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Is there a reason the prime factors of $|M_{24}|$ are all one less than the factors of $24$? Wikipedia says of the Mathieu group $M_{24}$, a $5$-transitive permutation group acting on $24$ points,
$$ |M_{24}|= 2^{10}\cdot3^3\cdot5\cdot7\cdot11\cdot23. $$
The prime factors $2,3,5,7,11,23$ are all one less than one of the factors $3,4,6,8,12,24$ of $24$.
Am I crazy? Is this a coincidence, or does it admit an explanation?
(I suppose it could be a combination of both: maybe the factors $11$ and $23$ for some $24$-related reason and $2,3,5,7$ because of the law of small numbers, for example.)
|
The fact that $M_{24}$ is 5-transitive forces its order to be
$$
|M_{24}| = 24 \cdot (24-1) \cdot (24-2) \cdot (24-3) \cdot (24-4) \cdot |H|,
$$
where $H$ is the subgroup fixing any ordered list of five distinct points. (In fact, we have $|H| = 48$.) It follows that $|M_{24}|$ is divisible by
\begin{align*}
24 \cdot \frac{24-1}{1} \cdot \frac{24-2}{2} \cdot \frac{24-3}{3} \cdot \frac{24-4}{4} & = 24 \cdot \left(\frac{24}{1}-1\right) \cdot \left(\frac{24}{2}-1\right) \cdot \left(\frac{24}{3}-1\right) \cdot \left(\frac{24}{4}-1\right) \\
& = 24 \cdot 23 \cdot 11 \cdot 7 \cdot 5.
\end{align*}
So the primes $5, 7, 11$, and $23$ arise "because" they're one less than divisors of $24$. The remaining primes, $2$ and $3$, appear in several places: as divisors of $24$ itself, as divisors of $|H|$, and in the denominators we introduced.
|
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"timestamp": "2023-03-29T00:00:00",
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|
modulus, floor, quotient $a = qM + r,\; q = \big\lfloor \frac{a}{M} \big\rfloor,\; 0 \leq r < M,\; r = a \bmod M$ are not self-consistent when $M < 0$ This question is different than other $\bmod M$ where $M < 0$ questions I've found because those don't specifically ask about the 5 relations I ask about below.
When dealing with modular arithmetic like $a\bmod M$ or division inside a floor function like $\big\lfloor \frac{a}{M} \big\rfloor$, I often use the following relations:
$$
\begin{align}
a &= qM + r &&(1)\\
q &= \Big\lfloor \frac{a}{M} \Big\rfloor &&(2)\\
0 &\leq r < M &&(3)
\end{align}
$$
which imply
$$0 \leq r = a - \Big\lfloor \frac{a}{M} \Big\rfloor M < M \qquad(4)$$
Sometimes, I also use
$$r = a \bmod M \qquad (5)$$
but I'm not sure if these relations are self-consistent in all cases, such as when $M$ is negative.
When $M < 0$, it is not possible to satisfy condition $(3)$ which is $0 \leq r < M < 0$. So does condition $(3)$ need to be changed?
For example, if we use $a = \pm 10$ and $M = \pm 3$, then solve for $q$ and $r$, we later find that $r$ does not satisfy $(3)$ when $M < 0$.
First, let's solve for $q = \big\lfloor \frac{a}{M} \big\rfloor$
$$ \begin{aligned} && M \rightarrow\\ &\;a\\ &\downarrow \end{aligned} $$
$3$
$-3$
$10$
$\big\lfloor \frac{10}{3} \big\rfloor = \big\lfloor 3.333\dots \big\rfloor = 3$
$\big\lfloor \frac{10}{-3} \big\rfloor = \big\lfloor -3.333\dots \big\rfloor = -4$
$-10$
$\big\lfloor \frac{-10}{3} \big\rfloor = \big\lfloor -3.333\dots \big\rfloor = -4$
$\big\lfloor \frac{-10}{-3} \big\rfloor = \big\lfloor 3.333\dots \big\rfloor = 3$
Now let's solve for $r = a - \Big\lfloor \frac{a}{M} \Big\rfloor M$
$$ \begin{aligned} && M \rightarrow\\ &\;a\\ &\downarrow \end{aligned} $$
$3$
$-3$
$10$
$10 - \Big\lfloor \frac{10}{3} \Big\rfloor 3 = 10 - 3 \cdot 3 = 1 \text{ which satisfies } (3)$
$10 - \Big\lfloor \frac{10}{-3} \Big\rfloor \cdot (-3) = 10 - (-4) \cdot (-3) = -2 \text{ which does not satisfy } (3)$
$-10$
$-10 - \Big\lfloor \frac{-10}{3} \Big\rfloor 3 = -10 - (-4) \cdot 3 = 2 \text{ which satisfies } (3)$
$-10 - \Big\lfloor \frac{-10}{-3} \Big\rfloor \cdot (-3) = -10 - 3 \cdot (-3) = -1 \text{ which does not satisfy } (3)$
As we can see, $r$ does not satisfy $(3)$ when $M < 0$. So what is the correct set of relations to use when $M < 0$?
|
Based on @BillDubuque's comment/Wikipedia link, I made a Desmos that explains 5 types of division and their corresponding quotient, remainder, and Modulo operation.
Summary: Modulo operation, mod(x, M), is defined in terms of remainder of integer division, which in turn, is defined in terms of quotient of integer division, which in turn, depends on the definition of integer division, of which there are several; therefore, there are several definitions of Modulo operation.
From programming, I had always thought of floored division and integer division being synonymous, but now I know floored division is just one possible definition of integer division.
For the particular case of floored division, for divisor $M \in \mathbb M_{\neq 0}$, we have
$$
\begin{cases}
0 \leq r < M &\text{if } M > 0\\
M < r \leq 0 &\text{if } M < 0
\end{cases}
$$
which can be consolidated as
$$0 \leq r < \lvert M \rvert$$
|
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|
How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful trick to evaluate the integral.
Noting that
$$I(1):= \int_{0}^{\infty} \frac{d x}{x^{4}-x^{2}+1} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{2}}{x^{4}-x^{2}+1}
$$
Combining them yields
\begin{aligned}
I(1)&=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\\&= \frac{1}{2}\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&= \frac{1}{2}\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
&= \frac{1}{2}\tan ^{-1}\left(x-\frac{1}{x}\right)_{0}^{\infty} \\
&= \frac{\pi}{2}
\end{aligned}
Later, I started to investigate the integrands with higher powers.
Similarly,
$$
I(2):= \int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{2}} \stackrel{x \mapsto \frac{1}{x}}{=}\int_{0}^{\infty} \frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}} d x
$$
By division, we decomposed $x^6$ and obtain $$
\frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}}=\frac{x^{2}+1}{x^{4}-x^{2}+1}-\frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}
$$$$
I(2)=\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x-\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}dx
$$
We can now conclude that $$I(2)=I(1)=\frac{\pi}{2} $$
My Question:
How about the integral $$\displaystyle I_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}$$ for any integer $n\geq 3$?
|
Too long for a comment
You could be interested by
$$\displaystyle I_{n}=\int_{0}^{t} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}=t \, F_1\left(\frac{1}{2};n,n;\frac{3}{2};e^{+\frac{i \pi }{3}} t^2,e^{-\frac{i \pi
}{3}} t^2\right)$$ where appears the Appell hypergeometric function of two variables and the roots of the quadratic in $x^2$.
I did not find any interesting reduction formula or any asymptotics for $t \to \infty$.
Nevertheless,
$$\displaystyle J_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}=\pi \,\frac {a_n}{2^{b_n}}$$ where the $a_n$ correspond to the sequence
$$\{1,1,9,21,25,483,2359,2907,115533,288915,363363,3674619,37326471,\cdots\}$$ and the $b_n$ the sequence
$$\{1,1,4,5,5,9,11,11,16,17,17,20,23,\cdots\}$$ None of these has been found in $OEIS$.
Edit
For
$$I_{m,n}=\int_0^\infty\frac{dx}{(x^m+1)^n}=\frac{\Gamma \left(1+\frac{1}{m}\right) \Gamma \left(n-\frac{1}{m}\right)}{\Gamma(n)} \qquad \text{if} \qquad \Re(m n)>1\land \Re(m)>0$$
gives the simple
$$I_{m,n+1}=\left(1-\frac{1}{m n}\right)I_{m,n}$$
|
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|
Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin{equation}\frac{1}{(-n)^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3} \end{equation}
so that $\Delta n = dn = -n-(n+1) = -2n-1$.
For $f(n) = 1/n^2$, $f'(n) = -2/n^3$, and
\begin{equation}
df = d(\frac{1}{n^2}) = f'(n)dn
\end{equation}
\begin{equation}
df = \frac{(2)(2n+1)}{n^3}
\end{equation}
Now, for very large $n$, $2n+1 \approx 2n$. Thus, \begin{equation}
df = \frac{4}{n^2}
\end{equation}
But, $4/n^2$ is not approximately equal to $2/n^3$ (required ans.) even if $n$ is very large.
So, please point out my mistake(s).
Thanks in advance (;
|
$x:=1/n$ and consider small $x$.
The expression reads
$x^2-\dfrac{x^2}{(1+x)^2}=$
$x^2(1-(1+x)^{-2})=$
$x^2(1-[1-2x+O(x^2)])=$
$2x^3+O(x^4)$, and we are done.
Used: Binomial expansion
|
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|
A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations and I have a rather complicated question question that I could not write as the sum of squares:
$$ 0\leq 6x^2y^2z^2 +19(x^3y^3+y^3z^3+z^3x^3)+27(x^4yz+xy^4z+xyz^4)-24(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3).$$
Thanks for any ideas that might help me clarify the issue.
|
Alternative proof:
(pqr method)
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
The desired inequality is written as
$$\frac{8}{27} \le \frac{p^3 r - 6pqr + q^3 + 8r^2}{q^3} = \frac{p^3 r - 6pqr + 8r^2}{q^3} + 1. \tag{1}$$
We split into two cases:
Case 1 : $p^3 r - 6pqr + 8r^2 \ge 0$
Clearly, (1) is true.
Case 2: $p^3 r - 6pqr + 8r^2 < 0$
Using $q^2 \ge 3pr$, it suffices to prove that
$$\frac{8}{27} \le \frac{p^3 r - 6pqr + 8r^2}{q \cdot 3pr} + 1$$
or
$$\frac{9p^3 - 35pq + 72 r}{27pq} \ge 0.$$
Using $r \ge \frac{4pq - p^3}{9}$ (3 degree Schur),
we have
$$9p^3 - 35pq + 72 r
\ge 9p^3 - 35pq + 72\cdot \frac{4pq - p^3}{9}
= p(p^2 - 3q) \ge 0.$$
We are done.
|
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|
Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluate $$I_1=\int_{0}^{\infty} \frac{\ln x}{x^{2}+1} d x \stackrel{x\mapsto\frac{1}{x}}{=} -I_1 \Rightarrow I_1= 0.$$
and then start to raise up the power of the denominator
$$I_n=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x .$$
In order to use differentiation, I introduce a more general integral
$$I_n(a)=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+a)^n} d x. $$
Now we can start with $I_1(a)$. Using $I_1=0$ yields
$$\displaystyle 1_1(a)=\int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x \stackrel{x\mapsto\frac{x}{a}}{=} \frac{\pi \ln a}{4 \sqrt a} \tag*{}$$
Now we are going to deal with $I_n$ by differentiating it by $(n-1)$ times
$$
\frac{d^{n-1}}{d a^{n-1}} \int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{a}\right)
$$
$$
\int_{0}^{\infty} \ln x\left[\frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{x^{2}+a}\right)\right] d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt a}\right)
$$
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+a\right)^{n}} d x=\frac{(-1)^{n-1} \pi}{4(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)
$$
In particular, when $a=1$, we get a formula for $$
\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x=\left.\frac{(-1)^{n-1} \pi}{4(n-1)!} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)\right|_{a=1}}
$$
For example, $$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{5}} d x=\frac{\pi}{4 \cdot 4 !}(-22)=-\frac{11 \pi}{48}
$$
and
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{10}} d x=\frac{-\pi}{4(9 !)}\left(\frac{71697105}{256}\right)=-\frac{1593269 \pi}{8257536}
$$
which is check by WA.
MY question
Though a formula for $I_n(a)$ was found, the last derivative is hard and tedious.
Is there any formula for $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$
|
Alternative solution:
Using the identity ($q > 0$)
$$
\frac{1}{q^n} = \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-qy}y^{n - 1}\mathrm{d} y,
$$
we have
\begin{align*}
I_n &= \int_0^\infty \frac{\ln x}{(x^2 + 1)^n}\mathrm{d} x\\
&= \int_0^\infty \left( \frac{1}{\Gamma(n)}\int_0^\infty e^{-y(1 + x^2)}y^{n - 1}\mathrm{d}y \right) \ln x\, \mathrm{d} x \\
&= \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-y} y^{n - 1}
\left(\int_0^\infty \mathrm{e}^{-y x^2}\ln x \, \mathrm{d} x\right) \mathrm{d} y\\
&= \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-y} y^{n - 1}
\left( - \frac{(\gamma + \ln 4)\sqrt \pi}{4\sqrt y} - \frac{\sqrt{\pi}\ln y}{4\sqrt y} \right) \mathrm{d} y \\
&= -\frac{(\gamma + \ln 4)\sqrt \pi}{4\Gamma(n)}\int_0^\infty \mathrm{e}^{-y} y^{n - 3/2}\mathrm{d} y - \frac{\sqrt{\pi}}{4\Gamma(n)}
\int_0^\infty \mathrm{e}^{-y} y^{n - 3/2}\ln y \, \mathrm{d}y\\
&= -\frac{(\gamma + \ln 4)\sqrt \pi}{4\Gamma(n)}\Gamma(n - 1/2) - \frac{\sqrt{\pi}}{4\Gamma(n)}\Gamma'(n - 1/2)\\
&= -\frac{(\gamma + \ln 4)\sqrt \pi}{4\Gamma(n)}\Gamma(n - 1/2) - \frac{\sqrt{\pi}}{4\Gamma(n)} \psi(n - 1/2)\Gamma(n - 1/2)
\end{align*}
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$,
and we have used
\begin{align*}
&\int_0^\infty \mathrm{e}^{-y x^2}\ln x \, \mathrm{d} x\\
=\,& \int_0^\infty \frac{1}{\sqrt y}\mathrm{e}^{-u^2}\ln \frac{u}{\sqrt y}\,\mathrm{d} u \\
=\,& \frac{1}{\sqrt y}\int_0^\infty \mathrm{e}^{-u^2}\ln u \,\mathrm{d} u - \frac{\ln y}{2\sqrt y}\int_0^\infty \mathrm{e}^{-u^2}\mathrm{d} u\\
=\,& - \frac{(\gamma + \ln 4)\sqrt \pi}{4\sqrt y} - \frac{\sqrt{\pi}\ln y}{4\sqrt y},
\end{align*}
and
$$\int_0^\infty \mathrm{e}^{-y} y^{n - 3/2}\ln y \, \mathrm{d}y = \Gamma'(n - 1/2) = \psi(n - 1/2)\Gamma(n - 1/2).$$
|
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|
n-rooks n-colors problem The problem goes as follows: $n$ colors are used to color the squares in a $n\times m$ chess board so that each color is used exactly $m$ times. Can you always place $n$ rooks on the board, so that the rooks don't attack each other and there is exactly one rook for every color?
Visual example for $8\times 8$:
What are the values of $n$ and $m$ where this is true? Particularly, what happens when $n=m$? Does this problem have a name? I encountered this problem here, but the comments were not helpful in my googling attempts.
Obviously, if $n>m$, it's impossible to place $n$ rooks, without them threatening each other. And also if the statement is true for some $(n,m)$ it's also true for $(n,m+1)$.
Here are some colorings that are impossible to satisfy
$$n=m=2$$
\begin{matrix}
1 & 2 \\
2 & 1 \\
\end{matrix}
$$n=m=3$$
\begin{matrix}
2 & 3 & 1 \\
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{matrix}
$$n=m=4$$
\begin{matrix}
4 & 4 & 1 & 3 \\
2 & 3 & 4 & 1 \\
2 & 4 & 1 & 2 \\
3 & 3 & 2 & 1 \\
\end{matrix}
$$n=m=5$$
\begin{matrix}
3 & 4 & 5 & 3 & 4 \\
1 & 2 & 4 & 1 & 2 \\
3 & 4 & 2 & 2 & 4 \\
1 & 5 & 2 & 1 & 5 \\
5 & 3 & 1 & 5 & 3 \\
\end{matrix}
|
Proof that this is impossible for $n=m$
The proof is disappointingly simple. You can just have every row be numbers $[1,n]$ in order, except the last one which is shifted one down.
Example for $n=4$:
\begin{matrix}
1 & 1 & 1 & 4 \\
2 & 2 & 2 & 1 \\
3 & 3 & 3 & 2 \\
4 & 4 & 4 & 3 \\
\end{matrix}
The proof that this is impossible is as follows: From the left $n-1$ columns, you will pick every number, except $x$. But because the bottom column is shifted, $x$ will not be on the only open row in the rightmost column. Thus $n=m$ is impossible.
As pointed out by Daniel Mathias, this can be easily extended for $n<m\le 2(n-1)$
\begin{matrix}
1 & 1 & 1 & 4 & 4 & 4 \\
2 & 2 & 2 & 1 & 1 & 1 \\
3 & 3 & 3 & 2 & 2 & 2 \\
4 & 4 & 4 & 3 & 3 & 3 \\
\end{matrix}
|
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|
How do I evaluate $ \int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x, \textrm{ where }0 \leq a\leq 1$? We tackle the integral by converting it into a Beta function by letting $y=\sin^2x$, then
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x &=\int_{0}^{1} \frac{y^{\frac{a}{2}} d y}{2 y^{\frac{1}{2}}(1-y)^{\frac{1}{2}}} \\
&=\frac{1}{2} \int_{0}^{1}y^ \frac{a-1}{2}(1-y)^{-\frac{1}{2}} d y \\
&=\boxed{\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right)}
\end{aligned}
$$
Also it can be expressed in terms of Gamma function by the property of Beta function.
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x &=\frac{\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{a}{2}+1\right)} \\
&=\boxed{\frac{\sqrt{\pi} \Gamma\left(\frac{a+1}{2}\right)}{a \Gamma\left(\frac{a}{2}\right)}}
\end{aligned}
$$
For example, $$
\begin{aligned}
&\int_{0}^{\frac{\pi}{2}} \sin x d x=\frac{1}{2} B\left(1, \frac{1}{2}\right)=1 \\
&\int_{0}^{\frac{\pi}{2}} \sqrt{\sin x} d x=\frac{2 \sqrt{\pi} \left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\\& \int_{0}^{\frac{\pi}{2}} \sqrt[3]{\sin x} d x=\frac{3 \sqrt{\pi} \Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)}\\&\int_{0}^{\frac{\pi}{2}} \sqrt[6]{\sin x} d x=\frac{1}{2} B\left(\frac{7}{12}, \frac{1}{2}\right)
\end{aligned}
$$
My Question
Is there any other closed form of the integral?
|
First, there is an antiderivative : for $0\leq x \leq \frac \pi 2$ $$I=\int \sin ^{a}( x)\, dx=-\cos (x) \,\,
_2F_1\left(\frac{1}{2},\frac{1-a}{2};\frac{3}{2};\cos ^2(x)\right)$$ In terms of summation
$$I=-\frac{1}{\Gamma \left(\frac{1-a}{2}\right)}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1-a}{2}\right)}{(2 n+1)\, \Gamma (n+1)}\,\cos ^{2 n+1}(x)$$
|
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|
Find $\alpha$ such that $b$ is in the column space of $A$ Let $A=\begin{bmatrix}
1 & 0 & -2 & 1\\
0 & 4 & -1 & 0\\
1 & 0 & -1 & 0\\
2 & 4 & -4 & 1
\end{bmatrix} $ and let $b= \begin{bmatrix}
1\\
-3\\
4\\
\alpha
\end{bmatrix}.$ Find $\alpha$ such that $b$ is in the column space of $A$.
I’ve found $rref(A)= \begin{bmatrix}
1 & 0 & 0 & -1\\
0 & 1 & 0 & -1/4\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}$ and thus Rank $A=3$.
Now, I know that the column space of $A$, $C(A)$, is the set of all linear combinations of the columns of $A$. And I know rank $A$. How can I relate all this information to find $\alpha$?
|
You have to find the RREF of the augemnted matrix
$A=\begin{bmatrix}
1 & 0 & -2 & 1&1\\
0 & 4 & -1 & 0&-3\\
1 & 0 & -1 & 0&4\\
2 & 4 & -4 & 1&\alpha
\end{bmatrix} $
And then see that for what value of $\alpha$ do you get rank of the augmented matrix to be $3$. (That is the last row is all zeroes). That would imply that you will have a solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given diagonals in a parallelogram. Find sides. Given a parallelogram with $d_1 = AC = 26$ cm, $d_2 = BD = 18$ cm and $\sin \displaystyle \angle AOD = \frac{12}{13}$.
Find $AB = a$ and $AD = b$.
What I did in order to solve it
*
*Using the formula for the area
$S = \frac{d_1d_2\sin \displaystyle \Phi}{2} = \frac{26 \times 18 \times 12/13}{2} = 216$
*
*Using Heron's formula for $\triangle BOC$ to find $b$
Defining $OC = a = 26 / 2 = 13$, $OB = b = 18 / 2 = 9$, $BC = c$
$p = \frac{a + b +c}{2} = \frac{13 + 9 + c}{2} = \frac{22 + c}{2}$
$S = \sqrt{p(p-a)(p-b)(p-c)}$
I believe when I find $b$, I will able to find $a$ too, but kinda stuck at that point.
|
You already noted that the diagonals are divided by half in the center. However, you're making it far too complicated by using the area.
Instead, take a look at the triangle $\triangle AOD$. You already know the length of two sides, $a=9$ and $c=13$ and their inclosed angle $\beta=\operatorname{arcsin} \frac{12}{13}\approx 67.38°$. With the help of the cosine theorem you get:
$$
b^2=a^2+c^2-2ac\cdot\cos \beta \\
b = \sqrt{81+169-2\cdot 9 \cdot 13\cdot\cos\left(\operatorname{arcsin} \frac{12}{13}\right)}=4\sqrt{10}\approx 12.65
$$
By using $\angle AOB = 90°-\angle AOD$ you can calculate the length of $a$ analogously within the triangle $\triangle AOB$.
|
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|
Elementary inequalities exercise - how to 'spot' the right sum of squares? I have been working through CJ Bradley's Introduction to Inequalities with a high school student and have been at loss to see how one could stumble upon the solution given for Q3 in Exercise 2c.
Question:
If $ad-bc=1$ prove that $a^2+b^2+c^2+d^2+ac+bd \geq \sqrt{3}$.
Solution:
Make the inequality homogenous by multiplying both sides by $ad-bc=1$. [That seems sensible.] Take everything onto one side so we now want to show
$$a^2+b^2+c^2+d^2+ac+bd -\sqrt{3}(ad-bc) \geq 0 \tag{1}.$$
[That's also a reasonable thing to do. The trouble is coming next...]
Now play around until you notice the left hand side can be written as
$$\frac{1}{4}(2a+c-\sqrt{3}d)^2 + \frac{1}{4}(2b+d+\sqrt{3}c)^2 \tag{2}.$$
[What??]
I played around for a fair while and didn't get to this. I have a suspicion that this question was created by reverse-engineering. What thought processes get you from (1) to (2), without knowing (2) beforehand? How can you get to the solution without pulling a rabbit out of a hat?
|
You could use the identity:
$$(ap+br+cs)^2+(aq+bs-cr)^2$$
$$=(a^2p^2+b^2r^2+c^2s^2+2abpr+2acps+2bcrs)+(a^2q^2+b^2s^2+c^2r^2+2abqs-2acqr-2bcrs)$$
$$=(a^2p^2+b^2r^2+c^2s^2+2abpr+2acps)+(a^2q^2+b^2s^2+c^2r^2+2abqs-2acqr)$$
$$=a^2(p^2+q^2)+(b^2+c^2)(r^2+s^2)+2ab(pr+qs)+2ac(ps-qr)$$
If $a^2=b^2+c^2$, dividing by $a^2$ leaves the four squares intact.
|
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|
Summation of cosines
Find the sum $\displaystyle\sum_{n=1}^\infty\left(\frac{2\cos n}{\sqrt{n}}-\frac{1}{\sqrt{n}}\right)\left(\frac{2\cos n}{n}\right)^3\left(\frac{2\cos n}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)$.
I tried in many ways, like applying Fourier series, GP formula, $\cos^3x$, conversions but nothing worked out.
|
Doing some algebraic manipulation we get that your sum $S$ is equal to
\begin{align*}
S &= 8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}\left(2\cos(n) -1 \right)\left(2\cos(n) +1 \right)\\
& =8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}\left(4\cos^2(n) -1 \right)\\
& = 32\sum_{n \ge 1}\frac{\cos^5(n)}{n^4} -8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}
\end{align*}
Now, expanding the cosines in the numerators we get $\cos^3(x) = \frac{3}{4}\cos(x) +\frac{1}{4}\cos(3x)$ and $\cos^5(x) = \frac{10}{16}\cos(x) + \frac{5}{16}\cos(3x) + \frac{1}{16}\cos(5x)$. This means that the sum can thus be written as:
\begin{align*}
S &= 2\left[10\sum_{n \ge 1}\frac{\cos(n)}{n^4} + 5\sum_{n \ge 1}\frac{\cos(3n)}{n^4} + \sum_{n \ge 1}\frac{\cos(5n)}{n^4} -3 \sum_{n \ge 1}\frac{\cos(n)}{n^4} -\sum_{n \ge 1}\frac{\cos(3n)}{n^4}\right]\\
& =2\left[7\sum_{n \ge 1}\frac{\cos(n)}{n^4} + 4\sum_{n \ge 1}\frac{\cos(3n)}{n^4} + \sum_{n \ge 1}\frac{\cos(5n)}{n^4} \right]
\end{align*}
But since we know from this answer that
$$
\sum_{n\ge 1}\frac{\cos(n\,x)}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac 1{48}x^4,\quad \text{for}\;x\in(0,2\pi)
$$
Plugging in $x = 1,3,5$ and simplifying gives the final result
\begin{align*}
S =\boxed{\frac{4 \pi^4}{15}- \frac{34 \pi^2}{3} + 40 \pi-\frac{239}{6}}
\end{align*}
|
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|
Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order and prove that $\frac{1}{\pi}\arctan(\frac{4}{3})$ is irrational.
by contradiction $\exists n$ s.t. $(\frac{3}{5}+\frac{4}{5}i)^n=1$
$(3+4i)^n=5^n$
when $n=2$, $(3+4i)^2=3+4i\pmod5$
stuck how prove that $(3+4i)^n$=$3+4i\pmod5$ use induction? if yes how in this case?
for second part of the question again by contradiction.
$\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$
$\phi=\arctan(\frac{4}{3})=\frac{\pi m}{n}$ how continue from here?
|
As you observed, $(3+4i)^2 = -7 + 24i \equiv 3 + 4i \pmod 5$. Therefore $$(3+4i)^n =(3+4i)^{n-2} (3+4i)^2 \equiv (3+4i)^{n-2} (3+4i) = (3+4i)^{n-1} \pmod 5$$
for all $n > 1$, so $(3+4i)^n \equiv 3+4i \pmod 5$ by induction. Hence $(3+4i)^n \neq 5^n$ for $n > 0$.
If you know some algebraic number theory, you can say the following. We can write $z = (3+4i)/5 = (2+i)/(2-i)$, and $2+i$ and $2-i$ are primes in $\mathbf{Z}[i]$, which is a UFD, so $z^n = 1$ is impossible by unique factorization.
I will leave the second part to you. Hint: write $(3+4i)/5$ in polar form.
|
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|
Differentiating $y = x - \frac2x + \frac3{x^2}$ Another easy question for you guys.
I'm differentiating the below to find the equation of the tangent at $(-3,-2)$
$$y = x - \dfrac{2}{x} + \dfrac{3}{x^2}$$
I simplified to:
$$ y = x - 2x^{-1} + 3x^{-2}$$
Then differentiated to get:
$$ \frac{dy}{dx} = 1 + 2x^{-2} - 6x^{-3}$$ or $$\frac{dy}{dx} = 1 + \frac{2}{x^2} - \frac{6}{x^3}$$
Placing $x = -3$ into this gives me $1$, and placing $m=1$ into $y=mx+c$ gives me $c = 1$.
Making the simple equation: $0 = x - y + 1$
However, I'm given the answer as: $0 = 13x - 9y + 21$
Where did I go wrong, I've studied it for longer than I'm willing to admit, have I made a stupid mistake somewhere?
