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 minimized. Find r.
My attempt so far
The total length of the wire is $y$ the length used for the square is $x$ therefore the perimeter of the square is $x$ the area of the square is then $\dfrac{x^2}{16}$, the length for the circle is $y-x$ and the circumference is therefore $y-x$
$$R=\dfrac{y-x}{2\pi}$$
Area of the circle is then $$\dfrac{(y-x)^2}{4\pi}$$
I need to minimize $$\dfrac{x^2}{16} + \dfrac{(y-x)^2}{4\pi}$$
In order to solve for $$r=\dfrac{y-x}{x}$$
Now I'm a little stuck. Thanks for any help.
|
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: $L^2 = (4x+2\sqrt{\pi}\sqrt{\pi}y)^2\leq (4^2+4\pi)(x^2+\pi y^2)\Rightarrow S \geq \dfrac{L^2}{16+4\pi}$. Equality occurs when $\dfrac{4}{x}=\dfrac{2\sqrt{\pi}}{\sqrt{\pi} y}=\dfrac{2}{y}\to x = 2y\to r = \dfrac{2\pi y}{4x}=\dfrac{\pi}{4}$
|
{
"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 it didn't work out
|
Notice, $$csc^2\theta=5\cot \theta+7$$
$$\cot^2\theta+1=5\cot \theta+7$$ $$\cot^2\theta-5\cot \theta-6=0$$
$$(\cot\theta -6)(\cot\theta+1)=0$$
$$\implies \cot\theta-6=0 \iff \tan \theta=\frac{1}{6}\iff \color{blue}{\theta=n\pi+\tan^{-1}\left(\frac{1}{6}\right)}$$
$$\implies \cot\theta+1=0 \iff \tan \theta=-1=-\tan\frac{\pi}{4} \iff \color{blue}{\theta=n\pi-\frac{\pi}{4}}$$
Where, $\color{blue}{n}$ is any integer
|
{
"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 &= n^2 - n + 1\\
B &= n^2 + n + 1 = A+2n\\
C &= n^2 + n + 3 = A+2n+2\\
A+B+C &= 3n^2+n+5
\end{align}
where A is the first element in row $n$.
For each shift of A to the right in the same row, A,B, and C increase by 2
and A+B+C increases by $6$.
Solving $3n^2+n+5 = 2149$ for n, we get $n = 26.567$ So we are in row $26$.
For the first element in row $26,
A = 651,\; B = 703,\; C = 705,$ and $A+B+C = 2059$.
Since $\dfrac{2149-2059}6 = 15$, we need to increase $A, B, $ and $C$ by $2\cdot15 = 30$.
So
\begin{align}
A &= 681\\
B &= 733\\
C &= 735
\end{align}
|
{
"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{3})\\
\theta&=& -\frac{5\pi}{3} + 5\pi n
\end{eqnarray}
|
$$\tan\frac\theta5=-\sqrt 3\iff\tan\frac\theta5=\tan\Bigl(-\frac\pi3\Bigr)\iff\frac\theta5\equiv -\frac\pi3\mod\pi\iff\theta\equiv -\frac{5\pi}3\mod 5\pi.$$
|
{
"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 degrees.
I tried to use this site for geometric sequences, How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?.
|
\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 +
\text{cis }8^\circ +
(\text{cis }8^\circ)^2 + \dots +
(\text{cis }8^\circ)^9 \right)\\
&= \text{cis }75^\circ
\dfrac{(\text{cis }8^\circ)^{10}-1}{\text{cis }8^\circ-1}\\
&= \text{cis }75^\circ
\dfrac{\text{cis }80^\circ-1}{\text{cis }8^\circ-1}\\
\end{align}
|
{
"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
$$f(-1)=\alpha(-1-\beta)(-1-1)(-1+2)=4$$ $$\implies \alpha(1+\beta)=\frac{4}{2}=2$$ $$\alpha+\alpha\beta=2\tag 1$$
$$f(2)=\alpha(2-\beta)(2-1)(2+2)=8$$ $$\implies \alpha(1+\beta)=\frac{8}{4}=2$$
$$2\alpha-\alpha\beta=2\tag 2$$ Now, adding (1) & (2), we get
$$\alpha+2\alpha=4\implies \alpha=\frac{4}{3}$$ Substituting the value of $\alpha$ in (1), we get $$\beta=\frac{2}{\frac{4}{3}}-1=\frac{1}{2}$$ Now, substitute these values in cubic polynomial, we get $$f(x)=\frac{4}{3}\left(x-\frac{1}{2}\right)(x-1)(x+2)$$ $$=\frac{2}{3}\left(2x-1\right)(x-1)(x+2)$$ $$=\frac{2}{3}(2x^3+x^2-5x+2)$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{f(x)=\frac{2}{3}(2x^3+x^2-5x+2)}}$$
|
{
"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 conclude $f(x) > 0, \forall x > \pi$.
|
{
"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 $\tan^{-1}\left(\frac{3-\tan B\tan C}{\tan B-\tan C}\right)$
I let the foot of perpendicular from A,B,C to opposite sides is D,E,F.Then
$$\tan B=\frac{AD}{BD},\tan C=\frac{AD}{CD}$$
$$\frac{3-\tan B\tan C}{\tan B-\tan C}=\frac{3-\frac{AD}{BD}\frac{AD}{CD}}{\frac{AD}{BD}-\frac{AD}{CD}}$$
I think this way i cannot get answer.Please help me getting the desired proof.I am stuck ...
|
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 $$H\equiv \left(\frac{0+a+\frac{a\tan C}{\tan B+\tan C}}{3}, \frac{0+0+\frac{a\tan B\tan C}{\tan B+\tan C}}{3} \right)\equiv \left(\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}, \frac{a\tan B\tan C}{3(\tan B+\tan C)} \right)$$ & the circumscribed center say $D$ can be calculated as $$D\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$
Hence, the slope of the line HD joining $H$ & $D$ with the BC (x-axis) is given as $$m=\frac{y_2-y_1}{x_2-x_1}$$ $$=\frac{\frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)}-\frac{a\tan B\tan C}{3(\tan B+\tan C)}}{\frac{a}{2}-\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}}$$ $$=\frac{3\tan B\tan C-3-2\tan B\tan C}{3\tan B+3\tan C-2\tan B-4\tan C}$$ $$=\frac{\tan B\tan C-3}{\tan B-\tan C}$$ Hence the angle of the line joining H & D with side BC is given as $$\tan \theta=|m|$$ $$\implies \theta=\tan^{-1}\left|\frac{\tan B\tan C-3}{\tan B-\tan C}\right|$$
or $$\bbox [5px, border:2px solid #C0A000]{\color{red}{\theta=\tan^{-1}\left|\frac{3-\tan B\tan C}{\tan B-\tan C}\right|}}$$
|
{
"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 $\cos 3x = \cos^3 x -3\cos x \sin^2 x$,
$$\begin{align*}
\cos 3x &= \cos^2 x - 3 \sin ^2 x\\
\cos^3 x -3\cos x \sin^2 x &= \cos^2 x - 3 \sin ^2 x\\
(\cos x)(\cos^2 x - 3 \sin ^2 x) &= \cos^2 x - 3 \sin ^2 x\\
\cos x &= 1&\text{ or }&& \cos^2 x - 3 \sin ^2 x &= 0\\
\cos x &= 1&\text{ or }&&\tan^2 x&=\frac13
\end{align*}$$
My formula of $\cos 3x$ may look non-standard, but that was because I did not remember it and instead got it by quickly expanding
$$\cos 3x + i\sin 3x = (\cos x + i\sin x)^3$$
|
{
"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:
$$\sum_{cyc}\frac{1}{a^3}=4(4^2-2\cdot(-4))+4(4)-24 = \color{red}{88}.$$
With a bit of experience, one may recognize $p(x)$ as the minimal polynomial of $\alpha=-\cos\frac{2\pi}{7}$, whose conjugates are $-\cos\frac{4\pi}{7}$ and $-\cos\frac{6\pi}{7}$. By this way, the problem is equivalent to proving a not-so-difficult trigonometric identity.
Another neat trick is the following: given $p(x)=\left(1-\frac{x}{a}\right)\left(1-\frac{x}{b}\right)\left(1-\frac{x}{c}\right)$,
$$ \log p(x) = \sum_{cyc}\log\left(1-\frac{x}{a}\right) = -\sum_{cyc}\left(\frac{x}{a}+\frac{x^2}{2a^2}+\frac{x^3}{3a^3}+\ldots\right) $$
hence $\sum_{cyc}\frac{1}{a^3}$ is minus three times the coefficient of $x^3$ in the Taylor series of $\log p(x)$ in a neighbourhood of $x=0$, or:
$$ \sum_{cyc}\frac{1}{a^3} = -\frac{1}{2}\left.\frac{d^3}{dx^3}\log p(x)\right|_{x=0}, $$
and we just need to evaluate:
$$\frac{\left(4+8 x-24 x^2\right)^3}{\left(1-4 x-4 x^2+8 x^3\right)^3}+\frac{3 (-4+24 x) \left(-4-8 x+24 x^2\right)}{\left(1-4 x-4 x^2+8 x^3\right)^2}-\frac{24}{1-4 x-4 x^2+8 x^3} $$
at $x=0$ to get $4^3+3\cdot 4^2-24=\color{red}{88}$ just as before.
|
{
"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 example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.
|
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
$\displaystyle (N+\sqrt{3})^2=4+2\sqrt{3}$
$\implies N^2+2\sqrt{3}\cdot N + 3=4+2\sqrt{3}$
$\implies (N^2-1)=2\sqrt{3}\cdot(1-N)$
$\implies N=-2\sqrt{3}-1$
or
$N=1$
But since $N>0$,
$\implies N=\boxed1$
|
{
"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
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
100\binom{100}{100}
$$
What I have tried:
$$\begin{align}
S &= 0\binom{100}{0}+1\binom{100}{1}+ \dotsb +99\binom{100}{99}+100\binom{100}{100} \\ \\
&=100\binom{100}{100}+99\binom{100}{99}+ \dotsb +1\binom{100}{1}+0\binom{100}{0}
\end{align}$$
and I'm stuck here, I don't know if it's true or not, any help will be appreciated.
|
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 binomial coefficients is well known - consider $(1+1)^{99}$
|
{
"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, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Rightarrow -2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)-\sin 3x=0$$
Unfortunately, I couldn't get further with this problem. Any help with this would be truly appreciated. Many thanks in advance!
|
\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) \\
& \sin\left(\frac{3x}{2}\right) \, \left(-2 \sin\left(\frac{x}{2}\right) - 2 \, \cos\left(\frac{3x}{2}\right) \right) = 0 \\
& \sin\left(\frac{3x}{2}\right) = 0 \quad \text{or} \quad \sin\left(\frac{x}{2}\right) = \cos\left(\frac{3x}{2}\right)
\end{align}
|
{
"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
$$
\lim _{ n\longrightarrow +\infty }{ \frac { \sqrt [ n ]{ n! } }{ n } } \quad =\quad \lim _{ n\longrightarrow +\infty }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (n-k) } } }{ n } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ n(1-\frac { k }{ n } ) } } }{ n } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty }{ \frac { \sqrt [ n ]{ { n }^{ n }\prod _{ k=0 }^{ n-1 }{ (1-\frac { k }{ n } ) } } }{ n } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty }{ \frac { n\sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (1-\frac { k }{ n } ) } } }{ n } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty }{ { \left( \prod _{ k=0 }^{ n-1 }{ \left( 1-\frac { k }{ n } \right) } \right) }^{ \frac { 1 }{ n } } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty }{ { e }^{ \frac { 1 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n } \right) } } } }
$$
Now we will evaluate that infinite sum:
$$
\underset { n\longrightarrow +\infty }{ lim } \frac { 1 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n } \right) } } \quad =\underset { n\longrightarrow +\infty }{ lim } \quad \frac { 1-0 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n } \right) } } \\ \qquad \qquad \qquad \qquad \qquad =\quad \int _{ 0 }^{ 1 }{ ln(1-x)\quad dx } \\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ x\ln { (1-x) } \right] }_{ 0 }^{ 1 }+\int _{ 0 }^{ 1 }{ \frac { x }{ 1-x } dx } \\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ x\ln { (1-x) } \right] }_{ 0 }^{ 1 }+{ \left[ -\ln { (1-x)-x } \right] }_{ 0 }^{ 1 }\\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ (x-1)\ln { (1-x)-x } \right] }_{ 0 }^{ 1 }\quad =\quad -1
$$
So your limit is: ${e}^{-1}$
|
{
"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 are no integer $n$ and Pythagorean
triple $(a, b, c)$ satisfying $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n$$
The foregoing question is from the 2005 Canada National Olympiad.
I need some help on part (b).
Part (a)
Examine the behaviour of the following function [that embodies the constraint $c^2=a^2+b^2$, $x$ can play the role of either $a$ or $b$] on $(0,c)$:
$$f(x) = \dfrac{c}{x} + \dfrac{c}{\sqrt{c^2-x^2}} \tag{1}$$
Now find the derivative: $$f'(x)=cx\left[\dfrac{1}{(c^2-x^2)^{3/2}}-\dfrac{1}{x^3}\right] \tag{2}$$
To find local extrema:
$$\begin{align}
f'(x)=0 &\implies x^3=(c^2-x^2)^{3/2} \\
&\implies x^6=(c^2-x^2)^3 &(\text{squaring}) \\
&\implies u^3=(k-u)^3 &(u=x^2,k=c^2) \\
&\implies 2u^3-3ku^2+3k^2u-k^3=0 \\
&\implies (2u-k)(u^2+-ku+k^2)=0 \tag{3}\\
\end{align}$$
So, solutions are: $$u=\dfrac{k}{2};u=\dfrac{k\pm\sqrt{k^2-4k^2}}{2}\notin\mathbb{R}$$
So take the only real solution as $$u=\dfrac{k}{2} \implies x^2=\dfrac{c^2}{2} \implies x=\dfrac{c}{\sqrt2}\quad(\text{since }x\in(0,c))$$
This is a local minima because $f$ is continuous on $(0,c)$ and $$\lim_{x\to0^+}{f(x)}=\lim_{x\to c^-}{f(x)}=+\infty$$
The minimum value is $$f\left(\dfrac{c}{\sqrt2}\right)=\sqrt2+\sqrt2=2\sqrt2 \tag{4}$$
This value is not achievable, because $\sqrt{2}$ is not a rational number. Hence,
$$[f(a)]^2 = \left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > (2\sqrt2)^2 = 8 \tag{5}$$
Part (b)
$$\begin{align}
\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n \implies c^2(a^2+b^2)=a^2b^2\cdot n \tag{6}
\end{align}$$
in a perhaps friendlier form.
I am not sure how to use this further.
|
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 $\frac{c}{a}+\frac{c}{b}=\frac{ca+cb}{ab}$. If this were an integer, we would have (in particular) $a\mid ca+cb$, so $a\mid cb$. But $a,b,c$ share no prime factor, so we would need to have $a=1$. But $1$ is not element of any Pythagorean triple.
|
{
"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
}
|
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 the circle that is intersected by the tangent/s.
For example,
Find the equation of the tangent/s of circle a with center $(5, 5)$ and a radius $3$, and intercepting the point $(15, 10)$.
