Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-... | From $$x-3\geq 0$$ follows:
$$x-1\geq 2$$
$$(x-1)^2\geq 4$$
$$x^2-2x+1\geq 4$$
$$x^2-2x\geq 3$$
$$x(x-2)\geq 3$$ so $$1\geq \frac{3}{x(x-2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find 'ordinary generating function': 1, 0, 2, 0, 3, 0, 4, 0, 5.... This was on my midterm yesterday, but I couldn't solve it.
Find an ordinary generating function: 1, 0, 2, 0, 3, 0, 4, 0, 5....
My answer was $(\frac{1}{1-x^2})^2$ because 1, 2, 3, 4, 5, 6, 7, ... is $\frac{1}{1-x^2}$. (now I feel like it just doesn'... |
We obtain
\begin{align*}
\color{blue}{1+2x^2+3x^4+\cdots}&=\sum_{n=0}^\infty (n+1)x^{2n}\\
&=\frac{1}{2x}\frac{d}{dx}\left(\sum_{n=0}^\infty \left(x^2\right)^{n+1}\right)\\
&=\frac{1}{2x}\frac{d}{dx}\left(\sum_{n=1}^\infty \left(x^2\right)^n\right)\\
&=\frac{1}{2x}\frac{d}{dx}\left(\frac{1}{1-x^2}-1\right)\\
&\,\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{... | Hint: $(2\cos \theta)^3-3(2\cos\theta)+1=2\cos3\theta+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A magic rectangle with the greatest size - original question I filled in a $3\times k$ rectangle with non negativ integers, such that the sum of the three numbers in each column is the same number $n$, and in each row all the numbers are different.
Find the maximum value of $k$.
If you try for $n=0,1,2,3,4,5,6$, you ge... | Well, here's a partial answer for you: the function $k(n)$ is obviously non-decreasing, the upper bound is $k={2\over3}n+1$, and this bound is sharp (that is, exact) when $n$ is divisible by 3.
The bound can be established as follows. The sum of all numbers in the table is obviously $k\cdot n$. On the other hand, if we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding area of a triangle using equation of a circle **Ignore notes I made they are stupid
Without a calculator
Question reads:
The diagram shows a sketch of the circle with equation $x^2 + y^2 = 5$.
The $y$-coordinate of point $A$ is $-1$.
The tangent to the circle at $A$ crosses the axes at $B$ and $C$ as shown... | Just going off the question alone and assuming the diagram is not drawn to scale, have some good information to off of.
Knowing the $y$ coordinate is $-1$, we can plug that into the equation of the circle to get the $x$ coordinate:
$x^2 + (-1)^2 = 5 \implies x=2$
This gives us a slope of the line $OA$ to be $\frac{-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For $f(x)f(y)+f(3/x) f(3/y)=2f(xy)$ choose correct choices Consider $f:\mathbb{R}^+ \to \mathbb{R}$ such that $f(3)=1$ and
$$f(x)f(y)+f(3/x) f(3/y)=2f(xy)\;\;\;\;\;\;\forall x,y \in \mathbb{R}^+$$ then choose the correct option(s):
(A) $f(2014)+f(2015)-f(2010)=100$
(B) $f$ is an even function
(C) $\frac{f(100)}{f(10)+... | $x$ and $y$ are always explicitly positive, and this will not be further noted.
$$
\begin{align}
2f(xy) &= f(x)f(y)+f(3/x) f(3/y) \tag{Definition} \label{def} \\
f(3) &= 1 \tag{Initial Condition}
\end{align}
$$
Following @ChristianF, we'll set variables to make the conditions more symmetric.
Considering the case where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove this inequality $Σ_{cyc}\sqrt{\frac{a}{b+3c}}\ge \frac{3}{2}$ with $a;b;c>0$ Let $a,b,c>0$. Prove $$\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge \frac{3}{2}$$
$A=\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}$
Holder: $A^2\cdot Σ_{cyc}\left(a^2\left(b+3c\right)\right)\g... | Your trying gets a wrong inequality.
Try $c=0$ and $a=b=1$.
Let $\frac{a}{b+3c}=\frac{x^2}{4}$, $\frac{b}{c+3a}=\frac{y}{4}$ and $\frac{c}{a+3b}=\frac{z^2}{4}$, where $x$, $y$ and $z$ are non-negative numbers.
Hence, the system
$$\begin{array}{l}4a-x^2b-3x^2c=0\\-3y^2a+4b-y^2c=0\\ -z^2a-3z^2b+4c=0\\ \end{array}$$
has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sid... | Notice that for any $x$ and and natural $n$ that $$(x-1)(x^{n-1} + ..... + x + 1) = (x^n + x^{n-1} +....... +x) - (x^{n-1} + x^{n-2} +.... +1) = x^n -1$$ so that $x-1|x^n - 1$ always.
Lemma: $x-1|x^n-1$ for natural $n$.
Now $d|n$ so let $m = \frac nd$ and let $y= x^d$.
Then $y-1|y^m -1$.
But $y-1 = x^d -1$ and $y^m -1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still... | By Newton's identities, with $e_k$ standing for the elementary symmetric polynomials and $p_k$ for the sum of $k^{th}$ powers, and using that $e_1=0\,$:
$$
\begin{align}
p_1 &= e_1 &&= 0\\
p_2 &= e_1p_1-2e_2 &&= -2e_2 \\
p_3 &= e_1p_2 -e_2p_1 + 3e_3 &&= 3e_3 \\
p_4 &= e_1p_3 - e_2p_2+e_3p_1-4e_4 &&= 2 e_2^2 \\
p_5 &= e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving $x^2+y^2=9$, $\arctan\frac{y+2}{x+2} + \arctan\frac{y-2}{x+4} =2\arctan\frac{y}{x}$ without graphing? The system of equations:
$$\begin{align}
x^2 + y^2 &= 9 \\[6pt]
\operatorname{arctan}\frac{y+2}{x+2} + \operatorname{arctan}\frac{y-2}{x+4} &=2\,\operatorname{arctan} \frac{y}{x}
\end{align}$$
I tried to in... | Hint:
$$\arctan\dfrac{y+2}{x+2}-\arctan\dfrac yx=\arctan\dfrac yx-\arctan\dfrac{y-2}{x+4}$$
$$\iff\dfrac{2x-2y}{x^2+y^2+2x+2y}=\dfrac{2x+4y}{x^2+y^2-2x+4y}$$
As $x^2+y^2=9$ $$\iff\dfrac{2x-2y}{9+2x+2y}=\dfrac{2x+4y}{9-2x+4y}$$
$$\iff8x^2+16y^2+54y=0\ \ \ \ (1)$$
As $x^2+y^2=9,$ WLOG $x=3\cos t,y=3\sin t$
Put these ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Partial Fractions Decomposition of $\frac{25s}{(s^2+16)(s-3)(s+3)}$ So this is the problem..
$$
\frac{25s}{(s^2+16)(s-3)(s+3)}
$$
So what I did was...
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {A}{s^2+16}+\frac {B}{s-3}+\frac{C}{s+3}
$$
then...
$$\begin{align}
25s &= A(s-3)(s+3)+B(s^2+16)(s+3)+C(s^2+16)(s-3) \\
25s &=... | $$\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {As+B}{s^2+16}+\frac {C}{s-3}+\frac{D}{s+3}$$
Use Heaviside method.
To find $C$, let $s=3$ and you get $C=1/2$
To find $D$, let $s=-3$ and you get $D=1/2$
Substitute for $C$ and $D$ in the RHS and subtract from LHS.
You will find $$\frac {25s}{(s^2+16)(s-3)(s+3)}-\frac {1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given:
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof th... | $$f(x)=e^x$$
$$f'(x)=e^x \tag 1$$ (I assumed you know this property of $e^x$. )
$$f''(x)=f'(x)=e^x$$
$$f^{(n)}(x)=e^x$$
$$f^{(n)}(0)=1$$
If we find the Taylor series for a function $f(x)$ is:
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$f(ix)=e^{ix}=\sum_{n=0}^{\infty} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Prove $n$ is prime. (Fermat's little theorem probably)
Let $x$ and $n$ be positive integers such that $1+x+x^2\dots x^{n-1}$ is prime. Prove $n$ is prime
My attempt:
Say the above summation equal to $p$ $$1+x+x^2\dots x^{n-1}\equiv 0\text{(mod p)}\\
{x^n-1\over x-1}\equiv0\\
\implies x^n\equiv1\text{ (as $p$ can't di... | If $x = 1$, it's obvious.
