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Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder.
| $$(x+1)^{2n+1}+x^{n+2}=(x+1)\{(x+1)^2\}^n+x^{n+2}=(x+1)(x^2+2x+1)^n+x^{n+2}$$
Now as $x^2+2x+1\equiv x\pmod{x^2+x+1}$
$$(x+1)(x^2+2x+1)^n+x^{n+2}\equiv (x+1)(x)^n+x^{n+2}\pmod{x^2+x+1}$$
$$\equiv x^n(x+1+x^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Quadratic equations and inequalities $\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$ and $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$ For every positive integer $n$, prove that
$$\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$$
Hence or otherwise, prove that $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$, where $[x]$ denotes the greatest integer not exceeding $x$.
This question was posed to me in class by my teacher. Since we are discussing quadratic equations. I can only imagine that this question is related to that topic.
Actually, by squaring the terms on both sides of the inequality, the first part of the question is solved easily. It is the second half that is causing me trouble.
Clearly we have to show, that if $x <\sqrt{4n+1} < x+1$ where $x$ is a natural number, then $x <\sqrt{n} + \sqrt{n+1} < x+1$, but how?
I am in high school, so please use techniques appropriate for my level.
| $$\sqrt n+\sqrt{n+1}>\sqrt{4n+1}\\n+2\sqrt{n(n+1)}+n+1>4n+1\\\sqrt{n(n+1)}>n\\n^2+n>n^2\\n>0\\\sqrt{4n+2}>\sqrt n+\sqrt{n+1}\\4n+2>2n+1+2\sqrt{n(n+1)}\\2n+1>2\sqrt{n(n+1)}\\4n^2+4n+1>4n^2+4n\\1>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
When is the sum of consecutive squares a prime? For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?
I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$
| Let our numbers be $a^2,(a+1)^2,\dots,(a+x-1)^2$. The sum of the squares is $a^2 x+2a(1+\cdots+(x-1))+(1^2+\cdots+(x-1)^2)$.
Note that $a^2 x$ is divisible by $x$, as is $2a(1+\cdots+(x-1))$. So we concentrate on the term $1^2+2^2+\cdots+x^2$. Call this number $N$. By the formula quoted in the post, we have
$$6N=(x-1)(x)(2x-1).$$
Suppose that $x\gt 6$. Then $N$ has a factor $d\gt 1$ such that $d\mid x$. We show that $d$ is a proper divisor of $a^2+(a+1)^2+\cdots +(a+x-1)^2$. This is because
$$a^2+(a+1)^2 +\cdots +(a+x-1)^2 \gt \left(\frac{x-1}{2}\right)^2,$$
and $\left(\frac{x-1}{2}\right)^2\gt x$ if $x\gt 6$. Thus all candidate $x$ are in the interval from $1$ to $6$.
We can rule out $x=4$, since in that case our sum of squares is even, and clearly cannot be $2$.
We can also rule out $x=5$, because if $x=5$ then $5$ divides $N$, and it is easy to verify that a sum of $5$ consecutive squares must be greater than $5$.
That leaves $x=1$ (no good), $x=2$, $x=3$, and $x=6$. For each of $x=2$, $x=3$, and $x=6$, we can produce examples of a sum of squares of $x$ consecutive integers which is prime. The simplest example for $x=3$ uses the consecutive integers $-1$, $0$, and $1$. The simplest example for $x=6$ uses $-2,-1,0,1,2,3$. There are also examples with all entries positive.
It is not known whether there are infinitely many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Associated Legendre functions special values I should prove that
$$P_n^n(\cos \theta)=(2n-1)!! \sin^n\theta$$
$$P_n^m(0)=\begin{Bmatrix} (-1)^{(m+n)/2}\displaystyle\frac{(n+m-1)!!}{(n-m)!!} & \mbox{ if }& n+m \text{ even}\\ 0 & \mbox{if}& n+m \text{ odd}\end{Bmatrix}$$
$P_n^m(x)$ is a associated Legendre functions
I don't know what way follow. I apreciatte any advice to solve it.
Thanks a lot!!!
| The associated Legendre polynomials $P_l^m$ (which are actually not polynomials for odd m :-)) are given by
$$P_l^m(x) = \frac{(-1)^m}{2^ll!}(1-x^2)^{m/2}[(x^2-1)^l]^{(l+m)}$$
Where $f^{(n)}$ stands for the $n^{th}$ derivative of $f$. In this formula, $x = cos \theta \in[-1,1]$. We derive
\begin{equation}
\begin{split}
P_n^n(x)&=\frac{(-1)^n}{2^nn!}(1-x^2)^{n/2}[(x^2-1)^n]^{(2n)}
\\ &=\frac{(-1)^n(2n)!}{2^nn!}(1-x^2)^{n/2}
\\ &=(-1)^n(2n-1)!!(1-x^2)^{n/2}
\end{split}
\end{equation}
or, in goniometric terms,
$$P_n^n(cos \theta) = (-1)^n(2n-1)!! \cdot sin^n \theta$$
Next,
\begin{equation}
\begin{split}
P_n^m(0)&=\frac{(-1)^m}{2^nn!}(1-x^2)^{m/2}[(x^2-1)^n]^{(n+m)} \left.\right|_{x=0}
\\ &=\frac{(-1)^m}{2^nn!}[(x^2-1)^n]^{(n+m)} \left.\right|_{x=0}
\\ &=\frac{(-1)^m}{2^nn!}\left[\sum_{k=0}^n \binom{n}{k}x^{2n-2k}(-1)^k\right]^{(n+m)} \left.\right|_{x=0}
\\ &=\begin{cases}
0 &\text{ if $n+m$ odd}
\\ \text {see below} &\text{ if $n+m$ even}
\end{cases}
\end{split}
\end{equation}
The latter is determined by the term with $2n-2k = n+m$, hence $k = \frac{n-m}{2}$, and can be evaluated as follows:
\begin{equation}
\begin{split}
P_n^m(0)&=\frac{(-1)^m}{2^nn!}\binom{n}{(n-m)/2}(n+m)!(-1)^{(n-m)/2}
\\ &= \frac{(-1)^{(n+m)/2}}{2^nn!}\frac{n!}{\frac{n-m}{2}!\frac{n+m}{2}!}(n+m)!
\\ &= (-1)^{(n+m)/2}\frac{(n+m)!}{(n-m)!!(n+m)!!}
\\ &= (-1)^{(n+m)/2}\frac{(n+m-1)!!}{(n-m)!!}
\end{split}
\end{equation}
| {
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"url": "https://math.stackexchange.com/questions/763924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Odd series convergence Prove that we have following inequality:
$1+ \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{397} > \frac{9}{4}$
Anybody can help me to figure it out?
| A silly approach:
$$
\sum_{n=0}^{198} \frac{1}{2n+1} > \int_0^{198} \frac{dx}{2x+1} = \frac{1}{2} \log 397 > \frac{1}{2} \log 361 = \log 19,
$$
$$
e^{9/4} < 3^{9/4} < 3^{10/4} = 9 \sqrt{3} < 9 \cdot 2 = 18.
$$
A more sensible approach:
Write
$$
\sum_{n=1}^{199} \frac{1}{2n-1} > \sum_{n=1}^{\large 2^7} \frac{1}{2n-1} > \frac{1}{2} \sum_{n=1}^{\large 2^7} \frac{1}{n}
$$
then use Cauchy condensation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/764846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving this linear second order Cauchy problem Here is my problem, I know that a function is a solution of this linear Cauchy problem
$$
\left\{
\begin{array}{rcl}
y'' &=& \frac{x^2+6}{4}y,\\
y(0)&=&0,\\
y'(0)&=& \frac{\sqrt{\pi}}{2}.
\end{array}
\right.
$$
and I want to find the solution of this problem with a simple method (I know the solution it is $x\mapsto \frac{\sqrt{\pi}}{2} e^{\frac{x^2}{4}}x$).
In fact this function is also a solution of
$$
\left\{
\begin{array}{rcl}
y'' -\frac{x}{2}y'-y&=&0,\\
y(0)&=&0,\\
y'(0)&=& \frac{\sqrt{\pi}}{2}.
\end{array}
\right.
$$
So here is my question what is the simplest method (and eventually what is the good differential equation) to get the solution. The general solution of this two differential equations involves the error function so I think that we need a method for this specific Cauchy problem. My idea was : since the solution is quite simple and has a power series, we may assume at first that the solution has a power series and derive from the equation this power series and then recognize the solution. But this problem comes from an difficult exercise and solving the differential equation is the last question and I think it is supposed to be easy (to find a solution of the ODE) in order to quickly finish the exercise (in fact the first question is "here is an integral with a parameter, show that it defines a $\mathcal{C}^2$ function", second question, "find an linear second order equation such that the previous integral with a parameter is a solution of the ODE", and last question "find an simple expression for the integral with a parameter").
| Apply the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf:
Let $y=e^\frac{x^2}{4}~u$ ,
Then $y'=e^\frac{x^2}{4}~u'+\dfrac{xe^\frac{x^2}{4}}{2}u$
$y''=e^\frac{x^2}{4}~u''+\dfrac{xe^\frac{x^2}{4}}{2}u'+\dfrac{xe^\frac{x^2}{4}}{2}u'+\biggl(\dfrac{x^2e^\frac{x^2}{4}}{4}+\dfrac{e^\frac{x^2}{4}}{2}\biggr)u=e^\frac{x^2}{4}~u''+xe^\frac{x^2}{4}~u'+\dfrac{(x^2+2)e^\frac{x^2}{4}}{4}u$
$\therefore e^\frac{x^2}{4}~u''+xe^\frac{x^2}{4}~u'+\dfrac{(x^2+2)e^\frac{x^2}{4}}{4}u=\dfrac{x^2+6}{4}e^\frac{x^2}{4}~u$
$e^\frac{x^2}{4}~u''+xe^\frac{x^2}{4}~u'-e^\frac{x^2}{4}~u=0$
$u''+xu'-u=0$
$u=x$ is a particular solution
$\therefore$ Let $u=xv$ ,
Then $u'=xv'+v$
$u''=xv''+v'+v'=xv''+2v'$
$\therefore xv''+2v'+x(xv'+v)-xv=0$
$xv''+(x^2+2)v'=0$
$xv''=-(x^2+2)v'$
$\dfrac{v''}{v'}=-x-\dfrac{2}{x}$
$\int\dfrac{v''}{v'}dx=\int\left(-x-\dfrac{2}{x}\right)dx$
$\ln v'=-\dfrac{x^2}{2}-2\ln x+c$
$v'=\dfrac{c_2e^{-\frac{x^2}{2}}}{x^2}$
$v=c_2\int\dfrac{e^{-\frac{x^2}{2}}}{x^2}dx$
$v=-c_2\int e^{-\frac{x^2}{2}}~d\left(\dfrac{1}{x}\right)$
$v=-c_2\biggl(\dfrac{e^{-\frac{x^2}{2}}}{x}-\int\dfrac{1}{x}d\biggl(e^{-\frac{x^2}{2}}\biggr)\biggr)$
$v=C_2\biggl(\dfrac{e^{-\frac{x^2}{2}}}{x}+\int e^{-\frac{x^2}{2}}~dx\biggr)$
$\dfrac{e^{-\frac{x^2}{4}}y}{x}=C_1+C_2\biggl(\dfrac{e^{-\frac{x^2}{2}}}{x}+\int_0^xe^{-\frac{x^2}{2}}~dx\biggr)$
$y=C_1xe^\frac{x^2}{4}+C_2\biggl(e^{-\frac{x^2}{4}}+xe^\frac{x^2}{4}\int_0^xe^{-\frac{x^2}{2}}~dx\biggr)$
$y(0)=0$ :
$C_2=0$
$\therefore y=C_1xe^\frac{x^2}{4}$
$y'=\dfrac{C_1(x^2+2)e^\frac{x^2}{4}}{2}$
$y'(0)=\dfrac{\sqrt\pi}{2}$ :
$C_1=\dfrac{\sqrt\pi}{2}$
$\therefore y=\dfrac{\sqrt\pi xe^\frac{x^2}{4}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove the inequality $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$
If $a,b,c\in\mathbb R^+$ prove that:
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$
| Apply Am-GM to the numerator of each fraction.
You get the statement of Nesbit inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sturm Liouville with periodic boundary conditions Background and motivation: I'm given the boundary value problem:
$$y''(x)+2y(x)=-f(x)$$
subject $y(0)=y(2\pi)$ and $y \, '(0)=y \, '(2\pi)$.
EDIT: These were not given to be zero !! Maybe this helps...
The text (Nagle Saff and Snider, end of Chapter 11 technical writing exercise) asks us to construct the Green's function for the problem. At the moment, I'm a bit stumped because there is no $\lambda$ in the given problem. Let me elaborate, if we were given:
$$ (py')'+qy+\lambda r y= 0 $$
where $p,p',q$ and $r$ were continuous, real-valued, periodic functions with period $2\pi$ then I think I'd be able to get started. I know the usual solutions then only fit the given boundary conditions for particular choices of $\lambda$. So, my initial observation is that $p=1$ is certainly continuous and periodic so we can set $p=1$.
*
*Question: what should I see as $q$ and $r$ for the problem stated at the start of this post? How can we massage the given problem into the standard form of Sturm Liouville?
I suppose it is important to note we must choose $r>0$ as it serves as the weight function in the inner product which is paired with the eigenspace of solutions for this problem.
Added: here is a picture of the problem from the text:
| The Green's function is the solution when $f(x)=\delta(x-x_s)$, where $x_s$ is some kind of point source position that forces the system. Let's suppose that $x_s\in(0,2\pi)$. For $x\neq x_s$, the delta function is zero, and so we solve the homogeneous equation
$$
\left|
\begin{array}{cc}
y'' + 2y = 0, & x<x_s\\
y'' + 2y = 0, & x>x_s
\end{array}
\right.
$$
And we'll worry about what happens right at $x=x_s$ in a bit. The homogeneous equations presented are solved by sines and cosines
$$
y(x) = \left\{
\begin{array}{cc}
A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x), & x<x_s\\
C\cos(\sqrt{2}x)+D\sin(\sqrt{2}x), & x>x_s\\
\end{array}
\right.
$$
Now the first boundary condition is that $y(0)=y(2\pi)$. For the left boundary, we use the left part of the piecewise $y$ above, and for the right boundary, we use the right part, so this reads
$$
A = C\cos(2\sqrt{2}\pi)+D\sin(2\sqrt{2}\pi)
$$
Doing the same with the derivative conditions gives
$$
B = D \cos \left(2 \sqrt{2} \pi \right)-C \sin \left(2 \sqrt{2} \pi \right)
$$
Now we need conditions to match the delta function. We expect the solution to be continuous at $x=x_s$, so
$$
A\cos(\sqrt{2}x_s)+B\sin(\sqrt{2}x_s) = C\cos(\sqrt{2}x_s)+D\sin(\sqrt{2}x_s)
$$
And then there is the appropriate "jump" condition on the derivative at $x=x_s$. We need $y$ to be discontinuous enough so that taking two derivatives will result in a negative delta function. This means that there must be a negative unit step discontinuity in the derivative:
$$
-\sqrt{2} A \sin \left(\sqrt{2} x_s\right)+\sqrt{2} B \cos\left(\sqrt{2} x_s\right)-1=\sqrt{2}D\cos\left(\sqrt{2} x_s\right)-\sqrt{2} C \sin \left(\sqrt{2} x_s\right)
$$
The above are four linear equation in the four unknowns $(A,B,C,D)$, which we can formulate as
$$
\left[
\begin{array}{cccc}
1 & 0 & -\cos(2\sqrt{2}\pi) & -\sin(2\sqrt{2}\pi) \\
0 & 1 & \sin \left(2 \sqrt{2} \pi \right) & -\cos \left(2 \sqrt{2} \pi \right)\\
\cos(\sqrt{2}x_s)&\sin(\sqrt{2}x_s)&-\cos(\sqrt{2}x_s)&-\sin(\sqrt{2}x_s) \\
-\sqrt{2} \sin \left(\sqrt{2} x_s\right)&\sqrt{2} \cos\left(\sqrt{2} x_s\right)&\sqrt{2}\sin \left(\sqrt{2} x_s\right)&-\sqrt{2}\cos\left(\sqrt{2} x_s\right)
\end{array}
\right]
\left[
\begin{array}{c}
A\\B\\C\\D
\end{array}
\right] =
\left[
\begin{array}{c}
0\\0\\0\\1
\end{array}
\right]
$$
Solving that system after a good bit of trigonometric simplifications gives
$$
A=-\frac{ \cos \left(\sqrt{2} (x_s-\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}
$$
$$
B=\frac{ \sin \left(\sqrt{2} (\pi -x_s)\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}
$$
$$
C=-\frac{ \cos \left(\sqrt{2} (x_s+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}
$$
$$
D=-\frac{ \sin \left(\sqrt{2} (x_s+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}
$$
Plugging into the piecewise form proposed for $y$ and doing more trigonometric simplifying,
$$
y(x) = \left\{
\begin{array}{cc}
-\frac{ \cos \left(\sqrt{2} (x-x_s+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}, & x<x_s\\
-\frac{ \cos \left(\sqrt{2} (x_s-x+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}, & x>x_s\\
\end{array}
\right.
$$
Notice these are the same except the role of $x$ and $x_s$ are switched between the two expressions. This allows us to write a more compact expression
$$
y(x)=-\frac{ \cos \left(\sqrt{2} (\pi-|x-x_s| )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}
$$
which is the Green's function for this problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Computing the Gaussian curvature of this surface $z=e^{(-1/2)(x^2+y^2)}$. Compute the Gaussian curvature of $z=e^{(-1/2)(x^2+y^2)}$. Sketch this surface and show where $K=0 $, $K>0$, and $K<0$.
So would the easiest way to do this question be to construct a parametrization $$\mathbf{x}(u,v)=(u, v, e^{-\frac{1}{2}(u^2+v^2)} )?$$
If so, I calculated the Normal to be $$\left( \frac{u}{u^2+v^2+e^{u^2+v^2}}, \frac{v}{u^2+v^2+e^{u^2+v^2}}, 1 \right).$$ Is that correct? Thanks
| The parametrization $$\sigma = (u,v, e^{-(x^2+y^2)/2})$$ would indeed be the simplest parametrization, but not for computational purposes.
So we can also let $x = r\cos\theta$ and $y = r\sin\theta$ and get,
$$\sigma = (r\cos\theta,r\sin\theta, e^{-r^2/2}).$$
After finding the normal vector, we get
$$\mathbf{N} = \frac{(e^{-r^2/2}r^2 \cos \theta, e^{-r^2/2}r^2 \sin\theta, r)}{|r|\sqrt{(e^{-r^2}r^2 +1)}}$$ which looks a lot like what you have got.
Find the component of the second fundamental, and we get (we will take $r > 0$ for now)
Therefore,
$$L = \left < (0,0,(r^2 - 1)e^{\frac{-r^2}{2}}), \mathbf{N} \right > = \frac{r(r^2 - 1)e^{\frac{-r^2}{2}}}{\sqrt{(e^{-r^2}r^2 +1)}}$$
$$M = \left < (-\sin \theta, \cos \theta,0 ), \mathbf{N} \right > = 0$$
$$N = \left < (-r\cos \theta, -r\sin \theta,0 ), \mathbf{N} \right > = -r^3e^{\frac{-r^2}{2}}$$
$$LN - M^2 = LN = \frac{-r^4(r^2 - 1)e^{-r^2}}{ \sqrt{(e^{-r^2}r^2 +1)} }$$
On the other hand,
$$E = \| (\cos \theta, \sin\theta, -re^{\frac{-r^2}{2}}) \|^2 = 1 + r^2 e^{-r^2}$$
$$F = \left < (-r\sin \theta, r\cos\theta, 0), (\cos \theta, \sin\theta, -re^{\frac{-r^2}{2}})\right > = 0$$
$$G = \| (-r\sin \theta, r\cos\theta, 0) \|^2 = r^2$$
$$EG - F^2 = EG = r^2(1 + r^2 e^{-r^2})$$
So FINALLY,
$$K = \frac{LN - M^2}{EG - F^2 } = \frac{LN}{EG} = \frac{1}{r^2(1 + r^2 e^{-r^2})} \frac{-r^4(r^2 - 1)e^{-r^2}}{ \sqrt{(e^{-r^2}r^2 +1)} } = \frac{r^2(1-r^2)e^{-r^2}}{(1+r^2 e^{-r^2})^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.
| Completing the square will yield
$$
x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}
$$
Normally, we will let $u=x-\frac{1}{2}$. However it can also be solved by letting $x-\frac{1}{2}=\frac{\sqrt3}{2}\sinh t$ and $dx=\frac{\sqrt3}{2}\cosh t\ dt$ which yields
$$
\begin{align}
\int \frac{dx}{\sqrt{x^{2}-x+1}}&=\int \frac{\frac{\sqrt3}{2}\cosh t\ dt}{\sqrt{\frac{3}{4}\sinh^2 t+\frac{3}{4}}}\\
&=\int \frac{\cosh t\ dt}{\sqrt{\cosh^2 t}}\\
&=\int \ dt\\
&=t+C
\end{align}
$$
where $\sinh t=\dfrac{2x-1}{\sqrt3}\;\Rightarrow\; t=\sinh^{-1}\left(\dfrac{2x-1}{\sqrt3}\right)$. Thus
$$
\int \frac{dx}{\sqrt{x^{2}-x+1}}=\sinh^{-1}\left(\dfrac{2x-1}{\sqrt3}\right)+C.
$$
As your book's solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Distance point on ellipse to centre I'm trying to calculate the distance of a certain point of an ellipse to the centre of that ellipse:
The blue things are known: The lengths of the horizontal major radius and vertical minor radius and the angle of the red line and the x-axis. The red distance is the desired result. It is not given where on the ellipse the point is. It can be anywhere on the ellipse. Is this problem possible? If so, in which can this be solved? Thanks in advance!
After reading Kaj Hansen's comment and trying a bit this is what I did, it still won't work though.
In a triangle, $tan(\theta)=\frac{\text{opposite side}}{\text{adjecent side}}$. The slope of a line is $\frac{\Delta y}{\Delta x}$. Therefor the slope of the red line is $\tan(\theta)$; the formula of the line is $y=\tan(\theta)\cdot x$.
