Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding $\cos ( 2 \sin^{-1}( \frac{5}{ 13} )) $ The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:
$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*}
... | Your mistake is when you squared $\sin2\theta$, it should be $4\sin^{2}\theta\cos^{2}\theta$ not $2\sin^{2}\theta\cos^{2}\theta$, but in case you want an easier method than going for a long method, here is one:
Just let $\displaystyle \theta=\sin^{-1}\left(\frac{5}{13}\right)$ so that we have $\sin \theta=\dfrac{5}{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\frac{\int_{0}^{\infty}(1+x^2)^{-2012}dx}{\int_{0}^{\infty}(1+x^2)^{-2011}dx}$
QUESTION: Evaluate $$\frac{\int_{0}^{\infty}(1+x^2)^{-2012}dx}{\int_{0}^{\infty}(1+x^2)^{-2011}dx}$$
MY ANSWER: I have got $2$ doubts on this; But let me write down my solution first..
First we solve for some general $\int_{0}^{... | Here is my approach which is to use Beta function:
$$I_n: = \int_{0}^{\infty} \frac{1}{\left(x^2+1\right)^n}~dx = \frac{1}{2}\cdot \int_{0}^{\infty} \frac{1}{\sqrt{t}\left(t+1\right)^n}~dt=\frac{1}{2}\cdot\mathrm{B} \left(\frac{1}{2}, n - \frac{1}{2}\right)$$$$= \frac{1}{2}\cdot\frac{\Gamma (1/2)\cdot \Gamma (n-1/2)}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$ For $x,y,z \geqslant 0.$ Proving$:$
$${\dfrac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\dfrac {{x}^{2}+yz}{6{y}^{2}
+6yz+6{z}^{2}}}}\quad \quad(\te... |
By Maple we have $$\sqrt{\frac{6(a^2+bc)}{b^2+bc+c^2}}\ge \frac{6a^2+4bc+ab+ac}{a^2+b^2+c^2+ab+bc+ca}$$
Note that $$\sum_{cyc} \frac{6a^2+4bc+ab+ac}{a^2+b^2+c^2+ab+bc+ca}=6$$
then we are done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Limit of $(x+1+\sqrt{(x+1)^2 +1})$ as $x\to-\infty$ This seemed easy at first, but then I confronted my result with the one I got from WolframAlpha and they are different:
$$
\lim_{x\to-\infty}(x+1+\sqrt{(x+1)^2 +1})
$$
WolframAlpha says the limit is equal to 0, though for me it seems to be negative infinity. I tried s... | $$
\lim_{x\to-\infty}\left(x+1+\sqrt{(x+1)^2 +1}\right)
$$
$$=\lim_{x\to-\infty}\left(x+1+|x+1|\sqrt{1+\dfrac1{(x+1)^2}}\right)$$
$$=\lim_{x\to-\infty}\left(x+1+|x+1|\left[1+\frac1{2(x+1)^2}\cdots\right]\right)$$
$$=\lim_{x\to-\infty}\left(x+1-(x+1)-(x+1)\left[\frac1{2(x+1)^2}\cdots\right]\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$ prove that
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$
My try: by given condition $a+b+c=ab+bc+ca$
we have $a+b+c\le a^2+b^2+c^2$
also using titu's lemma
$$\sum_{cyc}\frac{a^2}{b} +\fra... | We can use the following estimation.
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}.$$
The proof you can see here: https://artofproblemsolving.com/community/c6h296853
For hardest case $n=3$ we need to prove that:
$$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9}{a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving that no positive integer solutions exist for $x^2 = y^{119}+1$ The $119$ looks very frightening so I factored $119=7\times 17$. And I think it is enough to prove the result for $x^2 = y^7+1$.
I then tried to move the $1$ around, and factoring. On one hand, $y^7 = (x+1)(x-1)$, and $\gcd(x+1,x-1) = 1 \text{ or } ... | You have already found that $\gcd(x-1,x+1)=2$, from which it follows that
$$x-1=2^6a^7\qquad\text{ and }\qquad x+1=2b^7,$$
for some coprime $a,b\in\Bbb{Z}$ with $b$ odd, after changing the sign of $x$ if necessary. Then
$$b^{14}-(2a)^7=\left(\frac{x+1}{2}\right)^2-2(x-1)=\left(\frac{x-3}{2}\right)^2.\tag{1}$$
You have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the inverse function of $y=-\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$? How can the continuous random variable $x$ by isolated by itself on one side of the following equation
$$y = -\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$$
without resorting to a piece-wise equation?
$$ x = ?$$
Below is my init... | There are 2 cases:
*
*$x \geq 0$:
\begin{align*}
y&=-\frac{1}{2}ln(1-x^2)\\
-2y &= ln(1-x^2)\\
e^{-2y} &= 1-x^2\\
x^2 &= 1-e^{-2y}\\
x &= +\sqrt{1-e^{-2y}} \qquad :\text{since } x \geq 0
\end{align*}
*$x < 0$:
\begin{align*}
y&=\frac{1}{2}ln(1-x^2)\\
2y &= ln(1-x^2)\\
e^{2y} &= 1-x^2\\
x^2 &= 1-e^{2y}\\
x &= -\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Stuck on proof of $\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$ for $n \geq 3$ using induction I'm relatively familiar with induction, I'm just stuck on this step. I am currently taking Introduction to Abstract Math, and have taken Calculus I and II.
$P(n)$ is
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7... | $$ \sum\limits_{k=1}^{n+1}\frac 1 {k^{2}}$$ $$ =\sum\limits_{k=1}^{n}\frac 1 {k^{2}}+\frac 1 {(n+1)^{2}}$$ $$ \leq \frac 7 4 -\frac 1 n +\frac 1 {(n+1)^{2}}$$ $$ \leq \frac 7 4 -\frac 1 n +\frac 1 {n(n+1)}$$ $$ =\frac 7 4 -\frac 1 n +\left[\frac 1 n-\frac 1 {(n+1)}\right]$$ $$ \leq \frac 7 4 -\frac 1 {n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac... | The first mistake comes in your substitution of
$$t=y^{1-2}=\frac{1}{y}\implies t'= -\frac{1}{y^2}y'.$$
The next step after
$$\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)$$
should be
$$t'-\frac{2}{x^2-1}t=1-x.$$
Then the integrating factor is
$$\mu(x)=\exp\left(\int-\frac{2}{x^2-1}\,dx\right)=\frac{1+x}{1-x},$$
so the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$ I just came across the following question:
Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$
I solved it as follows:
$x=y=0$ obvious solution. If $xy\neq0$, let $d=gcd(x, y)$ and we write $x=da$, $y=db$, $a, b\in... | First of all,if $x=y$ then $x=y=0$ which does work. So,now assume $x \not =y$. Again if one of them is $0$, other one has to be also. So,from now on also assume that none of them is $0$
$\textbf{Case 1:}$ $x,y$ both are positive.
Then $(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)=16xy$
since obviously $x >y$,if $x \ge 3$ we have $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Solving the system $b+c+d=4$, $ad+bc=-8$, $a+b=5$, $cd=-8$ There is a given system of equations
\begin{align} ab+c+d&=\phantom{-}4 \\ ad+bc&=-8 \\ a+b&=\phantom{-}5 \\ cd&=-8 \end{align}
I have tried to simplify it by multiplying one equation by another or adding one to another or similar. Nothing came up.
Any hints wo... |
\begin{align} ab+c+d&=-4 \tag{1}\label{1} ,\\ ad+bc&=-8 \tag{2}\label{2},\\ a+b&=-5 \tag{3}\label{3},\\ cd&=-8 \tag{4}\label{4}.\end{align}
Substitution of
\begin{align}
a&=5-b
\tag{5}\label{5}
,\\
d&=-\frac8c
\tag{6}\label{6}
\end{align}
into \eqref{2} results in
\begin{align}
b&=\frac{40-8c}{c^2+8}
\tag{7}\label{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determinant of a certain Toeplitz matrix
Compute the following determinant
$$\begin{vmatrix} x & 1 & 2 & 3 & \cdots & n-1 & n\\ 1 & x & 1 & 2 & \cdots & n-2 & n-1\\ 2 & 1 & x & 1 & \cdots & n-3 & n-2\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ n-1 & n-2 & n-3 &... | Here are the first few determinants with the help of WA:
$$
\begin{array}{rl}
n & \text{determinant}
\\ 1 & x^2 - 1
\\ 2 & x^3 - 6 x + 4
\\ 3 & x^4 - 20 x^2 + 32 x - 12
\\ 4 & x^5 - 50 x^3 + 140 x^2 - 120 x + 32
\\ 5 & x^6 - 105 x^4 + 448 x^3 - 648 x^2 + 384 x - 80
\\ 6 & x^7 - 196 x^5 + 1176 x^4 - 2520 x^3 + 2464 x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Squares in the Complex plane In the complex plane, $z,$ $z^2,$ $z^3$ form, in some order, three of the vertices of a non-degenerate square. What are all the possible values of the area of the square?
