Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $g:[0,1]\rightarrow [0,1]$ such that $g(f(x))+g(x)=1$ Let $f:[0,1]\rightarrow [0,1]$ be a continuous function such that $f_\alpha= \frac{1-x}{1+\alpha x}$ , $\alpha \gt-1$.
Can you help me to find a function $g:[0,1]\rightarrow [0,1]$ such that $g(f(x))+g(x)=1$
| Consider a transform $g$ of the form $g(x) = \frac{a+bx}{c+dx}$. Then we must find $a, b, c, d$ satisfying
\begin{align*}
\frac{a + b\frac{1-x}{1+\alpha x}}{c + d\frac{1-x}{1+\alpha x}} + \frac{a + bx}{c + dx} = 1
\end{align*}
Or in other words
\begin{align*}
\frac{(a+b) + (a\alpha-b)x}{(c+d) + (c\alpha-d)x} = \frac{(c-a) + (d-b)x}{c + dx}
\end{align*}
which gives us the system of equations
\begin{align*}
a + b &= \lambda(c-a) \\
a\alpha - b &= \lambda(d-b) \\
c+d &= \lambda c \\
c\alpha - d &= \lambda d \\
\end{align*}
We have four equations for five variables ($\alpha$ is assumed to be known). In fact, one of the equations turns out to be redundant, giving us two free parameters in the end. We may solve for the positive roots of $c, d$ with a shameless application of your favorite computer algebra program as
\begin{align*}
c = \frac{\sqrt{\alpha+1}a + (a+b)}{\sqrt{\alpha+1}}, \quad d = \frac{(\sqrt{\alpha+1}-1)(\sqrt{\alpha+1}a + (a+b))}{\sqrt{\alpha+1}}
\end{align*}
Now, let's find sufficient conditions on $a, b$ such that $g([0, 1]) \subseteq [0,1]$. If we allow $a > 0$ and $b = 0$, then we have $c \ge a$ and $d > 0$. Clearly $g(x)\ge 0$. Also, for $0 \le x \le 1$,
\begin{align*}
g(x) = \frac{a}{c+dx} + \frac{b}{c+dx}x \le \frac{a}{c} \le 1
\end{align*}
And we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find integral solutions for surd problem
Problem. Find all integral $x$ and $y$ satisfying $$2\sqrt{6} + 5\sqrt{10} = \sqrt{x} + \sqrt{y}.$$
Squaring both sides, we get $$274 - x - y = 2\sqrt{xy} - 20\sqrt{60}.$$ Squaring again, we get $$-160\sqrt{15xy} \in \mathbb{Z},$$
which gives us $xy = 15k^2$. We know $2\sqrt{6} + 5\sqrt{10} < 2(2.5) + 5(3.2) = 21$. Hence $x, y < 441$. There are still quite a lot of cases to check. Is there a better way to do this?
Edit: I would like to find all possible solutions, and prove that those constitute all of them. Clearly $(24,250)$ and $(250,24)$ are solutions, but can we prove that there are no other?
| (1) Keeping in mind the linear independance of irrational quadratics we have
$$x=6X^2, y=10Y^2\Rightarrow 2\sqrt6+5\sqrt{10}=X\sqrt6+Y\sqrt{10}\Rightarrow X=2,Y=5$$ Hence $$(x,y)=(24,250)$$
(2) Applying conjugates in quadratic fields one has the system
$$\begin{cases}2\sqrt{6} + 5\sqrt{10} = \sqrt{x} + \sqrt{y}\\2\sqrt{6} - 5\sqrt{10} = \sqrt{x} - \sqrt{y}\end{cases}$$ From which it is immediate
$$4\sqrt6=2\sqrt x\Rightarrow x=24\\10\sqrt{10}=2\sqrt y\Rightarrow y=250$$
The only solution is $$(x,y)=(24,250)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{\frac{x^2-1}{x^2}}=\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2}\cdot\frac{x^2}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}$$
As you can see I'm going in circles. Can anyone give me a hint on how to start on this problem?
| Hint:
We can do partial fraction decomposition to show that
$$\frac x{x^2-1}=\frac12\left(\frac1{x-1}+\frac1{x+1}\right)$$
Then do the limit on each individual piece.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$ Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$
given that $a_0 = 0$, and $a_1 = 3$
My Attempt:
Get the characteristic equation and solve it.
For homogeneous equation
$x^2 -4x + 4 = 0$
$x = 2 $ or $ x = 2$
Hence, $a_n^h = (A+Bn)\cdot2^n $
Guess a particular solution: $n^22^nC$
$n^22^nC - 4(n-1)^22^{n-1}C + 4(n-2)^22^{n-2}C = 2^n$
$n^2C - 2(n-1)^2C + (n-2)^2C = 1$
$C= \frac12$
Hence, $a^p_n = \frac12n^22^n$
$a_n = a^p_n + a^h_n$
$a_n = (A+Bn)\cdot2^n + \frac12n^22^n$
$a_0 = 0 = A$
$a_1 = 3 = 2B + 1$
$B = 1$
Therefore, $a_n = (n+ \frac12n^2)2^n$
| No, $n^22^{n^2}C$ does not work. Try with $n^22^{n}C$. Then
$$2^n=a_n - 4a_{n-1} + 4a_{n-2} = n^22^{n}C-4(n-1)^22^{n-1}C+4(n-2)^22^{n-2}C=2^{n}\cdot 2C$$
which implies that $C=1/2$.
In general, if the r.h.s. is $r^n$ and $r$ is a solution of the characteristic polynomial of multiplicity $m$, then a particular solution has the form $n^m C r^n$.
Note that $n^i r^n$ is a solution of the homogeneous recurrence for $0\leq i<m$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2$
Let $x,y,z \in \mathbb{R}$ such that $x \geq y \geq z > 0$. Prove that $$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2.$$
I rearranged the inequality to get $$xy(x^2y)+yz(y^2z)+xz(z^2x) \geq xyz(x^2+y^2+z^2).$$ I then thought about using the rearrangement inequality but didn't see how to use it. How can we continue?
| We need to prove that $\sum\limits_{cyc}(x^3y^2-x^3yz)\geq0$ or
$$\sum\limits_{cyc}(z^3x^2+z^3y^2-2z^3xy)\geq\sum\limits_{cyc}(x^3z^2-x^3y^2)$$ or
$$\sum\limits_{cyc}z^3(x-y)^2\geq(xy+xz+yz)(x-y)(y-z)(z-x)$$
which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
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Incorrect method to find a tilted asymptote Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.
I would like to find out, though, what is wrong with the following incorrect way of finding the asymptote:
$\displaystyle y=\frac{x^2+x-6}{x+2}=\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}\approx\frac{x+1}{1}=x+1$, so $y=x+1$ is the slanted asymptote.
| The trouble is that the numerator in
$$\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}$$
stays large, so a small error in the denominator is magnified. Better to write
$$ \frac{1}{1+\frac{2}{x}} = 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots $$
and multiply; the infinite series indicated converges for $|x| > 2.$
$$ \left( x+1-\frac{6}{x} \right) \left( 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots \right) = x - 1 - \frac{4}{x} + \frac{8}{x^2} + \cdots $$
| {
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Describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$. I need to describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$.
After testing out some $n \in \mathbb{N}$, I came to the conclusion that $3^{n} - 2^{n}$ is divisible by $5$ iff $n$ is even - i.e., if $n$ is of the form $2k$ for $k \in \mathbb{N}$.
Here is my attempt so far:
$(\implies)$:
To show that if $n$ is even, then $3^{n}-2^{n}$ is divisible by $5$, we prove that $\forall k \in N$, $3^{2k}-2^{2k} \equiv \,0 \mod 5$ by induction on $k$:
*
*Basis Step: For $k = 1$, $3^{2(1)}-2^{2(1)}=3^{2}-2^{2}=9-4=5 \equiv\, 0 \mod 5 $.
*Suppose true for $\mathbf{k=m}$: $3^{2m}-2^{2m}\equiv \, 0 \mod 5 \, \implies \, 3^{2m}-2^{2m}=5l$, $l \in \mathbb{Z}$.
*Show true for $\mathbf{k = m+1}$: Consider $3^{2(m+1)}-2^{2(m+1)}\\ = 3^{2m+2}-2^{2m+2}\\ = 3^{2m}\cdot 3^{2}- 2^{2m}\cdot 2^{2}\\ = 9\cdot 3^{2m}-4 \cdot 2^{2m} \\= 5 \cdot 3^{2m} + 4\cdot 3^{2m} - 4 \cdot 2^{2m} \\ = 5(3^{2m}) + 4(3^{2m}-2^{2m}) \\ = 5(3^{2m})+4(5l)\, \text{(by the induction hypothesis)} \\ = 5(3^{2m} + 4l) \, \text{which is divisible by}\, 5.$
So, by induction, then the statement $3^{2k}-2^{2k} \equiv \, 0 \mod 5$ holds $\forall k \geq 1 \, \implies \, $ the statement $3^{n}-2^{n} \equiv \, 0 \mod 5$ holds $\forall n = 2k$, $k \geq 1$.
$(\Longleftarrow)$:
To show that $3^{n}-2^{n}$ divisible by $5$ $\implies$ $n$ even, we will show that $n$ odd $\implies$ $3^{n}-2^{n}$ is not divisible by $5$. Now, suppose the proposition is false. I.e., assume $\exists n \in \mathbb{N}$ for which $n$ is odd, but $3^{n}-2^{n}$ is divisible by $5$.
Since $n$ is odd, it is of the form $2k+1$, $k \in \mathbb{N}$.
So, we have that $3^{2k+1}-2^{2k+1} \equiv \, 0 \mod 5 \, \implies \, 3^{2k+1} - 2^{2k+1} = 5l$, $l \in \mathbb{N}$.
So, $\displaystyle \frac{3^{2k+1}}{5} - \frac{2^{2k+1}}{5} = l$.
At this point, I got stuck. I think my $(\implies)$ direction is fine, but I definitely need help on my $(\Longleftarrow)$ direction.
Could somebody please help me complete this proof?
Thank you.
| HINT: $$3^{2k+1} - 2^{2k+1} = 3\cdot 3^{2k} - 2 \cdot 2^{2k} = 3(3^{2k} - 2^{2k}) + 2^{2k}$$
Now can you see why the first summand is divisible by 5 and the second isn't?
| {
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Prove this trigonometry equation: $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
Prove that $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
I've tried writing $\sin 40^\circ$ as $\sin(40^\circ+10^\circ)$, then wrote $\sin(50^\circ+10^\circ)$ as $\sin 40^\circ \cos 10^\circ + \cos 40^\circ \sin 10^\circ$, but I don't know what to do next.
| As $\sin50^\circ=\cos(90-50)^\circ$
and $2\sin40^\circ\cdot\cos40^\circ=\sin(2\cdot40)^\circ$
Finally $\sin80^\circ=\cos(90-80)^\circ=?$
| {
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Solve the differential equation $4y=x^2+(y')^2$
Solve the differential equation $4y=x^2+(y')^2$.
My try:
$$4 y'=2x+2y'y''$$ hence $2y'=x+y'y''$. Let $y'=vx$, then $y''=v+xv'$, hence
$$2vx=x+vx(v+xv')$$
$$-\int {v\over (v-1)^2} dv=\int {1\over x} dx$$
For R.H.S. Let $u=v-1$
So :
$${1\over u}-\ln u=\ln x+c$$
$${1\over {v-1}}-\ln |v-1|=\ln |x|+c$$
By substitute $v={y'\over x}$ :
$${x\over {y'-x}}-\ln \left|{{y'-x}\over x}\right|=\ln |x| +c$$
that is:
$${x\over {y'-x}}-\ln \left|y'-x\right|=c$$
True ?
| $4y=x^2+(y')^2$
$\left(\dfrac{dy}{dx}\right)^2=4y-x^2$
$\dfrac{dy}{dx}=\pm\sqrt{4y-x^2}$
Let $u=\pm\sqrt{4y-x^2}$ ,
Then $y=\dfrac{u^2+x^2}{4}$
$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}+\dfrac{x}{2}$
$\therefore\dfrac{u}{2}\dfrac{du}{dx}+\dfrac{x}{2}=u$
$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{2u-x}{2}$
$\dfrac{du}{dx}=2-\dfrac{x}{u}$
Let $v=\dfrac{u}{x}$ ,
Then $u=xv$
$\dfrac{du}{dx}=x\dfrac{dv}{dx}+v$
$\therefore x\dfrac{dv}{dx}+v=2-\dfrac{1}{v}$
$x\dfrac{dv}{dx}=-v+2-\dfrac{1}{v}$
$x\dfrac{dv}{dx}=-\dfrac{v^2-2v+1}{v}$
$x\dfrac{dv}{dx}=-\dfrac{(v-1)^2}{v}$
$\dfrac{v}{(v-1)^2}~dv=-\dfrac{dx}{x}$
$\int\dfrac{v}{(v-1)^2}~dv=-\int\dfrac{dx}{x}$
$\ln(v-1)-\dfrac{1}{v-1}=-\ln x+c$
$(v-1)e^{-\frac{1}{v-1}}=\dfrac{C_1}{x}$
$\left(\dfrac{u}{x}-1\right)e^{-\frac{1}{\frac{u}{x}-1}}=\dfrac{C_1}{x}$
$(u-x)e^\frac{x}{x-u}=C_1$
$(\pm\sqrt{4y-x^2}-x)e^\frac{x}{x\mp\sqrt{4y-x^2}}=C_1$
$(x\pm\sqrt{4y-x^2})e^\frac{x}{x\pm\sqrt{4y-x^2}}=C$
| {
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Two spheres, triple integration, not their intersection
Two spheres, one of radius $1$ and one of radius $\sqrt{2}$, have centres that are $1$ unit apart. Find the volume of the smaller region that is outside one sphere and inside the other.
Can use either spherical or cylindrical coordinates. The $2$ spheres can be anywhere in three dimensional space. Apparently using the correct coordinates will lead into an easy integration. Been tossing around the two and haven't seen any of them working out to be easy.
| $x^2 + y^2 + z^2 = 2\\
x^2 + y^2 + (z-1)^2 = 1$
Would be an equation for two spheres with centers at (0,0,0) and (0,0,1) (1 unit apart), one of radius $\sqrt2$, and one of radius 1
Expand and subtract one from the other to find the the points of intersection.
$2z - 1 = 1\\
z = 1$
$x^2 +y^2 = 1$
Would you rather do this in spherical or cylindrical coordinates?
In cylindrical:
$x = r \cos \theta\\
y = r \sin \theta\\
z = z\\
dx\;dy\;dz = r \;dz\;dr\;d\theta$
Make these substitutions into the equations of the two spheres, and the circle formed by their intersection to find the limits of integration.
$z = \sqrt {2 - r^2}\\
z = \sqrt {1-r^2}+1\\
r = 1$
$\int_0^{2\pi}\int_0^{1}\int_{\sqrt {2-r^2}}^{\sqrt {1 - r^2}+1} r \;dz\;dr\;d\theta$
In spherical:
$x = \rho \cos \theta\sin\phi\\
y = \rho \sin \theta\sin\phi\\
z = \rho \cos\phi\\
dx\;dy\;dz = \rho^2 \sin\phi \;d\rho\;d\phi\;d\theta$
The limits tend to be harder to find in spherical, but the integrations tend to be easier.
Update________
The equations of the spheres, and the circle where the two intersect.
$x^2 + y^2 + z^2 = 2\\
x^2 + y^2 + (z-1)^2 = 1\\
z= 1, x^2 + y^2 = 1$
Substitute our new coordinate system.
$\rho^2 \sin^2\phi\cos^2\theta + \rho^2 \sin^2\phi\sin^2\theta + \rho^2 \cos^2\phi = 2\\
\rho^2 \sin^2\phi\cos^2\theta + \rho^2 \sin^2\phi\sin^2\theta + \rho^2 \cos^2\phi - 2\rho\cos\phi + 1 = 1\\
\rho\cos\phi= 1, \rho^2 \sin^2\phi\cos^2\theta + \rho^2 \sin^2\phi\sin^2\theta = 1$
Simplify:
$\rho^2 = 2\\
\rho^2 - 2\rho\cos\phi = 0\\
\rho\cos\phi= 1, \rho^2 \sin^2\phi = 1$
$\rho = \sqrt 2\\
\rho = 2\cos\phi\\
\frac {\rho\sin \phi}{\rho\cos \phi} = \tan \phi= 1\\
\phi = \frac {\pi}{4}$
Now, it may help to sketch these curves. Do it in 2D. Just look at the picture in the xz plane, and that should give you some idea of what is going on for $\rho$ and $\phi.$
\update______
$\int_0^{2\pi}\int_0^{\frac\pi4}\int_{\sqrt 2}^{2\cos\phi} \rho^2 \sin\phi \;d\rho\;d\phi\;d\theta$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How can I find the number of whole number solutions for $ x_1 + x_2 + x_3 + x_4 = 11 $? Find the number of whole number solutions for this equation
$ x_1 + x_2 + x_3 + x_4 = 11 $
where
$ x_1 \ge -2, x_2 \ge -1, x_3 \ge 0, x_4 \ge 5 $
| *$ x_1 + x_2 + x_3 + ...+x_k = n \to \left(\begin{array}{c}n+k-1\\ k-1\end{array}\right) $
where
$ x_1 \ge -2, x_2 \ge -1, x_3 \ge 0, x_4 \ge 5 $
$$ x_1 \ge -2 \to y_1=x_1 +2\ge 0\\
x_2 \ge -1 \to y_2=x_2 +1\ge 0\\
x_3 \ge 0\\
x_4 \ge 5 \to y_4=x_4 -5\ge 0$$
$$\color{red} {x_1+x_2+x_3+x_4=11\\(x_1+2)+(x_2+1)+(x_3)+(x_4-5)=11+2+1-5=9\\y_1+y_2+x_3+y_4=9} \to \left(\begin{array}{c}9+4-1\\ 4-1\end{array}\right)=\\\left(\begin{array}{c}12\\ 3\end{array}\right)=\frac{12.11.10}{3.2.1}=220$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Second partial derivative with $f(x,y)=x^3+5x^2y+y^3$ I have this problem:
I found $F_x=3x^2+10yx$ and $F_y=5x^2+3y^2$, then $D_uf=\frac{9}{5}x^2+6yx+4x^2$
$F_{x2}= \frac{58}{5}x + 6y, F_{y2}=6x+{24}{5}y$
$D_uf_2=\frac{174}{25}x+\frac{18}{5}y+\frac{32}{5}x+\frac{96}{25}y=>at(2,1)=>534/25$
The answer however is $\frac{774}{25}$
| Your mistake was at the first derivative, you forgot to add $\frac{12}5y^2$. See this full calculation:
$$f(x,y) = x^3+5x^2y+y^3$$
$$\nabla f = \left(3x^2+10xy \atop 5x^2+3y^2 \right) $$
$$\nabla f \cdot u = \left(3x^2+10xy \atop 5x^2+3y^2 \right)\cdot \left(\frac 3 5 \atop \frac 4 5\right) =\frac {29} 5x^2+6xy+ \frac{12}5y^2 $$
$$\nabla \left(\frac {29} 5x^2+6xy+ \frac{12}5y^2\right)\cdot u = \left( \frac{58}5x+6y\atop6x+\frac{24}5 y\right) \cdot \left(\frac 3 5 \atop \frac 4 5\right) $$
$$ = \frac{174}{25}x+\frac{18}{5}y+\frac{24}5x + \frac{96}{25}y$$
$$ = \frac{294}{25}x + \frac{186}{25}y
$$
Evaluating at $(2,1)$ gives:
$$\left(\frac{294}{25}x + \frac{186}{25}y\right)\left(2,1\right) =\frac{588}{25}+\frac{186}{25}=\frac{775}{25}$$
| {
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Why $(x-2)^2$ cannot be divided by $(x-2)(x-3)$ I know the question may seem ridiculous. My question is:
Why can't $(x-2)^2$ be divided by $(x-2)(x-3)$?
I know the answer is obvious if we do the division by hand. However, we also know that if remainder of $\frac{P(x)}{Q(x)}$ is zero, then $Q(x)$ is a divisor of $P(x)$.
Notice that:
$$(x-2)^2 = (x-3)(x-2)Q(x) + R(x).$$ If x =2 then
$$(2-2)^2 = (2-3)(2-2)Q(x) + R(x) \to R(x) = 0,$$ so $(x-2)^2$ can be divided by $(x-2)(x-3)$.
This happens because $R(x) = ax +b$, and it isn't a single number, like $b$. Is there is another reason for it?
Thanks.
| Let us have a polynomial $f(X)$. I claim that $X-a$ divides $f$ if and only if $a$ is a root of $f$.
From Euclidean division, we get that $f(X) = q(X)(X-a) + r(X)$, where $\deg r <\deg (X-a) = 1$ or $r = 0$.
Assume that $X-a$ divides $f$. Then $r = 0$, and $f(a)=q(a)(a-a) = 0$.
Now assume that $a$ is root of $f$. Then $0 = f(a) = q(a)(a-a)+r = r$, so $X-a$ divides $f$.
