Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$? Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$
Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?
I start with $AM-GM$
So $A \geq 8xyz$
Then I need to prove $xyz\geq 1$ right?
How to do that and if it's a wrong way, please help me to solve it !!!
| With Lagrange multipliers: consider $f(x,y,z)=(x^2+1)(y^2+1)(z^2+1)$ and
$$
L(x,y,z,t)=f(x,y,z)-t(x+y+z-3)
$$
Then you have
\begin{align}
\frac{\partial L}{\partial x}&=2x(y^2+1)(z^2+1)-t \\[1ex]
\frac{\partial L}{\partial x}&=2y(x^2+1)(z^2+1)-t \\[1ex]
\frac{\partial L}{\partial x}&=2z(x^2+1)(y^2+1)-t
\end{align}
These should vanish together, so we obtain, for instance,
$$
2x(y^2+1)(z^2+1)=2y(x^2+1)(z^2+1)
$$
that yields $xy^2+x=x^2y+y$, hence $xy(x-y)=x-y$. Thus either $xy=1$ or $x=y$. Similarly, $xz=1$ or $x=z$. Thus we have four cases to examine.
The case $x=y$ and $y=z$ yields $x=y=z=1$.
The case $xy=1$ and $xz=1$ yields $y=z$ and so $x=2,y=1/2,z=1/2$.
Since the function $f$ as well as the constraint are symmetric in $x,y,z$, the remaining two cases will provide $x=1/2,y=2,z=1/2$ and $x=1/2,y=1/2,z=2$.
We have $f(1,1,1)=8$ and $f(2,1/2,1/2)=125/16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$
I tried the most rudimentary method ie. differentiating wrt x and hoped to get at least 2 solutions for $f’(x)$, unfortunately I wasn’t able to find a solution for implying function is monotonous . Now whether it’s sis increasing or decreasing still remains a mystery, given that we don’t know what $a$ is.
What method can be used to solve this?
| This is actually a very devious inequality problem disguised as a calculus problem, and it took me very long to realise this. But eventually what gave it away was that calculating the second derivative of $f(x)$ would be a monstrous task, and that the $\sqrt{\dfrac{ax}{ax+8}}$ term seemed very suspicious (As we shall see, dividing both numerator and denominator by $ax$ would yield something very interesting and useful.)
First let us rename $x$ as $b, b \geq 0$. The case where $b=0$ is trivial, since we have $\dfrac{1}{\sqrt{1+a}} + \dfrac{1}{\sqrt{1+b}} + \sqrt{\dfrac{ab}{ab+8}} = \dfrac{1}{\sqrt{1+a}} +1 $. Thus, we may assume $a,b >0$. Let $a=2x,b=2y$, where $x, y >0$.
Thus, $$\dfrac{1}{\sqrt{1+a}} + \dfrac{1}{\sqrt{1+b}} + \sqrt{\dfrac{ab}{ab+8}} =\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+\dfrac{2}{xy}}}.$$
Let $2z=\dfrac{2}{xy} \Rightarrow xyz=1$. So really, this problem is equivalent to us proving the following inequality:
Inequality:
Let $x,y,z \in \mathbb{R^+}$, $xyz=1$. Prove that $1<\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} <2. $
Now, the lower bound is relatively easy, but the upper bound is a real killer.
Lower Bound:
Since $1+2x >1 , \sqrt{1+2x} < 1+2x \Rightarrow \dfrac{1}{\sqrt{1+2x}} > \dfrac{1}{1+2x}$. Thus it suffices for us to prove that:
\begin{align}
\sum_{\text{cyc}} \dfrac{1}{1+2x} \geq 1 & \iff \dfrac{(1+2y)(1+2z) + (1+2x)(1+2z) + (1+2x)(1+2y)}{(1+2x)(1+2y)(1+2z)} \geq 1 \\
& \iff 3+4(x+y+z) + 4(yz+xz+xy) \geq 9+2(x+y+z) + 4(yz+xz+xy) \\
& \iff 2(x+y+z) \geq 6 \\
& \iff x+y+z \geq 3 \\
\end{align}
Which follows immediately from AM-GM and the given condition.
Upper Bound:
The following proof borrowed an important idea introduced in Yufei Zhao's inequalities notes: Inequalities. Sometimes switching the roles of the constraint and the inequality can yield wonders!
We begin by letting $p=\dfrac{1}{\sqrt{1+2x}}, q=\dfrac{1}{\sqrt{1+2y}}, r=\dfrac{1}{\sqrt{1+2z}} $, and $p,q,r<1$. We have to prove that $xyz=1 \Rightarrow p+q+r <2$. Instead of proving this directly, we will prove the contrapositive: $p+q+r \geq 2 \Rightarrow xyz <1 \Rightarrow xyz \neq 1$. Next we carry out the following algebraic manipulations:
\begin{align}
p=\dfrac{1}{\sqrt{1+2x}} & \Rightarrow p^2=\dfrac{1}{1+2x} \\
& \Rightarrow \dfrac{1}{p^2} = 1+2x \\
& \Rightarrow x=\dfrac{1-p^2}{2p^2} = \dfrac{(1+p)(1-p)}{2p^2} < \dfrac{1-p}{p^2}\\
\end{align}
But
\begin{align}
p+q+r \geq 2 & \Rightarrow p \geq 2-q-r \\
& \Rightarrow 1-p \leq -1+q+r = qr-(1-q)(1-r) < qr \\
& \Rightarrow x < \dfrac{qr}{p^2}.
\end{align}
Similarly, we obtain that $y < \dfrac{pr}{q^2}$ and $z < \dfrac{pq}{r^2}$. Thus, $xyz <\dfrac{qr}{p^2} \cdot \dfrac{pr}{q^2} \cdot \dfrac{pq}{r^2} =1$, and we are done.
Final note: In fact, I believe that we have the following stronger upper bound:
$$\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq \sqrt{3}.$$
However, I have absolutely no clue on how to proceed. Perhaps someone else can come up with a nice idea?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute the integral $\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz$ Could you please help me to solve this integral. I have no idea on how to proceed. Thank you.
I think, i should rewrite $\frac{1}{z(z+2)^3}$ as a sum from a geometric series and then use the Cauchy integral formula, but I don´t see a trick..
$\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz=\oint\frac{\frac{1}{z}}{(z+2)^3}dz$
Then
$\oint\frac{\frac{1}{z}}{(z+2)^3}dz=\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2f(z)}{dz^2}\right]_{z=-2}$
$=\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2}{dz^2}(\frac{1}{z})\right]_{z=-2}$
$=\left({\pi i}\right)\left[{\frac{2}{(z)^3}}\right]_{z=-2}$
$=\color{blue}{-\frac{1\pi i}{4}}$
Is that correct?
*Solution (geometric progression)
$\int_{\delta_{B_1(-2)}} \frac{1}{z(z+2)^3}dz$
$\frac{1}{z(z+2)^3} = 2*(1-(1+\frac{z}{2}))^{-1}*\frac{1}{z(z+2)^3}$
$=\frac{2}{z(z+2)^3}*\frac{1}{1-(1+\frac{z}{2})}$
We know, that
$\frac{1}{1-(1+\frac{z}{2})}=\sum_{n=0}^∞ {(1+\frac{z}{2})^n}=1+(1+\frac{z}{2})^1+(1+\frac{z}{2})^2+(1+\frac{z}{2})^3+ ...)$
So we get
$\frac{2}{z(z+2)^3}*\frac{1}{1-(1+\frac{z}{2})}=\frac{2}{z(z+2)^3}*(1+(1+\frac{z}{2})^1+(1+\frac{z}{2})^2+(1+\frac{z}{2})^3+ ...)=$
$=\frac{4}{(z+2)^3}+\frac{z}{(z+2)^3}+\frac{1/2}{(z+2)}+\frac{1}{4}+\frac{2+z}{8}+...$
=>
$f(z)=1/2, z=-2
=>
2\pi if(z)=2\pi i*1/2=\pi i$
In this solution I have $\color{blue}{\pi i}$
And in the top solution I have another answer $\color{blue}{-\frac{\pi i}{4}}$.
What's my mistake?
| The only enclosed pole is a third-order one at $z=-2$, so the residue is $\tfrac12\lim_{z\to-2}\tfrac{d^2}{dz^2}[(z-2)^3f(z)]$. I leave you to evaluate this, then multiply by $2i\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are the following function 1st derivatives continuous at (0,0)? Apologies: I've had to rephrase this question quite a few times
Suppose the following function:
$$f(x,y)=\frac{2x^3y^2}{x^4+y^2}$$
$$f(0,0)=0$$
Show it is differentiable/not differentiable at $(0,0)$
I attempted this question first by applying the definition of differentiability which yielded the following expression:
$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}$$
If the above equals 0, then the function is infact differentiable at $(0,0)$. I can't seem to find a counterexample which proves that it is discontinuous so I attempted to prove its continuity by applying the $\epsilon-\delta$ definition of limits. Unfortunately, no matter whether I express the function in terms of polarcoordinates or not, I cannot bound the following
$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| < \epsilon$$
for all $\epsilon$.
I'd be greatly appreciative for a solution to this kind of problem!
| Credits to Kavi Rama Murthy for pointing out the $|2x^2y|<x^4+y^2$ inequality for me.
Note that:
*
*$(x^2-y)^2 \geq 0 \implies x^4 - 2x^2y + y^2 \geq 0 \implies x^4 + y^2 \geq 2x^2y$
*$(x-y)^2 \geq 0 \implies x^2 - 2xy + y^2 \geq 0 \implies x^2 + y^2 \geq 2xy \implies x^2+y^2 \geq xy \implies \sqrt{x^2+y^2} \geq \sqrt{xy}$
So it turns out the function is differentiable at (0,0). If $0 < \sqrt{x^2+y^2} < \delta$ then:
$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| = |\frac{(2x^2y)(\sqrt{xy})(\sqrt{xy})}{(x^4+y^2)\sqrt{x^2+y^2}}| = |\frac{2x^2y}{x^4+y^2}||\frac{\sqrt{xy}}{\sqrt{x^2+y^2}}||\sqrt{xy}| \leq (1)(1)\sqrt{xy} < \delta = \epsilon
$$
Thus the definition of limits is satisfied therfore $f(x,y)$ is differentiable at $(0,0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$
The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:
\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}
The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.
| If you do the substitution $x+1=t\iff x=t-1$ you only need do a (long) multiplication:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx=
\int \frac{\left((t-1)^2+t+2\right)\left((t-1)^3+7\right)}{t}dt=
$$
$$=\int \frac{\left(t^2-t+3\right)\left(t^3-3t^2+3t+6\right)}{t}dt=\int \frac{t^5-4 t^4+9 t^3-6 t^2+3 t+18}{t}dt=$$
$$=\int\left( t^4-4 t^3+9 t^2-6 t+3 +\frac{18}{t}\right)dt
$$
And now it's simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.
So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?
| In all four values. First, half angle formula
$$ \pm\sqrt{[1+\cos \left( \cot^{-1(...)})]/2\right)}$$
$$\pm \sqrt{\dfrac{1\pm\frac35 }{2}}$$
$$\pm \dfrac{2}{\sqrt 5}, \pm\dfrac{1}{\sqrt 5}.$$
So both answers correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Positive integer solution to $a^3=3b^2-2$ The equation $a^3=3b^2-2$ seem to only have one positive integer solution $(a,b)=(1,1)$, but I am unable to prove that.
What I did:
Google told me that the elliptic curve theory might help me, but I don't really know anything about it.
This thread in MO looks similar to what we have, but I was unable to apply it in my case.
| Let $a$ and $b$ be integers such that $a^3=3b^2-2$. First note that $a$ is not divisible by $3$, and that $a$ and $b$ are both odd. In the UFD $\Bbb{Z}[\sqrt{6}]$ we have the factorization
$$3a^3=9b^2-6=(3b+\sqrt{6})(3b-\sqrt{6}),$$
where the gcd of the two factors divides their difference $2\sqrt{6}$ as well as their product $3a^3$. As noted $a$ is odd, and so the two factors are coprime. The prime factorization $3=(3+\sqrt{6})(3-\sqrt{6})$ in $\Bbb{Z}[\sqrt{6}]$ shows that, after changing the sign of $b$ if necessary, we have
$$3b+\sqrt{6}=u(3+\sqrt{6})(c+d\sqrt{6})^3,\tag{1}$$
for some unit $u\in\Bbb{Z}[\sqrt{6}]$ and some integers $c$ and $d$. The unit group of $\Bbb{Z}[\sqrt{6}]$ is generated by $-1$ and $5+2\sqrt{6}$, meaning that $u=\pm(5+2\sqrt{6})^k$ for some integer $k$. Adjusting $c$ and $d$ if necessary, we may assume without loss of generality that $k\in\{-1,0,1\}$ and that we have the $+$-sign. Then we are left with solving
$$3b+\sqrt{6}=(5+2\sqrt{6})^k(3+\sqrt{6})(c+d\sqrt{6})^3,$$
for some integers $c$ and $d$, and some $k\in\{-1,0,1\}$. Expanding the product for each of the three values of $k$ and comparing coefficients of $\sqrt{6}$, we end up with the following three Diophantine equations:
\begin{eqnarray*}
1&=&-c^3+9c^2d-18cd^2+18d^3\\
1&=&c^3+9c^2d+18cd^2+18d^3\\
1&=&11c^3+81c^2d+198cd^2+162d^3\\
\end{eqnarray*}
These are three cubic Thue equations, for which there exist effective methods for finding all integral solutions. The first has the unique solution $(c,d)=(-1,0)$, the second has the unique solution $(c,d)=(-1,0)$, and the third has no integral solutions. These two solutions correspond to $b=-1$ and $b=1$, respectively, and this shows that these are the only integral solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Rewriting a Binomial expansion and expanding the rewritten version after making a convenience substitution to get a totally different answer I know that if I have a function $$f(x) = \frac{1}{x-1}$$ that if I perform a binomial expansion on I get the result $$f(x)\approx -1 -x -x^2-x^3-x^4 ...$$
However if I rewrite the function $$f(x) = \frac{1}{x-1} = \frac{1}{x(1-\frac{1}{x})} = \bigg[\big(1-\frac{1}{x}\big)\big(x\big) \bigg]^{-1}$$
Then we make the expansion of $\bigg[\big(1-\frac{1}{x}\big)\big(x\big) \bigg]^{-1}$ by using the subsitution $\frac{1}{x} = z$
We get $$(1-z)^{-1} = 1+z+z^2+z^3+z^4... = 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}+ \frac{1}{x^4}...$$
so that
$$x^{-1}(1-z)^{-1} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4}+ \frac{1}{x^5}...$$
This is completely different result!
Why is this the case?
| The expansion of $\frac 1 {x-1}$ you are using is valid only for $|x|<1$. When you change $x$ to $\frac 1 x$ and use the expansion you are assumung that $|x|>1$. Since there is no number with $|x|<1$ and $|x| >1$ you cannot use both of them for the same $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what is the circle with the smallest radius that contains the three points $(0,0)$ , $(1,1)$, and $(2,3)$? How to write a model that gives us a circle with the smallest radius that contain three points $(0,0)$ , $(1,1)$, and $(2,3)$?
I've tried to model this as following:
if $x, y$ be the location and $r$ is radius:
$ Min \ \ x^2+y^2 = r^2$
$ x^2+y^2 \le 0$
$ (x-1)^2+(y-1)^2 \le 1$
$ (x-2)^2+(y-3)^2 \le \sqrt 13$
| How about:
min $r^2$ subject to:
$(x-0)^2 + (y-0)^2 \leq r^2$
$(x-1)^2 + (y-1)^2 \leq r^2$
$(x-2)^2 + (y-3)^2 \leq r^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$
How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$
| You can start by substituting both sides into $\arccos()$ function:
$$\arccos(\cos(2 \arctan x)) = \arccos\Biggl(\frac{1-x^2}{1+x^2}\Biggl)$$ $$2 \arctan x =\arccos\Biggl(\frac{1-x^2}{1+x^2}\Biggl)$$
Then you can take the derivative of both sides: $$\begin{align} \ \frac{2}{1+x^2} &=\ \frac{1}{\sqrt{1-\Biggl(\frac{1-x^2}{1+x^2}\Biggl)^2}} \frac{4x}{(1+x^2)^2} \\
&=\ \sqrt{\frac{(1+x^2)^2}{4x^2}} \frac{4x}{(1+x^2)^2} \\ \end{align}$$
If you organize the right-hand side of the equation, you eventually get:
$$\bbox[yellow] {\frac{2}{1+x^2}=\frac{2}{1+x^2}}$$
This means the equation holds for all real $x$ values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$ Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.
Here is what I have done so far
$$ \cos x + \cos y = -\cos z $$
$$ (\cos x + \cos y)^{2} = (-\cos z)^{2} $$
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y = \cos^{2}z $$
likewise,
$$ \sin x + \sin y = -\sin z $$
$$ (\sin x + \sin y)^{2} = (-\sin z)^{2} $$
$$ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y = \sin^{2}z $$
from this, you get
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y + \sin^{2} x + \sin^{2} y + 2\sin x \sin y = \sin^{2}z + cos^{2}z $$
$$ 1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1 $$
$$ \cos x \cos y + \sin x \sin y = -\frac{1}{2} $$
$$ \cos(x-y) = -\frac{1}{2} $$
$$ x-y = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
likewise,
$$ \cos (x-z) = -\frac{1}{2} $$
$$ x-z = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
where do I go from here?
| Notice that $\begin{cases}\cos(x)+\cos(y)+\cos(z)=0\\\sin(x)+\sin(y)+\sin(z)=0\end{cases}\iff \underbrace{e^{ix}}_a+\underbrace{e^{iy}}_b+\underbrace{e^{iz}}_c=0$
We have $a+b+c=0\implies \bar a+\bar b+\bar c=0$
But since $|a|=|b|=|c|=1$ they verify $\bar a=\frac 1a$ and so on.
Therefore we get $$\frac 1a+\frac 1b+\frac 1c=0\implies \dfrac{ab+bc+ca}{abc}=0$$
Note that $|abc|=1$ so the numerator should be zero.
But then $$\overbrace{(a+b+c)^2}^{=0}=\overbrace{(a^2+b^2+c^2)}^{e^{i2x}+e^{i2y}+e^{i2z}}+2\overbrace{(ab+bc+ca)}^{=0}$$
And your sum is simply $0$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$
Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$
I couldn't proceed much, however I think I got the minimum and maximum case.
For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$
For maximum, we get $288$ with equality on $(-1,-1,-1,4,4).$
| This is the problem of China national in math olympiad here is my proof:
.
.
.
Wlog let $|e|=Max(|a|,|b|,|c|,|d|,|e|)$ so we have $(e+d)(e+a)>0$ (take note we are to find the min at first ) for it to be negative we have $2$ cases one is that all $3$ of $(b+c),(c+d),(a+b)$ are negative which we have from AM_GM that :
$$(-a-b)(-b-c)(-c-d)(d+e)(e+a)\le (\frac{-a-2b-2c-d}{3})^3(\frac{a+d+2e}{2})^2=(\frac{a+d+2e-10}{3})^3(\frac{a+d+2e}{2})^2 \le 512$$
so its done the second case is that only one of them is negative. Let $a+b$ be negative (the way we use AM_GM gives us this freedom to assume this) now we have
$$(-a-b)(b+c)(c+d)(d+e)(e+a)\le (\frac{2(c+d+e)}{5})^5=(\frac{2(5-a-b)}{5})^5\le(\frac{14}{5})^5< 512$$
so we are done here. now we want to find the Max which we do in the ofllowing way.
if all of $a+b,b+c,c+d$ are possitive we have from AM_GM that
$$S\le (\frac{2(a+b+c+d+e)}{5})^5=32<288.$$
the next case is the following . we have $a+b\le0$, $b+c\le0$, $c+d\ge0$, then by AM_GM
$$(-a-b)(-b-c)(c+d)(d+e)(e+a)\le (\frac{-a-2b-c}{2})^2(\frac{c+d+e+a}{2})^2(d+e)=(\frac{d+e-b-5}{2})^2(\frac{5-b}{2})^2(d+e)\le 288$$
the last case is :
$a+b\le0$, $b+c\ge0$, $c+d\le0$, then by AM_GM we have
$$S=(-a-b)(b+c)(-c-d)(d+e)(e+a)\le (\frac{-a-b-c-d}{2})^2(\frac{b+c+d+e+e+a}{3})^3=(\frac{e-5}{2})^2(\frac{5+e}{3})^3< 288$$
and done done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.
\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}
so $x > 100$
| Show this product is equal to:
$$\frac{4^{5000}}{\binom{10000}{5000}}$$
Then use the inequality about the central binomial coefficient:
$$\frac{4^n}{\sqrt{4n}}\leq\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}$$
Setting $n=5000$ this gives:
$$\frac{4^{5000}}{\sqrt{20000}}<\binom{10000}{5000}<\frac{4^{5000}}{\sqrt{15001}}$$
Or
$$122<\sqrt{15001}<\frac{4^{5000}}{\binom{10000}{5000}}<\sqrt{20000}<142$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin
First, I get the line which passes through point of contact of tangents from origin using result here which is :
$$ -12x-16y-2 \cdot 75 = 0$$
Or,
$$ 6x + 8y -75 =0 $$
Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:
$$ C(0,0) C(x,y) = (6x+8y-75)^2$$
This leads to:
$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$
$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$
For applying the result in this answer, then the formula
$$ a= 75 - 36, b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$
Or,
$$ a = 39, b= 11 , h=-48$$
$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$
However, the intended answer was:
$$ \tan \theta = \frac{1}{\sqrt{3} } $$
Where have I gone wrong?
| A much more sane and simpler way:
Observe the right triangle created by the line segment from origin to center, and origin to the point which is tangent to circle. Since, tangent and radius are perpendicular, we can apply good ol' trignometery easily:
$$ \sin \theta = \frac{r}{OC}= \frac{5}{10}= \frac{1}{2}$$
Hence, the $ x= \frac{\pi}{6}$ , which makes total angle between tangent as $ x = \frac{\pi}{3}$
However, all the other answer are beautiful in demonstrating the different way of approaching the same problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Using the ratio test on $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n^2}$ I already know that I should do the root test on this series due to the $n$th power, but I want to see if I can establish the result using the ratio test first. (Or would I always be stuck with only one kind of test)
I get something like this: $$\lim_{n \to \infty} \left|\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}\right|$$
But then I'm not sure how (or even if I could) simplify this. I've tried on Wolfram Alpha to make it simplify but the best I can get is numerical approximations that seem to converge to $e$
| First, note that
$$
\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}
= 1-\frac{1}{(n+1)^2} \tag{1}
$$
We will use that to make the "right" exponent appear, since we want to use the known limit $\lim_{m\to\infty} (1+\frac{u}{m})^m = e^u$.
Now, since $n^2=(n+1)^2-(2n+1)$, we can rewrite
$$\begin{align*}
\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
&= \left(1+\frac{1}{n}\right)^{2n+1}\cdot \frac{\left(1+\frac{1}{n+1}\right)^{(n+1)^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}
= \left(1+\frac{1}{n}\right)^{2n+1} \left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n^2} \\
&= \color{red}{\left(1+\frac{1}{n}\right)}\cdot \color{blue}{\left(1+\frac{2}{2n}\right)^{2n}}\cdot \color{green}{\left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2}} \tag{2}
\end{align*}$$
This is great! We know that
$$
\lim_{n\to\infty} \left(1+\frac{1}{n}\right) = \color{red}{1} \tag{3}
$$
and
$$
\lim_{n\to\infty} \left(1+\frac{2}{2n}\right)^{2n} = \color{blue}{e^{2}} \tag{4}
$$
and
$$
\lim_{n\to\infty} \left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2} = \color{green}{e^{-1}} \tag{5}
$$
so, combining (3), (4), and (5) into (2), we get
$$
\lim_{n\to\infty} \frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
= \color{red}{1} \cdot \color{blue}{e^{2}} \cdot \color{green}{e^{-1}} = \boxed{e}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| Prove by Induction that
$ 7| {4^{2^n} + 2^{2^n} + 1} , ∀ n ∈ N $
Base case:
$7 | 4^{2^1} + 2^{2^1} + 1, $
7|7⋅3
Which is true.