I tried doing $z=a+bi$ so $z^2=a^2-b^2+2ab i$ $\sqrt{(a^2-b^2-a)^2+(2ab-b)^2}=\sqrt{(a^3-3ab^2-a)^2+(3a^2-b^3-b)^2}$ but... | You have three choices, depending which pair of numbers is on the diagonal. The only two things I will use is that the lengths of sides is the same, and that the length of the diagonal is $\sqrt2$ times the length of the sides. Note that $z$ can't be any real number, since then $z$, $z^2$, and $z^3$ would be on the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In an acute angled triangle with angles $A,B$ and $C$, prove that $\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac6\pi$ In an Acute Angled Triangle Prove That : -
$$\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac{6}{\pi}$$
A , B , C represent the angles of the tri... | We can effectively use the graph of this function for this problem.
The graph of $\frac{\sin x}{x}$ for $x$ belongs to $[0,\pi]$
Assume the three points $(A,\frac{\sin A}{A}),(B,\frac{\sin B}{B}),(C,\frac{\sin C}{C})$ on the graph. The centroid of this triangle is:$$\left(\frac{\left(A+B+C\right)}{3},\frac{\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the unknowns from LCM HCF relation. Consider $2$ numbers $x,y$ such that $\frac{x+y}{lcm(x,y)}=\frac{7}{12}$, and it is given that hcf(x,y) is $4$.
How to find $\mathbf{x,y}$.
I have tried the question like this.
\begin{align*}
\frac{x + y}{(x\cdot y)/4} \Rightarrow \frac{4(x+y)}{x\cdot y} &= \frac{7}{12}\\
48x... | Cancelling $\,4\,$ from our gcd $\,(x,y)=4\,\Rightarrow\, \color{#c00}{(a,b)=1}$ thus $\,\color{#0a0}{(a\!+\!b,\,ab) = 1}\,$ by here.
So $\ \dfrac{a\!+\!b}{\color{#90f}{ab}}\! =\! \dfrac{7}{\color{#90f}{12}}\,$ are $\rm\color{#0a0}{reduced}$, thus $\,\color{#90f}{ab = 12},\,$ so $\,a,b\,$ are $\rm\color{#c00}{coprim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\fra... | By AM-GM $$x^z+y^z-(xy)^{\frac{z}{4}}\geq2\sqrt{(xy)^z}-(xy)^{\frac{z}{4}}=2\left((xy)^{\frac{z}{4}}-\frac{1}{4}\right)^2-\frac{1}{8}\geq-\frac{1}{8}.$$
The equality occurs for $x^z=y^z=\frac{1}{16},$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I find the maclaurin representation for $f(x)=\frac{\ln(2x+1)}{e^x-1}$? $f(x)=\frac{\ln(2x+1)}{e^x-1}$
I'm supposed to find the Maclaurin representation for $f(x)$ and then write down the first 4 terms. The question says I should use known the Maclaurin representation of elemental functions. Thus, I've tried usi... | This is a problem of composition of series
$$\log(2x+1)=2 x-2 x^2+\frac{8 x^3}{3}-4 x^4+\frac{32 x^5}{5}-\frac{32 x^6}{3}+\frac{128
x^7}{7}-32 x^8+O\left(x^9\right)$$
$$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{
x^7}{5040}+\frac{x^8}{40320}+O\left(x^9\right)$$ Now l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Roots and point of inflections Let $b$ and $c$ be the roots of a four degree polynomial. Also $x=b$ and $x=c$ are the real points of inflection of this four degree polynomial. If the other two roots of the polynomial be $a$ and $d$ where $a<b<c<d$ then prove that $$\int_{a}^{d}f(x)dx=0$$
My Attempt
I framed the equatio... | Shift the coordinate system so that the new origin is at $\left(\frac{b+c}{2},0\right)$; this obviously does not change the integral. We can then rephrase the problem as: suppose $\pm B = \pm\frac{b+c}{2}$ are two roots of a quartic and that these are also the $x$-coordinates of the points of inflection. If $A, D$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is this sum $\sum_{i=0}^{n-1}\sum_{j=0}^{i}x^i$? How to find this sum ?
$$\sum_{i=0}^{n-1}\sum_{j=0}^{i}x^i$$
without knowing this sum:
$$\sum_{i=1}^{n}ix^{i-1} = \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
| You have $$\sum_{i=1}^{n}ix^{i+1}=\sum_{i=0}^{n-1}(i+1)x^i$$
$$=\sum_{i=1}^{n-1} \underbrace{(1+1+...+1)}_{i+1\text{ times}}x^{i}$$
$$=\sum_{i=1}^{n-1}\sum_{j=0}^{i}x^i.$$
It is a geometric series so we have $$\sum_{i=0}^{n}x^{i}=\frac{1-x^{n+1}}{1-x}$$
Thus $\sum_{i=1}^{n}ix^{i-1}=\frac{d}{dx}\big(\frac{x^{n+1}-1}{x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +... | A concise Proof based on an earlier proof:
The given equation $(xz+1)(yz+1)=az^3+1$ can be rewritten as $az^2-xyz-(x+y)=0$. We shall show that for any solution $(x,y,z)$, we have $z \le a^2+2a. \ $
Note that $z \ | \ x+y$, therefore $z \le x+y. \ $ Treating $x, y$ as constants, the only positive solution for $z \ $ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find all functions $f:\mathbb N_0\to \mathbb N_0$ such that $f(a^2+b^2)=f(a)^2+f(b)^2$ I think the answer is that there are only 2 such functions: the zero function and the identity function, but I'm not able to prove it.
A few findings:
*
*$f(0)=0$ and thus $f(a^2)=f(a)^2$.
*If $f(1)=0$, then $f(2^n)=0$ for all $n... | With $u=f(1)$, let
$$ S=\{\,n\in\Bbb N_0\mid f(n)=nu\,\}.$$
Clearly,
*
*$0\in S$,
*$1\in S$,
*If two of $a,b,a^2+b^2$ are $\in S$, then so is the third
*If one of $a,2a^2$ is $\in S$, then so is the other
*If $a^2+b^2=c^2+d^2$ and three of $a,b,c,d$ are $\in S$, then so is the fourth
From these, we conclude by i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Let a, b, c be ints. $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an int, show that each of $\frac{ab}{c}, \frac{bc}{a}, \frac{ac}{b}$ is an int. Let a, b, c $\in \mathbb{Z}$ . If $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an integer, prove that each of $\frac{ab}{c}, \ \frac{bc}{a}, \ \frac{ac}{b}$ is an integ... | Let $x=\frac{bc}{a}, y=\frac{ca}{b}, z=\frac{ab}{c}$, then $x, y, z \in \Bbb{Q}$ and by condition $x+y+z=\alpha \in \Bbb{Z}$. It is easy to verify that $yz+zx+xy=a^2+b^2+c^2=\beta \in \Bbb{Z}$, $xyz=abc=\gamma \in \Bbb{Z}$. So $x, y, z$ are the rational roots of the monic polynomial $t^3-\alpha t^2+\beta t-\gamma=0$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
When are the curves $a =xy$ and $b = (1-x)(1-y)$ tangential? I've seen that the curves $xy = 1/4$ and $(1-x)(1-y) =1/4$ are tangential at the point $(x,y) = (1/2,1/2).$
Similarly I can see that the curves $xy = 1/16$ and $(1-x)(1-y) = 9/16$ are tangential at $(x,y) = (1/4,1/4).$
I therefore have the general conjecture ... | Your conjecture works if there is a specific relationship between values of $a$ and $b$.
First curve is $y = \frac{a}{x}$
Second curve is $y = 1 - \frac{b}{1-x}$
For your general conjecture to be true we should have only one point where both curves meet,
So, $\frac{a}{x} = 1 - \frac{b}{1-x}$ should have exactly one sol... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that for any $n$, $\sum_{k=0}^\infty \frac{1}{3^k}\binom{n+k \log(3)/\log(2)}{k}=c 2^n$ for some constant $c$ I stumbled upon this from playing around and it seems to be true but I don't know how to approach it.
| Let $0 < b < 2$, $0 < a < 2^{-b}$, and $t > 0$. We will evaluate the sum
$$\sum_{k=0}^\infty a^k \binom{bk + t}{k}$$
and will use the result to prove @Bartek's conjecture.