The formula of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. When I put the two formulas together I get
$$\frac{x^{2}}{a^{2}}+\frac{(\tan(\theta)\cdot x)^{2}}{b^{2}}=1$$
After a bit of rearranging:
$$x=\pm \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$
$$$$
$$y=\tan(\theta)\cdot x$$
$$y=\pm \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} $$
Now with the help of Pythagoras' theorem $c=\sqrt{a^{2}+b^{2}}$ the red line should be
$$\sqrt{\left ( \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}\right ) ^{2}+\left ( \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} \right )^{2}}$$
which can be simplified:
$$\sqrt{\frac{a^{2}\cdot b^{2} \cdot (\tan(\theta))^{2}+a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$
This, however, does not give the right answer. Let's try something:
$a=2$; $b=1$; $\theta=\frac{1}{2}\cdot \pi$ (The point is the point where the ellipse intersects with the minor radius)
$$\sqrt{\frac{2^{2}\cdot 1^{2} \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+2^{2}\cdot 1^{2}}{2^{2}\cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+1^{2}}}$$
$$\sqrt{\frac{4 \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+4}{4\cdot (\tan(\frac{1}{2} \pi))^{2}+1}}$$
But wait, $\tan(\frac{1}{2}\cdot \pi)$ is undefined. The formula cannot be filled in completely, which is a requirement. I need a formula that can be filled in for every value of $\theta$ on the domain $[0,\frac{1}{2}\cdot \pi]$
| I'm assuming that the centre is located on the origin.
Let us call the point where the red line meets the ellipse P. Let P have coordinates $(x_1, y_1) $. Notice that $tan\theta = y_1/x_1$. Let the horizontal semi-axis be $a$, meaning the ellipse is $2a$ wide. Similarly, let the vertical semi-axis be $b$.
Let the known blue angle be $\theta$.
Equation of an ellipse(centered at origin) =
$x^2/a^2 + y^2/b^2 = 1$
Since $(x_1,y_1)$ satisfies the equation, the equation becomes:
$x_1^2/a^2 + y_1^2/b^2 = 1$
But $tan\theta = y_1/x_1$ so $y_1^2 = x_1^2tan^2\theta$. Substituting this:
$x_1^2/a^2 + x_1^2tan^2\theta/b^2 = 1$
$x_1^2(1/a^2 + tan^2\theta/b^2) = 1$
$x_1^2(\frac{a^2tan^2\theta+b^2}{a^2b^2}) = 1$
$x_1^2= \frac{a^2b^2}{a^2tan^2\theta+b^2}$
Finding the value of $y_1^2$, we get:
$y_1^2 = \frac{a^2b^2}{a^2tan^2\theta+b^2}\times tan^2\theta$
$y_1^2 = \frac{a^2b^2tan^2\theta}{a^2tan^2\theta+b^2}$
Since we have assumed the centre to be the origin, the distance becomes:
$\sqrt{x_1^2+y_1^2}$
$\sqrt{\frac{a^2b^2}{a^2tan^2\theta+b^2}+\frac{a^2b^2tan^2\theta}{a^2tan^2\theta+b^2}}$
Since $1+tan^2\theta = sec^2\theta$, we have:
$\sqrt{\frac{a^2b^2sec^2\theta}{a^2tan^2\theta+b^2}}$
Addressing your problem, note that both tan and sec don't have continuous domains, however, both of them have cos as the denominator in one or the other representation. So, we can multiply both the numerator and denominator by $cos\theta$ to get rid of the discontinuous domain.
$\sqrt{\frac{a^2b^2sec^2\theta}{a^2tan^2\theta+b^2}}\times\sqrt{\frac{cos^2\theta}{cos^2\theta}}$
$\sqrt{\frac{a^2b^2}{a^2sin^2\theta+b^2cos^2\theta}}$
$\frac{ab}{\sqrt{a^2sin^2\theta+b^2cos^2\theta}}$
You could rationalize the denominator if you want. This formula works for $\theta = \pi/2$
Note: Even though we have assumed the ellipse to be centred at origin, the formula still works as any other ellipse with same a,b would be congruent to the discussed one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
} |
Prove that for infinitely many $n$, the numbers $(a+n)$,$(b+n)$,$(c+n)$ are mutually prime The problem is -
Let $a$, $b$ and $c$ be three distinct integers such that they are mutually prime. Show that for infinitely many $n$, the numbers $(a+n)$,$(b+n)$,$(c+n)$ are mutually prime.
I have been able to show that for infinitely many $n$ we can find $\gcd (a,b,c)$ $=$ $1$. The proof is very simple. Since by hypothesis the integers are distinct therefore there should be an ordering among them. Let $a$ $>$ $b$ $>$ $c$. Now choose $n$ to be such that $(a+n)$ is a prime. It is now easy to see that $\gcd (a,b,c)$ $=$ $1$.
| We prove that if $a$, $b$, and $c$ are distinct, then there are infinitely many $n$ such that $a+n$, $b+n$, and $c+n$ are pairwise relatively prime. We do not need to assume that $a$, $b$ and $c$ are pairwise relatively prime.
Without loss of generality we may assume that $a=0$ and $b$ and $c$ are positive, with $b\lt c$. Call the positive integer $n$ good if (i) $n$ is relatively prime to $b$, $c$, and $c-b$ and (ii) $n$ is divisible by any prime $\lt c$ that does not divide any of $b$, $c$, or $c-b$. There are infinitely many good $n$. We show that for any good $n$, the numbers $a+n$ (that is, $n$), $b+n$, and $c+n$ are pairwise relatively prime.
Suppose that $n$ and $b+n$ are not relatively prime. Then there is a prime $p$ that divides $n$ and $b+n$, and therefore that divides $b$. But this contradicts our choice of $n$. The same argument shows that $n$ and $c+n$ are relatively prime.
Finally, we show that $b+n$ and $c+n$ are relatively prime. If not, there is a prime $p$ that divides $b+n$ and $c+n$. It follows that $p$ divides $c-b$. So $p\lt c$, and therefore $p$ divides $n$. But then $p$ divides $b$ and $c$, contradicting the choice of $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.
Edited to add that it was a multiple choice question, with possible answers:
a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$
| Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.
$$\begin{align}
& \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\
= & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\
= & \sqrt{25-1} \\
= & 2\sqrt{6}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
$n$ is a square and a cube $a^2 = n = b^3\Rightarrow n\equiv 0,1\pmod{7}$
Verify that if an integer is simultaneously a square and a cube, then it must be either of the form ${7k}$ or ${7k +1}$.
I have no idea on how to proceed.
| Check the integers mod $7$. Which are squares and which are cubes:
$$0^2=0^3=0$$
$$1^2=1^3=1$$
$$2^2 = 4,\qquad 2^3=8\equiv 1$$
$$3^2 = 9 \equiv 2,\quad3^3=27\equiv 6$$
$$4^2 = 16 \equiv 2,\quad 4^3= 64 \equiv 1$$
$$5^2 = 25 \equiv 4,\quad 5^3= 125 \equiv 6$$
$$6^2 = 36 \equiv 1,\quad 6^3=216\equiv 6.$$
So only $0, 1, 2$ and $4$ are squares, only $0, 1$ and $6$ are cubes (mod $7$). The ones that are both give the wanted answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/774237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Limit as x approches infinity - Trouble with calculus I have this problem on a practice exam:
$\displaystyle \lim_{x\to\infty} 3x - \sqrt{9x^2+2x+1}$
We are dealing with L'hospitals rule, so when you plug $\infty$ in for $x$ you get
$\infty - \infty$. I multiplied by the conjugate to get:
$\displaystyle \frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}$
After applying L'hospitals a few times I was lost so I peeked at the solutions sheet and my teacher went from the above step straight to:
$\displaystyle \frac{-2}{3+3} = -\frac{1}{3}$
I'm kind of lost as to how she came to that answer. Can anyone explain to me how that works?
| This is to answer "what is your thought process" (I cannot read LAcarguy's mind so I cannot say what his thought process is). The trick of multiplying and dividing by the reciprocal of a leading term is a standard trick when we have a limits going to infinity. For instance suppose that we have the following limit, $$\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}.$$ Well note that the leading term of the numerator is $2x^2$ and the denominator is has leading term $5x^2$. So this suggests that $$\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}=\frac{2}{5}.$$ Now if we do this properly, we multiply numerator and denominator by $\frac{1}{x^2}$. This gives us,
$$\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}=\\
\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}\frac{1/x^2}{1/x^2}=\\
\lim_{n\to\infty}\frac{2+3/x+5/x^2}{5-3/x+100/x^2}\\ = \frac{2}{5}.$$ Now this is not the problem that you originally posed, so lettuce now look at that.
$$\lim_{n\to\infty}\frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}.$$ The long term behavior of the numerator is the same as the long term behavior of $-2x$, in particular the leading term. Now the denominator is a little more tricky. If we look at the long term behavior of $$\sqrt{9x^2+2x+1}$$, this will tell us what we want to know. But the long term behavior of the polynomial inside the radical is $$9x^2.$$ So then the long term behavior of the radical ,$$\sqrt{9x^2+2x+1}$$ is $$\sqrt{9x^2}=3x$$ as long as $x$ is nonnegative. Therefore the long term behavior of the denominator, $${3x + \sqrt{9x^2+2x+1}}$$ is $$3x+3x=6x.$$ Therefore, the long term behavior of the ratio $$\frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}$$ is $$\frac{-2x}{6x}=-\frac{1}{3}.$$ So we might think that $$\lim_{n\to\infty}\frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}=-\frac{1}{3}.$$ Well now if we do this properly, what we do is multiply the numerator and denominator by $1/x$, and we get the solution that LAcarguy gives.
| {
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"url": "https://math.stackexchange.com/questions/775073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the values of a and b? If the polynomial 6x4 + 8x3 - 5x2 + ax + b is exactly divisible by the polynomial 2x2 - 5, then find the values of a and b.
| You can just write $$6x^4 + 8x^3 - 5x^2 + ax + b=(2x^2-5)(Ax^2+Bx+C)$$ Expand the rhs to get $$6x^4 + 8x^3 - 5x^2 + ax + b=2Ax^4+2Bx^3+(2C-5A)x^2-5Bx-5C$$ Now identify the coefficients of the differents powers of $x$. Using $x^4,x^3,x^2$, this leads to $A=3$, $B=4$,$C=5$.
I am sure that you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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is 121 divides the my pattern for base 2? Is $121|2^{120}-1$? If yes, is there any online free calculation to check these type of values?
Advanced thanks to one and all!
-Richard Sieman
| You can rewrite your expression as$$2^{120}-1\\=(2^{60}+1)(2^{60}-1)\\=(2^{60}+1)(2^{30}+1)(2^{30}-1)\\=(2^{60}+1)(2^{30}+1)(2^{15}+1)(2^{15}-1)$$
so $$2^{120}-1\pmod{ 121}\\\equiv(2^{60}+1)(2^{30}+1)(2^{15}+1)(2^{15}-1)\pmod{ 121}$$
but since $2^{15}-1=32767 \equiv 97 \pmod{ 121}$
$$2^{120}-1\pmod{ 121}\\\equiv(2^{60}+1)(2^{30}+1)(2^{15}+1)(2^{15}-1)\pmod{ 121}\\\equiv(2^{60}+1)(2^{30}+1)\cdot97\cdot99\pmod{ 121}\\\equiv(2^{60}+1)\cdot44\cdot46\pmod{ 121}\\\equiv90\cdot88\pmod{ 121}\\\equiv55\pmod{ 121}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for the value of $$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \left(\frac{1}{n} \right)? $$
Unlike the other infinite series, this infinite series only converges conditionally.
| This is not an answer, but a useful way to transform the expression and link it to another, more simple sum:
$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\frac{\pi}{4}-\sum_{n=1}^{\infty} \left( \arctan \frac{1}{2n}-\arctan \frac{1}{2n+1}\right)$$
$$\arctan \frac{1}{2n}-\arctan \frac{1}{2n+1}=\arctan \frac{1}{4n^2+2n+1}$$
$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\frac{\pi}{4}-\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2+2n+1}$$
From this question (ananswered by the way) we have an identity:
$$\sum_{n=0}^{N} \arctan \frac{1}{n^2+n+1}=\arctan(N+1)$$
Which means
$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^2+n+1}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$
Considering:
$$(2n-1)^2+2n-1+1=4n^2-2n+1$$
We get another identity (separating even and odd terms):
$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^2+n+1}=\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2+2n+1}+\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2-2n+1}$$
And now we have:
$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2-2n+1}$$
The convergence of these two sums is slightly better than the original
If we take geometric mean of the last two expressions, it gives excellent convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 5
} |
Given $g(1) = 6$, $g'(1) = -1$, find $d/dx(2 g(x)/(x^2 + 1))$ when $x = 1$ Why is the answer $-7$? I plugged $1$ into the equation and I ended up with $12/2$ and got $6$. Can someone explain to me what I did wrong?
| Applying Quotient Rule, we find that:
\begin{align*}
\frac{d}{dx}\left[\frac{2g(x)}{x^2+1} \right]
&= \frac{(x^2+1)\frac{d}{dx}[2g(x)] - 2g(x)\frac{d}{dx}[x^2+1]}{(x^2+1)^2} \\
&= \frac{(x^2+1)[2g'(x)] - 2g(x)[2x]}{(x^2+1)^2} \\
&= \frac{2(x^2+1)g'(x) - 4xg(x)}{(x^2+1)^2} \\
\end{align*}
Hence, substituting $x = 1$, we obtain:
$$
\frac{2(1^2+1)g'(1) - 4(1)g(1)}{(1^2+1)^2}
= \frac{4(-1) - 4(6)}{4}
= -1 - 6 = -7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}\leq \frac {1}{8}$, $\alpha, \gamma\, \beta$ being angles of a triangle Prove $\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}\leq \frac {1}{8}$
I defined $f(x,y,z)=\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}$, and wanted to find max/min points under the constraint $\alpha+\beta+\gamma=\pi$.
What I reached, when using the Lagrange multipliers method is as follows:
$\alpha+\beta+\gamma=\pi$, and $\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\cos \frac{\gamma}{2}=\sin \frac{\alpha}{2}\cos \frac{\beta}{2}\sin \frac{\gamma}{2}=\cos \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}$
So obviously all points of the sort $(0,0,\pi), (\pi,0,0), (0,\pi,0)$ are fine, but I couldn't find the criticial points and extracting them from the Lagrange function.
Thanks in advance for any assistance!
| Using Algebra only,
$$2\sin\frac\alpha2\sin\frac\beta2=\cos\frac{\alpha-\beta}2-\cos\frac{\alpha+\beta}2$$
Now $\displaystyle\cos\frac{\alpha+\beta}2=\cdots=\sin\frac\gamma2$
Let $\displaystyle y=2\sin\frac\alpha2\sin\frac\beta2\sin\frac\gamma2$
$$\implies y=\left(\cos\frac{\alpha-\beta}2-
\sin\frac\gamma2\right)\sin\frac\gamma2\iff2\sin^2\frac\gamma2-\cos\frac{\alpha-\beta}2\sin\frac\gamma2+y=0$$ which is a Quadratic Equation in $\sin\dfrac\gamma2$
As $\gamma$ is real, so will be $\sin\dfrac\gamma2$
So, the discriminant $\displaystyle\cos^2\frac{\alpha-\beta}2-4\cdot2\cdot y$ must be $\ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the identity $\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B}=\sin B+\cos B$ I have worked on this identity from both sides of the equation and can't seem to get it to equal the other side no matter what I try.
$\displaystyle\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B} =\sin B+\cos B$
| What I would do here is to first convert $\tan B$ and $\cot B$ into $\frac{\sin B}{\cos B}$ and $\frac{\cos B}{\sin B}$, respectively.
$$\frac{\cos B}{1-\tan B}+\frac{\sin B}{1-\cot B}$$
$$=\frac{\cos B}{\left(1-\dfrac{\sin B}{\cos B}\right)}+\frac{\sin B}{\left(1-\dfrac{\cos B}{\sin B}\right)}$$
Simplify the denominators of both fractions.
$$\frac{\cos B}{\left(\dfrac{\cos B}{\cos B}-\dfrac{\sin B}{\cos B}\right)}+\frac{\sin B}{\left(\dfrac{\sin B}{\sin B}-\dfrac{\cos B}{\sin B}\right)}$$
$$=\frac{\cos B}{\left(\dfrac{\cos B - \sin B}{\cos B}\right)}+\frac{\sin B}{\left(\dfrac{\sin B -\cos B}{\sin B}\right)}$$
$$=\frac{\cos^2 B}{\cos B-\sin B}+\frac{\sin^2 B}{\sin B -\cos B}$$
$$=\frac{\cos^2 B}{\cos B-\sin B}+\frac{\sin^2 B}{-\left(\cos B -\sin B\right)}$$
$$=\frac{\cos^2 B}{\cos B-\sin B}-\frac{\sin^2 B}{\cos B-\sin B}$$
Add the two fractions
$$\frac{\cos^2 B-\sin^2 B}{\cos B-\sin B}$$
Factor the numerator using the difference of squares formula (which is $a^2-b^2=(a+b)(a-b)$)
$$\frac{(\cos B+\sin B)(\cos B-\sin B)}{\cos B-\sin B}$$
$$=\cos B+\sin B$$
$$=\sin B+\cos B \ \checkmark$$
$$\color{green}{\therefore \, \, \frac{\cos B}{1-\tan B}+\frac{\sin B}{1-\cot B}=\sin B+\cos B}$$
Hope I helped!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equilateral triange, sum... Just a short question:
In a triangle we have $\sum \left(\frac{a}{b+c}\right)^{2}$. Is the triangle equilateral?
I have derived
| For all real numbers $x,y,z$ we have that
$3(x^2+y^2+z^2)\geq (x+y+z)^2$. (It follows from expanding $\sum (x-y)^2 \geq 0$)
Applying this to your sum, we have that
$\sum \left(\frac{a}{b+c}\right)^2 \geq \frac{1}{3}\left(\sum\frac{a}{b+c}\right)^2$ with equality iff $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$.
Now Nesbitt's Inequality gives that $\sum\frac{a}{b+c}\geq\frac{3}{2}$, and so we have that
$\sum \left(\frac{a}{b+c}\right)^2\geq\frac{1}{3}\left(\frac{3}{2}\right)^2=\frac{3}{4}$.
We are given that we have equality, so we must have equality in each of the inequalities used. In particular, the equality condition for Nesbitt's Inequality is $a=b=c$ and hence the triangle is equilateral.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|
| The other answers are great but were pure speculation since Euler published how he solves quartics.
From Elements of Algebra by Euler section 4 chapter 15, his new method for resolving fourth order equations. Additionally, Google books has a full copy to download or read online here.
Suppose the root of the equation is of the form $x = \sqrt{p} + \sqrt{q} + \sqrt{r}$ where $p,q,r$ are the roots of an equation of degree three,
$$
z^3 -fz^2+gz-h=0
$$
and
\begin{align}
f&=p+q+r\tag{1}\\
g&=pq+pr+qr\tag{2}\\
h&=pqr\tag{3}
\end{align}
Now, he square $x$ and obtained
$$
x^2 = p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}
$$
From $(1)$, we can now write
$$
x^2-f=2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}\tag{4}
$$
Let's square $(4)$ again to obtain:
$$
x^4-2fx^2+f^2 = 4pq+4pr+4qr+8\sqrt{p^2qr}+8\sqrt{pq^2r}+8\sqrt{pqr^2}
$$
From $(2)$, we can now write
$$
x^4-2fx^2+f^2 -4g= 8\sqrt{pqr}(\sqrt{p}+\sqrt{q}+\sqrt{r})
$$
Using the identity first laid out for $x$ and $(3)$, we have
$$
x^4-2fx^2 -8x\sqrt{h}+f^2-4g= 0\tag{5}
$$
Euler says we don't need to worry about $yx^3$ because
for we shall afterwards shew, that every complete equation may be transformed into another, from which the second term is taken away.
Next, Euler goes onto factoring a quartic as a product of two real quadratics.
It is to be observed that the product of these three terms, or $\sqrt{pqr}$, must be equal to $\sqrt{h}=b/8$, and that if $b/8$ is positive, the product of the terms $\sqrt{p},\sqrt{q},\sqrt{r}$; must likewise be positive so that all variations that can be admitted are reduced to the four following:
\begin{align}
x &= \sqrt{p} + \sqrt{q} + \sqrt{r}\\
x &= \sqrt{p} - \sqrt{q} - \sqrt{r}\\
x &= -\sqrt{p} + \sqrt{q} - \sqrt{r}\\
x &= -\sqrt{p} - \sqrt{q} + \sqrt{r}
\end{align}
When $b/8$ is negative, we have
\begin{align}
x &= \sqrt{p} + \sqrt{q} - \sqrt{r}\\
x &= \sqrt{p} - \sqrt{q} + \sqrt{r}\\
x &= -\sqrt{p} + \sqrt{q} + \sqrt{r}\\
x &= -\sqrt{p} - \sqrt{q} - \sqrt{r}
\end{align}
In order to use Euler's method, you would first need to transform your equation such that the $x^3$ term is gone. Then you match up the coefficients of $f,g,h$ in your new polynomial with the general form $(5)$. That is, given a polynomial of the form
$$
x^4 - ax^2 - bx - c = 0
$$
you would set $a = 2f$, $b = 8\sqrt{h}$, and $-c = f^2 - 4g$. Then you can follow his book's examples which you will find in the second hyperlink.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 1
} |
To solve $ \frac {dy}{dx}=\frac 1{\sqrt{x^2+y^2}}$ How do we solve the differential equation $ \dfrac {dy}{dx}=\dfrac 1{\sqrt{x^2+y^2}}$ ?
| $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x^2+y^2}}$
$\dfrac{dx}{dy}=\sqrt{x^2+y^2}$
Apply the Euler substitution:
Let $u=x+\sqrt{x^2+y^2}$ ,
Then $x=\dfrac{u}{2}-\dfrac{y^2}{2u}$
$\dfrac{dx}{dy}=\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}-\dfrac{y}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}-\dfrac{y}{u}=u-\left(\dfrac{u}{2}-\dfrac{y^2}{2u}\right)$
$\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}-\dfrac{y}{u}=\dfrac{u}{2}+\dfrac{y^2}{2u}$
$\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}=\dfrac{u}{2}+\dfrac{y^2+2y}{2u}$
$(u^2+y^2)\dfrac{du}{dy}=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $\dfrac{dv}{dy}=2u\dfrac{du}{dy}$
$\therefore\dfrac{u^2+y^2}{2u}\dfrac{dv}{dy}=u^3+(y^2+2y)u$
$(u^2+y^2)\dfrac{dv}{dy}=2u^4+(2y^2+4y)u^2$
$(v+y^2)\dfrac{dv}{dy}=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$\dfrac{dv}{dy}=\dfrac{dw}{dy}-2y$
$\therefore w\left(\dfrac{dw}{dy}-2y\right)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$w\dfrac{dw}{dy}-2yw=2w^2+(4y-2y^2)w-4y^3$
$w\dfrac{dw}{dy}=2w^2+(6y-2y^2)w-4y^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{dy}=-\dfrac{1}{z^2}\dfrac{dz}{dy}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{dy}=\dfrac{2}{z^2}+\dfrac{6y-2y^2}{z}-4y^3$
$\dfrac{dz}{dy}=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Find the product $(1-a_1)(1-a_2)(1-a_3)(1-a_4)(1-a_5)(1-a_6)$ Let $1,$ $a_i$ for $1 \leq i \leq 6$ be the different roots of $x^7-1$.
