Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to show that this complex equation has 10 non real roots and how to express them
I did the first part successfully:
$$w^{12}=1= \cos 2\pi + i \sin 2\pi$$
$$w= \cos \frac{\pi k}{6} + i \sin \frac{\pi k}{6}$$
Where $k=0,1,2,3,4,5,6,7,8,9,10,11$
I struggled with this for a long time, I know that the two real roots are $z=2$ and $z=-2$
But how do I prove it and show it can be expressed as that. Please help.
Im new to these types of questions
I only know how to do this when its something like:
$$w^4=1$$
Where there is a the number $1$ or $-1$ and one sided equation. Because then I can write
$$w^4-1=0$$
$$(w-1)(w+1)(w^2+1)=0$$
| By writing the second equation as
$$\Bigl(\frac{z+2}{z}\Bigr)^{12}=1$$
it reduces to the first. Therefore the solutions are
$$\frac{z+2}{z}=\cos\frac{k\pi}{6}+i\sin\frac{k\pi}{6}\tag{$*$}$$
for $k=0$ or $k=6$ or $k=\pm1,\pm2,\ldots,\pm5$. Now $k=0$ gives
$$\frac{z+2}{z}=1$$
which is impossible, and $k=6$ gives
$$\frac{z+2}{z}=-1$$
which has the real solution $z=-1$. So this leaves ten non-real solutions. You can find them from $(*)$ and do a bit of simplification using trig functions in order to arrive at the given answers. Specifically, solve for $z$, rationalise the denominator and use half-angle formulae:
$$\eqalign{z
&=-\frac{2}{1-(\cos\frac{k\pi}{6}+i\sin\frac{k\pi}{6})}\cr
&=-2\frac{(1-\cos\frac{k\pi}{6})+i\sin\frac{k\pi}{6}}
{(1-\cos\frac{k\pi}{6})^2+(\sin\frac{k\pi}{6})^2}\cr
&=\cdots\cr
&=-\frac{(1-\cos\frac{k\pi}{6})+i\sin\frac{k\pi}{6}}{1-\cos\frac{k\pi}{6}}
\cr
&=-1-i\frac{2\sin\frac{k\pi}{12}\cos\frac{k\pi}{12}}{2\sin^2\frac{k\pi}{12}}
\ .\cr}$$
In fact, if you are familiar with complex exponentials you can write $(*)$ as
$$\frac{z+2}{z}=e^{k\pi i/6}$$
and the algebra becomes much easier.
Comment. Worth a thought: including the real solution, we have $11$ solutions. Can you explain why this is so, when your original equation involved polynomials of degree $12$? Why are there not $12$ solutions?
| {
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"url": "https://math.stackexchange.com/questions/996253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}$ I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.
$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$
$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$
$$\begin{align*}
(1)&\stackrel{(2)}{\iff} \frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\dots+\frac{2}{x(x+1)}=\frac{2011}{2013}\\\\
&\iff 2\left (\frac12 -\frac13+\frac13-\frac14+\frac14-\dots+\frac{1}{x}-\frac{1}{x+1}\right )=\frac{2011}{2013}\\\\
&\iff 1-\frac{2}{x+1}=\frac{2011}{2013}\\\\
&\iff x=2012
\end{align*}$$
| It surely isn't significantly different, but it can be done as follows:
The first step is the same:
$$
\sum_{n=1}^{x} n=\frac{x(x+1)}{2}=\binom{x+1}{2}\implies \frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\sum_{n=2}^{x}\frac{1}{\binom{n+1}{2}}=\sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}}
$$
Now, take a look at the following integral:
$$
\Delta(a,b)=\int_{0}^1 \left(1-t^{a}\right)^bdt
$$
With the substitution $t\to t^{\frac{1}{a}}$ we get:
$$
\Delta(a,b)=\int_{0}^1 \frac{1}{a}t^{\frac{1}{a}-1}\left(1-t\right)^b dt=\frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Gamma(b+1)}{\Gamma\left(\frac{1}{a}+b+1\right)}=\frac{\Gamma\left(\frac{1}{a}+1\right)\Gamma(b+1)}{\Gamma\left(\frac{1}{a}+b+1\right)}
$$
For $a>0$ and $b>-1$. Now, setting $a=\frac{1}{2}$ and $b=n$ we get:
$$
\Delta\left(\frac{1}{2},n\right)=\int_{0}^1 \left(1-\sqrt{t}
\right)^n dt=\frac{\Gamma\left(2+1\right)\Gamma(n+1)}{\Gamma\left(2+n+1\right)}=\frac{2!\cdot n!}{(n+2)!}=\frac{1}{\binom{n+2}{2}}
$$
And therefore:
$$
\sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}}=\sum_{n=1}^{x-1}\left(\int_{0}^1 \left(1-\sqrt{t}
\right)^n dt\right)=\int_{0}^1 \sum_{n=1}^{x-1}\left(1-\sqrt{t}
\right)^n dt=\int_{0}^1 \left(1-\sqrt{t}
\right)\frac{1-\left(1-\sqrt{t}
\right)^{x-1}}{1-\left(1-\sqrt{t}
\right)} dt=\int_{0}^1 \frac{\left(1-\sqrt{t}
\right)-\left(1-\sqrt{t}
\right)^{x}}{\sqrt{t}} dt
$$
Now, by making the substitution $t\to(1-t)^2$ we obtain:
$$
\sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}}=\int_{1}^0 -2(1-t)\frac{\left(1-(1-t)
\right)-\left(1-(1-t)
\right)^{x}}{(1-t)} dt=2\int_{0}^1 t-t^{x} dt=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=1-\frac{2}{x+1}
$$
And now we can determine $x$ as you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Finding a double integral $\int_1^\infty\int_0^\infty\frac{1}{(x^3+y^3)^3}\mathrm{d}x\ \mathrm{d}y=\frac{10\pi}{189\sqrt3}$ How do we prove that $$\int_1^\infty\int_0^\infty\dfrac{1}{(x^3+y^3)^3}\mathrm{d}x\ \mathrm{d}y=\dfrac{10\pi}{189\sqrt3}$$
I tried to expand and use partial fraction, but in vain. I don't have a clue what to do now. Please help me out. Thank you.
Please avoid using complex analysis, as I am not familiar with it.
| First, $$\int_0^\infty \frac{dx}{(x^3+y^3)^3} = \frac{1}{y^9}\int_0^\infty \frac{dx}{((x/y)^3+1)^3}= \frac{1}{y^8}\int_0^\infty \frac{dt}{(t^3+1)^3}= \frac{1}{y^8}\int_0^\infty \frac{\frac13u^{-2/3}du}{(u+1)^3}=$$
$$= \frac{1}{3y^8}\mathrm{B}\left(\frac13,3-\frac13\right)= \frac{1}{3y^8}\frac{5}{3}\frac{2}{3}\frac{\pi}{\sin\frac{\pi}{3}}=\frac{10\pi}{27\sqrt3}\frac{1}{y^8}.$$
Since $$\int_1^\infty\frac{dy}{y^8}=\frac17$$
we get the desired result $$\frac{10\pi}{189\sqrt3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
$ab$ divides $3^a+1$ and $3^b+1$ Find all positive integers $a,b$ such that $ab$ divides $3^a+1$ and $3^b+1$.
It is clear that $3$ cannot divide either $a$ or $b$, because $3$ doesn't divide $3^a+1$ or $3^b+1$.
$(a,b)=(1,1),(2,1),(1,2)$ all work. If $b=1$, the condition reduces to $a$ divides $3^a+1$ and $4$, so $a=1,2,4$, but this has already been covered.
[Source: Korean competition problem]
| Here is an unfinished idea. Maybe it's useful, maybe not - I'll post it and anyone is welcome to try to finish or adapt it.
Of course, a number $n$ divides both $x$ and $y$ if and only if $n$ divides $\gcd(x,y)$.
Assume without loss of generality that $b\geq a$, and write $b=qa+r$ for $0\leq r<a$. Observe that
$$\begin{align*}
\gcd(3^a+1,3^b+1)=\gcd(3^a+1,3^b-3^a)&=\gcd(3^a+1,3^{b-a}-1)\\
=\gcd(3^a+1,3^{b-a}+3^a)&=\gcd(3^a+1,3^{b-2a}+1)\\
&=\cdots\\
&=\gcd(3^a+1,3^r+(-1)^q)
\end{align*}$$
In fact by making an analogous argument, we always have
$$\gcd(3^a\pm1,3^b\pm 1)=\gcd(3^a\pm1,3^r\pm 1)\qquad \text{(signs not intended to match)}$$
Therefore, we can conclude
$$\gcd(3^a\pm 1,3^b\pm 1)=\gcd(3^a\pm1,3^{\gcd(a,b)}\pm 1)=\gcd(3^{\gcd(a,b)}\pm1,3^{\gcd(a,b)}\pm1)\\
=\begin{cases}
3^{\gcd(a,b)}\pm 1 & \text{if signs match},\\
2 & \text{if signs don't match}
\end{cases}$$
Suppose that $\gcd(3^a+1,3^b+1)=3^{\gcd(a,b)}\pm 1$. Then to say that this is divisible by $ab$ means
$$3^{\gcd(a,b)}\equiv\pm1\bmod ab$$
Thus, the order of $3$ in $(\mathbb{Z}/ab\mathbb{Z})^\times$ divides $2\gcd(a,b)$. Of course it also must divide $\varphi(ab)$... but that's not a contradiction yet.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
A Five Equations problem? If,
$$\begin{align*}
y+u+x+v&=0\\
z+y+v+u&=1\\
x+y+z+u&=5\\
z+u+v+x&=2\\
v+x+y+z&=4\,,
\end{align*}$$
What is the value of $xyzuv$?
| $$\begin{align*}
y+u+x+v&=0\tag1\\
z+y+v+u&=1\tag2\\
x+y+z+u&=5\tag3\\
z+u+v+x&=2\tag4\\
v+x+y+z&=4\tag5
\end{align*}$$
Adding these five equations, we get:
$$\begin{align*}
4(x+y+z+u+v)&=12\\
x+y+z+u+v&=3\tag{6}
\end{align*}$$
$$
\begin{align}
(6) - (1) &\implies z=3\\
(6) - (2) &\implies x=2\\
(6) - (3) &\implies v=-2\\
(6) - (4) &\implies y=1\\
(6) - (5) &\implies u=-1
\end{align}
$$
So, from these results,
$$
x\cdot y\cdot z \cdot u\cdot v = 2\times 1 \times 3 \times (-1) \times (-2)$$
$$
\therefore \boxed{\ x\,y\,z\,u\,v = 12\ }
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Proving this formula $1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt2$ I tried to prove this formula but I couldn't do.
$$1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt{2}$$
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large 1
+\sum_{n\ =\ 0}^{\infty}{1 \over \pi\pars{2n + 3/4}\pars{2n + 5/4}}}
=1 + {1 \over 4\pi}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{n + 3/8}\pars{n + 5/8}}
\\[5mm]&=1 + {1 \over 4\pi}\bracks{\Psi\pars{3/8} - \Psi\pars{5/8} \over 3/8 - 5/8}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function $\color{#000}{\bf 6.3.1}$.
Then
\begin{align}&\color{#66f}{\large 1
+\sum_{n\ =\ 0}^{\infty}{1 \over \pi\pars{2n + 3/4}\pars{2n + 5/4}}}
=1 + {1 \over \pi}\bracks{\Psi\pars{5 \over 8} - \Psi\pars{3 \over 8}}
\end{align}
With
Euler Reflection Formula $\color{#000}{\bf 6.3.7}$
$\ds{\pars{~\Psi\pars{1 - z} = \Psi\pars{z} + \pi\cot\pars{\pi z}~}}$ we'll get:
\begin{align}&\color{#66f}{\large 1
+\sum_{n\ =\ 0}^{\infty}{1 \over \pi\pars{2n + 3/4}\pars{2n + 5/4}}}
=1 + {1 \over \pi}\bracks{\pi\cot\pars{\pi\,{3 \over 8}}}
=1 + \cot\pars{3\pi \over 8}
\\[5mm]&=1 + {1 \over \tan\pars{135^{\circ}/2}}
=1 + {1 \over 1/\pars{\root{2} - 1}} =\color{#66f}{\large\root{2}}\,.\quad
{\tt\pars{~\mbox{See a proof below}~}}.
\end{align}
Also,
\begin{align}
-1&=\tan\pars{135^{\circ}}
= {2\tan\pars{135^{\circ} /2} \over 1 - \tan^{2}\pars{135^{\circ} /2}}
\quad\imp\quad
\tan^{2}\pars{135^{\circ} \over 2} - 2\tan\pars{135^{\circ} \over 2} - 1 = 0
\\[5mm]&\imp\quad\tan\pars{135^{\circ} \over 2}
={2 + \sqrt{8} \over 2} = 1 + \root{2} = {1 \over \root{2} - 1}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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recursive to explicit sequence I am trying to find the explicit formula for the following recursion:
$$a_{1}=3,\quad a_{n}=3- \frac{1}{a_{n-1}},\quad n \in \mathbb N,n>1$$
I tried in many ways but I cannot find any solution.
Can you help me?
| $$\dfrac{a_n - p}{a_n - q} = \dfrac{3- \dfrac{1}{a_{n-1}} - p}{3- \dfrac{1}{a_{n-1}} - q} = \dfrac{(3-p)a_{n-1} - 1}{(3-q)a_{n-1}-1} = \dfrac{3-p}{3-q}\dfrac{a_{n-1} -\dfrac{1}{3-p}}{a_{n-1} -\dfrac{1}{3-q}}$$
Take $p = \frac{1}{3-p}$ and $q = \frac{1}{3-q}$, i.e. take $p,q$ as the two different solutions of $x^2 - 3x + 1 = 0$
we get $$\dfrac{a_n - p}{a_n - q} = \left(\dfrac{3-p}{3-q}\right)^{n-1}\dfrac{a_1 - p}{a_1 - q}$$
Now solve for $a_n$ as function of $p,q,n$ and plug in the values of $p$ and $q$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate the integral $\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$? $$\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$$
The question is ask me to evaluate the integral but I have no idea how to start?
If there are any formulas required for this question, can you please list them ?
Thank you for any help!
| Try this
$$ t^2 = x^2 - 9, \quad t\,dt = x\,dx $$
so
$$ \int \frac{\sqrt{x^2-9}}{x^3}\,dx = \int \frac{\sqrt{x^2-9}}{x^4} x\,dx \\
= \int \frac{t}{(t^2+9)^2} t\,dt \\= \int t \frac{t}{(t^2+9)^2} dt $$
Now do integration by parts
$$ u = t, \quad dv = \frac{t}{(t^2+9)^2}\,dt $$
$$ du = dt, \quad v = -\frac{1}{2(t^2+9)}$$
$$ \int \frac{t^2}{(t^2+9)^2} dt = -\frac{t}{2(t^2+9)} + \int \frac{1}{2(t^2+9)}\,dt \\
= -\frac{t}{2(t^2+9)} + \frac{1}{2\sqrt{3}} \arctan{\frac{t}{\sqrt{3}}} + C \\
= -\frac{\sqrt{x^2-9}}{2x^2} + \frac{1}{2\sqrt{3}} \arctan{\frac{\sqrt{x^2-9}}{\sqrt{3}}} + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Unable to find the sum of a series I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + \frac{7}{9}$$
I also see that
$$a_2 = \frac{1}{9^2} + \frac{7^2}{9^2}$$
I know that in a geometric series the first term is $a$ and the second term is $ar$.
This allows me to see that
$$\left(\frac{1}{9}+\frac{7}{9}\right)r=\frac{1}{9^2}+\frac{7^2}{9^2}$$
which when solved for $r$ gives the value $\frac{25}{36}$.
Using the formula to find the sum of a geometric series $\frac{a}{1-r}$, I find that the sum is equal to $\frac{32}{11}$.
But this value is incorrect and the sum is actually $\frac{29}{8}$. How does one find that value?
| $$
\frac{\frac19}{1-\frac19}+\frac{\frac79}{1-\frac79}=\frac18+\frac72=\frac{29}{8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int\frac{1}{1+x^6} \,dx$ I came across following problem
Evaluate $$\int\frac{1}{1+x^6} \,dx$$
When I asked my teacher for hint he said first evaluate
$$\int\frac{1}{1+x^4} \,dx$$
I've tried to factorize $1+x^6$ as
$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing
$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps
I've also tried to reverse engineer the solution given by Wolfram Alpha
And I need to have terms similar to
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?
Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?
| With $1+x^6= (1+x^2)(x^4-x^2+1)$, decompose the integrand
\begin{align}
& \int \frac{dx}{1+x^6}
=\frac13\int \left( \frac{1}{1+x^2}+\frac12\frac{x^2+1}{x^4-x^2+1}- \frac32\frac{x^2-1}{x^4-x^2+1}\right)dx \\
&\hspace{15mm}=\frac13\int \frac{dx}{1+x^2}+\frac16\int \frac{d(x-\frac1x)}{(x-\frac1x)^2+1}dx - \frac12\int \frac{d(x+\frac1x)}{(x+\frac1x)^2-3} dx \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Proving $\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a.....}}}}-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a....}}}}=1$ I checked many values of this inequality, but I don't have the complete proving.
$$\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a.....}}}}-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a....}}}}=1$$ if $$a>1$$
| this is
$$
\lim U_n - \lim u_n
$$
where
$$
U_{n+1 } = \sqrt{a + U_n}
\\
u_{n+1 } = \sqrt{a - u_n}
$$
Provided that both limits exists, there are solutions of
$$
U^2 = a + U\\
u^2 = a - u
$$
the part $b^2 - 4ac $ is the same, so it remains only the $-\frac b {2a}$ part of both quadratic equations, giving
$$
U - u = \frac 12 - \left(-\frac 12\right) = 1
$$
Another 'proof':
\begin{align} \delta &=
\sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}- \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}
\\&=
\frac{a + \sqrt{a + \sqrt{a +\cdots}} - \left[a - \sqrt{a - \sqrt{a -\cdots}}\right]}
{\sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}+ \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}}
\\&=
\frac{ \sqrt{a + \sqrt{a +\cdots}} + \sqrt{a - \sqrt{a -\cdots}}}
{\sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}+ \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}}
\\&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Check my answer -ordinary differential equation I'm trying to solve the ODE $x^2y'+2xy-y^3=0$ or in other words $y'+\frac{2}{x}y=\frac{1}{x^2}y^3$
According to wikipedia (at least in hebrew), this is what's called a Bernoulli equation, that we can solve by using the transform $z=y^{1-3}=y^{-2}$ and then our differential equation will be $z'-\frac{4}{x}z=-\frac{2}{x^2}$
Note: The exact phrasing is - if $y'+p(x)y=q(x)y^n$, then use the transformation $z=y^{1-n}$ and you will get the linear equation $z'+(1-n)p(x)z=(1-n)q(x)$. It is not immediately apparent to me why this is true.
At any rate, I used the transformation, define $u=e^{\int \frac{-4}{x}dx}=e^{-4 \ln x}=x^{-4}$
$uz=\frac{z}{x^4}\int -\frac{2}{x^6}dx=-2\int x^{-6}=\frac{2}{5}x^{-5}+C$ and so $z=\frac{2}{5}x^{-1}+Cx^4$
Since $z=\frac{1}{y^2}$, we get that $y^2=\frac{1}{z}$ and so $y=\frac{1}{\sqrt{\frac{2}{5}x^{-1}+Cx^4}}$ or $y=-\frac{1}{\sqrt{\frac{2}{5}x^{-1}+Cx^4}}$
Is this correct? If so, why is the thing in the "note" correct? If it isn't, where did I go wrong?
| Your solution looks correct. Also, your equation for $z$ looks consistent with the 'note':
$$z' + (1-3)\frac{2}{x} z = (1-3)\frac{1}{x^2}$$
Added:
If $z = y^{1-n}$, then $z' = (1-n)y^{-n}y'$.
Thus
$$y' + p(x)y = q(x)y^n$$
$$\Leftrightarrow (1-n)y^{-n} [ y' + p(x)y ] = (1-n)y^{-n} [q(x)y^n]$$
$$\Leftrightarrow (1-n)y^{-n}y' + (1-n)p(x)y^{1-n} = (1-n)q(x)$$
$$\Leftrightarrow z' + (1-n)p(x)z = (1-n)q(x)$$
Personally, I never use this last form for Bernoulli equations. I just work through them starting with the substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arithmetic of a combinations formula I am trying to study, and I'm not quite sure how:
$$ \binom{5}{3} \cdot \binom{7}{3} = 350 $$
From my understanding the formula is
$$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $$
Therefore:
$$ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 $$
and
$$ \binom{7}{3} = \frac{7!}{3!(7-3)!} = 35 $$
Please help me understand, thank you.
