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$a+n^2$ is the sum of two perfect squares $a+n^2$ is the sum of two perfect squares for a given positive integer $a$ and all positive integers $n$.
Show that $a$ is a perfect square.
At first I thought of putting a bound on the difference between perfect squares up to a point, like maybe choosing $n=c$ so that one of the squares on the right hand side is the smallest, to try and show that if $a = x^2 + k< (x+1)^2$ that there would have to be two pairs of squares with the same difference, but I can't seem to work out the details
| The general solution of the equation
$$x^2+y^2=z^2+w^2$$ is given by the identity with four arbitrary parameters
$$(tX+sY)^2+(tY-sX)^2=(tX-sY)^2+(tY+sX)^2$$
Let $n$ be any integer so we have $$a+n^2=z^2+w^2$$Making
$$n=tY-sX\\z=tX-sY\\w=tY+sX$$ we have three equations with four unknowns which in general have infinitely many solutions. Any way we have
$$n^2=t^2Y^2+s^2X^2-2stXY\\z^2=t^2X^2+s^2Y^2-2stXY\\w^2=t^2Y^2+s^2X^2+2stXY$$ which implies $$z^2+w^2-n^2=t^2X^2+s^2Y^2+2stXY=(tX+sY)^2$$ Since $$a=z^2+w^2-n^2$$ we have
$$a=(tX+sY)^2$$ We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin
First, I get the line which passes through point of contact of tangents from origin using result here which is :
$$ -12x-16y-2 \cdot 75 = 0$$
Or,
$$ 6x + 8y -75 =0 $$
Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:
$$ C(0,0) C(x,y) = (6x+8y-75)^2$$
This leads to:
$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$
$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$
For applying the result in this answer, then the formula
$$ a= 75 - 36, b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$
Or,
$$ a = 39, b= 11 , h=-48$$
$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$
However, the intended answer was:
$$ \tan \theta = \frac{1}{\sqrt{3} } $$
Where have I gone wrong?
| Here is a simple solution: Find the center of the given circle, which is (6, 8). The line from the center of the circle to the point of contact of one of the tangents is perpendicular to the tangent. Use this fact to get the equation of a new circle, which is $x^2-6x+y^2-8x=0$. Find where both circles meet. Find the angle between the vectors, which are the two points where the circles meet using the formula:$$\text{Angle} = \arccos\left(\frac{x_1x_2 + y_1y_2}{\sqrt{(x_1^2+y_1^2) \cdot (x_2^2+y_2^2)}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013658",
"timestamp": "2023-03-29T00:00:00",
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If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.
| Note the beautiful identity $$y^2+z^2+t^2=\frac{1}{3}\left({(y+z+t)}^2+{(y-z)}^2+{(z-t)}^2+{(t-y)}^2\right)$$ thus we have $$x^2+\frac{1}{3}{(y+z+t)}^2+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=x(y+z+t)$$ $${\left (x-\frac{(y+z+t)}{2}\right)}^2+ \frac{{(y+z+t)}^2}{12}+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=0$$ which means $x=y=z=t=0$
(As its a sum of squares each term must be zero)
| {
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"timestamp": "2023-03-29T00:00:00",
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Expand $\frac{\Gamma\left(\frac{x}{2}\right)}{\Gamma\left(\frac{x-1}{2}\right)}$ at $x=\infty$ I used Wolfram Alpha for this problem and it gives me a "Puiseux expansion": $$\sqrt{x}-\frac{3 \sqrt{\frac{1}{x}}}{8}-\frac{7}{128}\left(\frac{1}{x}\right)^{3 / 2}-\frac{9\left(\frac{1}{x}\right)^{5 / 2}}{1024}+O\left(\left(\frac{1}{x}\right)^{3}\right)$$
which is exactly what I need.
My only question is how one would be able to obtain this expansion manually.
| Note that the expansion you gave is for $\Gamma(x)/\Gamma(x-1/2)$. Using the beta function and an appropriate change of integration variables, we obtain
\begin{align*}
&\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi }}\frac{{\Gamma \left( {\frac{x}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {\frac{{x + 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi }}B\left( {\frac{x}{2},\frac{1}{2}} \right) = \frac{{x - 1}}{{2\sqrt \pi }}\int_0^1 {\frac{{t^{x/2 - 1} }}{{\sqrt {1 - t} }}dt}
\\ &
= \frac{{x - 1}}{{\sqrt \pi }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - 2s} } }}ds} = \frac{{x - 1}}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - xs} s^{ - 1/2} \sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}} ds} .
\end{align*}
By the definition of the Nörlund numbers (cf. https://mathworld.wolfram.com/NorlundPolynomial.html),
$$
\sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}} = \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}s^n }
$$
for $|s|<\pi$. Thus, by Watson's lemma (cf. http://dlmf.nist.gov/2.3.ii),
\begin{align*}
\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} & \sim \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\int_0^{ + \infty } {e^{ - xs} s^{n - 1/2} ds} }
\\ &
= \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\Gamma \left( {n + \frac{1}{2}} \right)\frac{1}{{x^{n + 1/2} }}}
\\ &
= \sqrt {\frac{x}{2}} + \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} }}{{2\sqrt \pi n!}}\Gamma \left( {n + \frac{1}{2}} \right)\left( {\frac{{2n + 1}}{{n + 1}}B_{n + 1}^{(1/2)} + B_n^{(1/2)} } \right)\left(\frac{2}{{x}}\right)^{n+1/2}}
\end{align*}
as $x\to +\infty$. It is possible to simplify this result to
$$
\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} \sim \sqrt {\frac{x}{2}} + \sum\limits_{n = 0}^\infty {
\binom{ 1/2}{n + 1}B_{n + 1}^{(3/2)} \left(\frac{2}{{x}}\right)^{n+1/2}} .
$$
Note that this is a divergent asymptotic expansion. See http://dlmf.nist.gov/5.11.iii for more general results and references.
| {
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"url": "https://math.stackexchange.com/questions/4016560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| $ b - a = 4^{2^{k+1}} + 2^{2^{k+1}} - 4^{2^k}-2^{2^k}$
An intermediate step: notice $4=2^2$, so we have that $b-a = 2^{2^{k+2}} + 2^{2^{k+1}} - 2^{2^{k+1}}-2^{2^k} = 2^{2^{k+2}} -2^{2^k} $
Notice that each term is divisible by $2^{2^k}$, and $7$ does not divide $2^{2^k}$.
Now $\frac{b-a}{2^{2^k}} = 2^{2^{k+2} - 2^k} -2^{2^k - 2^k} = 2^{3 \times 2^k} - 2^0 = 8^{2^k} -1$
Now it is more of a bonus fact that $8^m -1$ is always divisible by $7$.
| {
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Solving $\tan(x) = \cos(x)$ I've been trying to solve the following trigonometric equation unsuccessfully. My intuition was to reduce the degree of the equation from $4$ to $2$ so I could solve it as a quadratic equation, but my attempt so far have proved unsuccessful. Any help would be appreciated.
$$\begin{align}
\tan(x) &= \cos(x) \tag1\\[4pt]
\frac{\sin(x)}{\cos(x)} &=\cos(x) \tag2\\[4pt]
\frac{\sin^2(x)}{\cos^2(x)} &=\cos^2(x) \tag3\\[4pt]
\frac{1-\cos^2(x)}{\cos^2(x)} &=\cos^2(x) \tag4\\[4pt]
1-\cos^2(x) &=\cos^4(x) \tag5\\[6pt]
\cos^4(x) + \cos^2(x) - 1 &= 0 \tag6
\end{align}$$
It's at this point that I get stuck
|
We could also construct the right triangle suggested by the equation $ \ \tan x \ = \ \cos x \ \ , $ where $ \ x \ $ is an angle (in the first quadrant). We may set the length of the side opposite $ \ x \ $ equal to $ \ 1 \ \ , $ and write the ratios $ \ \tan x \ = \ \frac{1}{y} \ \ $ and $ \ \cos x \ = \ \frac{y}{y^2} \ = \ \frac{1}{y} \ \ . $ The "Pythagorean" Theorem then leads us to $ \ y^4 \ = \ y^2 + 1 \ \ . $
If we substitute $ \ u \ = \ y^2 \ \ $ to write this equation as $ \ u^2 \ = \ u \ + \ 1 \ \ , \ $ then divide through by $ \ u \ \ $ (which is positive), we have one of the "Golden Ratio" relations, $ \ u \ = \ 1 \ + \ \frac{1}{u} \ \ , $ with $ \ u \ = \ \phi \ = \ \frac{1 + \sqrt5}{2} \ \approx \ 1.61803 \ \ . $ We thus have the value in the first quadrant for our equation, $$ \tan x \ \ = \ \ \cos x \ \ = \ \ \frac{1}{\sqrt{\phi}} \ \ \approx \ \ 0.7862 \ \ . $$
There is a second result in the "principal circle", since in the second quadrant we can have $ \tan x \ = \ \cos x \ = \ -\frac{1}{\sqrt{\phi}} \ \ . $ (There are no further solutions as $ \ \tan x \ $ and $ \ \cos x \ $ have opposite signs in the other two quadrants.) The general solution to our equation is then
$$ x \ \ = \ \ \arccos \left(\frac{1}{\sqrt{\phi}} \right) \ + \ 2n\pi \ \ , \ \ \left[ \ \pi \ - \ \arccos \left( \frac{1}{\sqrt{\phi}} \right) \ \right] \ + \ 2n\pi \ \ . $$
[Interestingly, the first quadrant angle is extremely close to $ \ \frac23 \ : \ \arccos \left(\frac{1}{\sqrt{\phi}} \right) \ \approx \ 0.66624 \ \ . \ ] $
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem solving complex equation with high power I have to solve :$$(i+z)^{16}-(7+7i)(i+z)^8+25i=0$$
I consider this equation like a quadratic equation and find its delta and solutions. But i can't solve two equations :$$(i+z)^8=4+3i$$
$$(i+z)^8=3+4i$$
I really need some helps to solve completely this equation. Thanks.
| If you don't want to use polar coordinates, the eighth roots can be found by repeated square root.
We get the square root
$$ (x+iy)^2 = a+ib $$
when
$$ x = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b} + \sqrt{\sqrt{a^2+b^2} - b} \right) $$
$$ y = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b} - \sqrt{\sqrt{a^2+b^2} - b} \right) $$
if the real parts $a,b$ are positive, so are the $x,y$ above are also positive. Switching $a,b$ to $b,a$ gives a different result.
| {
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Is the triangle equilateral? (resolve without using trigonometry) I have been asked the following question. Take a circle, and three arcs of magnitude $\dfrac{\pi}{3}$ (or you could say the corresponding centered angles are of $60$ degrees) each, that do not intersect with each other, say arc $AB$, arc $CD$ and arc $EF$.
For each of the three remaining arcs take the corresponding chords $BC$, $DE$ and $FA$. Let $M, N, P$ respectively be the middles of these three line segments. Is the triangle $MNP$ equilateral?
The person that asked me this question (a mathematician) did not know if the statement is true. I asked him where did he find it and he told me that his barber asked him. I tried to solve it using Euclidean geometry and I failed. I found a positive solution using trigonometry (I may present it here later).
I tried repeatedly to solve it again only using Euclidean geometry and I failed. Any idea? (I mean no analytic geometry or complex numbers, just to prove $MN=NP=PM$ using Euclidean geometry)
Edit : I will present the solution that I have, that is based on trigonometry.
Let $O$ be the center of the circle, $R$ its radius and a,b,c respectively the magnitudes of the angles BOC, DOE and FOA respectively.
Then $OM=R\cos\frac{a}{2}$, $ON=R\cos\frac{b}{2}$, $OP=R\cos\frac{c}{2}$.
The angle $MON$ is equal to $MON=\frac{a}{2}+\frac{\pi}{3}+\frac{b}{2}$ and taking into account that $a+b+c=\pi$ we get that $MON=\frac{5\pi}{6}-\frac{c}{2}$.
Thus $\cos(MON)=\cos(\frac{5\pi}{6}-\frac{c}{2})= \cos\frac{5\pi}{6}\cos\frac{c}{2}+\sin\frac{5\pi}{6}\sin\frac{c}{2}$
from which we deduce that
$$ \cos(MON)=
-\frac{\sqrt{3}}{2}\cos\frac{c}{2}+\frac{1}{2}\sin\frac{c}{2}$$
Since $\frac{c}{2}=\frac{\pi}{2}-(\frac{a}{2}+\frac{b}{2})$ we have that
$\sin\frac{c}{2}=cos(\frac{a}{2}+\frac{b}{2})
=\cos\frac{a}{2}\cos\frac{b}{2}-\sin\frac{a}{2}\sin\frac{b}{2}$.
We get $$ \sin\frac{c}{2}-\cos\frac{a}{2}\cos\frac{b}{2}=-\sin\frac{a}{2}\sin\frac{b}{2}. $$
Apply now the law of cosines in the triangle $MON$, in order to compute $MN^2$. Our main concern is to find a formula that is symmetric to any permutation of $a,b,c$, which will imlpy that the same formula is valid for $NP^2$ and $PM^2$.
$$ MN^2 = OM^2+ON^2-2 OM\cdot ON \cos(MON) \\
= R^2 \cos^2\frac{a}{2}+ R^2 \cos^2\frac{b}{2} -2 R\cos \frac{a}{2}R\cos\frac{b}{2} \cos(MON)\\
=R^2\Big(\cos^2\frac{a}{2}+ \cos^2\frac{b}{2} -2 \cos \frac{a}{2} \cos\frac{b}{2} \cos(MON)\Big)\\
=R^2\Big[\cos^2\frac{a}{2}+ \cos^2\frac{b}{2} -2 \cos \frac{a}{2} \cos\frac{b}{2}\big (-\frac{\sqrt{3}}{2}\cos\frac{c}{2}+\frac{1}{2}\sin\frac{c}{2}\big) \Big]\\
= R^2 \Big( \sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2}
-\cos \frac{a}{2}\cos\frac{b}{2}\sin\frac{c}{2}\Big)\\
=R^2 \Big( \sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}-1 +\sin^2\frac{c}{2}- \cos \frac{a}{2}\cos\frac{b}{2}\sin\frac{c}{2}\Big)\\
=R^2[-1+\sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}\\
\qquad\qquad\qquad\qquad\qquad\qquad
+\sin\frac{c}{2}(\sin\frac{c}{2}-\cos \frac{a}{2}\cos\frac{b}{2})\Big]\\
=R^2\Big[-1+\sqrt{3} \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}
-\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}\Big]
$$
It seems to me impossible to translate this trigonometric proof to a pure euclidean geometry proof. This is the reason that I made the post, for someone to suggest a pure proof based on euclidean geometry.
Edit3: As pointed out by @Blue the formula is simlpified to
$$MN^2= R^2(1+\sqrt{3}\cos\frac{a}{2} \cos\frac{b}{2}\cos \frac{c}{2}+
\sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}).$$
| It is not difficile but something tedious solve this problem without trigonometry. Let a circle of equation $x^2+y^2=r^2$ and put $A=(x_A,y_A),\cdots,F=(x_F,y_F)$. The triangles $\triangle{OAB},\triangle{OCD}$ and $\triangle{OEF}$ are clearly equilateral so we have $$\overline{AB}^2=r^2=(x_A-x_B)^2+(y_A-y_B)^2\Rightarrow 2(x_Ax_B+y_Ay_B)=r^2\space\space (1)$$ and similarly$$\overline{CD}^2=r^2\Rightarrow2(x_Cx_D+y_Cy_D)=r^2\space\space (2)\\\overline{EF}^2=r^2\Rightarrow2(x_Ex_F+y_Ey_F)=r^2\space\space (3)$$ Besides$$M=\left(\frac{x_B+x_C}{2},\frac{y_B+y_C}{2}\right)\\N=\left(\frac{x_D+x_E}{2},\frac{y_D+y_E}{2}\right)\\P=\left(\frac{x_A+x_F}{2},\frac{y_A+y_F}{2}\right)$$
It follows $$4\overline{PM}^2=(x_A+x_F-x_B-x_C)^2+(y_A+y_F-y_B-y_C)^2\\4\overline{PN}^2=(x_A+x_F-x_D-x_E)^2+(y_A+y_F-y_D-y_E)^2\\4\overline{NM}^2=(x_B+x_C-x_D-x_E)^2+(y_B+y_C-y_D-y_E)^2$$
We know by trigonometry that the request is true and then what we have to do is simply check the equalities $\overline{PM}=\overline{PN}=\overline{NM}$ using $(1),(2),(3)$ and the fact that $A,B,C,D,E,F$ are in the circle.(Easily we must have for example $4\overline{PM}^2-4\overline{PN}^2=0$ and $4\overline{PM}^2-4\overline{NM}^2=0$)
| {
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"timestamp": "2023-03-29T00:00:00",
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How many ways can $10$ different kids be distributed to $4$ distinct classes? I want to find the number ways to distribute 10 different kids to 4 different classes, such that each class has at least 2 kids.
My attempt: First distribute 8 kids to 4 classes such that each class has exactly 2 kids. Then there are 16 option to split the remaining 2 kids.
This is my result: $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}*16$ which is wrong. I don't have exact answer but I know it's either: $226800, 75600, 44100$.
Thanks.
| Note that there are 2 different size types for classes, $3,3,2,2$ or $4,2,2,2$. For the first one, choose firstly which classes will have $3$ kids, $\binom{4}{2}$, then distribute the kids, $\binom{10}{3} \cdot \binom{7}{3} \cdot \binom{4}{2} \cdot \binom{2}{2}$. Do the same for the other size type and you will arrive at $ \binom{4}{2} \cdot \binom{10}{3} \cdot \binom{7}{3} \cdot \binom{4}{2} \cdot \binom{2}{2} + \binom{4}{1} \cdot \binom{10}{4} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \binom{2}{2} = 226800$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration with trig substitution Trying to evaluate this using trig substitution:
$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$
Here's how I'm going about it, using $x = 5/7(\tan\theta)$
$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}\right)} \mathrm{d}\theta$$
$$=\int \frac {1}{(25\tan^2\theta + 25)} $$
$$=\int \frac {1}{25(\tan^2\theta + 1)}\mathrm{d}\theta $$
$$=\int \frac {1}{25\sec^2\theta}\mathrm{d}\theta $$
$$=\int \frac {\cos^2\theta}{25}\mathrm{d}\theta $$
$$=\frac{1}{50}(\theta + \sin\theta + \cos\theta) $$
To generalize for $x$, $\theta = \arctan(7x/5)$
$$\frac{1}{50}\left(\arctan\left(\frac{7x}{5}\right) + \sin\left(\arctan\left(\frac{7x}{5}\right)\right) + \cos\left(\arctan\left(\frac{7x}{5}\right)\right)\right) $$
$$\frac{1}{50} \left(\frac{7x}{5\left(\frac{49x^2}{25}+1\right)} + \arctan\left(\frac{7x}{5}\right)\right)$$
But taking the derivative of this gets me:
$$ \frac{35}{(49x^2 +25)^2}$$
Where is my mistake?
| We can calculate a more general integral of the form
$$I =\int\frac{1}{a^2x^2+b^2}\,\mathrm{d}x.$$
In your example $a=7$ and $b=5$. First of all do the non-trigonometrical substitution $u=ax/b$. That will give you
$$\int\frac{b}{a(b^2u^2+b^2)}\,\mathrm{d}x=\frac{1}{ab}\int\frac{1}{u^2+1}\,\mathrm{d}u.$$
You should be familiar with this integral. It's equal to $\arctan u$. Substituting back yields
$$I = \dfrac{\arctan(ax/b)}{ab}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $s_{n+1}=s_n^2-2$ diverges. Let $s_1=5$ and for $n\geq 1$, $s_{n+1}=s_n^2-2$. Do the following steps to show that $(s_n)$ diverges.
1.If $\lim s_n = a$ then $a^2 -2 = a$.
$a=\lim s_{n+1}=\lim {s_n^2-2}=\lim {s_n^2}-\lim (2)=(\lim {s_n})^2-\lim (2)=a^2-2$
2. If $\lim s_n = a$ then a has to be either $−1$ or $2$.
From (1), we have
$a=a^2-2$
$a^2-a-2=0$
$a=−1, 2$
3. Show that $\lim s_n$ can neither be $−1$ nor $2$. (Hint: write down the first few terms of $(s_n)$ to get some insights.)
$s_n\leq s_{n+1}$ for all $n$
Proof, by induction:
*
*$P_1$ is true; at $n=1$, $5=s_1\leq s_2=5^2-2=23$
*Suppose that $s_n\leq s_{n+1}$ for some $n$
*By the assumption we have $s_{n+1}=s_n^2-2\leq s_{n+1}^2-2=s_{n+2}$
As a result, $s_n$ is increasing, so $s_n\geq 5$ for all $n$. Now since $s_n\geq 5$, this implies that $\lim s_n \geq 5$ which implies that it cannot be $-1$ or $2$.
4. Combine these steps to show that $\lim s_n$ does not exist.
Is there any further step to do here or just combine the 2 and 3?
Are my answers correct, please? Thank you in advance.
| More convenient to begin with an $s_0.$
Suppose we find the real number $A>1$ for which
$$ s_0 = A + \frac{1}{A} $$
From $s_{n+1} = s_n^2 - 2$ we find
$$ s_1 = A^2 + \frac{1}{A^2} \; , \; \; $$
$$ s_2 = A^4 + \frac{1}{A^4} \; , \; \; $$
$$ s_3 = A^8 + \frac{1}{A^8} \; , \; \; $$
generally
$$ s_n = A^{2^n} + \frac{1}{A^{2^n}} \; . \; \; $$
This gets bigger and bigger, because $A> 1.$
With $ s_0 = 5 = A + \frac{1}{A} $ we find $A = \frac{5 + \sqrt{21}}{2} \approx 4.79.$
If we were to change the $s_0$ to something between $0$ and $2,$ most things would be the same but $A = \frac{s_0 + \sqrt{s_0^2 -4}}{2}$ is now a complex number of norm exactly one. So $s_n,$ which remains real, oscillates but always has absolute value less that two. Indeed, we have $s_n = 2 \cos \left(2^n T \right)$ where $T = \arctan \frac{\sqrt{4-s_0^2 }}{s_0} $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$ Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$
My attempt:
$$P=\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) = \frac{1}{2}\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)$$
$$\alpha = \cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$$
$$\Leftrightarrow \alpha \ge3.\sqrt[3]{\alpha} \Leftrightarrow \alpha^2\ge27\Leftrightarrow \alpha\ge3\sqrt{3}$$
$$\Rightarrow P=\frac{1}{2}\alpha\ge\frac{3\sqrt{3}}{2}$$
Hmm where I was wrong ?
| Let $p$ is the semi-perimeter, $R$ is circumradius and $r$ is radius.
We have
$$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C} = \frac{2R}{a} + \frac{2R}{b} + \frac{2R}{c} = \frac{4R+r}{2p}+\frac{p}{2r},$$
and
$$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2} = \sum \frac{r}{b+c-a} = \frac{4R+r}{p}.$$
The inequality become
$$p \geqslant 2\sqrt3 r.$$
Which is true by know inequality $$p \geqslant 3\sqrt 3 r.$$
| {
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"url": "https://math.stackexchange.com/questions/4035416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Bound on Truncated Alternating Harmonic Series I'm trying to prove the following inequality
$$\sum _{n=1}^{2N}\frac{(-1)^n}{n}+\log (2)<\frac{1}{4N+1}.$$
Most of the traditional inequalities I've seen for harmonic numbers aren't tight enough to prove this, unless I'm doing something wrong.
| As the taylor series of $\ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}x^n}{n}$, we have
$$\ln(2)=\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n}$$
then
\begin{align}
\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n} + \ln(2) &= \sum_{n=2N+1}^{+\infty} \frac{(-1)^{n-1}}{n} \\
&= \sum_{k=N}^{+\infty} \left( \frac{(-1)^{2k}}{2k+1}+\frac{(-1)^{2k+1}}{2k+2} \right) \\
&= \sum_{k=N}^{+\infty} \left( \frac{1}{2k+1}-\frac{1}{2k+2} \right) \\
&= \sum_{k=N}^{+\infty} \frac{1}{(2k+1)(2k+2)} \\
&= \sum_{k=N}^{+\infty} \frac{4}{(4k+2)(4k+4)} \\
&< \sum_{k=N}^{+\infty} \frac{4}{(4k+1)(4k+5)} =\sum_{k=N}^{+\infty} \left( \frac{1}{4k+1}-\frac{1}{4(k+1)+1} \right) = \frac{1}{4N+1}\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.