Then find the product:
$(1-a_1)(1-a_2)(1-a_3)(1-a_4)(1-a_5)(1-a_6)$
I don't know how to proceed.
| We have $$(x-1)(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)=(x-1)(1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}).$$ Hence we must have $$(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)=1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}.$$ This is because I am dividing the same monomial $x-1$ from both sides. Hence we have $$\prod_{i=1}^{6}(1-a_i)=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Elevator Probability Question There are four people in an elevator, four floors in the building, and each person exits at random. Find the probability that:
a) all exit at different floors
b) all exit at the same floor
c) two get off at one floor and two get off at another
For a) I found $4!$ ways for the passengers to get off at different floors, so $$\frac{4!}{4^4} \text{would be the probability} = \frac{3}{32}$$
For b) there are only four ways for them to all exit on the same floor, so $$\frac{4}{256} = \frac{1}{64}$$
For c) am I allowed to group the $4$ people so that I am solving for $2$ people technically? For two people there would be $12$ possibilities, and there are three ways to group the $4$ individuals, so $$\frac{12 \cdot 3}{256} = \frac{9}{64}$$
I'm not sure if I'm doing these right, can you please check? Thank you.
| Solutions using conditional probabilities:
Let's enumerate people $P_1, P_2, P_3, P_4$.
a) all exit at different floors
$P_1$ can pick any of 4 floors with probability $\frac{1}{4}$ each, so we have 4 cases. Because all 4 cases are equivalent we focus on just one. Now, given the floor $P_1$ exited at, $P_2$ can exit on any of the 3 floors left with probability $\frac{3}{4}$, then $P_3$ can exit on any of the 2 floors left with probability $\frac{2}{4}$, and finally, $P_4$ only one floor left with probability $\frac{1}{4}$. Thus, multiplying these conditional probabilities together:
$\sum_{floor=1,4}{\frac{1}{4} * \frac{3}{4} * \frac{2}{4}} * \frac{1}{4} = \frac{3}{4} * \frac{2}{4} * \frac{1}{4} = \frac{6}{64} = \boxed{\frac{3}{32}}$
b) all exit at the same floor
Again, $P_1$ can pick any of 4 floors with probability $\frac{1}{4}$ each, and then the rest of people have a probability $\frac{1}{4}$ to match that floor:
$\sum_{floor=1,4}{\frac{1}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4}} = (\frac{1}{4})^3 = = \boxed{\frac{1}{64}}$
c) two get off at one floor and two get off at another
Given $P_1$ picked a floor, we have 4 cases with probability $\frac{1}{4}$ each: $P_2$ picks the same floor with probability $\frac{1}{4}$, and the other two a different floor with probability ($\frac{3}{4})^2$:
$\sum_{floor=1,4}{\frac{1}{4} * (\frac{3}{4})^2} = \frac{1}{4} * (\frac{3}{4})^2 = = \boxed{\frac{9}{64}}$
Now, logical question would be what if it's not $P_2$ that exits on the same floor with $P_1$ but rather $P_3$ or $P_4$? There are $3!$ combinations that may happen with equal probabilities of $\frac{1}{3!}$, which would eliminate each other. But, to be complete here is the full formula:
$\sum_{floor=1,4}{\frac{1}{4} * \sum_{cases=1,3!}{\frac{1}{3!} * \frac{1}{4} * (\frac{3}{4})^2}} = \frac{1}{4} * (\frac{3}{4})^2 = = \boxed{\frac{9}{64}}$
The same reduction, of course, applied to a) and b).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
find the point $P$ such that the expression has minimum value Let $ABC$ be a triangle with sides
$$a,b,c.$$
Find a point $P$ inside the triangle such that
$$a(PA)^2+b(PB)^2+c(PC)^2$$
is minimum
| Let $I$ be the incenter of $ABC$. $I$ is the barycenter of the weighted points $(A;a)$,$(B;b)$ and $(C;c)$. This means that
$$
a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=\overrightarrow{0}\tag{1}
$$
Now
$$\eqalign{
a(PA)^2+b(PB)^2+c(PC)^2&=a(\overrightarrow{IA}-\overrightarrow{IP})^2
+a(\overrightarrow{IB}-\overrightarrow{IP})^2+
a(\overrightarrow{IC}-\overrightarrow{IP})^2\cr
&=a (IA)^2+b(IB)^2+c(IC)^2\cr
&\phantom{=}-2\underbrace{(a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC})}_0
\overrightarrow{IP}+(a+b+c)(IP)^2\cr
&=a (IA)^2+b(IB)^2+c(IC)^2+(a+b+c)(IP)^2
}
$$
Thus, the minimum is attained if and only if $IP=0$, that is, when $P$ is the incenter of the triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of the expression x^2 +y^2 subject to conditions Find the values of $x,y$ for which $x^2 + y^2$ takes the minimum value where $(x+5)^2 +(y-12)^2 =14$.
Tried Cauchy-Schwarz and AM - GM , unable to do.
| Any point satisfying $\displaystyle(x+5)^2 +(y-12)^2 =14$ can be expressed as $\sqrt{14}\cos\phi-5,\sqrt{14}\sin\phi+12$
$\displaystyle x^2 + y^2=14(\cos^2\phi+\sin^2\phi)+2\sqrt{14}(12\sin\phi-5\cos\phi)+5^2+12^2$
$\displaystyle=14+12^2+5^2+2\sqrt{14}(12\sin\phi-5\cos\phi)$
This will attain minimum if $\displaystyle12\sin\phi-5\cos\phi$ is minimum
Now set $12=r\sin\phi,5=r\cos\phi$ to find $\displaystyle12\sin\phi-5\cos\phi=13\sin\left(\phi-\arctan\frac5{12}\right)$
What is the minimum value of $\displaystyle\sin\left(\phi-\arctan\frac5{12}\right)?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluating Combination Sum $\sum{n+k\choose 2k} 2^{n-k}$ Evaluate $$\sum_{k=0}^n{n+k\choose 2k} 2^{n-k}$$
So im not really sure how to begin with this. I would imagine we start with dividing out $2^{n}$, but not really sure much past that
| The method used here is that of the generating function. Let $S_{n}$ be the series to be summed
\begin{align}
S_{n} = \sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k}.
\end{align}
The generating function and its reduction are as follows.
\begin{align}
\sum_{n=0}^{\infty} S_{n} \frac{t^{n}}{2^{n}} &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k} \ \frac{t^{n}}{2^{n}} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \binom{n+2k}{2k} \ 2^{-k} t^{n+k} \\
&= \sum_{k=0}^{\infty} \frac{t^{k}}{2^{k}} \cdot \sum_{n=0}^{\infty} \frac{(2k+1)_{n} t^{n}}{n!} \\
&= \sum_{k=0}^{\infty} \frac{t^{k}}{2^{k}} \cdot (1-t)^{-2k-1} \\
&= \frac{1}{1-t} \sum_{k=0}^{\infty} \left( \frac{t}{2(1-t)^{2}} \right)^{k}
= \frac{1-t}{1 - (5/2)t + t^{2}}.
\end{align}
Now,
\begin{align}
\frac{1-t}{1 - (5/2)t + t^{2}} &= \frac{1-t}{(1/2-t)(2-t)} = \frac{2}{3} \left[ \frac{1}{1-2t} + \frac{1}{2(1-t/2)} \right] \\
&= \sum_{n=0}^{\infty} \left[ \frac{2^{2n+1}+1}{3 \cdot 2^{n}} \right] \ t^{n}
\end{align}
and
\begin{align}
\sum_{n=0}^{\infty} S_{n} \frac{t^{n}}{2^{n}} = \sum_{n=0}^{\infty} \left[ \frac{2^{2n+1}+1}{3 \cdot 2^{n}} \right] \ t^{n}
\end{align}
which yields
\begin{align}
\sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k} = \frac{2^{2n+1}+1}{3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Finding determinant for matrix using upper triangle method Here is an example of a matrix, and I'm trying to evaluate its determinant:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
2 & 5 & -2 & 9 \\
3 & 7 & 0 & 1 \\
\end{pmatrix}
$$
When applying first row operation i get:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & -1 & -6 & 7 \\
0 & -2 & -6 & -2 \\
\end{pmatrix}
$$
Now, if I continue doing row operations until i get the upper triangle, the determinant will be 14 (which is said to be the correct one).
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & 0 & 2 & 3 \\
0 & 0 & 0 & -7 \\
\end{pmatrix}
$$
However, if I instead apply this certain operation, R4 --> (1/-2)R4...
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & -1 & -6 & 7 \\
0 & 1 & 3 & 1 \\
\end{pmatrix}
$$
...and then carry on with operations, I get a different final answer: The determinant will be 7 in this case!
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & 0 & -1 & 5 \\
0 & 0 & 0 & -7 \\
\end{pmatrix}
$$
Could someone explain that to me - is this operation illegal?
R4 --> (1/-2)R4
How so? Because i always tend to use it, just to simply things a little.
| If we multiply a certain row or column of a matrix $A$ by some scalar $\lambda$ then determinant of $A$ changes to $\lambda|A|.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/793217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What remainder does $34!$ leave when divided by $71$? What is the remainder of $34!$ when divided by $71$?
Is there an objective way of solving this?
I came across a solution which straight away starts by stating that
$69!$ mod $71$ equals $1$ and I lost it right there.
| Continuing in the line of Samrat Mukhopadhyay's answer, a method that is hardly any easier than actually computing $34!\pmod{71}$ by simply multiplying factor by factor:
By Wilson's theorem we know that $70!\equiv-1\pmod{71}$, from which it follows that
$$(34!)^2\times35\times36\equiv34!\times36!\equiv70!\equiv-1\pmod{71}.$$
Because $2\times35\equiv-1\pmod{71}$ and $2\times36\equiv1\pmod{71}$ we see that
$$(34!)^2\equiv-4\times(34!)^2\times35\times36\equiv4\pmod{71},$$
which shows that $34!\equiv\pm2\pmod{71}$.
Note that $-1$ is not a quadratic residue modulo $71$ as $71\equiv3\pmod{4}$. However $2$ is a quadratic residue because $71\equiv-1\pmod{8}$, and therefore $-2$ is not a quadratic residue modulo $71$. Some hand counting shows that the square-free part of $34!$ equals
$$3\times5\times11\times19\times23\times29\times31.$$
So the question is now whether this is a quadratic residue modulo $71$. By the law of quadratic reciprocity, and using the fact that $19\equiv3\pmod{8}$ and $23\equiv-1\pmod{8}$, we see that
\begin{eqnarray*}
\left(\frac{3}{71}\right)&=&-\left(\frac{-1}{3}\right)=1,\\
\left(\frac{5}{71}\right)&=&\left(\frac{1}{5}\right)=1,\\
\left(\frac{11}{71}\right)&=&-\left(\frac{5}{11}\right)=-\left(\frac{1}{5}\right)=-1,\\
\left(\frac{19}{71}\right)&=&-\left(\frac{14}{19}\right)=-\left(\frac{2}{19}\right)\left(\frac{7}{19}\right)=-\left(\frac{5}{7}\right)=-\left(\frac{2}{5}\right)=1,\\
\left(\frac{23}{71}\right)&=&-\left(\frac{2}{23}\right)=-1,\\
\left(\frac{29}{71}\right)&=&\left(\frac{13}{29}\right)=\left(\frac{3}{13}\right)=\left(\frac{1}{3}\right)=1,\\
\left(\frac{31}{71}\right)&=&-\left(\frac{9}{31}\right)=-1.
\end{eqnarray*}
We find an odd number of non-squares in the product above, which shows that it is not a quadratic residue modulo $71$, and hence $34!\equiv-2\equiv69\pmod{71}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/794470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Finding eigenvectors of a matrix I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$.
Here is how I find eigenvalues:
$$\begin{align*}
\det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix} \Bigg)\\
&= \det \Bigg(\begin{bmatrix} -\lambda&1&0 \\ 0&-\lambda&1 \\ -1&0&-\lambda \end{bmatrix} \Bigg)\\
&= -\lambda^3 - 1\\
\therefore \lambda =& -1
\end{align*}$$
Using eigenvalue that I found ($-1$), I want to find eigenvectors:
$$\begin{align*}
(A - \lambda I)\vec{V} =& 0\\
\Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\Bigg) \begin{bmatrix}x\\y\\z \end{bmatrix} =& \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\begin{bmatrix} x+y \\ y+z \\ -x+z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\end{align*}$$
But what I should do from now? What is really the eigenvectors? Does this means that I have unlimited eigenvectors and any number that satisfies three equations can be eigenvectors?
| Since your characteristic equation is:
$$
\lambda^3 = -1 \rightarrow \lambda = e^{\pi i + \frac{2n\pi}{3}i}
$$
and gives three distinct eigenvalues, there are exactly three eigenvectors only one of which has eigenvalue $\lambda = -1$.
$$
\begin{bmatrix} 1 & 1 & 0 \\
0 &1& 1 \\
-1 &0 & 1
\end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 0 \\
0 &1& 1 \\
0 &1 & 1
\end{bmatrix}
$$
Now the last two are degenerate (as we would expect) which gives:
$$
y = -z \\
x = -y = z \\
(z, -z, z) \rightarrow (1, -1, 1)
$$
So $\left\langle1, -1, 1\right\rangle$ or $\left\langle \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$ is the only eigenvector for $\lambda = -1$.
By only eigenvector, I mean that all eigenvectors for $\lambda = -1$ will be scalar multiples of the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding surface area of part of a plane that lies inside a cylinder??? I have a question::
Let $S$ be the part of plane $x+2y+3z=1$ that lies inside cylinder $x^2 + y^2 = 3$
They want me to find the surface area of S??
This is a way harder question than all my previous ones, and I think I should start by finding the intersection point of the plane and the cylinder:
$x+2y+3z-1 = x^2 + y^2 -3$
$x^2-x + y^2 - 2y -2 = 3z$
$x(x-1)+y(y-2)-2=3z$
Now I am stuck? Help me get the equation please!
| With the surface defined by $g(x,y,z)=x+2y+3z-1=0$ over the domain $(x,y)\in C=\{(a,b):a^2+b^2\le3\}$, use the formula:
\begin{align}
\text{Surface Area} &= \int \int_C \sqrt{\frac{g_x^2+g_y^2+g_z^2}{g_z^2}}\mathrm{d}x \mathrm{d}y\\
&=\int \int_C \sqrt{\frac{1^2+2^2+3^2}{3^2}}\mathrm{d}x \mathrm{d}y\\
&=\int \int_C \frac{\sqrt{14}}{3}\mathrm{d}x \mathrm{d}y\\
&=\frac{\sqrt{14}}{3}\int \int_C \mathrm{d}x \mathrm{d}y\\
&=\frac{\sqrt{14}}{3} \pi (\sqrt{3})^2\\
&=\pi \sqrt{14}
\end{align}
Note the final expression for the double integral was simply the area of the region in the $x$-$y$ plane that we were integrating over (a circle of radius $\sqrt{3}$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/798938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a - d) + x(b - e) = f - c \\
x^2(a - d) + x(b - e) + \frac{(b - e)^2}{4(a - d)} = f - c + \frac{(b - e)^2}{4(a - d)} \\
(x\sqrt{a - d} + \frac{b - e}{2\sqrt{a - d}})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
(a - d)(x + \frac{b - e}{2(a - d)})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
x + \frac{b - e}{2(a - d)} = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} \\
x = \pm\sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} - \frac{b - e}{2(a - d)} \\
$$
Then solving for $y$ is as simple as plugging $x$ into one of the equations.
$$
y = ax^2 + bx + c
$$
Is my solution for $x$ and $y$ correct? Is there a better way to solve for the intersection points?
| You should recognise a form of the quadratic formula:$$(a-d)x^2+(b-e)x+(c-f)=0$$
which gives $$x=\frac {-(b-e)\pm \sqrt {(b-e)^2-4(a-d)(c-f)}}{2(a-d)}$$
This is the same as yours except for a missing factor of $\frac 14$ under your square root, which you lost when you took the square root near the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange Multipliers
Find the extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange multipliers.
So I set it up:
$$
1 = 2x\lambda_1 + 2\lambda_2 \\
1 = -2y\lambda_1 \\
1 = \lambda_2
$$
Plug in for $\lambda_2$:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
$$
So we work with:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
1 = x^2 - y^2 \\
1 = 2x + z
$$
After some algebra I got $x = y$ as a solution but that's impossible because of the constraint $1 = x^2 - y^2$. What am I missing?
| Note: There is no extremas with the current situation...So I slightly changed one of the
"constraints" to:
$x^2 - 2y^2 = 1$, and to show you how to solve if using one parameter $LM$ method.
$x + y + z = x + y + (1 - 2x) = y - x + 1$, subject to: $x^2 - 2y^2 = 1$.
Define $f(x,y) = -x + y + 1$, and $g(x,y) = x^2 - 2y^2$. Then:
$\nabla f = \lambda\cdot \nabla g \to (-1,1) = (2x\lambda,-4y\lambda)$. Thus:
$-1 = 2x\lambda$
$1 = -4y\lambda$.
Add the above equations: $2\lambda(x - 2y) = 0 \to \lambda = 0$ or $x = 2y$. But $\lambda$ cannot be $0$, so $x = 2y$, and with $x^2 - 2y^2 = 1$ we get:
$4y^2 - 2y^2 = 1 \to 2y^2 = 1 \to y = \pm \dfrac{1}{\sqrt{2}}$. So $x = 2y = \pm \sqrt{2}$.
$f_{min} = f(\sqrt{2},\frac{1}{\sqrt{2}}) = \dfrac{\sqrt{2} - 1}{\sqrt{2}}$, and
$f_{max} = f(-\sqrt{2},-\frac{1}{\sqrt{2}}) = \dfrac{\sqrt{2} + 1}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/800575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find matrix determinant How do I reduce this matrix to row echelon form and hence find the determinant, or is there a way that I am unaware of that finds the determinant of this matrix without having to reduce it row echelon form given this is all I know and there exists no additional information.
$\left[
\begin{array}{ccc}
1+x & 2 & 3 & 4 \\
1 & 2+x & 3 & 4 \\
1 & 2 & 3+x & 4 \\
1 & 2 & 3 & 4+x \\
\end{array}
\right]$
| If you know how value of determinant is influenced by elementary row/column operations (see ProofWiki) then you could start by adding all other columns to the last one (which does not change the determinant) and the rest is relatively easy:
$$
\begin{vmatrix}
1+x & 2 & 3 & 4 \\
1 & 2+x & 3 & 4 \\
1 & 2 & 3+x & 4 \\
1 & 2 & 3 & 4+x
\end{vmatrix}=
\begin{vmatrix}
1+x & 2 & 3 & 10+x \\
1 & 2+x & 3 & 10+x \\
1 & 2 & 3+x & 10+x \\
1 & 2 & 3 & 10+x
\end{vmatrix}=
(x+10)
\begin{vmatrix}
1+x & 2 & 3 & 1 \\
1 & 2+x & 3 & 1 \\
1 & 2 & 3+x & 1 \\
1 & 2 & 3 & 1
\end{vmatrix}=
(x+10)
\begin{vmatrix}
x & 0 & 0 & 1 \\
0 & x & 0 & 1 \\
0 & 0 & x & 1 \\
0 & 0 & 0 & 1
\end{vmatrix}=(x+10)x^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
integral involving hypergeometric function $\int^1_0\frac{_2F_1(p,p;p+1;-\frac{1}{y})}{y}\,dy$ I arrived at the following result
$$\tag{1}\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{\Gamma(p)}{p}\int^1_0\frac{_2F_1(p,p;p+1;-\frac{1}{z})}{z}\,dz$$
where the exponential integral $E(z)$ is defined as
$$E(z)=\int^\infty_z \frac{e^{-t}}{t}\,dt$$
I have two questions
[1] Does (1) hold for all $p>0$ ?
[2] Is there a way to simplify or solve the integral on the right ?
| Let
\begin{align}
E_{1}(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \, dt
\end{align}
then the following two identities can be seen to be
\begin{align}
\int_{0}^{\infty} x^{n} E_{1}(ax) E_{1}(bx) \, dx &= - \frac{n!}{n+1} \left[ \frac{1}{a^{n+1}} \left\{ \ln\left(\frac{b}{a+b}\right) + \sum_{m=1}^{n} \frac{1}{m} \left( \frac{a}{a+b} \right)^{m} \right\} \right. \\
& \hspace{20mm} \left. + \frac{1}{b^{n+1}} \left\{ \ln\left(\frac{a}{a+b}\right) + \sum_{m=1}^{n} \frac{1}{m} \left( \frac{b}{a+b} \right)^{m} \right\} \right].
