Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Equations of the sides an equilateral triangle with centroid at the origin and one side is $x+y=1$
An equilateral triangle has its centroid at the origin and one side is $x+y=1$. Find the equations of the other sides.
My Attempt
$$
OD=\frac{1}{\sqrt{2}}\implies OC=\sqrt{2}\implies C=(-1,-1)\\
m_{AB}=m_1=-1\implies\tan60=\sqrt{3}=|\frac{m+1}{1-m}|\\
\sqrt{3}-m\sqrt{3}=m+1\quad\text{or}\quad m\sqrt{3}-\sqrt{3}=m+1\\
m(1+\sqrt{3})=\sqrt{3}-1\quad\text{or}\quad m(\sqrt{3}-1)=\sqrt{3}+1\\
m=\frac{\sqrt{3}-1}{\sqrt{3}+1}\quad\text{or}\quad m=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\
y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)\quad\text{or}\quad y+1=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x+1)
$$
But, my reference gives the solutions
$$y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)$$
and
$$y\pm1=3+\sqrt{3}(x-1)$$
So, what are the actual solution to the problem and the easiest way to solve it?
| An easy way to solve it is as follows:
Because the side $x+y=1$ makes the 135° angle with the x-axis and the triangle is equilateral, the other two sides make the 135° - 60° = 75° and 135° - 120° = 15° angles with the x-axis, respectively.
Therefore, their slopes are
$$ \tan(75^\circ) = 2+\sqrt{3}$$
$$ \tan(15^\circ) = 2-\sqrt{3}$$
and their equations passing the vertex C = (-1,-1) are
$$y+1=(2\pm \sqrt{3})(x+1)$$
Yours are the same after rationalizing the denominators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the convergence of the series $\sum \frac{n^{n-2}}{e^n n!}$ Show that the series $\sum_{n=1}^{\infty} \frac{n^{n-2}}{e^n n!}$ is convergent.
I tried to use root test but it yields 1 which makes the test indecisive . Any other approach ?
| As @robjohn pointed out in the comments theorem 4 from this answer suggest
$$
1 + \frac{1}{12\left(n + \frac{1}{2}\right)}
\le \frac{n! e^n}{n^n \sqrt{2 \pi n}}
\le 1 + \frac{1}{12\left(n - \frac{1}{3}\right)}.
$$
Simplifying the sums gives
$$
\frac{12n + 7}{12n + 6}
\le \frac{n! e^n}{n^n \sqrt{2 \pi n}}
\le \frac{12n - 3}{12n - 4}
$$
Now use $a \le b \le c \iff \frac{1}{c} \le \frac{1}{b} \le \frac{1}{a}$ and divide every term by $n^{2} \sqrt{2 \pi n}$ to obtain
$$
\frac{12n - 4}{(12n - 3) \sqrt{2 \pi} \cdot n^{\frac{5}{2}}}
\le \frac{n^{n - 2} e^{-n}}{n!}
\le \frac{12n + 6}{(12n + 7) \sqrt{2 \pi} \cdot n^{\frac{5}{2}}}
\le \frac{1}{\sqrt{2 \pi}} \cdot n^{-\frac{5}{2}}
$$
Now summing over all $n > 0$ and using the $p$-series and comparison test yields the convergence.
Using Stirlings approximation $n! \sim \sqrt{2 \pi n} \cdot n^n e^{-n}$ we have
$$
\sum_{n = 1}^{\infty} \frac{n^{n - 2} e^{-n}}{n!}
\sim \sum_{n = 1}^{\infty} \frac{n^{n - 2} e^{-n}}{\sqrt{2 \pi n} \cdot n^n e^{-n}}
= \sum_{n = 1}^{\infty} \frac{1}{\sqrt{2 \pi n} \cdot n^2}
= \sum_{n = 1}^{\infty} \frac{n^{-\frac{5}{2}}}{\sqrt{2 \pi}},
$$
which again converges because of the $p$-series test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The polynomial $x^3 + 2ax^2 + (2a^2 + b)x + c$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
The polynomial $$\large x^3 + 2ax^2 + (2a^2 + b)x + c$$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
I'm uncertain of how to prove this.
If $b$ is a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$(2a + b)x^2 + 2a^2x + (b^2 + c) = 0$$
which means the above polynomial has at least one root $\implies (a^2)^2 - (2a + b)(b^2 + c) \ge 0$
$\iff a^4 - 2ab^2 - 2ca - b^3 - bc \ge 0$
And $b$ is also a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$2ax^2 + (2a^2 + b)x + (b^3 + c) = 0$$
which means the above polynomial has at least one root $\implies (2a^2 + b)^2 - 4 \cdot 2a(b^3 + c) \ge 0$
$\iff 4a^4 + 4a^2b - 8ab^3 - 8ca + b^2 \ge 0$
But that's all I got.
| Hint.
If $b$ is a root then
$$
(x-b)(x-r_1)(x-r_2) = x^3+2ax²+(2a^2+b)x+c
$$
and comparing polynomials we get
$$
\left\{
\begin{array}{rcl}
b r_1 r_2 + c & = & 0\\
r_1 r_2 +b(r_1+r_2) & = & b(1-r_1)\\
r_1+r_2 + 2a+b & = & 0
\end{array}
\right.
$$
and after solving for $a,b,c$ we have
$$
a = \frac 12\left(1-r_1+r_2\pm\sqrt{1-r_1^2-r_2^2}\right)\\
b = -1\mp\sqrt{1-r_1^2-r_2^2}\\
c = r_1r_2\left(1\pm\sqrt{1-r_1^2-r_2^2}\right)
$$
so $a(r_1,r_2)c(r_1,r_2)$ have a surface over the circle $r_1^2+r_2^2\le 1$ represented respectively as follows:
The next step is the minima/maxima determination.
NOTE
In polar coordinates we have respectively
$$
\cases{
(a c)_1 = \frac{1}{4} \rho ^2 \sin (2 \theta ) \left(\left(\sqrt{1-\rho ^2}+1\right) (2-\rho (\sin (\theta )+\cos (\theta )))-\rho
^2\right)\\
\\
(a c)_2 = \frac{1}{4} \rho ^2 \sin (2 \theta ) \left(\left(\sqrt{1-\rho ^2}-1\right) \rho (\sin (\theta )+\cos (\theta )-2)-\rho ^2\right)
}
$$
now assuming $\max \frac{1}{4} \rho ^2 \sin (2 \theta )=\frac 14$ we have
$$
\max (a c)^2 \le 1.45
$$
with much algebraic effort we can conclude that $\max (a c)^2 \lt 0.64$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Zeros at the end of sum of factorials Find the number of zeros at the end of $15! + 16! + 17! + 18!$ ?
I know the method find the number of zeros at the end of x! where $x = { 15! , 16! , 17! ...}$ by dividing by number by $5,5^2, 5^3$ and so on .
| $1+16+16 \cdot 17+ 16 \cdot 17 \cdot 18 = 5185$. If the first non-zero digit of $15!$ is odd, then there will be just $3$ zeroes at the end from the $15!$. However, if it is even, there will be exactly one more zero. (hint: look at integers from $1$ to $9$)
Then we have to find:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 11 \cdot 12 \cdot 13 \cdot 14 \pmod {10}$$
and the result is an even number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
When does polynomial equation have a double root? I'm actually solving next task:
We have an polynomial: $x^n-ax^{n-1}+bx-1$.
And we need to find values of $a$ and $b$, such that we can divide polynomial by $(x-1)^2$ without remainder.
I tried equating polinomial to $0$ and got $a=x$ and $b=1/x$. But i'm sure there is at least one more solution. Please help me with it. Pointing to the material that would help would be greate.
| Given any polynomial $P(x)$, we have the general principle that
$(x - c)^2 \mid P(x) \tag 0$
if and only if
$(x - c) \mid P(x), \; (x - c) \mid P'(x), \tag 1$
for (0) implies
$P(x) = (x - c)^2Q(x) \tag 2$
for some polynomial $Q(x)$; then
$P'(x) = 2(x - c)Q(x) + (x - c)^2Q'(x), \tag 3$
and thus clearly
$(x - c) \mid P'(x), \tag 4$
and (1) binds. Going the other way, from (1) we may write
$P(x) = (x - c)R(x) \tag 5$
whence
$P'(x) = R(x) + (x - c)R'(x), \tag 6$
or
$R(x) = P'(x) - (x - c)R'(x); \tag 7$
since $(x - c) \mid P'(x)$ it follows that
$(x - c) \mid R(x), \tag 8$
and so we have
$R(x) = (x - c)S(x) \tag 9$
for some polynomial $S(x)$; now from (5),
$P(x) = (x - c)R(x)$
$= (x - c)(x - c)S(x) = (x - c)^2 S(x), \tag{10}$
and we see that (0) is in fact the case.
We apply this concept to
$P(x) = x^n - ax^{n - 1} + bx - 1, \tag{11}$
observing that
$(x - 1)^2 \mid P(x) \Longrightarrow P(1) = 0$ $\Longrightarrow -a + b = 1 - a + b - 1 = 0 \Longrightarrow a = b; \tag{12}$
thus (11) becomes
$P(x) = x^n - ax^{n - 1} + ax - 1, \tag{13}$
whence
$P'(x) = nx^{n - 1} - a(n - 1)x^{n - 2} + a; \tag{14}$
thus,
$P'(1) = 0 \Longrightarrow n - a(n - 1) + a = 0 \Longrightarrow n - an + a + a = 0$
$\Longrightarrow n + (2 - n)a = 0 \Longrightarrow a = \dfrac{n}{n - 2}; \tag{15}$
then
$P(x) = x^n - \dfrac{n}{n - 2}x^{n - 1} + \dfrac{n}{n - 2}x - 1, \tag{16}$
$P'(x) = nx^{n - 1} - \dfrac{n(n - 1)}{n - 2}x^{ - 2} + \dfrac{n}{n - 2}. \tag{17}$
We Check:
From (6),
$P(1) = 1 - \dfrac{n}{n - 2} + \dfrac{n}{n - 2} - 1 = 0; \tag{18}$
and from (7),
$P'(1) = n - \dfrac{n(n - 1)}{n - 2} + \dfrac{n}{n - 2} = \dfrac{n(n - 2)}{n - 2} - \dfrac{n(n - 1)}{n - 2} + \dfrac{n}{n - 2} = \dfrac{n^2 - 2n -n^2 + n + n}{n - 2} = 0. \tag{19}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sum of all entries In a Matrix of Linear Transformation So,few days before I had a Linear Algebra Test and we had following question :
Let $P_3[R]$ denote the vector space of all polynomials with degree $\le 3$. Define the linear Transformation $P_3[R]$ $\to$ $P_3[R]$ defined by :
$T(p(x))$ = $p(x)$ + $p'(x)$ + $p''(x)$ Then find the sum of all the matrix of T relative to the usual basis.
Now the basis for given vector space is :$\{1,x,x^2,x^3\}$ , Using this:
$T(1)$ = $1$
$T(x)$ = $x$ + $1$ +$0$ = $1.1$ + $1.x$ + $0.x^2$ + $0.x^3$
$T(x^2$) = $x^2$ + $2x$ + $2$ = $2.1$ + $2.x$ + $1.x^2$ + $0.x^3$
$T(x^3)$ = $x^3$ + $3x^2$ + $6x$ = $0.1$ + $6.x$ + $3.x^2$ + $1.x^3$
The Matrix is given by :
$$
\begin{pmatrix}
1 & 1 & 2 & 0\\
0 & 1 & 2 & 6 \\
0 & 0 & 1 & 3\\
0 & 0 & 0 & 1\\
\end{pmatrix}$$
Clearly sum of all entries in the matrix is $18$ but the answer given to me is $24$
Can anyone tell me what is the error in my solution ?
Thank you.
| Checking if your result is the good one is facilitated by the decomposition of $T$ into the sum of three operators : identity, derivation, derivation of order two :
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}+
\begin{pmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3\\
0 & 0 & 0 & 0\\
\end{pmatrix}+
\begin{pmatrix}
0 & 0 & 2 & 0\\
0 & 0 & 0 & 6 \\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}=\begin{pmatrix}
1 & 1 & 2 & 0\\
0 & 1 & 2 & 6\\
0 & 0 & 1 & 3\\
0 & 0 & 0 & 1
\end{pmatrix}$$
(see in particular how the diagonals are moving upwards).
This shows your result is good (entries' summation gives $18$).
It would have been $24$ if the operator had been with a supplementary term $p'''(x)$, bringing a $6$ in the upper right entry of the above matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How should I solve this definite Integral I need to solve the integral
$\displaystyle \int _{-1}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx$.
I thought I could use the identity $\arctan(x) + \operatorname{arccot}(x) = \dfrac{\pi}{2}$ to simplify the integral, But that gives the answer as $2\pi$ while the actual answer is given as:
$\pi$;
can anyone tell me how should I solve this definite integral?
Thank you.
| $\forall x\in\mathbb{R}^*,\,\arctan(x)+\arctan\left(\frac{1}{x}\right)=\mathrm{sgn}(x)\frac{\pi}{2}$, split the integral into two integrals :
$$ \int _{-1}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=\int _{-1}^{0} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx+\int _{0}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx $$
We then have $$\int _{0}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=\frac{3\pi}{2}$$ and $$\int _{-1}^{0} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=-\frac{\pi}{2}$$ so that $$ \int _{-1}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to calculate the limit $\lim_{n \to \infty}n\sin(2\pi\sqrt{1+n^2})$? The limit is: $$\lim_{n \to \infty}n\sin(2\pi\sqrt{1+n^2}),n\in \mathbb N$$
I wasn't able to convert it to any of the standard indeterminate forms so I thought of doing it analytically.
I realized that $\sin$ would only yield bounded real values which are periodic. This suggests some kind of oscillatory behaviour but I'm not sure how this could help me evaluate this limit.
Any hints or alternate approach?
| Firt note that
*
*$\sin 2\pi \sqrt{1+n^2} = \sin \left(2\pi \sqrt{1+n^2}- 2\pi n + 2\pi n\right) = \sin \frac{2\pi}{\sqrt{1+n^2} + n}$
Now, use $\lim_{x\to 0}\frac{\sin x}{x} = 1$ as follows
\begin{eqnarray*} n \sin \frac{2\pi}{\sqrt{1+n^2} + n}
& = & \frac{\sin \frac{2\pi}{\sqrt{1+n^2} + n}
}{\frac{2\pi}{\sqrt{1+n^2} + n}}\cdot \underbrace{\frac{2\pi n }{\sqrt{1+n^2} + n}
}_{\stackrel{n\to\infty}{\longrightarrow}\pi}\\
& \stackrel{n\to\infty}{\longrightarrow} & \pi
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Sum of increasing geometric series Sum of the series
$1 + \frac{1+3}{2!}+ \frac{1+3+3^2}{3!}+....... $
The series becomes $ \sum_{k=1}^{\infty} \frac {3^{k-1}}{k!}$.
How to calculate it's sum? Is it divergent due to the numerator?
| The numerator of the $k$th term: $1+3+9+\cdots+3^{k-1}=\dfrac{3^k-1}{2}$
The denominator of the $k$th term: $k!$
The sum:
$$\sum_{k=1}^\infty \dfrac{3^k-1}{2\left(k!\right)}=\dfrac{1}{2}\sum_{k=1}^\infty \dfrac{3^k}{k!}-\dfrac{1}{2}\sum_{k=1}^\infty \dfrac{1}{k!}=\dfrac{1}{2}\left(e^3-1\right)-\dfrac{1}{2}\left(e-1\right)=\dfrac{e^3-e}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $a$ and $b$ such that $\lim_{x\to25}\frac{\sqrt{x}-5}{ax+b} = \frac{1}{40}$ So I am given the following question:
Suppose
$$\lim _{x\to 25}\frac{\sqrt{x}-5}{ax+b} = \frac{1}{40}$$
Find $a$ and $b$.
I'm not exactly what to do from here, but what I did was multiplying $$\frac{\sqrt{x}-5}{ax+b}$$ by its conjugate
($\frac{ax-b}{ax-b}$), resulting in
$$\frac{100a}{625a^2-b^2} = \frac{1}{40}$$
Now I'm stuck and not sure what to do from here. Am I on a correct track?
| \begin{align}
\frac1{40}=\lim_{x\to25}\frac{\sqrt{x}-5}{ax+b}&=\lim_{x\to25}\frac{\sqrt{x}-5}{ax+b}\cdot\frac{\sqrt{x}+5}{\sqrt{x}+5}\\
\frac1{40}&=\lim_{x\to25}\frac{x-25}{(ax+b)(\sqrt{x}+5)}
\end{align}
Observe that by substituting $x=25$ into $\sqrt{x}+5$ we get $10$. Therefore, that missing factor of $4$ that we want (to make the denominator $40$) must come from $ax+b$. We also know that $x-25$ is a factor of $ax+b$. Thus, we can write
\begin{align}
ax+b&=4(x-25)\\
ax+b&=4x-100
\end{align}
Therefore, $\boxed{(a,b)=(4,-100)}$. To show that this is the solution, we can evaluate the limit:
\begin{align}
\lim_{x\to25}\frac{\sqrt{x}-5}{4x-100}&=\lim_{x\to25}\frac{\sqrt{x}-5}{4(x-25)}\cdot\frac{\sqrt{x}+5}{\sqrt{x}+5}\\
&=\lim_{x\to25}\frac{x-25}{4(x-25)(\sqrt{x}+5)}\\
&=\lim_{x\to25}\frac{1}{4(\sqrt{x}+5)}\\
&=\frac{1}{4(\sqrt{25}+5)}\\
&=\frac{1}{4(10)}\\
&=\frac{1}{40}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove $\sqrt{n!} \gt \frac{n}{2}$ $\forall$ $n \ge 1$ Prove that $$\sqrt{n!} \gt \frac{n}{2}$$ $\forall$ $n \ge 1$
My try:
I Tried using Induction:
$P(1)$ is True obviously.
Let $P(k)$ Be True Then we have
$$\sqrt{k!} \gt \frac{k}{2}$$
Now $$\sqrt{(k+1)!}=\sqrt{k+1}\sqrt{k!}\gt \sqrt{k+1}\frac{k}{2}$$
Now Since $x^2-x-1 \gt 0$ $\forall $ $x \ge 2$ We have
$$k^2 \gt k+1$$ $\forall$ $k \ge 2$
Hence
$$k \gt \sqrt{k+1}$$
Hence
$$\sqrt{(k+1)!} \gt \sqrt{k+1}\frac{k}{2}\gt \frac{1}{2}\sqrt{k+1}\sqrt{k+1}=\frac{k+1}{2}$$
Hence Proved
Is there any alternate approach?
| Assume $n \geq 2$,
Then, $$\frac{4}{3} < n$$
$$\Leftrightarrow n < 4(n-1)$$
$$\Leftrightarrow \frac{n}{4} < (n-1).$$
Then
$$\frac{n}{4} < (n-1) \leq (n-1)!$$
Multiply by $n$ and take the square root
$$ \frac{n}{2} < \sqrt{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a closed form for the integral $ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $ Is there a closed form for the integral
$$ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $$
where $ \displaystyle \mathrm{Li}_2(x) \;\;=\;\; \sum_{n=1}^\infty \frac{x^n}{n^2} $
a related sum is $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n H_n^{(2)}}{2n+1} $
| \begin{align}
I&=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=-\frac{\pi^3}{48}+2\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx\tag1\\
&=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1 x^{2n}\ dx\tag2\\
&=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\tag3\\
&=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\underbrace{\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}}_{\beta(3)=\frac{\pi^3}{32}}\tag4\\
&=\frac{5\pi^3}{48}-4\Im\sum_{n=1}^\infty\frac{(i)^nH_n}{n^2}\tag5\\
&=\frac{5\pi^3}{48}-4\left(-\frac{\pi}{16}\ln^22-\frac12\ln2\ G-\Im\operatorname{Li}_3(1-i)\right)\tag6\\
&=\frac{5\pi^3}{48}+\frac{\pi}{4}\ln^22+2\ln2\ G+4\Im\operatorname{Li}_3(1-i)
\end{align}
The related sum can be obtained from the the generating function
$$\sum_{n=1}^\infty H_n^{(q)}x^n=\frac{\operatorname{Li}_q(x)}{1-x}$$
set $q=2$ and replace $x$ with $-x^2$ we get
$$\sum_{n=1}^\infty (-1)^nH_n^{(2)}x^{2n}=\frac{\operatorname{Li}_2(-x^2)}{1+x^2}$$
now integrate both sides from $x=0$ to $1$
$$\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{2n+1}=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx$$
The index $n$ can start from $0$ as $H_{0}^{(2)}=0$
.
Explanation:
$(1)$ Integration by parts.
$(2)$ Using the identity $\arctan x\ln(1+x^2)=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}$.
$(3)$ $\int_0^1 x^{a-1}\ dx=\frac1a$
$(4)$ Using $H_n=H_{n+1}-\frac1{n+1}$
$(5)$ Using $\sum_{n=0}^\infty (-1)^n a_{2n+1}=\Im\sum_{n=1}^\infty i ^n a_n$
$(6)$ Using the generating function $\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$ .
Prove that $$\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$$
I think the best solution here is using right triangle . . . I have one too, but not pretty.
| Multiplying by $\cos(\pi/9)$, this is equivalent to
$$\sin\frac\pi 9+2\sin\frac{2\pi}9=\sqrt3\cos\frac\pi 9.\tag1$$
But
$$\sqrt3\cos\frac\pi 9-\sin\frac\pi9=
2\left(\sin\frac\pi3\cos\frac\pi 9-\cos\frac\pi3\sin\frac\pi9\right)
=2\sin\left(\frac\pi3-\frac\pi9\right)=2\sin\frac{2\pi}9$$
which proves $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find the maximum value of the expression $y=2(a-x)(x+\sqrt{x^2+b^2})$ The maximum value of the expression $y=2(a-x)(x+\sqrt{x^2+b^2})$
If we take derivative, then I am not getting anything, please guide how to proceed in such problems will be of great help. Thanks.
| Differentiation + some work:
$$
\begin{align}
y'
&=
-2\left(x+\sqrt{x^2+b^2}\right)
+2(a-x)\left(1+\frac{x}{\sqrt{x^2+b^2}}\right) \\
&=
-2\sqrt{x^2+b^2}\left(\frac{x+\sqrt{x^2+b^2}}{\sqrt{x^2+b^2}}\right)
+2(a-x)\left(\frac{\sqrt{x^2+b^2}+x}{\sqrt{x^2+b^2}}\right) \\
&=0
\end{align}
$$
this implies:
$$
\sqrt{x^2+b^2}=(a-x)
$$
and you can square both sides and work out the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$ I have this to solve :
Let $x,y,z>0$ such that $x+y+z=3$ then we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$
I try to use Jensen's inequality but the function $f(x)=\frac{x^2}{4x^3+3}$ is neither concave or convex on the interval $[0,3]$
I can't use Karamata's inequality too .
