Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Using Residue theorem to evaluate the integral Using Residue theorem to evaluate the integral:
$$\int_0^{\infty} \frac{x^2}{x^4 + 5x^2 +6}dx$$
I am using partial fraction to get:
$$\int_0^{\infty} \left( \frac{3}{x^2 +3} - \frac{2}{x^2+2} \right)dx$$
Then, next step, can someone show me how to use Residue theorem to evaluate the integral.
| You just use partial fraction expansion again, for each of the quadratic terms. We have
$$\frac{1}{x^2+3}=\frac{1}{i2\sqrt{3}}\left(\frac{1}{x-i\sqrt{3}}-\frac{1}{x+i\sqrt{3}}\right)$$
and
$$\frac{1}{x^2+2}=\frac{1}{i2\sqrt{2}}\left(\frac{1}{x-i\sqrt{2}}-\frac{1}{x+i\sqrt{2}}\right)$$
Can you finish from here?
SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO SEE THE REST OF THE ANSWER
First we note that the integrand is even and therefore we can write the integral of interest $I$ as $$I=\int_0^{\infty}\frac{x^2}{x^4+5x^2+6}dx=\frac12\int_{-\infty}^{\infty}\frac{x^2}{x^4+5x^2+6}dx$$ Next, we move to the complex plane and analyze the integral $$\oint_{C}\frac{z^2}{z^4+5z^2+6}dz$$where $C$ is the contour in the upper-half plane comprised of the real-line segment from $x=-R$ to $x=R$ and the semi-circle with radius $R$ and centered at the origin. Using the Residue Theorem, we have $$\oint_{C}\frac{z^2}{z^4+5z^2+6}dz=2\pi i \left(\frac{-2}{2i\sqrt{2}}+\frac{3}{2i\sqrt{3}}\right)=\pi(-\sqrt{2}+\sqrt{3}) $$ Notice that as as $R$ goes to infinity, the contribution from the integral over $C_R$ vanishes while the integral over the real line segment is equal to twice the integral of interest as $R\to \infty$. Therefore, we have $$I=\frac{\pi}{2}(\sqrt{3}-\sqrt{2})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| Note that,
$$
\lim_{x \to 0}\dfrac{\sqrt{x + a^2} - a}{x} = \lim_{x \to 0}\dfrac{1}{\sqrt{x + a^2} + a} = \dfrac{1}{2a}
$$
for $a > 0$. Thus,
\begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4} = \dfrac{\lim_{x \to 0}\dfrac{\sqrt{x + 25} - 5}{x}}{\lim_{x \to 0}\dfrac{\sqrt{x + 16} - 4}{x}} = \dfrac{1/10}{1/8} = \dfrac{4}{5}
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Gauss Jordan elimination method. Solve by Gauss-Jordan elimination method.
$$x + y + z = 2 \\
x + 3y + 3z = 0 \\
x + 3y + 5z = 2$$
Putting this in reduced row-echelon form:
$$\left(\begin{array}{ccc|r}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & -2 \\
0 & 0 & 1 & 1
\end{array}\right)
$$
So then
$\begin{pmatrix}
x \\ y \\ z
\end{pmatrix} =
\begin{pmatrix}
3 \\ -2 \\ 1
\end{pmatrix}$ and is
therefore consistent system.
Is my final answer correct?
| $$\begin{cases} x+y+z=2 \\ x+3y+3z=0 \\ x+3y+5z=2 \end{cases}\Rightarrow \begin{cases} x+y+z=2 \\ x+3y+3z=0 \\ 2z=2 \end{cases}\Rightarrow \begin{cases} x+y=1 \\ x+3y=-3 \\ z=1 \end{cases}\Rightarrow \begin{cases} x+y=1 \\ 2y=-4 \\ z=1 \end{cases}\begin{cases} x=3 \\ y=-2 \\ z=1 \end{cases}\\ \quad \quad \quad \quad \quad \quad \\ \quad \quad \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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algebra problem, Solve the equation
a nice problem: Solve the equation $$\left|2x-57-2\sqrt{x-55}+\frac{1}{x-54-2\sqrt{x-55}}\right|=|x-1|.$$
It's just for sharing a new ideas, thanks:)
| The left hand side of expression, by completing the square, is:
\begin{align*}
& \left|(\sqrt{x - 55})^2 - 2\sqrt{x - 55} + 1 + x - 3 + \frac{1}{(\sqrt{x - 55})^2 - 2\sqrt{x - 55} + 1}\right| \\
= & \left|(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} + x - 3\right|.
\end{align*}
Since $x \geq 55$, the expression inside the absolute symbol is positive, so is the one in the right hand side, so the equation can be written as
\begin{align*}
(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} + x - 3 & = x - 1 \\
(\sqrt{x - 55} - 1)^2 + \frac{1}{(\sqrt{x - 55} - 1)^2} & = 2 \tag{1}
\end{align*}
But by the algebraic-geometric inequality, the left hand side of $(1)$ is at least $2$, and the equality holds if and only if
$$(\sqrt{x - 55} - 1)^2 = 1$$
which gives $x = 55$ and $x = 59$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{1999} < \prod_{i=1}^{999}{\frac{2i−1}{2i}} < \frac{1}{44}$
Prove that $$\dfrac{1}{1999} < \prod_{i=1}^{999}{\dfrac{2i−1}{2i}} < \dfrac{1}{44}$$
from the 1997 Canada National Olympiad.
I have been able to prove the left half of the inequality using induction. Need help with the second part.
Prove $\prod_{i=1}^{n}{(1-\frac{1}{2i})} \ge \frac{1}{2n}$
Define $$p(n)=\prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \tag{1}$$
We wish to show $$p(n)\ge\frac{1}{2n},\quad\forall{n}\in\mathbb{Z^+} \tag{2}$$
For the base case, $n=1$, we have $p(1)=1-\frac{1}{2}=\frac{1}{2}\ge\frac{1}{2}$ which holds.
Assume the induction hypothesis (2) for $n$. Then for $n+1$, we can write:
$$\begin{align}
p(n+1) = \prod_{i=1}^{n+1}{\left(1-\frac{1}{2i}\right)} &= \left(1-\frac{1}{2(n+1)}\right) \cdot \prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \\
&\ge \left(1-\frac{1}{2(n+1)}\right) \cdot \frac{1}{2n} &(\text{by IH}) \\
&\ge \left(\frac{2n+1}{2(n+1)}\right) \cdot \frac{1}{2n} \\
&\ge \left(\frac{1}{2(n+1)}\right) \cdot \frac{2n+1}{2n} \\
&> \left(\frac{1}{2(n+1)}\right) &(\text{for }n\ge1)
\end{align}$$
This proves the induction step, establishing (2) for all positive $n$. Hence, we conclude
$$\boxed{\prod_{i=1}^{999}{\dfrac{2i−1}{2i}} = p(999) \ge \frac{1}{2\times999} = \frac{1}{1998} > \frac{1}{1999}}$$
Prove $\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} < \frac{1}{44}$
For this part, I have applied AM-GM to get an upper bound with a harmonic sum.
$$\left(\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)}\right)^{1/999} \le \frac{1}{999} \cdot \sum_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} \tag{3}$$
I can't see a way of evaluating the summation on the RHS.
| You have made the problem much hard than it is suppose to be.
For the left hand side, simply observe that $\frac{1}{2}>\frac{1}{3}$ and $\frac{3}{4}>\frac{3}{5}$ and so on... to $\frac{1997}{1998}>\frac{1997}{1999}$. Now, take the product of these inequalities to get the desired result.
For the other inequality, it is a similar trick except you need to observe that $P^2<\frac{1}{1999}<\frac{1}{1936}=\frac{1}{44^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating a function in a closed form Let $$g(x)=\lim_{n\rightarrow\infty}\sqrt{n}\int_0^x z^2e^{n(\cos^2z-1)}\ \mathsf dz$$
Evaluate $g$ in closed form. The answer is right here: http://math.nyu.edu/student_resources/wwiki/index.php/Advanced_Calculus:_2004_January:_Problem_2
However, in the last part of the answer. It says at singular $x=k\pi$, $$\int_{k\pi-\delta}^{k\pi} z^2e^{n(\cos^2z-1)}\ \mathsf dz\stackrel{\delta\to0}\longrightarrow \frac{1}{2}k^2\pi^{5/2}$$
Can someone please explain why this happens? Also, why does $$g(x)=\frac{\pi^{5/2}}6k(2k^2+1)$$ when $x=k\pi$. Thanks so much
| Actually, that limit for $x = k\pi$ that you're interested in is not as $\delta \to 0$ but as $n \to \infty$.
Heuristically that does follow from symmetry. It was shown in the line above that
$$
\int_{k\pi-\delta}^{k\pi+\delta} \sqrt{n}z^2e^{n(\cos^2z-1)}\ \mathsf dz\stackrel{n\to\infty}\longrightarrow k^2\pi^{5/2},
$$
and the dominant factor in the integrand, $e^{n(\cos^2 z-1)}$, is even about the point $z=k\pi$, so you would expect that
$$
\int_{k\pi-\delta}^{k\pi} \sqrt{n}z^2e^{n(\cos^2z-1)}\ \mathsf dz\stackrel{n\to\infty}\longrightarrow \frac{1}{2}k^2\pi^{5/2}.
$$
It still needs to be shown rigorously, though, but this is done just as it was done when the interval of integration was $(k\pi-\delta,k\pi+\delta)$:
$$
\begin{align}
&\int_{k\pi-\delta}^{k\pi} \sqrt{n}z^2e^{n(\cos^2z-1)}\ \, dz \\
&\qquad = \int_{-\sqrt{n}\delta}^0 \left(\frac{v}{\sqrt{n}}+k\pi\right)^2 e^{-v^2+O(1/n^2)}\,dv \\
&\qquad \to \int_{-\infty}^{0} (k\pi)^2 e^{-v^2}\,dv \\
&\qquad = \frac{1}{2}k^2\pi^{5/2}.
\end{align}
$$
For your second question, if $x = k\pi$ then
$$
\begin{align}
g(x) &= \lim_{n\to\infty} \int_0^{k\pi-\delta} \sqrt{n}z^2 e^{n(\cos^2 z-1)}\,dz + \lim_{n\to\infty} \int_{k\pi-\delta}^{k\pi} \sqrt{n}z^2 e^{n(\cos^2 z-1)}\,dz \\
&= \sum_{0<j\pi<x} j^2\pi^{5/2} + \frac{1}{2}k^2\pi^{5/2} \\
&= \sum_{0<j\pi < k\pi} j^2\pi^{5/2} + \frac{1}{2}k^2\pi^{5/2} \\
&= \pi^{5/2}\sum_{j=1}^{k-1} j^2 + \frac{1}{2}k^2\pi^{5/2} \\
&= \frac{\pi^{5/2}}{6} (k-1) k (2 k-1) + \frac{1}{2}k^2\pi^{5/2} \\
&= \frac{\pi^{5/2}}{6} k (2k^2+1).
\end{align}
$$
| {
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The sum of the lengths of the hypotenuse and another side of a right angled triangle is given. The sum of the lengths of the hypotenuse and another side of a right angled triangle is given.The area of the triangle will be maximum if the angle between them is:
$(A)\frac{\pi}{6}\hspace{1cm}(B)\frac{\pi}{4}\hspace{1cm}(C)\frac{\pi}{3}\hspace{1cm}(D)\frac{5\pi}{12}$
Let $a,b$ are sides of a right triangle other than hypotenuse.Then given that $\sqrt{a^2+b^2}+a$=constant=$k$
Area of triangle=$\frac{1}{2}ab=\frac{1}{2}a\sqrt{(k-a)^2-a^2}$
and then?
| Let $\theta$ be the angle between the hypotenuse & one side of length say $x$ of a right angled triangle such that $0<\theta<\frac{\pi}{2}$ then we have $$\text{hypotenuse}=x\sec \theta$$
$$\text{another side}=x\tan \theta$$
Now, according to the given condition, we have $$x\sec\theta+x=c\ \ \text{(any constant)}$$ $$x=\frac{c}{1+\sec\theta}$$ Now, the area say $A$ of right angled triangle is given as $$A=\frac{1}{2}(x)(x\tan \theta)=\frac{1}{2}x^2\tan \theta$$ Now, setting the value of $x$, we get $$A=\frac{1}{2}\left(\frac{c}{1+\sec\theta}\right)^2\tan \theta=\frac{c^2}{2}\frac{\tan \theta}{(1+\sec\theta)^2}$$ $$\frac{dA}{d\theta}=\frac{c^2}{2}\frac{(1+\sec\theta)^2(\sec^2\theta)-2\tan \theta(1+\sec\theta)(\sec\theta\tan \theta)}{(1+\sec\theta)^4}$$ For maximum value of $A$, we have $\frac{dA}{d\theta}=0$ hence $$\frac{c^2}{2}\frac{(1+\sec\theta)^2(\sec^2\theta)-2\tan \theta(1+\sec\theta)(\sec\theta\tan \theta)}{(1+\sec\theta)^4}=0$$ $$\sec\theta(1+\sec\theta)(\sec\theta+\sec^2\theta-2\tan^2\theta)=0$$ $$\sec\theta(1+\sec\theta)^2(2-\sec\theta)=0$$ $$\sec\theta \neq0$$ $$ 1+\sec\theta=0\iff \cos\theta=-1\iff \theta=\pi$$ but $0<\theta<\frac{\pi}{2}$ hence this solution is not acceptable. Now, $$2-\sec\theta=0\iff \sec\theta =2\iff $$ $$\theta=\frac{\pi}{3}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{option (C)}:\ \ \frac{\pi}{3}}}$$
| {
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Using right-hand Riemann sum to evaluate the limit of $ \frac{n}{n^2+1}+ \cdots+\frac{n}{n^2+n^2}$ I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:
\begin{align*}
\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\
&=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4}
\end{align*}
where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
| Notice, we have $$\lim_{n\to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\dots +\frac{n}{n^2+n^2}\right)=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}$$
$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}$$
Let, $\frac{r}{n}=x\implies \lim_{n\to \infty}\frac{1}{n}=dx\to 0$
$$\text{upper limit of x}=\lim_{n\to \infty }\frac{n}{n}=1$$
$$\text{lower limit of x}=\lim_{n\to \infty }\frac{1}{n}=0$$ Hence, using integration with proper limits, we get
$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}= \int_ {0}^{1}\frac{dx}{1+x^2}$$ $$=\left[\tan^{-1}(x)\right]_{0}^{1}$$
$$=\left[\tan^{-1}(1)-\tan^{-1}(0)\right]$$ $$=\left[\frac{\pi}{4}-0\right]=\frac{\pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does this "miracle method" for matrix inversion work? Recently, I answered this question about matrix invertibility using a solution technique I called a "miracle method." The question and answer are reproduced below:
Problem: Let $A$ be a matrix satisfying $A^3 = 2I$. Show that $B = A^2 - 2A + 2I$ is invertible.
Solution: Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, by power series expansion, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A + 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, compute the product $BB^{-1}$, and you'll find that, miraculously, this answer works!
I discovered this solution technique some time ago while exploring a similar problem in Wolfram Mathematica. However, I have no idea why any of these manipulations should produce a meaningful answer when scalar and matrix inversion are such different operations. Why does this method work? Is there something deeper going on here than a serendipitous coincidence in series expansion coefficients?
| Many (but not all!) things about scalar functions work with matrices through power series. Everything is easier when $A^n=A$ (I don't remember the name of this property) or the matrix is nilpotent ($A^n=0$). For the more numerically oriented people, I suggest "Computing matrix functions": http://dx.doi.org/10.1017/S0962492910000036
| {
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"timestamp": "2023-03-29T00:00:00",
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Application of Poincaré-Bendixson theorem Consider the system
$$x' = 3xy^2-x^2y \\
y' = 5x^2y - xy^2$$
Show that the system has no periodic solutions.
This is a tricky example. Linearization leads nowhere and I'm having a hard time constructing a Lyapunov function that does the trick. $V = 1/2(x^2+y^2)$ gives
$$V'(x,y) = 3x^2y^2-x^3y +5x^2y^2 -xy^3 = 8x^2y^2 -xy(x^2+y^2))$$
But this doesn't tell us much nice things about the origin. If anything, it looks as though the origin is repelling since small perturbations gives us that the $8x^2y^2$ term dominates the minus term. Maybe it's possible to show that there are no elliptical orbits somehow, but that doesn't exclude other, more exotic, periodic trajectories.
How to proceed...?
| I'll give the solution to your non-linear system $$\frac{dx}{dt}=3xy^2-x^2y$$
and $$\frac{dy}{dt}=5x^2y-xy^2$$
Divivding eq 2 by eq 1, we get $$\frac{dy}{dx}=\frac{5x^2y-xy^2}{3xy^2-x^2y}$$
Now, dividing by $x^3$ both the numenator and the denumenator we get that $$\frac{dy}{dx}=\frac{5\frac{y}{x}-(\frac{y}{x})^2}{3(\frac{y}{x})^2-\frac{y}{x}}$$
Let $y=x \cdot v$ then $\frac{dy}{dx}=x\frac{dv}{dx}+v$, thus the D.E transformed into a separable eq.
$$x\frac{dv}{dx}+v=\frac{5v-v^2}{3v^2-v}=\frac{5-v}{3v-1},$$ therefore
$$x\frac{dv}{dx}=\frac{5-v-3v^2+v}{3v-1}=\frac{5-3v^2}{3v-1}$$
$$ \Rightarrow \frac{3v-1}{5-3v^2}dv=\frac{dx}{x}$$
$$ \Rightarrow \int{\frac{3v-1}{5-3v^2}dv}=\int{\frac{dx}{x}}$$
Setting $
\sqrt {\frac{3}{5}} v= \sin \theta \Rightarrow \sqrt {\frac{5}{3}} \cos \theta d\theta = dv$, then
$$\int{\frac{3\sqrt {\frac{5}{3}} \sin \theta
-1}{5(1- \sin^2 \theta) }\sqrt {\frac{5}{3}} \cos \theta d\theta}=\int{\frac{dx}{x}}$$
so that $$\int{\frac{3\sqrt {\frac{5}{3}} \sin \theta
-1}{5(1- \sin^2 \theta) }\sqrt {\frac{5}{3}} \cos \theta d\theta}=\int {\tan \theta d\theta } - \sqrt {\frac{5}{3}} \frac{1}{5}\int {\sec \theta d\theta }
\\
= - \ln \left| {\cos \theta } \right| - \frac{1}{{\sqrt {15} }}\ln \left| {\sec \theta + \tan \theta } \right|
$$
Hence the solution is, $$
- \ln \left| {\cos \theta } \right| - \frac{1}{{\sqrt {15} }}\ln \left| {\sec \theta + \tan \theta } \right|
=\ln \left| {x } \right| +C$$ Finally, we may write the result as
$$\left| {\cos \theta } \right|\cdot
\left| {\sec \theta + \tan \theta } \right|^{{\textstyle{1 \over {\sqrt {15} }}}}
=\frac{k}{x}$$ where $k=\exp(-C)$.
| {
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Another integral $\int \frac{3 x^2+2 x+1}{ \left(x^3+x^2+x+2\right) \sqrt{1+\sqrt{x^3+x^2+x+2}}} \, dx$ Here is an indefinite integral that is similar to an integral I wanna propose for a contest. Apart from
using CAS, do you see any very easy way of calculating it?
$$\int \frac{1+2x +3 x^2}{\left(2+x+x^2+x^3\right) \sqrt{1+\sqrt{2+x+x^2+x^3}}} \, dx$$
EDIT: It's a part from the generalization
$$\int \frac{1+2x +3 x^2+\cdots n x^{n-1}}{\left(2+x+x^2+\cdots+ x^n\right) \sqrt{1\pm\sqrt{2+x+x^2+\cdots +x^n}}} \, dx$$
Supplementary question: How would you calculate the following integral using the generalization above? Would you prefer another way?
$$\int_0^{1/2} \frac{1}{\left(x^2-3 x+2\right)\sqrt{\sqrt{\frac{x-2}{x-1}}+1} } \, dx$$
As a note, the generalization like the one you see above and slightly modified versions can be wisely used for calculating very hard integrals.
| Let $$2+x+x^2+x^3=t^2\implies (1+2x+3x^2)dx=2tdt$$ $$\int \frac{2tdt}{t^2\sqrt{1+t}}$$ $$=2\int \frac{dt}{t\sqrt{1+t}}$$
Let $1+t=x^2\implies dt=2xdx$
$$=2\int \frac{2xdx}{(x^2-1)x}$$
$$=4\int \frac{dx}{x^2-1}$$
$$=4\int \frac{dx}{(x-1)(x+1)}$$
$$=4\frac{1}{2}\int \left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx$$
$$=4\frac{1}{2}\left(\int\frac{dx}{x-1}-\int\frac{dx}{x+1}\right)$$
$$=2\left(\ln|x-1|-\ln|x+1|\right)+c$$
$$=2\ln\left|\frac{x-1}{x+1}\right|+c$$
$$=2\ln\left|\frac{\sqrt{1+t}-1}{\sqrt{1+t}+1}\right|+c$$
$$=2\ln\left|\frac{\sqrt{1+\sqrt{1+2+x+x^2+x^3}}-1}{\sqrt{1+\sqrt{1+2+x+x^2+x^3}}+1}\right|+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the longest side of the triangle. The sides $a,b,c$ of a $\triangle ABC$ are in $GP$ whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.
I used law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
I took $a=a,b=\frac{2}{3}a,c=\left(\frac{2}{3}\right)^2a$
which gives $$\frac{\sin A}{\sin B}=\frac{\sin B}{\sin C}$$
I am stuck now,how to find the longest side $a$.
| If the side lengths are $l,\frac{2}{3}l,\frac{4}{9}l$ then the circumradius is given by:
$$ R=\frac{abc}{4\Delta}=\frac{\frac{8}{27}l^3}{l^2\sqrt{\left(1+\frac{2}{3}+\frac{4}{9}\right)\left(1-\frac{2}{3}+\frac{4}{9}\right)\left(1+\frac{2}{3}-\frac{4}{9}\right)\left(-1+\frac{2}{3}+\frac{4}{9}\right)}}$$
by Heron's formula, hence:
$$ R = \frac{24}{\sqrt{1463}} l $$
gives:
$$ l=\color{red}{\frac{7}{4}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions? This following Problem is from Pell equation chapters exercise
Let $y>3$ positive integer numbers, show that following diophantine equation
$$x^2+x+1=7^y\tag{1}$$ has no integer solutions.
I tried write the equation
$$(2x+1)^2+3=4\cdot 7^y$$
if $y=2k$ then we have
$$(2\cdot 7^k+2x+1)(2\cdot 7^k-2x-1)=3$$
this case has no integer.