For $x>1$, let $n$ be composite, say $n = pq$ with $p, q>1$. In that case, we have
$$
(1+x+\cdots+x^{n-1})(x-1) = x^n-1 = x^{pq}-1\\
= (x^p)^q-1\\
= (1+x+x^p+x^{2p}+\cdots + x^{(q-1)p})(x^p-1)
$$
Thus $x^p-1$ divides $x^n-1$, but is not equal to it. In other words, $\frac{x^n-1}{x^p-1}$ is an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What am I doing wrong while finding the Laurent series of $\frac{1}{\sin(z)}$? I'm trying to compute the Laurent series of $\frac{1}{\sin(z)}$ at $z_0=0$.
From what I've seen on the internet this is given as
$f(z)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+...$
My attempt:
$f(z)=\frac{1}{\sin(z)}$ can be rewritten as $... | Note the most direct way to obtain the expansion of $\dfrac1{\sin z}$ is division by increasing powers. I'll show how to obtain the first four terms:
Start from the expansion of $\sin z$ at order $7$:$$\frac1{\sin z}=\frac1{z-\dfrac{z^3}{6}+\dfrac{z^5}{120}-\dfrac{z^7}{5040}+o(z^7)}=\frac1z\cdot\frac1{1-\dfrac{z^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states
For positive real numbers $a,b,c$ prove that
$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$
After some algebraic wrangling we can get to the point where:
$(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥ 2... | Let's simplify the inequality:
$$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b) \Rightarrow \\
a^4+2a^2b^2+b^4\ge ((a+b)^2-c^2)(c^2-(a-b)^2) \Rightarrow \\
a^4+2a^2b^2+b^4\ge 2c^2(a^2+b^2)-(a^2-b^2)^2-c^4 \Rightarrow \\
c^4-2(a^2+b^2)c^2+2(a^4+b^4)\ge 0 \qquad (1)$$
It is a bi-quadratic inequality and its... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Edge length of an equilateral triangle if distances from a point $P$ to its vertices is given A point $P$ is located inside an equilateral triangle and is at a distance of 5, 12, and 13 from its vertices. Compute the edge length of the triangle.
The answer is $\sqrt{169 + 60\sqrt(3)}$.
If $s$ is the edge length of the ... |
Given $a,b,c$,
the equilateral triangle
with the internal point $P$,
located at distances $a,b,c$ from its vertices,
can be constructed as follows.
Start with $\triangle ABC$,
$|BC|=a$,
$|CA|=b$,
$|AB|=c$,
$\angle CAB=\alpha$,
$\angle ABC=\beta$,
$\angle BCA=\gamma$.
Construct a helper external equilateral triangle,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find an asymptotic for $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$ I need to find the asymptotic behavior of $$\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$$
as $N\to\infty$.
I found (using a computer) that this asymptotically will be equivalent to $\frac{1}{3}N^2$, but don't know how to prove it mathematically.... | Standard asymptotic calculus yields the estimate $$ \frac{1}{1-\cos x} = \frac{2}{x^2}+O(1)$$ hence $\displaystyle x\mapsto \frac{1}{1-\cos x} -\frac{2}{x^2}$ is bounded on a neighborhood of $0$, say $(0,\delta)$
Furthermore the function is continuous over $[\delta,\pi]$, thus bounded over $[\delta,\pi]$, hence bounded... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
My attempt:
Consider $\frac{1}{1 - z}$. Let $z = x + iy$, we know that $|z| < 1 \implies x, y < 1$. $$\frac{1}{1 - z} = \frac{1}{1 - x - iy} = \frac{1 - x + iy}{(1 - ... | $$\Re \Big(\frac{1}{1-z}\Big)={1\over 2}\Big(\frac{1}{1-z}+\frac{1}{1-\overline{z}}\Big)$$
$$={1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+z\overline{z}}$$
$$>{1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+1}$$
$$={1\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove that following recursively defined sequence converges and its limit is 1/2
\begin{cases}a_1 = 1\\ \\ \displaystyle a_{1+n} = \sqrt{\sum_{i=1}^n a_i}\end{cases}
Prove that $\{\frac{a_n}{n}\}$ is convergent and its limit is $\frac12$
My proof: We can recursively define relation between $a_{1+n}$ & $a_n$ as
$a_{... | You have shown that $a_{n+1}-a_{n-1} \gt \frac 12$ but that is a step of $2$ in the index. You therefore can only conclude that $a_{n+1} \gt a_1+\frac n4$ and the squeeze fails.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots... | It is maybe a bit easier to use the matrix determinant formula. If you rewrite your matrix as
$$
M = aI +UV^T
$$
where $UV^T$ is
$$
\begin{bmatrix}
(n-1) &0\\
(n-2) &0\\
\vdots&0\\
1 &0\\
0&1
\end{bmatrix}
\begin{bmatrix}
0&0&\cdots&1\\
(n-1)&(n-2)&\cdots&0
\end{bmatrix}
$$
Then, from
$$
\operatorname{det}({\mathbf{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
set G equal to what?
My attempts : i construct a matrix $A= \begin{bmatrix} 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&2&1&0&0&0&0\\ 0&0&0&2&0&0&0&0\\ 0&0&0&0&2&0&0&0\\ 0&0&0&0&0&2&0&0\\ 0&0&0&0&0&0&2&0\\ 0&0&0&0&0&0&0&2\end{bmatrix}$
Now we know that the characteristic polynomial of A is $(x − 1)^2
(x − 2)^6$
and the... | In the matrix $A$ we have an eigenvalue $2$ which has geometric multiplicity $4$ and one which has geometric multiplicity $1$ (the one of the jordan block $2-by-2$). Thus we have that $G=\{1,4\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sums of Nilpotent Matrices Let $A =$ diag$(a_1,a_2,…,a_n)$, where the sum of all $a_i$’s is zero.
Show that A is a sum of nilpotent matrices.
My idea:
$\begin{bmatrix}
2 & 0 \\
0 & -2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}$ + $\begin{bmatrix}
1 & -1 \\
1 & -1
\end{bmatrix}$
Where $\begin{bmatr... | My answer follows @JoséCarlosSantos' post.
It is enough to notice that for all $1 \leq i < j \leq n$, $$E_{i, i} -E_{j, j} = \left(E_{i, j}\right) + \left(- E_{j, i}\right) + \left(E_{i, i} -E_{i, j} + E_{j, i} - E_{j, j}\right)$$ and $$\left(E_{i, i} -E_{i, j} + E_{j, i} - E_{j, j}\right)^{2} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding all values in $\mathbb R$ for quadratic absolute value equation The question:
Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$
So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution s... | Knowing that $x^2+2x+2 > 0$
The equation is equivalent to
$$
x^2+2x+2 = \vert x^2-3x-4\vert
$$
which is equivalent to
$$
x^2+2x+2 =
\left\{\begin{array}{lcl}-x^2+3x+4 & \rightarrow & x = \{\frac{1\pm\sqrt{17}}{4}\}\\
x^2-3x-4 & \rightarrow & x = -\frac{6}{5}\end{array}\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one show that multiples of number does not have a zero in the decimal expansion?
Prove that there exists a number divisible by $5^{1000}$ not containing a single zero in its decimal notation.
The above question is taken from this site. It is question no. 88 in the list. I could not find any solutions for thi... | Idea of Approach
By induction, an using the fact that $5^{n}$ is a divisor of $10^n$ and $10^{n-1}$ leaves only a certain kind of remainder when divided by $5^n$.
Claim and proof
Claim : For all $n$, there exists an $n$ digit multiple of $5^n$ that does not contain any zeros.