Maybe brute force is the only way to solve it .
I try also to use the derivative but it becomes a little bit difficult .
In fact my idea was to use rearrangment inequality we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{x^3}{4x^3+3}+\frac{y^3}{4y^3+3}+\frac{z^3}{4z^3+3}$$
An use the inequality of Jensen's on $[0.8,1.2]$ with $f(x)=\frac{x^3}{4x^3+3}$
So it's a partial answer .
My question is how to complete my answer or can you provide an other answer ?
Thanks a lot for sharing your knowledge and your time .
| Let $\{x,y,z\}=\{a^2,b^2,c^2\},$ where $a\geq b\geq c>0$.
Thus, $a^2+b^2+c^2=3$ and by AM-GM, C-S, Rearrangement and AM-GM again we obtain:
$$\sum_{cyc}\frac{xy^2}{4y^3+3}=\sum_{cyc}\frac{xy^2}{2y^3+1+2y^3+2}\leq\sum_{cyc}\frac{xy^2}{3y^2+4\sqrt{y^3}}\leq$$
$$\leq\frac{1}{(3+4)^2}\sum_{cyc}xy^2\left(\frac{3^2}{3y^2}+\frac{4^2}{4\sqrt{y^3}}\right)=\frac{9}{49}+\frac{4}{49}\sum_{cyc}x\sqrt{y}=$$
$$=\frac{9}{49}+\frac{4}{49}\left(\sqrt{x}\sqrt{xy}+\sqrt{y}\sqrt{yz}+\sqrt{z}\sqrt{zx}\right)\leq\frac{9}{49}+\frac{4}{49}\left(a\cdot ab+b\cdot ac+c\cdot bc\right)=$$
$$=\frac{9}{49}+\frac{4b}{49}\left(a^2+ac+c^2\right)=\frac{9}{49}+\frac{4abc}{49}+\frac{4b(3-b^2)}{49}\leq\frac{9}{49}+\frac{4\cdot1}{49}+\frac{4\cdot2}{49}=\frac{3}{7}$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$
Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
| Expanding on @Dr.SonnhardGraubner's answer, since $\sin^2x=\frac{1-\cos 2x}{2}$ we have $$\sin^4x=\frac{1-2\cos 2x+\cos^22x}{4}=\frac{3-4\cos 2x+\cos4x}{8}$$ while $$\sin^6x=\frac{1-3\cos 2x+3\cos^22x-\cos^32x}{8}=\frac{10-15\cos 2x+6\cos4x-\cos6x}{32}.$$Thus $$\int(\sin^4x-\sin^6x)dx\\=\int\frac{2-\cos2x-2\cos4x+\cos6x}{32}dx=\frac{12x-3\sin2x-3\sin4x+\sin6x}{192}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Recursive sequence problem... Consider the recursive sequence
\begin{equation*}
\begin{split}
a_{n+1} = \frac{5}{6-a_n} \quad \textit{with} \quad a_1 = 4.
\end{split}
\end{equation*}
Prove that the sequence $(a_n)$ converges and find its limit, by working out the following steps.
1. First assume that the limit $L = \lim_{n\to \infty} a_n$ exists and find its possible values.
Let $L = \lim_{n\to\infty}a_n$. Then we get
\begin{equation*}
\begin{split}
L &= \lim_{n\to\infty}a_{n+1} \\
&= \lim_{n\to\infty} \frac{5}{6-a_n} \\
&= \frac{5}{6-L}.
\end{split}
\end{equation*}
So we have $L^2-6L+5 = 0 \Longleftrightarrow (L-1)(L-5) = 0$. So the possible values of $L$ are $L = 1$ or $L = 5$.
2. Starting with the initial value $a_1 = 4$, write down the first five entries in the sequence $(a_n)$. Can you see any pattern?
We have
\begin{equation*}
\begin{split}
a_2 &= \frac{5}{6-a_1} = \frac{5}{6-4} = \frac{5}{2} = 2.5 \\
a_3 &= \frac{5}{6-a_2} = \frac{5}{6-5/2} = \frac{10}{7} \approx 1.42857 \\
a_4 &= \frac{5}{6-a_3} = \frac{5}{6-10/7} = \frac{35}{32} = 1.09375 \\
a_5 &= \frac{5}{6-a_4} = \frac{5}{6-35/32} = \frac{160}{157} \approx 1.01091 \\
a_6 &= \frac{5}{6-a_5} = \frac{5}{6-160/157} = \frac{785}{782} \approx 1.00384.
\end{split}
\end{equation*}
3. Find the real valued function $f(x)$ defining the sequence, i.e. $a_{n+1} = f(a_n)$.
This is the question I'm having trouble with. Please help!
| Hints. For $3.$ $f(x)=\frac{5}{6-x}$ with $f'(x)=\frac{5}{(6-x)^2}>0$, i.e. the function is ascending. Now $a_2=f(a_1)=\frac{5}{2}<4=a_1$. Then $f(a_2)\leq f(a_1) \Rightarrow a_3\leq a_2$ and by induction
$$a_n\leq a_{n-1} \Rightarrow f(a_n)\leq f(a_{n-1}) \Rightarrow a_{n+1}\leq a_n$$
or the sequence is descending.
Let's show that it is also bounded. From $$x\in[1,5] \Rightarrow 1\leq x \leq 5 \Rightarrow 5\geq 6-x \geq 1 \Rightarrow 1\leq \frac{5}{6-x}\leq 5$$
or
$$x\in[1,5] \Rightarrow f(x)\in[1,5]$$
Now, $a_1\in[1,5] \Rightarrow a_2=f(a_1)\in[1,5]$ and, again by induction, $a_n\in[1,5]$. So, the sequence is bounded in monotone.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all $n\in\mathbb{Z^+}$ such that $3x^n+n(x+2)-3\ge nx^2$ for all $x\in\mathbb{R}$
Find all $n\in\mathbb{Z^+}$ such that $3x^n+n(x+2)-3\ge nx^2$ for all $x\in\mathbb{R}$
This question to me is tricky and I don't know where to start. I tried to substitute $n$ with multiple values like $1$ but I couldn't find a solution. For case $n=1$ I get that $4x-1\ge x^2$ which is not true for all $x$. The fact that $x$ can be any real value is troubling for me. I also tried to use induction to prove that some cases are incorrect but couldn't due to the fact that $x$ isn't a fixed value. Any help would be appreciated.
| The solutions to your problem are the even values of $n$.
Let $f_n(x)=3x^n+n(x+2)-3-nx^2$.
If $n$ is odd, then either $n=1$ in which case the leading monomial in $f_n(x)$
is $-x^2$, or $n\geq 3$ in which case the leading monomial in $f_n(x)$
is $3x^n$. In both cases, $f(x)\to -\infty$ when $x\to -\infty$ so $f$ cannot
be nonnegative.
So suppose now that $n$ is even. We will show that $f_n(x) \geq 0$.
For $n=2$, $f_2(x)=(x+1)^2 \geq 0$. For $n=4$, $f_4(x)=((x+1)^2)(3x^2-6x+5)\geq 0$. So we may assume $n\geq 6$.
If $x\leq -1$, we can write $x=-1-y$ where $y\geq 0$, and then
$$
\begin{array}{lcl}
f_n(x) &=& f_n(-1-y) \\
&=& 3(1+y)^n +n(1-y)-3-n(1+y)^2 \\
&=& 3(1+y)^n -ny^2-3ny-3 \\
&=& 3(\sum_{j=0}^n \binom{n}{j}y^j)-ny^2-3ny-3 \\
&=& \frac{n}{2}(3(n-2)+1)y^2+3(\sum_{j=3}^n \binom{n}{j}y^j) \geq 0
\end{array}
$$
If $x\in [-1,2]$, we have $3(1+x^n) \geq 0 \geq n(x^2-x-2)$ so $f_n(x)\geq 0$.
Finally, suppose that $x\geq 2$. We can then write $x=2+z$ with $z\geq 0$, and then
$$
\begin{array}{lcl}
f_n(x) &=& f_n(2+z) \\
&=& 3(2+z)^n +n(4+z)-3-n(2+z)^2 \\
&=& 3(2+z)^n -nz^2-3nz-3 \\
&=& 3(\sum_{j=0}^n \binom{n}{j}2^{n-j}y^j)-nz^2-3nz-3 \\
&=& 3(2^n-1)+3n(2^{n-1}-1)z+n((n-1)2^{n-3}-1)z^2+ 3(\sum_{j=3}^n \binom{n}{j}2^{n-j}y^j) \geq 0
\end{array}
$$
This finishes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof that $\frac{a_n}{3^n}$ is a Cauchy sequence that converges I'm trying to understand how to proof Cauchy sequence that converges.
Given that let $a_i$ be a sequence of number such that $a_i \in \{-1, 0,1,\}$ for each i.
Let $s_n$ be a sequence that $s_n = \frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ... + \frac{a_n}{3^n}$.
Proof:
Suppose $s_n$ converge to $s$, where $\lim s_n = s$. Let $\epsilon > 0$ then there exist $N \in \mathbb{N}$ such that
$|s_n - s| < \frac{\epsilon}{2}$. Then for all $n, m \geq N$, we have
$$|s_n - s_m| = |s_n - s + s - s_m| \leq |s_n - s| + |s - s_m|$$
$$= |\frac{a_n}{3^n} - s| + |\frac{a_m}{3^m} - s| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
Hence, we proved that $s_n$ is a Cauchy sequence. Therefore, the following sequence $s_n$ converges.
Can someone verify that if I did the proof correctly?
| Assume $n>m $ ,$s = \frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ...\frac{a_m}{3^m}+... + \frac{a_n}{3^n}+\cdots$
$$|s_n - s_m| = |s_n - s + s - s_m| \leq |s_n - s| + |s - s_m|\\=
|(\frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ...\frac{a_m}{3^m}+... + \frac{a_n}{3^n}) - s| + |s - (\frac{a_1}{3} + \frac{a_2}{3^n} + \frac{a_3}{3^3} + ...\frac{a_m}{3^m})|=\\
|\frac{a_{n+1}}{3^{n+1}}+\frac{a_{n+2}}{3^{n+2}}+\frac{a_{n+3}}{3^{n+3}}+\cdots|+|\frac{a_{m+1}}{3^{m+1}}+\frac{a_{m+2}}{3^{m+2}}+\frac{a_{m+3}}{3^{m+3}}+\cdots|$$ take over from here .
$$|\frac{a_{n+1}}{3^{n+1}}+\frac{a_{n+2}}{3^{n+2}}+\frac{a_{n+3}}{3^{n+3}}+\cdots|+|\frac{a_{m+1}}{3^{m+1}}+\frac{a_{m+2}}{3^{m+2}}+\frac{a_{m+3}}{3^{m+3}}+\cdots|\leq \\
|\frac{1}{3^{n+1}}+\frac{1}{3^{n+2}}+\frac{1}{3^{n+3}}+\cdots|+|\frac{1}{3^{m+1}}+\frac{1}{3^{m+2}}+\frac{1}{3^{m+3}}+\cdots|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why does this trick for solving this equation work? The question is to
solve the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
Now if we solve $ \left | 2x-2 \right | = \left | 4x-5 \right |$ first (that is, setting the denominators equal) we find that $x=\frac{3}{2},\frac{7}{6}.$ If we now go back to substitute these values of $x$ in LHS of the original equation, we get that $$\frac{ 4x\left ( \left ( 4x-5 \right )^{2}+3 \right ) + 12x\left ( \left ( 2x-2 \right )^{2}+3 \right ) }{\left ( \left ( 4x-5 \right )^{2}+3 \right )\left ( \left ( 2x-2 \right )^{2}+3 \right )} = 6.$$
Thus, we see that $x=\frac{3}{2},\frac{7}{6}$ both satisfy the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 6.$$
However, I don't know how this is relevant to the fact that $x=\frac{1}{2},\frac{7}{2}$ are solutions to the original equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
That is, what is the relationship between the respective solutions of $\text{original LHS}=6$ and $\text{original LHS}=1$?
Please tell me the mystery behind this process, and can we use this technique on other equations like this? Thank you.
| In problems like these, the first thing to do is to clear denominators. If we do so and expand everything, we will be left with
$$16x^4-100x^3+200x^2-175x+49=0.$$
If you substitute values (with smart guesses using the rational roots theorem), you will be able to find that $x=1/2$ and $x=7/2$ are solutions. Then by the Factor theorem , $(2x-1)$ and $(2x-7)$ are factors of the polynomial on the LHS. Then use long division to factor the LHS. You should obtain
$$(4x^2-9x+7)(2x-1)(2x-7)=0,$$
and the rest of the solution is easy to complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\sum _{n=1}^{\infty } \sin \left(\pi \sqrt{n^2+1}\right)$ How can one prove
$$\sum _{n=1}^{\infty } \sin \left(\pi \sqrt{n^2+1}\right)=-\frac{1}{2}\pi Y_1(\pi )-\int_0^{\infty } \exp ^{\frac{\pi}{2} \left(t-t^{-1}\right)} (\theta(2 \pi t)-1) \, dt$$
Here $\theta$ denotes Theta function of the third kind. Correpsonding cosine case is solved here but not so helpful. Any help will be appreciated.
| We can use the Poisson summation formula after a transformation to obtain the quoted result. Defining
\begin{equation}
I(a)=\sum _{n=1}^{\infty } \sin \left(\pi \sqrt{n^2+a^2}\right)
\end{equation}
we have
\begin{equation}
I'(a)=\pi a\sum _{n=1}^{\infty } \frac{\cos \left(\pi \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}
\end{equation}
For even functions, the Poisson summation formula reads
\begin{equation}
\sum _{n=1}^{\infty }f(n)=\sum _{k=1}^{\infty }\hat f(k)+\frac{1}{2}\left( \hat f(0)-f(0) \right)
\end{equation}
where
\begin{equation}
\hat f(k)=2\int_0^\infty f(t)\cos(2\pi kt)\,dt
\end{equation}
Using the tabulated integral (3.876.2) in G&R, with $a>0$,
\begin{equation}
\int_0^\infty \frac{\cos \left(\pi \sqrt{t^2+a^2}\right)}{\sqrt{t^2+a^2}}\cos(2\pi kt)\,dt=
\begin{cases}
K_0\left( \pi a\sqrt{4k^2-1} \right) & \text{for } k>1/4\\
-\frac{\pi}{2}Y_0\left(\pi a \right) & \text{for } k<1/4
\end{cases}
\end{equation}
Then
\begin{equation}
I'(a)=2\pi a\sum _{k=1}^{\infty } K_0\left( \pi a\sqrt{4k^2-1} \right) -\frac{\pi^2 a}{2}Y_0\left(\pi a \right)-\frac{\pi}{2}\cos \pi a
\end{equation}
From here, noticing that $I(0)=0$, we can integrate by exchanging summation and integration
\begin{align}
I(s)&=2\pi \sum _{k=1}^{\infty }\int_0^s K_0\left( \pi a\sqrt{4k^2-1} \right)a\,da -\frac{\pi^2}{2}\int_0^s Y_0\left(\pi a \right)a\,da-\frac{\pi}{2}\int_0^s \cos \pi a\,da\\
&=-2s\sum _{k=1}^{\infty }\frac{K_1\left( \pi s\sqrt{4k^2-1} \right)}{\sqrt{4k^2-1}}-\frac{\pi}{2}sY_1\left( \pi s \right)-\frac{1}{2}\sin \pi s
\end{align}
We chose an integral representation for the Bessel function DLMF
\begin{equation}
K_{1}\left(z\right)=\frac{z}{4}\int_{0}^{\infty}\exp
\left(-t-\frac{z^{2}}{4t}\right)\frac{\mathrm{d}t}{t^2}
\end{equation}
The above series reads
\begin{align}
-2s\sum _{k=1}^{\infty }\frac{K_1\left( \pi s\sqrt{4k^2-1} \right)}{\sqrt{4k^2-1}}&=-\frac{\pi s}{2}\sum _{k=1}^{\infty }\int_0^\infty e^{-t+\frac{\pi^2s^2}{4t}-\frac{\pi^2s^2k^2}{t}}\frac{\mathrm{d}t}{t^2}\\
&=-\frac{\pi s^2}{2}\int_0^\infty e^{-t+\frac{\pi^2s^2}{4t}}\left( \sum _{k=1}^{\infty }e^{-\frac{\pi^2s^2k^2}{t}} \right)\frac{\mathrm{d}t}{t^2}\\
&=-\frac{\pi s^2}{4}\int_0^\infty e^{-t+\frac{\pi^2s^2}{4t}}\left( f\left(\frac{\pi^2s^2}{t} \right)-1 \right)\frac{\mathrm{d}t}{t^2}\\
&=-\frac{s^3}{2}\int_0^\infty e^{\frac{\pi s}{2}\left( u-\frac{ 1}{u} \right)}\left(f\left( 2\pi su \right)-1 \right)\mathrm{d}u
\end{align}
Where $f(\pi^2 a^2/t)=\theta_3(0,e^{-\pi^2 a^2/t})$ where $\theta_3$ is the Theta function of the third kind DLMF. The latter expression was obtained by changing $t=\pi s/(2u)$.
Finally,
\begin{equation}
\sum _{n=1}^{\infty } \sin \left(\pi \sqrt{n^2+s^2}\right)=-\frac{s^3}{2}\int_0^\infty e^{\frac{\pi s}{2}\left( u-\frac{ 1}{u} \right)}\left(f\left( 2\pi su \right)-1 \right)\mathrm{d}u-\frac{\pi}{2}sY_1\left( \pi s \right)-\frac{1}{2}\sin \pi s
\end{equation}
which seems to be numerically correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the fastest way to calculate $(x+1)^3-(x-1)^3$?
What is the fastest way to calculate $(x+1)^3-(x-1)^3$?
My teacher gives me this question, and I use 5 minutes to calculate it, and there is an error. I want to know if there is a faster method to do it.
| $\mathbf{Method\ 1}:$ $$S=(x+1)^3 - (x-1)^3 = 2((x+1)^2 + (x-1)^2 + (x+1)(x-1))$$$$=2(2(x^2+1) + x^2 -1)$$ $$=2(3x^2 + 1)$$ $$=6x^2+2$$ $\mathbf{Method\ 2:}$ $$(x+1)^3 = \displaystyle\sum_{r=0}^3\binom 3rx^r$$ $$(x-1)^3=\displaystyle\sum_{r=0}^3(-1)^r\binom 3rx^r$$ Clearly in their addition only terms with even $r$ remain. Here, they are $r = 0,2$. So the sum is$$2\left(\binom 30 + \binom 32x^2\right)$$ $$=6x^2 + 2$$ In any case we have the solution $$\boxed{S = 6x^2+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve the following system of equations: $x^2 - 3x\sqrt{y - 2} + 3y = 9$ and $x^3 + 9(3y - 7)\sqrt{y - 2} = 3x$. $(x, y \in \mathbb R)$
Solve the following system of equations $(x, y \in \mathbb R)$ $$\large \left\{ \begin{align} x^2 - 3x\sqrt{y - 2} + 3y = 9\\ x^3 + 9(3y - 7)\sqrt{y - 2} = 3x \end{align} \right.$$
I have provided a solution below where there is an obsolete step for replacing $ -3 = \dfrac{x}{z}$ but who cares.
I hope there are other solutions which are better than mine.
| Let $\sqrt{y - 2} = z$ $(z \ge 0)$, we have that $\left\{ \begin{align} x^2 - 3zx + 3z^2 = 3\\ x^3 + 9z(3z^2 - 1) = 3x \end{align} \right.$
$$ \iff 3z^2 - 3zx + x^2 - 3 = 27z^3 + x^3 - 9z - 3x = 0$$
$$ \iff 3z^2 - 3zx + x^2 - 3 = (3z + x)(9z^2 - 3zx + x^2 - 3) = 0$$
$$ \iff 3z^2 - 3zx + x^2 - 3 = 6z^2(3z + x) = 0$$
$$ \iff 3z^2 - 3zx + x^2 - 3 = \left[ \begin{align} z\\ \dfrac{x}{z} + 3\\ \end{align} \right. = 0$$
If $z = 0 \iff y = 2$ and $x^2 - 3 = 0 \iff x = \pm \sqrt 3$
If $\dfrac{x}{z} = -3 \implies \left(-\dfrac{x}{z}\right) \cdot z^2 + \dfrac{x}{z} \cdot zx + x^2 + \dfrac{x}{z} = 0 \iff \dfrac{x}{z} \cdot (2zx - z^2 + 1) = 0$
$$ \iff 2zx - z^2 + 1 = 0 \iff 2z \cdot (-3z) - z^2 + 1 = 0 \iff z^2 = \dfrac{1}{7} \iff y = \dfrac{15}{7}$$
and $x = \dfrac{\mp 3}{\sqrt 7}$. We have that $(x, y) = \left\{(\pm \sqrt 3, 2), \left(\dfrac{\mp 3}{\sqrt 7}, \dfrac{15}{7}\right)\right\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Best game strategy in draws from a bin A bin has 2 white balls and 3 black balls. You play a game as follows: you draw balls one at a time without replacement. Every time you draw a white ball , you win a dollar, but every time you draw a black ball , you loose a dollar . You can stop the game at any time.Devise a strategy for playing this game which results in an expected profit.
According to my reasoning the best strategy is not to play the game at all : the expected value at every extraction remains the same and it's always negative $$E[X]=\left(\frac{2}{5}\right)*1 +\left(\frac{3}{5}\right)*(-1) =-\frac{1}{5}$$ So since I can treat it as a sum of expectations of random variables ,the best strategy is not to play…so the best expected profit is zero dollars right?
| Assume that we have $a$ white balls and $b$ blacks balls. We can choose between two things: to play or not to play. In the first case our profit is $0$. Now assume that we choose to play. With probability $\frac{a}{a + b}$ we gain one dollar and we are left with $a - 1$ balls and $b$ black balls. Next, with probability $\frac{b}{a + b}$ we lose one dollar and we are left with $a$ white balls and $b - 1$ black balls.