Now, having n=k, we assume that:
$7|4^{2^k} + 2^{2^k} + 1, ∀ k ∈ ℕ $
We have to prove that for n=k+1 that,
$7| 4^{2^{k+1}} + 2^{2^{k+1}} + 1, ∀ k ∈ ℕ $
$ 4^{2^{k+1}} + 2^{2^{k+1}} + 1 = 4^{2*2^k} + 2^{2*2^k} + 1 $
$ = {4^{2^k}}^2 + {2^{2^k}}^2 + 1 $
Put $ {4^{2^k}} = a, {2^{2^k}} = b $
$ ∴ {(4^{2^k})}^2 + {(2^{2^k})}^2 + 1 = a^2 + b^2 + 1 $
Also,
$ (a+b+1)(a+b-1) = a^2 + b^2 + 2ab - 1 $
Since $ 7|a+b+1 $,
$ 7| a^2 + b^2 + 2ab - 1 ...(1) $
$ ab-1 = {8^{2^k}} - 1 $
For n ∈ ℕ, $ 8^n = (7+1)^n $
= (Multiple of 7) + 1
(Theorem of Binomial Expansion)
$ ∴ 8^n -1 = 7p $
Where p ∈ ℕ if n ∈ ℕ
$ ∴ 7| ab - 1 ....(2) $
Subtracting Equation (2) from equation (1),
$ 7| a^2 + b^2 + 2ab - 1 -2*(ab-1) $
$ ∴ 7| a^2 + b^2 + 1 $
ie
$$7| 4^{2^{k+1}} + 2^{2^{k+1}} + 1, ∀ k ∈ ℕ $$
if
$7| 4^{2^{k}} + 2^{2^{k}} + 1, ∀ k ∈ ℕ $
Hence, by principle of Mathematical Induction, the above statement is proved to be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
Find the least square solution x
Find the least square solution $x$ for $Ax = b$ if
$$A =\left(\begin{array}{rrr} 2 & 0\\
1 & 3 \\
0 & 2\end{array}\right) \quad b =\left(\begin{array}{rrr} 7\\
0\\-1\end{array}\right) $$
My Solution:
I found $A^T A =\left(\begin{array}{rrr} 7 & -1\\
-1 & 8 \\
\end{array}\right) $
$$A^T b =\left(\begin{array}{rrr} 2\\
2\\
\end{array}\right) $$
$$\left(\begin{array}{rrr} 5 & -1\\
-1 & 5 \\
\end{array}\right) x = \left(\begin{array}{rrr} 2\\
2\\
\end{array}\right) $$
But I can't solve this equation to find $x$ as there is no solution.
My solution after finding $x$ was to then find this:
$$Ax = \left(\begin{array}{rrr} 2 & 0\\
1 & 1 \\
0 & 2\end{array}\right) \left(\begin{array}{rrr} x\\
x\\ \end{array}\right) = \left(\begin{array}{rrr} a\\
b\\c\end{array}\right) $$
and then use $b-Ax^ = \sqrt{1+0+1} - \sqrt{a^2 + b^2 +c^2} = \textrm{final answer}$
Where am I going wrong with my approach? I am stuck as I can't solve for $x$?
| Your first error is in computing $A^\top b$. We have
$$A^\top b = \begin{pmatrix} 2 & -1 & 0 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$$
So, we need to solve
$$\begin{pmatrix} 5 & -1\\
-1 & 5
\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2\\
-2
\end{pmatrix}.$$
I don't know where you went wrong specifically here, but there is a unique solution, as $A^\top A$ is invertible (compute its determinant to see this). We can use the old shortcut to compute the inverse of this matrix:
$$\begin{pmatrix} a & b \\ c & d\end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}.$$
In particular,
$$\begin{pmatrix} 5 & -1 \\ -1 & 5\end{pmatrix}^{-1} = \frac{1}{24}\begin{pmatrix} 5 & 1 \\ 1 & 5\end{pmatrix},$$
so
$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{24}\begin{pmatrix} 5 & 1\\1 & 5\end{pmatrix}\begin{pmatrix} 2\\-2\end{pmatrix} = \begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix}.$$
Therefore, the closest point to $(1, 0, -1)^\top$ in the columnspace of $A$ is
$$A\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix} =\begin{pmatrix} 2 & 0\\-1 & 1 \\0 & 2\end{pmatrix}\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{3}\end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ -\frac{2}{3} \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the coefficients of this polynomial. Given $f(x)=x^4 +ax^3 +bx^2 -3x +5$ , $ a,b\in \mathbb{R}$ and knowing that $i$ is a root of this polynomial. what are $a,b$?
My work until now: since all the coefficients are real, I now know that $i$ and $-i$ are roots of this polynomial, there must be two more roots. I tried to use vietta but it didn't really help, I get that $x_3+x_4=-a$ and $i(-i)x_3x_4=5$. What am I missing?
Appreciate any help.
| By long division,
$$
\require{enclose}
\begin{array}{r}
x^2+ax+(b-1)\\[-4pt]
x^2+1\enclose{longdiv}{x^4+ax^3+bx^2-3x+5}\\[-4pt]
\underline{x^4\phantom{\ +ax^3}+x^2\ }\phantom{-3x+5\ \ }\\[-2pt]
ax^3+(b-1)x^2-3x\phantom{\,+5\ \ }\\[-4pt]
\underline{ax^3\phantom{\,\,+(b-1)x^2}+ax}\phantom{\ +5}\\[-2pt]
(b-1)x^2-(a+3)x\ +5\\[-4pt]
\underline{(b-1)x^2\phantom{\ \qquad}+(b-1)}\\[-2pt]
-(a+3)x+(6-b)
\end{array}
$$
We must have that $-(a+3)x+(6-b)=0$. Therefore, $b=6\text{ and }a=-3$. Thus,
$$
\left(x^2+1\right)\left(x^2-3x+5\right)=x^4-3x^3+6x^2-3x+5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How fast does the sequence converge? We have the sequence $$a_1=\frac{9}{4} \quad \qquad a_{n+1}=2\frac{a_n^3-a_n^2+1}{3a_n^2-4a_n+1}$$ that converges to $\frac{9}{4}$.
I want to check how fast the convergence is.
Do we have to calculate $\lim_{n\rightarrow \infty}\frac{|a_{n+1}-9/4|}{|a_n-9/4|}$ ?
| Hint: Define
$$f(x)=2\frac{x^3 - x^2 + 1}{3 x^2 - 4 x + 1}$$
Prove the sequence $a_{n+1}=f(a_n)$ converge to $2$ with $a_1 =\frac{9}{4}$ as you said.
\begin{align}
a_{n+1} &= f(a_n) \\
&= 2\frac{(a_n-2)((a_n-2)^2+1)}{((a_n-2)+1)(3(a_n-2)+5)} \tag{1}\\
\end{align}
Denote $b_n = a_n-2$, we have $b_n \rightarrow 0$ when $n\rightarrow +\infty$ or $b_n=\mathcal{o}(1)$. And from (1) we have
$$b_{n+1}+2 = f(b_n+2)$$
or
\begin{align}
b_{n+1} &=f(b_n+2)-2 \\
&= \frac{b_n(b_n^2+1)}{(b_n+1)(3b_n+5)} -2 \\
&= \frac{2}{5}b_n(b_n^2+1)(1-b_n+\mathcal{O}(b_n^2))(1-\frac{3}{5}b_n+\mathcal{O}(b_n^2)) -2\\
&= \frac{4}{5}b_n^2-\frac{22}{5}b_n^3+\mathcal{O}(b_n^4)\\
&= \frac{4}{5}b_n^2(1+\mathcal{O}(b_n)) \tag{2}\\
\end{align}
Then,
$$\ln(b_{n+1})= \ln(\frac{4}{5})+ 2\ln(b_{n}) +\ln(1+\mathcal{O}(b_n)) $$
$$\iff (\ln(b_{n+1})-\ln(\frac{4}{5}))= 2(\ln(b_{n})-\ln(\frac{4}{5})) +\mathcal{O}(b_n) \tag{3}$$
We notice that from (2), we can deduce also that $b_{n+1}=\mathcal{O}(b_n^2)=\mathcal{o}(b_n)$ or $\mathcal{O}(b_{n+1})=\mathcal{o}(b_n)$. So,
$$(3)\iff (\ln(b_{n+1})-\ln(\frac{4}{5})+\mathcal{O}(b_{n+1}))= 2(\ln(b_{n})-\ln(\frac{4}{5})+\mathcal{O}(b_n)) $$
$$\iff \ln(b_{n})-\ln(\frac{4}{5})+\mathcal{O}(b_{n})= 2^{n-1}(\ln(b_{1})-\ln(\frac{4}{5})) =2^{n-1}\ln(\frac{5}{16}) $$
or
$$b_n=\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$
(because $b_n \rightarrow 0$ when $n \rightarrow +\infty$ then $\exp(\mathcal{O}(b_{n}))=1$)
Conclusion:
$$a_n \approx 2+\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$
$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?
If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.
|
Rotate $\triangle BCP$ counter-clockwise 60$^\circ$ around the point $B$ to $\triangle BAQ$ and connect $PQ$. Then, $BPQ$ is an equilateral triangle and $APQ$ is a right triangle due to $x^2+y^2=z^2$. Apply the cosine rule to $\triangle BPA$ to obtain the side $s$
\begin{align}
s^2 & = AP^2+BP^2 - 2AP\cdot BP\cos\angle APB \\
&= x^2 + y^2 - 2x y \cos 150^\circ\\
&= x^2 + y^2 +\sqrt3x y
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ How do I solve this example? I tried to point out the fastest growing term $2 ^ n$ and $3 ^ n$, but that doesn't seem to lead to the result. I know the limit is $0$ that's obvious, but I don't know how to work on it.
$$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$$
Or can I make a power estimate of a theorem on a tightened sequence, for example, that $2 ^ n$ / $13 ^ n$ (smaller) goes to zero and the other side $2 ^ n$ / $3 ^ n$ (bigger) also goes to zero?
$\lim_{n\to\infty} 2 ^ n$ / $13 ^ n$< $\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ < $\lim_{n\to\infty} 2 ^ n$ / $3 ^ n$
| It follows from the ratio test.
If $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = a < 1 \ $, then $\lim a_n = 0$.
$$ \dfrac{\frac{2^{n+1} \left((n+1)^4-1\right)}{4\cdot3^{n+1} + (n+1)^7}}{\frac{2^n(n^4-1)}{4\cdot3^n + n^7}} = \dfrac{2^{n+1} \left((n+1)^4-1\right)(4\cdot3^n + n^7)}{2^n(n^4-1)\left(4\cdot3^{n+1} + (n+1)^7\right)} \\ =\frac{2}{3} \left(\frac{n+1}{n}\right)^4 \left(\dfrac{1-\frac{1}{(n+1)^4}}{1-\frac{1}{n^4}}\right) \left(\dfrac{1+\frac{n^7}{3^n}}{1+\frac{(n+1)^7}{3^{n+1}}} \right) \longrightarrow \frac{2}{3} < 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $ \sum \frac{2^k}{k}$ is divisible by $2^M$ For each integer $M > 0$, there ${\bf exists}$ an $n$ such that
$$ \sum_{k=1}^n \dfrac{ 2^k}{k} $$
is divisible by $2^M$
${\bf try}$ Im struggling a bit to visualize this exercise. So, I tried to see for concrete number, for instance take $M=1$, then $n=2$ works: as
$$ 2 + \dfrac{2^1}{2} = 2^1 (1 + 2 )$$
Now, take $M=2$ and factor
$$ 2^{2} \underbrace{ \left( \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{2^4}{4} ... + \dfrac{2^{n-2} }{n} \right) }_{(*)}$$
now, we need to choose $n$ so that $(*)$ is an integer. Im unable to do so. Any help?
| Right so the "elementary answer." There are actually a couple that I've seen. One is cited by Gouvêa in his p-adic numbers book (Gouvêa, like me, also said the problem was too difficult to work out the answer from scratch).
The solution Gouvêa cites is Exercise 10.10 of D. P. Parent Exercises in Number Theory (1984).
The rough idea there (assuming I'm summarizing correctly) is
$$ 0 = \frac{1 - (1 - 2)^{2^n}}{2^n} = \sum_{k = 1}^{2^n} (-1)^{k+1}2^k \frac{1}{2^n}\binom{2^n}{k} \equiv \sum_{k=1}^{2^n} \frac{2^k}{k} \pmod{2^h} $$
for $n \ge h$. Plus there's some other work to compare this with the partial sums that are not powers of $2$.
The next two solutions I found cited by the OEIS (https://oeis.org/A087910). Namely the two solutions given for the 2002 Sydney University Mathematical Society Problems Competitions Problem 9.
Solution 1 Summary
If $n$ is even then
$$ 1 = (-1)^n = (1 - 2^n) = \sum_{k = 0}^n \binom{n}{k}(-2)^k. \tag{1} $$
Subtract 1 and divide by $n$ to get
$$ 0 = \sum_{k = 1}^n \frac1n \binom{n}{k}(-2)^k. $$
Then
$$ \frac1n \binom{n}{k} = \frac{(n-1)(n-2)\cdots(n-k+1)}{k!} = \frac{(-1)^{k-1}}{k} + n\frac{m_{n,k}}{k!} \tag{2}$$
for some integer $m_{n,k}$ (separate the term $(-1)(-2)\cdots(-k+1)$ from all the terms divisible by $n$).
By $(1)$ and $(2)$,
$$ \sum_{k = 1}^n \frac{2^k}{k} = n \sum_{k = 1}^n \frac{(-2)^km_{n,k}}{k!}$$
Then you use a well known fact that $v_2(k)! \le O(k)$. Thus
$$ v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) = v_2(n) + v_2\left( \sum_{k = 1}^n \frac{(-2)^km_{n,k}}{k!} \right) \ge v_2(n). $$
Now we can see that when $n = 2^k$ this tends to infinity.
Solution 2 Summary
First show that
$$ \sum_{k = 1}^n \frac{2^k}{k} = \frac{2^n}{n} \sum_{k = 0}^{n-1} \frac{1}{\binom{n-1}{k}}. $$
Next, use the well-known formula $$v_2(n!) = \sum_{i = 0}^\infty \left\lfloor \frac n{2^i} \right\rfloor$$ to get
$$ v_2\left( \binom{n}{k} \right) = v_2(n!) - v_2(k!) - v_2((n - k)!) = \sum_{i = 0}^\infty \left\lfloor \frac n{2^i} \right\rfloor - \left\lfloor \frac k{2^i} \right\rfloor - \left\lfloor \frac {n-k}{2^i} \right\rfloor. $$
Then by some analysis,
$$v_2\left( \binom{n-1}{k} \right) \le r \text{ if } 2^r + 1 \le n \le 2^{r + 1}. $$
So if $n$ is even and $2^r + 1 \le n \le 2^{r + 1}$ then
$$ v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) \ge n - r. $$
And the result follows.
You'll have to see the solutions I linked to if you want all the details, they wouldn't fit in one answer. I hope this gives you some appreciation for the $2$-adic logarithm approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
What is the formula for the nth number in this series. $1,3,7,9,11,13,17,19,21\dots$
Basically all the numbers that end in the digits $1,3,7,9$
I am working on a formula for approximating how many factors I have to test to find if a large number is prime. So for example to test the number $229,597$. How many possible factors will I have to check?
So the convention is to take the square root of $229,597$ which is approx $479$. Then I take $(\frac{479}{10}) \times (4) - 1$. I do this because out of every ten numbers there are 4 numbers that end in $1,3,7,9$. I subtract the 1 because prime numbers also have 1 as a factor.
So $(\frac{479}{10}) \times (4) - 1 = 190$ factors to check to see if $229,597$ is prime.
Then I look at the series $1,3,7,9,11,13,17\dots$ to start checking for possible factors. But how do I find the nth number in this series?
| One possible formula is $$f(n) = \frac{1}{4}\left(5(2n-1) + (-1)^{n+1} + 2 \cos \frac{n \pi}{2} - 2 \sin \frac{n \pi}{2}\right), \quad n = 1, 2, \ldots.$$ This corresponds to the recurrence relation
$$\begin{align}
f(n) &= f(n-1) + f(n-4) - f(n-5), \\
f(1) &= 1, \\
f(2) &= 3, \\
f(3) &= 7, \\
f(4) &= 9, \\
f(5) &= 11.
\end{align}$$
I don't know why you would need it, though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.
$$\frac{n(n+1)}{(n+3)^3}$$
Using the ratio test, then simplifying in stages:
$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+1)}$$
$$\frac{(n+1)(n+2)(n+3)^3}{(n+4)^3n(n+1)}$$
$$\frac{(n+2)(n+3)^3}{(n+4)^2(n+4)n}$$
$$\frac{(n+3)^3}{(n+4)^22n}$$
The concluding remark:
$$\frac{1}{2}\lim_{n \to \infty}\frac{(n+3)^3}{(n+4)^2n}$$
Hence the series is divergent, unless I went wrong somewhere?
| Without using asymptotic equivalence, an easy way is
\begin{equation}
\sum_{n=1}^\infty\left(\dfrac{n}{n+3}\times\dfrac{n+1}{n+3}\times\dfrac{1}{n+3}\right)\ge\sum_{n=1}^\infty\left(\dfrac14\times\dfrac24\times\dfrac{1}{n+3}\right)=\infty
\end{equation}
Another alternative solution is
\begin{equation}
\sum_{n=1}^\infty\dfrac{n(n+1)}{(n+3)^3}\ge\sum_{n=1}^\infty\dfrac{n^2}{(n+3n)^3}=\infty
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving equations involving two sets Out of curiosity, how would one go about solving an equation involving two sets? For example,
$$ \{1, 2, 3\} = \{a + b + c, a + b - c, a - b + c, a - b - c\} $$
An intuitive solution to this is $ \{a = 2, b = 0.5, c = 0.5\} $, but is there a specific process?
| Notice the set on the left has three distinct elements, but the set on the right has four representations so two of those represetations are of the same number.
So we have six options.
$a+b+c = a+b-c$ and $c = 0$ and we have the values $a+b$ and $a-b$. But that's only two different values (or fewer) and we have exactly $3$ so that's impossible.
$a+b + c = a-b+c$ and $b =0$ and have the values $a+c$ and $a-c$ and that's the same problem.
$a+b+c = a-b-c$ and $b+c= 0$ and $b=-c$ and we have the values $a, a+2b, a-2b$. We'll get back to this.
$a+b-c = a-b + c$ and $b-c=0$ and $b=c$ and we have the values $a+2b; a; a-2b$. That will have the same solutions as above.
$a+b-c=a-b-c$ and $b=0$ and we have the same problem we had earlier.
Or $a-b+c = a-b-c$ adn $c=0$ and we have the same problem from the very begining.
So either way we have $a,a+2b, a-2b = 1,2,3$.
$a\pm 2b, a , a\pm 2b$ are in arithmetic progression so we must have either $a-2b < a < a+2b$ and $a-2b = 1, a=2, a+2b=3$ or $a+2b < a < a-2b$ and $a+2b = 1, a=2, a-2b =3$
So we must have $a=2$ and $b =\pm 0.5$ and $c = \pm 0.5$ (so there are four sets of solutions.)
There isn't really any way to do this in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$p_{n}$ be the probability that $C+D$ is perfect square. Compute $\lim\limits_{n \to \infty}\left(\sqrt{n} p_{n}\right)$
Assume $C$ and $D$ are randomly chosen from $\{1,2, \cdots, n\}$. Let $p_{n}$ be the probability that $C+D$ is perfect square. Compute $$\lim\limits _{n \to \infty}\left(\sqrt{n} p_{n}\right)$$
My Approach: My assumptions were that $C,D$ can be equal and the order of $C,D$ are not considered [I don't know I assumed correct, but in the question it was not clear.]
The number of ways a number $k$ can be partitioned into $2$ parts is equal to $\Big\lfloor\frac{k}{2}\Big\rfloor$
Now with the numbers $\{1,2, \cdots, n\}$ all the numbers from $2$ to $2 n$ can be made by choosing $2$ numbers and adding them up. So number of perfect squares in this region are $\big\lfloor\sqrt{2 n}\big\rfloor-1$ ($-1$ because we cannot make $1$). So the probability of choosing $C, D$ such that $C+D$ is a perfect square is
$$p_n=
\frac{\displaystyle{\sum_{k=2}^{\big\lfloor\sqrt{2 n}\big\rfloor} \Bigg\lfloor\frac{k^{2}}{2}\Bigg\rfloor}}{n^{2}}
$$
Now I cant progress further.
Tell me if I am correct. Also are my assumptions correct ?
Edit: The answer of the limit is $\frac{4(\sqrt{2}-1)}{3}$
| $C$ and $D$ are chosen independently, so the number of pairs $(C,D)\in[n]^2$ with $C+D=k^2$ is
$$
\begin{cases}
k^2-1 &\text{if }2\leq k\leq \sqrt{n+1}\\
2n+1-k^2& \text{if }\sqrt{n+1}<k\leq\sqrt{2n}\\
0 & \text{otherwise}
\end{cases}
$$
So
\begin{align*}
\sqrt{n}p_n&=\frac{\displaystyle
\sum_{k=2}^{\lfloor\sqrt{n+1}\rfloor} (k^2-1)+\sum_{k=\lfloor\sqrt{n+1}+1\rfloor}^{\lfloor\sqrt{2n}\rfloor}(2n+1-k^2)}{n^{3/2}}\\
&\sim n^{-3/2}
\sum_{k=2}^{\lfloor\sqrt{n}\rfloor} k^2+n^{-3/2}\sum_{k=\lfloor\sqrt{n}\rfloor}^{\lfloor\sqrt{2n}\rfloor}(2n-k^2)\\
&\sim\int_0^1 x^2\,\mathrm{d}x+\int_1^{\sqrt{2}}(2-x^2)\,\mathrm{d}x=\frac43(\sqrt{2}-1)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing that the given sequence is bounded as follows Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.
Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows:
To show $x_{n+1}<x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that is s, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$
Back to the problem. It's clear that $A$ is bounded above by $\frac{1}{3^6}$ (Should I show this first?).
Then, to show that $A$ is bounded, it's suffices to show that $A$ is bounded below by $0$.
I show this one again by induction. Indeed, it's true for $n=1$.
Assume that it's true for $n=k$; that is $0 < x_k$. Then, $x_{k+1} = \frac{1}{3^{(k+1)+5}}
= \frac{1}{3^{k+5}} \cdot \frac{1}{3} > 0$. Hence, $0< x_n$ for all $n \in \Bbb Z^+$. Therefore, $A$ is bounded below by $0$. Thus, $A$ is bounded, as desired.
Now, I claim that $\sup A = \frac{1}{3^6}$ and $\inf(A) = 0$.
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.
For the supremum, let $M$ be an another upper bound of $A$. To show: $\frac{1}{3^6} \le M$. Suppose $M < \frac{1}{3^6}$. Then, by the Density Theorem, there exists $s \in \Bbb Q$ such that
$M < s < \frac{1}{3^6}$. Hence, $s \in A$, contradiction with the fact that $M$ is an upper bound of $A$.
Therefore, $\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.
Does those approach true?
| Using the "Archimedean property" , we get for every $\epsilon \gt 0 $ , there exists $m \in \mathbb{N}$ such that $\frac{1}{n} \lt \epsilon$ for every $n\ge m $ .
As, $3^{n+5}\gt n \implies \frac{1}{3^{n+5}} \lt \frac{1}{n} \lt \epsilon $ for all $n\ge m $
By definition of convergence of a sequence, sequence
$\{\frac{1}{3^{n+5}}\}$ converge to $0$.
So, the given sequence is bounded and $limsup=liminf=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof by induction: Inductive step struggles Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$
where $ n $ is a nonnegative integer.
Preforming the basis step where $ n $ is equal to 0
$$
1 = \frac{2^{1}+(-1)^{0}}{3\times2^{0}} = \frac{3}{3} = 1
$$
Now the basis step is confirmed.
