Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to c... | From sum and product, we have:$$a+b+ab+ab=0$$ $$a(b+1)=-b(a+1)$$$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\left(1-\dfrac{2a}{1+a}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1-\dfrac{2ab}{b(1+a)}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1+\dfrac{2b}{1+b}\right)^3+\left(1-\dfrac{2b}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Trig equation $\sin(x)+\sin(3x)=0$, the answers are given in a factored form $\sin(x)+\sin(3x)=0$
So to solve this I tried the following:
-First I transformed this expression from sum to product because it equals zero
$\sin(x)+\sin(3x)=0 => 2\sin(\frac{x+3x}{2})\cdot\cos(\frac{x-3x}{2})=0$
-Second part I got to
$2\si... | How I would have approached this problem,
$$\sin 3x = 3\sin x - 4\sin^3x$$
$$\sin x + \sin 3x = 4\sin x - 4 \sin^3x = 0 \implies \sin x(1-\sin^2 x) = 0 \implies \sin x \cos^2x = 0$$
$$\implies x = n\pi \text{ or } x = n\pi+\pi/2$$
$$\implies x = 2n\frac{\pi}{2} \text{ or } \frac{(2n+1)\pi}{2} \implies x = k\frac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Solving $\begin{cases} z_1z_2=10\operatorname{cis}(\frac{4\pi}5)\\\frac{z_1}{\overline{z_2}^2}=\frac2{25}\operatorname{cis}(\frac{\pi}5)\end{cases}$
Solve
$$\begin{cases}z_1z_2=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\\frac{z_1}{\overline{z_2}^2}=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}$$
o... | The line where you have $(*)$ should better be
$$
\left|\frac{z_1}{\overline{z_2}^{\,2}}\right|=
\frac{|z_1|}{\bigl|\overline{z_2}^{\,2}\bigr|}=\frac{r_1}{r_2^2}
$$
There is no reason for the equality before the one you mark with $(*)$.
The equality you mark with $({*}{*})$ is indeed wrong and it should be
$$
\frac{r_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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error in solutions? MIT single variable calculus - Simplifying formula for tangent line Question 1C-4 d) located here states:
Write an equation for the tangent line for the following questions:
d) $f(x) = \frac{1}{\sqrt x} $ at $x = a$.
By my workings:
$f'(a) = (\frac{-1}{2})a^\frac{-3}{2}$
Therefore, the equation ... | For the simplifcation:
$$ y = -\frac{1}{2}a^{-\frac{3}{2}}(x-a)+\frac{1}{\sqrt{a}} $$
$$ y = \frac{1}{2}a^{-\frac{3}{2}}\cdot a -\frac{1}{2}a^{-\frac{3}{2}} \cdot x + a^{-\frac{1}{2}} \\ \text{(multiplication, and definition of power and square root)} $$
$$ y = \frac{1}{2}a^{-\frac{1}{2}} + a^{-\frac{1}{2}} - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3... | Hint:
$$a^2+b^2+c^2=ab+bc+ca \Leftrightarrow $$
$$\Leftrightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ca \Leftrightarrow$$
$$\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0 \Leftrightarrow$$
$$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0 \Leftrightarrow a=b=c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ canc... | $$I = -\int \frac{1}{\sqrt{x^2-9}}\,dx$$
$\sqrt{bx^2 - a} \implies x = \sqrt{\frac{a}{b}}\sec \alpha$
$$x = 3\sec\alpha \implies \frac{dx}{d\alpha}=3\sec\alpha\tan\alpha$$
$$ -\int \frac{1}{\sqrt{\left(3\sec \alpha \right)^2-9}}3\sec\alpha\tan\alpha\,\,d\alpha$$
$$ -\int \frac{\sec\alpha \tan\alpha}{\sqrt{\sec^2\alpha-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram. Problem
Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.
My Solution
B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogr... | Hint:
since $\vec A+\vec B=\vec C+ \vec D$ we have:
$$
\frac{1}{2}(\vec A+\vec B)=\frac{1}{2}(\vec C+ \vec D)\iff \frac{1}{2}\vec A +\frac{1}{2}\vec B=\frac{1}{2}\vec C+\frac{1}{2}\vec D
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
prove $\sum_{cyc}\frac{1}{a(a+b)}\ge\frac{4}{ac+bd}$ Let $a,b,c,d$ be positives. Show that
$$\sum_{cyc}\dfrac{1}{a(a+b)}\ge\dfrac{4}{ac+bd}$$
Use Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{1}{a(a+b)} \cdot\sum_{cyc}a(a+b)\ge (1+1+1+1)^2$$
it suffient
$$\dfrac{16}{\sum\limits_{cyc}(a^2+ab)}\ge\dfrac{4}{ac+bd}$... | $$\Leftrightarrow \left ( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right )^2\cdot \dfrac{1}{4+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}}\ge \dfrac{4}{ac+bd}$$
$$\left ( \left (\dfrac{1}{a}+\dfrac{1}{b} \right )+\left (\dfrac{1}{c}+\dfrac{1}{d}\right )\right )^2 \cdot (ac)\ge 4 \left( \dfrac{c}{b}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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solving the inequality with three parameters Is there any way to show that the following inequality holds for the given function with constraints?
$\frac{(a x + y)^{y+1}}{a x (a x + y + 1)^y}\geq 1$ for $0.5 \leq a \leq 1$, $x >0,y \geq 0$. It can be easily checked that it saturates the inequality for $\lim_{x \rightar... | Assume WLOG that $a=1$ (otherwise replace $ax \mapsto x\,$).
$$
\begin{align}
\frac{(x + y)^{y+1}}{x (x + y + 1)^y}\geq 1 \;\;&\iff \;\; \frac{x+y+1}{x} \, \left(\frac{x+y}{x+y+1}\right)^{y+1} \ge 1 \\
&\iff\;\; \frac{x+y+1}{x}\,\left(1 - \frac{1}{x+y+1}\right)^{y+1} \ge 1
\end{align}
$$
The second factor on the LHS o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =... | HINT:
If calculus is not mandatory,
Method$\#1:$
Set $x=\tan A$
Method$\#2:$
$$\implies x^2y-2x+y=0$$
As $x$ is real, the discriminant must be $\ge0$
Method$\#3:$
If $x>0,$ $$\dfrac{2x}{x^2+1}=\dfrac2{x+\dfrac1x}$$
Now using AM-GM inequality, $$\dfrac{x+\dfrac1x}2\ge\sqrt{x\cdot\dfrac1x}=?$$
If $x<0,$ set $x=-u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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How does one expand $\sqrt{r^2+a^2}-r$ to $\frac{a^2}{2r}-\frac{a^4}{8r^3}+\frac{a^6}{16r^5}$? The problem and solution are attached as photos below.
I understand all parts of the solution except for the line above "By inspection..." where $V(r,\theta=0)$ is expanded. I tried a taylor expansion, taylor expansion + bin... | Taylor expansion is indeed the key here, but you have to be mindful of what variable you are expanding! Here, depending on how far $(r)$ you are away from the disk, the smaller the disk is to you and less its extended structure matters so $a$ is 'small' for your purposes. We then should Taylor expand around $a=0$.
You ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding all $ f : \mathbb R \to \mathbb R $ satisfying $ f \bigl ( f ( x ) f ( y ) \bigr ) + f ( x + y ) = f ( x y ) $ for all $ x , y \in \mathbb R $
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all reals $ x $ and $ y $:
$$ f \bigl ( f ( x ) f ( y ) \bigr ) + f ( x + y ) = f ( x y ) \text . $$
I... | $$ f(f(x)f(y))+f(x+y)=f(xy). \ \ \ (1)$$
Let $f(0)=c$.