But $y$ is odd number, How to prove equation (1) has no integer solutions for $x,y (y>3)$? Any help would be appreciated.
| An elementary solution [albeit not using Pellian methods] is reasonably simple to obtain.
As you show, there is no solution when $y$ is even. Now assume $y$ is odd. Observe that $7 \mid (x^2+x+1)$ implies $x \equiv 2\!\pmod{7}$ or $x \equiv 4\!\pmod{7}$.
In the first case, we substitute $x=7z+2$ into the equation and simplify to obtain
$$z(7z+5)=\bigl(7^\frac{y-1}{2}-1\bigr)\bigl(7^\frac{y-1}{2}+1\bigr).$$
Now our assumption of a positive solution with $y \ge 3$ implies there exist positive integers $a,b,c,d$ such that
\begin{align}
z &= ab, & 7^\frac{y-1}{2}-1 &= ac, \\
7z+5 &= cd, & 7^\frac{y-1}{2}+1 &= bd.
\end{align}
The conclusion follows fairly directly via algebraic manipulations.
The proof in the second case is nearly identical.
EDIT:
Here's a hint to get most of the way through the proof… Writing $k=(y-1)/2$, we have
$$
\frac{z}{7^k+1} = \frac{a}{d} = \frac{7^k-1}{7z+5}.
$$
Solving each relation for $7^k$ gives
$$
7^k = \frac{dz-a}{a} = \frac{7az+5a+d}{d},
$$
and solving the second relation for $z$ yields
$$
z = \frac{a(2d+5a)}{d^2-7a^2}.
$$
Since $z$ is an integer by hypothesis, there must be a prime $p$ dividing $\gcd(d,a)$, or $(d^2-7a^2) \mid (2d+5a)$, from which we can quickly deduce $p \mid 3a$.
| {
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"url": "https://math.stackexchange.com/questions/1403641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is the good substitution to make it integrable?Thanks in advance.
| It works, indeed. If we set $x=\frac{1}{2}\left(y+\sqrt{4+y^2}\right)$ we get:
$$ I = \int_{0}^{+\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2}\left(1+\frac{y}{\sqrt{y^2+4}}\right) $$
hence:
$$ I = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2} = \frac{\pi}{2a}$$
since the odd part of the previous integrand function clearly vanishes by symmetry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Trigonometric equations $\tan\theta-\sec\theta=\sqrt3$ For the following problem(s) I cannot get any answer(s). I would appreciate your help very much.
$$\tan { \theta -\sec { \theta } =\sqrt { 3 } } $$
TI get 30 degrees as the reference angle. What am I doing wrong because the answer is 210 degrees.
Link for my work since the post is not showing up.
Thanks for the help.
| $$\tan { \theta -\sec { \theta } =\sqrt { 3 } } \\ \frac { \sin { \theta } }{ \cos { \theta } } -\frac { 1 }{ \cos { \theta } } =\sqrt { 3 } \\ \sin { \theta -\sqrt { 3 } \cos { \theta =1 } } \\$$ divide both side to $2$
$$ \frac { 1 }{ 2 } \sin { \theta -\frac { \sqrt { 3 } }{ 2 } } \cos { \theta } =\frac { 1 }{ 2 } \\ \sin { \frac { \pi }{ 6 } \sin { \theta } -\cos { \frac { \pi }{ 6 } \cos { \theta =\frac { 1 }{ 2\\ } } } } \\ \cos { \left( \frac { \pi }{ 6 } +\theta \right) =-\frac { 1 }{ 2 } } \\ \frac { \pi }{ 6 } +\theta =\pm \arccos { \left( -\frac { 1 }{ 2 } \right)+2n\pi =\pm \left( \pi -\frac { \pi }{ 3 } \right) } +2n\pi \\ \theta =\pm \frac { 2\pi }{ 3 } -\frac { \pi }{ 6 } +2n\pi,n\in\Bbb Z\ $$
| {
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Given $a, b,c,d$ non-negative and $a+b+c+d=4$. Prove that $a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$
This is a problem on Mathlinks.ro and it have had no solution. So I hope it would have a nice answer and as simply as possible.
Given $a, b,c,d$ are non-negative numbers and $a+b+c+d=4$. Prove that $$a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$$
| Given the problem, we use
$$
\begin{eqnarray}
a &=& u^2\\
b &=& v^2\\
c &=& w^2\\
d &=& x^2
\end{eqnarray} \tag 1
$$
So the boundary condition is given by
$$
u^2 + v^2 + w^2 + x^2 = 4, \tag 2
$$
for any $u$, $v$, $w$ and $x$.
We define
$$
S(u,v,w,x) = u^6 v^2 + v^6 w^2 + w^6 x^2 + x^6 u^2 + 23 u^2 v^2 w^2 x^2.
\tag 3
$$
Whence
$$
\begin{eqnarray}
\frac{\partial S}{\partial u} &=& 6 u^5 v^2 + 2 x^6 u + 46 u v^2 w^2 x^2,\\
\frac{\partial S}{\partial v} &=& 6 v^5 w^2 + 2 u^6 v + 46 u^2 v w^2 x^2,\\
\frac{\partial S}{\partial w} &=& 6 w^5 x^2 + 2 v^6 w + 46 u^2 v^2 w x^2,\\
\frac{\partial S}{\partial x} &=& 6 x^5 u^2 + 2 w^6 x + 46 u^2 v^2 w^2 x.\\
\end{eqnarray} \tag 4
$$
We know that
$$
u = v = w = x = 1 \tag 5
$$
satisfies the boundary condition and we obtain
$$
\frac{\partial S}{\partial u} = \frac{\partial S}{\partial v}
= \frac{\partial S}{\partial w} = \frac{\partial S}{\partial x} = 54. \tag 6
$$
But as
$$
dx = - du - dv - dw, \tag 7
$$
we obtain
$$
d S = \left( \frac{\partial S}{\partial u}
- \frac{\partial S}{\partial x} \right) d u
+ \left( \frac{\partial S}{\partial v}
- \frac{\partial S}{\partial x} \right) d v
+ \left( \frac{\partial S}{\partial w}
- \frac{\partial S}{\partial x} \right) d w = 0. \tag 8
$$
So the "point" $(1,1,1,1)$ is an extreme value for $S$, and that value is $27$.
Another extreme "point " is $(0,0,0,2)$, and we obtain
$$
S(0,0,0,2) = 0 \tag 9
$$
So the extreme "point" $(1,1,1,1)$ is a maximum, whence
$$
u^6 v^2 + v^6 w^2 + w^6 x^2 + x^6 u^2 + 23 u^2 v^2 w^2 x^2 \le 27 \tag {10}
$$
Or going back to $(a,b,c,d)$, we get
$$
\bbox[16px,border:2px solid #800000] {
a^3 b + b^3 c + c^3 d + d^3 a + 23 a b c d \le 27. } \tag {11}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,c\in N$ and $\frac{b}{c}$is in its lowest form.Find the value of $a+b+c+abc$.
I tried to solve this question.Let $\left[\frac{3}{x}\right]=t$,therefore $\left[\frac{4}{x}\right]=5-t$.
$t\leq \frac{3}{x}<t+1$......(i) and $5-t\leq \frac{4}{x}<6-t$.....(ii)
Add (i) and (ii),we get $5\leq\frac{3}{x}+\frac{4}{x}<7$
$5\leq\frac{7}{x}<7$
Then i solved $5\leq\frac{7}{x}$ and $\frac{7}{x}<7$
and took the intersection of the two solution sets and got $x\in (1,\frac{7}{5}]$,so my $a+b+c+abc=48$ but the answer is given to be 20.What is wrong in my method.Please help me.
| Suppose that $x>1$, to get $5$ its needed to add two positive integers bcz $\displaystyle \forall x>1,\left(\lfloor \frac{3}{x}\rfloor < \lfloor \frac{4}{x}\rfloor \right)$.
So its obvious that only integer that satisfy the condition $(2,3)$ or $(1,4),$ However for the
Second couple its not find Corrosponding $x,$ bcz to have $\displaystyle \lfloor \frac{3}{x}\rfloor=1$ necessary $\displaystyle x>\frac{3}{2}$
and together with the values $\displaystyle \lfloor \frac{4}{x}\rfloor = 4\Rightarrow x<1$ Gives an empty set
Analyse Situation when $\displaystyle \lfloor \frac{3}{x}\rfloor = 2$ and $\displaystyle \lfloor \frac{4}{x}\rfloor = 3$
So we get $\displaystyle 2\leq \frac{3}{x}\leq 3$ and $\displaystyle 3\leq \frac{4}{x}\leq 4$
so solution $\displaystyle x\in \left(1,\frac{4}{3}\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}....\cos \alpha_{n}.$
Maximum value of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}.$ If it is given that
$\cot \alpha_{1}\cdot \cot \alpha_{2}\cdot \cot \alpha_{3}.......\cot \alpha_{n} = 1$ and $\displaystyle 0 \leq \alpha_{1},\alpha_{2},\alpha_{3},.......,\alpha_{n}\leq \frac{\pi}{2}.$
$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$
$$\displaystyle \frac{\sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}}{n}\geq \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}$$
So we get
$$\displaystyle \sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}\geq n\cdot \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}.......(1)$$
Similarly
$$\displaystyle \frac{\cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}}{n}\geq \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
So we get
$$\displaystyle \cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}\geq n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}.......(2)$$
Now Adding $(1)$ and $(2)\;,$ We get $$n\geq 2n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
Because above it is given that $$\sin \alpha_{1}\cdot \sin\alpha_{2}\cdot\sin \alpha_{3}.........\sin\alpha_{n}=\cos\alpha_{1}
\cdot\cos\alpha_{2}\cdot \sin\alpha_{3}......\cos\alpha_{n}$$
So we get $$\displaystyle \cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
My question is that can we solve it using any other short method? If yes then please explain it here.
Thanks.
| Apply AM-GM inequalities twice: Let $x_i = \cos \alpha_i \to 0 \leq x_i \leq 1 \to P = x_1x_2\cdots x_n \to P^2 = (1-x_1^2)(1-x_2^2)\cdots (1-x_n^2)\leq \left(\dfrac{n-(x_1^2+x_2^2+\cdots x_n^2)}{n}\right)^n\leq \left(\dfrac{n-n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}}{n}\right)^n= \left(1-\sqrt[n]{P^2}\right)^n\to \sqrt[n]{P^2} \leq 1-\sqrt[n]{P^2}\to \sqrt[n]{P^2} \leq \dfrac{1}{2}\to P^2 \leq 2^{-n}\to P \leq 2^{-\frac{n}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many 3 digit numbers can be formed using digits 1,2,3,4 and 5 such that the number is divisible by 6
How many $3$ digit numbers can be formed using digits $1,2,3,4$ and $5$ without repetition such that the number is divisible by $6$
First Approach:
A number is divisible by $6$ if it is divisible by $2$ and $3$.
Now the possible combinations I found are
$(1,3,2)$
$(3,1,2)$
$(2,3,4)$
$(3,2,4)$
$(4,3,2)$
$(3,4,2)$
$(3,5,4)$
$(5,3,4)$
total $8$ ways.
Second Approach
Case1:
unit digit can be filled in only two ways $(2,4)$ for nos $(3,2,4)$
Tens digit can be filled in $2$ ways
Hundred digit can be filled in $1$ ways
the required number is 2*1*2=4 ways
Case2:
unit digit can be filled in only one ways $(2)$ for nos $(1,2,3)$
Tens digit can be filled in $1$ ways
Hundred digit can be filled in $2$ ways
the required number is 2*1*1=2 ways
Case3:
unit digit can be filled in only one ways $(4)$ for nos $(3,4,5)$
Tens digit can be filled in $1$ ways
Hundred digit can be filled in $2$ ways
the required number is 2*1*1=2 ways
So, Total ways=8
Is there still a better way to solve this problem?
| A refinement of your first approach:
In numbers which end with $2$, the sum of the other two digits must be $4$ or $7$:
*
*$132$ the sum of the other two digits is 4
*$312$ the sum of the other two digits is 4
*$342$ the sum of the other two digits is 7
*$432$ the sum of the other two digits is 7
In numbers which end with $4$, the sum of the other two digits must be $5$ or $8$:
*
*$234$ the sum of the other two digits is 5
*$324$ the sum of the other two digits is 5
*$354$ the sum of the other two digits is 8
*$534$ the sum of the other two digits is 8
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\frac{n^n}{3^n} < n!$ for $n\ge6$ by induction How would I prove this using mathematical induction:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$.
Here is what I have tried:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$
Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true.
Assume for $k$ it's true so $\dfrac{k^k}{3^k} < k!$.
Now prove for $n=k+1$,
$\dfrac{(k+1)^{k+1}}{3^{k+1}} < (k+1)!$.
By assumption, we have that
$\dfrac{k^k}{3^k} < k!$.
Multiplying by $k+1$, we get
$$
\frac{k^k(k+1)}{3^k} < (k+1) k!
$$
and then
$$
\frac{k^k(k+1)}{3^k} < (k+1) !.
$$
I tried lots of things after this, but I am unable to figure out the proof.
| we have to show that $$(n+1)!>\frac{(n+1)^{n+1}}{3^{n+1}}$$ multiplying $$n!>\frac{n^n}{3^n}$$ by $$n+1$$ we get
$$(n+1)!>\frac{n^n(n+1)}{3^n}$$ this must be greater as $$\frac{(n+1)^{n+1}}{3^{n+1}}$$ this is true since $$\left(1+\frac{1}{n}\right)^n<3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410498",
"timestamp": "2023-03-29T00:00:00",
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Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
My attempt:
Let $z = x + iy$
$\displaystyle |z+1|=2 \iff | (x + iy)+1|=2 \iff |(x+1) +iy|=2 \iff (x+1)^2 + y^2 = 4$
Let $w = u + iv$
Now let $\displaystyle w = \frac{1}{z}$ hence we have that \begin{align}z &= \frac{1}{w} \\ &= \frac{1}{u + iv} \\ &= \frac{u-iv}{u^2 + v^2} \\ &= \frac{u}{u^2 + v^2} + i \big( - \frac{v}{u^2 + v^2} \big)\end{align}
From which we can deduce that $\displaystyle x = \frac{u}{u^2 + v^2}$ and $\displaystyle y = - \frac{v}{u^2 + v^2}$
and thus $$\displaystyle \bigg(\frac{u}{u^2 + v^2} +1\bigg)^2 + \bigg(- \frac{v}{u^2 + v^2}\bigg)^2 = 4$$
This is where I am stuck. I keep on messing up the simplification. Can someone please show me how to simplify this?
| Another approach, using inversive geometry: Observe that $f(z)$ almost produces the inversion of a figure, with respect to the unit circle; in fact, it produces the complex conjugate of the figure's inversion. However, in our case, the circle $|z+1| = 2$ is already symmetric with respect to the real axis, so $f(z)$ will in fact produce the figure's inversion.
Now, note that our input circle is tangent to the unit circle at $z = 1$, so the input circle's inversion, our output circle, will also be tangent to the unit circle at $z = 1$, but "on the inside." Since the input circle also intersects the real axis at $z = -3$, the output circle will intersect the real axis at $z = -1/3$, so the equation for the image must be
$$
\left|\, z-\frac{1}{3} \right| = \frac{2}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles? What are the dihedral angles in a disphenoid with four identical triangles, each having one edge of length $2$ and two edges of length $\sqrt{3}$? Tried to look it up, but couldn't find it...
| Here's a hedronometric approach ...
Let $a$ and $d$ be the (necessarily opposite) edges of length $2$. These edges are orthogonal to each other in space; moreover, their projections into a mutually-parallel plane form the diagonals of a square of side-length $\sqrt{2}$. This square is a "pseudo-face" of the tetrahedron, and has area $H = 2$.
The other pairs of opposite edges ---say, $b$ & $e$ and $c$ & $f$--- determine the remaining two pseudo-face areas $J$ and $K$. Although the tetrahedron's symmetry dictates that these areas must be equal, the values are a little harder to intuit than $H$. However, the Sum of Squares formula tells us that
$$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$
where $W$, $X$, $Y$, $Z$ are the areas of the tetrahedron's (ordinary) faces. In this case, by Heron's Formula, $W = X = Y = Z = \sqrt{2}$. We deduce that $J = K = \sqrt{2}$.
Finally, if we assume that faces $Y$ and $Z$ meet along edge $a$, bounding dihedral angle $A$, and that $Z$ and $X$ meet along edge $b$, bounding dihedral angle $B$, then the tetrahedral Law of Cosines says
$$\begin{align}
H^2 &= Y^2 + Z^2 - 2 Y Z \cos A \qquad\to\qquad \cos A = \,\,0 \qquad\to\qquad A = \frac{\pi}{2} \\[4pt]
J^2 &= Z^2 + X^2 - 2 Z X \cos B \qquad\to\qquad \cos B = \frac{1}{2} \qquad\to\qquad B = \frac{\pi}{3}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2.$
| Let $\omega = e^{2 \pi i/11}$, a primitive $11^{\text{th}}$ root of unity. We can assume that the circle is centered at the origin. We can also assume that $A_k$ is associated with the complex number $2 \omega^k$.
Let $p$ be complex number associated with the point $P$. Then
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \sum_{k = 0}^{10} |p - 2 \omega^k|^2.$
From the identity $z \cdot \overline{z} = |z|^2$,
\begin{align*}
\sum_{k = 0}^{10} |p - 2 \omega^k|^2 &= \sum_{k = 0}^{10} (p - 2 \omega^k)(\overline{p} - 2 \overline{\omega}^k) \\
&= \sum_{k = 0}^{10} (p \overline{p} - 2 \overline{\omega}^k p - 2 \omega^k \overline{p} + 4 \omega^k \overline{\omega}^k) \\
&= 11 p \overline{p} - 2p \sum_{k = 0}^{10} \overline{\omega}^k - 2 \overline{p} \sum_{k = 0}^{10} \omega^k + 4 \sum_{k = 0}^{10} \omega^k \overline{\omega}^k.
\end{align*}
The distance from $P$ to the origin is 3, so $11p \overline{p} = 11 \cdot |p|^2 = 11 \cdot 9 = 99$.
Since $\omega$ is a primitive $11^{\text{th}}$ root of unity, $\omega^{11} - 1 = 0$, which factors as
$(\omega - 1)(\omega^{10} + \omega^9 + \dots + \omega + 1) = 0.$
Since $\omega \neq 1$, we have $\omega^{10} + \omega^9 + \dots + \omega + 1 = 0$. Therefore,
$2 \overline{p} \sum_{k = 0}^{10} \omega^k = 0.$
Also, $|\omega| = 1$, so $\overline{\omega} = 1/\omega$, which means
$\sum_{k = 0}^{10} \overline{\omega}^k = 1 + \frac{1}{\omega} + \dots + \frac{1}{\omega^9} + \frac{1}{\omega^{10}} = \frac{\omega^{10} + \omega^9 + \dots + \omega + 1}{\omega^{10}} = 0.$
Finally, $\omega^k \overline{\omega}^k = \omega^k/\omega^k = 1$, so
$4 \sum_{k = 0}^{10} \omega^k \overline{\omega}^k = 4 \cdot 11 = 44.$
Therefore,
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = 99 + 44 = 143.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve this integral:$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx$ I occasionally found that $\displaystyle\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\dfrac{\pi}{2}\ln 2$.
I tried that
$$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^{\Large\frac{\pi}{2}}x \ \mathrm d(\ln \sin x)=-\int_0^{\Large\frac{\pi}{2}}\ln (\sin x)=\dfrac{\pi}{2}\ln 2$$
Then I tried another method
$$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^\infty\dfrac{\arctan x}{x(x^2+1)}\mathrm dx$$
I tried to expand $\arctan x$ and $\dfrac{1}{1+x^2}$, but got nothing, also I was confused that whether $\displaystyle\int_0^\infty$ and $\displaystyle\sum_{i=0}^\infty$ can exchange or not? If yes, on what condition?