Proof : Start with the base case : $5 , 25 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to find and characterise critical points of a polynomial? Find and characterise the critical points of: $f(x)=(2x^3-12x^2+18x-1)^5$
I differentiated to get:
$$\frac{df}{dx}=5(6x^2-24x+18)(2x^3-12x^2+18x-1)^4=30(x-3)(x-1)(2x^3-12x^2+18x-1)^4$$
But don't know how to factorise this further?
| If you set $$30(x-3)(x-1)(2x^3-12x^2+18x-1)^4=0$$ then the stationary points are at $$x-3=0\implies x=3\\x-1=0\implies x=1\\2x^3-12x^2+18x-1=0$$ and this cubic equation can be solved using Cardano's method.
To find the critical points, substitute each value of $x$ into $y=(2x^3-12x^2+18x-1)^5$ and the nature of them ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of multivariable function including absolute value I have some trouble with this limit:
$$\lim_{(x,y)\to (0,0)} \frac{x^3+y^2}{x^2+|y|}.$$
My first attempt to solve it was with polar coordinates but I couldn't find an expression which was independent of $\varphi$. Now I'm trying to solve it with the triangle ineq... | By Cauchy–Schwarz
$$|x^3+y^2|=|xx^2+|y||y||\le\sqrt{x^2+y^2}\sqrt{x^4+y^2}\le \sqrt{x^2+y^2}\sqrt{(x^2+|y|)^2}=\sqrt{x^2+y^2}(x^2+|y|)$$
then
$$\left|\frac{x^3+y^2}{x^2+|y|}\right|\le\sqrt{x^2+y^2}\to 0 \implies \frac{x^3+y^2}{x^2+|y|}\to 0$$
| {
"language": "en",
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"source": "stackexchange",
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If $z^{23}=1$ then evaluate $\sum^{22}_{z=0}\frac{1}{1+z^r+z^{2r}}$
If $z$ is any complex number and $z^{23}=1$ then evaluate $\displaystyle \sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}$
Try: From $$z^{23}-1=(z-1)(1+z+z^2+\cdots +z^{22})$$
And our sum $$\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}=\frac{1}{3}+\frac{1}{1+z+z^2}+\f... | If $z^{23}=1$ and $z\not=1$ (otherwise the sum is trivially equal to $23/3$) then $z^k$ is a primitive $23$-th root of unity for any $k=1,2\dots,22$ (note that $23$ is a prime number). Hence
$$\begin{align}
\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}&=\frac{1}{3}+\sum_{r=1}^{22}\frac{1}{1+z^r+z^{2r}}\\
&=\frac{1}{3}+\sum_{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integrate $\frac{1}{\sqrt{x^2+cx}}$ I am trying to compute the integral $\int\frac{1}{\sqrt{x^2+cx}}dx$.
To begin I completed the square of the denominator resulting in $$\int \frac{1}{\sqrt{(x+\frac{c}{2})^2-\frac{c^2}{4}}}dx$$
I then made the substitution $u=x+\frac{c}{2}$, which has left me with the integral $$\int ... | You've done it properly.
continuing from where you left off ;
$I =\displaystyle\int \frac{1}{\sqrt{u^2-\frac{c^2}{4}}}du$
$I =\displaystyle\int \frac{1}{\frac{c}2\sqrt{\frac{4u^2}{c^2}-1}}du$
let $ \frac{2u}{c} = \sec(t) \implies \frac 2c \,du = \sec(t)\tan(t)\,dt$
$I =\displaystyle\int\frac{1}{\sqrt{\sec^2(t)-1}}\sec... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7... This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges:
$$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\d... | The basic idea is good. But after you've proven that $\sum_{n=1}^N\frac1{2n+1}$ is smaller that the sum of the first $2N+1$ terms of your series, you're done. And all you need for that is that$$\frac1{2^2}+\frac1{4^2}+\cdots+\frac1{(2N)^2}<1.$$This is equivalent to$$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{N^2}<4,$$which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\lim_{x\to\infty}x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{1}{m+1}$ using the Mean Value Theorem Show that $\lim_{x\to\infty}x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{1}{m+1}$ using the Mean Value Theorem with m and x being whole numbers, for x greater or equal to 1 ... | When $x\to \infty$ since $x-1<x_0<x$ we also have that $x_0 \to \infty$ and
$$x-1<x_0<x \iff \frac{x-1}x<\frac{x_0}x<1$$
then by squeeze theorem $\frac{x_0}x\to 1$, thus
$$\frac{x^\frac{m}{m+1}}{(m+1)x_0^\frac{m}{m+1}}=\frac1{m+1} \left(\frac{x}{x_0}\right)^{\frac{m}{m+1}} \to\frac1{m+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove $\sum_{r=1}^{n-1}(n-r)^2\binom{n-1}{n-r}=n(n-1)2^{n-3}$ This is problem 2.38 from the book "Principles and Techniques in Combinatorics".
Prove $$\sum_{r=1}^{n-1}(n-r)^2\binom{n-1}{n-r}=n(n-1)2^{n-3}$$
I have have been trying different things on and off for two days and still can't derive RHS from LHS.
what I hav... | Another way:
$$
\eqalign{
& \sum\limits_{1\, \le \,r\, \le \,n - 1} {\left( {n - r} \right)^{\,2} \left( \matrix{
n - 1 \cr
n - r \cr} \right)} = \sum\limits_{1\, \le \,k\, \le \,n - 1} {k^{\,2} \left( \matrix{
n - 1 \cr
k \cr} \right)} = \cr
& = \sum\limits_{0\, \le \,k\, \le \,n - 1} {k^{\,2} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
| Note that $$\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}=\frac{3x^3 - 2x^2 +3x -2+\color{red}{5x^2-8x+6}}{3x^3 - 2x^2 +3x -2}=1+\frac{5x^2-8x+6}{3x^3 - 2x^2 +3x -2}$$ and $$3x^3-2x^2+3x-2=(3x-2)(x^2+1)$$ so $$\int_1^2\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\,dx=[x]_1^2+\int_1^2\frac{5x^2-8x+6}{(3x-2)(x^2+1)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Test $x_1=0, x_{n+1}=\frac{9+(x_n)^2}{6}$ for convergence and find its limit. Test $x_1=0, x_{n+1}=\frac{9+(x_n)^2}{6}$ for convergence and find its limit.
Note that $x_1=0, x_2=\frac{9+0²}{6}=\frac{9}{6}, x_3=\frac{9+9/6}{6}=\frac{17}{9}\dots$ with $x_1 < x_2 < x_3$. $(*)$
My plan for this is to show 1. that this rec... | Your conclusion that $(x_n)$ is bounded above by $3$ is unwarranted. Instead, I would do this by induction. Supposing that $x_n \le 3$, we have:
$$x_{n+1} \le \frac{9 + 3^2}{6} = 3$$
Now we conclude that $(x_n)$ is convergent and compute its limit as you have done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there another proof for Euler–Mascheroni Constant? Problem
Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent.
One Proof
This proof is based on the following inequality
$$\frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}$$
where $n=1,2,\cdots$, ... | Pictorial proof.
In the picture, take $n=11$. The red graph is $1/x$, so the area under the red graph from $1$ to $n$ is
$$
\int_1^n\frac{dx}{x} = \ln n.
$$ The area of the white rectangles under the graph is
$$
\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
$$
The difference is shown in green,
$$
\text{area}(\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Regular octagon inscribed in a square Problem: The corners of a 2 meter square are cut off to form a regular octagon. What is the length of the sides of the resulting octagon?
From the picture below, the octagon would form a right isosceles, specifically a right isosceles triangle on the corners. The sides of the octag... | Let $x$ be the length of your octagon (as in the left picture), and $c$ the length cut from one side of the square edge (which is the $x$ in the right picture).
Then you've correctly stated that $x = \sqrt{2}c$. Now you solve
$$
c + x + c = 2.
$$
This is rewritten as
$$
2c + x = 2c + \sqrt{2}c = (2 + \sqrt{2})c = 2.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Function as difference of convex functions I want to express the following function as the difference of two convex functions:
$f(x)=(x_2-2)^3+x_1^2-6x_1x_2+5x_1+max\{x_1^2,3-x_1^2\}+10$.
I have already seen answers that addressed these tasks, but I couldn't see a pattern in them. Is there an algorithmic approach to so... | Before beginning to answer your question I need to say that I do not know if the method I am about to propose is the best ; it is only how I would do right now, and I never had a class about this particular topic. But it is a method nonetheless.