Thanks,
|
Your calculation up to
$$\frac{dy}{dx} = 1 + \frac{2}{x^2} - \frac{6}{x^3}$$
is correct. However, placing $x=-3$ we get
$$m=1+\frac{2}{(-3)^2}-\frac{6}{(-3)^3}=\frac{13}{9}$$
and so the tangent's equation is given by
$$y - (-2) = \frac{13}{9}(x-(-3))$$
i.e.
$$y = \frac{13}{9}x + \frac{7}{3}$$
or equivalently,
$$9y - 13x - 21 = 0.$$
|
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|
Lyapunov function for a second order system involving trigonometric functions I am studying the stability of the following system:
\begin{aligned}
\dot{x}_{1} &= -x_{1}^{2} - \sin x_{2}\\
\dot{x}_{2} &= x_{1} - \frac{\cos x_{2}}{x_{1}}\\
\end{aligned}
The system itself has 2 equilibrium points:
\begin{equation}
\begin{aligned}
x_{1} &= -\frac{1}{\sqrt[4]{2}}\\
{x}_{2} &= -\frac{\pi}{4}\\
\end{aligned}
\qquad
\qquad
\text{and}
\qquad
\qquad
\begin{aligned}
x_{1} &= \frac{1}{\sqrt[4]{2}}\\
{x}_{2} &= -\frac{\pi}{4}\\
\end{aligned}
\end{equation}
Which I confirmed by analyzing the phase portrait of the system:
Through linearization, I determined that the left equilibrium point is an unstable focus and the right one is a stable focus according to the eigenvalues of the Jacobian matrix of the system evaluated in each equilibrium point.
\begin{equation}
\begin{aligned}
\lambda_{1} &= 1.2613 + 1.1124i\\
\lambda_{2} &= 1.2613 - 1.1124i\\
\end{aligned}
\qquad
\qquad
\text{,}
\qquad
\qquad
\begin{aligned}
\lambda_{1} &= -1.2613 + 1.1124i\\
\lambda_{2} &= -1.2613 - 1.1124i\\
\end{aligned}
\end{equation}
As far as I know, it is possible to make a change of coordinates and translate the stable equilibrium point to the origin. So, my goal consists in finding an appropriate Lyapunov function for determining the stability of the origin (after the change of coordinates), employing Lyapunov's second method.
Among others, here are some candidate functions I've tried so far without satisfactory results:
\begin{equation}
\begin{aligned}
V(x) &= x_{1}\sin x_{1} + \frac{1}{2}x_{2}^{2}\\
V(x) &= \frac{1}{2}x_{1}^{2} + (1-\cos x_{2})\\
V(x) &= x_{1}\sin x_{1} + (1-\cos x_{2})\\
\end{aligned}
\end{equation}
I got stuck trying to find a Lyapunov function due to the trigonometric functions involved in the original system. Could someone give me a suggestion or hint?
|
From the stable equilibrium point $\left(\frac{1}{\sqrt[4]{2}}, - \frac{\pi}{4}\right)$, the state variable transformation can be constructed as
$$
\begin{aligned}
y_{1} &= x_{1} - \frac{1}{\sqrt[4]{2}} \\
y_{2} &= x_{2} + \frac{\pi}{4} \\
\end{aligned}
$$
for which there is an inverse
$$
\begin{aligned}
x_{1} &= y_{1} + \frac{1}{\sqrt[4]{2}} \\
x_{2} &= y_{2} - \frac{\pi}{4} \\
\end{aligned}
$$
such that system model
$$
\begin{aligned}
\dot{x}_{1} &= - x_{1}^{2} - \sin(x_{2}) \\
\dot{x}_{2} &= x_{1} - \frac{\cos(x_{2})}{x_{1}} \\
\end{aligned}
$$
in the new coordinates has the form
$$
\begin{aligned}
\dot{y}_{1} &= - \left(y_{1} + \frac{1}{\sqrt[4]{2}}\right)^{2} - \sin\left(y_{2} - \frac{\pi}{4}\right) \\
\dot{y}_{2} &= y_{1} + \frac{1}{\sqrt[4]{2}} - \frac{\cos\left(y_{2} - \frac{\pi}{4}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} \\
\end{aligned}
$$
that can be expanded to
$$
\begin{aligned}
\dot{y}_{1} &= - y_{1}^{2} - 2^{\frac{3}{4}} y_{1} - \frac{1}{\sqrt{2}} - \sin\left(y_{2}\right) \cos\left(\frac{\pi}{4}\right) + \cos\left(y_{2}\right) \sin\left(\frac{\pi}{4}\right) \\
\dot{y}_{2} &= y_{1} + \frac{1}{\sqrt[4]{2}} - \frac{\cos\left(y_{2}\right) \cos\left(\frac{\pi}{4}\right) + \sin\left(y_{2}\right) \sin\left(\frac{\pi}{4}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} \\
\end{aligned}
$$
and then be simplified to
$$
\begin{aligned}
\dot{y}_{1} &= - \left(y_{1} + 2^{\frac{3}{4}}\right) y_{1} - \left[\cos\left(\frac{\pi}{4}\right) \frac{\sin\left(y_{2}\right)}{y_{2}}\right] y_{2} + \sin\left(\frac{\pi}{4}\right) \cos\left(y_{2}\right) - \frac{1}{\sqrt{2}} \\
\dot{y}_{2} &= y_{1} - \left[\frac{\sin\left(\frac{\pi}{4}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} \frac{\sin\left(y_{2}\right)}{y_{2}}\right] y_{2} - \frac{\cos\left(\frac{\pi}{4}\right) \cos\left(y_{2}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} + \frac{1}{\sqrt[4]{2}} . \\
\end{aligned}
$$
Next, the above modeling equations is represented in the nonlinear state-space format
$$
\begin{aligned}
\dot{\mathbf{y}} &= \mathbf{F}(\mathbf{y}) \mathbf{y} + \mathbf{B} \mathbf{u}(\mathbf{y}) \\
\begin{bmatrix}
\dot{y}_{1} \\
\dot{y}_{2} \\
\end{bmatrix} &= \begin{bmatrix}
- \left(y_{1} + 2^{\frac{3}{4}}\right) & - \frac{1}{\sqrt{2}} \frac{\sin\left(y_{2}\right)}{y_{2}} \\
1 & - \frac{1}{\sqrt{2}} \frac{1}{y_{1} + \frac{1}{\sqrt[4]{2}}} \frac{\sin\left(y_{2}\right)}{y_{2}} \\
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\
\end{bmatrix} + \begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix} \begin{bmatrix}
\frac{1}{\sqrt{2}} \cos\left(y_{2}\right) - \frac{1}{\sqrt{2}} \\
- \frac{1}{\sqrt{2}} \frac{1}{y_{1} + \frac{1}{\sqrt[4]{2}}} \cos\left(y_{2}\right) + \frac{1}{\sqrt[4]{2}} \\
\end{bmatrix} . \\
\end{aligned}
$$
Taking the limit of $\dot{\mathbf{y}}$ as $y_{1} \rightarrow 0$ and $y_{2} \rightarrow 0$, we have
$$
\begin{aligned}
\begin{bmatrix}
\dot{y}_{1} \\
\dot{y}_{2} \\
\end{bmatrix} &= \begin{bmatrix}
- 2^{\frac{3}{4}} & - \frac{1}{\sqrt{2}} \\
1 & - \frac{1}{\sqrt[4]{2}} \\
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\
\end{bmatrix} \\
\begin{bmatrix}
\dot{y}_{1} \\
\dot{y}_{2} \\
\end{bmatrix} &= \begin{bmatrix}
- a & - b \\
1 & - c \\
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\
\end{bmatrix} \\
\dot{\mathbf{y}} &= \mathbf{A} \mathbf{y} . \\
\end{aligned}
$$
Since this is a linear system, we postulate there exists a quadratic Lyapunov function $V(\mathbf{y}) = \mathbf{y}^{T} \mathbf{P} \mathbf{y}$, where $\mathbf{P}$ is a real symmetric positive-definite matrix. The derivative of V along the trajectories of $\dot{\mathbf{y}} = \mathbf{A} \mathbf{y}$ is given by
$$
\dot{V}(\mathbf{y}) = \mathbf{y}^{T} \left(\mathbf{P} \mathbf{A} + \mathbf{A}^{T} \mathbf{P}\right) \mathbf{y} = - \mathbf{y}^{T} \mathbf{Q} \mathbf{y}
$$
where $\mathbf{Q}$ is also a symmetric positive-definite matrix defined by
$$
\begin{aligned}
\mathbf{Q} &= 2 (a + c) (a c + b) \begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix} \\
\end{aligned}
$$
and $2 (a + c) (a c + b) = 9 \sqrt[4]{2} > 0$. Solving the Lyapunov equation
$$
\mathbf{P} \mathbf{A} + \mathbf{A}^{T} \mathbf{P} = - \mathbf{Q}
$$
then the unique solution, $\mathbf{P}$ is obtained
$$
\begin{aligned}
\mathbf{P} &= \begin{bmatrix}
c^2 + a c + b + 1 & a - b c \\
a - b c & a^2 + a c + b^2 + b \\
\end{bmatrix} \\
\end{aligned}.
$$
This approach can be used to establish the local stability for the stable equilibrium point.
|
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|
Derivative of $f(x) = \csc(x)\cot(x)$ using first principles. How to find derivative of $f(x) = \csc(x)\cot(x)$ using First principle of derivative?
I tried the following method.
$f(x) = \csc(x)\cot(x) = \dfrac{\cos(x)}{\sin^2(x)}$
Now using the limit,
$f'(x) = \lim\limits_{h\to0}\dfrac{\dfrac{\cos(x+h)}{\sin^2(x+h)} - \dfrac{\cos(x)}{\sin^2(x)}}{h}$
$f'(x) = \lim\limits_{h\to0}\dfrac{\cos(x+h)\sin^2(x) - \cos(x) \sin^2(x+h)}{h\sin^2(x+h)\sin^2(x)}$
Now, what should I do with this limit? I tried to apply the following identities,
*
*$\cos(A+B)= \cos(A)\cos(B) - \sin(A)\sin(B)$
*$\sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B)$
I also tried to change $\sin^2(x) $ into $\dfrac{1 - \cos(2x)}{2}$,
But all of these formulas seem not to work here. Can anyone guide me for the same?
|
Your $\sin x$ in the last denominator you found should be $\sin^2x$.
Act on the numerator:
$$
\cos(x+h)\sin^2x=\cos x\sin^2x\cos h-\sin^3x\sin h
$$
and
\begin{align}
\cos x\sin^2(x+h)
&=\cos x(\sin x\cos h+\cos x\sin h)^2 \\
&=\cos x\sin^2x\cos^2h+2\sin x\cos^2x\sin h\cos h+\cos^3x\sin^2h
\end{align}
Now subtract and collect the terms with $\sin h$ and those with $\cos h$:
\begin{align}
&(-\sin^3x-2\sin x\cos^2x\cos h+\cos^3x\sin h)\sin h \\
&+\cos x\sin^2x\cos h(1-\cos h)
\end{align}
Now you want
\begin{multline}
\lim_{h\to0}\frac{1}{\sin^2(x+h)\sin^2x}\Bigl((-\sin^3x-2\sin x\cos^2x\cos h+\cos^3x\sin h)\frac{\sin h}{h}\\+\cos x\sin^2x\cos h\frac{1-\cos h}{h}\Bigr)
\end{multline}
Since
$$
\lim_{h\to 0}\frac{\sin h}{h}=1,\qquad \lim_{x\to0}\frac{1-\cos h}{h}=0
$$
your limit evaluates to
$$
\frac{-\sin^3x-2\sin x\cos^2x}{\sin^4x}=-\frac{\sin^2x+2\cos^2x}{\sin^3x}
$$
|
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|
Infinite sums of squares $$\sum_{n=0}^{\infty} \frac {k^2(1-k)^2}{(n+k)^2(n+1-k)^2}$$
Here can anyone help me to solve this question,I can't think of any logic like telescopic, coefficient compare etc .
It would be helpful if anyone could provide the solution
|
The series diverges for integer $k$, Assuming $k \in \mathbb{R} \setminus \mathbb{Z}$, the series sums to $$\frac{k^2(1-k)^2}{(1-2k)^2}\left[\frac{\pi^2}{\sin^2(\pi k)} - \frac{2\pi\cot(\pi k)}{1-2k}\right]\tag{*1}$$
Let $\alpha = k$ and $\beta = 1 - k$, the sum at hand equals to
$$\begin{align}\verb/sum/
&= \alpha^2\beta^2\sum_{n=0}^\infty \frac{1}{(n+\alpha)^2(n+\beta)^2}\\
&= \frac{\alpha^2\beta^2}{(\alpha-\beta)^2}\sum_{n=0}^\infty\left(\frac{1}{n+\alpha} - \frac{1}{n+\beta}\right)^2\\
&= \frac{\alpha^2\beta^2}{(\alpha-\beta)^2}\sum_{n=0}^\infty\left(\frac{1}{(n+\alpha)^2} + \frac{1}{(n+\beta)^2} - \frac{2}{(n+\alpha)(n+\beta)}\right)\\
&= \frac{\alpha^2\beta^2}{(\alpha-\beta)^2}\sum_{n=0}^\infty\left[\frac{1}{(n+\alpha)^2} + \frac{1}{(n+\beta)^2} - \frac{2}{\beta - \alpha}\left(\frac{1}{n+\alpha} - \frac{1}{n+\beta}\right)\right]
\end{align}
$$
The four pieces inside above bracket can be combined into two pieces
$$\sum_{n=0}^\infty \left(\frac{1}{(n+\alpha)^2} + \frac{1}{(n+\beta)^2}\right)
= \sum_{n=0}^\infty \left(\frac{1}{(n+k)^2} + \frac{1}{(-(n+1)+k)^2}\right) = \sum_{n=-\infty}^\infty\frac{1}{(n+k)^2}$$
and
$$\sum_{n=0}^\infty \left(\frac{1}{n+\alpha} - \frac{1}{n+\beta}\right)
= \sum_{n=0}^\infty \left(\frac{1}{n+k} + \frac{1}{-(n+1)+k}\right) = \sum_{n=-\infty}^\infty\frac{1}{(n+k)^2}$$
where sums of the form $\sum\limits_{n=-\infty}^\infty (\cdots)$ is a short hand for the symmetric limit $\lim\limits_{p\to\infty}\sum\limits_{n=-p}^p (\cdots)$
Using following Mittag-Leffler expansion for cotangent and its derivative, $(*1)$ follows immediately.
$$\pi \cot(\pi x) = \sum_{n=-\infty}^\infty \frac{1}{x + n}
\quad\text{ and }\quad
\frac{\pi^2}{\sin^2(\pi x)} = \sum_{n=-\infty}^\infty \frac{1}{(x+n)^2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function for the number of strings following a certain pattern. Let $f_n$ denote the number of strings of length $n$ using letters $A,B,C$ and $D$ such that there are even number of $A's$; $1,2,3$ or $4 \ B's$; either $2$ or $5$ $C's$ and at least $1$ $D$.
So $f_0 = f_1 = f_2 = f3 = 0$, $f_4 = 12$ and $f_5 = 60$. To find the generating function for $f_n$.
I was trying to find a recurrence relation for $f_n$.
Its fine that there is at least one $D$.
But how to control how many $B's$ and $C's$ we have already used?
|
$\color{blue}{\text{Generating Function:}}$
I think that exponential generating functions is the best choice for your question.
*
*If there are even number of $A's$ , then E.G.F of $A$ is : $$1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...+\frac{x^{2k}}{(2k)!}+..=\frac{e^x + e^{-x}}{2}$$
*E.G.F of $B's$ : $$\bigg(\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\bigg)$$
*E.G.F of $C's$ : $$\bigg(\frac{x^2}{2!}+\frac{x^5}{5!}\bigg)$$
*E.G.F of $D's$ : $$\bigg(\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\bigg)=(e^x -1)$$
For any desired $n$ find the coefficient of $x^n$ and multiply it by $n!$ in the expansion of $$n![x^n]\bigg[\bigg(\frac{e^x + e^{-x}}{2}\bigg)\bigg(\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\bigg)\bigg(\frac{x^2}{2!}+\frac{x^5}{5!}\bigg)(e^x -1)\bigg]$$
$\color{red}{\text{EXPANSION:}}$
For example , for $n=5$ , $$5! \times \frac{1}{2}=60$$
|
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"url": "https://math.stackexchange.com/questions/4418411",
"timestamp": "2023-03-29T00:00:00",
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|
Solve $y’’ – 4y’ + 5y = 4e^{2x}\sin(x)$ using $\mathcal D$ operator Hello – I am working through the following question and get stuck at step 6. Could someone please advise in simple terms which I can hopefully understand. Thanks
$$y'' – 4y' + 5y = 4e^{2x}\sin(x)$$
Step one – Order equation so that differential operator is in front of the RHS of the equation
$\newcommand{\D}{\mathcal D}$
$$1 = \frac 1 {\D^2 – 4\D + 5} \cdot 4e^{2x}\sin(x)$$
Step two – move constant and exponential in front of the $\D$ operator
$$1 = 4e^{2x}\cdot \frac 1 {\D^2 – 4\D + 5}\cdot \sin(x)$$
Step three – calculate $a$
Because of $e^{2x}$, $a = 2$, and because of $\sin(x)$, $a = 2 + i$.
Step four – insert $a$ into the $\D$ operator and then calculate to see if it equals zero
\begin{align}
1 &= 4e^{2x}\cdot \frac 1 {(2+i)^2 – 4(2+i) + 5}\cdot \sin(x)
\\
&= 4e^{2x}\cdot \frac 1 {(4+4i+4-8-4i+5)}\cdot\sin(x)
\\
&= 4e^{2x}\cdot \frac 1 {(0)}\cdot \sin(x)
\end{align}
Step 5 – because there is a zero, note $a = 2$ therefore make it $\D+2$
\begin{align}
1 &= 4e^{2x}\cdot \frac 1 {(\D + 2)^2 – 4(\D + 2) + 5}\cdot \sin(x)
\\
&= 4e^{2x}\cdot\frac 1 {\D^2 + 1}\cdot\sin(x)
\end{align}
What do I do for step 6? Please explain in simply terms and assume my calculus knowledge is low.
|
For simplicity, we denote the particular solution as $y_p(t)$.
Then
$$
y_p(t) = {1 \over D^2 - 4 D + 5} \left( 4 e^{2 x} \sin x \right) =
4 {1 \over D^2 - 4 D + 5} \left( e^{2 x} \sin x \right)
$$
Thus, we have
$$
y_p(t) = 4 e^{2 x} {1 \over (D + 2)^2 - 4 (D + 2) + 5} \ \sin x
= 4 e^{2 x} {1 \over (D^2 + 4 D + 4) - (4 D + 8) + 5} \ \sin x
$$
Simplifying, we get
$$
y_p(t) = 4 e^{2 x} {1 \over D^2 + 1} \ \sin x =
4 e^{2 x} \left[ -{x \over 2} \cos x \right]
$$
(Using the standard formulas)
Hence,
$$
\boxed{y_p(t) = - 2 x e^{2 x} \ \cos x}
$$
Finally, I have added a reference for the general formula
$$
y_p(t) = - {x \over 2 a} \ \cos(a x)
$$
for a particular integral of $(D^2 + a^2) y = \sin(a x)$.
Proof for the Particular Integral of $(D^2 + a^2) y = \sin(a x)$
|
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|
Is this nice-looking inequality actually trivial?
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz)≥2(x+y+z)^2 $$
I tried to use the inequality
$$x^2+y^2+z^2≥xy+yz+xz$$
Then I got,
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz) ≥2(xy+yz+xz)^2≥2(x+y+z)^2\implies xy+yz+xz≥x+y+z$$
Which is correct, I think. Because, the degree of the polynomial $xy+yz+xz$ is greater than the degree of the polynomial $x+y+z$. But, I am not sure.
Is this inequality trivial and is my solution correct?
|
Let $S=x+y+z, T=xy+xz+yz$ and note that $S,T \ge 3$ by AMGM and $xyz=1$ and also $S^2 \ge 3T$ by C-S
The inequality is: $T(S^2-T) \ge 2S^2$
Case 1: $S \ge T$ then $S^2(T-2) \ge S^2 \ge T^2$ which rewrites to the required inequality
Case2: $S \le T$ then using $S^2-T \ge 2T$ we get $T(S^2-T) \ge 2T^2 \ge 2S^2$ so we are done also!
|
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|
MCQ about the product of $4$ consecutive odd numbers The product of four consecutive odd numbers must be …
(A) A multiple of 3, but not necessarily of 9.
(B) A multiple of 5 .
(C) A multiple of 7.
(D) A multiple of 9.
(E) A multiple of 3×5×7×9= 945.
Let $n\in \mathbb{Z}$, then
$(2n-1)(2n+1)(2n+3)(2n-3)$
$= (4n^2-1)(4n^2-9)$
$= 16n^4-40n^2+9$
$= 8n^2(2n^2-5)+9.$
For being a multiple of $9$, need $8n^2(2n^2-5)=9m$, for some suitable value of $m\in \mathbb{Z}$; or
$$2n^2-5=9m\cup n=3j, \text{for suitable} \,\,j\in \mathbb{Z}.$$
But, it gets more confusing to pursue further using my approach. Is my approach workable?
Want to add that if take case of showing not divisible by $5$, then : $x+9= 5k$, then $x= 5k-9$.
Hence, for $j, k\in \mathbb{Z}$, $$8n^2 = 5k-9\cup (2n^2-5)= 5j-9.$$
Both options are seemingly not possible, though not clear how to show theoretically the impossibility of both.
|
We use the fact that every number $n\neq 3$. can be written as $3k+1$ or $3k+2$. We chek this in:
$A=(2n-1)(2n+1)(2n+3)(2n+5)$
*
*$n=3k+1\Rightarrow 2n+1=6k+3\Rightarrow 3|A$
$2n+3=6k+5 \Rightarrow 5 \nmid A$
$2n+5=6k+7 \Rightarrow 7 \nmid A$
2): $n=3k+2$
$2n-1=6k \Rightarrow 3\mid A$
$2n+3=6k+7 \Rightarrow 7\nmid A$
$2n+1=6k+5\Rightarrow 5\nmid A$
The common point is that numbers 5 and 7 do not divide A except:
$n=3k, k=0 \Rightarrow 2n+5=5 \Rightarrow 5 \mid A$
$k=2, n=2k+3=7$
$k=4, 2k+5= 13$
$k=4, 2k+1=9$
So options B, C , D and E are exceptions and only option A can always be true.
Reply to comment: When we say every number it includes all primes and composites.
|
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|
How to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$ I want to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$.
Letting $y=f(x)$, we first note that $x=0 \iff y=0$. Next, we consider the case when $y \neq 0$:
$y=\frac{x}{1-x^2} \implies-x^2y+y-x=0 \implies x^2+\frac{x}{y}-1=0 \implies \left(x+\frac{1}{2y}\right)^2-\left(1+\frac{1}{4y^2} \right)=0$
, which can be simplified to:
$\left|x+\frac{1}{2y} \right|=\sqrt{1+\frac{1}{4y^2}} \quad (*)$
If $x+\frac{1}{2y}\gt 0$, then we have: $g_1(y)=x=\sqrt{1+\frac{1}{4y^2}}-\frac{1}{2y}$
If $x+\frac{1}{2y} \lt 0$, then we have: $g_2(y)=x=(-1)\cdot\sqrt{1+\frac{1}{4y^2}}-\frac{1}{2y}$
In its current state, the collection of $(y,x)$ that satisfies $(*)$ does not represent a function...because for a single $y$, there is more than one $x$ that satisfies $(*)$.
Without making reference to graphing software, how does one determine the appropriate inverse for $f$?
Is there a standard procedure? One might argue that I should just try $g_1 \circ f$, $f \circ g_1$, $g_2 \circ f$, and $f \circ g_2$ and see which yield the $I$...but it seems to me like this doesn't necessarily exhaust all possibilities. For example, what if the true inverse is chopped up and pasted together versions of $g_1$ and $g_2$, so as to avoid the doubled values. i.e. let $g_3(x) = \begin{cases}g_1(x) &\text{ if } x \in (a,b)\\ g_2(x) &\text{ if }x \notin (a,b)\end{cases}$
|
I will retain the notation found in the original question...starting from:
$$\left|x+\frac{1}{2y} \right|=\sqrt{1+\frac{1}{4y^2}}$$
This expression means that for any $y \in \mathbb R \setminus \{0\}$:
$x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y} \ \lor\ x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y} \quad (*)$
First suppose that $y \lt 0$. Assume that $x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$. If $y \lt 0$, then $-\frac{1}{2y} \gt 0$. Given that $\sqrt{\left(1+\frac{1}{4y^2}\right)} \gt 1$, we must have that $x \gt 1$.
Next, suppose that $y \gt 0$. Assume that $x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$. If $y \gt 0$, then $-\frac{1}{2y} \lt 0$. Given that $(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)} \lt -1$, we must have that $x \lt -1$.
Now, our original function $f$ is defined on the open interval $(-1,1)$. This means that the inverse function $f^{-1}$ must necessarily have an image of $(-1,1)$.
In order for $(*)$ to be upheld, we must then have the following:
(1) If $y \lt 0$, then $x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$
(2) If $y \gt 0$, then $x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$
Letting $x = f^{-1}(y)$, we have the following description of $f^{-1}$:
$$f^{-1}(y)=\begin{cases}(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}&\text { if } y\lt 0 \\0 &\text { if } y= 0\\\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}&\text { if } y\gt 0 \end{cases}$$
|
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|
Making a matrix with an unknown diagonalizable I'm stuck on a question about diagonalizability of matrices and need some help moving forward.I think I made some progress but I'm also not sure if its any good. Let $a \in \mathbb{R}$, $A$ is a matrix. For which values of $a$ is the matrix $A$ diagonalizable.
$$A:= \begin{pmatrix}
1 & 2 & 3 & 4\\
0 & 2 & a & 5\\
0 & 0 & 2 & 6\\
0 & 0 & 0 & 7 \\
\end{pmatrix} \in \mathbb{R^{4\times4}}$$
Firstly, if I'm not mistaken, any matrix whose charachteristic polynomial is a product of distinct linear factors can be diagonalized. So I simply tried to find the characteristic polynomial first.
$$ p_A(x)= \det(xE_n-A)=\det \begin{pmatrix}
x-1 & -2 & -3 & -4\\
0 & x-2 & -a & -5\\
0 & 0 & x-2 & -6\\
0 & 0 & 0 & x-7 \\
\end{pmatrix}= (x-2) \times \det \begin{pmatrix}
-2 & -a & -5 \\
x-2 & x-2 & -6 \\
0 & 0 & x-7 \\
\end{pmatrix}.$$
$$ (x-2) \times \det \begin{pmatrix}
-2 & -a & -5 \\
x-2 & x-2 & -6 \\
0 & 0 & x-7 \\
\end{pmatrix}=-a^2 x^3 + 2 a^2 x^2 - 9 a x^2 + 32 a x - 28 a - 2 x^3 + 23 x^2 - 66 x + 56.$$ Unless there is an error, this should be the characteristic polynomial. From here I just checked some simple cases like what if $a=0,a=1,a=2$ and found that if $a=2$ then $p_A(x)=-6x^3+13x^2-2x=(-x)(x-2)(6x-1).$ From here however I don't know how to generalize this for any other $a$ or if there is a way to check whether this is the only possible value for $a$.
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Since the matrix $A$ is triangular, the eigenvalues coincide with the diagonal elements of $A$. Therefore, it is given by $p(x)=(x-1)(x-2)^2(x-7)$. Since the only repeated eigenvalue is 2, we need to make sure that the geometric multiplicity of this eigenvalue is equal to 2 to make the matrix diagonalizable.