Then I started the inductive step where $ n = k $ is assumed true and I needed to prove $ n = k+1 $
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left(- \frac{1}{2}\right)^{k} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times2^{k+1}}
$$
Using the inductive hypothesis
$$
\frac{2^{k+1}+(-1)^{k}}{3\times2^{k}} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times 2^{k+1}}
$$
After this I am struggling here trying to get around to the end. I would appreciate any guidance.
| Use instead the perturbation method from Concrete Mathematics:
\begin{align}
S_{n} + (-1)^{n+1}\frac{1}{2^{n+1}}&= \sum_{k=0}^{n}(-1)^{k}\frac{1}{2^{k}} +(-1)^{n+1}\frac{1}{2^{n+1}}=1 -\frac{1}{2}\sum_{k=0}^{n}(-1)^{k}\frac{1}{2^{k}} \\
&= 1-\frac{1}{2}S_n
\end{align}
\begin{align}
\frac{3}{2}S_n&=1 - (-1)^{n+1}\frac{1}{2^{n+1}} \\
S_n &=\frac{2(1 - (-1)^{n+1}\frac{1}{2^{n+1}})}{3} = \frac{2^{n+2}-(-1)^{n+1}}{3\cdot2^{n+1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$ Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$
I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that
$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$
Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.
Update:
I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.
Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.
|
I hope following answer would help. The given series is an infinite arithmetic-geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How do I simplify $34\csc{\frac{2\pi}{17}}$? I have $$ 34\csc\left(\dfrac{2\pi}{17}\right)$$ is equal to $$\dfrac{136}{\sqrt{8-\sqrt{15+\sqrt{17} + \sqrt{34 + 6\sqrt{17} - \sqrt{34-2\sqrt{17} } + 2\sqrt{ 578-34\sqrt{17}} - 16\sqrt{34-2\sqrt{17}} } }}}.$$
I want to rationalize it, but I am not sure where to start. Could anyone provide an explanation on how to rationalize this, and what should the answer be?
| $$\csc^2\left(\frac{2\pi}{17}\right)=\frac{1}{17}\left[102+17\sqrt{17}-17\frac{\sqrt{34-2\sqrt{17}}}{2}-17\frac{\sqrt{34+2\sqrt{17}}}{2}+\sqrt{17}\cdot \sqrt{A}\right]$$
$$A=850+204\sqrt{17}-143{\sqrt{34+2\sqrt{17}}}-113{\sqrt{34-2\sqrt{17}}}$$
$$\tan^2\left(\dfrac{2\pi}{17}\right)=5+3\sqrt{17}+5\frac{\sqrt{34-2\sqrt{17}}}{2}+5\frac{\sqrt{34+2\sqrt{17}}}{2}-\sqrt{A}$$
$$A=850+204\sqrt{17}+161{\sqrt{34+2\sqrt{17}}}+127{\sqrt{34-2\sqrt{17}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$ I hit a wall on this question. Below are my steps
$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$
Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$
$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$
Use partial fraction decomposition,
$$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$
Solve for $A$ and $B$:
$$\begin{cases}
A=\frac{1}{-2a^{2}}
\\
B=\frac{1}{2a^{2}}
\end{cases}$$
Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$
Factoring out $a$ yields
$$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$
Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.
My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$
Feedback says my answer is wrong. Where did I mess up?
| $$I=\int\frac{-7x}{x^4-a^4}dx=-7\int\frac{x}{x^4-a^4}dx$$
now for ease lets let $u=x^2\Rightarrow dx=\frac{du}{2x}$ and $b=a^2$ so:
$$I=-\frac72\int\frac{du}{u^2-b^2}$$
now let $u=bv\Rightarrow du=b\,dv$ so:
$$I=-\frac{7b}{2}\int\frac{dv}{b^2(v^2-1)}=-\frac{7}{2b}\int\frac{dv}{v^2-1}$$
now this is a standard integral that you can solve with PFD or using a substitution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculation in Routh's theorem The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:
Option 1
The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.
That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.
Option 2
This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.
The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.
The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.
| The question asks to find an elegant proof of
$$ 1-\frac{x}{zx\!+\!x\!+\!1}-\frac{y}{xy\!+\!y\!+\!1}
-\frac{z}{yz\!+\!z\!+\!1}= \\
\frac{(xyz-1)^2}
{(zx\!+\!x\!+\!1)(xy\!+\!y\!+\!1)(yz\!+\!z\!+\!1)}. \tag{1} $$
In order to simplify algebraic manipulation define
$$ X:= zx+x+1,\quad Y:= xy+y+1,\quad Z:= yz+z+1. \tag{2}$$
Move all terms to the same side of the equation and
eliminate all denominators to get
$$ 0 = -X\,Y Z+x\, Y Z+X\,y\, Z+X\,Y z+(1-xyz)^2. \tag{3} $$
Define the polynomial expression
$$ A := X\,Y Z-x\, Y Z-X\,y\, Z-X\,Y z. \tag{4} $$
Proving equation $(3)$ and equation $(1)$ is equivalent to
proving $$ A = (1-xyz)^2.\tag{5} $$
One possible proof is to identity the homogeneous parts of
the degree six polynomial $\,A.\,$
For degrees $0$ and $6$ the only contribution is from the
first term of $\,A\,$ and thus
$$ A_0 = 1, \qquad A_6 = (xyz)^2. $$
For similar reasons,
$$ A_1 = (x+y+z)-x-y-z=0. $$
$$ A_5 = (xyz)^2(1/x+1/y+1/z) - (xyz)^2(1/z+1/x+1/y) = 0.$$
$$ A_2 = 2(zx+xy+yz)-x(y+z)-y(x+z)-z(x+y)=0. $$
$$A_4 = 2xyz (x+y+z)-xyz ((x+y)+(y+z)+(z+x))=0. $$
$$ A_3 \!=\! (4xyz\!+\!z^2x\!+\!x^2y\!+\!y^2z) \\
\!-\!xy(2z\!+\!x)\!-\!yz(2x\!+\!y)\!-\!zx(2y\!+\!z)
\!=\! -2xyz. $$
Putting all the parts together proves equation $(5)$.
This is essentially expanding the polynomial expression
$\,A\,$ and doing the same with $\,(1-xyz)^2.$
Another possible proof is to note that $\,A\,$ is a
polynomial in $\,x,y,z\,$ with maximum degree of each
variable being $2$. There are $3^3=27$ possible
monomials in $\,A.\,$ If equation $(5)$ can be proved
for at least $27$ generic values of $\,x,y,z\,$ then
the equation holds in general. Consider
$$ x=a/b,\;\; y=b/c,\;\; z=c/a,\;\;
x y z = 1, \\ d:=a+b+c,\quad (X,Y,Z) =
d\Big(\frac1b,\;\frac1c,\;\frac1a\Big), \\
A = \frac{d^2}{abc}(d-a-b-c) = 0 = (1-1)^2. $$
We can choose any nonzero values for $\,a,b,c\,$
which gives more than enough values of $\,x,y,z\,$
which proves equation $(5)$.
A simpler variation of this proof uses
$$ x = y = z =: w,\quad x y z = w^3,\quad X=Y=Z=1+w+w^2 =: W,
\\ A = W^3-3wW^2 = W^2(W-3w) = (W(1-w))^2 = (1-w^3)^2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.
Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.
Any help is appreciated!
| We must multiply the modulus by $\,6\,$ to balance the division by $6$, i.e.
$\qquad 6\mid a \,\Rightarrow\, a/6 \bmod 1000 = (a \bmod 6000)/6\ $ by the mod Distributive Law
$6000\!=\! 2^4\cdot 3\cdot 5^3$ whose totients $2^3,2,100\mid 2000\,$ so $\,\color{#c00}{7^{2000}\!\equiv 1}\bmod 2^4,3,5^3$ so also mod $6000,\,$ so $\bmod 6000\!:\ a\equiv 7^5 \color{#c00}{7^{2000}}-1 \equiv 7^5-1\equiv4806,\,$ so $\bmod 1000\!:\ a/6\equiv 4806/6 \equiv 801$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
What sequence is this $\prod_{i=1}^n \frac{2i}{2i + 1} = $? I'm trying to find a closed form solution for $\prod_{i=1}^n \frac{2i}{2i + 1} $, but I'm confused what kind of sequence this is. Does it have a closed form?
| Closed Form
$$
\begin{align}
a_n
&=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{1a}\\
&=\frac{\color{#C00}{2^nn!}\,\color{#090}{2^nn!}}{\color{#090}{(2n+1)!}}\tag{1b}\\[3pt]
&=\frac{4^n}{(2n+1)\binom{2n}{n}}\tag{1c}
\end{align}
$$
Bounds
$$
\begin{align}
\frac{\sqrt{n+1}\ a_n}{\sqrt{n}\ a_{n-1}}
&=\sqrt{\frac{n+1}{n}}\frac{2n}{2n+1}\tag{2a}\\
&=\sqrt{\frac{4n^3+4n^2}{4n^3+4n^2+n}}\tag{2b}\\[9pt]
&\le1\tag{2c}
\end{align}
$$
Therefore, $\sqrt{n+1}\ a_n$ is decreasing.
$$
\begin{align}
\frac{\sqrt{n+\frac34}\ a_n}{\sqrt{n-\frac14}\ a_{n-1}}
&=\sqrt{\frac{4n+3}{4n-1}}\frac{2n}{2n+1}\tag{3a}\\
&=\sqrt{\frac{16n^3+12n^2}{16n^3+12n^2-1}}\tag{3b}\\[12pt]
&\ge1\tag{3c}
\end{align}
$$
Therefore, $\sqrt{n+\frac34}\ a_n$ is increasing.
Since $\sqrt{n+1}\ a_n$ is decreasing and greater than $\sqrt{n+\frac34}\ a_n$, which is increasing, and their ratio tends to $1$, they tend to a common limit, $L$.
$$
\begin{align}
L
&=\lim_{n\to\infty}\sqrt{n+1}\prod_{k=1}^n\frac{2k}{2k+1}\tag{4a}\\
&=\lim_{n\to\infty}\frac{\sqrt{n+1}}{2n+1}\frac{4^n}{\binom{2n}{n}}\tag{4b}\\
&=\frac{\sqrt\pi}2\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: definition of $L$
$\text{(4b)}$: apply $(1)$
$\text{(4c)}$: Theorem $1$ from this answer
Thus $(2)$, $(3)$, and $(4)$ yield
$$
\bbox[5px,border:2px solid #C0A000]{\sqrt{\frac{\pi}{4n+4}}\le\prod_{k=1}^n\frac{2k}{2k+1}\le\sqrt{\frac{\pi}{4n+3}}}\tag5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$
applying Taylor's series.
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$
$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3- x^5+ o(x^5))}= \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 - x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2} \sim \frac{2}{3}$$
The suggested solution in my book is $2$. can someone indicate where I made mistakes?
| The OP's error has been pegged to an error with the logarithm of a difference being equated to the difference of logarithms. An alternative approach, which avoids all this, is to note that
$${\ln(1-\cos2x)\over\ln\tan2x}={\ln(2\sin^2x)\over\ln\displaystyle\left({2\tan x\over1-\tan^2x}\right)}={2\ln\sin x+\ln2\over\ln\sin x-\ln\cos x+\ln2-\ln(1-\tan^2x)}$$
Since $\ln\sin x\to-\infty$ as $x\to0^+$ while all the other terms tend to finite limits (i.e. $\ln2\to\ln2$, $\ln\cos x\to\ln\cos0=\ln1=0$, and $\ln(1-\tan^2x)\to\ln(1-\tan^20)=\ln1=0$), the requested limit is easily seen to equal $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate the limit $\lim\limits_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$ I have to calculate the limit $$\lim_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$$
I tried to use Sandwich Theorem like this $$\frac{3n^{2}}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}} \leq \Bigg(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\Bigg) \leq \frac{3n^{2}}{\sqrt[3]{(8n^{3}+2)}^{2}}$$
And for result I got that limit is $\frac{3}{4}$
Is this correct?
| You could have done it a bit faster and get more than the limit using generalized harmonic numbers.
$$S_n=\sum_{k=1}^{3n^2}\frac{1}{\left(8 n^3+2k\right)^{3/2}}=\frac 1{2\sqrt 2}\Bigg[H_{4 n^3+3 n^2}^{\left(\frac{3}{2}\right)}-H_{4
n^3}^{\left(\frac{3}{2}\right)} \Bigg]$$
Now, using the asymptotics
$$H_{p}^{\left(\frac{3}{2}\right)}=\zeta \left(\frac{3}{2}\right)-\frac 2{p^{1/2}}+\frac 1{2p^{3/2}}-\frac 1{8p^{5/2}}+O\left(\frac{1}{p^{9/2}}\right)$$ apply it twice and continue with Taylor series to get
$$S_n=\frac{3}{16 \sqrt{2}\, n^{5/2}}\Bigg[1-\frac{9}{16 n}+\frac{45}{128 n^2}-\frac{1713}{4096
n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I evaluate $\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx$? I need to calculate the following definite integral:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx.$$
The only thing that I've found is:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx = \int_{1/3}^3 \frac{\arctan \frac{1}{x}}{x^2 - x + 1} \; dx,$$
but it doesn't seem useful.
| Hint
$$\arctan{x}+\arctan{\dfrac{1}{x}}=\dfrac{\pi}{2}$$
so
$$\int_{\frac{1}{3}}^{3}\dfrac{\arctan{x}}{x^2-x+1}dx=I$$
let
$x=\dfrac{1}{u}$,then
$$I=\int_{\frac{1}{3}}^{3}\dfrac{\arctan{\frac{1}{x}}}{x^2-x+1}dx$$
so
$$2I=\dfrac{\pi}{2}\cdot\int_{\frac{1}{3}}^{3}\dfrac{1}{x^2-x+1}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.
Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.
| Look for perfect squares
$x^2 + y^2 + 1 \ge xy + x+y\iff $
$x^2 - 2xy+y^2 \ge -xy + x + y - 1\iff$
$(x-y)^2 \ge x+y - xy - 1$.
Now as the LHS is $\ge 0$ if we can show the RHS is $\le 0$ that be great. If not... well we may hit an inspiration on the way. We can always factor
$x+y - xy - 1 = x-xy + y-1 = x(1-y) + y-1) = x(1-y)-(1-y) = (x-1)(1-y)$.
Hmmm, no need for that to be negative but if it is positive then either $(x-1)$ and $1-y$ are both positive or both negative.
If both are positive then $x > 1$ and $y< 1$. So $x-y > x-1$ and $x-y > 1-y$ and so $(x-y)(x-y) > (x-1)(1-y)$.
And if both are negative then $x < 1<y$ and so $(x-y)^2 = (y-x)^2$ and $y-x > y-1>0$ and $y-x > 1-x>0$ and $(y-x)^2 > (y-1)(1-x) = (x-1)(1-y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 6
} |
Factorization $x^4+px^3+qx^2+r x +s=(x^2+a x +b)(x^2+\bar a x +\bar b)$ Question:
Under what condition, does the quartic polynomial with rational coefficients
$p$, $q$, $r$ and $s$ factorizes as
$$x^4+px^3+qx^2+r x +s= (x^2+a x +b)(x^2+\bar a x +\bar b) $$
with $a$, $b$ complex numbers, along with their conjugates $\bar a $, $\bar b$.
Examples:
$$x^4+2x^3+6x^2+2x+1=( x^2 +(1-i \sqrt3)x +1) (x^2 +(1+i \sqrt3)x +1) $$
$$x^4+2x^3+4x^2+2=( x^2 +(1+i)x +(1-i)) (x^2 +(1-i)x +(1+i)) $$
Note that the symmetry of coefficients leads to such factorization, as seen in the first example; but not exclusively so, as shown by the second example.
Is there any test on the coefficients $p$, $q$, $r$ and $s$ that can be carried out to determine the possibility of such factorization? I reviewed here the discriminate tests on the nature of roots for quartic equations and did not find anything applicable.
| Short answer:
Actually, the things I do, do not involve any special manipulation. It just contains simple algebra.
Let $a,b,c,d \in\mathbb R$ then we have,
$$\left(x^2+(a+bi)x+(c+di)\right)\left(x^2+(a-bi)x+(c-di)\right)=x^4+2ax^3+(a^2+b^2+2c)x^2+(2ac+2bd)x+(c^2+d^2).$$
$$x^4+px^3+qx^2+rx+s=x^4+2ax^3+(a^2+b^2+2c)x^2+(2ac+2bd)x+(c^2+d^2)$$
which follows
$$\begin{cases} p=2a \\ q=a^2+b^2+2c \\ r=2ac+2bd \\s=c^2+d^2\end{cases} $$
*
*If $d=0$, then the system of equation becomes extremely simple. I'll leave this case to you. Here we will work with $d≠0.$
*If $d≠0$, then $s-c^2≠0$. We have,
$$\begin{cases} a=\frac p2 \\b^2+2c=q-\frac{p^2}{4} \\pc+2bd=r \\c^2+d^2=s \end{cases}$$
$$\implies \begin{cases}\left( \frac{r-pc}{2d}\right)^2+2c-q+\frac{p^2}{4} =0 \\ c^2+d^2=s \end{cases}$$
Finally we get,
$$\frac{(r-pc)^2}{4(s-c^2)}+2c-q+\frac{p^2}{4} =0$$
$$\color {gold}{\boxed {\color{black}{8 c^3-4qc^2+(2pr-8s)c+(4qs-sp^2-r^2)=0.}}}$$
As it seems, we obtain the cubic equation with respect to $c$.
If all of the coefficients of the cubic equation are real numbers, then it has at least one real root. Therefore, we can always choose $c$ to be a real number.
If you accept $a,b,c,d$ as rational, then we can immediately use the Rational Root theorem.
For this, all we need is to find the factors of this expression $\color{red}{\dfrac{4qs-sp^2-r^2}{8}.}$
We're almost done. We have the last two restrictions.
After finding real $c$, then we must check the following two cases:
*
*$s-c^2≥0$
*$p^2-4q+8c≤0.$
If these conditions hold, then you can easily find the $a,b,d$ from the system of equations.
End of the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$ Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.
| $$(2r+1)^2+1\equiv8\cdot\dfrac{r(r+1)}2+2=8p+1\text{(say)}\equiv2\pmod8$$
$$(2r+1)^4+1=(8p+1)^2+1\equiv2\pmod{16}$$
Now $n^2-1=(2r+1)^2-1=8\cdot\dfrac{r(r+1)}2$
$\implies k\ge2\cdot3+2\cdot1+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Probabilities in a game John plays a game with a die. The game is as follows: He rolls a fair six-sided die. If the roll is a $1$ or $2$, he gets $0$ dollars. If he rolls a $3, 4$ or $5$, he gets $5$ dollars. If he rolls a $6$, he wins $X$ dollars where $X$ is a continuous random variable that is uniform on the interval $(10, 30)$. Let $Y$ be the amount of money John wins by playing the game.
(i) Compute the cumulative distribution function (cdf) of $Y$.
(ii) What is the probability that John rolled a $6$ given that he won less than $15$ dollars?
(iii) Compute $E(Y)$.
My attempt:
(i) If $Y > 5$ then $P(Y \leq y) = P(X \leq y)$. Then $P(Y \leq 5) = P(Y = 0) + P(Y = 5) = P(roll = 1 or 2) + P(roll = 3, 4 or 5) = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$.
Not sure if this is even correct.
(ii) $P(roll = 6 | Y < 15) = \frac{P(roll = 6 \land Y < 15)}{P(Y<15)}$
But by Bayes theorem $P(roll = 6 | Y < 15) = \frac{P(Y < 15 | roll = 6) \cdot P(roll = 6)}{P(Y<15)}$
We know $P(roll = 6) = \frac{1}{6}$.
Given that he rolled a $6$, the probability that he won less than $15$ dollars is equal to $P(X < 15) = \frac{15 - 10}{30 - 10} = \frac{1}{4}$
Lastly, $P(Y < 15) = 1 - P(Y \geq 15)$. However, since the only situation in which John wins $15$ dollars or more is if he rolls a $6$. We may infer that $P(Y \geq 15) = P(X \geq 15) = 1 - P(X < 15) = \frac{3}{4}$. So $P(Y < 15) = 1 - \frac{3}{4} = \frac{1}{4}$.
So $P(roll = 6 | Y < 15) = \frac{\frac{1}{4} \cdot \frac{1}{6}}{\frac{1}{4}} = \frac{1}{6}$
Not sure how to determine the other probabilities.
(iii) We can check $Y$ conditioned on $X = x$: $E(Y|X=x) = (0 \cdot \frac{1}{3}) + (5 \cdot \frac{1}{2}) + (x \cdot \frac{1}{6}) = \frac{5}{2} + \frac{x}{6} = \frac{x + 15}{6}$. But $E(E(Y|X)) = E(Y)$ by the law of iterated expectation.
Applying it we have $E(Y) = E(E(Y|X)) = E(\frac{X + 15}{6}) = \frac{1}{6} E(X) + \frac{15}{6}$
Since $X$ is uniform on $(10, 30)$ we know $E(X) = \frac{1}{2} (10 + 30) = 20$
Therefore $E(Y) = \frac{35}{6}$.
Is this correct? I have a feeling that my attempt for (i) is incorrect, and I am unsure about (ii) and (iii). Any assistance is much appreciated.
| Here is a simulation of a million iterations of your
process. Possibly except for conditional probabilities, it should
be accurate to a couple of decimal places. Also the ECDF
plot of from the simulation is very nearly the CDF of $Y,$
which you have not drawn.
set.seed(2021)
die =sample(1:6, 10^5, rep=T)
y = runif(10^5, 10, 30)
y[die<=2.1] = 0
y[die>=2.9 & die <=5.1] = 5
[1] 5.838549 # aprx E(Y)
35/6
[1] 5.833333 # exact
mean(die[y<15]==6)
[1] 0.04736204 # aprx P(Die = 6 | Y < 15)
hdr = "ECDF: Mixture of Discrete and Continuous"
plot(ecdf(y), ylab="ECDF", xlab="y", lwd=2, main=hdr)
At the 'jumps', the CDF takes the upper value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$ I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.
Do you have any ideas? :)
EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to
$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$
By expanding with 1:
$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$
$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$
$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$
And using the substitution: $t=\tan\left(x\right)$
$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$
$$t^2=\tan^2\left(x\right)$$
$$t^2=\frac{\sin^2x}{\cos^2x}$$
$$t^2=\frac{1-\cos^2x}{\cos^2x}$$
$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$
$$t^2+1=\frac{1}{\cos^2x}$$
Using it:
$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$
I don't think I got to the expected result but I can't seem to be able to find why…
| Following your substitution $t= \tan x$, integrate the resulting as follows
\begin{align}
&\int \frac1{1+\sin^4x}dx\\
=&\int \:\frac{t^2+1}{2t^4+2t^2+1}dt
=\int \frac{1+\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt\\
=&\frac{\sqrt2+1}{2\sqrt2} \int \frac{\sqrt2+\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt
-\frac{\sqrt2-1}{2\sqrt2} \int \frac{\sqrt2-\frac1{t^2}}{(\sqrt2t)^2 + \frac1{t^2}+1}dt\\
=&\frac{\sqrt2+1}{2\sqrt2} \int \frac{d(\sqrt2t-\frac1{t})}{(\sqrt2t-\frac1t)^2 + 2(\sqrt2+1)}dt
-\frac{\sqrt2-1}{2\sqrt2} \int \frac{d(\sqrt2t+\frac1{t})}{(\sqrt2t+\frac1t)^2 -2(\sqrt2-1)} dt\\
=&\frac{\sqrt{\sqrt2+1}}4\tan^{-1} \frac{t-\frac1{\sqrt2 t}}{\sqrt{\sqrt2+1}}
+\frac{\sqrt{\sqrt2-1}}4\coth^{-1} \frac{t+\frac1{\sqrt2 t}}{\sqrt{\sqrt2-1}}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find all positive $x,y\in\Bbb{Q}$ such that $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural numbers. What I did is, the product is also a natural number.
$xy+\frac{1}{xy}=m$ is also a natural number. Assuming $xy=z$, we get the equation $z^2-mz+1=0$
Hence, $z=\frac{m\pm \sqrt{m^2-4}}{2}$. Since z is rational $\sqrt{m^2-4}$ has to be rational. It is rational only when $m=2$ (not sure). So $z=xy=1$. So, $y=\frac{1}{x}$
Hence $2x$ and $2y$ are natural numbers.
I can't proceed further. Is my way correct? Any help please.
| If $x,y$ are rational $x = \frac pq$ and $y = \frac ab$ (assume in lowest terms) so
$\frac pq + \frac ba = \frac {pa + qb}{aq}\in \mathbb Z$
And $\frac ab +\frac qp = \frac {pa+qb}{pb}\in \mathbb Z$
So $q|pa+qb$ so $q|pa$ but $q$ and $p$ are relatively prime so $q|a$. Likewise $a|pa + qb$ so $a|qb$ so $a|q$ so $a= q$. And similarly $p = b$.
