Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
A generalization of arithmetic and geometric means using elementary symmetric polynomials Let $a_1, a_2, \ldots, a_n$ be positive real numbers. A while ago I noticed that if you form the polynomial
$$
P(x) = (x - a_1)(x-a_2) \cdots (x-a_n)
$$
then:
*
*The arithmetic mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{n-1}$.
*The geometric mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{0}.$
It looks like this can be extended: for any $0 \le i \le n-1$, let the $i$th mean be the number $m_i$ such that $(x - m_i)^n$ and $P(x)$ have the same coefficient of $x^i$. (Alternatively, one can define $m_i$ in terms of elementary symmetric polynomials.)
For instance, with three variables $x, y, z$ we get
\begin{align*}
m_0 &= \sqrt[3]{xyz} \\
m_1 &= \sqrt{\frac{xy + yz + zx}{3}} \\
m_2 &= \frac{x + y + z}{3}
\end{align*}
I am not sure what qualifies something as a "mean", but $m_1$ is symmetric and lies strictly between the min and the max, and it probably has other properties as well. Here's a more concrete question:
Question: Must it be true that $m_0 \le m_1 \le m_2 \le \cdots \le m_{n-1}$?
|
Denote, $(\overline{a}) = (a_1,\cdots,a_n)$ and $\displaystyle P(x) = \prod\limits_{k=1}^{n}(x - a_k) = x^n +\sum\limits_{k=1}^{n} (-1)^k\binom{n}{k}u_k(\overline{a})x^{n-k}$
where, $\displaystyle u_k(\overline{a}) = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n}a_{j_1}\cdots a_{j_k}}{\binom{n}{k}} = m_{n-k}^{k}$ (in your notation).
Note that $P(x)$ has $n$ real roots in the interval $\left[\min\limits_{i=1}^n\{a_i\},\max\limits_{i=1}^n\{a_i\}\right]$,
Thus $P^{(n-2)}(x) = \dfrac{n!}{2}(x^2 - 2u_1x + u_2)$ has two real roots in that interval, i.e., $u_1^2 \ge u_2$.
Similarly apply the same idea for the $(n-2)^{th}$ derivative of the polynomial with roots $\dfrac{1}{a_k}$, ($k = 1(1)n$)
We get, $\displaystyle \frac{1}{\binom{n}{2}}\sum\limits_{i<j}a_ia_j \le \frac{1}{\binom{n}{1}^2}\left(\sum\limits_{i=1}^{n} \frac{1}{a_i}\right)^2 \implies u_{n-1}^2 \ge u_nu_{n-2}$.
We show, $u_{k-1}(\overline{a})u_{k+1}(\overline{a}) \le u_k^2(\overline{a})$ for $k = 2,3,\cdots,n-1$,
We can prove this result by induction on $n$, suppose the inequality holds for any $n-1$ positive real numbers.
We have $\displaystyle P'(x) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}u_k(\overline{a})x^{n-k-1}\right)$
If the roots of $P'(x)$ are $b_k$, for $k=1,2,\cdots,n-1$, (which, are positive reals by M.V.T.).
and define $v_k = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n-1}b_{j_1}\cdots b_{j_k}}{\binom{n-1}{k}}$
Then $\displaystyle P'(x) = n\prod\limits_{k=1}^{n-1}(x - b_k) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}v_kx^{n-k-1}\right)$
Thus, $u_k = v_k$ for $k=1,2,\cdots,n-1$ and by the induction hypothesis on the numbers $(b_1,\cdots,b_{n-1})$ we get $u_k^2 \ge u_{k+1}u_{k-1}$ for $k = 2,\cdots,n-2$ and together with $u_{n-1}^2 \ge u_nu_{n-2}$ completes the induction.
If we take the $k$th inequality to the $k$th power and then multiply all these inequalities for $k=1,\cdots,r$, we get,
$u_1^2 u_2^4 u_3^6 \cdots u_{r-1}^{2r-2}u_{r}^{r-1}u_{r+1}^r \le u_1^2u_2^4\cdots u_r^{2r} \implies u_{r+1}^r \le u_{r}^{r+1}$ for $r=1, \cdots,n-1$
Thus, $u_r^{1/r} = m_{n-r}$ forms a non increasing sequence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1081075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
}
|
Calculate $\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \ldots $ $$
\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} \times\ldots$$
I already know a way to calculate it:
With $\cos{\frac{\pi}{4}} = \sqrt{\frac{1}{2}}$ and denote $\frac{\pi}{4} = x$. Observe that:
$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} = \sqrt{\frac{1+\cos{x}}{2}} = \cos{\frac{x}{2}} \\
\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} = \cos\frac{x}{4}$$
Thus it becomes
\begin{align}
P(n) &= \cos{\frac{x}{2^n}}\cos{\frac{x}{2^{n-1}}} \cdots \cos{x} \\
&= \frac{2\sin{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-2}}} \cdots\cos{\frac{x}{2}}}{2\sin{\frac{x}{2^{n-1}}}}
\end{align}
Taking in to account that $2\sin{x}\cos{x} = \sin{2x}$, we have
\begin{align}
P(n) &= \lim_{n \to \infty} \frac{\sin{2x}}{2^n\sin{\frac{x}{2^{n-1}}}} \\
&= \lim_{n\rightarrow \infty} \frac{\sin{2x}}{2x} \\
&= \frac{2\sin\frac{\pi}{2}}{\pi} \\
&= \boxed{\frac{2}{\pi}}
\end{align}
Now, I'm looking for another solution, please comment on.
|
Well probably the easiest way to derive this result is to employ the same method as it's creator, François Viète. He started with a circle of radius 1 and inscribed a square, whose area we will denote $A_0 = 2$. Now let $A_k$ be the area of the inscribed regular polygon with $2^{k+2}$ sides. Since $\lim_{k\to\infty}{A_k} = \pi$, we get that, on the one hand, $$\prod_{k=0}^{\infty}\frac{A_k}{A_{k+1}} = \frac{2}{\pi}.$$ On the other hand, by noticing that $\frac{A_0}{A_1} = \sqrt{\frac{1}{2}}$, $\frac{A_1}{A_2} = \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}$, etc. we get that $$\frac{2}{\pi} = \sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}...$$
The history of this product, and of Viète, is really quite interesting. This is, to the best of my knowledge, the first explicitly written infinite process in all of mathematics. It is really a corollary of the ideas of Archimedes, but Archimedes lacked the ability to write out an explicit formula for an infinite expression. Anyway, if you are interested, you should check out Viète's Variorum de rebus mathematicis responsorum liber VIII, which is freely available from Google. It's chock full of geometrical curiosities. This formula is in a note to Corollary 2 to Proposition 2 of Chapter 18.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 1
}
|
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
|
Let's square both of them:
\begin{align}
\sqrt{\frac{1}{2}}^2 &= \frac{1}{2}, \text{ while} \\
\left(\frac{\sqrt{2}}{2}\right)^2
&=\frac{\sqrt{2}^2}{2^2} \\
&=\frac{2}{4} \\
&= \frac{1}{2}
\end{align}
So they're both positive numbers, and their squares are the same, so they must be the same.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1082664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
}
|
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$ $a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$
I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next
|
As it was said, you can take $x=2^a$, $y=2^b$ and $z=2^c$ to obtain:
$$
x^3+y^3+z^3\ge x+y+z
$$
With $x,y,z>0$ and $xyz=1$. It is equivalent to:
$$
x^2(x-1)+y^2(y-1)+z^2(z-1)\ge 0 \iff \frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge 0
$$
Since $x^2$ and $x-1$ ar equally ordered, we might apply Chebychevs inequality to obtain:
$$
\frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge \frac{x^2+y^2+z^2}{3}\frac{(x-1)+(y-1)+(z-1)}{3}
$$
And to show that the two factors are greater than zero isn't difficult. (AM-GM)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1083277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Removing the root squares from this expression? I would like to understand how to remove the root squares from this expression:
$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$
How to do it?
|
Its all about rationalization,
\begin{align}
\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}} &= \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}
\cdot \frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(2+3+2\sqrt{6})-(5)}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2\sqrt{6}}\cdot\frac{\sqrt6}{\sqrt6}
\\[10pt]&=\frac{\sqrt{12}+ \sqrt{18} - \sqrt{30}}{12}
\\[10pt]&=\frac{2\sqrt{3} + 3\sqrt{2} - \sqrt{30}}{12}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1083442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
}
|
Formal proof of limit By using $\epsilon-\delta$ definition, prove that $$\lim_{x\rightarrow1}\frac{3}{x^2}=3$$
Here is what I have tried.
$$|\frac{3}{x^2}-3|=3|\frac{1}{x^2}-1|=3|\frac{1-x^2}{x^2}|=3\frac{|x-1||x+1|}{x^2}$$
So I have problem with how to elimate the denominator.
|
Given $\epsilon > 0$, choose $\delta = \text{min}\left(\frac{1}{2},\frac{\epsilon}{30}\right) \to |1|-|x| < |1-x|=|x-1| < \delta < \dfrac{1}{2} \to |x| > \dfrac{1}{2}$, and $|x+1| = |x-1+2| < |x-1|+|2| < \delta + 2 < \dfrac{1}{2} + 2 = \dfrac{5}{2} \to \left|\dfrac{3}{x^2} - 3\right| < 3\cdot |x-1|\cdot \dfrac{5}{2}\cdot \dfrac{1}{\left(\frac{1}{2}\right)^2}=30|x-1|< \epsilon$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1083599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Cauchy P.V. Of an improper inegral the poles are $x=+1,-1,i,-i$
we should take only the upper have of axis
so we should take residue of $1$ and $i$? right
in this problem the book took only $x= i$.
I don't know why !! please help
|
Denoting by $a$ and $b$ the relevant poles ($\Im a>0, \; \Im b>0 $), we have:
$$a= (1+i)/\sqrt{2}, \quad b= (-1+i)/\sqrt{2}$$
$$\bar{a}= (1-i)/\sqrt{2}, \quad \bar{b}= (-1-i)/\sqrt{2}$$
$$ a - \bar{a} = i\sqrt{2}, \quad a - b = \sqrt{2}, \quad a - \bar{b} = (1+i)\sqrt{2}$$
$$ b - \bar{b} = i\sqrt{2}, \quad b - a = -\sqrt{2}, \quad b - \bar{a} = (1-i)\sqrt{2}$$
$${\rm Res} \left(\frac{z^2+1}{z^4+1}; a\right) = \frac{i+1}{2i\sqrt{2}(1+i)}=\frac{1}{2i\sqrt{2}}$$
$${\rm Res} \left(\frac{z^2+1}{z^4+1}; b\right) = \frac{-i+1}{-2i\sqrt{2}(1-i)}=\frac{1}{2i\sqrt{2}}$$
So:
$$1/2\cdot {\rm p.v.}\int_{-\infty}^{\infty}\frac{x^2+1}{x^4+1}\; dx = \pi i\left({\rm Res} \left(\frac{z^2+1}{z^4+1}; a\right) +{\rm Res} \left(\frac{z^2+1}{z^4+1}; b\right)\right)= \frac{\pi}{\sqrt{2}}$$
Edit:
As per Ron Gordon's comment, as $\deg (x^4+1)-\deg (x^2+1)\geq 2$, improper integral
$$ \int_{-\infty}^{\infty}\frac{x^2+1}{x^4+1}\; dx = \int_{-\infty}^{0}\frac{x^2+1}{x^4+1}\; dx + \int_{0}^{\infty}\frac{x^2+1}{x^4+1}\; dx$$
already exists and it is finite, so its principal value doesn't need to be taken. Note that:
$$\int\frac{x^2+1}{x^4+1}\; dx = \left(\arctan (\sqrt{2}x+1) - \arctan (-\sqrt{2}x+1) \right)/\sqrt{2} + {\cal C},$$
gives a a direct way of computing the improper integral.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1085230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Complex numbers problem: $ |\frac 1z - \frac 14 | = \frac 14 $ How do you go about solving the following equation $ |\frac 1z - \frac 14 | = \frac 14 $ where $ z = a + bi $.
A hint is provided, and apprently the equation can be simplified to $ | {z-4\over z} | = 1 $ (don't understand how they did this).
|
We have:
$\left\lvert\dfrac1z-\dfrac14\right\rvert=\dfrac14$. Combine fractions:
$\left\lvert\dfrac{4-z}{4z}\right\rvert=\dfrac14$. Get rid of fractions (multiply by $\lvert4z\rvert$):
$\lvert4-z\rvert=\lvert z\rvert$. Let $z=a+bi$:
$\lvert4-a-bi\rvert=\lvert a+bi\rvert$.
$\sqrt{(4-a)^2+b^2}=\sqrt{a^2+b^2}$. Get rid of square root signs (they're ugly):
$(4-a)^2+b^2=a^2+b^2$. Subtract $b^2$:
$(4-a)^2=a^2$
Now, you can solve and get $a=2$. Thus, $a$ is $2$, and $b$ can be anything.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1085587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?
|
Set $y = 2\log\left(\frac{x}{\sin x}\right)$ and $z=\log\left(\frac{\tan x}{x}\right)$.
Exploting Weierstrass products, for every $x\in I=\left(0,\frac{\pi}{2}\right)$ we have:
$$ z-y = \sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{x^{2m}}{m\cdot \pi^{2m}}\left(\frac{4^m}{(2n-1)^{2m}}-\frac{3}{n^{2m}}\right), $$
so, given that $ \zeta(2m) = \sum_{n=1}^{+\infty} \frac{1}{n^{2m}},$
$$ z-y = \sum_{m=1}^{+\infty}\frac{x^{2m}\cdot\zeta(2m)}{m\cdot \pi^{2m}}(4^m-4).\tag{1} $$
Since every term in the RHS of $(1)$ is non-negative, it follows that $z > y$ for every $x\in I$.
So we have:
$$ \left(\frac{\sin x}{x}\right)^2+\frac{\tan x}{x} = e^z + e^{-y} > e^{y}+e^{-y} \geq 2. \tag{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1085780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Series acceleration with Fourier-Bessel series coefficients I was investigating methods for series acceleration and I found this identity:
$$e=\sum _{n=0}^{\infty } \left(\sqrt{\frac{\pi }{2}} (2 n+1)\right) I_{n+\frac{1}{2}}(1)$$
where $I$ is the modified Bessel function of the first kind.
Could you explain me where this identity comes from ?
I tried computing the Fourier-Bessel series for the exponential function but it did not work.
|
good work Jack but the Integral it is very difficult to resolve anyway i try to get the series of the Sinh(x)/x using series Bessel Fourier and we get a different seriesthan you give it above i think the following is it faster
$$\frac{\sinh (x)}{x}=\sum _{n=0}^{\infty } I_{n+\frac{1}{2}}(1) \left(\pi 2^{n-\frac{7}{2}} \left((-1)^n+1\right) (2 n+1) x^n \left(\frac{(-1)^{\left\lfloor \frac{n}{2}\right\rfloor +1} 4^{-\left\lfloor \frac{n}{2}\right\rfloor } x^{-2 \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)} \sec \left(\pi \left(n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) \, _3\tilde{F}_2\left(1,-\frac{n}{2}+\left\lfloor \frac{n}{2}\right\rfloor +\frac{1}{2},-\frac{n}{2}+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2,-n+\left\lfloor \frac{n}{2}\right\rfloor +\frac{3}{2};\frac{1}{x^2}\right)}{\Gamma \left(n-2 \left\lfloor \frac{n}{2}\right\rfloor \right)}+\frac{4 \sec (\pi n) \, _2\tilde{F}_1\left(-\frac{n}{2}-\frac{1}{2},-\frac{n}{2};\frac{1}{2}-n;\frac{1}{x^2}\right)}{\Gamma (n+2)}\right)\right)$$ but it is very complicate thanks anyway.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1086600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Integration of $x/\sqrt{x^2-4x}$ How can one integrate: $$\int \frac{x}{\sqrt{x^2-4x}}\,dx$$
I tried setting $x^2-4x$ as a $t$ (and changing $dx$ concordantly) but it didn't work....
|
$t=x^2-4x$ and $\frac 1 2 \,dt=(x-2)\,dx$, so
$$
\int\frac x {\sqrt{x^2-4x}} \,dx = \int \frac{x-2}{\sqrt{x^2-4x}}\,dx+\int\frac2{\sqrt{x^2-4x}}\,dx.
$$
Use the substitution to evaluate the first integral.
Then: $\underbrace{x^2-4x = (x^2-4x+4)}_{\text{completing the square}}-4=(x-2)^2-4=4\left(\dfrac{x-2}{2}\right)^2-4 = 4\left(\left(\dfrac{x-2}2\right)^2-1\right)$.
Then you can say
\begin{align}
\sec\theta& =\dfrac{x-2}2 \\[8pt]
\sec\theta\tan\theta\,d\theta & =\dfrac{dx}2 \\[8pt]
\left(\dfrac{x-2}2\right)^2-1 & =\sec^2\theta-1=\tan^2\theta.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1086941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Justify the solution of $x^2-x+\arctan{x}=0$ It is obvious that a solution of $x^2-x+\arctan{x}=0$ is $x=0$, but I would like you to show me how this can be derived more conceptually than by plugging in $0$, and how can one prove that such solution is unique.
|
like you conjectured, $x = 0$ is the only solution. i will first show that there are no positive solutions.
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1 - t^2 + {t^4 \over 1 + t^2} \right) dt = x -{1 \over 3}x^3 + \int_0^x {t^4 \over 1 + t^4}dt$$
for $x > 0, \arctan x > x - {1 \over 3}x^3,$ therefore
$$x^2 - x + \arctan x > x^2 + {1 \over 3}x^3 > 0$$ the last inequality establishes the claim for $x > 0$
now for the case $x < 0$
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1 - {t^2 \over 1 + t^2} \right) dt = x - \int_0^x {t^2 \over 1 + t^4}dt = x + \int_x^0 {t^2 \over 1 + t^4}dt$$
$$x^2 - x + \arctan x = x^2 + \int_x^0 {t^2 \over 1 + t^4}dt > x^2 > 0$$ the last inequality establishes the claim for $x < 0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1088529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
}
|
Prove convergence of series Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$
so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since
both series are convergent so the sum of it is also convergent but I'm not sure if it's legall
second series I'm not sure is like this $\displaystyle b_n=\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$
here I also split it into smaller parts namely $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n-3}+\sum_{n=1}^{\infty}\frac{1}{4n-1}-\sum_{n=1}^{\infty}\frac{1}{2n}$ and I prove that $\displaystyle \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}<\frac{1}{n^2} \cdot c $ for some const c so it implies that origin series is convergent or not ?
|
Consider the series $\displaystyle\sum_{m=1}^{\infty} a_m=\sum_{n\in A}(-1)^n\frac{1}{n^{1/3}-1}=\frac{1}{2^{1/3}-1}-\frac{1}{5^{1/3}-1}+\frac{1}{6^{1/3}-1}-\frac{1}{9^{1/3}-1}+\cdots$ $\hspace{.8 in}$and $\displaystyle\sum_{m=1}^{\infty}b_m=\sum_{n\in B}(-1)^n\frac{1}{n^{1/3}+1}=-\frac{1}{3^{1/3}+1}+\frac{1}{4^{1/3}+1}-\frac{1}{7^{1/3}+1}+\frac{1}{8^{1/3}+1}-\cdots$,
$\hspace{.4 in}$where $A=\{n\in\mathbb{N}, n\equiv1 \text{ or } n\equiv2 \pmod 4 \ \text { and } n> 1\}$ and
$\hspace{.8 in}B=\{n\in\mathbb{N}, n\equiv3 \text{ or } n\equiv4 \pmod 4 \}$.
Then both series converge by the Alternating Series Test; so if we let
$(T_m), (U_m), \text{ and }(S_m)$ be the partial sums for $\displaystyle\sum_{m=1}^{\infty} a_m, \displaystyle\sum_{m=1}^{\infty} b_m$, and the given series, respectively,
then $T_m\to T$ and $U_m\to U$ for some numbers $T$ and $U$.
Since $S_{2m}=T_m+U_m$ for every $m$, $\;\;S_{2m}\to T+U$;
and since $S_{2m+1}=S_{2m}+\frac{1}{(2m+2)^{1/3}\pm1}, \;\;S_{2m+1}\to T+U$.
Therefore $S_m\to T+U$, and the series converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1089885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.
I find it interesting but have trouble seeing the proof behind it. Could anyone could give me a formal proof based on this animation?
|
Method 1 (By $k^3=k\cdot k^2=k$ pieces of squares of side $k$)
$\displaystyle \quad (1^{3}+2^{3}+3^{3}+\ldots+n^{3} )cm^3
\\= 1 \cdot (1^{2} cm^2\times 1 cm)+2 \cdot (2^{2}cm^2\times 1 cm)+3 \cdot (3^{2} cm^2\times 1cm)+\ldots+n \cdot (n^{2}cm^2\times 1 cm)
\\=(1cm+2cm+3cm+\cdots+ncm)^2\times 1cm$
Method 2(By Induction)
The statement is true for $n=1$, now assume it is true for some $n$, then $$
\begin{aligned}
\sum_{k=1}^{n+1} k^{3}&=\sum_{k=1}^{n} k^{3}+(n+1)^{3} \\
&=\left(\sum_{k=1}^{n} k\right)^{2}+n(n+1)^{2}+(n+1)^{2} \\
&=\left(\sum_{k=1}^{n} k\right)^{2}+2(n+1)\left(\sum_{k=1}^{n} k\right)+(n+1)^{2} \\
&=\left[\sum_{k=1}^{n} k+(n+1)\right]^{2} \\
&=\left(\sum_{k=1}^{n+1} k\right)^{2}
\end{aligned}
$$
Method 3(By difference)
Considering the difference yields
\begin{aligned}\\ n^{3}-(n-1)^{3}&=3 n^{2}-3 n+1 \\
\therefore \sum_{k=1}^{n}\left[n^{3}-(n-1)^{3}\right]&=3 \sum_{k=1}^{n} n^{2}-3 \sum_{k=1}^{n} n+n \\ k^{4}-(k-1)^{4}&=4 k^{3}-6 k^{2}+4 k-1 \end{aligned}
Summing up yields
\begin{aligned}
\sum_{k=1}^{n}\left[k^{4}-(k-1)^{4}\right]&=4 \sum_{k=1}^{n} k^{3}-6 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k-n \\
n^{4}&=4 \sum_{k=1}^{n} k^{3}-6 \cdot \frac{n}{6} \cdot(n+1)(2 n+1)+4 \cdot \frac{n}{2}(n+1)-n \\
4 \sum_{k=1}^{n} k^{3}&=n^{4}+n \left( n+1\right)(2 n+1)-2 n(n+1)+n =[n(n+1)]^{2} \\
\sum_{k=1}^{n} k^{3}&=\left[\frac{n(n+1)}{2}\right]^{2}=(1+2+3+\cdots+n)^{2}
\end{aligned}
Wish you enjoy the explanation!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1089990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
}
|
How to solve $\tan x =\sin(x+45^{\circ})$? How do I solve $\tan x = \sin(x +45^{\circ})$?
This is how far I have come: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$
|
Set $x=y+45^\circ$ to get $\tan(y+45^\circ)=\sin(45^\circ+45^\circ+y)=\cos y$
Clearly, $y=0$ is a solution
Now $-1\le\cos y\le1\implies-1\le\tan(y+45^\circ)\le1$
If we consider $-180^\circ<y\le180^\circ,-45^\circ\le y+45^\circ\le45^\circ\iff-90^\circ\le y\le0\implies\cos y\ge0$
$\implies0\le\tan(y+45^\circ)\le1\implies0\le y+45^\circ\le45^\circ\iff-45^\circ\le y\le0$
$\implies\cos y\ge\cos45^\circ=\dfrac1{\sqrt2}$
$\implies\dfrac1{\sqrt2}\le\tan(y+45^\circ)\le1$
But as $\dfrac1{\sqrt2}>\dfrac1{\sqrt2+1}=\sqrt2-1=\tan15^\circ,$
$\implies15^\circ<y+45^\circ\le45^\circ\iff-30^\circ<y\le0$
$\implies\cos y\ge\cos30^\circ=\dfrac{\sqrt3}2$
If we follow this method, the range will continually shrink and eventually we shall reach at $0\le y\le0\iff y=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1090782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Ways to prove $ \int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$? I am wondering if we can show in a simple way that
$$
I=\int_0^1 \frac{\ln^2(1+x)}{x}dx = \int_1^2 \frac{\ln^2(t)}{t-1}dt = \frac{\zeta(3)}{4}.
$$
Because the end result is very simple, I suspect that there might be a fast way to prove this.
Can you prove it without using polylog identities? Complex analysis is allowed.
It may be easier to show the equivalent identity
$$
\sum_{k=1}^\infty \frac{(-1)^k H_k}{k^2} = -\frac{5 \zeta (3)}{8}
$$
I know you can do that one with the generating function of the harmonic numbers, but that gives a nasty expression with polylogs which I would like to avoid.
|
Here is a particularly efficient way to get to your Euler sum.
In this post I show that
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}.$$
Replacing $x $ with $-x$ gives
$$\ln^2 (1 + x) = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1} x^n}{n}.$$
So if we replace the term $\ln^2 (1 + x)$ with its above Maclaurin series expansion the integral becomes
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n} \int_0^1 x^{n - 1} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n^2}.$$
Making use of the following property for harmonic numbers
$$H_n = H_{n - 1} + \frac{1}{n},$$
the integral can be expressed as
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 2}^\infty \frac{(-1)^n}{n^3} = 2 \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n^3}.$$
For the sums, as you note
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3),$$
and
$$\sum_{n = 1}^\infty \frac{(-1)^n}{n^3} = - \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} = - \eta (3) = -(1 - 2^{1-3}) \zeta (3) = -\frac{3}{4} \zeta (3),$$
where $\eta (s)$ is the Dirichlet eta function, one finally has
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = -\frac{5}{4} \zeta (3) + \frac{3}{2} \zeta (3) = \frac{1}{4} \zeta (3),$$
as expected.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1091812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
}
|
$e^{\pi\sqrt N}$ is very close to an integer for some smallish $N$s. What about $\pi^{e\sqrt N}$? Heegner numbers (1, 2, 3, 7, 11, 19, 43, 67, 163 - let's use symbol $H_n$) are know for peculiar property that $e^{\pi\sqrt{H_n}}$ are almost integers:
$$e^{\pi \sqrt{19}} \approx 96^3+744-0.22$$
$$e^{\pi \sqrt{43}} \approx 960^3+744-0.00022$$
$$e^{\pi \sqrt{67}} \approx 5280^3+744-0.0000013$$
$$e^{\pi \sqrt{163}} \approx 640320^3+744-0.00000000000075$$
(given that they are all less than 200, it goes far beyond "chance" and "randomness")
Even stranger, related to the above:
$$19 = 3 \cdot 2 \cdot 3+1$$
$$43 = 7 \cdot 2 \cdot 3+1$$
$$67 = 11 \cdot 2 \cdot 3+1$$
$$163 = 27 \cdot 2 \cdot 3+1$$
and
$$96^3 =(2^5 \cdot 3)^3$$
$$960^3=(2^6 \cdot 3 \cdot 5)^3$$
$$5280^3=(2^5 \cdot 3 \cdot 5 \cdot 11)^3$$
$$640320^3=(2^6 \cdot 3 \cdot 5 \cdot 23 \cdot 29)^3$$
Related to this, I would like to know:
Are there natural numbers N (of fairly similar range, so, lets say under 500) that produce "almost" integers in the expression $\pi^{e\sqrt{N}}$?
If yes, do they have other interesting properties, like Heegner numbers do?
If not, all right, one more reason to appreciate Heegners. :)
|
Up to 100000, the 10 best $N$ such that $e^{\pi\sqrt{N}}$ is almost an integer. The error $\delta$ is given such that the nearest integer is at $10^{\delta}$ from the result.
$$
\begin{array}{c|c}
N & \delta \\\hline
163 & -12.12\\
4\cdot163 & -9.79\\
9\cdot163 & -8.01\\
58 & -6.75\\
16\cdot163 & -6.51\\
67 & -5.87\\
22905 & -5.61\\
95041 & -5.55\\
54295 & -5.37\\
25\cdot163 & -5.2\\
\end{array}
$$
As you can see, no $N$ beats 163 up to 100000. (For N = 4 x 163.)
For $\pi^{e\sqrt{N}}$, the behaviour is much more regular and you obtain :
$$
\begin{array}{c|c}
N & \delta \\\hline
66972 & -5.03 \\
85516 & -5.01 \\
53204 & -4.91 \\
46665 & -4.9 \\
50237 & -4.8 \\
93909 & -4.53 \\
52970 & -4.4 \\
10024 & -4.32 \\
84702 & -4.17 \\
6814 & -4.17 \\
\end{array}$$
So, it seems there is something strange in $e^{\pi\sqrt{N}}$ that makes that question interesting !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1092248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
}
|
Two-dimensional limit, is my approach correct? The limit is
$$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$$
As usual, I tried checking along particular paths, namely the axes and the curves $y=mx^n$ for various values of $n$, but to no avail; all the limits evaluate to $0$. I resorted to converting to polar coordinates, giving
$$\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}$$
In an attempt to apply the squeeze theorem, I found that trig expression in the numerator is bounded, with $|\cos^3t\sin t|\le\dfrac{3\sqrt3}{16}$, and that in the denominator I have
$$\begin{align*}
|r^2\cos^4t+\sin^2t|&\le r^2|\cos^4t|+|\sin^2t|\\
&\le r^2+1
\end{align*}$$
and so
$$\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}=\frac{3\sqrt3}{16}\lim_{r\to0^+}\frac{r^4}{r^4+r^2}=0$$
Is my reasoning valid? I am unsure about whether or not the bound for the denominator is correct.
|
$$ (x^2 - y)^2 \geq 0, $$
$$ x^4 - 2 x^2 y + y^2 \geq 0, $$
$$ x^4 + y^2 \geq 2 x^2 y. $$
$$ (x^2 + y)^2 \geq 0, $$
$$ x^4 + 2 x^2 y + y^2 \geq 0, $$
$$ x^4 + y^2 \geq -2 x^2 y. $$
$$ x^4 + y^2 \geq 2 x^2 |y|. $$
$$ \color{magenta}{ \left| \frac{x^2 y}{x^4 + y^2} \right| \leq \frac{1}{2} } $$
Your numerator is $x^3 y$ rather than $x^2 y;$ what happens now?
Some detail, not directly needed for your question but probably helpful anyway, at recent Continuity in $\mathbb R^n$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1093239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How can I show that $\lim_{n\to\infty} 2^n \left( \frac{n}{n+1} \right ) ^{n^2} = 0$? How to calculate limit of the following expression:
$$2^n \left( \frac{n}{n+1} \right ) ^{n^2} $$
I know that limit of this sequence is equal to zero, but how to show that?
|
$$\left(\frac{n}{n+1}\right)^{n^2}=\left(\frac{n+1}{n}\right)^{-n^2}=\left(1+\frac{1}{n}\right)^{-n^2}=\exp\left\{-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}$$
therefore
$$2^n\left(\frac{n}{n+1}\right)^{n^2}=\exp\{n\ln 2\}\cdot \exp\left\{-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}=\exp\left\{n\ln 2-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}=\exp\left\{n\underbrace{\left(\ln(2)-\underbrace{\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}}_{\to 1}\right)}_{\to \ln(2)-1<0}\right\}\longrightarrow 0 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1095083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
}
|
Finding the matrix of linear transformation What is the orthogonal projection on the line of equation $x = y$ of the point
$\begin{pmatrix} 3 \\ -1 \end{pmatrix}$? Assume this is a linear transformation.
The matrix for this linear transformation is $\begin{bmatrix} \frac 12 & \frac 12 \\ \frac 12& \frac 12 \end{bmatrix}$ since the perpendicular line from $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to the line of equation $x = y$ intersects that line at $\begin{pmatrix} \frac 12 \\ \frac 12 \end{pmatrix}$, as does the perpendicular line from $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. To determine the orthogonal projection of the point $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$, we multiply $\begin{bmatrix} \frac 12 & \frac 12 \\ \frac 12& \frac 12 \end{bmatrix}$$\begin{bmatrix} 3 \\ -1 \end{bmatrix} =$$\begin{bmatrix} 1 \\ 1 \end{bmatrix}.$
I am having hard time seeing:
the perpendicular line from $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to the line of equation $x = y$ intersects that line at $\begin{pmatrix} \frac 12 \\ \frac 12 \end{pmatrix}$
Wouldn't $x = y$ be intersected at $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$?
|
When your quote says "the perpendicular line from $(1,0)$ to the line of equation $x=y$", what is meant is the line through $(1,0)$ that is perpendicular to $x=y$.
It looks like you understood it as perpendicular to the vector $(1,0)$, but that is not how an orthogonal projection works.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1095303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$
I am also presuming once this formula is discovered, we can prove it by induction for any $n$.
|
$n^2 = \sum_{k=0}^{n-1} n = \sum_{k=0}^{n-1} ((n(n+2)-k)-(n(n+1)-k))$,
since $(n(n+2)-k)-(n(n+1)-k) = n$.
So we have
$n^2+\sum_{k=0}^{n-1} (n(n+1)-k) = \sum_{k=0}^{n-1} (n(n+2)-k)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1095581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
}
|
Calculate limit on series with nested sum I want to calculate the limit of following series:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{3^k} \cdot \frac{1}{2^{n-k}}$$
As far I could simply the series to:
$$\sum_{n=0}^{\infty} (\sum_{k=0}^{n} (\frac{1}{3})^k) \cdot (\sum_{k=0}^{n} 2^{n-k})$$
which would then allow me to use the geometric series:
$$\sum_{n=0}^{\infty} (\frac{1-(\frac{1}{3})^{n+1}}{\frac{2}{3}}) \cdot (\sum_{k=0}^{n}2^{n-k})$$
which can even be simplified further to:
$$\sum_{n=0}^{\infty} ((1-(\frac{1}{3})^n\cdot\frac{1}{3})(\frac{3}{2})) \cdot (\sum_{k=0}^{n}2^{n-k})$$
$$\sum_{n=0}^{\infty} (\frac{3}{2}-(\frac{1}{3})^n\cdot\frac{1}{2}) \cdot (\sum_{k=0}^{n}2^{n-k})$$
$$\sum_{n=0}^{\infty} (\frac{1}{2}(3-(\frac{1}{3})^n) \cdot (\sum_{k=0}^{n}2^{n-k})$$
This is as far as I know what to do. The solution by the way is:
$$\sum_{k=0}^{\infty} (\frac{1}{3})^k \sum_{k=0}^{\infty} (\frac{1}{2})^k$$ which could be simplified again with the geometric series.
However I have now idea what to do with $\sum_{k=0}^n2^{n-k}$ though we could write it as $\sum_{k=0}^n 2^n \sum_{k=0}^n 2^{-k}$ this would not make much sense as the former sum would converge against infinity.
|
Consider writing the sum as $\sum_{n=0}^{\infty}\dfrac{1}{2^n}\sum_{k=0}^{n}\dfrac{2^k}{3^k}=\sum_{n=0}^{\infty}\dfrac{1}{2^n}\dfrac{1-\dfrac{2^{n+1}}{3^{n+1}}}{1-\dfrac{2}{3}}=\sum_{n=0}^{\infty}\dfrac{1}{2^n}\dfrac{3^{n+1}-2^{n+1}}{3^n}=3\sum_{n=0}^{\infty}\dfrac{1}{2^n}-2\sum_{n=0}^{\infty}\dfrac{1}{3^n}=3(2)-2(\dfrac{3}{2})=3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1096394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:
Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
$$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\dfrac{1}{8}$$
My some idea: since
$$1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n=\dfrac{(n!)^n}{1!\cdot 2!\cdot 3!\cdots (n-1)!}$$
and it is well know
$$v_{5}(n!)=\lfloor \dfrac{n}{5}\rfloor+\lfloor\dfrac{n}{5^2}\rfloor+\lfloor\dfrac{n}{5^3}\rfloor+\cdots+\lfloor\dfrac{n}{5^k}\rfloor+\cdots=\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor$$
so
$$v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=n\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor-\sum_{i=1}^{n-1}\sum_{k=1}^{\infty}\left(\lfloor\dfrac{i}{5^k}\rfloor\right)$$
then I can't it.
Thank you
|
This seemed like a fun problem and wanted to prove it for general prime p. I use Legendre's formula directly which is,
$$v_p(n!) = \frac{n-s_p(n)}{p-1}$$
$s_p(n)$ being the sum of digits of $n$ when written in base $p$.
$$\frac{1}{n^2}v_p(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=\frac{1}{n^2}v_p\left(\dfrac{(n!)^n}{1!\cdot 2!\cdot 3!\cdots (n-1)!}\right)$$
This splits up by additivity to be,
$$\frac{1}{n}v_p(n!) - \frac{1}{n^2}\sum_{k=1}^{n-1} v_p(k!)$$
Now using Legendre's formula
$$\frac{1}{n}\frac{n-s_p(n)}{p-1} - \frac{1}{n^2}\sum_{k=1}^{n-1} \frac{k-s_p(k)}{p-1}$$
A little more algebra and replacing the sum over k with $\frac{n(n-1)}{2}$,
$$\frac{1}{p-1} \left( \frac{1}{2} + \frac{1}{2n}- \frac{s_p(n)}{n} + \frac{1}{n^2}\sum_{k=1}^{n-1} s_p(k)\right)$$
We can put a very cheap upper bound on $s_p(n)$ since the number of digits it has is $\lceil\log_p(n)\rceil$ and the largest a digit can be is $p-1$, so
$$s_p(n) \le \lceil\log_p(n)\rceil(p-1)$$
This higher bound still has a limit of 0 for the terms containing $s_p$, and thus squeezes out :
$$\frac{1}{2(p-1)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1097865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
|
Use the remainder theorem for polynomials.
$$
f(a) = 0 \implies f(x) = (x-a)p(x)
$$
Where $p(x)$ is reduced a degree. To find the $2$ you would trail solutions to find a root using a change of sign.
$\textbf{update}$
Since we have found one root we can assume that the we have an equation of the form
$$
(x+2)\left(ax^2+bx+c\right)
$$
Thus we have to compare the coefficents leading to
$$
a = 1\\
2a+ b = 4\\
2b + c = 1\\
2c = -6
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1098960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
}
|
Closed form solution for $\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}$. Let
$$
\text{S}_k = \sum_{n=1}^\infty\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{\ddots1+\cfrac{1}{1+n^2}}}},\quad\text{$k$ rows in the continued fraction}
$$
So for example, the terms of the sum $\text{S}_6$ are
$$
\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+n^2}}}}}}
$$
Using a symbolic computation software (Mathematica), I got the following interesting results:
$$
\begin{align}
\text{S}_4 &= \frac{\pi}{4}\left(\coth(\pi)+\sqrt{3}\coth(\sqrt{3}\pi)\right)-\frac{1}{2}\\
\text{S}_6 &= \frac{\pi}{4}\left(\sqrt{2}\coth\left(\sqrt{2}\pi\right)+\sqrt{\frac{4}{3}}\coth\left(\sqrt{\frac{4}{3}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_8 &= \frac{\pi}{4}\left(\sqrt{\frac{3}{2}}\coth\left(\sqrt{\frac{3}{2}}\pi\right)+\sqrt{\frac{7}{4}}\coth\left(\sqrt{\frac{7}{4}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_{10} &= \frac{\pi}{4}\left(\sqrt{\frac{5}{3}}\coth\left(\sqrt{\frac{5}{3}}\pi\right)+\sqrt{\frac{11}{7}}\coth\left(\sqrt{\frac{11}{7}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_{12} &= \frac{\pi}{4}\left(\sqrt{\frac{8}{5}}\coth\left(\sqrt{\frac{8}{5}}\pi\right)+\sqrt{\frac{18}{11}}\coth\left(\sqrt{\frac{11}{7}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_{14} &= \frac{\pi}{4}\left(\sqrt{\frac{13}{8}}\coth\left(\sqrt{\frac{13}{8}}\pi\right)+\sqrt{\frac{29}{18}}\coth\left(\sqrt{\frac{29}{18}}\pi\right)\right)-\frac{1}{2}.\\
\end{align}
$$
The numbers appearing at the first $\coth$ term are easy to guess: they are the famous Fibonacci numbers.
The numbers at the second $\coth$ term can also be guessed: they appear to be the Lucas numbers. Those are constructed like the Fibonacci numbers but starting with $2,1$ instead of $0,1$.
Hence:
Conjecture: $$\text{S}_{2k}=\frac{\pi}{4}\left(\sqrt{\frac{F_k}{F_{k-1}}}\coth\left(\sqrt{\frac{F_k}{F_{k-1}}}\pi\right)+\sqrt{\frac{L_k}{L_{k-1}}}\coth\left(\sqrt{\frac{L_k}{L_{k-1}}}\pi\right)\right)-\frac{1}{2}$$
I have verified this conjecture for many $k$'s and it always work out perfectly. To me this is quite amazing, but I am not able to verify the conjecture. Can anyone prove it?
Moreover, if true the conjecture implies that
$$\lim_{k\to\infty}\text{S}_{2k}=\frac{\sqrt{\varphi}\pi\coth\left(\sqrt{\varphi}\pi\right)-1}{2}$$
which is also very nice ($\pi$ and $\varphi$ don't meet very often).
|
This answer contains computer-assisted algebra bashing. Pencil wielding mathematicians be warned.
Let
$$f_k(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{\frac{^\ddots_1}{1+n^2}}}}$$
where there are $2k$ horizontal fraction bars. Then, we have
$$f_2(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{1}{1+n^2}}}}=\frac{\frac{1}2}{n^2+1}+\frac{\frac{3}2}{n^2+3}$$
$$f_{k+1}(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{\frac{1}{f_k(x)}-1}{n^2}}}}$$
where the first identity can be checked by Mathematica and the second is pretty obvious, if you think about undoing the top of the fraction, adding a bit in the middle, then redoing the top.
Next, suppose we have
$$f_k(n)=\frac{\frac{\alpha}2}{n^2+\alpha}+\frac{\frac{\beta}2}{n^2+\beta}$$
where $\alpha=\frac{F_{k}}{F_{k-1}}$ and $\beta=\frac{L_{k}}{L_k-1}$. This is true for $k=2$. We can prove it inductively in a very elegant way by substituting in Binet's formula to get $$\alpha=\frac{\varphi^n-(1-\varphi)^n}{\varphi^{n-1}-(1-\varphi)^{n-1}}$$
$$\beta=\frac{\varphi^n+(1-\varphi)^n}{\varphi^{n-1}+(1-\varphi)^{n-1}}$$
and then plugging the whole mess for $f_k$ into the recurrence relation, setting it equal to $f_{k+1}$ and letting FullSimplify look at it and shrug tell you this is equivalent to a certain rational expression equaling zero, which is, after multiplying out the denominator* $$2 \left(-2+\sqrt{5}\right) \left(\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}\right) e^{i k \pi } n+2 \left(-3+\sqrt{5}\right) \left(\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}\right) e^{i k \pi } n^3+\left(-1+\sqrt{5}\right) \left(\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}\right) e^{i k \pi } n^5=0$$
which looks pretty nasty, and even though Mathematica can't seem to handle it itself, we can solve it pretty easily; it comes down to showing that each coefficient vanishes; in particular we could factor out the following expression from each term
$$\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}$$
however, notice that $(-1+\sqrt{5})^2=6-2\sqrt{5}$ so the expression equals $$(6-2\sqrt{5})^{2k}-(6-2\sqrt{5})^{2k}=0.$$
This, along with the assumption that Mathematica's reductions suffice to complete the inductive proof that $$f_k(n)=\frac{\frac{\alpha}2}{n^2+\alpha}+\frac{\frac{\beta}2}{n^2+\beta}$$
with $\alpha=\frac{F_k}{F_{k-1}}$ and $\beta=\frac{L_k}{L_{k-1}}$.
Then, your conjecture follows immediately from the identity
$$\sum_{n=1}^{\infty}\frac{1}{n^2+c}=\frac{\pi\sqrt{c}\coth(\pi\sqrt{c})-1}{2c}.$$
*The denominator is a non-constant polynomial in $n$, which can only be $0$ at finitely many points for a fixed $k$. Thus this proof shows that the desired equality holds for all but finitely many points, and, since two rational functions agreeing at countably many points are clearly equal, this suffices and we can therefore safely ignore the denominator.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1099652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 2,
"answer_id": 1
}
|
Complex Solution I was looking at a simple complex problem and I came by this:
Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\frac{2}{x}$.
Substitute $y = -\frac 2x$ into $x^2 - y^2 = 3$, giving
$$x^2 - \frac{4}{x^2} = 3$$
Multiply by $x^2$:
$x^4 - 4 = 3x^2$
Factor
$$x^4 - 3x^2 - 4 = 0$$
$$(x^2 + 1) (x^2 - 4) = 0$$
$$(x^2+ 1) (x + 2) (x - 2) = 0$$
Given $xy = -2$
for $x = 2$ then $y = -1$ therefore $z = 2 - i$
for $x = -2$ then $y = 1$ therefore $z = -2 + i$
Now, I understand everything, but why do we ignore the root of $i$ and $-i$ for the root $x^2+1=0$? Is it because we are only looking for real roots? When plugging these two back into $(x+iy)=z$, I saw that it gave the same two solutions as the real roots.
|
Lets say you said let:$$z=x+yi$$but you assumed $x$ and $y$ themselves could also be complex numbers. Then, in effect, you are stating that:$$x=a+bi$$$$y=c+di$$where $a,b,c$ and $d$ are real numbers. This effectively means that you are saying:$$z=(a+bi)+(c+di)i=(a-d)+(b+c)i$$so you will be computing these extra components: $a,b,c$ and $d$ unnecessarily to get back to the same result.
You might as well here say that let:$$u=a-d$$$$v=b+c$$so that:$$z=u+vi$$ and solve for the real numbers $u$ and $v$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1103294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find minimum number of sums of $3$ out of $5$ whose vanishing implies all five to be zero Problem:
let $a,b,c,d,e$ be real numbers, now there are $\left(\binom{5}{3}=10\right)$ numbers
$$a+b+c,a+b+d,a+b+e,a+c+d,a+c+e,a+d+e,b+c+d,b+c+e,b+d+e,c+d+e$$
Question1:($\textbf{Jérémy Blanc has solve it}$)
Find the least k such that if $k$ out of these $10$ numbers are $0$,then $$a=b=c=d=e=0$$
Question 2:
let $n$ is give postive integer numbers,Assmue that $x_{1},x_{2},\cdots,x_{2n+1}$ be real numbers,now there are $\binom{2n+1}{n+1}$ numbers
$$x_{1}+x_{2}+\cdots+x_{n}+x_{n+1},x_{1}+x_{2}+\cdots+x_{n}+x_{n+2},\cdots,x_{n+1}+x_{n+2}+\cdots+x_{2n+1}$$
Find the least $k$ such that if $k$ out of these $\binom{2n+1}{n+1}$ numbers are $0$,then $$x_{1}=x_{2}=\cdots=x_{2n+1}=0$$
Use Jeremy ieda:
if $n=3$,then let
$a=-3,b=c=d=e=f=g=1$,then $\binom{6}{3}=20$,so I guess $k\ge 21$?
In general，so I guess
$$k\ge \binom{2n}{2}+1$$
such
$$x_{1}=n,x_{2}=x_{3}=\cdots=x_{2n+1}=-1$$
I guess is right? Then How prove it?
|
Answer: $k=7$
$(i)$ If $a=-2$ and $b=c=d=e=1$, then $\binom{4}{2}=6$ of the numbers are zero (all sums involving $a$). Hence, $k\ge 7$.
$(ii)$ In order to show that $k\le 7$, we assume that at least $7$ sums are zero and show that this implies that $a=b=c=d=e=0$.
In your $7$ sums you have $21$ letters, each one belonging to $\{a,b,c,d,e\}$. Hence, there is one letter which appear five times (at least), because $5\times 4=20<21$. We can assume that $a$ appears five times. There are exactly six sums involving $a$, and five of them are zero. Up to permutation, we can assume that
$$0=a+b+c=a+b+d=a+b+e=a+c+d=a+c+e$$
(the only sum with $a$ that I did not take is $a+d+e$). We find $b=c=d=e$ and $a=-2b$.
Since $7$ sums are zero, there is one of the sum which does not involve $a$, but only $b,c,d,e$, hence is equal to $0=3b$. So we have found $a=b=c=d=e=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1103788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.
Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1.$$
How to prove it by trigonometric method ?
Thank in advances.
|
We need $\displaystyle\sqrt3\sec\frac\pi5+\tan\frac\pi{30}=\cot\frac{2\pi}{15}=\tan\dfrac{11\pi}{30}$ as $\dfrac{2\pi}{15}+\dfrac{11\pi}{30}=\dfrac\pi2$
$\iff\displaystyle\sqrt3\sec\dfrac\pi5=\tan\dfrac{11\pi}{30}-\tan\dfrac\pi{30}$
$\iff\displaystyle\dfrac{\sqrt3}{\cos\dfrac\pi5}=\frac{\sin\left(\dfrac{11\pi}{30}-\dfrac\pi{30}\right)}{\cos\dfrac\pi{30}\cos\dfrac{11\pi}{30}}$
As $\sin\left(\dfrac{11\pi}{30}-\dfrac\pi{30}\right)=\sin\dfrac\pi3=\dfrac{\sqrt3}2,$
$\iff\displaystyle\cos\frac\pi5=2\cos\frac\pi{30}\cos\dfrac{11\pi}{30}=\cos\dfrac\pi3+\cos\frac{2\pi}5$
$\iff\displaystyle\cos\frac\pi5-\cos\frac{2\pi}5=\cos\dfrac\pi3=\dfrac12$
Now see Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1105022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Question about equivalent equations I am solving equations with radicals using this substitution:
$f(x)^{\frac{1}{3}}+g(x)^{\frac{1}{3}}=h(x)$
$f(x)+g(x)+3(f(x)g(x))^{\frac{1}{3}}(f(x)^{\frac{1}{3}}+g(x)^{\frac{1}{3}})=h(x)^{{3}}$
Now I substitute the expression $f(x)^{\frac{1}{3}}+g(x)^{\frac{1}{3}}$ with $h(x)$. Why after this substitution the new equation is not equivalent to the original??
|
To make things clear, let's denote $a = f(x), b = g(x), c = h(x)$, then
$$a^{1/3}+b^{1/3} = c$$
$$(a^{1/3}+b^{1/3})^3 = a + 3a^{2/3}b^{1/3} + 3a^{1/3}b^{2/3} + b = a+3a^{1/3}b^{1/3}(a^{1/3} + b^{1/3}) + b =$$
$$= a+3a^{1/3}b^{1/3}c + b = a+3(ab)^{1/3}c + b =c^3.$$
Therefore
$$(a+3(ab)^{1/3}c + b)^{1/3} = c = a^{1/3} + b^{1/3}$$
or
$$ a+ b =c^3-3(ab)^{1/3}c = c(c^2-3(ab)^{1/3}).$$
Because you are mixing the terms, you can get some messy equations, but it does not mean they are wrong.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1105306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Boolean algebra Simplification of "xy + x'z + yz" I'd like to simplify the following expression "xy + x'z + yz":
xy + x'z + yz = xy + z(x' +y)
= (xy + z)(xy + x' + y)
= (xy + z)(y(x + 1) + x')
= (xy + z) ( y + x')
What do I do after the last step?
|
By either drawing a Karnaugh map or recognizing that this is just the Consensus Theorem, observe that:
\begin{align*}
xy + x'z + yz
&= xy + x'z + (1)yz \\
&= xy + x'z + (x + x')yz \\
&= xy + x'z + (xyz + x'yz) \\
&= (xy + xyz) + (x'z + x'yz) \\
&= xy(1 + z) + x'z(1 + y) \\
&= xy(1) + x'z(1) \\
&= xy + x'z
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1108195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Determining conic section equation given foci and sum of distance to each point Disclaimer: This title was hard to formulate. Edits welcome.
Problem:
Given foci $$F_1 = (1,0)$$ $$F_2 = (3,0)$$
of a conic section, find the equation for all points $P$ that satisfy $$|PF_1| + |PF_2| = 6$$
My attempt:
I tried going about it algebraically. Letting $$P = (x,y)$$
I formulated
$$|PF_1| = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(x-1)^2 + y^2}$$
and likewise for $|PF_2|$ and then rewriting the above equation, but this turned out to be a mess, and even when I resorted to WolframAlpha, it turned out to be far uglier than I believe is intended.
I expect there is a more elegant solution here, but I'm not able to find it.
|
It's not extremely difficult, just take care not to have a square root everywhere:
$$
\sqrt{(x-1)^2+y^2}+\sqrt{(x-3)^2+y^2}=6 \\
\sqrt{(x-1)^2+y^2} = 6 - \sqrt{(x-3)^2+y^2} \\
(x-1)^2+y^2 = 36 + (x-3)^2+y^2 - 12\sqrt{(x-3)^2+y^2} \\
x^2-2x+1+y^2-36-(x^2-6x+9)-y^2 = -12\sqrt{(x-3)^2+y^2} \\
-12\sqrt{(x-3)^2+y^2} = 4x-44 \\
9\left[x^2-6x+9+y^2\right] = (x-11)^2 \\
8x^2-32x+9y^2=40 \\
8(x-2)^2+9y^2=72 \\
(x-2)^2/9+y^2/8=1 \\
\left(\frac{x-2}{3}\right)^2+\left(\frac{y}{2\sqrt{2}}\right)^2 = 1
$$
An ellipse with midpoint $(2,0)$ and axes $(3,2\sqrt{2})$ :
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1109508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$
Then
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
|
$$a_{n}=\left(\sqrt{n}-\sqrt{n-1}\right)-\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{\sqrt{n}+\sqrt{n-1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1109867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
}
|
Closed form of a recursive relation A sequence $\langle a_n\rangle$ is defined recursively by $a_1=0$, $a_2=1$ and for $n\ge 3$,
$$a_n=\frac 12 na_{n-1}+\frac 12n(n-1)a_{n-2}+(-1)^n\left(1-\frac n2\right).$$
Find a closed form expression for
$$f_n=a_n+2\binom n1a_{n-1}+3\binom n2a_{n-2}+\cdots+(n-1)\binom n{n-2}a_2+n\binom n{n-1}a_1.$$
I substituted $b_k=\binom n{n-k}a_k$ which reduces the given recursion to $$b_n=\frac {b_{n-1}}2+b_{n-2}+(-1)^n\left(1-\frac n2\right)\\ \Longrightarrow2b_n-b_{n-1}-2b_{n-2}=(-1)^n(2-n)\\ \Longrightarrow 2b_{n-1}-b_{n-2}-2b_{n-3}=-(-1)^n(3-n)\\ \Longrightarrow2b_{n-1}-b_{n-2}-2b_{n-3}=-(-1)^n(3-n)\\ \Longrightarrow 2b_{n-2}-b_{n-3}-2b_{n-4}=(-1)^n(4-n)$$ Adding the last four equations, we get $$2b_n+3b_{n-1}-2b_{n-2}-5b_{n-3}-2b_{n-4}=0$$ Now using the standard way of solving such recursions, we set $b_k=\lambda^k$, which gives $$2\lambda^4+3\lambda^3-2\lambda^2-5\lambda-2=0$$ We have to find $f_n=a_n+2\binom n1a_{n-1}+3\binom n2a_{n-2}+\cdots+(n-1)\binom n{n-2}a_2+n\binom n{n-1}a_1\\=b_n+2b_{n-1}+3b_{n-2}+\cdots+(n-1)b_2+nb_1\\=\lambda^n+2\lambda^{n-1}+3\lambda^{n-2}+\cdots+(n-1)\lambda^2+n\lambda$
This is where I got stuck. What should I do after this?
|
Let
$$ f(z) = \sum_{n\geq 0} b_n z^n.\tag{1}$$
Since:
$$ \sum_{n\geq 0}\left(1-\frac{n}{2}\right)(-1)^n z^n = \frac{2+3z}{2(1+z)^2}$$
the recursion on $b_n$ gives:
$$ f(z) = \frac{1}{2}z\,f(z)+z^2\, f(z)+\frac{2+3z}{2(1+z)^2},$$
hence:
$$ f(z) = \frac{2+3z}{(1-z)^2(2-3z^2)}.\tag{2}$$
We want to find:
$$ f_n = n b_1 + (n-1) b_2 + \ldots + 2 b_{n-1} + b_{n} \\= (n+1)(b_1+\ldots+b_n)-(1\cdot b_1 + 2\cdot b_2 +\ldots+ n\cdot b_n)\tag{3}$$
where $b_1+\ldots+b_n$ can be extracted from:
$$ [z^n]\frac{f(z)}{1-z} $$
and $1\cdot b_1 + \ldots + n\cdot b_n $ can be extracted from:
$$ [z^{n-1}]\frac{f'(z)}{1-z},$$
hence it suffices to provide a partial fraction decomposition for the RHS of $(2)$.
I bet you can finish from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1113635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Demonstration of the inequality of Cauchy-Schwarz After the demonstration of the inequality of Cauchy-Schwarz make by my professor, I still don't understand some steps of the demonstration.
To prove this inequality, my professor use the induction princile.
First, verify $P(1)$.
\begin{align}
\big(\sum^1{a_kb_k}\big)^2 &\le \big(\sum^1{a_k^2}\big)\big(\sum^1{b_k^2}\big) \\
\big(a_1b_1\big)^2 &= a_1^2b_1^2
\end{align}
We can also verify $P(2)$.
\begin{align}
\big(\sum^2{a_kb_k}\big)^2 &\le \big(\sum^2{a_k^2}\big)\big(\sum^2{b_k^2}\big) \\
\big(a_1b_1+a_2b_2\big)^2 &\le \big(a_1^2+a_2^2\big)\big(b_1^2+b_2^2\big) \\
a_1^2b_1^2+2a_1b_1a_2b_2+a_2^2b_2^2 &\le a_1^2b_1^2+a_1^2b_2^2+a_2^2b_1^2+a_2^2b_2^2 \\
2a_1b_1a_2b_2 &\le a_1^2b_2^2+a_2^2b_1^2 \\
0 &\le a_1^2b_2^2+a_2^2b_1^2 -2a_1b_1a_2b_2 \\
0 &\le \big(a_1b_1-a_2b_2\big)^2
\end{align}
If we suppose $P(n)$ to be true, we can verify $P(n+1)$.
\begin{align}
\big(\sum^{n+1}{a_kb_k}\big)^2 &\le \big(\sum^{n+1}{a_k^2}\big)\big(\sum^{n+1}{b_k^2}\big) \\
\end{align}
By the inequality of the triangle, we can write the follow statement.
$$\left|\sum^{n+1}{a_kb_k}\right| \le \left|\sum^{n}{a_kb_k}\right|+\left|a_{n+1}\right|\left|b_{n+1}\right|$$
So, we can assume,
$$\left|\sum^{n+1}{a_kb_k}\right| \le \sqrt{\sum{a_k^2}}\sqrt{\sum{b_k^2}}+\left|a_{n+1}\right|\left|b_{n+1}\right|$$
because we know,
$$\left|\sum{a_kb_k}\right| \le \sqrt{\sum{a_k^2}}\sqrt{\sum{b_k^2}}$$
If we define $A_1=\sqrt{\sum{a_k^2}}$, $B_2=\sqrt{\sum{b_k^2}}$, $A_2=\left|a_{n+1}\right|$ and $B_2=\left|b_{n+1}\right|$ we can write this inequation like this :
$$\left|\sum^{n+1}{a_kb_k}\right| \le A_1B_1+A_2B_2$$
And than, there, it's where I don't understand...
$$\left|\sum^{n+1}{a_kb_k}\right| \le \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}$$
And why we have to verify $P(2)$.
|
Actually, this expression is Cauchy-Schwarz's inequality itself. I'll give a proof and I hope you can follow. We have that $$(A_1+tB_1)^2 + (A_2+tB_2)^2 \geq 0$$ for all $t \in \Bbb R$, for being a sum of squares. This is equivalent to: $$(A_1^2+A_2^2)+2(A_1B_1+A_2B_2)t + (B_1^2+B_2^2)t^2 \geq 0.$$
This a polynomial in $t$ which is always positive, so $\Delta \leq 0$. Seeing this as $at^2+bt + c$, we have that $a = B_1^2+B_2^2, b = 2(A_1B_1+A_2B_2)$ and $c = A_1^2+A_2^2$, if this helps you seeing it. Then: $$\Delta = b^2-4ac = 4(A_1B_1+A_2B_2)^2-4(A_1^2+A_2^2)(B_1^2+B_2^2) \leq 0,$$ so that dividing by $4$ and taking roots, results $$|A_1B_1+A_2B_2| \leq \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1114201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
If gcd$(a, 4) = 2$ and gcd$(b, 4) =2$, then gcd$(a + b, 4) = 4$ If gcd$(a, 4) = 2$ and gcd$(b, 4) =2$, then gcd$(a + b, 4) = 4$
can someone help me solve this.
|
$$\gcd(a, 4) = 2 \rightarrow a=4k+2\\\gcd(b, 4) = 2 \rightarrow b=4q+2\\\gcd(a+b, 4) = \gcd((4k+2+4q+2,4)=\gcd(4k+4q+4,4)\\\gcd(4(q+k+1),4)=4 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1114721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n $$
and we know that:
$$\sum_{n=1}^\infty b_n$$
doesnt converge cause
$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$
doesnt converge
so from here my conclusion is that $\sum_{n=1}^\infty a_n$ doesnt converge
but i know the final answer is that it does converge
so what am i doing wrong?
|
$$
\begin{align}
&2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\\[16pt]
&=(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})\\[9pt]
&=\frac1{\sqrt{n}+\sqrt{n-1}}-\frac1{\sqrt{n+1}+\sqrt{n}}\\
&=\frac{\sqrt{n+1}-\sqrt{n-1}}{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}\\
&=\frac2{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n-1})}\\
&\le\frac1{4(n-1)^{3/2}}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{n=1}^\infty\left(2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\right)
&\le(2-\sqrt2)+\sum_{n=2}^\infty\frac1{4(n-1)^{3/2}}
\end{align}
$$
and the series converges by the $p$-test.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1114819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Problem with Lagrange multipliers I am asked to find local extrema of $f(x,y,z)=ax+by$ ($a,b$ non-zero and fixed) defined on $\{(x,y,z)\colon (x,y)\neq 0\}$ subject to
$$\left (R-\sqrt{x^2+y^2}\right)^2 + z^2 - r^2 = 0.$$
(here $0<r<R$ are fixed). Okay, let us define the auxiliary function by:
$$F(x,y,z) = ax+by - \lambda \left( \left(R-\sqrt{x^2+y^2}\right)^2 + z^2 - r^2 \right).$$
How to get rid of $\lambda$ from $F^\prime_x, F^\prime_y, F^\prime_z$? Can one please help me to find stationary points of $F$?
|
First we calculate $F_x'$, $F_y'$, and $F_z'$:
$$F_x'=a-\lambda\left(2\left(R-\sqrt{x^2+y^2}\right)\times\frac{-2x}{2\sqrt{x^2+y^2}}\right)=a+\frac{2\lambda x\left(R-\sqrt{x^2+y^2}\right)}{\sqrt{x^2+y^2}}$$
$$F_y'=b-\lambda\left(2\left(R-\sqrt{x^2+y^2}\right)\times\frac{-2y}{2\sqrt{x^2+y^2}}\right)=b+\frac{2\lambda y\left(R-\sqrt{x^2+y^2}\right)}{\sqrt{x^2+y^2}}$$
$$F_z'=-2\lambda z$$
To make these equations nicer, let's define $u:=\frac{2\left(R-\sqrt{x^2+y^2}\right)}{\sqrt{x^2+y^2}}$. Now, setting the derivatives to $0$, we have the following system of equations:
$$0=a+\lambda x u$$
$$0=b+\lambda y u$$
$$0=-2\lambda z$$
From the final equation, we have either $\lambda=0$ (in which case we have a solution if and only if we also have $a=b=0$ and so $f(x,y,z)=0$), or $z=0$. Since we have dealt with the former case, we'll assume from here that $z=0$.
Now multiplying the first equation by $y$, the second by $x$, and taking the difference, we get
$$0=ay+\lambda xyu-bx-\lambda xyu=ay-bx$$
so we have $y=\dfrac{b}{a}x$.
Finally, we can plug this into our constraint equation:
$$\begin{align}&0=\left (R-\sqrt{x^2+y^2}\right)^2 + z^2 - r^2
\\\implies&0=\left (R-\sqrt{x^2+\left(\frac{b}{a}\right)^2x^2}\right)^2 - r^2
\\\implies&r^2=\left (R-|x|\sqrt{1+\left(\frac{b}{a}\right)^2}\right)^2
\\\implies&\pm r=R-|x|\sqrt{1+\left(\frac{b}{a}\right)^2}
\\\implies&|x|=\frac{R\pm r}{\sqrt{1+\left(\frac{b}{a}\right)^2}}
\\\implies&x=\frac{\pm R\pm r}{\sqrt{1+\left(\frac{b}{a}\right)^2}}
\end{align}$$
Since we have $y=\frac{b}{a}x$,
$$y=\frac{\pm R\pm r}{\sqrt{1+\left(\frac{a}{b}\right)^2}}$$
Hence the critcal points are
$$\frac{a(\pm R\pm r)}{\sqrt{1+\left(\frac{b}{a}\right)^2}}+\frac{b(\pm R\pm r)}{\sqrt{1+\left(\frac{a}{b}\right)^2}}$$
Note in this final expression if we choose signs for $R$ and $r$ in the $ax$ term, we need to choose the same signs in the $by$ term.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1117793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
What is $\lim_{n\to\infty}\left(\frac{2}{5}\right)^{1/n}$? What is $$\lim\limits_{n\to\infty}\left(\frac{2}{5n}\right)^{\frac{1}{n}}$$ This is clearly equal to $\lim\limits_{n\to\infty}\left(\frac{2}{5}\right)^{\frac{1}{n}}\left(\frac{1}{n}\right)^{\frac{1}{n}}$. We know that $\lim\limits_{n\to\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}=1$. What is $\lim\limits_{n\to\infty}\left(\frac{2}{5}\right)^{\frac{1}{n}}$, though?
|
We have $$\left(\frac{2}{5n}\right)^{\frac{1}{n}} = e^{\frac{1}{n} \log \left(\frac{2}{5n}\right)}$$ and $$\frac{1}{n} \log \left(\frac{2}{5n}\right) = \frac{\log2 - \log5 - \log n}{n} \to 0$$ Thus $$\lim\limits_{n\to\infty}\left(\frac{2}{5n}\right)^{\frac{1}{n}} = 1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1118072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
All automorphisms of splitting fields Let $M \le \mathbb{C} $ be the splitting field of polynomial $ f(x)
\in \mathbb{Q}[x] $. Find all automorphisms of field $ M $ in cases:
1) $ f(x) = x^6 - 1 $
2) $ f(x) = x^{2011} - 1 $
1) In the first case I found all complex roots of the 6th degree of 1. It is $$ 1, \frac{1}{2} + i \frac{\sqrt{3}}{2}, -\frac{1}{2} + i \frac{\sqrt{3}}{2}, -1, -\frac{1}{2} - i \frac{\sqrt{3}}{2}, \frac{1}{2} - i \frac{\sqrt{3}}{2} $$
Hence splitting field is $ M = \mathbb{Q}(i\sqrt{3}) $ and $ \dim(\mathbb{Q}(i\sqrt{3}):\mathbb{Q}) = 2 $
All elements in $ M $ have the form $ a + b \cdot i\sqrt{3} $, where $ a, b \in \mathbb{Q} $
Let $f$ is our automorphism and $ f(i\sqrt{3}) = \alpha $.
$$ -3 = f(-3) = f((i\sqrt{3})^2) = \alpha^2 $$
And $ \alpha $ can be $ i\sqrt{3} $ or $ -i\sqrt{3} $.
If $ \alpha = i\sqrt{3} $ then it is identity automorphism and $f(a + b \cdot i\sqrt{3}) = a + b \cdot i\sqrt{3}$
If $ \alpha = -i\sqrt{3} $ then $f(a + b \cdot i\sqrt{3}) = a - b \cdot i\sqrt{3}$
And there are exactly 2 automorphisms. Is that correct solution?
2) But I have no solution in the second case. I have an idea to take one of the roots of 2011th degree of 1: $ \phi = \cos{\frac{2\pi}{2011}} + i\sin{\frac{2\pi}{2011}}$ and find degree of minimal irreducible polynomial for $ \phi $ for find $ n $ in form of element $ a = a_0 + a_1\phi + ...+a_n\phi^n $
And then write that $1 = f(1) = f(\phi^{2011}) = \alpha^{2011}$
But I have no finished solution...
Thanks for the help!
|
For $(2)$, we have that $2011$ is a prime number taking the thousand and eleventh root of unity, that is $\xi \neq 1$. Then $L = \mathbb{Q}[\xi]$ and as
$$\xi^{2010}+\xi^{2009}+\ldots +\xi^2 + \xi + 1 = 0$$
and $p(x) = x^{2010}+x^{2009}+\ldots +x^2 + x + 1$ is irreducible* over $\mathbb{Q}$ then $[L:\mathbb{Q}] = 2010$.
To find the automorphism notice that $\sigma \in Aut_{\mathbb{Q}} L $ is complete determined by $\sigma(\xi)$. That is the possibilies are $\{\xi,\xi^2\,\ldots,\xi^{2010}\}$.
(*)It's possible to prove for any $p$ prime that
$$x^{p-1} + x^{p-2} + \ldots + x + 1$$
is irreducible over $\mathbb{Q}$. Show that $$q(x+1) = (x+1)^{p-1}+ (x+1)^{p-2}+\ldots+ (x+1)+ 1$$
is irreducible by using Eiseinstein Criterion.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1119006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
So first we have two potentially problematic points which are $1,\infty$
We split the integral to $$\int_1^2 \frac{x\ln x}{x^4-1} dx + \int_2^\infty \frac{x\ln x}{x^4-1} dx$$
Now first I tried to study $I_1 = \int_1^2 \frac{x\ln x}{x^4-1} dx$.
I evaluated $\lim_{x\to 1^+} \frac{x\ln x}{x^4-1} = \frac{1}{4}$ so basically the function is "nice" at $x=1$ and we can evaluate the integral, but how exactly?
I tried various of methods without luck (integration by parts , substituting $t=\ln x$ , $(x^4 - 1) = (x^2-1)(x^2+1)$ , $(x^4-1) = (x-1)(1+x+x^2+x^3)$)
EDIT
Following the answer I got, I thought about:
Since $\frac{x\ln(x)}{x^4-1}$ is monotonically decreasing to $0$ then we can use the inequality:
$$\int_1^\infty \frac{x\ln(x)}{x^4-1} \le \sum_{n=1}^\infty \frac{n\ln(n)}{n^4-1} \le \sum_{n=1}^\infty \frac{n^2}{n^4-1} \le \sum_{n=1}^\infty \frac{n^2}{n^4} = \sum_{n=1}^\infty \frac{1}{n^2} \lt \infty$$
|
If the problem asks "Study the convergence", you're not required to find a closed form. Once you know the function is "nice" at $1$, you're done with that part: $I_1$ exists. Next look at $I_2$, and again don't worry about finding a formula for it, just whether or not it converges.
By the way, the integral from $1$ to $\infty$ happens to be $\pi^2/32$, but again you are not expected to find that. The integrals from $1$ to $2$ and $2$ to $\infty$ are much more complicated, involving the dilog function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1119289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !
|
(a - b)^2 + (b - c)^2 + (a - c)^2 = 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 2 - 2(ab + bc + ac). This is how we prove that ab + bc + ca <= 1.
Now:
If ab + bc + ac < -1/2, then 2(ab + bc + ac) < -1.
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2(a + b)(c) = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 1 + 2(ab + ac + bc) < 1 + (-1) = 0. Which is impossible.
(Edited).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1119879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Understanding how to estimate $\pi(x)$ based on Paul Erdos's proof of Bertrand's Postulate I am reading the 4th Edition of Proofs from the Book. I am not clear on how the proof behind Bertrand's postulate leads to the following statement on page 10 (of my edition):
From (2) one can derive with the same methods that $\prod\limits_{n<p\le 2n}p \ge 2^{\frac{1}{30}n}$ for $n \ge 4000$
Here's (2):
$4^n \le (2n)^{1+\sqrt{2n}} \cdot \prod\limits_{\sqrt{2n}< p \le \frac{2}{3}n} p \cdot \prod\limits_{n < p \le 2n} p$ for $n \ge 3$
After (2) comes the details that for completing the proof. I understand how this leads to the proof. I don't understand how this leads to the estimate for the product of primes
Here are the details presented:
Here's (1):
$\prod\limits_{p \le x}p \le 4^{x-1}$ for all real $x \ge 2$
Assuming that there are no primes between $n$ and $2n$, applying (1) to (2):
Substituting (1) into (2) we get:
$$4^n \le (2n)^{1+\sqrt{2n}}4^{\frac{2}{3}n}$$
Then applying $a+1 < 2^a$, the result becomes:
$$2n = (\sqrt[6]{2n})^6 < (\left\lfloor\sqrt[6]{2n}\right\rfloor + 1)^6 < 2^{6\left\lfloor\sqrt[6]{2n}\right\rfloor} \le 2^{6\sqrt[6]{2n}}$$
Then, for $n \ge 50$ ($18 < 2\sqrt{2n}$), combining the last two results gets:
$$2^{2n} \le (2n)^{3(1 + \sqrt{2n})} < 2^{\sqrt[6]{2n}(18+18\sqrt{2n})} < 2^{20\sqrt[6]{2n}\sqrt{2n}} = 2^{20(2n)^{\frac{2}{3}}}$$
While I follow all the details up to this point in the proof (I think), I don't follow how this gets us to:
From (2) one can derive with the same methods that $\prod\limits_{n<p\le 2n}p \ge 2^{\frac{1}{30}n}$ for $n \ge 4000$
If someone could provide details or an explanation about this last result, I would greatly appreciate it.
|
We can put (1) into (2) and we get$$4^{n}\leq\left(2n\right)^{1+\sqrt{2n}}\underset{\sqrt{2n}<p\leq\frac{2}{3}n}{\prod}p\underset{n<p\leq2n}{\prod}p\leq\left(2n\right)^{1+\sqrt{2n}}4^{\frac{2}{3}n-1}\underset{n<p\leq2n}{\prod}p\,\Longrightarrow$$
$$\frac{2^{2n-\frac{4}{3}n-\sqrt{2n}+1}}{n^{1+\sqrt{2n}}}\leq\underset{n<p\leq2n}{\prod}p.$$
Now, if $n$
is large enough (and $n\geq4000$
is large enough) holds$$\frac{1}{n^{1+\sqrt{2n}}}>\frac{1}{2^{n/2}}$$
so$$2^{2n-\frac{4}{3}n-\sqrt{2n}+1-\frac{n}{2}}\leq\underset{n<p\leq2n}{\prod}p$$
and trivially $$2n-\frac{4}{3}n-\sqrt{2n}+1-\frac{n}{2}=\frac{12n-8n-6\sqrt{2n}+6-3n}{6}=\frac{n-6\sqrt{2n}+6}{6}>\frac{n}{30}$$
if $n$ large enough, so we have $$2^{n/30}\leq\underset{n<p\leq2n}{\prod}p.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1120766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
|
A nice solution can be obtained by modifying the "exact" solution. The "exact" solution is
$$
f(x) = \frac{8-\pi}8 + \sqrt{x(1-x)}
$$
which has an arc length of $\frac{8+\pi}4$.
As such, I propose a solution of the form
$$
f(x) = \sqrt{x(1-x)}(1+g(x))
$$
where the "exact" solution uses $g(x)=(8-\pi)/(8\sqrt{x(1-x)})$. We want a solution similar to this, but with a finite value at $x=0$ and $x=1$. As such, I propose a simple modification.
$$
f(x) = \sqrt{x(1-x)}\left(1+\frac{A}{\sqrt{(x+B)(1+B-x)}}\right)
$$
Note that we recover the "exact" solution if $B=0$ and $A=\frac{8-\pi}8$. We can thus get arbitrarily close to this solution by selecting appropriate values for $A$ and $B$. Although a closed-form expression relating the two parameters isn't obvious, values can be chosen numerically. For example, for $B=0.0001$, we have $A\approx\frac{8-\pi}8+0.00058333971346\approx0.60788425801473$. For these, we have
$$
\int_0^1 \sqrt{1+f'(x)^2}dx\approx 2.78567 \approx \frac{8+\pi}4 + 2.67\times10^{−4}
$$
In this case, the expression works out to be
$$
f(x)=\sqrt{x(1-x)}\left(1+\frac{0.60788425801473}{\sqrt{(x+0.0001)(1.0001-x)}}\right)
$$
Note that this can also be expressed as
$$
f(x)=\sqrt{x(1-x)}\left(1+\frac{0.60788425801473}{\sqrt{x(1-x)+0.00010001}}\right)\tag{$\dagger$}
$$
Here is the graph of the $f(x)$ given in $(\dagger)$:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 12,
"answer_id": 1
}
|
Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta
I am having difficulty finding the $\delta$ value.
Here is what I have done so far:
What I want to show:
$$\forall \epsilon > 0, \exists \delta > 0 \; such \; that \; 0<\mid x -2 \mid < \delta \implies \mid x^4 - 2x^3 + x - 2\mid < \epsilon $$
After a lot of manipulation, I was able to turn the equation to the following:
$$\mid x^4 - 2x^3 + x - 2\mid \; = \; \mid x-2 \mid \mid x-2 \mid \mid x + 1 \mid \mid x + 1 \mid + 3\mid x + 1 \mid \mid x - 2 \mid$$
I already know that $ \mid x -2 \mid < \delta$. I must find an upper bound for $\mid x + 1 \mid$.
$$\mid x + 1 \mid = \mid x + 1 -3 + 3 \mid \; \le \; \mid x - 2 \mid + 3 $$
Choose $\delta = 1$, then
$$\mid x - 2 \mid < 1 \implies \mid x -2 \mid + 3 < 4$$
Therefore $\mid x + 1 \mid < 4$
This means:
$$\mid x^4 - 2x^3 + x - 2\mid < (\delta)(\delta)(4)(4) + 3(\delta)(4) = 16\delta^2 + 12\delta = \epsilon$$
This is where I get stuck. How do I transform $16\delta^2 + 12\delta = \epsilon$ into solving for $\delta$ in terms of $\epsilon$?
I am not sure if I am over thinking the whole question or if the above is even remotely correct. Please let me know any mistakes that I have made.
|
Here's a very easy and brief way to do it:
Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-2|<\delta$, then $|(x^4-2x^3+x+3)-5|<\epsilon$.
Now,
$$|(x^4-2x^2+x+3)-5|=|(x-2)(x^3+1)|.
$$ If $|x-2|<1$, that is, $1<x<3$, then $x^3+1<3^3+1=28$, and so
$$
|(x^4-2x^3+x+3)-5|=|x-2|(x^3+1)<28|x-2|.
$$ So if we take $\delta=\min\left\{1,\frac{\epsilon}{28}\right\}$, then
$$
0<|x-2|<\delta\to|(x^4-2x^2+x+3)-5|=|(x-2)|(x^3+1)<\frac{\epsilon}{28}\cdot 28=\epsilon.
$$ Thus, by the definition of a limit, $\lim_{x\to 2}(x^4-2x^3+x+3)=5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1124293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Existence of solution to Congruence relation $(x^2-2)(x^2-6)(x^2-3) \equiv 0\pmod p$ I'm taking the final exam in "Number Theory" tomorrow and stuck with:
Prove that $\,\,\forall p\in\mathbb{Z}_p\,$ the congruence relation: $$(x^2-2)(x^2-6)(x^2-3) \equiv 0\pmod p$$ has a solution.
Here $p$ is an arbitrary prime number.
Any hints? Ideas?
|
For $p=2,$
$$(x^2-2)(x^2-3)(x^2-6)\equiv x^2(x^2-1)x^2\pmod2$$
But $x^2(x^2-1)x^2$ is divisible by $x(x-1)$ which being a product of two consecutive integers is always divisible by $2$
$$(x^2-2)(x^2-3)(x^2-6)\equiv0\pmod2 $$ for all integer $x$
For $p=3,$
$$(x^2-2)(x^2-3)(x^2-6)\equiv (x^2+1)(x^2)x^2\pmod3$$
But $3\nmid(x^2+1),$ so we need $x^4\equiv0\pmod3\iff x\equiv0$
For prime $p>3,$
Using Legendre Symbol for prime $p$, $$\left(\frac6p\right)=\left(\frac3p\right)\cdot\left(\frac2p\right)$$
At least one of $\displaystyle\left(\frac6p\right),\left(\frac3p\right),\left(\frac2p\right)$ must be $=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1124845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
What are all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$? I am having some problems getting started with this problem, as I never had to deal with an inequality that was between two values with absolute values. Any help is appreciated. The problem is find all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$. I keep trying to find cases with $x < -2$ or $x \geq 1$, but that is not getting me anywhere.
|
If $4 < |x+2| + |x-1| < 5$ then $\ \pm (x+2) \pm (x-1) < 5$. Note that for an opposite choice of sign the inequality is never satisfied since $(x+2) - (x-1)=3$. So the inequality implies
$$ 4 < (x+2) + (x-1) < 5$$
or
$$4 < -(x+2) -(x-1) < 5$$
The inequality $ 4 < (x+2) + (x-1) < 5$ is equivalent to $4 < 2x+1<5 \iff 1.5 < x < 2$ and on this interval both $x+2$, $x-1$ are $>0$ so we have part of the solution.
The inequality $ 4 < -(x+2) -(x-1) < 5$ is equivalent to $4 < -2x-1<5 \iff -3 < x < -2.5$ and on this interval both $x+2$, $x-1$ are $<0$ so we have another part of the solution.