Now, $X-3$ does not divide $(X-2)^2$ since $(3-2)^2 \neq 0$, so $(X-3)(X-1)$ can't divide it either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Formula for monic quadratic polynomials over $\mathbb{Z}$ Let $p(x)$ be a monic quadratic polynomial over $\mathbb{Z}$. Show that,for any integer $n$, there exists an integer $k$ such that
$(p(n))(p(n+1))=p(k)$
| Let P(x)= x^2 + bx + c
Therefore, P(n)= n^2 + bn + c, for an arbitrary integer n
P(n+1) = (n+1)^2 + b(n+1) + c
= n^2 + 2n + 1 + bn + b + c
= P(n) + 2n + b + 1
Therefore, P(n)P(n+1) = P(n)*{P(n) + 2n + b + 1}
= {P(n)}^2 + 2nP(n) + bP(n) + P(n)
= {P(n)}^2 + 2nP(n) + n^2 + bn + bP(n) +c
= (P(n) + n}^2 + b{P(n) + n} + c
= k^2 + bk + c , where k = P(n) + n, (i.e) an integer,
= P(k)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Determine the elements of a triangle knowing relationships between lengths and angles A triangle's sides' lengths are three successive numbers (meaning b=a+1, c=a+2). The smallest angle is a half of the triangle's biggest angle.
Find the area of this triangle and its angles.
Solution:
Sides are 4, 5 and 6.
Area: $15\sqrt{7}/4$
Angles : $41° 24' 34"$ and $82°49'8"$
| Using the law of sines,
\begin{align}
\dfrac{\sin \theta}{a} &= \dfrac{\sin 2\theta}{a+2} \\
\dfrac{\sin \theta}{a} &= \dfrac{2 \sin(\theta) \cos(\theta)}{a+2} \\
\cos \theta &= \dfrac{a+2}{2a} \\
\end{align}
Using the law of cosines,
\begin{align}
a^2 &= (a+1)^2 +(a+2)^2 - 2(a+1)(a+2) \cos \theta \\
\cos \theta &= \dfrac{(a+1)^2 +(a+2)^2 - a^2}{2(a+1)(a+2)} \\
\cos \theta &= \dfrac{a^2+6a+5}{2(a+1)(a+2)} \\
\cos \theta &= \dfrac{(a+1)(a+5)}{2(a+1)(a+2)} \\
\cos \theta &= \dfrac{a+5}{2(a+2)} \\
\hline
\dfrac{a+2}{2a} &= \dfrac{a+5}{2(a+2)} \\
\dfrac{a+2}{a} &= \dfrac{a+5}{a+2} \\
1 + \dfrac 2a &= 1 + \dfrac{3}{a+2} \\
\dfrac 2a &= \dfrac{3}{a+2} \\
3a &= 2a + 4 \\
a &= 4
\end{align}
CHECKING
$a=4$
Smallest angle is $\theta$:
$\qquad \cos \theta = \dfrac{a+2}{2a} = \dfrac 34$
$\qquad \sin \theta = \dfrac{\sqrt 7}{4}$
$\qquad \theta \approx 41.41^\circ$
The 'middle' angle is $180^\circ - 3\theta$
$\qquad \cos(180^\circ - 3\theta) = -\cos 3\theta = \dfrac {9}{16}$
$\qquad \sin(180^\circ - 3\theta) = \sin 3\theta = \dfrac {5\sqrt 7}{16}$
$\qquad 180^\circ - 3\theta \approx 55.77^\circ$
Largest angle is $2 \theta$:
$\qquad \sin 2\theta
= 2 \; \sin \theta \; \cos \theta
= \dfrac{3 \sqrt 7}{8}$
$\qquad \cos 2\theta
= 2\cos^2 \theta - 1
= \dfrac 18$
$\qquad 2\theta \approx 82.82^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the values of $n$ such that $(1+\sqrt3i)^n$ is a real number
Find the values of $n$ such that $z^n=(1+\sqrt3i)^n$ is a real number.
My reasoning: The power will be real iff $\sin\arg z=0$. Since $\sin 0,\sin\pm\pi,\sin\pm2\pi,\dots=0$, $3\mid n$. Is it correct?
$$z=re^{ia}=2e^{i\pi/3}$$
$$z^n=2^n\left(\cos\frac{n\pi}3+i\sin\frac{n\pi}3\right)$$
$$\to\{n\mid n=3k, k\in\Bbb Z\}$$
| An alternate view for $a_{n} = (1+ \sqrt{3} i)^{n}$ is:
\begin{align}
a_{0} &= 1 \\
a_{1} &= 1 + \sqrt{3} i \\
a_{2} &= -2 + 2 \sqrt{3} i \\
a_{3} &= -8.
\end{align}
It is determined that $n \in {0, 3, \cdots}$. Consider $n \to 3n$ then
\begin{align}
a_{3n} &= (1 + \sqrt{3} i)^{3n} = [(1 + \sqrt{3} i)^{3}]^{n} = (-8)^n = (-1)^{n} \, 2^{3n}.
\end{align}
This yields the same result as that given by the proposer's solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
| $x+\frac 1x=1\\
(x+\frac 1x)^2=1\\
x^2 + 2 + (\frac 1x)^2 = 1\\
x^2 + (\frac 1x)^2 = -1\\
x^4 + (\frac 1x)^4 = -1\\$
$x^5+(\frac 1x)^5 = (x+\frac 1x)(x^4 - x^3(\frac 1x) + x^2(\frac 1x)^2 - x(\frac 1x)^3 + (\frac 1x)^4)\\
(x+\frac 1x)(x^4 + (\frac 1x)^4 - x^2 - (\frac 1x)^2 + 1 )\\
( 1 )((-1) - (-1) + 1) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I calculate the following trigonometric integrals with the methods of contour integrals?
How can I calculate the integrals $$\int_0^\pi \frac{dx}{a+b\sin(x)}$$
and $$\int_0^\pi \frac{dx}{a+b\cos(x)}$$
using contour integrals ?
I know the residue theorem , and my idea is to choose the line from $0$ to $\pi$ on the $x$-axis followed by the upper half-circle connecting the points $(\pi/0)$ and $(0/0)$.
But I do not know the singularities of the functions $f(z)=\frac{1}{a+b\sin(z)}$ and $g(z)=\frac{1}{a+b\cos(z)}$.
The other problem is that I do not know whether the integral over the half-circle can be calculated by hand.
Is there a way, or must I begin with a substitution, for example $t=\tan(\frac{x}{2})$ ?
| The integral over the sine is a bit tricky because the integration region cannot be extended to $2 \pi$, so if you want to used the residue theorem, a unit circle is out. Rather, we use a semicircle. Thus consider the complex integral
$$\oint_C \frac{dz}{b z^2+i 2 a z-b} $$
where $C$ is the unit semicircle in the upper half-plane along with the diameter along the real axis. Thus, the contour integral is equal to
$$\frac12 \int_0^{\pi} \frac{d\theta}{a+b \sin{\theta}} + \int_{-1}^1 \frac{dx}{b x^2+i 2 a x-b}$$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$. In this case, let's assume $a \gt b \gt 0$; the poles are at
$$z_{\pm}=-i \frac{a}{b} \pm i \sqrt{\frac{a^2}{b^2}-1} $$
Note that both poles are outside of the interior of the contour $C$; thus, the contour integral is zero. Accordingly,
$$\begin{align} \int_0^{\pi} \frac{d\theta}{a+b \sin{\theta}} &= -2 \int_{-1}^1 \frac{dx}{b x^2+i 2 a x-b} \\ &= -\frac{2}{b} \frac1{z_+-z_-} \int_{-1}^1 dx \left (\frac1{x-z_+} - \frac1{x-z_-} \right ) \\ &= - \frac{2}{i 2 \sqrt{a^2-b^2}} \left [\log{\left (\frac{1-z_+}{-1-z_+} \right )} - \log{\left (\frac{1-z_-}{-1-z_-} \right )} \right ]\\ &= - \frac{2}{i 2 \sqrt{a^2-b^2}} \left [\log{\left (\frac{1-z_+}{1+z_+} \right )} - \log{\left (\frac{1-z_-}{1+z_-} \right )} \right ]\\ &= - \frac{2}{i 2 \sqrt{a^2-b^2}} i 2 \left [ \arctan{\left ( \frac{a}{b} - \sqrt{\frac{a^2}{b^2}-1}\right )} - \arctan{\left ( \frac{a}{b} + \sqrt{\frac{a^2}{b^2}-1}\right )} \right ] \\ &= \frac{2}{\sqrt{a^2-b^2}} \arctan{\left (\frac{\sqrt{a^2-b^2}}{b} \right )} \end{align}$$
Things work out similarly when $b \lt a$. Accordingly, when $a \gt 0$ and $b \gt 0$,
$$\int_0^{\pi} \frac{d\theta}{a+b \sin{\theta}} = \begin{cases}\frac{2}{\sqrt{a^2-b^2}} \arctan{\left (\frac{\sqrt{a^2-b^2}}{b} \right )} & a \gt b \\ \frac{1}{\sqrt{b^2-a^2}} \log{\left (\frac{b+\sqrt{b^2-a^2}}{b-\sqrt{b^2-a^2}} \right )} & a \lt b \end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $a, b$ and $c$ from this system of linear equations? Find $a$, $b$ and $c$ from:
$$
\begin{align}
a+c &= 3 \\
b+a &= 2 \\
c+b &= -1 \\
\end{align}
$$
I tried the following way, the answer comes wrong:
$$(a+c) + (b+a) = 3+2$$
| $$(a+c)+(c+b)+(b+a)=3+2-1$$
$$2(a+b+c)=4$$
$$a+b+c=2$$
Then if $a+b+c=2$ and $a+c=3$ then $b=-1$
Then if $a+b+c=2$ and $a+b=2$ then $c=0$
Then if $a+b+c=2$ and $b+c=-1$ then $a=3$
| {
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"url": "https://math.stackexchange.com/questions/1977561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying logarithm of a product My textbook reads (without explanation, naturally):
$\log \prod_{y=0}^{9}{(1+\frac{1}{10x+y})} = \log{(1+\frac{1}{x})}$
Wondering how this was achieved...
| Notice that $1+\frac{1}{10x+y}=\frac{10x+y+1}{10x+y}$, so the product is: $$\prod_{y=0}^{9}{\left(1+\frac{1}{10x+y}\right)}=\frac{10x+1}{10x}\cdot\frac{10x+2}{10x+1}\cdot\frac{10x+3}{10x+2}\dots\frac{10x+10}{10x+9}.$$ Everything cancels out except the first denominator and the last numerator, giving $$\frac{10x+10}{10x}=\frac{x+1}{x}=1+\frac{1}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many integers $x$ satisfy: $-4How many integers are there in the set $\{x\in\Bbb R\mid-4<x^2+5x<14\}$?
The correct answer is four, but I only find $0$ and $1$ as integers in the set.
Is the answer wrong?
| This looks like a homework problem with the following intended solution:
Consider $-4<x^2+5x<14$ one inequality at a time.
(In each case, it may be helpful to draw a graph of the corresponding parabola.)
First, $x^2 + 5x > -4$ iff $(x+1)(x+4) = x^2 + 5x + 4 > 0$.
So, we have the points $x < -4$ or $x > -1$.
Second, $x^2 + 5x < 14$ iff $(x+7)(x-2) = x^2 + 5x - 14 < 0$.
So, we have the points $-7<x<2$.
Among integers, this means considering $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.
Within this set, and returning to the first inequality, which integers are less than $-4$?
Answer: $-6$ and $-5$.
Similarly, which integers are greater than $-1$?
Answer: $0$ and $1$.
Thus, the answer to your question is the four integers $\{-6, -5, 0, 1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What are the conditions on $a, b, c$ so that $x^3+ax^2+bx+c$ is bijective? I would like to find the conditions on $a$, $b$, $c$ so that function $$f(x)=x^3+ax^2+bx+c$$ is bijective.
I thought about resolving the equation
$$x^3+ax^2+bx+c=y$$
but I didn't succeed. And our math teacher told us that we cannot prove that a function is bijective by proving that this function is strictly increasing or decreasing.
Thanks for your help!
Marie
| Disclaimer: I started from a simple geometrical idea, but, unfortunately, it turned out messier than expected. This answer, therefore, just serves the purpose of showing that this can be done. Other answers are by far more elegant and are recommended over this one.
Let $f(x) = x^3+ax^2 + bx + c$. Since $y\mapsto y + c$ is bijective, we can assume that $c = 0$.
As others already noted, surjectivity is automatic, so we only need to deal with injectivity.
I will appeal to geometry at this point:
$f$ is not injective if and only if there is a line parallel to
$x$-axis that crosses the graph of $f$ at more than one point.
This tells us that:
$f$ is not injective if and only if there exists $y\in\Bbb R$ such
that $f(x) - y$ has at least two different real roots.
Since non-real roots of polynomials over $\Bbb R$ come in conjugate pairs,
$f$ is not injective if and only if there exists $y\in\Bbb R$ such
that $f(x) - y = (x-A)(x-B)(x-C)$ where $A,B,C$ are real
numbers, not all equal.
Expanding we get $$x^3 + ax^2 + bx - y = x^3 -(A+B+C)x +(AB+BC+AC)x - ABC$$ so the question boils down to finding real $A,B,C$ such that \begin{array}{c c}\begin{align}-(A+ B+C)&=a\\ AB+BC+AC &= b\end{align} & \tag{1} \end{array} and then we set $y = ABC$.
We can now conclude that
$f$ is injective if and only if $(1)$ has no real solutions or all the real solutions satisfy $A=B=C$.
Solving system $(1)$ in terms of $A$, we get $$B = \frac 1 2 (-a - A \pm \sqrt{- 3 A^2 - 2 a A +a^2 - 4 b})\\ C = -a - A - B\tag{2}$$
We can easily see that $(1)$ has no real solutions if and only if $B$ and $C$ from $(2)$ are not real for any real $A$, and that happens if and only if $- 3 A^2 - 2 a A +a^2 - 4 b<0,\ \forall A\in \Bbb R$.
From the discriminant of $- 3 A^2 - 2 a A +a^2 - 4 b$ as a polynomial in $A$, we finally get condition that $(1)$ has no real roots if and only if $a^2 < 3b$.
Assume now that there is a real solution to $(1)$ such that $A = B = C$. Substituting in $(1)$ we get necessary condition $A = B = C = -a/3$ and $a^2 = 3b$.
Now we need to show that $a^2 = 3b$ implies that all the real solutions to $(1)$ satisfy $A = B = C$. But this is almost immediate: we must have $$- 3 A^2 - 2 a A +a^2 - 4 b\geq 0$$ but the discriminant of $- 3 A^2 - 2 a A +a^2 - 4 b$ is $0$ (because $a^2 = 3b$), so we can conclude that $A = -a/3$ is the only real that satisfies this, and also we get that in that case $A = B = C$. Thus, $a^2 = 3b$ implies that $A=B=C=-a/3$ is the unique real solution to $(1)$.
Collecting the pieces, we get that $f$ is injective if and only if $a^2 \leq 3b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of the fraction
What common fraction is equivalent to $$\frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \text{ ?}$$
I didn't see how to relate the numerator to the denominator. We can't find the value of the numerator and denominator easily it seems, so what else should we try?
| \begin{align}
& \frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \\[12pt] = {} & \frac{ \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right) - \left( \dfrac 1 {2^3} + \dfrac 1 {4^3} + \dfrac 1 {6^3} + \cdots \right)}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots} \\[12pt]
= {} & \frac{ \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right) - \dfrac 1 {2^3} \cdot \left( \dfrac 1 {1^3} + \dfrac 1 {2^3} + \dfrac 1 {3^3} + \cdots \right)}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots} \\[12pt]
= {} & \frac{ \left( 1 - \dfrac 1 {2^3} \right) \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right)}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots} \\[12pt]
= {} & 1 - \dfrac 1 {2^3}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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A line L is tangent to two curves $x^2 + 1$ and $-2x^2 - 3$, find L's equation Some things I'd like to clarify beforehand:
From my understanding, for this situation to occur these conditions need to be fulfilled:
1) $f(a) = g(a)$
2) $f'(a) = g'(a)$
However, both of these conditions are not fulfilled.
What I first did:
$a^2 + 1 = -2a^2 - 3$
I get $3a^2 + 4 = 0$.
Am I on the right track here? The thing is my class doesn't allow calculators, which is why I believe I'm wrong as the equation would get too complicated following this logic.
Thanks.
| $f(x) = x^2 +1\\
g(x) = -2x^2 - 3$
both are tangent to the line $L: y = mx+b$
If the curve is tangent, when we subtract one from the other, we have a root of multiplicity. The discriminant in the quadratic formula is zero.
$f(x) = x^2 -mx +1-b\\
m^2 - 4(1-b) = 0$
$g(x) = -2x^2 -mx -3-b\\
m^2 - (-2)(-3-b)(4) = 0$
$4-4b = 24 + 8b\\
12b = -20\\
b = -\frac {5}{3}\\
m = \pm \sqrt {\frac {32}3}$
There are two lines:
$y_1 = 4\sqrt {\frac 23} x - \frac {5}{3}\\
y_2 = -4\sqrt {\frac 23} x - \frac {5}{3}$
Now it is probably wroth checking that each line is tangent to each curve
$y_1$ is tangent to $f(x)$ at $2\sqrt{\frac 23}$ and $g(x)$ at $2\sqrt{\frac 23}$
At the points of tangency $f(x) = y(x)$ and $f'(x) = y'(x).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of sequence formula I need help solving the following exercise:
The sequence $(a_n)_{n\in \mathbb N}$ is given by
$$a_0 = a_1 = 1 \quad \text{and} \quad a_n=2a_{n-1}+4a_{n-2} \quad \forall n \geq 2$$
Proof the explicit formula
$$a_n = \frac{1}{2}((1+\sqrt{5})^n + (1-\sqrt{5})^n) \quad \forall n\in \mathbb N$$
I think the way to go is by induction over n, however I didn't really know what do to after I plugged in everything. I couldn't get $n+1$ into the exponent. Can you guys give me a hint?
| The method presented by MrYouMath is called characteristic polynomial.
Another method is by using generating functions. E.g. $$f(x)=\sum_{n=0}^{\infty } a_nx^n=a_0+a_1x+\sum_{n=2}^{\infty } (2a_{n-1}+4a_{n-2})x^n=$$
$$a_0+a_1x+\sum_{n=2}^{\infty } 2a_{n-1}x^n + \sum_{n=2}^{\infty } 4a_{n-2}x^n=$$
$$a_0+a_1x+2x\sum_{n=1}^{\infty } a_{n}x^{n}+4x^2\sum_{n=0}^{\infty }a_{n}x^{n}=$$
$$a_0+a_1x-2xa_0+2xf(x)+4x^2f(x)=1-x+2xf(x)+4x^2f(x)$$
or $$f(x)=\frac{x-1}{4x^2+2x-1}=\frac{x-1}{4\left(x+\frac{1+\sqrt{5}}{4}\right)\left(x+\frac{1-\sqrt{5}}{4}\right)}=$$
$$\frac{1}{4(a-b)}\left( \frac{1-4b}{1-4xb} - \frac{1-4a}{1-4xa} \right)=\frac{1}{4(a-b)}\left( (1-4b)\sum_{n=0}^{\infty }4^nb^nx^n - (1-4a)\sum_{n=0}^{\infty }4^na^nx^n\right)$$
where $a=\frac{1+\sqrt{5}}{4}$ and $b=\frac{1-\sqrt{5}}{4}$. Then
$$a_n=\frac{1}{4(a-b)}\left((1-4b)4^nb^n -(1-4a)4^na^n \right)=$$
$$\frac{1}{2\sqrt{5}}\left( (1-1+\sqrt{5})(1-\sqrt{5})^n -(1 -1 -\sqrt{5})(1+\sqrt{5})^n \right)=\frac{1}{2}\left( (1+\sqrt{5})^n + (1-\sqrt{5})^n \right)$$
Here is another example for Fibonacci sequence.
| {
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Find the radius of convergence and convergence interval Find the radius of convergence and convergence interval
$$\sum^\infty_{n=1}\frac{n^5\;(x+8)^n}{9^n\; n^\frac{17}{3}}$$
My attempt:
I'm computing $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|$ and
I got the answer $|9(x+8)|$. So the radius of convergence $R=\frac{1}{9}$ and interval =$(\frac{-73}{9},\frac{-71}{9})\,$.
Am I right ?
| The series is a power series around $x = -8$, that is, it is of the form
$$ \sum_{n=1}^{\infty} a_n (x - (-8))^n $$
where $a_n = \frac{n^5}{9^n n^{\frac{17}{3}}} = \frac{1}{9^n n^{\frac{2}{3}}}$. The radius of convergence is given by $R = \lim_{n \to \infty} \frac{a_n}{a_{n+1}}$ (if it exists) and not by $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$! In our case, we have
$$ \frac{a_{n+1}}{a_n} = \frac{9^n n^{\frac{2}{3}}}{9^{n+1}(n+1)^{\frac{2}{3}}} = \frac{1}{9} \left( \frac{n}{n+1} \right)^{\frac{2}{3}} \to \frac{1}{9} $$
and so the radius of convergence is $R = 9$ and you are guaranteed that the series will converge in $(-17, 1)$. Now you also need to check the end points of the interval to determine the full domain of convergence.