Since the sum of convex functions is a convex function, it would be enough... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\int\frac{1}{(\sin x+a\sec x)^2}dx$
Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$
Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$
put $\tan x=t$ and $dx=\sec^2 tdt$
So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$
Could some help me how to solve above Integral ... | Hint: $\sin x = \dfrac{\tan x}{\sec x}$ and $\sec^2x = 1 + \tan^2x$.
$$\begin{align}
\dfrac1{(\sin x+a\sec x)^2} &= \dfrac1{\left(\dfrac{\tan x}{\sec x}+a\sec x\right)^2} \\
&= \dfrac{\sec^2x}{(\tan x + a\sec^2x)^2} \\
&= \dfrac{\sec^2x}{\tan^2x + 2a\tan x\sec^2x + a^2\sec^4x} \\
&= \dfrac{\sec^2x}{\tan^2x + 2a\tan x(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$ Show that:
$$ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$$
My try:
As we can see that the LHS is in form of $\cos 8 \theta$ which must be converted into $\cos \theta $ in order to solve the equation.
There are several ques... | Since $2(1+\cos x)=(2\cos x/2)^2$, $$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=\sqrt{2+\sqrt{2+2\cos 4\theta }}=\sqrt{2+2\cos 2\theta}=2\cos\theta,$$provided $\cos\theta,\,\cos 2\theta,\,\cos 4\theta>0$. In fact this implies $$\cos 2\theta =\sqrt{\frac{1+\cos 4\theta}{2}}\ge\frac{1}{\sqrt{2}},\,\cos\theta=\sqrt{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cominatorics (sticks and plus, generating function) If I should distribute 25 identical cookies to 10 children each children should get at least 1 cookie and maxium of 4 cookies. The problem should be solved using both inclusion-exclusion and generating function.
My solution with in-ex (Wrong):
$$x_{1} + x_{2} + x_{3}... | I find it a little clearer not to skip the first step, which is writing the generating function.
you just write down a series $A(x)$ of formal variables $x^k$,
where k denotes how many cookies each child becomes.
As specified k must be between 1 and 4.
Thus $$ A(x) = x+x^2+x^3+x^4 $$
As you have 10 children, the ge... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Bernoulli like inequality When does it hold that for $x\in (-1, 0), n>2$, $(1+x)^n<1+nx+\frac{n(n-1)}{2}x^2$? I know that for $n=3$ it's always true as
$$(1+x)^3=1+3x+\frac{3(3-1)}{2}x^2+x^3<1+3x+\frac{3(3-1)}{2}x^2$$
as $x^3<0$ for $x\in (-1, 0)$.
EDIT: I think that I found a proof by induction: I already proved the ... | (Too long for a comment)
Rewrite the inequality for $y\in(0,1)$
$$(1-y)^n<1-ny+\frac{n(n-1)}{2}y^2$$
while
$$(1-y)^n=1-ny+\frac{n(n-1)}{2}y^2+\sum_{k=3}^{n}{n\choose k}(-y)^{k}$$
So it is equivalent to prove
$$\sum_{k=3}^{n}{n\choose k}(-y)^{k}<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using Snake Oil Method to Evaluate Sum $$\sum_k \binom{n+k}{2k} \binom{2k}{k}\frac{(-1)^k}{k+1+m}$$
This is Problem 8 in "Basic Practice" Section of Concrete Maths by Knuth.
In the book, the answer comes out to be:
$$ (-1)^n \frac{m!n!}{(m+n+1)!}\binom{m}{n}$$
I want to know how to use generating functions to solve th... | Lets start by manipulating the binomial coefficients
\begin{eqnarray*}
\binom{2k}{k} \binom{n+k}{2k} = \binom{n+k}{k} \binom{n}{k}.
\end{eqnarray*}
Now two more tricks ....
\begin{eqnarray*}
\binom{n+k}{k} = [x^k]:(1+x)^{k+n} = [x^0] : \frac{(1+x)^{k+n}} {x^k} \\
\frac{1}{k+m+1} = \int_{0}^{1} y^{k+m} dy
\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $x$ if $\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$ if $x \lt 0$ Then Find value of $$\frac{(1-x^2)^{\frac{3}{2}}}{x^2}$$ if
$$ \cot^{-1} \left(\frac{1}{x}\right)+\cos^{-1}(-x)+\tan^{-1}(x)=\pi$$
My try:
Since $x \lt 0$ we have $$ \cot^{-1}\left(\frac{1}{x}\right)=\pi +\tan ^{-1}x$$
Also $$... | $$\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$$
Suppose $\tan^{-1}(x) = y \implies x = \tan y \implies \cot^{-1}(1/x) = y$
Let $\cos ^{-1}(-x) = z$
$ 2y + z = \pi \implies z = \pi - 2y$
Taking tangents,
$\tan(\pi - 2y) = \tan z \implies -\tan(2y) = \sin z/\cos z $
We get $-\tan 2y = \frac{\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Continuous antiderivative of $\frac{1}{1+\cos^2 x}$ without the floor function. By letting $u = 2x$ and $t = \tan \frac{u}{2}$, I found the continuous antiderivative of the function to be:
$$\int \frac{1}{1+\cos^2 x}dx\\= \int \frac{2}{3+\cos2x} dx\\ = \int \frac{1}{3+\cos u}du \\=\int \frac{\frac{2}{1+t^2}}{3+\frac{1-... | With Floor Function
Let $(1+\cos^2x )^{-1} = f(x)$. Now as you found,
$$\int \frac{dx}{1+\cos^2 x} = \int \frac{\sec^2 x }{2+\tan^2 x} dx = \frac{1}{\sqrt{2} } \arctan\left(\frac{\tan x}{\sqrt 2}\right)$$
The issue is that integral of a continuous function should be continuous. The one we found is discontinuous at all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$ How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$
I have already tried it myself: see here
but it is just not working out...
Thanks in advance!
| For n=1:
$$
\begin{align}
\sum_{i=1}^{1}\frac{1}{i\left(i+1\right)}&=\frac{1}{1(1+1)}=\frac{1}{2}\\
\frac{n}{n+1}&=\frac{1}{1+1}=\frac{1}{2}
\end{align}
$$
hence proved for n=1.
Assume it holds true for n=k, where k is any natural number. Therefore:
$$
\begin{align}
\sum_{i=1}^{k}\frac{1}{i\left(i+1\right)}&=\frac{k}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Gauss' law and a half-cylinder The question is:
A half cylinder with the square part on the $xy$-plane, and the length $h$ parallel to the $x$-axis. The position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$.
$S_1$ is the curved portion of the half-cylinder $z=(r^2-y^2)^{1/2}$ of length $h$... | "the boundaries of $y$ are $0$ and $r$"
There's your problem.
It should be $-r$ and $r$.
| {
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"url": "https://math.stackexchange.com/questions/2808159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$
What I did is
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}... | Don't rationalize like that; instead consider
\begin{align}
&\lim_{h\to0}\frac{1}{h}\left(
\frac{-5(x+h)}{2+\sqrt{x+h+3}}-\frac{-5x}{2+\sqrt{x+3}}\right)
\\[6px]
&\qquad=
\lim_{h\to0}\frac{
-10x-5x\sqrt{x+3}-10h-5h\sqrt{x+3}+10x+5x\sqrt{x+h+3}
}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})}
\\[6px]
&\qquad=
\lim_{h\to0}\frac{-10-5\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Tricky question on polynomials
For any real numbers $x$ and $y$ satisfying $x^2y + 6y = xy^3 +5x^2 +2x$, it is known that $$(x^2 + 2xy + 3y^2) \, f(x,y) = (4x^2 + 5xy + 6y^2) \, g(x,y)$$
Given that $g(0,0) = 6$, find the value of $f(0,0)$.