Any non-calculus based methods would be greatly appreciated!
|
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 $$\frac{|m(5)-5+c|}{\sqrt{m^2+(-1)^2}}=3$$
$$\frac{|5m-5+10-15m|}{\sqrt{m^2+1}}=3$$
$$|5-10m|=3\sqrt{m^2+1}$$
$$25+100m^2-100m=9m^2+9$$ $$91m^2-100m+16=0$$
On solving above quadratic equation for $m$, we get
$$m=\frac{50\pm 6\sqrt{29}}{91}$$ By substituting these values in the eq(1), we get $$m=\frac{50+6\sqrt{29}}{91}\implies c=10-15\left(\frac{50+6\sqrt{29}}{91}\right) =\frac{160- 90\sqrt{29}}{91}$$
$$m=\frac{50-6\sqrt{29}}{91}\implies c=10-15\left(\frac{50-6\sqrt{29}}{91}\right) =\frac{160+ 90\sqrt{29}}{91}$$
Hence, we get two tangent lines from the external point $(15, 10)$ to the given circle as follows $$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\left(\frac{50+6\sqrt{29}}{91}\right)x+\frac{160- 90\sqrt{29}}{91}}}$$ &
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\left(\frac{50-6\sqrt{29}}{91}\right)x+\frac{160+90\sqrt{29}}{91}}}$$
|
{
"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
}
|
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 $(1 - (-1 + \sqrt{-3})x)$ and $(1 - (-1 - \sqrt{-3})x)$
Course Notes (pg. 30)
Are these two expressions equivalent, or am I simply not factoring correctly? Thanks.
|
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 $$\displaystyle \left(2x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}i}{2}\right)^2 = \left(2x+\frac{1}{2}+\frac{\sqrt{3}i}{2}\right)\cdot \left(2x+\frac{1}{2}-\frac{\sqrt{3}i}{2}\right)$$
|
{
"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 not get answer.Please help..
|
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
$$f'(x)=\frac{1}{1+x^2}+\frac12\left(\frac{1}{1+x}+\frac{1}{1-x}\right) \tag 2$$
Next, we can avoid a series of integration by parts steps by making use of the following $3$ identities for $x^3$.
$$\begin{align}
x^3&=x(x^2+1)-x \tag 3\\\\
x^3&=(x^2-x+1)(1+x)-1 \tag 4\\\\
x^3&=-(x^2+x+1)(1-x)+ 1\tag 5
\end{align}$$
Then, using $(2)-(5)$ in $(1)$ yields
$$\begin{align}
F(x)&=2\int \frac{x(x^2+1)-x}{1+x^2} \,dx \\\\
&+\int \frac{(x^2-x+1)(1+x)-1}{1+x} \,dx \\\\
&+\int \frac{-(x^2+x+1)(1-x)+ 1}{1-x}\,dx\\\\
&=x^2-\log(1+x^2)\\\\
&+\frac13x^3-\frac12x^2+x-\log|1+x|\\\\
&-\frac13x^3-\frac12x^2-x-\log|1-x|\\\\
&-\log|1-x^4|+C
\end{align}$$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{\int \frac12 f'(x)d(x^4)=-\log|1-x^4|+C}$$
|
{
"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
}
|
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 roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?
|
$$
\begin{align}
\sum_{i=1}^6 \dfrac {x^i} {i!} &=\dfrac 1 {720} \cdot (x^6+6x^5+30x^4+120x^3+360x^2+720x+720= \\
&=\dfrac 1 {720} \cdot \{x^4(x+3)^2+20x^2(x+3)^2+x^4+180x^2+720x+720\}
\end{align}
$$
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 $C$.
|
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 increasing in Geometric Progression.
$$\therefore A < C < B $$
|
{
"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
}
|
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 to divide by the coefficient of $x^2$, I would get $1$ and $-1/3$, which are the correct answers. So, my question is, can all quadratics be solved in this way or was this just a fluke? If they can be, then why? If not, then is there any other quick and easy way of solving quadratics mentally? thanks for your help
|
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 factors, thus
we can factorize the expression as follows $$-3x^2+2x+1=-3(x-1)\left(x+\frac{1}{3}\right)$$
Hence, we have $$-3(x-1)\left(x+\frac{1}{3}\right)=0$$ or
$$(x-1)\left(3x+1\right)=0$$
Alternative Method Factorization can also be done by rearranging the terms as follows $$-3x^2+2x+1=0$$ $$-(3x^2-2x-1)=0$$
$$-(3x^2-3x+x-1)=0$$ $$-(\underbrace {3x^2-3x}+\underbrace{x-1})=0$$
$$-(3x(x-1)+(x-1))=0$$ $$(x-1)(3x+1)=0$$
|
{
"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:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$
(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
|
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 by letting the index $n$ run from $1$ to $99$ and replacing $\sqrt{100-n}$ with $\sqrt{n}$ since both cases will yield the same summands. Hence
$$\sum_{n=1}^{99} \Delta_n = \sqrt{2} \sum_{n=1}^{99}\sqrt{10 - \sqrt{n}} = \sqrt{2} \cdot b$$
thus
$$\frac{a-b}{b} = \frac{a}{b} - 1 = \sqrt{2} \implies \frac{a}{b} = \sqrt{2} + 1$$
|
{
"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 an even function, i.e. $p(x)=p(-x)$
*$q(x)q(-x) = 1$
then
\begin{align*}
\int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\frac{1}{2}\int_{-a}^{a}p(x)\,dx
\end{align*}
A reasoning is given below. But first let's check our example.
The current integral is symmetric around $x=0$
\begin{align*}
\int_{-3}^{3}\frac{x^8}{1+e^{2x}}
\end{align*}
with an even function $p(x)=x^8$ and a function $q(x)=e^{-2x}$ with
\begin{align*}
q(x)q(-x)=e^{-2x}e^{2x}=1
\end{align*}
Now recall that each function $f(x)$ can be uniquely written as sum of an odd function $f_o(x)$ and of an even function $f_e(x)$, since
\begin{align*}
f(x)&=f_o(x)+f_e(x)\\
f_o(x)&=\frac{1}{2}\left(f(x)-f(-x)\right)\\
f_e(x)&=\frac{1}{2}\left(f(x)+f(-x)\right)
\end{align*}
Let's define
\begin{align*}
f(x)=\frac{p(x)}{1+q(x)}
\end{align*}
with $f_e(x)$ it's even and $f_o(x)$ it's odd part.
Since the integral is symmetric around $x=0$ the odd part vanishes and we obtain
\begin{align*}
\int_{-a}^{a}f(x)\,dx&=\int_{-a}^{a}f_e(x)\,dx\\
&=\frac{1}{2}\int_{-a}^{a}\left(f(x)+f(-x)\right)\,dx\\
&=\frac{1}{2}\int_{-a}^{a}\left(\frac{p(x)}{1+q(x)}+\frac{p(-x)}{1+q(-x)}\right)\,dx\\
&=\frac{1}{2}\int_{-a}^{a}p(x)\frac{1+q(-x)+1+q(x)}{1+q(x)+q(-x)+q(x)q(-x)}\tag{1}\,dx\\
&=\frac{1}{2}\int_{-a}^{a}p(x)\frac{2+q(x)+q(-x)}{2+q(x)+q(-x)}\tag{2}\,dx\\
&=\frac{1}{2}\int_{-a}^{a}p(x)\,dx
\end{align*}
In (1) we use the fact that $p(x)$ is even and in (2) we use $q(x)q(-x)=1$.
$$ $$
In the current situation we obtain
\begin{align*}
\int_{-3}^{3}\frac{x^8}{1+e^{-2x}}\,dx&=\frac{1}{2}\int_{-3}^{3}\left(\frac{x^8}{1+e^{-2x}}+\frac{x^8}{1+e^{2x}}\right)\,dx\\
&=\frac{1}{2}\int_{-3}^{3}x^8\frac{1+e^{2x}+1+e^{-2x}}{1+e^{-2x}+e^{2x}+e^{-2x}e^{2x}}\,dx\\
&=\frac{1}{2}\int_{-3}^{3}x^8\frac{2+e^{-2x}+e^{2x}}{2+e^{-2x}+e^{2x}}\,dx\\
&=\frac{1}{2}\int_{-3}^{3}x^8\,dx
\end{align*}
Note: This technique can be found e.g. in Inside Interesting Integrals written by P.J. Nahin.
He applies this technique to the seemingly complicated integral
\begin{align*}
\int_{-1}^{1}\frac{\cos(x)}{1+e^{(1/x)}}\,dx
\end{align*}
which becomes easy if you know the trick.
|
{
"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 true for $n = 2$.
Then, assuming that it works for $n = m$, I took the case wherein $n = m+1\ldots$
Using this the left hand side of the equation is :
$$\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1} $$
and the right hand side would be
$$\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} }$$
How do I prove that the LHS = RHS??. I tried squaring both numerator and denominator, as well as using the properties of an arithmetic sequence but I havent been able to simplify the algebra.
|
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 $$\sum _{k=2}^{m}(\sqrt{a_{(k-1)}}+\sqrt{a_k})^{-1}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$
$$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{a_{(k-1)}-a_{k}}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$
$$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{a_1+(k-2)d-(a_1+(k-1)d)}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$
$$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{-d}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$
$$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}\tag 1$$
Now, setting $n=m+1$ in (1), we get $$\frac{1}{d}\sum _{k=2}^{m+1}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m+1-1}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}$$
$$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})+\frac{1}{d}(\sqrt{a_{(m+1)}}-\sqrt{a_{m}})=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}$$
$$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{d}(\sqrt{a_{(m+1)}}-\sqrt{a_{m}})$$
$$=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{d}\frac{(a_{(m+1)}-a_{m})}{\sqrt{a_{(m+1)}}+\sqrt{a_{m}}}$$
$$=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{d}\frac{d}{\sqrt{a_{(m+1)}}+\sqrt{a_{m}}}$$ $$=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{\sqrt{a_{(m+1)}}+\sqrt{a_{m}}}$$
Now, setting $a_{(m+1)}=a_m+d$ & simplifying, we get
$$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m-1}{\sqrt {a_1}+\sqrt{a_{m}}}$$ Which is true from (1), hence the equality holds for $n=m+1$
Hence, the equality holds for all positive integers $n\geq 1$
|
{
"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\int_0^{\frac{\pi}{2}}\frac{x \cos x}{\sin x+\sin^3 x} \, dx = \frac{\pi}{4} \log 2 + \int_0^{\frac{\pi}{2}} \frac{\log(1+\sin^2 x)}{2}\, dx \tag 1$$
$$$$
Now, let (for some $k$):
$$I(k) = \int_0^{\frac{\pi}{2}} \frac{\log(1+\sin^2 x + k \sin^2 x)}{2}\, dx \tag 2$$
In order to find $I(k)$:
$$\frac{\partial I}{\partial k} = \int_0^{\frac{\pi}{2}} \frac{\sin^{2}{x}\, dx}{1+\sin^2 x +k\, \sin^2 x} $$
$$\implies I'(k) = \int_0^{\frac{\pi}{2}} \frac{1}{k+1}\left(1-\frac{\sec^2 x}{1+(k+2)\tan^2 x}\right)\, \, dx $$
$$=\frac{x}{k+1}-\frac{\tan^{-1}{(\sqrt{k+2}\,\tan x)}}{(k+1)\, \sqrt{k+2}}\bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{2k+2}\left(1-\frac{1}{\sqrt{k+2}}\right)$$
Now just integrate (using Calculus 1 knowledge):
$$I(k)= \frac{\pi}{2}\left(\log{(k+1)}-\log\left(\frac{\sqrt{k+2}-1}{\sqrt{k+2}+1}\right)\right)+C=\frac{\pi}{2}\, \log\left(2\, (\sqrt{k+2}+1)^2\right)+C$$
$$\therefore I(k) = \frac{\pi}{2}\, \log{\left(2 (\sqrt{k+2}+1)^2\right)}+C \tag 3$$
From this step forward, we will solve for $C$:
To find $C$, substitute $k=-1$.
From $(2)$, we have:
$$\therefore I(-1) = 0$$
From $(3)$, we have:
$$I(-1) = \frac{3\,\pi}{2}\log 2+C$$
$$\implies C= - \frac{3\,\pi}{2}\log 2$$
$$\therefore I(0) = \int_0^{\frac{\pi}{2}} \log(1+\sin^{2}{x})\, dx = \frac{\pi}{2}\, \log\left(2\, (\sqrt{2}+1)^2\right) - \frac{3\,\pi}{2} \log 2$$
Now use $(1)$:
$$\therefore\int_0^{\frac{\pi}{2}}\frac{x \cos x\, dx}{\sin x +\sin^3 x} = \frac{\pi}{4}\,\log 2 + \frac{\pi}{4}\, \log\left(2\, (\sqrt{2}+1)^2\right) - \frac{3\,\pi}{4}\log 2$$
$$= \frac{\pi}{4}\log\left(\frac{3+2\,\sqrt{2}}{2}\right)$$
|
{
"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$ decomposition where $L$ is a lower triangular matrix with only $1s$ on the diagonal. But that's not what I'm getting. Here is what I did: \begin{align*} \begin{pmatrix} 3 & -6 & 9 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} & \begin{matrix} \xrightarrow{R_1 \rightarrow 1/3 R_1} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} \begin{matrix} \xrightarrow{R_2 \rightarrow R_2 + 2R_1} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ 0 & 3 & 4 \\ 0 & 1 & 5 \end{pmatrix} \\ & \begin{matrix} \xrightarrow{R_2 \rightarrow 1/3 R_2} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & \frac{4}{3} \\ 0 & 1 & 5 \end{pmatrix} \begin{matrix} \xrightarrow{R_3 \rightarrow R_3 - R_2} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & \frac{4}{3} \\ 0 & 0 & \frac{11}{3} \end{pmatrix} = U. \end{align*} This matrix is now in echelon form. The elementary matrices corresponding to the operations are: \begin{align*} E_1 & = \begin{pmatrix} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (R_1 \rightarrow 1/3 R_1) \quad E_2 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (R_2 \rightarrow R_2 + 2R_1) \\ E_3 &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (R_2 \rightarrow 1/3 R_2) \quad E_4 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} \qquad (R_3 \rightarrow R_3 - R_2) \end{align*} Hence we have $E_4 E_3 E_2 E_1 A = U$. Now \begin{align*} E_4 E_3 E_2 E_1 = M = \begin{pmatrix} \frac{1}{3} & 0 & 0 \\ \frac{2}{9} & \frac{1}{3} & 0 \\ - \frac{2}{9} & - \frac{1}{3} & 1 \end{pmatrix} \end{align*} is a lower triangular matrix. Now we can write \begin{align*} A = (E_4 E_3 E_2 E_1)^{-1} U = M^{-1} U = LU, \end{align*} with \begin{align*} M^{-1} = L = \begin{pmatrix} 3 & 0 & 0 \\ -2 & 3 & 0 \\ 0 & 1 & 1 \end{pmatrix} \end{align*} But why does my matrix $L$ not have $1s$ on the diagonal? I thought an $LU$-factorization of an invertible matrix is unique, and that in that case $L$ is an unit lower triangular matrix? Did I overlook something or made a mistake? I haven't interchanged any rows. Please help!
|
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 $2^n$.
Personally a tough problem. Hints Only Please!