We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$
Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{ This is what we need to prove}$$
$$
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing and is bounded sufficient?
| We you define the recursive formula you need to define $c_0 = \sqrt 2$
you don't need to define both $c_n$ and $c_{n+1}$. Defining $c_n = \sqrt{2 + c_{n-1}}$ when $n>1$ is enough. Or defining $c_{n+1}=\sqrt{2 + c_n}$ is enough.
We can streamline our prove that $c_n$ is increasing and bounded by $2$ in one fell induction step.
$0 \le \sqrt{2} \le c_0 < 2$.
And if $\sqrt {2} \le c_n < 2$ then $c_{n+1} =\sqrt {2 +c_n} > \sqrt {c_n^2 + c_n} > \sqrt{c_n^2} = c_n \ge \sqrt 2$.
And $c_{n+1}= \sqrt{2+c_n} < \sqrt{2 + 2} = 2$.
(Your inductions had no base cases!)
Then as we know it is bounded above and increasing $\lim_{n\to \infty} c_n=c$ exists.
And as $\lim_{n\to \infty} c_n = \lim_{n\to \infty} c_{n+1} = \lim_{n\to\infty} \sqrt {2 + c_n} = \sqrt{2 + \lim_{n\to\infty}{c_n}}$ so
$c = \sqrt {2+c} \ge 0$ so
$c^2 - c -2 =0$
$(c-2)(c+1) = 0$ so $c = 2$ or $c = -1$ but $c> 0$ so $c = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The angle between the pair of tangents drawn from a point P to the circle and its locus The angle between the pair of tangents drawn from a point P to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$ . Then the equation of the locus of the point P is
A- $x^2+y^2+4x−6y+4=0$
B- $x^2+y^2+4x−6y-9=0$
C- $x^2+y^2+4x−6y-4=0$
D- $x^2+y^2+4x−6y+9=0$
My approach is as follow
${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0 \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2} = 13{\sin ^2}\alpha - 9{\sin ^2}\alpha = {\left( {2\left| {\sin \alpha } \right|} \right)^2} = {R^2}$
${X^2} + {Y^2} = {R^2}$
$Y = mX \pm \sqrt {{{\mathop{\rm R}\nolimits} ^2} + {R^2}{m^2}} $ is the general equation of the tangent
$\Rightarrow Y - mX = \pm R\sqrt {1 + {m^2}} \Rightarrow {Y^2} + {m^2}{X^2} - 2mYX = {R^2} + {R^2}{m^2}$
$ \Rightarrow {m^2}\left( {{X^2} - {R^2}} \right) - 2mYX + {Y^2} - {R^2} = 0$
$\tan 2\alpha = \frac{{{m_1} + {m_2}}}{{1 - {m_1}{m_2}}} = \frac{{\frac{{2YX}}{{\left( {{X^2} - {R^2}} \right)}}}}{{1 - \frac{{\left( {{Y^2} - {R^2}} \right)}}{{\left( {{X^2} - {R^2}} \right)}}}} = \frac{{2YX}}{{{X^2} - {Y^2}}}$
$2\sin \alpha = R \Rightarrow \sin \alpha = \frac{R}{2} \Rightarrow \tan \alpha = \frac{R}{{\sqrt {4 - {R^2}} }}$
$\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{{2YX}}{{{X^2} - {Y^2}}} = \frac{{2\left( {y - 3} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2} - {{\left( {y - 3} \right)}^2}}}$
not able to proceed from here
|
The choices should alert you that a simpler method is possible.
On drawing a diagram, we see that $P$ is at a constant distance from center $O=(-2,3)$. You can now finish.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
| Recall the well-known fact: If $g(u)$ is a convex function on $\mathbb{R}^n$, then
$$g(u) \ge g(v) + \nabla g(v)^\mathsf{T}(u - v), \ \forall u, v \in \mathbb{R}^n.$$
Clearly, $f(x, y)$ is a convex function on $\mathbb{R}^2$. Thus, we have
$$f(x, y) \ge f(\tfrac{5}{12}, \tfrac{4}{3}) + \tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (x - \tfrac{5}{12})
+ \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (y - \tfrac{4}{3}) = \sqrt{10}.$$
Also, $f(\tfrac{5}{12}, \tfrac{4}{3}) = \sqrt{10}$.
The minimum of $f(x, y)$ is $\sqrt{10}$ achieved at $x = 5/12$ and $y = 4/3$.
Another way: Since $f(x, y)$ is a convex function on $\mathbb{R}^n$ and $\tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3})
= \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) = 0$, $f(x,y)$ achieves its minimum at $x = 5/12$ and $y = 4/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
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Find general solution of differential equation $x\frac{dy}{dx} + y = a \sin^2x.$ and $a \in \mathbb{R}$ The real number $a$ and the differential equation are given $$x\frac{dy}{dx} + y = a \sin^2x.$$
a) Find a general solution of the equation.
b) Find all $a \in \mathbb{R}$ that solutions of $y (x)$ of the differential equation
$$\lim_{x \to \infty}y (x) = 3.$$
Which of these solutions have limited functions?
So in example a) I got function $$y(x)=\frac{a(2x-\sin2x)+4C}{4x}$$
When I calculate limit of this function I got $\frac{a}{2}$, so $a = 6$.
Is this the only solution or is some different?
| $$\begin{align}x\frac{dy}{dx} + y &= a\sin^2(x) \\ x\frac{dy}{dx} + \frac{dx}{dx}y &= a\sin^2(x) \\ \frac{d}{dx}(xy) &= a\sin^2(x)\\ \therefore xy = \int a\sin^2x\ dx &= \frac{ax}{2} - \frac{a\sin(2x)}{4} + c\\ \implies y &= \frac{a}{2} - \frac{a\sin(2x)}{2x}+ \frac{c}{x}\end{align} $$
Therefore, as $ \lim_{x \to \infty} y = 3$ we have $\frac{a}{2} = 3 \implies a =6$.
| {
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"url": "https://math.stackexchange.com/questions/4049672",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.
Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.
| If you look at the inequality as a quadratic inequality with respect to the variable $x$, then we have
$$x^2+y^2+1 \geq xy+x+y$$
$$\implies x^2+y^2+1 -xy-x-y≥0$$
$$\implies x^2-x(y+1)+(y^2-y+1)≥0$$
$$\implies \left( x -\frac{y+1}{2}\right)^2-\left(\frac{y+1}{2}\right)^2+y^2-y+1≥0$$
$$\implies \left( x -\frac{y+1}{2}\right)^2+ \frac 34 (y - 1)^2≥0.$$
*
*Equality occurs if and only if, when $y=1$ and $x=\dfrac{y+1}{2}=\dfrac 22=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$
$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$
Find $x$.
I am not sure how to proceed.
Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?
Any help would be sincerely appreciated.
| This is just an empirical answer. Consider
$$f(n) = \sqrt{t+\sqrt{t+1 - \sqrt{t+2 + \sqrt{t+3 - \cdots}}}}$$
We need to find $f(1)$. But
\begin{align}
f(1) &= \sqrt{1+\sqrt{2 - f(3)}}\\
f(3) &= \sqrt{3+\sqrt{4 - f(5)}}\tag{Similarly moving on }\\
f(n) &= \sqrt{n+\sqrt{n+1 - f(n+2)}}\\
\end{align}
After squaring and rearranging, we get
$$f(n+2) = n+1 - (f(n)^2 - n)^2$$
I used the wolfram alpha recurrence solver to solve the recurrence and it gave me this:
$$\begin{array}{|c|c|c|c|}
\hline
n& 0 & 1 & 2 & 3 & 4 \\ \hline
g(n) & 0 & 1 & 1 & 2& 2\\ \hline
\end{array}$$
I don't know if this helps, but $x = 1$.
| {
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"url": "https://math.stackexchange.com/questions/4056039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$
Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$
Attempt:
First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...
Now I try to find the sum and I can't get to the solution $\frac{5}{2}$.
Let's try:
$$\sum_{k=2}^{\infty} \frac{5}{2^k}=5\sum_{k=2}^{\infty} \frac{1}{2^k}=5\sum_{k=2}^{\infty} 2^{-^k}=5\cdot2^{-2}\sum_{k=2}^{\infty}2^{-k}$$
I don't know how to work with the $2^{-k}$ in here, hope for your help^^
| $$\sum_{k=2}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} - (1 +\frac{1}{2}) = \frac{1}{1 - \frac{1}{2}} -\frac{3}{2} = \frac{1}{2}$$So the answer is $5\times\frac{1}{2} = \frac{5}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution to $y' + \frac xy = 1$ $y' + \frac xy = 1$
I've identified it as a homogeneous differential equation and transformed it to $xv' + v = 1-\frac 1v$ (where $v = \frac yx$) but am stuck here.
| $$
\begin{aligned}
y' + \frac{x}{y} = 1 &\Leftrightarrow
\left|
\begin{aligned}
y(x) &= z(x)\cdot x \\
y'(x) &= z'(x)x + z(x)
\end{aligned}
\right| \Leftrightarrow \\
&\Leftrightarrow z'x + z + \frac{x}{zx} = 1 \Leftrightarrow \\
&\Leftrightarrow z'x + z + \frac{1}{z} = 1 \Leftrightarrow \\
&\Leftrightarrow z' = -\frac{1}{x}\frac{z^2-z+1}{z}\Leftrightarrow \\
&\Leftrightarrow z' = -\frac{1}{x}\frac{z^2-z+1}{z}\Leftrightarrow \\
&\Leftrightarrow \frac{zdz}{z^2-z+1} = -\frac{dx}{x} \Leftrightarrow \\
&\Leftrightarrow \int\frac{zdz}{z^2-z+1} = -\int\frac{dx}{x}
\end{aligned}
$$
Left-hand side integral:
$$
\begin{aligned}
\int\frac{zdz}{z^2-z+1} &= \frac{1}{2}\int\left(\frac{2z-1}{z^2-z+1} + \frac{1}{z^2-z+1}\right)dz = \\
&= \frac{1}{2}\left(\int\frac{d(z^2-z+1)}{z^2-z+1}+\int\frac{dz}{\left(z-\frac{1}{2}\right)^2 +\left(\frac{\sqrt{3}}{2}\right)^2}\right) = \\
&= \frac{1}{2}\left(\text{ln}\left(z^2-z+1\right) + \frac{2}{\sqrt{3}}\arctan\left(\frac{2\left(z-\frac{1}{2}\right)^2}{\sqrt{3}}\right)\right).
\end{aligned}
$$
Right-hand side integral:
$$
-\int\frac{dx}{x} = -\text{ln}(x) + C.
$$
Finaly ($z = \frac{y}{x}$),
$$
\frac{1}{2}\text{ln}\left(\frac{y^2}{x^2}-\frac{y}{x}+1\right) + \frac{1}{\sqrt{3}}\arctan\left(\frac{2\left(\frac{y}{x}-\frac{1}{2}\right)^2}{\sqrt{3}}\right) = -\text{ln}(x) + C.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$? I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I have to integrate it. I would think that the $-x$ would equal to $\frac{-x^2}{2}$ but the answer on the back of my book is: $\displaystyle \frac{y^2}{2} - \frac{2}{5} y^{5/2} =\frac{x^3}{3}-x-\frac{54}{5}$
What is the particular solution given $y=4$ and $x=3$?
My Solution
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}\\
y(1-\sqrt y)\frac{dy}{dx}=\frac{1}{x}\cdot (x^3-x)\\
\int (y-y^\frac{3}{2}) \frac{dy}{dx}=\int{x^2-x}\\=\frac{y^2}{2}-\frac{2y^{5/2}}{5}=\frac{x^3}{3}-\frac{x^2}{2}+c\\
\frac{y^2}{2}-\frac{2y^{5/2}}{5}-\frac{x^3}{3}+\frac{x^2}{2}=c$$
| Okay, so you have to solve the equation:
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
The first step is to move things around, which you did properly.
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}$$
Now we can move the $dx$ to the other side and write the equation as,
$$y(1-\sqrt y)dy=\frac{1}{x}\cdot (x^3-x)dx$$
Simplifying and adding in the integration,
$$\int \left(y-y^\frac{3}{2}\right)dy= \int(x^2-1)dx$$
So, integrating both sides, we have,
$$\frac{y^{1+1}}{(1+1)}-\frac{y^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}=\frac{x^{2+1}}{(2+1)}-x+C\\
\frac{1}{2}y^2-\frac{2}{5}y^{\frac{5}{2}}=\frac{1}{3}x^3-x+C$$
Finally, plugging in $y=4$ and $x=3$, we have,
$$\frac{1}{2}(4)^2-\frac{2}{5}(4)^{\frac{5}{2}}=\frac{1}{3}(3)^3-3+C\\
-\frac{24}{5}=6+C\\ C=-\frac{54}{5}$$ or, we can make it a little cleaner by saying that there is some other constant, say $C_1$, such that, $$C_{1}=\frac{54}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
factor $1/(x\pm y)$ as $f(x)g(y)$ Is it possible to rewrite
$$\frac{1}{x\pm y}$$
as multiplication of an expression containing only $x$ and another containing only $y$ ($x$ and $y$ are real independent variables), i.e.
$$\frac{1}{x\pm y}\stackrel{?}{=}f(x)g(y)$$
If it is possible, what are $f(x)$ and $g(y)$?
| Sure it's possible. Take $A=\frac{a}{a+b}$ and take $B=0$. Then it should be trivial to show that $$\frac Aa + \frac Bb = \frac{\frac{a}{a+b}}{a} + \frac0b = \frac{1}{a+b}.$$
In fact, taking any value of $B$, you can set $A=a\cdot\left(\frac{1}{a+b} - \frac Bb\right)$ and get the equality
$$\frac{A}{a} + \frac Bb = \frac1{a+b}.$$
However, if you demand that $a,b, A, B$ are all integers, then the answer is no. A simple counterexample can be found by taking $a=b=1$. Then, $\frac{1}{a+b}=\frac12$, while $\frac Aa + \frac Bb = \frac A1 + \frac B1 = A + B$, and since $A$ and $B$ are integers, this means $\frac Aa + \frac Bb$ must also be an integer.
Since $\frac12$ is not an integer, it's clear that $\frac Aa + \frac Bb$ cannot be equal to $\frac1{a+b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
| $$\begin{align*}
I &= \int_0^\infty \frac{\ln x}{(x^2+1)^2} \, dx \\[1ex]
&= \int_0^1 \frac{\ln x}{(x^2+1)^2} \, dx + \int_1^\infty \frac{\ln x}{(x^2+1)^2} \, dx \\[1ex]
&= \int_0^1 \frac{1-x^2}{(x^2+1)^2} \ln(x) \, dx \tag{1} \\[1ex]
&= 2 \int_0^1 \frac{\ln(x)}{(x^2+1)^2} \, dx - \int_0^1 \frac{\ln(x)}{x^2+1} \, dx \tag{2} \\[1ex]
&= 2 \sum_{n=0}^\infty (-1)^n (n+1) \int_0^1 x^{2n} \ln(x) \, dx - \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n} \ln(x) \, dx \tag{3} \\[1ex]
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} - 2 \sum_{n=0}^\infty \frac{(-1)^n(n+1)}{(2n+1)^2} \tag{4} \\[1ex]
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} - \sum_{n=0}^\infty (-1)^n \left(\frac1{2n+1} + \frac1{(2n+1)^2}\right) \tag{5} \\[1ex]
&= - \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \\[1ex]
&= -\arctan(1) = \boxed{-\frac\pi4} \tag{6}
\end{align*}$$
*
*$(1)$ : substitute $x\mapsto\frac 1x$ in the latter integral
*$(2)$ : polynomial division
*$(3)$ : exploit the Maclaurin series of $\frac1{1-x}$
*$(4)$ : integrate by parts
*$(5)$ : partial fractions
*$(6)$ : recognize the Maclaurin series of $\arctan(x)$ evaluated at $x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
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Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?
This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!
| Use $\sin x \approx x$ in the denominator, it becomes $x^4$, next expand numerator complete up to $x^4$.Then use $e^{z}=1+z+z^2/2+O(z^3)$, $\cos z =1-z^2/2+z^4/24+O(z^6)$ when $|z|$ is very small
$$L=\lim_{x\to\infty} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}=\lim_{x\to 0} \frac{1+x^2+x^4/2+2(1-x^2/2+x^4/(24)-3}{x^4}$$.
$$L=\lim_{x \to 0} \frac{7x^4/12}{x^4}=\frac{7}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve the following inequality $\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$ Solve the following inequality with a real parameter a:
$\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$
I am not sure if my answer is correct. Please feel free to share your thoughts.
First, we need to multiply both sides of the given inequality by the LCD, which is $x^2$.
Since $x^2$ is always positive for all real numbers $x$, we don't have to change the direction of the inequality.
$x^2 \left[\frac{x-a}{x^2} + x \right] \ge 2x^2 \left(1 - \frac{a}{x}\right)$
$x-a+x^3 \ge 2x^2 - 2ax$
Isolate all terms on the left-hand side of the inequality.
$x^3-2x^2+(2a+1)x - a \ge 0$
Factor the left-hand side of the inequality. Since $a$ is a constant, it will always have a factor of $1$. Let us try to determine if $x-1$ is a factor using synthetic division.
$\begin{array}{cccccc} 1| & 1 & -2 & 2a+1 & -a \\ & & 1 & -1 & 2a \\ \hline & 1 & -1 & 2a & a \end{array}$
The result means that we can factor the left-hand side as
$(x-1) \left[x^2-(a+1)x+a\right] \ge 0.$
We can further factor the quadratic term on the left-hand side.
Note that we can express the inequality as
$(x-1) \left[x^2+(-a-1)x+a\right] \ge 0.$
The middle term is equal to the sum of $-a$ and $-1$, and the last term can be expressed as $(-a) * (-1)$. Hence, we can further express this as
$(x-1)(x-1)(x-a) \ge 0$
$(x-1)^2(x-a) \ge 0.$
Notice that for all real values of $x$, $(x-1)^2$ will always be positive. So, to make the left-hand side always greater than or equal to 0, the factor $x - a$ should be greater than or equal to $0$. That is
$(x-a) \ge 0.$
Therefore, we can now say that the solution to the inequality must be
$x \ge a.$
| Your argument is fine up to the point
$$
f(x) = x^3-2x^2+(2a+1)x - a \ge 0
$$
Then, your roots are wrong. So one needs a curve analysis. We have
$$
f´(x) = 3x^2-4x+(2a+1)
$$
so extreme points are $x = \frac13(2 \pm \sqrt{1 - 6 a})$. So for $a > \frac16$, there are no extremal points at all, which means $f(x)$ is rising everywhere. Since it is ugly to determine the (sole) root of $f(x)$ in this case, we can make a nice estimate. Notice
$$
f(x) = x(x-1)^2 +a(2x -1)
$$
which lets us give the bound that $f(x)$ is $\ge 0$ at least for $a \ge 0$ and $x\ge 0.5$. In fact, for large $a$, $x\ge 0.5$ becomes asymptotically exact.
This bound $x\ge 0.5$ is not that bad also for smaller positive $a$. The true values are that for $a = \frac16$ we have $x \ge 0.161$, and for $a = 0$ we obviously have $x \ge 0$.
Considering negative $a$, the behavior of $f(x)$ becomes more beasty. Taking the above extremal values gives us that the value of $f(x)$ at the minimum is
$\frac{a}{3} + \sqrt{1-6a} \frac{12 a -2}{27}+ \frac{2}{27} \le 0$ and at the maximum,
$\frac{a}{3} - \sqrt{1-6a} \frac{12 a -2}{27}+ \frac{2}{27} \ge 0$. Clearly, our third order polynomial will therefore have exactly one interval where $f(x) \ge 0$ about the maximum $\frac13(2 - \sqrt{1 - 6 a})$, and then $f(x)$ will stay positive again for $x > x_0$, where that $x_0$ can be estimated for negative $a$:
$$
f(x) = x(x-1)^2 +a(2x -1) \ge x(x-1)^2 +a(2x) = x [(x-1)^2 +2a]
$$
Hence $f(x) \ge 0$ at least for $x \ge x_0 = 1 + \sqrt{-2a}$ which indeed is a rather precise approximation. The same bound on $f(x)$ can be used to say, for $a < -\frac18$, that $f(x) \ge 0$ at least in the interval $x \in [1 - \sqrt{-2a} \quad , \quad 0.5]$. That holds true, as it is known that $f(x) \ge 0 $ on both ends of that interval, and since we know from the above discussion that there is only one continuous interval with positive $f(x)$ about the maximum.
Again, this is a rather precise approximation for more negative $a$.
Let's summarize these findings: We have, with rather precise approximation, that $f(x) \ge 0$ at least for:
$a \ge 0$ and $x\ge 0.5$;
$a \le 0$ and $x \ge 1 + \sqrt{-2a}$ ;
$a \le -\frac18$ and $x \in [1 - \sqrt{-2a} \quad , \quad 0.5]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can I find the limit of this fraction? ($0/0$ type) Suppose I have two complex numbers $g_0$ and $g_1$, when $g_0\rightarrow1$ and $g_1\rightarrow0$, how can I evaluate
$$
\sqrt\frac{g_1^2}{1-g_0^2}
$$
Intuitively I think that should be infinity, but I'm not pretty sure how can I justify that? Thanks!!
| Let
$g_1 = a$
and
$g_0 = 1-b$
so
$\sqrt\frac{g_1^2}{1-g_0^2}
=\sqrt\frac{a^2}{1-(1-b)^2}
=\sqrt\frac{a^2}{b(2-b)}
$
where
$a, b \to 0$.
If
$a=b$ this is
$\sqrt\frac{b^2}{b(2-b)}
=\sqrt\frac{b}{(2-b)}
\to 0
$.
If $a = b^{1/4}$ this is
$\sqrt\frac{b^{1/2}}{b(2-b)}
=\sqrt\frac{1}{b^{1/2}(2-b)}
\to \infty
$.
If $a = cb^{1/2}$ this is
$\sqrt\frac{c^2b}{b(2-b)}
=\sqrt\frac{c^2}{(2-b)}
\to \dfrac{c^2}{2}
$
so this can approach any positive value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots? Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots and has only integral parameters?
Since it has two different roots the it must be of degree at least $2$.
If it is of degree $2$, then it is of the form $(x-\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})$, but this type of polynomial has parameters which are not integers.
If it is of degree $4$ then we have the polynomial $(x-\frac{\sqrt{3}}{2})(x+\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})(x+\frac{\sqrt{2}}{3})$.
It remains to examine the case where it is of degree $3$. This is where I got stuck. I don't know how to examine that case. Could you please explain to me how to solve this question?
| The polynomial with integer coefficients with least degree that has $\frac{\sqrt{3}}{2}$ as a root is $4 x^2 - 3$.
Every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ as a root is a rational polynomial multiple of $4 x^2 - 3$.
Likewise, for $\frac{\sqrt{2}}{3}$, the polynomial is $9 x^2 - 2$.
Since these two polynomials are coprime, every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{2}}{3}$ as roots is a multiple of $(4 x^2 - 3)(9 x^2 - 2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What is $\arcsin(\cos{x})$ (Spivak Calculus exercise 15-18)? In the first part of the exercise I proved that $\sin(x+\frac{\pi}2)=\cos{x}$.