Sincerely thanks your help!
| $\begin{align} J&=\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx\tag1\\
&=\int_0^1\frac{\arctan x}{x(x^2+1)}\mathrm dx+\int_1^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align} J&=\int_0^1\frac{\arctan x}{x(x^2+1)}\mathrm dx+\int_0^1\frac{x\arctan\left(\dfrac{1}{x}\right)}{x^2+1}\mathrm dx\\
&=\left(\int_0^1\frac{\arctan x}{x}\mathrm dx-\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx\right)+\dfrac{\pi}{2}\int_0^1 \dfrac{x}{x^2+1}dx-\int_0^1\frac{x\arctan x}{x^2+1}\mathrm dx\\
&=\left(\int_0^1\frac{\arctan x}{x}\mathrm dx-2\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx\right)+\dfrac{\pi}{4}\ln 2
\end{align}$
In $(1)$ perform the change of variable $x=\dfrac{2y}{1-y^2}$,
$\begin{align}
J&=\int_0^1 \dfrac{(1-y^2)\arctan\left(\dfrac{2y}{1-y^2}\right)}{y(1+y^2}dy\\
&=2\int_0^1 \dfrac{(1-y^2)\arctan y}{y(1+y^2}dy\\
&=2\left(\int_0^1\frac{\arctan x}{x}\mathrm dx-2\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx\right)
\end{align}$
Therefore,
$\displaystyle \int_0^1\frac{\arctan x}{x}\mathrm dx-2\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx=\dfrac{J}{2}$
Therefore,
$\displaystyle J=\dfrac{J}{2}+\dfrac{\pi}{4}\ln 2$
Finally,
$\boxed{J=\displaystyle \dfrac{\pi}{2}\ln 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Range of a Rational Function How to find the Range of function $$f(x)= \frac{x^2-3x-4}{x^2 - 3x +4}$$
I tried to equate the expression to $y$, then cross multiplied
$$ y= \frac{x^2-3x-4}{x^2 - 3x +4}$$
$$ y(x^2 - 3x +4)= x^2-3x-4 $$
bought the terms to one side so it becomes a quadratic and made Discriminant to zero , but i cant seem to reach anywhere.. Any suggestions?
| Given $$\displaystyle y = \frac{x^2-3x-4}{x^2-3x+4}\Rightarrow yx^2-3yx+4y = x^2-3x-4$$
so we get $$\displaystyle (y-1)x^2-3(y-1)x+4(y+1) = 0$$
For real values of $y\;,$ Then given equation has real roots. So $\bf{Discriminant\geq 0}$
So $$9(y-1)^2-16(y-1)\cdot (y+1)\geq 0$$
So we get $$(y-1)\left[9(y-1)-16(y+1)\right]\geq 0$$
so $$(y-1)\left[-7y-25\right]\geq 0$$
So $$\displaystyle (y-1)(y+\frac{25}{7})\leq 0$$
So $$\displaystyle -\frac{25}{7}\leq y\leq 1\Rightarrow y\in \left[-\frac{25}{7}\;,1\right]$$
Now when $y=1\;,$ Then $$\displaystyle 1 = \frac{x^2-3x-4}{x^2-3x+4}\Rightarrow -4=+4\; \bf(Not\; Possible)$$
So we get $y\neq 1$
So Original Range $$\displaystyle -\frac{25}{7}\leq y< 1\Rightarrow y\in \left[-\frac{25}{7}\;,1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to solve differential equation $3p^2e^y-px+1=0$ ,$p =\frac{dy}{dx}$ How to solve differential equation
$$3p^2e^y-px+1=0$$
where $$p =\frac{dy}{dx}$$
I have tried to solve for p and for x, but i am not getting anywhere. Can someone help me with this
Thanks
| By using the quadratic formula we get
\begin{align*}
3p^2e^y-px+1&=0\\
3p^2-pe^{-y}x+e^{-y}&=0\\
p&=\frac{xe^{-y}\pm\sqrt{x^2e^{-2y}-12e^{-y}}}{6}\\
pe^y&=\frac{x\pm\sqrt{x^2-12e^{y}}}{6}\\
6pe^y-x&=\pm\sqrt{x^2-12e^{y}}
\end{align*}
Let $u=-6e^y+\frac{1}{2}x^2$, then we have
\begin{align*}
-u'&=\pm\sqrt{2u}\\
u^{-1/2}u'&=\mp\sqrt{2}\\
\end{align*}
Integrating respect to $x$
$$2u^{1/2}=\mp\sqrt{2}x+c\implies 4u=2(x+k)^2$$
Here $c,\;k$ are constants, then
$$-24e^y+2x^2=2(x+k)^2$$
$$-24e^y=4k x+2k^2$$
Finally we get
$$\boxed{\color{blue}{y=\log\left(-\frac{1}{6}kx-\frac{1}{12}k^2\right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x$
Show that $$\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x,\forall x\in \left ( 0;\frac\pi2 \right )$$
Firstly, I had use the differentiation of $f(x)=\left (\frac{\sin x}{x} \right )^3- \cos^{2}x$,but I think it very complex: $$f'(x)=3\left ( \frac{\sin x}{x} \right )^2\frac{x\cos x-\sin x}{x^2}+\sin2x$$ Or I can't prove it be positive in $ \left ( 0;\frac\pi2 \right )$
Secondly, I had use the convex function, but function $f(x)=\left (\frac{\sin x}{x} \right )^3-\cos^{2}x$ isn't convex or concave in all $ \left ( 0;\frac\pi2 \right )$
| Start with
$$
1-\cos(x)\ge0\tag{1}
$$
For $x\ge0$, successive integration yields
$$
x-\sin(x)\ge0\tag{2}
$$
and
$$
\frac{x^2}2-1+\cos(x)\ge0\tag{3}
$$
and
$$
\frac{x^3}6-x+\sin(x)\ge0\tag{4}
$$
and
$$
\frac{x^4}{24}-\frac{x^2}2+1-\cos(x)\ge0\tag{5}
$$
From $(4)$, we get
$$
\frac{\sin^3(x)}{x^3}\ge1-\frac{x^2}2+\frac{x^4}{12}-\frac{x^6}{216}\tag{6}
$$
From $(5)$, we get
$$
\cos(x)\le1-\frac{x^2}2+\frac{x^4}{24}\tag{7}
$$
Subtracting $(7)$ from $(6)$ yields
$$
\frac{\sin^3(x)}{x^3}-\cos(x)\ge\frac{x^4}{24}-\frac{x^6}{216}\tag{8}
$$
and for $|x|\le3$, $\frac{x^4}{24}-\frac{x^6}{216}\ge0$. Thus for, $x\in\left[0,\frac\pi2\right]$,
$$
\frac{\sin^3(x)}{x^3}\ge\cos(x)\ge\cos^2(x)\tag{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How many ways to write $2010$?
Let $ N$ be the number of ways to write $ 2010$ in the form $ 2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $ a_i$'s are integers, and $ 0 \le a_i \le 99$. An example of such a representation is $ 1\cdot10^3 + 3\cdot10^2 + 67\cdot10^1 + 40\cdot10^0$. Find $ N$.
I picked the biggest $a_1$ so: $a_1 = 2$, there are only two ways to form $2010$.
Take $a_1 = 1$ now. This opens up to a lot of possibilities.
Specific Casework should work:
Cases 1-1: $a_2 = 10$, then possibilities are: $a_1 = 1, a_0 = 0$ or $a_1 = 0, a_0 = 10$
Actually, I think a number-theoretic way is easier.
But still.
Case 1: $a_1 = 1$ then we must solve:
$100x + 10y + z = 1010$.
Since $0 \le x \le 10$, we can casework $x$ so that:
Case 1-1:$x = 0$. So that:
$10y + z = 1010 \implies z \equiv 0 \pmod{10}, z = 10k$ and $y = 101 - k$.
Hence, $(0, 101 - k, 10k)$. $\min{k} = 0 $ and we need to find the max of $k$. We must have, $101 - k \le 99$ and $10k \le 99$. This suggests, $k \le 9$.
Cases 1-2: $x=1$. So that:
$10y + z = 910 \implies z \equiv 0 \pmod{10}$ Again, $z = 10k$ and $y = 91 - k$. Giving a set of $(1, 91 - k, 10k)$.Here again, $\min{k} = 0$ and $10k \le 99$ so $k \le 9$.
I am conjecturing that since we are always increasing $x$ values, the value on the RHS will always be divisible by $10$.
$x = 9$ so that:
$10y + z = 110 \implies z = 10k$ and $y = 11 - k$, which again there are $9$ values.
Except if $x=10$ then there is: $10y + z = 10$ then $z = 10k$ and $y = 1 - k$. Then $k$ must be $1$.
So there are: $10(9) + 1 + 2 = 93$ solutions total.
This is just an attempt!
Bump: anybody have anything?
| The generating function for these representations is
$$
\frac{x^{100\cdot1000}-1}{x^{1000}-1}\cdot\frac{x^{100\cdot100}-1}{x^{100}-1}\cdot\frac{x^{100\cdot10}-1}{x^{10}-1}\cdot\frac{x^{100}-1}{x-1}=\frac{x^{100000}-1}{x^{10}-1}\cdot\frac{x^{10000}-1}{x-1}\;,
$$
which is also the generating function for representations of the form $b_1\cdot10+b_0$ with $0\le b_i\le9999$. Since this latter restriction is irrelevant for representations of $2010$, we are simply looking for the number of ways to represent $2010$ by tens and ones. We can have anything from $0$ to $201$ tens, so the number of such representations is $202$.
In response to Calvin Lin's answer: The cancellation in the generating functions corresponds to a bijection between the two forms of representation, with $b_1=100a_3+a_1$ and $b_0=100a_2+a_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$? \begin{align}
& \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt]
= {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)}
\end{align}
Then I use quadratic formula on numerator to factor it :
$a=2,b=-1,c=-1$
$$=\frac{2(x+2)(x-\frac{5}{2})}{(x-3)(2x+1)}$$
But apparently this can be factored further. What else can I do?
| I cannot comment, so I shall answer instead.
The first thing I tend to do with quadratics like is, is rather than use the quadratic formula, I would find the discriminant. If this is a square number, then you know your quadratic will factorise nicely. You can then either do quick trial and error, or use the formula to find the solutions and deduce the factorisation.
So in this case, b^2-4ac = 9, so we know it factorises. You know it will be (2x+a)(x+b), and it is then easy to finish from here.
Hope this helps :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What are solutions of the functional equation $f\big(f(x)\big)=-4f(x)+3x$? How to find all the functions $f:[0,\infty)\rightarrow [0,\infty)$ satisfying the functional equation
$$f\big(f(x)\big)=-4f(x)+3x\text?$$
I deduced $f(x) \le \frac 3 4 x.$
| Fix some $x$, and consider the sequence given by
$$T_0 = x$$
and
$$T_{k+1} = f(T_k)$$
for all $k \geq 0$.
The functional equation then tells us that the sequence $(T_n)$ satisfies the recurrence relation
$$T_{k+2}+4T_{k+1}-3T_k=0$$
The roots of the characteristic equation $\lambda^2+4\lambda-3$ are $\phi_1=-2+\sqrt{7}$ and $\phi_2=-2-\sqrt{7}$. All possible solutions to the recurrence are given by
$$T_n=A(x)\phi_1^n + B(x)\phi_2^n$$
where $A$ and $B$ are some values which depend only on $x$ and not on $n$.
Now since the codomain of $f$ is the non-negative reals, we must have that $T_n\geq 0$ for all $n$, and so
$$A\phi_1^n + B\phi_2^n\geq 0$$
for all $n$, or equivalently
$$A + B\phi^n \geq 0$$
for all $n$, where
$$\phi=\frac{\phi_2}{\phi_1}$$
We note that $\phi<0$ and that $|\phi|>1$, and so $|\phi|^n\to\infty$ as $n\to\infty$. In particular, $\phi^n$ takes arbitrarily large (in magnitude) positive and negative values.
Now if $B<0$ then we see that for large enough even $n$, we would have that $T_n<0$. Similarly, if $B>0$ then for large enough odd values of $n$ we would also have that $T_n<0$. Thus to have $T_n\geq 0$ for all $n$, we must have that $B=0$.
The solution to the recurrence then becomes
$$T_n=A\phi_1^n$$
for all $n$. Taking $n=0$ we see that $A=x$. Thus
$$T_n = x(-2+\sqrt{7})^n$$
for all $n$.
In particular,
$$f(x)=T_1=x(-2+\sqrt{7})$$
for all $x$, which we can check satisfies the functional equation.
Edit: Another approach
Another approach continues from your observation that
$$0x \leq f(x) \leq \frac{3}{4} x$$
for all $x$.
Suppose that $\lambda x \leq f(x)$ for all $x$. Then the functional equation gives us
$$-4f(x)+3x=f(f(x))\geq \lambda f(x) \geq \lambda^2 x$$
for all $x$, and so
$$ f(x) \leq \frac{3-\lambda^2}{4} x $$
for all $x$.
Similarly, if $f(x) \leq \lambda x$ for all $x$, then we get that
$$-4f(x)+3x \leq \lambda^2 x$$
for all $x$, and so
$$ f(x) \geq \frac{3-\lambda^2}{4} x $$
for all $x$.
(The above implicitly assumes that $\lambda \geq 0$, which will be the case in what we consider)
We see that if we define the sequence $(T_n)$ by $T_0=0$ and
$$T_{n+1}=\frac{3-T_n^2}{4}$$
then
$$T_{2k} x \leq f(x) \leq T_{2k+1} x$$
for all $x$.
We will now show that the sequence $(T_n)$ has a limit and that it is equal to $-2+\sqrt{7}$, which then solves the question.
The sequence is clearly bounded above by $\frac{3}{4}$. Also, since $\frac{3}{4} < \sqrt{3} $, the recurrence also shows that the sequence terms are all non-negative. (And hence are bounded below by $0$)
We now consider the odd and even indexed terms separately. Using the recurrence twice gives us that
$$T_{n+2}=\frac{39+6T_n^2-T_n^4}{64}$$
Consider the function $g(x)=\frac{39+6x^2-x^4}{64}$. (So that $T_{n+2}=g(T_n)$) We note that $64g^\prime(x)=12x-4x^3$, and so for non-negative values of $x$, $g(x)$ is increasing provided that $x \leq \sqrt{3}$. In particular, we see that if $x \leq -2+\sqrt{7}$ then $g(x) \leq g(-2+\sqrt{7}) = -2+\sqrt{7}$, and similarly, if $-2+\sqrt{7} \leq x \leq \sqrt{3}$ then $-2+\sqrt{7}\leq g(x)$.
We see that if $T_n \leq -2+\sqrt{7}$ then $T_{n+2} \leq -2+\sqrt{7}$, and that if $T_n \geq -2+\sqrt{7}$ then $T_{n+2} \geq -2+\sqrt{7}$. In particular, since $T_0=0\leq -2+\sqrt{7}$, and $T_1=\frac{3}{4} \geq -2+\sqrt{7}$, we have that
$$T_{2k}\leq -2+\sqrt{7}$$
and
$$T_{2k+1} \geq -2+\sqrt{7}$$
for all $k$.
Now consider $g(x)-x$. We have that
$$64(g(x)-x)=39-64x+6x^2-x^4=(x^2+4x-3)(x^2-4x+13)$$
which has the same sign as $(x^2+4x-3)$.
The roots of $x^2+4x-3$ are $-2+\sqrt{7}$ and $-2-\sqrt{7}$, and so for non-negative values of $x$ we see that $g(x) \geq x$ if and only if $x \geq -2+\sqrt{7}$.
Combining all of our observations above, we see that the even indexed terms form an increasing sequence of real numbers which is bounded above, and that the odd indexed terms form a decreasing sequence of real numbers which is bounded below. Thus both of the sequences $(T_{2k})$ and $(T_{2k+1})$ have a limit. If $\lambda$ is either of these limits, then we have that $g(\lambda) = \lambda$ by passing to the limit in the recurrence for these sequences, and so
$$(\lambda^2+4\lambda-3)(\lambda^2-4\lambda+13)=0$$
We know that $\lambda$ should be a non-negative real number, and so we get that $\lambda=-2+\sqrt{7}$. So the limit of both sequences is $-2+\sqrt{7}$.
This shows that $T_n\to -2+\sqrt{7}$ as $n \to \infty$, which completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$min\{ \sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{a+b}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}\}$ There is an Olympiad problem: $$a,b,c \in \mathbb{R}^+,M=\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{b}{a+c}}+\sqrt[4]{\frac{c}{a+b}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}},$$ find the minimum value of $M$.
I think $M$ is minimum when $a=b=c$, but I can't prove it.
Some idea to prove? Or disprove?
Maybe AM-GM work on this problem?
| Hint: As the expression is homogeneous, we may set $a+b+c=1$, say. Then
$$M = \sum_{cyc}\left( \sqrt[4]{\frac{a}{1-a}}+\sqrt{\frac{1-a}a}\right)$$
Now you are summing a convex function, so use Jensen's inequality to find the minimum...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lagrange Differential equation I have problems proceeding in solving the following differential equation $$xy' + y + (y')^2 = 0.$$
After solving for $\frac{dy}{dx}$ in the quadratic and using the substitution $u^2 = x^2 - 4y$ in the discriminant, I obtain $\frac{du}{dx} = 2 \frac{x}{u} -1$. Please how can I proceed?
| $$xy+y+y’^2=0$$
Let : $y’=p$
$$y=-xy’-y’^2=-xp-p^2$$
$$ \frac{dy}{dp} =-p\frac{dx}{dp}-x-2p$$
$$p=y’=\frac{dy}{dx}=\frac{dy}{dp} \frac{dp}{dx} =-p-(x+2p) \frac{dp}{dx}$$
$$2p=-(x+2p) \frac{dp}{dx}$$
$$2p\frac{dx}{dp}+x=-2p$$
The solution of this first order linear ODE is :
$$x=-\frac{2}{3}p+\frac{C}{2\sqrt{p}}$$
$$y=-xp -p^2= \frac{2}{3}p^2-\frac{C}{2}\sqrt{p} –p^2 =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p}$$
Finally, the solution of $xy+y+y’^2=0$ expressed on parametric form with parameter $p$ is :
$$\begin{cases}
x=-\frac{2}{3}p+ \frac{C}{2}\frac{1}{\sqrt{p} } \\
y =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p} \\
\end{cases}$$
If one want to find the explicit function $y(x)$ the parameter $p$ has to be eliminated from the system $\left(x(p),y(p)\right)$. This is possible, but will lead to complicated equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$? How to find the sum $\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+..+\frac{2015}{2015^4+2015^2+1}$ ?
I'm not being able to approach the problem.Hints please!
| You might want to prove by induction that the sum
$$\sum_{i=1}^n \frac{i}{i^4+i^2+1} = \frac{\frac{1}{2} n(n+1)}{n^2+n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Length of tangent common to two semicircles which are also tangent to a larger semicircle. In the diagram,the semicircles centered at ${P}$ and $Q$ are tangent to each other and to the large semicircle ,and their radii are $6$ and $4$ respectively.Line $LM $ is tangent to semicircles $P$ and $Q$ .Find $LM$
Efforts made:
I've been able to calculate the length of the segment between the points of tangency of the interior semicircles ,but i've not been able to find the lengths of the external segments.
This problem is meant to be solved by synthetic methods(geometric methods).
thanks in advance
| DISCALIMER : This is a brute force method using coordinate geometry. A better method should exists.
Take the center of the large circle be $(0, 0)$. Then $P = (-4, 0)$ and $Q = (6, 0)$.
Now let $y = mx + c$ denote the line extending $LM$ on both sides, where $m, c \in \mathbb{R}$ are constants such that both the systems of equations,
$$y = mx + c$$
$$(x - (-4))^2 + y^2 = 6^2$$
and
$$y = mx + c$$
$$(x - 6)^2 + y^2 = 4^2$$
have only one unique solution.
From the first system we can work out, by substitution first and then by considering discriminant second, one required condition:
$$(2mc + 8)^2 - 4(m^2 + 1)(c^2 - 20) = 0$$
$$20 m^2 + 8mc - c^2 + 36 = 0$$
and similarly from the second system,
$$(2mc - 12)^2 - 4(m^2 + 1)(c^2 + 20) = 0$$
$$20m^2 + 12mc + c^2 - 16 = 0$$
Subtract the twice the second condition from thrice the first to eliminate $mc$:
$$3(20 m^2 + 8mc - c^2 + 36) - 2(20m^2 + 12mc + c^2 - 16)= 0$$
$$20m^2 -5c^2 + 140 = 0$$
$$c^2 = 4m^2 + 28$$
Now consider the sum of $4$ times the first condition and $9$ times the second condition to eliminate the constant term:
$$4(20 m^2 + 8mc - c^2 + 36) + 9(20m^2 + 12mc + c^2 - 16) = 0$$
$$52m^2 + 28mc + c^2 = 0$$
$$(2m + c)(26m + c)$$
From which we get either $c = -2m$ or $c = -26m$.
Using the results in the previous two parts, we see that there a few possibilities for $m$. Clearly $c = -2m$ has no solutions because
$$(-2m)^2 = 4m^2 + 28$$
has no solution for $m$. Hence $c = -26m$. Therefore
$$(-26m)^2 = 4m^2 + 28$$
$$672m^2 = 28$$
$$m = \pm \frac{1}{2\sqrt{6}}$$
Since from our geometric definition of $m$ being the gradient of the negatively sloped $LM$-extended, $m$ must be negative. Hence
$$m = -\frac{1}{2\sqrt{6}}$$
which leaves $$c = \frac{13}{\sqrt{6}}$$
Thus, the line $LM$ extended has equation $y = -\frac{1}{2\sqrt{6}}x + \frac{13}{\sqrt{6}}$
Now use the equation of $LM$ extended and the equation of the large circle (with radius $\frac{6 + 6 + 4 + 4}{2} = 10$,
$$x^2 + y^2 = 10^2$$
to find the coordinates of $L$ and $M$.
Finding the distance between $L$ and $M$ then becomes trivial (Pythagoras' Theorem).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Straight lines divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$ Find the equation of straight lines which pass through $(7,1)$,and divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$
My attempt:
As the required line is dividing the circumference in the ratio of
$3:1$.Therefore,angle subtended by the required line on the center is $\frac{\pi}
{2}$
.But i could not find the equation of the lines.
I let the equation of line as $ax+by+c=0$ and it passes through $(7,1)$.So $7a+b+c=0$
Then i stuck.Please help me.
| HINT:
Let the equation of the line be $y=mx+c$ passing through the point $(7, 1)$ then we have $$1=m(7)+c$$ $$7m+c=1\tag 1$$
Substituting $y=mx+c$ in the equation of circle $x^2+y^2=100$, we get $$x^2+(mx+c)^2=100$$ $$(1+m^2)x^2+2mc x+c^2-100=0\tag 2$$
Let, the roots of the above equation be $x_1$ & $x_2$ then
$$x_1+x_2=-\frac{-2mc}{1+m^2}=\frac{2mc}{1+m^2}$$
$$x_1x_2=\frac{c^2-100}{1+m^2}$$
the points of intersection are $(x_1, y_1)$ & $(x_2, y_2)$
Now, the circumference $=2\pi\times 10=20\pi$ is divided in a ratio $3:1$ then the angle subtended by the small arc at the center $$=\frac{\text{arc length}}{\text{radius}}=\frac{5\pi}{10}=\frac{\pi}{2}$$ hence, the lines joining the points $(x_1, y_1)$ & $(x_2, y_2)$ to the center $(0, 0)$ will be normal to each other hence, we have $$m_1\times m_2=-1$$
$$\frac{y_1-0}{x_1-0}\times \frac{y_2-0}{x_2-0}=-1$$
$$x_1x_2+y_1y_2=0$$
$$x_1x_2+(mx_1+c)(mx_2+c)=0$$
$$(1+m^2)x_1x_2+2mc(x_1+x_2)+c^2=0$$
I hope you can take it from here to solve for the values of $m$ & $c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the limit of $\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}$? $$\lim_{n\to \infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)=? $$
I tried with the squeeze theorem, though I got the upper bound, but I couldn't find the lower bound. I also tried to solve it with the order limit theorem, but without any success. I guessed the result should be $\frac{1}{\sqrt{2}}$. How can I do this?
| Notice, we have $$\lim_{n\to \infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 1+\sqrt 3}+\frac{1}{\sqrt 3+\sqrt 5}+\cdots +\frac{1}{\sqrt {2n-1}+\sqrt {2n+1}}\right)$$
$$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt {2r-1}+\sqrt {2r+1}}\right)$$
$$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{\sqrt {2r+1}-\sqrt {2r-1}}{2r+1-(2r-1)}\right)$$
$$=\frac{1}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt {2r+1}-\sqrt {2r-1}}{\sqrt n}$$
$$=\frac{1}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt{2r}\left(1+\frac{1}{2r}\right)^{1/2}-\sqrt{2r}\left(1-\frac{1}{2r}\right)^{1/2}}{\sqrt n}$$
Using binomial expansion of $\left(1-\frac{1}{2r}\right)^{1/2}$ & neglecting higher power terms as $\left(\frac{1}{2r}\right)^2$, $\left(\frac{1}{2r}\right)^3,\ldots $
$$=\frac{\sqrt 2}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt r}{\sqrt n} \left[\left(1+\frac{1}{2}\frac{1}{2r}\right)-\left(1-\frac{1}{2}\frac{1}{2r}\right)\right]$$
$$=\frac{1}{\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt r}{\sqrt n} \left[\frac{1}{2r}\right]$$
$$=\frac{1}{2\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt {nr}}$$
$$=\frac{1}{2\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{\sqrt {\frac{r}{n}}}$$
$$=\frac{1}{2\sqrt 2}\int_{0}^{1}\frac{dx}{\sqrt {x}}$$
$$=\frac{1}{2\sqrt 2}[2\sqrt x]_{0}^{1}$$
$$=\frac{1}{2\sqrt 2}[2-0]$$
$$=\frac{1}{\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 2
} |
Diophantine-like equations So I was solving a problem and encountered a specific system of equations that I don't know if a solutions exists for it or not.
$$\begin{align}
4ny&=d^2-a^2\\
-4nx+4ny&=d^2-b^2\\
4nx&=d^2-c^2\\
\end{align}\tag1$$
where $0\leq x,y\leq n$ and $a,c,d\geq n$ and $0\leq b\leq n\sqrt{2}$ and $a,b,c,d,n\in \mathbb{N}$, I would appreciate it if any could suggest a hint as to how I should proceed. Or worded in a different way I want to solve the following system of equation with the same conditions on the variables
$$\begin{align}
b^2+c^2+d^2-3a^2&=-8n(x-y)\\
a^2+c^2+d^2-3b^2&=8n(x+y)\\
a^2+b^2+d^2-3c^2&=-8n(y-x)\\
a^2+b^2+c^2-3d^2&=-8n(x+y)
\end{align}\tag2$$
| I think you made a typo somewhere as $(1)$ and $(2)$ do not have equivalent solutions. Let's assume $(2)$ has no typos. Then what you can do is treat it as four equations in four unknowns $a,b,c,d$.
$$\begin{align}
b^2+c^2+d^2-3a^2&=-8n(x-y)\\
a^2+c^2+d^2-3b^2&=8n(x+y)\\
a^2+b^2+d^2-3c^2&=-8n(y-x)\\
a^2+b^2+c^2-3d^2&=-8n(x+y)
\end{align}\tag2$$
I get,
$$\begin{align}
a^2&=d^2-4n\,y\\
b^2&=d^2-4n(x+y)\\
c^2&=d^2-4n\,x\\
\end{align}\tag3$$
Then solve for this as three equations in three unknowns $x,y,d$ which yields,
$$\begin{align}
x&=\frac{a^2-b^2}{4n}\\
y&=\frac{-b^2+c^2}{4n}\\
d&=\sqrt{a^2-b^2+c^2}\\
\end{align}\tag4$$
If you substitute $(4)$ into $(2)$ you will see that it will solve it. But not if into $(1)$, so there must be a typo somewhere.
