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How to show that this difference of products is $O \left( \frac{1}{n^2} \right) $ Let $k \leq n$. Consider the following difference of products:
$$ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right)$$
For $n=1,2,3$, this is clearly $O \left( \frac{1}{n^2} \right) $ for $k=1,2,3$ respectively. How to show it for arbitrary $n$? And is it possible to give an estimate $C$ such that the difference is less than $\frac{C}{n^2}$?
| $$\begin{align}\prod_{i=1}^{k-1}{\left (1-\frac{i}{n+1} \right )} &= \prod_{i=1}^{k-1}{\left (1-\frac{i}{n} + \frac{i}{n^2} \right )} + O \left (\frac1{n^3} \right )\\ &= \prod_{i=1}^{k-1}{\left (1-\frac{i}{n} \right )} + \sum_{j=1}^{k-1}\frac{j}{n^2} \prod_{i=1,i \ne j}^{k-1}{\left (1-\frac{i}{n} \right )}+ O \left (\frac1{n^3} \right )\\ &=\prod_{i=1}^{k-1}{\left (1-\frac{i}{n} \right )} + \sum_{j=1}^{k-1}\frac{j}{n^2}+O \left (\frac1{n^3} \right )\\ &=\prod_{i=1}^{k-1}{\left (1-\frac{i}{n} \right )} + \frac{k(k-1)}{2 n^2}+O \left (\frac1{n^3} \right )\end{align}$$
Thus the difference is, when $k \lt n$,
$$\frac{k(k-1)}{2 n^2}+O \left (\frac1{n^3} \right )$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
| We have
$$\sqrt{x^2+5} - \sqrt{x^2-3} = 2
\sqrt{x^2+5} = 2+ \sqrt{x^2-3}\\
x^2+5 = 4+ x^2-3+4\sqrt{x^2-3}\\
4 = 4\sqrt{x^2-3}\\
1 = \sqrt{x^2-3}\\
1 = x^2 -3\\
x^2 = 4\\
x = \pm 2
$$
So that
$$\sqrt{x^2+5} + \sqrt{x^2-3} = 3 + 1 = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integrate a quotient with fractional power of a quadratic polynomial I need help finding the indefinite integral of
$$\int\,\frac{x}{(7x - 10 - {x^2})^{3/2}}\,\text{d}x\,.$$
| We have
$$
\int \frac{x}{(7x - 10 - x^2)^{3/2}}\,dx.
$$
Let $u=-x^2 +7x-10$, so that $du = (-2x+7)\,dx$ and $\dfrac{-du} 2 = \left( x - \dfrac 7 2 \right)\, dx$. Then the integral becomes
$$
\int\frac{x - \frac 7 2}{(7x-10+x^2)^{3/2}} \,dx + \frac 7 2 \int \frac{dx}{(7x-10+x^2)^{3/2}}
$$
The substitution above handles the first of these two integrlas; the one below handles the second:
\begin{align}
& -x^2+7x-10 = \overbrace{- (x^2-7x)-10 = -\left( x^2-7x + \frac{49}4 \right)^2 -10 + \frac{49}4}^{\text{completing the square}} \\[10pt]
= {} & -\left( x - \frac 7 2 \right)^2 + \left(\frac 3 2 \right)^2 = \left( \frac 3 2\right)^2 \left( 1 - \left( \frac 2 3 \left( x - \frac 7 2 \right) \right)^2 \right) \\[10pt]
= {} & \left( \frac 3 2 \right)^2 \left( 1 - \left( \frac{2x-7} 3 \right)^2 \right) = \left( \frac 3 2 \right)^2 (1-\sin^2\theta)
\end{align}
So
\begin{align}
\frac{2x-7} 3 & = \sin\theta \\[10pt]
\frac 2 3 \, dx & = \cos\theta\,d\theta
\end{align}
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
find the complex number $z^4$
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
Find $c+d$.
I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance between those two points.
$|z^2| = 25$ means since: $z^2 = a^2 - b^2 + 2abi$ that:
$625 = (a^2 - b^2)^2 + 4a^2b^2 = a^4 + b^4 + 2a^2b^2$
But that doesnt help much.
| If $z=5e^{i\theta}$, and we use $1+2i=\sqrt5e^{i\phi}=z_0$, then we have
$$|z^5-z^3z_0|=125|z^2-z_0|$$
as the thing to maximise, but this implies we need $z^2$ on the diameter through $z_0$, so we need $\theta=\frac{\phi+\pi}{2}$.
Then, $$z^4=625e^{4i\theta}=625e^{2i\phi+2i\pi}=625e^{2i\phi}=625\cos(2\phi)+625i\sin(2\phi)$$
Knowing that $\phi=\arctan(2)$ Provides a direct route to the numerical answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $x^2+y^2$ is constant for all values of $\theta$. Given that $x=3\sin \theta-2 \cos \theta$ and $y=3\cos \theta+2 \sin \theta$
i)Find the value of the acute angle $\theta$ for which $x=y$
ii)Show that $x^2+y^2$ is constant for all values of $\theta$.
My attempt,
$3\sin \theta-2 \cos \theta=3 \cos \theta+2 \sin \theta$
$\sin \theta=5 \cos \theta$
$\sin \theta-5 \cos \theta=0$
$\tan \theta-5=0$
$\theta=78.7$
How to solve question ii?
| hint: $x^2+y^2 = 9(\cos^2 \theta+\sin^2\theta)+4(\sin^2\theta + \cos^2\theta) = ....$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coefficient of binomial expansion The coefficient of $x^3$ is $4$ times the coefficient of $x^2$ in the new expansion of $(1+x)^n$. Find the value of $n$.
| Notice, we have
$$(1+x)^n=^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+\ldots +^nC_n.1^0.x^n$$ Given condition $$\color{red}{\text{coefficient of}\ x^3=4\times \text{coefficient of}\ x^2}$$ $$\implies ^nC_{3}=4\times ^nC_{2}$$ $$\frac{n!}{(n-3)!3!}=\frac{4\times n!}{(n-2)!2!}$$ $$\frac{1}{3}=\frac{4}{(n-2)}$$ $$n-2=12\implies n=14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}(xy+yz+zx)$
(D)$\frac{1}{2}(x+y+z)\sqrt{x^2+y^2+z^2}$
I tried applying Heron's formula but calculations are very messy and simplification is difficult.I could not think of any other method to find this area.Can someone assist me in solving this problem.
| There is such a triangle $\triangle\subset{\mathbb R}^3$ with vertices
$${\bf 0}=(0,0,0), \quad{\bf a}:=(x,y,0), \quad {\bf b}:=(0,y,z)\ .$$
Its area is given by
$${\rm area\,}(\triangle)={1\over2}\bigl|{\bf a}\times{\bf b}\bigr|={1\over2}\sqrt{y^2z^2+z^2x^2+x^2y^2}\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Minimum value of reciprocal squares I am bit stuck at a question. The question is : given: $x + y = 1$, $x$ and $y$ both are positive numbers. What will be the minimum value of: $$\left(x + \frac{1}{x}\right)^2 + \left(y+\frac{1}{y}\right) ^2$$
I know placing $x = y$ will give the right solution. Is there any other solution?
| We have $(x+1/x)^2+(y+1/y)^2=x^2+y^2+1/x^2+1/y^2+4=x^2+y^2+\frac{x^2+y^2}{x^2y^2 }+4\geq 1/2+16/2+4=25/2$. Here note that $x+y=1$ implies $xy\leq 1/4$ then $1/(xy)^2\geq 16$ and $x+y=1$ implies $x^2+y^2=1-2xy\geq 1-(x^2+y^2)$ so $x^2+y^2\geq 1/2$. For $x=y=1/2$ we have equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Proving an Inequality using a Different Method Is there another way to prove that: If $a,b\geq 0$ and $x,y>0$
$$\frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$$
using a different method than clearing denominators and reducing to $(ay-bx)^2 \ge 0$?
| $\frac{a^2}{x} + \frac{b^2}{y}
\ge \frac{(a+b)^2}{x+y}
$
Let
$u = a^2/x$
and
$v = b^2/y$,
so
$a = \sqrt{ux}$
and
$b = \sqrt{vy}$.
The inequality then becomes
$u+v
\ge \frac{(\sqrt{ux}+\sqrt{vy})^2}{x+y}
= \frac{ux+vy+2\sqrt{uxvy}}{x+y}
$
or
$ux+uy+vx+vy
\ge ux+vy+2\sqrt{uxvy}
$
or
$uy+vx
\ge 2\sqrt{uxvy}
$
or
$(\sqrt{uy}-\sqrt{vx})^2
\ge 0
$.
I don't know different this really is,
since it does clear the denominator,
but it looks different.
Anyway,
most inequalities usually
are equivalent to
$(something)^2 \ge 0
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Sums of Fourth Powers While fooling around on my calculator I found:
$$7^4 + 8^4 + (7 + 8)^4 = 2 * 13^4$$
$$11^4 + 24^4 + (11 + 24)^4 = 2 * 31^4$$
I'm intrigued but I can't explain why these two equations are true. Are these coincidences or is there a formula/theorem explaining them?
| I'll start at
Will Jagy's hint.
If
$c^2
=a^2+ab+b^2
$,
$\begin{array}\\
c^4
&=a^4+a^2b^2+b^4+2a^3b+2a^2b^2+2ab^3\\
&=a^4+3a^2b^2+b^4+2a^3b+2ab^3\\
\text{so}\\
2c^4
&=2a^4+6a^2b^2+2b^4+4a^3b+4ab^3\\
&=a^4+b^2+a^4+6a^2b^2+b^4+4a^3b+4ab^3\\
&=a^4+b^4+a^4+4a^3b+6a^2b^2+4ab^3+b^4\\
&=a^4+b^4+(a+b)^4\\
\end{array}
$
Yep.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving $\sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n}$ with induction Problem: Prove with induction that \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n} \end{align*} for every natural number $n \geq 2$.
Attempt at proof: Basic step: For $n = 2$ we have $1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ which is correct.
Induction step: Suppose the assertion holds for some natural number $n$, with $n > 2$. Then we now want to prove that it also holds for $n +1$, i.e. that \begin{align*} \sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} > \sqrt{n+1} \end{align*}
Now we have that \begin{align*} \sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} = \sum_{j=1}^n \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{n+1}} > \sqrt{n} + \frac{1}{\sqrt{n+1}} \end{align*} or \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{n+1}} > \frac{\sqrt{n} \sqrt{n+1} + 1}{\sqrt{n+1}} \end{align*}
Now I'm stuck, and I don't know how to get $\sqrt{n+1}$ on the right hand side. Help would be appreciated.
| Notice that :
$ \sqrt{n}\sqrt{n+1} + 1 > \sqrt{n}\sqrt{n} + 1 $
$\sqrt{n}\sqrt{n+1} + 1 > n + 1 $
$ \sqrt{n}\sqrt{n+1} + 1 > \sqrt{n+1}\sqrt{n+1} $
$ \frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}} > \sqrt{n+1}. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$
Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit?
Limit should be $L=4$
| Multiplying it by $$\frac{{1+\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\cdot\frac{2+\sqrt{4-x^2}}{{2+\sqrt{4-x^2}}}\ (=1)$$gives$$\begin{align}&\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (1-\sqrt{1-x^2})}{\sqrt{1-x}\ (2-\sqrt{4-x^2})}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ \color{red}{(1-\sqrt{1-x^2})}}{\sqrt{1-x}\ \color{blue}{(2-\sqrt{4-x^2})}}\cdot\frac{\color{red}{1+\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\cdot\frac{2+\sqrt{4-x^2}}{\color{blue}{2+\sqrt{4-x^2}}}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (2+\sqrt{4-x^2})\color{red}{(1-(1-x^2))}}{\sqrt{1-x}\ (1+\sqrt{1-x^2})\color{blue}{(4-(4-x^2))}}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (2+\sqrt{4-x^2})}{\sqrt{1-x}\ (1+\sqrt{1-x^2})}\\&=\frac{\sqrt 4\ (2+\sqrt 4)}{1\cdot (1+1)}\\&=4\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
last two digits of $14^{5532}$? This is a exam question, something related to network security, I have no clue how to solve this!
Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?
| Finding the last two digits necessarily implies $\pmod{100}$
As $(14^n,100)=4$ for $n\ge2$
Let use start with $14^{5532-2}\pmod{100/4}$ i.e., $14^{5530}\pmod{25}$
As $14^2\equiv-2^2\pmod{25}$
Now $2^5\equiv7,2^{10}\equiv7^2\equiv-1\pmod{25}$
$\implies14^{10}=(14^2)^5\equiv(-2^2)^5=-2^{10}\equiv-1(-1)\equiv1$
As $5530\equiv0\pmod{10},14^{5530}\equiv14^0\pmod{25}\equiv1$
Now use $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod {m\cdot c} $
$\displaystyle14^{5530}\cdot14^2\equiv1\cdot14^2\pmod{25\cdot14^2}$
As $100|25\cdot14^2,$
$\displaystyle14^{5530+2}\equiv14^2\pmod{100}\equiv?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Prove identity: $\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}$ Prove identity: $$\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}.$$
My work this far:
we take the left side
$$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac{1-\cos2\alpha}{2}}+\sqrt{\frac{1+\cos2\alpha}{2}}}=$$
then after some calulations I come to this
$$=\frac{\sqrt{2} +\sqrt{1-\cos2\alpha}-\sqrt{1+\cos2\alpha}}{\sqrt{2} +\sqrt{1-\cos2\alpha}+\sqrt{1+\cos2\alpha}}$$ but now I'm stuck...
| $\bf{My\; Solution::}$ Given $\displaystyle \frac{1+\sin \alpha-\cos \alpha}{1+\sin \alpha+\cos \alpha} = \frac{\left(1-\cos \alpha\right)+\sin \alpha}{\left(1+\cos \alpha\right)+\sin \alpha}$
$\displaystyle = \frac{2\sin^2 \frac{\alpha}{2}+2\sin \frac{\alpha}{2}\cdot \cos \frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}+2\sin \frac{\alpha}{2}\cdot \cos \frac{\alpha}{2}} = \tan \frac{\alpha}{2}$
Above We have used the formula
$\star 1 + \cos \alpha = 2\cos^2 \left(\frac{\alpha}{2}\right)$
$\star 1 - \cos \alpha = 2 \sin^2 \left(\frac{\alpha}{2}\right)$
$\star \sin \alpha = 2 \sin \left(\frac{\alpha}{2}\right)\cos \left(\frac{\alpha }{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Question about Euler's formula I have a question about Euler's formula
$$e^{ix} = \cos(x)+i\sin(x)$$
I want to show
$$\sin(ax)\sin(bx) = \frac{1}{2}(\cos((a-b)x)-\cos((a+b)x))$$
and
$$ \cos(ax)\cos(bx) = \frac{1}{2}(\cos((a-b)x)+\cos((a+b)x))$$
I'm not really sure how to get started here.
Can someone help me?
| $$\sin { \left( ax \right) } \sin { \left( bx \right) =\left( \frac { { e }^{ aix }-{ e }^{ -aix } }{ 2i } \right) \left( \frac { { e }^{ bix }-{ e }^{ -bix } }{ 2i } \right) } =\frac { { e }^{ \left( a+b \right) ix }-e^{ \left( a-b \right) ix }-{ e }^{ \left( b-a \right) ix }+{ e }^{ -\left( a+b \right) ix } }{ -4 } \\ =-\frac { 1 }{ 2 } \left( \frac { { e }^{ \left( a+b \right) ix }+{ e }^{ -\left( a+b \right) ix } }{ 2 } -\frac { { e }^{ \left( a-b \right) ix }+{ e }^{ -\left( a-b \right) ix } }{ 2 } \right) =\frac { 1 }{ 2 } \left( \cos { \left( a-b \right) x-\cos { \left( a+b \right) x } } \right) $$
same method you can do with $\cos { \left( ax \right) \cos { \left( bx \right) } } $
Edit:
$$\int { \sin { \left( ax \right) \sin { \left( bx \right) } } dx=\frac { 1 }{ 2 } \int { \left[ \cos { \left( a-b \right) x-\cos { \left( a+b \right) x } } \right] dx=\quad } } $$$$\frac { 1 }{ 2 } \int { \cos { \left( a-b \right) xdx } } -\frac { 1 }{ 2 } \int { \cos { \left( a+b \right) xdx= } } $$
now to order calculate $\int { \cos { \left( a+b \right) xdx } } $ write
$$t=\left( a+b \right) x\quad \Rightarrow \quad x=\frac { t }{ a+b } \quad \Rightarrow dx=\frac { 1 }{ a+b } dt\\ \int { \cos { \left( a+b \right) xdx=\frac { 1 }{ a+b } \int { \cos { \left( t \right) } dt=\frac { 1 }{ a+b } \sin { \left( t \right) = } } \frac { 1 }{ a+b } \sin { \left( a+b \right) x } } } +C\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?
| Hint:
Let, each side of the regular polygon $A_1A_2A_3\ldots A_n$ be $a$ then we have $$A_1A_2=a$$
$$A_1A_3=2a\sin\left(\frac{(n-2)\pi}{2n}\right)$$
$$A_1A_4=a-2a\cos\left(\frac{(n-2)\pi}{n}\right)$$
Now, we have
$$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$$
$$\frac{1}{a}=\frac{1}{2a\sin\left(\frac{(n-2)\pi}{2n}\right)}+\frac{1}{a-2a\cos\left(\frac{(n-2)\pi}{n}\right)}$$
Let, $\frac{(n-2)\pi}{2n}=\alpha$ then we have
$$1-\frac{1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$
$$\frac{1-2\cos2\alpha-1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$
$$8\sin^3\alpha-4\sin^2\alpha-4\sin\alpha+1=0$$
I hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Binomial expansion. Find the coefficient of $x$ in the expansion of $\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$.
I've used the way that my teacher teach me.
I've stuck in somewhere else.
$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6=\left(2-\frac{4}{x^3}\right)\left(x^6\left(1+\frac{2}{x^8}\right)^6\right)$
Can anyone teach me? Thanks in advance.
| In case you're interested, here's another way of doing this kind of question: remove fractional terms to the outside first...
$$(2-\frac{4}{x^2})(x+\frac{2}{x^2})^6=\frac{1}{x^{14}}(2x^2-4)(x^3+2)^6$$
So now you just need the coefficient of $x^{15}$ in the expansion of $$(2x^2-4)(x^3+2)^6$$ towards which the $2x^2$ term makes no contribution, and we just get $$-4\times\binom65\times 2=-48$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is
*
*Perpendicular to line $A$.
*Parallel to line $A$.
Attempting to find the parallel,
I tried $$y = -\frac{1}{2}x + \frac{1}{6}$$
$$y - (1) = -\frac{1}{2}(x-5)$$
$$Y = -\frac{1}{2}x - \frac{1}{10} - \frac{1}{10}$$
$$y = -\frac{1}{2}x$$
| Notice, in general, the equation of the line passing through the point $(x_1, y_1)$ & having slope $m$ is given by the point-slope form: $$\color{blue}{y-y_1=m(x-x_!)}$$
We have, equation of line A: $3x+6y-1=0\iff \color{blue}{y=-\frac{1}{2}+\frac{1}{6}}$ having slope $-\frac{1}{2}$
1.) slope of the line passing through $(5, 1)$ & perpendicular to the line A $$=\frac{-1}{\text{slope of line A}}=\frac{-1}{-\frac{1}{2}}=2$$ Hence, the equation of the line: $$y-1=2(x-5)$$ $$y-1=2x-10$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{2x-y-9=0}}$$
2.) slope of the line passing through $(5, 1)$ & parallel to the line A $$=\text{slope of line A}=-\frac{1}{2}$$ Hence, the equation of the line: $$y-1=-\frac{1}{2}(x-5)$$ $$2y-2=-x+5$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+2y-7=0}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$
Obtain Negative Reciprocal:
$${ m'=\frac{-10}{3}}$$
Get Midpoint fox X
$${ \frac{-6-4}{2} = -5 }$$
Get Midpoint for Y
$${ \frac{-0--3}{2} = \frac{3}{2} }$$
Make Point Slope Form:
$${ y = m'x +b = \frac{-10}{3}x + b}$$
Plugin Midpoints in Point Slope Form
$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$
Evaluate b
$${ b = \frac{109}{6}}$$
Get Equation and Simplify
$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$
Is the problem set wrong? What am I doing wrong?
| First you computed midpoint of $X$ wrongly.