This gives the following reccurence formula for $P_{a,b}$, the best expected profit we can get for $a$ white balls and $b$ black balls:
$$P_{a,b} = \max\left\{0, \frac{a}{a+b}\left(1 + P_{a-1,b}\right) + \frac{b}{a + b} \left(-1 + P_{a,b - 1}\right)\right\}$$
Now, if we have $a$ white balls and no black balls, then obviously the best we can do is to win $a$ dollars. On the other hand, if there is no white balls, then we should not play at all. I.e., $P_{a,0} = a, P_{0, b} = 0$.
Using these formulas, we obtain:
$$P_{0,0} = P_{0,1} = P_{0,2}=P_{0,3} = 0,$$
$$P_{1,0} = 1, P_{2,0} = 2,$$
$$P_{1,1} = \max\left\{0, \frac{1}{2}(1 + P_{0,1}) + \frac{1}{2}(-1 + P_{1,0})\right\} = \frac{1}{2}, $$
$$P_{2,1} = \max\left\{0, \frac{2}{3}(1 + P_{1,1}) + \frac{1}{3}(-1 + P_{2,0})\right\} = \frac{4}{3},$$
$$P_{1,2} = \max\left\{0, \frac{1}{3}(1 + P_{0,2}) + \frac{2}{3}(-1 + P_{1,1})\right\} = 0,$$
$$P_{1,3} = \max\left\{0, \frac{1}{4}(1 + P_{0,3}) + \frac{3}{4}(-1 + P_{1,2})\right\} = 0,$$
$$P_{2,2} = \max\left\{0, \frac{1}{2}(1 + P_{1,2}) + \frac{1}{2}(-1 + P_{2,1})\right\} = \frac{2}{3},$$
$$P_{2,3} = \max\left\{0, \frac{2}{5}(1 + P_{1,3}) + \frac{3}{5}(-1 + P_{2,2})\right\} = \frac{1}{5},$$
i.e., on average we can gain $1/5$ dollars in the inital game. And the strategy is as follows. Look at the numbers above. Assume that we are left with $a$ white balls and $b$ black balls. If $P_{a,b} > 0$, draw a ball, otherwise stop.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Proving that this function is not injective Let $f:\mathbb{R}^2\to\mathbb{R}^2$ given by $f(x,y)=((x+2)^2+y, 2x-3y-1)$
how do I prove that this function is not injective and thus not globally invertible?
I tried with $f(x,y)=f(u,v)$ but the system is quite impossible/hard for me.. is there another way?
| Find a case where $((x+2)^2 +y,2x -3y -1) = ((w+2)^2+ u, 2w-3u-1)$ but $(x,y)\ne (w,u)$.
Just spitballing:
If $x =0$ then we need
$(0+2)^2 + y = (w+2)^2 + u$ or
$y+4 = w^2 + 4w + 4 + u$ or $y = w^2 + 4w+u$
And $-3y-1 = 2w-3u-1$ so
$y = u-\frac 23w = w^2 + 4w + u$ so
$w^2 + \frac {14}3w=0$ If $w\ne 0$ then
$w=-\frac {14}3$.
$y = \frac {28}9 + u$ and $u$ can be anything.
So $f(0,\frac {28}9) = (2^2+\frac {28}9, -\frac {28}3-1) = (\frac {64}9,-\frac {31}3)$
And $f(-\frac {14}3,0) = ((-\frac {8}3)^2 + 0, -\frac {28}2-1)=(\frac {64}9, -\frac {31}3)$.
Thus it is not injective.
Probably more efficient ways to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate the value of $\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$ The Fibonacci sequence starts with 1, 1, 2, 3, 5, 8, 13, ... .(Start from the 3rd term,
each term is the sum of the two previous terms). Let $F_n$ be the $n$th term of this sequence. $S$ is defined as $S=\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$
Calculate the value of $S$
I have no idea how to solve this, hints aswell as solutions would be appreciated
Taken from the 2013 AITMO
| $$F_n=\frac{a^n-b^n}{\sqrt{5}},~ a+b=1, ~ab=-1,~ a,b=\frac{1\pm\sqrt{5}}{2}.$$
The required sim $$ s=\sum_{n=1}^{\infty} \frac{F_n}{2^n}= \frac{1}{\sqrt{5}} \sum_{n=1}^{\infty} \left ( \frac{a^n}{2^n}-\frac{b^n}{2^n} \right) =\frac{1}{\sqrt{5}}\left(\frac{a}{2-a}-\frac{b}{2-b}\right)= \frac{1}{\sqrt{5}}~\frac{2(a-b)}{4-2(a+b)+ab}=\frac{2 \sqrt{5}}{\sqrt{5}}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$ Consider S=$\{(x,y,z)\mid| 3x^2-4y+2y^2+2z-3=0\}$
How can I determine the volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$? That is the part of $S$ over the x-y axis.
My wrong approach was to integrate $z^2\pi=(\frac{1}{2}(3-2y^2+4y-3x^2))^2\pi$ from $z=0$ to the top $z=5/2$ but as you can see this thing isn't circular. Maybe using cylindrical coordinates?
Furthermore I am looking for equations to describe the bottom and the lateral surface of it.
For the bottom I have
$\{(x,y,z)\mid 3x^2-4y+2y^2-3\le 0, z=0\}$
And the surface:
$\{(x,y,z)\mid\frac{1}{2}(3-2y^2+4y-3x^2)=z,0<z<5/2\}$
| Rewrite the equation as
$$3x^2+2(y-1)^2+2z-5=0$$
Again, rewrite it in a standard elliptical form
$$\frac{x^2}{\frac{2}{3}(\frac52-z)}+\frac{(y-1)^2}{\frac 52 -z}=1$$
and note that the volume can be viewed as a stack of elliptical disks from 0 to 5/2 along the vertical direction. Use the area formula $\pi ab$ for an ellipse, where $a$ and $b$ are the major and minor axises, Then, the area of each disk at $z$ is
$$ \sqrt{\frac 23} \left(\frac52 -z\right) \pi$$
So, its volume integral can be simply expressed as
$$I= \pi\sqrt{\frac 23} \int_0^{5/2} \left( \frac 52 -z\right)dz
= \frac{25\pi}{8}\sqrt{\frac 23} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Convergence of series $x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$ Convergence of series $$S=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$$
When $x=\frac{1}{2}$
I used ratio test as:
$$a_{n}=\frac{\left(\frac{1}{2}\right)^{2n-1}}{2n-1}$$
Then
$$a_{n+1}=\frac{\left(\frac{1}{2}\right)^{2n+1}}{2n+1}$$
Then we get:
$$\frac{a_{n+1}}{a_n}=\frac{2n-1}{4(2n+1)}$$
Hence we get:
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0.25 \lt 1$$
Hence $\sum a_n$ Converges.
But how to find its value, i guess $\arctan x$ is not useful here
| Take derivative of $S$ to get $$S'=1+x^2+x^4+...=\frac {1}{1-x^2}$$
Integrating and you get $$S=(1/2)\ln\left|\frac {1+x}{1-x}\right|$$
At $x=1/2$ the result is $\frac {\ln 3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Combinatorics Probability: No Socks are a matched Pair? So suppose you have $16$ socks ($8$ distinct pairs) and you pull out $5$ of these without looking. What is the probability that none of the socks create a matched pair?
I know the answer is $8\over12$ by using a tree diagram, but I want to know how to do this with combinatorics, but I can't seem to get it.
We have $16\choose5$ ways of drawing $5$ socks from the $16$ socks, this is our sample space! But how do you calculate the numerator of this equation? I supposed it was $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8$, but obviously this yields a probability > $1$ and so I don't know how to proceed.
Can someone please demonstrate this to me? Thanks!
| You took order into account in your numerator but not your denominator.
Method 1: Without taking order into account.
It is true that you can select $5$ of the $16$ socks in $\binom{16}{5}$ ways. Notice that you have not taken the order of selection into account.
You then say that you can pick socks from five different pairs in $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8$ ways. However, this time you have taken order into account. For instance, if each pair of socks has a different color, choosing blue, black, grey, brown, and red in that order results in the same selection of five socks as choosing blue, brown, grey, black, and brown in that order. Since there are $5!$ orders in which you could pick the same socks, if we do not take order into account, the number of favorable cases is
$$\frac{1}{5!} \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 = \frac{16 \cdot 14 \cdot 12 \cdot 10 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 8 \cdot 14 \cdot 2 \cdot 8$$
Dividing by
$$\binom{16}{5} = \frac{16!}{5!11!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 8 \cdot 3 \cdot 14 \cdot 13$$
yields the probability
$$\Pr(\text{five different pairs}) = \frac{8 \cdot 14 \cdot 2 \cdot 8}{8 \cdot 3 \cdot 14 \cdot 13} = \frac{2 \cdot 8}{3 \cdot 13} = \frac{16}{39}$$
Method 2: Taking order into account.
We know that there are $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8$ favorable cases when order is taken into account.
If we also take the order of selection into account in counting all ways of selecting five of the sixteen socks, we obtain $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12$ possible selections.
Hence,
$$\Pr(\text{five different pairs}) = \frac{16 \cdot 14 \cdot 12 \cdot 10 \cdot 8}{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12} = \frac{10 \cdot 8}{15 \cdot 13} = \frac{2 \cdot 8}{3 \cdot 13} = \frac{16}{39}$$
Notice that these answers both agree with Sam's answer. Sam did not take order of selection into account, so there are $\binom{16}{5}$ of selecting five socks. For the favorable cases, Sam chose which five of the eight pairs of socks from which socks are drawn and one of the two socks from each of those pairs, giving
$\binom{8}{5}2^5$ favorable cases. Therefore,
$$\Pr(\text{five different pairs}) = \frac{\dbinom{8}{5}2^5}{\dbinom{16}{5}} = \frac{16}{39}$$
Personally, I prefer Sam's method, but I wanted to point out how you could correct your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Formula for $\prod\limits_{k=-\infty}^\infty \frac{1}{e}\left(1+\frac{1}{k+z}\right)^{k+z+\frac{1}{2}}$ Let $\displaystyle ~(-1)^z:=e^{i\pi z}~$ for $~z\in\mathbb{C} ~\land~\Re(z)\geq 0~$ .
$\displaystyle A(z):=\prod\limits_{k=0}^\infty \frac{1}{e}\left(1+\frac{1}{k+z}\right)^{k+z+\frac{1}{2}}~$ for $~z\in\mathbb{C}\setminus\mathbb{Z}~$ .
From that follows immediately $\displaystyle ~A(-z)\,A(z)=\prod\limits_{k=-\infty}^\infty \frac{1}{e}\left(1+\frac{1}{k+z}\right)^{k+z+\frac{1}{2}}~$ .
Question: $~$ How can we proof
$$A(-z)\,A(z)=\frac{1}{1-e^{-i2\pi z}} ~~, \hspace{5mm}\Re(z)\geq 0 \hspace{1cm} ?$$
Background:
Proof of the well-known Euler’s Infinite Product for the Sine using the Stirling formula.
$\displaystyle F(z) := \Gamma(z+1) = \left(\frac{z}{e}\right)^z\sqrt{2\pi z}\,A(z)$
Formal extension: $\enspace\displaystyle F(-z):=\left(-\frac{z}{e}\right)^{-z}\sqrt{-2\pi z}\,A(-z)$
We get $\enspace\displaystyle F(-z) = e^{i\pi\left(-z+\frac{1}{2}\right)}\left(\frac{z}{e}\right)^{-z}\sqrt{2\pi z}\,A(-z) = \frac{e^{i\pi\left(-z+\frac{1}{2}\right)}}{1-e^{-i2\pi z}}\left(\frac{e}{z}\right)^z\frac{\sqrt{2\pi z}}{A(z)} =
\frac{\sqrt{2\pi z}}{2A(z)\sin(\pi z)}\left(\frac{e}{z}\right)^z$
and therefore $\enspace\displaystyle F(-z)\,F(z) = \frac{\sqrt{2\pi z}}{2\,A(z)\sin(\pi z)}\left(\frac{e}{z}\right)^z \cdot \left(\frac{z}{e}\right)^z\sqrt{2\pi z}\,A(z) = \frac{\pi z}{\sin(\pi z)}~$ .
EDIT:
After a discussion with Diger, I think it's a problem to use the definition of $A(z)$ for $A(-z)$ . It comes from the interpretation of $(-1)^{-z}$ and $(\frac{1}{-1})^z$ . It makes sense to define:
$$A(z):=\prod\limits_{k=0}^\infty \frac{1}{e}\frac{(k+z+1)^{k+z+\frac{1}{2}}}{(k+z)^{k+z+\frac{1}{2}}}$$
EDIT 2: $~~$Solution
J. R. Quine, S. H. Heydari and R. Y. Song are discussing Zeta Regularized Products here.
On page 226, example $10$, is shown how my question could be answered.
| You can calculate $\log A(z)$ explicitly, by diferentiating with respect to $z$ two times and carrying out the summation.
You'll then obtain $$\frac{{\rm d}^2}{{\rm d}z^2} \, \log A(z) = \Psi'(z) - \frac{1}{z} - \frac{1}{2z^2} \\
\Longrightarrow \quad \log A(z) = \log \Gamma(z+1) + z - z\log z - \log \sqrt{z} + c_1 z + c_2 \, .$$
By comparing the limit value for $z \rightarrow \infty$ of the original sum (which is $0$), you will need $c_1=0$ and $c_2=-\log\sqrt{2\pi}$, so $$\log A(z) = \log \Gamma(z+1) + z - z\log z - \log \sqrt{2\pi z} \\ A(z) = \frac{\Gamma(z+1)}{\sqrt{2\pi z}} \left(\frac{e}{z}\right)^z \, .$$ I presume you can use this backwards to prove your equality. Note that the exponent $1/2$ is crucial in the sum/product. For any other value the sum/product diverges.
To be honest I still don't know what you precisely want :-(. Have you tried
\begin{align}
&\qquad \prod_{k=-N}^N \frac{1}{e} \, \frac{(k+z+1)^{k+z+1/2}}{(k+z)^{k+z+1/2}} \\
&=(1+N+z)^{N+z+1/2} (z-N)^{N-z+1/2} \frac{e^{-2N-1}}{z} \prod_{k=1}^N \frac{1}{z^2-k^2} \\
&=\left(1+\frac{z+1}{N}\right)^{N+z+1/2} N^{N+z+1/2} \cdot \left(1-\frac{z}{N}\right)^{N-z+1/2} N^{N-z+1/2} \\ &\quad \cdot (-1)^{1/2-z} \, \frac{e^{-2N-1}}{z} \cdot \frac{1}{N!^2} \prod_{k=1}^N \frac{1}{1-\left(\frac{z}{k}\right)^2} \\
&\sim \frac{(-1)^{1/2-z}}{2\pi z} \prod_{k=1}^\infty \frac{1}{1-\left(\frac{z}{k}\right)^2} \\
&= \frac{i e^{-i\pi z}}{2\sin(\pi z)} = \frac{1}{1-e^{i2\pi z}} \, ?
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt{2} +\sqrt{n}$ is irrational Suppose $\sqrt{2} + \sqrt{n}$ is rational,
so it can be represented like so: $\sqrt{2} + \sqrt{n} = {x \over y}$
then $\sqrt{2} = {{x \over y} - \sqrt{n}}$
so $\sqrt{2} = {x - y\sqrt{n}\over y}$
but $\sqrt{2}$ is irrational, which leads to a contradiction.
Does this proof make sense?
edit: fixed some mistakes
| The sum or product of rational numbers is rational.
If $\sqrt 2 + \sqrt n$ is rational, then so is $\frac{2 - n}{\sqrt 2 + \sqrt n} = \sqrt 2 - \sqrt n$.
Then $\frac{\left(\sqrt 2 + \sqrt n\right) + \left(\sqrt 2 - \sqrt n\right)}{2} = \sqrt 2$ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a^2+c^2>ab$ and $b^2>4c^2$ , for real x, show that $\frac{x+a}{x^2+bx+c^2}$ cannot lie between two limits If $a^2+c^2>ab$ and $b^2>4c^2$ , for real x, show that $\frac{x+a}{x^2+bx+c^2}$ cannot lie between two limits
My attempt is as follows:
$$y=\frac{x+a}{x^2+bx+c^2}$$
$$yx^2+byx+yc^2=x+a$$
$$yx^2+x(by-1)+yc^2-a=0$$
As x is real,so
$$D>=0$$
$$(by-1)^2-4y(yc^2-a)>=0$$
$$b^2y^2+1-2by-4y^2c^2+4ay>=0$$
$$(b^2-4c^2)y^2+2(2a-b)y+1>=0$$
As it is given $b^2>4c^2$, it means parabola is upwards, now this parabola will not lie between two limits if it does not cut x-axis at two distinct points.
So if $D<=0$ then parabola $(b^2-4c^2)y^2+2(2a-b)y+1>=0$ will not cut x-axis at two distinct points.
So lets calculate D for the equation $(b^2-4c^2)y^2+2(2a-b)y+1=0$
$$D=4(4a^2+b^2-4ab)-4(b^2-4c^2)$$
$$D=4(4a^2+4c^2-4ab)$$
$$D=16(a^2+c^2-ab)$$
But I am getting $D>0$ as $a^2+c^2>ab$
I am getting totally reversed result. What mistake am I doing here. Please help me.
| $x,y$ are reak.
Let $$y=\frac{x+a}{x^2+bx+c^2}~~~~(1)$$. Due to reality of $x$ we impose $B^2 \ge 4AC$ on the quadratic of $x$ from (1). We get
$$(b^2-4c^2) y^2 +y(4a-2b)+1 \ge 0, \forall ~ y\in R.~~~(2) $$
A quadratic $Az^2+Bz+C \ge 0, \forall Z \in R,~ then~. A>0$ and $B^2\le 4AC$ Applying this
to the quadratic (2). Hence, we get $$b^2-4c^2 >0~~~(3)$$ and $$4(2a-b)^2 \ge 4(b^2-4c^2) \implies (a^2+c^2) \ge ab. ~~~(4)$$ This means $y$ can take any real positive or negative.
This is also clear from the numerator of $y$ in (1) which is bound to have two real roots as $b^2 >4c^2$ at near thse roots $y$ will take any real value implying that $y$ is un-bounded not essentially lying between two values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3+b^3)$ for $a,b\in\mathbb{N}$. In other words, solutions to the Diophantine Equation:
$a^3+b^3=dc^3$ where $a,b,$ and $c$ are pairwise co-prime and $a,b,c>0$
Obviously there are some trivial solutions. If $c=1$, for example, $a$ and $b$ can be any integers, and $d$ can be chosen to simply be $a^3+b^3$.
By requiring that $a,b,$ and $c$ are pairwise co-prime and that $c\not=1$, we eliminate the trivial solutions, and what remains is of much more interest.
For $a,b,c,d\le20$, there are 5 solutions:
$4^3+5^3=7*3^3$
$2^3+7^3=13*3^3$
$1^3+8^3=19*3^3$
$3^3+5^3=19*2^3$
$1^3+19^3=20*7^3$
Under $100$ there are $16$ solutions, as found by Mathematica.
My question about this equation: Has it been studied previously? Are there infinitely many primitive solutions (which it seems like there are)? If so, can they be parametrized?
| There are infinitely many parametric solutions for arbitrary c.
We use simple identity below.
$$(c^3n+c^3-b)^3+b^3=c^3(n+1)(c^6n^2+2c^6n-3c^3nb+3b^2+c^6-3c^3b)$$
Let $a=c^3n+c^3-b, d=(n+1)(c^6n^2+2c^6n-3c^3nb+3b^2+c^6-3c^3b)$.
Hence we get the parametric solutions of $a^3+b^3=dc^3$.
We show the examples only for $c=2,3,4,5.$
$$(8n+7)^3+1^3 = 2^3(n+1)(64n^2+104n+43)$$
$$(27n+26)^3+1^3 = 3^3(3n+3)(243n^2+459n+217)$$
$$(64n+61)^3+3^3 = 4^3(n+1)(4096n^2+7616n+3547)$$
$$(125n+124)^3+1^3 = 5^3(n+1)(15625n^2+30875n+15253)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Bound for sum of fractional parts of $\sqrt{k}\, \alpha$ Let us consider the sum
$$C_\alpha(n)=\sum_{1\leq k\leq n} \left(\{ k\alpha\}-\frac{1}{2}\right)$$
where $\alpha$ is a positive irrational real number and $\{ x \}$ denotes the fractional part of $x$. The term $k \alpha$ is uniformly distributed modulo $1$. As reminded in this paper, Sierpinski showed that
$$C_\alpha(n)=o(n)$$
and stronger bounds were successively shown.
Now let us consider the sum
$$C_\alpha(n, 1/2)=\sum_{1\leq k\leq n} \left(\{ \sqrt{k} \,\alpha\}-\frac{1}{2}\right)$$
The term $(\sqrt{k}\, \alpha)$ is also uniformly distributed modulo $1$, and an $O(\sqrt{n})$ bound could be hypothesized, as cited in some comments of this previous question. I wonder whether this bound (or some stronger ones) can be formally proven, and whether there are references for this.
| First let us show that no stronger bound than $O(\sqrt{n})$ can hold. Consider a large $m$, and put
$$N_1 = N_1(m) = \biggl\lceil\biggl(\frac{m}{\alpha}\biggr)^2\biggr\rceil,\qquad N_2 = N_2(m) = \biggl\lfloor\biggl(\frac{m + \frac{1}{3}}{\alpha}\biggr)^2\biggr\rfloor\,.$$
Then $N_2 - N_1 \approx \frac{2m}{3\alpha^2} \approx \frac{2}{3\alpha}\sqrt{N_1} \approx \frac{2}{3\alpha}\sqrt{N_2}$, and for $N_1 \leqslant k \leqslant N_2$ we have $0 \leqslant \alpha\sqrt{k} \leqslant \frac{1}{3}$, so
$$C_{\alpha}(N_2,1/2) - C_{\alpha}(N_1,1/2) \leqslant -\frac{1}{6}(N_2 - N_1) \approx -\frac{1}{9\alpha}\sqrt{N_j}$$
for $j \in \{1,2\}$. Hence at least one of $C_{\alpha}(N_1(m),1/2) > \frac{1}{30\alpha}\sqrt{N_1(m)}$ and $C_{\alpha}(N_2(m),1/2) < -\frac{1}{30\alpha}\sqrt{N_2(m)}$ holds for all sufficiently large $m$.