I got that: $f(x) = g(2x) - g(x)$
Also,
$f(x) = a_0 + \sum_{k=1}^{n} a_k x^k$ and $g(x) = \sum_{k=1}^{n} \frac{a_k x^k}{2^k - 1}$ and $h(x) = g(2x) = \sum_{k=1}^{n} \frac{2^k a_k x^k}{2^k - 1}$
$g(1) = \sum_{k=1}^{n} \frac{a_k}{2^k - 1} = 0$
$g(2^{n+1}) = \sum_{k=1}^{n} \frac{2^k a_k 2^{nk}}{2^k - 1} = 0$
Realize that: $g(2^{n + 1}) = g(2 \cdot 2^{n}) = h(2^n). $ Hence the root of $h(x) = g(2x) \implies x = 2^n$.
Since $h(1/2) = g(1) = 0$, it follows $x= \frac{1}{2}$ is a root for $h(x) = g(2x)$.
So I have:
$g(x) = 0 \implies x = \{1, 2^{n+1} \}$
$g(2x) = 0 \implies x = \{\frac{1}{2}, 2^{n}\}$.
Now suppose $x = 2^{n} + i$. Then,
$g(2x) = g(2^{n+1} + 2i) > g(2^{n+1}) = 0$.
$g(x) = g(2^n + i)$
But I'm not sure what to do next?
|
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$ (the root of $g(x)$), one function must be "higher than" the other, so that $g(2x)-g(x)>0$ or $g(2x)-g(x)<0$. Conversely, on the right-most side, around $x=2^n$, the inverse is true. Hence the difference, in either case, changes sign somewhere in the interval $(1,2^n)$ which means that...
|
{
"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$$ I put $x^2-1=t$ but no benefit. Please guide me.
|
A couple of hints (not a full solution):
$x\to 2 \sin x$
\begin{align}
\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx&=\int_{\pi/6}^{\pi/2}\frac{\cot (x) \left(3-\cot ^2(x)\right)}{4 \sqrt{-12 \cos (2 x)+4 \cos (4 x)+9}}\,dx\\
\end{align}
now you may use $$\cot (x) \left(3-\cot ^2(x)\right)=\Big[\frac14(24 \sin (2 x)-16 \sin (4 x)) \csc ^2(x)\Big]-\Big[(-12 \cos (2 x)+4 \cos (4 x)+9) \cot (x) \csc ^2(x)\Big]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1403516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Triangle area inequalities; semiperimeter I've got stuck on this problem :
Proof that for every triangle of
sides $a$, $b$ and $c$ and area
$S$, the following
inequalities are true :
$4S \le a^2 + b^2$
$4S \le b^2 + c^2$
$4S \le a^2 + c^2$
$6S \le a^2 + b^2 + c^2$
The first thing that came to my mind was the inequality $S \le \frac 12 ab$. That is derived from the fact that $S = \frac 12ab \cdot \sin(\angle ACB)$ and $0 < \sin(\angle ACB) \le 1$.
Anyway, this wasn't enough to solve the problem. Some help would be well received.
Thanks!
|
Let $S$ denote the area of the triangle, so:
$$
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\leq 2bc\\
$$
And for $a,b,c\in {\mathbb{R}}^{+}$:
$$
{(a-b)}^{2}\geq 0\\
{(a-c)}^{2}\geq 0\\
{(c-b)}^{2}\geq 0
$$
We get:
$$
a^2+b^2\geq 2ab\\
a^2+c^2\geq 2ac\\
c^2+b^2\geq 2bc
$$
So we get:
$$
4S \leq a^2+b^2\\
4S\leq a^2+c^2\\
4S\leq c^2+b^2\\
$$
For the last one add the latter inequalities just obtained together:
$$
12S\leq a^2+b^2+a^2+c^2+c^2+b^2\\
12S\leq 2a^2+2b^2+2c^2\\
6S\leq a^2+b^2+c^2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1404335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Integrating $\frac{1}{(x^4 -1)^2}$ How to solve the the following integral? $$\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$$
Now Here $\bf{1^{st}}$ and $\bf{2^{nd}}$ can be Calculated Direct Formulas and for third put
$$\displaystyle \left(x^3-\frac{1}{x}\right) = t\;,$$ Then $$\displaystyle \left(3x^2+\frac{1}{x^2}\right)dx = dt$$
so we get $$\displaystyle I = \frac{3}{16}\ln \left|\frac{1+x}{1-x}\right|+\frac{3}{8}\tan^{-1}(x)-\frac{1}{4}\cdot \frac{x}{(x^4-1)}+\mathcal{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$$
Therefore the coordinates of $P$ is $(\frac{2}{9}, \frac{26}{9})$.
ii) Find the ratio in which $P$ divides the interval $(1,\frac{33}{9})$ and $(\frac{1}{3},3)$
Using the interval division formula, and letting the ratio be $k:l$, I got two equations, $\frac{k (1/3) + l(1) }{k+l}$ and $\frac{k (3) + l(33/9) }{k+l}$.
However when I simplified both, I ended up getting $7l+k=0$ on both sides, which left me stuck as I was not sure where I went wrong. Should I try with the ratio
$-k:l$?
|
Notice,let $m:n$ be the ration in which the given point $P\left(\frac{2}{9}, \frac{26}{9}\right)$ divides the line joining the points $\left(1, \frac{33}{9}\right)$ & $\left(\frac{1}{3}, 3\right)$ respectively
then the coordinates of the point P are calculated using division formula as follows $$P\equiv \color{red}{\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)}$$ Now, substituting the corresponding values we get
$$\left(\frac{2}{9}, \frac{26}{9}\right)\equiv \left(\frac{\frac{1}{3}\cdot m+1\cdot n}{m+n}, \frac{3\cdot m+n\cdot \frac{33}{9}}{m+n}\right)$$
Now, comparing the corresponding $x$ coordinates, we get $$\frac{\frac{1}{3}\cdot m+1\cdot n}{m+n}=\frac{2}{9}\iff m+7n=0$$ $$\frac{m}{n}=-\frac{7}{1}$$
Similarly, comparing the y-coordinate, we get
$$\frac{3\cdot m+n\cdot \frac{33}{9}}{m+n}=\frac{26}{9}\iff m+7n=0$$
$$\frac{m}{n}=-\frac{7}{1}$$
Negative sign indicates that the point $P$ divides the interval $\left(1, \frac{33}{9}\right)$ & $\left(\frac{1}{3}, 3\right)$ externally in a ratio $7:1$
Thus, we conclude that
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{green}{\text{Point P divides the given interval externally in a ratio}\ \ \color{blue}{7:1}}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1407592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Does the elliptic function $\operatorname{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\big|\frac{1}{2}\right)$ have a closed form? Given the complete elliptic integral of the first kind $K(k)$ for the modulus $k$,
can the elliptic function $$\text{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\bigg|\frac{1}{2}\right)$$ be expressed in closed form?
|
Assuming $1/2$ is $k$, as in Maple ... Maple
JacobiCN(2/3*EllipticK(1/2),1/2) evaluates to
$0.473058826656122429170671314726$.
From ISC we find that
this is a solution of
$$
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}\sqrt {1+2\,\sqrt [3]{6}}+2\,
\sqrt {1+2\,\sqrt [3]{6}}+14}}{2\sqrt [4]{1+2\,\sqrt [3]{6}}}}
$$
On the other hand, if $1/2$ is $m=k^2$, then in Maple we want
JacobiCN(2/3*EllipticK(1/sqrt(2)),1/sqrt(2)) which evaluates to
$0.43542054468233904782250442376$.
This is a solution of $-1+2Z+2Z^3-Z^4=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1407909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no idea.
|
Claim. $a+b+c\mid a^{2^n}+b^{2^n}+c^{2^n}$ for all $n\geq0$.
Proof. By induction: True for $n=0,1$ $\checkmark$. Suppose it's true for $0,\ldots,n$. Note that $$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-1}}c^{2^{n-1}}(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})$$
and that
$$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}})=(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2-(a^{2^n}+b^{2^n}+c^{2^n})$$
is divisible by $a+b+c$ by the induction hypothesis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
}
|
Calculate $\lim_{n\to\infty} (\frac{1}{{1\cdot2}} + \frac{1}{{2\cdot3}} + \frac{1}{{3\cdot4}} + \cdots + \frac{1}{{n(n + 1)}})$
Calculate
$$\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\infty}\left(\frac{X}{{n!(n + 1)}}\right).$$
How can I find $X$ and calculate limit?
|
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 \alpha + F_3 \cdot \sin \beta = 0
\end{cases}
\end{equation*}$
$\begin{equation*}
\begin{cases}
F_1 + F_2 \cdot \cos \alpha + F_3 \cdot \cos \beta = 0
\\
12 \cdot \sin \alpha = -13 \cdot \sin \beta
\end{cases}
\end{equation*}$
$\sin \alpha = - \frac {13}{12} \cdot \sin \beta$
$\cos^2 \alpha = 1 - \frac {169}{144} \cdot \sin^2 \beta=\frac {169}{144} \cdot \cos^2 \beta - \frac {25}{144}$
$ 5 + 12 \cdot \cos \alpha + 13 \cdot \cos \beta = 0$
$ 12 \cdot \cos \alpha =- 13 \cdot \cos \beta - 5$
$ 144 \cdot \cos^2 \alpha =(13 \cdot \cos \beta + 5)^2$
$169 \cdot \cos^2 \beta - 25 = 169 \cdot \cos^2 \beta + 130 \cdot \cos \beta + 25$
$130 \cdot \cos \beta = -50$
$\cos \beta = \frac {5}{13}, \beta < \pi$
$ \beta = 1.176$
$ \sin \alpha = -1$
$\alpha = \frac {3 \cdot \pi}{2}$
Answer:
$\angle \vec F_1 \vec F_2 = \frac {3 \cdot \pi}{2}, \angle \vec F_1 \vec F_3 = 1.176$
EDIT: Stupid mistake
|
{
"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$. Prove that
$$
\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+z)(1+x)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}
$$
Using AM-GM in the following form:
$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}{8} + \frac{1+z}{8} \ge \frac{3x}{4}$.
Now, this is where I get lost. Where was this expression derived from?
We conclude that
$\sum_{cyc}\frac{x^3}{(1+y)(1+z)} + \frac{1}{4}\sum_{cyc}(1+x) \ge \sum_{cyc}\frac{3x}{4} \rightarrow \sum_{cyc}\frac{x^3}{(1+y)(1+z)} \ge \frac{1}{4}\sum_{cyc}(2x-1) \ge \frac{3}{4}$
The equality holds for $x=y=z=1$
|
The AM-GM inequality for three terms can be written
$$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 rate of $3$ cm/s$^2$. Find the velocity and acceleration of particle at that instant.
Solution (Well at least my attempt at one)
$\mathbf{r}(t) = x\mathbf{i} - x^2\mathbf{j} + x^2\mathbf{k}$
$\mathbf{v}(t) = \left(\mathbf{i} - 2x\mathbf{j} + 2x\mathbf{k}\right)\dfrac{dx}{dt}$
\begin{align*}
\implies ||\mathbf{v}|| &=\left(\sqrt{1^2 +4x^2+4x^2}\right) \left|\dfrac{d x}{dt}\right|\\
&=(\sqrt{1+8x^2})\dfrac{dx}{dt}
\end{align*}
$|\frac{dx}{dt}|=\frac{dx}{dt}$ since the particle is always increasing
$\implies$ at the point $(1,\,-1,\,1),\,\, ||\mathbf{v}|| = 9$, so $dx/dt = 3$ and the velocity at that point is
$\mathbf{v}(t) = 3\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}$
Now acceleration is where I have a bit of trouble
$\mathbf{a}(t) =\dfrac{d^2x}{dt^2}(\mathbf{i} - 2x\mathbf{j} + 2x\mathbf{k}) + (\dfrac{dx}{dt})^2(- 2\mathbf{j} + 2\mathbf{k})$
But I don't exactly know what to do from here any help would be appreciated
|
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)^2}(3)-\dfrac{8(1)}{\sqrt{1+8(1)^2}}(3)(9)}{1+8(1)^2}\text{ cm/s}^2\\
&=\frac{9-72}{9}\text{ cm/s}^2\\
\frac{d^2x}{dt^2}&=-7\text{ cm/s}^2
\end{align*}
Hence, at the given instant,
\begin{align*}
\mathbf{a}&=\left(-7 \text{ cm/s}^2\right)(\mathbf{i}-2\mathbf{j}+2\mathbf{k} \text{ cm/s})+(9 \text{ cm/s})^2(-2\mathbf{j}+2\mathbf{k}\text{ cm}^{-1})\\
&=-7\mathbf{i}-4\mathbf{j}+4\mathbf{k}\quad\text{cm/s}^2
\end{align*}
|
{
"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{x^2}{1+x^2}}{3x^2}\\
&=\lim_{x\to0}\frac{\frac1{1+x^2}}{3}\\[3pt]
&=\frac13
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
transformation of uniformly distributed random variable f(x)=1/2pi into Y=cosx Let $X$ be a uniformly distributed function over $[-\pi;\pi]$. That is
$
f(x)=\left\{\begin{matrix}
\frac{1}{2 \pi} & -\pi\leq x\leq \pi \\
0 & otherwise
\end{matrix}\right.\\
$
Find the probability density function of $Y = \cos X$.
I figured out how to derive the final result using the change of variable method, but don't quite get why it needs to be multiplied by $2$.
I cannot get the result using the other cdf method at all.
Would you please help and clarify the steps of the both methods (variable change and cdf) for this type of the problem?
$$f(x)=y=cos(x) $$
$$f^{-1} (y) = \arccos(y)$$
$$f^{-1} (y)'= 1 / \sqrt{1-y^2}$$
$$fy(y)= \mathbf 2 \cdot \left(\frac{1}{2\pi}\right) \cdot \left|-1/\sqrt{1-y^2}\right|
= \frac 1\pi \cdot \sqrt{1-y^2}$$
using cdf method
\begin{align}
Fy(y) &= P(Y \leq y) = P(\cos x \leq y) = P(x \leq -\arccos y) + P(x \geq \arccos y) \\ \\
&= \int_{-\pi}^{-\arccos(y)} \frac{1}{2 \pi} \; dx + \int_{\arccos(y)}^{\pi} \frac{1}{2 \pi} dx
= \left[\frac{x}{2 \pi}\right]_{-\pi}^{-\arccos(y)}
+ \left[\frac{x}{2 \pi}\right]_{\arccos(y)}^{\pi} \\ \\
&= \frac{-\arccos(y)}{2 \pi} - \frac{-\pi}{2 \pi} + \frac{\pi}{2 \pi} -\frac{\arccos(y)}{2 \pi} \\ \\
&= \frac{-\arccos(y)}{2 \pi} + \frac{\pi}{2 \pi} - \frac{arccos(y)}{2 \pi} + \frac{\pi}{2 \pi} = 1- \frac{\arccos(y)}{ \pi} \\ \\
f(y) &= F'(y) = \left(1- \frac{\arccos(y)}{ \pi}\right)' = \frac{1}{\pi \sqrt{1-y^2}}
\end{align}
|
since we have $x$ belongs to $[-\pi,\pi]$ the $\arccos(x)$ can be written with two signs ($-$ & $+$) , that's why we have:
$$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 {\sqrt{1-x^2}}$$
That's it
Best regards
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1417683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find $\lim_{n \rightarrow \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$ $\lim_{n \rightarrow \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$
where the 2 inside the roots appear n times. For example if n = 2 :
$2^{2} \sqrt{2-\sqrt{2}}$
I discovered this. Has this been already developed/made before?
|
Since $$\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^2-n^2\pi^2}\\\\
&=\frac12\sum_{n=-\infty}^{\infty}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)
\end{align}$$
Then, using the trigonometric identity $\csc x=\cot(x/2)-\cot x$ reveals that
$$\begin{align}
\csc x&=\frac12\sum_{n=-\infty}^{\infty}\left(\frac{2}{x-2n\pi}-\frac{1}{x-n\pi}+\frac{2}{x+2n\pi}-\frac{1}{x+n\pi}\right)\\\\
&=\sum_{n=-\infty}^{\infty}\left(\frac{2}{x-2n\pi}-\frac{1}{x-n\pi}\right)\\\\
&=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{x-n\pi}
\end{align}$$
as was to be shown!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1419280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to calculate the limit that seems very complex.. Someone gives me a limit about trigonometric function and combinatorial numbers.