Anyway, what I described is the general method you can use. And there are infinitely many integers $a,b,c$ such that $d$ is also an integer, though you have to choose those that satisfy the additional constraints that you gave.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$. Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$.
When i solved this question,i got the answer $x<\frac{-28}{9}$ but the answer given in the book is $x\leq-3$.
This is how i solved.
$\sqrt{x^2-5x-24}-x-2>0\Rightarrow \sqrt{x^2-5x-24}>x+2$
$$\therefore x^2-5x-24>x^2+4x+4\Rightarrow9x<-28\Rightarrow x<\frac{-28}{9} \tag 1$$
and
$$x^2-5x-24\geq0\Rightarrow x\leq -3 \text{ or } x\geq 8 \tag 2$$
When taking the intersection of (1) and (2),we get $x<\frac{-28}{9}$. Where have i gone wrong? Please help me.
| Be careful when squaring inequalities.
The condition is indeed $\sqrt{x^2-5x-24}>x+2$, but this translates into two different sets of conditions:
\begin{align}
\textbf{Set 1}\quad
&\begin{cases}
x^2-5x-24\ge0\\
x+2<0
\end{cases}
\\[10px]
\textbf{Set 2}\quad
&\begin{cases}
x^2-5x-24\ge0\quad (*)\\
x^2-5x-24\ge(x+2)^2\\
x+2\ge0
\end{cases}
\end{align}
(The condition marked as $(*)$ is of course redundant.)
Set 1 is for the case squaring is not possible, but of course the desired inequality holds, because the radical exists and is non negative, while the left-hand side is negative.
Set 2 is for the case we can square, because we're assuming that also the right-hand side is non negative.
Where's the problem? You made an illicit squaring; it's like from $2>-3$ deduce, by squaring, that $4>9$. The inequality $a>b$ is equivalent to $a^2>b^2$ when $a,b\ge0$; but it's not in general equivalent when $a>0$ and $b<0$ (but it's true, of course, in this case).
Set 1 can be solved in an easy way, because the quadratic has roots $8$ and $-3$, and its solution set is $x\le-3$.
Set 2 becomes
\begin{cases}
9x+28\le 0\\
x+2\ge0
\end{cases}
which of course has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proof that there exists an integer $k$ such that $(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)k$ Without using the identity:
$$(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)(x^2+xy+x z+y^2+y z+z^2)$$ I am trying to find an argument to prove that for all $x,y,z$ there exists a nonzero integer $k$ such that:
$$(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)k$$
Proving $5$ divides the lefthand(LH) side expression is easy using FlT. How do I prove that $(x+y)(x+z)(y+z)$ divides the LH? Any hints?
| Using the factor theorem: the polynomial $P(x)$ is divisible by $x-a$ if an only if $P(a) = 0$.
Then $P(x)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $x+y$ given that
$$P(-y) = (-y+y+z)^5-(-y)^5-y^5-z^5 =z^5+y^5-y^5-z^5 = 0$$
Similarly, $P(x)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $x+z$ given that $P(-z)=0$.
Finally, $P(y)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $y+z$ given that $P(-z)=0$.
Then we conclude that $(x+y+z)^5-x^5-y^5-z^5$ is divisible by $(x+y)(x+z)(y+z)$.
Do you agree?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the possible real values of $(x,y)$ satisfying $\sin x+\cos x=\frac{y^2-2y+4}{y^2+2y+4}$ Find the possible real values of $(x,y)$ satisfying $\sin x+\cos x=\frac{y^2-2y+4}{y^2+2y+4}$.
I could not solve this question after trying many times.How can two variables be solved with one equation?Please help me.
Its answer is $(2n\pi+\frac{\pi}{4}\pm\arccos\frac{k}{\sqrt2},\frac{1+k\pm\sqrt{(k-3)(1-3k)}}{1-k});$ where $k\in [\frac{1}{3},\sqrt2]-\left\{1\right\}$ and $(2n\pi+\frac{\pi}{4}\pm\frac{\pi}{4},0)$
| HINT:
Let $u=\dfrac{y^2-2y+4}{y^2+2y+4}$
$\iff y^2(u-1)+2y(u+1)+4u-4=0$
As $y$ is real, the discriminant $4\{(u+1)^2-4(u-1)^2\}\ge0$
$\iff3u^2-10u+3\le0\iff (u-3)\left(u-\dfrac13\right)\le0\iff\dfrac13\le u\le3$
Now $\cos x+\sin x=\sqrt2\sin\left(\dfrac\pi4+x\right)\implies-\sqrt2\le\cos x+\sin x\le\sqrt2$
So, we need $\dfrac13\le\dfrac{y^2-2y+4}{y^2+2y+4}=\cos x+\sin x\le\sqrt2$
Now set $$\dfrac{y^2-2y+4}{y^2+2y+4}=\cos x+\sin x=k$$ with $\dfrac13\le k\le\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
$\bf{My\; try::}$ We can write $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldots ) = \frac{(1-x^3)(1)}{1-x^5}$$
So we can write it as $\displaystyle \frac{(1-x)(x^2+x+1)}{(1-x)(x^4+x^3+x^2+x+1)}$
So our Integral Convert into $\displaystyle \int_{0}^{1}\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$
Now How can I solve it, Help me
Thanks
| To expand the answer posted by @Arentino, we have
$$\begin{align}
S&=3\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+4)}\\\\
&=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k+4/5}\right)\\\\
&=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k}\right)+\frac15\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+4/5}\right)\\\\
&=\frac34+\frac15(\psi(9/5)-\psi(6/5)) \tag 1
\end{align}$$
Using the reflection and recurrence formulae of the digamma function in $(1)$ reveals that
$$\begin{align}
S&=\frac34+\frac15(\psi(9/5)-\psi(6/5))\\\\
&=\frac34+\frac15\left(\psi(4/5)+\frac54\right)-\frac15\left(\psi(1/5)+5\right)\\\\
&=\frac34+\frac15\left(\psi(4/5)-\psi(1/5)\right)+\left(\frac14-1\right)\\\\
&=\frac15 \pi \cot(\pi/5)\\\\\
&=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}
\end{align}$$
as expected!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Integral by the given $u$-substitution Original equation $$\int\sqrt{e^t-5}dt \qquad u=\sqrt{e^t-5}$$
This is what I've done so far$$\int{u}dt$$ $$du=\frac{e^t}{2\sqrt{e^t-5}}dt$$$$dt=\frac{2\sqrt{e^t-5}}{e^t}du$$$$\int{u}{\frac{2\sqrt{e^t-5}}{e^t}}du$$ From this point on what would i need to do?
| Try making a substitution of the inside function, that often works better. In this case, for the equation
$$\int\sqrt{e^t-5}dt$$
make the substitution $u=e^t$, $\,\,du = e^tdt$
$$=\int\frac{\sqrt{u-5}}{u}du$$
substitute $s = \sqrt{u-5}$, $\,\,ds = \frac{1}{2\sqrt{u-5}}$
$$=2\int\frac{s^2}{s^2 + 5}ds$$
$$=2\int\ \bigg(1-\frac{5}{s^2 + 5}\bigg)ds$$
$$=2\int1\,ds - 10\int \frac{ds}{s^2 + 5}$$
$$=2s - 2\int \frac{ds}{\frac{s^2}{5} + 1}$$
substitute $p = \frac{s}{\sqrt{5}}$, $\,\,dp = \frac{ds}{\sqrt{5}}$
$$=2s - 2\sqrt{5}\int \frac{dp}{p^2 + 1}$$
$$=2s - 2\sqrt{5}\tan^{-1}(p)$$
After a couple back-subtitutions we get
$$=2\sqrt{e^t - 5} - 2\sqrt{5}\tan^{-1}\bigg(\frac{\sqrt{e^t - 5}}{\sqrt{5}}\bigg) + C$$
I like solving the problem this way as it feels more linear... Instead of trying to find what trick to use you just keep messing with the integral, making it simpler until you notice that you are either near to or in a form you can easily integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex numbers - how to solve $(\sqrt{3}-i)z^6+16=0$ When $z = x + yi$ (or $a + bi$), I need to solve:
$$(\sqrt{3}-i)z^6+16=0$$
Here is how I started:
$(\sqrt{3}-i)z^6=-16$
$z^6=\frac{-16}{\sqrt{3}-i}$
$z=\sqrt[6]\frac{-16}{\sqrt{3}-i}$
In other cases I get a normal complex number under the root sign in the right side (in the form of $x+yi$) and then I represent this number in its trigonometric form and apply De Moivre's formula. But this case seems different... What should I do next?
| $$(\sqrt{3}-i)z^6+16 = 0 \Longleftrightarrow$$
$$(\sqrt{3}-i)z^6 = -16 \Longleftrightarrow$$
$$z^6 = \frac{-16}{\sqrt{3}-i} \Longleftrightarrow$$
$$z^6 = -4\sqrt{3}-4i \Longleftrightarrow$$
$$z^6 = 8e^{-\frac{5\pi}{6}i} \Longleftrightarrow$$
$$z^6 = 8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \left(8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i}\right)^{\frac{1}{6}} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{1}{6}\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{1}{6}\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{(-5+12k)\pi}{36}i}$$
With $k\in\mathbb{z}$ and $k=0-5$
With Eulers formula we found:
$$z = \sqrt{2}e^{\frac{(-5+12k)\pi}{36}i}=\sqrt{2}\left(\cos\left(\frac{(-5+12k)\pi}{36}\right)+\sin\left(\frac{(-5+12k)\pi}{36}\right)i\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\alpha:\mathbb{R}\to\mathbb{R}$ where $\alpha(x)=\frac{x^{3}}{x^{2}+1}$ is bijective Using the usual method that I was taught, in order to find that something is injective, I assume $f(x)=f(y)$ for some $x,y\in\mathbb{R}$, and then I demonstrate that $x=y$. This is not working out so well for me, using the following steps:
\begin{align}
\frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
x^3(y^2+1)&=y^3(x^2+1)\\
x^3y^2+x^3&=x^2y^3+y^3\\
x^3-y^3&=x^2y^3-x^3y^2\\
(x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
(x^2+xy+y^2)&=-x^2y^2\\
x^2+y^2&=-x^2y^2-xy\\
x^2+y^2&=-xy(xy+1)
\end{align}
And here is where I'm stuck. Am I assuming correctly that I should find that $x=y$ somehow?
| Assuming that $f(x) = f(y)$ and showing that $x = y$ is a correct method to show that $f$ is injective. You have one subtle error in your work.
$\begin{align} \frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
x^3(y^2+1)&=y^3(x^2+1)\\ x^3y^2+x^3&=x^2y^3+y^3\\
x^3-y^3&=x^2y^3-x^3y^2\\ (x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
(x^2+xy+y^2)&=-x^2y^2\\ \end{align}$
At the last step shown above, you divided both sides by $x-y$, which could potentially be $0$.
Continuing from above, you get that $f(x) = f(y)$ implies $$(x-y)(x^2+xy+y^2)=x^2y^2(y-x)$$, which can be rewritten as $$(x-y)(x^2y^2+x^2+xy+y^2) = 0$$
Here, there are two possibilities. Either $x-y = 0$, i.e. $x = y$ (which is what we want to show) or $x^2y^2+x^2+xy+y^2 = 0$. So you need to show that $x^2y^2+x^2+xy+y^2 = 0$ isn't possible unless the condition $x = y$ is also met.
This is easy to show since we can write $x^2y^2+x^2+xy+y^2$ as $(xy)^2 + \dfrac{3}{4}(x+y)^2 + \dfrac{1}{4}(x-y)^2$, which can only be $0$ if $xy = x+y = x-y = 0$, i.e. $x = y = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Parametric equations - stating values. A curve is defined by the parametric equations: $x=\cos2t, y=\sin2t, 0<t<π.$
a) Use parametric equations to find $\frac{dy}{dx}$. Hence find the equation of the tangent when $t=\frac{π}{8}$.
b) Obtain an expression for $\frac{d^2y}{dx^2}$ and hence show that: $\sin2t\left(\frac{d^2y}{dx^2}\right)+\left(\frac{dy}{dx}\right)^2=k$, where $k$ is an integer. State the value of $k$.
| Using $$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
So here $x=\cos 2t\;,$ Then $$\displaystyle \frac{dx}{dt} = -2\sin 2t$$ and $y=\sin 2t\;,$ Then $$\displaystyle \frac{dy}{dt} = 2\cos 2t$$
So $$\displaystyle \frac{dy}{dx} = -\frac{2\cos 2t}{2\sin 2t} = -\cot 2t\;,$$ Now $$\displaystyle \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{8}} = -\left[\cot 2t \right]_{t=\frac{\pi}{8}} = -1$$
Now $$\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right) = -\frac{d}{dx}(\cot 2t) =-\frac{d}{dt}(\cot 2t)\cdot \frac{dt}{dx} = 2\csc^2 2t\cdot \frac{1}{-2\sin 2t} $$
So $$\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{\sin^3 2t}$$
Now $$\displaystyle \sin 2t\cdot \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=-\frac{1}{\sin^2 2t}+\frac{\cos^2 2t}{\sin^2 2t} = -\frac{1}{\sin^2 2t}(1-\cos^2 2t)$$
So we get $$\displaystyle \sin 2t\cdot \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2 = -\frac{\sin^2 2t}{\sin^2 2t} = -1=k$$(Given)
So we get $$k=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Fundamental theorem of Calculus: $\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt$ Possible textbook mistake using the FTC we are supposed to evaluate
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
$$
and the answer, the textbook says, should be a constant. When I evaluate that derivative I get
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt = \frac{d}{dx} \int_{-x}^{0} \frac{1}{3+t^2} \ dt + \frac{d}{dx} \int_{0}^{x} \frac{1}{3+t^2} \ dt\\
= - \frac{d}{dx} \int_{0}^{-x} \frac{1}{3+t^2} \ dt + \frac{d}{dx} \int_{0}^{x} \frac{1}{3+t^2} \ dt\\
= - \frac{1}{3+(-x)^2} \cdot (-1) + \frac{1}{3+x^2}\\
= \frac{1}{3+x^2} + \frac{1}{3+x^2}\\
= \frac{2}{3+x^2}
$$
but I suppose that the textbook is expecting the answer $0$ (which for me is not correct since the chain rule, on the first term, makes the fraction not cancel). Is there really a mistake?
Thank you.
| $$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
= 1\frac{1}{3+x^2}+1\frac{1}{3+(-x)^2}=\frac{2}{3+x^2}$$
Note that $$
\frac{d}{dx} \int_{g(x)}^{f(x)} h(t) \ dt=f'(x)h(f(x))-g'(x)h(g(x))$$
| {
"language": "en",
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"source": "stackexchange",
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Prove that the probability that the sum of the digits used leaves the remainder 2 when divided by $4$ is $\frac{1}{4}.$ A number of 5 digits is written down at random.Prove that the probability that the sum of the digits used leaves the remainder 2 when divided by $4$ is $\frac{1}{4}.$
The sum of digits must be like 6,10,14 etc.But i am not sure how many digits we have to consider for summing.Please help me in solving this problem.
| While waiting for some smarter solution (if any), I'll give a direct count of how many 5-digit numbers are congruent to $2$ (mod $4$).
Let's start with 1-digit numbers (from $0$ to $9$): we have $3$ of them congruent to $0$ (they are $0$, $4$ and $8$), $3$ congruent to $1$ ($1$, $5$ and $9$), $2$ congruent to $2$ ($2$ and $6$) and $2$ congruent to $3$ ($3$ and $7$).
Consider now a 2-digit number $n$ (between $00$ and $99$), composed of two single digits $x$ and $y$. We have $n\equiv0$ if:
$$
\begin{align}
&x\equiv0,\ y\equiv0\quad (3\times3=9\ \hbox{cases})\cr
&x\equiv1,\ y\equiv3\quad (3\times2=6\ \hbox{cases})\cr
&x\equiv2,\ y\equiv2\quad (2\times2=4\ \hbox{cases})\cr
&x\equiv3,\ y\equiv1\quad (2\times3=6\ \hbox{cases})\cr
\end{align}
$$
for an overall total of $9+6+4+6=25$ cases. In the same way one can count $n\equiv1$ in $26$ cases, $n\equiv2$ in $25$ cases and $n\equiv3$ in $24$ cases.
We can repeat the same procedure for a 4-digit number $n$ (between $0000$ and $9999$), which is composed of two 2-digit numbers (we'll denote the sums of their digits by $x$ and $y$). We have $n\equiv 0$ if:
$$
\begin{align}
&x\equiv0,\ y\equiv0\quad (25\times25=625\ \hbox{cases})\cr
&x\equiv1,\ y\equiv3\quad (26\times24=624\ \hbox{cases})\cr
&x\equiv2,\ y\equiv2\quad (25\times25=625\ \hbox{cases})\cr
&x\equiv3,\ y\equiv1\quad (24\times26=624\ \hbox{cases})\cr
\end{align}
$$
so that we have a total of $2498$ cases with $n\equiv 0$. In a similar way we can compute a total of $2500$ cases with $n\equiv 1$, then $2502$ cases with $n\equiv 2$ and $2500$ cases with $n\equiv 3$.
The same process can be repeated for a 5-digit number: we must add the leftmost digit to a 4-digit number, taking into account that it cannot be zero. Between 1 and 9 there are 2 digits $\equiv0$, 3 digits $\equiv1$, 2 digits $\equiv2$ and 2 digits $\equiv3$. We can then form a 5-digit number congruent to $2$ in $2\times2502+3\times2500+2\times2498+2\times2500=22500$ different ways. So that the probability to pick one of them is $22500/90000=1/4$.
Notice that if we had allowed for a zero leftmost digit that probability would have been different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expected value of size of subset Given a set $S$ such that $|S|=n$,
A random item is chosen randomly from $S$, and being appended to a new set $T$.
This process is being repeated $n$ times (with repetition), what is the expected value of $|T|$ ?
| This question can also be answered by total enumeration using Stirling
numbers of the second kind. Observe that the expectation is given by
$$\mathrm{E}[X] = n^{-n} \sum_{k=0}^n k \times {n\choose k}
k! {n\brace k}.$$
Here we multiply the value of the random variable by the probability,
which is obtained from the fact that a sample with $k$ different
entries needs a choice of these entries from the $n$ possibles and a
partition of the $n$ slots into $k$ sets whose elements receive the
same value. All $k!$ permutations of the $k$ chosen values result in
a valid unique assignment.
We have
$${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}$$
by the species equation for set partititions
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields for the expectation
$$n^{-n} n! [z^n]
\sum_{k=0}^n k \times {n\choose k}
(\exp(z)-1)^k
\\ = n^{-n} n! [z^n] n
\sum_{k=1}^n {n-1\choose k-1}
(\exp(z)-1)^k
\\ = n^{-n} n! [z^n] n (\exp(z)-1)
\sum_{k=0}^{n-1} {n-1\choose k}
(\exp(z)-1)^k
\\ = n^{-n} n! [z^n] n (\exp(z)-1) \exp((n-1)z)
= n n^{-n} n! [z^n] (\exp(nz)-\exp((n-1)z))
\\ = n n^{-n} (n^n - (n-1)^n)
\\ = n \left(1 - \left(1-\frac{1}{n}\right)^n\right)
\sim n \left(1 - \frac{1}{e}\right).$$
Now we can also compute higher moments with this,
the variance for example.