The second part is asking what $\arcsin(\cos{x})$ is. Presumably, they want a rule for unrestricted $x$.
We have $\arcsin(\cos{x})=\arcsin(\sin(x+\frac{\pi}2))$. It was said that the domain of $\sin$ usually restricted to $\left[-\frac{\pi}2, \frac{\pi}2\right]$, hence I assume that $x+\frac{\pi}2 \in \left[-\frac{\pi}2, \frac{\pi}2\right]$, and this set is also the co-domain of the $\arcsin$ function.
Since $\sin$ is periodic with period $2\pi$, $\sin(x+\frac{\pi}2)=\sin(2k\pi+x+\frac{\pi}2)$, $2k\pi+x+\frac{\pi}2 \in \left[2k\pi-\frac{\pi}2, 2k\pi+\frac{\pi}2\right], k \in \mathbb{Z}$.
And we have that $\arcsin(\sin(2k\pi+x+\frac{\pi}2))=2k\pi+x+\frac{\pi}2$.
But the answer in the solutions manual is $\begin{cases} x-2k\pi+\frac{\pi}2, x \in \left[2k\pi-\pi, 2k\pi\right] \\ 2k\pi+\frac{\pi}2-x, x \in \left[2k\pi, 2k\pi+\pi\right] \end{cases}$
I can plot the graph and see how $2k\pi+x+\frac{\pi}2$ doesn't work, but where did they get this piecewise function?
| here's my solution based on @YvesDaoust's answer:
$\cos{x}=\sin\left({x}+\frac{\pi}2\right), x \in R$, hence $\arcsin(\cos(x)) = \arcsin\left(\sin\left({x}+\frac{\pi}2\right)\right)$
The co-domain of $\arcsin$ is $[-\frac{\pi}2,\frac{\pi}2]$, and we must restrict the domain of $\sin$ to the same interval, there are 2 cases, in therms of our $\sin(x+\frac{\pi}2)$:
Case 1: $x+\frac{\pi}2 \in [2k\pi-\frac{\pi}2, 2k\pi+\frac{\pi}2]$
Subtract $2k\pi$, $x+\frac{\pi}2-2k\pi \in [-\frac{\pi}2,\frac{\pi}2]$, moreover $\sin(x+\frac{\pi}2)=\sin(x+\frac{\pi}2-2k\pi)$, hence $\arcsin(\cos(x))=x+\frac{\pi}2-2k\pi$ when $x \in [2k\pi-\pi, 2k\pi]$.
Case 2: $x+\frac{\pi}2 \in [2k\pi+\frac{\pi}2, 2k\pi+\frac{3\pi}2]$
$\pi-(x+\frac{\pi}2) \in [\pi-2k\pi-\frac{3\pi}2, \pi-2k\pi-\frac{\pi}2]=[-2k\pi-\frac{\pi}2, -2k\pi+\frac{\pi}2]$
Add $2k\pi$ to get the desired interval:
$\pi-(x+\frac{\pi}2)+2k\pi = \frac{\pi}2+2k\pi-x \in [-\frac{\pi}2, \frac{\pi}2]$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving $\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$ I am trying to solve this equation
$$\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$$
By using Mathematica, I know that, the equation has unique solution $x=1$.
I tried to write the equation in the form
$$\sqrt{(x-1) (x+1) \left(2x^2-1\right)} + \sqrt{x^2\left(2x^2-1\right)} =4 x-3.$$
From here, I can not solve it. How can I solve it?
| Note that the domain of the equation is $x\ge1$ and
$$\sqrt{2x^4-x^2} - (2x^2-1)= \sqrt{2x^2-1} \left( x-\sqrt{2x^2-1}\right)
= -\frac{\sqrt{2x^2-1} (x^2-1)}{x+\sqrt{2x^2-1} }
$$
Then, factorize the equation as follows
\begin{align}
&\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}-(4 x-3)\\
=& \sqrt{(x^2-1) (2x^2-1)} +\left( \sqrt{x^2(2x^2-1)} - (2x^2-1)\right)+2(x^2-1)^2\\
=& \sqrt{(x^2-1) (2x^2-1)} -\frac{\sqrt{2x^2-1} (x^2-1)}{x+\sqrt{2x^2-1} } +2(x^2-1)^2 \\
=& \sqrt{(x^2-1) (2x^2-1)} \> \frac{x+\sqrt{2x^2-1}-\sqrt{x^2-1}}{x+\sqrt{2x^2-1} } +2(x^2-1)^2 \\
=& \sqrt{(x^2-1) (2x^2-1)} \> \frac{x+\sqrt{2x^2-1}-\sqrt{x^2-1}}{x+\sqrt{2x^2-1} } +2(x^2-1)^2\\
=& \sqrt{x^2-1}\cdot f(x)
\end{align}
where
$$f(x) = \sqrt{2x^2-1} \> \frac{x+\sqrt{2x^2-1}-\sqrt{x^2-1}}{x+\sqrt{2x^2-1} } +2\sqrt{x^2-1}>0
$$
Thus, $x=1$ is the sole solution.
| {
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"url": "https://math.stackexchange.com/questions/4079527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the maximal area between hyperbola and chord $AB$?
Suppose $A$ and $B$ are variable points on upper branch of hyperbola $\mathcal{H}:\;x^2-y^2=-4$ such that $AB = 1$. What is the maximal area between $AB$ and $\mathcal{H}$?
Clearly $y= \sqrt{x^2+4}$. If we set $A\big(t,\sqrt{t^2+4}\big)$ then $B\big(t+h, \sqrt{(t+h)^2+4} \big)$ for some positive $h$, such that $$h^2 +\Big(\sqrt{(t+h)^2+4} -\sqrt{t^2+4} \Big)^2=1$$
and we have to find a minimum value of
$$f(t) = \int_{t}^{t+h}\sqrt{x^2+4}\;dx$$
Notice that if $$F(x):= \int\sqrt{x^2+4}\;dx$$ then $$ F(x)= 2\ln\left(\left|\sqrt{x^2+4}+x\right|\right)+x\sqrt{\dfrac{x^2}{4}+1}$$
and things get complicated very quickly.
| Parametrize the hyperbola as $x=2\sinh t $, $y= 2\cosh t$ and let the endpoints of the chord be $A(2\sinh a, 2\cosh a)$, $B(2\sinh b, 2\cosh b)$. Then, the equation of chord $AB$ is
$$y= x\tanh\frac{a+b}2 +2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 }$$
and the area between the chord and hyperbola is
\begin{align}
K =&\int_{x_a}^{x_b}( y - \sqrt{x^2+4})dx\\
=&\int_a^b \left(2\tanh\frac{a+b}2 \sinh t +
2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 }
- 2 \cosh t \right)2\cosh t\>dt\\
=& 2\tanh\frac{a+b}2 (\sinh^2 b -\sinh^2a)
+2\sinh(b-a)(2-\cosh (a+b) )-2(b-a)
\end{align}
Moreover, $1=AB^2=(x_b-x_a)^2 + (y_b-y_a)^2 $ leads to
$$\cosh2b +\cosh 2a -2 \cos(a+b)=\frac14$$
$$\frac{d b}{da}
=\frac{\sinh2a}{\sinh(a+b)-\sinh2b}$$
Then, it is straightforward to verify that $\frac{d b}{da} = 1 $, $\frac{d K}{da} = 0 $ if $b=-a$, which leads to $\sinh b =-\sinh a= \frac14$ and the maximum area
$$K_{max}= \frac{\sqrt{17}}4-4\sinh^{-1}\frac14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Prove that $\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}+\sqrt{(x-1)^2+(y-2)^2}\geq8$ Prove that $\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}+\sqrt{(x-1)^2+(y-2)^2}\geq8$ for $x,y\in\mathbb{R}$. I used this Proof of an inequality about sequences to solve it and it worked, but I have no ideea how to interpret it geometrically. Any help, please? Thanks in advance.
| Let's $A(2,0), B(4,2),C(2,5), D(1,2)$ and $G(x,y)$ on the plane $Oxy$. The left hand side of the inequality is the sum $$\mathbf{S} =GA+GB+GC+GD$$
The inequality becomes: find $G$ such that the sum $\mathbf{S}$ is minimized.
We have
$$GA+GC \ge AC$$ $$GB+GD \ge BD$$
then $$S = GA+GB+GC+GD \ge AC+BD = \sqrt{2^2+5^2}+3 = \sqrt{29}+3$$
The sum is then minimized if $G$ is at $I$, the intersection of two lines: $AC$ and $BD$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $3^n \gt n^4 \ \forall n \gt 8$ Regarding this question, I found no answer that combines mine, so I want to know if my proof is valid as well:
Problem: Prove that:$\ 3^n \gt n^4 \ \forall n \ge 8$
Base step (n=8): is true because$$P(8): 6561 \gt 4096 \\$$
Inductive step ($P(n) \implies P(n+1)$): $\ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$
$$3^{n+1} = 3^n + 3 = 3^n + 3^n + 3^n = 3^n + 3^n + \frac{(n+1)^4}{3} \gt (n+1)^4 \\ \text{I found } \frac{(n+1)^4}{3} \text{ by dividing 3 on both sides of } 3^{n+1} \gt (n+1)^4 \text{ as soon as I suppose this to be true.}$$
| Rewrite the expression as $e^{n \log 3 - 4 \log n}$. This is your induction step. Now you need to show it for $n+1$.
$$
e^{(n+1)\log 3 - 4 \log (n+1)} = e^{n \log 3 - 4 \log n} \cdot e^{\log 3 - \log (1+\frac{1}{n})}
$$
The first term is greater that $1$ by the inductive argument. All is left to show that the second term is greater than $1$. This is true (again, using the definition of logarithm) for
$$
\frac{3n}{n+1} >e
$$
which is true for $n>\frac{e}{3-e} \approx 1.5$
| {
"language": "en",
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} |
Aproximation of the Normal Distribution by the Normal Density Function In Feller's introduction to probability the next lemma is stated:
"As $x\rightarrow \infty$
$\tag1 \frac{1-R(x)}{x^{-1}n(x)} \rightarrow 1$
Where $R(x)$ is the normal distribution and $n(x)$ is the normal density function. And more precisely:
$\tag1 (x^{-1}-x^{-3})n(x) < 1-R(x) < x^{-1}n(x)$
"
This lemma is shown by proving that the left and right hand side of the inequality are the integrals from $x$ to $\infty$ of $(1-3x^{-4})n(x)$ and $((1+x^{-2})n(x))$ and using the inequality:
$\tag1(1-3x^{-4})n(x) < n(x) < (1+x^{-2})n(x)$
The problem i have is in a generalization of this that comes as a problem of the book in the same chapter. The statement of the problem is:
"Prove that:
$\tag1 \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}-\frac{1}{x^3}+\cdots+(-1)^k\frac{1\cdot3\cdot7\cdots(2k-1)}{x^{(2k+1)}})}\rightarrow 1$ for $x>0$".
I've only tried to show that a similar inequality like the second one holds like in the lemma using the derivatives of $x^{-k}n(x)$ and $(x^{-k}-x^{-(k+2)})n(x)$ but i don't see exactly how to get the sum that is involved, so any help would be much appreciated.
| About the statement
$$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}-\frac{1}{x^3}+...+(-1)^k\frac{1*3*7*...*(2k-1)}{x^{(2k+1)}})}\xrightarrow{x \to +\infty} 1$$
In fact, it holds true, even for the general form like this one
$$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k})}\xrightarrow{x \to +\infty} 1 \qquad \forall n\ge 2, \forall (a_k)_{k=1,...,n} \in \mathbb{R}$$
The reason is
$$\frac{\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k}}{\frac{1}{x}}=1+\sum_{k=2}^n \frac{a_k}{x^{k-1}}\xrightarrow{x \to +\infty} 1$$
Then, you apply the lemma in Fella's book:
$$\frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k})} = \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x})}\times \frac{\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k}}{\frac{1}{x}} \xrightarrow{x \to +\infty} 1$$
PS: I provide another proof for Feller's lemma using the Hopital's rule
Proof:
$$\frac{1-R(x)}{x^{-1}n(x)} = \frac{\int_x^{+\infty}\exp\left(-\frac{t^2}{2} \right)dt}{\frac{1}{x}e^{-\frac{1}{2}x^2}}$$
Let's denote $$G(x) = \int_x^{+\infty}e^{-\frac{1}{2}t^2}dt$$ $$F(x) =\frac{1}{x}e^{-\frac{1}{2}x^2} $$
We have
$$\lim_{x \to +\infty}G(x) = 0$$
$$\lim_{x \to +\infty}F(x) = 0$$
Apply the Hopital's rule for $c = +\infty$
\begin{align}
\lim_{x \to +\infty}\frac{G(x)}{F(x)} &= \lim_{x \to +\infty}\frac{G(x)}{F(x)}\\
&= \lim_{x \to +\infty}\frac{G'(x)}{F'(x)}\\
&= \lim_{x \to +\infty}\frac{-e^{-\frac{1}{2}x^2}}{-e^{-\frac{1}{2}x^2 \left(1+ \frac{1}{x^2} \right)}}
&= \lim_{x \to +\infty}\frac{1}{1+ \frac{1}{x^2} } \\
&= 1
\end{align}
Q.E.D
| {
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Prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1 $ Does my proof make sense?
Problem: prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1 $
Base step (P(1)):$\ \ \ \ \ \ \ \ x-1 \ge x-1$
Hypotesis (P(n)): $ \ \ \ \ \ x^n -1 \ge (x-1)^n$
Thesis (P(n+1)):$ \ \ \ \ \ \ \ \ x^{n+1}-1 \ge (x-1)^{n+1}$
Induction step:
$$x^{n+1}-1 = x^n \cdot x -1 = x \cdot x^n -1 \ge x \cdot (x-1)^n \ge (x-1)^{n+1}$$
where: $x \cdot (\not x\not-\not1\not)^n \ge \not(\not x\not-\not1\not)^n \cdot (x-1)$ so $x \ge x-1$
| $(x-1)^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots+\binom{n}{n}x^n$ by the Binomial Theorem.
Evaluating the above at $x=1$ we see
$\sum\limits_{i=0}^n(-1)^{n-i}\binom{n}{i}=0, \sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}+\binom{n}{n}=0 \therefore (\color{red}{*})\sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}=-1 $ because $\binom{n}{n}=1$.
Let
$f(x)=(x-1)^n-x^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots +\binom{n}{n-1}(-1)^1x^{n-1}$,
$\frac{d}{dx}[f(x)]=\frac{d}{dx}[(x-1)^n-x^n]=n(x-1)^{n-1}-nx^{n-1}=n((x-1)^{n-1}-x^{n-1})\leq 0$ for $x\geq1$.
Note that $f(1)=-1$ by above $(\color{red}{*})$, and $f$ is decreasing, thus $f(x)\leq -1$ for $x\geq 1$.
$\therefore (x-1)^n=f(x)+x^n\leq -1+x^n$ for $x\geq 1, n\in \mathbb{N}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sin\angle HIO$ given $\triangle ABC$ with $(\overline{AB},\overline{BC},\overline{CA})=(8,13,15)$? In triangle $ABC, AB = 8, BC = 13$ and $CA = 15$.
Let $H, I, O$ be the orthocenter, incenter and circumcenter of triangle $ABC$ respectively. Find $\sin$ of angle $HIO$ .
My attempt: By using law of cosine I can see angle $A = 60^{\circ}$. I also observed that angle $BHC$, $BIC$ and $BOC$ equals $120^{\circ} $. How do I proceed after this?
|
Given $a=13$, $b=15$, $c=8$, we
can find semiperimeter, area, inradius and circumradius
ot the triangle:
\begin{align}
\rho&=\tfrac12(a+b+c)=18
,\\
S_{ABC}&=
\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
=
30\sqrt3
,\\
r&=\frac{S_{ABC}}\rho=\tfrac53\sqrt3
,\\
R&=\frac{abc}{4S_{ABC}}=
\tfrac{1}3\sqrt3
\end{align}
And use known expressions for $|OI|$, $|OH|$, and $|IH|$:
\begin{align}
|OI|&=\sqrt{R(R-2r)}=\sqrt{13}
,\\
|OH|&=\sqrt{R^2+2((r+2R)^2-\rho^2)}
=7
,\\
|IH|&=
\sqrt{(r+2R)^2+2r^2-\rho^2}
=\sqrt{13}
,\\
S_{OIH}&=
\tfrac14\sqrt{4|OI|^2|OH|^2-(|OI|^2+|OH|^2-|IH|^2)^2}
=\tfrac74\sqrt3
,\\
S_{OIH}&=\tfrac12|OI|\cdot|IH|\sin OIH
=\tfrac{13}2\sin OIH
,\\
\sin OIH&=
\frac{\tfrac74\sqrt3}{\tfrac{13}2}
=
\tfrac7{26}\sqrt3
.
\end{align}
| {
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Does $2x^5 - x^4 - 22x^3 - 23x^2 + 22x +24 = 0$ have exact solutions in radicals?
Does $2x^5 - x^4 - 22x^3 - 23x^2 + 22x +24 = 0$ have exact solutions in radicals?
A mysterious commenter said on Youtube this was the "easiest quintic equation of my life," and I'm suspecting a troll is afoot. In fact, I believe the opposite is true, this is unsolvable in radicals and you need to prove it.
Symbolab gave up.
Update: Aahaan now says "bro i just took a bunch of linear equations and multiplied them to get this. Now we can factor it into them." The similar equation
$$x^5 - 2x^4 - 22x^3 - 23x^2 + 22x +24 = 0$$
has roots $-2, 1$ and reduces to a straightforward after polynomial division, so there might've been a typo.
| Your polynomial $2x^5-x^4-22x^3-23x^2+22x+24$ has no solution in radical it's Galois group is the symmetric group if $5$ objects, on the other hard, $x^5-2x^4-22x^3-23x^2+22x+24$ factors into $(x-1)(x+2)(x^3-3x^2-17x-12)$
Since the polynomials are related in shape
$$2x^5-x^4-22x^3-23x^2+22x+24 = x^5-2x^4-22x^3-23x^2+22x+24 +(x^5+x^4)$$
$$2x^5-x^4-22x^3-23x^2+22x+24 = (x-1)(x+2)(x^3-3x^2-17x-12)+x^4(x+1)$$
There is no common linear factor, I believe it's a typo
| {
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Prove that $\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + O\left(\frac{1}{x^3}\right)$ for $x \to +\infty$ Want to prove that $$\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + O\left(\frac{1}{x^3}\right) \ \text{as}~~~ x \to +\infty$$
Wolfram says that it's true but I'm trying to find some formal prove of this equality.
Here's what i found about that:
We know that $\arctan(x) = \frac{\pi}{2} - \int_x^{+\infty}\frac{dt}{1 + t^2}$ (we can easily prove this by solving such integral). But what this one can give us for asymptotic assessment?
Seems like we have to prove that this integral equals to $-\frac{1}{x} + O\left(\frac{1}{x^3}\right)$, but how?
| To show $$\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + O\left(\frac{1}{x^3}\right) \ \text{as}~~~ x \to +\infty,$$
you have to prove (by definition)
$$\limsup_{ x\to +\infty}\left|\frac{\arctan(x)-\frac{\pi}{2} + \frac{1}{x}}{\frac{1}{x^3}}\right|<\infty.
$$
This can be easily done by calculating $\lim_{ x\to +\infty}\frac{\arctan(x)-\frac{\pi}{2} + \frac{1}{x}}{\frac{1}{x^3}}$ using L'Hôpital's rule. As you want to use $\arctan(x) = \frac{\pi}{2} - \int_x^{+\infty}\frac{dt}{1 + t^2}$, you get
$$\lim_{ x\to +\infty}\frac{\arctan(x)-\frac{\pi}{2} + \frac{1}{x}}{\frac{1}{x^3}}=\lim_{ x\to +\infty}\frac{- \int_x^{+\infty}\frac{dt}{1 + t^2} + \frac{1}{x}}{\frac{1}{x^3}} =\lim_{ x\to +\infty}\frac{\frac{1}{1 + x^2} - \frac{1}{x^2}}{-\frac{3}{x^4}}=\lim_{ x\to +\infty}\frac{1}{3}\frac{x^2}{1 + x^2}=\frac{1}{3}.
$$
| {
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If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
| Write $x=\tan A, y=\tan B, z=\tan C$.
By Vieta’s formula, we have
$$x+y+z=a,\ \ xy+yz+zx=0,\ \ xyz=-b.$$
Now you want to find the value of $$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2.$$
Note that
$$a^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx),$$
so $x^2+y^2+z^2=a^2$. Similarly,
$$0=(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$
implies $x^2y^2+y^2z^2+z^2x^2=2ab$. Hence
$$(1+x^2)(1+y^2)(1+z^2)=1+a^2+2ab+b^2=1+(a+b)^2.$$
| {
"language": "en",
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"answer_count": 5,
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?
| For the title question. The factorization is given by
$$
(ax-b)(bx-a)=abx^2-(a^2+b^2)x+ab.
$$
We can find it by multiplying out $(\rho_1x+\rho_2)(\rho_3x+\rho_4)$ and comparing coefficients. This is easier than computing a discriminant.
| {
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"source": "stackexchange",
"question_score": "4",
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Harmonic numbers as ratio of two Determinants Provide a proof to this interesting identity:
$$\frac{\begin{vmatrix} 1^0 & 1^2 & 1^3 & \cdots & 1^n \\ 2^0 & 2^2 & 2^3 & \cdots & 2^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^0 & n^2 & n^3 & \cdots & n^n \end{vmatrix}}{\begin{vmatrix} 1^1 & 1^2 & 1^3 & \cdots & 1^n \\ 2^1 & 2^2 & 2^3 & \cdots & 2^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^1 & n^2 & n^3 & \cdots & n^n \end{vmatrix}} = H_n \tag{1}\label{harmonic}$$
where $H_n$ is the nth Harmonic Number.
I accidently stumbled upon this observation while Programming. The inverse of a $(n+1)\times (n+1)$ Vandermonde Matrix (say $V_n$) with $ [\alpha_1 , \alpha_2, \alpha_3, \cdots, \alpha_n , \alpha_{n+1} ] = [0,1,2,\cdots,n-1,n] $, where we treat $0^0$ as $1$,
$$V_n = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 1^0 & 1^1 & 1^2 & \cdots & 1^n \\ 2^0 & 2^1 & 2^2 & \cdots & 2^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^0 & n^1 & n^2 & \cdots & n^n \end{bmatrix} $$
has some interesting properties.
The first row of $V_n^{-1}$ is $ [ 1 \ 0 \ 0 \cdots \ 0]$ . We also have this recurrence relation:
$$V_n^{-1}[m][0] = - \sum_{k=1}^n \frac{V_n^{-1}[m-1][0]}{k} \tag{2}\label{recurrence}$$
Using \eqref{recurrence} we find: $V_n^{-1}[1][0] = -H_n$.
We can also find $V_n^{-1}[1][0]$ using $ \ adj(V_n)/det(V_n)$ which gives $V_n^{-1}[1][0] = C_{01}/det(V_n)$ where $C_{01}$ is the Cofactor of $V_n[0][1]$. But it is same as the -ve of LHS of \eqref{harmonic}.
QCD
We can prove the same using Cramer's rule by Reframing the Arguments here a little. But None of these proofs are "nice" in a way of not using inverse of Vandermonde Matrix but only elementary arguments like Expansion of the determinants or Mathematical Inductions or something else. I've put exhaustive efforts but haven't succeded yet.