Later you evaluated $b$ wrongly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Calculate the volume bounded by the surfaces Calculate the volume of the solid bounded by the surfaces $$\begin{aligned}z&=4x^2+4y^2, \\ z&=x^2+y^2, \\z&=4.\end{aligned}$$
I made an equation of $4x^2+4y^2=4-x^2+y^2$ and solved it to get $x^2+y^2=\dfrac{4}{5}$.
Then I did a double integration
$$\displaystyle
\iint_{x^2+y^2\leq \frac{4}{5}} \left(4x^2+4y^2\right)-\left(4-x^2-y^2\right)dA,\
$$
did the subtraction and the changed to polar
$$\displaystyle
\int_0^{2\pi} \int_0^{\frac{4}{5}} \left(3r^2-4\right) r \,dr \,d\theta,\
$$
got a result of
$$
\displaystyle\int_0^{ 2\pi}\Bigg[ \left.\left(\frac{3}{4}r^4-2r^2\right) \right\vert_{r=0}^{\frac{4}{5}} \Bigg]d\theta
$$
Obviously this is not correct. Can you tell me where I have gone wrong?
Thanks in advance :)
| Three volumes.
Say $r^2 = x^2 + y^2$, cut through the volume gives the following image.
Cylinder #1
$$
V_{C_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^4 dz r = 16 \pi
$$
Paraboloid #1
$$
V_{P_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^{r^2} dz r = 8 \pi
$$
Cylinder #2
$$
V_{C_1} = \int_0^{2\pi} d\phi \int_0^1 d r \int_0^4 dz r = 4 \pi
$$
Paraboloid #2
$$
V_{P_1} = \int_0^{2\pi} d\phi \int_0^1 d r \int_0^{4 r^2} dz r = 2 \pi
$$
End volume is
$$
\Big( V_{C_1} - V_{P_1} \Big) - \Big( V_{C_2} - V_{P_2} \Big) = 6 \pi.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Branch cuts for the contour integral $\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx$ How can the branch cut be handled in the contour integral, for $|b| \leq 1, \, a > 1$,
$$\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx \quad ?$$
If $a=1$ can the value of the integral be shown to be zero?
Edit: The contour involves a branch cut from $-a + \sqrt{a^{2}-1}$ to $\infty$ along the $x$-axis, a branch cut from $- a - \sqrt{a^{2}-1}$ to $- \infty$ along the $x$-axis. There are two poles $- z_{b} = e^{i(\theta_{b} - \pi)}$ and $- \frac{1}{z_{b}} = e^{-i(\theta_{b}+\pi)}$ where $\theta_{b} = \cos^{-1}(b)$. The resulting proposed value is:
\begin{align}
\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx = \frac{2 \pi}{\sqrt{1-b^{2}}} \, \tan^{-1}\left[ \frac{\sqrt{1-b^{2}} \, (a - \sqrt{a^{2}-1})}{1 - b \, (a - \sqrt{a^{2}-1})} \right].
\end{align}
| The problem of the integrand is that, if we use the principal branches then the singular points somehow overlap, which makes it hard to deal with directly. So if you want to apply complex analysis technique, we need some modification.
Notice that the integral
$$ I = \mathrm{PV}\!\int_{-1}^{1} \frac{\log(x+a)}{(x+b)\sqrt{1-x^2}} \, dx $$
exists in principal value sense, due to the presence of the pole at $x = -b$. One way to circumvent this problem is to move the pole slightly away from the line of integration. So we introduce the following function:
$$ I(w) = \int_{-1}^{1} \frac{\log(x+a)}{(x+w)\sqrt{1-x^2}} \, dx. $$
Then we can express $I$ as
$$ I = \lim_{s \downarrow 0} \frac{I(b+is) + I(b-is)}{2}. \tag{1} $$
So it suffices to deal with $I(w)$ when $w$ lies outside the line $[-1, 1]$. Now choose a contour $C$ that winds $[-1, 1]$ counter-clockwise:
Using the principal square root, introduce the function $f(z) = i\sqrt{z-1}\sqrt{z+1}$. It is easy to see that the branch cut of $f$ is $[-1, 1]$ and that for $-1 < r < 1$,
$$ f(r + 0^+i) = -\sqrt{1-r^2}, \qquad f(r - 0^+i) = \sqrt{1-r^2}. $$
Then the integral is written as
$$ I(w) = \frac{1}{2} \oint_C \frac{\log(z+a)}{(z+w)f(z)} \, dz. $$
Now deform this contour as follows:
This can be done by enlarging the contour $C$ while wrapping around the singular points. (And it is possible since the integrand grows like $\mathcal{O}(\log R/ R^2)$ as the radius $R \to \infty$.) Then it follows that
\begin{align*}
I(w)
&= \pi i \int_{-\infty}^{-a} \frac{dx}{(x+w)f(x)} - \pi i \underset{z=-w}{\operatorname{Res}} \frac{\log(z+a)}{(z+w)f(z)} \\
&= \pi \int_{a}^{\infty} \frac{dx}{(x-w)\sqrt{x^2 - 1}} - \frac{\pi \log(a-w)}{\sqrt{-w-1}\sqrt{-w+1}}
\end{align*}
Plugging this to (1), we have
\begin{align*}
I &= \pi \int_{a}^{\infty} \frac{dx}{(x-b)\sqrt{x^2 - 1}} \\
&= \frac{2\pi}{\sqrt{1-b^2}} \left( \arctan\sqrt{\frac{1+b}{1-b}} - \arctan\sqrt{\frac{1+b}{1-b}\cdot\frac{a-1}{a+1}} \right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Alternative area of a triangle formula The problem is as follows:
There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$
Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumption that base is $c$ and went on to prove that height is $c \frac{\sin A \sin B}{\sin (A+B)}$. But I end up expressing height in terms of itself.. i.e. $h \equiv \frac{ch}{a\cos B + b \cos A}$. How do I prove this alternative area of triangle formula?
| Here is what I did:
$$\frac{1}{2}bh \equiv \frac{1}{2}c^2\frac{\sin A \sin B}{\sin (A+B)}$$
Assume that $b$ = $c$.
Then, $c\frac{\sin A \sin B}{\sin (A+B)} \equiv \frac{c}{\cot A + \cot B}$. But $\cot A \equiv \frac{b \cos A}{h}$ and $\cot B \equiv \frac{a\cos B}{h}$. Therefore $\frac{c}{\cot A + \cot B} \equiv \frac{ch}{a\cos B + b\cos A}$. Since $a\cos B + b\cos A \equiv c$, we have $h \equiv \frac{ch}{c}$. $h\equiv h$.
I realised this after I posted the question...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding the area between three curves in which one is a piecewise function Let $f(x)$ be a continuous function given by $$f(x)=\begin{cases} 2x,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |x|\leq1\\ x^2+ax+b, \ |x|>1\end{cases}$$
Find the area of the region in the third quadrant bounded by the curves $x=-2y^2$ and $y=f(x)$ lying on the left of the line $8x+1=0$.
What I did: First of all, lets find the values of $a$ and $b$. So $\lim_{x\to-1}f(x)=-2$ and $\lim_{x\to1}f(x)=2$. Thus, $1-a+b=-2$ and $1+a+b=2$. Solving, we get $a=2$ and $b=-1$.
So $$f(x)=\begin{cases} 2x,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |x|\leq1\\ x^2+2x-1, \ |x|>1\end{cases}$$
So $f(x),x=\frac{-1}{8},x=-2y^2$ intersect at $(\frac{-1}8,\frac{-1}{4})$.
So area would be- $$\int\limits^{\frac{-1}8}_{-1} 2x-\sqrt{\frac{-x}2}dx+\int\limits^{-1}_{-2}x^2+2x-1-\sqrt{\frac{-x}2}dx$$
Now putting $-x=t$, and solving we get $\frac{-185}{192}-3$. Area will be positive, so answer is $\frac{761}{192}$. But according to my textbook, answer is $\frac{257}{192}$. Please explain what mistake did I commit?
| It should be
$$\int\limits^{\frac{-1}8}_{-1} \left(-\sqrt{\frac{-x}2}-2x\right)\,dx+\int\limits^{-1}_{-2} \left( -\sqrt{\frac{-x}2}-(x^2+2x-1)\right) \,dx$$
The curve $x=-2y^2$ is on top in both of the two parts, so should be the first term. Also you are picking the branch of it where $y$ is negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can't orthogonally diagonalise this symmetric matrix. So I have a symmetric matrix A = $\begin{bmatrix} 2 & 2 & -4 \\ 2 & -1 & -2 \\ -4 & -2 & 2 \end{bmatrix}$ and I want to orthogonally diagonalise it. I know that there are eigenvalues $-2$ and $7$ (-2 with an algebraic multiplicity of two, and a basis $$\left\{{\pmatrix{1\\0\\2}},{\pmatrix{0\\1\\1}}\right\}$$ I, so far, haven't been able to find:
*
*the eigenvector corresponding to 7 (I know from online calculators that it is in fact $\begin{array}{c} 2 \\1\\-2 \end{array}$), but the matrix (A - -7$\lambda$) = $\begin{bmatrix} 2 & 2 & -4 \\ 2 & -1 & -2 \\ -4 & -2 & 2 \end{bmatrix}$ which row reduces down (as calculated online) to $$A = \begin{bmatrix}{} 1 & 0 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$ which gives $x = 2y = -z$, which leads to a vector $\begin{array}{c} 1 \\2\\-1 \end{array}$ or similar.
*how to correctly orthonormalize the vectors (by using Pythagoras geometrically, or by Gram-Schmidt?)
and therefore I can't orthogonally diagonalise A.
| If you know there are only eigenvalues $-2$ and $7$, then you can eliminate those with eigenvalue $-2$ and be left with the others. This is done by considering
$$
(A+2I) = \left[\begin{array}{ccc} 4 & 2 & -4 \\ 2 & 1 & -2 \\ -4 & -2 & 4\end{array}\right]
$$
The columns have to be eigevectors with eigenvalue $7$. There is only one independent column, which is what you expect:
$$
\left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right],\;\;\;\;
A \left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right]
= \left[\begin{array}{c}14 \\ 7 \\ -14\end{array}\right]
= 7\left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right]
$$
Now consider the case for eigenvalue $-2$:
$$
(A-7I) = \left[\begin{array}{ccc} -5 & 2 & -4 \\ 2 & -8 & -2 \\ -4 & -2 & -5\end{array}\right]
$$
The column space is two-dimensional; it's obviously not one-dimensional, and you can easily check that the columns are orthogonal to the eigenvector with eigenvalue $7$ given above. So everything checks. If you add the first and the third columns, you find that the following is an eigenvector with eigenvalue $-2$:
$$
\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right],
$$
and this one is orthogonal to the middle column. So an orthogonal basis of the eigenspace with eigenvalue $-2$ is
$$
\left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array},\right]
\left[\begin{array}{c} 1 \\ -4 \\ -1 \end{array}\right]\right\}
$$
If you normalize the column vectors, then you get the required orthogonal change of basis matrix $U$ such that $U^{-1}AU=D=\left[\begin{array}{ccc}7 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]$ is a diagonal matrix. The matrix $U$ is
$$
U = \left[\begin{array}{ccc}
2/3 & 1/\sqrt{2} & 1/\sqrt{18} \\
1/3 & 0 & -4/\sqrt{18} \\
-2/3 & 1/\sqrt{2} & -1/\sqrt{18}
\end{array}\right]
$$
The matrix $U$ is orthogonal, which means $U^{-1}=U^{T}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Line for set of three-dimensional vectors If there is a set for 3D vectors $v$ where
$ v \times \begin{pmatrix} -1 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -27 \\ 8 \end{pmatrix}$
is a line, what is this line's equation?
I'm not sure how to solve this, except for setting $v = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and maybe having a linear system of equations, but I don't see how I could use that to solve the problem.
| $
\newcommand{\i}{\hat{\mathbf{i}}}
\newcommand{\j}{\hat{\mathbf{j}}}
\newcommand{\k}{\hat{\mathbf{k}}}
\newcommand{\v}{\vec{\mathbf{v}}}
$
Let $v = \v$ and
$$
\v = \begin{bmatrix} a \\ b \\ c \end{bmatrix},
\quad
\i = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
\quad
\j = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix},
\quad
\k = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
$$
Then
$$
v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix}
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
$$
can be rewritten as
$$
\v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} =
\begin{bmatrix} a \\ b \\ c \end{bmatrix}
\times
\begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix}
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
$$
Let us expand vector product:
$$
\begin{aligned}
\v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix}
=
\begin{vmatrix}
\i & \j & \k \\
a & b & c \\
-1 & 1 & 4
\end{vmatrix}
=
\begin{vmatrix} b & c \\ 1 & 4 \end{vmatrix} \i-
\begin{vmatrix} a & c \\ -1 & 4 \end{vmatrix} \j+
\begin{vmatrix} a & b \\ -1 & 1 \end{vmatrix} \k
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
\end{aligned}
$$
Thus, we conclude
$$
\begin{cases}
4 b - c = 5 \\
4 a + c = 27 \\
a + b = 8
\end{cases}
\implies
\begin{cases}
4 b - c = 5 \\
4 a + 4b = 32 \\
a = 8 - b
\end{cases}
\implies
\begin{cases}
4 b - c = 5 \\
32 + 8b = 32 \\
a = 8 - b
\end{cases}
\implies
\begin{cases}
c = -5 \\
b = 0 \\
a = 8
\end{cases}
$$
Finally, we write
$$
\bbox[9px, border:3px solid #FF0000]{v = \begin{pmatrix} 8 \\ 0 \\ -5 \end{pmatrix}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
| Your solution is correct. To reach to required form See here,You allready Got this
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
Now multiply and divide your result by $2$, you will get
$$\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x}{10}+C\\
=\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x+e^x-e^x}{10}+C\\
=\frac{2e^x \sin^2 x-e^x -4e^x \sin x \cos x +5e^x}{10}+C\\
=\frac{e^x(2 \sin^2 x-1) -4e^x \sin x \cos x +5e^x}{10}+C\\
=-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
Where we used the identities
1) $2 \sin^2 x-1=-\cos 2x$
2)$2\sin x\cos x= \sin 2x $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review:
How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$?
Answer is $2.73$
Simplify into Standard Form:
$$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$
$$ a^2 = 1 $$
$$ b^2 = 2 $$
$$ c^2 = 5 $$
Hyperbola is Vertical:
$$ C(-2,1) ; y = 1 $$
$$ y = 1 + \sqrt5 = 3.24 $$
$$ y = 1 - \sqrt5 = 1.24 $$
Both answers don't match; What am I doing wrong?
| Please check the formulas as you have two foci. $x^2 - 4x - 4 - 2y^2 - 4y = 0\to (x-2)^2 -2(y+1)^2 =6 \to \dfrac{(x-2)^2}{(\sqrt{6})^2}-\dfrac{(y+1)^2}{(\sqrt{3})^2}=1\to a = \sqrt{6}, b = \sqrt{3}\to c = \sqrt{6+3} = 3\to F= (2\pm 3,-1)=(-1,-1), (5,-1)\to \text{ distance to x-axix = 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others
| As Zhanxiong answered, Taylor expansion always works.
Consider, for large value of $x$, $$A(x)=\big(1+\frac 1x\big)^x$$ $$\log\big(A(x)\big)=x\log\big(1+\frac 1x\big)=x\big( \frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+\cdots\big)=1-\frac{1}{2 x}+\frac{1}{3 x^2}+\cdots$$ $$A(x)=e \, e^{-\frac{1}{2 x}+\frac{1}{3 x^2}}$$ So $$A(n+1)-A(n)\approx e\Big(e^{-\frac{1}{2 (n+1)}+\frac{1}{3 (n+1)^2}} -e^{-\frac{1}{2 n}+\frac{1}{3 n^2}}\Big)$$ Now, using , for small $y$, $e^y=1+y+\frac{y^2}2+\cdots$ $$A(n+1)-A(n)\approx e\Big(\frac{1}{2 n^2}-\frac{17}{12 n^3}+\cdots\Big)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first partials to zero we have \begin{align*} \frac{\partial L}{\partial x} &= 2x + \lambda y^2 z^4 = 0 \qquad (1) \\ \frac{\partial L}{\partial y} &= 2y + 2 \lambda x y z^4 = 0 \qquad (2) \\ \frac{\partial L}{\partial z} &= 2z + 4 \lambda x y^2 z^3 = 0 \qquad (3) \\ \frac{\partial L}{\partial \lambda} &= xy^2 z^4 - 32 = 0 \qquad (4)
\end{align*}
Now I'm having a hard time solving this system for $x,y$ and $z$. Here is what I did so far. From $(1)$ and $(2)$ we get \begin{align*} \frac{2x}{y^2 z^4} = - \lambda \qquad \text{and} \qquad \frac{1}{xz^4} = - \lambda \end{align*} Thus $\frac{2x}{y^2 z^4} = \frac{1}{xz^4} $ or $y^2 = 2x^2$ after simplification. Also, from $(2)$ and $(3)$ we can deduce that \begin{align*} \frac{1}{xz^4} = - \lambda = \frac{2z}{4xy^2 z^3} \end{align*} so that $2y^2 = z^2$ after simplification. Now I used all this and substituted it into $(4)$. This gave me \begin{align*} x(2x^2) (4y^4) - 32 = 0 \end{align*} or (since $y^4 = 4x^4)$ \begin{align*} 8x^3 (4x^4) - 32 = 0 \end{align*}
This means that $32x^7 - 32 = 0$, so that $x = 1$. Then $y^2 = 2$, so that $y = \pm \sqrt{2}$. Then $z^2 = 4$, so that $z = \pm 2$. So I found the points $(x,y,z) = (1, \sqrt{2}, 2)$ and $(1, - \sqrt{2}, -2)$. They both give me the distance $\sqrt{x^2 + y^2 + z^2} = \sqrt{7}$, so I'm guessing they are equal? Is my reasoning correct?
| Another possibility. First,
$$\tag{$*$}
xy^2z^4 = 32 \Longrightarrow y^2 = \frac{32}{xz^4}.
$$
Notice that we can arbitrarily divide by $x,y$, or $z$ since any point with a coordinate equal to $0$ do not belong to your surface.
Now, consider the squared norm of a generic point $(x,y,z)$ and use $(*)$:
$$
x^2+y^2+z^2 = x^2 + \frac{32}{xz^4} + z^2 =: N(x,z).
$$
We want to minimize $N(x,z)$ over $\mathbb{R}^2$. Thus, looking for zeros of the gradient, we obtain the system
$$
\begin{cases}
0 = \frac{\partial N}{\partial x}(x,z) = 2x - \frac{32}{x^2z^4},\\
0 = \frac{\partial N}{\partial z}(x,z) = 2z - \frac{128}{xz^5},
\end{cases}
$$
that is solved by $x = 1$ and $z = \pm 2$. In particular, the minimum is $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)$, $(2)$ and $(3)$ together we reach
$$(a+b)(b+c)(c+a) \ge8abc.$$
Now, I need to show that $abc = 1$.
Again by AM-GM inequality we have
$$\frac{a+b+c}{3}\ge\sqrt[3]{abc}
\implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc}
\implies abc\le1$$
Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?
Other solutions to the above question are also welcomed.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $9uv^2-w^3\geq8$, which is a linear inequality of $v^2$
and the condition does not depend on $v^2$.
Thus it remains to prove our inequality for an extremal value of $v^2$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or
$x^3-3ux^2+3v^2x-w^3=0$ or $3v^2x=-x^3+3ux^2+w^3$.
Thus, the graph of $y=3v^2x$ and the graph of $y=-x^3+3ux^2+w^3$ have three common points.
Thus, an extremal value of $v^2$ holds for the case, when a line $y=3v^2x$ is a tangent line to the graph of $y=-x^3+3ux^2+w^3$, which happens for equality case of two variables.
Id est, it remains to prove our inequality for $b=a$ and the condition gives $c=\frac{\sqrt{a^4+3}-a^2}{a}$.
Thus, we need to prove that $2a\left(a+\frac{\sqrt{a^4+3}-a^2}{a}\right)^2\geq8$, which is
$a^4-4a+3\geq0$, which is AM-GM.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Tangent planes to $2+x^2+y^2$ and that contains the $x$ axis I need to find the tangent planes to $f(x,y) = 2+x^2+y^2$ and that contains the $x$ axis, so that's what I did:
$$z = z_0 + \frac{\partial f(x_0,y_0)}{\partial x}(x-x_0)+\frac{\partial f(x_0,y_0)}{\partial y}(y-y_0) \implies \\ z = 2 + x_0^2 + y_0^2 + 2x_0(x-x_0) + 2y_0(y-y_0) \implies \\ 2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0$$
So since the plane must contain the $x$ axis, its normal vector must have the form $(0,y,z)$. The normal vector fot the plane I found is:
$$(2x_0, 2y_0, -1)$$
so $x_0 = 0, y_0 = y_0$
our plane has the form:
$$2y_0y -z -y_0^2+2 = 0$$
but when I plot this graph for some values of $y_0$ I only get 1 tangent plane at $y_0\approx 1.5$
| So far you have worked out that the tangent plane to the surface at $(x_0,y_0,x_0^2 + y_0^2+2)$ has equation
$$
2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0
\tag{1}
$$
If this plane contains the $x$-axis, it contains all points $(x,y,z)$ with $y=z=0$. So the equation reduces to
$$
2x x_0 = x_0^2 + y_0^2 -2
\tag{2}
$$
Again, we are looking for pairs $(x_0,y_0)$ such that (2) is true for all $x$. This is different from looking for triples $(x,x_0,y_0)$ satisfying (2). If we substitute $x=\frac{x_0}{2}$ into (2) we get
$$
x_0^2 = x_0^2 + y_0^2 - 2 \implies y_0^2 = 2 \implies y_0 = \pm\sqrt{2}
$$
Now equation (2) reduces to $2x x_0 = x_0^2$. If we substitute $x=0$ we get
$$
x_0^2 = 0 \implies x_0=0
$$
Since the points are on the paraboloid, their $z$-coordinates satisfy
$$
z_0 = x_0^2 + y_0^2 + 2
= (0)^2 + \left(\sqrt{2}\right)^2 +2 = 4
$$
So there are two points on the surface for which the tangent plane contains the $x$-axis: $(0,\sqrt{2},4)$ and $(0,-\sqrt{2},4)$.
We can find the equations for the planes by substituting these points into (1).
The first gives
$$
2x(0) + 2y(\sqrt{2}) - z - 0^2 - 2 + 2 = 0
\implies z = 2 \sqrt{2} y
$$
The second is $z = -2\sqrt{2} y$.
Another way to think about this is to project into two dimensions.
It's equivalent to asking: Which lines in the $yz$-plane are tangent to $z=y^2+2$ and pass through the origin? The line through $(y_0,y_0^2+2)$ and the origin has slope $\frac{y_0^2 + 2}{y_0}$. The line through $(y_0,y_0^2+2)$ tangent to $z=y^2+2$ has slope $2y_0$. So
$$
\frac{y_0^2 + 2}{y_0} = 2y_0
\implies y_0^2 + 2 = 2y_0^2
\implies y_0^2 = 2
\implies y_0 = \pm\sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $(x^2+y^2+z^2)=2(x+z-1)$, then show that $x^3+y^3+z^3$ is constant and find its numeric value. I am trying to solve this question,
If $(x^2+y^2+z^2)=2(x+z-1)$, then show that $x^3+y^3+z^3$ is constant and find its numeric value.
I've tried this,
$$x^2-2x + z^2-2z + 2 + y^2 = 0$$
$$ (x-1)^2 + (z-1)^2 + y^2 = 0$$
The left hand side can only become $0$ if $x=1$, $y=0$ and $z=1$, so the only solution is $x=z=1$ and $y=0$, which gives $x^3 + y^3 + z^3 = 2$.
*
*Have I solved it correctly?
*And is there any other method to solve it?
| The OP wrote:
The left hand side can only become $0$ if $x=1$, $y=0$ and $z=1$, so the only solution is $x=z=1$ and $y=0$, which gives $x^3 + y^3 + z^3 = 2$.
I would add a justification that these values for $x$, $y$, and $z$ are unique.
We know that each of the three terms, $(x-1)^2$, $(z-1)^2$ and $y^2$ are positive if we are dealing with Real Numbers. So, if any of them were non-zero, then at least one of the other terms would have to be negative for the equation to equal $0$. If we are dealing with Real Numbers, that is not possible. So each term must equal $0$.
Once that justification is taken into account, including the assumption that ${x,y,z}\in \mathbb R$, then I think the OP has made a sound demonstration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows that
$$\int \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$$
$$=-\frac{1}{2} i \text{Li}_2\left(e^{2 i \tan ^{-1}(x)}\right)-\frac{1}{2} i \tan ^{-1}(x)^2+\tan ^{-1}(x) \log \left(1-e^{2 i \tan ^{-1}(x)}\right)-\frac{65}{256} \sin \left(2 \tan ^{-1}(x)\right)-\frac{23 \sin \left(4 \tan ^{-1}(x)\right)}{1024}-\frac{5 \sin \left(6 \tan ^{-1}(x)\right)}{2304}-\frac{\sin \left(8 \tan ^{-1}(x)\right)}{8192}+\frac{65}{128} \tan ^{-1}(x) \cos \left(2 \tan ^{-1}(x)\right)+\frac{23}{256} \tan ^{-1}(x) \cos \left(4 \tan ^{-1}(x)\right)+\frac{5}{384} \tan ^{-1}(x) \cos \left(6 \tan ^{-1}(x)\right)+\frac{\tan ^{-1}(x) \cos \left(8 \tan ^{-1}(x)\right)}{1024}$$
and this form doesn't look that nice.
Having given the nice form of the closed form I wonder if we can find a very nice and simple way of getting the answer. What do you think?
A supplementary question:
$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$
| What about Feynman's way? If we take:
$$ f(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{x(1+x^2)^5}\,dx $$
we have $f(0)=0$ and:
$$ f'(a)=\int_{0}^{+\infty}\frac{dx}{(1+x^2)^5 (1+a^2 x^2)} $$
that is a manageable integral through partial fraction decomposition / the residue theorem.
We have:
$$\begin{eqnarray*} f'(a) &=& \frac{\pi}{256(1+a)^5}\left(35+175 a+345 a^2+325 a^3+128 a^4\right)\\&=&\frac{\pi}{256(1+a)^5}\left(8-52(a+1)+138 (a+1)^2-187(a+1)^3+128(a+1)^4\right)\end{eqnarray*} $$
and it is not difficult to integrate such expression over $(0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?
| $$
\begin{align}
\int_0^1\frac{\sin^{-1}\left(\sqrt{x}\right)}{x^2-x+1}\,\mathrm{d}x
&=\int_0^1\frac{\frac12\cos^{-1}(1-2x)}{x^2-x+1}\,\mathrm{d}x\tag{1}\\
&=\int_{-1}^1\frac{\cos^{-1}(x)}{3+x^2}\,\mathrm{d}x\tag{2}\\
&=\frac12\int_{-1}^1\frac{\cos^{-1}(x)+\cos^{-1}(-x)}{3+x^2}\,\mathrm{d}x\tag{3}\\
&=\frac\pi2\int_{-1}^1\frac1{3+x^2}\,\mathrm{d}x\tag{4}\\
&=\frac\pi{2\sqrt3}\int_{-1/\sqrt3}^{1/\sqrt3}\frac1{1+x^2}\,\mathrm{d}x\tag{5}\\
&=\frac\pi{\sqrt3}\tan^{-1}\left(\frac1{\sqrt3}\right)\tag{6}\\
&=\frac{\pi^2}{6\sqrt3}\tag{7}
\end{align}
$$
Explanation:
$(1)$: $\sin^{-1}\left(\sqrt{x}\right)=\frac12\cos^{-1}(1-2x)$
$(2)$: substitute $x\mapsto\frac{1-x}2$
$(3)$: $\int_{-1}^1f(x)\,\mathrm{d}x=\frac12\int_{-1}^1(f(x)+f(-x))\,\mathrm{d}x$
$(4)$: $\cos^{-1}(x)+\cos^{-1}(-x)=\pi$
$(5)$: substitute $x\mapsto\sqrt3x$
$(6)$: integrate
$(7)$: simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
My Attempt:
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$
Replacing $x$ by $1-x$,we get
$$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$
Then I got stuck. Please help.
| Let $$\displaystyle I = -\int\frac{x-1}{\left(x+1\right)}\cdot \frac{1}{\sqrt{x^3+x^2+x}}dx = -\int\frac{(x^2-1)}{\left(x+1\right)^2\cdot \sqrt{x^3+x^2+x}}dx$$
So $$\displaystyle = -\int\frac{(x^2-1)}{(x^2+2x+1)\sqrt{x^3+x^2+x}}dx = -\int\frac{\left(1-x^{-2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dx$$
Now Let $$\displaystyle \left(x+\frac{1}{x}+1\right) = t^2\;,$$ Then $$\displaystyle \left(1-\frac{1}{x^2}\right)dx = 2tdt$$
So we get $$\displaystyle I = -2\int\frac{1}{t^2+1}dt = -2\tan^{-1}(t)+\mathcal{C} = -2\left[\frac{\pi}{2}-\cot^{-1}(t)\right]+\mathcal{C}$$
Above we have used the formula $$\bullet \; \displaystyle \tan^{-1}(x)+\cot^{-1}(x) = \frac{\pi}{2}.$$
So we get $$\displaystyle I = \int\frac{1-x}{(1+x)\sqrt{x^3+x^2+x}}dx= 2\cot^{-1}\left(\frac{x^2+x+1}{x}\right)+\mathcal{C'}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\int_{0}^{2\pi}\frac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$ Prove that $$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$$
My Attempt:
$$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\int_{0}^{2\pi}x^2\frac{\sin x}{8+\sin^2x}$$
I applied integration by parts,considering $x^2$ as first function and $\dfrac{\sin x}{8+\sin^2x}$ as second function
$$=x^2\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx-\int_{0}^{2\pi}2x\left(\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx\right)dx$$
but this calculates out to be zero because $$\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx=0$$
What has gone wrong?Please help me..
| Just a guideline:
Use geometric series to get
$$
I=\frac{1}{8}\sum_{n=0}^{\infty}\frac{(-1)^n}{8^n}\underbrace{\int_{0}^{2 \pi}x^2 \sin^{2n+1}(x)}_{I_n}
$$
Show that the inner integral is equal to (use integration by parts and find a recurion relation):
$$
I_n=2\pi ^2\frac{\sqrt{\pi }\Gamma (n+1)}{\Gamma(n+\frac{3}{2})}
$$
Use the Series expansion of $\text{arcsinh}(x)$ together with $\Gamma$-duplication formulas
To obtain
$$
I=-\frac{4\pi^2}{3} \text{arcsinh}\left(\frac{1}{2 \sqrt{2}}\right)
$$
which is after some algebra equal to $$-\frac{2 \pi^2}{3}\log(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| Don't take this answer too seriously, but I could not resist. I leave details to you to fill in.
First, you factor the denominator into second degree factors, as
$$
\begin{aligned}
x^4+3x^3+3x^2-3x+1&=\bigl(x^2+\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}+\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}\bigr)\\
&\qquad\times\bigl(x^2-\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}-\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}\bigr)
\end{aligned}
$$
Then, do a partial fraction decomposition. The ansatz is
$$
\begin{aligned}
\frac{x^2+1}{x^4+3x^3+3x^2-3x+1}&=\frac{ax+b}{x^2+\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}+\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}}\\
&\qquad+\frac{cx+d}{x^2-\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}-\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac{5}{4}}.
\end{aligned}
$$
A direct calculation shows that $a=c=0$ and
$$
b=\frac{1}{2}+\sqrt{\frac{\sqrt{37}}{22}-\frac{7}{44}}
\quad
\text{and}
\quad
d=\frac{1}{2}-\sqrt{\frac{\sqrt{37}}{22}-\frac{7}{44}}.
$$
What you then have to do is the usual completing the square in the denominator, and you will get some horrible arctans. I leave that to you.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:
$$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$
Then by expanding each $e^{nx}$ term individually:
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$
$$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$
$$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$
$$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$
So substituting $(2),(3),(4),(5)$ into $(1)$ i get:
$$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$
Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.
| Here's another way to derive it term by term:
$$y^{-1}=1+e^x\Rightarrow -y^{-2}y'=e^x=y^{-1}-1\Rightarrow y'=y^2-y$$
Now you can continue to differentiate with the minimum of fuss:
$$y''=2yy'-y'$$
$$y'''=2yy''+2y'y'-y''$$
$$y^{(4)}=2yy'''+6y''y'-y'''$$
And so on, evaluating at each step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$ Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$
I tried to solve it,i got range $\frac{\pi}{2}$ but the answer is ${\frac{-\pi}{2},\frac{\pi}{2}}$
$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt=\int_{-1}^{1}\frac{\sin x}{(t-\cos x)^2+\sin^2x}dt=\sin x\int_{-1}^{1}\frac{1}{(t-\cos x)^2+\sin^2x}dt$
$=\frac{\sin x}{\sin x}\left[\tan^{-1}\frac{t-\cos x}{\sin x}\right]_{-1}^{1}=\tan^{-1}\frac{1-\cos x}{\sin x}+\tan^{-1}\frac{1+\cos x}{\sin x}$
$=\tan^{-1}\tan\frac{x}{2}+\tan^{-1}\cot\frac{x}{2}=\frac{\pi}{2}$
But i am not getting $\frac{-\pi}{2}$.Have i done some wrong?Please enlighten me.
| $$f(x)=\int_{-1}^{1}\frac{\sin x}{t^2-2t\cos x+1}\ dt$$
$$=\int_{0}^{1}\frac{\sin x}{t^2-2t\cos x+1}+\frac{\sin x}{t^2+2t\cos x+1}\ dt$$
$$=2\sin x\int_{0}^{1}\frac{t^2+1}{(t^2+1)^2-4t^2\cos^2 x}\ dt$$
$$=2\sin x\int_{0}^{1}\frac{1+\frac {1}{t^2}}{\biggl(t+\frac {1}{t}\biggr)^2-4\cos^2 x}\ dt$$
$$=2\sin x\int_{0}^{1}\frac{1}{\biggl(t-\frac {1}{t}\biggr)^2+(2\sin x)^2}\ d\biggl(t-\frac {1}{t}\biggr)$$
$$=\biggl(\arctan \biggl(\frac{t-\frac {1}{t}}{2\sin x}\biggr)\biggr)_{0}^{1}$$
$$=\frac {\pi}{2} \text { for } \sin x>0$$
Or
$$=-\frac {\pi}{2} \text { for } \sin x<0$$
Hence Range of $$f(x)=\biggl\{-\frac {\pi}{2},\frac {\pi}{2}\biggr\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove an improper double integral is convergent I need to prove the following integral is convergent and find an upper bound
$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{1+x^2+y^4} dx dy$$
I've tried integrating $\frac{1}{1+x^2+y^2} \lt \frac{1}{1+x^2+y^4}$ but it doesn't converge
| Continuing from zhw.'s answer,
$$ \int_{0}^{+\infty}\frac{\sqrt{r}}{2(1+r^2)}\,dr = \int_{0}^{+\infty}\frac{u^2\,du}{1+u^4}=\frac{\pi}{2\sqrt{2}} $$
by the residue theorem, while:
$$ \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin\theta}}=\int_{0}^{1}\frac{du}{\sqrt{u(1-u^2)}}=\frac{1}{2\sqrt{2\pi}}\,\Gamma\left(\frac{1}{4}\right)^2$$
from the Euler beta function and the $\Gamma$ reflection formula. Hence we have:
$$ \iint_{(0,+\infty)^2}\frac{dx\,dy}{1+x^2+y^4}=\color{red}{2\sqrt{\pi}\cdot\Gamma\left(\frac{5}{4}\right)^2}=2.9123736927\ldots$$
A simple upper bound may be derived from:
$$ \int_{0}^{+\infty}\frac{u^2}{1+u^4}\,du \leq \int_{0}^{1}u^2\,du+\int_{1}^{+\infty}\frac{du}{u^2}=\frac{4}{3},$$
$$ \int_{0}^{1}\frac{2\,du}{\sqrt{1-u^4}}\leq\int_{0}^{1}\frac{2\,du}{\sqrt{1-u^2}}=\pi. $$
Another simple proof comes from:
$$ \int_{0}^{+\infty}\frac{dx}{1+y^4+x^2}=\frac{\pi}{2}\cdot\frac{1}{\sqrt{1+y^4}}$$
and obviously:
$$ \int_{0}^{+\infty}\frac{dy}{\sqrt{1+y^4}}\leq \int_{0}^{1}dy+\int_{1}^{+\infty}\frac{dy}{y^2}$$
giving $\color{red}{I\leq \pi}$. That can be improved through Cauchy-Schwarz inequality:
$$ \int_{0}^{+\infty}\frac{du}{\sqrt{1+u^4}}\leq \sqrt{\left(\int_{0}^{+\infty}\frac{1+u^2}{1+u^4}\,du\right)\cdot\left(\int_{0}^{+\infty}\frac{du}{1+u^2}\right)}$$
leads to:
$$ \color{red}{I} \leq \frac{\pi^2}{2\sqrt{2\sqrt{2}}}=2.93425\ldots\color{red}{< 3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| We have $$\cos\left(x\right)+\sin\left(x\right)\tan\left(\frac{x}{2}\right)=1\Leftrightarrow\cos\left(x\right)\cos\left(\frac{x}{2}\right)+\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\cos\left(\frac{x}{2}\right)
$$ and, using product to sum identity and the fact that cos is a even function$$\cos\left(x\right)\cos\left(\frac{x}{2}\right)=\frac{1}{2}\cos\left(\frac{x}{2}\right)+\frac{1}{2}\cos\left(\frac{3x}{2}\right)
$$ and again using product to sum identity we get $$\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\frac{1}{2}\cos\left(\frac{x}{2}\right)-\frac{1}{2}\cos\left(\frac{3x}{2}\right)
$$ and we have done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 6
} |
Rearrangement and Cauchy Let $a_1, \ldots, a_n$ be distinct positive integers. I want to prove that
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + \cdots + \frac{a_n}{n^2} \geq \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}.$$
I've been considering using Rearrangement and Cauchy Schwarz, but cannot make any progress
| Use the following: If $a_1, \ldots, a_n$ are distinct positive integers, then we can sort them according to their size $a_{n_1} \leq a_{n_2} \leq \ldots \leq a_{n_n}$ and get
$$k \leq a_{n_k} ~ \forall k = 1, \ldots, n.$$
Hence, using Cauchy Schwarz, we compute
\begin{align*}
\left( \sum_{k=1}^n \frac{1}{k} \right)^2 & = \left( \sum_{k=1}^n \frac{\sqrt{a_k}}{k} \cdot \frac{1}{\sqrt{a_k}}\right)^2 \leq \left( \sum_{k=1}^n \frac{a_k}{k^2} \right) \cdot \left( \sum_{k=1}^n \frac{1}{a_k} \right) = \left( \sum_{k=1}^n \frac{a_k}{k^2} \right) \cdot \left( \sum_{k=1}^n \frac{1}{a_{n_k}} \right)\\
& \leq \left( \sum_{k=1}^n \frac{a_k}{k^2} \right) \cdot \left( \sum_{k=1}^n \frac{1}{k} \right)
\end{align*}
Thus, dividing the inequality by $\sum_{k=1}^n 1/k$ yields
$$\frac{1}{1} + \ldots + \frac{1}{n} = \sum_{k=1}^n \frac{1}{k} \leq \sum_{k=1}^n \frac{a_k}{k^2} = \frac{a_1}{1^2} + \ldots + \frac{a_n}{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
My approach:
For $x=1$, $z$ is not divisible by $6$.
For $x=2$, $z$ is divisible by $6$.
For $x=3$, $z$ is not divisible by $6$.
For $x=4$, $z$ is divisible by $6$.
For $x=5$, $z$ is not divisible by $6$.
For $x=6$, $z$ is not divisible by $6$.
For $x=7$, $z$ is divisible by $6$.
For $x=8$, $z$ is divisible by $6$.
I could not identify the pattern in these questions. Also can this problem be solved with a better approach?
| Just a different way to highlight what has already been correctly noted, and to explain why the answer could be $66$, as stated in one of the initial comments by the author of the OP. To avoid confusion, I interpreted the problem as the question of how many values of $x$ lead to a positive $z$ that is divisible by $6$.
Let us start from $z=(100-x)(100+x)$, where the two factors $100-x$ and $100+x$ clearly have the same parity. First, we can observe that the only possibility that $z$ is divisible by $6$ occurs when at least one of these two factors is divisible by $6$. In fact, if any of the two factors is divisible by $3$ and not by $2$, the other factor is also not divisible by $2$, and so $z$ cannot be divisible by $6$.
Then, we can note that $100-x$ is divisible by $6$ only when $x=6j+4$ (with $j$ integer), whereas $100+x$ is divisible by $6$ only when $x=6j+2$.
If we limit the problem to solutions that give positive values of $z$ (although this is not specified in the OP), then $|x|$ can only assume integer values $<100$. It is not difficult to note that there are $16$ positive integers $<100$ of the form $x=6j+4$ (they are $4,10,16,22...94$) and $17$ positive integers $<100$ of the form $x=6j+2$ (they are $2,8,14,20...98$). So, there are $33$ positive integer values of $x$ that lead to a $z$ value that is divisible by $6$.
Lastly, because each of these solutions with a given positive $x$ corresponds to another symmetric solution with $-x$, we have to count other $33$ negative values of $x$. This finally leads to a total of $66$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find cartesian coordinates of the incenter $A(a_1,a_2)$, $B(b_1,b_2)$ and $C(c_1,c_2)$ form the triangle $ABC$. What are the cartesian coordinates of the incenter and why?
| I'll argue up to the formula, so the "why" will be clear. Let's use this diagram of $\triangle ABC$ with the Cartesian coordinates moved so that points $A$ and $B$ are on the $x$-axis and $C$ has a positive $y$-coordinate.
Point $I$ is the incenter, segments $ID$, $IE$, and $IF$ are perpendiculars from $I$ to sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ respectively. Since $I$ is the incenter, those perpendicular segments are radii of the incircle and have equal lengths, which we call $r$. The area of $\triangle ABC$ is of course half-times-base-times-height, with the base here $c$ and height $c_2$. We then get
$$\begin{align}
\operatorname{Area}(\triangle ABC)
&= \operatorname{Area}(\triangle ABI)+\operatorname{Area}(\triangle BCI)
+\operatorname{Area}(\triangle ACI) \\[2 ex]
&= \frac 12cr+\frac 12ar+\frac 12br \\[2 ex]
&= \frac{a+b+c}2r
\end{align}$$
Thus $r$, which is the $y$-coordinate of point $I$, is given by
$$\begin{align}
r&=\frac{2\operatorname{Area}(\triangle ABC)}{a+b+c} \\[2 ex]
&=\frac{2\cdot\frac 12c\cdot c_2}{a+b+c} \\[2 ex]
&=\frac{c\cdot c_2}{a+b+c}
\end{align}$$
If we look only at the $y$-coordinates, and consider $A$, $B$, $C$, and $I$ to be vectors, we get
$$I=\frac{aA+bB+cC}{a+b+c}$$
Note that this formula does not use the coordinate system directly and is symmetrically dependent on the triangle's vertices. Through an argument too tedious to give here, we get that the formula for $I$ is true in both $x$- and $y$-coordinates in any Cartesian coordinate system.
If you really want a formula with only the coordinates $a_1,a_2,b_1,b_2,c_1,c_2$, we replace $A$ with $(a_1,a_2)$ and $a$ with $\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}$ etc., and separate the $x$- and $y$-coordinates, to get
$$I=$$
$$\left(
\frac{a_1\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+b_1\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+c_1\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
{\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
,\right.$$
$$\left.
\frac{a_2\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+b_2\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+c_2\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
{\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}
+\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}
+\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| \begin{align}
1^2+2^2+...+&(n-1)^2 \quad\quad= \frac{n(n-1)(2n)}{6} = \frac{n^2(n-1)}{3} &< \frac{n^3}{3}
\\
1^2+2^2+...+&(n-1)^2+n^2 =\frac{n(n+1)(2n+1)}{6} =\frac{2n^3+3n^2+n}{6} =\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} &>\frac{n^3}{3}
\end{align}
Cleary our inequality is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}
$$
The result should be a number.
I try this:
$$
\frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} =
$$
$$
= \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}}
$$
what next?
| Let $s=a+b$ be our sum, where $a=\sqrt[3]{2+\sqrt{5}}$ and $b=\sqrt[3]{2-\sqrt{5}}$. Note that
$$s^3=a^3+b^3+3ab(a+b)=a^3+b^3+3abs.$$
Thus since $a^3+b^3=4$ and $ab=\sqrt[3]{-1}=-1$, we have
$s^3=4-3s$. This has the obvious root $s=1$ and no other real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\frac{x^2}{\sqrt{16-x^2}}$ using trig substitution During our integration of the following integral, using $x = 4 \sin \theta$
$$\int \frac{x^2}{\sqrt{16-x^2}} dx$$
We eventually come to the following point:
$$\int \frac{16 {\sin ^2 \theta} }{4 \sqrt{\cos ^2 \theta}} dx$$
At this point, we say that $\sqrt{\cos ^2 \theta} = \cos \theta$ and then complete the integral
But doesn't $\sqrt{\cos ^2 \theta} = |\cos \theta|$
What is going on?
| Let $$\displaystyle I = \int\frac{x^2}{\sqrt{16-x^2}}dx\;,$$ Let $x=4\sin \phi\;,$ Then $dx = 4\cos \phi d\phi$
So Integral $$\displaystyle I = \int\frac{16\sin^2 \phi}{4|\cos \phi|}\cdot 4\cos \phi d\phi = \pm 8\int 2\sin^2 \phi d \phi$$
Now Using $\bullet\; 2\sin^2 \phi = 1-\cos 2\phi$ and $\bullet\; \sin 2\phi = 2\sin \phi\cdot \cos \phi.$
so we get $$\displaystyle I = \pm 8\int (1-\cos 2\phi)d\phi = \pm 8\left[\phi-\frac{\sin 2\phi}{2}\right]+\mathcal{C} = \pm 8\left[\phi-\sin \phi \cdot \cos \phi\right]+\mathcal{C}$$
So we get $$\displaystyle I =\pm 8\left[\sin^{-1}\left(\frac{x}{4}\right)-\frac{x\sqrt{16-x^2}}{16}\right]+\mathcal{C} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Riemann integral on trigonometric functions I have to calculate Riemann integral of function $g:[0;\pi/4]\rightarrow\mathbb{R}$ (on interval $[0;\pi/4]$) given as $g(x)=\frac{\tan(x)}{(\cos(x)^2-4)}$. Function $g$ is continous on interval $[0;\pi/4]$ so it is enough to calculate $\int_0^{\frac{\pi}{4}}{\frac{\tan(x)}{(\cos(x)^2-4)}}$. How to do that?
| Hint: let $t=\tan(\frac{x}{2})$, then
$$\sin(x) = \frac{2t}{1+t^2}, \cos(x) = \frac{1-t^2}{1+t^2}$$
and
$$x=2\arctan(t) \implies dx = \frac{2 dt}{1+t^2}$$
so
$$\begin{aligned}
&\int \frac{\tan(x)}{(\cos(x)^2-4)} dx = \int \frac{\sin(x)}{\cos(x)(\cos(x)-2)(\cos(x)+2)} dx\\
&= \int \frac{2t}{1+t^2} \cdot \frac{1+t^2}{1-t^2} \cdot \frac{1}{\frac{1-t^2}{1+t^2} - 2} \cdot \frac{1}{\frac{1-t^2}{1+t^2} + 2} \cdot \frac{2 dt}{1+t^2}\\
&\ldots
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$.
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\frac{\pi}{8}$
Since $z^n=r^n\cdot e^{in\theta}$, $z^4$ can now be rewritten as $z^4=e^{i\cdot4\cdot\frac{\pi}{8}} \iff z=e^{i\frac{\pi}{8}}$ However, my answer file says this is wrong. Can anyone give me a hint on how to find $z$?
| We have $z^4=e^{3\pi i/2+2k\pi}$.
So $z=(e^{3\pi i/2+2k\pi})^{1/4}=e^{3\pi i/8+k\pi/2}$
As $e^{i\theta}=e^{i(\theta+2\pi)}$, the $4$ values of $z$ are $e^{3\pi i/8}, e^{7\pi i/8}, e^{11\pi i/8}, e^{15\pi i/8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $5\times 5$ determinant The following $5\times 5$ det. comes from a Russian book. I don't want to expand the det. rather than do some operations on it and extract the result.
Prove:
$$\begin{vmatrix}
-1 &1 &1 &1 &x \\
1& -1 &1 &1 &y \\
1& 1 & -1 & 1 &z \\
1& 1 & 1 & -1 & u\\
x& y & z & u &0
\end{vmatrix}= -4 [x^2+y^2+z^2 +u^2 - 2(xy+zx+zu+yz+yu+\color{red}{x}u)]$$
I have tried many things e.g $R_1 \leftrightarrow R_5+ x R_1$ and of course cyclic ($R_2 \leftrightarrow R_5+ y R_2$ etc) and then adding the last row to the second etc. resulting in:
$$\begin{vmatrix}
0 &y+x &z+x &u+x &x^2 \\
y+x&0 &z+y &u+y &y^2 \\
x+z& z+y & 0 &z+u & z^2\\
u+x& u+y &u+z & 0 &u^2 \\
x &y &z &u &0
\end{vmatrix}$$
but does not look that promising.
On the other hand if I add the the first column to the others I do get a lot of zeros, since the first row for example takes the form $-1, 0, 0, 0, x+1$. This looks more promising. But in either case I get stuck.
May I have some hints or answers?
Edit: The red letters as suggested by Peter.
| I am not sure if this is the most efficient way to do it but doing the following row opperations:
$R_2 \to R_2 + R_1, R_3 \to R_3 + R_1, R_4 \to R_4 + R_1, R_5 \to R_5 + xR_1$
Assuming i didnt make any silly mistakes (as is always the case with there types of problems)
$\begin{vmatrix}
-1&1&1&1&x\\
1&-1&1&1&y\\
1&1&-1&1&z\\
1&1&1&-1&u\\
x&y&z&u&0\\
\end{vmatrix}$ $\to$
$\begin{vmatrix}
-1&1&1&1&x\\
0&0&2&2&x+y\\
0&2&0&2&x+z\\
0&2&2&0&x+u\\
0&x+y&x+z&x+u&x^2\\
\end{vmatrix}$ $\to$
$-\begin{vmatrix}
0&2&2&x+y\\
2&0&2&x+z\\
2&2&0&x+u\\
x+y&x+z&x+u&x^2\\
\end{vmatrix}$
Now removing any one of the $2's$ from any row or column and expanding further into a linear combination of $3\times3$ matrices... Havent finished it but this is the most promising version I've tried
| {
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Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\ dx$. Therefore $$\begin{align}I&=\int\frac {\sqrt{2-x-x^2}}{x^2}\ dx\\&=\int \left(\frac tx\right)\left(\frac{2t\ dt}{-\frac4{x^3}+\frac 1{x^2}}\right)\\&=\int \frac{2t^2\ dt}{\frac{x-4}{x^2}}\\&=\int \frac{2t^2(1+t^2)\ dt}{{x-4}\over{2-x}}\\&=\int \frac{2t^2(1+t^2)(5+4t^2-\sqrt{8t^2+9})\ dt}{\sqrt{8t^2+9}-(8t^2+9)}\end{align}$$ Now I substituted $8t^2+9=z^2 \implies t^2=\frac {z^2-9}8 \implies 2t\ dt=z/4\ dz$. So, after some simplification, you get $$\begin{align}I&=-\frac1{512}\int (z^2-9)(z+1)(z-1)^2\ dz\end{align}$$ I didn't have the patience to solve this integration after all these substitutions knowing that it can be done (I think I have made a mistake somewhere but I can't find it. There has to be an $ln(...)$ term, I believe). Is there an easier way to do this integral, something that would also strike the mind quickly? I have already tried the Euler substitutions but that is also messy.
| Hint: $\sqrt{2-x-x^2}=\sqrt{\frac94-(x+\frac12)^2}$
Substitute $x+\frac12=t; dx=dt$
$$\int \frac{\sqrt{\frac94-t^2}}{(t-\frac12)^2}\ dt$$
Put $\frac32\sin\theta=t;\frac32\cos\theta d\theta=dt$
$$\int \frac{\frac32\cos\theta.\frac32\cos\theta d\theta}{(\frac32\sin\theta-\frac12)^2}$$
$$9\int \frac{\cos^2\theta d\theta}{(9\sin^2\theta+1-6\sin \theta)}$$
Now do $u=\tan\frac{\theta}2;du=\frac12\sec^2{\frac{\theta}2}d\theta$
$$\boxed{\sin\theta=\frac{2u}{1+u^2};\cos\theta=\frac{1-u^2}{1+u^2};d\theta=\frac{2du}{u^2+1}}$$
I guess you can take it from here?
| {
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Determining values of a coefficient for which a system is and isn't consistent. Given the system : \begin{array}{ccccrcc}
x & + & 2y & + & z & = & 3 \\
x & + & 3y & - & z & = & 1 \\
x & + & 2y & + & (a^2-8)z & = & a
\end{array}
Find values of $a$ such that the system has a unique solution, infinitely many solutions, or no solutions.
I begin by placing the system into an augmented matrix.
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
1 & 2 & a^2-8 & a \\
\end{array}
\right]
$
I then perform operations on the matrix.
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
1 & 2 & a^2-8 & a \\
\end{array}
\right]
$ $: (R_3-R_1)\rightarrow$ $\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
0 & 0 & a^2-9 & a-3 \\
\end{array}
\right]
$
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
1 & 3 & -1 & 1 \\
0 & 0 & a^2-9 & a-3 \\
\end{array}
\right]
$ $: (R_2-R_1; R_3/(a-3))\rightarrow$ $\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & -2 & -2 \\
0 & 0 & a+3 & 1 \\
\end{array}
\right]
$
$\displaystyle
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & -2 & -2 \\
0 & 0 & a+3 & 1 \\
\end{array}
\right]
$ $: (R_1-2R_2)\rightarrow$ $\displaystyle
\left[
\begin{array}{rrr|r}
1 & 0 & 5 & 3 \\
0 & 1 & -2 & -2 \\
0 & 0 & a+3 & 1 \\
\end{array}
\right]
$
Truth be told, at this point, I'm not too clear on how to progress (or if my approach is even ideal for this issue). I desire to find intervals across all of $a\in\mathbb{R}$ that explain where $a$ causes the system to have a unique, infinite, or inconsistent solution set.
| Just from the first reduction you can find:
$$
(a^2-9)z=a-3
$$
Now you simply have to discuss such equation ( the others does not contain the parameter). And you see that this equation has a solution only if $a^2-9 \ne0$ i.e $a \ne \pm 3$.
For $a=3$ the equation becomes: $ 0=0$ that is an identity and this means that the system has infinitely many solutions.
For $a=-3$ the equation becomes $ =-6$ and this means that there are no solutions for the equation and so for the system.
For all other values $a \ne \pm 3$ you find $ z= 1/(a+3)$ and back-substituting in the other equations you find the solutions $x, y$ of the system that, obviously, depend from the parameter.
Substituting $z=\dfrac{1}{a+3}$ in the first two equations we have:
$$
\begin{cases}
x+2y=3-\dfrac{1}{a+3}\\
x+3y=1+\dfrac{1}{a+3}
\end{cases}
$$
now subtracting the two equations:
$$
y=-2+\dfrac{2}{a+3}=\dfrac{-2(a+2)}{a+3}
$$
and substituting this value in the first equation we have:
$$
x-\dfrac{4(a+2)}{a+3}=3-\dfrac{1}{a+3}
$$
that gives
$$
x=3-\dfrac{1}{a+3}+\dfrac{4a+8}{a+3}=\dfrac{7a+16}{a+3}
$$
| {
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} |
Are there any $x$ for which $x ^2= (x-1)(x+1)$?
$x^2 = (x-1)(x+1)$.
Does anybody know if this is true for any numbers (specifically the larger the better)?
Because: $256^2 = 255 \cdot 257 + 1$
It's very very close, just an interesting thing I noticed when dealing with 2-byte integers in computer programming.
| This identity, $(x-1)(x+1) = x^2 - 1$, or more generally, $(x-a)(x+a)=x^2 - a^2$, occurs a lot in algebra. For example
\begin{align}
199 \times 201 = (200-1)(200+1) = 200^2 - 1^2 = 40000 - 1 &= 39999 \\
(\sqrt 7 - \sqrt 2)(\sqrt 7 + \sqrt 2) = 7 - 4 &= 3
\end{align}
You can also use it to solve the "sum-product problem". For example, to solve the system of equations: $$\text{$xy = -55 \quad$ and $\quad x+y = 6$.}$$
If $x+y = 6$, then $\frac 12(x+y) = 3$ is half-way between $x$ and $y$. That means there is some number, say $z$, such that $x = 3-z$ and $y = 3 + z$. We can now reason
\begin{align}
xy &= -55 \\
(3-z)(3+z) &= -55 \\
9 - z^2 &= -55 \\
-z^2 &= -64 \\
z^2 &= 64 \\
z &= 8 &\{\text{You could also have said $z = -8$.}\} \\
x = 3-8 &=-5 \\
y = 3+8 &= 11
\end{align}
| {
"language": "en",
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"answer_count": 3,
"answer_id": 2
} |
Where I'm wrong? Why my answer is different from the book? The question is to integrate
$$\int x\cos^{-1}x dx$$
Answer in my book
$$(2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+C$$
I'm learning single variable calculus right now and at current about integration with part. I'm confused in a problem from sometime. I don't know where I am wrong. Please have a look at the images.
Solution. $\newcommand{\dd}{\; \mathrm{d}}$
$$
\begin{align}
&\cos^{-1} x \int x \dd x - \int \left[\frac{-1}{\sqrt{1-x^2}}\cdot\frac{x^2}2\right] \dd x=\\
=&\cos^{-1}x \cdot \frac{x^2}2 + \int \frac{x^2}{2\sqrt{1-x^2}} \dd x2=\\
=&\frac{x^2}2 \cdot \cos^{-1}x + \frac12 \int \frac{x}{\sqrt{1-x^2}} \dd x =
\begin{vmatrix}x=\sin t\\ \dd x=\cos t\dd t\end{vmatrix} = \\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\sqrt{1-\sin^2t}} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\cos t} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \sin^2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{1-\cos2t}2 \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac12 \dd t -\frac 14 \int \cos2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 t - \frac18 \sin 2t + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin2\sin^{-1}x + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x) + C
\end{align}
$$
Please help. Thankyou in advance.
| With the help of Bernard's answer you can further transform your term
\begin{align}
\int x \cos^{-1}(x)dx&=\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x)\\
&=\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac 14 x\sqrt{1-x^2}+c \ \text{ but this does not equal}\\
&\neq (2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+c \\&=\frac{x^2\cos^{-1}x}2 - \frac14 \cos^{-1}x - \frac 14 x\sqrt{1-x^2}+c
\end{align}
so additionally to the fact that you can further simplify your answer the given answer in the book is wrong.
| {
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} |
Does the determinant of a matrix made up of column vectors being non-zero imply that the vectors are independent? Let's say we have 3 vectors and we make up a matrix where we depict the vectors as the columns of the matrix. If we calculate the determinant of the matrix and we get a non-zero number, does that mean that the vectors are linearly independent?
|
Let's say we have 3 vectors
$$
\left(
\begin{matrix}
a\\
b\\
c
\end{matrix}
\right),
\left(
\begin{matrix}
d\\
e\\
f
\end{matrix}
\right),
\left(
\begin{matrix}
g\\
h\\
i
\end{matrix}
\right)
$$
and we make up a matrix where we depict the vectors as the columns of the matrix
$$
\left(
\begin{matrix}
a & d & g\\
b & e & h\\
c & f & i\\
\end{matrix}
\right)
$$
If we calculate the determinant of the matrix and we get a non-zero number
$$
\left|
\begin{matrix}
a & d & g\\
b & e & h\\
c & f & i\\
\end{matrix}
\right|
=aei+dhc+gbf-gec-dbi-ahf\neq 0
$$
does that mean that the vectors are linearly independent?
Yes.
Look at the 6 summands that make up the determinant. Note how each one is the product of one element of each of the original vectors and also including one element from each "row". Let me illustrate
$$
\left(
\begin{matrix}
\color{red}{a}\\
\color{green}{b}\\
\color{blue}{c}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{d}\\
\color{green}{e}\\
\color{blue}{f}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{g}\\
\color{green}{h}\\
\color{blue}{i}
\end{matrix}
\right)
$$
and
$$\color{red}{a}\color{green}{e}\color{blue}{i}+\color{red}{d}\color{green}{h}\color{blue}{c}+\color{red}{g}\color{green}{b}\color{blue}{f}-\color{red}{g}\color{green}{e}\color{blue}{c}-\color{red}{d}\color{green}{b}\color{blue}{i}-\color{red}{a}\color{green}{h}\color{blue}{f}\neq 0$$
There seems to be a pattern to this. If you allow yourself for a moment to only care about the colors and substitute each summand by $\color{red}{R}\color{green}{G}\color{blue}{B}$, the result is clearly 0
$$\color{red}{R}\color{green}{G}\color{blue}{B}+\color{red}{R}\color{green}{G}\color{blue}{B}+\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B} = 3\color{red}{R}\color{green}{G}\color{blue}{B}-3\color{red}{R}\color{green}{G}\color{blue}{B}= 0$$
And that's exactly what happens when the vectors are linear dependent, which means all of them can be expressed as one vector multiplied with a scalar, solely for artistic reasons that vector shall be $(\color{red}{R},\color{green}{G},\color{blue}{B})^T$ which is equal to $(\color{red}{a},\color{green}{b},\color{blue}{c})^T$
$$
\left(
\begin{matrix}
\color{red}{a}\\
\color{green}{b}\\
\color{blue}{c}
\end{matrix}
\right)=\left(
\begin{matrix}
\color{red}{R}\\
\color{green}{G}\\
\color{blue}{B}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{d}\\
\color{green}{e}\\
\color{blue}{f}
\end{matrix}
\right) = x\left(
\begin{matrix}
\color{red}{R}\\
\color{green}{G}\\
\color{blue}{B}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{g}\\
\color{green}{h}\\
\color{blue}{i}
\end{matrix}
\right)=y\left(
\begin{matrix}
\color{red}{R}\\
\color{green}{G}\\
\color{blue}{B}
\end{matrix}
\right)
$$
Let the transformation begin
$$
\begin{array}{c|ccccccl}
& \color{red}{a}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{b}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{b}\color{blue}{i}&-\color{red}{a}\color{green}{h}\color{blue}{f}&\neq 0\\
\hline
\color{red}{a}=\color{red}{R}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{b}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{b}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{green}{b}=\color{green}{G}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{blue}{c}=\color{blue}{B}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{B}&-\color{red}{d}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{red}{d}=x\color{red}{R}& \color{red}{R}\color{green}{e}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{green}{e}=x\color{green}{G}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{blue}{f}=x\color{blue}{B}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}x\color{blue}{B}&-\color{red}{g}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}x\color{blue}{B}&\\
\color{red}{g}=y\color{red}{R}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}x\color{blue}{B}&\\
\color{green}{h}=y\color{green}{G}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}y\color{green}{G}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}y\color{green}{G}x\color{blue}{B}&\\
\color{blue}{i}=y\color{blue}{B}& \color{red}{R}x\color{green}{G}y\color{blue}{B}&+x\color{red}{R}y\color{green}{G}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}y\color{blue}{B}&-\color{red}{R}y\color{green}{G}x\color{blue}{B}&\\
\end{array}
$$
And that last line is (apart from the x&y's) exactly as the one above, which equalled zero
$$xy\color{red}{R}\color{green}{G}\color{blue}{B}+xy\color{red}{R}\color{green}{G}\color{blue}{B}+xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B} = 3xy\color{red}{R}\color{green}{G}\color{blue}{B}-3xy\color{red}{R}\color{green}{G}\color{blue}{B}= 0$$
Intuitively speaking, the linear dependency "smears" all 3 distinguishable a, d and g into one homogenous blob R. The same holds true for the other two rows. The scalar factors cancel out as well due to the construction of the determinant.
tl;dr; got color blind;
Checking vectors if they are linear dependent and assembling them into a matrix and checking if the determinant is 0 is the same thing.
| {
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} |
Rearange an expression I am learning induction. At one step I have to show that:
$$ 1 - \frac{1}{(1+n)} + \frac{1}{(n+1)(n+2)} $$
can be transformed to
$$ 1 - \frac{1}{n+2} $$
Theese are the steps for the transformation, but I cant understand them:
$$ 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)} = 1 - \frac{n+1}{(n+1)(n+2)} = 1 - \frac{1}{n+2} $$
Could you explain to me what happens there?
| The first step is as follows:
$$1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)}$$
Consider the fractional part of the first expression; this is:
$$- \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$
Notice that there is a common factor of $1/(n+1)$, so we will factor this out and obtain:
$$\frac{1}{(n+1)}\left(-1+\frac{1}{n+2}\right)$$
Write $-1=-\frac{n+2}{n+2}$. Then the expression inside the brackets becomes $\frac{(n+2) -1}{n+2}$. Multiplying this by $1/(n+1)$ gives $\frac{(n+2) -1}{(n+1)(n+2)}$.
Now $(n+2)-1=n+1$; but there's $(n+1)$ on both the numerator and the denominator, and therefore this fraction reduces to $1/(n+2)$. Putting it all together, this gives the string of equivalent expressions:
$$1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)} = 1 - \frac{n+1}{(n+1)(n+2)} = 1 - \frac{1}{n+2}$$
| {
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Mathematical Inequality Proof There are $4$ positive real numbers $\,a,b,c,d\,$ and one positive integer $\, p>0 $.
Prove:
$$
\left\lvert\sqrt[p]{a^p+b^p}-\sqrt[p]{c^p+d^p}\right\rvert^p
\leq \left\lvert a-c\right\rvert^p + \left\lvert b-d\right\rvert^p
$$
I have proved the cases $\,p=1$ and $\,p=2$. But the general case is difficult to prove.
Can anyone help me to finish the final prove for the general case? Thanks.
When $\,p=1,\,$
$LHS =\left\lvert (a+b)-(c+d)\right\rvert=\left\lvert (a-c)+(b-d)\right\rvert\leq \left\lvert a-c\right\rvert+\left\lvert b-d\right\rvert$=RHS.
When $\,p=2,\,$
$LHS =\left\lvert\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\right\rvert^2=\left(a^2+b^2\right)-2\sqrt{a^2+b^2}\sqrt{c^2+d^2}+\left(c^2+d^2\right)\,$
$RHS =a^2-2ac+c^2+b^2-2bd+d^2\,$
$RHS-LHS \geq 0$
if and only if $\,-2ac-2bd \geq -2\sqrt{a^2+b^2}\sqrt{c^2+d^2}$
if and only if $\,\big(ac+bd\big)^2\leq \big(a^2+b^2\big)\big(c^2+d^2\big)\,$
if and only if $\,2acbd\leq a^2d^2+b^2c^2\,$
if and only if $\,\big(ad-bc\big)^2\geq 0\,$ which is true and $\,RHS \geq LHS$
| The inequality is equivalent to
$$\left|(a^q+b^q)^{1/q}-(c^q+b^q)^{1/q}\right|\le \left(|a-c|^q+|b-d|^q\right)^{1/q}.$$
Up to a change of notation (|a-c|=x, |b-d|=y) and by symmetry, we only need to show
$$((x+a)^q+(y+b)^q)^{1/q}\le (x^q+y^q)^{1/q}+(a^q+b^q)^{1/q},$$
which is a simple case of the Minkowski Inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $8$ does not divide $x^2-1$, then $x$ is even; prove by contrapositive If $8$ does not divide $x^2-1$, then $x$ is even
proof by contrapositive
the contrapositive of this is :
if $x$ is odd, then $8$ divides $x^2-1$
proof by contrapositive:
Assume $x$ is odd
by definition of odd $∃k∈ℤ$ such that $x=2k+1$
Well, $x^2-1$ = $(2k+1)^2-1$
= $4k^2+4k$
= $4k(k+1)$
Therefore, $k(k+1)$ is an even integer
( i also know the definition of division : $∃m∈ℤ$ such that $x^2-1$ = $8m$)
$x^2-1$ = $4 · 2 · k(k+1)/2$
Therefore,
$8(k^2+1)/2$ is divisible by 8
| 8 does not divide $x^2-1$ is the same as
"if 8 divides $x^2-1$, then x is odd"
which is the same as:
$$8|(x^2-1)\rightarrow 2|(x+1)$$
but
$$x^2-1=(x-1)(x+1)$$
Then since $8=4*2$
Two divides both $x-1$ and $x+1$, and we proven our case
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $\sum_n n^6$ series? I went to OEIS but I didn't quite understand the formula for sum of this series. Could someone type it in normal shape and answer me?
and if it's no bother please do this for 7th, 8th, 9th and 10th powers?
| $$
\sum_{k=1}^n k^6 = \frac{1}{42} n (n+1) (2 n+1) (3 n^4+6 n^3-3 n+1)\\
\sum_{k=1}^n k^7 = \frac{1}{24} n^2 (n+1)^2 (3 n^4+6 n^3-n^2-4 n+2)\\
\sum_{k=1}^n k^8 = \frac{1}{90} n (10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3)\\
\sum_{k=1}^n k^9 = \frac{1}{20} n^2 (n+1)^2 (n^2+n-1) (2 n^4+4 n^3-n^2-3 n+3)\\
\sum_{k=1}^n k^{10} = \frac{1}{66} n (6 n^{10}+33 n^9+55 n^8-66 n^6+66 n^4-33 n^2+5)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?
| ${x^2+y^2\over x+y}=4 \implies x^2+y^2=4x+4y \implies x^2+y^2-4x-4y=0 \implies (x-2)^2+(y-2)^2=(2\sqrt{2})^2$ which is a circle with center $(2,2)$ and radius $2\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?
| You mixed up square roots and cube roots:
$$(1-\sqrt[2]{a})*(1+\sqrt[2]{a})=1-a$$
$$(1-\sqrt[3]{a})*(1+\sqrt[3]{a})≠1-a$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of the expression given below. Is $a,b,c$ are three different positive integers such that :
$$
ab+bc+ca\geq 107
$$
Then what is the minimum value of $a^3+b^3+c^3-3abc$.
I expanded this expression to $(a+b+c)((a+b+c)^2-3(ab+bc+ca))$ and tried to find the minimum value of $(a+b+c)$ by AM-GM inequalities but the problem is that the minimum value occurs at scenario where $a=b=c$, which is ommitted by assumptions of this question.
I'd appreciate some help.
| You can see $a^3+b^3+c^3-3abc=\frac{1}{2}[(a-b)^2+(a-c)^2+(c-b)^2](a+b+c)$
For $[(a-b)^2+(a-c)^2+(c-b)^2]$, since they are three different integers, this is greater or equal to $[(0-1)^2+(0-2)^2+(1-2)^2]=6$;
For $(a+b+c)$, $(a+b+c)=\sqrt{(a+b+c)^2}\geq \sqrt{3(ab+bc+ca)}\geq\sqrt{3*107}>17$. Since the're integers, we have $a+b+c\geq 18$.
Check for $a=5, b=6, c=7$, should be the minimum point.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I solve $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.
$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$
What is wrong with $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}-1+1}{\sqrt{1+x}-1} = \lim_{x \to 0} \frac{\sqrt{1+x}-1}{\sqrt{1+x}-1} + \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ = 1 + DIV?
| Notice, use rationalization as follows $$\lim_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$$
$$=\lim_{x\to 0}\frac{1+x-(1-x^2)}{1+x-1}\times \left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=\lim_{x\to 0}\frac{x^2+x}{x}\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=\lim_{x\to 0}(x+1)\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=(0+1)\left( \frac{\sqrt{1+0}+1}{\sqrt{1+0}+\sqrt{1-0}}\right)$$
$$=1\left(\frac{2}{2}\right)=\color{red}{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Flea on a triangle "A flea hops randomly on the vertices of a triangle with vertices labeled 1,2 and 3, hopping to each of the other vertices with equal probability. If the flea starts at vertex 1, find the probability that after n hops the flea is back to vertex 1."
Could someone provide a hint to help me start this?
| After hop $i$, if the flea is at vertex $1$, then it has $0$ probability to be at vertex $1$ next; if the flea is not at vertex $1$, then it has $0.5$ probability to be at vertex $1$ next.
Let $p_i$ be the probability of the flea at vertex $1$ after $i$ hops, and $q_i = 1-p_i$.
$$\begin{align*}
\begin{pmatrix}p_{i}\\q_{i}\end{pmatrix}
&= \begin{pmatrix}0&0.5\\1&0.5\end{pmatrix}\begin{pmatrix}p_{i-1}\\q_{i-1}\end{pmatrix}\\
\begin{pmatrix}p_0\\q_0\end{pmatrix} &= \begin{pmatrix}1\\0\end{pmatrix}
\end{align*}$$
Then simplifying the recurrence, for example with eigendecomposition,
$$\begin{align*}
\begin{pmatrix}p_n\\q_n\end{pmatrix}
&= \begin{pmatrix}0&0.5\\1&0.5\end{pmatrix} \begin{pmatrix}p_{n-1}\\q_{n-1}\end{pmatrix}\\
&= \begin{pmatrix}1&1\\-1&2\end{pmatrix} \begin{pmatrix}-0.5&0\\0&1\end{pmatrix} \begin{pmatrix}1&1\\-1&2\end{pmatrix}^{-1} \begin{pmatrix}p_{n-1}\\q_{n-1}\end{pmatrix}\\
&\vdots\\
&= \begin{pmatrix}1&1\\-1&2\end{pmatrix} \begin{pmatrix}-0.5&0\\0&1\end{pmatrix}^n \begin{pmatrix}1&1\\-1&2\end{pmatrix}^{-1} \begin{pmatrix}p_0\\q_0\end{pmatrix}\\
&= \frac13 \begin{pmatrix}1&1\\-1&2\end{pmatrix} \begin{pmatrix}(-0.5)^n&0\\0&1^n\end{pmatrix} \begin{pmatrix}2&-1\\1&1\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\\
&= \frac13 \begin{pmatrix}(-0.5)^n&1\\-(-0.5)^n&2\end{pmatrix} \begin{pmatrix}2&-1\\1&1\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\\
&= \frac13 \begin{pmatrix}2(-0.5)^n+1&-(-0.5)^n+1\\-2(-0.5)^n+2&(-0.5)^n+2\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\\
&= \frac13 \begin{pmatrix}2(-0.5)^n+1\\-2(-0.5)^n+2\end{pmatrix}\\
p_n
&= \frac13 + \frac23\cdot\frac1{(-2)^n}
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Based on a coordinate system centered in a sphere where is $M(x, y, z) = 6x - y^2 + xz + 60$ smallest? I am trying to work through a few examples in my workbook, and this one has me completely dumbfounded.
Suppose I have a sphere of radius 6 metres, based on a coordinate system centred in that sphere, at what point on the sphere will $M(x, y, z) = 6x - y^2 + xz + 60$ be smallest?
I have been racking my brain for ages, but I don't see how best to solve this.
If someone has any ideas, I would greatly appreciate your help.
Cheers
Tim
| Using a Lagrange Multiplier
You want to minimize
$$
M(x,y,z) = 6x -y^2 + xz + 60
$$
under the constraint
$$
R(x,y,z) = 6
$$
where
$$
R(x,y,z) = \sqrt{x^2+y^2+z^2}
$$
We define the Lagrange function
$$
\Lambda(x,y,z, \lambda) = M(x,y,z) + \lambda(R(x,y,z)-6)
$$
for some constant $\lambda$ and derive its gradient
$$
\mbox{grad } \Lambda =
(\partial_x \Lambda, \partial_y \Lambda,
\partial_z \Lambda, \partial_\lambda \Lambda)^\top
$$
with
\begin{align}
\partial_x \Lambda &=
6 + z + \frac{\lambda x}{\sqrt{x^2+y^2+z^2}} \\
\partial_y \Lambda &=
-2y + \frac{\lambda y}{\sqrt{x^2+y^2+z^2}} \\
\partial_z \Lambda &=
x + \frac{\lambda z}{\sqrt{x^2+y^2+z^2}} \\
\partial_\lambda \Lambda &=
\sqrt{x^2+y^2+z^2} - 6
\end{align}
A vanishing gradient leads to the system
$$
\left(
\begin{matrix}
-6 \\
0 \\
0
\end{matrix}
\right)
=
\left(
\begin{matrix}
\lambda/6 & 0 & 1 \\
0 & -2 + \lambda / 6 & 0 \\
1 & 0 & \lambda / 6
\end{matrix}
\right)
\left(
\begin{matrix}
x \\
y \\
z
\end{matrix}
\right)
$$
and the constraint $R(x,y,z) = 6$.
The second row implies $\lambda/6 = 2$.
Then the third row gives $x = -2z$ and the first row gives
$2x = -6 - z$ or $6 = 3z$ or $z = 2$ and thus $x = -4$.
To honour the constraint $y$ should be $y = \pm 4$.
So we found the critical points $(-4,\pm 4,2)^\top$ which are on the surface and have the value $M = 12$ (if I made no mistake :-)
Here are some visualizations:
Using Spherical coordinates
Another approach is to move to spherical coordinates
$$
\begin{align}
x &= r \cos \phi \sin \theta \\
y &= r \sin \phi \sin \theta \\
z &= r \cos \theta
\end{align}
$$
This leads to
$$
M(\phi,\theta) =
36 \left(
\cos \phi \sin \theta -
\sin^2 \phi \sin^2 \theta +
\cos \phi \sin \theta \cos \theta
\right) + 60
$$
A vanishing gradients leads to
\begin{align}
0
&= \partial_\phi M \\
&= - \sin \phi \sin \theta -
2 \sin \phi \cos \phi \sin^2 \theta -
\sin \phi \sin \theta \cos \theta \\
&= -\sin \phi \sin \theta(1 + 2 \cos \phi \sin \theta + \cos \theta) \\
0
&= \partial_\theta M \\
&= \cos \phi \cos \theta -
2 \sin^2 \phi \sin \theta \cos \theta +
\cos \phi (\cos^2 \theta - \sin^2 \theta) \\
&= \cos \phi \cos \theta - \sin^2 \phi \sin(2 \theta) +
\cos \phi \cos(2 \theta)
\end{align}
Again some visualizations.
Note: The images seem to be upside down. I will replace them later, after checking again.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Product $abc$ if $a$,$b$,$c$ $\in$ $\mathbb{R}$ and if $$a+\frac{1}{b}=\frac{7}{3}$$ $$b+\frac{1}{c}=4$$ $$c+\frac{1}{a}=1$$ Then find the value of $abc$
I multiplied the three equations with $bc$, $ca$ and $ab$ respectively we get
$$abc+c=\frac{7bc}{3}$$ $$abc+a=4ac$$ $$abc+b=ab$$ Again multiplying above three equations with $a$,$b$ and $c$ respectively we get
$$a(abc)+ac=\frac{7abc}{3}$$ $$b(abc)+ab=4abc$$ $$c(abc)+bc=abc$$ $\implies$
$$ac=(\frac{7}{3}-a)abc$$
$$ab=(4-b)abc$$ and $$bc=(1-c)abc$$ Multiplying above all three equations we get
$$a^2b^2c^2=(\frac{7}{3}-a)(4-b)(1-c)a^2b^2c^2$$ $\implies$
$$1=(\frac{7}{3}-a)(4-b)(1-c)$$ But from first set of equations $\frac{7}{3}-a=\frac{1}{b}$ and so on finally we get
$$1=\frac{1}{abc}$$ so $$abc=1$$
I feel this method is little long...can i have any other approach
| Hint: Multiplying the first three expressions we get $$abc+\left(a+\dfrac{1}{b}\right)+\left(b+\dfrac{1}{c}\right)+\left(c+\dfrac{1}{a}\right)+\dfrac{1}{abc}=\dfrac{28}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$ I was thinking of doing $\binom{8}{4}$ but not sure if right.
| Notice, the binomial expansion $$(2x-y+5)^8=^{8}C_0(2x-y)^85^0+\color{red}{^{8}C_1(2x-y)^75^1}+^{8}C_2(2x-y)^65^2+^{8}C_3(2x-y)^55^3+^{8}C_4(2x-y)^45^4+\dots +^{8}C_8(2x-y)^05^8$$
In the given term $x^3y^4$, the sum of powers $3+4=7$ hence in the above expansion there is only one term $^{8}C_1(2x-y)^75^1$ which has sum of powers $7$ hence using binomial expansion of $^{8}C_1(2x-y)^75^1$ as follows $$^{8}C_1(2x-y)^75^1=\color{blue}{5(^{8}C_1)}[^{7}C_0(2x)^7(-y)^0+^{7}C_1(2x)^6(-y)^1+^{7}C_2(2x)^5(-y)^2+^{7}C_3(2x)^4(-y)^3+\color{red}{^{7}C_4(2x)^3(-y)^4}\ldots +^{7}C_7(2x)^0(-y)^7]$$
Hence, the coefficient of $x^3y^4$ of $5(^{8}C_1)(^{7}C_4)(2x)^3(-y)^4$ is given as $$=5(^{8}C_1)(^{7}C_4)(2)^3(-1)^4$$
$$=5(8)(35)(8)(1)=\color{red}{11200}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Long Inequality problem for $a, b, c $ positive real numbers $$ \left( a+\frac{1}{b} -1\right) \left( b+\frac{1}{c} - 1\right) +\left( b+\frac{1}{c} -1\right) \left( c+\frac{1}{a} -1\right) +\left( c+\frac{1}{a} -1\right) \left( a+\frac{1}{b} -1\right) \geq 3$$
How we can prove the inequality above. Actually it take long time to prove it but I couldn't complete. How we prove it? . Thanks for help
| Maybe this could help:
If you denote $x=a+\frac{1}{b}>0, \, y=b+\frac{1}{c}>0, \, z=c+\frac{1}{a}>0$
you will have $x+y+z=a+1/a +b+1/b+c+1/c\ge 2+2+2=6$ and the equality is achieved iff $a=b=c=1$
You get $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)\ge 3$ which is equivalent to
$$xy+yz+zx-2(x+y+z)\ge 0$$
So you have to prove the last one, having in mind that $x,y,z>0$ and $x+y+z\ge 6$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Calculate $\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx $ Calculate
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx
$$
My try:
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx = \left| {x + 1 = {u^2}} \right| = 2\int {\frac{{(u - \sqrt { - 2 + {u^2}} )}}{{u + \sqrt { - 2 + {u^2}} }}du}
$$
I tried to do first Euler substitute:
$$\sqrt { - 2 + {u^2}} = {u_1} - u
$$
But it did not lead me to the goal. Any thoughts will be appriciated.
| $$\begin{eqnarray*}\int\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\,dx &=& \int\frac{2x-2\sqrt{x^2-1}}{2}\,dx\\ &=&\; C+\frac{x^2}{2}-\frac{x}{2}\sqrt{x^2-1}+\frac{1}{2}\log\left(x+\sqrt{x^2-1}\right).\end{eqnarray*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$.
$(3^{k+1}+7^{k+1}-2)$
$(3^k*3+7^k*7-2)$
But now I'm stuck...
| $(3^{k+1}+7^{k+1}-2)=3^k+7^k-2+2\cdot3^k+6\cdot7^k$
So, it is enough to show that $8$ divides $2\cdot3^k+6\cdot7^k=6\cdot3^{k-1}+6\cdot7^k$
Also, $3^{odd}=3$mod$(8)$ , $3^{even}=-1$mod$(8)$
$7^{odd}=-1$mod$(8)$, $7^{even}=1$mod$(8)$.
implies $6\cdot3^{k-1}+6\cdot7^k=0$mod$(8)$.
| {
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Prove by induction area of koch snowflake For all $n>=0$ prove that the area of a Koch snowflake is $a_n = a_0(\frac{8}{5}-\frac{3}{5}(\frac{4}{9})^n)$ where $a_0=\frac{\sqrt{3}}{4}$
I'm trying to get $P(n+1)$ from $P(n)$ but I'm not sure how to proceed.
For $P(n)$ I now have $\frac{2\sqrt{3}}{5}-\frac{3\sqrt{3}}{20}(\frac{4}{9})^n$.
The target is $P(n+1) = \frac{2\sqrt{3}}{5}-\frac{3\sqrt{3}}{20}(\frac{4}{9})^{n+1}$.
| Do you have to use induction?
Let's say that the flake after $j$ steps has $n_j$ sides of length $s_j$, now since you replace each side with four of length $s_j/3$ and thereby adding a equilateral triangle of side $s_j$ you will end up with $s_{j+1}$ = $s_j/3$ and $n_{j+1} = 4n_j$ and the area added in this step to be $n_j a_0 s_j^2/9$.
Now it's easy to see from this that $s_j = s_03^{-j}$, $n_j = n_04^j$, so the area added is
$\sum a_0 n_j s_j^2/9 = a_0\sum n_0 4^j s_0^2 3^{-2j}/9 = a_0n_0s_0^2\sum(4/9)^j/9$
Then it's just a matter of putting it into the formula for geometric series:
$a_0n_0s_0^2{1 - (4/9)^{N+1}\over1-4/9}/9 = a_0n_0s_0^2{1-(4/9)^{N+1}\over 5/9}/9$
This is however only the area added to the original triangle that has to be added:
$A_n = a_0s_0^2 + a_0n_0s_0^2( {1-(4/9)^{N+1}\over 5/9})/9$
Now we start with a triangle so $n_0=3$ and with unit length we get:
$A_n = a_0( 1 + 3{1-(4/9)^{N+1}\over 5/9}/9) = a_0(1 + 3/5 - {3\over 5}(4/9)^{N+1}) = a_0({8\over5} - {3\over 5}({4\over 9})^{N+1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x^2+y^2+xy=1\;,$ Then minimum and maximum value of $x^3y+xy^3+4\;,$ where $x,y\in \mathbb{R}$
If $x,y\in \mathbb{R}$ and $x^2+y^2+xy=1\;,$ Then Minimum and Maximum value of $x^3y+xy^3+4$
$\bf{My\; Try::} $Given $$x^2+y^2+xy=1\Rightarrow x^2+y^2=1-xy\geq 0$$
So we get $$xy\leq 1\;\;\forall x\in \mathbb{R}$$
and $$x^2+y^2+xy=1\Rightarrow (x+y)^2=1+xy\geq0$$
So we get $$xy\geq -1\;\;\forall x\in \mathbb{R}$$
So we get $$-1\leq xy\leq 1$$
$$\displaystyle f(x,y) = xy(x^2+y^2)+4 = xy(1-xy)+4 = (xy)-(xy)^2+4 = -\left[(xy)^2-xy-4\right]$$
So $$\displaystyle f(x,y) = -\left[\left(xy-\frac{1}{2}\right)^2-\frac{17}{4}\right] = \frac{17}{4}-\left(xy-\frac{1}{2}\right)^2$$
So $$\displaystyle f(x,y)_{\bf{Min.}} = \frac{17}{4}-\left(-1-\frac{1}{2}\right)^2 = 2\;,$$ which is occur when $xy=-1$
But I did not understand how can i calculate $f(x,y)_{\bf{Max.}}$
Plz Help me, Thanks
| You already showed that $$f(x,y) = xy(x^2+y^2)+4=xy(1-xy)+4,$$
which is a quadratic function in $xy$. It remains to find the exact range for $xy$. For which you showed that $xy\ge -1$, equality happens when $(x,y)=(1,-1)$ or $(x,y)=(-1,1)$.
The other restriction $xy\le1$, while correct, never reaches equality. Instead, note that
$$xy\le \frac{x^2+y^2}2=\frac{1-xy}2.$$
So $xy\le \frac13$, equality happens when $x=y=\pm\frac1{\sqrt3}$.
Now, let $t=xy$, then $-1\le t\le \frac13$, and
$$f =t-t^2+4.$$
Note that the derivative with respect to $t$ is $1-2t$ and is positive in the range $[-1,1/3]$. So $f_\min$ is at $t=-1$ and $f_\max$ is at $t=1/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Upper bound on $\ln(\frac{1}{1-x})$ for $0\leq x\leq 1/2$ Prove that $$\ln\left(\frac{1}{1-x}\right)\leq x+2x^2$$ for $0\leq x\leq 1/2$.
I thought about the Taylor series $\ln(1+x)=x-x^2/2+x^3/3-\ldots$. For small $x$, the values $1+x$ and $1/(1-x)$ are very close to each other, so the inequality should hold since in the Taylor expansion we have $-x^2/2$ while in the desired inequality we have $2x^2$. However, we need to prove the inequality up to $x\leq 1/2$, so something more is needed.
| There are many ways. You suggested series, so let's do it that way.
We have
$$\ln\left(\frac{1}{1-x}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots.$$
The part from $\frac{x^2}{2}$ on is $\le \frac{x^2}{2}\left(1+x+x^2+x^3+\cdots\right)$. Note that $1+x+x^2+\cdots=\frac{1}{1-x}$ and $\frac{1}{1-x}\le 2$ on our interval. That gives an inequality sharper than the proposed one.
| {
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"source": "stackexchange",
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Need what the final answer is How do I integrate:
$\int \frac{1}{(1+x\tan x)^2}dx$
Tried the following: substituting $\tan x = \frac{\sin x}{\cos x}$ and got stuck
Next tried $\tan x = \frac{1}{\cot x}$ ,took LCM and did substitution.
Don't know the final answer is correct or not
| your answer is correct
$$\int \frac{1}{(1+x\tan x)^2}dx =\dfrac{\tan x}{x\tan x+1}+Cst $$ and Substitute $$ \tan x = \dfrac{1}{\cot x}$$
\begin{align}
\int \frac{1}{(1+x\tan x)^2}dx&=\dfrac{\tan x}{x\tan x+1}+Cst\\
&=\dfrac{\dfrac{1}{\cot x}}{x\dfrac{1}{\cot x}+1}+Cst \\
&=\dfrac{1}{x+\cot x}+Cst \\
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Fundamental Identities I was given a task to prove:
$$\frac{\sec\theta}{\sec\theta\tan\theta} = \sec\theta(\sec\theta + \tan\theta)$$
Then I replaced them with their Ratio and Reciprocal Identities
\begin{align*}
\sec\theta & = \frac{1}{\cos\theta}\\
\tan\theta & = \frac{\sin\theta}{\cos\theta}
\end{align*}
so I came up with this:
$$\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta}} = \frac{1}{\cos\theta}\left(\frac{1}{\cos\theta} + \frac{\sin
\theta}{\cos\theta}\right)$$
then,
$$\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1}{\cos\theta}\left(1+\frac{\sin\theta}{\cos\theta}\right)$$
and I had the reciprocal,
$$\frac{1}{\cos\theta} \cdot \frac{\cos\theta}{\sin\theta} = \frac{1}{\cos\theta}\left(1+\frac{\sin\theta}{\cos\theta}\right)$$
and I don't know what to do next, can someone explain to me how? T_T
| Well the identity you mentioned is not correct:
Let $\theta = \frac{\pi}{4}$. Them your identity becomes $$\frac{\sec{\frac{\pi}{4}}}{(\sec{\frac{\pi}{4}})(\tan{\frac{\pi}{4}})}=1$$But the result you mentioned gives a different answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\ \epsilon\ C : |Z-1|+|Z+1|=4\}$
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=16$
$x^2+y^2=7$
$(x-0)^2+(y-0)^2=\sqrt7$
=This is a circle with center $0$ and radius $\sqrt7$
My answer is different from the correct answer given: "This is an ellipse with foci at $-1$ and $1$ passing through $2$"
I have no idea how to get to this answer. Could someone please help me here?
| Hint:
For brevity, let us write $R_1:=|z-1|, R_2:=|z+1|$, which are square roots of a quadratic polynomial.
Squaring the sum,
$$(R_1+R_2)^2=R_1^2+2R_1R_2+R_2^2=a^2.$$
Then
$$4R_1^2R_2^2=(a^2-R_1^2-R_2^2)^2
=a^4+R_1^4+R_2^4-2a^2R_1^2-2a^2R_2^2+2R_1^2R_2^2,$$
$$a^4+R_1^4+R_2^4-2a^2R_1^2-2a^2R_2^2-2R_1^2R_2^2=0,$$
$$a^4+(R_1^2-R_2^2)^2-2a^2R_1^2-2a^2R_2^2=0.$$
As $R_1^2-R_2^2$ simplifies to the square of a linear expression, you get a quadratic equation, i.e. a conic.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find powers of 3 mod 17 Show that powers of $3\ (\text{mod}\ 17)$ are $3,9,10,13,15,11,16,14,8,7,4,12,2,6,1$.
What is meaning of power of something, how do I proceed this question? Just provide me hint to get start in this problem.
| Let us try a very naive approach. I should add that this is precisely the method suggested in André Nicolas's comment.
You start by $3^0=1$ and multiply both sides by $3$ repeatedly
$$3^0 \equiv 1 \pmod{17}\\
3^1 \equiv 3 \pmod{17}\\
3^2 \equiv 9 \pmod{17}$$
The next result would be $3^3 \equiv 27 \pmod{17}$. But $27\equiv10\pmod{17}$.
$$
3^3 \equiv 10 \pmod{17}\\
3^4 \equiv 13 \pmod{17}\\
$$
We can also use $3^4 \equiv -4 \pmod{17}$ if it makes computation easier. In this case we either multiply $-4$ or $13$ by $3$. But in both case we get $-12 \equiv 5 \pmod{17}$ and $39 \equiv 5 \pmod{17}$.
$$
3^5 \equiv 5 \pmod{17}\\
3^6 \equiv 15 \pmod{17}\\
3^7 \equiv 11 \pmod{17}\\
3^8 \equiv 16 \pmod{17}\\$$
We continue this way until we get $1$ and the sequence starts to repeat. Or we could notice that $3^8\equiv-1\pmod{17}$, so from this point on we can repeat the same numbers we already have but with the minus sign.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question on Converting between Base Number Systems Question on Converting Between Base Numbers
The people of Jupiter use Base 13. Therefore, their numerals are 0,1,2,3,4,5,6,7,8,9,A,B,C.
The people on Saturn use Base 7. Therefore, their numerals are 0,1,2,3,4,5,6.
A person on Jupiter has A906BC Jupits on Jupiter. This same person has 65325406 Jupits in her bank account on Saturn.
Which bank account holds more money and by how much?
Thanks so much, really having trouble here with this and any help would be much appreciated!!!
| Jovian bank account: $10 \cdot 13^5 + 9 \cdot 13^4 + 6 \cdot 13^2 + 11 \cdot 13^1 + 12 = 3971148 \text{ Jupits}$
Saturnian bank account: $6 \cdot 7^7 + 5 \cdot 7^6 + 3 \cdot 7^5 + 2 \cdot 7^4 + 5 \cdot 7^3 + 4 \cdot 7^2 + 6 = 5586643 \text{ Jupits}$
The remaining parts should be easy for you.
| {
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The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$
I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not solve it further.Please help me.Thanks.
| Using $\displaystyle \bf{A.M\geq G.M}$
$$\displaystyle \frac{4\sin^2 x+\csc^2 x}{2}\geq \sqrt{4\sin^2 x\cdot \csc^2 x}\Rightarrow 4\sin^2 x+\csc^2 x\geq 4$$
and equality hold when $$4\sin^2 x=\csc^2 x$$
and $$\displaystyle \frac{\tan^2 x+\cot^2 x}{2}\geq \sqrt{\tan^2 x\cdot \cot^2 x}\Rightarrow \tan^2 x+\cot^2 x\geq 2$$
and equality hold when $$\tan^2 x= \cot^2 x$$
Now adding these two , We get $$\displaystyle 4\sin^2 x+\csc^2 x+\tan^2 x+\cot^2 x\geq 6$$
and here equality condition is satisfied
So we get $$4\sin^2 x=\csc^2 x$$ and $$\tan^2 x= \cot^2 x$$ where $0\leq x\leq 2\pi$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rank of a matrix of binomial coefficients This question arose as a side computation on error correcting codes.
Let $k$, $r$ be positive integers such that $2k-1 \leqslant r$ and let $p$ a prime number such that $r < p$. I would like to find the rank of the following $k \times k$ matrix with coefficients in the field $F_p$
$$
\begin{pmatrix}
\binom {r}{k} & \binom {r}{k+1} & \dotsm & \binom {r}{2k-1}\\
\vdots & \\
\binom {r}{2} & \binom {r}{3} & \dotsm & \binom {r}{k+1} \\
\binom {r}{1} & \binom {r}{2} & \dotsm & \binom {r}{k}
\end{pmatrix}
$$
where all binomial coefficients are taken modulo $p$. I conjecture the rank should be $k$ but I have no formal proof. I am aware of Lucas's theorem, but it didn't help so far.
| Not true, unfortunately.
Indeed, by adding the $2$nd row to the $1$st, $3$rd to $2$nd, ... , $k$th to $(k-1)$th etc, you end up with the matrix
$$
\begin{pmatrix}
\binom {r+1}{k} & \binom {r+1}{k+1} & \dotsm & \binom {r+1}{2k-1}\\
\vdots & \vdots & \ddots & \vdots \\
\binom {r+1}{2} & \binom {r+1}{3} & \dotsm & \binom {r+1}{k+1} \\
\binom {r}{1} & \binom {r}{2} & \dotsm & \binom {r}{k}
\end{pmatrix}
$$
which has the same rank.
In particular, for $r=p-1$ we get $\pmod p$
$$
\begin{pmatrix}
0 & 0 & \dotsm & 0\\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dotsm & 0 \\
* & * & \dotsm & *
\end{pmatrix}.
$$
I doubt this has rank $k$.
Moreover, it is easy to see with this argument that the rank is at most $p-r$.
Update: For $k\ge p-r$, the rank is exactly $p-r$. Sketch: make the above transformations with first $k-1$ rows, then with first $k-2$ rows, etc. Then do the same to columns. You will end up with
$$
\begin{pmatrix}
\binom {r+k-1}{k} & \binom {r+k}{k+1} & \dotsm & \binom {r+2k-2}{2k-1}\\
\vdots & \vdots & \ddots & \vdots \\
\binom {r+2}{3} & \binom {r+3}{4} & \dotsm & \binom {r+k+1}{k+2} \\
\binom {r+1}{2} & \binom {r+2}{3} & \dotsm & \binom {r+k}{k+1} \\
\binom {r}{1} & \binom {r+1}{2} & \dotsm & \binom {r+k-1}{k}
\end{pmatrix}
$$
Now $\pmod p$ the non-zero elements of this matrix form a lower-left triangle of size $p-r$ (I hope this is clear, as I don't know how LaTeX such thing). Hence, the rank is $p-r$.
It is also possible to do this for $k>p-r$, which gives only lower bound $2p-2r-k$ for the rank; not a lot of help, probably.
| {
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Solution to this recurrence? Is there exists a solution to this recurrence.
$$F(N,1) = N, N≥1$$
$$F(N,K) = \frac {1}{\lfloor\frac 1{F(N-1,K-1)} -\frac 1{F(N,K-1)}\rfloor} \;\;\;\;2≤K≤N$$
I tried to simplify the equation but i am not able to find F(1,2) and thus unable to proceed.
I am new to this site,so please let me know if i have not asked the question properly.
| The original recurrence relation is equivalent to
$$
\frac{1}{F(N,K)}
=
\frac{1}{F(N-1,K-1)}
-
\frac{1}{F(N,K-1)}.
$$
For $K = 2$, we have
\begin{align}
\frac{1}{F(N,2)}
&=
\frac{1}{F(N-1,1)}
-
\frac{1}{F(N,1)} \\
&=
\frac{1}{N-1}
-
\frac{1}{N}
\\
&=
\frac{1}{N\,(N-1)}.
\end{align}
Next, for $K = 3$, we have
\begin{align}
\frac{1}{F(N,3)}
&=
\frac{1}{F(N-1,2)}
-
\frac{1}{F(N,2)} \\
&=
\frac{1}{(N-1) \, (N-2)}
-
\frac{1}{N \, (N-1)}
\\
&=
\frac{N}{N \, (N-1) \, (N-2)}
-
\frac{N-2}{N \, (N-1) \, (N-2)}
\\
&=
\frac{2}{N\,(N-1)\,(N-2)}.
\end{align}
So we can guess,
(1)
\begin{align}
\frac{1}{F(N,K)}
&=
\frac{(K-1)!}{N\,(N-1)\,(N-2)\,(N-K+1)}.
\end{align}
Let us prove it by induction,
suppose (1) holds for $K = k$,
For $K = k+1$,
\begin{align}
\frac{1}{F(N,k+1)}
&=
\frac{1}{F(N-1,k)}
-
\frac{1}{F(N,k)} \\
&=
\frac{(k-1)!}{(N-1) \, (N-2) \, (N-k)}
-
\frac{(k-1)!}{N \, (N-1) \, (N-k+1)}
\\
&=
\frac{(k-1)! \, N }{ N \, (N-1) \, (N-2) \dots (N-k)}
-
\frac{(k-1)! \, (N-k)}{N \, (N-1) \, (N-k+1)\dots (N-k)}
\\
&=
\frac{k!}{N\,(N-1)\,(N-2)\,(N-k)},
\end{align}
which means (1) also holds for $K = k+1$.
So
$$
F(N,K) = \frac{N(N-1)\dots(N-K+1)}{(K-1)!} = N \, {N-1 \choose K-1}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The names of 8 students are listed randomly. What is the probability that Al, Bob, and Charlie are all in the top 4? My attempt at a solution: There are $C_{4,3}$ ways to arrange Al, Bob, and Charlie in the top 4, and $C_{5,4}$ ways to arrange the other 5 people. All of this is over $C_{8,4}$, giving $$\frac{C_{4,3}C_{5,4}}{C_{8,4}}=\frac{2}{7}$$ Is this correct?
| Here is a slightly different approach:
There are $\dbinom{8}{4}$ ways to select the top 4, and
there are $\dbinom{5}{1}$ ways to choose the top 4 including Al, Bob, and Charlie
$\;\;\;$(since 1 person out of the remaining 5 must be selected),
so the probability is $\displaystyle\frac{5}{\binom{8}{4}}=\frac{5}{70}=\frac{1}{14}$.
Alternate solution:
There are $8!$ ways to list the 8 people in order, and
there are $\binom{4}{3}$ ways to choose the places for Al, Bob, and Charlie in the top 4, $3!$ ways to arrange them in these places, and $5!$ ways to arrange the remaining 5 people in their spots;
so this gives a probability of $\displaystyle\frac{\binom{4}{3}3!5!}{8!}=\frac{4\cdot6}{6\cdot7\cdot8}=\frac{1}{14}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximizing $\sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta$ I need to maximize
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \tag{1}$$
where $\alpha, \beta \in [0, \frac{\pi}{2}]$.
With numerical methods I have found that
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \leq 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} - \sin^2 \frac{\alpha + \beta}{2}. \tag{2} $$
If $(2)$ is true then I can denote $x = \frac{\alpha + \beta}{2}$ and prove (using Cauchy inequality) that
$$ 2 \sin x \cos x - \sin^2 x \leq \frac{\sqrt{5}-1}{2}. \tag{3}$$
But is $(2)$ true? How do I prove it? Maybe I need to use a different idea to maximize $(1)$?
| Let $$f = \sin \alpha \cos \alpha + \sin \beta \cos \beta -\sin \alpha \sin \beta = \frac{1}{2}(\sin 2\alpha + \sin 2\beta) -\sin \alpha \sin \beta$$
$$\frac{\partial f}{\partial \alpha} = 0 \implies \cos 2\alpha - \sin \beta \cos \alpha = 0$$
$$\frac{\partial f}{\partial \beta} = 0 \implies \cos 2\beta - \sin \alpha \cos \beta = 0$$
From symmetry (also may not be difficult to show directly) $$\beta = \alpha$$
Finally, $$\alpha = \beta = \frac{1}{2} \tan^{-1} 2$$
Substituting in $f$, the maximum value is 0.61803398874989484820
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to 0} \frac{1-\cos (1-\cos (1-\cos x))}{x^{a}}$ is finite then the max value of $a$ is? I know the formula $\frac{1-\cos x}{x^{2}}=\dfrac{1}2$
But how do i use it here and tried l hospital and no use and doing l hospital twice will make it very lengthy how do i approach ?
| Using Taylor series: when $x\to 0$, we have
$\cos x = 1-\frac{x^2}{2} + o(x^2)$.
Step by step, we can use this as follows:
$$
1-\cos x = \frac{x^2}{2} + o(x^2)
$$
so that, plugging it back:
$$
\cos(1-\cos x) = 1-\frac{1}{2}\left(\frac{x^2}{2} + o(x^2)\right)^2 = 1-\frac{x^4}{8} + o(x^4)
$$
and therefore
$$
1-\cos (1-\cos x) = \frac{x^4}{8} + o(x^4).
$$
One more time:
$$
\cos(1-\cos (1-\cos x)) = \cos(\frac{x^4}{8} + o(x^4)) = 1 - \frac{1}{2}\cdot\frac{x^8}{64} + o(x^8)
$$
leading to
$$
\frac{1-\cos(1-\cos (1-\cos x))}{x^a} = \frac{\frac{x^8}{128}+o(x^8)}{x^a}
$$
Note that here, expanding to the second order $\cos x = 1+\frac{x^2}{2} + o(x^2)$ was enough for the "composition of Taylor series" to work; and that the whole approach relies on the fact that when $x\to0$, then $x^k\to 0$ (for $k=2$ and $k=4$), so that we have $\cos(\alpha x^k) = 1-\frac{(\alpha x^k)^2}{2}+o(x^{2k})$.
| {
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Number Theory: How to solve $x^2\equiv 4\pmod{143}$? I'm not sure how to solve this quadratic congruence:
$x^2 \equiv 4\pmod{143}$
Thanks!
So here's what I did, but the solution seems a bit long the way that I did it:
The four solutions are
1: $x\equiv 2\pmod{11}, x\equiv 2\pmod{13}$:
$x = 2 + 11k\implies 2+11k\equiv 2\pmod{13}\implies 11k\equiv 0\pmod{13}$
$\implies k\equiv 0\pmod{13}\implies k=13k\implies x=2+143k\implies x\equiv2\pmod{143}$.
2: $x\equiv -2\pmod{11}, x\equiv -2\pmod{13}$:
$x = -2 + 11k\implies -2+11k\equiv -2\pmod{13}\implies 11k\equiv 0\pmod{13}$
$\implies k\equiv 0\pmod{13}\implies k=13k\implies x=-2+143k\implies x\equiv-2 \equiv 141\pmod{143}$.
3: $x\equiv -2\pmod{11}, x\equiv 2\pmod{13}$:
$x=-2+11k\implies -2+11k\equiv 2\pmod{13}\implies 11k\equiv 4\equiv 121\pmod{13}$
$\implies k\equiv 11\pmod{13}\implies k=11+13i\implies x=119+143i\implies x\equiv 119\pmod{143}$.
4: $x\equiv 2\pmod{11}, x\equiv -2\pmod{13}$:
$x=2+11k\implies 2+11k\equiv -2\pmod{13}\implies 11k\equiv -4\equiv 22\pmod{13}$
$\implies k\equiv 2\pmod{13}\implies k=2+13i\implies x=24+143i\implies x\equiv 24\pmod{143}$.
So all incongruent solutions of $x^2\equiv 4\pmod{143}$ are:
$x\equiv 2, 24, 119, 141 \pmod{143}$.
| The four possible solutions are
$x\equiv 2\ (mod\ 11)$ and $x\equiv 2\ (mod\ 13)$
$x\equiv 2\ (mod\ 11)$ and $x\equiv -2\ (mod\ 13)$
$x\equiv -2\ (mod\ 11)$ and $x\equiv 2\ (mod\ 13)$
$x\equiv -2\ (mod\ 11)$ and $x\equiv -2\ (mod\ 13)$
The solutions $2$ and $143-2=141$ are easy to get, a bit more
difficult are the other solutions.
Following the numbers $2,13,24$ of the form $11k+2$, we see that $24$
is a solution.
Following the numbers $9,20,31,42,53,64,75,86,97,108,119$ of the form $11k-2$, we finally
find $119$ to be the last solution. Note, that $24+119=143$, so the
last solution could be found more quickly this way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first inequality
$$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$
second inequality:
$$\frac{a+b+c}{3} > \sqrt[3]{abc}$$
I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
| Late answer since it was tagged as duplicate there and this way of solving is neither here nor there.
\begin{eqnarray*} abc(a+b+c)
& \stackrel{GM-AM}{\leq} & \left(\frac{a+b+c}{3}\right)^3(a+b+c)\\
& = & 3\left(\frac{a+b+c}{3}\right)^4 \\
& \stackrel{x^4 \; is \; convex}{\leq} & 3\cdot \frac{1}{3}(a^4+b^4+c^4) = a^4+b^4+c^4
\end{eqnarray*}
Note that the inequality becomes sharp ("<"), if at least two of the numbers are different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How to find the inverse Laplace transform of this? I am looking for the inverse Laplace transform of $$\frac{1}{s-1}e^{-\sqrt{s}x}$$
This is for an introductory partial differential equations class so I am thinking that it should not be too hard. If it wasn't for the $s-1$ in the denominator and we had just $s$ instead, then the transform would simply be $$\textrm{erfc} \left(\frac{x}{2\sqrt{t}}\right) $$
Edit: Here is the problem from where that transformation originated.
| You could simply convolve the ILT with that of $1/(s-1)$, which is $e^t$. But I prefer using the definition of the ILT in the complex plane.
In this case, consider the contour integral
$$\oint_C dz \frac{e^{-x \sqrt{z}}}{z-1} e^{t z}$$
where $C$ is a Bromwich contour in the left half-plane with a detour along the negative real axis and around the branch point at $z=0$, as pictured below.
(The pole at $z=1$ is represented by the small disk.)
The contour integral is equal to, taking the limits as the outer radius goes to $\infty$
$$\int_{c-i \infty}^{c+i \infty} \frac{ds}{s+1} e^{-x \sqrt{s}} e^{s t} + \int_{\infty}^0 \frac{dy}{y+1} e^{-i x \sqrt{y}} e^{-t y} + \int_0^{\infty} \frac{dy}{y+1} e^{i x \sqrt{y}} e^{-t y}$$
The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=1$. Thus,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s+1} e^{-x \sqrt{s}} e^{s t} = e^{-x} e^t - \frac1{\pi} \int_0^{\infty} dy \, \frac{\sin{x \sqrt{y}}}{1+y} e^{-t y} $$
To evaluate the integral, sub $y=u^2$ and get
$$\int_0^{\infty} dy \, \frac{\sin{x \sqrt{y}}}{1+y} e^{-t y} = \int_{-\infty}^{\infty} du \frac{\sin{x u}}{u} e^{-t u^2} - \int_{-\infty}^{\infty} \frac{du}{1+u^2} \frac{\sin{x u}}{u} e^{-t u^2}$$
The first integral may be evaluated using Parseval's theorem:
$$\int_{-\infty}^{\infty} du \frac{\sin{x u}}{u} e^{-t u^2} = \frac1{2 \pi} \int_{-x}^{x} dk \, \pi \, \sqrt{\frac{\pi}{t}} e^{-k^2/(4 t)} = \pi \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )}$$
The second integral may be expressed as an integral of the first. Write that integral as
$$\begin{align}e^t \int_{-\infty}^{\infty} \frac{du}{1+u^2} \frac{\sin{x u}}{u} e^{-t (1+u^2)} &= e^t \int_t^{\infty} dt' \, e^{-t'} \, \int_{-\infty}^{\infty} du \frac{\sin{x u}}{u} e^{-t' u^2} \\ &= \pi\, e^t \int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} \end{align}$$
Now I will state the result here for now so the forest isn't lost for the trees:(*)
$$\int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} = e^{-t} \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )} + \frac12 e^x \operatorname{erfc}{\left (\sqrt{t}+\frac{x}{2 \sqrt{t}} \right )} - \frac12 e^{-x} \operatorname{erfc}{\left (\sqrt{t}-\frac{x}{2 \sqrt{t}} \right )} $$
and thus
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s-1} e^{-x \sqrt{s}} e^{s t} = e^{t-x} - \frac12 e^{t-x} \operatorname{erfc}{\left (\sqrt{t}-\frac{x}{2 \sqrt{t}} \right )} + \frac12 e^{t+x} \operatorname{erfc}{\left (\sqrt{t}+\frac{x}{2 \sqrt{t}} \right )}$$
ADDENDUM
I will now derive the result (*). Begin by integrating by parts:
$$\int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} = e^{-t} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )} - \frac{x}{2 \sqrt{\pi}} \int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')}$$
$$\underbrace{\int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')}}_{t'=1/u^2} = 2 \int_0^{1/\sqrt{t}} du \, e^{-1/u^2} e^{-x u^2/2} = 2 e^{-x} \int_0^{1/\sqrt{t}} du \, e^{-\left (\frac1{u} - \frac{x}{2} u \right )^2}$$
Now sub $v=\frac1{u} - \frac{x}{2} u$. Then
$$u = \frac1{x} \left (\sqrt{v^2+2 x}-v \right ) $$
$$du = \frac1{x} \left (\frac{v}{\sqrt{v^2+2 x}}-1 \right ) dv $$
and carrying out the substitution...
$$\int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')} = \frac{2 e^{-x}}{x} \int_{\infty}^{\sqrt{t} - \frac{x}{2 \sqrt{t}}} dv \, \left (\frac{v}{\sqrt{v^2+2 x}}-1 \right ) e^{-v^2}$$
The first term evaluates to
$$\begin{align}\int_{\infty}^{\sqrt{t} - \frac{x}{2 \sqrt{t}}} dv \, \frac{v}{\sqrt{v^2+2 x}} e^{-v^2} &= \frac12 \int_{\infty}^{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}}\right )^2} dw\, (w+2 x)^{-1/2} e^{-w}\\ &= \frac12 e^{2 x} \int_{\infty}^{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}}\right )^2} dw\, w^{-1/2} e^{-w} \\ &= e^{2 x} \int_{\infty}^{\sqrt{t} + \frac{x}{2 \sqrt{t}}} dy \, e^{-y^2} \\ &= -\frac{\sqrt{\pi}}{2} e^{2 x} \operatorname{erfc}{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}} \right )} \end{align}$$
The second term is obviously simpler:
$$\int_{\infty}^{\sqrt{t} - \frac{x}{2 \sqrt{t}}} dv \, e^{-v^2} = -\frac{\sqrt{\pi}}{2} \operatorname{erfc}{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}} \right )} $$
Therefore...
$$\int_t^{\infty} dt' \,t'^{-3/2} \, e^{-t'} e^{-x^2/(4 t')} =\frac{\sqrt{\pi}}{x} \left [e^{-x} \operatorname{erfc}{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}} \right )} - e^{x} \operatorname{erfc}{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}} \right )} \right ]$$
and...
$$\int_t^{\infty} dt' \, e^{-t'} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t'}} \right )} = e^{-t} \, \operatorname{erf}{\left (\frac{x}{2 \sqrt{t}} \right )} - \frac12 \left [e^{-x} \operatorname{erfc}{\left (\sqrt{t} - \frac{x}{2 \sqrt{t}} \right )} - e^{x} \operatorname{erfc}{\left (\sqrt{t} + \frac{x}{2 \sqrt{t}} \right )} \right ]$$
ADDENDUM II
A little manipulation puts the answer in a slightly nicer form:
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s-1} e^{-x \sqrt{s}} e^{s t} = e^{t} \cosh{x} - e^{t} \frac12 \left [ e^{x} \operatorname{erf}{\left (\sqrt{t}+\frac{x}{2 \sqrt{t}} \right )} - e^{-x} \operatorname{erf}{\left (\sqrt{t}-\frac{x}{2 \sqrt{t}} \right )}\right ]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $x= \sin(k- x)$? Is there a way to solve $x = \sin (k-x)$ without a computer, that is with a pocket calculator or pencil and paper?
| Let $f(x) = x - \sin(k-x)$. It is trivial that there is no zero when $|x| > 1$. We also know that $f(0) = -\sin(k) \ne 0$.
The problem with the Newton's method is that it will fail when $f'(x) = 0$.
Let $x-k=y$, $g(y) = f(y+k) = y + k + sin(y)$. Let's also constrain $k \in [-pi/2, pi/2]$.
Using the Taylor's series for $sin(y)$, $g(y) \approx y + k + y - y^3/6$.
Thus:
$$
g(y) = 0 \\
k + 2y - y^3/6 = 0 \\
y^3/6 - 2y - k = 0 \\
y^3 - 12y - 6k = 0
$$
I used the WolframAlpha to help solving this polinomial.
All three solutions can be seen here
You can notice that only one solution is between $[-1, 1]$. Take a look at this solution.
Let $w = 3k + \sqrt{-64+9 k^2} = 3k + i\sqrt{64-9 k^2}$. But, as $|w| = 8$, it can be rewritten as $w = 8 cis(\alpha)$, $\alpha = arg(w) = acos(3k/8)$.
Let $z = \sqrt[3]{w} = 2cis(\alpha/3)$.
Thus, our root is $y = \frac{2i(\sqrt{3}+i)}{z} - \frac{(1+i\sqrt{3})z}{2}$. Using the polar form for complex numbers, $y = 2cis\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2cis\left( \frac{\pi}{3} + \frac{\alpha}{3} \right)$.
But we already know that $y$ is a real number. So, let's get only the real part.
$$
y = 2cis\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2cis\left( \frac{\pi}{3} + \frac{\alpha}{3} \right) \\
y = 2\cos\left(\frac{2pi}{3}-\frac{\alpha}{3}\right) - 2\cos\left( \frac{\pi}{3} + \frac{\alpha}{3} \right) \\
y = -4 \sin\left( \frac{\pi}{6} - \frac{\alpha}{3} \right) \\
y = -4 \sin\left( \frac{\pi}{6} - \frac{cos^{-1}(3k/8)}{3} \right)
$$
Finally, $x = k -4 \sin\left( \frac{\pi}{6} - \frac{cos^{-1}(3k/8)}{3} \right)$, if $k \in [-\pi/2, \pi/2]$.
Let's check: $$
k=\pi/4, x = 0.38744, f(x)=-0.00009688 \\
k=\pi/3, x = 0.51073, f(x)=-0.00037297
$$
Notice that any value of $k$ can be mapped in the interval $[-\pi, \pi]$.
We still have to solve for $k \in [-pi, -pi/2)$ and $k \in (pi/2, \pi]$. But, notice that, for k in these intervals, we can just map $k$ to the interval $[-\pi/2, \pi/2]$ and invert the sign of $sin(x-k)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the coordinates of the points on the curve $y=2x^4 - 3x^2 + x - 7$ where the gradient is parallel to the line $y=3x$ I got that $\frac{dy}{dx}= 3$
(because of the gradient of the parallel line)
and then, I found the derivative of the equation of the curve to be $8x^3 -6x +1= 3$.
This simplifies as $8x^3 - 6x -2 =0$
From here, I simplified the equation and got $4x^3- 3x- 1= 0$
And I don't know how to solve this equation for $x$.
| Gradient of $y=3x$ is $3$.
So by differentiating, we have $\frac{dy}{dx}=8x^3-6x+1$
Now solve $8x^3-6x+1=3$.
Clearly $8x^3-6x+1=3\Rightarrow 8x^3-6x-2=0$.
$\Rightarrow 8x^3-6x-2=(x-1)(8x^2+8x+2)=0\Rightarrow x=1$.
Thus coordinates of the point $=(1,-7)$
| {
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"url": "https://math.stackexchange.com/questions/1484172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the maximum value of $ \sin x +\sin {2x} (0I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ \sin x +\sin {2x} (0<x<\frac{\pi}{2})$$
I have done my work here
$$f (x)= \sin x +\sin {2x} $$
$$f'(x)= \cos x +2\cos {2x}=\cos x + 4\cos^2 x-2 =0$$
$$\cos x=\frac{1}{8} \left(\sqrt{33}-1\right),\cos x=\frac{1}{8} \left(-\sqrt{33}-1\right)$$
Done...!
| $$f (x)= \sin x +\sin {2x} $$
$$f'(x)= \cos x +2\cos {2x}=\cos x + 4\cos^2 x-2 = 0 \implies
\cos x = \frac{-1 \pm \sqrt {33}}{8}$$
Since we are interested in the maxima only, discard the negative root. In general, $$x = 2k\pi \pm \cos^{-1}\frac{\sqrt{33}-1}{8} = 2k\pi \pm 0.93592946$$
However, in this case $$x = 0.93592946$$
Since $$f''(x) = -(\sin x + 4 \sin 2x)$$ verify that for the above value of $x$, you really have $f''(x) < 0$
In fact, $$f''(0.93592946) = -4.62524$$ and the maxima is $$f(0.93592946) = 1.76017$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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representation of a complex number in polar from
Write the following in polar form: $\frac{1+\sqrt{3}i}{1-\sqrt{3}i}$
$$\frac{1+\sqrt{3}i}{1-\sqrt{3}i}=\frac{1+\sqrt{3}i}{1-\sqrt{3}i}\cdot\frac{1+\sqrt{3}i}{1+\sqrt{3}i}=\frac{(1+\sqrt{3}i)^2}{1^2+(\sqrt{3})^2}=\frac{1+2\sqrt{3}i-3}{4}=\frac{-2+2\sqrt{3}i}{4}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$$
$|z|=r=\sqrt{(-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$
arg$z$=$tan^{-1}(\frac{\sqrt{3}}{2}\cdot -2)=-\frac{\pi}{3}$
acorrding to Wolfram $\theta=0$
| You forget the brackets in your wolfram formula, the correct one is this and as you can see your result is correct.
| {
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"question_score": "3",
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Use of property of ratio and factorization method Recently I have faced a problem related to algebraic ratio.I have tried much but I can not find any clue.Can anyone give me some hints.
Here is the problem.
I am given $$\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\neq0$$
I have to prove: $(a+b+c)(x+y+z)=ax+by+cz$
Please help me to solve the problem with clear explanation
| Define $k$ by $$\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}=k\neq0,$$
so $1/k\ne0$.
Now, finding the product, we have
$$\begin{align}(a+b+c)(x+y+z)=&\;ax+by+cz\\&+a(y+z)\\ &+b(x+z)\\&+c(x+y);\end{align}$$
thus we need to show that $a(y+z)+b(x+z)+c(x+y)=0$.
Since
$$\frac{x^2-yz}{k}\cdot(y+z)=a(y+z)\,$$
$$\frac{y^2-zx}{k}\cdot(x+z)=b(x+z)\,$$
$$\frac{z^2-xy}{k}\cdot(x+y)=c(x+y),$$
we obtain
$$\begin{align}a(y+z)+b(x+z)+c(&x+y)\\&=\frac1k((x^2-yz)(y+z)\\&\qquad+(y^2+zx)(x+z)\\&\qquad+(z^2+xy)(x+y))\\&=0.\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\int _0^{2\pi }(1-\cos t)(\frac{5}{4}-\cos t)^{-3/2}\,dt$
How to compute
$$ \int_{0}^{2\pi}\dfrac{1-\cos(t)}{\biggl(\dfrac{5}{4}-\cos(t)\biggr)^{\dfrac{3}{2}}} dt $$
I'm interested in more ways of computing this integral.
My thoughts:
I'm tired to use Bioche rules :
Let $x=\operatorname{tg}(\frac{t}{2})$ then $$\cos(t)=\dfrac{1-x^{2}}{1+x^{2}},\qquad dt=\dfrac{2}{1+x^{2}}dx$$
\begin{align}
\int_{0}^{2\pi}\dfrac{1-\cos(t)}{\biggl(\dfrac{5}{4}-\cos(t)\biggr)^{\dfrac{3}{2}}} dt&=\int_{0}^{2\pi}\dfrac{1-\dfrac{1-x^{2}}{1+x^{2}}}{\biggl(\dfrac{5}{4}-\dfrac{1-x^{2}}{1+x^{2}}\biggr)^{\dfrac{3}{2}}} \dfrac{2}{1+x^{2}}dx \\
&=\int_{0}^{2\pi}\dfrac{\frac{2x^2}{x^2+1}}{\left(\frac{9x^2+1}{4\left(x^2+1\right)}\right)^{\frac{3}{2}}}\dfrac{2}{1+x^{2}}dx\\
&=\int_{0}^{2\pi}\frac{4x^2}{\left(\frac{9x^2+1}{4\left(x^2+1\right)}\right)^{\frac{3}{2}}\left(x^2+1\right)^2}dx
\end{align}
| First we can see that the integral is twice the integral over $[0,\pi]$ and further using $\cos t=1-2\sin^2(t/2)$ and substitution $t=2x$ we get $$I=64\int_{0}^{\pi/2}\frac{\sin^2x}{(1+8\sin^2x)^{3/2}}\,dx=64\int_{0}^{\pi/2}\frac{\sin^2x}{(9-8\cos^2x)^{3/2}}\,dx$$ and this further leads to $$I=\frac{64}{27}\int_{0}^{\pi/2}\frac{1-\sin^2x}{(1-k^2\sin^2x)^{3/2}}\,dx$$ where $k^2=8/9$. It should now be anyone's guess that we are going to have to deal with elliptic integrals. The following uses the notation $$K(k) =\int_{0}^{\pi /2}\frac{dx}{\sqrt{1-k^2\sin^2x}},\,E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx$$ We can now write $$1-\sin^2x=\frac{k^2-1+1-k^2\sin^2x}{k^2}$$ and hence $$I=\frac{64(k^2-1)}{27k^2}\int_{0}^{\pi/2}(1-k^2\sin^2x)^{-3/2}\,dx+\frac{64}{27k^2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}=\frac{8}{3}K(k)-\frac{8}{27}\int_{0}^{\pi/2}(1-k^2\sin^2x)^{-3/2}\,dx$$ For the integral on right hand side note that $$\frac{d} {dx} \frac{\sin x\cos x} {\sqrt{1-k^2\sin^2x}}=\frac{\sqrt{1-k^2\sin^2x}} {k^2}-\frac{1-k^2}{k^2}\cdot(1-k^2\sin^2x)^{-3/2}$$ and thus integrating the above we get $$\int_{0}^{\pi/2}(1-k^2\sin^2x)^{-3/2}\,dx=9E(k)$$ and therefore we have the desired integral $$I=\frac{8}{3}(K(k)-E(k))$$
| {
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"url": "https://math.stackexchange.com/questions/1491193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determine co-ordinates of point?
How do you work out the co-ordinates of $P_{t+1}$ if you're given the co-ordinates of $P_0, P_1, P_2$ ? example (1,1);(2,5);(6,6)
The angle alpha remains constant between $(P_1, P_2)$ and $(P_2, P_{t+1})$
| Slope of $P_0P_1 = \frac{1-5}{1-2} = 4$
Slope of $P_2P_1 = \frac{6-5}{6-2} = \frac{1}{4}$
Slope of $P_{t+1}P_2 = \frac{y-6}{x-6} = \tan \alpha$
$\alpha=\tan^{-1}\left(\frac{4-\frac{1}{4}}{1+4\cdot \frac{1}{4}}\right) = \tan^{-1}\left(\frac{\frac{15}{4}}{2}\right) = \tan^{-1}\left(\frac{15}{8}\right)$
By the problem, $\frac{15}{8}=\left(\frac{|\frac{y-6}{x-6}-\frac{1}{4}|}{1+\frac{y-6}{x-6}\frac{1}{4}}\right)=\left(\frac{|4y-24-x+12|}{4x-24+y-6}\right)=\left(\frac{|4y-x-12|}{4x+y-30}\right)$
or, $60x+15y-450 = |32y-8x-96|$
$$=>60x+15y-450 = 32y-8x-96$$
or, $$68x-17y-354=0$$
$$\mathbf{OR}$$
$$=>60x+15y-450 =-(32y-8x-96)$$
or,$$52x+47y=546=0$$
From both lines applying distance formula, you will get 4 points. You will reject 2 points on account of their lying on that part of the st. line where the lines make angle of $180-\alpha$.
This pic is quite bad, still I was talking about 1,2,3 and 4. You have to reject 1 and 4.
Hope you can do this much.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the reflection that reflects in an arbitrary line y=mx+b How can I find the reflection that reflects in an arbitrary line, $y=mx+b$
I've examples where it's $y=mx$ without taking in the factor of $b$
But I want to know how you can take in the factor of $b$
And after searching through for some results, I came to this matrix which i think can solve my problems. But it doesn't seem to work.
$$
\begin{bmatrix} x' \\ y' \\ 1 \end{bmatrix} =
\begin{bmatrix}
\frac{1-m^2}{1 + m^2} & \frac{-2m}{1 + m^2} & \frac{-2mb}{1 + m^2} \\
\frac{-2m}{1 + m^2} & \frac{m^2-1}{1 + m^2} & \frac{2b}{1 + m^2} \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix} x \\ y \\ 1 \end{bmatrix}.
$$
The example I tried to use using this matrix is
the point $(0,8)$ reflected on $y=-\frac{1}{2}x+2$.
The result I get from that matrix is $[6.4,-0.6,0]$.
The actual answer should be $[-4.8, -1.6]$ , according to Geogebra
| Geometrical Approach:
In general (see derivation), when a given point $P(x_0, y_0)$ is reflected about the line: $y=mx+c$ then the co-ordinates of the point of reflection $P'(x', \ y')$ are calculated by the following formula
$$\color{blue}{(x', y')\equiv \left(\frac{(1-m^2)x_0+2m(y_0-c)}{1+m^2}, \frac{2mx_0-(1-m^2)y_0+2c}{1+m^2}\right)}$$
now, the point of reflection of $(0, 8)$ about the given line: $y=-\frac{1}{2}x+2$ is calculated by setting the corresponding values, $x_0=0, \ y_0=8, \ m=-\frac{1}{2}$ & $c=2$ as follows$$\left(\frac{(1-\left(-\frac{1}{2}\right)^2)(0)+2\left(-\frac{1}{2}\right)(8-2)}{1+\left(-\frac{1}{2}\right)^2}, \frac{2\left(-\frac{1}{2}\right)(0)-(1-\left(-\frac{1}{2}\right)^2)(8)+2(2)}{1+\left(-\frac{1}{2}\right)^2}\right)\equiv\left(\frac{-24}{5}, \frac{-8}{5}\right)\equiv \color{red}{(-4.8, \ -1.6)}$$
So the answer $[-4.8, -1.6]$ according to Geogebra is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that a matrix $A=\pmatrix{a&b\\c&d}$ satisfies $A^2-(a+d)A+(ad-bc)I=O$
Let $A=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
,a,b,c,d\in\mathbb{R}$ . Prove that every matrix $A$ satisfies the condition $$A^2-(a+d)A+(ad-bc)I=O .$$ Find $$ \begin{bmatrix}
a & b \\
c & -a
\end{bmatrix}^n .$$
For $a=1,b=2,c=3,d=4$ equality $A^2-(a+d)A+(ad-bc)I=O$ holds.
How to prove this equality for every $A=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$?
| The short answer is: The equality can be proofen by simply inserting A, performing a matrix multiplication and adding up the matrices. For the second part of your question, one may find a recursion formula.
The long answer including the calculation follows here:
Assuming $I=\begin{bmatrix}1& 0 \\ 0 &1\end{bmatrix}$ and $O = \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}$, the equality can be shown by plugging in $A$.
\begin{align}
A^2 &- (a+d)A&+&(ad-bc)I &=& O\\
\rightarrow \begin{bmatrix}a& b \\ c &d\end{bmatrix}\begin{bmatrix}a& b \\ c &d\end{bmatrix}&
-\begin{bmatrix}(a+d)a& (a+d)b \\ (a+d)c &(a+d)d\end{bmatrix}&+&\begin{bmatrix}ad-bc& 0 \\ 0 &ad-bc\end{bmatrix}&=& \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
Now we still need to do the matrix multiplication
\begin{align}
\begin{bmatrix}a& b \\ c &d\end{bmatrix}\begin{bmatrix}a& b \\ c &d\end{bmatrix} = \begin{bmatrix}a^2+bc& ab+bd \\ ac+dc &bc+d^2\end{bmatrix}
\end{align}
Now, we may add the matrices component-wise:
\begin{align}
\begin{bmatrix}a^2+bc& ab+bd \\ ac+dc &bc+d^2\end{bmatrix}&
-\begin{bmatrix}(a+d)a& (a+d)b \\ (a+d)c &(a+d)d\end{bmatrix}&+&\begin{bmatrix}ad-bc& 0 \\ 0 &ad-bc\end{bmatrix}&=& \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
\begin{align}
\rightarrow \begin{bmatrix}a^2+bc -(a+d)a+ad-bc& ab+bd -(a+d)b \\ ac+dc -(a+d)c&bc+d^2-(a+d)d+ad-bc\end{bmatrix}= \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
After rearranging the terms, we see that the quation holds.
\begin{align}
\rightarrow \begin{bmatrix}a^2-a^2 -ad+ad+bc-bc& ab-ab -ab-db \\ ac+dc -ac-cd&bc-bc+d^2-d^2+ad-ad\end{bmatrix}= \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}
\end{align}
The second part is about how this helps evaluating
$ \begin{bmatrix}a& b \\ c &-a\end{bmatrix}^n=(\tilde A)^n$. Here we defined $\tilde A= \begin{bmatrix}a& b \\ c &-a\end{bmatrix}$.
Using our equation we just proofed with d=-a, we get:
\begin{align}
(\tilde A)^2-(a+(-a))\tilde A+(-a^2-bc)I=&O\\
\rightarrow (\tilde A)^2+(-a^2-bc)I=&O
\end{align}
Multiplying this equation with $(\tilde A)^{n-2}$ gives the recursion formula
\begin{align}
(\tilde A)^n=(a^2+bc)(\tilde A)^{n-2}
\end{align}
This is useful since you may plug the right side back in yielding
\begin{align}
(\tilde A)^n=\begin{cases}(a^2+bc)^{n/2}I&\text{if } n \text{ even} \\
(a^2+bc)^{(n-1)/2}\tilde A &\text{if } n \text{ odd}
\end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} $ never zero? Here $n_i$ are integral numbers, and not all of them are zero.
It is natural to conjecture that similar statement holds for even more prime numbers. Namely,
$$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} + n_5 \sqrt{11} +n_6 \sqrt{13} $$ is never zero too.
I am asking because this is used in some numerical algorithm in physics
| Assume that:
$$ n_1\sqrt{2}+n_2\sqrt{3}+n_3\sqrt{5}+n_4\sqrt{7}=0.\tag{1} $$
That implies:
$$ (2n_1^2+3n_2^2)+2n_1 n_2 \sqrt{6} = (5n_3^2+7n_4^2)+2n_3 n_4 \sqrt{35}\tag{2} $$
as well as (just keep rearranging and squaring):
$$ (2n_1^2+3n_2^2-5n_3^2-7n_4^2)^2 = 24 n_1^2 n_2^2 + 140 n_3^2 n_4^2 - 8n_1 n_2 n_3 n_4 \sqrt{2\cdot 3\cdot 5\cdot 7} \tag{3}$$
but the last line implies $\sqrt{2\cdot 3\cdot 5\cdot 7}\in\mathbb{Q}$, which we know to be false.
The same approach works also in the second case.
| {
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"url": "https://math.stackexchange.com/questions/1496567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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solve: $z^3=\sqrt(3)-i$
Solve: $z^3=\sqrt(3)-i$
$r=\sqrt{\sqrt{3}^2+(-1)^2}=\sqrt{4}$
$\theta=tan^{-1}(\frac{-1}{\sqrt{3}})=\frac{-\pi}{6}$
0: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{-\pi}{18})+isin(\frac{-\pi}{18})]=1.24-0.96i$
1: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{11\pi}{18})+isin(\frac{11\pi}{18})]=-0.43-1.18i$
2: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{23\pi}{18})+isin(\frac{23\pi}{18})]=-0.8-0.96i$
Is it right? Wolfram has a much shorter way
| $$z^3=\sqrt{3}-i \Longleftrightarrow$$
$$z^3=\left|\sqrt{3}-i\right|e^{\arg\left(\sqrt{3}-i\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{\left(\sqrt{3}\right)^2+1^2}e^{\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{3+1}e^{-\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{4}e^{-\frac{\pi}{6}i} \Longleftrightarrow$$
$$z^3=2e^{-\frac{\pi}{6}i} \Longleftrightarrow$$
$$z=\left(2e^{\left(2\pi k-\frac{\pi}{6}\right)i}\right)^{\frac{1}{3}} \Longleftrightarrow$$
$$z=2^{\frac{1}{3}}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{6}\right)i} \Longleftrightarrow$$
$$z=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi k-\frac{\pi}{6}\right)i} $$
With $k\in\mathbb{Z}$ and $k:0-2$
So the solutions are:
$$z_0=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 0-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{-\frac{\pi}{18}i}\approx 1.2407-0.2187i$$
$$z_1=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 1-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{\frac{11\pi}{18}i}\approx -0.4309+1.1839i$$
$$z_2=\sqrt[3]{2}e^{\frac{1}{3}\left(2\pi \cdot 2-\frac{\pi}{6}\right)i}=\sqrt[3]{2}e^{-\frac{13\pi}{18}i}\approx -0.809-0.965i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $p=4k+1$ then $p$ divides $n^2+1$ I am stuck in one step in the proof that if $p$ is congruent to $1 \bmod 4$, then $p\mid (n^2+1)$ for some $n$.
The proof uses Wilson's theorem, $(4k)!\equiv -1 \pmod p$.
The part I am stuck is where it is claimed that $(4k)!\equiv (2k)!^2 \bmod p$. Why is this so?
| Take a typical small prime of the form $4k+1$, such as $17$. So $k=4$.
Then
$$16!=\left[(1)(2)(3)(4)(5)(6)(7)(8)\right]\left[(9)(10)(11)(12)(13)(14)(15)(16)\right].\tag{1}$$
We have $16\equiv -1\pmod{17}$, $15\equiv -2\pmod{17}$, $14\equiv -3\pmod{17}$, and so on down to $9\equiv -8\pmod{17}$.
It follows that $(16)(15)(14)(13)(12)(11)(10)(9)\equiv (-1)(-2)(-3)(-4)(-5)(-6)(-7)(-8)\pmod{17}$.
Our product has an even number of minus signs. It follows that
$$(16)(15)(14)(13)(12)(11)(10)(9)\equiv 8!\pmod{17}.$$
Thus from (1) we find that
$$16!\equiv [8!][8!]=(8!)^2\pmod{17}.$$
The same argument works for any prime of the form $4k+1$. It does not work for a prime of the form $4k+3$, for in that case we get an odd number of minus signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of the region bounded by the parabola $y = 2x^2$, the tangent line to this parabola at $(3, 18)$, and the $x$-axis. Is my answer correct?
$f(x) = 2x^2 \gets$ this is the parabola
$f(3) = 2 \times 9 = 18 \to$ the parabola passes through $A (3 ; 18$), so its tangent line does too.
$f'(x) = 4x \gets$ this is the derivative
…and the derivative is the slope of the tangent line to the curve at $x$
$f'(3) = 4 \times 3 = 12 \gets$ this is the slope of the tangent line to the curve at $x = 3$
Equation of the tangent line
The typical equation of a line is: $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept.
We know that the slope of the tangent line is $12$.
The equation of the tangent line becomes $y = 12x + b$.
The tangent line passes through $A (3 ; 18)$, so these coordinates must satisfy the equation of the tangent line.
$y = 12x + b$
$b = y - 12x \to$ I substitute $x$ and $y$ by the coordinates of the point $A (3 ; 18)$.
$b = 18 - 36 = - 18$
$\to$ The equation of the tangent line is $y = 12x - 18$.
Intersection between the tangent line to the curve and the $x$-axis: $\to$ when $y = 0$.
\begin{align}
y &= 12x - 18 \to \text{when } y = 0 \\
12x - 18 &= 0 \\
12x &= 18 \\
x &= 3/2
\end{align}
$\to$ Point $B (3/2 ; 0)$
Intersection between the vertical line passes through the point $A$ and the $x$-axis: $\to$ when $x = 3$.
$\to$ Point $C (3 ; 0)$
The equation of the vertical line is $x = 3$.
Area of the region bounded by the parabola $y = 2x^2$, the tangent line to this parabola at $(3; 18)$, and the $x$-axis.
$=$ (area of the region bounded by the parabola $y = 2x^2$ and the $x$-axis) minus (area of the triangle $ABC$)
$=$ [integral (from $0$ to $3$) of the parabola] minus $[(x_C-x_B)\cdot(y_A-y_C)/2]$
\begin{align}
&= \int_0^3 2x^2 dx -\frac{(x_C-x_B)(y_A-y_C}{2} \\
&= \left. \frac23 x^3 \right|_0^3 -\frac{(3-3/2)(18-0)}{2} \\
&= \frac23 \cdot 3^2 -(6/2 - 3/2)\cdot9 \\
&= \frac23 \cdot 27 -\frac32 \cdot 9 \\
&= 18 - \frac{27}{2} \\
&= \frac{36}{2} -\frac{27}{2} \\
&= \frac92 \text{ square units}
\end{align}
| $y=2x^2$ has derivative $y'=4x$. So the derivative will be positive for all $x>0$. This means the tangent line will intersect the x-axis at some point $x_a<x$ for a given x.
This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $x<x_a$ which is made up of only area under the parabola, and a region $x>x_a$ where area is from the area between the parabola and the tangent to $(3,18)$.
The equation of a tangent line at $(x_0,y_0)$ is $$\frac{y-f(x_0)}{x-x_0}=f'(x_0)$$
Or : $y=f'(x_0)(x-x_0)+f(x_0)$
The x intercept happens where $y=0$.
Requiring $y=0$ implies an x intercept of $x_c=\frac{-f(x_0)}{f'(x_0)}+x_0$
So from the above arguments with $x_0=3$:
$$A=\int_0^{x_c}2x^2dx+\int_{x_c}^32x^2-(12x-18)dx$$
But this can be simplified. The area under the tangent line is a triangle. So the the parabola can be integrated ignoring the tangent line, and then subtracting the area of the triangle.
From the above, we know $(x_0-x_c)=\frac{f(x_0)}{f'(x_0)}$, the base of the triangle. The height is just $f(x_0)$.
So the area of the triangle is $A_t=\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$
So:
$$A=\int_0^{x_0}f(x)dx-\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$$
and solve for $x_0=3$.
In this form, an expression can be found for $x_0$ which extremizes the area.
| {
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