Choose $x=y=0$, we can get
$$f(c^2)=0.\ \ \ (2)$$
Choose $y=0$, we can get
$$f(cf(x))+f(x)=c.\ \ \ (3)$$
Choose $y=\frac{x}{x-1}(x\neq 1)$, we can get
$$f\left(f(x)f\left(\frac{x}{x-1}\right)\right)=0(x\neq 1).\ \ \ (4)$$
When $c=0$, from equation $(3)$ we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists.
Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists.
I tried many different numbers and nothing is working but I don't know how to prov... | I assume you mean "difference of squares of positive integers".
Let $n$ be any positive integer, and $x,y$ positive integers such that $n = x^2 - y^2$. Then $n = (x - y)(x + y)$, so we get a factorization of $n$ into two integers. On the other hand, if we have a factorization $n = ab$, we can get $x$ and $y$ by $x = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\frac{-1+\sqrt{3}i}{2}$ is a cube root of 1? I know how to directly prove that complex number is one of the cube roots of 1. But the textbook solution also gives another solution as shown below,
However, I don't understand how it jumps to the conclusion that $\frac{-1+\sqrt{3}i}{2}$ is a root of the quadrat... | take $x=\frac{-1+\sqrt{3}i}{2}$ show that $x^3=1$
$$x=\frac12(-1+\sqrt{3}i)\\x^3=\frac{1}{8}((-1)^3+(\sqrt{3}i)^3+3(\sqrt{3}i)(-1)(-1+\sqrt{3}i))=\\
\frac18 (-1-3\sqrt3i+3\sqrt3i-9i^2)=\\\frac18(-1-9i^2)=\\\frac18(-1+9)=\\1$$
other method
$$x^3-1=(x-1)(x^2-x+1)=0\\x=1\\x^2-x+1=0\\\to (x-\frac12)^2-(\frac14)+1=0\\(x-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Logical suite and inequalities This is a related problem (see here)
We have the following inequalities :
For $n=3$ with $a,b,c$ real numbers the following inequality holds.
$$\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(c-b)^2+(c-a)^2+(a-b)^2\geq \sqrt{54}$$
For $n=4$ with $a,b,c,d$ real numbers
$$\frac{1... | To complete the answer of achille hui I will show that the lower bound
$$ \frac{n(n-1)}{2}\sqrt{2n}$$
is actually a minimum.
Let $z_j$ be the $n$ zeros of the Hermite polynomial of order n, then we have the so-called Stieltjes sums (see http://epubs.siam.org/doi/pdf/10.1137/0514028) :
$$ \sum_{j=1}^n \frac{1}{(z_k-z_j... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Which area is larger, the blue area, or the white area? In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
My Solution
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles h... | There was a mistake in my earlier solution. I correct that mistake here.
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - 2 {\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369403",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "16",
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"answer_id": 2
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Probability that a length $8$ password is made by $2$ digits and $6$ characters
You have $10$ digits $(0,1,2,...,9)$ and $26$ characters (from the
American alphabet).
Calculate the probability to automatic generate a 8-length password
with $2$ digits and $6$ characters.
I know the answer is $ p=\frac{\binom{8}{2}... | You could use $8 \choose 6$ instead of $8 \choose 2$ just fine. They are equal. In general ${n \choose r}={n \choose n-r}$ and your proof is a good one. For your second question, you should not use ${8 \choose 1}{8 \choose 3}{8 \choose 4}$. The first $8 \choose 1$ picks which character will be special. Having chos... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, ca... | $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \frac{x}{2}$$
Using componendo dividendo
$$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}+({\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}})}{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}-(\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}})} = \frac{x+2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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To prove $n\binom{2n-1}{n-1}$ is divisible by $n(2n-1)$
Prove that for natural $n \ge 2$ $$n\binom{2n-1}{n-1}$$ is divisible by $$n(2n-1)$$
We have $$n\binom{2n-1}{n-1}=n \frac{(2n-1)!}{(n-1)! \:n!}=n(2n-1)\frac{(2n-2)!}{(n-1)! \:n!}$$
Now it suffices to prove $\frac{(2n-2)!}{(n-1)! \:n!}$ is an integer
Now
$$\frac{(... | Because $$\frac{\binom{2n-1}{n-1}}{2n-1}=\frac{\binom{2n-2}{n-1}}{n}$$ and
$gcd(2n-1,n)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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Find the limit if it exists of $S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})$
Suppose that $S_0$ and A are positive numbers, let $$S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right)$$ with $n \geq 0 $.
(a)Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$
(b)Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
(c) Show that $s... | Hint You can simplify a) by applying AM-GM, obviously showing that each $S_n>0$ which is easy to show
$$\frac{1}{2}\left(S_n+\frac{A}{S_n}\right)\geq \sqrt{S_n \cdot \frac{A}{S_n}}=\sqrt{A}$$
For c) more details here and you already proved the sequence is decreasing and bounded below in b) and a).
For d) as Eugen sugge... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x... | I don't think there is a way to evaluate the limit without using the l'Hospital rule in some form. (Using Taylor series is basically the same, because they are obtained by calculating the derivatives.)
I would calculate the limit by applying l'Hospital once and then using known series expansions:
\begin{align}
&~~~~\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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On a determinant identity
Show that
$$\det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = (-1)^n(1+x-n)(1+x)^{n-1}$$
for all $n \in \mathbb{N}$.
I tried using induction but I didn't manage (using row/column swaps). Does anyone have ... | Subtract the first row from all of the other rows ... we have
\begin{eqnarray*}
\left[\begin{array}{cccccc}
-x & 1 & 1 & 1 &\cdots &1\\
1+x & -(x+1) & 0 & 0 &\cdots &0\\
1+x & 0 &-(x+1) & 0 &\cdots &0\\
1+x & 0 &0 & -(x+1) & \cdots &0\\
\vdots & \vdots &\vdots &\vdots &\ddots &\vdots & \\
1+x & 0 &0 & 0 & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is the determinant of the all one matrix minus the identity matrix n-1? Context (skippable)
I was asked (by a friend who is preparing for an exam) whether there was a special trick to compute the determinant of the following matrix. I didn't see anything beyond using the standard computations (like using "Gauss" to... | Here is a matrix (here with $n=10$) with columns that are eigenvectors of your (symmetric) matrix. Indeed, given any constants $\alpha, \beta,$ this shows a basis of eigenvectors for
$\alpha I_n + \beta \, 1_n$
Note that $P$ is not orthogonal, although the columns are pairwise orthogonal.
$$
P =
\left( \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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General solution of $\cos{4x}=\cos{2x}$ My attempt:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
\cos^2{2x}-\sin^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-1+\cos^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-\cos{2x}-1 &= 0 \\
(2\cos{2x}+1)(\cos{2x}-1) &= 0,
\end{align}
which leads to
$$\cos2x= -\frac{1}{2} \hspace{5mm} \text{or} \hspace{5mm} \co... | They are same answers.