So, we have that
$$A-2I=\begin{pmatrix}
-1 & 2 & 3 & 4\\
0 & 0 & a & 5\\
0 & 0 & 0 & 6\\
0 & 0 & 0 & 5 \\
\end{pmatrix} $$
One can see that if $a\ne0$, then this matrix has rank 3, which would mean that the kernel has dimension 1 and that the eigenvalue 2 has geometric multiplicity 1, which we do not want. Therefore, $a$ must be equal to 0.
|
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How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit
*
*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our results for both odd and even multiples $n$ of $x$, we can conclude that
$$
\lim _{n \rightarrow \infty} \displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(nx)+\cos ^{6}(nx)\right] \ln (1+\tan x) d x =\frac{5 \pi\ln 2}{64}
$$
*As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below as an answer:
$$
I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)\right] \ln (1+\tan x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}
$$
In order to evaluate the even case
$$\int_{0}^{\frac{\pi}{4}}\left[\sin^{6}(2 n x)+\cos^{6}(2 nx)\right] \ln (1+\tan x) d x $$
we first simplify
$\displaystyle \begin{aligned}\sin ^{6}(2 n x)+\cos ^{6}(2 n x) =& {\left[\sin ^{2}(2 n x)+\cos ^{2}(2 n x)\right]\left[\sin ^{4}(2 n x)-\sin ^{2}(2 n x) \cos ^{2}(2 n x)\right) } \\&\left.+\cos ^{4}(2 n x)\right] \\=& 1-3 \sin ^{2}(2 n x) \cos ^{2}(2 n x) \\=& 1-\frac{3}{4} \sin ^{2}(4 n x) \\=& 1-\frac{3}{8}(1-\cos 8 n x) \\=& \frac{1}{8}(5+3 \cos (8nx))\end{aligned} \tag*{} $
To get rid of the natural logarithm, a simple substitution transforms the integral into
$\begin{aligned}I &=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln (1+\tan x) d x \\& \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\&=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(\frac{2}{1+\tan x}\right) d x \\&=\frac{1}{8} \ln 2 \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x) )d x-I \\I &=\frac{\ln 2}{16} \int_{0}^{\frac{\pi}{4}}(5+3 \cos 8 n x) d x\\&=\frac{\ln 2}{16}\left[5 x+\frac{3}{8 n} \sin (8 n x)\right]_0^{\frac{\pi}{4} }\\ &=\frac{5 \pi}{64} \ln 2\end{aligned} \tag*{} $
My Question:
How can we deal with the odd one
$$\displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x ?$$
Can you help?
|
Similar to the even case, recognize
$$\sin ^{6}mx+\cos ^{6}mx = \frac58+\frac38 \cos4mx $$ and rewrite the integral as follows
\begin{align}
I_{2n+1}=&\int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x \\
=&\> \frac58\int_{0}^{\frac{\pi}{4}}\ln \overset{\frac\pi4-x \to x}{(1+\tan x) } d x
+ \frac38\int_{0}^{\frac{\pi}{4}}
\cos4(2n+1)x\ln\frac{\sqrt2\cos(\overset{\frac\pi4-x \to x}{\frac\pi4-x})}{\cos x}dx\\
=&\> \frac58\cdot \frac\pi8\ln2-
\frac34\int_{0}^{\frac{\pi}{4}}
\cos4(2n+1)x\ln \cos x\> \overset{ibp}{dx}\\
=&\> \frac{5\pi}{64}\ln2-\frac3{16(2n+1)}
\int_{0}^{\frac{\pi}{4}}\frac{\sin 4(2n+1)x\sin x}{\cos x}dx\tag1
\end{align}
Derive the integral below recursively
\begin{align}
K_m=&\int_{0}^{\frac{\pi}{4}}\frac{\sin 4m x\sin x}{\cos x}dx
= K_{m-1}+\frac{(-1)^{m-1}}{2m-1}=-\frac\pi4+\sum_{j=0}^{m-1} \frac{(-1)^j}{2j+1}\\
\end{align}
Then, evaluate $K_{2n+1}$ and plug into (1) to obtain
$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)
$$
In contrast, the even case $I_{2n}= \frac{5\pi}{64}\ln2$ is much simpler.
|
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|
Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$
Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.
I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2),$$
and $3 \le a+b+c$ (using $\frac{9}{a+b+c+6} \le \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$). But I didn't know how to deal with the $\sqrt{a}+\sqrt{b}+\sqrt{c}$.
Can anyone help me? Thank you so much.
|
Complement to @richrow's nice answer
Using $\sqrt{u} \ge \frac{2u}{1 + u}$ for all $u\ge 0$ (easy to prove using AM-GM),
we have
$$\sqrt a = \sqrt 2 \sqrt{a/2}
\ge \sqrt 2 \frac{2 \cdot a/2}{1 + a/2}
= \sqrt 2 \frac{2a}{2 + a} = 2\sqrt 2
- 4\sqrt 2 \frac{1}{2 + a}.$$
Thus, we have
$$\sqrt{a} + \sqrt{b} + \sqrt{c}
\ge 6\sqrt 2 - 4\sqrt 2 \left(\frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c}\right) = 2\sqrt 2$$
where we have used (see OP)
$$\frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c} - 1
= \frac{4 - ab - bc - ca - abc}{(2 + a)(2 + b)(2 + c)} = 0.$$
Also, when $a = b = 2, c = 0$, we have
$ab + bc + ca + abc = 4$ and $\sqrt{a} + \sqrt{b} + \sqrt{c} = 2\sqrt 2$.
Thus, the minimum is $2\sqrt 2$.
|
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|
Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $p^3+q^3+3pq-1=0$
A sequence of numbers $t_0,t_1,t_2,...$ satisfies $$t_{n+2} = pt_{n+1} + qt_n, ~~~n\geq0$$
where $p$ and $q$ are real. Throughout this question, $x,y,z$ are non-zero real numbers.
Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $$p^3+q^3+3pq-1=0~~~~~~(*)$$
Deduce that either either $p+q+1=0$ or $(p-q)^2+(p+1)^2+(q+1)^2=0$. Hence show that either $x=y=z$ or $x+y+z=0$
I have attempted this question, but despite arriving at (*), I don't believe I did it in the correct way.
Here are my workings:
When $n+2 = 3p, 3p+1, 3p+2$ we get three equations.
$$\ z=py+qx\\ x=pz+qy \\
y=px+qz$$
Adding all three equations, we get, $$x+y+z = p(x+y+z) + q(x+y+z) \\ \therefore x+y+z \not=0 \implies p+q=1$$
I then cubed both sides of this equation resulting in $p^3+q^3+3p^2+3pq^2=1 \implies p^3+q^3+3pq(p+q)-1=0 \implies p^3+q^3+3pq-1=0$
then we can proceed to factor this expression to get $p+q=1$ or the sum of the squared terms.
I suspect this approach is incorrect, but not entirely sure why it's incorrect. I would appreciate an explanation on the matter.
|
The approach looks correct to me, since all the steps either follow from previous step or from given information in the question. Here I provide another solution which might be of interest for you based on linear algebra.
(FYI I am currently revising for my linear algebra exam in a few weeks, so it is a cool revision exercise for me too ;)
Firstly, we can write the linear recurrence in matrix form:
$$\begin{pmatrix} t_{n+2} \\ t_{n+1} \end{pmatrix} = \begin{pmatrix} p&q\\1&0 \end{pmatrix}\begin{pmatrix}t_{n+1}\\t_n\end{pmatrix}$$
This is usually enough. However, in this question we want to analyse the terms of period $3$, so let's extend the vectors:
$$\underbrace{\begin{pmatrix}t_{n+3}\\ t_{n+2}\\ t_{n+1}\end{pmatrix}}_{\vec{v_{n+1}}}=\underbrace{\begin{pmatrix}p&q&0\\1&0&0\\0&1&0\end{pmatrix}}_A\underbrace{\begin{pmatrix}t_{n+2}\\t_{n+1}\\t_n\end{pmatrix}}_{\vec{v_n}}$$
(Notice how the third column is all $0$, showing how useless it is.)
Now, let's look at the recurrence every $3$ terms:
$$\vec{v_{n+3}}=A^3\vec{v_n}=\begin{pmatrix} p^3+2pq& p^2q+q^2& 0\\p^2+q&pq&0\\p&q&0\end{pmatrix}\vec{v_n}$$
Since we're given $(t_{3n},t_{3n+1},t_{3n+2})=(x,y,z)$ is constant, it means that the vector $\vec{v}=\begin{pmatrix}z\\y\\x\end{pmatrix}$ satisfies
$$\vec{v} = A\vec{v}$$
What does this remind you of? Eigenvalues! This equation precisely tells us that $1$ is a eigenvalue of $A$ (and $\vec{v}$ is the corresponding eigenvector), since $\vec{v}\neq\vec{0}$.
This means that
$$\begin{align*} \det(A - I_3) &= \begin{vmatrix}p^3+2pq-1&p^2q+q^2&0\\p^2+q&pq-1&0\\p&q&-1\end{vmatrix}\\&=-\begin{vmatrix}p^3+2pq-1&p^2q+q^2\\p^2+q&pq-1\end{vmatrix} \\&= \cdots \\&= p^3+q^3-3pq-1 \\&= 0&(\text{eigenvalue})\end{align*}$$
As desired, and the rest follows.
Hope this helps!
|
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|
Algebraic Proof help with real numbers Let $a$ and $b$ be distinct non-zero real numbers
Show that $-\frac{4ab}{(a-b)^2} = 1 - (\frac{a+b}{a-b})^2$
It's been 15 years since I have been in high school but I came across this problem and wanted to solve it.
$-\frac{4ab}{(a-b)^2} = -\frac{4ab}{a^2+b^2-2ab}$
And now I am stuck I assume I have to split this fraction up but I can't remember how to d that correctly. Any help will be appreciated.
|
Recall that
$(a+b)^2=a^2+2ab+b^2\tag{1}$$(a-b)^2=a^2-2ab+b^2\tag{2}$
Subtracting $(1)$ from $(2)$
$$4ab=(a+b)^2-(a-b)^2$$
$$\begin{align}-\dfrac{4ab}{(a-b)^2}&=-\dfrac{(a+b)^2-(a-b)^2}{(a-b)^2}\\&=-\left(\dfrac{a+b}{a-b}\right)+\dfrac{(a-b)^2}{(a-b)^2}\\&=1-\left(\dfrac{a+b}{a-b}\right)^2\qquad\blacksquare\quad\text{Q.E.D}\end{align}$$
|
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|
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?
|
HINT
A little bit different approach from the other users.
To begin with, notice that
\begin{align*}
\frac{\sqrt{(x - 1)(x + 3)}}{x + 2} & = \frac{\sqrt{x^{2} + 2x - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x^{2} + 4x + 4) - (2x + 4) - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x + 2)^{2} - 2(x + 2) - 3}}{x + 2}\\\\
& = \frac{\sqrt{[(x + 2) - 1]^{2} - 4}}{x + 2}\\\\
& = \frac{\sqrt{(x + 1)^{2} - 4}}{(x + 1) + 1}
\end{align*}
Now we can make the change of variable $u = x + 1$ and study the range of the transformed expression:
\begin{align*}
y = \frac{\sqrt{u^{2} - 4}}{u + 1} \Longleftrightarrow
\begin{cases}
y(u + 1) = \sqrt{u^{2} - 4}\\\\
|u| \geq 2
\end{cases} \Longleftrightarrow
\begin{cases}
(1 - y^{2})u^{2} - 2y^{2}u - y^{2} - 4 = 0\\\\
|u| \geq 2
\end{cases}
\end{align*}
Can you take it from here?
|
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|
Maclaurin series of $\frac{x^2}{1- x \cot x}$ I wonder if there is an explicit formula for the Maclaurin expansion of $\frac{x^2}{1 - x \cot x}$. We know an explicit formula for $1- x \cot x$.
Due to the continued fraction formula for $\tan x$, we know that all of the coefficients after the first are negative.
$\bf{Added:}$ I was lead to this question in trying to prove that in the Maclaurin expansion of $ \frac{x^2}{1 - x \cot x} + \frac{3}{5} ( 1 - x \cot x) - 2 $ all of the coefficients are positive. Since we have formulas for the expansion of the second term, we are interested in explicit formulas for the expansion of the first term.
$\bf{Added:}$ We have the continued fraction
$$\frac{x^2}{1 - x \cot x} = 3 - \frac{x^2}{ 5 - \frac{x^2}{ 7 - \cdots} } $$
following easily from the continued fraction for $\tan x$.
|
Using Bessel functions, we find
\begin{align*}
\frac{{x^3 }}{{1 - x\cot x}} & = 3 - x\frac{{J_{5/2} (x)}}{{J_{3/2} (x)}} = \frac{3}{2} + x\frac{{J'_{3/2} (x)}}{{J_{3/2} (x)}} = 3 - 2x^2 \sum\limits_{k = 1}^\infty {\frac{1}{{1 - (x/j_{3/2,k}^2 )^2 }}}
\\ & = 3 - 2x^2 \sum\limits_{k = 1}^\infty {\sum\limits_{n = 0}^\infty {\frac{1}{{j_{3/2,k}^{2n} }}x^{2n} } } = 3 - 2x^2 \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 1}^\infty {\frac{1}{{j_{3/2,k}^{2n} }}} } \right)x^{2n} } \\& = 3 - 2x^2 \sum\limits_{n = 0}^\infty {\sigma _{2n} \!\left( \tfrac{3}{2} \right)x^{2n} },
\end{align*}
where $j_{3/2,k}$ denotes the $k$th positive zero of $J_{3/2}$ and $\sigma_n$ is the Rayleigh function of order $n$. If $n\geq 1$, we can write
$$
\sigma _{2n} \!\left( {\tfrac{3}{2}} \right) = ( - 1)^{n - 1} 3 \cdot 2^{2n - 1} \frac{{V_{2n} }}{{(2n)!}},
$$
where $V_n$ is the $n$th van der Pol number. See this paper for properties of these numbers, including recurrence relations. In particular, your generating function is equation $(d)$ in Section $1$.
|
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|
Prove that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} = 0$ without substituting I found this possible solution:
Let $r^2=x^2 + y^2, x = r \cos(\theta)$ and $y=r\sin(\theta)$. Then:
$$
\begin{split}
\lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
&= \lim_{r \to 0} \frac{(r \cos(\theta))^2}{\sqrt{r^2}} \\
&= \lim_{r \to 0} \frac{r^2 \cos^2(\theta)}{r} \\
&= \lim_{r \to 0} r \cos^2(\theta)
\end{split}$$
Since $\lvert \cos^2(\theta)\rvert \leq 1$ and $\lim\limits_{r \to 0} r = 0$, we have:
$$
\lim_{r \to 0} r \cos^2(\theta)
= 0
= \lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
$$
I learned this substitution in this class.
First of all I would like to know if my solution is possible, but I would also would like to see a prove that does not use this kind of substitution.
|
You can consider that the following generalized limit:
$$\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}=0.$$
Observe that,
$$\begin{align}\frac{x^2}{\sqrt{x^2}}&=\frac{x^2}{|x|}=|x|\\
&≥\frac{x^2}{\sqrt{x^2 +y^2}}\\
&≥0\end{align}$$
This implies
$$\begin{align}0&=\lim_{(x,y) \to (0,0)}|x|\\
&=\lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
\end{align}$$
This also shows that
$$\begin{align}\lim_{(x,y) \to (0,0)}|x|&=\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}&\\
&=0.
\end{align}$$
where, $r\in\mathbb R$.
|
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|
How to prove the product of first n consecutive odd numbers is a square less than another square? I have observed for first few values of consecutive odd numbers, the result is always of the form:
$m^2 - n^2$, where $m$ and $n$ are two distinct positive integers. That is:
$$1\cdot 3\cdot 5 \cdot 7\cdots (2k-1) = m^2 - n^2$$
For example:
$
1\cdot 3 = 3 = 2^2 - 1^2 \\
1\cdot 3\cdot 5 = 15 = 4^2 - 1^2 \\
1\cdot 3\cdot 5\cdot 7 = 105 = 11^2 - 4^2 \\
1\cdot 3\cdot 5\cdot 7\cdot 9 = 945 = 31^2 - 4^2 \\
\vdots
$
But not sure, how to prove it.
Here is an attempt using induction:
Let it be true for some value of k, that is:
$1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1) = m^2 - n^2$
Then when $k$ takes the value of $k+1$, we have
$$\begin{align}
1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1)\cdot(2k + 1)
&= (m^2 - n^2)\cdot(2k + 1)\\
&= (m^2 - n^2)\cdot {(k+1)^2 - k^2}
\end{align}$$
and got stuck here.
Can you please suggest to proceed further or an altogether a different way of proving so or prove me wrong.
Thanks in advance.
|
Note that any product of two distinct odd numbers or two distinct even numbers can be written as a difference of squares. Let $p<q$ be both odd or both even. Then $(q-p)/2$ and $(q+p)/2$ are distinct integers, and
$$
((q+p)/2)^2 - ((q-p)/2)^2=(q^2/4+pq/2+p^2/4)-(q^2/4-pq/2+p^2/4)=pq.
$$
So this is a fairly weak condition, and one that the product of consecutive odd numbers certainly meets.
|
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|
Find extreme values of ellipse I have the curve equation
$$
Ax^2 + By^2 + Cxy = 1
$$
which represents ellipse with center in (0, 0) and rotated some angle.
How can I find max X and Y values (not semi-axes) on this ellipse? Points are marked on the picture.
There x-es and y-es can not be moved left or right side, so I cannot find dx/dy and dy/dx.
|
Another way, use polar co-ordinates $x=r \cos t, y= r \sin t$ to get
$r^2(t)=\frac{1}{A \cos^2 t+ B\sin ^2 t+C \sin t \cos t}$
Let $u(t)=1/r^2(t)$ $\implies u(t)-\frac{A+B}{2}= \frac{A-B}{2} \cos 2t + \frac{c}{2}\sin 2t= \sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}~ \sin(2t+k)$
$\implies \frac{A+B}{2}-\sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}\le u(t) \le \frac{A+B}{2}+\sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}.$
From here one can get $r_{min}$ and $r_{max}$.
|
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|
Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$ is equal to $1$.
I'm looking for other approaches/ideas to factorize the expression.
|
$xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=xy\left(x+y\right)+yz(y+\color{red}{x}-\color{red}x-z)-xz\left(x+z\right)=xy(x+y)+yz(y+\color{red}x)-yz(\color{red}x+z)-xz(x+z)=y(x+y)(x+z)-z(x+z)(x+y)=(x+y)(x+z)(y-z)$
|
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|
Let p be a prime. Prove that $x^4 + 4 y^4 = p$ has integer solutions if and only if $p = 5$. In this case, find the solutions for x and y. I have tried to factor
$$x^4 + 4 y^4 = p$$
using the Sophie Germain's Identity, as someone suggested in the comments, which yields:
$$((x+y)^2 + y^2)((x-y)^2 + y^2) = p$$
Since p is a prime, either
$$(x+y)^2 + y^2 = 1$$
or
$$(x-y)^2 + y^2 = 1$$
I do not know how to solve this for $x,y$. And I do not see how this will proves that p can only be 5. Furthermore, I do not know how to find the integer solutions for the equation:
$$x^4 + 4 y^4 = 5$$
|
The only prime ramified in the ring of Gauss integers $\mathbb Z[i]$ it is known to be $2=i^3((1+i)^2$ so we have since $i^4=1$ $$x^4+4y^4=(x^2+i2y^2)(x^2-i2y^2)=(x^2+(1+i)^2y^2)(x^2-(1+i)^2y^2)$$
then the factorization in $\mathbb Z[i]$ of $x^4+4y^4$ becomes
$$(x+i(1+i)y)(x-i(1+i)y)(x+(1+i)y)(x-(1+i)y)$$ which gives
$$(x-y+iy)(x+y-iy)(x+y+iy)(x-y-iy)=[(x-y)^2+y^2][(x+y)^2+y^2]$$ Consequently we must consider the two cases $(x-y)^2+y^2=1$ and $(x+y)^2+y^2=p$ and the correlative $(x+y)^2+y^2=1$ and $(x-y)^2+y^2=p$ which easily give the solution to the problem.
|
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|
Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\iff\tan^2\alpha+\tan\alpha-1=0$$ This equation has solutions $\left(\tan\alpha\right)_{1,2}=\dfrac{-1\pm\sqrt5}{2}$ but as $\alpha\in(0^\circ;45^\circ)\Rightarrow$ $\tan\alpha=\dfrac{\sqrt5-1}{2}$. Now $\sin\alpha=\dfrac{\sqrt5-1}{2}\cos\alpha$ and plugging into $\sin^2\alpha+\cos^2\alpha=1$ got me at $\cos^2\alpha=\dfrac{2}{5-\sqrt5}$
|
Consider a right triangle with acute angle $2\alpha$ and side lengths
\begin{align}
\text{opposite} &= 2\\
\text{adjacent} &= 1\\
\text{hypotenuse} &= \sqrt{2^2 + 1^2} = \sqrt{5}
\end{align}
Hence,
\begin{align}
\tan(2\alpha) &= \frac{2}{1} = 2\\
\cos(2\alpha) &= \frac{1}{\sqrt{5}}\\
\sin(2\alpha) &= \frac{2}{\sqrt{5}}
\end{align}
We can now compute
\begin{align}
C = \sin(3\alpha)\cos(\alpha) &= \cos(\alpha)\cdot(3\sin(\alpha) - 4\sin^3(\alpha))\\
& = 3\sin(\alpha)\cos(\alpha) - 4\sin^2(\alpha)\sin(\alpha)\cos(\alpha)\\
& = 3\frac{\sin(2\alpha)}{2} - 4\sin^2(\alpha)\frac{\sin(2\alpha)}{2}\\
& = \frac{\sin(2\alpha)}{2}(3 - 4\sin^2(\alpha))\\
& = \frac{\sin(2\alpha)}{2}(2(1 - 2\sin^2(\alpha)) + 1)\\
& = \frac{\sin(2\alpha)}{2}(2\cos(2\alpha) + 1)\\
& = \frac{\left(\frac{2}{\sqrt{5}}\right)}{2}\left(2\left(\frac{1}{\sqrt{5}}\right) + 1\right)\\
& = \boxed{\frac{2 + \sqrt{5}}{5}}
\end{align}
|
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|
Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$ How to show that
$$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$$
without evaluating each integral individually?
I came up with this problem while working on some problem where I found ( by comparing some results) that
$$\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{2n+1}=\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x$$
and by using $$\int_0^1 x^{2n}\ln(1-x)
\mathrm{d}x=-\frac{H_{2n+1}}{2n+1},$$
we have $$\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{2n+1}=-\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x.$$
|
Performing integration by parts,
\begin{align*}
\int_{0}^{1} \frac{\log(1-x)}{1+x^2} \, \mathrm{d}x
&= \underbrace{\left[ \log(1-x) \left(\arctan x - \frac{\pi}{4}\right) \right]_{0}^{1}}_{=0} - \int_{0}^{1} \frac{\frac{\pi}{4} - \arctan x}{1 - x} \, \mathrm{d}x.
\end{align*}
Now substituting $x = \frac{1-t}{1+t}$ and using the identity $\arctan\bigl(\frac{1-t}{1+t}\bigr) = \frac{\pi}{4} - \arctan t$ for $t > -1$, we get
\begin{align*}
\int_{0}^{1} \frac{\frac{\pi}{4} - \arctan x}{1 - x} \, \mathrm{d}x
&= \int_{0}^{1} \frac{\arctan t}{t(t+1)} \, \mathrm{d}t.
\end{align*}
This proves the desired identity.
|
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|
Calculate the integral $\int\limits_{0}^{8}\frac{dx}{x^2+\sqrt[3]{x}}$ My attempt:
$$\phi =\frac{1+\sqrt{5}}{2}, \; \bar{\phi }=\frac{1-\sqrt{5}}{2}\Rightarrow y^4-y^3+y^2-y+1=\left ( y^2-\phi y+1 \right )\left ( y^2-\bar{\phi }y+1 \right )$$
\begin{multline*}
\int\limits_{0}^{8}\frac{dx}{x^2+\sqrt[3]{x}}\overset{y=\sqrt[3]{x}}{=}\int\limits_{0}^{2}\frac{3y^2}{y^6+y}dy=\int\limits_{0}^{2}\frac{3y}{y^5+1}dy=\int\limits_{0}^{2}\frac{3ydy}{(y+1)(y^2-\phi y+1)(y^2-\bar{\phi }y+1)}=\\=3\int\limits_{0}^{2}\left ( \frac{A}{y+1}+\frac{By+C}{y^2-\phi y+1}+\frac{Dy+E}{y^2-\bar{\phi }y+1} \right )dy=\\=3\Bigg( A\ln(y+1)\Bigg|_0^{2}+\frac{B}{2}\ln(y^2-\phi y+1)\Bigg|_0^{2}+\left ( C-\frac{B}{2} \right )\int\limits_{0}^{2}\frac{dy}{y^2-\phi y+1}+\\+\frac{D}{2}\ln\left ( y^2-\bar{\phi }y+1 \right )\Bigg|_{0}^{2}+\left ( E-\frac{D}{2} \right )\int\limits_{0}^{2}\frac{dy}{y^2-\bar{\phi }+1} \Bigg)=\\=3\Bigg( A\ln 2+\frac{B}{2}\ln (5-2\phi )+\left ( C-\frac{B}{2} \right )\frac{2}{\sqrt{4-\phi ^2}}\arctan \left ( \frac{2y+\phi}{\sqrt{4-\phi^2}} \right )\Bigg|_0^{2}+\\+\frac{D}{2}\ln \left ( 5-2\bar{\phi} \right )+\left ( E-\frac{D}{2} \right )\frac{2}{\sqrt{4-\bar{\phi}^2}}\arctan \left ( \frac{2y+\bar{\phi}}{\sqrt{4-\bar{\phi}^2}} \right )\Bigg|_{0}^2 \Bigg)
\end{multline*}
Where
$$\left ( A,B,C,D,E \right )=\left ( -\frac{1}{5},\frac{1-\sqrt{5}}{10},\frac{1-\sqrt{5}}{10},\frac{1+\sqrt{5}}{10},\frac{1+\sqrt{5}}{10} \right )$$
Wolfram writes such an answer through a hypergeometric function:
$$\int\frac{dx}{x^2+\sqrt[3]{x}}=\frac{3}{2}x^{2/3}{}_2F_1\left ( \frac{2}{5},1;\frac{7}{5};-x^{5/3} \right )+C$$
How do you get the same answer and calculate a definite integral through this answer?
|
There is an antiderivative (have a look here) which can be simplified and write
$$I=\int\frac{dx}{x^2+\sqrt[3]{x}}$$
$$I=\frac{3 \log \left(x^{2/3}+\frac{1}{2} \left(\sqrt{5}-1\right)
\sqrt[3]{x}+1\right)}{4 \sqrt{5}}+\frac{3}{20} \log \left(x^{2/3}+\frac{1}{2}
\left(\sqrt{5}-1\right) \sqrt[3]{x}+1\right)-\frac{3 \log
\left(x^{2/3}-\frac{1}{2} \left(1+\sqrt{5}\right) \sqrt[3]{x}+1\right)}{4
\sqrt{5}}+\frac{3}{20} \log \left(x^{2/3}-\frac{1}{2} \left(1+\sqrt{5}\right)
\sqrt[3]{x}+1\right)-\frac{3}{5} \log \left(\sqrt[3]{x}+1\right)-\frac{3}{5}
\sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)} \tan ^{-1}\left(\frac{-4
\sqrt[3]{x}+\sqrt{5}+1}{\sqrt{10-2 \sqrt{5}}}\right)-\frac{3}{10} \sqrt{10-2
\sqrt{5}} \tan ^{-1}\left(\frac{4 \sqrt[3]{x}+\sqrt{5}-1}{\sqrt{2
\left(5+\sqrt{5}\right)}}\right)$$
So, no problem for the definite integral.
|
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Integrate $\frac{x}{\sin 2x}$ \begin{align}\int_{{\pi}\over5}^{{3\pi}\over10}\frac{x}{\sin2x}\,dx\end{align}
This integral came up while learning integration using Leibnitz rule. What has been tried is taking the integral as:
\begin{align}I(a)=\frac{1}{2}\int_{\pi\over5}^{3\pi\over10}\frac{\arctan (a \tan x)}{\sin x\cos x}\,dx\end{align} which gives $I'(a)$ as:
\begin{align}I'(a)=\frac{1}{2}\int_{2\pi\over5}^{3\pi\over5}\frac{dx}{\cos^2x(1+a^2\tan^2x)}=\frac{1}{2a}\arctan{\frac{a\tan \frac{3\pi}{10}-a\tan{\pi\over5}}{1+a^2\tan{\pi\over5}\cot{\pi\over5}}}\end{align}
\begin{align}=\frac{1}{2a}\arctan{\frac{a(\tan \frac{3\pi}{10}-\tan{\pi\over5})}{1+a^2}}\end{align}
Proceeding after this seems uncertain.
|
A simpler approach using the following property of definite integral:
$$\int_{a}^ {b} f(x) dx=\int_{a} ^{b} f(a+b-x)dx$$
\begin{align}
I&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{{\pi \over 5}+{3 \pi \over 10}-x}{\sin\left({2\left({\pi \over 5}+{3 \pi \over 10}-x\right)}\right)}\: dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{\frac{\pi}{2}-x}{\sin{2\left(\frac{\pi}{2}-x\right)}}\:dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{\frac{\pi}{2}}{\sin{2x}}\:dx -\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\
&=\frac{\pi}{2}\int_{\pi \over 5}^{3 \pi \over 10}\frac{dx}{\sin{2x}} -I\\
\implies 2I &=\frac{\pi}{2}\int_{\pi \over 5}^{3 \pi \over 10} \csc{2x}\: dx\\
&=\frac{\pi}{2}\frac{1}{2} \left[\ln\left|\tan\left(\frac{2x}2\right)\right|\right]_{\pi \over 5}^{3 \pi \over 10}\\
&=\frac{\pi}{4} \left(\ln\left|\tan\left(\frac{3\pi}{10}\right)\right|-\ln\left|\tan\left(\frac{\pi}{5}\right)\right|\right)\\
\implies I&=\frac{\pi}{8} \ln\left|\frac{\tan\left(\frac{3\pi}{10}\right)}{\tan\left(\frac{\pi}{5}\right)}\right|\\
&\approx 0.250901
\end{align}
|
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Spivak, Ch. 13 "Integrals", Prob. 2: Show $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding unique number such that $L(f,P)\leq \int_0^b x^4 \leq U(f,P)$? Consider the problem (Spivak, Chapter 13 "Integrals", problem 2) of showing that $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding the unique number $\int_0^b x^4$ such that
$$L(f,P)\leq \int_0^b x^4 \leq U(f,P)$$
where $L(f,P)$ the lower sum of $f$ for partition $P$ on $[0,b]$, and $U(f,P)$ is the upper sum of $f$ for partition $P$ on $[0,b]$.