I have tried expressing $f(x,y)$ in terms of $g(x,y)$. But seems that some t... | Note that:
$$x^2y + 6y = xy^3 +5x^2 +2x \Rightarrow y=\frac{x(y^3 +5x +2)}{x^2+6}.$$
Hence:
$$\begin{align}(x^2 + 2xy + 3y^2) \, f(x,y) &= (4x^2 + 5xy + 6y^2) \, g(x,y) \Rightarrow \\
\lim_{x,y\to 0} \frac{f(x,y)}{g(x,y)}&=\lim_{x,y\to 0} \frac{(8x+5y)^2+71y^2}{16((x+y)^2+2y^2)}\\
\frac{f(0,0)}{g(0,0)}&=\lim_{x,y\to 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find real part of $\frac{1}{1-e^{i\pi/7}}$ How can you find
$$\operatorname{Re}\left(\frac{1}{1-e^{i\pi/7}}\right).$$
I put it into wolframalpha and got $\frac{1}{2}$, but I have no idea where to begin. I though maybe we could use the fact that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2},$$ where $\bar{z}$ is the conjugate of ... | Try this,
$$z=\frac{1}{1+e^{\frac{i\pi}{7}}}=\frac{1}{1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}}$$
$$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{(1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7})\times(1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7})}$$
$$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{(1+\cos\frac{\pi}{7})^2+\sin^2\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also,
$$\int_{0... | $$\int_0^1 x^m\ln x\,dx=\int_0^1 ye^{-(m+1)y}\,dy=-\frac1{(m+1)^2}.$$
Then
$$\int_0^1\frac{\ln x}{1+x}\,dx=\sum_{m=0}^\infty(-1)^m
\int_0^1 x^m\log x\,dx=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(m+1)^2}
=-\frac{\pi^2}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding maximum of $A=\frac a{2+bc}+\frac b{2+ca}+\frac c{2+ab}$
Let $a,b,c\ge 0$ satisfy $a^2+b^2+c^2=2$. Find maximum of $$A=\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}.$$
I see $\max A=1$ and it occurs when $(a,b,c)=(1,1,0)$ and its permutation. So I will prove this inequality:$$\frac{a}{2+bc}\le \frac{a}{a+b+c} ... | It is true that $2(2+2bc)=(1+1)(a^2+(b+c)^2)\ge (a+b+c)^2$, but it doesn't prove your inequality because $2 \le a+b+c$ is not proved. Actually, it is not true. Try $a=b=c=\sqrt{\frac{2}{3}}$.
Instead, $2+bc\ge a+b+c$ is equivalent to$$2-b-c+bc\ge a$$or$$(2-b-c+bc)^2\ge a^2$$or$$b^2 c^2 - 2 b^2 c + 2b^2 - 2 b c^2 + 6 bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Proof involving generating function The following is part of a proof that the number of ways of associating a product with $n$ terms (different ways of inserting parentheses) is $$
a_1 = 1,\ a_n = \frac{1}{n} \binom{2n-2}{n-1},$$
and the relationship$$
a_{n+1} = a_1a_n + a_2a_{n-1} + a_3a_{n-2} + \cdots + a_na_1$$
is a... | Note that
$$
\begin{align}
f(x)
&=\sum_{n=1}^\infty\color{#C00}{a_n}x^n\tag1\\
&=\color{#C00}{a_1}x+\sum_{n=2}^\infty\color{#C00}{\sum_{k=1}^{n-1}a_ka_{n-k}}x^n\tag2\\
&=x+\left(\sum_{n=1}^\infty a_nx^n\right)^2\tag3\\[9pt]
&=x+f(x)^2\tag4
\end{align}
$$
Explanation:
$(1)$: $f$ is the generating function for $a_n$
$(2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Maximize $f(x,y)=xy$ subject to $x^2-yx+y^2 = 1$
Use Lagrange multipliers method to find the maximum and minimum values of the function
$$f(x,y)=xy$$
on the curve
$$x^2-yx+y^2=1$$
Attempt:
First I set let $g(x,y)=x^2-xy+y^2-1$ and set $$\nabla f=\lambda\nabla g$$
so
$$(y,x)=\lambda(2x-y,2y-x)$$
then
$$\begin{c... | As a check, not using Lagrange multipliers:
$x^2-xy+y^2 =1$.
1) Minimum.
$(x+y)^2 -3xy =1.$
$3xy= (x+y)^2 -1 ;$
Minimum of $f(x,y) =-(1/3).$
2) Maximum.
$(x-y)^2 +xy =1;$
$xy = 1- (x-y)^2;$
Maximum of $f(x,y) = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Why are the solutions of the $3\times 3$ system like that? Consider the problem:$$
\min \quad -x_1^2-4x_1x_2-x_2^2\\ \text{s.t.} \quad x_1^2 + x_2^2 = 1$$
The KKT system is given by\begin{align*}
x_1 (-1 + v) + 2 x_2 &= 0 \tag{1} ,\\ x_2 (-1 + v) + 2 x_1 &= 0 \tag{2},\\ x_1^2 + x_2^2 &= 1 \tag{3}
\end{align*}
The solut... | Using polar coordinates,
$$\begin{aligned} x_1 &= \cos (\theta)\\ x_2 &= \sin (\theta)\end{aligned}$$
we obtain the unconstrained $1$-dimensional maximization problem in $\theta$
$$\text{maximize} \quad 1 + 2 \sin (2\theta)$$
Differentiating the objective and finding where the derivative vanishes, we obtain $\cos(2\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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The length of a segment constructed from an equilateral triangle and some auxiliary points In the diagram, $CE=CF=EF,\ EA=BF=2AB$, and $PA=QB=PC=QC=PD=QD=1$, Determine $BD$.
I tried working this question as :
$$\angle ACB = \angle DQB = 12^\circ$$
So, by the cosine rule for $\triangle QBD$,
$$q^2 = b^2 + d^2 - 2bd\... | Let $EF=5a$ and $CD\cap EF=\{K\}.$
Thus, $DK\perp AB$, $AB=a$ and by the Pythagoras theorem we obtain: $$AC=\sqrt{AK^2+CK^2}=\sqrt{\frac{AB^2}{4}+\frac{3EF^2}{4}}=a\sqrt{19}.$$
Now, $$\measuredangle ADC=\frac{1}{2}\left(2\measuredangle PAD+2\measuredangle PCD\right)=$$
$$=\frac{1}{2}\left(180^{\circ}-\measuredangle APD... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Determine the image of the map Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by \begin{equation*}f: \begin{pmatrix}x \\ y\end{pmatrix}\rightarrow \begin{pmatrix}u \\ v\end{pmatrix}=\begin{pmatrix}x(1-y) \\ xy\end{pmatrix}\end{equation*}
I want to determine th eimage $f(\mathbb{R}^2)$.
$$$$
We have that
\begi... | $f$ is not linear. Hence, you can't assume that the image of $f$ is a subspace of $\mathbb R^2$.
Hint: Consider
$$
f(x,y)=x\begin{pmatrix}1-y\\y\end{pmatrix}.
$$
You see that the image of $f$ contains all lines through $\begin{pmatrix}1-y\\y\end{pmatrix}$ for some $y\in\mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1... | $$\sqrt2-\sum_{cyc}\frac{a}{1+a^2}=\sum_{cyc}\left(\frac{\sqrt2}{3}-\frac{a}{1+a^2}\right)=\sum_{cyc}\left(\frac{\sqrt2}{3}-\frac{a}{1+a^2}+\frac{1}{2\sqrt2}\left(\frac{2}{3}-\frac{a^2}{1+a^2}\right)\right)=$$
$$=\sum_{cyc}\frac{(a-\sqrt2)^2}{2\sqrt2(1+a^2)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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In $\triangle ABC$, $(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = $?
In $\triangle ABC$,
$$(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = \text{?}$$
I tried solving this by cosine rule but it is becoming too long. Any short step solution for this?
| Solving this problem by Law of Cosines isn't short but not that unmanageable.
The key is the observation of pattern within the expression:
$$(b^2-c^2)\cot A = \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R} = \frac{R}{abc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]$$
This has the form $\phi(a,b,c) - \phi(b,c,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove generating function of $a_n = (b+1)^n - b^n$ The problem is prove that $$\frac{x}{(1-bx)(1-bx-x)}$$
is the generating function for the sequence $$a_n = (b+1)^n - b^n$$
I tried separating the fraction as $$x\frac{1}{1-bx}\frac{1}{1-x(b+1)}$$
which yields the generating functions $$x\sum_{n=0}^\infty (bx)^n\sum_{n=... | There is no mistake with your approach. The calculation is just, let's say slightly more cumbersome.
We obtain
\begin{align*}
\color{blue}{x\sum_{k=0}^\infty}&\color{blue}{ (bx)^k\sum_{l=0}^\infty ((b+1)x)^l}\\
&=x\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}b^k(b+1)^l\right)x^n\tag{1}\\
&=x\sum_{n=0}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^... | Alternatively, using modular arithmetic:
$$\begin{align}n&\equiv 0,1,2 \pmod 3 \\
n^2&\equiv 0,1 \pmod 3\\
4n^2&\equiv 0,1 \pmod 3\\
n^4&\equiv 0,1 \pmod 3\\
n^4-4n^2&\equiv 0 \pmod 3.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $[0,1)$? Consider the following series for $x \in [0,1)$:
$$
\sum \frac{x^n}{1+x^n}
$$
I figured that it converges, by using the ratio test:
$$
\left| \frac{a_{n+1}}{a_n} \right| = \frac{x^{n+1}}{1+x^{n+1}} \cdot \frac{1+x^{n}}{x^n} = \frac{x+x^{n+1}}{1+x^... | The convergence is not uniform. For $x_n = \sqrt[n+1]{1- \frac1n}$ we have:
$$
\sup_{x \in [0,1)} \left|\sum_{k=1}^\infty \frac{x^k}{1+x^k} - \sum_{k=1}^n\frac{x^k}{1+x^k}\right| = \sup_{x \in [0,1)} \sum_{k=n+1}^\infty\frac{x^k}{1+x^k}
\ge \frac{x_n^{n+1}}{1+x_n^{n+1}}
= \frac{1-\frac1n}{2-\frac1n} \xrightarrow{n\to\i... | {
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"url": "https://math.stackexchange.com/questions/2829732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$
Prove for every $a, b, c \in\mathbb{R^+}$, given that $abc=1$ :
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$$
I tried using AM-GM or AM-HM but I can't f... | By AM-GM $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2-3(a+b+c+1)=\sum_{cyc}\left(a^2+\frac{2a}{b}+\frac{1}{b^2}-3a-1\right)\geq$$
$$\geq\sum_{cyc}(2a-1+2+a^2b^2-3a-1)=\sum_{cyc}(a^2b^2-a)=$$
$$=\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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The number of incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$ I was asked to find the number of the incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$, and the fact $2019=673\times 3$ is given.
My attempt: Given congruence is equivalent to
$$
\begin{cases}
x^4-14x^2+36\equiv 0\pmod{673}\\
x^4-14x^2+3... | Playing around with numbers (what follows is $\bmod 673$):
$\sqrt{13}\equiv \sqrt{2032}\equiv 4\sqrt{127}\equiv 4\sqrt{800}\equiv 80\sqrt{2}\equiv 80\sqrt{675}\equiv (-146)\sqrt{3}\equiv (-146)\sqrt{676}\equiv (-146)×(\pm 26)\equiv \pm 242$.
So the roots for $x^2$ are $249, -235 \bmod 673$, which can be checked for the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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A problem with definite integral for trigonometric functions I need to evaluate $ \displaystyle \int\limits_{0}^{2\pi} \frac{dx}{5-3\cos x}$
The answer in the book is $\frac{\pi}{2}$.
What i did is $\displaystyle \int \frac{dx}{5-3\cos{x}} = \dfrac{\arctan {2 \tan{\frac{x}{2}}}}{2}$ , So evaluating from $0$ to $2\pi$ g... | The function mapping $x$ into $\frac{1}{2}\tan(2x)$ is increasing and differentiable on each connected component of its domain, but not everywhere. And we require that change of variables are given by one-to-one maps (at least essentially, with respect to some measure). A correct way to handle your integral is
$$\begin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $x^2+y^2+z^2=14^n$ holds for distinct integers $x,y,z$ for every natural $n$? Prove that for all natural numbers $n$, there exist distinct integers $x, y, z$ for which,
$x^2+y^2+z^2=14^n$
How to prove this using mathematical induction?
Some context:
A related question asks for solutions of $x^2 +... | For $n=1$ we have
$$14=1^2+2^2+3^2$$
For $n=2$ we have
$$14^2=196=4^2+6^2+12^2$$
If $14^n=x^2+y^2+z^2$ then
$$14^{n+2}=14^2(x^2+y^2+z^2)=(14x)^2+(14y)^2+(14z)^2$$
so you can do induction on even and odd numbers separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Number of different sums with k numbers from {1, 5, 10, 50}
Say we have $k$ numbers, each of which belongs to the set $S = \{1, 5, 10, 50\}$
How many different sums can be created by adding these numbers?
If $k = 1$, the are four different sums.
Also, if $k = 2$, there are ten:
$$\begin{align} 1 + 1 = 2 \quad 1 +... | We can calculate the number of solutions with some algebra. We represent multiples of $\{1,5,10,50\}$ by generating functions (in $x$) and count the number of occurrences in $y$.
We calculate the number of sums with $k$ members from $\{1,5,10,50\}$ as the coefficient of $[y^k]$ evaluated at $x=1$:
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt... | Although I prefer an algebraic approach, here is a "non-algebraic" way by directly determining $A$. Let
*
*$C =A^3 \Rightarrow C^2=\begin{bmatrix}7& 6\\-6&-5\end{bmatrix}$
Looking at the pattern of the entries comparing $A^6$ with $A^3$ we find:
*
*the entries in the first row go $3$ up
*the entries in the s... | {
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"url": "https://math.stackexchange.com/questions/2838190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt:
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed a... | Easiest way is to draw a graph $$f(x) = {x^4+11x^2+1\over -x(x^2+1)}$$
and we want to find for which $k$ line $y=k$ cuts graph of $f$ for positive $x$ four times. If you do some calculus you see that $f$ has local minimum $-6,5$ for positive $x$ and local maximum $-6$ so the finally answer is for $$-6,5<k<-6$$
If you... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there are no non-zero integers $m$ and $n$ such that $m^2 = 180 n^4$ I made the following:
$(m^2)^{1/2} =(n^4)^{1/2} 180^{1/2} $
Then $|m| = (n^2) 180^{1/2} $ but $180^{1/2} = 6 (5^{1/2})$ isn't an integer number. Then $m$ is an integer number if and only if $n= a (5^{1/2})$, $a$ is any integer number. Th... | For $m^2 = 180 n^4$, assuming there is an integer solution, implies that $m^2$ is wholly divisible by $180 = 2\cdot 2\cdot 3\cdot 3\cdot 5$.
That is, $\frac{m^2}{2\cdot 2\cdot 3\cdot 3\cdot 5} = n^4$
Now, factors of $m$ must appear as an even quantity of each in $m^2$. So if we divide $m^2$ by a single factor of $5$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\frac{x^5}{5!}- \frac{x^7}{7!}+ \frac{x^9}{9!}+\dots>0$ holds for $x>0$ Show that:
$\sin x> x- \dfrac{x^3}{6} ~~\forall ~x>0$
Attempt:
Let $y = \sin x - x+\dfrac{x^3}{6}$
We have to prove that y is increasing for $x >0$
$y' = \cos x -1 + \dfrac{x^2}{2}$
$y'' =-\sin x+x$
$y''>0$ for all $x>0$.
$\implies y'$ is incr... | Consider any alternating series $a_1-a_2+a_3...$ with $a_n >0$ and $a_n$ decreasing to $0$. If the series is absolutely convergent then we can group the terms as $(a_1-a_2)+(a_3-a_4)+...$ and the sum is greater than $a_1-a_2$ because the other terms are non-negative. Your question is a special case of this. Attempt 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating indefinite integrals.
If $$\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\dfrac{d}{b}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$ where $b,d$ are relatively prime find $b+d$.
My solution:
$$\displaystyle\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\int\dfrac{1}{x}\sqrt{\dfrac{x^3}{a^3-x^3}}dx$$
Then let $x^3=t\im... | Since you already have the form you must get the answer into then why not get some hints from it.
$$\int \frac {\sqrt x}{\sqrt {a^3-x^3}} dx=\int \frac {\sqrt x}{a^{\frac 32}\sqrt {1-\left(\frac {x^{3/2}}{a^{3/2}}\right)^2 }} dx$$
Let $$\frac {x^{3/2}}{a^{3/2}}=t$$ hence $$dt=\frac 32\cdot \frac {\sqrt x}{a^{3/2}}dx$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the value of $a$ by evaluating the limit $L=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}=\frac{1}{60}$
For $a\in\mathbb R-\{-1\}$
$$L=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}=\frac{1}{60}$$
find the value of $a$.
My attempt:
... | $L=\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+...+(na+n)]}=$
$= \lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{\frac{1}{n^a}(\frac{n+1}{n})^{a-1}\frac{1}{n}[n^2a +\frac{n(n+1)}{2}]} =$
$= \lim_{n\to\infty}\frac{(\frac{1}{n})^a+(\frac{2}{n})^a+...+1^a}{(\frac{n+1}{n})^{a-1}[na +\frac{(n+1)}{2}]} =$
$= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What does is mean $Y=\min \{X_1,X_2\}$? $Y=\min \{X_1,X_2\}$ and $Z=\max \{X_1,X_2\}$?
Let's determine $X_1$ distribution to be $\operatorname{Bin}(2,\frac{1}{2})$ and the distribution of $X_2$ to be $U(1,2,3)$. Also $X_,X_2$ are independent.
After some calculations:
*
*$P(X_1=0)=\frac{1}{4}$
*$P(X_1=1)=\frac{1}{2... | You have not said what the joint distribution of $X_1,X_2$ is. But if we assume they are independent, although you didn't say that, then the joint distribution is as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 1/12 & 1/12 & 1/12 \\
1 & 1/6 & 1/6 & 1/6 \\
2 & 1/12 & 1/12 & 1/12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability on cards involving two conditions . $\text{Problem (i) :}$
When $4$ cards are drawn out of $52$ cards, what is the probability of only $2$ cards being same value and the rest being different ?
MY WORK:
The probability I found is :
$$\frac{\binom{13}{1}\times \binom{4}{2}\times \binom{12}{2}\times \bino... |
$$\frac{\binom{13}{1}\times \binom{4}{2}\times \binom{12}{2}\times \binom{4}{1}\times \binom{4}{1}}{\binom{52}{4}}$$
$\checkmark$ Select a rank and two suits for it, and select two other ranks, and a suit for each.
$$\frac{\binom{2}{1}\times \binom{13}{1}\times \binom{2}{2}\times \binom{12}{1}\times \binom{2}{1}\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of the tangent lines to the ellipse having a given angular coefficient Find the equations of the tangent lines to the ellipse $E : x^2/a^2 + y^2/b^2 − 1 = 0$ having a given angular coefficient $m ∈ R$.
| Here's a go at it:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Differentiate implicitly.
$$ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 $$
Assuming angular coefficient is the slope.
$$ \frac{dy}{dx} = m $$
$$ \frac{x}{a^2} + \frac{y}{b^2}m = 0 $$
$$ x = - y \frac{a^2}{b^2}m $$
$$ x^2 = y^2 \frac{a^4}{b^4}m^2 $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient ex... | In this precise case, you can notice that $x^3$ is an obvious factor. You are left with: $(x^3+1)/(x+1)$ Using the remarkable identity $A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+...+B^{n-1})$ with $A=x$ and $B=-1$ gives $q(x)=x^3(x^2-x+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Proof by Induction: If $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$
If $x_1,x_2,\dots,x_n$ are positive real numbers and if $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$
There is a step in which I am confuse. My proof is as follows (it must be proven using induction).
By induction, for $n=1$ the... | In fact， we may prove the more stonger one
Let $x_1,x_2,\cdots,x_n$ be positive numbers. Then
$$\frac{1}{n}(x_1+x_2+\cdots+x_n)\geq \sqrt[n]{x_1x_2\cdots x_n}.\tag1$$
Put $x_1x_2\cdots x_n=1$ into $(1)$. You will get yours.
Inductive Proof
Obviously, $(1)$ holds for $n=1$, which is trivial.
Notice that, from $(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.... | While $x$ is large then $4x^2-7\sim4x^2$ hence
$$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty} \frac{5x+9}{3x+2-|2x|}=\lim_{x\to-\infty} \frac{5x+9}{3x+2+2x}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | The following solution may be the most essential solution...
Denote $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}, ~~~x \in (0,1).$$
Since $$f''(x)=\dfrac{3-x}{4x^{5/2}}>0,~~~\forall x \in (0,1),$$ hence $f(x)$ is a convex function over $(0,1)$. Notice that $a^2, b^2 \in (0,1).$Therefore, $$a+b+\frac{1}{a}+\frac{1}{b}=f(a^2)+f(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 6
} |
$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ It is known that
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?
Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\co... | No, this can't be correct. Remember that $\sin^2x+\cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$\sin a=\frac35\qquad\cos a=\frac45$$
$$\sin b=\frac5{13}\qquad\cos b=\frac{12}{13}$$
and thus
$$\cos(a+b)=\cos a\cos b-\sin a\sin b=\frac45\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Positive integer solutions to $a^3 + b^3 = c^4$ Let $n$ and $m$ be positive integers. We know that
$n^3 + m^3 = n^3 + m^3$
Multiply both sides by $(n^3 + m^3)^3$; on the LHS you distribute, and on the RHS you use power addition rule;
$(n^4 + nm^3)^3 + (m^4 + mn^3)^3 = (n^3 + m^3)^4$
So we get an infinite array of solut... | There are many other solutions $(a,b,c)$. First, start with arbitrary integers $u,v>0$ (you can add the condition $\gcd(u,v)=1$ so as to produce non-overlapping of infinite families of solutions). Then, write
$$u^3+v^3=\prod_{r=1}^k\,p_r^{t_r}\,,$$
where $p_1,p_2,\ldots,p_k$ are pairwise distinct prime natural number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Evaluate $(1+i)^{(1-2i)}$ Find all the values of $(1+i)^{(1-2i)}$ and show that there are small values as we wish (else from $0$) and big values as we wish
\begin{align}
(1+i)^{(1-2i)}&=e^{\ln(1+i)^{(1-2i)}}
=e^{(1-2i)\ln(1+i)}
\\&=e^{(\ln\sqrt{2}+i(\frac{\pi}{2}+2\pi k))(1-2i)}
\\&=e^{\ln\sqrt{2}-2\ln\sqrt{2}*i+i(\fr... | There should only be one value.$$(1+i)^{1-2i}=(\sqrt 2e^\dfrac{i\pi}{4})^{1-2i}=\sqrt 2e^\dfrac{i\pi}{4}(0.5e^\dfrac{-i\pi}{2})^i=\sqrt 2e^\dfrac{\pi}{2}e^{i(\dfrac{\pi}{4}+\ln 0.5)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$
For $x,y,z>0,$ I have to prove that
$$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$$
I tried to use $$ \sqrt {\frac {1}{3} \sum\... | Let $\frac{2x}{y+z}=a$, $\frac{2y}{x+z}=b$ and $\frac{2z}{x+y}=c$.
Thus, $ab+ac+bc+abc=4$ and we need to prove that
$$\sum_{cyc}\frac{1}{(a+1)^2}+\frac{3}{8}abc\geq\frac{9}{8}.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.
Hence, the condition does not depend on $u$.
In another hand, by AM-GM $$4=a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by ... | After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= \frac{-1\pm \sqrt7}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$ I've got one integration question which I first felt was not a hard nut to crack. But, as I proceeded, difficulties arose. This is the one:
$\displaystyle\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$
I went ahead simplifying the... | It can be seen easily that for $H\left( x \right)=\sqrt{{{x}^{2}}-3x+2}$
$$\begin{align}
& {{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}=\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}, \\
& {{\left( xH\left( x \right) \right)}^{\prime }}=\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}, \\
& {{\left( H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi... | Solution
Notice that
$$\sin x \pm\cos x=\sqrt{2}\sin\left(x\pm\frac{\pi}{4}\right),~~~\forall x \in \mathbb{R}$$
Therefore,
\begin{align*}
&\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\\
=&\sqrt{2}\sin\left(\frac{\pi}{20}+\frac{\pi}{4}\right)+\sqrt{2}\sin\left(\frac{3\pi}{20}-\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
Linear Transformations between 2 non-standard basis of Polynomials
If
$$
A = \begin{pmatrix} 1 & -1 & 2 \\
-2 & 1 &-1 \\ 1 & 2 & 3 \end{pmatrix}
$$
is the matrix representation of a
linear transformation $T : P_3(x) \to P_3(x)$ with respect to
bases $\{1-x,x(1-x),x(1+x)\}$ and $\{1,1+x,1+x^2\}$. Find T.... | What you did is fine, but now you have to compute $T[\alpha+\beta x+\gamma x^2]$ for arbitrary $\alpha,\beta,\gamma\in\mathbb R$. In order to do that, solve the equation$$\alpha+\beta x+\gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$\left\{\begin{array}{l}a-b+2c=\alpha\\-2a+b-c=\beta\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
My solution to inhomogeneous $\frac{d^2y}{dx^2} + y = \sin{x}$ does not conform to my book's solution! I need help with the solution of this particular equation:
$$\frac{d^2y}{dx^2} + y = \sin{x}$$
Due to me having to go to work, I cannot display all my work in mathjax, my shift starts in 5 min...but my solution is: ... | Since you mention variation of parameters, I'll summarize the method here. Hopefully you can retrace your steps to find the mistake.
From the fundamental solution
$$ y_h(x) = c_1\cos x + c_2\sin x $$
we then seek a particular solution of the form
$$ y_p(x) = u_1(x)\cos x + u_2(x)\sin x $$
Plugging in the solution and f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The diophantine equation $5\times 2^{x-4}=3^y-1$ I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5\times 2^{x-4}-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
| Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 \cdot 16 = 81 - 1. $ Write this as
$5 \cdot 16 \cdot 2^x = 81 \cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 \cdot 2^x - 80 = 81 \cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Number theory question with floor function
Define $[a]$ as the largest integer not greater than $a$. For example, $\left[\frac{11}3\right]=3$. Given the function
$$f(x)=\left[\frac x7\right]\left[\frac{37}x\right],$$
where $x$ is an integer such that $1\le x\le45$, how many values can $f(x)$ assume?
A. $1$ B. $3$ ... | First, $f(x) \le \left[\frac{x}{7}\cdot\frac{37}{x}\right] = 5$, and $f(x) \ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.
Second, $f(x)=1$ implies $\left[\frac{x}{7}\right]=1$ and $\left[\frac{37}{x}\right]=1$. The first condition implies $7 \le x \le 13$, while the latter requires $19 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$ I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases,
$C-1, \, x \in [2, \infty) $
So, $ x-1 + x-2 = 1 \Rightarrow x= 2$
$C-2, \, x \in [1, 2) $
$x-1 - x + 2 = 1 \Rightarrow 1 =1 \Rightarrow x\in [1,2) $
$C-3, \, x \in (- \infty, 1)$
$ - x + 1 - x... | By direct way we need to distinguish three cases
*
*$x<1 \implies |x-1| + |x-2| = 1-x+2-x=1 \implies -2x=-2\implies x=1$
*$1\le x<2 \implies |x-1| + |x-2| = x-1+2-x=1 \implies 1=1$
*$x\ge 2 \implies |x-1| + |x-2| = x-1+x-2=1 \implies 2x=4\implies x=2$
therefore $1\le x\le 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and... | Try Jensen's Inequality with $\varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$\left(\frac{1}{b-a}\int_a^b(x-p)\,dx\right)^{\!2}\le \frac{1}{b-a}\int_a^b(x-p)^2\,dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Is there a clever way to compute $\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} dx$ I'm wondering how to compute $\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $\int \frac{w^3}{(w^3+1)^3} dw$ what is quite unpleasant as Wolframalpha ... | Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:
We will require two basic properties. The first one is the Euler reflection formula:
$$\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin z \pi}$$
The second is $\Gamma(z+1)=z\Gamma(z)$. So,
$$\Gamma\left(\frac{5}{3}\right) \Gamma\left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{n\rightarrow \infty} ( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +....+ \arctan \frac {1}{2n^2})$
Calculate
$$\lim_{n\rightarrow \infty} \left( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +....+ \arctan \frac {1}{2n^2}\right)$$
My answer: i know that $$ \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right)... | $$\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{2}{1+4n^2-1}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$$
Therefore,
$$\arctan \left( \frac{1}{2n^2}\right) = \arctan (2n+1) - \arctan(2n-1) $$
Can you perform the telescoping sum now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given a matrix $A$ find $A^n$. $A=$$
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} $
Find $A^n$.
My input:
$A^2= \begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix} $
$A^3 ... | The minimal polynomial of $A$ is $(x-1)^2$ so for any entire function $f$ we have
$$f(A) = f(1)P + f'(1)Q$$
for some $P, Q$ polynomials in $A$.
Plugging in $f \equiv 1$ and $f(x) = x$ we get $P = I$ and $Q = A-I$.
Therefore for $f(x) = x^n$ we have
$$A^n = 1^n\cdot I + n1^{n-1}\cdot (A-I) = I + n(A-I) = \begin{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find an equivalent to $\sum_{k=n}^\infty \frac{1}{k!}$ I would like to find an equivalent of $\sum_{k=n}^\infty \frac{1}{k!}$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.
$$n!\sum_{k=n}^\infty \frac{1}{k!}=1+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+... | See this post: Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=1}^{n-1} \frac{1}{k!}$, so the fractional part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=n}^{\infty} \frac{1}{k!}$.
Hence, an equivalent expression is $\frac{\{(n-1)!e\}}{(n-1)!}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating the integral $\int \frac1{(2+3\cos x)^2}\mathrm dx$ Please someone give me an idea to evaluate
this: $$\int \frac1{(2+3\cos x)^2}\mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $\cos^2x$ does not work, so help me here.
| You can use the following approach:
$$\int{\frac{dx}{(2+3\cos x)^2}}=\int{\frac{dx}{(2\cos^2{\frac{x}2}+2\sin^2{\frac{x}2}+3\cos^2{\frac{x}2}-3\sin^2{\frac{x}2})^2}}=$$
$$=\int{\frac{dx}{(5\cos^2{\frac{x}2}-\sin^2{\frac{x}2})^2}}=\int\frac{dx}{\cos^4{\frac x2}(5-\tan^2{\frac x2})^2}=\left[t=\tan\frac{x}{2}\right]=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How to find the sum of $1+(1+r)s+(1+r+r^2)s^2+\dots$? I was asked to find the geometric sum of the following:
$$1+(1+r)s+(1+r+r^2)s^2+\dots$$
My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:
$$1+(s+rs+\dots)... | You expanded the brackets, but did not actually group:
$$ 1 + (1+r)s + (1+r+r^2)s^2 + \dotsb
= 1 + (s+rs \color{red}{{} + \dotsb}) + (s^2+rs^2+r^2s^2\color{red}{{} + \dotsb}). $$
Here is the way to group:
\begin{align*}
& 1+(1+r)s+(1+r+r^2)s^2+\dotsb \\
&= 1+(\color{red}{s}+\color{green}{rs})+(\color{red}{s^2}+\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Let $m$ be the largest real root of the equation $\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4$ find $m$
Let $m$ be the largest real root of the equation $$\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4.$$ Find $m$.
do we literally add all the fractions or do we do ... | HINT:
Write the LHS as $$\left(\frac x{x-3}-1\right)+\left(\frac x{x-5}-1\right)+\left(\frac x{x-17}-1\right)+\left(\frac x{x-19}-1\right)$$ and this gives a constant term of $-4$. Equating this with the RHS, we have$$x\left(\frac1{x-3}+\frac1{x-5}+\frac1{x-17}+\frac1{x-19}\right)=x(x-11)$$ and note that $3,5$ are 'sym... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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 $... | (I won't change the notation although it's not 100% correct)
Hint: Rewrite the basis of $W_1$ as
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=a\left(\begin{array}{c}
1\\
0\\
-2\\
-1
\end{array}\right)+b\left(\begin{array}{c}
1\\
-1\\
0\\
0
\end{array}\right)\right\} $$
A basis of $W_2$ is
$$W... | {
"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": 1
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Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
| I try to calculate $\tan 1 <1.57<\frac{\pi}{2}$
$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}+o(x^9)$
$|\tan x-x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}|<\frac{62x^9}{2835},x\in[0,1]$
So:$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}+\frac{62... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"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.