To show the $O(\sqrt{n})$ bound, put $M = \lfloor \alpha \sqrt{n}\rfloor$. There are $O(\sqrt{n})$ values $k$ for which $\alpha\sqrt{k} \geqslant M$, the corresponding terms hence contribute at most $O(\sqrt{n})$. For $1 \leqslant m \leqslant M$, the sum over those $k$ for which $m-1 \leqslant \alpha\sqrt{k} < m$ is bounded by a constant (depending on $\alpha$), and there are $M = O(\sqrt{n})$ such sums, hence we have an overall $O(\sqrt{n})$ bound.
It remains to show that the sums mentioned above are indeed uniformly bounded. For $m \geqslant \alpha + 1$, put
$$x = \biggl(\frac{m-1}{\alpha}\biggr)^2,\qquad y = \biggl(\frac{m}{\alpha}\biggr)^2\,.$$
The sum we want to estimate is
$$\sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)\,,$$
where $f(t) = \alpha \sqrt{t} - \bigl(m - \frac{1}{2}\bigr)$. Since $f$ is (strictly) increasing, we have
\begin{align}
\sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)
&< \int_{\lceil x\rceil}^{\lceil y\rceil} f(t)\,dt \\
&= \int_x^y f(t)\,dt + \int_{y}^{\lceil y\rceil} f(t)\,dt - \int_x^{\lceil x\rceil} f(t)\,dt
\end{align}
and
\begin{align}
\sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)
&> \int_{\lceil x\rceil - 1}^{\lceil y\rceil - 1} f(t)\,dt \\
&= \int_{x}^{y} f(t)\,dt + \int_{\lceil x\rceil - 1}^{x} f(t)\,dt - \int_{\lceil y\rceil - 1}^y f(t)\,dt\,.
\end{align}
Also $\lvert f(t)\rvert \leqslant \frac{1}{2}$ for $x \leqslant t \leqslant y$, and $f(y+1) - f(y) = \alpha (\sqrt{y+1} - \sqrt{y}) = \frac{\alpha}{\sqrt{y+1} + \sqrt{y}} < \frac{\alpha}{2\sqrt{y}} = \frac{\alpha^2}{2m} \leqslant \frac{1}{2}\alpha^2$, $f(x) - f(x-1) = \frac{\alpha}{\sqrt{x} + \sqrt{x-1}} \leqslant \frac{\alpha}{\sqrt{x}} = \frac{\alpha^2}{m-1} \leqslant \alpha^2$. Thus $\lvert f(t)\rvert \leqslant \frac{1}{2} + \alpha^2$ for $\lceil x\rceil - 1 \leqslant t \leqslant \lceil y\rceil$ and
$$\biggl\lvert \sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k) - \int_x^y f(t)\,dt\biggr\rvert < 1 + 2\alpha^2\,.$$
The integral is easily evaluated,
$$\int_x^y f(t)\,dt = \frac{2\alpha}{3}\bigl(y^{3/2} - x^{3/2}\bigr) - \bigl(m - \tfrac{1}{2}\bigr)(y-x) = \frac{2m^2 - 2m + \frac{2}{3}}{\alpha^2} - \frac{2m^2-2m + \frac{1}{2}}{\alpha^2} = \frac{1}{6\alpha^2}\,,$$
whence
$$\biggl\lvert \sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)\biggr\rvert \leqslant 1 + 2\alpha^2 + \frac{1}{6\alpha^2}$$
for $m \geqslant \alpha + 1$.
For $m < \alpha + 1$, i.e. $m \leqslant \lceil \alpha\rceil$ there is of course a common bound (these are $\lceil\alpha\rceil$ fixed sums, independent of everything except $\alpha$), say $B_{\alpha}$, so $K_{\alpha} := \max\:\{B_{\alpha}, 1 + 2\alpha^2 + 1/(6\alpha^2)\}$ is a bound holding uniformly for all $m$.
Note that irrationality of $\alpha$ played no role here, only $\alpha > 0$ was used (and for $\alpha < 0$ little would change).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $| |y-x| - |z-y| | \leq |x-z|$ How to prove equation of the form $| |y-x| - |z-y| | \leq |x-z| x,y, z \in \Re$ ?
There are 10(?) cases:
(1) $x>y>z$ : $||y-x|-|z-y|| \le x-z$
(2) $x>z>y$ : $|y-x|-z+y \le x-z$
(3) $y>x>z$ : $|y-x-|z-y|| \le x-z$
(4) $y>z>x$ : $|y-x-|z-y||\le |x-z|$
(5) $z>x>y$ : $||y-x|-z+y|\le|x-z|$
(6) $z>y>x$ : $|y-x-z+y|\le|x-z|$
(7) $x=y>z$ : $|x-x|-|z-x|\le|x-z|$
(8) $x=z>y$ : $|y-x|-|x-y| \le |x-x|$
(9) $y=z>x$ : $|z-x|-|z-z| \le |x-z|$
(10)$x=y=z$ : $||x-x|-|x-x||\le |x-x|$
(1) $||y-x|-|z-y||\le x-z \to ||x-y|-|z-y||\le x-z \to |x-y-z+y|\le x-z \\ \to |x-z|\le x-z \to x-z \le x-z$ proved
(2) $|y-x|-z+y\le x-z \to |y-x|+y\le x \to |x-y|+y\le x \to x \le x$ proved
(3) $|y-x-|z-y|| \le x-z \to |y-x-|y-z|| \le x-z \to |y-x-y+z| \le x-z \to \\|-x+z|\le x-z \to |z-x|\le x-z \to |z-x| \le x-z \to |x-z| \le x-z \to \\ x-z\le x-z $ proved
(4) $|y-x-|z-y|| \le |x-z| \to |y-x-|y-z||\le z-x \to |-x+z| \le z-x \to \\ |-x+z| \le z-x \to |z-x| \le z-x \to z-x \le z-x $ proved
(5) $||y-x|-z+y|\le z-x \to |x-y-z+y|\le z-x \to |x-z|\le z-x \to z-x\le z-x$ proved
(6) $|y-x+y-z| \le |x-z| \to |2y-x-z|\le|x-z| \to $ still no idea
(7) $|0-|z-x||\le|x-z| \to |z-x|\le|x-z|$ proved
(8) $|y-x|-|x-y|\le|x-x| \to 0 \le 0$ proved
(9) $|z-x|-|z-z| \le |x-z| \to |z-x|\le|x-z|$ proved
(10)$||x−x|−|x−x||≤|x−x| \to 0\le0$ proved
any ideas how to proceed? what could i do next?
| By triangle inequality:
$$|y-x|=|y-z+z-x|\leq |x-z|+|z-y|$$
and
$$|z-y|=|y-x+x-z|\leq |y-x|+|x-z|.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove a reduction formula for: $\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}}$
For $n > 1\in \Bbb N$, prove that:
$$
J_n = \int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = \\
{1\over na}\left(x^{n-1}\sqrt{ax^2 + bx + c} - {b\over 2}(2n-1)J_{n-1} - c(n-1)J_{n-2}\right)
$$
I've been working on this for a while without any success. I've first tried to use the fact that:
$$
\int {P_n(x)dx\over \sqrt{ax^2 + bx + c}} = Q_{n-1}(x)\sqrt{ax^2 + bx + c} + \int {\lambda dx \over \sqrt{ax^2 + bx + c}}
$$
where $Q_{n-1}(x)$ is a polymonial of $n-1$ degree at max and coeeficients for $Q_{n-1}(x)$ and $\lambda$ are yet to be determined. Using polynomials from the problem statement we get:
$$
\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = Ax^{n-1}\sqrt{ax^2 + bx + c} + \int{\lambda dx \over \sqrt{ax^2 + bx + c}}
$$
Differente both parts of the equality, after which we need to find the coefficients:
$$
\frac{x^n dx}{\sqrt{ax^2 + bx + c}} = {d(Ax^{n-1}\sqrt{ax^2 + bx + c})\over dx} + {\lambda \over \sqrt{ax^2 + bx + c}} = \\
A(n-1)x^{n-2}\sqrt{ax^2 + bx + c} + Ax^{n-1}\frac{2ax + b}{2\sqrt{ax^2 + bx + c}} + {\lambda dx \over \sqrt{ax^2 + bx + c}}$$
Multiplying both sides by $\sqrt{ax^2 + bx + c}$ and skipping some algebraic transformation I was indeed able to get that:
$$
A = {1\over na}
$$
However, the term $\lambda$ appears to be equal to $0$. Which yields:
$$
J_n = {x^{n-1}\over na}\sqrt{ax^2 + bx + c}
$$
And this approach doesn't seem to lead anywhere. I've then tried a different technique. Let $b = 2b_0$, then:
$$
aJ_{n+2} = \int \frac{ax^{n+2}dx}{\sqrt{ax^2 + 2b_0x+c}}\\
2b_0J_{n+1} = \int \frac{2b_0x^{n+1}dx}{\sqrt{ax^2 + 2b_0x+c}}\\
cJ_n = \int \frac{cx^{n}dx}{\sqrt{ax^2 + 2b_0x+c}}
$$
Summing left and right parts we get:
$$
aJ_{n+2} + 2b_0J{n+1} + cJ_n = \\
\int \frac{(ax^2 + 2b_0x + c)x^n dx}{\sqrt{ax^2 + 2b_0x + c}}= \\
\int {x^n\sqrt{ax^2 + 2b_0x + c}} dx
$$
Integration by parts yields:
$$
u = \sqrt{ax^2 + 2b_0x + c}\\
du = {2ax + 2b_0\over 2\sqrt{ax^2 + 2b_0x + c}}dx\\
dv = x^n\\
v = {x^{n+1}\over n+1}\\
\int {x^n\sqrt{ax^2 + 2b_0x + c}} dx = uv - \int vdu = \\
= {x^{n+1}\over n+1} \sqrt{ax^2 + 2b0x + c} - {J_{n+2}\over n+2} - {J_{n+1}\over 2(n+1)}
$$
Which seems to be "the other way round". The question is what technique do I use to prove what's stated in the question section? Hopefully, I didn't make typos in the body. Thank you!
| Hint:
$$\dfrac{d(x^m\sqrt{ax^2+bx+c})}{dx}$$
$$=mx^{m-1}\sqrt{ax^2+bx+c}+\dfrac{x^m(2ax+b)}{2\sqrt{ax^2+bx+c}}$$
Now let $x^3(2ax+b)=(2x^2+dx+e)(ax^2+bx+c)-ce$
Find $d,e$ by comparing the coefficients of $x,x^2,x^3$
Integrate both sides of the first relationship wrt $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{n}+\sqrt{n+2}]$ for all integers $n\ge 1$ This is the full question
Prove that of the two equations $$ [\sqrt{n}+\sqrt{n+1}] = [\sqrt{n}+\sqrt{n+2}] \\ [\sqrt[3]{n}+\sqrt[3]{n+1}] = [\sqrt[3]{n}+\sqrt[3]{n+2}]$$ the first one holds for every positive integer $n$, but the second one does not. Note: $[x]$ denotes the greatest integer function/floor function of $x$.
For the second equation, using python I was able to find two values of $n$ between $1$ and $100$ for which the equation is false. They were $n=15$ and $n=42$. Is there a way to actually calculate these values directly from the equation. Also, I don't have any ideas for proving the first equation. Any hints?
| Assume that there is an integer $m$ such that
$$\sqrt{n}+\sqrt{n+1}<m<\sqrt{n}+\sqrt{n+2}.$$
Squaring quickly proves that
$$\lceil(\sqrt{n}+\sqrt{n+1})^2\rceil=4n+2$$
and
$$
\lceil(\sqrt{n}+\sqrt{n+2})^2\rceil=4n+4.
$$
These imply that $m^2$ must be equal to either $4n+2$ or $4n+3$, both of which are impossible by considering quadratic residues modulo $4$.
It seems to me that for all $n\ge2$ we have
$$\lfloor(\root3\of{n}+\root3\of{n+1})^3\rfloor=8n+3$$ and
$$\lfloor(\root3\of{n}+\root3\of{n+2})^3\rfloor=8n+7.$$ Whenever there is a perfect cube in between (allowing equality at the upper limit), there will be a failure with the cube roots. When $n=15$ we have $8n+3=123<5^3<127=8n+7$. When $n=42$ we have
$8n+3=339<7^3=8n+7$.
The cubes of even numbers are divisible by eight, and won't produce more examples. On the other hand for odd numbers $m$ we have $m^3\equiv m\pmod8$. Therefore the cubes of integers $m\equiv5,7\pmod8$ do give more examples. Indeed, with $m=13$ we have $m^3=8\cdot274+5$, and we get the next counterexample when $n=274$. The next counterexample comes at $n=421$ as
$8\cdot421+3<15^3=8\cdot421+7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$ $$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$$
I tried using this substitute:
$$(\tan x=u),(sin2x= \frac{2\tan x}{1+\tan^2x} =\frac{2u}{u^2+1}), (dx=\frac{1}
{u^2+1}du)$$
and after long answer I get:
$$\frac{6+3\sqrt{3}}{6+4\sqrt{3}}.\ln(\frac{\tan x+2
-\sqrt{3}}{\tan x+2+\sqrt{3}})$$
and now I can not use $$F(\frac\pi2)-F(\frac\pi6)$$ because $$\tan(\frac\pi2)=\infty$$
| I would do $u=\tan x$ too, and then your integral becomes$$\int_{1/\sqrt3}^\infty\frac3{t^2+4t+1}\,\mathrm dt.$$But$$\int\frac3{t^2+4t+1}\,\mathrm dt=\frac{\sqrt3}2\left(\log \left(-t+\sqrt{3}-2\right)-\log\left(t+\sqrt{3}+2\right)\right)$$and therefore\begin{multline}\int_{1/\sqrt3}^\infty\frac3{t^2+4t+1}\,\mathrm dt=\\=\lim_{t\to\infty}\frac{\sqrt3}2\left(\log\left(-t+\sqrt{3}-2\right)-\log\left(t+\sqrt{3}+2\right)\right)-\\-\frac{\sqrt3}2\left(\log\left(-\frac1{\sqrt3}+\sqrt{3}-2\right)-\log\left(\frac1{\sqrt3}+\sqrt{3}+2\right)\right)=\\=\frac{\sqrt3}2\log\left(\frac{5+3\sqrt3}2\right).\end{multline}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}= \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0} \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=
\lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
| Only ${x\rightarrow 0^+}$ is possible here. Let $x=y^2$, then
$$L =\lim_{x \rightarrow 0^+} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}= \lim_{y \rightarrow 0} \frac{1-\sqrt{\cos y^2}}{1-\cos y} \lim_{y \rightarrow 0} \frac{1-\sqrt{1-y^4/2}}{1-(1-y^2/2)}= \lim_{y \rightarrow 0} \frac{1-(1-y^4/4)}{1-(1-y^2/2)}$$ $$\implies L= \lim_{y \rightarrow 0}\frac{y^4/4}{y^2/2}=\lim_{y \rightarrow 0} \frac{y^2}{2}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
What are the roots of this equation? (Quadratic Equation) $\frac{x}{\sqrt{1-x^2}}=A \Rightarrow \frac{x^2}{1-x^2}=A^2$
$\Rightarrow x^2=A^2(1-x^2)=A^2-A^2x^2$
$\Rightarrow x^2+A^2x^2=A^2$
$\Rightarrow x^2(1+A^2)=A^2$
$\Rightarrow x^2(1+A^2)-A^2=0$
I have tried this;
$\Delta=b^2-4ac=-4(1+A^2)(-A^2)=4(1+A^2)(A^2)$
$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{\pm2A\sqrt{1+A^2}}{{2(1+A^2)}}$
$x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$
But I got nothing.
| So $x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$.
That's it. Go home. Eat lunch. You are done.
Well, not entirely. $\sqrt{1-x^2}$ must not equal $0$ and $1-x^2\ge 0$. Need to check that this holds for $x_{1,2}$.
$\sqrt{1-x^2} = 0 \iff x^2 = 1\iff x=\pm 1$. And $1-x^2 \ge 0 \iff x^2 \le 1 \iff |x| \le 1$. So we must show $|x_{1,2}| < 1$.
$|A| = \sqrt {A^2} < \sqrt {1+A^2}$. So $|A|\sqrt{1+A^2} < \sqrt{1+A^2}^2 =1+A^2$
So $|x_{1,2}| = \frac {|A|\sqrt{1+A^2}}{1+A^2} < \frac {1+A^2}{1+A^2} = 1$.
Also you can simplify $x_{1,2} = \frac {\pm A}{\sqrt{1+A^2}}$ but that is not required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}$ using the definition of $e$
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}.$
I know that $e^x=\lim\limits_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n,$ so I need to somehow convert the limits to this form. I also noticed that $\left(1+\frac{1}{n+1}\right)=\left(\frac{n+2}{n+1}\right)=\left(\frac{n(n+2)}{(n+1)^2}\right)\cdot\left(1+\frac{1}{n}\right).$
Thus, the limit can be rewritten as $$\lim\limits_{n\to\infty}\dfrac{\left(1+\frac{1}{n}\right)^{n^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}\cdot\left(\dfrac{(n+1)^2}{n(n+2)}\right)^{(n+1)^2}\\
=\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^{-2n}\cdot\left(1+\dfrac{1}{n}\right)^{-1}\cdot\left(1+\dfrac{1}{n^2+2n}\right)^{n^2+2n}\cdot\left(1+\dfrac{1}{n^2+2n}\right)$$
$$=\dfrac{1}{e^2}\cdot(1)\cdot e\cdot(1)=\dfrac{1}{e}$$
| Your work is correct. Just make sure you know how to prove that $e=\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$ There's a nice geometric proof using areas under the curve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
What are the solutions for the following equation? I have the following equation:
$\log_{2x}4x+\log_{4x}16x=4$
What are the solutions of this equation?
This is what I did:
Firstly, I applied the following conditions:
$2x>0 \Rightarrow x>0$
$2x \ne 1 \Rightarrow x \ne \dfrac{1}{2}$
$4x>0 \Rightarrow x>0$
$4x \ne 1 \Rightarrow x \ne \dfrac{1}{4}$
$16x > 0 \Rightarrow x > 0$
Considering all of these at once, I have: $x \in (0, + \infty) \setminus \bigg\{\dfrac{1}{2}, \dfrac{1}{4} \bigg \}$
Next, I wrote the equation like so:
$\log_{2x}(2x*2) + \log_{4x}(4x*4)=4$
$\log_{2x}2x + \log_{2x}2 + \log_{4x}4x+\log_{4x}4 = 4$
$1 + \log_{2x}2 + 1 + \log_{4x}4 = 4$
$\log_{2x}2 + \log_{4x}4 =2$
$\log_{2x}2 + 2\log_{4x}2=2$
$\log_{2x}2 + \dfrac{2}{\log_{2}4x}=2$
$\log_{2x}2 + \dfrac{2}{\log_2 2 + \log_2 2x}=2$
$\log_{2x}2+\dfrac{2}{1+\dfrac{1}{\log_{2x}2}}=2$
Then I used the substitution $\log_{2x}2=t$
$t+ \dfrac2{1+ \dfrac{1}{t}}=2$
$...$
$t^2+t-2=0$
$(t+2)(t-1)=0$
In the case that $t=-2$,
$\log_{2x}2=-2$
$(2x)^{-2}=2$
$\dfrac{1}{4x^2}=2$
$x^2= \dfrac{1}{8}$
$x = \pm \dfrac{1}{2 \sqrt{2}}$, but because of our conditions: $x = \dfrac{1}{2 \sqrt{2}}$
And in the case that $t=1$,
$\log_{2x}2=1$
$2x=2$
$x=1$
So, in the end I got: $x \in \bigg \{ \dfrac{1}{2 \sqrt{2}}, 1 \bigg \}$. The only problem with this is that it is completly wrong...
My textbook says that the correct answer should be: $\bigg [ \dfrac{1}{2 \sqrt{2}}, 2 \bigg ]$. So you see I'm not even close, I need an interval of values, not just $2$ solutions like I got. So, where did I make mistakes and what should I do to get the right answer?
| As mentioned in the comments, your answer is perfectly correct and the textbook is the one which is wrong. You can check this by substituting the values in. Another way to see that it certainly cannot be an interval of solutions is by noticing that the equation $$\log_{2x}2+\log_{4x}4=2,\tag{1}$$
which is equivalent to the given one, has a LHS which changes with $x$, and on no interval $[a,b]$ is it constant. So it is absurd that there can be an interval on which the LHS is constant at $2$.
I will say though that in questions involving the logarithm like this one, a smart general approach is to convert everything to the same base. So starting from $(1)$, write
$$\frac{1}{1+\log_2(x)}+\frac{2}{2+\log_2(x)}=2,$$
and then substitute $y=\log_2(x)$ to solve for $y$ and hence for $x$. This is slightly cleaner than your solution, but of course yours is correct too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$ Show that
$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$
cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.
My attempt is as follows:
There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-
First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=0$
Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=\pi$
Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.
| If $\frac{3 + \cos x}{\sin x} = k, k \in \mathbb R$, then $3 + \cos x = k \sin x \Rightarrow 9 + 6 \cos x + \cos^2 x = k^2 - k^2 \cos^2 x$, thus $(k^2+1)\cos^2 x + 6 \cos x + (9 - k^2) = 0$. Let $u = k^2$. Now for no value of $\cos x$ to exist, the discriminant must be less than $0$:
$$6^2 - 4(u+1)(9-u) < 0 \Rightarrow 4(u+1)(9-u) > 36$$
$$\Rightarrow -u^2 + 8u + 9 > 9 \Rightarrow u(-u+8) > 0 \Rightarrow 0 < u < 8$$
where in the last step, a sketch of the quadratic shows that it is concave up, hence the direction of the inequality.
