Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Minimize the area of a wire divided into a circle and square.
A wire is divided into two parts. One part is shaped into a square, and the other part is shaped into a circle. Let r be the ratio of the circumference of the circle to the perimeter of the square when the sum of the areas of the square and circle is minimi... | Let $L$ be the length of the wire, $x, y$ be the side of the square and the radius of the circle, $S$ be the sum of the areas of the circle and the square. Then: $S = x^2 + \pi y^2, 4x+2\pi y = L$. By the Cauchy-Schwarz inequality: $L^2 = (4x+2\sqrt{\pi}\sqrt{\pi}y)^2\leq (4^2+4\pi)(x^2+\pi y^2)\Rightarrow S \geq \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the trigonometric equation $\csc^2 \theta= 5 \cot \theta + 7$ Solve the given equation. Let k be any integer.
$$\csc^2 θ = 5 \cot θ + 7$$
I just need the first step or two please. I tried converting it:
$$\frac{1}{\sin^2 θ} = \frac {5\cosθ}{\sinθ} + 7$$
Then I tried a number of different ways to simplify it but ... | Notice, $$csc^2\theta=5\cot \theta+7$$
$$\cot^2\theta+1=5\cot \theta+7$$ $$\cot^2\theta-5\cot \theta-6=0$$
$$(\cot\theta -6)(\cot\theta+1)=0$$
$$\implies \cot\theta-6=0 \iff \tan \theta=\frac{1}{6}\iff \color{blue}{\theta=n\pi+\tan^{-1}\left(\frac{1}{6}\right)}$$
$$\implies \cot\theta+1=0 \iff \tan \theta=-1=-\tan\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
| Looking at the first element in each row, we see
row A B C A+B+C
1 1 3 5 9
2 3 7 9 19
3 7 13 15 35
4 13 21 23 57
5 21 31 33 85
6 31 43 45 139
Using finite differences, we find that
\begin{align}
A &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
How do you solve for θ in the equation $\tan \frac{\theta}{5} + \sqrt{3} = 0$ $$\tan \frac{\theta}{5} + \sqrt{3} = 0$$
Alright so the $\frac{\theta}{5}$ is confusing me.
Would it be wrong to do
\begin{eqnarray}
\tan \frac{\theta}{5}&=&-\sqrt{3}\\
\frac{\theta}{5}&=&\tan^{-1}(-\sqrt{3})\\
\theta&=& 5\tan^{-1}(-\sqrt... | $$\tan\frac\theta5=-\sqrt 3\iff\tan\frac\theta5=\tan\Bigl(-\frac\pi3\Bigr)\iff\frac\theta5\equiv -\frac\pi3\mod\pi\iff\theta\equiv -\frac{5\pi}3\mod 5\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$ The number
$\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$
is expressed in the form $r \, \text{cis } \theta$, where $0 \le \theta < 360^\circ$. Find $\theta$ in degr... | \begin{align}
\text{cis }75^\circ +
\text{cis }83^\circ +
\text{cis }91^\circ + \dots +
\text{cis }147^\circ
&= \text{cis }75^\circ \left(
1 +
\text{cis }8^\circ +
\text{cis }16^\circ + \dots +
\text{cis }72^\circ \right)\\
&= \text{cis }75^\circ \left(
1 +
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Unable to find the f(x)
Find the Cubic in $x$ which vanishes when $x=1$ and $x=-2$ and has values $4$ and $8$ when $x=-1$ and $x=2$ resprectively.
I have proceeded like $P(x)=(x-1)(x+2)f(x)$ but I unable to find $f(x)$ to satisfy for $x=-1,2$.
| Since, cubic polynomial vanishes at $x=1$ & $x=-2$ hence, $(x-1)$ & $(x+2)$ are the factors.
Now, let the third root be $\beta$ then the cubic polynomial is given as $$f(x)=\alpha(x-\beta)(x-1)(x+2)$$ Now, we have
$$f(-1)=\alpha(-1-\beta)(-1-1)(-1+2)=4$$ $$\implies \alpha(1+\beta)=\frac{4}{2}=2$$ $$\alpha+\alpha\beta=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}$ Proving that $$\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}, \qquad\forall x>\pi$$
| Hint: say $$f(x) = \sin(x) - \frac{x(\pi^2 - x^2)}{\pi^2+x^2}$$ now $$f'(x) = \cos(x) + \frac{x^4 + 4\pi^2x^2 - \pi^4}{x^4 + 2\pi^2x^2 + \pi^4} = \cos x + 1 + \frac{2\pi^2\left ( x^2 - \pi^2 \right )}{\left ( x^2 + \pi^2 \right )^2}$$ then it is clear that $f'(x) > 0, \forall x > \pi$ and $f(\pi) = 0$ hence we can conc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Line joining the orthocenter to the circumcenter of a triangle ABC is inclined to BC at an angle $\tan^{-1}(\frac{3-\tan B\tan C}{\tan B-\tan C})$ Show that the line joining the orthocenter to the circumscribed center of a triangle ABC is inclined to BC at an angle $\tan^{-1}\left(\frac{3-\tan B\tan C}{\tan B-\tan C}\r... | Let $a$ be the length of side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$ Hence, the ortho-center will be $$H\equiv \left(\frac{0+a+\frac{a\tan C}{\tan B+\tan C}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
help with trigonometric equations How do I solve this?
$$\cos3x=\cos^2x-3\sin^2x$$
| Using $\cos 3x = \cos^3 x -3\cos x \sin^2 x$,
$$\begin{align*}
\cos 3x &= \cos^2 x - 3 \sin ^2 x\\
\cos^3 x -3\cos x \sin^2 x &= \cos^2 x - 3 \sin ^2 x\\
(\cos x)(\cos^2 x - 3 \sin ^2 x) &= \cos^2 x - 3 \sin ^2 x\\
\cos x &= 1&\text{ or }&& \cos^2 x - 3 \sin ^2 x &= 0\\
\cos x &= 1&\text{ or }&&\tan^2 x&=\frac13
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the value below
Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$
It's just for sharing a new ideas, thanks:)
| If $a,b,c$ are the roots of $p(x)=8x^3-4x^2-4x+1$ then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are the roots of
$$ q(x) = x^3-4x^2-4x+8 $$
so:
$$ \frac{1}{a^3} = \frac{4}{a^2}+\frac{4}{a}- 8 $$
and:
$$\sum_{cyc}\frac{1}{a^3}=4\sum_{cyc}\frac{1}{a^2}+4\sum_{cyc}\frac{1}{a}-24$$
then Viète's theorem applied to $q(x)$ gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar exam... | Let $\displaystyle N=\sqrt{4+2\sqrt{3}}-\sqrt{3}$
$=\sqrt{{(\sqrt{3})}^2+1^2+2\cdot\sqrt{3}\cdot1}-\sqrt{3}$
$=(\sqrt{3}+1-\sqrt{3})$
$=\boxed1$
Aliter: If you want to use polynomials, you can see that
$\displaystyle (N+\sqrt{3})^2=4+2\sqrt{3}$
$\implies N^2+2\sqrt{3}\cdot N + 3=4+2\sqrt{3}$
$\implies (N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
... | You can compute the sum directly, using
Note that $$r\cdot \binom nr=r\cdot \frac {n!}{r!\cdot(n-r)!}=n\cdot\frac {(n-1)!}{(r-1)!\cdot(n-r)}=n\cdot\binom {n-1}{r-1}$$
This gives you a factor $100$ you can extract from every term leaving the sum from $\binom {99}0$ to $\binom {99}{99}$. The sum of such a complete set of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the general solution of $\cos(x)-\cos(2x)=\sin(3x)$ Problem:
Find the general solution of $$\cos(x)-\cos(2x)=\sin(3x)$$
I tried attempting this by using the formula$$\cos C-\cos D=-2\sin(\dfrac{C+D}{2})\sin(\dfrac{C-D}{2})$$
Thus, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Right... | \begin{align}
& -2 \, \sin\left(\frac{x}{2}\right) \, \sin\left(\frac{3x}{2}\right) = \sin(3x) \\
& \sin(3x) = 2 \, \sin\left(\frac{3x}{2}\right) \, \cos\left(\frac{3x}{2}\right) \\
& -2 \, \sin\left(\frac{x}{2}\right) \, \sin\left(\frac{3x}{2}\right) = 2 \, \sin\left(\frac{3x}{2}\right) \, \cos\left(\frac{3x}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\lim\limits_{n\to+\infty}\frac{\sqrt[n]{n!}}{n}$ I tried using Stirling's approximation and d'Alambert's ratio test but can't get the limit. Could someone show how to evaluate this limit?
| This is an alternate solution for those who don't know stirling approximation yet like me
$$
\lim _{ n\longrightarrow +\infty }{ \frac { \sqrt [ n ]{ n! } }{ n } } \quad =\quad \lim _{ n\longrightarrow +\infty }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (n-k) } } }{ n } } \\ \qquad \qquad \qquad \qquad =\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Let $(a, b, c)$ be a Pythagorean triple. Prove that $\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2$ is greater than 8 and never an integer.