\end{align}
When $a=b=1$ this becomes
\begin{align}
\int_{0}^{\infty} x^{p} E_{1}^{2}(x) \, dx &= \frac{\Gamma(p+1)}{2^{p} \, (p+1)} \, \sum_{m=0}^{\infty} \frac{1}{2^{m} (m+p+1)} .
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} = 0 \\ $
It seems that x = 0 is a critical point.
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' = -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?
| Observe that:
$\\ \\ \displaystyle \sin(3\gamma)=\sin(2\gamma)\cos(\gamma)+\sin(\gamma)\cos(2\gamma)=2\sin(\gamma)\cos^2(\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=2\sin(\gamma)(1-\sin^2\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=3\sin(\gamma)-4\sin^3(\gamma)\Rightarrow \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)$
\begin{equation}
\sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)
\end{equation}
Do it $\displaystyle \gamma=\frac{\phi}{3^k}$:
\begin{equation}
\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Multiplying by $\displaystyle 3^{k-1}$:
\begin{equation}
3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Applying summation on both sides of equality, we will have:
$\\ \\ \displaystyle \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\sum_{k=1}^{n}\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right) =\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\Rightarrow \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\\ \\$
Take the limit:
\begin{equation*}
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right)=\frac{1}{4}\left(\phi-\sin(\phi)\right)
\end{equation*}
On the other hand, using the inequality $ \displaystyle \sin x \leq x $ and using the infinite geometric progression formula, it follows that:
$\\ \displaystyle \sin\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi}{3^{k}}\Rightarrow \sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{3k}} \Rightarrow 3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{2k+1}}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \sum_{k=1}^{\infty} \frac{\phi^3}{3^{2k+1}}=\frac{\phi^3}{3}\sum_{k=1}^{\infty} \frac{1}{3^{2k}}=\frac{\phi^3}{3\times 8}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \frac{1}{4}\left(\phi-\sin(\phi)\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \phi-\frac{\phi^3}{6}\leq \sin(\phi) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 9,
"answer_id": 2
} |
How to find determinant of this matrix? Is there a manual method to find $\det\left(XY^{-1}\right)$ ?
Let $$X=\left[ {\begin{array}{cc} 1 & 2 & 2^2 & \cdots & 2^{2012} \\
1 & 3 & 3^2 & \cdots & 3^{2012} \\
1 & 4 & 4^2 & \cdots & 4^{2012} \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
1 & 2014 & 2014^2 & \cdots & 2014^{2012} \\ \end{array} } \right], $$
$$Y=\left[ {\begin{array}{cc}\frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016} \\
2 & 3 & 4 & \cdots & 2014 \\
2^2 & 3^2 & 4^2 & \cdots & 2014^{2} \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
2^{2012} & 3^{2012} & 4^{2012} & \cdots & 2014^{2012} \\ \end{array} } \right] $$.
Thanks in advance.
| Not a solution, but a possible step in the right direction (too long for a comment):
Following Cramer's rule, we note that the solution $\vec y = (y_1,\dots,y_{2014})^T$ to
$$X \,\vec y = \pmatrix{
\frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016}
}^T$$
Will satisfy
$$
y_1 = \det(Y^T)/\det(X) = [\det(XY^{-1})]^{-1}
$$
With that in mind: if there is another way to solve this equation, it will probably be easier than actually computing the determinant.
Note also that $X$ is a Vandermonde matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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} |
Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$ Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq
2xy(x+y+xy)$ for $x,y \in \mathbb{R}^+$.
I started by multiplying everything out on the RHS to get the equivalent statement
\begin{align*}
x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq
2 x^2 y + 2 x y^2 + 2 x^2 y^2.
\end{align*}
I've tried numerous methods of attacking this problem, but I haven't had much luck yet. The closest I seem to have been able to get is to use the identity $(x-y)^2\geq 0$ to deduce that
\begin{align*}
x^3+x^3y^2\geq x^3+xy^2=x(x^2+y^2)\geq 2x^2y,
\end{align*}
and similarly,
\begin{align*}
y^3x^2+y^3\geq x^2y+y^3=y(x^2+y^2)\geq 2xy^2.
\end{align*}
But now I am left with having to prove that $x^2+y^2\geq 2x^2y^2$, which is obviously false for $x,y \in \mathbb{R}^+$. I'm thinking I'm giving up too much trying to make the identity $(x-y)^2\geq 0$ work for this problem.
I also tried writing everything as a polynomial in $x$ and showing that the resulting polynomial is positive for $x>0$, but this gets extremely messy. Does anyone have any ideas? Thanks in advance!
| By $AM-GM :$
$$x^3+x^3y^2\ge 2x^3y$$
$$x^2+y^2\ge2xy$$
$$y^3 +x^2y^3\ge2xy^3$$
Add :
$$\text{LHS}\ge2xy(x^2+y^2+1)$$
WLOG Assume $x\ge y\ge 1$
Use rearrangement inequality on same inequality :
$$x^2+y^2+1\ge x+y+xy$$
By transitivity...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Examples of quasigroups with no identity elements If you scroll to the bottom of this page, there is a table claiming quasigroups have divisibility but not identity (in general).
What would be some examples of quasigroups without an identity element?
| The Cayley tables:
$$\begin{array}{c|ccccccc}
\ast & 1 & 2 & 3 & 4 & 5 & 6 & 7\\ \hline
1 & 1 & 3 & 2 & 5 & 4 & 7 & 6\\
2 & 3 & 2 & 1 & 6 & 7 & 4 & 5\\
3 & 2 & 1 & 3 & 7 & 6 & 5 & 4\\
4 & 5 & 6 & 7 & 4 & 1 & 2 & 3\\
5 & 4 & 7 & 6 & 1 & 5 & 3 & 2\\
6 & 7 & 4 & 5 & 2 & 3 & 6 & 1\\
7 & 6 & 5 & 4 & 3 & 2 & 1 & 7
\end{array}
\qquad\qquad
\begin{array}{c|ccccccc}
\circ & 1 & 2 & 3 & 4 & 5 & 6 & 7\\ \hline
1 & 1 & 7 & 6 & 5 & 4 & 3 & 2\\
2 & 7 & 2 & 5 & 6 & 3 & 4 & 1\\
3 & 6 & 5 & 3 & 7 & 2 & 1 & 4\\
4 & 5 & 6 & 7 & 4 & 1 & 2 & 3\\
5 & 4 & 3 & 2 & 1 & 5 & 7 & 6\\
6 & 3 & 4 & 1 & 2 & 7 & 6 & 5\\
7 & 2 & 1 & 4 & 3 & 6 & 5 & 7
\end{array}
$$
show two different Steiner quasigroups of order $7$. None of them have left nor right identities. Such quasigroups of order $m>1$ exist for $m\equiv_6 1$ or $m\equiv_6 3$.
Another example of finite quasigroup (order $4$) with no element $e$ such that
$$ex=x=xe$$
for all $x$ is:
$$\begin{array}{c|cccc}
\bullet & 1 & 2 & 3 & 4\\ \hline
1 & 3 & 2 & 1 & 4\\
2 & 2 & 1 & 4 & 3\\
3 & 1 & 4 & 3 & 2\\
4 & 4 & 3 & 2 & 1\\
\end{array}
$$
of course a latin square as well.
Last but not least. In part two of this article you'll see a classical example of quasigroup over an infinite set, with no identity, having its roots and motivations not in abstract algebra but in geometry. The first time I saw this one I thought it was pretty different from any other quasigroup I had seen before and definitely not trivial...hope you like it!
I would like to make clear that we call quasigroup a groupoid (magma) with left and right division and cancellation laws, while we call loop a quasigroup with identity element.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Algebra Manipulation Contest Math Problem The question was as follows:
The equations $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ have two roots in common. Compute the product of these common roots.
Because $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ it means that $x^3+Ax+10=x^3+Bx^2+50$
Take $x^3+Ax+10=x^3+Bx^2+50$ and remove $x^3$ from both sides, you get $Ax+10=Bx^2+50$ or $Bx^2-Ax+40=0$
By the quadratic equation, we get $\frac {A \pm \sqrt {(-A)^2 - 4*40B}}{2B}=\frac {A \pm \sqrt {A^2 - 160B}}{2B}$
This gives us two answers: $\frac {A + \sqrt {A^2 - 160B}}{2B}$ and $\frac {A - \sqrt {A^2 - 160B}}{2B}$
$\frac {A + \sqrt {A^2 - 160B}}{2B} * \frac {A - \sqrt {A^2 - 160B}}{2B}=\frac {A^2 - {A^2 - 160B}}{4B^2}$
This simplifies as $\frac {160B}{4B^2}=\frac{40}{B}$
$\frac{40}{B}$ is an answer, but in the solutions, they expected an integer answer. Where did I go wrong?
| Let $a,b$ be the roots.
Then $a,b$ are roots of
$$x^3+Ax+10=0$$
The sum of the three roots of this polynomial is negative the coefficient of $x^2$, thus $0$. It follows that the third root is $-(a+b)$.
As the product of the three roots is $-10$ we get
$$ab(a+b)=10$$
Now let $c$ be the third root of
$$x^3+Bx^2+50=0$$
Then
$$ab+ac+bc =0$$
or
$$ab+c(a+b)=0$$
and
$$abc=-50$$
Replacing $a+b=\frac{10}{ab}$ we get
$$(ab)^2+10c =0$$
$$abc=-50$$
Multiply the first of these two equations by $ab$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$ Question:
Prove that
$$\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk=\frac{1}{16\pi}\Gamma^4\left(
\frac{1}{4}\right)$$
My attempt
Start by the transformation
$$k \to \frac{2\sqrt{k}}{1+k}$$
Hence we have
$$\int^{1}_0 K\left(\frac{2\sqrt{k}}{1+k}\right)\,\frac{1}{\sqrt{k}(1+k)}dk$$
Now we use that
$$K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right)$$
So we have
$$\int^1_0 \frac{K(k)}{\sqrt{k}}\,dk=2\int^1_0 K(k^2)\,dk$$
[1] I have no idea how to solve the last integral?
[2] Should I use another approach to solve the integral ?
By definition we have
$$K(k) = \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,$$
| A simple method for the last integral:
\begin{align}
\int \limits_0^1 \frac{\operatorname{K}(k)}{\sqrt{k}} \, \mathrm{d} k &= \int \limits_0^1 \int \limits_0^{\pi/2} \frac{\mathrm{d} \phi}{\sqrt{k(1-k^2 \sin^2(\phi))}} \, \mathrm{d} k \stackrel{\text{Tonelli}}{=} \int \limits_0^{\pi/2} \int \limits_0^1 \frac{\mathrm{d} k}{\sqrt{k(1-k^2 \sin^2(\phi))}} \, \mathrm{d} \phi \\
&\!\!\!\!\!\!\stackrel{k = \frac{\sin(\theta)}{\sin(\phi)}}{=} \int \limits_0^{\pi/2} \int \limits_0^\phi \frac{\mathrm{d} \theta \, \mathrm{d} \phi}{\sqrt{\sin(\theta)\sin(\phi)}} = \frac{1}{2} \int \limits_0^{\pi/2} \int \limits_0^{\pi/2} \frac{\mathrm{d} \theta \, \mathrm{d} \phi}{\sqrt{\sin(\theta)\sin(\phi)}} = \frac{1}{2} \left[\int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{\sin(t)}}\right]^2 \\
&= \frac{1}{8} \operatorname{B}^2\left(\frac{1}{4},\frac{1}{2}\right) = \frac{1}{8} \left(\frac{\operatorname{\Gamma}\left(\frac{1}{4}\right) \operatorname{\Gamma}\left(\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{3}{4}\right)}\right)^2 \stackrel{\Gamma\text{-reflection}}{=} \frac{1}{8} \left(\frac{\operatorname{\Gamma}^2\left(\frac{1}{4}\right) \operatorname{\Gamma}\left(\frac{1}{2}\right) \sin\left(\frac{\pi}{4}\right)}{\pi}\right)^2 \\
&= \frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{16 \pi} \, .
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
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Integer solutions to $x^x=122+231y$ How can I find the integer solutions to the following equation (without a script or trial and error)?
$$x^x=122+231y$$
| By the Chinese Remainder Theorem, to know the value of $x^x$ modulo $231$ we only need to know it modulo $3,7,11$.
The value of $x^y$ modulo $n$ depends on the values of $x$ modulo $n$ and $y$ modulo $\phi(n)$.
So if $n$ is prime, the value of $x^x$ modulo $n$ depends on $x$ modulo $n(n-1)$.
A computation gives that
$x^x \equiv 122 \pmod 3 \iff x^x \equiv 2 \pmod 3 \iff x \equiv 5 \pmod 6$,
$x^x \equiv 122 \pmod 7 \iff x^x \equiv 3 \pmod 7 \iff x \equiv 5,31 \pmod {42}$
So far we know that we must have $x \equiv 5 \pmod {42}$.
Out of the $28$ solutions modulo $110$ to $x^x \equiv 122 \pmod {11}$, only $9$ are congruent to $1$ modulo $2$, and those are $1,5,15,23,25,45,67,75,89$ modulo $110$.
So modulo $2310$, we have $9$ solutions to $x^x = 122 \pmod {231}$, and they are $x \equiv 5,89,551,845,1013,1475,1895,1937,2105 \pmod {2310}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do I find the sum of the infinite geometric series? $$2/3-2/9+2/27-2/81+\cdots$$
The formula is $$\mathrm{sum}= \frac{A_g}{1-r}\,.$$
To find the ratio, I did the following:
$$r=\frac29\Big/\frac23$$
Then got:
$$\frac29 \cdot \frac32= \frac13=r$$
and $$A_g= \frac23$$
Then I plug it all in and get:
$$\begin{align*}
\mathrm{sum} &= \frac23 \Big/ \left(1-\frac13\right)\\
&= \frac23 \Big/ \left(\frac33-\frac13\right)\\
&= \frac23 \Big/ \frac23\\
&= \frac23 \cdot \frac32\\
&= 1\,.
\end{align*}$$
But the real answer is $\frac12$.
What did I do wrong?
| $$\frac{2}{3}-\frac{2}{9}+\frac{2}{27}-\frac{2}{81}+...$$
$$\frac{2}{3}+\frac{2}{27}+\cdots - (\frac{2}{9}+\frac{2}{81}+\cdots)$$
$$2(\dfrac{\frac{1}{3}}{1-\frac{1}{9}}) - 2(\dfrac{\frac{1}{9}}{1-\frac{1}{9}})$$
$$2.(\frac{3}{8}) - 2(\frac{1}{8})$$
$$\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
How to solve $\int \frac{\,dx}{(x^3 + x + 1)^3}$?
How to solve $$\int \frac{\,dx}{(x^3 + x + 1)^3}$$ ?
Wolfram Alpha gives me something I am not familiar with. I thought that the idea was using partial fractions because $x^3$ and $x$ are bijections, there must be a real root but it seems that Wolfram Alpha is using numerical methods to approximate the root so it's not a "nice" number. I can't thing of any substitution which could help me nor any formula to transform this denominator. The formula $\int u\,dv = uv - \int v\,du$ also yields a more complicated integral:
$$ \,dv = \,dx \implies v = x \\
u = \frac{1}{(x^3 + x + 1)^3} \implies du = -3\frac{3x^2 + 1}{(x^3 + x + 1)^4} \,dx \\
\int \frac{\,dx}{(x^3 + x + 1)^3} = \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x}{(x^3 + x + 1)^4} \,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x \pm 2x \pm 3}{(x^3 + x + 1)^4}\,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + 3x + 3}{(x^3 + x + 1)^4}\,dx + 3\int \frac{- 2x - 3}{(x^3 + x + 1)^4}\,dx \\
= \frac{x}{(x^3 + x + 1)^3} + 9\int \frac{\,dx}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx \\
I = \int \frac{\,dx}{(x^3 + x + 1)^3} \implies \\
-8I = \frac{x}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx$$
Is this one really as complicated as Wolfram Alpha "tells" me or is there some sort of "trick" which can be applied?
| The integral is a rational function so one ought to be able to apply the method of partial fractions, however, the real root of $x^3+x+1$ is irrational without a nice expression, this makes it difficult to apply partial fractions, however some progress can be made.
Consider first what kinds of functions we get from integrating partial fractions.
1) Logarithms. These come from linear factors of the denominator.
2) Inverse tangent functions. These come from quadratic factors of the denominator.
3) Rational functions. These come form the occurrence of repeated factors in the denominator, as is the case with this integral.
Now to find the expressions for terms of the first two types we need the factorization of the denominator. However the term of the third type can be calculated arithmetically without factoring the denominator.
Thus we want to find an expression of the form
$$\int \frac{P}{Q}=\frac{P_1}{Q_2}-\int \frac{P_2}{Q_2}$$
Where $Q_2$ contains has the same factors as $Q$ only with out repetition.
And where we have $Q=Q_1Q_2$, Now $Q_1=(Q,Q^{\prime})$ is calculated arithmetically
$Q_2$ is then found by division. In our case $Q=(x^3+x+1)^3$ and $Q_2= x^3+x+1$ and
$Q_1=(x^3+x+1)^2$.
It remains to find $P_1$ and $P_2$.
By differentiating the above expression we get the formula,
$$P=P_1^{\prime}Q_2-P_1H+P_2Q_1$$ where
$$H=\frac{Q_1^{\prime}Q_2}{Q_1}$$ which is easily seen to be a polynomial.
In our case $$H=\frac{Q_1^{\prime}Q_2}{Q_1}=\frac{2(x^3+x+1)
(3x^2+1)(x^3+x+1)}{(x^3+x+1)^2}=6x^2+2$$
Now we set
$$P_1=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$$ and
$$P_2=Gx^2+Hx+K$$
Thus we set this in our formula and obtain a system of equations. The first few equations give, This gives $G=0$ and $H=A$ and $K=2B$ we can simplify a little writing,
\begin{equation*}
\begin{split}
1=&(5Ax^4+4Bx^3+3Cx^2+2Dx+E)(x^3+x+1)\\
&-(Ax^5+Bx^4+Cx^3+Dx^2+Ex+F)(6x^2+2)\\
&+(Ax+2B)(x^6+2x^4+2x^3+x^2+2x+1)\\
\end{split}
\end{equation*}
After successive elimination we reduce to
$$119A+162B=0$$
$$13A+50B=6$$
which finally gives us the solution, so we have ultimately,
$$\int \frac{dx}{(x^3+x+1)^3}=\frac{-486x^5+ 357x^4-810 x^3-315x^2+312x-448}{1922(x^3+x+1)^2}+
\int \frac{-486x+714}{1922(x^3+x+1)}dx.$$
Now to go further we need a root for the equation. There is only one real root so we would expect a logarithmic and an arctan term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Get variables with Matrix I try to get the variables for this equation:
$$\begin{cases}
6x_1 + 4x_2 + 8x_3 + 17x_4 &= -20\\
3x_1 + 2x_2 + 5x_3 + 8x_4 &= -8\\
3x_1 + 2x_2 + 7x_3 + 7x_4 &= -4\\
0x_1 + 0x_2 + 2x_3 -1x_4 &= 4
\end{cases}$$
So i started with:
$$ \begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
3 & 2 & 5 & 8 & -8 \\
3 & 2 & 7 & 7 & -4\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}$$
Then I continued: $$
\begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
-6 & -4 & -10 & -16 & 16 \\
-6 & -4 & -14 & -14 & 8\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}
$$
And began to count:
$$
\begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
0 & 0 & -2 & -1 & -4 \\
0 & 0
& -6 & 3 & -12\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}
$$
And finally I transformed it to:$$
\begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
0 & 0 & -2 & -1 & -4 \\
0 & 0
& -6 & 3 & -12\\
0 & 0 & 0 & 0 &0\\
\end{pmatrix}
$$
What did i wrong? I think the last row cannot be correct because:
$$(0, 0 ,0 , 0) \neq 0x_1 + 0x_2 + 2x_3 -1x_4 = 4$$
How should i solve it? Thanks
| $$ \begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
0 & 0 & -2 & \color{red}{-1} & -4 \\
0 & 0 & -6 & 3 & -12\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}$$
should be $$ \begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
0 & 0 & -2 & \color{green}{+1} & -4 \\
0 & 0 & -6 & 3 & -12\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\
\ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\
\Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\
\mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\
\end{array}
$
help me what you please
| $$
\displaylines{\mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x - x - 1}}{{x^2 }} = \frac{1}{2} \cdots \left( 1 \right) \cr}
$$
$$
\displaylines{ \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{t^2 }}{{te^t - e^t + 1}}} \right) = 2 \cdots \left( 2 \right) \cr}
$$
$$\displaylines{ \mathop {\lim }\limits_{_{x \to 1} } \frac{{x - 1 - \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{_{t \to 0} } \frac{{e^t - t - 1}}{{te^t - e^t + 1}} \cr
= \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{e^t - t - 1}}{{t^2 }} \times \frac{{t^2 }}{{te^t - e^t + 1}}} \right) \cr
= \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{e^t - t - 1}}{{t^2 }}} \right) \times \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{t^2 }}{{te^t - e^t + 1}}} \right) \cr
= \frac{1}{2} \times 2 = 1 \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
In the Quadratic Formula, what does it mean if $b^2-4ac>0$, $b^2-4ac<0$, and $b^2-4ac=0$? Concerning the Quadratic Formula:
What does it mean if $b^2-4ac>0$, $b^2-4ac<0$, and $b^2-4ac=0$?
| Since one has the term $\sqrt{b^2-4ac}$ in solution of the quadratic formula, if $b^2-4ac>0$, then the equation has two real solutions (since $\sqrt{b^2-4ac}$ is real, and $b$ and $2a$ are also real). When $b^2-4ac<0$, then the quadratic equation has two complex solutions (since $\sqrt{b^2-4ac}$ is complex imaginary). If $b^2-4ac=0$, then $\sqrt{b^2-4ac}=0$, implying that the solution is $x=\dfrac{-b}{2a}$ (can you see why?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove $a_n=1+\frac{a_{n-1}}{1+a_{n-1}}$ increasing There is a homework question in Calculus-1 course:
Calculate the limit of $\{a_n\}$: $$a_1=1,\ a_n=1+\frac{a_{n-1}}{1+a_{n-1}}$$
I think the key points are bounded and increasing, and I have proved that $$a_n\in(1, 2)$$
If I knew it's increasing then $$a=1+\frac{a}{1+a}\Rightarrow\lim a_n=\frac{\sqrt5+1}{2}$$
My question is How to Prove it's increasing?