At $x = 1$ we get the series
$$ \sum_{n=1}^{\infty} \frac{1}{n^{\frac{2}{3}}} $$
which diverges by the $p$ criterion.
At $x = -17$ we get the series
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\frac{2}{3}}} $$
which converges using the alternate series test. Thus, the full interval of convergence is $[ -17, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
I want to find a pair of integers $X$, $Y$ which satisfy $X^2 - 2Y^2=1$ such that $X > Y > 50$
Find a pair of integers $X, Y$, which satisfy $X^2 - 2Y^2 = 1$, such that $X > Y > 50$.
I have started by finding a pair of much smaller integers that work: $X(1) = 3$ and $Y(1) = 2$.
When I looked up a solution it was as follows:
$$\begin{array}
&X(2) = 3 × 3 + 4 × 2 = 17, & Y(2) = 2 × 3 + 3 × 2 = 12 \\
X(3) = 3 × 17 + 4 × 12 = 99, & Y(3) = 2 × 17 + 3 × 12 = 70
\end{array}$$
So $X = 99$ and $Y = 70$ is such a pair.
But what is the method used? I tried searching for similar questions, but didn't find an answer to this.
| $$X_{i+1}^2-2Y_{i+1}^2 = (3X_i + 4Y_i)^2 - 2(2X_i + 3Y_i)^2 =$$ $$ = (9-2\cdot4)X_i ^ 2 + (12-12)X_i \cdot Y_i + (16-2\cdot9)Y_i ^ 2= $$ $$= X_{i}^2-2Y_{i}^2 = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
${\sqrt{2x+1}=1+\sqrt{x}}$ — I dont know if the solution is correct. Help?
${\sqrt{2x+1}=1+\sqrt{x}}$
${2x+1=1+2\sqrt{x}+x}$
${x=2\sqrt{x}}$
${x*\frac{1}{x^{1/2}}=2}$
${\sqrt{x}=2}$
${x=4}$
| You are mostly correct but there are two conditions you didn't take into account which you should have.
$\sqrt{2x + 1}=1+\sqrt{x}$ Note, this implies $2x + 1 \ge 0$ i.e $x \ge -\frac 12$.
$2x +1 = 1 + 2\sqrt{x} + x$ Now, we have "lost" that assumption. It is possible that we will end up with some extraneous answers where $x < -\frac 12$. As it turns out that isn't an issue and it doesn't happen but it could have. (As we still have $\sqrt{x}$ that implies $x \ge 0$ so $x < -\frac 12$ is impossible).
$x = 2\sqrt{x}$
$x/x^{1/2} = 2$ Here you divided by $x^{1/2}$ in the assumption $x^{1/2} \ne 0$. You can not make that assumption. You must consider the possibility that $x^{1/2}$.
So say: Case 1: If $\sqrt{x} = 0$, then $x= 0$ and we have $0 = 2\sqrt{0}$ which consistent so $x = 0$ is a possible answer.
But if $\sqrt{x} \ne 0$ then
$x/x^{1/2} = 2$
$x^{1/2} = 2$
$x =4$
So $x=4$ is the only other possible solution . So $x = 0$ or $x = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\
&= u \ln\sin(u) - \int \ln\sin(u) du
\tag{1}
\label{eq:20161030-1}
\end{align}
\begin{align}
\int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} - \mathrm{e}^{-iu}}{i2} \right) du \\
&= \int \ln\left(\mathrm{e}^{iu} - \mathrm{e}^{-iu} \right) du \,- \int \ln(i2) du \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \int \ln\mathrm{e}^{iu} du \,-\, u\ln(i2) \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln2 \,-\, ui\frac{\pi}{2}
\tag{2}
\label{eq:20161030-2}
\end{align}
To evaluate the integral above, let $y=\mathrm{e}^{-i2u}$
\begin{equation}
\int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy
= -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u}
\tag{3}
\label{eq:20161030-3}
\end{equation}
Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to (x), and apply limits,
\begin{align}
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx
&= \sin^{-1}(x)\ln(x) + \sin^{-1}(x)\left(\ln2 + i\frac{\pi}{2}\right) \\
&- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_0^1 \\
&= \frac{\pi}{2}\ln2
\end{align}
I would be interested in seeing other solutions.
| $$\int_{0}^{1} \frac{\sin^{-1}x}{x} dx
= \int_{0}^{1} \int_{0}^{1} \frac{\sqrt{1-x^2}}{1-x^2+x^2y^2} dy\ dx \overset{x=\sin t}
=\int_0^1 \frac{\pi}{2(1+y)}dy=\frac{\pi}{2}\ln2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
} |
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.
$0^2 + 0 + 4 = 4 \\
1^2 + 1 + 4 = 6 \\
2^2 + 2 + 4 = 10 \\
3^2 + 3 + 4 = 5 \\
4^2 + 4 + 4 = 2 \\
5^2 + 5 + 4 = 1 \\
6^2 + 6 + 4 = 2 \\
7^2 + 7 + 4 = 5 \\
8^2 + 8 + 4 = 10 \\
9^2 + 9 + 4 = 6 \\
10^2 + 10 + 4 = 4 $
But this process is very long one. I am seeking a short method by which I can solve such problem.
Now suppose that it is reducible, then the factors should be linear. So how can we solve the problem based on this kind of observation? Any help will be great. Thanks.
| ${\rm mod}\ 11\!:\,\ 0\equiv x^2\!+\!x\!+\!4\equiv x^2\!+\!12x\!+\!4\equiv (x\!+\!6)^2\!+1\,\Rightarrow\,(x\!+\!6)^2\equiv -1.$
But $\,-1\,$ is not a square, else by little Fermat: $\left[ a^2\equiv -1\right]^5\Rightarrow\, 1\equiv -1.$
Remark $ $ We implicitly applied Euler's Criterion to show $-1$ is not a square $\!\bmod 11.\,$ More generally we can use Legendre or Jacobi symbols or the generalized Euler Criterion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solutions to $a^2+b^2+(ab)^2=c^2$ In a comment to the question $n^2+(n+1)^2+n^2\cdot(n+1)^2$ is a perfect square it is proved that:
$(1)\quad b=a+1$ give integer solutions to
$a^2+b^2+(ab)^2=c^2$ for all $a\in\mathbb N$.
In the answers to the question $\forall m\in\mathbb N\exists n>m+1\exists N\in\mathbb N:m^2+n^2+(mn)^2=N^2$ it is proved that:
$(2)\quad b=2a^2$ give integer soulutions for all $a\in\mathbb N$.
Conjecture:
For each $a\in\mathbb N$ there are integer solutions to
$a^2+b^2+(ab)^2=c^2$ that are neither of the types $(1)$ or $(2)$
above.
Prove the conjecture or give a counter-example. It is tested for all $a<1000$.
| The Pell type equation $$ x^2 - (1 + A^2) y^2 = A^2 $$ has an infinite set of solutions $(x,y).$ The first three predictable types are
$$
\left(
\begin{array}{c}
A \\
0
\end{array}
\right),
$$
$$
\left(
\begin{array}{c}
A^2 - A + 1 \\
A - 1
\end{array}
\right),
$$
$$
\left(
\begin{array}{c}
A^2 + A + 1 \\
A + 1
\end{array}
\right).
$$
After that, an infinite sequence of $(x,y)$ as column vectors may be found by multiplying by the generator of the (oriented) automorphism group of the binary quadratic form $ x^2 - (1 + A^2) y^2. $ That matrix is
$$ M =
\left(
\begin{array}{cc}
2A^2 + 1 & 2 A^3 + 2 A \\
2A & 2 A^2 + 1
\end{array}
\right).
$$
The answer you write as $b = 2 a^2$ comes up as
$$ M =
\left(
\begin{array}{cc}
2A^2 + 1 & 2 A^3 + 2 A \\
2A & 2 A^2 + 1
\end{array}
\right)
\left(
\begin{array}{c}
A \\
0
\end{array}
\right) =
\left(
\begin{array}{c}
2A^3 + A \\
2 A^2
\end{array}
\right).
$$
The bad news is that, for a fixed $A,$ there may be others. These others show up as later entries in the Pell sequences for smaller values of $A.$ I do not have a genuinely sensible two-dimensional description for all answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Congruence involving prime numbers Given the function $f(x)= x^{\frac{p(p-1)}{2}}-1.$ Also let $p$ be an odd prime number. If $\epsilon$ is a number $\pmod{p}$ such that
$f(\epsilon) \equiv 0 \pmod{p}$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(\epsilon)\equiv 0 \pmod{p^2}$$
| Let $f(x) = x^{p(p-1)/2} - 1$. Suppose we have an integer $a$ such that $f(a) \equiv 0 \pmod{p}$.
First, suppose $p \mid a$. Then, $f(a) = a^{p(p - 1)/2} - 1 \equiv -1 \pmod{p}$, a contradiction.
Then, suppose $p \nmid a$. By Euler's Theorem, $a^{\phi(p^2)} = a^{p(p - 1)} \equiv 1 \pmod{p^2}$. Treating this as a quadratic congruence $(a^{p(p-1)/2})^2 \equiv 1 \pmod{p^2}$, we see that $a^{p(p - 1)/2} \equiv \pm 1 \pmod{p^2}$.
If $a^{p(p - 1)/2} \equiv -1 \pmod{p^2 }$, we imply $a^{p(p - 1)/2} \equiv -1 \pmod{p}$, so $a^{p(p - 1)/2} - 1 = f(a) \equiv -2 \not \equiv 0 \pmod{p}$, a contradiction. Thus, we conclude that $a^{p(p - 1)/2} \equiv 1 \pmod{p^2}$, or $f(a) \equiv 0 \pmod{p^2}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Combinatoric Identities Ho to prove the following two identities?
I cannot see the trick:
(a) $\binom{-x}{k}=(-1)^k\binom{x+k-1}{k}$
(b) $\binom{k+x}{2k+1}=-\binom{k-x}{2k+1}$
| Note that $(2k+1)-(k+x)-1=k-x$, the upper number of $\binom{k-x}{2k+1}$. Thus, you can negate the upper index:
$$\binom{k+x}{2k+1}=(-1)^{2k+1}\binom{(2k+1)-(k+x)-1}{2k+1}\;,$$
Therefore:
${k+x \choose 2k + 1}=-{k-x \choose 2k + 1} $
$=(-1)^{2k+1}\binom{(2k+1)-(k+x)-1}{2k+1}\;$
$=(-1)^{2k+1}\binom{2k-(k+x)}{2k+1}$
$=(-1)^{2k+1}\frac{(2k-(k+x)).(2k-(k+x) -1).(2k-(k+x)-2) ... (2k-(k+x)-2k) }{(2k+1)!}$
$=\frac{-1(2k-(k+x)).-1(2k-(k+x) -1).-1(2k-(k+x)-2) ... -1(2k-(k+x)-2k) }{(2k+1)!}$
$=\frac{1}{(2k+1)!}.((k+x)-2k).((k+x) -2k +1).((k+x)-2k +2) ... ((k+x)-2k+2k)$
$=\frac{1}{(2k+1)!}.(\frac{(k+x)!}{((k+x)-2k-1)!})$
$=\frac{1}{(2k+1)!}.(\frac{(k+x)!}{((k+x)-(2k +1))!})$
$=\frac{(k+x)!}{(2k+1)! . ((k+x)-(2k +1))!}$
$=\binom{k+x}{2k+1}$
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$
$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$
Now I do the nested radicals formula and I get $254\sqrt{41}$ which is none of those where did I mistaked?
| Considering all positive values, $45\pm4\sqrt{41}$ can be written as $45\pm4\sqrt{41}=(\sqrt{41}\pm2)^2$, thus, the given expression can be simplified as follows: $$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}=\left(\sqrt{45+4\sqrt{41}}\right)^3-\left(\sqrt{45-4\sqrt{41}}\right)^3=\left(\sqrt{\left(\sqrt{41}+2\right)^2}\right)^3-\left(\sqrt{\left(\sqrt{41}-2\right)^2}\right)^3=\left(\sqrt{41}+2\right)^3-\left(\sqrt{41}-2\right)^3$$
Now recall, $a^3-b^3=(a-b)(a^2+ab+b^2)$. Thus, $$\left(\sqrt{41}+2\right)^3-\left(\sqrt{41}-2\right)^3=\left((\sqrt{41}+2)-(\sqrt{41}-2)\right)\left((\sqrt{41}+2)^2+(\sqrt{41}+2)(\sqrt{41}-2)+(\sqrt{41}-2)^2\right)=\left(\sqrt{41}+2-\sqrt{41}+2\right)\left(45+4\sqrt{41}+41-4+45-4\sqrt{41}\right)=4(45+41-4+45)=508$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
| $z=\tan(\theta/2)$, $\tan(\theta)=\frac{2z}{1-z^2}$, $\mathrm{d}\theta=\frac{2\,\mathrm{d}z}{1+z^2}$, $\cos(\theta)=\frac{1-z^2}{1+z^2}$, $\sin(\theta)=\frac{2z}{1+z^2}$
$z=\frac{\sec(\theta)-1}{\tan(\theta)}=\frac{-1+\sqrt{x^2+2x+2}}{x+1}$
$$
\begin{align}
\int\frac{\sqrt{x^2+2x+2}}{x}\,\mathrm{d}x
&=\int\frac{\sec^3(\theta)}{\tan(\theta)-1}\,\mathrm{d}\theta\\
&=\int\left(\frac{1+z^2}{1-z^2}\right)^3\frac{1-z^2}{z^2+2z-1}\frac{2\,\mathrm{d}z}{1+z^2}\\
&=\int\left(\frac{1+z^2}{1-z^2}\right)^2\frac{2\,\mathrm{d}z}{z^2+2z-1}\\
&=\int\frac{2z^4+4z^2+2}{(z+1+\sqrt2)(z+1-\sqrt2)(z-1)^2(z+1)^2}\,\mathrm{d}z\\
&=\int\left(\frac1{(z-1)^2}-\frac1{z-1}-\frac1{(z+1)^2}+\frac1{z+1}+\frac{\sqrt2}{z+1-\sqrt2}-\frac{\sqrt2}{z+1+\sqrt2}\right)\mathrm{d}z\\
&=-\frac2{z^2-1}+\log\left(\frac{z+1}{z-1}\right)+
\sqrt2\log\left(\frac{z+1-\sqrt2}{z+1+\sqrt2}\right)+C
\end{align}
$$
Now back substitute for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to find the value of $\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$ How to find the exact value of the following limit?
$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$
| $$\cos x^2=1-\frac{x^4}{2}+o(x^4)$$
so
$$e^{x\cos x^2} = e^{x-\frac{x^5}{2}+o(x^5)} = e^xe^{-\frac{x^5}{2}+o(x^5)}$$
therefore
$$\frac{e^{x\cos x^2}-e^x}{x^5} = e^x\frac{e^{-\frac{x^5}{2}+o(x^5)}-1}{x^5} \sim e^x\frac{-\frac{x^5}{2}}{x^5} \xrightarrow[x\to0]{}-\frac{1}{2}$$
using $e^u-1\sim u$ when $u\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}$ We have
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{2^{7/3}}+\frac{\sqrt{3}}{2^{4/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag1$$
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-2}=\frac{3}{2}-\frac{\ln(\sqrt[3]{2}-1)}{2^{5/3}}+\frac{\sqrt{3}}{2^{2/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag2$$
with the first discussed in this post. (Update) Courtesy of the answers below, we also have,
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}=\frac{4}{3}+\frac{\ln\big((4^{1/4}-3^{1/4})\sqrt{2+\sqrt{3}}\big)}{3^{5/4}\sqrt{2}}+\frac{\sqrt{2}}{3^{5/4}}\arctan\big(3^{1/4}\sqrt{2+\sqrt{3}}\big)\tag3$$
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-3}=\frac{4}{3}-\frac{\ln\big((4^{1/4}-3^{1/4})\sqrt{2+\sqrt{3}}\big)}{3^{3/4}\sqrt{2}}+\frac{\sqrt{2}}{3^{3/4}}\arctan\big(3^{1/4}\sqrt{2+\sqrt{3}}\big)\tag4$$Walpha gives a closed-form for $(3)$, but uses two log functions and two arctans. The two answers below show it can be simplified further similar to $(1),(2)$.
| Yes. The logarithms have the same denominator. So factoring this out gives a simple difference of logs: namely the simple log of a quotient, which itself can be simplified to $$2\ln\left(\frac12+\frac{\surd3}2-\sqrt{2\surd3}\right)$$(if my calculation is right). A similar method will simplify the inverse cotangents to a single inverse cotangent, by use of the formula $$\alpha+\beta=\operatorname{arcot}\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}$$(exactly or to within $\pm\pi$).
(Revised) According to my (corrected) calculation, Wolfram's expression for formula 3 can be written as $$\frac43+\frac4{3a}\arctan \frac{2a+a^3}4-\frac2{3a}\ln\frac{2+2a+a^2}4,$$where $a:=\sqrt{2\surd3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$ - my answer is wrong (why?) Need to find
$$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$$
One thing I use is
$$\lim_{x \to 1}\frac{x^{1/m}-1}{x^{1/n} - 1} = n/m$$
$$ \lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}} = 3\lim_{x \to 1}\frac{x^{1/2} - x^{1/3}}{(1-x^{1/2})((1-x^{1/3})} = 3\lim_{x \to 1}\frac{x^{2/3} (x^{1/6}-1)^2 - 2x^{1/3}+2x^{5/12}}{(1-x^{1/2})((1-x^{1/3})}= 3\lim_{x \to 1}\frac{x^{1/3} (x^{1/6}-1)^2 }{(1-x^{1/2})((1-x^{1/3})} + \frac{2x^{8/12}(x^{1/12} - 1)^2+4x^{5/24}-4x^{4/24}}{(1-x^{1/2})((1-x^{1/3})} + ... = 3*(1/6 + 2*(3/12*2/12) + 4*(3/24*2/24)... = 3(1/6 + 1/12 + 1/24+...) = 3(\frac{1/6}{1-1/2}) = 1 $$
Answer in my book is 1/2.
Important note I'm in the part of book where integration and differntiation is not covered.
| It's likely there was a typo, and the intended problem was
$$\lim_{x\to1}\left({3\over1-x^{1/2}}-{2\over1-x^{1/3}} \right)$$
In general, if you are taking a limit of ${A\over1-x^{1/2}}-{B\over1-x^{1/3}}$, Abdallah Hammam's trick of letting $x=t^6$ is a good way to go:
$${A\over1-x^{1/2}}-{B\over1-x^{1/3}}={A\over1-t^3}-{B\over1-t^2}={1\over1-t}\left({A\over1+t+t^2}-{B\over1+t} \right)={(A-B)+(A-B)t-Bt^2\over(1-t)(1+t+t^2)(1+t)}$$
Since there is a $1-t$ in the denominator, you need for the numerator to go to $0$ as $t$ goes to $1$ in order for the limit to exist. That is, you need
$$(A-B)+(A-B)-B=0$$
which is to say, you need $2A=3B$ in order to have a limit. If, indeed, the problem meant to have $A=3$ and $B=2$ (instead of $A=B=3$), the last line is
$${1+t-2t^2\over(1-t)(1+t+t^2)(1+t)}={(1-t)(1+2t)\over(1-t)(1+t+t^2)(1+t)}={1+2t\over(1+t+t^2)(1+t)}$$
which tends to $3/(3\cdot2)=1/2$ as $t\to1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to estimate this sum? I encounter a problem and I just can't figure out a useful way to estimate it.I have tried taylor expansion but it doesn't work.
I want to know the big O-estimate of $n$ in this formula.
$$\sum_{k=2}^{n}\frac{\cos\frac{\pi}{n+1}}{\cos\frac{\pi}{n+1}-\cos\frac{k\pi}{n+1}}$$
| Since $\cos(x)=1-x^2/2+O\!\left(x^4\right)$
$$
\begin{align}
\frac{\cos\left(\frac\pi{n+1}\right)}{\cos\left(\frac\pi{n+1}\right)-\cos\left(\frac{k\pi}{n+1}\right)}
&=\frac{1-\frac{\pi^2}{2(n+1)^2}+O\left(\frac1{n^4}\right)}{\left(1-\frac{\pi^2}{2(n+1)^2}\right)-\left(1-\frac{k^2\pi^2}{2(n+1)^2}\right)+O\!\left(\frac{k^4}{n^4}\right)}\\
&=\frac{2(n+1)^2}{\left(k^2-1\right)\pi^2}+O\!\left(1\right)\\[4pt]
&=\frac{2n^2}{\left(k^2-1\right)\pi^2}+O\!\left(1+\frac{n}{k^2}\!\right)
\end{align}
$$
Therefore
$$
\sum_{k=2}^n\frac{\cos\left(\frac\pi{n+1}\right)}{\cos\left(\frac\pi{n+1}\right)-\cos\left(\frac{k\pi}{n+1}\right)}=\frac{3n^2}{2\pi^2}+O(n)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find x and y coordinates based on the given distance? The problem says: Find the point (coordinates $(x,y)=~?$) which is symmetrical to the point $(4,-2) $ considering the given equation $y=2x-3$
I have found the perpendicular line-slope $y=-~\frac{1}{2}x$ and the
intersection point which is shown in the graph $\left(\frac{6}{5},\frac{-3}{5}\right)$
I'm somehow unable to find the $x$ and $y$ I have found the distance based on the point and point distance equation:$$d(p_1,p_2)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{\left(\frac{6}{5}-4\right)^2+\left(-\frac{3}{5}+2\right)^2}= \frac{7\sqrt{5}}{5}$$
$$d_1=d_2=d(p_1,p_2)$$
So now I know the distance, the min. distance to the unknown point is the same.