Answer: $ 10\cdot 35 = 350$.
| $$\binom53=\frac{5!}{3!(5-3)!}=\frac{5!}{3!2!}=\frac{1\cdot2\cdot3\cdot4\cdot5}{1\cdot2\cdot3\cdot1\cdot2}=\frac{4\cdot5}{1\cdot2}=\frac{20}2=10$$
$$\binom73=\frac{7!}{3!(7-3)!}=\frac{7!}{3!4!}=\frac{7!}{4!3!}=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}{1\cdot2\cdot3\cdot4\cdot1\cdot2\cdot3}=\frac{5\cdot6\cdot7}{1\cdot2\cdot3}=\frac{5\cdot6\cdot7}6=5\cdot7=35$$
$$\binom53\binom73=10\cdot35=350$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Understanding $ \frac {e^x}{x^2} $ as $ x \to -\infty $ Given the limit of this function
$$
\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x} =
\begin{bmatrix}
\frac {\infty}{\infty}
\end{bmatrix} = \lim_{x\to -\infty} \frac {e^x (1 + \frac{x^2}{e^x} + \frac{1}{e^x})} {x^2(1 + \frac{1}{x} \sin x)}
$$
Since $ \lim_{x\to -\infty} e^x = 0$ and $ \lim_{x\to -\infty} x^2 = +\infty$ , I'd give the result $ \frac{0}{+\infty} = 0$ but it is not correct.
What am I mistaking? Have I made any wrong assumption?
| Arithmetic of limits:
$$\lim_{x\to -\infty}\frac{e^x}{x^2}=\lim_{x\to-\infty}e^x\cdot\lim_{x\to -\infty}\frac1{x^2}=0\cdot 0= 0$$
And about the original limit:
$$\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x}=\lim_{x\to -\infty}\frac{1+\frac{e^x}{x^2}+\frac1{x^2}}{1+\frac{\sin x}x}=\frac{1+0+0}{1+0}=1$$
Observe that in the limit, $\;x^2\;$ rules, not $\;e^x\;$ .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove complex equation Prove the following $$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$$
for any integer n greater than 0. DO NOT USE ....
I believe that I can use mathematical induction. Any help would be greatly appreciated.
| And here's a proof by induction.
It is true for $n=0$
when it is
$\frac1{1-z}
=
\frac1{1-z}
$.
Suppose it is true for $n$.
We want to show that
it is true for $n+1$,
or that
$\frac{1}{z-1}*\frac{1}{z^{n+1}}
= \dfrac{1}{z-1} - \sum_{k=1}^{n+1}\frac{1}{z^k}
$.
Using the induction hypothesis that
$\frac{1}{z-1}*\frac{1}{z^{n}}
= \dfrac{1}{z-1} - \sum_{k=1}^{n}\frac{1}{z^k}
$
$\begin{array}\\
\dfrac{1}{z-1} - \sum_{k=1}^{n+1}\frac{1}{z^k}
&=\dfrac{1}{z-1} - \left(\sum_{k=1}^{n}\frac{1}{z^k}+\frac1{z^{n+1}}\right)\\
&=\left(\dfrac{1}{z-1} - \sum_{k=1}^{n}\frac{1}{z^k}\right)-\frac1{z^{n+1}}\\
&=\left(\frac{1}{z-1}*\frac{1}{z^{n}}\right)-\frac1{z^{n+1}}
\text{ (induction hypothesis used here)}
\\
&=\frac{z-(z-1)}{(z-1)z^{n+1}}\\
&=\frac{1}{(z-1)z^{n+1}}\\
\end{array}
$
and the result is true for $n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help me evaluate this equation in Scientific Notation How do i solve this using Scientific Notation?
$$ \frac{3\times 10^8}{4\times 10^{-5}} $$
I have been trying for hours now! Help please
| First note that
$$ \frac{x^a}{x^{-b}}= \frac{x^a}{\frac{1}{x^{b}}}=x^ax^b =x^{a+b}, \quad x\not =0$$
So now
$$ \frac{3\times 10^8}{4\times 10^{-5}}= \frac{3}{4}\times \frac{10^8}{10^{-5}}= \frac{3}{4}\times 10^8\times 10^5 $$
$$= 0.75\times 10^{8+5}= 0.75\times 10^{13}=7.5\times 10^{12}$$
| {
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"url": "https://math.stackexchange.com/questions/1016977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.
|AB| = a, P is midpoint of segment AB
| Let $|AB| = a$, $|BC|=a$, $|CA|=a.$ By Pythag we know $|PC|=\sqrt{a^2 - (\frac{a}{2})^2}$ and therefore the area of the equilateral triangle $ABC$ is $\frac{a\left(\sqrt{a^2 - (\frac{a}{2})^2}\right)}{2}$. Now if $ABC$ is an equilateral triangle then the angle between $AB$ and $BC$ is $\pi/3$.
So we work out the area between the edge of the circle and the triangle $ABC$, there are 2 of these but they are equal. To do this we find the area of one of the segments of the circle $AB$, $BC$ which is $\frac{\pi a^2}{6}$ and then minus the area of the equilateral triangle, so we get:
\begin{equation}
X=\frac{\pi a^2}{6}-\frac{a\left(\sqrt{a^2 -(\frac{a}{2})^2}\right)}{2}
\end{equation}
So the total area in the $ABC$ area (the blue and the white circles combined) is:
\begin{equation}
\begin{aligned}
Y=& \frac{a\left(\sqrt{a^2 - (\frac{a}{2})^2}\right)}{2} +2X \\ =&
\frac{\pi a^2}{3} - \frac{a\left(\sqrt{a^2 -(\frac{a}{2})^2}\right)}{2}
\end{aligned}
\end{equation}
Now we have that area we just need to minus the area of the white circles inside it. The two semi-cicles on the line $AB$ are of same size and therefore we can just minus the area of the full circle equal to $\frac{\pi a^2}{16}$. Now we just need the area of the full circle at the top, call it $Z$. Then the blue area is equal to $Y - \frac{\pi a^2}{16} - Z$. Unfortunately at the moment I can't work out the area $Z$ but I thought i'd still share the rest of this answer and hopefully you can get it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding a closed form of $\cos^n(x)$ in terms of $\cos$
By definition we have $$\cos(x):= \frac{1}{2}\left(e^{ix}+e^{-ix} \right) $$
Using this definition and applying the binomial expansion to the right hand side I would get $$\cos^n(x)= \left(\frac{1}{2}\right)^n\sum_{k=0}^n\left(\begin{matrix} n \\ k\end{matrix}\right) e^{ixk} e^{(-ix)(n-k)} = \left(\frac{1}{2} \right)^n \sum_{k=0}^n\left(\begin{matrix} n \\ k\end{matrix}\right) e^{(2k-n)ix}$$
which looks promising, but I can't continue from there, apparently to this Wikipedia entry here there is a much more elegant form. I do of course with the above already obtain the $\cos(2k-n)$ part but there is still the $i\sin(2k-n)$ part present. Namely: $$\cos^n(x) = \frac{1}{2^n}\sum_{k=0}^n \left(\begin{matrix} n \\ k\end{matrix}\right)\left(\cos((2k-n)x)+i \sin((2k-n)x)\right) $$
It feels like an invalid argument to say that $\cos^n(x)$ is real and therefore the $i \sin(2k-n)$ part will vanish.
| For the power of trigonometric function we have the following identities:
\begin{eqnarray}
\cos^{2n}x &=& \frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos[2(n-k)x] \tag 1\\
\cos^{2n+1}x &=& \frac{1}{4^n}\sum_{k=0}^{n}\binom{2n+1}{k}\cos[(2n+1-2k)x]\tag 2\\
\sin^{2n}x &=& \frac{1}{2^{2n}}\binom{2n}{n}+\frac{(-1)^n}{2^{2n-1}}\sum_{k=0}^{n}(-1)^k\binom{2n}{k}\cos[2(n-k)x] \tag 3
\\
\sin^{2n+1}x &=& \frac{(-1)^n}{4^n}\sum_{k=0}^{n-1}\binom{2n+1}{k}\sin[(2n+1-2k)x] \tag 4\\
\end{eqnarray}
We prove the identity (1). We have
$$
\cos x=\frac{(\cos x+i\sin x)+(\cos x-i\sin x)}{2}.
$$
Put $\cos x+i\sin x=z$ so that $\cos x-i\sin x=z^{-1}$. Then we have
$$
\cos^{2n}x=\left (\frac{z+z^{-1}}{2}\right)^{2n}=\frac{1}{2^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}z^{2n-k}.
$$
Further
$$
2^{2n}\cos^{2n}x=\sum_{k=0}^{n-1}\binom{2n}{k}z^{2(n-k)}+\binom{2n}{n}+\sum_{k=n+1}^{2n}\binom{2n}{k}z^{2(n-k)}.
$$
In the second sum put $n-k=-(n-k')$. Then this sum is rewritten as
$$
\sum_{k'=n-1}^{0}\binom{2n}{2n-k'}z^{-2(n-k')}=\sum_{k=0}^{n-1}\binom{2n}{k}z^{-2(n-k)}
$$
and so
$$
2^{2n}\cos^{2n}x=\sum_{k=0}^{n-1}\binom{2n}{k}\left[z^{2(n-k)}+z^{-2(n-k)}\right]+\binom{2n}{n}.
$$
However, $z^{2(n-k)}+z^{-2(n-k)}=2\cos[2(n-k)x]$, and therefore
$$
\cos^{2n}x =\frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos[2(n-k)x].
$$
Replacing in (1) $x$ by $\frac{\pi}{2}-x$, we get formula (2). Formulas (3) and (4) are deduced as (1) and (2).
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I prove $\pi ^2=\sum_{n=0}^{\infty }\frac{1}{(2n+1+\frac{a}{3})^2}+\frac{1}{(2n+1-\frac{a}{3})^2}$ Proving this formula
$$
\pi^{2}
=\sum_{n\ =\ 0}^{\infty}\left[\,{1 \over \left(\,2n + 1 + a/3\,\right)^{2}}
+{1 \over \left(\, 2n + 1 - a/3\,\right)^{2}}\,\right]
$$
if $a$ an even integer number so that
$$
a \geq 4\quad\mbox{and}\quad{\rm gcd}\left(\,a,3\,\right) = 1
$$
| Here is another solution. Noting
$$\sum_{n=-\infty}^\infty f(n)= -\sum_{j=1}^k \operatorname*{Res}_{z=j}\pi \cot (\pi z)f(z) $$
we have
\begin{eqnarray}
&&\sum_{n=0}^{\infty}\left[{1 \over \left(2n + 1 + a/3\right)^{2}}
+{1 \over \left(2n + 1 - a/3\right)^{2}}\right]\\
&=&\sum_{n=-\infty}^{\infty}{1 \over \left(2n + 1 + a/3\right)^{2}}=\frac{1}{4}\sum_{n=-\infty}^{\infty}{1 \over \left(n + \frac{a+3}{6}\right)^{2}}\\
&=&-\frac14\text{Res}\left(\frac{\pi\cot(\pi z)}{\left(z + \frac{a+3}{6}\right)^{2}},z=-\frac{a+3}{6}\right)\\
&=&\frac{\pi^2}{4}(1+\tan^2(\frac{a\pi}{6}))\\
&=&\pi^2.
\end{eqnarray}
This becase $a$ is even and $\gcd(a,3)=1$ and $\tan(\frac{a\pi}{3})=\pm\sqrt 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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problematic limit with squares I have problem with calculation such limit : $\displaystyle \lim_{n\to \infty} {\frac{10^{\sqrt{n+1}-\sqrt{n}}-1}{2^{\frac{1}{\sqrt{n}}}-1}}$
I only know the answer from wolfram that it's $\frac{1}{2}+\frac{\ln{5}}{\ln{4}}$
| We need to understand that both $$x = \sqrt{n + 1} - \sqrt{n} = \frac{1}{\sqrt{n + 1} + \sqrt{n}}$$ and $y = 1/\sqrt{n}$ tend to $0$ as $n \to \infty$ and we aslo know that $$\lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a$$ for $a > 0$. Hence we have $$\begin{aligned}l &= \lim_{n \to \infty}\frac{10^{\sqrt{n + 1} - \sqrt{n}} - 1}{2^{1/\sqrt{n}} - 1}\\
&= \lim_{n \to \infty}\frac{10^{\sqrt{n + 1} - \sqrt{n}} - 1}{\sqrt{n + 1} - \sqrt{n}}\cdot\frac{\sqrt{n + 1} - \sqrt{n}}{1/\sqrt{n}}\cdot\frac{1/\sqrt{n}}{2^{1/\sqrt{n}} - 1}\\
&= \lim_{x \to 0^{+}}\frac{10^{x} - 1}{x}\cdot\lim_{n \to \infty}\frac{\sqrt{n + 1} - \sqrt{n}}{1/\sqrt{n}}\cdot\lim_{y \to 0^{+}}\frac{y}{2^{y} - 1}\\
&= \frac{\log 10}{\log 2}\cdot\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{n + 1} + \sqrt{n}}\\
&= \frac{\log 10}{\log 2}\cdot\lim_{n \to \infty}\dfrac{1}{\sqrt{1 + \dfrac{1}{n}} + 1}\\
&= \frac{\log 10}{2\log 2}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of fractions with square roots inequality What is the greatest integer $n$ such that
$n \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{2014}}$?
| For positive integer $k$, we have $$2\sqrt{k} < \sqrt{k+1} + \sqrt{k} < 2\sqrt{k+1}$$
Taking reciprocals and rationalising the middle term, we get
$$\frac1{2\sqrt{k+1}}< \sqrt{k+1}-\sqrt k < \frac1{2\sqrt k}$$
The middle term telescopes and hence sums easily. Thus we can have tight bounds on both sides:
$$\sum_{k=1}^{2014}\frac1{\sqrt{k}} = 1+\sum_{k=1}^{2013}\frac1{\sqrt{k+1}} < 1+2\sum_{k=1}^{2013} (\sqrt{k+1}-\sqrt{k})= 2\sqrt{2014}-1 \approx 88.75$$ and
$$\sum_{k=1}^{2014} \frac1{\sqrt k} > 1+\frac1{\sqrt2}+2\sum_{k=3}^{2014}(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{2015} - 2\sqrt3+1+\frac1{\sqrt2} \approx 88.02 $$
so you have your integer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Laplace Transform solution verification: $\ddot{y} + 2y = 2e^t\implies \frac13\cos(\sqrt{2}t)-\frac{2}{3\sqrt{2}}\sin(\sqrt{2}t)+\frac23e^t\,\text{?}$ Does
$$\ddot{y} + 2y = 2e^t\quad y(0)=1,\dot{y}(0)=0$$
Give
$$\frac13\cos(\sqrt{2}t)-\frac{2}{3\sqrt{2}}\sin(\sqrt{2}t)+\frac23e^t\,\text{?}$$
This is what I have got, and it seems to go back and forth just fine. I just wanted a quick verification. Thank you very much sirs, and madams.
| I am not sure if you used partial fractions and table to solve the problem so I will post a solution using the inverse Mellin transform.
The Laplace transform of the ODE is
\begin{align}
s^2Y(s) - sy(0) - \dot{y}(0) + 2Y(s) &= 2\int_0^{\infty}e^{t}e^{-st}dt\\
Y(s)(s^2 + 2) - s &= \frac{2}{s-1}\\
Y(s) &= \frac{2}{(s-1)(s^2+2)}+\frac{s}{s^2+2}
\end{align}
Then by the inverse Mellin transform, we have that the poles are at $s=\pm \sqrt{2}i,1$
\begin{align}
y(t) &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\biggl[\frac{2e^{st}}{(s-1)(s^2+2)}+\frac{se^{st}}{s^2+2}\biggr]ds\\
&=\sum\text{Res}\\
&=\lim_{s\to 1}(s-1)\frac{2e^{st}}{(s-1)(s^2+2)}+\lim_{s\to \sqrt{2}i}(s-\sqrt{2}i)\frac{2e^{st}}{(s-1)(s^2+2)}\\
&+\lim_{s\to -\sqrt{2}i}(s+\sqrt{2}i)\frac{2e^{st}}{(s-1)(s^2+2)} + \lim_{s\to \sqrt{2}i}(s-\sqrt{2}i)\frac{se^{st}}{s^2+2}\\
&+\lim_{s\to -\sqrt{2}i}(s+\sqrt{2}i)\frac{se^{st}}{s^2+2}\\
&= \frac{2e^t}{3}-\frac{2}{3}\cos(t\sqrt{2})-\frac{\sqrt{2}}{3}\sin(t\sqrt{2})+\cos(t\sqrt{2})\\
&=\frac{2e^t}{3}+\frac{1}{3}\cos(t\sqrt{2})-\frac{\sqrt{2}}{3}\sin(t\sqrt{2})
\end{align}
which is of no surprise the solution you obtained.
| {
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Functional Equation: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $(x+y)(f(x)-f(y))=(x-y)f(x+y)$ Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$
My attempt:
If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.
Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon>0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$.
Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!
| $f(x)$ must be of the form $ax^2 + bx$.
Letting $x = 1$ and $y = 0$ in the equation, we find $f(0) = 0$.
Now define $g(x) = f(x)/x$ for $x \ne 0$. Taking $f(0) = 0$ for granted, the functional equation can be rewritten as
$$(x-y)g(x + y) = xg(x) - yg(y), \qquad x, y, x+y \ne 0.$$
Substituting $1$ for $y$, we find
$$(x-1)g(x+1) = xg(x) - g(1), \qquad x \ne -1, 0.$$
Now substituting $x + 1$ for $x$ and $-1$ for $y$, we find
$$
\begin{align*}
(x+2)g(x) &= (x+1)g(x+1) + g(-1) &\quad \text{($x \ne -1,0$)}\\
(x-1)(x+2)g(x) &= (x+1)(x-1)g(x+1) + g(-1)(x-1) \\
(x^2 +x - 2)g(x) &= (x+1)[xg(x) - g(1)] + g(-1)(x-1) \\
-2g(x) &=-g(1)(x+1) +g(-1)(x-1).
\end{align*}
$$
The last relation is in fact true for all $x \ne 0$, including $x = -1$.
This proves that the function $g(x)$ is linear, say $g(x) = ax + b$. Thus $f(x) = xg(x) = ax^2 + bx$. The relation $f(x) = ax^2 + bx$ is valid for all $x$, including $x = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Verifying $\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$
Verify: $$\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$$
My solution:
$$
\begin{align}\sec^2x+\tan^2x&=\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\cdot\frac{1-\sin^2x}{1-\sin^2x}\\
&=\frac{1-\sin^4x}{\cos^2x\cdot\cos^2x}\\
&=\frac{1-\sin^4x}{\cos^4x}\\
&=\frac{1}{\cos^4x}-\frac{\sin^4x}{\cos^4x}\\
&=\sec^4x-\sin^4x\sec^4x\\
&=\sec^4x(1-\sin^4x)\\
\end{align}$$
Is it incorrect to multiply in $1-\sin^2x$ like in the fourth equality?
| It's okay because $1-\sin^2 x=\cos^2 x$ so it can only become $0$ if $\cos x$ is zero, wich can be excluded due to the domain of the equality, thus no division by zero can occur and you may proceed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$
$$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$
Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?
| Multiply all equations by the product of the denominators ($(a-\omega)(b-\omega)(c-\omega)$ in (1), $(a-\omega^2)(b-\omega^2)(c-\omega^2)$ in (2) etc.), to get rid of the fractions.
The first two equations then simplify to (remember $\omega^3=1$):
$$ 0=(ab+ac+bc)+(2abc-1)\omega^2 $$
$$ 0=(ab+ac+bc)+(2abc-1)\omega $$
Because $\omega\neq\omega^2$, these can only both be true if $2abc-1=0$. In that case both equations become
$$ab+ac+bc=0.$$
If I replace the questionmark in your equation (3) by $X$, the equation becomes (after multiplying with the three terms):
$$X=\frac{(b+1)(c+1)+(a+1)(c+1)+(a+1)(b+1)}{(a+1)(b+1)(c+1)}$$
Expanded:
$$X=\frac{ab+bc+ac+2a+2b+2c+3}{abc+ab+ac+bc+a+b+c+1}$$
Use the fact that $ab+ac+bc=0$ and $2abc-1=2$:
$$X=\frac{2a+2b+2c+3}{a+b+c+\frac{3}{2}}=2$$
There must be a nicer way to show this, considering the nice symmetry in the problem, but I can't see that now.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding a polynomial mod 5: What did they do here? I'm reading through this solution for the problem below.
The given solution is this:
What I don't quite understand is how they got say from $\frac{1}{3}(X-2)(X-4)$ to be 'equal' to $2(X-2)(X-4)$.
