Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Calculate the sum of $\left\lfloor \sqrt{k} \right\rfloor$ I'm trying to calculate for $n\in \mathbb{N}$ the following sum :
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor$.
I tried putting in the first terms, which gave me
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=(1+2+3+\cdots+n)+\left\lfloor \sqrt{2} \right\rfloor+\left\lfloor \sqrt{3} \right\rfloor+\left\lfloor \sqrt{5} \right\rfloor+\cdots+\left\lfloor \sqrt{n^2-1} \right\rfloor$
$\iff \sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=\frac{n(n+1)}{2}+\left\lfloor \sqrt{2} \right\rfloor+\left\lfloor \sqrt{3} \right\rfloor+\left\lfloor \sqrt{5} \right\rfloor+\cdots+\left\lfloor \sqrt{n^2-1} \right\rfloor$.
I've been trying to somehow find a pattern between the different integer parts of the irrational numbers just like I did with the integers but I fail to success.
Is there a trick to use here or is my take wrong ?
Thank you.
| Put
$$k= j^2+i$$
so that$j,i$ shall span
$$
k \in \left[ {1,n^2 } \right]\quad \Rightarrow \quad \left\{ \begin{array}{l}
j \in \left[ {1,n} \right] \\
i \in \left[ {0,\left( {j + 1} \right)^2 - j^2 } \right) = \left[ {0,2j} \right] \\
\end{array} \right.
$$
and clearly it is
$$
\left\lfloor {\sqrt k } \right\rfloor = j
$$
and then the sum becomes
$$
\begin{array}{l}
S(n) = \sum\limits_{k = 1}^{n^2 } {\left\lfloor {\sqrt k } \right\rfloor }
= \sum\limits_{j = 1}^{n - 1} {\sum\limits_{i = 0}^{2j} j } + \sum\limits_{j = n}^n {\sum\limits_{i = 0}^0 j } = \\
= \sum\limits_{j = 1}^{n - 1} {j\sum\limits_{i = 0}^{2j} 1 } + \sum\limits_{j = n}^n {j\sum\limits_{i = 0}^0 1 }
= \left( {\sum\limits_{j = 1}^{n - 1} {\left( {2j + 1} \right)j} } \right) + n
= \left( {2\sum\limits_{j = 1}^{n - 1} {j^2 } + \sum\limits_{j = 1}^{n - 1} j } \right) + n = \\
= 2\frac{{n\left( {2n - 1} \right)\left( {n - 1} \right)}}{6} + \frac{{n\left( {n - 1} \right)}}{2} + n = \\
= \frac{{2n\left( {2n - 1} \right)\left( {n - 1} \right) + 3n\left( {n - 1} \right) + 6n}}{6} = \\
= \frac{{4n^3 - 3n^2 + 5n}}{6} \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $x+y+z=4$, $x^2+y^2+z^2=14$, $x^3+y^3+z^3=34$
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x+y+z=4 \\
x^2+y^2+z^2=14 \\
x^3+y^3+z^3=34
\end{split}
\end{equation}$$
My work:
I found out that $$xy+yz+xz=1$$ and $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=22$$
After this I'm stuck. Any help is greatly appreciated.
EDIT This not a duplicate. I'm looking for a detailed solution and not a solution just by inspection. Also I thought of a new idea. Maybe e should consider a cubic polynomial whose roots are $x,y,z$
| There is always the way of a direct computation. Substituting $z=4-x-y$, the other two equations are $f(x,y)=0$ and $g(x,y)=0$, and the resultant of $f$ and $g$ with respect to $y$ yields
$$
(x + 1)(x - 2)(x - 3)=0.
$$
So we have $x=-1$ or $x=2$ or $x=3$. To be more precise,
\begin{align}
f(x,y)& =x^2 + xy - 4x + y^2 - 4y + 1,\\
g(x,y)& = - x^2y + 4x^2 - xy^2 + 8xy - 16x + 4y^2 - 16y + 10.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
find the Laurent series expansion of $f(z) = \frac{1}{(z + 1)(z +3)}$ for the region $0 \lt |{z + 1}| \lt 2$ I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$
Using partial fration f(z) can written as:
$$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3}$$
since we have from Maclurin's series,
$$\sum_0^\infty \frac{1}{1+z} = (-z)^n for |z| <1$$
so ,
$$f(z) = \frac{1}{2} [\sum (-1)^n z^n - \frac{1}{1 + (z + 2)}]$$
How to proceed from here?
| It should rather be
$$f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \frac{1}{1+\frac{(z+1)}{2}}$$
$$\implies f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \left(\frac{z+1}{2}\right)^k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $f(x)=\cos x-\sin x$ is bounded Show that $$f(x)=\cos x-\sin x$$ is bounded.
We know that $$-1\le\cos x\le1\\{-1}\le\sin x\le1,$$ so I tried to subtract these two inequalities, but it seems that it doesn't work as we get: $$0\le\cos x-\sin x\le0$$
I also noticed: $f(x)=\cos x-\sin x\ge\cos x-(-1)=\cos x+2$ and $f(x)=\cos x-\sin x\ge\cos x - 1$.
| You can write
$$\begin{cases}-1\le\cos x\le1\\{-1}\le\sin x\le1\end{cases}\\
\implies \begin{cases}-1\le\cos x\le1\\{1}\ge-\sin x\ge -1\end{cases}
$$
We see that it is not possible to sum these inequalities directly from side by side.
We can try the following way to subtract the two inequalities.
Let $a≥x≥b \wedge c≥y≥d$, then we have
\begin{align*}&\begin{cases} a≥x≥b\\ c≥y≥d \end{cases}\\
\implies &\max x-\min y ≥x-y≥ \min x-\max y \\
\implies & a-d≥x-y≥b-c \end{align*}
Hence you get,
$$-2\le \cos x-\sin x\le 2$$
However, you can also write
$${1}\ge-\sin x\ge -1\\
\implies
2\ge1+\cos x\ge\cos x-\sin x\ge \cos x-1\ge-2\\
\implies -2\le \cos x-\sin x\le 2$$
But, more rigorous way is:
$$-\sqrt {a^2+b^2}≤a\cos x+b\sin x≤\sqrt {a^2+b^2}$$
Because, $\cos x-\sin x=-2$ and $\cos x-\sin x=2$ is not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove $ \int_{0}^{\infty} {x}^{a - 1} \sin \left(\frac{\pi}{2} a - b x\right) \frac{r}{{x}^{2} + {r}^{2}} d x = \frac{\pi}{2} {r}^{a - 1} {e}^{- br} $ I am attempting to calculate the following integral from Whittaker and Watson chapter 6 using the residue theorem. Show that if $a \in \left(0 , 2\right)$, $b > 0$, and $r > 0$, then
\begin{align}
\int_{0}^{\infty} {x}^{a - 1} \sin \left(\frac{\pi}{2} a - b x\right) \frac{r}{{x}^{2} + {r}^{2}} d x = \frac{\pi}{2} {r}^{a - 1} {e}^{- b r} \\
\end{align}
During the chapter, we proved that if $Q \left(z\right)$ is a rational function s.t. ${z}^{a} Q \left(z\right) \to 0$ when $z \to 0$ or $z \to \infty$, then
\begin{align}
\int_{C} {\left(- z\right)}^{a - 1} Q \left(z\right) \mathrm{dz} = 2 \pi i \sum_{\zeta}^{} {\text{Res}}_{z = \zeta} {\left(- z\right)}^{a - 1} Q \left(z\right) \\
\end{align}
and finally that
\begin{align}
\int_{0}^{\infty} {x}^{a - 1} Q \left(x\right) \mathrm{dx} = \csc \left(a \pi\right) \sum_{\zeta}^{} {\text{Res}}_{z = \zeta} {\left(- z\right)}^{a - 1} Q \left(z\right) \\
\end{align}
Attempt: I am not entirely sure how I should proceed with this integral. I assume that I must first show that
\begin{align}Q \left(x\right) = \frac{\sin \left(\frac{\pi}{2} a - b x\right) r}{{x}^{2} + {r}^{2}}\end{align}
will satisfy the necessary asymptotic conditions. I know
\begin{align}
{z}^{a} Q \left(z\right) & = {z}^{a} \sin \left(\frac{\pi}{2} a - b z\right) \frac{r}{{z}^{2} + {r}^{2}} \\
& = {z}^{a} \frac{1}{2 i} \left({e}^{\frac{\pi i}{2} a - i b z} - {e}^{\frac{- \pi i}{2} a + i b z}\right) \frac{r}{{z}^{2} + {r}^{2}} \\
\end{align}
such that when $z = x + i y$ and $| z {|}^{2} > {r}^{2}$
\begin{align}
| {z}^{a} Q \left(z\right) | \le \frac{| z {|}^{a}}{2} \left({e}^{b y} - {e}^{- b y}\right) \frac{r}{| z {|}^{2} + {r}^{2}} \\
\end{align}
Taking $x = 0$ and allowing $y \to \infty$ causes the above bound to diverge, so I am not sure how to proceed. Am I to use a different theorem for integration. It appeared to me that this integral was specifically designed to be addressed with the above method, for it was the only such problem to feature an ${x}^{a - 1}$ term just like the theorem.
In the meantime, I have computed the residues of the function ${\left(- z\right)}^{a - 1} Q \left(z\right)$.
\begin{align}
{\text{Res}}_{z = i r} {\left(- z\right)}^{a - 1} Q \left(z\right) & = {\text{Res}}_{z = i r} {\left(- z\right)}^{a - 1} \frac{\sin \left(\frac{\pi}{2} a - b z\right) r}{{z}^{2} + {r}^{2}} \\
& = \lim_{z \to i r} {\left(- z\right)}^{a - 1} \frac{\sin \left(\frac{\pi}{2} a - b z\right) r}{\left(z + i r\right) \left(z - i r\right)} \left(z - i r\right) \\
& = {\left(- i r\right)}^{a - 1} \frac{\sin \left(\frac{\pi}{2} a - b i r\right)}{2 i} \\
\end{align}
and likewise
\begin{align}
{\text{Res}}_{z = - i r} {\left(- z\right)}^{a - 1} Q \left(z\right) & = \lim_{z \to - i r} {\left(- z\right)}^{a - 1} \frac{\sin \left(\frac{\pi}{2} a - b z\right) r}{z - i r} \\
& = - {\left(i r\right)}^{a - 1} \frac{\sin \left(\frac{\pi}{2} a + b i r\right)}{2 i} \\
\end{align}
When $a$ is even,
\begin{align}
{\text{Res}}_{z = i r} {\left(- z\right)}^{a - 1} Q \left(z\right) + {\text{Res}}_{z = - i r} {\left(- z\right)}^{a - 1} Q \left(z\right) & = \frac{- {\left(i r\right)}^{a - 1}}{2 i} \left[\sin \left(\frac{\pi}{2} a - b i r\right) - \sin \left(\frac{\pi}{2} a + b i r\right)\right] \\
& = \frac{- {\left(i r\right)}^{a - 1}}{2 i} \sin \left(- b i r\right) \cos \left(\frac{\pi}{2} a\right) \\
& = \frac{{\left(i r\right)}^{a - 1}}{4} \cos \left(\frac{\pi}{2} a\right) \left[{e}^{b r} - {e}^{- b r}\right] \\
\end{align}
I feel that calculating the residue at the origin will be tedious and might not be necessary.
| Let contour $C=C_1+C_2+C_3+C_4$, where $C_1$ is the positive real axis, from $\epsilon$ to $\infty$. $C_2$ is large semi-circle on the upper half plane, with radius $R$, $C_3$ is negative real axis, from $-\infty$ to $-\epsilon$. $C_4$ is small semi-circle on the upper half plane, with radius $\epsilon$.
On $C_1$,
$$I_1=\int_0^\infty z^{a-1}\frac{r}{z^2+r^2}e^{ibz}dz$$
On $C_2$, $z=R e^{i\theta}$
$$|I_2|\le \frac{2\pi r}{R^{2-a}}\to0~~\text{as}~~R\to\infty$$
On $C_3$,
$$\begin{align} I_3&=\int_{-\infty}^0 z^{a-1}\frac{r}{z^2+r^2}e^{ibz}dz,~~~z=e^{i\pi}x\\
\\
I_3&=e^{i(a-1)\pi}\int_0^\infty x^{a-1}\frac{r}{x^2+r^2}e^{-ibx}dx\\
\\
I_3&=-e^{ia\pi}\int_0^\infty x^{a-1}\frac{r}{x^2+r^2}e^{-ibx}dx \end{align}$$
On $C_4$, $z=\epsilon e^{i\theta}$
$$|I_4|\le \frac{\pi}{2r}\epsilon^a\to0~~\text{as}~~\epsilon\to0$$
$$I_1+I_2+I_3+I_4=2\pi i\cdot \text{Res}(z=ir)$$
$$\text{Res}(z=ir)=\lim_{z\to ir} (z-ir)\frac{z^{a-1}r}{z^2+r^2}e^{ibz}=\frac{r^{a-1}e^{-br}}{2i}e^{i\frac{\pi}{2}(a-1)}=-\frac{r^{a-1}e^{-br}}{2}e^{i\frac{\pi}{2}a}$$
$$\begin{align}\int_0^\infty z^{a-1}\frac{r}{z^2+r^2}\left(e^{ibz}-e^{ia\pi-ib z}\right)dz&=-i\pi \cdot r^{a-1}e^{-br}e^{i\frac{\pi}{2}a}\\
\\
\int_0^\infty z^{a-1}\frac{r}{z^2+r^2}\left(e^{-i\left(\frac{\pi}{2}a-bz\right)}-e^{i\left(\frac{\pi}{2}a-bz\right)}\right)dz&=-i\pi \cdot r^{a-1}e^{-br}\\
\\
-2i\int_0^\infty z^{a-1}\frac{r}{z^2+r^2}\sin\left(\frac{\pi}{2}a-bz\right)dz&=-i\pi \cdot r^{a-1}e^{-br}\\
\\
\int_0^\infty z^{a-1}\frac{r}{z^2+r^2}\sin\left(\frac{\pi}{2}a-bz\right)dz&=\frac{\pi}2 \cdot r^{a-1}e^{-br}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Obtaining $\zeta(4)$ from specific derived expression I started from the expansion of cot(x) by the Mittag-Leffler theorem, where:
$$\cot(x) = \frac{1}{x} + \sum_{k = 1}^{\infty} \frac{2x}{x^{2} - k^2 \pi^2}.$$
After splitting up $\cot(x)$, taking the derivative of both sides, and some algebra, I eventually was left with the following expression:
$$-1 = 6\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} + 4x^2
\left [ \sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 - 4x^2\sum_{k = 1}^{\infty} \frac{1}{({x^{2} - k^2 \pi^2})^2} $$
where $x = 0$ gives $\zeta(2)$. I am now trying to obtain the exact value for $\zeta(4)$ through this expression. To do this, I am attempting to once again take the derivative and then set $x = 0$. My work is shown below.
$$-1 = 6\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} + 4x^2
\left [ \sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 - 4x^2\sum_{k = 1}^{\infty} \frac{1}{({x^{2} - k^2 \pi^2})^2} \Longrightarrow $$
$\Longrightarrow 0 = -12x\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} + 8x\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 + 8x^2\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] \times \frac{d}{dx}\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] - 8x \left [\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} \right ] \left [ \displaystyle\sum_{k = 1}^{\infty} \frac{-4x}{(x^{2} - k^2 \pi^2)^3} \right ] \Longrightarrow
$
Dividing by $x$:
$\Longrightarrow0 =-12\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} + 8\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 + \\8x\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] \times \frac{d}{dx}\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] - 8x \left [\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} \right ] \left [ \displaystyle\sum_{k = 1}^{\infty} \frac{-4}{(x^{2} - k^2 \pi^2)^3} \right ] \Longrightarrow
$
Setting $x = 0$:
$\Longrightarrow 0 = -12\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(- k^2 \pi^2)^2} + 8\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{- k^2 \pi^2} \right ] ^2 \Longrightarrow
$
Multiplying everything by $\pi^4$:
$\Longrightarrow 0 = -12\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(- k^2 )^2} + 8\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{- k^2 } \right ] ^2 \Longrightarrow
$
$\Longrightarrow 0 = 3\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(k^2 )^2} - 2\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{ k^2 } \right ] ^2 \Longrightarrow
$
Solving for ζ(4):
$\Longrightarrow 0 = 3\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(k^2 )^2} - 2\left [ \frac{\pi^2}{6} \right ] ^2 \Longrightarrow
$
$\Longrightarrow \displaystyle\sum_{k = 1}^{\infty} \frac{1}{k^4 } = \frac{\pi^4}{54}$
However, my answer is wrong. The correct result would be $\frac{\pi^4}{90}$. Any help on this would be greatly appreciated.
| Expanding the internal passages:
\begin{align}
\frac{\text{d}}{\text{d}x} \left(4x^2\left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2\right) & = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 + 4x^2 \frac{\text{d}}{\text{d}x} \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2
\\\\
& = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 + 8x^2 \sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\frac{\text{d}}{\text{d}x}\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}
\\\\
& = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 -16x^3 \left(\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right) \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}
\end{align}
Also
\begin{align}
\frac{\text{d}}{\text{d}x} \left(-4x^2 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}\right) & = -8x \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} - 4x^2 \frac{\text{d}}{\text{d}x}\left(\sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}\right)
\\\\
& = -8x \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} + 16x^3 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^3}
\end{align}
When you divide by $x$, it remains
$$0 = -12 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} + 8\left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2- 8 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}$$
Setting $x = 0$ then
$$0 = -12 \sum_{k = 1}^{+\infty} \frac{1}{(k^2\pi^2)^2} + 8\left[\sum_{k = 1}^{+\infty} \frac{1}{-k^2\pi^2}\right]^2 -8 \sum_{k = 1}^{+\infty} \frac{1}{(k^2\pi^2)^2} $$
Summing the extreme terms and multiplying by $\pi^4$:
$$0 = -20 \sum_{k = 1}^{+\infty} \frac{1}{(k^2)^2}+ 8\left[\sum_{k = 1}^{+\infty} \frac{1}{-k^2}\right]^2$$
Now
$$\sum_{k = 1}^{+\infty} \frac{1}{k^4} = \frac{8}{20} \frac{\pi^4}{36} = \frac{2\pi^4}{360} = \frac{\pi^4}{90}$$
As wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Alternating sum of binomial coefficients $\sum_{k=0}^{49}(-1)^k \binom{99}{2k} = -2^{49}$ How can you prove that $$\sum_{k=0}^{49} \binom{99}{2k}(-1)^k = -2^{49}?$$
A more general formula seems to be $\sum_{k=0}^{n} \binom{2n + 1}{2k}(-1)^k$
For $n = 0:$ it equals $2^0$
For $n = 1: -2^1$
For $n = 2: -2^2$
For $n = 3: 2^3$
For $n = 4: 2^4$
For $n = 5: -2^5$
For $n = 6: -2^6$
For $n = 7: 2^7$
etc.
| Use binomial theorem as
$$F(x)=\frac{(1+x)^{2n+1}+(1-x)^{2n+1}}{2}=\sum_{k=0}^{n} {2n+1 \choose 2k} x^{2k}$$
Then take$x=i$, to get
$$\sum_{k=0}^{n} {2n+1 \choose 2k} (-1)^{2k}=F(i)=\frac{(1+i)^{2n+1}+(1-i)^{2n+1}}{2}=2^{n+1/2}\cos[(2n+1)\pi/4]n$$
where we have used $(1\pm i)=\sqrt{2}[\cos(\pi/4)\pm i\sin(\pi/4)]$ and demoivre's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving definite integral $\int_{0}^{1} \tan^{-1}(1-1/x)dx$ $$\int_{0}^{1} \tan^{-1}\left(1-\frac1x\right)dx$$
Here's what I have done so far. (the answer is given as $-\pi/4$)
Let
$$
I = \int_{0}^{1}\tan^{-1}\left(1-\frac1x\right)dx = \int_{0}^{1}\tan^{-1}\left(\frac{x-1}x\right)dx.
$$
Since, $\int_{0}^{1} f(x)dx = \int_{0}^{1} f(1-x)dx$ one has\begin{align}
I &= \int_{0}^{1}\tan^{-1}\left(1-\frac1{1-x}\right)dx\\
& = \int_{0}^{1}\tan^{-1}\left(\frac x{x-1}\right)dx\\
& = \int_{0}^{1}\frac\pi2-\cot^{-1}\left(\frac x{x-1}\right)dx\\
& = \int_{0}^{1}\frac\pi2-\tan^{-1}\left(\frac{x-1}x\right)dx\\
& = \frac\pi2 - I,
\end{align}
Hence, $I = \dfrac\pi4$. The given answer is $-\dfrac\pi4$. Where have I gone wrong?
| Noticing that
$$
\tan \left[\tan ^{-1}\left(\frac{x-1}{x}\right)+\frac{\pi}{4}\right]=\frac{\frac{x-1}{x}+1}{1-\frac{x-1}{x}}=2 x-1
$$
For $x\in (0,1)$, $$
\tan ^{-1}\left(\frac{x-1}{x}\right)+\frac{\pi}{4}=\tan ^{-1}(2 x-1)
$$
Integrating both sides from $0$ to $1$ yields
$$
\begin{aligned}
\int_{0}^{1} \tan ^{-1}\left(1-\frac{1}{x}\right) d x+\frac{\pi}{4} &=\int_{0}^{1} \tan ^{-1}(2 x-1) d x \stackrel{2x-1\mapsto x}{=} \frac{1}{2} \int_{-1}^{1} \tan ^{-1} x d x=0
\end{aligned}
$$
Hence
$$\boxed{\int_{0}^{1} \tan ^{-1}\left(1-\frac{1}{x}\right) dx =-\frac{\pi}{4} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Solving $y' = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}$ without tricks I would like to solve the equation:
$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy - x^2 - 2} \tag{1}$
From that, we have:
$y'(y^2 + 2xy - x^2 - 2) = y^2 - 2xy - x^2 + 2 \implies $
$\frac{(2x + 2yy')(x+y) - (x^2 + y^2 + 2)(1 + y')}{(x+y)^2} = 0 \implies $
$(\frac{x^2 + y^2 + 2}{x + y})' = 0 \implies $
$\frac{x^2 + y^2 + 2}{x + y} = C$
Where the last part agrees with the solution in the book.
The "trick" is that I first looked at the solution in the book, differentiated it, and then applied here the process in reverse, so it looks like a solution. I was obviously not happy with that approach, so I looked online for a better solution. Unfortunately, I could not see a generic pattern in neither of the two existing solutions [solution 1, solution 2] I found.
Does there exists a more standard solution to the equation?
Are the two existing solutions common, and are not considered as "tricks"? Why would they be less of a trick than the one I described above?
Thanks!
| $$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy - x^2 - 2} $$
$$\dfrac {dy}{y^2 - 2xy - x^2 + 2} = \dfrac{dx}{y^2 + 2xy - x^2 - 2}$$
$$\dfrac {d(x+y)}{-y^2 +x^2} = \dfrac{d(x-y)}{- 2xy +2}$$
Substitute $u=x+y; v=x-y$:
$$\dfrac {du}{2uv} = \dfrac{dv}{v^2-u^2 +4}$$
$$ 2uv\dfrac{dv}{du}=v^2-u^2 +4$$
$$ u\dfrac{dv^2}{du}=v^2-u^2 +4$$
This is a first order linear DE.
$$ us'-s=-u^2 +4$$
Where $s=v^2=(x-y)^2$.
I only used substitutions to transform the DE into a first order linear DE that can be easily solved with well-known methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show inequality for positive real numbers If $x,y$ are positive real numbers then we have that $$ \frac{1}{\sqrt{x+y}}<\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}$$ right? But how can we show that?
I have tried the following but I don't think that this is the way we should go.
\begin{align*}\frac{1}{\sqrt{x + y}} < \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}}=\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}} & \iff \left(\frac{1}{\sqrt{x + y}}\right)^2 < \left(\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}}\right)^2 \\ & \iff \frac{1}{x + y} <\frac{x+y+2\sqrt{xy}}{xy} \\ & \iff xy<(x+y)(x+y+2\sqrt{xy}) \\ & \iff xy<x^2+xy+2x\sqrt{xy}+xy+y^2+2y\sqrt{xy}\\ & \iff 0<x^2+2(x+y)\sqrt{xy}+xy+y^2\end{align*}
| $f(x) = 1/\sqrt{x}$ is strictly convex on $(0, \infty)$, so that
$$
f\left( \frac{x+y}{2}\right) < \frac 12 \left( f(x) + f(y) \right) \, ,
$$
for $0 < x < y$, this gives the stronger inequality
$$
\frac{1}{\sqrt{x+y}}<\frac{1}{2\sqrt 2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right) \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Double integral $\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$ I'm trying to evaluate this double integral:
$$I=\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$$
Here is the closed-form:
$$\frac{91π^4}{11520}+\frac{21}{32} ζ(3)-\frac{7}{8} ζ(3)\log{2}-\text{Li}_4 (\frac{1}{2})+\frac{π^2}{24}\log^2{2}-\frac{π^2}{16}\log{2}-\frac{1}{24}\log^4{2}$$
My try: because of its symmetry we can rewrite it as:$$2I=\int_{0}^{1}\int_{0}^{1}\frac{(x^2y^3+x^3y^2)\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$$
Then use the substitution: $x\to \frac{1}{x}$ and $y\to \frac{1}{y}$, now it becomes:
$$I=\int_{1}^{\infty}\int_{1}^{\infty}\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dxdy$$
Seem I could get rid of the power of $x$ and $y$ from numerator, but after splitting:$$I=\int_{1}^{\infty}\frac{\log{(x)}}{x(1-x^2)}dx\int_{1}^{\infty}\frac{\log{(y)}}{(1-y^2)(x^2y^2-1)}dy$$
The latter integral gives (Mathematica):$$\fbox{$-\frac{-2 x \text{Li}_2\left(x^2\right)+8 x \text{Li}_2(x)+2 x \log ^2(-x)-2 x \log ^2(x)+\pi ^2}{8 \left(x^2-1\right)}\text{ if }\Im(x)\neq 0\lor -1<\Re\left(\frac{1}{x}\right)<1$}$$
Now, I don't know how to process further. Maybe I misdirected at the beginning. Can you guys help me with this? Thank you very much.
| It is nice to have arrives at $$I=\int_{1}^{\infty}\int_{1}^{\infty}\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dx\,dy$$
$$J=\int\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dx$$
$$\frac{2 \left(y^2-1\right)^2}{\log (y)}\,J=-y^2 \text{Li}_2(-x y)-y^2 \text{Li}_2(x y)-\text{Li}_2(1-x)+\text{Li}_2(-x)+y^2
\log ^2(x)-$$ $$y^2 \log (x) \log (1-x y)-y^2 \log (x) \log (x y+1)-\log ^2(x)+\log
(x) \log (x+1)$$ Integrated between $1$ and $\infty$ the rhs is
$$\frac{1}{12} \left(6 y^2 \text{Li}_2\left(y^2\right)+\pi ^2 \left(1-2 y^2\right)+12
y^2 \log (y) (\log (y)+i \pi )\right)$$ So, it remains to compute
$$I=\frac 1 {24}\int_1^\infty \frac{\log (y) \left(6 y^2 \text{Li}_2\left(y^2\right)+\pi ^2 \left(1-2
y^2\right)+12 y^2 \log (y) (\log (y)+i \pi )\right)}{\left(y^2-1\right)^2}\,dy$$
At this point, I am stuck !
Numerically integrated, this gives
$$I=0.072175120285406568206969486373268731903370185258386\cdots$$ which is exactly the value of
$$\frac{91π^4}{11520}+\frac{21}{32} ζ(3)-\frac{7}{8} ζ(3)\log(2)-\text{Li}_4 (\frac{1}{2})+\frac{π^2}{24}\log^2(2)-\frac{π^2}{16}\log(2)-\frac{1}{24}\log^4(2)$$
Edit
If we start with
$$\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}\,dx\,dy$$
$$J=\int \frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}\,dx$$ gives
$$\frac{2 \left(1-y^2\right)^2}{y^2 \log (y)}\,J=y (\text{Li}_2(1-x)+\text{Li}_2(-x))-\text{Li}_2(-x y)+\text{Li}_2(x y)+$$ $$\log (x)
\left(y \log (x+1)-2 \tanh ^{-1}(x y)\right)$$ Usingg the bounds, it remains to compute
$$\int_0^1 \frac{y^2 \log (y)}{2 \left(y^2-1\right)^2}\left(-\frac{\pi ^2 y}{4}-\text{Li}_2(-y)+\text{Li}_2(y)\right)\,dy$$
Even if it looks simpler, at this point, I am still stuck !
| {
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"url": "https://math.stackexchange.com/questions/4522660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$? I'm having difficulties solving $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$. I guess I'm missing a crucial point. The solution is $x=\frac{49}{4}$. What are the steps?
| As the user Dark Magician said, we can use the substitution $t=\sqrt{x}$. We get the following from using the quadratic formula:
\begin{align}
(t+7)(t-1)&=\frac{105}{4}\\
t^2-t+7t-7&=\frac{105}{4}\\
t^2+6t-7-\frac{105}{4}&=0\\
t^2+6t-\frac{133}{4}&=0\\
4t^2+24t-133&=0\\
t&=\frac{-24\pm\sqrt{24^2-4\times4\times(-133)}}{2\times4}\\
t&=\frac{-24\pm\sqrt{576+2128}}{8}\\
t&=\frac{-24\pm\sqrt{2704}}{8}\\
t&=-3\pm\frac{13}{2}\\
t&=-\frac{19}{2}\ or\ \frac{7}{2}
\end{align}
We know that $t$ is nonnegative, as it is an arithmetic square root, so
$$t=\frac{7}{2}$$
$$x=t^2=\frac{49}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Excircles of a right triangle
Given a $\triangle ABC$ such that the $A$ excircle has a radius $r_A$ and the $B$ excircle has a radius $r_B$. Prove that if $r_A + r_B = AB$ then $\triangle ABC$ is a right triangle.
I called $D$ the point where the $A$ excircle touches the side $BC$, $E$ the point where the $B$ excircle touches the side $AC$ and $F$ the point where the $C$ excircle touches the side $AB$.
Then I noticed that $AE = BD$ and that in order to prove the given statement I need to show that $BF = r_B$ and $AF = r_A$ but I don't know how to use these facts.
| The formula for the ex-radius is
$ r_A = \dfrac{ \Delta } {s - a} $
$ r_B = \dfrac{ \Delta }{ s- b} $
where $ \Delta $ is the area of the triangle, and $s = \frac{1}{2}(a + b + c) $
Given in the problem
$ r_A + r_B = AB = c $
Therefore,
$ \Delta \left( \dfrac{1}{s-a} + \dfrac{1}{s - b} \right) = c $
so that
$ \Delta ( 2 s - a - b ) = c (s - a)(s - b) $
But $2 s - a - b = c $, therefore
$ \Delta = (s - a)(s - b) $
From Heron's formula for the area,
$ \Delta = \sqrt{ s (s - a) (s - b)(s - c) }$
Therefore
$ s (s - a)(s - b) (s - c) = (s - a)^2 (s - b)^2 $
so that
$ s (s - c) = (s - a)(s - b) $
Expanding
$ s^2 - s c = s^2 - s (a + b) + a b $
Hence,
$ s (a + b - c) = a b $
But, $ s = \dfrac{1}{2}(a + b + c ) $
Therefore,
$ (a + b + c) (a + b - c) = 2 a b $
so that
$(a + b)^2 - c^2 = 2 a b $
And, from this,
$ a^2 + 2 a b + b^2 - c^2 = 2 a b $
Therefore,
$ c^2 = a^2 + b^2 $
Hence, $\triangle ABC $ is a right triangle at $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of two independent Gaussian r.v.s. given that one is negative Let $X \sim \mathcal N(0, \sigma_x^2)$ and $Y \sim \mathcal N(0, \sigma_y^2)$ be two independent r.v.s.
Is there a simple solution for $p(X+Y = a ~|~ X \leq 0)$?
I'm aware that without the conditioning on $X \leq 0$ we have that $Z = X+Y \sim \mathcal N(0, \sigma_x^2 + \sigma_y^2)$, but for me it seems that conditioning on $X \leq 0$ makes the problem considerably harder.
Here's what I have so far:
\begin{equation}
\begin{split}
p(X+Y=a ~|~ X \leq 0) &= \frac{p(X+Y=a, X \leq 0)}{P(X \leq 0)} \\
&= 2\int_{-\infty}^0 p(X+Y=a, X=b) db \\
&= 2 \int_{-\infty}^0 p(Y=a-b) ~ p(X=b) db \\
& \quad\quad\quad\quad \vdots \quad \text{(following the proof for the non-conditioned case)}\\
&= 2 ~ \frac{1}{\sqrt{2 \pi (\sigma_x^2 + \sigma_y^2)}} \exp \left(- \frac{a^2}{2 (\sigma_x^2 + \sigma_y^2)} \right) \int_{-\infty}^0 \frac{1}{\sqrt{2 \pi \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}} \exp \left(- \frac{\left(b - \frac{\sigma_x^2 a}{\sigma_x^2 + \sigma_y^2}\right)^2}{2 \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}} \right) db \\
&= 2 ~ p(X+Y=a) \int_{-\infty}^0 \frac{1}{\sqrt{2 \pi \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}} \exp \left(- \frac{\left(b - \frac{\sigma_x^2 a}{\sigma_x^2 + \sigma_y^2}\right)^2}{2 \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}} \right) db \\
&= 2 ~ p(X+Y=a)~p \left(W \geq \frac{\sigma_x a}{\sigma_y \sqrt{\sigma_x^2 + \sigma_y^2}} \right), \quad\quad W \sim \mathcal N(0,1)\\
\end{split}
\end{equation}
I'm trying to simplify this further such that its entropy can be 'neatly' computed.
| In your third equality line, the integrand is zero as these are absolutely continuous random variables. Dont you get an answer of zero just from that?
| {
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Why $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ results in pythagorean triples? As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2*0)^2+(0^2-1)^2=(0^2+1)^2$
$(2*1)^2+(1^2-1)^2=(1^2+1)^2$
$(2*2)^2+(2^2-1)^2=(2^2+1)^2$
$(2*0)^2+(0-1)^2=(0+1)^2$
$(2*1)^2+(1-1)^2=(1+1)^2$
$(2*2)^2+(4-1)^2=(4+1)^2$
$0^2+1^2=1^2$
$2^2+0^2=2^2$
$4^2+3^2=5^2$
$0+1=1$
$4+0=4$
$16+9=25$
$1=1$
$4=4$
$25=25$
| Here's a visual proof:
(This space intentionally left blank.)
| {
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A is an n by n matrix such that $A^3=2I$. A is an n by n matrix such that $A^3 = 2I$.
(1) Is $A$ invertible? If it is, then what is $A^{-1}$?
(2) Is $A+I$ invertible? If it is, then what is $(I+A)^{-1}$
The original version of the question is not in English, so maybe there are some grammar mistakes, sorry in advance.
| A matrix $B$ that satisfies $AB=BA = I$ is an inverse of $A$.
Since $A ({1 \over 2} A^2) = I$ we see that $A^{-1} = {1 \over 2} A^2$.
Suppose $x^3 = 2$, then we see that
${1 \over x+1} = {1 \over x} (1-{1 \over x} + {1 \over x^2} - {1 \over x^3} + {1 \over x^4} - {1 \over x^5} +\cdots) $, and using $x^3 = 2$ we get
$=1+(-1+{1 \over x} - {1 \over x^2}) (1-{1 \over 2} + {1 \over 2^2} - \cdots)$ and using ${x^3 \over 2} =1$ we get
$= {1 \over 3} (1-x+x^2)$.
(Check by multiplying by $1+x$.)
This suggests trying $B={1 \over 3} (I-A+A^2)$ as an inverse for $I+A$.
As above, if we multiply $(I+A)B$ we get $I$ hence
$(I+A)^{-1} = {1 \over 3} (I-A+A^2)$.
Addendum: Another approach to computing $(A+I)^{-1}$.
We want to find a polynomial $p$ such that $(x+1) p(x)$ is a non zero constant. Since $x^3=2$ we posit that a degree two polynomial will suffice. Hence we are looking for $a,b$ such that $(x+1) (1 + ax+bx^2)$ is a constant.
Expanding (and using $x^3=2$) gives
$a=-1, b=1$ which is the same as above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Relationship between e and sine I am reading a physics text about scattering but I am stuck with the maths.
I think the solution might be related to Euler's formula where:
$$\frac{e^{i\theta} - e^{-i\theta} }{2i} = \sin\theta$$
The text says that:
$$I(Q) = \bigg|\sum_{n=0}^{N-1}b e^{ikna\sin\theta}\bigg|^2$$
and
$$\sum_{n=0}^{N-1} u^{n} = \frac{1-u^N}{1-u}$$
where $u = e^{ika\sin\theta}$ and they say that this means:
$$I(Q) = b^2\frac{\sin^2(\tfrac12kNa\sin\theta)}{\sin^2(\tfrac12ka\sin\theta)}$$
Can someone show this?
Thanks
| So first, we have the identities
$$|x+iy|^2 = x^2 + y^2 \text{ for } x,y\in\mathbb{R}$$
$$\left|\frac{x}{y}\right| = \frac{|x|}{|y|}$$
$$1-\cos(x) = 2\sin^2(x/2)$$
$$\cos^2(x) + \sin^2(x) = 1$$
$$\sum_{j=0}^{N-1} r^j = \frac{1-r^N}{1-r}$$
Also for any $x$,
$$
\begin{align*}
\left|1-e^{ix}\right| &= |1-\cos(x)-i\sin(x)|^2 \\
&= (1-\cos(x))^2 + \sin^2(x) \\
&= 1 - 2\cos(x) + \cos^2(x) + \sin^2(x) \\
&= 2 - 2\cos(x) \\
&= 4\sin^2(x/2)
\end{align*}
$$
Let $\phi = ka\sin(\theta)$
Then
$$
\begin{align*}
I &= b^2\left|\sum_{n=0}^{N-1} e^{i\phi n}\right|^2 \\
&= b^2\left|\frac{1-e^{i\phi N}}{1-e^{i\phi}}\right| \\
&= b^2 \frac{|1-e^{i\phi N}|}{|1-e^{i\phi}|} \\
&= b^2 \frac{\sin^2(N\phi/2)}{\sin^2(\phi/2)} \\
&= b^2 \frac{\sin^2(\frac{1}{2}Nka\sin(\theta))}{\sin^2(\frac{1}{2}ka\sin(\theta))}
\end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the value of $\sqrt{\Delta}$ for the given question? If it is given that $\alpha , \beta, \gamma$ are the roots of $x^3+x^2-2x+1=0$, then find the value $\sqrt{\Delta}$, where $$
\Delta=\left|\begin{array}{ccc}
\gamma^2 & \beta^2 & 2 \beta \gamma-\alpha^2 \\
2 \alpha \gamma-\beta^2 & \alpha^2 & \gamma^2 \\
\alpha^2 & 2 \alpha \beta-\gamma^2 & \beta^2
\end{array}\right|
$$
I tried expanding the determinant and it comes out to be,
\begin{align*}
\Delta &= \alpha^{6} - 6 \alpha^{4} \beta \gamma + 2 \alpha^{3} \beta^{3} + 2 \alpha^{3} \gamma^{3} + 9 \alpha^{2} \beta^{2} \gamma^{2} - 6 \alpha \beta^{4} \gamma - 6 \alpha \beta \gamma^{4} + \beta^{6} + 2 \beta^{3} \gamma^{3} + \gamma^{6} \\
&= (\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma)^2
\end{align*}
But I don't know how to proceed from here, also I believe that there should be another way to do it, as this expansion is way too tedious to do. Any help is appreciated.
Edit: I managed to get the answer $\sqrt{-7}$ using Newton's identities.
| By confort put $a,b,c$ for the three roots. You do have
$$a+b+c=-1\\ab+ac+bc=-2\\abc=-1\tag1$$ Besides $$(a+b+c)^3=a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc\\(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\tag2$$ From $(2)$ and $(1)$
$$-1=a^3+b^3+c^3+3a^2(-1-a)+3b^2(-1-b)+3c^2(-1-c)-6$$ then $$2(a^3+b^3+c^3)=-3(a^2+b^2+c^2)-5=-3((-1)^2+4)-5=-20$$ Finally, using your
$\Delta=(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma)^2$
we get
$$\Delta=(a^3+b^3+c^3-3abc)^2=(-10+3)^2=49$$
| {
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Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$ I tried using integration by parts and could get to this point:
\begin{align}
\int \sqrt{x^2 + a^2}\mathrm{d}x &= x\sqrt{x^2+a^2} - \int x\frac{2x}{\sqrt{x^2 + a^2}}\mathrm{d}x
\\\\ &= x\sqrt{x^2+a^2} -2\int \sqrt{x^2 + a^2}\mathrm{d}x + \int \frac{2a^2}{\sqrt{x^2 + a^2}}\mathrm{d}x
\end{align}
then, I don't know how to get:
$$\int \frac{2a^2}{\sqrt{x^2 + a^2}}\mathrm{d}x$$
| W.l.o.g. we can assume that $a>0$ (because, otherwise one can replace $a$ with its absolute value)
Setting
$$
x=a\tan(u),
$$
we have
$$
dx=a\sec^2(u)du,
$$
so we get
\begin{eqnarray}
\int\sqrt{x^2+a^2}dx
&=&\int a\sec^2(u)\sqrt{a^2\tan^2(u)+a^2}du\\
&=&\int a\sec^2(u)\sqrt{a^2(1+\tan^2(u))}du\\
&=&\int a\sec^2(u)\sqrt{a^2\sec^2(u)}du\\
&=& a^2\int \sec^3(u)du
\end{eqnarray}
Using integration by parts, we have
\begin{eqnarray}
\int \sec^3(u)du
&=&\int \sec(u)[\tan(u)]’du\\
&=& \sec(u)\tan(u)-\int \tan(u)]\sec(u)]’du\\
&=&\sec(u)\tan(u)-\int\tan(u)\sec(u)\tan(u)du\\
&=&\sec(u)\tan(u)-\int\sec(u)\tan^2(u)du\\
&=&\sec(u)\tan(u)-\int\sec(u)[\sec^2(u)-1]du\\
&=&\sec(u)\tan(u)+\int\sec(u)du-\int\sec^3(u)du\\
&=&\sec(u)\tan(u)+\ln|\sec(u)+\tan(u)|+2B-\int\sec^3(u)du
\end{eqnarray}
where $B$ is an arbitrary constant. Therefore
$$
\int\sec^3(u)du=\frac12\sec(u)\tan(u)+\frac12\ln|\sec(u)+\tan(u)|+B.
$$
Since
$$
\tan(u)=\frac{x}{a},
$$
we have
$$
\sec(u)=\frac1a\sqrt{x^2+a^2}.
$$
Hence
$$
\int\sqrt{x^2+a^2}dx=\frac12x\sqrt{x^2+a^2}+\frac12a^2\ln|x+\sqrt{x^2+a^2}|+C
$$
where
$$
C=B-\frac12a^2\ln(a).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate this improper integral $\int_{-\infty}^\infty x^2 e^{-x^2}\cos x \, dx$? How to evaluate
$$
\int_{-\infty}^{\infty} x^2 e^{-x^2} \cos x \, dx?
$$
My Attempt:
Since the integrand is an even function, we have
$$
\int_{-\infty}^{\infty} x^2 e^{-x^2} \cos x \, dx = 2 \int_0^{\infty} x^2 e^{-x^2} \cos x \, dx. \tag{1}
$$
Since $-1 \leq \cos x \leq 1$ and $x^2 e^{-x^2} > 0$ for all $x$ such that $0 \leq x < \infty$, we have
$$
-x^2 e^{-x^2} \leq x^2 e^{-x^2} \cos x \leq x^2 e^{-x^2},
$$
which implies
$$
-\int_0^{\infty} x^2 e^{-x^2} \, dx \leq \int_0^{\infty} x^2 e^{-x^2} \cos x \, dx \leq \int_0^{\infty} x^2 e^{-x^2} \, dx,
$$
and hence
$$
-2 \int_0^{\infty} x^2 e^{-x^2} \, dx \leq 2 \int_0^{\infty} x^2 e^{-x^2} \cos x \, dx \leq 2 \int_0^{\infty} x^2 e^{-x^2} \, dx.
$$
Now if we can show that
$$
\int_0^{\infty} x^2 e^{-x^2} \, dx = 0,
$$
then would obtain
$$
\int_{-\infty}^{\infty} x^2 e^{-x^2} \cos x \, dx = 0.
$$
[ Refer to (1) above.]
Am I right? If so, then what next? How to proceed?
Or, is there another better way around this problem?
| How to solve!
The actual result is different:
$$ \int_{-\infty}^{\infty} x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x = \frac{\sqrt{\pi}}{4 \cdot \sqrt[4]{e}} \approx 0.345097 $$
The easiest way to solve the problem would be to form one od the antiderivatives:
(You can easily do this by relating the trigonometric functions to the exponential functions:
$$ \cos(x) = \frac{e^{x \cdot \mathrm{i}} + e^{-x \cdot \mathrm{i}}}{2} $$
$$ \int x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x = \dfrac{\mathrm{e}^{-\frac{1}{4}}\cdot\left(\sqrt{{\pi}} \cdot \operatorname{erf}\left(x+\frac{\mathrm{i}}{2}\right)+\sqrt{{\pi}} \cdot \operatorname{erf}\left(x-\frac{\mathrm{i}}{2}\right)\right)}{16}+\dfrac{\mathrm{e}^{-x^2}\cdot\left(\sin\left(x\right)-2 \cdot x \cdot \cos\left(x\right)\right)}{4} $$
or
$$ \int x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x =
\dfrac{\frac{\mathrm{e}^{-\frac{1}{4}}\mathrm{i}\cdot\left(\frac{\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{2},\frac{\left(\mathrm{i}-2x\right)^2}{4}\right)\left(\mathrm{i}-2x\right)}{4\sqrt{\left(\mathrm{i}-2x\right)^2}}-\frac{\mathrm{i}\operatorname{\Gamma}\left(\frac{3}{2},\frac{\left(\mathrm{i}-2x\right)^2}{4}\right)\left(\mathrm{i}-2x\right)^3}{\left(\left(\mathrm{i}-2x\right)^2\right)^\frac{3}{2}}-\operatorname{\Gamma}\left(1,\frac{\left(\mathrm{i}-2x\right)^2}{4}\right)\right)}{2}+\frac{\mathrm{e}^{-\frac{1}{4}}\mathrm{i}\cdot\left(\frac{\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{2},\frac{\left(-2x-\mathrm{i}\right)^2}{4}\right)\left(-2x-\mathrm{i}\right)}{4\sqrt{\left(-2x-\mathrm{i}\right)^2}}-\frac{\mathrm{i}\operatorname{\Gamma}\left(\frac{3}{2},\frac{\left(-2x-\mathrm{i}\right)^2}{4}\right)\left(-2x-\mathrm{i}\right)^3}{\left(\left(-2x-\mathrm{i}\right)^2\right)^\frac{3}{2}}+\operatorname{\Gamma}\left(1,\frac{\left(-2x-\mathrm{i}\right)^2}{4}\right)\right)}{2}}{2} $$
(calculated by Integralrechner, wich shows a step by step solution too)
or
$$ \int x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x = \frac{(\frac{\sqrt{\pi} \cdot \operatorname{erf}(\frac{2 \cdot x + \mathrm{i}}{2})}{\sqrt[4]{e}} - \frac{\mathrm{i} \sqrt(\pi) \operatorname{erfi}\frac{1}{2} + \mathrm{i} \cdot x)}{\sqrt[4]{e}} - 2 e^{-x \cdot (x + \mathrm{i})} \cdot (2 x + e^{2 \cdot \mathrm{i} \cdot x} (2 \cdot x + \mathrm{i}) - \mathrm{i}))}{16} + constant $$
(calculated by Wolfram|Alpha)
or
...
Now you have to form limes with from x to infinity and x to -infinity of you're formula and you will get you're sulution! (as shown above)
$$ \int_{-\infty}^{\infty} x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x = \lim_{{x} \to {\infty}} (\dfrac{\mathrm{e}^{-\frac{1}{4}}\cdot\left(\sqrt{{\pi}} \cdot \operatorname{erf}\left(x+\frac{\mathrm{i}}{2}\right)+\sqrt{{\pi}} \cdot \operatorname{erf}\left(x-\frac{\mathrm{i}}{2}\right)\right)}{16}+\dfrac{\mathrm{e}^{-x^2}\cdot\left(\sin\left(x\right)-2 \cdot x \cdot \cos\left(x\right)\right)}{4}) - \lim_{{x} \to {-\infty}} (\dfrac{\mathrm{e}^{-\frac{1}{4}}\cdot\left(\sqrt{{\pi}} \cdot \operatorname{erf}\left(x+\frac{\mathrm{i}}{2}\right)+\sqrt{{\pi}} \cdot \operatorname{erf}\left(x-\frac{\mathrm{i}}{2}\right)\right)}{16}+\dfrac{\mathrm{e}^{-x^2}\cdot\left(\sin\left(x\right)-2 \cdot x \cdot \cos\left(x\right)\right)}{4})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
How to find the number of integral solutions of a given equation analytically? I was given $$\tan^{-1} x+\cot^{-1} y= \tan^{-1} 3$$
and asked to find the number of integral solutions.
I was able to find two, which are $(1,2) ; (2,7)$ which was obtained by solving the given equation by taking the tangent of both sides and writing $\cot y \\$ as $ \tan \frac{1}{y}$
after which I used trial and error.
However, this doesn't avoid the possibility of missing solutions.
What's the analytical method of solving such problems?
I have come across this question here on solving equations using Simon's favourite factoring technique but, I don't really get that. Hence I'm asking here.
| According to your description of the problem, you got at least this:
$$
\frac{x+y^{-1}}{1-xy^{-1}} = 3
$$
Multiply numerator and denominator of the left side with $y$:
$$
\frac{xy+1}{y-x} = 3
$$
which means
$$
xy+1 = 3y-3x
$$
or
$$
(x-3)(y+3)=-10
$$
Now you only have to check in how may ways $-10$ can be expressed as product of two numbers. Then you get the candidates of the possible solutions.
$$
(x-3,y+3) \in\{(-10,1),(-5,2),(-2,5),(-1,10),(1,-10),(2,-5),(5,-2),(10,-1)\}
$$
That is
$$
(x,y) \in\{(-7,-2),(-2,-1),(1,2),(2,7),(4,-13),(5,-8),(8,-5),(13,-4)\}
$$
However, we have only shown that this is a necessary condition. We also have to account for the fact that we would have got the same necessary condition if we added an integer multiple of $\pi$ to one of the sides of the original equation. So we can expect that some of the candidates do not fulfill the original equation, but make the left side and the right side differ by an integer multiple of $\pi.$
If we use the convention $\tan^{-1}x\in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $\cot^{-1}y\in \left(-\frac{\pi}{2},\frac{\pi}{2}\right]\setminus\{0\}$, we get
\begin{eqnarray}
\tan^{-1}(-7)+\cot^{-1}(-2) &=& \tan^{-1}(3) -\pi \\
\tan^{-1}(-2)+\cot^{-1}(-1) &=& \tan^{-1}(3) -\pi \\
\tan^{-1}(1)+\cot^{-1}(2) &=& \tan^{-1}(3) \\
\tan^{-1}(2)+\cot^{-1}(7) &=& \tan^{-1}(3) \\
\tan^{-1}(4)+\cot^{-1}(-13) &=& \tan^{-1}(3) \\
\tan^{-1}(5)+\cot^{-1}(-8) &=& \tan^{-1}(3) \\
\tan^{-1}(8)+\cot^{-1}(-5) &=& \tan^{-1}(3) \\
\tan^{-1}(13)+\cot^{-1}(-4) &=& \tan^{-1}(3)
\end{eqnarray}
So we have $6$ solutions.
But if we use the convention $\cot^{-1}y\in(0,\pi)$, we get
\begin{eqnarray}
\tan^{-1}(-7)+\cot^{-1}(-2) &=& \tan^{-1}(3) \\
\tan^{-1}(-2)+\cot^{-1}(-1) &=& \tan^{-1}(3) \\
\tan^{-1}(1)+\cot^{-1}(2) &=& \tan^{-1}(3) \\
\tan^{-1}(2)+\cot^{-1}(7) &=& \tan^{-1}(3) \\
\tan^{-1}(4)+\cot^{-1}(-13) &=& \tan^{-1}(3) +\pi\\
\tan^{-1}(5)+\cot^{-1}(-8) &=& \tan^{-1}(3) +\pi\\
\tan^{-1}(8)+\cot^{-1}(-5) &=& \tan^{-1}(3) +\pi\\
\tan^{-1}(13)+\cot^{-1}(-4) &=& \tan^{-1}(3)+\pi
\end{eqnarray}
and we have only $4$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
volume of the region in the first octant bounded by the coordinate planes and the planes The problem requires me to find the volume of the region in the first octant bounded by the coordinate planes and the planes $x+z=1$, $y+2z=2$, and here is my setup:
$$
\begin{aligned}
&\phantom{\implies}x+z=1\\
&\implies z=1-x\\
\end{aligned}
$$
$$
\begin{aligned}
&\phantom{\implies}y+2z=2\\
&\implies y=2-2z=2-2(1-x)=2x\\
\end{aligned}
$$
But the solution in the notes is as follows:
$$
\begin{aligned}
V&=\int_0^1\int_0^{1-x}\int_0^{2-2x}dydzdx\\
&=\int_0^1\int_0^{1-x}\left[y\right]_0^{2-2x}dzdx\\
&=\int_0^1\int_0^{1-x}(2-2x)-(0)dzdx\\
&=\int_0^1\int_0^{1-x}2-2xdzdx\\
&=\int_0^1(2-2x)\left[z\right]_0^{1-x}dx\\
&=\int_0^1(2-2x)[(1-x)-(0)]dx\\
&=\int_0^1(2-2x)(1-x)dx\\
&=\int_0^12x^2-4x+2dx\\
&=\left[\frac{2}{3}x^3-\frac{4}{2}x^2+2x\right]_0^1\\
&=\left[\frac{2}{3}(1)^3-\frac{4}{2}(1)^2+2(1)\right]-\left[\frac{2}{3}(0)^3-\frac{4}{2}(0)^2+2(0)\right]\\
&=\frac{2}{3}\\
\end{aligned}
$$
So I wonder where the upper limit of $y$, which is $2-2x$, comes from.
| The region is given by the inequalities $x,y,z\geq 0$, $x+z\leq 1,y+2z\leq 2$. It is pyramid with a rectangular base of sides $1$ and $2$, which lays in the $xy$-plane, and of height $1$ because the apex is at $(0,0,1)$. By elementary geometry the volume of this pyramid is $V=\frac{B\cdot h}{3}=\frac{(1\cdot 2)\cdot 1}{3}=\frac{2}{3}$.
Such volume can be evaluated also as an iterated integral in at least two different ways:
*
*The slices are perpendicular to the $z$-axis: $z\in [0,1]$, $(x,y)\in [0,1-z]\times [0,2-2z]$,
$$\begin{aligned}
V&=\int_{z=0}^1\int_{x=0}^{1-z}\int_{y=0}^{2-2z}dydxdz\\
&=\int_{0}^1(1-z)(2-2z)\,dz=2\int_{z=0}^1(1-z)^2\,dz\\
&=2\int_{t=0}^1t^2 dt=2\Big[\frac{t^2}{3}\Big]_0^1=\frac{2}{3}.
\end{aligned}
$$
*The slices are perpendicular to the $x$-axis: $x\in [0,1]$, $z\in [0,1-x]$, and $y\in [0,2-2z]$,
$$\begin{aligned}
V&=\int_{x=0}^1\int_{z=0}^{1-x}\int_{y=0}^{2-2z}dydzdx\\
&=\int_{x=0}^1\int_{z=0}^{1-x}(2-2z)dzdx\\
&=\int_{x=0}^1(2(1-x)-(1-x)^2)dx\\
&=\int_{t=0}^1(2t-t^2)dt=\Big[t^2-\frac{t^2}{3}\Big]_0^1=\frac{2}{3}.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $y = \frac{{a{x^2} - 3x + 5}}{{5{x^2} - 3x + a}}$ If $y = \frac{{a{x^2} - 3x + 5}}{{5{x^2} - 3x + a}}$, then number of possible integral values of 'a' for which y may be capable of all values $\forall x \in R$ is
(1) 4
(2) 5
(3) 6
(4) none of these
My approach is as follow $5{x^2} - 3x + a > 0 \Rightarrow 9 - 20a < 0 \Rightarrow a > \frac{9}{{20}}$
For this case we get the following range that $x\in \mathbb R$, but not able to approach
| Cross-multiply and simplify to get: $$(5y-a)x^2+(3-3y)x+(ay-5)=0\tag{1}$$
The statement “y may be capable of all values…” suggests that $(1)$ has at least one real solution of $x$ for all $y$. Thus, the discriminant of $(1)$ must always be non-negative $\forall y, a\in\mathbb R$: $$(3-3y)^2-4(ay-5)(5y-a)\geq 0$$ which simplifies to $$(9y^2+9-18y)-4(5ay^2-(25+a^2)y+5a)\geq 0$$ or $$(9-20a)y^2+(82+4a^2)y+(9-20a)\geq0$$ $\forall y\in \mathbb R$. But a quadratic polynomial in y always being greater than $0$ when the domain of $y$ is all reals, means that the discriminant of THAT polynomial must be negative. Thus, $$(82+4a^2)^2-4(9-20a)^2\lt 0$$ or $$(82+4a^2)^2-(18-40a)^2\lt 0$$$$\implies (4a^2+40a+64)(4a^2-40a+100)\lt 0$$$$\implies (a^2+10a+16)(a^2-10a+25)\lt 0$$$$\implies (a^2+10a+16)(a-5)^2\lt 0$$ Now, the second factor is always non-negative by virtue of being a perfect square. This means that $a$ must satisfy $(a^2+10a+16)\lt0\implies a\in (-8,-2)$ which gives us $a=-3,-4,-5,-6,-7$. Now to just cap it off, we check the boundary values whether $a=5,-2,-8$ satisfy the conditions and then we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
prove that $|x^y - y^x| > 2$
Prove that for any integers $x\neq y ,x,y>2, |x^y-y^x|>2$.
I know that the function $f(x) = \dfrac{\ln x}{x}$ is increasing for $x<e$ and decreasing for $x > e$. So $|x^y-y^x| = x^y - y^x \ge 1$ if $x < y$. I was thinking of proving something involving ratios (e.g. $f(x)/f(y)$ for $x < y$). I could possibly consider the case where x and y are consecutive and I could assume WLOG that $y>x$ so that it suffices to prove $x^y > y^x+2$ for $y>x>2.$ The inequality seems to be fairly weak; even for the smallest possible values of x and y, $y=4,x=3,$ we have $x^y = y^x+17.$ I think the sequence $\dfrac{x^{x+1}}{(x+1)^x} = x(1-\dfrac{1}{x+1})^x$ is an increasing function of x, and one could prove this using derivatives.
Source: A PuMAC 2008 problem.
| Letting $f(x)=x^{1/x}$ then for integers $2<x<y,$ $$\begin{align}y^x-x^y&=f(x)^{xy}-f(y)^{xy}\\&=(f(y)-f(x))\sum_{k=0}^{xy-1} f(x)^k f(y)^{xy-1-k}\\&>xyf(y)^{xy-1}(f(y)-f(x)).\end{align}$$
the last step because $f$ is decreasing on $[3,\infty)$ and thus $f(x)>f(y).$
Now, $f(x)-f(y)=f'(c)(x-y)$ for some $c\in(x,y).$ And $$f'(x)=\frac{1-\ln x}{x^2}x^{1/x}=\frac{1-\ln x}{x^2}f(x).$$
Since $\frac{\ln x-1}{x^2}$ is decreasing and positive for $x\geq e^{3/2}<5$ and the same is true for $x^{1/x},$ the largest value for $|f'(c)|$ is $|f'(x)|$ if $x\geq 5.$
Adding that $f(y)^{xy-1}=\frac{y^x}{f(y)}$ you get:
$$\left|y^x-x^y\right|> (y-x)xy\cdot y^{x}\frac{\log x-1}{x^2}\frac{f(x)}{f(y)}>(y-x)y^x(\log x-1)$$
When $5\leq x<y$ are integers, $y-x\geq 1,$ $y^x>2$ and $\log x>\frac32.$ (You actually have $y^x\geq 6^5,$ so this would give you a lower bound of $3888$ when $x\geq 5.$)
So you are left with the cases $x=3$ and $x=4,$ which can be proved seperately. You can easily show that you only need to check a finite number of values $y$ in those cases, because $3^y>y^4$ for all but finitely many $y,$ and $4^y>y^5$ for all but finitely many $y.$
Looking at the graph for $f''(x),$ it seems like $|f'(x)|$ is decreasing for $x\geq 1.$ Above, I used a shorthand to avoid computing $f''$ directly, but if true, the rest of the argument works for $x=3$ and $4,$ too.
Then the lower bound for $x\geq 3$ is $(\log(x)-1)x^{x+1}>6.$
We still need $x\geq 3$ to get $f(x)>f(y).$
| {
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"url": "https://math.stackexchange.com/questions/4550484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Problem regarding solving logarithm without a calculator I have a question regarding the answer provided from this question, which is solving $\log$ without a calculator.
This was one of the solutions and I am having a hard time understanding it.
$$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$ Let us apply it for computing the logarithm of $\frac 53 =1.666667$; this would correspond to $x=\frac 14$. So $$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \frac{1}{5120}+\cdots\right)=\frac{3923}{7680}\approx 0.5108$$ Now, $$\frac {166}{100}=\frac 53\times\frac{249}{250}$$ So $$\log(1.66)=\log \left(\frac53\right)+\log \left(\frac{249}{250}\right)=\log \left(\frac53\right)+\log \left(1-\frac{1}{250}\right)\approx\log \left(\frac53\right)-\frac{1}{250}$$ $$\log(1.66)\approx 0.5108-0.0040=0.5068$$ while the exact value would be ... the same.
If you need less accuracy, just truncate the first part, that is to say
$$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \cdots\right)=\frac{49}{96}$$ $$\log(1.66)=\frac{49}{96}-\frac{1}{250}=\frac{6077}{12000}\approx 0.5064$$
And my first question is how does the equation $\frac{166}{100} = \frac{5}{3} \times \frac{249}{250} $ came out.
Second question is how does $\log(1- \frac{1}{250})$ becomes $-\frac{1}{250}$.
Thank you for your help.
| There is formula the solver knew very well and was comfortable with.
$\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......$
From this they conclude two things
$\ln \frac{1+x}{1-x}=\ln(1+x) - \ln (1-x) = (x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......) - (-x -\frac {x^2}2 - \frac {x^3}3- \frac {x^4}4 -......)=$
$2(x +\frac {x^3}3 + \frac {x^5}5 + ...... )$
And
If $|x|$ is very very small then all the $\frac {x^k}k$ terms are negligibly small and $\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + .... \approx x$.
They figure that as $ 1.66$ is close to $\frac 53$ and we can use $\frac 53 = \frac {1+\frac 14}{1-\frac 14}$ and use the second formula to approximate $\ln \frac 53$. (Although why they didn't they use $\frac 53 = 1 + \frac 23$ and use the first formula, isn't clear to me.)
But as they didn't want $\ln \frac 53$ but $\ln 1.66$. However they figured that as $ 1.66$ is smaller but very close to $\frac 53$ there must be some very small $e$ so that $1.66 = \frac 53(1-e)$.
If so, to solve $\ln 1.66=\ln\frac 53 (1-e)$ we get
$\ln \frac 53 + \ln (1-e)=$
$\ln \frac {1+\frac 14}{1-\frac 14} + \ln (1-e)$.
By the second formula we can figure out what $\ln \frac {1+\frac 14}{1-\frac 14}$ is.
And by the third approximation we can assume $\ln (1-e) \approx e$.
So what is $e$?
Well $1.66 = \frac 53(1-e)\implies e =\frac 1{250}$.
But they didn't actually do it that way. They did:
$1.66 =(1.66\cdot \frac 35)\times \frac 53 =$
$\frac {166\cdot 3}{500}\times \frac 53 =$
$\frac {83\cdot 3}{250}\times \frac 53=$
$\frac {249}{250}\times \frac 53=$
$\frac 35(1-\frac 1{250})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Need Help Proving An Infinite Series Relating The Gamma Function And Riemann Zeta In a paper entitled 'On Question 330 Of Professor Sanjana' (by Ramanujan), Ramanujan goes on to define a function $f(p)$ as follows
$$f(p) = \frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{5+p} + ...$$
$$= \frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{\Gamma\left(\frac{p+2}{2}\right)^2}$$
And then goes on to say that
$$\ln{f(p)}= \ln\left(\frac{\pi}{2}\right) - p\ln(2) + \frac{p^2}{2} \Bigl(1-\frac{1}{2}\Bigr)\zeta(2) - \frac{p^3}{3} \Bigl(1-\frac{1}{2^2}\Bigr)\zeta(3) + ...$$
In another post, here they go on to explain that this can be proven using the fact that
$$\Gamma(p+1)=e^{-\gamma p}\prod_{n=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}$$
I tried to do so but got stuck figuiring out what to do next. Here is my attempt:
$$\frac{\Gamma(p+1)}{\Gamma\left(\frac{p+2}{2}\right)^2} = \frac{e^{-\gamma p}\prod_{n=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}}{\left(e^{-\frac{\gamma p}{2}}\prod_{n=1}^\infty\left(\frac{n}{n+\frac{p}{2}}\right)e^{p/2n}\right)^2}$$
$$= \prod_{n=1}^{\infty}\frac{\left(\frac{n}{n+p}\right)}{\frac{n^2}{\left(n+\frac{p}{2}\right)^2}} = \prod_{n=1}^{\infty}\frac{2n+p}{2n(n+p)}$$
therefore
$$\frac{2^{p+1}}{\pi} f(p) = \prod_{n=1}^{\infty}\frac{2n+p}{2n(n+p)}$$
$$\ln\left(\frac{2^{p+1}}{\pi}\right) + \ln(f(p)) = \sum_{n=1}^{\infty}\ln\left(\frac{2n+p}{2n(n+p)}\right)$$
$$ = \sum_{n=1}^{\infty}\ln\left(2n+p\right) - \ln\left(2n^2+np\right)$$
Here is where I'm not sure what to do next. I know I need to us the series expansion for $\ln(1 \pm x)$, so what do I do? I would appreciate hints and/or solutions. Thanks!
| If another approach is allowed, I would propose the following. First, we note that
$$S(p)=\frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{5+p} + ...=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k-1)!!}{(2k)!!}$$
$$=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k-1)!!(2k)!!}{\big((2k)!!\big)^2}=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k)!}{2^{2k}(k!)^2}$$
Next, we use the fact that $\int_0^{2\pi}e^{ikx}=2\pi$ only if $k=0$, otherwise - zero. Therefore, opening the brackets,
$$\int_0^{2\pi}\sin^{2k}xdx=\frac{1}{2^{2k}i^{2k}}\int_0^{2\pi}(e^{ix}-e^{-ix})^{2k}dx=\frac{2\pi}{2^{2k}}\binom{2k}{k}=\frac{2\pi}{2^{2k}}\frac{(2k)!}{(k!)^2}$$
and
$$S(p)=\frac{1}{2\pi}\sum_{k=0}^\infty\frac{1}{2k+1+p}\int_0^{2\pi}\sin^{2k}xdx=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{2k+1+p}\int_0^{\pi/2}\sin^{2k}xdx$$
Using $\displaystyle \frac{1}{2k+1+p}=\int_0^1y^{2k+p}dy$ and changing the order of summation and integration
$$S(p)=\frac{2}{\pi}\int_0^1dy\int_0^{\pi/2}dx\sum_{k=0}^\infty\sin^{2k}(x)\,y^{2k+p}=\frac{2}{\pi}\int_0^1y^pdy\int_0^{\pi/2}\frac{dx}{1-y^2\sin^2x}$$
Using the substitution $t=\tan x$
$$=\frac{2}{\pi}\int_0^1y^pdy\int_0^\infty\frac{dt}{(1-y^2)t^2+1}=\int_0^1\frac{y^p}{\sqrt{1-y^2}}dy=\frac{1}{2}\int_0^1\frac{t^\frac{p-1}{2}}{\sqrt{1-t}}dt$$
$$=\frac{1}{2}B\Big(\frac{p+1}{2};\frac{1}{2}\Big)=\frac{\sqrt\pi}{2}\frac{\Gamma\Big(\frac{p+1}{2}\Big)}{\Gamma\Big(\frac{p+2}{2}\Big)}=\frac{\sqrt\pi}{2}\frac{\Gamma\Big(\frac{p+1}{2}\Big)\Gamma\Big(\frac{p+2}{2}\Big)}{\Gamma\Big(\frac{p+2}{2}\Big)^2}$$
Now, using the duplication formula for gamma function $\Gamma(z)\Gamma\big(z+\frac{1}{2}\big)=2^{1-2z}\sqrt\pi\,\Gamma(2z)$ and the main property $\Gamma(z+1)=z\Gamma(z)$
$$S(p)=\frac{\sqrt\pi}{2}\frac{p}{2}2^{1-p}\sqrt\pi\frac{\Gamma(p)}{\Gamma\Big(\frac{p+2}{2}\Big)^2}=\frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{\Gamma\left(\frac{p+2}{2}\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $0$ is a regular value of $F$
How can i show that $0$ is a regular value of $F:\mathbb{R}^{3} \rightarrow \mathbb{R}^2$ given by
$$F(x,y,z)=(x^3+y^3+z^3-1,z-xy)$$
I compute the diferential of $F$ obtaining:
$$dF=\pmatrix{ 3x^2 & -y\cr 3y^2 & -x\cr 3z^2 &1}$$
and i know that if $det[(dF)(dF)^t] \neq 0$ then $F$ is a submersion but compute this determinant and see the points where it is $0$ its hard.
Any hint or alternatively way to show that i will be very grateful.
| To solve this problem, you have to play with the equations $x^3+y^3+z^3 = 1$ and $z=xy$.
Let $Z = F^{-1}(0)$.
Let $(x,y,z)\in Z$, and consider the matrix of $dF_{(x,y,z)}$ in the canonical basis, which is
$$
\mathrm{Mat}_{\mathrm{can}}(dF_{(x,y,z)}) = \begin{pmatrix}3x^2 & 3y^2 & 3z^2 \\ -y & -x & 1 \end{pmatrix} = \begin{pmatrix}3x^2 & 3y^2 & 3x^2y^2 \\ -y & -x & 1 \end{pmatrix},
$$
since on $Z$, $z=xy$.
(Note that this is the transpose of what you have stated: indeed, $dF_{(x,y,z)}$ goes from $\Bbb R^3$ to $\Bbb R^2$, not the converse.)
Let us show that this matrix has rank $2$.
First of all, $x$ and $y$ cannot be simultaneously vanishing: By contradiction, if $x=y=0$, then $x^3+y^3+z^3-1=0$ gives $z=1$, while $z-xy=0$ gives $z=0$.
Without loss of generality, let us assume that $x\neq 0$.
Consider the determinant of the matrix formed by the first and last columns:
$$
\begin{vmatrix}3x^2 & 3x^2y^2 \\ -y & 1\end{vmatrix} = 3x^2 + 3x^2y^3 = 3x^2(1+y^3).
$$
It vanishes if and only if $y^3=-1$, that is if and only if $y=-1$.
Assume by contradiction that $y=-1$.
Then $z-xy=0$ leads to $z=-x$. Hence, $x^3+y^3+z^3-1=0$ leads to $-2=0$, which is obviously false.
If follows that $y\neq -1$ and that $dF_{(x,y,z)}$ has rank $2$.
Hence, $(x,y,z)$ is a regular point of $F$.
This being true for all $(x,y,z)$ in $Z$, $0$ is a regular value for $F$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the series $\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$ I am trying to find the value of
$$\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$$
and using the formula for $\binom{n}{r}$, I have simplified it to
$$\sum\limits_{x=0}^{\infty} \frac{(x+4)(x+3)(x+2)(x+1)}{24\cdot 2^x}$$
but I have no idea how to proceed. It's question 9 from this link:
https://www.nanomath.org/sample.pdf
| The problem is: Evaluate $\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$
Let $$S(n) = \sum\limits_{x=0}^{\infty} \binom{n+x}{x} \frac{1}{2^x}$$
So we have $$S(4) = \sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$$
$$S(3) = \sum\limits_{x=0}^{\infty} \binom{3+x}{x} \frac{1}{2^x}$$
$$S(4) - S(3) = \sum\limits_{x=0}^{\infty} \left(\binom{4+x}{x} - \binom{3+x}{x}\right) \frac{1}{2^x}$$
$$ = \frac{1}{2}\sum\limits_{x=1}^{\infty} \binom{3+x}{x-1} \frac{1}{2^{x-1
}}$$
$$ = \frac{1}{2}\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x} =\frac{S(4)}{2}$$ by change of variable.
So $S(4) = 2S(3) = 4S(2) = 8S(1) = 16S(0)$ where
$$S(0) = \sum\limits_{x=0}^{\infty} \binom{0+x}{x} \frac{1}{2^x} = 2$$
So $$S(4) = 32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but since $(\sin x + \cos x) \in\left(0, \sqrt2\right]$ the maximum possible value of $(\sin x + \cos x)(6 - \sin x)$ can be $(6+ \dfrac{\sqrt2}{2}) * \sqrt2 = 1+6\sqrt2$ which is greater than $9$.
My second attempt was to find the derivative, but I couldn't find its roots.
| $\begin{equation} \begin{split}
(\sin x + \cos x)(6 - \sin x) & = 6 \cos x + 6 \sin x - \sin^2 x - \sin x \cos x \\ &= 6 \cos x + 6 \sin x - \dfrac{1}{2} (1 - \cos(2x) ) - \dfrac{1}{2} \sin(2x)\\
&= 6 \cos x + 6 \sin x + \dfrac{1}{2} (\cos(2x) - \sin(2x) ) - \dfrac{1}{2} \\
&\le 6 \sqrt{2} + \dfrac{1}{2} \sqrt{2} - \dfrac{1}{2} = 8.692388 \\ &\lt 9 \end{split} \end{equation} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $? Latest Edit
We are glad to see there are 4 alternative solutions which give the same closed form to the integral:
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$
where $\psi$ denotes the Digamma Function.
In the post, we found that
$$\int_0^{\infty} \frac{\ln x}{a^2+t^2} d x =\frac{\pi \ln a}{2 a }$$
Now I want to generalise the integral as
$$
I_n=\int_0^{+\infty} \frac{\ln t}{\left(a+t^2\right)^n} d t
$$
where $n\in N$.
Replacing a by $\sqrt{a}$ and differentiating both sides w.r.t. $a$ by $n$ times yields
$$
\begin{aligned}
&J(a)=\int_0^{\infty} \frac{\ln t}{a+t^2} d t=\frac{\pi}{4 \sqrt{a}} \ln a\\
&\frac{d^n}{d a^n}(J(a))=\frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) \\ & (-1)^n n ! \int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)} d t= \frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)\\& \boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{(-1)^n \pi}{4 n !} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)}
\end{aligned}
$$
By Wolfram-alpha, we have
$$
\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)=(-1)^n a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)\cdots (*)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
In particular, when $a=1$, we have
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} \left(\frac{1}{2}\right)_n\left(\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
For example,
$$$$
\begin{aligned}
\int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right)
=-\frac{23 \pi}{96}
\end{aligned}
$$
$$
My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?
| Assuming that $n$ is a positive integer, using hypergeometric functions$$I_n=\int \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=t \log (t) \, _2F_1\left(\frac{1}{2},n;\frac{3}{2};-t^2\right)-t \,
_3F_2\left(\frac{1}{2},\frac{1}{2},n;\frac{3}{2},\frac{3}{2};-t^2
\right)$$
$$J_n=\int_0^\infty \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=-\frac{\sqrt{\pi }}{4}\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}\,\left(H_{n-\frac{3}{2}}+2 \log (2)\right)$$
Asymptotically
$$J_n=-\frac{\sqrt{\pi }}{4 \sqrt{n}}(\log (n)+\gamma +2 \log (2))+O\left(\frac{1}{n^{3/2}}\right)$$
Trying to simplify what is given in other answers
$$\frac{d^n}{d a^n}\left(\frac{\log( a)}{\sqrt{a}}\right)=\sqrt \pi \frac{ \left(\log (a)-H_{n-\frac{1}{2}}-2 \log
(2)\right)}{a^{n+\frac{1}{2}}\,\,\Gamma \left(\frac{1}{2}-n\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Confusion: When can I preform operation of infinity in limit (without using the explanation of Epsilon Delta Definition) Question 1:
$$\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}$$
Knowing this:
$$\lim_{x\to\infty}\frac{1}{x}$$
The denominator of question 1 is $\infty$, therefore, $\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}} = 0$
My confusion: Why could we assume that $\sqrt{\infty+1}+\sqrt{\infty} = \infty$ since there aren't any algebraic operation on inf and negative inf? e.g. This operation is wrong $\lim_{x\to\infty}x-x = \infty - \infty$
Question 2:
$$\lim_{x\to 0^{-}}\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}$$
Knowing that $\lim_{x\to 0^{-}}\frac{1}{x}$ and $\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}$ is continuous.
Using this theorem
$$\frac{1+2^{\lim_{x\to 0^{-}}\frac{1}{x}}}{3+2^{\lim_{x\to 0^{-}}\frac{1}{x}}}$$
$$\frac{1+2^{-\infty}}{3+2^{-\infty}}$$
since $2^{-\infty} = \frac{1}{2^{\infty}}$, Therefore
$$\frac{1+2^{-\infty}}{3+2^{-\infty}} = \frac{1}{3}$$
My confusion: Why could we assume that $2^{-\infty} = \frac{1}{2^{\infty}}$?
| I usually substitute $-\frac{1}{M}$ for $x$, in case $x\rightarrow 0^{-}$ and $\frac{1}{M}$ in case $x\rightarrow 0^{+}$ and then let $M\rightarrow\infty$. For example,
$$
\lim_{x\rightarrow 0^{-}}\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}=\lim_{M\rightarrow\infty}\frac{1+2^{-M}}{3+2^{-M}}=\lim_{M\rightarrow\infty}\frac{2^{M}+1}{3.2^{M}+1}=\lim_{y\rightarrow\infty}\frac{y+1}{3y+1}.
$$
It reduced to a rational function limit problem.
| {
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"url": "https://math.stackexchange.com/questions/4568102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C = \sec A + \sec A - \tan A$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{{{\sec }^2}A - {{\tan }^2}A}}{{\sec A + \tan A}}} \right) \Rightarrow \sqrt 3 \sin C = \sec A + \left( {\frac{1}{{\sec A + \tan A}}} \right)$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{1}{{\cos A}} + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{{1 + \sin A + 1 - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
$\sqrt 3 \sin C = \frac{{2 + \sin A - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
Not able to proceed further
| Another approach:
Assume $0\le x\lt1$, then:
$$4x^2+1\ge4x\implies 4-4x+x^2\ge 3-3x^2\implies \frac {(2-x)^2}{1-x^2}\ge3\implies \frac {2-x}{\sqrt {1-x^2}}\ge \sqrt3;$$
notice the equality holds when $x=\frac{1}{2}.$ Now, we have:
$$\sqrt3 \sin C=\frac{2}{cos A}-\frac{\sin A}{\cos A}=\frac {2-\sin A}{\sqrt {1-\sin^2 A}}.$$
Hence, by putting $x=\sin A$, we come to the conclusion that $\sin C=1$ and $\sin A=\frac {1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the length PD in the figure below. For reference:
Find the length PD in the figure below.
ABCDEF is a regular hexagon
AB = 10cm
AQE is a sector( center at F)
PD is tangent of sector
My progresss:
$\triangle DEH:10.sin60^o = \frac{DF}{2} \therefore
DF = 10\sqrt3$
$\triangle DFQ :DF^2 = DQ^2+10^2 \implies DQ = \sqrt{300-100} \therefore DQ = 10\sqrt2 $
I can't find PQ
| We can, without loss of generality, set up a coordinate system so that the center of the hexagon is at (0, 0) and the top and bottom are parallel to the x-axis. Then A= (-10 sin(30), -10 cos(30)), B= (10 sin(30), -10 cos(30)), C= (10, 0), D= (10 sin(30), 10 cos(30)), E= (-10 sin(30), 10 cos(30)), and F= (-10, 0).
The circle with center at F and radius 10 is given by $(x+ 10)^2+ y^2= 100$ and any line through D is $y= m(x- 10 sin(30))+ 10 cos(30)= m(x- 5)+ 5\sqrt{3}$. For the circle, $2(x+ 10)+ 2yy'= 0$ so $y'= -\frac{x+ 10}{y}$. A line tangent to the circle at (x, y) must have slope $m= -\frac{x+ 10}{y}$.
We have $y= \frac{x+ 10}{y}(x- 5)+ 5\sqrt{3}= \frac{x^2+ 5x- 50}{y}+ 5\sqrt{3}$. The point at which the line DP is tangent to the circle must satisfy
$y= \frac{x+ 10}{y}(x- 5)+ 5\sqrt{3}= \frac{x^2+ 5x- 50}{y}+ 5\sqrt{3}$ and $(x+ 10)^2+ y^2= 100$. Solve those two equations for x and y, the coordinates of the line of tangency.
Once you have that, you can find the equation of the line through point D and this point of tangency, then find the coordinates of P, where the line intersects $y= -10 cos(30)= -5\sqrt{30}$ and so find the distance between D and P.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$z^4=\bar z$ complex number equation Find all $z\in\mathbb{C}$ that satisfy $z^4=\bar z$ and display them on the complex plane.
I started with the exponential forms of the two and got to $r^3e^{i4\varphi}=e^{-i\varphi}$. Is it wrong to assume, that $r=1$ and $4\varphi =-\varphi \Rightarrow \varphi=0$?
| An alternative route
Let $z=x+yi$. Then, $$(x+yi)^4=x-yi.$$ Expansion on the left side yields $$x^4-6x^2y^2+y^4+(4x^3y-4xy^3)i=x-yi$$. This gives us two equations:
$$x^4-x-6x^2y^2+y^4=0$$ and $$y+4x^3y-4xy^3=0$$
The bottom equation simplifies when you divide by $y$ (assuming $y\ne 0$):
$$1+4x^3-4xy^2=0.$$
This is quadratic in $y$: $y^2=\frac{-4x^3-1}{-4x}.$
Substituting this into the original top equation:
$$x^4-x-6x^2\frac{-4x^3-1}{-4x}+\frac{(-4x^3-1)^2}{(-4x)^2}=0.$$
Assuming $(-4x)^2\ne 0$, we can remove the denominators:
$$16x^6-6x^2(-4x^3-1)+(-4x^3-1)^2=0.$$
Simplifying:
$$32x^6+24x^5+8x^3+6x^2+1=0.$$
From this, you can find approximate values of $x$ and subsequently, using the formula above for $y$ in terms of $x$, approximate values of $y$.
| {
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"url": "https://math.stackexchange.com/questions/4571874",
"timestamp": "2023-03-29T00:00:00",
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Prove that: $\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}$
Let $x$, $y$ and $z$ be positive numbers. Prove that:
$$\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}.$$
This problem is similar to very many contest problems, but I think it's hard enough.
I tried to use Holder, C-S, AM-GM and more, but without any success.
For example, by Holder $$\sum_{cyc}\sqrt{\frac{4x^2+y^2}{3x^2+yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{4x^2+y^2}{3x^2+yz}}\right)^2\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}{\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}}\geq$$
$$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(4x^2+y^2)(kx+my+z)\right)^3}{\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}},$$ but I did not find non-negatives $k$ and $m$, for which the inequality $$4\left(\sum\limits_{cyc}(4x^2+y^2)(kx+my+z)\right)^3\geq45\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3$$ is true.
Thank you!
| (WARNING: this proof is very bad)
First, define two sequences $a_i$ and $b_i$ where $a_1=4x^2+y^2,a_2=4y^2+z^2,a_3=4z^2+x^2$ and $b_1=\frac{1}{3x^2+yz},b_2=\frac{1}{3y^2+xz},b_3=\frac{1}{3z^2+xy}$.
Since we have two three term sequences, we may now use Holder's inequality:
$$\sqrt{4x^2+y^2+4y^2+z^2+4z^2+x^2}=\sqrt{5x^2+5y^2+5z^2}$$
and
$$\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}$$
Thus, by Holder's inequality:
$$\left(\sqrt{5x^2+5y^2+5z^2}\right)\left(\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}\right) \ge \frac{3\sqrt{5}}{2}$$
Since equality holds when the two sequences are proportional, we may set $x=y=z$ and get:
$$\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}$$
which is the minimum value of this inequality, we may then say that since:
$$\left(\sqrt{5x^2+5y^2+5z^2}\right)\left(\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}\right)=\sqrt{\frac{4x^2+y^2}{3x^2+yz}} + \sqrt{\frac{4y^2+z^2}{3y^2+xz}} + \sqrt{\frac{4z^2+x^2}{3z^2+xy}}$$
This holds to be true.
| {
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Show that $\prod_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$ Show that
$$\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$$
Context:
Inspired by this question, I considered the product of the areas of every regular odd-gon inscribed in a circle of area $1$. This is $\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi} \sin{\frac{2\pi}{2k+1}}=0.18055...$ which I guess does not have a closed form.
On a whim, I divided this number by the Kepler-Bouwkamp constant, $\prod\limits_{k=1}^\infty \cos{\frac{\pi}{k+2}}=0.11494...$, and numerical calculation suggests that the result is $\pi/2$.
My attempt:
I used the double angle formula for sine to get $\prod\limits_{k=1}^\infty \frac{2k+1}{\pi}\sin{\left(\frac{\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{2k+2}\right)}$ but this seems just as intractable as the original form of the product.
| Let
$$K = \prod_{k=3}^\infty \frac1{\cos\left(\frac\pi k\right)}$$
As Bouwkamp points out under First method in this paper,
$$\cos(x) = \frac{\sin(2x)}{2x} \cdot \frac x{\sin(x)} \tag{1}$$
Then
$$\begin{align*}
\frac1K &= \prod_{k=3}^\infty \cos\left(\frac\pi k\right) \\[1ex]
&= \prod_{k=3}^\infty \frac{\sin\left(\frac{2\pi}k\right)}{\frac{2\pi}k} \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{1}\\[1ex]
&= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}} \cdot \prod_{k=2}^\infty \frac{\sin\left(\frac{2\pi}{2k}\right)}{\frac{2\pi}{2k}}\right] \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{2}\\[1ex]
&= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}}\right] \cdot \frac2\pi
\end{align*}$$
where in $(2)$ we split the product over even and odd indices.
| {
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Find all solutions for $\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$ I've been struggling with a quite difficult problem the past few days. This is my class's 'Exponentials and Logarithms' unit and my peers and I are stumped. Here it is:
Find all integer solutions to this system for x and y: $\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$.
So far, I've tried to get y isolated, giving me $y=(x^{\frac{1}{4}}-5)^5$ and $y=\pm \sqrt{(x^{\frac{1}{2}}-13)^5}$, but I don't know what to do next. Could someone please give me guidance? Thanks!
| Given
$$\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$$
then let $t = x^{1/4}$ and $u = y^{1/5}$ to obtain
$$ \begin{cases} t + u = 5 \\ t^2 + u^2 = 13 \end{cases}. $$
Square the first equation, and use the second, to obtain
$$25 = t^2 + u^2 + 2 u t = 13 + 2 u t $$
gives $u t = 6$. Now use this in the first equation to obtain $ t^2 - 5 t + 6 = (t-3)(t-2) = 0$. Choosing $t = 3$ leads to $u = 2$, and finally $ x = 3^4$ and $ y = 2^5$. A second set can be obtained by setting $t = 2$ and $u = 3$ which gives $x = 2^4$ and $y = 3^5$.
| {
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Find a plane M such that $l_1\in M (l_1$ is contained in $M$), and $M$ is parallel $l_2$. $l_{1} :
\left\lbrace
\begin{array}{lcl}
2 x + 3 y + 4 z = 7 && \mbox{ } \\
4 x + y - 2 z = -1 && \mbox{ } \\
\end{array}
\right.$
$l_{2} : \frac{x-2}{20}=\frac{y-2}{10}=\frac{z +1 }{5}$
I have to find a plane M such that $l_1\in M (l_1$ is contained in $M$), and $M$ is parallel $l_2$.
My attempt :
$\begin{pmatrix}
i&j&k\\
2&3&4\\
4&1&-2\\
\end{pmatrix}=-10\hat{i}+20\hat{j}-10\hat{k} \implies a=-10,b=20,c=-10$
$
\left\lbrace
\begin{array}{lcl}
2 x + 3 y + 4 z = 7 && \mbox{ } \\
4 x + y - 2 z = -1 && \mbox{ } \\
\end{array}
\right. \implies x=z-1,y=-2z+3$.
Denote $z=0 \implies x=-1,y=3$
Plan's equation is $ax+by+cz+d=0$ , subtitue $a,b,c,x,y,z \implies d=-70$
$\vec{l_2}=(2,2,-1)+s(20,10,5)$
I dont how to solve $M$ is parallel $l_2$ , any help is welcome .
| You get the direction vector of $l_1: v_1 = (-1,2,-1)$
and the direction vector of $l_2: v_2=(4,2,1)$
You can just cross them to get the normal vector of M. Since both $l_1,l_2$ are $\perp$ to normal vector of M. So
$N_m=v_1\times v_2 = (4,-3,-10)$
Then M is $4x-3y-10z=d$, find d and you are done. (Use any point on $l_1$)
| {
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Arrange $m$ H's and $n$ T's to have exactly $k$ H-runs In an arrangement of $m$ H's and $n$ T's, an uninterrupted sequence of one kind of symbol is called a run. (For example: HHHTHHTTTH is a sequence of $6$ H's and $4$ T's which opens with an H-run of length $3$, followed by a T-run of length $1$,an H-run of length $2$, a T-run of length $3$ and an H-run of length $1$). Find the number of arrangements of $m$ H's and $n$ T's in which there are exactly $k$ H-runs.
My solution goes like this :
We consider $k$ boxes where we will put $m$ H's. Now, between these boxes we have $k-1$ different boxes where we place only T's. We must place at least one T in all those $k-1$ boxes. If the boxes where H is contained are named as $x_1,x_2,\dots,x_k$, then $x_1+\dots+x_k=m$ and each $x_i\geq 1$. If the boxes where T is contained are named $y_1,y_2,\dots,y_{k-1}$ then $y_1+\dots+y_{k-1}=k-1\space \text{or}\space k\space \text{or}\space k+1,...\text{or}\space n$ and each $y_i\geq 1$. Now, the number of solutions of $x_i$ is $\binom{m-1}{k-1}$. Now, $y_i$ we have different cases and they are all mutually exclusive, so the total number of solutions are $\binom{k-2}{k-2}+\binom{k-1}{k-2}+\dots +\binom{n-1}{k-2}$. Now, the total number of ways to have $k$ H-runs is $\binom{m-1}{k-1}[\binom{k-2}{k-2}+\binom{k-1}{k-2}+\dots+\binom{n-1}{k-2}].$
Is the above solution correct? Is it valid? If not, where is it going wrong? There may be some posts relating to this topic on this site but I can't seem to find it either...
| As you observed, the number of $T$-runs is either $k - 1$, $k$, or $k + 1$. Let's consider cases.
Case 1: $k$ $H$-runs and $k - 1$ $T$-runs.
For this to occur, there must be a run of at least one $T$ in each of the $k - 1$ spaces between runs of at least one $H$. If $x_i$ is the number of $H$s in the $i$th run, then
$$x_1 + x_2 + \cdots + x_k = m \tag{1}$$
is an equation in the positive integers. As you found, equation $1$ has
$$\binom{m - 1}{k - 1}$$
solutions in the positive integers. If $y_j$ is the number of $T$s in the $j$th run, then
$$y_1 + y_2 + \cdots + y_{k - 1} = n \tag{2}$$
is an equation in the positive integers. The number of solutions of equation $2$ in the positive integers is
$$\binom{n - 1}{k - 2}$$
Hence, there are
$$\binom{m - 1}{k - 1}\binom{n - 1}{k - 2}$$
sequences of $m$ $H$s and $n$ $T$s with $k$ $H$-runs and $k - 1$ $T$-runs.
Case 2: $k$ $H$-runs and $k$ $T$-runs
Runs of at least one $H$ must alternate with runs of at least one $T$. The number of ways of distributing the $H$s is the number of solutions of equation $1$ in the positive integers, which is
$$\binom{m - 1}{k - 1}$$
The number of ways of distributing the $T$s is the number of solutions of the equation
$$y_1 + y_2 + \cdots + y_k = n \tag{3}$$
in the positive integers. Equation $3$ has
$$\binom{n - 1}{k - 1}$$
solutions in the positive integers. Since the first run either begins with an $H$ or with a $T$, the number of sequences of $m$ $H$s and $n$ $T$s which have $k$ $H$-runs and $k$ $T$-runs is
$$\binom{2}{1}\binom{m - 1}{k - 1}\binom{n - 1}{k - 1}$$
Case 3: $k$ $H$-runs and $k + 1$ $T$-runs
For this to occur, there must be a run of at least one $H$ in each of the $k$ spaces between runs of at least one $T$.
The number of ways of distributing the $H$s is the number of solutions of equation $1$ in the positive integers, which is
$$\binom{m - 1}{k - 1}$$
The number of ways of distributing the $T$s is the number of solutions of the equation
$$y_1 + y_2 + \cdots + y_{k + 1} = n \tag{4}$$
in the positive integers. Equation $4$ has
$$\binom{n - 1}{k}$$
solutions in the positive integers. Hence, the number of sequences of $m$ $H$s and $n$ $T$s with $k$ runs of $H$s and $k + 1$ runs of $T$s is
$$\binom{m - 1}{k - 1}\binom{n - 1}{k}$$
Total: Since these three cases are mutually exclusive and exhaustive, the number of sequences of $m$ $H$s and $n$ $T$s which have exactly $k$ $H$-runs is
$$\binom{m - 1}{k - 1}\binom{n - 1}{k - 2} + \binom{2}{1}\binom{m - 1}{k - 1}\binom{n - 1}{k - 1} + \binom{m - 1}{k - 1}\binom{n - 1}{k}$$
This expression can be simplified using Pascal's identity.
\begin{align*}
\binom{m - 1}{k - 1}\binom{n - 1}{k - 2} & + \binom{2}{1}\binom{m - 1}{k - 1}\binom{n - 1}{k - 1} + \binom{m - 1}{k - 1}\binom{n - 1}{k}\\
& \qquad = \binom{m - 1}{k - 1}\left[\binom{n - 1}{k - 2} + 2\binom{n - 1}{k - 1} + \binom{n - 1}{k}\right]\\
& \qquad = \binom{m - 1}{k - 1}\left[\binom{n - 1}{k - 2} + \binom{n - 1}{k - 1} + \binom{n - 1}{k - 1} + \binom{n - 1}{k}\right]\\
& \qquad = \binom{m - 1}{k - 1}\left[\binom{n}{k - 1} + \binom{n}{k}\right]\\
& \qquad = \binom{m - 1}{k - 1}\binom{n + 1}{k}
\end{align*}
Addendum: In the comments, you suggested the formula
$$\binom{m - 1}{n - 1}\sum_{a = k - 1}^{n} \binom{a - 1}{k - 2}\binom{n - a + 1}{1}$$
where the term in front of the summation is the number of ways of distributing the $m$ $H$s into $k$ runs, the term $\binom{a - 1}{k - 2}$ is the number of ways of distributing $a$ $T$s to the $k - 1$ runs of tails between successive runs of heads, and $\binom{n - a + 1}{1}$ is the number of ways of distributing the remaining $n - a$ $H$s to the spaces before the first $H$ or after the last $H$ without restriction.
Each of these formulas is correct. The first term was discussed above. The number of ways of distributing $a$ $T$s to $k - 1$ runs is the number of solutions in the positive integers of the equation
$$y_1 + y_2 + \cdots + y_{k - 1} = a$$
which is
$$\binom{a - 1}{k - 2}$$
The number of ways of distributing the remaining $n - a$ heads to the spaces to the left of the first head or to the right of the last head without restriction is the number of solutions of the equation
$$y_0 + y_k = n - a$$
in the nonnegative integers, which is
$$\binom{n - a + 2 - 1}{2 - 1} = \binom{n - a + 1}{1}$$
Observe that
$$\sum_{a = k - 1}^{n} \binom{a - 1}{k - 2}\binom{n - a + 1}{1}$$
is the number of ways of selecting $k$ elements from the $(n + 1)$-element set $\{0, 1, 2, \ldots, a - 1, a, \ldots, n\}$, where we select $k - 2$ elements which are less than $a - 1$, select $a - 1$, and then select one of the $n - (a - 1) = n - a + 1$ elements larger than $a$, where $a$ can vary from $k - 1$ to $n$. Hence,
$$\binom{m - 1}{n - 1}\sum_{a = k - 1}^{n} \binom{a - 1}{k - 2}\binom{n - a + 1}{1} = \binom{m - 1}{n - 1}\binom{n}{k + 1}$$
as required. Therefore, the formula you wrote in the comments is correct.
That said, the answer I wrote above is easier to work with since there are fewer terms and is easier to simplify since the simplifications rely on Pascal's identity rather than a pivot point.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the minimum value of all products of two values is less than $-\frac{1}{n}$.
Let $x_1, x_2, \dots, x_n$ be $n$ real numbers, where $n \geq 3$ is an odd natural number. We know that:
$$x_1^2 + x_2^2 + \dots + x_n^2 = 1$$
and:
$$x_1^3 + x_2^3 + \dots + x_n^3 = 0$$
Prove that the minimum value of a product of type $x_ix_j$, with $0 < i < j < n +1$ is less than $\frac{-1}{n}$.
My thoughts: Let's consider two points $P$ and $Q$ in the $n$-dimensional topological space, one defined by $P(x_1, x_2, \dots, x_n)$ and the other by $Q(x_1^2, x_2^2, \dots, x_n^2)$. Obviously, from the first hypothesis, $P$ is located on the boundary of the $n - 1$-hypersphere. While, from the second condition, the scalar product $\vec{OP} \cdot \vec{OQ} = 0$, therefore $OP \perp OQ$ (perpendicularity defined in the $n$-space). From here, I want to find a $n$-geometrical condition for the collection of products, make their sum, and impose som condition on the smallest term.
| Denote $a = \underset{1\le i\le n}{\min}{x_i}$ and $b = \underset{1\le i\le n}{\max}{x_i}$, then
$$\begin{align}
\sum_{i=1}^n (x_i^3-a^3)(x_i^3-b^3)\le0 &\Longleftrightarrow \sum_{i=1}^n x_i^6-\underbrace{\left( \sum_{i=1}^n x_i^3 \right)}_{=0}(a^3+b^3)+na^3b^3\le 0 \\
&\Longleftrightarrow a^3b^3\le -\frac{1}{n}\sum_{i=1}^n x_i^6 \tag{1} \\
\end{align}$$
Applying the Holder's inequality for $(p,q)= \left(3,\frac{3}{2} \right)$
$$\left(\sum_{i=1}^n \left(x_i^2\right)^{3} \right)^{1/3}\left(\sum_{i=1}^n 1^{3/2} \right)^{2/3} \ge \sum_{i=1}^n x_i^2 =1$$
$$\Longleftrightarrow \sum_{i=1}^n x_i^6 \ge \frac{1}{n^2} \tag{2}$$
From $(1),(2)$, we deduce that:
$$a^3b^3\le -\frac{1}{n^3} \Longleftrightarrow \color{red}{ab\le -\frac{1}{n}}$$
The equality occurs if and only if $n=2k$ and for example $x_1=...=x_k = -\frac{1}{\sqrt{n}}$ and $x_{k+1}=...=x_{n} = \frac{1}{\sqrt{n}}$.
Hence, the minimum value of a product of type $x_ix_j$ is less than $-\frac{1}{n}$ as $n$ is odd number.
Q.E.D
| {
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Any way to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}$? I was solving a radical equation $x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$. I deduced it to $\sqrt{x } + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}.$
Answer is $\frac1{24}$.
The first equation has two solutions however the latter one has only one solution. I lossed up one solution but still, I'm interested in solving it.
I wish if someone could help me in solving any of the above equations.
Here's my work:
$$\begin{align}x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} &= 2\\x + \frac{1}{2}\left(2 \sqrt{x}\sqrt{x+1} + 2\sqrt{x+1}\sqrt{x+2} + 2\sqrt{x}\sqrt{x+2}\right) &=2\tag{1}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (x + x + 1 + x + 2)\Big]& = 2\tag{2}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 2\\2x + \Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 4\\(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 & = x + 7\\\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} & = \sqrt{x+7}\end{align}$$
Moving from $(1)$ to $(2)$, I used $2(ab + bc + ca) = (a+b+c)^2 - (a^2+b^2+c^2)$.
In general, is it possible to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+k}$ by hand?
| Given the equation
$$ \sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7} $$
consider it in the form
$$ \sqrt{x} + \sqrt{x+1} = \sqrt{x+7} - \sqrt{x+2} $$
then by squaring both sides the result is
$$ \sqrt{x (x+1)} = 4 - \sqrt{(x+2)(x+7)}. $$
Now, with this equation written as $4 = \sqrt{x (x+1)} + \sqrt{(x+2)(x+7)}$ then, by squaring both sides,
$$ \sqrt{x (x+1) (x+2) (x+7)} = x^2 + 5 x -1. $$
Again squaring both sides leads to
$$ x (x+1) (x+2)(x+7) = x^4 + 10 x^3 + 23 x^2 - 10 x + 1 = x(x+1)(x+2)(x+7) - 24 x + 1$$
which gives the equation $$24 x = 1 \hspace{5mm} \text{and the result} \, x = \frac{1}{24}. $$
| {
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Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1.$
Let $n$ be a positive integer and let $z$ be a root of $x^n-x-1.$ Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1$.
Solution. Write $z = re^{i\theta}, r > 0$ as $0$ is obviously not a root. Then $$1+2r\cos \theta = |z+1|^2-r^2 = r^{2n} - r^2.$$ We need to show that $$2(r\cos\theta - 1/r \cos \theta) = 2 \operatorname{Re}(re^{i\theta} - 1/r e^{-i\theta}) > 1/r^2 - 1.$$ We know that $2r\cos \theta = r^{2n}-r^2 - 1\Rightarrow \cos \theta = \dfrac{r^{2n}-r^2-1}{2r}$ from above. So we need to show that $$r^{2n}-r^2 - 1 - \dfrac{r^{2n}-r^2-1}{r^2} > 1/r^2 - 1\Leftrightarrow r^{2n+2} - r^4 - r^2 - r^{2n} + r^2 + 1 - 1 + r^2>0 \\ \Longleftrightarrow r^{2n+2} - r^{2n} - r^4 + r^2 = r^2(r^{2n} - r^{2n-2} -r^2 + 1).$$
I don't think I've made any algebraic mistakes, so I was wondering how I can show the last quantity is positive?
| We have to prove that $$r^{2n}+1>r^{2n-2}-r^2\tag{1}$$ which is easy to see since $r^{2n}>r^{2n-2}$ (when $r>1$, that is) and so the inequality is proven for $r>1$. When $r<1$ we have $r^{2n-2}<r^2$ so the right side of $(1)$ is negative and the left side is positive. Therefore your inequality is true.
| {
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how to solve $\int\sqrt{1-16cos^2(x)}\ dx$ How can we solve this integral?
$\int\sqrt{1-16cos^2(x)}\ dx$
I tried to make some substitutions, but I failed to solve it.
For example, I took the substitution
$u=4cos(x)$
$dx=-\frac{du}{\sqrt{16-u^2}}$
hence, the integration will be
$\int-\frac{\sqrt{1-u^2}}{\sqrt{16-u^2}}du$
then I tried the substitution
$z=\sqrt{1-u^2}$
$du=-\frac{z}{\sqrt{1-z^2}}dz$
hence, the integration will be
$\int\frac{z^2 dz}{\sqrt{1-z^2}\sqrt{15+z^2}}$
But I do not know how to complete from this step..
| You can easily represent the antiderivative using the indefinite elliptic integrals of the second kind (Jacobi-Form) and complex numbers:
$$
\begin{align*}
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{1 - 16 \cdot \left(1 - \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{1 - 16 + 16 \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{-15 + 16 \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{-\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{(-1) \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{-1} \cdot \sqrt{\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \mathrm{i} \cdot \sqrt{\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{1 \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{\frac{15}{15} \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{15} \cdot \sqrt{\frac{1}{15} \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \sqrt{15} \cdot \int_{}^{} \sqrt{1 - \frac{16}{15} \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \sqrt{15} \cdot \left( \int_{0}^{x} \sqrt{1 - \frac{16}{15} \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x + c \right)\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \sqrt{15} \cdot \operatorname{E}\left( x, ~\frac{16}{15} \right) + c\\
\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= c + \sqrt{15} \cdot \operatorname{E}\left( x, ~\frac{16}{15} \right) \cdot \mathrm{i}\\
\Re\left( \int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x \right) &= \Re(c)\\
\Im\left( \int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x \right) &= \Im(c) + \sqrt{15} \cdot \operatorname{E}\left( x, ~\frac{16}{15} \right)\\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Right triangle with its perimeter and median and altitude A right triangle $ABC$ is given with $\measuredangle ACB=90^\circ$. If the perimeter of the triangle is $72$ and the difference between the lengths of the median and the altitude to the hypotenuse is $7$, find the area.
Let $CD=x$ $(x>0)$, then $CM=x+7$. By the Pythagorean theorem, it follows that $$MD=\sqrt{(x+7)^2-x^2}=\sqrt{14x+49}.$$
For the perimeter as $AM=MB=CM=x+7$ we have $$P_{ABC}=2x+14+a+b=72\Rightarrow a+b=58-2x,$$ where $BC=a,AC=b$.
We have to find $S_{ABC}=\dfrac{2(x+7)x}{2}=x(x+7)=x^2+7x$. I can't see how to do that.
On the other hand, if the sides of the triangle are $a,b$ and $c$, then $$\begin{cases}a+b+c=72\\\dfrac{c}{2}-\dfrac{ab}{c}=7\\a^2+b^2=c^2\end{cases}$$ and we only need to solve it for $\dfrac{ab}{2}$. I cannot do that as well. Thanks!
| Let $a = BC$, $b = AC$, $c = AB$ and $S$ the area of the triangle $ABC$ then $S = \dfrac{a b}{2}$.
*
* We have $a + b + c = 72$ then $a + b = 72 - c$ so :
$$(72 - c)^2 = (a + b)^2 = a^2 + b^2 + 2 a b = c^2 + 4 S$$
We deduce that :
$$4 S = (72 - c)^2 - c^2 = 5184 - 144 c \quad (1)$$
* We have :
$$S = \dfrac{1}{2} CD \times c = \dfrac{1}{2} (CM - 7) \times c$$
then :
$$4 S = (2 CM - 14) \times c = (c - 14) \times c = c^2 - 14 c \quad (2)$$
From $(1)$ and $(2)$ :
$$c^2 - 14 c = 5184 - 144 c$$
then :
$$c^2 + 130 c - 5184 = 0$$
The discriminant is $\Delta = 37636 = 194^2$ then :
$$c = \dfrac{-130 + 194}{2} = 32 \text{ or } c = \dfrac{-130 - 194}{2} = -162$$
We deduce that $c = 32$ because $c \geq 0$ then :
$$S = \dfrac{1}{4}(5184 - 144 c) = 144$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that: $\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$
Prove that:
$$\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$$
This inequality is the result of an integral representation/inequality.
I lost access to the article that mentioned this inequality. Now I want to prove it myself. Using series is the standard way. However, this involves heavy calculations, so it's almost useless. We know that,
$$\ln 28=2\ln 2+\ln 7$$
From here, it is necessary to evaluate the $\ln2$ and $\ln 7$ numbers independently. More work is required.
*
*If you've seen this inequality before, please share the solution with us. Presumably this will be the Integral representation.
*If you have made any solution that requires as little computation as possible, please share it with us.
*The question is open to all solutions involving elementary techniques or advanced techniques.
Less computational solutions are preferred.
| For the left inequality:
We have
$$\ln 28 = 5 \ln 2 - \ln(8/7)
\ge 5\ln 2 - 1/7$$
where we have used $\ln(1 + u) \le u$ for all $u \ge 0$.
From
$$0 \le \int_0^1 \frac{x^2(1 - x)^3}{1 + x}\,\mathrm{d} x = 8\ln 2 - \frac{83}{15},$$
we have
$\ln 2 \ge \frac{83}{120}$.
(Note: Alternatively,
we may use $\ln \frac{1+x}{1-x}
\ge 2x + \frac23 x^3$ ($3$-th order Taylor approximation) to get
$\ln 2 \ge \frac{56}{81}$ by letting $x = 1/3$.)
Thus, we have
$$\ln 28 \ge 5 \cdot \frac{83}{120} - 1/7 = \frac{557}{168}.$$
Let $A = \sqrt[3]{36}$ and $B = 10/3$. Then $A < B$. We have
$$A = B - \frac{B^3 - A^3}{B^2 + BA + A^2} \le B - \frac{B^3 - A^3}{3B^2} = \frac{743}{225}.$$
(Note: $\sqrt[3]{36} = 3\sqrt[3]{1 + 1/3} \le 3(1 + 1/9) = 10/3$ by Bernoulli inequality.)
Since $\frac{743}{225} < \frac{557}{168}$, we have $36^{1/3} < \ln 28$.
$\phantom{2}$
For the right inequality:
Using $37 \cdot 27 = 999$, we have
$$\sqrt[3]{37} = \sqrt[3]{999/27} = \frac{10}{3}\sqrt[3]{999/1000}
\ge \frac{10}{3} \cdot \frac{3 \cdot 999/1000}{2\cdot 999/1000 + 1} = \frac{4995}{1499}$$
where we have used
$$u^{1/3} = \frac{u}{\sqrt[3]{u \cdot u \cdot 1}} \ge \frac{u}{(u+u+1)/3} = \frac{3u}{2u + 1}.$$
It suffices to prove that
$$\ln 28 < \frac{4995}{1499}.$$
I don't have a nice approach to deal with it.
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$X$ has pdf $f(x)=c(x+1)$ Find pdf of $Y=X^2$.
Let $X$ be continuous random variable with pdf $f(x)=c(x+1)$ for
$-2<x<1$. Determine the pdf of $Y=X^2$.
I actually asked this question earlier but the second part of the question didn't make sense. I just wanted to verify if my solution for this question is correct.
$\int_{-2}^{1} c(x+1)dx=\frac{3}{2}c$
$c=\frac{2}{3}$
For $0<y<1$
$$P(X\leq \sqrt{y})=\int_{-\sqrt{y}}^{\sqrt{y}} 2/3 (x+1)dx=\frac{4\sqrt{y}}{3}$$
Taking the derivative, I get
$f_{y}(y)=\frac{2}{3\sqrt{y}}, 0<y<1$
Similarly, for $1<y<4$ the cdf is
$$\int_{-2}^{\sqrt{y}} 2/3(x+1)dx = \frac{y}{3}+2\sqrt{y}$$
Again, taking the derivative I get
$f_y(y)=\frac{1}{3}+\frac{1}{\sqrt{y}}$
So finally:
$$f_Y(y) =
\begin{cases}
\frac{2}{3\sqrt{y}}, & \text{0<y<1} \\[2ex]
\frac{1}{3}+\frac{1}{\sqrt{y}}, & \text{1<y<4}
\end{cases}$$
| As @Alex said, $c(x+1)$ becomes both negative, positive value on $-2 < x < 1$, which makes $f$ not a pdf.
To make $f$ a pdf, you should separate $c$ within the range with proper $c$'s.
i.e. $c_1(x+1)$ in ($-2 < x < -1$), $c_2(x+1)$ in ($-1 < x < 1$)
Let's ignore the negative value problem and consider the integration parts.
Let's keep in mind that the support of $x$ is $(-2, 1)$
\begin{align}
F_Y(y) &= P(Y \leq y) \\
&= P(X^2\leq y)\\
&= P(- \sqrt y \leq X \leq \sqrt y)
\end{align}
For $0 < y < 1$,
\begin{align}
-1 < - \sqrt y \leq X \leq \sqrt y < 1
\end{align}
So, the interval is contained in the support.
Thus,
\begin{align}
F_Y(y) = \int_{-\sqrt y}^{\sqrt y} f_X(x) dx
\end{align}
For $1 < y < 4$,
\begin{align}
1 < \sqrt y < 2 \text{ and} -2< -\sqrt y < -1 \\
\end{align}
\begin{align}
-2 < -\sqrt y \leq X < 1 < \sqrt y < 2
\end{align}
So, the interval $[- \sqrt y, \sqrt y]$ is not contained in the support.
Thus, we should get the intersection of $[- \sqrt y, \sqrt y]$ and support(=$(-2, 1)$) as integration interval which is
\begin{align}
[- \sqrt y, 1)
\end{align}
Thus,
\begin{align}
F_Y(y) = \int_{-\sqrt y}^{1} f_X(x) dx
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
| Let us make the substitution $y := \sqrt{1-x^2}$; then we get the system of polynomial equations
\begin{align*}
x^2 + y^2 & = 1, \\
x(1+y) & = y.
\end{align*}
From here, if we solve the second equation for $y$, $y = \frac{x}{1-x}$, and substitute into the first equation, it is straightforward to get the polynomial equation
$$x^4 - 2x^3 + x^2 + 2x - 1 = 0.$$
However, let us go back to the original system of two equations in $(x, y)$. Here, the first equation is the unit circle, and the second equation is equivalent to $(x-1)(y+1) = -1$, so it is a hyperbola with center $(1, -1)$. We see that both curves are symmetric about the line $y = -x$; therefore, if $(x, y)$ is a solution to the system of equations, then so is $(-y, -x)$. This implies that if $\alpha$ is a root of the quartic polynomial, then so is $-\frac{\alpha}{1 - \alpha} = \frac{\alpha}{\alpha - 1}$.
From here, we can hypothesize that if $K$ is the splitting field of the quartic over $\mathbb{Q}$, then there might be an element $\sigma \in \operatorname{Gal}(K / \mathbb{Q})$ which permutes the roots according to $\alpha \mapsto \frac{\alpha}{\alpha - 1}$. If this is the case, then it is easy to see that $\sigma^2 = \operatorname{id}$ since it fixes each root of the quartic; however $\sigma \ne \operatorname{id}$ since the only fixed point of $\alpha \mapsto \frac{\alpha}{\alpha - 1}$ is 2, and 2 is not a root of the quartic. Then setting $L$ to be the fixed field of $\sigma$, we will have $\operatorname{Tr}^K_L(\alpha) = \alpha + \frac{\alpha}{\alpha - 1} = \frac{\alpha^2}{\alpha - 1}$. This suggests searching for a polynomial which will have $\frac{\alpha^2}{\alpha - 1}$ as a root, and hopefully finding the roots of this polynomial will be simpler because $L$ has smaller degree over $\mathbb{Q}$.
However, from this point, we fairly quickly see that
$$\left( \frac{\alpha^2}{\alpha-1} \right)^2 - 2 \left( \frac{\alpha^2}{\alpha-1} \right) - 1 = \frac{\alpha^4 - 2\alpha^2(\alpha-1) - (\alpha-1)^2}{(\alpha-1)^2} = \\
\frac{\alpha^4 - 2\alpha^3 + \alpha^2 + 2\alpha - 1}{(\alpha-1)^2} = 0.$$
But then, that means that $\frac{\alpha^2}{\alpha-1} = 1 \pm \sqrt{2}$. Another application of the quadratic formula to the equation $\alpha^2 - (1 \pm \sqrt{2}) \alpha + (1\pm\sqrt{2}) = 0$ gives the roots of the quartic. (Then from here, two of the roots of the quartic will be complex, and one of the real roots will not satisfy the original equation because it does not satisfy $\sqrt{1-x^2} = \frac{x}{1-x} \ge 0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Does the sequence of positive roots of $x^n+x^{n-1}+\cdots+x-1=0$ converge? For a positive integer $n$, let $a_n$ denote the unique positive real root of the equation
$$x^n+x^{n-1}+\cdots+x-1=0.$$
Then what can we say about the sequence $\{a_n\}$?
We can observe that all $a_n<1$ so if $\lim a_n$ exists, let's say at $L$, then $L \le 1$. Also
\begin{align}
&&x^n+x^{n-1}+\cdots+x-1 &= 0 \\
\implies&&
x^{n+1}-1 &= 2(x-1) \\
\implies&& x^{n+1} &= 2x-1 \\
\implies&& a_n^{n+1} &= 2a_n-1.
\end{align}
Taking limit both side we get $0=2L-1$, i.e.
$L=\frac12$ but how to show that the limit always exists?
| Clearly this unique positive $x$ lies in the interval $\big(\frac12,1\big)$. Also,
$$
x^n+\cdots+x-1=0 \quad\Longrightarrow\quad (x-1)(x^n+\cdots+x)-x+1=0
\quad\Longrightarrow\quad x^{n+1}-2x+1=0.
$$
Since $x$ depends on $n$, we denote it as $x_n$.
If we set $x_n=\frac12+u_n$, then $u_n\in \big(0,\frac12\big)$, and $u_n$ satisfies
$$
f(u_n)=\Big(\frac12+u_n\Big)^{n+1}-2u_n=0.
$$
Clearly, $f(0)>0$, while $f\big(\frac1{n+1}\big)<0,$ for $n$ sufficiently large, since
$$
f\Big(\frac1{n+1}\Big)=\Big(\frac12+\frac1{n+1}\Big)^{n+1}-\frac{2}{n+1}
\\=2^{-n-1}\Bigg(\Big(1+\frac{2}{n+1}\Big)^{n+1}-\frac{2^{n+2}}{n+1}\Bigg)
<2^{-n-1}\Big(e^2-\frac{2^{n+2}}{n+1}\Big)<0,
$$
and $\frac{2^{n+2}}{n+1}\to\infty$. Thus, for $n$ large enough, $f(u)$ vanishes for some $u\in \big(0,\frac1{n+1}\big)$.
Thus $0<u_n<\frac{1}{n+1}$,
and hence, for $n$ sufficiently large
$$
\frac{1}{2}<x_n<\frac{1}{2}+\frac{1}{n+1}
$$
which implies that $\lim_{n\to\infty}x_n=\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$
Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$
This is my textbook contest exercise.
$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$
$$x=\big(\sqrt [4]{2-x}+1\big)^4$$
I know that
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
So, we will have some cube powers which introduce still $\sqrt [4]{\cdot}$ roots.
This makes more complicated the equation. How can I find the minimal polynomial using a more simple way?
| Let $x=u^4,\;2-x=v^4$ with the restriction $u,v≥0$, then we have:
$$\begin{align}&\begin{cases}u-v=1\\u^4+v^4=2\end{cases}
\\\implies &v^4+(v+1)^4=2\end{align}$$
Then, using the key substitution $v+\frac 12=y$ leads to:
$$\begin{align}&\left(y-\frac 12\right)^4+\left(y+\frac 12\right)^4=2\\
\implies &16y^4+24y^2-15=0\end{align}$$
Finally, substitute $y^2=z$, you obtain:
$$16z^2+24z-15=0$$
The solution is completed by reversing the steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral $\int\ln(1−x)dx$ by parts I have an exercise that explicitly asks for the indefinite integral of the following equation using integral by parts.
$$
\int\ln(1-x)\,dx
$$
I tried to define: $ u = \ln(1-x) $ so,
$$
\frac{\,du}{\,dx} = \frac{-1}{1-x}\\
\,du = \frac{-\,dx}{1-x}
$$
and, $ \,dv = \,dx $ resulting in $ v = x $ and finally applying: $ \int{u\,dv} = uv - \int{v\,du} $
$$
\begin{align}
\int\ln(1-x)\,dx&= x \ln(1-x) -\int x \, \frac{-\,dx}{1-x}\\
&=x\ln(1-x) - \int \frac{-x}{1-x} \,dx\\
&=x\ln(1-x) - \int \frac{1-x-1}{1-x} \,dx\\
&=x\ln(1-x) - \int \left(\frac{1-x}{1-x} - \frac{1}{1-x} \right)\,dx\\
&=x\ln(1-x) - \int \left(1 - \frac{1}{1-x} \right)\,dx\\
&=x\ln(1-x) - \left(\int{1\,dx} - \int{\frac{1}{1-x}}\,dx\right)\\
&=x\ln(1-x) - x -\left(- \int{\frac{1}{1-x}}\,dx\right)\\
&=x\ln(1-x) - x + \int{\frac{1}{1-x}}\,dx\\
&=x\ln(1-x) - x + \ln(1-x) + c\\
&=(x + 1)\ln(1-x) - x + c
\end{align}
$$
that does not match the answer in the book:
$(x - 1)\ln(1-x) - x + c$
I have found this other solution that I believe is not valid considering that the exercise asks to utilize integral by parts.
| The mistake was to consider that $\int{\frac{\,dx}{1-x}}$ is simply $\ln(1-x)$ when the correct path of continuing is using the substitution method resulting in $-\ln(1 - x) + c$:
$ t = x - 1 $
$ \frac{\,dt}{\,dx} = -1$
$ \,dt = -\,dx$
$ \,dx = -\,dt$
$$
\int{\frac{-\,dt}{t}} = -1\int{\frac{\,dt}{t}} = -1\ln(1-x) + c
$$
| {
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"url": "https://math.stackexchange.com/questions/4595973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}dx=\frac{\pi^2}{4}$? I am trying to show that
$$\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}dx=\frac{\pi^2}{4}$$
Context: I was working on another question ("Attempt $2$") and miscopied an integral, so I was trying to evaluate the integral in my question here. Anyway, now I'm intrigued by this integral, because my computer strongly suggests that it has a closed form, $\frac{\pi^2}{4}$.
My attempt: I have tried substituting $u=\cos{x}$ or $u=\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}$, and the half-angle tangent substitution, but they do not seem to work.
(Ideally, there is an elementary solution, but any solution would be appreciated.)
| One can show that $\sin(x)/\sqrt{5/4-\cos(x)}$ increases from $0$ to $1$ for $x$ going from $0$ to $\pi/3$ and decreases from $1$ to $0$ for $x$ going from $\pi/3$ to $\pi$. Then the integral can be separated into
\begin{align}
I&=\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx\\
&=\int_0^{\pi/3} \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx+\int_{\pi/3}^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx
\end{align}
We define
\begin{equation}
u=\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}
\end{equation}
or $u^2(5/4-\cos{x})=1-\cos^2x$.
Solving this equation for $\cos x$,
*
*for $0\le x\le\pi/3$, we have $\cos x=(u^2+\sqrt{u^4-5u^2+4})/2$
*for $\pi/3\le x\le\pi$, we have $\cos x=(u^2-\sqrt{u^4-5u^2+4})/2$.
Substituting $u$ into the integral, it comes
\begin{align}
I&=\int_0^1\arcsin(u)\frac{\sqrt{5-2 u^{2}-2 \sqrt{u^{4}-5 u^{2}+4}}}{\sqrt{u^{4}-5 u^{2}+4}}\,du+
\int_0^1\arcsin(u)\frac{\sqrt{5-2 u^{2}+2 \sqrt{u^{4}-5 u^{2}+4}}}{\sqrt{u^{4}-5 u^{2}+4}}\,du\\
&=\int_0^1\frac{\arcsin(u)}{\sqrt{u^{4}-5 u^{2}+4}}
f(u)\,du
\end{align}
where $f(u)=\sqrt{5-2 u^{2}-2 \sqrt{u^{4}-5 u^{2}+4}}+\sqrt{5-2 u^{2}+2 \sqrt{u^{4}-5 u^{2}+4}}$.
Now, it is possible to simplify:
\begin{align}
\left[f(u)\right]^2&=10-4u^2+2\sqrt{5-2 u^{2}-2 \sqrt{u^{4}-5 u^{2}+4}}\sqrt{5-2 u^{2}+2 \sqrt{u^{4}-5 u^{2}+4}}\\
&=4(4-u^2)
\end{align}
and
\begin{equation}
\sqrt{\frac{4(4-u^2)}{u^{4}-5 u^{2}+4}}=\frac{2}{\sqrt{1-u^2}}
\end{equation}
Then
\begin{align}
I&=2\int_0^1\frac{\arcsin(u)}{\sqrt{1-u^2}}\,du\\
&=\left.\arcsin^2(x)\right|_0^1\\
&=\frac{\pi^2}{4}
\end{align}
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Converting a differential equation to a hypergeometric equation I would like to transform the following differential equation into a hypergeometric equation:
${z^2}\frac{{{d^2}W}}{{d{z^2}}} + z\frac{{{d^{}}W}}{{d{z^{}}}} + \left( {{A^2} - {B^2}\left( {1 - {C_{}}z + D\frac{{z\,}}{{{{\left( {1 - z} \right)}^{}}}}} \right) - \frac{F}{{{{\left( {1 - z} \right)}^2}}}} \right)W = 0.$
Here A, B, C, D and F are constants.
I tried by letting I tried by letting $W = {z^p}{\left( {1 - z} \right)^q}y.$
After replacing it into the above equation, I ran into trouble trying to "fit" the values of p and q so that the y term in the resulting equation would be independent of z. I thought it was a straightforward thing to do. However, the values that I got for p and q did not reduce the equation to a hypergeometric one. Obviously, I am missing something simple here.
Any advice will be most appreciated.
| Using
\begin{align}
w &= x^p \, (1-x)^q \, y \\
w' &= x^p \, (1-x)^q \, \left(y' + \left(\frac{p}{x} - \frac{q}{1-x}\right) \, y \right) \\
w'' &= x^p \, (1-x)^q \, \left(y'' + \left(\frac{2 p}{x} - \frac{2 q}{1-x}\right) \, y' + \left(\frac{p(p-1)}{x^2} - \frac{2 p q}{x(1-x)} + \frac{q(q-1)}{(1-x)^2} \right) \right)
\end{align}
then the equation
$$ x^2 \, w'' + x \, w' + \left(a + b \, x + \frac{c \, x}{1-x} - \frac{d}{(1-x)^2} \right) \, w = 0 $$
can be transformed as seen in the following. The differential equation can be placed into the form
$$x (1-x) \, w'' + (1-x) \, w' + \left((-a+b+c) + \frac{a-d}{x} - \frac{d}{1-x} \right) \, w = 0.$$
From here, and with the substitutions, this equation can be reduced to
$$x (1-x) \, y'' + ((2p+1) - (2p+2q+1) \, x) \, y' + f(x) \, y = 0,$$
where
$$ f(x) = -(p+q)^2 - a + b + c + \frac{p^2 + a - d}{x} + \frac{q(q-1)-d}{1-x}. $$
This yields the equations $q(q-1) - d=0$ and $p^2 + a -d = 0$. Now that the equation is further reduced it is determined that the resulting form is
$$x (1-x) \, y'' + ((2p+1) - (2p+2q+1) \, x) \, y' - ((p+q)^2 + a - b - c) \, y = 0.$$
Comparing this to the desired hypergeometric equation, namely,
$$ x (1-x) \, y'' + (\gamma - (\alpha + \beta + 1) \, x) \, y' - \alpha \, \beta \, y = 0 \\
y(x) = {}_{2}F_{1}(\alpha, \beta; \gamma; x)$$
then, by using $\alpha \beta = (p+q)^2 + a - b - c$ and $\alpha + \beta = 2p + 2q$,
$$ y(x) = {}_{2}F_{1}(\phi_{1}, \phi_{2}; 2p+1; x),$$
where $\phi_{1} = p+q+\sqrt{c+b-a}$ and $\phi_{2} = p+q-\sqrt{c+b-a}$.
The $w(x)$ form is
$$ w(x) = x^p \, (1-x)^q \, {}_{2}F_{1}(\phi_{1}, \phi_{2}; 2p+1; x), $$
where
\begin{align}
& p^2 + a -d = 0 \\
& q(q-1) - d = 0 \\
& \phi_{1} = p+q+\sqrt{c+b-a} \\
& \phi_{2} = p+q-\sqrt{c+b-a}.
\end{align}
Solving for $p$ and $q$, equating the coefficients in general, and making the necessary reductions should lead to the desired solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluating the limit $\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$ How can I evaluate the following limit?
$$\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$$
To solve this limit, I divided the numerator and denominator by $x^5$ and I got
$$\lim_{x\to\infty}\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x$$
But I realized when $x\to\infty$ the expression is still indeterminate.
I tried changing the $x=\frac {1}{y}$ variable to be able to apply Lophital's rule, but I couldn't get a useful result.
| $$\lim_{x\to\infty}\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x$$
Arrived at this point, you can use geometric series:
$$\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x\sim\left(\left(1+\frac {1}{x}\pi+\frac {1}{x^5}e\right)\left(1-\frac {1}{x}e-\frac {1}{x^5}\pi \right)\right)^x\sim \left(1 +\frac{\pi-e}{x}\right)^x \to e^{\pi-e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $3^n + 4^n + 5^n$ is divisible by $12$ for all positive odd integers $n$ by induction I am trying to prove by induction that $\sum_{i=3}^{5} i^n$ is divisible by $12$ for all positive odd integers $n$. I heard that the solution might involve normal induction, where you assume that it works for the lowest positive odd number (in this case, $1$) and then try to deduce $P(n+2)$ for the next odd number.
I am not sure how to go about this, so any help would be greatly appreciated. I was thinking that if we define $n$ to be odd, then we could try to show that $n+2$ works using the fact that $n+2$ is also odd. However, I am not sure if this is the correct approach or how to proceed.
Could someone please help me with this problem?
Thank you!
| We set $n = 2k - 1$ and carry an induction over $k$. For the base case, $k = 1$, $n = 1$ and $3 + 4 + 5 = 12$, which is clearly divisible by 12. Assume true for $k$, and consider the case for $k + 1$. We have.
\begin{align*}
3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1} & = 9\cdot 3^{2k-1} + 16\cdot 4^{2k-1} + 25\cdot 5^{2k-1}\\
& = 8\cdot 3^{2k-1} + 15\cdot 4^{2k-1} + 24\cdot 5^{2k-1} + (3^{2k-1} + 4^{2k-1} + 5^{2k-1}).
\end{align*}
From our induction hypothesis, we know $3^{2k-1} + 4^{2k-1} + 5^{2k-1} \equiv 0 \ \ \ (\text{ mod } 12 \ )$. Also, $8\cdot 3^{2k-1}$, $15\cdot 4^{2k-1}$ and $24\cdot 5^{2k-1}$ are all $\equiv 0 \ \ \ (\text{ mod } 12 \ )$ implying $3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1}$ is divisible by 12.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to LHS but did not succeed in solving by this means.
Please provide a solution.
| Another way.
$$\sum_{cyc}\frac{a^2+bc}{b+c}-a-b-c=\sum_{cyc}\left(\frac{a^2+bc}{b+c}-\frac{b+c}{2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2-c^2}{b+c}=\frac{1}{2}\sum_{cyc}\frac{a^2-b^2-(c^2-a^2)}{b+c}=$$
$$=\frac{1}{2}\sum_{cyc}(a^2-b^2)\left(\frac{1}{b+c}-\frac{1}{a+c}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a+b)}{(a+c)(b+c)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Let $f(x)=1+2\cos x +3 \sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Let $f(x)=1+2\cos x +3\sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Solution given as in book:
since $f(x)+f(x+\pi)=2 \implies \dfrac{1}{2}f(x)+\dfrac{1}{2}f(x+\pi)=1$
and after comparing with original equation we get $a=\dfrac{1}{2}, b=\dfrac{1}{2}, c=\pi$
My doubt: Isn't there other way to approach this problem as it is not obvious to do $f(x)+f(x+\pi)=2$ because this kind of step does not come to mind immediately.
| Note that $3\sin x+2\cos x=\sqrt{13}\sin (x+\alpha)$ where $\tan \alpha=\frac{2}{3}$.
Now we get $$a+b+\sqrt{13}(a\sin(x+\alpha)+b\sin(x+c+\alpha))=1$$
for all $x\in \mathbb R$.
Now if we set $x=-\alpha$ and $x=\pi-\alpha$, then $a+b=1$ and $\sin c=0$ so $c=k\pi$ for some integer $k$.
Plugging $x=\frac{\pi}{2}-\alpha$ gives $a+b\cos k\pi=0$ so $k$ is odd and $a-b=0$. Therefore $a=b=\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinitely integrate it. But the closed form are in terms $\text{Si}(x)$ and $\text{Ci}(x)$. I would highly appreciate if there's any method or unique substitution to evaluate this Integral.
| $$I=\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\int_{0}^{\infty}\frac{\left|\cos\left ( \frac{x}{2}-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( \frac{x}{2}+\frac{\pi}{4} \right ) \right| }{x}dx$$
$$=\frac{1}{\sqrt2}\int_0^\infty\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}dx=\sqrt 2\int_0^\infty\frac{\sin x}{x}\frac{1}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}dx$$
Using Lobachevsky' integral
$$I=\sqrt 2\int_0^{\pi/2}\frac{dx}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}=\sqrt 2\int_0^{\pi/2}\frac{dx}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}$$
$$\sqrt 2\int_0^{\pi/2}\frac{dx}{\sqrt2\cos\frac{x}{2}+\sqrt2\sin\frac{x}{2}}=\sqrt2\int_0^{\pi/4}\frac{dx}{\sin(x+\frac{\pi}{4})}=\sqrt2\int_{\pi/4}^{\pi/2}\frac{dx}{\sin x}=\sqrt 2\ln\tan\frac{x}{2}\,\bigg|_{\pi/4}^{\pi/2}$$
$$=-\sqrt2\ln\tan\frac{\pi}{8}=-\sqrt 2\ln\sqrt\frac{1-\frac{1}{\sqrt2}}{1+\frac{1}{\sqrt2}}=\frac{1}{\sqrt 2}\ln\frac{\sqrt2+1}{\sqrt2-1}=\sqrt2\ln(\sqrt2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded. Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded.
What I've tried: Since
$\frac{1}{2^4}+\frac{1}{3^4}\leq \frac{1}{2^4}+\frac{1}{2^4}=\frac{1}{2^3}$
$\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}\leq \frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}=\frac{1}{4^3}$
and so on. So the series in question is less than $1+\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{8^3}...$. If I can show the latter is bounded I'm done. But I can't so maybe there is and easier solution?
| \begin{equation}
\sum_{k=1}^{\infty}\frac{1}{k^4}
\leq \sum_{k=1}^{\infty}\frac{1}{k^2}
= 1 + \sum_{k=1}^{\infty}\frac{1}{\left(k + 1\right)^2}
\leq 1+\int_{1}^{\infty}\frac{1}{x^2} dx
= 1 + 1
= 2
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question to first order PDE with Characteristics Method We have the given problem
\begin{align}
-2yu_{x} + u_{y} + 2yu &= 2y \tag 1 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 2
\end{align}
where $x > 0, y \in \mathbb{R^*}$. We can modify the formula $(1)$ to get
\begin{align}
-2yu_{x} + u_{y} &= 2y - 2yu \\
\implies -2yu_{x} + u_{y} &= 2y(1-u) \\
\implies -u_{x} + \frac{u_{y}}{2y} &= 1-u \\
\implies u_{x} - \frac{u_{y}}{2y} &= u-1
\end{align}
so our problem would be
\begin{align}
u_{x} - \frac{u_{y}}{2y} &= u-1 \tag 3 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 4
\end{align}
The method of characteristics gives
$$\dfrac{dx}{1} = -2y dy = \dfrac{du}{u-1}$$
Edited after @MatthewCassell suggestion
Now, we have
\begin{align}
\dfrac{dx}{1} = -2y dy \implies
C_{1} &= y^2+x
\end{align}
and
\begin{align}
\dfrac{dx}{1} = \dfrac{du}{u-1} \implies u &= C_{2} e^x + 1 \\
\implies C_{2} &= \frac{u-1}{e^x}
\end{align}
The general solution of the PDE expressed in the form of implicit equation
$$C_2=G(C_1)$$
So $$u=e^x G(x+y^2)+1$$
Now, using the initial condition $u(1,y)=1+e^{-1-2y^2}$ gives
$$1+e^{-1-2y^2} = e G(1+y^2)+1 \implies G(1+y^2) = \frac{e^{-1-2y^2}}{e}$$
I get something strange (from what I've seen at least). Is this possible?
| Solution
\begin{align}
-2yu_{x} + u_{y} &= 2y - 2yu \\
\implies -2yu_{x} + u_{y} &= 2y(1-u) \\
\implies -u_{x} + \frac{u_{y}}{2y} &= 1-u \\
\implies u_{x} - \frac{u_{y}}{2y} &= u-1
\end{align}
so our problem would be
\begin{align}
u_{x} - \frac{u_{y}}{2y} &= u-1 \tag 3 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 4
\end{align}
The method of characteristics gives
$$\dfrac{dx}{1} = -2y dy = \dfrac{du}{u-1}$$
Now, we have
\begin{align}
\dfrac{dx}{1} = -2y dy \implies
C_{1} &= y^2+x
\end{align}
and
\begin{align}
\dfrac{dx}{1} = \dfrac{du}{u-1} \implies u &= C_{2} e^x + 1 \\
\end{align}
The general solution of the PDE expressed in the form of implicit equation
$$C_2=G(C_1)$$
So $$u=e^x G(x+y^2)+1$$
Now, using the initial condition $u(1,y)=1+e^{-1-2y^2}$ gives
$$1+e^{-1-2y^2} = e G(1+y^2)+1 \implies G(1+y^2) = \frac{e^{-1-2y^2}}{e}$$
Then,
\begin{align}
G(1+y^2) &= e^{-2-2y^2}\\
&=e^{-2(1+y^2)}
\end{align}
So we can see that
\begin{align}
G(x+y^2) &= e^{-2(x+y^2)}\\
\implies u&=e^x \cdot e^{-2(x+y^2)} +1\\
&=e^{-x-2y^2} +1
\end{align}
which satisfies the PDE and boundary condition
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$, find $\frac{r}{R}$ in terms of $x, c, y$ This problem comes from a previous JEE Advanced examination, the problem is as follows:
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$ where $c$ is the third side, find the ratio of the inradius to the circumradius in terms of $x, c, y$
This is indeed a pretty challenging and tricky problem. Upon initial examination, spamming algebra seems to not lead anywhere. I found a solution using some angle chasing, which I'll share below, please share your own approaches as well!
Let $a+b=x$ and $ab=y$
We know that:
$$x^2-c^2=y$$
$$(a+b)^2-c^2=ab$$
$$a^2+b^2-c^2=-ab$$
Dividing by $2ab$ and exploiting the Law of Cosines, we can say that:
$$\frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}$$
Now, this implies that $\angle C=120^\circ$
Now, using the Law of Sines, we get:
$$\frac{1}{2R}=\frac{\sin{C}}{c}$$
$$\frac{c}{2R}=\frac{\sqrt3}{2}$$
$$R=\frac{c}{\sqrt3}$$
We also know that:
$$\Delta= \frac{1}{2}ab\sin{C}$$
$$\Delta=\frac{\sqrt{3}y}{4}$$
Further, we can note that:
$$s=\frac{a+b+c}{2}$$
$$s=\frac{x+c}{2}$$
Also, $r=\frac{\Delta}{s}$, therefore:
$$\frac{2\sqrt{3}y}{4(x+c)}=r$$
$$r=\frac{\sqrt{3}y}{2(x+c)}$$
Finally:
$$\frac{r}{R}=(\frac{\sqrt{3}y}{2(x+c)})(\frac{\sqrt{3}}{c})$$
$$\frac{r}{R}=\frac{3y}{2c(x+c)}$$
| From Law of Cosine, replace cosine with accurate haversine.
$c^2 = a^2 + b^2 - 2ab×\cos(C) = (a-b)^2 + 4ab×hav(C)$
Let $\displaystyle z = ab×hav(C) = \frac{(c-(a-b)(c+(a-b))}{4}$
$→ z = (s-a)(s-b)$
$→ (a\,b-z) = s(s-c)$
$Δ = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(ab-z)\,z}$
If c is the smallest side, (a-b) is exact, RHS area formula is very accurate.
Back to the oringal problem, without "angle chasing"
$\displaystyle z = \frac{c^2 - (a-b)^2}{4}
= \frac{(x^2 - y) - (x^2 - 4y)}{4}
= \frac{3y}{4}$
$\displaystyle Δ = \sqrt{(ab-z)\,z}
= \sqrt{\left(y-\frac{3y}{4}\right) \left(\frac{3y}{4}\right)}
= \frac{\sqrt{3}\,y}{4}$
$\displaystyle \frac{r}{R}
= \frac{Δ}{s} ÷ \frac{a\,b\,c}{4Δ}
= \frac{4Δ^2}{s\,a\,b\,c}
= \frac{3y^2/4} {\left( \frac{x+c}{2} \right)y\,c}
= \frac{3y}{2c(x+c)}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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What is the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true in $\mathbb{R}$? The question arose from the following proof:
$f(x)=x^3+x+1$ is injective:
Let $a,b\in \mathbb{R}$ and $f(a)=f(b)$:
$f(a)=f(b)\Leftrightarrow a^3+a+1=b^3+b+1\Leftrightarrow a^3+a=b^3+b \Rightarrow a=b$
Is $a^3+a=b^3+b\Rightarrow a=b$ mathematically rigorous enough? I don't know the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true?
| Since
$\quad a^3+a=b^3+b \text{ implies } (a = 0 \land b = 0) \; \displaystyle{\Large \lor} \; \frac{a^2+1}{b^2+1} = \frac{b}{a}$
we can write, with $a \ne 0$
$$a^3+a=b^3+b \; \text{ iff } \; \frac{a^2+1}{b^2+1} = \frac{b}{a}$$
Examining the fraction you must have that both $a$ and $b$ are of the same sign.
Again, examining the fraction, assuming $0 \lt a \lt b$ leads to a contradiction.
Similarly, assuming any of
$\quad 0 \lt b \lt a\;$ or
$\quad b \lt a \lt 0\;$ or
$\quad a \lt b \lt 0$
leads to a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that $(1+1/x)\tan^{-1} x > 1$, when $x> 0$ Monotonicity of a function is an itneresting method to prove inequalities. However, one needs to construct a suitable function $f(x)$ in the given domain. In this, out of several choices one particular form of $f(x)$ may suceed. Here, we prove that $$(1+1/x)\tan^{-1} x>1, ~\text{when}~ x > 0. \tag{$*$}$$
Out of several options, let us take $$f(x)=\tan^{-1}x -\frac{x}{1+x}.$$
$$\implies f'(x)=\frac{1}{1+x^2}-\frac{1}{(1+x)^2}=\frac{2x}{(1+x^2)(1+x)^2}\ge 0, \forall~ x\ge 0.$$
So, $f(x)$ is an increasing function in $[0,\infty)$, so $f(0)\ge 0$. Hence we get
$$\tan^{-1}x -\frac{x}{1+x}\ge 0 \implies (1+x)\tan^{-1}x\ge x.$$
Finally for $x>0$, we have
$$\implies (1+1/x)\tan^{-1} x >1.$$
It will be interesting to know other options/methods to prove $(*)$.
| Dividing both sides by $(1+1/x)$,
$$
\arctan x > \frac x{x+1}
$$
Then we have
$$
\begin{align}
x > \tan \frac x{x+1}
\end{align}
$$
Applying inequalities about $\sin$ and $\cos$ i.e. $\sin \theta < \theta$ and $\cos \theta > 1 - 1/2\theta^2, \forall \theta \in (0,1)$,
$$
\begin{align}
\tan \frac x{x+1} & = \frac{\sin \frac x{x+1}}{\cos \frac x{x+1}} \\
&< \frac{\frac x{x+1}}{1 - \frac12(\frac x{x+1})^2} \\
&< \frac{\frac x{x+1}}{1 - \frac x{x+1}} \\
&= x
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Proving that $\lim_\limits{x \to \infty} x\cdot \ln\left(1 + \frac{1}{x}\right) = 1$ using Taylor series We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.
To do this, when we expand $\ln(1 + \frac{1}{x})$, we should get a series like $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$, multiplying by $x$ would make it $1 + \frac{a_1}{x} + \frac{a_2}{x^2} + ...$ which is $1$ if $x \to \infty$.
The Taylor expansion of $f(x) = \ln(1 + \frac{1}{x})$ at some $a$ is $\sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6}(x-a)^3 + ...$.
$f'(a) = -\frac{1}{a^2 + a}$, $f'(a) \cdot (x-a) = - \frac{x-a}{a^2 + a}$
$f''(a) = \frac{2a + 1}{(a^2 + a)^2}$, $f''(a) \cdot (x-a) = \frac{(2a + 1)(x-a)}{(a^2 + a)^2}$
So the Taylor series begins like this:
$\ln(1 + \frac{1}{x}) = \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ...$
Since the first term doesn't involve $x$ and we want it to be $\frac{1}{x}$, we would like it to be $0$ and the next term to be $\frac{1}{x}$. To make the first term $0$, $a$ needs to be approaching infinity.
$\lim_\limits{a \to \infty} \left( \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right) = \lim_\limits{a \to \infty} \left(- \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right)$.
Now, this isn't in the form $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$ that we wanted.
Do we need to pick some other $a$?
I found https://math.stackexchange.com/a/1071689/1095885 which says
When x is very large, using Taylor
$\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$
The series $\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{x^3} + ...$, which would be perfect, seems to be a Laurent series though.
Edit: When substituting $u = \frac{1}{x}$ and taking the Taylor series of $\ln(1 + u)$, why don't we have to use the chain rule?
| Using Taylor series here is like the proverbial killing a fly with a sledgehammer. Set $h = 1/x$, so the condition $x \to \infty$ is the same as $h \to 0^+$ and
$$
x\ln\left(1+\frac{1}{x}\right) = \frac{\ln(1+h)}{h} = \frac{\ln(1+h)-\ln 1}{h}.
$$
Therefore
$$
\lim_{x \to \infty} x\ln\left(1+\frac{1}{x}\right) =
\lim_{h \to 0^+} \frac{\ln(1+h)-\ln 1}{h},
$$
which shows your calculation follows from calculating the derivative of $\ln x$ at $x = 1$ (using $h \to 0$, not just $h \to 0^+$):
$$
\lim_{h \to 0} \frac{\ln(1+h)-\ln 1}{h} = \left.\frac{d}{dx}\right|_{x=1}\ln x = \ ??
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2- \frac{2}{\sqrt{n+1}}$
Let $i\in\mathbb{N}$ $$\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{n+1}}$$
A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?
| We first observe that
$$\frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}} = \frac{2}{\sqrt{n(n+1)}(\sqrt{n} + \sqrt{n+1})}
< \frac{2}{\sqrt{n(n+1)}\cdot 2[n(n+1)]^{\frac{1}{4}}} = \frac{1}{(n^2+n)^{\frac{3}{4}}}$$
and then use this to argue
\begin{align*}
\displaystyle\sum_{i=1}^{n} \frac{1}{(i^2+i)^\frac{3}{4}} > \displaystyle\sum_{i=1}^{n} \frac{2}{\sqrt{i}} - \frac{2}{\sqrt{i+1}} = 2-\frac{2}{\sqrt{n+1}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given positive real numbers $x_1,x_2,\ldots,x_n$ such that $\sum_{k=1}^{n}{\frac{1}{x_k+1}}=n-2$ , how to prove the following inequality? Given positive real numbers $x_1,x_2,\ldots,x_n$ such that $\sum_{k=1}^{n}{\frac{1}{x_k+1}}=n-2$ , prove that
$$
\sum_{i=1}^{n}{\frac{1}{x_i}}+\frac{n(n-2)}{\sum_{i=1}^{n}{x_i}}\geqslant(n-1)(n-2)
$$
This problem was initially asked on the site 'zhihu'. The case $n=3$ has been proved by 'SOS' method, and I've given a $pqr$ method proof. How to prove (or falsify) this generalised inequality?
I've tried to set $a_i=1/x_i$ , then $\sum_{k=1}^{n}{\frac{a_k}{a_k+1}}=n-2$ and the inequality is equivalent to
$$
nA+(n-2)H\geqslant(n-1)(n-2)
$$
where $A$ is the Arithmetic Mean among $a_i$ , and H is the Harmonic Mean among $a_i$ , and maybe we can apply some theorems relevant to these means.
| Let $$\sum_{i=1}^{n}{\frac{1}{x_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{x_i}}<(n-1)(n-2)$$ for some values of $x_i$, $x_i=ka_i,$ where $k>0$ and
$$\sum_{i=1}^{n}{\frac{1}{a_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{a_i}}=(n-1)(n-2).$$
Thus, $$\frac{1}{k}(n-1)(n-2)<(n-1)(n-2),$$ which gives $k>1$ and we obtain:
$$n-2=\sum_{i=1}^{n}{\frac{1}{x_i+1}}=\sum_{k=1}^{n}{\frac{1}{ka_i+1}}<\sum_{k=1}^{n}{\frac{1}{a_k+1}},$$ which is a contradiction because we'll prove now that $$\sum_{k=1}^{n}{\frac{1}{a_k+1}}\leq n-2$$ for any positives $a_i$ such that $$\sum_{i=1}^{n}{\frac{1}{a_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{a_i}}=(n-1)(n-2).$$
Indeed, let $f(x)=\frac{1}{1+x}$ and $g(x)=f'\left(\frac{1}{\sqrt{x}}\right)=-\frac{x}{(1+\sqrt{x})^2}.$
Thus,
$$g''(x)=\frac{3}{2\sqrt{x}(1+\sqrt{x})^4}>0,$$ which says that $g$ is a strictly convex on $(0,+\infty).$
Now, let $\sum\limits_{i=1}^na_i=constant$.
Thus, $\sum\limits_{i=1}^n\frac{1}{a_i}=constant$ and by the Vasc's EV Method Corollary 1.6 it's enough to prove that $\sum\limits_{i=1}^nf(a_i)\leq n-2$
for equality case of $n-1$ variables.
Id est, it's enough to prove that:
$$\frac{n-1}{x+1}+\frac{1}{y+1}\leq n-2,$$ where $x$ and $y$ are positives such that
$$\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}=(n-1)(n-2)$$ or
$$\frac{n-1}{x+\frac{(n-1)(n-2)}{\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}}}+\frac{1}{y+\frac{(n-1)(n-2)}{\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}}}\leq\frac{(n-2)\left(\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}\right)}{(n-1)(n-2)}$$ or
$$(x-y)^2x(2nxy+(x-y)^2)\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $\int_{0}^{1}\frac{K(x)\ln\left(1-x^2\right)}{\sqrt{1-x^2}}\text{d}x,\int_{0}^{1}\frac{xK(x)^2\ln\left(1-x^2\right)}{\sqrt{1-x^2}}\text{d}x$ I am recently interested in integrals containing an elliptic integral $K(x)$, which is defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2} }\text{d}t$ for $|x|<1$ and $x$ is the elliptic modulus.
Then I come across the following two integrals
$$
\begin{aligned}
&\int_{0}^{1} \frac{K(x)\ln\left ( 1-x^2 \right ) }{
\sqrt{1-x^2} }\text{d}x=-\frac{\Gamma\left(\frac14\right)^4}{24},\\
&\int_{0}^{1} \frac{xK(x)^2\ln\left ( 1-x^2 \right ) }{
\sqrt{1-x^2} }\text{d}x=-\frac{\pi^4}{4}\,_4F_3\left ( \frac12,\frac12,\frac12,\frac12;1,1,1;1 \right ).
\end{aligned}
$$
by estimation. Where generalized hypergeometric function($\,_pF_q$) is used. For the first one, I write
$$
\begin{aligned}
\int_{0}^{1} \frac{K(x)\ln\left ( 1-x^2 \right ) }{
\sqrt{1-x^2} }\text{d}x
&=\frac{\pi}{2} \frac{\mathrm{d}}{\mathrm{d}n} \frac{\sqrt{\pi}\,\Gamma\left ( \frac{n+2}{2} \right ) }{
\Gamma\left ( \frac{n+3}{2} \right ) }\,_3F_2
\left ( \frac12,\frac12,\frac12;1,\frac{3+n}{2} ;1 \right )\Bigg|_{n=-1}\\
&=\frac{\pi}{2}\left [ -\pi\ln(2)\cdot\frac{\pi}{\Gamma\left ( \frac34 \right )^2 }
+\pi \frac{\mathrm{d}}{\mathrm{d} n} \,_3F_2
\left ( \frac12,\frac12,\frac12;1,\frac{3+n}{2} ;1 \right )\Bigg|_{n=-1}\right ] .
\end{aligned}
$$
Edit: We just need to prove
$$
\,_3F_2
\left ( \frac12,\frac12,\frac12;1,\frac{3+n}{2} ;1 \right )
\Bigg|_{n=-1}
=-\frac{(\pi-3\ln(2))\Gamma\left ( \frac14 \right )^4 }{12\pi^3}.
$$
Or we equivalently write($H_n$ denotes harmonic numbers)
$$
\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^3 }{
(n!)^3}H_n=\frac{(\pi-3\ln(2))\Gamma\left ( \frac14 \right )^4 }{6\pi^3}.
$$
Which has been evaluated here.
Question: Can we verify these closed-forms? Thanks for reaching my hand.
| For the first integral we can start by substituting $1-x^2\to x^2$ followed by expanding the denominator into power series.
$$I=\int_0^1 \frac{K(x)\ln (1-x^2)}{\sqrt{1-x^2}}dx=2\int_0^1 \frac{K'(x)\ln x}{\sqrt{1-x^2}}dx$$
$$=2\sum_{k=0}^\infty \frac{\binom{2k}{k}}{2^{2k}}\int_0^1 x^{2k}K'(x)\ln x\,dx$$
We have:
$$\int_0^1 x^{2k}K'(x)dx=\frac{\pi}{4}\frac{\Gamma\left(k+\frac12\right)^2}{\Gamma\left(k+1\right)^2}$$
$$\Rightarrow \int_0^1 x^{2k}K'(x)\ln x\,dx=\frac{\pi}{4}\frac{\Gamma\left(k+\frac12\right)^2\left(\psi\left(k+\frac12\right)-\psi\left(k+1\right)\right)}{\Gamma\left(k+1\right)^2}$$
Therefore we desire to find:
$$I=\frac{\pi^2}{2}\sum_{k=0}^\infty \frac{\binom{2k}{k}^3}{\left(2^{2k}\right)^3}\left(\psi\left(k+\frac12\right)-\psi\left(k+1\right)\right)$$
And using $(*)$ from here, gives us:
$$I=-\frac{\pi^3}{6} {_2F_1}{\left( {\frac{1}{2},\frac{1}{2};1;\frac{1}{2}} \right)^2} = -\frac{\Gamma\left(\frac14\right)^4}{24}$$
Note that:
$$\binom{2k}{k}=\frac{2^{2k}}{\sqrt \pi}\frac{\Gamma\left(k+\frac12\right)}{\Gamma\left(k+1\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}{b(a+c+d)} + \frac{c^4}{c(a+b+d)} + \frac{d^4}{d(a+b+c)} \ge \frac{(a^2 + b^2 + c^2 + d^2)^2}{a(b+c+d) \ + \ b(a+c+d) \ + \ c(a+b+d) \ + \ d(a+b+c)}$
Using the condition that $ab + bc + cd + da = 1$, the RHS results in $\frac{(a^2+b^2+c^2+d^2)^2}{2+2(ac+bd)}$.
Nothing more seems to help. Opening up the numerator or using AM-GM on it doesn't work and I have no idea on how to turn this result in a $\frac{1}{3}$ fraction.
| Continuing from where you left off, note the condition is equivalent to $(a+c)(b+d)=1$. Then reduce the numerator using
$$
a^2+b^2+c^2+d^2 = (a+c)^2+(b+d)^2-2(ac+bd)\ge 2(a+c)(b+d)-2(ac+bd) =2-2(ac+bd)
$$
You now need to show $2[1-(ac+bd)]^2/(1+ac +bd)\ge 1/3$, which is equivalent to $ ac+bd\le 1/2$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Prove $a+b+c \geq \frac{3+\sqrt{7}}{2}$ for $a,b,c \in [0, \frac32]$ with $a^2+b^2+c^2+abc=4$
Let $a,b,c \in \left[0;\frac{3}{2}\right]$ and $a^2+b^2+c^2+abc=4$. Prove that
$$a+b+c \geq \dfrac{3+\sqrt{7}}{2}.$$
Source: This is a math problem that my teacher gave me $3$ months ago (the submission deadline has expired). My teacher wrote a book and sent me to test the difficulty of the problem.
My attempt: I have converted $a$ to $b,c$, used to trigonometric conversion, but all failed.
Related problem (the same source, with the same conditions): https://artofproblemsolving.com/community/c6h2975620p26673082
Please give me a suggestion! Thank you!
| Rewriting the inequality as :
$$\frac{1}{2}\left(-bc+\sqrt{b^2c^2-4\left(b+c\right)^{2}+8bc+16}\right)+b+c \geq \dfrac{3+\sqrt{7}}{2}$$
We make :
$$b+c=\operatorname{constant}=C$$
The function :
$$g(bc)=-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}$$
Is increasing as $bc$ increases with $-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}=a\geq 0$
So if $b=0$ with the constraint we have an equality as $c=1.5$
Done .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving y(t) from a differential equation Given the initial problem
$$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0, y(1) =3 $$
I got the solution to be
$$\frac{t^2}{4y^4}-\frac{3}{2y^2} = -\frac{53}{324} $$
But how do I write it in terms of $y(t)$
| Letting $P:=\frac{t}{2y^4}$ and $Q:=\frac{-t^2+3y^2}{t^5}$ then $P_y=Q_t$ so the non linear DE is exact and we have the implicit solution $\frac{-3}{2y^2}+\frac{t^2}{4y^4}=k$ for an arbitrary constant $k$. We have for $y,k\not=0$ that $-\frac{3}{2y^2}+\frac{t^2}{4y^4}=k$ give $t^2-6u-4ku^2=0$ with $u:=y^2$. Completing square we have $(u+\frac{3}{4k})^2=\frac{1}{4}(\frac{t^2}{k}+\frac{9}{4k^2})$. Squaring both sides we find $u+\frac{3}{4k}=\pm\frac{1}{2}\sqrt{\frac{4kt^2+9}{4k^2}}$ and then $y^2=\pm\frac{\sqrt{4kt^2+9}}{4k}-\frac{3}{4k}$. Hence $y=\pm\frac{1}{2}\sqrt{\frac{-3}{k}\pm\frac{\sqrt{4kt^2+9}}{k}}$. Hence, we have the four cases for to study. Invoking the condition $y(1)=3$ we find only $k=\frac{-53}{324}$ with the combination the sign $(+,-)$. I think that you can complete the details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Natural log : Has this proof jumped a step? This proof in Nevanlinna and Paatero's Introduction to Complex Analysis (1969) is meant to show that if $$c_n=\left(1+\frac{z}{n}\right)^n$$ $(n=1,2,\dots,$ and $z = x + iy)$ then $$e^x = \liminf_ {n}|c_n|$$ ($|.|$ denotes the absolute value (corrected from "the real part").
How do they get from Step $2$ to Step $3$? It looks as if the log function has just been ignored, but they're too reputable for that!
Expanding $$|c_n|=\left\lvert1+\frac{x}{n}+i\frac{y}{n}\right\rvert^n
=\bigg(\left(1+\frac{x}{n}\right)^2+\left(\frac{y}{n}\right)^2\bigg)^\frac{n}{2}$$
so $$\ln|c_n|=\frac{n}{2}\ln\bigg(1+2\frac{x}{n}+\frac{x^2+y^2}{n^2}\bigg)\tag 1$$
$$=\frac{n}{2}\bigg(\frac{2x}{n}+\left[\frac{1}{n^2}\right]\bigg) = x + \left[\frac{1}{n}\right]\tag 2$$
($[.]$ meaning roughly "of the order of").
So $$\liminf_{n}\ln|c_n|=x\tag 3$$ and $$|c_n|\rightarrow e^x$$ QED.
| This may be the least troublesome derivation.
For $t > 0,$
$$
\log(1 + t) = \int_1^{1 + t}\frac{du}u < \int_1^{1 + t}du = t,
$$
and
$$
\log(1 + t) = \int_1^{1 + t}\frac{du}u >
\int_1^{1 + t}\frac{du}{1 + t} = \frac{t}{1 + t} =
t - \frac{t^2}{1 + t} > t - t^2,
$$
so
$$
0 < t - \log(1 + t) < t^2.
$$
We don't need the detailed inequality, only the consequential
estimate:
$$
\log(1 + t) = t + O(t^2).
$$
This gives us:
\begin{align*}
\log\left(1 + \frac{2x}n + \frac{x^2 + y^2}{n^2}\right) & =
\log\left(1 + \frac{2x}n + O\left(\frac1{n^2}\right)\right) \\
& = \left(\frac{2x}n + O\left(\frac1{n^2}\right)\right) +
O\left(\left(\frac{2x}n + O\left(\frac1{n^2}\right)\right)^2\right)
\\ & =
\frac{2x}n + O\left(\frac1{n^2}\right) + O\left(\frac1{n^2}\right)
\\ & = \frac{2x}n + O\left(\frac1{n^2}\right),
\end{align*}
whence:
$$
\frac{n}2\log\left(1 + \frac{2x}n + \frac{x^2 + y^2}{n^2}\right) =
x + O\left(\frac1n\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factor $1-64m^6$ under $\mathbb{Q}$ Factor $1-64m^6$ under $\mathbb{Q}$. We can employ the well known formula: $a^3-b^3 = (a-b)\cdot(a^2+ab+b^2)$.
$1-64m^6 = (1)^3-(4m^2)^3 = (1-4m^2)(1+4m^2+16m^4)=(1+2m)(1-2m)(1+4m^2+16m^4)$.
However, the answer is provided as $(1+2m)(1-2m)(1-2m+4m^2)\cdot(1+2m+4m^2)$.
I cannot understand how it is possible to simplify $(1+4m^2+16m^4)$ to $(1-2m+4m^2)(1+2m+4m^2)$?
| $$1+4m^2+16m^4$$
$$=1+2\cdot 4m^2+16m^4-4m^2$$
$$=(1+4m^2)^2-(2m)^2$$
$$=(1-2m+4m^2)(1+2m+4m^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!
| Hint for $a,b,c,x,y,z\in[1,\infty)$:
Let $x,y,z\in[1,\infty) $ then we have the inequality (tangent line method):
$$f\left(x\right)=\frac{x^{2}y+2}{y+z}\geq f'(1)(x-1)+f(1)$$
remark that again the inequality of the tangent line method is greater than the following inequality :
$$f\left(x,y,z\right)=\frac{\frac{3}{2}\left(\left(2x-1\right)y+2\right)}{x+z+y}$$
Then :
$$f\left(a,b,c\right)+f\left(b,c,a\right)+f\left(c,a,b\right)\geq 4.5$$
Then $a=b=c=1$ is the minimum we are done .
To show the intermediate inequality we have :
$$\sum_{cyc}\frac{\left(2x-1\right)y\left(2y-z-x\right)}{2\left(x+z\right)\left(x+y+z\right)}\geq 0\tag{I}$$
Because we use :
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-\frac{4.5}{a+b+c}\geq 0$$
$I$ can be shown using Sum of Square method .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Confusion while factoring quadratics. I recently factored the quadratic $3x^2+4x-4$, the results where $x = -2$; $x = 2/3$. My question is why is it incorrect to write the factors of the quadratic as $(x+2)(x-2/3)$? The correct way to write it is $(x+2)(3x-2)$, but why? What is the difference in $(x-2/3)$ and $(3x-2)$, both when $x$ is $2/3$ result in the y being zero.
| Observe that
\begin{align*}
3x^2 + 4x - 4 & = 3x^2 - 2x + 6x - 4\\
& = x(3x - 2) + 2(3x - 2)\\
& = (x + 2)(3x - 2)\\
& = 3(x + 2)\left(x - \frac{2}{3}\right)\\
& \neq (x + 2)\left(x - \frac{2}{3}\right)\\
& = x^2 + \frac{4}{3}x - \frac{4}{3}
\end{align*}
For two polynomials to be equal, they must be equal at each value of $x$ within their domains.
Let $f(x) = 3x^2 + 4x - 4$. Then $f(0) = -4$.
Let $g(x) = (x + 2)\left(x - \frac{2}{3}\right)$. Then $g(0) = -4/3 \neq -4 = f(0)$, so the polynomials $3x^2 + 4x - 4$ and $(x + 2)(x - 2/3)$ are not equal.
The graphs of $f$ and $g$ are shown below.
Your confusion stems from the fact these two polynomials have the same roots.
They have the same roots since the two polynomials differ by a scaling factor that can be canceled when you solve for the roots.
\begin{align*}
3x^2 + 4x - 4 & = 0\\
3(x + 2)\left(x - \frac{2}{3}\right) & = 0\\
(x + 2)\left(x - \frac{2}{3}\right) & = 0
\end{align*}
\begin{align*}
x + 2 & = 0 & \text{or} & & x - \frac{2}{3} & = 0\\
x & = -2 & & & x & = \frac{2}{3}
\end{align*}
| {
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Proving $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$ using matrices I am trying to show that $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$. I have calculated the minimal polynomial $x^4-14x^2+9$ and I know that its basis as a $\mathbb{Q}$-vector space is $\{1,
\sqrt{2}, \sqrt{5}, \sqrt{10}\}$. But I'm not sure how to construct an invertible matrix to express the basis as rational polynomials. I'm thinking of Fly by Night's comment here
| As Mariano notes in the comments, there's a much quicker argument using degrees, but here's the matrix argument. (A nice bonus of this argument is that you can explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$.)
Let $\alpha = \sqrt{2} + \sqrt{5}$. Note that $\alpha \in \mathbb{Q}(\sqrt{2}, \sqrt{5}),$ so immediately we have $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{5})$. Suppose there exists an invertible rational $4 \times 4$ matrix $A$ such that
$$\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix} = A \begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix}.$$
Then, we have
$$\begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix} = A^{-1}\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix},$$ which would prove that each of the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ can be expressed as a $\mathbb{Q}$-linear combination of powers of $\alpha$, and thus we must have $\mathbb{Q}(\sqrt{2}, \sqrt{5}) \subseteq \mathbb{Q}(\alpha)$. Since we already have inclusion the other way, we would then be done.
So, we just need to find such an $A$. Well, there's only one real candidate for what $A$ can be: the $i$th row of $A$ must be $\begin{pmatrix}a & b & c & d \end{pmatrix}$, where $a + b\sqrt{2} + c\sqrt{5} + d\sqrt{10}$ is the unique way to write $\alpha^{i}$ as a $\mathbb{Q}$-linear combination of the basis elements $\{1, \sqrt{2}, \sqrt{5}, \sqrt{10}\}.$ (This is unique because $\alpha^{i} \in \mathbb{Q}(\sqrt{2}, \sqrt{5}).$)
Let's compute:
$$\alpha = 0 \cdot 1 + 1 \cdot \sqrt{2} + 1 \cdot \sqrt{5} + 0 \cdot \sqrt{10},$$
$$\alpha^{2} = 7 + 2\sqrt{10} = 7 \cdot 1 + 0 \cdot \sqrt{2} + 0 \cdot \sqrt{5} + 2 \cdot \sqrt{10},$$
$$\alpha^{3} = 17\sqrt{2} + 11\sqrt{5} = 0 \cdot 1 + 17 \cdot \sqrt{2} + 11 \cdot \sqrt{5} + 0 \cdot \sqrt{10},$$
$$\alpha^{4} = 89 + 28\sqrt{10} = 89 \cdot 1 + 0 \cdot \sqrt{2} + 0 \cdot \sqrt{5} + 28 \cdot \sqrt{10}.$$
So, our candidate for $A$ is
$$A = \begin{pmatrix} 0 & 1 & 1 &0 \\ 7 & 0 & 0 & 2 \\ 0 & 17 & 11 & 0\\ 89 & 0 & 0 &28 \end{pmatrix}.$$ A quick computation shows that this matrix has nonzero determinant, hence is invertible. At this point, we are already done, but we can actually invert $A$ to explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$,
A more laborious computation (just kidding, I used a calculator) finds the inverse to be
$$A^{-1} = \begin{pmatrix} 0 & \frac{14}{9} & 0 & - \frac{1}{9} \\ -\frac{11}{6} & 0 & \frac{1}{6} & 0 \\ \frac{17}{6} & 0 & -\frac{1}{6} & 0\\ 0 & -\frac{89}{18} & 0 & \frac{7}{18} \end{pmatrix}.$$
Since we have
$$\begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix} = A^{-1}\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix},$$ it follows that
$$1 =\frac{14}{9} \alpha^{2} - \frac{1}{9} \alpha^{4},$$
(Look! It's the minimal polynomial!)
$$\sqrt{2} = -\frac{11}{6}\alpha + \frac{1}{6} \alpha^{3},$$
$$\sqrt{5} = \frac{17}{6}\alpha - \frac{1}{6} \alpha^{3},$$
$$\sqrt{10} = -\frac{89}{18}\alpha^{2} + \frac{7}{18}\alpha^{4}.$$
| {
"language": "en",
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"source": "stackexchange",
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sum of 100 terms of logarithmic expression
Calculate value of
$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$
My try :: $\displaystyle x^4+4y^4$
$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$
$\displaystyle =\sum^{100}_{k=1}\ln\bigg[\frac{(1+2(2k+1)+2(2k+1)^2)(1-2(2k+1)+2(2k+1)^2)}{(1+2(2k)+2(2k)^2)(1-2(2k)+2(2k)^2)}\bigg]$
How can I decompose that complex expression into partial fractions?
| Let's do bit by bit.
$$\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}$$
$$=\frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}}$$
Now,
$$\ln\Big( \frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}} \Big)$$
$$=\ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$
Now bring the summation.
$$\sum^{100}_{k=1} \ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$
$$=\Big[\ln\Big( (1+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 1^4+\frac{1}{64} \Big)\Big] + \Big[ \ln\Big( (2+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 2^4+\frac{1}{64} \Big) \Big] + \Big[ \ln\Big( (3+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 3^4+\frac{1}{64} \Big) \Big] + \dots + \Big[ \ln\Big( (100+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 100^4+\frac{1}{64} \Big) \Big]$$
Hope you can continue after this as it's now simplified
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Random Variables as Sums of Different Random Variables for $R = \sqrt{X^2 + Y^2}$ I am trying to understand the derivation of the PDF of $$R = \sqrt{X^2 + Y^2}$$
I am looking at some derivations and some derivations say that if $X$ and $Y$ are independent, then we can say that $$f_{X,Y} = f_X f_Y$$ which is just the law of independence for joint distributions. I don't really understand how we can break up $R$ into $2$ marginals which are both just the marginals of $X$ and $Y$.
For reference, I am trying the simplest example, where $X$ and $Y$ are uniform random variables from $0$ to $1$. I'll attach some of my work just so maybe someone can help me move forward too.
WORK THUS FAR
Let $R = \sqrt{X^2 + Y^2}$. Then, $$P(R \leq r) = P(\sqrt{X^2 + Y^2} \leq r) = P(X^2 + Y^2 \leq r^2)$$
Then, we know that the bounds are on the unit circle so we are solving:
$$F_R = \int_{-r}^{r}\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}f_{XY}dxdy$$
I'm stuck here.
Thanks.
| Now let me suppose that $X$ and $Y$ are valued in $\mathbb{R^+} = (0, \infty)$ and $S \colon = X/Y$.
Then we have
$$
\begin{cases}
X = \frac{RS}{\sqrt{1+S^2}} \\
Y = \frac{R}{\sqrt{1+S^2}}
\end{cases}
$$
Denote the density of $(X, Y)$ and $(R, S)$ by $f(x, y)$ and $g(r, s)$ respectively.
We can prove (maybe has been proved in the probability course) that
$$
g(r, s) = \left.
f\left( \frac{rs}{\sqrt{1+s^2}}, \frac{r}{\sqrt{1+s^2}} \right) \middle/
\left|
\begin{pmatrix}
r_x & r_y \\
s_x & s_y
\end{pmatrix}
\right|
\right.
$$
where $r_x = \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}$ and it is similar to calculate9 $r_y, s_x, s_y$. Inserting into the above equation yields that
$$
g(r, s) = \left.
f\left( \frac{rs}{\sqrt{1+s^2}}, \frac{r}{\sqrt{1+s^2}} \right) \frac{r}{1+s^2}\right.
$$
Then by integrating on $s$ the marginal density $g(r)$ of $R$ is given the following formula.
$$
g(r) = \int_0^\infty \left.
f\left( \frac{rs}{\sqrt{1+s^2}}, \frac{r}{\sqrt{1+s^2}} \right) \frac{r}{1+s^2}
\right.\ ds
$$
If we add the assumption that $X$ and $Y$ are indenpent which implies that $f(x, y) = f_X(x) f_y(y)$. Substituting $f$ by $f_Xf_Y$ simplify the formula.
$$
g(r) = \int_0^\infty \left.
f_X\left( \frac{rs}{\sqrt{1+s^2}}\right)f_Y\left(\frac{r}{\sqrt{1+s^2}} \right) \frac{r}{1+s^2}
\right.\ ds
$$
Actually I do think the formula is ugly which makes me suspend that if there is any wrong above.
Update: I have veirfied the formula and it should be correct.
For uniform, $f_X(x) = I(0 < x < 1)$ and $f_Y(y) = I(0 < y< 1)$.
Hence
$$
g(r) = \int_0^\infty I\left(0<\frac{rs}{\sqrt{1+s^2}}<1\right) I\left(0<\frac{r}{\sqrt{1+s^2}}<1\right) \frac{r}{1+s^2}\ ds
=\int_D \frac{r}{1+s^2} ds
$$
where $D$ is the joint of $D_1 = \{s: 0<\frac{rs}{\sqrt{1+s^2}}<1\}$ and $D_2 = \{0<\frac{r}{\sqrt{1+s^2}}<1\}$.
Case 1: $0<r<1, D = \mathbb{R}^+, g(r) = \frac\pi2 r$;
Case 2: $1 \le r < \sqrt2, D = (\sqrt{r^2-1}, 1/\sqrt{r^2-1}), g(r) = r(\tan 1/\sqrt{r^2-1} - \tan \sqrt{r^2-1})$;
Case 3: $\sqrt 2 \le r, D = \phi, g(r) = 0$.
| {
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Evaluate $\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$ (from MIT Integration Bee 2023) This is from the final round of MIT Integration Bee 2023.
$$\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$$
The given answer is $7\pi$.
I found a way to do it using contour integrals (see answer below), but this is not a calculation I can finish within 4 minutes (the time limit in the competition). I am still looking for other elegant methods, possibly without using contour integrals.
| $$A=\frac{\sin (2 x)\, \sin (3 x)\, \sin (5 x) \,\sin (30 x) }{\sin (x) \,\sin (6 x) \,\sin (10 x) \,\sin (15 x) }$$ Simplifying $$A=-1-2 \cos (2 x)+2 \cos (6 x)+2 \cos (8 x)$$ Squaring and simplifying again
$$A^2=7+8 \cos (2 x)-2 \cos (4 x)-8 \cos (6 x)-8 \cos (8 x)-4 \cos (10
x)+$$ $$2 \cos (12 x)+4 \cos (14 x)+2 \cos (16 x)$$
$$\int A^2\, dx=7 x+4 \sin (2 x)-\frac{1}{2} \sin (4 x)-\frac{4}{3} \sin (6
x)-\sin (8 x)-$$ $$\frac{2}{5} \sin (10 x)+\frac{1}{6} \sin (12
x)+\frac{2}{7} \sin (14 x)+\frac{1}{8} \sin (16 x)$$
Edit (detailed steps for the simplication)
Use the half-angle formulae for $\sin(2x)$, $\sin(30x)$ in numerator and for $\sin(6x)$, $\sin(10x)$ in denominator. This gives
$$A=\frac {\cos (x)\, \cos (15 x) }{\cos (3 x)\, \cos (5 x) }$$
Using product to sum and let $x=\frac t 2$
$$A=\frac{\cos (14 x)+\cos (16 x)}{\cos (2 x)+\cos (8 x)}=\frac{\cos (7 t)+\cos (8 t)}{\cos (t)+\cos (4 t)}$$
Now, $t=\cos ^{-1}(z)$ and multiple angle formulae to make
$$A=\frac{128 z^8+64 z^7-256 z^6-112 z^5+160 z^4+56 z^3-32 z^2-7z+1 } {8 z^4-8 z^2+z+1 }$$
Factorization
$$A=\frac{ (z+1) (2 z-1) \left(4 z^2-2 z-1\right) \left(16 z^4+8 z^3-16 z^2-8
z+1\right)} {(z+1) (2 z-1) \left(4 z^2-2 z-1\right) }$$
Simplify and back to $t$
$$A=16 \cos ^4(t)+8 \cos ^3(t)-16 \cos ^2(t)-8 \cos (t)+1$$
Now, using
$$\cos^2(t)=\frac 12(1+\cos(2t))$$
$$\cos^3(t)=\frac 14 (3 \cos (t)+\cos (3 t))$$
$$\cos^4(t)=\frac{1}{8} (3+4 \cos (2 t)+\cos (4 t))$$
Replace, make $t=2x$ to find the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$ Prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$
It is clear, that:
*
*$x^2 \cos^2 (x) \to 0, x \to \frac{\pi}{2}$, since cosine goes to 0.
*$x^2 \cos^2 (x) \to 0, x \to 0$, since $x$ goes to 0.
Of course, we can split $(0, \frac{\pi}{2})$ into $A_1 = (0, \frac{\pi}{4}]$, $ A_2 = [\frac{\pi}{4}, \frac{\pi}{3}]$ and $A_3 = [\frac{\pi}{3}, \frac{\pi}{2})$
(sorry for intersection, just don't want to write limits. It doesn't affect anything)
Then we take max and min values from $A_i$ and evaluate max value at $x^2$ and min value at $\cos^2 x$:
$A_1$: $\max(A_1)^2 = \frac{\pi^2}{16} \in (0, 1); \cos^2 \min(A_1) = 1 \implies \max(A_1)^2 \cos^2 \min(A_1) < 2$
$A_2$: $\max(A_2)^2 = \frac{\pi^2}{9} \in (1, 2) ; \cos^2 \min(A_2) = \frac{1}{2} \implies
\max(A_2)^2 \cos^2 \min(A_2) < 2$
$A_3$: $\max(A_3)^2 = \frac{\pi^2}{4} \in (2, 3) ; \cos^2 \min(A_3) = \frac{1}{4} \implies
\max(A_3)^2 \cos^2 \min(A_3) < 2$
Actually, here we can see that upper bound is even lower (not bigger that 1).
Unfortunately, I am not satisfied with this writing (somehow I am sure there are much better proofs).
But what are they? Thank you.
| I shall give you the Hint:
*
*Assume that $f(x)=x^2\cdot \cos^2x$, and then prove, $$f'(x)=2x\cos x(\cos x-x\sin x)$$
*Find the maxima of $f(x)$ between $\left[0,{\pi\over2}\right]$. As $f(0)=f({\pi\over2})=0$ so there must be peak between $\left[0,{\pi\over2}\right]$. Hence, $f'(x)=0$ has the maxima. And for your inequality to be true maxima$<2$
| {
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Find all $x$ for that $x^2 + (x+1)^2$ is a square How to find all natural $x$ for that $x^2 + (x+1)^2$ is a perfect square?
| Suppose $x^2 + (x+1)^2 = y^2$. We can rewrite it as $(2x+1)^2 + 1 = 2y^2$ or $(2x+1)^2 - 2y^2 = -1$.
If $z=2x+1$ then we have $z^2 - 2y^2 = -1$. This is Pell's equation.
Wikipedia article shows how to solve it.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve a cyclic quintic in radicals? Galois theory tells us that $\frac{z^{11}-1}{z-1} = z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be $\zeta^1,\zeta^2,\ldots,\zeta^{10}$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group $[2,4,8,5,10,9,7,3,6,1]$ we can partition into the five subgroups of conjugate pairs $[2,9]$,$[4,7]$,$[8,3]$,$[5,6]$,$[10,1]$.
Now put $q_1 = \zeta^2+\zeta^9$,$q_2 = \zeta^4+\zeta^7$,$q_3 = \zeta^8+\zeta^3$,$q_4 = \zeta^5+\zeta^6$,$q_5 = \zeta^{10}+\zeta^1$. So if we can solve the quintic $(q - q_1)(q - q_2)(q - q_3)(q - q_4)(q - q_5) = q^5 + q^4 - 4q^3 - 3q^2 + 3q + 1 = 0$ we would just be left to solve a few quadratic equations.
Now pari/gp tells me this quintic has the cyclic group C(5):
? polgalois(x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1)
%1 = [5, 1, 1, "C(5) = 5"]
I've worked through examples of solving quadratic and cubic equations based on the galois group but when it comes to this quintic I'm completely stumped so any advice would be tremendously helpful! Thanks.
Edit: Thanks to Robin Chapman the problem is reduced significantly. Let $\omega$ be a primitive 5th root of unity (which is easy to express in radicals), it only remains to express $(q_1 + \omega q_2 + \omega^2 q_3 + \omega^3 q_4 + \omega^4 q_5)^5$ in terms of rationals and powers of $\omega$ (and then everything can be substituted back and solved easily). We know this is possible because the term is fixed by the quintics galois group, how to actually perform this evades me but I will try to find a way.
| The solution of a solvable quintic,
$$x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$$
can be given by,
$$x = \frac{1}{5}\left(-a+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$
and the $z_i$ are the roots of the quartic. However, there is also a form where one takes a fifth root only once.
$p=11$
Let,
$$x^5 +x^4 −4x^3 −3x^2 +3x+1 = 0$$
The five roots $x_k$ for $k=0,1,2,3,4$ are,
$$x_k = \frac{-1}{5}\left(\frac{1}{\beta_k^{-1}}+\frac{1}{\beta_k^0}+\frac{11}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$
where,
$$a=\tfrac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10(5+\sqrt{5})}\right)$$
$$b=\tfrac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10(85+31\sqrt{5})}\right)$$
$$\beta_k = \zeta_5^k\,\left(\frac{ab}{11}\right)^{1/5}$$
$$\zeta_5 = e^{2\pi\,i/5}$$
$p=31$
As discussed in this post,
$$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$
$$x_k = \frac{1}{5}\left(\frac{1}{\beta_k^{-1}}-\frac{1}{\beta_k^0}+\frac{31}{\beta_k^1}+\frac{a}{\beta_k^2}+\frac{b}{\beta_k^3} \right)$$
where,
$$a=\tfrac{31}{4}\left(11+5\sqrt{5}+\sqrt{-10(25-11\sqrt{5})}\right)$$
$$b=\tfrac{31}{4}\left(-1-5\sqrt{5}+\sqrt{-10(1525-\sqrt{5})}\right)$$
$$\beta_k = \zeta_5^k\,\left(\frac{ab}{31}\right)^{1/5}$$
and so on for other prime $p=10n+1$.
| {
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| $\dfrac{k+1}2-\dfrac{k-1}2=1 \implies\dfrac{k(k+1)}2-\dfrac{(k-1)k}2=k$
$\implies \sum_{k=1}^n\Bigg(\dfrac{k(k+1)}2-\dfrac{(k-1)k}2\Bigg)=\sum_{k=1}^nk\implies\dfrac{n(n+1)}2-\dfrac{1(1-1)}2=\sum_{k=1}^nk$
$\implies\sum_{k=1}^nk=\dfrac{n(n+1)}2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it possible to 'split' coin flipping 3 ways? When flipping a coin to make important decisions in life you can flip once to choose between 2 possible outcomes. (Heads I eat cake, Tails I eat chocolate!)
You can also flip twice to choose between 4 outcomes. (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads)
Can you use a coin to choose evenly between three possible choices? If so how?
(Ignoring slight abnormalities in weight)
| Here's my method. Like voltrevo's method the expected number of tosses is also only $2\frac{2}{3}$.
In order to explain this easily, lets say there are three people named p1, p2 and p3. If the coin comes up heads on the first flip then p3 looses and p1 and p2 flip to decide who wins. However, if it comes up tails on the first flip then you keep flipping until you get either two tails in a row or two heads. If it's two tails then p3 wins, and if it's two heads then p1 and p2 flip to decide who wins again.
To see why this works, lets first find the probability that p3 loses. This can happen in the following ways where something like TTH represents tails followed by tails followed by heads. The probability of each case is also shown next to the case:
H = $\frac{1}{2^1}$
THH = $\frac{1}{2^3}$
THTHH = $\frac{1}{2^5}$
THTHTHH = $\frac{1}{2^7}$
THTHTHTHH = $\frac{1}{2^9}$
...
In each new case we see that two new flips must be added, and so the probability of any one of these cases happening (ie the probability that p3 looses) can be given by the following geometric sum:
$$\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n+1}$$
Solving this sum:
$$\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n+1}=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n}=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\boxed{\frac{2}{3}}$$
From this we see that the probability of p3 losing is $\frac{2}{3}$ which means he has a $\frac{1}{3}$ chance of winning, and there is a $\frac{2}{3}$ chance of either p1 or p2 winning. So by having p1 and p2 flip an additional time whenever p3 loses they split this $\frac{2}{3}$ chance among each other giving each person a $\frac{1}{3}$ chance of winning.
Finally, we can make a probability model for the number of tosses in a given game. Here is a table which represents the number of tosses (x), the cases which yielded them and the probability of obtaining them (p(x)).
$$\boxed{
\begin{matrix}
x & 1 & 2 & 3 & 4 & 5 & 6 & ...\\
Case & N/A & H,TT & N/A & THH, THTT & N/A & THTHH,THTHTT & ... \\
p(x) & 0 & \left ( \frac{1}{2^1}+\frac{1}{2^2} \right ) & 0 & \left ( \frac{1}{2^3}+\frac{1}{2^4} \right ) & 0 &\left ( \frac{1}{2^5}+\frac{1}{2^6} \right ) & ...
\end{matrix}
}$$
You can see that all of the even numbers have two cases and these two cases correspond to P3 loosing and P3 winning respectively. Analyzing the pattern we can make a general formula for the probability of the game ending in x flips where x is an even integer:
$$p(x)=\left ( \frac{1}{2^{x-1}}+\frac{1}{2^x} \right )$$
Using this formula, we can find the expected number of tosses.
$$\mathbb{E}(x)=\sum_{x=0}^{\infty}\left(2x\right)\left(\frac{1}{2^{\left(2x-1\right)}}+\frac{1}{2^{2x}}\right)=\sum_{n=0}^{100}\frac{n}{4^{n-1}}+\sum_{n=0}^{100}\frac{n}{2^{2n-1}}$$
To find the value of these sum, we will first consider the general case of the geometric series and take the derivative of both sides.
$$\frac{d}{dx}\sum_{n=0}^{\infty}x^n=\frac{d}{dx}\frac{1}{1-x}$$
$$\boxed{\sum_{n=0}^{\infty}nx^{\left(n-1\right)}=\frac{1}{\left(1-x\right)^2}}$$
You can then see that the above two sums are just special cases of this formula and so we get that:
$$\mathbb{E}(x)=\sum_{n=0}^{100}\frac{n}{4^{n-1}}+\sum_{n=0}^{100}\frac{n}{2^{2n-1}} = \frac{16}{9}+\frac{8}{9}=\boxed{2\frac{2}{3}}$$
So, in summary, this method should take less than 3 flips on average to evenly split a coin toss among three people.
| {
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Cartesian Equation for the perpendicular bisector of a line
Find the Cartesian equation for the perpendicular bisector of the line joining A(2,3) and B(0,6)
How do I do this?
Thank you!
| Notice, the slope of the line joining the given points $(2, 3)$ & $(0, 6)$
$$=\frac{6-3}{0-2}=\frac{-3}{2}$$ Now, the slope of the perpendicular bisector $$=\frac{-1}{\text{Slope of line}\space AB}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}$$ Now, the perpendicular bisector will pass through the mid-point of the line joining points $(2, 3)$ & $(0, 6)$ i.e. $\left(\frac{2+0}{2}, \frac{3+6}{2}\right)\equiv\left(1, \frac{9}{2}\right)$ Hence, the cartesian equation of the perpendicular bisector having slope $\frac{2}{3}$ & passing through the point $\left(1, \frac{9}{2}\right)$ is given as $$y-\frac{9}{2}=\frac{2}{3}(x-1)$$ $$\implies 6y-27=4x-4$$ $$\implies \color{blue}{4x-6y+23=0}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving separable differential equation Seems straight-forward but I've been unable to get it right. Here are my steps:
$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$
$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x$$
$$-\sqrt{28-2y} = x + c$$
$$(28-2y) = (x+c)^2$$
$$y = -1/2x^2 - cx - c^2/2 + 14$$
$$c = {-2, 10}$$
$$\Rightarrow y = -1/2x^2 - 10x - 36$$
I've checked it over many countless times but for the life of me I can't figure out why it won't work. I've tried plugging the result back into the original equation and it seems to me like it checks out if you take the negative of the square root..
| So what we want here is a particular solution to our ODE given our condition: $y(-4) = -4$
$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$
$$\Rightarrow \dfrac{dy}{dx} = \sqrt{-2y(x) + 28}$$
$$\Rightarrow \int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$u = 28-2y$
$du = -2dx$
$dx = -\dfrac{1}{2}du$
$$\Rightarrow -\dfrac{1}{2}\int {1 \over \sqrt{u} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$$\Rightarrow -\dfrac{1}{2}\int {u^{-\dfrac{1}{2}}} \hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$$\Rightarrow -\dfrac{1}{2}2u^{\dfrac{1}{2}} = x$$
$$\Rightarrow -\sqrt{28-2y} = x + c$$
$$\Rightarrow \sqrt{28-2y} = -c - x,~~y(-4) = -4$$
$$\Rightarrow \sqrt{28-2(-4)} = -c - (-4)$$
$$\Rightarrow \sqrt{36} = -c + 4$$
$$\Rightarrow 6 = -c + 4$$
$$\Rightarrow c = -2$$
$$\Rightarrow \sqrt{28-2y}^{2} = (-c-x)^2$$
$$\Rightarrow 28-2y = (-(-2)-x)$$
$$\Rightarrow 28-2y = (2-x)^{2}$$
$$\Rightarrow 28-2y = 4-4x+x^{2}$$
$$\Rightarrow 2y = 28-4+4x-x^{2}$$
$$\Rightarrow y(x) = \dfrac{28-4+4x-x^{2}}{2}$$
$$\Rightarrow y(x) = \dfrac{28}{2}-\dfrac{4}{2}+\dfrac{4x}{2}-\dfrac{x^{2}}{2}$$
$$\Rightarrow y(x) = 14-2+2x-\dfrac{1}{2}x^{2}$$
$$\Rightarrow y(x) = -\dfrac{1}{2}x^{2}+2x+12.$$
Hence,
$~~~~~~~~~~~~~~~~~~~~~~~~y(x) = -\dfrac{1}{2}x^{2}+2x+12$
is our particular solution found to our original first-order seperable linear ordinary differential equation. $\blacksquare$
I hope this helped out, and hopefully I did not make any mistakes to cause any type of confusion
here.
| {
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Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $ Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $
Please suggest an approach for this task.
| This is similar to what has been said by Branimir, but shows how we can extend the result to
$$\sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}.$$
We can build up the result from the identities
$${1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1)$$
$${1 \over k(k+1)(k+2)} = {1 \over 2} \left( {1 \over k(k+1)} - { 1 \over (k+1)(k+2)} \right),$$
$${1 \over k(k+1)(k+2)(k+3)} =
{1 \over 3} \left( {1 \over k(k+1)(k+2)} - { 1 \over (k+1)(k+2)(k+3)} \right), \quad \textrm{ etc...}$$
Write $S_1 = \sum_{k=1}^n {1 \over k(k+1)},$
$S_2 = \sum_{k=1}^n {1 \over k(k+1)(k+2)},$ etc
Summing for $S_1$ using (1) all terms on RHS cancel to get the classic
$$S_1 = \sum_{k=1}^n {1 \over k(k+1)} = 1 – {1 \over n+1} = {n \over n+1}.$$
We then sum the series for $S_2$ using this result obtained for $S_1,$ and so on.
| {
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For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$
| First a kooky Lemma:
For $x > 1$ then $\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$
[$\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$ iff
$1 - \frac 1 x + 1 + \frac 3 x + 2\sqrt{(1 - \frac 1 x)(1 + \frac 3 x)} > 4$ iff
$2 + \frac 2 x + 2\sqrt{(1 + \frac 2 x -\frac 2 {x^2}} > 4$ iff
$1 + \frac 2 x -\frac 2 {x^2} > 1$ iff
$\frac 2 x -\frac 2 {x^2} > 0$ ]
================
So now to answer your question:
Let $n^2 \le x < (n+1)^2$ or in other words $n^2 \le x \le n^2 + 2n$.
It's obvious:
$2n < \sqrt x + \sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = \{2n, 2n + 1\}$.
Likewise $2n = \sqrt{4n} < \sqrt{4x + 2} \le \sqrt{4n^2 + 8n + 2} < \sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $\lfloor \sqrt {4x + 2} \rfloor = \{2n, 2n + 1\}$.
====================
Case 1:
$n^2 \le x < x + 1 \le n^2 + n < n^2 + n + \frac 1 4 = (n + \frac 1 2)^2$
Then $ \sqrt x + \sqrt {x + 1} < n + \frac 1 2 + n + \frac 1 2 < 2n + 1$
So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n$.
Likewise $ \sqrt{4x + 2} \le \sqrt{4n^2 + 4(n -1) + 2} < \sqrt{4n^2 + 4n + 1} = 2(n + \frac 1 2) = 2n + 1$.
So $\lfloor \sqrt {4x + 2} \rfloor = 2n = \lfloor \sqrt x + \sqrt {x + 1} \rfloor$.
================
Case 2:
$n^2 + 2n + 1 \ge x + 1 > x \ge n^2 + n$.
Then $\sqrt{4x + 2} \ge \sqrt{4n^2 + 4n + 2} = 2(n + \frac 12) = 2n + 1$.
So $\lfloor \sqrt {4x + 2} \rfloor = 2n + 1$.
Now $\sqrt x + \sqrt {x + 1} \ge \sqrt{n^2 + n} + \sqrt{n^2 + n + 1} = \sqrt{n^2 + n + \frac 1 4 - \frac 1 4} + \sqrt{n^2 + n + \frac 1 4 + \frac 3 4}$
$= (n + \frac 1 2)[\sqrt{1 - \frac 1 {4(n + \frac 1 2)^2}} + \sqrt{1 + \frac 3 {4(n + \frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + \frac 1 2)2 = 2n + 1$.
So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n + 1 = \lfloor \sqrt{4x +2} \rfloor$.
| {
"language": "en",
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"source": "stackexchange",
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Simple Proof by induction: $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:
*
*Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid
*Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.
And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
| Here's an all-purpose technique for proving your kind of induction problems.
You are trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever n is a non-negative integer.
You have already demonstrated the base case, so here's the inductive step.
Assuming it is true for n, is it also true for n+1?
That means to say: is $(n+1)^3 + (n+2)^3 + (n+3)^3$ divisible by 9?
Well, if $[(n+1)^3 + (n+2)^3 + (n+3)^3] -[n^3 + (n+1)^3 + (n+2)^3]$ (the difference of the inductive step and the formula) is divisible by 9, so is $(n+1)^3 + (n+2)^3 + (n+3)^3$.
Why? A bit of modular arithmetic here.
If (x mod 9) - (0 mod 9) = (0 mod 9) then x = (0 mod 9). This means that if the difference is divisible by 9, and one of the numbers in our subtraction is divisible by 9, the other number must be too.
I'm not going to do the algebra on here, but you can verify that $[(n+1)^3 + (n+2)^3 + (n+3)^3] -[n^3 + (n+1)^3 + (n+2)^3]$ simplifies into $9n^2+27n+27$ which you can rewrite as $9(n^2+3n+3)$. Hence the difference is divisible by 9, and so is our inductive step.
Q.E.D.
| {
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How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.
The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.
EDIT: I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked svenkatr and Bill Dubuque solution), hence I am removing the combinatorics tag.
| Use Gauss' trick for summing a sequence: Combine the sum term-by-term with its reversed-order self, noting that the binomial coefficients are symmetric.
$$S = \sum_{r=0}^n r { n \choose r } = \sum_{r=0}^n (n-r) {n \choose n-r }$$
So,
$$\begin{eqnarray}
2 S &=& \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose n-r} \right)\\\\
&=& \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose r} \right)\\\\
&=& \sum_{r=0}^n \left( n {n\choose r} \right)\\\\
&=& n \sum_{r=0}^n {n\choose r}\\\\
&=& n 2^n \hspace{0.1in} \text{, by familiar identity}
\end{eqnarray}$$
To use a specific example, say, $n= 4$. The "trick" is deal to add the sum to its reverse, so let's look at those sums ...
$$\begin{align}
\sum_{r=0}^4 r\binom{4}{r} &\quad=\quad 0 \binom{4}{0} + 1 \binom{4}{1} + 2 \binom{4}{2} + 3\binom{4}{3} + 4\binom{4}{4} \quad= S \\
\sum_{r=0}^4(4-r)\binom{4}{4-r} &\quad=\quad 4 \binom{4}{4} + 3 \binom{4}{3} + 2 \binom{4}{2} + 1 \binom{4}{1} + 0\binom{4}{0} \quad= S
\end{align}$$
The fact that the binomial coefficients are "symmetric" says that each such coefficient in the first sum matches the one below it in the second sum. Adding the sums "vertically" (that is, term-by-term), we have
$$\sum_{r=0}^4 \left( r + (4-r) \right) \binom{4}{r} \quad=\quad 4\binom{4}{0} + 4\binom{4}{1} + 4\binom{4}{2} + 4\binom{4}{3} + 4\binom{4}{4} \quad=\quad 2 S$$
The trick has given us a common multiplier ($4$, which is to say $n$) that we factor-out:
$$n \sum_{r=0}^4 \binom{4}{r} \quad=\quad 4\left(\;\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} \;\right) \quad=\quad 2 S$$
Finally, because we know the sum of binomial coefficients is an appropriate power of $2$ (why?), this gives
$$4\cdot 2^4 = 2 S \qquad\to\qquad S = 4\cdot 2^3 = n\cdot 2^{n-1}$$
| {
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Number of digits from $1$ to $n$ Let $n$ be a natural number $k$ digits. Show that the quantity $Q$ of digits required to write the natural numbers from $1$ to $n$ is:
$Q = k(n+1) - \underbrace{111\ldots11}_{k\textrm{ digits}}$
| As this question is now 14 hours old, I will give an answer.
First sum the total number of digits of all the natural numbers $ \le k-1.$ Call this sum $S.$
There are $9 \times 10^{i-1}$ natural numbers of length $i,$ so the total number of digits of natural numbers of length $i$ is $9i \times 10^{i-1}.$ Hence
$$ S=9(1+2.10 + 3.10^2 + \cdots + (k-1)10^{k-2}) = (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}.$$
Now the first natural number with $k$ digits is $10^{k-1}$ and so there are $n- 10^{k-1}+1$ natural numbers from $10^{k-1}$ to $n$ inclusive. Each of these numbers has $k$ digits and so the number of digits required to write the natural numbers from $1$ to $n$ is
$$ k( n- 10^{k-1}+1) + S = k( n- 10^{k-1}+1) + (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}$$
$$ = k(n+1) - \frac{10^k-1}{9}.$$
| {
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What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| I tried a geometry proof for this and thought it was worth sharing it here.
I am unable to draw the figure. I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures.
Consider the triangle $ABC$. Let the angle bisector of $A$ meet $BC$ at $D$.
By angle bisector theorem, we get
$\frac{AB}{AC} = \frac{DC}{DB}$. Hence, we get $\frac{b}{c} = \frac{DC}{DB}$.
Add one to both sides, we get
$\frac{b+c}{c} = \frac{DC+DB}{DB} = \frac{a}{DB}$.
Rearranging we get $\frac{c}{DB} = \frac{b+c}{a} = \cot(\frac{A}{2})$.
A similar argument (or using the fact that $\frac{c}{DB} = \frac{b}{DC}$) yields, $\frac{b}{DC} = \frac{b+c}{a} = \cot(\frac{A}{2})$.
Now, we are almost done. (Note that if $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$, the above is true.) However, we still need to prove that for no other possibility this is satisfied.
To prove the above. Draw a perpendicular to $AB$ at $B$ and similarly draw a perpendicular to $AC$ at $C$. Let these two will intersect $AD$ at $D'$ and $D''$. Now note that $BD' = BD$ since $\frac{BD}{AB} = \cot(\frac{A}{2}) = \frac{BD'}{AB}$ (Since $D'BA$ is a right angled triangle). Similarly $CD'' = CD$.
But this is not possible unless $D' = D$ or $D'' = D$, which implies $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$
Again, I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures. The argument I hope however gives you the idea behind the method.
| {
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Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$ I am quite a baffled now, I am not getting by how it can be written that :
\begin{align*}
&\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ \\
&= \log_{10} \tan 40^\circ + \log_{10} \tan 41^\circ + \log_{10} \tan 42^\circ + \log_{10} \tan 43^\circ + \cdots + \log_{10} \tan 50^\circ
\end{align*}
Is it even valid ? If yes,how ?
| HINT: Observe that $$\log_{10} \tan{40} + \log_{10} \tan{50} = \log_{10} \sin{40} - \log_{10} \cos{40} + \log_{10} \sin{50} - \log_{10} \cos{50}$$ Now use the fact that $\sin(90 - x) =\cos{x}$ and see that they cancel out.
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AM-GM application How to show that
$$ \displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9 $$
$a_1,a_2,a_3$ are all of same algebraic sign.
| WLOG we may assume $a_{1}\leq a_{2}\leq a_{3}$.
$$\dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{1}$$
It is not hard to see that $a_{1},a_{2},a_{3}$ and $\dfrac{1}{a_{1}},\dfrac{1}{a_{2}},\dfrac{1}{a_{3}}$ are oppositely sorted
Using Rearrangement Inequality we find
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\dfrac{a_{3}}{a_{1}}\geq \dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{2}$$
$$\dfrac{a_{1}}{a_{3}}+\dfrac{a_{2}}{a_{1}}+\dfrac{a_{3}}{a_{2}}\geq \dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{3}$$
Add the 3 inequalities to get the required result.
Another way would be to apply AM-GM since on expanding the RHS we get the terms of the form $$\dfrac{x}{y}+\dfrac{y}{x}$$ which is obviously $\geq 2$
There will be 3 such pairs + the result from (1) gives us the RHS$\geq 9$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove a number is composite How can I prove that $$n^4 + 4$$ is composite for all $n > 5$?
This problem looked very simple, but I took 6 hours and ended up with nothing :(. I broke it into cases base on quotient remainder theorem, but it did not give any useful information.
Plus, I try to factor it out:
$$n^4 - 16 + 20 = ( n^2 - 4 )( n^2 + 4 ) - 5\cdot4$$
If a composite is added to a number that is a multiple of $5$, is there anything special? A hint would suffice.
Thanks,
Chan
| You can factor out $n^{4}+4$ algebraically by finding the four roots of $n^{4}+4=0$.
Since $n^{4}+4=0\Leftrightarrow n^{4}=4e^{i\pi }$, we have
$$\begin{eqnarray*}
n &=&4^{1/4}e^{i (\pi +2k\pi)/4}\quad k=0,1,2,3 \\
&& \\
n &=&\sqrt{2}e^{i\pi /4 }=1+i\quad \left( k=0\right)
\\[2ex]
n &=&\sqrt{2}e^{i 3\pi /4 }=-1+i\quad \left( k=1\right)
\\[2ex]
n &=&\sqrt{2}e^{i 5\pi/4 }=-1-i\quad \left( k=2\right)
\\[2ex]
n &=&\sqrt{2}e^{i 7\pi/4}=1-i\quad \left( k=3\right).
\end{eqnarray*}$$
Now combining the complex conjugates factors, we get
$$\begin{eqnarray*}
n^{4}+4 &=&\left( n-1-i\right) \left( n+1-i\right) \left( n+1+i\right)
\left( n-1+i\right) \\
&=&\left( \left( n+1-i\right) \left( n+1+i\right) \right) \left( \left(
n-1-i\right) \left( n-1+i\right) \right) \\
&=&\left( n^{2}+2n+2\right) \left( n^{2}-2n+2\right).
\end{eqnarray*}$$
Note: for $n>1$, $n^2+2n+2>5$ and $n^2-2n+2>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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having trouble with limit of integral How do I solve the following?
$$\lim_{x\to 0} \int_0^1 \cos\left(\frac{1}{xt}\right)\, dt$$
| Here's another take. Since $\cos \left(\frac{1}{xt}\right)$ is an even function of $x$, we can take the limit from the right or from the left, and the result is the same. So assume $x > 0$. Then apply the $u = 1/(xt)$ substitution as in joriki's answer, followed by integration by parts in the opposite direction from joriki's answer. Rewriting the result in terms of the sine integral produces the output from Wolfram Alpha mentioned by Eivind and Sivaram:
$$ \lim_{x \to 0} \left(\frac{1}{x} \int_0^{1/x} \frac{\sin u}{u} du - \frac{\pi}{2x} + \cos \left(\frac{1}{x}\right)\right).$$
Now, apply the following known series expansion for the sine integral $\int_0^x \frac{\sin u}{u} du$ (valid for large values of $x$):
$$\int_0^x \frac{\sin u}{u} du = \frac{\pi}{2} -\frac{\cos x}{x} \left(1 - \frac{2!}{x^2} + \frac{4!}{x^4} \pm \cdots \right) - \frac{\sin x}{x} \left(\frac{1}{x} - \frac{3!}{x^3} \pm \cdots \right).$$
After simplifying, we are left with
$$\lim_{x \to 0} \left(-\cos \left(\frac{1}{x}\right) \left(- 2! x^2 + 4! x^4 \pm \cdots \right) - \sin\left(\frac{1}{x}\right) \left(x - 3! x^3 \pm \cdots \right) \right).$$
Since $\cos \left(\frac{1}{x}\right)$ and $\sin \left(\frac{1}{x}\right)$ are both bounded by $-1$ and $1$, the limit is $0$.
In addition, we can see that the dominant term in the limit is $- x\sin\left(\frac{1}{x}\right)$, which is clearly bounded by $|x|$, as conjectured by Sivaram in the comments.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$ $\displaystyle\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}\,\mathrm{d}x}$
I used partial fractions, solved $A = 2, C = 3$.
$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$
\begin{align*}
&8x^4+15x^3+16x^2+22x+4\\
&\quad = A(x+1)^2(x^2+2)+B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2
\end{align*}
Substitute in $x=0$ to get $4=A(1)(2)$, so $A = 2$
$$6x^4+11x^3+10x^2+14x = B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2$$
Substitute in $x=-1$ to get
$$6-11+10-14 = C(-1)(1+2)$$
so $-9=-3C$, thus $C=3$.
Leaving me what I have below:
Which brings me to where I am currently stuck.
$$6x^4 +8x^3 +10x^2+8x = B(x)(x+1)(x^2+2) + (Dx + E) (x) (x+1)^2$$
Is the next best move to use substitution to solve for $B$?
| Below I show how the Heaviside cover-up method generalizes to handle nonlinear denominators. With numerator $\rm\:f(x)\ =\ 8x^4+15x^3+16x^2+22x+4\:,\: $ the undetermined partial fraction is
$$\rm\frac{f(x)}{x(x+1)^2(x^2+2)}\ =\ \frac{a}{x}\, + \,\frac{b\, (x+1) + c}{(x+1)^2}\, + \,\frac{d\, x+e}{x^2+2}$$
To find $\,\rm d\,x+e\,$ in the $\rm\: x^2+2\ $ fraction, clear denominators and collect factors of $\rm\: x^2 + 2\: $
$$\rm f(x)\ \ =\ \ x\ (x+1)^2\ (d\, x +\: e)\, +\, (x^2+2)\ g(x)\ ,\quad some\ \ g(x) \in \mathbb Q[x]$$
Evaluating this $\rm\bmod\, x^2 + 2,\ $ i.e.$\:$ iteratively applying the rewrite rule $\rm\ x^2 \to -2\,\ $ yields
$$\rm - 8\, x + 4\ =\: -(4\, d + e)\, x\, +\, 2\, d - 4\, e\ \iff \ d=2,\ e=0 $$
Notice this method amounts to ignoring ("covering") all the undetermined partial fractions having denominator different (i.e. coprime) from the current denominator $\rm\:p(x) = x^2+2\:$ (i.e. having different roots) then evaluating what remains at the roots of $\rm\,p(x)\,$ or, equivalently, evaluating it $\rm\bmod p(x).\,$ To avoid computing inverses $\rm\bmod p(x)\:$ we scale to clear denominators before evaluating. This is simply the higher-degree analog of the classical Heaviside method - where covering up and evaluating at $\rm\: x = r\:$ is equivalent to evaluating modulo $\rm\:x-r\:$.
Using the same method we can solve for the numerator of the $\rm\ (x+1)^2\,$ fraction
$$\rm f(x)\ =\ x\, (x^2+2)\, (b\, (x+1) + c)\, + \, (x+1)^2\ h(x)\ ,\quad some\ \ h(x) \in\mathbb Q[x]$$
Evaluating it mod $\rm\, (x+1)^2,\, $ i.e. iteratively applying rewrite rule $\rm\, x^2 \to\, -2\ x - 1\, $ yields
$$\rm 3\ x - 6\ =\ (5\, c - 3\, b)\, x\, +\, 2\, c - 3\, b\ \iff\ c = 3,\ b = 4 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/23484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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What is the remainder of $16!$ is divided by 19? Can anyone share me the trick to solve this problem using congruence?
Thanks,
Chan
| A more explicit answer would go like this:
By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$
$$18! = (18\cdot 17)(16!)$$
Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$
But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$
Then it follows that $(18\cdot 17)(16!)$ is congruent to $((-1)\cdot (-2))(16!)$ is congruent to $-1 \pmod{19}$
Simplifying, we get, $2\cdot(16!)$ is congruent to $-1 \pmod{19}$. But we need the remainder of $16!$, not $2\cdot 16!$. and $-\frac12$ isn't a valid answer.
Then also notice that $18$ is congruent to $-1 \pmod{19}$. Then $2\cdot 16!$ is congruent to $18 \pmod{19}$ by the fact that if $a$ is congruent to $b \pmod{p}$, then $b$ is congruent to $a \pmod{p}$
So we have that $2\cdot 16!$ is congruent to $18 \pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \pmod{19}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ $$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$$
$$\frac{dy}{dx}=\frac{5}{6}x^{4}-\frac{3}{10x^{4}}$$
squaring this $$=\frac{25}{36}x^{8}+\frac{9}{100x^{8}}$$
Plugging into the formula $$ds=\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}$$
$$\int_{1}^{2}\sqrt{1+\frac{25}{36}x^{8}+\frac{9}{100x^{8}}}$$
Is this correct so far? And how would I go about evaluating this integral.
| As Américo Tavares pointed out, you are missing the term $-1/2$ due to the crossterm when squaring $d y/ dx$. The integral you want to solve is
$$\int_1^2 dx \,\sqrt{\frac{1}{2}+ \frac{25 x^8}{36} + \frac{9}{100 x^8}}
= \int_1^2 dx \, \sqrt{\frac{(9+25 x^8)^2}{900 x^8}}$$
$$ = \int_1^2 dx \,\frac{9+25 x^8}{30 x^4} = \left[-\frac{1}{10 x^3} + \frac{x^4}{6} \right]_{x=1}^2 = \frac{1261}{240}. $$
I hope every step is reproducible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of Sum over Integer Lattice Does the sum
$$\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)$$
converge pointwise or even uniformly for $\varepsilon < |{\bf x}| < 1-\varepsilon$?
| It seems to converge conditionally, with convergence if you sum over a sphere centered at the origin and then let the radius of the sphere increase without bound. Let ${\bf x}=(0,0,d)$ and ${\bf z}=(x,y,z)$ (sorry for reusing $x$). Then $$\begin{align}\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)&=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left(\frac{1}{x^2+y^2+(z-d)^2}-\frac{1}{x^2+y^2+z^2}\right) \\ &=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \frac{(z-d)^2-z^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}\\ &=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \frac{-2dz+d^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}\end{align}$$
Now we add together the contributions from the points $(x,y,z)$ and $(x,y,-z)$ so the sum is over the upper half space $$\begin{align}&=\sum \frac{-2dz+d^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}+\frac{2dz+d^2}{(x^2+y^2+(z+d)^2)(x^2+y^2+z^2)} \\ & =\sum \frac{-8d^2z^2+d^2(x^2+y^2)}{(x^2+y^2+(z-d)^2)(x^2+y^2+(z+d)^2)(x^2+y^2+z^2)}\end{align}$$
Far from the origin, where convergence is determined, this goes down at least as fast as $\frac{1}{r^4}$ and there are $r^2$ points in a spherical shell at a given radius, so we have a sum of $\frac{1}{r^2}$, which converges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to evaluate an expresssion $$\frac{(z x^{-1} y)^5 y^5}{x^{-4} z^{-4}}$$
How would I evaluate this if $x = 10$, $y = -3$ and $z = 3$ ? I would like a step-by-step solution to help me fully understand it.
| First, use the fact that $x^{-n}=\frac{1}{x^n}$ to find $$\frac{(z x^{-1} y)^5 y^5}{x^{-4} z^{-4}}=(zx^{-1}y)^5 \cdot \frac{y^5}{x^{-4}z^{-4}}=\left(\frac{zy}{x}\right)^5\cdot x^4z^4y^5 $$ Next, $$\left(\frac{zy}{x}\right)^5\cdot x^4z^4y^5=\frac{z^5y^5}{x^5}\cdot x^4z^4y^5=z^9y^{10}x$$ Can you finish from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?
Thanks,
| $$\begin{aligned}
2x^3+7x-4 \equiv 0 ( mod 5)\\
2x^3+2x-4 \equiv 0 ( mod 5)\\
2(x-1)(x^2+x+2) \equiv 0 ( mod 5)
\end{aligned}$$
Now you have two solutions $x-1 \equiv 0 (mod5)$ or $x^2+x+2 \equiv 0 (mod5)$.
You can continue and verify if $x \equiv 1 ( mod 5)$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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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.