Beside an elementary proof, any other perspective on the Result would be helpful and appreciated.
| As @achillehui suggested, we can consider the Vandermonde determinant
$$
P(x) = \begin{vmatrix}
x^0 & x^1 & x^2 & \cdots & x^n \\
1^0 & 1^1 & 1^2 & \cdots & 1^n \\
2^0 & 2^1 & 2^2 & \cdots & 2^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n^0 & n^1 & n^2 & \cdots & n^n
\end{vmatrix}
= \prod_{k=1}^n (x-k) \cdot \prod_{j<k} (j-k) \, .
$$
$P$ is a polynomial, its constant term is exactly the denominator of your fraction, and its coefficient of $x^1$ is the negative of the numerator:
$$
\frac{\begin{vmatrix}
1^0 & 1^2 & 1^3 & \cdots & 1^n \\
2^0 & 2^2 & 2^3 & \cdots & 2^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n^0 & n^2 & n^3 & \cdots & n^n
\end{vmatrix}}{\begin{vmatrix}
1^1 & 1^2 & 1^3 & \cdots & 1^n \\
2^1 & 2^2 & 2^3 & \cdots & 2^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n^1 & n^2 & n^3 & \cdots & n^n
\end{vmatrix}}
= \frac{-P'(0)}{P(0)} \, .
$$
The logarithmic derivative of a product is the sum of the logarithmic derivatives of each factor:
$$
\frac{P'(x)}{P(x)} = \sum_{k=1}^n \frac{1}{x-k}
$$
and this gives the desired result:
$$
-\frac{P'(0)}{P(0)} = \sum_{k=1}^n \frac{1}{k} = H_n \, .
$$
In the same way one can show that more generally
$$
\frac{\begin{vmatrix}
x_1^0 & x_1^2 & x_1^3 & \cdots & x_1^n \\
x_2^0 & x_2^2 & x_2^3 & \cdots & x_2^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_n^0 & x_n^2 & x_n^3 & \cdots & x_n^n
\end{vmatrix}}{\begin{vmatrix}
x_1^1 & x_1^2 & x_1^3 & \cdots & x_1^n \\
x_2^1 & x_2^2 & x_2^3 & \cdots & x_2^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_n^1 & x_n^2 & x_n^3 & \cdots & x_n^n
\end{vmatrix}}
= \sum_{k=1}^n \frac{1}{x_k}
$$
for any non-zero distinct numbers $x_1, \ldots, x_n$.
| {
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Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?
| $(1+x^2+x^4+x^6)^n = (1+x^4)^n(1+x^2)^n$
the $r^{th}$ term in first series: ${n \choose r} x^{4r}$
the $l^{th}$ term in second series: ${n \choose l} x^{2l}$
so we have $4r+2l = 12 \implies r=0, l=6; r=1, l=4; r=2, l=2; r=3, l=0$
So we have coefficient: ${n \choose 0}{n \choose 6} +{n \choose 1}{n \choose 4} + {n \choose 2}{n \choose 2} + {n \choose 3}{n \choose 0}$
| {
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Finding integer solutions to the system $xy=6(x+y+z)$, $x^2+y^2=z^2$ How do you go about solving a system of equations like below for integer solutions?
$$xy=6(x+y+z)$$
$$x^2+y^2=z^2$$
Would you first try and list out a number of Pythagorean triples, then try and see which ones when multiplied are divisible by 6 (by 2 and 3)? Since the other side is just the sum of the 3 of them left.
I mean is there an elegant way to find these triplets, apart from a brute-force or any exhaustion method? Or even using Graphs perhaps? And why is there not infinite solutions to this? (there are actually 6 triplets).
P.S. As a bonus, replace the first equation above with:
$$x^2 + (z-510)x + 1020p = 0$$
while in the 2nd equation:
$$y=1020p/x$$
where $p$ is Prime.
Mathematica gives me only 2 solutions for $p$ for this. But how??
| This is the first way that came to my mind. Maybe there is a shorter way.
$$(x+y)^2=z^2+12z+12(x+y)$$
$$t^2-12t-(z^2+12z)=0, ~t=x+y$$
$$\Delta_{\text{half}}=36+z^2+12z=(z+6)^2$$
$$t_{1,2}=6±|z+6|=6±z+6=12±z$$
$$x+y=12±z$$
$$xy=6(x+y+z)=6(z+12±z)$$
If $x+y=12-z$, then $xy=72$.
Check all integers $x,y$ such that $xy=72$.
If $x+y=12+z$, then
$$xy=6(x+y+z)=6(12+2z)=12(z+6)$$
Finally, we have
$$(x-y)^2=(x+y)^2-4xy=(12+z)^2-48(z+6)=z^2-24z-144=(z-12)^2-288$$
$$(z-12)^2-(x-y)^2=288$$
$$(z-12-x+y)(z-12+x-y)=288=2^5×3^2$$
Then, check all possible factors of $288$:
If $m=±1,±2 \cdots ±288$, then
$$\begin{cases} z-x+y-12=\frac{288}{m} \\z+x-y=m \\x+y=12+z \end{cases}$$
| {
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Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.
I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-3)^2}=3PA+PB$$
Where $P$ is a generic point on $x^2+y^2=4$ and $A(1,0),B(0,3)$.
So the problem essentially means which point on the circle $x^2+y^2=4$ minimizes $3PA+PB$.
I am really struggling to find the geometrical notion of this and hence unable to solve.
| Nice problem! Here is a solution using geometry.
For the moment, lets forget $P$ is lying on the circle $x^2+y^2=4$. Call the origin $O=(0,0)$. $A=(1,0), B=(0,3)$. So $OA=1, OB=3$ and $\angle BOA = 90^\circ$. Draw the quadrilateral $OAPB$.
Note that we have to minimize
$$3\cdot PA+1\cdot PB=OB\cdot PA + OA\cdot PB$$
We know from Ptolemy's inequality that
$$OB\cdot PA + OA\cdot PB \ge AB \cdot OP$$
with equality only when $O,A,P,B$ lie on a circle.
Thus the desired minimum is attained when $P$ is the intersection of circle with diameter $AB$, call it $S_2$ and $S_1:x^2+y^2=4$.
The value of this minimum is $AB\cdot OP $. Since $AB=\sqrt{10}$ and $OP$ is radius of $S_1$, this minimum value is $2\sqrt{10}$.
Though coordinates of $P$ are not needed, they can be quickly found as intersection of common chord of the two circles and the circle $S_1$. Equation for common chord is $x+3y=4$ hence ordinate of $P$ satisfies $5y^2-12y+6=0$ whose only geometrically valid solution is
$$y_P=\frac{6-\sqrt{6}}{5} \Rightarrow x_P=\frac{2+3\sqrt{6}}{5}$$
which matches other answers and WA.
| {
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Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
| With the equation
$$x^3 + y^3 = x^2 + y^2 + 42xy \tag{1}\label{eq1A}$$
you've already handled the cases of $x = 0$ or $y = 0$. For the non-zero cases, have
$$\gcd(x,y) = d, \; x = de, \; y = df, \; \gcd(e, f) = 1 \tag{2}\label{eq2A}$$
Then \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned}
(de)^3 + (df)^3 & = (de)^2 + (df)^2 + 42(de)(df) \\
d^3(e^3 + f^3) & = d^2(e^2 + f^2 + 42ef) \\
d(e + f)(e^2 - ef + f^2) & = e^2 + f^2 + 42ef
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This shows that $e^2 - ef + f^2$ must divide the right side, so using this and $e^2 + f^2 \equiv ef \pmod{e^2 - ef + f^2}$, gives
$$e^2 + f^2 + 42ef \equiv ef + 42ef \equiv 43ef \pmod{e^2 - ef + f^2} \tag{4}\label{eq4A}$$
Due to $\gcd(e,f) = 1 \implies \gcd(ef, e^2 - ef + f^2) = 1$, then
$$e^2 - ef + f^2 \mid 43 \tag{5}\label{eq5A}$$
Since $d$ is the $\gcd$, it'll be positive. If both $e$ and $f$ are negative, then there's no solution (since the left side of \eqref{eq1A} will be negative & the right side would be positive). Thus, either $e$ and $f$ are both positive or one is negative with the other being positive.
Either case gives $e^2 - ef + f^2 \gt 0$ so, from \eqref{eq5A}, this means $e^2 - ef + f^2 = 1$ or $e^2 - ef + f^2 = 43$. For it being equal to $1$, we must have $e$ and $f$ being positive and, since $e^2 - 2ef + f^2 = (e - f)^2 = 1 - ef \ge 0 \implies ef = 1$, then $e = f = 1$ is the only possible solution. This gives in \eqref{eq3A}
$$d(1 + 1)(1 - 1 + 1) = 1 + 1 + 42 \implies d = 22 \tag{6}\label{eq6A}$$
This shows $(x, y) = (22, 22)$ is a solution, as wimi's question comment states. Next, consider the other case of
$$e^2 - ef + f^2 = 43 \implies e^2 - (f)e + (f^2 - 43) = 0 \tag{7}\label{eq7A}$$
Treating $f$ as being constant, then solving for $e$ using the quadratic formula, gives a discriminant of $f^2 - 4(f^2 - 43) = 172 - 3f^2$, which must be $\ge 0$ so $\lvert f \rvert \le 7$, and is a perfect square. Trying the various integer possibilities shows that $f \in \{\pm 1, \pm 6, \pm 7\}$. Using these values to check $e = \frac{f \pm \sqrt{172 - 3f^2}}{2}$, then $d$ from \eqref{eq3A} and $(x,y)$ from \eqref{eq2A} (which I'll leave to you to do), results in the remaining non-zero solutions of $(x, y)$ are $\{(-6, 1)$, $(1, -6)$, $(1, 7)$, $(7, 1)\}$, as Christian Blatter's question comment indicates.
| {
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When we express $\sin x - \cos x$ as $A \sin (x+c)$, how many solutions are there for $c \in [0, 2\pi)$? This is a problem from problem set 1 of MIT OCW 18.01SC:
express $\sin x - \cos x$ in the form $A \sin (x+c)$.
Their solution is $\sqrt{2} \sin (x - \frac{\pi}{4})$
I found two solutions (for $c \in [0, 2\pi]$):
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
Are there two solutions or is there something amiss with the second solution above?
Here is how I got the solutions:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be written in the form $A \sin(x+c)$.
We have two equations in two unknowns (A and c) and so we can solve for these variables. Square both sides of each equation:
$$A^2 \cos^2 c = 1$$
$$A^2 \sin^2 c = 1$$
Sum the two equations
$$A^2(\cos^2 x + \sin^2 x) = 2$$
$$A^2 = 2 \Rightarrow A = \pm \sqrt{2}$$
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
$$\cos c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
$$\sin c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{5\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{7\pi}{4}$, which is the same as $-\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
$$\cos c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{3\pi}{4} \text{ or } c = \frac{5 \pi}{4}$$
$$\sin c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{3 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{3\pi}{4}$.
| We can use sum to product formula to confirm your solution:
\begin{align}\sin x - \cos x = \sin x - \sin (\frac{\pi}{2}- x)=2\cos \frac{\pi}{4}\sin(x-\frac{\pi}{4})\end{align}\begin{align}=\sqrt 2\sin(x-\frac{\pi}{4})=-\sqrt 2\sin(\frac{\pi}{4}-x)\\=-\sqrt 2\sin(\pi-(\frac{\pi}{4}-x))=-\sqrt 2\sin(x+\frac{3\pi}{4})\end{align}
So, you are correct if $c \in[0,2\pi]$, the textbook is correct if $A>0$.
| {
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Show that $\lim \sqrt{1-\frac{1}{n}} = 1$ by definition Is my attempt correct?
Proof. Let $\varepsilon > 0$. Take $N > \frac{1}{\varepsilon}$ and let $n \geq N$. Then
$\begin{align}\displaystyle\left\lvert \sqrt{1- \frac{1}{n}} -1 \right\rvert
&= \left\lvert \frac{\sqrt{n-1}}{\sqrt{n}} -1 \right\rvert\\\\
&= \left\lvert \frac{\sqrt{n-1} - \sqrt{n}}{\sqrt{n}} \right\rvert\\\\
&= \left\lvert \frac{(\sqrt{n-1} - \sqrt{n})(\sqrt{n-1} + \sqrt{n})}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{(n-1)-n}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{-1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \frac{1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})}\\\\
& < \frac{1}{\sqrt{n}(\sqrt{n})}= \frac{1}{n} \leq \frac{1}{N} < \varepsilon\end{align}$
| One way to shorten this proof would be to use this lemma which says that for $x,y\in[0,\infty)$, we have $$\left|\sqrt{x}-\sqrt{y}\right|\leq\sqrt{\big||x|-|y|\big|}\leq\sqrt{|x-y|}$$
Then your proof would go $$\left|\sqrt{1-\frac{1}{n}}-\sqrt{1}\right|\leq\sqrt{\left|1-\frac{1}{n}-1\right|}=\frac{1}{\sqrt{n}}\leq\frac{1}{\sqrt{N}}<\varepsilon$$
for $1/N<\varepsilon^2$ and $n\geq N$.
Don't forget to mention the Archimedean Property, but otherwise, nice job :)
| {
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How to find $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$? For numbers $a,b,c$ we know that $\frac{a-c}{b+c}+\frac{b-a}{c+a}+\frac{c-b}{a+b}=1$ What is the value of $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$ ?
$1)3\qquad\qquad2)4\qquad\qquad3)6\qquad\qquad4)2$
Because it is a multiple choice question at first I tried to find $a,b,c$ so that each fraction be equal to $\frac13$ and their sum be $1$. but unfortunately I couldn't find such numbers.
I think adding the fractions by finding common denominator is not a good idea because the numerator will be so ugly. but I can rewrite it as:
$$\frac{a}{b+c}-\frac c{b+c}+\frac b{c+a}-\frac{a}{c+a}+\frac{c}{a+b}-\frac{b}{a+b}=1$$
But I can't continue from here.
| $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} = \frac{a+b+c}{b+c} - \frac{c}{b+c} +\frac{b+c+a}{c+a} -\frac{a}{c+a}+\frac{c+a+b}{a+b} - \frac{b}{a+b}$
$= \frac{a}{b+c} + 1 - \frac{c}{b+c}+ \frac{b}{c+a} + 1 -\frac{a}{c+a} + \frac{c}{a+b} + 1 - \frac{b}{a+b}$
$= \frac{a-c}{b+c}+1 + \frac{b-a}{c+a}+1 + \frac{c-b}{a+b} +1$
$ = 4$
| {
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Solving multivariable 4 grade equation Find the integer solutions of the following equation:
$$8x^4+y^4=6x^2y^2$$
I tried to reconfigure the equation using the notable identity $(a-b)^2=a^2+b^2-2ab$, but due to the numeric coeficients I was not able
| You can factor
$8x^4-6x^2y^2+y^4$ as
$9x^4-6x^2y^2+y^4-x^4=(3x^2-y^2)^2-(x^2)^2=(4x^2-y^2)(2x^2-y^2)$
So the equation becomes
$(4x^2-y^2)(2x^2-y^2)=0$
At least one of these two factors has to be equal to 0
But $2x^2=y^2$ is impossible, because doubling a square changes the exponent of $2$ present in the prime factorization of the square from even to odd, which makes it a non-square integer. So:
$4x^2=y^2$
$y=\pm 2x$
The solutions are in the form $(k;\pm2k)$
| {
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Is $5^2-3^2=16$ the only example for $(P_n)^2 - (P_{n-1})^2 \bmod 6 ≠ 0$ Is $5^2-3^2=16$ the only example for $(P_n)^2 - (P_{n-1})^2 \bmod 6 ≠ 0$ where $P$ is an element of the consecutive list of prime numbers (without $2$): ${3,5,7,11,13,17...P_n}$
As a web developer and a self math learner, out of curiosity I have created a script to find the difference between a prime number its previous prime number on a consecutive list.
From the first $100000$ primes, Starting with $5^2-3^2=16$, only $5^2-3^2=16$ was not divisible by $6$.
Is it proven to be the case (Yes being divisible by $6$) for all the remaining differences, and if so what makes only $5^2-3^2=16$ act different?
The only thing that stands out for me is that $5$ is the only prime ending in $5$, I wish I could post my own ideas for what I think might be the cause, but I really don't have a clue, so any help is appreciated.
Attached you can see a screen shot of the first results:
| Yes. First, we claim that every prime $>3$ is congruent to $\pm 1 \bmod{6}$. This is easy to see since a prime bigger than $3$ cannot be congruent to $0,2,3,4 \bmod{6}$ since it would then be divisible by either $2$ or $3$, which is impossible.
Next, assume that $p_{n-1}>3$. Then,
$$p_{n-1} \equiv \pm 1 \pmod{6} \implies p_{n-1}^2 \equiv 1 \pmod{6}$$
Similarly, since $p_n>p_{n-1}>3$, we have $p_n^2 \equiv 1 \pmod{6}$. This shows that $p_n^2-p_{n-1}^2 \equiv 0 \pmod{6}$. Now, the only possibility where this does not hold where $p_n$ and $p_{n-1}$ are odd primes, is when $p_{n-1}=3$ and hence, $5^2-3^2=16$ is the only exception to the observation.
| {
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Are the Sylow $p$-subgroups of $S_4$ also Sylow $p$-subgroups of $S_5$? Here is my thinking:
$|S_4| = 4! = 1 \times 2 \times 3 \times 4 = 2^3 \times 3$.
$|S_5| = 5! = 1 \times 2 \times 3 \times 4 \times 5 = 2^3 \times 3 \times 5$.
Since $2^3$ is the maximal power of $2$ which divides the order of $S_4$ and $S_5$ and since $S_4$ is a subgroup of $S_5$, we know that the Sylow $2$-subgroups of $S_4$ are also Sylow $2$-subgroups of $S_5$.
Similarly, since $3$ is the maximal power of $3$ which divides the order of $S_4$ and $S_5$ and since $S_4$ is a subgroup of $S_5$, we know that the Sylow $3$-subgroups of $S_4$ are also Sylow $3$-subgroups of $S_5$.
On the other hand, if we wanted to check whether the Sylow $p$-subgroups of $S_4$ are also Sylow $p$-subgroups of $S_6$ we would see the following.
$|S_4| = 4! = 1 \times 2 \times 3 \times 4 = 2^3 \times 3$.
$|S_6| = 6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 2^3 \times 3 \times 6 = 2^4 \times 3^2$.
Since $6$ can be decomposed into $2 \times 3$, the maximal powers of $2$ and $3$ are no longer the same in $S_4$ and $S_6$ and so the Sylow $p$-subgroups of $S_4$ are not also $p$-subgroups of $S_6$.
Is this correct?
| Hint: Embed $S_4\hookrightarrow S_5$ and then compose with the inclusion $P\hookrightarrow S_4$. Once you've a subgroup of the right order, it's a Sylow subgroup!
| {
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Prove $\sum_{n=1}^{\infty}(\sqrt{n^2+3}-\sqrt{n^2-1})=\infty$ I want to prove $$\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty$$
If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\neq0$, I can prove it.
In fact, however, $$\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=0$$
As a solution to finding a sum of series, I only know one that uses a partial sum or a progression of differences. How can I apply it to this proof?
I would appreciate it if you could tell me the solution.
| $\sqrt{n^2+3}-\sqrt{n^2-1}=$
$=\dfrac{\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\left(\sqrt{n^2+3}+\sqrt{n^2-1}\right)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=$
$=\dfrac{n^2+3-(n^2-1)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=\dfrac4{\sqrt{n^2+3}+\sqrt{n^2-1}}>$
$>\dfrac4{\sqrt{n^2+3n^2}+\sqrt{n^2}}=\dfrac4{\sqrt{4n^2}+\sqrt{n^2}}=$
$=\dfrac4{3n}>\dfrac1n\;,\quad$ for all $\;n\in\mathbb{N}\;.$
Hence,
$\sqrt{n^2+3}-\sqrt{n^2-1}>\dfrac1n\;,\quad$ for all $\;n\in\mathbb{N}\;.$
Since $\;\displaystyle\sum_{n=1}^\infty\dfrac1n=+\infty\;,\;$ by using the comparison test, we get that $\;\displaystyle\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=+\infty\;.$
| {
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Converting equation into short Weierstrass form I have the curve $y^2 + 4y = x^3 + 3x^2 −x + 1$. I need to find a transformation of the form $X=x+a$ and $Y=y+b$
that turns this curve into the standard form of the elliptic curve: $$Y^2=X^3+AX+B.$$
I completed the square for the left side and now have $(y+2)^2= x^3 + 3x^2 −x + 5$, but I'm not sure what to do next to fix the right side. Any help would be great!
| Hint: Observe that
\begin{align}
&& y^2+4y&=(y+2)^2-4 \\
&\text{and}& \\
&&\qquad x^3+3x^2&=(x+1)^3-3x-1
\end{align}
Can you continue?
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Determine the largest and smallest of the rational numbers $x$, $y$, $z$ if $x^2+4y^2+z^2-4xy-4yz+2xz-2x+2z+2=0$
Determine what number out of three rational numbers $x$, $y$, $z$ is the biggest and smallest one if following equality is true:
$$x^2+4y^2+z^2-4xy-4yz+2xz-2x+2z+2=0$$
I think it may be somewhat connected with full squares extraction
$$(x-2y)^2+(z+1)^2-2z(2y-x)-2x+1=0$$
Or
$$(z-2y)^2+(x-1)^2-2x(2y-z)+2z+1=0$$
Or
$$(x+z)^2-4y(y-x-z)-2(x-z-1)=0$$
But i am still not quite sure, what to do next
| Let's start by rewriting the equality as
$$
(x+z-2y)^2 + 2 = 2(x-z)
$$
Now complete the square on the LHS first by adding $2(x+z-2y)$ to both sides:
$$
(x+z-2y+1)^2 + 1 = 2(x-z) + 2(x+z-2y) = 4(x-y)
$$
And again complete the square by subtracting $2(x+z-2y)$:
$$
(x+z-2y-1)^2 + 1 = 2(x-z) - 2(x+z-2y) = 4(y-z)
$$
Since LHS is positive in all 3 equations, you can now determine the ordering between the 3 variables: $x > y >z$
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Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\dfrac{x-13}{x-14}-\dfrac{x-15}{x-16}=-\dfrac{1}{12}.$$
For $x\ne14$ and $x\ne 16$ by multiplying the whole equation by $$12(x-14)(x-16)$$ we get: $$12(x-16)(x-13)-12(x-14)(x-15)=-(x-14)(x-16).$$ This doesn't look very nice. Can we do something else at the beginning? $$x-14=(x-13)-1\\x-15=(x-14)-1\\..?$$
| It's easier if you subtract $1$ from each term on the LHS, i.e. $$-\frac{1}{12}=\frac{1}{x-14}-\frac{1}{x-16}=\frac{-2}{(x-14)(x-16)},$$ which is equivalent to $$(x-14)(x-16)=24.$$ This can be rearranged to $(x-10)(x-20)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \right) \right) \, dx$ How can the following improper integral be evaluated?
$$\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \right) \right) \, dx$$
or alternatively:
$$\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx$$
Note: I am only really interested in the imaginary component of the second integral.