How did they go from one step to the other? I am not quite understanding this.
| All my arithmetic is modulo 5:
$5(X-2)(X-4)=0$
Any of ${1,2,3,4}$ when multiplied by $3$ is non-zero, so
$\frac{5}{3} (X-2)(X-4)=0 \Rightarrow$
$ \frac{6}{3} (X-2)(X-4)- \frac{1}{3} (X-2)(X-4)=0$
Hence $ \frac{1}{3} (X-2)(X-4)= 2(X-2)(X-4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof by Induction Divisibility. $6^n-5n+4$ is divisible by 5 for all positive integers $n$.
$n >=1$
Prove By Induction
My attempt is as follows:
$n=1$
$6^1-5(1) +4$
$=5$, Therefore 5 is divisible by 5 so $n=1$ is true
Assume its true for $n=k$
consider $n=k+1$
$6^k-5k+4=5.x$
I am stuck here would appreciate some assistance.
| Assume $6^n+5n+4$ is a multiple of $5$, so
$$
6^n+5n+4=5x
$$
for some integer $x$.
Now you want to do the inductive step; since $6^{n+1}$ seems to be the toughest term, we isolate $6^n$ from the previous identity:
$$
6^n=5x-5n-4
$$
and recall that $6^{n+1}=6\cdot6^n$; then
\begin{align}
6^{n+1}+5(n+1)+4
&=6\cdot6^n+5(n+1)+4\\
&=6(5x-5n-4)+5(n+1)+4\\
&=5(6x-6n+n+1)-24+4\\
&=5(6x-5n+1-4)\\
&=5(6x-5n-3)
\end{align}
is a multiple of $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of obtaining a double six in at least two throws The question: A pair of fair dice is thrown 10 times. What is the probability of obtaining a double six in at least two throws?
My attempt: Let X denote the total number of double sixes obtained. On any throw, the probability of obtaining a double six is $\frac 1{36}$ and thus, the probability of not obtaining a double six is $\frac {35}{36}$.
$P(X=1) = 10(\frac 1{36})(\frac {35}{36})^9$ since the double six can be obtained in any of the 10 throws.
$P(X=0) = (\frac {35}{36})^{10}$
So $P(X \ge 2) = 1 - P(X=1) - P(X=0) = 1 - (\frac {10}{36})(\frac{35}{36})^9 - (\frac{35}{36})^{10}$
However, the school's solution says that $P(X \ge 2) = 1 - (\frac {1}{36})(\frac{35}{36})^9 - (\frac{35}{36})^{10}$
Which is correct? And why?
| Your answer is correct. The school’s solution is in error.
$\begin{align}
\mathsf P(X\geq 2) & = 1 - \mathsf P(X\leq 1) & \text{By the Law of Complements}
\\ & = 1 - {10\choose 0}p^0(1-p)^{10} - {10\choose 1}p^1(1-p)^9 & \text{because it is a binomial distribution}
\\ & = 1- \frac{35^{10}}{36^{10}}-\frac{10\cdot 35^9}{36^{10}}
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx$
Calculate the definite integral
$$
I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx
$$
given that $a>b>0$.
My Attempt:
If we replace $x$ by $C$, then
$$
I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC
$$
Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives
$$
\begin{align}
\cos C &= \frac{a^2+b^2-c^2}{2ab}\\
a^2+b^2-2ab\cos C &= c^2
\end{align}
$$
From here we can use the formula $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ to transform the integral to
$$
\begin{align}
I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC
\end{align}
$$
Is my process right? If not, how can I calculate the above integral?
| Using the formula $$a^2+b^2-ab\cos C = c^2\;,$$ Now If $C\leftrightarrow x\;,$ Then $a^2+b^2-2ab\cos x= c^2$
So Integral Convert into $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\mathrm dC\;,$$ Now Using Sin formula $\displaystyle \frac{\sin C}{c} = \frac{\sin A}{a}.$
So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}\mathrm dC\;,$$ Now $A+B+C = \pi\;,$ Then $\mathrm dC = 0-\mathrm dA-\mathrm dB$
Now when $C\rightarrow 0,$ Then $A\rightarrow \pi$ and $B\rightarrow 0$
When $C\rightarrow 0,$ Then $A\rightarrow 0$ and $B\rightarrow 0$
So $$\displaystyle I = \int_{\pi}^{0}\frac{\sin^2 A}{a^2}(-\mathrm dA)+\int_{0}^{0}\frac{\sin^2 A}{a^2}(-\mathrm dB)$$
So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}\mathrm dA = \frac{2}{2a^2}\int_{0}^{\frac{\pi}{2}}\left[1-\cos 2A\right]\mathrm dA = \frac{\pi}{2a^2}$$
which agrees with the other solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this diophantine equation $x^2+y^2+z^3=n$ always have integer solution
show that: For any postive ineteger $n$,then the equation
$$n=x^2+y^2+z^3$$
always have integer solution
My idea: such as $n=1$,then we have
$$1=0^2+0^2+1^3$$
$$2=0^2+1^2+1^3$$
$$3=1^2+1^2+1^3$$
$$4=2^2+0^2+0^3$$
$$5=1^2+2^2+0^3$$
$$6=1^2+2^2+1^3$$
$$7=2^2+2^2+(-1)^3$$
$$8=0^2+0^2+2^3$$
$$9=1^2+0^2+2^3$$
$$10=1^2+1^2+2^3$$
$\cdots\cdots\cdots$
But for general $n$, How prove it?
| Here’s a proof using a solution I found here.
Write $n$ in the form $8^m\cdot s$, for an integer $m\ge0$ and with $s$ not divisible by $8$. This can always be done. Note that an integer $s$ that is not a multiple of $8$ can be written in one of the following three forms: $2k+1$ (if $s$ is odd), $4k+2$ (if $s$ is even, but not a multiple of $4$), or $8k+4$ (if $s$ is even and a multiple of $4$).
First, find an expression for $s$ in the form $a^2+b^2+c^3$ as follows. (I haven’t checked these details.)
If $s=2k+1$, let $a=k^2-k-1$, $b=k^3-3k^2+k$, and $c=-k^2+2k$.
If $s=4k+2$, let $a=2k^3-2k^2-k$, $b=2k^3-4k^2-k+1$, and $c=-2k^2+2k+1$
If $s=8k+4$, let $a=k^2-2k-1$, $b=k^3+k+2$, and $c=-k^2-1$
Now observe that if $s$ is the sum of two squares and a cube, so is $8s$, because $8(a^2+b^2+c^3)=(2a+2b)^2+(2a-2b)^2+(2c)^3$. Inductively, if $s$ is the sum of two squares and a cube, then so is $8^ms$ for any nonnegative integer $m$.
We have already shown that $s$ is the sum of two squares and a cube, so $n=8^ms$ is as well, completing the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$ I realize that is a basic math problem, but I am still having problems with it.
The expression $$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$$ equals one of the following:
*
*$2\sqrt{3}-1$
*$\dfrac{6}{11}$
*$4\sqrt{3}$
*$12$
How do I simplify this?
| $$\begin{align}\frac{\sqrt3}{2\sqrt3+1}+\frac{\sqrt3}{11}&=\frac{\sqrt3(2\sqrt3-1)}{(2\sqrt3+1)(2\sqrt3-1)}+\frac{\sqrt3}{11}\\&=\frac{6-\sqrt3}{11}+\frac{\sqrt3}{11}\\&=\frac{6}{11}\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1039261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Modulo: Calculating without calculator?? Calculate the modulo operations given below (without the usage of a calculator):
$101 \times 98 \mod 17 =$
$7^5 \mod 15 =$
$12^8 \mod 7 =$
$3524 \mod 63 =$
$−3524 \mod 63 =$
Ok with calculator I have no problem with it. However, I need to learn how can I compute them without calculator.
| By definition of modular arithmetic, $3524 \pmod{63}$ is the remainder when $3524$ is divided by $63$.
To find $-3524 \pmod{63}$, multiply your answer for $3524 \pmod{63}$ by $-1$. If you want a positive residue, add $63$ to this result.
For the product $101 \cdot 98 \mod{17}$, use the theorem that if $a \equiv b \pmod{n}$ and $c \equiv d \pmod{n}$, then $ac \equiv bd \pmod{n}$.
Since $101 = 5 \cdot 17 + 1$, $101 \equiv 16 \pmod{17}$. Since $98 = 5 \cdot 17 + 13$, $98 \equiv 13 \pmod{17}$. Thus,
$$101 \cdot 98 \equiv 16 \cdot 13 \equiv 208 \equiv 4 \pmod{17}$$
since $208 = 12 \cdot 17 + 4$.
However, you can simplify the calculations further if you use residues with absolute value at most $n/2$.
Since $101 = 6 \cdot 17 - 1$, $101 \equiv -1 \pmod{17}$. Since $98 = 6 \cdot 17 - 4$, $98 \equiv -4 \pmod{17}$. Thus,
$$101 \cdot 98 \equiv -1 \cdot -4 \equiv 4 \pmod{17}$$
which agrees with our previous result.
For $12^8 \pmod{7}$, observe that $12 \equiv 5 \pmod{7}$, so $12^8 \equiv 5^8 \pmod{7}$.
If $p$ is a prime modulus and $k \neq 0$, then $k^{p - 1} \equiv 1 \pmod{p}$. Hence, $5^6 \equiv 1 \pmod{7}$, so
$$12^8 \equiv 5^8 \equiv 5^65^2 \equiv 5^2 \equiv 4 \pmod{7}$$
For $7^5 \pmod{15}$, reduce modulo $15$ after you calculate each power. For instance,
$$7^2 \equiv 49 \equiv 4 \pmod{15}$$
so
$$7^3 \equiv 7 \cdot 7^2 \equiv 7 \cdot 4 \equiv 28 \equiv -2 \pmod{15}$$
Since you know the residues of $7^2 \pmod{15}$ and $7^3 \pmod{15}$, you can multiply their residues to find the residue of $7^5 = 7^2 \cdot 7^3$ modulo $15$. If you want a positive residue, add a suitable multiple of $15$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$? How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$
I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result.
| $$\int \frac{dx}{x^{2}\sqrt{x^2-1}}\\
y=\frac{1}{x}\Rightarrow dy=-\frac{dx}{x^{2}},\sqrt{x^{2}-1}=\sqrt{\frac{1}{y^{2}}-1}=\frac{\sqrt{1-y^{2}}}{y}\\
\int \frac{dx}{x^{2}\sqrt{x^2-1}}=\frac{1}{2}\int \frac{-2ydy}{\sqrt{1-y^{2}}}
$$$$=\frac{\sqrt{1-y^{2}}}{2}+c\\
\int_{1}^{\infty }\frac{dx}{x^{2}\sqrt{x^2-1}}=1\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 8
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How many words can be written with $aabbbccdd$ such that no two equal letters are adjacent? I'm trying to count this using the principle if inclusion-exclusion. I've done the following:
*
*Counting the number of permutations of $aabbbccdd$.
*
*$9!$
*
*Counting the number of permutations of $aabbbccdd$ without repetitions.
*$\displaystyle\frac{9!}{2!3!2!2!}$
*
*Now I guess I need need to count the number of permutations in which all letters are together which is $4!$.
*$\displaystyle\frac{9!}{2!3!2!2!}-4!$
*
*Now counting the number of permutations in which $a,c,d$ are adjacent, I guess that this number is $\displaystyle \frac{3\cdot 7!}{3!2!2!}$.
*$\displaystyle\frac{9!}{2!3!2!2!}-4!-\frac{3\cdot 7!}{3!2!2!}$
*
*Now counting the number of permutations in which we have permutations of 2 $b$'s together and then the number of permutations which have $3$ $b$'s together, I have $3\cdot 2/2!$, then $\displaystyle\frac{3 \cdot 7!}{2!2!2!}+\frac{6!}{2!2!2!}$
*$\displaystyle\frac{9!}{2!3!2!2!}-4!-\frac{3\cdot 7!}{3!2!2!}-\frac{3 \cdot 7!}{2!2!2!}-\frac{6!}{2!2!2!}$
*
*Now I know that are intersections in there. But I'm unable to think about something to count them, I guess that I should add the number of intersections to the result:
*$\displaystyle \frac{9!}{2!3!2!2!}-\left( 4!-\frac{3\cdot 7!}{3!2!2!}-\frac{3 \cdot 7!}{2!2!2!}-\frac{6!}{2!2!2!}+ \cap_n \right)$
Is this correct? If so, how can I count the intersections? I guess that the actual formula would be expressed as:
$$\displaystyle \frac{9!}{2!3!2!2!} \cap (4! \cup \frac{3\cdot 7!}{3!2!2!} \cup \frac{3 \cdot 7!}{2!2!2!} \cup \frac{6!}{2!2!2!})$$
| We begin by writing arbitrary words containing only $a$, $a$, $c$, $c$, $d$, $d$.
There are ${6!\over2!2!2!}=90$ such words .
$6$ of these words contain the forbidden pairs $a^2$, $c^2$, and $d^2$. We then need the three $b$'s to separate these pairs. Makes $6$.
There are ${4!\over2!}=12$ words containing $a^2$, $c^2$, and $d$, $d$ in any order. In $6$ of these the $d$'s are paired as well. It follows that there are $3\cdot(12-6)=18$ words containing exactly two forbidden pairs. For each such word we need two $b$'s to separate the forbidden pairs, and there are $5$ slots left for the third $b$. Makes $18\cdot 5=90$.
There are ${5!\over2!2!}=30$ words containing $a^2$, and $c$, $c$, $d$, $d$ in any order. $2\cdot6$ of these words contain exactly one of $c^2$ and $d^2$, and another $6$ contain both $c^2$ and $d^2$. It follows that there are $3\cdot(30-12-6)=36$ words containing exactly one forbidden pair. We need one $b$ to separate this pair, and there are $6$ slots left for the remaining two $b$'s. Makes $36\cdot{6\choose2}=540$.
There are $90-6-18-36=30$ words containing no forbidden pair. This allows for $7$ slots where the $b$'s can be written in. Makes $30\cdot{7\choose3}=1050$.
Adding up the obtained counts gives a grand total of $$N=1686$$ allowed words from $a$, $a$, $b$, $b$, $b$, $c$, $c$, $d$, $d$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding a generalized form for this series While i was just playing around with series i came across this one,
$$
S = \sum_{k=1}^\infty[\frac{k}{k-\frac{1}{2}}+\frac{k-\frac{1}{2}}{k}-\frac{k+\frac{1}{2}}{k} - \frac{k}{k-\frac{1}{2}}]
$$
Which simplified nicely into
$$
S=\sum_{k=1}^{\infty}\frac{1}{4k^3-k} = \ln(4)-1
$$
Then i wondered the results would be if i changed the $\frac{1}{2}$ into $\frac{1}{n}$
I simplified the initial summation to
$$
S_n = \sum_{k=1}^{\infty}\frac{2}{n^3 k^3 - nk}
$$
Then i found a few of the results of the summations,
$S_2 = \ln(4)-1$
$S_3 = \ln(3)-1$
$S_4 = \frac{\ln(8)}{2}-1$
$S_5 = \ln(\frac{5}{8} + \frac{\sqrt{5}}{8})(\sqrt{5}+1)-\sqrt{5}\ln(\frac{5}{8}-\frac{\sqrt{5}}{8}) + \frac{1}{10}[\ln(\frac{5}{8}-\frac{\sqrt{5}}{8}) + 4\ln(10)-10]$
$S_6 = \frac{\ln(432)}{6}-1$
My interest is piqued and i would love to see a generalized form of what $S_n$ would yield and how one would go about finding this.
| Thanks to Raymond Manzoni's answer, i have found the generalized form
Since
$$\psi(1+x) = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k(k+x)}$$
$$
\psi(1+\frac{1}{n})+\psi(1-\frac{1}{n})+2\gamma = \sum_{k=1}^{\infty}\frac{1}{nk(k+\frac{1}{n})}-\frac{1}{nk(k-\frac{1}{n})}
$$
$$
= \sum_{k=1}^{\infty}\frac{-2k}{n^2k^4-k^2} = \sum_{k=1}^{\infty}\frac{-2}{n^2 k^3-k} \therefore
$$
$$
S_n=\sum_{k=1}^{\infty}\frac{2}{n^3k^3-nk} = -\frac{\psi(1+\frac{1}{n})+\psi(1-\frac{1}{n})+2\gamma}{n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Generating sequence I have these two sequences $ a,b $
$$
\begin{array}{c|lcr}
n & \text{a} & \text{b} \\
\hline
1 & 1 & 0 \\
2 & 2 & 1 \\
3 & 2 & 0 \\
4 & 3 & 2 \\
5 & 3 & 1 \\
6 & 3 & 0\\
7 & 4 & 3 \\
8 & 4 & 2 \\
9 & 4 & 1 \\
10 & 4 & 0 \\
11 & 5 & 4 \\
\vdots & \vdots & \vdots
\end{array}
$$
what should I do to write a function using $a$ and $b$ to get $n$? Can you please show me some direction? Thanks.
| $$n = {a+1\choose2} - b$$
How do we get that?
Given $a$, we conclude
$$
1+2+\cdots+a \geq n > 1+2+\cdots+(a-1)
$$
where
$$
1+2+\cdots+a = {a+1\choose2}
$$
by a "well known" formula. Now, looking at $b$, we see that all we have to do is subtract $b$ from ${a+1\choose2}.$
There you are.
| {
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How to prove the matrix inequality $\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}+\det{(B)}|$ Let $A,B\in M_{2}(R)$ be two square matrices so that
$AB=BA$, prove the inequality
$$\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}|$$
I only know prove this
$$\sqrt{a^2-ab+b^2}+\sqrt{a^2+ab+b^2}\ge |a+b|,a,b\in R$$
because
$$\sqrt{x}+\sqrt{y}\ge\sqrt{x+y}$$
so
$$LHS\ge\sqrt{2(a^2+b^2)}\ge (a+b)$$
But for matrices I can't prove it. Thank you
| I'm late to the party and I haven't time to cook up a full solution, but here are some observations.
As $A$ commutes with $B+\epsilon I$ for every real $\epsilon$, by continuity, it suffices to prove the inequality for invertible $B$. Let $X=B^{-1}A$. Divide both sides of the inequality by $|\det(B)|$, the inequality becomes
$$
\sqrt{\det{(X^2-X+I)}}+\sqrt{\det{(X^2+X+I)}}\ge |\det{(X)}+1|.
$$
When $X$ has two real eigenvalues $x,y$, the inequality becomes
$$
\sqrt{(x^2 - x + 1)(y^2 - y + 1)} + \sqrt{(x^2 + x + 1)(y^2 + y + 1)}
\ge |1 + xy|.\tag{1}
$$
When $X$ has a pair of conjugate complex eigenvalues $z$ and $\bar{z}$, the inequality becomes
$$
|z^2 - z + 1| + |z^2 + z + 1| \ge 1 + |z|^2.\tag{2}
$$
Hence the problem boils down to proving $(1)$ and $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Why four roots to this equation: $(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$
$$(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$$
I figured the roots are $0$, $1$, $-1$, and $9$. But why?
| Consider this system of polynomial equations:
$$\begin{align}
a+b+c &= 2 \\
a^3 &= 7x + 1 \\
b^3 &= 8 + x - x^2 \\
c^3 &= x^2 - 8 x - 1
\end{align}$$
Using the method of resultants (via the oh-so-convenient Resultant[] function in Mathematica), we can eliminate $a$, $b$, $c$ to get
$$x^3\;(x-1)^6\;(x+1)^3\;(x - 9 )^3 \;=\; 0$$
Therefore, the only candidate values for $x$ are, as you have previously determined, $0$, $1$, $-1$, and $9$. Sometimes the method of resultants introduces extraneous factors, so we should double-check. As it turns out (assuming the $1/3$ power indicates the real cube root of a real number), each of the four candidates is in fact a valid solution. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Solving recurrence relations? How do I solve $$ a_{n} = a_{n-1} + n, a_{0} = 1$$??
I solved for n=1 thru n=5:
1: 2 = a0 + 1
2: 4 = a0 + 1 + 2 = a0 + 3
3: 7 = a0 + 3 + 3 = a0 + 6
4: 11 = a0 + 6 + 4 = a0 + 10
5: 16 = a0 + 10 + 5 = a0 + 15
I'm failing to see THE pattern. I see that for each iteration from n=1, the number you add to a_0 (=1) increments by one from the last iteration in the sequence... like n=2, it's the a_0 + 1 we started with + 2 makes a_0 + 3, and for n=3, we have a_0 + 6 which is the last one + 3, for n=4 it's +4, etc.... but that's still reliant upon previous iterations, not a direct solution without having to actually iterate through them. How do I extract the solution? Isn't there an actual WAY to go about it?
| One clear pattern is that the right-hand side of your equations is always $a_0$
plus some number. The numbers to add to $a_0$ follow the sequence
$1, 3, 6, 10, 15, 21, \ldots .$
Now, many sequences can be analyzed by finding out something about the rate
at which the terms increase (the percentage increase, not the value added each time).