So $x = \frac 1y$.
So we must have $x + \frac 1y = x + x = 2x$ and $y + \frac 1x = 2y$ are natural numbers.
Whoo boy.
Okay. Case one $x = 1 \in \mathbb Z$ then $y = \frac 11 = 1$ and $x = y = 1$ are obviously solutions.
Case 2: $x \in \mathbb Z$ but $x > 1$. Then $y=\frac 1x \not \in \mathbb Z$. But $2y=\frac 2x\in \mathbb N$ so $x|2$ so $x = 2$ and $y= \frac 12$ is a solutions ($2+ \frac 1{\frac 12} = 4$ and $\frac 12 + \frac 12 = 1$).
Case 3: $x\not \in \mathbb Z$ then $2x \in \mathbb Z$ so $x = \frac m2$ for some odd $m$. But then $y =\frac 2m$ and $2y=\frac 4m$ is an integer so $m|4$. But $m$ is odd so $m =1$.
So those are the only three possibilities.
$x = y = 1; x = 2; y =\frac 12$ and $x = \frac 12$ and $y = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
nth term of $2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$ What is the nth term of the sequence: $$2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$$
in terms of $s_{n-1}$.
I have tried for some time to calculate this but to no avail. Hopefully somebody with a better understanding of the world of sequences could help/point me in the right direction!
| In my answer to this question, I detailed the steps for solving a first-order rational difference equation such as
$${ a_{n+1} = \frac{ma_n + x}{a_n + y} }=m+\frac{x-m y}{a_n+y}$$
For your case $m=2$, $x=1$ and $y=0$. So, using the initial condition,
$$a_n=\frac{\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n } { \left(1+\sqrt{2}\right) \left(1-\sqrt{2}\right)^n+\left(\sqrt{2}-1\right)\left(1+\sqrt{2}\right)^n}$$
Edit
In the documentation of sequence $A000129$ in $OEIS$, there is superb formula given by Peter Luschny in year $2018$. It write
$$a_n=\frac 1{\sqrt{2}}\, e^{\frac{i \pi n}{2}}\,\sinh \left(n \cosh ^{-1}(-i)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here? Am I going correct?
| For AM-GM, $-2xy \le 2|xy| \le x^2+y^2$ so $2xy\ge -x^2-y^2$.
$$5(x^2+y^2)\le 6x^2+2xy+6y^2 =9.$$
Then $x^2+y^2\le 9/5$. The maximum is reached when $x=-y=\pm\sqrt{9/10}$.
You can also get the min value of $x^2+y^2$. Because $2xy\le 2|xy|\le x^2+y^2$
$$7(x^2+y^2)\ge 6x^2+2xy+6y^2=9.$$
Then $x^2+y^2\ge 9/7$. The minimum is reached when $x=y=\pm\sqrt{9/14}$.
I think the shape $6x^2+2xy+6y^2=9$ is an ellipse with major/minor axes given by the lines $x=\pm y$. The axes have length $2\sqrt{9/5}$ and $2\sqrt{9/7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$ How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$
What I tried
Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamma \left(n+\frac{3}{4}\right)=\Gamma\left(\frac{3}{4}\right)\prod_{k=0}^{n-1}\left(k+\frac{3}{4}\right).$$
Then
$$\lim_{n\to\infty}\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
Since
$$\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\prod_{k=1}^n \frac{4k(4k-2)}{(4k-1)^2}$$
for all $n\in\mathbb{N}$, it follows that
$$\prod_{k=1}^\infty \frac{4k(4k-2)}{(4k-1)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
But note that this actually gives an interesting Wallis-like product:
$$\frac{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdots}{3\cdot 3\cdot 7\cdot 7\cdot 11\cdot 11\cdots}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
I'm stuck at the Wallis-like product, though.
| I suppose you could do it the cheap way and use Stirling's approximation:
$$n! \sim \sqrt{2\pi n} (n/e)^n$$ implies $$\Gamma^4(n+3/4) \sim 4\pi^2 \frac{(n-1/4)^{4n+1}}{e^{4n-1}},$$ and $$\Gamma^2(2n+1) \sim 2\pi \frac{(2n)^{4n+1}}{e^{4n}};$$ hence $$2^{4n} \frac{\Gamma^4(n+3/4)}{\Gamma^2(2n+1)} \sim \pi \left(1 - \frac{1}{4n}\right)^{4n+1} e,$$
and the rest is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$ Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.
I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.
So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.
I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.
| Well putting $x=\sinh^2(a)$
We have to show (using $\cosh^2(a)-\sinh^2(a)=1$):
$$\cosh(a)\geq 1+\frac{1}{2}\sinh^2(a)-\frac{1}{8}\sinh^4(a)$$
Or :
$$\sinh^6\Big(\frac{a}{2}\Big)(\cosh(a)+3)\geq 0$$
Wich is clearly obvious !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find polynomials $M_1(x)$ and $M_2(x)$ such that $(x+1)^2M_1(x) + (x^2 + x + 1)M_2(x) = 1$ I have been trying to solve this with no success. Could you suggest me a solution?
| Here is an alternate basic approach based upon coefficient comparison of polynomials. We are looking for polynomials $M_1(x), M_2(x)$ so that the identity
\begin{align*}
\left(x^2+2x+1\right)M_1(x)+\left(x^2+x+1\right)M_2(x)=1\tag{1}
\end{align*}
is valid.
We start with some observations:
*
*Since the right-hand side of (1) is a polynomial of degree zero, the left-hand side has also to be a polynomial of degree zero.
*We are looking for polynomials $M_1$ and $M_2$ which have smallest possible degree. We observe setting $M_1(x)=1$ and $M_2(x)=-1$ is not appropriate since we get rid of the square terms but not of the linear terms.
*Ansatz: We start with linear polynomials
\begin{align*}
M_1(x)=ax+b\qquad M_2(x)=cx+d
\end{align*}
with unknown coefficients $a,b,c,d\in\mathbb{R}$, multiply out and make a comparison of coefficients of LHS and RHS of (1).
We denote with $[x^n]$ the coefficient of $x^n$ of a polynomial and obtain from (1)
\begin{align*}
[x^3]:\qquad\qquad &a+c&=0\\
[x^2]:\qquad\qquad&(2a+b)+(c+d)&=0\\
[x^1]:\qquad\qquad&(a+2b)+(c+d)&=0\\
[x^0]:\qquad\qquad&b+d&=1\\
\end{align*}
Coefficient comparison of $[x^3]$ gives: $c=-a$. Putting this in the other three equations results in
\begin{align*}
a+b+d&=0\\
2b+d&=0\\
b+d&=1\\
\end{align*}
from which we easily find $a=-1, b=-1, c=1, d=2$.
We finally obtain
\begin{align*}
\color{blue}{M_1(x)=-x-1\quad M_2(x)=x+2}
\end{align*}
in accordance with the other given answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Induction with divisibility QUESTION: Prove that $16 \mid 19^{4n+1}+17^{3n+1}-4$ for all $n \in \mathbb{N}$
This is what I have so far but I'm not sure where to go from here.
PROOF:
Let $16 \mid 19^{4n+1}+17^{3n+1}-4$ equal $S(n)$.
Base case, let $n=0$.
\begin{align*}
16 &\mid 191+171-4\\
= 16 & \mid 32\\
= 2
\end{align*}
Therefore, $S(0)$ is true.
Using induction hypothesis, suppose $19^{4k+1}+17^{3k+1}-4$ is divisible by $16$ for all $n \in \mathbb{N}$.
Claim,
$16 \mid 19^{4k+1}+17^{3k+1}-4$,
that is $19^{4k+1}+17^{3k+1}-4=16m$, whereby $m$ is a multiple of $16$.
The above equation simplifies into,
\begin{align*}
16 & \mid 19^{4n+5}+17^{3n+4}-4\\
16 & \mid 19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4
\end{align*}
photo of my working out
| Q. Show that $16$ $|$ $19^{4n+1} +17^{3n+1} -4$ $\forall$ $ n \in \mathbb N$.
First of all, simplify the expression. Write $19=16+3$ and $17=16+1$, then use binomial theorem to expand it. That way you'll get a simpler expression to use induction.
$16$ $|$ $19^{4n+1} +17^{3n+1} -4 \implies$ $16$ $|$ $3^{4n+1}+1-4 \implies 16$ $|$ $3(3^{4n}-1)$
$\implies 16$ $|$ $3^{4n}-1$.
Method 1 to proceed:
Now here it's easier to just use the fact that $x-y$ $|$ $x^n-y^n$, because it will prove that $ 16$ $|$ $3^{4n}-1$ directly, without induction. ( As $16$ $|$ $3^4 -1$ $|$ $3^{4n}-1$. )
But if you wish to use induction only, then here it is:
Method 2 to proceed:
Let $P(n):16$ $|$ $3^{4n}-1$.
$P(1), P(2)$ are true.
Let $P(i)$ be true for $i=1,2,\cdots,k$
$16$ $|$ $3^{4n}-1 \implies 16$ $|$ $(3^{4n}-1)(3^4+1) \implies 16$ $|$ $3^{4n+4}-1 +3^{4}(3^{4n-4}-1)$
And we know $16 $ $|$ $(3^{4n-4}-1)$ (As $P(n-1)$ is true).
$\implies 16$ $|$ $3^{4n+4}-1 \implies P(n+1)$ is true, and hence $P(n)$ is true for all natural numbers $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$p = x^2 + y^2$ with $p$ prime and $y$ even, show that $x$ and $y/2$ are quadratic residues mod $p$
Let $p$ be a prime which can be written as $p = x^2 + y^2$ for positive integers $x$ and $y$. Also assume that $y$ is an even integer.
Show that $x$ and $y/2$ are quadratic residues mod $p$.
My attempt:
The case $p = 2$ is trivial.
For $p > 2$ we know that $p \equiv 1$ mod $4$. The integer $y$ is even. Since $p = x^2+y^2$ and $y^2 \equiv 0 $ mod $4$, we have that $x^2 \equiv 1$ mod $4$.
From this point I would like to use the legendre symbol $\left(\frac{x}{p}\right)$ to get further, but I don't get anywhere. Any tips?
| Use properties of the Jacobi symbol.
Firstly, note that if $p>2$, then $p\equiv 1\pmod 4$. Since $x$ and $p$ are odd we have
$$
\left(\frac{x}{p}\right)=(-1)^{(x-1)(p-1)/4}\left(\frac{p}{x}\right)=\left(\frac{p}{x}\right)=\left(\frac{x^2+y^2}{x}\right)=\left(\frac{y^2}{x}\right)=1,
$$
so $x$ is a quadratic residue.
For $y$ we consider two cases: $y\equiv 2\pmod 4$ (meaning that $p\equiv 5\pmod 8$) or $y\equiv 0\pmod 4$ (meaning that $p\equiv 1\pmod 8$).
In the first case we can just do the same
$$
\left(\frac{y/2}{p}\right)=(-1)^{(y/2-1)(p-1)/4}\left(\frac{p}{y/2}\right)=\left(\frac{p}{y/2}\right)=\left(\frac{x^2+y^2}{y/2}\right)=\left(\frac{x^2}{y/2}\right)=1.
$$
For the second case let $y=2^kz$ where $z$ is odd and $k\ge 2$. Since $p\equiv 1\pmod 8$ we have $\left(\frac{2}{p}\right)=1$, so the previous argument can be modified in the following way:
$$
\left(\frac{y/2}{p}\right)=\left(\frac{2^{k-1}z}{p}\right)=\left(\frac{z}{p}\right)=(-1)^{(z-1)(p-1)/4}\left(\frac{p}{z}\right)=\left(\frac{p}{z}\right)=\left(\frac{x^2+y^2}{z}\right)=\left(\frac{x^2}{z}\right)=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Taylor series about 1 I was trying to solve the Taylor series about one for $\dfrac{x^2}{2 - x}$ but my answer seems to be wrong. I got $T(x) = \frac{1}{n!}(x-1)^n$.
| We are looking for a representation
\begin{align*}
\sum_{n=0}^\infty a_n(x-1)^n
\end{align*}
We obtain
\begin{align*}
\color{blue}{\frac{x^2}{2-x}}&=\frac{(x-1+1)^2}{1-(x-1)}\\
&=\left((x-1)^2+2(x-1)+1\right)\sum_{n=0}^\infty (x-1)^n\\
&=\sum_{n=0}^\infty(x-1)^{n+2}+2\sum_{n=0}^\infty(x-1)^{n+1}+\sum_{n=0}^\infty(x-1)^n\\
&=\sum_{n=2}^\infty(x-1)^{n}+2\sum_{n=1}^\infty(x-1)^{n}+\sum_{n=0}^\infty(x-1)^n\\
&\,\,\color{blue}{=1+3(x-1)+4\sum_{n=2}^\infty(x-1)^n}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Area of $(x-3)^2+(y+2)^2<25: (x,y) \in L_1 \cap L_2$ Two lines $(L_1,L_2)$ intersects the circle $(x-3)^2+(y+2)^2=25$ at the points $(P,Q)$ and $(R,S)$ respectively. The midpoint of the line segment $PQ$ has $x$-coordinate $-\dfrac{3}{5}$, and the midpoint of the line segment $RS$ has $y$-coordinate $-\dfrac{3}{5}$.
If $A$ is the locus of the intersections of the line segments $PQ$ and $RS$, then the area of the region $A$ is:
What I've done:
Consider $L_1: y= ax+b$. The midpoint of the chord $PQ$ is $(-\dfrac{3}{5}, -\dfrac{3a}{5}+b)$. Now, using the property that the midpoint of a chord of a circle and the center of the circle $(3,-2)$ are perpendicular we have: $\dfrac{-\dfrac{3a}{5}+b-(-2)}{-\dfrac{3}{5}-(3)} *a = -1$
$$\implies b= \dfrac{3a^2-10a+18}{5a}$$ This means we can eliminate one variable and write the equation of $L_1: y= ax+ \dfrac{3a^2-10a+18}{5a}$
From this form of $L_1$ we can get the value of the minimum value of the $y$-intercept by differentiating $b$. Let $b= f(a) = \dfrac{3a^2-10a+18}{5a}, f'(a) = \dfrac{3a^2-18}{5a^2}, f'(a)=0 \implies a = \pm \sqrt{6}$. I just found this hoping we will get bounds on the y-intercept of $L_1$.
Now lets do the same process for $L_2$.
Consider $L_2: y= cx+d$. The midpoint of the chord $RS$ is $(\dfrac{-\dfrac{3}{5} -d}{c},-\dfrac{3}{5})$. Now, using the property that the midpoint of a chord of a circle and the center of the circle $(3,-2)$ are perpendicular we have: $\dfrac{-\dfrac{3}{5}+2}{\dfrac{-\dfrac{3}{5} -d}{c}-3} *c = -1$
$$\implies d= \dfrac{7c^2-15c-3}{5}$$ This means we can eliminate one variable and write the equation of $L_2: y= cx+ \dfrac{7c^2-15c-3}{5}$
From this form of $L_2$ we can get the value of the minimum value of the $y$-intercept by differentiating $d$. Let $d= f(c) = \dfrac{7c^2-15c-3}{5}, f'(c) = \dfrac{14c-15}{5}, f'(c)=0 \implies c = \dfrac{15}{14}$. I just found this hoping we will get bounds on the y-intercept of $L_2$.
Along with all this, we can set bounds when the line segment is just about to leave the circle (tangent to the circle).
What I can visualize:
Let $X$ = union of all line segments $PQ$.
Let $Y$ = union of all line segments $RS$.
Every point in the intersection of $X$ and $Y$ is a candidate intersection point of the lines $L_1$ and $L_2$. So A = $X \cap Y$
Edit 1: I saw the equation of $L_1$ varying as $a$ varies on DESMOS and think the boundry of the union of all line segments PQ might be an outer circle.
| EDIT (Original answer at the end).
I want to show how the envelope of chords $RS$ (or $PQ$) can be obtained without calculus and without coordinates (see figure below).
Let's start with a chord $AB$ of a circle of centre $O$. For any point $M$ on that chord, we can construct a line $RS$ passing through $M$ and perpendicular to $OM$. We want to find the envelope of all those lines, i.e. the curve which is tangent to all the lines $RS$ as $M$ varies on $AB$.
Consider then another point $M'$ on $AB$ and its associated line $R'S'$. Let $P$ be the intersection of $RS$ and $R'S'$, $T$ the tangency point of $RS$ with the envelope, $T'$ the tangency point of $R'S'$ with the envelope. As $M'$ approaches $M$, both $T'$ and $P$ approach $T$.
But the circle through $OMM'$ also passes through $P$ (because $\angle PMO=\angle PM'O=90°$) and this circle, as $M'\to M$, tends to the circle through $O$ tangent to $AB$ at $M$. Hence $T$, which is the limiting position of $P$, is the intersection of that circle with line $RS$. Moreover, $OT$ is a diameter of that circle.
Now that we know how to construct point $T$ on $RS$,
let's also construct line $HK$, parallel to $AB$ at a distance from it equal to the distance of $O$ from $AB$. If $J$ is the projection of $T$ on it, then $TJ=TO$, because line $CH$ joining the midpoints of the legs of a trapezoid is the arithmetic mean of bases $OK$ and $TJ$.
It follows that point $T$ has the same distance from $O$ and from line $HK$. Hence its locus (which is the envelope) is a parabola, having $O$ as focus and $HK$ as directrix.
ORIGINAL ANSWER.
What you need is the envelope $\gamma_1$ formed by lines $L_1$ and the envelope $\gamma_2$ formed by lines $L_2$.
As the equation of $L_1$ is $y=\left(x+{3\over5}\right)a-2+{18\over5a}$, differentiating w.r.t. $a$ we get: $x+{3\over5}-{18\over5a^2}=0$, which can be solved for $a$:
$$
a^2={18\over 5x+3}.
$$
Inserting this into the equation of $L_1$ we get (after some algebra):
$$
25(2+y)^2={72}(5x+3),
$$
which is the desired envelope $\gamma_1$ (a parabola).
Repeating the same process for $L_2$ we can find the equation of $\gamma_2$ (another parabola):
$$
y=-{5\over28}(3-x)^2-{3\over5}.
$$
Lines $L_1$ and $L_2$ are tangent to their envelope, hence
the area you want to compute is that external to both parabolas but inside the circle.
To compute the area, one has to find the coordinates of the upper intersection $A$ between $\gamma_1$ and the circle,
of the left intersection $C$ between $\gamma_2$ and the circle
and of the intersection $B$ between $\gamma_1$ and $\gamma_2$
lying inside the circle:
$$
A=\left(\frac{4}{5},\frac{6}{5} \sqrt{14}-2\right),\quad
C=\left(3-\frac{6}{5}\sqrt{14},-\frac{21}{5}\right),\quad
B=\left(\frac{1}{5} \left(29-12 \sqrt{7}\right),\frac{2}{5} \left(6
\sqrt{7}-23\right)\right).
$$
Integrating along $y$ the area can then be computed as:
$$
\begin{align}
area&=\int_{y_B}^{y_A}
\left[\left(\frac{5}{72}(y+2)^2-\frac{3}{5}\right)-\left(3-\sqrt{25-(y+2)^2}\right)\right] \, dy \\
&+\int_{y_C}^{y_B}
\left[\left(-\frac{2}{5} \sqrt{-35 y-21}+3\right)-\left(3-\sqrt{25-(y+2)^2}\right)\right] \, dy \\
&=\frac{25 \pi }{4}+4 \sqrt{14 \left(9-4 \sqrt{2}\right)}-\frac{3004}{75}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all odd functions of the form $f(x) = \frac{ax + b}{x + c}$ I am working through a pure maths book and am stuck on odd and even functions.
Let $f(x) = \frac{ax + b}{x + c}$ where x, a, b, c are real and $x \ne \pm c$. Show that if $f$ is an even function then $ac = b$. Deduce that if $f$ is an even function then $f(x)$ must reduce to the form $f(x) = k$, where k is constant. Find all odd functions of the form $\frac{ax + b}{x + c}$
I have solved the first part. I am fairly sure of the second part. But I cannot solve the third part.
My calculations are as follows:
If $f(x)$ is even $f(x) = f(-x)$
$\implies \frac{ax + b}{x + c} = \frac{-ax + b}{-x + c}$
$\implies (ax + b)(c-x)= (x+c)(b-ax)$
$ac = b$
$\implies \frac{ax + b}{x + c} = \frac{ax + ac}{x + c} = \frac{a(x + c)}{x + c} = a$
So I am assuming that the a is the k to which the question refers.
Now, if $\frac{ax + b}{x + c}$ is odd, $f(-x) =-f(x)$. So
$\frac{-ax + b}{-x + c} = \frac{-(ax + b)}{x + c}$
$\implies (b-ax)(c+x)= (c-x)(-ax-b)$
$\implies 2bc=2ax^2 \implies bc=ax^2 \implies x = \sqrt\frac{bc}{a}$
but this does not lead to the answer in the book, which is $f(x)=\frac{k}{x}$
| You need $$ax^2 -bc =0 $$ for all $x$. That is, $ax^2-bc$ should be the zero polynomial, i.e. $a=0$ and $bc=0$. If $b=0$, then $$f(x) =0$$ If $c=0$, then $$f(x)=\frac bx $$ for any $b\in \mathbb R$. But setting $b=0$ gives the first case, so a general solution is $$f(x)=\frac kx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
College Algebra: Find real-valued closed formulas for the trajectory $x(t+1)=Ax(t)$ ?? Hey so I have this problem on my webwork that I do not understand:
The problem says to find real-valued closed formulas for the trajectory:
$x(t+1)=Ax(t)$ where
$A=\begin{bmatrix} -0.8 & 0.6 \\ -0.6 & -0.8 \end{bmatrix}$ and $\overrightarrow{x}(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$
My thinking is that I have to make the dynamical system $x_{k+1} = Ax_k$
I solved for the eigenvalues $\lambda_{1} = \frac{-4}{5} + \frac35i$ with its respective eigenvector $\begin{bmatrix} -i \\ 1 \end{bmatrix}$
and $\lambda_{2} = \frac{-4}{5} - \frac35i$ with its respective eigenvector $\begin{bmatrix} i \\ 1 \end{bmatrix}$
I made a formula $x_k = 1(\frac{-4}{5} + \frac35i)^k\begin{bmatrix} -i \\ 1 \end{bmatrix} + 0(\frac{-4}{5} - \frac35i)^k\begin{bmatrix} i \\ 1 \end{bmatrix}$ which evaluates to
$x_k = 1(\frac{-4}{5} + \frac35i)^k\begin{bmatrix} -i \\ 1 \end{bmatrix}$
but the answer is looking for a vector with 2 rows and 1 column.. what am I doing wrong??
| Hint: $A$ is rotation matrix for angle $\theta = \pi + \arccos 0.8 \approx 3.785$ radians, which is approximate $216.9^{\circ}$. So $x(1)$ is $x(0)$, but rotated by this angle, $x(2)$ is $x(0)$ rotated by $2\theta$ and so on. You can write in as
\begin{align}
x(k) = A^k x(0) &= \begin{pmatrix}
\cos{k(\pi + \arccos 0.8)} & -\sin{k(\pi + \arccos 0.8)}\\
\sin{k(\pi + \arccos 0.8)} & \cos{k(\pi + \arccos 0.8)}
\end{pmatrix} x(0) \\&=
\begin{pmatrix}
(-1)^k cos{(k\arccos 0.8)} & - (-1)^k\sin{(k\arccos 0.8)} \\
(-1)^k\sin{(k\arccos 0.8)} & (-1)^k \cos{(k\arccos 0.8)}
\end{pmatrix} x(0) \\&=
(-1)^{k} \begin{pmatrix}
\cos{(k\arccos 0.8)} & -\sin{(k\arccos 0.8)} \\
\sin{(k\arccos 0.8)} & \cos{(k\arccos 0.8)}
\end{pmatrix} x(0)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let X be uniform on $[0, 10]$. Let $Y$ be exponential with $E(Y ) = 5$. Find $P(X < Y )$ From the given information above. I'm able to derive that
$f(x) = \cfrac{1}{10}$ from $0\leq x \leq 10$
and
$f(y) = \cfrac{1}{5}e^{\cfrac{-1}{5}y}$ for $0 < y$
I think since they're asking for P(X < Y) it's safe to assume that X and Y are independent (is this a fair assumption?) so $f(x,y) = \cfrac{1}{50}e^{\cfrac{-1}{5}y}$ for $0 \leq x \leq 10$ and $0 < y$
so $P(X < Y) = \int_0^{\infty}\int_0^y\cfrac{1}{50}e^{\cfrac{-1}{5}y}dxdy = \cfrac{1}{2}$
Have I understood and done this problem correctly?
| $$P(X<Y)=\int_{[0,10]\times[0,\infty)}1_{x\le y}f_{X,Y}(x,y)d(x,y)$$
Now, since $X$ and $Y$ are independent, the joint density function is the product $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. Using partial integration, one finds
$$\begin{align*}
\int_{[0,10]\times[0,\infty)}1_{x\le y}f_{X,Y}(x,y)d(x,y) &=\int_0^\infty \int_0^{10} 1_{x\le y}\frac{1}{10}\frac{1}{5}e^{-\frac{1}{5}y}dxdy\\
&=\int_0^\infty\int_0^{min(10,y)} \frac{1}{10}\frac{1}{5}e^{-\frac{1}{5}y}dxdy\\
&=\int_0^{10}\frac{y}{10}\frac{1}{5}e^{-\frac{1}{5}y}dy+\int_{10}^\infty\frac{10}{10}\frac{1}{5}e^{-\frac{1}{5}y}dy\\
&=\frac{1}{50}\Big[-5ye^{-\frac{1}{5}y}\Big]_{y=0}^{y=10}-\frac{1}{50}\int_0^{10}-5e^{-\frac{1}{5}y}dy+e^{-2}\\
&=-e^{-2}+-\frac{1}{2}(e^{-2}-1)+e^{-2}\\
&=\frac{1}{2}(1-e^{-2})
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $P=\sqrt{a^2-2ab+b^2}+\left(\frac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\frac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$ is rational Show that the number $$P=\sqrt{a^2-2ab+b^2}+\left(\dfrac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\dfrac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$$ is a rational number ($P\in\mathbb{Q})$ if $a\in\mathbb{Q},b\in\mathbb{Q},a>0,b>0$ and $a\ne b$. Find the value of $P$ if $a=1.1$ and $b=1.22$.