In the course of solving this problem, there is a step in which one must prove
$$\frac{1}{6n^5}\left [ 6n^5-15n^4+10n^3-n \right ] < 1\tag{1}$$
and
$$\frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ]>1\tag{2}$$
My question is how to show these two inequalities are true.
Below is context on the steps leading up to the necessity of proving the inequalities $(1)$ and $(2)$.
We start with an equally spaced partition of $[0,b]$, $P_n=\{t_0,...,t_n\}$ where $t_i-t_{i-1}=\frac{b}{n}$.
In each partition subinterval, we have
$$m_i=\inf\{f(x): t_{i-1}\leq x \leq t_i \}=t_{i-1}^4$$
$$M_i=\sup\{f(x): t_{i-1}\leq x \leq t_i \}=t_{i}^4$$
Let's consider $L(f,P)$ specifically
$$L(f,P)=\sum_{i=1}^n m_i(t_i-t_{i-1})=\sum_{i=1}^n t_{i-1}^4(t_i-t_{i-1})$$
$$=\sum_{i=1}^n \left ( \frac{b(i-1)}{n} \right )^4\cdot \frac{b}{n}$$
$$= \frac{b^5}{n^5}\sum_{i=1}^n (i-1)^4$$
$$= \frac{b^5}{n^5}\sum_{i=0}^{n-1} i^4$$
$$= \frac{b^5}{n^5} \left [ \frac{(n-1)^5}{5} +\frac{(n-1)^4}{2}+ \frac{(n-1)^3}{3} -\frac{n-1}{30} \right ]$$
After a bit of algebra we reach
$$L(f,P)=\frac{b^5}{5}\frac{1}{6n^5}\left [ 6n^5-15n^4+10n^3-n \right ]$$
I'd like to show that $(1)$ is $<1$ so that I can assert that $$L(f,P)<\frac{b^5}{5}\tag{3}$$
Similarly,
$$U(f,P)=\sum_{i=1}^n M_i(t_i-t_{i-1})$$
$$=\sum_{i=1}^n t_i^4 (t_i-t_{i-1})$$
$$=\sum_{i=1}^n \left ( \frac{bi}{n} \right )^4\cdot \frac{b}{n}$$
$$=\frac{b^5}{n^5} \sum_{i=1}^n i^4 $$
$$=\frac{b^5}{n^5} \left [ \frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30} \right ]$$
$$=\frac{b^5}{5} \cdot \frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ]$$
I'd like to show that $$\frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ] > 1$$
so I can assert that $$U(f,P)>\frac{b^5}{5}\tag{4}$$
Once I can assert $(3)$ and $(4)$, I have
$$L(f,P)<\frac{b^5}{5}<U(f,P)\tag{5}$$
for all partitions $P$. Since It is also true that $U(f,P)-L(f,P)<\epsilon$ for any $\epsilon>0$ (by definition of $f$ being integrable), we can conclude that there is only one number satisfying $(5)$, and since $\int_0^b f$ definitely satisfies it by definition of $f$ being integrable, it must be the unique number. Hence $\int_0^b f=\frac{b^5}{5}$.
|
Since those 2 inequalities derive from equally spaced partition of $[0,b], P_n = \{t_0, t_1, \ldots, t_n \},$ where $t_i-t_{i-1}=\frac{n}{b}$, it must be true that $n \geq 1$.
Let's show that
\begin{equation}
\frac{1}{n^5}(n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6}) > 1.
\end{equation}
By doing some algebra we get:
\begin{equation}
1+\frac{5}{2n}+\frac{5}{3n^2}-\frac{1}{6n^4} > 1 \Leftrightarrow \frac{5}{2n}+\frac{5}{3n^2}-\frac{1}{6n^4} > 0.
\tag{2}
\label{eq:2}
\end{equation}
Since $n \geq 1$, we can multiply \eqref{eq:2} with $6n^4$. Now, we just have to prove that $15n^4+10n^2-1>0, n \geq 1$.
If we define the function $g(n) = 15n^4+10n^2-1 , n \geq 1$ we have:
\begin{equation}
g(1) = 15+10-1>0 \\
g'(n) = 50n^3+20n>0 \Rightarrow g \uparrow
\end{equation}
So $g(n)>0 \ \forall n \geq 1.$ We can work using the same method to prove the other inequality. Hope that helps !
|
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|
How to solve long integration by partial fraction decomposition problems faster? Some problems are just too time consuming for short exam times what is the fastest way to solve problems like this one for example $$\int \frac{5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}dx$$
|
If you're going for a purely partial fraction decomposition approach, then I'm afraid there isn't much you can do to expedite the process. Partial fraction decomposition is just a naturally tedious and long-winded thing to do.
$$\frac {5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}=\frac A{x-2}+\frac {Bx+C}{x^2-2x+2}+\frac {Dx+E}{(x^2-2x+2)^2}\tag1$$
Clear the fraction to get
$$5x^4-21x^3+40x^2-37x+14=A(x^2-2x+2)^2+(Bx+C)(x-2)\times\ldots\\(x^2-2x+2)+(Dx+E)(x-2)\tag2$$
There is a slightly faster way to obtain most of the coefficients than expanding the right-hand side and comparing the values. First, set $x=2$ and observe that the second and third terms of the right-hand side vanish.
$$12=4A\qquad\implies\qquad A=3\tag3$$
Note that another way to find $A$ would've been to take the limit of the left-hand side of $(1)$ ignoring the $x-2$ term$$A=\lim\limits_{x\to2}\frac {5x^4-21x^3+40x^2-37x+14}{(x^2-2x+2)^2}=3$$This is essentially the same as the way I've shown above, just in less steps.
Next, set the quadratic equal to zero and observe how the first and second terms vanish. To do so, no need to actually solve the quadratic, observe that $x^2=2x-2$ and replace all occurrences of even powers with that.
$$5(2x-2)^2-21x(2x-2)+40(2x-2)-37x+14=(Dx+E)(x-2)$$
After much simplification, and replacing any resulting $x^2$ terms with $2x-2$ after expanding, you should arrive
$$x-2=(Dx+E)(x-2)\qquad\implies\qquad(D, E)=(0, 1)\tag4$$
Unfortunately, I do not know of a way to solve for $B$ and $C$ with a single substitution. Replace $A$, $D$, and $E$ in $(2)$ with the values derived from $(3)$ and $(4)$ and simplify.
$$(2x-1)(x-2)(x^2-2x+2)=(Bx+C)(x-2)(x^2-2x+2)$$
Clearly, $B=2$ and $C=-1$. Putting everything together gives
$$\frac {5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}=\frac 3{x-2}+\frac {2x-1}{x^2-2x+2}+\frac 1{(x^2-2x+2)^2}$$
|
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Probability regarding random marbles chosen from a bag I am unsure if my thought process for these questions is correct.
Susie has a bag of marbles containing 3 Red, 7 Green, and 10 Blue marbles.
*
*What is the probability of picking 5 marbles and getting at least one red marble? Calculate the probability (a) with replacement, and (b) without replacement.
What I have tried so far is that with replacement, it would be $$1- \left(\frac{17}{20}\right)^5 = 0.556$$
For without replacement, I tried $$1- \left(\frac{17}{20} \times \frac{16}{19} \times \frac{15}{18} \times \frac{14}{17} \cdot \frac{13}{16}\right) = 0.601$$
*Calculate the probability of picking $8$ marbles and getting exactly $4$ green and $4$ blue (a) with replacement and (b) without replacement.
What I have tried for this so far is that with replacement, it would be $$\left(\frac{7}{20}\right)^4 \cdot \left(\frac{10}{20}\right)^4 \cdot 8^2 = 0.066$$
For without replacement, I am unsure so I just assumed it would follow the same as 1(b) based off of 1(a).
|
You solved both parts of the first problem correctly.
For the second part of the first problem, where you are selecting without replacement, you could also express the answer in the form
$$\Pr(\text{at least one red}) = 1 - \Pr(\text{no reds}) = 1 - \frac{\dbinom{17}{5}}{\dbinom{20}{5}}$$
where the denominator counts the number of ways of selecting five of the $20$ marbles in the bag and the numerator of the subtracted term counts the number of ways of selecting $5$ of the $20 - 3 = 17$ non-red marbles in the bag.
Susie has a bag of marbles containing $3$ red, $7$ green, and $10$ blue marbles. Calculate the probability of picking $8$ marbles and getting exactly $4$ green and $4$ blue with replacement.
Since the selections are made with replacement, the probability of selecting a particular color is the same for each outcome. We can use the multinomial distribution. The probability of selecting exactly $r$ red, $g$ green, and $b$ blue marbles from $3$ red, $7$ green, and $10$ blue marbles when $n = r + g + b$ marbles are selected from the bag with replacement is
$$\Pr(r, g, b) = \binom{n}{r, g, b}\left(\frac{3}{20}\right)^r\left(\frac{7}{20}\right)^g\left(\frac{10}{20}\right)^b$$
Substituting $8$ for $n$, $0$ for $r$, $4$ for $g$, and $4$ for $b$ yields
$$\Pr(r, g, b) = \binom{8}{0, 4, 4}\left(\frac{3}{20}\right)^0\left(\frac{7}{20}\right)^4\left(\frac{10}{20}\right)^4$$
Susie has a bag of marbles containing $3$ red, $7$ green, and $10$ blue marbles. Calculate the probability of picking $8$ marbles and getting exactly $4$ green and $4$ blue without replacement.
Since the marbles are selected without replacement, we can use the multivariate hypergeometric distribution. The probability of selecting exactly $r$ red, $g$ green, and $b$ blue marbles when $n = r + g + b$ marbles are selected from the $3$ red, $7$ green, and $10$ blue marbles in the bag without replacement is
$$\Pr(r, g, b) = \frac{\dbinom{3}{r}\dbinom{7}{g}\dbinom{10}{b}}{\dbinom{20}{n}}$$
Can you finish?
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|
${f}\begin{pmatrix}x \\y \\\end{pmatrix}$=$\begin{pmatrix}x^2-y^2 \\2xy \end{pmatrix}$ is differentiable at each point I need help with understanding some steps of a task. I tried to solve the uncertainty by myself by using "Approach zero" but the problem is, that I am not able to write these kind of functions in "Approach zero". Maybe you could already help me out by showing how its done.
Show that the function $\mathbf{f}:\mathbb{R^2}\to\mathbb{R^2}$ with $$\mathbf{f}\begin{pmatrix}x \\y \\\end{pmatrix}=\begin{pmatrix}x^2-y^2 \\2xy \\\end{pmatrix}$$ is differentiable at each point in $\mathbb{R^2}$ and calculate its derivative.
My first question is how to type that in "Approach Zero" because I am 100% sure that I could find some posts dealing with this question using approach.
Here is the solution that I don't quite understand:
We now that $\frac{\partial(f_1,f_2)}{\partial(x,y)}\begin{pmatrix}x \\y \\\end{pmatrix}=\begin{pmatrix}2x&-2y \\2y&2x \\\end{pmatrix}$. Now we want to show, that this matrix is the total derivative d$\mathbf{f[x]}$. So we have to show that $\mathbf{f}$ is differentiable at each point with d$\mathbf{f[x]}$ being its derivative , which is the case, if $\lim\limits_{\mathbf{h}\to 0}\frac{\mathbf{f(x+h)-f(x)}-\frac{\partial(f_1,f_2)}{\partial(x,y)}\begin{pmatrix}x \\y \\\end{pmatrix}\mathbf{h}}{|\mathbf{h}|}=0$ But what is the solution doing:
|$\mathbf{f(x+h)-f(x)}-\frac{\partial(f_1,f_2)}{\partial(x,y)}(\mathbf{x})\mathbf{h}|=|\begin{pmatrix}h_1^2-h_2^2 \\2h_1h_2 \\\end{pmatrix}|=O(|\mathbf{h}|^2)$ for $\mathbf{h}\to O$ and therefore $\mathbf{f}$ is differentiable.
My question is why? Where does this whole thing come from. Why aren't we using this instead $\lim\limits_{\mathbf{h}\to 0}\frac{\mathbf{f(x+h)-f(x)}-\frac{\partial(f_1,f_2)}{\partial(x,y)}\begin{pmatrix}x \\y \\\end{pmatrix}\mathbf{h}}{|\mathbf{h}|}$. How does a substraction between a vector and a matrix even work? And where does this whole thing $|\begin{pmatrix}h_1^2-h_2^2 \\2h_1h_2 \\\end{pmatrix}|=O(|\mathbf{h}|^2)$ even come from? What even is this $O$ and why are we allowed to say that $\mathbf{f}$ is totally differentiable?
Is there anyone who could help me out? I would be very grateful.
|
Here is a sketch of a solution: Let $\mathbf{x}=[x, y]^\intercal$ and $\mathbf{h}=[h, k]^\intercal$.
Notice that
$$(x+h)^2-(y+k)^2=x^2-y^2+2xh-2yk+h^2-k^2$$
and
$$2(x+h)(y+k)=2xy+2xk+2yk+2hk$$
Thus
\begin{align}
\mathbf{f}(\mathbf{x}+\mathbf{h})&=\begin{pmatrix}x^2-y^2\\ 2xy\end{pmatrix} + \begin{pmatrix} 2xh-2yk\\ 2xk+2yh\end{pmatrix}+ \begin{pmatrix} h^2-k^2\\ 2hk\end{pmatrix} \\
&=\mathbf{f}(\mathbf{x})+\begin{pmatrix}2x & -2y\\ 2y & 2x\end{pmatrix}\begin{pmatrix}h
\\k\end{pmatrix} + \begin{pmatrix} h^2-k^2\\ 2hk\end{pmatrix}
\end{align}
Set
$$\mathbf{r}(\mathbf{h})=\begin{pmatrix} h^2-k^2\\ 2hk\end{pmatrix}$$
Then
\begin{align}
\frac{\|\mathbf{r}(\mathbf{h})\|}{\|\mathbf{h}\|}=\sqrt{\frac{(h^2-k^2)^2+4h^2k^2}{h^2+k^2}}=\sqrt{\frac{(h^2+k^2)^2}{h^2+k^2}}=\sqrt{h^2+k^2}\xrightarrow{\|\mathbf{h}\|\rightarrow0}0
\end{align}
The rational is based on the meaning of differentiation. A function $\mathbf{f}$ is differentiable at a point $\mathbf{x}_0$ there is a linear transform $A_{\mathbf{x}_0}$ such that
\begin{align}
\mathbf{f}(\mathbf{x}_0+\mathbf{h})=\mathbf{f}(\mathbf{x}_0)+A_{\mathbf{x}_0}\mathbf{h} + \mathbf{r}(\mathbf{x}_0;\mathbf{h})\tag{1}\label{one}
\end{align}
where
\begin{align}
\lim_{\mathbf{h}\rightarrow\mathbf{0}}\frac{\|\mathbf{r}(\mathbf{x_0};\mathbf{h})\|}{\|\mathbf{h}\|}=0\tag{2}\label{two}
\end{align}
In the OP, the function $\mathbf{f}$ is independent of the point $\mathbf{x}$. This is not a surprise in this case since $\mathbf{f}$ is a quadratic function.
|
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|
Conjecture: $\frac1\pi=\sum_{n=0}^\infty\left((n+1)\frac{C_n^3}{2^{6n}}\sum_{k=0}^n(-1)^k{n\choose k}{\frac{(n-k)(k-1)}{(2k-1)(2k+1)}}\right)$ Let $C_n$ denote the $n$-th Catalan number defined by
$${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}=\prod \limits _{k=2}^{n}{\frac {n+k}{k}}\quad \left(n\geqslant 0\right).}$$
Next, we define the sequence
$${\displaystyle A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}}.$$
I've numerically managed to verify that
\begin{equation}\tag{1}\label{pi}
{\displaystyle \sum _{n=0}^{\infty }(n+1)A_{n}={\frac {1}{\pi}}}.
\end{equation}
Is it possible to prove the relation in (\ref{pi})? If yes, then how could we go about it?
Also, is this series already known or studied in the literature? If yes, then any references will be highly appreciated. Thanks!
|
$$\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}=\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}$$
$$ A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}=\frac{\Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi \Gamma (n+2)^2}$$
$$(n+1)A_n=\frac{(n+1) \Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi \Gamma (n+2)^2}$$
$$S_p=\sum_{n=0}^p(n+1)A_n=\frac 1 \pi \frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}$$ Now, Stirling approximation
$$\log \left(\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}\right)=\log(p+1)+2\log \left(\Gamma \left(p+\frac{3}{2}\right)\right)-2\log \left(\Gamma \left(p+{2}\right)\right)$$
$$\log \left(\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}\right)=-\frac{1}{4 p}+\frac{1}{4 p^2}+O\left(\frac{1}{p^3}\right)$$
$$\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}=1-\frac{1}{4 p}+\frac{9}{32 p^2}+O\left(\frac{1}{p^3}\right)$$
$$S_p=\frac 1 \pi \left(1-\frac{1}{4 p}+\frac{9}{32 p^2}+O\left(\frac{1}{p^3}\right) \right)$$
|
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|
Integral with Weierstrass substituion, limits gone wrong I'm trying to evaluate the following integral with a Weierstrass substitution:
$$
\int_{\pi/3}^{4\pi/3} \frac{3}{13 + 6\sin x - 5\cos x} \text{d}x
$$
This comes out to about 0.55 when evaluated numerically. I thought it would be possible to rewrite the integral using the substitution $t = \tan\frac{x}{2}$ like so:
$$
\text{Upper limit: }\frac{4\pi}{3} \to \tan\frac{4\pi}{6} \to -\sqrt{3}
$$
$$
\text{Lower limit: }\frac{\pi}{3} \to \tan\frac{\pi}{6} \to \sqrt{3}/3
$$
$$
\int_{\sqrt{3}/3}^{-\sqrt{3}} \frac{3(1+t^2)}{2(3t+1)^2 + 6} \cdot \frac{2}{1 + t^2} \text{d}x
$$
However, evaluating this numerically now gives -1.26, so it's clear that this method is wrong. I am sure that
$$
\frac{3}{13 + 6\sin x - 5\cos x} = \frac{3(1+t^2)}{2(3t+1)^2 + 6}
$$
So something must have gone wrong changing the limits, but I've never come across this problem before. I thought the integral may be improper but plotting $\frac{3}{13 + 6\sin x - 5\cos x}$ on a graph shows it has no asymptotes or any other obvious problems in the range of integration.
Any help would be greatly appreciated :)
|
The Weierstrass substitution is valid for $x\in(-\pi,\pi)$ or $x\in(\pi,3\pi)$, but not for an interval containing a neighborhood of $\pi$. As $x$ tends to $\pi$ from below, $z$ tends to $+\infty$. As $x$ tends to $\pi$ from above, $z$ tends to $-\infty$.
$$
\begin{align}
&\int^{4\pi/3}_{\pi/3}\frac3{13+6\sin(x)-5\cos(x)}\,\mathrm{d}x\\
&=\int^{\pi}_{\pi/3}\frac3{13+6\sin(x)-5\cos(x)}\,\mathrm{d}x
+\int^{4\pi/3}_{\pi}\frac3{13+6\sin(x)-5\cos(x)}\,\mathrm{d}x\tag{1a}\\
&=\int_{1/\sqrt3}^\infty\frac3{13+6\frac{2z}{1+z^2}-5\frac{1-z^2}{1+z^2}}\frac{2\,\mathrm{d}z}{1+z^2}
+\int_{-\infty}^{-\sqrt3}\frac3{13+6\frac{2z}{1+z^2}-5\frac{1-z^2}{1+z^2}}\frac{2\,\mathrm{d}z}{1+z^2}\tag{1b}\\
&=\int_{1/\sqrt3}^\infty\frac{3\,\mathrm{d}z}{4+6z+9z^2}
+\int_{-\infty}^{-\sqrt3}\frac{3\,\mathrm{d}z}{4+6z+9z^2}\tag{1c}\\
&=\int_{\tan^{-1}\left(\frac{3+\sqrt3}3\right)}^{\pi/2}\frac{\mathrm{d}u}{\sqrt3}
+\int_{-\pi/2}^{\tan^{-1}\left(\frac{-9+\sqrt3}3\right)}\frac{\mathrm{d}u}{\sqrt3}\tag{1d}\\[3pt]
&=\frac1{\sqrt3}\tan^{-1}\left(\frac{60-24\sqrt3}{13}\right)\tag{1e}
\end{align}
$$
Explanation:
$\text{(1a)}$: break the integral into ranges where $z$ is increasing
$\text{(1b)}$: $z=\tan(x/2)$, $\sin(x)=\frac{2z}{1+z^2}$, $\cos(x)=\frac{1-z^2}{1+z^2}$, $\mathrm{d}x=\frac{2\,\mathrm{d}z}{1+z^2}$
$\text{(1c)}$: simplify
$\text{(1d)}$: if $\sqrt3z+1/\sqrt3=\tan(u)$, then $4+6z+9z^2=3\sec^2(u)$ and $\mathrm{d}z=\frac1{\sqrt3}\sec^2(u)\,\mathrm{d}u$
$\text{(1e)}$: $\tan^{-1}\left(\frac{3+\sqrt3}3\right)\gt\frac\pi4$ and $\tan^{-1}\left(\frac{9-\sqrt3}3\right)\gt\frac\pi4$, and
$\phantom{\text{(1e): }}$$\tan\left(\pi-\tan^{-1}\left(\frac{3+\sqrt3}3\right)-\tan^{-1}\left(\frac{9-\sqrt3}3\right)\right)=\frac{60-24\sqrt3}{13}$
|
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|
Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\textrm{ and }$ $b+c$ are positive real numbers.
First of all, let’s ‘kill’ the term $\sin x$ and the square of $\cos x$ by identity and double angle formula.
$\displaystyle \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x =& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b) \cos ^{2} x+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b)\left(\frac{1+\cos 2 x}{2}\right)+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[\frac{a-b}{2} \cos 2 x+\left(b+c+\frac{a-b}{2}\right)\right] d x \\\stackrel{2x\mapsto x}{=} & \frac{1}{2} \int_{0}^{\pi} \ln \left[\frac{a-b}{2} \cos x+\left(\frac{a+b+2 c}{2}\right)\right] d x\end{aligned}\tag*{} $
By my post, I found, by Feynman’s Integration Technique, that
$\displaystyle \int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right),\tag*{} $
where $\left|\frac{c}{b}\right|\geq 1$.
$\displaystyle I=\frac{\pi}{2} \ln \left[\frac{\frac{a+b+2 c}{2}+\sqrt{\left(\frac{a+b+2 c}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}}}{2}\right]\tag*{} $
Simplifying gives the result
$\displaystyle \begin{aligned}I&=\frac{\pi}{2} \ln \left[\frac{a+b+2 c+2 \sqrt{(a+c)(b+c)}]}{4}\right] \\&=\frac{\pi}{2} \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)^{2} \\&=\pi \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)\end{aligned}\tag*{} $
|
This isn't really an answer as such, but as a general comment, the next trick you should know is how to "peel off" the logarithm. Let's guess that for some functions $g$ and $h$:
$$\int \log f(x)\,dx = h(x) \log f(x) + g(x)$$
Taking the derivative of both sides:
$$\log f(x) = h'(x) \log f(x) + h(x) \frac{f'(x)}{f(x)} + g'(x)$$
This is clearly going to have a solution if $h'(x) = 1$, that is, $h(x) = x$. So:
$$\int \log f(x)\,dx = x \log f(x) + g(x)$$
where:
$$g(x) = - \int x \frac{f'(x)}{f(x)}\,dx$$
In your case:
$$\int \log \left(b \cos x + c\right)\,dx = x \log \left(b \cos x + c\right) + g(x)$$
where:
$$g(x) = \int \frac{x \sin x}{\cos x + c/b}\,dx$$
This works for any elementary field extension, not just the logarithm function. You can think of this as integration by parts, but I find the Risch algorithm interpretation easier to apply.
|
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|
Log rules for calculating joint entropy This question is probably not so hard for you.
Why is the entropy equal to:
$$
H(x,y)=2\log_2(5)-\frac{8}{25}\log_2(2)-\frac{6}{25}\log_2(3),
$$
for the following joint distribution?
$$
p(x,y) =
\frac{1}{25}
\begin{bmatrix}
1&1&1&1&1\\
2&1&2&0&0\\
2&0&1&1&1\\
0&3&0&2&0\\
0&0&1&1&3
\end{bmatrix}
$$
My answer is:
$$
H(x,y)=\frac{11}{25}\log_2(\frac{1}{25})-\frac{8}{25}\log_2(\frac{2}{25})-\frac{6}{25}\log_2(\frac{3}{25}),
$$
and the only thing I could think of applying here is $\log(\frac{1}{x})=-\log(x)$ and $\log_2(2)=1$.
Which is useful for $-\log_2(\frac{1}{25})=\log_2(25)=\log_2(5^2)=2\log(5)$.
Hope someone can help. Its from this book: https://link.springer.com/book/10.1007/978-0-387-79234-7.
|
As far as I understand you need help with logarithms simplification? Because I think your calculations are almost (possible typo) ok:
My answer is: $H(x,y)=\frac{11}{25}\log_2(\frac{1}{25})-\frac{8}{25}\log_2(\frac{2}{25})-\frac{6}{25}\log_2(\frac{3}{25}).$
I thing you miss minus sine $H(x,y)=-\frac{11}{25}\log_2\frac{1}{25}-\frac{8}{25}\log_2\frac{2}{25}-\frac{6}{25}\log_2\frac{3}{25}.$ Sincethe joint Shannon’s entropy is given by $$H(X,Y)=-\sum_{x\in X}\sum_{y\in Y}p(x,y)\log p(x,y)$$
while I'm referring to notation from your book Information Theory
and Network Coding, Raymond W. Yeung p. 43. So let's simplify logarithms. But at first notice there are many equivalents ways to do this so you (and potential reader) do not have to stick with mine. Since $$-\frac{11}{25}\log_2\frac{1}{25} = \frac{11}{25}\log_2 25 = \frac{22}{25}\log_2 5 \\ -\frac{8}{25}\log_2\frac{2}{25}= -\frac{8}{25}\log_2 2 + \frac{8}{25}\log_2 25 = -\frac{8}{25} + \frac{16}{25}\log_2 5 \\ -\frac{6}{25}\log_2\frac{3}{25} = -\frac{6}{25}\log_2 3 +\frac{6}{25}\log_2 25= -\frac{6}{25}\log_2 3 +\frac{12}{25}\log_2 5. $$ I'm using dwo facts $$\log \xi/\eta =\log \xi -\log \eta \qquad \& \qquad \log \xi^{\eta}=\eta \log \xi. $$ At the end sum up everything $$\frac{22}{25}\log_2 5 -\frac{8}{25} + \frac{16}{25}\log_2 5 -\frac{6}{25}\log_2 3 +\frac{12}{25}\log_2 5 = 2 \log_2 5 -\frac{8}{25} -\frac{6}{25}\log_2 3 $$ what is the expected answer.
|
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|
Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}D\sec^2\theta d\theta
= \frac{1}{D^3}\int_{-\infty}^\infty \cos^2 \theta d\theta$$
$$=\frac{1}{D^3}[\frac{\theta}{2} + \frac{\sin {2\theta}}{4} + C]_{-\infty}^\infty$$
Put $\theta = \arctan{\frac{x}{D}}$;
$$=\frac{1}{D^3}[\frac{\arctan{\frac{x}{D}}}{2} + \frac{\sin {(2\arctan{\frac{x}{D})}}}{4} + C]_{-\infty}^\infty$$
We get the integral as $\frac{\pi}{2D^3}$. Since $\lim_{x \to +\infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = \frac{-\pi}{2}$
I feel something isn't right here. Can anyone point the mistake please. Thank you very much.
|
Keep it easy: for any $A>0$
$$ f(A) = \int_{0}^{+\infty}\frac{dx}{x^2+A}\stackrel{x\mapsto z\sqrt{A}}{=} \frac{1}{\sqrt{A}}\int_{0}^{+\infty}\frac{dz}{z^2+1}=\frac{\pi}{2\sqrt{A}}.\tag{1}$$
By differentiating with respect to $A$ we get:
$$ -\frac{\pi}{4 A\sqrt{A}} = -\int_{0}^{+\infty}\frac{dx}{(x^2+A)^2} \tag{2}$$
and by replacing $A$ with $D^2$ we have:
$$ \int_{0}^{+\infty}\frac{dx}{(x^2+D^2)^2} = \color{red}{\frac{\pi}{4 D^3}}.\tag{3}$$
|
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|
Distribution of amount of green balls drawn Urn of type A contains 4 green and 3 blue balls. Urn of type B contains 4 blue and 5 green balls.
