Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How can I simplify this expression
I know that I cannot just do $\frac{x^{-1}}{x^{-3}}$ and $\frac{y^{-1}}{y^{-3}}$ and get $x^{2}+y^{2}$ but how can i factor either the numerator or denominator to get something to cancel out?
| $\dfrac{x^{-1}+y^{-1}}{x^{-3}+y^{-3}}=\dfrac{x^3y^3(x^{-1}+y^{-1})}{x^3y^3(x^{-3}+y^{-3})}= \dfrac{x^{2}y^3+x^3y^{2}}{x^{3}+y^{3}}=\dfrac{(x^2y^2)(x+y)}{x^{3}+y^{3}}=\dfrac{(x^2y^2)(x+y)}{(x+y)(x^2-xy+y^2)}=\dfrac{(x^2y^2)}{(x^2-xy+y^2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving Diophantine equation without brute force I am having trouble trying to solve this Diophantine equations:
$$
^3=4^2+4−3
$$
I was wondering if anyone could help me find the integer solutions to these and any advice of techniques to use?
Thank you!
| Another way: You can rewrite the equation:
$$x^3+8=4y^2+4y+5$$
Or $$(x+2)(x^2-2x+4)=4y^2+4y+5$$
Let $(x+2;x^2-2x+4)=d (d\in N^+)$
Note that : $x^2-2x+4=x(x+2)-4(x+2)+12\Rightarrow d\mid 12$
We will prove $4y^2+4y+5\not\mid12$. Assume $4y^2+4y+5\mid12$, then we have $4y^2+4y+5=12k(k\in Z)$
$\Leftrightarrow (2y+1)^2=12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
prove that the triangle is isosceles
In a $\triangle ABC$, If
$\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \... | With transformation $R_3 \longrightarrow R_3- \left(\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{B}{2}\right)+\tan\left(\dfrac{C}{2}\right)\right)R_1$
we get $\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Confusion with Summation notation I need to compute the value of this:
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)$$
the $N$ is a gaussian noise with mean=2 and standard deviation=2.
The question:
is this equivalent to:
$$
\frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z... | It could be either one. It is more likely to be the latter; at least I myself would interpret it as
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)=\frac{1}{3}\left(\left(\frac{z_{1}}{f} +\frac{z_{2}}{f} +\frac{z_{3}}{f}\right) + N(2,2)\right),$$
because the former would more likely be written with ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determinant of block matrix with equal diagonals
Let $$Q=\begin{bmatrix}A&B\\-B&A\end{bmatrix}$$ where $A,B\in \mathbb{R}^{n\times n}$. Prove that $$\det(Q)=\det(A^2+B^2)$$
Since $A$ and $B$ do not commute, I cannot use Schur's formula. Also, $A$ and $B$ may not be invertible. Any comment or response is appreciated.... | The trick, if $A$ and $B$ were commutative, is to notice that
$$\begin{bmatrix}A&B\\-B&A\end{bmatrix} \cdot \begin{bmatrix}A&-B\\B&A\end{bmatrix}=\begin{bmatrix}A^2+B^2&0\\0&A^2+B^2\end{bmatrix},$$
which I will refer to as $Q\cdot Q^\top = D$. It is true that $\det(M)=\det(M^\top)$ for any real matrix, so we obtain $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
... | Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
$$\binom{9}{0}+\binom{9}{2}+\binom{9}{4}+\binom{9}{6}+\binom{9}{8}=1+36+126+84+9=256$$
The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 3
} |
Circle problem in which i wanted to find value of $PQ$ Find a value of $PQ$:
Let the radius of bigger circle be $R$, and that of the smaller circles be $r_1$ and $r_2$.
Hence we can find value of $R = 10$.
I don't know how to proceed further to find the Value of $r_1$ and $r_2$.
|
Using the power of the point $O$
with respect to circle $O_1$
we have
\begin{align}
|OT_2|\cdot|OT_1|&=|OP|^2
,\\
\text{or }\quad
R(R-2r_1)&=(R-(|AD|-r_1))^2
,\\
r_1&=
\sqrt{2R(2R-|AD|)}-(2R-|AD|)
\tag{1}\label{1}
.
\end{align}
Similarly,
\begin{align}
r_2&=\sqrt{2R\cdot|AD|}-|AD|
\tag{2}\label{2}
.
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \f... | Consider how $\left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor$ depends on the binary representation of $n$. Dividing by $2^k$ shifts the digits to the right by $k$. There are then two cases:
*
*If the $k^{th}$ binary digit of $n$ is $0$, then adding $1/2$ will not cause a carry, so $\left\lfloor \frac{n}{2^k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is co... | The inequality after "Adding both we get:" is true, but it is not the whole story. You have lost information here, which means that not every solution to this inequality is a solution to both of the constituent inequalities.
Here is a simpler example: suppose we seek the range of solutions to
$$0<x<2$$ and $$1<x<3$$
Ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate using Riemann sums $\int_{\pi/2}^{3\pi/2}(4\sin 3x - 3 \cos 4x)dx$ I have to evaluate this integral using Riemann sums. I've already tried:
$$\int_{\pi/2}^{3\pi/2}(4\sin 3x - 3\cos 4x)dx = [\delta x = \frac{\pi}{n}, x_i = \frac{\pi}{2} + \frac{i\pi}{n}] =$$$$= \lim_{n \rightarrow \infty}\sum_{i = 1}^n(4\sin(3... | To evaluate the Riemann sums use the identities:
$$\sum_{k=1}^n \sin kx = \frac{\sin \frac{nx}{2} \sin \frac{(n+1)x}{2}}{\sin \frac{x}{2}}, \quad \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2} \cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}, $$
which can be derived as the imaginary and real parts of the geometric sum $\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tr... | Let $\sqrt{x+y}=a\ge0,\sqrt{x+3}=b\ge0$
$$(b^2-3)(a+b)=a^2-b^2$$
If $a+b=0,a=b=0$
Else $b^2-3=a-b\iff a=b^2+b-3=x+b\iff x=a-b\ \ \ \ (1)$
Squaring we get
$$x+y=x^2+x+3+2bx$$
$$\iff y=x^2+3+2x\sqrt{x+3}$$
The value of $x,y$ must satisfy $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Algebraic equation I mean if we change interval of "x" to:
$k>1$ and $x\in(\sqrt[3]{k},\sqrt[5]{k^3})$.
Define the $B(x)$ as a polynomial as follows;
$B(x)=-2x^7-3x^6k+2k^2x^4-k^3x^3+2k^3x^2+3xk^4-2xk^3+k^4$
I want to check that whether $B(x)$ is positive?
if not, for which values of $k$ it is positive?
Do we have t... | Let $y:=\sqrt[3]k$ and solve the equation
$$x^7+3x^6-2x^5-\left(y^3+1\right)x^4+\left(2y^3-3\right)x^2+1=0.$$
We easily draw
$$y=\sqrt[3]{\frac{x^7+3x^6-2x^5-x^4-3x^2+1}{x^4-2x^2}}$$ which is the zero set.
That curve is asymptotic to $x=y$ and $x=\sqrt2$, and there are two roots which satisfy $1<x<y$ for every value o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prime numbers $pFind the prime numbers $p<q<r$ such that $r^2-q^2-p^2$ is a perfect square.
I think the only solution is (2,3,7) but i cannot prove it.
The equation would be $r^2-q^2-p^2=k^2$ equivalently
$q^2+p^2+k^2=r^2$ which is somehow a classical diophantine equation (of the pythagorean quadruples) which have a ... | Your guess is correct: $(2,3,7)$ is the unique solution. There is a simple proof using two basic and easy to verify facts:
*
*if $m$ is odd, then $m^2\equiv 1\pmod 4$;
*if $m$ is not divisible by $3$, then $m^2\equiv 1\pmod 3$.
We proceed as follows.
Write $r^2-p^2-q^2=k^2$ with an integer $k$.
If we had $p>2$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$... | Is easy to prove the last step of Andreas:
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} =\\
\frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac1{20} (\log(16/5) + \sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)))\\
\simeq 0.07756
$$
which is exactly as given in the question: $\frac14\log(5) + \frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$ without expansions in limits
Evaluate $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$
One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{\log_e (1+x)^{1/x}} = e^... | Write $f(x) = \frac{1}{x}\log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,
\begin{align*}
\lim_{x\to0} \frac{e^{f(x)} - e + \frac{e}{2}x}{x^2}
&= \lim_{x\to0} \frac{e^{f(x)}f'(x) + \frac{e}{2}}{2x} \\
&= \lim_{x\to0} \frac{e^{f(x)}f''(x) + e^{f(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proof that there exist only 17 Wallpaper Groups (Tilings of the plane) I had a professor who once introduced us to Wallpaper Groups. There are many references that exist to understand what they are (example Wiki, Wallpaper group).
The punchline is
$$There \,\, are \,\, exactly \,\, 17 \,\, wallpaper \,\, groups \,\,(... | Sketch of the proof: Let $\Gamma \le {\rm Iso}(\Bbb R^2)$ be a wallpaper group. Then $\Gamma$ has a normal subgroup isomorphic
to $\Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $\Bbb Z^2$ by conjugation.
We obtain a faithful representation
$$
F \hookrightarrow {\rm Aut}(\Bbb Z^2)\cong GL_2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:
*
*$x^2-1=(x+1)(x-1)$
*$\sqrt{4x-4}=2\sqrt{x-1}$
*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}... | Set $x-1=t^2$, which you can because the limit is for $x\to1^+$. Then the limit becomes
$$
\lim_{t\to0^+}\frac{t\sqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
\lim_{t\to0^+}\frac{\sqrt{t^2+2}+t}{2+t(t^2+2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$
Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$
My attempt:
Firstly, since, $\Bbb Q(\sqrt{2}+3^{1/3}) \subseteq \Bbb Q(\sqrt{2},3^{1/3})$ , I computed $(\sqrt{2}+3^{1/3})^{-1} = 6-4\sqrt{2}+4(3^{1/3})+3(3^{2/3})-3(\sqrt{2}3^{1/3})-2(\sqrt{2... | Let $\alpha = \sqrt2+ \sqrt[3]{3}$.
First isolate $\sqrt[3]{3}$ and raise everything to the third power:
$$\sqrt[3]{3} = \alpha - \sqrt{2} \implies 3 = (\alpha - \sqrt{2})^3 = \alpha^3-3\alpha^2\sqrt{2}+6\alpha - 2\sqrt{2}$$
Then isolate $\sqrt{2}$
$$\sqrt{2}(3\alpha^2+2) = \alpha^3+6\alpha-3$$
so $$\sqrt{2} = \frac{\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Lower Bound on the Sum of Reciprocal of LCM While reading online, I encountered this post which the author claims that
\begin{align}
S(N, 1):=\sum_{1\le i, j \le N} \frac{1}{\text{lcm}(i, j)} \geq 3H_N-2
\end{align}
and $S(N, 1) \geq H_N^2$ where $H_N$ is the partial harmonic sum.
The prove of the latter inequality is... | Sketch proof for 1: assume that there is $c>0$ absolute constant s.t for each $1 \leq a \leq N^{\frac{1}{4}}$ ($N$ high enough), there are at least $c\frac{N^2}{a^2}$ pairs $N \leq k,l \leq 2N$ with $gcd(k,l)=a$; then as $lcm(k,l)gcd(k,l)=kl$,
$s(N):=\sum_{N \leq k, l \leq 2N} \frac{1}{\text{lcm}(k, l)} =\sum_{N \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the Diophantine equation $24x^4-5y^4=z^2$ I want to solve $24x^4-5y^4=z^2$ in integers not all zero, and to fix ideas, I want to find them such that x,y are coprime.
I've tried plugging in small values of $x$ and $y$ and they don't return a square value for $24x^4-5y^4$. So I guess that no (nontrivial) solutions ... | Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives
\begin{align}
z^2 &= 24x^4-5y^4\\
&\equiv -y^4\mod 4
\end{align}
Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2\equiv y^4\equiv 0\mod 4$, so $2|z,y$. Then setting $\overline{y}=\frac{y}{2}$ and $\overline{z}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\gcd(d^{\large 671}\! +\! 1, d^{\large 610}\! −\!1),\ d = \gcd(51^{\large 610}\! +\! 1, 51^{\large 671}\! −\!1)$ Let $(a,b)$ denote the greatest common divisor of $a$ and $b$.
With $ \ d = (51^{\large 610}\! + 1,\, 51^{\large 671}\! −1)$
and $\ \ x \,=\, (d^{\large 671} + 1,\, \ d^{\large 610} −1 )$
fi... | First note: $\gcd(a^m \pm 1, a+1)=\gcd((a^{m}\pm 1)-(a^{m}+a^{m-1}),a+1) = \gcd(a^{m-1}\mp 1, a+1)$
And via induction $\gcd(a^{m}+1, a+1) = \gcd(2, a+1)$ if $m$ is even. $\gcd(a^{m} - 1,a+1) =\gcd(0, a+1) = a+1$ if $m$ is even. (And the opposite results if $m$ is odd).
so if we let $a= 51^{61}$.
Then $\gcd(51^{610} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equival... | $ \frac{2}{a^2+b^2} \leq \frac{1}{ab} , \frac{2}{c^2+d^2} \leq \frac{1}{cd}$
$ \Rightarrow \frac{2(a^2+b^2+c^2+d^2)}{(a^2+b^2)(c^2+d^2)} \leq \frac{1}{ab} + \frac{1}{cd}$
$ \Rightarrow \frac{a^2+b^2+c^2+d^2}{2} \cdot (\frac{4}{(a^2+b^2)(c^2+d^2)} - 1)\leq \frac{1}{ab} + \frac{1}{cd} - \frac{a^2+b^2+c^2+d^2}{2}$
also... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing f... | Here is a brute force solution. Letters $x,y,z$ stand for nonzero integers.
Case 1. Permutations of $(x,x,x)$.
$(1,1,1)$
$(2,2,2)$
$2$ combinations, one permutation of each, $2^3=8$ ways to assign $\pm$ signs, so $2\cdot1\cdot8=\boxed{16}$ solutions.
Case 2. Permutations of $(x,x,y)$, $x\ne y$.
$(1,1,2)$
$(1,1,3)$
$(1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$. I want to show:
If $c^2+8 \equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$, so it should be ... | $$2c-7\equiv2c-7+c^2+8\pmod p\equiv(c+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Trigonometric residue integral The trigonometric complex integral is
$$
\int_0^{2\pi} \frac 1{(2+\sin \theta)^2} d\theta.
$$
My answer is below, and it is correct because I verified this numerically on WolframAlpha: https://www.wolframalpha.com/input/?i=%5Cint_0%5E(2pi)+1%2F%7B(2%2B%5Csin+%5Ctheta)%5E2%7D+d%5Ctheta
| I applied the substitution $z=e^{i\theta}$ and then used the residue theorem to obtain
\begin{align*}
\int_0^{2\pi} \frac 1{(2+\sin \theta)^2} d\theta &= \int_{|z|=1} \frac 1{(2+\frac 1{2i}(z-\frac 1z))^2} \frac{dz}{iz} \\
&= 4i\int_{|z|=1} \frac z{(z^2+4iz-1)^2} dz \\
&= 4i \cdot 2\pi i \operatorname{Res}_{z=(-2+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Complex sum of sine and cosine functions
By using the complex representations of sine and cosine, show that $$\sum_{m=0}^n\sin m\theta =\frac{\sin\frac{n}{2}\theta\sin\frac{n+1}{2}\theta}{\sin\frac{1}{2}\theta}$$
So I am not too sure how to go about this proof. I tried to substitute sine for its complex representatio... |
Based upon the finite geometric series formula and
$\sin \theta=\frac{1}{2i}\left(\exp(i\theta)-\exp(-i\theta)\right)$
we obtain:
\begin{align*}
&\color{blue}{\frac{\sin\frac{n}{2}\theta\sin\frac{n+1}{2}\theta}{\sin\frac{1}{2}\theta}}\\
&\qquad=\frac{\frac{1}{2i}\left(\exp\left(i\frac{n}{2}\theta\right)-\exp\left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof for $\frac{\cos^6 x}{\cos 6x}=1$ I tried some transformations but nothing is bringing me to 1.
$$\frac{\cos^6 x}{\cos 6x} = 1$$
I tried something like $\left(\cos^2 x\right)^3$, then $\left(2\cos^2x-1\right)^3$. And, for $\cos 6x$, to transform into $\cos(2(3x))$ or $\cos(3 (2x))$. But it's not going well.
| Let $\cos^2x=t$.
Thus, since $$\cos3\alpha=4\cos^2\alpha-3\cos\alpha,$$ we obtain:
$$\cos^6x=\cos6x$$ or
$$t^3=2(4\cos^3x-3\cos x)^2-1$$ or
$$t^3=2t(4t-3)^2-1$$ or
$$31t^3-48t^2+18t-1=0$$ or
$$31t^3-31t^2-17t^2+17t+t-1=0$$ or
$$(t-1)(31t^2-17t+1)=0.$$
1. $t=1$.
Thus, $$\cos^2x=1$$ or $$\sin{x}=0$$ or $$x=\pi k,$$ where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum_{n=1}^\infty\frac{k^2}{k^2-2k+5}$, $\sum_{n=1}^\infty\frac{6\cdot 2^{2n-1}}{3^n}$, and $\sum_{n=1}^\infty\frac{2^n+4^n}{e^n}$ I have three series questions:
1) $$\sum_{n=1}^{\infty} \frac{k^2}{k^2 - 2k + 5}$$
I'm going to use this theorem:
So $$\lim_{x\to \infty} \frac{x^2}{x^2-2x+5}$$
So $$\li... | In both cases you can use the necessary condition for convergence of a series:
If
$$\sum_{n=0}^\infty a_n$$
converges, then
$$a_n\to 0$$
(the same is true if the sum starts at some integer different from zero).
This implies that if $a_n\to 0$ is not true (whether because the limit is a non zero number, or infinity, or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving difference between two functions is small when $x$ is small... I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| \leq 10^{-4} \implies \bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}$$
My attempt at a... | For $|x|<\epsilon=10^{-4}$, by the mean value theorem $$\left|\frac{e^x-1}{x}\right|=e^y\ge e^{-\epsilon}\ge1-\epsilon,$$ and $$1+\frac{x}{2}+\frac{x^2}{6}\ge1-\frac{\epsilon}{2},$$ so it follows that $$\left|\frac{x}{e^x-1}\frac{1}{1+\frac{x}{2}+\frac{x^2}{6}}\right|\le\frac{1}{(1-\epsilon)^2}\le2\le\frac{4!}{e^\epsil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determining a marginal of a joint categorical distribution I have a set of $n$ discrete variables $\mathcal{A} = \{ a_{1}, \dots, a_{n} \}$, where every $a_{i} \in \mathcal{A}$ can take on the values in $ \{ 0, 1, \dots, m \}$ and $n,m \in \mathbb{N}$. I wish to create a joint measure -- an unnormalised probability di... | Here's how you prove this is a probability distribution. I have removed the upper limits of the summations; this is allowed since all subsequent terms are zero.
\begin{align}
\sum_{j=0}^m P(a_i=j)
&=P(a_i=0)+mP(a_i\neq 0)
\\&=\frac{1}{K} \sum_{k\ge 0} k! \binom{n-1}{k} \binom{m}{k}+m\cdot\frac1K\sum_{k\ge 0}k!\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = \frac{-130}{13}=-10$.
I... | You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $c\in[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$ \lim\limits_{x\rightarrow +\infty} \left(\frac{2}{\pi} \arctan x \right)^x$ and $\lim_{x\rightarrow 3^+} \frac{\cos x \ln(x-3)}{\ln(e^x-e^3)}$?
I got stuck on two exercises below
$$
\lim\limits_{x\rightarrow +\infty} \left(\frac{2}{\pi} \arctan x \right)^x \\
\lim_{x\rightarrow 3^+} \frac{\cos x \ln(x-3)}{\ln(e... | Without L'Hospital
$$y=\left(\frac{2}{\pi} \arctan (x) \right)^x\implies \log(y)=x \log\left(\frac{2}{\pi} \arctan (x) \right) $$
Now, by Taylor for large values of $x$
$$\arctan (x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$\frac{2}{\pi} \arctan (x) =1-\frac{2}{\pi x}+\frac{2}{3 \pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$
and
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac... | Following your professor's advice, let $A=\frac{x+y}{2}$, $B=\frac{x-y}{2}$. Then $$x=A+B\\y=A-B$$So the LHS of your equation becomes $$\sin(A+B)-\sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2\cos A\sin B$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Nice way of showing the locus of $\left|z-\frac{1}{z}\right| = 2$ is the union of 2 intersecting circles. For visualisation, the following code was run in Mathematica.
ContourPlot[{Abs[x + I y - 1/(x + I y)] == 2}, {x, -2.42, 2.42}, {y, -2.42, 2.42}, Axes -> True]
Here is the locus plot generated by the above code.
I ... | $$\begin{align}
\left|z-{1\over z}\right|=2
&\iff|z-1||z+1|=2|z|\\
&\iff((x-1)^2+y^2)((x+1)^2+y^2)=4(x^2+y^2)\\
&\iff(x^2+y^2+1-2x)(x^2+y^2+1+2x)=4(x^2+y^2)\\
&\iff(x^2+y^2+1)^2-4x^2=4(x^2+y^2)\\
&\iff(x^2+y^2-1)^2-4x^2=0
\end{align}$$
On the other hand, the equation for the locus of two circles of radius $r$, centered... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum the first n^3 how provide with using mathematical induction? I having next formula and need solving with helps mathematical induction.
$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$
My next steps in solving the task:
a)I check the validity of this assertion for n = 1
$\sum_{i=1}^1 1^3 = \biggl(\frac{1(1+1)}... | You need to show that
$$
\sum_{i=1}^{n+1} i^3 = \left( \frac{(n+1)(n+2)}{2}\right)^2,
$$
using as an hypothesis that
$$
\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2}\right)^2.
$$
So,
\begin{align*}
\sum_{i=1}^{n+1} i^3=& \sum_{i=1}^n i^3 + (n+1)^3= \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3\\
=& (n+1)^2\left( \frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right).$ Find
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right)$$
Choices:
| 1
B: $\frac{1}{2}$
C: $\frac{1}{4}$
D: 0
My attempt is:
The question is directly equal to
$$\int_0^1\left(e^{(1+x)^2} - \frac{3e^{(1+3x)}}{2\sqrt{1+3x}}\right)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Need help for integration I need help with finding the following integral
$$\int_0^{+\infty} \frac{x^a\,dx}{(1+x)^2(A+x^a)}, $$ where $A>0$ and $a \in \mathbb{R}\,.$
This is my first trial.
\begin{align}
\int_0^\infty \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)} \\
&= \int_0^1 \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+... | I give you a hint to solve your integral analytical:
First you split the integral in two parts:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, \
dx=\int_0^{\infty } \frac{1}{(1+x)^2} \, dx-\int_0^{\infty } \
\frac{A}{(1+x)^2 \left(A+x^a\right)}\, dx$$
The first integral is trivial:
$$\int_0^{\infty } \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Analytical solution to $\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2} \cdot \sqrt{b^2-x^2}$ Can't find beauty analytical solution to such equation :
$$\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2}\cdot\sqrt{b^2-x^2}$$
Assuming $a,\ b \in \mathbb{N},\ a \le b,\ x \in \mathbb{R}$
Is it possible to find a solution f... | So, we have $$\dfrac1{\sqrt{a^2-x^2}}+\dfrac1{\sqrt{b^2-x^2}}=1$$
WLOG $a^2-x^2=\sin^4t, b^2-x^2=\cos^4t$
$\implies b^2-a^2=\cos2t$
$$4a^2-4x^2=(2\sin^2t)^2=(1+a^2-b^2)^2$$
$$\iff4x^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all integral solutions to $xy^2 + (x^2 + 1)y + x^4 + 1 = 0$. I have an equation $$xy^2 + (x^2 + 1)y + x^4 + 1 = 0$$ and I need to find a set of solutions for it in integers. I tried to solve it using Maple with the isolve () command, but it doesn’t give anything. This suggests that either the Maple algorithms cann... | This answer makes use of the rational root theorem a lot. Let $x$ and $y$ be integers such that
$$xy^2 + (x^2 + 1)y + x^4 + 1 = 0.\tag{1}$$
We first make a few simple observations.
*
*If $x=0$ then $y=-1$, and if $y=-1$ then $x=0$.
*There are no other integral solutions with $|x|\leq2$.
*Equation $(1)$ shows that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\fr... | Draw a right triangle whose tangent is $x$. Label the opposite side $x$ and the other leg $1$. Now find the length of the hypotenuse; it's $\sqrt{x^2 + 1}$.
Then, compute the cosine of the complementary angle. You are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Why is this limit evaluated like so? Question:
If
$$\lim_{x \to 0}{\frac{-1 + \sqrt{(\tan x - \sin x) + \sqrt{(\tan x - \sin x) + \sqrt{(\tan x - \sin x) + \cdots \infty}}}}{-1 + \sqrt{x^3 + \sqrt{x^3 + \sqrt{x^3 + \cdots \infty}}}}} = \frac{1}{k}$$
Then find the value of $k$.
The way I approached the problem was by s... | Okay. when we look at these continued roots....lets look at the one in the numerator, the logic will hold for the one in the denominator.
First the roots will not be defined as real functions if $x<0$
$\lim_\limits{x\to 0} \tan x - \sin x = 0$
$\tan x - \sin x > 0$ for all $0<x<\frac {\pi}2$
When x is small and positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Showing matrix $\left[\begin{smallmatrix} 4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3 \end{smallmatrix}\right]$ is positive definite $$\begin{bmatrix}
x_1 & x_2 & x_3
\end{bmatrix}\begin{bmatrix}
4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3
\end{bmatrix}\begin{bmatrix}
x_1\\x_2\\x_3
\end{bmatrix} = \\4x_1^2 + x_1x_2 + x_3x_... | Note that we can show that a matrix is positive definite by looking at its $n$ upper left determinants. Note that for the matrix
$$\begin{bmatrix}
4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3
\end{bmatrix}$$
We have
$$\begin{vmatrix}
4
\end{vmatrix} = 4$$
and
$$\begin{vmatrix}
4 & 1\\
1 & 2
\end{vmatrix} = 7$$
and lastly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $ substitution I have this integral.
$$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $$
But I am unable to do it right and I just don't know where is the problem in my steps.
My steps:
Complete the square
$$\int \frac{1}{{2-(x+1)^2}} \, \mathrm{d}x $$
$$\frac{1}{2}*\int \frac{1}{{... | Use the u-substation $u=\frac{x+1}{\sqrt{2}}$ and then do a partial fraction decomposition on the fraction that you will get:
$$
\int\frac{1}{1-2x-x^2}\,dx=\int\frac{1}{2-(x+1)^2}\,dx=\\
\frac{\sqrt{2}}{2}\int\frac{1}{1-\left(\frac{x+1}{\sqrt{2}}\right)^2}\frac{d}{dx}\left(\frac{x+1}{\sqrt{2}}\right)\,dx=
\frac{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why do I get different results when testing for increasing/decreasing intervals of a function here? I have the function $f(x)=6+\frac{6}{x}+\frac{6}{x^2}$ and I want to find the intervals where it increases or decreases. The problem is that when I find $f'(x)=0$, which becomes $x=-2$. Once I put $-2$ on a number line, ... | Notice that
$$ f^\prime(x)=-\frac{6}{x^2} -\frac{12}{x^3}=-\frac{6}{x^3}(x+2)$$
So $f^\prime$ is undefined at $x=0$ and $f$ has slope $0$ at $x=-2$.
You should therefore check for increasing/decreasing on the intervals $(-\infty,-2),\,(-2,0)$ and $(0,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the partial sum formula of $\sum_{i=1}^n \frac{x^{2^{i-1}}}{1-x^{2^i}}$ Given next series:
$$\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \frac{x^4}{1 - x^8} + \frac{x^8}{1 - x^{16}} + \frac{x^{16}}{1 - x^{32}} + ... $$
and $|x| < 1$. Need to derive $S_n$ formula from series partial sums.
I could only find that $... | Hint:
Use method of cancellation
$$\frac {x^2} {1-x^4} = \left[\frac 1 {1-x^2} - \frac 1 {1-x^4}\right]$$
$$\frac {x^4} {1-x^8} = \left[\frac 1 {1-x^4} - \frac 1 {1-x^8}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}$ when $x^2+y^2=1$ I have to prove the inequality
$$
\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}
$$
when $x^2+y^2=1$, using Cauchy-Schwarz Inequality.
The RHS is equal to $\frac{1... | For $xy\geq0$ by C-S $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{3}{2+x^2y^2}+\frac{1}{1+xy}=$$
$$=\frac{4}{\frac{4(2+x^2y^2)}{3}}+\frac{1}{1+xy}\geq\frac{(2+1)^2}{\frac{4(2+x^2y^2)}{3}+1+xy}.$$
Id est, it's enough to prove that
$$\frac{9}{\frac{4(2+x^2y^2)}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or
$$(1-2xy)(1+8xy)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sum_{k=1}^{n}(5^{2k}-5^{2k-1})=\frac{5^{2n+1}-5}{6}$, for n = 1,2,3, ... Show that
$\sum_{k=1}^{n}(5^{2k}-5^{2k-1})=\frac{5^{2n+1}-5}{6}$, for n = 1,2,3, ...
n = 1
$LHS = 5^2 - 5 = 20$
$RHS =\frac{ 5^{3} - 5}{6} = \frac{120}{6} = 20$
n = p
$LHS_{p} = (5^{2(1)}-5^{2(1)-1}) + (5^{2(2)}-5^{2(2)-1})+....+ (5^{... | Hint:
$$5^{2k}-5^{2k-1}=5^{2k-1}(5-1)=4\cdot5^{2k-1}$$
So, we have
$$\sum_{k=1}^n4\cdot5^{2k-1}$$ which is a geometric Series
Use Induction proof dealing with geometric series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lyapunov's theory for differential equations Use Lyapunov's theory to discuss stability of the origin for
$$x' = -x^3 - xy^2 + xy^5$$
$$y' = -y^3 + x^2y - 3x^2y^4$$
$$L(x,y) = ax^2 + by^2$$
In class, we define $L(x,y)$ as being positive definite. Then we determine if the origin is a critical point. The we take the deri... | Given the system
$\dot x = -x^3 - xy^2 + xy^5, \tag 1$
$\dot y = -y^3 + x^2y - 3x^2y^4, \tag 2$
it is easily seen that $(0, 0)$ is a critical point; we seek a Lyapunov function of the form
$L(x,y) = ax^2 + by^2; \tag 3$
we have
$\dot L = 2ax \dot x+ 2by \dot y$
$= 2ax(-x^3 - xy^2 + xy^5) + 2by(-y^3 + x^2y - 3x^2y^4)$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that the simple continued fraction of $\sqrt{d}$ has period length 1 iff $d=a^2+1$ Given that I know if $d$ is an integer that $\sqrt{d}=[\alpha_0,\bar{\alpha_1},...\bar{\alpha_n},\bar{2\alpha_0}]$. I want to show that $\sqrt{d}$ has period length 1 if and only if $d=a^2+1$, for some natural number a.
Here is w... | $$\sqrt{a^2}\lt\sqrt{a^2+1}\lt\sqrt{a^2+2a+1}$$
$$\therefore a\lt\sqrt{a^2+1}\lt a+1\implies\left\lfloor\sqrt{a^2+1}\right\rfloor=a$$
Hence,
$$\begin{align}
\sqrt{d}
&=\sqrt{a^2+1}\\
&=a+\sqrt{a^2+1}-a\\
&=a+\frac1{\left(\frac1{\sqrt{a^2+1}-a}\right)}\\
&=a+\frac1{a+\sqrt{a^2+1}}\\
&=a+\frac1{a+\sqrt{d}}\\
&=a+\frac1{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inscribed circle in a right-angled triangle
In right-angled $\triangle ABC$ with catheti $a = 11\,\text{cm}, b=7\,\text{cm}$ a circle has been inscribed. Find the radius and altitude from $C$ to the hypotenuse.
I found that the hypotenuse is $c = \sqrt{170}$ and the radius is $r = \frac{a+b-c}{2}=\frac {18 - \sqrt{17... | I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.
Since $\triangle ABC$ is a right triangle, and since
$\triangle AHC$ and $\triangle HBC$ each share one angle with $\triangle ABC$,
the three triangles are similar. We have
$$
\frac{CH}{AC} = \frac{BC}{AB}
$$
and (equivalently)
$$
\frac{CH}{BC} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Intriguing Limit Prove that:
$$L=\lim_{n\to\infty} \frac {\sqrt 2 n^{\left(n-\frac 12\right)}}{n!}\left(\frac {(2\sqrt[n] {n} -1)^n}{n^2}\right)^{ \frac {n\left(n-\frac 12\right)}{\ln^2 n}}=\sqrt {\frac {e}{\pi}}$$
My method:
Properties I am going to use :
1)Stirling's approximation:$$n!\sim\sqrt {2\pi n} \left(\fra... | We can start with
$$
\log\left(n^{1/n}\right)=\frac{\log(n)}n
$$
and use the power series for $e^x$ to get
$$
2n^{1/n}-1=1+2\frac{\log(n)}n+\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{3n^3}+O\!\left(\frac{\log(n)^4}{n^4}\right)
$$
The power series for $\log(1+x)$ yields
$$
\begin{align}
\log\left(2n^{1/n}-1\right)
&=\overbr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way? I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42
\end{equation*}... | Outstanding solution due to Cornel Valean.
Recall the generating function $\displaystyle-\ln(1+x)\ln(1-x)=\sum_{n=1}^\infty x^{2n}\frac{H_{2n}-H_n}{n}+\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2},$ where if we multiply both sides by $\ln(1+x)/x$ and use the simple fact $\displaystyle\int_0^1x^{2n-1}\ln(1+x)\ dx=\frac{H_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now,... | Maybe it would be easier with prettier numbers? If you were to solve the equation $$x^2+200x+10000=0,$$ you’d recognize that $$(x+100)^2 = x^2+200x+10000$$ which would simplify things greatly, quickly leading to the solution $x=-100$.
If, however, your equation does not match the $(x+a)^2 = x^2 + 2ax + a^2$ formula per... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Minimum possible values of $\frac{x^2+y^2+z^2+1}{xy+yz+z}$ and $\frac{x^2+y^2+z^2+1}{xy+y+z}$ Let $k$ and $m$ be the minimum possible values of $$\frac{x^2+y^2+z^2+1}{xy+yz+z} \quad \text{and} \quad \frac{x^2+y^2+z^2+1}{xy+y+z}$$ respectively where $x,y,z$ are non-negative real numbers. What is the value of $km+k+m$?
I... | Let $U = [0,\infty)^3$ and $V = [0,\infty)^3 \times (0,\infty)$.
We will parameterize $U$ by elements in $V$ through following map:
$$V \in (X,Y,Z,W) \mapsto (x,y,z) = \left(\frac{X}{W},\frac{Y}{W},\frac{Z}{W}\right) \in U$$
The minimum $k$ we seek equals to
$$k = \min_{(x,y,z)\in U}\left\{\frac{x^2+y^2+z^2+1}{xy+yz+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$?
Prove the identity
\begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
for each nonnegative integer $n$.
To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^... | We will start off with the basis step using the base value of $n = 0$.
Basis step
$\prod_{i=0}^{0} (1 + q^{2^i}) = \dfrac{1 - q^{2^{n + 1}}}{1 - q}$
=> $1 + q^{2^0} = \dfrac{1 - q^{2^{0 + 1}}}{1 - q}$
=> $1 + q = \dfrac{1 - q^2}{1 - q}$
=> $(1 + q)(1 - q) = 1 - q^2$
=> $1 - q^2 = 1 - q^2$
The basis step is hereby prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when... | $\sqrt{x^2+25}$ is the distance between $(x,0)$ and $(0,5)$, while $\sqrt{(x-12)^2+16}$ is the distance between $(x,0)$ and $(12,-4)$. Hence we need to find $x$ such that the sum of these two distances reaches the minimum. That is exactly the point that the line through $(12,-4)$ and $(0,5)$ meets the $x$-axis. Therefo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Uniform convergence of $\sum_{n=0} ^\infty \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1}$
I am given:
$$ f(z)=\sum_{n=0} ^\infty \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1}$$
And asked to show that this complex function series converges uniformly for $|z|\leq \frac{1}{3}$. I am also asked t... | Observe for $|z|\leq \frac{1}{3}$ We have:
$$|f_n|=\left| \frac{(-1)^n}{2n+1} \left(\frac{2z}{1-z^2}\right)^{2n+1} \right| =
\frac{1}{2n+1} \left|\left(\frac{2z}{1-z^2}\right)\right|^{2n+1} $$
By reverse triangle inequality we get:
$$|f_n|\leq \left|\left(\frac{2|z|}{|1|-|z^2|}\right)^{2n+1} \right|=\left|\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For which $k\in \Bbb N$ does $x^2+x+1$ divide $ (x + 1)^{2k}+x^{2k} + 1$? I tried many attempt to solve this problem but failed every time.
The polynomial $x^{2k} + 1 + (x + 1)^{2k}$ is not divisible by $x^2 + x + 1$. Find the value of $k \in \mathbb{N}$
My question is:
can this be solved with using remainder/factor th... | You said you haven't learned number theory yet.
I don't know if you have learned about complex roots of unity.
Here is a proof where you just need to know the binomial expansion and mathematical induction.
Let $f_n(x)=(x+1)^n+x^n+1.$
I claim $f_{6s}(x)=q_1(x)(x^2+x+1)+3,$
$ f_{6s+2}(x)=q_2(x)(x^2+x+1),$ and
$f_{6s+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the solution of the equation $xyp^2 + (3x^2 - 2y^2)p - 6xy=0$, where $p = \frac{dx}{dy}$ What is the solution of the equation $xyp^2 + (3x^2 - 2y^2)p - 6xy=0$, where $p = \frac{dx}{dy}$
I was trying to solve it by dividing the whole equation by $xy$ and then integrate it
$\frac{dy}{dx}[\frac{dy}{dx} + (\frac{... | Using quadratic formula, we have
$$p=\frac{2y^2 - 3x^2\pm\sqrt{(3x^2-2y^2)^2+24x^2y^2}}{2xy}$$
$$=\frac{2y^2 - 3x^2\pm(3x^2+2y^2)}{2xy}. $$
That is:
$$\frac{dy}{dx}=p=\frac{2y}{x} \text{ or } \frac{-3x}{y}. $$
In both cases, we can easily solve the ODE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
not every matrix is sum of two idempotent matrices i'm reading the proof of theory show that the matrix ring $M_{n}(R)$ over $R$ contains an element which cannot be written as sum of two idempotents. ($R$ is a ring with $1$, and $n$ is a positive integer greater than $1$)
Lemma: Let $a$ be an element of $R$ with $a^{2}... |
Why we got $4R=0$
Because $\begin{bmatrix}0&4r\\0&0\end{bmatrix}=r4\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ for any $r$, so $4R=\{0\}$.
And compare the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$?
Just write out $a^{2}= a + ae + ea — 2e$:
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing volume inside a ball and outside a cylinder I try to calculate the volume inside the ball $\ x^2 + y^2 + z ^2 = 4 $ the outside the cylinder $\ x^2+y^2=2x $ using double integral.
Since the shape is symmetric I chose $\ z = 0 $ as bottom limit and the ball is the upper limit. I try to convert to polar form a... | Let $I = \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$
$I = \frac{-1}{2}\int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ d(4-r^2) \ d \theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-r^2)^{3/2}\big]^{2cos\theta}_0d\theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-4cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to
$$100!\times\left(1+\sum_{n=1}^{10... | Write the sum like
$$\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\frac{(-1)^n n}{n!} + \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=1}^{100}\frac{(-1)^n n}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that the characteristic polynomial is the same as the minimal polynomial
Let $$A =\begin{pmatrix}0 & 0 & c \\1 & 0 & b \\ 0& 1 & a\end{pmatrix}$$
Show that the characteristic and minimal polynomials of $A$ are the same.
I have already computated the characteristic polynomial
$$p_A(x)=x^3-ax^2-bx-c$$
and I kn... | Compute:
$$A^2 = \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix}.$$
So, we just need to show that $A^2, A, I$ are linearly independent. Clearly $A$ is not a multiple of $I$, so we just need to show there is no solution to the equation
$$A^2 = pA + qI \iff \begin{pmatrix} 0 & c & ac \\ 0 & b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x =... | Edit: This is now obsolete - it applies to the original version of the question, without the condition "in simplest form":
It's clear that $a+b+c$ is not determined, since $$\frac{2a+\sqrt{4b}}{2c}=\frac{a+\sqrt b}c$$but $$2a+4b+2c\ne a+b+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me pleas... | Let $x^2+6x+7=y$.
Thus, $y^2+6y+7=x$, which gives
$$x^2+6x+7-(y^2+6y+7)=y-x$$ or
$$(x-y)(x+y+7)=0.$$
Thus, $x=y,$ which does not give real solutions, or
$$y=-x-7,$$ which gives
$$x^2+6x+7=-x-7.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Show that $2 - \sqrt{2}$ is irrational I suppose $2 - \sqrt{2} $ is rational.
so $$2- \sqrt{2} = {a/b} $$
where a,b are integers and gcd(a,b) = 1.
$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$
$$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$
$$\text{Step 3. }2b^2 = (2k)^2$$
Step 3 since $a... | Yet another proof: $0<2-\sqrt{2}<1$, so $(2-\sqrt{2})^n\to 0$. Note that $(2-\sqrt{2})^n=a_n-b_n\sqrt{2}$ for some integers $a_n,b_n$ by binomial theorem, so $$0<(2-\sqrt{2})^n=(a_n-2b_n)+b_n(2-\sqrt{2}).\tag{1}$$
If $2-\sqrt{2}$ were rational, say $=p/q$, $p,q\in\mathbb{N}$, then by equation (1) $(2-\sqrt{2})^n$ must... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Why does $ \lim_{x \to 2} \frac{x^2-4}{x-2} =4 $ if x cannot be 2? I know that $ \lim_{x \to 2} \frac{x^2-4}{x-2} $ is evaluated as follows :-
$$ \lim_{x \to 2} \frac{x^2-4}{x-2} \\ = \lim_{x \to 2} \frac{(x+2)(x-2)}{x-2} \\ = \lim_{x \to 2} x+2 \\ = 2+2 \\ = 4 $$
By looking at the function $ \frac{x^2 - 4}{x - 2} $, I... | Limit doesn't have to do with the existence of function at that point. In the context of the given problem,
$$f(x) = \frac{x^2-4}{x-2}= \frac{(x-2)(x+2)}{x-2}$$
We study the behavior of $f$ as $x$ approaches a certain point. For example let's study the RHL by considering the value of $f$ at some points:
$$f(2.01) = 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
| Note that $(1+ 2 \sqrt3)^2 = 1 + 4 \sqrt 3 + 4 \cdot \sqrt 3^2 = 13 + 4 \sqrt 3$. The expression is now easy to simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
prove $\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$ this series was evaluated by Cornel Valean here using series manipulation.
I took a different path as follows:
using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
multiply both sides by $\ln^3x/x$ ... | For a slight variation on a theme.
As
$$\int_0^1 x^{n - 1} \ln^2 (1 - x) \, dx = \frac{H^2_n}{n} + \frac{H^{(2)}_n}{n},$$
for a proof of this result, see here, we can write the sum as
\begin{align}
\sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{1}{n^3} \cdot \frac{H^2_n}{n}\\
&= - \sum_{n = 1}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Quickest way to find characteristic polynomial from a given matrix Find the Rational form of
$$
A=\begin{pmatrix} 1&2&0&4\\ \:\:\:4&1&2&0\\ \:\:\:0&4&1&2\\ \:\:\:2&0&4&1\end{pmatrix}
$$
I don't wanna the solution, instead I would like to know a quickest way to calculate $\det (\lambda I-A)$.
$$
\begin{vmatrix}
x-1 &... | Add columns $\,2,3,4\,$ to the first one (this is a circulant matrix), and get:
$$\det(t I-A)=\begin{vmatrix}t-7&-2&0&-4\\
t-7&t-1&-2&0\\
t-7&-4&t-1&-2\\
t-7&0&-4&t-1\end{vmatrix}\stackrel{R_i-R_1}=\begin{vmatrix}t-7&-2&0&-4\\
0&t+1&-2&4\\
0&-2&t-1&2\\
0&2&-4&t+3\end{vmatrix}=$$$${}$$
$$=(t-7)\begin{vmatrix}
t+1&-2&4\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expr... | It's not actually a full answer but, in my view, interesting generalization.
Note that the problem is a special case of the following statement.
Proposition. Let $\beta$ be a symmetric bilinear form in $n$-dimensional euclidean space $\left(V, \langle\cdot, \cdot\rangle\right)$ and $\lambda_1\geq \lambda_2\geq\ldots\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1... | Hint
$$p(x)=(1 + x + x^2 +\cdots+x^n)^{n-1}=\left(\frac{x^{n+1}-1}{x-1}\right)^{n-1}$$ $$ p(x)(x-1)^{n-1}\equiv(x^{n+1}-1)^{n-1}$$
and remember that $p(x)$ has degree equal to $(n+1)(n-1)-(n-1)=n^2-n$. So,
$$p(x)=a_0+a_1x+...+a_{n^2-n}x^{n^2-n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17 $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equi... | Since $x^2+5x+6=(x+2)(x+3)$,\begin{align}x^2+5x+6\equiv187\pmod{187}&\iff(x+2)(x+3)\equiv0\pmod{187}\\&\iff(11\mid x+2\vee11\mid x+3)\wedge(17\mid x+2\vee17\mid x+3).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2... | The result can be proved using AM $\geq$ GM.
\begin{align}
\sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}&=\sqrt[2m+1]{\prod_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}}}\leq\frac{1}{2m+1} \sum_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}} \\
&=\frac{1}{2m+1} \left[\sum_{i=0}^{m-1} \left(\frac{1+x^{i+1}}{1+x^{i}}+\frac{1+x^{2m-i+1}}{1+x^{2m-i}} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
$\int_{0}^{\pi /2} (\cos x)^{2.5} \mathrm dx$ $$\int_{0}^{\pi /2} (\cos x)^{2.5} \mathrm dx$$
I tried changing to sine and substituting, but it's getting complex. Help.
| It is a difficult problem because you face elliptic integrals.
$$\int\cos ^{\frac{5}{2}}(x)\,dx=\frac{2}{5} \left(3 E\left(\left.\frac{x}{2}\right|2\right)+\sin (x) \cos^{\frac{3}{2}}(x)\right)$$ making
$$\int_0^{\frac \pi 2}\cos ^{\frac{5}{2}}(x)\,dx=\frac{3}{5} \sqrt{2} \left(2
E\left(\frac{1}{2}\right)-K\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $ Compute $\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $. I tried to apply the first mean value theorem for definite integrals and then apply the squeeze theorem, but it didn't work. The answer given ... | We should start by noting that
$$
0 \leq \int_n^{2n}\frac{x}{1+x^5}dx\leq \int_n^{2n} \frac{x}{1+x^4} dx=\left[\frac 12 \arctan(x^2)\right]_n^{2n}=\frac 12 (\arctan(2n)-\arctan n) \to 0.
$$
This justifies that we can apply L'Hospital's rule getting
$$
\lim n^3 \int_n^{2n}\frac{x}{1+x^5}\,dx = \lim \frac{\int_n^{2n}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What property was used in this sine transformation? I have this expression:
$$
ψ(χ) = A\sin^3(\frac{πχ}{α})
$$
And somehow the book i read equalizes the previous equation to this one:
$$
ψ(χ) = \frac{A}{4}[3\sin(\frac{πχ}{α}) - \sin(\frac{3πχ}{α})]
$$
What trigonometric identity was used to make this possible?
| By DeMoivre's identity,
$\cos(3x)+i\sin(3x)=(\cos(x)+i\sin(x))^3=$
$\cos^3(x)+3i\cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-i\sin^3(x)$.
Grouping the imaginary terms together, we see that $\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)$.
We can rewrite $3\cos^2(x)\sin(x)$ as $3\sin(x)(1-\sin^2(x))$, which gives us $\sin(3x)=3\sin(x)-4\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the maximum value of $\sum_{cyc}\frac{1}{\sqrt{a^2 + b^2}}$ where $a, b, c > 0$ and $abc = a + b + c + 2$.
$a$, $b$ and $c$ are positives such that $abc = a + b + c + 2$. Caculate the maximum value of $$\large \frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}}$$
This proble... | So, by Cauchy-Schwarz, we have $a^2+b^2 \geq (a+b)^2/2$, so we just have to prove that
$\displaystyle \sum_{cyc} \frac{1}{a+b} \leq 3/4$ (It's obvious that this maximum occurs when $a=b=c=2$, I will prove it.)
Letting, $a=2k,b=2l,c=2m$ we have to prove that $\displaystyle \sum_{cyc} \frac{1}{k+l} \leq 3/2$ subject to t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Minimal Rook Difference Grids In the below grid all 18 orthogonal differences are distinct, with a difference of 18 missing.
Could the highest number be 18? The resulting graph would have valence 4, making it an Eulerian Graceful graph with edges(mod 4)=2. Rosa (1967) proved Eulerian Graceful graphs must have edges... | I found a new optimal solution for the $3 \times 5$ rook graph:
\begin{array}{|c|c|c|c|c|}
\hline
5 & 46 & 45 & 28 & 0 \\
\hline
31 & 7 & 1 & 44 & 11 \\
\hline
40 & 21 & 13 & 6 & 42 \\
\hline
\end{array}
Also I investigated the queen graph.
$qdg(2,2)=6$, so the $2 \times 2$ queen graph is grace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This s... | A nice generalization proved here:
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=(2a+1)\beta(2a+2)-\frac{\ln(2)|E_{2a}|}{(2a)!}\left(\frac{\pi}{2}\right)^{2a+1}$$
$$-\frac12\left(\frac{\pi}{2}\right)^{2a+1}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1),$$
where $E_r$ is the Euler n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
| Also, we can use SOS:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}=\sum_{cyc}\frac{a-b-(c-a)}{(b+c)^2}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{(b+c)^2}-\frac{1}{(a+c)^2}\right)=\sum_{cyc}\frac{(a-b)^2(a+b+2c)}{(a+c)^2(b+c)^2}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int_{-\infty}^{\infty} \arctan(e^x) \arctan(e^{-x})dx=\frac{7}{4}\zeta(3)$
How to prove that $$\int_{-\infty}^{\infty} \arctan(e^x) \arctan(e^{-x})\rm dx=\frac{7}{4}\zeta(3)?$$
This integral appeared while attempting to solve an unanswered question, namely:
Closed form of :$ \int_{-\infty}^{\infty}\arctan\... | Here is another slight variation on a theme.
Let
$$I = \int_{-\infty}^\infty \arctan (e^x) \arctan (e^{-x}) \, dx.$$
Setting $u = e^x$ gives
\begin{align}
I &= \int_0^\infty \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u}\\
&= \int_0^1 \arctan u \arctan \left (\frac{1}{u} \right ) \frac{du}{u} + \int_1^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, becaus... | The OP mentioned the solution given in the book, so here we express that logic.
$\quad \{ x \in \Bbb R \, | \, \sqrt{x^2+1} - 2x+ 1 \gt 0\} =$
$\quad\quad\quad \{ x \in (-\infty, \frac{1}{2}] \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \; \bigcup $
$\quad\quad\quad\{ x \in [\frac{1}{2},\frac{4}{3}] \; | \; \sqrt{x^2+1} - 2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$ The sum of series is
$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$
I used to solve this problem as its $n$th term
$${1\over (2n+1)(2n+3)}$$
Now how can I proceed??
| Hint : The $n$ th term of your series is $$ a_n= \dfrac{1}{2}\left ( \dfrac{1}{2n+1}- \dfrac{1}{2n+3}\right)$$
expanding your series you get $$ \dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}...\right)$$
as you see the terms cancel out each other the only term left is $\dfrac{1}{3}$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^... | To prove the inequality you were required to prove, note
$\dfrac1{n+1}+\dfrac1{(n+1)^2}<\dfrac1{n+1}+\dfrac1{(n+1)^2}+\dfrac1{(n+1)^3}+...=\dfrac{\dfrac1{1+n}}{1-\dfrac1{n+1}}=\dfrac1n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
If $u=(1+\cos t)(1+\cos 2t)-\sin t\sin 2t$ and $v=\sin t(1+\cos 2t)+\sin 2t(1+\cos t)$, then $u^2+v^2=4(1+\cos t)(1+\cos 2t)$
If
$$u = (1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta \qquad v = \sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)$$
then show that
$$u^2 + v^2 = 4(1+\cos\theta)(1+\co... | Assigning $a$, $b$, $c$, $d$ as
$$a := 1+\cos\theta \qquad b := 1+\cos2\theta \qquad c := \sin\theta \qquad d := \sin 2\theta$$
we can write
$$u = ab - c d \qquad v = bc + ad$$
When squaring and adding, the $-2abcd$ will cancel with $2abcd$, leaving
$$u^2 + v^2 = a^2 b^2 + c^2 d^2 + b^2 c^2 + a^2 d^2 = \left(a^2+c^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Prove the inequality $b\le \sqrt{a^2 + b^2}$ How do i prove $b\le \sqrt{a^2 + b^2}$ ?
What i think is :
$\begin{align}
a^2& \gt 0,\, b^2 \gt 0 \\
b &= \sqrt{b^2}\\
&\le\sqrt{a^2+b^2} \\
\therefore b &\le \sqrt{a^2 + b^2}
\end{align}$
Am i right? Please do a correction.
| It is not true that $b =\sqrt {b^{2}}$. You have to say $b \leq |b| =\sqrt {b^{2}} \leq \sqrt {a^{2}+b^{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expanding the Polynomial Using Taylor Series? Expand the polynomial
$$f(x) = x^3-2x^2-3x+5$$
In power of $(x-2)$
This might be a simple question but all we have to do to solve this question is to expand using the Taylor series with $x =2 $ , as opposed to the Maclaurin at $x=0$.
$f(x) = f(a) + \frac{f'(a)(x-a)^1}{1!}+... | Just like for the change of basis for numbers, you can use Horner's algorithm:
divide the polynomial by $x-2$, then divide again the quotient by $x-2$, and so on until you obtain a constant quotient. The coefficients of the successive powers of $x-2$ are the successive remainders (including the final quotient).
Illu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . .
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that
$$a^2 + b^2 + c^2=d$$
then $a+b+c+d$ equals?
$$(A)\ 0 \quad (B)\ 1 \quad (C)\ 2 \quad (D)\ 4$$
Here is what I tried: assu... | You made a slight error, this needs to be $5x^2$ and not $2x^2$.
So you have this equation $$3a^2+a(6x-1)+(5x^2-3x)=0$$
This equation has discriminant $\Delta=-24x^2+24x+1$
The way to solve this for Diophantine equations is to introduce an integer $m$ such that $\Delta=m^2$ (because we need the square root to be an int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit:
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$
I tried limiting it with $$n\c... | Converting this
to a Riemann sum,
$\begin{array}\\
\frac{1}{n}\sum_{k=1}^n (a+\frac{k}{n})^2
&\to \int_0^1 (a+x)^2dx\\
&= \int_a^{a+1} x^2dx\\
&=\dfrac{x^3}{3}|_a^{a+1}\\
&=\dfrac{a^3+3a^2+3a+1-a^3}{3}\\
&=\dfrac{3a^2+3a+1}{3}\\
&=a^2+a+\frac13\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Any suggestions on a closed form of $\sum_{a=1}^{N} \left({\lfloor{\frac{N+1}{a}}\rfloor+\lfloor{\frac{N-1}{a}}\rfloor}\right)$ Now the sum is approximated by $$\sum_{a=1}^{N} \left({\left\lfloor{\frac{N+1}{a}}\right\rfloor + \left\lfloor{\frac{N-1}{a}}\right\rfloor}\right) \approx 2 \sum_{a=1}^{N} \left\lfloor{\frac{N... | A first useful point to consider is that
$$
\eqalign{
& \left\lfloor {{{N + 1} \over a}} \right\rfloor + \left\lfloor {{{N - 1} \over a}} \right\rfloor = \cr
& = {{N + 1} \over a} + {{N - 1} \over a} - \left\{ {{{N + 1} \over a}} \right\} - \left\{ {{{N - 1} \over a}} \right\} = \cr
& = {{2N} \over a} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{... | Quick solution using Lifting the Exponent Lemma:
Note that $3^{7^n} + 5^{7^n} - 1 \equiv -4^{7^n} - 2^{7^n} - 1 \equiv -(4^{7^n} + 2^{7^n} + 1) \equiv -\dfrac{8^n-1}{2^n-1} \pmod{7^{n+1}}$.
Now, by LTE, we have $\nu_7(8^{7^n} - 1) = \nu_7(8-1) + \nu_7(7^n) = n+1$, and $2^{7^n} \equiv 2^{7^n \pmod{6}} \equiv 2^{1^n} \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find : $\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$ Evaluate :
$I=\int_0^{b}\frac{\ln x}{x^2-1}dx$ where $b>1$
My attempt :
I use integral by part : $u=\ln x$ $dv=\frac{1}{x^2-1}$
Then : $u'=\frac{1}{x}$ and $v=\frac{\ln \frac{x-1}{x+1}}{2}$
So :
$I=[\frac{\ln x\ln\frac{x-1}{x+1}}{2}]_0^{b}$
$+\int_0^{b}\frac{\ln (... | $
\begin{align}
\int_0^b\dfrac{\ln x}{x^2-1}~dx&=\int_0^1\dfrac{\ln x}{x^2-1}~dx+\int_1^b\dfrac{\ln x}{x^2-1}~dx
\\&=\dfrac{\pi^2}{8}+\int_1^b\frac{\ln x}{x^2\left(1-\dfrac{1}{x^2}\right)}~dx
\\&=\dfrac{\pi^2}{8}+\sum\limits_{n=0}^\infty\int_1^b\frac{\ln x}{x^{2n+2}}~dx
\\&=\dfrac{\pi^2}{8}-\sum\limits_{n=0}^\infty\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
LU Decomposition of $A$ I am currently working with a practice problem in which we are asked to find the $LU$ factorization of some matrix $A$, given below:
$$
A= \begin{pmatrix}
1 & 2 & 3 \\
2 & 4 & 7 \\
3 & 6 & 10 \\
3 & 6 & 10 \\
\end{pmatrix}
$$
In the first step, I added $-2$ of the first row ... | Using notation from here, you obtained that
$$L_{24}(-1)L_{23}(-1)L_{14}(-3)L_{13}(-3)L_{12}(-2)A = U$$
so
$$A = L_{12}(-2)^{-1}L_{13}(-3)^{-1}L_{14}(-3)^{-1}L_{23}(-1)^{-1}L_{24}(-1)^{-1}U = L_{12}(2)L_{13}(3)L_{14}(3)L_{23}(1)L_{24}(1)U$$
where $L = L_{12}(2)L_{13}(3)L_{14}(3)L_{23}(1)L_{24}(1)$ is lower triangular w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sum_\text{cyc}\frac{1}{3-ab}\le\frac{3}{2}$
Let $a,b,c>0$ such that $a^2+b^2+c^2=1$. Prove that $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le\frac{3}{2}$$
By Am/Gm we have: $$ {1\over 3-ab}\leq {2\over 6-a^2-b^2} = {2\over 5+c^2}<{2
\over 5}$$
so
$$\sum_{cyc} {2\over 5+a^2} < {6\over 5} <{3\over 2}$$
So ... | Hint.-Because of $a^2+b^2+c^2=1$, each $a,b,c$ must be in the interval $[-1,1]$ then one has for positive values
$$\frac13\le\frac{1}{3-ab}\le\frac12\\\frac13\le\frac{1}{3-ac}\le\frac12\\\frac13\le\frac{1}{3-bc}\le\frac12$$ Thus the sum is less or equal than $\dfrac32$.
If some of $a,b,c$ is negative, the fraction is l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}... | When you want to find the value of $~x~$ inside a logarithm function, first you have to make both side in term of logarithm first (convert four into a logarithm term) and then inverse it to find $~x~$ :
$\log_{3}(x) = 4(1)$ $\Leftarrow$ convert $~1~$ into logarithm base $~3~$
$\log_{3}(x) = 4.\log_{3}(3)$ $\Leftarrow$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Is every prime contained in the set $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Is every prime contained in the set
$\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Let $p\notin\{2,5\}$ be a prime number. Then $\frac{1}{p}$ has a decimal period of length at most $p-1$. Denote $\fr... | A prime number $p$ divides $10^k-1$ if and only if $10^k \equiv 1 \bmod p$. But if $p \nmid 10$ then $10^{p-1} \equiv 1 \bmod p$ by Fermat's little theorem.
Therefore $p$ divides $10^{p-1}-1$ for all prime numbers $p \not\in \{ 2,5 \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum whe... | Let $(x-a)^2+y^2=k^2,$ where $k\geq0.$
Thus, the equation $$(x-a)^2+4\left(1-\frac{x^2}{9}\right)=k^2$$ has real roots.
We have
$$\frac{5}{9}x^2-2ax+a^2+4-k^2=0,$$ which gives
$$a^2-\frac{5}{9}(a^2+4-k^2)\geq0$$
and $$k\geq2\sqrt{1-\frac{1}{5}a^2},$$ as your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Why do the digits of a number squared follow a similar quotient? I realized that, any number $k$ with $n$ digits, and when $k$ is squared, i.e $k^2$ will have $2n-1$ or $2n$ digits.
Per example, let $k = 583$, thus $n = 3$ and the digits of $583^2$ are $2n$.
But I started thinking, how can I narrow the result to be mor... | If $k^2$ has $2n$ digits, then it is true that $10^{2n-1} \leq k^2 < 10^{2n}$, so we have $10^{n-1}\sqrt{10} = \sqrt{10^{2n-1}} \leq k < \sqrt{10^{2n}} = 10^n$.
If $k^2$ has $2n-1$ digits, then it is true that $10^{2n-2} \leq k^2 < 10^{2n-1}$, so we have $10^{n-1} = \sqrt{10^{2n-2}} \leq k < \sqrt{10^{2n-1}} = 10^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^... | $f(z)=z^5+2z^3+3 \Rightarrow f'(z)=5z^4+6z^2>0 \Rightarrow f(z)$ is an increasing function. So $f(z)=0$ can have at most oone real root and it is $z=-1.$
Take $$-3=z^5+2z^3 \Rightarrow 3=|z^5+2z^3|\le |z|^5+2|z|^3 \le 3.$$
This proves that at least one root of this equation is s.t. |z|=1.
Other roots cannot not satisfy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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.