My try: $$P=\left|a-b\right|+\dfrac{a-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}:\dfrac{b\sqrt{a}+\sqrt{b}\left(a-\sqrt{ab}\right)}{a-\sqrt{ab}}=\\=|a-b|+\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a-\sqrt{ab}}{b\sqrt{a}+a\sqrt{b}-\sqrt{b^2a}}.$$
| \begin{align}
P&=\sqrt{a^2-2ab+b^2}+\left(\dfrac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\dfrac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)=\\
&=|a-b|+\dfrac{\sqrt{ab}}{\sqrt a-\sqrt b}:\dfrac{b\sqrt a+a\sqrt b-b\sqrt a}{a-\sqrt{ab}}=\\
&=|a-b|+\dfrac{\sqrt{ab}}{\sqrt a-\sqrt b}:\dfrac{a\sqrt b}{\sqrt a\left(\sqrt a-\sqrt b\right)}=\\
&=|a-b|+\dfrac{\sqrt{ab}}{\sqrt a-\sqrt b}\cdot\dfrac{\sqrt a\left(\sqrt a-\sqrt b\right)}{a\sqrt b}=\\
&=|a-b|+\dfrac{a\sqrt b\left(\sqrt a-\sqrt b\right)}{a\sqrt b\left(\sqrt a-\sqrt b\right)}=\\
&=|a-b|+1\;.
\end{align}
Since $\;a\;$ and $\;b\;$ are rational numbers, also $\;P\;$ is a rational number.
If $\;a=1.1\;$ and $\;b=1.22\;,\;$ the value of $\;P\;$ is
$\begin{align}
P&=|a-b|+1=|1.1-1.22|+1=|-0.12|+1=\\
&=0.12+1=1.12
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Weird "hidden answer" in $2\tan(2x)=3\text{cot}(x)$ The question is
Find the solutions to the equation $$2\tan(2x)=3\cot(x) , \space 0<x<180$$
I started by applying the tan double angle formula and recipricoal identity for cot
$$2* \frac{2\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)}$$
$$\implies 7\tan^2(x)=3 \therefore x=\tan^{-1}\left(-\sqrt\frac{3}{7} \right)$$
$$x=-33.2,33.2$$
Then by using the quadrants
I was lead to the final solution that $x=33.2,146.8$ however the answer in the book has an additional solution of $x=90$, I understand the reasoning that $\tan(180)=0$ and $\cot(x)$ tends to zero as x tends to 90 however how was this solution found?
Is there a process for consistently finding these "hidden answers"?
| $$\frac{4\tan(x)}{1-\tan^2(x)} = \frac{3}{\tan(x)}$$
$$\frac{4\tan(x)}{1-\tan^2(x)} - \frac{3}{\tan(x)} = 0$$
$$\frac{4\tan^2(x)-3[1-\tan^2(x)]}{\tan(x)[1-\tan^2(x)]} = 0$$
$$\frac{7\tan^2(x)-3}{\tan(x)[1-\tan^2(x)]} = 0$$
You focused in the fact that the equation is satisfied when the numerator is zero, i.e., $7\tan^2(x)-3=0$, but the equation is also satisfied when $\tan(x)\to\infty$ (when the denominator itself tends to infinity).
$$\lim_{\tan(x)\to\infty} \frac{7\tan^2(x)-3}{\tan(x)[1-\tan^2(x)]} = \lim_{\tan(x)\to\infty} \frac{\tan^2(x)\left[7-\frac{3}{\tan^2(x)}\right]}{\tan^2(x)\left[\frac{1}{\tan(x)}-\tan(x)\right]} = \lim_{\tan(x)\to\infty} \frac{7-\frac{3}{\tan^2(x)}}{\frac{1}{\tan(x)}-\tan(x)} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Prove that $\sum_{n=0}^{\infty} \frac{n}{n+1}x^{n+1}=\frac{x}{1-x} + \ln(1-x)$ Prove that $\sum_{n=0}^{\infty} \frac{n}{n+1}x^{n+1}=\frac{x}{1-x} + \ln(1-x)$
Right off the bat, I noticed that $\frac{x}{1-x}=x\frac{1}{1-x}= x \sum_{n=0}^{\infty}x^n$, and I know that $\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}$. If somehow I can express $\ln(1-x)$ in terms of $\ln(1+x)$, then I'm set. Can I do something like $\ln(1-x)=\ln(1+(-x))=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x)^n}{n}$?
What's next? Should I add these 2 together? Can someone give me an exact answer? Also, the radius of convergence is supposed to be $(-1, 1)$ right?
| Know:
\begin{equation}
\frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n} \implies \frac{1}{1-x} - 1 = \sum_{n=1}^{\infty}x^{n}, \label{first series}
\end{equation}
which holds for $|x| < 1$, and
\begin{equation}
\log(1+x) = -\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n} \implies \log(1-x) = -\sum_{n=1}^{\infty}\frac{1}{n}x^{n}, \label{second series}
\end{equation}
also valid for $|x| < 1$. If we now look at our series of interest,
\begin{equation}
f(x) = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n+1},
\end{equation}
we can use an index substitution $k = n+1 \iff n = k-1$ so that
\begin{align}
f(x) &= \sum_{k=1}^{\infty}\frac{k-1}{k}x^{k}\\
&= \sum_{k=1}^{\infty}\left(1-\frac{1}{k}\right)x^{k},\\
\end{align}
and under the condition that our series converges, we can use the linearity of the series operator to show
\begin{equation}
f(x) = \sum_{k=1}^{\infty}x^{k} + \left(-\sum_{k=1}^{\infty}\frac{1}{k}x^{k}\right),
\end{equation}
provided both the individual series converge (which they do). Notice that on the right-hand side of the above equation, the leftmost series is equal to $1/(1-x) - 1$ (from \eqref{first series}), and the rightmost series is equal to $\log(1-x)$ (from \eqref{second series}). Therefore
\begin{align}
f(x) &= \frac{1}{1-x} - 1 + \log(1-x)\\
&= \frac{x}{1-x} + \log(1-x).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?
| Your approach is also fine. In the following it is convenient to denote with $[x^k]$ the coefficient of $x^k$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{12}]}&\color{blue}{\left(\frac{1-x^8}{1-x^2}\right)^n}\\
&=[x^{12}](1-x^8)^n\sum_{j=0}^{\infty}\binom{-n}{j}\left(-x^2\right)^j\tag{1}\\
&=[x^{12}]\left(1-\binom{n}{1}x^8\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^{2j}\tag{2}\\
&=\left([x^{12}]-n[x^4]\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^{2j}\tag{3}\\
&\,\,\color{blue}{=\binom{n+5}{6}-n\binom{n+1}{2}}\tag{4}
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion.
*In (2) we expand $(1-x^8)^n$ up to terms of $x^8$, since other terms do not contribute. We also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.
*In (4) we select the coefficients of $x^k$ accordingly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors
| A solution proceeds as follows: Denote $t:=(1/x)+(1/y)$. Observe that $$t^3=\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\cdot t.$$ For the given equation, you may divide $x^3y^3$ on both sides, which gives $$1=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}.$$ Subtracting these two formulas, we can obtain $$t^3-1=\frac{3(t-1)}{xy}.$$ If $t=1$, this formula holds definitely. If $t\neq 1$, then $$t^3-1=\frac{3(t-1)}{xy}\implies t^2+t+1=\frac{3}{xy}\implies\frac{1}{x}\cdot\frac{1}{y}=\frac{t^2+t+1}{3}.$$ If you regard $u:=1/x$ and $v:=1/y$, then $u$ and $v$ are solutions to the quadratic equation $$z^2-tz+\frac{t^2+t+1}{3}=0.$$ Since $x$ and $y$ are real numbers, both $u$ and $v$ are also real numbers. This implies that the discriminant of this quadratic equation is non-negative. Thus $$\Delta=t^2-\frac{4(t^2+t+1)}{3}=-\frac{(t+2)^2}{3}\geq 0, $$ hence $t=-2$ if $t\neq 1$.
In general, $t=-2$ or $t=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving global minimum by lower bound of 2-variable function $f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y$ I would like to prove that the following function $f :\mathbb{R}^2\to\mathbb{R}$ has a global minimum:
$f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y=(x^2+y)^2-4(x^2+2x+2y)$
$f$ has strict local minimum at
$f(1,3)=-20$
I think that what I need to show is that $-20$ is a lower bound of this function, and then conclude that's a global minimum, but I didn't manage to do so.
Please advise.
Thank you.
| You can find the stationary points:
\begin{align}
\frac{\partial f}{\partial x}&=4x^3+4xy-8x-8 \\[6px]
\frac{\partial f}{\partial y}&=2x^2+2y-8
\end{align}
At a critical point $y=4-x^2$ and also
$$
x^3+x(4-x^2)-2x-2=0
$$
that is, $x=1$, that implies $y=3$.
Since clearly the function is upper unbounded on the line $y=0$, we just need to show it is lower bounded. Conjecturing that the stationary point is a minimum, we have $f(1,3)=-20$, we need to see whether $f(x,y)\ge-20$.
Now let's try completing the square in
$$
y^2+2(x^2-4)y+x^4-4x^2-8x+20
$$
Since $(x^2-4)^2=x^4-8x^2+16$, we have
$$
f(x,y)+20=(y+x^2-4)^2+4x^2-8x+4=(y+x^2-4)^2+4(x-1)^2
$$
which is everywhere nonnegative, so we proved that $f(x,y)\ge-20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
the integral PDF $\sqrt {{2 \over \pi }} \int_{t\, = \,0}^\infty {e^{\, - \,{1 \over 2}\left( {v^{\,2} /t^{\,2} + 2t} \right)} dt\,} $ In my answer to this post I came to the conclusion that the PDF of the volume of a parallelepiped with normal distributed coordinates,
having one vertex at the origin, is
$$
p_t (v) = \sqrt {{2 \over \pi }} \int_{t\, = \,0}^\infty
{e^{\, - \,{1 \over 2}\left( {v^{\,2} /t^{\,2} + 2t} \right)} dt\,}
$$
with the corresponding CDF
$$
P_{\,t} (v) = \int_{t = 0}^\infty
{\,\,t\;e^{\, - \,\,t} \,{\rm erf}\left( {{v \over {\sqrt 2 \,t}}} \right)dt\,}
$$
where $0 \le v$.
I wonder whether these two integrals may have an interesting expression in terms of known functions.
| In terms of the Meijer G function
$$p_t (v) =\frac{\sqrt{2}}{\pi }\,\, G_{0,3}^{3,0}\left(\frac{v^2}{8}|
\begin{array}{c}
0,\frac{1}{2},1
\end{array}
\right)$$
$$P_t (v) =\frac{2}{\pi }\,\, G_{1,4}^{3,1}\left(\frac{v^2}{8}|
\begin{array}{c}
1 \\
\frac{1}{2},1,\frac{3}{2},0
\end{array}
\right)$$
Edit
In terms of series, for small values of $v$
$$p_t (v)=\sqrt{\frac{2}{\pi }}-v+$$ $$\frac{v^2 \left(-2 \sqrt{2} \log (v)-2 \sqrt{2} \gamma
+\sqrt{2}+\sqrt{2} \log (8)+\sqrt{2} \psi
^{(0)}\left(-\frac{1}{2}\right)\right)}{4 \sqrt{\pi
}}+$$ $$\frac{v^3}{6}+O\left(v^4\right)$$
$$P_t (v)=\sqrt{\frac{2}{\pi }} v-\frac{v^2}{2}+\frac{v^3 (-6 \log (v)-9 \gamma +11+\log
(8))}{18 \sqrt{2 \pi }}+O\left(v^4\right)$$
For large values of $v$, using $v=t^3$
$$p_t (v)=\frac{4t}{\sqrt{3}}\,e^{-\frac{3 t^2}{2}} \left(1+\frac{5}{18 t^2}-\frac{35}{648 t^4}+\frac{665}{34992
t^6}+O\left(\frac{1}{t^8}\right)\right)$$
$$P_t (v)=1-\frac{34}{9 \sqrt{3}}e^{-\frac{3 t^2}{2}}\left(\frac{18 t^2}{17}+1-\frac{35}{612 t^2}+\frac{1925}{33048
t^4}+O\left(\frac{1}{t^6}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4098056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximize the value of $(a,b) + (b,c) + (a,c)$ given is $a + b + c = 5n$; $a,b,c,n$ are positive integers
$(a,b)$ stands for the gcd of $a$ and $b$
Maximize the value of $(a.b) + (b,c) + (a,c)$
So I found that since $a \geq(a,x)$, that $5n \geq (a.b) + (b,c) + (a,c)$
And for $n=0 $ mod 3 , we can set $a=b=c$, and find the maximum. I've been trying to do the cases $n$=1 and 2 mod 3 separately, but I haven't made any progress with those cases.
| We are going to explore how large the fraction $\frac{\gcd(a,b) + \gcd(b,c) + \gcd(c,a)}{a+b+c}$ can be, where $a,b,c$ are any positive integers such that all of them are not equal. (We forget about the sum being $5n$).
Case $1: a=b$.
Notice $\gcd(a,b) + \gcd(b,c) + \gcd(c,a) = a + \gcd(a,c) + \gcd(c,a)$.
If $a$ is a multiple of $c$ then we get $kc + 2c = (k+2)c$ which when compared to $2a+c = (2k+1)c$ gives a ratio of at most $\frac{4}{5}$.
If $c$ is a multiple of $a$ then we get $a + 2a = 3a$ which when compared to $2a+c = a(2+k)$ gives a ratio of at most $\frac{3}{5}$.
If neither number is a multiple of another then $a + \gcd(a,c) + \gcd(c,a) \leq a + a/2 + c/2 = \frac{3a}{2} + c/2 $ which when compared to $2a+c$ gives a ratio of at most $\frac{3}{4}$
If no number is equal to another without loss of generality $a<b<c$ and now notice:
$\gcd(a,b) + \gcd(b,c) + \gcd(c,a) \leq b/2 + c/2 + a < \frac{2}{3}(a+b+c)$
So the best ratio possible is $\frac{4}{5}$ with equality if and only if $a=b=2c$, in other words if the solution is of form $(n,2n,2n)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{\infty} \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}\>dx$ How to Integrate
$$ I = \int_{0}^{\infty} \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \>dx\approx -0.295512 $$
Mathematica returns a result that does not match numerically with the integral approximation :
$$ \frac{i C}{4} -\frac{\pi}{4}\sqrt{4-3i} +\frac{3 \pi^2}{32} +\pi\left(\frac{1}{4}-\frac{i}{8}\right)\coth^{-1}(\sqrt{2}) \approx -0.048576 + 0.43823\,i $$
Where C denotes Catalan's Constant
Motivation
I was able to find a closed form for
$$ \int_{0}^{1} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx $$
using double infinite sums. Upon plotting the function within the integral i saw that it could be integrated from $0$ to ${\infty}$ .
Attempts
Number 1
I tried to Split the integral as
$$ \int_{0}^{1} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx + \int_{1}^{\infty} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx $$
and use the Taylor Series for $\tan^{-1}(x^2) $ when $|x| >1$
Number 2
I tried using partial fractions as
$$ \frac{1}{x^4-1} = \frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^2+1)} $$
$$ \frac{1}{x^2(x^4-1)} = \frac{1}{2(x^2+1)}-\frac{1}{x^2}-\frac{1}{4(x+1)} + \frac{1}{4(x-1)} $$
from which I obtained
$$\int_{0}^{\infty} \frac{\tan^{-1}(x^2)}{2(x^2+1)}dx = \frac{\pi^2}{16} $$
$$ \int_{0}^{\infty} \frac{\pi}{8(x^2+1)} dx= \frac{\pi^2}{8} $$
but was unable to proceed further.
Number 3
A number of basic integration techniques such as U-Sub and Integration by parts.
Number 4
Using the same technique i used to evaluate the same integral but from $0$ to $1$
I will continue to try , but for now I find myself to be stuck.
Q - Is there a closed form for I? If the solution is easy and i am missing something , could you provide hints instead?
Thank you for your help and time.
| Note
$$\frac1{x^4-1} =\frac1{2(x^2-1)} - \frac1{2(x^2+1)}\\
\frac1{x^2(x^4-1)} =\frac1{2(x^2-1)} + \frac1{2(x^2+1)}-\frac1{x^2}
$$
and rewrite the integral as
\begin{align}
I &= \int_{0}^{\infty} \left( \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \right)dx\\
&=\frac{\pi^2}{16}+ \frac12\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx
- \int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx
+ \frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\tag1
\end{align}
where $\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx=\frac{\pi^2}8$, $
\int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx=\frac{\pi}{\sqrt2}$ and
\begin{align}
&\frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx
=\int_0^{1} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\\
\overset{IBP}=&-\int_0^1 \frac x{1+x^4} \ln \frac{1-x}{1+x}dx
\overset{x\to \frac{1-x}{1+x}}=-\int_0^\infty\frac {\ln x}{1+6x^2+ x^4}dx\\
=&-\frac1{4\sqrt2}\left( \int_0^\infty\frac{\ln x}{x^2+ (\sqrt2-1)^2}dx -\int_0^\infty\frac{\ln x}{x^2+ (\sqrt2+1)^2}dx \right)\\
=&-\frac1{8\sqrt2}\left(\frac{\ln(\sqrt2-1)}{\sqrt2-1} - \frac{\ln(\sqrt2+1)}{\sqrt2+1} \right)
\end{align}
Substitute above results into (1) to obtain
$$I= \frac{\pi^2}8-\frac\pi{\sqrt2}\left(1+\frac{\ln(\sqrt2-1)}{8(\sqrt2-1)} - \frac{\ln(\sqrt2+1)}{8(\sqrt2+1)} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the Integral $\int_{-\infty}^\infty e^{-x^2}\cos\big(2x^2\big)\,\mathrm dx$ How to compute the integral
$$\int_{-\infty}^\infty e^{-x^2}\cos\big(2x^2\big) dx$$
I am wondering if there's a nice closed form of this elegant integral. I have tried to compute this integral using some substitutions, Laplace and Mellin transforms, however it doesn't seem to get or transform to something more simplified.
Any approach (including complex analysis) is most welcomed.
Thanks.
EDIT $\textbf{1}$: @heropup has provided a beautiful answer, however I would be much more happy if there's another nice way to prove the same. Thanks.
| The first thing that comes to my mind is to do something like this: $$e^{-x^2} \cos (2x^2) = e^{-x^2} \frac{e^{2x^2 i} + e^{-2x^2 i}}{2} = \frac{1}{2} \left( e^{-(1-2i)x^2} + e^{-(1+2i)x^2} \right),$$ then use the fact that $$\int_{x=-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}$$ to evaluate $$\int_{x=0}^\infty e^{-zx^2} \, dx = \frac{\sqrt{\pi}}{2 \sqrt{z}}, \quad \Re(z) > 0.$$ Then we obtain $$\int_{x=-\infty}^\infty e^{-x^2} \cos (2x^2) \, dx = \frac{\sqrt{\pi}}{2} \left((1-2i)^{-1/2} + (1+2i)^{-1/2}\right) = \sqrt{\frac{1 + \sqrt{5}}{10} \pi}.$$
In case that last step is unclear, one simply writes
$$\frac{1}{\sqrt{1 - 2i}} + \frac{1}{\sqrt{1 + 2i}} = \frac{\sqrt{1 + 2i} + \sqrt{1 - 2i}}{\sqrt{5}}.$$ Then because $$1 \pm 2i = \sqrt{5} \left(\frac{1}{\sqrt{5}} \pm \frac{2}{\sqrt{5}} i\right) = \sqrt{5} e^{\pm i \theta}$$ where $\theta = \tan^{-1} 2$, we have $$\sqrt{1 + 2i} + \sqrt{1 - 2i} = 5^{1/4}(e^{i\theta/2} + e^{-i\theta/2}) = 2 \cdot 5^{1/4} \cos \frac{\theta}{2} = 2 \cdot 5^{1/4} \sqrt{\frac{1 + \cos \theta}{2}} \\ = 2 \cdot 5^{1/4} \sqrt{\frac{1 + 1/\sqrt{5}}{2}}.$$ The rest is algebra. I believe such a computation should be accessible to a student of complex analysis--indeed, a student of high school trigonometry and complex number arithmetic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$ $$\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$$
I can reduce it to the general term,
$$\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)2^r}$$
I don't know how to go about this any further though. I also ran this in python and the sum is exceeding $1.5$ for $10,000$ terms, which is weird since it should converge to $1.5$, so it makes me doubt if the general term I've written is correct.
| Consider $$f(x)=\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)}x^r=\sum_{r=1}^\infty \left(\frac{1}{r+1}-\frac{1}{r+2}+\frac{2}{r} \right)x^r$$
$$f(x)=\frac{\log (1-x)}{x^2}+\frac{1}{x}-\frac{\log (1-x)}{x}-2 \log
(1-x)-\frac{1}{2}$$
$$f\left(\frac{1}{2}\right)=\frac 3 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluate:- $\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
Evaluate:- $\dfrac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
What I Tried:- Let $a = 4 , b = \sqrt{15} , c = 6, d= \sqrt{35}$ . Then I get :-
$$\rightarrow \frac{[a + b]^{3/2} + [a - b]^{3/2}}{[c + d]^{3/2} - [c - d]^{3/2}}$$
Now I can put the formulas $(a^3 + b^3)$ and $(c^3 - d^3)$
$$\rightarrow \frac{[(a + b)^{1/2} + (a - b)^{1/2}][(a + b) - \sqrt{a^2 - b^2} + (a - b)]}{[(c + d)^{1/2} - (c - d)^{1/2}][(c + d) + \sqrt{c^2 - d^2} + (c - d]}$$
$$\rightarrow \frac{7[(a + b)^{1/2} + (a - b)^{1/2}]}{13[(c + d)^{1/2} - (c - d)^{1/2}]}$$
From here, I do not know how to proceed. Can anyone help me?
| $$\sqrt{8+2\sqrt{15}}=\sqrt5+\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why can we use $\cos^2t+\sin^2t=1$ to eliminate $t$ from $x=5\cos t$, $y=2\sin t$? Given the parametric equations $x=5 \cos t$ and $y=2 \sin t$, I want to eliminate the parameter.