There are three urns of type A and two urns of type B. We pick one urn at random (out of 5) and we draw 3
balls (with a single draw) out of the chosen urn. Let X denote the number of green balls drawn. Determine
the distribution of X.
I tried the following approach
$
P(X=N)=\frac{3}{5}\binom{3}{N}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\binom{3}{N}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{7}{3}}
$
where $N \in \{ 0,1,2,3 \}$
However sum of probabilities is greater than 1:
$
\sum_{N=0}^{3} \frac{3}{5}\binom{3}{N}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\binom{3}{N}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{7}{3}} = \frac{283}{105}
$
I have a problem with understanding how to calculate the probability of obtaining N green balls from a single draw.
|
I think you may have some unneeded terms. You want $\frac35$ times some hypergeometric probability and $\frac25$ times some hypergeometric probability as you seem to have found. But then you included $\binom{3}{N}$ which I do not understand. You also had a $7$ instead of a $9$ in the bottom right denominator.
Perhaps $$\mathbb P(X=N)=\frac{3}{5}\frac{\binom{4}{N}\binom{3}{3-N}}{\binom{7}{3}}+\frac{2}{5}\frac{\binom{5}{N}\binom{4}{3-N}}{\binom{9}{3}}$$ might sum to $1$.
|
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|
Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment says.
This identity has a special feature, that the form of the expression maintains from LHS to RHS.
Also, for addition, we can expanse this identity to:
$$ (a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2. $$
or:
$$ X=xz-Cyw, Y=axw+a'yz+BYw. \\ (ax^2+Bxy+a'Cy^2)(a'z^2+Bzw+aCw^2)=aa'X^2+BXY+CY^2 $$
, from the answer of @Will Jagy.
This can be proved by various solutions, for example, just multiplying out this identity, or with trigonometric functions, or with the imaginary number "$i$".
I want you to prove this identity with more solutions.
|
The identity is trivially satisfied if $\,a=b=0\,$ or $\,c=d=0\,$. Otherwise $\,a^2+b^2 \ne 0\,$ and $\,\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1\,$, so there exists an angle $\,\alpha\,$ such that $\,\frac{a}{\sqrt{a^2+b^2}} = \sin \alpha\,$, $\,\frac{b}{\sqrt{a^2+b^2}} = \cos \alpha\,$. Similarly, define $\,\gamma\,$ such that $\,\frac{c}{\sqrt{c^2+d^2}} = \sin \gamma\,$ and $\,\frac{d}{\sqrt{c^2+d^2}} = \cos \gamma\,$.
After dividing by $\,(a^2+b^2)(c^2+d^2) \ne 0\,$, the identity can then be written as:
$$
(\sin\alpha\sin\gamma \pm \cos\alpha\cos\gamma)^2 + (\sin\alpha\cos\gamma \mp \cos\alpha\sin\gamma)^2 = 1
\\ \iff\quad \cos^2 (\alpha \mp \gamma) + \sin^2 (\alpha \mp \gamma) = 1 \quad\quad
$$
|
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|
Proving inequality using root of unity Let $\omega$ be a complex cube root of unity. It can be shown that if $a,b,c \in \mathbb{R}$, then $$(a+b+c)(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)= a^3+b^3+c^3-3abc$$
I was wondering if this could be extended to prove the AM/GM inequality for $n=3$. All that needs to be done is to prove that the left expression is non-negative, but I cannot find a way.
|
Proof of AM-GM:
Using $\omega^3=1, \omega^2+\omega+1=0$, We know that $(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)=a^2+b^2+c^2-ab-bc-ca$, then
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ge 0$
If $a,b,c>0$, then $a+b+c>0$ and $\frac{a^2+b^2}{2}\ge ab$ etc give $a^2+b^2+c^2\ge ab+bc+ca$
Hence $\frac{a^3+b^3+c^3}{3}\ge abc \implies \frac{x+y+z}{3} \ge (xyx)^{1/3}.$
|
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|
Compute $\int_0^{\frac\pi2} \frac1{\sin(x+\frac\pi3)\sin(x+\frac\pi6)} dx$ Compute the following definite integral:
$$\int_0^{\frac\pi2} \frac1{\sin(x+\frac\pi3)\sin(x+\frac\pi6)} dx$$
I know that I can rewrite the denominator as shown below:
$$\int_0^{\frac\pi2} \frac1{\frac12\left(\cos(\frac\pi6) - \cos(2x+\frac\pi2)\right)} dx$$
$$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 2\sin(2x)} dx$$
And then use the substitution
$$\sin(2x) = \frac{2\tan(x)}{1+\tan^2(x)}$$
$$dx=\frac1{1+\tan^2(x)}dt$$
which results in having to compute
$$\int_0^\infty \frac4{\sqrt{3}t^2 + 4t + \sqrt{3}} = 2\ln(3)$$
However, is there a simpler (or quicker) way to solve this integral? Thank you!
|
Here is some trick:
$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 2\sin(2x)} dx$=$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 4\sin(x)\cos(x)} dx$=$\int_0^{\frac\pi2} \frac{4\sqrt{3}\sec^2{x}}{\sec^2{x}(3 + 4\sqrt{3}\sin(x)\cos(x))} dx$=$\int_0^{\frac\pi2} \frac{4\sqrt{3}}{(3(1+\tan^2{x}) + 4\sqrt{3}\tan{x})} d(\tan{x})$
=$-\int_0^{\frac\pi2} \frac{4}{1-(2+\sqrt{3}\tan{x})^2} d(2+\sqrt{3}\tan{x})$.
Now, it becomes trivial integral, and the result is $2\log{3}$ as desired.
|
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|
Proving $\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$ using $\epsilon$-$\delta$ definition I want to prove that $$\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$$ but I am not sure if my proof is valid because I did some algebraic manipulation. Can you please verify my proof?
Given $ \epsilon \gt 0 $, Choose $ \delta = \min\{3, \sqrt{2\epsilon}\} $
Suppose $ 0 \lt x - 2 \lt \delta $
Check:
$$\begin{align}
\left|\frac{x - 2}{\sqrt{x^{2} - 4}} + 1 - 1\right| &= \left| \frac{x - 2}{\sqrt{x^{2} - 4}}\right|\\
&= \frac{x - 2}{\sqrt{x^{2} - 4}} \\
&= \frac{x - 2}{\sqrt{x + 2}\sqrt{x - 2}} \\
&= \frac{\left(x - 2\right)^{1}}{\sqrt{x + 2}\cdot\left(x - 2\right)^{0.5}} \\
&= \frac{\sqrt{x - 2}}{\sqrt{x + 2}}
\end{align}$$
$$0 \lt x - 2 \lt 3\quad\Rightarrow\quad 4 \lt x + 2 \lt 7
\quad\Rightarrow\quad 2 \lt \sqrt{x + 2} \lt \sqrt{7}$$
$\Rightarrow$
$$\frac{\sqrt{x - 2}}{\sqrt{x + 2}} \lt \frac{\sqrt{\delta}}{2} \le \epsilon\tag*{$\blacksquare$}$$
|
You should have $\delta<4\varepsilon^2$, so that $\frac{\sqrt\delta}2\leqslant\varepsilon$. So, take $\delta=\min\left\{3,4\varepsilon^2\right\}$.
|
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|
Find all the real values of $a,b,c,e,d,f$ for which $A$ is diagonalizable $\newcommand{\span}{\text{span}}$
The matrix in question is:
$$A=\begin{pmatrix}
2&0&0&0\\
a&-1&0&0\\
b&c&-1&0\\
d&e&f&2
\end{pmatrix}$$
This was on a test that I had and I got it wrong because I needed to add more conditions, what I did was:
Since A is a upper triangular matrix the eigenvalues of A are the values on the diagonal so $$\lambda_1=\lambda_4=2 \qquad \lambda_2=\lambda_3=-1$$
For $$\lambda_1=\lambda_4=2$$ we have:
$(A-2I)=\begin{pmatrix}
2&0&0&0\\
a&-1&0&0\\
b&c&-1&0\\
d&e&f&2
\end{pmatrix}-\begin{pmatrix}
2&0&0&0\\
0&2&0&0\\
0&0&2&0\\
0&0&0&2
\end{pmatrix}=
\begin{pmatrix}
0&0&0&0\\
a&-1&0&0\\
b&c&-1&0\\
d&e&f&0
\end{pmatrix}$
Since we need that the geometric multiplicity equals the algebraic multiplicity (which is $2$) we need to have 2 linearly independent vectors on each solution for the eigenvalues, so by making $a=b=c=d=e=f=0$ we get:
$$\begin{pmatrix}
0&0&0&0\\
0&-1&0&0\\
0&0&-1&0\\
0&0&0&0
\end{pmatrix}$$
The will give us the eigenspace $E_{2}=\span\left(\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right)$
By a similar process for $\lambda_2=\lambda_3=-1$ we get $E_{-1}=\span\left(\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\right)$
What are the remaining conditions, or values, that I need on $a,b,c,d,e,f$ for $A$ to be diagonalizable. Any help would be really appreciated.
|
Ummm. A matrix is diagonalizable if and only if the minimal polynomial is squarefree.
Your characteristic polynomial is $ (\lambda +1)^2 (\lambda - 2)^2.$ Both factors occur in the minimal polynomial with exponent either $1$ or $2.$ We need both exponents $1,$ so that the minimal poly becomes $ (\lambda +1) (\lambda - 2)= \lambda^2 - \lambda - 2.$
This happens when
$$ A^2 - A - 2I = 0 $$
calculating
$$
A^2 - A - 2 I =
\left(
\begin{array}{cccc}
0&0&0&0 \\
0&0&0&0 \\
ac&-3c&0&0 \\
ae+bf+3d& cf&0&0
\end{array}
\right)
$$
|
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|
Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.
By multiplying both sides by $(x+1)^2$ we get,
$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$
$$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$
But I can't factor the third degree polynomial.
|
$x^2 + \left(\frac{x}{x+1}\right)^2 = \frac{5}{4}$
Let $y=-x\qquad ⇒ y^2 + \left(\frac{y}{1-y}\right)^2 = \frac{5}{4}$
Let $z=\frac{x}{x+1}\qquad ⇒ z^2 + \left(\frac{z}{1-z}\right)^2 = \frac{5}{4}$
By inspection, $(x=1)$ is a solution.
$(x=1) → (y=-1) → (z=-1)$
Solve $z=-1$, back for x, we have $\;\left(x=-\frac{1}{2}\right)$
Original formula can be rewritten as quartic polynomial.
With 2 known roots, it reduced to simple quadratics.
|
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|
Conditions for $A\cos^2(x)+B\sin^2(x)+C\sin(x)\cos(x)+D\cos(x)+E\sin(x)+F = 0$ to have (a) real root(s) I am doing research on Linear Algebra and encounter a function in the form like below:
$$A\cos^2(x)+B\sin^2(x)+C\sin(x)\cos(x)+D\cos(x)+E\sin(x)+F = 0$$
What are the conditions that the coefficients of the equation need to satisfy for it to have (a) real root(s)? I am just wondering if there are some known results for the problem.
|
The conic section given in Zarrax's answer,
$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$
can be reordered into a simple quadratic in terms of either $x$ or $y$:
$$Ax^2 + (Cy + D)x + (By^2 + Ey + F) = 0$$
$$By^2 + (Cx + E)y + (Ax^2 + Dx + F) = 0$$
Plugging the coefficients into the quadratic formula, we get:
$$x = \frac{-(Cy + D) \pm \sqrt{(Cy + D)^2 - 4A(By^2 + Ey + F)}}{2A}$$
$$y = \frac{-(Cx + E) \pm \sqrt{(Cx + E)^2 - 4B(Ax^2 + Dx + F)}}{2B}$$
In order for a real solution to exist, both discriminants must simultaneously be non-negative, i.e.,
$$(Cy + D)^2 - 4A(By^2 + Ey + F) \ge 0$$
$$(Cx + E)^2 - 4B(Ax^2 + Dx + F) \ge 0$$
We can add these together to get:
$$(Cy + D)^2 - 4A(By^2 + Ey + F) + (Cx + E)^2 - 4B(Ax^2 + Dx + F) \ge 0$$
$$C^2y^2 + 2CDy + D^2 - 4ABy^2 - 4AEy - 4AF + C^2x^2 + 2CEx + E^2 - 4ABx^2 - 4BDx - 4BF \ge 0$$
$$C^2(x^2 + y^2) - 4AB(x^2 + y^2) + (2CE - 4BD)x + (2CD - 4AE)y - 4AF - 4BF + D^2 + E^2 \ge 0$$
But any solution must also lie on the unit circle $x^2 + y^2 = 1$, so this simplifies slightly to:
$$C^2 - 4AB + (2CE - 4BD)x + (2CD - 4AE)y - 4AF - 4BF + D^2 + E^2 \ge 0$$
$$(2CE - 4BD)x + (2CD - 4AE)y \ge 4(AB + AF + BF) - (C^2 + D^2 + E^2)$$
Now, define the function:
$$f(\theta) = (2CE - 4BD)\cos(\theta) + (2CD - 4AE)\sin(\theta)$$
Recall that
$$\alpha \cos(\theta) + \beta \sin(\theta) = \sqrt{\alpha^2 + \beta^2} \sin(\theta + \tan^{-1}(\frac{\alpha}{\beta}))$$
So the amplitude of the sine wave defined by $f$ is:
$$\sqrt{(2CE - 4BD)^2 + (2CD - 4AE)^2}$$
$$\sqrt{4C^2E^2 - 16BCDE + 16B^2D^2 + 4C^2D^2 - 16ACDE + 16A^2E^2}$$
$$2 \sqrt{C^2E^2 - 4BCDE + 4B^2D^2 + C^2D^2 - 4ACDE + 4A^2E^2}$$
$$2 \sqrt{C^2(D^2 + E^2) + 4(A^2E^2 + B^2D^2) - 4(A + B)CDE}$$
Thus, a necessary condition for the original equation to have real solutions is:
$$2 \sqrt{C^2(D^2 + E^2) + 4(A^2E^2 + B^2D^2) - 4(A + B)CDE} \ge 4(AB + AF + BF) - (C^2 + D^2 + E^2)$$
There's probably a way to simplify this further, though.
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|
How do I solve a recurrence relation of the type $a_{n+1} = K(n)a_{n} + P(n)a_{n-1}$ where $K(n)$ and $P(n)$ are rational functions? I've been stuck on a problem for days now and was able to boil it down to this recurrence relation. I researched a bit about ways to solve it -- generating functions, characteristic polynomials, homogenous and non-homogenous recurrence relations -- but couldn't find a method of tackling this specific one. Any tips, solutions, or suggestions would be welcomed. Thanks.
Edit: As made clear in the comments, treating the functions as any rational function is too general to solve. The specific problem I am solving requires $K(n) = (2n+5)/(n+2)$ and $P(n) = (n+1)(n+3)^2/(n+2)$
Edit #2: As requested, $a_1 = 5 ; a_2 = 33 ; a_3 = 168$
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$a_{n+1} = \dfrac{2n+5}{n+2} a_{n} + \dfrac{(n+1)(n+3)^2}{n+2} a_{n-1}$
Rewriting $\,2n+5 = (n+3)^2 - (n+2)^2\,$:
$$
\quad\quad a_{n+1} = \frac{(n+3)^2 - (n+2)^2}{n+2} a_{n} + \frac{(n+1)(n+3)^2}{n+2} a_{n-1}
\\ \iff\quad\quad a_{n+1} + (n+2) a_n = \frac{(n+3)^2}{n+2} \big(a_n + (n+1) a_{n-1}\big)
$$
With $\,b_n = a_{n} + (n+1) a_{n-1}\,$, this telescopes to:
$$
\require{cancel}
b_{n+1} = \frac{(n+3)^2}{n+2} \,b_n = \frac{(n+3)^2}{\bcancel{n+2}} \,\frac{(n+2)^{\bcancel{2}}}{n+1} \, b_{n-1} = \dots = (n+3)\,\frac{(n+3)!}{2} \, b_0
$$
Once $\,b_n\,$ is determined, what's left to solve is the simpler recurrence:
$$
a_{n} + (n+1) a_{n-1} = b_n \quad\iff\quad \frac{a_n}{(n+1)!} = -\frac{a_{n-1}}{n!} + \frac{b_n}{(n+1)!}
$$
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|
if $~f(x)=x^2+10x+20~$, find the number of solutions of $f(f(f(f(x))))=0$ or $f^4(x)=0$ where $f^n$ means f is composed to itself n times. The answer given is 2. I tried to bash the question by $~f^3(x)=\alpha~$ where $~\alpha~$ is the root. Then find the solution, and keep on continuing. It is a long process, so I ditched the idea. I'm looking for a short and elegant solution.
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To do it the very hard way
$a^2 + 10a + 20 = k$ was solutions $a =\frac{-10\pm\sqrt{100-4(20-k)}}2=-5\pm \frac {\sqrt{20 + 4k}}2=-5\pm \sqrt{5+k}$.
So if $f^4(f^3(x)) = 0$ we have solutions $f^3(x)=-5\pm \sqrt{5}$.
To solve $f^3(x) =f(f^2(x))=-5\pm \sqrt{5}$ we have solutions $f^2(x) = -5\pm\sqrt{5 + (-5\pm\sqrt 5)}=-5\pm \sqrt{\pm \sqrt 5} =-5\pm \sqrt{\sqrt 5}=-5\pm \sqrt[4] 5$.
To solve $f^2(x) =f(f(x)) = -5\pm \sqrt[4] 5$ we have solutions $f(x) = -5\pm\sqrt{5 -(-5\pm \sqrt[4]5)} = -5\pm \sqrt[8] 5$ (definitely seeing a pattern here).
And finally to solve $f(x) = -5\pm \sqrt[8]5$ we have solutions $x = -5\pm \sqrt{5+(-5\pm \sqrt[8]5)} =-5\pm \sqrt[16]5$. Two solutions.
(And by induction the solution solution to $f^n(x)=0$ is $x = -5 \pm \sqrt[2^n]5$)
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|
Evaluate $\lim\limits_{n\to\infty}\big(\frac{n}{2}+\min\limits_{x\in\mathbb{R}}\sum_{k=0}^n{\cos(2^k x)}\big)$ What is the exact value of the following limit?
$$L=\lim_{n\to\infty}\left(\frac{n}{2}+\min_{x\in\mathbb{R}}\sum_{k=0}^n{\cos(2^k x)}\right)$$
Experimenting on desmos suggests the following claims:
$\sum_{k=0}^n{\cos{(2^k x)}}$ is minimized when $x\approx \left(2m\pm\dfrac{2}{3}\right)\pi,m\in\mathbb{Z}$, with the approximation approaching equality as $n\to\infty$
$L\approx -0.704$
I do not know how to prove these claims.
(This question was inspired by another question.)
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Long comment. I believe this type of question is extremely hard, if not impossible by the current technology, to answer. However, let me share some observation.
Define $\varphi_n$ by the $n$th sum:
$$ \varphi_n(x) = \frac{n}{2} + \sum_{k=0}^{n} \cos(2^k x) $$
We will consider the behavior of $\varphi_n (x)$ near $x = \frac{2\pi}{3}$.
\begin{align*}
\varphi_n\left(\frac{2\pi}{3} + (-1)^n \frac{x}{2^n} \right)
&= \frac{n}{2} + \sum_{k=0}^{n} \cos\left(\frac{2^{k+1}\pi}{3} + (-1)^n \frac{x}{2^{n-k}}\right) \\
&= \frac{n}{2} + \sum_{k=0}^{n} \left[ -\frac{1}{2}\cos\left(\frac{x}{2^{n-k}}\right) + (-1)^{n-k+1}\frac{\sqrt{3}}{2}\sin\left(\frac{x}{2^{n-k}}\right) \right] \\
&= -\frac{1}{2} + \sum_{k=0}^{n} \left[ \sin^2\left(\frac{x}{2^{n-k+1}}\right) + (-1)^{n-k+1}\frac{\sqrt{3}}{2}\sin\left(\frac{x}{2^{n-k}}\right) \right] \\
&= -\frac{1}{2} + \sum_{j=0}^{n} \left[ \sin^2\left(\frac{x}{2^{j+1}}\right) + (-1)^{j+1}\frac{\sqrt{3}}{2}\sin\left(\frac{x}{2^{j}}\right) \right] \\
&= -\frac{1}{2} + 2 \sum_{j=0}^{n} \sin\left(\frac{x}{2^{j+1}}\right)\sin\left(\frac{x}{2^{j+1}}+(-1)^{j+1}\frac{\pi}{3}\right).
\end{align*}
Using this, define $\psi(x)$ by
$$ \psi(x) = -\frac{1}{2} + 2 \sum_{j=0}^{\infty} \sin\left(\frac{x}{2^{j+1}}\right)\sin\left(\frac{x}{2^{j+1}}+(-1)^{j+1}\frac{\pi}{3}\right). $$
Since $\varphi_n \bigl( \frac{2\pi}{3} + (-1)^n \frac{x}{2^n} \bigr)$ converges locally uniformly to $\psi(x)$ on $\mathbb{R}$, it follows that
$$\inf_{x\in\mathbb{R}} \psi(x) \geq \limsup_{n\to\infty} \left( \min_{x\in\mathbb{R}} \varphi_n(x) \right). $$
A numerical calculation suggests that $L = \inf \psi$ with an approximate value
$$\inf \psi \approx -0.70399210451640656752$$
at $x \approx 0.66123108104874561312$. Unfortunately, all the inverse symbolic calculators I tried could not identify this value. My gut is also telling that $L$ has no elementary closed-form, but again, this is rather a bold claim.
For fun, I included the graph of $\psi$:
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|
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
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Here $x$ is common, hence we can "cancel" it to get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19}$ & $x^2+1$
By your method, let this be $f(x) = (x^2+1)g(x)+ax+b$
At $x=i$, we get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19} = 0 + ai+b$
At $x=-i$, we get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19} = 0 - ai+b$
LHS in both can be calculated by Geometric Progression.
We will end up with $2$ Simultaneous Equations having complex co-efficients.
Solving that will give the required reminder.
Alternately, we know that $x^4 = 1$ when $x=i$ or $x=-i$.
With that, we can reduce the given Equations:
$x^1+x^{2}+x^{3}+x^{0}+ \cdots +x^{3} = 0 + ai+b$
$x^1+x^{2}+x^{3}+x^{0}+ \cdots +x^{3} = 0 - ai+b$
$(+i)+(-1)+(-i)+(+1)+ \cdots +(-i) = 0 + ai+b$
$(-i)+(-1)+(+i)+(+1)+ \cdots +(+i) = 0 - ai+b$
$(0i)+(-1) = 0 + ai+b$
$(0i)+(-1) = 0 - ai+b$
Both give $(a,b)=(0,-1)$
Plugging into the original, before we did the "cancellation", we get $0x^2-x = -x$
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|
How to find the maximum area of a quadrilateral, when three of the sides add up to 24? If there's a quadrilateral ABCD, and the sides AB + BC + CD add up to 24, how can I find the maximum area of the quadrilateral formed, what length should the four sides of the quadrilateral be, and what should the interior angles be?
I think Lagrange multiplier needs to be used for such a case, but I'm not sure how to start. I think the constraint will be x + y + z = 24, but I'm not sure how to represent the area function. Any help is greatly appreciated!
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Denote the four sides of the quadrilateral by $w = DA$, $x = AB$, $y = BC$, $z = CD$.
Let $d = AC$ be the length of one diagonal. This splits the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ACD$.
Use Heron's formula to find the area of the two triangles.
$$A = \frac{1}{4}\sqrt{(x + y + d)(-x + y + d)(x - y + d)(x + y - d)} + \frac{1}{4}\sqrt{(w + z + d)(-w + z + d)(w - z + d)(w + z - d)}$$
But we're given that $x + y + z = 24$, so:
$$A = \frac{1}{4}\sqrt{(x + y + d)(-x + y + d)(x - y + d)(x + y - d)} + \frac{1}{4}\sqrt{(w + z + (24 - x - y))(-w + (24 - x - y) + d)(w - (24 - x - y) + d)(w + (24 - x - y) - d)}$$
This gives us a formula for the area in terms of three sides of the quadrilateral ($w$, $x$, and $y$) and the diagonal $d$. Some work will needed to be done to find the maximum, but at least now we know what to maximize.
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|
Sum of all real roots of $\frac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \frac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$?
The sum of real roots of $\dfrac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \dfrac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$ is____
How do I proceed with this type of problem.
Cross multiplying and segregating the coefficient is a cumbersome process
My putting in desmos.com my answer is $«6»$.
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Cross-multiplying gives:
$$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (x^2 + 3x + 10)(5x^2 - 7x + 19)$$
$$9x^4 + 15x^3 + 36x^2 - 27x^3 - 45x^2 - 108x + 51x^2 + 85x + 204 = 5x^4 - 7x^3 + 19x^2 + 15x^3 - 21x^2 + 57x + 50x^2 - 70x + 190$$
$$9x^4 - 12x^3 + 42x^2 - 23x + 204 = 5x^4 + 8x^3 + 48x^2 - 13x + 190$$
$$4x^4 - 20x^3 - 6x^2 - 10x + 14 = 0$$
$$x^4 - 5x^3 - \frac{3}{2}x^2 - \frac{5}{2}x + \frac{7}{2} = 0$$
There, that wasn't too bad. Now the hard part: finding the roots.
The Rational Root Theorem tells us that all rational roots are in $\pm\lbrace 1, 7, \frac{1}{2}, \frac{7}{2} \rbrace$. Unfortunately, it turns out that none of those actually are roots, so let's take another path and try a quadratic factorization:
$$f(x) = (x^2 + Ax + B)\left(x^2 - (A + 5)x + \frac{7}{2B}\right)$$
where the coefficients in the second factor are set to make the cubic ($-5$) and constant ($\frac{7}{2}$) coefficients correct. Expanding the product:
$$f(x) = x^4 - (A + 5)x^3 + \frac{7}{2B}x^2 + Ax^3 - A(A + 5)x^2 + \frac{7A}{2B}x + Bx^2 - (A + 5)Bx + \frac{7}{2}$$
$$f(x) = x^4 - 5x^3 + \left(\frac{7}{2B} - A^2 - 5A + B \right)x^2 + \left(\frac{7A}{2B} - AB - 5B\right)x + \frac{7}{2}$$
Solving for the $x^2$ and $x$ coefficients is a bit difficult, but it turns out that $A = B = 1$ works, so:
$$f(x) = (x^2 + x + 1)\left(x^2 - 6x + \frac{7}{2}\right) = 0$$
The first factor has a negative discriminant (-3) and thus no real solutions, but the second one has a positive discriminant (22).
We don't actually have to find the roots, just use the fact that the two roots of a quadratic $ax^2 + bx + c = 0$ add up to $\frac{-b}{a}$.
So the answer to your question is indeed 6.
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|
Prove the sequence of three real numbers
If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then prove that $a,b,c$ are terms in $G.P$
My work:
I assumed that they are in $G.P$ and so assumed $b=ak$ and $c=ak^2$ for some arbitrary $k$. After that I expanded both sides of the equality and got the same results so that means the equality is true. But here's the twist!
Suppose the question was
If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then the terms $a,b,c$ are in $G.P$ or $A.P$ or $H.P\:\:?$
Then what should be one's approach. Will expanding work here$?$
Any help is greatly appreciated.
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We can think of the problem in terms of polynomials too. Assume that $a, b, c$ are the roots of the cubic polynomial '$ P(x) = x^3 + ux^2 + vx + w$. We get
$$-u = a+b+c$$
$$v = ab+bc+ca$$
$$-w = abc$$
Substituting these in the original condition, we have
$$v^3 = (-w)(-u)^3 = wu^3$$ or
$$ w = \frac{v^3}{u^3}$$
$$P(x) = x^3 + ux^2 + vx +\frac{v^3}{u^3} = (x^3 + \frac{v^3}{u^3}) + (ux^2 + vx) $$
$$= (x + \frac{v}{u})(x^2 + \frac{v^2}{u^2} - \frac{v}{u}x) + ux(x+\frac{v}{u}) $$
This shows $\frac{-v}{u}$ is a root of the P(x). The product of three roots is $-w = \frac{-v^3}{u^3}$. So the product of the other two roots is $\frac{v^2}{u^2}$. We get that square of one root is product of other two and they must be in a G.P.
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|
How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta}$$
Adding two versions together yields
$$
\begin{aligned}
2 I &=2 a \int_{0}^{\pi} \frac{d \theta}{a^{2}-b^{2} \cos ^{2} \theta} \\
&=4 a\int_0^{\frac{\pi}{2}} \frac{\sec ^{2} \theta}{a^{2} \sec ^{2} \theta-b^{2}} d \theta \quad \textrm{( By symmetry )}\\
&=4 a{\int_{0}^{\infty}} \frac{d t}{\left(a^{2}-b^{2}\right)+a^{2} t^{2}} \\
&=\frac{4}{\sqrt{a^{2}-b^{2}}}\left[\tan^{-1} \left(\frac{at}{\sqrt{a^{2}-b^{2}}}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\frac{\pi}{\sqrt{a^{2}-b^{2}}}}
$$
Are there any other methods to deal with the integral?