We first compute the expectation of $X(X-1)$, getting
$$\mathrm{E}[X(X-1)] =
n^{-n} \sum_{k=0}^n k (k-1) \times {n\choose k}
k! {n\brace k}
\\ = n^{-n} n! [z^n] n (n-1)
\sum_{k=2}^n {n-2\choose k-2}
(\exp(z)-1)^k
\\ = n^{-n} n! [z^n] n (n-1) (\exp(z)-1)^2
\sum_{k=0}^{n-2} {n-2\choose k}
(\exp(z)-1)^k
\\ = n^{-n} n! [z^n] n (n-1) (\exp(z)-1)^2 \exp((n-2)z).$$
Extracting coefficients we obtain
$$n (n-1) n^{-n} (n^n - 2(n-1)^n + (n-2)^n)
= n(n-1) \left(1-2\left(1-\frac{1}{n}\right)^n
+\left(1-\frac{2}{n}\right)^n\right)
\\ \sim
n(n-1) \left(1-\frac{2}{e}+\frac{1}{e^2}\right).$$
We thus get for the variance
$$\mathrm{Var}[X] = \mathrm{E}[X^2] - \mathrm{E}[X]^2
\\ = n(n-1) \left(1-2\left(1-\frac{1}{n}\right)^n
+\left(1-\frac{2}{n}\right)^n\right)
\\ + n\left(1-\left(1-\frac{1}{n}\right)^n\right)
- n^2\left(1-\left(1-\frac{1}{n}\right)^n\right)^2
\\ = n^2 \left(\left(1-\frac{2}{n}\right)^n
- \left(1-\frac{1}{n}\right)^{2n}\right)
\\ + n \left(\left(1-\frac{1}{n}\right)^n
- \left(1-\frac{2}{n}\right)^n \right).$$
The asymptotic expansion then yields
$$\mathrm{Var}[X] \sim n \left(\frac{1}{e}-\frac{2}{e^2}\right).$$
Addendum. We need the second term rather than the constant term in
the asymptotic expansion of the factor on $n^2.$
We get for the first term in the difference
$$\exp\left(n\log\left(1-\frac{2}{n}\right)\right)
= \exp\left(-\sum_{q\ge 1} \frac{2^q}{n^{q-1} q}\right)
\\ = \sum_{k\ge 0} \frac{(-1)^k}{k!}
\left(\sum_{q\ge 1} \frac{2^q}{n^{q-1} q}\right)^k.$$
This yields for the constant term
$$\sum_{k\ge 0} \frac{(-1)^k}{k!} 2^k = \frac{1}{e^2}.$$
We get for the next term
$$\frac{1}{n}
\sum_{k\ge 0} \frac{(-1)^k}{k!}
{k\choose 1} 2\times 2^{k-1}
= -\frac{2}{n}
\sum_{k\ge 1} \frac{(-1)^{k-1}}{(k-1)!} 2^{k-1}
= -\frac{2}{n} \frac{1}{e^2}.$$
For the second term in the difference we have
$$\exp\left(2n\log\left(1-\frac{1}{n}\right)\right)
= \exp\left(-2\sum_{q\ge 1} \frac{1}{n^{q-1} q}\right)
\\ = \sum_{k\ge 0} \frac{(-2)^k}{k!}
\left(\sum_{q\ge 1} \frac{1}{n^{q-1} q}\right)^k.$$
This time we get for the constant term
$$\sum_{k\ge 0} \frac{(-2)^k}{k!} = \frac{1}{e^2}$$
and for the next term
$$\frac{1}{n}
\sum_{k\ge 0} \frac{(-2)^k}{k!}
{k\choose 1} \frac{1}{2} \times 1^{k-1}
= -\frac{1}{n}
\sum_{k\ge 1} \frac{(-2)^{k-1}}{(k-1)!}
= -\frac{1}{n} \frac{1}{e^2}.$$
The difference then yields for the variance
$$n^2 \left(-\frac{1}{n} \frac{1}{e^2}\right)
+ n\left(\frac{1}{e}-\frac{1}{e^2}\right)$$
which is the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
How to Compute $\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}$
How to compute
$$\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=-\dfrac{3}{2}$$
I'm interested in more ways of computing limit for this expression.
My Thoughts
at least i tried to use L'Hospital's rule but with no luck
\begin{align}
&\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\lim _{x\to \:0}\dfrac{2x\left(\dfrac{\cos \left(x^2\right)}{\sin \left(x^2\right)+1}-1\right)}{\dfrac{2\arcsin \left(x\right)}{\sqrt{1-x^2}}-2x}\\
&=\lim _{x\to \:0}\dfrac{2\left(\dfrac{\cos \left(x^2\right)\left(\sin \left(x^2\right)+1\right)-2x^2\left(\cos ^2\left(x^2\right)+\sin ^2\left(x^2\right)+\sin \left(x^2\right)\right)}{\left(\sin \left(x^2\right)+1\right)^2}-1\right)}{2\left(\dfrac{x\arcsin \left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{1}{1-x^2}-1\right)}
\end{align}
Note this limit was taken from competition mathematics so i can't go with L'Hopital because each time i use it i got big terms that i have to derivative for the next time
| You must use power series development at order $4$:
*
*$\sin x= x+o(x^2)$, hence $\sin x^2=x^2+o(x^4)$
*$\ln(1+u)=u-\dfrac{u^2}2+o(u^2) $, hence
$$\ln(1+\sin x^2)=\ln\bigl(1+ x^2+o(x^4)\bigr)=x^2-\frac{x^4}2+o(x^4)$$
*$\arcsin x=x+\dfrac12\dfrac{x^3}3+o(x^4)$, hence $(\arcsin x)^2=x^2+\dfrac{x^4}3+o(x^4)$
Grouping all the results we get:
$$\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\frac{-\dfrac{x^4}2+o(x^4)}{\dfrac{x^4}3+o(x^4)}=-\frac32+o(1)\to-\frac32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the cube roots of $-11-2i$. How do I find the roots of $\sqrt[3]{ - 11 - 2i}$ ?
Tried to use Moivre's theorem, but can not find the solutions by using the polar form:
$z_k=\sqrt{5}[\cos(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})+i.\sin(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})]$, for $k=0, 1$ and $2$.
Also resorted to $(a+bi)^3=-11-2i$ and did not have success, because I could not solve
$a³-3ab² = -11$ and $3a²b - b³ = -2$.
Solutions at the end of the book are $1+2i$ and $-\frac{1+2\sqrt{3}}{2}+\frac{\sqrt{3}-2}{2}i$ and $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.
| Let's try to find a solution $z=a+bi$ with $a,b \in \mathbb Z$.
We must have $-11=a^3-3ab^2 = a(a^2-3b^2)$ and so $a=\pm 1$ or $\pm 11$.
We must have $-2=3a^2b - b^3 = b(3a^2-b^2)$ and so $b=\pm 1$ or $\pm 2$.
Testing all cases, we see that $a=1$ and $b=2$ works.
Having found one solution $z=1+2i$, the others are $z\omega$ and $z\omega^2$, where $\omega$ is a primitive cubic root of unity:
$$
\omega = \dfrac{-1+\sqrt{3} \, i}{2}
\qquad
\omega^2 = \bar \omega = \dfrac{-1-\sqrt{3} \, i}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
And now:
$$\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x} = \infty$$
What I'm doing wrong? I know that answer is $\sqrt{3}$, but where is my mistake?
| $$
\underbrace{\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x}}_\text{I have no idea what you did here!}
$$
Let's recall some simple algebra:
$$
\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x}
= \overbrace{\frac{x(\cdots\cdots)}{x(\cdots\cdots)} = \frac{\sqrt{3+x}+\sqrt{3-x}}{2} \vphantom{\frac\int\int} }^\text{Cancel the $x$s.}
$$
Now you have
$$
\lim_{x\to0} \frac{\sqrt{3+x}+\sqrt{3-x}} 2 = \frac{2\sqrt 3} 2 = \sqrt 3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Showing that the sequence $a_n=[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}]-\ln(n)$ converges How do you show that
the sequence
$$a_n=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}\right)-\ln(n)$$ is convergent.
Give a convincing argument, that $a_n$ converges to a number called gamma. Using a picture if you want. Remember the anti-derivative of $\ln(x)$ is $\frac 1x$
I am not sure how to give a convincing argument that this sequence converges to $\gamma$ I never heard of this number gamma before.
| $a_{n+1} - a_n = \dfrac{1}{n+1} - \ln(n+1)+\ln n = \dfrac{1}{n+1} - \ln\left(1+\dfrac{1}{n}\right)= \dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}-\ln\left(1+\dfrac{1}{n}\right)= \dfrac{x}{1+x} - \ln(1+x)=1-\dfrac{1}{x+1} -\ln(1+x)=f(x), 0< x = \dfrac{1}{n}<1\Rightarrow f'(x) = \dfrac{1}{(x+1)^2}-\dfrac{1}{x+1}<0\Rightarrow a_{n+1}-a_n = f(x) < f(0) = 0\Rightarrow a_{n+1} < a_n\Rightarrow \{a_n\} \text{ is a strictly decreasing sequence, and further it is bounded below by $0$ because } a_n = \dfrac{1}{n} + \left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n-1} - \ln n\right) > \dfrac{1}{n} + 0 = \dfrac{1}{n} > 0, \text{ thus it converges .}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Tricky trig question from GRE
Please evaluate
\begin{align}
1-\sin^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
What I've tried so far is to use Pythagorean identity and I got
\begin{align}
\cos^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
If \begin{align}
\arccos \frac{\pi}{12}=y
\end{align}
then
\begin{align}
\cos y =\frac{\pi}{12}
\end{align}
and here I can't continue because the answers are in such form:
*
*(A) $\sqrt{\frac{1-\cos\frac\pi{24}}{2}}$
*(B) $\sqrt{\frac{1-\cos\frac\pi{6}}{2}}$
*(C) $\sqrt{\frac{1+\cos\frac\pi{24}}{2}}$
*(D) $\frac\pi6$
and one more is missing, but that's pdf's issue
| Once you get to this stage:
$\cos^2 (\arccos \frac{\pi}{12})$,
Use the facts that $\cos(\arccos x) = x$ and $\cos^2 y = (\cos y)^2$ to get to the final answer.
In this case, that's simply $(\frac{\pi}{12})^2 = \frac{\pi^2}{144}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
| A possible short proof comes from Parseval's identity.
It is well known that the Fourier sine series of the $\text{sign}$ function over $(-\pi,\pi)$ is given by:
$$ \text{sign}(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} \tag{1}$$
hence by Parseval's identity:
$$ 2\pi = \int_{-\pi}^{\pi} 1\,dx = \frac{16}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\tag{2}$$
and the claim readily follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all conditions of a and b for when matrix has determinant zero Find all conditions on a and b for which A =
\begin{bmatrix} a & 0 &a^2 \\ 2 & a & 3a \\ a & -1 & ba+b\end{bmatrix}
has determinant 0.
| Subtract $a$ times first column from third to obtain
$$\begin{pmatrix}a&0&0\\2&a&a\\a&-1&ba+b-a^2\end{pmatrix} $$
which has determinant $0$ iff the original matrix has detrminant $0$.
If $a=0$, the first row is zero, hence $\det A=0$.
If $a\ne 0$, any linear combinations of the columns that results in zero must not use the first column. Hence we need only check if the second and third column are linearly dependent. Under the premise that $a\ne 0$, this is the case iff $-1=ba+b-a^2$. To solve this for $b$ we'd like to divide by $a+1$, hence have to distinguish cases: If $a=-1$, we obtain $ba+b-a^2=-1$ for all $b$; if $a\ne -1$ we can solve as $b=\frac{a^2-1}{a+1}=a-1$.
Thus
$\det A=0$ iff $a=0$ or $a=-1$ or $b=a-1$.
| {
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"url": "https://math.stackexchange.com/questions/1456081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm. Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm.
I let two non-parallel sides and one of the parallel sides as $k$(shorter one).I know that area of the trapezium is $\frac{1}{2}\times $sum of parallel sides $\times$ height.But in this question,neither height is given nor longest side is given.How should i formulate the equation of the area?
| Notice, let the unknown side be $x$ which is parallel to one of three equal sides each of length $k$ then the normal distance between the parallel sides can be determined as follows (using pytagorean theorem in a right triangle)
$$\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}$$
the area of the trapezium is given as
$$A=\frac{1}{2}(x+k)\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}\tag 1$$
differentiating (1) w.r.t. $x$, we get
$$\frac{dA}{dx}=\frac{1}{2}(x+k)\frac{-(x-k)}{4\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}+\frac{1}{2}\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}$$
$$=\frac{1}{8}\frac{x^2-k^2+4k^2-4\left(\frac{x-k}{2}\right)^2}{\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}=\frac{1}{4}\frac{(-x^2+kx+2k^2)}{\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}$$
Again differentiating w.r.t. $x$, we get
$$\frac{d^2A}{dx^2}=\frac{x^3-3kx^2-7k^2x+5k^3}{16\left(k^2-\left(\frac{x-k}{2}\right)^2\right)^{3/2}}$$
Now, for maximum or minimum, setting $\frac{dA}{dx}=0$, we get
$$\frac{1}{4}\frac{(-x^2+kx+2k^2)}{\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}=0$$
$$x^2-kx-2k^2=0$$
$$x=\frac{-(-k)\pm\sqrt{(-k)^2-4(1)(-2k^2)}}{2(1)}=\frac{k\pm 3k}{2}$$
But, side $k>0$ hence, we accept $$x=\frac{k+3k}{2}=2k$$
It can be checked that $\frac{d^2A}{dx^2}=-\frac{13}{6\sqrt 3}<0$ at $x=2k$. Hence, the area is maximum at $x=2k$
Hence, substituting $x=2k$, the greatest possible area is $$A_{\text{max}}=\frac{1}{2}(2k+k)\sqrt{k^2-\left(\frac{2k-k}{2}\right)^2}$$
$$=\frac{3k}{2}\sqrt{k^2-\frac{k^2}{4}}=\frac{3k}{2}\sqrt{\frac{3k^2}{4}}$$
$$\color{red}{A_{\text{max}}=\frac{3\sqrt 3\ k^2}{4}\ \text{sq. cm.}}$$
| {
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"url": "https://math.stackexchange.com/questions/1457865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
Split that:
$$\frac{\sin x\cdot \cos x}{x^2\cdot\sin 2x\cdot \cos x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
In the left side, we can cancel the $\cos x$ and also apply $\frac{\sin x}{x} = 1$ once:
$$\frac{1}{x\cdot\sin 2x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
That was probably a bad idea, since $x \cdot \sin2x$ will definitely be $0$... But anyway, let's keep going with the right side. There, we can apply the identity $\frac{\sin x}{x} = 1$ again:
$$\frac{1}{x\cdot\sin 2x} - \frac{1}{x\cdot\sin 2x\cdot \cos x}$$
Hey, I could get rid of the $\sin 2x$ on the left side if I multiply and divide by $2x$... the same on the right side:
$$\frac{1}{2x^2} - \frac{1}{2x^2\cdot \cos x}$$
Looking pretty, but sadly that's not going anywhere. What can I do?
| first :
$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^3}=\frac{-1}{2}$$
proof :$\lim_{x\to 0} \frac{\tan ^nx - \sin ^m x}{x^r}=?$without l'Hôpital's rule.
now :
$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}=\lim_{x\to 0}\frac{\sin x-\tan x}{x^3}.\frac{2x}{2\sin 2x}=?$$
since :
$$\lim_{x\to 0}\frac{2x}{\sin 2x}=1$$
so :
$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}=\frac{-1}{2}.\frac{1}{2}=\frac{-1}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the derivative of the function $f(x)=x/(x^2+1)$ at a point a $$\frac{x}{x^2+1}$$
I have to find the derivative and set a point by myself.
So I just set $x=a$ ( is it correct?)
and using the limit,
$$ \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{\frac{a+h}{(a+h)^2 + 1} - \frac{a}{a^2+1}}{h}$$
therefore, the answer is $$\frac{ 1-a^2}{(a^2+1)^2}$$
is it correct?
| Your answer is correct.
Notice, $$f(x)=\frac{x}{x^2+1}$$
$$\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$$
$$=\lim_{h\to 0}\frac{(x+h)(x^2+1)-x((x+h)^2+1)}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{x^3+hx^2+x+h-x^3-2hx^2-h^2x-x}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{h-hx^2}{h((x+h)^2+1)(x^2+1)}$$
$$=\lim_{h\to 0}\frac{1-x^2}{((x+h)^2+1)(x^2+1)}$$$$=\frac{1-x^2}{((x+0)^2+1)(x^2+1)}=\frac{1-x^2}{(x^2+1)^2}$$
Now, substituting arbitrary value $x=a$
$$f'(a)=\frac{1-a^2}{(a^2+1)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do we know that $x^2 + \frac{1}{x^2}$ is greater or equal to $2$? For one problem, we were supposed to know that:
$$x^2 + \frac{1}{x^2}\geq 2.$$
How do you deduce this instantly when looking at the expression above?
| $x^2 + \frac{1}{x^2} = \frac{x^4 + 1}{x^2}$
$ = \frac{x^4 - 2x^2 + 1}{x^2} + 2$
$ = \frac{(x^2 - 1)^2}{x^2} + 2$
$ \geq 0 + 2 = 2$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
| By replacing $x$ with $\arctan t$,
$$ I = \int_{0}^{\pi/2}\frac{dx}{4\sin^2 x+5\cos^2 x} = \int_{0}^{+\infty}\frac{dt}{4t^2+5}=\color{red}{\frac{\pi}{4\sqrt{5}}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to multiply radicals? The problem is $$0 = -(\sqrt{7 -2 b} - 2) * -\sqrt{7-2 b} - 2) $$
the original question was this
$$-2 = \sqrt{7 -2 b} - \sqrt{2 b + 3}$$
and than i found out the you had to isolate one radical version so i substracted $$-(\sqrt{7 -2 b} - 2)$$ and than you get $$(-\sqrt{7 -2 b} -2)^2 = (-\sqrt{2 b +3})^2 $$
i got stuck after trying to square the side with $$(-\sqrt{7 -2 b} -2)^2$$ can anyone tell me how you're supposed to square/multiply them? I am trying to perform multiplication and ultimately try to solve the equation...if anyone knows how to solve the equation the way i have shown above please help me as i have been stuck with this equation for long time
| $$\begin{align}(x + y)^2 &= (x + y)(x+y) \\ &= x(x +y) + y(x +y) \\ &= (x^2 + xy) + (xy + y^2) \\ &= x^2 + 2xy + y^2\end{align}$$
Now in your case $x = -\sqrt{7 - 2b}$ and $y = -2$, so
$$\begin{align}(-\sqrt{7 - 2b} + -2)^2 &= (-\sqrt{7 - 2b})^2 + 2(-\sqrt{7 - 2b})(-2) + (-2)^2\\&=(7 - 2b) + 4\sqrt{7 - 2b} + 4\\&=11 - 2b + 4\sqrt{7 - 2b} \end{align}$$
Which is as far as you can take it.
| {
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Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:
*
*the denominator has to be one,
*the numerator has to be a multiple of the denominator or
*the numerator and denominator have to be the same.
Obviously, with the given conditions case 3 is eliminated since $a \neq b$.
For case 1, if $b=1$, then $a=1$ which is a contradiction to the given condition that says $a \neq b$.
For case 2, I would think of examples. For example if $b=2$ then a multiple is $4$, so $a=4$. Then we have:
$$\frac{a}{b}+\frac{b}{a}=\frac{4}{2}+\frac{2}{4}=\frac{10}{4}$$
which is not an integer, but how would i proceed to show this case for all integers.
| If $a,b$ are not integers, nothing stops $x+\frac1x$ being an integer. In fact, that happens infinitely many times as the function is continuous.
Let $d=gcd(a,b)$ and $a=xd, b=yd$ with $gcd(x,y)=1$. Then
$$\frac{a}{b}+\frac{b}{a}=\frac{x^2+y^2}{xy}.$$
We will show that if $p$ is a common prime divisor of $x^2+y^2$ and $xy$ then $p=2$. Indeed,
$$(x+y)^2=x^2+y^2-2xy\text{ and }(x-y)^2=x^2+y^2-2xy$$
are both multiples of $p$. Thus $p$ divides both $x+y$ and $x-y$, and so $p$ divides $2x$ and $2y$. Since $gcd(x,y)=1$, $p$ divides $2$ and $p$ must be $2$.
It follows that $x^2+y^2$ and $xy$ are both even. From $xy$ is even, we see that either $x$ or $y$ is even. That together with $x^2+y^2$ is even, we see that both $x$ and $y$ are even. It contradicts the fact that $gcd(x,y)=1$.
Thus $x^2+y^2$ and $xy$ don't have a common prime factor. $\dfrac{x^2+y^2}{xy}$ is an integer only if $xy=1$, equivalently $x=y=\pm1$, or $a=b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the maximun $n\in N^{+}$, such $x^n+y^n+z^n$ is divisible by $x+y+z$ A prime number $p$ and integers $x,y,z$ with $0<x<y<z<p$ are given. Show that if the numbers $x^3,y^3,z^3$ give the same remainder when divided by $p$ , Find the maximun $n\in N^{+}$, such $x^n+y^n+z^n$ is divisible by $x+y+z$
when $n=2$ case be see Poland 2003
| I think you have to edit your post.
Are you sure you don’t ask about the maximum $n$ such that all of the $x^m+y^m+z^m$ with $m> n$ are not divisible by $x+y+z$?
In your hyperlink, Poland 2003, it is not question of maximum n but just for $n=2$.
See at the following example: $0<3<5<6<7$ and $3^3, 5^3, 6^3$ have the same residue, $-1$ modulo $7$.
In fact, $3^2+5^2+6^2= 5(3+5+6)$, this is true.
But you have also $3^4+5^4+6^4=143(3+5+6)$ so $n=2$ is not a maximum as required. And it is not proved in any way that $n=4$ is.
| {
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"timestamp": "2023-03-29T00:00:00",
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square root of $\frac{2+\sqrt{3}}{4}$
Find the square root of $\frac{2+\sqrt{3}}{4}$.
Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?
| $\frac{2+\sqrt{3}}{4}$ is a positive number so it has a real square root.
Notice first that:
$$\sqrt{\dfrac{2+\sqrt{3}}{4}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$$
so what you have to find is $\sqrt{2+\sqrt{3}}$. A technique is to try to find what are $a,b\in\mathbb{R}$ where $2+\sqrt{3}=(a+b)^2$ so that $2+\sqrt{3}=|a+b|$.
You have $$(a+b)^2=a^2+b^2+2ab$$.
The technique in such cases is to try to get $a^2+b^2=2$ and $2ab=\sqrt{3}$.
Notice that $b=\frac{\sqrt{3}}{2a}$
If you put it in $a^2+b^2=2$ you'll have an equation with one variable $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?