Recall that for any $\alpha \in \mathbb{C}$, the function $z \mapsto z^\alpha := e^{\alpha \log z}$, where $\log$ denotes the principal branch of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the radius of the circle in complex plane given by $|z - i| = 3|z +2i|$ where $z$ is a point on the circle. I tried to arrange the following into the standard equation of the circle i.e $|z - z'| = r$ where $r$ is the radius, $z'$ is the centre and $z$ is a point on the circle.
I rearranged the following equation ... | $|z - i| = 3|z +2i|$
$x^2+(y-1)^2=9(x^2+(y+2)^2)$
$x^2+y^2-2y+1=9x^2+9y^2+36y+36$
$8x^2+8y^2+38y+35=0$
$x^2+y^2+\frac{19}{4}y+\frac {35}{8}=0$
$x^2+(y+\frac{19}{8})^2=\frac{81}{64}$
Circle is centered at $(0;-\frac{19}{8})$ with radius $r=\frac{9}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many words of length $n ≥ 4$ in the alphabet ${1, 2, 3}$ contain at least two $1$’s, at least two $2$’s, and at least two $3$’s? How many words of length $n ≥ 4$ in the alphabet ${1, 2, 3}$ contain at least two $1$’s, at least two $2$’s, and at least two $3$’s?
My approach involves the incluson-exclusion principle.... | Inclusion-exclusion yields:
\begin{align}
&\quad\binom{3}{0}3^n\\
&-\binom{3}{1}\left(\binom{n}{0}2^{n-0}+\binom{n}{1}2^{n-1}\right)\\
&+\binom{3}{2}\left(\binom{n}{0}1^{n-0}+2\binom{n}{1}1^{n-1}+2\binom{n}{2}1^{n-2}\right)\\
&-\binom{3}{3}\left(\binom{3}{0}[n=0]+\binom{3}{1}[n=1]+2\binom{3}{2}[n=2]+3!\binom{3}{3}[n=3]... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the degree of finite field extension
Compute the degree $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}]$.
By the Degree Theorem we have $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}(\sqrt{3 + \sqrt{3}})] \cdot [\mathbb{Q}(\sqrt{3 ... | Here's a way to do it.
First of all, consider both $[\mathbb{Q}(\sqrt{3+\sqrt{3}} ): \mathbb{Q}]$ and $[\mathbb{Q}(\sqrt{3} ): \mathbb{Q}]$. The first is $4$ as you have stated and the second is $2$, so this gives you the hint that $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}]$ is a multiple of $4$.
I want to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explanation in a step of the summation $\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}}$ I have the solution to the above equation but I wanted to know the explanation for it.
$$\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}} = 2^n \frac{1}{ 2^{n+1}}$$
and
$$2^n\frac{1}{2^{n+1}}=\frac{1}{2} \qq... | We have an inequality since $$\frac{1}{2^n+i}\geq \frac{1}{2^{n+1}}$$
for $1\leq i\leq 2^{n}.$ Thus $$\sum_{i=1}^{2^{n}}\frac{1}{2^n+i}\geq\sum_{i=1}^{2^n}\frac{1}{2^{n+1}}=\frac{1}{2^{n+1}}\sum_{i=1}^{2^n}1=2^n\frac{1}{2^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$... | $\ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C= \ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|{x^2+2x+3}\right|-\ln{2}}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C=\frac{1}{2}\left(\ln{\left|{x^2+2x+3}\right|}- \sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion.... | This is what I tried after reading your answers, please correct me if there are mistakes!$$cos^3+sin^3=1$$ $$cos^3+sin^3=cos² +sin²$$ $$cos^3-cos²+sin^3-sin²=0$$ $$cos²(cos-1) + sin²(sin-1)=0$$ $$(1-sin²)(cos-1)+(1-cos²)(sin-1)=0$$ $$ (1+sin)(1-sin)(cos-1)+(1+cos)(1-cos)(sin-1)=0 $$ $$(1+sin)(1-sin)(cos-1)-(cos-1)(1-co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $1+n+n^2+n^3$ is a perfect square, then $n=1$ or $n=7$ I want to prove that if $1+n+n^2+n^3$ is a perfect square then $n=1$ or $n=7$.
I managed to prove that $1+n+n^2+n^3=(n^2+1)(n+1)$ and that $(n^2+1,n+1)$ is either $1$ or $2$.
I found out that it could not be $1$, and then $\frac{1}{2}(n^2+1,n+1)=1$.
From here I ... | Note that the LHS is a 4-term GP with $a = 1, r = n$. The sum is
$${1(1 - n^4) \over (1 - n)} = {n^4 - 1 \over n-1}$$
So, the equation becomes,
$${n^4 - 1 \over n-1} = y^2$$
This is the Nagell–Ljunggren Equation whose general form is:
$${x^n - 1 \over x - 1} = y^q$$
where, $x > 1, y > 1, n > 2, q ≥ 2$.
Known solutions ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6} \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3)... | Let $x-3=y$, then
$$L=\lim_{y\to 0} \frac{\tan\frac{y}{y+6} (\sin (9\sin y)}{\sin y \sin[(y+3)^3-27]}=\lim_{y\to 0}\frac{\sin(9\sin y)}{\sin y} \frac{\tan\frac{y}{y+6}}{\sin(y(y^2+9y+27)}=\lim_{y \to 0}\frac{\sin(9\sin y)}{y} \lim_{y \to 0} \frac{\frac{1}{y}\tan\frac{y}{y+6}}{(y^2+9y+27)}$$
As $\tan z \to z, \sin z \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Exponential recurrence Let $f_0 = a, f_1 = b$.
Define $f_n = f_{n-1}$ for odd $n$ and
$$f_{n} = f_{n/2}+f_{n/2 - 1}$$
How do I solve this recurrence? Normally I would use generating functions but the fraction is screwing me up.
| (I will assume that "$f_n=f_{n-1}$ for odd $n$" includes $n=1$, so that $f_1=f_0$; if that's not the case, a similar method will apply but it gets more annoying.)
Define $F(x) = \sum_{n=0}^\infty f(n)x^n$ and $F_2(x) = \sum_{n\text{ even}} f(n)x^n$. Since $f_{n+1}=f_n$ when $n$ is even, we have $F(x) = F_2(x) + xF_2(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $2\cot4x = \cot2x - \tan2x$ How to prove the trigonometry equation is an identity?
$$2\cot4x = \cot2x - \tan2x$$
Thank you in advance.
Thank you for the comments and hints. I got an answer after many tries. ;)
Below is my answer. Thank you.
$2cot4x = cot2x - tan2x$
$2\frac{1}{tan 4x} = cot2x - tan2x$
$2\f... | Thank you for the comments and hints. I got an answer after many tries. ;)
Below is my answer.
$$\begin{align}
2\cot4x = \cot2x - \tan2x \\
2\cdot\frac{1}{\tan 4x} = \cot2x - \tan2x\\
2\cdot\frac{1}{\frac{2 \tan 2x}{1 - \tan^2 2x}} = \cot2x - \tan2x\\
2\cdot\frac{1 - \tan^2 2x}{2 \tan 2x} = \cot2x - \tan2x\\
\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5... | Just another way.
Consider
$$y=\frac{x^4 + px^3 + 5x^2 + 5x + q } {(x+1)(x-3) }$$ and perform the long division. You will have
$$y=-\frac{q}{3}+\left(\frac{2 q}{9}-\frac{5}{3}\right) x-\left(\frac{7
q}{27}+\frac{5}{9}\right) x^2+\frac{1}{81} x^3 (-27 p+20 q-15)+\frac{1}{243}
x^4 (54 p-61 q-96)+\cdots$$
So,
$$-27 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Two urns with balls
We consider two urns that contain blue balls and red balls. Urn $1$ has $9$ blue balls and $1$ red one. Urn $2$ has $8$ blue balls and $2$ red ones. We now draw one ball from each urn.
(1) Find the probability of at least one ball being red.
(2) Find the probability of exactly one ball being red.
(... | *
*$$\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}+\frac {1}{10}\times\frac{2}{10}=1-\frac {9}{10}\times\frac{8}{10}$$
*$$\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}$$
*$$P( U_1| R)=\frac {P(U_1\cap R)} {P(R)}= \frac{\frac {1}{10}\times\frac{8}{10}}{\frac {1}{10}\times\frac{8}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| By AM-GM $$b+c=\sqrt{(b+c)^2}=\sqrt{b^2+2bc+c^2}=\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{2}{3}bc}\leq$$
$$\leq\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{b^2+c^2}{3}}=\sqrt{\frac{4}{3}(b^2+bc+c^2)}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve the diffrential equation $\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0$ A solution curve of the differential equation $\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0$, $x>0$ passes through the point (1,3). Find the solution curve.