Thus $0 < k^2 < 8$. This implies there are no values of $x$ such that $-2 \sqrt{2} < \frac{3 + \cos x}{\sin x} < 2 \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
how many 3 digit numbers are there such that at least one of its digit is 6? I know the answer using
Total-none of it is $6 = 900−648=252$
My doubt is that if we you it like
when unit digit is $6 = 9\cdot10\cdot1 = 90$
when tens place is $6 = 9\cdot1\cdot10 = 90$
when hundredth place is $6 = 1\cdot10\cdot10 = 100$
So total $= 90 + 90 + 100 = 280$
why my answer is not the same
| If exactly one digit is a $6,$ there are $1\cdot 9\cdot 9+8\cdot1 \cdot 9+8\cdot 9 \cdot 1=225$ possibilities.
If exactly two digits are a $6,$ there are $9+9+8=26$ possibilities.
If exactly three digits are $6,$ there is only one possibility.
Thus there are $252$ possibilities in total.
Just to clear things up,
If the units digit is $6,$ then you should have $8\cdot9\cdot1$ possibilities, not what you got.
If the tens digit is $6,$ then you should have $8\cdot1\cdot9=81$ possibilities.
If the hundreds digit is $6,$ then there are $1\cdot9\cdot9$ possibilities. This is because you cannot include $0$ as the hundreds digit and you must subtract $1$ from $10$ since the other digits cannot be $6.$ Also you need to add the possibilities where there is more than one $6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Cubic roots of the equation $x^3-x-2=0$ If $\alpha,\beta,\gamma$ are the cubic roots of the equation $x^3-x-2=0$, then find the value of $\alpha^5+\beta^5+\gamma^5$.
One of the root is real and two roots are imaginary.
I get three equation.
$\alpha+\beta+\gamma=0$;
$\alpha\beta+\beta\gamma+\gamma\alpha=-1$
$\alpha\beta\gamma=2$
But not able to get the requisite result.
| Note:
$$a^3=a+2 \Rightarrow a^5=a^3+2a^2=2a^2+a+2$$
Hence:
$$a^5+b^5+c^5=2(a^2+b^2+c^2)+(a+b+c)+6=\\
2((a+b+c)^2-2(ab+bc+ca))+6=\\
2(0^2-2(-1))+6=10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating the exponential of an upper triangular matrix
Find the matrix exponential $e^A$ for
$$ A = \begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix}.$$
I think we should use the proberty
If $AB = BA$ then $e^{A+B} = e^A e^B$.
We can use that
$$\begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}
+\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{bmatrix}$$
Both matrices obviously commute. But I dont know how to calculate the exponential of
$$\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
\end{bmatrix}.$$
Could you help me?
| You should decompose your matrix like this
$$\begin{equation}
\begin{pmatrix}
2 & 1 & 1 \\
0 & 2 & 1\\
0 & 0 & 2
\end{pmatrix}
=
\begin{pmatrix}
2 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\end{pmatrix}
+
\begin{pmatrix}
0 & 1 & 1\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}
\end{equation}
$$
The left one commutes with the right one and the right one is nilpotent. So it is easy to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Find the general solution of $\theta$ for which the following quadratic equation is the square of a linear function. Find the general solution of $\theta$ for which the quadratic equation
$$\left(\sin\theta\right)x^2+(2\cos\theta)x+\dfrac{\cos\theta+\sin\theta}{2}$$ is the square of a linear function.
$$D=0$$
$$4\cos^2\theta-2\sin\theta\left(\sin\theta+\cos\theta\right)=0$$
$$2\cos^2\theta-\sin^2\theta-\sin\theta\cos\theta=0$$
$$2\cos^2\theta-2\sin\theta\cos\theta+\sin\theta\cos\theta-\sin^2\theta=0$$
$$2\cos\theta(\cos\theta-\sin\theta)+\sin\theta(\cos\theta-\sin\theta)=0$$
$$(\cos\theta-\sin\theta)(2\cos\theta+\sin\theta)=0$$
$$\tan\theta=1 \text { or } \tan\theta=-2$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=\tan(2\pi-\tan^{-1}(2))$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=n\pi+2\pi-\tan^{-1}(2)$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=\pi(n+2)-\tan^{-1}(2)$$
$$\sin\theta\ne 0$$
$$\theta\ne m\pi \text { where $m \in$ I }$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { can't be integral multiple of $\pi$ as } \theta=\dfrac{\pi(4n+1)}{4}$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { is the valid solution }$$
$$\theta=\pi(n+2)-\tan^{-1}(2) \text { cannot be the integral multiple of $\pi$ as $\tan^{-1}(2)$ is not the integral multiple of $\pi$ } $$
$$\theta=\pi(n+2)-\tan^{-1}(2) \text { is the valid solution }$$
Hence $\theta=\pi(n+2)-\tan^{-1}(2) \text { or } \theta=n\pi+\dfrac{\pi}{4}$
But actual answer is $\theta= 2n \pi+\dfrac{\pi}{4} \text{or } \theta =(2n+1)\pi - \tan^{-1}(2) \text { where $n \in$ I}$
I tried to find out the mistake but didn't get any breakthrough. What am I missing.
| Your problem is that the discriminant tells you when there is a double root, not when the polynomial is a square. When $\sin \theta$ is negative the polynomial factorises in the form $-(ax+b)^2$ and is the negative of a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Computing the integral $\int_{0}^{\theta}\frac{x^{a}}{(bx^{2}+c)^{5}}dx$ How to compute the following integral:
\begin{equation}
\int_{0}^{\theta}\frac{x^{a}}{(bx^{2}+c)^{5}}dx,
\end{equation}
where $b,c,\theta>0$ and $a>2$?
Any hints appreciated.
| $\int_0^\theta\dfrac{x^a}{(bx^2+c)^5}~dx$
$=\int_0^{\theta^2}\dfrac{(\sqrt x)^a}{(bx+c)^5}~d(\sqrt x)$
$=\dfrac{1}{2}\int_0^{\theta^2}\dfrac{x^\frac{a-1}{2}}{(bx+c)^5}~dx$
$=\dfrac{1}{2}\int_0^\frac{b\theta^2}{c}\dfrac{\left(\dfrac{cx}{b}\right)^\frac{a-1}{2}}{(cx+c)^5}~d\left(\dfrac{cx}{b}\right)$
$=\dfrac{c^\frac{a-9}{2}}{2b^\frac{a+1}{2}}\int_0^\frac{b\theta^2}{c}\dfrac{x^\frac{a-1}{2}}{(x+1)^5}~dx$
$=\dfrac{c^\frac{a-9}{2}}{2b^\frac{a+1}{2}}\int_0^\frac{\frac{b\theta^2}{c}}{\frac{b\theta^2}{c}+1}\dfrac{\left(\dfrac{x}{1-x}\right)^\frac{a-1}{2}}{\left(\dfrac{x}{1-x}+1\right)^5}~d\left(\dfrac{x}{1-x}\right)$
$=\dfrac{c^\frac{a-9}{2}}{2b^\frac{a+1}{2}}\int_0^\frac{b\theta^2}{b\theta^2+c}x^\frac{a-1}{2}(1-x)^\frac{7-a}{2}~dx$
$=\dfrac{c^\frac{a-9}{2}}{2b^\frac{a+1}{2}}B\left(\dfrac{b\theta^2}{b\theta^2+c};\dfrac{a+1}{2},\dfrac{9-a}{2}\right)$ (according to https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{-3x+1}{x^2-6x-16}>0$ and $y=-2/x +1$ find the interval of y The known info $\frac{-3x+1}{x^2-6x-16}>0$
so, i find that : x not 8 nor -2. And $x \geq 1/3$
$y=-2/x +1$
For y i find that
$y > -5$ and y not 3/4 or 2.
Based on that. So interval for $ y = -5 < y < 3/4$
is it right?
What is the interval of y?
| It is obviously that for $x\to \infty $ we have a negative sign for the function $f(x) =\frac{-3x+1}{x^2-6x-16}$ which change the sign at every critical point, that is at $8,1/3$ and $-2$, since each if has an odd degree. So $x\in(-\infty,-2)\cup ({1\over 3},8)$.
Now since $x={2\over 1-y}$ we have to solve:
$${2\over 1-y}<-2\;\;\;{\rm and}\;\;\;{1\over 3}<{2\over 1-y}<8$$
From the first one we get $1>y-1$ so $\boxed{y<2}$ and from the second $1-y<6$ so $\boxed{y>-5}$ and $1<4-4y$ so $\boxed{ y<{3\over 4}}$.
Thus the final result is $y\in (-5,{3\over 4})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Place the counters on a $4\times 4$ grid such that no column or row has the same color counters
There is a $4\times 4$ grid posted on the wall. Find the number of ways of placing two identical red counters and two identical blue counters on four different squares of the grid such that no column or row has two counters of the same color.
My solution: place the first blue counter $B_1$: there are 16 ways. Next, place $B_2$: there are 9 ways.
After, I split into cases:
-The first case: $R_1$ is placed not coinciding with the cross of $B_1$ or $B_2$.
-The second case: $R_1$ coincides with exactly one cross of $B_1$ or $B_2$.
-The third case: $R_1$ coincides with the crosses of $B_1$ and $B_2$.
I find my solution is quite long and complicated. Therefore, I want to refer to another solution. Please everyone help me.
| Your approach is fine!
As regards the blue counters we have $16$ ways for $B_1$ and $9$ for $B_2$. Now we may assume that $B_1$ is on square $(1,1)$ and $B_2$ on square $(2,2)$ and we have 3 cases for the red counters:
1) if $R_1$ is on $(2,1)$ or on $(1,2)$, then $9$ places are left for $R_2$.
2) if $R_1$ is on $(3,1), (3,2),(4,1), (4,2),(1,3), (2,3),(1,4), (2,4)$, then $8$ places are left for $R_2$.
3) if $R_1$ is on $(3,3), (3,4),(4,3), (4,4)$, then $7$ places are left for $R_2$.
Hence the number of ways is equal to
$$\frac{16\cdot 9}{2}\cdot \frac{2\cdot 9+8\cdot 8+4\cdot 7}{2}=3960.$$
By using the same strategy, we find the number of ways for a grid $n\times n$:
$$\frac{n^2\cdot(n-1)^2}{2}\cdot \frac{2\cdot(n-1)^2+4(n-2)\cdot((n-1)^2-1)+(n-2)^2\cdot((n-1)^2-2)}{2}$$
which reduces to
$$\frac{n^2\cdot(n-1)^2\cdot(n^4-2n^3-3n^2+8n-2)}{4}.$$
For $n\geq 1$ the numbers are $0, 2, 198, 3960, 33800, 180450, 715302, 2306528$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $4x \equiv 7 \pmod{15}$ and $3x \equiv 5 \pmod{16}$ (different exercises) This is how I solved each:
$$4x \equiv 7\pmod {15} \\
4x - 7 \equiv 0\pmod{15} \\
4x-7 = 15k \Leftrightarrow 4x-15k= 7 \\
$$
$$15 = 4*3+3 \\
4=3*1+1\\
3=3*1+0$$
$$1 = 4-3*1 \\
3 = 15-4*3 \\
1 = 4 - (15-4*3)*1 \\
1 = 4-15+4*3 \\
1 = 4*4-15*1 \\
$$
$$7 = (7*4)*4-(7*1)*15$$
$7*4*4 = 112$ which is congruent with 7 in mod 15. Yet my book says the answer is 13. What went wrong?
The second one:
$$3x - 5 \equiv 0 \pmod{16} \\
3x-5=16k\\
3x-16k=5 \\$$
$$16=3*5+1\\
3=1*3+0$$
$$1 = 16-3*5 \\
5=16*5-3*5*5$$
Yet $3*5*5-5$ is not congruent with zero in mod 16. What went wrong?
| You arrived at $7=28(4)-7(15)$, which means $7\equiv 4(28)-7(15)\equiv4(28)\equiv4(\boxed{13})$ (mod 15).
Similarly, $5=16(5)-3(25)$, means $5\equiv-3(25)\equiv-3(-7)\equiv3(\boxed{7})$ (mod 16).
You made a mistake in the second part because you forgot to take into account the negative sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How do I solve this: I was doing my homework and I stumbled over this particular exercise. I would've known how to solve it, if it had been the same thing under square root in both cases.
$$
\lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)
$$
| $$\left({\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}}\right)\frac{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}\\=\frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$
so evaluate
$$\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}=\lim_{n\to\infty} \frac{2n}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}+\\\lim_{n\to\infty} \frac{3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$
by which
$$\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}=\lim_{n\to\infty} \frac{2n}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$
where the second term is zero because the numerator is a constant while the denominator contains $\sqrt{4n^2}$. Therefore
\begin{align}\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}&=\lim_{n\to\infty} \frac{2n}{{2n\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+2n\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}\\&=\lim_{n\to\infty} \frac{1}{{\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}\\&=\frac{1}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding the limits of integration for $\iint\limits_A \frac{1}{(1 + x^2 + y^2)^2} dxdy$ where A is one loop of lemniscate $(x^2 + y^2)^2 = x^2 - y^2$ Work so far: Switch to polar coordinates
$A \rightarrow r^2 = \cos^2\theta - \sin^2\theta \rightarrow r^2 = \cos2\theta$
So $A$ is $r^2 = \cos 2 \theta$
And the integral becomes
$$\iint\limits_A \frac{r}{(1 + r^2)}drd\theta$$
Now I want to find the limits of integration without actually graphing the curve.
So
$$2rdr = -2\sin(2\theta)$$
$$\frac{dr}{d\theta} = \frac{-\sin(2\theta)}{r}$$
Critical points for $\theta$ come out to be $0$ and $\frac{\pi}{2}$
Now I know how to solve the integral using u substitution but I'm not sure about the limits.
| In polar coordinates the curve is determined by
$$r^2=\cos2\theta,
$$
which has real solutions for $r $ if and only if:
$$
\cos2\theta\ge0 \implies -\frac\pi4\le\theta\le\frac\pi4\text { or }\frac{3\pi}4\le\theta\le\frac{5\pi}4.$$
As the problem asks to find the area of a single loop it suffices to consider only the first interval. We have:
$$\iint\limits_A \frac{r}{(1 + r^2)^2}drd\theta=
\int\limits_{-\frac\pi4}^{\frac\pi4}d\theta
\int\limits_0^\sqrt {\cos2\theta}\frac{rdr}{(1 + r^2)^2}.
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the 4 roots of the equation: $z^{4}+4=0$ Find the 4 roots of the equation: $z^{4}+4=0$
Note: I can't use the form: $e^{i\theta}$
My attempt:
Note we have: $z=(-4)^{\frac{1}{4}}$, consider $w=-4$, then $w^{1/4}=z$ this implies $w=z^4$.
Consider the polar form of $z$ and $w$:
$|w|=r=4$,
$Arg(w)=\pi=\theta$
Then,
$$w=4\cos\pi+i\sin\pi$$
$$z=p\cos\phi+i\sin\phi$$
This implies: $w=z^4$ iff $4=p^4$ and $4\phi=\pi+2k\pi$ iff $p=4^{1/4}$ and $\phi=\frac{\pi+2k\pi}{4}$ with $k=0,1,2,3$
Then the roots are:
$$z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$
with $k=0,1,2,3$
is correct this?
| It is $$(z^2)^2+(2^2)^2=0$$ or $$a^2+b^2=0$$ and this is $$(a-bi)(a+bi)=0$$ where $$a=z^2,b=2^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How should I solve the equation $\sqrt{x-\frac 1x}+\sqrt{1-\frac 1x}=x$ I could square both sides of the equation, but that ends up giving me a cubic to solve. What I need is a beginning approach to solve such questions, not the whole answer.
Thanks
| Note that $x = 0$ is clearly not a solution. Observe that:
\begin{align*}
&\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x \\
&\Rightarrow \left(\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right)\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\
&\Rightarrow \left(\left(x - \frac{1}{x}\right) - \left(1 - \frac{1}{x}\right)\right) = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\
&\Rightarrow x - 1 = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\
&\Rightarrow \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} = 1 - \frac{1}{x}
\end{align*}
Adding both together yields:
\begin{align*}
&\left(\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right) + \left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) = x + \left(1 - \frac{1}{x}\right) \\
&\Rightarrow 2\sqrt{x - \frac{1}{x}} = \left(x - \frac{1}{x}\right) + 1
\end{align*}
We substitute $y = \sqrt{x - \frac{1}{x}}$, and we see that:
\begin{align*}
2y = y^2 + 1 \Rightarrow (y - 1)^2 = 0 \Rightarrow y = 1
\end{align*}
It remains to solve $\sqrt{x - \frac{1}{x}} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Setting up the volume $\iiint_{?}^{?}dV$ Let $$S = \{ (x,y,z) | x=a+b, y = b+c, z = -b, ac-b^2\ge0, c\ge0, a\ge0\}$$ and
$$x^2+y^2+z^2\le 1$$
Compute the volume of S.
My work:
$$V=\iiint_R 1 dV = \int\limits_{-1}^{1}\int\limits_{-\sqrt{(1-b^2)-x^2}}^{\sqrt{(1-b^2)-x^2}}\int\limits_{-b}^{\sqrt{1-x^2-y^2}} 1. dzdydx$$
Am i right ???
how to use spherical ?? or cartesian is better here?
Next try:
eliminating a,b,c we get the planes $ x+z\ge0 $ and $ y+z\ge0$, and the hyperboloid $xy+(x+y)z\ge0$ with $x^2+y^2+z^2\le 1$
How to set up now ?
$$V=\int\limits_{?}^{?}\int\limits_{?}^{?}\int\limits_{?}^{?} 1. dzdydx$$
| Substitute $a=x+z, c=y+z,b=-z$.
The inequalities are $$ac-b^2=xy+yz+xz\geq0\\ \ c=y+z\geq 0\\\ a=x+z\geq 0\\ x^2+y^2+z^2\leq 1$$.
The surface $xy+yz+xz=0$ consists of two cones whose vertices touch each other at the origin. Both cones are symmetric around the line $x=y=z$. All three coordinate axes lie on the cone.
The intersection of the plane $x+z=0$ with the cone $xy+yz+xz=0$ is the $y$ axis since $xy+yz+xz=(x+z)y+xz$ is zero whenever $x=z=0$.
Since the plane $x+z=0$ does not slice through the interior of either cone, if one point in the cone is in the region $x+z\geq 0$ then the whole cone is in the region.
As it turns out, one of the two cones is in both the region $x+z\geq 0$ and the region $y+z\geq 0$.
All that remains is to find the intersection of this cone with the sphere $x^2+y^2+z^2\leq 1$.
This will be done with a volume integral in spherical coordinates. The region to be integrated over is $r=[0,1]$, $\theta=[0,2\pi]$, $\phi=[0,\cos^{-1}(\frac{1}{\sqrt{3}})]$.
The upper bound of $\cos^{-1}(\frac{1}{\sqrt{3}})$ is the angle between any coordinate axis and the line $x=y=z$.
$$\int_0^1\int_0^{2\pi}\int_0^{\cos^{-1}(\frac{1}{\sqrt{3}})}r^2\sin\phi \ \mathrm{d}\phi\mathrm{d}\theta \mathrm{d}r=\frac{2}{3} \left(1-\frac{1}{\sqrt{3}}\right)\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $1+\sin(x)+\sin(2x)+\sin(3x)=\cos(x)+\cos(2x)+\cos(3x)$ I'm trying to solve the following trigonometric equation
$$1+\sin(x)+\sin(2x)+\sin(3x)=\cos(x)+\cos(2x)+\cos(3x)$$
but I can't get any further than
$\iff 1 + 2\sin(2x)\cos(x) + \sin(2x) = 2\cos(2x)\cos(x) + \cos(2x)$
$\iff 1 + \sin(2x)(2\cos(x) + 1) = \cos(2x)(2\cos(x) + 1)$
$\iff 1 + \sin(2x)(2\cos(x) + 1) = \cos(2x)(2\cos(x) + 1)$
$\iff 1 = (\cos(2x)-\sin(2x))(2\cos(x) + 1)$
Could someone point me out in the right direction?
| Let us use the Swiss army knife of trigonometry by setting $t = \tan(\frac{x}{2})$. One gets
\begin{align}
\sin{x} &= \frac{2t}{1+t^2} \\
\sin{2x} &= 2\sin{x}\cos{x} = \frac{4t(1-t^2)}{(1+t^2)^2}\\
\sin{3x} &= 3\sin{x}- 4\sin^3{x} = \frac{6t}{1+t^2} - \frac{32t^3}{(1+t^2)^3} \\
\cos{x} &= \frac{1-t^2}{1+t^2} \\
\cos{2x} &= \cos^2{x} - \sin^2{x} = \frac{t^4-6t^2 + 1}{(1+t^2)^2}\\
\cos{3x} &= 4\cos^3{x} - 3 \cos{x} = \frac{(1-t^2)^3}{(1+t^2)^3} - \frac{3(1-t^2)}{1+t^2}
\end{align}
Substituting in your equation, and up to possible errors, one finally gets the equation
$$
t^6 + 2t^5 -3t^4 -8t^3 +11t^2 +6t -1 = 0
$$
which has two real roots and four complex roots. According to WolframAlpha, the real roots are
$0.136187$ and $- 0.548975$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $1 - 3xy + x^3 + y^3 = 0$ I am trying to solve the equation
$$
1 - 3xy + x^3 + y^3 = 0
$$
in the real numbers. I can see several (infinitely many) solutions:
*
*$x=y = 1$
*$x+ y = -1$
I am trying to show that these are all of them.
For example, if $x = y$, then $x^2(-3+2x) = -1$ forcing $x = y = 1$
I can't seem to make much progress with the case where $x\neq y$ and showing that $x + y = 1$ in that case. (I am pretty sure that this must be true.)
| Worth memorizing:
$$ x^3 + y^3 + z^3 - 3xyz = (x+y+z) \left(x^2 + y^2 + z^2 - yz - zx - xy \right) $$
Another step is possible, in that
$$ x^2 + y^2 + z^2 - yz - zx - xy = \frac{1}{2} \; \left((x-y)^2 + ( y-z)^2 + ( z-x)^2 \right) $$
The cubic is zero, for real numbers $x,y,z,$ when either
$x+y+z = 0$ or $x=y=z.$ In your case, take $z=1$
More solutions are possible over the complexes, in that the quadratic further factors as
$$ (x+\omega y + \omega^2 z)(x + \omega^2 y + \omega z) $$
where $\omega$ is a primitive cube root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Is $d(x,y)=|x^3-y^3|$ metric on $\mathbb R$? What about property $2$:
$d(x,y)=0$ iff $x=y$
$|x^3-y^3|=0$
$x^3-y^3=0$
$(x-y)(x^2+xy+y^2)=0$
If $x-y=0$ then $x=y$.