$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom{n}{1}\cos\frac{2}{n^2}+\binom{n}{2}\cos\frac{3}{n^2}\cdots\binom{n}{n}\cos\frac{n+1}{n^2}}+1\right)^n$
when $n$ is big enough, $\displaystyle 0\leqslant\frac{n+1}{n^2}\leqslant \frac\pi 2$ I tried $\displaystyle \frac{x}{1+x}\leqslant\sin x\leqslant x$
Use $\sin x\leqslant x$, I got $I\leqslant e$, But use another inequality I got nothing.
I don't know the answer is $e$ or not. Who can help me. Thanks.
|
With rough (but safe) approximations, we can avoid the use of trigonometric stuff, that is $$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}$$
where $\displaystyle \sin\left(\frac{k}{n^2}\right)$ behaves like $\displaystyle \frac{k}{n^2}$ and $\displaystyle \cos\left(\frac{k}{n^2}\right)$ behaves like $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1421920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
}
|
How to evaluate this infinite product ? (Fibonacci number) Let $F_n$ be Fibonacci numbers.
How to evaluate
$$\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}$$
But how to prove it?
Thank in advances.
|
The product is indeed equal to $\frac{1}{3}$.
We can use Binet's formula, which states that
$$F_n=\frac{1}{\sqrt{5}}\left(\phi^n - \left(-\frac{1}{\phi}\right)^n\right)$$
for all $n$ where $\phi=\frac{1+\sqrt{5}}{2}$
We can plug this directly into the product which we want to evaluate, but our life becomes slightly easier if we first simplify $F_{n+1}^2-F_{n-1}^2$ using other means.
We note that the addition formula for Fibonacci numbers tells us that $$F_{m+n}=F_{m+1}F_n + F_{m}F_{n-1}$$
and that in particular
$$F_{2n}=F_n(F_{n+1}+F_{n-1})$$
We can then write $F_{n+1}^2-F_{n-1}^2$ as
$$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})=F_n(F_{n+1}+F_{n-1})=F_{2n}$$
as noted by Henning Makholm in the comments on the question.
The product then becomes
$$\prod_{k=2}^\infty \frac{F_{2n}-1}{F_{2n}+1}$$
Using Binet's formula, the terms in the product become
$$\frac{F_{2n}-1}{F_{2n}+1}=\frac{\frac{1}{\sqrt{5}}\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)-1}{\frac{1}{\sqrt{5}}\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)+1}=\frac{\phi^{4n}-\sqrt{5}\phi^{2n}-1}{\phi^{4n}+\sqrt{5}\phi^{2n}-1}$$
Noting that $\phi^2-\frac{1}{\phi^2}=\sqrt{5}$, we see that this can be factorised as
$$\frac{\left(\phi^{2n}-\phi^2\right)\left(\phi^{2n}+\phi^{-2}\right)}{\left(\phi^{2n}-\phi^{-2}\right)\left(\phi^{2n}+\phi^2\right)} = \frac{\left(\phi^{2n-2}-1\right)\left(\phi^{2n+2}+1\right)}{\left(\phi^{2n+2}-1\right)\left(\phi^{2n-2}+1\right)}=\frac{a_{n-1}b_{n+1}}{a_{n+1}b_{n-1}}$$
where we define
$$a_n=\phi^{2n}-1$$
and
$$b_n=\phi^{2n}+1$$
The product then becomes
$$\prod_{k=2}^\infty \frac{a_{k-1}b_{k+1}}{a_{k+1}b_{k-1}}$$
which telescopes, and we see that the product is equal to
$$\lim_{n\to\infty} \frac{a_1 a_2 b_n b_{n+1}}{a_n a_{n+1} b_1 b_2}$$
It is straightforward to see that
$$\lim_{n\to\infty} \frac{b_n}{a_n} = 1$$
so that the product is then equal to the value
$$\frac{a_1 a_2}{b_1 b_2}=\frac{(\phi^2-1)(\phi^4-1)}{(\phi^2+1)(\phi^4+1)}=\frac{(\phi^2-1)^2}{\phi^4+1}=\frac{\phi^2}{3\phi+2+1}=\frac{1}{3}$$
as desired.
(Since $\phi^2=\phi+1$, and so $\phi^4=\phi^2+2\phi+1=3\phi+2$)
|
{
"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
}
|
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 it true that $k\cdot p_x + m$ is always prime when $k\cdot p_x + m < p_{x+1} ^2$ and $\gcd(k,m)=1$ where $k$ is a positive even number?
|
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. I suspect that the counterexamples come very easily if we consider Fermat primes, though I can't prove this right now.
EDIT 2: If we consider $p = 2^k + 1$, then $p^2 = 2^{2k} + 2\cdot 2^k + 1$ so $n = 2^{2k-1} + 2^k$. Then, letting $k = 1$, we have $n + p = 2^k(2^{k-1} + 1) + 2^k + 1 = (2^k + 2)(2^{k-1} + 1) - 1 = 2(2^{k-1} + 1)^2 - 1$. My guess is that many numbers of this form are composite. Maybe if we set up some sort of Pell equation $2a^2 - 1 = b^2$ some good things will happen?
Lastly, if $p = 11$, then $m = 61$, and if $k = 4$ we have $105$ divisible by $5$ but $105 < 121$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1424861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
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)}}$$
I have no idea for this question, does anyone can help me to answer this?
|
For $a=0$ and $b=c=1$ we have a value $2$.
We'll prove that it's a minimal value.
Indeed, let $a\neq0$, $a=x$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, by AM-GM we obtain: $$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$$
and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1424950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to get formula for sums of powers? Assuming I have Bernoulli numbers:
$B = [\frac{1}{1},\frac{1}{2},\frac{1}{6},\frac{0}{1},-\frac{1}{30}, \frac{0}{1}, \frac{1}{42}, ...]$
How can I get the coefficients of the sums of powers formulas?
For example the sum of squares is $(1/3)n^3 + (1/2)n^2 + 1/6n$
|
A vey-well rememberable scheme is perhaps the following.
Use the vector of Bernoulli-numbers in the first row of a little scheme and multiply through each column (the result per column is in the last row of the table):
$$ \begin{bmatrix}\frac{1}{1}&\frac{1}{2}&\frac{1}{6}&\frac{0}{1}&-\frac{1}{30}& \frac{0}{1}& \frac{1}{42}& ... \\
*x^4 & *x^3&*x^2&*x& & & & \\
/4 & /3& /2&/1& & & & \\
*1 & *3& *3&*1& & & & \\ \hline
1/4x^4 & 1/2x^3& 1/4x^2& 0x& & & &
\end{bmatrix}$$
Sum of cubes is $1/4x^4 + 1/2x^3+ 1/4x^2$
$$ \begin{bmatrix}\frac{1}{1}&\frac{1}{2}&\frac{1}{6}&\frac{0}{1}&-\frac{1}{30}& \frac{0}{1}& \frac{1}{42}& ... \\
*x^5 & *x^4&*x^3&*x^2&*x & & & \\
/5 & /4& /3&/2&/1 & & & \\
*1 & *4& *6&*4&*1 & & & \\ \hline
1/5x^5 & 1/2x^4& 1/3x^3& 0x^2&-1/30x
\end{bmatrix}$$
sum-of-$4$'th powers is $ 1/5x^5 + 1/2x^4 + 1/3x^3 -1/30x $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1426663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
Solve $\vert x-2\vert+2\vert x-4\vert\leq \vert x+1\vert$ I was helping someone with abolute values and inequalities and found this question.
What is the easiest way to solve this?
The only thing I thought of is to add the L.H.S and graph it with the R.H.S to answer the questoin is there simpler way to deal with this?
Thank you
|
An alternate non case-wise approach that involves more calculation with larger numbers:
Look for equality first. Squaring preserves the equality (although it may introduce extraneous solutions which can be ruled out at the end).
$$
\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+64
&= x^2+2x+1\\
4|(x-2)(x-4)|
&=-4x^2+38x-67\\
16x^4-192x^3+832x^2-1536x+1024
&=16x^4-304x^3+1980x^2-5092x+4489\\
112x^3-1148x^2+3556x-3465
&=0\\
16x^3-164x^2+508x-495
&=0\\
(4x-9)(2x-5)(2x-11)
&=0\\
\end{align}
$$
where the final factorization uses the rational root theorem. Checking, $\frac{9}{4}$ does not give equality, but both $\frac{5}{2}$ and $\frac{11}{2}$ do. Now check the direction of inequality on $\left(-\infty,\frac52\right)$, $\left(\frac52,\frac{11}{2}\right)$, and $\left(\frac{11}{2},\infty\right)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1428567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Use Proof of Induction to prove $\sum_{k=1}^{2n} (-1)^k k = n$ Base Case:
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n\\
(-1)^1 (1) + (-1)^2(2) &=&1 \\
1=1
\end{eqnarray*}
Inductive Step: For this step we must prove that
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n \Rightarrow \sum_{k=1}^{2(n+1)} (-1)^k k = n+1 \\
\sum_{k=1}^{2(n+1)} (-1)^k k = \sum_{k=1}^{2n} (-1)^k k + \sum_{k=1}^{2} (-1)^k k
\end{eqnarray*}
We know that,
\begin{eqnarray*}
\sum_{k=1}^{2n} (-1)^k k = n
\end{eqnarray*}
and from the base case we conclude,
\begin{eqnarray*}
\sum_{k=1}^{2} (-1)^k k = 1
\end{eqnarray*}
Therefore in we have $n+1 = n+1$. Is this a method that could be utilized?
|
Basically, yes. There is a bit of a snag with your second summation though: it should be $k = 2n + 1$ and $k = 2n + 2$. So we have:
\begin{align*}
\sum_{k=1}^{2(n + 1)} (-1)^k k
&= \sum_{k=1}^{2n} (-1)^k k + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) \\
&= n + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) &\text{by the induction hypothesis}\\
&= n - (2n + 1) + (2n + 2)\\
&= n + 1\\
\end{align*}
as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1430210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be:
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-2)^2}$ (first of all I don't get why that is?).
But in this exercise, will $\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64}$ be equal to $\frac A {x-2} + \frac B {(x-2)^2} + \frac C {x^2+2x+4} + \frac D {(x^2+2x+4)^2}$?
Thanks in advance.
|
Let's look at this a bit differently. As a first stage, we want to decompose the numerator $x^5+x^4+3x^3-8x^2+28x+48$ into components $p(x)(x^2+2x+4)^2+q(x)(x-2)^2$ so that we have $$\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}$$
We note by comparing degrees that $q(x)$ has degree at most $3$, so is of the form $(Cx+D)(x^2+2x+4)+Ex+f$ and $p(x)$ is of degree at most $1$ and can be written in the form $A(x-2)+B$. Then the expression becomes $$\frac A{x-2}+\frac B{(x-2)^2}+\frac {Cx+D}{x^2+2x+4}+\frac {Ex+F}{(x^2+2x+4)^2}$$
Now we are trying to match a numerator of degree $5$ which has $6$ independent coefficients, and this expression has a matching six constants to be determined. The numerator might be kinder than that, but in principle we cannot simplify this general form, where the degree of the numerator in the partial fractions is one less than the relevant degree in the denominator, unless we can use some special feature of the problem at hand.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1430739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
If $ax+by+cz=k$,prove that the minimum value of $x^2+y^2+z^2$ is $\frac{k^2}{a^2+b^2+c^2}$ If $ax+by+cz=k$,prove that the minimum value of $x^2+y^2+z^2$ is $\frac{k^2}{a^2+b^2+c^2}$.
I know that this problem can be solved by Cauchy Schwartz.How can i find its minimum value using multivariable calculus or by other methods.Please help me.
|
$$(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}$$
This is in fact an elemental proof of Cauchy–Schwarz_inequality on $\mathbb{R}^3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1431655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Show that there exist infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$ Let $(a_n)$ be a sequence defined by: $a_o=2, a_1=4, a_2=11$ and $\forall n \geq 3$, $$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).$$
So, we can have
$$a_n-na_{n-1}+n\cdot 3^{n-1}=3^n.$$
This can be written as
$$a_n-3^n=n(a_{n-1}-3^{n-1}).$$
So,
$$a_n-3^n=n!,$$
i.e.
$$a_n=3^n+n!.$$
Now $a_n-1=3^n-1+n!$.
Note here that $3^{2^k}-1$ is divisible by $2^{k+2}$ (you can prove this by induction on $k$). Also, $n!$ is divisible by $i$ at least when $n\ge i$.
Thus, we know that $a_i-1$ is divisible by $2^{2015}$ at least when $i=2^{2015+m}$ where $m=0,1,2,\cdots$.
Hence, it follows from this that there are infinitely many $i$ such that $a_i-1$ is divisible by $2^{2015}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1432801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving a simple rational equation $(\frac{6x}{6-x})^2+x^2=400$ Clearly we could multiply both sides of
$$\left(\dfrac{6x}{6-x}\right)^2+x^2=400$$
by $(6-x)^2$ which leads to a degree 4 polynomial equation, which we can solve using the bi-quadratic formula. Moreover, we could approximate the solutions using Newtons method. However, I have a feeling there is a much more graceful way to solve this. Any help is appreciated.
|
The quartic (or bi-quadratic) polynomial for $x$ can be factored into two quadratic polynomials with coefficients in $\mathbb{Z}[\sqrt{109}]$, so that the closed-form expression for roots is not as terrible as it can be generally.
Clearing fractions one gets:
$$ x^4 - 12x^3 - 328x^2 + 4800x - 14400 = 0 $$
Then notice that:
$$ x^4 - 12x^3 - 328x^2 + 4800x - 14400 = (x^2 - Ax + 6A)(x^2 - Bx + 6B) $$
provided $A+B = 12$ and $AB = -400$.
It follows that the roots for $x$ may be found by solving the respective quadratic equations with $A = 6 + 2\sqrt{109}$ and $B = 6 - 2\sqrt{109}$.
More explicitly:
$$ x = 3 - \sqrt{109} \pm \sqrt{82 + 6\sqrt{109}}\;,\; 3 + \sqrt{109} \pm \sqrt{82 - 6\sqrt{109}} $$
A crude estimate shows that the expressions under the radicals are all positive, and thus the four roots are indeed real roots.
How was the reducibility of the polynomial recognized?
Initially I saw that setting $y = \frac{6x}{x-6}$ leads to a system of equations:
$$ x^2 + y^2 = 20^2 $$
$$ (x - 6)(y - 6) = 36 $$
Geometrically this asks for the intersection of a circle of radius $20$ centered at the origin and a "right" hyperbola with center $(6,6)$. Since both curves are symmetric about the line $y = x$, it follows that when $(x,y)$ is a point of intersection, so too is $(y,x)$.
Algebraically this amounts to saying that $x \mapsto \frac{6x}{x-6}$ is an involution which pairs up the roots $x$. A bit of scratchwork showed that the pairing of roots in this way results in factors $x^2 - Ax + 6A$ and $x^2 - Bx + 6B$ where $A,B$ are roots of $w^2 - 12w - 400 = 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1434452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Change a 3D plane to XY plane I have a 3D cloud of points from witch I determine an "average" plane.
The plane equation is Ax + By + Cz + D =0;
I would like to re-compute all the points to obtain their new position considering that "average" plane is the new XY plane.
I suppose I have to do first a Z translation of -D/C and then a rotation.
Can someone give me the [3*3] rotation matrix?
|
$\vec{n}=(A,B,C)$ is a vector perpendicular to your average plane, forming an angle $\theta$ with the $z$-axis.
Consider the vector
$\vec{n}'=(B,-A,0)$, which is perpendicular to the plane formed by $\vec{n}$ and $z$-axis: a rotation of angle $\theta$ around $\vec{n}'$ will carry $\vec{n}$ along the $z$-axis and is thus what you are looking for. The rotation matrix is then:
$$
\left(
\begin{array}{ccc}
\frac{B^2}{A^2+B^2}+\frac{\left(1-\frac{B^2}{A^2+B^2}\right) C}{\sqrt{A^2+B^2+C^2}} &
-\frac{A B \left(1-\frac{C}{\sqrt{A^2+B^2+C^2}}\right)}{A^2+B^2} &
-\frac{A}{\sqrt{A^2+B^2+C^2}} \\
-\frac{A B \left(1-\frac{C}{\sqrt{A^2+B^2+C^2}}\right)}{A^2+B^2} &
\frac{A^2}{A^2+B^2}+\frac{\left(1-\frac{A^2}{A^2+B^2}\right) C}{\sqrt{A^2+B^2+C^2}}
& -\frac{B}{\sqrt{A^2+B^2+C^2}} \\
\frac{A}{\sqrt{A^2+B^2+C^2}} & \frac{B}{\sqrt{A^2+B^2+C^2}} &
\frac{C}{\sqrt{A^2+B^2+C^2}}
\end{array}
\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1435018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
|
Our old friend the golden ratio, or rather its enhancement by $1$, namely $a=1+\phi$, along with its reciprocal, is lurking behind this problem.
Let $a$ and $\bar a$ be respectively the larger and smaller root of $x^2-3x+1=0$. We have $a\bar a=1$ and $a+\bar a=3$. It's easy to show that $a^3+\bar a^3=18$. Also,$$a^{11}+\bar a^{11}=(a^5+\bar a^5)(a^6+\bar a^6)-(a+\bar a).$$After breaking $a^5+\bar a^5$ down to $(a^3+\bar a^3)[(a+\bar a)^2-2]-(a+\bar a)$, and $a^6+\bar a^6$ to $(a^3+\bar a^3)^2-2$, we get the answer$$a^{11}+\bar a^{11}=(18\times7-3)(18^2-2)-3=39,\!603.$$(Here we are ignoring the complex roots.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1435075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
}
|
Solving the absolute value equation $2-3|x-1| = -4|x-1|+7$ $$2-3|x-1| = -4|x-1|+7$$
This is an example from my text book, and I do not understand how they got the answers.
Solution: (this is the solution in my textbook)
Isolate the absolute value of expression on one side
Add $4|x-1|$ to both sides $ \rightarrow 2+|x-1|=7$
Subtract $2$ from both sides $ \rightarrow |x-1|=5$
If the absolute value of an expression is equal to $5$, then the expression is equal to either $-5$ or $5$.
$$x-1= -5, \ x-1=5$$
$$\implies x= -4, \ \implies x= 6.$$
I don't understand where the $4|x-1|$ went or what happened to it.
|
This is simply due to the fact that we can add or subtract anything to both sides of an equation and preserve the equality. Maybe this will help you understand it
\begin{align}
2-\color{red}{3|x-1|} &= \color{red}{-4|x-1|}+7 \\
2-\color{red}{3|x-1|} + \underbrace{\color{blue}{4|x-1|}}_{\text{Added}} &= \color{red}{-4|x-1|}+7+ \underbrace{\color{blue}{4|x-1|}}_{\text{Added}}\\
2+(\color{blue}{4}-\color{red}{3})|x-1| &= (\color{blue}{4}-\color{red}{4})|x-1| + 7\\
2+1\cdot|x-1|&= 0\cdot|x-1|+7\\
2+|x-1| &= 7\\
2-\color{blue}{2}+|x-1| &= 7-\color{blue}{2}\\
|x-1| &= 5.
\end{align}
I hope it makes sense now.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1438692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\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.$$ The result is now straightforward.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1439732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
How to find the root of this polynomial equation of third degree? x^3 -3x + 4 = 0
How do I find roots of the above equation?
I have tried plugging in values of x but it is not satisfying the equation above.
|
The full solution to the cubic is:
$$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 = 0^3 + \frac{0-3(4)}{6} = \frac{-12}{6} = -2$$
$$u = \sqrt{(-2)^2 + (-1)^3} = \sqrt{3}$$
$$x = (\sqrt{3}-2)^{\frac{1}{3}} + (\sqrt{3}-2)^{\frac{1}{3}}$$
$$\therefore x= 2 (\sqrt{3}-2)^{\frac{1}{3}}$$
There are also two complex solutions, but to get these simply divide your original polynomial by the solution we get, yielding a quadratic. An application of the quadratic formula will yield the other two solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1440367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Proving a progression inequality Prove that $$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}.$$
The case $n=2$ is easily verified ($\frac{9}{8}\leq \frac{5}{4}-\frac{1}{21}$). Now for the induction step:
$$\sum_{j=1}^{n+1}\frac{1}{j^3}=\frac{1}{(n+1)^3}+\sum_{j=1}^{n}\frac{1}{j^3}\leq \frac{1}{(n+1)^3}+\frac{5}{4}-\frac{4}{15n^2}$$ and now what we have to verify is that $$\frac{1}{(n+1)^3}+\frac{5}{4}-\frac{4}{15n^2}\leq \frac{5}{4}-\frac{4}{15(n+1)^2}.$$
This is equivalent to $$\frac{1}{(n+1)^3}\leq \frac{4}{15}\cdot\frac{2n+1}{n^2(n+1)^2}$$ $$\frac{4}{15}\geq\frac{n^2}{(n+1)(2n+1)}$$ which is true because we have equality in $n=2$ and the function on the RHS is decreasing.
Actually, the most natural way to proceed is to write $S_n\leq \frac{5}{4}-\frac{A}{n^s}$ and to go the above steps to find $A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1440859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How is it possible that $\int\frac{dy}{(1+y^2)(2+y)}$ = $\frac{1}{5}\int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} $? Suppose we have a fraction
$$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} ?$$
How are they using partial fractions to do this?
|
Setting
$$\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$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1442739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solving $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$? How to solve the definite integral $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$?
Funnily enough, this is actually a problem that showed up during my real analysis course. At first glance, the problem seemed solvable by using knowledge from basic elementary calculus, but it turned out not to be so simple. I initially tried the substitution $u=\sqrt{x}$, but that lead me nowhere...
Let $u=x^{\frac{1}{2}}$, $du=\frac{1}{2}x^{-\frac{1}{2}}$.
$\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$ =
$2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}~du$.
EDIT: I have tried an additional substitution to the above, namely $u= \frac{\sin t}{\sqrt{2}}$.
$$
\begin{align*}
2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}~du&=
2\int_0^{\frac{\pi}{2}} \frac{\sin^4 t}{4} (1-\sin^2 t)^{\frac{3}{2}}\cdot \frac{\cos t}{\sqrt{2}}~dt\\
&=\frac{1}{2\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^4t\cos^4t~dt.
\end{align*}
$$
Hopefully my calculations aren't wrong...
|
Note that there $x$ and $(\frac 12 - x)$ is symmetric about $x= \frac 14$. Thus try to substitute $y = \frac 14 -x$. Then the integral is
$$- \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.$$
From here one can use trigonometric substitution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1443182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
|
Since $x \geq \lfloor x \rfloor$ for all $x$, you have $x^2 - 19\lfloor x \rfloor + 88 \geq x^2 - 19x + 88$ for all $x$. Hence the graph of $x^2 - 19\lfloor x \rfloor + 88$ lies above that of $x^2 - 19x + 88$. So the function $x^2 - 19\lfloor x \rfloor + 88$ can only be zero where the function $x^2 - 19x + 88 = (x - 11)(x - 8)$ is less than or equal to zero, namely the interval $[8,11]$.
One can then solve $x^2 - 19\lfloor x \rfloor + 88 = 0$ separately on the four ranges in $[8,11]$ where $\lfloor x \rfloor$ is constant: the intervals $[8,9),[9,10],[10,11)$ and the singleton $x = 11$. You get the equations $x^2 - 64 = 0$, $x^2 - 83 = 0$, $x^2 - 102 = 0$, and $x^2 - 121 = 0$ respectively, leading to the four solutions $8, \sqrt{83}, \sqrt{102},$ and $11$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1444045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Given area of square $= 9+6\sqrt{2}$ Without calculator show its length in form of $(\sqrt{ c}+\sqrt{ d})$ $\sqrt{9+6\sqrt{2}}$ to find length
But how do I express the above in the form of $\sqrt{c} + \sqrt{d}$.
|
Notice, let the side of the square be $a$ then its area is given as $$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)$$
Using $a^2+b^2+2ab=(a+b)^2$, we get $$a^2=(\sqrt 6+\sqrt 3)^2$$
hence, side of square $$a=\color{red}{\sqrt6+\sqrt 3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1445691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Is there an easier way to solve this integral problem? I have the following integral:
$$F(t) = \int_0^b (\sqrt{2 + t} - 2) dt$$
I need to find a positive value of $b$ such that $F(t) = 0$.
Going through the integration steps and using the Fundamental Theorem of Calculus, I end up with
\begin{align}
0 & = (\frac{2}{3}\sqrt{(2 + t)}^3 - 2t) |_0^b \\
& = (\frac{2}{3}\sqrt{(2 + b)}^3 - 2b) - (\frac{2}{3}\sqrt{2}^3) \\
\Rightarrow \frac{4\sqrt{2}}{3} & = \frac{2}{3}\sqrt{(2 + b)}^3 - 2b
\end{align}
But isolating $b$ here is very complicated, and WolframAlpha gives a very ugly solution. It's been a while since I took calculus, but I feel like I'm missing something obvious.
|
If you change variable $b=x^2-2$, the equation becomes $$ \frac{2 x^3}{3}-2 x^2+4\big(1-\frac{ \sqrt{2}}{3}\big)=0$$ which can be solved with radicals using Cardano method.
But, there is one "obvious" root $x_1=\sqrt 2$ (remember that this corresponds to $b=0$ which is a trivial solution of the problem). So, what is left is a quadratic equation $$\frac{2 x^2}{3}+\left(\frac{2 \sqrt{2}}{3}-2\right) x-\big(2 \sqrt{2}-\frac{4}{3}\big)=0$$ the roots of which being $$x_2=\frac{1}{2} \left(3-\sqrt{2}-\sqrt{3+6 \sqrt{2}}\right)$$ $$x_3=\frac{1}{2} \left(3-\sqrt{2}+\sqrt{3+6 \sqrt{2}}\right)$$ The root $x_2$ must be discarded since it is negative and the only one left is then $x_3$; so, after simplifications, $$b=x_3^2-2=\frac{1}{2} \left(3+\sqrt{48 \sqrt{2}-39}\right)\approx 4.18711$$ which is quite close to Shailesh's answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1447653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Maximising cardinality of a set given the mean. Given a set of distinct, positive integers of mean $x$ and largest element $y$ how can we maximise the cardinality of the set?
|
For a corresponding set of $n$ elements, there must be a partition of $nx-y$ into $n-1$ unique addends. If it exists, you've found the maximum cardinality - $n$. Finding the maximum value requires three cases:
*
*$y=x$, the cardinality is $1$ (set $\{x\}$)
*$y < 2x$, the cardinality is $\lfloor \frac{1}{2} (\sqrt{4 x^2-8 x y+4 x+4 y^2+4 y+1}-2 x+2 y-1) \rfloor$
*$y \geq 2x$, cardinality is$\lfloor \frac{1}{2} (\sqrt{4 x^2+4 x-8 y+1}+2 x+1)) \rfloor$
This gives you $5$ for $x=3, y=5$ or $8$ for $x=5, y=11$ (the set for the latter is $\{1, 2, 3, 4, 5, 6, 8, 11\}$). The sets themselves can be found trivially.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1451567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
A given polynomial equation of 5 degree has three equal roots . Given equation is $x^5-10a^3x^2+b^4x+c^5=0$ which has 3 equal roots. What I know is that since its a 5th degree equation therefore it must have 5 roots out (of which 3 are equal).
Aim is to establish the relationship between the constants $a,b$ and $c$.
Options given are:
1) $6a^5+c^5=0$.
2) $b^4=15a^4.$
I have to find which one is correct out of the two, any help?
|
Question : I have to find which one is correct out of the two, any help?
Answer : Actually, both the options are correct. I am giving the explanation.
Let $~f(x) = x^5 – 10a^3x^2 + b^4x + c^5~$.
$$~⇒ f'(x) = 5x^4 – 20a^3x + b^4~~~~ ⇒ f'' (x) = 20x^3 – 20a^3~$$
Since $~x = \alpha~$ be a root that is repeated three times, so
$$f''(\alpha) = 0,~~~~~~ f'(\alpha) = 0,~~~~~~ f(\alpha) = 0$$
$$⇒ 20\alpha^3 – 20a^3=0,~~~~~~ 5a^4 – 20a^3\alpha + b^4 = 0,~~~~~~ \alpha^5 – 10a^3\alpha^2 + b^4\alpha +c^5 = 0$$
$$⇒ \alpha = a,~~~~~~ 5a^4 – 20a^3\alpha + b^4 = 0,~~~~~~ \alpha^5 – 10a^3\alpha^2 + b^4\alpha +c^5 = 0$$
$$⇒ b^4 = 15a^4, ~~~~~~c^5 + ab^4 – 9a^5 = 0\qquad (\text{using the result}~~\alpha = a )~~~~$$
$$⇒ b^4 = 15a^4, ~~~~~~c^5 + 15a^5 – 9a^5 = 0\qquad (\text{using the value}~~b^4 = 15a^4)$$
$$⇒ b^4 = 15a^4, ~~~~~~6a^5 + c^5 = 0~.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1455008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
How do I transform, with long division, this polynomial into the answer given? I need to transform:
$ \frac {x^5}{(x-2)(x+2)(x^2+4)} $
into
$ \frac{-2x}{x^2+4} + x + \frac{1}{x-2} + \frac{1}{x+2} $
How can I solve it?
Thanks.
|
First, you should notice the difference of squares formula (twice) in the denominator. $(x-2)(x+2)=x^2-4$ Then $(x^2-4)(x^2+4)=x^4-16$ so we have $\frac {x^5}{x^4-16}=x+\frac {16x}{x^4-16}$ The rest is a partial fraction decomposition of the second term.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1455985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Without L'Hopital,
$$\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 \sin\frac{x}{2}} + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
On the left one, it would be great to have $\frac{x}{2} \cdot \frac{x}{2}$. To do so, I will multiply and divide by $\frac{1}{4}$:
$$\frac{\frac{x}{2}\cdot \frac{x}{2}}{\frac{1}{4}\cdot-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}} + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
Then we use the identity $\frac{x}{\sin x} = 1$:
$$-2 + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
On the right side, I'd like to have $\frac{x}{2}$ there, so I will multiply and divide by $\frac{1}{2}$:
$$-2 + \frac{\frac{x}{2}\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}\cdot\frac{1}{2}}$$
We use the identity $\frac{x}{\sin x} = 1$:
$$-2 + \frac{\sin x}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}}$$
Hmm... we could perform the addition I suppose:
$$\frac{\sin x - 2(-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2})}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}}$$
Simplify:
$$\frac{\sin x + (4\cdot-2\sin\frac{x}{2}\cdot-2\frac{1}{2})}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}} = \frac{\sin x +8\sin\frac{x}{2}}{\sin\frac{x}{2}}$$
Then
$$\frac{\sin x}{\sin\frac{x}{2}} + \frac{8\sin\frac{x}{2}}{\sin\frac{x}{2}} = \frac{\sin x}{\sin\frac{x}{2}} + 8$$
I'm close! The answer should be $-4$, but I don't know what to do with
$$\frac{\sin x}{\sin\frac{x}{2}} + 8$$
So basically I need to know what to do with $\frac{\sin x}{\sin\frac{x}{2}}$, but I also included my whole procedure just in case I've been doing it wrong all along.
|
Perhaps a simpler approach is as follows
\begin{align}
L &= \lim_{x \to 0}\frac{x^{2} + x\sin x}{-1 + \cos x}\notag\\
&= \lim_{x \to 0}\frac{x^{2} + x\sin x}{-1 + \cos x}\cdot\frac{1 + \cos x}{1 + \cos x}\notag\\
&= \lim_{x \to 0}\frac{x^{2} + x\sin x}{-1 + \cos^{2} x}\cdot(1 + \cos x)\notag\\
&= -2\lim_{x \to 0}\frac{x^{2} + x\sin x}{\sin^{2} x}\notag\\
&= -2\lim_{x \to 0}\left(\frac{x^{2}}{\sin^{2}x} + \frac{x}{\sin x}\right)\notag\\
&= -2(1 + 1) = -4\notag
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1457217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
What is the number of terms in this series $1+0.5+0.25+\ldots+0.5^{n−1}$? And what is the sum?
What is the number of terms in this series $1+0.5+0.25+\ldots+0.5^{n−1}$? And what is the sum?
I stumped by this question, but have a feeling the answer's really obvious:S
Am I supposed to get an actual number as the answer?
Thanks for the help!
|
use the geometric series
$$\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$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1463145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$ Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.
$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \frac{\alpha}{2}\right ) \cos \left ( \frac{\alpha}{2}\right ) + \sin \left ( \frac{\beta}{2}\right ) \cos \left ( \frac{\beta}{2}\right )}{\cos^2 \left ( \frac{\alpha}{2} \right) - \sin ^2 \left ( \frac{\beta}{2} \right )}$$
But I know that if $\alpha$ and $\beta$ are angles in a triangle, then this expression should simplify to
$$\tan \left ( \frac{\alpha + \beta}{2} \right )$$
I can see that the denominator becomes $$\cos \left ( \frac{\alpha + \beta}{2} \right ) $$
But I cannot see how the numerator becomes
$$\sin \left ( \frac{\alpha + \beta}{2} \right )$$
What have I done wrong here?
|
$$\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}).$$
So, you get the conclusion.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1465321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red?}}{=}}\{a+b\sqrt2+c\sqrt 3:a,b,c\in \Bbb Q\}$
But from my last question it seems that we need to check this for multiplicative closure, $(a+b\sqrt2 + c\sqrt3)(d+e\sqrt 2 + f\sqrt 3)=ad+ae\sqrt 2+ af\sqrt 3 + 2eb+bf\sqrt 3+cd\sqrt 3+ ce\sqrt2\sqrt3+ 3cf$
So we actually find I believe that:
$\Bbb Q(\sqrt2,\sqrt3)=\{a+b\sqrt 2+c\sqrt 3+ d\sqrt2\sqrt3:a,b,c,d\in \Bbb Q\}$?
I believe that $\Bbb Q(\sqrt 2 +\sqrt 3)\overset{\huge{\color\red ?}}=\{a+b(\sqrt 2+ \sqrt 3):a,b\in \Bbb Q\}$
So let's verify multiplicative closure $(a+b\sqrt 2+ b\sqrt 3)(c+d\sqrt 2+ d\sqrt 3)$
$$=ac+ad\sqrt 2+ad\sqrt 3+bc\sqrt2+2bd+bd\sqrt3+bc\sqrt3+bd\sqrt2\sqrt3+3bd$$
$$=(ac+3bd+2bd)+(ad+bc)\sqrt 2 + (ad+bd+bc)\sqrt 3 +(bd)\sqrt2\sqrt3$$
Since $\sqrt2$ and $\sqrt 3$ and $\sqrt2\sqrt3$ don't share common coefficients, they are linearly independent and hence $\Bbb Q(\sqrt2+\sqrt 3)=\Bbb Q(\sqrt2,\sqrt3)$(from above, i.e we have deduced that:) $\Bbb Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt2+c\sqrt3:a,b,c\in\Bbb Q\}$.
Is that the correct way to show this? Also this means we have:
$$\begin{matrix}&&\left(\Bbb Q(\sqrt{2}+\sqrt{3})=\Bbb Q(\sqrt2,\sqrt3)\right)\\\\&{\huge\diagup}&&{\huge\diagdown}\\\Bbb Q(\sqrt2)&&&&\Bbb Q(\sqrt 3)\end{matrix}$$
|
Just to add to completeness of the other answers let me find minimal polynomial of $\alpha = \sqrt 2 + \sqrt 3$ over $\mathbb Q$ directly.
\begin{align}
\alpha = \sqrt 2 + \sqrt 3 &\implies \alpha - \sqrt 2 = \sqrt 3\\
&\implies \alpha^2 -2\alpha\sqrt 2 + 2 = 3\\
&\implies \alpha^2 - 1 = 2\alpha\sqrt 2\\
&\implies \alpha^4 - 2\alpha^2 + 1 = 8\alpha^2\\
&\implies \alpha^4 - 10\alpha^2 + 1 = 0
\end{align}
Thus, $\alpha$ is a root of $p(x) = x^4 - 10x^2 + 1$, so if we show that it is irreducible over $\mathbb Q$, it will follow that it is minimal polynomial. By Gauss's lemma, since $p$ is primitive, it is enough to prove that it is irreducible over $\mathbb Z$. Assume the contrary, that $p = fg$ for non-invertible $f, g\in\mathbb Z[x]$. Neither $f$ nor $g$ can be of degree $0$ since $p$ is primitive, nor of degree $1$ since $p$ has no rational roots by Rational root theorem. So we have $\deg f = \deg g = 2$ and thus can write $$ p(x) = (x^2 + ax + b)(x^2 + cx + d),\quad a,b,c,d\in\mathbb Z $$ By expanding we get $$x^4 - 10x^2 + 1 = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad + bc)x + bd$$ or equivalently
\begin{align}
a + c &= 0\\
b + d + ac &= -10\\
ad + bc &= 0\\
bd &= 1
\end{align}
Now, we note that $b = d = \pm 1$ and that neither cases yield integer solutions of above system and thus arrive at contradiction that $p$ is reducible.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1466935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Solving a first order IVP I want to solve the following initial value problem:
$$(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 $$
But the initial condition suggests that $(1+x) \gt 0$. Thus we have
$$ 2\sqrt{y} = -\ln(1+x) +C~~~~~~(*)$$ Applying the initial condition we get $C=2$ so that the solution becomes $$ 2\sqrt{y} = -\ln(1+x) + 2 \,\,\text{or}\,\, y = ( -\frac{1}{2} \ln(1+x) +1)^2~. $$
My question is this: From (*), one could also first find $$ y = (-\frac{1}{2} \ln(1+x) + \frac{C}{2})^2 $$ and applying the initial condition, yields
$$ 1 = \frac{C^2}{4} \implies C = \pm 2$$
Now choosing $C=2$, we obtain $y$ as before. But choosing $C =-2$ yields a different solution. How do we know which $C$ to choose? How do we reconcile the two approaches? Which approach is better?
|
You have to remember that the square root of a number $y$ is a non-negative number. Hence when you take $C=-2$, $-\frac{1}{2} \ln(1+x) + \frac{C}{2}$ is negative in the neighborhood of $x=0$.
Consequently the square root of $$y(x) = (-\frac{1}{2} \ln(1+x) -2)^2 $$ is not $\sqrt{y(x)} = -\frac{1}{2} \ln(1+x) -2$ (which is non-positive in the neighborhood of $0$) but the opposite $$\sqrt{y(x)} = \frac{1}{2} \ln(1+x) +2$$
But then $y(x)$ is no more a solution of the original IVP.
Conclusion: the IVP has only one solution with the initial condition $y(0)=1$ which corresponds to $C=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1468380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
obtain a three-term asymptotic solution In the limit as $\epsilon \to \infty$, obtain a three-term asymptotic solution to the roots of the following equation.
$$\epsilon x^3+x^2-2x+1=0$$
What I've done so far
I've assumed $x=O(\epsilon^r)$ as $\epsilon \to \infty$ therefore,
$\epsilon x^3=O(\epsilon^{1+3r})$
$x^2=O(\epsilon^{2r})$
$x=O(\epsilon^{r})$
$1=O(\epsilon^{0})$
I'm not too sure what needs to be done from here. any help would be much appreciated.
|
Outline (rather messy: is there a more elegant approach?):
*
*First, prove that you must have $x=x(\epsilon)\xrightarrow[\epsilon\to\infty]{} 0$.
*Then, you can get the expansion one term at a time...
Now, the second bullet (up to order two) below -- if I have not screwed up.
$$
1-2x + x^2+\epsilon x^3 = 1+\epsilon x^3 + o(1)
$$
so you need $\epsilon x^3 \operatorname*{\sim}_{\epsilon\to\infty} -1$. I.e., $x = -\frac{1}{\epsilon^{1/3}} + o\left(\frac{1}{\epsilon^{1/3}}\right)$.
Write $x = -\frac{1}{\epsilon^{1/3}} + h$, so that $x^3 = -\frac{1}{\epsilon} + \frac{3h}{\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)$. Plugging it back,
$$
1-2x + x^2+\epsilon x^3 = 1 -1 + 3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}} -2h + x^2 = 3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}} + o(h\epsilon^{1/3})
$$
To get that equal to $0$, this implies in particular $3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}}=o(1)$, so that
$$
h = \frac{-2}{3\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)
$$
i.e.
$$
x = -\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)
$$
Now, for the final step, write
$$
x = -\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g
$$
with $g=o\left(\frac{1}{\epsilon^{2/3}}\right)$.
$$
1-2x + x^2+\epsilon x^3 = 1 -2(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g)+\left(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g\right)^2+\epsilon\left(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g\right)^3
$$
and expand (possibly with Taylor expansions to the right order if this helps). To satisfy the equation, this must be $o(1)$: with this constraint, you will get $g$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1468686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2}x-1)}+\frac{x+\sqrt{2}}{2\sqrt{2}(x^2+\sqrt{2}x+1)}$$
BTW, are there any easier method to solve this integral? As seen on the answer, it seems not very possible. ;)
|
I am giving a hint.
$$\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)$$
Now, write $x^2+\frac 1{x^2}=\left(x-\frac 1x\right)^2+2$ and $x^2+\frac 1{x^2}=\left(x+\frac 1x\right)^2-2$, for the respective integrals, substitute for $x+\frac 1x=z$ and $x-\frac 1x=u$.
Now, you have $\left(1-\frac 1{x^2}\right)dx=dz$ and $\left(1+\frac 1{x^2}\right)dx=du$
Now,proceed from here, it is easy for you now.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1470540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\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}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels like nothing simple fits. Next I noticed we can normalise: restricting $a+b+c=1$ it can be made to look like this:
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
Another idea is to let $x=a/b,y=b/c, z=c/a:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{1}{x(y+1)}}+\sqrt{\frac{1}{y(z+1)}}+\sqrt{\frac{1}{z(x+1)}}\right)\leq \sqrt{x+y+z}$$
But I can't see where to go from here.
|
After using C-S $\left(\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq(a+b+c)\sum\limits_{cyc}\frac{1}{b+c}$ we'll obtain something obvious:
$$\sum_{cyc}(3a^4c^2+3a^3b^3+a^4bc-2a^3b^2c-2a^3c^2b-3a^2b^2c^2)\geq0.$$
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1471811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
}
|
Finding Pythagorean triplet given the hypotenuse I have a number $c$ which is an integer and can be even or odd. It is the hypotenuse of a right angled triangle. How can I find integers $a,b$ such that
$$ a^2 + b^2 = c^2 $$
What would be the complexity of the calculation?
|
You can find one or more triples, given only $C$ if $GCD(A,B,C)=2$ or a perfect square, which includes all primitives and such as $(27,36,45)$. You cannot find those that are otherwise, such as $3,5,6,7,8,10$ times a primitive, e.g. $(45,24,51)=3*(15,8,17)$.
We solve $C=m^2+n^2$ for $n$, and look for positive integers
$$n=\sqrt{C-m^2}\text{ where }{\lceil \sqrt{\frac{C}{2}}\space\rceil\le m\le \lfloor\sqrt{C}\rfloor}$$.
If we want to find one or more triples for $C=145$, then we have
$$m_{min}=\lceil \sqrt{\frac{145}{2}}\space\rceil=9\qquad m_{max}=\lfloor\sqrt{145}\rfloor=12$$
If we try $9\le m \le 12$, we find $n\in\mathbb{N}$ only for $\sqrt{145-9^2}=8$ and $\sqrt{145-12^2}=1$
$$f(9,8)=(17,144,145)\qquad f(12,1)=(143,24,145)$$
For $C=1105$, there are $4$ triples; for $C=29$, there is $1$ triple; for $C=31$, there are no triples. If $C=20$, we find $f(4,2)=(12,16,20)=2^2*f(2,1)=2^2*(3,4,5)$. Try the numbers and see for yourself.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1473201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of n such that l is non zero finite real number. Find l Problem :
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of $n$ such that $l$ is non zero finite real number. Find value of $l$.
My approach :
$$\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$$
$$ \Rightarrow \lim_{x \to 0} \frac{x^n}{-2\sin(\frac{\sin x+x}{2})\sin(\frac{\sin x-x}{2})} =l $$
Please suggest how to move further , will be of great help thanks.
|
We have $cos x= \sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}x^{2j}$ hence \begin{align}
\frac{x^n}{\cos \sin x -\cos x}&= \frac{x^n}{\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}(\sin^{2j}x-x^{2j})}\\
&= \frac{x^n}{\sum_{j=1}^{\infty}\frac{(-1)^j}{(2j)!}(\sin^{2j}x-x^{2j})}\\
&=\frac{1}{\color{blue}{\frac{(-1)^1}{2!}}\frac{\sin^{2}x-x^{2}}{x^n}+\frac{1}{4!}\frac{\sin^{4}x-x^{4}}{x^n}+...}\\
\end{align}
Now note that $$\sin^2x-x^2=\color{red}{-\frac13}x^4+\frac{2 x^6}{45}+O\left(x^8\right)$$ and $$\sin^4x-x^4=-\frac{2 x^6}{3}+\frac{x^8}{5}+O\left(x^{10}\right)$$ This indicates that choosing $n=4$ we will have a leading non-zero term in denumerator (coming from $(\sin^2x-x^2)/x^4$) making for a limit of $6$. Choosing $x>4$ say $x=6$, endows the second term a nonzero limit but the first term will be diverging making for a zero limit. Hence your choice is only $n=4$ and in this case the limit is $$\frac{1}{\color{blue}{\frac{(-1)^1}{2}}\color{red}{\frac{-1}{3}}}$$
|
{
"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
}
|
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 the origin, or all of $R^3$
I understand that for the solution set of a linear system Ax=0 to be a single point through the origin only, Ax=0 has only the trivial solution.
Also that a line through the origin means Ax=0 has one free variable.
And a plane through the origin means Ax=0 has 2 free variables.
I can't figure out how to accurately describe b to reflect each of these, I think I'm confusing myself. Would someone mind walking through this step by step with me? Thanks.
|
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
$
\begin{bmatrix}
1&1&b\\
0&b-1&1-b\\
b-1&0&1-b
\end{bmatrix}
$
$\xrightarrow{(b-1)R1}$
$
\begin{bmatrix}
b-1&b-1&b(b-1)\\
0&b-1&1-b\\
b-1&0&1-b
\end{bmatrix}
$
$
\xrightarrow{R3-R1}
$
$
\begin{bmatrix}
b-1&b-1&b(b-1)\\
0&b-1&1-b\\
0&1-b&(1-b)^2
\end{bmatrix}
$
$
\xrightarrow{R3+R2}
$
$
\begin{bmatrix}
b-1&b-1&b(b-1)\\
0&b-1&1-b\\
0&0&(1-b)(2-b)
\end{bmatrix}
$
So if the $3,3$ entry is nonzero the matrix is in reduced echelon form and so the system has unique solution.
The conclusion is: $b=1$ gives $2$ free variables, $b=2$ gives one free variable, any other value of $b$ gives unique solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1486212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
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 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Apart from simply multiplying $A$ with $B$ and generating a $12$ variable system of equations, is there any simpler way of finding $B$ ?
|
Your matrix
$$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\\2&2&1\\2&1&2\end{bmatrix}
$$
$$
(AA^T)^{-1}=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\-2&4&0\\-2&0&4\end{bmatrix}
$$
$$
B=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\1&2&-2\\-1&2&2\\1&-2&2\end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1487338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
The rule for evaluating limits of rational functions by dividing the coefficients of highest powers I have a Limit problem as below:
Connor claims: " $\lim_{x\to \infty} \left(\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}\right) = 3$ because my high school calculus teacher told us the limit of ratio of polynomials is always the quotient of the coefficients of the highest power terms"
If correct, show in detail how to use algebra and the limit theorems to evaluate this limit and get the same answer
If wrong,
1, use algebra and limit theorems to correctly evaluate the limit and
2, write a paragraph that explain why Connor shouldn't expect the rule he remember from high school to work in this particular problem
I am able to correctly evaluate the limit ( lim = 0) but since my first language is not English, I don't understand what Connor claimed and how to explain it. Can anyone help? Thanks in advance
|
$$\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}-\frac{1}{x^3}}\right)=$$
Te expressions $\frac{3}{x^3},\frac{7}{x^2},\frac{6}{x},-\frac{1}{x^3},-\frac{2}{x^3},\frac{1}{x}$ all tend to zero as $x$ approaches $\infty$:
$$\frac{0+0+0}{2+0-0-0}=\frac{0}{2}=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1489105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
}
|
Integral : $\int \frac{\sqrt{9-x^2}}{x^2}$ I have the following integral :
$$\int \frac{\sqrt{9-x^2}}{x^2} = -\frac{\sqrt{9-x^2}}{x} - \arccos (\frac{x}{3})$$
But it seems like I must use $\arcsin$... Is there a difference between the two ?
|
First off the sign on the $\arccos(\frac{x}{3})$ is wrong.
I guess you could use the constant of integration to solve this. From the definitions of the inverse trig functions we know that
$\arccos x = \frac{\pi}{2} - \arcsin x$
So your antiderivative can be written as
\begin{equation}
-\frac{\sqrt{9 - x^2}}{x^2} + \arccos(\frac{x}{3}) = -\frac{\sqrt{9 - x^2}}{x^2} + \frac{\pi}{2} - \arcsin(\frac{x}{3})
\end{equation}
Since $-\frac{\sqrt{9 - x^2}}{x^2} + \frac{\pi}{2} - \arcsin(\frac{x}{3})$ and $-\frac{\sqrt{9 - x^2}}{x^2} - \arcsin(\frac{x}{3})$ have the same derivative(the constant $\frac{\pi}{2}$ does nothing) it means that they are both antiderivatives of your original function. This is why a $ + C$ is added at the end of the antiderivative's general form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1489950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Finding the locus in terms of a and b Find the locus of the point equidistant from $(a+b,b-a)$ and $(a-b,a+b)$
Okay so I don't know how do I find the locus of the point in terms of $ a$ & $b $.
|
Let $(x,y)$ be a point on the locus. Then:
$$(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$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1490259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
Velocity of a train A train travels with a velocity that doubles every hour. In 3 hours, it has travelled 350km. What was the trains velocity in the second hour?
A. 225 km/h
B. 175 km/h
C. 150 km/h
D. 100 km/h
Answer is D. Can someone explain?
|
Assuming that the intent is to solve this using linear algebra, here are the steps
$$ 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$$
Which can be rewritten as
$$x+y+z=350$$
$$2x-y+0z=0$$
$$0x+2y-z=0$$
The augmented matrix of this linear system is
$$\left[\begin{array}{ccc|c}
1 & 1 & 1 & 350 \\
2 & -1 & 0 & 0 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 1 & 1 & 350 \\
0 & -3 & -2 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & -3 & -2 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & -7 & 0 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & 1 & 0 & 100 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 0 & 0 & 50 \\
0 & 1 & 0 & 100 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 0 & 0 & 50 \\
0 & 1 & 0 & 100 \\
0 & 0 & 1 & 200 \\
\end{array}\right]$$
Which implies that
$$ x =50\ \mathrm{km}$$
$$ y =100\ \mathrm{km}$$
$$ z =200\ \mathrm{km}$$
Since $y$ represents the distance that the train traveled in the second hour, the answer then is $100 \frac{\mathrm{km}}{h}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1490717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$ $$\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}$$
WolframAlpha says this can be shown as $\sqrt[3]{3}$, but I can't find out how to do this.
|
Let $t=\sqrt[3]3$. Then, we have $t^3=3$, so
$$\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.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1490854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
geometric inequality involving lengths of hypotenuses In the diagram, $AD = AB + AC$ and all these lengths are positive. The following inequality holds:
$$DG < 2(BE + CF).$$
The constant 2 is sharp. I am able to prove the weaker inequality $DG < BE + 3 CF$ where $CF \ge BE$ but I am not able to prove the stronger inequality. Could someone please suggest an approach? I am not looking for a complete answer.
|
Using $a=AB$, $b=AC$, and $a+b=AD$, this is equivalent to the following
Lemma: For $a,b\gt0$,
$$
\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}}+\frac{b^2}{\sqrt{b^2+1}}\right)
$$
Substituting $a\mapsto at$ and $b\mapsto bt$, multiplying by $\frac2t$, and integrating from $0$ to $1$ yields
$$
\sqrt{(a+b)^2+1}-1\lt2\left(\sqrt{a^2+1}-1+\sqrt{b^2+1}-1\right)
$$
$\square$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1491673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
What is the maximum value of $ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}<x<\frac{2\pi}{3})$$
I have done my work here
$$f'(x)=\frac{2 \left(2 \sqrt{3} \cos (2 x)+\sqrt{3} \cos (4 x)-8 \sin ^3(x) \cos (x)\right)}{3 \left(\sin (x)+\sqrt{3} \cos (x)\right)^2}=0$$
I tried to solve problems but I couldn't make further progress
|
$$f(x) = \frac{1}{3}\cdot\frac{\sin(3x)}{\sin\left(x+\frac{\pi}{3}\right)} $$
is a negative function on the interval $\left(\frac{\pi}{3},\frac{2\pi}{3}\right)$, while $f\left(\frac{\pi}{3}\right)=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1493858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
|
$$\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)(x-3)}<0$$
The above inequality implies the following:
$$x\not = 1,x\not = -2,x\not = 3,x\not = -\frac{3}{4}$$
Now check the values of $x$ in the following intervals: $(-\infty,-2),(-2,-\frac{3}{4}),(-\frac{3}{4},1),(1,3),(3,+\infty)$
You will find that the inequality holds in the interval $(-2,-\frac{3}{4})\cup(1,3)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1495145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Solving line integral without using Green’s theorem Good morning everyone, I am a new user here and I have the following problem that I am dealing with since yesterday but I cannot find the correct answer. Here is the problem
Calculate
$$\oint_C (x^2-2xy)dx+(x^2y+3)dy$$
where $C$ is the closed curve of the region bounded by $y^2=8x$ and $x=2$.
Okay, using Green’s theorem, I got
$$\oint_C (x^2-2xy)dx+(x^2y+3)dy=\int_{-4}^4\int_{y^2/8}^{2}(2xy+2x)dxdy=\frac{128}{5}$$
but without using the theorem, I got
The the region bounded by $y^2=8x$ and $x=2$ intersect at $(2,-4)$ and $(2,4)$, hence
The line integral along $x=2$ equals
$$\int_{x=2}^{2}(x^2-2xy)dx+(x^2y+3)dy=0\tag{*}$$
and the line integral along $y^2=8x$ equals
$$\begin{align}\int_{y=4}^{-4}\left(\left(\frac{y^2}{8}\right)^2-2\left(\frac{y^2}{8}\right)y\right)d\left(\frac{y^2}{8}\right)+\left(\left(\frac{y^2}{8}\right)^2y+3\right)dy&=\int_{y=4}^{-4}\left(\frac{5y^5}{256}-\frac{y^4}{16}+3\right)dy\\&=\frac{8}{5}\end{align}$$
So, where is my mistake? Could someone here help me out? Thanks for your comments and answers, appreciate it.
|
Use parametrisation. That is the trick to do line integral problems.
Take $y=t$ and $x=\frac{t^2}{8}$ in Path- I and $x=2,dx=0$ in Path-II
So your line integral $$\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}[(\frac{t^5}{256}-\frac{t^4}{16}+\frac{t^5}{64}+3)dt]$$
$$=[-\frac{t^5}{64}]_{4}^{-4}$$ since the other functions are odd functions, so their integrals are $0$
$$=\frac{128}{5}$$
And $$\oint_2=\int_{-4}^{4}(4y+3)dy$$
$$=0$$ since the function is odd.
So your answer is $\oint=\frac{128}{5}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1496536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Definite Integrals with Substitution. How would you use the substitution: $x=1+sin\theta$ to evaluate $\int_{0}^{π/2}\frac{cos\theta}{(1+sin\theta)^3}d\theta$.
Furthermore what would you when changing the limits, since its a definite integral.
|
HINT:
$$\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}{u^3}\right)\text{d}u=$$
$$\int_{1}^{2}\left(u^{-3}\right)\text{d}u=$$
$$\left[-\frac{1}{2u^2}\right]_{1}^{2}=$$
$$-\frac{1}{2}\left[\frac{1}{u^2}\right]_{1}^{2}=$$
$$-\frac{1}{2}\left(\frac{1}{2^2}-\frac{1}{1^2}\right)=$$
$$-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{1}\right)=$$
$$-\frac{1}{2}\left(\frac{1}{4}-1\right)=$$
$$-\frac{1}{2}\left(-\frac{3}{4}\right)=$$
$$--\frac{1}{2}\cdot\frac{3}{4}=$$
$$\frac{1}{2}\cdot\frac{3}{4}=$$
$$\frac{1\cdot 3}{2\cdot 4}=$$
$$\frac{3}{8}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1496672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solving linear system equation How would I solve this linear system equation?
$$\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 & -1 & 0 & 4 \\
0 & -1 & 0 & 3 & 6 \\
0 & 3 & 0 & -9 & 1 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{cccc|c}
2 & 1 & -1 & 0 & 4 \\
0 & -1 & 0 & 3 & 6 \\
0 & 0 & 0 & 0 & 19 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}\right]$$
This system is inconsistent because we have
$$0w+0x+0y+0z=19$$
Therefore this system has no solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1497495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Evaluate $P(1)/P(-1)$
The polynomial $f(x)=x^{2007}+17x^{2006}+1$ has distinct zeroes $r_1,\ldots,r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j+\dfrac{1}{r_j}\right)=0$ for $j=1,\ldots,2007$. Determine the value of $P(1)/P(-1)$.
Let $P(x) = a_{2007}x^{2007} + ... + a_0$.
$r_j + \frac{1}{r_j} = \frac{r_j^2 + 1}{r_j}$
So we need the map: $x + \frac{1}{x} \to x$.
We need a $u \to x$. Where $u(x + 1/x) = x$.
The inverse function is: $u = \frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4})$ [USED WOLFRAM ALPHA]
So,
$P(x) = \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4}) \bigg)^{2007} + 17 \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4}) \bigg)^{2006} + 1$
But letting $x= 1$ or $-1$ makes it so that the answer is imaginary!
|
For first, $\frac{1}{r_i}+r_i = \frac{1}{r_j}+r_j$ is equivalent to $(r_i-r_j)(r_i r_j - 1)=0$. Since $f(x)$ is not a palyndromic polynomial, the set of numbers $r_i+\frac{1}{r_i}$ is exactly the set of roots of $P$.
Assuming that $P$ is a monic polynomial, Vieta's formulas give:
$$ P(1) = \prod_{i=1}^{2007}\left(1-r_i-\frac{1}{r_i}\right),\qquad P(-1)=-\prod_{i=1}^{2007}\left(1+r_i+\frac{1}{r_i}\right)\tag{1} $$
hence it follows that:
$$ \frac{P(1)}{P(-1)} = \prod_{i=1}^{2007}\frac{r_i^2-r_i+1}{r_i^2+r_i+1}=\prod_{i=1}^{2007}\frac{r_i-1}{r_i+1}\cdot\prod_{i=1}^{2007}\frac{r_i^3-1}{r_i^3+1}\tag{2} $$
where the first term of the RHS can be easily computed from $f(1)$ and $f(-1)$, and the whole expression just depends on the values of $f$ over the sixth roots of unity, due to $r_i^2+r_i+1 = (r_i-\omega^2)(r_i-\omega^4)$, where $\omega = \exp\left(\frac{2\pi i}{6}\right)$. Given $(2)$, it follows that:
$$ \frac{P(1)}{P(-1)} = \frac{f(\omega)\cdot f(\omega^5)}{f(\omega^2)\cdot f(\omega^4)}=\frac{17\omega^2\cdot (-17\omega)}{(2-17\omega)\cdot(2+17\omega^2)}=\color{red}{\frac{1}{\frac{297}{578}-\frac{i\sqrt{3}}{2}}}.\tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1500547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
If $\gcd(a,b,c)=1$ and $a^2 + b^2 = c^2$, prove $12|abc$ I know from $a^2+b^2=c^2$ that $a = st$, b = $\frac{s^2-t^2}{2}$, c = $\frac{s^2+t^2}{2}$ , where $s>t>1$, where $s$ and $t$ are both odd, but how do I make use of this information?
|
$s,t$ odd means $s^2\equiv t^2\equiv 1\mod 8$, hence $b$ is divisible by $4$. If neither $s$ nor $t$ is divisible by $3$, then $s^2\equiv t^2\equiv 1\mod 3$ and $3|b$. Otherwise $3|a$, in both cases $12|abc$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1503864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plane $x+y+z=1$
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plain $x+y+z=1$
Hi i was doing this question but i'm not sure i was right. Does this make sense?
Let
$$f(x) = x^2+y^2 = 1$$
$$f(y) = x+y+z=1, \quad \text{our constraint}$$
$$ \nabla f(x) = \lambda \nabla g(x) $$
$f_x = 2x = \lambda$ ..... (1)
$f_y = 2y = \lambda$ ...... (2)
$f_z = 0 = \lambda$
(1) = (2)
$2x = 2y = \lambda = 0 $
$x=y$
Sub into f(x)
$$f(x) = x^2 + x^2 = 1$$ $$f(x) = 2x^2 = 1$$ $$f(x) = x = \pm \frac{1}{\sqrt{2}} $$ $$ \therefore \, y = \pm \frac{1}{\sqrt{2}}$$
Points: $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}
\right), \quad \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$
Sub these points into g
$$g\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, z \right) = \frac{2}{\sqrt{2}} + z = 10 $$
$$ z = 1 - \frac{2}{\sqrt{2}} $$
$$g\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, z \right) = -\frac{2}{\sqrt{2}} + z = 10 $$
$$ z = 1 + \frac{2}{\sqrt{2}} $$
So highest and lowest points are....
$$\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1 - \frac{2}{\sqrt{2}} \right) \quad lowest$$
$$\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1 + \frac{2}{\sqrt{2}} \right) \quad highest$$
|
By inequality $2x^2+2y^2\geq (x+y)^2$ we indeed have $-\sqrt2 \leq x+y \leq \sqrt{2}$ and to maximize/minimize $z$ we need to minimize/maximize $x+y$ so your solution until $x=y=\pm{1\over\sqrt2}$ is correct.
However how do you get $z$ looks unreasonable as $x+y+z=1$ but your solution, for example ${1\over\sqrt2}+{1\over\sqrt2}+{1\over\sqrt8}$ does not equal $1$.
You should simply calculate $z=1-x-y=1+{\sqrt2}$ for maximum and $1-\sqrt2$ for minimum.
There are two major mistakes after a look at your way of finding z:
(1) $g(x,y,z)=1\neq10$ as stated in the question. I do not know where the $10$ comes from.
(2) $10-{2\over \sqrt2}$ is not $8\over\sqrt2$. You cannot calculate irrational numbers like this.
As to answer the comment, what Matthew is talking about is the general way of solving multiple constraints.
We want to maximize $h(x,y,z)=z$ using constraint $g(x,y,z)=x+y+z=1$ and $f(x,y,z)=s^2+y^2=1$. Let the first multiplier be $\lambda$ and the second be $u$
Then we have the system:
$0=\lambda+2ux=\lambda+2uy$
$1=\lambda$
$x^2+y^2=1$
$x+y+z=1$
By solving the equations we will find the value of $z$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1504489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Can all differential equations be solved using power series? To elaborate, does the differential equation have to be of some form to be solvable by power series. More specifically, if I wanted to solve this equation by power series would I be able to?
$$
(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)=0
$$
Also, $n$ is a constant in the stated equation.
|
$(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}^{\infty} k(k-1)a_k x^{k-2}
=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
$
so
$\begin{array}\\
(1-x^2)y''(x)
&=(1-x^2)\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}\\
&=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
-x^2\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}\\
&=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
-\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k+2}\\
&=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
-\sum_{k=2}^{\infty} (k-1)ka_{k} x^{k}\\
&=2a_2+6a_3x
+\sum_{k=2}^{\infty} ((k+1)(k+2)a_{k+2}-(k-1)ka_{k}) x^{k}\\
\end{array}
$.
Adding these all up
$\begin{array}\\
0
&=(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)\\
&=n(n+1)(a_0+a_1x)-3(a_1x)+2a_2+6a_3x
+\sum_{k=2}^{\infty} x^k\left( n(n+2)a_k-3ka_{k}+((k+1)(k+2)a_{k+2}-(k-1)ka_{k})\right)\\
&=n(n+1)a_0+2a_2+x(n(n+1)a_1-3a_1+6a_3)
+\sum_{k=2}^{\infty} x^k\left( (n(n+2)-3k-k(k-1))a_k+(k+1)(k+2)a_{k+2})\right)\\
&=n(n+1)a_0+2a_2+x(n(n+1)a_1-3a_1+6a_3)
+\sum_{k=2}^{\infty} x^k\left( (n(n+2)-k(k+2))a_k+(k+1)(k+2)a_{k+2}\right)\\
\end{array}
$
Therefore,
assuming no errors
($P(\text{no errors}) < 1-1/e)$),
$0
=n(n+1)a_0+2a_2
$,
$0
=(n(n+1)-3)a_1+6a_3
$,
and
$0
= (n(n+2)-k(k+2))a_k+(k+1)(k+2)a_{k+2}
$.
These recurrences
show how to compute
the $a_k$
in terms of
$a_0$ and $a_1$.
Note that
$y(x)$ is the sum of
two independent series,
one with even exponents
and one with odd.
The last recurrence has
an obvious problem
at $k=n$,
so LutzL's suggestion
of multiplying the series
by $x^{\alpha}$
might have to be done.
I'll leave it at this.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1510076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
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 contradiction.
Assume that $y^2 \gt 2$, so there exists $z \gt 0$ such that $y^2 − 4z \gt 2$.
Then:
$$ (y − z)^2 = y^2 − 2zy + z^2 > y^2 − 4z > 2 \tag{1} $$
Therefore $y − z$ is also an upper bound of $A$ and $y$ cannot be the least upper bound.
If $y^2 \lt 2$, so there exists $c \gt 0$ such that $y^2 + 5c \lt 2$ and $c^2 \lt c$.
Then:
$$ (y + c)^2 = y^2 + 2cy + c^2 < y^2 + 5c < 2. \tag{2} $$
Therefore also $y + c \in A$ and $y$ cannot be the least upper bound.
I apologize for this basic question. The inequalities labelled 1 and 2 appear to be true to me, but I can't see why they are true logically. I don't see the "then" logical link.
|
The expressions (1) and (2) might be easier to follow if they are expanded to
$$\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{align}$$
Note, both expansions use the fact that $y\le2$ (i.e., $2$ is an upper bound for the set $A$); the second makes explicit use of the condition $c^2\lt c$. The key idea is the insertion of $0$, in the form $-4z+4z$ and $5c-5c$, respectively.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1511364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Prove $\frac{a+b}{\sqrt{c}}+\frac{a+c}{\sqrt{b}}+ \frac{b+c}{\sqrt{a}} \geq 2(\sqrt{a} + \sqrt{b} +\sqrt{c})$ Could anyone advise me on how to prove this inequality:
$$\dfrac{a+b}{\sqrt{c}}+\dfrac{a+c}{\sqrt{b}}+ \dfrac{b+c}{\sqrt{a}} \geq 2(\sqrt{a} + \sqrt{b} +\sqrt{c}),$$
where $a,b,c $ are any positive real numbers.
Do I use the AM-GM inequality somewhere?
Thank you.
|
From AM-GM you have
$$ \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{b}}+ \dfrac{b+c}{\sqrt{a}} \right) \geq 6( \sqrt{a} + \sqrt{b} +\sqrt{c} )$$
Simplifying gives the desired inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1512074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
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 irreducible in $R$ if whenever $r = ab$ with $a,b\in R$, at least one $a$ or $b$ must be a unit in $R$.
Then let $2 = xy$. Then taking the norm, we have $4 = N(x)N(y)$.
Then we need to show that $N(x)=1$ and $N(y) = 4$ or $N(y)=1$ and $N(x) = 4$.
Suppose $x = a + b\sqrt{-n}$. Then assume $N(x) = 2$. Then $x = a + b\sqrt{-n}$ implies $N(x)= 2 = a^2 + nb^2$ . Since $n \ge 3$, then equation only makes sense if $b = 0$ and $a = \sqrt2$ . So we have a contradiction, so $N(x) $ = $4$ and $N(y)$ must be $1$, so that $y$ is a unit.
So $2$ is irreducible.
Now for $\sqrt{-n}$, suppose $\sqrt{-n} = xy$. Then taking the norm, we have, $n = N(x)N(y)$. Then let $\sqrt{-n} = xy = (a + b\sqrt{-n})(c + d\sqrt{-n})$. Then $n = (a^2 + nb^2)(c^2 + nd^2)$ .
Can someone please help me for $\sqrt{-n}$ ? I am stuck.
Thank you!
|
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$. Consequently $c=0$ and $a=d=\pm 1$. Which again gives a trivial factorization.
Note: In the second case $d \neq 0$, otherwise $n$ will not be square free.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1512399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
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 and minimum values of $|\vec{PA}||\vec{PB}|$ are $M$ and $m$ respectively then prove that the value of $M^2+m^2=34$
My Attempt:
Let the position vector of $P$ be $x\hat{i}+y\hat{j}$.Then $\vec{PA}=(1-x)\hat{i}-y\hat{j}$ and $\vec{PB}=(-1-x)\hat{i}-y\hat{j}$
$\vec{OA}=\hat{i},\vec{OB}=-\hat{i}$
$(\vec{PA}.\vec{PB})+3(\vec{OA}.\vec{OB})=0$ gives
$x^2-1+y^2-3=0$
$x^2+y^2=4$
$|\vec{PA}||\vec{PB}|=\sqrt{(1-x)^2+y^2}\sqrt{(-1-x)^2+y^2}=\sqrt{(x^2+y^2-2x+1)(x^2+y^2+2x+1)}$
$|\vec{PA}||\vec{PB}|=\sqrt{(5-2x)(5+2x)}=\sqrt{25-4x^2}$
I found the domain of $\sqrt{25-4x^2}$ which is $\frac{-5}{2}\leq x \leq \frac{5}{2}$.Then i found the minimum and maximum values of $\sqrt{25-4x^2}$ which comes out to be $M=5$ and $m=0$.So $M^2+m^2=25$
But i need to prove $M^2+m^2=34$.Where have i gone wrong?Please help me.Thanks.
|
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 $\displaystyle \theta =0,2\pi.$
and $\displaystyle \max\left[f(\theta)\right] = \sqrt{25-0} = 5$ at $\displaystyle \theta =\frac{\pi}{2}\;\ \frac{3\pi}{2}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1514094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
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}$$
Im stuck here !
|
I'm not a math major so I'm not sure about the formalism but does this work?
If a!=b,
Then a=b+d ; where d is a non zero integer.
Plug in to the expression (a^3+a)-(b^3+b) :-
3bd^2+3db^2+d^3.
This has only one real root which is d=0, which is a contradiction, so the two expressions cannot be equal.
Too informal? Correct or incorrect, math majors?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1516745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Integral $\sqrt{1+\frac{1}{4x}}$ $$\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}\ $$
How would one get this answer by integrating manually?
|
Assume $x>0.$ Let $u=\sqrt{x},$ then $u^{2}=x,$ $dx=2udu,$ and $1+\frac{1}{4x%
}=1+\frac{1}{4u^{2}}$
\begin{equation*}
\int \sqrt{1+\frac{1}{4x}}dx=\int \sqrt{1+\frac{1}{4u^{2}}}(2u)du=\int \sqrt{%
(2u)^{2}+1}du.
\end{equation*}
Now, substitution trigonometric, $2u=\tan \theta ,$ $du=\frac{1}{2}\sec
^{2}\theta d\theta ,$ and $\sqrt{(2u)^{2}+1}=\sec \theta ,$ hence
\begin{equation*}
\int \sqrt{(2u)^{2}+1}du=\int (\sec \theta )\frac{1}{2}\sec ^{2}\theta
d\theta =\frac{1}{2}\int \sec ^{3}\theta d\theta .
\end{equation*}
Then, we use the well-known integral
\begin{equation*}
\int \sec ^{3}\theta d\theta =\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}%
\ln \left\vert \sec \theta +\tan \theta \right\vert +2C
\end{equation*}
it follows that
\begin{eqnarray*}
\int \sqrt{1+\frac{1}{4x}}dx &=&\frac{1}{4}\sec \theta \tan \theta +\frac{1}{%
4}\ln \left\vert \sec \theta +\tan \theta \right\vert +C \\
&=&\frac{1}{4}(\sqrt{(2u)^{2}+1})(2u)+\frac{1}{4}\ln \left\vert \sqrt{%
(2u)^{2}+1}+2u\right\vert +C \\
&=&\frac{1}{2}u\sqrt{1+4u^{2}}+\frac{1}{4}\ln \left\vert \sqrt{(2u)^{2}+1}%
+2u\right\vert +C \\
&=&\frac{1}{2}\sqrt{x}\sqrt{1+4x}+\frac{1}{4}\ln \left\vert \sqrt{1+4x}+2%
\sqrt{x}\right\vert +C.\ \ \ \ \blacksquare
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1516828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
}
|
Can someone explain the integration of $\sqrt{v²+\tfrac14}$ to me? I am currently trying to integrate this root:
$$\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}$$
However, I just can't get my head around it. I have absolutely no idea where that $\operatorname{arsinh}$ is coming from.
My take on this would have been:
$$f = \left(v^2+\frac{1}{4}\right)^{\frac{1}{2}}\implies F = \frac{2}{3}\cdot\frac{1}{2v}\left(v^2+\frac{1}{4}\right)^{\frac{3}{2}}$$
|
People don't seem comfortable going directly to the hyperbolic trig functions, but that is simplest here.
$$ 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{1}{4}} \; dv = \frac{1}{4} \cosh^2 t \; dt} $$
Next we want the double "angle" formulas. In general,
$$ \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y $$
$$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y $$
so
$$ \cosh 2 t = \cosh^2 t + \sinh^2 t = 2 \cosh^2 t - 1, $$ and
$$ \color{red}{ \cosh^2 t = \frac{1 + \cosh 2t}{2}}. $$
$$ \color{magenta}{ \sqrt{v^2 + \frac{1}{4}} \; dv = \frac{1 + \cosh 2t}{8} \; dt} $$
The integral $dt$ is
$$ \color{green}{ \frac{t}{8} + \frac{\sinh 2t}{16}.} $$
We have $$ t = \operatorname{arsinh} 2v. $$ We already found
$$ \cosh t = 2 \sqrt{v^2 + \frac{1}{4}} $$ and
$$ \sinh t = 2 v, $$ whence
$$ \sinh 2t = 2 \sinh t \cosh t = 8 v \, \sqrt{v^2 + \frac{1}{4}} $$
Put them together, the integral $dv$ is
$$ \color{green}{ \frac{ \operatorname{arsinh} 2v}{8} + \frac{ v \, \sqrt{v^2 + \frac{1}{4}}}{2}.} $$
I looked on wikipedia, they say there is some preference for arsinh over both arcsinh and argsinh. News to me.
|
{
"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
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.