What is the easiest way to find the symmetrical $x$ and $y$ coordinates?
(This is a high school problem)
| Alternative rapid way: calling $x_0$ and $y_0$ the coordinates of the symmetric point, the simplest method is to note that the differences in the $x $- and $y $- coordinates between the searched point and $( \frac{6}{5}, -\frac{3}{5})$ must equal those between this last point and $(4,-2) $. Thus:
$$ 4-\frac{6}{5} = \frac{6}{5} -x_0$$
$$ y_0 - \left(-\frac{3}{5}\right) = -\frac{3}{5} - \left(-2\right) $$
Now solve these equations to find the coordinates.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$
I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x$. But maybe the questions doesn't require solving the integral.
Apparently the $\lim_{n \to \infty}(\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x)^n$ should be $\frac{1}{2}$ for the question to make sense. That's all I know.
| Let $I(n)$ be given by the integral
$$\begin{align}
I(n)&=\int_0^1 \frac{1}{1+x^n}\,dx \tag 1\\\\
\end{align}$$
Then, expanding the integrand of the integral on the right-hand side of $(1)$ in the Taylor series $ \frac{1}{1+x^n}=\sum_{k=0}^\infty (-1)^kx^{nk}$ reveals
$$\begin{align}
I(n)&=\sum_{k=0}^\infty \frac{(-1)^k}{nk+1}\\\\
&=1+\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} \tag 2
\end{align}$$
Next, using the fact that $\log(2)= \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$ and that $\frac{\pi^2}{12}=-\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$, we can write the series in $(2)$ as
$$\begin{align}
\sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} &=-\log(2)+\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k+1/n}-\frac1k\right)\\\\
&=-\log(2)-\frac1n \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1/n)}\\\\
&=-\log(2)-\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\frac1n\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k(k+1/n)}-\frac{1}{k^2}\right)\\\\
&=-\log(2)+\frac{\pi^2}{12n}+O\left(\frac1{n^2}\right) \tag 3
\end{align}$$
Using $(1)-(3)$ yields
$$I(n)=1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right) \tag 4$$
Next, using $(4)$, we can write
$$\begin{align}
\left(I(n)\right)^n&=e^{n\log\left(1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right)\right)}\\\\
&=e^{-\log(2)+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)}\\\\
&=\frac12 \left(1+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)\right)
\end{align}$$
Finally, we have
$$\begin{align}
\lim_{n\to \infty}\left(n\left(\left(I(n)\right)^n-\frac12\right)\right)&=\lim_{n\to \infty}\left(\frac{\pi^2}{24}-\frac{\log^2(2)}{4}+O\left(\frac1n\right)\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^2}{24}-\frac{\log^2(2)}{4}}
\end{align}$$
And we are done!
| {
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Fraction Decomposition I have the following problem:
Suppose $x + y + z = 0$. Show that $$\frac{x^5 + y^5 + z^5}{5}= \frac{x^3 + y^3 + z^3}{3} \times \frac{x^2 + y^2 + z^2}{2}$$ and $$\frac{x^7 + y^7 + z^7}{7}= \frac{x^2 + y^2 + z^2}{2} \times \frac{x^5 + y^5 + z^5}{5}$$
I thought this was a fun problem to tackle, but I've been stuck on this for hours, and I can't seem to get anywhere. I've tried expanding, but no matter what I do, I just get a huge mess.
I also cannot find how to incorporate the initial assumption ($x + y + z = 0$) anywhere. Have I missed something?
| The first identity does not involve much algebra, if you write $z = -(x+y)$:
$$ z^5 + (x^5+y^5) = -5x^4y-10x^3y^2-10x^2y^3-5xy^4$$
where the $(-x^5-y^5)$ cancels out in the binomial exapnsion.
A similar thing happends in the cubes term.
So the identity reads
$$
-x^4y-2x^3y^2-2x^2y^3-xy^4 = (-x^2y-xy^2)\frac{(x^2+y^2+x^2+2xy+y^2)}{2}
$$
and of the six terms on the right, two pairs add together, and you get the left side.
| {
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Determine the last digit of the following number. Problem: Determine the last digit of the following number $$\underbrace{7^{7^{7^{7...}}}}_{1001\text{ }7's}.$$
My Attempt: By Euler's theorem $7^4\equiv 1\pmod {10}\Rightarrow 7^7\equiv 3\pmod {10}.$ After this I am clueless.
Edit $1$: So I tried taking even and odd number of sevens and this is what I observed $$7^{7}\equiv 3\pmod{10}.$$ $$7^{7^{7}}\equiv 7\pmod{10}.$$ $$7^{7^{7^{7}}}\equiv 7^{7^{7*7^6}}\equiv \left(7^{7^{7}}\right)^{7^6}\equiv 7^{7^6}\equiv 3\pmod{10}.$$ Thus for an odd number of sevens it seems that the remaineder will be $7$ whereas for an even number of sevens the remainder will be $3.$
Edit $2$: We need to find $7^7\equiv 1\pmod 4$ therefore $7^{7^{7}}\equiv 7^{4k+1}\equiv 7\pmod {10}.$ Similarily $7^{7^{7^{7}}}\equiv 7^{7^{4k+1}}\equiv 7^7\equiv 3\pmod {10}.$ Also observe that $7^{7^{7}}\equiv 7^{2k+1}\equiv 3\pmod 4.$ If we have $5$ sevens $7^{7^{7^{7^7}}}\equiv 7^3\equiv 3\pmod{10}.$
| As you observed, by Euler's theorem, the exponent works modulo $4$. And $7$ happens to be congruent to $-1 (\bmod{4})$. The exponent is a tower of $1000$ $7$'s, which, modulo $4$, is $-1$ raised to an odd power. So modulo $10$, your tower of $1001$ $7's$ is $\equiv 7^{-1} \equiv 3 (\bmod{10}).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of solutions $(x,y)$ in non-negative integers such that $ax+by\leq ab$, where $a$ and $b$ are positive integers. Problem: Find the number of solutions $(x,y)$ in nonnegative integers such that $ax+by\leq ab$, where $a$ and $b$ are positive integers.
My Attempt: I observed that this is equivalent to counting the number of lattice points on and under the line $ax+by= ab.$ So I know that the number of solutions must be at least $$\frac{(a+1)(b+1)}{2}$$ since there are $(a+1)(b+1)$ points in the rectangle $[0,b]*[0,a].$ However, I am having trouble in counting the points on the line $ax+by=ab.$ Any ideas/hints will be much appreciated.
| So far, you should agree that the number of solutions $(x,y)$ is
$$
\frac{(a+1)(b+1) + B}{2}, \tag{1}
$$
where $B$ is the number of points on the boundary $ax + by = ab$. So we need to count $B$.
Write $a = da'$, $b = db'$, where $d = \gcd(a,b)$, so that $\gcd(a',b') = 1$.
Our equation becomes
\begin{align*}
(da') x + (db') y &= d^2 a' b'\\
\iff a' x + b' y &= d a' b' \tag{2}
\end{align*}
Mod $a'$, $b'y \equiv 0$ so $a' \mid y$.
Likewise mod $b'$, $a'x \equiv 0$ so $b' \mid x$.
So if we write $x = b'x'$ and $y = a' y'$, the number of solutions to (2) is equal to the number solutions $(x',y')$ to
$$
a'b'x' + a'b'y' = da'b'
$$
i.e. just $x' + y' = d$.
There are $d+1$ solutions to this since $x'$ and $y'$ are nonnegative.
So $B = d + 1 = \gcd(a,b) + 1$ in (1), and the final answer is
$$
\boxed{\frac{(a+1)(b+1) + \gcd(a,b) + 1}{2}.}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there an identity for $\arctan(x+y)$? Ive tried looking on the internet and I can't seem to find any identities for $\arctan(x+y)$. I was wondering if anyone knows any?
Thanks!
| $$\arctan(x+y) + \arctan(x-y) = \arctan\left(\frac{2 x}{1-x^2+y^2}\right) \tag 1 $$
is valid because take tan on both sides using
$$ \tan( x+y)=\frac{\tan x + \tan y}{1- \tan x \tan y} $$
(1) holds. Similarly
$$\arctan(x+y) - \arctan(x-y) = \arctan\left(\frac{2 y}{1+x^2-y^2}\right) \tag 2 $$
Taking sum
$$2 \arctan(x+y) = \arctan\left(\frac{2 x}{1-x^2+y^2}\right)+\arctan\left(\frac{2 y}{1+x^2-y^2}\right)$$
If $z= x+iy , u = $ real part of $ z^2 $
$$ 2 \arctan(x+y) = \arctan\left(\frac{2 x}{1-u}\right)+\arctan\left(\frac{2 y}{1+u}\right)$$
for whatever use it could be put to..
| {
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Help with question on limits It is given that $\lim _{ x\rightarrow 0 }{ \frac { f(x) }{ { x }^{ 2 } } =a } $ and $\lim _{ x\rightarrow 0 }{ \frac { f(1-\cos x) }{ g(x)\sin^2x } = b }$, where $b \neq 0$, then find $\lim _{ x\rightarrow 0 }{ \frac { g(1-\cos2x) }{ x^4 } }$.
My Approach:
Since $\sin^2x + \cos^2x = 1 \implies \sin^2x = (1 - \cos x)(1+\cos x)$.
$$\therefore \lim _{x \rightarrow 0} \frac{f(1-\cos x)}{g(x)(1-cos x)(1+cos x)}=b$$
$$\implies\lim _{x \rightarrow 0} \frac{f(1-\cos x)}{(1-cos x)^2} \times \lim _{x \rightarrow 0} \frac{(1-\cos x)}{g(x)(1+\cos x)}=b$$
$$\implies a\times \lim _{x \rightarrow 0} \frac{(1-\cos x)}{g(x)(1+\cos x)}=b$$
Now, since $a$ can't be equal to $0$, because if $a = 0 \implies b=0$, which is not possible as per the question, thus,
$$\implies \lim _{x \rightarrow 0} \frac{(1-\cos x)}{g(x)(1+\cos x)}=\frac{b}{a}$$
After this, I am not able to proceed further. How should I get to the answer?
| You are on the right track.
Method 1.
One may recall that, by the Taylor series expansion, as $u \to 0$, $$ \cos u=1-\frac{u^2}2+o(u^2) $$ giving, as $x \to 0$, $$ \cos 2x=1-2x^2+o(x^2) $$ and $$ \cos (1-\cos 2x)=1-2x^4+o(x^4). \tag1 $$ Now you have obtained, as $u \to 0$, $$ \frac{(1-\cos u)}{g(u)(1+\cos u)} \to \frac{b}{a} \tag2 $$ then replace $u$ with $1-\cos 2x$, as $x \to 0$ in $(2)$ to get using $(1)$, $$ \frac{(1-\cos (1-\cos 2x))}{g(1-\cos 2x)(1+\cos (1-\cos 2x))}=\frac{2x^4+o(x^4)}{g(1-\cos 2x)(2-2x^4+o(x^4))} \tag3 $$ giving, as $x \to 0$,
$$ \frac{x^4}{g(1-\cos 2x)} \sim \frac ba\tag4 $$
obtaining then the desired limit.
Method 2.
One may use
$$
1-\cos 2x=2\sin^2 x
$$ giving
$$
1-\cos(1-\cos 2x)=1-\cos(2\sin^2 x )=2\sin^2(\sin^2x)
$$ then, making $x \to 0$ in $(3)$ (on the left hand side of) and writting $$
\begin{align} &\frac{(1-\cos (1-\cos 2x))}{g(1-\cos 2x)(1+\cos (1-\cos 2x))}
\\\\&=\frac{\sin^2(\sin^2x)}{(\sin^2x)^2}\cdot \frac{(\sin^2x)^2}{(x^2)^2}\cdot \frac{2}{1+\cos(2\sin^2 x )}\cdot\frac{x^4}{g(1-\cos 2x)} \tag5
\end{align}$$ we get $(4)$ again using $\displaystyle \lim_{u\to 0}\frac{\sin u}{u}=1$.
| {
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Can this be solved without resorting to graphical method? I need to find the points of intersection of a circle with radius $2$ and centre at $(0,0)$ and a rectangular hyperbola with equation $xy=1$. As per the topic statement is there any way to solve this without the graphical method. I have tried setting the $y$ values equal but I cant solve the resulting equation for $x $.
| The equation of the circle is $x^2 + y^2 = 4$ and the equation of the hyperbola is $xy=1$
So the point of intersection would be a common solution to
$xy =1$
$x^2 + y^2 = 4$
so
$y = 1/x$
$x^2 + \frac 1{x^2} = 4$
$x^4 +1 = 4x^2$
$x^4 - 4x^2 + 1 = 0$
$x^2 = \frac {4 \pm \sqrt {12}}2$
$x^2 = 2 \pm \sqrt 3$
$x = \pm \sqrt{2 \pm \sqrt{3}}$
$y = 1/x = \pm \frac 1{\sqrt{2 \pm \sqrt{3}}}$
$= \pm \frac 1{\sqrt{2 \pm \sqrt{3}}}\frac {\sqrt {2\mp \sqrt {3}}}{\sqrt {2\mp\sqrt{3}}} $
$=\pm \frac{\sqrt {2\mp \sqrt {3}}}{\sqrt {4-3}}=\pm {\sqrt {2\mp \sqrt {3}}}$
So there are four points: $(\sqrt{2 + \sqrt{3}},{\sqrt{2 - \sqrt{3}}});(\sqrt{2 - \sqrt{3}},{\sqrt{2 +\sqrt{3}}});(-\sqrt{2 + \sqrt{3}},-{\sqrt{2 - \sqrt{3}}});(-\sqrt{2 - \sqrt{3}},-{\sqrt{2 + \sqrt{3}}});$
| {
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maximum value of $xy+yz+zx.$ given $x+2y+z=4$ if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
| $$xy+yz+zx+y^2 = (x+y)(y+z) \le \frac{(x+2y+z)^2}{4} = 4$$
$$\Rightarrow xy+yz+zx \le 4-y^2 \le 4$$
Equality occurs when $y=0$, $x=z=2$
| {
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Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
| There is a much simpler way involving no contrived derivatives, binomial expansions, etc., that unfortunately does not easily generalize to any other quadratic reciprocal.
If we notice that $x^2+2x+2$ occurs in the factorization of a certain binomial:
$$ x^8 - 16 = \left( x^4 - 4 \right) \left( x^4 + 4 \right) = \left( x^4 - 4 \right) \left( x^2 - 2x + 2 \right) \left( x^2 + 2x + 2 \right) = \left( x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8 \right) \left( x^2 + 2x + 2 \right) $$
Rearranging:
$$ \frac{1}{x^2 + 2x + 2} = \frac{x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8}{x^8 - 16} $$
We can bring the right-hand side to the form of a simple geometric series:
$$ \frac{1}{x^2 + 2x + 2} = \frac{\frac{8}{16} - \frac{8}{16} x + \frac{4}{16} x^2 - \frac{2}{16} x^4 + \frac{2}{16} x^5 - \frac{1}{16} x^6}{\frac{16}{16} - \frac{x^8}{16}} = \frac{\frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6}{1 - \frac{x^8}{16}} = \left( \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6 \right) \left( 1 + \frac{x^8}{16} + \frac{x^{16}}{256} + \cdots \right) $$
Clearly, there is no overlap between the terms when multiplying out, so if we write only the coefficients, we see that they group together in sets of eight:
$$ \frac{1}{x^2 + 2x + 2} = \left[ \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, 0, -\frac{1}{8}, \frac{1}{8}, -\frac{1}{16}, 0, \frac{1}{16}, -\frac{1}{32}, \frac{1}{32}, 0, -\frac{1}{64}, \frac{1}{64}, -\frac{1}{128}, 0, \cdots \right] $$
in an elegant pattern!
| {
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About the diophantine equation $y^3=8x^6+2x^3y-y^2$
How I can solve the equation $y^3=8x^6+2x^3y-y^2$ in integers?
I made the substition $x^3=z$ and got the equation $8z^2+2yz-y^3-y^2=0$. So I decided to apply the general formula for quadratic equations and thus I got $$z=\frac{-2y\pm \sqrt{32y^3+36y^2}}{16}=\frac{-y\pm y\sqrt{8y+9}}{8}.$$
So in order to have $z\in \mathbb{Z}$, $8y+9$ must be a perfect square, let's say $8y+9=k^2$, with $k\in \mathbb{Z}$. From this we get $8y=(k+3)(k-3)$, but what could I do next? Also, is there another approach, like an appropriate factorization, to solve this equation? Thanks in advance.
| Well, after a couple of days thinking I think the diophantine equation can be solved in a different way. First of all, asumme that $y=0$, so the equation becomes $8x^6=0$, then $x=0$. Now, let's consider $y\neq 0$, we note that the LHS is divisible by $y$ and thus the RHS also has to be divisible by $y$, so we get that $y\mid 8x^6$. The idea is to prove that $\gcd(x,y)=1$. We can write the equation in the form $$(2x^3+y^2+y)(4x^3-y)=4x^3y^2.$$
Let's suppose $\gcd(x,y)=d>1$, so there exists a prime number $p$ such that $p\mid x$ and $p\mid y$. Set $v_{p}(x)=\alpha>0$ and $v_{p}(y)=\beta>0$. First, let's assume that $p$ is odd, then $v_{p}(4x^3y^2)=3\alpha+2\beta$.
On the other hand, $v_{p}((2x^3+y^2+y)(4x^3-y))=v_{p}(2x^3+y^2+y)+v_{p}(4x^3-y)$, but $v_{p}(2x^3+y^2+y)=\min\{v_{p}(2x^3), v_{p}(y^2), v_{p}(y)\}=3\alpha$ or $\beta$ if $3\alpha\le \beta$ or if $3\alpha>\beta$, respectively. Also $v_{p}(4x^3-y)=\min\{v_{p}(4x^3), v_{p}(y)\}=3\alpha$ or $\beta$ if $3\alpha\le \beta$ or if $3\alpha>\beta$, respectively. In summary, we deduce that
$$v_{p}((2x^3+y^2+y)(4x^3-y))=
\begin{cases}
6\alpha &\text{if } 3\alpha\le \beta \\
2\beta &\text{if } 3\alpha>\beta
\end{cases}
$$
So we conclude that $6\alpha=3\alpha+2\beta$ or $2\beta=3\alpha+2\beta$. In the first case we get $\beta\le 0$, and in the second case we get $\alpha=0$, contradiction in both cases. Now, Let's suppose $p=2$. In this case we get $v_{p}(4x^3y^2)=3\alpha+2\beta+2$, and after an analogous reasoning we deduce that
$$v_{p}((2x^3+y^2+y)(4x^3-y))=
\begin{cases}
6\alpha+3 &\text{if } 3\alpha+1< \beta \\
3\alpha+\beta+1 &\text{if }\, 3\alpha+1=\beta \\
2\beta &\text{if } 3\alpha+1>\beta
\end{cases}
$$
So we have either $6\alpha+3=3\alpha+2\beta+2$ or $3\alpha+\beta+1=3\alpha+2\beta+2$ or $2\beta=3\alpha+2\beta+2$. In any case we arrive to a contradiction. In conclusion such $p$ doesn't exists and therefore $\gcd(x,y)=1$. Using this and the fact that $y\mid 8x^6$ we get that $y\mid 8$. Therefore, the possible values for $y$ are $\{\pm1, \pm2, \pm4, \pm8\}$.
A routine check using the possible values of $y$ gives us solutions only for $y=-1$ and $y=2$. In the first case we get $x=0$, and in the second case we get $x=1$. Hence, all the integer solutions for the equation are: $$(x,y)=(0,0), (0,-1), (1,2).$$
| {
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"url": "https://math.stackexchange.com/questions/2028099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that at least one of the expressions does not exceed $\sqrt[3]{3}$
Let $m,n > 1$ be positive integers. Prove that at least one of the numbers $$\sqrt[n]{m} \ , \ \sqrt[m]{n}$$ does not exceed $\sqrt[3]{3}$.