But if $x^2+xy+y^2=0$ then $x$ does not equal to $y$.
| Note that $$x^2+xy+y^2= (x+\frac {y}{2} )^2 + \frac {3}{4}y^2$$
Thus $$x^2+xy+y^2=0 \iff x=y=0$$
Therefore there is no problem with your metric.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}$ is a multiple of 3 if it is an integer I'm trying to solve the following problem by induction but I'm getting stuck.
For positive integers $a$ and $b$, define $$f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}.$$ If $f(a,b)$ is an integer, prove that it is a multiple of 3.
Proof by induction on a:
Base case: $a=1$
$$f(1,b) = \frac{1}{b} + \frac{b}{1} + \frac{1}{b} = \frac{2}{b} + b$$
Since $f(1,b)$ is an integer than $\frac{2}{b}$ must be an integer and $b\in\{1,2\}$ . Then $f(1,b) = 3$.
Inductive hypothesis: Assume for some integer $k \ge 1,$ $$f(k,b) = \frac{k}{b} + \frac{b}{k} + \frac{1}{kb}$$ is an integer and is a multiple of 3.
I want to show that $f(k+1,b)$ is also a multiple of 3. Then
$$f(k+1,b) = \frac{k+1}{b} + \frac{b}{k+1} + \frac{1}{(k+1)(b)} = \cdots.$$
And this is where I get stuck. I know there are other ways to solve this problem but I wanted to try it by induction. Hope someone can help with this! Thanks.
| I don't see how to solve the problem by induction but there's a fairly simple solution without induction. Note that
$$f(a, b)= \frac{a^2+b^2+1}{ab}.$$
If either $3 \vert a$ or $3 \vert b$, then the numerator is not a multiple of $3$, but the denominator is, so the fraction can't be an integer.
If neither $a$ nor $b$ is a multiple of $3$, then $a^2 \equiv b^2 \equiv 1 \pmod{3}$ so the numerator is a multiple of $3$ and the denominator is not, so if the fraction is an integer it must be a multiple of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Invariant to inversions 2
Let $a,b,c,d,e,f>0$ satisfying
$a+b+c+d+e+f=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}$
. Prove
$$ab+bc+cd+da+ac+bd+ae+be+ce+de+af+bf+cf+df+ef+10\sqrt{abcdef}\geq25$$
Attempt: Prove that if $k>10$ then the inequality
$$ab+bc+cd+da+ac+bd+ae+be+ce+de+af+bf+cf+df+ef+k\sqrt{abcdef}\geq15+k$$is not always true.
Also, I tried
$$a+b+c+d+e+f\geq\frac{36}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}}=\frac{36}{a+b+c+f+e+f}.$$
Also, $(a+b+c+d+e+f)^2$ appears $ab+ac+ad+...+ec+ed+ef.$
| Let $a+b+c+d+e+f= const$ and $a^2+b^2+c^2+d^2+e^2+f^2=const.$
Thus, by the Vasc's EV Method : https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf (corollary 1.8 (b))
it's enough to prove our inequality for equality case of five variables.
Now, let $b=c=d=e=f$ and $a=xb.$
Thus, the condition gives $$a+5b=\frac{1}{a}+\frac{5}{b}$$ and we need to p-rove that
$$5ab+10b^2+10\sqrt{ab^5}\geq25$$ or
$$x+2+2\sqrt{\frac{5x+1}{x+5}}\geq\frac{5x(x+5)}{5x+1}$$ or
$$\sqrt{\frac{5x+1}{x+5}}\geq\frac{7x-1}{5x+1}$$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the $n^{th}$ derivative of $y=\frac {x}{x^2+a^2}$ Find the $n^{th}$ derivative of $y=\dfrac {x}{x^2+a^2}$.
My Attempt:
$$y=\dfrac {x}{x^2+a^2}$$
$$y=x.(x^2+a^2)^{-1}$$
Differentiating both sides with respect to $x$
$$y_{1}=x.((-1).(x^2+a^2)^{-2}.2x)+(x^2+a^2)^{-1}$$
$$y_{1}=(-1).(x^2+a^2)^{-2}.2x^2+(x^2+a^2)^{-1}$$
| $$y=\frac{x}{x^2+a^2}=\frac{1}{2} \left( \frac{1}{x-ia}+\frac{1}{x+ia} \right)$$
The nth derivative of $(x+b)^{-1}$ is given as
$$D^n (x+b)^{-1}=(-1)^n n!(x+b)^{-n-1}$$
Using this
$$D^n y= (-1)^n n!~ \Re (x+ia)^{-n-1}=(-1)^n n! (x^2+a^2)^{(-n-1)/2} \cos[(n+1) \tan^{-1}(a/x)].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is what I did:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )=0$$
$$\lim\limits_{x \to \infty} x\bigg (\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + \sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-a-\dfrac{b}{x} \bigg )=0$$
So, we'd have something like:
$$\infty \cdot(2-a) = 0$$
$(*)$If $a \in (-\infty, 2)$ $\Rightarrow$ $(2-a) > 0$, which means:
$$\infty \cdot(2-a) = \infty$$
$(*)$If $a\in (2, +\infty)$ $\Rightarrow$ $(2-a) < 0$, which means:
$$\infty \cdot(2-a) = -\infty$$
So the only option left for the limit to have any chance of being true is:
$$a=2$$
which would result in the indeterminate form:
$$\infty \cdot (2-a) = \infty \cdot 0$$
And now that we have $a=2$, we need to find the value of $b$ for the limit to be true. Substituting $a$ with $2$ in the initial limit we get:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x-b)=0$$
Since $b$ is a constant we can pull it out of the limit and get:
$$b=\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x)$$
And this is where I got stuck. I tried a bunch of methods and tricks for finding this limit and I got nowhere. That leads me to think that either I made some mistake/mistakes along the way, or I simply don't know how to solve this limit. So, how can I find $b$?
| HInt
$$F=\lim_{x\to\infty}{\sqrt{x^2+px+q}-x}=\lim_{h\to0}\dfrac{\sqrt{1+ph+qh^2}-1}h$$
Rationalize the numerator to find
$$F=\lim\dfrac{1+ph+qh^2-1}{h(?)}=\dfrac p{1+1}$$
Set $p=q=1$
and $p=q=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate $\sin\alpha$
Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate
$\sin\alpha$
My proof:
$\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}\\
\frac{\sin\left(\alpha-\frac{\pi}{4}\right)}{\cos\left(\alpha-\frac{\pi}{4}\right)}=\frac{1}{3}\\3\sin\left(\alpha-\frac{\pi}{4}\right)=\cos\left(\alpha-\frac{\pi}{4}\right)\\\sin^2\left(\alpha-\frac{\pi}{4}\right)+9\sin^2\left(\alpha-\frac{\pi}{4}\right)=1\\\sin\left(\alpha-\frac{\pi}{4}\right)=\pm\frac{1}{\sqrt{10}}\\
\sin\left(\alpha-\frac{\pi}{4}\right)=\sin\alpha\cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cos\alpha=\frac{\sqrt{2}}{2}\sin\alpha-\frac{\sqrt2}{2}\cos\alpha=\frac{\sqrt2}{2}\left(\sin\alpha-\cos\alpha\right)=\pm\frac{1}{\sqrt{10}}\\\sin\alpha-\cos\alpha=\pm\frac{1}{\sqrt{5}}\\\sin\alpha=\pm\frac{1}{\sqrt{5}}+\cos\alpha$
I have no idea how to determine $\sin\alpha$
| Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$ and $-\arctan(x)=\arctan(-x),$
$$\alpha=n\pi+\dfrac\pi4-\arctan\dfrac13=n\pi+\arctan\dfrac{1+\left(-\dfrac13\right)}{1-1\left(-\dfrac13\right)}=n\pi+\arctan\dfrac12$$ where $n$ is any integer
$$\sin\alpha=(-1)^n\sin\left(\arctan\dfrac12\right)$$
If $\arctan\dfrac12=y,\tan y=\dfrac12,$
$$\dfrac{\sin y}1=\dfrac{\cos y}2=\pm\sqrt{\dfrac{\sin^2y+\cos^2y}{2^2+1^2}}$$
Now as $0<\arctan\dfrac12=y<\dfrac\pi2,\sin y>0,\cos y>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can we give a closed form expression for $\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}$? Let $a>0$ and $x\in(-1,1)$. Can we give a closed form expression for $\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}$? Note that the series is convergent. There should be an expression in terms of the cotangent.
My goal is to find a (sharp) upper bound $c(x)$ of the series.
| Observe
\begin{align}
\sum^\infty_{k=-\infty}\frac{1}{a+(x+k)^2} =&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\
&+\sum^\infty_{k=1} \frac{1}{a+(k+(1+x))^2}+\sum^{\infty}_{k=1} \frac{1}{a+(k+(1-x))^2}\\
\leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\
&\ +\int^\infty_0\frac{dk}{a+(k+(1+x))^2}+ \int^\infty_{0} \frac{dk}{a+(k+(1-x))^2}\\
\leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\
&\ +\frac{1}{\sqrt{a}}\left(\tan^{-1}\left( \frac{\sqrt{a}}{1+x}\right)+\tan^{-1}\left( \frac{\sqrt{a}}{1-x}\right)\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding a partial derivative with two independent variables and two dependent variables The following problem is from the 7th edition of the book "Calculus and Analytic Geometry" by Thomas and Finney. It is problem 5b in section 16.5. Below is my attempt to solve it. However, I am getting a different answer than the book. Where did I go wrong?
Problem:
Find
$$ \left( \frac{\partial w}{\partial y} \right)_z $$
at the point $(w, x, y, z) = (4, 2, 1, -1)$ if
$$ w= x^2y^2 + yz - z^3 $$
and
$$ x^2 + y^2 + z^2 = 6 .$$
Answer:
The first step in the process is to eliminate the variable $x$ from the first equation.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\
w &= x^2y^2 + yz - z^3 \\
w &= (6 - y^2 - z^2 )y^2 + yz - z^3 \\
\left( \frac{\partial w}{\partial y} \right)_z &= (6 - y^2 - z^2)(2y) + y^2(-2y) + z \\
\left( \frac{\partial w}{\partial y} \right)_z &= (6 - y^2 - z^2)(2y) - 3y^3 + z \\
\left( \frac{\partial w}{\partial y} \right)_z ( 4, 2, 1, -1 ) &= ( 6 - 1^1 - 1^2)(2(1)) - 3(1^3) - 1 \\
\left( \frac{\partial w}{\partial y} \right)_z ( 4, 2, 1, -1 ) &= ( 4)(2) - 3 - 1 \\
\left( \frac{\partial w}{\partial y} \right)_z ( 4, 2, 1, -1 ) &= 4 \\
\end{align*}
However, the book's answer is $5$.
| You have
$$\left( \frac{\partial w}{\partial y} \right)_z = (6 - y^2 - z^2)(2y) + y^2(-2y) + z \\
\left( \frac{\partial w}{\partial y} \right)_z = (6 - y^2 - z^2)(2y) - 3y^3 + z \\$$
but $$y^2(-2y)=-2y^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Infinite prime cylinders? Define a prime $n$-circle as a circular sequence of
$n$ distinct natural numbers such that adjacent elements sum to a prime
(including $n^{\textrm{th}}$ $+$ $1^{\textrm{st}}$).
For example:
$$6, 1, 2, 5, 8, 9, 4, 7 \;.$$
This is a variation on the prime circles counted in
the integer sequence
A051252,
in that I am not insisting that the numbers be drawn from
$\{1,2,\ldots,n\}$. (The above circle misses $3$.)
Define a prime $n$-cylinder as an aligned stacking of prime
circles, such that adjacent numbers vertically also sum to a prime.
(However, no wrap-around requirement from top to bottom.)
Further, every circular rung of the cyclinder should be a
distinct permutation—no repeats allowed (also disallowing
reversals).
For example, here is a prime octagonal-cylinder consisting of
four prime circles:
And here displayed as a matrix:
$$
\left(
\begin{array}{cccccccc}
7 & 4 & 3 & 8 & 9 & 2 & 5 & 6
\\
16 & 15 & 8 & 9 & 4 & 3 & 2 &
1 \\
7 & 4 & 9 & 8 & 3 & 2 & 15 &
16 \\
6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
\\
\end{array}
\right)
$$
Q. Do there exist infinitely tall prime $n$-cylinders,
for each $n$?
The answer should
be the same whether the cyclinder is
infinite just in one direction, or bi-infinite, extending infinitely
in both directions.
Although it seems relatively easy to add rungs in a greedy fashion,
I don't see how to prove that this approach can extend infinitely.
For example, here is how one might get "stuck."
Suppose you are adding the last number $x$ to the
top prime circle. The three numbers adjacent to $x$ (left, right, below)
could be $1$, $3$, and $5$. But the only number $x$ such
that each of $\{x+1, x+3, x+5 \}$ is prime is $x=2$, which may already
have been used in that top circle.
| Let's denote the numbers in rung $j$ by $a(i,j)$ where $i$ runs from $1$ to $n$. So the conditions for a infinite (in one direction) cylinder are:
*
*A. $a(1,j) + a(n,j)$ is prime for $j \ge 1$
*B. $a(i,j) + a(i-1,j)$ is prime for $i = 2 \dots n$, for $j \ge 1$
*C. $a(i,j) + a(i,j-1)$ is prime for $i=1 \dots n$, for $j \ge 2$
Clearly $n$ must be even and the numbers in each rung must alternate between even and odd.
If we have constructed up to rung $k$ then we can construct most of rung $k+1$ as follows:
$a(i,k+1) = a(i+1,k)$ for $i=1 \dots n-1$
This meets conditions B and C on rung $k+1$, up to $i=n-1$. To meet the remaining conditions we just need to choose $a(n,k+1)$ so that:
*
*$a(n,k+1) + a(n-1,k+1)$ is prime
*$a(n,k+1) + a(n,k)$ is prime
*$a(n,k+1) +a(1,k+1)$ is prime
and also $a(n,k+1) \gt \max(a(i,k))$ for $i=1..n$ to prevent a repeat. And since $a(n-1,k+1)=a(n,k)$, the three constraints above in fact reduce to two.
Since we can make $a(n,k+1)$ as large as we like, we simply find two primes $p$ and $q$ such that $p,q \gt 2\max(a(i,k))$ and $p-q=a(n-1,k+1)-a(1,k+1)$, and then set
$a(n,k+1) = p - a(n-1,k+1) = q-a(1,k+1)$
As an example, an infinite cylinder with initial rung $\left(
\begin{array}{cccccccc}
6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
\\
\end{array}
\right)$ could be
$\left(
\begin{array}{cccccccc}
6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
\\
1 & 2 & 5 & 8 & 9 & 4 & 7 & 12
\\
2 & 5 & 8 & 9 & 4 & 7 & 12 & 17
\\
5 & 8 & 9 & 4 & 7 & 12 & 17 & 24
\\
8 & 9 & 4 & 7 & 12 & 17 & 24 & 29
\\
. & . & . & . & . & . & . & .
\\
\end{array}
\right)$
... and if you want a doubly infinite cylinder you can do the same trick in the other direction ...
$\left(
\begin{array}{cccccccc}
. & . & . & . & . & . & . & .
\\
33 & 28 & 25 & 6 & 1 & 2 & 5 & 8
\\
28 & 25 & 6 & 1 & 2 & 5 & 8 & 9
\\
25 & 6 & 1 & 2 & 5 & 8 & 9 & 4
\\
6 & 1 & 2 & 5 & 8 & 9 & 4 & 7
\\
1 & 2 & 5 & 8 & 9 & 4 & 7 & 12
\\
2 & 5 & 8 & 9 & 4 & 7 & 12 & 17
\\
5 & 8 & 9 & 4 & 7 & 12 & 17 & 24
\\
8 & 9 & 4 & 7 & 12 & 17 & 24 & 29
\\
. & . & . & . & . & . & . & .
\\
\end{array}
\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What's the visualization of $z=\sec\theta+i\cdot\csc\theta$ in complex-plane? What would be the graph of $z=\sec\theta+i\cdot\csc\theta$ in complex plane? Since $z=\cos\theta+i\cdot\sin\theta$ is a circle.
| With $x=sec \theta$ adn $y=csc \theta $ you get $$\frac {1}{x^2} + \frac {1}{y^2}=1$$ or $$y=\pm \sqrt{\frac {x^2}{x^2-1}}$$
The graph consists of four branches with two vertical asymptotes at $x=\pm 1$ and two horizontal asymptotes at $y=\pm 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim\limits_{x \rightarrow \infty}(\frac{2}{\pi}\arctan(x))^{\frac{x^2}{1+2x}}$ I have to calculate next limit: $$\lim\limits_{x \rightarrow \infty}\left(\frac{2}{\pi}\arctan(x)\right)^{\frac{x^2}{1+2x}}$$
So far I got to this point $$e^{\lim\limits{x\rightarrow\infty}\frac{x^2}{1+2x}\frac{2\arctan(x)-\pi}{\pi}}$$
When I start to calculate this limit on $e$ I then came to this $$\frac{1}{\pi} \lim\limits_{x\rightarrow\infty}\frac{x^2(2\arctan(x)-\pi)}{1+2x}$$
And also I must not use L'Hôpital's rule for this one.
Any help?
| We are to compute
\begin{align*}
\lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right).
\end{align*}
With the change of variable $u=1-\dfrac{2}{\pi}\tan^{-1}x$, then
\begin{align*}
\lim_{x\rightarrow\infty}\dfrac{x^{2}}{1+2x}\log\left(\dfrac{2}{\pi}\tan^{-1}x\right)&=\lim_{u\rightarrow 0^{+}}\dfrac{\cot^{2}\dfrac{\pi}{2}u}{1+2\cot\dfrac{\pi}{2}u}\log(1-u)\\
&=\lim_{u\rightarrow 0^{+}}\dfrac{2}{\pi}\cdot\dfrac{\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u}\cdot\dfrac{\cos^{2}\dfrac{\pi}{2}u}{\sin\dfrac{\pi}{2}u+2\cos\dfrac{\pi}{2}u}\cdot\dfrac{1}{u}\cdot\log(1-u).
\end{align*}
Note that
\begin{align*}
\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\cdot\log(1-u)=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{u}\int_{0}^{u}\dfrac{1}{1-t}dt=-\lim_{u\rightarrow 0^{+}}\dfrac{1}{1-\eta_{u}}=-1,
\end{align*}
where $\eta_{u}$ is in between $u$ and $0$, chosen by Mean Value Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$ in $\Delta \text{ABC}$ I am attempting the following trigonometry problem. Given that in an acute angled triangle $\Delta \text{ABC}$, the following equalities hold true
$$\cos^2(A)+\cos^2(B)+2\sin(A)\sin(B)\cos(C)=\dfrac{15}{8}\\
\cos^2(B)+\cos^2(C)+2\sin(B)\sin(C)\cos(A)=\dfrac{14}{9}$$
Find the value of $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$.
My Attempt:
Let the unknown quantity be $x$. Then we have, by adding all the terms.
$$2\sum_{cyc}\cos^2(A)-2\sum_{cyc}\sin(A)\sin(B)\cos(A+B)=\dfrac{15}{8}+\dfrac{14}{9}+x$$
Also simplifying the second summation term as follows, we get $$\sin(A)\sin(B)\cos(A+B)=\dfrac{\sin(2A)\sin(2B)}{4}-(1-\cos^2(A))(1-\cos^2(B))$$
I'm not sure how to proceed further. Any hints are appreciated. Even hints to other possible pathways to the solution are welcome. Thanks
| Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\cos^2C+\cos^2A+2\sin A\sin B\cos C$$
$$=1+\cos(C+A)\cos(C-A)+...$$
$$=1-\cos B(\cos C\cos A+\sin C\sin A-2\sin C\sin A)$$
$$=1-\cos B\cos(C+A)$$
$$=1+\cos^2B$$
$\implies\cos^2B=\dfrac{15}8-1,\sin B=+\sqrt{1-\cos^2B}=?$
Similarly we can find $\cos^2A$ and hence $\sin A$
Use Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ to find $\cos C$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$
Let $k$ be equal to:
$$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$
I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is there any algorithm that I should know to express the expressions as squares?
| You have $\sqrt{A \pm \sqrt{B}}=\sqrt{\frac{A+C}{2}}\pm \sqrt{\frac{A-C}{2}}$ where $C=\sqrt{A^2-B}$
So we can write $\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+1}{2}}+\sqrt{\frac{5-1}{2}}=\sqrt{3}+\sqrt{2}$
and $\sqrt{5-2\sqrt{6}}=\sqrt{5-\sqrt{24}}=\sqrt{\frac{5+1}{2}}-\sqrt{\frac{5-1}{2}}=\sqrt{3}-\sqrt{2}$
So substituting we get $k=-\sqrt{3}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=-6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What's the coefficent of $x^{20}$ in $(x^3+x^4+x^5+...)^5$? Only hint is needed. I know about Binomial/Multinomial expansion but I got stuck on this series; it doesn't look like anything I've solved before. I already searched for any hint/formula and couldn't find one, any help is appreciated.
| Observe that
$$
(x^3+x^4+\cdots+x^n+\cdots)^5=x^{15}(1+x+x^2+\cdots)^5=\frac{x^{15}}{(1-x)^5}\\ =\frac{x^{15}}{3!}\frac{d^4}{dx^4}\frac{1}{1-x}= \frac{x^{15}}{4!}
\left(1+x+x^2+\cdots\right)^{(4)}\\=\frac{x^{15}}{4!}
\left(\frac{4!}{0!}+\frac{5!}{1!}x+\frac{6!}{2!}x^2+\frac{7!}{3!}x^3+\frac{8!}{4!}x^4+\frac{9!}{5!}x^5+\cdots\right)
$$
Hence coefficient of $x^{20}$ is
$$
c_{20}=\frac{9!}{4!5!}
$$
Note that
$$
\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}, \quad
\frac{d}{dx}\frac{1}{(1-x)^2}=\frac{2}{(1-x)^3}, \quad
\frac{d}{dx}\frac{1}{(1-x)^3}=\frac{3!}{(1-x)^4}, \quad
\frac{d^n}{dx^n}\frac{1}{1-x}=\frac{(n-1)!}{(1-x)^n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Paradox in indefinite integral Today,I met a problem about calculating such an indefinite integral$$ I=\int \frac{1}{(1+x^2)(1+x^{2019})}$$ My thought is as follows $$
\begin{aligned} I & = \int \frac { 1 } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = \int \frac { 1 } { \left( 1 + x ^ { - 2 } \right) \left( 1 + x ^ { - 2019} \right) } \left( - \frac { 1 } { x ^ { 2 } } \right) d x \\ & = \int \frac { - x ^ { 2019 } } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = I - \int \frac { 1 } { 1 + x ^ { 2 } } d x \end{aligned}
$$ No doubt,it is wrong.However, I don’t know how to illustrate this paradox?
| Let $f_n$ denote an antiderivative of $\frac{1}{(1+x^2)(1+x^n)},\,n:=2019$ so $\int\frac{-x^ndx}{(1+x^2)(1+x^n)}=f_n^\prime(1/x)+C$. So you've actually shown $f_n(x)=f_n(1/x)-\arctan x+C_n$ for some $C_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Possible solutions for complex equations I have trouble solving this complex equation:
If $$z^4 +z^3 + z^2 + z + 1 = 0 $$ and $$u = z + z^{-1}$$ find all the possible values for u . I have tried substituting u into the equation in two different ways, finding that $$ u = \frac{-1}{z^3 + z^2}$$ and $$ u = \frac{z^4 -1}{z^3 - z} $$
but I'm not sure where to go from here. How many possible values should I expect to find for u? Is it possible to find the values of z with the information I have?
| We have
\begin{eqnarray*}
u &=& z+\frac{1}{z} \\
u^2 &=& z^2+\frac{1}{z^2}+2. \\
\end{eqnarray*}
So
\begin{eqnarray*}
u^2+u-1=0 \\
u = \frac{-1 \pm \sqrt{5}}{2}. \\
\end{eqnarray*}
Now you just need to solve the quadratic
\begin{eqnarray*}
z+\frac{1}{z}= \frac{-1 \pm \sqrt{5}}{2}. \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\frac{1}{5}\big((4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8\big)$ is the sum of 3 consecutive squares
Prove that for every positive integer $n$, the number
$$
\large \frac{(4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8}{5}
$$
can be expressed as a sum of squares of three consecutive integers.