I've attempted multiple methods, all of which seeming unsuccessful, however, I believe contour integration may be the solution to the second integral above, which would also easily allow me to get the integral I'm interested in.
| Partial solution
We can present the second integral as a series (imaginary part of the integral forms a series)
$$I= \int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx=\int_{0}^{\infty} \frac{(x \text{coth} (x) - 1)(2\pi-ix)}{x^2 ((2 \pi)^2 + x^2)} \, dx$$
$$x\coth x=1+2\sum_{k=1}^{\infty}\frac{x^2}{\pi^2k^2+x^2}$$
$$I=2\int_{0}^{\infty}dx\frac{2\pi-ix}{(2\pi)^2+x^2}\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2+x^2}=\Re I+i\Im I=I_1+I_2$$
$$I_1=2\pi\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (4 \pi^2 + x^2)}dx=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\frac{1}{\pi^2k^2+x^2}dx$$$$=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{dx}{\pi^2k^2-4\pi^2}$$
$$=4\pi\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2-4\pi^2}\Bigl(\frac{1}{2\pi}-\frac{1}{\pi k}\Bigr)\frac{\pi}{2}=2\pi^2\sum_{k=1}^{\infty}\frac{1}{2\pi^2k(2\pi+\pi k)}=\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$$
$$I_1=\frac{1}{2\pi}\sum_{k=1}^{\infty}\biggl(\frac{1}{k}-\frac{1}{k+2}\biggr)=\frac{3}{4\pi}$$
$$I_2=-2i\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\,\frac{x}{\pi^2k^2+x^2}dx$$
For $k\neq2$
$$I_2=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{xdx}{\pi^2k^2-4\pi^2}$$
$$=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{d(x^2)}{(2\pi)^2(1+\frac{x^2}{(2\pi)^2})}-\frac{d(x^2)}{(\pi k)^2(1+\frac{x^2}{(\pi k)^2})}\Bigr)\frac{1}{\pi^2k^2-4\pi^2}$$ $$=-\frac{4i}{\pi^2}\sum_{k\neq2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$
For $k=2$
$$I_2=-2i\int_{0}^{\infty}\frac{xdx}{((2\pi)^2+x^2)^2}=-i\int_{0}^{\infty}\frac{dt}{((2\pi)^2+t)^2}=-\frac{i}{4\pi^2}$$
$$I=I_1+I_2=\frac{3}{4\pi}-\frac{4i}{3\pi^2}\log2-\frac{i}{4\pi^2}-\frac{4i}{\pi^2}\sum_{k>2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the equation of the locus of the mid point of AB as m varies I am working through a pure maths book as a hobby. This question puzzles me.
The line y=mx intersects the curve $y=x^2-1$ at the points A and B. Find the equation of the locus of the mid point of AB as m varies.
I have said at intersection:
$mx = x^2-1 \implies x^2 - mx - 1=0$
Completing the square:
$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$
$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$
$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$
x-coordinates for points of intersection are
$x= \frac{-\sqrt(4+m^2) +m}{2}$ and $x= \frac{\sqrt(4+m^2)+m}{2}$
$\implies$ x-co-ordinate of P is mid-way between the two above points, namely $\frac{2m}{2} = m$
So $x=m, y = mx = m^2\implies y = x^2$
But my book says $y=2x^2$
| If $x_1$ and $x_2$ are x-coordinates of intersection points, x-coordinate of midpoint,
$x_m = \frac{x_1 + x_2}{2} = \frac{m}{2}$
$y = mx \implies y_m = 2 x_m^2$ so locus of midpoint is $y = 2 x^2$.
Your approach is correct but note that you can also get to it quickly using Vieta's formula for quadratic equation.
If $ax^2+bx+c = 0, x_1 + x_2 = - \frac{b}{a}, \ x_1 \ x_2 = \frac{c}{a}$
(where $x_1$ and $x_2$ are roots of the quadratic)
Here we have, $x^2 - mx - 1=0 \implies x_1 + x_2 = m \ $ so $x_m = \frac{m}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to show a sequence of functions is increasing How does one show that for all $x\geq 0$ that the following sequence of functions is increasing where $f_n$ is defined by
$$f_n(x)= x\left(1+\frac{x^2}{n}\right)^n$$
using the fact that for all $y\geq 0$
$$\frac{y}{y+1}\leq \ln(1+y) \leq y$$
I have been trying different ways including trying to show $\frac{f_{n+1}}{f_n}$ is positive but I’m obviously missing the trick.
| Proof only using AM$-$GM: if $x\geq 0$ then proving $f_{n+1}(x)\geq f_{n}(x)$ is equivalent to prove $$\left(1+\frac{x^2}{n}\right)^{n}$$
is an incresing sequence. Using AM$-$GM inequality we have that
\begin{align*}
\left(1+\frac{x^2}{n}\right)^{\frac{n}{n+1}}&=\sqrt[n+1]{1\cdot\underbrace{\left(1+\frac{x^2}{n}\right)\cdot\left(1+\frac{x^2}{n}\right)\cdots\left(1+\frac{x^2}{n}\right)}_{n~\text{times}}}\\
\\
&\leq\frac{1+\overbrace{\left(1+\frac{x^2}{n}\right)+\left(1+\frac{x^2}{n}\right)+\cdots+\left(1+\frac{x^2}{n}\right)}^{n~\text{times}}}{n+1}\\
&=1+\frac{x^2}{n+1}
\end{align*}
and raising both sides to $n+1$ gives
$$\left(1+\frac{x^2}{n}\right)^{n}\leq\left(1+\frac{x^2}{n+1}\right)^{n+1}$$
In fact we have the strict inequality, since $1\neq 1+\frac{x^2}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of a Triangle in $\mathbb{R}^4$ I find this question quite tricky, and I don't know if this kind of treatment is right.
I was asked about the area of a triangle formed by the points: $A:(1,2,-3,3)$; $B:(3,-6,-4,2)$; and $C:(-3,-16,-4,0)$. The only way I could make a reason out of this would be getting all the $2\times2$ matrices within the matrix M whose columns are AB and AC, getting their determinants, and finally dividing it by two. Is this procedure right? I will write the processes below:
Firstly, get vectors AB and AC:
$AB = (2,-8,-1,-1)$
$AC = (-4,-18,-1,-3)$
Secondly, construct matrix M:
\begin{equation}
M_{4,2}=
\begin{pmatrix}
2 & -4 \\
-8 & -18 \\
-1 & -1 \\
-1 & -3
\end{pmatrix}
\end{equation}
Thirdly, get all the $2\times2$ determinants within M:
\begin{equation}
S_{2}=
\begin{vmatrix}
2 & -4 \\
-8 & -18
\end{vmatrix}
+
\begin{vmatrix}
2 & -4 \\
-1 & -1
\end{vmatrix}
+
\begin{vmatrix}
2 & -4 \\
-1 & -3
\end{vmatrix}
+
\begin{vmatrix}
-8 & -18 \\
-1 & -1
\end{vmatrix}
+
\begin{vmatrix}
-8 & -18 \\
-1 & -3
\end{vmatrix}
+
\begin{vmatrix}
-1 & -1 \\
-1 & -3
\end{vmatrix}
\end{equation}
Forthly, dividing by two, and getting the absolute value (Area=$|\frac{S_{2}}{2}|$):
Area$=43$
| As @Unit commented, Herons formula might be an easier bet, we have:
$$\overrightarrow{AB}=\begin{pmatrix}3-1\\-6-2\\-4+3\\2-3\end{pmatrix}=\begin{pmatrix}2\\-8\\-1\\-1\end{pmatrix}\Rightarrow\left|\overrightarrow{AB}\right|=\sqrt{2^2+8^2+1^2+1^2}=\sqrt{70}$$
$$\overrightarrow{BC}=\begin{pmatrix}-6\\-10\\0\\-2\end{pmatrix}\Rightarrow\left|\overrightarrow{BC}\right|=\sqrt{6^2+10^2+2^2}=\sqrt{140}$$
$$\overrightarrow{AC}=\begin{pmatrix}-4\\-18\\-1\\-3\end{pmatrix}\Rightarrow\left|\overrightarrow{AC}\right|=\sqrt{4^2+18^2+1^2+3^2}=\sqrt{350}$$
We can say that the Area of the triangle is:
$$A_{abc}=\sqrt{S\left(S-|\overrightarrow{AC}|\right)\left(S-|\overrightarrow{BC}|\right)\left(S-|\overrightarrow{AB}|\right)}$$
where:
$$S=\frac{|\overrightarrow{AC}|+|\overrightarrow{AB}|+|\overrightarrow{BC}|}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve over the positive integers: $7^x+18=19^y.$ Solve over the positive integers:
$$7^x+18=19^y.$$
Progress:-
I first took $\mod 7,$ so we get $5^y\equiv 4 \mod 7$ since $5$ is a primitive root of $7$ and $5^2\equiv 4\mod 7.$ So we get $y\equiv 2\mod 6.$
And then took $\mod 9$
So we get $7^x\equiv 1\mod 9.$ Since residues of $7^x$ are $\{7,9,1\}.$
We get $x\equiv 0\mod 3.$
Then I couldn't get any progress, I tried Zsigmondy,etc stuff and also noticed $7^x-1=19(19^{y-1}-1)$
Any hints? Thanks in advance.
| Excluding the obvious solution $(x, y) = (0, 1)$, an argument mod $7$ shows that $y$ is even (which you have done). Similarly, an argument mod $19$ shows that $x$ is a multiple of $3$.
Thus we may look at the elliptic curve $Y^2 = X^3 + 18$. A computer algebra system such as Sage can be used to find all integral points: $(X, Y) = (7, \pm 19)$.
Therefore the only remaining solution is $(x, y) = (3, 2)$.
There might be elementary solutions, but this is the most efficient way of solving it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Circle passing through two points and tangent to a line
"Find the equation of the circle passing through the origin $(0,0)$, the point $(1,0)$ and tangent to the line $x-2y+1=0$."
What I have done:
The equation of a circle with radius $R$ and center $(x_0,y_0)$ is $(x-x_0)^2+(y-y_0)^2=R^2$. Since the circle passes through $(0,0)$ and $(1,0)$ it must be $$\tag{1} x_0^2+y_0^2=R^2$$ $$ \tag{2} (1-x_0)^2+y_0^2=R^2$$ and since it is tangent to the line $x-2y+1=0$ the distance from the center to this line must equal the radius $R$ hence (using the fact that the distance $d$ from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is $d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$) we have $$R=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}=\frac{|1x_0-2y_0+1|}{\sqrt{1^2+(-2)^2}}=\frac{|x_0-2y_0+1|}{\sqrt{5}} \tag{3}$$
From (1) and (2) we have
$$x_0^2 + y_0^2 = (1 - x_0)^2 + y_0^2 \Rightarrow x_0 = \dfrac{1}{2} \tag{4}$$
Equating (1) , (3) and (4) we have
$$\dfrac{1}{4}+y_0^2 = \dfrac{(\frac{3}{2}-2y_0)^2}{5} \Rightarrow y_0 = -3 \pm \sqrt{10}$$
So there are two solutions,
$$R^2 = \dfrac{1}{4}+(-3+\sqrt{10})^2 \Rightarrow (x - \dfrac{1}{4})^2+(y + 3 - \sqrt{10})^2 = \dfrac{1}{4} + (-3 + \sqrt{10})^2$$
and
$$R^2 = \frac{1}{4} + (-3 - \sqrt{10})^2 \Rightarrow (x - \frac{1}{4})^2 + (y + 3 + \sqrt{10})^2 = \dfrac{1}{4} + (-3 - \sqrt{10})^2$$.
Now, when I plot these solutions they appear to be wrong, so I would like to know which mistake(s) I have made.
| There is a very neat way of doing it if you are familiar with the idea of family of circles.
Equation of family of circles passing through $O(0, 0)$ and $A(1, 0)$ is given by
$$(x-0)(x-1)+(y-0)(y-0)+ky=0$$
where k is the parameter.
On simplification we get
$x^2+y^2-x+ky=0$
So the centre of above circle is $\left(\frac{1}{2},-\frac{k}{2}\right)$ and radius is $\frac{\sqrt{1+k^2}}{2}$
Since the line $x-2y+1=0$ is tangent to above circle so perpendicular dropped from centre to the line must be equal to the radius of the circle
$$\frac{\sqrt{1+k^2}}{2}=\frac{|1-2(-\frac{k}{2})+1|}{\sqrt{1^2+2^2}}$$
Solving for k we get $k=8\pm5\sqrt3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n = \frac{1}{e}$? I tried to solve this limit: $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n$.
My approach was to re-write it as $\lim\limits_{n \to +\infty} \bigg(\dfrac{n}{n+2} + \dfrac{1}{n+2}\bigg)^n$, and since $\dfrac{n}{n+2}$ tends to 1 and $\dfrac{1}{n+2} \sim \dfrac{1}{n}$ as $n \to +\infty$, I figured the solution would be $e$, as $\lim\limits_{n \to +\infty} \bigg(1+\dfrac{1}{n}\bigg)^n = e$.
I suppose I've done something wrong, since by plotting the function I noticed the solution is $\dfrac{1}{e}$.
Where is my error?
| You must know that $\left(1+\frac{1}{n}\right)^n \xrightarrow{n\rightarrow \infty} e$. So, we are going to play wiht it.
$\left(\frac{n+1}{n+2}\right)^n=\left[\left(\frac{n+2}{n+1}\right)^n\right]^{-1}= \left[\left(1+\frac{1}{n+1}\right)^n\right]^{-1}=\left[\left(1+\frac{1}{n+1}\right)^{(n+1)}\cdot\left(1+\frac{1}{n+1}\right)^{(-1)}\right]^{-1}=\left[\left(1+\frac{1}{n+1}\right)^{(n+1)}\right]^{-1}\cdot\left(1+\frac{1}{n+1}\right)$.
Now changing $m=n+1$ we have:
$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)$.
Now taking limits when $n$ tends to $+\infty$, as $m=n+1$, is the same as taking limits when $m$ tends to $+\infty$. Then, as $\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty}1$:
$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty} \frac{1}{e}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding sum of the roots of $4(x-\sqrt x)^2-7x+7\sqrt x=2$
Find sum of the roots of $$4(x-\sqrt x)^2-7x+7\sqrt x=2$$
By substituting $t=x-\sqrt x$ we have $4t^2-7t-2=0$
$$4t^2-8t+t-2=0$$
$$(4t+1)(t-2)=0$$
So we get $x-\sqrt x=2$ Hence $x=4$.
Or $x-\sqrt x=-\frac14$ then $x-\sqrt x+\frac14=0$ and $(\sqrt x-\frac12)^2=0$ and $x=\frac14$
So sum of the roots is $4+\frac14=\frac{17}{4}\quad$ Or $\quad 4+\frac14+\frac14=\frac92$ (adding $\frac14$ twice)?
| We are asked to find the sum of roots of this function:
$$4(x-\sqrt{x})^2-7x+7\sqrt{x}=2 \Leftrightarrow \\ 4(x-\sqrt{x})^2-7x+7\sqrt{x}-2=0$$
$\text{As you are saying by substituting } t=x-\sqrt{x} \text{ (1)}\text{ ,we have:}$
$$4t^2-7t-2=0 \Leftrightarrow (4t+1)\cdot(t-2)=0$$
$\text{And the solution we get from the previous equation are:}$
$$\boxed{t=-\frac 1 4 \text{ or } t=2 \text{ (2)}}$$
$\text{And now by replacing the varialbe } t \text{ we have that:} $
$$x-\sqrt{x}=-\frac 1 4 \text{ (3) or }x-\sqrt{x}=2 \text{ (4)}$$
$\text{Let } u=\sqrt{x} \Rightarrow u^2=x ,u \ge 0 $
$\text{We demand } u\ge 0 \text{ because the function:} f(x)=\sqrt{x} \text{ with a domain: }$ $D=[0,+\infty) \text{ it has a range: } R_f=[0,+\infty)$
$\text{For the equation (3) we have:}$
$$u^2-u-2=0 \Rightarrow u_{1}=2 \text{ or } u_2=-1$$
$\text{The second root we found we have to reject it because we demand } u\ge0 $
$\text{So we can accept only } u_1=2$
$\text{For the equation (4) we have:}$
$$u^2-u+\frac 1 4=0, \Delta=0 \Rightarrow u_3=\frac {b \pm \sqrt{\Delta}} a= \frac 1 2 $$
$\text{Now by replacing the variable u we have that the roots are:}$
$$\boxed{x_1=4 \text{ or } x_2=\frac 1 4}$$
$\text{And now when it comes to the sum of roots we have:}$
$$\boxed{S=x_1+x_2=4+\frac 1 4=\frac {17} {4}}$$
As you can see i didn't add twice the number $\displaystyle\frac 1 4$.It may be called "doulbe root" but as you can see above the quadratic formula give us one solution when $\Delta=0$
| {
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"url": "https://math.stackexchange.com/questions/4135580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$? (Note: This question is an offshoot of this closely related one.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers likely needs no introduction.
The initial question is as is in the title:
If $p^k m^2$ is an odd perfect number with special prime $p$, then is there a constant $D$ such that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$
(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)
MY ATTEMPT
Since $\gcd(p^k,\sigma(p^k))=1$, we know that
$$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}=\frac{D(m^2)}{s(p^k)},$$
where $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
Now, let $D > 0$ be a constant such that
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}.$$
This is equivalent to
$$D > \frac{p^k(m^2 - p^k)}{\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{\sigma(p^k)\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{2p^k m^2}$$
$$= \frac{\sigma(p^k)}{2} - \frac{p^k \sigma(p^k)}{2 m^2} = \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)}.$$
Using mathlove's answer to a closely related question, we have the lower bound
$$\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375.$$
Hence, we have
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p - 1)} - \frac{p^k}{3375} = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$
Let
$$f(k) = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$
Then the first derivative
$$\frac{\partial f}{\partial k} = \frac{(3373p + 2)p^k \log(p)}{6750(p - 1)}$$
is positive for $p \geq 5$. This implies that $f$ is an increasing function of $k$.
Therefore, since $k \geq 1$, we obtain
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2}.$$
But we know that $p$ is at least $5$, since $p$ is the special prime satisfying $p \equiv 1 \pmod 4$. Therefore, we have
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2} \geq \frac{2024}{675} = 2.9985\overline{185}.$$
Here are my final questions:
Does this mean that we can take $D=3$ in the inequality
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$
If $D=3$ does not work, then what value of $D$ works?
| This is a partial answer.
If $D=3$, then $p=5$ and $k=1$ both hold.
Suppose that $D=3$.
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{3}$$
$$2 > I(m^2)=\frac{\sigma(m^2)}{m^2} > \frac{p^k \bigg(1 - \dfrac{p^k}{m^2}\bigg)}{3}$$
$$\implies 6 > p^k \bigg(1 - \frac{p^k}{m^2}\bigg) \geq p^k \bigg(1 - \frac{2}{3375}\bigg) = \frac{3373p^k}{3375}$$
since
$$\frac{\sigma(m^2)}{p^k} \geq 3375 \implies \frac{p^k}{m^2} \leq \frac{2}{3375}.$$
But the inequality
$$\frac{3373p^k}{3375} < 6$$
only holds when $p = 5$ and $k = 1$. QED
Hence, we may not take $D=3$, as it will not make the inequality hold in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that all elements of sequence $a_{n}=\frac{ \left(1+\sqrt{n^4-n^2+1}\right)^{n} + \left(1-\sqrt{n^4-n^2+1}\right)^{n}}{2^{n}}$ are integers. $\textbf{PROBLEM}$:
Prove or disprove that all elements of sequence $a_{n}=\frac{ \left(1+\sqrt{n^4-n^2+1}\right)^{n} + \left(1-\sqrt{n^4-n^2+1}\right)^{n}}{2^{n}}$ are integers.
$\textbf{MY THOUGHTS}$:
First thing to note is that $${\forall n \in \mathbb{N}~~~~~ \exists k \in \mathbb{Z}:~~~~~ \left( n^4-n^2 \right) = 4k}$$
Secondly, according to Binomial theorem, $$ a_{n}=\frac{\sum\limits_{i=0}^{\lfloor \frac n2 \rfloor} {n \choose 2i} \cdot \left(n^4-n^2+1 \right)^i}{2^{n-1}}$$
I've tried to prove statement by induction on $n$, using facts above, but didn't manage to do it.
I also thought about using the fact that $$\begin{cases} \left(1+\sqrt{n^4-n^2+1}\right) + \left(1-\sqrt{n^4-n^2+1}\right) = 2\\ \left(1+\sqrt{n^4-n^2+1}\right) \cdot \left(1-\sqrt{n^4-n^2+1}\right)= n^2 - n^4\end{cases} $$
So, we could think about $\left(1+\sqrt{n^4-n^2+1}\right)$ and $\left(1-\sqrt{n^4-n^2+1}\right)$ as roots of $x^2 -2x + n^2 -n^4$.
In that case we can try to solve equivalent problem: Prove that $ \frac{x_1^n+x_2^n}{2^n}$ is integer if $x_1$ and $x_2$ are roots of $x^2 -2x + n^2 -n^4$.
| Let $x_1$ and $x_2$ denote $\left(1 + \sqrt{n^4 - n^2 + 1} \right)$ and $\left(1 - \sqrt{n^4 - n^2 + 1} \right)$. We want to prove that $\frac{x_1^n+x_2^n}{2^n}$ is integer.
Proof can be obtained by induction on n.
Firstly, let's fix $x_1$ and $x_2$. By that I mean: choose any fixed $n = n_0$ and let $x_1 = \left(1 + \sqrt{n_0^4 - n_0^2 + 1} \right)$ and $x_2 = \left(1 - \sqrt{n_0^4 - n_0^2 + 1} \right)$. Now we can think of $x_1$ and $x_2$ as of constants.
No matter how one will chose $n_0$ above, for $n=0$ and $n=1$ statement is true.
Let it be true $\forall n \le k$.
As mentioned in comments, $\forall k \ge 2 \in \mathbb{N},~~~~~ x_1^{k+1} + x_2^{k+1} = \left(x_1+x_2\right)\left(x_1^k+x_2^k\right) - x_1x_2\left(x_1^{k-1} + x_2^{k-1}\right)$. Using facts from section "My thoughts" one can easily see that this decomposition immediately proves induction step (no matter how we choose $n_0$).
So, for any chosen $n_0$ we completed proof by induction. So, to prove original statement for any $n$ put $n_0=n$.
In this way we obtained proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5}$, then $\dfrac{a}{b} \leq \dfrac{91}{150}$ and if $\dfrac{a}{b}<\dfrac{3}{5}$, then $\dfrac{a}{b} \geq \dfrac{89}{150}$.
Hence the maximum value of $\dfrac{a}{b}$ is $\dfrac{91}{150}$. And $\dfrac{a}{b}$ is maximum when $b$ is the smallest.
So, I thought $150$ would be the smallest value of $b$.
But my friend said that the smallest value of $b$ would be $32$ as $\left\lvert \dfrac{19}{32}- \dfrac{3}{5}\right\rvert= \dfrac{1}{160}$ works actually. Though he said that he found this intuitively.
So, how to solve the problem with proper procedure?
| Once you get that $\frac{a}{b}$ is between $\frac{89}{150}$ and $\frac{91}{150}$, you may simply consider the continued fractions of these rational numbers:
$$ \frac{89}{150}=[0;1,1,2,5,1,1,2] $$
$$ \frac{91}{150}=[0;1,1,1,1,5,2,2] $$
In particular we are looking for a fraction of the form $[0;1,1,2,\alpha]$ with $\alpha > [5;1,1,2]=\frac{28}{5}$. In order to have the smallest denominator the most effective choice is just $\alpha=6$, and
$$ [0;1,1,2,6]=\frac{19}{\color{blue}{32}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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} |
Locus of point M such that its projections onto two fixed lines are always at the same distance. A point moves so that the distance between the feet of the perpendiculars drawn from it to the lines $ax^2+2hxy+by^2=0$ is a constant $c$. Prove that the equation of its locus is
$$4(x^2+y^2)(h^2-ab)=c^2(4h^2+(a-b)^2).$$
What I have done: Suppose $P$ is any point on the locus, $A$ and $B$ be the feet of the perpendiculars, and $O$ is the origin. Then I want to show that
$$OP=\frac{AB}{\sin AOB}=\frac{c}{\sin \theta}$$
where $\theta$ is the angle between the two lines.
Note that the arc $PAO$ is a semi-circle, since $\angle PAO=90$. Similarly the arc $PBO$ is a semi-circle since $\angle PBO=90$. Also $\angle APB=\angle AOB$, as it is the angle of the same arc.
Now, $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. From here I can find $\sin \theta$ and $OP=\sqrt{x^2+y^2}$. Putting all these in,
$$OP=\frac{c}{\sin \theta}$$
will give me the equation of the locus. The problem is how to show that
$$OP=\frac{AB}{\sin AOB}.$$
| Please note that $OAPB$ is cyclic quadrilateral and $OP$ is the diameter of the circle. Also both lines pass through origin.