So, how can we eliminate the parameter here? Supposedly, all we need to do according to the solution is to recall this trig identity:
$$
\cos ^{2} t+\sin ^{2} t=1
$$
Then from the parametric equations we get,
$$
\cos t=\frac{x}{5} \quad \sin t=\frac{y}{2}
$$
Then, using the trig identity from above and these equations we get,
$$
1=\cos ^{2} t+\sin ^{2} t=\left(\frac{x}{5}\right)^{2}+\left(\frac{y}{2}\right)^{2}=\frac{x^{2}}{25}+\frac{y^{2}}{4}
$$ and should thus conclude that we haven an ellipse.
However, I don't understand why this works. We just took some equation (trig identity) and plugged something in - how do we know that this is equal to our parametric equations? I mean we could have taken any other formula and plug in values and make a completely different conclusion - it seems very arbitrary to me.
| This is because $\left(\frac{x}{5}\right)^{2} + \left(\frac{y}{2}\right)^{2}$ matches the form of $\cos^{2}t + \sin^{2}t$. But before this, let's understand what $\cos^{2}t + \sin^{2}t = 1$ mean.
We know that the center-radius (standard) form of a circle of radius $r$ with its center at the origin has the equation $$x^{2} + y^{2} = r^{2}.$$
Dividing both sides by $r^{2}$,
$$\begin{align*}\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} &= 1 \\ \left(\frac{x}{r}\right)^{2} + \left(\frac{y}{r}\right)^{2} &= 1.\end{align*}$$
But we know that $\cos t =\frac{x}{r}$ and $\sin t = \frac{y}{r}$ where $t$ is the angle from the positive $x$-axis. By substitution, we get the equation $$\cos^{2}t + \sin^{2}t = 1 \tag{1}.$$
We will now go to ellipses. Circles as special cases of ellipses where both $x$ and $y$ are scaled by a factor of $r$. However, ellipses that are not circles have different scaling factors for $x$ and $y$. If the scaling factor for $x$ and $y$ are $a$ and $b$, the equation will be $$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} = 1 \tag{2}.$$
As $(1)$ is similar to $(2)$, we can equate the terms to each other.
\begin{align*}\left(\frac{x}{a}\right)^{2} &= \cos^{2}t &\qquad \left(\frac{y}{b}\right)^{2} &= \sin^{2}t \\ \frac{x}{a} &= \cos t &\qquad \frac{y}{b} &= \sin t \\ x &= a\cos t &\qquad y &= b \sin t.\end{align*}
This is why it works. I can't think of anything aside from this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Fourth root of a $3\times 3$ matrix I have solved this question by generalizing my assumption about some properties of the given matrix. However, it is not a rigorous proof and I haven't fully understood the background/meaning of it.
Here's the question. Find the matrix X satisfying
$ X^4=
\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0\\
\end{bmatrix}
$
I found that
$(X^4)^n=\begin{bmatrix}3^n&0&0\\0&3^n&3^{n-1}\\0&0&0\\\end{bmatrix}$
Therefore I assumed
$X^n=\begin{bmatrix}3^\frac{n}{4}&0&0\\0&3^\frac{n}{4}&3^\frac{n-1}{4}\\0&0&0\\\end{bmatrix}$
Although I am on the entry-level in linear algebra, any approach to this question is welcomed, I will try my best to understand it.
| Note that $X^4$ is diagonalizable with eigenspaces
\begin{align}
E_3(X^4) &= \langle (1,0,0), (0,1,0) \rangle, \\
E_0(X^4) &= \ker(X^4) = \langle (0,1,-3)\rangle.
\end{align}
Hence, after a change of basis the problem translates to finding $Y$ with
$$
Y^4 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0\end{pmatrix}.
$$
Since $Y^4$ is singular, $Y$ needs to be singular as well and since $\ker(Y)\subseteq\ker(Y^4)$ we know that $\ker(Y)=\ker(Y^4)=\langle(0,0,1)\rangle$, so the third column of $Y$ must be zero.
Since the dimensions of the kernels of $Y$ and $Y^4$ are equal, the dimensions of the images have to be equal as well and we get $\operatorname{im}(Y)=\operatorname{im}(Y^4)=\langle (1,0,0), (0,1,0)\rangle$.
Hence, we have
$$
Y = \begin{pmatrix} * & * & 0 \\ * & * & 0 \\ 0 & 0 & 0\end{pmatrix},
$$
where $Z=(\begin{smallmatrix}*&*\\*&*\end{smallmatrix})$ is any solution to $Z^4=3I$, where $I$ is the $2\times 2$ identity matrix. Such matrices are of the form
$$
Z = U^{-1} \begin{pmatrix} \eta_1 & 0 \\ 0 & \eta_2\end{pmatrix} U,
$$
where $\eta_1,\eta_2$ are any (possibly equal) solutions to $\eta^4=3$ and $U$ is an invertible $2\times 2$ matrix, so $U=(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ with $ad-bc\neq 0$.
Putting things together, we first get
\begin{align}
Z &= \frac{1}{ad-bc} \begin{pmatrix}d&-b\\-c&a\end{pmatrix}\begin{pmatrix} \eta_1 & 0 \\ 0 & \eta_2\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix} \\
&= \frac{1}{ad-bc} \begin{pmatrix} ad\eta_1-bc\eta_2 & bd(\eta_1-\eta_2)\\ac(\eta_2-\eta_1) & ad\eta_2-bc\eta_1\end{pmatrix}.
\end{align}
Then
$$
Y = \frac{1}{ad-bc} \begin{pmatrix} ad\eta_1-bc\eta_2 & bd(\eta_1-\eta_2) & 0 \\ ac(\eta_2-\eta_1) & ad\eta_2-bc\eta_1 & 0 \\ 0 & 0 & 0\end{pmatrix},
$$
and changing back to the original basis finally
$$
X = \frac{1}{ad-bc} \begin{pmatrix} ad\eta_1-bc\eta_2 & bd(\eta_1-\eta_2) & \frac 1 3 b d (\eta_1-\eta_2) \\ ac(\eta_2-\eta_1) & ad\eta_2-bc\eta_1 & \frac 1 3 (ad\eta_2-bc\eta_1)\\ 0 & 0 & 0\end{pmatrix}.
$$
The constraints are $ad-bc\neq 0$ and $\eta_i^4=3$ for $i=1,2$.
The solution obtained by you (with the correction of Henry) is obtained by choosing $\eta_1=\eta_2=3^{\frac 1 4}$ and $U=I$, so $a=d=1$, $b=c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $f(x,y) =\sqrt{16-x^2-y^2}$ is continuous using $\epsilon$, $\delta$? I want to prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that
$$(x-x_0)^2 + (y-y_0)^2 < \delta^2 \Rightarrow \left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| < \epsilon.$$
I note that $|x-x_0| \leq \delta$ and $|y-y_0| \leq \delta$.
What I've tried:
$$\left| \sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| = \frac{|x^2-x_0^2 + y^2-y_0^2|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} }$$
$$= \frac{|(x-x_0)(x+x_0) + (y-y_0)(y+y_0)|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} } \leq
\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} }$$
$$\leq
\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x_0^2-y_0^2} }$$
But then I do not know how to continue.
| HINT
You are on the right track! Notice that
\begin{align*}
\begin{cases}
|x + x_{0}| = |(x - x_{0}) + 2x_{0}| \leq |x - x_{0}| + 2|x_{0}|\\\\
|y + y_{0}| = |(y - y_{0}) + 2y_{0}| \leq |y - y_{0}| + 2|y_{0}|
\end{cases}
\end{align*}
Consequently, if we set $k^{-1} = \sqrt{16 - x^{2}_{0} - y^{2}_{0}}$, then we have that
\begin{align*}
k(|x - x_{0}||x + x_{0}| + |y - y_{0}||y + y_{0}|) \leq 2k\delta^{2} + 2k\delta(|x_{0}| + |y_{0}|) := \varepsilon
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction on $S(n,k)$ Consider $$S(n,k):= 1^{k}+\ldots+n^{k}.$$
I have to find a formula for $$S(n,3).$$
After trying for some n values I found out that
$$S(1,3)=1^{3}=1^{2}$$
$$S(2,3)=1^{2} +2^{2}=3^{2}$$
$$S(3,3)=1^{2} +2^{2} +3^{3}=6^{2}$$
$$S(4,3)=1^{2} +2^{2} +3^{3}+4^{3}=10^{2}$$
$$S(5,3)=1^{2} +2^{2} +3^{3}+4^{3}+5^{3}=15^{2}$$
Which led to the conjecture that $$S(n,3)=\left(1+\ldots+n\right)^{2}$$
I know for a fact that $$S(n,1)=\frac{n^{2}}{2}+\frac{n}{2}$$
Therefore,
$$S(n,3)=\left(\frac{n^{2}}{2}+\frac{n}{2}\right)^2$$
But when trying to prove the formula using induction I get stuck on the inductive step.
$$S(k+1,3)=1^{3}+\ldots+k^{3}+(k+1)^3=\left(\frac{(k+1)^{2}}{2}+\frac{k+1}{2}\right)^2$$
I don't know how to proceed. I have tried manipulating the RHS, but it's been fruitless.
| Assuming that $S(k,3)=(\frac{k^2}{2}+\frac{k}{2})^2$ from the induction step, $$S(k+1,3)=1^3+\dots k^3 +(k+1)^3 = S(k,3)+(k+1)^3 = \frac{k^4+2k^3+k^2}{4}+(k+1)^3=\frac{k^4+2k^3+k^2+4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{(k^4+k^3)+(5k^3+5k^2)+(8k^2+8k)+(4k+4)}{4}=\frac{(k+1)(k^3+5k^2+8k+4)}{4}=\frac{(k+1)((k^3+k^2)+(4k^2+4k)+(4k+4))}{4}=\frac{(k+1)^2(k^2+4k+4)}{4}=\frac{((k+1)(k+2))^2}{4}=(\frac{(k+1)^2+(k+1)}{2})^2$$ which is what we were looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove by induction $\sum\limits_{k=0}^{n}( k+1)( n-k+1) =\binom{n+3}3$ Prove by induction :
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\binom{n+3}{3}$$
induction basis: $n=0$
$$\displaystyle \sum _{k=0}^{n}( 0+1)( 0-0+1) =1=\binom{3}{3}$$
For $n+1$:
\begin{aligned}
\sum _{k=0}^{n+1}( n+1+1)( n+1-(n+1)+1) & =\sum _{k=0}^{n}( k+1)( n-k+1) +( n+2)(( n+1) -( n+1) +1)\\
& =\binom{n+3}{3} +( n+2)\\
\end{aligned}
I had to stop here because I realize there is a mistake ...
Unfortunately, I didn't succeed in many ways.
| You set up the induction correctly by getting the base case and stating your induction hypothesis.
For the induction step:
$\sum\limits_{k=0}^{n+1}(k+1)((n\color{red}{+1})-k+1)$ You seem to have forgotten this red $\color{red}{+1}$.
So, we have $\sum\limits_{k=0}^{n+1}(k+1)((n\color{red}{+1})-k+1) = \sum\limits_{k=0}^n(k+1)(n-k+1)\color{red}{+\sum\limits_{k=0}^n(k+1)}+(n+2)$.
This should hopefully get you back on track.
The first summation simplifies to $\binom{n+3}{3}$ per induction hypothesis, the second summation simplifies to $\binom{n+2}{2}$ recognizing it as the $n+1$'st triangular number. So we have $\binom{n+3}{3}+\binom{n+2}{2}+\binom{n+2}{1}=\binom{n+3}{3}+\binom{n+3}{2}=\binom{n+4}{3}=\binom{(n+1)+3}{3}$ using Paschal's identity to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4125660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\cos ^{-1} x-\cos ^{-1} y$ $$
\cos ^{-1} x-\cos ^{-1} y=\left\{\begin{array}{l}
\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) ; \text { if }-1 \leq x, y \leq 1 \quad \text{and} \quad x \leq y \\
-\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) ; \text { if }-1 \leq y \leq 0,0<x \leq 1 \quad \text{and} \quad x \geqslant 1
\end{array}\right.
$$
I'm having some issues proving for different cases, this is what I tried so far:
Let $\cos ^{-1} x=\alpha, \quad \cos ^{-1} y=\beta \quad \Longrightarrow \quad x=\cos \alpha, y=\cos \beta$
$$
\begin{aligned}
\cos (\alpha-\beta) &=\cos \alpha \cos \beta+\sin \alpha \sin \beta \\
&=\cos \alpha \cos \beta+\sqrt{1-\cos ^{2} \alpha} \sqrt{1-\cos ^{2} \beta} \\
&=\left(x y+\sqrt{1-\cos ^{2} \alpha} \sqrt{1-\cos ^{2} \beta}\right)
\end{aligned}
$$
$$
\begin{aligned}
\therefore \alpha-\beta &=\cos ^{-1} x-\cos ^{-1} y \\
&=\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)
\end{aligned}
$$
| $\cos^{-1}z$ will be real $\iff-1\le z\le1$
Now as $0\le\cos^{-1}x,\cos^{-1}y\le\pi,$
$$-\pi<\cos^{-1}x-\cos^{-1}y\le\pi$$
$$\cos^{-1}(xy+\sqrt{(1-x^2)(1-y^2)})\text{ will be }=\cos^{-1}x-\cos^{-1}y$$
$$\iff\cos^{-1}x-\cos^{-1}y\ge0$$
$$\iff\dfrac\pi2-\sin^{-1}x\ge\dfrac\pi2-\sin^{-1}y\iff\sin^{-1}x\le\sin^{-1}y\iff x\le y$$ as $\sin^{-1}x$ is an increasing function
Similarly, $$\cos^{-1}(xy+\sqrt{(1-x^2)(1-y^2)})\text{ will be }=\cos^{-1}y-\cos^{-1}x\iff
y\le x$$
See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ | Since $|\sin t| \leq |t|$ for all $t\in\mathbb{R}$,
$$
|x^2 y \sin(xy^2)| \leq |x|^3 |y|^3 \leq \frac{x^6+y^6}{2}\,.
$$
the second inequality by the AM-GM inequality. This means
$$
\frac{|x^2 y \sin(xy^2)|}{(x^4+y^4)\sqrt{x^2+y^2}}
\leq \frac{1}{2} \frac{(x^6+y^6)}{(x^4+y^4)\sqrt{x^2+y^2}}
\leq \frac{1}{2}\frac{2\max(x^6,y^6)}{\max(x^4,y^4)\sqrt{\max(x^2,y^2)}}
= \max(|x|,|y|)
$$
and $\lim_{(x,y)\to (0,0)} \max(|x|,|y|) = 0$, so you can conclude by the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Recursive geometric series to explicit formula I'm trying to figure out how to convert this recursive formula $A_n = \frac{8}{9} A_{n-1} + \frac{6}{5} \left(\frac{20}{9}\right)^n $ into $A_n = 4 \left(\frac{8}{9}\right)^n + 2 \left(\frac{20}{9}\right)^n$ and I'm not sure how to approach it.
The first thing I thought of was to try to find a pattern by substituting a previous term into the next one
\begin{align*}
A_n & = \frac{8}{9} A_{n-1} + \frac{6}{5} \left(\frac{20}{9}\right)^n \\
A_{n+1} & = \frac{8}{9} \left( \frac{8}{9} A_{n-1} + \frac{6}{5} \left(\frac{20}{9}\right)^n\right) + \frac{6}{5} \left(\frac{20}{9}\right)^{n+1} \\
& = \left(\frac{8}{9}\right)^2 A_{n-1}+ \frac{6}{5}\frac{8}{9}\left(\frac{20}{9}\right)^n + \frac{6}{5} \left(\frac{20}{9}\right)^{n+1} \\
& = \left(\frac{8}{9}\right)^2 A_{n-1}+ \frac{6}{5}\left(\frac{20}{9}\right)^n \left(\frac{8}{9}+\frac{20}{9}\right)
\end{align*}
But that didn't seem to work, although it's possible I simplified incorrectly (or not in the way you're supposed to)
--
Actually, I think I forgot a crucial piece of information: $A_0 = 6$
| $$A_{n+1}=aA_n+bc^n.$$
Try a solution of the form $dc^n$. You have
$$dc^{n+1}=adc^n+bc^n,$$ giving $$d=\frac b{c-a}.$$
Now by subtraction,
$$A_{n+1}-dc^{n+1}=aA_n+bc^n-dc^{n+1}$$
is
$$(A_{n+1}-dc^{n+1})=a(A_n-dc^n),$$
which is easy to solve for
$$A_n-dc^n$$ as it is a geometric progression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$ $$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
I tried $$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\leq\frac{n}{\sqrt{n^2+n+1}}$$
But$$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+n}}\right)^n=\lim_{n\to\infty}\left(1+\frac2n\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac2n)}=\frac1e$$
And in the same way,i got $$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+1}}\right)^n=\lim_{n\to\infty}\left(1+\frac1n+\frac{1}{n^2}\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac1n+\frac{1}{n^2})}=\frac{1}{\sqrt e}$$
So i only got $$\frac1e\leq I\leq\frac{1}{\sqrt e}$$
Could someone help me get the value of $I$. Thanks!
| I use the Binomial expansion repeatedly without using logs, as in Tavish's answer. I just want to show that such a method works for this question. The method is long, but it is neither difficult to understand nor ugly.
Write $\ t = n^2 + n.\ $ Then we have
\begin{align}\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\ldots+\frac{1}{\sqrt{n^2+n+n}}\\
\\
= \frac{1}{\sqrt{t+1}}+\frac{1}{\sqrt{t+2}}+\ldots+\frac{1}{\sqrt{t+n}}\\
\\
=\frac{1}{\sqrt{t}}\left( \frac{1}{\sqrt{1+\frac{1}{t}}} + \frac{1}{\sqrt{1+\frac{2}{t}}} +\ldots + \frac{1}{\sqrt{1+\frac{n}{t}}} \right)\\
\\
=\frac{1}{\sqrt{t}}\left( \left(1+\frac{1}{t}\right)^{-\frac{1}{2}} + \left(1+\frac{2}{t}\right)^{-\frac{1}{2}} +\ldots + \left(1+\frac{n}{t}\right)^{-\frac{1}{2}} \right)\\
\\
=\frac{1}{\sqrt{t}}\left(\left[ 1+\left(-\frac{1}{2}\right)\left(\frac{1}{t}\right) + \ldots \right] + \left[ 1+\left(-\frac{1}{2}\right)\left(\frac{2}{t}\right) + \ldots \right] +\ldots + \left[ 1+\left(-\frac{1}{2}\right)\left(\frac{n}{t}\right) + \ldots \right] \right)\\
\\
=\frac{1}{\sqrt{t}}\left( 1\times n + \left(-\frac{1}{2}\right)\times\left(\frac{1}{t}\right)\times \left( \displaystyle\sum_{i=1}^n i \right) + O\left(\frac{1}{n}\right) \right)\\
\\
=\frac{1}{\sqrt{n^2+n}} \left( n + \frac{\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)n(n+1)}{n^2+n} + O\left(\frac{1}{n}\right) \right) \\
\\
=\frac{n-\frac{1}{4}+O\left(\frac{1}{n}\right)}{\sqrt{n^2+n}}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right)\left( \sqrt{n^2+n} \right)}{n^2+n}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right) \times n \times \left( \sqrt{1+\frac{1}{n}} \right)}{n^2+n}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right) \times \left( \left(1+\frac{1}{n}\right)^{\frac{1}{2}} \right)}{n+1}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right) \times \left(1+\left(\frac{1}{2}\right) \left(\frac{1}{n}\right) + O\left(\frac{1}{n^2}\right) \right)}{n+1}\\
\\
=\frac{n+\frac{1}{4}+O\left(\frac{1}{n}\right)}{n+1} \\
\\
=1 - \left(\frac{3}{4}\right)\frac{1}{n+1} + O\left(\frac{1}{n^2}\right). \\
\\
\end{align}
Therefore,
\begin{align}
I=\displaystyle\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n\\
\\
=\displaystyle\lim_{n\to\infty}\left(\ \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right) + O\left(\frac{1}{n^2}\right)\ \right)^n\\
\\
=\displaystyle\lim_{n\to\infty}\left(\ \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^n + \underset{\Large{\to 0 \text{ as } n\to \infty}}{\underbrace{\displaystyle\sum_{k=1}^n \binom{n}{k} \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^{n-k} O\left(\frac{1}{\left(n^2\right)^k}\right)}}\ \right)\\
\\
=\displaystyle\lim_{n\to\infty}\left(\ \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^{n+1}\ \cdot\ \underset{\Large{\to 1 \text{ as } n\to \infty}}{\underbrace{\left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^{-1}}} \right)\\
\\
=\displaystyle\lim_{(n+1)\to\infty}\left(\ \left(1 + \left(-\frac{3}{4}\right)\frac{1}{n+1}\right)^{n+1} \right)\\
\\
=e^{-\frac{3}{4}}.\\
\end{align}
Within the proof, I actually left out a couple of details for the sake of brevity. For example, getting from line $5$ to line $6$ isn't immediately obvious, but is in fact not too difficult to prove.
For the final step, see here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 0
} |
Using definite integral to find the limit. I would like someone to verify this exercise for me. Please.
Find the following limit:
$\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\right)$
$=\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+2n}\right)$
$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n+k}$
$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n\left(1+\frac{k}{n}\right)}$
$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n}\left(\dfrac{1}{1+\frac{k}{n}}\cdot\dfrac{1}{n}\right)$
$=\displaystyle\int_{1+0}^{1+2} \frac{1}{x} \,dx$
$=\displaystyle\int_{1}^{3} \frac{1}{x} \,dx$
$=\big[\ln|x|\big] _{1}^3$
$=\ln|3|-\ln|1|$
$=\ln(3)-\ln(1)$
$=\ln(3)$
| \begin{align}S_n&=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\\
&=\sum_{k=1}^{2n}\frac{1}{n+k}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}+\underbrace{\sum_{k=n+1}^{2n}\frac{1}{n+k}}_{l=k-n}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}+\sum_{k=1}^{n}\frac{1}{2n+l}\\
&=\sum_{k=1}^{n}\left(\frac{1}{n+k}+\frac{1}{2n+k}\right)\\
&=\frac{1}{n}\sum_{k=1}^{n}\left(\frac{1}{1+\frac{k}{n}}+\frac{1}{2+\frac{k}{n}}\right)\\
\end{align}
Therefore,
\begin{align}\lim_{n\rightarrow \infty}S_n&=\int_0^1 \left(\frac{1}{1+x}+\frac{1}{2+x}\right)dx\\
&=\Big[\ln(1+x)+\ln(2+x)\Big]_0^1\\
&=\ln 2+\ln 3-0-\ln 2\\
&=\boxed{\ln 3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to show that: $(-1)^nn+2+\sum_{k=1}^{n}\frac{(-1)^k}{k+1}(2)_{k+1}S_{n}^{k+1}H_{k+1}=B_{n-1}$ I was observing these formulas concerning $B_n$, Bernoulli numbers
I was able to conjecture this formula:
$$(-1)^nn+2+\sum_{k=1}^{n}\frac{(-1)^k}{k+1}(2)_{k+1}S_{n}^{k+1}H_{k+1}=B_{n-1}\tag1$$
Where $H_n$ is Harmonic numbers
$S_{n}^m$ is Stirling number of second kind
$(n)_m$ is Pochhammer
Does anyone know how to prove it?
| We seek to prove the identity
$$B_n = \sum_{k=0}^{n} \frac{(-1)^k H_{k+1} (k+2)!
{n+1\brace k+1}}{k+1} + (-1)^{n+1} (n+1).$$
The sum is
$$(n+1)! [z^{n+1}]
\sum_{k=0}^{n} (-1)^k \left(1+\frac{1}{k+1}\right)
H_{k+1} (\exp(z)-1)^{k+1}.$$
With $\exp(z)-1 = z+\cdots$ the coefficient extractor enforces the upper
limit of the sum and we get
$$(n+1)! [z^{n+1}]
\sum_{k\ge 0} (-1)^k \left(1+\frac{1}{k+1}\right)
(\exp(z)-1)^{k+1} [w^{k+1}] \frac{1}{1-w} \log\frac{1}{1-w}.$$
Note that the term in $w$ starts at $w$. We get for the first piece
$$-(n+1)! [z^{n+1}] \exp(-z) \log \exp(-z)
\\ = (n+1)! [z^n] \exp(-z) = (-1)^n (n+1).$$
We see that this cancels the extra term from the initial closed form.