I think it's better the following way.
$$4x=2x+2\pi k$$ or
$$4x=-2x+2\pi k,$$
where $k\in\mathbb Z$, which gives else the same answer:
$$\left\{\frac{\pi k}{3}|k\in\mathbb Z\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $0.a_1b_1a_2b_2a_3b_3 \ldots = 0.21742615\ldots$ is irrational, where $(a_i,b_i)$ are the solutions of $x^2-3y^2=1$
Let $S$ be the set of all ordered pairs of positive integers $(x,y)$ such that $x^2-3y^2 = 1$. Let $(a_1,b_1),(a_2,b_2),\ldots$ be the elements of $S$ listed in ascending order, i.e., $a_i < a... | Hints: If $k$ is rational then its decimal expansion repeats after some point. Let $p$ be the period, and let $N=10^p$. Then there is a finite set $F$ of pairs such that
$$
a_i,b_i\in\{(r+sN^n)/(N-1)\mid(r,s)\in F,\,n\geq0,\,s>0\}
$$
for $i$ sufficiently large. For some fixed $(r,s),(t,u)\in F$ with $s,u>0$, there ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int{\frac{1}{x^2-x+1}dx}$ Original question was to integrate $$\int{\frac{(x+1)}{x^2-x+1}dx}$$
But I was able to break it into 2 parts: $$\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}+\frac{3}{2}\int{\frac{1}{x^2-x+1}dx} $$
The first part $\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}$ could be easily integrated by s... | Hint : $x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many natural numbers $n$ not greater than $10000$ are there such that $2^n - n^2$ is divisible by $7$? How many natural numbers $n$ not greater than $10000$ are there such that $2^n - n^2$ is divisible by $7$?
Please help me solve this question!
| Hint. Since $2^3\equiv 1 \pmod {7}$, consider $n=3q+r$ with the remainder $r=0,1,2$.
If $r=0$, then $1\leq q \leq 3333$ and
$$0\equiv 2^n - n^2=1-9q^2\equiv 1-2q^2\implies q\equiv 2, q\equiv 5\pmod{7}$$
Therefore the number of $n=6q$ is equal to
$$\left\lfloor\frac{3333-2}{7}\right\rfloor+1+\left\lfloor\frac{3333-5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Reorder the terms of the series This problem comes from a Brazilian book of Real Analysis.
Explicitly perform a reordering of the terms of the series $$1 -\frac{1}{2} + \frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots $$ so that its sum becomes zero.
It suggests to use the fol... | Its not actually four terms. Perhaps, to make the hint more obvious:
$$\begin{align}0&<1\\0&>1-\frac12-\frac14-\frac16-\frac18\\0&<1-\frac12-\frac14-\frac16-\frac18+\frac13\\0&>1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac1{10}-\frac1{12}-\frac1{14}-\frac1{16}\end{align}$$
If you rearrange it so that the partial sums... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Tartaglia triangle for real field? Let's say i need to find the coefficient of the 2nd term in a binomia, we can use pascal's triangle. This works only if the power is a natural number.
What if we want to know it for a real power binomial? We have something for that?
| We can derive a Pascal triangle as follows:
We expand
\begin{align*}
(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k\qquad\qquad n=0,1,2,\ldots
\end{align*}
and obtain the coefficients
\begin{array}{lcccccccccccccccc}
(1+x)^0\qquad\qquad&&&&&1&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\color{lightgrey}{0}&&\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$ The question is to evaluate $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.
| Set $\cos 2x = \sin^2\theta$. Then
$$ \cos x = \sqrt{\frac{1+\sin^2\theta}{2}} \qquad \text{and} \qquad \left|\frac{dx}{d\theta}\right| = \frac{\sin\theta}{\sqrt{1+\sin^2\theta}}. $$
So we have
$$ \int_{0}^{\frac{\pi}{4}} \cos^{3/2}(2x) \cos x \, dx
= \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sin^4\theta \, d\theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How many positive integers less than $1000$ divisible by $3$ with sum of digits divisible by $7$?
How many positive integers less than $1000$ are divisible by $3$ with their sum of digits being divisible by $7$?
Well, I got Answer: $28$.
$a+b+c = 7k$ and $a,b,c$ are multiples of 3 so $a+b+c$ is a multiple of both $7... | A number is divisible by $3$ iff the sum of its digits (in the decimal representation) is divisible by $3$. Since $3$ and $7$ are coprime, the problem boils down to finding the number of non-zero triples $(a,b,c)$ such that $a,b,c\in\{0,1,2,\ldots,9\}$ and $21\mid a+b+c$. $a+b+c$ is at most $27$, hence the problem is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
I... | If a is divisible by 3, then 3 divides a, and therefore a(a^2 + 2).
Otherwise, $a = 3k ± 1$, and 3 divides $(a^2 + 2)$ = $((3k ± 1)^2 + 2)$ = $((9k^2 ± 6k + 1) + 2)$ = $(9k^2 ± 6k + 3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $ \int_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} dt=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$ I've already known that
$$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4... | As noted, this integral is reducible to an elliptic integral and the form of the proposed solution may suggest that its modulus could be a singular value. Indeed, from the form
\begin{align}
I&=2\int_0^\infty \frac{\mathrm{d}u}{\sqrt {u^4+21u^2+112}} \\
&=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {u^2+\frac{21}{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
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Find $a_n$, given $a_1=1$ and $a_{n+1} =a_n/4+3/4$ The sequence is defined of rational numbers $a_1,a_2,a_3,...$. Let $a_1=1$ and $a_{n+1}=a_n/4+3/4$. What are the numbers $a_1,a_2,a_3,...$ ?
What I've done:
$a_1=1$
for $n=1$ we have $a_{n+1}=a_2=a_1/4+3/4=1$
for $n=2$ we have $a_{n+1}=a_3=a_2/4+3/4=1$
Therefore the se... | Claim: for every $n \in \mathbb{N}$; we have: $a_n=1$ .
Proof by induction: suppose that the assertion holds for $k=n$;
i.e. $a_n=1$ ; then one can see that
$$a_{n+1}=\dfrac{a_n}{4}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1;$$
so the assertion also holds for $k=n+1$.
Let $A \in \mathbb{R}$ be arbitrery, and let $a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to solve the equation $\log_2(x-9)+\log_{(2x-18)}6=3$.
Solve the following equation. $$\log_2(x-9)+\log_{(2x-18)}6=3.$$
I tried this way,
\begin{align}
\log_2(x-9)+\log_{(2x-18)}6 & \ = \ 3\\
\log_2(x-9)+\log_{2(x-9)}6 & \ = \ 3\\
\log_2(x-9)+\frac{\log_{2}6}{\log_2(2(x-9))} & \ = \ 3\\
\log_2(x-9)+\fra... | Now, $x=9+2^{1+\sqrt{4-\log_26}}$ or $x=9+2^{1-\sqrt{4-\log_26}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence of series $(\frac{1}{3})^{2}+(\frac{1.2}{3.5})^{2}+(\frac{1.2.3}{3.5.7})^{2}+...$ $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$
I am not able to find a general equation and that's creating problem for me as I can't proceed ... | Yes it converges of course! Use $$\frac{n}{2n+1}<\frac{1}{2}$$
We obtain:
$$\left(\frac{1}{3}\right)^{2}+\left(\frac{1.2}{3.5}\right)^{2}+\left(\frac{1.2.3}{3.5.7}\right)^{2}+...<\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Show $x=0$ and $x=1$ are the only integer solutions I'm trying to show that the only solutions to $x^2-x+1=y^2$ are when $x=0$ and $x=1$. All I can think of is completing the square gives $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}$, which is clearly not a perfect square. Does this suffice? Should I use induction?
| Multiply by $4$ to obtain $$(2x-1)^2+3=4y^2$$ or $$4y^2-(2x-1)^2=3$$ or $$(2y-2x+1)(2y+2x-1)=3$$
And the factors are in some order either $3,1$ or $-3, -1$
The sum of the two factors is $4y=4, -4$ respectively, so $y=\pm 1$
The difference between the two factors is $4x-2=\pm 2$ so $x=0,1$
All four combinations actually... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or ... | Without congruences, subtraction, or factoring a difference of common powers:
Inductive assumption:
For some $n$, there is a $k$ such that
$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1} = 19k $
i.e.