I tried it in two ways:
$$a_{n+1}-a_n=1+\frac{a_n}{1+a_n}-a_n=\frac{-a^2_n+a_n+1}{1+a_n}$$
But how to prove that $-a^2_n+a_n+1>0$?
Another way is
$$\frac{a_{n+1}}{a_n}=\frac{1}{a_n}+\frac{1}{1+a_n}=\frac{1+2a_n}{a_n+a^2_n}$$
But how to prove that $1+2a_n>a_n+a^2_n$?
This is not a proof question which means $\frac{\sqrt5+1}{2}$ is not a known result.
Thank you!
| Consider the sequence
\begin{align}
a_{n} = 1 + \frac{a_{n-1}}{1+a_{n-1}}
\end{align}
where $a_{1} = 1$.
Let $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1-\sqrt{5}$. It is seen that $\alpha > \beta$ and $\beta^{n} \rightarrow 0$ as $n \rightarrow \infty$. Now, the terms of $a_{n}$ are $a_{n} \in \{ 1, 3/2, 8/5, \cdots \}$ which are seen to be the Fibonacci numbers, and in general
\begin{align}
a_{n} = \frac{F_{2n}}{F_{2n-1}}.
\end{align}
Since $\sqrt{5} F_{n} = \alpha^{n} - \beta^{n}$ then
\begin{align}
a_{n} = \frac{\alpha^{2n} - \beta^{2n}}{\alpha^{2n-1} - \beta^{2n-1}} = \alpha \ \frac{1 - \left(\frac{\beta}{\alpha} \right)^{2n}}{1 - \left(\frac{\beta}{\alpha} \right)^{2n-1}} .
\end{align}
Taking the limit as $n \rightarrow \infty$ leads to
\begin{align}
\lim_{n \rightarrow \infty} a_{n} = \alpha = \frac{1+\sqrt{5}}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Volume between cylinder and plane Problem: Find the volume bounded by $z = y^2, x =0, y =0, z =9-x$.
My working:
$z$ goes from $y^2$ to $9-x$ so these are the limits of integration.
Work out the points of intersection of $9-x$ and $y^2$. When $y=0$, $9-x=0$ and $x=9$. So $x$ goes from 0 to 9. When $x=0$, $y^2 = 9$ so $y=3$ (take the positive one). So $y$ goes from 0 to 9.
Then evaluate
\begin{align}
\int_{x=0}^{x=9} \int_{y=0}^{y=9} \int_{z=y^2}^{z=9-x} dz dy dz &= \int_{x=0}^{x=9} \int_{y=0}^{y=9} y^2 - 9 + x dy dx
\\
&= \int_{x=0}^{x=9} 18+3x dx
\\
&= \frac{567}{2}
\end{align}
My textbook says the answer is $\frac{324}{5}$. What have I done wrong?
| Alternately the set up is:
$\displaystyle \int_{0}^3 \int_{0}^{9-y^2} \int_{y^2}^{9-x} 1dzdxdy = \dfrac{324}{5}$ (already checked)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The differential equation: $ \arctan (y) = \arctan(x)+C .$ I solved the equation and stalled. Help with decision please.
$$(1+y^2)\,dx=(1+y^2)\,dy \iff \int \frac{dx}{1+x^2} = \int\frac{dy}{1+y^2} $$
Transformed expression for the table of integrals.
$$ \arctan (y) = \arctan(x)+C $$
Prompt how to further transform expression.(Find the general solution)
| The differential set
$$
(1+y^{2}) \,dx = (1+x^{2})\, dy
$$
can be integrated as seen by
\begin{align}
\int \frac{dy}{1+y^{2}} = \int \frac{dx}{1+x^{2}}
\end{align}
and leads to
\begin{align}
\tan^{-1}(y) = \tan^{-1}(x) + c
\end{align}
or
\begin{align}
y = \tan(\tan^{-1}(x) + c).
\end{align}
Now using
\begin{align}
\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)}
\end{align}
the general result becomes
\begin{align}
y(x) = \frac{x+c_{1}}{1 - c_{1} x}
\end{align}
since $\tan(\tan^{-1}(x)) = x$ and $c_{1} = \tan(c)$ is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Compute an integral with residue theorem Using residue theorem, compute the following integral:
$$
\int_{0}^{2\pi}\frac{\left( 1+2\cos t\right) ^{n}\cos\left( nt\right)
}{5+4\cos t}\operatorname*{dt}.
$$
Or a source with a solution.
| \begin{align} \int_{0}^{2 \pi} \frac{(1+2 \cos t)^{n} \cos (nt)}{5 + 4 \cos t} \ dt &= \text{Re} \int_{0}^{2 \pi} \frac{(1+2 \cos t)^{n} e^{int}}{5 + 4 \cos x} \ dt \\ &= \text{Re} \int_{0}^{2 \pi}\frac{(1+ e^{it} + e^{-it})^{n}e^{int}}{5 + 2(e^{it} + e^{-it})} \ dt \\ &= \text{Re} \int_{|z|=1} \frac{(1+z+z^{-1})^{n}z^{n}}{5+2z+2z^{-1}} \frac{dz}{iz} \\ &= \text{Re} \frac{1}{i} \int_{|z|=1} \frac{(z^{2}+z+1)^{n}}{2z^{2}+5z+2} \ dz \\ &= \text{Re} \frac{1}{i} \int_{|z|=1} \frac{(z^{2}+z+1)^{n}}{(2z+1)(z+2)} \ dz\end{align}
Only the pole at $z=- \frac{1}{2}$ is inside of the unit circle.
Therefore,
\begin{align} \int_{0}^{2 \pi} \frac{(1+2 \cos t)^{n} \cos (nt)}{3 + 2 \cos t} \ dt &= \text{Re} \frac{1}{i} 2 \pi i \ \text{Res} \Big[\frac{(z^{2}+z+1)^{n}}{(2z+1)(z+2)}, - \frac{1}{2}\Big] \\ &= 2 \pi \ \text{Re} \lim_{z \to - \frac{1}{2}} \frac{1}{2} \frac{(z^{2}+z+1)^{n}}{z+2} \\ &= \pi \ \text{Re} \frac{(\frac{1}{4} - \frac{1}{2}+1)^{n}}{\frac{3}{2}} \\ &= \frac{2 \pi}{3} \left(\frac{3}{4} \right)^{n} \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all the asymptote of $1-x+\sqrt{2+2x+x^2}$ I find
$$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=2,$$ but i'am stuck when $x\rightarrow-\infty$ how to find that $y=-2x$ is an oblique asymptote. Any idea?
| General Method
Consider a curve given by the equation $f(x, y) = 0$, where $f(x, y)$ is a polynomial of degree $n$ (in $x$ and $y$). If it contains the term $y^n$ (with some non-zero coefficient), then it has no asymptote parallel to the $y$-axis. Otherwise, equate the total coefficient of the highest degree term of $y$ to 0 and solve (if possible). This gives all asymptotes parallel to the $y$-axis.
Example
$x^2 - xy - y + 1 = 0$
The degree of the equation is $2$, and $y^2$ is not present, which means the curve might have an asymptote parallel to the $y$-axis. The highest degree term of $y$ is $y$, and its total coefficient is $-(x + 1)$ ("total" here meaning we include not only constant coefficients, but also any $x$-term that is multiplied with $y$). Equating this to zero, $-(x + 1) = 0 \Rightarrow x = -1$ is an asymptote parallel to the $y$-axis.
Similarly, we can find the asymptotes parallel to the $x$-axis as well, but that is unnecessary as the procedure for finding the "oblique" asymptotes does this job as well. Suppose there are asymptotes of the form $y = mx + c$. Substitute $y = mx + c$ in $f(x, y)$ to get a polynomial in $x$. Equate the coefficients of the highest and next highest degree terms of $x$ to zero. This gives two equations in $m$ and $c$ (in general). If these can be solved simultaneously, each solution pair $(m, c)$ corresponds to an asymptote $y = mx + c$.
Example
$x^2 - xy - y + 1 = 0$
Substituting $y = mx + c$:
$x^2 - x(mx + c) - (mx + c) + 1 = 0$
Now, we do not have to expand the whole LHS (though it's simple enough in this case). It suffices to find the coefficients of the highest and next highest degree terms. The highest degree term is $x^2$ with coefficient $(1 - m)$. So $1 - m = 0 \Rightarrow m = 1$. The next highest degree term is $x$, with coefficient $-c - m$, and equating this to zero, with $m = 1$, we get $-c - 1 = 0 \Rightarrow c = -1$. Thus, $y = x - 1$ is the other asymptote.
Application
Applying this to the problem at hand, first we rewrite the equation in polynomial form:
$y = 1 - x + \sqrt{2 + 2x + x^2} \Rightarrow\\
x + y - 1 = \sqrt{2 + 2x + x^2} \Rightarrow$
$(x + y - 1)^2 = 2 + 2x + x^2$.
The degree of is $2$, and there is a $y^2$ term, so there is no asymptote parallel to the $y$-axis. Now, let $y = mx + c$ be an asymptote (not parallel to the $y$-axis). Substituting $y = mx + c$ in the equation:
$(x + mx + c - 1)^2 = 2 + 2x + x^2$
Without expanding the whole LHS, we can see that the highest degree term will be $x^2$ with coefficient $(m + 1)^2$, and on the RHS, $x^2$ has coefficient $1$. So $(m + 1)^2 = 1 \Rightarrow m + 1 = \pm 1 \Rightarrow m = 0, -2$.
The next highest degree term is $x$, with coefficient $2(m + 1)(c - 1)$ on the LHS, and $2$ on the RHS, so $2(m + 1)(c - 1) = 2 \Rightarrow (m + 1)(c - 1) = 1$.
For $m = 0$, $c - 1 = 1 \Rightarrow c = 2$, and for $m = -2$, $1 - c = 1 \Rightarrow c = 0$.
Thus the two asymptotes are $\boxed{y = 2}$ and $\boxed{y = -2x}$.
Explanation
How does this work? Recall that we can think of asymptotes intuitively as "tangents at infinity". That means we may able to find asymptotes by finding the tangents at infinity. This is exactly what the above procedure does. An oblique asymptote of the form $y = mx + c$ "intersects" (or "touches") the curve $f(x, y) = 0$ "at infinity". As it is not parallel to the $y$-axis, this happens as $x \to \infty$.
To find the "intersection" of the curve $f(x, y) = 0$ with the asymptote $y = mx + c$, we proceed as usual by substituting the equation of the line in the equation of the curve. This results in an $n^{\text{th}}$ degree polynomial equation in $x$ alone (as $f(x, y)$ is a polynomial of degree $n$, in $x$ and $y$). Let this be
$a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0$.
As this is an asymptote, it must intersect the curve at only infinity, so that $\infty$ is the only root of the equation. Replace $x$ with $1/z$, and the equation becomes
$\dfrac{a_n}{z^n} + \dfrac{a_{n-1}}{z^{n-1}} + \ldots + a_0 = 0 \Rightarrow\\
a_0 z^n + a_1 z^{n - 1} + \ldots + a_{n - 1}z + a_n = 0
$
As $x \to \infty$, $z = 1/x \to 0$. But for $0$ to be a root of the equation, the constant term $a_n$ must be $0$. Then the equation becomes
$a_0 z^n + a_1 z^{n - 1} + \ldots + a_{n - 1}z = 0 \Rightarrow\\
a_0 z^{n - 1} + a_1 z^{n - 2} + \ldots + a_{n - 1} = 0
$
Again, as $0$ is the only root, we can equate $a_{n - 1}$ to $0$ as well. And so on.
Thus, $a_n = a_{n - 1} = \ldots = 0$.
Note: This is an undergraduate level explanation, and not a rigorous analytic proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/830229",
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Show that it doesn't exist any of natural number $ n = 4m + 3$ that $ n= x^2+y^2 $ for any natural x and y Show that it doesn't exist any of natural number $ n = 4m + 3$ that $ n= x^2+y^2 $ for any natural x and y
Show that every prime number in form $ p=4m+1 $ could be showed as $ p = x^2+y^2$ (x and y are natural)
I checked it and it
| *
*It can be: $$x=2n+1 \text{ and } y=2k+1 \ \text{ OR } \ x=2n \text{ and } y=2k \ \text{ OR } x=2n \text{ and } y=2k+1 \text{ OR } x=2n+1 \text{ and } y=2k$$
You can check each case and you will see that it cannot be $x^2+y^2=4m+3$
For example,at the first case you will get:
$$x^2+y^2=(2n+1)^2+(2k+1)^2=4n^2+4n+1+4k^2+4k+1=4(n^2+n+k^2+k)+2=4m+2$$
| {
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Understanding 2012 AMC 12B #23
Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that
$P(Q(x))$ has zeros at $x=-23$, $-21$, $-17$, and $-15$, and $Q(P(x))$ has zeros
at $x=-59$,$-57$,$-51$ and $-49$. What is the sum of the minimum values of
$P(x)$ and $Q(x)$?
I found a solution here, and I was able to follow it up until this sentence:
Since Q is monic, the roots of $Q(x) = a + \sqrt {b}$ are "farther"
from the axis of symmetry than the roots of $Q(x) = a - \sqrt {b}$.
Thus, we have
$Q( - 23) = - 54 + \sqrt {b}$
$Q( -21) =- 54 - \sqrt {b},$
or substituting
$16 - d = - 54 + \sqrt {b}$
$ 4 - d = - 54 - \sqrt {b}$
What does it mean by the roots are farther from the axis of symmetry, and how did they arrive at that? And from that, how does one get the mentioned equations?
| Claim: If $ \alpha, \beta$ are the roots of $ Q(x) = a + \sqrt{b}$ and $\gamma , \delta$ are the roots of $Q(x) = a - \sqrt{b}$, then
$$|\alpha - \beta | > | \gamma - \delta|.$$
Proof: This follows by considering the graph of the monic quadratic $y=Q(x)$ and the lines $ y = a + \sqrt{b}$ and $ y = a - \sqrt{b}$. Use the fact that $ a + \sqrt{b} > a - \sqrt{b}$. $_\square$
The axis of symmetry is the line $ x = \frac{\alpha + \beta} { 2} = \frac{ \gamma + \delta} {2}$. Hence, from the above inequality $\alpha, \beta$ are further away from this axis, than $\gamma, \delta$.
| {
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A sum regarding prime factorization Prime factorization of $n$ is $n = p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_k^{a_k}$
Let $f(n) = \left((p_1^{a_1}+1)(p_2^{a_2}+1)(p_3^{a_3}+1)\cdots(p_k^{a_k}+1)\right)$
I want to find the value of $$\sum_{n=1}^{N}f(n)$$
For example if $N=6$, then the answer is $(1^{1}+1)+(2^{1}+1)+(3^{1}+1)+(2^{2}+1)+(5^{1}+1)+(2^{1}+1)(3^{1}+1) = 32$
Can I find the value without factorizing all the numbers?
$$
1=1 \\
2=2 \\
3=3 \\
4=2^2 \\
5=5 \\
6=2\cdot3 \\
7=7 \\
8=2^3 \\
9=3^2 \\
10=2\cdot5
$$
So for $N=10$ answer is $(1+1)+(2+1)+(3+1)+(2^2+1)+(5+1)+(2+1)(3+1)+(7+1)+(2^3+1)+(3^2+1)+(2+1)(5+1) = 77$
So the sum sums over all $n=1$ to $N$. Same term is not added $n$ times.
| Let $F(N) = \sum_{n=1}^N f(n)$ where $f(n) = \sum_{d|n, \gcd(d,n/d) = 1} d$ as was noted in the comments by Hagen von Eitzen.
Now changing the order of summation and noting that $n$ is of the form $n = n'd$, we have
$$F(N) = \sum_{d=1}^N \sum_{\gcd(d,n') = 1, 1 \le n' d \le N} d,$$
which we can also write (by replacing $n'$ by $n$) in the form
$$F(N) = \sum_{\gcd(n,d) = 1, 1 \le n d \le N} d$$
where the latter sum now runs over both $n$ and $d$.
Now consider the sum
$$\sum_{1 \le n d \le N} d = \sum_{g=1}^\infty \sum_{\gcd(n,d) = g, 1 \le n d \le N} d.$$
By replacing $n$ by $gn'$ and $d$ by $gd'$, we get that this is equal to
$$\sum_{g=1}^\infty g \sum_{\gcd(n',d') = 1, 1 \le n' d' \le \left\lfloor \frac{N}{g^2} \right\rfloor} d'= \sum_{g=1}^\infty g F\left(\left\lfloor \frac{N}{g^2} \right\rfloor\right).$$
Moving the terms with $g=2,\dots$ to the other side we get the recurrence relation
$$F(N) = \sum_{1 \le nd \le N} d - \sum_{g=2}^{\sqrt{N}} g F\left(\left\lfloor \frac{N}{g^2} \right\rfloor\right).$$
Here the first term is simply $\sum_{d=1}^N d \left\lfloor \frac{N}{d} \right\rfloor$. This can be turned into a pretty fast algorithm by using dynamic programming. See Project Euler for many similar problems utilizing this kind of idea.
| {
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$AM+GM+HM+RMS=66$ For two reals $a,b\in\mathbb{R}$ Arithmetic mean(AM), Geometric Mean(GM), Harmonic Mean(HM) and Root mean square (RMS) all are integers and
$AM+GM+HM+RMS=66$
Find all such $a,b$
I have assumed $a+b=u$ and $ab=v$ and written all the means in the terms of $u,v$ and used trial and error after some analysis. But this is long method. Can there be a shorter method?
| $a+b=2u,ab=v^2 \implies AM=u,GM=v.HM=\dfrac{2v^2}{2u}=\dfrac{v^2}{u},RMS=\sqrt{2u^2-v^2}$
$HM$ is integer ,$\implies u=tq^2, t $ is not a perfect square number $ v=tqp \implies RMS=\sqrt{2t^2q^4-t^2q^2p^2}=tq\sqrt{2q^2-p^2} \implies tq^2+tqp+tp^2+tq\sqrt{2q^2-p^2}=66 \implies t(q^2+pq+p^2+q\sqrt{2q^2-p^2})=66 $
$u \ge v \implies q\ge p $,if $q=1 \implies p=1 ,t= \dfrac{33}{2}$, it is not possible .
$q\ge2 \implies (q^2+pq+p^2+q\sqrt{2q^2-p^2})> 10, q^2+pq+p^2+q\sqrt{2q^2-p^2}$is mono increasing function for $q,p,\implies t=1,2,3,6$
look at $\sqrt{2q^2-p^2}$, there is a general possible solution $p=q \implies \sqrt{2q^2-p^2}=q$, but in this case, $4tq=66$ which is not possible also.
$t=1,(q^2+pq+p^2+q\sqrt{2q^2-p^2})=66, q_{max}=5$
$q=5,p=1$, there is no other possibility.
$q=4$,but $\sqrt{2q^2-p^2}$ can't be a integer when $p=3,2,1$
$q=3$,but $\sqrt{2q^2-p^2}$ can't be a integer when $p=2,1$
$q=2$,but $\sqrt{2q^2-p^2}$ can't be a integer when $p=1$
when $t \ge 2, \implies q \le 4$, but we already know there is no solutions for all the cases.
the only solution is $u=25,v=5 \iff a+b=50,ab=25 \implies 25\pm 10\sqrt{6}$
| {
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Find the value of a + b + c + d Let $a$ and $b$ be the roots of the equation: $x^2 - 10cx - 11d = 0$ where $c$ and $d$ be the roots of $x^2 - 10ax - 11b = 0$. Find the value of $a+b+c+d$, assuming that they all are distinct.
I first tried making an equation with roots $(a+b)$ and $(c+d)$ to get the sum of the roots, however I wasn't able to solve this question using that method as the answer which I got was in terms of the variables itself.
I also tried placing $a$ into the first equation and $c$ into the second to cancel out a common term ($-10ac$), but after cancelling, I got: $(a^2 - c^2 - 11d + 11b = 0)$. Now I don't know how to move ahead.
| My Solution ::
Given $a,b$ are the roots of $x^2-10cx-11d=0.$
So $$ a+b = 10c \tag{1}$$
and
$$ab = -11d \tag{2} $$
and $c,d$ are the roots of $x^2-10ax-11b=0.$
So $$c+d=10a\tag{3}$$
and
$$cd=-11b \tag{4}$$
So $$a+b+c+d = 10(a+c). \tag{5}$$
Now $$\frac{a+b}{c+d} = \frac{10c}{10a}=\frac{c}{a}\Rightarrow a^2+ab=c^2+cd\Rightarrow a^2-11d=c^2-11b$$
So $$(a^2-c^2)=-11(b-d)\Rightarrow (a+c)\cdot(a-c)=-11(b-d) \tag{6}$$
Now $(1)-(3)$, we get $$a+b-c-d=10c-10a\Rightarrow (b-d) = 11(c-a)=-11(a-c)$$
Now putting $(b-d) = -11(a-c)$ in eqn. $(6)$, we get $$(a+c)\cdot(a-c)=121(a-c)$$
So $(a-c)\cdot(a+c-121) = 0$. Now $a\neq c$, because $a,b,c,d$ are distinct real numbers.
So $a+c=121$. Put into eqn. $(5)$.
We get
$$
a+b+c+d = 10(a+c) = 10\cdot 121 = 1210\Rightarrow (a+b+c+d) = 1210.
$$
| {
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"question_score": "5",
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Fair die being rolled repeatedly A fair die is rolled repeatedly, and let $X$ record the number of the roll when the 1st $6$ appears. A game is played as follows. A player pays \$1 to play the game. If $X\leq 5$ , then he loses the dollar. If $6 \le X \le 10$, then he gets his dollar back plus \$1. And if $X > 10$, then he gets his dollar back plus \$2 . Is this a fair game? If not, whom does it favour?