Attempt. Since
$$
(m - 1)^2 + m^2 + (m + 1)^2 = 3m^2 + 2, \forall m \in \mathbb Z^+, m \ge 1,
$$
it suffices to show that
$$
(4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} = 15m^2 + 2, \,\,\text{for some}\,\,\, m \in \mathbb Z^+,
$$
$$\implies \sum_{p = 1}^{n - 1}[(4 + \sqrt{15})^{2p} + (4 - \sqrt{15})^{2p}] = 15m^2, \forall m \in \mathbb Z^+, m \ge 1$$
But this couldn't use any mathematical induction, since $m$ hasn't been known to be on a sequence for $n = 1, 2, 3, 4, \cdots$
| Note that
$$
\frac{(4+\sqrt{15})^{2n}+(4-\sqrt{15})^{2n}+8}{5}
=\frac{\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2+10}{5}=
\frac{\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2}{5}+2
$$
So it suffices to show that
$$
\frac{\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2}{5}=3m^2
$$
for some $m\in\mathbb N$,
or
$$
\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2=15m^2
$$
But, it is clear (inductive proof) that if
$$
(4+\sqrt{15})^{n}=a+b\sqrt{15}, \quad a,b\in \mathbb N,
$$
then
$$
(4-\sqrt{15})^{n}=a-b\sqrt{15}, \quad a,b\in \mathbb N,
$$
and hence
$$
\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2=(2b\sqrt{15})^2=15\cdot4b^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Taylor expansion of $\exp(\sin x)$ around $c=0$
Consider the Taylor's expansion around $c=0$
and find the first 4 terms for the function $\exp(\sin x)$.
I have done this but I'm not sure if is correct.
$$e^{\sin x}=\left(1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\right)^{x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots+(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}}$$
| We may avoid the use of the definition, by taking the composition of the expansions of $e^x$ and $\sin(x)$ at $x=0$:
$$\begin{align}
e^{\sin(x)}&=e^{x-\frac{x^3}{3!}+o(x^4)}\\
&=1+\left(x-\frac{x^3}{3!}+o(x^4)\right)
+\frac{1}{2!}\left(x-\frac{x^3}{3!}+o(x^4)\right)^2
\\&\qquad+\frac{1}{3!}\left(x+o(x^2)\right)^3+\frac{1}{4!}\left(x+o(x^2)\right)^4+o(x^4)\\
&=1+x-\frac{x^3}{6}+\frac{x^2}{2}-\frac{x^4}{6}+\frac{x^3}{6}
+\frac{x^4}{24}+o(x^4)\\
&=1+x+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$\omega$ satisfies $a \omega^3 + b \omega^2 + c \omega + d = 0$, Prove that $ |\omega| \leq \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$ If $\omega$ is a complex number that satisfies $a \omega^3 + b \omega^2 + c \omega + d = 0$, where $a,b,c,d$ are positive real numbers, prove that $|\omega| \leq \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$
This just seems so simple and elegant yet I can't wrap my head around. Was trying to tackle from vieta's formula but nothing fruitful yet.
Another idea was to make the polynomial $\omega^3 + \frac{b}{a} \omega^2 + \frac{c}{b} \omega + \frac{d}{c} = 0$, and prove some contradiction if $\omega > \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$. It might be closer to what we want.
| Let $r=\max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$. To simplify the problem, use the substitution $\omega =rz$,
$$
ar^3z^3+br^2z^2+crz+d=0,\\
az^3+\frac{b}{r} z^2+ \frac{c}{r^2} z+\frac{d}{r^3}=0,\\
z^3+B z^2+ C z+D=0, D\leq C\le B\le 1.
$$
Now, consider
$$
f(z)=(1-z)(z^3+B z^2+ C z+D)+z^4=(1-B)z^3+(B-C)z^2+(C-D)z+D.
$$
For $|z|=1$, $f(z)\leq |1-B|+|B-C|+|C-D|+D=1$, so $|z^3f(1/z)|\leq 1\times 1=1$.
By maximum modulus principle, $|z^3f(1/z)|\le 1$ for all $|z|\le 1$. That means for $|z|>1$, $|f(z)/z^3|\le 1$, $|f(z)|\le |z|^3$ Therefore,
$$
|(1-z)(z^3+B z^2+ C z+D)|=|f(z)-z^4|\ge -|f(z)|+|z|^4\\
\ge |z|^4-|z^3|=|z|^3(1-|z|)>0
$$
for $|z|>1$, so for $(1-z)(z^3+B z^2+ C z+D)=0$, $|z|\leq 1$. As a result,
$$
|\omega|\le |r||z|\le r.
$$
For more information, see this document.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $m \equiv 5\mod 10 $ prove that $1991 \mid 12^m + 9^m + 8^m + 6^m$ I tried to find the remainder of each one of $12$,$9$,$8$ and $6 \mod 5$ and then combine them but I didn’t get the answer
| From
$$m \equiv 5 \pmod{10}\iff m=10\cdot k +5=5\cdot(2k+1)\tag{1}$$
and
$$12^m + 9^m + 8^m + 6^m=\\
3^m \cdot 4^m + 3^m\cdot 3^m + 2^m \cdot 4^m + 2^m \cdot 3^m=\\
3^m(4^m+3^m)+2^m(4^m+3^m)=\\
(3^m+2^m)\cdot(4^m+3^m)\overset{(1)}{=}\\
(243^{2k+1}+32^{2k+1})\cdot(1024^{2k+1}+243^{2k+1})=...$$
From the well known
$a^{2k+1}+b^{2k+1}=(a+b)\cdot\left(a^{2k}+a^{2k-1}\cdot(-b)+a^{2k-2}\cdot(-b)^2+...+b^{2k}\right)$
$$...=(243+32)\cdot Q1 \cdot (1024+243)\cdot Q2=\\
275 \cdot 1267 \cdot Q1\cdot Q2=
2^5 \cdot \color{red}{11} \cdot 7 \cdot \color{red}{181} \cdot Q1\cdot Q2=\\
\color{red}{1991}\cdot 2^5 \cdot 7\cdot Q1\cdot Q2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve $x^2+5x+24=0$ mod $36$
Solve $x^2+5x+24=0$ mod $36$
So I have $x^2+5x+24\equiv 0$ mod $2^2\cdot 3^2$
SO I believe I need $x^2+x\equiv 0$ mod $2$ and $x^2+2x\equiv 0$ mod $3$
Which I determined that $0,1$ both work.
Then $f^\prime(x)=2x+5$
$f^\prime(0)=5$ and $f^\prime(1)=7$. So by hensel lemma there are solutions mod $4$
I'm not sure what to do now, am I supposed to lift to solutions mod $2^2$ and mod $3^2$?
I have $f(0+2t)=f(0)+2tf^\prime(0)\equiv 0 $ mod $4$
is $f(2t)=0+2tf^\prime(0)=2t(5)\equiv 0 $ mod $4$
then $10t\equiv 0$ mod $4$
which says $t=0$ which seems fine because $f(0)\equiv 0$ mod $4$ but I'm not sure I'm doing this right.
Im not sure how to deal with hensel lifting on numbers which aren't prime powers.
| You could solve this without Hensel lifting.
$x^2+5x+24\equiv0\bmod36\implies x^2+x\equiv0\bmod4$ and $x^2+5x+6\equiv0\bmod9$.
This means $x(x+1)\equiv0\bmod4$ and $(x+2)(x+3)\equiv0\bmod9$.
Note that $4|x(x+1)\implies 4|x$ or $4|x+1$, and $9|(x+2)(x+3)\implies 9|x+2$ or $9|x+3$,
because it is not possible for $2|x$ and $2|x+1$ or $3|x+2$ and $3|x+3$.
Anyway, can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that there are always three naturals $a, b, c \in (n^2, (n + 1)^2): c \mid (a^2 + b^2); \forall n \in \mathbb Z^+, n > 1$.
Prove that there are always three pairwise distinct naturals $$\large a, b, c \in \left(n^2, (n + 1)^2\right): c \mid (a^2 + b^2); \forall n \in \mathbb Z^+, n > 1$$
Here's an attempt of mine. Let $$a - x = b - y = c - z = n^2\implies x, y, z \in (0, 2n + 1)$$
It is necessitated to be sufficient to show that
$$(n^2 + z) \mid [(n^2 + x)^2 + (n^2 + y)^2] \iff (n^2 + z) \mid [(x - z)^2 + (y - z)^2]$$
But it can't think of anything anymore. Perhaps I could use apagogical arguments but in the cases of three changeable positive integers, that might not be disproven easily.
| There isn't much to this other than trial and error, formulating a guess, and verifying that it works. So it's a good question for you to hone your skills on. Avoid peeking too much.
Hint: Play with small cases, find a pattern.
For $ n = 2$, the only possibilities are $ 5 \mid 6^2 + 7^2$ and $5 \mid 6^2 + 8^2$.
If you really want a (potential) value:
$ c = n^2 + 1$ works. (Or we would really like it given the above. Of course, it could be $ c = n^2 + n -1$ too)
Continue with other cases given the above guess.
For $n=3$, we have $ 10 \mid 11^2 + 13^2$, $ 10 \mid 12^2 + 14 ^2 $.
Note: We can quickly verify that for $n=3$, $c = n^2 + n - 1=11$ would not work, so our previous guess about the value of $c$ might be accurate.
Note: The only other solution is $ 13 \mid 10^2 + 15^2$. This "seems" out of place, so we will ignore it for now.
If you really want another (potential) value:
Notice that $ 6 = 2^2 + 2, 12 = 3^2 + 3$. So we hope that $n^2 + n$ will work.
And if you must have the answer, the above cases - $5\mid 6^2 + 8^2, 10 \mid 12^2 + 14^2$ - yield the family:
$(a,b,c) = (n^2 + n, n^2+n+2, n^2 + 1)$.
$a^2 + b^2 = c\times 2(n^2+2n+2)$
Of course, there could be other families to consider too. For example, $ 5 \mid 6^2 + 7^2, 10 \mid 11^2 + 13^2 $ yields the family
$(a,b,c) = (n^2 + 2, n^2+n+1, n^2 + 1)$
$a^2+b^2 = c \times (2n^2 + 2n + 5)$
Note that this is the family that your original working suggests, which occurs when $ z = 1, y-z = 1, x-z = n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $xy\frac{dy}{dx} = \sqrt{x^2 - y^2 - x^2y^2 - 1}$ Can anyone please help me solving the below
$$
xy\frac{dy}{dx} = \sqrt{x^2 - y^2 - x^2y^2 - 1}
$$
If there would have been + instead of -, It would have been easier.
I tried 2 ways as mentioned below :
1. Putting $x^2 = u, y^2 = v$
2. Squaring both sides.Then, dividing both sides by $x^2y^2$
But didn't able to go much further.
Any hint would be appreciated. Thanks in advance.
| Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\therefore2x^2y\dfrac{dy}{du}=\sqrt{u-y^2-uy^2-1}$
$2uy\dfrac{dy}{du}=\sqrt{u-y^2-uy^2-1}$
Let $v=y^2$ ,
Then $\dfrac{dv}{du}=2y\dfrac{dy}{du}$
$\therefore u\dfrac{dv}{du}=\sqrt{u-v-uv-1}$
Let $w=\sqrt{u-v-uv-1}$ ,
Then $w^2=u-1-(u+1)v$
$v=\dfrac{u-1}{u+1}-\dfrac{w^2}{u+1}$
$\dfrac{dv}{du}=\dfrac{2}{(u+1)^2}+\dfrac{w^2}{(u+1)^2}-\dfrac{2w}{u+1}\dfrac{dw}{du}$
$\therefore\dfrac{2u}{(u+1)^2}+\dfrac{uw^2}{(u+1)^2}-\dfrac{2uw}{u+1}\dfrac{dw}{du}=w$
$\dfrac{2uw}{u+1}\dfrac{dw}{du}=\dfrac{uw^2}{(u+1)^2}-w+\dfrac{2u}{(u+1)^2}$
$w\dfrac{dw}{du}=\dfrac{w^2}{2(u+1)}-\dfrac{(u+1)w}{2u}+\dfrac{1}{u+1}$
This belongs to an Abel equation of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What's the size of an angle in a triangle with sides $\sin(x), \cos(x),$ and $\tan(x)$?
Imagining a scalene triangle with sides $\sin(x), \cos(x)$ and $\tan(x)$, how would you find angle $x$ if it was between $\cos(x)$ and $\sin(x)$ when $0<x<\frac{\pi}{2}$?
I tried using the law of cosines but it lead nowhere and honestly haven't gone very far.
$$\cos(x)=\frac{\sin^2(x)+\cos^2(x)-\tan^2(x)}{2\sin(x)\cos(x)}\\
\cos(x)=\frac{1-\tan^2(x)}{\sin(2x)}$$
| From the law of cosines, we have that $$c^2 = a^2 + b^2 - 2ab\cos(C)$$
In this case, we have that $c = \tan(x), a = \cos(x), b = \sin(x)$, and $C = x$.
Plugging those in, we get $$\tan^2(x) = \cos^2(x) + \sin^2(x) - 2\cos(x)\sin(x)\cos(x)$$
After making the simplification $\cos^2(x) + \sin^2(x) = 1$ and multiplying by $\cos^2(x)$, the result is $$\sin^2(x) = \cos^2(x)-2\cos^4(x)\sin(x)$$
Making the substitution $u = \sin(x)$, we get $$2u^{5}-4u^{3}+2u^{2}+2u-1=0$$
This is a quintic equation with no closed form for the roots. However, WolframAlpha says the relevant root is approximately $0.463$. $x$ is then the $\arcsin$ of this, which means that $$x \approx 0.481$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to:
$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$
Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\
\sqrt[3]{6+x} &= x \\
x^3 &= 6+x \\
x^3-2x^2+2x^2-4x+3x-6 &= 0 \\
(x-2)(x^2+2x+3) &= 0 \\
x &= 2
\end{align}
$$
| So the equation is as follows
$$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}$$
Notice that when you repetedly put back $x$ in the RHS, you get the following infinite sum.
$$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+{\cdots}}}}}}}$$
From here, you can replace the part from the first cube root with $x$ to get
$$x=\sqrt[3]{6+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
If $x\geq 1$, then $x^{x^{\cos(x)}}\geq (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$ Hi it's a curious problem that I really don't understand :
Let $x\geq 1$ then we have :
$$x^{x^{\cos(x)}}\geq (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$$
It's very curious for me because the accuracy of the inequality increase when $x$ increase
But clearing the expression or taking both side logarithm we get :
$$x^{\cos(x)}\ln(x)\geq \ln(1-\sqrt{x}+x)(x+1)^{\cos(x)}$$
But it doesn't help at all to explain this mystery .
It seems that there is similarity between the graphic of $f(x)=x^n$ $n$ a natural number and $g(x)=x^{x^{\cos(x)}}- (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$
I think furthermore that our function is quasiperiodic
but it's too fuzzy as definition .
That's all for me If you have nice ideas it would be great .
Thanks a lot for sharing your time and knowledge .
| Alternative solution:
It suffices to prove that
$$x^{\cos x}\ln x \ge (x+1)^{\cos x}\ln (1 - \sqrt{x} + x)$$
which is written as
$$ \ln x
\ge \left(1 + \frac{1}{x}\right)^{\cos x}\ln (1 - \sqrt{x} + x). $$
It suffices to prove that
$$ \ln x
\ge \left(1 + \frac{1}{x}\right)\ln (1 - \sqrt{x} + x). $$
It suffices to prove that
$$ x\ln x
\ge (x+1)\ln (1 - \sqrt{x} + x). $$
Let $f(x) = x\ln x - (x+1)\ln (1 - \sqrt{x} + x)$. We have
\begin{align}
f'(x) &= \ln \frac{x}{1 - \sqrt{x} + x} - \frac{x-1}{2(1 - \sqrt{x} + x)\sqrt{x}}\\
&\ge \frac{2(\frac{x}{1 - \sqrt{x} + x}-1)}{\frac{x}{1 - \sqrt{x} + x}+1}
- \frac{x-1}{2(1 - \sqrt{x} + x)\sqrt{x}}\\
&= \frac{(2\sqrt{x} - 1)(\sqrt{x} - 1)^3}{2(1 - \sqrt{x} + x)\sqrt{x}(2x + 1 - \sqrt{x})}\\
&\ge 0
\end{align}
where we have used $\ln u \ge \frac{2(u-1)}{u+1}$ for all $u \ge 1$ (Note:
$\frac{\mathrm{d}}{\mathrm{d} u}(\ln u - \frac{2(u-1)}{u+1}) = \frac{(u-1)^2}{u(u+1)^2}$).
Also, $f(1) = 0$. Thus, $f(x)\ge 0$ for all $x\ge 1$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Compute the $PV\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$ Problem :
Evaluate the closed form of : $PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$
Wolfram alpha give me :
$I=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx=\sqrt{3}-\coth^{-1}{\sqrt{3}}$
But i can't get by my try as following :
$\cos (4x)=8\cos^{4} x-8\cos^{2} x+1$
And
$\cos (3x)=4\cos^{3} x-3\cos x$
And I know that
$PV\displaystyle\int_0^{\frac{π}{3}}\frac{1}{\cos (3x)}dx=0$
So I need to find
$J=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos^{2} x}{\cos (3x)}dx=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\sin^{2} x}{\cos (3x)}dx$
Now take $y=\cos x$
$J=\displaystyle\int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{4x^{3}-3x}dx$
From here I don't know how I complete
| $$\mathrm{PV}\int_0^{\frac \pi3}\frac{\cos 4x}{\cos 3x}\mathrm{d}x=\lim_{\epsilon\to 0^+}\left(\int_0^{\frac \pi 6-\epsilon}\frac{\cos 4x}{\cos 3x}\mathrm{d}x+\int_{\frac \pi 6+\epsilon}^{\frac \pi 3}\frac{\cos 4x}{\cos 3x}\mathrm{d}x\right) $$
and you can find the antiderivative
$$\int \frac{\cos 4x}{\cos 3x}\mathrm{d}x=\int\frac{8\cos^4 x-8\cos^2 x+1}{\cos x(4\cos^2 x-3)}\mathrm{d}x=\int\frac{8\cos^4 x-8\cos^2 x+1}{\cos^2 x(4\cos^2 x-3)}\cos x\,\mathrm{d}x$$
by substituting $u=\sin x$ which transforms the integral to
$$\int\frac{8u^4-8u^2+1}{4u^4-5u^2+1}\mathrm{d}u=\int\left(2+\frac{2u^2-1}{(u-1)(u+1)(2u-1)(2u+1)}\right)\mathrm{d}u $$
and you continue with partial fractions. After you find the antiderivative, plug in the limits of integration, and take limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Am I computing this conditional expectation correctly? Let $f_X(t) = (5/6 - t^2)\mathsf 1_{(-1,1)}(t)$ be the density of $X$. Then according to this post, we may compute
$$
\mathbb E[X\mid X>k] = \frac1{1-F(k)}\int_k^1 tf_X(t)\ \mathsf dt
$$
We have $1-F(k) = \int_k^1 (5/6 - t^2)\ \mathsf dt = \frac{1}{6} \left(2 k^3-5 k+3\right)$
and
$$
\int_k^1 t(5/6-t^2)\ \mathsf dt = \frac{1}{12} \left(3 k^4-5 k^2+2\right),
$$
hence
$$
\mathbb E[X\mid X>k] = \frac{\frac{1}{12} \left(3 k^4-5 k^2+2\right)}{\frac{1}{6} \left(2 k^3-5 k+3\right)},
$$
so that
$$
\mathbb E[X\mid X>k] = \frac{1}{2} \left(2 k^3-5 k+3\right) \left(3 k^4-5 k^2+2\right).
$$
| There is no $\omega$ in your question.. but there is one in the other post, you formula should be :
$$E[X\mid X>k] = \frac1{1-F(k)}\int_k^\infty tf_X(t)\ \mathsf dt$$
Let $-1 \leq k \leq 1$ (the other cases are trivial)
$$F(k)=P(X<k)=\int_{-\infty}^k f_X(t)\ \mathsf dt=\frac{5}{6}(k+1)-\left[\frac{k^3}{3}+\frac{1}{3}\right]$$
$$E\left[X 1_{X>k}\right]=\int_{-\infty}^\infty tf_X(t)1_{t>k}\ \mathsf dt=\int_{k}^1 {t\left(\frac{5}{6}-t^2\right)}\ \mathsf dt=\frac{5}{6}\left(\frac{1}{2}-\frac{k^2}{2}\right)-\left[\frac{1}{4}-\frac{k^4}{4}\right]$$
You can finish by using the following formula
$$E\left[X |X>k\right]=\frac{E\left[X 1_{X>k}\right]}{P(X>k)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of $P = \frac{y + z}{x}$
Let $x,y,z > 0$ such that $3x(x+y+z)=yz$ . Find the minimum value of $$P=\frac{y+z}{x}$$
$$3x(x+y+z)=yz\Leftrightarrow 3(x+y)(x+z)=4yz$$
Or $$3(\frac{x}{y}+1)(\frac{x}{z}+1)=4$$
Let $a=x/y, b=x/z ( a,b>0)$
We have: $$4=3(a+1)(b+1)\ge3 \cdot 2 \sqrt a \cdot 2 \sqrt b=12 \sqrt {ab} \rightarrow \sqrt {ab}\le \frac 1 3$$
And $$P=\frac{y+z}{x}=\frac{1}{a}+\frac{1}{b}\ge \frac{2}{\sqrt {ab}}\ge 6$$
But by WA $Min_P=6+4\sqrt 3$. Is my solution is wrong?
| Your solution is wrong because in your solution equality occurs when $a=b=1$ which is not possible because $x=y=z$ doesn't satisfy the constraints. Instead, you should use method of lagrange multipliers. Let $m = \frac{y}{x}$ and $n = \frac{z}{x}$. Then, lagrangian
$$L = m+n-\lambda(3(1+m+n)-mn)$$
(the constraint is obtained by dividing the given equation by $x^2$ on both sides)Equations formed after differentiating $L$ with respect to $m, n$ and $\lambda$ are $$1 =\lambda(3-n)$$
$$1 = \lambda(3-m)$$
$$3(1+m+n)=mn$$
From first two equations, $m=n$, using it in third equation, we get,
$$m^2-6m-3=0$$
$$\implies m = 3+2\sqrt{3}$$
Required minimum value is $6+4\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of $ab+bc+ca$ depending on given constraints We have to minimize the function $ab+bc+ca$ on the given condition that $a^2+b^2+c^2=1$.