For the Fibonacci sequence, for example, we find that the ratio of terms approaches
$\tfrac12(1+\sqrt5)$ (the "golden ratio") as the sequence continues to infinity.
If you look at the rate at which the numbers increase in this sequence, you may notice that the rate is actually slowing down.
In fact,
$$\frac31 = 3,\ \frac63 = 2,\ \frac{10}{6} = \frac53,
\ \frac{15}{10} = \frac32,\ \frac{21}{15} = \frac75,$$
and so forth.
You may notice that the ratios steadily decrease; the pattern is a
little more obvious if we do not reduce the fractions quite to the lowest terms:
$$\frac31 = \frac31,\ \frac63 = \frac42,\ \frac{10}{6} = \frac53,
\ \frac{15}{10} = \frac64,\ \frac{21}{15} = \frac75.$$
That is,
$$\frac{a_n - a_0}{a_{n-1} - a_0} = \frac{n+1}{n-1}.$$
But the value of $a_n - a_0$ is just the cumulative effect of all these increases:
$$\begin{eqnarray}
a_n - a_0 &=& (a_1 - a_0) \cdot
\frac{a_2 - a_0}{a_1 - a_0} \cdot
\frac{a_3 - a_0}{a_2 - a_0} \cdot
\frac{a_4 - a_0}{a_2 - a_0} \cdot \cdots \cdot
\frac{a_{n-1} - a_0}{a_{n-2} - a_0} \cdot
\frac{a_n - a_0}{a_{n-1} - a_0} \\
&=& 1 \cdot \frac31 \cdot \frac42 \cdot \frac53 \cdot \cdots
\cdot \frac{n-1}{n-3} \cdot \frac{n}{n-2} \cdot \frac{n+1}{n-1}.
\end{eqnarray}$$
Now notice that each numerator from $3$ through $n-1$ inclusive (that is, all but the
last two) cancels the denominator of the fraction after the next in this product.
That is, all but the first two denominators end up being canceled.
We end up with
$$ a_n - a_0 = 1 \cdot \frac11 \cdot \frac12 \cdot {n} \cdot (n+1)
= \frac{n(n+1)}{2}. $$
That was a very unconventional way of looking at this sequence, but when we come to
a new problem that we are completely unfamiliar with, we may try many different things.
The way people usually solve a sequence like this (for the first time)
is more like the following:
Sometimes it is useful not to combine the numbers too much. So instead of writing
$$\begin{eqnarray}
a_1 &=& a_0 + 1, \\
a_2 &=& a_0 + 1 + 2 = a_0 + 3, \\
a_3 &=& a_0 + 3 + 3 = a_0 + 6, \\
a_4 &=& a_0 + 6 + 4 = a_0 + 10,
\end{eqnarray}$$
and so forth, you could write
$$\begin{eqnarray}
a_1 &=& a_0 + 1, \\
a_2 &=& a_0 + 1 + 2, \\
a_3 &=& a_0 + 1 + 2 + 3, \\
a_4 &=& a_0 + 1 + 2 + 3 + 4,
\end{eqnarray}$$
and so forth. A pattern then emerges:
$$ a_n = a_0 + 1 + 2 + 3 + \cdots + n. $$
Now there is a famous story that says a 10-year-old boy named Karl Friedrich Gauss
came up with a trick for adding these numbers quickly.
A variation of Gauss's method is to write the sum twice, but reverse the order
of the numeric terms on the second writing:
$$\begin{eqnarray}
a_n &=& a_0 + 1 + 2 + 3 + \cdots + n, \\
a_n &=& a_0 + n + (n - 1) + (n - 2) + \cdots + 1.
\end{eqnarray}$$
Now we can add the two values on the left side, and add the two equal values
on the right side term by term, and obtain a new equation,
$$\begin{eqnarray}
2a_n &=& 2a_0 + (n + 1) + ((n - 1) + 2) + ((n - 2) + 3) + \cdots + (1 + n) \\
&=& 2a_0 + \underbrace{(n+1) + (n+1) + (n+1) + \cdots + (n+1)}_{n} \\
&=& 2a_0 + n(n+1).
\end{eqnarray}$$
It is easy to find $a_n$ for any $n$ if you provide the value
of $a_0$ in this equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiation with the quotient rule I have the question:
Given that $$y=\frac{e^x}{\sqrt{1+2x}}$$
Show that $$\frac{dy}{dx} = \frac{2xe^x}{\sqrt{(1+2x)^3}}$$
I've done the question but I got $2xe^x(\sqrt{(1+2x)^3})$. I feel like this is too similar to the sheet's answer, so am I wrong or is the sheet printed wrong?
Working:
Let $u = e^x$, $v = 1+2x$, $w = \sqrt{v}$.
Using the chain rule, $$\frac{dw}{dx} = \frac{dw}{dv} * \frac{dv}{dx}$$
So $$\frac{dw}{dx} = 2(\frac{1}{2}v^{-\frac{1}{2}}) = \frac{1}{\sqrt{v}} = \frac{1}{\sqrt{1+2x}}$$
And $\frac{du}{dx}$ is clearly $e^x$.
Therefore, using the quotient rule, $$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})- e^x(\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\sqrt{1+2x}-\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\frac{2x}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{\frac{2xe^x}{\sqrt{1+2x}}}{(1+2x)^2}$$
Using the rule
$$\frac{x^m}{x^n} = x^{m-n}$$
We get
$$\frac{2xe^x}{(1+2x)^{-\frac{3}{2}}}$$
$$=2xe^x(\sqrt{1+2x})^3 $$
Have I done this correctly?
| Using Quotient rule:
$$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})-e^x(\frac{1}{\sqrt{1+2x}})}{1+2x}.$$ You can simplify now.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Can an explicit formula be found for a bijection $f \colon \mathbb{N} \to \mathbb{Q}$? Let $\mathbb{N}$ be the set of all natural numbers, and let $\mathbb{Q}$ denote the set of all rational numbers.
Then of course there is a bijection $f \colon \mathbb{N} \to \mathbb{Q}$ since $\mathbb{Q}$ is a countably infinite set.
Can we find an explicit formula for $f$?
| If you're looking for an explicit function, here's a sequence that maps $\mathbb{N}$ to $\mathbb{Q}^+$, which is almost what you want. Define $f$ as follows:
$$f(1) = 1, \quad f(n + 1) = \frac{1}{\lfloor f(n) \rfloor + 1 - \{ f(n) \}}$$
where $\lfloor x \rfloor$ is the integer part of $x$, and $\{ x \} = x - \lfloor x \rfloor$ is the fractional part. This creates a sequence that hits every rational number.
To show it's a bijection, we relate this function to a binary tree. Miscellaneous facts needed for the proof:
*
*$f(n) \in \mathbb{Q}^+$
*For any real $x$, $y$, if $\lfloor x \rfloor - \lfloor y \rfloor = \{ x \} - \{ y \}$, then $x = y$.
*The Euclidean algorithm works.
First, we show injectivity. Assume there are some $a$ and $b$ in $\mathbb{N}$, with $a < b$, such that $f(a) = f(b)$. If $a \ne 1$, then:
$$
\begin{align*}
\frac{1}{\lfloor f(a - 1) \rfloor + 1 - \{ f(a - 1) \}} &= \frac{1}{\lfloor f(b - 1) \rfloor + 1 - \{ f(b - 1) \}} \\
\lfloor f(a - 1) \rfloor - \{ f(a - 1) \} &= \lfloor f(b - 1) \rfloor - \{ f(b - 1) \} \\
\lfloor f(a - 1) \rfloor - \lfloor f(b - 1) \rfloor &= \{ f(a - 1) \} - \{ f(b - 1) \} \\
f(a - 1) &= f(b - 1)
\end{align*}
$$
Thus, we can descend until we hit $1$. There is some $n > 1$ such that $f(n) = f(1) = 1$. Then $\frac{1}{\lfloor f(n - 1) \rfloor + 1 - \{ f(n - 1) \}} = 1$, implying $\lfloor f(n - 1) \rfloor = \{ f(n - 1) \}$. Because the former is in $\mathbb{N} \cup \{ 0 \}$, and the latter is in $[0, 1)$, both sides must be zero, forcing $f(n - 1) = 0 \notin \mathbb{Q}^+$.
To show surjectivity, we introduce the Stern-Brocot tree. It's a binary tree with root $1$, and if a node has value $\frac{a}{b}$, its left and right children have values $\frac{a}{a+b}$ and $\frac{a + b}{b}$, respectively. If we want to find $\frac{p}{q}$ in the tree, and $p > q$, we look for $\frac{p - q}{q}$, which will have a right child of $\frac{p}{q}$. Similarly, if $p < q$, we look for $\frac{p}{q - p}$. Since this process always terminates at $\frac{1}{1}$, which is definitely in the tree, it must contain all of $\mathbb{Q}^+$.
How did that help? It turns out that if we index the nodes of the tree (breadth-first, left-to-right), this is exactly $f(n)$. To prove it, we show that it satisfies the same recurrence relations.
Claim: $f(1) = \frac{1}{1}$. Proof: trivial.
Claim: $f(n) = \frac{a}{a + b} \implies f(n + 1) = \frac{a + b}{b}$. Proof: plug-and-chug.
Claim: $f(n) = \frac{a}{b} \implies f(2n) = \frac{a}{a + b}$. We proceed by induction. Checking $n = 1$ is easy. For the recursive case, let $s$ such that $b = aq + a - s$ and $0 \le s < a$.
$$
\begin{align*}
f(n) &= \frac{a}{b} \\
&= \frac{a}{aq + a - s} \\
&= \frac{1}{q + 1 - \frac{s}{a}} \\
\frac{1}{ \lfloor f(n - 1) \rfloor + 1 - \{ f(n - 1) \} } &= \frac{1}{\lfloor q + \frac{s}{a} \rfloor + 1 - \{ q + \frac{s}{a} \} } \\
\end{align*}
$$
Thus, $f(n - 1) = q + \frac{s}{a} = \frac{aq + s}{a}$. (Just copy the steps from the injectivity argument.) Now apply the inductive hypothesis, use the theorem above, and then explicitly compute one more iteration:
$$
\begin{align*}
f(n - 1) &= \frac{aq + s}{a} \\
f(2n - 2) &= \frac{aq + s}{(aq + s) + a} \\
f(2n - 1) &= \frac{(aq + s) + a}{a} = q + 1 + \frac{s}{a} \\
f(2n) &= \frac{1}{ \lfloor f(2n - 1) \rfloor + 1 - \{ f(2n - 1) \} } \\
f(2n) &= \frac{1}{ (q + 1) + 1 - \frac{s}{a} } \\
f(2n) &= \frac{a}{ aq + a + a - s } \\
f(2n) &= \frac{a}{a + b} \\
\end{align*}
$$
So the $i$th node has value $f(i)$, showing that the tree is injective and the function surjective.
| {
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"url": "https://math.stackexchange.com/questions/1052517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that there exists a point $c$ such that $f''(c)\ge 4$
Let $f:[0,1/2]\to\mathbb{R}$ be a twice differentiable function such that $f(0) = 0$ and $f'(0) = 0$. Also $f(1/2)\ge 1/2$. Prove that there exists a $c\in[0,1/2]$ such that $f''(c)\ge4$.
I have tried different approached. By Lagrange's MVT twice we can prove the that there is a $c$ such that $f''(c)\ge 2$.
One of my second approaches involves taking two cases $c\le1/4$ and $c>1/4$. Another approach was to consider $g(x) = f(x) + f(x+1/4)$ or $g(x) = f(x)-f(x-1/4)$.
| We have 3 data points, $f(0) = f(0+\delta) = 0$ and $f(\tfrac{1}{2}) \geq \tfrac{1}{2}$. Using 3 points, we can try to estimate the 2nd derivative.
For any 3 data points, $a \mapsto f(a), b \mapsto f(b), c \mapsto f(c)$, the interpolating quadratic polynomial should be:
$$ f(x) \approx f(a)\frac{(x-b)(b-c)(c-x)}{(a-b)(b-c)(c-a)}
+ f(b)\frac{(a-x)(x-c)(c-a)}{(a-b)(b-c)(c-a)}
+ f(c)\frac{(a-b)(b-x)(x-a)}{(a-b)(b-c)(c-a)}
$$
Notice the cancellation, so we can just cross out the factors in numerator and denominator:
$$ f(x) \approx \frac{(x-b)(c-x)}{(a-b)(c-a)}
+ \frac{(a-x)(x-c)}{(a-b)(b-c)}
+ \frac{(b-x)(x-a)}{(b-c)(c-a)}
$$
This is just for any 3 points, there is a parabola going through it.
https://stackoverflow.com/questions/717762/how-to-calculate-the-vertex-of-a-parabola-given-three-points
In our case: $a=0,b = 0 + \delta, c = \tfrac{1}{2}$:
$$ f(x) \approx
0\cdot \frac{(x-\delta)(\tfrac{1}{2}-x)}{(-\delta )\tfrac{1}{2}}
+ 0\cdot \frac{(-x)(x-\tfrac{1}{2})}{(-\delta)(\delta-\tfrac{1}{2})}
+ (\tfrac{1}{2} + \epsilon) \cdot \frac{(\delta-x)x}{(\delta-\tfrac{1}{2})(\tfrac{1}{2})}
\approx \boxed{(2+\epsilon)x^2 }
$$
with $\epsilon > 0$. In this case the 2nd derivative should be something like: $f''(x) \approx 4 + \epsilon$.
The Mean Value Theorem should tell us this point with $f''(c) = 4$ is a topological feature and can't be removed.
See section 4 of Osculating curves: around the Tait-Kneser Theorem by Ghys, Tabachnikov and Timorin.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integers $a$ for which $x^2-x+a$ divide $x^{13}+x+90$. Find all integers $a$ for which $x^2-x+a$ divide $x^{13}+x+90$.
The answer is $a=2$.
| If $a$ is negative or zero, then the quadratic has two real roots.
But we can easily check that the other polynomial has derivative everywhere positive
and hence only one real root.
So $a$ must be positive.
If $x^2-x+a$ divides $x^{13}+x+90$, then $x^{13}+x+90=f(x)(x^2-x+a)$,
where $f(x)$ is a polynomial with integer coefficients.
Let $x=0$, we see that $a$ must divide $90$. Let $x=1$, we see that it must divide $92.$
Hence it must divide $92-90=2.$ So the only possibilities are $1$ and $2.$
Suppose $a=1$, then putting $x=2$, we have that $3$ divides $2^{13}+92$ but $2^{\text{odd}}$ is congruent to $2 \mod 3,$ so $2^{13}+92$ is congruent to $1 \mod 3.$
So $a$ cannot be $1.$
To see that $a=2$ is possible, we write
$(x^2-x+2)(x^{11}+x^{10}-x^9-3x^8-x^7+5x^6+7x^5-3x^4-17x^3-11x^2+23x+45)=x^{13}+x+90$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a, b, c$ be positive real numbers such that $a + 2b + 3c = 26$ and $a^2 + b^2 + c^2 = 52$. Find the largest possible value of $a$.
I used the Cauchy Schwarz inequality $(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$ as follows:
$a + 2b + 3c = 26$ is given; adding $a$ to both sides gives $2a + 2b + 3c = 26+a$. Then, we have $x = 2$, $y=2$, $z=3$. Putting this all into the inequality, we get:
$(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$
$(2a+2b+3c)^2 \leq (a^2+b^2+c^2)(2^2+2^2+3^2)$
$(26+a)^2 \leq (52)(17)$
$(26+a)^2 \leq 884$
$26 + a \leq \sqrt{884}$
$a \leq 2\sqrt{221} - 26 \approx 3.73$
However, the listed answer gives $a \leq \frac{26}{7} \approx 3.71$. What am I doing wrong?
| The issue is that you are asked to find the least upper bound.
All that you have done is to find an upper bound. You still have yet to show that it is indeed the least upper bound.
In your CS inequality, in order for equality to hold, we must have
$$ \frac{a}{2} = \frac{b}{2} = \frac{ c}{3} , a = 2 \sqrt{221} - 26 $$
However, you can check that the initial conditions would not hold.
| {
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Solving $3x\equiv 4\pmod 7$ I'm trying to learn about linear congruences of the form ax = b(mod m). In my book, it's written that if $\gcd(a, m) = 1$ then there must exist an integer $a'$ which is an inverse of $a \pmod{m}$. I'm trying to solve this example:
$$3x \equiv 4 \pmod 7$$
First I noticed $\gcd(3, 7) = 1$.
Therefore, there must exist an integer which is the multiplicative inverse of $3 \pmod 7$.
According to Bezout's Theorem, if $\gcd(a, m) = 1$ then there are integers $s$ and $t$ such that $sa+tm=1$
where $s$ is the multiplicative inverse of $a\pmod{m}$.
Using that theorem:
$\begin{align}7 = 3\cdot2 +1\\7 - 3\cdot2 = 1 \\-2\cdot3 + 7 = 1\end{align}$
$s=-2$ in the above equation so $-2$ is the inverse of $3 \pmod{7}$.
The book says that the next step to solve $3x \equiv 4 \pmod{7}$ is to multiply $-2$ on both sides.
By doing that I get:
$\begin{align}-2\cdot3x \equiv -2\cdot4 \pmod 7\\-6x\equiv -8 \pmod 7\end{align}$
What should I do after that?
I am working on this problem for hours.
Thanks :)
| $$\begin{align}
3x\equiv4\pmod{7} & (\text{Original equation})\\3x\equiv -3\pmod{7} &(\text{Replaced 4 with -3(by subtracting 7)})\\x\equiv-1\pmod{7}& (\text{Divide each side by 3})\\ x\equiv6\pmod{7} &(\text{replaced -1 with 6 (by adding 7))}
\end{align}$$
P.S.- The reason you can add or subtract $7$ is one of the properties of $\pmod{7}$. You can add or subtract multiples of $7$ to the number in front of the $mod$ without effecting the equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}X = \begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
2 & 1 & 1 \\
\end{pmatrix}
$$
When I try to peform Gaussian elimination to get the reduced form of A, I always get a row of zeroes, e.g:
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}
$$ R_3 - R_1 \to R_3 $$
I get
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
What can I conclude from the fact that I got a zeroes row?
Does this help solving the problem?
| Hint : let there exist $X_{4 \times 3}$ matrix satisfiy the equaltiy then make a contradiction whit entries of $X$
| {
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Finding $\tan \pi/8$ from $\sqrt{1+i}$. I want to prove that $\tan \pi/8 = \sqrt{2} - 1$ using $\sqrt{1+i}$ in some way.
Write: $\sqrt{1+i} = a+bi$, and let's find $a$ and $b$. We have: $$1+i = a^2+b^2 + 2abi,$$ so $a^2+b^2 = 1$ and $ab = 1/2$. This already seems wrong to me, because we know that $|1+i| = \sqrt{2} \implies |\sqrt{1+i}| = \sqrt[4]{2},$ regardless of what root we choose. Let's pretend everything is ok. Substituition gives $$a^2 + \frac{1}{4a^2} = 1 \implies a^4 - a^2 + \frac{1}{4} = 0 \implies a^2 = \frac{1}{2} \implies a = \pm \frac{\sqrt{2}}{2},$$ and $$b = \pm\frac{1}{2\frac{\sqrt{2}}{2}} = \pm\frac{\sqrt{2}}{2} = a.$$
And writing $1+i = re^{i((\pi/4)+2k\pi)}$, we get the roots are given by $\sqrt[4]{2}e^{i((\pi/8)+k\pi)}$.
Now, $z = x+iy \implies \tan \arg(x+iy) = y/x$ would give us $\tan \pi/8 = 1$!
Can someone give me a light here? It must be simple, but I'm failing to make sense of this. Thanks.
Yes, I know better ways to find $\tan \pi/8$. I want to do it this way.
| Another path could start with
$$\displaystyle \tan(2x)=\frac{2\tan x}{1-\tan^2 x},$$
putting $x=\pi/8$, using $\displaystyle \tan(\pi/4)=1$, then solving easily
$$
\frac{2y}{1-y^2 }=1
$$ to get
$$
\displaystyle \tan(\pi/8)=\sqrt{2}-1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\sin\theta +\sin3\theta =0$
Solve the equation by first using a Sum-to-Product Formula.
$$\sin\theta +\sin3\theta =0$$
Steps I took:
$$\begin{align}0&=2\sin\frac { \theta +3\theta }{ 2 } \cos\frac { \theta -3\theta }{ 2 }\\
0&=2\sin\frac { 4\theta }{ 2 } \cos\frac { -2\theta }{ 2 } \\
0&=2\sin2\theta \cos(-\theta)\\
0&=2\sin2\theta \cos\theta\\
0&=2(2\sin\theta \cos\theta )\cos\theta\\
0&=4\sin\theta \cos^2\theta \\
0&=4\sin\theta \frac { 1+\cos2\theta }{ 2 } \\
0&=4\sin\theta \frac { 1+1-2\sin^2\theta }{ 2 } \\
0&=4\sin\theta \frac { 2-2\sin^{ 2 }\theta }{ 2 } \\
0&=\frac { 8\sin\theta -8\sin^{ 2 }\theta }{ 2 }\end{align}$$
| As $\sin(-\theta)=-\sin\theta,$
$$\sin3\theta=-\sin\theta=\sin(-\theta)$$
$$3\theta=m\pi+(-1)^m(-\theta)$$ where $m$ is any integer
Check if $m$ is even $(=2r)$(say) and $m$ is odd $(=2r+1)$(say) one by one
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a formula for all integers $x$ such that $5x-1$ is divisible by $13$ and $19x-12$ is divisible by $23$ Find a formula for all integers $x$ such that $5x-1$ is divisible by $13$ and $19x-12$ is divisible by $23$.