I thought about doing a proof by contradiction. That is, assume this is not the case and so both exceed $\sqrt[3]{3}$. Then, $\sqrt[n]{m} > \sqrt[3]{3}$ and $\sqrt[m]{n} > \sqrt[3]{3}$. How do we continue?
| Without loss of generality, assume that $m\ge n$. Then $$\sqrt[m]{n} \le \sqrt[n]{n}$$
We just want to show that $$\sqrt[n]{n} \le \sqrt[3]{3}, $$
which is equivalent to
$$n^3 \le 3^n.$$
This is not hard by, for example, proof by induction.
When $n=2$ or $3$, this is true, and when $n > 3$, $$n^3 = (n-1)^3 + 3(n-1)^2+ 3(n-1) + 1 \le (n-1)^3 + (n-1)^3 + 3(n-1) + 1 = 2(n-1)^3 + 3n -2 < 2(n-1)^3 + (n-1)^3 = 3(n-1)^3 \le 3*3^{n-1} = 3^n$$
If you want to use calculus, it is to prove that $$x^{\frac{1}{x}}$$ is decreasing when $x\ge e$, which is also not hard.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$. Exercise
Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$.
I've posted my solution below. In case someone has a more clever solution, feel free to post it!
(TBH, I was surprised that there was no question on Math.SE regarding this equation!)
| The result is essentially the fact that $1+x+ \cdots + x^{n-1} = {1-x^n \over 1-x}$ (for $x \neq 1$), or equivalently that
$(1-x)(1+x+ \cdots + x^{n-1}) = 1 -x ^n$ for any $x$.
If $a=b$, or $a=0$ then the result is true.
Suppose $a \neq 0$ and $a \neq b$. Then letting $x={b \over a}$ in the above
and multiplying the result by $a^n$ yields the desired result.
To see the above result, note that
$(1-x) (1+x+ \cdots + x^{n-1}) = (1+x+ \cdots + x^{n-1})-(x+x^2+ \cdots + x^{n}) = 1- x^n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Calculating $\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)$ I've managed to solve it by rewriting the expression as
$$\frac{1 - \frac{\sqrt{x^2 +x + 1}}{x e^{\frac{1}{x}}} }{ \frac{1}{x e^{\frac{1}{x}}} }$$
then applying L'Hospital's rule.
This took up one whole page and was very hairy, even after substituting $t = \sqrt{x^2+x+1}$. I'm wondering if there's a simpler way. A friend suggested substituting $e = (1+\frac{1}{x})^x$, but that's a bit suspicious.
In both cases, the answer is $\frac{1}{2}$, as confirmed by my computer.
| Using the standard limit
$$
\lim_{t\to 0}\frac{e^t-1}{t}=1
$$
we have
$$
1=\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\lim_{x\to\infty}(xe^{1/x}-x).
$$
Now rewrite your limit as
$$
\lim_{x\to\infty}(xe^{1/x}-x+x-\sqrt{x^2+x+1})=1+\lim_{x\to\infty}(x-\sqrt{x^2+x+1}).
$$
To calculate the last limit we rewrite again
\begin{align}
&\lim_{x\to\infty}(x-\sqrt{x^2+x+1})=\lim_{x\to\infty}\frac{(x-\sqrt{x^2+x+1})(x+\sqrt{x^2+x+1})}{x+\sqrt{x^2+x+1}}=\\
&=\lim_{x\to\infty}\frac{x^2-(x^2+x+1)}{x+\sqrt{x^2+x+1}}=
\lim_{x\to\infty}\frac{-x-1}{x+\sqrt{x^2+x+1}}=\lim_{x\to\infty}\frac{-1-\frac1x}{1+\sqrt{1+\frac1x+\frac{1}{x^2}}}=-\frac{1}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Constructing a cubic given four points
Question: Is there an easier way to solve this problem?
Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.
My Attempt: I started off with the general cubic $ax^3+bx^2+cx+d=f(x)$ and manually plugged in each point to get the following system:$$\begin{align*} & 27a+9b+3c+d=2\\ & 64a+16b+4c+d=4\\ & 125a+25b+5c+d=-3\\ & 216a+36b+6c+d=8\end{align*}\tag1$$
Solving the system with the handy matrix gives the solutions as $a=\frac 92,b=-\frac {117}2,c=245,d=-328$. Thus, $f(0)=-328$.
Even though I (think) solved the problem correctly, this method seems a bit "bulky" especially when everything becomes a higher degree. So I'm wondering if there is a quicker way to evaluate this kind of problem.
| You can directly find out the polynomial $f$ by considering it according as the $x$-values available:
Let $f(x)=a_0+a_1(x-3)+a_2(x-3)(x-4)+a_3(x-3)(x-4)(x-5)$ for real constants $a_0,a_1,a_2,a_3$. Note that we don't need to take into account the value $x=6$ as this is already a cubic polynomial.
Then, $f(3)=2\Rightarrow a_0=2$
$f(4)=4\Rightarrow a_1=2$
$f(5)=-3\Rightarrow a_2=-\frac{9}{2}$
$f(6)=8\Rightarrow a_3=\frac{9}{2}$
Thus, $f(x)=2+2(x-3)-\frac{9}{2}(x-3)(x-4)+\frac{9}{2}(x-3)(x-4)(x-5)$.
I think the calculations are pretty simple this way as you have chosen $f$ to be such.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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n such that $|\sin (\sqrt{n+1})-\sin \sqrt n|< \lambda$
I broke $|\sin (\sqrt{n+1})-\sin \sqrt n|$ as $|2cos(\frac{\sqrt{n+1}+\sqrt n}{2})sin(\frac{\sqrt{n+1}-\sqrt n}{2})|$.I am facing trouble in proving that it is less than some number for some n.Please help me in his regard.Thanks.
| Note that for $0< x < \pi/2$, $\sin x < x$.
Also, for all $n > 0$, we have:
$$
\frac{1}{4n} + 1 + n > n+1 \\
\left( \frac{1}{2\sqrt{n}}+\sqrt{n}\right)^2 > n+1 > 0 \\
\frac{1}{2\sqrt{n}} +\sqrt{n}> \sqrt{n+1} \\
\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}
$$
Also it is obvious that
$$
\left|\cos \left( \frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\right| \leq 1
$$
So for any $\lambda > 0$ and sufficiently large $n$,
$$
|\sin \sqrt{n+1}-\sin\sqrt{n} |< \lambda$$
Thus the answer is C.
| {
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"url": "https://math.stackexchange.com/questions/2033066",
"timestamp": "2023-03-29T00:00:00",
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Find Power Series representation of the function $f(x) = {x\over 2x^2 + 1}$?
Find Power Series representation of the function $f(x) = \dfrac x{2x^2 + 1}$?
I'm not sure how to tackle this...I'm supposed to find interval of convergence.
| A variation: From the geometric series
\begin{align*}
\frac{1}{1-x}=\sum_{n=0}^\infty x^n\qquad\qquad\qquad |x|<1
\end{align*}
we obtain
\begin{align*}
\frac{x}{2x^2+1}&=\frac{x}{1-\left(-2x^2\right)}\\
&=x\sum_{n=0}^\infty (-2x^2)^n\qquad\qquad &|-2x^2|&<1\\
&=\sum_{n=0}^\infty (-2)^nx^{2n+1} &|x|&<\frac{1}{\sqrt{2}}\\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
| Using the hint of Mark Bennet you can write
$$ \dfrac{1}{(1-x)^3(1+x)(1+x+x^2)} = \dfrac{1}{(1-x)(1-x)(1+x)(1-x)(1+x+x^2)} = \dfrac{1}{(1-x)(1-x^2)(1-x^3)}. $$
Now, notice that for each fixed $j = 1, 2, 3$ we have:
$$ \frac{1}{1 - x^j} = 1 + x^j + x^{2j} + \ldots$$
Denote by $c_n$ the $n$-th coeficient of your rational function. Then, multiplying the known power series $\frac{1}{1-x^j}$ for $j = 1, 2, 3$ we discover
$$ c_n = \# \{(n_1, n_2, n_3); n_j \geq 0, n_1 + 2n_2 + 3n_3 = n\}$$
Now, you have a combinatorial problem. As Nil pointed out, there is a related post on this kind of problem. I'm following the ideas there.
First of all, notice that for any fixed $n$ the non-negative solutions of $x + 2y = n$ are described by first choosing some $0 \le 2y \le n$ and then noticing that $x$ is fixed after $y$ is chosen. Thus, you have $\lfloor \frac{n}{2} \rfloor + 1$ solutions.
For the problem we are interested in, first we choose $0 \le 3n_3 \le n$. There are $\lfloor \frac{n}{3} \rfloor + 1$ choices for this. Fixed this first choice, you have exactly $\lfloor \frac{n - 3n_3}{2} \rfloor + 1$ choices for the pair $(n_1, n_2)$. Thus, you can write:
$$ c_n = \sum_{z=0}^{\lfloor \frac{n}{3} \rfloor} \left( \lfloor \frac{n - 3z}{2} \rfloor + 1 \right) = 1 + \lfloor \frac{n}{3} \rfloor + \sum_{z=0}^{\lfloor \frac{n}{3} \rfloor} \lfloor \frac{n - 3z}{2} \rfloor.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Here's another approach for the record.
*
*The equation $x+ \frac{1}{x}=\alpha$ where $\alpha \in [-2,2]$ can be solved as follows. Identify $\alpha$ is $2\cos (\theta)$, and observe that letting $z=e^{i\theta}$,from the definition of $\cos (\theta)$, we have
$$z+ \frac{1}{z}=2\cos(\theta)=\alpha.$$
*Also, from the definition of $\cos$,
$$z^k + \frac{1}{z^k} = 2\cos (k \theta).$$
*Now back to our question. Let $\alpha = \frac{1+\sqrt{5}}{2}$. Then $\cos(\theta) = \frac{1+\sqrt{5}}{4}$. This gives us $\theta = \pm \frac{\pi}{5} \mod 2\pi$. Therefore the answer to the problem is $2\cos (400\pi)=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
| Let $N$ be such that $\sum \limits _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \epsilon$. Notice that
$$\sum _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {(2^{N+m})^5} = \sum _{m \ge 0} \frac {2^{N+m}} {(2^{N+m})^5} = \sum _{m \ge 0} \frac 1 {(2^{N+m})^4} = \\
\frac 1 {2^{4N}} \sum _{m \ge 0} \frac 1 {2^{4m}} = \frac 1 {2^{4N}} \frac 1 {1 - \frac 1 {16}} = \frac 1 {15} \frac 1 {16^{N-1}} .$$
It follows that in order to have the inequality that we have begun with, it is sufficient to impose that $\frac 1 {15} \frac 1 {16^{N-1}} \le \epsilon$, which means $16^{N-1} \ge \frac 1 {15 \epsilon}$, whence $N-1 \ge \log_{16} \frac 1 {15 \epsilon} = -\log_{16} (15\epsilon) = -\frac 1 4 \log_2 (15 \epsilon)$, so $N = \left[ 1 -\frac 1 4 \log_2 (15 \epsilon) \right] + 1$, where $[x]$ denotes the integer part of $x$ (the "floor" function, as some call it).
Now just choose $\epsilon = 10^{-3}$ (because you want the first 2 decimal places to be exact) in order to get $N = \left[ 1 -\frac 1 4 \log_2 (15 \cdot 10^{-3}) \right] + 1 = 3$. This means that
$$\sum _{n=1} ^{2^3-1} \frac 1 {n^5} = 1.\color{red}{03}6849\dots$$
has the first 2 decimal places correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Lagrange multiplier on unit sphere for $f(x,y,z)=x^2+2y^2+3z^2$
Find the maximum and minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ in the region $D=\{(x,y,z)\in \mathbb R^3| x^2+y^2+z^2=1\}$. And find a unit vector at which the maximum and minimum are attained respectively.
Attempt:
I know I need to proceed by Lagrange multiplier method, but I am not sure how to proceed after a step
we will get the equations as $$x=\lambda x$$$$y=\frac{\lambda}{2} y$$$$z=\frac{\lambda}{3} z$$$$x^2+y^2+z^2=1$$
Now how to solve them?
| We have: $x^2 = 1 - y^2 - z^2 \implies f(x,y,z) = 1 - y^2 - z^2 + 2y^2 + 3z^2= 1+y^2+2z^2\ge 1$, and this is the minimum of $f$ which is achieved when $y = z = 0, x = \pm 1$. For the max, we have $f_y = 2y, f_z = 4z$, and $f_y = 0 = f_z \implies y = 0 = z$ which is the only critical point in the domain $D = [-1,1]\times [-1,1]= \{(y,z): -1 \le y, z \le 1\}$ of $f$. Thus evaluating the values of $f$ at the boundary points which are $(y,z) = (\pm 1, 0), (0,\pm 1)$, we have the max is $3$ when $x = 0, y = 0, z = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047014",
"timestamp": "2023-03-29T00:00:00",
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how to show $a_{n}=[\frac{(2n)!!}{(2n-1)!!}]^2 \frac{1}{2n+1}$ converges? Question: $\displaystyle{a_{n} =
\left[{\left(2n\right)!! \over \left(2n - 1\right)!!}\,\right]^{2}
{1 \over 2n + 1}\,,\quad\mbox{prove}\ a_{n}}$ converges.
My thought: I want to prove {$a_{n}$} is an increasing sequence and it has an upper bound. I've figured out $$a_{n} = (\frac{2\cdot 2 }{1\cdot 3})\cdot(\frac{4\cdot4}{3\cdot5})\dots (\frac{(2n-2)(2n-2)}{(2n-3)(2n-1)})(\frac{(2n)^2}{2n-1})(\frac{1}{2n+1})$$ And, $$\frac{a_{n+1}}{a_{n}}=\frac{(2n)^2}{(2n-1)(2n+1)}\frac{(2n+2)^2}{(2n+1)}\frac{(2n-1)}{(2n)^2}(\frac{2n+1}{2n+3}) = \frac{(2n+2)^2}{(2n+1)(2n+3)} \gt 1$$ So it is increasing.
My problem: I'm stuck at proving it has a upper bound. Maybe there are some inequalities that can be used here?
In addition, I've searched this question and find this sequence converges to $\pi/2$ and it is known as Wallis Formula.But here, I only want to show it converges and I'm not going to find its limit.
Thanks for your time!
| $$
\begin{align}
& a_m=\left(\frac{(2m)!!}{(2m-1)!!}\right)^2 \frac{1}{2m+1} = \prod_{n=1}^{m}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) \Rightarrow \\
& \log(a_m)=\sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \\
& \text{Let}\space f(x) = -\log(x) \space\colon x\gt1 \space\Rightarrow f'(x) = -\frac{1}{x} \space\Rightarrow f''(x) = \frac{1}{x^2} \space\gt0 \space\Rightarrow f(x) \space\text{convex function} \\[8mm]
& \text{(1).}\space f\left(\frac{x_1+x_2}{2}\right)\le\frac{f(x_1)+f(x_2)}{2}\Rightarrow \\
& \quad -\log\left(\frac{(2n-1)+(2n+1)}{2}\right) \le \frac{-\log(2n-1)-\log(2n-1)}{2} \Rightarrow \\
& \quad -2\log(2n) \le -\log(2n-1)-\log(2n-1) \Rightarrow 0\le 2\log(2n)-\log(2n-1)-\log(2n-1) \\
& \quad \sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \ge 0 \Rightarrow \log(a_m) \ge 0 \Rightarrow \quad\color{red}{a_m \ge 1} \\[8mm]
& \text{(2).}\space f(x) \ge f(y)+f'(y)\,(x-y) \Rightarrow \\
& \quad -\log(2n) \ge -\log(2n-1)-\frac{2}{2n-1} \Rightarrow 0 \ge \log(2n)-\log(2n-1)-\frac{2}{2n-1} \quad\space\& \\
& \quad -\log(2n) \ge -\log(2n+1)+\frac{2}{2n+1} \Rightarrow 0 \ge \log(2n)-\log(2n+1)+\frac{2}{2n-1} \quad\Rightarrow \\
& \quad 0 \ge 2\log(2n)-\log(2n-1)-\log(2n+1)-\frac{2}{2n-1}+\frac{2}{2n+1} \Rightarrow \\
& \quad 2\log(2n)-\log(2n-1)-\log(2n+1) \le \frac{2}{2n-1}-\frac{2}{2n+1} = \frac{4}{4n^2-1} \Rightarrow \\
& \quad \sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \le \sum_{n=1}^{m}\frac{4}{4n^2-1} \Rightarrow \\
& \quad \lim_{m\rightarrow\infty} \log(a_m) \le \sum_{n=1}^{\infty}\frac{4}{4n^2-1} = 2 \Rightarrow \quad\color{red}{\lim_{m\rightarrow\infty} a_m\le e^2} \\
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$
| {
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} |
Solving an integral How to obtain an explicite formula for the following integral?
$$f_m(x_m)= \int_0^\infty\cdots \int_0^\infty x_1(x_1+x_2) \cdots (x_1+x_2+\cdots+x_m) e^{-\frac{(x_1+x_2+\cdots+x_m)^2}{2}} dx_1 dx_2 \cdots d x_{m-1},$$
where $m\geq 2$. We know that $f_2(x_2)= \int_{x_2}^\infty e^{-\frac{y^2}{2}} d y$, $f_3(x_3)= \frac{3}{2}\left(-x_3 e^{-\frac{x_3^2}{2}}+(1+x_3^2) f_2(x_3)\right)$, $f_4(x_4)= \frac{5}{16}\left(-2x_4(x_4^2+5) e^{-\frac{x_4^2}{2}}+ 2(x_4^4+6x_4^2+3) f_2(x_4)\right)$ and etc.
| It is convenient to consider the function $g_n(x) = f_{n+2}(x)$ with $n \geq 0$. Before the computation, we introduce some notations:
*
*$\Delta^n(a) = \{ (x_1, \cdots, x_n) \in \Bbb{R}^n : 0 < x_1 < \cdots < x_n < a \}$.
*$I_n(x)$ for $n \geq 0$ is defined recursively by $I_0 (x) = 1$ and $I_{n+1}(x) = \int_{0}^{x} t I_n(t) \, dt$. It is easy to check that for $x > 0$,
$$ I_n(x) = \frac{x^{2n}}{2^n \cdot n!} = \int_{\Delta^n(x)} t_1\cdots t_n \, dt_1 \cdots dt_n. $$
Now let us return to the original problem and compute $g_n(x)$. With the substitution
\begin{align*}
y_1 &= x_1 \\
y_2 &= x_1 + x_2 = y_1 + x_2 \\
\vdots & \qquad \qquad \vdots \\
y_{n+1} &= x_1 + \cdots + x_{n+1} = y_n + x_{n+1}
\end{align*}
and $y = y_{n+1}$, we have
\begin{align*}
g_n(x)
&= \int_{0}^{\infty} \int_{\Delta^n (y_{n+1})} y_1 \cdots y_n y_{n+1}(y_{n+1} + x) e^{-\frac{(y_{n+1}+x)^2}{2}} \, dy_1\cdots dy_{n+1} \\
&= \int_{0}^{\infty} I_n(y) y(y + x) e^{-\frac{(y+x)^2}{2}} \, dy \\
&= \frac{1}{2^n \cdot n!} \int_{0}^{\infty} y^{2n+1}(y + x) e^{-\frac{(y+x)^2}{2}} \, dy \\
&= \frac{2n+1}{2^n \cdot n!} \int_{0}^{\infty} y^{2n} e^{-\frac{(y+x)^2}{2}} \, dy.
\end{align*}
So in principle we can compute every $g_n(x)$ by applying integration by parts several times.
| {
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"url": "https://math.stackexchange.com/questions/2048021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculate $\sum\limits_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$
Calculate the sum of the series $$\sum_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$$
I have tried partial fraction decomposition.
$$\sum_{n=1}^\infty\frac{1}{4(n+2)}- \sum_{n=1}^\infty\frac{1}{4(n+4)}-\sum_{n=1}^\infty\frac{1}{2(n+4)^2}$$
Is this correct? What is the sum?
| First note that your series has the same convergence as $\sum \frac{1}{n^3}$ by the limit comparison test. And the latter series converges absolutely by the $p$-series test or the integral test, so therefore so does your series.
Next, to find the value the sum converges to, trying partial fractions, we get
$$\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\frac{1}{n+2}-\frac{1}{4}\frac{1}{n+4}-\frac{1}{2}\frac{1}{(n+4)^2}$$
If we break it into three series as you have attempted, then all three are divergent, and difference of divergent is indeterminate, so we cannot proceed. Instead, treat the first two terms together, and note that $\sum_{n=1}^\infty \frac{1}{n+2}-\frac{1}{n+4}$ is a telescoping series, its sum will converge to $1/3+1/4$, the uncanceled parts of the first two terms. For the final term, rememeber by the Basel problem $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ so by reindexing we have $\sum_{n=5}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{(n+4)^2}=\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1$.