I think that that it is not a fair game because it solely depends on whether the first number is $6$ which is a $\frac{1}{6}$ chance. But I don't know how to prove this further.
| The probability the number is $X$ is $(\dfrac{5}{6})^{X-1}\cdot\dfrac{1}{6}$
Therefore the expected gains of a game is
$$\sum_{X=6}^{\infty}(\dfrac{5}{6})^{X-1}\cdot\dfrac{1}{6}+2\sum_{X=11}^\infty(\dfrac{5}{6})^{X-1}\cdot\dfrac{1}{6}=\dfrac{1}{6}(\sum_{X=6}^\infty(\dfrac{5}{6})^{X-1}+2\sum_{X=11}^\infty(\dfrac{5}{6})^{X-1})=$$
$$\dfrac{1}{6}(18-(1+\frac{5}{6}+((\frac{5}{6})^2\dots(\frac{5}{6})^4)+2(1+\frac{5}{6}+((\frac{5}{6})^2\dots(\frac{5}{6})^9)=$$
$$\frac{1}{6}(18-\frac{4651}{1296}-2(\frac{50700551}{10077696}))\approx0.72$$ so no, it isn't worth it to play.
| {
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Check entry extended ternary Golay code The extended ternary Golay code is the linear $[12,6,6]$-code with the following generator matrix:
$$
C=\left(
\begin{array}
&1&0&0&0&0&0&0&1&1&1&1&1\\
0&1&0&0&0&0&1&0&1&2&2&1\\
0&0&1&0&0&0&1&1&0&1&2&2\\
0&0&0&1&0&0&1&2&1&0&1&2\\
0&0&0&0&1&0&1&2&2&1&0&1\\
0&0&0&0&0&1&1&1&2&2&1&0
\end{array}
\right)
$$
But if I look at the check-sum (the sum of the entries) of the second row, it is equal to $8$ and this is not $0\pmod 3$. What is going on? This code is supposed to be the extension of the ternary Golay code.
| Posting my comment as an answer lest this question gets stuck in the unanswered queue.
It is a possible explanation that the intended generator matrix for the code $C$ was
$$
G=\left(
\begin{array}
&1&0&0&0&0&0&0&1&1&1&1&1\\
0&1&0&0&0&0&2&0&1&2&2&1\\
0&0&1&0&0&0&2&1&0&1&2&2\\
0&0&0&1&0&0&2&2&1&0&1&2\\
0&0&0&0&1&0&2&2&2&1&0&1\\
0&0&0&0&0&1&2&1&2&2&1&0
\end{array}
\right)
$$
It is easy to see that $GG^T=0$, and as $G$ clearly has rank six, it follows that it generates a self-dual code. So we can use $G$ as a ternary check matrix as well.
Because $1^2\equiv2^2=1\pmod3$ it follows that the Hamming weights of all the words in $C$ are multiples of three. Because no set of three columns is linearly dependent (this takes some convincing as it splits into many cases, but they all are quite straightforward), the minimum Hamming distance of $C$ is at least six. It follows from the uniqueness (up to equivalence) of the Golay codes that $C$ must be equivalent to the Golay code.
So the original code is equivalent to the ternary Golay code. The seventh symbol of every word is negated. It depends on your definititions whether that is killjoy or not.
| {
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Prove this identity... $$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$
This is what I've done:
$$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{\sin x}{1+\cos x}$$
I have no idea what to do next.
edit:
Solution:
$$\frac{\sin2\frac{x}{2}}{1+\cos2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}cos\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=$$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}$$
| By your estimate we have $$\frac{\sin x}{1+\cos x}=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2\cos^2\frac{x}{2}}=\tan\frac{x}{2}.$$
| {
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Solving $x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$. I have problem with this equation:
$$x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$$
Any ideas on beatiful solving?
| Assume $x+6=a$ and $x-6=b$
So LHS= $(a+b)/2-\sqrt{ab}=1/2 * (\sqrt a-\sqrt b)^2$.
RHS= $b^2/2a$.
Hence, $b/\sqrt a=\sqrt a- \sqrt b$
Hence $b=a-\sqrt{ab}$
Therefore $\sqrt {ab}=a-b=12.$ Now, put $a=12+b$ and solve $b(12+b)=144$.
Hope this helps. Correct me if I'm wrong!
| {
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Probability Help (die problem) A die is rolled 20 times. How many different sequences
a) each number 1-6 is rolled exactly three times
My Answer: (20 choose 6)*(3 choose 1)
b) each number 1-6 are each rolled exactly once in the first six rolls?
My Answer: (20 choose 6)*(6 choose 1)
c) each number rolled is at least as big as the number that was rolled directly before it?
ex: 111222333344455566666
My Answer: No idea
| a) I am considering it to mean atleast 3 times, which can be obtained by an exponential generating function:
\begin{align*}
G(x) &= \left(\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\right)^6 \\
\left[\frac{x^{20}}{20!}\right]G(x) &= \frac{11}{414720}\cdot 20! = 64530097632000
\end{align*}
b) has been answered : $$6!\, 6^{14} = 56422198149120$$
c) Use a program to generate the sequence : $$6,21,56,126,252,462,792,1287,2002,\ldots$$
Looking up in OEIS gives A000389.
Hence, for $n$ dice throws, the number of valid sequences are:
\begin{align*}
a_n &= \binom{n+5}{5} \\
\implies a_{20} &= \binom{25}{5} = 53130
\end{align*}
| {
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Product in terms of $n$ of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$ What is the following product in terms of $n$?
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$$
Thank you.
| $$\frac{1}{2}\frac{3}{4}\frac{5}{6} \dotsb \frac{2n-1}{2n}=\frac{1\mathbf{2}3\mathbf{4}5 \dotsb (2n-1)\mathbf{2n}}{(2 \cdot 4 \cdot 6 \dotsb 2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$
| {
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Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$
Prove $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$.
I tried to do this in two ways, I'm not sure about CMVT and I have a problem with the other way.
Using Cauchy's MVT:
RHS:
$\sin x \le x \implies \frac {\sin x}{x}\le 1$
So define:
$f(x)=\sin x, \ g(x)=x$ then from CMVT:
$\dfrac {f(\frac {\pi} 2)}{g(\frac {\pi} 2)}=\dfrac {f'(c)} {g'(c)}=\cos c$
and from the fact that $c$ is between $0$ and $\pi/2 \implies \cos c \le 1$.
LHS: In the same manner but here I run into some trouble:
$\frac2\pi x \le \sin x\implies \frac {2x}{\pi\sin x}\le 1$
So:
$\dfrac {f(\frac {\pi} 2)}{g(\frac {\pi} 2)}=\dfrac {f'(c)} {g'(c)}\implies\frac {1}{\sin {\frac {\pi}{2}}}=\frac {2}{\pi \cos c}$
Here actually
$\frac {1}{\sin {\frac {\pi}{2}}}=1$ so it's also $\le 1$
Is it correct to use CMVT like this ?
The other way:
We want to show: $f(x)=\sin x - x < 0$ and $g(x)=\frac {2x}{\pi}-sinx <0 $ by deriving both it's easy to show that the inequality stands for $f$ but for $g$ it isn't so obvious that $g'(x)=\frac {2}{\pi}-\cos x$ is negative. In fact for $x=\frac {\pi} 2$ it's positive. Please help figure this out.
This is the same The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$ but all the answers there are partial or hints and I want to avoid convexity.
Note: I can't use integrals.
| For any $x \in (0,\frac{\pi}{2})$, consider the expression
$$\frac{\sin x - \sin 0}{x - 0} - \frac{\sin\frac{\pi}{2} - \sin x}{\frac{\pi}{2}- x}
= \frac{\sin x}{x} - \frac{1-\sin x}{\frac{\pi}{2} - x}\tag{*1}
$$
Apply MVT on for the first term on $[0,x]$ and the second term on $[x,\frac{\pi}{2}]$, we can find two numbers $y, z$ such that
$$0 < y < x < z < \frac{\pi}{2}\quad\text{ and }\quad \frac{\sin x}{x} = \cos y \;\land\; \frac{1-\sin x}{\frac{\pi}{2} - x} = \cos z$$
Since $\cos t$ is strictly decreasing on $[0,\frac{\pi}{2}]$, we have $\cos y > \cos z$ and hence
$$\begin{align}\frac{\sin x}{x} - \frac{1-\sin x}{\frac{\pi}{2} - x} > 0
&\iff
\left(\frac{\pi}{2} - x \right)\sin x - x \left(1 - \sin x\right) > 0\\
&\iff
\sin x > \frac{2x}{\pi}\end{align}$$
You may wonder how I arrive the expression in $(*1)$. Geometrically,
*
*$\displaystyle\;\frac{\sin x}{x}\;$ is the slope of $\sin x$ over $[0,x]$.
*$\displaystyle\;\frac{1-\sin x}{\frac{\pi}{2} - x}\;$ is the slope over $[x,\frac{\pi}{2}]$.
This proof works because the slope $\cos x$ is decreasing on $[0,\frac{\pi}{2}]$.
This is sort of equivalent to $\sin''(x) = -\sin x < 0$. In certain sense, this is really a proof with convexity hiding under the carpet.
| {
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Basic Trigonometry Question If $\cos{(A-B)}=\frac{3}{5}$ and $\sin{(A+B)}=\frac{12}{13}$, then find $\cos{(2B)}$.
Correct answer = 63/65.
I tried all identities I know but I have no idea how to proceed.
| From
$$\cos{(A-B)}=\frac{3}{5}\Rightarrow \sin(A-B)=\sqrt{1-(3/5)^2}=\frac{4}{5}$$
and from
$$\sin{(A+B)}=\frac{12}{13}\Rightarrow \cos(A+B)=\sqrt{1-(12/13)^2}=\frac{5}{13}$$
then
$$\cos(2B)=\cos((A+B)-(A-B))=$$
$$=\cos(A+B)\cos(A-B)+\sin(A+B)\sin(A-B)=$$
$$=\frac{5}{13}\frac{3}{5}+\frac{12}{13}\frac{4}{5}=\frac{63}{65}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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how to solve this question of polynomials
Given the polynmial is exactly divided by $x+1$, when it is divided by
$3x-1$, the remainder is $4$. The polynomial leaves remainder $hx+k$ when divided by
$3x^2+2x-1$. Find $h$ and $k$.
This is the question which is confusing me.. i have done this question like this:
$p(x) = g1(x)(x+1) +0 $
$p(x) = g2(x)(3x-1) +4 $
$p(x) = g3(x)(3x+2x-1)+hx+k $
$=> p(x) = g3(x)(x+1)(3x-1) + hx+k $
now putting the value of $x$ in each $p(x) $
$p(-1) = 0 \tag{1}$
$p(1/3) = 4 \tag{2}$
$p(-1,1/3) = hx+k \tag{3}$
from equation $(2)$ and $(3)$
$4=hx+k\tag{4}$
now putting the value of $x$ in equation $(4)$.
$h(-1)+k=4 => -h+k=4 \tag{5}$
$h(1/3)+k=4 => h+3k=12 \tag{6}$
now adding equations $(5)$ and $(6)$
$ h+3k-h+k = 4+12 =>4k=16 => k=4$
now puting the value of of $k$ in equation $(5)$
$h=0$
therefore my answer is $h=0,k=4$ but the answers are$ h=3, k=3$. please help me to sort out my problem
| The error in your solution is that p(−1)=0 implies h(−1)+k=0, not 4,Right
| {
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Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a^2}-\sqrt{b} < \sqrt{a^2+b}$, but plugging this in ends up with a non-sense lower-bound of $-\sqrt{b}$ even though the target number is positive.
\begin{equation*}
\big(\sqrt{a^2}-\sqrt{b} \big) - a < \sqrt{a^2+b}-a
\end{equation*}
How can I obtain some positive lower-bound?
| By the mean value theorem,
$$\sqrt{1 + x} - 1 = f(1 + x) - f(1) = x f'(c)$$
where $f$ is square root, $f'$ is its derivative,
and $c$ is some point in $[1, 1+x]$.
We need a lower bound and $f'$ is decreasing,
so $c$ is at worst $1 + x$ and we obtain
$$x f'(c) ≥ xf'(1 + x) = \frac{x}{2\sqrt{1 + x}}.$$
Backtrack: you want to underestimate $\sqrt{a^2 + b} - a$,
which is $a(\sqrt{1 + b/a^2} - 1)$
and we can let $x$ be $b/a^2$:
$$\sqrt{a^2 + b} - a ≥ a\frac{b/a^2}{2\sqrt{1 + b/a^2}} = \frac{b}{2\sqrt{a^2+b}}.$$
For an upper bound, $c$ is at best $1$ so
$$\sqrt{a^2 + b} - a ≤ \frac{b}{2a}$$
which is trivial (cf. completing the square).
| {
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What technique turns $\frac{x}{(x-2)^2(x+1)}$ into $-\frac{1}{9x+9}+\frac{1}{9x-18}+\frac{2}{3(x-2)^2}$? I found this:
Let's rewrite the integrand so that it's easier to integrate:
$$\dfrac{x}{(x-2)^2(x+1)} = -\dfrac{1}{9x+9}+\dfrac{1}{9x-18}+\dfrac{2}{3(x-2)^2}$$
I see how the expressions are equal, but I don't know how to get from the left to the right. There must be a technique that is used; could you tell me what the technique is called so I can read more about it or maybe you could explain it to me?
Thanks for your help.
| The technique that is being used is called partial fraction decomposition. You'll get "tons" of hits if you Google the phrase.
As applied to this particular question, first note that
$$\dfrac{x}{(x-2)^2(x+1)} = -\dfrac{1}{9x+9}+\dfrac{1}{9x-18}+\dfrac{2}{3(x-2)^2}$$
$$= -\frac 19\cdot \frac 1{x + 1} + \frac 19 \cdot \frac{1}{x-2} + \frac 23\cdot \frac 1{(x-2)^2}\tag{1}$$
Now, back to the original fraction; let's go through how it is decomposed:
$$\dfrac{x}{(x-2)^2(x+1)} = \frac A{x+1} + \frac B{x-2} + \frac C{(x-2)^2}\tag{2}$$ Now the objective is to solve for $A, B, C$.
We now find that $$A(x-2)^2 + B(x+1)(x-2) + C(x+1) = x \tag{3}$$ $(3)$ is what we'd get if we multiplied each side of equation $(2)$ by the denominator of the right-hand side.
We can then solve for $A, B, C$ by expanding the left hand side of $(3)$, and equating coefficients from left and right, and obtaining a system of three equations in three unknowns. But in this case, another method works quite nicely.
Evaluate $(3)$ at $x = 2 \implies 3C = 2 \iff C= \frac 23$.
Evaluate $(3)$ at $x = -1 \implies 9A = -1\iff A =-\frac 19$.
Now, solve for $B$ using the found values of $A, C$, setting $x = 0$, say. Then evaluate. If you do this, you'll obtain $B = \frac 19$.
Now, replace $A, B, C$ in the right-hand side of equation $(2)$. Compare this with the right-most expression of equation $(1)$.
| {
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Prove that $\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}$ is a strictly decreasing function. This is part of an actuarial science problem. Unfortunately, the official solution of this problem takes the derivative of
$$\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}\text{, } \quad x \geq 0\text{.}$$
and shows that it is always $\leq 0$. However, this does not at all show that the function is strictly decreasing.
I'm trying to prove this myself. If I assume $x > y$, I want to show that
$$\dfrac{0.5x^2 + x + 1}{x^2 + x + 1} < \dfrac{0.5y^2 + y + 1}{y^2 + y + 1}\text{.}$$
Needless to say, this does not look clean if I were to "work backwards."
Any suggestions?
| We can write the function as $\dfrac{0.5x^2+x+1}{x^2+x+1} = 1 - \dfrac{0.5x^2}{x^2+x+1} = 1 - \dfrac{0.5}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}$.
You can easily see that $1+\dfrac{1}{x}+\dfrac{1}{x^2}$ is strictly decreasing, so $\dfrac{0.5}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}$ is strictly increasing.
Therefore, $1 - \dfrac{0.5}{1+\dfrac{1}{x}+\dfrac{1}{x^2}} = \dfrac{0.5x^2+x+1}{x^2+x+1}$ is strictly decreasing.
| {
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How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1} \frac{1}{m^2+(N-m)^2}$$
$$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$
but it doesn't work.
| If the sum were finite, then we could get a contradiction as follows. Breaking it up into 4 sums depending on whether or not $m$ and $n$ are even, we have
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2} = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n)^2} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m-1)^2+(2n)^2} $$
$$+ \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n-1)^2} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m-1)^2+(2n-1)^2} $$
Note that each of the last three sums is greater than the first due to the denominators of each term being smaller. Thus we have
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2} > 4 \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n)^2} $$
But factoring out the 4 from the denominator we see the right hand side is the same as the left. Hence we get a contradiction and the sum is not finite.
| {
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"source": "stackexchange",
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} |
Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?
| As $x \to\infty$, $A\to x$ and $B\to x$. Therefore the expression would tend to $0$.
| {
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For the polynomial For the polynomial, -2 is a zero. $h(x)= x^3+8x^2+14x+4$. Express $h(x)$ as a product of linear factors.
Can someone please explain and help me solve?
| Okay, the first thing to do is polynomial division (or synthetic division, whichever you prefer).
Since -2 is a 0, we know that $(x+2)$ is a factor of $h(x)$. We then divide $h(x)$ by $(x+2)$
Dividing: $$\frac{x^3 + 8x^2 + 14x + 4}{x+2} = x^2 + 6x + 2$$
So $h(x)$ becomes: $$(x+2)*(x^2+6x+2)$$
Now you must use the Quadratic formula to find the root of $x^2 + 6x +2$
$$x = \frac{-6 +/- \sqrt{28}}{2}$$
$$x = -3 +/- \sqrt{7}$$
So $h(x)$ is $$(x+3+\sqrt{7})(x+3-\sqrt{7})(x+2)$$
| {
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What's the intuition behind the 2D rotation matrix? Can anyone offer an intuitive proof of why the 2D rotation matrix works?
http://en.wikipedia.org/wiki/Rotation_matrix
I've tried to derive it using polar coordinates to no avail.
| There's two ways to think of it. At this stage, I think seeing both might be helpful.
Hard way: Let $(x,y) \in \Bbb R^2$ be represented by polar coordinates $(r, \varphi)$. I mean the relations $x = r \cos \varphi, y = r \sin \varphi$. So, let $(x_\theta, y_\theta)$ be the point after a rotation of $\theta$. Clearly, we have: $$\begin{align} (x_\theta, y_\theta) &= (r \cos(\varphi + \theta), r \sin (\varphi + \theta)) \\ &= (r(\cos \varphi \cos \theta - \sin \varphi \sin \theta), r(\sin \varphi \cos \theta + \cos \varphi \sin \theta)) \\ &= (r \cos \varphi \cos \theta - r \sin \varphi \sin \theta, r \sin \varphi \cos \theta + r \cos \varphi \sin \theta) \\ &= (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta) \end{align}$$
To find the matrix, in the standard basis, see that $(1,0)$ goes to $(\cos \theta, \sin \theta)$, and $(0,1)$ goes to $(- \sin \theta, \cos \theta)$. So, the matrix is: $$\begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} $$
Easy way: Let $R_\theta$ be the rotation. See that rotations are linear mappings, hence, it suffices to know the effect of the transformation on a basis, let's say... $((1,0), (0,1))$. Drawing, it is easy to see that $R_\theta (1,0) = (\cos \theta, \sin \theta)$ and $R_\theta (0, 1) = (- \sin \theta, \cos \theta)$. This way, the matrix is: $$\begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$
Finally, we can give the transformation: $$(R_\theta (x,y)) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}$$
Maybe you don't agree with me on what point of view is easier, but if you get a clear idea of both, I'm happy.
| {
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Basic induction proof methods so we're looking to prove $P(n)$ that
$$1^2+2^3+\cdots+n^3 = (n(n+1)/2)^2$$
I know the basis step for $p(1)$ holds.
We're going to assume $P(k)$
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
And we're looking to prove $P(k+1)$
What I've discerned from the internet is that I should be looking to add the next term, $k+1$, to both sides so...
$$1^3+2^3+\cdots+k^3 + (k+1)^3=(k(k+1)/2)^2 + (k+1)^3$$
now I saw some nonsense since that we assumed $p(k)$ we can use it as a definition in our proof, specifically on the left hand side
so since
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
then
$$(k(k+1)/2)^2 + (k+1)^3 = (k(k+1)/2)^2 + (k+1)^3$$
and we have our proof
OK so far thats wrong
so far ive figured this.
$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+1)+1)/2)^2$$
Then
$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+2)/2)^2$$
using the definition
$$(k(k+1)/2)^2 + (k+1)^3 = ((k+1)((k+2)/2)^2$$
$$(k^2+k/2)^2 + (k^2+2k+1)(k+1) = (k^2+3k+2/2)^2$$
$$(k^4+k^2/4)+(k^2+2k^2+k+k^2+2k+1)= (k^4+9k^2+4/4)$$
Where should I go from here? It doesn't possibly look like these could equate, I'll keep going though
| An example proof:
Let $P(k)$ denote the statement that $\sum\limits_{i = 1}^k i^3 = \left( \frac{k(k + 1)}{2} \right)^2$.
We wish to prove that for all $k \in \mathbb Z$ such that $k > 0$, $P(k)$.
We will prove this by induction on $k$.
In the base case, $k = 1$, and we have that $1^3 = 1 = \left(\frac{1(1 + 1)}{2}\right)^2$.
Now, for the inductive case, we suppose that for some $k$, $P(k)$ holds.
Then, as we have assumed $P(k)$, we have that $\sum\limits_{i = 1}^k i^3 = \left(\frac{k(k + 1)}{2} \right)^2$.
Adding $(k + 1)^3$ to both sides, we get that $\sum\limits_{i = 1}^{k + 1} i^3 = \left(\frac{k(k + 1)}{2} \right)^2 + (k + 1)^3$.
Now, the right-hand side is equal to, by factoring, $(k + 1)^2 \left(\frac{k^2}{4} + (k + 1)\right)$.
We can then rewrite this as $(k + 1)^2 \left(\frac{k^2 + (4(k + 1))}{4}\right)$, which then becomes $\frac{(k + 1)^2(k^2 + 4k + 4)}{4}$, or $\frac{(k + 1)^2(k + 2)^2}{4}$.
This is equal to $\left(\frac{(k + 1)(k + 2)}{2}\right)^2$, and so we have that
$$
\sum\limits_{i = 1}^{k + 1} i^3 = \left(\frac{(k + 1)(k + 2)}{2}\right)^2,
$$
which is precisely $P(k + 1)$.