Hello. I am working on a review sheet for my test tomorrow and I am stuck on this question.
I found earlier that all integers $x$ such that $19x-12$ is divisible by $23$ is
$x\equiv 20 \pmod{23}$, but I am not sure if this helps.
| Yes, $19x\equiv 12\pmod{23}$ is equivalent to $x\equiv 20\pmod{23}$. Similarly, $5x\equiv 1\pmod{13}$ is equivalent to $x\equiv 8\pmod{13}$.
So we want to solve the system of congruences $x\equiv 8\pmod{13}$, $x\equiv 20\pmod{23}$. In general to solve systems of linear congruences we use the machinery of the Chinese Remainder Theorem. But for a "small" problem like ours, one can find the answer by playing around. We want $20+23k$ to have remainder $8$ on division by $13$. We can try $k=0$ to $12$, or use more machinery.
We want $7+10k\equiv 8\pmod{13}$ or equivalently $10k\equiv 1\pmod{13}$. The inverse of $10$ modulo $13$ is $4$, so we want $k=4$. Thus the general solution is $20+4\cdot 23\pmod{(13)(23)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
General Solution for a non-Homogeneous Recurrence
What is the general solution to the recurrence:
$$x(n + 2) = x(n + 1) + x(n) + n - 1$$
where $n\geq 1$ with $x(1) = 0, x(2) = 1$?
It's a question on a practice exam I'm reviewing and I'm not quite sure why my solution is incorrect.
So I went through the characteristic equation and found the constants, and the solution to the homogeneous part of this question is:
$$x(n) = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{\sqrt{5}}$$
This is the Fibonacci expression, right? So, what do I do with the non-homogeneous part of the expression? My professor mentioned something about solving them separately but I'm not sure what to do with it now.
| Solution Using Method of Annihilators
Let $E$ denote the "shift" operator, and let $\langle x_n \rangle$ denote the sequence $x_1, x_2, x_3, \ldots$
Then our recurrence can be re-written as:
$$x_{n+2} - x_{n+1} - x_n = n - 1 \iff (E^2-E-1)\langle x_n\rangle = \langle n-1\rangle$$
The annihilator for $n-1$ is $(E-1)^2$, we have:
$$(E-1)^2(E^2-E-1)\langle x_n\rangle = (E-1)^2\langle n-1\rangle=\langle 0\rangle$$
So, we've now converted our problem from a non-homogenous problem to a homogeneous problem; the only difficulty is that we have also changed the recurrence from a $2$nd order recurrence to a $4$th order recurrence. Thus, we need to compute two more initial conditions:
\begin{align}
x_2 &= 1\\
x_3 &= 3
\end{align}
We can easily move from the annihilator operator form to the characteristic of this fourth-order recurrence:
$$(r-1)^2(r^2-r-1) \implies r=1, 1, \frac{1+\sqrt 5}{2}, \frac{1-\sqrt 5}{2}$$
Thus, our general solution has the form:
$$x_n = A + Bn + C\left(\frac{1+\sqrt 5}{2}\right)^n + D\left(\frac{1-\sqrt 5}{2}\right)^n$$
Using our four initial conditions, we find:
\begin{align}
x_1 &= 0 = A + B + \left(\frac{1+\sqrt 5}{2}\right)C + \left(\frac{1-\sqrt 5}{2}\right)D\\
x_2 &= 1 = A + 2B + \left(\frac{1+\sqrt 5}{2}\right)^2C + \left(\frac{1-\sqrt 5}{2}\right)^2D\\
x_3 &= 1 = A + 3B + \left(\frac{1+\sqrt 5}{2}\right)^3C + \left(\frac{1-\sqrt 5}{2}\right)^3D\\
x_4 &= 3 = A + 4B + \left(\frac{1+\sqrt 5}{2}\right)^4C + \left(\frac{1-\sqrt 5}{2}\right)^4D\\
\end{align}
This is a system of linear equations; using any method, we arrive at the solution $A = 0, B=-1, C=1, D = 1$.
Thus, the solution is:
$$x_n =\left(\frac{1+\sqrt 5}{2}\right)^n + \left(\frac{1-\sqrt 5}{2}\right)^n - n$$
For what it's worth, the first several values of the sequence are:
1 | 0
2 | 1
3 | 1
4 | 3
5 | 6
6 | 12
7 | 22
8 | 39
9 | 67
| {
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To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}
Similarly
\begin{equation*}
(\cos\theta + \sec\theta )^2 \ge 4 \tag{2}
\end{equation*}
Adding equation (1) and (2)
\begin{equation*}
(\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8
\end{equation*}
What is wrong?
| Using Cauchy–Schwarz inequality:
$$(\sin x + \csc x)^2+(\cos x + \sec x)^2\\=[(\sin x + \csc x)^2+(\cos x + \sec x)^2][\sin^2 x+ \cos^2 x]\\\geq \mid (\sin x + \csc x)\sin x+(\cos x + \sec x)\cos x \mid^2\\=\mid (\sin^2 x +1)+(\cos^2 x +1) \mid^2\\=\mid 1+1+1 \mid^2=9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the linear space of Eigenvectors associated with a certain Eigenvalue? The following matrix $A$ has $\lambda=2$ and $\lambda=8$ as its eigenvalues
$$ A = \begin{bmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{bmatrix}$$
let $P$ be the orthogonal matrix that diagonalizes $A$. in other words, $A= PDP^T$. you can check that
$$ P = \begin{bmatrix} (-√2)/2 & (-√6)/6 & (-√3)/3 \\ (√2)/2 & (-√6)/6 & (√3)/3 \\ 0 & (√6)/3 & (√3)/3 \end{bmatrix} $$
then the linear space of eigenvectors associated to the eigenvalue $\lambda=2$ is generated by?
*
*The last two columns of the matrix $P$
*The last two rows of matrix $P$
*The first two columns of the matrix $P$
*The first two rows of the matrix $P$
*The last column of the matrix $P$
I Don't know how to solve this, I tried doing it but it got really messy and I didn't get any answer that made sense. If someone could please explain how to do it or how to solve it, that would be great.
Thank You
| Since matrix $P$ is orthogonal ($P^\top = P^{-1}$) one can rewrite $A = PDP^\top$ as
$$
D = P^{-1}A(P^{-1})^\top = P^\top AP
$$
After multiplying one would obtain (there is a typo in your $P$ matrix)
$$
D =
\begin{bmatrix}
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
-\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & \sqrt{\frac{2}{3}} \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}
\end{bmatrix}
\begin{bmatrix}
4 & 2 & 2 \\
2 & 4 & 2 \\
2 & 2 & 4
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
0 & \sqrt{\frac{2}{3}} & \frac{1}{\sqrt{3}}
\end{bmatrix} =
\begin{bmatrix}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 8
\end{bmatrix}
$$
Note that $A = PDP^\top$ could be rewritten as $AP = PD$ which means that $P$ columns are eigenvectors of $A$ and corresponding eigenvalues are on $D$ diagonal:
$$
A\mathbf{p}_i = d_{ii} \mathbf{p}_i, \qquad P = \begin{bmatrix}
\mathbf{p}_1 & \mathbf{p}_2 & \mathbf{p}_3
\end{bmatrix}
$$
Because $d_{11} = d_{22} = 2$ the eigenspace corresponding to $\lambda = 2$ is formed by $\mathbf{p}_1$ and $\mathbf{p}_2$
| {
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Approximate matrix by a rank 2 matrix using singular values I only understand the singular value decomposition process. Do I apply it to this matrix?
\begin{bmatrix} 0 & 0 & \pi \\ 0 & e & 0 \\ 1&0&0 \end{bmatrix}
What is the idea behind matrix approximation using this process?
Applying SVD, we have $\begin{bmatrix} 0 & 0 & \pi \\ 0 & e & 0 \\ 1&0&0 \end{bmatrix}^T\begin{bmatrix} 0 & 0 & \pi \\ 0 & e & 0 \\ 1&0&0 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & e^2 & 0 \\ 0&0&\pi^2 \end{bmatrix}$
How to proceed?
| Decomposition
By agreement, the SVD orders the singular values. The decomposition here is
$$
\mathbf{A}
=
%
\left(
\begin{array}{ccc}
0 & 0 & \pi \\
0 & e & 0 \\
1 & 0 & 0 \\
\end{array}
\right)
%
= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{ccc}
\pi & 0 & 0 \\
0 & e & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{array}
\right).
%
$$
The approximation through order $k$ uses the first $k$ singular values. In your case, $k=2$ and the approximation is
$$
\mathbf{A}_{2}
=
%
\left(
\begin{array}{ccc}
0 & 0 & \pi \\
0 & e & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
%
= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{ccc}
\pi & 0 & 0 \\
0 & e & 0 \\
0 & 0 & \color{red}{0} \\
\end{array}
\right)
%
\left(
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{array}
\right).
%
$$
The red $0$ shows that you have discarded $\sigma_{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$ Find the set of complex numbers $z$ which satisfy
$$\left\lvert\frac{z-3}{z+3}\right\rvert=2\text.$$
I need help on that one. Thank you.
| Take square both sides then we get
$$|z-3|^2=4|z+3|^2.$$
Evaluate it (using the fact $|z|^2=z\overline{z}$ and $2\operatorname{Re}z=z+\overline{z}$) then
$$3|z|^2+30\operatorname{Re}z+27=0$$
Let $z=x+iy$. From above equation, we get
$$x^2+10x+y^2+9=0$$
and is equivalent to $(x+5)^2+y^2=4^2$
Details: From $|z-3|^2=4|z+3|^2$, we get
$$(z-3)\overline{(z-3)}=4(z+3)\overline{(z+3)}$$
and is equivalent to
$$(z-3)(\overline{z}-3)=4(z+3)(\overline{z}+3).$$
Expand the above equality then
$$z\overline{z}-3(z+\overline{z})+9=4(z\overline{z}+3(z+\overline{z})+9)$$
Note that $z\overline{z}=|z|^2$ and $z+\overline{z}=2\operatorname{Re}(z)$ (you can check it easily - take $z=x+iy$ and calculate it) so
$$3|z|^2+30\operatorname{Re}z+27=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An exercise from Knuth's book - Proving a formula by induction I would like to find a formula for this sum:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}
$$
The answer given (Knuth's book, The Art of Computer Programming Volume 1, Third Edition, Section 1.2.1, Exercise 11) is $$ \frac{(-1^n)(n+1)}{4(n+1)^2+1}$$
Please give me an approach or a solution to get this sum.
I tried to solve by pairing the consecutive terms. The last and the second last terms will correspond to $(2n+1)$ and $(2n-1)$ respectively. But this approach couldn't yield much.
The second part is to prove this sum by induction, i.e we've to prove this equation:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4} = \frac{(-1^n)(n+1)}{4(n+1)^2+1}
$$
I approached it like this:
The sum is correct for $k=1$( and $k=0$). (k is a non-negative integer)
Assuming that the sum is correct for $k = 0,1,..,n$
we'll prove that its correct for $k=n+1$.
The sum for $k=n+1$ will be equal to: the sum for k=n + the last term, i.e. $$ \frac{(-1)^n(n+1)}{4(n+1)^2+1} + \frac{(-1)^{n+1}(2(n+1)+1)^3}{(2(n+1)+1)^4+4}$$
This should then be proved equal to $$\frac{(-1)^{n+1}((n+1)+1)}{4((n+1)+1)^2+1}$$
which again I'm not able to do. So, please help in proving this also.
P.S.: Someone please create a 'knuth' or 'taocp' tag for mathematical questions of this book.
|
This is my proof.
We can represent
$$\frac{a^3}{a^4+4}$$ as a $$\frac{a-1}{2((a-1)^2+1) } + \frac{a+1}{2((a+1)^2+1) }$$
After that we easily can proof this two summas.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Probability of ultimate extinction? Need to show that an infinite series is less than $1$ I have the following probability generating function for a branching process -
$$G_n(s) = \frac{n}{n+1} + \sum_{r=1}^{\infty}\frac{n^{r-1}}{(n+1)^{r+1}}s^r$$
It says in a book that extinction is certain for this process. Which implies that the mean of the probability generating function must be $\le1$.
I.e. $G_n'(1) \le 1$
But
$$G_n'(1) = \sum_{r=1}^{\infty}\frac{n^{r-1}}{(n+1)^{r+1}}$$
and I don't see how to show this is less than or equal to $1$?
| If
$G_n'(s)
=\sum_{r=1}^\infty \frac{n^{r-1}}{(n+1)^{r+1}}\frac{d}{dx}\left[s^r\right]
= \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}s^{r-1}
$,
then
$G_n'(1)
= \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}
= \frac1{n(n+1)}\sum_{r=1}^\infty\frac{rn^{r}}{(n+1)^{r}}
= \frac1{n(n+1)}\sum_{r=1}^\infty rz^r
$,
where
$z = \frac{n}{n+1}
$.
Since
$\sum_{r=1}^\infty rz^r
= \frac{z}{(1 - z)^2}
$,
$\begin{array}\\
G_n'(1)
&=\frac1{n(n+1)}\frac{z}{(1 - z)^2}\\
&=\frac1{n(n+1)}\frac{\frac{n}{n+1}}{(1 - \frac{n}{n+1})^2}\\
&=\frac{\frac{1}{(n+1)^2}}{\frac{1}{(n+1)^2}}\\
&=1\\
\end{array}
$
More generally,
$\begin{array}\\
G_n'(s)
&= \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}s^{r-1}\\
&= \frac1{sn(n+1)}\sum_{r=1}^\infty\frac{rn^{r}}{(n+1)^{r}}s^{r}\\
&= \frac1{sn(n+1)}\sum_{r=1}^\infty r\left(\frac{ns}{n+1}\right)^r\\
&= \frac1{sn(n+1)}\sum_{r=1}^\infty rz^r
\quad \text{ where } z=\frac{ns}{n+1}\\
&= \frac1{sn(n+1)}\frac{z}{(1-z)^2}\\
&= \frac1{sn(n+1)}\frac{\frac{ns}{n+1}}{(1-\frac{ns}{n+1})^2}\\
&= \frac1{(n+1)^2}\frac{1}{(1-\frac{ns}{n+1})^2}\\
&= \frac{1}{(n+1-ns)^2}\\
&= \frac{1}{(1+n(1-s))^2}\\
\end{array}
$
Note that if
$1+n(1-s)
= 0$,
or
$s
=1+\frac1{n}
$,
then
$G_n'(s)
= \infty
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove with Cauchy's limit definition ($\epsilon, \delta$) that $\lim_{x \rightarrow 0} \frac{x^2-8}{x-8}=1$ Prove with Cauchy's limit definition ($\epsilon, \delta$) that $$\lim_{x \rightarrow 0} \frac{x^2-8}{x-8}=1$$
Got really troubled with the proper technique of solving this.
Any assistance will be much appreciated!
| For any $\epsilon > 0$, choose $\delta = \text{min}\left(1,\frac{7\epsilon}{2}\right)$, then: if $0 < |x| < \delta$ then $\left|\dfrac{x^2-8}{x-8} - 1\right| = \left|\dfrac{x^2-x}{x-8}\right| \leq \dfrac{|x^2-x|}{8-|x|} < \dfrac{|x^2-x|}{7} < \dfrac{|x^2| + |x|}{7} < \dfrac{|x|+|x|}{7} = \dfrac{2|x|}{7} < \dfrac{2}{7}\cdot \dfrac{7\epsilon}{2} = \epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072421",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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$x^2-y^2=2s$, s cannot be an odd integer How can we prove that if $x^2-y^2=2s$ holds, s cannot be an odd integer. What theorem in number theory should we use?
| You don't need any special theorem, just reason through it:
If $x^2 - y^2$ is even, that means either both $x$ and $y$ are even or they're both odd.
If both $x$ and $y$ are even, let's say $x = 2m$ and $y = 2n$. Then $x^2 - y^2 = 4m^2 - 4n^2 = 4(m^2 - n^2)$ and $s = 2(m^2 - n^2)$, meaning that it's even.
If both $x$ and $y$ are odd, let's say $x = 2m + 1$ and $y = 2n + 1$. Then $x^2 - y^2 = (4m^2 + 4m + 1) - (4n^2 + 4n + 1) = 4(m^2 + m - n^2 - n)$ and $s = 2(m^2 + m - n^2 - n)$, meaning that it's even again.
| {
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"url": "https://math.stackexchange.com/questions/1075128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$ Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution
$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$
$$x^2+1=\frac{u^2-2}{u^2-1}$$
$$2xdx=\frac{-2u}{(u^2-1)^2}$$
$$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}dx}$$
$$\int{\frac{1}{\frac{-1}{u^2-1}u\frac{u^2-2}{u^2-1}}\left(\frac{-u}{(u^2-1)^2}\right)du} $$
$$\int{\frac{1}{(u^2-2)}du} $$
$$\frac{\log \left(\sqrt{2}-x\right)-\log \left(x+\sqrt{2}\right)}{2 \sqrt{2}}$$
$$\frac{\log \left(\sqrt{2}-\sqrt{x^2-1}/x\right)-\log \left(\sqrt{x^2-1}/x+\sqrt{2}\right)}{2 \sqrt{2}}$$
However at mathematica I get $$\frac{\log \left(-3 x^2-2 \sqrt{2} \sqrt{x^2-1} x+1\right)-\log \left(-3 x^2+2 \sqrt{2} \sqrt{x^2-1} x+1\right)}{4 \sqrt{2}}$$
Where is the error?
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\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
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\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\overbrace{\color{#66f}{\large%
\int{1 \over \root{x^{2} - 1}\pars{x^{2} + 1}}\,\dd x}}
^{\ds{\dsc{x} = \dsc{\cosh\pars{t}}}}\ =\
\int{1 \over \underbrace{\root{\cosh^{2}\pars{t} - 1}}_{\dsc{\sinh\pars{t}}}\ \bracks{\cosh^{2}\pars{t} + 1}}\,\sinh\pars{t}\,\dd t
\\[5mm]&=\int{\dd t \over \cosh^{2}\pars{t} + 1}
=\int{\sech^{2}\pars{t}\,\dd t \over 1 + \sech^{2}\pars{t}}
=\int{\sech^{2}\pars{t}\,\dd t \over 2 - \tanh^{2}\pars{t}}
\\[5mm]&={1 \over 2\root{2}}
\int\bracks{{1 \over \root{2} - \tanh\pars{t}}
+{1 \over \root{2} + \tanh\pars{t}}}\sech^{2}\pars{t}\,\dd t
\\[5mm]&={\root{2} \over 4}\,
\ln\pars{1 + \tanh\pars{t}/\root{2} \over 1 - \tanh\pars{t}/\root{2}}
={\root{2} \over 2}\,\,{\rm arctanh}\pars{\tanh\pars{t} \over \root{2}}
\\[5mm]&={\root{2} \over 2}
\,\,{\rm arctanh}\pars{\root{\cosh^{2}\pars{t} - 1} \over \root{2}\cosh\pars{t}}
=\color{#66f}{\large{\root{2} \over 2}
\,\,{\rm arctanh}\pars{\root{x^{2} - 1} \over \root{2}x}}
+ \mbox{a constant}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Closed form of $\int_{0}^{\eta}\cos nt\log\left(\frac{\cos(t/2)+\sqrt{\cos^2(t/2) -\cos^2(\eta/2)}}{\cos(\eta/2)}\right) dt$ I am reading a paper (sorry, no e-copy) with a number of infinite series, in which each term of the series is an integral of a complicated transcendental function like the one in the title.
There are no fewer than a dozen of these infinite sums of integrals in the paper.
For instance, the authors define
$$
I_1(\eta) = \sum_{n=2,3,4}^{\infty} \dfrac{n}{n^2-1} \int_{0}^{\eta} \cos nt \log \left( \dfrac{\cos(t/2) + \sqrt{\cos^2(t/2) - \cos^2(\eta/2)}}{\cos(\eta/2)}\right) dt
$$
where $0 \leq \eta < \pi$. The authors then assert that
$$
I_1(\eta) = \frac{\pi}{8} \left( (2\cos\eta -2) \log \sin \frac{\eta}{2} + \frac{1}{2}\cos\eta -\frac{1}{2}\right)
$$
without any proof or derivation (the results are simply listed in an Appendix).