I am not able to proceed as I am ... | $$\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0,$$
$$\left( {{{\left( {x + 2} \right)}^2} + y\left( {x + 2} \right)} \right)dy-y^2 dx=0$$
After substituting $x+2$ with $t$, we get
$$ (t^2+yt)dy-y^2dt=0$$
After multiplying with $t^{-2}$, we get
$$(1+\frac{y}{t})\frac{dy}{dt}-(\frac{y}{t})^2=0$$
... | {
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"url": "https://math.stackexchange.com/questions/3878056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $12(ab+ba+ac) <7a^2+15b^2+18c^2$ holds for all positive numbers. Prove that $12(ab+ba+ac) < 7a^2+15b^2+18c^2$ holds for all positive numbers.
I tried completing the square, but that solution would suggest that inequality holds for all real numbers. Inequalities between means did not work for me either.
$$12(... | The completing to square works:
We need to prove that:
$$18c^2-12(a+b)c+7a^2+15b^2-12ab>0$$ or
$$9c^2-6(a+b)c+\frac{7}{2}a^2+\frac{15}{2}b^2-6ab>0$$ or
$$(3c-a-b)^2+\frac{5}{2}a^2+\frac{13}{2}b^2-8ab>0,$$
which is true by AM-GM:
$$\frac{5}{2}a^2+\frac{13}{2}b^2-8ab\geq2\sqrt{\frac{5}{2}a^2\cdot\frac{13}{2}b^2}-8ab=\lef... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$
if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find maximum value of $Q$ and minimum value of $P$
This is actually a question made up myself ,so i don,t know whether a nice solution exists!?.
Finding P seems easier:AM-GM results $$... | Let $x=y=z=\frac 13$ then $Q \leqslant \frac{4913}{64}.$ We will show that it's a maximum value, or
$$(5x+5y+7z)(5z+5x+7y)(5y+5z+7x) \geqslant \frac{4913}{64}(2x+y+z)(2y+z+x)(2z+x+y).$$
Let
$$\left\{\begin{aligned} & a = 2x+y+z\\& b = 2y+z+x \\& c = 2z+x+y\end{aligned}\right. \Rightarrow \left\{\begin{aligned} & x = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$ $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
Can we apply Stolz-Cesaro?
$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{... | The limit is $+\infty$.
You made an error in the first step. Applying Stolz-Cesaro, you should rather consider:
\begin{eqnarray*} \frac{\frac 1{\sqrt{n+1}}}{\ln{(n+1)}-\ln n}
& = & \frac 1{\sqrt{(n+1)}}\cdot \frac 1{\ln \left(1+\frac 1 n\right)}\\
& = & \frac n{\sqrt{(n+1)}}\cdot \underbrace{\frac{\frac 1n}{\ln \left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885831",
"timestamp": "2023-03-29T00:00:00",
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Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$ Can this question have solutions for N>3:
Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb N... | We have:
$a + b = u^2$, $c + d = v^2$
$a + c = w^2, b + d = x^2$
$a + d = y^2, b + c = z^2$
$a + b + c + d = k^2$
Let:
$u=m^2-n^2$ & $v =2mn$
$w=p^2-q^2$ & $x=2pq$
$y=r^2-s^2$ & $z=2rs$
We apply the condition:
$m^2+n^2=p^2+q^2=r^2+s^2=k^2$
Hence, $u^2+v^2=w^2+x^2=y^2+z^2=k^2$
Since:
$2a=u^2+w^2-z^2$
$2b=x^2+z^2-v^2$
$2... | {
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"url": "https://math.stackexchange.com/questions/3887416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)... | The two points i and -i are distance 2 apart.
The origin is distance one from one of them and one from the other so it satisfies the equation.
All the points on the line between those two points satisfy the equation.
Any other point got on the line between these two has the sum of the distances to those two points bein... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Double integral with a change of variables where $u = \frac{y}{x^3}$ and $v = \frac{y}{x}$. Sketch the planar region D which is bounded by the following four curves and located in the positive quadrant: $y = x^3$, $y = 10x^3$, $y = x$ and $y = 2x$.
Use transformation given by $u=\frac{y}{x^3}$ and $v=\frac{y}{x}$.
Eval... | Solving wrt to $u,v$ we get
$$\left\{x=\sqrt{\frac{v}{u}},y=v \sqrt{\frac{v}{u}}\right\}$$
Jacobian is
$$J=\left(
\begin{array}{cc}
-\frac{\sqrt{\frac{v}{u}}}{2 u} & -\frac{1}{2} \left(\frac{v}{u}\right)^{3/2} \\
\frac{\sqrt{\frac{v}{u}}}{2 v} & \frac{3 \sqrt{\frac{v}{u}}}{2} \\
\end{array}
\right)$$
and $\det J=-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3891298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding convergence of series using Ratio Test I have this series but Im not too sure whether my calculations are correct
$$ \sum_{n=1}^\infty \left( \frac{(3n)!}{(n!)^3}\right) $$
$$ \left( \frac{(3(n+1)!}{((n+1)!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$
$$ \left( \frac{(3(n+1)(3n)!}{((n+1)^3(n!)^3}\righ... | There is a mistake in your first step, indeed by ratio test we obtain
$$ \frac{(3n+3)!}{((n+1)!)^3} \frac{(n!)^3}{(3n)!}=\frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} $$
since $(3(n+1))!=(3n+3)!$.
To avoid ratio test we can use that
*
*$(3n)! =3n\cdot(3n-1)\cdot \ldots \cdot (n+1) \cdot n!\ge (n+1)^{2n} \cdot n! \ge n^{2n} \cd... | {
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help integrating I'm really struggling to integrate this.
$$\int_0^t \sqrt{\cos^2\left(1\right)\sinh^2\left(u\right)+4\sin^2\left(\dfrac{1}{2}\right)\cosh^2\left(\dfrac{u}{2}\right)} \,\mathrm{d}u$$
This is to find the arc length of a plane curve. An online integral calculator gives me such a complicated result with $\... | Hint:
Set $a={{\cos }^{2}}\left( 1 \right),b={{\sin }^{2}}\left( \frac{1}{2} \right)$
and using $\sinh \left( u \right)=2\sinh \left( \frac{u}{2} \right)\cosh \left( \frac{u}{2} \right)$ the integral becomes:
$$\begin{align}
& =\int{\sqrt{4a{{\sinh }^{2}}\left( \frac{u}{2} \right){{\cosh }^{2}}\left( \frac{u}{2} \r... | {
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Sketching a set of $Re(z^2(1+i))$ on the Complex plane. So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.
I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies Re(z^2) = a^2 - b^2$$
Similarly:
$$Re(z^2(1+i)) \im... | Some facts:
We require the real part to be less than zero, which means $(1 + i)z^2$ must be located on the left half of the complex plane. In terms of the argument, this means $(1+i) z^2$ must have an argument between $\pi/2$ and $3\pi/2$, plus some integer multiple of $2\pi$.
Squaring a complex number doubles its arg... | {
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Find $\sqrt{a}+\sqrt{b}+\sqrt{c}$ only in terms of $p$. If $a^{2} x^{3}+b^{2} y^{3}+c^{2} z^{3}=p^{5}, a x^{2}=b y^{2}=c z^{2}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{p},$ find $\sqrt{a}+\sqrt{b}+\sqrt{c}$
only in terms of $p$.
I tried simple factorization like shifting terms here and there but didnt get any... | We are given
$$a^2x^3+b^2y^3+c^2z^3=p^5\quad (1)$$
$$ax^2=by^2=cz^2=t\ (\mathrm{let})\quad (2)$$
$$\frac 1 x +\frac 1 y +\frac 1 z =\frac 1 p\quad (3)$$
Squaring $(2)$, we get
$$a^2x^4= b^2y^4= c^2=z^4=t^2$$
$$a^2x^3=\frac t x ; b^2y^3=\frac t y ; c^2z^3=\frac t z$$
Substituting these values in $(1)$, we get
$$\frac t ... | {
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Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
... | A value of $x$ satisfying the equation
$$
x=-\sqrt{x+1}
$$
must be nonpositive since $\sqrt{x+1}$ is nonnegative.
Squaring both sides is legal but of the two solutions you found, one of them is positive so doesn't satisfy the given equation, hence should not excluded from the domain.
Thus the domain is $[-1,r)\cup(r,... | {
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Solving $2^{16^x} = 16^{2^x}$
Find $x$ if $2^{16^x} = 16^{2^x}$.
If $x = 0$ we have $2=16$.
$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.
$256$ or $2^8$ difference. I can't see anything useful in figuring this out.