Putting it all together we have
$$\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\left(\sum_{n=1}^\infty\frac{1}{n+2}-\frac{1}{n+4}\right)-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(n+4)^2} \\=\frac{1}{4}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{2}\left(\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1\right) = \frac{7}{48} -\frac{\pi^2}{12}+\frac{205}{288}=\frac{247}{288}-\frac{\pi^2}{12}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve a system of equations for real $x,y,z$ Solve the following system of equations in $x,y,z \in \Bbb R$:
$$ (x+y)^3 = z $$
$$ (x+z)^3 = y$$
$$ (y+z)^3 = x$$
I found the solutions $(0,0,0)$, $(\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$ and $(-\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}})$, but I'm not sure if my proof is correct. Also I was wondering if there is a more elegant solution than the one I found (described below).
First, I took the first two equations and solved for $x$, getting
$$ x = \sqrt[3]{z} - y$$
$$ x = \sqrt[3]{y} - z$$
Then substituting $z = c^3$ and $y = b^3$ (since we are working in $\Bbb R$, these values of $b$ and $c$ exist and are unique), we get
$$c - b^3 = b - c^3$$
and thus
$$c + c^3 = b + b^3$$
Since the function $f(x) = x^3 + x$ is one-to-one and onto, this implies $b=c$. By symmetry, this means $a=b=c$. Since $f(x) = x^3$ is one-to-one and onto, this implies $x=y=z$. Plugging this into the original equation gives
$$8x^3 = x$$
which has the solutions $x = 0, \pm \frac{1}{2\sqrt{2}}$. Hence the three solution pairs I found.
Is this solution ok? Is there an alternate, more elegant approach (that perhap more obviously uses the symmetry between $x$, $y$ and $z$?)
| This problem is very pretty:
Let's suppose that $x\geq y\geq z$ (any order will let you the same, you can check that). Because of this,
$$x+y\geq x+z\geq z+y$$
$$\Rightarrow (x+y)^3\geq (x+z)^3\geq (z+y)^3$$
But that means that:
$$z\geq y\geq x$$
So from that $x=y=z$ then for any equation you will have $8x^3=x$. From this last equation you have $x=0$ is solution or $x=\pm \frac{1}{2\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
limit of $\frac{x^2}{x^2+y^2}$ To prove $\frac{x^2}{x^2+y^2}$ has no limit at $(0,0)$ we can take $y=kx$ and therefore:
$$\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2+y^2}=\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2+k^2x^2}=\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2(1+k^2)}=\lim_{(x,kx)\to (0,0)}\frac{1}{(1+k^2)}$$
For different $k$ we will get different limits, so there is no limit at $(0,0)$
But what about the limit at $(3,3)$ intuitively there is a limit which is $\frac{1}{2}$ but if we look at:
$$\lim_{(x,kx)\to (3,3)}\frac{x^2}{x^2+y^2}$$ we get
$$\lim_{(x,kx)\to (3,3)}\frac{1}{(1+k^2)}=?$$
Where am I getting it wrong?
| The error is that you assume $y=kx\to3$, which is only true for $k=1$. This is not the case for the limit to $3$, but it is true when $(x,y)\to(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub? I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've worked these problems before, I usually end up performing a u-substitution after the trig sub, but I'm not sure where to do that or even if this is the right strategy for this integral.
$\int\frac{x^5}{(36x^2+1)^{3/2}}dx=\int\frac{x^5}{((6x)^2+1)^{3/2}}dx$
$x=\frac{1}{6}\tan\theta, dx=\frac{1}{6}\sec^2(\theta) d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\tan^2(\theta)+1)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\sec^2\theta)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{\sec(\theta)}d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^5(\theta)}*\cos(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$
| Hint: Change $$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$$ to $$\frac{1}{6^6}\int\sin(\theta)\tan^4(\theta)d\theta$$ then go to
$$\frac{1}{6^6}\int\sin(\theta)(\sec^2(\theta) - 1)^2d\theta$$ then use a u substitution of $u = \cos\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve system of equations involving square roots How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect.
$$
\left\{
\begin{array}{c}
\sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\
\sqrt{1 - x_1} + \sqrt{1 - x_2} + \sqrt{1 - x_3} + \sqrt{1 - x_4} = 2\sqrt{3}
\end{array}
\right.
$$
| Denote
$$\sqrt{1+x_i}=a_i,~~~\sqrt{1-x_i}=b_i, ~~~(i=1,2,3,4)$$
It's clear that $$a_i,b_i \geq 0,$$and$$a_i^2+b_i^2=2.\tag1$$
Moreover, the system of equations could be rewritten as $$\sum_{i=1}^4a_i=2\sqrt{5},~~~~\sum_{i=1}^4b_i=2\sqrt{3}.\tag2$$
From $(1)$, we have $$\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2=8.$$
From $(2)$, we have $$\left(\frac{1}{4}\sum_{i=1}^4a_i\right)^2+\left(\frac{1}{4}\sum_{i=1}^4b_i\right)^2=2.\tag3$$
But by the inequality $A_n \leq Q_n$, we have
$$\left(\frac{1}{4}\sum_{i=1}^4a_i\right)^2+\left(\frac{1}{4}\sum_{i=1}^4b_i\right)^2 \leq \frac{1}{4}\sum_{i=1}^4a_i^2+\frac{1}{4}\sum_{i=1}^4b_i^2=\frac{1}{4}\left(\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2\right)=2.\tag4 $$
Form $(3)$, we see that the equality holds in $(4)$. Recall the condition for equality in $A_n \leq Q_n$, which is $$a_1=a_2=a_3=a_4,~~~b_1=b_2=b_3=b_4.$$
Thus, we may obtain $$a_i=\frac{\sqrt{5}}{2},~~~b_i=\frac{\sqrt{3}}{2}.$$
As a result, $$x_i \equiv\frac{1}{4},$$ where $i=1,2,3,4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An interesting result in ratio and proportions If
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then
$$
\frac{a+c}{b+d}=k
$$
Also
$$
\frac{a^2}{b^2}=\frac{c^2}{d^2}=k^2
$$
And
$$
\left( \frac{a+c}{b+d} \right)^2=k^2
$$
Also
$$
\frac{a^2 + c ^2}{b^2+d^2}=k^2
$$
Hence
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
$$
My question is... if the reverse is true. That is...
if
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
$$
Then can we assume that :
$$
\frac{a}{b}=\frac{c}{d}
$$
??
| It is clear that the truth of $$ \frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}$$
does not change if we swap $b\leftrightarrow d$, but the truth of $\frac ab=\frac cd$ may well change
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all the possible values of $(a,b,c,d)$. Find all quadruples of real numbers $(a,b,c,d)$ satisfying the system of equations
$(b+c+d)^{2010}=3a$
$(a+c+d)^{2010}=3b$
$(a+b+d)^{2010}=3c$
$(a+b+c)^{2010}=3d$
I tried to find the solutions using hit and trial and by using some logic also I find two solutions which are $(0,0,0,0)$ and $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$
I think these are the only solutions that exist.
| If $(a,b,c,d)$ satisfies the equations, then we may well assume that $a\leq b\leq c\leq d$.These are non-negative because an even power of a real number is always non-negative. It follows that $$ b+c+d\geq a+c+d\geq a+b+d\geq a+b+c$$ since $x$ and hence $x^{2010}$ is increasing for $x\geq 0$, we have that $$3a=(b+c+d)^{2010}\geq (a+c+d)^{2010}\geq (a+b+d)^{2010}\geq (a+b+c)^{2010}=3d.$$Thus, we conclude that $a=b=c=d$ and all the equations take the form $(3a)^{2010}=3a$, so $a=0$ or $3a=1$. Finally, all solutions are $(0, 0, 0 ,0)$ and $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The limit of a function with sum of two roots. I need to find the limit of the following function:
$$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$
I derived it to the form of:
$$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqrt[3]{(x^3+x^2)^2}+\sqrt[3]{(x^3+x^2)^4}\right)},$$
hoping that I'd be able to simplify something but still have $\frac{0}{0}$ and don't see how to do that.
| Let me change the sign on $x$, I find it easier to think about.
the expression is then
$$\sqrt{x^2-2x}+\sqrt[3]{x^2-x^3}$$
Now investigate the two limits
$$\sqrt{x^2-2x}-x\rightarrow -1$$
and
$$x+\sqrt[3]{x^2-x^3}$$ and this latter is equal
to
$$\frac{x^2}{x^2-x\sqrt[3]{x^2-x^3}+(\sqrt[3]{x^2-x^3})^2}$$
for which I get the limit of $\frac{1}{3}$ (assuming I haven't made a mistake, (a big assumption)).
Ok so I get the limit $-\frac{2}{3}$ in agreement with the book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \frac {a^c}{a^a}=\frac{b^b}{b^c}$
$ \frac {a^c \cdot{b^c}}{a^a} =b^b\qquad \text{(i)}$
Similarly, taking the next two terms we obtain,
$b^b=\frac{b^a \cdot c^a}{c^c}\qquad \text{(ii)}$
I tried to solve the two equations obtained to get to the desired statement but I couldn't. Is the way adopted correct or is there another way to reach the desired answer. Please help me proceed with this question
| It is given that $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\tag1$$
Now $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log a+\log b}{b-c+c-a} \,\,\,\,\,\,\,\text {(by Addendo)}$$
$$=\frac{\log ab}{b-a} \tag2$$
So from $(1)$ and $(2)$, we get that $$\frac{\log c}{a-b}=\frac{\log ab}{b-a}$$
$$\implies ab =\frac {1}{c}\tag3$$
From equation $(i)$ established by you in the question and $(3) $, we get $$\frac {a^c \cdot{b^c}}{a^a} =b^b$$
$$\implies a^c \cdot b^c=a^a \cdot b^b$$
$$\implies \frac {1}{c^c}=a^a \cdot b^b$$
$$\implies a^a \cdot b^b \cdot c^c= 1$$
Hope this helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I derive the polar representation of Bernoulli's lemniscate, $r^2 = a^2\cos 2\theta$? If I start with the formula $(x^2+y^2)^2 -a(x^2-y^2) = 0$,
Converting to polar coordinates gives
$r^2 - a(r^2\cos^2 \theta - r^2 \sin^2 \theta) = 0$.
Applying the identity $ \cos^2 \theta - \sin^2 \theta = \cos 2\theta$,
I find that
$r^2 - ar^2(\cos 2\theta) = 0$
How do I derive $r^2 = a^2\cos 2\theta$ from here?
| You forgot to square the first term
$$
x^2+y^2 = r^2\implies (x^2+y^2)^2 = (r^2)^2 = r^4
$$
| {
"language": "en",
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"source": "stackexchange",
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Find the integral part of the product $\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdots \frac{2016}{2015}.$
Find the integral part of the following number $$T = \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dfrac{8}{7} \cdots \dfrac{2016}{2015}.$$
We can show that $T = 2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$, since $$\int_0^{\frac{\pi}{2}} \sin^{2n+1}(x) dx = \dfrac{2}{3} \cdot \dfrac{4}{5} \cdot \dfrac{6}{7} \cdots \dfrac{2n}{2n+1},$$ but how do we calculate the integral part of $2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$?
| We may prove the inequality mentioned by achille hui in the comments without resorting to Stirling's approximation. For large values of $n$, we have:
$$ \frac{(2n)!!}{(2n-1)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^{-1} \tag{1} =2\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^{-1}$$
and since $\left(1-\frac{1}{2k}\right)^{2}$ is close to $\left(1-\frac{1}{k}\right)$, that is the term of a telescopic product,
$$ \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 = 4\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)^{-1} \prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)\tag{2} $$
and since $\prod_{k\geq 2}\left(1-\frac{1}{(2k-1)^2}\right)=\frac{\pi}{4}$ by Wallis product,
$$ \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 = \pi n \prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)\tag{3} $$
where:
$$\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)\approx \exp\sum_{k>n}\frac{1}{4k(k-1)} = \exp\left(\frac{1}{4n}\right) \tag{4}$$
by "telescopic luck" again. Here $\approx$ is actually a $\leq$. With similar arguments one may prove
$$ \exp\left(\frac{1}{8n}-\frac{1}{96n^3}\right)\leq\frac{1}{\sqrt{\pi n}}\cdot\frac{(2n)!!}{(2n-1)!!}\leq \exp\left(\frac{1}{8n}\right) \tag{5}$$
and since $\sqrt{1008\cdot \pi}=56.273\ldots$, $\color{blue}{\large 56}$ is the correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Computing an integral by making it into a complex integral. I am computing an integral
$$\int_0^{2\pi} \frac{\sin^2(\theta)}{5-3\cos(\theta)}$$
I introduced the substitution $z=e^{i \theta}$ which implies $\frac{dz}{zi}=d\theta$. Using the complex analysis for $\sin$ and $\cos$ and this substitution we can write them as
\begin{align}
\sin(\theta)= \frac{z-z^{-1}}{2i} && \cos(\theta)= \frac{z+z^{-1}}{2}
\end{align}
Thus we can see that $\sin^2(\theta)= \frac{z^2-z^{-2}-2}{-4}$. from there we have where $C$ is the unit circle,
$$\int _C \frac{\frac{z^2-z^{-2}-2}{-4}}{5-3/2(z+z^{-1})} \frac{dz}{iz}$$
Where we may simplify by multiplying the top and bottom by 4 and distributing the $z$ from the $dz$ term
$$\frac{1}{i}\int_C \frac{z^2-z^{-2}-2}{20z-6z^2+6}dz$$
Then we separate this into two integrals
$$\int_C \frac{z^2-z^{-2}-2}{20z-6z^2+6}dz= \int_C \frac{z^2-2}{20z-6z^2+6}dz+ \int_C \frac{-z^{-2}}{20z-6z^2+6}dz$$
Now we have to evaluate the two integrals but because our $C$ is the unit circle and the polynomial in the denominator has its roots outside the unit circle, we know the first integral will be an integral of an analytic function, thus it is zero. Does this seem acceptable so far? I would really appreciate comments.
| Your approach is reasonable, but there were a few careless/typographical errors along the way. For example, since $\sin(z)=\frac{z-z^{-1}}{2i}$, then $\sin^2(z)=\frac{z^2+z^{-2}-2}{-4}\ne \frac{z^2-z^{-2}}{-4}$.
METHODOLOGY $1$: MODIFY THE FORM OF THE INTEGRAND
However, there is a much more efficient way forward. Proceeding, note that we can write
$$\frac{\sin^2(\theta)}{5-3\cos(\theta)}=\frac59+\frac13\cos(\theta)-\frac{16}{9}\left(\frac{1}{5-3\cos\theta)}\right)$$
Then, we have
$$\begin{align}
\int_0^{2\pi}\frac{\sin^2(\theta)}{5-3\cos(\theta)}\,d\theta&=\frac{10\pi}{9}-\frac{16}{9}\oint_{|z|=1}\frac{1}{5-3\left(\frac{z+z^{-1}}{2}\right)}\,\frac{1}{iz}\,dz\\\\
&=\frac{10\pi}{9}-\frac{16}{9}\oint_{|z|=1}\frac{1}{5-3\left(\frac{z+z^{-1}}{2}\right)}\,\frac{1}{iz}\,dz\\\\
&=\frac{10\pi}{9}-\frac{32i}{9}\oint_{|z|=1}\frac{1}{(3z-1)(z-3)}\,dz\\\\
&=\frac{10\pi}{9}-\frac{32i}{9}(2\pi i)\text{Res}\left(\frac{1}{(3z-1)(z-3)},z=1/3\right)\\\\
&=\frac{10\pi}{9}-\frac{8\pi}{9}\\\\
&=\frac{2\pi}{9}
\end{align}$$
METHODOLOGY $2$: BRUTE FORCE
$$\begin{align}
\int_0^{2\pi}\frac{\sin^2(\theta)}{5-3\cos(\theta)}\,d\theta&=\oint_{|z|=1}\frac{\left(\frac{z-z^{-1}}{2i}\right)^2}{5-3\left(\frac{z+z^{-1}}{2}\right)}\,\frac{1}{iz}\,dz\\\\
&=\frac{-i}{2}\oint_{|z|=1}\frac{(z^2-1)^2}{z^2(3z-1)(z-3)}\,dz\\\\
&=\pi\text{Res}\left(\frac{(z^2-1)^2}{z^2(3z-1)(z-3)}, z=0,1/3\right)\\\\
&=\pi\left(\frac{10}{9}-\frac89\right)\\\\
&=\frac{2\pi}{9}
\end{align}$$
as expected.
Note that the residue at $0$ can be evaluated as
$$\lim_{z\to 0}\frac{d}{dz}\left(\frac{(z^2-1)^2}{(3z-1)(z-3)}\right)=\lim_{z\to 0}\left(\frac{4z(z^2-1)}{(3z-1)(z-3)}-\frac{(z^2-1)^2(6z-10)}{(3z-1)^2(z-3)^2}\right)=\frac{10}{9}$$
| {
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Proving no integer solution for $x^3+y^3+4=z^3$ I've been trying to shoe that there are no integer solutions for the equation:
$$x^3+y^3+4=z^3$$
What I've tried out so far is to prove by contradiction and to get to a contradiction using modulus.
Let us assume that there are $x,y,z\in\mathbb{Z}$ so that they solve the equation above. therefore:
$$x^3+y^3+4=x^3+3x^2y+3y^2x+y^2-3x^2y-3y^2x+4=(x+y)^3-3xy(x+y)+4=z^3$$
$$4-3xy(x+y)=z^3-(x+y)^3$$
What I did next is I checked all the possibilities for $x,y,z$ to be odd/even: $x,y,z\equiv 0 \mod2$, $x,z\equiv 0 \mod2, y\equiv 1 \mod2$ and so on...
All cases give a contradiction except for the case where $x,y$ are odd and $z$ is even.
Help?
| Hint: Try the equation mod $9$. The cubes mod $9$ are $0,\pm1$.
Solution:
Write the equation as $x^3+y^3+(-z)^3=-4$. The range of the LHS mod $9$ is $-3,\dots,3$ and so is never $-4$.
| {
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$\frac{2}{3}$ of the people on Weird Island tell the truth all the time and the rest lie all the time
$\frac{2}{3}$ of the people on Weird Island tell the truth all the
time and the rest lie all the time. You are sitting in a room with no
windows and two people come in from outside.
Person 1 says: "It is raining outside"
Person 2 says: "Person 1 is telling the truth"
What is the probability that it is raining outside?
My progress:
| According to Bayes theorem.
$P(E1 = truth) = \frac{1}{2}$
$P(E2 = lie) = 1- \frac{1}{2}$
$P(E2 = lie) = \frac{1}{2}$
$P\left(\frac{A}{E1}\right) = $people speaks truth $= \frac{2}{3}$
$P\left(\frac{A}{E2}\right) = $people speaks lie $= 1- \frac{2}{3}$
$P\left(\frac{A}{E2}\right) = $people speaks lie $= \frac{1}{3}$
$Probability = \frac{\frac{A}{E1} * E1}{\frac{A}{E1} * E1 + \frac{A}{E2} * E2}$
$$= \frac{\frac{2}{3} * \frac{1}{2}}{\frac{2}{3} * \frac{1}{2} + \frac{1}{3} * \frac{1}{2}}$$
$= \frac{1}{3} * \frac{2}{1}$
$= \frac{2}{3}$
| {
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What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$? What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$?
How will I be able to find the answer to such questions? Is there any special technique to find the answer to such problems? Thank you.
| $18 = 2\cdot 3^2$. We can find the power of a small prime in a large factorial by successive division to find base divisibility, then divisibility by squares, etc. So the multiplicity of powers of $2$ in $50!$, $v_2(50!),$ is
$$ v_2(50!) = \left\lfloor\frac{50}{2}\right\rfloor + \left\lfloor\frac{50}{4}\right\rfloor + \left\lfloor\frac{50}{8}\right\rfloor + \cdots = 25+12+6+3+1 = 47$$
and similarly $v_2(25!)=22$, $v_3(50!)=16+5+1 = 22$ and $v_3(25!)=8+2 =10$, so
$$v_2\left(\frac{50!}{25!25!}\right) = 47-2\cdot22=3 \\
v_3\left(\frac{50!}{25!25!}\right) = 22-2\cdot 10=2 $$
and only $2$ available powers of $3$ means that $v_{18}\left(\frac{50!}{25!25!}\right)=1$ - the highest power of $18$ dividing the given expression is $18^1=18$.
| {
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Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but giving it to wolfram alpha gives five complex set of answers. Another idea that looks to be nice is this:
$\sum\limits_{}^{cyc}\frac{1}{xy+z-1}=\sum\limits_{}^{cyc}\frac{x}{x^2-x+4}$
Maybe it gives the answer but it is to hard to calculate. Any hints?
| Edit : Since the OP changes some signs, this answer also changes the signs.
You have already noticed that
$$\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}$$
This answer shows that using that $x,y,z$ are the solutions of $t^3-2t^2+\frac 12t-4=0$ enables us to have a simpler form and to find the value easily.
We have that
$$x+y+z=2,\quad xyz=4$$
and that
$$xy+yz+zx=\frac 12\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{2^2-3}{2}=\frac 12$$
So, we know that $x,y,z$ are the solutions of $$t^3-2t^2+\frac 12t-4=0,$$
i.e.
$$(t-1)(t^2-t+4)-\frac{9}{2}t=0,$$
i.e.
$$\frac{t}{t^2-t+4}=\frac{2}{9}(t-1)$$
since $t^2-t+4\not=0$.