$20(50^n)+18(12^n) = 19k$
For this $n$, add $980(50^n) + 198(12^n)$ to both sides.
The left-hand side will turn into the desired $(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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A equation about determinant I read from the Wiki that:
When A = D and B = C, the blocks are square matrices of the same order and the following formula holds (even if A and B do not commute)
$$\det\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}=\det(A-B)\det(A+B).$$
How to prove this relation when $A$ and $B$ do not ... | Substitute the row $i$ by $R_i-R_{i+n}$ for each $i\in \{1,\dots,n\}$, where $n$ is the dimension of $A$ and $B$. We get
$$\det\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}=\det\begin{pmatrix}
A-B & B-A\\
B & A
\end{pmatrix}.$$
Now, consider the matrix $\begin{pmatrix}
A-B & B-A\\
B & A
\end{pmatrix}$ and substitute the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the cosine of $\theta$ is $\frac{1 - t^2}{1 + t^2}$. Where am I going wrong? I've recently picked up the book Mathematics and it's History by John Stillwell due to a recent curiosity in the history of math.
I started doing one of the exercises in the book and got a little stumped (I'll admit, I'm quite rust... | You never say what $t$ is. I suppose that it is the slope of the line. Since it is a line passing through $(-1,0)$ and $(\cos\theta,\sin\theta)$, you have $t=\frac{\sin\theta}{1+\cos\theta}$. So,$$t^2=\frac{1-\cos^2\theta}{(1+\cos\theta)^2}$$and therefore\begin{align}\frac{1-t^2}{1+t^2}&=\frac{(1+\cos\theta)^2-1+\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{n \to \infty}\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right)=?$ What is the value of limit
$$\lim_{n \to \infty}\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right)$$
If... | Hint:
$$
\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right) \leq \left( \frac{\sqrt[n]a}{n}+\frac{\sqrt[n]{a^2}}{n}+\frac{\sqrt[n]{a^3}}{n}+\cdots+\frac{\sqrt[n]{a^n}}{n}\right)=\sqrt[n]{a}.\frac{a-1}{n\left(\sqrt[n]{a}-1\right)}
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
I have reached:
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$.
How do I continue from here? I am unable to find a suita... | $y = 1 +\frac{2x}{x^2 + x + 1} = 1 + \frac{2}{1+x+1/x}$, $x\ne 0.$
A) Let $x\gt 0:$
1) $x+1/x \ge 2$
AM-GM: $x+1/x \ge 2\sqrt{x×1/x}$
$y_{max} = 1 + \frac{2}{1+2} = 5/3$.
B) Let $x \lt 0:$
2) $x+1/x \le -2$
$y_{min} = 1 + \frac{2}{1- 2} = 1+ (-2) = -1.$
Since
$y = \frac{x^2 + 3x + 1}{x^2+x+1}$ is continuous on $\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ where $n$ is a positive integer.
We have \begin{align*}\lfloor(1+\sqrt{3})^{2n-1}\rfloor &= (\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}\\&=\dfrac{(4+2\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(4-2... | Set $x_n=\frac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\frac{(2-\sqrt{3})^n}{1-\sqrt{3}}$. Show that $x_1, x_2$ are integer and $x_{n+2}=4x_{n+1}-x_{n}$ holds, thus by induction $x_n$ is an integer for all $n \in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (Click here.) $$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots\to~?$$
I tried like below
$$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots=\\\sum_{n=1}^{\infty}\frac{1}{1^2+2^2+3^2+\dots+n^2}=\\\sum_{n=1}^{\infty}\frac{1}{\fr... | Starting from
$$S=6 \sum _{n=1}^{\infty } \left(\frac{1}{n+1}-\frac{4}{2 n+1}+\frac{1}{n}\right)$$
I changed the index $n\to n+1$
$$S=6 \sum _{n=0}^{\infty } \left(\frac{1}{n+2}-\frac{2}{n+\frac{3}{2}}+\frac{1}{n+1}\right)$$
Then I applied an amazing property of
digamma function that can be found here
and I got
$$S=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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solve equation with cos and powers
Solve the following equation.
$$3^{x^2-6x+11}=8+\cos^2\frac{\pi x}{3}.$$
Here is the problem that I am struggling with it
I tried to take a logarithm of both side but I kinda stuck
can someone help me with this?
Thanks
| Add 1 to each side of the equation to get
$$3^{x^2-6x+11} + 1 = 9 + \cos^2 \frac {\pi x} {3}$$
Take $\log_3$ of both sides to get
$$x^2-6x+11 = 2 + \log_3 \cos^2 \frac {\pi x} {3}$$
The maximum of $\cos^2 \frac {\pi x} {3} = 1$, thus $\log_3 1 = 0$ , so we have
$$x^2 - 6x + 11 = 2$$
Bringing the 2 to the right hand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ende... | $$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}=
\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} + \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}}=
\lim_{x\rightarrow \infty} \frac{ax-bx}{|x+\frac{a}{2}| + |x+\frac{b}{2}|}
$$
this is
$$\lim_{x\rightarrow +\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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$66$ points in $100$ shots. I just received a probability problem from a friend via a text and by the time I took to read it, I was sent a solution as well - which is confusing.. The question goes like this..
A person shoots basketball 100 times. First time he scores a point and second time he misses it. For the foll... | Here is a slightly shorter version of Zubin Mukerjee's solution, with a (I believe) simpler argument for the edge cases $P(1,N)$ and $P(N-1,N)$. The arguments are effectively the same up to that point.
We denote the probability that we have a score of $x$ after $N$ shots as $P(x,N)$.
The problem states that
$$ P(1,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 5
} |
Showing $ \mathbb{Q}[\sqrt[4]{2}]$ is a field Here i want to prove
\begin{align}
\mathbb{Q}[\sqrt[4]{2}] = \{ a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8} \mid a,b,c,d\in\mathbb{Q}\}
\end{align}
this is a field. To do that i want to prove
\begin{align}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}}... | Through the magic of upper triangulation and backsubstitution...
\begin{align*}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} &= e + f \sqrt[4]{2} + g \sqrt[4]{4} + h \sqrt[4]{8} \\
\iff 1 &= (a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8})(e + f \sqrt[4]{2} + g \sqrt[4]{4} + h \sqrt[4]{8}) \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Prove by induction: $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ I want to prove that $2^{n+2} +3^{2n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$ using proof by induction.
Attempt
Prove true for $n = 1$
$2^{1+2} + 3^{2(1) +1} = 35$
35 is divisible by 7 so true for $n =1$
Induction step: Ass... | \begin{eqnarray*}
2^{k+3}+3^{2k+3} =(9-7)2^{k+2} +9 \times 3^{2k+1}= 9(\color{blue}{2^{k+2}+3^{2k+1}})-\color{blue}{7} \times2^{k+2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
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How many committees of $2$ women and $3$ men can be formed if two of the men refuse to serve on the committe together? From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?
This is ... | From a set of 5 women you always take two women.
From a set of two men you take one or none of them
Then you take 2 or 3 from a set of 5 men. So:
\begin{align}
P=\binom{5}{2}\binom{2}{1}\binom{5}{2} +\binom{5}{2}\binom{2}{0}\binom{5}{3} \\
P=\underbrace{\binom{5}{2}}_\text{Two women}(\underbrace{\binom{2}{1}\binom{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Explicit form of $\sin(x-y/2)$ I want to know if there exist and how to evaluate the explicit form of a curve given by $$y=\sin(x-y/2)$$
My intuition tells me there should exist one, because deivative of y w.r.t. x is always finite:
$$y'(x) = \frac{2\cos(x - y/2)}{2 + \cos(x - y/2)}$$
but I have no idea how tho evalua... | First, solve for $x$ when $|x|\le\frac{\pi+1}2$, where the function is invert-able:
$$f(y)=\arcsin(y)+\frac12y=x$$
By Taylor expanding,
\begin{align}f(y)&=\frac12y+\sum_{k=1}^\infty\binom{2k}k\frac{2ky^{2k-1}}{4^k(2k-1)^2}\\&=\frac32y+\frac16y^3+\frac3{40}y^5+\dots\end{align}
Substitute $y=a_0x+a_1x^3+a_2x^5+\dots$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$... | Here's a non-induction approach, provided for information only:
$$\begin{align}
&1^2+3^2+5^2+\cdots+(2n-1)^2\\
&=\sum_{r=1}^n(2r-1)^2\\
&=\sum_{r=1}^n\binom {2r-1}2+\binom {2r}2\\
&=\sum_{s=1}^{2n}\binom s2\\
&=\binom {2n+1}3
&&=\frac {(2n+1)(2n)(2n-1)}{1\cdot 2\cdot 3}\\
& &&=\frac 13n(2n-1)(2n+1)\\
& && =\frac 13 n(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Prove that the perpendicular from the origin upon the straight line Prove that the perpendicular drawn from the origin upon the straight line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ bisects the distance between them.
My Attempt:
Equation of the line joining the points $(c\cos ... | Let $O$ be the origin.
Midpoint $M$ of $A$ and $B$:
$x_M =(1/2)c(\cos\alpha +\cos\beta)$;
$y_M =(1/2)c(\sin\alpha + \sin\beta)$;
Line $l_1$ joining points $O$(origin) and $M$ has slope:
$m: = \dfrac{y_M}{x_M}$.
Line $l_2$ joining $A$ and $B$ has slope:
$m':= \dfrac{\sin\beta - sin\alpha}{\cos\beta - \cos\alpha}$.
Rema... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Arithmetic and Geometric progression in 3 numbers Suppose
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$
$$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally $$a=b=c=\frac{1}{2}.$$... | $2b=a+c,\;
a+b+c=\dfrac{3}{2}$
give $3b=\dfrac{3}{2}\to b=\dfrac{1}{2}$
and $a+c=1\to c=1-a$
Substitute in $b^4=a^2c^2$ and get
$a^2(1-a)^2=\dfrac{1}{16}$
$a(1-a)=\pm \dfrac{1}{4}$
$a (1-a)=\dfrac{1}{4}$ gives $a=b=c=\dfrac{1}{2}$ you have already found
$a(1-a)=-\dfrac{1}{4}$ gives
$\color{red}{a=\dfrac{1}{2} \left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
Question:
Find the values of the constants $a$,$b$ and $c$ such that the ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
What I've done so far:
Substitute $(-1,2)$ in $4x^2+y^2+ax+by... | Your equation $(1)$ is correct:
$$a - 2 b - c = 8 \tag{1}$$
For the tangency condition, you're thinking a little too hard. If the ellipse is tangent to the $x$-axis, then the equation should have a double-root for $x$ when $y=0$; that is, its discriminant must vanish:
$$4 x^2 + a x + c = 0 \quad\to\quad \text{discrimin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$? $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e... | Hint: Use the Taylor approximation for sine and the exponential function.
$$\sin(u)=u-u^3/6+O(u^5)$$
$$\exp(u)=1+u+u^2/2 + O(u^3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find the remainder when $f(x)$ divided by $(x^2 + x + 1)(x+1)$. When a polynomial $f(x)$ is divided by $x^2 + x + 1$ and $(x+1)^2$, the remainder are $x+5$ and $x-1$ respectively. Find the remainder when $f(x)$ is divided by $(x^2 + x + 1)(x+1)$.
First, I let the remainder be $Ax^2 + Bx +C$, then I try to find the valu... | We know that
$$f(x)=Q_1(x)(x^2 + x + 1)+x+5=Q_2(x)(x+1)^2+x-1$$
for some polynomials $Q_1$ and $Q_2$.
Moreover, for some polynomial $Q_3$,
$$f(x)=Q_3(x)(x^2 + x + 1)(x+1)+Ax^2+Bx+C.$$
Hence
$$f(-1)=-1-1=A(-1)^2+B(-1)+C.$$
Now consider the complex zeros $w_1$ and $w_2$ of $x^2 + x + 1$. Then
$$f(w_1)=w_1+5=Aw_1^2+Bw_1+C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Minimum value of an expression in two variables $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}$$
$x$ and $y$ are positive real numbers, what is the minimum value?
I have tried finding the minimum value of the first expression and the third expression, with the first expression being $\frac{7... | let $$f(x,y)=\sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2+y^2-\sqrt{2}xy}+\sqrt{50+y^2-10y}$$
then solve
$$f_x=1/2\,{\frac {2\,x-7\,\sqrt {2}}{\sqrt {{x}^{2}-7\,\sqrt {2}x+49}}}+1/2
\,{\frac {2\,x-\sqrt {2}y}{\sqrt {{x}^{2}+{y}^{2}-\sqrt {2}xy}}}
=0$$
and
$$f_y=1/2\,{\frac {2\,y-\sqrt {2}x}{\sqrt {{x}^{2}+{y}^{2}-\sqrt {2}xy}}}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Show that no GCD exists for $4$ and $2 \cdot (1 + \sqrt{-3})$ in $\{ x + y \sqrt{-3} | x, y \in \mathbb{Z}\}$ I need to show that there is no greatest common divisor for $4$ and $2 \cdot(1 + \sqrt{-3})$ in the ring $R := \{ x + y \cdot\sqrt{-3} \,\,|\,\, x, y \in \mathbb{Z}\}$.
I found that both $2$ and $(1 + \sqrt{-3}... | Let the common divisor of $4$ and $2(1+\sqrt{-3})$ be $x+y\sqrt{-3}$. Then we have
\begin{align}
(x+y\sqrt{-3})(a+b\sqrt{-3})&=4,\\
(x+y\sqrt{-3})(c+d\sqrt{-3})&=2(1+\sqrt{-3}),
\end{align}
where $x,y,a,b,c,d\in\mathbb{Z}$. Multiplying $x-y\sqrt{-3}$ on both sides of the equations, one obtains
\begin{align}
a&=\frac{4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve system of equation, $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, $\sqrt{y-4}+\sqrt{z-4}=4\sqrt3$ and $ \sqrt{x-9}+\sqrt{z-9}=4\sqrt3$ . $$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ I tried somthing,like go to the power of two , and change of ... | So here are the 3 equations:
$$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$
As suggested by transcenmental,
$\sqrt{x-1}-\sqrt{y-1}=4\sqrt 3 - 2 \sqrt{y-1}$ and multiplying with the first eq. gives
$$
x - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$ I need to show that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$. My attempted method was showing that as the distance between $(x,y)$ and $(0,0)$ approaches $0$, so does the function in the limit. I tried to show $\lvert{\frac{x^2y^3}{x^... | Hint: Where defined, $\frac{1}{x^2 + y^2}\leq \frac{1}{y^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ How to prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ ? I have tried AM-GM but with no effect.
| By Cauchy inequality we have:
$$(a^2 + 1)(1+b^2) \ge (a + b)^2$$
$$(a^2 + 1)(1+c^2) \ge (a + c)^2$$
$$(c^2 + 1)(1+b^2) \ge (c + b)^2$$
Multiply these and take square root and thus the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Evaluate $5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5$
Evaluate
$$5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5.$$
The options are $5^6$, $5^7$, $5^8$, $5^9$, $5^{10}$.
I'm new to this site. I came across this question in an Olympiad foundation site. I have no idea how to solve it. Can I get the answe... | $$a=5+4.5+4.5^2+4.5^3+4.5^4+4.5^5\to \times 5\\5a=25+4.5^2+4.5^3+4.5^4+4.5^5+4.5^6$$ now $5a-a= ?$
$$\quad{5a-a=(25-5)-(4.5)+(4.5^2-4.5^2)+(4.5^3-4.5^3)+(4.5^4-4.5^4)+(4.5^5-4.5^5)+4.5^6\\\to 5a-a=4.5^6 \\4a=4.5^6\\a=5^6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Conditions necessary for a function to contain multiple attractive fixed points Given a function $f(x)$, what are the necessary conditions for $f(x)$ to contain multiple attractive fixed points?
Another key condition is I want the construction of the fixed point $x^*$ to be repeated application of $f$: $\lim\limits_{x ... | All attracting fixed points must be fixed points first of all. So, the function $f(x)$ must have multiple fixed points, or solutions for $f(x^*)=x^*$. If function $f(x)$ is differentiable in open neighbourhoods of those fixed points $\{x^*\}$ and $|f'(x^*)|<1$, then those points are also attracting. Let's use these sta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $\overline{(2+\sqrt{3})}^{17} = \overline{2 - \sqrt{3}}$ in $\mathbb{Z}[\sqrt{3}]/(17)$ I am trying to solve the following problem:
Show that $\overline{(2+\sqrt{3})}^{17} = \overline{2 - \sqrt{3}}$ and $\overline{(2-\sqrt{3})}^{17} = \overline{2 + \sqrt{3}}$ in $\mathbb{Z}[\sqrt{3}]/(17)$, being $(17)$ the p... | Without answering your actual questions, here is another approach to the problem.
We have $\mathbb Z[\sqrt 3]/(17)\simeq \mathbb Z[x]/(x^2-3,17)\simeq \mathbb Z/(17)[x]/(x^2-3)$. We are looking to simplify $(2+x)^{17}$ in this ring. But because $17$ is prime, all the binomial coefficients $\binom{17}{i}$ with $0<i<17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving that a certain function approaches a certain limit at a certain point using epsilon-delta method (Proof Check) I have to prove that,
$\lim_{x\to 1}$ $\frac{x}{x+1}$ = $\frac{1}{2}$
using delta-epsilon method.
My answer: Let $\epsilon$ > $0$ and let $\delta$ = min( 1 , 2$\epsilon$)
Then, $\left|\frac{x}{x+1} -... | Depending on where you are in your studies/class, more justification around some of these conclusions could be helpful for the sake of clarity. For example, writing:
$$\Big|\frac{x}{x+1} - \frac{1}{2}\Big| = \Big|\frac{2x}{2(x+1)} - \frac{x+1}{2(x+1)}\Big| = \Big|\frac{x-1}{2(x+1)}\Big| = \frac{1}{2}\Big|x-1\Big|\Big|\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why is $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ rational? For this question,
Show the following irrational-looking expressions are actually rational numbers.
(a) $\sqrt{4+2\sqrt{3}}-\sqrt{3}$, and
(b) ...
I solved it as follows:
$$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\
x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\
(x+\sqrt{3})... | $\sqrt{4+2\sqrt{3}}=\sqrt{1+2\sqrt{3}+3}=\sqrt{(1+\sqrt{3})^2}=1+\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e. $$\frac{\sqrt{6-x}-2}{\... | Just another way:
Let $\sqrt{6-x}-2=h,\sqrt{3-x}-1=k\implies h\to0,k\to0$
and $$6-4-4h-h^2=x=3-1-2k-k^2\implies-h(4+h)=-k(k+2)\implies\dfrac hk=\dfrac{k+2}{h+4}$$
$$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{h\to0,k\to0}\dfrac hk=\lim_{h\to0,k\to0}\dfrac{k+2}{k+4}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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p - Factorization modulo n I learned that ${n!} = p^{e}n_0$ where $e = \sum\limits_{k = 1}^\infty \lfloor\frac{n}{p^{k}}\rfloor$ where n is the number, p is the prime, $n_0$ is the product of factors of n! divided by $p^{e}$. Does it has something to do with this problem? Or Im referring to different thing?
Prove that:... | Case 1 : $k=\lfloor \frac{n}{p} \rfloor=0$ , so we have $\frac{n!}{p^0} = (-1)^0 * 0! *(n-0*p)! \mod p $ => $ n! = n! \mod p$ which is true.
Case 2 : $k=\lfloor \frac{n}{p} \rfloor=1$ , so we have $\frac{p!}{p} = (-1)^1 *1! *(0)! \mod p $ => $ (p-1)! = -1 \mod p$ which is true Wilson's Theorem
Assume its true for $n$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multiplicative group modulo $2^n$ I remember learning that the multiplicative group modulo $2^n$, namely the group $\mathbb{Z}_{2^n}^\times$of integers coprime with $2^n$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_{2^{n-2}}$, which is due to Gauss.
It boils down to proving that the multiplicative order of $3$ in th... | We want to show $$3^{2^{n-3}}\ne 1\mod 2^n$$ and $$3^{2^{n-2}}\equiv 1\mod 2^n$$
for $n\ge 3$
For $n=3$ , we have $3^{2^{n-3}}=3\ne 1\mod 8$ and $3^{2^{n-2}}=9\equiv 1\mod 8$
hence the claim is true for $n=3$
For $n=4$; we have $3^{2^{n-3}}=9\ne 1\mod 16$ and $3^{2^{n-2}}=81\equiv 1\mod 16$ , hence the claim is true fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with d... | By AM-GM for positive variables:
$$\sum_{cyc}\frac{b+c}{a}\geq2\sum_{cyc}\frac{\sqrt{bc}}{a}\geq6\sqrt[3]{\prod_{cyc}\frac{\sqrt{bc}}{a}}=6.$$
AM-GM it's the Jensen inequality for $f(x)=\ln{x}$ because $f''(x)=-\frac{1}{x^2}<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Does the following limit exist? (a)$ \lim_{(x,y) \to (0,0)} f(x,y) = \frac{x^2+y}{x^2+y^2}$? Does the following limit exist? If exist, evaluate (a)$ \lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}$ $(b)\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}$?
for (a)
approaching (0,0) using $y=0$ we have
$$\lim_{(x,y)
\to (0,0)} \fra... | we use $$x=\frac{1}{n},y=\frac{\sqrt{n}}{n}$$ and plug in this in your function
$$f(x,y)=\frac{x^2+y}{x^2+y^2}$$ we get
$${\frac {{n}^{2}+\sqrt {n}}{\sqrt {n} \left( n+1 \right) }}$$
and we obtain $$\lim_{n\to \infty}{\frac {{n}^{2}+\sqrt {n}}{\sqrt {n} \left( n+1 \right) }}=\infty$$ therefore the limit doesn't exist
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde ... | multiplying your given inequality by $$xyz>0$$ we get
$$x^2+y^2+z^2\geq xy+xz+zy$$
and this is equivalent to
$$(x-y)^2+(y-z)^2+(z-x)^2\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Solving systems of equation with unknown How do i solve this system?
$$
\left\{
\begin{array}{c}
ax+y+z=1 \\
x+ay+z=a \\
x+y+az=a^2
\end{array}
\right.
$$
Ive reduced to this form. How should i continue to get infinitely many solutions, no solution and unique?
$$ \left[
\begin{array}{ccc|c}
1&a&1&a\\
... | hint:$$\left\{
\begin{array}{c}
ax+y+z=1 \\
x+ay+z=a \\
x+y+az=a^2
\end{array}
\right.$$sum of the equations :
$$x(a+2)+y(a+2)+z(a+2)=1+a+a2\\x+y+z=\frac{1+a+a^2}{a+2}\\$$now $$ax+y+z=1-(x+y+z=\frac{1+a+a^2}{a+2})\\(a-1)x=1-\frac{1+a+a^2}{a+2}\\(a-1)=\frac{a+2-1-a-a^2}{a+2}\\x=\frac{1-a^2}{(a+2)(a-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Relationship between Pythagorean Theorem and leg-hypotenuse lengths With a standard right triangle, we have the legs $a$ and $b$ and the hypotenuse $c$, where typically $a$ is the shorter leg. I'm curious about the fact that
$$a + b > c$$
but
$$a^2 + b^2 = c^2$$
This discussion seems to be a simple proof -- assuming t... | In $\triangle ABC$ with $\angle C$ denoting the interior angle at the vertex $ C$ and the sides' lengths being $a,b,c$ we have the Cosine Law: $$c^2=a^2+b^2-2ab\cdot \cos \angle C$$ which is very ancient. Therefore $$(a+b)^2=a^2+b^2+2ab=$$ $$=a^2+b^2-2ab \cdot \cos \angle C+2ab(1+\cos \angle C)=$$ $$=c^2+2ab (1+\cos \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a+b+c=0$ prove that $ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $$
I made this question as a more difficult (higher degree) version of this question. My idea... | Let's define
$$S_n=a^n+b^n+c^n$$
and consider the generating function
$$F(t)=\sum_{n=0}^\infty S_nt^n=\frac1{1-at}+\frac1{1-bt}+\frac1{1-ct}.$$
Using $a+b+c=0$ gives
$$(1-at)(1-bt)(1-ct)=1+pt^2+qt^3$$
for some $p$ and $q$, and
$$F(t)=\frac{3+pt^2}{1+pt^2+qt^3}.$$
From this we expand as a power series
$$F(t)=(3+pt^2)(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $-1$is a root for $ax^2+bx−3$, find $a^2+b^2$
Given: -1 is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$.
Find: the numeric value for $a^2+b^2$.
Background: question asked in an entrance exam (Colégio Militar 2005).
My attempt: the other root is $3/a$ and by substitution we can easily fin... | Starting from $a-b=3$, which is odd, this implies the positive primes cannot be both odd. So one of them is $2$, the other is an odd prime. As $a-b>0$, it's $b$ which is equal to $2$, so $a=3+b=5$, and $a^2+b^2=29$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving the cubic $x^3-x^2-2x+1 = 0$ Solving the cubic $x^3-x^2-2x+1 = 0$.
Using the Cubic Formula I get the following three solutions.
$x_1 = \frac{1}{3} - \frac{1}{3}\left(\frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3} - \frac{7}{3}\frac{1}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong -1.2469796... | I'm sorry, but I'm not entirely sure if there's a way make a transformation to get the appropriate trigonometric roots. However, there is another way to derive its roots using Euler's formula.
Starting with your cubic$$y=x^3-x^2-2x+1$$First, make the substitution $x=t+t^{-1}$. Expanding and collecting in $t$, we find ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the ratio of the areas
I'm having a trouble in solving this problem. How am I supposed to approach the problem?
| Here's a "coordinate geometry" method: Set up a Cartesian coordinate system so that A is (0, 0), B is (1, 0) and R is (0, -1). Then Q is (1, -1). Each interior angle of a regular hexagon is 120 degrees so angles FAR and BAS are both 120- 90= 30 degrees. Then F is $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is ... | This is only true for odd integers $n=2k+1$:
$$2^{2k+1}+1 = 2^{2k+1}-(-1)^{2k+1} = (2+1) (2^{2k}-2^{2k-1}\pm\cdots+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$? How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$
My Try :
$$x^2+x-2=(x-1)(x+2)$$
$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\s... | HINT: write
$$\frac{2(x-3+\sqrt{x(x+3)})(x-3-\sqrt{x(x+3)})}{(x-1)(x+2)(\sqrt{x}+\sqrt{x+3}+3)(x-3-\sqrt{x(x+3)}}$$
and this is equal
$$\frac{-9(x-1)}{(x-1)(x+2)(\sqrt{x}+\sqrt{x+3}+3)(x-3-\sqrt{x(x+3)})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the differential equations: $$y' = {y \over {\sin x}}+\tan{x \over 2}$$
I was trying to do this by substitution $u=y/x $ and it did not work and also with
$$y' - {y \over {\sin x}} = 0$$
$${dy \over dx} = {y \over {\sin x}} $$
$${\ln(y) = \ln\left|\tan\frac{x}{2}\right|+c } $$
$${y = c\cdot\tan\frac{x}{2} } $$
bu... | I don't think there "too many equations"
$${ y=C\left( x \right) \tan \frac { x }{ 2 } }\\ y'=C'\left( x \right) \tan { \frac { x }{ 2 } } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } }{ 2 } \\ C'\left( x \right) \tan { \frac { x }{ 2 } } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Sum of sixth power of roots of $x^3-x-1=0$ Question:
Find the sum of sixth power of roots of the equation $x^3-x-1=0$
My First approach:
Let $S_i$ denote the sum of $i^{th}$ power of roots of the given equation.
Now, multiplying given equation by $x^3$ , putting each of the three roots and adding the three formed equat... | Squaring $x^3=x+1$, we have
$$
x^6=x^2+2x+1
$$
Furthermore, we know that the coefficient of $x^2$ in $x^3-x-1$ is the negative of the sum of the roots, and the coefficient of $x$ is the sum of the products of pairs of distinct roots.
Therefore, we need to find the sum of the squares of the roots, which is
$$
\left(\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac... | $$(x+1)^2=x^2+2x+1=x^2-x+3x -3+4=(x-1)(\underbrace{x+3}_\text{quotient})+\underset{\strut{\rlap{\text{remainder}}\qquad}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Given $\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0$, $\{a,b\}\subset \mathbb R^+$, show the roots are real and within defined intervals
Given: $\{a,b\}\subset \mathbb R^+$, $P(x)=\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0.$ (correction: last term is $\frac{1}{x{\color{red}+}b})$
Show: all roots of $P(x)$ are real, one in... | I am using calculus. Please tell me if it is not allowed in the exam, then I can post an alternate solution.
Case $1$ : $P(x) = \frac 1{x} + \frac 1{x-a} + \frac 1{x-b} = 0$ (This was done before the edit, but is included for completeness)
$P(x)$ is a continuous function on $\mathbb R \backslash\{0,a,b\}$. Without loss... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$ Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
I already tried to use the Sqeeze theorem on it, but I just was not able to find some reasonable upper series for it, only lower:
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]... | $$\lim_{n \to \infty} (\sqrt[3]{n^3+1}-\sqrt{n^2+1})=$$
$$=\lim_{n\to \infty}((\sqrt[3]{n^3+1}-n)-(\sqrt{n^2+1}-n))=$$
Use the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1}), k\ge 2, k\in\mathbb Z, a,b\in\mathbb R$$
with, e.g., $a=\sqrt[3]{n^3+1}$, $b=n$, $k=3$, etc.
$$=\lim_{n\to \infty}\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How to compute the following limit$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$ I am trying to find
$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$
where $p>0$. I have tried to factor out as
$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =... | We can use the expansion
$$(1+y)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} \cdot y^n.$$
Taking $y = \frac{1}{x^p}, \ \alpha = \frac{1}{p}$, we get
$$\left( 1 + \frac{1}{x^p} \right)^{\frac{1}{p}} = 1 + \frac{1}{p x^p} + o \left( \frac{1}{x^{p}} \right).$$
Similarly
$$\left(1+\frac{1}{x^{p+1}} \right)^{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Nonlinear first order differential equation (exact)
Find an implicit solution to the initial value problem $$(x+y)^2+(2xy+x^2-1)\dfrac{dy}{dx} = 0 \\ y(1) = 1$$
Attempted solution
The equation can be written as $M(x,y)+N(x,y)y'=0$. We try to find find the potential function $\psi(x,y)$: $$\psi_x = M(x,y) = (x+y)^2 \... | Conventionally an implicit solution is of the form $\psi(x,y) =0$. To this end, you should set $$C=-\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute $ \lim_{n \to \infty} \frac{5^n}{5^n + 3^n} $ How can I compute the next limit of sequences:
$$ \lim_{n \to \infty} \frac{5^n}{5^n + 3^n} $$
It is $\frac{\infty}{\infty}$ indeterminate when evaluating in $n$, but using L'Hopital never eliminates the $5^n, 3^n$ terms. But because the denominator is greater than ... | $$\lim_{n \to \infty} \frac{5^n}{5^n + 3^n} = \lim_{n \to \infty} \frac{5^n}{5^n\left(1 + \left(\frac{3}{5}\right)^n\right)} = \lim_{n \to \infty} \frac{1}{1 + \left(\frac{3}{5}\right)^n} = 1,$$
since $\left(\frac{3}{5}\right)^n$ goes to $0$ as $n$ goes to $+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Trigonometry Problems Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac{1}{2}$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
My s... | \begin{align}
{\sin2x\over\sin2y}+{\cos2x\over\cos2y}=
{}& {2\sin x\cos x\over2\sin y\cos y}+{\cos^2x-\sin^2x\over\cos^2y-\sin^2y} \\[10pt]
= {} &
{3\sin y\cos y\over2\sin y\cos y}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y} \\[10pt]
= {} &
{3\over2}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y}.
\end{align}
On the other... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -... | Use the fact that
$$
a^3-b^3=(a-b)(a^2+ab+b^2)
$$
to get that
\begin{align}
\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}
&=\lim_{x\to 3}\left(\frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}\times
\frac{(x+5)^{2/3}+2(x+5)^{1/3}+4}{(x+5)^{2/3}+2(x+5)^{1/3}+4}
\right)\\
&=\lim_{x\to 3}\left(\frac{\cos (\pi x)+1}{x-3}\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prime counting function formulas Are there any elementary (including floor, ceiling, mod) representations of the prime counting function. Or one without an integral.
| I made this elementary formula, although it really has no practical importance:
for $n > 3$
$\pi (n) = 2 + \sum \limits_{m=4}^{n} \left\lfloor \frac{\displaystyle 1}{\displaystyle 1+ \displaystyle\sum_{p=2}^{\left\lfloor \sqrt{m} \right\rfloor}{\left( \left\lfloor\frac{m}{p}\right\rfloor- \left\lfloor\frac{m-1}{p}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Concerning this integral $\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}$ Motivation from this question $(1)$
$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{1\over 2}\right){x^s\over 1-x}\mathrm dx=F(s)\tag1$$
setting $s=0$ then $F(0)=-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over... | Use
\begin{eqnarray*}
H_v= \int_0^1 \frac{1-x^v}{1-x} dx.
\end{eqnarray*}
We have
\begin{eqnarray*}
\int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{x^s}{2}\right)\frac{ dx}{1-x} &=& \int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{1}{2} +\frac{1-x^s}{2}\right)\frac{ dx}{1-x} \\
&=& F(0)+ \int_0^1 \frac{1-x^s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2}... | Here's a way
to get bounds on $x$
for a generalized version of this.
We want to solve
$x
=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor
$.
Upper bound:
If
$a = \max(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\le\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? Mor... | Notice that $7^6\cdot 7^{-6}=7^{6-6}=7^0=1$
Notice also that $7^6\cdot\frac{1}{7^6}=\frac{7^6}{7^6}=1$
So, we learned that $7^6\cdot 7^{-6}=7^6\cdot\frac{1}{7^6}$.
Remembering that $x\cdot a=x\cdot b$ for nonzero $x$ implies that $a=b$ by cancelling this tells us that $7^{-6}=\frac{1}{7^6}$
In general the following pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 1
} |
Exponential Generating Functions With Coefficients Given the exponential generating function, $B(x)=\sum_{n\geq 0}b_n\frac{x^n}{n!}=b_0+b_1\frac{x}{1!}+b_2\frac{x^2}{2!}+\cdots$, I want to find a formula for the generating function
\begin{equation*}
\tilde{B}(x)=b_1+b_0\frac{x}{1!}+b_3\frac{x^2}{2!}+b_2\frac{x^3}{3... | You can bisect a generating function into its even and odd parts with the formulae
$$ \sum_{n \text{ even}} b_n \frac{x^n}{n!} = \frac{B(x) + B(-x)}{2} $$
and
$$ \sum_{n \text{ odd}} b_n \frac{x^n}{n!} = \frac{B(x) - B(-x)}{2}. $$
You can shift the terms in an exponential generating function by integrating or different... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)... | $$\sqrt[3]{26+15\sqrt3}=\sqrt[3]{8+3\cdot4\cdot\sqrt3+3\cdot2\cdot3+3\sqrt3}=\sqrt[3]{(2+\sqrt3)^3}=2+\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate: $\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}$ Evaluate: $\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}$
My Attempt:
$$=\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}$$
$$=\lim_{y\to x} \dfrac {y\cos x-x\cos y}{\cos x\cos y \cdot (y-x)}$$
| $\lim_\limits{y\to x} \frac {y\sec y - x\sec x}{y-x}$ is the definition of $\frac {d}{dx} x\sec x$
But since you requested a solution to this limit that was algebraic, here goes:
$\frac {y\cos x - x\cos y}{(y-x)\cos x\cos y}\\
\frac {y\cos x - y\cos y + y \cos y- x\cos y}{(y-x)\cos x\cos y}\\
\frac {y(\cos x - \cos y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximum and minimum values of function
For any number $c$ we let $f_c(x)$ be the smaller of the two numbers
$(x-c)^2$ and $(x-c-2)^2$. Then we define $g(c)=\int_0^1f_c(x)dx$.
Find the maximum and minimum values of $g(c)$ if $-2\le c \le 2$.
So the maximum/minimum values are either $g(\pm2)$ (the endpoints) or $g(... | Hint $f_c(x)$ depends only on $x-c$, in fact $f_c(x) = f_0(x-c)$, hence
$$g(c)= \int_0^1 f_0(x - c) d x = \int_{-c}^{1-c}f_0(t) d t$$
Edit: As ${f}_{0}$ is continuous, it follows that $g$ has the
derivative
$${g'} \left(c\right) =-{f}_{0} \left(1-c\right)+{f}_{0} \left({-c}\right) = \min \left({d}^{2} \left(c , {-2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $\int_0^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx$ converges I have to show that the above improper integral is convergent and I'm stuck. But this is my work so far:
*
*Split it: $$\int_0^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx = \int_0^\frac{\pi}{2} \frac{e^{\sin x}-1}{\sqrt{x}x}dx + \int_\frac{\pi}{2}^\inf... | \begin{align*}
\frac{e^{x}-1}{x^{3/2}}=\frac{1}{x^{1/2}}+\sum_{k=2}\frac{1}{k!}x^{k-3/2}
\end{align*}
we know that
\begin{align*}
\int_{0}^{\pi/2}\frac{1}{x^{1/2}}dx<\infty
\end{align*}
and
\begin{align*}
\sum_{k=2}\frac{1}{k!}\int_{0}^{\pi/2}x^{k-3/2}dx=\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}x^{k-(1/2)}\bigg|_{x=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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