Therefore, the set of solutions is $(-3,-2.5) \cup (1.5,2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1126165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
What is $\int \frac{1}{\sqrt{25y^2-10y-3}}dy$ $= \int \dfrac{1}{\sqrt{(5y-1)^2-4}}dy$
$=\int \dfrac{1}{\sqrt{u^2-4}}\dfrac{du}{5}, \quad U$ substitution
$=\int \dfrac{1}{10\cos(\theta)} 2\cos(\theta) d\theta, \quad$ Trig substitution
$= \dfrac{1}{5} \theta$
$= \dfrac{\cos^{-1}\frac{\sqrt{(5y-1)^2-4}}{2}}{5} +C$
Where did I go wrong?
work after u substitution $\frac{1}{5}\int \sec \theta d\theta \\ =\frac{1}{5} \ln \left | \sec \theta + \tan \theta\right | $
* New Answer* $\\
\frac{1}{5}\ln \left | \frac{5y-1}{2} +\frac{\sqrt{(5y-1)^2 -4}}{2}\right | +C$
|
The problem is in the 3rd line. $$\sqrt{4\sin^2\theta-4} \neq 2\cos \theta$$
Try $x=2\cosh\theta$ or $x=2\sec\theta$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1127251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Calculation of determinant of an arrowhead matrix Is there any easier way to make sure the determinant of the following $n \times n$ matrix is $n$?
$$\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & 0 &\cdots & 1
\end{vmatrix} = n$$
I figured it with a smaller dimension and it indeed produces the determinant that is the size of dimension. I tried to do a cofactor expansion with the first row, and each term produces the determinant of $1$ and if you sum them up, then the total determinant will be $n$. But the sign change for each cofactor is confusing, and it is not easily seen that each cofactor term is actually positive $1$.
|
We know that for $n=2$ the statement is true. We'll prove the rest by induction.
We can compute the determinant based on the last row. The result is the sum of two determinants
$\ \ \ \begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & & 0 \\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 & 1
\end{vmatrix}=(-1)^{n+1}
\begin{vmatrix}
-1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 0 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & 0 & \cdots & 0 \\
0 & 0 & 1 & 0 & & 0 \\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 & 0
\end{vmatrix}
+\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & & 0 \\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 & 1
\end{vmatrix},$
where the second term is of the form of the original determinant. By hypothesis the value of this second determinant is $n-1$.
From here it would be enough to show that the determinant of the first term is $-1$ for even $n$'s and $+1$ for odd $n$'s. For $n=2 \text { and } 3$ this is true.
In the general case we can do the first term based on the last row. Surprisingly, by hiding the last but one column and the last row we get a determinant of the form of its predecessor. This completes the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1128330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
Alternative matrix representation for translation The ''usual" way to write translation for $\textbf{v}\in \mathbb{R}^2$ is with the following $3\times3$ matrix
$$ \left( {\begin{array}{ccc}
1 & 0 & x_{0} \\
0 & 1 & y_{0} \\
0 & 0 & 1 \\
\end{array} } \right) \cdot \left( {\begin{array}{c}
x \\
y \\
1 \\
\end{array} } \right)= \left( {\begin{array}{c} x+x_{0}\\y+y_{0}\\1\end{array}}\right)$$
But also it seems to be possible to write this transformation with the following $2\times2$ matrix:
$$ \left( {\begin{array}{cc}
1 & \frac{x_{0}}{y} \\
\frac{y_{0}}{x} & 1 \\
\end{array} } \right) \cdot \left( {\begin{array}{c}
x \\
y \\
\end{array} } \right)= \left( {\begin{array}{c} x+x_{0}\\y+y_{0}\end{array}}\right)$$
I didn't find a source using this as a version of translation, what are the problems associated with this representation?
|
I want to answer my own question in response to the answer given above. In the answer Marc van Leeuwen says that ''matrix representing linear transformation should not depend on the input"
Why exactly? What properties of linear transformations prevent that?If we find a transformation that depends on the input but satisfies $A(\alpha \textbf{v}+\beta \textbf{u})=\alpha A(\textbf{v})+\beta A(\textbf{u})$ isn't that automatically a linear transformation?
If we try it on the above transformation we see where the problem comes from:
Let $\textbf{u} = \left( {\begin{array}{c} u^1\\u^2\end{array}}\right)$ and $\textbf{v} = \left( {\begin{array}{c} v^1\\v^2\end{array}}\right)$ Then from the definition of $A$ we have:
$$A(\alpha \textbf{v}+\beta \textbf{u})=A\cdot \Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} \\ \alpha u^{2}+\beta v^{2} \end{array}} \Bigg) = \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{\alpha u^2+\beta v^2}\\
\frac{y_{0}}{\alpha u^1 + \beta v^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} \\ \alpha u^{2}+\beta v^{2} \end{array}} \Bigg) $$
$$=\Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} + x_{0} \\ \alpha u^{2}+\beta v^{2} + y_{0} \end{array}} \Bigg)$$
On the other hand
$$\alpha A(\textbf{v})+\beta A(\textbf{u}) = \alpha \cdot \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{ u^2}\\
\frac{y_{0}}{u^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} u^{1} \\ u^{2} \end{array}} \Bigg)+\beta \cdot \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{ v^2}\\
\frac{y_{0}}{v^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} v^{1} \\ v^{2} \end{array}} \Bigg)$$
$$=\Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} + (\alpha + \beta)x_{0} \\ \alpha u^{2}+\beta v^{2} + (\alpha + \beta)y_{0} \end{array}} \Bigg)$$
So we proved that $A(\alpha \textbf{v}+\beta \textbf{u})\neq \alpha A(\textbf{v})+\beta A(\textbf{u})$ thus the transformation provided by the matrix $A$ is not linear.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1128777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\begin{align}
& \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\
& \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\
& e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\
& e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\
& e^{-\infty} = 0\\
\end{align}
$$
Edit #1:
Continuing after the mistake of the $\tan(x)$:
$$\begin{align}
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\
\end{align}
$$
|
\begin{align*}
\left(\frac{\sin x}x\right)^{\frac1{x^2}}
&=\left[1+\left(\frac{\sin x}x-1\right)\right]^{\frac1{x^2}}\\
&=\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{\frac1{x^2}}\\
&=\left[\underbrace{\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{[x/(\sin x-x)]}}_{=:A(x)}\right]^{\frac{\sin x-x}{x^3}}
\end{align*}
Next observe that
$\bullet\sin x=x-\frac{x^3}{6}+o(x^3) \Longrightarrow\frac{\sin x-x}{x^3}\stackrel{x\to0}{\longrightarrow}-\frac16$;
$\bullet\lim_{\xi\to\infty}\left(1+\frac1\xi\right)^{\xi}=e \Longrightarrow A(x)\stackrel{x\to0}{\longrightarrow} e \;\;(\mbox{in fact} \;\;x/(\sin x-x)\stackrel{x\to0}{\longrightarrow}\infty)$
thus we immediately get
$$
\lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=e^{-\frac16}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Area of an isosceles triangle where the tangents of some angles are in geometric progression In $\triangle ABC$, $AB=BC$ and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $\overline{BE}=10$. The values of $\tan CBE$, $\tan DBE$, and $\tan ABE$ form a geometric progression, and the values of $\cot DBE$, $\cot CBE$, and $\cot DBC$ form an arithmetic progression. What is the area of $\triangle ABC$?
I'm pretty stuck on this problem. My first thought was to convert cotangent into tangent, but that got messy very quickly. To find the area of $\triangle ABC$, we would need to use the base and the height, or $BD$ and $AE$. I don't know how to go about finding those lengths. Any helps would be great!
|
Here is a picture:
You know that $\triangle ABC$ is isosceles, and $D$ is the midpoint of the base across from $B$. Further, $\triangle BDE$ is a right triangle.
Let $\theta = \angle DBE$ and $\phi = \angle DBC$, so $\tan \angle CBE = \tan (\theta - \phi)$ and $\tan \angle ABE = \tan(\theta+\phi)$. Applying sum and difference formulas, we get
$$ \tan \angle CBE = \frac{\tan\theta-\tan \phi}{1+\tan\theta\tan \phi} = \frac{x-y}{1+xy} $$
$$ \tan \angle ABE = \frac{\tan\theta+\tan \phi}{1-\tan\theta\tan \phi} = \frac{x+y}{1-xy} $$
where $x=\tan\theta $ and $y=\tan \phi$. For a geometric progression, we need
$$ \frac{\frac{x-y}{1+xy} }{x} = \frac{x}{ \frac{x+y}{1-xy} }$$
Simplifying,
$$ x^2 = \frac{x^2-y^2}{1-x^2y^2}$$
$$ x^4 = 1 $$
So the only valid solution is $x=1$, or $\angle DBE = \frac{\pi}{4}$.
Looking to the cotangents, $\cot\angle DBE = 1$, $\cot\angle CBE = \frac{1+y}{1-y}$, and $\cot\angle DBC = \frac{1}{y}$ must form an arithmetic progression, so
$$ \frac{1+y}{1-y} - 1 = \frac{1}{y} - \frac{1+y}{1-y} $$
Simplifying,
$$ 3y^2-2y-1=0$$
So $y = \frac{1}{3},-1$. Clearly, $\frac{1}{3}$ is the only solution that makes sense.
Now, the area of $\triangle ABC = \overline{BD}^2 \tan\angle DBC = \left(\frac{10}{\sqrt{2}}\right)^2 y = \frac{50}{3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1132236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$
After that I tried using the formula:
$$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$
to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.
Any help would be appreciated!
|
Use the linearisation formulae. The expression can be rewritten as:
$$2^2\Bigl(2\sin\frac{\pi}{16}\,\sin\frac{7\pi}{16}\Bigr)\Bigl(2\sin\frac{3\pi}{16}\,\sin\frac{5\pi}{16}\Bigr)=2^2\cos\frac{3\pi}{8}\cos\frac{\pi}{8}=2\cos\frac{\pi}{4}=\sqrt2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
}
|
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I can apply the standard limit laws.
|
$$\lim_{x\rightarrow \infty }\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\\\lim_{x\rightarrow \infty }\frac{(x+1)^{x+1}}{(x+2)^{x+1}}*\frac{x}{x+2}=\\\lim_{x\rightarrow \infty }(\frac{x+1}{x+2})^{x+1}\frac{x}{x+2}=\\(1-\frac{1}{x+2})^{x+1}* \lim_{x\rightarrow \infty }\frac{x}{x+2}=\\e^{-1}*\lim_{x\rightarrow \infty }\frac{x}{x+2}\\e^{-1}*1\\=\frac{1}{e}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
}
|
Jordan matrices with the only eigenvalue $1$. So, I need to list all of the Jordan matrices of a $4x4$ matrix with the only eigenvalue $1$. If the only eigenvalue is 1, then there can't be any other value on the diagonal, correct? So am I wrong in thinking that only Jordan matrix would be this? $$ J_{1,n} =\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0&1&1&0\\0&0&1&1\\0&0&0&1 \end{bmatrix}$$
|
Read egreg's comment and complete. Some of the possibilities are:
$$\begin{align}&4=4\;\;\rightarrow&\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{pmatrix}\\{}\\
&4=3+1\;\;\rightarrow&\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\\{}\\
&4=2+2\;\;\rightarrow&\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&1\end{pmatrix}\end{align}$$
Can you see the relation? $\;4=n+k+r=$ one block of size $\;n\;$ and one of size $\;k\;$ and one of size $\;r\;$ (and etc.), and thus, by the comment quoted before, there are as many possibilities as partitions of $\;4\;$ , which are five.
Complete now according to the other partitions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1136759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve $\int_{0}^{t}\cos x\sin (t-x)dx$ I want to solve the following integral $$\int_{0}^{t}\cos x\sin (t-x)dx$$
What I tried to do is integration by parts twice, but it just brought me straight back to $$\int_{0}^{t}\cos x\sin (t-x)dx=\int_{0}^{t}\cos x\sin (t-x)dx$$ Not very informative indeed.
Is there any way to do this without resorting to trigonometric identities?
|
We have
\begin{align}
\int_0^t\cos\left(x\right)\sin\left(t-x\right)\:dx,\tag{1}
\end{align}
and since $\sin\left(t-x\right)=\sin\left(t\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(t\right)$,we may re-write this integral as
\begin{align}
&\int_0^t \cos\left(x\right)\left[\sin\left(t\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(t\right)\right]\:dx\tag{2}\\
&=\sin\left(t\right)\int_0^t \cos^2\left(x\right)\:dx-\cos\left(t\right)\int_0^t\sin\left(x\right)\cos\left(x\right)\:dx,\tag{3}
\end{align}
and because of the reduction formula
\begin{align}
\int\cos^{n}\left(x\right)\:dx=\frac{\cos^{n-1}\left(x\right)\sin\left(x\right)}{n}+\frac{n-1}{n}\int\cos^{n-2}\left(x\right)\:dx,
\end{align}
we have
\begin{align}
&=\sin\left(t\right)\left[\frac{\cos\left(x\right)\sin\left(x\right)}{2}\bigg|_0^t+\frac{1}{2}\int_0^t\:dx\right]-\cos\left(t\right)\left[\frac{1}{2}\sin^2\left(x\right)\right]_0^t\tag{4}\\
&=\frac{\cos\left(t\right)\sin^2\left(t\right)}{2}+\frac{\sin\left(t\right)t}{2}-\frac{\cos\left(t\right)\sin^2\left(t\right)}{2}\tag{5}\\
&=\boxed{\frac{t\sin\left(t\right)}{2}.}\tag{6}
\end{align}
Another way to look at this problem is via convolution such that
\begin{align}
\left(f\ast g\right)\left(t\right)=\int_0^t f\left(t-x\right)g\left(x\right)\:dx,\tag{7}
\end{align}
and in this case our $f\left(t-x\right)=\sin\left(t-x\right)$ and $g\left(x\right)=\cos\left(x\right)$. Therefore, this is simply
\begin{align}
\sin\left(x\right)\ast\cos\left(x\right)=\cos\left(x\right)\ast\sin\left(x\right)=\int_0^t\sin\left(t-x\right)\cos\left(x\right)\:dx\\=\int_0^t \cos\left(t-x\right)\sin\left(x\right)\:dx.\tag{8}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1137015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me that:
$$\frac{x}{(x-1)^2} = \frac{1}{(x-1)} + \frac{1}{(x-1)^2}$$
How to intuitively think about this partial fraction expansion? I've seen some examples but suddenly these conter intuitive examples opo out and I get confused. I can check that his is true but I couldn't find this expansion by myself
Then:
$$\int \frac{x}{(x-1)^2} dx = \int \frac{1}{(x-1)} + \frac{1}{(x-1)^2}dx = \ln (x-1) + (x-1)^{-1}$$
Then:
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+\frac{x}{(x-1)^2}dx = \frac{x^2}{2} + 2x + 3[\ln(x-1)+(x-1)^{-1}]$$ but wolfram alpha gives another answer. What I did wrong?
|
Here's a way to find the partial fraction expansion:
\begin{align}
\frac{x}{(x-1)^2} &= \frac{A}{x-1} + \frac{B}{(x-1)^2}\\
(x-1)^2\frac{x}{(x-1)^2} &= (x-1)^2\frac{A}{x-1} + (x-1)^2\frac{B}{(x-1)^2}\\
x &= A(x-1) + B
\end{align}
You can turn this into two linear equations and solve, or you can do the following:
*
*plug in $x = 1$ to get $1 = B$.
*Take derivatives on both sides to get
$$
1 = A
$$
*Plug $x = 1$ into this equation (which does nothing, but it's part of a pattern) to get $1 = A$.
In general, you plug in $x = c$ for this kind of equation [polynomial over a power of $x-c$], and for each derivative, and doing so will give you the coefficients one after another.
Is there an intuition for these partial fraction expansions? Yes, a slight one ...but it'll mostly come after you take Complex Analysis, alas.
Also: in integrating $1/(x-1)^2$, you have a sign error.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expression and then manipulating, and I also thought of integer solutions(none exist). How do I solve it then?
|
First, by expanding the four equations, we note that the system of equations is equivalent to
$$ \begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 4 & 2 & 4 & 4 & 8 & 2 & 4 & 4 & 8 & 4 & 8 & 8 \\
1 & 3 & 3 & 9 & 3 & 9 & 9 & 27 & 3 & 9 & 9 & 27 & 9 & 27 & 27 \\
1 & 4 & 4 & 16 & 4 & 16 & 16 & 64 & 4 & 16 & 16 & 64 & 16 & 64 & 64
\end{bmatrix}
\begin{bmatrix}
abcd \\
abc \\
abd \\
ab \\
acd \\
ac \\
ad \\
a \\
bcd \\
bc \\
bd \\
b \\
cd \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
14 \\
29 \\
52 \\
83
\end{bmatrix}.
$$
Using row reduction, we see that this is equivalent to the matrix equation
$$ \begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
abcd \\
abc \\
abd \\
ab \\
acd \\
ac \\
ad \\
a \\
bcd \\
bc \\
bd \\
b \\
cd \\
c \\
d
\end{bmatrix} =
\begin{bmatrix}
7 \\
3 \\
4 \\
0
\end{bmatrix}.
$$
This resolves into the following system of equations:
$$ \begin{aligned}
abcd = 7 \\
abc + abd + acd + bcd = 3 \\
ab + ac + ad + bc + bd + cd = 4 \\
a + b + c + d = 0.
\end{aligned}
$$
Finally, we note that $(a+5)(b+5)(c+5)(d+5)$ is equal to
$$abcd + 5(abc + abd + acd + bcd) + 25(ab + ac + ad + bc + bd + cd) + 125(a + b + c + d) + 625.$$
So $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 2
}
|
Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\ &=
\lim _{n\to \infty}\frac{n^2\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\&=
\lim_{n\to \infty}\frac{\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{\sqrt{3+\frac{1}{n^2}}}\end{align}$
Which looks like is equal to $\infty$ because of the $n^3\cdot 1$ but it's wrong.
What am I doing wrong here?
|
$$\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}\cdot\frac{\frac1{\sqrt n}}{\frac1{\sqrt n}}=\frac{\sqrt{1+\sqrt{1+\frac1{n^2}}}}{\sqrt{3+\frac1n}}\xrightarrow[n\to\infty]{}\sqrt\frac23$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Solve: $ab+bc+ca\mid (a+b+c)^2$ I couldn't make any progress on this problem, can anyone help?
I found it's the same as:
Find all integers $a,b,c$ such that $ab+bc+ca$ divides $a^2+b^2+c^2$.
I found a solution $a=-b=1$, and $c$ any integer.
Any more solutions?
|
Since $(a+b+c)^2 = 2(ab+ac+bc)+(a^2+b^2+c^2)$, if
$$ (a+b+c)^2 = k(ab+ac+bc), \tag{1}$$
then $k\geq 2$, and for $k=2$ we have only the trivial solution $(a,b,c)=(0,0,0)$.
Assuming $k=3$, we have:
$$ a^2+b^2+c^2 = ab+ac+bc \tag{2}$$
and by the Cauchy-Schwarz inequality $(2)$ holds only for $a=b=c$.
Assuming $k=4$ we have the parametric solution:
$$ (a,b,c) = (m^2,n^2,(n+m)^2) \tag{3}$$
so there are plenty of solutions, and even more can be computed by Vieta jumping.
Markov triples are deeply related.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$
$$p = \cos\theta \implies dp = -\sin\theta d\theta$$
$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$
$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$
$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$
But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.
Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?
|
$$
\begin{aligned}
\text { Let } y &=\sqrt{\frac{x}{x+1}}, \text { then } y^{2}=\frac{x}{x+1}=1-\frac{1}{x+1} \text{ and } \quad 2 y d y =\frac{1}{(x+1)^{2}} d x=\left(1-y^{2}\right)^{2} d x \\
I &=\int y \frac{2 y}{\left(1-y^{2}\right)^{2}} d y \\
&=2 \int\left(\frac{y}{1-y^{2}}\right)^{2} d y \\
&=2 \int\left[\frac{1}{2}\left(\frac{1}{1-y}-\frac{1}{1+y}\right)\right]^{2} d y \\
&=\frac{1}{2}\left[\frac{d y}{(1-y)^{2}}+\int \frac{d y}{(1+y)^{2}}-\int \frac{2}{(1-y) x(1+y} d y\right] \\
&=\frac{1}{2}\left[\frac{1}{1-y}-\frac{1}{1+y}-\int\left(\frac{1}{1-y}+\frac{1}{1+y}\right) d y\right]\\
&=\frac{\sqrt{\frac{x}{x+1}}}{1-\frac{x}{x+1}}+\ln \left|\frac{1+\sqrt{\frac{x}{x+1}}}{1-\sqrt{\frac{x}{x+1}}}\right|+C \\
&=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left[\left(1+\sqrt{\frac{x}{x+1}}\right)^{2}(x+1)\right]+C\\
&=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left|\left(1+2 \sqrt{\frac{x}{x+1}}+\frac{x}{x+1}\right)(x+1)\right|+C\mid \\
&=(x+1) \sqrt{\frac{x}{x+1}}+\ln \left|2 x+1+2(x+1) \sqrt{\frac{x}{x+1}}\right|+C
\end{aligned}
$$
$$*********$$
Because $\dfrac{x}{x+1}>0 \Leftrightarrow x<-1$ or $x>0$, we can simplify the answer by cases:
A. When $x\geq 0,$ as $\sqrt{(x+1)^{2}}=|x+1|=x+1$, we have $$
I=\sqrt{x(x+1)}+2 \ln (\sqrt{x}+\sqrt{x+1})+C
$$
B. When $x<-1$, as $\sqrt{(x+1)^{2}}=|x+1|=-(x+1)$ , we have$$
I=-\sqrt{x(x+1)}+\ln | 2 x+1-2 \sqrt{x(x+1)}|+C
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
}
|
From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
|
Replacing 1 with $\cos\frac2{x}{2}$ + $\sin\frac{x}{2}$, $\cos$$x$ with $\cos\frac2{x}{2}- \sin\frac{x}{2}$ and $\sin$$x$ with $2\sin\frac{x}{2}$$\cos\frac{x}{2}$ the numerator and denominator simplify to $2\sin^2\frac{x}{2}$ and $2\cos\frac{x}{2}\sin\frac{x}{2}$ and hence the fraction simplifies to $\tan\frac{x}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1144072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
How can I prove that $2/9$x$ and $y$ are real numbers.
Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?
Then can I use that to prove that for any natural number $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$
|
1) to show $x^4+y^4 < 8$, you already have a good solution by squaring $x^2+y^2<2+xy$.
2) By AM-GM and Cauchy Schwarz, we get
$$\left(x^4+y^4+\frac{x^4+y^4}2\right)(3) \ge (x^4+y^4+x^2y^2)(1^2+1^2+(-1)^2) \ge (x^2+y^2-xy)^2 > 1$$
hence $x^4+y^4 > \frac29$
For the more general case, you could use Power Means in addition to the above, i.e. for $n>2$:
$$\sqrt[2^n]{\frac{x^{2^n}+y^{2^n}}2} \ge \sqrt[4]{\frac{x^4+y^4}2} > \frac1{\sqrt3} \implies x^{2^n}+y^{2^n} > \frac2{3^{2^{n-1}}}$$
which is a tighter bound than your inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1145506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
System of Exponential Equations in $x$ and $y$ Three of the elements in the solution set of the simultaneous system $x^{x+y} = y^4, y^{x+y} = x$ are ordered pairs of integers $(x, y)$. Find these ordered pairs.
I found the trivial solution at $(1,1)$, but was unable to find the other two solutions. Could I get help with the others? Thanks!
|
$$1) \quad x^{x+y} = y^4$$
$$2) \quad y^{x+y}=x$$
$$x^{x+y}=(y^{x+y})^{x+y}=y^{(x+y)^2}=y^4\implies (x+y)^2=4\implies x+y =\pm 2$$
Else $(x+y)^2$ need not equal $4$ if $y=-1,0,1$ as Meelo has commented(but $(x+y)^2$ need be even for $y=-1$ case.)
Clearly they have designed this so$x=y=0$ and $(-1,-1)$ solutions are inconsistent.
$$x^2=y^4\implies(x,y)=(\pm n^2,\pm n),n\in\Bbb Z$$ of these consider $(\pm 4, \pm 2)$: $x+y=2$ has only the soltuion here $(4,-2)$ only solution. We can see from $2)$, $y^{x+y}=x$, here ${-2}^2 = 4$ holds. Thus this is a solution.
$$x+y=-2,x^{-2}=y^4\implies (\pm1,1)$$ are these consistent with $y^{x+y}=x$, only $(1,1)$ is.
Consider the options of $y=-1,0,1$, we have already found $y=1$ and ruled out $y=0$, consider $y=-1$:
$$y=-1\implies x^{x-1}=1,-1^{x+y}=x, x=1,$$
$$(x,y)=(1,-1)$$
Solutions $(1,\pm1),(4,-2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Method of characteristics Given the equation
$y^2 u_x + x u_y = \sin(u^2)$ with
initial condition $u(x,0) = x$,
determine the values of $u_x$, $ u_y$, $u_{xx}$, $u_{yy}$, $u_{xy}$, $u_{yx}$ on the $x$-axis.
I tried the following:
$$\begin{cases}
\frac{dx}{dt} = y^2 \\\\
\frac{dy}{dt} = x \\\\
\frac{dz}{dt} = \sin(z^2)
\end{cases}$$
However, the first two coupled ODEs are really difficult to solve. So I assume there is an easier way to do this.
|
Hint:
$$\begin{align}
\frac{dx}{dt} &= y^{2} \ \ \ (1)\\
\frac{dy}{dt} &= x \ \ \ \ \ (2)\\\\
(2) \implies \frac{d}{dt} \bigg( \frac{dy}{dt} \bigg) &= \frac{d^{2}y}{dt^{2}} \\
&= \frac{dx}{dt} \\
&= y^{2} \\
&= (1)
\end{align}$$
EDIT:
We have
$$y'' = y^{2}$$
Multiply through by $y'$ and integrating
$$\begin{align}
y' y'' &= y' y^{2} \\
\implies \frac{d}{dt} \bigg( \frac{1}{2} y'^{2} \bigg) &= y' y^{2} \\
\implies \frac{1}{2} y'^{2} &= \frac{y^{3}}{3} + C_1 \\
\implies y'^{2} &= \frac{2y^{3}}{3} + C_1 \\
\implies y' &= \pm \sqrt{\frac{2y^{3}}{3} + C_1} \\
\end{align}$$
Separating and integrating
$$\int \frac{dy}{\sqrt{\frac{2y^{3}}{3} + C_1}} = \pm \int dt$$
Can you take it from there?
EDIT 2:
From our PDE we have
$$y^{2}u_x + xu_y = \sin(u^{2})$$
Separating into parts, we have
$$\frac{dx}{y^{2}} = \frac{dy}{x} = \frac{du}{\sin(u^{2})}$$
From this we find two equations, take
$$\frac{dy}{dx} = \frac{x}{y^{2}} \ \ \ \ \ (*) \\
\frac{du}{dy} = \frac{\sin(u^{2})}{x} \ \ \ \ \ (**)$$
From $(*)$ we find
$$\begin{align}
y^{2}dy &= xdx \\
\implies x &= \sqrt{\frac{2y^{3}}{3} + C_1} \\
\end{align}$$
Hence, using $(**)$ we see
$$\begin{align}
\frac{du}{\sin(u^{2})} &= \frac{dy}{\sqrt{\frac{2y^{3}}{3} + C_1}} \\
\end{align}$$
However, the $u$ integration is ridiculous and can't be done in terms of elementary functions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $(n+1)^{n-1}How would one prove that $$(n+1)^{n-1}<n^n \ \forall n>1$$
I have tried several methods such as induction.
|
My try to prove $\forall n>1 : (n+1)^{n-1} < n^n$:
$$(n+1)^{n-1} < n^n \Leftrightarrow (n+1)^{n} < n^n (n+1) \Leftrightarrow \left(\frac{n+1}{n}\right)^n < n+1$$
Now apply induction on this result:
Base case: $n = 2 \Rightarrow \left(\frac{3}{2}\right)^2 < 3 \Rightarrow 2.25 < 3$ OK
Induction hypothesis: assume $\left(\frac{n+1}{n}\right)^n < n+1$ true for all $n \leq k$
Induction step: we prove that $\left(\frac{n+1}{n}\right)^n < n+1$ is true for $n = k+1$
$$\left(\frac{k+2}{k+1}\right)^{k+1} < k+2 \Leftrightarrow \left(\frac{k+2}{k+1}\right)^k < k+1$$
Now we have that $\left(\frac{k+1}{k}\right)^k > \left(\frac{k+2}{k+1}\right)^k$ and because of the induction hypothesis, we have proven by induction that $\forall n>1 : \left(\frac{n+1}{n}\right)^n < n+1$ and thus that $\forall n>1 : (n+1)^{n-1} < n^n$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
}
|
Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$ $$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$
Can someone help me to solve it?
result of online calculator: 3/5
|
Hint:
$$\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}} = \frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\cdot\frac{\sqrt{x+1}+\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-3}}\cdot\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+1}+\sqrt{x-2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Number of solutions of a variable matrix Given matrix:
$$\left(\begin{array}{ccc|c}
c & c & 1-c & 1\\
c & c^{2} & 1-c^{2} & 1\\
2c & c+c^{2} & 2-2c & c+2
\end{array}\right)$$
row reduced to:
$$\left(\begin{array}{ccc|c}
c & 1 & 0 & 1\\
0 & c-1 & 1-c & 0\\
0 & 0 & c+1 & 1
\end{array}\right)$$
question asks to find the values of 'c' for which the matrix has:
*
*infinite solutions
*one solution
*no solution
my answer:
the column vectors are linearly independent thus the Rank[A|b] = Rank[A] = n = 3
then for all c ≠ 1 we have one solution
And for c = 1 we have no solutions.
Thanks,
|
let us look at the cases separately.
in case $c = 0,$ you have the system $\pmatrix{0&0&1&|&1\\0&0&1&|&1\\0&0&2&|&2}$ which is and has infinitely many solutions.
in case $c = 1,$ you have the system $\pmatrix{1&1&0&|&1\\1&1&0&|&1\\2&2&0&|&3}$ which has no solutions.
in case $c \ne 0, c \neq 1$ we can reduce
$\pmatrix{c&c&1-c&|&1\\c&c^2&1-c^2&|&1\\2c&c+c^2&2-2c&|&c+2} \rightarrow
\pmatrix{c&c&1-c&|&1\\0&c^2-c&c-c^2&|&0\\0&-c+c^2&0&|&c} \rightarrow
\pmatrix{c&c&1-c&|&1\\0&c^2-c&c-c^2&|&0\\0&0&c-c^2&|&c}
$
has a unique solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make a mistake in assuming that $2^{4n+2} + 3^{n+2}$ is divisble by 13? In induction we have to prove that if something is divisible/whatever by something then n+1 is divisible too which I did
|
Your proof isn't wrong per se, but the wording is a bit confusing (and I would probably end up taking a point or two off if I were grading it). Here's how I would write it:
For $n=0$, $$2^{4\cdot0+2}+3^{0+2} = 2^2 + 3^2 = 13.$$
Now assume that $13|(2^{4n+2}+3^{n+2})$ for some integer $n\geqslant0$. Then $2^{4n+2}+3^{n+2}=13k$ for some integer $k$, and
$$2^{4(n+1)+2}+3^{(n+1)+2} = 2^4\cdot2^{4n+2}+3\cdot3^{n+2}=16(2^{4n+2}+3^{n+2})-13\cdot3^{n+2}=13(16k - 3^{n+2}) $$
You could also factor it into
$$3(2^{4n+2}+3^{n+2}) + 13\cdot2^{4n+2}= 13(3k + 2^{4n+2}).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1147659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
Identity with Harmonic and Catalan numbers Can anyone help me with this.
Prove that
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
The left side is equal to $$2\log(C(x))=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)$$ where $C(x)=\sum_{n=0}^{\infty}C_n x^n$ is the generating function of the Catalan numbers.
|
We start with the key identity:
$$\sum\limits_{j=1}^{n-k} \frac{(-1)^{j-1}}{j}\binom{n}{k+j} = \binom{n}{k}(H_n - H_k)$$
which can be proved elementarily by induction on $n$ or otherwise.
Thus we have for our special case: $\displaystyle \binom{2n}{n}(H_{2n} - H_n) = \sum\limits_{j=1}^{n} (-1)^{j-1}\frac{1}{j}\binom{2n}{n+j}$
The series then decomposes to two parts:
$$\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \underbrace{\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n}- H_n)\frac{z^n}{n}}_{=A} - \frac{1}{2}\underbrace{\sum\limits_{n=1}^{\infty} \binom{2n}{n} \frac{z^n}{n^2}}_{ = B}$$
Evaluation of $A$:
$$\begin{align}\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n}- H_n)\frac{z^n}{n} & = \sum\limits_{n=1}^{\infty} \sum\limits_{j=1}^{n} (-1)^{j-1}\frac{1}{j}\binom{2n}{n+j}\frac{z^n}{n} \\ &= \sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}}{j}\sum\limits_{n=j}^{\infty} \binom{2n}{n+j}\frac{z^n}{n} \\ &= \sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}z^{j}}{j}\sum\limits_{m=0}^{\infty} \binom{2j+2m}{2j+m}\frac{z^m}{j+m} \\ &= \int_0^z \frac{1}{z}\sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}z^{j}}{j}\sum\limits_{m=0}^{\infty} \binom{2j+2m}{2j+m} z^{m} \,\mathrm{d}z \tag{*}\\ &= \int_0^z \frac{1}{z}\sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}z^j}{j}.\frac{4^j}{\sqrt{1-4z}(1+\sqrt{1-4z})^{2j}} \,\mathrm{d}z\\&= \int_0^z \frac{1}{z\sqrt{1-4z}}\log \left(1+\frac{4z}{(1+\sqrt{1-4z})^2}\right) \,\mathrm{d}z \\ &= -\int_0^z \frac{1}{z\sqrt{1-4z}}\log \left(\frac{1+\sqrt{1-4z}}{2}\right) \,\mathrm{d}z\end{align}$$
Consider, $\displaystyle f(z) = \log \left(\frac{1+\sqrt{1-4z}}{2}\right)$, then $\displaystyle f'(z) = \frac{1}{2z} - \frac{1}{2z\sqrt{1-4z}}$
Continuing the computation with the substitution:
$$\begin{align}& = \int_0^z f(z)\left(2f'(z) - \frac{1}{z}\right)\,\mathrm{d}z \\&= \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) - \int_0^z \frac{1}{z}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)\,\mathrm{d}z\end{align}$$
Since,
$$\displaystyle \begin{align} \sum\limits_{n=1}^{\infty} \binom{2n}{n}z^n = \frac{1}{\sqrt{1-4z}} &\implies \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n} = -2\log \left(\frac{1+\sqrt{1-4z}}{2}\right)\\ & \implies \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n^2} = -2\int_0^z\frac{1}{z}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)\,\mathrm{d}z =B \end{align}$$
And also, $\displaystyle C(z) = \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n+1} = \frac{1}{z}\int_0^z \frac{1}{\sqrt{1-4z}}\,\mathrm{d}z = \frac{1-\sqrt{1-4z}}{2z} = \frac{2}{1+\sqrt{1-4z}}$
Hence, we get our final result:
$$\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) = \log^2 (C(z))$$
In $(*)$ we used the fact that: $\displaystyle \sum\limits_{m=0}^{\infty} \binom{p+2m}{p+m}z^m = \frac{2^p (1+\sqrt{1-4z})^{-p}}{\sqrt{1-4z}}$, for integers $p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1148203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
}
|
What is the coefficient of $x^{10}$ in $\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2$? I did (parcially) the following exercise:
There are $10$ identical gift boxes. Each one must be colored with a unique color and there are the colors red, blue, green and yellow. It's possible to color a maximum of $2$ boxes with red, and a maximum of $3$ colors with blue. Write the ordinary generating function associated with the problem and find the number of ways to color 10 boxes.
I've managed to find the generating function:
$$(1+x+x^2)(1+x+x^2+x^3)(1+x+x^2+\dots)^2=\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2\\ \frac{(1-x^3)(1-x^4)}{(1-x)^4}$$
But for calculating the number of colored boxes, I'd have:
$$\frac{(1-x^3)(1-x^4)}{(1-x)^4}=[1-x^3][1-x^4] \left[ \frac{1}{(1-x)} \right]^4$$
The expansion of $\left[\frac{1}{(1-x)}\right]^4$ is:
$$\left[\frac{1}{(1-x)}\right]^4=\sum_{j=0}^\infty {4+j-1 \choose j} y^k$$
But this doesn't seem too revealing. I don't know how to proceeed the counting in this exercise.
|
This is mechanized in Maple:
coeftayl(((-x^3+1)/(1-x)*((-x^4+1)/(1-x)))/(1-x)^2, x = 0, 10);
$$ 102. $$
See coeftayl for info.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1150280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Non vanishing of an infinite product I need to prove that the infinite product
$$\prod_n \left(1-\frac{1} {(a^n+1)^2} \right)^{\frac{a^n}{n}} $$
with $a$ an integer $\geq 3$,
converges to a real number $L$ such that $0<L<1$.
It's immediate to see that $L<1$, as we have $\left(1-\frac{1} {(a^n+1)^2} \right)^{\frac{a^n}{n}}<1$, but I didn't found a way to say that the limit is not zero.
Thanks for your help.
|
Since $a\ge3$,
$$
\frac{1} {(a^n+1)^2}\le\frac{1}{16}\quad\forall n\ge1.
$$
Let $C>0$ be such that $\log(1-x)\ge-C\,x$ if $0<x\le1/16$. For any $N>1$
$$\begin{align}
\log\Biggl(\prod_{n=1}^N \Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)^{\frac{a^n}{n}}\Biggr)&=\sum_{n=1}^N\frac{a^n}{n}\log\Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)\\
&\ge-C\sum_{n=1}^N\frac{a^n}{n}\,\frac{1}{(a^n+1)^2}\\
&\ge-C\sum_{n=1}^N\frac{1}{a^n}\\
&=-\frac{C}{a-1}.
\end{align}$$
Thus
$$
\prod_{n=1}^N \Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)^{\frac{a^n}{n}}\ge e^{-\tfrac{C}{a-1}}>0.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1150601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove: $\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Let $a;b;c>0$. Prove that :
$\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
I think:
$\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Leftrightarrow \frac{1}{a}-\frac{b+c}{a^2+ab}+\frac{1}{b}-\frac{c+a}{b^2+ac}+\frac{1}{c}-\frac{a+b}{c^2+ab}\geq 0\Leftrightarrow \frac{(a-b)(a-c)}{a^3+abc}+\frac{(b-c)(b-a)}{b^3+abc}+\frac{(c-a)(c-b)}{c^3+abc}\geq 0$
And then i don't know what i should do next ! :(
|
WLOG,you can assume $a\ge b\ge c>0$, then
$$\frac{(b-c)(a-b)}{b^3+abc}\le\frac{(a-c)(b-c)}{c^3+abc}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1151632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Proving the following set of real numbers is a field Show that the following set $A$ of real numbers under addition and multipication is a field:
$A = {a + b\sqrt{2} : a,b \ \text{rational}}$
I am not sure if I am right but here is what I have thus far:
Closure: Let $a_{1} + b_{1}\sqrt{2}$, $a_{2} + b_{2}\sqrt{2}\in \mathbb{R}$. Then $(a_{1} + b_{1}\sqrt{2})+ (a_{2} + b_{2}\sqrt{2}) = (a_{1} + a_{2}) + (b_{1} + b_{2})\sqrt{2}\in A$.
Now, $(a_{1} + b_{1}\sqrt{2})\times (a_{2} + b_{2}\sqrt{2}) = (a_{1}\times a_{2} + 2\times b_{1}\times b_{2}) + (a_{1}\times b_{2} + b_{1}\times a_{2})\sqrt{2}\in A$
So both addition and multipication are closed on $A$
Associativity: As addition and multipication are associative on $\mathbb{R}$ it follows from Restriction of Operation Associativity that they are also associative on the set $A$.
Commutativity: As addition and multipication are commutative on $\mathbb{R}$ it follows from Restriction of Operation Commutativity that they are also commutative on the set $A$.
Identity: We have $(a + b\sqrt{2}) + (0 + 0\sqrt{2}) = (a + 0) + (b + 0)\sqrt{2} = a + b\sqrt{2}$ and similarly for $(0 + 0\sqrt{2}) + (a + b\sqrt{2})$. So, $(0 + 0\sqrt{2})$ is the identity for addition on $A$. Now, for multipication we have $(a + b\sqrt{2})(1 + 0\sqrt{2}) = (a\times 1 + 2\times b\times 0) + (b\times 1 + a\times 0)\sqrt{2} = a + b\sqrt{2}$ and similarly for $(1 + 0\sqrt{2})(a + b\sqrt{2})$. So, $(1 + 0\sqrt{2})$ is the identity for multipication on $A$.
Inverses: $(a + b\sqrt{2}) + (-a + (-b)\sqrt{2}) = (a - a) + (b - b)\sqrt{2} = 0 + 0\sqrt{2}$ and similarly for $(-a + (-b)\sqrt{2}) + (a + b\sqrt{2})$. So, $(-a + (-b)\sqrt{2})$ is the inverse of $(a + b\sqrt{2})$ for addition on $A$.\
For product inverse consider the difference of squares: $(a + b\sqrt{2})(a - b\sqrt{2}) = a^{2} - 2b^{2}$ which leads to $(a + \sqrt{2})(\frac{a - b\sqrt{2}}{a^{2} - 2b^{2}}) = 1 = 1 + 0\sqrt{2}$. So, demonstrating that the product inverse of $(a + b\sqrt{2})$ is $\frac{a}{a^{2} - 2b^{2}} - \frac{b\sqrt{2}}{a^{2} - 2b^{2}}$ as $a,b$ are rational, it follows that so are $\frac{a}{a^{2} - 2b^{2}}$ and $\frac{b}{a^{2} - 2b^{2}}$. So, the product inverse $(a + b\sqrt{2})\in A$
Distributivity: We have the Real Multipication Distributes over addition, so by Restriction of Operation Distrbutibity, multipication is distributive over addition on $A$.
|
Hints:
After realizing that $\;0=0+0\cdot\sqrt2\;,\;\;1=1+0\cdot\sqrt2\;$, we see all the axioms of a field that are inherited to subsets are fulfilled in $\;A\;$ since the operations used are exactly the same as the ones in $\;\Bbb R\;$ .
The only thing thus that is left to show is closedness of operations and existence of multiplicative inverse. For the first ones you already were commented enough, as as for the last one think of
$$a,b\in\Bbb Q\;,\;\;a\neq 0\;\;or\;\;b\neq 0\implies\frac1{a+b\sqrt2}=\frac1{a+b\sqrt2}\cdot\frac{a-b\sqrt2}{a-b\sqrt2}$$
and now do the maths in the rightmost expression and use the fact the rationals is a field.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1157407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
|
${1\over x}+{1\over y}={1\over z}$ if and only if it has the form
$${1\over a(a+b)d}+{1\over b(a+b)d}={1\over abd}\quad\text{with}\quad \gcd(a,b)=1$$
(all variables being understood as positive integers).
The "if" part is trivial to verify. The "only if" part comes by setting $x=ga$ and $y=gb$ with $g=\gcd(x,y)$, which gives
$${1\over x}+{1\over y}={a+b\over gab}$$
and noting that $a+b$ is relatively prime to both $a$ and $b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1166999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
}
|
Integrating $1/{(1-x)}$ Given $y = 1 - x = -1 \cdot (x-1) $ with $x \in (0,1)$ then
\begin{equation} \int \frac{1}{y} \: \mathrm{d}x = \int \frac{1}{1-x} \: \mathrm{d}x = \ln(1-x) + C
\end{equation}
but also
\begin{equation} \int \frac{1}{y} \: \mathrm{d}x = - \int \frac{1}{x-1} \: \mathrm{d}x = -\ln(x-1) + C
\end{equation}
which are not equal. What am I missing?
|
Your first attempt is not correct:
Given the integral $$\int \frac{1}{1-x} \,dx$$ let's put $1-x = u,$ and so $\,du = (1-x)'\,dx = -dx \iff dx = -du.$
Upon substitution, $$\int \frac{1}{1-x} \,dx = -\int \frac{1}{u} \,du = -\ln|u| + C = -\ln|1-x| + C = -\ln|x-1| + C$$
Since $x \in (0, 1)$, we can write the answer as $$- \ln(1-x)+ C\tag{$\dagger$}$$ $\dagger:\,$ We can drop the absolute value sign because $(1-x)>0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1168010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Geometric proof of $\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$ It is well-known that $$\sin{20^\circ}\sin{40^\circ}\sin{80^\circ}=\frac{\sqrt{3}}{8}$$
It follows that $$\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$$
But how to prove this by geometry?
Thank you.
|
Another Trigonometric & algebraic proof:
If $\sin3x=\sin3A,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x+\sin3A=0$
$\implies\prod_{r=1}^3\sin x_r=-\dfrac{\sin3A}4$
Again $\sin3x=\sin3A\implies3x=n180^\circ+(-1)^n3A$ where $n$ is any integer
$\implies x=n60^\circ+(-1)^nA$
If we choose even $n=2m,x=120^\circ m+A$
$\implies\sin A\sin(120^\circ+A)\sin(240^\circ+A)=-\dfrac{\sin3A}4$
Now use $\sin(180^\circ-y)=\sin y$ to find $\sin(120^\circ+A)=\sin[180^\circ-(60^\circ-A)]=\sin(60^\circ-A)$
and
use $\sin(180^\circ+y)=-\sin y$ to find $\sin(240^\circ+A)=\sin[180^\circ+(60^\circ+A)]=-\sin(60^\circ+A)$
Here $A=20^\circ$
You can also choose odd $n$ and use the last formula to derive the same identity
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1171625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
If $n$ is an odd integer prove that $n - 2^k$ is divisible by $3$ So let $n$ be a odd integer. Show that $n - 2^k$ is divisible by $3$ if $k$ is SOME SPECIFIC positive integer. $k \ge 0$. So there only has to exist one. For example:
$$7 - 2^2 = 3$$ is divisible by $3$
The approach is modular arithemetic, but it is hard since,
$$2 \equiv 2 \pmod{3}$$
$$n \equiv p \pmod{3}$$
It is hard to combine these? What should I do?
|
So if $n$ is an odd integer, you have to prove that there exists some integer exponent $k$ such that $3|(n - 2^k)$? Trouble is, it depends on what $n$ is. There are either infinitely many solutions or none at all.
Let's look at this modulo 6: there are three possibilities: $n \equiv 1, 3 \textrm{ or } 5 \pmod 6$. For the powers of 2, we have $2^k \equiv 2 \textrm{ or } 4 \pmod 6$ for $k > 0$. We want $n - 2^k \equiv 0 \textrm{ or } 3 \pmod 6$.
But how to combine all these? It's actually quite easy: we just subtract! Just be sure to mind the wrap-around.
For example, if $n \equiv 1 \pmod 6$ and $2^k \equiv 4 \pmod 6$, then $n - 2^k \equiv 1 - 4 \equiv 3 \pmod 6$.
But what if $n \equiv 3 \pmod 6$? No $k$ can work: if $2^k \equiv 2 \pmod 6$, then $n - 2^k \equiv 3 - 2 \equiv 1 \pmod 6$, and if $2^k \equiv 4 \pmod 6$, then $n - 2^k \equiv 3 - 4 \equiv 5 \pmod 6$. Don't forget $k = 0$, but that's of no help here: $n - 1 \equiv 3 - 1 \equiv 2 \pmod 6$.
The solution for $n \equiv 5 \pmod 6$ should be easy enough for you to figure out now.
One of the other answerers mentioned a requirement for $n - 2^k$ to be positive, but you didn't. I don't know the rules of the contest, so I assume negative values are perfectly acceptable. For example, $-9 \equiv 3 \pmod 6$ since $(-2) \times 6 + 3 = -12 + 3 = -9$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1173691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
|
If the equation out this
$$(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})\geq2(1+\frac{a+b+c}{\sqrt[3]{abc}})$$
We can rewrite the equation:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
writing
$\frac{a+b}{c}=\frac{a+b+c}{c}-1$
$\frac{b+c}{a}=\frac{a+b+c}{a}-1$
$\frac{c+a}{b}=\frac{a+b+c}{b}-1$
Inequality can be written as:
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
Then by AG
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{3(a+b+c)}{\sqrt[3]{abc}}=\frac{2(a+b+c)}{\sqrt[3]{abc}}+\frac{(a+b+c)}{\sqrt[3]{abc}}\geq\frac{2(a+b+c)}{\sqrt[3]{abc}}+3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1178979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
Prove formally that $\frac {n^2 + 2}{3n^3 - 5n}\to 0$ as $n \to \infty$. I'm reviewing some Sequences notes from a Mathematics Analysis course I'm taking. I'm finding the beginning of the formal proof below confusing. Some clarity on the following questions would be much appreciated.
Questions
*
*Why is $n\ge3$ noted? If it is to make $\frac {n^2+2}{3n^3 -5n}$ positive, why not use $n\ge2$?
*How does one arrive at $\frac {n^2+2}{3n^3 -5n} \lt \frac {n^2+2}{2n^3}$? Where does $\frac {n^2+2}{2n^3}$ come from?
Formal Proof
|
If $n \geqslant 3$ then $n^3 > 5n$ which implies $3n^3 -5n > 2n^3$. This is not true if $n = 2$.
Therefore,
$$\frac{n^2+2}{3n^3 -5n} < \frac{n^2+2}{2n^3}.$$
We also have $2 < 2n^2$ for $n \geqslant 2$ which implies $n^2 + 2 < n^2 +2n^2 = 3n^2$.
Hence if $n \geqslant 3$,
$$\frac{n^2+2}{3n^3 -5n} < \frac{3n^2}{2n^3} = \frac{3}{2n} .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1179314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Closed form of $ \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x } } \log { (\cos { x } ) }\ dx }$ Does there exists a closed form of$$ \displaystyle \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x } } \log { (\cos { x } ) }\ dx }$$
If exists can someone find a way to tackle this integral and provide a closed-form of it. Many similar integrals have closed form and I believe this one, too.
|
Substitute $t = \sqrt{\tan x}$
\begin{align}
I=\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\cos x})\ dx=J(1)\\
\end{align}
where $J(a)= -\int_0^\infty \frac{t^2 \ln(1+t^4) \tan^{-1}(a^2t^2)}{1+t^4}dt$
\begin{align}
J’(a)=&-\int_0^\infty \frac{2at^2 \ln(1+t^4)}{(1+t^4)(1+a^4t^4)}dt\\
=& \ \frac{2a}{1-a^4}\int_0^\infty
\left(\frac{\ln(1+t^4)}{1+t^4}- \frac{\ln(1+t^4)}{1+a^4t^4} \right)dt
\end{align}
Let $K(a,b)=\int_0^\infty \frac{\ln(1+b^4t^4)}{1+a^4t^4}dt$
\begin{align}
K_b’(a,b)=&\int_0^\infty \frac{4b^3 t^4}{(1+a^4t^4)(1+b^4t^4)}dt
=\frac{\sqrt2\pi b^2}{a(a+b)(a^2+b^2)}
\end{align}
Then
\begin{align}
K(a,b)= &\int_0^b K_b’(a,s)ds
= \frac\pi{\sqrt2 a}\left(\frac12\ln\frac{a^2+b^2}{a^2}
+\ln\frac{a+b}a-\tan^{-1}\frac ba\right)\\
J’(a)=&\ \frac{2a}{1-a^4}\left[K(1,1)-K(a,1) \right]\\
=& \frac{\sqrt2\pi}{1-a^4}
\left[a\left(\frac32\ln2-\frac\pi4 \right)-\frac12 \ln\frac{1+a^2}{a^2} -\ln\frac{1+a}{a}+\cot^{-1}a \right]
\end{align}
and
\begin{align}
I=&\ J(1)=\int_0^1 J’(a)da\\
=& -\sqrt2\pi \bigg[
\left(\frac32\ln2-\frac\pi4 \right)\int_0^1 \frac{1-a}{1-a^4}da
+\frac12\int_0^1 \frac{\ln\frac{1+a^2}{2a^2}}{1-a^4}da\\
&\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>
+\int_0^1 \frac{\ln\frac{1+a}{2a}}{1-a^4}da
+\int_0^1 \frac{\frac\pi4-\cot^{-1}a}{1-a^4}da\bigg]\\
=& -\sqrt2\pi \bigg[
\left(\frac32\ln2-\frac\pi4 \right)\left(\frac\pi8+\frac14\ln2\right)
+\frac12\left(\frac12G+\frac{3\pi^2}{32} +\frac\pi{8}\ln2 \right)\\
&\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>
+\left(\frac12G+\frac{\pi^2}{24} -\frac\pi{16}\ln2\right)
+\left(-\frac14G-\frac{\pi^2}{64}\right)\bigg]\\
=&-\frac\pi{\sqrt2}\left(G+\frac{\pi^2}{12}+\frac\pi4\ln2+\frac34\ln^22 \right)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1180401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
}
|
Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give me some hints? Thanks in advance!
|
use this well known identity
$$(a+b)^3=a^3+b^3+3ab(a+b)$$,so we Let
$$x=\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}$$
then we have
$$x^3=14+3\sqrt[3]{49-50}\cdot x=-3x+14$$
so $$x=2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
}
|
Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$
Here's what I did to solve this exercise:
$$\frac{a+b}2 \ge \sqrt{ab} \implies \frac12 \ge \sqrt{ab} \implies \frac14\ge ab \implies ab\le\frac14$$
$$\left(1+\frac1a\right)\left(1+\frac1b\right) = 1+\frac1b+\frac1a+\frac1{ab} \ge 9 \Rightarrow \frac{a+b+1}{ab}\ge8 \Rightarrow \frac2{ab}\ge8 \Rightarrow 2\le8ab \Rightarrow ab\le\frac14$$
I came to a known point, but I'm not sure that this is the right form. If there is any other more clear proof please show it to me.
|
If the inequality$$\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2$$is used, then we can deduce that $a^2+b^2\ge2ab$ must hold. This can be seen from squaring the terms on the LHS and simplifying the resulting expression.
Now, $$ab\le\frac12(a^2+b^2)=\frac{(a+b)^2-2ab}{2}=\frac12-ab$$This implies that $$2ab\le\frac12$$which in turn gives the first inequality$$ab\le\frac14$$
For the second inequality, we have
$$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=1+\frac{a+b+1}{ab}=1+\frac{2}{ab}$$But we just showed that $ab\le\frac14$ which means that $\frac{1}{ab}\ge 4$. Using this, we find $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge1+(2)(4)=9$$and that completes the proof for the second inequality!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1185465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
Positive numbers inequality Let $ x_1, x_2, \ldots, x_n \; \in \; \mathbb{R}_{+} $ so that $ x_1 \geq x_2 \geq \cdots \geq x_n $
We also know that:
$$ \frac{1}{2^{x_1}} + \frac{1}{2^{x_2}} + \cdots + \frac{1}{2^{x_n}} = 1 $$
Prove the following inequality:
$$ \frac{1}{3^{x_1}} + \frac{2}{3^{x_2}} + \cdots + \frac{n}{3^{x_n}} \geq \frac{n + 1}{2n} $$
I've tried to prove it several times but I haven't succeeded. I would really appreciate if you can help me.
|
Following Michael Hardy's comment, the problem is equivalent to showing
$$\sum_{k=1}^n k\cdot p_k^{\log_2 3}\ge \frac{n+1}{2n}$$
if $0<p_1\le \cdots\le p_n$ and $p_1+\cdots+p_n=1$. This can be proven as follows:
\begin{align*}
&\sum_{k=1}^n k\cdot {p_k}^{\log_2 3}\\
&\ge \sum_{k=1}^n \frac{n+1}{2}\cdot {p_k}^{\log_2 3}\\
&= \frac{n+1}{2} \sum_{k=1}^n {p_k}^{\log_2 3}\\
&\ge \frac{n+1}{2} n\left(\frac1n\right)^{\log_2 3}\\
&\ge \frac{n+1}{2} n\left(\frac1n\right)^2\\
&=\frac{n+1}{2n}.
\end{align*}
1st line to 2nd line: For $1\le k\le \lfloor\frac{n+1}{2}\rfloor$,
$$
k {p_k}^{\log_2 3}+(n+1-k) {p_{n+1-k}}^{\log_2 3}
\ge \frac{n+1}{2} {p_k}^{\log_2 3}+\frac{n+1}{2} {p_{n+1-k}}^{\log_2 3}.
$$
3rd line to 4th line: apply Jensen's inequality to a convex function $f(x)=x^{\log_2 3}$:
$$
\sum_{k=1}^{n}\frac1n f(p_k) \ge f(\frac1n(p_1+\cdots+p_n)).
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1185564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right?
|
\begin{array}{l}
f(x) = \sqrt {7x + \sqrt {7x + \sqrt {7x} } } \\
{f^2}(x) = 7x + \sqrt {7x + \sqrt {7x} } \\
{f^2}(x) - 7x = \sqrt {7x + \sqrt {7x} } \\
{\left( {{f^2}(x) - 7x} \right)^2} = 7x + \sqrt {7x}
\end{array}
Now I'm doing calculus using chain rule:
\begin{array}{l}
2\left( {{f^2}(x) - 7x} \right)\left( {2f(x)f'(x) - 7} \right) = 7\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right)\\
\sqrt {7x + \sqrt {7x} } \left( {2f(x)f'(x) - 7} \right) = \frac{7}{2}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right)\\
2f(x)f'(x) = \frac{7}{2}\frac{1}{{\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 7\\
2f(x)f'(x) = 7\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)\\
f(x)f'(x) = \frac{7}{2}\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)
\end{array}
No needs to factor it out:
\begin{array}{l}
f'(x) = \frac{7}{2}\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)\frac{1}{{f(x)}}\\
f'(x) = \frac{7}{2}\left( {\frac{1}{{2\sqrt {7x + \sqrt {7x} } }}\left( {1 + \frac{1}{2}\frac{1}{{\sqrt {7x} }}} \right) + 1} \right)\frac{1}{{\sqrt {7x + \sqrt {7x + \sqrt {7x} } } }}
\end{array}
But if somebody wanted to:
$$f'(x) = \left( {\frac{{14\sqrt x + \sqrt 7 }}{{8\sqrt {7{x^2} + x\sqrt {7x} } }} + \frac{7}{2}} \right)\frac{1}{{\sqrt {7x + \sqrt {7x + \sqrt {7x} } } }}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1190891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
Can anyone tell me if this is correct? Suppose that the temperature of a metal plate is given by $T(x; y) = x^2 +2x+y^2$, for points $(x, y)$ on the elliptical plate defined by $x^2 + 4y^2 <= 24$.
Find the maximum and minimum temperatures on the plate.
This is what i have done so far.
Finding critical point:
$T(x)=2x+2$, $T(y)=2y$. Equating to $0$, $x=-1$, $y=0$.
Critical point is $(-1,0)$ and is a minimum.
On the boundary, $ x^2 + 4y^2 = 24$
$g(x,y)=x^2 + 4y^2$
$g(x)=2x, g(y)=8y$
$2x+2=A2x$---------(1)
$2y =A8y$---------(2)
$x^2+4y^2=24$------(3)
When solving from equation 1 and 3 im getting $ x=-1,y=|(23/4)^{0.5}|,$ and $x=|24^{0.5}|, y=0$ and when from eqn 2 and 3 im getting $x=|24^{0.5}|,y=0, A= 0.25 , x=-4/3, y=|(50/9)^{0.5}|$.
Is this correct? am getting different values when using equation $1$ and $2$.
|
For minimum, point $(-1,0)$ should be correct.
Although I didn't understand your approach for finding maximum, maximum temperature will occur at the boundaries.
$x^2 + 4y^2 = 24$
$y^2 = \frac{24-x^2}{4}$
Substituting this value into the equation for temperature yields $T(x)= \frac{3x^2+8x+24}{4}$
As $x$ ranges from $-\sqrt{24} \space to \space \sqrt{24}$.
$x=\sqrt{24} $ gives $33.798$ as the maximum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1191826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
For a random variable $X$ such that $P(aI've worked on the following problem and have a solution (included below), but I would like to know if there are any other solutions to this problem, particularly more elegant solutions that apply well known inequalities that I've overlooked.
QUESTION: Suppose we have a random variable s.t. $P(a<X<b) =1$ where $0 < a < X < b$ , $a$ and $b$ both positive constants.
Show that $$E(X)E\left(\frac{1}{X}\right) \le \frac{(a+b)^2}{4ab}$$
Hint: find constant c and d s.t. $\frac{1}{x} \le cx+d$ when $a<x<b$, and argue that then we shall have $E(\frac{1}{X}) \le cE(X)+d$
MY SOLUTION: For a line $cx+d$ that cuts through $\frac{1}{X}$ at the points $x=a$ and $x = b$, it's easy to show that $ c = - \frac{1}{ab} $ and $d = \frac{a+b}{ab} $,
$$ E\left(\frac{1}{X}\right) \le - \frac{1}{ab} E(X) + \frac{a+b}{ab} $$
$$ abE\left(\frac{1}{X}\right) + E(X) \le (a+b) $$
and because both sides of the inequality are positive, it follows that:
$$ \left(abE\left(\frac{1}{X}\right) + E(X)\right)^2 \le (a+b)^2 $$
$$ (ab)^2E\left(\frac{1}{X}\right)^2 + 2abE\left(\frac{1}{X}\right)E(X) + E(X)^2 \le (a+b)^2 $$
Now, for the LHS, we can see that
$2abE\left(\frac{1}{X}\right)E(X) \le (ab)^2E\left(\frac{1}{X}\right)^2 + E(X)^2$
because
$0 \le (ab)^2E\left(\frac{1}{X}\right)^2 - 2ab\,E\left(\frac{1}{X}\right)E(X) + E(X)^2 = \left(ab\,E\left(\frac{1}{X}\right) - E(X)\right)^2 $
So,
$$ 4ab\,E\left(\frac{1}{X}\right)E(X) \le (ab)^2E\left(\frac{1}{X}\right)^2 + 2ab\,E\left(\frac{1}{X}\right)E(X) + E(X)^2 \le (a+b)^2 $$
and therefore:
$$ E\left(\frac{1}{X}\right)E(X) \le \frac{(a+b)^2}{4ab} $$ Q.E.D.
Thanks for any additional solutions you might be able to provide. Cheers!
|
For any number $x$ such that $0<a\le x\le b$, we have $x-a\ge0$ and $b-x\ge0$, so
$$ (x-a)(b-x)\ge0,$$
which rearranges into the equivalent form
$$ {ab\over x} + x\le a+b,\tag{1}$$
which is the inequality you obtained earlier. You can slickify the rest of your proof as follows. Let $X$ be a random variable with $0<a\le X\le b$. Use $(1)$ to get:
$$m+n\le a+b,$$
where $m:=abE(1/X)$ and $n:=E(X)$ are both positive. Then
$$4abE(1/X)E(X) = 4mn=(m+n)^2-(m-n)^2\le(m+n)^2\le(a+b)^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1193262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Evaluating $\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right)$ $$\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) = \;?$$
I have been trying to see if it can be written as sum of two telescope terms but it looks tricky. Any help ?
|
$$\sum\limits_{n=1}^k \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) =1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2k} + \sum\limits_{n=k}^{2k-1} \frac{1}{2n+1}$$
Can you evaluate the limits of the two series on the right-hand side?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1193654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
Why is $\sqrt{3}=[1;1,2,1,2,\dots]$?
Why is $\sqrt{3}=[1;1,2,1,2,\dots]$ ?
$\displaystyle[1;1,2,1,2,\dots]=1+\frac{1}{[1;2,1,2,\dots]}=1+\frac{1}{1+\frac{1}{2+\frac{1}{[1;2,1,2,\dots]}}}$
If I set $x=[1;2,1,2,\dots]$ then;
$\frac{x+1}{x}=\frac{5x+2}{3x+1}$
with solutions $x_{1,2}=\frac{1\pm\sqrt{3}}{2}$
what did go wrong ?
|
But then $\sqrt{3} = 1+1/x_1$. That's because $$\frac{1}{x_1}=\frac{2}{1+\sqrt 3} = \frac{2(\sqrt{3}-1)}{2} = \sqrt{3}-1$$
The other approach is to define:
$$y = 1+ \dfrac{1}{1+\dfrac1{1+y}}$$
Then you get $y=1+\frac{y+1}{y+2}=\frac{2y+3}{y+2}$ which reduces to $y^2=3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1195526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Determining convergence of $\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$ I have the following infinite series:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$$
I want to examine its convergence. First thing that came to my mind was "unfolding" $e^{\frac{1}{n}}$ and see what will happen:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}=\sum_{n=1}^{+\infty}(\sum_{k=0}^{+\infty}\frac{1}{n^kk!}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}=\sum_{n=1}^{+\infty}(\sum_{k=3}^{+\infty}\frac{1}{n^kk!})^{\frac{1}{2}}$$
Now it kinda "looks" convergent but I'm clueless about how to prove it. Any hints?
|
We have
$$\sum_{n=1}^{\infty} \sqrt{\sum_{k=3}^{\infty} \frac{1}{n^k k!}}$$
We can use the Cauchy condensation test to study the convergence of
$$\sum_{n=1}^{\infty} 2^n \sqrt{\sum_{k=3}^{\infty} \frac{1}{2^{nk} k!}} = \sum_{n=1}^{\infty} 2^n \sqrt{\frac{1}{2^{2n}} \sum_{k=3}^{\infty} \frac{1}{2^{n(k-2)} k!}}$$
$$\sum_{n=1}^{\infty} \sqrt{\sum_{k=3}^{\infty} \frac{1}{2^{n(k-2)} k!}} \leq \sum_{n=1}^{\infty} \sqrt{\sum_{k=1}^{\infty} \frac{1}{2^{nk} k!}} = \sum_{n=1}^{\infty} \sqrt{e^{2^{-n}} - 1}$$
The final series converges by the ratio test:
$$\lim \limits_{n \to \infty} \sqrt{ \frac{e^{2^{-(n+1)}} - 1}{e^{2^{-n}} - 1} }$$
This limit goes to $0/0$, so use L'Hopital's on the inside:
$$\lim \limits_{n \to \infty} \sqrt{\frac{- 2^{-(n+1)} e^{2^{-(n+1)}} \log 2}{-2^{-n} e^{2^{-n}} \log 2}} = \frac{1}{\sqrt{2}} < 1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1200163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} + 3
\\ -3 &= \sqrt{36x^2-36x^4}
\\ (-3)^2 &= (\sqrt{36x^2-36x^4})^2
\\ 0 &= -36x^4 + 36x^2 - 9
\\ \frac{0}{9} &= \frac{9(-4x^4 + 4x^2 - 1)}{9}
\\ 0 &= -4x^4 + 4x^2 - 1
\\ \end{align*}
$$
I've considered the rational root theorem at this point, through which I find possible roots to be $x = \{ \pm1, \pm\frac{1}{2}, \pm\frac{1}{4} \}$.
Clearly I've made an error, as the actual root is $x = -\dfrac{1}{\sqrt{2}}$.
|
$0(-1)=(-4x^4+4x^2-1)(-1) \\
0=4x^4-4x^2+1 \\
0=(2x^2-1)^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1202949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Eigenvectors of a complex matrix Given the following matrix
$\begin{pmatrix}
0 & 1-i & 0\\
1+i & 0 &1-i\\
0& 1+i &0\\
\end{pmatrix}$
I have found the Eigenvalues $0, 2,-2$. But I have no idea how to calculate the corresponding Eigenvectors and I failed with Gaussian method. What could you recommend?
Thanks in advance!
|
For the eigenvalue $-2$:
\begin{align}
\begin{bmatrix}
2 & 1-i & 0\\
1+i & 2 &1-i\\
0& 1+i &2\\
\end{bmatrix}
&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
1+i & 2 &1-i\\
0& 1+i &2\\
\end{bmatrix} && R_1\gets \tfrac{1}{2}R_1
\\[6px]&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
0 & 1 &1-i\\
0& 1+i &2\\
\end{bmatrix} && R_2\gets R_2-(1+i)R_1
\\[6px]&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
0 & 1 &1-i\\
0& 0 &0\\
\end{bmatrix} && R_3\gets R_3-(1+i)R_2
\\[6px]&\to
\begin{bmatrix}
1 & 0 & i\\
0 & 1 &1-i\\
0& 0 &0\\
\end{bmatrix} && R_1\gets R_1-\tfrac{1-i}{2}R_2
\end{align}
Thus the equations can be written
$$
\begin{cases}
x_1=-ix_3\\
x_2=(i-1)x_3
\end{cases}
$$
so an eigenvector is
$$
\begin{bmatrix}
-i\\
i-1\\
1
\end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1203497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
}
|
How to solve $\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$? Here is my question
$$\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$$
I have tried it by substituting $x$ = $\frac{1}{t}$. I got the answer $0$ but the correct answer is $\pi log(2)$. Any suggestion would be appreciated.
|
Approach 1:
\begin{align}
I&=\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\
&=\int_0^{1} \frac{\log(x+\frac{1}{x})}{1+x^2}dx+\int_1^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\
&=2\int_0^{1} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\
&=2\int_2^{\infty} \frac{\log(u)}{u\sqrt{u^2-4}}du\\
\end{align}
using $u=x+\frac{1}{x}$ in the last step. Now set $u=2\sec y$ to obtain
\begin{align}
I&=\int_0^{\pi/2} \log(2\sec y)dy\\
&=\frac{\pi}{2}\log2-\int_0^{\pi/2} \log(\cos y)dy\\
&=\pi\log2
\end{align}
using this in the last step.
Approach 2:
$x \to \tan x$
\begin{align}
I&=-\int_0^{\pi/2} \log(\sin x\cos x)dx\\
&=-\int_0^{\pi/2} \log(\sin x)dx+\int_0^{\pi/2} \log(\cos x)dx\\
&=-2\int_0^{\pi/2} \log(\cos x)dx\\
&=\pi \log 2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1205543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 2nd sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 3rd sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 4th sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 5th sample space is $\dfrac{1}{2}$
The final result is:
$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$
Is this reasoning correct? Are there any simpler ways to solve this problem?
|
You are probably familiar with "tree diagrams" to solve these types of problems. For the first pick you have a (2/5) chance to remove an even, and a (3/5) chance to remove an odd.
When you remove an odd first, there are only (2/4) odds left. Meanwhile, if you remove an even first, there are now (3/4) odds left.
Finally (2/5)(3/4)+(3/5)(2/4) = (6/20)+(6/20) = (12/20) = (3/5)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1206670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
Integral of rational function with trigonometric functions $$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}
$$
I saw this problem online and it looked like an interesting/difficult problem to try and tackle. My attempt so far is to use tangent half-angle substitution.
Let $t= \tan^2 (\frac{x}{2})$, then $dt= \frac{1}{2} \sec^2 (\frac{x}{2})\;dx.$ With algebra we get that $dx= \frac{2}{1+t^2}dt,$ $\sin x = \frac{2t}{1+t^2},$ and $\cos x = \frac{1-t^2}{1+t^2}.$ Therefore we get the following:
$$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4} = \int \frac{\frac{2}{1+t^2}}{(\sqrt{\frac{1-t^2}{1+t^2}}+\sqrt{\frac{2t}{1+t^2}})^4}dt = 2\int \frac{1}{1+t^2}\cdot \frac{(1+t^2)^2}{(\sqrt{1-t^2}+ \sqrt{2t})^4}dt
$$
$$
= 2\int \frac{1+t^2}{(\sqrt{1-t^2}+ \sqrt{2t})^4}dt
$$
I expanded the denominator, but only made things worse. According to wolfram alpha, the solution is in terms of elementary functions. Can you give me any clues or hints on how to evaluate this beast. Thank you!
|
Multiplying top and bottom of the original integral by $\sec^2x$ yields
$$\int\frac{\sec^2xdx}{(1+\sqrt{\tan x})^4}$$
$$u=\tan x,du=\sec^2xdx$$
$$\int\frac{du}{(1+\sqrt u)^4}$$
$$u=t^2,du=2tdt$$
$$\int\frac{2tdt}{(1+t)^4}=\int\frac{(2t+2-2)dt}{(1+t)^4}=\int\frac{2dt}{(1+t)^3}-\int\frac{2dt}{(1+t)^4}=$$
$$\frac{2}{3(1+t)^3}-\frac{1}{(1+t)^2}=\frac{2}{3(1+\sqrt{\tan x})^3}-\frac1{(1+\sqrt{\tan x})^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1207121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Showing Function is Continuous Let $f: \mathbb{R} \backslash \{2\} \to \mathbb{R}$ be the function given by $f(x) = \frac{2x^2+x-10}{3x-6}$.
Let $g: \mathbb{R} \to \mathbb{R}$ given by:
$$g(x)=\begin{cases}{f(x)} & \text{if } x \neq 2 \\ 3 &\text{if } x = 2 \end{cases}$$
Prove that $g$ is continuous for all $x \in \mathbb{R}$.
Workings:
Suppose $\epsilon > 0$ and let $\delta =$ ______
Then in $|x-2| < \delta$, we have
$|g(x) - g(c)| = \left|\frac{2x^2+x-10}{3x-6} - 3\right|$
$= \left|\frac{2x+5}{3} - 3\right|$
$= \left|\frac{2x+5}{3} - \frac{9}{3}\right|$
$= \left|\frac{2x-4}{3}\right|$
$= \frac{2}{3}\left|x-2\right|$
$< \frac{2}{3}\delta$
Now I'm not too sure what to do. Any help will be appreciated.
|
You just have to calculate the 2 limits :
$$l_1 = \lim_{x \rightarrow 2; x > 2} g(x) \text{ and } l_2 = \lim_{x \rightarrow 2; x < 2} g(x)$$
and show that $l_1 = l_2 = g(2) = 3$. But $$l_1 = \lim_{x \rightarrow 2; x > 2} g(x) = \lim_{x \rightarrow 2; x > 2} f(x) = \lim_{x \rightarrow 2; x > 2} \frac{2x^2 + x - 10}{3x-6} = \lim_{x \rightarrow 2; x > 2} \frac{(2x^2 + x - 10)'}{(3x-6)'} = \lim_{x \rightarrow 2; x > 2} \frac{4x+1}{3} = 3$$ (using L'Hopital's rule). And similarly $l_2 = 3$. So the function is continuous.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1207904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Differential Equations (how to proceed) $$y^2+\frac{dy}{dx}=4$$ $y=0$, when $x=\ln 2$
$$\frac{dy}{dx}=-y^2+4$$
$$\frac{\frac{dy}{dx}}{-y^2+4}=1$$
$$\int \frac{\frac{dy}{dx}}{-y^2+4}dx=\int 1dx$$
$$-\frac{1}{4}\ln (-y+2)+\frac{1}{4}\ln (y+2)=x+c_1$$
How to proceed to solve y?
|
$$ \begin{align}x - \ln 2 = \int_{\ln 2}^x dx &= \int_0^y\frac{dy}{4-y^2}\\& = \frac 14\int_0^y \left(\frac{1}{2-y} + \frac 1{2+y}\right)\\&=\frac 14 \left(\ln(2+y) - \ln(2-y)\right)\\&=\frac 14\ln\left(\frac{2+y}{2-y}\right)\end{align}$$
that is $$\frac{2 + y}{2-y}=\frac{e^{4x}}{16}\to \frac{y}{2} = \frac{e^{4x}-16}{e^{4x} + 16} $$ finally we have $$ y = \frac{2(e^{4x}-16)}{e^{4x} + 16}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1210862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Simplify $\left(\sqrt{\left(\sqrt{2} - \frac{3}{2}\right)^2} - \sqrt[3]{\left(1 - \sqrt{2}\right)^3}\right)^2$ I was trying to solve this square root problem, but I seem not to understand some basics.
Here is the problem.
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2$$
The solution is as follows:
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2 = \Bigg(\frac{3}{2} - \sqrt{2} - 1 + \sqrt{2}\Bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \frac{1}{4}$$
Now, what I don't understand is how the left part of the problem becomes:
$$\frac{3}{2} - \sqrt{2}$$
Because I thought that $$\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2}$$
equals to $$\bigg(\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2\bigg)^{\frac{1}{2}}$$
Which becomes $$\sqrt{2} - \frac{3}{2}$$
But as you can see I'm wrong.
I think that there is a step involving absolute value that I oversee/don't understand.
So could you please explain by which property or rule of square root is this problem solved?
Thanks in advance
|
Nicely put question.
You are right about the absolute value missing somewhere. Indeed, we have:
$$\sqrt{x^2} = |x|.$$
In your case, we have
$$\sqrt{\left(\sqrt{2}-\frac{3}{2}\right)^2}=\left|\sqrt{2}-\frac{3}{2}\right|.$$
But $\sqrt{2}-\frac{3}{2}$ is negative, so the absolute value "chooses" the positive version of this, that is,
$$\left|\sqrt{2}-\frac{3}{2}\right| = -\left(\sqrt{2}-\frac{3}{2}\right)=\frac{3}{2}-\sqrt{2}.$$
I hope this helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1212000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Does $a \leq b + c $ imply $a^2 \leq (b+c)^2 + (b-c)^2$? Givens
$$
a \leq b + c
$$
or
$$
a^2 \leq b^2 + c^2+2bc
$$
Can we prove that?:
$$
a^2 \leq (b+c)^2 + (b-c)^2 = 2b^2 + 2c^2
$$
|
$$(b-c)^2\geq0$$
therefore $$b^2+c^2\geq 2bc$$
Hence $$2b^2+2c^2\geq b^2+c^2+2bc\geq a^2$$
But the problem is that the very first inequality $b+c\geq a$ doesn't tell us if $a$ is non-negative. It is possible that $b+c\geq a$ but $a$ is very negative like $a=-\sqrt{(b^2+c^2)+1}$, while $b,c\geq0$. Then $b+c\geq0\geq a$ but $a^2=b^2+c^2+1>b^2+c^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1213281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Error in approximating the sum I am watching one of the online probability courses and in one of the lectures, the professor simplifies the sum:
$$A = \sum_{j=0}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}$$
in the following way: $A \approx \int_{0}^{1}x^kdx=\frac{1}{k+1}$
The problem is that in my opinion the simplification should work in the following way:
$$A = \frac{1}{N+1} \cdot \sum_{j=0}^{N}\frac{j^k}{N^k} $$ and then if one will make a substitution $x = \frac{j}{N}$, then $$A \approx \frac{1}{N+1} \cdot \int_{0}^{1}x^kdx$$ and doing completely the same I will end up with $$A = \frac{1}{N+1} \cdot \frac{1}{k+1}$$
I can not figure out where is my problem here.
|
SalmonKiller gave a good point and I shall not repeat. However, one could notice that $$A_{k,N} = \sum_{j=1}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}=\frac{ H_N^{(-k)}}{(N+1)N^k}$$ where appear the harmonic numbers. Expanding as series for large values of $N$, we then have $$A_{k,N}=\left(\frac{1}{k+1}+\frac{\frac{1}{2}-\frac{1}{k+1}}{N}+O\left(\left(\frac{1}{N}\right)^2\right)\right)+ \frac{\zeta (-k)}{N^k} \left(\frac{1}{N}-\frac{1}{N^2}+O\left(\left(\frac{1}{N}\right)^3\right)\right)$$ So, if $k>1$, $$A_{k,N} \approx\frac{1}{k+1}+\frac{\frac{1}{2}-\frac{1}{k+1}}{N}=\frac{1}{k+1}\Big(1-\frac 1N\Big)+\frac 1{2N}$$
Let us try with small numbers $A_{5,10}=\frac{803}{4000}$ while the approximation gives $\frac{800}{4000}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1213708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Let $p$=prime and $\sqrt{x}+\sqrt{y}<\sqrt{2p}$ Let $p$ be a fixed odd prime. Let $x,y\in \mathbb{Z}_+$ such that $\sqrt{x}+\sqrt{y}<\sqrt{2p}$. Prove that $$\sqrt{x}+\sqrt{y}\le \sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}.$$
Any ideas at all? This seems extremely difficult to do using elementary methods.
Note: It is from the 2015 Moldova TST (IMO selection test). The original problem was: Let $p$ be a fixed odd prime and $x,y\in \mathbb{Z}_+$. Find the minimum positive value of $\sqrt{2p}-\sqrt{x}-\sqrt{y}$.
EDIT! : This is apparently an old IMO Shortlist problem which Moldova recycled for their TST. I note that my reformulation of it was correct; see here for a solution: See here.
|
Let's suppose that the problem is not true and there exists $x,y$ such that
$$\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}<\sqrt{x}+\sqrt{y}<\sqrt{2p}\qquad (1)$$
First, note that if $x+y\le p$ then $$\sqrt{x}+\sqrt{y}\le\sqrt{x}+\sqrt{p-x}\le\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}$$
Since $\sqrt{x}+\sqrt{p-x}$ is concave function and maximal at $x=p/2$. Squaring both sides of $(1)$ gives
$$p+\sqrt{p^2-1}<x+y+2\sqrt{xy}<2p$$
Now, because we showed $x+y>p$, let $\epsilon=p-\sqrt{p^2-1}$ then it is enough to prove that there is no integer $n, m$ with $m<p$ and $$m-\epsilon<2\sqrt{n}<m\qquad (2)$$ Now, suppose that $m$ is even and there exists $k$ such that $2k=m$. Then we get $k<p/2$, also $n<k^2$, which is $n\le k^2-1$ and $$2k-\epsilon <2 \sqrt{n}\le 2\sqrt{k^2-1}$$
$$2(k-\sqrt{k^2-1})<p-\sqrt{p^2-1}$$
However, since $f(x)=\sqrt{x}-\sqrt{x-1}$ is strictly decreasing function for positive $x$, therefore
$$\begin{align}2(k-\sqrt{k^2-1})& > 2\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}-1}\right)\\& =p-\sqrt{p^2-4}\\& > p-\sqrt{p^2-1}\end{align}$$
And this is contradiction.
Now we suppose that $m$ is odd and there exists $k$ such that $m=2k+1$. Also we get $4n < 4k^2+4k+1$, so $n \le k^2+k$. Also, from $2k+1-\epsilon<2\sqrt{n}\le2\sqrt{k^2+k}$,$$2k+1-2\sqrt{k^2+k}<p-\sqrt{p^2-1}$$
However, this cannot be true, because
$$\begin{align}2k+1-2\sqrt{k^2+k}& = 2k+1-\sqrt{(2k+1)^2-1}\\& =m-\sqrt{m^2-1}\\& > p-\sqrt{p^2-1}\end{align}$$
Therefore, this is contradiction and there exists no integers $m,n$ with $(2)$, and it follows that there exists no integers $x,y$ satisfying $(1)$. Proved!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1214870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.