Hence, we get
$$\begin{align}\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}&=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}\\\\&=\frac{2}{9}(y-1)+\frac{2}{9}(x-1)+\frac{2}{9}(z-1)\\\\&=\frac{2}{9}(x+y+z-3)\\\\&=\color{red}{-\frac{2}{9}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Problem with an identity involving fractions. In a proof concerning normal distributions, the following step is a little difficult to understand:
$$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma_Y} e^{-{(z-x-\mu_Y)^2 \over 2\sigma_Y^2}} \frac{1}{\sqrt{2\pi}\sigma_X} e^{-{(x-\mu_X)^2 \over 2\sigma_X^2}} \, dx$$
$$= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sqrt{\sigma_X^2+\sigma_Y^2}} \exp \left[ - { (z-(\mu_X+\mu_Y))^2 \over 2(\sigma_X^2+\sigma_Y^2) } \right] \frac{1}{\sqrt{2\pi} \frac{\sigma_X\sigma_Y}{\sqrt{\sigma_X^2+\sigma_Y^2}}} \exp \left[ - \frac{\left(x-\frac{\sigma_X^2(z-\mu_Y)+\sigma_Y^2\mu_X}{\sigma_X^2+\sigma_Y^2}\right)^2}{2\left(\frac{\sigma_X\sigma_Y}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)^2} \right] \, dx$$
The part that I'm not sure about is why
$$ e^{-{(z-x-\mu_Y)^2 \over 2\sigma_Y^2}} e^{-{(x-\mu_X)^2 \over 2\sigma_X^2}} = \exp \left[ - { (z-(\mu_X+\mu_Y))^2 \over 2(\sigma_X^2+\sigma_Y^2) } \right] \exp \left[ - \frac{\left(x-\frac{\sigma_X^2(z-\mu_Y)+\sigma_Y^2\mu_X}{\sigma_X^2+\sigma_Y^2}\right)^2}{2\left(\frac{\sigma_X\sigma_Y}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)^2} \right] $$
In other words, why
$$ -{(z-x-\mu_Y)^2 \over 2\sigma_Y^2}-{(x-\mu_X)^2 \over 2\sigma_X^2} = - { (z-(\mu_X+\mu_Y))^2 \over 2(\sigma_X^2+\sigma_Y^2) } - \frac{\left(x-\frac{\sigma_X^2(z-\mu_Y)+\sigma_Y^2\mu_X}{\sigma_X^2+\sigma_Y^2}\right)^2}{2\left(\frac{\sigma_X\sigma_Y}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)^2} $$
Any help is appreciated.
| To prove that
$$
-{(z-x-\mu_Y)^2 \over 2\sigma_Y^2}-{(x-\mu_X)^2 \over 2\sigma_X^2} = - { (z-(\mu_X+\mu_Y))^2 \over 2(\sigma_X^2+\sigma_Y^2) } - \frac{\left(x-\frac{\sigma_X^2(z-\mu_Y)+\sigma_Y^2\mu_X}{\sigma_X^2+\sigma_Y^2}\right)^2}{2\left(\frac{\sigma_X\sigma_Y}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)^2}
$$
we could first change the signs and multiply by $2$. Then we could multiply by $\sigma_X^2\sigma_Y^2(\sigma_X^2+\sigma_Y^2)$ to have the LHS
$$
(z-x-\mu_Y)^2\sigma_X^2(\sigma_X^2+\sigma_Y^2)+(x-\mu_X)^2\sigma_Y^2(\sigma_X^2+\sigma_Y^2)
$$
and RHS
$$
(z-\mu_X-\mu_Y)^2\sigma_X^2\sigma_Y^2+\left[x(\sigma_X^2+\sigma_Y^2)-\sigma_X^2 z+\sigma_X^2\mu_Y-\sigma_Y^2\mu_X\right]^2
$$
and it is now relatively simple to check term by term that the two sides are equal. For instance we find $z^2\sigma_X^2(\sigma_X^2+\sigma_Y^2)$ on the LHS and $z^2\sigma_X^2\sigma_Y^2+\sigma_X^4 z^2$ on the RHS, so those match up. As another example $-2z\mu_Y\sigma_X^2(\sigma_X^2+\sigma_Y^2)$ from LHS is found as $-2z\mu_Y\sigma_X^2\sigma_Y^2-2z\mu_Y\sigma_X^4$ on the RHS, so those also match up. The term $-2z\mu_X\sigma_X^2\sigma_Y^2$ stemming from the first term on the RHS is cancelled by the second term where we find $2\sigma_X^2 z\sigma_Y^2\mu_X$, so this letter combination not found on the LHS is cancelled on the RHS anyway. Checking the other terms is similar and they have a nice repetitive pattern in the reasoning behind the matching up process.
| {
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What is the number of integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$? What is the number of non negative integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$ ?
I tried it as :
${x_{3}}+{x_{4}}={12} - 2{x_{1}}-2{x_{2}}$
Now, finding the solutions of ${x_{1}}+{x_{2}}$
*
*${x_{1}}+{x_{2}} = 0 => 1$ Solution
*${x_{1}}+{x_{2}} = 1 => 2$ Solutions
*${x_{1}}+{x_{2}} = 2 => 3$ Solutions
*${x_{1}}+{x_{2}} = 3 => 4$ Solutions
*${x_{1}}+{x_{2}} = 4 => 5$ Solutions
*${x_{1}}+{x_{2}} = 5 => 6$ Solutions
*${x_{1}}+{x_{2}} = 6 => 7$ Solutions
And, now respectively finding for ${x_{3}}+{x_{4}}$
*
*${x_{3}}+{x_{4}} = 12 => 13$ Solutions
*${x_{3}}+{x_{4}} = 10 => 11$ Solutions
*${x_{3}}+{x_{4}} = 8 => 9$ Solutions
*${x_{3}}+{x_{4}} = 6 => 7$ Solutions
*${x_{3}}+{x_{4}} = 4 => 5$ Solutions
*${x_{3}}+{x_{4}} = 2 => 3$ Solutions
*${x_{3}}+{x_{4}} = 0 => 1$ Solution
Then, Multiplying respective numbers
$1.13 + 2.11 + 3.9 + 4.7 + 5.5 + 6.3 + 7.1 = 140$ Solutions
I don't have answer for it. Am i right here ?
| Here is another solution, using generating formulas:
Let the power of $x$ represent the value of $x_i$.
Then we have $$(1+x^2+x^4+\cdots)\times (1+x^2+x^4+\cdots)\times(1+x+x^2+\cdots)\times (1+x+x^2+\cdots)$$
$$=\left(\frac{1}{1-x^2}\right)^2\cdot\left(\frac{1}{1-x}\right)^2$$
$$=\frac{1}{(1-x^2)^2(1-x)^2}$$
$$=\frac{1}{1-2x-x^2+4x^3-x^4-2x^5+x^6}.$$
Now we want to find the coefficient of $x^{12}$. By (arduous or computer aided) long division, we get the same answer, $140$.
By the way, one easy way to do long division here is to type the expression into Wolfram Alpha with the phrase "taylor series for." Then it will give you the option to calculate more terms.
| {
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Solutions for diophantine equation $3^a+1=2^b$ I am looking for solutions for the diophantine equation
$3^a+1=2^b$
where $a\in \Bbb N$ and $b\in \Bbb N$.
Is there a power of $3$ that gives a power of $2$ when you add $1$?
Two solutions are easy to find:
*
*$3^0+1=2^1 \rightarrow 1+1=2$
*$3^1+1=2^2 \rightarrow 3+1=4$
But I'm looking for other solutions (solutions where $a>1$).
I believe that there is no other solution, but how can you proof this conjecture?
More general:
How can you find solutions for
$p_1^a+n=p_2^b$
where $p_1$ and $p_2$ are prime and $a,b,n\in \Bbb N$?
| Another solution: consider the powers of $3$$\pmod 8$. We have $3^1\equiv 3\pmod 8$ and $3^2\equiv 1\pmod 8$. So by induction it follows that $3^{2k+1}\equiv 3\pmod 8$ and $3^{2k}\equiv 1\pmod 8$, for $ k\in\Bbb {N} $. So $3^a+1$ would be congruent with $4$ or $2$$\pmod 8$. Therefore we deduce that if $3^a+1=2^b $, then $ b\le 2$.
If $ b=2$, then $ a=1$.
If $ b=1$, then $ a=0$. Hence all the solutions are the pairs $(a, b)=(0,1), (1,2) $.
| {
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How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint
Replace $d$ by a variable $x$; make use of the fact that the sum of the roots of a fourth-degree polynomial is equal to the coefficient of $x^3$
but I didn't get that.
| As you already noted the determinant is a polynomial of 4th degree in $d$.
This polynomial is zero if you replace $d$ by $a$, $b$, or $c$ (we have two identical rows).
Moreover the coefficient of $d^3$ is zero which implies that the sum of the roots is zero. Hence the fourth root is $-(a+b+c)$.
Finally, by Vandermonde's determinant, the coefficient of $d^4$ is
$$\det\begin{pmatrix}1& a& a^2\\1& b& b^2\\1& c& c^2
\end{pmatrix}=(a-b)(b-c)(c-a).$$
Therefore the required determinant is
$$(a-b)(b-c)(c-a)\cdot (d-a)(d-b)(d-c)(d+a+b+c).$$
| {
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Prove that $\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$ Show that
$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$$
(where $\eta(n)$ is the Dirichlet eta function, and A is the Glaisher-Kinkelin constant).
I try:
$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\eta(n)-\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)$$
We use this series $$\ln{{\pi\over 2}}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\eta(n)$$ to simplify to
$$-\ln{\left({A^6\over 2^{1\over6}\sqrt{\pi}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)$$
We have
$$\ln{\left[\prod_{k=1}^{\infty}\left({k\over k+1}\right)^{(-1)^{k+1}}\right]}=\sum_{n=1}^{\infty}{(-1)^n\eta(n)\over n}$$
We can't use this to apply on the above series.
I just wonder, is there an infinite product for
$$\ln{\left[F(k)\right]}=\sum_{n=1}^{\infty}{(-1)^n\eta(n)\over 1+n}$$
Can anyone please give a hand here? Thank you.
| Another approach is to use a common integral representation of the Dirichlet eta function, along with an integral representation of the Hurwitz zeta function.
Specifically, we can use $$\eta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \, dx , \quad \text{Re}(s) >0,\tag{1}$$ and $$\zeta(s,z) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-zx}}{1-e^{-x}} \, dx \tag{2}, \quad (\text{Re}(s) >1, \, \text{Re}(z) >0). $$
From $(2)$ it follows that
$$ \begin{align} \zeta(s)-\zeta \left(s, \frac{3}{2} \right) &= \frac{1}{\Gamma(s)} \int_{0}^{\infty}\frac{x^{s-1}}{1-e^{-x}} \left(e^{-x}-e^{-(3/2)x} \right) \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}}{1-e^{-2u}} \left(e^{-2u} - e^{-3u} \right) \, du \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}e^{-2u}} {1+e^{-u}} \, du \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}e^{-u}} {1+e^{u}} \, du, \quad \text{Re}(s) >0. \end{align}$$
And using $(1)$, we get $$ \begin{align} &\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n(n+1)} \, \eta(n) \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n+1} }{n(n+1)}\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{x^{n-1}}{1+e^{x}} \, dx \\ &= \int_{0}^{\infty} \frac{1}{1+e^{x}} \frac{1}{x^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!} \, \frac{x^{n+1}}{n+1} \, dx \\ &= \int_{0}^{\infty} \frac{1}{1+e^{x}} \frac{1}{x^{2}} \left(e^{-x}+x-1 \right) \, dx \\ &= \lim_{s \to -1^{+}} \int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \left(e^{-x}+x-1 \right) \, dx \\ &= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left( \zeta(s) - \zeta \left(s, \frac{3}{2} \right)\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \\&= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left(\zeta(s) - \zeta \left(s, \frac{1}{2} \right) +2^{s}\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \tag{3} \\&= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left(\zeta(s) - \left(2^{s}-1\right)\zeta (s) +2^{s}\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \tag{4} \\ &= \lim_{s \to -1^{+}} \left[\left(2^{1-s}-1 \right) \Gamma(s) \zeta(s) + \Gamma(s) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s)\right] \\&= \lim_{s \to -1^{+}} \left[ \Gamma(s) + \Gamma(s+1) \eta(s+1) -2\Gamma(s) \eta(s) \right] \tag{5}. \end{align}$$
Expanding about $s=-1$, we get
$$ \begin{align} &\lim_{s \to -1^{+}} \left[ \Gamma(s) + \Gamma(s+1) \eta(s+1) -2\Gamma(s) \eta(s) \right] \\ &= \small \lim_{s \to -1^{+}} \Big[\left(- \frac{1}{s+1} + \gamma-1 + \mathcal{O}(s+1) \right) + \left(\frac{1}{s+1} - \gamma + \mathcal{O}(s+1) \right) \Big(\frac{1}{2} +\eta'(0)(s+1) \\ &+ \small \mathcal{O}\left( (s+1)^{2} \right) \Big) - \small 2 \left(- \frac{1}{s+1} + \gamma-1 + \mathcal{O}(s+1) \right) \left(\frac{1}{4} + \eta'(-1)(s+1) + \mathcal{O}\left((s+1)^{2} \right) \right) \Big] \\ &= \lim_{s \to -1^{+}} \left[\eta'(0) + 2 \eta'(-1) - \frac{1}{2} + \mathcal{O}\left((s+1) \right) \right] \\&= \frac{1}{2} \, \ln \left(\frac{\pi}{2} \right) +2 \big( 4 \ln(2) \zeta(-1) -3 \zeta'(-1) \big) -\frac{1}{2} \tag{6} \\&= \frac{1}{2} \, \ln \left(\frac{\pi}{2} \right) +2 \left(-\frac{1}{3} \, \ln(2) -3 \left(\frac{1}{12} - \ln(A) \right) \right) - \frac{1}{2} \tag{7} \\&= \ln \left(\frac{\sqrt{\pi} \, A^{6}}{ 2^{7/6} e} \right) . \end{align}$$
$(3)$ $\zeta(s,a) = \zeta(s,a+1) + a^{-s}$
$(4)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (11)
$(5)$ $\eta(s) = (1-2^{1-s}) \zeta(s)$
$(6)$ http://mathworld.wolfram.com/DirichletEtaFunction.html (11)
$(7)$ https://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant
To expand $\Gamma(s+1)$ in a Laurent series about $s=-1$, use the identity $\Gamma(s+1) = \frac{\Gamma(s+2)}{s+1}$.
| {
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Prove $\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac{x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1$ for $a+b+c=m$
If $a+b+c=m$, prove that:
$$\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac {x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1.$$
My Attempt:
$a+b+c=m$
$a=m-b-c$
$b=m-a-c$
$c=m-a-b$,
L.H.S.$=\frac {x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}}+\frac {x^{2b}}{x^{2b}+x^{a+b}+x^{b+c}}+\frac {x^{2c}}{x^{2c}+x^{b+c}+x^{a+c}}$.
Now, how should I move on?
| Hint:
$$
\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} = \frac {x^{a}}{x^{a}+x^{m-b-a}+x^{m-c-a}} = \frac {x^{a}}{x^{a}+x^{b}+x^{c}}
$$
| {
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"source": "stackexchange",
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} |
How prove $\sum\limits_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le ab+bc+ca$ Let acute-angled triangle $ABC$,and $AB=c,BC=a,AC=b$,show that
$$\sum_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le ab+bc+ca$$
I try use AM-GM
$$\sum_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le\sum_{cyc}\dfrac{(a^2+b^2-c^2)+(a^2-b^2+c^2)}{2}=a^2+b^2+c^2$$
But this following not hold,because
$$a^2+b^2+c^2\ge ab+bc+ac$$
| Let $a^2+b^2-c^2=z^2$, $a^2+c^2-b^2=y^2$ and $b^2+c^2-a^2=x^2$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that $\sum\limits_{cyc}\sqrt{(x^2+y^2)(x^2+z^2)}\geq2(xy+cz+yz)$,
which after using C-S $\sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+yz$
gives $\sum\limits_{cyc}(x^2-xy)$, which is $\sum\limits_{cyc}(x-y)^2\geq0$.
Done!
| {
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How to quickly solve $y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx$? I'm currently trying to solve $$y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx,$$ which is a GRE math subject test problem. I was able to get the answer ($\sqrt{2}$) by breaking up the integral $$\int_{-\pi/4 }^{\pi/4 } \cos x \ dx+\int_{-\pi/4 }^{\pi/4 } \sqrt{1+x^2}\sin^3x\cos^3x \ dx.$$ Then I used $$1+x^2=(z-x)^2$$ to get $$t=\frac{z^2-1}{2z}$$ as well as $$\sin x = \frac{2z}{1+z^2}, \ \ \cos x = \frac{1-z^2}{1+z^2}, \ \ dx = \frac{2 \ dz}{1+z^2}.$$ Note that $$z=\tan(x/2).$$ I then plugged these into $$\int_{-\pi/4 }^{\pi/4 }\sqrt{1+x^2}\sin^3x\cos^3x \ dx$$ to get something rather ugly
\begin{align*}
&\int_{-\tan{\pi/8}}^{\tan{\pi/8}}\sqrt{1+ \frac{z^2-1}{2z}^2 } \cdot\left( \frac{2z}{1+z^2}\right)^3\cdot\left(\frac{1-z^2}{1+z^2}\right)^3\cdot\left(\frac{2}{1+z^2}\right)dz \\
&= \int_{-\tan{\pi/8}}^{\tan{\pi/8}} \frac{16 z^3 (1 - z^2)^3 \sqrt{\frac{(z^2 - 1)^2}{4 z^2} + 1}}{(z^2 + 1)^7}dz.
\end{align*}
Since this is an odd function, the solution is $0$ and thus $$y=\int_{-\pi/4 }^{\pi/4 } \cos x \ dx + 0 =\sqrt{2}.$$ Even though I was able to get to the answer, I'd imagine on the math GRE subject test, I would not have had enough time to figure all this out. Are there any tricks for quickly solving definite integrals like this?
| If you noticed that $\sqrt{1+x^2}\sin^3 x \cos^3 x$ was an odd function, you would immediately get that $$\displaystyle\int_{-\pi/4}^{\pi/4}\sqrt{1+x^2}\sin^3 x \cos^3 x\,dx = 0,$$
and thus, the given integral simplifies to just $y = \displaystyle\int_{-\pi/4}^{\pi/4}\cos x\,dx$.
| {
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Showing that $f(x) = x^3$ is injective? This is my attempt. Is it right? Is there a simpler way? Is there a way which relies on less background knowledge?
A function is injective iff $f(x) = f(y) \implies x = y$. This is equivalent to $x \not = y \implies f(x) \not = f(y)$. We will prove this latter statement by contradiction:
Suppose there are two numbers $x$, $x+a$ with $a >0$ and $x \not = x+a$ but $x^3 = (x+a)^3$. Expanding the RHS, subtracting $x^3$, and dividing both sides by $a>0$ yields $0 = 3x^2 + 3xa + a^2$. Taking this last equation as a quadratic in $x$, it has a real solution iff the discriminant is non-negative. But the discriminant is $9a^2 - 12a^2 = -3a^2$, so there are no real $x$ which satisfy $x^3 = (x+a)^3$.
| Is there a way which relies on less background knowledge?
Yes, there is (probably the most elementary one, which is based essentially in the monotonicity of multiplication):
Claim 1. If $x\neq 0$, $y\neq 0$ and $x^3=y^3$ then $x$ and $y$ have the same sign.
Proof: Assume that $x^3=y^3$ with $x>0$ and $y<0$. Then
$$0<x^3=y^3=(-1)^3(-y)^3<0,$$
which is a contradiction. $\square$
Claim 2: $f(x)=x^3$ is injective.
Proof: Note that $x^3=xxx\neq 0$ for all $x\neq 0$ and $0^3=0$. So, if $f(x)=x^3$ is not injective then there exist two real numbers $x$ and $y$ such that:
*
*$x\neq 0$, $y\neq0$
*$x\neq y$
*$x^3=y^3$
Without loss of generality, we can assume that $x<y$. It follows from Claim 1 that $0<x<y$ or $0>y>x$. Thus
$$x^3=xxx<yyy=y^3$$
which is a contradiction. $\square$
Remark. Note that this proof relies only in the following elementary facts:
*
*$a<b \text{ and }c>0\quad \Longrightarrow \quad ac<bc$
*$a<b \text{ and }c<0\quad \Longrightarrow \quad ac>bc$
*$ab=0\quad \Longrightarrow \quad a=0\text{ or }b=0$
*$-a=(-1)a\text{ and } -(-a)=a$
*$ab=ba$
*$0^2=0$
*$(-1)^2=1$
*$-1<0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Find the remainder of $51!$ when divided by $61$? Find the remainder of $51!$ when divided by $61$ ?