Thus, we have shown that $P(1)$ holds and that for all $k \in \mathbb Z$ such that $k > 0$, $P(k) \implies P(k + 1)$, and so, by the principle of mathematical induction, $P(k)$ holds for all $k$.
| {
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Let $N$ and $M$ be two digit numbers. Then the digits of $M^2$ are those of $N^2$, but reversed.
Let $N$ be a two digit number and let $M$ be the number formed from $M$ by reversing $N$'s digits. The digits of $M^2$ are precisely those of $N^2$, but reversed.
$Proof$:
Since $N$ is a two digit number, we can write $N = 10a + b$ where $a$ and $b$ are the digits of $N$. Since $M$ is formed from $N$ by reversing digits, $M = 10b + a$.
$N^2 = (10a + b)^2 = 100a^2 + 20ab + b^2 $. The digits of $N^2$ are $a^2, 2ab, b^2$.
$M^2 = (10b + a)^2 = 100b^2 + 20ab + a^2$. The digits of $M^2$ are $b^2, 2ab, a^2$, exactly the reverse of $N^2$.
This proposition is false. Let $N$ be $15$. That means the proof above is not correct, but I can't see where exactly.
| however there exists an infinite sequence of such numbers: we have
$12^2=144$
$102^2=10404$
$1002^2=1004004$
$10002^2=100040004$
$\dots\ \dots\ \dots$
and the same with digits reversed:
$21^2=441$
$201^2=40401$
$2001^2=4004001$
$20001^2=400040001$
$\dots\ \dots\ \dots$
| {
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Solve $a$ and $b$ for centre of mass in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Given ellipse:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
What length do $a$ and $b$ have to be so the centre of mass is $S(4;2)$?
I've tried steps to solve the equation to $$y=b\sqrt{1-\frac{x^2}{a^2}}$$ and integrate
$$A=b\int_0^a{\sqrt{1-\frac{x^2}{a^2}}}$$
But I'm not achieving a satisfying result. There must be an easier way . Enlighten me please
|
We can also use Pappus' (Second) Centroid Theorem, which states that the volume of a solid of revolution is equal to the area of the region revolved about the axis of symmetry times the circumference of the circular path "swept out" by the centroid of the region,
$$ V \ = \ A \ \cdot \ 2 \pi \ \overline{r} \ \ . $$
The area of the quarter-ellipse (in the diagram for the problem) for which we wish to locate its centroid is $ \ \frac{1}{4} \cdot \pi \ ab \ $ .
If we revolve this region about the $ \ x-$ axis, we obtain a solid of revolution which is half of a "prolate spheroid", an ellipsoid with one of its semi-axes having length $ \ a \ $ and two with length $ \ b \ $ . Its volume is then $ \ \frac{1}{2} \cdot \ \frac{4 \pi}{3} \ ab^2 \ $ . The circumference of the centroid's path (in blue) is $ \ 2 \pi \cdot \ \overline{y} \ $ . By Pappus' Theorem, we conclude
$$ \frac{2 \pi}{3} \ ab^2 \ = \ \frac{\pi}{4} \ ab \ \cdot \ 2 \pi \cdot \ \overline{y} \ \ \Rightarrow \ \ \overline{y} \ = \ \frac{4}{3 \pi} \ b \ \ . $$
By similar reasoning, revolving the quarter-ellipse about the $ \ y-$ axis generates half of an "oblate spheroid" with one semi-axis of length $ \ b \ $ and two with length $ \ a \ $ ; the circumference of the centroid path (in green) is $ \ 2 \pi \cdot \ \overline{x} \ $ . Hence,
$$ \frac{2 \pi}{3} \ a^2b \ = \ \frac{\pi}{4} \ ab \ \cdot \ 2 \pi \cdot \ \overline{x} \ \ \Rightarrow \ \ \overline{x} \ = \ \frac{4}{3 \pi} \ a \ \ . $$
The Centroid Theorems have a direct connection to the moment integrals used to compute the coordinates of a centroid, although the means to describe them in this way was not available to Pappus.
Answering the question for the problem is then a matter of solving
$$ 4 \ = \ \frac{4}{3 \pi} \ a \ \ \ \text{and} \ \ \ 2 \ = \ \frac{4}{3 \pi} \ b \ \ , $$
as also shown in other posted answers here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Identities and formulas There are so many identities like $\sin2θ$, $\cos2θ$, $\tan2θ$, $\sin(θ/2)$, $\cos(θ/2)$ and $\tan(θ/2)$. there are other formulas too like $\cos(α-β)$, $\sin(α-β)$ etc and yes the sum and product formulas of trigonometric function...
I am stuck in memorizing all. Is there any simple trick to memorize these formulas and identities?
| For the multiple angle stuff it suffices to remember De Moivre's formula:
$$
(\cos x + i \sin x)^n = \cos (nx) + i \sin (nx).\,
$$
With it you easily get
$$
\begin{align}(\cos x + i \sin x)^2 &= \cos^2 x + 2i\sin x \cos x - \sin^2 x = (\cos^2 x - \sin^2 x) + i(2 \sin x \cos x)\\ &= \cos(2x) + i \sin (2x)\end{align}
$$
and then
$$
\cos^2 x - \sin^2 x = \cos(2x) \\
2\sin x \cos x = \sin(2x) .
$$
by comparing real and imaginary parts...
Sums and differences can be written in Matrix Form:
$$
\begin{align} & {} \quad \left(\begin{array}{rr} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right) \left(\begin{array}{rr} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{array}\right) \\[12pt] & = \left(\begin{array}{rr} \cos\alpha\cos\beta - \sin\alpha\sin\beta & -\cos\alpha\sin\beta - \sin\alpha\cos\beta \\ \sin\alpha\cos\beta + \cos\alpha\sin\beta & -\sin\alpha\sin\beta + \cos\alpha\cos\beta \end{array}\right) \\[12pt] & = \left(\begin{array}{rr} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{array}\right). \end{align}
$$
| {
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$13\mid4^{2n+1}+3^{n+2}$ How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?
| $$4^{2n+1}+3^{n+2}=16^n\cdot 4+3^n\cdot 9\\16^n\cdot4+3^n\cdot 9\equiv3^n\cdot4+3^n\cdot 9\pmod {13}\\3^n(4+9)\equiv3^n\cdot13\equiv0\pmod{13}$$
This can also be solved with induction,for $n=0$
$$4+3^2=13$$
Assume it holds for $n=k$
$$4^{2k+1}+3^{k+2}$$
Prove it holds for $n=k+1$
$$4^{2k+3}+3^{k+3}=16\cdot4^{2k+1}+3\cdot3^{k+2}=16\cdot4^{2k+1}+16\cdot3^{k+2}-13\cdot3^{k+2}=16(4^{2k+1}+3^{k+2})-13\cdot3^{k+2}$$
by the inductive hypothesis $4^{2k+1}+3^{k+2}$ is divisible by $13$ because of that the whole expression is
| {
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"url": "https://math.stackexchange.com/questions/859523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to Show the following converges to $e^{\frac{t^2}{2}}$ How to prove that $$\lim_{n\to\infty}\left[\left[e^{t\sqrt{\frac{1-p}{np\vphantom{()}}}}-1-t\cdot\sqrt{\frac{1-p}{np}}-\frac{1}{2}t^2\left(\frac{1-p}{np}\right)\right]\cdot p \\+\left[e^{-t\sqrt{\frac{p}{n(1-p)}}}-1+t\cdot\sqrt{\frac{p}{n{(1-p)}}}-\frac{1}{2}\cdot t^2\left(\frac{p}{n(1-p)}\right)\right](1-p)+\left(1+\frac{t^2}{2n}\right)\right]^n$$ converges to $e^{\frac{t^2}{2}}$ . where $0 \leq p\leq 1$ and $t$ is a parameter which can take any real value
| Reacll that: $e^{x}=1+x+\frac{x^{2}}{2}+o(x^{2})$ for small $x$ which corresponds to large $n$. Taking $x=t\sqrt{\frac{1-p}{np}}$ and $x=-t\sqrt{\frac{p}{n(1-p)}}$ we see that the first two terms are $o(x^{2})$. Notice also that as $n\to\infty$ we have $(1+\frac{\frac{t^{2}}{2}}{n})^{n}\to e^{\frac{t^{2}}{2}}$. So we have:
$((1+\frac{\frac{t^{2}}{2}}{n})+o(t^{2}(\frac{1-p}{np}))+o(t^{2}\frac{p}{n(1-p)}))^{n}=(1+\frac{\frac{t^{2}}{2}}{n})^{n}+o(t^{2}\frac{1-p}{p})+o(t^{2}\frac{p}{1-p})$
which tends to $e^{\frac{t^{2}}{2}}$.
To make the notation simpler let $u=t\sqrt{\frac{1-p}{p}}$ and $v=t\sqrt{\frac{p}{n(1-p)}}$. Then the limit we are interested in is:
$\lim_{n\to\infty}((e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p+(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)+(1+\frac{\frac{t^{2}}{2}}{n}))^{n}$
Taking exponentials it suffices to show the following tends to $\frac{t^{2}}{2}$:
$\lim_{n\to\infty} n\ln((e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p+(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)+(1+\frac{\frac{t^{2}}{2}}{n}))$
$=\lim_{n\to\infty}\bigg(\frac{\ln\bigg(\frac{(e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p}{\frac{u^{2}}{n}}\frac{u^{2}}{n}+\frac{(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)}{\frac{v^{2}}{n}}\frac{v^{2}}{n}+(1+\frac{\frac{t^{2}}{2}}{n})\bigg)}{(\frac{(e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p}{\frac{u^{2}}{n}}\frac{u^{2}}{n}+\frac{(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)}{\frac{v^{2}}{n}}\frac{v^{2}}{n}+(\frac{\frac{t^{2}}{2}}{n}))}\bigg)\cdot$
$\lim_{n\to\infty}\bigg({(\frac{(e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p}{\frac{u^{2}}{n}}u^{2}+\frac{(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)}{\frac{v^{2}}{n}}v^{2}+(\frac{t^{2}}{2}))}\bigg)$
Notice that as $x\to0$ then $\frac{e^{x}-1-x-\frac{x^{2}}{2}}{x^{2}}\to0$ and $\frac{\ln(1+x)}{x}\to1$ which corresponds to $n$ tending to $\infty$. Notice that the final limit is a product of two limits. I had to separate them so that it would fit properly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generalization of Bernoulli's Inequality Is it possible to generalize Bernoulli's Inequality to $(x+y)^n \geq x + ny$ provided $x+y \geq 0 $ and $x \geq 1$ and $n$ is a positive natural number?
I was thinking that the proof follows by induction: In base case, have $n=1$. Then we have that $(x+y)^n = (x+y)^1 = x+y \geq x+y$. Now, inductively assume that $(x+y)^k \geq x + ky$. Then $(x+y)^{k+1} = (x+y)(x+y)^k \geq (x+y)(x+ky)$ by our inductive hypothesis and as $(x+y) \geq 0$. Now, "foiling,' we see that $(x+y)(x+ky) = x^2 + kxy + xy + ky^2 \geq x^2 + (k+1)xy$. As $x \geq 1$, we have that $1/x \leq 1$, so $x^2 + (k+1)xy \geq (1/x)(x^2+(k+1)xy) = x + (k+1)y$. Then by transitivity of order, we have $(x+y)^{k+1} \geq x + (k+1)y$, closing induction.
Thus, $(x+y)^n \geq x + ny$ for all positive natural n.
Is this valid? Did I do anything that was incorrect/invalid? I am just curious because I have not found this generalization anywhere, and am wondering if somehow it is not a generalization at all.
| There is one step that is not (always) valid. The inequality
$$x^2 + (k+1)xy \geqslant \frac{1}{x}(x^2+(k+1)xy)$$
only holds for $x+(k+1)y \geqslant 0$, if $x > 1$. If $x+(k+1)y < 0$ and $x > 1$, you have a strict inequality in the other direction,
$$x^2 + (k+1)xy < \frac{1}{x}(x^2+(k+1)xy).$$
However, in that case, the desired inequality follows from
$$(x+y)^{k+1} \geqslant 0 > x + (k+1)y.$$
For $x+ky \geqslant 0$ (for $x+ky < 0$, we have the sharper $(x+y)^k \geqslant 0 > x+ky$), the Bernoulli inequality yields a sharper bound per
$$(x+y)^k = x^k\left(1+\frac{x}{y}\right)^k \geqslant x^k\left(1+k\frac{x}{y}\right) = x^k + kx^{k-1}y = x^{k-1}(x+ky),$$
which may explain why your inequality is not mentioned in the literature.
| {
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$
So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.
I do not know what to do next. Please help. Thank you.
| Use quadratic reciprocity.
$$4(a^2+ab+b^2)=(2a+b)^2+3b^2$$
so if
$$a^2+ab+b^2\equiv 0 \pmod p$$ then
$$(2a+b)^2\equiv -3b^2 \pmod p$$ and $-3$ is a quadratic residue so
$$\left(\frac{-3}{p}\right)=1.$$
However by reciprocity,
$$\left(\frac{-3}{p}\right)=\left(\frac{p}{-3}\right)=\left(\frac{2}{3}\right)=-1$$
| {
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"url": "https://math.stackexchange.com/questions/867413",
"timestamp": "2023-03-29T00:00:00",
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Surface area of sphere $x^2 + y^2 + z^2 = a^2$ cut by cylinder $x^2 + y^2 = ay$, $a>0$ The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.
The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.
We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:
$$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$
where $F$ is the equation of the sphere.
Plugging in the expressions and simplifying ($z \ge 0)$, we get:
$$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$
Converting to polar coordinates, we have:
$$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$
Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I going wrong?
| I know what you r doing rong, I solved this a week ago, the same way you did.
You forgot that $\sqrt{\sin^2\theta} = |\sin\theta|$, not $\sin\theta$
$ $
$ $
This is how you might have done
$$ 2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta}\dfrac{a}{\sqrt{a^2-r^2}}\cdot r \cdot drd\theta$$
$ $
$$= 2\int_{-\pi/2}^{\pi/2}\left[ -a\sqrt{a^2-r^2}\right]_0^{a\cos\theta}d\theta$$
$ $
$$= 2\int_{-\pi/2}^{\pi/2}- a^2\sqrt{\sin^2\theta}-\left(-a^2 \right)d\theta$$
$ $
$$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2\sin\theta d\theta$$
MISTAKE !!!!
$ $
INSTEAD $$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2|\sin\theta| d\theta$$
$ $
$$= 4\int_{0}^{\pi/2}a^2- a^2\sin\theta d\theta$$
$ $
$$= 4a^2 \left[1- \sin\theta \right] _{0}^{\pi/2}$$
$ $
$$= a^2\left(2\pi-4 \right)$$
| {
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modulo group defined by an algebraic relation I am asked if $\{n, n^{2}, n^{3}\}$ forms a group under multiplication modulo $m$ where $m = n + n^{2} + n^{3}.$
As an example we see that $\{2, 4, 8\}$ does form a group modulo $14,$ with identity $8,$ but am stuck starting the proof for the general case. Thanks in advance.
| We work modulo $m=n+n^2+n^3$. Since $m|n^4-n$, we have that $n^4=n$. This implies* $n^3=1$, so $n^3$ is the identity. Also, when $x,y\in\{n,n^2,n^3\}=G$, we have $xy\in G$, because $xy=n^an^b=n^{a+b}=n^c$, where $c-1=a+b-1\mod 3$. Take for example $x=n^2$ and $y=n^3$. Then, $xy=n^5=n^4\cdot n=n\cdot n=n^2$. So in the exponent, we may subtract $3$ when it is at least $4$, making it the modulo $3$ remainder. (Note that we should take $n^3$ instead of $n^0$ here, since $n^3$ is already in our group.)
*this is not always true, but since we don't actually use that $n^3=1$ but only $n^4=n$, everything is fine.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rational number trivial case Let $a,b,c$ denote rational numbers, such that $(a+b\sqrt[3]2+c\sqrt[3]4)^3$ is also rational. Prove that at least two of the numbers $a,b,c$ must be zero. Actually I confused of the beginning steps for this proof. Should I apply the strategies of contradictory proofs
| Since
$$(a+bx+cx^2)^3\equiv (6bc^2+3a^2c+3ab^2)x^2+(6ac^2+6b^2c+3a^2b)x+(4c^3+2b^3+a^3+12abc)\pmod{(x^3-2)}$$
if $(a+b\sqrt[3]{2}+c\sqrt[3]{2})^3$ belongs to $\mathbb{Q}$ then
$$(2bc^2+a^2c+ab^2)=0, \qquad (2ac^2+2b^2c+a^2b)=0,\tag{1}$$
hence:
$$ a^2 b^2 c^2 = (a^2c+ab^2)(ac^2+b^2 c) = a^3 c^3 + 2a^2 b^2 c^2 + ab^4 c,$$
$$ a^3 c^3 + a^2 b^2 c^2 + ab^4 c = 0,$$
$$ ac (a^2 c^2 + a c b^2 + b^4) = 0.$$
Since the discriminant of $y^2+y+1$ is negative, the last equation implies $a=0$ or $c=0$. If we plug this identities back into $(1)$, we get that the only possibilities are that at least two variables among $\{a,b,c\}$ are zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate: $\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} - x - 2$ Find the following limit $$\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} - x - 2$$
How do I find this limit? If I had to guess I'd say it converges to $-2$ but the usual things like L'Hôpital or clever factorisation don't seem to work in this case.
| Take aside the $-2$, for the moment. You can use the identity
$$
a^3-b^3=(a-b)(a^2+ab+b^2),
$$
with $a=\sqrt[3]{x^3-1}$ and $b=x$. Then
\begin{align}
\lim_{x\to\infty}(\sqrt[3]{x^3-1}-x)&=
\lim_{x\to\infty}
\frac{
(\sqrt[3]{x^3-1}-x)
(\sqrt[3]{(x^3-1)^2}+x\sqrt[3]{x^3-1}+x^2)
}{\sqrt[3]{(x^3-1)^2}+x\sqrt[3]{x^3-1}+x^2}\\
&=
\lim_{x\to\infty}\frac{x^3-1-x^3}{\sqrt[3]{(x^3-1)^2}+x\sqrt[3]{x^3-1}+x^2}
\end{align}
and you should be able to continue from here. Then put in $-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum area of a rectangle inscribed in a semi - circle with radius r.
A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.
My Try:
Let length of the side be $x$,
Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.
Then the area function is
$$A(x) = 2x\sqrt{r^2-x^2}$$
$$\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\
&=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$
setting $A'(x) = 0$,
$$\implies x^2 +2x -r^2 = 0$$
Solving, I obtained:
$$x = -1 \pm \sqrt{1+r^2}$$
That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.
Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.
Thanks!
|
Eqn of circle: x^2 + y^2 = r^2
|
area of rectangle = a = 2xy
==> a^2 = 4 x^2 y^2
==> a^2 = 4 x^2 (r^2 - x^2)
==> a^2 = 4 (x^2 r^2 - x^4)
|
differentiating:
==> 2a da/dt = 4 (2x r^2 - 4x^3)
since, da/dt=0
==> 2 x r^2 = 4 x^3
==> x = r/sqrt(2)
|
second derivative test:
==>4 (2 r^2 - 12 x^2)
==>8 r^2 - 24 r^2
==>(-16) r^2
==> -ve
therefore, maximum area for x= r/sqrt(2)
|
finding y:
==>y^2 = r^2 - x^2
==>y^2 = r^2 - (r^2)/2
==>y=r/sqrt(2)
| {
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Find the remainder of the polynomial division $p(x)/(x^2-1)$ for some $p$ Let $f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5. $
Without using long division (which would be horribly nasty!), find the remainder when $f(x)$ is divided by $x^2-1$.
I'm not sure how to do this, as the only way I know of dividing polynomials other than long division is synthetic division, which only works with linear divisors. I thought about doing $f(x)=g(x)(x+1)(x-1)+r(x)$, but I'm not sure how to continue. Thanks for the help in advance.
| Plug in $1$ and $-1$ to get two values of $r(x)$ which is linear. From there you can get what $a,b$ are in $ax+b.$
Since
$$f(x)=g(x)(x+1)(x-1)+r(x)$$
we have
$$ f(1)=g(1)(1+1)(1-1)+r(1)=r(1)=-10$$
$$ f(-1)=g(1)(-1+1)(-1-1)+r(-1)=r(-1)=16$$
We know the remainder is of degree $1$, so
$r(x)=ax+b$
and now we know,
$$r(1)=ax+b=a+b=-10$$
$$r(-1)=ax+b=-a+b=16$$
so, solve
$$a+b=-10$$
$$-a+b=16$$
which yields, $a=-13$ $b=3$, so
$$r(x)=-13x+3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Cyclic Group Presentation Show that the the group with presentation $$\langle x, y\ \mid\ x^2=y^2x^2y,\ (xy^2)^2=yx^2, \ yx^{-1}y^2=x^7\rangle $$ is cyclic of order 24.
This presentation was obtained using the Todd-Coxeter process for a subgroup of index 2 in the group presented in problem 476854.
| $$x^2=y^2x^2y$$
$$x=x^{-1}y^2x^2y=y^{-1}(yx^{-1}y^2)x^2y$$
$$x=y^{-1}x^9y$$
then we can say that;
$$y^2x^2y=x^2=y^{-1}x^{18}y$$
$$y^2x^2=y^{-1}x^{18}$$
$$y^3=x^{16}$$
Now, $$(xy^2)^2=yx^2$$
$$xy^2xy^2=yx^2$$
$$xy^2x(y^3)=yx^2y=y^3x^2y^2=x^{18}y^2$$
$$y^2x^{17}=x^{17}y^2$$
$$y^5x=xy^{5}$$
As a last step;
$$x^2=y^2x^2y$$
$$x^{18}=x^2y^3=y^3x^2=(y^5x^2)y=x^2y^6$$
$$x^2y^3=x^2y^6$$
$$e=y^3$$
Thus, we can say that $x,y^2$ will commute with each other so will $x,y^4=y$. And $x=y^{-1}x^9y=x^9\implies x^8=e$. From that point you can easily conclude that $G$ is an cyclic group of order $24$.