I have a feeling $x$ must be a fraction. I know only know about logarithm. I have... | If you take the $\log_2$ of both sides you end up with
\begin{align*}
\log_2(2^{16^x}) &= 16^x \log_2(2) = 16^x = 2^x\cdot8^x\\
\log_2(16^{2^x}) &= 2^x\log_2{16} = 2^x\cdot4
\end{align*}
From this you get that $2^x\cdot 4 = 2^x \cdot 8^x \implies 8^x = 4 \implies x =\log_{8}(4) = 2/3.$
($\log_8(4) = \log_8(2^2) = 2\log... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$
To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgus... | Factorize as follows
\begin{align}x^{11} + x + 1
&= \frac{ x^{11}(x-1) + (x+1)(x-1)}{x-1}\\
&= \frac{x^{12}-x^{11}+x^2-1}{ x-1}
= \frac{(x^{12}-1 )-x^2(x^{9}-1)}{ x-1}\\
&= \frac{(x^{3}-1)(x^9+x^6+x^3+1)-x^2(x^{3}-1)(x^6+x^3+1)}{ x-1}\\
&= \frac{x^{3}-1}{x-1}(x^9+x^6+x^3+1-x^{8}-x^5-x^2)\\
&= (x^2+x+1)(x^9-x^8+x^6-x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove or disprove $\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\left(\frac{k}{p}\right)\equiv 0\pmod{ p^2}$ let $p=8k+1$ prime,prove or disprove
$$\sum_{k=1}^{p-1}\dfrac{(-1)^k}{k}\left(\dfrac{k}{p}\right)\equiv 0\pmod {p^2}\tag{1}$$
where $\left(\dfrac{k}{p}\right)$ is Legendre symbol
It seem this Notes on Wolstenholme’s Theorem... | This is what I thought of. It is not finished but it is a start.
Observe that $$\frac{1}{p}\binom{p}{k}=\frac{1}{p}\cdot\frac{p!}{k!\cdot(p-k)!}\equiv\frac{(p-1)!}{k!\cdot-k\cdot...\cdot-(p-1)}\equiv\frac{(p-1)!}{(p-1)!\cdot k}\cdot(-1)^{p-k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}$$
So $$\frac{1}{p}\binom{p}{k}\equiv\frac{(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function I am trying to derive the Hardy-Ramanujan asymptotic formula
$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$
from Radmacher's formula for the partition function $p(n)$ given by
$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}... | Using the leading term of Radmacher's formula, we have
\begin{align*}
p(n) & \sim \frac{1}{{\pi \sqrt 2 }}\left[ {\frac{\mathrm{d}}{{\mathrm{d}x}}\frac{{\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {x - \frac{1}{{24}}} \right)} } \right)}}{{\sqrt {x - \frac{1}{{24}}} }}} \right]_{x = n}
\\ &
= \frac{{4\sqrt 3 }}{{24n ... | {
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"source": "stackexchange",
"question_score": "5",
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Values of $a$ such that $x^5-x-a$ has quadratic factor I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.
My Attempt
Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:
$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\e... | This is essentially an elliptic curve.
There might be elementary methods, but there are also computer algebra systems that can (in many cases) solve this kind of diophantine equations.
We may rewrite the equation as: $m^2n^2 = 5n^6 + 4n^2$.
If we write $y = 5mn$ and $x = 5n^2$, then it becomes $y^2 = x^3 + 20x$.
Now we... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Number of partitions of an integer with extra conditions In how many ways can you write number 15 as a sum of positive integers if number 1 can appear 3 times at most in the sum and number 3 can only appear even number of times?
I've tried doing this as $$(1+t+t^2+t^3)(1+t^2+t^4+t^6+t^8+t^{10}+t^{12}+t^{14})(1+t^6+t^{1... | Via inclusion-exclusion, where $p_n$ is the number of unrestricted partitions of $n$:
\begin{align}
& p_{15}
- \left[p_{15-4\cdot1} + (p_{15-1\cdot3} - p_{15-2\cdot3}) + (p_{15-3\cdot3} - p_{15-4\cdot3}) + p_{15-5\cdot3} \right]
+ \left[(p_{15-4\cdot1-1\cdot3} - p_{15-4\cdot1-2\cdot3}) + p_{15-4\cdot1-3\cdot3} \right]... | {
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"source": "stackexchange",
"question_score": "1",
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Prove by induction $n!<4n^2+3$ How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?
Here´s my try:
The equation is valid for every $n\leq4$.
n=5: n!=5!=120>103
n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$
Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.
Any ... | Your work is mostly fine, but you can perfect your redacting.
What you want to show is that the set $A=\{n\in\mathbb N\mid n!<4n^2+3\}=\{0,1,2,3,4\}$
For that you need to verify the inequality for the a few numbers which you certainly did, but it needs to appear on your solution.
$\begin{cases}
n=0&0!=1<4\times 0^2+3=3... | {
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Evaluating limits using a series I'm trying to use a Taylor series centered at $0$ to evaluate this limit:
$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$
I rewrote the function as its Maclaurin series:
$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$
In expanded form:
$$\sum_{k=1}^\infty... | $$e^{-\dfrac2{x^3}}=1-\dfrac2{x^3}+\dfrac{\left(-\dfrac2{x^3}\right)^3}{2!}+\cdots=1-\dfrac2{x^3}+O\left(\dfrac1{x^6}\right)$$
Alternatively, Set $-\dfrac2{x^3}=h$ to find
$$=-4\cdot2\lim_{h\to0^-}\dfrac{e^h-1}h$$
| {
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"question_score": "3",
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Balkan MO :Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$. Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.
This problem was posted 2 times in MSE, but was never solved. Can someone help me?
I am supposed to use orders and mods
| Another approach:
Due to Euler's criterion:
$A=3^{\frac{7-1}2=3}+1 \equiv 0 \ mod (7)$
$3\times 3^{q-1}+1\equiv 0 \ mod (7)$
These two relations give $p=q=3$
In this case $A=28$.
We check this with RHS, i.e $B=11^p+17^q$:
$11\equiv 4 \mod (7)$ , $\rightarrow 11^p \equiv 4^p \mod (7)=7a+4^p$
$17 \equiv 3 \ mod (7)$, $\r... | {
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prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.
When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so
$$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac... | Equivalent problem: Let $0 < y < \pi$. Prove that $\frac{\sin y}{y} \ge \frac{\pi^2 - y^2}{\pi^2 + y^2}$.
It suffices to prove that, for all $y$ in $[0, \pi]$,
$$\sin y - y \frac{\pi^2 - y^2}{\pi^2 + y^2} \ge 0.$$
Denote LHS by $g(y)$. We have the following result. The proof is given at the end.
Fact 1: If $y\in (0, \p... | {
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| Like Barry Cipra,
replacing $\sin x=s,\cos x=c,\cos2x=2c^2-1,\sin2x=2sc$
$$3(2sc)+4(3c^2-1)-2c+6s-6=0$$
$$\iff4c^2-c(1-3s)+3s-5=0$$
$$\implies c=\cdots=-1,\dfrac{5-3s}4$$
If $\cos x=c=-1, x=(2n+1)\pi$
Else $4\cos x+3\sin x=5$
Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$
See also: Find $4\cos\... | {
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"answer_id": 4
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Induction proof done, need to simplify $\sqrt[4]{(n+1)^5}\ln(1 + \frac{1}{(n+1)^2}) \leq 4 \sqrt[4]{\frac{n+1}{n}}$ Prove that:
$$\sum_{j=1}^{n} \sqrt[4]{j^5}\ln(1+\frac{1}{j^2}) \leq 4 \sqrt[4]{n}, n \in \mathbb{N} $$
I got everything right and came up with an inequality:
$$\sqrt[4]{(n+1)^5} \ln\left(1 + \frac{1}{(n+1... | As regards your final step, just notice that $\ln(1+x)\leq x$.Therefore it suffices to show that
$$\sqrt[4]{(n+1)^5}\cdot \frac{1}{(n+1)^2} \leq 4 \sqrt[4]{\frac{n+1}{n}}$$
that is
$$\sqrt[4]{n}\leq 4 (n+1)^{\frac{1}{4}+2-\frac{5}{4}}=4(n+1)$$
which trivially holds.
| {
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Find the solution to cauchys problem I'm learning differential equation and not sure how to solve the following problem:
Find the solution for the cauchy's problem:
$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$
$y(0)=0$
$y'(0)=1$
What I tried so far:
$\lambda^2-6\lambda+9=0 \Leftrightarrow (\lambda-3)(\lambda-3)=0$
$y(x)=c_... | To begin with, I would notice that
\begin{align*}
y'' - 6y' + 9y & = (y'' - 3y') - (3y' - 9y) = (y' - 3y)' - 3(y' - 3y)
\end{align*}
Thus if we make the change of variable $u = y' - 3y$, one gets that
\begin{align*}
u' - 3u = e^{3x}\left(\frac{x^{2}+2}{x^{2}+1}\right) & \Longleftrightarrow (ue^{-3x})' = \frac{x^{2}+2}{... | {
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Prove that ${a_{1} +2a_{2} +3a_{3}+\cdots+np\ a_{n}}_{p} \ =\frac{np}{2}( 1+p)^{n}$ In expansion of ${\ \left( 1+x+x^{2} +\cdots+x^{p}\right)^{n} =a_{0} +a_{1} x+a_{2} x^{2} +\cdots+a_{np}} x^{np}$,
prove that ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p} \ =\frac{np}{2}( 1+p)^{n}$.
My attempt:
${\ \left( 1+x+x^{2} +\... | $$(1+x+x^2+...+x^p)^n=(a_0+a_1x+a_2x^2+....+a_{np}x^{np})$$
D.w.r.t. $x$ on both sides
$$n(1+x+x^2+...+x^p)^{n-1}(1+2x+3x^2+4x^3+...+px^{p-1})=a_1+2a_2x+a_3x^2+a_4x^3+...+np a_{np} x^{np-1}$$
Put$x=1$ to get
$$n (1+p)^{n-1}\frac{p(p+1)}{2}=a_1+2a_2+3a_3+...+np a_{np}=\frac{1}{2}np(1+p)^n.$$
| {
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"question_score": "2",
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Is $f(n)=\sqrt{n}$ the only function from $\Bbb{N}_0$ to $[0,\infty)$, with $f(100)=10$ and $\sum_{k=0}^n\frac{1}{f(k)+f(k+1)}=f(n+1)$?
Let $f:\mathbb{N_0} \rightarrow \left[0,\infty\right)$ be a function such that
a) $f(100)=10$
b)$\dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}=f(n+1)$ for al... | For $n\geq 1$,
\begin{align}
f(n+1) &= \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}\\
& = f(n)+\dfrac{1}{f(n)+f(n+1)}\\
\implies f(n+1)^2&=1+f(n)^2 =n+f(1)^2\\
\implies f(n) &= \sqrt{n-1+f(1)^2}.
\end{align}
Further, we have $$10 = f(100)=\sqrt{99+f(1)^2}\implies f(1)=1.$$
Finally, we also de... | {
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"url": "https://math.stackexchange.com/questions/3933243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$x^3 + y^3 = p^2$ over the integers $x^3 + y^3 = p^2$ has a solution over the integers for some three digit prime p. Find all p that satisfy.
The first thing I did was factorize the left hand side, getting $(x+y)(x^2 - xy + y^2) = p^2$
I then considered the case $x^2 - xy + y^2 = p^2$, which gave me $p^2 = 3y^2 -3y +1$... | Consider the linear factor first instead; this allows you to directly express $y$ in terms of $x$. From
$$p^2=x^3+y^3=(x+y)(x^2-xy+y^2),$$
it follows that $x+y=p^k$ for some $k\in\{0,1,2\}$, and $x^2-xy+y^2=p^{2-k}$. Then $y=p^k-x$, so we can eliminate $y$ to get the quadratic equation
$$3x^2-3p^kx+p^{2k}-p^{2-k}=0.$$
... | {
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Maximize the function $f(x) = (1-x) (1+x)^3$ Maximize the function $f(x) = (1-x)(1+x)^3.$
I took the first derivative of $f(x)$ and got $(1+x)^2 (2-4x)$
From there I got $2$ critical points of: $-1, \frac{1}{2}$
Then $I$ took the second derivative of $f(x)$ ($-12x^2 -12x$) and plugged in the critical points to get $0$ ... | Since $f(x) $ is non-negative only when $ |x| \leqslant 1$, then we only need to care when $|x|\leqslant 1$.
With that, we know that $ 1-x$ and $ 1 + x$ are non-negative numbers. Thus, we can apply the AM-GM inequality:
$$ AM\left(1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) \geqslant GM \left (1 - x , \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Basis of Image in a GF(5) matrix with variables consider $A_\alpha := \begin{pmatrix} \alpha & 3 & 4\\ 3 & 1 & 3 \\ 4 & 2 & \alpha \end{pmatrix} \in \mathbb{Z_5}^{3\times3}$. Determine, dependant on $\alpha$, a Basis and the dimension of the Image of the standardinterpretation $F_{A_\alpha}$. What is the dimension of t... | Your row reduction is indeed correct. Your steps from there are problematic. We have the reduced matrix
$$
\pmatrix{
a+1 & 0& 0\\
3& 1& 3\\
0& 0& a+4}.
$$
It is easy to see that for $\alpha \notin \{1,4\}$, further row-reduction brings us to the identity so that $A$ is invertible. In the case that $\alpha = 1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a$,$b$,$c$ be real numbers. Let $a$,$b$,$c$ be real numbers such that $a^{4}+b^{4}+c^{4}\leq27$. Prove the inequality:
$$\sum_{cyc}\frac{1}{\sqrt{a^{2}+1}}\geq \frac{3}{2}$$
My solution:
$\frac{1}{\sqrt{a^{2}+1}}\geq\frac{1}{2}\implies a^{2}\leq 3\implies a^{4}\leq 9$.
Similary : $b^{4}\leq9$ and $c^{4}\leq9$.
The... | Let $x_a = \sqrt{a^2 + 1}$, and similarly for $x_b$ and $x_c$. Then by RMS-HM inequality on $x_a, x_b, x_c$,
$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}}
&\le \sqrt{\frac{x_a^2+x_b^2+x_c^2}3} \\
&= \sqrt{\frac{(a^2+1)+(b^2+1)+(c^2+1)}3}\\
&= \sqrt{\frac{a^2+b^2+c^2}3 + 1}\tag1\\
\end{align*}$$
By RMS-AM inequa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$ Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1... | $$I=\int\frac{x}{x^2+(1-x)^2}\,dx=\int \frac{x}{2x^2-2x+1}\,dx=\int \frac{x}{2(x-a)(x-b)}\,dx$$
$$I=\frac 1{2(a-b)}\int \Big[\frac a{x-a}-\frac b{x-b} \Big]\,dx$$
$$I=\frac 1{2(a-b)}\Big[a \log(|x-a|)-b\log(|x-b|)\Big]$$ Now $a=\frac{1-i}2$, $b=\frac{1+i}2$,
$$I=\frac {1-i}4\left(\log
\left(x-\frac{1+i}{2}\right)+i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$
This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\a... | Here is perhaps a better approach. Note $\arg\left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ implies
$$\frac{z-8}{z-2}=re^{i \frac{\pi}{2}}\implies
\frac{z-8}{z-2}= - \frac{\bar z-8}{\bar z-2}\implies |z|^2 -5(z+\bar z )+16=0
$$
which is $|z-5|^2=9$, or $|z-5|=3$, i.e. a circle of center $(5,0)$ and radius $3$.
Note that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations? First, let us recall some trigonometric values
$\sin(0)=0$
$\sin(\pi/6)=\frac{1}{2}$
$\sin(\pi/3)=\frac{\sqrt{3}}{2}$
$\sin(\pi/2)=1$
$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$
$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Here, we can obse... | It is not so bad !
As @Doug M commented, it is one of the roots of
$$8x^3 - 4x^2 - 4x + 1 = 0$$
Using the trigonometric method for three real roots,we have
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6} \left(1+2 \sqrt{7} \sin \left(\frac{1}{3} \csc ^{-1}\left(2
\sqrt{7}\right)\right)\right)$$
Otherwise, using radi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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The limit of a series I need help with this exercise!
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$
I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
But I dont understand why!
So following that then,
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_... | This is what happens when you rush into learning calculus!
The quantity $S(n) = 1^4 + 2^4 + \cdots + n^4$ can be summed in a variety of ways. One such way is to guess that it grows like a fifth degree polynomial, then use Lagrange interpolation and induction. Another way might be to notice that
$$\sum_{i=0}^{n-1} (i+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$ I don’t know how to begin solving. Can I get a hint?
My failed attempt
Let $t =\tan (x/2)$
Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve
| $$\int_0^\tfrac{\pi}{2} \frac{1 + 2\cos x}{(2 + \cos x)^2}{\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\frac{/\sin^2 x}{/\sin^2 x} {\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{\csc^2 x + 2\cot x \csc x}{(2\csc x + \cot x)^2}{\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{-1}{(2\csc x + \cot x)^2}{\rm d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Proof by induction, $2^{n} > n^{2} - 2$ So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.