My Try :
By Wilson's theorem $60! ≡ −1\pmod{61}$
Then, I can write
$(60)(59)(58)(57)(56)(55)(54)(53)(52)51!≡−1\pmod{61}$
$(−1)(−2)(−3)(−4)(−5)(−6)(−7)(−8)(−9)51!≡−1\pmod{61}$
$(362880)51!\equiv1\pmod{61}$
How can I proceed after this OR Is there any other approach ?
| Continued from your working without evaluating $9!$ explicitly:
$$51!9!\equiv 1 \mod 61$$
$$51! 2(3)(4)(5)(6)(7)(8)(9) \equiv 1 \mod 61$$
Since $2(5)(6)=60$,
$$51! (60)(3)(4)(7)(8)(9) \equiv 1 \mod 61$$
$$51! (-1)(3)(4)(7)(8)(9) \equiv 1 \mod 61$$
$$51! (3)(4)(7)(8)(9) \equiv -1 \mod 61$$
Since $9(7)=63$,
$$51! (3)(4)(8)(63) \equiv -1 \mod 61$$
$$51! (3)(4)(8)(2) \equiv -1 \mod 61$$
Since $2(4)(8)=64$,
$$51! (3)(64) \equiv -1 \mod 61$$
$$51! (3)(3) \equiv -1 \mod 61$$
$$51! (9) \equiv -1 \mod 61$$
Let's do Euclidean algorithm to compute $9^{-1} \mod 61$:
$$61=9(6)+7$$
$$9=7+2$$
$$7=3(2)+1$$
Hence $$1=7-3(2)=7-3(9-7)=4(7)-3(9)=4(61-9(6))-3(9)=4(61)-27(9)$$
$$9^{-1}\equiv -27 \mod 61$$.
Hence $$51!\equiv 27 \mod 61$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077580",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to find $3^{3^{3^{\dots}}}\pmod{100}$? I can show that $3^{3^{3^n}}\equiv7\pmod{10}$ since
$3^1\equiv3\pmod{10}$
$3^2\equiv9\pmod{10}$
$3^3\equiv7\pmod{10}$
$3^4\equiv1\pmod{10}$
Thus, it reduces to $3^{(3^{3^n}\mod4)}$. I can then notice that
$3^1\equiv3\pmod4$
$3^2\equiv1\pmod4$
Reducing it down to $3^{(3^{(3^n\mod2)}\mod4)}=3^{(3^1\mod4)}=3^3\equiv7\pmod{10}$
However, this is tedious and not capable of solving the following problem:
$$3^{3^{3^{\dots}}}\pmod{100}$$
where the power tower keeps going up until the value becomes fixed for all further power towers. How would I take this problem?
To clarify a bit, we take the exponents at the top and work our way down. For example,
$$3^{3^3}=3^{27}\ne(3^3)^3$$
For clearer notation, $3^{3^{3^{\dots}}}=3\uparrow(3\uparrow(3\uparrow(\dots(3\uparrow n)\dots)))$, and we take as many $\uparrow$'s required such that for all $n,k\in\mathbb N$ we have
$$3\uparrow(3\uparrow(3\uparrow(\dots(3\uparrow n)\dots)))\equiv3\uparrow(3\uparrow(3\uparrow(\dots(3\uparrow k)\dots)))\pmod{100}$$
In Knuth's up arrow notation.
| We know that
$$3^2 \equiv 1 \pmod{4} \\
3^{20} \equiv 1 \pmod{25}$$
with the last following from Euler Theorem. Therefore
$$3^{20} \equiv 1 \pmod{100}$$
The problem then reduces to finding the powers of $3 \pmod{20}$.
Again
$$3^2 \equiv 1 \pmod{4} \\
3^4 \equiv 1 \pmod{5} \\$$
Therefore $3^4 \equiv 1 \pmod{20}$.
We thus have
$$3^{3^{3 ^{...}}} \equiv 3^{3^{3 ^{...}} \pmod{20}} \equiv 3^{3^{3 ^{...} \pmod{4}}} \pmod{100} \equiv 3^{3^{3 }} \pmod{100}\\
\equiv 3^{27}\equiv 3^{7} \pmod{100}$$
which is easy to calculate.
| {
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"url": "https://math.stackexchange.com/questions/2077664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Mathematical Induction on Fibonacci numbers I have ben stuck on this for a while.
Let $F(N)$ be the Fibonacci numbers with $F(1)=F(2)=1$. Show that $4(-1)^n + 5(F(N))^2$ is a square for all integers $N$.
| This doesn't prove it inductively, so if you specifically need an inductive proof, this wouldn't work.
Instead, this uses the closed form for the Fibonacci sequence, which is that $F(N)=\dfrac{\alpha^N-\beta^N}{\sqrt{5}}$, where $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}=\frac{-1}{\alpha}$.
The expression $4\cdot(-1)^N+5(F(N))^2$ becomes
$$4\cdot(-1)^N+5\left(\dfrac{\alpha^N-\beta^N}{\sqrt{5}}\right)^2
=4\cdot(-1)^N+\alpha^{2N}-2\alpha^N\beta^{N}+\beta^{2N}.$$
Since $\beta=\frac{-1}{\alpha}$, $2\alpha^N\beta^N=2\cdot(-1)^N$ and so our expression becomes $$\begin{align}
4\cdot(-1)^N+\alpha^{2N}-2(-1)^{N}+\beta^{2N}= \\
\alpha^{2N}+2(-1)^N+\beta^{2N}= \\
\alpha^{2N}+2\alpha^N\beta^N+\beta^{2N}= \\
(\alpha^N+\beta^N)^2
\end{align}$$ which is a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral involving $\sqrt{\tan x}$ $$\int \dfrac{1}{1+\sqrt{\tan x}}\quad dx$$
$$\int \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\quad dx$$
It is difficult for me to solve this integration.
| $\displaystyle \mathcal{I} = \int\frac{1}{1+\sqrt{\tan x}}dx = \int \frac{\sqrt{\cot x}}{1+\sqrt{\cot x}}dx$
substitute $\cot x= t^2$ and $\displaystyle dx = -\frac{1}{1+t^4}dt$
$\displaystyle \mathcal{I}= -\int\frac{t}{(1+t)(1+t^4)}dt = -\frac{1}{2}\int\frac{\bigg((1+t^4)+(1-t^4)\bigg)t}{(1+t)(1+t^4)}dt$
$\displaystyle = -\frac{1}{2}\int\frac{t}{1+t}dt-\frac{1}{2}\int\frac{(t-t^2)(1+t^2)}{1+t^4}dt$
$\displaystyle = -\frac{1}{2}\int \frac{(1+t)-1}{1+t}dt-\frac{1}{2}\int \frac{t+t^3-(t^2-1)-t^4-1}{1+t^4}dt$
$\displaystyle =-\frac{t}{2}+\frac{1}{2}\ln|t+1|-\frac{1}{4}\int\frac{2t}{1+t^4}-\frac{1}{2}\int\frac{t^3}{1+t^4}dt+\frac{1}{2}\int \frac{t^2-1}{1+t^4}dt-\frac{1}{2}t+\mathcal{C}$
all integrals are easy except $\displaystyle \mathcal{J} = \int\frac{t^2-1}{1+t^4}dt = \int\frac{1-t^{-2}}{\left(t+t^{-1}\right)^2-2}dt = \int\frac{(t-t^{-1})'}{(t-t^{-1})^2-2}dt$
| {
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Prove something similar to a variant of Cauchy-Shwarz inequality Cauchy-Shwarz Inequality is:
$$(a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 \leq (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2)$$
However, it can be manipulated as:
$$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2}$$
I'm tasked with proving the following inequality:
$$\sqrt{(a_1 + b_1 + \cdots + z_1)^2 + (a_2 + b_2 + \cdots + z_2)^2 + \cdots + (a_n + b_n \cdots + z_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2} + \cdots + \sqrt{z_1^2 + z_2^2 + \cdots + z_n^2}$$
I've proved both Cauchy-Shwarz and its manipulation, but am lost when it come to the inequality right above. Hints and/or solutions are welcome.
| Hint:
Replacing $b_i$ by $-b_i$ one can transform
$$
\sqrt{(a_1-b_1)^2+\cdots+(a_n-b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}
$$
into
$$
\sqrt{(a_1+b_1)^2+\cdots+(a_n+b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}
$$
Then,
\begin{align*}
\sqrt{(a_1+b_1+c_1)^2+\cdots+(a_n+b_n+c_n)^2}&=\sqrt{(a_1+(b_1+c_1))^2+\cdots+(a_n+(b_n+c_n))^2}\\
&\leq\sqrt{a_1^2+\cdots+a_n^2}+\sqrt{(b_1+c_1)^2+\cdots+(b_n+c_n)^2}.
\end{align*}
Applying the result again, to the second radical, we get
\begin{align*}
\sqrt{a_1^2+\cdots+a_n^2}&+\sqrt{(b_1+c_1)^2+\cdots+(b_n+c_n)^2}\\
&\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}+\sqrt{c_1^2+\cdots+c_n^2}.
\end{align*}
Now, use induction to complete the proof.
| {
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Circumradius of triangle ABC where H is orthocentre and $(AH)(BH)(CH) = 3 $ and $(AH)^2 + (BH)^2 + (CH)^2 = 7 $ Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC. Given $(AH)(BH)(CH) = 3$
and $ (AH)^2 + (BH)^2 + (CH)^2 = 7 $
Then what is the circumradius of triangle?
I know that AH = 2RcosA and so but I get stuck after two or three steps.
| As you wrote, we have
$$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C$$
Now using that
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1\tag1$$
(the proof for $(1)$ is written at the end of this answer)
we get
$$\frac{7}{4R^2}+2\cdot\frac{3}{8R^3}=1,$$
i.e.
$$(R+1)(2R-3)(2R+1)=0$$
to have
$$\color{red}{R=\frac 32}$$
Let us prove $(1)$.
We have
$$\begin{align}\cos 2A+\cos 2B+\cos 2C&=2\cos(A+B)\cos(A-B)+\cos 2C\\\\&=-2\cos C\cos(A-B)+2\cos^2C-1\\\\&=-2\cos C(\cos(A-B)-\cos C)-1\\\\&=-2\cos C\ (\cos(A-B)+\cos(A+B))-1\\\\&=-4\cos A\cos B\cos C-1\end{align}$$
from which we have
$$(2\cos^2A-1)+(2\cos^2B-1)+(2\cos^2C-1)=-4\cos A\cos B\cos C-1$$
| {
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Find the power series of rational function How to find the power series of rational functions of type, e.g.
$$\frac{1}{1-6x+12x^2},\frac{6x}{1-6x+12x^2}$$
where denominator can't be factorized over the real domain.
Is there a way to use the method of undetermined coefficients (by completing the square), or is it necessary to use complex domain and trigonometry?
| Here is an approach based upon the geometric series expansion (see the comment from @YvesDaoust).
We obtain
\begin{align*}
\frac{1}{1-6x+12x^2}&=\frac{1}{1-6x(1-2x)}\\
&=\sum_{k=0}^\infty (6x)^k(1-2x)^k\\
&=\sum_{k=0}^\infty (6x)^k\sum_{j=0}^k\binom{k}{j}(-2x)^j\\
&=1+6x+24x^2+72x^3+144x^4\\
&\qquad+\color{grey}{0}x^5-1728x^6-10368x^7-41472x^8-\cdots
\end{align*}
It is convenient to use the coefficient of operator $[x^n]$ to extract the coefficient of $x^n$.
We get
\begin{align*}
[x^n]\frac{1}{1-6x+12x^2}&=[x^n]\sum_{k=0}^\infty (6x)^k\sum_{j=0}^k\binom{k}{j}(-2x)^j\\
&=\sum_{k=0}^n6^k[x^{n-k}]\sum_{j=0}^k\binom{k}{j}(-2x)^j\tag{1}\\
&=\sum_{k=0}^n6^k\binom{k}{n-k}(-2)^{n-k}\\
&=(-2)^n\sum_{k=0}^n\binom{k}{n-k}(-3)^k\tag{2}
\end{align*}
and find this way a summation formula for the coefficient of the power series.
Comment:
*
*In (1) we use the linearity of the coefficient of operator and the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.
*In (2) we select the coefficient of $x^{n-k}$.
Note: Since the zeros of $f(x)=1-6x+12x^2$ are non-real, we don't expect to get a closed formula of the binomial sum formula in real numbers only. In fact the following is valid (see e.g. formula 1.70 in this paper by R. Sprugnoli).
\begin{align*}
\sum_{k=0}^n\binom{n-k}{k}z^k
=\frac{1}{\sqrt{1+4z}}
\left(\left(\frac{1+\sqrt{1+4z}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{1+4z}}{2}\right)^{n+1}\right)
\end{align*}
Replacing the index $k$ with $n-k$ gives
\begin{align*}
\sum_{k=0}^n\binom{k}{n-k}z^{n-k}
=\frac{1}{\sqrt{1+4z}}
\left(\left(\frac{1+\sqrt{1+4z}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{1+4z}}{2}\right)^{n+1}\right)
\end{align*}
This expression evaluated at $z=-\frac{1}{3}$ and multiplied with $(-2)^n$ gives the coefficient of the power series and we obtain
\begin{align*}
(-2)^n\sum_{k=0}^n\binom{k}{n-k}(-3)^k
=\frac{i\sqrt{3}}{2}
\left(\left(1-\frac{i}{\sqrt{3}}\right)^{n+1}-\left(1+\frac{i}{\sqrt{3}}\right)^{n+1}\right)3^n
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the last two digits in the decimal representation (base 10) of 17^20 My attempt is in this image. But I want the answer to this question by using modulo congruence method since the method used by me involves tedious calculations.
| The units digit of $17^1$ is equal to $7$.
The units digit of $17^2$ is equal to $9$.
The units digit of $17^3$ is equal to $3$.
The units digit of $17^4$ is equal to $1$.
Calculating,
$\quad 17^4 = (17^2)(17^2) = (289)(289) \equiv (-11)(-11) \equiv 21 \pmod{100}$
Continuing,
$\quad 21^2 \equiv 41 \pmod{100}$
$\quad 21^3 \equiv 61 \pmod{100}$
$\quad 21^4 \equiv 81 \pmod{100}$
$\quad 21^5 \equiv \;\,1 \pmod{100}$
So,
$\quad 17^{20} = (17^4)^5 \equiv 21^5 \equiv 1 \pmod{100}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
There exists $n$ such that there exists an n-digit number M for which $M(10^n + 1)$ is a perfect square. Show that there exists a positive integer $n$ greater than 3 such that there exists an $n$-digit positive integer $M$ for which $M(10^n + 1)$ is a perfect square.
| As I explained in the comments, this is impossible if $10^n+1$ is squarefree. On the other hand, if $p^2$ divides $10^n+1$ for some prime $p$, then $M_0=\frac{10^n+1} {p^2}$ is an integer, and $M_0(10^n+1)=\left(\frac{10^n+1} {p}\right)^2$ is a perfect square. $M_0$ does not necessarily have $n$ digits; but we can always multiply it by some power of $4$ or $9$ in such a way that the product $M$ has $n$ digits, and then $M$ will satisfy the conditions.
So all we have to do is prove that there is some $n$ for which $10^n+1$ is divisible by the square of a prime number $p$; $p$ cannot be $2$, $3$ or $5$, so we will try $p=7$. We want to have
\begin{align}10^n \equiv -1 \pmod{49}.\end{align}
By Euler's theorem
$$10^{42}\equiv 1 \pmod{49},$$
and thus
$$(10^{21}+1)(10^{21}-1)\equiv 0 \pmod{49}.$$
To conclude that $49|10^{21}+1$ it suffices to prove that it does not divide $10^{21}-1$. We have
\begin{align}10^{21}\equiv (10^{3})^7\equiv 10^3\equiv 3^3\equiv 27\equiv -1\pmod 7, \end{align}and thus $7\not |10^{21}-1$, which implies that $49\not |10^{21}-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions for $a^2+b^2+c^2=d^2$ I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $x\in\mathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
| Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$\bigg(b^2-a^2,2ab,2cd,c^2+d^2\bigg)$$
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the limit of $6^n(2-x_n)$ where $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ with $n$ roots
Let $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ where the expression in the RHS has $n$ roots.
Find the following limit: $\lim \limits_{n\to \infty}6^n(2-x_n)$
My approach: I had two approaches. The first one was the following: I showed that $x_n$ is increasing with upper bound which is equal 2 then by Weierstrass theorem its convergent with limit $2$. But we cannot deduce from that the limit of $6^n(2-x_n)$ is zero because the last expression is uncertainty.
The second one was that $x_{n+1}^3=6+x_n$ then $x_{n+1}^3-8=x_n-2$. From the last equation we get: $(x_{n+1}-2)(x_{n+1}^2+2x_{n+1}+4)=x_n-2$. I tried work out with this but I have stuck.
Would be very grateful for hints or solutions.
| takes $f(x)= \sqrt[3]{x+6}$ then see that $x_{n+1} = f(x_n)$
and $$|f'(x) |= \frac{1}{3\sqrt[3]{(x+6)^2}} < \frac{1}{3\sqrt[3]{36}} <\frac{1}{3\sqrt[3]{2^3\times 3}}=\frac{1}{6\sqrt[3]{3}}$$
so that
for any $x,y $ one has
$$|f(x)-f(y)| = |\int_x^yf'(s)ds|\le \frac{1}{6\sqrt[3]{3}}|x-y|$$
show then by induction using this inequality and the fact that $f(2)=2$
$$|x_n-2| \le \frac{1}{6^n\sqrt[3]{3^n}}|x_0-2|$$
so $$6^n| x_n-2|\le \frac{1}{\sqrt[3]{3^n}}|x_0-2| \to 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding the Smith Normal Form of an integer matrix I want to put the following integer matrix into Smith Normal Form:
$$\begin{pmatrix} -9 & 6 \\ 5 & -2 \\ 6 & 3 \end{pmatrix}$$
I have done this and found the answer to be $$\begin{pmatrix} 1 & 0 \\ 0 & 3 \\ 0 & 0\end{pmatrix}$$
Could someone verify whether this is correct or not?
| $$\begin{pmatrix}-9 & 6 \\ 5 & -2 \\ 6 & 3\end{pmatrix}\to\begin{pmatrix}2 & -5 \\ 6 & -9 \\ 3 & 6\end{pmatrix}\to \begin{pmatrix}2 & -5 \\ 0 & 6 \\ 1 & 11\end{pmatrix}\to\begin{pmatrix}1 & 11 \\ 0 & 6 \\ 0 & -27\end{pmatrix}\to\begin{pmatrix}1 & 0 \\ 0 & 6 \\ 0 & 3\end{pmatrix}\to\begin{pmatrix}1 & 0 \\ 0 & 3 \\ 0 & 0\end{pmatrix}$$
Rearrange rows and columns, reduce, rearrange again, reduce (twice), rearrange and reduce.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rotation Matrix of rotation around a point other than the origin In homogeneous coordinates, a rotation matrix around the origin can be described as
$R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix}$
with the angle $\theta$ and the rotation being counter-clockwise.
A translation amongst $x$ and $y$ can be defined as:
$T(x,y) = \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix}$
As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. The formula of this operations can be described in a simple multiplication of
$T(x,y) * R * T(-x,-y) \qquad (I)$
I find this to be counter-intuitive. In my understanding, it should be
$T(-x,-y) * R * T(x,y) \qquad (II)$
The two formulations are definitely not equal.
The first equation yields
$E1 = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & -x\cdot\cos(\theta)+y\cdot\sin(\theta)+x\\\sin(\theta) & \cos(\theta) & -x\cdot\sin(\theta)-y\cdot\cos(\theta)+y \\ 0&0&1\end{bmatrix}$
The second one:
$E2 = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & x\cdot\cos(\theta)-y\cdot\sin(\theta)-x\\\sin(\theta) & \cos(\theta) & x\cdot\sin(\theta)+y\cdot\cos(\theta)-y \\ 0&0&1\end{bmatrix}$
So, which one is correct?
| Your first formula is correct. Remember, the point to which this is applied appears on the RIGHT:
$$
T(x,y) * R * T(-x,-y) (P)
$$
So to evaluate the expression above, we first translate $P$ by $(-x, -y)$, then rotate the result, then translate back. Let's see what happens when $P$ is the point $(x, y, 1)$. That amounts to evaluating the following product:
\begin{align}
f((x, y))
&= \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix}
\begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix}
\begin{bmatrix}1&0&-x\\ 0& 1&-y\\0&0&1\end{bmatrix}
\begin{bmatrix}x\\ y\\1\end{bmatrix}\\
&= \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix}
\begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix}
\begin{bmatrix}0\\ 0\\1\end{bmatrix}\\
&= \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix}
\begin{bmatrix}0\\ 0\\1\end{bmatrix}\\
&= \begin{bmatrix}x\\ y\\1\end{bmatrix}\\
\end{align}
as expected: the point $(x, y)$ remains fixed by this composite transformation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
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