$, given that $x^3-3xy^2=3$ and $y^3-3x^2y=3\sqrt{2}$ Let $x,y$ be reals that satisfy the equations:
$$\begin{align*}
x^3-3xy^2&=3 \\
y^3-3x^2y&=3\sqrt{2}
\end{align*}$$
Find the value of $x^2+y^2.$
The terms in the above equations are similar to the expansion $(x\pm y)^3=x^3\pm y^3\pm3xy(x\pm y)$. But subtracting the two equations leads to $$x^3+3x^2y-3xy^2-y^3=(x-y)^3+6xy(x-y)=(x-y)\left((x-y)^2+6xy\right)=3-3\sqrt{2}$$
I also tried solving for $x$ and $y$ using the substitution $y=xt$,
$$\sqrt{2}x^3-3\sqrt{2}x^3t^2=x^3t^3-3x^3t\implies t^3+3\sqrt{2}t^2-3t-\sqrt{2}=0 $$
Both of these approaches don't seem to be particularly useful.
| $\begin{array}\\
(x+iy)^3
&=x^3+3ix^2y+3i^2xy^2+i^3y^3\\
&=x^3+3ix^2y-3xy^2-iy^3\\
&=x^3-3xy^2+i(3x^2y-y^3)\\
&=3-3i\sqrt{2}\\
&=3(1-i\sqrt{2})\\
&=3\sqrt{3}(1/\sqrt{3}-i\sqrt{2/3})\\
&=3\sqrt{3}(1/\sqrt{3}-i\sqrt{2/3})\\
&=3\sqrt{3}(\cos(a)-i\sin(a))
\qquad \cos(a)=1/\sqrt{3}\\
&=3\sqrt{3}e^{ia}\\
x+iy
&=\sqrt{3}e^{ia/3}\\
x^2+y^2
&=|x+iy|^2\\
&=3\\
\end{array}
$
In general, if
$x^3-3xy^2=u,
y^3-3x^2y=v
$
then
$\begin{array}\\
(x+iy)^3
&=x^3+3ix^2y+3i^2xy^2+i^3y^3\\
&=x^3+3ix^2y-3xy^2-iy^3\\
&=x^3-3xy^2+i(3x^2y-y^3)\\
&=u-iv\\
(x-iy)^3
&=u+iv\\
(x^2+y^2)^3
&=(x+iy)^3(x-iy)^3\\
&=(u+iv)(u-iv)\\
&=(u^2+v^2)\\
x^2+y^2
&=(u^2+v^2)^{1/3}\\
\end{array}
$
Here
$u=3, v=3\sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Length of the shadow of a tower at 4:00 pm compared to the length of the shadow at 10:00 am Consider a location on the Earth where the Sun is overhead at noon. Compared to its shadow at 10:00 am, the shadow of a tower at 4:00 pm would be
(1) twice longer
(2) three times longer
(3) four times longer
(4) eight times longer
Let $h$ be the height of the tower. Let $x,y$ be the lengths of the shadow at 10:00 am and 4:00 pm, respectively. If $\theta$ is the angle of elevation of the Sun from the ground at 10:00 am, then the corresponding angle of elevation at 4:00 pm should be $\theta/2$, as the Sun is overhead at noon (12:00 noon).
Now,
\begin{align}
\tan\theta=\frac{h}{x}, \quad \tan\frac{\theta}{2}=\frac{h}{y}.
\end{align}
Also,
\begin{align}
\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}=\frac{2\cdot\frac{h}{y}}{1-\frac{h^2}{y^2}}.
\end{align}
So, we have
\begin{align}
\frac{h}{x}=\frac{2\cdot\frac{h}{y}}{1-\frac{h^2}{y^2}}\\
\implies\frac{1}{x}=\frac{2}{y-\frac{h^2}{y}}.
\end{align}
What to do after that?
| You would compare the two after you calculate it.
Here is another solution:
$\frac{y}{x} = tan(60) = \sqrt 3$.
Therefore, $y = \sqrt 3 \times x$.
$\frac{z}{x}=tan(30)=\frac{1}{\sqrt3}$
Therefore, $z = \frac{x}{\sqrt 3}$.
$\frac{y}{z}=\frac{x \times \sqrt 3}{\frac{x}{\sqrt 3}} = 3$.
Therefore, the shadow will be 3 times longer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve this equation $12^x-5^y=19$ positive integers Find all $x,y$ be positive integers,such
$$12^x-5^y=19$$
I found $(x,y)=(2,3)$ is solution,maybe have other,so I consider case $x,y>3$ and $\pmod 9$,since
$$12^x\equiv 0\pmod 9,x\ge 2$$
then $5^y\equiv -1\pmod 9$,
$5^1\equiv 5 \pmod 9$
$5^2\equiv 7 \pmod 9$
$5^3\equiv 8 \pmod 9$
$5^4\equiv 4 \pmod 9$
$5^5\equiv 2 \pmod 9$
$5^6\equiv 1 \mod 9\Rightarrow y\equiv 3 \mod 6\Rightarrow y$ is odd.
| To show that there are no solutions with $x \ge 3$, let's consider the equation modulo $27$. When $x\ge 3$, the equation becomes $-5^y \equiv 19 \pmod{27}$, or $5^y \equiv 8 \pmod{27}$. This calculation is intimidating, but rewarding: $5$ has multiplicative order $18$ modulo $27$, and so we are able to conclude $y \equiv 15 \pmod{18}$. In particular, $y \equiv 6 \pmod 9$.
To make use of this, we look at the factors of $5^9-1$: two reasonably-sized ones are $19$ and $31$. We want factors of $5^9-1$ because when $d \mid 5^9-1$, knowing that $y \equiv 6 \pmod 9$ tells us that $5^y \equiv 5^6 \pmod d$.
*
*Taking the equation mod $19$, we get $12^x - 5^6 \equiv 0 \pmod{19}$, or $12^x \equiv 7 \pmod{19}$. This is true when $x \equiv 4 \pmod 6$.
*Taking the equation mod $31$, we get $12^x - 5^6 \equiv 19 \pmod{31}$, or $12^x \equiv 20 \pmod{31}$. This is true when $x \equiv 2 \pmod {30}$.
But these two contradict each other, so there are no solutions (with the assumption $x\ge3$ we started from).
We could also have combined the last two steps into a calculation modulo $19\cdot31 = 589$, where $12^x - 5^6 \equiv 19 \pmod{589}$ has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I prove algebraically that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$? I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering:
$$\sqrt{6}-\sqrt{2} =
\sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } = \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-\sqrt{6×2}-\sqrt{6×2}+2}=\sqrt{ 8-2\sqrt{12}}=\sqrt{ 8-4\sqrt{3}}$$
But how can I do it the other way, also for other similar problems?
I tried multiple problem solving sites, but none could simplify $\sqrt{8-4\sqrt{3}}$ further then $2\sqrt{2-\sqrt{3}}$
| In the general case let solve: $$\sqrt{x\pm y}=\sqrt{a}\pm\sqrt{b}$$
Since LHS exsits and is positive we will also assume $x\ge y$ and $a\ge b$.
By squaring expressions we get the system $\begin{cases}x+y=a+b+2\sqrt{ab}\\x-y=a+b-2\sqrt{ab}\end{cases}\iff \begin{cases}x=a+b\\y=2\sqrt{ab}\end{cases}$
Solving for $a,b$ we get $\begin{cases}a=\frac 12 x+\frac 12\sqrt{x^2-y^2}\\b=\frac 12 x-\frac 12\sqrt{x^2-y^2}\end{cases}$
And there you have it:
$$\sqrt{\vphantom{|}x\pm y}=\sqrt{\frac 12 x+\frac 12\sqrt{x^2-y^2}}\pm\sqrt{\frac 12 x-\frac 12\sqrt{x^2-y^2}}$$
Applying for $x=8$ anf $y=4\sqrt{3}$ we get $\sqrt{x^2-y^2}=\sqrt{64-16\times 3}=\sqrt{16}=4$
So $\sqrt{8-4\sqrt{3}}=\sqrt{4+2}-\sqrt{4-2}=\sqrt{6}-\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find minimum of the function using AM-GM Problem: Find the minimum of the function $f(x,y)=x + \frac{8}{y(x-y)}$, where $x>y>0$ using AM-GM.
My attempt:
$$f(x,y)=2\cdot \frac{x+\frac{8}{y(x-y)}}{2} \ge 2 \sqrt{\frac{8x}{y(x-y)}}$$
But here nothing cancels and I couldn't get a constant. I also tried partial fraction, but didn't work either.
$$f(x,y)=x+\frac{8}{xy}-\frac{8}{\color{red}{{x(y-x)}}}$$
I think the issue is with the denominator in red. Any hint please?
| First off $x= \frac{y+ (x-y)}{2} >= \sqrt{y(x-y)}$
AM-GM was used for the first time.
So now we have $x+ \frac{8}{\sqrt{y(x-y)}} >= x+ \frac{8}{x}$
Then one uses AM-GM inequality for the second time to conclude that
$x+ \frac{8}{x} >= 2\sqrt{x*\frac{8}{x}} = 4\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $x,y,z \in \mathbb{R}$ are such that $x^2+x=y^2$, $y^2+y=z^2$ and $z^2+z=x^2$, prove that $(x-y)(y-z)(z-x)=1.$ If $x,y,z$ are real numbers that not equal to 0 such that $x^2+x=y^2$, $y^2+y=z^2$ and $z^2+z=x^2$ prove that $(x-y)(y-z)(z-x)=1$.
I solve this problem by adding each two equations, but I'm interested can we solve this problem by subtracting equations? By subtracting we get $(x-y)(x+y+1)=(y-z)(y+z)$.
| Adding the three equations we get
$$x+y+z=0.$$
So then, if we rewrite the equations as $y^2-x^2=x$, $z^2-y^2=y$, etc., we find $(y+x)(y-x)=x$, $(z+y)(z-y)=y$, $(x+z)(x-z)=z$, which in view of the above relation we can rewrite as $(x-y)z=x$, $(y-z)x=y$, $(z-x)y=z$. Multiplying these three equations, and using that $xyz \neq 0$, we get the desired conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,\mathrm{d}y$; how do you avoid using a complex substitution? $\newcommand{\d}{\mathrm{d}}$The given exam question - I provide the beginning for context:
Let: $$I=\int\frac{1}{(b^2-y^2)\sqrt{c^2-y^2}}\d y$$Where $b,c\gt0$, and employ the substitution $y=\frac{cx}{\sqrt{x^2+1}}$ to show that: $$I=\int\frac{1}{b^2+(b^2-c^2)x^2}\d x$$And hence evaluate: $$\int_1^{\sqrt{2}}\frac{1}{(3-y^2)\sqrt{2-y^2}}\d y$$And: $$\int_{1/\sqrt{2}}^1\frac{y}{(3y^2-1)\sqrt{2y^2-1}}\d y$$
This I managed straightforwardly. The next part:
By means of a suitable substitution, evaluate: $$\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\d y$$
I ordinarily wouldn't have a clue about evaluating that integral, as I am not very well practised in integration (which is why I am seeking out such exercises) so I assumed they wanted us to use the same ideas. My attempt is based on the context of the question at the start, but I shouldn't have needed a complex variable substitution.
$$\begin{align}I:=&\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\d y\\\overset{y\mapsto1/y}{=}&\int_1^{\sqrt{2}}\frac{y}{(3-y^2)\sqrt{2-y^2}}\d y\end{align}$$I optimistically employ the same substitution $y=\frac{\sqrt{2}x}{\sqrt{x^2+1}}$ to get: $$\begin{align}I=&\,\sqrt{2}\int_1^\infty\frac{x}{(x^2+3)\sqrt{x^2+1}}\d x\\\overset{x\mapsto1/x}{=}&\sqrt{2}\int_0^1\frac{1}{(1+3x^2)\sqrt{1+x^2}}\d x\end{align}$$
Now, I decided that instead of trying for perhaps a very long time to come up with a similar substitution to the one at the beginning (the problem being that we have $b^2+y^2$ instead of $b^2-y^2$) I decided to enforce negative signs by introducing a complex variable substitution - which will cause some weird problem.
Let $x=\frac{iz}{\sqrt{z^2+1}}$. Then $x^2(z^2+1)=-z^2,\,z^2(x^2+1)=-x^2,\,z=\frac{-ix}{\sqrt{x^2+1}}$ (the minus sign comes from pluggin $z$ back in and seeing which of $\pm i$ works) and $\frac{\d x}{\d z}=\frac{i}{(z^2+1)\sqrt{z^2+1}}$ and we get: $$\begin{align}I&=i\sqrt{2}\int_0^{-i/\sqrt{2}}\frac{1}{1-2z^2}\d z\\&\overset{z=-\frac{iu}{\sqrt{2}}}{=}\int_0^{1}\frac{1}{1+u^2}\d u\\&=\frac{\pi}{4}\end{align}$$
However, although this isn't really complex integration, I am extremely certain we were not supposed to do it this way (since the syllabus is for pre-university mathematics) so how were we supposed to do it?
Many thanks.
| Yes, the same kind of substitution can be used here. The substitution $t=\frac{y}{\sqrt{2y^2-1}}$ transforms the integral in question to
$$\int_1^\infty\frac{\mathrm{d}t}{t^2+1}=\tan^{-1}t\Big|_1^\infty=\frac{\pi}{4} $$
More generally, this substitution can be used to evaluate integrals of the form
$$\int\frac{\mathrm{d}x}{\left(x^2+p\right)^k\sqrt{c\left(x^2+q\right)}} $$
(See Gradshteyn & Ryzhik, Eighth edition, p. 94 for more details.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$ I am trying to show that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000}
=\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}.
$$
It seems there is some sort of generalizable pattern here, so I will verify it for smaller numbers:
$$
\begin{align*}
\text{Say, }n=4 \hspace{35pt} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}&=\frac{1}{3}+\frac{1}{4}\\
\frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}&=\frac{4}{12}+\frac{3}{12}\\
\frac{7}{12}&=\frac{7}{12}
\end{align*}
$$
So, my guess on the general formula is
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n/2+1}+\ldots+\frac{1}{n-1}+\frac{1}{n}.
$$
This really seems like I am getting somewhere, but how can I finish off the proof? Is induction viable?
| Let $$S(n) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n}.$$
Then notice if we add $$T(n) = 2 \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n}\right)$$ to $S(n)$, all the negative terms become positive and we get a nice sum:
$$S(n) + T(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n} = H(2n),$$ where $H(n) = 1 + 1/2 + \cdots + 1/n$ is the $n^{\rm th}$ harmonic number.
But notice that $T(n)$ is itself a harmonic number: just distribute the $2$: $$T(n) = \frac{2}{2} + \frac{2}{4} + \cdots + \frac{2}{2n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} = H(n),$$ so they are actually the same. Therefore,
$$S(n) = H(2n) - H(n).$$ And from here it is easy to see that
$$S(n) = \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n}\right) - \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right) \\
= \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n},$$
which proves the claim.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x^4+y^4+z^4=1$ Find the minimum value of $\sum_{cyc}\frac{x^3}{1-x^8}$. Let $x,y,z$ be positive reals such that $x^4+y^4+z^4=1$. Find the minimum value of
$$\sum_{cyc}\frac{x^3}{1-x^8}$$
First I tried Jensen. Let $$f(x)=\frac{x^3}{1-x^8}$$
Then $f$ is convex on $\mathbb R^+$, using Jensen $$\frac{1}{3}\sum_{cyc}f(x)\ge f\left(\frac{x+y+z}{3}\right)$$
but we don't have enough information about $x+y+z$.
By the way my guess for the minimum value is at $x=1/\sqrt[4]{3}$
| Consider the function $$f(x)=\frac{1}{1-x^8}$$
And note that $f$ is convex on $[0,1[$ hence it lies above its tangent line, $$f(x)\ge f'(a)(x-a)+f(a)$$
set $a=1/\sqrt[4]{3}$
$$\frac{1}{1-x^8}\ge \frac{9}{8}\sqrt[4]{3}x$$
Meaning $$\frac{x^3}{1-x^8}\ge \frac{9}{8}\sqrt[4]{3}x^4$$
Now cycle sum to finish.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a point of the plane which distance to a line minimal is Let be $P \subseteq R^3$ (P is a plane)and $X_1, X_2 \subseteq R^3$ two lines given by:
$P = \Bigg\langle \begin{pmatrix} 2 \\ 0 \\ 1\end{pmatrix},\begin{pmatrix} 0 \\ 2 \\ 1\end{pmatrix} \Bigg\rangle$,
$X_1 = \Bigg\langle \begin{pmatrix} 2 \\ 0 \\ 1\end{pmatrix}\Bigg\rangle$,
$X_2 = \Bigg\{\begin{pmatrix} 0 \\ -2 \\ 5\ \end{pmatrix} +s \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\Bigg \}$, where $s \in \mathbb{R}$
(a) Let be $T_s=(0,-2+2s,5+s) \in X_2$. Determine a point $T_s^{*} \in P $ that $||T_s-T_s^{*}|| = min||T_s-T||$
as I understand the problem $T_s^{*}$is a point of the plane $P$ which distance to $T_s \in X_2$ minimal is. I guess that I have to work with some kind of orthogonal projection but I still don't have no idea how does it work. Thank you very much for your help in advance.
| A bit late answer but I think worth mentioning it because it doesn't need any Gram-Schmidt:
Note that the direction vector $\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$ of $X_2$ is also a spanning vector of $P$. Hence, the line $X_2$ is parallel to $P$.
The line $X_2$ passes through point $(0 , -2 , 5)$. So, you need to find the orthogonal projection of this point onto $P$.
Since $P$ contains point $(0,0,0)$, you only need to project $\begin{pmatrix} 0 \\ -2 \\ 5\ \end{pmatrix}$ onto a normal vector of $P$ in order to find the component of $\begin{pmatrix} 0 \\ -2 \\ 5\ \end{pmatrix}$ which is normal to $P$.
A normal vector of $P$ is quickly found to be, for example, $\begin{pmatrix} -1 \\ -1 \\ 2\ \end{pmatrix}$. Hence, the position vector of such a point you are looking for is
$$\begin{pmatrix} 0 \\ -2 \\ 5\ \end{pmatrix}- \frac{\begin{pmatrix} 0 \\ -2 \\ 5\ \end{pmatrix}\cdot\begin{pmatrix} -1 \\ -1 \\ 2\ \end{pmatrix}}{\left|\begin{pmatrix} -1 \\ -1 \\ 2\ \end{pmatrix}\right|^2}\begin{pmatrix} -1 \\ -1 \\ 2\ \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 1\ \end{pmatrix}$$
| {
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Writing the recurrence $O_t=-\frac1{T_w}\sum_{i=t-T_p}^{t-1}O_i-\frac1{T_i}\sum_{i=1}^{t-T_p-1}O_i+B_t$ in terms of its initial value I want to write the following solely in terms of its initial value $O_1$
$$
O_t = - \frac{1}{T_w} \sum_{i=t-T_p}^{t-1} O_{i}
- \frac{1}{T_i} \sum_{i=1}^{t-T_p-1} O_i
+ B_t
$$
where
*
*$T_w , T_i$ are non zero constants
*$T_p$ is positive integer
*$B_t$ is some function of t
*Define $O_i = 0 , \; B_i = 0\quad \forall \; i < 1$
I'd expanded the first 3 terms in order to find pattern , we may compare the blue terms .
I found a pattern , like the coefficients have the form $a_{n+1} = a_n + a_n^2 $ but I don't have a close form for it . Furthermore , there seems to be other patterns as well .
Context : above is actually a model for supply chain , with $\frac{1}{T_w} ,\frac{1}{T_i} $ as proportional gain (they are 2 indepedent feedback controllers) , you don't see the error because I've already simplified the expression . I want to do this because I do have the analytic expressions for $O_1$ , so this is the first step to prepare for optimization .
| Hint.
Considering first $T_p = 3$ from
$$
\left\{
\begin{array}{rcl}
\frac{o_0+o_1+o_2}{T_w}+o_3 &=& b_3 \\
\frac{o_1+o_2+o_3}{T_w}+o_4&=&b_4 \\
\frac{o_1}{T_i}+\frac{o_2+o_3+o_4}{T_w}+o_5 &=& b_5\\
\frac{o_1+o_2}{T_i}+\frac{o_3+o_4+o_5}{T_w}+o_6 &=& b_6\\
\end{array}
\right.
$$
we can represent a matrix formula as
$$
M_1 O_{t+T_p} + M_2O_t=B_{t+T_p}
$$
with
$$
\cases{
O_t = (o_{-1},o_0,o_1,o_2)^T\\
M_1 = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
\frac{1}{T_w} & 1 & 0 & 0 \\
\frac{1}{T_w} & \frac{1}{T_w} & 1 & 0 \\
\frac{1}{T_w} & \frac{1}{T_w} & \frac{1}{T_w} & 1 \\
\end{array}
\right)\\
M_2 = \left(
\begin{array}{cccc}
0 & \frac{1}{T_w} & \frac{1}{T_w} & \frac{1}{T_w} \\
0 & 0 & \frac{1}{T_w} & \frac{1}{T_w} \\
0 & 0 & \frac{1}{T_i} & \frac{1}{T_w} \\
0 & 0 & \frac{1}{T_i} & \frac{1}{T_i} \\
\end{array}
\right)
}
$$
and as $M_1$ is invertible we have
$$
O_{t+T_p} + M_1^{-1}M_2O_t=M_1^{-1}B_{t+T_p}
$$
NOTE
$$
M_1^{-1} = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
-\frac{1}{T_w} & 1 & 0 & 0 \\
\frac{1-T_w}{T_w^2} & -\frac{1}{T_w} & 1 & 0 \\
-\frac{(T_w-1)^2}{T_w^3} & \frac{1-T_w}{T_w^2} & -\frac{1}{T_w} & 1 \\
\end{array}
\right)
$$
now assuming $O_t = 0$ we have
$$
O_{t+T_p} = \left(
\begin{array}{c}
b_3 \\
b_4-\frac{b_3}{T_w} \\
b_5+\frac{b_3 (1-T_w)}{T_w^2}-\frac{b_4}{T_w} \\
b_6-\frac{b_3 (T_w-1)^2}{T_w^3}+\frac{b_4 (1-T_w)}{T_w^2}-\frac{b_5}{T_w} \\
\end{array}
\right)
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\sqrt[3]{9\sqrt3-11\sqrt2}$ Simplify $$\sqrt[3]{9\sqrt3-11\sqrt2}$$
How can we actually simplify this radical?
| Let $x = \sqrt[3]{9\sqrt 3 -11 \sqrt 2}$. Cubing both sides gives $x^3 = 9\sqrt 3 -11 \sqrt 2$.
Per the suggestions in the comments, assume $x = a\sqrt 3 + b \sqrt 2$. Then:
$$x^3 = (a\sqrt 3 + b \sqrt 2)^3$$
$$x^3 = (a\sqrt 3)^3 + 3(a\sqrt 3)^2(b \sqrt 2) + 3(a\sqrt 3)(b \sqrt 2)^2 + (b \sqrt 2)^3$$
$$x^3 = a^3(\sqrt 3)^3 + 3a^2b(\sqrt 3)^2(\sqrt 2) + 3ab^2(\sqrt 3)(\sqrt 2)^2 + b^3(\sqrt 2)^3$$
$$x^3 = 3a^3\sqrt 3 + 9a^2b \sqrt 2 + 6ab^2 \sqrt 3 + 2 b^3 \sqrt 2$$
$$x^3 = (3a^3 + 6ab^2)\sqrt 3 + (9a^2b + 2b^3)\sqrt 2$$
If we can find $a$ and $b$ such that $3a^3 + 6ab^2 = 9$ and $9a^2b + 2b^3 = -11$, then it will solve the original equation.
From $3a^3 + 6ab^2 = 9$, we get $b = \pm \sqrt{\frac{3 - a^3}{2a}}$. Plugging this into the other equation gives:
$$9a^2(\pm \sqrt{\frac{3 - a^3}{2a}}) + 2(\pm \sqrt{\frac{3 - a^3}{2a}})^3 = -11$$
which, with a bunch of algebra that I won't show here, simplifies to:
$$64a^6 - 73a^3 + 9 = 0$$
Let $u=a^3$. Then $64u^2 - 73u + 9 = 0$. Applying the quadratic formula gives:
$$u = \frac{73 \pm \sqrt{3025}}{128} = \frac{73 \pm 55}{128}$$
Which has the solutions $u=1$ or $u=\frac{9}{64}$.
If $u=1$, then $a=1$, so $b=\pm 1$.
If $u=\frac{9}{64}$, then $a=\frac{\sqrt[3]9}{4}$, and $b=\pm\sqrt[6]{\frac{680943}{2^{15}}}$.
That gives 4 possible combinations for $a$ and $b$, and it just so happens that $a=1$ and $b=-1$ satisfies the original equation. Therefore,
$$x = \sqrt 3 - \sqrt 2$$
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find number of positive integer solutions to the equation $a+b+c=15$ where a,b are even and c is odd This is equivalent to:
Coefficient of $x^{15}$ in $(1+x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
Or
Coefficient of $x^{14}$ in $(1+x^2+x^4+\cdots+x^{14})^3$
Is this approach correct?
| It's already been pointed out that the OP has an error in that $a,b,c$ are supposed to be positive, not non-negative.
But I would like to point out that the algebra is a little simpler if we use the constraints $a,b,c \ge 1$ without taking into account our knowledge that $a,b,c \le 15$. This gives the generating function
$$\begin{align}
f(x) &= (x^2 + x^4 + x^6 + \dots)^2 (x + x^3 + x^6 + \dots) \\
&= x^5 (1 + x^2 + x^4 + \dots)^3 \\
&= x^5 \left( \frac {1}{1-x^2} \right)^3 \\
&= x^5 (1-x^2)^{-3} \\
&= x^5 \sum_{i=0}^{\infty} \binom{3+i-1}{i} x^{2i}
\end{align}$$
From the last equation we see that the coefficient of $x^{15}$ is $\binom{7}{5} = 21$.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
How to find $x$ such that $\frac{1}{2^{p-1}} -\frac{1}{a^p}>x$? Let $a, p\in\mathbb{R}$ be such that $a\ge 2$ and $p>1$. Consider the quantity
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}.$$
Since $a\ge 2$, it seems clear to me that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>0.$$
My question is: it is possible to find a positive real number $x$ (not depending on $a$) such that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>x?$$
Could someone please help me in finding that?
Thank you in advance!
| $$\frac{1}{2^{p-1}} -\frac{1}{a^p}\geq \frac{1}{2^{p-1}} -\frac{1}{2^p}$$
So if you want strict inequality, we simply have:
$$\frac{1}{2^{p-1}} -\frac{1}{a^p} > \frac{1}{2}\left(\frac{1}{2^{p-1}} -\frac{1}{2^p}\right)$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
provided $f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}$, what is f(x)? $$
f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}
$$
Here is one of the solution from my workbook:
$$
\begin{array}{l}
f( x+1) \ =\ \lim _{n\rightarrow \infty }\left[\left( 1+\frac{2+x}{n-2}\right)^{\frac{n-2}{2+x}}\right]^{n\left(\frac{2+x}{n-2}\right)} =\ e^{2+x} =e^{1+( x+1)}\\
f( x) \ =\ e^{1+x}
\end{array}
$$
but is a bit preplexing for me :(
| $$y=\left(\frac{n+x}{n-2}\right)^{n}=\left(\frac{(n-2)+(x+2)}{n-2}\right)^{n}=\left(1+\frac{x+2}{n-2}\right)^{n}$$ Taking logarithms
$$\log(y)=n \log\left(1+\frac{x+2}{n-2}\right)\sim n \frac{x+2}{n-2}=\frac{n}{n-2}(x+2)\sim x+2$$ So
$$y=f(x+1)\sim e^{x+2} \implies f(x)=e^{x+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$
Find $x$ in the equation $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$.
My attempt:
$$\log\left(x\right)\ -\ \frac{1}{2}\ \log\left(x-\frac{1}{2}\right)\ =\ \log\left(x+\frac{1}{2}\right)\ -\ \frac{1}{2}\log\left(x\ +\ \frac{1}{8}\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(\frac{2x-1}{2}\right)\ =\ \log\left(\frac{2x+1}{2}\right)\ -\ \frac{1}{2}\log\left(\frac{8x+1}{8}\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ +\ \frac{1}{2}\log\left(2\right)\ =\ \log\left(2x+1\right)\ -\ \log\left(2\right)\ -\ \frac{1}{2}\log\left(8x+1\right)\ +\ \frac{1}{2}\log\left(8\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ =\ \log\left(2x+1\right)\ -\ \frac{1}{2}\log\left(8x+1\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ -\ \log\left(2x+1\right)\ +\ \frac{1}{2}\log\left(8x+1\right)=0$$
$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2(8x+1)\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right) = 0$$
$$\log\left(\frac{x^{2}\left(8x+1\right)}{(2x-1)(4x^2 + 4x + 1)}\right)=\ 0\ \tag{1}$$
I'm not getting any idea what to do with this further. Also, the original equation has only $x = 1$ as a solution but the equation $(1.)$ has $2$ solutions $x = 1; x= -1/3$.
I think there might be a simpler solution to this equation rather than my cumbersome one.
Not sure what's the mistake.
|
$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
Note that $$\log\left(x^2\right)=t\kern.6em\not\kern-.6em\implies2\log(x)=t,$$ but that $$2\log(x)=t\iff\log\left(x^2\right)=t \;\text{ and }\;x>0.$$ Neglecting the implicit condition $x>0$ was how, in the above step, you introduced the extraneous solution $x= -\frac13.$
In general, for $n\in\mathbb Z,$ $$n\log x\not\equiv \log x^n;$$ instead, $$n\log |x|\equiv \log |x|^n.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximal and minimal eigenvalues of a symmetric tridiagonal Toeplitz matrix
Given $m \times m$ symmetric tridiagonal Toeplitz matrices $$M=\begin{pmatrix}
4 & 1 & & \\ 1 & 4 & \ddots & \\ & \ddots & \ddots & 1\\ & & 1 & 4\end{pmatrix}, \qquad A=\begin{pmatrix} 2 & -1 & & \\ -1 & 2 & \ddots & \\ & \ddots & \ddots & -1\\ & & -1 & 2\end{pmatrix} $$ determine the maximal and minimal eigenvalues of $$F = M + h^{-\alpha} A$$ with $\alpha \in \mathbb{Z}, h=\frac{1}{m+1}$ when $\mu_j = 4 + 2\cos(jh\pi)$ are the eigenvalues of $M$ and $\eta_j = 2 - 2\cos(jh\pi)$ are the eigenvalues of $A$ with $j=1,\dots, m$.
I guess it is $\lambda_j=4+2\cos(jh\pi) + h^{-\alpha}(2-2\cos(jh\pi))$ but I cannot show it. To determine the eigenvalues of
$$M=\begin{pmatrix}
4 & 1 & & \\
1 & 4 & \ddots & \\
& \ddots & \ddots & 1\\
& & 1 & 4 \\
\end{pmatrix}$$
I used the ansatz $\det(M-\mu_jI)=\det(J-(\mu_j-4)I)$ with
$$J=\begin{pmatrix}
0 & 1 & & \\
1 & 0 & \ddots & \\
& \ddots & \ddots & 1\\
& & 1 & 0 \\
\end{pmatrix}$$
for which the eigenvalues are $\sigma_j=2\cos(jh\pi)$ and I hope $\mu_j=4+2\cos(jh\pi)$ is correct! The eigenvalues of $A$ were already given. The maximal eigenvalue of $M$ is $\mu_1$ but the maximal eigenvalue of $A$ is $\eta_m$.
I need to show that the spectral condition $\kappa(F)=\frac{\lambda_{max}}{\lambda_{min}}$ in first approximation of $h$ (means with the big O notation) is $$\kappa(F)\sim \begin{cases}
3-6h^{-\alpha}, \alpha \leq -1\\
1, \alpha=0 \text{ (done})\\
\frac{2}{3}h^{-1}, \alpha=1 \\
\frac{4}{6+\pi^2}h^{-2}, \alpha=2 \\
\frac{4}{\pi^2}h^{-2}, \alpha\geq 3
\end{cases}$$
Consider $\alpha \leq -1$, then it is $$\kappa(F)=\frac{\lambda_{\max}}{\lambda_{\min}}=\frac{4+2h^{-\alpha} +2\cos(\pi h)(1-h^{-\alpha})}{4+2h^{-\alpha} -2\cos(\pi h)\,(1-h^{-\alpha})} \sim \frac{4+2h^{-\alpha} +2(1-h^{-\alpha})}{4+2h^{-\alpha} -2\,(1-h^{-\alpha})} $$
since $$cos(h\pi)=1-O(h^2)$$ with $\alpha \leq -1$. I cannot show that $\kappa(F)=3-6h^{-\alpha}$ for $\alpha\leq -1$...
| According to the notation $$M=4I+J,\quad A=2I-J$$ Therefore
$$ M+h^{-\alpha}A =(4+2h^{-\alpha})\,I +(1-h^{-\alpha})\,J$$ Every eigenvalue of $M+h^{-\alpha}A$ is thus of the form $$ [4+2h^{-\alpha}]+(1-h^{-\alpha})\, \lambda\qquad (*)$$ where $\lambda$ is an eigenvalue of $J.$ The eigenvalues of $J$ are known, and are listed in the question. That's actually a nontrivial fact which follows from the trigonometric identity
$$2\cos \theta\,\sin k\theta= \sin (k+1)\theta+ \sin (k-1)\theta$$ for $\theta $ satisfying $\sin(m+1)\theta =0.$
In the question it is assumed that $\alpha\in \mathbb{N}.$ I will consider $\alpha>0.$
Hence $h^{-\alpha}>1$ and by $(*)$
the maximal eigenvalue $\lambda_{\max}$ corresponds to the minimal eigenvalue $\sigma_{\min}$ of $J,$ and the minimal one $\lambda_{\min}$ to the maximal eigenvalue $\sigma_{\max}$ of $J.$
We have
$$\sigma_{\max}=2\cos{\pi\over m}=2\cos \pi h,\qquad \sigma_{\min}=2\cos{\pi(m-1)\over m}=-\sigma_\max$$
The ratio $\lambda_{\max}/\lambda_{\min}$ is thus equal
$$\displaylines{{4+2h^{-\alpha} -2\cos {\pi h}\,(1-h^{-\alpha}) \over 4+2h^{-\alpha} +2\cos {\pi h}\,(1-h^{-\alpha}) }={2+4h^\alpha +2\cos {\pi h}\,(1-h^{\alpha}) \over 2+ 4h^\alpha -2\cos {\pi h}\,(1-h^{\alpha})}\\ =
{4+2h^\alpha-2(1-\cos\pi h)(1-h^\alpha )\over 6h^\alpha+2(1-\cos\pi h)(1-h^\alpha )}}$$
Applying $$2(1-\cos x)={x^2}+O(x^4),\qquad x\to 0,$$ gives
$$ \displaylines{4+2h^\alpha-2(1-\cos\pi h)(1-h^\alpha )=4+2h^\alpha -\pi^2h^2+O(h^{2+\alpha})+O(h^4)\\
=\begin{cases}4+2h^\alpha +O(h^2) & \alpha<2\\
4+(2-\pi^2)h^2+O(h^4) &\alpha=2\\
4 -\pi^2h^2+O(h^\alpha) &\alpha>2
\end{cases}}
$$
Next
$$ \displaylines{6h^\alpha+2(1-\cos\pi h)(1-h^\alpha )=6h^\alpha +\pi^2h^2+O(h^{2+\alpha})+O(h^4)\\
=\begin{cases}6h^\alpha +O(h^2) & \alpha<2\\
(6+\pi^2)h^2+O(h^4) &\alpha=2\\
\pi^2h^2+O(h^\alpha) &\alpha>2
\end{cases}}$$
Therefore
$${\lambda_\max\over \lambda_\min}=\begin{cases}
\displaystyle {2\over 3}h^{-\alpha} +O(1) &\alpha <2\\
\displaystyle {4\over 6+\pi^2}h^{-2}+O(1) & \alpha=2\\
\displaystyle {4\over \pi^2}h^{-2} +O(1) & \alpha>2
\end{cases}$$
| {
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$\forall a\ge 2, \exists n\ge 1$ such that $2n+1\mid a+n!$ Let $a\ge 2$ be an integer. Prove that $\exists n\ge 1$ such that $$2n+1\mid a+n! \text{ or }a-n!$$
I think we'll construct a solution for $n$ as a function of $a$. But I have no idea what that function looks like. I just took care of some trivial cases.
First if $a$ is a factorial number i.e. $a=k!$ then set $n=k$
If $a$ is of the form $k!+2k+1$ then set $n=k$ and observe that $$a-n!=2k+1$$
Note that if $a\equiv 1$ or $2$ $\pmod 3$ then set $n=1$,so we only have to deal with multiples of $3$ now .
| As you've already noted, $n = 1$ works with either $a \equiv 1 \pmod{3}$ (since $3 \mid a - 1$) or $a \equiv 2 \pmod{3}$ (since $3 \mid a + 1$). Also, as aschepler's question comment indicates, $n = 4$ works where $a$ has just one factor of $3$. This is because with $a = 3i$, then if $i \equiv 1 \pmod{3}$, we get $a + 4! = 3(i + 8) \equiv 0 \pmod{9}$, while if $i \equiv 2 \pmod{3}$, then $a - 4! = 3(i - 8) \equiv 0 \pmod{9}$.
This just leaves where $a$ is an integral multiple of $9$. As Dan's question comment indicates, $n = 20$ works for $a = 9$. Consider next if $a$ has any other odd factors, say $j \gt 1$. If so, then $2n + 1 = 9j \; \to \; n = \frac{9j - 1}{2}$ works since $3(3j) \mid a$, with $3 \neq 3j$, plus $3 \lt n$ and $3j \lt n$, so $3(3j) \mid n!$, which means $2n + 1 \mid a + n!$ (as well as $2n + 1 \mid a - n!$).
The final case to consider is with $a = 9(2^k)$ for $k \ge 1$. Then for $k = 1$ let $b = a + 1$, and for $k \ge 2$ let $b = a - 1$. Then $b \equiv 3 \pmod{4}$, so there's at least one prime $p \equiv 3 \pmod{4}$ where $p \mid b$. Let $p = 2n + 1 \; \to \; n = \frac{p - 1}{2}$. Then as shown in If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$, we get that $n! \equiv \pm 1 \pmod{p}$. Thus, since $a \equiv \pm 1 \pmod{p}$, we have that either $2n + 1 \mid a + n!$ or $2n + 1 \mid a - n!$.
This concludes the proof since all cases for $a$ have been handled.
| {
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Show that $\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ=\cos7^\circ$
Show that $$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ=\cos7^\circ$$
NOTE: I have seen the other questions and solutions for this problem. I have a particular question if my idea has a potential.
I decided to rearrange the LHS (without a particular reason, it just felt right to me) as follows $$\begin{align}(&\sin47^\circ-\sin11^\circ)+(\sin61^\circ-\sin25^\circ)\\&=2\cos\frac{47^\circ+11^\circ}{2}\sin\frac{47^\circ-11^\circ}{2}+2\cos\frac{61^\circ+25^\circ}{2}\sin\frac{61^\circ-25^\circ}{2}\\&=2\cos29^\circ\sin18^\circ+2\cos43^\circ\sin18^\circ\\&=2\sin18^\circ(\cos29^\circ+\cos43^\circ)\\&=4\sin18^\circ\cos36^\circ\cos7^\circ\end{align}$$
If this won't work, what is the intuition that leads to the appropriate rearranging?
| $$4\sin18^\circ\cos36^\circ\cos7^\circ=\frac{4\sin18^\circ\cos18^\circ\cos36^\circ\cos7^\circ}{\cos18^\circ}=\frac{2\sin36^\circ\cos36^\circ\cos7^\circ}{\cos18^\circ}=\\
\frac{\sin72^\circ\cos7^\circ}{\cos18^\circ}=\cos7^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find all integer values of $m$ such that the equations $x+y+\sqrt{4-x^2}+\sqrt{4-y^2}=4$ and $x\sqrt{4-x^2}+y\sqrt{4-y^2}=m+2$ have a solution.
Consider the system of equation $\left\{ \begin{aligned} x + y + \sqrt{4 - x^2} + \sqrt{4 - y^2} &= 4\\ x\sqrt{4 - x^2} + y\sqrt{4 - y^2} &= m + 2 \end{aligned} \right. (x; y \in \mathbb R)$. How many integer values of $m$ are there such that there exists a solution to the above system?
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Of course, it's definitely not $$x\sqrt{4 - x^2} + y\sqrt{4 - y^2} \le \dfrac{(x + \sqrt{4 - x})^2 + (y + \sqrt{4 - y^2})^2}{2}$$, nor is it $$\left(x\sqrt{4 - x^2} + y\sqrt{4 - y^2}\right)^2 \le (x^2 + y^2)[(4 - x^2) + (4 - y^2)]$$
Let's just try to solve this the typical way.
Consider the function $f(z) = z + \sqrt{4 - z^2}$. The domain and codomain of the function are respectively $[-2; 2]$ and $[-2; 2\sqrt 2]$.
(This is an unnecessary step, but I'm going to include it anyway.
Since $f(x) + f(y) = 4$, this means $$f(x), f(y) \in [4 - 2\sqrt 2; 2\sqrt 2] \iff x, y \in \left[2 - \sqrt 2 - 2\sqrt{\sqrt 2 - 1}; 2\right]$$ This is just to further strengthen the domains of $f(x)$ and $f(y)$ specifically. Strengthen's definitely not the right word.)
Consider function $g(z) = z\sqrt{4 - z^2}$. For $\displaystyle x, y \in \left[2 - \sqrt 2 - 2\sqrt{\sqrt 2 - 1}; 2\right]$, the codomain of $g(x)$ and $g(y)$ are $[10 - 8\sqrt{2}; 2]$. Therefore, $g(x) + g(y) \in [20 - 16\sqrt{2}; 4]$, right?
If only it was that easy.
How about this? Let $\left\{ \begin{aligned} x = 2\cos a, \sqrt{4 - x^2} &= 2\sin a\\ y = 2\cos b, \sqrt{4 - y^2} &= 2\sin b \end{aligned} \right.$, (since $x^2 + (\sqrt{4 - x^2})^2 = y^2 + (\sqrt{4 - y^2})^2 = 4$). Afterwards, the system of equation could be rewritten as $$\left\{ \begin{aligned} \sin\left(\dfrac{a + b}{2} + \dfrac{\pi}{4}\right)\cos\left(\dfrac{a - b}{2}\right) &= \dfrac{\sqrt 2}{2} &&(1)\\ \sin(a + b)\cos(a - b) &= \dfrac{m + 2}{4} &&(2) \end{aligned} \right.$$
Combining $(1)$ and $(2)$, we have $\sin(a + b)\left[\dfrac{1}{\sin^2\left(\dfrac{a + b}{2} + \dfrac{\pi}{4}\right)} - 1\right] = \dfrac{m + 2}{4}$.
Let $\dfrac{a + b}{2} = z$ and function $$f(z) = 2\sin z\cos z\left[\dfrac{2}{(\sin z + \cos z)^2} - 1\right] \implies f'(z) = 2 \times \left[\dfrac{2(\cos z - \sin z)}{(\cos z + \sin z)^3} - \cos 2z\right]$$
How would one go about actually solving $f'(z) = 0$?
Hmmm~ well, it's obvious that $\cos z - \sin z = 0 \iff z = n\pi + \dfrac{\pi}{4}, \forall n \in \mathbb Z$ is one of the subset of roots.
Actually, let's graph the function.
(seeing that the $y$-intercept extends to $-\infty$) How about we don't graph the function?
Aside from the apparent fact that $z\sqrt{4 - z^2} = \dfrac{(z + \sqrt{4 - z^2})^2 - 4}{2}$, I don't know what to do next. Also, it's past midnight already. So as always, thanks for reading, (and even so if you could help), and have a great tomorrow, everyone~
By the way, the options were $1, 2, 3$ and $4$, and there are seven integers between $20 - 16\sqrt{2}$ and $4$, so my first attempt would have led me to a dead end.
| $$\begin{align}
x + y + \sqrt{4-x^2} + \sqrt{4-y^2} &= 4 \tag 1\\
x\sqrt{4-x^2} + y\sqrt{4-y^2} &= m + 2 \tag 2
\end{align}$$
Obviously, $x,y\in[-2,2]$, that is we can represent $x$ and $y$ as
$$\begin{align}
x &= 2\sin a \\
y &= 2\sin b
\end{align}$$
which transforms (1) and (2) into
$$\begin{align}
\sin a + \sin b + \cos a + \cos b &= 2 \tag{1a}\\
\sin a \cos a + \sin b \cos b &= m + 2 \tag{2a}\\
\end{align}$$
where $a,b\in[-\pi/2, \pi/2]$ because the square roots are non-negative, and similar must apply to $\cos$.
We can rewrite (2a) as
$$\sin 2a + \sin 2b = 4+2m \tag{2b}$$
which constrains $m$ to
$$
-3\leqslant m\leqslant -1 \implies m\in\{-3,-2,-1\}
$$
Now let's try to find $a$, $b$ for these $m$:
*
*$m=-1 \implies a=b=\pi/4$ because the only way to get the right-hand side of (2b) to equal 2 is to have both sines equal to 1. However for (1a) this means $4\cdot 1/\sqrt 2 = 2$. Thus, $m=-1$ has no solution.
*$m=-3 \implies a=b=-\pi/4$ because the only way to get the right-hand side of (2b) to equal -2 is to have both sines equal to −1. However for (1a) this means $0 = 2$. Thus, $m=-1$ has no solution.
*$m=-2 \implies \sin2a + \sin2b = 0$ which has solutions $a=-b$ and $a + b=\pm\pi/2$, yielding the sub-cases:
*
*$a=-b$ cancels the sines from (1a) and yields $\cos a+ \cos b=2$, i.e. $a=b=0$, i.e. $x=y=2$.
*$a =\pm\pi/2 -b$ is left as an exercise :-)
As the question just asks for the number of $m$'s such that the equations posess solution(s), the correct answer is "1"; no need to work out the last sub-case as it won't change the number of $m$'s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Criteria for $3 \times 3$ matrix to positive definite Here it is said that a $2\times 2$ matrix $A$ is positive definite if and only if $tr(A) >0$ and $det(A)>0$. This will not work if $A$ is $3\times 3$. But is there any way to enforce the positive definiteness of the matrix $A$ via the trace and determinant of $A$, if $A$ is of size $3\times 3$?
| No, it is not possible. Take any two real numbers $a$ and $b$ such that $a^2+b^2=1$ and the matrix
$$
A=
\begin{pmatrix}
1 & a & b \\
a & a^2+1 & a b \\
b & a b & -a^2+2
\end{pmatrix}.
$$
Then $\operatorname{tr}(A)=4$, $\det(A)=1$, and $A$ is definite positive.
But if you take two real numbers $a$ and $b$ such that $a^2+b^2=5$ and you define
$$
B=
\begin{pmatrix}
1&a&b\\
a&a^2-1&ab\\
b&ab&-a^2+4
\end{pmatrix},
$$
then $\operatorname{tr}(B)=4$, $\det(B)=1$, but $B$ is not definite positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find $ \int \frac{dx}{(x^2 + a^2)^3} $ I would like to find the anti-derivative
$$\displaystyle \int \dfrac{dx}{(x^2 + a^2)^3 }$$
My attempt:
By substitution: $ x = a \tan(\theta) \Rightarrow dx = a \sec^2(\theta) d\theta$
Then the integral becomes
$$\displaystyle \int \dfrac{ a \sec^2(\theta) d\theta }{a^6 \sec^6(\theta) } $$
And this reduces to
$$\dfrac{1}{a^5} \displaystyle \int \cos^4(\theta) d\theta $$
Now, $\cos^4(\theta) = ( \cos^2(\theta) )^2 = \dfrac{1}{4} (1 + \cos(2 \theta) )^2 = \dfrac{1}{4} ( 1 + 2 \cos(2 \theta) + \cos^2(2 \theta) ) $
So this reduces to
$$ \cos^4(\theta) = \dfrac{1}{4} + \dfrac{1}{2} \cos(2 \theta) + \dfrac{1}{8} (1 + \cos(4 \theta) ) $$
and these are integrable easily, but I want to relate their integrals to the original variable $x$.
So, now I have the anti-derivative as
$$ \dfrac{3}{8} \theta + \dfrac{1}{4} \sin(2 \theta) + \dfrac{1}{32} \sin(4 \theta) $$
Since $ x = a \tan(\theta) $, then $ \cos(\theta) = \dfrac{a}{\sqrt{x^2 + a^2}}$, and $\sin(\theta) = \dfrac{x}{\sqrt{x^2 + a^2}} $
Hence
$$\sin(2 \theta) = \dfrac{ 2 a x }{x^2 + a^2} $$
$$\cos(2 \theta) = \dfrac{a^2 - x^2}{x^2 + a^2} $$
And finally,
$$\sin(4 \theta) = \dfrac{4 a x (a^2 - x^2) }{ (x^2 + a^2)^2} $$
Hence, my integral becomes
$$ \displaystyle \int \dfrac{dx}{(x^2 + a^2)^3} = \dfrac{1}{a^5}\left( \dfrac{3}{8} \tan^{-1}\left(\dfrac{x}{a}\right) + \dfrac{1}{2} \dfrac{a x }{x^2 + a^2} + \dfrac{1}{8} \dfrac{ a x (a^2 - x^2) }{ (x^2 + a^2)^2 } \right) + C $$
where $C$ is an arbitrary constant of integration.
And my question is: Are my approach and derivation correct?
| Alternatively, integrate by parts to avoid the trig substitution
\begin{align}
I_3=&\int \frac{1}{(x^2 + a^2)^3 }dx
=\int \frac{1}{4a^2x^3}d\left(\frac{x^4}{ (x^2 + a^2)^2}\right)
= \frac{x}{4a^2 (x^2 + a^2)^2}+\frac3{4a^2}I_2\\
I_2=&\int \frac{1}{(x^2 + a^2)^2}dx
=\int \frac{1}{2a^2x}d\left(\frac{x^2}{ x^2 + a^2}\right)
=\frac{x}{2a^2 (x^2 + a^2)}+\frac1{2a^2}I_1\\
I_1=&\int \frac{1}{x^2 + a^2}dx= \frac1a \tan^{-1}\frac xa
\end{align}
which leads to
$$I_3
= \frac{x}{4a^2 (x^2 + a^2)^2}+ \frac{3x}{8a^4 (x^2 + a^2)}+
\frac{3}{8a^5}\tan^{-1}\frac xa +C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Eigenvalue and spectral condition
Let $A=\begin{pmatrix} 1& 1 \\ a^2 &1 \end{pmatrix} \text{ with } a\in (0,\frac{1}{2}]$. Show $$cond_2(A)=||A||_2 \cdot ||A^{-1}||_2\leq 4(1-a^2)^{-1}$$ by first showing $||A||^2_2\leq||A||_1||A||_{\infty}$.
$||A||^2_2$ is the maximal eigenvalue of $A^TA$ and $||A||_1||A||_{\infty}=2\cdot2 = 4$. Prove $||A||^2_2\leq||A||_1||A||_{\infty}=4$ by contradiction: Suppose $\lambda_{max} >4$, then $$4<\lambda_{\ast}=\frac{\sqrt{a^8+2a^4+4a^2+1}+a^4+3}{2} \iff 5<\sqrt{a^8+\underbrace{2a^4}_{\leq \frac{1}{2}}+\underbrace{4a^2}_{\leq 1}+1}+\underbrace{a^4}_{\leq 1} \\ \leq \sqrt{4}+1 = 3$$ contradiction!
So I know $||A||^2_2\leq 4 \Rightarrow ||A||_2\leq 2$ but I need to show $cond_2(A) = \underbrace{||A||_2}_{\leq 2}||A^{-1}||_2\leq 4(1-a^2)^{-1}$. I don't know why I needed to show $||A||^2_2\leq||A||_1||A||_{\infty}$ first? How does it help? I still need the maximal eigenvalue of $(A^{-1})^TA^{-1}$. The eigenvalues are $$\lambda_1=\frac{\sqrt{a^4-2a^2+5}-a^2+1}{2\sqrt{a^4-2a^2+5}-4} \\ \lambda_2=\frac{\sqrt{a^4-2a^2+5}-a^2-1}{2\sqrt{a^4-2a^2+5}+4}$$ What am I supposed to do now? Thanks for any help!
| $0 \lt 1-a^2 = \det\big(A)=\lambda_1\cdot \lambda_2 = \sigma_1\cdot \sigma_2$
i.e. the product of singular values gives the modulus of the determinant (and in particular when the determinant positive it is equal to the product of singular values)
proof: using polar decomposition
$\det\big(A\big) =\det\big(UP\big)=\det\big(U\big)\det\big(P\big)=\pm \det\big(P\big) = \pm(\sigma_1\cdot\sigma_2)$
now I assume you know that $||A||_2 = \sigma_1$ and $||A^{-1}||_2=\frac{1}{\sigma_2}$ so we may write
$\text{cond}_2(A)=||A||_2 \cdot ||A^{-1}||_2 = \frac{\sigma_1}{\sigma_2}=\frac{\sigma_1^2}{\sigma_1\sigma_2}\lt \frac{\sigma_1^2+\sigma_2^2}{\sigma_1\sigma_2}=\frac{\big \Vert A\big \Vert_F^2}{\det\big(A\big)}= \frac{3+a^4}{1-a^2}\lt \frac{4}{1-a^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate/asymptotic of the sum $\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$ I am trying to evaluate the sum and its asymptotic limit as $N \rightarrow \infty$ of
$$\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$$
where the upper limit can be either $L \left({N}\right) = {L}_{1} \left({N}\right) = \left\lfloor{\frac{N}{6} - \frac{1}{2}}\right\rfloor$ or $L \left({N}\right) = {L}_{2} \left({N}\right) = \left\lfloor{\frac{1}{2} \left\lceil{\sqrt{N + 2}}\right\rceil}\right\rfloor-1$.
It looks like I should count the number of even and odd terms of $$\left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$$
The case for ${L}_{2} \left({N}\right)$ when $N \rightarrow \infty$ looks to be $\sim \sqrt[4]{N}$. The similar case for ${L}_{1} \left({N}\right) \sim \alpha\, N$ for some constant $\alpha$.
My literature search has not revealed any number theory equivalents nor anything for the asymptotic limits. I suspect that the sum equivalent of the sum of the number of divisors or $\sum_{n = 1}^{x} \left\lfloor{\frac{N}{n}}\right\rfloor$ has been broken down into the even and odd floor function results, but again I am not finding any literature on this subject.
These sums arise from computing the exact number of factorable quartics of the form $\left({x+a}\right)^{2} \left({x+b}\right) \left({x+c}\right)$ where $a, b, c$ are integer roots.
| I get
$ \frac{N}{2}(1-2\ln(2))+O(L(N))
$.
So we want
$m \le N/(2a+1) \lt m+1
$
or
$m/N \le 1/(2a+1) \lt (m+1)/N
$
or
$N/m \ge 2a+1 > N/(m+1)
$
or
$(\frac{N}{m}-1)/2 \ge a \gt (\frac{N}{m+1}-1)/2
$
or
$\frac{N-m}{2m} \ge a \gt \frac{N-m-1}{2(m+1)}
$.
The number of these is about
$\begin{array}\\
\frac{N-m}{2m} -\frac{N-m-1}{2(m+1)}
&=\frac{(N-m)(m+1)-(N-m-1)(m)}{2m(m+1)}\\
&=\frac{(N-m)m+(N-m)-(N-m)m+m}{2m(m+1)}\\
&=\frac{N}{2m(m+1)}\\
\end{array}
$
so the sum is about
$\begin{array}\\
s(N)
&=\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}
\\
&\approx \sum_{m=1}^{N/(2L(N))} \frac{N}{2m(m+1)}(-1)^m\\
&= \frac{N}{2}\sum_{m=1}^{f(N)} \frac{(-1)^m}{m(m+1)}
\qquad f(N)=N/(2L(N))\\
&= \frac{N}{2}\sum_{m=1}^{f(N)} (-1)^m(\frac1{m}-\frac1{m+1})\\
&= \frac{N}{2}\left(\sum_{m=1}^{f(N)} \frac{(-1)^m}{m}-\sum_{m=1}^{f(N)} \frac{(-1)^m}{m+1}\right)
\\
&= \frac{N}{2}\left(\sum_{m=1}^{f(N)} \frac{(-1)^m}{m}-\sum_{m=2}^{f(N)+1} \frac{(-1)^{m-1}}{m}\right)\\
&= \frac{N}{2}\left(\sum_{m=1}^{f(N)} \frac{(-1)^m}{m}+\sum_{m=2}^{f(N)+1} \frac{(-1)^{m}}{m}\right)\\
&= \frac{N}{2}\left((-1)+\sum_{m=2}^{f(N)} \frac{(-1)^m}{m}+\sum_{m=2}^{f(N)} \frac{(-1)^{m}}{m}+\frac{(-1)^{f(N)+1}}{f(N)+1}\right)\\
&= \frac{N}{2}\left((-1)+2\sum_{m=2}^{f(N)} \frac{(-1)^m}{m}+\frac{(-1)^{f(N)+1}}{f(N)+1}\right)\\
&\approx \frac{N}{2}\left((-1)+2(1-\ln(2))+\frac{(-1)^{f(N)+1}}{f(N)+1}\right)\\
&\approx \frac{N}{2}(1-2\ln(2))+\frac{(-1)^{f(N)+1}N}{2(f(N)+1)}\\
&\approx \frac{N}{2}(1-2\ln(2))+\frac{(-1)^{f(N)+1}N}{2N/(2L(N))}\\
&= \frac{N}{2}(1-2\ln(2))+(-1)^{f(N)+1}L(N)\\
&= \frac{N}{2}(1-2\ln(2))+O(L(N))\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/4455284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x$ Latest Edit
As suggested by Mr Claude Leibovici, I go further to investigate the integral in general,
$$I_{m}:=\int_{0}^{\frac{\pi}{2}} x^{m} \ln (\cos x) d x$$
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^m$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$$\begin{aligned}
I_m&=-\int_{0}^{\frac{\pi}{2}} x^{m} \ln 2 d x+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !} \int_{0}^{\frac{\pi}{2}} x^{m} \cos (2 n x) d x\\&= -\frac{\pi^{m+1}\ln 2}{(m+1) 2^{m+1}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} J(m, n)\end{aligned}
$$
Next we need a reduction formula on $J(m,n)$.
Applying integration by parts twice, we obtain
$$
\begin{aligned}J(m, n)&=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m} d(\sin 2 n x)\\ &= -\frac{m}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m-1} \sin 2 n x d x\\&=\frac{m}{4 n^{2}} \int_{0}^{\frac{\pi}{2}} x^{m-1} d(\cos 2 n x)\\&=\frac{(-1)^{n} \pi^{m-1} m}{2^{m+1} n^{2}} -\frac{m(m-1)}{4 n^{2}} J (m-2, n)\end{aligned}
$$
By the reduction formula of $J(m,n)$, we can find $I_m$ by plugging $J(m,n)$.
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^2$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$\displaystyle I=-\underbrace{\int_{0}^{\frac{\pi}{2}} x^{2} \ln 2 d x}_{\frac{\pi^{3} \ln 2}{24}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \underbrace{ \int_{0}^{\frac{\pi}{2}} x^{2} \cos (2 n x) d x}_{J_n}\tag*{} $
Integrating by parts twice yields
$\displaystyle \begin{aligned}J_{n} &=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{2} d(\sin 2 n x) \\&=\frac{1}{2 n}\left[x^{2} \sin 2 n x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{n} \int_{0}^{\frac{\pi}{2}} x \sin 2 n x d x \\&=\frac{1}{2 n^{2}} \int_{0}^{\frac{\pi}{2}} x d(\cos 2 n x) \\&=\left[\frac{1}{2 n^{2}} x \cos 2 n x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2 n^{2}} \int_{0}^{\frac{\pi}{2}} \cos 2 n x d x \\&=\frac{\pi}{4 n^{2}} \cos n \pi-\frac{1}{2 n^{2}}\left[\frac{\sin 2 n x}{2 n}\right]_{0}^{\frac{\pi}{2}} \\&=\frac{\pi}{4 n^{2}} \cos n \pi\end{aligned}\tag*{} $
We can now conclude that
$\displaystyle \begin{aligned}I &=-\frac{\pi^{3} \ln 2}{24}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cdot \frac{\pi}{4 n^{2}} \cos n \pi \\&=-\frac{\pi^{3} \ln 2}{24}+\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^{n}}{n^{3}} \\&=-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{1}{n^{3}} \\&=\boxed{-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \zeta(3)}\end{aligned}\tag*{} $
Is there any other method to evaluate $I$?
| Using $(*)$, I think that we could even make the problem more general using the incomplete gamma function or the exponential integral function
$$\int x^p \cos(2nx)\,dx=-\frac{1}{2} x^{p+1} (E_{-p}(-2 i n x)+E_{-p}(2 i n x))$$
$$K_p=\int_0^{\frac \pi 2} x^p \cos(2nx)\,dx=$$ $$-i^{p+1} 2^{-p-2} n^{-p-1} \left(\Gamma (p+1,-i n \pi )-(-1)^p \Gamma (p+1,i n \pi
)+\left((-1)^p-1\right) \Gamma (p+1)\right)$$ which simplifies a lot if $p$ is even
$$K_{2m}=i\,(-1)^{m+1} \,2^{-2 (m+1)}\,n^{-(2m+1)}\,(\Gamma (2 m+1,-i n \pi )-\Gamma (2 m+1,i n \pi ))$$
So, as you found
$$K_2=\frac{\left(\pi ^2 n^2-2\right) \sin (\pi n)+2 \pi n \cos (\pi n)}{8 n^3}=(-1)^n\frac{\pi }{4 n^2}$$
$$K_4=(-1)^n \frac{\pi \left(\pi ^2 n^2-6\right)}{8 n^4}\qquad \qquad K_6=(-1)^n \frac{3 \pi \left(\pi ^4 n^4-20 \pi ^2 n^2+120\right)}{64 n^6}$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_2=-\frac{\pi }{4}\zeta (3)$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_4=-\frac{\pi}{8} \left(\pi ^2 \zeta (3)-6 \zeta (5)\right)$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_6=-\frac{3\pi}{64} \left(\pi ^4 \zeta (3)-20 \pi ^2 \zeta (5)+120 \zeta (7)\right)$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_8=-\frac{\pi }{64} \left(\pi ^6 \zeta (3)-42 \pi ^4 \zeta (5)+840 \pi ^2 \zeta (7)-5040
\zeta (9)\right)$$
Edit
In fact, no much problem if $p$ is odd
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_1=-\frac{7 }{16}\zeta (3)$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_3=-\frac{3}{128} \left(8 \pi ^2 \zeta (3)-31 \zeta (5)\right)$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_5=-\frac{5}{512} \left(8 \pi ^4 \zeta (3)-96 \pi ^2 \zeta (5)+381 \zeta (7)\right)$$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} K_7=-\frac{7}{4096} \left(16 \pi ^6 \zeta (3)-480 \pi ^4 \zeta (5)+5760 \pi ^2 \zeta (7)-22995 \zeta
(9)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solving the integral of $\cos^2x\sin^2 x$ Solving the integral of $\cos^2x\sin^2 x$:
My steps are: $(\cos x\sin x)^2=\left(\frac{\sin(2x)}{2}\right)^2$. Now we know that
$$\sin^2(\alpha)=\frac{1-\cos (2x)}{2}\iff\left(\frac{\sin(2x)}{2}\right)^2=\frac 14\sin^2(2x)=\frac 14\cdot \frac{1-\cos (4x)}{2}=\frac 18(1-\cos (4x))$$
Rewriting all the steps:
$$\int \sin ^2\left(x\right)\cos ^2\left(x\right)dx=\frac{1}{8}\left(x-\frac{1}{4}\sin \left(4x\right)\right)+k, \,\, k\in \Bbb R \tag 1$$
Is there another method to solve this integral (1)?
| Integrate by parts
\begin{align}
\int \cos^2x\sin^2 x \ dx =& -\frac1{16}\int\tan 2x\ d(\cos^2 2x)= -\frac1{16}\tan2x\cos^22x+\frac18 x+C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{k=1}^{n} a_k \left(1-\frac{k^2}{n^2}\right) \to 1$
Problem : Given $a_n$ which satisfies $\displaystyle\sum a_n = 1$, prove that
$$\lim_{n\to\infty}\sum_{k=1}^na_k\left(1-\frac{k^2}{n^2}\right) = 1.$$
My Attempt
From $\displaystyle\sum a_n = 1$, I changed this problem to show $\displaystyle\lim\sum\frac{a_kk^2}{n^2}\to 0$.
From given condition, I know that $a_n \to 0$ and it means that $a_n$ is bounded.
So, for arbitrary large $n$, $|a_n|<1$, and I tried to bound an original sum
$$\left|\sum\frac{a_kk^2}{n^2}\right|\le \frac{1}{n^2}\left|\sum a_kk^2\right|\le \frac{1}{n^2}\sum k^2$$
but I failed because the right term goes to infinity.
Thanks for help.
| The standard approach to tacking abelian-type results like this is to consider $s_n = \sum_{k=1}^{n} a_k$ and write the sum in terms of $(s_n)$ using summation by parts:
\begin{align*}
\sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right)
&= \sum_{k=1}^{n} (s_k - s_{k-1}) \left( 1 - \frac{k^2}{n^2} \right) \\
&= \sum_{k=1}^{n} s_k \left( 1 - \frac{k^2}{n^2} \right) - \sum_{k=1}^{n} s_{k-1} \left( 1 - \frac{k^2}{n^2} \right) \\
&= \sum_{k=1}^{n-1} s_k \left( 1 - \frac{k^2}{n^2} \right) - \sum_{k=1}^{n-1} s_{k} \left( 1 - \frac{(k+1)^2}{n^2} \right) \\
&= \sum_{k=1}^{n-1} s_k \frac{2k+1}{n^2}.
\end{align*}
Then by using $\sum_{k=1}^{n-1} \frac{2k+1}{n^2} = 1 - \frac{1}{n^2}$, the difference $d_n$ between the sum $\sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right)$ and the limit candidate $1$ is bounded by
\begin{align*}
d_n := \left| \sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right) - 1 \right|
&= \left| \sum_{k=1}^{n-1} (s_k - 1) \frac{2k+1}{n^2} + \frac{1}{n^2} \right| \\
&\leq \sum_{k=1}^{n-1} \left| s_k - 1 \right| \frac{2k+1}{n^2} + \frac{1}{n^2}.
\end{align*}
Now, for an arbitrary $\varepsilon > 0$, choose $N$ such that $|s_k - \varepsilon| < \varepsilon$ whenever $k \geq N$. Then for $n > N$, by splitting the last sum into two parts, one over $k < N$ and the other over $k \geq N$, $d_n$ is further bounded by
\begin{align*}
d_n
&\leq \sum_{k=1}^{N-1} \left| s_k - 1 \right| \frac{2k+1}{n^2}
+ \sum_{k=N}^{n-1} \left| s_k - 1 \right| \frac{2k+1}{n^2}
+ \frac{1}{n^2} \\
&\leq \left( \max_{k < N} |s_k - 1| \right) \left( \sum_{k=1}^{N-1} \frac{2k+1}{n^2} \right)
+ \varepsilon \left( \sum_{k=N}^{n-1} \frac{2k+1}{n^2} \right)
+ \frac{1}{n^2} \\
&\leq \left( \max_{k < N} |s_k - 1| \right) \frac{N^2}{n^2}
+ \varepsilon
+ \frac{1}{n^2}.
\end{align*}
Letting limsup as $n\to\infty$, this bound reduces to
\begin{align*}
\limsup_{n\to\infty} d_n
&\leq \varepsilon
\end{align*}
However, since the left-hand side does not depend on $\varepsilon$, we may let $\varepsilon \to 0^+$ to obtain
$$ \limsup_{n\to\infty} \left| \sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right) - 1 \right| = 0. $$
Therefore the desired assertion follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$ So, my friend has challenged me to solve the following integral
$\displaystyle \tag*{} I=\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$
I started by doing $x^2=t$, so
$\displaystyle \tag{1} I=\frac 12 \int_0^\infty \frac{\arctan (x) \ln(1+x^2)}{x(x^4+x^2+1)} \mathrm dx$
Now, I first tried to solve this generalized integral:
$\displaystyle \tag*{} \mathcal L(m,n,a) = \int_0^\infty \frac{\arctan(mx)\ln(1+n^2x^2)}{x(a^2+x^2)} \ \mathrm dx$
We have $\mathcal L(0,1,a)=0$ and $\mathcal L(1,0,a)=0$. We now differentiate $\mathcal L(m,n,a)$ w.r.t $m$ first and then w.r.t $n$. We get:
$\displaystyle \tag{2} \begin{align} \partial m \ \partial n \ \mathcal L(m,n,a) &= \int_0^\infty \frac{2nx^2}{(a^2+x^2)(1+m^2x^2)(1+n^2x^2)} \ \mathrm dx \\\\
&= \frac{\pi n}{(m+n)(am+1)(an+1)} \\\\
&= \frac{\pi}{2(am+1)(an+1)} , \ \ \text{via symmetry} \end{align}$
And now taking the double integeral gives
$\displaystyle \tag{3} \mathcal L(1,1,a) = \int_0^1\int_0^1 \frac{\pi}{2(am+1)(an+1)} \ \mathrm dm \ \mathrm dn$
Using the elementary result:
$\displaystyle \tag*{} \int_0^1 \frac{1}{ay+1} \ \mathrm dy = \frac{\log(a+1)}{a}$
We find the value of $(3)$ as
$\displaystyle \tag*{} \mathcal L(1,1,a) = \frac{\pi \log^2(a+1)}{2a^2}$
Finally, we conclude that
$\displaystyle \tag*{}\int_0^\infty \frac{\arctan(x)\ln(1+x^2)}{x(a^2+x^2)} \ \mathrm dx={\color{Red} { \frac{\pi \log^2(a+1)}{2a^2}}}:= {\color{Blue}{ f(a)}}$
Now by completing square and performing, we have:
$\displaystyle \tag*{} \begin{align}\frac{1}{x^4+x^2+1} &=\frac{1}{\left(x^2+\frac 12\right)^2+\frac 34 }\\ &= \frac{1}{\left(x^2+\frac 12 - \frac{i \sqrt 3}{2}\right)\left(x^2+\frac 12 + \frac{i \sqrt 3}{2}\right)} \\ &= \frac{2}{i \sqrt 3} \left(\frac{1}{\left(x^2+\frac 12 - \frac{i \sqrt 3}{2}\right)}-\frac{1}{\left(x^2+\frac 12 + \frac{i \sqrt 3}{2}\right)}\right)\end{align} $
Adding all the pieces together, we get
$\displaystyle \tag{4} I=\frac{1}{i \sqrt 3} \left(f\left(\sqrt{\frac{1-i\sqrt3}{2}}\right) - f\left(\sqrt{\frac{1+i\sqrt3}{2}}\right)\right)$
My question is to simplify $(4)$ (if all my steps are correct) and also is there any short ways (methods others than contour integration) to destroy this integral?
Thanks :).
| Note that $a_\pm =\sqrt{\frac{1\pm i\sqrt3}{2}}=\frac{\sqrt3\pm i}2$
and
\begin{align}
I=&\ \frac 12 \int_0^\infty \frac{\arctan x\ln(1+x^2)}{x(x^4+x^2+1)} dx\\
=& \ \frac{1}{2\sqrt 3 i} \int_0^\infty \frac{\arctan x\ln(1+x^2)}{x(x^2+a_-^2)}-\frac{\arctan x\ln(1+x^2)}{(x^2+a_+^2)}\ dx\\
= & \ \frac\pi{4\sqrt3 i}\bigg(\frac{\ln^2(a_- +1)}{a_-^2}- \frac{\ln^2(a_+ +1)}{a_+^2} \bigg)\\
=& \ \frac\pi{4\sqrt3 i}\bigg(\frac{\ln(a_- +1)}{a_-}- \frac{\ln(a_++1)}{a_+} \bigg)\bigg(\frac{\ln(a_- +1)}{a_-}+\frac{\ln(a_+ +1)}{a_+} \bigg)\\
=& \ \frac\pi{4\sqrt3 i}\cdot\frac{\sqrt3 \pi -6\ln(2+\sqrt3)}{12i} \cdot \frac{6\sqrt3\ln(2+\sqrt3)+\pi}{12}\\
= &\ \frac\pi{16}\ln^2(2+\sqrt3)-\frac{\pi^2}{48\sqrt3}\ln(2+\sqrt3)-\frac{\pi^3}{576}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Proof $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition I'm trying to prove that $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition. Here is my proof so far:
My effort: Let $ \epsilon \gt 0 $, Choose $ \delta = min\{1, \}.$ Suppose $ 0 \lt \left|x - 3\right| < \delta.$
Check: $ \left|\frac{x-3}{x^2 - 5x + 6} - 1\right| = \left|\frac{x - 3}{\left(x - 2\right)\left(x - 3\right)} - 1\right| = \left|\frac{1}{x-2} - 1\right| = \left|\frac{1 - x + 2}{x - 2}\right| = \left|\frac{-x + 3}{x - 2}\right| = \frac{\left|x-3\right|}{\left|x-2\right|}.$
I want to find a lower bound $ M \gt 0$ such that $ \left|x - 2\right| \ge M $ so that I can write $ \frac{\left|x - 3\right|}{\left|x - 2\right|} \le \frac{\left|x -3\right|}{M}$.
I know that $ -1 \lt x-3 \lt 1 $ and therefore $ 0 \lt x - 2 \lt 2 $, but I can't choose $ 0 $ to be the lower bound because it needs to be positive. What can I do?
| $|\frac{x-3}{x^2-5x+6}-1|<\epsilon$
$ \Leftrightarrow$ $\frac{|x-3||x-3|}{|x-3||x-2|}<\epsilon$
$ \Leftrightarrow$$\frac{|x-3|}{|x-2|}<\epsilon$
let's put $|x-3|<\frac{1}{2}$ so $\frac{1}{|x-2|}<2$,so finally;
$|x-3|<\frac{\epsilon}{2}$ and $|x-3|<\frac{1}{2}$(our assumption),so we can put $\theta = min(\frac{\epsilon}{2},\frac{1}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Solve the following equation: $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Since real solutions are to be found, the domain of $x$ is $[-3; 5]$.
I immediately found that $15+2x-x^2$ can be factored to $(5-x)(x+3)$, this gave:
$\sqrt{x+3}+\sqrt{5-x}-2\sqrt{(5-x)(x+3)}=-4$
Squaring both sides was too complicated so I tried to substitute $a=\sqrt{5-x}$ for $a \in [0; \sqrt{8}]$ and $b=\sqrt{x+3}$ for $b \in [0; \sqrt{8}]$, this gave the new multivariable equation: $a+b-2ab=-4$
Another thing I noticed was $a^2+b^2=8$, together with the above equation, I got this system of equations: $\left\{ \begin{array}{l}
a + b - 2ab = - 4\\
{a^2} + {b^2} = 8
\end{array} \right.$
. Solving this by elimination is quite difficult for me.
This problem needs to be solved using algebra so I wonder how do I continue with this or are there any better way to solve this algebraically?
|
Squaring both sides was too complicated so I tried to substitute
$a = \sqrt{5−x}, ~b = \sqrt{x + 3}.$
This gave the new multivariable equation: $~a+b−2ab=−4$.
$a + b - 2ab = -4.$
$a^2 + b^2 = 8.$
This is nice work, so far.
Subtracting the 1st equation above, from the 2nd equation above gives
$(a^2 + b^2 + 2ab) - (a + b) = 8 - (-4) = 12$.
Let $u = (a + b).$
Then, $u^2 - u = 12 \implies u \in \{4,-3\}.$
However, by the constraints, $~~a,b~~$ must each be positive.
Therefore, $u = -3$ must be rejected.
Therefore, $a + b = u = 4.$
This implies that $2ab = 8 = a^2 + b^2 \implies (a - b)^2 = 0.$
Therefore, $a = b = 2,~$ since (again), $~~a,b~~$ must each be positive.
This forces $x = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
I need $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
$((2207^2 -2)^2 - 2)^2 -2)^2 ... - 2)$
quite big
But I realized I didn't have to simplify it, I just have to check if
$((((((2207^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2$ is divisible by 3
| Thank you all for those who answered. Different amazing solutions!
But I'd still share my final solution, this is to be done as fast as possible so there's no proving needed and algebraic/arithmetic manipulation is probably the desired/suggested solution (not that mine is the optimal one, I just find this easier).
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
What is needed: $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
Or simply, this can be done under mod 3:
$((((((((((-3)^2 - 2)^2 - 2)^2 - 2)^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2 $
The innermost is just $(-3)^2 - 2 \equiv 1~$ mod 3,
and then it can even be done mentally continuing the process $(1)^2 - 2 \equiv -1~$ mod 3, squaring: $(-1)^2$ and getting the same, $(-1)^2 - 2 \equiv -1~$ mod 3 .
Recognizing the pattern it's $-1 \equiv 2$ mod 3 all the way. So it isn't divisible by 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
complex integral $\int_0^{2\pi} (\cos x)^n \;dx$ I tried to solve following integral $$\int_0^{2\pi} (\cos x)^n \;dx$$
by saying $z = e^{ix}$ we get $dx=\frac{-i \;dz}{z}$ and $\cos x = \frac{1}{2}\left( z + \frac{1}{z}\right)$
$$\int_0^{2\pi} (\cos x)^n \;dx = \frac{-i}{2^n}\oint_{|z|=1} \frac{1}{z}\left( z + \frac{1}{z}\right)^n \; dz = 2\pi i \left( \frac{-i}{2^n} \right) \cdot Res(f(z) = \frac{z^2 + 1} {z^{n+1}}; z_0 = 0)$$
$$= \frac{\pi}{2^{n-1} n!} \left[ z^2 + 1\right]^{(n)}(0)$$
if $n=0$ we get $\frac{\pi}{2^{-1}} = 2\pi$, if $n=1$ we get $0$, if $n=2$ we get $\frac{\pi}{2!} = \frac{\pi}{2}$ and for $n = 3$ and $n > 3$ we get $0$.
But if I use integral solver for $n=4$ the result is $\frac{3\pi}{4}$ and for $n=2$ we get $\pi$. Why residue theorem doesn't work here? Or have I done some stupid mistake somewhere and I can't spot it?
| The problem is that the $z^2+1$ in the numerator should be raised to the same power as the $2z$ in the denominator.
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_0^{2\pi}\cos^n(x)\,\mathrm{d}x
&=\int_{|z|=1}\left(\frac{z+\frac1z}2\right)^n\,\frac{\mathrm{d}z}{iz}\tag1\\
&=\frac{2\pi}{2^n}\Res_{z=0}\left(\frac1z\left(z+\frac1z\right)^n\right)\tag2\\[3pt]
&=[n\text{ is even}]\frac{2\pi}{2^n}\binom{n}{n/2}\tag3
\end{align}
$$
$(1)$: convert to complex integral
$(2)$: apply residue theorem
$(3)$: compute the residue; $[\dots]$ are Iverson brackets
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the power series in powers of $x$ for the function Find the power series in powers of $x$ for the function
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt$$
Solution Verification Requesteed
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt \implies f'(x) = \frac{x^2}{1-x^4}$$
Now, we proceed with partial fraction decomposition. We obtain
$$f'(x) = \frac{x^2}{1-x^4} = - \frac{x^2}{x^4-1} = -x^2 \cdot \frac{1}{(x^2+1)(x+1)(-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+1} + \frac{D}{x-1}$$
We do partial fractions and obtain $A=0, B= -\frac{1}{2}, C = \frac{1}{4}, D = - \frac{1}{4}$. Now, we see that
$$-x^2 \cdot \frac{1}{x^4-1} = - \frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} - \frac{1}{4(x-1)}$$
$$f'(x) = \frac{x^2}{x^4-1} = \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} + \frac{1}{4(x-1)}$$
$$f'(x)= \frac{1}{2} \cdot \frac{1}{1-(-x^2)} - \frac{1}{4} \cdot \frac{1}{1-(-x)} + \frac{1}{4} \cdot \frac{1}{-(1-x)}$$
$$f'(x)= \frac{1}{2} \sum_{n=0}^\infty (-1)^nx^{2n} - \frac{1}{4} \sum_{n=0}^\infty (-1)^n x^n - \frac{1}{4} \sum_{n=0}^\infty x^n$$
We can integrate each sum term-by-term to obtain
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} - \frac{1}{4} \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - \frac{1}{4} \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} $$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} - \frac{1}{4} \Big[ \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} + \sum_{n=0}^\infty \frac{x^{n+1}}{n+1}\Big]$$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} + \Big[(x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4}) \Big]$$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} + 2 \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}$$
I'd simplify it further into one series but that dang $(-1)^n$ in the first sum keeps things from being nice and intuitive.
UPDATE
$$f'(x) = x^2 \cdot \frac{1}{1-(x^4)} = x^2 \sum_{n=0}^\infty x^{4n} = \sum_{n=0}^\infty x^{4n+2}$$
$$\implies f(x) = \sum_{n=0}^\infty \frac{x^{4n+3}}{4n+3}$$ as Jack D'Aurizio has concluded.
| Way faster:
$$ \frac{t^2}{1-t^4}=\sum_{n\geq 0} t^{4n+2} $$
for any $t\in(-1,1)$, so by termwise integration
$$ \int_{0}^{x}\frac{t^2}{1-t^4}\,dt = \sum_{n\geq 0}\frac{x^{4n+3}}{4n+3} $$
for any $x\in(-1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Sub-blocks of matrix quadratic equations Suppose the following matrix quadratic equation has at least one real solution for $X$:
$$
\begin{bmatrix}
A & a \\
0 & 1\\
\end{bmatrix} X^2 +
\begin{bmatrix}
B & b \\
0 & 0\\
\end{bmatrix} X +
\begin{bmatrix}
C & c \\
0 & 0\\
\end{bmatrix} = 0,
$$
where $A$, $B$ and $C$ are real square matrices, $a$, $b$ and $c$ are real vectors, and $X$ is a square matrix the same size as $\begin{bmatrix}
A & a \\
0 & 1\\
\end{bmatrix}$.
Now consider the following matrix quadratic equation for $Y$:
$$A Y^2 + B Y + C = 0,$$
where $Y$ is a square matrix the same size as $A$.
Must this have a real solution for $Y$? If not, must this at least hold generically over all $A,B,C,a,b,c$ for which the first equation has a real solution for $X$?
Note: In the case in which $A,B,C,a,b,c$ are all scalars, the solutions to the original equation are as follows:
Either:
*
*$X_{1,1}$ is a solution to $A X_{1,1}^2+B X_{1,1}+C=0$, $X_{2,1}=X_{2,2}=0$, $X_{1,2}=-\frac{c}{AX_{1,1}+B}$. OR
*$X_{1,1}=-\frac{C c}{B c-b C}$, $X_{1,2}=-\frac{c^2}{B c-b C}$, $X_{2,1}=-\frac{C^2}{B c-b C}$, $X_{2,2}=\frac{C c}{B c-b C}$.
In the former case, my claim holds. In the latter case, there is no guarantee that it does (it seems).
| When $A=B=0$ but $C\neq 0$,
$A Y^2 + B Y + C = 0$ fails to have a solution.
However, letting $A,B,C$ be $1$-by-$1$ square matrices,
you can set $A=B=0\;$; $C=1\;$; $a=1\;$; $b=-1\;$; $c=0\;$; $X=\begin{bmatrix}
0 & 0 \\
1 & 0\\
\end{bmatrix}$
satisfying $\begin{bmatrix}
A & a \\
0 & 1\\
\end{bmatrix} X^2 +
\begin{bmatrix}
B & b \\
0 & 0\\
\end{bmatrix} X +
\begin{bmatrix}
C & c \\
0 & 0\\
\end{bmatrix} = 0$.
$\\$
@cfp If you want a more generic-looking example, $A=\begin{bmatrix} -2 & 0 \\ 0 & 1\\ \end{bmatrix}$, $B=0$, $C=\begin{bmatrix} 0 & 0 \\ 0 & 1\\ \end{bmatrix}$, the value of $a$ doesn't matter, $b=\begin{bmatrix} 0 \\ -1\\ \end{bmatrix}$, $c=\begin{bmatrix} 0 \\ 0\\ \end{bmatrix}$, $X=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0\\ \end{bmatrix}$.
You can check that you cannot solve for $Y$ where $AY^2+C=0$, which is equivalent to $Y^2=\begin{bmatrix} 0 & 0 \\ 0 & -1\\ \end{bmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$, $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and I cant simplify the determinant in a good way with row operations.
I also tried solving $z$ from $g=1$. $z=1-x-y$.
$f(x,y,1-x-y)=xy-x^2y-xy^2$ restricted to $h(x,y)=2x^2+2y^2-2x-2y+2xy+1$ with Lagrange multiplier but I made no progress there either as the partial derivatives got too messy.
| The Lagrange multiplier method is effective for this problem. First of all observe that the case $x=y=z$ is inadmissible. By symmetry we may assume that $z\neq x$ and $z\neq y.$ We have to solve for
$$\nabla(xyz)=\lambda \nabla(x+y+z)+\mu\,\nabla(x^2+y^2+z^2) $$ i.e.
\begin{eqnarray*}
yz &=& \lambda +2\mu x\\
xz &=& \lambda +2\mu y \\
xy &=& \lambda +2\mu z
\end{eqnarray*}
Subtracting the third equation from the first two ones gives
\begin{eqnarray*}y(z-x)=-2\mu(z-x)\\
x(z-y)=-2\mu(z-y)
\end{eqnarray*}
Therefore $x=y.$ We thus get $$2x+z=1,\quad 2x^2+z^2=1$$
This leads to two solutions $x=y=0,$ $z=1$ and $x=y={2\over 3},$ $z=-{1\over 3}.$ Thus
$$M=0,\qquad m={2\over 3}\cdot {2\over 3}\cdot \left (-{1\over 3}\right )=-{4\over 27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$ I want to show that:
$$\lim_{(x,y)\to(0,0)} (1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} = 1$$
Direct substitution doesn't work, because of the denominator of $\frac{1}{x^2 + y^2}$.
Usually these problems are solved by bounding the following expression:
$$\left|(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} -1 \right| = \left| \frac{1 - (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}} }{ (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}}} \right|$$
I don't know how to continue.
| Hint
Take the natural logarithm and use that $z-\frac{z^2}2\le \ln(1+z) \le z$.
Additionally we can switch to radial coordinates $x=r\cos\theta,\, y=r\sin\theta$ and let $r\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How to solve such fraction differential equation? Here's my first-order differential equation:
$$
(x^3 - 2xy^2)dx + 3yx^2dy = xdy - ydx
$$
I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.
| $$(x^3 - 2xy^2)dx + \color {red}{ 3yx^2dy} = xdy - ydx$$
$$(x^3 - 2xy^2)dx +\color {red}{ 2yx^2dy+yx^2dy} = xdy - ydx$$
Rearrange terms:
$$(x^3dx+x^2ydy) - 2xy(ydx - xdy )= xdy - ydx$$
$$x^2(xdx+ydy) = (xdy - ydx)(1-2xy)$$
$$xdx+yd y = (1-2xy)d \left (\dfrac {y}{x}\right)$$
$$d(x^2+y^2) = 2(1-2xy)d \left (\dfrac {y}{x}\right)$$
Divide by $x^2+y^2$ and substitute $u=x^2+y^2$ and $v=\dfrac yx$:
$$\dfrac 12 \dfrac {du}{u} =\left (\dfrac 1u-\dfrac {2v}{v^2+1}\right)dv$$
$$ \dfrac {du}{dv} = 2-\dfrac {4uv}{v^2+1}$$
$$ u' +\dfrac {4v}{v^2+1}u=2$$
This is a first order linear DE. Try to integrate by integrating factor method:
$$\mu (v)=\exp \int \dfrac {4v }{v^2+1}dv$$
$$\mu (v)=\exp \int \dfrac {2}{v^2+1} dv^2=(v^2+1)^2$$
The DE becomes:
$$ (u(v^2+1)^2)'=2(v^2+1)^2$$
$$ u(v^2+1)^2=2\left(\dfrac {v^5}{5}+2\dfrac {v^3}{3}+v\right)+C$$
Unsubstitute $u$ and $v$:
$$u=x^2+y^2 \text { and }v=\dfrac yx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
With $x_1^2+3x_2^2+2x_1x_2=32$, find max value of $|x_1-x_2|$
With $x_1^2+3x_2^2+2x_1x_2=32$, find the maximum of $|x_1-x_2|$.
I have tried with AM-GM, but can't solve it. With $d = |x_1-x_2|$ and $d^2 = 2x_1^2 + 4x_2^2 - 32$, then I wonder how to do next.
Thank a lot for helping!
| $(x_1+x_2)^2+2(x_2)^2=32.$
We are finding the value of $\max(|x_1-x_2|)=\max(|x_1+x_2-2x_2|)$.
Let $x_1+x_2=X, x_2=Y.$
Then, we can rewrite the problem:
If $X^2+2Y^2=32$, Find the maximum of $|X-2Y|$.
Let $\sin(A)=\dfrac{X}{4\sqrt{2}}.$
Then, $\cos(A)=\dfrac{Y}{4}.$
So, the problem changes:
Find the maximum of $|4\sqrt{2}\sin{A}-8\cos{A}|.$
Let $\sin(B)=-\dfrac{\sqrt{2}}{\sqrt{3}}, \cos(B)=\dfrac{1}{\sqrt{3}}$.
Then, $|4\sqrt{2}\sin{A}-8\cos{A}| = |4\sqrt{6}(\cos{B}\sin{A}+\sin{B}\cos{A})| =|4\sqrt{6}\sin(A+B)| \leq 4\sqrt{6}.$
So, the answer will be $4\sqrt{6}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485747",
"timestamp": "2023-03-29T00:00:00",
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Evaluate the definite integral $I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}$ I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omissions), and (2) if there are any better/faster ways of evaluating this integral. Thanks!
Problem:
Evaluate the definite integral
$$I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}.$$
Attempted Solution:
We first note that the integrand is an even function, thus
$$I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)} = \frac{1}{2}\int_{-\pi}^{\pi} \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}. $$
Let $z=e^{i\theta} \implies d\theta = \frac{dz}{iz}$, then we have
$$\sin\theta =\frac{e^{i\theta} - e^{-i\theta}}{2i} =\frac{z-z^{-1}}{2i} \implies \sin^2\theta = \frac{z^2-2+z^{-2}}{-4} $$
and
$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z+z^{-1}}{2}. $$
We can now substitute such that for the unit circle $\gamma(\theta) = e^{i\theta}, \;\theta\in[-\pi,\pi]$, we have
\begin{align*}
\frac{1}{2}\int_{-\pi}^{\pi} \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)} &= \frac{1}{2}\oint_\gamma \frac{(-\frac{1}{4})(z^2-2+z^{-2})}{10-3(z+z^{-1})}\frac{dz}{iz} \\
&= \frac{i}{8} \oint_\gamma \frac{z^2-2+z^{-2}}{z(10-3z-3z^{-1})} dz \\
&= \frac{i}{8} \oint_\gamma \frac{z^4 - 2z^2 + 1}{z^2(10z-3z^2 -3)} dz \\
&= -\frac{i}{24} \oint_\gamma \frac{z^4-2z^2 + 1}{z^2(z^2-\frac{10}{3}z + 1)} dz \\
&= -\frac{i}{24} \oint_\gamma \frac{z^4-2z^2 + 1}{z^2(z-3)(z-\frac{1}{3})} dz
\end{align*}
It's clear that our integrand has a pole of order 2 at $z=0$, and two simple poles at $z=3$ and $z=1/3$. Using the Residue Theorem, we can evaluate this integral as $2\pi i$ times the sum of the residues at $z=0$ and $z=1/3$, disregarding $z=3$ since this pole is outside of the curve $\gamma$.
At the simple pole $z=1/3$, we calculate
\begin{align*}
\text{Res}(1/3) &= \lim_{z\to\frac{1}{3}} (z-\frac{1}{3})\left(-\frac{i}{24} \frac{z^4-2z^2 + 1}{z^2(z-3)(z-\frac{1}{3})} \right) \\
&= -\frac{i}{24}\lim_{z\to\frac{1}{3}} \left( \frac{z^4-2z^2 + 1}{z^2(z-3)} \right) \\
&= \frac{i}{9}
\end{align*}
The pole at $z=0$ is of order 2, therefore we calculate
\begin{align*}
\text{Res}(0) &= \lim_{z\to 0}\frac{d}{dz} z^2 \left(-\frac{i}{24} \frac{z^4-2z^2 + 1}{z^2(z-3)(z-\frac{1}{3})} \right) \\
&= -\frac{i}{24} \lim_{z\to 0}\frac{d}{dz} \left( \frac{z^4-2z^2 + 1}{(z-3)(z-\frac{1}{3})} \right) \\
&= -\frac{i}{24} \lim_{z\to 0} \frac{(4z^3-4z)(z-3)(z-\frac{1}{3})-(z^4-2z^2+1)(2z-\frac{10}{3})}{(z-3)^2(z-\frac{1}{3})^2}\\
&= \frac{-5i}{36}
\end{align*}
Finally, we calculate
$$
I=2\pi i\Big(\text{Res}(0) + \text{Res}(\frac{1}{3})\Big) = 2\pi i\Big(\frac{-5i}{36} + \frac{i}{9}\Big) = \frac{\pi}{18}
$$
|
(2) if there are any better/faster ways of evaluating this integral.
Note that $$\int_a^b f(x)dx=\int_a^bf(a+b-x)dx$$ Also,
$\sin(\pi-x)=\sin x, \ \cos(\pi-x)=-\cos x.$ Thus,
$$I=\int_0^{\pi}\frac{\sin^2\theta}{10-6\cos\theta}d\theta=
\int_0^{\pi}\frac{\sin^2\theta}{10+6\cos\theta}d\theta$$ Thus,
$$2I=20\int_0^{\pi}\frac{\sin^2\theta}{100-36\cos^2\theta}d\theta=
20\int_0^{\pi}\frac{\sin^2\theta}{64+36\sin^2\theta} d\theta$$
$$\implies I=10\int_0^ {\pi}\frac{1}{36}\left(1-\frac{64}{64+36\sin^2\theta}d\theta\right)$$$$=\frac{5\pi}{18}-\frac{160}{9}\int_0^{\pi}\frac{d\theta}{64+36\sin^2\theta}$$$$= \frac{5\pi}{18}-\frac{160}{9}\int_0^{\pi}\frac{\sec^2\theta d\theta}{64+100\tan^2\theta}$$$$=\frac{5\pi}{18}-\frac{320}{9}\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta d\theta}{64+100\tan^2\theta}= \frac{5\pi}{18} -\frac{2\pi}{9}=\frac{\pi}{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$ A\sin\alpha + B\cos\alpha=C$ solution confusion The solution of equation:
$$ A\sin\alpha + B\cos\alpha=C$$
according to python is:
$$\alpha=2\arctan\left[\frac{A \pm \sqrt{A^2+B^2-C^2}}{B+C}\right]$$
My question is why two answers or in otherwords in which situations we choose the postive and which the negative ? Can we get rid of the plus minus sign to outside of the function inside the $\arctan$ to outside?
| Starting with:
$$Asin(\alpha) + Bcos(\alpha) = C,$$
divide by $cos(\alpha)$ on both sides:
$$Atan(\alpha) + B = Csec(\alpha).$$
Then, square both sides:
$$A^2tan^2(\alpha) + 2ABtan(\alpha) + B^2 = C^2sec^2(\alpha).$$
Apply a variant of the Pythagorean identity to the RHS to get:
$$A^2tan^2(\alpha)+2ABtan(\alpha) + B^2 = C^2(tan^2(\alpha)+1).$$
Regroup to get quadratic in $tan(\alpha)$:
$$(A^2 - C^2)[tan(\alpha)]^2+(2AB)[tan(\alpha)] + (B^2-C^2) = 0.$$
Apply quadratic formula and simplify:
$$tan(\alpha) = \frac{-2AB\pm\sqrt{4A^2B^2 - 4(A^2-C^2)(B^2 - C^2)}}{2(A^2 - C^2)}$$
$$ = \frac{-AB\pm \sqrt{A^2B^2 - A^2B^2+A^2C^2+B^2C^2-C^4}}{A^2-C^2}$$
$$ = \frac{-AB\pm C\sqrt{A^2+B^2-C^2}}{A^2-C^2}.$$
Hence:
$$\alpha = arctan\bigg(\frac{-AB\pm C\sqrt{A^2+B^2-C^2}}{A^2-C^2}\bigg).$$
So the plus and minus comes from the quadratic formula and represents two arguments provided $A^2+B^2\neq C^2$ (i.e. that the discriminant is nonzero). If it is zero of course there is one argument for arctan. The $\pm$ can't pull out. The quadrant in which $\alpha$ lies depends on the sign of the numerator and denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Existence of Simsons Line using complex numbers Let $ABC$ be a triangle and $P$ is a point on its circumcircle. $X,Y$ and $Z$ are the feets of the perpendicular from $P$ on $BC, CA$ and $AB$, respectively. Prove that points $X, Y, Z$ are collinear.
Now proving this using angle chasing isn't hard. But I want to prove it using complex numbers.
Set $O=0$ where $O$ is the center of the circumcircle of $ABC$ ($(ABC)$ is the unit circle) and denote the corresponding complex number of every point with its lowercase letter. Here is the formula for each feet,
$$x=\frac{1}{2}(b+c+p-bc\bar p), \quad y= \frac{1}{2}( a+c+p-ac\bar p),\quad z= \frac{1}{2}( a+b+p-ab\bar p)$$
We know $x,y,z$ are collinear iff $x-y/x-z$ is real. But $$\frac{x-y}{x-z}=\frac{b-a-c\bar p(b-a)}{c-a-b\bar p(c-a)}=\frac{(b-a)(1-c\bar p)}{(c-a)(1-b\bar p)}$$ Is that even real?
| Noting $\;\overline{p}=1/p \;$ we have
$$\frac{(b-a)(1-c\bar p)}{(c-a)(1-b\bar p)}=\frac{(b-a)(p-c)}{(c-a)(p-b)} =\frac{a-b}{a-c}\div \frac{p-b}{p-c}$$
which is real since $A,B,C,P \;$ are concylic. $\; \blacksquare$
Alternatively, it also suffices to verify
$$\frac{\left(\frac{1}{b}-\frac{1}{a}\right) \left(\frac{1}{p}-\frac{1}{c}\right)}{\left(\frac{1}{c}-\frac{1}{a}\right) \left(\frac{1}{p}-\frac{1}{b}\right)}= \cdots = \frac{(b-a)(p-c)}{(c-a)(p-b)}.$$
| {
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"url": "https://math.stackexchange.com/questions/4490608",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to prove that for any $k$ between $2$ and $n-2$ we have $\binom{n}{k} \ \ge \binom{n}2$ I am trying to show that for any $k$ from $2$ to $n-2$ we have $\binom{n}{k} \ \ge \binom{n}2$.
What I tried
I started from the fact that $\binom{n}{k} \le \binom{n}{k+1}$ if $k \le \frac{n-1}{2}$ and by using the symmetry of binomial coefficients we also have this for $k \ge\frac{n+1}{2}$, but I don't know how to proceed from here on out and tie this to $k$ being between $2$ and $n-2$.
Any help would be greatly appreciated.
| Proof by induction:
(1) Base case: $n=6$, easy to verify, $2<k=3<6-2$,
$$~~~~~~~~~~~~~\binom{6}{3}>\binom{6}{2},~~\text{True.}$$
(2) Assume it is true for $n=m$, where $m>6$ and $m\in \mathbb{N^+}$, so we have:
$$\binom{m}{k}>\binom{m}{2}$$
(3) Induction step: when $n=m+1$,
$$\begin{align}
\binom{m+1}{k}&=\frac{(m+1)!}{k!(m+1-k)!}=\frac{m+1}{m+1-k}\binom{m}{k}>\frac{m+1}{m+1-k}\binom{m}{2}\\
\\
&=\frac{m+1}{m+1-k}\cdot\frac{m!}{2!(m-2)!}=\frac{m-1}{m+1-k}\cdot\frac{(m+1)!}{2!(m-1)!}=\frac{m-1}{m+1-k}\binom{m+1}{2}
\end{align}$$
since $k>2,~~\rightarrow~~m-1>m+1-k$, therefore:
$$\frac{m-1}{m+1-k}\binom{m+1}{2}>\binom{m+1}{2}$$
So we have:
$$\binom{m+1}{k}>\binom{m+1}{2}$$
Proof is completed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?
If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?
I solved this problem as follow,
Assuming $\sqrt[3]{2009}=\alpha$ , we have $x=\frac12(\alpha-\frac1{\alpha})$ and $$(x+\sqrt{1+x^2})^3=\left[\frac12(\alpha-\frac1{\alpha}) +\sqrt{1+\frac14(\alpha^2+\frac1{\alpha^2}-2)}\right]^3=\left[(\frac{\alpha}2-\frac1{2\alpha}) +\sqrt{\frac{\alpha^2}4+\frac1{4\alpha^2}+\frac12)}\right]^3=\left(\frac{\alpha}2-\frac1{2\alpha}+\left|\frac{\alpha}2+\frac1{2\alpha}\right|\right)^3=\alpha^3=2009$$
I'm wondering, is it possible to solve this problem with other approaches?
| Here's a rather elegant approach that involves pattern matching. Let $x=\sinh t$. Then
$$(x+\sqrt{1+x^2})^3 = (\sinh t + \cosh t)^3 = e^{3t}$$
And from comparing definitions
$$\sinh t = \frac{1}{2}(e^t-e^{-t})$$
$$x = \frac{1}{2}\left(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}}\right)$$
clearly, $e^t=\sqrt[3]{2009}$, thus
$$e^{3t} = 2009$$
| {
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"url": "https://math.stackexchange.com/questions/4496913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Analytical Solution to a nonlinear system of equations Is it possible to analytically solve for x,y,z in the following system, and if so how would one go about it?
$a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x = 0$
$a_{1} (z - 2y + x) - y + 2y^{3} - a_{2}y = 0$
$a_{1} (x - 2z + y) - z + 2z^{3} - a_{2}z = 0$
I've tried numerous methods such as cyclic permutations of the three equations and substitution but none seem to work, hence any advice would be greatly appreciated.
| Let $\,s = x+y+z\,$ then the first equation can be written as:
$$
\begin{align}
0 &= a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x
\\ &= a_{1} (s - 3x) - x + 2x^{3} - a_{2}x
\\ &= 2x^{3} - (3 a_1 + a_2 + 1) x + a_{1} s
\end{align}
$$
The same equation is satisfied by $\,y\,$ and $\,z\,$. Assuming $\,x,y,z\,$ are distinct, they are the roots of the same depressed cubic, so their sum must be $\,s = 0\,$, and the equations reduce to:
$$
2t^{3} - (3 a_1 + a_2 + 1) t = 0 \quad\quad \Big|\quad t = x, y, z
$$
Therefore the non-trivial distinct solutions of the system are permutations of:
$$
x = \sqrt{\frac{3a_1+a_2+1}{2}} \;,\quad y = - \sqrt{\frac{3a_1+a_2+1}{2}}\;, \quad z = 0
$$
This leaves the case where pairs of $\,x,y,z\,$ are equal to be solved separately.
[ EDIT ] The following covers the remaining case where two solutions are equal. Assume WLOG that $\,y=z\,$, then the last equation becomes redundant, and the system reduces to:
$$
\begin{cases}
\begin{align}
2x^{3} - (a_2+1) x - 2a_{1} (x-y) &= 0
\\ 2y^{3} - (a_2 + 1)y + a_{1} (x-y) &= 0
\end{align}
\end{cases}
$$
It can be seen "by inspection" that a solution is $\,x=y\,$ where $\,x\,$ is a root of $\,2x^{3} - (a_2+1) x = 0\,$, which gives the identical triplets $\,x = y = z \in \left\{0, \pm \sqrt{\dfrac{a_2+1}{2}}\right\}\,$ as solutions.
To find all solutions, the first equation gives $\,y\,$ as a cubic in $\,x\,$, then substituting into the second equation gives an equation of degree $\,9\,$ in $\,x\,$ alone. We know from the previous step that $\,2x^{3} - (a_2+1) x\,$ must be a factor, so that leaves a sextic in $\,x\,$, which in the end turns out to be a cubic in $\,x^2\,$ (courtesy WA):
$$
x^2 (a_2 - 2 x^2 + 1)^2 + 6 a_1 x^2 (a_2 - 2 x^2 + 1) - 2 a_1^2 (a_2 - 6 x^2 + 1) - 6 a_1^3 = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Interpreting the ratio test for $\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $ We want to find the limit of this.
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $$
I would have done it as follows:
$$\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = \lim_{n\to\infty}\frac{\frac{n+1}{(n+1)^2+n}}{\frac{n}{n^2+n}} =\lim_{n\to\infty} \frac{n+1}{(n+1)^2+n} \cdot \frac{n^2+n}{n} = \lim_{n\to\infty}\frac{n^3+2n^2+n}{n^3+3n^2+n} \\
=\lim_{n\to\infty} \frac{n^3 \cdot \bigl(1+\frac{2}{n} + \frac{1}{n^2} \bigr)}{n^3\cdot\big(1+\frac{3}{n} + \frac{1}{n^2} \bigr)} = \frac{1}{1} = 1$$
According to the ratio test, the series converges if $\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| <1$ and it diverges if it's $> 1$.
But since we get $1$ here, the series converges towards that value, no? But according to the ratio test, we can't make a statement about the series of the limit of $\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = 1$
What am I misunderstanding here?
| Will you allow me a proof without the ratio test which I think you
didn't use it in the correct way. We are interested in the limit of the
sum $s_{n}=n[\dfrac{1}{n^{2}+1}+\dfrac{1}{n^{2}+2}+....+\dfrac{1}{n^{2}+n}]$.
Notice that $\dfrac{1}{n^{2}+1}<\dfrac{1}{n^{2}}$ and likewise .... till
$\dfrac{1}{n^{2}+n}<\dfrac{1}{n^{2}}$. Hence the sum in brackets is less or equal to:
$\dfrac{n}{n^{2}}$=$\dfrac{1}{n}$.
Therefore $s_{n}\leq n\dfrac{1}{n}=1$. Also the sum in brackets is $\geq$ than $\dfrac{n}{n^{2}+n}$, which is $\dfrac{1}{n+1}$. Multiplied by $n$ we get $s_{n}\geq\,\dfrac{n}{n+1}$.Thus the sequence $s_{n}$ is such that $\dfrac{n}{n+1}\,\leq\,s_{n}\leq\,1$ and certainly converges to $1$!! Therefore the series converges to $1$.!!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$ To evaluate the integral
$$I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$$
I define $t=-1+\sqrt{\frac{4}{x}-3}$, then we have $x=\frac{4}{(t+1)^2+3}$ and
$$dx=\frac{-8(t+1)}{(t^2+2t+4)^2}dt$$
So we get
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\int_0^\infty \frac{\sqrt{t}(t+1)}{(t^2+2t+4)^2}\ dt$$
Again, let $s=\sqrt{t}$, then $dt=2sds$ and the integral becomes
$$8\int_0^\infty \frac{s(s^2+1) 2s}{(s^4+2s^2+4)^2}\ ds=16\int_0^\infty \frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}\ ds$$
Next, I write the integrand as
$$\frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}=\frac{1}{s^4+2s^2+4}-\frac{s^2+4}{(s^4+2s^2+4)^2}$$
so I have to evaluate
$$I_1=\int_0^\infty \frac{1}{s^4+2s^2+4}\ ds\ and\ I_2=\int_0^\infty \frac{s^2+4}{(s^4+2s^2+4)^2}\ ds$$
But when I evaluate the integral $I_1$, I get a weird result as follows:
$$I_1=\int_0^\infty \frac{1}{(s^2+2)^2-(\sqrt{2}s)^2}\ ds
=\int_0^\infty \frac{1}{(s^2+\sqrt{2}s+2)(s^2-\sqrt{2}s+2)}\ ds=\int_0^\infty \left(\frac{\frac{1}{4\sqrt{2}}s+\frac{1}{4}}{s^2+\sqrt{2}s+2}+\frac{\frac{-1}{4\sqrt{2}}s+\frac{1}{4}}{s^2-\sqrt{2}s+2}\right)\ ds$$
and if we separate two parts and evaluate the integrals, we would get two divergent improper integrals. So how can I find $I_1$?(Hope that the method is elementary and without complex analysis.)
Another question: I'm a beginner at learning complex analysis. I conjecture that we can evaluate the integral $I$ in complex analysis (or maybe not worked). Hope everybody can give me some hints or solutions with the method in complex analysis.
| *
*Continue with
\begin{align}
I=&\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx\\
=&\ 8\int_0^\infty \frac{\sqrt{t}(t+1)}{\left(t^2+2t+4\right)^2}dt\overset{x=\sqrt{\frac t2}}
=2\sqrt2\int_0^\infty \frac{2+\frac1{x^2}}{\left(x^2+\frac1{x^2}+1\right)^2}\overset{x\to\frac1x}{dx}\\
=& \ 3\sqrt2 \int_0^\infty \frac{1+\frac1{x^2}}{\left(x^2+\frac1{x^2}+1\right)^2}dx\overset{x-\frac1x\to x }
=3\sqrt2 \int_{-\infty}^\infty \frac{1}{\left(x^2+3\right)^2}dx= \frac\pi{\sqrt6}
\end{align}
*Alternatively, let
$$y= \sqrt{-1+\sqrt{\frac{4}{x}-3}}\implies
x=\frac{4}{y^4+2y^2+4}$$
and it is much simpler integrating in $y$ instead
$$I= \int_0^1 y(x)dx =\int_0^\infty x(y)dy
= \int_0^\infty \frac{4}{y^4+2y^2+4} dy=\frac\pi{\sqrt6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How many six-letter "words" can be formed from the alphabet $\{a-z\}$, if a "word" contains at least one vowel and at least one consonant?
How many six-letter “words” can be formed from the alphabet $\{a – z \}$?
*
*A “word” for this question must have at least one vowel $\{a, e, i, o, u\}$ and have at least one consonant (letters not in $\{a, e, i, o, u\}$ are considered as consonants).
What I have tried is $26^6−21^6$, but this does not account for the "at least one consonant case".
| I think that exponential generating functions can be used for this question. If there must be at least one consonant and at least one vowel , then there can be at most $5$ consonant or vowel.
*
*E.G.F for consonant : $$\bigg(21x+21^2\frac{x^2}{2!}+21^3\frac{x^3}{3!}+21^4\frac{x^4}{4!}+21^5\frac{x^5}{5!}\bigg)$$
*E.G.F for vowels : $$\bigg(5x+5^2\frac{x^2}{2!}+5^3\frac{x^3}{3!}+5^4\frac{x^4}{4!}+5^5\frac{x^5}{5!}\bigg)$$
Now find the coefficient of $\frac{x^6}{6!}$ or find $x^6$ and multiply it by $6!$ in the expansion of $$\bigg(21x+21^2\frac{x^2}{2!}+21^3\frac{x^3}{3!}+21^4\frac{x^4}{4!}+21^5\frac{x^5}{5!}\bigg)\bigg(5x+5^2\frac{x^2}{2!}+5^3\frac{x^3}{3!}+5^4\frac{x^4}{4!}+5^5\frac{x^5}{5!}\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\binom{n}{0}\binom{n+1}{0}+\binom{n}{1}\binom{n+1}{1}+\ldots+\binom{n}{n-1}\binom{n+1}{n-1}+\binom{n}{n}\binom{n+1}{n}$ Here's a question that arose when I was trying to solve a probability problem in my textbook. I'm trying to calculate the sum:
$$\binom{n}{0}\binom{n + 1}{0} + \binom{n}{1}\binom{n + 1}{1} + \binom{n}{2}\binom{n + 1}{2} + \ldots + \binom{n}{n - 1}\binom{n + 1}{n - 1} + \binom{n}{n}\binom{n + 1}{n}$$
But I'm not sure how to get started with evaluating this. Could anyone help? Any hints would be well-appreciated.
| Let's use the binomial theorem to find:
$$(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n$$
and
$$(x + 1)^{n + 1} = \binom{n + 1}{0}x^{n + 1} + \binom{n + 1}{1}x^n + \binom{n + 1}{2}x^{n - 1} + \cdots + \binom{n + 1}{n}x + \binom{n + 1}{n + 1}$$
Let's look at the coefficent of the term $x^{n + 1}$ in the product of these two.
$x^{n + 1}$ can be formed by multiplying $\binom{n}{0}x^0$ with $\binom{n + 1}{0}x^{n + 1}$, $\binom{n}{1}x^1$ with $\binom{n+1}{1}x^{n}$ a.s.o.
In other words the coefficient of $x^{n + 1}$ in $(x + 1)^{n + (n + 1)}$ is what you're looking for.
So, answer $${\binom{2n + 1}{n + 1}}$$
PS. This is a very frequently used technique to solve such type of problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\sum_{cyc} \frac{\cos(\frac{\alpha+\beta}{2})}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}=2$ when $\alpha+\beta+\gamma=\pi$ I saw this problem in a math magazine:
Let $\alpha,\beta$ and $\gamma$ be the angles of a triangle, so that
$$
\alpha+\beta+\gamma=\pi
$$
Show that
$$
\frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\cos\left(\frac{\alpha+\gamma}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
I tried to rewrite $\alpha+\beta$ with $\pi-\gamma$ so that
$$
\cos\left(\frac{\alpha+\beta}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)= \cos \frac{\pi}{2}\cos \frac{\gamma}{2}+ \sin\frac{\pi}{2}\sin \frac{\gamma}{2}= \sin\frac{\gamma}{2}
$$
Now the above equation equals
$$
\frac{\sin\frac{\gamma}{2}}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\sin\frac{\beta}{2}}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\sin\frac{\alpha}{2}}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
But now I don't have ideas for continuation. Any hints to tackle this?
| First off the bat, your formula in the first line is wrong, it should be, as per my comment, $$\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right) = \cos \frac{\pi}{2}\cos \frac{\gamma}{2}\color{purple}+\sin\frac{\pi}{2}\sin \frac{\gamma}{2}=\sin\frac{\gamma}{2}$$
I’ll take an alternative route. Expand $\cos\left(\dfrac{\alpha+\beta}{2}\right)$ as $\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}-\sin \dfrac{\alpha}{2}\sin \dfrac{\beta}{2}$ so that $$\dfrac{\cos\left(\dfrac{\alpha+\beta}{2}\right)}{\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}}=\dfrac{\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}-\sin \dfrac{\alpha}{2}\sin \dfrac{\beta}{2}}{\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}}$$$$=1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}$$
So summing similar terms we get $$LHS=3-\left(\tan\frac{\alpha}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}\right)\tag{1}$$
Now, $$\alpha+\beta+\gamma=\pi$$$$\implies\frac{\alpha}{2}+\frac{\beta}{2}=\frac{\pi}{2}-\frac{\gamma}{2}$$ so that taking tangents and expanding using the formula gives us $$\frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}=\tan\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)=\frac{1}{\tan\frac{\gamma}{2}}.\tag{2}$$ Rearranging gives us $$ \tan\frac{\alpha}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}=1$$ so using $(1)$ and $(2)$, we get our desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the right algebra to show they are indeed equivalent.
Note I ran into this problem, trying to do: $2\cos(\pi/8)+2\sin(\pi/8)$, where
$$2\cos(\pi/8)=\sqrt{2+\sqrt{2}},$$
$$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}.$$
|
$2\cos(\pi/8)=\sqrt{2+\sqrt{2}}$
$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}$
Start from your work,
$$2\cos(\pi/8)=2\sin(\frac{\pi}2-\frac{\pi}8)=2\sin(\frac{3\pi}8)$$
So
$$S=2\cos(\pi/8)+2\sin(\pi/8)=2\sin(\frac{3\pi}8)+2\sin(\frac{\pi}8)$$
Use formula:
$$\sin(x)+\sin(y)=2\sin({\frac{x+y}{2}})\cos(\frac{x-y}2)$$
We have
$$S=4\sin(\frac{\pi}4)\cos(\frac{\pi}8)=2\sqrt{2}\cos(\frac{\pi}8)=\sqrt{4+2\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
} |
Solve $(x^2+y^2+2z^2)^{1/2}=\cos z$ for $z$ near $(0,1,0)$
Can the equation $(x^2+y^2+2z^2)^{1/2}=\cos z$ be solved uniquely for $z$ in terms of $x$ and $y$ near $(0,1,0)$?
Put $F(x,y,z)=(x^2+y^2+2z^2)^{1/2}-\cos z$. Then $F(0,1,0)=0$,
$$
\nabla F(x,y,z)=(x(x^2+y^2+2z^2)^{-1/2},y(x^2+y^2+2z^2)^{-1/2},2z(x^2+y^2+2z^2)^{-1/2}+\sin z)
$$
and $\nabla F(0,1,0)=(0,1,0)$. Hence by the implicit function theorem, $y$ is a function of $x$ and $z$ near $(0,1,0)$.
But we cannot conclude the solvability of $z$ from this. Since for all $z$ sufficiently close to $0$, $F(x,y,z)=0$ if and only if $F(x,y,-z)=0$, $z$ cannot be solved as a function of $x$ and $y$. Am I correct? How to make this rigorous?
| For this kind of problems, you need to use the multidimensional Taylor series taking directional derivatives.
Truncated to the second order, this gives
$$\frac{1}{2}x^2+y+\frac{3 }{2}z^2-1=0 \tag 1$$ Checking at $x=\frac 1{10}$, $y=\frac 9{10}$, we have
$$\sqrt{2 z^2+\frac{41}{50}}-\cos (z)=0$$ which, expanded, gives
$$0=\frac{1}{10} \left(\sqrt{82}-10\right)+\left(\frac{1}{2}+5
\sqrt{\frac{2}{41}}\right) z^2+O\left(z^4\right)$$ that is to say
$$z^2=\frac{41 \left(10-\sqrt{82}\right)}{5 \left(41+10 \sqrt{82}\right)}=0.0589$$
while $(1)$ gives
$$z^2=\frac{19}{300}=0.0633$$
Expanded to the next order
$$\left(-\frac{x^2 y}{2}+x^2+y-1\right)+\left(\frac{5}{2}-y\right) z^2=0$$ which, for the used values would give
$$z^2=\frac{189}{3200}=0.0591$$ the exact solution being $z^2=0.0604$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
A question from Integration Bee 2022: Evaluate $\int^2_{1/2}\ln(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})})dx$ I am trying to evaluate the integral $$\int^2_{\frac{1}2}\ln\left(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})}\right)dx$$ by separating it as $$\int^2_{\frac{1}2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^2_{\frac{1}2}\ln\ln\left(\left(x-\frac{1}{2}\right)^2+4\right)dx$$
I think the substitution $x=1/u$ should be applied for the first item but I don't know how to go on. Please help. Thank you. (The final answer is $-\frac{3}{2}\ln2$)
| Separately
\begin{align}
\hspace{-10mm}\int^2_{1/2}\ln\ln(x^2-x+\frac{17}{4})dx\overset{x-\frac12\to x}
= \int^{3/2}_{0}\ln\ln(x^2+4)\ dx \\
\end{align}
\begin{align}
\int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)\overset{x\to \frac1x}{dx}
= &\ \frac12 \int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)(1+\frac1{x^2})\overset{x-\frac1x\to x}{dx}\\
=&\ \frac12 \int^{3/2}_{-3/2} \ln \frac12 + \ln\ln(x^2+4)\ dx \\
= &-\frac32 \ln2 +\int^{3/2}_{0} \ln\ln(x^2+4)\ dx \\
\end{align}
Together
$$\int^2_{1/2}\ln\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})}dx=-\frac32 \ln2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Upper bound of the sum $\sum_{n=1}^\infty \tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)$ as $x\to \infty$ I am stuck at finding a Big-O upper bound of the following sum as $x\to \infty$ $$\sum_{n=1}^\infty \tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)$$
I tried: Since $$ \sum_{n=1}^\infty\tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)=\sum_{n=1}^\infty O\left(\frac{x}{n^2}\right) $$
Now we have for $x_0>0$ and $M>0$ $$ \sum_{n=1}^\infty\tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)=\left(M \sum_{n=1}^\infty \frac{1}{n^2}\right)x $$
Now since $\sum_{n=1}^\infty\frac{1}{n^2}$ converges so can we write the following ? $x\to \infty$ $$ \sum_{n=1}^\infty\tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)=O(x) $$
Thnak you for your valuable time and comments.
| To get the leading asymptotics term, we can switch from summation to integration. As @Jack D'Aurizio mentioned, we are getting $O(1)$ term while evaluating $S_2$, therefore we can only try to catch the leading term. As was shown above, the leading term is $O(\sqrt x)$.
Indeed, implementing the program, denoting for a while $x=N\to\infty$ and dropping the lower terms
$$-S(N)=-\sum_{n=1}^\infty \arctan\left(\frac{N}{1-(N-n)^2}\right)\sim-\int_0^\infty\arctan\frac{N}{1-(n-N)^2}dn$$
$$\sim\int_0^\infty\arctan\frac{N}{(n-N)^2}dn$$
We do not care about the accurate evaluation near $n=N$: as was mentioned, it gives only the contribution $\sim O(1)$.
Integrating by part,
$$-S(N)\sim\int_0^\infty\frac{2Nn(n-N)}{N^2+(n-N)^4}dn=2N^2\int_{-N}^\infty\frac{tdt}{N^2+t^4}+2N\int_{-N}^\infty\frac{t^2dt}{N^2+t^4}$$
Making some changes and dropping vanishing at $N\to0$ terms,
$$\sim N^2\int_{N^2}^\infty\frac{dx}{N^2+x^2}+2N\int_{-\infty}^\infty\frac{t^2}{N^2+t^4}dt-2N\int_N^\infty\frac{t^2}{N^2+t^4}dt$$
It is easy to see that only second term is divergent at $N\to\infty$, therefore
$$-S(N)\sim 2N\int_{-\infty}^\infty\frac{t^2}{N^2+t^4}dt=2\sqrt N\int_{-\infty}^\infty\frac{t^2}{1+t^4}dt=\sqrt N\int_0^1(1-t)^{-\frac{1}{4}}t^{-\frac{3}{4}}dt$$
$$\boxed{\,\,-S(x)\sim\sqrt x\, B\Big(\frac{1}{4};\frac{3}{4}\Big)=\sqrt 2\,\pi\,\sqrt x\quad\text{at}\,\,x\to\infty\,\,}$$
We can also check that at
$\displaystyle x=100\quad -S(100)=34.22;\,\,\sqrt 2\pi\,\sqrt {100}=44.43$
$\displaystyle x=300\quad -S(300)=76.07;\,\,\sqrt 2\pi\,\sqrt {300}=76.95$
$\displaystyle x=500\quad -S(500)=98.44;\,\,\sqrt 2\pi\,\sqrt {500}=99.34$
$\displaystyle x=1000\,\, -S(1000)=139.6;\,\,\sqrt 2\pi\,\sqrt {1000}=140.5$, etc.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
Furthermore, $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\
&=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\
&=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3)
\end{aligned}
$$
and $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$
| Method. $(1)$, Use Mittag-Leffler's series for $\cot(x)$
$$\begin{align}
I&=\int_0^\frac{\pi}{2}x\ln(\sin(x))~dx=\frac{1}{2}\int_0^\frac{\pi}{2}\ln(\sin(x))~dx^2\\
\\
&=-\frac{1}2 \int_0^\frac{\pi}{2}\frac{x^2}{\tan(x)}~dx=-\frac{1}2\int_0^\frac{\pi}{2}x^2\left(\frac{1}x+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2} \right)~dx
\end{align}$$
Next, integrate them term by term and you will get the answer.
Method. $(2)$, Construct two bases, as mentioned by @Quanto:
$$\begin{align}
I_1&=\int_0^\frac{\pi}{2}x\left[\ln(\sin(x))+\ln(\cos(x))\right]dx=\int_0^\frac{\pi}{2}x\ln\left( \frac{1}{2}\sin(2x) \right)dx\\
\\
I_2&=\int_0^\frac{\pi}{2}x\left[\ln(\sin(x))-\ln(\cos(x))\right]dx=\int_0^\frac{\pi}{2}x\ln(\tan(x))dx
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Finding all functions satisfying $f(x f(x+y))+f(f(y) f(x+y))=(x+y)^{2}$ Determine all functions $f:\mathbb{R} \to \mathbb{R}$ such that $$f(x f(x+y))+f(f(y) f(x+y))=(x+y)^{2}, \forall x,y \in \mathbb{R} \tag1)$$
My approach:
Let $x=0$, we get
$$f(0)+f\left((f(y))^2\right)=y^2$$
$\Rightarrow$
$$f\left((f(y))^2\right)=y^2-f(0)\tag2 $$
Let us assume $f(0)=k \ne 0$
Put $y=0$ above, we get
$$f(k^2)=-k$$
Also put $y=-x$ in $(1)$, we get
$$f(kf(x))+f(kf(-x))=0, \forall x \in \mathbb{R}$$
Put $x=0$ above we get
$$f(k^2)=0$$
$\Rightarrow$
$f(k^2)$ has two different images $0,-k$ which contradicts that $f$ is a function. Hence $k=0 \Rightarrow f(0)=0$.
So from $(2)$ we get:
$$f\left((f(y))^2\right)=y^2 \cdots (3)$$
Now put $y=0, x=f(x)$ in $(1)$, and use the fact $f(0)=0$,we get
$$f\left((f(x))^2\right)=(f(x))^2$$
Since $x$ is dummy variable, we get $$f\left((f(y))^2\right)=(f(y))^2 \cdots (4)$$
From $(3),(4)$, we get $$f(x)=\pm x$$
I just want to ask, is my approach fine? If not where is the flaw? Also other approaches are welcomed.
|
$f: \Bbb{R \to R}, f(xf(x+y))+f(f(y)f(x+y))=(x+y)^2.$
My attempt was to show that $f(0)=0$.. And I did it.
\begin{align}
P(0, y): \; & f(0)+f(f(y)^2)=y^2. \\
P(0, 0): \; & f(0)+f(f(0)^2)=0. \\
& \text{let } f(0)=k. \\
\ \\
\Rightarrow \; & k+f(f(y)^2)=y^2, k+f(k^2)=0. \\
\ \\
& \text{let } f(a)^2=f(b)^2. \\
P(0, a): \; & k+f(f(a)^2)=a^2. \\
P(0, b): \; & k+f(f(b)^2)=b^2. \\
\therefore \; & f(a)^2=f(b)^2 \Leftrightarrow a^2=b^2. \\
\ \\
P(k, -k): \; & f(kf(0))+f(f(-k)f(0))=0. \\
\Rightarrow \; & f(k^2)+f(kf(-k))=0, f(kf(-k))=k. \\
\ \\
P(-k, 0): \; & f(-kf(-k))+f(kf(-k))=k^2. \\
\Rightarrow \; & f(-kf(-k))=k^2-k. \\
P(k, -2k): \; & f(kf(-k))+f(f(-2k)f(-k))=k^2. \\
\Rightarrow \; & f(f(-2k)f(-k))=k^2-k. \\
\ \\
\Rightarrow \; & f(-kf(-k))=f(f(-2k)f(-k)). \\
\Rightarrow \; & f(-kf(-k))^2=f(f(-2k)f(-k))^2. \\
\ \\
\therefore \; & k^2f(-k)^2=f(-2k)^2f(-k)^2, k^2=f(-2k)^2. \\
\Rightarrow \; & f(0)^2=f(-2k)^2, 4k^2=0, k=0. \\
\therefore \; & f(0)=0. \\
\ \\
\ \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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.