Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Euclidean Algorithm for polynomials I know how to use the extended euclidean algorithm for finding the GCD of integers but not polynomials. I can't really find any good explanations of it online. The question here is to find the GCD of
m(x) = $\ x^3+6x+7 $ and n(x) = $\ x^2+3x+2 $.
I try to use it the same way as for integers but don't really get anywhere and just get huge lines without ever reducing it and getting closer to finding the GCD.
| $$ \left( x^{3} + 6 x + 7 \right) $$
$$ \left( x^{2} + 3 x + 2 \right) $$
$$ \left( x^{3} + 6 x + 7 \right) = \left( x^{2} + 3 x + 2 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 13 x + 13 \right) $$
$$ \left( x^{2} + 3 x + 2 \right) = \left( 13 x + 13 \right) \cdot \color{magenta}{ \left( \frac{ x + 2 }{ 13 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x - 3 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x - 3 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ x + 2 }{ 13 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{2} - x + 7 }{ 13 } \right) }{ \left( \frac{ x + 2 }{ 13 } \right) } $$
$$ \left( x^{2} - x + 7 \right) \left( \frac{ 1}{13 } \right) - \left( x + 2 \right) \left( \frac{ x - 3 }{ 13 } \right) = \left( 1 \right) $$
$$ \left( x^{3} + 6 x + 7 \right) = \left( x^{2} - x + 7 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$
$$ \left( x^{2} + 3 x + 2 \right) = \left( x + 2 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x + 1 \right) } $$
$$ \left( x^{3} + 6 x + 7 \right) \left( \frac{ 1}{13 } \right) - \left( x^{2} + 3 x + 2 \right) \left( \frac{ x - 3 }{ 13 } \right) = \left( x + 1 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Recurrence relation $a_n = -a_{n-1}+n -1 , a_0=7$ $a_n = -a_{n-1} +n -1 , a_0=7$
I am trying to find an explicit formula for this recurrence relation by using backward iteration.
So far I've ended up with this and I don't know what to do to find an explicit formula for this.
$a_n = -a_{n-1}+n -1 , a_0=7$
$= -a_{n-1} +n -1$
$= a_{n-2} +1$
$= -a_{n-3} +n -2$
$= a_{n-4} +2$
$= - a_{n-5} +n-3$
$= a_{n-6} +3$
......
| I think you already discovered this pattern:
$$\begin{array}\\
a_n &=& a_{n-0} +0\\
&=& a_{n-2} +1\\
&=& a_{n-4} +2\\
&=& a_{n-6} +3\\
&\ldots
\end{array}$$
So for even $n$, $n=2m,$ we have
$$\begin{array}\\
a_{2m} &=& a_{2m-0} +0\\
&=& a_{2m-2} +1\\
&=& a_{2m-4} +2\\
&=& a_{2m-6} +3\\
&\ldots\\
&=& a_{2m-2m} +m\\
&=& a_{0} +m\\
&=& 7 +m\\
\end{array}$$
and for odd $n$, $n=2m+1$, we have
$$\begin{array}\\
a_{2m+1} &=& a_{2m+1-0} +0\\
&=& a_{2m+1-2} +1\\
&=& a_{2m+1-4} +2\\
&=& a_{2m+1-6} +3\\
&\ldots\\
&=& a_{2m+1-2m} +m\\
&=& a_{1} +m\\
&=& -7 +m\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that $A^2+B^2+C^2=D^2$ In an orthogonal parallellepipide (cuboid) with sides $a,b,c$, a plane bisects it in halv through three side-diagonals of the cuboid, leaving a tetrahedral with three of it's sides being right triangles. Denote these right triangle sides by $A,B,C$ and denote the fourth side by $D$. Show that $A^2+B^2+C^2=D^2.$
My method starts with putting the cuboid in a coordinate system like this below.
Fixing the vectors $\vec{u}=(a,0,0)-(0,0,-c) = (a,0,c)$ and $\vec{v}=(a,b,-c)-(0,0,-c)=(a,b,0).$
Now, the area of the fourth triangle is $$D=\frac{1}{2}|\vec{u}\times\vec{v}|= \frac{1}{2}\left(\begin{vmatrix}
0 & c \\
b & 0
\end{vmatrix}-\begin{vmatrix}
a & c \\
a & 0
\end{vmatrix}+\begin{vmatrix}
a & 0 \\
a & b
\end{vmatrix}\right)=-\frac{bc}{2}+\frac{ac}{2}+\frac{ab}{2}.$$
Now, by the figure it follows that $A^2=\frac{b^2c^2}{4}, \ B^2=\frac{a^2b^2}{4}$ and $C^2=\frac{a^2c^2}{4}$ so $$A^2+B^2+C^2=\frac{b^2c^2}{4}+\frac{a^2b^2}{4}+\frac{a^2c^2}{4}.$$
But Squaring $D$ does not give me $\frac{b^2c^2}{4}+\frac{a^2b^2}{4}+\frac{a^2c^2}{4}.$ This method should work!
**
Note: I don't want to know other methods, I want to know why the above doesn't work.
**
| The result you want to prove has a name : it is called De Gua's theorem.
In fact :
$$\vec{u}\times\vec{v}=\begin{pmatrix}
-bc\\ \ \ ac\\ \ \ ab
\end{pmatrix}$$
giving:
$$D=\frac{1}{2}\|\vec{u}\times\vec{v}\|=\frac12\sqrt{(bc)^2+(ac)^2+(ab)^2}$$
instead of
$$-\frac{bc}{2}+\frac{ac}{2}+\frac{ab}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of $ \sin (a+ \frac{b}{2}(2n+1) ) - \sin (a+ \frac{b}{2} ) $? What is the value of $$ \sin \bigg(a+\displaystyle \frac{b}{2}(2n+1) \bigg) - \sin \bigg(a+\displaystyle \frac{b}{2} \bigg) ? $$
I already tried the sum-to-product rules and did not go so far with that. It is supposed to equal $$ 2 \sin \bigg( \frac{bn}{2} \bigg) \cos \bigg( \frac{2a+bn+b}{2} \bigg) .$$
| Take d = $ \frac{bn}{2}$ and c = $\frac{2a+bn+b}{2}$
So your equation reduces to
$$ \sin \big(c+d \big) - \sin \big(c-d \big) $$
which is $$ 2\cos (c)\sin(d) $$ = $$ 2\cos (\frac{2a+bn+b}{2})\sin(\frac{bn}{2}) $$
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum\limits_{n=2}^\infty \frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$
Convergence of $$\sum_{n=2}^\infty \frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$$
Well, I have tried almost everything. D'Alembert's criterion doesn't work because the limit is 1. I have tried to bound $\frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$ in several ways but every bound diverges. Any hint?
P.S: I know this series converges to $\frac{1}{2}$
| You can use Gauss test (or Rabee's test, the latter is weaker but still suffices), I will use the Gauss test:
If $u_n >0$ and satisfies
$$\frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(\frac{1}{n^r})$$ for some $r>1$, then $\sum u_n$ converges iff $h>1$.
In your case, $$u_n = \frac{1\cdot 3\cdots (2n-3)}{2^n n!} = \frac{(2n-2)!}{2^{2n-1}n!(n-1)!}$$ we easily obtain
$$\frac{u_n}{u_{n+1}} = \frac{2n+2}{2n-1} = 1+\frac{3}{2n-1} = 1+\frac{3}{2n}+O(\frac{1}{n^2})$$
so $\sum u_n$ converges.
Alternatively, you can use the Stirling formula
$$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$ but I think this is an overkill.
The value of the sum can be derived from the following series (after integrating and shifting terms):
$$\sum_{n=0}^\infty \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve: $\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}} =4 $ I tried:
$\sqrt{x +14} = t$
$\implies \sqrt{x+3 + t} + \sqrt{x+3 - t} = 4$
But I think it is the hard way. Any ideas?
| The domain gives $x\geq-3$ and $(x+3)^2\geq x+14$.
Now, our equation it's
$$\left(\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}}\right)^2 =16 $$ or
$$2x+6+2\sqrt{(x+3)^2-x-14}=16$$ or
$$\sqrt{x^2+5x-5}=5-x.$$
Now, we get also $5-x\geq0$, for which we obtain
$$x^2+5x-5=x^2-10x+25$$ or
$$x=2,$$ which is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve the differential equation $ 2xy' = y+ \frac{3}{2}x^2 $ Question: Solve the differential equation below
$$ 2xy' = y+ \frac{3}{2}x^2 $$
Normal $y'= dy/x$ doest not work I think.
Should I start it with substitution? $$ y' =u'x+u $$ $$y=\frac{u}{x} $$
But that method leads me to nothing, how I should do it?
| $$2xy'=y+\frac { 3 }{ 2 } x^{ 2 }\\ 2xy'-y=\frac { 3 }{ 2 } x^{ 2 }\\ 2xy'-y=0\\ 2x\frac { dy }{ dx } =y\\ 2\int { \frac { dy }{ y } } =\int { \frac { dx }{ x } } \\ \ln { \left| y \right| =\ln { \left| Cx \right| } } \\ { y }^{ 2 }=Cx\\ y=C\sqrt { x } \\ y=C\left( x \right) \sqrt { x } \\ { y }^{ \prime }={ C }^{ \prime }\left( x \right) \sqrt { x } +\frac { C\left( x \right) }{ 2\sqrt { x } } \\ 2x\left( { C }^{ \prime }\left( x \right) \sqrt { x } +\frac { C\left( x \right) }{ 2\sqrt { x } } \right) =C\left( x \right) \sqrt { x } +\frac { 3 }{ 2 } { x }^{ 2 }\\ 2x\sqrt { x } { C }^{ \prime }\left( x \right) =\frac { 3 }{ 2 } { x }^{ 2 }\\ { C }^{ \prime }\left( x \right) =\frac { 3\sqrt { x } }{ 4 } \\ { C }\left( x \right) =\frac { 1 }{ 2 } x\sqrt { x } +{ C }_{ 1 }\\ y=C\left( x \right) \sqrt { x } =\sqrt { x } \left( \frac { 1 }{ 2 } x\sqrt { x } +{ C }_{ 1 } \right) =\frac { { x }^{ 2 } }{ 2 } +{ C }_{ 1 }\sqrt { x } \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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What is the 3rd Derivative of Cos(x) using this Derivative Formula? There is a general formula for the derivative of a function:
$$\frac{d^n}{dx^n}f(x)=\lim_{\epsilon\to0}\frac{1}{\epsilon^n}\sum_{j=0}^n{((-1)^j\frac{\Gamma(n+1)}{j!\Gamma{(n+1-j)}}f(x-j\epsilon))}$$
Where $\Gamma(x) $ is the Gamma function
I tried using the formula to evaluate the 3rd derivative of $\cos(x)$, but I get confused quickly. It would be very appreciated if someone could show a step by step solution to this problem.
I'm totally aware the answer is $\sin(x)$, but what's the process to get to that solution?
| We want
$$ \lim_{h \to 0} \frac{\cos{(x+3h)}-3\cos{(x+2h)+3\cos{(x+h)}-\cos{x}}}{h^3}. $$
Then
$$ \cos{(x+3h)}-\cos{(x+2h)} = -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{5h}{2}\right)} \\
-\cos{(x+2h)}+\cos{(x+h)} = 4\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} \\
\cos{(x+h)}-\cos{x} = -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{h}{2}\right)},
$$
then
$$ -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{5h}{2}\right)} + 2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} = -\left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + 2h \right)} \\
2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{h}{2}\right)} = \left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + h \right)}, $$
and finally
$$ -\left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + 2h \right)} + \left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + h \right)} = \left( 2\sin{\left(\frac{h}{2}\right)}\right)^3\sin{\left( x + \frac{3h}{2} \right)}. $$
Then
$$ \lim_{h \to 0} \frac{1}{h^3}\left( 2\sin{\left(\frac{h}{2}\right)}\right)^3 = 1, $$
and the other term tends to $\sin{x}$, so the whole thing converges to $\sin{x}$. Exactly the same argument works for any number of derivatives, as can be shown by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $\lim\limits_{x \to x_0} f(x) = \lim\limits_{x \to x_0} g(x) = +\infty$ prove that $\lim\limits_{x \to x_0} \frac{f(x)+g(x)}{f^2(x)+g^2(x)}=0$
Consider the functions $f, g : \mathbb{R} \rightarrow \mathbb{R}$ for which holds that $\lim\limits_{x \to x_0} f(x) = \lim\limits_{x \to x_0} g(x) = +\infty$ for some $x_0 \in \mathbb{R}$. Prove that $\lim\limits_{x \to x_0} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}=0$
Is the following proof correct? For some $x \to x_0$ holds that $f(x)>0$ and $g(x)>0$
So it is true that $2f(x)g(x)\geq 0 \\ \Rightarrow (f(x)+g(x))^2 \geq f^2(x)+g^2(x) \\ \Rightarrow \dfrac{1}{f^2(x)+g^2(x)} \geq \dfrac{1}{(f(x)+g(x))^2} \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \dfrac{f(x)+g(x)}{(f(x)+g(x))^2}=\dfrac{1}{f(x)+g(x)}$
But $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}>0$ hence $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}=\Bigg|\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}\Bigg|$
Thus $\Bigg|\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}\Bigg| \geq \dfrac{1}{f(x)+g(x)} \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \dfrac{1}{f(x)+g(x)} \text{OR} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq -\dfrac{1}{f(x)+g(x)} \\ \Rightarrow \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \lim_{x \to x_0}\dfrac{1}{f(x)+g(x)} \text{OR} \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq -\lim_{x \to x_0}\dfrac{1}{f(x)+g(x)} \\ \Rightarrow \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq 0 \text{OR} \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq 0 \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} = 0$
| $\bigg|\dfrac{f(x)+g(x)}{f(x)^2+g(x)^2}\bigg|=\bigg|\dfrac{r(x)\big(\cos \theta(x)+\sin\theta(x)\big)}{r(x)^2}\bigg|\le\dfrac{2}{r(x)}\to 0$
Since $r(x)=\sqrt{f(x)^2+g(x)^2}\to+\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of solutions to $a + 10b + 20c = 100$ with $a,b,c$ non-negative integers
How many solutions are there to the equation $$a + 10b + 20c = 100$$ where $a,b,c$ are non-negative integers
The problem I'm asking is another way of phrasing "How many ways can you break $100$ dollars into $1$, $10$, and $20$ dollar bills", or any multitude of variants.
Solutions can, for example, be $(a,b,c) = (100, 0, 0)$, $(10, 7, 1)$, $(0, 6, 2)$, etc.
I wrote a quick program which gave me a total of $36$ solutions, but I'd like to have a better understanding as to where this answer comes from without something resembling brute-force. Preferably, I'd like to see how to use generating functions to solve this, as that was the method I first tried (and failed) with.
The solution should be the coefficient of the 100th-power term in the product:
$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$
I can get this coefficient from the following:
$$\lim_{x \to 0} \frac{1}{100!} \frac{d^{100}}{dx^{100}}(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$
For whatever reason, applying product rule fails (or I fail to use it correctly), as I get $3$. So, assuming $|x| < 1$, this equals:
$$\lim_{x \to 0} \frac{1}{100!} \frac{d^{100}}{dx^{100}} \frac{1}{(1-x)(1-x^{10})(1-x^{20})}$$
However, to much dismay, this is not such a simple thing to find derivatives of. Any advice?
| Er, ..., calculate carefully?
$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...) \\
= \dots +30 x^{98} + 30 x^{99} + 36 x^{100} + 36 x^{101} + ...$$
Another way: Suppose there are $n_k$ ways to write $k$ as a non-negative integer weighted sum of $\{1,10\}$. Then the total number of ways to write $k$ as a non-negative integer weighted sum of $\{1,10,20\}$ is $N = n_0 + n_{20} + n_{40} + n_{60} + n_{80} + n_{100}$, where $n_0$ counts the number of ways we finish by adding five $20$s, $n_{20}$ the number of ways we finish by adding four $20$s, ..., and $n_{100}$ the number of ways we finish by adding no $20$s.
*
*Now $n_0 = 1$ because $0 \cdot 1 + 0 \cdot 10$ is the only way to write $0$ as a non-negative integer weighted sum of $\{1,10\}$.
*And $n_{20} = 3$, depending on whether there are zero, one, or two $10$s in the sum, the rest being made up with $1$s.
*And $n_{40} = 5$, for the same reason : zero to four $10$s and the remainder made of $1$s.
*Finishing, $n_{60} = 7$, $n_{80} = 9$, and $n_{100} = 11$.
Finally, $N = 1 + 3 + 5 + 7 + 9 + 11 = 36$. That the number of ways is controlled by the $10$s and $20$s, with the $1$ making up the rest, is hinted in the snippet of the series in $x$, above. There is only one way to make totals up to $9$ from $\{1,10,20\}$. There are two ways for totals in $[10,19]$, four ways in $[20,29]$, six ways in $[30,31]$, nine ways in $[40,41]$, where the increment in the number of ways in each range of ten totals increases by $1$ each time we pass a multiple of $20$ (counting that we could either include or exclude one more $20$ in our sums).
Another way: You don't have to multiply out the full power series. You can calculate with
$$( 1 + x+ x^2 + \cdots + x^{99} + x^{100})(1 + x^{10} + \cdots + x^{100})(1+x^{20}+\cdots+x^{100})$$
always discarding any term with power greater than $100$. If you do this from right to left, you essentially perform the calculation described above.
| {
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"url": "https://math.stackexchange.com/questions/2482032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of solutions of $3\sin^2 x+\cos^2 x+\sqrt{3} \sin x+\cos x+1=\sqrt{3}\sin x\cos x$ Find Number of solutions of
$$3\sin^2 x+\cos^2 x+\sqrt{3} \sin x+\cos x+1=\sqrt{3}\sin x\cos x$$ in $\left[0 \:\: 10\pi\right]$
My Try:
The given equation is
$$2+2\sin^2 x+2\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\sin (2x)$$ $\implies$
$$3+2\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\sin (2x)+\cos (2x)$$
Any idea here?
| Expand out the difference (left side - right side) as
$$ F(x) = 4-2\, \cos^2 \left( x \right) +\sqrt {3}\sin \left(
x \right) +\cos \left( x \right) -\sqrt {3}\sin \left( x \right) \cos
\left( x \right)
$$
If $\sin(x) = s$, $\cos(x) = \pm \sqrt{1-s^2}$, and
$$ F(x) = 2 + 2 s^2 + \sqrt{3} s \pm (1-\sqrt{3} s) \sqrt{1-s^2}$$
In order for that to be $0$, we'd need
$$ \sqrt{1-s^2} = \mp \frac{2 s^2 + \sqrt{3} s + 2}{1-\sqrt{3} s} $$
and thus $$1 - s^2 = \dfrac{(2s^2 + \sqrt{3} s + 2)^2}{(1-\sqrt{3} s)^2}$$
so that
$$ 7\,{s}^{4}+2\,\sqrt {3}{s}^{3}+6\,\sqrt {3}s+9\,{s}^{2}+3 = 0$$
This quartic has positive discriminant ($209088$), so its roots are either all real or all non-real. In this case they turn out to be all non-real.
| {
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"url": "https://math.stackexchange.com/questions/2490246",
"timestamp": "2023-03-29T00:00:00",
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find the $x$ : $(\cos 3x+\cos 4x)(\cos 3x +\cos x)=\frac{1}{4}$ find the $x$ :
$$(\cos 3x+\cos 4x)(\cos 3x +\cos x)=\frac{1}{4}$$
My try :
$$\cos x +\cos y= 2\cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$$
So :
$$\left(2\cos \left(\frac{7x}{2}\right)\right) \cos \left(\frac{-x}{2}\right)\left(2\cos \left(\frac{4x}{2}\right)\right) \cos \left(\frac{2x}{2}\right)=\frac{1}{4}$$
$$\cos \left(\frac{7x}{2}\right) \cos \left(\frac{x}{2}\right)\cos (2x) \cos (x)=\frac{1}{16}$$
Now what ?
| Let $\cos{x}=t$.
Thus, we get
$$(4t^3-3t+8t^4-8t^2+1)(4t^3-3t+t)=\frac{1}{4}$$ or
$$(2t+1)(4t^2+2t-1)(16t^4-8t^3-16t^2+8t+1)=0.$$
$$16t^4-8t^3-16t^2+8t+1=0$$ we can solve by the following way
$$\left(4t^2-t-\frac{3}{2}\right)^2-5\left(t-\frac{1}{2}\right)^2=0.$$
I think the rest is smooth.
I think the following way a bit of better.
After using of your work we need to solve
$$16\sin\frac{x}{2}\cos\frac{x}{2}\cos{x}\cos2x\cos\frac{7x}{2}=\sin\frac{x}{2}$$ or
$$2\sin4x\cos\frac{7x}{2}=\sin\frac{x}{2}$$ or
$$\sin\frac{15x}{2}+\sin\frac{x}{2}=\sin\frac{x}{2}$$ or
$$\sin\frac{15x}{2}=0.$$
Now, we need to delete all roots of $\sin{\frac{x}{2}}=0$ and to write the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Ratio of Trigonometric Functions tanx/tany = 1/3; sin2x/sin2y = 3/4 where 0 < x,y< pi/2. What is the Value of
tan2x/tan2y
I couldn't find the appropriate trig identities to use for this problem, and what relationship for sin and tan that we can exploit.
| $$\frac{\tan x}{\tan y}=\frac{1}{3};\;\frac{\sin 2x}{\sin 2y}=\frac{3}{4}$$
set $\tan x =t;\;\tan y=u$
$$\sin 2x=\frac{2t}{1+t^2};\;\sin 2y=\frac{2u}{1+u^2}$$
$$\frac{t}{u}=\frac{1}{3}\to u=3t$$
$$\frac{2t}{1+t^2}\frac{1+u^2}{2u}=\frac{3}{4}$$
$$\frac{2t}{1+t^2}\frac{1+9t^2}{6t}=\frac{3}{4}$$
$$8t(1+9t^2)=18t(1+t^2)\to 8t(1+9t^2)-18t(1+t^2)=0\to 2t(4+36t^2-9-9t^2)=0$$
$$\;t=\pm\sqrt{\frac{5}{27}};\;u=\pm 3\sqrt{\frac{5}{27}}=\pm\sqrt{\frac{5}{3}}$$
it $t=0$ then $u=0$ which is impossible since $u$ is the denominator of the fraction
if $t=\pm\sqrt{\frac{5}{27}};\;x=\pm\arctan\sqrt{\frac{5}{27}}+k\pi,\;\forall k\in\mathbb{Z}$
$u=\pm\sqrt{\frac{5}{3}};\;y=\pm\arctan \sqrt{\frac{5}{3}}+h\pi ,\;\forall h\in\mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding upper bound of Riemann's sum I have
$$f(x) = -x^2 + 2x + 5,\ x \in [1, 3]$$
Since $f(1) = 6$ and $f(3) = 2$, $f$ is decreasing.
$$\overline{S_n} = \sum_{i=0}^{n-1} f(\frac{i}{n})\Delta x_i,\ \Delta x_i = \frac{b - a}{n} = \frac{2}{n}$$
$$\overline{S_n} = \sum_{i=0}^{n-1} \left(-\left(\frac{i}{n}\right)^2 + 2\frac{i}{n} + 5\right)\frac{2}{n}$$
$$\overline{S_n} = \frac{2}{n}\left(\frac{-1}{n^2}\sum_{i=0}^{n-1} i^2 + \frac{2}{n}\sum_{i=0}^{n-1} i + \sum_{i=0}^{n-1} 5\right)$$
$$\overline{S_n} = \frac{2}{n}\left(\frac{-1}{n^2}\frac{n(n-1)(2n - 1)}{6} + \frac{2}{n}\frac{n(n-1)}{2} + 5n\right)$$
$$\overline{S_n} = \frac{2}{n}\left(\frac{-(n-1)(2n - 1)}{6n} + n - 1 + 5n\right)$$
$$\overline{S_n} = \frac{2}{n}\left(\frac{-(2n^2 - 3n + 1)}{6n} + 6n -
1\right)$$
$$\overline{S_n} = \frac{2}{n}\left(\frac{-2n^2}{6n} + \frac{3n}{6n} - \frac{1}{6n} + 6n -
1\right)$$
$$\overline{S_n} = \frac{2}{n}\left(\frac{17n}{3} - \frac{1}{2} - \frac{1}{6n}\right)$$
$$\overline{S_n} = \frac{34}{3} - \frac{1}{n} - \frac{1}{3n^2}$$
But the answer is supposed to be
$$\overline{S_n} = \frac{28}{3} + \frac{4}{n} - \frac{4}{3n^2}$$
Where is my mistake?
| Daniel Fischer's comment to an answer that can be accepted:
$$\overline{S_n} = \sum_{i=0}^{n-1} f(\frac{i}{n})\Delta x_i,\ \Delta x_i = \frac{b - a}{n} = \frac{2}{n}$$
You're evaluating the function in points $i/n$, which are between $0$ and $1$, so your result has no correspondence to the integral from $1$ to $3$. You should instead evaluate in points $a + i\frac{b-a}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder of the division Find the remainder when $f(x)=x^{2017}-1$ is divided by $(x^2+1)(x^2+x+1)$
I have expanded the function and factorized it. The remainder I had found was $-x^3-2x^2-3$
I am not sure if I was right ! So can anyone tell me the other way to solve it ..?
| $$x^{2017}-1=x((x^3)^{672}-1)+x-1\equiv x-1\pmod{x^3-1}\equiv x-1\pmod{x^2+x+1}$$
and $$x^{2017}-1=x((x^4)^{504}-1)+x-1\equiv x-1\pmod{x^4-1}\equiv x-1\pmod{x^2+1}$$
Luckily both remainders are same and $$(x^2+x+1,x^2+1)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $0 = A\sin^2(\alpha) + B\sin(2\alpha) - C$? I am facing a problem from physics class involving a projectile motion which can be described with such an given equation:
$$
h = -\frac{1}{2} \frac{g}{v_{0}^2 \cos^2\alpha} d^2 + \frac{\sin\alpha}{\cos\alpha} d + y.
$$
The goal is to find the minimum value of (rearranging above equation)
$$
v_0(\alpha) = \frac{d}{\cos\alpha} \cdot \sqrt{\frac{1}{2} \cdot \frac{g}{\tan\alpha \cdot d + y - h}}.
$$
This involves finding solutions to $v_0'(\alpha) = 0$. I was able to find the derivative ($t := d\cdot \tan\alpha +y-h$):
$$
v_0'(\alpha) = \frac{\tan \alpha}{\cos\alpha \cdot \sqrt{t}} - \frac{d}{2\cos^3\alpha \cdot (\sqrt{t})^3}
$$
Because of the condition $v_0'(\alpha) = 0$ this simplifies to (it is know that the solution is around $50^\circ$)
$$
0 = \sin\alpha - \frac{d}{2\sin\alpha \cos\alpha \cdot (d\cdot \tan\alpha + y - h)}
$$
or
$$
0 = d\sin(2\alpha)\tan\alpha + (y-h)\sin(2\alpha) - d.
$$
or
$$
0 = 2d\sin^2(\alpha) + (y-h)\sin(2\alpha) - d
$$
or with some constants
$$ 0 = 2d\sin^2(\alpha) + B\sin(2\alpha) - d. $$ How can I find the
solutions from here?
| You can expand $\sin2\alpha$:
$$
2d\sin^2\alpha+2B\sin\alpha\cos\alpha-d\sin^2\alpha-d\cos^2\alpha=0
$$
that becomes
$$
d\sin^2\alpha+2B\sin\alpha\cos\alpha-d\cos^2\alpha=0
$$
If $d\ne0$, we cannot have $\cos\alpha=0$ as a solution, so we can divide by $\cos^2\alpha$, leading to
$$
d\tan^2\alpha+2B\tan\alpha-d=0
$$
so
$$
\tan\alpha=\frac{-B\pm\sqrt{B^2+d^2}}{d}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Explicit formula for summation How can I find an explicit formula for the summation
$$\sum_{i=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor+1}\binom{n}{2i-1}\left(\frac{1}{6}\right)^{i}\left(\frac{5}{6}\right)^{n-(2i-1)}$$
Wolfram Alpha comes up with
$$-\frac{\left(60+31\sqrt{6}\right)\left[\left(\frac{5}{6}-\frac{1}{\sqrt{6}}\right)^{n}-\left(\frac{5}{6}+\frac{1}{\sqrt{6}}\right)^{n}\right]}{372+120\sqrt{6}},$$ but I have no idea how it came up with it.
| The sum can be rewritten as
$$\sum_{\substack{0\le k \le n,\\ \color{red}{k \text{ odd}}}} \binom{n}{k}\left(\frac{1}{6}\right)^{\frac{k+1}{2}}\left(\frac{5}{6}\right)^{n-k}
=\frac{1}{\sqrt{6}}\sum_{k=0}^n \frac{1-(-1)^k}{2}\binom{n}{k}\left(\frac{\sqrt{6}}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}\\
=
\frac{1}{2\sqrt{6}}
\left[\left(\frac{5+\sqrt{6}}{6}\right)^n
-\left(\frac{5-\sqrt{6}}{6}\right)^n
\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove this inequality $\frac{n}{3^n-2^n}<\left(\frac{2}{5}\right)^{n-1}$ Let $n\ge 3$. How to prove show that
$$\dfrac{n}{3^n-2^n}<\left(\dfrac{2}{5}\right)^{n-1}\tag1$$
or
$$3^n\cdot 2^{n-1}-2^{2n-1}-5^{n-1}\cdot n>0,n\ge 3\tag2$$
It seem right:see wolfram
when I do this problem found it:How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$
if this inequality (1) has prove it,then
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<1+\dfrac{2}{5}+\sum_{k=3}^{+\infty}\dfrac{2^{k-1}}{5^{k-1}}=\dfrac{5}{3}$$
Question: How to prove inequality (1) or (2)
| $(2)$ can be proved by induction. $(2)$ is equivalent to
$$3^n\cdot 2^{n-1}>2^{2n-1}+5^{n-1}\cdot n\quad \quad (3)$$
It is easy to check the cases $n=3,4,5,6$.
Suppose $(3)$ is true for $n=k$. Then
$$3^{k+1}\cdot 2^k = 6(3^k\cdot 2^{k-1})>6(2^{2k-1}+5^{k-1}\cdot k)=\frac{3}{2}\cdot 2^{2k+1}+\frac{6}{5}\cdot 5^{k}\cdot k > 2^{2k+1}+\frac{6}{5}\cdot 5^{k}\cdot k$$
Provided $k>5$, $\frac{6}{5}k>k+1$, so the statement holds for all $n>5$ by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Euclidean algorithm and GL2(Z) How can I use the Euclidean algorithm to prove the matrices
$$
\begin{pmatrix}
1&1\\
0&1\\
\end{pmatrix}
\begin{pmatrix}
0&1\\
1&0\\
\end{pmatrix}
$$
are generating GL(2,Z)?
| Denote the first matrix u, and the second v. You have, then that
$
u^n=
\left[ {\begin{array}{cc}
1 & n \\
0 & 1\\
\end{array} } \right], n\in \mathbb{Z}
$
and $v^{2n}=I_2$ (you can prove it by induction). You can also show that you can generate all the matrices of the form $\left[ {\begin{array}{cc}
0 & 1 \\
1 & n\\
\end{array} } \right],\left[ {\begin{array}{cc}
1 & 0 \\
n & \pm1\\
\end{array} } \right]$.
Now, if we have a matrix in $\left[ {\begin{array}{cc}
a & b \\
c & d\\
\end{array} } \right]\in GL_2(\mathbb{Z})$ , then write $a=q\cdot b +r$.
Now we can put:
$$
\left[ {\begin{array}{cc}
a & b \\
c & d\\
\end{array} } \right]\cdot \left[ {\begin{array}{cc}
0 & 1 \\
1 & -r\\
\end{array} } \right]= \left[ {\begin{array}{cc}
b & a-br \\
c' & d'\\
\end{array} } \right]=\left[ {\begin{array}{cc}
b & r \\
c' & d'\\
\end{array} } \right]
$$
Continue to multiply with the corresponding q. Eventually, you will get the matrix $
\left[ {\begin{array}{cc}
gcd(a,b) & 0 \\
c'' & d''\\
\end{array} } \right]
$ But as we know, $det\left[ {\begin{array}{cc}
a & b \\
c & d\\
\end{array} } \right]=ad-bc=\pm1$. So we can write a linear combination of $a,b$ that is equal to 1, so $gcd(a,b)=1$. So the matrix we got is:
$\left[ {\begin{array}{cc}
1 & 0 \\
c'' & d''\\
\end{array} } \right]$ and it's determinant is still $\pm 1$, so,
$d''=\pm1$. But we showed that we can generate this matrix, so we finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $m/(x^m - 1) - n/(x^n - 1)$ as $x\rightarrow 1$ without l'Hôpital's rule? Let $m,n\in\mathbb N$. How can I solve this limit without using l'Hopital's rule, please?
$$\lim_{x\to1}\left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right)$$
| \begin{align}\lim_{x\to1}\left(\frac m{x^m-1}-\frac n{x^n-1}\right)&=\lim_{x\to1}\frac{m(x^n-1)-n(x^m-1)}{(x^m-1)(x^n-1)}\\&=\lim_{x\to1}\frac{m(x-1)(x^{n-1}+\cdots+1)-n(x-1)(x^{m-1}+\cdots+1)}{(x^m-1)(x^n-1)}\\&=\lim_{x\to1}\frac{m(x^{n-1}+\cdots+1)-n(x^{m-1}+\cdots+1)}{(x-1)(x^{m-1}+\cdots+1)(x^{n-1}+\cdots+1)}\\&=\frac1{mn}\lim_{x\to1}\frac{m(x^{n-1}+\cdots+1)-n(x^{m-1}+\cdots+1)}{x-1}.\end{align}By definition of derivative, the previous limit is the derivative at $1$ of the function$$x\mapsto m(x^{n-1}+x^{n-2}+\cdots+1)-n(x^{m-1}+x^{m-2}+\cdots+1),$$which is\begin{align}m\frac{(n-1)n}2-n\frac{(m-1)m}2&=\frac{mn^2-mn-nm^2+nm}2\\&=\frac{mn(n-m)}2.\end{align}Therefore, your limit is $\dfrac{n-m}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $x_0=\sqrt{2}+\sqrt{3}+\sqrt{6}$ and $x_{n+1}=\frac{x_n^2-5}{2(x_n+2)}$, how can I find $x_n$?
Let $x_0=\sqrt{2}+\sqrt{3}+\sqrt{6}$ and
$$\forall n \in \mathbb{N} : x_{n+1}=\dfrac{x_n^2-5}{2(x_n+2)}$$
Then $x_n=?$
My Try :
$$x_{n+1}(2x_n+4)=x_n^2-5 \\2x_{n+1}x_n+4x_{n+1}=x_n^2-5\\x_n^2-2x_{n+1}x_n-(4x_{n+1}+5)=0$$
So we have :
$$x_n=\dfrac{2x_{n+1}\pm\sqrt{2x_{n+1}^2+4(4x_{n+1}+5)}}{2}$$
Now what ?
| A crucial observation is that the substitution $x_n=a_n-2$ brings the given recurrence relation into a familiar one, namely
$$ a_{n+1} = \frac{a_n^2-1}{2a_n} $$
which is associated with the duplication formula for the cotangent function. In particular, $a_0=-\cot\theta$ implies $a_n=-\cot(2^n\theta)$ and $x_n=-2-\cot(2^n\theta)$. Since $a_0=\cot\frac{\pi}{24}$ we have
$$ x_n = -2+\cot\left(\frac{\pi\cdot2^n}{24}\right) $$
and our sequence keeps oscillating between $-2-\frac{1}{\sqrt{3}}$ and $-2+\frac{1}{\sqrt{3}}$ from $x_3$ on.
Anyway, for any $x_0\in\mathbb{R}$ the given sequence is never converging, since, as remarked in the comments, it is Newton's method applied to a quadratic polynomial with a negative discriminant ($x^2+4x+5$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
How do I find k so that this rank is true? $$
\begin{bmatrix}
3 & 2 & -2 \\
1 & 1 & -1 \\
-1 & 2 & k \\
\end{bmatrix}
$$
Find the values k for which rank(A) = 3
From what I have tried myself, I have brought it down to row-echelon form where it looks something like this:
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & k+2 \\
\end{bmatrix}
$$
Through various row operations, but I seem to be stuck on whether or not I'm doing the right thing. Thanks!
To start:
$$
\begin{bmatrix}
3 & 2 & -2 \\
1 & 1 & -1 \\
-1 & 2 & k \\
\end{bmatrix}
$$
Interchanging Rows 1 and 2.
$$
\begin{bmatrix}
1 & 1 & -1 \\
3 & 2 & -2 \\
-1 & 2 & k \\
\end{bmatrix}
$$
then, Row 2 minus 3 times Row 1 and Row 3 plus Row 1.
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & -1 & 1 \\
0 & 3 & k-1 \\
\end{bmatrix}
$$
then, multiply Row 2 by (-1)
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 3 & k-1 \\
\end{bmatrix}
$$
Finally, Row 3 minus 3 times Row 2.
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & k+2 \\
\end{bmatrix}
$$
| $\begin{align*}
\det(A)&=3(k+2)-2(k-1)-2(2+1)\\
&=k+2
\end{align*}$
So $\det(A)=0 \iff k=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Using EGFs to solve the recurrence relation $a_n=n a_{n-1}+(n+1)!$ I have to solve the recurrence relation
$$a_n = na_{n-1} + (n + 1)!,\qquad a_0 = 1.$$
I am struggling to finish the problem. I have attached my work. Can you please help me finish?
$$\begin{eqnarray*}A(x)&=&\sum_{n\geq 0}\frac{a_n}{n!}x^n = a_0+\sum_{n\geq 1}\frac{a_n}{n!}x^n=a_0+\sum_{n\geq 1}\left(na_{n-1}+(n+1)!\right)\frac{x^n}{n!}\\&=&a_0+\sum_{n\geq 1}a_{n-1}\frac{x^n}{(n-1)!}+\sum_{n\geq 1}(n+1)x^n = a_0 + x \sum_{n\geq 1}a_{n-1} \frac{x^{n-1}}{(n-1)!}+\sum_{n\geq 1}(n+1)x^n\\&=&a_0+ x\,A(x)+\underbrace{\sum_{n\geq 1}(n+1)x^n}_{\text{I know that }(n+1)x^n = \frac{d}{dx}x^{n+1}.}\end{eqnarray*}$$
| By differentiating a geometric series and simplifying we get
$$ a_0+\sum_{n\geq 1}(n+1)x^n = 1+\frac{d}{dx}\left(\frac{x^2}{1-x}\right)=\frac{1}{(1-x)^2},$$
$$ A(x) = x\,A(x)+\frac{1}{(1-x)^2},\qquad A(x)=\frac{1}{(1-x)^3}\stackrel{\text{stars and bars}}{=}\sum_{n\geq 0}\binom{n+2}{2}x^n $$
hence
$$ a_n = \binom{n+2}{2}n! = \color{blue}{\frac{1}{2}(n+2)!}. $$
Let's check it, too:
$$ na_{n-1}+(n+1)! = \frac{1}{2}n(n+1)!+(n+1)! = \frac{n+2}{2}(n+1)!\stackrel{\color{green}{\checkmark}}{=}\frac{1}{2}(n+2)!.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
| If you go on to rewrite $2x=A(x+2)+B$ as $2x=Ax+(2A+B)$, then you can match up the coefficients of the polynomials (of $x$) on each side. You must have
$$2=A\qquad\text{and}\qquad 0=2A+B$$
Now solve for $A$ and $B$. The first equation (obviously!) says $A=2$. When you plug this into the second, obtaining $0=2\cdot2+B$, you get $B=-4$.
Incidentally, where you write "$2x={A\over x+2}+{B\over(x+2)^2}$," the left hand side of the equation should be $2x\over(x+2)^2$, not just $2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 7
} |
leading terms in sequence What is the leading term of the following sequences as $n \to \infty$
(a) $n^{-1}- (n^{2}+1)^{-1/2}$
(a) $7^{1/n^2}- 2(7^{2/n^2})+7^{3/n^2}$
| We expand the expression in a Taylor series around $n=0$ and extract the most significant term when $n\rightarrow \infty$
We obtain
\begin{align*}
\color{blue}{n^{-1}-\left(n^2+1\right)^{-\frac{1}{2}}}
&=\frac{1}{n}-\frac{1}{n}\left(1+\frac{1}{n^2}\right)^{-\frac{1}{2}}\\
&=\frac{1}{n}-\frac{1}{n}\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}\frac{1}{n^{2j}}\tag{1}\\
&=\frac{1}{n}-\frac{1}{n}\left(1+\left(-\frac{1}{2}\right)\frac{1}{n^2}
+\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\frac{1}{n^4}+\cdots\right)\\
&=\frac{1}{2n^3}-\frac{3}{4n^5}-\cdots\\
&\color{blue}{\sim \frac{1}{2n^3}}
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion.
We obtain
\begin{align*}
\color{blue}{7^{1/n^2}}&\color{blue}{-2\cdot7^{2/n^2}+7^{3/n^2}}\\
&=e^{\frac{1}{n^2}\ln 7}-2e^{\frac{2}{n^2}\ln 7}+e^{\frac{3}{n^2}\ln 7}\tag{2}\\
&=\left(1+\frac{\ln 7}{n^2}+\frac{1}{2}\cdot\frac{\ln^2 7}{n^4}+\cdots\right)
-2\left(1+\frac{2\ln 7}{n^2}+\frac{1}{2}\cdot\frac{4\ln^2 7}{n^4}+\cdots\right)\tag{3}\\
&\qquad +\left(1+\frac{3\ln 7}{n^2}+\frac{1}{2}\cdot\frac{9\ln^2 7}{n^4}+\cdots\right)\\
&=\left(\frac{1}{2}-4+\frac{9}{2}\right)\frac{\ln^2 7}{n^4}+\cdots\\
&\color{blue}{\sim \frac{\ln^2 7}{n^4}}
\end{align*}
Comment:
*
*In (2) we use the representation $a^b=e^{b\ln(a)}$.
*In (3) we use the exponential series expansion $$e^x=\sum_{j=0}^\infty\frac{x^j}{j!}=1+x+\frac{1}{2}x^2+\cdots$$
Author's example:
Let's verify the author's calculation. We recall the cosine series expansion at $x=0$ is
\begin{align*}
\sum_{j=0}^\infty(-1)^j\frac{x^{2j}}{(2j)!}
\end{align*}
It is also convenient to use the Landau Big-O notation.
We obtain
\begin{align*}
\color{blue}{1-\cos\left(\frac{2n^{1/2}+3}{3n^2+1}\right)}
&=1-\cos\left(\frac{\frac{2}{3}n^{-3/2}+n^{-2}}{1+\frac{1}{3}n^{-2}}\right)\tag{5}\\
&=1-\cos\left(\left(\frac{2}{3}n^{-3/2}+O(n^{-2})\right)\left(1+O(n^{-2})\right)\right)\tag{6}\\
&=1-\cos\left(\frac{2}{3}n^{-3/2}+O(n^{-2})\right)\tag{7}\\
&=1-\left(1-\frac{1}{2}\left(\frac{2}{3}n^{-3/2}+O(n^{-2})\right)^2+O(n^{-6})\right)\\
&=\frac{1}{2}\cdot\frac{4}{9}n^{-3}+O(n^{-7/2})\\
&\color{blue}{\sim \frac{2}{9}n^{-3}}
\end{align*}
confirming the author's result.
Comment:
*
*In (5) we factor out $3n^2$ as preparation for a geometric series expansion.
*In (6) we do the series expansion noting that
$$\frac{1}{1+\frac{1}{3}n^{-2}}=1-\frac{1}{3}n^{-2}+\frac{1}{9}n^{-4}-\cdots=1+O(n^{-2})$$
*In (7) we observe that terms are swallowed by $O(n^{-2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum distance from the point Find the minimum distance from the point $ \ (5,0) \ $ to the curve $ \ y=\sqrt x+2 \ $.
Answer:
Let $ \ (x,y) \ $ be the closest point on $ \ y=\sqrt x+2 \ $ from $ (5,0) $ .
Then the distance is given by
$ d(x,y)=\sqrt{(x-5)^2+y^2} \ $
We will minimize the function $ \ g(x,y)=(x-5)^2+y^2 \ $
replacing $ y \ \ by \ \ \sqrt x+2 \ $ , we get
$ g(x)=(x-5)^2+(\sqrt x+2)^2 \ $
The extreme points \ are
$ f'(x)=0 \\ 2(x-5)+\frac{\sqrt x+2}{\sqrt x} =0 \\ 2x \sqrt x-9 \sqrt x+2=0 $
This becomes complicated . I am unable to calculate the closest point.
Help me out
| write your last equation as $$2x-10+1+\frac{2}{\sqrt{x}}=0$$ and Isolate the term with the square root:
$$\frac{2}{\sqrt{x}}=-2x+9$$ squaring gives
$$\frac{4}{x}=(-2x+9)^2$$
can you finish?
factorizing the last equation gives $$- \left( x-4 \right) \left( 4\,{x}^{2}-20\,x+1 \right) =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim\limits_{x \to π/2}\frac{2x\sin(x) - π}{\cos x}$ without l'hospital There is a problem that is easy to solve with L'hospital but we are required to solve it without it, but I could not find the answer. x* sinx part is especially confusing me, because other examples I have solved did not include such part.
$$\lim_{x \to π/2}\frac{2x\sin(x) - π}{\cos x}$$
| Substitute $y=\frac\pi 2 - x$. Then
$$\lim_{x \to π/2}\frac{2x\sin x - π}{\cos x} = \lim_{y\to0}\frac{2(\frac\pi 2-y)\cos y - π}{\sin y}$$
and the function under the limit is
$$\frac{2(\frac\pi 2-y)\cos y - π}{\sin y}
= \frac{(\pi-2y)\cos y - π}{\sin y} \\
= \frac{\pi(\cos y - 1)-2y\cos y}{\sin y} \\
= \pi\frac{\cos y-1}{\sin y}-\frac{2y}{\sin y}\cos y \\
= \pi\frac{\color{red}{\cos y}-\color{blue}1}{\sin y}-2\frac y{\sin y}\cos y \\
= \pi\frac{\color{red}{\cos(2\cdot\frac y2)}-\color{blue}1}{\sin(2\cdot\frac y2)}-2\frac y{\sin y}\cos y \\
= \pi\frac{\color{red}{(\cos^2\frac y2-\sin^2\frac y2)}-\color{blue}{(\sin^2\frac y2+\cos^2\frac y2)}}{2\sin\frac y2\cdot\cos\frac y2}-2\frac y{\sin y}\cos y \\
= \pi\frac{-2\sin^2\frac y2}{2\sin\frac y2\cdot\cos\frac y2}-2\frac y{\sin y}\cos y \\
= -\pi\frac{\sin\frac y2}{\cos\frac y2}-2\frac y{\sin y}\cos y \\
= -\pi\tan\frac y2-2\frac y{\sin y}\cos y
$$
We know (...?) that
$$\tan 0 = 0$$
$$\cos 0 = 1$$
and
$$\lim_{y\to 0}\frac y{\sin y} = 1$$
so the limit sought is
$$\lim_{y\to 0} \left(-\pi\tan\frac y2-2\frac y{\sin y}\cos y\right) = -\pi \cdot 0 - 2\cdot 1 \cdot 1 = \boxed{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve system of three equations I got these equations as a system
$2x+3y-2z=2$
$3x-y+2z=-1$
$7x+16y-12z=11$
And I cannot figure out how to solve this. I have tried addition, tried to get rid of z and only solve for x and y but always end up with $0=0$
| $$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
3 & -1 & 2 & -1 \\
7 & 16 & -12 & 11\\
\end{array}\right)$$
Add $-\frac{3}{2}$ times the first row to the second, and $-\frac{7}{2}$ times the first row to the third:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
0 & \frac{11}{2} & -5 & 4\\
\end{array}\right)$$
Now add the second row to the third:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
0 & 0 & 0 & 0\\
\end{array}\right)$$
Remove the third row:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
\end{array}\right)$$
Move the third column to the right side (all of the numbers there will be multipled by $-1$), and name it as $p$ (so $z=p$)
$$\left(\begin{array}{cc|cc}
2 & 3 & 2 & 2 \\
0 & \frac{-11}{2} & -4 & -5 \\
\end{array}\right)$$
Now add $\frac{6}{11}$ times the second row to the first:
$$\left(\begin{array}{cc|cc}
2 & 0 & \frac{-2}{11} & \frac{-8}{11} \\
0 & \frac{-11}{2} & -4 & -5 \\
\end{array}\right)$$
Now the equations:
$$2x=-\frac{2}{11}-\frac{8}{11}p$$
$$-\frac{11}{2}y=-4-5p$$
$$z=p$$
So:
$$x=-\frac{1}{11}-\frac{4}{11}p$$
$$y=-\frac{8}{11}-\frac{10}{11}p$$
$$z=p$$
In vector form:
$$\begin{pmatrix} x \\ y \\ z\end{pmatrix}=\frac{1}{11}\begin{pmatrix} -1 \\ -8 \\ 0\end{pmatrix} + \frac{p}{11} \begin{pmatrix} -4 \\ -10 \\ 11\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
General solution of $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $ Find general solution of the equation $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $.
My approach:
Squared on both sides, formed a quadratic equation in $\cos\theta$ and finally got two solutions for theta,
$$\theta = 2n\pi \pm \frac{\pi}{6}$$
$$\theta = 2n\pi \pm \frac{\pi}{3}$$
But the answer given in my book is $$2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$$
Pretty strange
can anyone help me?
| In general, to solve $a \cos x +b\sin x=c$. Divide both sides by $\sqrt{a^2+b^2}$ and then let $\sin \alpha =\frac{a}{\sqrt{a^2+b^2}}$ to get
$$\sin (x+\alpha)=\frac{c}{\sqrt{a^2+b^2}}$$
Now solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proof $|x+y|+|x-y| \geq |x| + |y|$ Proof that for all $x,y \in \mathbb{R}$:
$$|x+y|+|x-y| \geq |x| + |y|$$
My (failed) attempt:
$$|x + y| \leq |x| + |y|$$
$$|x - y| \leq |x| + |y|$$
$$|x + y|+|x - y| \leq 2(|x| + |y|)$$
Thank you
| We have,
$$\color{red}{(|x+y|+|x-y|)^2 = 2(x^2+y^2) +2|x^2-y^2|}$$ Whence,
\begin{align}&2|x^2-y^2| \geq -x^2-y^2+2|x| |y|= -(|x|-|y|)^2\\
&\Longleftrightarrow(x^2+y^2) +2|x^2-y^2|\geq 2|x||y|\\
&\Longleftrightarrow 2(x^2+y^2) +2|x^2-y^2|\geq x^2+y^2+2|x||y|\\
&\Longleftrightarrow (|x+y|+|x-y|)^2 \geq (|x| + |y|)^2\\
&\Longleftrightarrow|x+y|+|x-y| \geq |x| + |y|
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_{0}^{\infty} \frac{x^{-\mathfrak{i}a}}{x^2+bx+1} \,\mathrm{d}x$ using complex analysis? We were told today by our teacher (I suppose to scare us) that in certain schools for physics in Soviet Russia there was as an entry examination the following integral given
$$\int\limits_{0}^{\infty} \frac{x^{-\mathfrak{i}a}}{x^2+bx+1} \,\mathrm{d}x\,,$$
where $a \in \mathbb{R}$, $b \in [0,2)$, and $\mathfrak{i}$ is the imaginary unit. And since we are doing complex analysis at the moment, it can, according to my teacher, be calculated using complex methods.
I was wondering how this could work? It seems hard to me to find a good curve to apply the residue theorem for this object, I suppose. Is there a trick to compute this integral?
| $\newcommand{\Res}{\text{Res}}$
Firstly define:
\begin{align}
f(z) = \frac{1}{z^2+bz+1}
\end{align}
Secondly define:
\begin{align}
g(z)= (-z)^{-ia}f(z)
\end{align}
We use the Principal Log to define $(-z)^{-ia}$. The main reason for the minus sign is that I want to work with the principal Log. Consider now the keyhole contour $K_R$ consisting of $C_R \cup C_R^+ \cup C_R^-$. The circle part is $C_R$ with radius $R$, and $C_R^+$ is the segment that connects $0$ to $R$ on the right half plane from above and $C_R^-$ the same as $C_R^+$ but from below.
So our contour looks like the one below:
Note that $\int_{C_R} g(z)dz \to 0$ as $R\to \infty$ (why?). Now the poles of this function $g(z)$ is at $z_1= -\frac{1}{2}b - i\ \sqrt[]{1-\frac{1}{4}b^2}$ and $z_2 = -\frac{1}{2}b + i\ \sqrt[]{1-\frac{1}{4}b^2}$. Note that for large enough $R$ both poles will be in the area enclosed by our contour.
On $C_R^+$ we have:
\begin{align}
\lim_{R\to \infty} \int_{C_R^+} g(z) dz = e^{-a\pi}\int^{\infty}_0\frac{x^{-ia}}{x^2+bx+1} dx
\end{align}
Similarly on $C_R^-$ we have:
\begin{align}
\lim_{R\to \infty} \int_{C_R^-} g(z) dz = -e^{a\pi}\int^{\infty}_0\frac{x^{-ia}}{x^2+bx+1} dx
\end{align}
So:
\begin{align}
\lim_{R\to\infty} \int_{K_R} g(z) dz &= (-e^{a\pi} + e^{-a\pi} )\int^\infty_0 \frac{x^{-ia}}{x^2+bx+1} dx \\
&=-2\sinh(a\pi) \int^\infty_0 \frac{x^{-ia}}{x^2+bx+1} dx
\end{align}
On the other hand we have by the Residue theorem:
\begin{align}
\lim_{R\to\infty} \int_{K_R} g(z) dz = 2\pi i\left( \Res_{z=z_1}g(z)+\Res_{z=z_2}g(z)\right)
\end{align}
Let's calculate the residues:
\begin{align}
\Res_{z=z_1} g(z) = \frac{\left(\frac{1}{2}b + i\ \sqrt[]{1-\frac{1}{4}b^2}\ \right)^{-ia}}{-2i \ \sqrt[]{1-\frac{1}{4}b^2} }
\end{align}
The other one:
\begin{align}
\Res_{z=z_2} g(z) = \frac{\left(\frac{1}{2}b - i\ \sqrt[]{1-\frac{1}{4}b^2}\ \right)^{-ia}}{2i \ \sqrt[]{1-\frac{1}{4}b^2} }
\end{align}
Define $\beta:=\frac{1}{2}b + i\ \sqrt[]{1-\frac{1}{4}b^2}$. So:
\begin{align}
\lim_{R\to\infty} \int_{K_R} g(z) dz = \pi \frac{\bar\beta^{-ia}-\beta^{-ia}}{\sqrt[]{1-\frac{1}{4}b^2}} = 2\pi \frac{\bar\beta^{-ia}-\beta^{-ia}}{ \sqrt[]{4-b^2}}
\end{align}
This means:
\begin{align}
\int^{\infty}_0 \frac{x^{-ia}}{x^2+bx+1}dx &= \frac{2\pi}{-2\sinh(a\pi)} \frac{\bar\beta^{-ia}-\beta^{-ia}}{ \sqrt[]{4-b^2}}
\end{align}
Note that $\beta = \exp(i\arccos(b/2))$. So $\beta^{-ia}=\exp(a\arccos(b/2))$ and $\bar\beta^{-ia}=\exp(a\arccos(b/2)).$ Substituting $\beta$ gives us the final result:
\begin{align}
\int^{\infty}_0 \frac{x^{-ia}}{x^2+bx+1}dx = \color{red}{\frac{2\pi\sinh(a\cdot\arccos(b/2))}{\sinh(a\pi) \ \sqrt[]{4-b^2}} }
\end{align}
This integral is real.
Remark
It may be interesting to note the following. For positive $x$ one has: $x^{-ia}=e^{-ia\text{Log}(x)}=e^{-ia\ln(x)}=\cos(a\ln(x))-i\sin(a\ln(x))$. So we have:
\begin{align}
\int^\infty_0 \frac{x^{-ia}}{x^2+bx+1} dx = \int^\infty_0 \frac{\cos(a\ln(x))}{x^2+bx+1} dx - i\int^\infty_0 \frac{\sin(a\ln(x))}{x^2+bx+1} dx
\end{align}
Since our integral is real we get the following identities for free:
\begin{align}
\color{blue}{\int^\infty_0 \frac{\cos(a\ln(x))}{x^2+bx+1} dx = \frac{2\pi\sinh(a\cdot\arccos(b/2))}{\sinh(a\pi) \ \sqrt[]{4-b^2}}}
\end{align}
And:
\begin{align}
\color{blue}{\int^\infty_0 \frac{\sin(a\ln(x))}{x^2+bx+1} dx = 0 }
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "95",
"answer_count": 3,
"answer_id": 0
} |
What is the probability of rolling 3 different pairs with seven dice? for example one outcome can be this :
(1,1,2,2,3,3,4)
My answer is different compared to the book
This is how i solve it:
$$\frac{\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}}{6^7} $$
The book solves it like this:
$$\frac{\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{6}{3}\binom{3}{1}}{6^7}=0.135 $$
am i wrong?
isn't this outcome :(1,1,2,2,3,3,4) different compared to this :(2,2,1,1,3,3,4) ?
| Here is a justification of the book's answer:
Choose the three values for the pairs from the six possible values on the dice in $\binom{6}{3}$ ways. Choose the value of the singleton from the remaining three values in $\binom{3}{1}$ ways. For the three pairs, there are $\binom{7}{2}$ ways for the smallest value to appear on two of the seven dice, $\binom{5}{2}$ ways for the middle value to appear on two of the other five dice, and $\binom{3}{2}$ ways for the largest value to appear on two of the other three dice. The remaining die shows the singleton value. Hence, there are
$$\binom{6}{3}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2}$$
favorable outcomes.
In your attempt, you count each favorable outcome six times, once for each way you designate one of the three pairs as your first pair, another pair as your second pair, and the remaining pair as your third pair. The order in which the pairs are selected does not matter. Notice that
$$6\binom{6}{3}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2} = \binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine a basis for a solution set of a linear system I'm studying, in this class we are given practice problems (but no answer sheet, go figure!). So my question is:
Is my answer correct? Please point out anything I might have gotten wrong.
Determine a basis for the solution set of the linear system:
$$\left\{\begin{array}{l}
3x + 4y - 2z =0\\
2x + 3y + z =0\\
-2x -2y + 6z =0\\
\end{array}\right.$$
My solution:
Then the augmented matrix is:
\begin{align*}
\left[ \begin{array}{ccc|c}
3 & 4 & -2 &0\\
2 & 3 & 1 &0\\
-2 & -2 & 6 &0\\
\end{array}
\right]
\end{align*}
Reduced Row Echelon Form
\begin{align*}
\left[ \begin{array}{ccc|c}
1 & 0 & -10 &0\\
0 & 1 & 7 &0\\
0 & 0 & 0 &0\\
\end{array}
\right]
\end{align*}
This means that any vector $\vec x$ can be represented as
\begin{align*}
\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} =
\begin{bmatrix}
10z\\
-7z\\
z\end{bmatrix}=z
\begin{bmatrix}
10\\
-7\\
1
\end{bmatrix}
\end{align*}
Where $z$ is any scalar.
This means that the linear system represents a one-dimensional subspace and its basis is:
\begin{align*}\left(\begin{bmatrix}
10\\
-7\\
1
\end{bmatrix}\right)
\end{align*}
| Your computations are correct, and so is your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$.
If $z\in C$ satisfies $|z^4+z^3-z-1| = |z^4-z^3+z-1|.$ Then minimum value of $|z-1-2i|$ and maximum value of $|z^2-1-2i|$ is . . .
$\bf{Attempt}$ from $|z^4+z^3-z-1| = |z^4-z^3+z-1|$
$$|(z+1)(z^3-1)| = |(z-1)(z^3+1)|$$
$$\bigg|\frac{z-1}{z+1}\bigg|=\bigg|\frac{z^3-1}{z^3+1}\bigg|=k$$
Could some help me how to solve it, thanks
| You've gone just a bit astray.
First of all, since you don't know what $z$ is, it's dangerous to divide by $|z+1|,$ as it could well be equal to $0.$ In fact, $z=-1$ is a solution to the given equation, so that is entirely reasonable!
Next, you haven't quite finished factoring. Observing that $z=1$ is clearly a root of $z^3-1=0,$ then $z-1$ is a factor of $z^3-1.$ In particular, $$z^4+z^3-z-1=(z+1)(z-1)\left(z^2+z+1\right),$$ and similarly, $$z^4-z^3+z-1=(z-1)(z+1)\left(z^2-z+1\right).$$ Now, the given equation becomes $$|z+1||z-1|\left|z^2+z+1\right|=|z-1||z+1|\left|z^2-z+1\right|,$$ or equivalently, $$|z+1||z-1|\left(\left|z^2+z+1\right|-\left|z^2-z+1\right|\right)=0.$$ Thus, we can conclude that $z\in\{1,-1\}$ or that $\left|z^2+z+1\right|=\left|z^2-z+1\right|.$ It is easily verified that $z^2+z+1$ and $z^2-z+1$ are never both equal to $0,$ and neither is equal to $0$ when $z\in\{1,-1\}.$ Thus, assuming that the given equation holds and that $z\notin\{1,-1\},$ we have that $$\left|z^2+z+1\right|=\left|z^2-z+1\right|,\tag{1}$$ and you could divide through by one side of the equation safely.
However, we still don't have to do so. Since the modulus function is non-negative, then $(1)$ holds if and only if $$\left|z^2+z+1\right|^2=\left|z^2-z+1\right|^2\tag{2}$$ holds. Recall now the following facts (which hold for any $u,v,w\in\Bbb C$):
*
*$|w|^2=w\cdot\overline w,$ where $\overline w$ is the complex conjugate of $w.$ In particular, if $w=x+iy$ for $x,y\in\Bbb R,$ then $|w|^2=x^2+y^2.$
*$\overline{u+v}=\overline u+\overline v.$
*$\overline{u\cdot v}=\overline u\cdot\overline v.$
*For any $\alpha\in\Bbb R,$ we have $\overline{\alpha}=\alpha.$
*Applying facts 2-4 shows that $\overline{u-v}=\overline u-\overline v.$
*$\overline{\left(\overline w\right)}=w.$
*The real part of $w$ is $\Re(w)=\frac12\left(w+\overline w\right).$
With these facts at our disposal, we see that $$\begin{eqnarray}\left|z^2+z+1\right|^2 &=& \left(z^2+z+1\right)\cdot\overline{(z^2+z+1)}\\ &=& \left[\left(z^2+1\right)+z\right]\cdot\left[\overline{(z^2+1)}+\overline z\right]\\ &=& \left|z^2+1\right|^2+\overline z\cdot\left(z^2+1\right)+z\cdot\overline{\left(z^2+1\right)}+|z|^2\\ &=& \left|z^2+1\right|^2+|z|^2+2\Re\left(\overline z\cdot\left(z^2+1\right)\right),\end{eqnarray}$$ and similarly that $$\left|z^2-z+1\right|^2=\left|z^2+1\right|^2+|z|^2-2\Re\left(\overline z\cdot\left(z^2+1\right)\right),$$ whence we see that $(2)$ is equivalent to $$\Re\left(\overline z\cdot\left(z^2+1\right)\right)=0.\tag{3}$$ From there, $$\begin{eqnarray}0 &=& 2\Re\left(\overline z\cdot\left(z^2+1\right)\right)\\ &=& \overline z\cdot\left(z^2+1\right)+z\cdot\overline{\left(z^2+1\right)}\\ &=& \overline z\cdot\left(z^2+1\right)+z\cdot\left(\left(\overline z\right)^2+1\right)\\ &=& |z|^2\cdot z+z+|z|^2\cdot\overline z+\overline z\\ &=& \left(|z|^2+1\right)\cdot z+\left(|z|^2+1\right)\cdot\overline z\\ &=& 2\left(|z|^2+1\right)\Re(z),\end{eqnarray}$$ so since $|z|^2+1\ne 0,$ then we have that $(3)$ is equivalent to $$\Re(z)=0.\tag{4}$$
Thus, letting $A:=\{1,-1\}\cup\{\alpha i:\alpha\in\Bbb R\},$ the given equation holds if and only if $z\in A.$
At this point, we need only minimize $|z-1-2i|$ and maximize $\left|z^2-1-2i\right|$ among these points.
Suppose that $z=\alpha i$ for some real $\alpha.$ Observe that $z$ minimizes $|z-1-2i|$ if and only if $z$ minimizes $$|z-1-2i|^2=|-1+i(\alpha-2)|^2=(-1)^2+(\alpha-2)^2=(\alpha-2)^2+1.$$ Since $(\alpha-2)^2\ge0$ for all real $\alpha,$ then $\alpha=2$ clearly minimizes $(\alpha-2)^2+1.$ Thus for $z\in A,$ the minimum value of $|z-1-2i|$ occurs at some point $z\in\{1,-1,2i\}.$ I leave it to you to determine said minimum value.
Similarly, if $z=\alpha i$ for some real $\alpha,$ then $z$ maximizes $\left|z^2-1-2i\right|$ if and only if $z$ maximizes $$\left|z^2-1-2i\right|=(\alpha^2+1)^2+4.$$ Is this possible?
| {
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prove that $2\sqrt5 +\sqrt{11}$ is irrational how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational?
I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields
$\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
| Here's a more advanced approach, with some common details to the above.
A monic polynomial with integer coefficients:
$$ (x^2 - (2\sqrt 5 + \sqrt{11})^2)(x^2 - (2\sqrt 5 - \sqrt{11})^2) = x^4 - 62x^2 +81 $$
Now $x = 2\sqrt 5 + \sqrt{11}$ is a root of this polynomial, but by the Rational Roots Thm., any rational root would be an integer divisor of $81$.
Thus one only needs to verify that $2\sqrt 5 + \sqrt{11}$ is not an integer. A simple hand computation (or calculator computation) shows this positive number lies strictly between $7$ and $8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minima and maxima of a trig function let $f(x) = \sqrt {4\sin^4x - \sin^2x\cos^2x + 4\cos^4x}$
what is the product of the minimum and maximum value of the function?
I tried turning them all to $\sin x$
$-\sin^2x\cos^2x = \sin^4x-\sin^2x ; 4\cos^4x = (1-\sin^2x)^2 = 4sin^4x - 4sin^2x +4 $
which gives us :
$9\sin^4x - 2\sin^2x +4$
then i dont know how to proceed
EDIT:
The equation above is wrong, i got if you try solving for the ones i have given :
$9\sin^4x - 5\sin^2x +4$
then I tried making $sin^2x = a$ (Thanks @Greninja for the info)
where in the the min is when $a = 5/18$ and the value is 47/36,
then the max is when $a=1$ and the value is 8
Min x Max = $12\sqrt{94}$ which looks far fetched
I dont know where I went wrong with my solution Please help.
And yes, I have been told that the answer is $\sqrt{7}$
| The expression you should get is $\sqrt{9(\sin x ) ^4-9(\sin x ) ^2+4}$
$$\sqrt{9(\sin^4 x -\sin^2 x) +4}$$
$$\sqrt{9((\sin^2 x - \frac{1}{2})^2 - \frac{1}{4})+4}$$
$$\sqrt{9(\sin^2 x-\frac{1}{2})^2+ \frac{7}{4}}$$
Now using range and domain of $sin^2 x $ get your answer
Edit1: now you would get the desired answer.
| {
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Number which is simultaneously sum of 2 and 3 squares Is there positive integer $m$ such $m=x_1^2+x_2^2$ and $m=y_1^2+y_2^2+y_3^2$ where $x_i, y_j$ are nonzero integers. I have tried by hand for the ten natural numbers but I was not able to find such $m$.
Would be thankful for help.
| Alternatively you could use a formula to generate examples; for example
$$m=\left(a^2+b^2+c^2\right)^2+\left(a^2-c^2\right)^2=\left(a^2+c^2\right)^2+\left(2cb\right)^2+\left(a^2+b^2-c^2\right)^2$$
originating from the identity
$$\left(a^2+b^2+c^2\right)^2-\left(a^2+b^2-c^2\right)^2=\left(2ca\right)^2+\left(2cb\right)^2$$
and using $$\left(a^2+c^2\right)^2-\left(a^2-c^2\right)^2=\left(2ca\right)^2$$
to substitute for $\left(2ca\right)^2$
| {
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Eigenvalue Decomposition Error for the following matrix. Let us assume that $$A=\left[\begin{array}{rrr}2&-1&-1\\-1&2&-1\\-1&-1&2\end{array}\right].$$
Then, eigenvalues of $A$ is $0, 3, 3$.
For $\lambda_1=0$, I could find its eigenvector as $\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$.
For $\lambda_1=3$, I could find its eigenvector as $\left[\begin{array}r0\\ 1\\ -1\end{array}\right]$ and $\left[\begin{array}r0\\ -1\\ 1\end{array}\right]$.
Thus, I set $S=\left[\begin{array}{rrr}1&0&0\\1&1&-1\\1&-1&1\end{array}\right]$ and $\Lambda=\begin{bmatrix}0&0&0\\0&3&0\\0&0&3\end{bmatrix}$.
The problem is $S\times \Lambda \times S^{-1}$ does not give $A$.
By matlab, $S$ is given as $\left[\begin{array}{rrr}0.5774&0.7634&0.2895\\0.5774&-0.6325&0.5164\\0.5774&-0.1310&-0.8059\end{array}\right]$
Using $S$ given by matlab, $S\times \Lambda \times S^{-1}$ gives $A$.
The textbook said the $i$th column of $S$ must be the eigenvector corresponding the $i$th eigenvalue of $A$.
I think this problem happens because two lambdas are same as $3$.
How can I find correct $S$?
| Directly:
$$|xI-A|=\begin{vmatrix}
x-2&1&1\\
1&x-2&1\\
1&1&x-2\end{vmatrix}=(x-2)^3+2-3(x-2)=x^3-6x^2+9x\implies$$
the eigenvalues are $\;0,\,3\;$, the last one of multiplicity two, and a basis for the corresponding eigenspaces is
$$\left\{\,\begin{pmatrix}1\\1\\1\end{pmatrix}\,,\;\begin{pmatrix}1\\\!-1\\0\end{pmatrix}\,,\;\begin{pmatrix}1\\0\\\!-1\end{pmatrix}\,\right\}$$
Try to take it from here now.
| {
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Show that $(ab' - a'b)^2 + 4(ah' - a'h)(bh' - b'h)$ is a perfect square Reference: A Course of Pure Mathematics (ed. 3) page 38
Show that if $a,a',b,b',h,h'$ are rational and all the values of $x$ and $y$ given by
$ax^2 + 2hxy + by^2 = 1$, $a'x^2 + 2h'xy + b'y^2=1$
are rational, then
*
*$(h - h')^2 - (a-a')(b-b')$ is a perfect square
*$(ab' - a'b)^2 + 4(ah' - a'h)(bh' - b'h)$ is a perfect square
My attempt:
$$ax^2 + 2hxy + by^2 = 1 \\\implies ax^2 + 2hxy + by^2 - (a'x^2 + 2h'xy + b'y^2) = 1 -1 \\\implies (a-a')x^2 + 2(h-h')xy + (b-b')y^2 = 0 \\\implies \frac{1}{y^2}[(a-a')x^2 + 2(h-h')xy + (b-b')y^2] = 0 \\\implies (a-a')\frac{x^2}{y^2} + 2(h-h')\frac{x}{y} + (b-b') = 0$$ Assuming $a-a' \neq 0$, $$\\\implies \frac{x}{y} = \frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'} $$
$\frac{x}{y}$ is always rational if $(h-h')^2 - (a-a')(b-b')$ is a perfect square and vice versa. This proves (1).
$$ax^2 + 2hxy + by^2 = 1$$
$$\implies a (\frac{x}{y})^2 + 2h \frac{x}{y} + b = \frac{1}{y^2}$$
$$\implies a \cdot [\frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'}]^2 + 2h \cdot [\frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'}] + b = \frac{1}{y^2}$$
Now, if $(h-h')^2 - (a-a')(b-b')$ is a perfect square, then $\frac{x}{y}$ is rational and if $$a \cdot [\frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'}]^2 + 2h \cdot [\frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'}] + b$$ is a perfect square, then $y$ must be rational. And if $\frac{x}{y}$ and $y$ is rational then $x$ is rational. Therefore it seems to me that if $a-a' \neq 0$ and $(h-h')^2 - (a-a')(b-b')$ is a perfect square and $$a \cdot [\frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'}]^2 + 2h \cdot [\frac{-(h-h')\pm \sqrt{(h-h')^2 - (a-a')(b-b')}}{a-a'}] + b$$ is a perfect square, then all values of $x$ and $y$ that satisfy the equations are rational.
| Update: Here's a new proof, constructed via resultant theory. I think this is probably what the original poser wanted.
Let $z = x/y$, then
$$(a - a')z^2 + 2(h - h')z + (b - b') = 0$$
$$y^2 = \frac{1}{az^2 + 2hz + b}$$
Define
$$s = 2(a'b - ab')(a - a') + 4(ah' - a'h)(h - h')$$
$$t = (ab' - a'b)^2 + 4(ah' - a'h)(bh' - b'h)$$
Then one can verify that $y$ satisfies a quartic equation
$$ty^4 + sy^2 + (a - a')^2 = 0$$
The solutions to this equation are given by
$$y = \pm\sqrt{-\frac{s}{4t} + \frac{\sqrt{(a - a')^2/t}}{2}} \pm \sqrt{-\frac{s}{4t} - \frac{\sqrt{(a - a')^2/t}}{2}}$$
Since all solutions of $y$ are rational, then
$$\sqrt{\frac{(a - a')^2}{t}}$$
is rational. Hence $\sqrt{t}$ is rational.
Original proof: I assume that in this problem, $x,y$ range over all complex numbers. If complex solutions are ignored, then we can construct counterexamples in which all real solutions are rational, but the second value is negative, hence not a perfect square.
Let $z = x/y$. Suppose that
$$(h - h')^2 - (a - a')(b - b') \neq 0$$
Then there exists two rational roots to the equation
$$(a - a')z^2 + 2(h - h')z + (b - b') = 0$$
Let us call them $z_1,z_2$. Correspondingly, there are four solutions to the original system:
$$\left\{\begin{aligned}x &= \pm\frac{1}{\sqrt{az^2 + 2hz + b}}\\y &= \pm\frac{z}{\sqrt{az^2 + 2hz + b}}\end{aligned}\right.$$
As you have noted, in each case $az^2 + 2hz + b$ must be a perfect square. Suppose that
$$\left\{\begin{aligned}az_1^2 + 2hz_1 + b &= t_1^2\\az_2^2 + 2hz_2 + b &= t_2^2\end{aligned}\right.$$
We find that
$$2h = \frac{(t_2^2 - t_1^2) - a(z_2^2 - z_1^2)}{z_2 - z_1}$$
$$b = t_1^2 - az_1^2 - 2hz_1$$
Now if we write
$$(a - a')z^2 + 2(h - h')z + (b - b') = u(z - z_1)(z - z_2)$$
where $u = a - a'$ is a rational number, then
$$a'z^2 + 2h'z + b' = az^2 + 2hz + b - u(z - z_1)(z - z_2)$$
Matching coefficients of $z$ on both sides, and we obtain expressions for $a',h',b'$.
Plug all the above relations into
$$(ab' - a'b)^2 + 4(ah' - a'h)(bh' - b'h)$$
and simplify the result (this will be very messy, I did it with Mathematica), and it will turn out to be a square of a rational number.
In the above we have assumed that
$$(h - h')^2 - (a - a')(b - b') \neq 0$$
If instead we have
$$(h - h')^2 - (a - a')(b - b') = 0$$
then we can assume
$$\left\{\begin{aligned}h - h' &= k\\a - a' &= p\\b - b' &= q\end{aligned}\right.$$
Then $q = k^2/p$, and
$$\left\{\begin{aligned}h' &= h - k\\a' &= a - p\\b' &= b - k^2/p\end{aligned}\right.$$
Plug these relations into the value, simplify the result, and it will turn out to be a square.
| {
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AM>HM Problem $\frac{1}{n+1}+...+\frac{1}{3n+1}>1$ I am having difficulty solving one of the problems from "Problems in Mathematical Analysis I" - W. J. Kaczor;M. T. Nowak .
It's a problem 1.2.5 b), and it goes like this:
1.2.5. For $n \in \mathbb{N}$, verify the following claims:
$$\tag{b} \qquad \dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3} + \ldots + \dfrac{1}{3n + 1} \, > \, 1$$
In solutions it says: "Use the arithmetic-harmonic mean inequality!"
I tried to apply it on whole inequality but got:
\begin{align}
& \dfrac{\frac{1}{n+1}+\ldots+\frac{1}{3n+1}}{2n}\, >\, \dfrac{2n}{n+1+\ldots+3n+1} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{8n^2}{2n(n+1+3n+1)} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{2n}{2n+1}
\end{align}
| We can use C-S by another way.
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}=\frac{1}{2n+1}+\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{3n+2-k}\right)\geq$$
$$\geq\frac{1}{2n+1}+\sum_{k=1}^n\frac{(1+1)^2}{n+k+3n+2-k}=\frac{1}{2n+1}+\frac{2n}{2n+1}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\int\limits_0^{\pi/2} {1 \over 1+8\sin^2(\tan x)}\ dx$ I’m trying to evaluate$$\int\limits_0^{\pi/2}dx\,\frac 1{1+8\sin^2(\tan x)}$$And made the substitution $u=\tan x$. Therefore$$I=\int\limits_0^{\infty}dx\,\frac 1{(1+x^2)(1+8\sin^2x)}$$Through some manipulations, you arrive at$$I=\int\limits_0^{\infty}dx\,\frac 1{(1+x^2)(5-4\cos 2x)}$$However, I’m confused what to do after that. The answer key started with$$\int\limits_0^{\infty} dx\,\frac 1{(1+x^2)(5-4\cos 2x)}=\int\limits_0^{\infty}dx\,\frac 1{1+x^2}\left(\frac 13+\frac 23\sum\limits_{k\geq1}\frac {\cos 2kx}{2^k}\right)$$However, I’m confused where they got the infinite sum from. Any ideas?
| It is just a geometric series in disguise. You may consider that
$$ \sum_{k\geq 0}\frac{\cos(2kx)}{2^k}=\text{Re}\sum_{k\geq 0}\left(\frac{e^{2ix}}{2}\right)^k = \text{Re}\left(\frac{2}{2-e^{2ix}}\right) = 2\,\text{Re}\left(\frac{2-e^{-2ix}}{5-4\cos(2x)}\right)$$
equals $\frac{4}{5-4\cos(2x)}$. In particular
$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{dx}{(1+x^2)(5-4\cos(2x))}&=&\frac{\pi}{6}+\frac{2}{3}\sum_{k\geq 1}\frac{1}{2^k}\int_{0}^{+\infty}\frac{\cos(2kx)}{1+x^2}\,dx\\&=&\frac{\pi}{6}+\frac{2}{3}\sum_{k\geq 1}\frac{\pi}{2^{k+1}e^{2k}}=\color{red}{\frac{\pi}{6}\cdot\frac{2e^2+1}{2e^2-1}}.\end{eqnarray*}$$
| {
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Integral and trig substitution I have a problem with this integral:
$$\int \frac{dx}{\sqrt{4x+x^2}}$$
I tried to transform it like this:
$$\int \frac{dx}{\sqrt{4x+x^2}}=\int \frac{dx}{\sqrt{x^2+4x+4-4}}=\int \frac{dx}{\sqrt{(x+2)^2-4}}$$
$u=x+2$ , $du=dx$ then
$$\int \frac{dx}{\sqrt{(x+2)^2-4}}=\int \frac{du}{\sqrt{u^2-4}}$$
Unfortunately I have no idea what I should do now. I think that I can use trig substitution, but I do not know how I can do it.
| As suggested, we use $u=2\cosh z$, $du=2\sinh z\ dz$. Hence
$$\int \frac{du}{\sqrt{u^2-4}}=2\int \frac{\sinh z}{\sqrt{4\cosh^2 z-4}}dz=\int \frac{\sinh z}{\sqrt{\cosh^2 z-1}}$$
Now, recall the relation $\cosh^2 z-\sinh^2 z=1$; we have that $\sinh^2 z=\cosh^2 z-1$, so
$$\int \frac{\sinh z}{\sqrt{\cosh^2 z-1}}=\int \frac{\sinh z}{\sinh z}dz=\int dz=z+c$$
Let's come back to the original variable: $z=\cosh^{-1} \frac{u}{2}$, and $u=x+2$. So
$$\int \frac{dx}{\sqrt{4x+x^2}}=\cosh^{-1}\left(\frac{x+2}{2}\right)+c$$
Remembering that $\cosh^{-1} y=\ln\left(y+\sqrt{y^2-1}\right)$
$$\int \frac{dx}{\sqrt{4x+x^2}}=\ln\left(\frac{x+2}{2}+\sqrt{\frac{(x+2)^2}{4}-1}\right)+c=\ln\left(\frac{1}{2}\left(x+2+\sqrt{x^2+4x}\right)\right)+c=$$
$$=\ln \left(\frac{1}{2}\right)+\ln\left(x+2+\sqrt{x^2+4x}\right)+c$$
For all $c\in\mathbb{R}$.
Now $\ln \left(\frac{1}{2}\right)$ is a costant, so we can incorporate it in the $c$. So finally
$$\int \frac{dx}{\sqrt{4x+x^2}}=\ln\left(x+2+\sqrt{x^2+4x}\right)+c$$
If you don't know hyperbolic functions, notice that they're defined by
$$\cosh z:=\frac{e^z+e^{-z}}{2}$$
$$\sinh z:=\frac{e^z-e^{-z}}{2}$$
| {
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Find the area between a cardioid and a straight line with polar equations I have to find the area between a line and a cardioid, given by $\rho_1 = 8 + 8 \sin(\theta)$ and $\rho_2 = 4/\hspace{-0.5mm}\sin(\theta)$.
First I found when both are positive, which is always for the first one, and only from $0$ to $2\pi$ for the second one. Then I graph them both and now I have to find the intersections, but I get $8+\sin(\theta) = 4/\hspace{-0.5mm}\sin(θ)$, which is $\sin^2(\theta) + 1 = 1/2$, and I have no idea how to solve that.
Then, for the area, I know it's the top part of the graphic, so let's say they intersect in $\theta=a$ and $\theta=b$, so I have $$\int_{a}^{b} (8+8\sin\theta)^2 - (4/\hspace{-0.5mm}\sin\theta)^2 d\theta$$ for the cardioid is over the line. Is that correct? I'm not supposed to solve it, just to propose it.
And that's all, I hope you can help me. Thank you!
| The general formula for the area between two polar curves is $$\frac{1}{2}\int_{\theta_1}^{\theta_2}\left(r_1^2-r_2^2\right)\mathrm{d}\theta,$$ where $r_1$ is the "outer" curve (the cardioid in this case) and $r_2$ is the "inner" curve (the line in this case). You were mostly on the right track, aside from the $1/2$ out front.
Finding the points of intersection of the two polar curves can often be, as you found, the trickiest part of setting up the integral. In this case, since $r_1=8+8\sin\theta$ and $r_2=4/\hspace{-0.5mm}\sin\theta$, we have
\begin{align}
8+8\sin\theta&=4/\hspace{-0.5mm}\sin\theta \\
\hspace{2.5cm}8\sin\theta+8\sin^2\theta&=4 \\
2\sin^2\theta+2\sin\theta-1&=0 \\
2x^2+2x-1&=0 &&\text{Let $x=\sin\theta$} \\
\end{align}
\begin{align}
x&=\frac{-2\pm\sqrt{2^2-4(2)(-1)}}{2(2)} \\
x&=\frac{-2\pm\sqrt{12}}{4} \\
x&=-\frac{1}{2}\pm\frac{\sqrt{3}}{2} \\
\sin\theta&=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}
\end{align}
We can see from this result that $\theta_1=\arcsin\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\right)$. Unfortunately, since the domain of the $\arcsin$ function is only $[-1,1]$, $-\frac{1}{2}-\frac{\sqrt{3}}{2}$ is outside the domain of the $\arcsin$ function. We can correct for this by taking $\theta_2=\pi-\theta_1$, since it's clear from graphing the two functions that that's the angle we want. Therefore, the final form of the integral is
$$\frac{1}{2}\int_{\arcsin\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}^{\pi-\arcsin\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}\left[\Big(8+8\sin\theta\Big)^2-\left(\frac{4}{\sin\theta}\right)^2\right]\mathrm{d}\theta$$
Numerical evaluation shows that this integral is approximately equal to $204.16$.
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Evaluating $\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$ The question is to evaluate $$\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$$
I tried using leibnitz rule
$$F'(a)=\int_0^{\pi /2} \frac{ -\sin a}{(1+\cos a \cos x)}dx$$ Now I used the substitution $\tan(x/2)=t$ to get $$-2 \sin a \int_0^1 \frac{ dt}{1+t^2 +\cos a (1-t^2)} $$ which can be rewritten as
$$-2\frac{\sin a} {1- \cos a}\int_0^{1} \frac{ dt}{t^2 +\frac{1+ \cos a}{1-\cos a}} $$ which evaluates to $-a$.i am not sure where I went wrong.Any ideas?
| Using Feynman’s Technique Integration by differentiating w.r.t. $a$,
$$I(a)=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos a \cos x)}{\cos x} d x,$$
we obtain $$
I^{\prime}(a)=\int_{0}^{\frac{\pi}{2}} \frac{-\sin a}{1+\cos a \cos x} d x
$$
Letting $t=\tan \dfrac{x}{2}$ yields $$
\begin{aligned}
I^{\prime}(a) &=-2 \sin a \int_{0}^{1} \frac{d t}{(1+\cos a)+(1-\cos a) t^{2}} \\
&=-2 \sin a \int_{0}^{1} \frac{d t}{2 \cos ^{2} \frac{a}{2}+2 t^{2} \sin ^{2} \frac{a}{2}} \\
&=-2 \cos \frac{a}{2} \int_{0}^{1} \frac{d\left(t \sin \frac{a}{2}\right)}{\left(t \sin \frac{a}{2}\right)^{2}+\cos ^{2} \frac{a}{2}} \\
&=-2\left[\tan ^{-1}\left(t \cdot \tan \frac{a}{2}\right)\right]_{0}^{1} \\
&=-2 \tan ^{-1}\left(\tan \frac{a}{2}\right) \\
&=-a
\end{aligned}
$$
Now we can get $I(a)$ by integrating $I^{\prime}(y)$ w.r.t $y$ from $\dfrac{\pi}{2} $ to $a$.
$$
I(a)=\int_{\frac{\pi}{2}}^{a} I^{\prime}(y) d y=\int_{\frac{\pi}{2}}^{a}-yd y=\boxed{\frac{\pi^{2}}{8}-\frac{a^{2}}{2}}
$$
For example, $$
\int_{0}^{\frac{\pi}{2}} \frac{\ln \left(1+\frac{1}{2} \cos x\right)}{\cos x} d x=\frac{5 \pi^{2}}{72}
$$
:|D Wish you enjoy my solution!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2545972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Modular arithmetic - How to find failure case theoretically I am unable to find the wrong intersection point of two modular classes theoretically, for the below question:
If $n$ is an odd integer, and $n^2 + 2n$ and $n$ have the same residue $\pmod 8$. What can be the remainder on division by $8$?
Residues of $n \mod 8$, for odd $n$, are: $\{1,3,5,7\}$. And of $2n \mod 8$ the residues are: $(2,6,2,6)$, and of $n^2 \mod 8$ are found by simply finding $r^2 \mod 8$:$(1,1,1,1)$, or $\{1\}$. $n^2 + 2n$ have the possible residues found by adding up the corresponding values of residues for $2n \pmod 8$ and $ n^2 \pmod 8$: $(3,7,3,7)$, or the set: $\{3,7\}$. So, $n$ values can be $8k+3, 8k+7, \forall k \in \mathbb {Z}$. So, the remainders $\mod 8$ can be : $3, 7$.
But, there is a flaw in this approach: you cannot figure out that $8k+3$ case will not work before-hand. One need see this case manually, with $n \pmod 8$ yielding residue $3$ and $n^2 + 2n \pmod 8$ yielding residue (using residue rather than actual value of $8k+3$):
$3^2 + 6 => 15 \pmod 8 => 7$.
This is also substantiated by a value of $n=19$ that falls in the class $8k+ 3$, with the value of $n^2 + 2n = 399 \equiv 7 \pmod 8$. However, $19 \equiv 3\pmod 8$.
Addendum Based on the answers given, have modified the answer to :
Need to equate the solutions given (theoretically, or algebraically) for the two modular classes:
$n^2 + 2n \equiv n\pmod 8 => n + 2 \equiv 1\pmod 8 => n \equiv -1 \pmod 8$.
Addendum -2 My approach for modification is to theoretically factor in the residue(remainder) from the start, as follows for the OP case, with $k$ representing not $n$ but the residue.
$n^2 \mod 8$ is given for odd $n$ as: $(2k+1)^2 \mod 8 => (4k^2 + 4k + 1) \mod 8 $. On adding $2n \mod 8$ to it, get: $((4k^2 + 4k + 1) + (4k+2)) \mod 8 => (4k^2 + 8k + 3) \mod 8$. This should be equal to $n \pmod 8$.
$4k^2 + 8k + 3 \equiv 2k+1 \pmod 8 => 2k^2 + 3k + 1 \equiv 0 \pmod 8 => 2k(k+1) + 1.(k+1) \equiv 0 \pmod 8 => (2k+1)(k+1) \equiv 0 \pmod 8$.
Only answer for integer value of $k$ (residue) possible is: $-1$ in modulo $8$ system
| I think that you're overthinking the question.
As you said in your question, $n^2 \equiv 1 \pmod 8$. This means that $2n+1 = n \pmod 8$. Solving the equation $2n+1=n$ gives $n=-1$. However, since this is modulo $8$, the general form is $8n-1$.
We can check this works by substitution. $(8n-1)^2 + 2(8n-1) = (64n^2-16n+1) + (16n-2)$ $= 8(8n^2) - 1 \equiv -1 \pmod 8$, so the expression $n^2$+$2n$ has the same residue of $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Solve over integers: $x(x+1) = 2y(y+1)$ Solving $x(x+1) = 2y(y+1)$
I've got this :
y = $\frac{1}{2} \sqrt{2x^2+2x+1}$ so $2x^2+2x+1 = n^2$
so $x \equiv 0 \pmod 4$ or $x \equiv -1 \pmod 4$ and I totally stuck...
| Solving it as quadratic equation in terms of $x$ we have:
$$x^2 + x -2y^2 - 2y = 0$$
$$x= \frac{-1 \pm \sqrt{1 + 8y^2 + 8y}}{2}$$
Now we have that $8y^2 + 8y + 1 = m^2$, as $x$ is an integer. Again solving for $y$ we have:
$$y = \frac{-8 \pm \sqrt{64 + 32m^2 - 32}}{16}$$
So we must have that $2(m^2 + 1)$ is a square. So we get $n^2 - 2m^2 = 2$. This Diophantene Equation has infinitely many solutions. For example:
$$n_k + m_k\sqrt{2} = (10 + 7\sqrt{2})(3 + 2\sqrt{2})^k$$
generates infinitely many. Now note that the solutions for $n$ are even, but never divisible by $4$, so we have $m^2 + 1 = 2s^2$, where $s$ is odd. So:
$$y = \frac{-8 \pm 8s}{16} \in \mathbb{Z}$$
$$x = \frac{-1 \pm \sqrt{2s^2 - 1}}{2} \in \mathbb{Z}$$
So we have that each solution of $n^2 - 2m^2 = 2$ generates a solution of $x(x+1)=2y(y+1)$, so there are infinitely many of them. Some examples are: $(x,y) = (3,2),(20,14),(119,84),(696,492) \dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $a$ if $(a+3)x^2-4x+2<0$ $\forall x\in [-2,1]$ Find $a$ if $(a+3)x^2-4x+2<0 \qquad \forall x\in [-2,1]$
My Attempt:
$a+3=0$ $\implies -4x+2<0$
$$2x-1>0$$
$$x>\dfrac {1}{2}$$
$\implies a \neq -3$.
| Let $f(x)=(a+3)x^2-4x+2$
1) Let $a+3>0 \Leftrightarrow a>-3$.
Then $f(-2)<0$ and $f(1)<0$
$f(-2)<0 \Leftrightarrow 4(a+3)+10<0$, then $a<-\frac{11}{2}$
$f(-1)<0 \Leftrightarrow a+3-2<0$, then $a<-1$
2) Let $a+3>0$ and let $x_0=\frac{4}{2(a+3)}$
Then $f(-2)<0$ and $f(1)<0$ and $x_0\not\in[-2;1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Question about difference of powers related to FLT I am wondering about differences of powers and some idea about it. It is related to FLT in that if the idea is correct something good could be done with it. I will give an example for prime $3$ but it could easily work for any odd prime. My question is: is the idea correct?
Let the following equation hold:
$$x^3 - y^3 = 3y^2(x - y) + 3y(x - y)^2 + (x - y)^3 = z^3$$
for some integers $x,y,z$.
We can obtain this by means of the binomial expansion of $((x - y) + y)^3 = x^3$ and subtracting $y^3$ to get $z^3$.
Rewrite the equation as follows(by recursively factoring out $x - y$):
$$(x - y)(3y^2 + (x - y)(3y + (x - y))) = z^3$$
We will prove: $p \mid x - y \implies p^3 \mid x - y$ for any prime other than $3$:
Assume $\gcd(x,y,z) = 1$,
Let $p \ne 3$ be some prime for which $p \mid x - y$,
$\quad\quad p \mid x - y \implies p \mid z \implies p^3 \mid z^3$
So we get:
$\quad\quad (x - y)(3y^2 + (x - y)(3y + (x - y))) \equiv z^3 \equiv 0 \pmod{p^3}$
Suppose now $p^3 \nmid x - y$, then:
$\quad\quad 3y^2 + (x - y)(3y + (x - y)) \equiv 0 \pmod{p}$
And so:
$\quad\quad (x - y)(3y + (x - y)) \equiv -3y^2 \pmod{p}$
$\quad\quad \implies p \mid 3y^2$ which is impossible because of $p \ne 3$ and $\gcd(x,y,z) = 1$.
We may conclude that for every prime $p \ne 3$, $p \mid x - y$ we have $p^3 \mid x - y$
The same idea should work for prime-divisors of $x - z$.
| Assuming we have $z^3=x^3-y^3=(x-y)(x^2+xy+y^2)$ and then
\begin{align}
\gcd(x-y,x^2+xy+y^2) &= \gcd(x-y,x^2+xy+y^2)\\
&= \gcd(x-y,2xy+y^2)\\
&= \gcd(x-y,3xy).
\end{align}
But because you assume also $\gcd(x,y,z)=1$, it follows $\gcd(x-y,x^2+xy+y^2)=\gcd(x-y,3)$.
In other words, $x-y$ and $x^2+xy+y^2$ have no common factor apart from $3$, and so of course, if you find prime $p \mid x-y$, $p\neq 3$ then indeed $p^3 | x-y$.
But that is not new, same applies for generic $n$. If you have
$$z^n=x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})$$ then you can show similarly as before (assuming $\gcd(x,y,z)=1$ again) that
$$
\gcd(x-y,x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})=\gcd(x-y,n)
$$
and so if there is a prime $p$ such that $p\nmid n$ and $p|x-y$, then it follows $p^n\mid x-y$.
So the idea is correct, question is whether it is of any use, which I do not know.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Centroid of a Trapezoid using double integrals
So, I need to find the center of mass from that trapezoid using double integration knowing the mass function: $\delta(x, y) = 1 + 2x + y$
What I've found so far is the boundaries are $y = -x + 3$, $y = 1$, $x$-axis, $y$-axis
But then, I confused about what is the next step, since I need to use the mass function and use the double integration to find the answer.
How do I find the centroid using double integration and the given mass function?
| There is one mistake in the other answer, because the total mass should be computed using the function $\delta$.
Let $M$ be the total mass of the given shape. Then,
$$
M = \int_0^1 \left( \int_0^{3-y} \delta(x,y)\, dx\right)dy.
$$
First, we shall compute the inner integral:
$$
\begin{align*}
\int_0^{3-y} \delta(x,y)\, dx &= \int_0^{3-y} (1+2x+y)\, dx \\
&= \left. \frac{(1+2x+y)^2}{4} \right|_0^{3-y}\\
&= \frac{(7-y)^2}{4} - \frac{(1+y)^2}{4} \\
&= \frac{48 - 16y}{4} \\
&= 12 - 4y.
\end{align*}
$$
So,
$$
\begin{align*}
M &= \int_0^1 (12-4y)\, dy \\
&= \left. \frac{(12-4y)^2}{-8} \right|_0^1 \\
&= -8 - (-18) \\
&= 10.
\end{align*}
$$
Hence, the total mass is $M = 10$.
I have written out the rest of the computation as well, so that you can check your calculations.
Next, we want to find the coordinates of the centroid, $(\bar{x},\bar{y})$. To find $\bar{x}$, consider an infinitesimal mass element at the point $(x,y)$. It has mass $\delta(x,y)\, dx\, dy$. We weight this infinitesimal mass by the position coordinate $x$, and then integrate over the whole shape. This value divided by the total mass will give $\bar{x}$. Similarly, for $\bar{y}$, we weight the infinitesimal mass element $\delta(x,y)\, dx\, dy$ by $y$ and perform the integration as before. Mathematically,
$$
\begin{align*}
M \bar{x} &= \int_0^1 \left( \int_0^{3-y} x (1+2x+y)\, dx \right) dy \\
M \bar{y} &= \int_0^1 \left( \int_0^{3-y} y (1+2x+y)\, dx \right) dy.
\end{align*}
$$
For $\bar{x}$, the inner integral evaluates as:
$$
\begin{align*}
\int_0^{3-y} x (1+2x+y)\, dx &= \left. \frac{(1+y)x^2}{2} + \frac{2x^3}{3} \right|_0^{3-y} \\
&= \frac{(1+y)(3-y)^2}{2} + \frac{2(3-y)^3}{3} \\
&= \frac{(3-y)^2(15-y)}{6}.
\end{align*}
$$
For $\bar{y}$ the inner integral was already computed earlier, and it evaluates to:
$$
\int_0^{3-y} y (1+2x+y)\, dx = y(12-4y).
$$
So,
$$
\begin{align*}
M\bar{x} &= \int_0^1 \frac{(3-y)^2(15-y)}{6} \, dy \\
&= \int_0^1 \frac{135 - 99y + 21y^2 - y^3}{6}\, dy \\
&= \left. \frac{45y}{2} - \frac{33y^2}{4} + \frac{7y^3}{6} - \frac{y^4}{24} \right|_0^1 \\
&= \frac{45}{2} - \frac{33}{4} + \frac{7}{6} - \frac{1}{24} \\
&= \frac{123}{8}
\end{align*}
$$
and
$$
\begin{align*}
M\bar{y} &= \int_0^1 y(12-4y) \, dy \\
&= \int_0^1 (12y - 4y^2)\, dy \\
&= \left. 6y^2 - \frac{4y^3}{3} \right|_0^1 \\
&= 6 - \frac{4}{3} \\
&= \frac{14}{3}.
\end{align*}
$$
Therefore,
$$
\begin{align*}
\bar{x} &= \frac{123}{80} \\
\bar{y} &= \frac{7}{15}.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$ (without calculus) I'm trying to prove the following proposition (I'm not supposed to use calculus):
$$\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$$
(I'm assuming that $0 \notin \Bbb{N}$)
This is what I've tried so far:
$\sum_{i=1}^{n} \frac{n+i}{i+1} = \sum_{i=1}^{n}[ \frac{n}{i+1} + \frac{i}{i+1} ] = [\sum_{i=1}^{n} \frac{n}{i+1}] + \sum_{i=1}^{n} \frac{i}{i+1} = n[\sum_{i=1}^{n} \frac{1}{i+1}] + \sum_{i=1}^{n} \frac{i}{i+1}$
Let $j= i +1$ . So now we have:
$\sum_{i=1}^{n} \frac{n+i}{i+1} = n[\sum_{j=2}^{n+1} \frac{1}{j}] + \sum_{j=2}^{n+1} \frac{j-1}{j}$
$= n[\sum_{j=2}^{n+1} \frac{1}{j}] + \sum_{j=2}^{n+1} \frac{j}{j} - \sum_{j=2}^{n+1} \frac{1}{j}$
$= n[\sum_{j=2}^{n+1} \frac{1}{j}]- \sum_{j=2}^{n+1} \frac{1}{j} + \sum_{j=2}^{n+1} 1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + [\sum_{j=1}^{n+1} 1 ] -1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n+1 -1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n$
Therefore proving the following inequality is the same as proving the original:
$$ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n(n-1)$$
If $n=1$ then $ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n = 1$ and $1 + n(n-1) = 1$
Now I assume that $n>1$:
$ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n(n-1) \iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n^2 - n$
$\iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] \le 1 + n^2 - 2n$
$\iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] \le (n-1)^2$
$\iff \sum_{j=2}^{n+1} \frac{1}{j} \le \frac{(n-1)^2}{n-1}$
$\iff \sum_{j=2}^{n+1} \frac{1}{j} \le n-1$
$\iff 1 + \sum_{j=2}^{n+1} \frac{1}{j} \le n$
$\iff \sum_{j=1}^{n+1} \frac{1}{j} \le n$
And here is where I'm stuck. I know that an upper bound for the harmonic sum can be found using an integral test but I'm not supposed to use calculus. Is there a discrete way of proving that $\sum_{j=1}^{n+1} \frac{1}{j} \le n$ for all natural $n \ge 2$ ? Or maybe another way of proving the proposition without ending up with the harmonic sum?
| It's slightly easier to show $\sum_{j=2}^{n+1} \frac{1}{j} \le n-1$, so I'll do that instead (as you note, they're equivalent). In particular, note that every term on the left hand side is no larger than $\frac{1}{2}$:
$$\sum_{j=2}^{n+1} \frac{1}{j} \le \sum_{j=2}^{n+1} \frac{1}{2} \\
\leq \frac{1}{2}n.$$
So, all you need to do is show that $\frac{1}{2} n \leq (n-1)$ for $n \geq 2$, which follows immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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To show that $x^2+y^2\propto xy$. Let $x$ and $y$ are variables satisfying $3x-4y\propto\sqrt{xy}$. Then show that $x^2+y^2\propto xy$, where $y∝x$ implies $y$ directly proportional to $x$.
My attempt... $3x-4y=k\sqrt{xy}$ for some real $k$. This implies $(3x-4y)^2=k^2xy$ implies $9x^2+16y^2=(k^2+24)xy$ implies $x^2+y^2=((k^2+24)x-7y)y/9$. So i have to assert now that $[(k^2+24)x-7y]\propto x$. But I cant proceed from here. Please help me to solve this.
| Let $t = \sqrt{\frac{x}{y}}$. Observe that $3t - \frac{4}{t} = k$, or $3t^2 -kt - 4 = 0$, giving $t = \frac{k \pm \sqrt{k^2 + 12}}{6}$, that is $t$ is constant.
On the other hand, $\frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x} = t^2 + t^{-2}$ is also constant, and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the function $\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}$ Continuous at $(0,0)$ The function is defined piece-wise with $f(0,0) = 0$, so I took the limit $(x,y) \rightarrow (0,0)$
I proceeded like this:
$$\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2} \leq \frac{|y|(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2} \leq \frac{|y|(x^4+4x^2y^2+y^4)}{(x^2+y^2)^2}
$$
$$= \frac{|y|(x^2+y^2)^2+|y|2x^2y^2}{(x^2+y^2)^2} = |y| + \frac{|y|2(xy)^2}{(x^2+y^2)^2} \leq |y| + \frac{|y|2(\frac{x^2+y^2}{2})^2}{(x^2+y^2)^2} = |y| + \frac{|y|}{2} \rightarrow0$$
Hence by the sandwich/squeeze theorem, the original function also goes to $0$
Is this answer ok?
| Your proof is correct but here is shorter way.
Since $|x|,|y|\le \|(x,y)\|$
we get $$\left|\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\right|=\left|\frac{y(x^4+4x^2y^2-y^4)}{\|(x,y)\|^4}\right|\le 6\frac{\|(x,y)\|^5}{\|(x,y)\|^4} =\|(x,y)\|\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability of 6 die rolls
If two fair $6$-sided dice are tossed six times, find the probability that
the sixth sum obtained is not a repetition.
The solution given to me is not very helpful in explaining the steps, and my attempt at it is far away from the final answer. This question appears to be different from the standard die questions I am used to. Intuition and explanation would be appreciated.
The Solution: $2*\frac{1}{36}(\frac{35}{36})^5+2*\frac{1}{18}(\frac{7}{18})^5+2*\frac{1}{12}(\frac{11}{12})^5+2*\frac{1}{9}(\frac{8}{9})^5+2*\frac{5}{36}(\frac{31}{36})^5+\frac{1}{6}(\frac{5}{6})^5\approx0.5614$
| If you toss two dice, the resulting sum can be anywhere between 2 and 12.
$$\begin{array}{c|cccccc}
&1&2&3&4&5&6\\
\hline
1&2&3&4&5&6&7\\
2&3&4&5&6&7&8\\
3&4&5&6&7&8&9\\
4&5&6&7&8&9&10\\
5&6&7&8&9&10&11\\
6&7&8&9&10&11&12\\
\end{array}$$
For me, one helpful mental trick here is considering the probability that the first roll of the dice is not a repetition. This must be the same as the probability that the last roll of the dice is not a repetition, because order does not matter.
So what we do is we consider each possible sum the first dice might have. For example, the sum might be 6. Out of the 36 outcomes in this table, five outcomes result in 6, and thirty-one outcomes result in something other than 6. Hence the probability of getting a 6 on the first roll of dice, then never getting a 6 again is:
$$\frac{5}{36} \cdot (\frac{31}{36})^5$$
We can compute the probability that the first roll of the dice results in $k$, then the next five rolls all result in something other than $k$. If we do this for all sums in the range $k=2\ldots 12$ and add up the probabilities, we will have our result.
The answer is:
$$P = \underbrace{\frac{1}{36} \left(\frac{36-1}{36}\right)^5}_{k=2} + \underbrace{\frac{2}{36} \left(\frac{36-2}{36}\right)^5}_{k=3} + \ldots +
\underbrace{\frac{5}{36} \left(\frac{36-5}{36}\right)^5}_{k=6} +\underbrace{\frac{6}{36} \left(\frac{36-6}{36}\right)^5}_{k=7} + \underbrace{\frac{5}{36} \left(\frac{36-5}{36}\right)^5}_{k=8} + \ldots +\underbrace{\frac{2}{36} \left(\frac{36-2}{36}\right)^5}_{k=11}+ \underbrace{\frac{1}{36} \left(\frac{36-1}{36}\right)^5}_{k=12}$$
In each case, we have a term that looks like $\frac{r}{36}\left(\frac{36-r}{36}\right)^5$, where $r$ is the number of times the given sum appears in the table. It's the probability of getting that sum on the first roll, then getting some other sum in each of the five subsequent rolls.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Value of $a$ such that the equation has integral roots If $(a+b+c)x^2+2(a-b)x +a+b-8=0$ has integral roots and $a,b,c \in R^+ ,c>b-3a>0$,then the value of $a$ is ----
I am aware that a condition for integral roots is that the discriminant must be a perfect square but I couldn't deduce anything from it.Any ideas?Thanks.
| Rewrite as
$$\tag1 x^2+px+q = 0$$
with $$ p=\frac{2(a-b)}{a+b+c},\qquad q=\frac{a+b-8}{a+b+c}.$$
As $p$ is the negative sum and $q$ is the product of the roots (Vieta), both must be integers.
From the given inequality,
$$-p=\frac{(a+b)+(b-3a)}{a+b+c}\begin{cases}<\frac{a+b+c}{a+b+c}=1\\
>\frac{a+b}{a+b+c}>0 \end{cases}.$$
But there is no such integer.
| {
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How to find area of a circle? I have the equation in polar coordinates:
$$r=\cos(\phi)-\sin(\phi)$$
I just moved to
Cartesian coordinates and drew a graph:
$$\left(x-\frac{1}{2}\right)^2+\left(y-\left(-\frac{1}{2}\right)\right)^2=\left(\sqrt{\frac{1}{2}}\right)^2$$
And now I should find the area of this circle using integral. What I need to do?Please help
| If you are going to do the algebra to change the coordinate system, you might as well finish the job algebraically.
$(x-h)^2 + (y-k)^2 = r^2$
and $A = \pi r^2 \implies A = \frac {\pi}{2}$
other options.
$r = \cos \phi - \sin \phi\\
r = \sqrt 2 (\cos (\phi + \frac {\pi}{4}))$
let $\theta = \phi + \frac {\pi}{4}$
$r = \sqrt 2 \cos \theta$
Which you might recognize as a circle of radius $\frac {\sqrt 2}{2}$
Caculus in polar coordinates
$A = \int_a^{b} \frac 12 r^2 \ d\phi$
The tricky part, $r\ge 0$ so you must find limits for $\phi$ such that this is true.
$\cos\phi \ge \sin\phi\\
\phi \in [-\frac {3\pi}{4},\frac {\pi}{4}]$
$\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \frac 12 (cos\phi - \sin\phi)^2\ d\phi\\
\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \frac 12 (cos^2\phi + \sin^2\phi - 2cos\phi\sin\phi)\ d\phi\\
\frac 12 (\phi + \frac 12 \sin^2 \phi)|_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}\\
\frac {\pi}{2}$
And finally integrate in Cartesian.
$(x-\frac {1}{2})^2 + (y+\frac {1}{2})^2 = \frac 12\\
y = \pm \sqrt {\frac 12 - (x-\frac {1}{2})^2} - \frac {\sqrt 2}{2}\\
\int_{\frac 12 - \frac {\sqrt 2}{2}}^{\frac 12 + \frac {\sqrt 2}{2}} (\sqrt {\frac 12 - (x-\frac {1}{2})^2} - \frac {\sqrt 2}{2}) - (-\sqrt {\frac 12 - (x-\frac {1}{2})^2} - \frac {\sqrt 2}{2})\ dx\\
\int_{\frac 12 - \frac {\sqrt 2}{2}}^{\frac 12 + \frac {\sqrt 2}{2}} 2\sqrt {\frac 12 - (x-\frac {1}{2})^2}\ dx\\
x - \frac 12 = \frac {\sqrt 2}{2} \sin t\\dx = \frac {\sqrt 2}{2}\cos t\ dt\\
\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \cos^2 t\ dt\\
\frac 12 (t + \sin t\cos t)|_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\\
\frac {\pi}{2}$
| {
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Integrate $\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \,dx$
Integrate $$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}dx$$
I solved the question by using the identity $\cos^4(x)+\sin^4(x) = \frac{1}{4}(\cos4x+3)$ and the substitution $u=\cos4x +3$, which turned it into a relatively familiar integral (see my answer below).
However, I'm pretty sure there are easier ways I am missing, so please feel free to post alternative answers.
There is a similar question here.
Problem Source: James Stewart Calculus, 6E
| $u = \cos 2t \implies du = -2\sin 2t dt$
$$\begin{align}\dfrac{-1}2\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}\dfrac{du}{\sin 2t} =& \dfrac{-1}4\int \frac{1}{\sin^4x + \cos^4x}du &\\=& \dfrac{-1}2\int \frac{1}{(\cos^2x + \sin^2x)^2 +(\cos^2x - \sin^2x)^2 } &\\=& \dfrac{-1}2\int \frac{1}{1+u^2 }du = \dfrac{-1}{2}\tan^{-1}(\cos 2x) + C\end{align}$$
| {
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How do you find the values that $x$ can take by squaring? $$|x+4| \cdot |x-4| = x+4$$
How do you find the values that $x$ can take by squaring?
How I've tried
$$|x+4|^2 \cdot |x-4|^2 = (x+4)^2$$
and
$$|x^2+16| \cdot |x^2+16| = x^2+16$$
I think I went wrong.
| Given
$$
|x+4||x-4|=x+4,
$$
and we want to find $x$, we observe that
$$
|x+4||x-4|=|(x+4)(x-4)|=|x^2+4x-4x-16|=|x^2-16|.
$$
There are two cases:
*
*$x^2-16\geq 0$. This happens when $x\geq 4$ or $x\leq -4$. When $x^2-16\geq 0$, $|x^2-16|=x^2-16$ since both are nonnegative. Therefore, you want to solve $x^2-16=x+4$. Now, use the quadratic formula, you can get the solutions (but you'll need to check them against the inequalities above).
*$x^2-16<0$. This happens when $-4<x<4$. When $x^2-16<0$, then $|x^2-16|=-(x^2-16)=16-x^2$. This is true since $x^2-16<0$ but the absolute value makes it positive by multiplying by $-1$. Again, you need to use the quadratic formula to solve $16-x^2=x+4$ (and again, discard any values that you get that violate $-4<x<4$).
By other's observations on the positivity, you only need to consider the case where $x\geq -4$, but that only slightly simplifies this approach.
| {
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Deriving least squares coefficients for curve of form $y=a/(x^2+b)$ I am trying to derive the coefficients $a$ and $b$ for a least squares curve of the form $y=\frac{a}{x^2+b}$
To start I have written $$S=\sum_i(y_i-y)^2=\sum_i\left(y_i-\frac{a}{x_i^2+b}\right)^2=\sum_i\left(y_i^2-\frac{2a y_i}{x_i^2+b}+\frac{a^2}{(x_i^2+b)^2}\right) \\S=\sum_iy_i^2-2a\sum_i\frac{y_i}{x_i^2+b}+a^2\sum_i\frac{1}{(x_i^2+b)^2}$$
Now the way I have seen it done for the line $y=a+bx$ is they minimize $S$ with respect to both $a$ and $b$, applying that for this curve gives the two constraint equations:
$$0=a\sum_i\frac{1}{(x_i^2+b)^2}-\sum_i\frac{y_i}{x_i^2+b}\\
0=-a\sum_i\frac{1}{(x_i^2+b)^3}+\sum_i\frac{y_i}{x_i^2+b}$$
Summing these together we find $$0=a\left(\sum_i\frac{1}{(x_i^2+b)^2}-\sum_i\frac{1}{(x_i^2+b)^3}\right)$$
Since $a\neq 0$ we must have $$\sum_i\frac{1}{(x_i^2+b)^2}=\sum_i\frac{1}{(x_i^2+b)^3}$$ Which seems to imply $$(x_i^2+b)^2=(x_i^2+b)^3$$ and unfortunately there doesnt appear to be a constant $b$ that solves this equation. Did I make a mistake somewhere or is this the result?
| Starting with a rather silly sentence : since the model is nonlinear with respect to $b$, by the end, you will need a nonlinear regression and, as usual, you will need reasonable starting guesses for the parameters.
You can get these estimates in a preliminary step
$$y=\frac{a}{x^2+b} \implies \frac 1y=\frac {x^2} a+\frac ba$$ So, defining for each $(x_i,y_i)$ data points, new variables $t_i=x_i^2$ and $z_i=\frac 1{y_i}$, you have $$z=\alpha\, t+\beta$$ and, from the corresponding linear regression, you will get, as estimates $a=\frac 1 \alpha$ and $b=\frac \beta \alpha$.
If you do not want to use nonlinear regression, you can reduce the problem to a nasty equation in $b$.
Using Rohan's expressions, let us define
$$S_1=\sum \frac{y_i}{x_i^2+b} \qquad S_2=\sum \frac{1}{(x_i^2+b)^2}$$
$$S_3=\sum \frac{y_i}{(x_i^2+b)^2} \qquad S_4= \sum \frac{1}{(x_i^2+b)^3}$$ Eliminating $a$, you then need to to find the zero of function $$f(b)=S_2\, S_3-S_1 S_4$$ Since you have the estimate, you could use it as the starting point of Newton method with analytical derivatives since $$f'(b)=S_2S'_3+S'_2S_3-S_1S'_4-S'_1S_4$$ where will appear other summations (I let you the task of writing them - they are not difficult). The iterative process will converge very fast. So $b$ and then $a$.
Edit
For illustration purposes, consider the following data set
$$\left(
\begin{array}{cc}
x & y \\
0 & 46 \\
1 & 46 \\
2 & 45 \\
3 & 43 \\
4 & 41 \\
5 & 38 \\
6 & 36 \\
7 & 33 \\
8 & 30 \\
9 & 28
\end{array}
\right)$$ The preliminary step leads to $z=0.000176905 t+0.0216373$ from which the estimates $a=5652.76$ and $b=122.310$.
Starting with this guess, the iterates of Newton method will be
$$\left(
\begin{array}{cc}
n & b_n \\
0 & 122.31000 \\
1 & 121.84307 \\
2 & 121.85174
\end{array}
\right)$$ from which $a=5635.22$. A nonlinear regression would have be providing exactly the same values for the parameters.
| {
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"question_score": "1",
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Sum of fifth power of the roots of equation $x^3-x^2+1=0$ The equation $x^3-x^2+1=0$ has three roots $\alpha$, $\beta$ and $\gamma$. Find the value of $\alpha^5 + \beta^5 + \gamma^5$
I tried it this way:
$x^3=x^2-1$
$\alpha + \beta + \gamma = 1$
$\alpha \cdot \beta \cdot \gamma = -1$
$\alpha \cdot \beta + \beta \cdot \gamma + \alpha \cdot \gamma = 0$
So, $\alpha^3=\alpha^2-1$
$\alpha^5=\alpha^4-\alpha^2$
And similarly for $\beta$ and $\gamma$
Now I did add them but I am unable to find something useful in it.
| $$\alpha^2+\beta^2+\gamma^2=(\underbrace{\alpha+\beta+\gamma})^2-2(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})=?$$
$$\alpha^4+\beta^4+\gamma^4=(\underbrace{\alpha^2+\beta^2+\gamma^2})^2-2(\underbrace{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2})$$
Now $$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\underbrace{\alpha\beta+\beta\gamma+\gamma\alpha})^2-2\underbrace{\alpha\beta\gamma}(\underbrace{\alpha+\beta+\gamma})$$
| {
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How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
$$...=...$$
$$2=2$$
$$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$
$$\sqrt2=\sqrt2\text{ hence proved.}$$
Cheers in advance :)
| This one is a little bit round about.
$\sin \frac {\pi}{12} = \sin (\frac {\pi}{3} - \frac {\pi}{4})$ by angle addition rules
and
$\sin \frac {\pi}{12} = \sqrt {\frac {1-\cos \frac {\pi}{6}}{2}}$ by the half angle rules.
$\sin (\frac {\pi}{3} - \frac {\pi}{4}) = \frac {\sqrt 6 - \sqrt 2}{4}$
$\sqrt {\frac {1-\cos \frac {\pi}{6}}{2}} = \frac {\sqrt {2-\sqrt 3}}{2}$
$\frac {\sqrt 6 - \sqrt 2}{4} = \frac {\sqrt {2-\sqrt 3}}{2}$
similarly
$\cos \frac {\pi}{12} = \cos (\frac {\pi}{3} - \frac {\pi}{4}) = \sqrt {\frac {1+\cos \frac {\pi}{6}}{2}}\\
\frac {\sqrt 6 + \sqrt 2}{4} = \frac {\sqrt {2+\sqrt 3}}{2}$
$2\cos \frac{\pi}{12} - 2\sin \frac {\pi}{12} = \frac {\sqrt 6 + \sqrt 2}{2} - \frac {\sqrt 6 - \sqrt 2}{2} = \sqrt 2 = \sqrt {2+\sqrt 3} - \sqrt {2-\sqrt 3}$
| {
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Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$. I'm looking for help with (b) and (c) specifically. I'm posting (a) for completeness.
(a) Show convergence for $a_n=\sqrt{n+1}-\sqrt{n}$ towards $0$ and test $\sqrt{n}a_n$ for convergence.
(b) Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$.
(c) For which $\alpha\in\mathbb{Q}_+$ does $n^\alpha b_n$ converge?
I'm pretty sure I solved (a). I have proven the convergence of $a_n$ by using the fact that $$\sqrt{n}<\sqrt{n+1}\leq\sqrt{n}+\frac{1}{2\sqrt{n}}$$ which holds true since $$(\sqrt{n}+\frac{1}{2\sqrt{n}})^2=n+1+\frac{1}{4n}\geq n+1\,.$$
This gives us $$0<\sqrt{n+1}-\sqrt{n}\leq\frac{1}{2\sqrt{n}}$$
and after applying the squeeze theorem with noting that $\frac{1}{2\sqrt{n}}\longrightarrow0$ we can tell that also $a_n\longrightarrow0$.
Now $x_n=\sqrt{n}a_n=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$.
We have \begin{align*}\sqrt{n}(\sqrt{n+1}-\sqrt{n})&=\sqrt{n}\sqrt{n+1}-\sqrt{n}\sqrt{n}\\&=\sqrt{n(n+1)}-n\\&=\sqrt{n^2+n}-n\\&=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\\&=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}\\&=\frac{n}{\sqrt{n^2+n}+n}\\&=\frac{n}{n\sqrt{1+\frac{1}{n}}+n}\\&=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\end{align*}
and hence since the harmonic sequence $\frac{1}{n}$ converges towards 0 we have $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$
| Write $$b_n=\frac{\sqrt[k]{1+\frac{1}{n}}-\sqrt[k]{1}}{\frac{1}{\sqrt[k]{n}}}=\frac{\sqrt[k]{1+\frac{1}{n}}-\sqrt[k]{1}}{\frac{1}{{n}}}\frac{1}{n^{1-1/k}}$$
The first part is a difference quotient, hence $$\lim_{n\to \infty} b_n= f'(1)\lim_{n\to \infty}n^{1/k-1}$$
where $f(x)=\sqrt[k]{x}$. We can use calculus to compute $f'(x)=\frac{1}{k}x^{(\frac{1}{k}-1)}$ and hence $f'(1)=\frac{1}{k}$, a finite constant. Hence the limit will be zero if $\lim_{n\to \infty}n^{1/k-1}=0$, which is zero for $k>1$. This allows us to also calculate that $b_nn^\alpha$ converges, provided $\alpha+\frac{1}{k}-1\le 0$.
Note that this works even for $k$ not an integer.
| {
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System of congruent equations Given the following system, how do you prove that there only is one solution congruent 8?
$$3x+7y \equiv 2 (\text{mod } 8)$$
$$4x+5y \equiv 7 (\text{mod } 8)$$
My main idea have been to solve the equation, then to come to the solution that there only is one solution, however I get flawed results to say the least. Here is my method:
$$\left\{
\begin{array}{ll}
3x+7y \equiv 2 (\text{mod } 8)
\\4x+5y \equiv 7 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+14y \equiv 4 (\text{mod } 8)
\\8x+10y \equiv 14 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+6y \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
$$
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+18 \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
-2x\equiv -14 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
2x\equiv 6 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
Which then gives:
$$
\left\{
\begin{array}{ll}
x=4t+3
\\y=4s+3
\end{array}
\right.$$
$t$ and $s$ being arbitary whole numbers
So solution given congruent 8 is:
$$x\equiv 3 (\text{mod } 8) \text{ or } x\equiv 7 (\text{mod } 8)$$
$$y\equiv 3 (\text{mod } 8) \text{ or } y\equiv 7 (\text{mod } 8)$$
However when you put that into the original equation, the answer is wrong. So where did i do wrong...
| Don't multiply equations by $2$: as $2$ is not coprime to $8$ you're not getting equivalent equations.
One way: given:
$$\begin{align}3x+7y\equiv 2 \pmod 8\\4x+5y\equiv 7\pmod 8\end{align}$$
Take away the 1st equation from the 2nd:
$$\begin{align}3x+7y\equiv 2 \pmod 8\\x-2y\equiv 5\pmod 8\end{align}$$
Take away 3 times the 2nd equation from the 1st one:
$$\begin{align}13y\equiv -13 \pmod 8\\x-2y\equiv 5\pmod 8\end{align}$$
Divide the first equation by $13$, which is coprime to $8$:
$$\begin{align}y\equiv -1 \pmod 8\\x-2y\equiv 5\pmod 8\end{align}$$
Add the first equation, multiplied by 2, to the 2nd:
$$\begin{align}y\equiv -1\equiv 7 \pmod 8\\x\equiv 3\pmod 8\end{align}$$
Bonus reading about what I've actually been (partially) doing to the equations: https://en.wikipedia.org/wiki/Smith_normal_form
| {
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$p(x) = x^3 + 2x^2 + kx − 2$ zeroes at $a$, $\frac{1}{a}$ and $b$ for what value of k? For what values of $k$ has $p(x) = x^3 + 2x^2 + kx − 2$ zeroes at $a$, $\frac{1}{a}$ and $b$?
The options I'm given are $k=-7$, $k=1$, $ k=0$ and $k=2$.
| The product of the roots is $2$ and also $a(1/a)b$. So $b = 2$. So $2$ is a root. What $k$ makes $p(2) = 0$?
| {
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Elementary proof for $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$ I am searching for an elementary proof of the AM-GM inequality in three variables:
$\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$
The inequality of the geometric mean vs the arithmetic mean of two variables can be proven elementarily via
$x - 2 \sqrt{xy} + y \geq 0$
whenever $x,y > 0$. I am searching for a proof that uses a similar technique based on elementary arithmetic.
| An equivalent form is
$$x^3+y^3+z^3\ge 3xyz\tag{*}$$
for $x$, $y$, $z\ge0$. Without loss assume that $0\le x\le y\le z$.
Then $y=x+a$ and $z=x+a+b$ with $a$, $b$, $x\ge0$.
Also
$$x^3+y^3+z^3-3xyz=x^3+(x+a)^3+(x+a+b)^3-3x(x+a)(x+a+b)
=3xa^2+3xab+3xb^2+2a^3+3a^2b+3ab^2+b^3\ge0.$$
Then (*) follows.
| {
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I don't know how to solve this integral
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$
What I've done so far is:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$
So
$$x^2+x-4 = A(x^2-4)+Bx$$
And I did this so I could figure out the value of A and B but I am having a rough time trying to calculate A because of the $x^2$.
I've also seen that the solution solves this exercise by separating the fraction's numerator like this:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{x^2}{x(x^2+4)}dx+\int \frac{x}{x(x^2+4)}dx + \int \frac{-4}{x(x^2+4)}dx $$
When do I know that I have to use this? Is what I though correct? If yes, how do I continue it?
Thank you very much. Please, let me know if something is not very clear in my question.
Agapita.
| Your partial fraction decomposition should be
$$\frac{x^2+x-4}{x(x^2+4)} = \frac{A}{x}+\frac{Bx+C}{x^2+4}. \tag{1}$$
Multiplying both sides by $x(x^2+4)$ results in
$$x^2+x-4 = (A+B)x^2+Cx+4A. \tag{2}$$
Consequently, $4A=-4 \implies A=-1$, $C=1$ and $A+B = 1 \implies B =2$.
After partial-fraction decomposition, the integral can be computed using standard transformations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sums and harmonic series I found the solution of series on Wolfram Alpha
http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n
$ \sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = \sum\limits_{k=1}^{n} \frac{1}{(2k+1)(2k+2)} = \frac{1}{2} \left(H_{n+\frac{1}{2}} - H_{n+1} -1 + \text{ln}(4)\right)$
Can someone tell how to prove this in the form of Harmonic numbers?
| This is not an answer to the specific question but a long comment which gives a derivation of a much simpler formula for the sum in question.
Let
$$s = \sum _{k=1}^n \left(\frac{1}{2 k+1}-\frac{1}{2 k+2}\right)$$
Adding and subtracting a sum of even terms we get for $s$
$$\begin{array}
&=& \sum _{k=1}^n \left(\frac{1}{2 k}+\frac{1}{2 k+1}\right)-\sum _{k=1}^n \left(\frac{1}{2 k}+\frac{1}{2 k+2}\right)\\
=& (\frac{1}{2}+\frac{1}{3}+ ... + \frac{1}{2n+1})-\frac{1}{2}(1+\frac{1}{2}+...+\frac{1}{n}) -\frac{1}{2}(\frac{1}{2}+\frac{1}{3}+...+ \frac{1}{n+1})\\
=&(H_{2n+1}-1) -\frac{1}{2}H_n -\frac{1}{2}(H_{n+1}-1)\\
=&H_{2n+1} -\frac{1}{2}(H_n +H_{n+1})-\frac{1}{2}
\end{array}
$$
Those who wish can simplifiy this further using the relation $H_{n+1}=H_n + \frac{1}{n+1}$.
| {
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"question_score": "1",
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Find the Laurent series in the given domain: I'm stuck on the following question: (it is from 'the fundamentals of complex analysis')
Find the Laurent series for $\frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$
Im working towards a geometric series in the following way:
$\frac{z+1}{z(z-4)^3}$=$\frac{1}{(z-4)^3}(1+\frac{1}{z})$
Then I tried rewriting the $\frac{1}{z}$ part since eventually for a geometric series we want something like $A*\frac{1}{1-\frac{z-4}{4}}$ since we know $|\frac{z-4}{4}|<1$
$\frac{1}{z}=-\frac{1}{4}*\frac{1}{1-\frac{z-4}{4}}=-\frac{1}{4}\sum_{j=0}^{\infty}(\frac{z-4}{4})^j$
So all together we'd obtain $\frac{1}{(z-4)^3}-\frac{1}{4}\sum_{j=0}^{\infty}(\frac{z-4}{4})^j$
However the solution manual says:
Can someone please help me to spot where I go wrong? Or are they just different answers that express the same?
Thanks in advance
|
Can someone please help me to spot where I go wrong?
There is a mistake when you write
$$
\frac{1}{z}=-\frac{1}{4}*\frac{1}{1-\frac{z-4}{4}}
$$ a correct version is rather
$$
\frac{1}{z}=\frac{1}{4}\cdot\frac{1}{1+\frac{z-4}{4}}.
$$
| {
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Producing two functions from a rotated parabola that graph the parabola in question The following is my attempt at splitting a rotated parabola into two real functions of one real variable (the $x$-coordinate). Let $P$ be the parabola, $V = (x_V, y_V)$ the vertex of $P$, $\alpha$ the angle between the $x$-axis and the line perpendicular to the axis of symmetry of $P$ and growing in counter-clockwise direction. Finally, let $g(t) = \beta t^2$ define the "geometry" of $P$.
Now, the line perpendicular to the axis of symmetry of $P$ and passing through $V$ (call it $\ell$) is given by equation
$$
y = (x - x_V) \tan \alpha + y_V.
$$
Let $x$ be given. We wish to calculate two values $y_1$ and $y_2$ such that $(x, y_1)$ and $(x, y_2)$ are located at the parabola's "lower" and "upper" parts. To do that, I start from $(x, 0)$, go up or down until I reach $\ell$ at point $B$, travel through the line segment $BT$, and finally travel through $TF$. If we fix $V$, $T$ becomes
$$
T(t) = (x_V + t \cos \alpha - \beta t^2 \sin \alpha, y_V + t \sin \alpha + \beta t^2 \cos \alpha).
$$
Next, we set $x = x_V + t \cos\alpha - \beta t^2 \sin \alpha$ and solve for $t$
$$
-\beta \sin(\alpha)t^2 + \cos(\alpha)t + (x_V - x) = 0,
$$
which has solutions
$$
\begin{aligned}
t &= \frac{-\cos\alpha \pm \sqrt{\cos^2\alpha - 4(-\beta\sin\alpha)(x_V - x)}}{-2\beta\sin\alpha} \\
&= \frac{\cos\alpha \pm \sqrt{\cos^2\alpha + 4\beta\sin\alpha(x_V - x)}}{2\beta\sin\alpha}.
\end{aligned}
$$
Next, we need to find out $|VB|$:
$$
\begin{aligned}
|VB| &= \sqrt{(x - x_V)^2 + (\tan(\alpha)x - \tan(\alpha)x_V + y_V - y_V)^2} \\
&= \sqrt{(x - x_V)^2 + (\tan(\alpha)x - \tan(\alpha)x_V)^2} \\
&= \sqrt{(x - x_V)^2 + \tan^2(\alpha)(x - x_V)^2} \\
&= \sqrt{(x - x_V)^2 (1 + \tan^2(\alpha))^2} \\
&= (x - x_V)\sqrt{1 + \tan^2 \alpha}.
\end{aligned}
$$
Finally, the desired $y_1$ or $y_2$ are given by
$$
\begin{aligned}
y &= y_S + \sqrt{(|VB| - t)^2 + (\beta t^2)^2} \\
&= y_S + \sqrt{(|VB| - t)^2 + \beta^2 t^4},
\end{aligned}
$$
where $y_S = \tan(\alpha)(x - x_V) + y_V$.
Question: is this calculation correct? Does it work for all $\alpha$, $\beta$, $V$? (Except $\alpha = n\pi$.)
| The form of the solution will depend on the form in which you put the parabola. Any parabola (or circle or ellipse or hyperbola) in the plane can be expressed in the form $By^2 + Ax^2 + Cx + Dy + Exy + F = 0$. In the case of parabolas of interest here, that do not open up or down, $B$ will not be zero, so without loss of generality we can divide out by $B$ and assume the parabola is of the form $y^2 + Ax^2 + Cx + Dy + Exy + F = 0$. The problem is relatively easy to solve in this form. First, separate all terms with a $y$:
$y^2 + (Ex + D)y = -Ax^2 - Cx - F.$
Then complete the square in $y$ on the LHS:
$y^2 + (Ex + D)y + (\frac{Ex+D}{2})^2 = (\frac{E^2}{4} -A)x^2 + (\frac{ED}{2} - C)x + (\frac{D^2}{4} - F)$
That is,
$(y + \frac{Ex+D}{2})^2 = (\frac{E^2}{4} -A)x^2 + (\frac{ED}{2} - C)x + (\frac{D^2}{4} - F)$
so that
$ y + \frac{Ex+D}{2} = \pm\sqrt{(\frac{E^2}{4} -A)x^2 + (\frac{ED}{2} - C)x + (\frac{D^2}{4} - F)}$
or
$ y = -\frac{Ex+D}{2} \pm\sqrt{(\frac{E^2}{4} -A)x^2 + (\frac{ED}{2} - C)x + (\frac{D^2}{4} - F)}$
with the "upper" curve for $y$ being given by the choice of + in the $\pm$ and the lower by -. (Aside: and there is no solution when the expression inside the square root is negative. If $E^2 < 4A$, that clearly happens whenever $x$ is sufficiently negative or sufficiently positive. On the other hand, we know a rotated parabola either "opens to the right" or "opens to the left", so there will at least be two solutions for all sufficiently positive or all sufficiently negative $x$. Hence, we conclude that if the original curve was a parabola, then in fact $E^2 \geq 4A$.)
Anyhow, to use this solution in your case, I suggest you convert your parabola into this form. In particular, you can start with the upward-opening parabola with vertex $(a, b)$ which has equation $u - b = \beta(v-a)^2$ and then simply rotate about the origin by an angle $\alpha$, e.g. by substituting $u = x\cos\alpha - y\sin\alpha$ and $v = x\sin\alpha + y\cos\alpha$. You can arrange that the rotated vertex ends up at $(x_V, y_V)$ by taking $a = x_V\cos\alpha - y_V\sin\alpha$ and $b = x_V\sin\alpha + y_V\cos\alpha$.
I recognize that this is not a direct answer to the question of whether the expressions you list are correct, but I hope it will still be helpful, given that the original question of what are the expressions for the "upper and lower branches" of the rotated parabola is much simpler starting from the general quadratic equation in the plane as expressed in this solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\left(\frac{-2}{p}\right)$
Prove that for $p\ge 3$, a prime: $$\left(\frac{-2}{p}\right) = \begin{cases}
1 & p\equiv1,3\pmod{8}\\
-1 & p\equiv-1,-3\pmod{8}
\end{cases}$$
I already now that:
$$\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p^{2}-1}{8}}=(-1)^{\frac{(p-1)(p^{2}-1)}{8}}=(-1)^{\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\frac{p-1}{4}}$$
And I'm trying to show when the exponent is even (for the $1$ case).
We have:
*
*$\frac{p-1}{4} = 2k \implies p- 1 = 8k \implies p = 8k + 1$. So we can infer that if $p\equiv 1\pmod{8}$ then $\left( \frac{-2}{p} \right)=1$
*$\frac{p-1}{2}$ - doesn't yield anything new to us (I think)
*$\frac{p+1}{2} = 2t \implies p+1 = 4t \implies p = 4t-1$
but I don't really see how to infer the other option $(p \equiv 3 \pmod{8})$
| $p=4t-1=4(t-1)+3$, so $p=3$ mod 4. Modulo 8 there are two possibilities:
1) $p=3$ (mod 8)
2) $p=7$ (mod 8)
By plugging in the numbers, 2) doesn't work, and 1) work
Edit:
There seems to be a mistake, remember that $(-1)^\frac{p-1}{2}(-1)^\frac{p^2-1}{8}=(-1)^{\frac{p-1}{2}+\frac{p^2-1}{8}}=(-1)^\frac{(p-1)(p+5)}{8}$
| {
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How to prove the value of an trigonometric term to zero? Given that , in $\triangle ABC$ , $AC \neq BC$ . We have to prove that , $\dfrac{BC\cos C-AC\cos B}{BC\cos B-AC\cos A}+\cos C=0$
My trying:
As $AC \neq BC \Rightarrow b \neq a \Rightarrow a=c \Rightarrow A=C$ . So for the following term , $$ \\ $$
\begin{align}
\frac{BC\cos C-AC\cos B}{BC\cos B-AC\cos A}+\cos C
&= \frac{a\cos C-b\cos B}{a\cos B-b\cos A}+\cos C = \frac{a\cos A-b\cos B}{a\cos B-b\cos A}+\cos A \\[1ex]
&= \frac{a\cos A-b\cos B+a\cos A\cos B=b{\cos }^2A}{a\cos B-b\cos A}
\end{align}
Then I got stuck and have no clue how to go ahead . Can anyone please help me to solve this problem ? It will be of great help .
| Let's write $BC=a$, $AC=b$ and $AB=c$, and the opposite angles $\widehat{BAC}=\alpha$, $\widehat{ABC}=\beta$, $\widehat{ACB}=\gamma$, so your equation becomes
$$
\frac{a\cos\gamma-b\cos\beta}{a\cos\beta-b\cos\alpha}+\cos\gamma=0
$$
The sine law tells us that, if $R$ is the circumradius,
$$
a=2R\sin\alpha
\qquad
b=2R\sin\beta
$$
so we can rewrite the equation as
$$
\frac{\sin\alpha\cos\gamma-\sin\beta\cos\beta}{\sin\alpha\cos\beta-\sin\beta\cos\alpha}+\cos\gamma=0
$$
or
$$
\sin\alpha\cos\gamma(1+\cos\beta)-\sin\beta(\cos\beta+\cos\alpha\cos\gamma)=0
$$
Since $\beta=\pi-(\alpha+\gamma)$, we easily see that $\cos\beta+\cos\alpha\cos\gamma=\sin\alpha\sin\gamma$ and the equation becomes
$$
\sin\alpha\cos\gamma(1+\cos\beta)-\sin\alpha\sin\beta\sin\gamma=0
$$
or (as $\sin\alpha\ne0$)
$$
\cos\gamma+\cos\beta\cos\gamma-\sin\beta\sin\gamma=0
$$
that is, $\cos\gamma+\cos(\beta+\gamma)=0$ and therefore $\cos\gamma=\cos\alpha$.
So the stated equality implies the angles at $A$ and $C$ are equal. Your triangle must be isosceles on the base $AC$.
Tracing the steps back, we set that the given equality holds for $\alpha=\gamma$.
| {
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Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice factorization of $2^{3x+1}$
Any help specifically on how to treat $2^{3x+1}$ would be appreciated.
| For natural $x$ we have
$$4^{3x+1}+2^{3x+1}=4\cdot64^x+2\cdot8^x+1=4(64^x-1)+2(8^x-1)+7\equiv0\pmod{7}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something.
Hints and answers appreciated. Sorry if this is a duplicate.
| Set $x=-y\to+\infty$
$$\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=\frac{y^2-\sqrt{y^4+1}}{-y^3-\sqrt{y^6+1}}=-\frac{y^2-\sqrt{y^4+1}}{y^3+\sqrt{y^6+1}}\frac{y^2+\sqrt{y^4+1}}{{y^2+\sqrt{y^4+1}}}=\frac{1}{(y^3+\sqrt{y^6+1})({y^2+\sqrt{y^4+1})}}\to0$$
| {
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How to solve for $m/n$ given that $(2x^{\frac{1}{6}})^{m} = (2x^{\frac{1}{8}})^{2n}$ for some constant $x$? I was solving a problem when I got stuck on a particular step, I don't know how to proceed further.
$$(2x^{\frac{1}{6}})^{m} = (2x^{\frac{1}{8}})^{2n}$$
I want to make the bases equal, so I can compare the exponents (i.e., the ratio of $m$ to $n$). Here $x$ is constant.
I tried but ended up making $x$ terms equal but then 2's exponents (in the base) became different.
Here is the original question:
If $m$ and $n$ are the smallest positive integers satisfying the
relation $$\left(2\, \mathrm{cis} \frac\pi6\right)^m = \left(4\, \mathrm{cis} \frac\pi4\right)^n$$ then find the value of $m+n$.
where $\mathrm{cis}\, \theta \equiv \cos \theta + i \sin\theta$
| Solving for $x$:
$$(2x^{1/6})^m=(2x^{1/8})^{2n}$$
$$2^mx^{m/6}=2^{2n}x^{2n/8}$$
$$x^{m/6}=2^{2n-m}x^{n/4}$$
$$x^{m/6-n/4}=2^{2n-m}$$
$$x=2^{\frac{2n-m}{m/6-n/4}}$$
$$x=2^{\frac{48n-24m}{4m-6n}}$$
$$x=2^{\frac{24n-12m}{2m-3n}}$$
$$x=2^{\frac{8*3n-6*2m}{2m-3n}}$$
$$x=2^{-\frac{6*2m-8*3n}{2m-3n}}$$
$$x=2^{-\frac{6*2m-6*3n-2*3n}{2m-3n}}$$
$$x=2^{\frac{6n}{2m-3n}-6}$$
$$x=\frac{1}{2^6}2^{\frac{6n}{2m-3n}}$$
The ratio of $m$ and $n$:
$$(2x^{1/6})^m=(2x^{1/8})^{2n}$$
$$m\log(2x^{1/6})=2n\log(2x^{1/8})$$
$$\frac{m}{n}=2\log_{2x^{1/6}}(2x^{1/8})$$
Or:
$$\frac{m}{n}=2\frac{\log(2)+\frac{1}{8}\log(x)}{\log(2)+\frac{1}{6}\log(x)}$$
$$\frac{m}{n}=2\frac{6\log(2)+\frac{6}{8}\log(x)}{6\log(2)+\log(x)}$$
$$\frac{m}{n}=2\frac{6\log(2)+(1-\frac{2}{8})\log(x)}{6\log(2)+\log(x)}$$
$$\frac{m}{n}=2\frac{6\log(2)+\log(x)-\frac{2}{8}\log(x)}{6\log(2)+\log(x)}$$
$$\frac{m}{n}=2\left(1-\frac{\frac{2}{8}\log(x)}{6\log(2)+\log(x)}\right)$$
$$\frac{m}{n}=2-\frac{1}{2}\frac{\log(x)}{6\log(2)+\log(x)}$$
The origial question:
$$2^m \text{cis}(\frac{\pi m}{6})=4^n \text{cis}(\frac{\pi n}{4})$$
$$2^{m-2n}=\text{cis}(\pi (\frac{n}{4}-\frac{m}{6}))$$
The left side is real, so the right side must be real aswell:
$$\sin(\pi (\frac{n}{4}-\frac{m}{6}))=0$$
$$\cos(\pi (\frac{n}{4}-\frac{m}{6}))=2^{m-2n}$$
For the first equation:
$$\pi (\frac{n}{4}-\frac{m}{6})=k \pi$$
$$\frac{n}{4}-\frac{m}{6}=k$$
$$6n-4m=24k$$
For the second, the right side must be 1:
$$m-2n=0$$
So:
$$6n-8n=24k$$
$$-2n=24k$$
$$n=12k$$
$$m=24k$$
But the $\cos$ must be positive, so the smallest possible $k$ value is $2$ (you get $-1$ as the value of the $\cos$ with $k=1$, so the next integer ($k=2$) will give $1$): $n=24$, $m=48$, so $n+m=72$.
| {
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Why $r$ is equal to $1/\phi$ in the System of Equations I have this system
$$\begin{align*} & \cos\theta\left(r+\frac 1r\right)=-2\\ & r^2+\frac 1{r^2}+4\cos^2\theta=10\end{align*}$$
I solved for $r$ and $\cos\theta$ by first squaring the first equation$$r^2+\frac 1{r^2}=\frac 4{\cos^2\theta}-2$$So$$\frac 4{\cos^2\theta}+4\cos^2\theta=12\quad\implies\quad\cos\theta=\frac {\pm1\pm\sqrt5}2$$
Question:
However after this, I'm not sure which value of the cosine I should take. The result used is $\cos\theta=1/\phi$, but I'm not sure why the other three solutions were discarded. The full solutions are$$\cos\theta=\frac 1{\phi}\quad\qquad\quad r=\phi+\sqrt \phi$$where $\phi=\tfrac 12\left(1+\sqrt5\right)$.
For context, I'm working out this problem which was solved by Ron Gordon. I'm having a hard time figuring out why he chose $1/\phi$ and $\phi+\sqrt{\phi}$.
| Starting from
$$ \frac{4}{\cos^2\theta} + 4\cos^2\theta - 12 = 0 $$
You get
$$ \cos^4\theta - 3\cos^2\theta + 1 = 0 $$
which gives
$$ \cos^2\theta = \frac{3\pm\sqrt{5}}{2} $$
However $\cos^2\theta \le 1$, so we must take the negative sign. Furthermore you want $\cos\theta < 0$ since
$$ \cos\theta \left(r + \frac{1}{r}\right) = -2 < 0 $$
This leaves
$$ \cos\theta = -\sqrt{\frac{3-\sqrt{5}}{2}} = \frac{1-\sqrt{5}}{2} = -\frac{1}{\varphi} $$
which means
$$ r + \frac{1}{r} = 2\varphi $$
or
$$ r = \varphi \pm \sqrt{\varphi} $$
both roots are equally valid as they're reciprocals of each other, and $r$ and $r^{-1}$ are interchangeable.
| {
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Find determinant Find determinant of A.
$${det}A =\left( \begin{array}{cc}
1 & 1 & 1 & 1 & 1\\
a & b & c & d & x\\
a^2 & b^2 & c^2 & d^2 & x^2\\
a^3 & b^3 & c^3 & d^3 & x^3\\
a^4 & b^4 & c^4 & d^4 & x^4\end{array} \right)$$
I was thinking that I should reduce ones on top, but then I would be stuck with this:
$${det}A =\left( \begin{array}{cc}
b-a & c-a & d-a & x-a\\
b^2-a^2 & c^2-a^2 & d^2-a^2 & x^2-a^2\\
b^3-a^3 & c^3-a^3 & d^3-a^3 & x^3-a^3\\
b^4-a^4 & c^4-a^4 & d^4-a^4 & x^4-a^4\end{array} \right)$$
Any suggestions?
| Now, notice that
*
*$(b-a)$ is a factor of the first column,
*$(c-a)$ a factor of the second column,
*$(d-a)$ a factor of the third, and,
*$(x-a)$ a factor of the fourth.
You will again get a row of $1$s first up.
| {
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Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
By taylor polynomials we get: $e^{x^4}=1+x^4+\frac{x^8}{2}+\mathcal{O}(x^{12})$
$\sin(x^2)=x^2-\frac{x^6}{6}+\mathcal{O}(x^{10})$
$\cos(x^3)=1-\frac{x^6}{2}+\mathcal{O}(x^{12})$
so putting these together:
$$ \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)} = \frac{x^2+x^6+\frac{x^{10}}{2}-x^2+\frac{x^6}{6}-\mathcal{O}(x^{10})}{\frac{x^6}{2}-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}$$
Now I am not too familiar with the Big-Oh notation for limits so I am stuck here.
How does arithmetic work with them, can I simplify the oh's in the numerator and can I take $x$'s out?
| $\mathcal{O}(x^n)$ stands for an “unnamed” function $f(x)$ such that there exists a positive constant $M$ so that $|f(x)|\le M|x^n|$ for $x$ sufficiently close to $0$. In particular, $f(x)$ can be a function such that $\lim_{x\to0}f(x)/x^n$ is finite.
As a consequence, $\mathcal{O}(x^n)$ is also $\mathcal{O}(x^m)$ if $m\le n$: if you can find $M$ as before, then
$$
|f(x)|\le M|x^n|=M|x^m|\,|x^{n-m}|\le M|x^m|
$$
as soon as $|x|<1$.
Also needed here is that, if $f(x)$ is $\mathcal{O}(x^n)$, then $x^kf(x)$ is $\mathcal{O}(x^{n+k})$.
By the triangle inequality, the sum of two functions that are $\mathcal{O}(x^n)$ is also a function that is $\mathcal{O}(x^n)$.
For your numerator you can write
$$
e^{x^4}=1+x^4+\mathcal{O}(x^8)
$$
and so
$$
x^2e^{x^4}=x^2+x^6+\mathcal{O}(x^{10})
$$
Therefore
$$
x^2e^{x^4}=x^2+x^6-x^2+\frac{x^6}{6}+\mathcal{O}(x^{10})=\frac{7}{6}x^6+\mathcal{O}(x^{10})
$$
The denominator is
$$
1-1+\frac{1}{2}x^6+\mathcal{O}(x^{12})=\frac{1}{2}x^6+\mathcal{O}(x^{10})
$$
Finally, you can prove that if $n>m$ and $f(x)$ is $\mathcal{O}(x^n)$, then $f(x)/x^m$ is $\mathcal{O}(x^{n-m})$.
You may want to use the “small o” notation; roughly speaking, $o(x^n)$ means you disregard terms with order larger than $x^n$. In this case
$$
x^2e^{x^4}-\sin(x^2)=
x^2(1+x^4+o(x^4))-\left(x^2-\frac{1}{6}x^6+o(x^6)\right)=\frac{7}{6}x^6+o(x^6)
$$
and
$$
1-\cos(x^3)=1-1+\frac{1}{2}x^6+o(x^6)
$$
so your limit becomes
$$
\lim_{x\to0}\frac{\frac{7}{6}x^6+o(x^6)}{\frac{1}{2}x^6+o(x^6)}
=
\lim_{x\to0}\frac{\frac{7}{6}+o(x^0)}{\frac{1}{2}+o(x^0)}
$$
and you're done because $\lim_{x\to0}o(x^0)=0$.
| {
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How to display $x^5 -1$? Exercise:
Write $x^5-1$ as product of real polynomials of rate (max) 2.
I would start as $(x-1)(x^4+ x^3+x^2+x+1)$, but then I have no idea how to continue. Till now I had only even terms in parenthesis and I managed. Now I cannot get the solution. Please help, because I really have no idea how to continue. Is there maybe an algorithm for solving this?
| The roots of $x^5-1$ are $1$, $e^{\pm\frac{2\pi i}5}$, and $e^{\pm\frac{4\pi i}5}$. Therefore,\begin{align}x^5-1&=(x-1)\left(x-e^{\frac{2\pi i}5}\right)\left(x-e^{\frac{-2\pi i}5}\right)\left(x-e^{\frac{4\pi i}5}\right)\left(x-e^{\frac{-4\pi i}5}\right)\\&=(x-1)\left(x^2-2\cos\left(\frac{2\pi}5\right)x+1\right)\left(x^2-2\cos\left(\frac{4\pi}5\right)x+1\right).\end{align}
| {
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If in ,$r=1$,$R=3$ and $s=5$ then find value of $a^2+b^2+c^2$ If in $\triangle ABC$,$r=1$,$R=3$ and $s=5$, then find value of $a^2+b^2+c^2$
(where $r$=inradius,$R$=circumradius and $s$=semi-perimeter)
The answer given is $24$
| You know that $$r=\frac{Δ}{s}$$ Hence we get $$Δ=5$$
Also $$R=\frac{abc}{4Δ} $$ From this we get $$ abc=60$$
We also have a formula that
$$\frac{4(s-a)(s-b)(s-c)}{abc}=\frac{r}{R}$$
Hence, on substituting values we get
$$(5-a)(5-b)(5-c)=5$$
Which reduces to
$$125-50s+5\sum_{cyc} ab -abc = 5$$
Further on reducing we get
$$\sum_{cyc} ab=38$$
Now
$${(\frac{a+b+c}{2})}^2=s^2$$
$$\frac{a^2+b^2+c^2+2\sum ab}{4}=25$$
$$a^2+b^2+c^2=100-76=24$$
Hence $\sum a^2 =24$
| {
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Solve $\frac{1}{x-1}-\frac{4}{x-2}+\frac{4}{x-3}-\frac{1}{x-4}<\frac{1}{30} \forall x\in R$ I plugged this into WolframAlpha. It gave the factorized form as:
$$\frac{(x-7)(x-6)(x+1)(x+2)}{(x-4)(x-3)(x-2)(x-1)}>0$$
Once we've reached this point, I know how to find the solution for $x$. But, currently, I am having trouble converting the given question expression into the factorized form above. I started off strong with:
$$(\frac{1}{x-1}-\frac{1}{x-4})+(\frac{4}{x-3}-\frac{4}{x-2})<\frac{1}{30}$$
which simplifies to:
$$\frac{-3}{(x-1)(x-4)}+\frac{12}{(x-3)(x-2)}<\frac{1}{30}$$
which simplifies to:
$$\frac{3x^2-15x+10}{(x-1)(x-2)(x-3)(x-4)}<\frac{1}{90}$$
and now this is a dead end! As you can see, I got to the denomiator, but the numerator is not what I expected it to be. I could transfer $1/30$ to the LHS but that would bring the deadly fourth power into the numerator. If this was a competitive exam question, I would not be sitting in the hall factorizing fourth powers...
Is there a simpler method to factorize the LHS? Or is there an altogether different approach to solve this question?
| To solve the inequality, leonbloy already gave you the best suggestion.
To reconstruct WolframAlpha answer, doesn't seem that there be a shorter way other
than the plain
$$
\eqalign{
& {{x^{\,2} - 5x - 2} \over {\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} - {1 \over {30}} = \cr
& = {{30x^{\,2} - 150x - 60 - \left( {x^{\,4} - 10x^{\,3} + 35x^{\,2} - 50x + 24} \right)} \over {30\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} = \cr
& = - {{x^{\,4} - 10x^{\,3} + 5x^{\,2} + 100x + 84} \over {30\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} = \cr
& = - {{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 6} \right)\left( {x - 7} \right)} \over {30\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)}} \cr}
$$
where the factorization in the last passage is to be obtained by Ruffini's rule.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\iint_{\Bbb R^2}\frac{dxdy}{(1+x^2+xy+y^2)^\alpha}$ I need to determine for what values of $\alpha$ the following improper integral converges: $$\iint_{\Bbb R^2}\frac{dxdy}{(1+x^2+xy+y^2)^\alpha}$$
Any hints on how to change variable?
| $x^2+xy+y^2$ is a quadratic form associated to $\left(\begin{smallmatrix}1 & \frac{1}{2}\\\frac{1}{2}& 1\end{smallmatrix}\right)$, having eigenvalues $\frac{1}{2}$ and $\frac{3}{2}$. By enforcing the substitution $x=\frac{A+B}{\sqrt{2}}, y=\frac{A-B}{\sqrt{2}}$ the original integral is converted into
$$ I(\alpha)=\iint_{\mathbb{R}^2}\frac{dA\,dB}{\left(1+\frac{3}{2}A^2+\frac{1}{2}B^2\right)^\alpha}=\frac{2}{\sqrt{3}}\iint_{\mathbb{R}^2}\frac{du\,dv}{\left(1+u^2+v^2\right)^{\alpha}} $$
and by switching to polar coordinates:
$$ I(\alpha)=\frac{4\pi}{\sqrt{3}}\int_{0}^{+\infty}\frac{\rho}{(1+\rho^2)^{\alpha}}\,d\rho=\color{red}{\frac{2\pi}{\sqrt{3}(\alpha-1)}} $$
as soon as $\color{red}{\alpha>1}$.
Remark: an explicit diagonalization (in order to find the correct substitution) is not really needed. If $Ax^2+Bxy+Cy^2=q(x,y)$ is a positive definite quadratic form,
$$ \forall \alpha>1,\qquad \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+q(x,y))^\alpha}=\frac{2\pi}{(\alpha-1)\sqrt{4AC-B^2}} .$$
| {
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Using induction, prove that $({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$. Question: Using the principle of mathematical induction, prove that for every integer $n\geq 1, ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.
My attempt:
Let $P(n): ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$ ($n\geq 1).$
BASE CASE:
$P(1): ({3^2}^1 -1)$ is divisible by $2^{1+2}$ but not by $2^{1+3}$ , that is, $8$ is divisible by $8$ but not by $16$, which is true.
$\therefore P(1)$ is true
INDUCTION HYPOTHESIS:
Let $P(n)$ be true for $n=k$.
$\therefore P(k)$ is true, that is, $({3^2}^k -1)$ is divisible by $2^{k+2}$ but not by $2^{k+3}$.
$\therefore$ Let ${3^2}^k -1=\lambda.2^{k+2}$
$\implies {3^2}^k=1+\lambda.2^{k+2}$
INDUCTIVE STEP:
$P(k+1):({3^2}^{k+1} -1)$ is divisible by $2^{k+1+2}$ but not by $2^{k+1+3}$
We have to prove that $P(k+1)$ is true
${3^2}^{k+1}-1={3^{{2^k}2}}-1=(1+\lambda.2^{k+2})^2-1=1+2\lambda.2^{k+2}+\lambda^2.2^{2(k+2)}-1=\lambda.2^{k+3}+\lambda^2.2^{k+3+k+1}=2^{k+3}(\lambda+\lambda^2.2^{k+1})$
$\therefore {3^2}^{k+1}-1$ is divisible by $2^{k+3}$
My problem: How do I prove that the number is not divisible by $2^{n+3}$? What should I add to the induction hypothesis part and the inductive step. I am not allowed to use congruence. I am allowed to use the properties of numbers and their division. I thought of doing it by proving that the number leaves a particular remainder on dividing by $2^{n+2}$ but then realized that it would be very cumbersome since the remainder can range from $1$ to $2^{n+3}-1$.
Please help.
| Extract a factor $2\lambda$ from your final bracket. Since you have raised the exponent by $1$ you need to prove divisibility by $2^{k+3}$ but not $2^{k+4}$. You should be able to show that $\lambda$ is odd, as is the remaining factor.
| {
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Integrate $\int \frac{\sin(3x)}{1-\cos^2(\frac{3x}{2})}dx$ Integrate $$\int \frac{\sin(3x)}{1-\cos^2(\frac{3x}{2})}dx$$
My attempt,
Let $u=1-\cos^2(\frac{3x}{2})$. Then
$$du=-2\cdot\cos\left(\frac{3x}{2}\right)\cdot -\sin\left(\frac{3x}{2}\right)\cdot \frac{3}{2}=3\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)\,dx$$
How should I continue? Thanks in advance.
| Recall that $$\cos^2(x)=\frac{1+\cos(2x)}{2}\implies 1-\cos^2\left(\frac{3x}{2}\right)=1-\frac{1+\cos(3x)}{2}=\frac{1-\cos(3x)}{2}$$
Then you could try a substitution like $u=\cos\left(3x\right)$.
Another way to do this: notice that $1-\cos^2\left(\frac{3x}{2}\right)=\sin^2\left(\frac{3x}{2}\right)$ and $\sin(3x)=\frac{\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)}{2}$, and then much in the integrand cancels out.
| {
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Find $\sum_{n=1}^{\infty}$ $\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$
MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
= Lim$_{k\rightarrow\infty}$$\sum_{k=1}^{n}\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$
$\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$=
$\frac{1}{6\left(n+1\right)}$-$\frac{2}{\left(n+2\right)}$+$\frac{9}{\left(n+3\right)2}$-$\frac{8}{\left(n+4\right)3}$
I don't think i can telescope it, And i don't know any other method
| This is one way to rewrite the summand into a form suitable for telescoping by hand.
Let $p_k(n) = \prod_{\ell=1}^k (n+\ell)$. The series at hand has the form
$$\mathcal{S} \stackrel{def}{=} \sum_{n=1}^\infty \frac{n^2}{(n+1)(n+2)(n+3)(n+4)} =
\sum_{n=1}^\infty \frac{n^2}{p_4(n)}$$
Notice we have the identity
$$\frac{1}{p_{k+1}(n)} = \frac{1}{k} \frac{(n+k+1)-(n+1)}{p_{k+1}(n)} = \frac{1}{kp_k(n)} - \frac{1}{kp_k(n+1)}$$
Repeat apply this identity to the summand of above series, we find
$$\begin{align}
\frac{n^2}{p_4(n)} &= \frac{n^2}{3p_3(n)} - \frac{n^2}{3p_3(n+1)}
= \color{red}{\frac{2n-1}{3p_3(n)}} + \color{green}{\frac{(n-1)^2}{3p_3(n)}} - \frac{n^2}{3p_3(n+1)}\\
\color{red}{\frac{2n-1}{3p_3(n)}} &= \frac{2n-1}{6p_2(n)} - \frac{2n-1}{6p_2(n+1)}
= \color{orange}{\frac{1}{3p_2(n)}} + \color{blue}{\frac{2n-3}{6p_2(n)}} - \frac{2n-1}{6p_2(n+1)}\\
\color{orange}{\frac{1}{3p_2(n)}} &= \color{magenta}{\frac{1}{3p_1(n)}} - \frac{1}{3p_1(n+1)}
\end{align}$$
Combine these, we can rewrite the summand as $\displaystyle\;\frac{n^2}{p_4(n)} = g(n) - g(n+1)$
where
$$\begin{align}
g(n)
&=\color{magenta}{\frac{1}{3p_1(n)}} + \color{blue}{\frac{2n-3}{6p_2(n)}} + \color{green}{\frac{(n-1)^2}{3p_3(n)}}\\
&= \frac{\color{magenta}{2(n+2)(n+3)} + \color{blue}{(2n-3)(n+3)} + \color{green}{2(n-1)^2}}{6(n+1)(n+2)(n+3)}\\
&= \frac{6n^2+9n+5}{6(n+1)(n+2)(n+3)}
\end{align}$$
The series is now a telescoping one and
$$\mathcal{S} = \lim_{p\to\infty} \sum_{n=1}^p \frac{n^2}{p_4(n)} = \lim_{p\to\infty}( g(1) - g(p+1)) = g(1) = \frac{6+9+5}{6\cdot 2\cdot 3 \cdot 4} = \frac{5}{36}$$
| {
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Find $\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right).$ This is what my prof did:
Rewrite as $$x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\frac{x}{e}\left(1-e\cdot e^{-x\ln{\left(1+\frac{1}{x}\right)}}\right).$$
Here he uses that
$$\ln{(1+t)}=t-\frac{t^2}{2}+O(t^3),$$
so
$$\frac{x}{e}\left(1-e\cdot e^{-x\left(\frac{1}{x}-\frac{1}{2x^2}+O(\frac{1}{x^3})\right)}\right)=\frac{x}{e}\left(1-e^{\frac{1}{2x}+O\left(\frac{1}{x^3}\right)}\right).$$
So far so good. Now, he uses that $e^t=1+t+O(t^2)$ so that the expression aove becomes
$$\frac{x}{e}\left(1-\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\right)=-\frac{1}{2e}+O\left(\frac{1}{t}\right)$$
and now he lets $x\rightarrow\infty$ to get the answer. However, when he introduces the expansion of $e^t$, how does he get $O(1/x^2)$?
It is my understanding that I should plug in $t=\frac{1}{2x}+O\left(\frac{1}{x^3}\right)$ everywhere in $1+t+O(t^2),$ doing that I get
$$1+\frac{1}{2x}+O\left(\frac{1}{x^3}\right)+O\left(\left(\frac{1}{2x}+O\left(\frac{1}{x^3}\right)\right)^2\right),$$
Squaring that last ordo-term, the largest power of $x$ is $1/x^2$ and the smallest is $1/x^6.$ Why choose $1/x^2$?
| $$O\left(\left(\frac{1}{2x}+O\left(\frac{1}{x^3}\right)\right)^2\right)=O\left(\frac{1}{4x^2}+O\left(\frac{1}{x^4}\right)+O\left(\frac{1}{x^6}\right)\right)=O\left(\frac{1}{x^2}\right)$$
because you let $x\to \infty$ and $\frac{1}{x^6}$ approaches $0$ much faster than $\frac{1}{x^2}$, so you choose $\frac{1}{x^2}$, the "slowest of the penguins". Here you can estimate $\frac{1}{x^6}$ by $\frac{1}{x^2}$.
Compare it with choosing $O\left(x^2\right)$ when you could choose between $x^2$ and $x^6$: if you let $x\to 0$, then $x^6$ goes much faster, so you only need to consider the terms of orders of $x^2$. Here you would estimate $x^6$ by $x^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Seat friends at a dinner table I have the following question:
At a squared table I seat $8$ friends of mine.
At each side, I place two friends.
Always two friends know each other.
What's the probability that no friend knows it's sided, neighbor?
I tried to solve it by:
*
*Calculating all different seating possibilities: $4!$ $\Rightarrow 24$
*And writing down all options on paper where no side neighbors know the other side neighbor. For that, I got 9 options.
So I guess my final result is $$\frac{9}{4!}\Rightarrow 0.375$$
$37.5\%$ that none of the side neighbors knows the other side neighbor.
Is this correct? How would I calculate the count of all possible seating options where no side neighbor knows each other?
Thanks
| A slight improvement on user8734617's method:
Exactly $\frac{1}{7}$ of the possible arrangements have a fixed pair $A_1,B_1$ seated together. To see this, given any arrangement with the pair seated together, fix the position of $A_1$ and rotate the other $7$ around the table, giving $6$ other arrangements with $A_1,B_1$ not seated together. This method produces all possible arrangements around the table, so each arrangement with $A_1,B_1$ seated together corresponds to $6$ other arrangements, meaning $A_1,B_1$ sit together with probability $\frac{1}{7}$.
By the same method, of those arrangements with a fixed pair $A_1,B_1$ seated together, $\frac{1}{5}$ have another fixed pair $A_2,B_2$ also seated together, so two fixed pairs sit together with probability $\frac{1}{7 \cdot 5}$; similarly, three fixed pairs sit together with probability $\frac{1}{7 \cdot 5 \cdot 3}$ and four with probability $\frac{1}{7\cdot5\cdot3\cdot1}$.
Then we apply inclusion-exclusion: our answer is $$\begin{align}
&1 - \binom{4}{1}\cdot\frac{1}{7} + \binom{4}{2}\cdot\frac{1}{7 \cdot 5} - \binom{4}{3}\cdot\frac{1}{7 \cdot 5 \cdot 3} + \binom{4}{4}\cdot\frac{1}{7\cdot5\cdot3\cdot1} \\
&= 1 - \frac{4}{7} + \frac{6}{7\cdot5} - \frac{4}{7 \cdot 5 \cdot 3} + \frac{1}{7 \cdot 5 \cdot 3} \\
&= \frac{3}{7} + \frac{6}{7\cdot5} - \frac{3}{7 \cdot 5 \cdot 3} \\
&= \frac{3}{7} + \frac{5}{7\cdot5} \\ &= \frac{4}{7}
\end{align}$$
| {
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Whats the result of double integral $\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$ Whats the result of double integral
$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$
I was trying to get this $$\int_{0}^{1}\sqrt{1+x^2}\arctan {\frac{1}{\sqrt{1+x^2}}}dx=\int_{0}^{1}(1+x^2)\frac{1}{\sqrt{1+x^2}}\arctan {\frac{1}{\sqrt{1+x^2}}}dx$$
$$=\int_{0}^{1}(1+x^2)\int_{0}^{1}\frac{1}{1+x^2+y^2}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$
So far I don't know what to do next.any helps are to be grateful.
| Changing the variable $u=1+x^2$, integrating for y, and reversing the change, you can get:
$$I = \int(\sqrt{(x^2+1)}\arctan(\frac 1 {\sqrt{(x^2+1)}}), x = 0..1 ) = 0.819755$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Solution to a 4th order polynomial equation as Stillwell does (1989) I have a 4th order polynomial equation,
\begin{equation}
x^4 + px^2 + qx + r = 0,
\end{equation}
which is of the same form as the beginning to a general solution for 4th order polynomials detailed by Stillwell's book, Mathematics and its history, First Edition (1989). Stillwell rewrites this as,
\begin{equation}
\left(x^2 + p\right)^2 = px^2 - qx + p^2 - r
\end{equation}
and asserts that for any $y$,
\begin{equation}
\left(x^2 + p + y\right)^2 = \left(p+2y\right)x^2 - qx + \left(p^2 - r + 2py + y^2\right).\tag{*}
\end{equation}
Then, he says that
The quadratic $Ax^2 + Bx + C$ on the right-hand side will be a square
if $B^2-4AC=0$, which is a cubic equation for $y$. We can therefore
solve for $y$ and take the square root of both sides of the equation
for $x$, which then becomes quadratic and hence also solvable. The final result is a formula for $x$ using just square and cube roots of rational functions of the coefficients.
I don't want to copy Stillwell's textbook verbatim, and I more-or-less understand his solution to the cubic equation, so I will just put the solution here, as it's not available online. Given a general cubic equation,
\begin{equation}
x^3 + ax^2 + bx + c = 0,
\end{equation}
it holds true that a linear change of variable, $x = y - a/3$ gives
\begin{equation}
y^3 = py + q,
\end{equation}
which is then solved by,
\begin{equation}
y = \sqrt[3]{\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}} + \sqrt[3]{\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^2 - \left(\frac{p}{3}\right)^3}}.
\end{equation}
So, after doing all this, I still can't figure out how to recover a solution for $x$. However, I have a source (a part of a computer code) that claims the equation that must be solved is
\begin{equation}
\left(x^2+p+y\right)^2 = ax^2+bx+c,
\end{equation}
which is of the same form as equation (*). My source does this by re-writing $ax^2+bx+c$ as a square, then "solving the equation" to obtain 2 results:
\begin{align}
x^2 + \left(-\sqrt{a}\right)x + \left(-\frac{b}{2\sqrt{a}}+p+y\right) &= 0\\
x^2 + \left(\sqrt{a}\right)x + \left(\frac{b}{2\sqrt{a}}+p+y\right) &= 0
\end{align}
If I can figure out the nitty-gritty details behind how these solutions came about, I would be set, but as it is I have spent several days grinding away at Stillwell's solutions without any progress. Please help!!
| I'll try to add some details to what he says.
We can write
$$\left(x^2 + p + y\right)^2 = Ax^2+Bx +C\tag{*}$$
where $A=p+2y,B=-q,C=p^2-r+2py+y^2$.
The right-hand side will be a square $\left(\sqrt Ax+\frac{B}{2\sqrt A}\right)^2$ if $B^2-4AC=0$, i.e.
$$8y^3+20py^2+(16p^2-8r)y-q^2+4p^3-4pr=0$$
which is a cubic equation for $y$.
We can therefore solve for $y$ and take the square root of both sides of the equation for $x$
$$\left(x^2 + p + y\right)^2 = \left(\sqrt Ax+\frac{B}{2\sqrt A}\right)^2$$
to have
$$x^2+p+y=\pm\left(\sqrt Ax+\frac{B}{2\sqrt A}\right)$$
which then becomes quadratic and hence also solvable.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The value of the expression $x^4-8x^3+18x^2-8x+2$ when $x=\cot\frac{\pi}{12}$ The value of the expression $x^4-8x^3+18x^2-8x+2$ when $x=\cot\frac{\pi}{12}$
I know that $\cot\frac{\pi}{12}=2+\sqrt3$,but putting this value and evaluating is a long process.Isn't there a short and intuitive method to solve this?
| Hint:
$$\cot15^\circ+\tan15^\circ=\dfrac2{\sin(2\cdot15^\circ)}=?$$
$$\cot15^\circ\tan15^\circ=?$$
So, $\cot15^\circ,\tan15^\circ$ are the roots of $$t^2-4t+1=0$$
Now we can straightaway divide $$x^4-8x^3+18x^2-8x+2$$ by $$x^2-4x+1$$ to find the remainder.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What will be area of shaded region, in two rectangles, where 2 vertex are common? In a figure, there are two rectangles $ABCD$ and $DEBG$, each of lengh $7\ cm$ and width $3\ cm$. The area of shaded region. in $cm^2$ is approximately ?
Options given : $12$, $10$, $8$, $4$
|
Here, notice that $\Delta DEK$ and $\Delta BCK$ are congruent. If we find $|DK|$, we are done. By Pythagoras Theorem, $(7-a)^2 = a^2+9 \implies a = \frac{20}{7}$. Therefore, $|DK| = 7-\frac{20}{7} = \frac{29}{7}$ and shaded area is $3 \cdot \frac{29}{7} = \frac{87}{7} \approx12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$X^2 + X =A$ with $X, A\in \text{Mat}_{2,2} (\mathbb{R})$ . Show that there exists a solution $X$ for a given $A$ Prove the following statement:
There exists an $\epsilon > 0$ such that the following holds:
If $A = (a_{ij}) \in \text{Mat}_{2,2} (\mathbb{R})$ a matrix with $(|a_{ij}|) < \epsilon$ for $i,j \in \{1,2\}$, then the following equation
$$X^2 + X = A$$
has a solution $X \in \text{Mat}_{2,2} (\mathbb{R})$
My Idea on how to solve this:
Let $X = \begin{bmatrix}
v& w \\ x & y
\end{bmatrix}$.
Therefore $X^2 + X = \begin{bmatrix}
v^2 + v + w x& v w + w y + w\\ v x + x y + x & w x + y^2 + y
\end{bmatrix} = \begin{bmatrix}
a_0 & a_1 \\ a_2 & a_3\end{bmatrix}$
Lets now define the function $$ 0=h(v,w,y,x,a_0,a_1,a_2,a_3) =\begin{cases} v^2 + v + w x - a_0 \\ v w + w y + w - a_1\\v x + x y + x -a_2
\\w x + y^2 + y-a_3
\end{cases} $$
We can now calculate the derivative of $h$:
$$dh = \begin{bmatrix}
2v + 1 & x & 0 & w & -1&0&0&0\\
w& v+y+1& w& 0& 0&-1&0&0\\
x & 0&x&v +1 & 0&0&-1&0
\\0&x&2y+1&w& 0&0&0&-1
\end{bmatrix}$$
The idea now would be to apply the implicit function theorem and show that there exists an $X$ which solves this equation. I am not sure though if this approach is correct.
Last but not least.. this question comes from an analysis sheet, so I assume one should use the methods of analysis to solve it.
Is my approach the correct way? And how does one proceed from here?
Feel free to use another approach.
Thank you for your time.
| My way of attacking this problem uses both some algebra, and some analysis; also, there is nothing in the following which requires we restrict ourselves to $M_2(\Bbb R)$, the set of $2 \times 2$ real matrices; everything we say applies equally to $M_n(\Bbb R)$.
First, some algebra: the equation
$X^2 + X = A \tag 1$
has a solution $X$ for some $A$ if and only if the equation
$X^2 + X + \dfrac{1}{4} I = \dfrac{1}{4}I + A \tag 2$
has the same solution $X$ as well; also, since
$(X + \dfrac{1}{2}I)^2 = X^2 + X + \dfrac{1}{4}I, \tag 3$
we may write (2) as
$(X + \dfrac{1}{2}I)^2 = \dfrac{1}{4} I + A, \tag 4$
which has a solution precisely when there exists a matrix $C$ such that
$C^2 = \dfrac{1}{4}I + A, \tag 5$
in which event we may take
$X + \dfrac{1}{2} I = C, \tag 6$
or
$X = C - \dfrac{1}{2}I. \tag 7$
The reader informed in the machinations of high-school algebra will recognize the above process as completing the square of the polynomial $X^2 + X$.
The preceding discussion indicates that we need to resolve the question of the existence of $C$ satisfying (5) when the entries of $A$ are sufficiently small. To this end we turn to the tools of analysis. Consider the mapping
$B \mapsto B^2 \tag 8$
defined in some neighborhood of $\frac{1}{2}I$; for any linear map $H$ we have
$(B + H)^2 = B^2 + HB + BH + H^2, \tag 9$
whence
$(B + H)^2 - B^2 - (HB + BH) = H^2, \tag{10}$
whence
$\Vert (B + H)^2 - B^2 - (HB + BH) \Vert = \Vert H^2 \Vert \le \Vert H \Vert^2; \tag{11}$
since the right-hand side of (11) is of the from $\Vert H \Vert o(\Vert H \Vert)$, we find that the derivative $DB^2$ at any $B$ is the linear map
$DB^2(H) = HB + BH; \tag{12}$
with
$B = \dfrac{1}{2}I, \tag{13}$
(12) becomes
$DB^2(H) = H \left (\dfrac{1}{2}I \right ) + \left (\dfrac{1}{2}I \right ) H = H, \tag{14}$
that is, $DB^2$ is the identity mapping when (13) binds. We may now invoke that powerful tool of analysis, the inverse function theorem, to infer that there is a neighborhood $U$ of $\frac{1}{4}I$ and a continuously differentiable function
$S:U \to M_n(\Bbb R), \; S \left ( \dfrac{1}{4} \right ) = \dfrac{1}{2}, \tag{15}$
satisfying, for every $E \in U$,
$(S(E))^2 = E; \tag{16}$
that is, $S$ is an effective square-root on $U$.
The hypothesis placed upon $A$, that $\vert a_{ij} \vert < \epsilon$ for all $i, j$, allows us to conclude that for $\epsilon$ sufficiently small
$\dfrac{1}{4} I + A \in U; \tag{17}$
thus we may take
$C = S \left (\dfrac{1}{4}I + A \right ); \tag{18}$
then
$C^2 = \dfrac{1}{4}I + A, \tag{19}$
and with $X$ as in (7) we have a solution to (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2609524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$ and avoid extra solutions while squaring Solve the equation,
$$
\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x
$$
My Attempt:
$$
\cos\Big[ \sin^{-1}x+\sin^{-1}(1-x) \Big]=x\\
\cos\big(\sin^{-1}x\big)\cos\big(\sin^{-1}(1-x)\big)-\sin\big(\sin^{-1}x\big)\sin\big(\sin^{-1}(1-x)\big)=x\\
\sqrt{1-x^2}.\sqrt{2x-x^2}-x.(1-x)=x\\
\sqrt{2x-x^2-2x^3+x^4}=2x-x^2\\
\sqrt{x^4-2x^3-x^2+2x}=\sqrt{4x^2-4x^3+x^4}\\
x(2x^2-5x+2)=0\\
\implies x=0\quad or \quad x=2\quad or \quad x=\frac{1}{2}
$$
Actual solutions exclude $x=2$.ie, solutions are $x=0$ or $x=\frac{1}{2}$.
I think additional solutions are added because of the squaring of the term $2x-x^2$ in the steps.
So, how do you solve it avoiding the extra solutions in similar problems ?
Note: I dont want to substitute the solutions to find the wrong ones.
| The domain gives
$$-1\leq x\leq1$$ and $$-1\leq1-x\leq1,$$ which gives $$0\leq x\leq1,$$
which says that the answer is $$\left\{\frac{1}{2},0\right\}.$$
I think it's better after your third step to write
$$\sqrt{2x-x^2}=\sqrt{1-x^2}$$ or $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding a recursive relation from a differential equation. We consider the following differential equation: $$(1-x)\psi'(x) - 1 = 0$$
We then define: $\psi(x) = \sum_{n = 0}^{\infty}a_nx^n$.(power series)
Plugging it into the formula we get to this step: $$\sum_{n=0}^{\infty}a_nn(x^{n-1}-x^{n}) - 1= 0$$
How can we then identify the recursive relation between the summands thus determine $a_{n+1}$ ?
| Assuming $\psi(x) = a_0 + a_1 x+a_2 x^2+\ldots$, then it follows
\begin{align}
(1-x)\psi'(x) =&\ (1-x)(a_1 + 2a_2x+\ldots + na_n x^{n-1}+\ldots)\\
=&\ a_1+ (2a_2-a_1)x + (3a_3-2a_2)x^2+\ldots + ((n+1)a_{n+1}-na_n)x^n+\ldots
\end{align}
So, we have
\begin{align}
(1-x)\psi'(x)-1 = (a_1-1) + (2a_2-a_1)x + \ldots + ((n+1)a_{n+1}-na_n)x^n+\ldots = 0
\end{align}
We know that in order for a power series to be identicially zero, then it must be that all of its coefficients are zeros. Hence we have
\begin{align}
a_1-1 = 0, 2a_2-a_1=0, \ldots, (n+1)a_{n+1}-n a_n = 0.
\end{align}
So the recursion formula is
\begin{align}
a_{n+1} = \frac{n}{n+1}a_n \ \ \ \text{ with } \ \ \ a_1=1.
\end{align}
Solving the recursion, we see that
\begin{align}
a_n = \frac{n-1}{n}a_{n-1} = \frac{n-1}{n}\frac{n-2}{n-1}a_{n-2}= \frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{1}{2}a_1 = \frac{1}{n}a_1 = \frac{1}{n}.
\end{align}
Thus, the solution is given by
\begin{align}
\psi(x) = a_0+\sum^\infty_{n=1} \frac{x^n}{n} = a_0 - \log(1-x)
\end{align}
for some constant $a_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the greatest and least values of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$ Find the upper and lower limit of
$$
(\sin^{-1}x)^2+(\cos^{-1}x)^2
$$
My Attempt:
$$
\frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&\quad(\cos^{-1}x)^2\leq\pi^2\\
0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4}
$$
Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.
Why am I not getting the lower limit in my approach ?
How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?
| $$(\arcsin x)^2+(\arccos x)^2=(\arccos x+\arcsin x)^2-2\arccos x\cdot \arccos x\leq \left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{4}.$$
Equality hold when $x=0$ or $\displaystyle x=1$
Similarly using $$\frac{(\arcsin x)^2+(\arccos x)^2}{2}\geq \bigg(\frac{\arcsin x+\arccos x}{2}\bigg)^2$$
so $$(\arcsin x)^2+(\arccos x)^2\geq \frac{\pi^2}{8}$$
equality hold when $\displaystyle x=\frac{1}{\sqrt{2}}$.
And at $x=-1$ value of $\displaystyle (\arcsin x)^2+(\arccos x)^2=\frac{5\pi^2}{4}$
$$\Rightarrow \frac{\pi^2}{8}\leq (\arcsin x)^2(\arccos x)^2\leq \frac{5\pi^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$.
My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here. Any help?
| You may solve this problem using geometry or you may use Lagrange Multipliers method.
You have already found $$(x-1)^2+(y+1)^2=4$$ and you want to find a point on this circle which maximizes $$x^2+y^2$$.
Graphing the first circle and finding the largest possible circle $$x^2+y^2= R^2$$ to have a point on the circle $$ (x-1)^2+(y+1)^2=4$$ leads to the point with $ x=1+\sqrt 2$ and $y=-1-\sqrt 2$.
That implies $x^2= y^2 =3+2\sqrt 2 $ which in turns gives $$x^2+y^2= 2(3+2\sqrt 2) =11.6568.. $$
The Lagrange Multipliers Methods provides the same results.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Specific steps in applying the Chinese Remainder Theorem to solve modular problem splitting modulus I am trying to get an idea of how the Chinese Remainder Theorem (CRT) can be used to finish up this problem, in which the problem
$$7^{30}\equiv x\pmod{ 100}$$
is attempted by splitting the modulus into relatively prime factors $25$ and $4,$ arriving at
$$\begin{align}
7^{30}&\equiv1\pmod4\\
7^{30}&\equiv-1\pmod{25}
\end{align}$$
I understand that the CRT may be called upon because $m=\prod m_i,$ and we have the same $7^{30}$ value on the LHS, but I don't know how to carry it out.
The question was touched upon in this post as the second entry:
How do I efficiently compute $a^b \pmod c$ when $b$ is less than $c.$ For instance $5^{69}\,\bmod 101.$
However, I don't see this particular point clearly worked out, perhaps because it is a multi-pronged question.
Following this presentation online, this seems to be the verbatim application of the CRT without any added concepts or shortcuts:
From @gimusi's answer (upvoted):
$$\begin{cases}
x \equiv 7^{30} \pmod4\\
x\equiv 7^{30} \pmod{25}
\end{cases}$$
rearranged into
\begin{cases}
x \equiv 1 \pmod4\\
x\equiv -1 \pmod{25}
\end{cases}
Given the general form of the equations above as $x\equiv a_i \pmod {m_i},$ the CRT states $x\equiv a_1 b_1 \frac{M}{m_1}+a_2 b_2 \frac{M}{m_2}\pmod M$ with $M=\prod m_i,$ and with
$$b_i =\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}.$$
The inverse of $\frac{M}{m_i}$ is such that $\frac{M}{m_i}\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}\equiv 1.$
Calculating the components:
$$\begin{align}
a_1&=1\\
a_2&=-1\\
M&=4\times 25 =100\\
\frac{M}{m_1} &= \frac{100}{4}=25\\
\frac{M}{m_2} &= \frac{100}{25}=4\\
b_1 &= \left(\frac{M}{m_1}\right)^{-1} \pmod 4 = (25)^{-1}\pmod 4 =1\\
b_2 &= \left(\frac{M}{m_2}\right)^{-1} \pmod {25}= (4)^{-1} \pmod{25}=19
\end{align}$$
Hence,
$$x=1\cdot 25 \cdot 1 + (-1)\cdot 4 \cdot 19 = -51 \pmod{100}\equiv 49.$$
| $$7^{30}\equiv x\pmod{ 100}$$
Could be solved easily without Chinese Remainder Theorem. Note that $7^4=2401 \equiv 1\pmod {100} $ Thus $$ 7^{30} = 7^{28}\times 49 \equiv 49 \pmod {100}$$
Solving the system with Chinese Remainder Theorem requires finding a linear combination of $25$ and $4$ to equal 1.
Such a combination is $$ 1= 1(25) -6(4) $$
Therefore the answer to the system is $$ x\equiv (1)(1)(25) +(-1)(-6)(4) \pmod {100}$$
That is $$ x\equiv 49 \pmod {100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solving $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$? (1983 AIME problem 3)
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?
I know it is a messy/bad idea, but I first started off by squaring both sides and moving everything to one side to get $$x^4 + 36x^3 + 384x^2 + 1080x + 900 - 4x^2 - 72x - 180 = x^4 + 36x^3 + 380x^2 + 1008x + 720 .$$
And by (generalisation) of Vieta's formula, the product of the real roots should be $\frac{720}{1} = 720$, but that is wrong, and I don't understand why.
| $$x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$$
let $x^2+18x+30=u$
so
$$u= 2 \sqrt{u+15}$$
$$u^2=4(u+15)$$
$$u^2-4u-60=0$$
$$(u+6)(u-10)=0$$
so
$$({\color{Red} {x^2+18x+36}})({\color{Blue} {x^2+18x+20}})=0$$
the roots of $({\color{Red} {x^2+18x+36}})$ do not satisfy the original equation especially in the term $\sqrt{x^2 + 18x + 45}$ so the product of real roots will be $20$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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If $A$ is $2\times 2$ skew orthogonal with $A^TA=-I$ and $\det A=1$ then $\operatorname{tr} A=0$ $A$ is skew orthogonal if $A^TA=-I$ in $\mathbb{Z_p}$ for $p>2$. The general form of $2\times 2$ characteristic polynomial: $x²-(\operatorname{tr} A)x+\det A$. It is given: $\det A=\pm 1$
If $\det A=-1$ then $\operatorname{tr} A= 0,1,-1$. But if $\det A=1$, then $\operatorname{tr} A$ must be $0$ only. What is the reason behind this?
| Edited 27/01: As the OP has warned me privately, there was a mistake in a sign in the case $\det(A)=-1$. This makes a new type of solution to appear.
This answer includes also the case with $\det(A)=-1$.
Let $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$. Then $A^T=\begin{pmatrix}a & c \\ b & d\end{pmatrix}$ and
$$AA^T=\begin{pmatrix}a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2\end{pmatrix}=-I,$$
1) If $\det(A)=1$ then we have the identities
$$\begin{align*}
a^2+b^2&=-1 & (1)\\
c^2+d^2&=-1 & (2)\\
ac+bd&=0 & (3)\\
ad-bc&=1 & (4)
\end{align*}$$ in $\mathbb{Z}_p$.
Multipying (3) by $c$ and (4) by $d$ we find $ac^2+bcd=0$, $ad^2-bcd=d$, so
$$ac^2+ad^2=d\implies a(c^2+d^2)=d\implies -a=d$$
by (2) and therefore trace$(A)=a+d=-d+d=0$.
Observe that this proof is valid over any commutative unital ring, we have not used that it is $\mathbb{Z}_p$ in any way (nor that it is a field!).
2) If $\det(A)=-1$ then we have similar identities, changing (4) with $ad-bc=-1 \ (4')$. Then similar computations give us $a=d$ and $b=-c$.
Therefore $$A=\begin{pmatrix} a & b \\ -b & a\end{pmatrix}$$
with $a,b$ such that $a^2+b^2=-1$. Observe that this implies $AA^T=-I$, $\det(A)=-1$. Then trace$(A)=2a$, but this does not imply that trace$(A)\in\{-1,0,1\}$!
For example, we can pick $b=0$, $a^2=-1$. By quadratic reciprocity, such an $a$ exists in $\mathbb{Z}_p$ with $p$ prime if and only if $p\equiv 1(\text{mod}4)$. We can pick for example $p=13$. Then $8^2\equiv -1(\text{mod}{13})$, so
$$A=\begin{pmatrix} 8 & 0 \\ 0 & 8\end{pmatrix}$$
satisfifes the hypotheses in $\mathbb{Z}_{13}$ but trace$(A)\equiv 16\equiv 3\not\equiv 0,\pm1(\text{mod}13)$.
Similarly, in $\mathbb{C}$ we can pick $a=i$ to get trace$(A)=2i\neq 0,\pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to figure out this definite integral How to calculate this definite integral?
$$\int_0^1\frac{x\cos x}{\sqrt{1+x^2}}\mathrm{d}x$$
I've been thinking about this for a few days but still have no ideas.
| I got an answer with Kampé de Fériet Function.
\begin{aligned}
&\int_0^1{\frac{x\cos x}{\sqrt{1+x^2}}}\text{d}x\\
&=\int_0^{\ln \left( 1+\sqrt{2} \right)}{\text{sinh}x\cos \left( \text{sinh}x \right)}\text{d}x \ \ \ \ \ (x\rightarrow \mathrm{sinh}x)
\\
&=\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n \right) !}}\int_0^{\ln \left( 1+\sqrt{2} \right)}{\text{sinh}^{2n+1}x\text{d}x}
\\
&=-\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}{\frac{\Gamma \left( n+1 \right)}{\Gamma \left( 2n+1 \right) \Gamma \left( n+\frac{3}{2} \right)}}\\
&\ \ \ -\sqrt{\frac{\pi}{2}}\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{\left( -1 \right) ^n2^k}{\Gamma \left( 2n+1 \right) \Gamma \left( \frac{3}{2}+k \right) \Gamma \left( \frac{3}{2}+n \right) \Gamma \left( \frac{3}{2}+k+n \right)}}}\ \ \ \ \ (\text{with the help of MMA})
\\
&=-\frac{\pi}{2}\mathbf{L}_{-1}\left( 1 \right) -\sqrt{\frac{\pi}{2}}\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{\left( -1 \right) ^n2^k}{\Gamma \left( 2n+1 \right) \Gamma \left( \frac{3}{2}+k \right) \Gamma \left( \frac{3}{2}+n \right) \Gamma \left( \frac{3}{2}+k+n \right)}}}
\\
&=-\frac{\pi}{2}\mathbf{L}_{-1}\left( 1 \right) -\frac{4\sqrt{2}}{\pi}\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{\left( 1 \right) \!_n\left( 1 \right) \!_k}{\left( 1 \right) \!_n\left( \frac{1}{2} \right) \!_n\left( \frac{3}{2} \right) \!_n\left( \frac{3}{2} \right) \!_k\left( \frac{3}{2} \right) \!_{n+k}}}}\frac{\left( -\frac{1}{4} \right) \!^n}{n!}\frac{2^k}{k!}
\end{aligned}
where $\mathbf{L}_{\nu}\left( z \right)$ is modified Struve function and by the definition of Kampé de Fériet Function
$$\displaystyle \mathbf{F}_{o,m,n}^{r,p,q}\left( \left. \begin{array}{c}
\mathbf{A}_r;\mathbf{a }_p;\mathbf{a }_{q}^{'}\\[5pt]
\mathbf{B}_o;\mathbf{b }_m;\mathbf{b }_{n}^{'}\\
\end{array} \right|x,y \right) =\sum_{i,j=0}^{\infty}{\frac{\prod_{s=1}^r{\left( A_s \right) \!_{i+j}}\prod_{k=1}^p{\left( a _k \right) \!_i}\prod_{\ell =1}^q{\left( a _{\ell}^{'} \right) \!_j}}{\prod_{v=1}^o{\left( B_v \right) \!_{i+j}}\prod_{t=1}^m{\left( b _t \right) \!_i}\prod_{u=1}^n{\left( b _{u}^{'} \right) \!_j}}}\frac{x^i}{i!}\frac{y^j}{j!}$$
where $\mathbf{a }_p=\alpha _1,\alpha _2,...,\alpha _p.$ Hence we have
$$\int_0^1{\frac{x\cos x}{\sqrt{1+x^2}}}\text{d}x=-\frac{\pi}{2}\mathbf{L}_{-1}\left( 1 \right) -\frac{4\sqrt{2}}{\pi}\mathbf{F}_{1,3,1}^{0,1,1}\left( \left. \begin{array}{c}
1;1\\
\dfrac{3}{2};1,\dfrac{1}{2},\dfrac{3}{2};\dfrac{3}{2}\\
\end{array} \right|-\frac{1}{4},2 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| Let $x=1+y$ with $y\to0$ and use
$$(1+y)^a=1+ay+o(y)$$
$$\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}=\frac{1+\frac{1}{5}y-1+o(y)}{1+ \frac{1}{6}y -1+o(y)}=\frac{\frac{1}{5}y+o(y)}{\frac{1}{6}y+o(y)}=\frac{\frac{1}{5}+o(1)}{\frac{1}{6}+o(1)}\to\frac65$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 3
} |
How do I solve this fractional indices equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$? Solve the equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$
I thought that the best way of approaching this would be to rewrite everything using $3$ as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields:
$$\frac{3^{5x+2}}{3^{2(1-x)}}=\frac{3^{3(4+3x)}}{3^6}$$
Equating the exponents of each numerator:
$$ 5x+2=3(4+3x) $$
$$ 5x+2=12+9x $$
$$ -4x=10 $$
$$ x = \frac{10}{-4}=-\frac{5}{2} $$
Doing this for the denominator yields a different value of $x$:
$$ 2(1-x)=6 $$
$$ 2-2x=6 $$
$$ -2x=4 $$
$$ x = -2 $$
Why is that I'm obtaining two different values of $x$?
Further to this, the solution in the book states the answer as $x=-3$, what am I doing wrong?
| You cannot just equate the numerator and denominator here, because they are not coprime.
Use that $$\frac{3^{5x+2}}{3^{2(1-x)}}=3^{5x+2-2(1-x)}$$ and $$\frac{3^{3(4+3x)}}{3^6}=3^{3(4+3x)-6}$$
then solve
$$5x+2-2(1-x)=3(4+3x)-6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
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