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One more
$$I=\int_{0}^{\pi} \frac{dx}{a+b \cos x}=\int_{0}^{\pi}\frac{dx}{a+b(2\cos^2(x/2)-1)}, a>b.$$
Use $t=x/2$, then
$$I=\int_{0}^{\pi/2} \frac{2dt}{a-b+2b \cos^2 t}=\int_{0}^{\pi/2} \frac{2\sec^2t ~dt}{(a-b)\sec^2 t+2b}=2\int_{0}^{\pi/2} \frac{\sec^2 t~dt}{(a+b)+(a-b)\tan^2 t}$$
Let $\tan t=u$, then
$$I=\frac{2}{a-b}\int_{0}^{\infty}\frac{du}{u^2+\frac{a+b}{a-b}}=\frac{2}{\sqrt{a^2-b^2}} \tan^{-1} u|_{0}^{\infty}=\frac{\pi}{\sqrt{a^2-b^2}}.$$
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|
Find the limit of $\frac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$ Find the limit $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$$
For the numerator we have $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac{1}{3}\right)^{n+1}}{1-\frac13}=\dfrac32-\dfrac12\dfrac{1}{3^n}.$
By analogy, $1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}=\dfrac54-\dfrac14.\dfrac{1}{5^n}$
So we have $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}=\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}}=\lim_{n\to\infty}{\left(\dfrac{3^{n+1}-1}{2.3^n}\cdot\dfrac{4.5^n}{5^{n+1}-1}\right)}$$ What am I supposed to do next?
My initial mistake was that I thought $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac13\right)^n}{1-\frac13}$. How could we actually see (and show) that the terms in the sum(s) are not $n$, but $n+1$?
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How could we actually see (and show) that the terms in the sum(s) are not $n$, but $+1$
The sum is $$1+\frac13+…\frac{1}{3^n}=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+…+\frac{1}{3^n}$$$$=\sum_{i=0}^n 3^{-n}.$$ The index $i$ starts from the value $\textbf 0$ (and not $1$) and ends at $n$. So the number of terms is $n+1$.
What am I supposed to do next?
From $$\lim_{n\to\infty}\left(\dfrac{3^{n+1}-1}{2\cdot 3^n}\cdot\dfrac{4.5^n}{5^{n+1}-1}\right)$$ you divide the numerator and denominator of the first part by $3^n$ and the numerator and denominator of the second part by $5^n$. $$\lim_{n\to \infty}\left(\frac{3-\frac{1}{3^n}}{2}\cdot\frac{4}{5-\frac{1}{5^n}}\right)$$ Note that $$\lim_{n\to \infty}\frac{1}{3^n}= \lim_{n\to \infty}\frac{1}{5^n}=0$$ so the limit is $$\frac{3}{2}\cdot\frac{4}{5}=\frac65.$$
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then
*
*$ \ \displaystyle{ \alpha<-2 }$ ;
*$ \ \displaystyle{ 3<\alpha<7 }$ ;
*it is impossible ;
*none of (a)-(c).
$$$$
I have done the following :
The value $\alpha$ is real and such that the equation $x^2+2(\alpha-1)x-\alpha+7=0$ has two different real negative solutions.
The solutions of the equation are given from the quadratic formula \begin{align*}x_{1,2}&=\frac{-2(\alpha-1)\pm \sqrt{[2(\alpha-1)]^2-4\cdot 1\cdot (-\alpha+7)}}{2}\\ & =\frac{-2(\alpha-1)\pm \sqrt{4(\alpha^2-2\alpha-1)-4\cdot (-\alpha+7)}}{2}\\ & =-(\alpha-1)\pm \sqrt{2(\alpha^2-2\alpha-1)-2\cdot (-\alpha+7)} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-4\alpha-2+2\alpha-14} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}\end{align*}
So that we have two different solutions the discriminant must be non-zero.
So that we have two negative solutions, it must hold $-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}<0$.
So that we have real solutions the expression under the square root must be non negative.
The expression under the square root has the sign of the coefficient of $x^2$, i.e. positive, outside the roots.
We have that \begin{equation*}2\alpha^2-2\alpha-16=0 \Rightarrow \alpha_{1,2}=\frac{1}{2}\pm \frac{\sqrt{33}}{2}\end{equation*}
So we have that the expression under the square root if $\alpha<\frac{1}{2}- \frac{\sqrt{33}}{2}$ and if $\alpha>\frac{1}{2}+ \frac{\sqrt{33}}{2}$.
Is my attempt correct so far? Now do we check if the first two intervals of $\alpha$ can hold for all these conditions? Or how do we continue? Or is there a better way to solve that exercise ?
|
You made a slight algebra error.
$$a = 1; b = 2(\alpha - 1); c=-\alpha + 7$$
$$x = \frac{-2(\alpha - 1) \pm \sqrt{(2(\alpha - 1))^2 - 4(-\alpha + 7)}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4(\alpha^2 - 2\alpha + 1) + 4(\alpha - 7)}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4\alpha^2 - 8\alpha + 4 + 4\alpha - 28}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4\alpha^2 - 4\alpha - 24}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4}\sqrt{\alpha^2 - \alpha - 6}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4}\sqrt{\alpha^2 - \alpha - 6}}{2}$$
$$x = -\alpha + 1 \pm \sqrt{\alpha^2 - \alpha - 6}$$
For the roots to be real, the radicand must be nonnegative. For the roots to be distinct, the radicand must be nonzero. Thus,
$$\alpha^2 - \alpha - 6 > 0$$
$$(\alpha + 2)(\alpha - 3) > 0$$
$$\alpha < -2 \text{ or } \alpha > 3$$
But we also need both roots to be negative. This is equivalent to the greater of the two roots being negative.
$$-\alpha + 1 + \sqrt{\alpha^2 - \alpha - 6} < 0$$
$$\sqrt{\alpha^2 - \alpha - 6} < \alpha - 1$$
But $\sqrt{\alpha^2 - \alpha - 6} > 0$, so by transitivity we must also have $\alpha - 1 > 0$, or $\alpha > 1$. This rules out the $\alpha < -2$ possibility from earlier, so at this point we know $\alpha > 3$. Anyhow, since both sides of the inequality are positive, we can square it without changing their order.
$$\alpha^2 - \alpha - 6 < (\alpha - 1)^2$$
$$\alpha^2 - \alpha - 6 < \alpha^2 - 2\alpha + 1$$
$$-6 < -\alpha + 1$$
$$-7 < -\alpha$$
$$7 > \alpha$$
Therefore, $3 < \alpha < 7$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Question about continuity and the Epsilon-Delta definition So, I' currently working through an example:
Show that $\lim \limits_{x \to 0}\frac{x^3}{x-4}\cos{(\frac{1}{x})}=0$ using the $\epsilon-\delta$ definition.
So far I have the following working:
For any $\epsilon>0$, we must find a $\delta>0$ such that if $|x-0|<\delta$ then:
$|\frac{x^3}{x-4}\cos{(\frac{1}{x})}-0|<\epsilon, \qquad \forall x \in(-\delta,0)\cup(0,\delta)$
I know that the cosine function is bound by $[-1,1]$, so we can say:
$|\frac{x^3}{x-4}\cos{(\frac{1}{x})}-0|=|\frac{x^3}{x-4}||cos{(\frac{1}{x})}|\leq|\frac{x^3}{x-4}|<\epsilon$
Now the trouble I'm having is developing an upper bound for $|\frac{x^3}{x-4}|$. My suspicion is that since we a looking for a small neighbour hood of $\delta$; if I bind $\delta\leq1$ then $|x^3|$ should also be bounded the same way implying: $-1<x<1$.
My gut is telling me that $\delta=\epsilon$ is going to be the road I need to go down, but I want to prove it. Are there any insights I'm missing? Should I be placing an lower bound on $\frac{1}{x-4}$ by putting a positive upper bound on $x-4$?
Thanks for reading.
|
The first answer is correct. But this one goes (i believe), more in your original direction, but without restricting $\delta$.
In this first part $x<4$ :
For $f = |\frac{x^3}{x-4}|$, We call $f_+ = |\frac{x^3}{x-4}|$ for $x>0$ and $f_- = |\frac{x^3}{x-4}|$ for $x<0$ (and both are $0$ on the rest). We have $f = f_+ + f_-$ for $x<4$.
*
*For $x < 4$, $|x-4| = 4-x$,
*$|x^3| = x^3$ for $x>0$ and $|x^3| = -x^3$ for $x<0$.
*Therefore : $|\frac{x^3}{x-4}| = \frac{x^3}{4-x}$ for $x >0$ and $|\frac{x^3}{x-4}| = \frac{-x^3}{4-x}$ for $x<0$.
*Hence, $f_+ = \frac{x^3}{4-x}$ and $f_- = -\frac{x^3}{4-x}$
Notice that $4-x > 1$ and therefore $f_+ = x^3\cdot\frac{1}{4-x} < x^3 \cdot 1$ for $x>0$ and $f_- = -x^3\cdot\frac{1}{4-x} < -x^3 \cdot 1$
We therefore have a bound $f < |x^3|$ for all $x$, because the signs just flip and cancel for $x>4$ and the cases are analoguous.
You should have no issue fixing $\delta$ now. But here it is, just in case :
Set $\delta = \varepsilon^{1/3}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$
This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve the problem. I've tried the following: $$\begin{align} 4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ})&=16\sin^2(12^{\circ})\cos^2(12^{\circ})+8\sin^2(12^{\circ})
\cos(12^{\circ})\\
\\
&=8\sin^2(12^{\circ})\cos(12^{\circ})\Big(2\cos(12^{\circ} ) + 1\Big)\end{align}$$
As you can see, I was trying to simplify the expression so that it'll contain only $\sin(12^{\circ})$ and $\cos(12^{\circ})$, since I thought by unifying the angles I would have a bigger chance of solving it. However, I couldn't find a way to make any further progress. Can someone show me the way?
|
I prove: $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) - 1=0$ without using special values at $18^\circ$ or $36^\circ$
Start:
$$\begin{align} \text{LHS} &=4\sin^2(24^{\circ})+\frac{2\sin(24^{\circ})\sin(24^{\circ})}{\cos(12^\circ)} - 1\\
\\
&=4\sin^2(24^\circ)\cdot\frac{\cos(12^\circ)+\cos(60^\circ)}{\cos(12^\circ)} -1\\
\\
&=\frac{8\sin^2(24^\circ)\cos(24^\circ)\cos(36^\circ)}{\cos(12^\circ)}-1\\
\\
&=\frac{4\sin(24^\circ)\sin(48^\circ)\cos(36^\circ)}{\cos(12^\circ)}-1\\
\\
&=8\sin(12^\circ)\sin(48^\circ)\cos(36^\circ)-1\\
\\
&=4\left(\cos(36^\circ)-\frac{1}2 \right)\cos(36^\circ)-1\\
\\
&=4\cos^2(36^\circ)-2\cos(36^\circ)-1\\
\\
&=\frac{\left(4\cos^2(36^\circ)-2\cos(36^\circ)-1\right)\left(\cos(36^\circ)+1\right)}{\cos(36^\circ)+1}\\
\\
&=\frac{4\cos^3(36^\circ)-3\cos(36^\circ)+2\cos^2(36^\circ)-1}{\cos(36^\circ)+1}\\
\\
&=\frac{\cos(108^\circ)+\cos(72^\circ)}{\cos(36^\circ)+1}\\
\\
&=\frac{-\cos(72^\circ)+\cos(72^\circ)}{\cos(36^\circ)+1}\\
\\
&=0
\end{align}$$
|
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|
The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th term becomes $(n-1) \cdot n$ and so it's summation becomes $\frac{n(n+1) (2n+1)}{6} - \frac{n(n+1)}{2}$ but it's the same summation as above but the results are different what am I doing wrong?
|
In first case. ${\color{green}S_n}=\frac{n(n+1)(n+2)}{6}+\frac{n(n+1)}{2}={\color{red}{\frac{n(n+1)(n+2)}{3}}}$
In second case, $S'_{n+1}=\frac{(n+1)(n+2)(2n+3)}{6}-\frac{(n+1)(n+2)}{2}={\color{red}{\frac{n(n+1)(n+2)}{3}}}={\color{green}S_n}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $0\le ab+bc+ca-abc\le2$ A question asks
Let $a,b,c$ be non-negative reals such that $$a^2+b^2+c^2+abc=4$$
Prove that$$0\le ab+bc+ca-abc\le2$$
The lower bound can be proven as follows:
If $a,b,c>1$, then $$a^2+b^2+c^2+abc>4$$
This implies that at least one of $a,b,c$ is less than or equal to $1$, and we can assume WLOG that $a\le1$. Then,
$$ab+bc+ca-abc=a(b+c)+bc(1-a)\ge0$$
For the upper bound, I had the idea of treating $a^2+b^2+c^2+abc=4$ as a quadratic in $a$ but don't see any way to proceed from there. Please only give me hints!
|
If one of $a,b,c$ is zero or two of them are equal, the inequality holds (almots) trivially.
Consider the function $ f(a,b,c) = ab+bc+ca-2abc $, with $a,b,c\ge 0, g(a,b,c) = 1$, where $g(a,b,c) = a^2 + b^2 + c^2 + abc$. The domain is a compact subset of $\mathbb{R}^3$ so $f$ must admit maximum. Assume the maximum is strictly larger than $2$, then it is taken in the interior of the domain, and $a,b,c$ are pairwisely distinct. Thus we can apply Lagrange's multiplier to get
\begin{equation}
\nabla f + \lambda \nabla g = 0.
\end{equation}
Notice that $\partial_a g = 2a + bc > 0$, so we have
\begin{equation}
\frac{a+c -ac}{2b+ac} = \frac{b+c-bc}{2a+bc},
\end{equation}
or
\begin{equation}
(a-b)(2a+2b+2c-2ac-2bc+c^2) = 0
\end{equation}
and two similar equations. Since we assumed $a,b,c$ are pairwisely distinct, we yield
\begin{equation}
a^2 - 2ac- 2ab = b^2 - 2bc -2ba
\end{equation}
or $a+b-2c = 0$. Then we have $a = b = c$ for sure, a contradiction.
|
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|
Find the limit of the sequence $a_{n+1}=a_n + \frac{1}{2^n a_n}$ with $a_1=1$ Find the limit of the sequence $a_{n+1}=a_n + \frac{1}{2^n a_n}$ with $a_1=1$.
I could only estimate the upper and lower bound for $a_n$. My result is
$$\sqrt{3} \leq \lim_{n\to \infty} a_n \leq \sqrt{\frac{79}{24}}$$.
Is there any way to compute $\lim_{n\to \infty} a_n$, or find the non-linear equality it satisfies.
B.T.W., I find the limit value depends on the initial value $a_1=1$.
|
Write $L = \lim_{n \to \infty} a_n$. Squaring gives
$$a_{n+1}^2 = a_n^2 + \frac{1}{2^{n-1}} + \frac{1}{2^{2n} a_n^2}$$
which gives, ignoring the last term
$$L^2 \ge 1 + \sum_{k=0}^{\infty} \frac{1}{2^k} = 3$$
which I presume is where your lower bound of $\sqrt{3}$ comes from. Since $a_n$ is monotonically increasing, we have $a_n^2 \ge 1$ and so $\frac{1}{a_n^2} \le 1$, which gives
$$L^2 \le 3 + \sum_{k=1}^{\infty} \frac{1}{2^{2k}} = 3 + \frac{1}{3} = \frac{10}{3}$$
which gets us an upper bound of $\sqrt{ \frac{10}{3} } = \sqrt{ \frac{80}{24} }$, slightly worse than the one you obtained. In particular $a_n$ is both monotonic and bounded so the limit actually exists. By monotonicity this is an upper bound on the entire series, so we have $a_n^2 \le \frac{10}{3}$, which gives a stronger lower bound
$$L^2 \ge 3 + \sum_{k=1}^{\infty} \frac{1}{2^{2k} \frac{10}{3}} = 3 + \frac{1}{10}.$$
We can improve the lower and upper bounds starting from the exact value of any particular term $a_k$ of the sequence, as follows. Starting the recurrence from this term gives
$$L^2 \ge a_k^2 + \sum_{i=k-1}^{\infty} \frac{1}{2^i} = a_k^2 + \frac{1}{2^{k-2}}$$
as a lower bound, and since $a_n$ is monotonically increasing we have $a_n \ge a_k$ for $n \ge k$ so we get
$$L^2 \le a_k^2 + \frac{1}{2^{k-2}} + \sum_{i=k} \frac{1}{2^{2i} a_k^2} = a_k^2 + \frac{1}{2^{k-2}} + \frac{1}{3 \cdot 4^{k-1} a_k^2}.$$
As above this upper bound lets us strengthen the lower bound again, to
$$L^2 \ge a_k^2 + \frac{1}{2^{k-2}} + \frac{1}{3 \cdot 4^{k-1} \left( a_k^2 + \frac{1}{2^{k-2}} \right)}.$$
The first few terms are $a_1 = 1, a_2 = \frac{3}{2}, a_3 = \frac{5}{3}, a_4 = \frac{209}{120}$. Applying the above bounds to $a_3 = \frac{5}{3}$ to keep the arithmetic simple gives
$$\frac{25}{9} + \frac{1}{2} + \frac{3}{424} \le L^2 \le \frac{25}{9} + \frac{1}{2} + \frac{3}{400}$$
which gives $L \in [1.81242 \dots, 1.81253 \dots]$, and using larger terms would get us exponentially closer to $L$, but I doubt there's anything nice to say as far as closed forms.
|
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|
Two different answers for diameter $x$ I am solving this question shown below:
Two circles embedded between two parallel lines whose distance is $x$ and three other circles with indicated diameters touch both the bigger ones. Find $x$
My try:
I used the following facts:
Fact $1:$ If two circles touch externally, distance between their centers is sum of radii.
Fact $2:$ If two circles touch internally, distance between centers is absolute difference of radii.
Fact $3:$ The distance between parallel tangents is Diameter.
Based on these, I have got this figure below:
$O$ and $M$ are centers of Bigger circles.$P,Q,R$ are centers of smaller ones.
By Cosine rule in $\Delta OPM, \Delta OPQ$, we have:
$$\cos \theta=\frac{\left(\frac{x}{2}+4\right)^2+12^2-\left(\frac{x}{2}-4\right)^2}{2(12)\left(\frac{x}{2}+4\right)}=\frac{\left(\frac{x}{2}+4\right)^2+\left(\frac{x}{2}+6\right)^2-10^2}{2\left(\frac{x}{2}+4\right)\left(\frac{x}{2}+6\right)}$$
This gives $$x=12\sqrt{5}$$
Also by cosine rule in $\Delta OMR, OQR$, we get
$$\cos \alpha=\frac{\left(\frac{x}{2}+\frac{9}{2}\right)^2+144-\left(\frac{x}{2}-\frac{9}{2}\right)^2}{2\left(\frac{x}{2}+\frac{9}{2}\right)(12)}=\frac{\left(\frac{x}{2}+\frac{9}{2}\right)^2+\left(\frac{x}{2}+6\right)^2-\frac{441}{4}}{2\left(\frac{x}{2}+\frac{9}{2}\right)\left(\frac{x}{2}+6\right)}$$
This gives $$x=12\sqrt{7}$$
Where i went wrong?
|
As mentioned in a comment, OP's error was in assuming that point $M$ was on segment $OQ$, an impossibility due to the vertical asymmetry of the figure.
There's probably a slick solution here involving inversive geometry, but here's one that follows OP's trigonometric lead:
We take the large circles $\bigcirc O$ and $\bigcirc O'$ to have radius $r$, and the smaller circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ (with $\bigcirc A$ between $\bigcirc B$ and $\bigcirc C$) to have radii $a$, $b$, $c$. Define $\beta := \angle OAB$, $\gamma := \angle OAC$, $\beta' := \angle O'AB$, $\gamma' := \angle O'AC$.
The Law of Cosines in $\triangle OAB$, $\triangle OAC$, $\triangle O'AB$, $\triangle O'AC$ tells us
$$\begin{align}
\cos\beta &= \frac{-(r+b)^2+(a+b)^2+(r+a)^2}{2(a+b)(r+a)}=\frac{a^2 + a b + a r - b r}{(a + b) (r+a)} \\[4pt]
\cos\gamma&=\frac{a^2 + a c + a r - c r}{(a + c) (r+a)} \\[4pt]
\cos\beta' &= \frac{a^2 + a b - a r + b r}{(a + b) (r - a)} \\[4pt]
\cos\gamma' &= \frac{a^2 + a c - a r + c r}{(a + c) (r - a)}
\end{align}$$
Now, since $\beta+\gamma=\angle BAC=\beta'+\gamma'$, we have $\cos(\beta+\gamma)=\cos(\beta'+\gamma')$, so that
$$\cos\beta\cos\gamma-\sin\beta\sin\gamma = \cos\beta'\cos\gamma' - \sin\beta'\sin\gamma'$$
With a couple of rounds of squaring, we can replace sines with cosines, and then substitute the above expressions. (A computer algebra system really helps with this process.) When the dust settles, we find
$$r^2 = \frac{a^2 (a b + a c + 2 b c)^2}{4 b c (a^2 - b c)} \qquad\to\qquad
r = \frac{2\bar{a}+\bar{b}+\bar{c}}{2\bar{a}\,\sqrt{\bar{b}\bar{c} -\bar{a}^2}}\qquad\left(\bar{x}:=\frac1x\right)$$
For OP's specific case, $a=6$, $b=4$, $c=9/2$, so that $r=29/2$. $\square$
We can also consider the Law of Cosines in $\triangle OAO'$
$$|OO'|^2 = |OA|^2 + |O'A|^2 - 2 |OA||O'A|\cos(\beta'-\beta) $$
to determine that (ignoring an extraneous solution)
$$|OO'|^2 = \frac{a^2 (a b + a c + 2 b c) (a b + a c - 2 b c)}{b c (a^2 - b c)}= \frac{(\bar{b}+\bar{c}+2\bar{a})(\bar{b}+\bar{c}-2\bar{a})}{\bar{a}^2(\bar{b}\bar{c} - \bar{a}^2)}$$
This is helpful for drawing a figure. (One can also calculate $\angle AOO'$, $\angle BOA$, $\angle COA$ to determine accurate placement of the smaller circles. Frankly, I just eye-balled that. :)
Note that
$$\frac{4r^2}{|OO'|^2} = \frac{\bar{b}+\bar{c}+2\bar{a}}{\bar{b}+\bar{c}-2\bar{a}}=\frac{\mu+\bar{a}}{\mu-\bar{a}} \qquad\left(\mu:=\frac12(\bar{b}+\bar{c})\right)$$
This suggests that there might be something interesting going on in the underlying geometry; perhaps something that helps side-step the rather arduous algebraic manipulations above. Investigation is left as an exercise to the reader.
|
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Contradictions when do integral $\unicode{x222F} ~f(x,y,z)~dS $, given $f(tx,ty,tz)=t^4 f(x,y,z)$ and $f_{xx}+f_{yy}+f_{zz}=x^2+y^2+z^2$ If $f(tx,ty,tz)=t^4 f(x,y,z)$ and $f_{xx}+f_{yy}+f_{zz}=x^2+y^2+z^2$, then compute the surface integral on the unit sphere $x^2+y^2+z^2=1$:
$$\unicode{x222F} ~f(x,y,z)~dS $$
One particular solution is $f_p=\frac{1}{12}(x^4+y^4+z^4)$,
Another particular solution is $f_p=\frac{1}{4}(x^2y^2+y^2z^2+z^2x^2)$.
Any solutions of the form $f=f_p+h$ where $h$ is a harmonic function and homogenous with degree $4$ (for example, $h=x^4-6x^2y^2+y^4$) will satisfy the given conditions.
Take solution $f=f_p+h$ by Mean Value Theorem, the integral on the harmonic function $h$ equals the value $4\pi h(0,0,0)$. Due to $h$ is homogeneous of degree $4$, we have $h(0,0,0)=0$, so the integral equals
$$I=\unicode{x222F} ~f_p~dS=\int_0^{2\pi} \int_0^\pi f_p(\theta, \phi) \sin(\phi)d\phi d\theta$$
But the problem is this integral seems depends on the choice of $f_p$
If let $f_p=\frac{1}{12}(x^4+y^4+z^4),~~I=\frac{7\pi^2}{64}$
If let $f_p=\frac{1}{4}(x^2y^2+y^2z^2+z^2x^2),~~I=\frac{11\pi^2}{128}$
Do I make some mistakes?
|
Both integrals should have a value of $\dfrac\pi5$.
Using the first choice of $f_p$:
$$\iint\limits_{x^2+y^2+z^2=1} \frac{x^4+y^4+z^4}{12} \, dS \\ = \int_0^{2\pi} \int_0^\pi \frac{\cos^4(\theta)\sin^4(\phi) + \sin^4(\theta)\sin^4(\phi) + \cos^4(\phi)}{12} \sin(\phi) \, d\phi \, d\theta$$
Now,
$$\begin{align*}
I &= \int_0^{2\pi} \int_0^\pi \cos^4(\theta) \sin^5(\phi) \, d\phi \, d\theta \\[1ex]
&= \left(\int_0^{2\pi} \cos^4(\theta)\,d\theta\right) \left(\int_0^\pi \sin^5(\phi) \, d\phi\right) \\[1ex]
&= \left(\int_0^{2\pi} \frac{3 + 4\cos(2\theta) + \cos(4\theta)}8 \, d\theta\right) \left(\int_0^\pi \sin(\phi) (1 - \cos^2(\phi))^2 \, d\phi\right) \\[1ex]
&= \frac{3\pi}4 \times \frac{16}{15} = \frac{4\pi}5
\end{align*}$$
The integrals of $\sin^4(\theta)\sin^4(\phi)$ and $\cos^4(\phi)$ have the same value, so the surface integral is $\dfrac{I+I+I}{12} = \dfrac\pi5$.
Using the second choice of $f_p$:
$$\iint\limits_{x^2+y^2+z^2=1} \frac{x^2y^2 + x^2z^2 + y^2z^2}4 \, dS \\ = \int_0^{2\pi} \int_0^\pi \frac{\cos^2(\phi)\sin^2(\phi) + \cos^2(\theta)\sin^2(\theta)\sin^2(\phi)}4 \sin(\phi) \, d\phi \, d\theta$$
Now,
$$\begin{align*}
I_1 &= \int_0^{2\pi} \int_0^\pi \cos^2(\phi)\sin^3(\phi) \, d\theta \, d\phi \\[1ex]
&= 2\pi \int_0^\pi \cos^2(\phi) (1 - \cos^2(\phi)) \sin(\phi) \, d\phi \\[1ex]
&= 2\pi\times\frac4{15} = \frac{8\pi}{15} \\[2ex]
I_2 &= \int_0^{2\pi} \int_0^\pi \cos^2(\theta) \sin^2(\theta) \sin^5(\phi) \, d\phi \, d\theta \\[1ex]
&= \left(\int_0^{2\pi} \left(\frac{\sin(2\theta)}2\right)^2 \, d\theta\right) \left(\int_0^\pi(1-\cos^2(\phi))^2\sin(\phi)\,d\phi\right) \\[1ex]
&= \frac14 \left(\int_0^{2\pi} \frac{1-\cos(4\theta)}2 \, d\theta\right) \left(\int_0^\pi(1-\cos^2(\phi))^2\sin(\phi)\,d\phi\right) \\[1ex]
&= \frac14\times\pi\times\frac{16}{15} = \frac{4\pi}{15}
\end{align*}$$
Then the surface integral is again $\dfrac{I_1+I_2}4 = \dfrac\pi5$.
|
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|
$a^2 + b^2 = c^2$, $ab = 6d^2$, $(a,b) = 1$ has no solutions $(a,b,c,d) \in \mathbb N^4$
Prove that $$a^2 + b^2 = c^2$$
$$ab = 6d^2$$
$$\gcd(a,b) = 1$$
has no solutions $(a,b,c,d) \in \mathbb N^4$.
Put $a = p^2 - q^2, b = 2pq, c= p^2 + q^2$ where $(p,q) = 1$, and $p - q \equiv 1 \bmod 2$ (i.e., one is odd, the other is even.) Then,
$$pq(p+q)(p-q) = 3d^2$$
$\{p,q,p+q,p-q\}$ are pairwise coprime. I thank Thomas Andrews for his thoughts and help with a similar (now deleted) question, which helped me make progress on this problem.
The following cases arise:
*
*$p = 3x_1^2, q = x_2^2, p+q = x_3^2, p-q = x_4^2$.
$3x_1^2 + x_2^2 = x_3^2$ gives $x_2^2 = x_3^2 \bmod 3$. $3x_1^2 = x_4^2 + x_2^2$ gives $x_4^2 + x_2^2 = 0 \bmod 3$, forcing $x_2^2 = x_3^2 = x_4^2 = 0 \bmod 3$. So, $3 \mid q$, which contradicts $(p,q) = 1$.