So if $(x, y)$ is the coordinates of $P$, radius of the circle is $ \displaystyle r = \frac{\sqrt{x^2 + y^2}}{2}$.
If angle between two lines is $\theta$, the angle subtended by chord $AB$ on the center is $2\theta$. We then have,
$ \displaystyle AB = 2 r \sin \theta \implies \sin\theta = \frac{c}{ \sqrt{x^2+y^2}}$ (as $AB = c$)
Now if the equations of lines are $l_1: a_1 x + b_1 y = 0$ and $l_2: a_2 x + b_2 y = 0$
Multiplying both equations and equating to $ax^2 + 2hxy+b y^2 = 0$,
$a_1a_2 = a, \ b_1b_2 = b, \ a_1b_2 + a_2b_1 = 2h \tag1$
Also $ \displaystyle m_1 = - \frac{a_1}{b_1}, m_2 = - \frac{a_2}{b_2}$
So, $ \displaystyle \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\frac{a_1 b_2 - a_2b_1}{a_1a_2 + b_1b_2}\right| = \left|\frac{ \sqrt{(a_1 b_2 + a_2b_1)^2 - 4 a_1 a_2 b_1 b_2}}{a_1a_2 + b_1b_2}\right|$
With help of equation $(1)$, $ \displaystyle \tan \theta = \left|\frac{ \sqrt{4h^2 - 4 ab}}{a + b}\right|$
Using $ \displaystyle \sin^2 \theta = \frac{\tan^2\theta}{1+\tan^2\theta} \ $, $ \ \displaystyle \frac{c^2}{x^2+y^2} = \frac{4h^2 - 4ab}{4h^2 + (a-b)^2}$
EDIT: seeing Jean Marie's nice diagrams, I realize adding an example to go with my answer may help visualize the situation better. I do not have such nice tools with me so my diagram is basic Desmos :)
Before we get to the example, using the formula we derived, it is easy to see that the locus is a circle centered at the origin. Now why origin? Simply because the question uses pair of lines intersecting at the origin. In other words, if the lines intersected at a different point, that point will be the center of the circle.
Let's take pair of lines given by $x^2 - 5xy + 4y^2 = 0$. So we have $a = 1, b = 4, h = - \frac{5}{2}$.
The lines are $x - y = 0, x - 4y = 0$.
So using the formula we derived earlier, the locus of the point in this example is,
$ \displaystyle x^2 + y^2 = \frac{34 c^2}{9}$. See the Desmos diagram for $c = 1$. From any point on this circle, if we draw perpendicular to the given pair of lines, the distance between the feet of both perpendicular will be $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Sufficient criteria for specific constrained extrema I'm trying to solve the next problem using lagrange multipliers:
Find the distance from the origin to the curve $z^2=x^2+y^2$, $x-2z = 3$.
We define the Lagrangian
$$
\mathcal{L}(x,y,z,\lambda,\mu) = x^2+y^2+z^2 + \lambda(x^2+y^2-z^2)+\mu(x-2z-3)
$$
and found two solutions $(1,0,-1,-\frac{1}{3},-\frac{4}{3})$ and $(-3,0,-3,-3,-12)$. So the two candidates for extrema are $(1,0,-1)$ and $(-3,0,-3)$. Graphically it is obvious that they are minimum and maximum respectively. Now the problem arises when we try to prove it using some criteria. I tried using the Hessian of the distance square function, but it's not useful because the candidates are not critical point of the distance square. The other possible method that I'm aware at my level of math is to study the bordered hessian but one of the determinant for both of the point is zero, so the criteria is not useful. Is there some criteria (not too advanced) that could be useful for this problem?
| The surface $z^2 = x^2 + y^2$ is a cone that forms both above $z = 0$ and below it, and z-axis is the axis of cone. The plane $x-2z = 3$ is parallel to y-axis. It is easy to visualize where the maximum and minimum distance to origin occur. Here is a simple approach -
The curve is given by intersection of $z^2=x^2+y^2, x - 2z = 3$. The projection in XY-plane is,
$(x-3)^2 = 4z^2 = 4 (x^2+y^2) \implies 3x^2 + 6x + 4y^2 = 9$
So the intersection curve is, $\dfrac{(x+1)^2}{4} + \dfrac{y^2}{3} = 1 \ $ and $ \ z = \dfrac{x-3}{2}$
Parametrize the curve as $ \ r(\theta)= (-1 + 2 \cos\theta, \sqrt3 \sin\theta, -2 + \cos\theta), \ 0 \leq \theta \leq 2\pi$.
So if distance from origin to a point on the curve is $d$,
$d^2 = x^2 + y^2 + z^2 = (-1 + 2 \cos\theta)^2 + 3 \sin^2\theta + (-2 + \cos\theta)^2$
$ = 8 + 2 \cos^2\theta - 8 \cos\theta = 2 \cdot (2 - \cos\theta)^2$
So we can clearly see that $|d|$ is maximum when $\cos\theta = -1 \implies \theta = \pi$ and it is minimum when $\cos\theta = 1 \implies \theta = 0 \ (\text{ or } 2\pi)$
That leads to maximum distance of $3 \sqrt2$ and minimum distance of $\sqrt2$. You can also plug in value of $\theta$ in $r(\theta)$ to find points where maximum and minimum distance from origin occur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the equations of circles passing through $(1, -1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
Find the equations of circles passing through $(1,-1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
The point of intersection of the lines is $(\frac25,-\frac{11}5)$
If we want this point of intersection to be $(0,0)$ (because the given lines are perpendicular) then define the new $X=x-\frac25,Y=y+\frac{11}5$, where $x,y$ are the original coordinates.
So, the point $(1,-1)$ in the new system becomes $(\frac35,\frac65)$. Since it's in the first quadrant, so, the center of the circle will be $(r,r)$ where $r$ is the radius of the circle.
So, $(\frac35-r)^2+(\frac65-r)^2=r^2\implies r=\frac35,3$
So, the equations of circles are $$(X-\frac35)^2+(Y-\frac35)^2=\frac9{25}\implies(x-1)^2+(y+\frac85)^2=\frac9{25}$$
and $$(X-3)^2+(Y-3)^2=9\implies(x-\frac{17}5)^2+(y-\frac45)^2=9$$
Is this correct?
| Geometric approach.
The centers of all inscribed circles are located on a bisector line,
which equation is $y=7x-5$.
Construct auxiliary inscribed circle, say centered at $O=(2,9)$
with the radius $r=8$.
The line $XP$ intersect this circle at
points $D(2,1)$ and $E(\tfrac{42}5,\tfrac{49}5)$
and we have two scaling coefficients,
\begin{align}
k_1&=\frac{|XP|}{|XE|}
=\tfrac3{40}
,\\
\text{and }\quad
k_2&=\frac{|XP|}{|XD|}
=\tfrac38
,
\end{align}
so we can find the centers and the radii of sought circles through $P$
as
\begin{align}
O_1&=X+(O-X)\cdot k_1
=(\tfrac{13}{25},-\tfrac{34}{25})
,\\
r_1&=r k_1=\tfrac35
,\\
O_2&=X+(O-X)\cdot k_2
=(1,2)
,\\
r_2&=r k_2=3
.
\end{align}
Then the circle equations would be
\begin{align}
(x-\tfrac{13}{25})^2+(y+\tfrac{34}{25})^2&=\tfrac9{25}
\\
\text{and }\quad
(x-1)^2+(y-2)^2&=9
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Power series of a function around a point ≠ 0 Using the power series $ \sum_{k \geq 0}z^k = \frac{1}{1-z} $ for $ |z|<1 $, where it's centered around $0$. How would the series look like if someone wanted to let, for example $ z_0 = \frac{1}{2}$?
Also, how would one compute the power series of $ \frac{1}{1+z^2} $ around $z_0 = 1$?
I know that, centered around 0 we can manipulate the geometric series to obtain $ \frac{1}{1+z^2} = \sum_{k \geq 0} (-1)^k z^{2k} $. However, I want to write it on the form $ \sum_{k \geq 0} c_k (z-1)^k $.
| Just manipulate the fraction
$$
\dfrac{1}{1-z} = \dfrac{1}{1-z+1/2-1/2} = \dfrac{1}{1/2-z+1/2} = \dfrac{1}{1/2-(z-1/2)} = \dfrac{1}{\dfrac{1}{2} \Big(1 - 2(z-1/2) \Big)}.
$$
And
$$
\dfrac{1}{\dfrac{1}{2} \Big(1 - 2(z-1/2) \Big)} = 2 \sum_{k=0}^{\infty} \Big(2 \, (z - 1/2) \Big)^k = \sum_{k=0}^{\infty} 2^{k+1} \Big(z - \dfrac{1}{2}\Big)^k,
$$
with
$$
| 2 (z-1/2) | < 1, \\
|(z-1/2) | < 1/2.
$$
Since I like GEdgar's answer, I will write most of the details of his answer.
You want to manipulate something like this
$$
\dfrac{1}{1-z},
$$
not this
$$
\dfrac{1}{1+z^2}.
$$
Easy, use partial fractions
$$
\dfrac{1}{1+z^2} = \dfrac{-i}{2}\dfrac{1}{(z-i)} + \dfrac{i}{2}\dfrac{1}{(z+i)},
$$
and manipulate
$$
\dfrac{-i}{2}\dfrac{1}{(z-i)} = \dfrac{-i}{2}\dfrac{1}{(z-i + 1 -1)} = \dfrac{-i}{2(-1-i)} \dfrac{1}{ \Bigg[ 1- \dfrac{z-1}{-1-i} \Bigg]} \\
= \dfrac{-i}{2(-1-i)} \sum_{k=0}^{\infty} \Big(\dfrac{z-1}{-1-i}\Big)^k.
$$
Using the polar form of $1/(-1-i) = e^{i\, 3\pi/4}/ \sqrt{2} $
$$
\dfrac{-i}{2(-1-i)} \sum_{k=0}^{\infty} \Big(\dfrac{z-1}{-1-i}\Big)^k = \dfrac{-i}{2} \dfrac{1}{\sqrt 2} e^{i\, 3\pi/4} \sum_{k=0}^{\infty} \Bigg(\dfrac{1}{\sqrt 2} e^{i\, 3\pi/4} \Bigg)^{\!k} (z-1)^k .
$$
Doing some calculations, the other fraction is
$$
\dfrac{i}{2}\dfrac{1}{(z+i)} = \dfrac{i}{2} \dfrac{1}{\sqrt 2} e^{-i\, 3\pi/4} \sum_{k=0}^{\infty} \Bigg(\dfrac{1}{\sqrt 2} e^{-i\, 3\pi/4} \Bigg)^{\!k} (z-1)^k .
$$
Joining everything, the power series of $1/(1+z^2)$ around $z_0=1$ is
$$
\dfrac{1}{1+z^2} = \dfrac{i}{2 \sqrt 2} \sum_{k=0}^{\infty} \bigg(\dfrac{1}{\sqrt 2}\bigg)^{\!k} \bigg(\exp \Big(\!-i\, \dfrac{3\pi(k+1)}{4} \Big) - \exp \Big(i\, \dfrac{3\pi(k+1)}{4} \Big) \bigg) (z-1)^k .
$$
Simplifying both exponentials
$$
\exp \Big(\!-i\, \dfrac{3\pi(k+1)}{4} \Big) - \exp \Big(i\, \dfrac{3\pi(k+1)}{4} \Big) = -2\, i \sin \Big(\dfrac{3\pi(k+1)}{4} \Big)
$$
The result is
$$
\dfrac{1}{1+z^2} = \sum_{k=0}^{\infty} \bigg(\dfrac{1}{\sqrt 2}\bigg)^{\!k+1}\! \sin \Big(\dfrac{3\pi(k+1)}{4} \Big) (z-1)^k,
$$
with
$$
|z-1| < \dfrac{1}{\sqrt 2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ where $x$ is a real number.
Background: Doing Olympiad question and got one from the book.
Attempt:
Let $3^x$ be $u$.
\begin{align*}
3^{2x+1} + 4 \cdot 3x - 15 &= 0 \\
3^{2x} \cdot 3^1 + 4 \cdot 3^x - 15 &= 0 \\
3^{2x} \cdot 3 + 4 \cdot 3^x - 15 &= 0 \\
3 \cdot 3^{2x} + 4 \cdot 3^x - 15 &= 0 \\
3u^2 + 4u - 15 &= 0
\end{align*}
Factoring the equation we get
$$ u=\frac{5}{3} $$
(We eliminate $u=-3$ as $x$ is real.)
$$ 3^x=\frac{5}{3} $$
Taking $\log$ both sides
$$ x \log 3 = \log\left(\frac{5}{3}\right) $$
Now I want to know how to further solve this. Also is there any easier way to solve this?
| It depends. It seems as if you already have a short solution and you may consider taking base-$3$ logarithm to make a shorter one, but most of the time, logarithms are of base-$e$, which is exactly what you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Mean independent A random variable $X$ is said to be mean independent of another random variable $Y$
if its conditional expectation given $Y$ is equal to its unconditional expectation, that is, $E[X \mid Y]=E(X)$.
My question is, for $\phi$ Borel function we can say that $E[\phi(X) \mid Y]=E[\phi(X)]$ ?
Thanks.
| Let $R$ be the result of rolling a fair die.
Define the random variable $X$ by
$$
X=(R\;\text{mod}\;3)
\qquad\qquad\qquad\qquad\;\;\,
$$
and define the random variable $Y$ by
\begin{cases}
Y=0&\text{if}\;R\in\{2,3,5,6\}
\qquad\qquad\;\;\;\,
\\[4pt]
Y=1&\text{if}\;R\in\{1,4\}\\
\end{cases}
Then we get
$$
\left\lbrace
\begin{align*}
&E(X)
=
\frac{1}{3}{\,\cdot\,}0
+
\frac{1}{3}{\,\cdot\,}1
+
\frac{1}{3}{\,\cdot\,}2
=
1
\qquad
\\[4pt]
&E(X|Y=0)
=
\frac{1}{2}{\,\cdot\,}0
+
\frac{1}{2}{\,\cdot\,}2
=
1
\\[4pt]
&E(X|Y=1)
=
1{\,\cdot\,}1
=
1
\\[4pt]
\end{align*}
\right.
$$
so $X$ is mean independent of $Y$, but
$$
\left\lbrace
\begin{align*}
&E(X^2)
=
\frac{1}{3}{\,\cdot\,}0^2
+
\frac{1}{3}{\,\cdot\,}1^2
+
\frac{1}{3}{\,\cdot\,}2^2
=
\frac{5}{3}
\\[4pt]
&E(X^2|Y=0)
=
\frac{1}{2}{\,\cdot\,}0^2
+
\frac{1}{2}{\,\cdot\,}2^2
=
2
\\[4pt]
&E(X^2|Y=1)
=
1{\,\cdot\,}1^2
=
1
\\[4pt]
\end{align*}
\right.
$$
so $X^2$ is not mean independent of $Y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many five-digit numbers can be made from the digits $0, 1, 2, 3, 4$, provided that each digit can be repeated twice? The 10,000th place can be filled in $4$ different ways, the 1000th place can be filled in $5$ different ways, the 100th place can be filled in $4$ different ways, seeing the double repetition of numerical values, and the 10th and 1st places must be filled $(5-2)=3$ different ways each. Thus, the total number of possible such 5-digit numbers is $4 \cdot 5 \cdot 4 \cdot 3 \cdot 3 = 720$.
However, i found the following solution:
$$1) 1\cdot5!/1=120$$
$$2) 5\cdot5!/2=300$$
$$3) 10\cdot5!/4=300$$
$1)$ understandable. The number of all possible choices with zero in any position.
$1) 2)$ already incomprehensible. In theory, you need to calculate how many numbers there will be so that there is no zero in front. After all, such numbers are invalid. And somehow calculate, the number of options with repeating numbers $2$ times. Could you explain this to me?
| $\color{red}{HINT=}$There are three situation such that zero is used once , zero is used two times and zero is not used.
Zero is not used : Our generaitng function is $(1+ x+ \frac{x^2}{2})^4$ , then find the coefficient of $x^5$ and multiply it by $5!$
Zero is used once= Our generating function is $(1+ x+ \frac{x^2}{2})^4$ , then find the coefficient of $x^4$ and multiply it by $4!$. After that select one place for $0$ among $4$ suitable place.
Zero is used two times = Our generating function is $(1+ x+ \frac{x^2}{2})^4$ , then find the coefficient of $x^3$ and multiply it by $3!$. After that select two place for $0$ among $3$ suitable place.
$\color{blue}{OR}$,
More concisely, Lets think all of arrangement where zero may be leadind digit such that $(1+ x+ \frac{x^2}{2})^5$ , then find the coefficient of $x^5$ and multiply it by $5!$.
Now we should subtract the cases where zero is leading digit such that $(1+ x+ \frac{x^2}{2})^4 \times (1+x)$ , then find the coefficient of $x^4$ and multiply it by $4!$.
At last , subtract these two cases . Then you will find the number of $5$ digits numbers where it does not start with $0$ and each terms repeat at most $2$ times.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Algebra question to find all ordered pairs to a set of equations Find all ordered pairs of real numbers ($x,y$) for which
$$(1+x)(1+x^{2})(1+x^{4})=1+y^{7}$$
$$(1+y)(1+y^{2})(1+y^{4})=1+x^{7}$$
I am not sure on how to solve this. There is a clear symmetry in the problem. Also on expanding the LHS, we will get the GP
$$1+x+x^2+x^3+x^4+x^5+x^6+x^7=1+y^7$$ and similarly for the second equation. I tried manipulating the two equations and factorise them but I was unsuccesful. Any help would be much appreciated. Thanks a lot
| Observe that
*
*If $ x = y$, then we have the solutions $ (0, 0), (-1, -1)$.
*We will show that there are no solutions with $ x \neq y$.
*If $ x = 0$, then from the first equation $ y = 0$.
*If $ x > 0, y < 0 $ then $ 1 + x + x^2 + \ldots + x^7 > 1 > 1 + y^7 $, so no solutions. Likewise for $ x < 0, y > 0$.
*If $x, y > 0$ then notice for for real $x$, $ (1+x)(1+x^2)(1+x^4) > 1 + x^7$. Hence, we have $(1+x)(1+x^2)(1+x^4) = 1 + y^7 < (1+y)(1+y^2)(1+y^4 ) = 1 + x^7 $ which is a contradiction.
*If $ x , y < 0$, then multiplying the first equation by $ 1 - x \neq 0$ and the second by $ 1 - y \neq 0$, and subtracting the two, we get
$$ y^ 8 - x^8 = (y - x) + (y^ 7 - x^7) + xy (x^6 - y^6)$$
$\qquad$ If $x < y < 0 $, then the LHS is negative, but each term on the RHS is positive. Hence contradiction. Likewise if $ y < x < 0 $.
Note: A very common misconception when solving symmetric system of equations is thinking that all solutions to $x = f(y), y = f(x)$ are of the form $ y = x$.
In fact, what we truly require is $ f(f(x)) = x, y = f(x)$.
This is most obvious when looking at self-inverse functions like $f(x) = x, f(x) = -x, f(x) = \frac{1}{x}$. However, it's not about the function, but about the "self-inverse" point, namely $f(x) = f^{-1} (x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$$
This is my attempt.
We have
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}=\frac{a}{a+\sqrt{(a+b+c)a+bc}}+\frac{b}{b+\sqrt{(a+b+c)b+ca}}+\frac{c}{c+\sqrt{(a+b+c)c+ab}}=\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}=1-\frac{\sqrt{(a+b)(a+c)}}{a+\sqrt{(a+b)(a+c)}}+1-\frac{\sqrt{(b+c)(b+a)}}{b+\sqrt{(b+c)(b+a)}}+1-\frac{\sqrt{(c+a)(c+b)}}{c+\sqrt{(c+a)(c+b)}}$$
| We have for $y>1$ and $a\geq b$ and $a\geq c$: $$p=\left(\frac{\left(abc\right)^{y}}{a^{y}+b^{y}+c^{y}}\right)^{\frac{1}{y}}\leq bc$$
So we have with $ap=q$:
$$ f(a)+f(b)+f(c)=\frac{a}{a+\sqrt{2013a+\frac{q}{a}}}+\frac{b}{b+\sqrt{2013b+\frac{q}{b}}}+\frac{c}{c+\sqrt{2013c+\frac{q}{c}}}$$
The function :
$$g(x)=\frac{x}{x+\sqrt{2013x+\frac{a}{x}p}}$$
Is concave for $x\geq 20$ and assumptions of the OP on $a,b,c$ .
So we apply Jensen's inequality to get :
$$f(a)+f(b)+f(c)\leq f(a)+2f\left(\frac{2013-a}{2}\right)$$
Now $y\to\infty$ and we obtain
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{2013-a}{\frac{\left(2013-a\right)}{2}+\sqrt{\frac{2013\left(2013-a\right)}{2}+bc\cdot\frac{2a}{2013-a}}}\leq 1$$
This inequality is not hard.
| {
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"url": "https://math.stackexchange.com/questions/4171770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Computing $\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$ I have this limit as my question to solve:
$$\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$$
My procedure:
$$\lim_{x\to-1}\frac{(2x+\sqrt{3-x})(2x-\sqrt{3-x})}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{4x^2+x-3}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{(x+1)(x-\frac{3}{4})}{x(x+1)(2x-\sqrt{3-x})}$$
$(x+1)$ is the zero factor and has to be eliminated.
$$\frac{-1-\frac{3}{4}}{-1(2(-1)-\sqrt{3-(-1)})}=-\frac{7}{16}$$
But when I checked my answer, I found it's wrong and the right answer is $-\frac{7}{4}$.
What was my mistake here?
| It's because $\displaystyle4x^2+x-3=(x+1)(4x-3)$ rather than $\displaystyle(x+1)\left(x-\frac34\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$ While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried $x=\sin(u)$ then $\tan(u/2)=t$ and got
$$I=\int_0^{\pi/2}\cot(u)\ln(\sin u)\ln(1-\sin u)\ln(1+\cos u)\mathrm{d}u$$
$$=\int_0^1\frac{1-t^2}{t(1+t^2)}\ln\left(\frac{2t}{1+t^2}\right)\ln\left(\frac{(1-t)^2}{1+t^2}\right)\ln\left(\frac{2}{1+t^2}\right)\mathrm{d}t.$$
Any idea?
Thanks
| Bonus: Differentiate both sides of $\int_0^1 x^{2n-1}\ln(1-x)dx=-\frac{H_{2n}}{2n}$, we get
$$\int_0^1 x^{2n-1}\ln(x)\ln(1-x)dx=\frac{H_{2n}}{4n^2}+\frac{H_{2n}^{(2)}}{2n}-\frac{\zeta(2)}{2n}.$$
Multiply both sides by $\frac{2n\choose n}{4^nn}$ then sum up from $n=1$ to $\infty$, we obtain
$$\frac14\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}}{4^n n^3}+\frac12\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}-\zeta(2)\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n n^2}=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{{2n\choose n}x^{2n}}{4^n n}\right)dx$$
$$=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\left(2\ln\left(\frac{2}{1+\sqrt{1-x^2}}\right)\right)dx$$
$$=2\ln(2)\int_0^1\frac{\ln(x)\ln(1-x)}{x}dx-2\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}dx$$
$$=2\ln(2)\zeta(3)-2I.$$
The sum $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$ is calculated by Jorge above and the sum $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}}{4^n n^3}$ is calculated here. Collecting these two results along with using $\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n n^2}=\zeta(2)-2\ln^2(2)$, we get
$$I = -2 \pi \Im \left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-\frac12\text{Li}_4\left(\frac{1}{2}\right)+\frac{281 }{64}\zeta(4)+\frac{7}{16} \ln (2)\zeta (3)-\frac{5}{8} \ln ^2(2)\zeta(2)-\frac{1}{48} \ln ^4(2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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$\int_0^{\infty} \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}} dx = \frac{\pi}{2}$ and more general results? Problem: evaluate that $\int_0^{\infty} \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}} dx = \frac{\pi}{2}$ and prove if a more general case, $\int_0^{\infty} \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}} \frac{\sin \frac{x}{5}}{\frac{x}{5}} ... \frac{\sin \frac{x}{2n+1}}{\frac{x}{2n+1}} dx = \frac{\pi}{2}$ holds?