The options are:
*
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
*$$\frac{1-x^2}{1-y^2}$$
*$$\frac{1-y^2}{1-x^2}$$
I tried differentiating the entire expression:
$$\begin{align}
\frac{-2x}{2\sqrt{1-x^2}}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx} &= a(1-\frac{dy}{dx}) \\
a + \frac{x}{\sqrt{1-x^2}} &= \frac{dy}{dx}(a-\frac{y}{\sqrt{1-y^2}}) \\
\frac{dy}{dx} &= (a+\frac{x}{\sqrt{1-x^2}})(a-\frac{y}{\sqrt{1-y^2}})^{-1}
\end{align}$$
I am not sure how to proceed now. Is there a better approach? Is it advisable to use my approach?
| It looks like there's supposed to be a way to eliminate $a$ from the formula.
Let's try to tackle the term $a+\frac{x}{\sqrt{1-x^2}}$ first, using the fact that
$$a(x-y) = \sqrt{1-x^2} + \sqrt{1-y^2}.$$
In order to avoid dealing with a lot of fractions, I would
multiply $a + \frac{x}{\sqrt{1-x^2}}$ by the factor $x-y$ from the equation
and by the denominator $\sqrt{1-x^2}$.
Then
\begin{align}
\left(a + \frac{x}{\sqrt{1-x^2}}\right)(x-y)\sqrt{1-x^2}
&= a(x-y)\sqrt{1-x^2} + x(x - y) \\
&= (\sqrt{1-x^2} + \sqrt{1-y^2}) \sqrt{1-x^2} + x^2 - xy \\
&= 1 - x^2 + \sqrt{1-y^2}\sqrt{1-x^2} + x^2 - xy \\
&= 1 - xy + \sqrt{1-x^2}\sqrt{1-y^2} \\
\end{align}
Next, work out the value of
$$
\left(a - \frac{y}{\sqrt{1-y^2}}\right)(x-y)\sqrt{1-y^2}
$$
in a similar manner to get an expression in $x$ and $y$ only.
Compare the two results and see if that allows you to simplify your expression to one of the desired answers.
| {
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"timestamp": "2023-03-29T00:00:00",
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The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})?$ What is the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots (1+x^{100})?$
I manually expanded $(1+x)(1+x^2)(1+x^3)...(1+x^{10})$ and calculated the coefficient of $x^9$ as $8$ but i dont know how to solve it without expanding.Please help me.
| In every factor of your product, you will need to pick either the $1$ or the $x^i$, so immediatly we see that for all factors with $i > 9$ we need to pick the $1$, otherwise we will get $x^k$ with $k > 9$. So we can just look at the product up till $(1 + x^9)$. Now we look at the number of ways to pick $1$ and $x^i$ here such that we end up with $x^9$. Because all our coefficients are $1$, we do not need to worry about those, and count only the number of ways.
To simplify this, assume you always pick the $1$ if you don't write them down, then all possible combinations are:
$x^9, x^1\cdot x^8, x^2\cdot x^7, x^3\cdot x^6, x^4\cdot x^5, x^1\cdot x^2\cdot x^6, x^1\cdot x^3\cdot x^5, x^2\cdot x^3\cdot x^4$ which gives us $8$ as the final result.
It is not hard to see that this is equivalent to summing numbers $1, \ldots 9$ to $9$ using every number only once, and then counting the number of summations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric equation: $\cos3x+\cos x-\cos2x=0$ $\cos3x+\cos x-\cos2x=0$ find general solution
My answer:
$\cos 3x+\cos x-\cos2x=0$
$\cos2x(2\cos x-1)=0$
$\cos2x=0$ or $2\cos x-1=0$
When $\cos2x=0$
$x=(n+1)\pi/2$
When $\cos x=1/2$
Solving...$x=\pi/3$
Then $x=2n\pi\pm \pi/3$
Did I go wrong?
Is it correct?
| $$\cos(x)-\cos(2x)+\cos(3x)=0\Longleftrightarrow$$
$$1-2\cos(x)-2\cos^2(x)+4\cos^3(x)=0\Longleftrightarrow$$
$$\left(2\cos(x)-1\right)\cdot\left(2\cos^2(x)-1\right)=0\Longleftrightarrow$$
$$2\cos(x)-1=0\Longleftrightarrow \vee 2\cos^2(x)-1=0\Longleftrightarrow$$
$$2\cos(x)=1\Longleftrightarrow \vee 2\cos^2(x)=1\Longleftrightarrow$$
$$\cos(x)=\frac{1}{2}\Longleftrightarrow \vee \cos^2(x)=\frac{1}{2}\Longleftrightarrow$$
$$\cos(x)=\frac{1}{2}\Longleftrightarrow \vee \cos(x)=\pm\frac{1}{\sqrt{2}}\Longleftrightarrow$$
$$\cos(x)=\frac{1}{2}\Longleftrightarrow \vee \cos(x)=\frac{1}{\sqrt{2}} \vee \cos(x)=-\frac{1}{\sqrt{2}} \Longleftrightarrow$$
$$x=\frac{\pi}{3}+2\pi n_1 \vee x=\frac{5\pi}{3}+2\pi n_2 \vee x=\frac{\pi}{4}+2\pi n_3 \vee x=\frac{7\pi}{4}+2\pi n_4$$
$$\vee x=\frac{3\pi}{4}+2\pi n_5 \vee x=\frac{5\pi}{4}+2\pi n_6$$
With $n_1,n_2,n_3,n_4,n_5,n_6 \in \mathbb{Z}$
| {
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How many numbers less than $1000$ with digit sum to $11$ and divisible by $11$ How many positive (integers) numbers less than $1000$ with digit sum to $11$ and divisible by $11$?
There are $\lfloor 1000/11 \rfloor = 90$ numbers less than $1000$ divisible by $11$.
$N = 100a + 10b + c$ where $a + b + c = 11$ and $0 \le a, b, c \le 9$
I got $\binom{13}{2} - 9 = 69$ solutions.
| $N=a+10b+100c$. $a-b+c=0$ or $a-b+c=11$. Also we have $a+b+c=11$. We get two cases:
*
*$a-b+c=0,\;a+b+c=11$ from which we get $2a+2b=11$; impossible.
*$a-b+c=11,\;a+b+c=11$ from which we get $a+c=11,\; b=0$. They are $209, 308, 407, 506, 605, 704, 803, 902$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Checking primality for $2 \uparrow \uparrow n + 3 \uparrow \uparrow n$ Is there a clever way to test primality for
$$2 \uparrow \uparrow n \quad + \quad 3 \uparrow \uparrow n$$
where $n \gt 3$? (Not surprisingly, I got stuck after that)
For $n \le 3$ we get:
$$
\begin{matrix}
2 \uparrow \uparrow 0 \quad + \quad 3 \uparrow \uparrow 0 & = & 1 + 1 & = & 2 & (\text{prime}) & (\text{...}) \\
2 \uparrow \uparrow 1 \quad + \quad 3 \uparrow \uparrow 1 & = & 2 + 3 & = & 5 & (\text{prime}) & (\text{yes}) \\
2 \uparrow \uparrow 2 \quad + \quad 3 \uparrow \uparrow 2 & = & 2^2 + 3^3 & = & 31 & (\text{prime}) & (\text{okay}) \\
2 \uparrow \uparrow 3 \quad + \quad 3 \uparrow \uparrow 3 & = & 2^{2^2} + 3^{3^3} & = & 7625597485003 & (\text{prime}) & (\text{hmm...}) \\
\end{matrix}
$$
Example:
One simple way to rule out primality is to check if the last digit of the sum is $5$. Now we know that the last digit of $2^k$ and $3^k$ follow a pattern such that for $2^k$ with $k \equiv 0 \pmod{4}$ the last digit is $6$ and for $3^k$ with $k \equiv 3 \pmod{4}$ the last digit is $7$. From this we can state that for $n \gt 2$ the last digit of $2 \uparrow \uparrow n$ will be $6$ and for $n \gt 1$ the last digit of $3 \uparrow \uparrow n$ will be $7$, which means we can't rule out primality this way.
| For $n$ large enough ($n \ge 6$) ,$ 2\uparrow \uparrow n \equiv 2197 \pmod {4423}$ and $3 \uparrow \uparrow n \equiv 2226 \pmod {4423}$, and so $2 \uparrow \uparrow n + 3\uparrow \uparrow n$ is always a multiple of $4423$.
Assuming the limit of $a \uparrow \uparrow n$ modulo $p$ is "random" in $\Bbb F_p^*$, you can expect to get $2\uparrow \uparrow n +3\uparrow \uparrow n \to 0 \pmod p$ with probability about $1/p$.
The sum over all primes of $1/p$ diverges, so we can expect that there are infinitely many such primes. $4423$ is the smallest such prime.
So you are left with the cases $n=4$ and $n=5$ to inverstigate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Generalizations of the pentagonal number theorem Euler's pentagonal number theorem (see also the original paper and review by Jordan Bell) states
$$
\prod_{n=1}^\infty (1 - q^n) = \sum_{k=-\infty}^{\infty} (-1)^{k} q^{(3k^2 - k)/2},
$$
where $k \, (3k - 1)/2$ is the $k$th pentagon number.
It seems that we can generalize this to higher polygonal numbers, that is, for $t \ge 2$, we have
$$
(1-q)\prod_{n=1}^\infty(1-q^{nt - 1})(1-q^{nt})(1-q^{nt + 1})
=
\sum_{k=-\infty}^{\infty}
(-1)^k q^{k \, (t \, k + 2-t)/2},
$$
where $k \, (t \, k + 2 - k)/2$ is the $(t+2)$-gonal number.
Examples
For $t = 2$, it becomes,
\begin{align}
(1-q)^2 (1-q^2) (1-q^3)^2 (1-q^4) \cdots = 1 + 2 \sum_{k=1}^\infty (-1)^k q^{k^2}
\\
=1 - 2 \, q + 2 \, q^4 - 2 \, q^9 + 2 \, q^{16} - 2 \, q^{25} + 2 \, q^{36}
- 2 \, q^{49} + \cdots
\end{align}
For $t = 3$, we recover Euler's pentagonal number theorem
$$
(1-q) (1-q^2) (1-q^3) \cdots = 1 - q - q^2 + q^5 + q^7 - q^{12} - q^{15} + q^{22} + \cdots.
$$
For $t = 4$, we have
\begin{align}
(1-q) \cdot (1-q^3) (1 - q^4) (1 - q^5)
\cdot (1 - q^7) (1 - q^8) (1 - q^9) \cdots \\
= 1 - q - q^3 + q^6 + q^{10} - q^{15} - q^{21} + q^{28} + q^{36} + \cdots.
\end{align}
For $t = 5$, we have
\begin{align}
(1-q) \cdot (1-q^4) (1 - q^5) (1 - q^6)
\cdot (1 - q^9) (1 - q^{10}) (1 - q^{11}) \cdots \\
= 1 - q - q^4 + q^7 + q^{13} - q^{18} - q^{27} + q^{34} + q^{46} + \cdots.
\end{align}
I suppose that this is an established theorem. Can any one tell me more about this? Is it some special case of some general theorem? Or, are there general proofs, and further extensions? Thank you!
| It turns out that it follows directly from the Jacobi triple product identity:
\begin{align}
\prod_{n=0}^\infty
(1 - x^{2n+2}) (1 + x^{2n+1} z) \left(1 + x^{2n+1}z^{-1}\right)
=
\sum_{n=-\infty}^{\infty}
x^{n^2} \, z^n.
\tag{1}
\end{align}
With $x = q^{t/2}$ and $z = -q^{1-t/2}$, the left hand side becomes
\begin{align}
& \prod_{n=0}^\infty
\left(1 - q^{tn+t}\right)
\left(1 - q^{tn+t/2+1-t/2} \right)
\left(1 - q^{tn+t/2-1+t/2}\right)
\\
&=
\prod_{n=0}^\infty
\left(1 - q^{t(n+1)}\right) \left(1 - q^{tn+1} \right) \left(1 - q^{t(n+1)-1}\right)
\\
&=
(1-q) \prod_{n=1}^\infty
(1 - q^{tn}) (1 - q^{tn+1} ) (1 - q^{tn-1}).
\tag{2}
\end{align}
The right hand side becomes $(-1)^n \, q^{tn^2/2+n-nt/2}$, which proves the desired identity.
Proof
I shall take the chance to give a direct proof adapted from Andrews' proof for the triplet product identity, see also this post.
We first state two identities that can be proved by combinatorics.
\begin{align}
\prod_{i=0}^\infty (1 - u^i \, v)
&= \sum_{k = 0}^\infty
\frac{ (-1)^k u^{k(k-1)/2} v^k }{ (1 - u) (1 - u^2) \cdots (1 - u^k)},
\tag{3}
\\
\prod_{i=1}^\infty \frac{1}{(1 - u^i \, v)}
&= \sum_{k = 0}^\infty
\frac{ (u \, v)^k }{ (1 - u) (1 - u^2) \cdots (1 - u^k)},
\tag{4}
\end{align}
Using Eq. (3) for $u = q^t, v = q$, we get
\begin{align}
\prod_{n=0}^\infty (1-q^{tn+1})
&=
\sum_{k = 0}^\infty
\frac{ (-1)^k \, q^{t k (k - 1)/2 + k} }
{(1-q^t) (1 - q^{2t}) \cdots (1 - q^{kt}) }
\\
&=
\sum_{k = 0}^\infty
\frac{ (-1)^k \, q^{t k (k - 1)/2 + k} (1 - q^{(k+1)t}) (1 - q^{(k+2)t}) \cdots }
{(1-q^t) (1 - q^{2t})\cdots (1 - q^{kt})
(1 - q^{(k+1)t}) (1 - q^{(k+2)t}) \cdots }
\\
&=
\frac{ 1 } { \prod_{n=1}^\infty (1-q^{nt}) }
\sum_{k = 0}^\infty (-1)^k
q^{t k (k - 1)/2 + k} \left[(1 - q^{(k+1)t}) (1 - q^{(k+2)t}) \cdots \right]
\end{align}
At this point we note that if the product in the square brackets
starts from $k \le -1$, it will encounter a factor $(1 - q^0) = 0$
and thus contributes nothing to the sum. This means that it is harmless to extend the sum to negative $k$.
\begin{align}
\prod_{n=0}^\infty (1-q^{tn+1})
&=
\frac{ 1 } { \prod_{n=1}^\infty (1-q^{nt}) }
\sum_{k = -\infty}^{+\infty} (-1)^k
q^{t k (k - 1)/2 + k} \left[(1 - q^{(k+1)t}) (1 - q^{(k+2)t}) \cdots \right].
\end{align}
Now use again Eq. (3) for the product in the square brackets with $u = q^t, v = q^{(k+1)t}$, we get
\begin{align}
\prod_{n=0}^\infty (1-q^{tn+1})
&=
\frac{ 1 } { \prod_{n=1}^\infty (1-q^{nt}) }
\sum_{k = -\infty}^{+\infty}
(-1)^k \, q^{t \, k (k - 1)/2 + k}
\sum_{l = 0}^\infty
\frac{ (-1)^l q^{t \, l(l-1)/2 + (k+1)\, l \, t}}
{(1 - q^t) \, \cdots (1 - q^{lt}) } \\
&=
\frac{ 1 } { \prod_{n=1}^\infty (1-q^{nt}) }
\sum_{l = 0}^\infty
\frac{ q^{l \, (t - 1)} }
{(1 - q^t) \, \cdots (1 - q^{lt}) }
\sum_{k = -\infty}^{+\infty}
(-1)^{k + l} \, q^{t \, (k+l)^2/2 - t(k+l)/2 + (k+l)},
\end{align}
where we have changed the order of summation in the second step.
The inner sum over $k$ can be done in terms of $m = k+l$ from $-\infty$ to $\infty$.
The outer sum over $l$ can be converted to a product by Eq. (4) with $u = q^t, v = q^{-1}$, so
\begin{align}
\prod_{n=0}^\infty (1-q^{tn+1})
&=
\frac{ 1 } { \prod_{n=1}^\infty (1-q^{nt}) }
\frac{1}{ \prod_{j = 1}^\infty (1-q^{jt-1}) }
\sum_{m = -\infty}^{+\infty}
(-1)^m \, q^{t \, m^2/2 - t m/2 + m}.
\end{align}
Multiplying the two products on the denominator yields the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the integer values of c Find all possible positive integer values of c that $\frac {a^2+b^2+ab} {ab-1}$=c can take in $\mathbb N$.
I know that the solutions are c=7 and c=4 but I don't know how to prove this with Vieta Jumping method.
| THIS IS VIETA JUMPING. I RARELY LIE ABOUT MATHEMATICS.
$$ \frac{x^2 + xy + y^2}{xy-1} = c, $$
$$ x^2 + xy + y^2 = c xy - c, $$
$$ x^2 + (1-c)xy + y^2 = -c. $$
We want integers with $$ x^2 + (1-c)xy + y^2 = -c $$ and $c \geq 3,$ because otherwise the quadratic form on the left is positive definite or semidefinite.
There are no solutions with $xy \leq 0$ in any case. we take $x,y > 0.$
THIS IS THE PART WHERE VIETA ROOT JUMPING IS EXPLICIT!!!!!!!
We are going to look for solutions that minimize $x + y.$ We can replace $x$ by
$$ \color{blue}{x' = (c-1)y - x}. $$
We see that $x' < x$ so $x' + y < x+y,$ unless $2 x \leq (c-1)y. $
We can replace $y$ by
$$ \color{blue}{y' = (c-1)x - y}. $$
We see that $y' < y$ so $x + y' < x+y,$ unless $2 y \leq (c-1)x. $
SEE. VIETA ROOT JUMPING. RIGHT THERE ABOVE.
If there are any integer solutions for a particular $c,$ then there are solutions satisfying the inequalities
$$ 2 x \leq (c-1)y, $$
$$ 2 y \leq (c-1)x. $$
I'll call these the Hurwitz inequalities and Hurwitz lines. Article in 1907 in German.
For $c=4$ we get $(2,2)$
For (c=7) we get $(2,1), (1,2)$
Let's see: for those who know calculus, the Hurwitz lines intersect this branch of the hyperbola in the minimum value of $x$ and the minimum value of $y.$ What happens when $c \geq 8?$
Well, as you can see in the picture, the branch passes through a point very near $(1,1)$ when $x=y,$ to be specific
$$ x = y = \sqrt {\left(1 + \frac{3}{c-3} \right)}, $$
slightly larger then $1.$
However, as soon as $x=2,$ we find the $y$ value along the lower part smaller than $1,$
$$ y = (c-1) - \sqrt {c^2 - 3 c - 3} = \frac{c+4}{ (c-1) + \sqrt {c^2 - 3 c - 3}} $$
This thing has limit $1/2$ as $c \rightarrow \infty.$ $c=8, 0.91723,$ $c=9, 0.85857$
Not a coincidence: $c=7, x=2 \rightarrow y= 1, $ $c=4, x=2 \rightarrow y= 2. $
So that is the proof, for $c \geq 8$ there are no integer lattice points on the arc of the hyperbola within
$$ 2 x \leq (c-1)y, $$
$$ 2 y \leq (c-1)x. $$
You can draw graphs for $c = 3,5,6,$ you will see that the arc of the hyperbola between the Hurwitz lines does not contain any lattice points.
I put lots of detail at Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the derivative for a product of two polynomial functions? In my problem, I am attempting to find $f'(x)$ when $f(x)=(5x^2-2x+8)(4x^2+7x-3)$. For my work I have:
\begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt]
= {} & 80x^3 + 73x^2+6x+62
\end{align}
But when I plugged my original equation [$f(x)=(5x^2-2x+8)(4x^2+7x-3)$] into an online derivative calculator to check my answer, it comes out as:
$$=80x^3+81x^2+6x+62\ldots\text{ ?}$$
Can anyone spot where I am going wrong (if I am)?
| \begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt]
= {} & 80x^3 + 73x^2+6x+62
\end{align}
In the penultimate line of calculation, the 8th term from left should be $$4x^2 \times (-2) = -8x^2$$ instead of $(-16x^2)$. Correct that and then the answers will match.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Linear algebra Plane and line I have a plane with points: $ A = (-5, 1, 3), B = (-2, 6, -3), C = (-4, 3, 4)$
I also have a line with points: $ D = (-2, 7, 6), E = (-1, 8, -2)$
The vector equation form of the line:
$$
\left\{
\begin{array}{c|}
\begin{pmatrix}
-2\\
7\\
6
\end{pmatrix}
+ r \begin{pmatrix}
1\\
1\\
8
\end{pmatrix}
\end{array}
r \in \Bbb R
\right.
$$
The vector equation form of the plane:
$$
\left\{
\begin{array}{c|}
\begin{pmatrix}
-5\\
1\\
3
\end{pmatrix}
+ s
\begin{pmatrix}
3\\
5\\
-6
\end{pmatrix}+t
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}
\end{array}
s,t \in \Bbb R
\right.
$$
I am supposed to compute the intersection between these two.
What I know about intersection is that if I have a linear equation with no solution, then there is no intersection. If i have a Linear equation with one solution, then there is one intersection. If i have a linear equation with infinitely many solutions, then the line is contained within the plane.
I the answer is that the line is contained within the plane. I know the answer, but how is the solution squired is beyond me.
| Hint: We have that $\vec{x}$ lies in the intersection iff:
$\vec{x}=\begin{pmatrix}
-2\\
7\\
6
\end{pmatrix}
+ r \begin{pmatrix}
1\\
1\\
8
\end{pmatrix}$ and $\vec{x}=\begin{pmatrix}
-5\\
1\\
3
\end{pmatrix}
+ s
\begin{pmatrix}
3\\
5\\
-6
\end{pmatrix}+t
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}$.
This leads to: $$\begin{pmatrix}
-2\\
7\\
6
\end{pmatrix}
+ r \begin{pmatrix}
1\\
1\\
8
\end{pmatrix}=\begin{pmatrix}
-5\\
1\\
3
\end{pmatrix}
+ s
\begin{pmatrix}
3\\
5\\
-6
\end{pmatrix}+t
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}$$ This is a system of 3 linear equations with 3 unknowns. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of trigonometric relationship of angle bisector I have to prove the following equation:
$$CD=\frac{2ab\cos\frac{C}{2}}{a+b}$$ where $CD$ is the angle bisector of $C$ in $\Delta ABC $.