*$p = x_1^2, q = 3x_2^2, p+q = x_3^2, p-q = x_4^2$.
In this case, we get $x_1^2 = x_3^2 = x_4^2 = 1 \bmod 3$. I haven't been able to finish this part.
*$p = x_1^2, q = x_2^2, p+q = 3x_3^2, p-q = x_4^2$.
We have $x_1^2 + x_2^2 = 3x_3^2 = 0 \bmod 3$, forcing $x_1^2 = x_2^2 = 0 \bmod 3$. This contradicts $(p,q) = 1$.
*$p = x_1^2, q = x_2^2, p+q = x_3^2, p-q = 3x_4^2$.
$x_1^2 + x_2^2 = x_3^2 \bmod 3$, and $x_1^2 = x_2^2 \bmod 3$. Since $x_3^2 = 0$ or $1 \bmod 3$, we have $x_1^2 = x_2^2 = x_3^2 = 0 \bmod 3$, contradicting $(p,q) = 1$.
I need help with case ($2$) only, assuming the other cases are correctly done. Thank you!
|
For case 2, we could use the trick of Fermat's Infinite Descent.
Observe that $$\left(\dfrac{x_3+x_4}{2}\right)^2+\left(\dfrac{x_3-x_4}{2}\right)^2=x_1^2$$, and $3|x_3-x_4$. So we let $\dfrac{x_3+x_4}{2}=m^2-n^2, x_3-x_4=2mn, x_1=m^2+n^2$, where $m,n$ are positive integers and $(m,n)=1$. Then we have $x_3=m^2+2mn-n^2$. On the other hand, we have $x_3^2-x_1^2=3x_2^2$, which becomes
$$4mn(m^2-n^2)=3x_2^2$$
So $x_2$ is even, let it be $2l$. Then the equation becomes
$$mn(m+n)(m-n)=l^2$$
Now we see that $(p,q,d),(m,n,l)$ are solutions of $xy(x+y)(x-y)=3z^2$. If we assume $(p,q,d)$ is the smallest solution (in terms of the sum), which clearly should exist, then we will produce $(m,n,l)$ which is even smaller, contradiction.
|
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|
Given $\left|e^{i\frac{\phi}{2}}\left(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}\right)\right|$ show that it's equal to $2|\sin{\frac{\phi}{2}}|$ Given $\left|e^{i\frac{\phi}{2}}\left(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}\right)\right|$ I want to show that it's equal to $2\left|\sin{\frac{\phi}{2}}\right|.$
My work.
$|e^{i\frac{\phi}{2}}(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}})|=|(\cos\frac{\phi}{2}+i\sin{\frac{\phi}{2}})(2i\sin{\frac{\phi}{2}})|=|\sin\phi i - 2\sin^2{\frac{\phi}{2}}|$.
How from here go on to show that it's equal to $2|\sin{\frac{\phi}{2}}|?$
|
Continuing from $|i\sin\phi - 2\sin^2{\frac{\phi}{2}}|$ we have $$\sqrt{\sin^2\phi + 4\sin^4\frac{\phi}{2}} = \sqrt{4\sin^2\frac{\phi}{2}\cos^2\frac{\phi}{2} + 4\sin^4\frac{\phi}{2}} = \sqrt{4\sin^2\frac{\phi}{2}(\cos^2\frac{\phi}{2} + \sin^2\frac{\phi}{2}}) = \sqrt{4\sin^2\frac{\phi}{2}} = 2|\sin \frac{\phi}{2}|$$
|
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|
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
|
$$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{\sin^2\left(\dfrac{3\pi}{4}+\alpha\right)}{\cos^2\left(\dfrac{3\pi}{4}+\alpha\right)}=\frac{\left(\frac{-\sqrt{2}}{2}\cos\alpha+\frac{\sqrt{2}}{2}\sin\alpha\right)^2}{\left(\frac{-\sqrt{2}}{2}\cos\alpha-\frac{\sqrt{2}}{2}\sin \alpha\right)^2}=\frac{\frac{1}{2}\cos^2\alpha+\frac{1}{2}\sin^2 \alpha-\frac{1}{2}\sin{\alpha}\cos{\alpha}}{\frac{1}{2}\cos^2 \alpha+\frac{1}{2}\sin^2\alpha+\frac{1}{2}\sin{\alpha}\cos{\alpha}}=\frac{1-\sin{2\alpha}}{1+\sin{2\alpha}}$$ Use the addition formulas for trigonometric functions.
|
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|
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
|
Let $g(x)=Ax^2+Bx+C$,if $A>0$, then $g(x)\ge 0$
Case 1: $\forall x \in \Re$, if $B^2\le 4AC$.
Case 2: For $x \in [p,q]$, if the position of the min (bottom) of $g(x)$, $x_0=-\frac{B}{2A}$ should be s.t $x_0 \le p$ or $x_0\ge q$.
Here $$f(x)=4x^2-2ax+a^2-5a+4>0$$ in $x\in (0,2)$ can be met in both the cases.
Case 1: $4a^2 \le 16(a^2-5a+4) \implies 3a^2-20a+16 \ge 0 \implies (x-x_1)(x-x_2)\ge 0$ where $x_1=\frac{2}{3}(5-\sqrt{13}) \approx 0.92), x_2=\frac{2}{3}(5+\sqrt{13}) \approx 5.73$. So $$a\le \frac{2}{3}(5-\sqrt{13})~ \text{or}~ a\ge \frac{2}{3}(5+\sqrt{13}).....(1)$$
Case 2: $x_0=a/4$, $x_0<0 \implies a\le 0$ and $x_0\ge 2 \implies a\ge 8.$
So in this case $$a\le 0 ~\text{or}~ a\ge 8........(2)$$.
Interestingly, the final domain of $a$ needs to be the union of (1) and (2).
Lastly, $f(x)> 0, ~\text{ONLY}~\forall x \in(0,2)$, when $a\in (-\infty,0)\cup (8,\infty).$
Otherwise, $f(x)>0, \forall x \in \Re $, when $a \in (-\infty,\frac{2}{3}(5-\sqrt{13}) \cup (\frac{2}{3}(5+\sqrt{13}), \infty)$
|
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|
What is the inverse for: $x^2+4x+5,x\ge-2$ What is the inverse for: $x^2+4x+5,x\ge-2$
I have tried to solve it using the quadratic formula:
$x^2+4x+5=y$
$x^2+4x+5-y=0$
$x=-2\pm\sqrt{y-1}$
$y\ge1$
The inverse should be: $x=-2+\sqrt{y-1}$
I don't understand why it sould be $+\sqrt{y-1}$ and not $-\sqrt{y-1}$
Can someone explain why it is positive and not negative?
|
Try plugging in a number. Since we assume $x\geq -2$, plugging in $x=3$ gives us $f(3)=26$. We want $f^{-1}(26)=3$. This only happens when $f^{-1}(y)=-2+\sqrt{y-1}$. In case we were to take $x<-2$, the inverse would have been $-2-\sqrt{y-1}$. This phenomena occours since $x^2+4x+5$ is not globally invertible, and hence has two inverses depending on the region.
|
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|
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$using implicit differentiation
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ using implicit differentiation
My workings:
$y=\tan^{-1} x$,
$x= \tan y$
$\frac{d}{dx} (x) = \frac{d}{dx} \tan y$
$1 = \sec^2 y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$
How do I carry on from here?
|
Use the basic trigonometric identity
$$
y'=\cos^2y=\frac{\cos^2y}1=\frac{\cos^2y}{\cos^2y+\sin^2y}=\frac1{1+\tan^2y}=\frac1{1+x^2}
$$
|
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|
Probability of getting white balls from urns QUESTION: An urn $A$ contains $3$ white balls and $4$ black balls. $2$ balls are drawn from urn $A$ and without seeing the colour and we introduce them in urn $B$ (which was empty). Later, a ball is drawn randomly from each urn.
a) Find out the probability that the ball drawn from urn $A$ is white.
b) Find out the probability that the two balls drawn are white.
BELOW IS MY SOLUTION. Is it correct?
a) Let $w$ the event of drawing a white ball and $b$ the event of drawing a black ball. The probabilities for the events of two drawings from urn $A$ are:
$\begin{array}{lll}
P(\{ ww \}) &=& \displaystyle \frac{3}{7} \cdot \frac{2}{6} = \frac{1}{7}, \quad P(\{ wb \}) = \displaystyle \frac{3}{7} \cdot \frac{4}{6} = \frac{2}{7}, \\
P(\{ bw \}) &=& \displaystyle \frac{4}{7} \cdot \frac{3}{6} = \frac{2}{7}, \quad P(\{ bb \}) = \displaystyle \frac{4}{7} \cdot \frac{3}{6} = \frac{2}{7} \\
\end{array}$
Let $W_A$ the event of drawing a white ball from urn $A$ after $2$ drawings. By total probability theorem:
$\begin{array}{lll}
P\left( W_A \right) & = & \displaystyle P\left( \{ ww \} \right) \cdot P\left(W_A | \{ ww \} \right) + P\left( \{ wb \} \right) \cdot P\left(W_A | \{ bn \} \right) + \\
&& P\left( \{ bw \} \right) \cdot P\left(W_A | \{ bw \} \right) + P\left( \{ bb \} \right) \cdot P\left(W_A | \{ bb \} \right) \\
& = & \displaystyle \frac{1}{7} \cdot \frac{1}{5} + \frac{2}{7} \cdot \frac{2}{5} + \frac{2}{7} \cdot \frac{2}{5} + \frac{2}{7} \cdot \frac{3}{5} = \frac{1+4+4+6}{35}=\frac{15}{35} = \frac{3}{7} \\
\end{array}$
Consequently, the probability that the ball drawn from urn $A$ is white is $\displaystyle \frac{3}{7}$.
b) Let be the following events:
$\begin{array}{lll}
W_B & = & \{ \text{Get white ball from urn $B$ after $2$ drawings} \} \\
W_A \cap W_B & = & \{ \text{Get $2$ white balls after $2$ drawings} \} \\
\end{array}$
By total probability theorem:
$\begin{array}{lll}
P\left( W_A \cap W_B \right) & = & \displaystyle P\left( \{ bb \} \right) \cdot P\left(W_A \cap W_B | \{ bb \} \right) + P\left( \{ bn \} \right) \cdot P\left(W_A \cap W_B | \{ bn \} \right) + \\
&& P\left( \{ nb \} \right) \cdot P\left(W_A \cap W_B | \{ nb \} \right) + P\left( \{ nn \} \right) \cdot P\left(W_A \cap W_B | \{ nn \} \right) \\
& = & \displaystyle \frac{1}{7} \cdot \frac{1}{5} \cdot \frac{2}{2} + \frac{2}{7} \cdot \frac{2}{5} \cdot \frac{1}{2} + \frac{2}{7} \cdot \frac{2}{5} \cdot \frac{1}{2} + \frac{2}{7} \cdot \frac{3}{5} \cdot \frac{0}{2} = \frac{1+2+2}{35}=\frac{5}{35} = \frac{1}{7} \\
\end{array}$
The probability that the two balls drawn are white is $\displaystyle \frac{1}{7}$.
|
Given the solutions, we are hinted that the procedure of choosing $1$ or $2$ white balls does not affect the answer.
$$ \frac {3 \choose 1} {7 \choose 1} = {3 \over 7} $$
$$ \frac {3 \choose 2} {7 \choose 2} = {1 \over 7} $$
This makes sense.
Let's think to a deck with red an black cards. It does not matter how much we shuffle, cut or re-shuffle before we randomly turn two cards down side up and reveal the color.
|
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|
find two primes less than $500$ that divide $20^{22}+1$
Find two primes less than $500$ that divide $20^{22}+1$.
Note that $20^2+1=401$ divides the required number (since for any integer a, if k is odd, then $a+1$ divides $a^k+1$). Also, any prime dividing the given number must be congruent to $1$ modulo $4$ since $-1$ would be a quadratic residue modulo any such prime. Primes like $17$ and $13$ don't work. $20\equiv 7\mod 13, 20\equiv 3\mod 17,$ and in both cases $20$ is a primitive root with respect to the corresponding modulus. So it might be that we'll need to find a large prime. $a^k+1 = (a+1)(a^{(k-1)} - a^{(k-2)}+\cdots + 1),$ where $a = 20^2, k=11.$ I'm not sure if there's some way to factor $a^{(k-1)} -a^{(k-2)}+\cdots + 1.$
|
I claim that the two primes are $89$ and $401$. The $401$ is kind of obvious because $20^{22}=(20^2)^{11}\equiv (-1)^{11}\equiv -1\pmod{401}$. The $89$ is a bit more subtle. Note that $2^{44}\pmod{89}$ is just $\left(\frac{2}{89}\right)$, the Legendre symbol. It's easy to calculate by quadratic reciprocity that this is $1$. I'm too lazy to compute $5^{22}\pmod{89}$ the smart way, so just reduce: $5^{22}\equiv 5\cdot 125^7\equiv 5\cdot 36^7\equiv 5\cdot 36\cdot 1296^3\equiv 2\cdot 50^3\equiv -1\pmod{89}$. Then $20^{22}\equiv 5^{22}\cdot 2^{44}\equiv -1\pmod{89}$.
|
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|
Converse to Jensen's inequality for $1/x$ on a positive bounded interval? Consider the function $f(x) = 1/x$ on the interval $I = [a, b]$, where $0 < a \leq b$.
By Jensen's inequality, we have for any $\{x_j \}_{j=1}^n \subset I$,
$$
f(\overline{x}) \leq \frac{1}{n} \sum_{i=1}^n f(x_i).
$$
Above, $\overline{x} = (1/n)\sum_{j=1}^n x_j$.
Is there a converse inequality of the form
$$
\frac{1}{n} \sum_{i=1}^n f(x_i) \leq C f\Big(\frac{1}{n}\sum_{i=1}^n x_i\Big),
$$
where $C$ may depend on $a, b$?
One thing I thought of is a second-order Taylor expansion.
By Taylor expansion, there are $\xi_i$ in the interval between $x_i, \overline{x}$ such that
$$
\frac{1}{n} \sum_i f(x_i) = f(\overline{x}) + \frac{1}{2n} \sum_{i=1}^n f''(\xi_i)
(x_i - \overline{x})^2.
$$
Additionally, we have that $f''(\xi) = \tfrac{2}{\xi^3}$, so that the remainder is
$$
\frac{1}{n} \sum_{i=1}^n \frac{(x_i - \overline{x})^2}{\xi_i^3}
\leq
\frac{1}{n} \sum_{i=1}^n \max \Big\{
\frac{(x_i - \overline{x})^2}{x_i^3},
\frac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}.
$$
I struggled to bound this remainder in terms of $C(a, b)~f(\overline{x})$, though.
|
We want to determine an upper bound for the function
$$
h(x_1, \ldots, x_n) = \frac{1}{n^2} \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k}
= \sum_{k, l=1}^n \frac{x_k}{x_l}
$$
on the hypercube $[a, b]^n$. $h$ is convex in each variable, so that the maximum of $h$ is attained at a corner point of the hypercube. Therefore it suffices to compute
$$
\begin{align}
h(\underbrace{a, \ldots, a}_{m},\underbrace{b, \ldots, b}_{n-m})
&= \frac{1}{n^2}\bigl(ma - (n-m)b\bigr)\left( \frac m a + \frac{n-m} {b}\right) \\
&= \left( \frac mn \right)^2 + \frac mn\left( 1-\frac mn \right)\left( \frac a b + \frac b a\right) + \left( 1-\frac mn \right)^2 \\
&= \dots \\
&= \frac{(a+b)^2}{4ab} - \frac{(a-b)^2}{4ab}\left(2 \frac mn - 1\right)^2 \, .
\end{align}
$$
The sharp upper bounds are therefore
$$
h(x_1, \ldots, x_n) \le \frac{(a+b)^2}{4ab}
$$
if $n$ is even, and
$$
h(x_1, \ldots, x_n) \le \frac{(a+b)^2}{4ab} - \frac{1}{n^2}\frac{(a-b)^2}{4ab}
$$
if $n$ is odd.
The best upper bound which is independent of $n$ is $C = \frac{(a+b)^2}{4ab}$.
|
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|
Angle Between Two Vectors Discrepancy Here is something that is bugging me: Consider the vectors
$$ \mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k} \qquad \text{ and } \qquad \mathbf{w} = 2\mathbf{j} - \mathbf{k}. $$
We have two different formulas for the angle $\theta$ between these vectors:
$$ \theta = \cos^{-1}\left(\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|} \right) \qquad \text{ and } \qquad \theta = \sin^{-1}\left(\frac{|\mathbf{v} \times \mathbf{w}|}{|\mathbf{v} || \mathbf{w}|} \right) .$$
Well:
$$|\mathbf{v}| = 6, \qquad |\mathbf{w}| = \sqrt{5}, \qquad \mathbf{v} \cdot \mathbf{w} = -12, \qquad \text{ and } |\mathbf{v} \times \mathbf{w}| = 6. $$
But:
$$\cos^{-1}\left(-\frac{2}{\sqrt{5}} \right) \ne \sin^{-1}\left( \frac{1}{\sqrt{5}} \right). $$
So, what gives?
|
Both functions only give one specific inverse, but others exist. If you remove the inverse functions and convert it to a simple system of equations, you actually have a valid solution.
|
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|
A question on the conditions under which $\left|\begin{smallmatrix}x+a&b&c\\a&x+b&c\\a&b&x+c\end{smallmatrix}\right|=A$ is non singular The question is if
$$
\begin{vmatrix}
x+a & b & c \\
a & x+b & c \\
a & b & x+c \\
\end{vmatrix}=A
$$
find the conditions under which it's nonsingular
on performing some ERTs
I got
$$
\begin{vmatrix}
0 & -x & 0 \\
x & x & -x \\
a+b & b & x+c \\
\end{vmatrix}=A
$$
so $$-x[x(x+c)+x(a+b))] \ne 0$$
which means
$$-x^3+x^2(a+b+c)\ne 0$$
which implies $x \ne a+b+c$
however, my book says the answer is $x \ne 0$ and $-(a+b+c)$
where am I going wrong
|
You made a sign error in the computation of the determinant; it should be
$-x^3 - x^2(a+b+c)$. The solutions of $-x^3 - x^2(a+b+c) = 0$ are then $x = 0$ and $x = -(a+b+c)$.
|
{
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|
Solution the ode $x^2y''+4xy'+2y=f(x)$ I'm trying to find a solution for the second order ode $x^2y''+4xy'+2y=f(x)$, where $f\in\mathcal{C}^1(\mathbb{R})$.
I already found that the solution for the homogeneous part is equal to
$y_h(x)=\frac{C_1}{x^2}+\frac{C_2}{x}$. But now I'm stuck trying to get the nonhomogeneous solution for this equation. I tried substituting $y=f(x)$ but this leads me to nowhere.
Any help is greatly appriciated.
|
If you say that $y = f(x)$ you can solve it like this (homogeneous):
step 2: Assume a solution to this Euler-Cauchy equation will be proportional to for some constant $λ$. Substitute $y(x) := x^{λ}$ into the differentialequation.
$$
\begin{align*}
f(x) &= x^{2} \cdot y'' + 4 \cdot x \cdot y' + 2 \cdot y\\
y &= x^{2} \cdot y'' + 4 \cdot x \cdot y' + 2 \cdot y \quad\mid\quad \text{step 2}\\
0 &= x^{2} \cdot {x^{λ}}'' + 4 \cdot x \cdot {x^{λ}}' + x^{λ}\\
0 &= x^{2} \cdot \frac{\operatorname{d}^{2}}{\operatorname{d}x^{2}}x^{λ} + 4 \cdot x \cdot \frac{\operatorname{d}}{\operatorname{d}x}x^{λ} + x^{λ} \quad\mid\quad \frac{\operatorname{d}^{2}}{\operatorname{d}x^{2}}x^{λ} := λ \cdot (λ - 1) \cdot x^{λ - 2} \text{ and } \frac{\operatorname{d}}{\operatorname{d}x}x^{λ} = λ \cdot x^{λ - 1}\\
0 &= x^{2} \cdot λ \cdot (λ - 1) \cdot x^{λ - 2} + 4 \cdot x \cdot λ \cdot x^{λ - 1} + x^{λ}\\
0 &= λ^{2} \cdot x^{λ} + 3 \cdot λ \cdot x^{λ} + x^{λ}\\
0 &= (λ^{2} + 3 \cdot λ + 1) \cdot x^{λ} \quad\mid\quad\text{solve for } λ \text{ with assuming } x \ne 0 \\
0 &= (λ^{2} + 3 \cdot λ + 1) \cdot x^{λ} \quad\mid\quad\text{Zero product theorem}\\
0 &= (λ^{2} + 3 \cdot λ + 1) \cdot x^{λ} \quad\mid\quad\text{solve for } λ \\
0 &= λ^{2} + 3 \cdot λ + 1 \quad\mid\quad \text{use the pq formula}\\
λ &= - \frac{3}{2} \pm \sqrt{\frac{5}{4}}\\
λ &= - \frac{3}{2} \pm \frac{\sqrt{5}}{2}\\
\\
y_{1} &= \mathrm{c}_{1} \cdot x^{- \frac{3}{2} + \frac{\sqrt{5}}{2}}\\
y_{2} &= \mathrm{c}_{2} \cdot x^{- \frac{3}{2} - \frac{\sqrt{5}}{2}}\\
\\
y &= y_{1} + y_{2} = \mathrm{c}_{1} \cdot x^{- \frac{3}{2} + \frac{\sqrt{5}}{2}} + \mathrm{c}_{2} \cdot x^{- \frac{3}{2} - \frac{\sqrt{5}}{2}}
\end{align*}
$$
If you want to to it with a nonhomogeneous function...
step 3: Try $t := \log(t) \Rightarrow x := e^{t}$:
$$
\begin{align*}
f(x) &= x^{2} \cdot y'' + 4 \cdot x \cdot y' + 2 \cdot y\\
y &= x^{2} \cdot y'' + 4 \cdot x \cdot y' + 2 \cdot y \quad\mid\quad \text{step 2}\\
y &= e^{2 \cdot t} \cdot y'' + 4 \cdot e^{t} \cdot y' + 2 \cdot y\\
y(t) &= \frac{\operatorname{d}^{2}}{\operatorname{d}t^{2}}y(t) + 3 \cdot \frac{\operatorname{d}}{\operatorname{d}t}y(t) + 2 \cdot y(t)\\
0 &= \frac{\operatorname{d}^{2}}{\operatorname{d}t^{2}}y(t) + 3 \cdot \frac{\operatorname{d}}{\operatorname{d}t}y(t) + y(t)\\
0 &= \frac{\operatorname{d}^{2}}{\operatorname{d}t^{2}}e^{λ \cdot t} + 3 \cdot \frac{\operatorname{d}}{\operatorname{d}t}e^{λ \cdot t} + e^{λ \cdot t}\\
0 &= λ^{2} \cdot e^{λ \cdot t} + 3 \cdot λ \cdot e^{λ \cdot t} + e^{λ \cdot t}\\
0 &= (λ^{2} + 3 \cdot λ + 1) \cdot e^{λ \cdot t} \quad\mid\quad \text{Zero product theorem}\\
0 &= λ^{2} + 3 \cdot λ + 1 \quad\mid\quad \text{use the pq formula}\\
λ &= - \frac{3}{2} \pm \sqrt{\frac{5}{4}}\\
λ &= - \frac{3}{2} \pm \frac{\sqrt{5}}{2}\\
\\
y_{1}(t) &= \mathrm{c}_{1} \cdot e^{(- \frac{3}{2} + \frac{\sqrt{5}}{2}) \cdot t}\\
y_{2(t)} &= \mathrm{c}_{2} \cdot e^{(- \frac{3}{2} - \frac{\sqrt{5}}{2}) \cdot t}\\
\\
y &= x^{- \frac{3}{2} - \frac{\sqrt{5}}{2}} \cdot (\mathrm{c}_{1} + \mathrm{c}_{2} \cdot x^{\sqrt{5}})
\end{align*}
$$
|
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|
Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried relating this to well-known sums but that didn't work out either.
I would love some help with this.
|
The series appears to take the form
$$ f(x) = 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n $$
which can be seen in the form:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n} \, x^n - \sum_{n=1}^{\infty} \frac{(n+1) \, x^n}{n} \\
&= 2 + \frac{d^2}{d x^2} \, \sum_{n=1}^{\infty} \frac{x^{n+2}}{n} - \frac{d}{dx} \, \sum_{n=1}^{\infty} \frac{x^{n+1}}{n} \\
&= 2 + \frac{d^2}{d x^2} \, (- x^2 \, \ln(1-x) ) - \frac{d}{d x} \, (- x \, \ln(1-x) ) \\
&= \cdots \\
&= 2 + \frac{3 x - 2 x^2}{(1-x)^2} - \ln(1-x) \\
&= \frac{2 - x}{(1-x)^2} - \ln(1-x).
\end{align}
A faster way is:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n \\
&= 2 + \sum_{n=1}^{\infty} \frac{n (n+2) + 1}{n} \, x^n \\
&= 2 + \sum_{n=1}^{\infty} \left( (n+2) + \frac{1}{n} \right) \, x^n \\
&= \sum_{n=0}^{\infty} (n+2) \, x^n + \sum_{n=1}^{\infty} \frac{x^n}{n} \\
&= \frac{2-x}{(1-x)^2} - \ln(1-x).
\end{align}
|
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|
Write $ z = \frac{(1-i)^3(√3+i)}{4i}$ to polar form Write the complex number in polar form:
$$ z = \frac{(1-i)^3(\sqrt 3+i)}{4i}$$
So my try goes as follows:
\begin{align}
\frac{(1−i)^3(\sqrt 3+i)}{4i} &=
\frac{(1−i)^3(\sqrt 3+i) \times -4i}{16}\\& =
\frac{(1-3i-3+i)(\sqrt 3+i)\times-4i}{16}\\& =
\frac{(-2-2i)(\sqrt 3+i)\times-4i}{16} \\&=
\frac{(-1-i)(\sqrt 3+i)\times-4i}{8}\\& =
\frac{(4i - 4)(\sqrt 3+i)}{8} \\&=
\frac{(i-1)(\sqrt 3+i)}{2} \\&=
\frac{(\sqrt 3 \times i-1-\sqrt 3-i)}{2}\\& =
\frac{-(\sqrt 3+1)}{2} + \frac{(\sqrt 3-1)i} {2}\end{align}
Polar form:
$$r =\sqrt {\left(\frac {-1+\sqrt 3)}{2}\right)^2 + \left(\frac{\sqrt 3-1}{2}\right)^2} = \sqrt{2} $$
$$\tan(v) = \frac{(\sqrt 3-1)} { -(1+\sqrt 3)} \iff v = \arctan(\sqrt 3-2) $$
put it equal to $K$
$$z = \sqrt{2}(\cos K, i \sin K), \\ K = \arctan(\sqrt 3-2)$$
To me, this doesn’t seem like a clean answer, since this question could potentially be on an upcoming exam. Is there something I am missing, like a better approach?
|
The beat way to change to polar form is to change individual terms to polar form first
So
$$(1-i)^3=\left(\sqrt{2}e^{-\frac{\pi i}{4}}\right)^3=2\sqrt{2}e^{-\frac{3\pi i}{4}}$$
$$\sqrt{3}+i=2e^{\frac{\pi i}{6}}$$
and
$$4i=4e^{\frac{\pi i}{2}}$$
Which means that
$$z=\sqrt{2}e^{-\frac{13\pi i}{12}}$$
|
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|
Solve the system $ \left\lbrace \begin{array}{ccc} \sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\ x+y &=& 2xy \end{array}\right. $ I have to solve the following system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\
x+y &=& 2xy
\end{array}\right.
$$
I've showed that it is equivalent to
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 \\
x+y &=& 2xy
\end{array}\right. \text{ or } x=y= 1
$$
And i can't conclude nothing frome this system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 \\
x+y &=& 2xy
\end{array}\right.
$$
|
To solve the second equation system.
From $x+y=2xy$. We have $y=\frac{x}{2x-1}$.
Then use $\sqrt{1-x} = 2 + \sqrt{1-y}$ (I use the second equation, x=y=1 is excluded.)
$y-x-4 = 4 \sqrt{1-y}$
$\frac{x}{2x-1}-x-4 = 4 \sqrt{1-\frac{x}{2x-1}}$
$\frac{-2x^2-6x+4}{2x-1} = 4 \sqrt{\frac{x-1}{2x-1}}$
$(x^2+3x-2)^2 = 4(x-1)(2x-1)$
$x^4+6x^3-3x^2=0$
$x = -3 \pm 2 \sqrt{3}, 0$
rejected unsuitable roots, we have $x = -3 - 2 \sqrt{3}, y = -3 + 2 \sqrt{3}$
and $x=0, y=0$ but it does not satisfact first equation.