Let $(a, b, c)$ be a Pythagorean triple, i.e. a triplet of positive integers with $a^2 + b^2 = c^2$.
a) Prove that $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > 8$$
b) Prove that there ... | First note that if $n$ were a square of rational number, then it would be square of integer. So if $(\frac{c}{a}+\frac{c}{b})^2$ were an integer, so would be $\frac{c}{a}+\frac{c}{b}$. By dividing by common factor, we can assume that triple $(a,b,c)$ is primitive, i.e. no two numbers share a prime factor.
Now $\frac{c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Tangent of a circle While I was preparing for $AIME$, I ran into a question regarding circles and their tangents(on a coordinate plane). I've read several posts regarding similar question types but I haven't found a systematic way of finding tangent points on a circle given the equation of a circle and a point outside ... | Notice, let the equation of the tangent be $y=mx+c$ then satisfying this equation by the point $(15, 10)$ , we get $$10=m(15)+c$$ $$15m+c=10$$
$$c=10-15m\tag 1$$
Now, the length of perpendicular from the center $(5, 5) $ to the tangent $y=mx+c$ must be equal to the radius $3$ of the circle, hence we have $$\frac{|m(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to factor $4x^2 + 2x + 1$? I want to know how to factor $4x^2 + 2x + 1$? I found the roots using quadratic equation and got $-1 + \sqrt{-3}$ and $-1 - \sqrt{-3}$, so I thought the factors would be $(x - (-1 + \sqrt{-3}))$ and $(x - (-1 - \sqrt{-3}))$
However, according to MIT's course notes, the factors are $(1 - (... | Given $$\displaystyle 4x^2+2x+1 = \underbrace{(2x)^2+1^2+\left(\frac{1}{2}\right)^2}+1-\underbrace{\left(\frac{1}{2}\right)^2}$$
So we get $$\displaystyle \left(2x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}i}{2}\right)^2 = \left(2x+\frac{1}{2}+\frac{\sqrt{3}i}{2}\right)\cdot \left(2x+\frac{1}{2}-\frac{\sqrt{3}i}{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Integarate $\frac{1}{2}f'(x)$ wrt $x^4$ where $f(x)=\tan^{-1}x+\ln\sqrt{1+x}-\ln\sqrt{1-x}$ Integarate $\frac{1}{2}f'(x)$ wrt $x^4$ where $f(x)=\tan^{-1}x+\ln\sqrt{1+x}-\ln\sqrt{1-x}$
I tried: $\int \frac{1}{2}f'(x) \, d(x^4)=\int \frac{1}{2}f'(x)\cdot 4x^3 \, dx$. I integrated it by parts but it goes lengthy and could... | We are given $f(x)=\arctan(x)+\frac12 \log(1+x)-\frac12 \log(1-x)$ and are asked to find $F(x)$ where $F(x)$ is given by
$$F(x)=\int \frac12 f'(x)4x^3\,dx \tag 1$$
We can proceed directly by first determining the derivative of $f$. To that end, we find that
$$f'(x)=\frac{1}{1+x^2}+\frac12\left(\frac{1}{1+x}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no... | $$
\begin{align}
\sum_{i=1}^6 \dfrac {x^i} {i!} &=\dfrac 1 {720} \cdot (x^6+6x^5+30x^4+120x^3+360x^2+720x+720= \\
&=\dfrac 1 {720} \cdot \{x^4(x+3)^2+20x^2(x+3)^2+x^4+180x^2+720x+720\}
\end{align}
$$
It can be easily proved that $x^4+180x^2+720x+720 > 0$ by using the derivative. Therefore, there are no real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 1
} |
Olympiad-like Inequality Problem Let
$$ A := \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times\cdots\times \frac{2013}{2014};$$
let
$$ B := \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times\cdots \times \frac{2012}{2013};$$
and let
$$ C := \frac{1}{ \sqrt{2014} }. $$
Find the relations between $A, B,$ and... | Let fractions in $A:= a_1, a_2, a_3, \ldots, a_{1007} $
Let fractions in $B:=b_1, b_2, b_3, \ldots , b_{1006} $
But $ a_1 < b_1, a_2 < b_2, a_3 < b_3 \ldots a_{1006} < b_{1006}, a_{1007} < 1$
Multiplying all these, $ A < B $
Observe that $ A \times B = \frac{1}{2014} = C^2 $
So $A, C, B $ are positive numbers forming i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Quadratics question To solve $-3x^2 +2x +1=0$, I'd normally break the middle term and then factorise. But I was wondering if there was a way to skip the factorising step? The factors I'd use in place of the middle term would be $3$ and $-1$. If I were to flip the sign of each, I would get $-3$ and $1$. And then, were I... | Notice, we have $$-3x^2+2x+1=0$$
Finding the roots of $-3x^2+2x+1=0$ by quadratic rule as follows
$$x=\frac{-2\pm\sqrt{(2)^2-4(-3)(1)}}{2(-3)}$$
$$x=\frac{-2\pm4}{-6}$$ $$x=\frac{-2+4}{-6}\ \ \vee \ \ x=\frac{-2-4}{-6}$$
$$x=1\ \ \vee \ \ x=-\frac{1}{3}$$
Then, $(x-1)$ & $\left(x+\frac{1}{3}\right)$ will be the factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$... | Rewrite the sum as
$$\frac{\sum_{n=1}^{99} \sqrt{10+\sqrt{n}}}{\sum_{n=1}^{99} \sqrt{10-\sqrt{n}}} = \frac{a}{b}$$
then let
$$\Delta_n = \sqrt{10+\sqrt{n}} - \sqrt{10-\sqrt{n}}$$
By squaring $\Delta_n$ and simplifying we get
$$\Delta_n = \sqrt{2} \sqrt{10 - \sqrt{100-n}}$$
Now summing all those $\Delta_n$s can be done ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 6
} |
Evaluating $\int_{-3}^{3}\frac{x^8}{1+e^{2x}}dx$ $$\int_{-3}^{3}\frac{x^8}{1+e^{2x}}dx$$
I can't find solution for this task, can someone help me?
| The answer from @OlivierOloa presents a really cool trick and I'd like to add some information around it. This might help to apply this technique to similar expressions.
The following holds true. If the integral has the form:
\begin{align*}
\int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx
\end{align*}
with
*
*$p(x)$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Need help with an arithmetic sequence proving question It is given that $a_1, a_2, a_3, \ldots ,a_n$ are consecutive terms of an Arithmetic progression. I have to prove that
$$\sum_{k=2}^n (\sqrt{a_{(k-1)}} + \sqrt{a_k} )^{-1} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n} }$$
Using Mathematical induction I showed that it is tru... | Let $d$ be the common difference of A.P. then general $n$th term is given as $$a_n=a_1+(n-1)d$$
Now, assuming the equality holds for $n=m$, then we get $$\sum _{k=2}^{m}(\sqrt{a_{(k-1)}}+\sqrt{a_k})^{-1}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$
$$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{a_{(k-1)}-a_{k}}=\frac{m-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ My Calc 2 teacher wasn't able to solve this:
$$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$
Can someone help me solve this?
| First Start with Basic Integration by Parts:
$$ \int_0^{\frac{\pi}{2}}\dfrac{x \cos x}{\sin{x}+\sin^3 x}\, dx = -\frac{\pi}{4}\log 2 - \int_0^{\frac{\pi}{2}} \left( \log{\sin{x}} - \frac{\log(1+\sin^2 x)}{2}\, dx \right) $$
$$$$
Since,
$$\int_0^{\frac{\pi}{2}} \log\sin x\, dx = -\frac{\pi}{2} \log 2 $$
$$\therefore\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
LU-factorization: why can't I get a unit lower triangular matrix? I want to find an $LU$-factorization of the following matrix: \begin{align*} A = \begin{pmatrix} 3 & -6 & 9 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} \end{align*} This matrix is invertible (the determinant is $33$), so I should be getting a $LU$ decompos... | Uniqueness is only for such $L$ that $L_{ii} = 1$ for all $i$. To achieve this, move the scaling factors to $U$ like this:
$$A = LU = (LS)(S^{-1}U) = \tilde L \tilde U$$
Here you need
$$S = \pmatrix{\frac13&&\\&\frac13&\\&&1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the root is less than $2^n$
A polynomial $f(x)$ of degree $n$ such that coefficient of $x^k$ is $a_k$. Another constructed polynomial $g(x)$ of degree $n$ is present such that the coefficeint of $x^k$ is $\frac{a_k}{2^k-1}$. If $1$ and $2^{n+1}$ are roots of $g(x)$, show that $f(x)$ has a positive root less than... | Warning: This is just a wordy hint... :
Assuming $a_0=0$, notice that $g(2x)$ is a horizontally scaled version of $g(x)$, which means that both functions have the same sign in $(\dfrac 12,2^n)$ and $(1,2^{n+1})$ respectively, so they have the same sign in $(1,2^n)$. On the left-most side of the interval, around $x=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2... | A couple of hints (not a full solution):
$x\to 2 \sin x$
\begin{align}
\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx&=\int_{\pi/6}^{\pi/2}\frac{\cot (x) \left(3-\cot ^2(x)\right)}{4 \sqrt{-12 \cos (2 x)+4 \cos (4 x)+9}}\,dx\\
\end{align}
now you may use $$\cot (x) \left(3-\cot ^2(x)\right)=\Big[\frac14(24 \sin (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Triangle area inequalities; semiperimeter I've got stuck on this problem :
Proof that for every triangle of
sides $a$, $b$ and $c$ and area
$S$, the following
inequalities are true :
$4S \le a^2 + b^2$
$4S \le b^2 + c^2$
$4S \le a^2 + c^2$
$6S \le a^2 + b^2 + c^2$
The first thing that came to my mind was the inequal... | Let $S$ denote the area of the triangle, so:
$$
S = \frac{ab\sin{(\gamma)}}{2} = \frac{cb\sin{(\alpha)}}{2} = \frac{ac\sin{(\beta)}}{2}
$$
And as for $\theta \in[0,\pi]$: $0\leq\sin{\theta}\leq1$, we get as you said:
$$
S\leq\frac{ab}{2}\\
S\leq\frac{ac}{2}\\
S\leq\frac{bc}{2}\\
$$
So:
$$
4S \leq 2ab\\
4S\leq 2ac\\
4S\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\frac{1}{(x^4 -1)^2}$ How to solve the the following integral? $$\int{\frac{1}{(x^4 -1)^2}}\, dx$$
| Let $$\displaystyle I = \int\frac{1}{(x^4-1)^2}dx = \frac{1}{4}\int \left[\frac{-3}{(x^4-1)}+\frac{3x^4+1}{(x^4-1)^2}\right]dx$$
$$\displaystyle I = \frac{1}{4}\int\left[\frac{-3}{2(x^2-1)}+\frac{3}{2(x^2+1)}+\frac{3x^2+\frac{1}{x^2}}{\left(x^3-\frac{1}{x}\right)^2}\right]dx$$
Now Here $\bf{1^{st}}$ and $\bf{2^{nd}}$ c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Linear Functions: Division of an interval The lines $y= 4x + 2$ and $x+2y=6$ intersect at point $P$.
i) Find the coordinates of point $P$.
$$y= 4x + 2\ldots(1)$$
$$x+2y=6 \ldots(2)$$
Sub ($1$) into ($2$)
$$x+2(4x + 2) =6$$
$$9x=2$$
$$x=2/9$$
sub $x = 2/9$ into ($1$) to find $y$
$$y=26/9$$
Therefore the coordinates of ... | Notice,let $m:n$ be the ration in which the given point $P\left(\frac{2}{9}, \frac{26}{9}\right)$ divides the line joining the points $\left(1, \frac{33}{9}\right)$ & $\left(\frac{1}{3}, 3\right)$ respectively
then the coordinates of the point P are calculated using division formula as follows $$P\equiv \color{red}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Does the elliptic function $\operatorname{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\big|\frac{1}{2}\right)$ have a closed form? Given the complete elliptic integral of the first kind $K(k)$ for the modulus $k$,
can the elliptic function $$\text{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\bigg|\frac{1}{2}\right)$$... | Assuming $1/2$ is $k$, as in Maple ... Maple
JacobiCN(2/3*EllipticK(1/2),1/2) evaluates to
$0.473058826656122429170671314726$.
From ISC we find that
this is a solution of
$$
Z^4-2Z^3-6Z+3=0
$$
which may be written
$$
{\frac { \left( 1+2\,\sqrt [3]{6} \right) ^{3/4}+\sqrt [4]{1+2\,
\sqrt [3]{6}}-\sqrt {-2\,\sqrt [3]{6}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no i... |
Claim. $a+b+c\mid a^{2^n}+b^{2^n}+c^{2^n}$ for all $n\geq0$.
Proof. By induction: True for $n=0,1$ $\checkmark$. Suppose it's true for $0,\ldots,n$. Note that $$a^{2^{n+1}}+b^{2^{n+1}}+c^{2^{n+1}}=(a^{2^n}+b^{2^n}+c^{2^n})^2-2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})^2+4a^{2^{n-1}}b^{2^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
Calculate $\lim_{n\to\infty} (\frac{1}{{1\cdot2}} + \frac{1}{{2\cdot3}} + \frac{1}{{3\cdot4}} + \cdots + \frac{1}{{n(n + 1)}})$
Calculate
$$\lim_{n\to\infty} \left(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{n(n + 1)}\right).
$$
If reduce to a common denominator we get
$$\lim _{n\to... | This won't get you far. Try $\frac{1}{n(n + 1)} = \frac{n + 1 - n}{n(n+1)} = \frac{1}{n} - \frac{1}{n + 1}$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Triangle of forces Forces equal to $5P$, $12P$ and $13P$ acting on a particle are in equilibrium ;find ,by geometric construction and by calculation ,the angles between their directions?
I have an problem that, With three forces how can I use Lami's rule?
| let $\vec x\parallel\vec F_1, \vec x \perp \vec y$
$\vec x:$
$\angle \vec F_2=\alpha$, $\angle \vec F_3=\beta$
$\vec y:$
$\angle \vec F_2=\frac {\pi}{2} - \alpha$, $\angle \vec F_3=\frac {\pi}{2} - \beta$
$\begin{equation*}
\begin{cases}
F_1 + F_2 \cdot \cos \alpha + F_3 \cdot \cos \beta = 0
\\
0 + F_2 \cdot \sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trouble understanding inequality proved using AM-GM inequality I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I'm lost on the very first line of the solution.
Let $x,y,z$ be positive real numbers such that $xyz =1$. P... | The AM-GM inequality for three terms can be written
$$a+b+c\ge3\sqrt[3]{abc}\ .$$
Just substitute your three terms on the LHS into this. That is, take
$$a=\frac{x^3}{(1+y)(1+z)}\ ,\quad b=\frac{1+y}8\ ,\quad c=\frac{1+z}8\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Vector Functions of One Variable Question
A particle moves along the curve of the intersection of the cylinders $y=-x^2$ and $z=x^2$ in the direction in which $x$ increases. (All distances are in cm.) At the instant when the particle is at the point $(1,\,-1,\,1)$ its speed is $9$ cm/s, and the speed is increasing at a... | Notice that, at the given instant:
$x=1$ cm and
\begin{align*}
\frac{d^2x}{dt^2}&=\frac{d}{dt}\left(\frac{dx}{dt}\right)\\
&=\frac{d}{dt}\left(\frac{v}{\sqrt{1+8x^2}}\right)\qquad\text{where }v=||\mathbf{v}||\\
&=\frac{\sqrt{1+8x^2}\dfrac{dv}{dt}-\frac{8x}{\sqrt{1+8x^2}}\frac{dx}{dt}v}{1+8x^2} \\
&=\frac{\sqrt{1+8(1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
| Using L'Hôpital and $\lim\limits_{x\to0}\frac{\tan(x)}x=1$,
$$
\begin{align}
\lim_{x\to0}\left(\frac1{x\tan^{-1}(x)}-\frac1{x^2}\right)
&=\lim_{x\to0}\left(\frac{x}{x^3}-\frac{\tan^{-1}(x)}{x^3}\right)\cdot\lim_{x\to0}\frac{x}{\tan^{-1}(x)}\\
&=\lim_{x\to0}\frac{1-\frac1{1+x^2}}{3x^2}\cdot1\\
&=\lim_{x\to0}\frac{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
transformation of uniformly distributed random variable f(x)=1/2pi into Y=cosx Let $X$ be a uniformly distributed function over $[-\pi;\pi]$. That is
$
f(x)=\left\{\begin{matrix}
\frac{1}{2 \pi} & -\pi\leq x\leq \pi \\
0 & otherwise
\end{matrix}\right.\\
$
Find the probability density function of $Y = \cos X$.
I f... | since we have $x$ belongs to $[-\pi,\pi]$ the $\arccos(x)$ can be written with two signs ($-$ & $+$) , that's why we have:
$$fy(y)=\frac 1{ 2\pi} |\frac{-1} {\sqrt{1-x^2}}| + \frac 1 {2\pi} |\frac 1 {\sqrt{1-x^2}}|$$
So
$$fy(y)=2* \frac 1 {2\pi} * |\frac 1 {\sqrt{1-x^2}}|$$
Finally,
$$fy(y)= \frac 1 \pi \frac 1 {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{n \rightarrow \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$ $\lim_{n \rightarrow \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$
where the 2 inside the roots appear n times. For example if n = 2 :
$2^{2} \sqrt{2-\sqrt{2}}$
I discovered this. Has this been already developed/made before?
| Since $$\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}$$ is the side length of the regular $2^{n+1}$-gon inscribed in a radius $1$ circle, the sum of the $2^{n+1}$ sides will converge to the circumference of that circle. Thus $$\lim_{n\rightarrow\infty}2^{n+1}\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}=2\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Series of $\csc(x)$ or $(\sin(x))^{-1}$ In some cases I found that $$\csc(x)= \lim\limits_{k\rightarrow \infty}\sum_{n=-k}^{k}(-1)^{n}\frac{1}{x-n\pi}$$
Is anything to prove or disprove that?
| Recall that the sine function can be represented as the infinite product
$$\sin x=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right) \tag 1$$
Taking the logarithmic derivative of $(1)$, we obtian
$$\begin{align}
\cot x&=\frac1x+2x\sum_{n=1}^{\infty}\frac{1}{x^2-n^2\pi^2}\\\\
&=\sum_{n=-\infty}^{\infty}\frac{x}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to calculate the limit that seems very complex.. Someone gives me a limit about trigonometric function and combinatorial numbers.
$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom... | With rough (but safe) approximations, we can avoid the use of trigonometric stuff, that is $$l=\lim_{n\to \infty } \, \left(1+\frac{n+2}{2n^2}\right)^n=\lim_{n\to \infty } \, \exp\left(\frac{ n+2}{2n}\right)=\sqrt{e}$$
where $\displaystyle \sin\left(\frac{k}{n^2}\right)$ behaves like $\displaystyle \frac{k}{n^2}$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
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How to evaluate this infinite product ? (Fibonacci number) Let $F_n$ be Fibonacci numbers.
How to evaluate
$$\prod_{n=2}^\infty \left(1-\frac{2}{F_{n+1}^2-F_{n-1}^2+1}\right)\text{ ?}$$
It seem like that
$$\prod_{n=2}^\infty \left(1-\frac{2}{F_{n+1}^2-F_{n-1}^2+1}\right)=\frac{1}{3}$$
But how to prove it?
Thank in adva... | The product is indeed equal to $\frac{1}{3}$.
We can use Binet's formula, which states that
$$F_n=\frac{1}{\sqrt{5}}\left(\phi^n - \left(-\frac{1}{\phi}\right)^n\right)$$
for all $n$ where $\phi=\frac{1+\sqrt{5}}{2}$
We can plug this directly into the product which we want to evaluate, but our life becomes slightly eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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If $p_x ^2 = 2n+1$ then $n$ is always even? If an odd prime squared $p_x ^2$ is written in the form $2n+1$, is $n$ always even?
Furthermore, is it true that $k\cdot p_x + n$ is always prime when $k\cdot p_x + n < p_{x+1} ^2$ and $\gcd(k,n)=1$ where $k \in \mathbb{N}$ and $n$ is as above?
Lastly, for $p_x ^2 = 2m-1$, is... | Any odd number squared is of the form $4k+1$, so that $p^2 - 1 = 4k$ for some integer $k$. If $p^2 - 1 = 2m$ then of course $m$ would be even (because $2m$ is divisible by 4).
Take $p = 5, k = 3, n = 1$. Then $16 < 25$ and $16$ is not prime.
EDIT: $p = 17, n=144, k=1$. So $161 < 289$ and $161 = 7\cdot23$ is not prime. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the smallest value of $\sqrt[5]{\frac{abc}{b+c}} + \sqrt[5]{\frac{b}{c(1+ab)}} + \sqrt[5]{\frac{c}{b(1+ac)}}$ Let $a\ge0$ and $b,c>0$, we need to find the smallest value of the expression
$$S=\sqrt[5]{\frac{abc}{b+c}} + \sqrt[5]{\frac{b}{c(1+ab)}} + \sqrt[5]{\frac{c}{b(1+ac)}}$$
I have no idea for this question, ... | For $a=0$ and $b=c=1$ we have a value $2$.
We'll prove that it's a minimal value.
Indeed, let $a\neq0$, $a=x$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, by AM-GM we obtain: $$S=\sum_{cyc}\sqrt[5]{\frac{x}{y+z}}\geq\sum_{cyc}\sqrt{\frac{x}{y+z}}=$$
$$=\sum_{cyc}\frac{2x}{2\sqrt{x(y+z)}}\geq\sum_{cyc}\frac{2x}{x+y+z}=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to get formula for sums of powers? Assuming I have Bernoulli numbers:
$B = [\frac{1}{1},\frac{1}{2},\frac{1}{6},\frac{0}{1},-\frac{1}{30}, \frac{0}{1}, \frac{1}{42}, ...]$
How can I get the coefficients of the sums of powers formulas?
For example the sum of squares is $(1/3)n^3 + (1/2)n^2 + 1/6n$
| A vey-well rememberable scheme is perhaps the following.
Use the vector of Bernoulli-numbers in the first row of a little scheme and multiply through each column (the result per column is in the last row of the table):
$$ \begin{bmatrix}\frac{1}{1}&\frac{1}{2}&\frac{1}{6}&\frac{0}{1}&-\frac{1}{30}& \frac{0}{1}& \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Solve $\vert x-2\vert+2\vert x-4\vert\leq \vert x+1\vert$ I was helping someone with abolute values and inequalities and found this question.
What is the easiest way to solve this?
The only thing I thought of is to add the L.H.S and graph it with the R.H.S to answer the questoin is there simpler way to deal with this?
... | An alternate non case-wise approach that involves more calculation with larger numbers:
Look for equality first. Squaring preserves the equality (although it may introduce extraneous solutions which can be ruled out at the end).
$$
\begin{align}
(x-2)^2+4|(x-2)(x-4)|+4(x-4)^2
&=(x+1)^2\\
x^2-4x+4+4|(x-2)(x-4)|+4x^2-32x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Use Proof of Induction to prove $\sum_{k=1}^{2n} (-1)^k k = n$ Base Case:
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n\\
(-1)^1 (1) + (-1)^2(2) &=&1 \\
1=1
\end{eqnarray*}
Inductive Step: For this step we must prove that
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n \Rightarrow \sum_{k=1}^{2(n+1)} (-1)^k k = n+1 ... | Basically, yes. There is a bit of a snag with your second summation though: it should be $k = 2n + 1$ and $k = 2n + 2$. So we have:
\begin{align*}
\sum_{k=1}^{2(n + 1)} (-1)^k k
&= \sum_{k=1}^{2n} (-1)^k k + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) \\
&= n + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) &\text{by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be:
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-... | Let's look at this a bit differently. As a first stage, we want to decompose the numerator $x^5+x^4+3x^3-8x^2+28x+48$ into components $p(x)(x^2+2x+4)^2+q(x)(x-2)^2$ so that we have $$\frac{x^5+x^4+3x^3-8x^2+28x+48}{(x^2+2x+4)^2(x-2)^2}=\frac {p(x)}{(x-2)^2}+\frac {q(x)}{(x^2+2x+4)^2}$$
We note by comparing degrees that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ax+by+cz=k$,prove that the minimum value of $x^2+y^2+z^2$ is $\frac{k^2}{a^2+b^2+c^2}$ If $ax+by+cz=k$,prove that the minimum value of $x^2+y^2+z^2$ is $\frac{k^2}{a^2+b^2+c^2}$.
I know that this problem can be solved by Cauchy Schwartz.How can i find its minimum value using multivariable calculus or by other metho... | $$(x^2+y^2+z^2)(a^2+b^2+c^2)-(ax+by+cz)^2 = (ay-bx)^2+(bz-cy)^2+(cx-az)^2 \ge 0$$
so $$x^2+y^2+z^2\ge\frac{(ax+by+cz)^2}{a^2+b^2+c^2}=\frac{k^2}{a^2+b^2+c^2}$$
This is in fact an elemental proof of Cauchy–Schwarz_inequality on $\mathbb{R}^3$
| {
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"url": "https://math.stackexchange.com/questions/1431655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that there exist infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$ Let $(a_n)$ be a sequence defined by: $a_o=2, a_1=4, a_2=11$ and $\forall n \geq 3$, $$a_n = (n+6)a_{n-1}-3(2n+1)a_{n-2}+9(n-2)a_{n-3}$$
Show that there exist infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$
| $$a_n = (n+6)a_{n-1}-3(2n+1)a_{n-2}+9(n-2)a_{n-3}$$
can be written as
$$a_n-(n+3)a_{n-1}+3(n-1)a_{n-2}=3\left(a_{n-1}-(n+2)a_{n-2}+3(n-2)a_{n-3}\right)$$
So, we can have
$$a_n-(n+3)a_{n-1}+3(n-1)a_{n-2}=-3^{n-1}.$$
This can be written as
$$a_n-na_{n-1}+n\cdot 3^{n-1}=3\left(a_{n-1}-(n-1)a_{n-2}+(n-1)\cdot 3^{n-2}\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a simple rational equation $(\frac{6x}{6-x})^2+x^2=400$ Clearly we could multiply both sides of
$$\left(\dfrac{6x}{6-x}\right)^2+x^2=400$$
by $(6-x)^2$ which leads to a degree 4 polynomial equation, which we can solve using the bi-quadratic formula. Moreover, we could approximate the solutions using Newtons met... | The quartic (or bi-quadratic) polynomial for $x$ can be factored into two quadratic polynomials with coefficients in $\mathbb{Z}[\sqrt{109}]$, so that the closed-form expression for roots is not as terrible as it can be generally.
Clearing fractions one gets:
$$ x^4 - 12x^3 - 328x^2 + 4800x - 14400 = 0 $$
Then notice t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Change a 3D plane to XY plane I have a 3D cloud of points from witch I determine an "average" plane.
The plane equation is Ax + By + Cz + D =0;
I would like to re-compute all the points to obtain their new position considering that "average" plane is the new XY plane.
I suppose I have to do first a Z translation of -D/... | $\vec{n}=(A,B,C)$ is a vector perpendicular to your average plane, forming an angle $\theta$ with the $z$-axis.
Consider the vector
$\vec{n}'=(B,-A,0)$, which is perpendicular to the plane formed by $\vec{n}$ and $z$-axis: a rotation of angle $\theta$ around $\vec{n}'$ will carry $\vec{n}$ along the $z$-axis and is t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| Our old friend the golden ratio, or rather its enhancement by $1$, namely $a=1+\phi$, along with its reciprocal, is lurking behind this problem.
Let $a$ and $\bar a$ be respectively the larger and smaller root of $x^2-3x+1=0$. We have $a\bar a=1$ and $a+\bar a=3$. It's easy to show that $a^3+\bar a^3=18$. Also,$$a^{11}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the absolute value equation $2-3|x-1| = -4|x-1|+7$ $$2-3|x-1| = -4|x-1|+7$$
This is an example from my text book, and I do not understand how they got the answers.
Solution: (this is the solution in my textbook)
Isolate the absolute value of expression on one side
Add $4|x-1|$ to both sides $ \rightarrow 2+|x-1... | This is simply due to the fact that we can add or subtract anything to both sides of an equation and preserve the equality. Maybe this will help you understand it
\begin{align}
2-\color{red}{3|x-1|} &= \color{red}{-4|x-1|}+7 \\
2-\color{red}{3|x-1|} + \underbrace{\color{blue}{4|x-1|}}_{\text{Added}} &= \color{red}{-4|x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
| Observe $$(x+1)(x^2 - x + 1) = x^3 + 1,$$ so that $$f(x) = \frac{x^2+x+1}{x^2-x+1} = 1 + \frac{2x}{x^2-x+1} = 1 + \frac{2x(x+1)}{x^3+1}.$$ Then consider the geometric series expansion $$\frac{1}{1 - (-x^3)} = 1 + (-x^3) + (-x^3)^2 + (-x^3)^3 + \cdots = 1 - x^3 + x^6 - x^9 + \cdots, \quad |x| < 1.$$ The result is now ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the root of this polynomial equation of third degree? x^3 -3x + 4 = 0
How do I find roots of the above equation?
I have tried plugging in values of x but it is not satisfying the equation above.
| The full solution to the cubic is:
$$x = (q + u)^{\frac{1}{3}} + (q - u)^{\frac{1}{3}} + p$$
$$u = \sqrt{q^2 + (r-p^2)^3}$$
$$p = -\frac{b}{3a}$$
$$q = p^3 + \frac{bc-3ad}{6a^2}$$
$$r = \frac{c}{3a}$$
Here we have $a=1$, $b=0$, $c=-3$, and $d=4$. Hence:
$$r = \frac{-3}{3} = -1$$
$$p = -\frac{0}{3} = 0$$
$$q = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a progression inequality Prove that $$1+ \frac{1}{2^3} + \cdot \cdot \cdot + \frac{1}{n^3} < \frac{5}{4} $$
I got no idea of how to approach this problem.
| Another way to tackle these problems is by proving a stronger statement using induction.
For example, in this case you might prove that for $n\geq 2$ $$\sum_{j=1}^n\frac{1}{j^3}\leq \frac{5}{4}-\frac{4}{15n^2}.$$
The case $n=2$ is easily verified ($\frac{9}{8}\leq \frac{5}{4}-\frac{1}{21}$). Now for the induction step:... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How is it possible that $\int\frac{dy}{(1+y^2)(2+y)}$ = $\frac{1}{5}\int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} $? Suppose we have a fraction
$$I=\int\frac{dy}{(1+y^2)(2+y)}$$
How is it possible that
$$5I = \int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} ?$$
How are they using partial... | Setting
$$\frac{1}{(1+y^2)(2+y)}=\frac{A}{y+2}+\frac{By+C}{1+y^2}$$
gives
$$1=A(1+y^2)+(y+2)(By+C),$$
i.e.
$$0y^2+0y+1=(A+B)y^2+(C+2B)y+A+2C$$
Then, solve the following system :
$$0=A+B,\quad 0=C+2B,\quad 1=A+2C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$? How to solve the definite integral $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$?
Funnily enough, this is actually a problem that showed up during my real analysis course. At first glance, the problem seemed solvable by using knowle... | Note that there $x$ and $(\frac 12 - x)$ is symmetric about $x= \frac 14$. Thus try to substitute $y = \frac 14 -x$. Then the integral is
$$- \int_{\frac{1}{4}}^{-\frac 14} (\frac 14 -y)^\frac{3}{2} (\frac 12 + 2y)^{\frac 32} dy = 2^{\frac 32} \int_{-\frac 14}^{\frac 14} (\frac 1{16} - y^2)^{\frac 32} dy.$$
From here ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
| Since $x \geq \lfloor x \rfloor$ for all $x$, you have $x^2 - 19\lfloor x \rfloor + 88 \geq x^2 - 19x + 88$ for all $x$. Hence the graph of $x^2 - 19\lfloor x \rfloor + 88$ lies above that of $x^2 - 19x + 88$. So the function $x^2 - 19\lfloor x \rfloor + 88$ can only be zero where the function $x^2 - 19x + 88 = (x - 11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given area of square $= 9+6\sqrt{2}$ Without calculator show its length in form of $(\sqrt{ c}+\sqrt{ d})$ $\sqrt{9+6\sqrt{2}}$ to find length
But how do I express the above in the form of $\sqrt{c} + \sqrt{d}$.
| Notice, let the side of the square be $a$ then its area is given as $$a^2=9+6\sqrt 2$$
$$a^2=9+2(3)\sqrt 2$$ $$a^2=9+2(\sqrt 3\sqrt 3)\sqrt 2$$
$$a^2=9+2(3)\sqrt 2$$ $$a^2=9+2(\sqrt 3\sqrt 2)(\sqrt 3)$$
$$a^2=6+3+2\sqrt{6}\sqrt 3$$ $$a^2=(\sqrt 6)^2+(\sqrt 3)^2+2(\sqrt{6})(\sqrt 3)$$
Using $a^2+b^2+2ab=(a+b)^2$, we get... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there an easier way to solve this integral problem? I have the following integral:
$$F(t) = \int_0^b (\sqrt{2 + t} - 2) dt$$
I need to find a positive value of $b$ such that $F(t) = 0$.
Going through the integration steps and using the Fundamental Theorem of Calculus, I end up with
\begin{align}
0 & = (\frac{2}{3}\s... | If you change variable $b=x^2-2$, the equation becomes $$ \frac{2 x^3}{3}-2 x^2+4\big(1-\frac{ \sqrt{2}}{3}\big)=0$$ which can be solved with radicals using Cardano method.
But, there is one "obvious" root $x_1=\sqrt 2$ (remember that this corresponds to $b=0$ which is a trivial solution of the problem). So, what is l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Maximising cardinality of a set given the mean. Given a set of distinct, positive integers of mean $x$ and largest element $y$ how can we maximise the cardinality of the set?
| For a corresponding set of $n$ elements, there must be a partition of $nx-y$ into $n-1$ unique addends. If it exists, you've found the maximum cardinality - $n$. Finding the maximum value requires three cases:
*
*$y=x$, the cardinality is $1$ (set $\{x\}$)
*$y < 2x$, the cardinality is $\lfloor \frac{1}{2} (\sqrt{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A given polynomial equation of 5 degree has three equal roots . Given equation is $x^5-10a^3x^2+b^4x+c^5=0$ which has 3 equal roots. What I know is that since its a 5th degree equation therefore it must have 5 roots out (of which 3 are equal).
Aim is to establish the relationship between the constants $a,b$ and $c$.
Op... |
Question : I have to find which one is correct out of the two, any help?
Answer : Actually, both the options are correct. I am giving the explanation.
Let $~f(x) = x^5 – 10a^3x^2 + b^4x + c^5~$.
$$~⇒ f'(x) = 5x^4 – 20a^3x + b^4~~~~ ⇒ f'' (x) = 20x^3 – 20a^3~$$
Since $~x = \alpha~$ be a root that is repeated three ti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I transform, with long division, this polynomial into the answer given? I need to transform:
$ \frac {x^5}{(x-2)(x+2)(x^2+4)} $
into
$ \frac{-2x}{x^2+4} + x + \frac{1}{x-2} + \frac{1}{x+2} $
How can I solve it?
Thanks.
| First, you should notice the difference of squares formula (twice) in the denominator. $(x-2)(x+2)=x^2-4$ Then $(x^2-4)(x^2+4)=x^4-16$ so we have $\frac {x^5}{x^4-16}=x+\frac {16x}{x^4-16}$ The rest is a partial fraction decomposition of the second term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Without L'Hopital,
$$\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$$
That's
$$\frac{x^2+x\cdot \sin x}{-1+\left(1-2\sin^2\frac{x}{2}\right)} = \frac{x^2+x\cdot \sin x}{-2\sin^2\frac{x}{2}}$$
Let's split this
$$\frac{x\cdot x}{-2\cdot\sin\frac{x}{2}\cdot ... | Perhaps a simpler approach is as follows
\begin{align}
L &= \lim_{x \to 0}\frac{x^{2} + x\sin x}{-1 + \cos x}\notag\\
&= \lim_{x \to 0}\frac{x^{2} + x\sin x}{-1 + \cos x}\cdot\frac{1 + \cos x}{1 + \cos x}\notag\\
&= \lim_{x \to 0}\frac{x^{2} + x\sin x}{-1 + \cos^{2} x}\cdot(1 + \cos x)\notag\\
&= -2\lim_{x \to 0}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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What is the number of terms in this series $1+0.5+0.25+\ldots+0.5^{n−1}$? And what is the sum?
What is the number of terms in this series $1+0.5+0.25+\ldots+0.5^{n−1}$? And what is the sum?
I stumped by this question, but have a feeling the answer's really obvious:S
Am I supposed to get an actual number as the answer... | use the geometric series
$$\frac{1}{1-x}=1+x+x^2+x^3+.....x^n$$
$$\frac{1}{1-x}-1=x+x^2+x^3+.....x^n$$
$$\frac{1}{1-x}-1=x(1+x+x^2+x^3+.....x^{n-1})$$
$$\frac{x}{1-x}=x(1+x+x^2+x^3+.....x^{n-1})$$
hence
$$\frac{1}{1-x}=(1+x+x^2+x^3+.....x^{n-1})$$
$|x|<1$
plug $x=0.5$
$$\frac{1}{1-0.5}=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$ Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.
$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \f... | $$\sin\alpha + \sin\beta = 2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$
$$\cos\alpha + \cos\beta = 2\cos(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$
So, you get the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red... | Just to add to completeness of the other answers let me find minimal polynomial of $\alpha = \sqrt 2 + \sqrt 3$ over $\mathbb Q$ directly.
\begin{align}
\alpha = \sqrt 2 + \sqrt 3 &\implies \alpha - \sqrt 2 = \sqrt 3\\
&\implies \alpha^2 -2\alpha\sqrt 2 + 2 = 3\\
&\implies \alpha^2 - 1 = 2\alpha\sqrt 2\\
&\implies \alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Solving a first order IVP I want to solve the following initial value problem:
$$(1+x)dy + \sqrt{y}dx =0,\,\, y(0)=1~.$$
I notice that the equation is separable, hence,
$$ \int \frac{1}{\sqrt{y}} dy = - \int \frac{1}{1+x} dx $$
so that $$ 2\sqrt{y} = -\ln|1+x| +C $$
But the initial condition suggests that $(1+x) \gt... | You have to remember that the square root of a number $y$ is a non-negative number. Hence when you take $C=-2$, $-\frac{1}{2} \ln(1+x) + \frac{C}{2}$ is negative in the neighborhood of $x=0$.
Consequently the square root of $$y(x) = (-\frac{1}{2} \ln(1+x) -2)^2 $$ is not $\sqrt{y(x)} = -\frac{1}{2} \ln(1+x) -2$ (whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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obtain a three-term asymptotic solution In the limit as $\epsilon \to \infty$, obtain a three-term asymptotic solution to the roots of the following equation.
$$\epsilon x^3+x^2-2x+1=0$$
What I've done so far
I've assumed $x=O(\epsilon^r)$ as $\epsilon \to \infty$ therefore,
$\epsilon x^3=O(\epsilon^{1+3r})$
$x^2=O(\ep... | Outline (rather messy: is there a more elegant approach?):
*
*First, prove that you must have $x=x(\epsilon)\xrightarrow[\epsilon\to\infty]{} 0$.
*Then, you can get the expansion one term at a time...
Now, the second bullet (up to order two) below -- if I have not screwed up.
$$
1-2x + x^2+\epsilon x^3 = 1+\epsil... | {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2... | I am giving a hint.
$$\int \frac {dx}{1+x^4}=\frac 12\left(\int\frac {1+x^2}{1+x^4}dx-\int\frac {x^2-1}{1+x^4}dx\right)\\=\frac 12\left(\int\frac {1+\frac 1{x^2}}{x^2+\frac 1{x^2}}dx-\int\frac {1-\frac 1{x^2}}{x^2+\frac 1{x^2}}dx\right)$$
Now, write $x^2+\frac 1{x^2}=\left(x-\frac 1x\right)^2+2$ and $x^2+\frac 1{x^2}=\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels l... | After using C-S $\left(\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq(a+b+c)\sum\limits_{cyc}\frac{1}{b+c}$ we'll obtain something obvious:
$$\sum_{cyc}(3a^4c^2+3a^3b^3+a^4bc-2a^3b^2c-2a^3c^2b-3a^2b^2c^2)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding Pythagorean triplet given the hypotenuse I have a number $c$ which is an integer and can be even or odd. It is the hypotenuse of a right angled triangle. How can I find integers $a,b$ such that
$$ a^2 + b^2 = c^2 $$
What would be the complexity of the calculation?
| You can find one or more triples, given only $C$ if $GCD(A,B,C)=2$ or a perfect square, which includes all primitives and such as $(27,36,45)$. You cannot find those that are otherwise, such as $3,5,6,7,8,10$ times a primitive, e.g. $(45,24,51)=3*(15,8,17)$.
We solve $C=m^2+n^2$ for $n$, and look for positive integers... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of n such that l is non zero finite real number. Find l Problem :
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of $n$ such that $l$ is non zero finite real number. Find value of $l$.
My approach :
$$\lim_{x \to 0} \frac{x^n}{\cos(\s... | We have $cos x= \sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}x^{2j}$ hence \begin{align}
\frac{x^n}{\cos \sin x -\cos x}&= \frac{x^n}{\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}(\sin^{2j}x-x^{2j})}\\
&= \frac{x^n}{\sum_{j=1}^{\infty}\frac{(-1)^j}{(2j)!}(\sin^{2j}x-x^{2j})}\\
&=\frac{1}{\color{blue}{\frac{(-1)^1}{2!}}\frac{\sin^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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homogeneous linear systems, finding scalar I'm struggling with finding a solution to this problem.
For what values of b is the solution set of this linear system:
$x_{1} + x_{2} + bx_{3} = 0$
$x_{1} + bx_{2} + x_{3} = 0$
$bx_{1} + x_{2} + x_{3} = 0$
equal to the origin only, a line through the origin, a plane through t... | If we perform row reduction,
$
\begin{bmatrix}
1&1&b\\
1&b&1\\
b&1&1
\end{bmatrix}
$
$\xrightarrow[R3-R1]{R2-R1}$
$
\begin{bmatrix}
1&1&b\\
0&b-1&1-b\\
b-1&0&1-b
\end{bmatrix}
$
We see that if $b=1$, the second and third row are zero, and the system is reduced to the plane $x_1+x_2+x-3=0$.
If $b\ne1$, we can continue
... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding right inverse matrix Given a $3\times 4$ matrix $A$ such as
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
\end{pmatrix}
,$$
find a matrix $B_{4\times 3}$ such that
$$AB =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
... | Your matrix
$$A=\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}$$
has rank=$3$ and you can see that the square matrix $AA^T$ is invertible. Now note that $AA^T(AA^T)^{-1}=I$ so the matrix $B=A^T(AA^T)^{-1}$ is a right inverse of $A$ (but it is not the unique).
in this case we have:
$$
AA^T=\begin{bmatrix}4&2&2\... | {
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"question_score": "3",
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The rule for evaluating limits of rational functions by dividing the coefficients of highest powers I have a Limit problem as below:
Connor claims: " $\lim_{x\to \infty} \left(\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}\right) = 3$ because my high school calculus teacher told us the limit of ratio of polynomials is always t... | $$\lim_{x\to \infty} \left(\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}\right)=$$
The leading term in the denominator of $\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}$ is $x^3$. Divide the numerator and denominator by this:
$$\lim_{x\to \infty} \left(\frac{\frac{6}{x}+\frac{7}{x^2}+\frac{3}{x^3}}{2+\frac{1}{x}-\frac{2}{x^2}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Integral : $\int \frac{\sqrt{9-x^2}}{x^2}$ I have the following integral :
$$\int \frac{\sqrt{9-x^2}}{x^2} = -\frac{\sqrt{9-x^2}}{x} - \arccos (\frac{x}{3})$$
But it seems like I must use $\arcsin$... Is there a difference between the two ?
| First off the sign on the $\arccos(\frac{x}{3})$ is wrong.
I guess you could use the constant of integration to solve this. From the definitions of the inverse trig functions we know that
$\arccos x = \frac{\pi}{2} - \arcsin x$
So your antiderivative can be written as
\begin{equation}
-\frac{\sqrt{9 - x^2}}{x^2} + \ar... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding the locus in terms of a and b Find the locus of the point equidistant from $(a+b,b-a)$ and $(a-b,a+b)$
Okay so I don't know how do I find the locus of the point in terms of $ a$ & $b $.
| Let $(x,y)$ be a point on the locus. Then:
$$(x-(a+b))^2+(y-(b-a))^2=(x-(a-b))^2+(y-(a+b))^2$$
Expanding:
$$x^2-2x(a+b)+(a+b)^2+y^2-2y(b-a)+(b-a)^2=x^2-2x(a-b)+(a-b)^2+y^2-2y(a+b)+(a+b)^2$$
$$-2y(b-a)+2y(a+b)=2x(a+b)-2x(a-b)$$
$$4ya=4xb$$
$$y=\frac{b}{a}x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Velocity of a train A train travels with a velocity that doubles every hour. In 3 hours, it has travelled 350km. What was the trains velocity in the second hour?
A. 225 km/h
B. 175 km/h
C. 150 km/h
D. 100 km/h
Answer is D. Can someone explain?
| Assuming that the intent is to solve this using linear algebra, here are the steps
$$ x =\mbox{distance traveled in the first hour}$$
$$ y =\mbox{distance traveled in the second hour}$$
$$ z =\mbox{distance traveled in the third hour}$$
This problem is expressed mathematically as
$$x+y+z=350$$
$$2x=y$$
$$2y=z$$
Which c... | {
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"source": "stackexchange",
"question_score": "1",
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How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$ $$\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$$
I have simplified above to:
$$\frac{3-\sqrt[3]{3}}{(\sqrt[3]{3}+1)(\sqrt[3]{3}-1)}$$
What is equal to:
$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}$$
WolframAlpha says this can be shown as... | Let $t=\sqrt[3]3$. Then, we have $t^3=3$, so
$$\frac{1}{t-1}-\frac{2}{t+1}=\frac{3-t}{t^2-1}=\frac{t(3-t)}{t(t^2-1)}=\frac{t(3-t)}{3-t}=t.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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geometric inequality involving lengths of hypotenuses In the diagram, $AD = AB + AC$ and all these lengths are positive. The following inequality holds:
$$DG < 2(BE + CF).$$
The constant 2 is sharp. I am able to prove the weaker inequality $DG < BE + 3 CF$ where $CF \ge BE$ but I am not able to prove the stronger inequ... | Using $a=AB$, $b=AC$, and $a+b=AD$, this is equivalent to the following
Lemma: For $a,b\gt0$,
$$
\sqrt{(a+b)^2+1}-1\lt2\left(\sqrt{a^2+1}-1+\sqrt{b^2+1}-1\right)
$$
Proof: Since $2ab\le a^2+b^2$, we have $(a+b)^2\le2\left(a^2+b^2\right)$. Therefore,
$$
\frac{(a+b)^2}{\sqrt{(a+b)^2+1}}\lt2\left(\frac{a^2}{\sqrt{a^2+1}}+... | {
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What is the maximum value of $ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}<x<\frac{2\pi}{3})$$
I have done my work... | $$f(x) = \frac{1}{3}\cdot\frac{\sin(3x)}{\sin\left(x+\frac{\pi}{3}\right)} $$
is a negative function on the interval $\left(\frac{\pi}{3},\frac{2\pi}{3}\right)$, while $f\left(\frac{\pi}{3}\right)=0$.
| {
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"source": "stackexchange",
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What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
| $$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$$
$$\frac{(x+1)}{(x-1)(x+2)}-\frac{(x)}{(x-1)(x-3)}>0$$
$$\frac{1}{(x-1)}\left(\frac{(x+1)}{(x+2)}-\frac{(x)}{(x-3)}\right)>0$$
$$\frac{1}{(x-1)}\left(\frac{(x+1)(x-3)-x(x+2)}{(x+2)(x-3)}\right)>0$$
$$\frac{x^2-2x-3-x^2-2x}{(x-1)(x+2)(x-3)}>0$$
$$\frac{4x+3}{(x-1)(x+2)(... | {
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Solving line integral without using Green’s theorem Good morning everyone, I am a new user here and I have the following problem that I am dealing with since yesterday but I cannot find the correct answer. Here is the problem
Calculate
$$\oint_C (x^2-2xy)dx+(x^2y+3)dy$$
where $C$ is the closed curve of the region bound... | Use parametrisation. That is the trick to do line integral problems.
Take $y=t$ and $x=\frac{t^2}{8}$ in Path- I and $x=2,dx=0$ in Path-II
So your line integral $$\oint=\oint_{I}+\oint_{II}$$
Now $$\oint_1=\int_{4}^{-4}[(\frac{t^4}{64}-\frac{t^3}{4})\cdot \frac{t}{4}dt+(\frac{t^5}{64}+3)dt]$$
$$=\int_{4}^{-4}[(\fra... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Definite Integrals with Substitution. How would you use the substitution: $x=1+sin\theta$ to evaluate $\int_{0}^{π/2}\frac{cos\theta}{(1+sin\theta)^3}d\theta$.
Furthermore what would you when changing the limits, since its a definite integral.
| HINT:
$$\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos(x)}{(1+\sin(x))^3}\right)\text{d}x=$$
Substitute $u=\sin(x)+1$ and $\text{d}u=\cos(x)\text{d}x$ and
$
\begin{cases}
x=0 \Longrightarrow u=\sin(0)+1=1\\\
x=\frac{\pi}{2} \Longrightarrow u=\sin\left(\frac{\pi}{2}\right)+1=2 \\
\end{cases}
$:
$$\int_{1}^{2}\left(\frac{1}... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solving linear system equation How would I solve this linear system equation?
$$\begin{cases}
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7
\end{cases}$$
First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.
| $$2w+x-y=4$$
$$3z-x=6$$
$$-2y-x+9z+4w=7$$
Can be rewritten to
$$2w+x-y+0z=4$$
$$0w-x+0y+3z=6$$
$$4w-x-2y+9z=7$$
$$0w+0x+0y+0z=0$$
So now we have
$$\left[\begin{array}{cccc|c}
2 & 1 & -1 & 0 & 4 \\
0 & -1 & 0 & 3 & 6 \\
4 & -1 & -2 & 9 & 7 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{cccc|c}
2 & 1... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $P(1)/P(-1)$
The polynomial $f(x)=x^{2007}+17x^{2006}+1$ has distinct zeroes $r_1,\ldots,r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j+\dfrac{1}{r_j}\right)=0$ for $j=1,\ldots,2007$. Determine the value of $P(1)/P(-1)$.
Let $P(x) = a_{2007}x^{2007} + ... + a_0$.
$r_j + \frac... | For first, $\frac{1}{r_i}+r_i = \frac{1}{r_j}+r_j$ is equivalent to $(r_i-r_j)(r_i r_j - 1)=0$. Since $f(x)$ is not a palyndromic polynomial, the set of numbers $r_i+\frac{1}{r_i}$ is exactly the set of roots of $P$.
Assuming that $P$ is a monic polynomial, Vieta's formulas give:
$$ P(1) = \prod_{i=1}^{2007}\left(1-r_... | {
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If $\gcd(a,b,c)=1$ and $a^2 + b^2 = c^2$, prove $12|abc$ I know from $a^2+b^2=c^2$ that $a = st$, b = $\frac{s^2-t^2}{2}$, c = $\frac{s^2+t^2}{2}$ , where $s>t>1$, where $s$ and $t$ are both odd, but how do I make use of this information?
| $s,t$ odd means $s^2\equiv t^2\equiv 1\mod 8$, hence $b$ is divisible by $4$. If neither $s$ nor $t$ is divisible by $3$, then $s^2\equiv t^2\equiv 1\mod 3$ and $3|b$. Otherwise $3|a$, in both cases $12|abc$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plane $x+y+z=1$
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plain $x+y+z=1$
Hi i was doing this question but i'm not sure i was right. Does this make sens... | By inequality $2x^2+2y^2\geq (x+y)^2$ we indeed have $-\sqrt2 \leq x+y \leq \sqrt{2}$ and to maximize/minimize $z$ we need to minimize/maximize $x+y$ so your solution until $x=y=\pm{1\over\sqrt2}$ is correct.
However how do you get $z$ looks unreasonable as $x+y+z=1$ but your solution, for example ${1\over\sqrt2}+{1\o... | {
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"source": "stackexchange",
"question_score": "3",
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Can all differential equations be solved using power series? To elaborate, does the differential equation have to be of some form to be solvable by power series. More specifically, if I wanted to solve this equation by power series would I be able to?
$$
(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)=0
$$
Also, $n$ is a constant in... | $(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)
=0
$
I'll try
$y(x)
=\sum_{k=0}^{\infty} a_k x^k
=a_0+a_1x+\sum_{k=2}^{\infty} a_k x^k
$.
$y'(x)
=\sum_{k=1}^{\infty} ka_k x^{k-1}
=\sum_{k=0}^{\infty} (k+1)a_{k+1} x^{k}
$
so
$xy'(x)
=\sum_{k=0}^{\infty} (k+1)a_{k+1} x^{k+1}
=a_1x+\sum_{k=2}^{\infty} ka_{k} x^{k}
$.
$y''(x)
=\sum_{k=2... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Why two Inequalities are true $A$ is a subset of real numbers.
Consider the set $A$ of all real
numbers $x$ such that $x^2 \le 2$.
This set is nonempty and bounded from above, for example by 2. Call the $\sup(A),\; y$. Then $y^2 = 2$, because the other possibilities
$y^2 \lt 2$ and $y^2 \gt 2$ both lead to a c... | The expressions (1) and (2) might be easier to follow if they are expanded to
$$\begin{align}
(y-z)^2&=y^2-2yz+z^2\\
&=y^2-4z+4z-2zy+z^2\\
&=y^2-4z+z(z+2(2-y))\\
&\ge y^2-4z\\
&\gt2
\end{align}$$
and
$$\begin{align}
(y+c)^2&=y^2+2cy+c^2\\
&\lt y^2+2cy+c\\
&=y^2+5c-5c+2cy+c\\
&=y^2+5c-2c(2-y)\\
&\le y^2+5c\\
&\lt2
\end{... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a+b}{\sqrt{c}}+\frac{a+c}{\sqrt{b}}+ \frac{b+c}{\sqrt{a}} \geq 2(\sqrt{a} + \sqrt{b} +\sqrt{c})$ Could anyone advise me on how to prove this inequality:
$$\dfrac{a+b}{\sqrt{c}}+\dfrac{a+c}{\sqrt{b}}+ \dfrac{b+c}{\sqrt{a}} \geq 2(\sqrt{a} + \sqrt{b} +\sqrt{c}),$$
where $a,b,c $ are any positive real number... | From AM-GM you have
$$ \frac{a}{\sqrt{b}} +\frac{a}{\sqrt{b}} +\frac{b}{\sqrt{a}} \geq 3 \sqrt{a} $$
and
$$ \frac{a}{\sqrt{c}} +\frac{a}{\sqrt{c}} +\frac{c}{\sqrt{a}} \geq 3 \sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3\left ( \dfrac{a+b}{\sqrt{c}}+\dfrac{a+c}{\sqrt... | {
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Show $\sqrt{-n}$ is irreducible in $R = \mathbb{Z}[\sqrt{-n}]$
Let $R = \mathbb{Z}[\sqrt{-n}]$ where $n$ is a square free integer greater or equal than $3$. Prove that $2$ and $\sqrt{-n}$ are irreducible in $R$.
Proof. Recall that if $R$ is an integral domain, and $r \in R$ and is not a unit, then $r$ is called irred... | If $b\neq 0$, then $a^2+nb^2 \geq n$. The equality occurs only when $a=0$ and $b=\pm 1$. In which case the factorization you have is the trivial one $\sqrt{-n}=\pm\sqrt{-n}(c+d\sqrt{-n})$. Thus $c=\pm 1,d=0$.
If $b=0$, then $n=(ac)^2+(ad)^2n$, using the same reasoning as above, we can conclude that $ac=0$ and $ad=1$. C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the maximum and minimum values of $|\vec{PA}||\vec{PB}|$ are $M$ and $m$ respectively then prove that the value of $M^2+m^2=34$ Given three points on the $xy$ plane on $O(0,0),A(1,0)$ and $B(-1,0)$.Point $P$ is moving on the plane satisfying the condition $(\vec{PA}.\vec{PB})+3(\vec{OA}.\vec{OB})=0$
If the maximum a... | You have almost Got it. Here $x^2+y^2 = 4\;,$ Then Parametric Coordinats are
$x=2\cos \theta\;\;,y=2\sin \theta$ and Here $f(x) = \sqrt{25-4x^2}.$
So we get $f(\theta) = \sqrt{25-16\cos^2 \theta}\;,$ Now Using $0\leq \cos^2 \theta \leq 1$
So $\displaystyle \min\left[f(\theta)\right] = \sqrt{25-16} = 3$ at $\displaysty... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
| I'm not a math major so I'm not sure about the formalism but does this work?
If a!=b,
Then a=b+d ; where d is a non zero integer.
Plug in to the expression (a^3+a)-(b^3+b) :-
3bd^2+3db^2+d^3.
This has only one real root which is d=0, which is a contradiction, so the two expressions cannot be equal.
Too informal? Corre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating m... | Assume $x>0.$ Let $u=\sqrt{x},$ then $u^{2}=x,$ $dx=2udu,$ and $1+\frac{1}{4x%
}=1+\frac{1}{4u^{2}}$
\begin{equation*}
\int \sqrt{1+\frac{1}{4x}}dx=\int \sqrt{1+\frac{1}{4u^{2}}}(2u)du=\int \sqrt{%
(2u)^{2}+1}du.
\end{equation*}
Now, substitution trigonometric, $2u=\tan \theta ,$ $du=\frac{1}{2}\sec
^{2}\theta d\theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Can someone explain the integration of $\sqrt{v²+\tfrac14}$ to me? I am currently trying to integrate this root:
$$\sqrt{v^2+\frac{1}{4}}$$
According to several integration calculators on the web it is:
$$\frac{\operatorname{arsinh}(2v)}{8} +\frac{v\sqrt{v^2+\tfrac{1}{4}}}{2}$$
However, I just can't get my head around ... | People don't seem comfortable going directly to the hyperbolic trig functions, but that is simplest here.
$$ v = \frac{1}{2} \sinh t $$
$$ v^2 + \frac{1}{4} = \frac{1}{4} \cosh^2 t $$
$$ \sqrt{v^2 + \frac{1}{4}} = \frac{1}{2} \cosh t $$
$$ dv = \frac{1}{2} \cosh t \; dt $$
$$ \color{blue}{ \sqrt{v^2 + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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