I have tried the following approach (but I don't think I'm right )
$$\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)
$$
We know that all squares are greater than equal to 0 and hence: $\left(a+b+c\right)^2\geq0$
$$\therefore a^2+b^2+c^2+2\left(ab+bc+ca\right) \geq0$$
$$\Rightarrow 1+2\left(ab+bc+ca\right) \geq0$$
$$\Rightarrow 2\left(ab+bc+ca\right)\geq -1$$
$$\Rightarrow \left(ab+bc+ca\right)\geq -1/2$$
As we can see I have reached a definite answer but my question is since there is a given constraint on $a$,$b$ and $c$ we can't say for sure that this value will be greater than or equal to -1/2, or can we?
| You're doing it right. Another way of looking at what you're doing:
You have
$$ab + bc + ac = {1 \over 2} (a + b + c)^2 - {1 \over 2}(a^2 + b^2 + c^2)$$
Under the constraint that $a^2 + b^2 + c^2$, you therefore have
$$ab + bc + ac = {1 \over 2} (a + b + c)^2 - {1 \over 2}$$
You're trying to minimize this quantity. Since ${1 \over 2} (a + b + c)^2 \geq 0$, it's smallest when $a + b + c = 0$, in which case you have
$$ab + bc + ac = -{1 \over 2}$$
This is the minimum possible value of $ab + bc + ac$, and it is achieved for any $(a,b,c)$ for which $a^2 + b^2 + c^2 = 1$ and $a + b + c = 0$. There are many such $(a,b,c)$ as they correspond to the intersection of the sphere $a^2 + b^2 + c^2 = 1$ and the plane through the origin with equation $a + b + c = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
$x^{4\log x} = \frac{x^{12}}{{10}^8} $ all sum of x? ${10}^8 = \frac{x^{12}}{x^{4\log x}}$
${10}^8 = x^{12 - 4\log x}$
$x > 0$
$x \neq 1$
by guessing, x = 10 is true. but how sum of all x?
| Assuming you are wondering how to systematically get all $x$ for which the original expression is true, you can take logs of both sides (base 10 log) to obtain
\begin{align*}
8 = \log x^{12-4\log x} = (12-4\log x) (\log x) = 12\log x - 4(\log x)^2.
\end{align*}
Organizing a bit gives the equivalent statement,
\begin{align*}
(\log x)^2 - 3\log x + 2 = 0.
\end{align*}
The quadratic formula now gives
\begin{align*}
\log x = \frac{3\pm \sqrt{9-8}}{2} = \{2,1\}.
\end{align*}
The zeros of the function $f(x) = (\log x)^2-3\log x + 2$ are then $x = 10$ or $100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does $f{(x)} = a x^4 - x^3 + ax - a$ have exactly $1$ positive root given $a>0$? Let $$f{(x)} = a x^4 - x^3 + ax - a$$ where $a>0$.
We know that it contains $4$ roots. Also, by the Descartes' rule of signs, we know that
*
*There are either $1$ or $3$ positive roots.
*There is exactly $1$ negative root.
I observed from plotting the function that it may contain exactly one positive root.
Therefore, can we prove/disprove that the function always contain a pair of conjugate complex roots?
Equivalently, can we prove/disprove that it has exactly $1$ positive root?
| Let $f(x)=0$.
Thus, $$a=\frac{x^3}{x^4+x-1}.$$
Now, by the Descartes' rule of signs we see that $x^4+x-1$ has an unique positive root $x_0$.
Thus, for $x>x_0$ we have $\frac{x^3}{x^4+x-1}>0$ and for $0<x<x_0$ we have $\frac{x^3}{x^4+x-1}<0$.
Now, $$\left(\frac{x^3}{x^4+x-1}\right)'=-\frac{x^2(x^4-2x+3)}{(x^4+x-1)^2}<0$$ for any $x>x_0$,
$$\lim_{x\rightarrow x_0^+}\frac{x^3}{x^4+x-1}=+\infty,$$
$$\lim_{x\rightarrow+\infty}\frac{x^3}{x^4+x-1}=0,$$ and $\frac{x^3}{x^4+x-1}$ is a continuous on $(x_0,+\infty),$ which says that the equation
$$a=\frac{x^3}{x^4+x-1}$$ or $$f(x)=0$$ has an unique positive root.
By the same way we see that the equation $f(x)=0$ has an unique negative root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \cos 2\theta \ln\left(\frac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$ $$\int \cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$$
My attempt is as follows:-
$$\ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)=t\tag{1}$$
$$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\cdot\dfrac{\left(\cos\theta-\sin\theta\right)^2-(-\sin\theta-\cos\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)^2}=\dfrac{dt}{d\theta}$$
$$\dfrac{2}{\cos2\theta}=\dfrac{dt}{d\theta}$$
Let's calculate $\cos2\theta$ from equation $1$
$$\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}=e^t$$
$$\dfrac{1+\tan\theta}{1-\tan\theta}=e^t$$
Applying componendo and dividendo
$$\dfrac{2}{2\tan\theta}=\dfrac{e^t+1}{e^t-1}$$
$$\dfrac{e^t-1}{e^t+1}=\tan\theta$$
$$\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$\cos2\theta=\dfrac{(e^t+1)^2-(e^t-1)^2}{(e^t+1)^2+(e^t-1)^2}$$
$$\cos2\theta=\dfrac{4e^t}{2(e^{2t}+1)}$$
$$\cos2\theta=\dfrac{2e^t}{e^{2t}+1}\tag{2}$$
So integral will be
$$\dfrac{1}{2}\cdot\int \left(\dfrac{2e^t}{e^{2t}+1}\right)^2dt$$
$$\dfrac{1}{2}\cdot\int \dfrac{4e^{2t}}{(1+e^{2t})^2}$$
$$e^{2t}+1=y$$
$$2e^{2t}=\dfrac{dy}{dt}$$
$$2e^{2t}dt=dy
$$\int \dfrac{dy}{y^2}$$
$$-\dfrac{1}{y}+C$$
$$-\dfrac{1}{1+e^{2t}}+C$$
$$-\dfrac{1}{1+e^{\ln\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2}}+C$$
$$-\dfrac{1}{1+\dfrac{1+\sin2\theta}{1-\sin2\theta}}+C$$
$$-\dfrac{1-\sin2\theta}{2}+C$$
$$\dfrac{\sin2\theta}{2}+C'$$
And this should be actually wrong because if we differentiate the result, it will give $\cos2\theta$, but integrand is $\cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)$
What am I missing here, checked multiple times, but not able to get the mistake. Any directions?
| You've done nearly all your computations correctly, save for one critical error: after your substitution $$e^t = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta},$$ with $$dt = 2 \sec 2\theta \, d\theta,$$ and $$\cos 2\theta = \frac{2}{e^t + e^{-t}},$$ your integrand should be $$\int \color{red}{\cos 2\theta} \color{blue}{\log \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta}} \, \color{purple}{d\theta} = \frac{1}{2} \int \color{red}{\frac{2}{e^t + e^{-t}}} \cdot \color{blue}{t} \cdot \color{purple}{\frac{2}{e^t + e^{-t}} \, d t} = \frac{1}{2} \int \left(\frac{2}{e^t + e^{-t}}\right)^2 t \, dt.$$ You are missing that extra factor $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Solve complex equations with absolute value $|z|^2+z|z|+\bar{z} = 0$ I literally have no idea on how to solve equations like this:
$$|z|^2+z|z|+\bar{z} = 0$$
Even if I try to rewrite it as
$$(x^2+y^2)+(x+iy)\sqrt{x^2+y^2}+(x-iy) = 0$$
how should I proceed?
| Looking at imaginary parts:
$$y\sqrt{x^2+y^2}-y=0,$$
so $y=0$ or $\sqrt{x^2+y^2}-1=0$.
In the case $y=0$, you then just have the equation $2x^2+x=0$, so $x=0$ or $-\frac{1}{2}$, so $z=0$ or $z=-\frac{1}{2}$. But we find if $z=-\frac{1}{2}$, the real part is not zero, so we reject this solution.
In the case $\sqrt{x^2+y^2}=1$, we have $|z|=1$, so the equation becomes $1+z+\overline{z}=0$, which implies $x=-\frac{1}{2}$, so $y=\pm\sqrt{1-x^2}=\pm\frac{\sqrt{3}}{2}$ and $z=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.
Therefore, the solutions are $z=0,-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $\int \dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx$ $$\int \dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx$$
My attempts is as follows:-
Integrating by parts:-
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(-x\sin x(x\sin x+\cos x)+x\cos x(x\cos x-\sin x)\right)}{(x\cos x-\sin x)^2(x\sin x+\cos x)^2}dx$$
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(-x^2\sin^2x-x\sin x\cos x-x\cos x\sin x+x^2\cos^2x\right)}{(x\cos x-\sin x)^2(x\sin x+\cos x)^2}dx$$
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(x^2\cos2x-x\sin 2x\right)}{(x^2\cos x\sin x+x\cos^2x-x\sin^2x-\sin x\cos x)^2}dx$$
$$\dfrac{1}{(x\cos x-\sin x)(x\sin x+\cos x)}\cdot\dfrac{x^3}{3}-\dfrac{1}{3}\int\dfrac{x^3\left(x^2\cos2x-x\sin 2x\right)}{(\cos x\sin x(x^2-1)+x\cos 2x)^2}dx$$
I thought numerator and denominator in the second would be quite similar, but its not, so I was not able to proceed from here.
Any directions?
| The integrand is $\frac{x\cos x}{x\sin x+\cos x}-\frac{-x\sin x}{x\cos x-\sin x}$. In the numerator of the first fraction, add and subtract the $\sin x$ and in the second -- $\cos x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$ Hi I am stuck solving this problem:
'Employ a change of variables $z=\frac{y}{x}$ to solve the differential equation: $$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$$
Enter the result $y(3)$.
I have done the substitution and then integrated by I cannot get an explicit expression for y, which leaves me confused as to how I can find $y(3)$.
Please help!
| Starting from
$$(x+y)\frac{dy}{dx} -(4x+y)=0$$
rewrite as
$$\frac{dy}{dx}=\frac{4x+4}{x+y}=\frac{4+\frac xy}{1+\frac xy}$$
then with the substitution $z=\frac{y}{x}$ we have $\frac{dy}{dx}=z+x\frac{dz}{dx}$ so
$$z+x\frac{dz}{dx}=\frac{4+z}{1+z} \implies x\frac{dz}{dx}=\frac{4-z^2}{1+z}$$
which we can rewrite as
$$\frac{1+z}{4-z^2}dz=\frac{1}{x}dx$$
where a partial fraction decomposition of the LHS forms
$$\frac{1+z}{4-z^2}=\frac{-1}{4(z+2)}-\frac{3}{4(z-2)}$$
hence
$$-\frac{1}{4}\left(\frac{1}{z+2}+\frac{3}{z-2}\right)dz=\frac{1}{x}dx$$
whereby integrating both sides forms
$$-\frac{1}{4}\Big(\ln|z+2|+3\ln|z-2|\Big)=\ln|x|+C$$
or
$$\ln|z+2|+3\ln|z-2|=-4\ln|x|+C$$
which can be rewritten as
$$(z+2)(z-2)^3=Cx^{-4} \implies C=x^4(z+2)(z-2)^3$$
therefore substituting $z=\frac{y}{x}$
$$C=x^4\left(\frac{y}{x}+2\right)\left(\frac{y}{x}-2\right)^3$$
the initial/boundary condition of $y(1)=2$ forms
$$C=1(4)(0)=0$$
hence since $x\neq 0$ we see that
$$(y+2x)(y-2x)^3=0$$
which implies
$$y=-2x,\quad y=2x$$
where $y=-2x$ cannot be the solution since $y(1)=2$. Therefore, $y=2x$ is the solution and $y(3)=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Two rectangles: The $1$st has twice the perimeter of the $2$nd and the $2$nd has twice the area of the $1$st. How can this be solved using just algebra, where the first rectangle has sides $a$ and $b$, and the second rectangle has sides $c$ and $d$?
These are the two equations that follow:
$$a + b = 2(c + d)$$
and
$$2ab = cd$$
| Let $a,b,c,d$ be as described. Note that this implies
$$a,b,c,d>0$$
else we wouldn't have a rectangle
Now
$$ab=A_1=\frac{1}{2}A_2=\frac{1}{2}cd$$
$$2a+2b=P_1=2P_2=4c+4d$$
This gives us the set of equations
$$2ab=cd$$
$$a+b=2(c+d)$$
Now, this has an infinite number of solutions as we have four unknowns and two equations. Solving for $a$ and $c$ gives us
$$a=\frac{d (b-2 d)}{4 b-d}$$
$$c=\frac{2 b (b-2 d)}{4 b-d}$$
Since these are positive, we know
$$b>2d$$
$$d< 4b$$
which implies $b>2d$. However, we can also write this as
$$b<2d$$
$$d>4b$$
which implies $d>4b$. Thus, one infinite family of solutions might be $b=3d$ which gives us
$$a=\frac{d}{11}$$
$$b=3d$$
$$c=\frac{6d}{11}$$
$$d=d$$
Then for any $d>0$ we get
$$A_1=ab=\frac{6d^2}{22}$$
$$A_2=cd=\frac{6d^2}{11}$$
$$P_1=2a+2b=\frac{68 d}{11}$$
$$P_2=2c+2d=\frac{34 d}{11}$$
which satisfy the original problem. For example, $a=1, b=33, c=6,$ and $d=11$ would give areas of $33$ and $66$ and perimeters of $68$ and $34$, respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve in $\mathbb{R^{3}}$ : $\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$ Solve in $\mathbb{R^{3}}$ the following system :
$$\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$$
My try but I think not complete :
The system equivalent :
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}$$
Then from first equation we have :
$$x+1=\frac{20}{y+1}$$
So third equation give :
$$z+1=3\frac{y+1}{4}$$
Second equation $\implies $
$(y+1)^{2}=16$ then $y=3,-5$
This mean $x=\frac{17}{3}$ or $x=-6$ also $z=3$ or $z=-3$
I'm correct or no ??
| Your method of solution is correct. However, you've made a few trivial mistakes in computing the values of $\ x\ $ and $\ z\ $ when $\ y=3\ $, and the value of $\ z\ $ when $\ y=-5\ $, as John Omielan's answer demonstrates.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Nature of the critical point $(0,0)$ of the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$ Consider the function $$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$ The point $(0,0)$ is a critical point. Observe,
\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\
f_{yy} & = 4, f_{yy}=4
\end{align*}
So, in order to determine the nature of the above critical point, we need to check the Hessian at $(0,0)$ which is $0$ and hence the test is inconclusive. $$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$So, I tried to see the function on slices like $y=0$ and $y=x$ but nothing worked. So please suggest me how do I find the nature of the critical point in this case?
| You might note that your function factors as
$$(x^2-y)(x^4-2y).$$
So there are easy to find regions in the $xy$-plane where the function is positive an negative. Close to the origin and between the curves $y=x^2$ and $y=x^4/2$, the function is negative. This suggests trying the limit along the curve $y=x^3$. It's not too horrible to analyse
$$f(x,x^3) = 3x^6-2x^5 -x^7$$
around $x=0$ to see that it's negative there.
Comparing with the curve given by $y=0$, we get a saddle point at $(0,0).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \frac{1}{(x^4-1)^2}dx$ without partial fraction decomposition The substitution $x^2=\sec\theta$ doesn't seem to lead anywhere. I know the key is to manipulate it into a form
$$\displaystyle \int \dfrac{\text{d}\left(x^a\pm\frac{1}{x^a}\right)}{f\left(x^a\pm\frac{1}{x^a}\right)}$$
But I fail to do so.
| $$\int\dfrac{dx}{(x^4-1)^2}=\int\dfrac{x^3}{(x^4-1)^2}\cdot\dfrac1{x^3}dx$$
$$=\dfrac1{x^3}\cdot\int \dfrac{x^3}{(x^4-1)^2} dx+\dfrac34\int\dfrac{dx}{x^4(x^4-1)}$$
We can use $$\int\dfrac{dx}{x^4(x^4-1)}=\int\dfrac{x^4-(x^4-1)}{x^4(x^4-1)}dx$$
and $\displaystyle\int\dfrac2{x^4-1}=\int\dfrac{x^2+1-(x^2-1)}{x^4-1}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability that two plates with numbers ranging from $0001-9999$ have at least two common digits in them. I would calculate it as this $\frac{10^4-9^4-9^4+8^4}{10^4}$. But it may be incorrect Because I summing the $10^4-9^4$ and $9^4-8^4$. Is it the correct?
| Let's consider "common digits" to mean the digit appears in both plate numbers in any order. Let's find the probability that the two plates have no common digit.
If the first plate has one distinct digit, this can happen in any of 9 ways ($0000$ is not possible), so there are $9^4-1$ ways to choose the numbers for the second plate.
If the first plate has two distinct digits, there are 14 possible orders for the two digits: $\dfrac{4!}{3!1!}+\dfrac{4!}{2!2!}+\dfrac{4!}{1!3!}$. There are $\dbinom{10}{2}$ ways to choose the two digits. There are $8^4$ ways to choose a four digit number for the second plate without those two digits. We have to subtract off the number of ways to choose the first plate with no zeros and the second plate with all zeros, since that is a forbidden number: $14\dbinom{9}{2}$.
If the first plate has three distinct digits, there are $\dfrac{4!}{2!1!1!}+\dfrac{4!}{1!2!1!}+\dfrac{4!}{1!1!2!} = 36$ ways to order the numbers, and $\dbinom{10}{3}$ ways to choose them. Then there are $7^4$ ways to choose the number for the second plate without any of the three digits from the first plate. However, this overcounts when the second plate is $0000$, so we have to subtract $36\dbinom{9}{3}$.
Finally, if the first plate has four distinct digits: $\dbinom{10}{4}4!6^4-\dbinom{9}{4}4!$
Complement of the total probability that there are no common digits (which means at least one shared digit):
$$1-\dfrac{9(9^4-1)+14\dbinom{10}{2}8^4-14 \dbinom{9}{2}+36\dbinom{10}{3}7^4-36\dbinom{9}{3}+24\dbinom{10}{4}6^4-24\dbinom{9}{4}}{(10^4-1)^2} = \dfrac{8,937,985}{11,108,889} \approx 80\%$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solutions to $X^3 = I_2$ Show that there are an infinite number of solutions to $X^3 = I_2$ in $M_2(\mathbb{Q})$.
$\operatorname{det} (X) = 1$ because $X^3 = I_2$ and $\operatorname{det} (X) \in \mathbb{Q}$
$X^2 = X^{-1}$ and, using Cayley-Hamilton $X^2 = \operatorname{tr}(X) X - I_2$ so $$ X^{-1} = \operatorname{tr}(X)X - I_2 $$ but when i try to solve this equation the solution it gives do not verify $X^3 = I_2$. What am I doing wrong?
| Cayley-Hamilton gives us $X^2-\text{tr}(X)X+I = 0$, from which we find $X(X-\text{tr}(X)I) = -I$, and hence $X^{-1} = \text{tr}(X)I-X$. Letting
\begin{align*}
X = \left( \begin{matrix}
a & b \\
c & d
\end{matrix} \right)
\end{align*}
we get the following system of equations
\begin{align*}
a^2+bc =& d \\
b(a+d) =& -b \\
c(a+d) =& -c \\
d^2+bc =& a
\end{align*}
Which yields either $X = I$ or
\begin{align*}
X = \left( \begin{matrix}
a & b \\
c & -1-a
\end{matrix} \right)
\end{align*}, where $bc = -1-a-a^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.