My attempt:
I've started by rearranging the equation in terms of $\cos \frac{C}{2}$ ,then by squaring and subtracting $\sin^2 \frac{C}{2}$ from both sides, yielding thus:
$$\cos^2 \frac{C}{2}-sin^2\frac{C}{2} =\frac{(a+b)^2 \cdot CD^2}{(2ab)^2}- \ \sin^2\frac{C}{2}\tag{1}$$
$$\cos C =\frac{(a+b)^2 \cdot CD^2}{(2ab)^2} - \frac{(s-a)(s-b)}{ab}\tag{2}$$
Now i express $ \cos C $ in terms of $a,b,c$ ,then I simplify by multiplying for $ab$ both sides and I multiply out $(s-a)(s-b)$ , getting now:
$$2(b^2+a^2-c^2)=\frac{(a+b)^2 \cdot CD^2}{4ab}-(ab+bc-b^2+c^2-a^2)\tag{3}$$
By simplifying this and rearranging for $CD^2$ I get:
$$CD^2=\frac{(a^2+b^2-c^2 +ab+ bc)(ab)}{(a+b)^2}\tag{4}$$
Finally by replacing $CD^2=ab-(abc^2)/(a+b)^2\tag{5}$
and doing all the algebraic manipulation i get the final result that
$a+c=2a$ which is clearly wrong...And here i am asking for humble help on math.stackexchange
| HINT:
$$(s-a)(s-b)=\frac{ (-b^2-a^2+c^2+2 a b)}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the half-range Fourier series expansion of $f(x) = \cos(x)$ I am stuck on the problem of calculating the half-range Fourier series expansion of $$f(x) = \cos(x),$$ $$0 < x < \frac{\pi}{2}$$
I am at the point where I have calculated the definite integral of $b_n$.
$$b_n = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}} \cos(x)\sin(nx) dx = \frac{4}{\pi} \cdot \frac{n - \sin(\frac{\pi n}{2})}{(n^2 - 1)}$$
According to [this][1] Wikipedia link, the correct answer for $b_n$ is
$$b_n = \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} \cos(x)\sin(nx) dx = \frac{4}{\pi} \cdot \frac{n((-1)^n + 1)}{(n^2 - 1)}.$$
Would someone please explain to me how to go from
$$\frac{n - \sin(\frac{\pi n}{2})}{(n^2 - 1)}$$
to $$\frac{n((-1)^n + 1)}{(n^2 - 1)}.$$
I understand that, for example, $$\cos(n\pi)$$ can be written as $$(-1)^n,$$
because $\cos(n\pi)$ has alternating $1s$ and $0s$ as $n$ increases.
However, for $\sin(\frac{n\pi}{2})$ the alteration is $1$, $0$, $-1$, $0$. I have been stuck on this problem for hours now.
I will greatly appreciate any explanation :)
[1]: https://en.wikipedia.org/wiki/Half_range_Fourier_series
| for $n=1,5,9,....$ the $\sin(\frac{n\pi}{2})=1$
$$\frac{n-\sin(\frac{n\pi}{2})}{n^2-1}=\frac{n-1}{n^2-1}=\frac{1}{n+1}$$
for $n=2,4,6,8....$ all terms equal zero
for $n=3,7,11,...$ the $\sin(\frac{n\pi}{2})=-1$
$$\frac{n-\sin(\frac{n\pi}{2})}{n^2-1}=\frac{n+1}{n^2-1}=\frac{1}{n-1}$$
now add the $\frac{1}{n+1}$ with $\frac{1}{n-1}$ to get what you want
$$\frac{1}{n+1}+\frac{1}{n-1}=\frac{2n}{n^2-1}=\frac{n((-1)^n + 1)}{(n^2 - 1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The limit of $(1-\cos^{1/3} x)/(1-\cos x^{1/3})$ as $x\to 0$ I have a problem to find answer for this limit.
$$\lim_{x\to 0}\frac{1-{\cos^{1/3} x}}{1-\cos(x^{1/3})}$$
I did it like this $$\dfrac{\dfrac{\sin x}{3×\sqrt[3]{\cos x²}}}{\sin \sqrt[3]x×\dfrac 1{3×\sqrt[3]{x²}}}$$
| Applying l'Hopital's rule:
$$\lim_{x\to 0} \frac{1-\sqrt[3]{\cos(x)}}{1-\cos\left(\sqrt[3]{x}\right)}=$$
$$\lim_{x\to 0} \frac{\frac{\text{d}}{\text{d}x}\left(1-\sqrt[3]{\cos(x)}\right)}{\frac{\text{d}}{\text{d}x}\left(1-\cos\left(\sqrt[3]{x}\right)\right)}=$$
$$\lim_{x\to 0} \frac{\frac{\sin(x)}{3\cos^{\frac{2}{3}}(x)}}{\frac{\sin\left(\sqrt[3]{x}\right)}{3x^{\frac{2}{3}}}}=$$
$$\lim_{x\to 0} \frac{x^{\frac{2}{3}}\cdot\cos^{-\frac{2}{3}}(x)\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}=$$
$$\left(\lim_{x\to 0}\cos^{-\frac{2}{3}}(x)\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(\lim_{x\to 0}\frac{1}{\cos^{\frac{2}{3}}(x)}\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(\frac{1}{\cos^{\frac{2}{3}}(0)}\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(\frac{1}{1}\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\left(1\right)\cdot\left(\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}\right)=$$
$$\lim_{x\to 0}\frac{x^{\frac{2}{3}}\cdot\sin(x)}{\sin\left(\sqrt[3]{x}\right)}$$
Now l'Hopital's rule again, and you'll find the answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
| $$z^5+z^4+z^3+z^2+z+1 = 0 \Longleftrightarrow$$
$$(z+1)(z^2-z+1)(z^2+z+1) = 0 \Longleftrightarrow$$
$$z+1=0 \vee z^2-z+1=0 \vee z^2+z+1=0 \Longleftrightarrow$$
$$z=-1 \vee z=\frac{1\pm\sqrt{-3}}{2} \vee z=\frac{-1\pm\sqrt{-3}}{2} \Longleftrightarrow$$
$$z=-1 \vee z=\frac{1\pm i\sqrt{3}}{2} \vee z=\frac{-1\pm i\sqrt{3}}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 4
} |
Remainder of $7^{7^{7}}$ divided by $32$ The problem is to find remainder of $7^{7^{7}}$ divided by $32$ the quickest way.
My attempt:
$32 = 8.4$
Mod 8: $7^{7^{7}}\equiv(-1)^{7^{7}}\equiv-1$
Mod 4: $7^{7^{7}}\equiv(-1)^{7^{7}}\equiv-1$
Hence in mod $32$: $7^{7^{7}}\equiv(-1)$ mod $8.(-1)$mod$ 4 (?) \equiv 7.3\equiv21 $
I am quite worried about the last step. I can't justify it.
Is there a theorem related to (?) or
Is there a better and correct way to do this ?
| Fact: $a^{2^{k-2}}\equiv 1\pmod{2^k}$ for all $k\ge 3$, odd $a$.
Proof: We use induction. $a^2\equiv 1\pmod{8}$ for all odd $a$.
If $a^{2^{m-2}}\equiv 1\pmod{2^m}$, we want to prove $a^{2^{m-1}}\equiv 1\pmod{2^{m+1}}$.
I.e. if $a^{2^{m-2}}=2^mk+1$, we want to prove $\left(2^mk+1\right)^2\equiv 1\pmod{2^{m+1}}$.
$$\left(2^mk+1\right)^2\equiv 2^{2m}k+2^{m+1}k+1\equiv 1\pmod{2^{m+1}}$$
So $7^{8}\equiv 1\pmod{32}$. Then $$7^{7^7}\equiv 7^{7^7\pmod{8}}\equiv 7^{(-1)^7}\equiv 7^{-1}\pmod{32}$$
$$\equiv \frac{1}{7}\equiv \frac{5}{7\cdot 5}\equiv \frac{5}{3}\equiv \frac{-27}{3}\equiv -9\equiv 23\pmod{32}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How does $\sqrt{y^2 + y^2} = 1$ give $y = -1/\sqrt{2}$ here? I'm studying this problem:
and I cannot see the last step, how we go from $$\sqrt{y^2 + y^2} = 1$$ to $$y = -1/\sqrt{2}$$
When I try to solve this, I end up with:
$$y^2 + y^2 = 1^2 = 1$$
$$2y^2 = 1$$
$$y^2 = 2$$
$$y = \sqrt{2}$$
| There are basically two interpretations here: either solve $\sqrt{y^2+y^2}=1$ to get $-\frac{1}{\sqrt{2}}$.
Or, the other interpretation is: They didn't need to solve the equation $\sqrt{y^2+y^2}=1$. They just made the observation that $$\cos\left(\frac{5\pi}{4}\right)=\sin\left(\frac{5\pi}{4}\right)=-\frac{1}{\sqrt{2}}$$
To see this, recall that the sine or cosine of $45^{\circ}$ or $\frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}$.
So $\sin\left(\frac{5\pi}{4}\right)=\sin\left(\frac{\pi}{4}+\pi\right)$ (a translation in the negative $x$ direction).
By 'they', I mean the person who wrote the paper.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
2nd degree matrix equation Let $X$ be a matrix with 2 rows and 2 columns.
Solve the following equation:
$$ X^2 = \begin{pmatrix}
3 & 5\\
-5 & 8
\end{pmatrix} $$
Here is what I did:
Let $ X = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} $. After multiplying I got the following system:
$$ \left\{\begin{matrix}
a^2 + bc = 3\\
ab + bd = 5\\
ac + cd = -5\\
d^2 + bc = 8
\end{matrix}\right. $$
At this point I got stucked.
If you know how to solve this please help me! Thank you!
| A nonsingular $n \times n$ matrix $M$ will have some square roots that are polynomials in $M$ of degree $\le n-1$. Thus in this case we can look for solutions of the form $X = s M + t I$. By the Cayley-Hamilton theorem,
a matrix satisfies its characteristic polynomial: in this case the characteristic polynomial is $p(x) = x^2 - 11 x + 49$, and $M^2 - 11 M + 49 I = 0$. Thus $s M + t I$ will be a square root of $M$ if
$(s x + t)^2 - x$ is a multiple of $p(x)$. In this case
$$(s x + t)^2 - x - s^2 p(x) = (11 s^2 + 2 s t - 1) x - 49 s^2 + t^2$$
so we want
$$ \eqalign{11 s^2 & + 2 s t - 1 = 0\cr
-49 s^2 & + t^2 = 0\cr}$$
The solutions are
$$ \eqalign{ s &= 1/5, t = 7/5 \cr
s &= -1/5, t = -7/5\cr
s &= i/\sqrt{3}, t = -7 i/\sqrt{3}\cr
s &= -i/\sqrt{3}, t = 7 i/\sqrt{3}\cr}$$
corresponding to
$$ X = \pmatrix{2 & 1\cr -1 & 3\cr},\ \pmatrix{-2 & -1\cr 1 & -3\cr},\
\pmatrix{-4i/\sqrt{3} & 5i\sqrt{3}\cr -5i/\sqrt{3} & i/\sqrt{3}},\
\pmatrix{4i/\sqrt{3} & -5i\sqrt{3}\cr 5i/\sqrt{3} & -i/\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge? Does the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots $ converge or diverge?
| \begin{align}
& \underbrace{1+\frac{1}{2}-\frac{1}{3}}_{\ge 1} + \underbrace{\frac{1}{4}+\frac{1}{5}-\frac{1}{6}}_{\ge 1/4} + \underbrace{\frac{1}{7}+\frac{1}{8}-\frac{1}{9}}_{\ge 1/7} + \underbrace{\frac 1 {10} + \frac 1 {11} - \frac 1 {12}}_{\ge 1/10} + \cdots \\[10pt]
\ge {} & 1 + \frac 1 4 + \frac 1 7 + \frac 1 {10} + \frac 1 {13} + \cdots \\[10pt]
\ge {} & \frac 1 3 + \frac 1 6 + \frac 19 + \frac 1 {12} + \frac 1 {15} + \cdots \\[10pt]
= {} & \frac 1 3 \left( 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots \right) = \infty.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $p$ be prime not equal to 2 or 5. Show $p^2+1$ or $p^2-1$ is divisible by 10. I can do half the proof but can not think of a way to finish. If $p$ is prime then both $p-1$ and $p+1$ are even. In cases where either $p-1$ or $p+1$ is divisible by 5 that implies $(p-1)(p+1)=p^2-1=10n$ for some positive integer $n$. I am not sure if I am on the right track with this.
| If $p \equiv 3$ or $7 \pmod{10}$, then $p^2 \equiv 9 \pmod{10}$, so $p^2 + 1 \equiv 0 \pmod{10}$ as desired. But if $p \equiv 1$ or $9 \pmod{10}$, then $p^2 \equiv 1 \pmod{10}$, so $p^2 - 1 \equiv 0 \pmod{10}$ as desired.
Examples: $3^2 + 1 = 10$, $7^2 + 1 = 50$, $11^2 - 1 = 120$, $19^2 - 1 = 360$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
How do I prove that the inequality $\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $ holds for every $x,y>0$ Prove that for every $x,y>0$, the following inequality holds. Begin by proving the right side first and assume that $(\sqrt { x } -\sqrt { y } )^2\ge 0$
$$\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $$
What I did thus far:
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\sqrt { xy } \le \frac { x+y }{ 2 } $
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy }$
3) Then, $\frac { x+y }{ 2 } \ge \sqrt { xy } $
4) Therefore, $\sqrt { xy } \le \frac { x+y }{ 2 } $
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\frac { 2xy }{ x+y } \le \sqrt { xy } $
At this point, I get stuck. I don't know what do to prove the other side of the inequality. I also don't know if I am going about this proof the right way.
Hints are much more appreciated than the actual solution.
Edit: (added this after getting advice from answers below)
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy } $
3) By taking the reciprocal of both sides, we get: $\frac { 1 }{ x+y } \le \frac { 1 }{ 2\sqrt { xy } } $
4) Then, $(\sqrt { xy } )\frac { 1 }{ x+y } \le (\sqrt { xy } )\frac { 1 }{ 2\sqrt { xy } } \Rightarrow \frac { \sqrt { xy } }{ x+y } \le \frac { \sqrt { xy } }{ 2\sqrt { xy } } $
5) Then, $(2\sqrt { xy } )\frac { \sqrt { xy } }{ x+y } \le (2\sqrt { xy } )\frac { \sqrt { xy } }{ 2\sqrt { xy } } $
6) Therefore, $\frac { 2xy }{ x+y } \le \sqrt { xy } $
QED
| Since $x,y>0$, you can divide by any of them (or by their square roots) on both sides of the inequality. Thus,
$$\frac{2xy}{x+y}\le\sqrt{xy} \quad \Leftrightarrow \quad \frac{2\sqrt{xy}}{x+y}\le1 \quad \Leftrightarrow \quad 2\sqrt{xy}\le x+y.$$
You have already proved the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule
Find the following limit
$$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x\right)^\frac1x=e,\lim_{x\to0}x^x=1$ and equivalent infinitesimals at $x\to0$ such as $x\sim\sin x,a^x\sim1+x\ln a,(1+x)^a\sim1+ax$. This is what I did so far:
$$\begin{align}\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\lim_{x\to0}\frac{\left(\frac{3-(1-\cos x)}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(\frac{3-\frac12x^2}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^x-1}{x^3}\end{align}$$
I used fact that $x\sim\sin x$ at $x\to0$. After that, I tried to simplify $\left(1-\frac16x^2\right)^x$. Problem is because this is not indeterminate, so I cannot use infinitesimals here. Can this be solved algebraically using fundamental limits, or I need different approach?
| We can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\frac{1}{x^{3}}\left(\left(\frac{2 + \cos x}{3}\right)^{x} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{3}}\left(\exp\left(x\log\left(\frac{2 + \cos x}{3}\right)\right) - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{3}}\cdot x\log\left(\dfrac{2 + \cos x}{3}\right)\cdot\dfrac{\exp\left(x\log\left(\dfrac{2 + \cos x}{3}\right)\right) - 1}{x\log\left(\dfrac{2 + \cos x}{3}\right)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \log\left(\dfrac{2 + \cos x}{3}\right)\cdot 1\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \log\left(1 + \frac{\cos x - 1}{3}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \dfrac{\cos x - 1}{3}\cdot\dfrac{\log\left(1 + \dfrac{\cos x - 1}{3}\right)}{\dfrac{\cos x - 1}{3}}\notag\\
&= -\frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\notag\\
&= -\frac{1}{6}\notag
\end{align}
We have used the following standard limits $$\lim_{x \to 0}\frac{\exp(x) - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Using Hensel's Lifting Lemma to Solve $x^2 + x + 34 \equiv 0 \pmod{81}$ As in the title, I'm trying to solve $$x^2 + x + 34 \equiv 0 \pmod{81}.$$
Let $f(x) = x^2 + x + 34$ throughout.
I'm using Hensel's lemma, but it's a bit dense and I'm not sure my interpretation is correct. My method involves checking for solutions through mod 3, mod 9, mod 27 and mod 81, and checking the conditions on $f$ and $f'$ each time. But should I check the conditions for mod 81 too? Or just substitute my penultimate solutions into the original congruence? More specifically, in the penultimate step (mod 27) I get the solutions $$x \equiv 4 \pmod{27}, \quad x \equiv 13 \pmod{27}, \quad x \equiv 22 \pmod{27}.$$
Now, is all that remains to substitute $x = 4,13,22$ into the original equation? None of them are solutions and thus by Hensel's lemma there are no solutions?
| You probably can't use Hensel's Lemma:
we need $f'(x) = 2x + 1 \not \equiv 0 \mod 3$ so we need $x \equiv 0, 2 \mod 3$ and that just isn't likely to solve $x^2 + x + 7 + 27 \equiv 0 \mod 3^k$
But $x^2 + x + 34 \equiv x^2 - 80x + (27 + 7) \equiv$
$x^2 -80 x + 40^2 - 40^2 + (27+7) \equiv $
$(x-40)^2 -(27 + 13)^2 + (27 + 7) \equiv $
$(x-40)^2 - 26*27 - 13^2 + (27 + 7)\equiv $
$(x - 40)^2 - 25*27 - 169 +7\equiv $
$(x-40)^2 - 27 - 7+7 \equiv (x-40)^2 - 27 \mod 81$
$\implies (x-40)^2 \equiv 27 \mod 81$.
Which means $27 + 81k$ is a perfect square for some $k$.
So $27(3k + 1)$ is a perfect square. But as $3\not |3k + 1$ that is impossible.
So there are no solutions.
.....
(Or in hindsight: $x^2 + x + 34 \equiv 0 \mod 81 \iff$
$x^2 + x \equiv -27 - 7 \mod 81\iff$
$4x^2 + 4x \equiv -4*27 - 28 \equiv -5*27 - 1\equiv 27 -1 \mod 81\iff$
$4x^2 + 4x+1 = (2x + 1)^2 \equiv 27 \mod 81$
Same contradiction.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find limit "1 ^ infinity" $\displaystyle \lim_\limits{x \to \frac{\pi}{4}}
\left[\frac{1 + \sin(\pi/4 - x)\sin(2x)}{1 + 0.5\cos(2x)}\right]^{\cot^3(x - \frac{\pi}{4})}$
I've tried to use $\exp$, but found very difficult to simplify the expression.
| $$\displaystyle \lim_\limits{x \to \frac{\pi}{4}}
\left(\frac{1 + \sin\left(\frac{\pi}{4} - x\right)\sin(2x)}{1 + \frac{1}{2}\cos(2x)}\right)^{\cot^3(x - \frac{\pi}{4})}=$$
$$\displaystyle \lim_\limits{x \to \frac{\pi}{4}}
\left(\frac{2+\sqrt{2}(\cos(x)-\sin(x))\sin(2x)}{2+\cos(2x)}\right)^{-\tan^3\left(\frac{\pi}{4}+x\right)}=$$
$$\lim_{x \to \frac{\pi}{4}}
\exp\left(\ln\left(\left(\frac{2+\sqrt{2}(\cos(x)-\sin(x))\sin(2x)}{2+\cos(2x)}\right)^{-\tan^3\left(\frac{\pi}{4}+x\right)}\right)\right)=$$
$$\lim_{x \to \frac{\pi}{4}}
\exp\left(-\ln\left(\frac{2+\sqrt{2}(\cos(x)-\sin(x))\sin(2x)}{2+\cos(2x)}\right)\tan^3\left(\frac{\pi}{4}+x\right)\right)=$$
$$\exp\left(\lim_{x \to \frac{\pi}{4}}
-\ln\left(\frac{2+\sqrt{2}(\cos(x)-\sin(x))\sin(2x)}{2+\cos(2x)}\right)\tan^3\left(\frac{\pi}{4}+x\right)\right)=$$
$$\exp\left(
-\left(\lim_{x \to \frac{\pi}{4}}\ln\left(\frac{2+\sqrt{2}(\cos(x)-\sin(x))\sin(2x)}{2+\cos(2x)}\right)\tan^3\left(\frac{\pi}{4}+x\right)\right)\right)=$$
$$\exp\left(
-\left(\lim_{x \to \frac{\pi}{4}}\sin^3\left(x+\frac{\pi}{4}\right)\csc^3\left(\frac{\pi}{4}-x\right)\ln\left(\frac{\sqrt{2}\sin(2x)(\cos(x)-\sin(x))+2}{2+\cos(2x)}\right)\right)\right)=$$
$$\exp\left(
-\left(\lim_{x \to \frac{\pi}{4}}\csc^3\left(\frac{\pi}{4}-x\right)\ln\left(\frac{\sqrt{2}\sin(2x)(\cos(x)-\sin(x))+2}{2+\cos(2x)}\right)\right)\right)=$$
$$\exp\left(-\left(-\frac{3}{2}\right)\right)=e^{--\frac{3}{2}}=e^{\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of powers of sine
Find $\displaystyle \sum_{n=1}^{89} \sin^6(n) = \frac{m}{n}$
Let $x = \sin(n)$ and let $y = \cos(n)$.
Since $\cos(n) = \sin(90 - n)$ it follows that
$= \sin^6(1) +\sin^6(1) + ... + \sin^6(45) + \cos^6(1) + \cos^6(2) + ... + \cos^6(44)$.
EDIT:
$x^6 + y^6 = (x^2 + y^2)(x^4 - (xy)^2 + y^4)$.
$(x^2 + y^2)^2 = x^4 + y^4 + 2(xy)^2 \implies x^4 + y^4 - (xy)^2 = (x^2 + y^2)^2 - 3(xy)^2$
Hence,
$x^6 + y^6 = (x^2 + y^2)\bigg((x^2 + y^2)^2 - 3(xy)^2 \bigg)$
Since $x = \sin(n)$ and $y= \cos(n)$, it really follows that:
$x^6 + y^6 = (1)(1 - 3(xy)^2)) = 1 - 3(xy)^2$.
Since $\sin(n)\cos(n) = \frac{\sin(2n)}{2}$, let $t = \frac{\sin(2n)}{2}$
So then,
$x^6 + y^6 = 1 - \frac{3}{4} t^2$, also $\sin^6(45) = \frac{1}{8}$.