Therefore the remaining term must give the Bernoulli numbers:
$$(n+1)! [z^{n+1}]
\sum_{k\ge 0} (-1)^k \frac{1}{k+1}
(\exp(z)-1)^{k+1} [w^{k+1}] \frac{1}{1-w} \log\frac{1}{1-w}.$$
Differentiate to get
$$n! [z^n] \exp(z)
\sum_{k\ge 0} (-1)^k
(\exp(z)-1)^{k} [w^{k+1}] \frac{1}{1-w} \log\frac{1}{1-w}
\\ = n! [z^n] \exp(z)
\sum_{k\ge 0} (-1)^k
(\exp(z)-1)^{k} [w^k] \frac{1}{w} \frac{1}{1-w} \log\frac{1}{1-w}.$$
This is
$$n! [z^n] \exp(z) \frac{1}{1-\exp(z)} \exp(-z) \log\exp(-z)
\\ = n! [z^n] \frac{z}{\exp(z)-1} = B_n$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The sum of five different positive integers is 320. The sum of greatest three integers in this set is 283. The sum of five different positive integers is $320$. The sum of the greatest three integers in this set is $283$. The sum of the greatest and least integers is $119$. If $x$ is the greatest integer in the set, what is the positive difference between the greatest possible value and the least possible value of $x$?
I obtained the equations
$$\begin{align}x+b+c+d+e &= 320\\
x+b+c &=283\\
x+e &= 119\\
d+e &= 37\\
b+c+d &= 201\end{align}$$
How to proceed after this?
| $e\ge1,d>e\implies d\ge2$ and similarly $c\ge3, b\ge4$. Thus $x\le320-(1+2+3+4)=310$ from the first equation, $x\le283-(3+4)=276$ from the second and $x\le 119-1=118$ from the third. Since all three of these inequalities must be satisfied, the maximum value of $x$ is $118$. Can you verify by coming with with the values of the rest of the variables?
Suppose the minimum possible value of $x$ is $m$, then the highest possible sum of $x,b,c,d,e$ is $m+(m-1)+(m-2)+(m-3)+(m-4)=5m-10\ge320\implies m\ge66$. Similarly the second equation gives $m+(m-1)+(m-2)=3m-3\ge283\implies m\ge96$ and the third equations gives $m+(m-4)\ge119\implies m\ge62$. All three give $m\ge96$. Can you check that for $x=96$ we can find the values of other variables satisfying the equations? That is not the case: the problem occurs while choosing the value of $d,e$. We have $d=x-82,e=119-x$ so $d>e\implies 2x>201\implies x\ge101$. Does $101$ work?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do I find all values $x$ such that a vector is a linear combination of a nonempty set of vectors in vector space $\mathbb{R^3}$? In vector space $\mathbb{R^3}$.
Find all values $x$ such that $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \}$.
My solution:
I used the equation:
$r_1\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}$ + $r_2\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$ + $r_3\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$ = $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$
which can be represented with the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
1&0&1&2x^2\\
1&1&2&-3x\\
3&1&4&1\\
\end{array}
\right] $$
which has an RREF of:
$$ \left[
\begin{array}{ccc|c}
1&0&1&2x^2\\
0&1&1&-3x-2x^2\\
0&0&0&1-4x^2+3x\\
\end{array}
\right] $$
I understood this as the system could only be consistent iff:
$1-4x^2+3x = 0$
So then, I solved for $x$ which yields $x = 1,\frac {-1}{4} $.
Is it correct to say, then, that $1$ and $\frac {-1}{4}$ are all the values of $x$ that will make $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \}$?
| Yes. You approach is correct. Also (if you are allowed) you can remove the vector $\begin{bmatrix}1\\2\\4\end{bmatrix}$ by writing
$$
\text{span}\left\{\begin{bmatrix}1\\1\\3\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\4\end{bmatrix}\right\}
=
\text{span}\left\{\begin{bmatrix}1\\1\\3\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}\right\}
$$
since
$$
\begin{bmatrix}1\\2\\4\end{bmatrix}=
\begin{bmatrix}1\\1\\3\end{bmatrix}
+
\begin{bmatrix}0\\1\\1\end{bmatrix}
.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\cos\alpha=\frac13,$ find the value of $7\cos(180^\circ-\alpha)-2\sin(90^\circ+\alpha)$ If $\cos\alpha=\dfrac13,$ find the value of $$7\cos(180^\circ-\alpha)-2\sin(90^\circ+\alpha)$$
Let $A=7\cos(180^\circ-\alpha)-2\sin(90^\circ+\alpha)=-7\cos\alpha-2\cos\alpha=-9\cos\alpha$ or when $\cos\alpha=\dfrac13\Rightarrow A=-9.\dfrac13=-3.$ The given answer in my book is $3$. Am I wrong?
| $7 \cos(\pi-\alpha) = -7 \cos(\alpha) = -7/3$
$2 \sin(\pi/2 + \alpha) = 2 \cos (\alpha) = 2/3$
So you get $-7/3-2/3 = -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$
My question:
Let $a,b,c$ be positive real numbers satisfy $a+b+c=3.$
Prove that $$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}.$$
I have tried to change the LHS to $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}$$
And using Cauchy–Schwarz inequality for it
$$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}\geq \frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}$$
Then because $$ab+ca+ca\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3,$$
$$\frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}\geq \frac{9}{3+ac^2+ba^2+cb^2}$$
Finally, I can't prove $ac^2+ba^2+cb^2\leq 3$ $$ $$
I look forward to your help, thank you!
| Remarks: My proof of (1) is not nice. Hope to see a nice proof of it.
My proof:
Since the inequality is cyclic, WLOG, assume that $c = \min(a, b, c)$.
We split into two cases:
Case 1: $c \ge 1/5$
Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\mathrm{LHS} \ge \frac{(a + b + c)^2}{a(b + c^2) + b(c + a^2) + c(a + b^2)}.$$
It suffices to prove that
$$ab + bc + ca + a^2b + b^2c + c^2 a \le 6 \tag{1}$$
which is true (the proof is given at the end).
Case 2: $c < 1/5$
Using $c^2 \le c < 1/5$, we have
$$\mathrm{LHS} \ge \frac{a}{b + c} + \frac{b}{1/5 + a^2} \ge \frac{a}{3 - a} + \frac{3 - a - 1/5}{1/5 + a^2}.$$
It suffices to prove that
$$\frac{a}{3 - a} + \frac{3 - a - 1/5}{1/5 + a^2} \ge \frac32$$
or (after clearing the denominators)
$$25a^3 - 35a^2 - 53a + 75 \ge 0$$
which is true (actually for all $a\ge 0$).
(Hint: Using AM-GM, we have $a^3 + \frac94 a \ge 3a^2$. The rest is easy.)
We are done.
Proof of (1):
We split into two cases:
(1) If $a < 1$ or $b < 1$, using the well-known inequality $a^2b + b^2c + c^2a + abc \le \frac{4}{27}(a + b + c)^3$, it suffices to prove that $$ab + bc + ca + 4 - abc \le 6$$ which is written as $$(1 - a)(b - 1)(b + a - 2) + (ab - a - b)(a + b + c - 3) \ge 0$$ which is true (using $a + b > 2$).
(2) If $a, b \ge 1$, let $a = 1 + u$ and $b = 1 + v$ for $u, v \ge 0$, and the inequality is written as $$-u^3 - 3u^2v + v^3 + u^2 + uv + v^2 \ge 0.$$ From $c = 3 - a - b = 1 - u - v$ and $c \ge 1/5$, we have $u + v \le 4/5$. Using $-u^3 \ge - u^2 \cdot \frac45$, it suffices to prove that $$-u^2\cdot \frac45 - 3u^2v + v^3 + u^2 + uv + v^2 \ge 0$$ or $$(1/5 - 3v)u^2 + v^3 + uv + v^2 \ge 0.$$
If $1/5 -3v \ge 0$, the inequality is true.
If $1/5 -3v < 0$, let $g(u) := (1/5 - 3v)u^2 + v^3 + uv + v^2$. Then $g(u)$ is concave. Note that $0 \le u \le 4/5 - v$. Also, we have $g(0) \ge 0$ and $g(4/5 - v) = \frac{-250v^3 + 625v^2 - 180v + 16}{125} \ge 0$. Thus, $g(u) \ge 0$ on $[0, 4/5 - v]$. The desired result follows.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Showing that $\mathrm{Var}(X)=\mathrm{Var}(Y)$ Let $X$ and $Y$ be continuous random variables with the following PDF $(b>a)$:
Show that $\mathrm{Var}(X)=\mathrm{Var}(Y)$.
Here is my attempt:
\begin{align*}
\mathrm{Var}(X)&=\mathbb{E}(X^2)-(\mathbb{E}(X))^2 = \int_1^2 bx^2 \, dx + \int_2^4 ax^2 \, dx - \left(\int_1^2 bx \, dx + \int_2^4 ax \, dx\right)^2\\
&=\frac{7b+56a}{3}-\frac{9b^2}{4}-18ba-36a^2\\
\mathrm{Var}(Y)&=\mathbb{E}(Y^2)-(\mathbb{E}(Y))^2 = \int_2^4 ay^2\, dy - \int_4^5 by^2\, dy - \left(\int_2^4 ay\, dy - \int_4^5 by\, dy\right)^2\\
&=\frac{56a-61b}{3}-36a^2+54ab-\frac{81b^2}{4}
\end{align*}
| $Var(X)$ is correct but there are mistakes in signs while calculation $Var(Y)$.
\begin{align*}
\mathrm{Var}(X)&=\mathbb{E}(X^2)-(\mathbb{E}(X))^2 = \int_1^2 bx^2 \, dx + \int_2^4 ax^2 \, dx - \left(\int_1^2 bx \, dx + \int_2^4 ax \, dx\right)^2\\
&=\frac{7b+56a}{3}- \left(6a+\frac{3b}{2}\right)^2\\
\mathrm{Var}(Y)&=\mathbb{E}(Y^2)-(\mathbb{E}(Y))^2 = \int_2^4 ay^2\, dy + \int_4^5 by^2\, dy - \left(\int_2^4 ay\, dy + \int_4^5 by\, dy\right)^2\\
&=\frac{56a+61b}{3}- \left(6a+\frac{9b}{2}\right)^2
\end{align*}
$ \displaystyle Var(Y) - Var(X) = 18b + \left(6a+\frac{3b}{2}\right)^2 - \left(6a+\frac{9b}{2}\right)^2$
$ \displaystyle = 18b - 36 ab - 18b^2$
Now given the pdf, we also note that $2a + b = 1 \implies 2a = 1 - b$
So, $ \displaystyle Var(Y) - Var(X) = 18b - 18 b (1-b) - 18b^2 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find explicit form of following: $a_n=3a_{n-1}+3^{n-1}$ I wanted to find the explicit form of a recurrence relation , but i stuck in nonhomogenous part.
Find explicit form of following: $a_n=3a_{n-1}+3^{n-1}$ where $a_0=1 , a_1 =4,a_2=15$
My attempt:
For homogeneous part , it is obvious that $c_13^n$
For non-homogenouspart = $3C3^n=9C3^{n-1}+3^n \rightarrow 9C3^n=9C3^{n}+3 \times3^n$ , so there it not solution.
However , answer is $n3^{n-1} + 3^n$ . What am i missing ?
| Let $A(z)=\sum_{n \ge 0} a_n z^n$ be the ordinary generating function. The recurrence relation and initial condition imply that
\begin{align}
A(z)
&= a_0 + \sum_{n \ge 1} a_n z^n \\
&= 1 + \sum_{n \ge 1} (3a_{n-1} + 3^{n-1}) z^n \\
&= 1 + 3z \sum_{n \ge 1} a_{n-1} z^{n-1} + z \sum_{n \ge 1} 3^{n-1} z^{n-1} \\
&= 1 + 3z \sum_{n \ge 0} a_n z^n + z \sum_{n \ge 0} (3z)^n \\
&= 1 + 3z A(z) + \frac{z}{1-3z}.
\end{align}
Solving for $A(z)$ yields
\begin{align}
A(z)
&= \frac{1+z/(1-3z)}{1-3z} \\
&= \frac{1-2z}{(1-3z)^2} \\
&= \frac{2/3}{1-3z} + \frac{1/3}{(1-3z)^2} \\
&= \frac{2}{3}\sum_{n\ge 0}(3z)^n + \frac{1}{3}\sum_{n\ge 0}\binom{n+1}{1}(3z)^n \\
&= \sum_{n\ge 0}\left(\frac{2}{3}+\frac{1}{3}(n+1)\right)3^n z^n \\
&= \sum_{n\ge 0}(n+3)3^{n-1} z^n.
\end{align}
Hence $a_n=(n+3)3^{n-1}$ for $n \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove there is only one main class latin square of order 4. I'm learning about latin squares and orthogonal latin squares. My question is how can I prove there's only one main class latin square of order 4? I did this one
$$\array{0&1&2&3\\1&2&3&0\\2&3&0&1\\3&0&1&2}$$
which is the normalized one and there's shouldn't be any other latin square orthogonal to this one but I don't get how to formally prove it. Thanks for the help.
| I think you've written the wrong thing in the title. There are two main classes of Latin square of order $4$:
$$\begin{array}{|cccc|} \hline 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ \hline \end{array}
\qquad
\begin{array}{|cccc|} \hline 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \\ 3 & 4 & 1 & 2 \\ 2 & 3 & 4 & 1 \\ \hline \end{array}$$
You can prove they're not in the same main class by counting $2 \times 2$ subsquares (intercalates): the one on the left has $12$ intercalates, and the one on the right has $4$ intercalates.
As we'll see below, there is:
*
*one main class of sets of 2 orthogonal Latin squares of order $4$ (2-MOLS(4)), and
*one main class of sets of 3 orthogonal Latin squares of order $4$ (3-MOLS(4)).
The Latin square on the left above belongs to a set of 3 mutually orthogonal Latin squares (or as in PM 2Ring's comment). We can prove the one on the right has no orthogonal mate (it's the answer to this question), so it doesn't belong to any sets of orthogonal Latin squares.
Since the Latin squares are small, we can find the orthogonal mates of the Latin square on the left manually. Since we can permute the symbols in an orthogonal mate to obtain another orthogonal mate, we can assume the first row is $(1,2,3,4)$. Afterwards, the cell $(2,1)$ can be filled in $2$ ways without violating the Latin property nor the orthogonal property. Once this decision is made, the remainder of the Latin square is determined (from the Latin property or the orthogonal property)---you can fill it in like a sudoku.
$$\begin{array}{|cccc|} \hline 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ \hline \end{array}
\qquad
\begin{array}{|cccc|} \hline \bf 1 & \bf 2 & \bf 3 & \bf 4 \\ \bf \color{blue} 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 \\ \hline \end{array}
\qquad
\begin{array}{|cccc|} \hline \bf 1 & \bf 2 & \bf 3 & \bf 4 \\ \bf \color{blue} 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ \hline \end{array}
$$
So there is only one main class of sets of 3 orthogonal Latin squares.
To check there is only one main class of sets of 2 orthogonal Latin squares, we need to check the two we incidentally found above are in the same main class:
$$\left\{\begin{array}{|cccc|} \hline 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ \hline \end{array}, \qquad
\begin{array}{|cccc|} \hline \bf 1 & \bf 2 & \bf 3 & \bf 4 \\ \bf \color{blue} 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 \\ \hline \end{array}\right\}
$$
$$\left\{\begin{array}{|cccc|} \hline 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ \hline \end{array},
\qquad
\begin{array}{|cccc|} \hline \bf 1 & \bf 2 & \bf 3 & \bf 4 \\ \bf \color{blue} 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ \hline \end{array}\right\}
$$
To check this, we take the first pair and swap the rows 3 and 4, then swap the columns 3 and 4, then swap the symbols 3 and 4, which gives the second pair above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
| Suppose $1+x+x^2+x^3...=S$ where $S \in R$.
If $x=0$ it is trivial, so suppose $x \neq 0$. Subtract each side by 1 and divide both sides by $x$.
This leaves: $S=(S-1)x^{-1}$. Solving for $x$ yields: $S*x=S-1$ which simplifies to $S(x-1)=-1$ which then yields $1+x+x^2+x^3...=(1-x)^{-1}$
Note domain restrictions of $|x|<1$ applies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Solving $(x-2)^{x^2-6x+8} >1$ Solving this equation:
$(x-2)^{x^2-6x+8} >1$, by taking log on base $(x-2)$ both the sides, I get the solution as $x>4$. My work: Let $(x-2)>0$
$$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1 =0\implies x<2, or, x>4$$
But this doesn't appear to be the complete solution for instance $x=5/2$ is also a solution. I would like to know how to solve it completely.
| $(x-2)^{x^2-6x+8}\gt 1$
If
$x^2 -6x + 8=0$ $\implies$ $(x-2)^{x^2-6x+8}=1$
$x\ne4$ and $x\ne2$
At $x=3$
$(x-2)^{x^2-6x+8}=1$
$x\in (2,3)\cup (4,\infty)$
For $x=\frac{5}{2}$
$(0.5)^{-0.75}=1.68179....$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
For how many integer values of $m$ parabola $y=(m-2)x^2+12x+m+3$ passes through only three quadrants?
For how many integer values of $m$ the graph of parabola
$y=(m-2)x^2+12x+m+3$ passes through only three quadrants?
$1)0\qquad\qquad2)7\qquad\qquad3)12\qquad\qquad4)\text{infinity}$
If the parabola $ax^2+bx+c=0$ passes through three quadrant I think we should have $\frac ca>0$ and $\Delta>0$:
$$\frac{m+3}{m-2}>0\Rightarrow m\in(-\infty,-3] \cup(2,+\infty)$$
$$\Delta'>0\Rightarrow 36-(m-2)(m+3)>0\Rightarrow m^2+m-42<0\Rightarrow m\in [-7,6]$$
Also for $m=2$ we have $y=12x+5$ and it passes through three quadrants.
So $m$ can be $-7,-6,\cdots,-3, $ or $2,3,4,5,6$ . so there are $10$ possible values for $m$ but this isn't in the options. What am I missing?
| One mistake is that
$$ m^2+m-42<0$$
gives $m\in (-7,6)$ instead of $m\in [-7, 6]$. So $\{-7, 6\}$ are excluded.
If one excludes also $m=2$ (which does not give a parabola), then there are only $7$ possible choices: $\{-6, -5, -4, -3, 3, 4, 5\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\measuredangle C=120^\circ$ and two altitudes $AH$ and $BD$ are altitudes of $\triangle ABC$ and $\measuredangle ACB=120^\circ$. If $S_{\triangle HCD}=\dfrac{15\sqrt3}{4},$ find the area of $\triangle ABC$.
$$S_{\triangle HCD}=\dfrac12\cdot CH\cdot CD\cdot\sin\measuredangle HCD=\dfrac{\sqrt3}{4}CH\cdot CD=\dfrac{15\sqrt{3}}{4}\ \implies CH\cdot CD=15$$
On the other hand $$S_{\triangle ABC}=\dfrac12\cdot AC\cdot BC\cdot\sin\measuredangle ACB=\dfrac{\sqrt3}{4}AC\cdot BC=?$$
I noted that $ABDH$ is inscribed, because $\measuredangle ADB=\measuredangle AHB=90^\circ$, so $$AC\cdot CD=BC\cdot CH.$$ I am stuck here. Thank you in advance!
| Observe,
$$\angle ACH=\angle BCD=60^{\circ}\implies \angle CAH=CBD=30^{\circ}$$
In right-triangle $ AHC$,
$$\sin \angle CAH=\frac{CH}{AC}\implies AC=2\cdot CH $$
In right-triangle $ BDC$,
$$\sin \angle CBD=\frac{CD}{BC}\implies BC=2\cdot CD $$
Since you have arrived at $CH\cdot CD=15$,
$$\begin{align*}
\text{area}(\triangle ABC)&=\frac{1}{2}\cdot AC\cdot BC\cdot \sin \angle ABC \\
&=\frac{1}{2}\cdot (2\cdot CH)\cdot (2\cdot CD)\cdot \sin 120^{\circ}\\
&=\boxed{15\sqrt{3}}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n$
Show that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n.$
I'm not very certain how to proceed on this problem. I know $x^2+y^2=(x+y)^2-2xy,$ so we essentially have $x+y=3n, (x+y)^2-3xy=7n$ for positive integers $x, y, n.$ However, this doesn't really help. I've also tried writing it as a fraction and doing some algebraic manipulations, but I haven't gotten anywhere either. May I have some help? Thanks in advance.
| Because $x$ and $y$ are exchangeable, I represent them as
$$ x=m+d \qquad y=m-d
$$
Then
$$ x+y=3n \qquad \to \qquad 2m = 3n \qquad \to \qquad 7n= 14/3m \tag 1$$
$$ x^2+y^2-xy = 7n \qquad \to \\
2m^2+2d^2 - (m^2-d^2)= 7n \qquad\to \\
m^2+3d^2 = 14/3m \qquad\to \tag 2 $$
$$ m^2-14/3m + (7/3)^2 = (7/3)^2-3d^2 \qquad\to $$
$$ m= 7/3 \pm \sqrt{ (7/3)^2-3d^2} \tag 3 $$
The integer resp half-integer values of the term $7/3 \pm \sqrt{ (7/3)^2-3d^2}$ can be enumerated for $d \in \{0,1/2,1,3/2,...\}$ and all for $d \gt 1 $ become imaginary. From that only $d=0$ and $d=1/2$ are integer or half integer and lead to the solutions:
d m x=m+d y=m-d
--------------------------------------- for 7/3 + sqrt(...)
0 4.66666666667 4+2/3 4+2/3
0.5 4.50000000000 5 4 <---- the single integral solution
1 3.89680525327 <fractional>
--------------------------------------- for 7/3 - sqrt(...)
0 0 "trivial"
0.5 0.166666666667 <fractional>
1 0.769861413392 <fractional>
Conclusion: The only integer-solutions are $(x,y)=(y,x)=(5,4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Generalized Method to find nth power of matrix in $P^n = 5I - 8P$
Question: Let $I$ be an identity matrix of order $2 \times 2$ and $P = \begin{bmatrix}
2 & -1 \\
5 & -3 \\
\end{bmatrix}$. Then the value of $n ∈ N$ for which $P^n = 5I - 8P$ is equal to _______.
Answer: 6
Question Source: JEE Mains $18^{th}$ March Shift-2 2021
By characteristic Equation: $P^2-(\operatorname {Tr}(P))P+(\det (P))I=0$ we can find $n=6$ by hit and trial. i.e. multiplying equation by P gives
$P^3=(\operatorname {Tr}(P))P^2-(\det (P))P = (\operatorname {Tr}(P))[(\operatorname {Tr}(P))P+(\det(P))I]$
And going on solving for $P^6$. But is there a generalized method because it can't be done for high values of $n$ (eg:$100$)?
| We can diagonalize the matrix $P$ because it has unique eigenvalues.
One of the benefits of this is in finding matrix powers.