You need to square twice to get rid of all square roots and resulting a 4th order equation. Luckily it is solvable.
|
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|
Solving $\frac{dy}{dx} = \frac{xy+3x-2y+6}{xy-3x-2y+6}$ I'm stuck with this problem...
$$\frac{\operatorname{d}y}{\operatorname{d}x} = \frac{xy+3x-2y+6}{xy-3x-2y+6}$$
I have tried separating variables in the following way:
$$\frac{\operatorname{d}y}{\operatorname{d}x} = \frac{x(y+3)2(-y+3)}{y(x-2)3(-x+2)}$$
$$\left(\frac{y+3 \cdot 2(-y+3)}{y}\right)\operatorname{d}y = \left(\frac{x-2 \cdot 3(-x+2)}{x}\right)\operatorname{d}x$$
$$\left(\frac{-y+9}{y}\right) \operatorname{d}y = \left(\frac{-2x+4}{x}\right) \operatorname{d}x$$
Then integrating both sides I end up with
$$y=2x+\ln\left(\frac{y^9}{x^4}\right)+C$$
I'd appreciate very much if someone give me the right answer because I don't know if my solution is correct or not. Thanks in advance
|
$\begin{align}\frac{xy+3x-2y+6}{xy-3x-2y+6}&=\frac{6x+xy-3x-2y+6}{xy-3x-2y+6}\\
&=\frac{6x+x(y-3)-2(y-3)}{x(y-3)-2(y-3)}\\
&=\frac{6x+(x-2)(y-3)}{(x-2)(y-3)}\\
&=\frac{6x}{(x-2)(y-3)}+1\\
\end{align}$
So, your non-linear ODE is
$$y'=\frac{6x}{(x-2)(y-3)}+1$$
or
$$(y-3)(y'-1)=\frac{6x}{x-2}.$$
If we make a change of variable $u=y-3$ then $y'=u'$ and hence we have
$$uu'-u=\frac{6x}{x-2}.$$
I don't now how to solve it. WA couldn't solve too. https://www.wolframalpha.com/input?i=yy%27-y%3D6x%2F%28x-2%29
I need to unlock pro-wa?
|
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|
Finding $\lim_{x\to 0}\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}$ How to evaluate the following limit without L'Hôpital's rule
?$$\lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)$$
My attempt:
$$\begin{align}L &= \lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)\tag1\\& = \lim_{x\to 0}\left(\frac{x^2}{(\sin^{-1}(x))^2x^2} - \frac1{x^2}\right)\tag2\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot \frac1{x^2} - \frac1{x^2}\tag3\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag4\\& =\left(1\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag5\\& =\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2} \tag6\\& =0\tag7\end{align}$$
But later I realized, $\lim\limits_{x\to a} f(x)\ g(x) = \lim\limits_{x\to a} f(x)\lim\limits_{x\to a} g(x) $ if and only if both $\lim\limits_{x\to a} f(x)$ and $\lim\limits_{x\to a} g(x)$ are well defined and finite. Thus moving from $(3)$ to $(4)$ is absolutely wrong.
I don't have any more ideas about the problem.
|
$$\lim_{x \rightarrow 0} \left( \frac{1}{\arcsin(x)^2} - \frac{1}{x^2} \right) = \lim_{x \rightarrow 0} \left( \frac{x^2 - \arcsin(x)^2}{\arcsin(x)^2 x^2} \right)$$
Taylor series of $ \arcsin(x) = x + x^3/6 + O(x^5) $ at $x \rightarrow 0$, in this way
$$\lim_{x \rightarrow 0} \left( \frac{x^2 - x^2 - x^4/3 - x^6/36}{(x + x^3/6)^2x^2} \right) = \lim_{x \rightarrow 0} \left( \frac{- x^4/3 - x^6/36}{x^4 + x^6/3 + x^8/36} \right) = \lim_{x \rightarrow 0} \left( \frac{- x^4/3}{x^4} \right) = - \cfrac{1}{3}$$
|
{
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|
How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.
I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exapansions and factoring seem terrible. Maybe there is a good way for generalization.
The question is, can we generalize these factorisations?
$$(a+b)^n-(a^n
+b^n)$$
for all odd integer $n$?
|
A generalisation for non-negative integer $n$ is provided in this answer. The expansion there can be written as
\begin{align*}
&\,\,\color{blue}{(a+b)^n-\left(a^n+b^n\right)}\\
&\quad\,\,\color{blue}{=ab(a+b)\sum_{k=1}^{\lfloor n/2\rfloor}\left(\binom{n-k}{k}+\binom{n-k-1}{k-1}\right)(-ab)^{k-1}(a+b)^{n-2k-1}}\tag{1}\\
\end{align*}
where we use the binomial identity
$
2\binom{n-k}{k}-\binom{n-k-1}{k}=\binom{n-k}{k}+\binom{n-k-1}{k-1}
$.
We check formula (1) by taking $n=7$. We obtain
\begin{align*}
&\color{blue}{(a+b)^7-\left(a^7+b^7\right)}\\
&\qquad\quad=ab(a+b)\sum_{k=1}^{3}\left(\binom{7-k}{k}+\binom{6-k}{k-1}\right)(-ab)^{k-1}(a+b)^{6-2k}\\
&\qquad\quad=ab(a+b)\left(\left(\binom{6}{1}+\binom{5}{0}\right)(a+b)^4
-\left(\binom{5}{2}+\binom{4}{1}\right)ab(a+b)^2\right.\\
&\qquad\qquad\qquad\qquad\quad\left.+\left(\binom{4}{3}+\binom{3}{2}\right)(ab)^2\right)\\
&\qquad\quad=ab(a+b)\left(7(a+b)^4-14(ab)(a+b)^2+7(ab)^2\right)\\
&\qquad\quad=7ab(a+b)\left((a+b)^4-2(ab)(a+b)^2+(ab)^2\right)\\
&\qquad\quad=7ab(a+b)\left((a+b)^2-ab\right)^2\\
&\qquad\quad\,\,\color{blue}{=7ab(a+b)\left(a^2+ab+b^2\right)^2}
\end{align*}
in accordance with OPs derivation.
|
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|
The number of distinct powers of x that appear in the expansion of $(x^{7}+x^{11}+x^{14})^{10}$ is
The number of distinct powers of x that appear in the expansion of
$(x^{7}+x^{11}+x^{14})^{10}$ is
Let the power of $x^7$ is a, $x^{11}$ is b and $x^{14}$ is c
then by multinomial a+b+c=10
so we have to find the number of distinct values of $7*a+11*b+14*c= 7*(a+2c)+11*b$
how do I proceed after this?
|
First of all, notice that the answer to the question is equal to that of $(1+x^4+x^7)^{10}$, and then the answer is equal to the number of distinct sums of the form $4a+7b$, where $0\le a+b\le10$. If $a+b=k$, it is easy to notice that $4a+7b$ produces $k+1$ numbers, hence, we already have $1+2+...+11=66$ numbers. However some of these numbers are repeated while being produced. In other words;
$$4a+7b=4c+7d\implies 4(a-c)=7(d-b)\implies a-c=7.$$
Now, we have;
$$a=7, c=0\implies (d,b)=(4,0),(5,1),(6,2),or\; (7,3)$$
$$a=8, c=1\implies (d,b)=(4,0),(5,1), or\;(6,2)$$
$$a=9, c=2\implies (d,b)=(4,0), or\;(5,1)$$
$$a=10, c=3\implies (d,b)=(4,0).$$
Therefore, the answer is $66-10=56.$
|
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|
Stationary points of a cubic function If t is a positive constant, find the local maximum and minimum values of the function
$f(x) = (3x^2 - 4)\left(x - t + \frac{1}{t}\right)$
and show that the difference between them is $\frac{4}{9}(t + 1/t)^3$.
Find the least value of this difference as $t$ is varied.
My attempt:
\begin{align*}
f(x) & = (3x^2 - 4)\left(x - t + \frac{1}{t}\right)\\
f'(x) & = (3x^2 - 4)(1) + 6x\left(x - t + \frac{1}{t}\right)\\
\end{align*}
Stationary points when $f'(x) = 0$.
$$0 = 9x^2 - 6x\left(t - \frac{1}{t}\right) - 4$$
Find values for $x$ using quadratic formula gives
$$x = \frac{\left(t - \frac{1}{t}\right) \pm \sqrt{t^2 + \frac{1}{t^2}}}{3}$$
I am not sure if this is right so far, but if it is ... it seems very complicated to work out corresponding values for $f(x)$ and then find the difference between them.
|
Note that you can factor the derivative as
$$ \frac{\rm d}{{\rm d}x} f = \tfrac{1}{t} ( 3x -2 t) ( 3 t x + 2) = 0 $$
which gives you the $x$ values for the extrema
$$ \begin{gathered} x = \frac{2 t}{3} & x = -\frac{2}{3 t} \end{gathered} $$
For the next part, you can do a variable substitution with $ u = t + \frac{1}{t}$ since the answer you are looking for is in terms of $u$.
$$ f = (x - \sqrt{ u^2-4}) ( 3 x^2 - 4) $$
and
$$ \begin{gathered} x = \frac{\sqrt{u^2-4}-u}{3} & x = \frac{\sqrt{u^2-4}+u}{3} \end{gathered} $$
At this point plug in the extrema $x$ into $f$ and simplify. Take the difference between the two results.
|
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|
Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$x^3+y^3 = 4(x+y)$
$x^3+y^3 = 8$
Then:
$8 = 4(x+y)$
$x+y = 2$
However if we replace that value in $(x+y)^2$ we have:
$(x+y)^2 = 2^2 =4$
$(x+y)^2 = x^2+y^2+2xy = 5 + 2 = 7$
As you can see $4 \neq 7$, what is happening?
|
Let us call $a := x + y$ and $b := xy$. Then the proposed system of equations is equivalent to
\begin{align*}
\begin{cases}
b = 1\\\\
a^{2} - 2b = 5\\\\
a^{3} - 3ab = 8
\end{cases} \Longleftrightarrow
\begin{cases}
b = 1\\\\
a^{2} = 7\\\\
a^{3} - 3a = 8
\end{cases} & \Longleftrightarrow
\begin{cases}
b = 1\\\\
a = \pm\sqrt{7}\\\\
a = 2
\end{cases}
\end{align*}
which clearly has no solutions. I think this answers your question about what is happening.
Hopefully this helps!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all the solutions for $f\left(x\right) =2f\left(\frac{1}{x}\right)-\frac{2x^{2}-1}{x^{2}+1}$ The function is $f:\left(0,\infty\right)\rightarrow\mathbb{R}$, I tried to do it like that, first I saw that: $$f\left(x\right) =2f\left(\frac{1}{x}\right)-\frac{2x^{2}-1}{x^{2}+1}$$ and then decided to try to put $\frac{1}{x}\in\left(0,\infty\right) $ and came out with: $$f\left(\frac{1}{x}\right) =2f\left(x\right)-\frac{2\frac{1}{x^{2}}-1}{\frac{1}{x^{2}}+1}
=2f\left(x\right)-\frac{\frac{2-x^{2}}{x^{2}}}{\frac{1+x^{2}}{x^{2}}}
=2f\left(x\right)-\frac{2-x^{2}}{1+x^{2}}$$ i used it at the last equation and it came out that: $$f\left(x\right)=\frac{5-4x^{2}}{3\left(1+x^{2}\right)}$$ now i get $f\left(1\right)=\frac{1}{6}$ but i know it should equal 1/2... I have now idea how to move forward, and why it doesn't work, any tips?
|
$$f(x)=2f\left(\frac 1x\right)-\frac{2x^2-1}{x^2+1}\\f\left(\frac 1x\right)=2f(x)-\frac{2-x^2}{x^2+1}$$ Adding and simplifying we have
$$f(x)+f\left(\frac 1x\right)=1$$ It follows
$$\boxed{f(x)= \frac{1}{x^2+1}}$$ and it is easy to verify that in fact
$$f(x)=2f\left(\frac 1x\right)-\frac{2x^2-1}{x^2+1}$$
|
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|
Inequalities showing monotonicity imply another inequality I came across a proof that shows that $\lim\limits_{n \rightarrow \infty} (1+ \frac{1}{n})^{n\cdot (\frac{1}{n+1} +\dots + \frac{1}{2n})} =2$
First it is shown that $((1+\frac{1}{n})^{n+1})_{n=1}^{\infty}$ is monotonically decreasing and $((1+\frac{1}{n})^n)_{n=1}^{\infty}$ is monotonically increasing, thus
$$(1 + \frac{1}{n})^n \leq (1 + \frac{1}{n+1})^{n+1} \leq \dots \leq (1+ \frac{1}{n+k})^{n+k}$$ for $((1+\frac{1}{n})^n)_{n=1}^{\infty}$ and
$$(1+\frac{1}{n})^{n+1} \geq (1+ \frac{1}{n+1})^{n+2} \geq \dots \geq (1 + \frac{1}{n+k})^{n+k+1}$$ for $((1+\frac{1}{n})^{n+1})_{n=1}^{\infty}$
I understand the proof up until this point. But then it says that those two inequalities yield
$$
\left(1+\frac1{n+k}\right)^{\frac{n}{n+1}}\le\left(1+\frac1n\right)^\frac{n}{n+k}\le\left(1+\frac1{n+k}\right)
$$
How do I obtain this result, is there something obvious I am missing?
|
By starting from
$$
\left(1 + \frac1n\right)^{n} \le \left(1 + \frac1{n+k}\right)^{n+k}
$$
and raising both sides to the power $\frac1{n+k}$, we get
$$
\left(1 + \frac1n\right)^{\frac{n}{n+k}} \le \left(1 + \frac1{n+k}\right). \tag{1}
$$
Also, by starting from
$$
\left(1 + \frac1{n+k}\right)^{n+k+1} \le \left(1 + \frac1n\right)^{n+1}
$$
and raising both sides to the power $\frac{n}{(n+1)(n+k+1)}$, we get
$$
\left(1 + \frac1{n+k}\right)^{\frac{n}{n+1}} \le \left(1 + \frac1n\right)^{\frac{n}{n+k+1}}. \tag{2}
$$
Finally, we have
$$
\left(1 + \frac1n\right)^{\frac{n}{n+k+1}} \le \left(1 + \frac1n\right)^{\frac{n}{n+k}} \tag{3}
$$
just because the exponent is bigger on the right.
To get the inequality we want, chain together $(2)$, $(3)$, and $(1)$.
|
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|
How to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $? I need to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $. I've checked numerically that this is true, but I haven't been able to prove it.
I've tried trigonometric substitutions. Let $\tan u= t:$
$$\int_0^1 (1+t^2)^{\frac 7 2} dt = \int_0^{\frac{\sqrt 2}{2}} (1+\tan^2 u )^{\frac 9 2} du = \int_0^{\frac{\sqrt 2}{2}} \sec^9 u \ du = \int_0^{\frac{\sqrt 2}{2}} \sec^{10} u \cos u\ du = \int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du$$
Now let $\sin u = w$. Then:
$$\int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du = \int_0^{\sin {\frac{\sqrt 2}{2}}} \frac {1}{(1-w^2)^5} dw.$$
This last integral is solvable using partial fraction decomposition, but even after going through all the work required I'm not really sure how to compare it with $\frac 7 2$, because of that $\sin {\frac {\sqrt{2}}{2}}$ term, which is not easy to compare.
|
Alternative proof:
Clearly, we have
$\sqrt{1 + t^2} \le 1 + t^2/2$. Thus, we have
$$(1 + t^2)^{7/2} \le (1 + t^2)^3 (1 + t^2/2) = \frac12t^8 + \frac52 t^6 + \frac92 t^4 + \frac72 t^2 + 1.$$
Thus, we have
\begin{align*}
\int_0^1 (1 + t^2)^{7/2} \, \mathrm{d} t
&\le \int_0^1 \left( \frac12t^8 + \frac52 t^6 + \frac92 t^4 + \frac72 t^2 + 1\right)\,\mathrm{d} t \\
&= \frac{1}{18} + \frac{5}{14}
+ \frac{9}{10} + \frac{7}{6} + 1\\
& < \frac72.
\end{align*}
We are done.
|
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|
Extracting sequence from generating function I was provided the following generating function, and was unsure how to use it. I have never seen an example where the function “involved” itself.
The generating function is
$F(z)^8$
Where
$$F(z)=z+z^6 F(z)^5+z^{11} F(z)^{10}+z^{16} F(z)^{15}+z^{21} F(z)^{20}$$
Any help appreciated
|
A method that will take a while but gives the first few terms for the equation
$$ F(x) = x + x^6 F(x)^5 + x^{11} F(x)^{10} + x^{16} F(x)^{15} + x^{21} F(x)^{20} $$
is to let $F(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots$.
First notice that the right-hand side does not have a constant term. This leads to $F(x)$ being $F(x) = x + a_{2} x^2 + \cdots$. Also notice that $a_2 = a_{3} = \cdots = a_{10} = 0$ and further reduces $F(x)$ to $F(x) = x + x^{11} + \cdots$
This leads to
\begin{align}
x + x^{11} + a_{21} \, x^{21} + \cdots &= x + x^{11} (1 + 5 \, x^{10} + \cdots) + x^{21} (1 + 10 x^{10} + \cdots) + \cdots \\
&= x + x^{11} + (5 + 1) \, x^{21} + \cdots.
\end{align}
Continuing gives the result mentioned by Gary in the comments.
|
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|
How to solve the differential equation $y'=\frac{y-xy^2}{x+x^2y}$ $$y'=\frac{y-xy^2}{x+x^2y}$$
This is the equation I want to solve.
My idea is to substitute $xy$ with $u,$ $u=xy$ and $\frac{du}{dx}=xy'+y.$
So, the equation becomes
$y'=\frac{y(1-u)}{x(1+u)}.$
What I am struggling with is how to deal with the $x$ and $y$ that are left in the equation.
|
Solve
\begin{gather*}
\boxed{y^{\prime}-\frac{y-x y^{2}}{x +x^{2} y}=0}
\end{gather*}
Using $u=yx$ the above becomes
\begin{align*}
u' &= \frac{2 u}{x \left(u +1\right)}
\end{align*}
This is separable. Integrating gives
\begin{align*}
\left(\frac{u +1}{u}\right)\mathop{\mathrm{d}u}&= \left(\frac{2}{x}\right)\mathop{\mathrm{d}x}\\
\int \left(\frac{u +1}{u}\right)\mathop{\mathrm{d}u}&= \int \left(\frac{2}{x}\right)\mathop{\mathrm{d}x}
\end{align*}
Which gives
\begin{align*}
u +\ln \left(u \right) = 2 \ln \left(x \right)+c_{1}
\end{align*}
Replacing $u \left(x \right)$ in the above solution by $yx$ gives
\begin{align*}
x y+\ln \left(x y\right)-2 \ln \left(x \right)-c_{1} = 0
\end{align*}
This is implicit solution. If you want explicit, you'll have to deal with LambertW special function.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How resolve $y''+y=\sin(x)$ by power series around the point $x_0=0$ The ODE I tried to solve is:
$$y''+y=\sin(x)$$
using the power series method around the point $x_0=0$, with the conditions:
$$y(0)=0,\qquad y'(0)=0$$
Let
\begin{align}
y(x) &= \sum_{n=0}^{\infty} a_{n} \, x^n \\
y^{'}(x) &= \sum_{n=0}^{\infty} (n+1) \, a_{n+1} \, x^n \\
y^{''}(x) &= \sum_{n=0}^{\infty} (n+1)(n+2) \, a_{n+2} \, x^n.
\end{align}
then the differential equation gives
\begin{align}
\sum_{n=0}^{\infty} (n+1)(n+2) \, a_{n+2} \, x^n + \sum_{n=0}^{\infty} a_{n} \, x^n &= \sin(x) \\
\sum_{n=0}^{\infty} ( (n+1)(n+2) \, a_{n+2} + a_{n} ) \, x^n &= \sum_{n=0}^{\infty} \frac{(-1)^n \, x^{2n+1}}{(2 n +1)!}.
\end{align}
I got stuck in this last line, there I see that the sums on both sides are not very identical. So I want to know what change of variable I could apply to the index $n$ in order to find the values of $a_n$. Thanks.
|
The difficulty as stated in the work of the proposed problem the left-hand side should be broken into even and odd index to match the right-hand side. Doing so gives the desired recurrence equations and thus solution.
This solution takes a similar but different path. Let
$$ y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n $$
which leads to
\begin{align}
y(x) &= \sum_{n=0}^{\infty} a_{n} \, x^n \\
y^{'}(x) &= \sum_{n=0}^{\infty} (n+1) \, a_{n+1} \, x^n \\
y^{''}(x) &= \sum_{n=0}^{\infty} (n+1)(n+2) \, a_{n+2} \, x^n.
\end{align}
Now, by using
$$ \sin(x) = \frac{e^{i x} - e^{-i x}}{2 i} = \sum_{n=0}^{\infty} \frac{i^{n-1} \, (1 - (-1)^n)}{2 \, n!} \, x^n, $$
the differential equation becomes
\begin{align}
y^{''} + y &= \sin(x) \\
\sum_{n=0}^{\infty} \left( (n+1)(n+2) \, a_{n+2} + a_{n} \right) \, x^n &= \sum_{n=0}^{\infty} \frac{i^{n-1} \, (1-(-1)^n)}{2 \, n!} \, x^n
\end{align}
which gives the recurrence
$$ a_{n+2} = - \frac{a_{n}}{(n+1)(n+2)} + \frac{i^{n-1} \, (1-(-1)^n)}{2 \, (n+2)!}. $$
From this is can be seen that when $n$ is even the term with $1 - (-1)^n$ is zero and leads to
$$ a_{2n} = - \frac{a_{0}}{(2 n)!}. $$
The odd values take the form
\begin{align}
a_{1} &= a_{1} \\
a_{3} &= \frac{1}{3!} - \frac{a_{1}}{3!} \\
a_{5} &= - \frac{2}{5!} + \frac{a_{1}}{5!} \\
a_{7} &= \frac{3}{7!} - \frac{a_{1}}{7!} \\
\cdots &= \cdots
\end{align}
This leads to
$$ y(x) = - a_{0} \, \cos(x) - a_{1} \, \sin(x) - \frac{x \, \cos(x)}{2}. $$
Applying the initial conditions $y(0) = y^{'}(0) = 0$ gives
$$ y(x) = \frac{\sin(x) - x \, \cos(x)}{2}. $$
|
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|
Better algorithm to find numerical value of one over squareroot Description: To find the value of $1/\sqrt{a}$, basically we find the root $x^{\star}$ of the function $f(x)$
$$
f(x) = x^2 - \dfrac{1}{a}
$$
One of the ways to find it is by using Newton's method and iterating from an initial value $x_0$.
$$
x_{k+1} = x_{k} - \dfrac{f(x_{k})}{f'(x_k)} = \dfrac{1}{2}\left(x_k + \dfrac{1}{ax_k}\right)
$$
$$
\lim_{k\to \infty}x_{k} = \dfrac{1}{\sqrt{a}}
$$
This is known as Newton's method which has quadratic convergence, for some positive constant $M$:
$$
\left|x_{k+1} - \dfrac{1}{\sqrt{a}}\right| \le M \cdot \left|x_{k} - \dfrac{1}{\sqrt{a}}\right|^2
$$
Question: Is there a better numerical method, which convergence is higher than Newton's method, to find $x^{\star} = \dfrac{1}{\sqrt{a}}$?
EDIT: I thought about using Taylor expansion until second order, once Newton's method is made using first-order expansion:
$$ f(x) = f(x_k) + (x-x_k) \cdot f'(x_k) + \dfrac{(x-x_k)^2}{2} \cdot f''(x_k) + \dfrac{(x-x_k)^3}{6} \cdot f'''(\varepsilon(x)) $$
And then find the value of $x_{k+1}$ that
$$0 = f(x_k) + (x_{k+1} - x_k) \cdot f'(x_k)+ \dfrac{(x_{k+1}-x_k)^2}{2} \cdot f''(x_k)$$
Using the function $f(x) = x - \dfrac{1}{ax}$ we rewrite
$$x_{k+1}^2 - \left(3+ax_k^2\right) \cdot x_k \cdot x_{k+1} + 3x_k^2 = 0$$
Which solution is
$$x_{k+1} = \dfrac{x_k}{2}\left(3ax_k^2 - \sqrt{\left(3+ax_{k}^2\right)^2 - 12}\right)$$
As there's a square-root, it may be hard to compute and then we make a second-order Taylor approximation around $\frac{1}{\sqrt{a}}$ of this square-root:
$$\sqrt{\left(3+ax^2\right)^2 - 12}\approx 6x\sqrt{a} - ax^2 -3$$
Once $\sqrt{a} \approx \dfrac{1}{2}\left(y+\dfrac{a}{y}\right) = \dfrac{1}{2}\left(\dfrac{1}{x} + ax\right)$ then
$$\sqrt{\left(3+ax^2\right)^2 - 12}\approx 6x \cdot \dfrac{1}{2}\left(\dfrac{1}{x} + ax\right) -ax^2 - 3 = 2ax^2$$
$$\boxed{x_{k+1} = \dfrac{x_k(3-ax_k^2)}{2}}$$
Which is a simple function of polynomials.
But I have no idea about the convergence of using this iteration function.
Its order would be $3$ cause we used second-order taylor expansion of $f$, but using linear approximations of the square root around $1/\sqrt{a}$ seems it's bellow $3$.
*
*Then, how about other functions like $f=x^4-a^{-2}$ and using the same tricks?
Motivation: There's an algorithm known as Fast Inverse Square Root (more details here) which starts from a good estimative (using logarithm and bit manipulation), but it uses Newton's iteration to refine the value at final step.
|
Householder's method of order $d$ to solve $f(x) = 0$ gives
$$x \to H_d(x) = x + d \frac {\left(1/f\right)^{(d-1)}(x)}{\left(1/f\right)^{(d)}(x)}$$
with rate of convergence $d + 1$.
Using the (wx)Maxima computer algebra system, for
$$f(x) = x^{-2} - a$$
I get
$$\begin{aligned}
H_1(x) &= \frac{(3 - a x^2)x}{2} \\
H_2(x) &= \frac{(3 + a x^2)x}{3 a x^2 + 1} \\
H_3(x) &= \frac{a^2x^4 + 6ax^2 + 1}{4ax(ax^2+1)} \\
H_4(x) &= \frac{x(a^2x^4 + 10ax^2 + 5)}{5a^2x^4 + 10ax^2 + 1} \\
\end{aligned}$$
Suppose the target precision is $N$ accurate digits.
The last iteration giving $N$ accurate digits needs $N/(d+1)$ accurate digits as input, so there are around $\log_{d+1} N$ iteration steps needed.
Assume the cost of $D$-digit digit multiplication (and hence division) is given by some function $M(D)$ (for example $M(D) = D^{\log_2 3}$ when using Karatsuba algorithm), and assume that multiplication and division by small integers, and addition and subtraction of $D$-digit numbers, are insignificant in comparison. Then the cost depends on whether the number of digits of $a$ is small or large. Many calculations can be reused/shared to reduce number of multiplications.
For small $a$:
$$\begin{aligned}
\operatorname{cost}(H_1) = 2 (M(N/1) + M(N/2) + M(N/4) + \cdots + M(1)) &< \frac{2}{\log 2} M(N) \log(N) \\
\operatorname{cost}(H_2) = 3 (M(N/1) + M(N/3) + M(N/9) + \cdots + M(1)) &< \frac{3}{\log 3} M(N)\log(N) \\
\operatorname{cost}(H_3) = 4 (M(N/1) + M(N/4) + M(N/16) + \cdots + M(1)) &< \frac{4}{\log 4} M(N) \log(N) \\
\operatorname{cost}(H_4) = 4 (M(N/1) + M(N/5) + M(N/25) + \cdots + M(1)) &< \frac{4}{\log 5} M(N) \log(N) \\
\end{aligned}$$
showing that higher orders improve overall cost, at least when $N$ is large enough that additions etc really are insignificant.
For $a$ with $N$ digits:
$$\begin{aligned}
\operatorname{cost}(H_1) = 3 (M(N/1) + M(N/2) + M(N/4) + \cdots + M(1)) &< \frac{3}{\log 2} M(N) \log(N) \\
\operatorname{cost}(H_2) = 4 (M(N/1) + M(N/3) + M(N/9) + \cdots + M(1)) &< \frac{4}{\log 3} M(N) \log(N) \\
\operatorname{cost}(H_3) = 5 (M(N/1) + M(N/4) + M(N/16) + \cdots + M(1)) &< \frac{5}{\log 4} M(N) \log(N) \\
\operatorname{cost}(H_4) = 5 (M(N/1) + M(N/5) + M(N/25) + \cdots + M(1)) &< \frac{5}{\log 5} M(N) \log(N) \\
\end{aligned}$$
Reference: https://en.wikipedia.org/wiki/Householder%27s_method
(wx)Maxima source code:
f(x) := x^(-2) - a;
H(d, x) := factor(ratsimp(x + d * (diff(1/f(x), x, d - 1)) / (diff(1/f(x), x, d))));
H(1, x);
H(2, x);
H(3, x);
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.