I confess I am stumped as to how the authors arrived at their results. Any advice or help would be greatly appreciated.
| $\newcommand{\s}{\sigma}$$\newcommand{\e}{\eta}$$\newcommand{\al}[1]{\begin{align}#1\end{align} }$As suggested by Norbert Schuch in the comments, it suffices to check that the derivatives with respect to $\eta$ agree. This simplifies our task somewhat by getting rid of one of the logarithms in the integrand.
Starting from the expression derived by JJacquelin, we have for the case $0 \leq \e < \pi/2$:
$$
\al{
I_1'(\e) &= -\frac{\tan \frac \e 2}{8} \int_0^\e d\e \frac{ \cos \frac{t}{2} \left[2+ \cos t+2 \cos t \ln (2-2\cos t)\right]}{ \sqrt{\cos^2 \frac t 2 - \cos^2 \frac \e 2}}
\\&= -\frac{\tan \frac \e 2}{8} \int_0^\e d\e\frac{ \cos \frac{t}{2} \left[2 + (1 - 2 \sin^2 \frac t 2)+2 (1 - 2 \sin^2 \frac t 2) \ln (4 \sin^2 \frac t 2)\right]}{ \sqrt{\sin^2 \frac \e 2 - \sin^2 \frac t 2}}
\\&= -\frac{\tan \frac \e 2}{8} \int_0^1 dx\frac{ 2\left[3 - 2 x^2 \sin^2 \frac \e 2+2 (1 - 2 x^2 \sin^2 \frac \e 2) \ln (4 x^2 \sin^2 \frac \e 2)\right]}{ \sqrt{1 - x^2}}
\\&\equiv -\frac{\tan \frac \e 2}{4} I
}
$$
Here I put $\sin \frac t 2 = x \sin \frac \e 2 $, and there is no boundary term because the integrand vanishes at $t = \e$.
Now we abbreviate $\s = \sin \frac \e 2$ for clarity. We have
$$
I = A+B+C+D+E+F,
$$
with
$$
\al{
&A=\int_0^1 dx \frac{3}{\sqrt{1-x^2}}=\frac{3\pi}{2}
\\&B=\int_0^1 dx \frac{-2 \s^2 x^2}{\sqrt{1-x^2}}= -\frac \pi 2 \s^2
\\&C=\int_0^1 dx \frac{2 \ln (4\s^2)}{\sqrt{1-x^2}}= \pi \ln (4\s^2)
\\&D=\int_0^1 dx \frac{4 \ln x}{\sqrt{1-x^2}}= 4 \int_0^{\pi/2} dt \ln \sin t = -\pi \ln 4
\\&E=\int_0^1 dx \frac{-4x^2 \s^2 \ln(4\s^2)}{\sqrt{1-x^2}}= -\pi \s^2 \ln (4\s^2)
\\&F=\int_0^1 dx \frac{-4x^2 \s^2 (2 \ln x)}{\sqrt{1-x^2}}= -8 \s^2 \int_0^{\pi/2} dt \sin^2 t\ln \sin t = \pi \s^2 (\ln 4-1).
}
$$
Therefore,
$$
I = \pi(1-\s^2) \left[ \frac{3}{2} + 2 \ln \s \right] = \pi \cos^2 \frac \e 2 \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right].
$$
This gives
$$
I_1'(\e) = -\frac \pi 4 \sin \frac \e 2 \cos \frac \e 2 \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right] = -\frac \pi 8 \sin \e \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right].
$$
This agrees with the derivative of the desired result.
The case $\pi/2 \leq \eta \leq \pi$ can be handled similarly; the only thing that changes is that now we cannot assume that $\cos^2 \frac t 2 > \cos^2 \frac \e 2$ everywhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Upperbound this difference between two log expressions I have the difference between the following log expressions and I am trying to bound the difference,
$$F=
\log \left(1+ \left(2+\frac{1}{\sqrt{2}}\right)^2 x^2\right) -
\log \left(1+ \left(1-\frac{1}{\sqrt{2}}\right)^2 x^2\right) $$
Can I say that
$$F \leq \log\left(1+ \frac{ \left(2+\frac{1}{\sqrt{2}}\right)^2}{\left(1-\frac{1}{\sqrt{2}}\right)^2}\right)$$
Is this the tightest bound that exists?
Thanks in advance.
| $f(a) =\log (1+ax^2) - \ln a\Rightarrow f'(a) = \dfrac{x^2}{1+ax^2}- \dfrac{1}{a}< 0\Rightarrow \text{ since } b=\left(2+\dfrac{1}{\sqrt{2}}\right)^2 > \left(1-\dfrac{1}{\sqrt{2}}\right)^2=a\Rightarrow f(a) < f(b)\text{ and} \ln\left(\dfrac{b}{a}\right) \leq \dfrac{b}{a}-1 \Rightarrow \text{the inequality holds and the tighter bound is} \dfrac{b}{a}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$
Prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$.
Let $\theta$ be a root of $f(x)$. Compute $\frac{1}{\theta}$ and $(2 + \theta^2)^{-1} $ in $\mathbb Q[\theta ]$.
\begin{array}{l}
f\left( \theta \right) = \theta ^3 + 3\theta - 1 = 0 \\
\Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\
\Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta \ne 0} \right) \\
\end{array}
\begin{array}{l}
\frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta \ne 0} \right) \\
\Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2 \\
\Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1} = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2 \notin Q} \right) \\
\Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1} = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta \ne 1} \right) \\
\end{array}
But I can't show that $f$ is irreducible.
| Let $$g(x)=f(x+1)=x^3+3x^2+3x+1+3x+3-1=x^3+3x^2+6x+3$$
Now we can apply Eisenstein's Criterion (with $p=3$) to find that $g(x)$ is irreducible in $\mathbb{Q}[x]$, so $f(x)$ is also irreducible in $\mathbb{Q}[x]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
solution verification here is solution of my old question but i can't see it would someone explain to me the principal idea and what he wants to show
Solution
from $u_n=\sqrt{n}\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),$ To $u^2_n=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right),$
and from
$u^2_n= \frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)$
To $u^2_n=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)$
and if we've that $u_n^2$ convergent is that means $u_n$ also convergent
i just need to see the principal idea of what he did Thanks
I think the principal idea of solution is that he want to show $u_n^2$ convergent and bounded then tell that $u_n$ is convergent too
Any help would be appreciated
| Note first that
$$
1-\frac{1}{k}+\frac{1}{4k^2}
=\left(1-\frac{1}{k}\right)\left(1+\frac{k}{k-1}\cdot\frac{1}{4k^2}\right)=
\left(1-\frac{1}{k}\right)\left(1+\frac{1}{4(k-1)k}\right).
$$
Next,
$$
\frac{n}{4}\prod_{k=2}^n\left(1-\frac{1}{k}\right)=\frac{n}{4}\prod_{k=2}^n
\frac{k-1}{k}=\frac{1}{4}.
$$
Then
$$
\prod_{k=2}^n\left(1+\frac{1}{4k(k-1)}\right)=\exp\left(\sum_{k=2}^n
\log\left(1+\frac{1}{4k(k-1)}\right)\right)\le\exp\left(\sum_{k=2}^n
\frac{1}{4k(k-1)}\right),
$$
and
$$
\sum_{k=2}^n
\frac{1}{k(k-1)}=\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{1}{n}<1.
$$
Thus
$$
\prod_{k=2}^n\left(1+\frac{1}{4k(k-1)}\right)\le \exp\left(\frac{1}{4}\right)
$$
and hence $\{u_n^2\}$ is upper bounded and increasing, and thus convergent, and so is $\{u_n\}$, as $u_n>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The point of contact of between two circles and common tangent at this point. A large circle and a small circle have equations $x^2+y^2+2x-4y-27=0 $ and $x^2+y^2-12x+10y+43=0$ respectively.
a) Show that the two circles externally touch at a single point and find the point of contact.
b) Establish the equation of the common tangent at this point.
Answers:
a) Proof and $(3,-2)$
b) $y=x-5$
I need explanation provided please.
| $\mathcal{C_1}: (x+1)^2 + (y-2)^2 = (4\sqrt{2})^2\Rightarrow C_1 = (-1,2), r_1 = 4\sqrt{2}$.
$\mathcal{C_2}: (x-6)^2 + (y+5)^2 = (3\sqrt{2})^2\Rightarrow C_2 = (6,-5), r_2 = 3\sqrt{2}$.
$\overline{C_1C_2} = \sqrt{7^2+7^2} = 7\sqrt{2} = r_1+r_2\Rightarrow \mathcal{C_1}, \mathcal{C_2} $ are touching each other as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proof that sum of first $n$ cubes is always a perfect square I know that
$$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$$
What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first $n$ cubes always a perfect square.
| Let's prove this quickly by induction.
If needed I will edit this answer to provide further explanation.
To prove: $$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2}\right)^2$$
Initial case $n=1$: $$\sum_{i=1}^1 i^3=1^3=\left(\frac{2}{2}\right)^2 = \left(\frac{1(1+1)}{2}\right)^2$$
Given for induction: $$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2}\right)^2$$
To show in induction step: $$\sum_{i=1}^{n+1} i^3=\left(\frac{(n+1)(n+2)}{2}\right)^2$$
Induction step:
$$
\sum_{i=1}^{n+1} i^3=(n+1)^3+\sum_{i=1}^{n} i^3\stackrel{!}{=}
(n+1)^3+\left(\frac{n(n+1)}{2}\right)^2
= \left(\frac{4(n+1)^3+(n(n+1))^2}{2^2}\right)
= \left(\frac{4(n+1)(n+1)^2+n^2(n+1)^2}{2^2}\right)
= \left(\frac{(n+1)^2(4(n+1)+n^2)}{2^2}\right)
= \left(\frac{(n+1)^2(n^2+2\cdot2n+2^2)}{2^2}\right)
= \left(\frac{(n+1)(n+2)}{2}\right)^2
$$
In $(!)$ we used the equation holding for $n$.
If you are not familiar with the principle of induction, take a look at this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How can prove that $\sum_{n=1}^{\infty }\frac{\zeta (2n)}{4^{n-1}}(1-\frac{1}{4^n})=\frac{\pi }{2}$
$$\zeta (2)(1-\frac{1}{4})+\frac{\zeta (4)}{4}(1-\frac{1}{4^2})+\frac{\zeta (6)}{4^2}(1-\frac{1}{4^3})+...=\frac{\pi }{2}$$
The WolframAlph couldn't recognize the closed-form which is $\pi/2$ when I gave the series,
so I used the WolframAlph again to compute many terms of infinite series.
I think that the WolfarmAlph cannot say the value is $\pi/2$,So we need to prove it.
| Since,
$\begin{align}\sum_{k=1}^\infty\zeta(2k)\,x^{2k} &= \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{x^{2k}}{n^{2k}}\\
&=\sum_{n=1}^\infty\frac{\frac{x^2}{n^2}}{1-\frac{x^2}{n^2}}\\
&=\sum_{n=1}^\infty\frac{x^2}{n^2-x^2}\\
&=-\frac{x}{2}\sum_{n=1}^\infty\left(\frac{1}{x+n} + \frac{1}{x-n}\right)\\
&=-\frac{x}{2}\left(\pi\cot(\pi x)-\frac1x\right)\\
&=\frac{1}{2}\left(1-\pi x\cot(\pi x)\right)
\end{align}$
Plug in the particular cases $x = \dfrac{1}{2}$ and $x = \dfrac{1}{4}$ in the series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$ $$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$
I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugate $\left(\sqrt{4x^2+5x} + \sqrt{4x^2+x}\right)$, then I get $\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}$, but I can conclude nothing out of it.
| Note that $$\sqrt{4x^2+5x}-\sqrt{4x^2+x} = \frac{4x}{\sqrt{4x^2+5x}+\sqrt{4x^2+x}}$$
And the denominator is between $2x+2x=4x$ and $(2x+\frac{5}{4})+(2x+\frac{1}{4})=4x+\frac{3}{2}$. So the limit be $1$.
Alternatively, show that:
$$\lim \left(2x+\frac{5}{4}-\sqrt{4x^2+5x}\right) = 0$$
and
$$\lim \left(2x+\frac{1}{4}-\sqrt{4x^2+x}\right) = 0$$
Then again deduce that the limit of the difference must be $0$, so the limit you are seeking is $\frac{5}{4}-\frac 14=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Maclaurin series of $\arctan(x)$ up to degree $4$ How can I find the Maclaurin series up to degree 4 for:
$$\arctan(x)$$
Calculating the derivatives becomes complex very quickly.
Is there a special expansion for $\arctan$ like there is for $\cos(x)$ and $\sin(x)$?
| If by “bgtan” you mean the arctangent, implicitly defined by
$$
\tan\arctan x=x
$$
with values in $(-\pi/2,\pi/2)$, then the Taylor expansion at $0$ (also known as MacLaurin expansion) is the famous Gregory series:
$$
\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dotsb
$$
which converges for $x\in(-1,1]$.
Computing it by derivatives is not really difficult. Consider $f(x)=\arctan x$; then
$$
f'(x)=\frac{1}{1+x^2}
$$
and so
$$
f''(x)=-\frac{2x}{(1+x^2)^2}
$$
The third derivative is
$$
f'''(x)=-2\frac{(1+x^2)^2-4x(1+x^2)}{(1+x^2)^4}=
-2\frac{1-4x+x^2}{(1+x^2)^3}
$$
You don't need to compute the fourth derivative, because $f(x)=-f(-x)$, so all derivatives of even order are zero.
Now
$$
f(0)=0,\quad
f'(0)=1,\quad
f''(0)=0,\quad
f'''(0)=-2\quad
f''''(0)=0
$$
so
$$
\arctan x=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+
\frac{f''''(0)}{24}+o(x^4)=x-\frac{x^3}{3}+o(x^4)
$$
A different approach would be to notice that
$$
\frac{1}{1+x^2}=1-x^2+x^4+o(x^4)
$$
because of geometric series. Integrating between $0$ and $x$ gives
$$
\arctan x=\int_0^x (1-t^2+t^4+o(t^4))\,dt=
x-\frac{x^3}{3}+\frac{x^5}{5}+o(x^5)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Diophantine equation $1 + \sum_{j=1}^{n-1}\left(j \prod_{k=1}^j x_k\right) = \prod_{j=1}^n x_j$ What are the positive solutions $(x_1,x_2,\ldots,x_n)$ for the Diophantine equation:
$$1 + \sum_{j=1}^{n-1}\left(j \prod_{k=1}^j x_k\right) = \prod_{j=1}^n x_j$$
| It seems the following.
It can be easily checked by induction, that the equation has a simple solution $x_k=k$.
I investigated the equation for small values of $n$. It seems that its solutions can be obtained when we consequently consider admissible by divisibility values for $x_1,\dots,x_{n-1}$, branching our consideration. For instance, by this way we obtain
$n=1$. $1=x_1$
So $x_1=1$.
Solutions:
*
*$(1)$
$n=2$. $1+x_1=x_1x_2$.
So $x_1|1$ and $x_1=1$, $x_2=2$.
Solutions:
*
*$(1,2)$
$n=3$. $1+x_1+2x_1x_2=x_1x_2x_3$.
So $x_1|1$ and $x_1=1$. Then $2+2x_2=x_2x_3$.
So $x_2|2$ and $x_2=1$ or $x_2=2$.
Suppose that $x_2=1$. Then $x_3=4$.
Suppose that $x_2=2$. Then $x_3=3$.
Solutions:
*
*$(1,1,4)$
*$(1,2,3)$
$n=4$. $1+x_1+2x_1x_2+3x_1x_2x_3=x_1x_2x_3x_4$.
So $x_1|1$ and $x_1=1$. Then $2+2x_2+3x_2x_3=x_2x_3x_4$.
So $x_2|2$ and $x_2=1$ or $x_2=2$.
Suppose that $x_2=1$. Then $4+3x_3=x_3x_4$.
So $x_3|4$ and $x_3=1$ or $x_3=2$ or $x_3=4$.
Suppose that $x_3=1$. Then $x_4=7$.
Suppose that $x_3=2$. Then $x_4=5$.
Suppose that $x_3=4$. Then $x_4=4$.
Suppose that $x_2=2$. Then $3+3x_3=x_3x_4$.
So $x_3|3$ and $x_3=1$ or $x_3=3$.
Suppose that $x_3=1$. Then $x_4=6$.
Suppose that $x_3=3$. Then $x_4=4$.
Solutions:
*
*$(1,1,1,7)$
*$(1,1,2,5)$
*$(1,1,4,4)$
*$(1,2,1,3)$
*$(1,2,3,4)$
One may write a program which continues this consideration for larger $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find Minimum Value of: $P=x^2+y^2+2xy+2-\frac{1}{xy}$ Given: $x,y>0$ and $x^2y-x+xy^2-y-3xy=0$
Find Minimum Value of:
$P=x^2+y^2+2xy+2-\frac{1}{xy}$
I found $\min P =\frac{71}{4}$ at $x=y=2$ but I cant prove that
Could some one help me ?
| $x^2y-x+xy^2-y-3xy=0 , x>0, y>0$
$xy(x+y)-(x+y)-3xy=0$
Let $\begin{array}{|l} {x+y=a} \\ {xy=b} \end{array} \Rightarrow a>0, b>0$
Then $ab-a-3b=0$
$b(a-3)=a \Rightarrow a-3>0 \Rightarrow a>3$
$b=\frac{a}{a-3 }$
x and y are roots of $t^2-at+b=0$
$D=a^2-4b\ge 0$
$a^2-\frac{4a}{a-3 }\ge 0$ |.$(a-3)>0$
$a^3-3a^2-4a\ge 0$
$a(a^2-3a-4)\ge 0$ |:$a>0$
$a^2-3a-4\ge 0$
$(a-4)(a+1)\ge 0$ |:$(a+1)>0$
$a-4\ge 0 \Rightarrow a\ge 4$
$P=x^2+y^2+2xy+2-\frac{1}{xy }=(x+y)^2-\frac{1}{xy }+2=a^2-\frac{1}{b }+2=$
$=a^2-\frac{a-3}{a }+2= a^2+\frac{3}{a }+1$
$a^2+\frac{3}{a }+1\ge \frac{71}{4 } \Leftrightarrow 4a^3-67a+12\ge 0 \Leftrightarrow (a-4)(4a^2+16a-3)\ge 0$
DONE ($4a^2+16a-3\ge 0$ if $a\ge 4$)!
| {
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Prove by induction that $\sum_{k=0}^{n}(-1)^{n+k} k^{2} = \frac{n(n+1)}{2}$ Prove by mathematical induction that
$\forall n \in \mathbb{N}:~~~~ \sum_{k=0}^{n}(-1)^{n+k} k^{2} = \frac{n(n+1)}{2}$
Step 1: Show true for $n = 1$:
LHS: $(-1)^{(1+0)} \cdot 0^{2} + (-1)^{(1+1)} \cdot 1^{2} = 1$
RHS: $\frac{0(0+1)}{2} + \frac{1(1+1)}{2} = 1$
Step 2: Show "if true for $n = p$, then true for $n = p + 1$":
So we start by the LHS of $n = p + 1$ and try to get to the RHS of it, by using the equality for $n = p$.
The first step involves breaking out the highest term in the sum.
$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (-1)^{(2p+2)} (p+1)^{2} + \sum_{k=0}^{p}(-1)^{(p+k)} k^{2} $
Replacing the last term with the equality for $n = p$.
$(-1)^{2p+2} (p+1)^{2} + \sum_{k=0}^{n}(-1)^{p+k} k^{2} = (-1)^{2p+2} (p+1)^{2} + \frac{p(p+1)}{2}$
Since 2p+2 is an even number, the -1 becomes a 1:
$(-1)^{2p+2} (p+1)^{2} + \frac{p(p+1)}{2} = (p+1)^{2} + \frac{p(p+1)}{2}$
All that is required now is to show that this equals to $\frac{(p+1)(p+2)}{2}$, but there seems to be a $(p+1)$ too many in the first term for it to work.
Question 1: Where have I gone wrong in the above attempt?
Question 2: Is it more productive to attempt to prove
$\forall n \in \mathbb{N}:~~~~ \sum_{k=0}^{n}(-1)^{n+k} k^{2} = \sum_{k=1}^{n} k$
instead?
| In this step you have made an error:
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (-1)^{(2p+2)} (p+1)^{2} + \sum_{k=0}^{p}(-1)^{(p+k)} k^{2}$$
This should be
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (-1)^{(2p+2)} (p+1)^{2} + \sum_{k=0}^{p}(-1)^{(p+1+k)} k^{2}.$$Notice the extra $+1$ in the exponent of $(-1)$ in the last sum. We can remove it by factoring out a $-1$ to get
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (p+1)^{2} - \sum_{k=0}^{p}(-1)^{(p+k)} k^{2}.$$
By the assumption this is
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (p+1)^{2} -\frac{p(p+1)}{2}.$$
Now we can easily work it out:
$$(p+1)^2 - \frac{p(p+1)}{2} = (p+1)(-\frac{p}{2} + p +1)= (p+1)\frac{p+2}{2}$$
as desired.
| {
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Number of sequences of $0$s, $1$s, and $2$s with length $n$ such that there is a $0$ somewhere between every pair of $2$s Let $a(n)$ be the number of sequences with length $n$ which consists the digits $0,1,2$ such that between every two occurrences of $2$ there is an occurrence of $0$ (not necessarily next to the $2$'s). An example for good sequence is $0102102$. An example for bad sequence is $01212$.
A. Find a recursion formula for $a(n)$
B. Find an explicit expression for $a(n)$.
My try:
let $x(n)$ be the number of sequences with length $n$ that if we add $2$ at the end of sequence it is still a good sequence with length $n+1$.
Now, if the last digit in the sequence was $0$, then we have to look on the number of good sequences with length $n-1$, hence $\displaystyle a(n-1)$.
If the last digit was $1$, then we now need to look on $x(n-1)$ and work with the same manipulation. This way we get the recursion $x(n)=a(n-1)+x(n-1)$. By induction I can get a recursion formula for $x(n)$, but I don't know how to find a formula for $a(n)$ or find an explicit expression for $a(n)$.
Any help will be appreciated, thank you!
| Consider the following:
\begin{align*}
r(n) &:= \text{number of sequences of 0s and 1s} = 2^n \\
s(n) &:= \text{number of sequences of 0s and 1s with at least one $0$} = 2^n - 1 \\
\end{align*}
The corresponding (ordinary) generating functions are
\begin{align*}
R(x) &= \sum_{n \ge 0} 2^n x^n = \frac{1}{1 - 2x} \\
S(x) &= \sum_{n \ge 0} (2^n - 1) x^n = \frac{1}{1 - 2x} - \frac{1}{1 - x}
= \frac{x}{(1-2x)(1-x)}.
\end{align*}
The generating function $A(x)$ of $a(n)$ satisfies, by casework on the number $k$ of $2$s in the sequence:
\begin{align*}
A(x)
&= \underbrace{R(x)}_{k=0} + \underbrace{R(x) x R(x)}_{k=1} + \sum_{k \ge 2} R(x) (x S(x))^{k-1} x R(x) \\
&= R(x) + \sum_{k \ge 1} R(x)^2 S(x)^{k-1} x^{k} \\
&= R(x) + \frac{x R(x)^2}{1 - x S(x)} \\
&= \frac{1}{1 - 2x} + \frac{x / (1 - 2x)^2}{1 - \frac{x^2}{(1-x)(1-2x)}} \\
&= \frac{1}{1 - 2x} + \frac{x(1-x)}{(1-2x) \left[ (1-2x)(1-x) - x^2 \right]} \\
&= \frac{1 - 3x + x^2}{(1 - 2x)(1 - 3x + x^2)} + \frac{x - x^2}{(1-2x)(1 - 3x + x^2)} \\
&= \frac{1 - 2x}{(1 - 2x)(1 - 3x + x^2)} \\
&= \boxed{\frac{1}{1 - 3x + x^2}}.
\end{align*}
Now that we have the generating function, we realize that $a(n)$ satisfies a much simpler recurrence:
$$
a(n) = 3 a(n-1) - a(n-2)
$$
which follows directly from $A(x) (1 - 3x + x^2) = 1$.
This recurrence can be explained combinatorially as well, and if we had bothered
to be a bit more clever we might have found the recurrence in the first place and saved some algebra.
Conclude either from the recurrence or from searching the first few values in oeis
that in fact
$$
a(n) = f(2n+2)
$$
where $f(k)$ is the $k$th Fibonacci number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$
If $x>0$, $y>0$, $z>0$ and $xyz = 1$ then
$$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$$
I tried using $\displaystyle x = \frac{a}{b},y = \frac{b}{c}$ and $\displaystyle z = \frac{c}{a}$ substitution,
$\displaystyle \sum_{cyc} \frac{\sqrt{a^2+b^2}}{b} \le \sqrt{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \\ \displaystyle \iff 6+ 2\sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le \sum_{cyc}\frac{a^2+b^2}{b^2} + 4\sum_{cyc}\frac{b}{a}$
So it would suffice if we showed $\displaystyle \sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le 2\sum_{cyc}\frac{b}{a}$
Squaring again does not lead anywhere nice. Is the last inequality true ? If not is there a better way of showing the result ?
| Consider the function $f(t) = \sqrt2t - \sqrt{t^2+1}-\left(\sqrt2-\frac1{\sqrt2} \right)\log t$ for $t > 0$. Note that $f(t) \ge 0 \implies f(x)+f(y)+f(z) \ge 0 $ which gives the desired inequality.
$$$$
Now $f'(t) = \dfrac{-2t^2+2t\sqrt{2(t^2+1)}-\sqrt{2(t^2+1)}}{2t\sqrt{t^2+1}}$ so $f'(1)=0$. Further,
$$f''(t) = \frac{(t^2+1)^{3/2}- \sqrt2 t^2}{\sqrt 2 t^2 (t^2+1)^{3/2}} > 0$$
so $f'(t)> 0$ for $t> 1$ and $f'(t) < 0$ for $t < 1$. Hence $f(1)=0$ is the minimum, $f(t) \ge 0 \quad \forall t > 0$ and the inequality holds.
P.S.: @user1537366 when the inequality is separable, i.e. of form $g(x)+g(y)+g(z)+... \ge 0$ and a constraint which can be also expressed as $h(x)+h(y)+\dots = 0$, it is worthwhile to check if for a suitable constant $k$, we can get $f(t) = g(t)+k\,h(t) \ge 0$.
Often $k$ can be determined by setting $f'(t^*)=0$ where $t=t^*$ is when you expect equality.
| {
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I have a Hard time solving this system of nonlinear equations $x^2+y^2-z^2=20$
$x^4+y^4-z^4=560$
$x^3+y^3+z^3=3xyz$
I know the fact that if $x^3+y^3+z^3=3xyz$ then $x+y+z=0$ (coming from Euler's identity) and first equation can be written as
$(x+y-z)^2-2(xy-xz-yz)=20$ and since $x+y=-z$ then $(xy-xz-yz)=2z^2-10$.
And the second equation becomes
$(x^2+y^2-z^2)^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$.
Since we know $x^2+y^2-z^2=20$, therefore $20^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$ or
$(x^2y^2-2x^2z^2-2y^2z^2)=80$.
Now I am stuck! Help please!
| Hint
Using $z=-x-y$, the first equation write $2 x y+20=0$ and so $y=-\frac{10}{x}$. Now the second equation write, after minor simplifications, $$x^2+\frac{100}{x^2}=29$$ in which you recognize a quadratic in $x^2$.
I am sure that you can take from here.
| {
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If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be? If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be?
I have seen such problems many times but no specific method to solve them.
I hope I find the general rule to solve them here.
| We have $$2^x = 9\implies x\ln(2)=\ln(2^x)=\ln(9)\implies x = \frac{\ln(9)}{\ln(2)}= 2 \frac{\ln(3)}{\ln(2)}$$
and
$$4^y = 27\implies y = \frac{\ln(27)}{\ln(4)}= \frac{3}{2}\frac{\ln(3)}{\ln(2)}$$
So
$$\frac {x+2y}{2y-4x} =\frac {2 \frac{\ln(3)}{\ln(2)}+3\frac{\ln(3)}{\ln(2)}}{3\frac{\ln(3)}{\ln(2)}-8\frac{\ln(3)}{\ln(2)}} =-1$$
| {
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Show that $\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$
Show that
$$\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$$
Playing around with the expression, I found a proof which I will post as an answer.
I'm asking this question because I would like to see if there are alternative solutions which are perhaps faster / more direct / elementary / elegant / methodical / insightful etc.
| Here is my solution:
Let $\alpha = \sqrt[3]{3\sqrt{21} + 8}$ and $\beta = \sqrt[3]{3\sqrt{21} - 8}$.
Then
$$ \alpha\beta = \sqrt[3]{(3\sqrt{21} + 8)(3\sqrt{21} - 8)} = \sqrt[3]{(3\sqrt{21})^2 - 8^2} = \sqrt[3]{189 - 64} = \sqrt[3]{125} = 5
$$
and
$$
\alpha^3 - \beta^3 = (3\sqrt{21} + 8) - (3\sqrt{21} - 8) = 16
$$
Now
$$ (\alpha - \beta)^3 = \alpha^3 - 3\alpha^2\beta + 3\alpha\beta^2 - \beta^3 = (\alpha^3 - \beta^3) - 3\alpha\beta (\alpha - \beta) = 16 - 15(\alpha - \beta)
$$
so $\alpha - \beta$ is a root of the polynomial
$$
x^3 + 15x - 16 = (x-1)(x^2 + x + 16).
$$
The part $x^2 + x + 16$ does not have any real roots since its discriminant is $-1 - 4\cdot 16 < 0$.
So $\alpha - \beta$ is a root of $x-1$ and thus
$$
\alpha - \beta = 1
$$
| {
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"question_score": "4",
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Evaluate $ \int_0^1 \sqrt{x}\sin(x)dx $ to accuracy 0.001. Evaluate
$$
\int_0^1 \sqrt{x}\sin(x)dx
$$
to accuracy 0.001.
By definition, there exists an N such that for n > N,
$$
\left| \int_0^1 \sqrt{x}\sin(x)dx - \sum\limits_{i=1}^n \sqrt{x^{*}_i}\sin(x^{*}_i) \Delta x\right| <0.001.
$$
But from there I do not know how to proceed. Are there any theories regaurding approximation that would be helpful?
| Using @Mohammad Ali Baydoun: ideea, the integral lies between the increasing sequence
$$ \int_0^1 \sqrt{x}\cdot (x - \frac{x^3}{6}) dx, \int_0^1 \sqrt{x}\cdot (x - \frac{x^3}{6}+ \frac{x^5}{120} - \frac{x^7}{7!})\,dx, \ldots $$
and the decreasing sequence
$$ \int_0^1 \sqrt{x}\cdot x\, dx, \int_0^1 \sqrt{x}\cdot ( x- \frac{x^3}{6}+ \frac{x^5}{120})\,dx, \ldots $$
These are sequences of rational numbers, easy to evaluate, that converge to the same limit $\int_0^1 \sqrt{x} \sin x dx$.
Note that the inequalities between $\sin x$ and its odd and even order Taylor approximations are valid for all positive $x$, something that does not follow right away from Leibniz estimate for alternating series with decreasing absolute values of terms. See the picture below:
The fifth and the seventh order Taylor approximation already give the correct fourth decimal :
$$\frac{608527}{1670760} < \int_0^1 \sqrt{x} \sin x dx< \frac{2557}{7020}$$
The approximation on both sides
$$\sqrt{x}\cdot (x - \frac{x^3}{6}+ \frac{x^5}{120} - \frac{x^7}{7!})\le \sqrt{x} \sin x \le \sqrt{x}\cdot (x - \frac{x^3}{6}+ \frac{x^5}{120} )$$ on $[0,1]$ has an error less than $\frac{1}{7!} = \frac{1}{5040}< \frac{1}{5000} = 0.0002$ so the estimates for the integrals are apriori better than $0.0002$ since the interval has length $1$. It turns out that the estimates are better than $0.0001$, as we see from evaluating both of the fractions.
| {
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If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem :
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$
Solution :
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other
Thanks
| The equations are satisfied by $f(x) = x^3 + 1$. For:
$$
f(x) f\left(\frac{1}{x}\right)
= (x^3 + 1) \left(\frac{1}{x^3} + 1\right)
= 1 + x^3 + \frac{1}{x^3} + 1
= f(x) + f \left(\frac{1}{x}\right)
$$
and $f(4) = 4^3 + 1 = 65$.
So
$f(6) = 6^3 + 1 = 217$.
| {
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| I think it may be done the folowing way:
$x^3+4x^2+x−6=x^3+2x^2+2x^2+4x-3x-6=x^2(x+2) + 2x(x+2)-3(x+2)=(x+2)(x^2+2x-3)$
| {
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How can I evaluate $\int_0^{\pi/2}\frac{x\cos{x}}{3\sin^2x+1}dx$ and $\int_0^{\pi/2}\frac{x\cos{x}}{\sin^2x+3}dx$? I do not find the closed form of the following integrals$$\int_0^{\pi/2}\frac{x\cos{x}}{3\sin^2x+1}\mathrm dx$$
$$\int_0^{\pi/2}\frac{x\cos{x}}{\sin^2x+3}\mathrm dx$$
On the other side, I find
$$\int_0^{\pi/2}\frac{x(1+\sin^2x)\cos{x}}{(3\sin^2x+1)(\sin^2x+3)}\mathrm dx=-\frac{\sqrt3}{24}\ln(3)\ln{(2-\sqrt3)}$$
| Integrating by parts, we get
\begin{align*}
\int_{0}^{\pi/2} \frac{x\cos x}{3\sin^{2}x + 1} \, dx
&= \left[ \frac{x}{\sqrt{3}} \arctan(\sqrt{3}\sin x) \right]_{0}^{\pi/2} - \frac{1}{\sqrt{3}} \int_{0}^{\pi/2} \arctan(\sqrt{3}\sin x) \, dx \\
&= \frac{\pi^{2}}{6\sqrt{3}} - \frac{1}{\sqrt{3}} \int_{0}^{\pi/2} \arctan(\sqrt{3}\sin x) \, dx.
\end{align*}
Now by noting the identity
$$ \int_{0}^{\pi/2} \arctan(r\sin x) \, dx = 2\chi_{2}\left(\frac{\sqrt{1+r^{2}} - 1}{r} \right), $$
where $\chi_{2}$ is the Legendre chi function of order 2, it follows that
$$ \int_{0}^{\pi/2} \frac{x\cos x}{3\sin^{2}x + 1} \, dx
= \frac{\pi^{2}}{6\sqrt{3}} - \frac{2}{\sqrt{3}} \chi_{2}\left( \frac{1}{\sqrt{3}} \right). $$
There are only a handful of cases where the exact value of $\chi_{2}(z)$ are known. And unfortunately $z = 3^{-1/2}$ is not the case. Similarly,
$$ \int_{0}^{\pi/2} \frac{x\cos x}{\sin^{2}x + 3} \, dx
= \frac{\pi^{2}}{12\sqrt{3}} - \frac{2}{\sqrt{3}} \chi_{2}(2-\sqrt{3}). $$
| {
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Solution to $\sqrt{x^2-5}+3>|x-1|$ I tried many ways to solve this but I just can't figure it out...
$$\sqrt{x^2-5}+3>|x-1|$$
| first the domain of the inequality is $(-\infty, -\sqrt 5] \cup [\sqrt 5, \infty).$
so we will deal with the absolute value sign by breaking the function into two pieces:
on $x \le -\sqrt 5$ we need to solve $\sqrt{x^2 - 5} + 3 > 1 - x$ and on $x ge \sqrt 5$ solve
$\sqrt{x^2 - 5} + 3 > x - 1$
first for $x \le -\sqrt 5:$
$ \sqrt{x^2 - 5} > -x -2 $ implies $x^2 - 5 > x^2 + 4x + 4$ which says that $x < -9/4.$
now for $x \ge \sqrt 5:$
$ \sqrt{x^2 - 5} > x -4 $ implies $x^2 - 5 > x^2 -8x + 16$ which says that $x > 21/8$ but for $x = 21/8$ the right hand side $x - 4 < 0.$ that is $x = 21/8$ solves
$\sqrt{x^2 - 5} > 4 - x$ instead.
combining the two the solution is $$-\infty < x < -\dfrac{9}{4} \text{ or } \sqrt 5 < x < \infty$$
p.s. every day you learn a new thing.
| {
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Is there a linear operation such that $F(1,1,1) = (1,2,3),F(1,2,3) = (1,4,9),F(2,3,4) = (1,8,27)$? The exercise asks me verify if there exists a linear operator $F$ such that:
$$F(1,1,1) = (1,2,3)\\F(1,2,3) = (1,4,9)\\F(2,3,4) = (1,8,27)$$
First I tried to write a vector $(x,y,z)$ as a linear combination of $(1,1,1),(1,2,3),(2,3,4)$:
$$(x,y,z) = a(1,1,1) + b(1,4,9) + c(1,8,27)$$
then we have:
$$x = a + b + 2c\\y = a + 2b + 3c\\z = a + 3b + 4c$$
which is a system that has determinant $0$, therefore there isn't a way to represent a vector $(x,y,z)$ as a linear combination of $(1,1,1),(1,2,3),(2,3,4)$.
If it were possible to write the vector $(x,y,z)$ I would then apply $F$ to both sides and find the linear operator. But I can't do it since that vectors do not form a basis. But my reasoning does not logically prove that such linear operator does not exsists.
By the way, my book also says that this is impossible because $(2,3,4)$ should be equal to $(2,6,12)$. Why?
| The vectors $v_1=(1,1,1)$, $v_2=(1,2,3)$ and $v_3=(2,3,4)$ are not linearly independent; Gaussian elimination gives
\begin{align}
\begin{bmatrix}
1 & 1 & 2 \\
1 & 2 & 3 \\
1 & 3 & 4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 2 & 2
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\end{align}
which says that $v_3=v_1+v_2$.
Since $(1,2,3)+(1,4,9)=(2,6,12)\ne(1,8,27)$ no linear map exists with the requested property.
Had the problem said $F(2,3,4)=(2,6,12)$, then infinitely many linear maps would exist: complete $\{(1,1,1),(1,2,3)\}$ to a basis and send the third basis vector to any vector.
In the RREF above, the dominant columns (those with a leading $1$) are the first and the second column. The coefficients in the third column give exactly the coefficients to use for getting it as a linear combination of the dominant columns to its left.
Change the third column to represent the vector $(1,0,-1)=2v_1-v_2$; Gaussian elimination is
\begin{align}
\begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 0 \\
1 & 3 & -1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 & -1 \\
0 & 2 & -2
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}
\end{align}
and indeed we find the right coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it in the form:
$$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$
Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 < \frac{1}{n} < \ldots < \frac{n}{n}$.
Can you explain to me to reach this conclusion?
| Hints : First look at the coefficient $\frac{1}{n}$ in front of the expression inside brackets. The $1$ at the top of the fraction is $1-0$, where $1$ and $0$ are the end points of the interval $[0,1]$. The $n$ at the bottom would be the length of a subinterval of $[0,1]$ when you partition $[0,1]$ into $n$ equal parts.
Then look at each individual term inside brackets. For example,
$$
\frac{1}{1+(\frac{3}{n})^2}
$$
would be the value of the function $\frac{1}{1+x^2}$ when $x$ is $3$ times the length of a subinterval of length $\frac{1}{n}$ added from the starting point $0$.
Can you now figure out that this is the Riemann sum you had to recognize ? What the expression really represents is a sum of areas of rectangles of base $\frac{1}{n}$ and height $\frac{1}{1+(\frac{i}{n})^2}$. Here you approach the area under the curve $\frac{1}{1+x^2}$ over $[0,1]$ by taking the right end points of the subintervals of an equally distributed partition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Proving that the series 1 + ... + $1 / \sqrt{x}$ < $2 \sqrt{x}$ Proving that the series 1 + ... + $1 / \sqrt{x}$ < $2 \sqrt{x}$
I am doing it by simple induction adding
$1/\sqrt{x+1}$ to both sides, but I can't find a way to manipulate this expression and find that the new series is $< 2 \sqrt{x+1}$.
Can someone show me the correct process?
I failed many times and the furthest I've gotten has been to prove that it's $< 2 \sqrt{x+1} + 1$ which is close, but not enough.
| I suppose $x$ is an integer and you series is
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{n}}$$
For $n=1,$ our inequity is obviously true.
Suppose it is true for $n=k.$ Then
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}<2\sqrt{k}$$
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}<2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}$$ Note that
$$\sqrt{k+1}-\sqrt{k}=\dfrac{1}{\sqrt{k+1}+\sqrt{k}}>\dfrac{1}{2\sqrt{k+1}}.$$ Hence $$\dfrac{1}{2\sqrt{k+1}}+\sqrt{k}<\sqrt{k+1}$$ and therefore
$$1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dots+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}<2\sqrt{k+1}$$ Therefore by MI, our result is true for all $n\in\mathbb{N}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integral $\int \sqrt{x^4+x^3} \, dx$ $$
\int \sqrt{x^4+x^3} \, dx?
$$
Using the binomial method and by setting $\frac{1}{x}+1=t$, I get to solve
$$
\int \frac{-t^{\frac{3}{2}}}{(t^2-1)^4} dt?
$$
It is on degree $\frac 32$. How to solve this it seems a bit hard and impossible.
| $$\int \sqrt{x^4 + x^3} \;dx = \int x\sqrt{x^2 + x}\;dx = \frac{1}{2} \int (2x+1)\sqrt{x^2+x}\;dx - \frac{1}{2} \int \sqrt{x^2 + x} \;dx \\
= \frac{1}{2} \int (2x+1)\sqrt{x^2+x}\;dx - \frac{1}{2} \int \sqrt{\left( x + \frac{1}{2} \right)^2-\frac{1}{4}} \;dx $$
For the first term, substitute $t = x^2 + x$. For the second term, substitute $x + \frac{1}{2} = \frac{1}{2}\cosh u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Limit of Ratio of Chebyshev Polynomials I have been trying to compute the limit
$$\lim_{n\to\infty}{{U_n(x)^2}\over{U_{n-1}(x)^2+U_n(x)^2}}$$
where $U_n(x)$ is the $n$-th Chebyshev polynomial of the second kind and $x\ge 1$.
Using software I have been able to compute these limits exist when $x$ a half-integer, but I would like to have an explicit formula of the limit as a function of $x$ (at least for $x$ a half-integer).
| First thing we should note is:
$$\lim_{n\to\infty}\frac{U_n(x)^2}{U_{n-1}(x)^2+U_n(x)^2}=\lim_{n\to\infty}\frac{1}{\left(\dfrac{U_{n-1}}{U_n}\right)^2+1}$$
So, we look at $\displaystyle\lim_{n\to\infty}\frac{U_{n-1}}{U_n}$. Wikipedia gives the formula:
$$U_n=\frac{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}{2\sqrt{x^2-1}}$$.
Thus, the limit becomes:
$$\lim_{n\to\infty}\frac{\frac{\left(x+\sqrt{x^2-1}\right)^{n}-\left(x-\sqrt{x^2-1}\right)^{n}}{2\sqrt{x^2-1}}}{\frac{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}{2\sqrt{x^2-1}}}=\lim_{n\to\infty}\frac{\left(x+\sqrt{x^2-1}\right)^{n}-\left(x-\sqrt{x^2-1}\right)^{n}}{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}$$
Now let's call $a=x+\sqrt{x^2-1}$. Then $\frac{1}{a}=x-\sqrt{x^2-1}$, so the limits is now:
$$\lim_{n\to\infty}\frac{a^n-a^{-n}}{a^{n+1}-a^{-(n+1)}}$$
This depends on whether $a$ is $>$, $=$ or $<1$:
I'll do $|a|<1$ first:
We multiply the denominator and the numerator by $a^{n+1}$:
$$\lim_{n\to\infty}\frac{a^{2n+1}-a}{a^{2n}-1}$$
Here $a^{2n+1},a^{2n}\to0$ as $n\to\infty$, so the limit is equal to $a$.
In this case, the original limit is equal to $\frac{1}{a^2+1}=\frac{1}{2\left(x+\sqrt{x^2-1}\right)}=\frac{x-\sqrt{x^2-1}}{2}$.
The case where $|a|<1$ is very similar to this one.
The two remaining cases are $a=1$ and $a=-1$.
When $a=-1$, the denominator equals $2$ and the numerator is 0, so the limit does not exist and neither does the original one for values of $x$ for which $a=-1$, which is only $x=-1$.
The only case left is $a=1$ which gives $\frac{0}{0}$ which is undefined, so the limit is also undefined. This is for $x=1$.
| {
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"url": "https://math.stackexchange.com/questions/1109517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$
Then
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
| Write $$a_n = \frac {4} {(2 \sqrt {n} + \sqrt {n - 1} + \sqrt {n + 1}) (2n + 2 \sqrt {n^2 - 1})} \sim \frac {1} {4 n \sqrt {n}}.$$ Then, $$\sum a_n \sim \frac {1} {4} \zeta (\frac {3} {2}).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve system of 3 equations $x+y+z=0$
$x^2+y^2+z^2=6ab$
$x^3+y^3+z^3=3(a^3+b^3)$
this is what i reasoned out so far;
$xyz=a^3+b^3$
$x^2+zx+z^2=3ab$
$y^2+zy+z^2=3ab$
$x^2+xy+y^2=3ab$
$y^2=3ab+zx$
$x^2=3ab+zy$
$z^2=3ab+xy$
I'd prefer a hint rather than a full answer - and how should I solve this kind of systems?
| If we set $z=-x-y$, then we obtain two equations in $x,y$, namely
\begin{align*}
3ab - x^2 - xy - y^2 & = 0,\\
a^3 + b^3 + x^2y + xy^2 & = 0.
\end{align*}
For $b\neq 0$ the first equation here gives $a=\frac{x^2+xy+y^2}{3b}$, so that the second equation becomes
$$
(3b^2 + 3bx + 3by + x^2 + xy + y^2)(3b^2 - 3bx + x^2 + xy + y^2)(3
b^2 - 3by + x^2 + xy + y^2)=0.
$$
Each factor is a quadratic equation, say , in $y$, which can be solved easily. Since there should not be a full answer, I suggest for you to consider the case $b=0$ (for $a=b=0$ we obtain $x=y=z=0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
| The absolute least value you can get is a rectangle topped by a half circle (the circle has the best area to arc length ratio of any shape) with a total arc length of $2 \big(1 - \frac{\pi}{8}\big) + \frac{\pi}{2} \approx 2.78539$. If you use Fourier approximation, you can come arbitrarily close to this limit. (I assume the fun of this challenge is to find an arbitrarily "low-term" function.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
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"answer_id": 6
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For what positive integer values $b,d$ does $(b^2-d)\mid(b^2-1)?$ hold? I am curious about the answer to the following questions: And hope that you can help me
For what positive integer values $b, d$ does
$$(b^2-d)|(b^2-1)?$$
hold?
Is it correct that the only solutions are those of the form $(b,1)$? The following leads to an affirmative answer:
The above expression implies $(b+\sqrt d)(b-\sqrt d)\mid(b^2-1)$. This means that either $b+\sqrt d$ or $b-\sqrt d$ divides $b^2-1$.
Using synthetic division in solving $\frac{b^2-1}{b+\sqrt d}$ and $\frac{b^2-1}{b-\sqrt d}$ I get both the remainer $d-1$. Since we want $d-1=0$ we have $d=1$. So the solution is $(b,d)=(b,1)$.
Thanks in advance.
| This is only a partial answer to your question.
Since we have the identity
$$\frac{b^2 - 1}{b^2 - d} = \frac{b^2 - d}{b^2 - d} + \frac{d - 1}{b^2 - d} = 1 + \frac{d - 1}{b^2 - d}$$
then your question is equivalent to
For what positive integer values $b, d$ does
$$b^2 - d \mid d - 1$$
hold?
Now, $b^2 - d \mid d - 1$ implies that $b^2 - d \leq d - 1$, which implies that $b^2 \leq 2d - 1$.
Unfortunately, this does not seem to narrow things that much.
| {
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$x^2+3$ has two zeros over ${\Bbb F}_p$ provided that $x^2+x+1\in{\Bbb F}_p[x]$ has two? The following is an exercise in abstract algebra:
If $p=1\pmod{3}$, then $x^2+x+1\in\Bbb{F}_p[x]$ has two zeros. Prove in this case that $-3$ is a quadratic residue mod $p$.
Showing that $(x^2+x+1)\mid (x^p-x)$, one can get the first part done. But I don't see how the second part, namely, $x^2+3\in\Bbb{F}_p[x]$ has two zeros, can be deduced from the first one. Could one come up with some hints of how to get the first part? (I was thinking one might relate $x^2+3$ to $x^2+x+1$ to make an argument about the zeros of $x^2+3$. But I don't see how I can go on.)
| The map $f\colon x\mapsto 2x+1$ maps roots of $x^2+x+1$ to roots of $x^2+3$. Indeed, $$f(x)^2+3=(2x+1)^2+3=4(x^2+x+1)=0$$ whenever $x^2+x+1=0$.
On the other hand, if $2$ is invertible, then the map $g\colon x\mapsto \frac12(x-1)$ maps roots of $x^2+3$ to roots of $x^2+x+1$. Indeed, $$g(x)^2+g(x)+1=\frac14(x-1)^2+\frac12(x-1)+1=\frac14(x^2+3)=0$$ whenever $x^2+3=0$.
Since $f$ and $g$ are inverse maps (if $2$ is invertible), they show that the sets of roots of $x^2+3$ and $x^2+x+1$ are in bijection, hence of the same cardinality.
Of course, the claim is trivially true for $\mathbb F_2$ (though it would be wrong for $\mathbb F_4$, which has two roots of $2^2+x+1$ and only one root of $x^2+3$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Reference request: Controllable and Observable form for transform function I came across some online material a year ago that claimed that a the ABCD matrix of a transfer function $$G(z) = \frac{b_1 z+b_2}{z^2+a_1z + a_2}$$ can be directly computed from the coefficients of the transfer function
Can someone provide a reference to some material that contains the proof to this concept?
| Consider the generic transfer function
$$
H(s) = \frac{Y(s)}{X(s)} = \frac{b_0s^n + \cdots + b_ns^0}{s^n + a_1s^{n-1} + \cdots + a_n}
$$
Then the state space model is
\begin{align}
\dot{\mathbf{q}} &= \mathbf{Aq} + \mathbf{B}u\\
y &= \mathbf{Cq} + Du
\end{align}
where
\begin{align}
\mathbf{A} &=
\begin{bmatrix}
-a_1 & 1 & 0 & \cdots & 0\\
-a_2 & 0 & 1 & 0 & \vdots\\
\vdots & \vdots & & \ddots & 0\\
-a_{n-1} & 0 &\cdots & & 1\\
-a_n & 0 & 0&\cdots & 0
\end{bmatrix}\\
\mathbf{B} &=
\begin{bmatrix}
b_1 - a_1b_0\\
b_2 - a_2b_0\\
\vdots\\
b_{n-1} - a_{n-1}b_0\\
b_n - a_nb_0
\end{bmatrix}\\
\mathbf{C} &= \begin{bmatrix}1 & 0 & \cdots & 0\end{bmatrix}\\
D &= b_0
\end{align}
You can go through the full derivation here.
| {
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How does $(x+3)^2 - 2^2$ become $(x+1)(x+5)$? I don't understand how $(x+3)^2 - 2^2$ can be transformed to equal $(x+1)(x+5)$. A short demonstration and/or reference to math rules would be very kind.
| Yet another way of looking at it is to look at the roots.
$(x+3)^2-2^2=0$, so $(x+3)^2=2^2$. Taking square roots, we get $x+3=2$ or $x+3=-2$. Solving for x, we get $x=-1$ or $x=-5$. That means the quadratic is of the form $a(x+1)(x+5)$ where $a$ is the coefficient of your $x^2$ term, which in this case is 1.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for any integer, $n^2 + 5$ is not divisible by $4$. So I got that there is two cases: odd or even.
If odd then say $n^2$ is $(2k+1)^2 = 4k^2 + 4k + 1.$
then $4k^2 + 4k + 1 + 5$ would need to be divisible by 4 and I don't know where to go from there.
| When $n$ is even,$n^2$ should also be even.Then $n^2+5$ should be odd,so it cannot be divided by 4.
When $n$ is odd,say $n=2k+1$,then $n^2+5=(2k+1)^2+5=4k^2+4k+6$.Since $4k^2$ and $4k$ can be divided by 4,this simply is about whether 6 can be divided by 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
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To find all odd integers $n>1$ such that $2n \choose r$ , where $1 \le r \le n$ , is odd only for $r=2$ For which odd integers $n>1$ is it true that $2n \choose r$ where $1 \le r \le n$ is odd only for $r=2$ ? I know that $2n \choose 2$ is odd if $n$ is odd but I want to find those odd $n$ for which the only value of $r$ between $1$ and $n$ such that $2n \choose r$ is odd , is $2$ . I am wanting to know this as for such an $n$ , if $a^{2n}=a , \forall a \in R $ where $R$ is a ring then $R$ is commutative .Indeed $-a=(-a)^{2n}=a^{2n}=a$ so $2a=0 , \forall a \in R$ ; then since $a^2,a$ commute , so in the expansion of $(a^2+a)^{2n}$ , since $2n \choose r$ is even for all $r$ between $3$ and $2n-3$ , and also ${2n \choose 1 }={2n \choose 2n-1}=2n$ is even , these terms vanish due to $2a=0 , \forall a \in R$ , so
$(a^2+a)^{2n}=(a^2)^{2n}+ {2n \choose 2}(a^2)^{2n-2}.a^2+{2n \choose 2n-2}(a^2)^2.a^{2n-2}+a^{2n}=a^2+ a^{4n-2}+a^{2n+2}+a=a^2+a^{2n-1}+a^3+a$
so $0=(a^2+a)^{2n}-(a^2+a)=a^{2n-1}+a^3=a^{2n-1}-a^3$ that is $a^{2n-1}=a^3 , \forall a \in R$ , so
$a^4=a.a^3=a.a^{2n-1}=a^{2n}=a , \forall a \in R$ , thus R is commutative ; so please help to find all odd
integers $n>1$ such that $2n \choose r$ , where $1 \le r \le n$ , is odd only for $r=2$ . Thanks in advance
| Let $2 n = n_{0} + n_{1} 2 + \dots + n_{k} 2^{k}$ be the base $2$ expansion of $2 n$, so all $n_{i} \in \{0, 1\}$. We have $n_{0} = 0$, $n_{1} = 1$, and may assume $n_{k} = 1$, that is, $2^{k-1} \le n < 2^{k}$.
Similarly, let $1 \le r = r_{0} + r_{1} 2 + \dots + r_{k-1} 2^{k-1} \le n$, with $n_{i} \in \{0, 1\}$.
Then by Lucas' Theorem
$$
\binom{2n}{r}
\equiv
\prod_{i=0}^{k} \binom{n_{i}}{r_{i}}
\equiv
\prod_{i=0}^{k-1} \binom{n_{i}}{r_{i}}\pmod{2}.
$$
So the left hand side is even when there is an $i$ such that $n_{i} = 0$ and $r_{i} = 1$. You want this to happen for all $2 < r \le n$.
If there is $1 < i < k$ such that $n_{i} = 1$, then $r = 2^{i} \le 2^{k-1} \le n$, and
$$
\binom{2n}{r} \equiv 1 \pmod{2}.
$$
So the required condition is that for all $1 < i < k$ we have $n_{i} = 0$, that is, $2 n = 2 + 2^{k}$, that is, $n = 1 + 2^{k-1}$ for some $k > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$ I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$
where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS:
$$
\begin {align}
a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\
&=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\
&=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\
&=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc]
\end{align}$$
where $$s=\frac{a+b+c}{2}$$
I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
But I'm stuck there. Any help please?
| HINT:
Use Law of sines $a=2R\sin A$ etc.
Now Double-Angle Formula $2\sin A\cos A=\sin2A$
Then Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Finally use $\triangle=\dfrac{abc}{4R}$ (Proof)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding all values of $p$ for which $\operatorname{div}\vec{F} = 0$ Here is the full question:
Let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ and let $r = \|\vec{r}\|$. Let $\vec{F} = r^p\vec{r}$. Find all values of $p$ for which div $\vec{F} = 0$
I'm a bit confused on this as it seems that at any point in the vector field, $r^p$ cannot equal zero for any value of $p$ unless $r = 0$, then the value of $p$ would be irrelevant.
So, if I were to start finding div $\vec{F}$, the first thing I would need to do it find:
$$\operatorname{div}\vec{F} = \frac{\partial}{\partial x} \left(x\sqrt{x^2+y^2+z^2}^p\right) + \frac{\partial F_2}{\partial x} + \frac{\partial F_3}{\partial x}$$
And looking just at the first partial:
$$\frac{\partial}{\partial x}\left(x\sqrt{x^2+y^2+z^2}^p\right) = 2x^2\left(p\sqrt{x^2+y^2+z^2}^{p-1}\right) + \sqrt{x^2+y^2+z^2}^p$$
In order to get div $\vec{F} = 0$, $F_1$, $F_2$, and $F_3$ must go to zero when taking the partials or be zero, however, if I did the above correctly, they will not go to zero, regardless of the value of $p$.
Am I completely missing the mark here?
Thank you very much for your time and assistance.
Best,
Eric
| Observe that in taking one of the partial derivatives we find that
\begin{eqnarray*}
\frac{\partial}{\partial x} x \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}} & = & \frac{p}{2} x \cdot (2x) \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}-1} + \left (x^2 + y^2 + z^2
\right )^{\frac{p}{2}} \\
& = & px^2 \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}-1} + ||\textbf{r}||^p.
\end{eqnarray*}
When we sum all three components we find that
\begin{eqnarray*}
\nabla \cdot \textbf{F} & = & p(x^2 + y^2 + z^2) \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}-1} + 3||\textbf{r}||^p \\
& = & p\left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}} + 3||\textbf{r}||^p \\
& = & p||\textbf{r}||^p + 3||\textbf{r}||^p
\end{eqnarray*}
and this quantity equals zero only when $p = -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$ Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$.
My attempt:
I tried multiplying the whole thing but that didn't help at all. So, I tried to manipulate the triangle inequality and bring out the given form but that didn't help too. I am out of ideas now. Please help. Thank you.
| Note
$$(a+1)(b+1)(c+1)=1+abc+a+b+c+ab+bc+ca.$$ So it suffices to show $$abc+a+b+c<2,$$
or $$abc+a+b+c-2=abc+a+b+c-1-ab-bc-ca<0.$$
This is the same as
$$(a-1)(b-1)(c-1)<0.$$
The only problematic possibility is $a>1>b>c.$ In this case,
$$1=ab+bc+ca>b+bc+c>b+c>a,$$
a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Expand $(a−b)^2 +(b−c)^2 +(a−c)^2$ , and hence prove that $a ^2 +b^2 +c^2 ≥ ab+bc+ac$. I would like some advice on how I can prove that $$a^2 +b^2 +c^2 ≥ ab+bc+ac.$$
I have completed the first part of the question that asks for an expansion:
$$ (a−b)^2 +(b−c)^2 +(a−c)^2$$
$$(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)$$
$$ 2a^2+2b^2+2c^2-2ab-2bc-2ac $$
$$ 2(a^2+b^2+c^2-ab-bc-ac)$$
If I have done the expansion incorrectly, corrections in this aspect would also be helpful.
| Hint: $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq .....0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can I quickly find the determinant of this matrix $$
\begin{vmatrix}
14 & 2 & 1 & 3\\
31 & 4 & 5 & 6\\
26 & 3 & 7 & 4\\
10 & 1 & 3 & 2\\
\end{vmatrix}
=
\begin{vmatrix}
5\cdot2+1+3 & 2 & 1 & 3\\
5\cdot4+5+6 & 4 & 5 & 6\\
5\cdot3+7+4 & 3 & 7 & 4\\
5\cdot1+3+2 & 1 & 3 & 2\\
\end{vmatrix}$$$$
=
\begin{vmatrix}
5\cdot2 & 2 & 1 & 3\\
5\cdot4 & 4 & 5 & 6\\
5\cdot3 & 3 & 7 & 4\\
5\cdot1 & 1 & 3 & 2\\
\end{vmatrix}
+
\begin{vmatrix}
1 & 2 & 1 & 3\\
5 & 4 & 5 & 6\\
7 & 3 & 7 & 4\\
3 & 1 & 3 & 2\\
\end{vmatrix}
+
\begin{vmatrix}
3 & 2 & 1 & 3\\
6 & 4 & 5 & 6\\
4 & 3 & 7 & 4\\
2 & 1 & 3 & 2\\
\end{vmatrix}
$$
However I am not able to proceed beyond. The answer given is zero. Is there any simple determinant property that I am not able to guess?
| The two last determinants in your right side are zero as they both have two equal columns, so you're left with:
$$\begin{vmatrix}
14 & 2 & 1 & 3\\
31 & 4 & 5 & 6\\
26 & 3 & 7 & 4\\
10 & 1 & 3 & 2\\
\end{vmatrix}= \begin{vmatrix}
5\cdot2 & 2 & 1 & 3\\
5\cdot4 & 4 & 5 & 6\\
5\cdot3 & 3 & 7 & 4\\
5\cdot1 & 1 & 3 & 2\\
\end{vmatrix}= 5\begin{vmatrix}
2 & 2 & 1 & 3\\
4 & 4 & 5 & 6\\
3 & 3 & 7 & 4\\
1 & 1 & 3 & 2\\
\end{vmatrix}=0
$$
Of course, much easier, short and clear, after you wrote your first equality sign, is to remark that the determinant is zero directly as
$$C_1=5C_2+C_3+C_4\;,\;\;C_i=i-\text{th column}$$
so that the matrix's rank isn't full and etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Fractions in Questions and Answers
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