For your example, we have eigenvalues as
$$\lambda_{1,2} = \frac{1}{2} \left(\mp\sqrt{5}-1\right)$$
The eigenvectors are
$$v_{1,2} = \begin{pmatrix} \dfrac{1}{10} \left(5\mp\sqrt{5}\right) \\ 1\end{pmatrix}$$
We can now write $P = V D V^{-1}$ as
$$\begin{pmatrix}
\dfrac{1}{10} \left(5-\sqrt{5}\right) & 1 \\
\dfrac{1}{10} \left(\sqrt{5}+5\right) & 1 \\
\end{pmatrix} \begin{pmatrix}
\dfrac{1}{2} \left(-\sqrt{5}-1\right) & 0\\ 0 & \dfrac{1}{2} \left(\sqrt{5}-1\right)
\end{pmatrix}\begin{pmatrix}
\dfrac{1}{2} \left(1-\sqrt{5}\right) & 0 \\
0 & \dfrac{1}{2} \left(\sqrt{5}+1\right) \\
\end{pmatrix}$$
Now $P^n = V D^n V^{-1}$ is
$$\begin{pmatrix}
\dfrac{1}{10} \left(5-\sqrt{5}\right) & 1 \\
\dfrac{1}{10} \left(\sqrt{5}+5\right) & 1 \\
\end{pmatrix} \begin{pmatrix}
\left(\dfrac{1}{2} \left(-\sqrt{5}-1\right)\right)^n & 0\\ 0 & \left(\dfrac{1}{2} \left(\sqrt{5}-1\right)\right)^n
\end{pmatrix}\begin{pmatrix}
\dfrac{1}{2} \left(1-\sqrt{5}\right) & 0 \\
0 & \dfrac{1}{2} \left(\sqrt{5}+1\right) \\
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Solve $3xdy = y(1+x\sin(x) - 3y^3 \sin(x))dx$ Solve $3xdy = y(1+x\sin(x) - 3y^3 \sin(x))dx$
Mt Attempt
$$3xy' = y(1+x\sin(x) - 3y^3 \sin(x)) \rightarrow 3xy'-y(1+x\sin(x)) = -3y^3\sin(x)$$
$$y'-y\frac{(1+x\sin(x))}{3x}= -\frac{y^3\sin(x)}{x}$$
To me, this is the Bernoulli form. Hence let $u = y^{1-3} = y^{-2}$ and $u' = -2y^{-3}y'$
Divide by $y^3$
$$\frac{1}{y^3}y'-\frac{1}{y^2}\frac{(1+x\sin(x))}{3x}= -\frac{\sin(x)}{x}$$
$$\frac{-1}{2}u'-u\frac{(1+x\sin(x))}{3x}= -\frac{\sin(x)}{x}$$
$$u'+u\frac{2(1+x\sin(x))}{3x}= \frac{2\sin(x)}{x}$$
The integrating factor and several steps become very messy, so it is probably wrong somewhere here. Is this the right approach?
| Your approach is correct, but you have made a mistake. The correct differential equation would be:
$$y'-y\frac {1+x\sin x}{3x}=-\frac {y^{\mathbf 4} \sin x}{x}$$
This is indeed the Bernoulli form, so the procedure ahead is clear.
The IF would not come out to be messy or complicated, in fact it is of the form $x e^{-\cos x}$, which when multiplied with $\frac {\sin x}{x}$ gives an elementary integral solvable by substituting $t=\cos x$..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $x$ if $\cot(x)=\csc(12^{\circ})-\sqrt{3}$ Find $x$ in degrees if $\cot(x)=\csc(12^{\circ})-\sqrt{3}$
My attempt:
$$\cot (x)=\frac{1}{\sin (12^{\circ})}-2 \sin \left(60^{\circ}\right)$$
$$\Rightarrow \cot x=\frac{1-2 \sin (12^{\circ}) \sin (60^{\circ})}{\sin \left(12^{\circ}\right)}$$
$$\Rightarrow \cot x=\frac{1-\cos 48^{\circ}+\cos 72^{\circ}}{\sin \left(12^{\circ}\right)}$$
Now let $, \theta=12^{\circ},s=\sin(\theta)$, then we get
$$\cot x=\frac{1-\cos (4 \theta)+\cos (6\theta)}{\sin (\theta)}$$
Converting to rational function in $s$, we get
$$\cot x=\frac{-32 s^{6}+40 s^{4}-10 s^{2}+1}{s}$$
| This is a very similar yet a little different question compared to this problem, and the answer by @albert chan paves the way for a similar solution to this problem.
$$ \sin(12°) = \sin(30°-18°) = \sin(30°)\cos(18°) - \cos(30°)\sin(18°)$$
$$= {1\over2} (\cos(18°) - \sqrt3 \sin(18°))$$
Converting $\sin$ to $\csc$ and rationalizing the denominator:
$$\csc(12°) = \left({2 \over \cos(18°) - \sqrt3 \sin(18°)}\right) \left({\cos(18°) + \sqrt3 \sin(18°) \over \cos(18°) + \sqrt3 \sin(18°)}\right)$$
$$= {2(\cos(18°) + \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} $$
Using $\cot(x)\sin(x)=\cos(x)$ for numerator and Pythagorean trig identity for denominator:
$$= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) + \sqrt3)
$$
Let $s=\sin(18°)$, using multiple angles formula and the fact that $s≠1$:
$$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s -1 = (s-1)(4s^2+2s-1)^2 = 0 $$
$$\implies 4s^2+2s-1 = 0 \implies {2s \over 1-4s^2} = 1$$
$$\fbox{$ \cot(x) = (\cot(18°) + \sqrt3) - \sqrt3 = \cot(18°)$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$?
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$
Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac x2)}{1+\tan^2(\frac{x}2)}=-\frac35\quad$. by testing the options we can find out $\tan(\frac x2)=-3$ works (although by solving the quadratic I get $\tan(\frac x2)=-\frac13$ too. $-3$ isn't the only possible value.)
I wonder is it possible to solve the question with other (quick) approaches?
| This is the same way to go as in the OP, maybe combining the arguments looks simpler. Let $t$ be $t=\tan(x/2)$ for the "good $x$" satisfying the given relation. Then $\displaystyle \sin x=\frac {2t}{1+t^2}$, so
$$
4=
\frac{1-\sin x}{1+\sin x}
=
\frac{(1+t^2)-2t}{(1+t^2)+2t}
=\left(\frac{1-t}{1+t}\right)^2\ .
$$
This gives for $(1-t)/(1+t)$ the values $\pm 2$, leading to the two solutions $-3$ and $-1/3$ mentioned in the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
If $f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}$, then $\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx =\,$? This is a question from a practice workbook for a college entrance exam.
Let
$$f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}.$$
Find
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx.$$
While I know that computing $f(f(x))$ is an option, it is very time consuming and wouldn't be practical considering the time limit of the exam. I believe there must be a more elegant solution.
Looking at the limits, I tried to find useful things about $f(\frac{1}{7}+\frac{6}{7}-x)$
The relation I obtained was that $f(x) + f(1-x) = 12/12 = 1$. I don't know how to use this for the direct integral of $f(f(x)).$
| In a comment after @Math Lover's elagant answer, I told that nobody tried the change of variable $f(x)=y$. So, I tried for the fun of it
$$I=\int f[f(x)]\,dx=\int \Big[ \frac 1{12}+f(x)-\frac 12 f^2(x)+\frac 13 f^3(x)\Big]\,dx$$
Let $f(x)=y$. Solving the cubic with the hyperbolic method for only one real root
$$x=\frac{1}{2}-\sqrt{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2-4
y}{\sqrt{3}}\right)\right)$$
$$dx=\frac{4 \cosh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2-4
y}{\sqrt{3}}\right)\right)}{\sqrt{48 (y-1) y+21}}\,dy$$ So,
$$20480\,I=-10240 \sqrt{3} \sinh (t)+2700 \cosh (2 t)+1350 \cosh (4 t)+15 \cosh (8 t)+12 \cosh (10 t)$$ where $$t=\frac{1}{3} \sinh ^{-1}\left(\frac{2-4 y}{\sqrt{3}}\right)$$ Now, for $y$, the upper and lower bounds are $\frac{3223}{4116}$ and $\frac{893}{4116}$ and then for $t$, they are $$\mp \frac{1}{3} \sinh ^{-1}\left(\frac{1165}{1029 \sqrt{3}}\right)$$ Because of the symmetry in $t$, all cosines disappear and the result for the definite integral is just
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f[f(x)]\,dx =\sqrt{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{1165}{1029 \sqrt{3}}\right)\right)$$ whic, numerically is $0.3571428571428571428571429$; its reciprocal is $2.8=\frac {14}5$ so the value of $\frac 5{14}$.
For those who are curious, this took me close to one and half hour.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Proving upper bound on a term involving binomial coefficients I am trying to show:
$$\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}\leq \frac{1}{n^2}$$ for $k\in \{2,3,\cdots,n-1\}$ for all $n$
I observe that numerically that is true but analytically, I am not able to look at the term on LHS as a term of a binomial expansion or something like that to get an upper bound. Any ideas?
| Sketch of a proof:
Fact 1: Let $x, y$ be real numbers with $x \ge 6$ and $3 \le y \le x - 2$. Then
$$\ln \frac{\Gamma(x + 1)}{\Gamma(y + 1)\Gamma(x - y + 1)}
+ (2x - 2y + 2)\ln(1 - y/x)
+ 2y\ln \frac{y - 1}{x} + 2\ln x \le 0.$$
The proof of Fact 1 is given at the end.
By Fact 1, when $n\ge 6$ and $3 \le k \le n - 2$, we have
$$\ln \binom{n}{k}
+ (2n - 2k + 2)\ln(1 - k/n)
+ 2k\ln \frac{k - 1}{n} + 2\ln n \le 0.$$
Thus, the desired inequality is true.
When $n \ge 6$ and $k = 2, n - 1$, the desired inequality is verified directly.
When $n \le 5$, the desired inequality is verified directly.
We are done.
Proof of Fact 1:
Denote the LHS by $F(x, y)$.
We have
\begin{align*}
\frac{\partial^2 F}{\partial y^2}
&= - \psi'(y + 1) - \psi'(x - y + 1)\\
&\qquad + {\frac {2{x}^{2}y - 2x{y}^{2} - 4{x}^{2} + 4xy - 2{y}^{2} + 2x + 2y - 2}{ \left( x - y \right) ^{2} \left( y - 1 \right) ^{2}}}\\[5pt]
&\ge - \frac{1}{y + 1} - \frac{1}{(y + 1)^2}
- \frac{1}{x - y + 1} - \frac{1}{(x - y + 1)^2}\\[8pt]
&\qquad + {\frac {2{x}^{2}y - 2x{y}^{2} - 4{x}^{2} + 4xy - 2{y}^{2} + 2x + 2y - 2}{ \left( x - y \right) ^{2} \left( y - 1 \right) ^{2}}}\\[6pt]
&= \frac{G}{(y + 1)^2(x - y + 1)^2(x - y)^2(y - 1)^2}\\
&\ge 0
\end{align*}
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$, and
\begin{align*}
G &= {x}^{4}{y}^{3}-3\,{x}^{3}{y}^{4}+3\,{x}^{2}{y}^{5}-x{y}^{6}+2\,{x}^{3}
{y}^{3}-6\,{x}^{2}{y}^{4}+6\,x{y}^{5}-2\,{y}^{6}\\
&\qquad -3\,{x}^{4}y+10\,{x}^{
3}{y}^{2}-11\,{x}^{2}{y}^{3}+2\,x{y}^{4}+2\,{y}^{5}-6\,{x}^{4}+18\,{x}
^{3}y\\
&\qquad -18\,{x}^{2}{y}^{2}+6\,x{y}^{3}-2\,{y}^{4}-11\,{x}^{3}+30\,{x}^{2
}y-23\,x{y}^{2}+4\,{y}^{3}\\
&\qquad -6\,{x}^{2}+12\,xy-2\,{y}^{2}-2\,x+2\,y-2,
\end{align*}
and we have used
$\psi'(u) \le \frac{1}{u} + \frac{1}{u^2}$ for all $u > 0$, and we have used $G \ge 0$.
Hint for the proof of $G \ge 0$: With the substitutions $x = 6 + t, \ y = \frac{1}{1 + s}\cdot 3 + \frac{s}{1 + s}\cdot (x - 2)$ for $t, s \ge 0$, we have
$(1 + s)^6G$ is a polynomial in $s, t$ with non-negative coefficients.
Thus, $F(x, y)$ is convex with respect to $y$. Also,
$F(x, 3) \le 0$ and $F(x, x-2) \le 0$.
Thus, $F(x, y) \le 0$ for all $x \ge 6$ and $3 \le y \le x - 2$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine the Galois group of the splitting field of $f(x) = x^4+x+t \in F_2(t)[x]$ I used Gauss's lemma to show that the polynomial is irreducible since it is irreducible in $F_2[t,x]$, used the derivative GCD test to conclude that the polynomial is separable, and finally started by taking the quotient $F_2(t)[x]/(f(x))$ which gives a degree $4$ extension that contains at least $2$ roots: $\bar{x}, \bar{x}+1$ (using $a^2+b^2= (a+b)^2$). However, from here, I can't seem to find other roots or clearly show the non-existence of any other roots to have to take another quotient by a degree $2$ factor to find that the Galois group is $D_4$ (thought of as a subgroup of $S_4$). It seems that all subgroups of $S_4$ of order $2^n$ are transitive so that doesn't help.
Let me know if you can think of any ways to finalize the computation of the Galois group.
| Let $\alpha$ be a root of $f=X^4+X+t$ in an algebraic closure of $k=\mathbb{F}_2(t)$. If $j$ is a primitive third root of $1$ (so $j^2+j+1=0$), the roots of $f$ are $\alpha,\alpha+1,\alpha+j,\alpha+j+1$ and the splitting field of $f$ is $L=k(\alpha,j)$.
Let us prove that $j\notin k(\alpha)$. Otherwise, $j=a+b\alpha+c\alpha^2+d\alpha^3$.
Then $j^2=a^2+b^2\alpha^2+c^2(\alpha+t)+d^2(\alpha^3+t\alpha^2)$ since $\alpha^4=\alpha+t$ and $\alpha^6=\alpha^2\alpha^4=\alpha^2(\alpha+t)$.
Thus
$0=j^2+j+1=(a^2+a+tc^2+1)+(b+c^2)\alpha+(b^2+c+td^2)\alpha^2+(d+d^2)\alpha^3$.
In particular, $b=c^2$, so $0=(a^2+a+tc^2+1)+(c^4+c+td^2)\alpha^2+(d+d^2)\alpha^3$.
Now $d^2+d=0$, so $d=0$ or $1$. If $d=1$, we would get $c^4+c+t=0$, which is not possible since $f$ is irreducible. So $d=0$ and $c^4+c=0=c(c+1)(c^2+c+1)$. Since $X^2+X+1$
is irreducible over $k$ (easy), we get $c=0$ or $1$.
If $c=0$, we get $a^2+a+1=0$, which is not possible since $X^2+X+1$ is irreducible over $k$.
Hence $c=1$, and $a^2+a+t+1=0$. But $X^2+X+(t+1)$ is irreducible over $k$ , so we get a contradiction.
Finally, $j\notin k(\alpha)$, $[k(\alpha)(j):k(\alpha)]=2$ and $[L:k]=8$. Now, the subextension $k(\alpha)/k$ is non Galois (since the splitting field of $f$ has degree $8$) of degree $4$, so the Galois group has a non normal subgroup of order $2$. The only group of degree $8$ satisfying this property is the dihedral group $D_4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $x= \left ( 6, 2, -3 \right ),$ how to find the coordinate $x$ up to the basis $V$ In the vector space $\mathbb{R}^{3},$ given two systems of vectors
$$U= \left \{ u_{1}= \left ( 4, 2, 5 \right ), u_{2}= \left ( 2, 1, 3 \right ), u_{3}= \left ( 3, 1, 3 \right ) \right \}$$
$$V= \left \{ v_{1}= \left ( 5, 2, 1 \right ), v_{2}= \left ( 6, 2, 1 \right ), v_{3}= \left ( -1, 7, 4 \right ) \right \}$$
Proved that $U$ and $V$ are two bases of $\mathbb{R}^{3}.$
Source: StackMath/@haidangel_ in.edit
In the edited part, I gave a bonus question:
Given $x= \left ( 6, 2, -3 \right ).$ How to find the coordinate $x$ up to the basis $V.$
Now I have two approaches but I don't know which one is true ? I need to the help.
First approach. Consider the linear combination
$$\alpha_{1}v_{1}+ \alpha_{2}v_{2}+ \alpha_{3}v_{3}= x$$
This is equivalent to the matrix equation
$$\begin{bmatrix} 5 & 6 & -1\\ 2 & 2 & 7\\ 1 & 1 & 4 \end{bmatrix}\begin{bmatrix} \alpha_{1}\\ \alpha_{2}\\ \alpha_{3} \end{bmatrix}= \begin{bmatrix} 8\\ 2\\ -3 \end{bmatrix}$$
To find the solution, consider the augmented matrix.
Applying elementary row operations, we obtain
$$\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 2 & 2 & 7 & 2\\ 1 & 1 & 4 & -3 \end{array} \right ]\xrightarrow{R_{3}\leftrightarrow R_{2}}\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 1 & 1 & 4 & -3\\ 2 & 2 & 7 & 2 \end{array} \right ]\xrightarrow{2R_{2}- R_{3}}\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 1 & 1 & 4 & -3\\ 0 & 0 & 1 & -8 \end{array} \right ]$$
$$\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 1 & 1 & 4 & -3\\ 0 & 0 & 1 & -8 \end{array} \right ]\xrightarrow{6R_{2}- 25R_{3}- R_{1}}\left [ \begin{array}{rrr|r} 1 & 0 & 0 & 174\\ 1 & 1 & 4 & -3\\ 0 & 0 & 1 & -8 \end{array} \right ]$$
It follows that the solution is $\alpha_{1}= 174, \alpha_{3}= -8, \alpha_{2}= -3- \alpha_{1}- 4\alpha_{3}= -145.$ We obtain
$$\left [ x \right ]_{V}= \begin{bmatrix} 174\\ -145\\ -8 \end{bmatrix}$$
Second approach. By the coordinate transformation equation
$$\left [ x \right ]_{V}= P_{V\rightarrow E}\cdot\left [ x \right ]_{E}= \left ( P_{E\rightarrow V} \right )^{-1}\cdot\left [ x \right ]_{E}= \begin{bmatrix} 5 & 6 & -1\\ 2 & 2 & 7\\ 1 & 1 & 4 \end{bmatrix}^{-1}\begin{bmatrix} 6\\ 2\\ -3 \end{bmatrix}= \begin{bmatrix} 176\\ -147\\ -8 \end{bmatrix}$$
| In the first approach, you wrote $8$ in place of $6$. Otherwise, both approaches are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Geometric reason why this determinant can be factored to (x-y)(y-z)(z-x)? The determinant $\begin{vmatrix}
1 & 1 &1 \\
x & y & z \\
x^2 & y^2 &z^2 \\
\end{vmatrix}$ can be factored to the form $(x-y)(y-z)(z-x)$
Proof:
Subtracting column 1 from column 2, and putting that in column 2,
\begin{equation*}
\begin{vmatrix}
1 & 1 &1 \\
x & y & z \\
x^2 & y^2 &z^2 \\
\end{vmatrix}
=
\begin{vmatrix}
1 & 0 &1 \\
x & y-x & z \\
x^2 & y^2-x^2 &z^2 \\
\end{vmatrix}
\end{equation*}
$
= z^2(y-x)-z(y^2-x^2)+x(y^2-x^2)-x^2(y-x)
$
rearranging the terms,
$
=z^2(y-x)-x^2(y-x)+x(y^2-x^2)-z(y^2-x^2)
$
taking out the common terms $(y-x)$ and $(y^2-x^2)$,
$
=(y-x)(z^2-x^2)+(y^2-x^2)(x-z)
$
expanding the terms $(z^2-x^2)$ and $(y^2-x^2)$
$
=(y-x)(z-x)(z+x)+(y-x)(y+x)(x-z)
$
$
=(y-x)(z-x)(z+x)-(y-x)(z-x)(y+x)
$
taking out the common term (y-x)(z-x)
$
=(y-x)(z-x) [z+x-y-x]
$
$
=(y-x)(z-x)(z-y)
$
$
=(x-y)(y-z)(z-x)
$
Is there a geometric reason for this?
The determinant of this matrix is the volume of a parallelopiped with sides as vectors whose tail is at the origin and head at x,y,z coordinates being equal to the columns(or rows) of the matrix.$^{[1]}$
So is the volume of this parallelopiped equals $(x-y)(y-z)(z-x)$ in any obvious geometric way?
References
[1] Nykamp DQ, “The relationship between determinants and area or volume.” From Math Insight. http://mathinsight.org/relationship_determinants_area_volume
| Subtracting a multiple of another column (or row) to an
existing column (or row) does not change the determinant.
$$
\begin{vmatrix}
1 & 1 & 1 \\
x & y & z \\
x^2 & y^2 & z^2 \\
\end{vmatrix} $$
$$=
\begin{vmatrix}
1 & 0 &0 \\
x & y-x & z-x \\
x^2 & y^2-x^2 &z^2-x^2 \\
\end{vmatrix}$$
$$=
\begin{vmatrix}
1 & 0 &0 \\
0 & y-x & z-x \\
0 & y^2-x^2 &z^2-x^2 \\
\end{vmatrix}$$
$$={(y-x)(z-x)
\begin{vmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & y+x &z+x \\
\end{vmatrix} }
$$
$$=(y-x)(z-x)
\begin{vmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & y+x & z-y \\
\end{vmatrix}
$$
$$=(y-x)(z-x)(z-y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
mean and variance formula for negative binomial distribution The equation below indicates expected value of negative binomial distribution. I need a derivation for this formula. I have searched a lot but can't find any solution. Thanks for helping :)
$$
E(X)=\sum_{x=r}^\infty x\cdot \binom {x-1}{r-1} \cdot p^r \cdot (1-p)^{x-r} =\frac{r}{p}
$$
I have tried:
\begin{align}
E(X) & =\sum _{x=r} x\cdot \binom{x-1}{r-1} \cdot p^r \cdot (1-p)^{x-r} \\[8pt]
& = \sum_{x=r}^\infty x \cdot \frac{(x-1)!}{(r-1)! \cdot ((x-1-(r-1))!} \cdot p^r \cdot (1-p)^{x-r} \\[8pt]
& = \sum_{x=r}^\infty \frac{x!}{(r-1)!\cdot ((x-r)!} \cdot p^r \cdot (1-p)^{x-r} \\[8pt]
\Longrightarrow & \phantom{={}} \sum_{x=r}^\infty r\cdot \frac{x!}{r!\cdot (x-r)!}\cdot p^r \cdot (1-p{)}^{x-r} \\[8pt]
& = \frac{r}{p} \cdot \sum_{x=r}^\infty \frac{x!}{r!\cdot (x-r)!}\cdot p^{r+1}\cdot (1-p)^{x-r}
\end{align}
If the power of $p$ in the last equation were not $r + 1,$ I can implement Newton Binomial. So It will be true. But I am stuck here.
| If you want to continue that derivation instead of using linearity of expectation on a sum of i.i.d. geometric random variables, then you can follow this; however, doing it this way is much more complicated than the method using the i.i.d. variables.
When you arrive at the step $\operatorname{E}(X) = \sum_{x\geq r} r \binom{x}{r} p^r (1 - p)^{x - r}$, we can use this fact about power series:
$$
\frac{1}{(1 - z)^{r + 1}} = \sum_{n\geq r} \binom{n}{r}z^{n-r}, \quad \text{for }\lvert z\rvert < 1.
$$
If this fact is unfamiliar to you, then you can derive it from the geometric series $\frac{1}{1 - z} = \sum_{n\geq 0} z^n$ by differentiating both sides $r$ times and dividing by $r!$. Of course, we are tacitly assuming that $p \neq 0$ in order to use this. Otherwise, the event that we want to occur $r$ times could not occur at all!
It follows that
$$\begin{align*}
\operatorname{E}(X) &= r p^r\sum_{x\geq r} \binom{x}{r} (1 - p)^{x - r} \\
&= rp^r \cdot \frac{1}{\big(1 - (1 - p)\big)^{r + 1}} \\
&= rp^r \cdot \frac{1}{
p^{r + 1}} \\
&= \frac{r}{p}
\end{align*}$$
We can do something similar for the variance using the formula
$$\begin{align*}
\operatorname{Var} X &= \operatorname{E}\big(X^2\big) - \big(\operatorname{E}(X)\big)^2 \\
&= \operatorname{E}\big(X(X + 1)\big) - \operatorname{E}(X) - \big(\operatorname{E}(X)\big)^2.
\end{align*}$$
This means that
$$\begin{align*}
\operatorname{Var} X &= \sum_{x\geq r} x (x + 1)\binom{x - 1}{r - 1} p^r (1 - p)^{x - r} - \frac{r}{p} - \frac{r^2}{p^2} \\
&= \sum_{x\geq r} r (r + 1)\binom{x + 1}{r + 1} p^r (1 - p)^{x - r} - \frac{r p + r^2}{p^2} \\
&= r(r + 1)p^r \sum_{x\geq r+1} \binom{x}{r + 1} (1 - p)^{x - (r + 1)} -\frac{r p + r^2}{p^2} \\
&= r(r + 1)p^r \cdot \frac{1}{\big(1 - (1 - p)\big)^{r + 2}} -\frac{r p + r^2}{p^2} \\
&= \frac{r^2 + r}{p^2} - \frac{rp + r^2}{p^2} \\
&= \frac{r (1 - p)}{p^2}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits