Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all $n \in \mathbb{N}$ such that ${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}$
Find all $n \in \mathbb{N}$ such that $${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}.$$
I know that if $2n+1$ were prime number, then
$${{2n}\choose{n}} = \frac{(2n) (2n-1) \cdots (n+1)}{n!} \equiv \frac{(-1)(-2)\cdots(-n)}{n!} = (-1)^n \... | Consider $N=(2n+1)$, an odd number. If it is a prime number, then with the argument in the OP we have the congruence $\binom {2n}n=(-1)^n$ modulo $N$.
In case $N$ is not a prime we have to divide first in the above "scheme". I will give an example. For $n=10$, $N=21=3\cdot 7$, we have to compute modulo $21$
$$
\binom{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Let $W_{1},W_{2}$ be sub-spaces of $\mathbb{R}^{4}$, find a subspace $W_{3}$ s.t $W_3\subset W_{2}$ and $W_{1}\oplus W_{3}=W_{1}+W_{2}$ Let $W_{1},W_{2}$ be linear sub-spaces of $\mathbb{R}^{4}$.
$W_{1}=\text{sp}\{(1,2,3,4),(3,4,5,6),(7,8,9,10)\}$
$W_{2}=\text{sp }\{(x,y,z,w)| \ x+y=0\}$
Find a linear subspace of $... | HINT
We should write as a span (check again you derivation by RREF)
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=s\left(\begin{array}{c}
1\\
1\\
1\\
1
\end{array}\right)+t\left(\begin{array}{c}
0\\
1\\
2\\
3
\end{array}\right)\right\}$$
and we can also easily find that
$$W_{2}=\left\{ \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determining the null space of the matrix
Determine the null space of the matrix:$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$$
My try:
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}_{R_2\rightarrow R_2-2R_1\\R_3\rightarrow R_3-R_1}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}_{R_3\... | Right Nullspace
If vector $\alpha = [x,y]$ is in the right nullspace of A then
$$\begin{bmatrix}
1 & -1 \\
2 & 3 \\
1 & 1 \\
\end{bmatrix}[x,y]^T = \begin{bmatrix} 0 \\ 0 \\ 0
\end{bmatrix}$$
This gives
$$x - y = 0$$
$$2x + 3y = 0$$
$$x + y = 0$$
First and third equation tells us that $x = y = -x$, i.e. the only solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is
For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
I use AM-GM Inequality to do this
$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$... | Your answer is right!
Indeed, we'll prove that $\sqrt{14}$ is a maximal value.
Let $a=x^2$ and $b=3y^2$, where $x$ and $y$ are positives.
Thus, $2a^2+7ab+3b^2=2x^4+21x^2y^2+27y^4$ and we need to prove that
$$x^2+3xy+6y^2\leq\sqrt{14}$$ or
$$(x^2+3xy+6y^2)^2\leq2(2x^4+21x^2y^2+27y^4)$$ or
$$(x-y)^2(x^2+6y^2)\geq0,$$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d... | hint: $x+y = (x+2) + (y-2)$ , and use $(A+B)^2 = A^2 + 2AB +B^2, A = x+2,B = y - 2$ to expand the numerator and simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$ Find the $n$-th derivative of
$$f(x)=\frac{x}{\sqrt{1-x}}$$
First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansio... | $x(1-x)^{-\frac 12}$
Generalized binomial theorem.
$x(1-x)^{-\frac 12} = x(1^{-\frac {1}{2}} + \frac {-1}{ 2}1^{-\frac {3}{2}}(-x) + \frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}1^{-\frac {5}{2}}(-x)^2+\frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}\frac {-5}{2}\frac {1}{3}1^{-\frac {5}{2}}(-x)^3\cdots$
$x(1-x)^{-\frac 12} = x + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Counting problem - fault in my reasoning. The problem is as follows:
The dean of science wants to select a committee consisting of mathematicians and physicists to discuss a new curriculum. There are $15$ mathematicians and $20$ physicists at the faculty; how many possible committees of $8$ members are there, if there ... | You made a calculation error.
$$\frac{20 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 29 \cdot 28}{8!} = \color{red}{1161160}$$
Let's examine your numerator:
$$20 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 29 \cdot 28 = 20 \cdot \frac{15!}{10!} \cdot \frac{29!}{27!} = \binom{20}{1} \cdot \binom{15}{5} \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
multidimensional chain rule, example Given $f:\mathbb{R}^3\to\mathbb{R}$, $f(u,v,w)=uv+vw-uw$ and
$g:\mathbb{R}^2\to\mathbb{R}^3$, $g(x,y)=(x+y, x+y^2, x^2+y)$, I have to give $D(f\circ g)(x,y)$
I calculated like this:
$Df(u,v,w)=(v-w, u+w, v-u)$ and
$Dg(x,y)=\begin{pmatrix} 1&1&2x\\ 1&2y& 1\\ 0&0&0\end{pmatrix}$
Furth... | $Dg(x,y)$ should be a matrix of $3\times 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Derive the following identity $1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.
Count the elements of the following set
$$A=\{(x,y,z): 1\leq x,y,z \leq n+1, z>\max\{x,y\}\}.
$$
From this derive the following identity:
$$1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
In the same manner find the formula for... | One method is this: When n= 0, the sum is 0. When n= 1, the sum is 1. When n= 2, the sum is 1+ 4= 5. When n= 3, the sum is 1+ 4+ 9= 14. When n= 4 the sum is 1+ 4+ 9+ 16= 30. When n= 5 the sum is 1+ 4+ 9+ 16+ 25= 55, etc.
Obviously, at each step we are just adding the next square so the "first differences" are: n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculate, $f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)$
If $$f(x)=\frac{4^x}{4^x+2}$$
Calculate,
$$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots
f\bigg(\frac{1996}{1997}\bigg)$$
My Attempt:
I was... | Given $f(x)=\dfrac{4^x}{4^x+2}$
From that we get $f(1-x)=\dfrac{2}{4^x+2}$
First let us take the last term $f\left(\dfrac{1}{1997}\right)$
Notice that $f\left(\dfrac{1}{1997}\right)=f\left(1-\dfrac{1}{1997}\right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$f\left(\dfrac{1}{1997}\right)+f\left(\dfrac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Distribution function of $X_1 + X_2$ given probability distributions of $X_1$ and $X_2$ Suppose I have a probability distribution for random variable $X_1$ given by:
$$
P(x) =
\begin{cases}
1/a, & \text{if} \;\; 0 \leq x \leq a\\
0, & \text{otherwise}
\end{cases}
$$
and $X_2$ with probability distribution:
$$
P(x) =
\b... | These are both uniform random variables if they independent we can write it like this. First $X_{1} \sim U(0,a)$ $X_{2} \sim U(0,b)$ . Also note the proof assumes that $ a \leq b$ The pdf of a uniform random variable $X \sim U(a,b)$ is given by
$$ f_{X}(x) =\begin{align}\begin{cases} \frac{1}{b-a} & \textrm{ for } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\s... | In these cases, in order to avoid mistakes with the sign, we can let $y=-x\to \infty$ to obtain
$$\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}=\lim_{y\to \infty} \frac{-7y}{\sqrt {4y^2-7y}+2y}=-\frac 74$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. T... | Hint:
$$2x^2-3xy-2y^2=2x^2-4xy+xy-2y^2=2x\underbrace{(x-2y)}+y\underbrace{(x-2y)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
(Modular Arithmetic) Congruences With Exponents I am trying to find the following:
The least positive integer $n$ for which
$3^n \equiv 1$ (mod $7$)
And hence $3^{100}$ (mod $7$).
The least positive integer $n$ for which
$5^n \equiv 1$ (mod $17$) or $5^n \equiv -1$ (mod $17$)
And hence evaluate $5^{243}$ (mod $17$).
E... | Regarding congruences with exponents - the cool thing about them is you can raise 'each side' to some power, or multiply them by a common factor and it remains true. It's often useful to show that $x^n\equiv \pm 1 \pmod k $. Regarding your first question, we know $3\equiv 3\pmod 7 $ so $3^2\equiv 2\pmod 7$ since $9\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What is the radius of the black circle tangent to all three of these circles? The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the ... | Thank you for your answers that helped me make a solution
$M_{red}=\dbinom{0}{3/2}$ , $r_{red}=\dfrac{3}{2}$ , $M_{blue}=\dbinom{0}{2}$ , $r_{blue}=2$, $M_{green}=\dbinom{2}{3/2}$ , $r_{green}=\dfrac{5}{2}$
$M_{black}=M=\dbinom{x}{y}$ , $r_{black}=r$
$length(v)=\dfrac{v}{\sqrt{v_{x}^{2}+v_{y}^{2}}}$ , $unitvec(v)=\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 4,
"answer_id": 0
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Last digit of sequence of numbers
We define the sequence of natural numbers
$$
a_1 = 3 \quad \text{and} \quad a_{n+1}=a_n^{a_n},
\quad
\text{ for $n \geq 1$}.
$$
I want to show that the last digit of the numbers of the sequence $a_n$ alternates between the numbers $3$ and $7$. Specifically, if we symbolize wit... | First, you should prove that $a_n$ is odd for all $n$ : this follows from the fact that $a_1$ is odd, and if $a_n$ is odd, then $a_{n+1}=a_n^{a_n}$ is an odd number multiplied with itself some number of times, so it is odd.
Let us notice something more. First, $a_1 = 3 \equiv 3 \mod 4$. Next, if $a_n \equiv 3 \mod 4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $p(x) = x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of $f$ around $x=1$, then $1$ is a local minimum
Consider $f:\mathbb{R} \to \mathbb{R} \in C^4$. Show that $p(x) =
x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of order $4$ of $f$ around $x=1$, then
$1$ is a local minimum.
I'm not sure how to proceed. I know t... | By the Remainder theorem of the Taylor polynomials we have $$ f(x)=P(x) + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$
$$ = (x-1)^4 + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$
$$ = (x-1)^4 \big[1+\frac {f^{(5)}(\xi)}{5!} (x-1)\big]$$
Let $x$ be close enough to $1$ such that $$ \big| \frac {f^{(5)}(\xi)}{5!} (x-1)\big|<1/2$$
Then $$ \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$
Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.
| In this case, since the equation is linear and given the "clue" that the coefficient of $x^2$ is non-zero, then consider $f(x) = a + b x + c x^2$. This yields the possible polynomial solutions, but not necessarily all possible solutions.
\begin{align}
x^2 &= (a + b x + c x^2) - 2 \, \left( a + \frac{b x}{2} + \frac{c x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Limit as $(x,y,z)\to (0,0,0)$ of $f(x,y,z) = \dfrac{xy+yz+xz}{\sqrt{x^2+y^2+z^2}}$ To find this limit, I converted to spherical coordinates and rewrote:
$$\lim_{r\to 0} \dfrac{r^2(\sin^2\theta \cos\phi \sin \phi + \sin\theta \cos \theta \sin \phi + \sin\theta \cos \theta \cos \phi)}{r} = 0$$
Is this method alright? Our... | Notice that $x,y$ and $z$ are less and equal than $\sqrt{x^2+y^2+z^2}$ so that
$$ |xy+yz+xz| \leq 3\left(\sqrt{x^2+y^2+z^2}\right)^2.$$
Hence, by a straightforward application of sandwich theorem, there holds that the aforementioned limit equals zero, since
$$ 0\leq \lim_{(x,y,z)\rightarrow (0,0,0)}\left|\dfrac{xy+yz+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Tough Divisibility Problem
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B \equiv 12 \pmod{19}$$
$$20000 + 1000A + 100 + 30 + B \equiv 12 \pmod{19}$$
$$ 5 + 12A + 5 + 11 + B \equiv 12 \pmod{19}$$
$$ 21+ 12A+ B \equiv 12 ... | You have some errors, which I will fix:
$$\begin{align}\overline{2A13B}=20000 + 1000A + 100 + 30 + B &\equiv 12 \pmod{19} \Rightarrow \\
(19\cdot 1052+12)+(19\cdot 52+12)A+(19\cdot 5+5)+(19\cdot 1+11)+B&\equiv 12 \pmod{19} \Rightarrow \\
12+12A+5+11+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+19\cdot 1+9+B&\equiv 12 \pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Finding all complex roots of polynomial Given $\ i $ is a root for the polynom $\ p(x) = x^4 +2x^3 +3x^2 +2x +2 $ find all the roots of $\ p(x) $ in the $\ \mathbb C $ field.
$\ x^4 + 2x^3 + 3x^2 + 2 \ / \ x-i $
I get confused everytime I try to divide $\ p(x) $ by $\ x -i $. I end up with this residual: $\ 2+x^3i ... | As the coefficients are real, when $a$ is a root so $\bar{a}$ is a root too. Then
$$\dfrac{x^4 +2x^3 +3x^2 +2x +2}{x^2+1}=x^2+2x+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding Jordan canonical form of a matrix given the characteristic polynomial I am trying to find the Jordan canonical form of a matrix $A$ given its characteristic polynomial. Suppose $A$ is a complex $5\times 5$ matrix with minimal polynomial $X^5-X^3$. The end goal of the problem is to find the characteristic polyno... | The minimal polynomial is $m_A(x) = x^3(x-1)(x+1)$ and the characteristic polynomial is the same up to sign change $\chi_A(x) = -x^3(x-1)(x+1)$.
We can assume that $A$ is equal to its Jordan form, which is:
$$A = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
When The curvature is maximum of $x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$ QUESTION
Find Where The Curvature has an extremum ... | Perhaps not a complete answer, but below you will find my solution to the problem, which leads to the same answer as yours, so we are either both wrong or the answer sheet is wrong.
Parameterize your curve as
$$ x(t) = a\cos^4 t, \quad y(t) = a\sin^4 t. $$
Then
\begin{align}
\dot x(t) &= -4a\cos^3 t \sin t,\\
\ddot x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding the limit of a recursive complex sequence I am currently struggling with the following exercise:
Let $z_0 = x_0+iy_0 \ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = \frac{1}{2} \left( z_n+\frac{1}{z_n} \right)$$
for $n \ge 0$. Show that if $x_{0} > 0$ then $\lim... | The way I see it is that you got two recursions going on
\begin{equation}
z_n = x_n + iy_n
\end{equation}
So
\begin{equation}
x_{n+1} + iy_{n+1}
=
\frac{1}{2}
(x_n + iy_n + \frac{1}{x_n + iy_n})
\end{equation}
The right hand side is
\begin{equation}
\frac{1}{2}
(x_n + iy_n + \frac{1}{x_n + iy_n})
=
\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express a matrix $A$ as the sum of a symmetric and a skew symmetric matrix I'm trying to express
$$A = \left[ \begin{array} { r r r } { 2 } & { - 2 } & { - 4 } \\ { - 1 } & { 3 } & { 4 } \\ { 1 } & { - 2 } & { - 3 } \end{array} \right]$$
as the sum of a symmetric and a skew symmetric matrix.
So far I have tried this:
... | HINT
Recall that for any square matrix $A$
$$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$$
and
$$\left(\frac{A+A^T}{2}\right)^T=\frac{A^T+A}{2}=\frac{A+A^T}{2}$$
$$\left(\frac{A-A^T}{2}\right)^T=\frac{A^T-A}{2}=-\frac{A-A^T}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verify Elementary Number Theory Proof (Romanian Olympiad) Can you check that there are no errors in this proof I wrote?
Here's the problem (from the 2002 Romanian Olympiad):
Let $p, q$ be two distinct primes. Prove that there are
positive integers $a, b$ so that the arithmetic mean of all the divisors of the number $n ... | Your proof is fine,
but the magic values
of $a$ and $b$
bother me.
So,
I'm going to try
to find all solutions.
Spoiler:
I don't,
but do give some conditions.
The sum of the divisors
of $p^aq^b$ is
$\sigma(p^aq^b) =\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)}
$
and the number of divisors is
$d(p^aq^b) =(a+1)(b+1)$.
So the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } ... | Because for $a > 0$ we know $\sqrt a = \frac a{\sqrt a}$ so $\sqrt{2 -x^2} = \frac {2-x^2}{\sqrt{2-x^2}}$ (assuming that $2 - x^2 > 0$ which must be the case as you have had $\sqrt{2 - x^2}$ in the denominator elsewhere in the equation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve $2a_{n-2} = a_n + a_{n-1}$ using generating function Need to solve:
$$2a_{n-2} = a_n + a_{n-1}$$
with: $a_0 = 0$ and $a_1=1$
I get:
$$f(x) = \frac{2x^3-x^2-x}{2x^2-x-1}$$ so I tried to scompose the denominator and I get:
$$f(x) = \frac{2x^3-x^2-x}{(x-1)(x+\frac{1}{2})}$$ now I think I have to use partial fraction... | I think you committed a typo in calculations. From:
$$a_n=2a_{n-2}-a_{n-1}$$
we should have:
$$f(x)=\sum\limits_{n=0}\color{blue}{a_{n}}x^{n}=
x+\sum\limits_{n=2}a_nx^n=
x+\sum\limits_{n=2}\left(2a_{n-2}-a_{n-1}\right)x^n=\\
x+2x^2\left(\sum\limits_{n=2}a_{n-2}x^{n-2}\right)-x\left(\sum\limits_{n=2}a_{n-1}x^{n-1}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Show $\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx = 2\pi e^{2/3}$ Real Methods Taken from the post: The Integral that Stumped Feynman?
I want to know if the integral:
$$\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin ... | Start with $\tan \left(\frac{x}{2} \right)$$=2\tan \left( \frac{t}{2} \right)$ $$I=2\int_0^{\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=8e^{5/8}\int_0^{\pi}\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)\frac{dt}{5-3\cos t} $$ Using $$ \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Roll a die probability question
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.
I have tried to solve this question this wa... | There are six possible outcomes for each of the five throws, so your denominator should be $6^5$.
For the favorable cases, we must have a sequence of five throws in which three of the outcomes are 1, 2, or 3, two of the outcomes are 4 or 5, and two of the outcomes are a six. There are $\binom{5}{2}$ ways to select whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Definite integral of $x\sin^n x$ from $0$ to $\pi/2$ How to find
\begin{equation*}
\int_0^{\pi/2} x\sin^n x dx
\end{equation*}
where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives
\begin{equation*}
\frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} (\sin y+\cos y)^n dy+\frac{1}{2^{n/2}}\int_{-\pi/4}^{\p... | Assuming that $n$ is odd ($n=2m+1$), we may consider the triangle wave
$$ t(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}\,\sin((2n+1)x)$$
and the fact that, by symmetry,
$$ \int_{0}^{\pi/2}x\sin^n(x)\,dx = \frac{1}{4}\int_{0}^{2\pi}t(x)\sin^n(x)\,dx. $$
On the other hand the Fourier sine series of $\sin^n(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Sum of series $\frac{ 4}{10}+\frac{4\cdot 7}{10\cdot 20}+\frac{4\cdot 7 \cdot 10}{10\cdot 20 \cdot 30}+\cdots \cdots$
Sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots $$
Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot... | For a complex number $z$ such that $|z|<1$, let
$$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\,z^n\,.$$
Then, the required sum $S:=\displaystyle\sum_{n=1}^\infty\,\prod_{k=1}^n\,\left(\frac{3k+1}{10k}\right)$ satisfies
$$S=f\left(\frac{3}{10}\right)-1\,.$$
We claim that $f(z)=(1-z)^{-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
| $$7\times 2^k+1=n^2\iff7\times2^k=(n-1)(n+1)$$
$n$ must be odd so, $7\times 2^{k-2}=(\frac{n-1}{2})(\frac{n+1}{2})$ and $\gcd(\frac{n-1}{2},\frac{n+1}{2})=1$, since $\gcd(\frac{n-1}{2},\frac{n+1}{2})$ divides $\frac{n+1}{2}-\frac{n-1}{2}=1$.
This gives two possibilities:
*
*$\frac{n-1}{2}=7\Rightarrow n=225\Rightarr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the substitution of variables allowed in proofs? EDIT: I later realized no one saw that substituting $\sqrt{x} = x$ is not valid because they are not equal, you would have to assume $x$ as a completely different variable (i.e. $a$) and not the same one because they have no relation to each other. However, the solut... | $x^2 > 0$ we have
$2y \le \frac {y^2}{x^2} + x^2 \iff yx^2 \le \frac{x^2 + x^4}2$.
Now do the substition. Substitute $y = \sqrt a$ and $x^2 = \sqrt b$.
EXCEPT we don't know that $y$ is positive.
Let $y = \pm \sqrt a$ and $x^2 = \sqrt b$ so $y^2 = a \ge 0$ and $b = x^4 > 0$
And we know $\sqrt ab \le \frac {a+b}2$ for... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$ is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$.
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$... | given hyperbola is $$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$ now differentiate the equation with respect to x to get the slope of tangent at that point $$\implies {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4} = 1 \right)\implies \frac{2}{9}(x- 3)- \frac{1}{2}(y- 2){dy\over dx}= 0 \\ \implies{dy\over dx}= ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit Problem
If there exists $a \in \mathbb{R}$ such that:
$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$
has limit in $-2$. If such $a$ exists what is limit in $-2$ ?
Attempt to solve
My idea was first to try factorize ... | Since the denominator at $x=-2$ is equal to $0$ it is necessary that also the numerator at $x=-2$ is equal to $0$ that is
$$3(-2)^2+a(-2)+a+3 =0 \implies a=15 $$
and therefore
$$\lim_{x \rightarrow -2}\frac{3x^2+15x+18}{x^2+x+2}=\lim_{x \rightarrow -2}\frac{3(x+2)(x+3))}{(x-1)(x+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
The pdf of $\tan X$ when $X$ is Uniform Let $X$ be a random variable with pdf
$$ f(x) = \begin{cases} \frac{1}{\pi}, & \text{if } -\frac{\pi}{2} < x <\frac{\pi}{2} \\ 0, & \text{elsewhere} \end{cases} $$
Find the pdf of $Y = \tan(x)$.
My attempt:
$$F_Y(y) = P(Y\le y) = P(\tan(x)\le y) = P(x\le \tan^{-1}(y))=F_X(\tan^{... | \begin{align}F_Y(y) &= P(Y\le y) \\&= P(\tan(\color{red}X)\le y) \\&= P(X\le \tan^{-1}(y))\\&=F_X(\tan^{-1}y) \\&= \frac{1}{\pi}\left({\tan^{-1}y}+\frac{\pi}2 \right)\end{align}
$$ f_Y(y) = \frac{1}{\pi} \frac{1}{1+y^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$
I would like to find the minimum of
$$f(x)=\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n,$$
where $n$ is a natural number.
I know there is possible by derivate, but
$$f'(x)=n \left(\left(... | Let $A = \dfrac{1+\sin^2x}{\sin^2x}, B = \dfrac{1+\cos^2x}{\cos^2x}$. Then $A+B = 2+\dfrac{1^2}{\sin^2x}+\dfrac{1^2}{\cos^2x}\ge 2+\dfrac{(1+1)^2}{\sin^2x+\cos^2x}=6$. Thus: $A^n+B^n\ge \dfrac{(A+B)^n}{2^{n-1}}= \dfrac{6^n}{2^{n-1}}=2\cdot 3^n$. Equality occurs when $\sin x = \cos x$ or $x = \dfrac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Difficulty Finding $A^k$ Let $A=
\begin{bmatrix}
1& -1 & 1\\
0 & 1 & 1 \\
0 & 0 & 1\\
\end{bmatrix}$. Compute $A^k$.
My attempt
I'm trying to compute $A^k$ using this approach as follows:
$$
A=I+N=
\begin{bmatrix}
1& 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}+
\begin{bmatrix}
0& -1 & 1\\
0 & 0 & 1 \\
0 & 0 &... | You have written $A=I+N$, and you know that $N^3$ (and hence all higher powers) are zero. If $X$ and $Y$ are two matrices that commute with each other, then you can still use the bionomial theorem:
$$(X+Y)^n=\sum_{i+j=n}\binom{n}{i} X^i Y^j$$
Because $I$ commutes with $N$, and because all the higher powers of $N$ vani... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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The minimum value of perimeter of the triangle Let there be $\triangle ABC$ having integral side lengths such that angle $$ \angle A=3 \angle B $$ Then find the minimum value of its perimeter.
I did this with trigonometry and got an equation as: $$ 2p = 4b[2\cos^3( \theta) + \cos^2(\theta) - \cos(\theta)] $$ where $b$ ... | My try:
$$\frac{a}{\sin3\beta}=\frac{b}{\sin\beta}$$
$$\sin3\beta=3\sin\beta-4\sin^3\beta=\frac{a}{b}\sin\beta$$
$$3-4\sin^2\beta=\frac{a}{b}$$
$$\sin^2\beta=\frac14(3-\frac ab)\tag{1}$$
$$\frac{b}{\sin\beta}=\frac{c}{\sin4\beta}$$
$$\frac{\sin4\beta}{\sin\beta}=\frac{2\sin2\beta \cos2\beta}{\sin\beta}=4\cos\beta (1-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930594",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Discontinuity of $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ at $(x,y) = (0,0)$ Let $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ , when $(x,y) \neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x \rightarrow 0$, to get the limit to be $0$.
But I ... | Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -\frac{x^2}{2}+x^n$ where $n >0$,then we have
\begin{align}\lim_{x \to 0}\frac{2(x^3+(-\frac{x^2}{2}+x^n)^3)}{2x^n} &= \lim_{x \to 0} \frac{(x^3+(-\frac{x^2}{2}+x^n)^3)}{x^n} \\
&= \lim_{x \to 0} \left[ x^{3-n}+\left(-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The tangent at $P$ to $y = x^2 - x^3$ meets the curve again at $Q$. Show that locus of midpoint of $PQ$ is $y=1-9x+28x^2-28x^3$
The tangent at a variable point $P$ of the curve $y = x^2 - x^3$ meets it again at $Q$. Show that the locus of the middle point of $PQ$ is $$y = 1 - 9x + 28x^2 - 28x^3$$
My approach
$$y^\pr... | Let $P(h,h^2-h^3),Q(k,k^2-k^3), h\ne k$
Now the gradient of $PQ$ is $$=\dfrac{k^2-k^3-(h^2-h^3)}{k-h}=k+h-(k^2+kh+h^2)$$
The gradient of the tangent at $P$ is $$=2h-3h^2$$
$\implies2h-3h^2=k+h-(k^2+kh+h^2)$
$\iff2h^2-(k+1)h+k-k^2=0$
$$\implies h=\dfrac{k+1\pm\sqrt{(k+1)^2+8(k^2-k)}}4=\dfrac{k+1\pm(3k-1)}4=k,\dfrac{1-k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\lef... | Eqn 1
$$(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$$
$$(-\frac{6}{x^2})+(-\frac{6}{x^3})+(\frac{6}{x^3})(1+\frac{1}{x^2}+\frac{2}{x})$$
$$(-\frac{6}{x^2})+ (-\frac{6}{x^3}) + \frac{6}{x^3}+\frac{6}{x^5}+\frac{12}{x^4}$$
$$\frac6{x^2}[\frac{1}{x^3}+\frac{2}{x^2}-1]$$
Eqn 2
$$\frac{6}{x^3}(1+\frac{1... | {
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"url": "https://math.stackexchange.com/questions/2935936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Condition for primeness of a degree 2 Solinas Prime Given $2^{2k+1}+2^{k+1}+1$ is prime, this necessarily imply $2k+1$ is a prime.
This result is assumed true in Primality tests for $2^p \pm 2^{(p+1)/2}+1$ using elliptic curves by YU TSUMURA in Proceedings of the American mathematical society vol. 139 num. 8 August 201... | We begin with the following identity applied on $x=2^k$.
$$4x^4+1=(2x^2+2x+1)(2x^2-2x+1).$$
This gives that
$$\left(2^{2k+1}+2^{k+1}+1\right)\left(2^{2k+1}-2^{k+1}+1\right)=2^{4k+2}+1.$$
Assume for contradiction that $2k+1$ is not prime, but our first factor is prime. Let $p$ be a prime dividing $2k+1$. We have that, a... | {
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"url": "https://math.stackexchange.com/questions/2937965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}... | Continuing from yours:
$$I=\int\frac{1}{\sqrt{(x-a)(x-b)}}dx$$
now if we look at this:
$$(x-a)(x-b)=x^2-(a+b)x+ab=\left(x-\frac{a+b}{2}\right)^2-\frac{a^2+b^2}{2}$$
and so we can rewrite our integral as:
$$I=\int\frac{1}{\sqrt{(x-\frac{a+b}{2})^2-\frac{a^2+b^2}{2}}}dx$$
now let $u=x-\frac{a+b}{2}$ so $dx=du$
and the in... | {
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"url": "https://math.stackexchange.com/questions/2944173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose t... | A different appraoch:
Note that
\begin{align}
\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} &= \underbrace{\frac{n(n+1)(2n+1)}{6}}_{\displaystyle=\sum_{k=1}^n k^2 }-2\,\underbrace{\frac{n(n+1)}{2}}_{\displaystyle=\sum_{k=1}^{n}k}+n=\sum_{k=1}^n(k^2-2k+1)\in\Bbb Z
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 3
} |
Find value of n such the function has local mimima at x=1. If $f$ is defined by
$$f(x)=(x^2-1)^n(x^2+x-1)$$
then $f$ has a local minimum at $x=1$, when
*
(i) $n=2$
(ii) $n=3$
(iii) $n=4$
(iv) $n=5$
Multiple options are correct.
The given answer is $n=2$ and $n=4$.
I tried putting derivative equal to zero and ... | The second derivative test is inconclusive for $n > 2$.
As a fall back, you can use the first derivative test.
Let $n > 1$ be a positive integer.
Computing $f'(x)$, we get the factored form
$$f'(x)=(x^2-1)^{n-1}g(x)$$
where
$$g(x)= (2n+2)x^3+(2n+1)x^2-(2n+2)x-1$$
Noting that $g(1)=2n$, it follows that $g(x) > 0$ ne... | {
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"source": "stackexchange",
"question_score": "2",
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why using $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to factorize a polynomial degree 2 dose not always work i tried to factorize $5x^3-11x^2+2x$ so i took out $x$ and used $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the roots 2 and $\frac{1}{5}$ but to my surprise multiplying the roots like so $x(x-2)\cdot(x-\frac{1}{5})$ produces... | It’s related to the derivation of the quadratic formula, which is done by completing the square. So, we know the formula.
$$ax^2+bx+c = 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
In reality, you aren’t solving for the original equation, which is $ax^2+bx+c = 0$. You’re solving for $x^2+\frac{b}{a}x+\frac{c}{a} = 0... | {
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Where does this derivation of the Fourier Series for csc(x) go wrong? In this post, the following derivation for the Fourier series of csc(x) is given:
\begin{align}
\csc x
&= \dfrac{1}{\sin x}\\
&= \dfrac{2i}{e^{ix}-e^{-ix}}\\
&= \dfrac{2ie^{-ix}}{1-e^{-2ix}}\\
&= 2ie^{-ix}\sum_{n\geq0}e^{-2inx}\\
&= 2i\sum_{n\geq0}e... | The imaginary part actually seems to vanish — except at $x=0$ where it's infinite.
The Fourier cosine series of $\delta(x-\epsilon)$ is given by
$\frac{1}{\pi} + \sum_{n=1}^{\infty} \frac{2}{\pi} \cos n\epsilon \cos nx.$
Letting $\epsilon \to 0$ gives
$\delta(x) = \frac{1}{\pi} + \sum_{n=1}^{\infty} \frac{2}{\pi} \cos ... | {
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"answer_count": 1,
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Integrate $\int\frac{x^3+1}{x^4+x^3+x^2+x}\,dx$ How do I integrate $\displaystyle\int\dfrac{x^3+1}{x^4+x^3+x^2+x}\,\mathrm dx$?
I tried by splitting the equation in two parts like $\dfrac{x^3}{x^4+x^3+x^2+x}$ and $\dfrac1{x^4+x^3+x^2+x}$ and then cancelling out the $x$ terms from the first part and then trying to integ... | Hint:
$$\dfrac{x^3+1}{x^4+x^3+x^2+x}=\dfrac{1}{x}-\dfrac{1}{x^2+1}$$
| {
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Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$.
Now, consider a cubic polynomial function with 3 distinct real roots,
$f(x)=A(x-a)(x-b)(x-c)$
It ... | Translating and scaling, you can assume that $a=0$ and $b=1$. Then
$$f(x)=x^3-(c+1)x^2+cx$$
$$f'(x)=3x^2-2(c+1)x+c$$
The roots of $f'$ are
$$\frac{2c+2\pm\sqrt{4c^2-4c+4}}6=\frac{c+1\pm\sqrt{c^2-c+1}}3$$
Note that the discriminant is increasing for $c>1/2$. Then, since $c>b=1$, we have
$$\frac{c+1+\sqrt{c^2-c+1}}3>\fra... | {
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"source": "stackexchange",
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3 digit numbers whose one digit is the average of the other two
Question: How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
How can this be solved with an arithmetic sequence?
I can see that any two digits must be same parity to produce the even su... | Can be done by simple trial and error
(Note that plus sign is only used to distinguish the triplets of digits that can be used to form the desired number)
$0+2,4,6,8+1,2,3,4$
$1+3,5,7,9+2,3,4,5$
$2+4,6,8+3,4,5$
$3+5,7,9+4,5,6$
$4+6,8+5,6$
$5+7,9+6,7$
$6+8+7$
$7+9+8$
In total we get 20 possible triplets of digits that c... | {
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"url": "https://math.stackexchange.com/questions/2953392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ if and only if $a,b,c$ are the lengths of a right angled triangle Here is a problem from the book: Everything connected to Pithagoras
It is well known that in every right angled triangle $ABC$, $a^2+b^2=c^2$. Howevee, there are some more complicated equations as well. Here is one:
Prove... | Hint:
$$
2(a^4b^4+b^4c^4+c^4a^4)-(a^8+b^8+c^8)\\
=(a^2+b^2+c^2)(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2).
$$
Edit: I arrived at the above by guessing. My intuition is that: the given identity $2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ is symmetric in its info about $a$, $b$, and $c$, therefore, the conclusion has to be symmetric ... | {
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"source": "stackexchange",
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How can I simplify the derivative further to match the correct answer? I've been stuck on trying to simplify my expression in order to match the correct answer but I can't seem to get the correct solution. Guidance towards the proper steps would be greatly appreciated!
The original question:
Differentiate and simplify... | To remove some visual clutter, I'll define $a:=x^3-1$ and $b:=x^3+1$. So, your solution looks like:
$$\frac{2x^2}{\sqrt[3]{a^2}\;\sqrt[3]{b^4}} \tag{1}$$
Note that $\sqrt[3]{b^4}=\sqrt[3]{b^3 b} = b\sqrt[3]{b}$, so that $(1)$ becomes
$$\frac{2x^2}{b\;\sqrt[3]{a^2}\;\sqrt[3]{b}}\tag{2}$$
From here, we want to rationaliz... | {
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"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to 4} \frac{x^4-4^x}{x-4}$, where is my mistake? Once again, I am not interested in the answer. But rather, where is/are my mistake(s)? Perhaps the solution route is hopeless:
Question is: evaluate $\lim_{x \to 4} \frac{x^4 -4^x}{x-4}$.
My workings are:
Let $y=x-4$. Then when $x \to 4$, we have that ... | You can break up this limit under certain circumstances.
$\lim_\limits{x\to a} (f(x) + g(x)) = \lim_\limits{x\to a} f(x) + \lim_\limits{x\to a}g(x)$
You can do it if $\lim_\limits{x\to a} f(x), \lim_\limits{x\to a}g(x)$ both exist and are finite.
But if $\lim_\limits{x\to a} f(x) = \infty$ and $\lim_\limits{x\to a}g(x)... | {
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"url": "https://math.stackexchange.com/questions/2957072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Constructing the graph from parametric equations
$$x(\theta)=|\cos4\theta|\cos\theta$$
$$y(\theta)=|\cos 4\theta|\sin\theta$$
$$0\le \theta\le2\pi$$
I have to graph the equation.
Now, I have no idea as the parameter $\theta$ cannot be eliminated. At most I found that $x^2+y^2=\cos^24\theta$. Now, suppose I let $\thet... | $\theta$ can be eliminated with $\tan\theta=\dfrac yx$, then
$$\cos\theta=\frac x{\sqrt{x^2+y^2}},
\\\sin\theta=\frac y{\sqrt{x^2+y^2}},
\\\cos4\theta=8\frac{x^4}{(x^2+y^2)^2}-8\frac{x^2}{x^2+y^2}+1.$$
Then
$$x^2+y^2=\left(8\frac{x^4}{(x^2+y^2)^2}-8\frac{x^2}{x^2+y^2}+1\right)^2,$$
or
$$(x^2+y^2)^5=(x^2+y^2)^2-8x^2y^2.... | {
"language": "en",
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"source": "stackexchange",
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Show that the exact values of the square roots of $z=1+i$ are... Show that the exact values of the square roots of $z=1+i$ are...
$w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}$
$w_1=-\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$
My attempt
Let $z=1+i\in \mathbb{C}$. Then
$r=|z|=\sqrt{2}$
... | Well, you could cheat and just do
$$
w_0^2=\frac{1+\sqrt{2}}{2}-\frac{-1+\sqrt{2}}{2}+2i\sqrt{\frac{1+\sqrt{2}}{2}\cdot\frac{-1+\sqrt{2}}{2}}=1+i
$$
and note that $w_1=-w_0$.
Without cheating, you did quite well: $|1+i|=\sqrt{2}$, so
$$
1+i=\sqrt{2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)=
\sqrt{2}\left(\cos... | {
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"url": "https://math.stackexchange.com/questions/2959857",
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"source": "stackexchange",
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Compute the following limit if they exist: $\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$ Compute the following limit if they exist:
$\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$
Attempt:
Given $\epsilon > 0$. Choose $\delta = \text{Not sure yet}$. Then $||(x, y, z) - (0,0,0)|| = \sqrt... | Well, we have $$2\sqrt{x^2+y^2} \le 2\sqrt{x^2+y^2+z^2} $$
and you can pick $\delta = \frac{\epsilon}2$.
| {
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Is the matrix $A = I - \frac{J}{n+1}$ idempotent? I am supposed to figure out if the following statement is false or true.
If $I$ is the $n \times n$ identity matrix, and $J$ is an $n \times n$ matrix consisting entirely of ones, then the matrix $$A = I - \frac{J}{n+1}$$ is idempotent (i.e., $A^{2} = A$).
I underst... | The expression $A = I - \frac{J}{n+1}$ is matrix addition, where $\frac{J}{n+1}$ represents scalar multiplication of the matrix $J$ by $\frac{1}{n+1}$. Since the matrix $J$ consists completely of $1$s, each entry of the scaled matrix $\frac{J}{n+1}$ is $\frac{1}{n+1}$. Here is the matrix addition:
$$ I - \frac{J}{n+1} ... | {
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How can I solve this inequality using convexity? Given $a, b, c\ge 0$ and $x, y, z> 0$ and $a + b + c = x + y + z$.
Show that $$a ^ 3 / x ^ 2 + b ^ 3 / y ^ 2 + c ^ 3 / z ^ 2 \ge a + b + c$$ prove inequality using convexity
| By Holder
$$\left(\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\right)(x+y+z)^2\geq(a+b+c)^3$$ and we are done!
Also, since $f(x)=x^3$ is a convex function on $[0,+\infty),$ by Jensen we obtain:
$$\sum_{cyc}\frac{a^3}{x^2}=(x+y+z)\sum_{cyc}\frac{x}{x+y+z}\left(\frac{a}{x}\right)^3\geq$$
$$\geq(x+y+z)\left(\sum_{cyc}\... | {
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Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{... | $$\sqrt[3]{n^3+1}-\sqrt{n^2-1}=\sqrt[3]{n^3+1}-n-\sqrt{n^2-1}+n
\\=\frac1{(n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2}+\frac1{\sqrt{n^2-1}+n}$$
and both terms are decreasing. The supremum is with $n=1$,
$$x_1=\sqrt[3]2.$$
| {
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"url": "https://math.stackexchange.com/questions/2968028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ when $n \rightarrow \infty$? Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ ? How to solve such type of questions considering $n \rightarrow \infty$?
| Let $n = m^2$.
$a = \ln (\sqrt n)! = \ln 1 + \ln 2 + \cdots + \ln m$
$b = \ln \sqrt {n!} = \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m + \ln (m+1) + \ln (m+2) + \cdots + \ln m^2\}$
$a - b = \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m\}
- \frac{1}{2} \{\ln (m+1) + \cdots + \ln (m+m) + \cdots + \ln m^2\}$
$a - b < \fr... | {
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"source": "stackexchange",
"question_score": "1",
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Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}... | Denote $p(n):=(1+\frac{1}{1^3})(1+\frac{1}{2^3})(1+\frac{1}{3^3})...(1+\frac{1}{n^3})$.
Claim:
$$p(n)\leq3-\frac2{n^2},\,\forall n\geq2.$$
For $n=2$, we have $\frac94\leq3-\frac12$.
Then suppose $p(n)\leq3-\frac2{n^2}$ for some $n$. We see
$$\eqalign{
p(n+1)&=p(n)(1+\frac1{(n+1)^3})\cr
&\leq3-\frac{2}{n^2}+\frac3{(n... | {
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"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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is there are any short cut method to find the determinant of A? Find the determinant of A
$$A=\left(\begin{matrix}
x^1 & x^2 & x^3 \\
x^8 & x^9 & x^4 \\
x^7 & x^6 & x^5 \\
\end{matrix}\right)$$
My attempts : By doing a laplace expansion along the first column i can calculate,but it is a long process My qu... | Use Gaussian elimination, while keeping track of changes to the determinant.
$$ \begin{aligned}
\left\lvert \begin{matrix} x^1 & x^2 & x^3 \\ x^8 & x^9 & x^4 \\ x^7 & x^6 & x^5 \end{matrix} \right\rvert
&= x^6 \left\lvert \begin{matrix} 1 & 1 & 1 \\ x^7 & x^7 & x^1 \\ x^6 & x^4 & x^2 \end{matrix} \right\rvert \\
&= x^6... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Find the value of an integer $a$ such that $ a^2 +6a +1 $ is a perfect square. I was able to solve this but it required me using hit and trial at one step. I was wondering if i could find a more solid method to solve it.
p.s. this is the first time im asking a question here so sorry if i couldn't construct the question... | $$
\\(a+1)^2=a^2+2\cdot a+1\leq a^2+6\cdot a + 1 <a^2+6\cdot a+9=(a+3)^2
\\(a+1)^2\leq a^2+6\cdot a+1\leq(a+2)^2
\\=>
$$
if $\>a^2+6\cdot a + 1\>$ is a perfect square if and only if $\>a^2+6\cdot a + 1=(a+1)^2\>$ or $\>a^2+6\cdot a + 1=(a+2)^2\>=>\>a=0\>$ or $\>a=-6$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the matrix A in Ax=b I am watching the linear algebra course of MIT presented by Gilbert Strang and in Lecture 13 he solves the equation $Ax = b$. This is not the usual "find $x$", instead $x$ and $b$ are given and we need to find the matrix $A$ The link is at https://www.youtube.com/watch?v=l88D4r74gtM&t=841s ... | It's again a linear system, with unknowns living in a vector space, precisely the $3\times 1$ column vectors. Write $A=[a_1\ a_2\ a_3]$; then you know that
$$
\begin{cases}
2a_1 = b & (c=0,\;d=0) \\[4px]
3a_1+a_2=b & (c=1,\;d=0) \\[4px]
2a_1+a_3=b & (c=0,\;d=1)
\end{cases}
$$
This immediately entails that $a_3=0$, $a_1... | {
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Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n <... | For monotonic behavior:
$a_{n+1}-a_n=\frac{(n+1)^2+3}{(n+2)^2}-\frac{n^2+3}{(n+1)^2}=\frac{(n+1)^4+3(n+1)^2-n^2(n+2)^2-3(n+2)^2}{(n^2+3n+2)^2}=\frac{2(n^2-n-4)}{(n^2+3n+2)^2}.$
Observe that for $n \geq 3$, the numerator is always positive, so for all $n \geq 3$, the sequence will be an increasing sequence (so ultimatel... | {
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If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_... | $$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})
=F_n(F_{n+1}+F_{n-1}).$$
Also $F_{n+1}=F_n+F_{n-1}\equiv F_{n-1}\pmod{F_n}$, so
$F_{n+1}+F_{n-1}\equiv 2F_{n-1}\pmod{F_n}$. If $F_n$ is even,
then so is $F_{n+1}+F_{n-1}$.
Similarly
$$F_{n+1}^3-F_{n-1}^3=F_n(F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2)$$
and
$$F_{n+1}^2+F... | {
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Prove $\lim_{x\to 4}\sqrt{x^3}=8$ $\lim_{x\to 4}\sqrt{x^3}=8$
My attempt:
Given $\epsilon>0$, choose $\delta=\min(1, \frac{\epsilon}{61})$
$|\sqrt{x^3}-8|=|\frac{(x-4)(x^2+4x+16)}{\sqrt{x^3}+8}|=|x-4||x^2+4x+16||\frac{1}{\sqrt{x^3}+8}|$
If $|x-4|<1$ then $-5<x<5$
So, $|x^2+4x+16|<61$
And $|\frac{1}{\sqrt{x^3}+8}|<1$
So... | The proof is good. You could avoid some steps by factoring
$$
\sqrt{x^3}-8=(\sqrt{x}-2)(x+2\sqrt{x}+4)
$$
and observe that, if $|x-4|<1$, then $|x+2\sqrt{x}+4|<5+2\sqrt{5}+4<15$ and you just need to take $|\sqrt{x}-2|<\frac{\varepsilon}{15}$; since
$$
|\sqrt{x}-2|=\left|\frac{x-4}{\sqrt{x}+2}\right|<|x-4|
$$
when $x>3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+... | Answer in the image. Hopefully it is legible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Real and imaginary part of $\tan(a+bi)$ When dealing with complex trigonometric functions, it is quite natural to ask how the real/imaginary part of $\tan(a+bi)$ can be expressed using $a$ and $b$.
Of course, since $\tan z$ and $\tanh z$ are tightly linked for complex variables, we could derive the real/imaginary part ... | We want to express $\tan(a+bi)$ in the form
$$\tan(a+bi)=A(a,b)+B(a,b)i,$$
the two functions $A(a,b)$ and $B(a,b)$ are what we are looking for.
We have
\begin{align*}
\tan(a+bi)&=\frac{\sin(a+bi)}{\cos(a+bi)}\\
&\overset{(1)}=\frac{\sin a\cos(bi)+\cos(a)\sin(bi)}{\cos a\cos(bi)-\sin a\sin(bi)}\\
&\overset{(2)}=\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Probability that the circumference of circle is larger than $\pi$
The joint probability density function of the random variables $X$ and
$Y$ is given by $f(x,y)=x+y$ for $0 \leq x,y \leq 1$ and $f(x,y)=0$
otherwise. Consider the circle centered at the origin and passing
through the point $(X,Y)$. What is the probabili... | \begin{align}
P(X^2 + Y^2 \le \frac14) &=\int_0^{\frac12} \int_0^{\frac{\pi}2}r(\cos(\theta)+\sin(\theta) \cdot r \, \, dr d\theta \\
&= \int_0^\frac12r^2 \, dr \int_0^\frac{\pi}2 \cos(\theta) + \sin(\theta) \, d\theta\\
&=\frac13\left( \frac12\right)^3 [ \sin(\theta) - \cos(\theta)]_0^\frac{\pi}2 \\
&=\frac13\left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$ For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$
Since the power of the right hand side is 4, $p(x)$ has to be 2.
So I assumed a solution of: $p(x)=ax^2+bx+c$ and then i put it in $p(p(x))+p(x)... | The polynomials
$$Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$$
and
$$ax^5+bx^4+cx^3+dx^2+ex+f$$
are equal for all $x$ if and only if
$$A=a\\B=b\\C=c\\D=d\\E=e\\F=f$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\bi... | Start off with the identity $\sin^2(x)+\cos^2(x)=1$. Rearrange this to find $\sin(x)=\pm\sqrt{1-\cos^2(x)}$. Substitute this into your equation and then rearrange to find $\sqrt{1-\cos^2(x)}$. You should be able to take it from there.
Note that it's not an identity that $\cos^2(x)=\sin(x)+\sin^3(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Error in calculation of number of ways to put $20$ identical balls in $4$ labelled boxes if each box contains at most $18$ balls The number of ways to put $20$ identical balls in $4$ labelled boxes in such a way that each box contains at most $18$ balls is,
$$(a)~~\binom{24}{4}-16~~(b)~~\binom{24}{4}-10~~(c)~~\binom{... | We need the coefficient of $t^{20}$ in the expansion $(1+t+\cdots +t^{18})^4$.
We have
\begin{align*}
(1+t+\cdots +t^{18})^4 &= \left(\dfrac{1-t^{19}}{1-t}\right)^4 \\
&= (1-t^{19})^4(1-t)^{-4} \\
&= (1-4t^{19} + \cdots )\sum_{n=0}^\infty \binom{n+3}{n}t^n
\end{align*}
and hence equals $\binom{23}{20} - 4\binom{4}{1} =... | {
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"url": "https://math.stackexchange.com/questions/2993718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To prove: $\frac {\log (1+x)}{(1+x)} = \sum_{k=1}^∞ (-1)^{k+1}H_kx^k$ For the open interval $(-1,1)$ we have:
$$\log (1+x) = x - \frac {1}{2}.x^2 + \frac {1}{3}.x^3 - ……$$
$$(1+x)^{-1} = 1 - x + x^2 - x^3 + ……$$
Multiplying the two series and collecting together the coefficients of like powers of $x$ we have:
$$\frac {... | $$
\\\frac{\log{(1+x)}}{1+x}=\sum_{n=1}^{+\infty}{\frac{(-1)^{n+1}}{n}x^n}\sum_{m=0}^{+\infty}{(-1)^{m}x^m}=
\\=\sum_{n=1}^{+\infty}\sum_{m=0}^{+\infty}{\frac{(-1)^{n+m+1}}{n}x^{n+m}}=
\\=\sum_{n=1}^{+\infty}(-1)^{n+1}H_nx^n
\\k={n+m}=>n=1,\;2,\;3,\dots k=>
\\H_k=\sum_{n=1}^{k}{\frac{1}{n}}
$$
Q.E.D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The $1997$ IIT JEE problem
Let $S$ be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$
This problem appeared in IIT JEE $1997$ (re-exam).I really do not ... | $a^2 = p^2 + s^2$
$ b^2 = (1 – p)^2 + q^2$
$c^2 = (1 – q)^2 + (1 – r)^2$
$ d^2 = r^2 + (1 – s)^2$
$$\implies a^2 + b^2 + c^2 + d^2 = p^2 + (1 - p)^2 + q^2 + (1 – q)^2 + r^2 + (1 – r)^2 + s^2 + (1 – s)^2$$
Where $p$, $q$, $r$, $s$ all vary in the interval $[0, 1]$.
Now consider the function
$y^2 = x^2 + (1 – x)^2, 0 ≤ x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limits - Calculating $\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule
Calculate $\displaystyle\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule.
I'm messing around with this limit. I've tried using substitution for $x^a -1$, but it d... | You can use Bernoulli's ineq (which can be proved by am-gm and by continuity argument, so minimal calculus knowledge i think) and the squeeze theorem. You can assume wlog that $a\geq 1$, or otherwise, if $0<a<1$, take $X=x^a$, so
$$\lim_{x\to 1}\frac{x^a-1}{x-1}=\lim_{X\to 1}\frac{X-1}{X^{\frac1a}-1}=\frac{1}{\lim_{X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help calculating $\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$ I need some help calculating this limit:
$$\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$$
I know it's equal to 1 but I have no idea how to get there. Can anyone give me a tip? I can't use l'Ho... | HINT
Use that
$$ \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} =\left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)\frac{ \sqrt{x + \sqrt{x}}+ \sqrt{x - \sqrt{x}} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }=$$
$$=\frac{ x + \sqrt{x}- x + \sqrt{x} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$ In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{\left(\frac{x}{1}\rig... | \begin{multline}
\int_0^\infty \frac{\sin(x)}{x^3}dx = \int_0^1 \frac{\sin(x)}{x^3}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx \\> \int_0^1 \frac{x/2}{x^3}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx = \frac{1}{2}\int_0^1 \frac{1}{x^2}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx = \infty
\end{multline}
The integral diverges.
| {
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"url": "https://math.stackexchange.com/questions/3000733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integer solutions of a variable coefficient polynomial I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 \leq a \leq b-1$
I am looking for integer solutions to this equation. Short of wr... | Say $b \ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 \mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2\mid b(b^2+2)$$
If $b$ is odd then $\gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2\mid b^2+2 \implies 2b^2+b+2\leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Definition of $\binom{\frac{1}{2}}{1}$? How interpret the combination
$\binom{\frac{1}{2}}{n}$ when $n$ is a positive integer ?
| The definition is
$$
\binom{x}{n}=\frac{x(x-1)\dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
\beg... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $n$ is an integer , find all the possible values for $(8n+6,6n+3)$ I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
| $(1)$ A euclidean sequence is $\ \overbrace{8n\!+\!6,\,6n\!+\!3,\,2n\!+\!3,\,{-}\color{#c00}6}^{\Large a_{k-1} -\, j\ a_k\ =\ a_{k+1}}\,$ so the gcd is
$$(2n\!+\!3,\,\color{#c00}{2\cdot 3}) = (2n\!+\!3,\color{#c00}2)(2n\!+\!3,\color{#c00}3) = (3,2)(2n,3) = (n,3)\qquad\qquad $$
$(2)$ A euclidean sequence is $\ 2n^2\!... | {
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"url": "https://math.stackexchange.com/questions/3005287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)=\frac{\pi^3}{32}-2G\ln2$ How to prove
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$
where $G$ is the Catalan's constant.
Attempt
For the first sum,
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\... | \begin{align}
S&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}\\
&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\
&=2\Im\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n^2}+\sum_{n=1}^\infty\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Integral of $1/(x^2 + y^2)^2$ under the parabola $y = x^2 - x - 1$ I'm trying to show that
\begin{equation}
\int \limits_{y < x^2 - x - 1} \!\!\!\! \frac{1}{(x^2 + y^2)^2}\, dxdy = \pi,
\tag{1}
\end{equation}
and more generally that
\begin{equation}
\int\limits_{y < ax^2 + bx + c} \!\!\!\! \frac{1}{(x^2 + y^2)^2}\, ... | We can factor the parabola into the form $y=a(x-A)(x-B)$. If $A=B$ or $A$ and $B$ are not real numbers, out integral will diverge. If $AB<0$, then polar substitution converts the integral into the form of $\iint\frac{dr}{r^3}d\theta$. The limits of integration are from $r=0$ to $\frac{\sin(\theta)-b\cos(\theta)+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that 2 heads do not come consecutively.
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?
Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{H... | Given that a fair coin is tossed $10$ times. So, the possible outcomes are $2^{10}$
Now to find the probability that no two heads appear consecutively, then the possible outcomes are
$$\begin{matrix} Heads & Tails & Total \\ 0 & 10 & \dbinom{11}{0} \\ 1 & 9 & \dbinom{10}{1} \\ 2 & 8 & \dbinom{9}{2} \\ 3 & 7 & \dbinom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{x\to\pi/2}\left(\frac{1-\sin x}{(\pi-2x)^4}\right)(\cos x)(8x^3 - \pi^3)$ using algebra of limits Let $$
F(x) = \left(\frac{1-\sin x}{(\pi-2x)^4}\right)(\cos x)(8x^3 - \pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $\pi/2$.
How can we find the limit using algebra of limits?
The limit of $\dfrac... | Make life simpler using $x=y+\frac \pi 2$ which makes
$$\lim_{x\to\frac \pi2}\left(\frac{1-\sin (x)}{(\pi-2x)^4}\right)\cos (x)(8x^3 - \pi^3)=\lim_{y\to 0}\frac{\left(4 y^2+6 \pi y+3 \pi ^2\right) \sin (y) (\cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$\frac{\left(4 y^2+6 \pi y+3 \pi ^... | {
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"url": "https://math.stackexchange.com/questions/3013994",
"timestamp": "2023-03-29T00:00:00",
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Calculate $\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$ $$\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$$ should be calculated using complex numbers I think, the Wolfram answer is :
$ \frac{1}{3} (e^x + 2 e^{-x/2} \cos(\frac{\sqrt{3}x}{2})) $
How to approach this problem?
| We have that by $f(x)=\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$
$$f'(x)=\frac{d}{dx}\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}$$
$$f''(x)=\frac{d}{dx}\sum\limits_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line ... | Your method is also really good. Consider the two points $A(-1,2)$ and $B(2,5)$. We need to find the point $(x,x+2)$ on the line $y=x+2$ for which the sum $AD+BD$ is minimum (see the graph below).
$\hspace{3cm}$
You reflect the point $A$ over the line $y=x+2$ to find the point $C(0,1)$. The line through $C$ and $B$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros? Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros?
\begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \\
&= -(1-x+x^2)+x^3(1-x+x^2)\\
&=(x... | The derivative has a single real root, namely $1$. The other two factors never vanish on $\mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-\infty$ and $\infty$ are both $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the complex roots of $x^3-1$? What are the complex roots of $x^3-1$?
Work I've done so far:
I've set $x = a + bi$. Since $x^3-1=0$, I set $x^3 = (a+bi)^3=1$.
This gives me the following:
(1) $(-ab^2 + a^3) + (2ab^2 + 2a^2b + a^2b - b^3)i$
Which means that I set $(-ab^2 + a^3) = a(a^2-b^2)= 1$ which is also equ... | You can just use difference of cubes, which gets the answer much more quickly:
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
Applying it here, you get
$$x^3-1 = (x-1)(x^2+x+1) = 0$$
The first factor obviously gives the real root of $x = 1$, so solve for the second factor. It should be pretty straightforward.
As for your error, you ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding magnitude of a complex number
$$z = \dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
My attempt:
$$\dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = \dfrac{4+12i}{20} = \dfrac{4}{20}+\dfrac{12}{20}i = \dfrac{1}{5} + \dfrac{3}{5}i$$
Now taking its magnitude and we have that
$$|z| = \sqrt{\biggr (\dfrac 1 5 \biggr ) ^2 +\biggr (\dfrac 3 5... | It is better to use $$\left|\dfrac ab\right|=\dfrac{|a|}{|b|}$$
$|2+2i|=\sqrt{2^2+2^2}=2\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
if such three condition find the $\sin{\angle OPA}$ Convex quadrilateral $ABCD$,and circumcenter is $O$,if Point $P$ lie on sides $AD$,and such
$$\dfrac{AP}{PD}=\dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=\dfrac{3}{2}CD$$find $\sin{\angle OPA}$
I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get v... | I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-\sqrt{1 - a^2}, -a)$ and $D(\sqrt{1 - a^2}, -a)$
Using the ratio formula $P(\frac{3\sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = \frac{3}{13}\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\... | (i)
$$(\sec^2x + \tan^2x)(\csc^2x + \cot^2x)$$
=>
$$(\sec^2x\csc^2x + \tan^2x\csc^2x + \sec^2x\cot^2x + \tan^2x\cot^2x)$$
=>
$$(\sec^2x\csc^2x+ \frac{\sin^2x}{\cos^2x\sin^2x} + \frac{\cos^2x}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{1}{\cos^2x} + \frac{1}{\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{\sin^2x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
find closed form for following double integral containing radicals question:
To prove :
$\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}=\dfrac{3\left(\Gamma{\dfrac{1}{3}}\right)^6}{2^{\dfrac{17}{3}}\pi^2}$
i tried attempting it by transforming into p... | This double integral is first transformed into a single integral of a complete elliptic integral:
\begin{align}
I&=\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}\\
&\stackrel{x=\cos t}{=}\int_0^1\frac{dy}{\sqrt{1-y^2}}\int_{0}^{\pi/2}\frac{dt}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving $E[\max(X^2,Y^2)]\le 1+\sqrt{1-\rho^2}$ The question is
Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $\rho$. Show that $E[\max(X^2,Y^2)]\le 1+\sqrt{1-\rho^2}$.
My attempt:
$$E[\max(X^2,Y^2)]=
E\left(\frac{X^2+Y^2-|X^2-Y^2|}{2}\right)=
1-\frac{E|X^2-Y^2|}{2}$... | Actually, the correct expression is
$$\max(X^2,Y^2)=\frac{1}{2}\left[X^2+Y^2\color{red}+|X^2-Y^2|\right]\tag{1}$$
Now simply note that
\begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\\&=E\left[\sqrt{(X+Y)^2(X-Y)^2}\right]
\\&\le \sqrt{E\left[(X+Y)^2(X-Y)^2\right]}\tag{2}
\\&\le\sqrt{E\,[(X+Y)^2]E\,[(X-Y)^2]}\tag{3}
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using Vieta's formula to evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}{x_1}$ Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,
$$x_1 + x_2 + x_3 = -\frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = \frac{c}{a},$$
$$x_1x_2x_3 = -\frac{d}{a},$$
How would one evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}... | Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= \frac{r}{s}+\frac{s}{t}+\frac{t}{r}$ and $q=\frac{s}{r}+\frac{t}{s}+\frac{r}{t}$. Then,
$$p+q=\frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
$$pq=3+\frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
That i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The li... | We have that
$$\frac{\sqrt{(n-1)!}}{\prod_{k=1}^n\big(1+\sqrt{k}\big)}=\frac1{\sqrt n}\prod_{k=1}^n\frac{\sqrt{k}}{1+\sqrt{k}}\le \frac1{\sqrt n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
prove there exist postive integers $a,b$ such $p^2|a^2+ab+b^2$ Problem 1: Let prime $p\equiv 1\pmod 3$.show that:there exist postive integers $a\le b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triple... | Alternate solution via Thue's Lemma
Proposition. Let $p\equiv 1\pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1\equiv 0 \pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $\pmod p$, hence
$$
u^2 \equiv -3 \pmod p
$$
for some integer $u$. Setting $u\equiv 2c+1\pmod p$, this becomes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it possibile to obtain the sum of the following series without using hypergeometric functions? I know that the following series:
$$
\sum_{n=1}^{+\infty}\frac{ (n!)^2}{(2n)!}
$$
converges. If I plug it in Wolphram Alpha, I can see that its sum is
$$
\frac{1}{27} \left(9 + 2 \sqrt{3} \pi\right).
$$
Is it possibile to... | I started from the fact that
$\Gamma(n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt{\pi}$
}
Divide both side by $n!$ and express $\frac{(n!)^2}{(2n)!}$
We get
$\frac{(n!)^2}{(2n)!}=\frac{n!}{4^n \Gamma(n+\frac{1}{2})}\sqrt\pi=\frac{\Gamma(n+1)}{4^n \Gamma(n+\frac{1}{2})}\Gamma(\frac{1}{2})$
Introduce $\beta$ function and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
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