N.Q.E.D
| {
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A transcendental number from the diophantine equation $x+2y+3z=n$ Let $\displaystyle n=1,2,3,\cdots.$ We denote by $D_n$ the number of non-negative integer solutions of the diophantine equation
$$x+2y+3z=n$$
Prove that
$$
\sum_{n=0}^{\infty} \frac{1}{D_{2n+1}}
$$
is a transcendental number.
| We have that
$$
\displaystyle D_{n} = \left \lfloor \frac {(n+3)^2}{12} \right \rceil
$$
where $\lfloor x \rceil$ is the nearest integer to $x$. You can find a detailed proof here (first pdf link, p.50).
It gives
$$
D_{2n+1} = \left \lfloor \frac {(n+2)^2}{3} \right \rceil.
$$
As Jack D'Aurizio explained, we deduce that
$$
\sum_{n=0}^{\infty} \frac{1}{D_{2n+1}}=\frac{3}{4} + \frac{\pi \sqrt{3}}{6} + \frac{\pi^2}{18}
$$
Since $\sqrt{3}$ is algebraic, this sum is transcendental as $\pi$ is.
Likewise, we obtain
$$
\sum_{n=0}^{\infty} \frac{1}{D_{n}}= \frac{135}{36} - \frac{\pi \sqrt{3}}{6} + \frac{\pi^2}{18} +\frac{\pi \sqrt{3}}{3}\tan \left( \frac{\pi \sqrt{3}}{6} \right),
$$
$$
\sum_{n=0}^{\infty} \frac{(-1)^n}{D_{n}}=\frac{9}{4} - \frac{\pi \sqrt{3}}{2} + \frac{\pi^2}{18} +\frac{\pi \sqrt{3}}{3}\tan \left( \frac{\pi \sqrt{3}}{6} \right).
$$
For these two last sums, it seems rather difficult to prove they are transcendental, as we may suspect they are.
| {
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Trigonometric Identities help How do you solve this? I can't figure out what I should do.
$$\sin ^4\left(A\right)+\cos ^2\left(A\right)=\cos ^4\left(A\right)+\sin ^2\left(A\right)$$
Also, why is this equal zero? Can someone explain how that simplifies to be zero?
$$\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}$$
| Note that $\sin^4 A-\cos^4 A=(\sin^2 A-\cos^2 A)(\sin^2 A+\cos^2 A)=\sin^2 A-\cos^2 A$.
For the other question, we have $\frac{1}{\cos x}-1=\frac{1-\cos x}{\cos x}$ and $\frac{1}{\cos x}+1=\frac{1+\cos x}{\cos x}$. When we divide, the $\cos x$ cancel, and we get $\frac{1-\cos x}{1+\cos x}$.
| {
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Find the maximun value of the expression $P=\sum \sqrt[3]{\frac{a^{2}+a}{a^{2}+a+1}}$ Let $a,b,c$ be positive real numbers such that $abc\leq 1$ .Find the maximun value of the expression
$P=\sqrt[3]{\frac{a^{2}+a}{a^{2}+a+1}}+\sqrt[3]{\frac{b^{2}+b}{b^{2}+b+1}}+\sqrt[3]{\frac{c^{2}+c}{c^{2}+c+1}}$
| Holder's inequality or Power Means will give
$$ \sum_{cyc} \frac{a^2+a}{a^2+a+1} \ge \frac19\left(\sum_{cyc} \sqrt[3]{\frac{a^2+a}{a^2+a+1}} \right)^3$$
with equality iff $a=b=c$. We claim that the LHS reaches a maximum of $2$ exactly when $a=b=c=1$, and show the same below.
This is equivalent to showing
$$ \sum_{cyc} \left(1- \frac1{a^2+a+1} \right) \le 2 \iff \sum_{cyc} \frac1{a^2+a+1} \ge 1 $$
As $abc=1$ for the minimum, there exists positive reals $x, y, z$ s.t. $a = \frac{yz}{x^2}, b = \frac{zx}{y^2}, c = \frac{xy}{z^2}$. So we can homogenise and rewrite as
$$\sum_{cyc} \frac{x^4}{x^4+x^2yz+y^2z^2} \ge 1$$
Using Cauchy Schwarz on the LHS, it is sufficient to show
$$(x^2+y^2+z^2)^2 \ge \sum_{cyc} x^4 + xyz (x+y+z)+\sum_{cyc} x^2y^2$$
$$\iff \sum_{cyc} x^2y^2 \ge xyz \sum_{cyc} x \iff \sum_{cyc} x^2(y-z)^2 \ge 0$$
Equality holds iff $x=y=z$, i.e. when $a=b=c=1$, so that is when we have the maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving inequality $3^{n^2} > (n!)^4$ Prove that $3^{n^2} > (n!)^4$ for all positive integers $n$.
I tried to use induction on this problem but failed to do so. I instead tried to prove $3^{2n+1}>(n+1)^4$, but couldn't come up with the solution.
| The induction-step should be pretty easy:
*
*Assume $\displaystyle3^{n^2}>(n!)^4$
*Prove $\displaystyle3^{(n+1)^2}>(n+1)!^4$:
*
*$\displaystyle3^{(n+1)^2}=3^{n^2+2n+1}$
*$\displaystyle3^{n^2+2n+1}=3^{n^2}3^{2n+1}$
*$\displaystyle3^{n^2}3^{2n+1}>(n!)^43^{2n+1}$
*$\displaystyle(n!)^43^{2n+1}=\frac{(n+1)!^4}{(n+1)^4}3^{2n+1}$
*$\displaystyle\frac{(n+1)!^4}{(n+1)^4}3^{2n+1}=(n+1)!^4\frac{3^{2n+1}}{(n+1)^4}$
So all you have left to prove by induction is $3^{2n+1}>(n+1)^4$:
*
*Assume $\displaystyle3^{2n+1}>(n+1)^4$
*Prove $\displaystyle3^{2(n+1)+1}>(n+2)^4$:
*
*$\displaystyle3^{2(n+1)+1}=3^{2n+3}$
*$\displaystyle3^{2n+3}=3^{2n+1}3^2$
*$\displaystyle3^{2n+1}3^2>(n+1)^43^2$
*$\displaystyle(n+1)^43^2=9(n^4+4n^3+6n^2+4n+1)$
*$\displaystyle9(n^4+4n^3+6n^2+4n+1)=9n^4+36n^3+54n^2+36n+9$
*$\displaystyle9n^4+36n^3+54n^2+36n+9>n^4+8n^3+24n^2+32n+16$
*$\displaystyle n^4+8n^3+24n^2+32n+16=(n+2)^4$
| {
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"answer_count": 2,
"answer_id": 0
} |
if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good partial fractions methods by hand?
because I take an hour to solve this problem.
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you
| Note that $\left(1-x^2\right)\left(1-x^3\right)\left(1-x^4\right)$ divides $\left(1-x^{12}\right)^3$ and $\displaystyle\left(1-z\right)^{-3}=\sum_{n=0}^{\infty}{n+2 \choose 2}z^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
solving integral with complex analysis I have problems with understanding of the evaluation of this integral below. It has been a long a time ago since I had complex analysis.
where $a = (1-\sqrt y )^2$ and $b = (1+\sqrt y )^2$.
Now my question are: How do I substitute in the last 2 steps, how can I calculate the residuals and how do I get from there to answer $1/(1-y)$.
I hope you guys can help me. Thanks a lot!
Van
| Note that
$$
\sqrt{(b-x)(x-a)}=\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{b+a}{2}\right)^2}\tag{1}
$$
Therefore, we can substitute
$$
x=\frac{b+a}{2}-\frac{b-a}{2}\cos(\theta)\quad\text{so that}\quad\sqrt{(b-x)(x-a)}=\frac{b-a}{2}\sin(\theta)\tag{2}
$$
Furthermore, by the definition of $y$,
$$
\frac{b+a}2=1+y\qquad\text{and}\qquad\frac{b-a}2=2\sqrt{y}\tag{3}
$$
Thus, applying $(2)$ and $(3)$,
$$
\begin{align}
&\int_a^b\frac1x\frac1{2\pi xy}\sqrt{(b-x)(x-a)}\,\mathrm{d}x\tag{4}\\[9pt]
&=\frac1{2\pi y}\int_0^\pi\frac{4y\sin^2(\theta)}{\left(1+y-2\sqrt{y}\cos(\theta)\right)^2}\,\mathrm{d}\theta\tag{5}\\
&=\frac1{4\pi y}\int_{-\pi}^\pi\frac{4y\sin^2(\theta)}{\left(1+y-2\sqrt{y}\cos(\theta)\right)^2}\,\mathrm{d}\theta\tag{6}\\
&=\frac1{4\pi}\int_{-\pi}^\pi\frac{-\left(e^{i\theta}-e^{-i\theta}\right)^2}{\left(1+y-\sqrt{y}\left(e^{i\theta}+e^{-i\theta}\right)\right)^2}\frac{\mathrm{d}e^{i\theta}}{ie^{i\theta}}\tag{7}\\
&=-\frac1{4\pi i}\oint_{|z|=1}\frac{\left(z-\frac1z\right)^2}{\left(1+y-\sqrt{y}\left(z+\frac1z\right)\right)^2}\frac{\mathrm{d}z}{z}\tag{8}\\
&=-\frac1{4\pi i}\oint_{|z|=1}\frac{\left(z^2-1\right)^2}{\left((1+y)z-\sqrt{y}\left(z^2+1\right)\right)^2}\frac{\mathrm{d}z}{z}\tag{9}\\
&=-\frac1{4\pi i}2\pi i\left(\frac1y-\frac{1+y}{y(1-y)}\right)\tag{10}\\[9pt]
&=\frac1{1-y}\tag{11}
\end{align}
$$
Explanation:
$\:\ (5)$: apply $(2)$ and $(3)$
$\:\ (6)$: the integrand is even; double the domain of integration and divide by $2$
$\:\ (7)$: write $\sin(\theta)$ and $\cos(\theta)$ in terms of $e^{i\theta}$
$\:\ (8)$: substitute $z=e^{i\theta}$
$\:\ (9)$: multiply numerator and denominator by $z^2$
$(10)$: residue at $z=0$ is $\frac1y$ and the residue at $z=\sqrt{y}\ $ is $-\frac{1+y}{y(1-y)}$
Note that $(8)$ is the final integral in $(\mathrm{A}.13)$.
Computation of the Residues
The singularities are at $z=0$ and $z=\sqrt{y}$ and $z=\frac1{\sqrt{y}}$.
Since the singularity at $z=0$ is simple, multiply by $z$ and evaluate at $z=0$ to get a residue of $\frac1y$.
For $0\lt y\lt1$, the singularity at $z=\frac1{\sqrt{y}}$ is outside $|z|=1$.
For the singularity at $\sqrt{y}$, substitute $z\mapsto z+\sqrt{y}$ and look near $0$:
$$
\begin{align}
&\frac{\left(z^2+2z\sqrt{y}+y-1\right)^2}{\left((1+y)(z+\sqrt{y})-\sqrt{y}\left(z^2+2z\sqrt{y}+y+1\right)\right)^2}\frac1{z+\sqrt{y}}\\
&=\frac{(1-y)^2-4\sqrt{y}(1-y)z+O(z^2)}{\left((1-y)z-\sqrt{y}z^2\right)^2}\left(\frac1{\sqrt{y}}-\frac{z}{y}+O(z^2)\right)\\
&=\frac{1-\frac{4\sqrt{y}}{1-y}z+O(z^2)}{z^2\left(1-\frac{2\sqrt{y}}{1-y}z+O(z^2)\right)}\frac1{\sqrt{y}}\left(1-\frac{z}{\sqrt{y}}+O(z^2)\right)\\
&=\frac1{z^2\sqrt{y}}\left(1-\frac{z}{\sqrt{y}}-\frac{2\sqrt{y}}{1-y}z+O(z^2)\right)\\
&=\frac1{z^2}\left(\frac1{\sqrt{y}}-\frac{1+y}{y(1-y)}z+O(z^2)\right)\tag{12}
\end{align}
$$
The residue at $\sqrt{y}$ is the coefficient of $\frac1z$, that is $-\frac{1+y}{y(1-y)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Intuitive ways to get formula of cubic sum Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$
I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$$
But how to prove it? Please help me. Grazie!
| Here's a proof by induction
I came up with
about two years ago.
Let
$s_k(n)
=\sum_{i=1}^n i^k
$.
Want to show that
$s_3(n) = (s_1(n))^2
$.
If you want to
prove by induction
that $a(n) = b(n)$,
there are two possibilities:
(1) Show
$a(0)-b(0) = 0$
and
$a(n)-b(n) = 0
\implies
a(n+1)-b(n+1) = 0$;
(2) show
$a(0) = b(0)$
and
$a(n) = b(n)
\implies
a(n+1)-a(n)
=b(n+1)-b(n)
$.
These are,
of course,
equivalent,
but I find that (2)
is often easier
since the math
is simpler
in the initial parts.
In this case,
$s_3(n+1)-s_3(n)
=(n+1)^3
=n^3+3n^2+3n+1
$
and
$(s_1(n+1))^2-(s_1(n))^2
=(s_1(n)+n+1)^2-(s_1(n))^2\\
=s_1^2(n)+2(n+1)s_1(n)+(n+1))^2-(s_1(n))^2\\
=2(n+1)s_1(n)+n^2+2n+1\\
$
so we want to show that
$n^3+3n^2+3n+1
=2(n+1)s_1(n)+n^2+2n+1
$
or
$n^3+2n^2+n
=2(n+1)s_1(n)
$
or
$n(n^2+2n+1)
=2(n+1)s_1(n)
$
or
$n(n+1)^2
=2(n+1)s_1(n)
$
or
$s_1(n)
=\dfrac{n(n+1)}{2}
$.
This can either
be assumed known
or, in turn,
proved by induction using
$\dfrac{(n+1)(n+2)}{2}
-\dfrac{n(n+1)}{2}
=\dfrac{n^2+3n+2-(n^2+n)}{2}
=\dfrac{2n+2}{2}
=n+1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
} |
taylor series of $\ln(1+x)$? Compute the taylor series of $\ln(1+x)$
I've first computed derivatives (up to the 4th) of ln(1+x)
$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$
But this doesn't seem to be correct. Can anyone please explain why this doesn't work?
The supposed correct answers are: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$
$$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$
| You got the general expansion about $x=a$. Here we are intended to take $a=0$. That is, we are finding the Maclaurin series of $\ln(1+x)$.
That will simplify your expression considerably. Note also that $\frac{(n-1)!}{n!}=\frac{1}{n}$.
The approach in the suggested solution also works. We note that
$$\frac{1}{1+t}=1-t+t^2-t^3+\cdots\tag{1}$$
if $|t|\lt 1$ (infinite geometric series). Then we note that
$$\ln(1+x)=\int_0^x \frac{1}{1+t}\,dt.$$
Then we integrate the right-hand side of (1) term by term. We get
$$\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$
precisely the same thing as what one gets by putting $a=0$ in your expression.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 4,
"answer_id": 0
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A closed form for $\int_{0}^{\pi/2}\frac{\ln\cos x}{x}\mathrm{d}x$? The following integrals are classic, initiated by L. Euler.
\begin{align}
\displaystyle \int_{0}^{\pi/2} x^3 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^4}{64} \ln 2-\frac{3\pi^2}{16} \zeta(3)+\frac{93}{128} \zeta(5),
\\ \int_{0}^{\pi/2} x^2 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^3}{24} \ln 2-\frac{\pi}{4} \zeta(3),
\\ \int_{0}^{\pi/2} x^1 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3),
\\ \int_{0}^{\pi/2} x^0 \ln\cos x\:\mathrm{d}x & = -\frac{\pi}{2}\ln 2.
\end{align}
We may logically consider the case when the first factor of the integrand is $\displaystyle x^{-1} = \frac 1x $ leading to the following non classic convergent integral.
$$ \int_{0}^{\pi/2} \frac{\ln\cos x}{x}\:\mathrm{d}x \qquad (*)$$
I do not have a closed form for this integral.
My question is does someone have some references/results about $(*)$?
| Other series expansions can be found by noticing that:
$$\int\limits_{0}^{\frac{\pi}{2}} \frac{\log{\left(\cos{\left(x \right)} \right)}}{x}\, dx = -\frac{\pi}{2}\int\limits_{0}^{1}\int\limits_{0}^{1} \tan{\left(\frac{\pi x y}{2} \right)}\, dx\, dy \tag{1}$$
which enables us to use the series expansion of the tangent for $-\pi/2<x<\pi/2$:
$$\tan{\left(x \right)} = \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n - 1} \cdot 2^{2 n} x^{2 n - 1} \left(2^{2 n} - 1\right) B_{2 n}}{\left(2 n\right)!} \tag{2}$$
to obtain:
$$\begin{aligned}
\int\limits_{0}^{\frac{\pi}{2}} \frac{\log{\left(\cos{\left(x \right)} \right)}}{x}\, dx &= \sum_{n=1}^{\infty} \frac{\left(- \pi^{2}\right)^{n} \left(4^{n} - 1\right) B_{2 n} \int\limits_{0}^{1}\int\limits_{0}^{1} \left(x y\right)^{2 n - 1}\, dx\, dy}{\left(2 n\right)!}\\
&=\sum_{n=1}^{\infty} \frac{\left(- \pi^{2}\right)^{n} \left(4^{n} - 1\right) B_{2 n}}{4 n^{2} \left(2 n\right)!}\\
&=\sum_{n=1}^{\infty} \frac{\left(- 1 + \frac{1}{2^{2 n}}\right) \zeta\left(2 n\right)}{2 n^{2}}\\
&=\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\left(- 1 + \frac{1}{2^{2 n}}\right)}{2 n^{2}k^{2 n}}\\
&=\frac{1}{2}\sum_{k=1}^{\infty} \left(\operatorname{Li}_{2}\left(\frac{1}{4 k^{2}}\right) - \operatorname{Li}_{2}\left(\frac{1}{k^{2}}\right)\right)\\
&=-\frac{1}{2}\sum_{k=1}^{\infty} \operatorname{Li}_{2}\left(\frac{1}{\left(2 k - 1\right)^{2}}\right)
\end{aligned} \tag{3}$$
where $B_{n}$ is a Bernoulli number, $\zeta$ is the Riemann zeta function, and $\operatorname{Li}_{2}$ is a polylogarithm of order 2 (aka dilogarithm or Spence's function). Unlike this answer from the op, on this occasion we do not obtain poly-Stieltjes constants but rather a higher order generalisation of sorts to the dilogarithm. The final expression in terms of the dilogarithm can be calculated to arbitrary precision in Python's Sympy package and agrees with the numerical evaluation of the integral $−0.941237674287746$ at $1e20$ terms.
More generally, following the above method it can be shown that:
$$\begin{aligned}
\int\limits_{0}^{1} x^{- s} \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}\, dx &= - \sum_{n=1}^{\infty} \frac{\left(1 - 4^{- n}\right) \zeta\left(2 n\right)}{n \left(2 n - s + 1\right)}\\
&=- \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \frac{1}{n \left(2 n - s + 1\right)\left(2 k - 1\right)^{2 n}}
\end{aligned} \tag{4}$$
and in particular, by using:
$$\sum_{n=1}^{\infty} \frac{z^{2 n}}{n \left(2 n - 1\right)} = \left(1 - z\right) \log{\left(1 - z \right)} + \left(z + 1\right) \log{\left(z + 1 \right)} \tag{5}$$
the special case with $s=2$ can be expressed in closed form in terms of the poly-Stieltjes constants:
$$\int\limits_{0}^{1} x^{- 2} \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}\, dx = - \sum_{n=1}^{\infty} \frac{1}{n \left(2 n - 1\right)} \\- \sum_{k=1}^{\infty} \left(\left(1 - \frac{1}{2 k + 1}\right) \log{\left(1 - \frac{1}{2 k + 1} \right)} + \left(1 + \frac{1}{2 k + 1}\right) \log{\left(1 + \frac{1}{2 k + 1} \right)}\right) \tag{6}$$
$$\int\limits_{0}^{1} x^{- 2} \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}\, dx = - \log{\left(\pi \right)} +\frac{\gamma_1(0,\frac{1}{2})}{2} - \frac{\gamma_1(0,-\frac{1}{2})}{2} \tag{7}$$
where the limit as $z$ tends to 1 in $\left(5\right)$ was used to derive:
$$\sum_{n=1}^{\infty} \frac{1}{n \left(2 n - 1\right)} = 2 \log{\left(2 \right)} \tag{8}$$
the product formula for the cosine was used to derive:
$$\log{\left(\frac{\pi}{4}\right)}=\sum_{k=1}^{\infty} \log{\left(1 - \frac{1}{\left(2 k + 1\right)^{2}} \right)} \tag{9}$$
and the poly-Stieltjes constants are defined as:
$$\gamma_k(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) \tag{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
} |
Summation of Infinite Geometric Series Determine the sum of the following series:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} $$
My work:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} = \sum_{n=1}^{\infty } \frac{-1}{7} (\frac{3}{7})^{n-1}$$
$$\sum_{n=1}^{\infty } ar^{n-1} = \frac{a}{1-r} = \frac{\frac{-1}{7}}{1-\frac{3}{7}} = -\frac{1}{4}$$
Why does this not work?
Sorry for the incorrect initial post!!!
Edit: -3 changed to (-3)
| $$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{3^n}{7^n}=-\frac{1}{3} \sum_{n=1}^{\infty} \left ( \frac{3}{7} \right )^n=-\frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{3}{7} \right )^n+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\frac{3}{7}}+\frac{1}{3} \\ =-\frac{1}{3}\frac{7}{7-3}+\frac{1}{3}=-\frac{7}{12}+\frac{4}{12}=\frac{-1}{4}$$
EDIT:
$$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{(-3)^n}{7^n}=-\frac{1}{3} \sum_{n=0}^{\infty} \frac{(-3)^n}{7^n}+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\left (\frac{-3}{7} \right )}-\frac{1}{3}=-\frac{1}{3} \frac{1}{1+\frac{3}{7}}+\frac{1}{3}=-\frac{1}{3} \frac{7}{10}+\frac{1}{3}=-\frac{7}{30}+\frac{10}{30}=\frac{1}{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits