Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Does this vector lie in the vector subspace $U$?
Given are the vectors $v_{1}=\begin{pmatrix} 1\\ 3\\ 5
\end{pmatrix}, v_{2}=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix},
v_{3}=\begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}$ from
$\mathbb{R}^{3}$.
The vector subspace $U = \text{span}\left\{v_{1},v_{2},v_{3}\right\}$
Does ... | Equivalently, we want a+ 4b+ 6c= 2, 3a+ 5b+ 4c= 4, and 5a+ 6b+ 2c= 5. Subtract the second equation from twice the third: 7a+ 7b= 6. Subtract the first equation from three times the third: 14a+ 14b= 13. Multiplying 7a+ 7b= 6 by two, 14a+ 14b= 12. a and b cannot satisfy both 14a+ 14b= 13 and 14a+ 14b= 12. There are n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve that logarithmic inequality? $$\log_{1\over31} (4x-5)^2 > \log_{1\over31} (5x-7)^2$$
$\begin{cases}
x\neq {7\over 5} \\
x\neq {5\over 4} \\
(4x-5)^2<(5x-7)^2
\end{cases}$
$(4x-5)^2-(5x-7)^2<0, \quad (4x-5-(5x-7))(4x-5+5x-7)<0, \quad (2-x)(x-{4\over3})<0, \quad x \in \left(-\infty; {4\over3}\right) \cup \... | $\log_b$ with basis less than $1$ is decreasing. Therefore the equality is equivalent to
$$(4x-5)^2>(5x-7)^2\iff 9x^2-30x+24=3(3x^2-10x+8)>0.$$
Use the Rational root theorem to find $2$ is a root, hence by Vieta's relations, $4/3$ is the other root. The solutions are the $x$ outide of the interval of the roots:
$$(-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\... | HINT:
Like Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$,
$$\sum_{k=1}^n\dfrac{a^{k+1}}{k+1}\binom nk=\dfrac1{n+1}\sum_{k=1}^n\binom{n+1}{k+1}a^{k+1}$$
Now $$\sum_{k=-1}^n\binom{n+1}{k+1}a^{k+1}=(a+1)^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Five Consecutive Integers divisible by a square greater than 1 I have the task of finding 5 consecutive
integers of the
form {x, x+1, x+2, x+3, x+4, x+5}
where each number in the sequence
is divisible by a square k greater than 1..
I tried to write a simple JAVA program
to find a sequence of that form by checking
i... | Chinese remainder theorem:
$$\begin{align}x&\equiv 0\pmod{4}\\
x&\equiv -1\pmod{9}\\
x&\equiv -2\pmod{25}\\
x&\equiv -3\pmod{49}\\
x&\equiv -5\pmod{121}
\end{align}$$
That's gonna be ugly.
If you only need five, then you can ignore the last line and you get $$x\equiv 29348\pmod{4\cdot 9\cdot 24\cdot 49}.$$ Then $x,x+4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Why is there another transformation matrix for the bases of the image and of the preimage of this mapping? There is the following Matrix:
\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix}
At a point it is needed to calculate the determinant of the matrix. In the official solution it is written:
$det\begin{pmatrix... | Substracting the first column from the second and third, you get
$\begin{vmatrix}
1 & 1&1 \\
a&b &c \\
a^{2} & b^{2} &c^{2}
\end{vmatrix}=\begin{vmatrix}
1 & 0&0 \\
a&b-a &c-a \\
a^{2} & b^{2}-a^{2} &c^{2} -a^{2}
\end{vmatrix}=$
$=(b-a)\cdot (c-a)\begin{vmatrix}
1 & 0&0 \\
a&1 &1 \\
a^{2} & b+a &c+a
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the number $\;'c'\;$ that satisfies the conclusion of Rolle's theorem? I have a problem
What is the number $\;'c'\;$ that satisfies the conclusion of Rolle's theorem for the function $$f(x)=(x^2-1)(x-2)\;\;\text{in}\;(1,\;2]$$
I've tried$$f'(x)= 3x^2-4x-1$$
we know, to fiind 'c' $$f'(x)=0$$
So, $$\Rightarrow 3... | Solve quadratic equation correctly
$$3x^2-4x-1=0\\\Rightarrow \dfrac{4\pm\sqrt{16-4(3)(-1)}}{2(3)}\\\Rightarrow x=\dfrac {2\pm \sqrt 7}{3}\\\Rightarrow x=\dfrac {2+ \sqrt 7}{3}\;\text{and}\; x=\dfrac {2- \sqrt 7}{3}\\x=\dfrac {2+ \sqrt 7}{3}=1.55\in (1,\;2]\\ x=\dfrac {2- \sqrt 7}{3}=-.215\notin(1,\;2]$$
Hence,
$\; c=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{j=0}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k} = 0$ when $n+k$ is odd An integral led me to a power series with these coefficients:
$$a_k = \sum_{j=k}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k}$$
I strongly suspect that the series should have $a_k = 0$ w... | Suppose we seek to verify that
$$a_{n,k} = \sum_{j=k}^n \left(-\frac{1}{2}\right)^j
{n\choose j} {n+j\choose j} {j\choose k}$$
is zero when $n+k$ is odd. We have
$${n+j\choose j} {j\choose k}
= \frac{(n+j)!}{n! k! (j-k)!}
= {n+k\choose k} {n+j\choose n+k}$$
and obtain for the sum
$${n+k\choose k} \sum_{j=k}^n \left(-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the value of the $X$?
With the details, what could be the value of X?
| Let $d(x, y)$ be the distance of two vectors $x, y \in \mathbb{R}^2$. Let $p$ be the point where the two triangles touch. We aim to find $||p||$. $d((15/2, 0), p) = 15/2$ and $d((0, 10), p) = 5$. Let $p = (p_1, p_2)$. Then
\begin{align*}
\sqrt{(15/2 - p_1)^2 + p_2^2} = 15/2, \ \sqrt{p_1^2 + (10-p_2)^2} = 5
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Put identical coins into identical boxes In how many ways can I put 200 (identical) coins into 3 identical boxes?
I started count it, and I saw that the possibilities goes this way:
$200, 0, 0$
$199, 1, 0$
$198, 2, 0$
$198, 1, 1$
$197, 3, 0$
$197, 2, 1$
ect.
which mean I can count it this way: $2 \cdot 1 + 2 \cdot 2 +... | See OEIS sequence A001399. You are asking for the value of $a_{200}$ where $a_n$ is the number of partitions of the number $n$ into at most $3$ parts, or equivalently, the number of partitions of $n$ into parts of size at most $3$, that is, the number of ways we can write $n$ as a sum of $1$s, $2$s, and $3$s without re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove that $(ab+ac+bc)\sum\limits_{cyc}\frac{1}{(a-7b)^2}\geq\frac{1}{4}$
Let $a$, $b$ and $c$ be non-negative numbers such that $\prod\limits_{cyc}(a-7b)\neq0$. Prove that:
$$(ab+ac+bc)\left(\frac{1}{(a-7b)^2}+\frac{1}{(b-7c)^2}+\frac{1}{(c-7a)^2}\right)\geq\frac{1}{4}$$
I think this inequality is very interesting... | Hints:
1)Put : $A=a$,
$AB=b$,
$AC=c$
With this following substitution you can eliminate a variable
We obtain $$(B+C+BC)(\frac{1}{(1-7B)^2}+\frac{1}{(B-7C)^2}+\frac{1}{(C-7)^2})$$
2)Try to prove this :
$(x+\alpha+\frac{1}{x+\beta}+\frac{x+\alpha}{x+\beta})(\frac{1}{(1-7(x+\alpha))^2}+\frac{1}{((x+\alpha)-7\frac{1}{x+\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Consider the family of functions of domain $\mathbb{R}\setminus \{a\}$, $\frac{2x^2+x-3}{x-a}$.Find a so that there are no vertical asymptotes
Consider the family of functions of domain $\mathbb{R}\setminus
\{a\}$ defined by $f(x) = \frac{2x^2+x-3}{x-a}$.
Determine $a$ so that $f$ has no vertical
asymptotes.
I kn... | Note that $$f(x) = \frac{2x^2+x-3}{x-a} = \frac{(2x+3)(x-1)}{x-a} = \frac{2(x+1.5)(x-1)}{x-a}.$$ Thus if $a=-1.5$ or $a=1$, then $f(x)$ simplifies to a linear polynomial over the domain $\mathbb{R} \setminus \{a\}$, and linear polynomials do not have vertical asymptotes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\cot(\pi/14)-4\sin(\pi/7)=\sqrt7$ Show that $\cot(\pi/14)-4\sin(\pi/7)=\sqrt7$.
This problem is from G.M. 10/2016 and I can't solve it. I tried with an isosceles triangle with angles $3\pi/7, 3\pi/7$ and $\pi/7$ and I tried to find a relation between the sides of the triangle but I couldn't find anything. I ... | Let $a=\frac{\pi}{7},c=\cos a$.
This answer uses that $x=c$ is a root of
$$8x^3-4x^2-4x+1$$
The proof is written at the end of this answer.
Multiplying the both sides of
$$8c^3-4c^2-4c+1=0$$
by $2$ gives
$$16c^3-8c^2-8c+2=0,$$
i.e.
$$(16c^2-24c+9)+(16c^3-24c^2+9c)-7+7c=0,$$
i.e.
$$(1+c)(16c^2-24c+9)=7(1-c)$$
Multiplyin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Using the simmetry of the function $t \rightarrow 1/t $ to change limits of an integral. I was looking at this very famous answer and unfortunately I could not get through even the first step:
\begin{align}
& 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty}... | In the integral
$$\int \limits _1 ^\infty \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \Bbb dt$$
make the change $u = \frac 1 t$. This will turn it into
$$\int \limits _1 ^0 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}?$ True or false?
If $x \geq 0, y \geq 0, z \geq 0 $ and $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}.$
I want to know if there is a way to demonstrate this conditional inequality.
I know I can make a connection with two properties known in a ... | Let $(a,b,c)=(\lambda x,\lambda y,\lambda z)$, where $\lambda(x+y+z)=\frac32$. Then $a+b+c=\frac32$, and we have
$$a^2+b^2+c^2+2abc=\lambda^2(x^2+y^2+z^2+2\lambda xyz),$$
which is increasing in $\lambda$. Hence to show $\lambda\geq1$ it suffices to prove
$$a+b+c=\frac32\implies a^2+b^2+c^2+2abc\geq1.$$
We now use the n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the closed form for a recurrence relation I'm having trouble finding a closed form for a geometric recurrence relation where the term being recursively multiplied is of the form (x+a) instead of just (x).
Here's the recursive sequence:
$a_{n} = 4a_{n-1} + 5$ for $n \geq 1$ with the initial condition $a_{0} = 2... | Finish the distributing the operators:
$$\begin{array} {rll}
%
a_0 &= 2 \\
%
a_1 &= 4\cdot 2 + 5 \\
%
a_2 &= 4\cdot (4\cdot 2 + 5) + 5 &= 4^2\cdot 2 + 4\cdot 5 + 5 \\
%
a_3 &= 4 \cdot (4^2\cdot 2 + 4\cdot 5 + 5) + 5 &= 4^3 \cdot 2 + 4^2\cdot 5 + 4 \cdot 5 + 5 \\
%
a_4 &= 4 \cdot (4^3 \cdot 2 + 4^2\cdot 5 + 4 \cdot 5 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove that: $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a^2+3bc}}+\frac{b}{\sqrt{b^2+3ac}}+\frac{c}{\sqrt{c^2+3ab}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$$
I tried Cauchy-Schwarz:
$$\left(\sum\limits_{cyc}\frac{a}... | Using the substitutions $(a, b, c) \to (a^2, b^2, c^2)$, the inequality becomes
$$\sum_{\mathrm{cyc}}\frac{a^2}{\sqrt{a^4+3b^2c^2}} \le \frac{9(a^4+b^4+c^4)}{2(a^2+b^2+c^2)^2}.$$
Using the Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
\mathrm{LHS}^2 &= \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the gcd is 1. I need to show that for $x$ odd the $\gcd\bigg(x,\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$. Im trying it doing it pairwise. I already show that $\gcd\bigg(\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$, but I dont have an idea how to show that $\gcd\bigg(x,\dfrac{x^2-1}{2}\bigg)=1$ and $\gcd\bigg(... | If $p|x$ then $p|x^2$ and $p\not |x^2 \pm 1$ so $p\not|\frac{x^2 \pm 1}k$.
So $x$ and $\frac{x^2 \pm 1}2$ have no prime factors in common. So $\gcd(x, \frac{x^2 -1}2) = \gcd(x,\frac{x^2 +1}2) = 1$.
And $\gcd(\frac{x^2-1}2,\frac{x^2 + 1}2) = \gcd(\frac{x^2 -1}2, \frac{x^2 +1}2 - \frac{x^2-1}2) = \gcd(\frac{x^2-1}2, 1) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$? Let
$$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$
Find the maximum of the value of
$$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$
where
$$x_{7}=x_{1},\quad x_{8}=x_{2},\quad x_{9}=x_{3}\,.$... | Let $x_1=a,\ x_2=b,\ x_3=c,\ x_4=d,\ x_5=e,\ x_6=f.$
Objective function is
$$Z(a,b,c,d,e,f) = abcd + bcde + cdef + defa + efab + fabc.$$
Let maximize $Z(a,b,c,d,e,f)$ using Lagrange mulptiplyers method for the function
$$F(a,b,c,d,e,f,\lambda) = abcd + bcde + cdef + defa + efab + fabc + \lambda(1-a-b-c-d-e-f).$$
Equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Surface area of a sphere with integration of disks Why it is not correct to say that the surface area of a sphere is:
$$
2 \int_{0}^{R} 2\pi r \text{ } dr
$$
In my mind we are summing up the perimeters of disks from $r=0$ to $r=R$, so by 1 integration, we would have $\frac{1}{2}$ of the surface area of the sphere.
I kn... | $$\text{Area will be}~~: 2 \int_{0}^{R} 2\pi x ~ ds$$
Where $ds$ is width of strip bounded by circles of radius $x$ and $x+dx$ situated at height $y$. Also $ds \neq dr$ it's tilted in $y$ direction too. Only horizontal projection of $ds$ is $dr$.What you have done is valid for Disk See image below.
$$(ds)^2=(dx)^2+(dy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
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Find sum binomial coefficients $$
\sum_{k>=1}^{\infty} {2N \choose N-k}k
$$
How to find this sum?
I know that the answer is $ \frac{1}{2}N{2N \choose N}$
But it is very interesting to know the solution :)
| Our first goal in dealing with this sum is to get rid of the $k$. The standard approach is to rewrite something like $\binom{n}{k} \cdot k$ as $\frac nk \binom{n-1}{k-1} \cdot k$, or $n \binom{n-1}{k-1}$. Here, the bottom index doesn't match the extra factor, but we can make it so with a little extra work:
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}$ Consider this double sums $(1)$
$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}\tag1$$
Where $H_n$ is the n-th harmonic
An attempt:
Rewrite $(1)$ as
$$\sum_{k=... | We seek to show that
$$\sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}.$$
This is
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} \frac{k+1}{n+1} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}$$
or
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1}
\sum_{j=0}^k \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Calculate the volume of the solid bounded laterally. How do i find the volume bounded below by the plane $xy$ and bounded above by $x^2+y^2+4z^2=16$ and laterally by the cylinder $x^2+y^2-4y=0$.
Since when i change to polar coordinates $x^2+y^2-4y=0$. is equal to $4sin(\theta)$.
And for the limits. $z=\frac{\sqrt{16-x... | You have some mistake in the limits of integration. Using the symmetry of the solid around the $y-z$ plane ( see the figure), we can take for $\theta$ the values between $0$ and $\frac{\pi}{2}$ and duplicate the integral, so the limits becomes:
$$
0<\theta<\frac{\pi}{2} \qquad 0<r<4\sin \theta \qquad 0<z< \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$n^{n-1}-1$ is a multiple of $k$
Find the number of integers $k$ with $2 \leq k \leq 1000$ satisfying the following property:
*
*For every positive integer $n$ relatively prime to $k$, $n^{n-1}-1$ is a multiple of $k$.
Let $k = 2^{\alpha_1}3^{\alpha_2} \cdots p_n^{\alpha_n}$ be the prime decomposition of $k$. Then... | If $\gcd(n,k)=1$ then $\gcd(n+k,k)=1$. So:
$$1\equiv (n+k)^{n+k-1}\equiv n^{n+k-1}=n^{n-1}n^k\pmod{k}$$
and hence $n^k\equiv 1\pmod{k}$ for all $n$ relatively prime to $k$.
Now if $0<n<k$ with $\gcd(n,k)=1$, then:
$$1\equiv (k-n)^{k-n-1} \equiv (-1)^{k-n-1} n^kn^{-(n-1)}n^{-2}\pmod{k}$$
But $n^k\equiv n^{-(n-1)}\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Show $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ I know there are various methods showing that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, but I want to know how to derive it from letting $t\rightarrow 0^{+}$ for the following identity:
$$\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{\pi}{t}\frac{1+e^... | Note that we have
$$\sum_{n=-\infty}^\infty \frac{1}{t^2+n^2}=\frac1{t^2}+2\sum_{n=1}^\infty\frac{1}{t^2+n^2}$$
Therefore, using $\sum_{n=-\infty}^\infty \frac{1}{t^2+n^2}=\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$, we find that
$$\sum_{n=1}^\infty\frac{1}{t^2+n^2}=\frac12\left(\frac\pi t \frac{1+e^{-2\pi t}}{1-e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Expansion of this expression Let $x$ be a real number in $\left[0,\frac{1}{2}\right].$ It is well known that
$$\frac{1}{1-x}=\sum_{n=0}^{+\infty} x^n.$$
What is the expansion or the series of the expression $(\frac{1}{1-x})^2$?
Many thanks.
| Or do by brute force multiplication and gathering terms:
$$\frac{1}{(1-x)^2}=$$
$$
(1+x+x^2+x^3+x^4+\ldots)(1+x+x^2+x^3+x^4+\ldots)\\
=(1+x+x^2+x^3+x^4+\ldots)\\
+x+x^2+x^3+x^4+\ldots\\
\quad +x^2+x^3+x^4+\ldots\\
\quad \quad +x^3+x^4+\ldots\\
=1+2x+3x^2+4x^3+\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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What is wrong with this way of solving trig equations? Let's suppose I have to find the values of $\theta$ and $\alpha$ that satisfy these equations:
*
*$\cos^3 \theta$ = $\cos \theta $
*$3\tan^3 \alpha = \tan \alpha$
on the interval $[0; 2 \pi]$.
If I try to solve, for instance, the first equation like this:
$$\co... | The equation $x^3=x$ has three solutions: you can write it as
$$
x^3-x=0
$$
so
$$
x(x-1)(x+1)=0
$$
and the roots are $0$, $1$ and $-1$.
You cannot “divide by $x$”, which is the mistake you make when you “divide by $\cos\theta$”.
Thus your equation becomes
$$
\cos\theta=0
\quad\text{or}\quad
\cos\theta=1
\quad\text{or}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that for every natural number $n$ and for every real numbers Prove that for every natural number $n$ and for every real numbers $x\neq \dfrac {k\pi}{2^t }$ $(t=0,1,....n;)$ (where $k$ is any integer)
$$\dfrac {1}{\sin 2x}+\dfrac {1}{\sin 4x}+.....+\dfrac {1}{\sin 2^n x}=\cot x-\cot 2^nx$$
| Since
\begin{align}
\frac{1}{\sin 2x}=\frac{2{\cos}^2x-\cos2x}{2\sin x\cos x}=\cot x-\cot 2x,
\end{align}
then, if we regard $x$ as $2x$
\begin{align}
\frac{1}{\sin 4x}=\frac{2{\cos}^22x-\cos 4x}{2\sin 2x \cos 2x}=\cot 2x-\cot 4x.
\end{align}
Similarly, repeat the above step until $n$,
we have
\begin{align}
LHS=\cot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^... | A differential approach follows with by use of $\delta = x D = x \frac{d}{dx}$ and integration. Since $\delta x^n = n x^n$ then
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, x^n}{n!} &= \delta^{2} \, e^{x} = x(x+1) \, e^{x}
\end{align}
now, by integration,
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, t^{n+1}}{(n+1)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Calculate the rotation area when $y = \cos(\frac{\pi x}{6})$, $\left| x \right| \leq 3$ rotates around the x-axis. Question
Calculate the rotation area when $y = \cos(\frac{\pi x}{6})$, $\left| x \right| \leq 3$ rotates around the x-axis.
Attempt at solution
$$ A = 2\pi \int_{-3}^{3} f(x) \sqrt{1+ f'(x)^2} dx$$
Which ... | $$ A = 2\pi \int_{-3}^{3} \cos \left(\frac{\pi x}{6}\right) \sqrt{ 1 + \frac{1}{36} \pi^2 \sin^2\left(\frac{\pi x}{6}\right)} dx$$
By substitution $u = \sin \left(\dfrac{\pi x}{6}\right)$ and $du = \dfrac{\pi }{6}\cos \left(\dfrac{\pi x}{6}\right)dx$, we get:
$$ A = 2\pi\dfrac{6}{\pi} \int_{-1}^{1} \sqrt{ 1 + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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One particular equation involving floor function I've got this equation to solve
$ \left\lfloor{\frac {x + 1}{x+7}}\right\rfloor = \left\lfloor{\frac {x }{x+3}}\right\rfloor $
It's not very hard to solve it if using properties of rational functions but I wonder if it can be solved using only arithmetics. Anyone got id... | If $x \ge 0$ then $0 < \frac {x+1}{x+7} < 1$ and $0 \le \frac x{x +3} < 1$ so $[\frac{x+1}{x+1}] = [\frac x{x+3}] = 0$. So all $x \ge 0$ are solutions.
If $x < 0$ is a matter of solving $[\frac {y - 1}{y-7}] = [\frac y{y- 3}]$ for $y = - x > 0$.
If $y = 1$ then $[\frac {y - 1}{y-7}] = 0 \ne [\frac y{y- 3}] = [- 1/2] =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
| $$\cos (4\theta)=2\cos^2 (2\theta)-1=2(1-2\sin^2 \theta)^2-1=8\sin^4 \theta-8\sin^2 \theta+1$$
$$4\cos 2\theta = 4(1-2\sin^2 \theta)=4-8\sin^2 \theta$$
$$\cos 4\theta-4\cos 2\theta= (8\sin^4 \theta-8\sin^2 \theta+1)-(4-8\sin^2 \theta)=8\sin^4 \theta-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Need direction for solving integral
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$
I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There... | HINT:
$$\frac{x^2}{5-x^2}=-\frac{-x^2+5-5}{5-x^2}=-1+\frac{5}{5-x^2}=-1+\frac{A}{\sqrt{5}-x}+\frac{B}{\sqrt{5}+x}\\A(\sqrt{5}+x)+B(\sqrt{5}-x)=5$$
Now when you solve for $A,B$ instead of $x$ plug in $\sin(x)$ [$\cos^2x+4=5-\sin^2 x$]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to prove $\sqrt{x(3x+y)} + \sqrt{y(3y+z)} +\sqrt{z(3z+x)} \le 2(x+y+z)$ Prove $\sqrt{x(3x+y)} + \sqrt{y(3y+z)} +\sqrt{z(3z+x)} \le 2(x+y+z)$ using triangle inequality
To answer the comment: I'm pretty stumped on the question, i know i can substitute some variables in to make it simpler but from there im clueless,... | By AM-GM
$$\sum_{cyc}\sqrt{x(3x+y)}=\sqrt{\sum_{cyc}(3x^2+xy+2\sqrt{xy(3x+y)(3y+z)}}=$$
$$=\sqrt{\sum_{cyc}\left(3x^2+xy+2\sqrt{x(3y+z)\cdot y(3x+y)}\right)}\leq$$
$$\leq\sqrt{\sum_{cyc}(3x^2+xy+x(3y+z)+y(3x+y))}=2(x+y+z).$$
Done!
The proof by triangle inequality (Minkowski).
$$2(x+y+z)=\frac{1}{4}\sum_{cyc}(4x+3x+y)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $2^p+p^2$ is prime for $p=3$ only I do know that all prime numbers larger than $3$ can be expressed as $3k + 1$ and or $3k + 2$. Plugging those in I still see no solution.
EDIT: $p$ can only be a prime number.
| So you know that all primes larger than 3 are of the form $3k + 1$ or $3k + 2$.
Do you know which of those forms $2^p$ is? Obviously $2 = 3 \times 0 + 2$. Then $4 = 3 \times 1 + 1$, $8 = 3 \times 2 + 2$, $16 = 3 \times 5 + 1$, $32 = 3 \times 10 + 2$, etc. In short, $2^n \equiv 2 \pmod 3$ if $n$ is odd, so if $p$ is a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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How to evaluate this limit??
Evaluate: $$\ f(x)= \lim_{n\rightarrow \infty}\left( \dfrac{n^n(x+n)\left( x+\dfrac{n}{2}\right)\left( x+\dfrac{n}{3}\right)... \left( x+\dfrac{n}{n}\right)}{n!(x^2+n^2)\left( x^2+\dfrac{n^2}{4}\right)\left( x^2+\dfrac{n^2}{9}\right)...\left( x^2+\dfrac{n^2}{n^2}\right)}\right)^{\dfrac{x}{... | We solve it using Riemann sum. You can write it as $$\left(\frac {(\frac {x}{n}+1)\cdots (\frac {nx}{n}+1)}{((\frac{x}{n})^2+1)\cdots ((\frac {nx}{n})^2+1)}\right)^{x/n}$$
Now take logs and use the fact of Riemann sum. Put $\dfrac {1}{n}=\mathrm dt,\dfrac {r}{n}=t$ in numerator and denominator where $r$ varies from $1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A,B <<1$ I need to evaluate the definite integral $$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for various}\ A,B \text{; with}\ A,B<<1.$$
Wolfram Alpha provides the followi... | Note that the integral of interest fails to converge if $\sqrt{A^2+B^2}\ge 1$. So, we restrict $A$ and $B$ such that $\sqrt{A^2+B^2}< 1$.
Then, we can write
$$\begin{align}
\int_0^{2\pi}\frac{1}{1+A\sin(\theta)+B\cos(\theta)}\,d\theta&=\int_0^{2\pi}\frac{1}{1+\sqrt{A^2+B^2} \cos(\theta-\arctan(A/B)}\,d\theta\\\\
&=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proving three lines are concurrent
In the triangle $ABC$ the segment $AD$ is altitude. $M$ is the midpoint of $BC$ and $N$ is the reflection of $M$ in $D$. The circumcircle of $ANM$ meets $AB$ at $P$ and $AC$ at $Q$. Show that $AN, BQ, CP$ are concurrent.
All I have tried to do is angle chasing and hopefully get the ... | One can also solve it without trigonometry:
First, show that $AP \cdot AB = QA \cdot CA$. This can be done by producing $AD$ until it intersects the circle at $L$ for the second time. Then $\angle \, LPA = \angle \, BDA = 90^{\circ}$ so triangles $LPA$ and $BDA$ are similar and this latter fact implies the ratio $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ We have to find the integration of
$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$
In this I tried to do substitution of $x=e^t$
After that got stuck .
| May be this will be a more elementary solution.
Let us start with a the substitution $x=2y$. Then
\begin{gather*}
I = \int_{0}^{\infty}\dfrac{\ln(x)}{x^2+2x+4}\, dx = \int_{0}^{\infty}2\dfrac{\ln(2y)}{4y^2+4y+4}\, dy =\\[2ex] \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln(2)}{y^2+y+1}\, dy + \dfrac{1}{2}\int_{0}^{\infty}\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
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Find the probability that $m^2-n^2$ is divisible by $4$ Two integers are $m$ , $n$ chosen at random with replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$ .
my solution : Two integers can be chosen (with replacement ) in $9\times 9=81$ ways .
$m^2-n^2 $ is divisib... | We will use the following three things:
1)
$(x^2-y^2) = (x-y)(x+y)$
2) Odd + Odd = Even
Odd - Odd = Even
Even + Even = Even
Even - Even = Even
Odd + Even = Even + Odd = Odd
Odd - Even = Even - Odd = Odd
3)
Even $\cdot$ Even = Multiple of $4$
Odd $\cdot$ Odd = Odd $\not = $ Multiple of $4$
Hence if $x,y $ both ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$ Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$
My Attempt:
$$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$$
$$\sin \dfrac {180}{10} - \sin \dfrac {3\times 180}{10}$$
$$\sin 18^\circ - \sin 54^\circ$$
Now,
Let $A=18^\circ$.
$$5A=90^... | $\cos 2A=1-2\sin^2 A$, then substitute $x=\sin A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$ I can't solve this problem. How to solve it?
The Problem is
"Find the constant term in the expansion of $
\left({{x}^{2}\mathrm{{+}}{1}}\right){\left({{x}\mathrm{{+}}\frac{1}{x}}\right)}^{\mathrm{10}}
$"
| Let $c_k$ be the coefficient of $\left(x+\dfrac{1}{x}\right)^{10}$ before $x^k$. Then
$$(x^2+1)\left(x+\dfrac{1}{x}\right)^{10} = (x^2+1)\sum\limits_{-\infty}^{+\infty} c_kx^k = \sum\limits_{-\infty}^{+\infty} c_kx^{k+2} + \sum\limits_{-\infty}^{+\infty} c_kx^{k} = \sum\limits_{-\infty}^{+\infty} (c_{k-2} + c_k)x^{k}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} ... | I think it means that the variables are reals.
By your identity we obtain $$\left(\sum_{cyc}\frac{a+b}{a-b}\right)^2=\sum_{cyc}\frac{(a+b)^2}{(a-b)^2}-2\geq0$$
Thus, $$\sum_{cyc}\frac{(a+b)^2}{(a-b)^2}\geq2$$ or
$$\sum_{cyc}\left(\frac{(a+b)^2}{(a-b)^2}-1\right)\geq-1$$ or
$$\sum_{cyc}\frac{4ab}{(a-b)^2}\geq-1$$ or
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Triangle - length of the sides - proof a, b and c are the lengths of the sides of a triangle. Prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
Let $\gamma$ be the angle between sides a and b. then:
$$a^2 + b^2 - 2ab\cos(\gamma) = c^2$$
Hence we need to prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
$$2a^2+2b^2 \ge a^2+b^2 -... | The squared length of the median through $C$ is exactly $\frac{1}{4}\left(2a^2+2b^2-c^2\right)$ and that gives $a^2+b^2\geq \frac{1}{2}c^2$ as a straightforward consequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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What the set $I$ of invertible elements in $\mathbb{Z}_{9}$? $5.2 = 1$
The set of invertible elements in $\mathbb{Z}_{9}$ is $I = \{1,2,5\}$.
Q1: Is it correct?
I was also asked to show that it is not isomorphic to $S_{3}$
How can I show that it is not isomorphic? Thank you.
| $$\begin{array}{c|c}
\times&0&1&2&3&4&5&6&7&8\\\hline
0&0&0&0&0&0&0&0&0&0\\\hline
1&0&\color{red}1&2&3&4&5&6&7&8\\\hline
2&0&2&4&6&8&\color{red}1&3&5&7\\\hline
3&0&3&6&0&3&6&0&3&6\\\hline
4&0&4&8&3&7&2&6&\color{red}1&5\\\hline
5&0&5&\color{red}1&6&2&7&3&8&4\\\hline
6&0&6&3&0&6&3&0&6&3\\\hline
7&0&7&5&3&\color{red}1&8&6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help calculating a math limit Can anyone help with this limit?
\begin{equation*}
\lim_{x \rightarrow 4}
\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16(x-4)^2}
\end{equation*}
I've tried a variable change of \begin{equation*} y=\sqrt{x} \end{equation*} but this didn't help.
| The variable change is a good idea to begin with, but I'd use $2y=\sqrt{x}$
$$
\lim_{y\to 1}
\frac{16\sqrt{4y^2-2y}-12\sqrt{2}\,y^2-4\sqrt{2}}{16(4y^2-4)^2}=
\frac{4\sqrt{2}}{256}
\lim_{y\to 1}
\frac{4\sqrt{2y^2-y}-(3y^2+1)}{(y^2-1)^2}
$$
You can do this with a Taylor expansion or by “rationalizing”: note that
$$
16(2y... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\l... | Put $t=x^2$
and
$$\frac {1-t}{1+t}=u^2$$
or
$$t=\frac{1-u^2}{1+u^2} $$
$$=\frac {2}{1+u^2}-1$$
and
$$dt=\frac{-4u}{(1+u^2)^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Decide if the series $\sum\frac{4^{n+1}}{3^{n}-2}$ converges or diverges and, if it converges, find its sum
Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.
Is this how you would show divergence attempt:
For $n \in [1,\infty), a_n = \frac{4^{n... | Assume that the sequence converges for some $n$. Then:
$$\lim_{n \to \infty} \left| \ \frac{4^{n+2}}{3^{n+1}-2} \cdot \frac{3^n-2}{4^{n+1}}\ \right| <1$$
$$\Rightarrow \lim_{n \to \infty} \left| \ \frac{4^{n+2}}{3^{n+1}} \cdot \frac{3^n}{4^{n+1}} \ \right| < 1$$
$$\Rightarrow \lim_{n \to \infty} \left| \ 4 \cdot \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\int \sqrt{a- x \over x} dx$.
$$ \int \sqrt{a- x \over x} dx$$
Substituting $u = \sqrt{a - x}$,
$$\int \sqrt{a- x \over x} dx = -2\int {u^2 \over \sqrt{a - u^2}} du$$
Now, for $\sqrt{a}\sin t = u$,
$$-2\int {a\sin^2 t \over \sqrt{a} \cos t} \ \ dt \ \ \sqrt{a }\cos t = -a\int(1 - \cos 2t) dt = a(t - \cos t... | You didn't go wrong - actually, we have that
$$\arcsin\sqrt{x \over a} = \frac{\pi}{2} - \arcsin\sqrt{a -x \over a}$$
as $$\left(\sqrt{x \over a} \right)^2 + \left(\sqrt{a -x \over a} \right)^2=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine the number of permutations of $ \ \{1,2,3,4,5,6,7,8,9,10\} \ $ that have exactly 3 numbers in their natural position Determine the number of permutations of $ \ \{1,2,3,4,5,6,7,8,9,10\} \ $ that have exactly 3 numbers in their natural position. $$ $$ Is it $ \ \ \begin{pmatrix} 10 \\ 3 \end{pmatrix} \time... | You need choose $3$ elements that stay on their positions in $\binom{10}{3} = 120$ and multiply it by the number $!7$ of permutations of special kind called derangement on remaining $7$ elements. There are many ways to compute number $!k$ (see link), particularly it is known to be equal to $\frac{k!}{e}$ rounded to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How do I compute the taylor series for $(1+x+x^2)^\frac{1}{x}$ I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..
| Note that $$f(x) = (1 + x + x^{2})^{1/x} = \exp\left(\frac{\log(1 + x + x^{2})}{x}\right) = \exp\left(\frac{\log(1 - x^{3}) - \log(1 - x)}{x}\right)$$ so that we can get the series $$\frac{\log(1 - x^{3}) - \log(1 - x)}{x} = \sum_{n = 1}^{\infty}\frac{x^{n - 1} - x^{3n - 1}}{n} = 1 + \frac{x}{2} - \frac{2x^{2}}{3} + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is $x$ in the formula? What is $x$ (assume it's integer) in $512p+ 1 = x^3$, where $p$ is a prime number.
Attempt:
$$a^3 - b^3 = (a-b) (a^2 + ab+b^2) \Longrightarrow 512p = x^3 -1 = (x-1)(x^2+x+1).$$
Here I got stuck. Am I suppose to plug in $p$ and try it one by one?
What about $16p + 1 = x^3$??
| You are on the right track. Since $512p+ 1 = x^3$, we know that $x^3$ is odd, so that $x$ must be odd.
When you have factored it into $512p = x^3 -1 = (x-1)(x^2+x+1)$, since $x$ is odd, $x^2+x+1$ is odd, so $x-1$ must be some multiple of 512. However, if $x-1$ is not exactly 512, $p$ won't be a prime. So we have $x-1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Point On a Plane Closest to a Point
*
*Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Find the point in this plane that is closest to $(0,3,-1)$.
*Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Let $\ell$ be the line cont... | Problem 1.
The equation of the plane containing the points $(−3,4,−2)$, $(1,4,0)$ and $(3,2,−1)$ is given by
\begin{align*}
\begin{vmatrix}
x & y & z & 1 \\
-3 & 4 & -2 & 1 \\
1 & 4 & 0 & 1 \\
3 & 2 & -1 & 1
\end{vmatrix}
&= 0
\end{align*}
or, $x + 2y - 2z - 9 = 0$
The vector normal to this plane is given by $(1, 2, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Why can we see if a binary number is divisible by 3 when we look at the $1$'s position Why is a binary number $z=b_n\dots b_1b_0$ divisible by $3$ iff the amount of $1$'s on the even position $b_{2n}$ minus the amount of $1$'s on odd positions $b_{2n+1}$is divisible by $3$?
So meaning $11000_b$ $(=24)$ is divisible by ... | Even powers of $2$ are congruent to $1$ modulo $3$. Odd powers of $2$ are congruent to $2$ modulo $3$.* So suppose that in the binary representation of some number $k$, there are $m$ $1$'s in the even positions, and $n$ $1$'s in the odd positions. Then,
\begin{align}
k & \equiv m+2n \mod 3 \\
& \equiv m+2n-3n \mod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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factorization of polynomials in $ \mathbb F_7$ I am not sure in finding a factorization of $f(X) = 6 x^6 +1 \in \mathbb F_7$.
I got $f(X) =6\cdot (x-1)\cdot(x+1)\cdot(x-2)\cdot(x+2)\cdot(x-3)\cdot(x+3) $
Is that correct?
| In $\Bbb{F}_7$ one has on one hand
$$\begin{align}
6x^6+1&=6(x^6-1)\\
&=6(x^2-1)(x^4+x^2+1)\\
&=6(x-1)(x+1)(x^4+x^2+1)
\end{align}$$
On the other hand
$$(x-2)(x+2)(x-3)(x+3)=(x^2-4)(x^2-2)=x^4-6x+8=x^4+x^2+1$$
So the factorisation is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Angle bisector in a right angled triangle In a right angled triangle, the legs adjacent to the right angle are equal to $a$ and $b$. Prove that the length of the bisector (of the right angle) is equal to $$\frac{a\cdot b\cdot \sqrt{2}}{a+b}.$$
While approaching this question, I was very puzzled as to how I would end up... | In a General Triangle
Using the Law of Sines, we get that side $c$ is broken up as follows by the angle bisector:
Using the Law of Cosines we get
$$
\left(\frac{ac}{a+b}\right)^2+d^2-\frac{2acd}{a+b}\cos(D)=a^2\tag{1}
$$
and
$$
\left(\frac{bc}{a+b}\right)^2+d^2+\frac{2bcd}{a+b}\cos(D)=b^2\tag{2}
$$
Multiply $(1)$ by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | It is sufficed prove that: $3(x^2 +y^2 +z^2) \geq (x+y+z)^2$
$\Leftrightarrow \frac{1}{2}(x+y-2z)^2 +\frac{3}{2}(x-y)^2 \geq 0$
Which is obvious.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 12
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Every complex number has 2 square roots - Rudin I am using Baby Rudin for self study and as many of you are aware, it is one thing to follow Rudin's calculations to verify that his proof is correct. It is quite another to work backwards to see how he came up with his proof in the first place. I find that unless I do th... | I figured it out:
$a^2 - b^2 = u$ and $2ab = v$
$b = \frac{v}{2a}$
$a^2 - \frac{v^2}{4a^2} - u = 0$
$4a^4 -4ua^2 - v^2 = 0$
By the quadratic formula:
$a^2 = \frac{4u \pm \sqrt{16^2 - 4(4)(-v^2)}}{8} = \frac{4u \pm \sqrt{16u^2 + 16v^2}}{8} = \frac{4u+4\sqrt{u^2 + v^2}}{8} = \frac{u \pm |w|}{2}$
So that $a = \sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Limit of product of mixed root indexes I am unable to recover the limit of this function.
$$
\lim_{x \to +\infty} \left (\frac{1}{\sqrt{x}} \times \sqrt[3]{x+1} \right )
$$
I have tried many ways of solving it. These are the two simplest:
*
*Variable substitution
Using $y = \sqrt[3]{x+1} $, with which I got a comm... | Multiplying and dividing by $x^{1/3}$ we get $\frac{1}{\sqrt{x}}(x+1)^{1/3}=\frac{x^{1/3}}{\sqrt{x}}(\frac{x+1}{x})^{1/3}$. The factor $(\frac{x+1}{x})^{1/3}$ tends to $1$ as $x \to \infty$. The factor $\frac{x^{1/3}}{\sqrt{x}}$ equals $x^{-1/6}$ that tends to $0$ as $x \to \infty$. So the limit you're looking for is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Four points, sine, Taylor's series I am creating a program that finds the cubic equation $y = ax^3+bx^2+cx+d$ through four points; I have already found how to calculate the coefficients, but I am also attempting to use a Taylor’s series to approximate a sine equation $A\sin(Bx+C)+D$ through roughly the same points base... | Note that the Taylor expansion you need is
$$
\sin x = \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}
= x - x^3/6 + \Theta\left(x^5\right)
$$
Hence
$$
A\sin(Bx+C) + D
\approx D + A\left[ (Bx+C) - (Bx+C)^3/6 \right]
$$
and now expand the brackets and do arithmetic. The next term in the series would be $x^5$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct... | If you divide both sides by $2^n$, you will get :
$$\forall k\ge1,\,\frac{a_k}{2^k}-\frac{a_{k-1}}{2^{k-1}}=\frac1{2^{k-1}}$$
Now, cancellation occurs when summing up ($k$ from $1$ to $n$) :
$$\frac{a_n}{2^n}-a_0=\sum_{k=1}^n\frac1{2^{k-1}}=\frac{1-\frac1{2^n}}{1-\frac12}=2-\frac1{2^{n-1}}$$
that is :
$$a_n=2^n\left(4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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The inequality $\frac{3\sqrt{3}}{2}\frac{27(A+B)(B+C)(C+A)}{8(A+B+C)^3}\ge \sin{A}+\sin{B}+\sin{C}$ In $\triangle ABC$, show that
$$\dfrac{3\sqrt{3}}{2}\dfrac{27(A+B)(B+C)(C+A)}{8(A+B+C)^3}\ge \sin{A}+\sin{B}+\sin{C}$$
My attempt: Since $A+B+C=\pi$, it suffices to show that
$$\dfrac{3\sqrt{3}}{2}\cdot \dfrac{27(\pi-A)... | Not a full solution just yet, but we can go like this:
$$\sin A + \sin B + \sin C = 4 \sin\frac{A+B}2\sin\frac{B+C}2\sin\frac{C+A}2.$$
So if we let $x= \frac{B+C}2,y=\frac{C+A}2,z=\frac{A+B}2$, we need to show that
$$xyz\ge \lambda \sin x\sin y\sin z$$
for $x,y,z\in(0,\pi/2)$, and $ x+y+z=\pi$ and some constant $\lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using th... | We can proceed as follows
\begin{align}
L&=\lim_{x\to 0}\frac{\log\cosh(x^{2}-cx)}{x^{2}}\notag\\
&=\lim_{x\to 0}\frac{\log\cosh(x^{2}-cx)}{\cosh(x^{2}-cx)-1}\cdot\frac{\cosh(x^{2}-cx)-1}{x^{2}}\notag\\
&=\lim_{x\to 0}\frac{2\sinh^{2}((x^{2}-cx)/2)}{x^{2}}\notag\\
&=\lim_{x\to 0}2\cdot\frac{\sinh^{2}((x^{2}-cx)/2)}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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Show that there are no positive integer solutions to $x^4-y^4=z^4$ Also, Show that the only solution in relatively prime positive integers to $x^4+y^4=2z^4$ is $x=y=z=1$.
Here I tried a similar contradiction proof to the proof for $x^4+y^4=z^4$, but this doesn't work the same way.
| We show that if there exist $x,y,z \in \mathbb N$ with $x^2=y^4+z^4$ then there exist $x',y',z'\in \mathbb N$ with $x'^2=y'^4+z'^4$ and $x'<x.$ This implies that, if $S=\{x\in \mathbb N:\;\exists y,z \in \mathbb N\;(x^2=y^4+z^4)\}$ is not empty, then $S$ is a non-empty subset of $\mathbb N$ with no least member, which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if: The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if:
$1$. $c=a\sqrt {1+m^2}$
$2$. $c=\pm a\sqrt {1+m^2}$
$3$. $c^2=\pm a\sqrt {1+m^2}$
$4$. $\textrm {None}$
My Attempt:
The tangent to circle $x^2+y^2=a^2$ at point $(x_1,y_1)$ is given by :
$$xx_1+yy_1=a^2$$
Now, what s... | A possible approach:
1) Calculate distance from (0,0) to line $y = mx + c$.
The perpendicular line passing through the origin is
$y = - (1/m)x$.
Point of intersection: $- (1/m)x = mx + c$.
$- x = m^2x + mc$
$x ( m^2 + 1) + mc = 0$;
$x = (-mc)/(m^2 +1)$;
Corresponding $y$:
$y = - (1/m) (-mc) /(m^2 + 1)$;
$y = c/(m^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix.
| One option is use the projector onto the $\color{red}{null}$ space
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A^{*}}\right)}} = \mathbf{I}_{4} - \mathbf{A}\mathbf{A}^{+}
$$
The target matrix and pseudoinverse are
$$
\mathbf{A} =
\left(
\begin{array}{cccc}
1 & 2 & 1 & 3 \\
1 & 3 & 2 & 4 \\
2 & 5 & 3 & 7 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\dfrac{1}{2\pi i}\int_{C}\dfrac{e^{z}}{z^{3}-1}dz = \sum_{n=0}^{\infty}\dfrac{1}{(3n+2)!}$ a) Use residue at infinity to solve the problem:
If $C$ is a circle $C(0,2)$ traversed in the counter clockwise direction, then $$\dfrac{1}{2\pi i}\int_{C}\dfrac{e^{z}}{z^{3}-1}dz = \sum_{n=0}^{\infty}\dfrac{1}{(3n+2)!}$$
b) B... | We first note that for a simple pole we have
$$\mathrm{Res}(f/g,z_0) = \frac{f(z_0)}{g'(z_0)}$$
The sum of the residues
$$\sum \mathrm{Res}(f,z) =\frac{e}{3}+\dfrac{e^{e^{2\pi i/3}}}{3e^{4\pi i/3}}+\dfrac{e^{e^{4\pi i/3}}}{3e^{8\pi i/3}}$$
Note that
$$\dfrac{e^{e^{2\pi i/3}}}{e^{4\pi i/3}}+\dfrac{e^{e^{4\pi i/3}}}{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Hermitian matrices in an arbitrary inner product I was wondering if anyone can give me any hints for how to prove the following proposition:
Proposition: Let $V = F^n$ and let $A\in M_{n\times n}(F)$. Prove that $\langle A^* x,y \rangle = \langle x, A y \rangle$.
The problem that I am having is how to prove this for ... | This is true only for $V$ with the standard inner product.
Counter-example:
Assume $V=\mathbb{R}^2$ and we define the inner product in the following way:
$$\langle x | y\rangle = x^T \begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y$$
Lets take for simplicity $x = \begin{pmatrix} 1\\0 \end{... | {
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"url": "https://math.stackexchange.com/questions/2257106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Floors Complicated Proof Problem:
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
I have my proof down below. Its a little complicated though.
This is not a duplicate of the same question on another page on math.stackexcha... | Let $n=\lfloor x \rfloor$ and $m=\lfloor y \rfloor$. Consider $4$ cases:
Case 1: $x \in [n,n+\dfrac{1}{2})$ and $y \in [m,m+\dfrac{1}{2})$
Case 2: $x \in [n+\dfrac{1}{2},n+1)$ and $y \in [m,m+\dfrac{1}{2})$
Case 3: $x \in [n,n+\dfrac{1}{2})$ and $y \in [m+\dfrac{1}{2},m+1)$
Case 4: $x \in [n+\dfrac{1}{2},n+1)$ and $y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the Differential equation $x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$ Solve the Differential equation $$x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$$
i reduced the equation as
$$x^3\frac{dy}{dx}=y^3\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right)$$ $\implies$
$$\frac{x^3}{y^3}\frac{dy}{dx}=\left(1+\sqrt{1+\left(\frac... | Well, let $\text{y}\left(x\right)=x\cdot\text{r}\left(x\right)$, which gives $\text{y}'\left(x\right)=x\cdot\text{r}'\left(x\right)+\text{r}\left(x\right)$:
$$x^3\cdot\text{y}'\left(x\right)=\text{y}\left(x\right)^3+\text{y}\left(x\right)^2\cdot\sqrt{x^2+\text{y}\left(x\right)^2}\space\Longleftrightarrow\space$$
$$x^3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Given $A^2$, find all $A$ that satisfy Find the set of all matrices $A\in M_{n\times n}(\mathbb R)$ satisfying $A^2=\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix}$
I have been given this problem. Do I just need to take $A=\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}$ and solve for the variab... | Let $$A=\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}$$ as OP suggests. Now clearly,
$$A^4=0$$ and thus the minimal polynomial of $A$ divides $x^4$. However the minimal polynomial has degree $\leq 3$ and therefore the minimal polynomial divides $x^3$. Thus we have
$$A^3=0$$ and so
$$\begin{pmatrix} a &b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that limit = 0. So the question is as in the title.
$$\lim_{(x,y,z)\to (0,0,0)}\frac{\sin(xyz)}{\sqrt{(x^2+y^2+z^2)}}$$
I decided to prove it from the definition.
so $L = 0$
$$\left\lvert\frac{\sin(xyz)}{\sqrt{(x^2+y^2+z^2)}}-0 \right \rvert \leq \frac {1}{\sqrt{(x^2+y^2+z^2)}} $$
and then trnasform it to this
... | Use $|\sin t|\le|t|$. Then
$$\left|\frac{\sin(xyz)}{\sqrt{x^2+y^2+z^2}}\right|
\le\left|\frac{xyz}{\sqrt{x^2+y^2+z^2}}\right|.$$
If we let $r=\sqrt{x^2+y^2+z^2}$ then $|xyz|\le r^3$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Interesting integral: $\int_0^1{\frac{nx^{n-1}}{x+1}}dx$ Find the value of $$\int_0^1{\frac{nx^{n-1}}{x+1}}dx.$$
I had no luck while integrating it. I also tried differentiating w.r.t n but still couldn't reach anywhere. Need help.
| We can write the integral as
$$
F(n) = \int_0^{\;1} {\frac{{nx^{n - 1} }}{{1 + x}}dx} = \int_0^{\;1} {\frac{1}{{1 + x}}dx^n } = \int_0^{\;1} {\left( {1 - \frac{x}{{1 + x}}} \right)dx^n } = 1 - \int_0^{\;1} {\frac{x}{{1 + x}}dx^n }
$$
$F(n+1)$ will be
$$
F(n + 1) = \int_0^{\;1} {\frac{{\left( {n + 1} \right)x^n }}{{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove that $x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}$
Let $p > 3$ be a prime, and let $x,y$ be integers such that $\gcd(x,y) = 1$. Prove that $$x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}.$$
I thought about expanding $(x+y)^p$ using the binomial theorem, but I didn't see how that would help because the modulus is an alg... | Not true, you can use polynomial modular arithmetic, i will give an example for what i am talking about :
when $p=3$ we have $(x+y)^3=x^3+3x^2 y+3x y^2+y^3 $, and we want to find the remainder module $x^2+x y+y^2$ , the leading expression in the bigger polynomial is $x^3$ which means we need to multiply the smaller pol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determining the first return distribution of 2-states Markov chain
A Markov chain has the transition matrix $$
P= \begin{pmatrix}
1-a & a \\
b & 1-b \\
\end{pmatrix}
$$
Find the first return distribution $f_{00}^n=\Pr\{X_1\neq 0,...,X_{n-1} \neq0,X_n=0\mid X_0=0\}$.
My attempt:
For $n=0... | Draw the state transition diagram and find out the possible transitions.
The following is the state transition diagram:
Note that $f_{00}^{(1)}=p_{00}^{(1)}=1-a$.
\begin{eqnarray*}
f_{00}^{(1)}&=&p_{00}^{(1)}=1-a\\
f_{00}^{(2)}&=&p_{01}^{(1)}p_{10}^{(1)}=a\cdot b\\
f_{00}^{(3)}&=&p_{01}^{(1)}p_{11}^{(1)}p_{10}^{(1)}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose that $\cos a=-\frac{8}{17}$ for $\pi < a < 3\pi/2$ find $\sin 2a$ Suppose that $\cos A= -\frac{8}{17}$ for $\pi < A < 3\pi/2$ find $\sin 2A$. how is this question solved and in what steps ?
| Tools
$$
\sin 2\theta = 2\cos\theta \sin \theta
\tag{1}
$$
$$
\sin \theta = \pm \sqrt{ 1 - \cos^{2} \theta }
$$
Data
Let the angle
$$
\theta = \arccos \left( -\frac{8}{17} \right)
$$
then the sine is
$$
\sin \theta = -\frac{15}{17}
$$
Solution
$$
\sin 2\theta = 2 \cos\theta \sin \theta =
2
\left( -\frac{8}{17}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $... | HINT: we have:
$$\frac{3\sin\left(\text{A}\right)}{\sin\left(\text{B}\right)}=\frac{2\cos\left(\text{B}\right)}{\cos\left(\text{A}\right)}\space\Longleftrightarrow\space3\sin\left(\text{A}\right)\cos\left(\text{A}\right)=2\cos\left(\text{B}\right)\sin\left(\text{B}\right)\space\Longleftrightarrow\space$$
$$\frac{3\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to prove this property involving the floor function? How does one prove that $$\lfloor x \rfloor = \lfloor x/2 \rfloor + \lfloor (x+1)/2 \rfloor $$
where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$?
I tried to do this by using inequalities but end up with $\lfloor x \rfloor + 1/2$ ... | I here assume that $x$ is a positive real. We then have that $x = n + r$, where $n \in \mathbb{N}$ and $r$ is a real number with $r \in [0;1)$. We have that $\lfloor x \rfloor = n$.
Either $n$ or $n+1$ is even. If $n$ is even, then $\lfloor x/2 \rfloor = \frac{n}{2}$ and $\lfloor (x+1)/2 \rfloor = \lfloor (n+1)/2 \rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Domain of $f(x)=\sqrt{\lfloor x\rfloor-1+x^2}$ I drew the number line and tested with different values, getting the correct domain $(-\infty,-\sqrt3)\cup[1,\infty)$. However, how do I solve this faster by manipulating the function?
| Let $g(x) = x^2 + \lfloor{x}\rfloor - 1$.
Then the domain of $f$ is $\{x \in \mathbb{R} \mid g(x) \ge 0\}$.
Note that for all $x \in \mathbb{R}$, we have $x-1 < \lfloor{x}\rfloor \le x$, hence for all $x \in \mathbb{R}$,
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&\ge x^2 + (x-1) - 1\\[4pt]
&=x^2 + x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the value of $A$ for the inequality if $x,y,z$ are positive reals and
$$S=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)} \le A$$ find $A$
i have taken L.C.M getting
$$S=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ dividing by $xyz$ we get
$$S=\frac{2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{xyz\le... | For $x=y=z\rightarrow0^+$ we get $\sum\limits_{cyc}\frac{x}{(x+y)(x+z)}\rightarrow+\infty$,
which says that the needed value of $A$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove Ramanujan's formula for nested cubic roots $\sqrt[3]{{a}+b\sqrt[3]{r}}$ Ramanujan found that
$$\begin{align*} & \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}\\ =&\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9}-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9}+\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9}
\end{align*}$$
for arbitrary $... | Denote
$$c_1=\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9},\>\>\>
c_2=-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9},\>\>\>
c_3=\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9}
$$
and it is straightforward to verify that
\begin{align}
c_1c_2c_3 &=- \frac{1}{9}(m-n)(2m+n)(m+2n)\\
c_1^3+c_2^3+c_3^3 &= \frac{1}{3}(m^3+6m^2n+3mn^2-n^3)\\
c_1^3c_2^3+c_2^3c_3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $ \prod_{n=1}^{80}n^{k-n} \pmod{83}$
Evaluate $\displaystyle \prod_{n=1}^{80}n^{80-n} \pmod{83}$.
Attempt:
The product is \begin{align*}\prod_{n=1}^{80}n^{80-n} &= 1^{79} \cdot 2^{78} \cdot 3^{77} \cdots 79^1 \cdot 80^0\\&\equiv -2^{78} \cdot 3^{77} \cdot 4^{77} \cdots 41^{77}\\&\equiv -2 \cdot (1 \cdot 2 \c... | For $\prod_{n=1}^{82} n^{80-n} \mod 83$, we could pair each $n$ with $83-n$:
$$ n^{80-n} (83-n)^{n-3} \equiv n^{80-n} (-n)^{n-3} \equiv (-1)^{n-3} n^{77}$$
so that
$$ \prod_{n=1}^{82} n^{80-n} \equiv \prod_{n=1}^{41} (-1)^{n-3} n^{77} \mod 83$$
$\prod_{n=1}^{41} (-1)^{n-3}$ has $20$ factors of $-1$ (one for each even $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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To prove $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}$ To prove $$P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}$$
i have written $P$ as
$$P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{... | We need to prove that $$\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}.$$
Indeed, by induction for $n=1$ we get an equality.
Now, $$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}\leq\frac{4^n}{\sqrt{3n+1}}\cdot\frac{2(2n+1)}{n+1}.$$
Thus, it remains to prove that
$$\frac{4^n}{\sqrt{3n+1}}\cdot\frac{2(2n+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =... | From where you are:
You obtained a quadratic function in $\sin(\theta)$. Perform the substitution $u =\sin(\theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$\Rightarrow u_{1,2} = \frac{- 2 \pm \sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = \frac{- 2 \pm \sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a point on an ellipsoid Find a point on the ellipsoid $x^2+4y^2+z^2=9$ where the tangent plane is perpendicular to the line with parametric equations \begin{align}x&=2+2t\\y&=1+2t\\z&=3-t\end{align}
The answer to this question is: $$\left(\frac{6\sqrt{13}}{13}, \frac{6\sqrt{13}}{13}, -\frac{3\sqrt{13}}{13}\righ... | The normal to the plane must be parallel to the given line, hence
$$(x,4y,z)=\lambda(2,2,-1).$$
Combining with the equation of the ellipsoid,
$$x^2+4y^2+z^2=9,$$
we get
$$2^2\lambda^2+4\frac1{2^2}\lambda^2+(-1)^2\lambda^2=6\lambda^2=9,$$ and solve for $\lambda$.
$$(x,y,z)=\pm\sqrt{\frac32}\left(2,\frac12,-1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$ and : $\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$then : $x+y+z =?$ let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$
and :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$$
then :
$$x+y+z =?$$
my try :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}... | It should be $x=4$, $y=6$, $z=7$ and the answer to the original question is 17.
You can use the method described here for obtaining the Engel expansion of a number.
The Engel expansion for $x$ is written $x=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\cdots$
As described in the wiki, the non-decreasing values $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Probability of a quarter final combination There are three Spanish teams, two German, one Italian, one French, and one English. What is the probability that Germany plays itself? What is the probability that Spain plays itself? What is the intersection of this probability?
The first question I think is $1/28$ because $... | We can list the teams in some order, say alphabetically. There are seven ways we can match a team with the first team on the list. This leaves six teams. There are five ways to match a team with the first team remaining on the list. This leaves four teams. There are three ways to match a team with the first team r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse element of
Given the extension $ \dfrac{\mathbb Q[x]}{\langle x^3-5\rangle}: \mathbb Q$ and $\alpha =1+x^2+\langle x^3-5\rangle \in \dfrac{\mathbb Q[x]}{\langle x^3-5 \rangle}$, find $\alpha^{-1}$.
Can somebody give a hint? or an answer? please?
| A general element of this ring looks like $a + bx + cx^2$. Suppose this were the inverse of $1+x^2$. Then:
\begin{align*}
1 &= (1+x^2)(a+bx+cx^2)\\
&= a + bx + (a+c)x^2 + bx^3 +cx^4
\end{align*}
In this ring, $x^3 = 5$ so:
\begin{align*}
1 &= a + bx + (a+c)x^2 +5b + 5cx \\
&=(a+5b) + (b+5c)x + (a+c)x^2.
\end{align*}
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $n^5+n^4+1$ is composite for $n>1.$
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.
This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was t... | Building upon OP's idea:
I tried multiplying it by $n+1$ or $n-1$ and then tried factorising it
$$
\begin{align}
(n-1)(n^5+n^4+1) = \color{red}{n^6} \color{blue}{- n^4 + n} \color{red}{- 1} &= \color{red}{(n^3-1)(n^3+1)} \color{blue}{-n(n^3-1)} \\
&= (n^3-1)(n^3-n+1) \\
&= (n-1)(n^2+n+1)(n^3-n+1)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How can we prove the equality of the following determinants? Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?
$\det(A-xI)$=\begin{vmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
... | Recall that if an $n\times n$ matrix $B$ is obtained from an $n\times n$ matrix $A$ by adding a scalar multiple of one row of $A$ to another row, then $\det(B)=\det(A)$.
In your problem the second matrix is obtained from the first by adding rows $2$ through $n$ to row $1$. So the determinants are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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PDF of $\frac{X}{1+X^2}$ in terms of the PDF of $X$ Given a r.v. $X$ with a know probability density $g$, I would like to find the probability density of $\frac{X}{1+X^2}$. Below I write my calculations:
denote
$f(X) := \frac{X}{1+X^2}$
Now $f$ is not monotone so we can't apply the known theorem, we proceed by finding... | Using the CDF
Let
$$
G(x)=\int_{-\infty}^xg(t)\,\mathrm{d}t\tag{1}
$$
be the CDF of $X$ and $Y=\frac{X}{X^2+1}$. Then the CDF of $Y$ is
$$
\begin{align}
F(\lambda)
&=P(y\lt\lambda)\\
&=P\left(\frac{x}{x^2+1}\lt\lambda\right)\\
&=\left\{\begin{array}{}
P\left(x\lt\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-P\left(x\lt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solving a rational inequality with an unknown using cases So the question is,
For what values of x is $\dfrac{1}{1 + x}> -1 $ ?
Now, one way to do it is write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator:
$\dfrac{1}{1 + x}> -1 \implie... | Shortly $${ \frac { 1 }{ 1+x } }>-1\\ \frac { 1 }{ 1+x } +1>0\\ \frac { x+2 }{ x+1 } >0\\ \frac { \left( x+1 \right) \left( x+2 \right) }{ { \left( x+1 \right) }^{ 2 } } >0\\ \left( x+1 \right) \left( x+2 \right) >0\\ x\in \left( -\infty ;-2 \right) \cup \left( -1;+\infty \right) \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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The distance from the incenter to an acute vertex of a right triangle I'm seeking an alternative proof of this result:
Given $\triangle ABC$ with right angle at $A$. Point $I$ is the intersection of the three angle lines. (That is, $I$ is the incenter of $\triangle ABC$.) Prove that
$$|CI|^2=\frac12\left(\left(\;|B... | Let $AB = c, AC = b, BC = a$, and let $IT\perp AC $ at $I$. Then, $CI$ can be written as $$\begin{split}CI^2 &= CT^2 + IT^2 \\&= \left(\frac{a+b-c}{2}\right)^2+\left(\frac{b+c-a}{2}\right)^2\\&= \frac{2a^2+2b^2+2c^2-4ac}{4}\\&=\frac{(a-c)^2+b^2}{2} \\&=\frac{(BC-AB)^2+AC^2}{2}\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$
Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{... | You assume the expression to be true for $n=m$ and then prove it is true for $n=m+1$.
Expression:
\begin{gather}
a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}\\
\\
\text{For } n = m \text{ assume true}\\
\\
a_{m+1}= \frac{3a_m}{(2m+2)(2m+3)} \\
a_{m+1}= \frac{3}{(2m+2)(2m+3)}\cdot \frac {3! \cdot 3^{m-1}} {(2m+1)!}\\
a_{m+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prob. 24, Chap. 5 in Baby Rudin: For $\alpha>1$, let $f(x) = (x+\alpha/x)/2$, $g(x) = (\alpha+x)/(1+x)$ have $\sqrt{\alpha}$ as their only fixed point Here is Prob. 24, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
The process described in part (c) of Exercise 22 can of course ... | For $f:$
Suppose $x_0>0$ and $a>0$.Let $x_{n+1}=(x_n+a/x_n)/2.$ Assume $x_0\ne \sqrt a \;$( because if $x_0=\sqrt a$ then $x_n=\sqrt a$ for all $n,$ which is a trivial case). Then we have:
(I). $x_1^2>a$ so $x_1>\sqrt a.$
(II). If $x_n^2>a$ then $x_n>x_{n+1}>\sqrt a.$ So by (I), $n\geq 1\implies x_n>x_{n+1}>\sqrt a.$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Compute $gcd\left(1714, 1814\right)$ using Euclidean Algorithm So I know the answer for this is $2$, but based on my own work, I can't get to that solution. I haven't done a gcd before where $b>a$. I thought I could just flip the numbers and use the same method but that didn't seem to work. Here's what I have so far, w... | written as a continued fraction:
$$ \frac{ 1814 }{ 1714 } = 1 + \frac{ 100 }{ 1714 } $$
$$ \frac{ 1714 }{ 100 } = 17 + \frac{ 14 }{ 100 } $$
$$ \frac{ 100 }{ 14 } = 7 + \frac{ 2 }{ 14 } $$
$$ \frac{ 14 }{ 2 } = 7 + \frac{ 0 }{ 2 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve $x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $
Find the real roots of the equation $$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $$ $$(0<a<\frac{1}{4})$$
My attempt,
$(x^2+2ax+\frac{1}{16}+a)^2=a^2+x-\frac{1}{16}$
I did an expansion which becomes $x^4+4ax^3(4a^2+2a+\frac{1}{8})x^2+(4a^2+\frac{1}{4}... | Hint
$$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2-a^2+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2=a^2-\frac{1}{16}-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2+(x+a)=a^2+x-\frac{1}{16}+\sqrt{a^2+x-\frac{1}{16}}$$
Now call:
$$y=\sqrt{a^2+x-\frac{1}{16}}\to y^2=a^2+x-\frac{1}{16}$$
and then
$$(x+a)^2+(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Can we prove the divergence of harmonic series like this? Let $S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots$.
Or $S=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots)+(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots)$.
$\implies S=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{... | Yes, we can prove the divergence of the harmonic series like that, but one must at the beginning state something along the lines of "suppose the harmonic series is convergent".
Once that assumption is introduced, it is a valid proof by contradiction. Since all terms in the harmonic series are positive, if it were conve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to show that $\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)?$ Given that:
Where $\alpha >0$
$$\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)\tag1$$
Setting... | $\displaystyle J=\int_{-\alpha}^{\alpha} \arccos\left(\dfrac{x}{\alpha}\right)\ln\left(\alpha+x\right)dx$
Perform the change of variable $y=\dfrac{x}{\alpha}$,
$\begin{align} J&=\alpha\int_{-1}^{1} \arccos\left(x\right)\ln\left(\alpha+\alpha x\right)dx\\
\end{align}$
Perform the change of variable $y=\arccos x$,
$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Limit of function calculations I must solve limit of next function:
$$\lim_{x\to \infty}\frac{2x^3+x-2}{3x^3-x^2-x+1}$$
Does my calculations are proper? If not where is my mistake?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =... | Put $y=1/x $ and compute
$$\lim_ {y\to 0^+}\frac { 2+y^2-2y^3 }{3-y -y^2 +y^3}=2/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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General Solution for a PDE I would like to know if someone could provide me the solutions for the next PDE:
\begin{align*}
f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)
\end{align*}
I empirically found the following particular solution given by $f(x,y) = \displaystyle\frac{x+y}{\sqrt{x^{2}+y^{2}}}$. Any contr... | One can use method of characteristics $\gamma = \left(x(t),y(t)\right)$:
$$f_x + f_y = \dfrac{y-x}{x^2+y^2}\,f
$$
$$
x' = y'=1, \qquad \frac{df}{dt} = \dfrac{y-x}{x^2+y^2}\,f
$$
Solving system of ODEs above get
$$
x = t+a,\quad y = t+b = x + c, \implies c = y-x \quad - \quad \operatorname{const}\\
f' = \frac{f}{2t^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.