I start of with the "basecase" for n = 3:
RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$
Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where ... | The inductive step uses $2k^2-4\ge(k+1)^2-2$, or equivalently $(k-1)^2\ge4$, which holds for $k\ge3$. With quadratic inequalities, it's always wise to complete the square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Dividing $x^{81}+x^{48}+2x^{27}+x^{6}+3$ by $x^{3}+1$ Letting $f(x)=x^{81}+x^{48}+2x^{27}+x^{6}+3$, and we seek to divide $f(x)$ by $x^{3}+1$, or just any $x^3-a$. Is subbing $y=x^3$ allowed in this kind of division? I've experimented doing so and so far it seems correct.
| Even without the substitution, the problem will succumb to
polynomial long division.
Therefore, the only issue is how much perspiration you can replace with inspiration.
As already discussed, setting $y = x^3$ reduces the problem to
$$\text{Divide} ~g(y) = (y^{27} + y^{16} + 2y^9 + y^2 + 3) ~\text{by}~ (y+1). \tag1$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perf... | There's a typo in the first step. @mjw caught it. Note that you're applying the quadratic formula on $p^{1/3}$, so $$p^{1/3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}, \text{ if }c \ne 0.$$
Another missing link in your proof is your lack of argument about the claimed irrationality of $± \sqrt{b^2 - 4ac} - 2cp^{2/3}$.
As yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number,... | In the first step he observed that there are $2n$ minus signs, so they cancel out, and he multiplied top and bottom by $(n-1)!$. In the next step he rearranged the factors in the numerator:
$$\begin{align*}
&\frac12\cdot\frac32\cdot\ldots\cdot\frac{2n-3}2(n-1)!\\
&\qquad=\frac12\cdot\frac32\cdot\ldots\cdot\frac{2n-3}2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$ I am going around in circles on this system
$$x^2 - yz = 1\\
y^2 - xz = 2\\
z^2 - xy = 3$$
I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get:
$x^2 + y^2 + z^2 - yz - xz - xy =6$. And also t... | You're on the right track but made an algebra error (cf. $y - z$). These are quadratic equations, so there are two solutions:
$$\left\{x\to -\frac{5}{3 \sqrt{2}},y\to \frac{1}{3 \sqrt{2}},z\to \frac{7}{3
\sqrt{2}}\right\}$$
$$\left\{x\to \frac{5}{3 \sqrt{2}},y\to -\frac{1}{3 \sqrt{2}},z\to -\frac{7}{3
\sqrt{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $9x \equiv 21 \bmod 30$
Solving $9x \equiv 21 \bmod 30$
My approach:
Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}
So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$... | $9x\equiv 21\pmod{30}$ means $9x=30k+21$ or $3x=10k+7$ or $3x\equiv 7\pmod{10}$ or $21x\equiv 49\pmod{10}$ or $\color{red}{x\equiv 9\pmod{10}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Identity with the floor function I'm struggling to complete proof of the following identity:
$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.
Then
\begin{align}
\frac{... | hint
Observe that
$$\frac{m-n+1}{2}=\frac{m+n+1}{2}-n$$
So, you just need to prove that
$$\lfloor a\rfloor +\lfloor a+\frac 12\rfloor=2a$$
where $ a=\frac{m+n}{2}$.
There are two cases
$$a\in \Bbb N \implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=a+a$$
$$a\notin \Bbb N \implies a+\frac 12\in \Bbb N$$
$$\implies \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Question on binomial theorem: Find the term independent of $x$ in the expansion of $(2+x)\left(2x+\frac{1}{x}\right)^5$
Find the term independent of $x$ in the expansion of $$(2+x)\left(2x+\frac{1}{x}\right)^5$$
I am only able to do this with one binomial, but when there are two I do not totally understand the proces... | You can make life a bit easier in first writing
$$(2+x)\left(2x+\frac 1x\right)^5 = \frac 1{x^5}\color{blue}{(2+x)(2x^2+1)^5}$$
Now, you see you need only calculate the coefficient of $x^5$ in $(2+x)(2x^2+1)^5$. This is easy since $(2x^2+1)^5$ contains only even powers of $x$ but you are looking for $x^5$. Hence using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x... | $$f(x)=\left(x^3+4 x^2+x-6\right) q(x)+a x^2+b x+c=\\=(x^2+2 x-3) (x+2) q(x)+a x^2+b x+c\tag{1}$$
$$r(x)=a x^2+b x+c$$
We know that
$$\frac{f(x)}{x+2}$$
gives remainder $r=1$, which means that $f(-2)=1$
then we know that
$$\frac{f(x)}{x^2+2 x-3}$$
gives remainder $r=x+2$, which means that
$$f(x)=p(x)(x^2+2x-3)+x+2=p(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that the product of 4 consecutive positive integers is divisible by 24 How can I prove that prove that the product of $4$ consecutive positive integers is divisible by $24$, ie for any positive integer $n$ : $n(n+1)(n+2)(n+3)$
is divisible by $24$.
I've noticed that:
$24$ = $2^3 * 3$
$n(n+1)(n+2)(n+3)$ is divisib... | Notice that among $4$ consecutive numbers exactly one is divisible by 4 and one more divisible by 2 (say $n$ and $n+2$ or $n+1$ and $n+3$).
Notice also that among $3$ consecutive numbers exactly one is divisible by 3.
So the number is divisible by $3\cdot 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\fra... | You have a mistake in:
$$
\frac{-2n-1}{n^2+n}=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
$$
It should be:
$$
\frac{-2n-1}{n^2+n}=\frac{-\frac{2}{n}-\frac{1}{n^2}}{1+\frac{1}{n}}
$$
However, it would be better to take a factor of $n$:
$$
\frac{-2n-1}{n^2+n}=\frac{-2-\frac{1}{n}}{n+1}
$$
Hence, you will find that the limit wil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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For what values of $x_0$, $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$ converges? $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$
For what values of $x_0$ does this sequence converge?
If $(x_n) \to L$, then assuming $L>0$, I have
$L=\lim_\limits{n \to \infty} x_n=... | $x_n > 0 $ for all $n$
$x_{n} \ge \sqrt{6}$ for all $n \ge 1$ by the AM-GM inequality
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}=x_{n-1}$ when $x_{n-1} = \sqrt{6}$
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}<x_{n-1}$ when $x_{n-1} > \sqrt{6}$
Basically its going decrease till $x_n = \sqrt{6}$
Therefore, it converges to $\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate integral $ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx $ using the Dirichlet integral I try to calculate the integral:
$$ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx, $$
using the Dirichlet integral
$$ \int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\... | $$\int\frac{x-\sin (x)}{x^3}\,dx=\int\frac{1}{x^2}\,dx-\int\frac{\sin (x)}{x^3}\,dx=$$
$$=-\frac{1}{x}-\int\frac{\sin (x)}{x^3}\,dx$$
by parts
$$\int\frac{\sin (x)}{x^3}\,dx=-\frac{\sin (x)}{2 x^2}-\int -\frac{\cos (x)}{2 x^2} \, dx=$$
again by parts
$$=-\frac{\sin (x)}{2 x^2}-\left(\frac{\cos (x)}{2 x}-\int -\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Die and coin variance of random variable question. Die and coin. Roll a die and flip a coin. Let $Y$ be the value
of the die. Let $Z = 1$ if the coin shows a head, and $Z = 0$ otherwise. Let
$X = Y + Z$. Find the variance of $X$.
My work:
$E(Y) = 1 \cdot \cfrac{1}{6} + 2 \cdot \cfrac{1}{6} + 3 \cdot \cfrac{1}{6} + 4 ... | $E(X^2)\ne E(Y^2+Z^2)$
$E(X^2)=E(Y^2+Z^2+2YZ)=E(Y^2)+E(Z^2)+2E(YZ)=\dfrac{91}{6}+\dfrac{1}{2}+2\cdot\dfrac{7}{2}\cdot \dfrac{1}{2}=\dfrac{47}{3}+\dfrac{1}{2}=\dfrac{97}{6}$
$V(X)=E(X^2)-E(X)^2=\dfrac{97}{6}-16=\dfrac{97-96}{6}=\dfrac{1}{6}$
An alternate way:
Since X and Y are independent events covariance term vanishes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate this limit without L'Hôpital's rule.
Calculate $$\lim_{x\to0}\frac{(x+32)^{1/5}-2}{x}$$ without L'Hôpital's rule.