So it's a rather well known results to prove Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?
and I use a computer program to validate that the general case does appear to hold, but not sure how to prove it..
| These are Borwein Integrals and the behavior of exactly equaling $\pi/2$ stops at $2n+1 = 15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q^2)^2$
which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.
| $$\color{magenta}{a=3p^2-2pq-3q^2, \quad b=-p^2-6pq+q^2, \quad c=p^2+q^2}$$
or, same thing
$$\color{green}{a=3x^2-2xy-3y^2, \quad b=-x^2-6xy+y^2, \quad c=x^2+y^2}$$
When $p,q$ are coprime, the primes that can still divide $\gcd(a,b,c)$ are $2$ and $5.$ The proof of all (primitive, integral) solutions is just showig that these don't matter.
When $x,y$ are both odd, all three of $a,b,c$ are divisible by $2,$ and we need to worry about whether half the triple is represented by the given parametrization. Well
taking
$$ p = \frac{x-y}{2} \; , \; \; q = \frac{x+y}{2} \; , \; \; $$
$$ 3 p^2 - 2pq -3q^2 = \frac{1}{2} \left( -x^2 - 6 xy + y^2 \right) $$
$$ - p^2 - 6pq +q^2 = \frac{-1}{2} \left( 3x^2 - 2 xy -3 y^2 \right) $$
$$ p^2 +q^2 = \frac{1}{2} \left( x^2 + y^2 \right) $$
$5$ is the other possibility . This happens when $$2x+y \equiv x - 2y \equiv 0 \pmod 5.$$
Taking
$$ p = \frac{2x+y}{5} \; , \; \; q = \frac{x-2y}{5} \; , \; \; $$
$$ 3 p^2 - 2pq -3q^2 = \frac{1}{5} \left( 3x^2 - 2 xy -3 y^2 \right) $$
$$ - p^2 - 6pq +q^2 = \frac{1}{5} \left( -x^2 - 6 xy + y^2 \right) $$
$$ p^2 +q^2 = \frac{1}{5} \left( x^2 + y^2 \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
The example I like to show is solving
$$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes,"
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 + 23 uv + 37 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
32 u^2 + 61 uv + 18 v^2 \\
18 u^2 -25 uv -11 v^2 \\
-11 u^2 + 3 uv + 32 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
38 u^2 + 45 uv + 4 v^2 \\
4 u^2 -37 uv -3 v^2 \\
-3 u^2 + 31 uv + 38 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
29 u^2 + 63 uv + 22 v^2 \\
22 u^2 -19 uv -12 v^2 \\
-12 u^2 -5 uv + 29 v^2
\end{array}
\right)
$$
For all four recipes,
$$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
How to find local maxima of $\frac{x^2+2x-3}{x^2+1}$ quickly?
What is the value of relative maximum of the function
$f(x)=\dfrac{x^2+2x-3}{x^2+1}$ ?
$1)-1+\sqrt5\qquad\qquad2)1+\sqrt5\qquad\qquad3)-1+\sqrt3\qquad\qquad4)1+\sqrt3$
It is a problem from a timed exam so I'm looking for the fastest approaches. Here is my approach:
First I tried taking derivative of the function and equate it to zero, but I realized by doing that I should solve a third degree equation because the degree of numerator is $2$. So I decided to write the fraction as:
$$f(x)=\dfrac{x^2+1+2x-4}{x^2+1}=1+\frac{2x-4}{x^2+1}$$
Hence it is easier to take derivative:
$$f'(x)=0\Rightarrow f'(x)=\dfrac{2x^2+2-4x^2+8x}{x^2+1}=0$$So we have $-x^2+4x+1=0\Rightarrow x=2\pm\sqrt5$. And from here by using Wolfram Alpha I got maximum is $\sqrt5-1$ at $2+\sqrt5$ and minimum is $-1-\sqrt5$ at $2-\sqrt5$.
But this approach takes much time (specially at the end) and it is not suitable for these kinds of exams. So can you please solve this problem differently?
| $$\frac{2x-4}{x^2+1}=u$$
$$\begin{align} &\implies ux^2-2x+(u+4)=0\\
&\implies \Delta=1-u(u+4)≥0 \\
&\implies u^2+4u-1≤0 \\
&\implies (u+2)^2-5≤0 \\
&\implies |u+2|≤\sqrt 5 \\
&\implies \max \left\{u\right\}=\sqrt 5-2 \\
&\implies \max \left\{f(x)\right\}=\sqrt 5-1.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$ I found a proof that $\lim \sqrt[n]{1^2+2^2+...+n^2}=1$ by $\varepsilon$-language, but I think it's quite complicated and not sure that it's correct.
My question is:
1- Is my proof correct?
2- Is there another simpler proof in the sense of $\varepsilon$-language?
Please help me. Thanks.
Solution:
Let $a_n=\sqrt[n]{1^2+2^2+...+n^2}-1$. Then $a_n>0$ and:
$$ (a_n+1)^n= 1^2+2^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}$$
By the binomial theorem we have:
$ (a_n+1)^n = 1 + na_n+\frac{n(n-1)}{2}a_n^2+\frac{n(n-1)(n-2)}{6}a_n^3+\frac{n(n-1)(n-2)(n-3)}{24}a_n^4+\cdots+a_n^n $
Since $a_n>0$, then $(a_n+1)^n>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$ and therefore:
$$\dfrac{n(n+1)(2n+1)}{6}>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$$
This is equivalent to
$$(n+1)(2n+1)>\dfrac{(n-1)(n-2)(n-3)}{4}a_n^4 \Leftrightarrow a_n^4<\dfrac{4(n+1)(2n+1)}{(n-1)(n-2)(n-3)}$$
$$\Rightarrow a_n^4< \dfrac{8(n+1)^2}{(n-2)^2(n-3)}=\dfrac{8}{n-3}\Big(1+\dfrac{3}{n-2}\Big)^2<\dfrac{128}{n-3}.$$
Thus: $a_n<\sqrt[4]{\dfrac{128}{n-3}}$. For every $\varepsilon >0$, take $N>3+\dfrac{128}{\varepsilon^4}$, then for all $n\ge N$,
$$ n-3 \ge N -3 > \dfrac{128}{\varepsilon^4} $$
Thus: $ \varepsilon^4 > \dfrac{128}{n-3} \Rightarrow \varepsilon > \sqrt[4]{\dfrac{128}{n-3}} >a_n $.
Hence, $\lim a_n=0$ or equivalently, $\lim \sqrt[n]{1^2+2^2+...+n^2} =1$.
| Note that $\sum_{k=1}^{n} k^a \sim \frac{n^{a+1}}{a+1}$
because $$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} (k/n)^a =\int_{0}^{1} x^a dx=\frac{1}{a+1}$$
So $$L=\lim_{n \to \infty} \left [\sum_{k=1}^{n} k^2 \right]^{1/n}=\lim_{n \to \infty} (n^3/3)^{1/n}= \lim_{n\to \infty} \exp[\frac{1}{n} \ln (n^3/3)]=\lim_{n \to \infty} \exp[\frac{1}{n}(3 \ln n-\ln3)]=e^{0}=1. $$
| {
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"url": "https://math.stackexchange.com/questions/4180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:
If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
S \space$ then if $a = b\cdot c\space$ prove that $c \in S$
My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\space$ and $b = 1 + 3 \cdot y\space$ and $x = y + j\space$ for some $j >= 1$
If $a = bc \implies c \mid a \space \And \space b\mid a$
Now $b \mid a \implies (1 + 3y)\mid (1 + 3x) \implies (1 + 3y) \mid (1 + 3(y + j)) \implies (1 + 3y) | (1 + 3y + 3j) \implies (1 + 3y) | (1 + 3y) + 3j$
Since $1 + 3y \mid (1 + 3y) \implies (1 + 3y) \mid 3j$ and we keep that.
Now we know that
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = \frac{1+3(y + j)}{1 + 3y} = \frac{1 + 3y + 3j}{1 + 3y} = \frac{1 + 3y}{1 + 3y} + \frac{3j}{1+3y} = 1 + \frac{3j}{1 + 3y}$$
But we have shown that $1 + 3y \mid 3j \implies 3j = k(1 + 3y)\space$ where $1 + 3y$ is a positive integer.
Hence $$c = 1 + \frac{k(1 + 3y)}{1 + 3y} = 1 + k$$
But now I am stuck because I need to show that $k$ is a multiple of $3$ to finish the proof and I am not sure how to do that.
I know that $$k = 3\frac{j}{1 + 3y}$$ but the $\frac{j}{1 + 3y}$ is not an integer I think so I a messing up somewhere.
Update:
Following the comment of @Infinity_hunter
Let $c = r + 3k\space$ for some $r >= 1$. We have:
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = 1 + \frac{3j}{1 + 3y}$$
This was shown earlier.
But
$$c = r + 3k = 1 + \frac{3j}{1 + 3y} $$
We know that $$\frac{3j}{1 + 3y}$$ since we have shown that $1 + 3y \mid 3j$ so let $\frac{3j}{1 + 3y} = p$
So we have:
$c = r + 3k = 1 + p \implies c = r - 1 - p + 3k \implies c = (r - 1 - p) + 3k$
And I am not sure how to progress from here.
| Since $a,b\in S$ we have $a=3x+1$ and $b=3y+1$ for some integers $x,y$. So $$3x+1 = (3y+1)c$$ and thus $$c = 3\underbrace{(x-yc)}_k +1 = 3k+1 \implies c\in S$$
| {
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"url": "https://math.stackexchange.com/questions/4183499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $a$ & $b$ , s.t. $a+b=0$ and $ab=-3$? I've a problem in this question :
In the polynomial identity $x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)$ , find $ab$ $? $
MY APPROACH
We have : $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$
Now according to the Problem : $$x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)=(x^2+1)(x^4-x^2+1)$$ Or $$x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1=x^4-x^2+1$$ $$(a+b)x^3+(ab+2)x^2+(a+b)x=-x^2$$ In order to satisfy this , $a+b=0$ and $ab+2=-1$
Hence , $$ab=-3$$
but what'll be the value of $a$ & $ b$ , which'll satisfy these conditions $?$
| Since, $a+b=0$
$\implies a=-b$
And, $ab=-3$
$\implies -b^2=-3$
$\implies b=\pm\sqrt{3}$
So, $a+b=0$
$\implies a\pm\sqrt{3}=0$
$\implies a=\mp\sqrt{3}$
$\therefore (a,b)=(\pm\sqrt{3},\mp\sqrt{3})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can i solve this integral by differentiating under integral sign or by any other method $$\int_{1}^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}}$$
where $\Im(a)\neq0$ and $\Re\ge-1$
Here's how i tried to solve
$$I(a)=\int_{1}^{\infty}\frac{dx}{(x+a)\sqrt{x-1}}=\frac{\pi}{\sqrt{a+1}}$$
Differentiating wrt $a$
$$I'(a)=\int_{1}^{\infty}\frac{dx}{(x+a)^2\sqrt{x-1}}=\frac{\pi}{2(a+1)^{3/2}}$$
$$I''(a)=\int_{1}^{\infty}\frac{dx}{(x+a)^3\sqrt{x-1}}=\frac{3\pi}{8(a+1)^{5/2}}$$
$$I'''(a)=\int_{1}^{\infty}\frac{dx}{(x+a)^4\sqrt{x-1}}=\frac{5\pi}{16(a+1)^{7/2}}$$
By continuing in similar way how can i obtain the general expression for it's $n$th derivative
| Upon differentiating, you should notice that
$$(n-1)! \int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \frac{(2n-3)\cdot(2n-5) \cdots1}{2\cdot2\cdots2}\frac{\pi}{(a+1)^{n-\frac{1}{2}}}$$
where there are $(n-1) \; 2's$ in product in the denominator.
Now the RHS can be manipulated a bit by multiplying and dividing by $(2n-2) \cdot (2n-4) \cdots 2 = 2^{n-1}\cdot (n-1)!$, which gives us
$$(n-1)! \int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \frac{(2n-2)!}{2^{n-1}\cdot2^{n-1}\cdot {(n-1)}!}\frac{\pi}{(a+1)^{n-\frac{1}{2}}}$$
which finally gives us
$$\int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \binom{2(n-1)}{n-1}\frac{\pi}{4^{n-1}(a+1)^{n - \frac{1}{2}}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is my mistake in finding this pythagorean triplet? Since Project Euler copyright license requires that you attribute the problem to them, I'd like to add that this is about question 9 there.
I am trying to solve this problem on only two brain cells and can't figure out what am I doing wrong. Here is the system for $(a, b, c) \in \mathbb{N}^3$,
\begin{align*}
a^2 +b^2 &= c^2 \\
a+ b + c &= 1000 \\
a &< b < c
\end{align*}
Here is my approach,
\begin{align*}
a + b + c &= 1000 \\
a + b &= 1000 - c &&\text{Subtract } c\\
a^2 + b^2 + 2ab &= 1000^2 + c^2 - 2000c &&\text{Square both sides}\\
c^2 + 2ab &= 1000^2 + c^2 - 2000c &&\text{Since }a^2 +b^2 = c^2\\
2ab &= 1000^2 - 2000c &&\text{Subtract } c^2\\
\frac{ab}{500} &= 1000 - 2c &&\text{Divide } 1000\\
2c &= 1000 - \frac{ab}{500} &&\text{Rearrange}\\
\end{align*}
Now let $a=5,b=200$,
\begin{align*}
2c &= 1000 - 2\\
2c &= 998 \\
c &= (998 \div 2) = 499 \\
\end{align*}
But certainly these values do not work. I can't see why.
| $a = p^2 - q^2\\
b = 2pq\\
c = p^2 + q^2$
For any $(p,q:p>q), (a,b,c)$ will be a pythagorean tripple
$p^2 - q^2 + 2pq + p^2 + q^2 = 1000\\
2p^2 + 2pq = 1000\\
p(q+p) = 500$
$p$ is a factor of $500$
If $p$ is too small, $q$ is too big, and $p^2 - q^2 < 0$
And $p$ is too big, $q< 0$
$\sqrt{250}<p<\sqrt{500}$
$p = 20,$ works
Our triple is $(375,200,425)$ but we need to reorder to meet one of the criteria.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How do I have to use this assumption?
Let be $V$ a vector space over $\mathbb{R}$ with dimension 4. Let be
$T$ a linear operator in $V$. In case that a base $\mathcal{B}$ of $V$
exists, such that the associated matrix of $T$ relative to the basis
$\mathcal{B}$ is: $$ \left [ T \right ]_{\mathcal{B}}=\begin{pmatrix}
1 & a & 0 & 0\\ 0 & 1 & b & 0\\ 0 & 0 & 1 & c\\ 0 & 0 & 0 & 1
\end{pmatrix} $$ with $a^{2}+b^{2}+c^{2} \neq 0$, determine if T is
diagonalizable.
My solution:
Let be $A=\left [ T \right ]_{\mathcal{B}}$
First we compute $det(A-tI)$ to find the eigenvalues of A, that is
\begin{align*}
det(A-tI)=(1-t)^{4}
\end{align*}
Then, we have an unique eigenvalue $\lambda=1$ with algebraic multiplicity $\mu_{1}=4$. Now, we have to find the eigenspace of the eigenvalue $\lambda=1$, so we have to solve this:
\begin{align*}
\begin{pmatrix}
0 & a & 0 & 0\\
0 & 0 & b & 0\\
0 & 0 & 0 & c\\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0
\end{pmatrix}
\end{align*}
Thus,
\begin{align*}
a \cdot x_2&=0& & &x_2=0 & &\\
b \cdot x_3&=0& &\Longrightarrow &x_3=0 & &\wedge& &x_1=r\\
c \cdot x_4&=0& & &x_3=0 & &
\end{align*}
Therefore, the eigenspace is:
\begin{align*}
S_{1}=\left \{ r \begin{pmatrix}
1\\
0\\
0\\
0
\end{pmatrix} : \forall r \in F \right \}
\end{align*}
So, $\dim(S_{1})=1 \neq \mu_{1}=4$ $\Longrightarrow$ $T$ is not diagonalizable.
My questions are:
*
*How do I have to use the assumption that $a^2+b^2+c^2\neq 0$?
*Is my solution correct?
(I have already asked this question here but in this post I put my complete solution and a new question about the assumption. I would really really appreciate your help.)
| Note that $a^2 + b^2 + c^2 = 0$, for $a, b, c \in \mathbb{R}$, only holds if $a = b = c = 0$.
Your error comes in this step:
\begin{align*}
a \cdot x_2&=0& & &x_2=0 & &\\
b \cdot x_3&=0& &\Longrightarrow &x_3=0 & &\wedge& &x_1=r\\
c \cdot x_4&=0& & &x_3=0 & &
\end{align*}
Because in this step, you've assumed that $a, b, c \neq 0$ (you also made a typo, since the last line should be $x_4 = 0$, not $x_3$). If one of the parameters is 0, then the corresponding $x$ value is actually free (e.g. if $a = 0$, then $x_2 = s \in \mathbb{R}$).
So the dimension of the eigenspace of $A$ is equal to one more than the number of parameters that is equal to zero. As long as at least one of $a, b, c$ is non-zero, your argument is correct and $T$ is not diagonalisable. If $a = b = c = 0$, then the eigenspace of $A$ is $\mathbb{R}^4$ and $T$ is diagonalisable (and, in fact, if $a = b = c = 0$ then $A$ is the identity matrix which is not just diagonalisable but already diagonal).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we obtain the relation between $x,y$ in terms of $a,b$ for which $ (a+x-y)(x+y)>b $? I have this condition
$$ (a+x-y)(x+y)>b ,$$
where $x,y$ are two positive parameters and $a,b$ are fixed positive numbers.
Are we able to obtain other simpler alternative relations between $x,y$ in terms of $a,b$ under which the given condition holds?
| Rearranging the expression:
\begin{align*}
(a+x-y)(x+y) &> b\\
a(x+y)+(x-y)(x+y) &> b\\
ax+ay+x^2-y^2 &> b\\
(x^2+ax)-(y^2-ay) &> b\\
(x^2+ax+a^2/4-a^2/4)-(y^2-ay+a^2/4-a^2/4) &> b\\
(x^2+ax+a^2/4)-a^2/4-(y^2-ay+a^2/4)+a^2/4 &> b\\
(x+a/2)^2-(y-a/2)^2 &> b\\
\frac{(x+a/2)^2}{b}-\frac{(y-a/2)^2}{b} &> 1,\text{ if $b>0$}\\
\end{align*}
So the relationship is that $(x,y)$ are points outside the horizontal hyperbola with center at $(-a/2,a/2)$ and semi-axis $\sqrt{b}$, if $b>0$ o inside the vertical hyperbola with center at $(-a/2,a/2)$ with semi-axis $\sqrt{-b}$, if $b<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)=3(a^2+b^2+c^2+(a+b+c)^2)^2$ in another way? How do I prove the following identity without expanding both sides directly.
$$2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)\\=3(a^2+b^2+c^2+(a+b+c)^2)^2$$
I expanded both sides directly and it is true. However, I was hoping there could be another way to prove it, like that of Candido's identity which could be proved using diagrams.
| You are trying to prove that $F=G$ where $F$ and $G$ are symmetric homogeneous 4th degree polynomials in three variables. Fully expanded, such a polynomial is in the form $$A\left(a^4+b^4+c^4\right)+B\left(a^3(b+c)+b^3(a+c)+c^3(a+b)\right)\\+C\left(a^2bc+b^2ac+c^2ab\right)+D\left(a^2b^2+a^2c^2+b^2c^2\right)$$
Imagine the above equals your $F$, and you do not yet know $A,B,C,D$. Substitute four points of the form $(a,b,c)$. You would get four linear equations in $A,B,C,D$. As long as you are not very unlucky, these four equations are independent and linear algebra will tell you the values of $A,B,C,D$.
But the same could be said with $G$ on the right side. And if your solutions for $A,B,C,D$ match, then you may conclude $F=G$.
Really, as long as you trust that you pick four triples $(a,b,c)$ that lead to independent equations, as long as you verify that your left side equals your right side at those four triples, it establishes that $F=G$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
| I wanted to offer a different approach, using a bit of algebra.
$$\begin{align}2^n+5^n+56 &\equiv \\ &\equiv2^n+5^n+54+2\\
&\equiv 2^n+5^n+2\\
&\equiv 10^n-1+2^{2n}+2\times 2^n+1\\
&\equiv2^{2n}+2\times 2^n+1\\
&\equiv \left(2^n+1\right)^2\\
&\equiv 0\thinspace \thinspace \thinspace(\text{mod 9}) \end{align}$$
Because, $$\begin{align}2^n+1\equiv 0 \thinspace \thinspace \thinspace (\text{mod 3})\end{align}$$
where, $~n\equiv 1 \thinspace \thinspace \thinspace (\text{mod 2}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Find the area of the right triangle with a right triangle I came up with a problem myself:
What is the area/sides of the largest right triangle?
Question: Point $D$ is the midpoint of $AB$ and point $E$. It is also known that $\triangle AEF$ is also a right triangle. The goal is to find the area and sides of the largest right triangle.
My attempts:
*
*Complete the two right triangles such that they become rectangles, let $AGBC$ and $AEFH$ be the rectangles, assume that point $I$ be the point such that $BI$ is perpendicular to $AH$, thus $\angle GAH=\angle GBI$, this may be able to be solved by similar triangles.
*Rotate $\triangle AED$ such that $AD$ overlap with $BD$ and $EBD$ is a triangle.
*Let $G$ be a point such that $BG$ is perpendicular to $EF$, that may possible be done using similar triangles.
|
Let
\begin{align}
|BC|&=a=6
,\quad
|AC|=b=8
,\quad
|AB|=|BE|=|BF|=c=10
,\\
|AD|&=|BD|=|CD|=\tfrac c2=5
,\\
\angle BCA&=
\angle FEA=
\angle DGC=
\angle DHE=
90^\circ
.