Hence,
$S = \frac{1}{8} + (1 - 3[\sin^2(1)\cos^2(1) + ... + \sin^2(44)\cos^2(44)])$.
$ = \frac{1}{8} + 1 - \frac{3}{4}[\sin^2(2) + \sin^2(4) + .. + \sin^2(88)]$
$ = \frac{1}{8} + 1 - \frac{3}{4} [ \sin^2(2) + \cos^2(2) + \sin^2(4) + \cos^2(4)+ ... + \sin^2(44) + \cos^2(44)]$
$ = \frac{1}{8} + 1 - \frac{3}{4} [22 ] = \frac{1}{8} + 1 - \frac{33}{2} = \frac{1 + 8 - 132}{8} = \frac{-123}{8}$.
But that is still not right?
| Let $$S = \sum^{89}_{n=1}\sin^6(n)\;,$$ as you have used $\sin(90^\circ-n) = \cos (n)$.
So we get $$S = \sum^{89}_{n=1}\sin^6(90^{\circ}-n) = \sum^{89}_{n=1}\cos^6(n)\;,$$ as you have used $\sin(90-n) = \cos (n)$.
Now Add these two equation, We get $$2S = \sum^{89}_{n=1}\left[\sin^6(n)+\cos^{6}(n)\right] = \sum^{89}_{n=1}\left[1-3\sin^2 (n)\cdot \cos^2(n)\right]$$
Above we have uesd the formula If $\bullet\; \sin^2 x+\cos^2 x+(-1) =0\;,$ Then $$\sin^6 x+\cos^6 x-1 = 3\cdot \sin^2 x\cdot \cos^2 x\cdot -1$$
So we get $$\sin^6 x+\cos^6 x= 1-3\sin^2 x \cdot \cos^2 x.$$
So we get $$2S = \sum^{89}_{n=1}1-\frac{3}{4}\sum^{89}_{n=1}\sin^2(2n) = 89-\frac{3}{4}T$$
Where $$T = \sum^{89}_{n=1}\sin^2(2n)=2\sum^{44}_{n=1}\sin^2(n)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there a difference between $(x)^{\frac{1}{n}} $ and $\sqrt[n]{x}$? Is there a difference between $(x)^{\frac{1}{n}}$ and $ \sqrt[n]{x}$ ?
I'm confused with this topic. Any ideas or examples ?
If $(x)^{\frac{1}{n}} = \sqrt[n]{x}$
Consider $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . Is it the same if I write $ x=\frac{-b \pm(b^2-4ac)^{\frac{1}{2}}}{2a}$ ?
| There is no difference. In the same way as I write $e^x$ when $x$ is simple enough, and $\exp(x)$ otherwise, I also write $\sqrt{x}$ when $x$ is simple enough, and $(x)^{1/2}$ otherwise.
For example, I write
$$
\left(\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}\right)^{1/2}
\qquad\text{and not the ugly}\qquad
\sqrt{\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}}
$$
Just as I write
$$
\exp\left(\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}\right)
\qquad\text{and not the illegible}\qquad
e^{\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$
$|(1+i\sqrt{3})(1+i)|=8$
Using Moivre's theorem on the LHS:
$$\sqrt{(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)}=2\sqrt{2}\left(\cos\frac{\phi+2k\pi}{2}+i\sin\frac{\phi+2k\pi}{2}\right),k=0,1$$
This is not correct (checked for $\phi=\pi/3$).
How to prove this equation?
| Hint:
Since you can write
$(1+i\sqrt{3})(1+i)e^{i \phi}=2\sqrt{2}e^{i \phi} e^{i {7 \pi \over 12}}$, so
you can concentrate on showing
$(1+i\sqrt{3})(1+i)=2\sqrt{2} e^{i {7 \pi \over 12}}$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
min polynomial of $\sqrt{6+2\sqrt{5}}$ min polynomial of $\sqrt{6+2\sqrt{5}}$ over $\mathbb{Q}$
I found $\alpha^4 -12\alpha^2 +16 = 0$ but wolframalpha does not agree with me. furthermore, would anything change is it was over $\mathbb{R}$?
| Since $\sqrt{6+ 2\sqrt{5}}$ is a real number, if it were "over R" then the minimum polynomial would be $x- \sqrt{6+ 2\sqrt{5}}$!
The way I would do this problem would be to note that $(x- \sqrt{6+ 2\sqrt{5}})(x+ \sqrt{6+ 2\sqrt{5}})= x^2- (6+ 2\sqrt{5})= (x^2- 6)- 2\sqrt{5}$.
Then $((x^2- 6)- 2\sqrt{5})((x^2- 6)+ 2\sqrt{5})= (x^2- 6)^2- 20= x^4- 12x^2+ 36- 20= x^4- 12x^2+ 16$.
Notice the use of $(a- b)(a+ b)= a^2- b^2$ to get rid of the square roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Where am I going wrong when trying to solve this system of equations using Gaussian Elimination? $$3x-y+z=5 \\ 2x+y-z=1 \\ x-y+z=2 \\ 4x+4y+z=3$$
Steps I took:
$$\left[\begin{array}{rrr|r}
3 & -1 & 1 & 5 \\
2 & 1 & -1 & 1 \\
1 & -1 & 1 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$\Rightarrow { R }_{ 1 }={ R }_{ 1 }+{ R }_{ 2 } = \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
2 & 1 & -1 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$\Rightarrow { R }_{ 2 }={ R }_{4}-2{ R }_{ 2 } = \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
0 & 2 & 3 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right]$$
$$ \Rightarrow { R }_{ 3 }=4{ R }_{3}-{ R }_{ 4 } = \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
0 & 2 & 3 & 1 \\
0 & -8 & 15 & 5\\
4 & 4 & 1 & 3
\end{array}\right]$$
At this point I have no idea what to do to get solve this system of equations. To be honest, I don't even know how to handle Gaussian Elimination with $4$ equations with $3$ unknowns. I would like an explanation as to what I need to do (in layman terms, please) and then point me in the right direction.
| $\left[\begin{array}{rrr|r}
3 & -1 & 1 & 5 \\
2 & 1 & -1 & 1 \\
1 & -1 & 1 & 2\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{ 1 }={ R }_{ 1 }+{ R }_{ 2 }\Rightarrow
\left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
2 & 1 & -1 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{ 2 }={ R }_{4}-2{ R }_{ 2 }
\Rightarrow \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
0 & 2 & 3 & 1 \\
1 & -1 & 4 & 2\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{ 3 }=4{ R }_{3}-{ R }_{ 4 }
\Rightarrow \left[\begin{array}{rrr|r}
5 & 0 & 0 & 6 \\
0 & 2 & 3 & 1 \\
0 & -8 & 15 & 5\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{ 1 }={ R }_{1}/5 \text{ and } { R }_{ 2 }={ R }_{2}/2
\Rightarrow \left[\begin{array}{rrr|r}
1 & 0 & 0 & 6/5 \\
0 & 1 & 3/2 & 1/2 \\
0 & -8 & 15 & 5\\
4 & 4 & 1 & 3
\end{array}\right],
{ R }_{ 4 }={ R }_{4}-4{ R }_{ 1 }
\Rightarrow \left[\begin{array}{rrr|r}
1 & 0 & 0 & 6/5 \\
0 & 1 & 3/2 & 1/2 \\
0 & -8 & 15 & 5\\
0 & 4 & 1 & -9/5
\end{array}\right],
{ R }_{ 4 }={ R }_{4}-4{ R }_{ 2 }\text{ and }{ R }_{ 3 }={ R }_{3}+8{ R }_{ 2 }
\Rightarrow \left[\begin{array}{rrr|r}
1 & 0 & 0 & 6/5 \\
0 & 1 & 3/2 & 1/2 \\
0 & 0 & 27 & 9\\
0 & 0 & -5 & -19/5
\end{array}\right]$
Now you can stop, since you got a contradiction you got $27x_3=9\Rightarrow x_3=1/3$ and $-5x_3=-19/5\Rightarrow x_3=-19/25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).
How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
| Permit me to contribute an algebraic proof that does not use Beta
functions.
Suppose we seek to verify that
$$S_n = \sum_{k=0}^n \frac{(-1)^k}{2k+1} {n\choose k} =
\frac{2^{2n} (n!)^2}{(2n+1) (2n)!}.$$
We have by inspection that
$$S_n = \sum_{k=0}^n \mathrm{Res}(f(z); z=k)$$
where $$f(z) =
(-1)^n n! \frac{1}{2z+1} \prod_{q=0}^n \frac{1}{z-q}.$$
This is because
$$\mathrm{Res}(f(z); z=k) =
(-1)^n \frac{n!}{2k+1} \prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\= (-1)^n \frac{n!}{2k+1} \frac{1}{k!} (-1)^{n-k} \frac{1}{(n-k)!}
= \frac{(-1)^k}{2k+1} {n\choose k}.$$
Now with $f(z)$ being rational we must have
$$S_n = - \mathrm{Res}(f(z); z=-1/2)
- \mathrm{Res}(f(z); z=\infty).$$
From $z=-1/2$ we get including the sign
$$- (-1)^n \frac{n!}{2} \prod_{q=0}^n \frac{1}{-1/2-q}
= - (-1)^n n! 2^n \prod_{q=0}^n \frac{1}{-1-2q}
= n! 2^n \prod_{q=0}^n \frac{1}{2q+1}
\\ = n! 2^n \times \frac{2^n n! }{(2n+1)(2n)!}
= \frac{2^{2n} (n!)^2}{(2n+1) (2n)!}.$$
For the residue at infinity we get including the sign
$$\mathrm{Res}_{z=0} \frac{1}{z^2}
(-1)^n n! \frac{1}{2/z+1} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = \mathrm{Res}_{z=0} \frac{1}{z}
(-1)^n n! \frac{1}{z+2} \prod_{q=0}^n \frac{z}{1-qz} = 0.$$
Collecting the two contributions we obtain
$$\frac{2^{2n} (n!)^2}{(2n+1)!}$$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Mathematical Induction Divisibility Problem Prove that if $n \ge 1$ is a positive integer, then $13^n − 6^n$ is divisible by $7$.
In proving the $n = k+1$ case, I get to $133k + 6^k\cdot13 - 6\cdot13^k = 7M$, where $M$ is a positive integer. $133k$ is divisible by $7$. How do I show the $6^k\cdot13 - 6\cdot13^k$ part is divisible by $7$?
| $13^{k+1} - 6^{k+1} = 13^k* 13 - 6^k * 6 = 13^k(7 + 6) - 6^k*6 = 7*13^k + 6*13^k - 6^k*6 = 7*13^k + 6(13^k - 6^k) = 7*13^k + 6(7*M) = 7*(13^k + k)$ is a multiple of 7.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use mathematical induction to show that $ H_{2^n} \geq 1+ \frac{n}{2} $ An Inequality for Harmonic Numbers.
The harmonic numbers $H_j, j=1,2,3,...,$ are defined by $$H_j = 1 + \cfrac{1}{2}+\cfrac{1}{3}+...+\cfrac{1}{j}$$
Use mathematical induction to show that $$ H_{2^n} \geq 1+ \frac{n}{2} $$
whenever $n$ is a nonnegative integer.
BASIS STEP: $P(0)$ is true, because $H_{2^0}=H_1=1 \geq 1+\dfrac{0}{2}$
INDUCTIVE STEP: The inductive hypothesis is the statement that $P(k)$ is true, that is, $H_{2^k} \geq 1 +\dfrac{k}{2}$, where $k$ is an arbitrary nonnegative integer. We must show that if $P(k)$ is true, then $P(k+1)$, which states that $H_{2^{k+1}} \geq 1 +\dfrac{k+1}{2}$, is also true. So, assuming the inductive hypothesis, it follows that
$$H_{2^{k+1}} = 1+ \frac{1}{2} + \frac{1}{3} + ...+ \frac{1}{2^k} + \frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}}$$
$$=H_{2^k}+\frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}}$$
$$\geq (1+\frac{k}{2})+ \frac{1}{2^k+1}+...+ \frac{1}{2^{k+1}} \qquad ...(?)$$
$$\geq (1+\frac{k}{2})+2^k \cdot \frac{1}{2^{k+1}}\qquad ...(??)$$
$$\geq (1+\frac{k}{2}) + \frac{1}{2}$$
$$=1+\frac{k+1}{2} $$
I don't understand what is going on at lines $(?)$ and $(??)$, why did it change from $=H_{2^k}$ to $\geq (1+\dfrac{k}{2})$ can somebody explain it to me?
| The first inequality is the inductive hypothesis.
As for the second, note that for all $j \in \{ 1, \dots 2^k\}$, $$\frac{1}{2^k+j} \ge \frac{1}{2^{k+1}}$$
So that
$$\sum_{j=1}^{2^k} \frac{1}{2^k+j} \ge \sum_{j=1}^{2^k} \frac{1}{2^{k+1}} = 2^k \frac{1}{2^{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$P$ is a prime number not equal to $2$ or $3$. Show that $p - 6$ is a quadratic residue $\pmod p$ whenever... $P$ is a prime number not equal to $2$ or $3$.
Show that $p - 6$ is a quadratic residue $\pmod p$ whenever $p \equiv 1, 5, 7, 11\pmod {24}$
And show that $p - 6$ is a quadratic non - residue $\pmod p$ whenever $p\equiv 13, 17, 19, 23\pmod {24}$
I know that $p - 6$ is a quadratic residue if there exists a $b$ such that $p - 6\equiv b^2\pmod p$
So for $p = 11$ we get $11 - 6\equiv b^2\pmod {11}$ and $b = 4$ satisfies this.
Similarly,
For $p = 1$ : $b = 0$
For $p = 5$ : $b = 2$
For $p = 7$ : $b = 1$
And for the quadratic non - residue:
Take $p = 13$ then we see that none of $1^2, 2^2, 3^2, \ldots, 12^2$ are congruent to $13 - 6 = 7 \pmod {13}$. So it is a quadratic non - residue. This is also the case for $17, 19, 23$.
What I do not understand is: Where does the number $24$ come from? And how do I use it in the explanation above, or is this not necessary?
| I.e. prove: if $p\ge 5$, then $\left(\frac{-6}{p}\right)=1\iff p\equiv \{1,5,7,11\}\pmod{24}$. Use Quadratic Reciprocity.
$$\left(\frac{-6}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)=1$$
$$\iff \begin{cases}\begin{cases}p\equiv 1\pmod{4}\\p\equiv \pm 1\pmod{8}\\p\equiv \pm 1\pmod{12}\end{cases}\\\text{or}\\\begin{cases}p\equiv 1\pmod{4}\\p\equiv \{3,5\}\pmod{8}\\p\equiv \pm 5\pmod{12}\end{cases}\\\text{or}\\\begin{cases}p\equiv 3\pmod{4}\\p\equiv \{3,5\}\pmod{8}\\p\equiv \pm 1\pmod{12}\end{cases}\\\text{or}\\\begin{cases}p\equiv 3\pmod{4}\\p\equiv \pm 1\pmod{8}\\p\equiv \pm 5\pmod{12}\end{cases}\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$
If one has the result it is not difficult to verify it by induction, but how can I solve it without induction ?
| We have :
$$
\begin{split}
\require{cancel}
\dfrac{1\cdot2\cdot\bcancel{3\cdot4\cdot5\cdots k}}{\bcancel{3\cdot4\cdot5\cdots k}\cdot k+1\cdot k+2}&=\dfrac{1\cdot2}{k+1\cdot k+2}\\&=\dfrac{2}{k+1}-\dfrac{2}{k+2}
\end{split}
$$
So the summation $S_j=\sum_{k=0}^{j-1}\dfrac{1\cdot2\cdots k}{3\cdot4\cdots k+2}$ is a telescoping sum and is equal to:
$$
\begin{split}
S_j&=\sum_{k=0}^{j-1}\left(\dfrac{2}{k+1}-\dfrac{2}{k+2}\right)
\\&=2\left(1-\dfrac{1}{j+1}\right)
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $x=y$, given $\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$ Given $x,y\in(0,\frac{\pi}{2}]$, $\alpha>0$ is a constant satisfying $0<x+y+\alpha<\pi$,
Also it is known that
$$\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$$
How can we prove $x=y$ "elegantly"? I am stuck on this problem for several days. Please help.
| $$\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$$
or, $$\frac{1}{\sin x}-\frac{1}{\sin y}=\frac{1}{\sin(y+\alpha)}-\frac{1}{\sin(x+\alpha)}$$
or, $$\frac{\sin y-\sin x}{\sin x \sin y}=\frac{\sin(x+\alpha)-\sin(y+\alpha)}{\sin(y+\alpha)\sin(x+\alpha)}$$
or, $$-\frac{2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})}{\sin x \sin y}=\frac{2\cos(\frac{x+y}{2}+\alpha)\sin(\frac{x-y}{2})}{\sin(y+\alpha)\sin(x+\alpha)}$$
or, $$2\sin(\frac{x-y}{2})\left(\frac{\cos(\frac{x+y}{2})}{\sin x \sin y}+\frac{\cos(\frac{x+y}{2}+\alpha)}{\sin(y+\alpha)\sin(x+\alpha)}\right)=0$$
It is very easy to say $x=y$ from here. I hope you can do this last step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to Find the value of a trigonometric function if other very complicated trigonometric equation is given?
Q)If $\operatorname{sin}\alpha+\operatorname{cos}\alpha=\frac{\sqrt7}2$, $0 \lt \alpha \lt\frac{\pi}{6}, then \operatorname{tan}\frac{\alpha}2 $ is:(1)$\sqrt{7}-2$(2)$(\sqrt{7}-2)/3$(3)$-\sqrt{7}+2$(4)$(-\sqrt{7}+2)/3$
i did this question by two ways but both the was were getting so complex however on solving them completely which took $4.25$ pages the answers were ridiculously different.
$I^{st}$ Method:
squaring both sides $$(\operatorname{sin}\alpha+\operatorname{cos}\alpha)^2=\frac{7}4=1+ \operatorname{sin}2\alpha=\frac 74\implies\operatorname{sin}2\alpha=\frac 34\implies\operatorname{cos}2\alpha=\frac{\sqrt{7}}{4}$$
$$\text{hence, }\operatorname{tan}2\alpha=\frac{3}{\sqrt{7}}\implies 3tan^2\alpha+2\sqrt{7}tan \alpha -3=0\implies tan\alpha=\frac{\surd7 \pm 4}3 $$
no from this if i want to calculate the value of $\tan\alpha/2 $ then you can think how different the answer could be!
$II^{nd} $ Method
$$1+2\sin\alpha\cos\alpha=\frac74\implies \sin^2\alpha\cos^2\alpha=\frac{9}{64}$$
$$64\sin^4\alpha-64\sin^2\alpha+9=0\implies \sin^2\alpha=\frac{64\pm 64\sqrt{7}}{2\times 64}$$
even from this you can have the idea how different the answer will be so different because the value of $$\tan\alpha=\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}$$.
| If $t=\tan\fracα2$ then
$$
\cosα = \frac{1-t^2}{1+t^2}\text{ and }\sinα = \frac{2t}{1+t^2}
$$
so that
$$
a=\cosα+\sinα=\frac{1-t^2+2t}{1+t^2}\iff (1+a)t^2 - 2t +a-1=0\\
\iff ((a+1)t-1)^2 -1 + (a^2-1) \\
\iff t=\frac{1\pm\sqrt{2-a^2}}{a+1}=\frac{a-1}{1\mp\sqrt{2-a^2}}
$$
and with $a=\frac{\sqrt7}2$
$$
t=\frac{2a-2}{2\mp\sqrt{8-(2a)^2}}=\frac{\sqrt7-2}{2\pm 1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a summation inequality with induction The exact question:
Prove:
$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)$
I have looked at similar problems but still don't understand how to prove this inequality by induction. So far I have this:
Induction basis:
Let n=1
$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}} = \frac{1}{\sqrt{1}} = 1 > 2(\sqrt{1+1}-1) = ~.828$
$1>.828$
So it proves the inequality true when n=1.
Now i really don't know how to continue even with all the examples i have browsed through. One of them i came across showed that the induction hypothesis should let P(n) equal the equation above and do something with P(n+1). I am not looking for the answer I just need help on how to continue with the problem. What other steps are necessary for me to complete this proof by induction.
| The induction step assumes
$$\sum_{k = 1}^n \frac{1}{\sqrt{k}} > 2\sqrt{n + 1} - 2.$$
Using it, it follows that
\begin{align}
& \sum_{k = 1}^{n + 1} \frac{1}{\sqrt{k}} \\
= & \sum_{k = 1}^n \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{n + 1}} \\
> &2\sqrt{n + 1} - 2 + \frac{1}{\sqrt{n + 1}} \\
= & 2 \sqrt{n + 2} - 2 + \left[2(\sqrt{n + 1} - \sqrt{n + 2}) + \frac{1}{\sqrt{n + 1}}\right] \\
= & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\
> & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\
= & 2 \sqrt{(n + 1) + 1} - 2,
\end{align}
finishes the induction step.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing the basis of a subspace of matrices with the given nullspace. Compute the dimension and find a basis for the subspace of matrices $H ⊂ M_{3×5} (\mathbb{R})$
such that $[0,2,−3,0,1]^T$ is in the nullspace of A ∈ H.
I have said that, if we want $[0,2,−3,0,1]^T$ to be in the nullspace of A, then we must be able to write,
$$c_1\vec{v_1}+...+c_k\vec{v_k} = [0,2,−3,0,1]^T $$
where $S = \{\vec{v_1},...,\vec{v_k}\}$ is a basis for the nullspace of A.
so we require
$$V\mathbb{\vec{c}} = [0,2,−3,0,1]^T $$ to be consistent. where $V$ is a matrix whose columns are the elements of S.
How can I move on from here?
I feel as though this is wrong in the first place. Could someone give me push please.
| $$\{\left( \begin{array}{ccc}
0 & -1 & 1 & 0 & 5\\
1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\end{array} \right),\left( \begin{array}{ccc}
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\end{array} \right),\left( \begin{array}{ccc}
0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 1\\
1 & 0 & 0 & 0 & 0\end{array} \right)
\left( \begin{array}{ccc}
0 & 0 & 0 & 0 & 0\\
0 & -1 & 1 & 0 & 5\\
0 & 0 & 0 & 1 & 0\end{array} \right)\left( \begin{array}{ccc}
1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 1\end{array} \right),\left( \begin{array}{ccc}
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & -1 & 1 &0 & 5\end{array} \right)\}$$
so $Dim(H) =6$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
} |
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.