My attempt: I first rationalized the expression to get $$\left(\frac{(x+32)^{1/5}-2}{x}\right)\left(\frac{(x+32)^{1/5}+2}{(x+32)^{1/5}+2}\right)=\frac{x+28}{x((x+32)^{1/5}+2)}$$ How should I get... | If you want detailed proof:
$\dfrac{(2^{5}(\frac{x}{32}+1))^{\frac{1}{5}}-2}{x}=\dfrac{2(\frac{x}{32}+1)^\frac{1}{5}-2}{x}$
$(1+x)^n=1+nx+\dfrac{n(n-1)}{2}\cdot x^2.....$
$\lim_{x\to0}=\dfrac{2\bigg(1+\frac{1}{5}\frac{x}{32}+\dfrac{\frac{1}{5}(\frac{1}{5}-1)}{2}\cdot\bigg(\frac{x}{32}\bigg)^2.....\bigg)-2}{x}=\lim_{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$ For $ n\geq2$ let $ a_1, a_2, \ldots a_n$ be positive real numbers such that
$$ (a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right) \leq \left(n + \frac {1}{2}\right)^2.
$$
Prove that $ \max(a_1, a_2,... | Update:
Just realized we can prove directly without taking derivative.
WLOG assume $a_1 \ge a_2 \ge \cdots \ge a_n$, and $a_1 > 4a_n$ then
$$\begin{split} & & \left(\sum a_i \right) \left( \sum \frac{1}{a_i} \right) \\ & = & n + \left(\frac{a_1}{a_n}+\frac{a_n}{a_1}\right) + \sum_{1 < i<j<n} \left( \frac{a_i}{a_j} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\fra... | Your mistake is a sign error:$$\begin{align}\frac{d}{dx}x^3(1-x^2)^{3/2}&=x^2(1-x^2)^{1/2}[3(1-x^2)+x\tfrac32(-2x)]\\&=x^2(1-x^2)^{1/2}[3-3x^2-3x^2].\end{align}$$You seem to have mistaken that last $-$ sign for a $+$. Everything after that is error-carried-forward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
How to simplify $(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha})\cdot \cos\alpha \cdot \cos\beta$? There is this problem:
$$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{1}{\tan\alpha -\tan\beta}$$
I started as $$\left(\frac{\sin\alpha - \sin\beta}{\cos... | The proposed identity is not true (even though after the correction).
Indeed, take $\alpha = \pi/4$ and $\beta = 0$. Then we obtain:
\begin{align*}
\left(\frac{\sin(\pi/4) - \sin(0)}{\cos(0) - \cos(\pi/4)}\right)\times\cos(\pi/4)\cos(0) = \frac{\sqrt{2}}{2 - \sqrt{2}}\times\frac{\sqrt{2}}{2} = \frac{1}{2-\sqrt{2}}
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complete square values of quadratic formula For Which values of $x \in [0,n]$ the polynomial $P(x)=ax^2+bx+c$ $\ $ gives a complete square value.
For instance the polynomial $P(x)=3x^2+5x+7$ on the interval $x \in [0,100]$ has only two values for $x=3 \rightarrow P(3)=49 $ and for $x=54 \rightarrow P(54)=9025 $.
I wa... | $y^2=3x^2+5x+7$.
$12y^2=36x^2+60x+84=(6x+5)^2+59$.
$u^2-3v^2=-59$ where $u=6x+5$, $v=2y$.
Given one solution of $u^2-3v^2=-59$, and a fundamental solution of $u^2-3v^2=1$, you can get infinitely many solutions of $u^2-3v^2=-59$, and from those infinitely many solutions of $y^2=3x^2+5x+7$.
Now a fundamental solution of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?
| $$\frac{2 \tan x}{1-\tan ^2 x}+\tan x=^*\frac{2}{\sqrt{3}}$$.
$$\tan x=-1.45424, \tan x=0.35178, \tan x =2.25716$$
$L=\tan x \cot 2x$ can be written$^{**}$ as $\frac{1}{2} \left(1-\tan ^2 x\right)$ thus possible values are $$L_1=-0.557406,L_2=0.438125,L_3=-2.04739$$
$$(^*)\quad \tan x=u \to u^3-\frac{2 u^2}{\sqrt{3}}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{... | Let $S:=1+2+4+\dots+2^{n-2}$.
Then, since it's a geometric series, we have $S=\dfrac{1-2^{n-1}}{1-2}=2^{n-1}-1\implies S+1=2^{n-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Integrating this expression over two spheres I'd like to analytically perform the following integral:
$$ I=\int\limits_V d^3\textbf x\,d^3\textbf y \frac{1}{(\textbf x-\textbf y)^2}, $$
where the volume $V$ of the integration is such that: $|\textbf x|\in [0,A]$ and $|\textbf y|\in [0,B]$, i.e. spherical for $x$ and $y... | Let's start with $I=\int\limits_{V_{x}}\int\limits_{V_y} d^3\textbf x\,d^3\textbf y \frac{1}{(\textbf x-\textbf y)^2}$$=\int\limits_{V_{y}}(\int\limits_{V_x} d^3\textbf x\, \frac{1}{(\textbf x-\textbf y)^2})d^3\textbf y$$=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is my claim true about the maximum and minimum of the given function? I have this three-variable function
$$f(x,y,z)=x^9 \cos (2 x y+x z)+\left(5 x^4+1\right) \cos (x y+x z)+\left(5 x^4+3\right)\\ \qquad+\left(4 x^3+1\right) \cos (x y+2 x z)+x^3 \cos (x z)+\left(2 x^2+7\right) \cos (2 x y+2 x z)\\ \qquad+\left(x^2+5\ri... | Yes, considering that the variables only take positive values, setting all the co-sines to 1 will provide an upper bound and setting all the co-sines equal to -1 will provide a lower bound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that for every $n$ there are integers $a,b$ such that $X_n = a^2 + 2b^2$ We have a sequence $X_0, X_1,...$ which is determined by $X_0=1, X_1=3$ and $X_n = 6X_{n-1} - X_{n-2}$
Show that for every $n$ there are integers $a,b$ such that
$X_n = a^2 + 2b^2$
I thought of changing $a^2 + 2b^2 = (a + \sqrt 2 i)(a - \sqrt... | Let $(a_n,b_n)$ be solutions to Pell's equation $a_n^2-2b_n^2=\pm1$.
The fundamental solution is $1^2-2\times1=(1+\sqrt2)(1-\sqrt2)=-1$,
so solutions are $a_n^2-2b_n^2=(a_n+\sqrt2b)(a_n-\sqrt2b)=(1+\sqrt2)^n(1-\sqrt2)^n=(-1)^n$, where
$a_n+\sqrt2b_n=(1+\sqrt2)^n=(a_{n-1}+\sqrt2b_{n-1})(1+\sqrt2)=(a_{n-1}+2b_{n-1})+(a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3... | I used Sage cell to calculate this.
var("X")
F.<x> = NumberField(X^5 - 2)
y = x/(x^2 + 1) + x^2/(x^4 + 1) + x^3/(x + 1) + x^4/(x^3 + 1)
print(y)
print(y.minpoly())
Output:
-44/765*x^4 + 122/765*x^3 + 79/765*x^2 + 233/765*x + 1336/765
x^5 - 1336/153*x^4 + 1556/51*x^3 - 8182/153*x^2 + 7210/153*x - 12806/765
In the outp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$
Prove that $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$$
My attempt: I simplified $\sqrt{4x^2-nx}+2x$ by conjugation $$\sqrt{4x^2-nx}+2x \left(\frac{\sqrt{4x^2-nx}-2x}{\sqrt{4x^2-nx}-2x}\right) = -\frac{nx}{\sqrt{4x^2-nx}-2x} $$ I thought about substit... | $$\lim_{x\rightarrow-\infty}-\frac{nx}{\sqrt{4x^2-nx}-2x} = \lim_{y\rightarrow \infty}\frac{ny}{\sqrt{4y^2+ny}+2y} = \lim_{y\rightarrow \infty}\frac{n}{\sqrt{4+n/y}+2} = \frac{n}{4}$$
Alternatively,
$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \lim_{y \to 0^-} \frac{-\sqrt{4-ny}+2}{y}$
Note the negative sign as $y$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For every natural number n, prove that the vector $x = (n, n + 1, n (n + 1)) ∈ \Bbb R^3$ has as a norm a natural number, that is, $\|x\| \in \Bbb N$.
For every natural number $n$, prove that the vector $x = (n, n + 1, n (n + 1))$ ∈ $\mathbb{R}^3$ has as a norm a natural number, that is, $\|x\| \in \mathbb{N}$.
I made... | \begin{align*}
\sqrt{n^2+(n+1)^2+n^2(n+1)^2}&=\sqrt{(n^2+1)(n+1)^2+n^2}\\
&=\sqrt{(n^2+1+2n)(n+1)^2+n^2-2n(n+1)^2}\\
&=\sqrt{(n+1)^4-2n(n+1)^2+n^2}\\
&=\sqrt{[(n+1)^2-n]^2}\\
&=(n+1)^2-n\\
&=n^2+n+1.
\end{align*}
This is obviously a natural number for $n\in\mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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.