\end{align}
Then $\triangle EDH\sim\triangle CDG$
and
\begin{align}
\frac{|EH|}{|CG|}&=
\frac{|DE|}{|DC|}
=\frac{|DC|+|CE|}{|DC|}
\tag{1}\label{1}
,\\
|CG|&=\frac{ab}c
\tag{2}\label{2}
\end{align}
and we can express $|EH|$ in terms of $|CE|$:
\begin{align}
|EH| &= \frac{ab(2|CE|+c)}{c^2}
\tag{3}\label{3}
,
\end{align}
where $|CE|$ can be found from $\triangle BED$
with the help of [Stewart's theorem]
%(https://en.wikipedia.org/wiki/Stewart%27s_theorem):
\begin{align}
|BE|^2\cdot|CD|+
|BD|^2\cdot|CE|
&=(|CD|+|CE|)(|BC|^2+|CD|\cdot|CE|)
\tag{4}\label{4}
,\\
|CE|&=
\frac1c\,(-a^2+\sqrt{a^4+b^2 c^2})
\tag{5}\label{5}
,
\end{align}
hence
\begin{align}
|EH|&=
\frac{ab}{c^3}\,(b^2-a^2+2\sqrt{a^4+b^2 c^2})
\tag{6}\label{6}
\end{align}
and the area of $\triangle AFE$ is
\begin{align}
[AFE]&=\tfrac12|EH|\cdot|AF|
=
\frac{ab}{c^2}\,(b^2-a^2+2\sqrt{a^4+b^2 c^2})
\\
&=
\tfrac1{25}(336+96\sqrt{481})
\approx 97.65777
\tag{7}\label{7}
\end{align}
The two other sides of $\triangle AFE$ can then be found as
\begin{align}
u&=\sqrt{2c^2 +2\sqrt{c^4-[AFE]^2}}
=\tfrac15(14\sqrt{37}-2\sqrt{13})
\tag{8}\label{8}
\\
v&=\sqrt{2c^2 -2\sqrt{c^4-[AFE]^2}}
=\tfrac15(14\sqrt{13}+2\sqrt{37})
\tag{9}\label{9}
.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the size of the segment joining the foot of the perpendiculars of a scalene triangle The sides of a scalene triangle measure 13, 14, and 15 units. Two outer bisectors of different angles are drawn and the third vertex is drawn perpendicular to these bisectors. Calculate the size of the segment joining the foot of these perpendiculars.(Answer:21)
My progress ..I thought of using sine theorem, cosine theorem and Pythagoras but it will be very complicated. There is probably a simpler solution
$D ~é ~excentro \therefore AD ~é~ bissetriz \triangle ABC\\
Teorema Bissetriz: \frac{BK}{AB}=\frac{KC}{AC}\rightarrow\frac{14-KC}{13}=\frac{KC}{15}\\
\therefore KC = 7,5 ~e~BK = 6,5\\
\triangle ABK\sim \triangle AML: Razão~Semelhança=\frac{13}{6,5} = 2\\
\therefore LM = \frac{6,5}{2}=3,25\\
De~forma~análoga: LN = 3,75\\
\triangle AHB: M(ponto~médio)\rightarrow HM = MB = 6,5\\
\triangle ACI:N(ponto ~médio) \rightarrow NI = NC = 7,5\\
\therefore \boxed{\color{red}x = 6,5+3,25+3,75+7,5 = 21 }$
| The way you had started does lead to a solution, although it is a little calculative. Note that:
$$AH=13 \cos \left(\frac {\beta}{2}\right)$$
$$AI=15 \cos \left(\frac {\gamma}{2} \right)$$
Also, $\angle AHI=\alpha +\frac {\beta+\gamma}{2}=90°+\frac {\alpha}{2}$.
Now, let $HI=x$. From the cosine law on $\Delta HAI$, we have:
$$\cos (\angle AHI)=\cos\left(90°+\frac {\alpha}{2}\right)=-\sin \frac {\alpha}{2}=\frac {AH^2+HI^2-x^2}{2 AH \cdot AI}$$
Thus, $$x=\sqrt {AH^2+HI^2-2 AH \cdot AI \sin \frac {\alpha}{2}} {\tag 1}$$
Now, use the formulae:
$$\cos \frac {\alpha}{2}=\sqrt {\frac {s(s-a)}{bc}}$$
$$\cos \frac {\beta}{2}=\sqrt {\frac {s(s-b)}{ac}}$$
$$\cos \frac {\gamma}{2}=\sqrt {\frac {s(s-c)}{ab}}$$
Here $a=14$, $b=15$ and $c=13$, while $s=\frac {a+b+c}{2}=21$.
Hence, all the half-angle trig values are known, and can be substituted back in $(1)$ to obtain the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integers $ a $ and $ b $ such that $7^a-3(2^b)=1$ I am looking for all positive integers satisfying the equation
$$7^a-3(2^b)=1$$
I used the fact that $ (1,1) $ is a solution and got
$$(7^a-7)=3(2^b-2)$$
I thaught about using Gauss Theorem in vain.
Thanks in advance for any idea.
| An argument mod $7$ shows that $b+2$ is divisible by $3$, then let $b=3n-2.$
We can take the two cases $a=2m, a=2m+1.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet\ a=2m$
Let $X=3\cdot2^{n}, Y=6\cdot7^{m}$, then we get
$Y^2 =X^3 + 36.$
According to LMFDB, this elliptic curve has integral solutions $(X,Y)=(-3,\pm 3), (0,\pm 6), (4,\pm 10), (12,\pm 42).$
From $(12,\pm 42)$ we get $(m,n)=(1,2) \implies (a,b)=(2,4).$
$\bullet\ a=2m+1$
Let $X=21\cdot2^{n}, Y=294\cdot7^{m}$, then we get
$Y^2 =X^3 + 12348.$
According to LMFDB, this elliptic curve has integral solutions $(X,Y)=(-14,\pm 98), (-3,\pm 111), (21,\pm 147), (37,\pm 251), (42,\pm 294), (378,\pm 7350), (11802,\pm 1282134).$
From $(42,\pm 294)$ we get $(m,n)=(0,1) \implies (a,b)=(1,1).$
Hence there are only integral solutions $(a,b)=(1,1),(2,4).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(1+A)^n \ge (1+a_1)\cdot(1+a_2)\cdots(1+a_n) \ge (1+G)^n$ $A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that
*
*$(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
*if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$
My try (Ques -1): To prove $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)=1+\sum{a_1}+\sum{a_1}\cdot{a_2}+\sum{a_1}\cdot{a_2}\cdot{a_3}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$
Consider $a_1,a_2,\ldots,a_n$ be $n$ positive real numbers and then apply AM$\ge$GM
$\frac{(a_1+a_2+\cdots+a_n)}{n}\ge(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$ imples $\sum{a_1}\ge n\cdot{G}$ because $G=(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$
again consider $(a_1\cdot{a_2}),(a_1\cdot{a_3}),\ldots({a_1}\cdot{a_n}),(a_2\cdot{a_3}),(a_2\cdot{a_4})\ldots,(a_2\cdot{a_n}),(a_3\cdot{a_4})\ldots(a_{n-1}\cdot{a_n})$ be $\frac{n(n-1)}{2!}$ positive real numbers.
Then Applying AM$\ge$GM , we get,
$\frac{\sum{a_1}\cdot{a_2}}{\frac{n(n-1)}{2!}}$ $\ge$ $\bigl(a_{1}^{n-1}\cdot{a_{2}^{n-1}}\cdots{a_{n}^{n-1}}\bigl)^\frac{2!}{n(n-1)}$ implies $\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$
Similary , $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.
Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge 1+nG+\frac{n(n-1)}{2!}G^2+\frac{n(n-1)(n-2)}{3!}G^3+\cdots+G^n$
Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge (1+G)^n$
To prove $(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$
Consider $(1+a_1),(1+a_2),\ldots,(1+a_n)$ be $n$ positive real numbers and then applying AM$\ge$GM
$\frac{(1+a_1)+(1+a_2)+\cdots+(1+a_n)}{n} \ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$
$\frac{n+a_1+a_2+\cdots{a_n}}{n}\ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$
$(1+A)^n\ge(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)$
My try (Ques -2): To prove $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$
Consider $(k+a_1),(k+a_2)\ldots(k+a_n)$ be $n$ positive real numbers and applying AM$\ge$GM
$\frac{(k+a_1)+(k+a_2)+\cdots(k+a_n)}{n}\ge[(k+a_1)\cdot(k+a_2)\cdots(k+a_n)]^\frac{1}{n}$
$(k+A)^n\ge (k+a_1)\cdot(k+a_2)\cdots(k+a_n)$
To prove $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$
$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)=k^n+k^{n-1}\sum{a_1}+k^{n-2}\sum{a_1\cdot{a_2}}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$
now
$\sum{a_1}\ge n\cdot{G}$,
$\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$
$\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.
Therefore
$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge k^n+nk^{n-1}G+\frac{n(n-1)}{2!}k^{n-2}G^2+\cdots+G^n$
Therefore, $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge (k+G)^n$
My Ques :
*
*Have I solved the questions correctly?
*How to prove $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$
Thanks
| I think there is a much faster way.
*
*$\log(k+x)$ is concave, hence by Jensen's inequality $\log(k+A)\geq\frac{1}{n}\sum_{m=1}^{n}\log(k+a_m)$. Exponentiation leads to $(k+A)^n \geq \prod_{m=1}^{n}(k+a_m)$.
*Let $b_m=\log(a_m)$ or, equivalently, $a_m=e^{b_m}$. We have that $\log(k+e^x)$ is convex, since $\frac{d^2}{dx^2}\log(k+e^x)=\frac{ke^x}{(k+e^x)^2}$. This leads to $(k+G)^n\leq \prod_{m=1}^{n}(k+a_m)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion.
It is given that $(x^2 +y +2t +3k)^{10}$. What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion. For example the terms are $x^2y^4t^3k^2$ etc.
I said that let $(x^2 +2t)$ be $"a"$ and $(y +3k)$ be $"b"$. Then, $(x^2 +y +2t +3k)^{10}=(a+b)^{10}$.
If we are looking for $x^2t^3$, then the exponential of $a$ must be $4$. Then, the exponential of $b$ is $6$.
Then, the coefficient of $x^2t^3$ is $4 \times 2^3=32$. Now, we have $(y +3k)^6$ as $b$. If we find the sum of the all coefficients in $(y +3k)^6$, we can handle the question.
The sum of the all coefficients in $(y +3k)^6$ is $4^6$.
Then, answer is $32 \times 4^6$
Is my solution correct ?
| As an alternative, we can use the multinomial expansion formula:
$$(x_1+x_2+\cdots +x_m)^n=\sum_{i_1+i_2+\cdots +i_m=n} {n\choose i_1,i_2,...,i_m} \prod_{t=1}^m x_t^{i_t}, \\
\quad \text{where} \quad {n\choose i_1,i_2,...,i_m}=\frac{n!}{i_1!i_2!\cdots i_m!}\\
$$
$$ (x^2 +y +2t +3k)^{10}=\sum_{i_1+i_2+i_3+i_4=10} {10\choose i_1,i_2,i_3,i_4} x^{2i_1}y^{i_2}(2t)^{i_3}(3k)^{i_4}$$
We want to have $i_1=1,i_3=3$:
$$2^33^6{10\choose 1,0,3,6}+2^33^5{10\choose 1,1,3,5}+2^33^4{10\choose 1,2,3,4}+\\
2^33^3{10\choose 1,3,3,3}+2^33^2{10\choose 1,4,3,2}+2^33^1{10\choose 1,5,3,1}+2^3{10\choose 1,6,3,0}= \\
2^3\cdot \left[\frac{10!}{3!6!}(3^6+1)+\frac{10!}{3!5!}(3^5+3)+\frac{10!}{2!3!4!}(3^4+3^2)+3^3\frac{10!}{3!3!3!}\right]=\\
2^3\cdot \left[613,200+1,239,840+1,134,000+453,600\right]=\\
27,525,120.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which prime $p$, is the group units $R^*$ cyclic? I'm studying for my qualifying exam and found the following question in the question bank of previous years paper:
Let $p$ be a prime and $\mathbb{F}_p$ the field with $p$ elements. Set $R=\mathbb{F}_p[x]/(x^3)$. For which prime $p$ is the group of units $R^*$ cyclic?
What I've got so far is that: $R$ is local ring and any element in $R$ with non-zero constant term will be a unit. I have proven it for $p=2$, but I don't know if it will be true for any other primes. Also, I have proven that if $r=a\overline{x}^2+b\overline{x}+c$ is the generator of the cyclic group $R^*$ then $b\not=0$. I tried to look at the powers of $r$ hoping that I might be able find some element in $R^*$ which can't be expressed as $r^n$, for some $n$, but it got complicated and now I'm completely lost.
| First, note that if $ax^2 + bx + c$ has $c \neq 0,$ then $\gcd(x^3, ax^2 + bx + c) = 1$ in $\mathbb{F}_p[x]$ (since the only prime factor of $x^3$ is $x,$ which won't divide $ax^2+bx+c$ for $c\neq 0$) and so $ax^2+bx+c$ is invertible mod $x^3.$ It's easy to see that nothing else can be invertible mod $x^3$ from a similar gcd argument. If $p=2$ this ring $R$ thus consists only of $x^2+x+1, x^2 + 1,$ and $x+1.$ It is cyclic with generator $x+1,$ since $(x+1)^2 = x^2 + 1$ and $(x+1)^3 = (x^2+1)(x+1) = x^2 + x + 1$ in $R.$
Suppose that $R$ had cyclic group of units, and let it be generated by $ax^2 + bx + c.$
Note that, mod $x^3,$ we find
$$(ax^2 + bx + c)^n = ((ax^2+bx)+c)^n = c^n + nc^{n-1}(ax^2+bx) + \binom{n}{2}c^{n-2}(ax^2+bx)^2 = c^n + nc^{n-1}(ax^2 + bx) + \binom{n}{2}c^{n-2}b^2x^2.$$
This lets us immediately rule out $b \neq 0,$ since otherwise we'd never generate a polynomial like $x+1$ with nontrivial linear term.
Now, as $n$ varies, we need $(c^n, nc^{n-1}b, \binom{n}{2}c^{n-2}b^2 + nc^{n-1}a)$ to run through all possible triples $(x,y,z)$ in $\mathbb{F}_p$ where $x \neq 0.$
The value of $c^n$ depends only on $n$ mod $p-1,$ since $c$ is fixed. Similarly, $n, \binom{n}{2}$ only depend on the value of $n$ mod $p$ for $p > 2,$ so the entire triple $(c^n, nc^{n-1}b, \binom{n}{2}c^{n-2}b^2 + nc^{n-1}a)$ depends only on $n$ mod $p(p-1).$ Thus, there are only $p^2 - p$ distinct possibilities for this triple at most.
How many $(x,y,z)$ triples with $x \neq 0$ are there, though? Well, there's $p^2(p-1) > p(p-1).$ So, it's not cyclic if $p > 2$.
(This breaks for $p=2$ since, as indicated above, $\binom{n}{2}$ mod $2$ cannot be determined just by knowing $n$ mod $2,$ since i.e. $\binom{3}{2} = 3$ is odd but $\binom{1}{2} = 0$ is even. This is because $\binom{n}{2} = n(n-1)/2$ and division by $2$ only makes since modulo an odd prime.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the value of $n\in\mathbb{Z}$ in the equation $16^n-25\times 12^{n-1}+9^n=0$ If $n$ is satisfy the equation $16^n-25\times 12^{n-1}+9^n=0$. Find the value of $n$.
Here is my work:
$$16^n-2\times12\times 12^{n-1}+9^n=12^{n-1}$$
$$(4^n-3^n)^2=12^{n-1}$$
For $n\in\mathbb{Z}^+\cup\{0\}$ we can see the LHS is an odd number and RHS is odd only for $n=1$. But for $n\in\mathbb{Z}^-$ we have fractions in both sides of the equation. I don't know how to check if there are answers for negative values of $n$.
| The original expression $16^n-25\times 12^{n-1}+9^n=0$ when taken $\dfrac{1}{12}$ out common reduces to $\dfrac{1}{12}(4\cdot 3^n-3 \cdot 4^n)(3^{n+1}-4^{n+1})=0$
This results in 2 cases:
Case 1:
$(3^{n+1}-4^{n+1})=0 \implies 3^{n+1}=4^{n+1}$
This is one possible when both have power $0$
So $n+1=0 \implies n=-1$
Case 2:
$ (4\cdot 3^n-3 \cdot 4^n)=0 \implies 4 \cdot 3^n=3 \cdot 4^n $
If you try to see deep you will notice that when $n=1$ only then the equation is satisfied.
So $n=-1,1$
Edit: In case you failed here is other solution-
Take $a=4^n, b=3^n$ and yay there you get quadratic $12(a-b)^2=ab$. Just solve this and get your answer
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $ \frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z) $ Show that for positive reals $x,y,z$ the following inequality holds and that the constant cannot be improved
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z)
$$
Background: I was digging through some old correspondence and found a letter from a very young me to Professor Love at the University of Melbourne. I had apparently ask via a letter (yes it was back when we wrote letters) how one could prove the above inequality (my version had $1/\sqrt{3}$ in it). He kindly wrote back but without a full proof. I just found the correspondence today and thought that this was a good question for this site.
Based on his letter and my old writings you can transform the above inequality as follows. First note that
$$
\text{g.l.b.}f(x,y,z) = \text{g.l.b.}f(x,z,y) = k \quad (say)
$$
where g.l.b is the greatest lower bound and $f(x,y,z)$ is the function
$$
\left(\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}}\right)\bigg/(x+y+z).
$$
and so
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq 2k (x+y+z)
$$
Thus we need to prove that for positive reals $x,y$ and $x$ the following is true and tight:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \sqrt{2} (x+y+z)
$$
One approach to prove this (used by Prof. Love) was to apply Hölder's inequality but this unfortunately only gives:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \frac{2}{\sqrt{3}} (x+y+z)
$$
Geometric View: From a geometric view point this inequality can be viewed as stating that the perimeter of $\Delta PQR$ is not less than $\sqrt{2}$ times the sum of the the three edge-lengths of the box of sides $x,y,$ and $z$ and the points $P,Q$ and $R$ are three corners of the box that are not adjacent to each other.
I suspect that this is a "well known" inequality in the right circles but it is still not known to me. Thought that is was a nice problem for lovers of inequalities.
| When $x=y=z$, lhs=$\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{x+y+z}{\sqrt{2}}$=rhs, which means that $\sqrt{2}$ is best constant.
| {
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"url": "https://math.stackexchange.com/questions/4215933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the system of differential equations and plot the curves given the initial conditions. We are given the following system of ordinary differential equations:
$$\dot{x} = x - 4y \quad \dot{y} = x - 2y -4.$$
Thus, the slope of the trajectories is given by
$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{x - 2y -4}{x-4y}.$$
We have the initial conditions $x(0) = 15$ and $y(0) = 5$.
How do I solve this system? The book does not give the solution but claims that we can solve for the trajectories. It does not suggest a method. It also asks what happens for other initial conditions.
I tried treating each separate equation as a linear equation and solved them via integrating factors, but that of course resulted in an integral equation on the right-hand side. I was not sure if that was the right way to go about it.
| Here is an approach, from the second equation, we have
$$x = y' + 2 y + 4, ~~\mbox{so}~~ x' = y'' + 2 y'$$
Substituting these two into the first equation
$$ y'' + y'+2 y = 4$$
For initial conditions, we are given $y(0) = 5$ and use the second equation to find $y'(0) = x(0) - 2 y(0) - 4 = 1$.
This DEQ and ICs lets you find $y$, which you then use to find $x$.
We can also set this up and solve a non-homogeneous system using many methods
$$ Y' = A Y + g = \begin{pmatrix} 1 & -4 \\ 1 &- 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}+ \begin{pmatrix} 0 \\ - 4 \end{pmatrix}$$
You should get
$$x(t)= -\frac{3 e^{-\frac{t}{2}} \sin \left(\frac{\sqrt{7} t}{2}\right)}{\sqrt{7}}+7 e^{-\frac{t}{2}} \cos \left(\frac{\sqrt{7} t}{2}\right)+8 \\y(t)= \frac{5 e^{-\frac{t}{2}} \sin \left(\frac{\sqrt{7} t}{2}\right)}{\sqrt{7}}+3 e^{-\frac{t}{2}} \cos \left(\frac{\sqrt{7} t}{2}\right)+2$$
You can also look at a phase portrait
Here is that same plot with a huge number of plot points
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Color ball drawer probability question There is a ball drawer.
Seven color balls will be drawn with the same probability ($1/7$).
(black, blue, green, yellow, white, pink, orange)
If Anson attempts $9$ times,
what is the probability that he gets all $7$ different color balls?
My work:
I separate the answer to $3$ ways.
*
*$7$ attempts -> done (get $7$ colors)
*$8$ attempts -> done (get $7$ colors)
*$9$ attempts -> done (get $7$ colors)
Therefore, my answer is $$\frac{9C7 + 8C7 + 7C7}{7^7 \cdot (7!)}$$
However, I don't know it is correct or not.
| Since there are seven choices for each of the nine balls Anson selects, there are $7^9$ possible sequences of colors.
Method 1: If each color appears among the nine balls, there are two possibilities:
*
*One color is selected three times and each of the other colors is selected once.
*Two colors are each selected twice and each of the other colors is selected once.
One color is selected three times and each of the other colors is selected once: There are seven ways to select the color which appears three times, $\binom{9}{3}$ ways to select the three positions occupied by that color, and $6!$ ways to arrange the remaining six colors in the remaining six positions. There are
$$\binom{7}{1}\binom{9}{3}6!$$
such cases.
Two colors are each selected twice and each of the other colors is selected once: There are $\binom{7}{2}$ ways to select the two colors which each appear twice, $\binom{9}{2}$ ways to select two positions for the selected color which appears first in an alphabetical list, $\binom{7}{2}$ ways to select two positions for the other selected color, and $5!$ ways to arrange the remaining five colors in the remaining five positions. There are
$$\binom{7}{2}\binom{9}{2}\binom{7}{2}5!$$
such cases.
Therefore, the number of favorable cases is
$$\binom{7}{1}\binom{9}{3}6! + \binom{7}{2}\binom{9}{2}\binom{7}{2}5!$$
so the probability that all seven colors are selected is
$$\Pr(\text{all seven colors selected}) = \frac{\dbinom{7}{1}\dbinom{9}{3}6! + \dbinom{7}{2}\dbinom{9}{2}\dbinom{7}{2}5!}{7^9}$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $7^9$ possible sequences of colors. From these, we must exclude those sequences in which one or more colors is missing.
There are $\binom{7}{k}$ ways to select which $k$ colors are missing and $(7 - k)^9$ sequences of colors which can be formed with the remaining colors. Thus, by the Inclusion-Exclusion Principle, the number of favorable cases is
\begin{align*}
& \sum_{k = 0}^{7} (-1)^k\binom{7}{k}(7 - k)^9\\
& \qquad = 7^9 - \binom{7}{1}6^9 + \binom{7}{2}5^9 - \binom{7}{3}4^9 + \binom{7}{4}3^9 - \binom{7}{5}2^9 + \binom{7}{6}1^9 - \binom{7}{7}0^9
\end{align*}
Hence, the probability that each color appear is
\begin{align*}
& \Pr(\text{all seven colors selected})\\
& \qquad = \frac{7^9 - \dbinom{7}{1}6^9 + \dbinom{7}{2}5^9 - \dbinom{7}{3}4^9 + \dbinom{7}{4}3^9 - \dbinom{7}{5}2^9 + \dbinom{7}{6}1^9 - \dbinom{7}{7}0^9}{7^9}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$ has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking for a simialr series solution. I suspect is is of the form
$$x = \sum_{k=0}^\infty \binom{3k+2}{k} \frac{b^{3k+2} }{(3)k+2} $$
| Too long for a comment.
Looking for truncated series expansions, you can directly solve for the roots solving the cubic using the trigonometric method.
The solutions are given by
$$x_k=\frac{2 \cos \left(\frac{2 \pi }{3}k-\frac{1}{3} \cos
^{-1}\left(-\frac{3\sqrt{3}}{2} b^{3/2}\right)\right)}{\sqrt{b}\sqrt{3}}\qquad \text{with} \qquad k=0,1,2$$
We have
$$x_1=b+b^4+3 b^7+12 b^{10}+55 b^{13}+273 b^{16}+1428 b^{19}+7752
b^{22}+O\left(b^{25}\right)$$ and the coefficients corrrespond to sequence $A001764$ in $OEIS$ that is to say
$$x_1=\sum_{n=0}^\infty \frac{ \binom{3 n}{n}}{2 n+1}b^{3 n+1}$$ which is the one you obtained using Lagrange inversion theorem.
Similarly,
$$x_0=\frac{1}{\sqrt{b}}-\frac{b}{2}-\frac{3 b^{5/2}}{8}-\frac{b^4}{2}-\frac{105
b^{11/2}}{128}-\frac{3 b^7}{2}-\frac{3003 b^{17/2}}{1024}-6
b^{10}+O\left(b^{23/2}\right)$$
$$x_2=-\frac{1}{\sqrt{b}}-\frac{b}{2}+\frac{3 b^{5/2}}{8}-\frac{b^4}{2}+\frac{105
b^{11/2}}{128}-\frac{3 b^7}{2}+\frac{3003 b^{17/2}}{1024}-6
b^{10}+O\left(b^{23/2}\right)$$ already given by @Somos.
You also have $x_0+x_1+x_2=0$ and $x_0x_1x_2=-1$ also as mentioned by @Somos.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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