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How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}}$, and then observe $t$ is a root of $f(x) = x^3 + 3x - 36 = 0$. Also observe that $f(3) = 0$. Now $f(x)$ has only one real root, since $f'(x) = x^2+3$ has no real solution. So $t =3$, as desired.
But I couldn't find any direct/algebraic way of fiddling it for a while. I guess morever this is true: $\left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}+\left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}} = t$, which can possibly be proved in the same indirect way but I don't know how to prove this by direct manipulation.
One way maybe is to write $z_+ = \left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, $z_- = \left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, and writing $t = z_+ + z_-$, observe $z_+z_- = -1$, so $(x-z_+)(x-z_-) = x^2 - tx -1$, and then with Fundamental Theorem of Symmetric Polynomials, maybe calculating $(z_+^3 - z_-^3)^2$ and $(z_+^3 + z_-^3)$ in terms of $t$ would help ?
| By direct calculation we have the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac).$
Then set $a=\sqrt[3]{-18+\sqrt{325}},b=\sqrt[3]{-18-\sqrt{325}}$ and $c=3$. Note that $ab=-1$ and $a^3+b^3=-36$, so $a^3+b^3+c^3=3abc$.
Since $(a-b)^2+(b-c)^2+(b-c)^2>0\Longleftrightarrow a^2+b^2+c^2-ab-bc-ac>0$ holds for any three distinct real numbers, we have that $a+b+c=0$, which means $$\sqrt[3]{-18+\sqrt{325}}+\sqrt[3]{-18-\sqrt{325}}=-3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \to 0}\frac{\frac{\cos2x }{2x}}{\frac{\sin 2x}{2x}}\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \frac{\cos 2x}{2x} \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \frac{{\cos^2 (x)}-{sin^2 (x)}}{2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{\cos^2(x)}{2x}-\frac{sin^2 x}{2x}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1-\sin^2 x}{2x}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin^2 x}{2x}-\frac{sin x}{2}\right) \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin x}{2}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-2\frac{\sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\sin x\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
| Hint: $$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))=\lim_{x \to 0} {\cos (2x)\over \sin 2x}\tan (x) =\lim_{x \to 0} {\cos (2x)\over 2\sin x\cos x}{\sin x\over \cos x}$$
$$=\lim_{x \to 0} {\cos (2x)\over 2\cos^2 x} = {1\over 2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it true that there aren't any three different numbers $x,y,z$ such that $x^3+x \equiv y^3+y \equiv z^3+z \pmod p $? Let $p$ be a prime number. Is it true that there aren't any three different numbers $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p $$
with $x -y, y-z, z-x$, each of them cannot be divided by $p$ ?
If not, what are the conditions of $p$ so that the statement is true for prime number $p$ ?
I tried with $p=3,7$ and both of them are correct, so I think that $p \equiv 3 \pmod 4$ may satisfy the statement.
My other attempt: Assume by contradiction, there exist $x,y,z$ such that $$x^3+x \equiv y^3+y \equiv z^3+z \pmod p$$
with $x -y, y-z, z-x$, each of them cannot be divided by $p$. Then $$x^2+xy+y^2 \equiv y^2+yz+z^2 \equiv z^2+zx+x^2 \pmod p$$
thus $$x+y+z \equiv 0 \pmod p.$$
Here I am stuck. How can I solve this problem ?
(Sorry for my English)
| Solutions exist for all primes $p\ge5,p\neq7$.
As the OP observed, we have the Vieta relation $x+y+z=0$ as $x,y,z$ are the zeros of the cubic
$$
P(T)=T^3+T+c=(T-x)(T-y)(T-z)
$$
in the field $\Bbb{F}_p$. Here $-c=-xyz$ is the shared value of $x^3+x,y^3+y$ and $z^3+z$ (treated as elements of $\Bbb{F}_p$ turning congruences into equalities).
The relation $z=-x-y$ takes care of the quadratic term, and
we are well placed to take advantage of the degree of freedom to select $c$ any which way we wish. Let's concentrate on the linear term! Expanding $(T-x)(T-y)(T+x+y)$ tells us that
$$
(T-x)(T-y)(T+x+y)=T^3-T(x^2+xy+y^2)-xyz,
$$
so we want to be able to choose distinct elements $x,y\in\Bbb{F}_p$ such that $x^2+xy+y^2=-1$.
This is possible whenever $p>3$.
Assume first that $p\equiv1\pmod3$. In this case there is a primitive cubic root of unity $\omega\in\Bbb{F}_p$. It satisfies the equation
$$
\omega^2+\omega+1=0.
$$
And that relation gives us the factorization
$$
a^2+ab+b^2=(a-\omega b)(a-\omega^2b).
$$
So we can select any two numbers $c,d\in\Bbb{F}_p$ such that $cd=-1$. Then the linear system
$$
\left\{\begin{array}{lcl}
a-\omega b&=&c\\
a-\omega^2b&=&d
\end{array}\right.
$$
has a unique solution $(a,b)$. After all, its determinant is $\omega-\omega^2\neq0$.
Then assume that $p\equiv-1\pmod3$. In this case $\omega$ only exists in the extension field $\Bbb{F}_{p^2}$. But, in that case we are dealing with the norm map
$$
N:\Bbb{F}_{p^2}\to\Bbb{F}_p, a-b\omega\mapsto (a-b\omega)(a-b\omega^2)=a^2+ab+b^2.
$$
By elementary properties of finite fields the norm is surjective, and takes each non-zero value in $\Bbb{F}_p$ exactly $p+1$ times. In particular, there are $p+1$ pairs $(a,b)$ such that $a^2+ab+b^2=-1$.
The above argument did not concern the possibility that some of $x,y,z$ may be equal (i.e. $P(T)$ has a multiple root for the resulting $c$). If $x=y$, then $x^2+xy+y^2=3x^2$. If $-1/3$ is a quadratic residue, we need to rule out two possible values of $x$. If $x=-y-x$ then $y=-2x$, and again $3x^2=-1$. Finally, if $y=-y-x$ then $x=-2y$ we need to rule the solutions of $3y^2=-1$.
At most six pairs $(x,y)$ were ruled out. If $p>7$ then in the first case the number of pairs $(c,d)$ such that $cd=-1$ is high enough to leave some solutions. All the cases where we had repetitions among $\{x,y,-x-y\}$ lead to the presence of a square root of $-3\in\Bbb{F}_p$, so the second case of $p\equiv-1\pmod 3$ is not affected.
The claim follows.
It may be worth noting that $p=7$ fails precisely because all the solutions of $a^2+ab+b^2=-1$, namely $(a,b)\in\{(1,3),(3,1),(3,3),(4,4),(4,6),(6,4)\}$ lead to repetitions among $\{a,b,-a-b\}$. None of the six solutions of $cd=-1$ work!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the sequence $a_n =\frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$\sqrt{(n^2-1)}/n$.
$\sqrt{(n^2-1)}/n= |n|\sqrt{1-1/n^2}/n$ if $n$ is positive than $\sqrt1$ else $-\sqrt1$;
bound: $\sqrt{(n^2-1)}/n \le \sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $\frac{\sqrt{(n+1)^2-1}}{n+1} \ge \frac{\sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = \frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
| $$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it without them, it would be better for me. Thanks!
| The first digit of $11^{11}=$ the first digit of $1^{11}=$ $1$.
The first digit of $12^5$ equals the first digit of $2^5=$ the first digit of $32=2$.
$\text{(The first digit of $12^{10}) =$ (The first digit of $12^5)^2 = 4$ }$.
$\text{(The first digit of $12^{12}) =
$ (The first digit of $12^{10}) \times($The first digit of $12^2) = 6$ }$.
$\text{(The first digit of $13^{4}) =$ (The first digit of $3^4) = 1$ }$.
$\text{(The first digit of $13^{12}) =$ (The first digit of $(13^4)^3) = 1$ }$.
$\text{(The first digit of $13^{13}) =$ (The first digit of $13^{12} \times 3) = 3$ }$.
$\text{(The first digit of $11^{11} + 12^{12} + 13^{13}) =$
(The first digit of $1+6+3) = 0$ }$.
Hence $11^{11} + 12^{12} + 13^{13}$ is a multiple of $10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\int \frac{4x^2-3x+2}{4x^2-4x+3}$ with partial fractions I'm trying to understand how my textbook solved this problem but something seems a bit off.
First thing to do was to perform long division since the degree of the numerator is not less than the degree of the denominator, such a division yielded:
$$1 + \frac{x-1}{4x^2-4x-3}$$
Calculating this integral, we got to the point where it seemed plausible to do a substitution $u = 2x - 1$
$$x + \frac{1}{2} \int\frac{\frac{1}{2}(u + 1) -1}{u^2+2}du$$
Now, here's the tricky part, somehow the next simplification was:
$$x + \frac{1}{4} \int\frac{(u -1)}{u^2+2}du$$
My question is how did he do this?
| HINT
We have that
$$\frac{4x^2-3x+2}{4x^2-4x+3}=\frac{4x^2-4x+3+x-1}{4x^2-4x+3}=1+\frac{x-1}{4x^2-4x+3}=$$
$$=1+\frac18\frac{8x-4-4}{4x^2-4x+3}=1+\frac18\frac{8x-4}{4x^2-4x+3}-\frac12\frac{1}{4x^2-4x+3}$$
$$=1+\frac18\frac{8x-4-4}{4x^2-4x+3}=1+\frac18\frac{8x-4}{4x^2-4x+3}-\frac12\frac{1}{2+(2x-1)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sqrt 5$ can be expressed as a polynomial in $e^{2\pi i/5}$ over $\Bbb Z$ Question from a Qualifying Exam:
*
*Show that $\sqrt 5$ can be expressed as a polynomial in $e^{(\frac{2\pi i}{5})}$ over $\Bbb Z$
*If in a field the equation $x^2-5$ has no solution then $x^5-1$ also has no non-trivial solution.
I am unable to show how to find the polynomial
Please givesome hints
| To your first question: Here is a high-faluting answer. If $p$ is any odd
prime number (i.e., any prime number $>2$), then the Gauss
sum is defined to be the
number
\begin{equation}
g\left( 1;p\right) :=\sum_{n=0}^{p-1}e^{2\pi in^{2}/p}.
\end{equation}
Gauss proved that
\begin{equation}
g\left( 1;p\right) =
\begin{cases}
\sqrt{p}, & \text{if }p\equiv1\operatorname{mod}4;\\
i\sqrt{p}, & \text{if }p\equiv3\operatorname{mod}4
\end{cases}
\end{equation}
(and this has been re-proven many times since Gauss; see a post by David
Speyer on
SBSeminar
for my favorite proof, although he denotes $g\left( 1;p\right) $ by
$g\left( \zeta\right) $ and defines it somewhat differently).
Applying this to $p=5$, we obtain $g\left( 1;5\right) =\sqrt{5}$ (since
$5\equiv1\operatorname{mod}4$). Hence,
\begin{align*}
\sqrt{5} & =g\left( 1;5\right) =\sum_{n=0}^{4}e^{2\pi in^{2}/5}=e^{2\pi
i\cdot0^{2}/5}+e^{2\pi i\cdot1^{2}/5}+e^{2\pi i\cdot2^{2}/5}+e^{2\pi
i\cdot3^{2}/5}+e^{2\pi i\cdot4^{2}/5}\\
& =z^{0^{2}}+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}},\qquad\text{where
}z=e^{2\pi i/5}.
\end{align*}
This is, of course, a polynomial in $e^{2\pi i/5}$ over $\mathbb{Z}$. Hence,
your first question is answered.
To your second question: Let $K$ be a field. We shall show that if $x^{2}-5$
has no solution in $K$, then $x^{5}-1$ has no non-trivial solution in $K$.
Indeed, let us prove the contrapositive: Let us prove that if $x^{5}-1$ has a
non-trivial solution in $K$, then $x^{2}-5$ has a solution in $K$.
So we assume that $x^{5}-1$ has a non-trivial solution in $K$. Fix such a
solution, and denote it by $z$. Thus, $z^{5}-1=0$ but $z\neq1$.
Inspired by the above answer to the first question, we set $w=z^{0^{2}
}+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}}$. We shall now prove that
$w^{2}-5=0$.
Indeed, $z-1\neq0$ (since $z\neq1$). Hence, we can cancel $z-1$ from the
equality $\left( z-1\right) \left( z^{4}+z^{3}+z^{2}+z+1\right)
=z^{5}-1=0$. We thus obtain $z^{4}+z^{3}+z^{2}+z+1=0$, so that $z^{4}
=-z^{3}-z^{2}-z-1$. Also, from $z^{5}-1=0$, we obtain $z^{5}=1$, thus
$z^{8}=z^{3}$ and $z^{9}=z^{4}$ and $z^{16}=z^{11}=z^{6}=z$. Hence,
\begin{align*}
w & =\underbrace{z^{0^{2}}}_{=z^{0}=1}+\underbrace{z^{1^{2}}}_{=z^{1}
=z}+\underbrace{z^{2^{2}}}_{=z^{4}}+\underbrace{z^{3^{2}}}_{=z^{9}=z^{4}
}+\underbrace{z^{4^{2}}}_{=z^{16}=z}\\
& =1+z+z^{4}+z^{4}+z=1+2z+2z^{4}.
\end{align*}
Squaring this equality, we find
\begin{align*}
w^{2} & =\left( 1+2z+2z^{4}\right) ^{2}=1+4z+4z^{2}+4z^{4}
+8\underbrace{z^{5}}_{=1}+4\underbrace{z^{8}}_{=z^{3}}\\
& =1+4z+4z^{2}+4z^{4}+8+4z^{3}=5+4\underbrace{\left( z^{4}+z^{3}
+z^{2}+z+1\right) }_{=0}=5.
\end{align*}
In other words, $w^{2}-5=0$. Hence, $x^{2}-5$ has a solution in $K$ (namely,
$w$). This answers the second question.
| {
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Matrix equation where some entries are solution to a polynome
Let $z$ be a solution to $z^2+z+1=0$. Find a solution to
\begin{bmatrix}1&1&1&3\\1&1&1&-1\\1&z&z^2&0\\1&z^2&z&0\end
{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end
{bmatrix} =\begin{bmatrix}9\\1\\0\\0\end{bmatrix}
This an old exam question in my linear algebra course. I initially just tried solving the equation, but setting $z$ and $z^2$ in the matrix becomes messy quickly when I try to row reduce. The solution is elegant, but I don't understand it;
1) The equation $z^2+z+1=0$ ensures that every vector of the form
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end
{bmatrix} = \begin{bmatrix}a\\a\\a\\b\end{bmatrix}
solves the last two equations. It therefore suffices to find such $a$ and $b$ that the first two equations are also satisified. These read $3a+3b=9$ and $3a-b=1$. This forces $a=1$ and $b=2$ and
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end
{bmatrix} = \begin{bmatrix}1\\1\\1\\2\end{bmatrix} is indeed a solution
I don't understand 1), the rest follows naturally I guess. Could someone elaborate on why 1) is true? Where do the 3 $a$s and 1 $b$ come from?
| $$\begin{bmatrix}1&1&1&3\\1&1&1&-1\\1&z&z^2&0\\1&z^2&z&0\end
{bmatrix}\begin{bmatrix}a\\a\\a\\b\end
{bmatrix} =\begin{bmatrix}3a+3b\\3a-b\\a(1+z+z^2)\\a(1+z+z^2)\end{bmatrix}=\begin{bmatrix}3a+3b\\3a-b\\0\\0\end{bmatrix}$$
We just have to solve for $3a+3b=9$ and $3a-b=1$.
We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.
| {
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Is $h:x\to \exp\left[-\frac{x^2}{2}\right]\left(\exp\left[\frac{x^4}{24n}\right]-1\right)$ increasing? I struggling to show this function ins increasing on $\mathbb{R}^+$
$$h:x\to \exp\left[-\dfrac{x^2}{2}\right]\left(\exp\left[\dfrac{x^4}{24n}\right]-1\right) \qquad n\in \mathbb{N}^*$$
With the derivative
I've found
$$h'(x)=xe^{-\frac{x^2}{2}}\left[\left(\dfrac{x^2}{6n}-1\right)e^{\frac{x^4}{24n}}+1\right]$$
$h'(x)\ge 0\iff \left(\dfrac{x^2}{6n}-1\right)e^{\frac{x^4}{24n}}+1\ge 0$
and now I'm blocked...
| We have that
$$h'(x)=-xe^{-\frac{x^2}{2}}\left(e^{\frac{x^4}{24n}}-1\right)+\frac{x^3}{6n}e^{-\frac{x^2}{2}}e^{\frac{x^4}{24n}}
=-xe^{-\frac{x^2}{2}}\left(\left(1-\frac{x^2}{6n}\right)e^{\frac{x^4}{24n}}-1\right)$$
then we need to prove that
$$\left(\frac{x^2}{6n}-1\right)e^{\frac{x^4}{24n}}+1 \ge 0 $$
and for $\frac{x^2}{6n}-1<0 \iff x<\sqrt{6n}$ it is equivalent to
$$\left(1-\frac{x^2}{6n}\right)e^{\frac{x^4}{24n}}-1\le 0 $$
and by $y=\frac{x^2}{6n}\in (0,1)$
$$(1-y)e^{\frac32ny^2}-1\le 0 $$
which doesn't hold for $n=2$ indeed
$$f(y)=(1-y)e^{3y^2}-1 \implies f'(y)=e^{3y^2}(-6y^2+6y-1)=0 \implies y_{1,2}=\frac12\pm\frac{\sqrt 3}6$$
and $f(y_1)>0$.
| {
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Prove a geometric sequence a, b, c from the arithmetic progression $1/(b-a)$, $1/2b$, $1/(b-c)$ The given task is:
The following forms an arithmetic sequence: $$\frac{1}{b-a}, \frac{1}{2b}, \frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ \frac{1}{2b}-\frac{1}{b-a}=\frac{1}{b-c}-\frac{1}{2b} \iff \frac{a+b}{a-b}=\frac{b+c}{b-c}$$ and that I need to prove that $$\frac{b}{a}=\frac{c}{b},$$ but between the two I just make it more and more complicated.
| We have
$$
\frac1b = \frac{1}{b-a} + \frac1{b-c}\\
(b-a)(b-c) = (b-c + b-a)b\\
b^2 - ab - bc + ac = 2b^2-bc - ab\\
ac = b^2
$$
and we are done.
| {
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"answer_id": 0
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Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$
Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$
If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and
$$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq {3\over 8}$$
Since $f(x)=\Big({1\over 1+x}\Big)^3$ is convex we get, by Jensen,: $$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq 3f({x+y+z\over 3})$$
Unfortunately, since $f$ is decreasing we don't have $f({x+y+z\over 3}) \geq f(1) = {1\over 8}$.
Some idea how to solve this?
| By Holder $$\left(\sum_{cyc}\frac{a^3}{(a+b)^3}\right)^2\sum_{cyc}1\geq\left(\sum_{cyc}\sqrt[3]{\left(\frac{a^3}{(a+b)^3}\right)^2\cdot1}\right)^3=\left(\sum_{cyc}\frac{a^2}{(a+b)^2}\right)^3.$$
Thus, it's enough to prove that
$$\frac{\left(\sum\limits_{cyc}\frac{a^2}{(a+b)^2}\right)^3}{3}\geq\frac{9}{64}$$ or
$$\sum\limits_{cyc}\frac{a^2}{(a+b)^2}\geq\frac{3}{4}.$$
Now, by C-S
$$\sum\limits_{cyc}\frac{a^2}{(a+b)^2}=\sum\limits_{cyc}\frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a+b)^2(a+c)^2}.$$
Thus, it's enough to prove that
$$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a+b)^2(a+c)^2,$$
which is true even for all reals $a$, $b$ and $c$.
Indeed, the last inequality is symmetric inequality by degree four,
which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )
it's enough to prove the last inequality for equality case of two variables and since
it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
$$(a-1)^2(a+3)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Textbook Proposition on Product of Real Analytic Functions
Let
\begin{align*}
\sum\limits_{j=0}^\infty a_j (x-c)^j && \sum\limits_{j=0}^\infty b_j (x-c)^j \\
\end{align*}
be two power series with intervals of convergence $\mathcal{C}_1$ and $\mathcal{C}_2$ centered on at $c$. Let $f_1(x)$ be the function defined by the first series on $\mathcal{C}_1$ and $f_2(x)$ the function defined by the second series on $\mathcal{C}_2$. Then, on their common domain $\mathcal{C} = \mathcal{C}_1 \cap \mathcal{C}_2$, it holds that
\begin{align*}
f(x) \cdot g(x) &= \sum\limits_{m=0}^\infty \sum\limits_{j+k=m} (a_j \cdot b_k) (x-c)^m \\
\end{align*}
Proof: Let
\begin{align*}
A_N &= \sum\limits_{j=0}^N a_j (x-c)^j & B_N &= \sum\limits_{j=0}^N b_j (x-c)^j \\
\end{align*}
be respectively, the Nth partial sums of the power series that define $f$ and $g$. Let
\begin{align*}
D_N &= \sum\limits_{m=0}^N \sum\limits_{j+k=m} (a_j \cdot b_k) (x-c)^m &
R_N &= \sum\limits_{j=N+1}^\infty b_j (x-c)^j \\
\end{align*}
We have:
\begin{align*}
D_N &= a_0 B_N + a_1 (x-c) B_{N-1} + \cdots + a_N (x-c)^N B_0 \\
&= a_0 (g(x) - R_N) + a_1 (x-c) (g(x) - R_{N-1}) + \cdots + a_N (x-c)^N (g(x) - R_0) \\
\end{align*}
[snip]
I've transcribed the above from my Real Analysis textbook and stopped at the part I'm having trouble understanding. How do you get from the first definition of $D_N$ to the second one? They seem like different equations. I've tried to do sequence manipulations to get from one definition to the other, but I've not had success.
| You can actually prove this by induction
Works for $N = 1$
\begin{eqnarray}
D_1 &=& a_0b_0 + (a_0b_1 + a_1 b_0)(x - c) \\ &=& a_0 [b_0 + b_1(x - c)] + a_1[b_0(x - c)] \\
&=& a_0 B_1 + a_1 B_0
\end{eqnarray}
Assume that it works for $N - 1$
$$
\sum_{m = 0}^{N - 1}\sum_{j + k = m}a_j b_k(x - c)^m = a_0 B_{N - 1}
+ a_1(x - c) B_{N - 2} + \cdots + a_{N-1}(x - c)^{N-1}B_0 \tag{1}
$$
Now let's prove it for $N$
\begin{eqnarray}
\sum_{m = 0}^{N}\sum_{j + k = m}a_j b_k(x - c)^m &=& \sum_{m = 0}^{N - 1} \sum_{j + k = m}a_j b_k(x - c)^m + \sum_{j + k = N}a_j b_k(x - c)^N \\
&\stackrel{(1)}{=}& a_0 B_{N - 1}
+ a_1(x - c) B_{N - 2} + \cdots + a_{N-1}(x - c)^{N-1}B_0 \\
&& + [a_0 b_N + a_1 b_{N - 1} + \cdots + a_N b_0](x - c)^N \\
&=& a_0 \left[ B_{N - 1} + b_N(x - c)^N \right] + a_1(x - c) \left[ B_{N - 2} + b_{N - 1}(x - c)^{N - 1} \right] \\
&& + \cdots + a_{N}(x - c)^{N}[b_0] \\
&=&a_0 B_N + a_1(x - c) B_{N - 1} + \cdots + a_{N}(x - c)^{N}B_0 \tag{2}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
value of $k$ in binomial expression If $\displaystyle \binom{404}{4}-\binom{4}{1}\cdot \binom{303}{4}+\binom{4}{2}\cdot \binom{202}{4}-\binom{4}{3}\cdot \binom{101}{4}=(101)^k.$ Then $k$ is
Iam trying to simplify it
$\displaystyle \frac{(404)!}{4!\cdot (400)!} -4\cdot \frac{(303)!}{4!\cdot (299)!}+6\cdot \frac{(202)!}{(198)!\cdot 4!}-4\cdot \frac{(101)!}{4!\cdot (97)!}$
but i did not understand how do i find $(101)$ as a factor in that expression
may be some other way to calculate it
please Help me to solve it
| Keep on simplifying:
$$\displaystyle \frac{(404)!}{4!\cdot (400)!} -4\cdot \frac{(\color{red}{303})!}{4!\cdot (299)!}+6\cdot \frac{(202)!}{(198)!\cdot 4!}-4\cdot \frac{(101)!}{4!\cdot (97)!}=\\
\displaystyle \frac{404\cdot 403\cdot 402\cdot 401}{24} - \frac{303\cdot 302\cdot 301\cdot 300}{6}+\frac{202\cdot 201\cdot 200\cdot 199}{4}-\frac{101\cdot 100\cdot 99\cdot 98}{6}=\\
101\cdot \left[ 403\cdot 67\cdot 401 - 302\cdot 301\cdot 150+201\cdot 100\cdot 199-50\cdot 33\cdot 98\right]=\\
101\cdot [1030301
]=101\cdot 101^3=101^4 \Rightarrow k=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Easier way to find eigenvalues of Matrices? I am trying to find eigenvalues for this matrix,
A =
$\begin{bmatrix}
3 & 2 & -3 \\
-3 & -4 & 9 \\
-1 & -2 & 5 \\
\end{bmatrix}$
I find the characteristic equation here:
$(\lambda I - A)
=
\begin{bmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{bmatrix}$
The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$\begin{vmatrix}
\lambda I - A \\
\end{vmatrix}
=
\begin{bmatrix}
\lambda - 3 & -2 & 3 & \lambda -3 & -2 \\
3 & \lambda + 4 & -9 & 3 & \lambda + 4 \\
1 & 2 & \lambda - 5 & 1 & 2 \\
\end{bmatrix}$
= $(1)(\lambda + 4)(3)$
$+ (2)(-9)(\lambda - 3)$
$+ (\lambda - 5)(3)(-2)$
$- (\lambda - 3)(\lambda + 4)(\lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 18 - 18$
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 36$
= $((\lambda + 4)(3) -(\lambda -3)(\lambda -5)) + (-18)(\lambda -3) + (-6)(\lambda - 5) - 36$
= $(\lambda - 3)((\lambda + 4)(3) -(\lambda - 5)) -18 + (-6)(\lambda - 5) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 60$
= $(\lambda - 5)(\lambda - 3)(3\lambda + 12 -1 - 60)$
= $(\lambda - 5)(\lambda - 3)(3\lambda - 49)$
I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?
| For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $\lambda^2 - T \lambda + D,$ where $T$ is the trace and $D$ is the determinant.
more in a few minutes.
For 3 by 3, it is $$ \lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3 $$
Here $\sigma_1$ is the trace and $\sigma_3$ is the determinant. The middle one, $\sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal.
With your original matrix, we have
$$
\sigma_2 =
\left|
\begin{array}{rr}
3 & 2 \\
-3 & -4 \\
\end{array}
\right| +
\left|
\begin{array}{rr}
3 & -3 \\
-1 & 5 \\
\end{array}
\right| +
\left|
\begin{array}{rr}
-4 & 9 \\
-2 & 5 \\
\end{array}
\right| = -6 + 12 -2 = 4
$$
We have the familiar trace $\sigma_1 = 4,$ and the determinant $\sigma_3 = 0$ by that Rule of Sarrus.
$$ \lambda^3 - 4 \lambda^2 + 4 \lambda - 0 = \lambda (\lambda- 2 )^2 $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ \lambda^4 - \sigma_1 \lambda^3 + \sigma_2 \lambda^2 - \sigma_3 \lambda + \sigma_4. $$
Once again $\sigma_1$ is the trace and $\sigma_4$ the determinant. This time we have two middle terms. Now $\sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $\sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can we find minimum radius of circle which contains $\arctan^2(x)+\arctan^2(y)=a$? How can we find minimum radius of circle which contains $\arctan^2(x)+\arctan^2(y)=a$, where I think $a<\pi/2$?
For example I have some plots from WolframAlpha and I see it depends on $a$.
But have no idea how to find maximum radius because this is not simple implicit function.
| Seems like the smallest circle is always tangent at $x=0$ or $y=0$. So you can put $\arctan(x)^2=a$ and solve $x=\tan\sqrt a$. This would be the radius.
Edit: let's make it rigorous.
Let's maximize $x^2+y^2$ with the constraint $\arctan(x)^2+\arctan(y)^2=a$. By symmetry we can work in the positive quadrant. With Lagrange multipliers we find
$$
\left(\frac{\arctan(x)}{1+x^2},\frac{\arctan(y)}{1+y^2}\right)
= \lambda (x,y),
$$
therefore (unless $x=0$ or $y=0$, which are easily seen to be stationary points) we have
$$
\frac{\arctan(x)}{(1+x^2)x} = \frac{\arctan(y)}{(1+y^2)y}.
$$
We want to show that the only possibility is $x=y$. The we need an argument to say that this is a minimum and not a maximum.
As a matter of fact, the function $\frac{\arctan(x)}{(1+x^2)x}$ is decreasing, so this shows that $x=y$ are the only solutions. Its derivative is in fact
$$
\begin{split}
\frac{d}{dx}\frac{\arctan(x)}{(1+x^2)x} &=
\frac{2}{x
\left(x^2+1\right)^2}-\frac{2 \arctan(x)}{x^2
\left(x^2+1\right)}-\frac{4 \arctan(x)}{\left(x^2+1\right)^2} \\
&< \frac{2}{x
\left(x^2+1\right)^2}-\frac{2 \arctan(x)}{x^2
\left(x^2+1\right)} \\
&= -2 \frac{(1+x^2)\arctan(x)-x}{(x+x^3)^2} < 0.
\end{split}
$$
You may rightfully wonder why is $(1+x^2)\arctan(x)-x>0$? Well, it vanishes at $0$ and its derivative is $2x\arctan(x)>0$.
So, why is $x=y$ a minimum? Well, we just need to compare it with the point at $y=0$.
The values of $x^2+y^2$ at the two points are respectively $2(\tan\sqrt{a/2})^2$ and $(\tan\sqrt a)^2$. Since the function $f(t)=(\tan\sqrt t)^2$ is convex and vanishes at $0$, it is superadditive, implying that $2f(a/2)\leq f(a)$.
Bonus
I'd like to add one little piece, related to proving that $(1+x^2)\arctan(x)-x$ is increasing. The insight is that it is of the form $\frac{g(x)}{g'(x)}-x$ with $g$ concave. Then it's derivative is $-\frac{g(x)g''(x)}{g'(x)^2} \geq 0$. I don't know if this general detail might be useful to someone.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find $f,g$ s.t. $f\circ g=\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ 10 & 4 & 5 & 7 & 8 & 9 & 2 & 6 & 3 & 1\end{pmatrix}.$
Let $f$ and $g$ be permutations such that
$$f \circ f = id,$$
$$g \circ g = id,$$
and
$$f\circ g =\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
10 & 4 & 5 & 7 & 8 & 9 & 2 & 6 & 3 & 1
\end{pmatrix}.$$
Find $f$ and $g$.
I can solve it by a lot of guess work, but I wonder if there is some general method.
| A permutation $f$ is an involution if $f\circ f=id$.
As you know, any permutation can be written as a product of disjoint cycles; your permutation is $(1\ 10)(2\ 4\ 7)(3\ 5\ 8\ 6\ 9)$. In order to write an arbitrary permutation as a product of two involutions, it suffices (since disjoint permutations commute) to write a cycle of arbitrary length as a product of two involutions. A permutation is an involution if it's a product of disjoint cycles of length $2$. If you experiment a little with multiplying involutions, you might discover the following pattern:
$$(1\ 2)\circ(2\ 3)=(1\ 2\ 3)\tag3$$
$$(1\ 2)(3\ 4)\circ(2\ 3)=(1\ 2\ 4\ 3)\tag4$$
$$(1\ 2)(3\ 4)\circ(2\ 3)(4\ 5)=(1\ 2\ 4\ 5\ 3)\tag5$$
$$(1\ 2)(3\ 4)(5\ 6)\circ(2\ 3)(4\ 5)=(1\ 2\ 4\ 6\ 5\ 3)\tag6$$
etc. So a cycle of any length can be obtained by multiplying two involutions. In particular, replacing $1,2,3$ by $2,4,7$ in $(3)$ we get
$$(2\ 4\ 7)=(2\ 4)\circ(4\ 7);$$
replacing $1,2,4,5,3$ by $3,5,8,6,9$ in $(5)$ we get
$$(3\ 5\ 8\ 6\ 9)=(3\ 5)(9\ 8)\circ(5\ 9)(8\ 6)=(3\ 5)(8\ 9)\circ(5\ 9)(6\ 8);$$
and of course
$$(1\ 10)=(1\ 10)\circ id;$$
so
$$(1\ 10)(2\ 4\ 7)(3\ 5\ 8\ 6\ 9)=(1\ 10)(2\ 4)(3\ 5)(8\ 9)\circ(4\ 7)(5\ 9)(6\ 8).$$
I.e., you can take
$$f=(1\ 10)(2\ 4)(3\ 5)(8\ 9),\ g=(4\ 7)(5\ 9)(6\ 8).$$
Of course there are other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Evaluating $1-\frac12-\frac13+\frac14+\frac15+\frac16-\cdots$
What is the value of $$S=1-\frac12-\frac13+\frac14+\frac15+\frac16-\cdots$$ where the sign alternates over the triangular numbers?
To start, it is easy to prove convergence. The sum of each set of reciprocals (e.g. $\{1/2,1/3\}$ and $\{1/4,1/5,1/6\}$) can be written as $$\sum_{i=1}^n\frac1{\frac{n(n-1)}2+i},\quad n=1,2,\cdots$$ and it can be shown that it is monotonically decreasing (and tending towards zero) since \begin{align}\sum_{i=1}^n\frac1{\frac{n(n-1)}2+i}>\sum_{i=1}^{n+1}\frac1{\frac{n(n+1)}2+i}&\impliedby \sum_{i=1}^n\frac1{\frac{n(n-1)}2+i}-\sum_{i=1}^n\frac1{\frac{n(n+1)}2+i}>\frac2{n^2+3n+2}\\&\impliedby \sum_{i=1}^n\frac 1{(n^2-n+2i)(n^2+n+2i)}>\frac1{2n(n+1)(n+2)}\\&\impliedby \frac n{(n^2-n+2\cdot1)(n^2+n+2\cdot1)}>\frac1{2n(n+1)(n+2)}\\&\impliedby (n-2)(n^2+n+2)<2n^2(n+2)\\&\impliedby n^3+5n^2+4>0\end{align} Is there a closed-form expression for the value of $S$?
| In other terms we want to evaluate
$$ \sum_{n\geq 1}(-1)^{n+1}\left(H_{n(n+1)/2}-H_{n(n-1)/2}\right)=\int_{0}^{1}\sum_{n\geq 1}(-1)^{n+1}\frac{x^{n(n+1)/2}-x^{n(n-1)/2}}{x-1}\,dx$$
where the theory of modular forms ensures
$$ \sum_{n\geq 0} x^{n(n+1)/2} = \prod_{n\geq 1}\frac{(1-x^{2n})^2}{(1-x^n)}=\prod_{n\geq 1}\frac{1-x^{2n}}{1-x^{2n-1}} $$
but I do not see an easy way for introducing a $(-1)^n$ twist in the LHS. On the other hand, the Euler-Maclaurin summation formula ensures
$$ H_n = \log n + \gamma + \frac{1}{2n} - \sum_{m\geq 2}\frac{B_m}{m n^m} $$
in the Poisson sense. Replacing $n$ with $n(n\pm 1)/2$,
$$ H_{\frac{n(n+1)}{2}}-H_{\frac{n(n-1)}{2}} = \log\left(\tfrac{n+1}{n-1}\right)+\tfrac{2}{(n-1)n(n+1)}-\sum_{m\geq 2}\tfrac{2^m B_m}{m n^m}\left(\tfrac{1}{(n-1)^m}-\tfrac{1}{(n+1)^m}\right) $$
then multiplying both sides by $(-1)^n$ and summing over $n\geq 2$:
$$ \sum_{n\geq 2}(-1)^n\left(H_{\frac{n(n+1)}{2}}-H_{\frac{n(n-1)}{2}} \right)=\\=5\left(\log(2)-\tfrac{1}{2}\right)-\sum_{m\geq 2}\tfrac{2^m B_m}{m}\sum_{n\geq 2}(-1)^n\left(\tfrac{1}{n^m(n-1)^m}-\tfrac{1}{n^m(n+1)^m}\right)
\\=5\left(\log(2)-\tfrac{1}{2}\right)-\sum_{m\geq 2}\frac{2^m B_m}{m}\left[\frac{1}{2^m}-2\sum_{n\geq 2}\frac{(-1)^m}{n^m(n+1)^m}\right]$$
where the innermost series is a linear combination of $\log(2),\zeta(3),\zeta(5),\ldots$ by partial fraction decomposition. This allows a reasonable numerical approximation of the original series and a conversion into a double series involving $\zeta(2a)\zeta(2b+1)$. I am not sure we can do better than this, but I would be delighted to be proven wrong.
Playing a bit with functions, a nice approximation of $\sum_{n\geq 0}(-1)^n x^{n(n+1)/2}$ over $[0,1]$ is given by $\frac{1}{x+1}-x^2(1-x)^2$, so the value of the original series has to be close to $\log(2)-\frac{1}{6}$. A better approximation of the function is $\frac{1}{x+1}-x^2(1-x)^2+\frac{3}{4}x^4(1-x)\left(\frac{4}{5}-x\right)$, leading to the following improved approximation for the series: $\log(2)-\frac{53}{300}$. A further refinement,
$$ g(x)=\sum_{n\geq 0}(-1)^n x^{n(n+1)/2} \approx \frac{1+x+2x^2}{1+2x+5x^2}$$
leads to $\color{red}{S\approx\frac{\pi+3\log 2}{10}}$. It might be interesting to describe how I got this approximation. $g(0)$ and $g'(0)$ are directly given by the Maclaurin series, while $\lim_{x\to 1^-}g(x)=\frac{1}{2}$ and $\lim_{x\to 1^-}g'(x)=-\frac{1}{8}$ can be found through $\mathcal{L}(f(e^{-x}))(s)$. $g(x)$ is convex and decreasing on $(0,1)$ and any approximation of the
$$ \frac{1+ax+(1+a)x^2}{1+(1+a)x+(2+3a)x^2}$$
kind with $a$ in a suitable range matches such constraint and the values of $g$ and $g'$ at the endpoints of $(0,1)$. We still have the freedom to pick $a$ in such a way that the derived approximation is both simple and accurate enough - I just picked $a=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
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If $\frac {a}{3^{x-1}}=\frac{b}{3^{y+2}}=\frac{c}{3^{z-1}}=\frac 15\;$ then which of the following equals $a×b×c$? The problem is:
If $\frac {a}{3^{x-1}}=\frac{b}{3^{y+2}}=\frac{c}{3^{z-1}}=\frac 15,\;$ then which of the following equals $a×b×c$ ?
A) $\frac {1}{375}$
B) $\frac{1}{125}$
C) $\frac{27}{125}$
D) $\frac{3}{125}$
E) $\frac{27}{5}$
I think the question is wrong.
My counterexample:
Let $x=m,\; y=m-3,\; z=m.$
Then $a=b=c=\frac{3^{m-1}}{5}.$
So, $a×b×c=\frac{3^{3m-3}}{125},\; m\in\mathbb{R}.$
Am I right?
| Yes some information is missing, indeed we have that
$$\frac {a}{3^{x-1}}=\frac{b}{3^{y+2}}=\frac{c}{3^{z-1}}=\frac 15$$
then
$$abc=\frac{3^{(x+y+z)}}{125}$$
then we need a condition for $t=x+y+z\in \mathbb R$, since $3^{t}$ can assume any value $\in(0,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of these variables. Five real numbers $a_1, a_2, a_3, a_4\;\text{and}\; a_5\;$ are such that
$$\sqrt{a_1- 1} + 2\sqrt{a_2- 4}+3\sqrt{a_3- 9}
+4\sqrt{a_4- 16} + 5 \sqrt{a_4- 25} =\frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
| For real $\sqrt{a-b^2},$ we need $a-b^2\ge0$
and for $b>0,$ and as $\sqrt{a-b^2}\ge0$ by AM-GM inequality,
$$\dfrac{(\sqrt{a-b^2})^2+(b)^2}2\ge b\sqrt{a-b^2}$$
the equality will occur if $\sqrt{a-b^2}=b$
$$\implies\dfrac{(\sqrt{a_1-1})^2+1^2+\cdots+(\sqrt{a_5-5^2})^2+5^2}2=\dfrac{a_1+a_2+a_3+a_4+a_5}2 \ge \sqrt{a_1-1}+\cdots$$
| {
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The number of integral solution of $\alpha+\beta+\gamma+\delta$=18 such that.. Question
The number of integral solution of the equation $$\alpha+\beta+\gamma+\delta=18$$, with the conditions:
$1\leq\alpha\leq5$; ${-2}\leq\beta\leq4$; $0\leq\gamma\leq5$ and $3\leq\delta\leq9$ is $k$. Find $k$.
$$Attempt$$
I tried to do to it by giving the minimum number required to the variables to generate new variable to whom then nil can also be given and hence the equation I wrote is $$x+y+z+v=16$$ But, now, I had a problem in adjusting the higher limit for every variable for accounting each of them. Afterwards, I don't know what to do.
*Any hints or suggestions?*
Thanks for helping me!
| Here is a proof using the method of generating functions. By adjusting the constraints we may write the problem as finding the number of integral solutions to
$$
\alpha'+\beta'+\gamma'+\delta'=16
$$
with $0\le\alpha'\leq 4, 0\leq \beta'\leq 6, 0\leq \gamma'\leq 5, 0\leq\delta'\leq 6$. The number of solutions is then the coefficient in the expansion of
$$
\begin{align}
F(x)&=(1+x+x^2+x^3+x^4)(1+x+\dotsb+x^6)^2(1+x+\dotsb+x^5)\\
&=\frac{1-x^5}{1-x}\frac{(1-x^7)^2}{(1-x)^2}\frac{1-x^6}{1-x}\\
&=\frac{x^{25} - x^{20} - x^{19} - 2 x^{18} + x^{14} + 2 x^{13} + 2 x^{12} + x^{11} - 2 x^7 - x^6 - x^5 + 1}{(1-x)^4}.
\end{align}
$$
For the coeffficient extraction use the fact that
$$
\sum_{n=0}^\infty\binom{n+3}{3}x^n=(1-x)^{-4}
$$
and linearity of coefficient extraction to get that the number of solutions is
$$
\binom{16+3}{3}-\binom{16-5+3}{3}-\binom{16-6+3}{3}-2\binom{16-7+3}{3}+\binom{16-11+3}{3}+2\binom{16-12+3}{3}+\\2\binom{16-13+3}{3}+\binom{16-14+3}{3}
$$
| {
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Integer Solutions of the Equation $u^3 = r^2-s^2$ The question says the following:
Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.
The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $\gcd(r,s) = 1$and $r+s \equiv 1 \pmod {2}$
In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.
I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.
Any ideas?
| We can find one or more Pythagorean triples for any odd leg $\ge 3$using a function of $(m,A)$:
$$\text{We can let }n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le \frac{A+1}{2}$$
This is useful in finding $x^3+B^2=C^2$ because we can plug in any odd cube $(A)$ an make a small finite search of values based on $A$. We demonstrate with the first odd cube.
$$m_{min}=\lceil\sqrt{27}\space\rceil=\lceil 5.196152423\rceil=6\qquad \qquad m_{max}=\frac{A+1}{2}=\frac{28}{2}=14$$
Testing for $6\le m\le 14$, we find integers for $(m,n)=(6,3)\text{ and }(14,13)$
$$\text{For }(6,3)\quad A=6^2-3^2=27\quad B=2*6*3=36\quad C=6^2+3^2=45\qquad (3,36,45)$$
$$\text{For }(14,13), A=14^2-13^2=27, B=2*14*13=364, C=14^2+13^2=365\quad (3,364,365)$$
Here are triples generated by the same algorithm(s) for other cubes
$$\text{For }5^3=125\qquad f(15,10)=(5,300,325)$$
$$\text{For }7^3=343\qquad f(28,21)=(7,1176,1225$$
$$\text{For }9^3=729\quad f(45,36)=(9,3240,3321)\qquad f(123,120)=(9,29520,29529)$$
$$\text{For }11^3=1331\qquad f(66,55)=(11,7260,7381)$$
| {
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Prove that $\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0. $
Let $n$ be a positive integer. Prove that
$$
\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0.
$$
I am trying to solve this by using induction on $n$. I have proven the sum to be zero in the case $n=1$. Assuming that the sum is zero for $n=m$ ($m$ is a positive integer), how do I prove that it implies that the sum is zero for $n=m+1$?
Can I get some hints?
| Starting from
$$\sum_{k=0}^{2n} {2n+k\choose k} {2n\choose k}
\frac{(-1)^k}{2^k} \frac{1}{k+1}$$
we get
$$\frac{1}{2n} \sum_{k=0}^{2n} {2n+k\choose k+1} {2n\choose k}
\frac{(-1)^k}{2^k}
= \frac{1}{2n} \sum_{k=0}^{2n} {2n+k\choose 2n-1} {2n\choose k}
\frac{(-1)^k}{2^k}
\\ = \frac{1}{2n} \sum_{k=0}^{2n} {2n\choose k}
\frac{(-1)^k}{2^k} [z^{2n-1}] (1+z)^{2n+k}
\\ = \frac{1}{2n} [z^{2n-1}] (1+z)^{2n}
\sum_{k=0}^{2n} {2n\choose k}
\frac{(-1)^k}{2^k} (1+z)^{k}
\\ = \frac{1}{2n} [z^{2n-1}] (1+z)^{2n}
\left(1-\frac{1}{2} (1+z)\right)^{2n}
\\ = \frac{1}{2^{2n+1}n} [z^{2n-1}] (1+z)^{2n} (1-z)^{2n}
= \frac{1}{2^{2n+1}n} [z^{2n-1}] (1-z^2)^{2n} = 0.$$
The last step is zero by inspection since we are extracting a
coefficient on an odd power from a polynomial where all the powers are
even, and we have the claim.
| {
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How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see the pattern of multiple occurrences of $(x-a)$, $(x-b)$ I can't see any way to simplify the fraction further, so I go on to simplify the expression by multiplying by $\sqrt{x-a}+\sqrt{x-b}$:
$$\begin{align*}
(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}&=(a-b)(\sqrt{x-a}+\sqrt{x-b})\\
&=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}
\end{align*}$$
Now I have the following:
$$(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
Simplifying the RHS as I was out of ideas at that point:
$$x\sqrt{x-a}-a\sqrt{x-a}+x\sqrt{x-b}-b\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
I noticed that all one of the common factors $\sqrt{x-a},\sqrt{x-b}$ so I tried to isolate them and factor them out -- that is, all $\sqrt{x-b}$ terms on one side and $\sqrt{x-a}$ terms on the other.
$$\sqrt{x-b}(x-a)=\sqrt{x-a}(2a-b-x)$$
I tried to then square both sides, but that led to quite a mess.
$$(x-b)(x^2-2ax+a^2)=(x-a)(4a^2-4ab+2bx-4ax+b^2+x^2)$$
I'm afraid to even begin trying to simplifying this. I'm convinced I'm going about it in the wrong way.
The $a>b$ hint is interesting, but I have no clue what implication it may have here.
I think the $(x-a)\sqrt {x-a}$ patterns may mean something, perhaps I could do something with $a\sqrt a=\sqrt{a^3}$, but at this point it is probably a dead end.
I appreciate any help.
| Hint: Define $$u=\sqrt{x-a}\\w=\sqrt{x-b}$$therefore $${w^3+u^3\over u+w}=w^2-u^2$$which yields to $$2u^3=uw^2-u^2w$$one answer is $u=0$ or $x=a$ which is valid. The others can be found by solving $$2u^2=w^2-uw$$or $$u^2+uw=a-b$$by substituting we obtain $$x-a+\sqrt{(x-a)(x-b)}=a-b$$
| {
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Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$\frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}\longrightarrow\frac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
| 1) $n(1+1/n )^{1/3} = $
$\dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$\dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$\small{\dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$\small{=\dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)\dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$\lim_{n \rightarrow \infty}\dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$\small{-\lim_{n \rightarrow \infty}(1/n^2)\dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$\small{\lim_{n \rightarrow \infty}(1/n^2)×}$
$\small{\lim_{n \rightarrow \infty}\dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$\small{0 \cdot (1/3)=0.}$
| {
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Closed form of this type $\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}$ Given that, $$\sum_{j=0}^{\infty}\frac{2^j\left(j-\frac{1}{3}\right)^3\left(j^2+j-1\right)}{(2j+1)(2j+3){2j \choose j}}=A\tag1$$
We have $A=2\pi+12+\frac{1}{3}?$
We can generalize the above $(1)$:
$$\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}=F(n)\tag2$$
I believe the closed for $(2)$ is
$$F(n)=(-1)^{n}(2n+1)-(-1)^n (2n-1)\cdot \frac{\pi}{2}$$
How can we show that the proposed $(2)$ is correct?
| From the expansion of $\arcsin^2 t$
\begin{equation}
\arcsin^2 t=\sum_{p=0}^\infty \frac{2^{2p}t^{2p+2}}{(2p+1)(p+1)\binom{2p}{p}}
\end{equation}
we can obtain by differentiation
\begin{equation}
2\frac{\arcsin t}{\sqrt{1-t^2}}=\sum_{p=0}^\infty \frac{2^{2p+1}t^{2p+1}}{(2p+1)\binom{2p}{p}}
\end{equation}
Multiplying the above identity by $t$ and integrating, one obtains
\begin{equation}
2\int_0^x\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\sum_{p=0}^\infty \frac{2^{2p+1}x^{2p+3}}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation}
We choose $x=\sqrt{y}/2$ and divide the identity by $y^{3/2}$ to write
\begin{equation}
\frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\sum_{p=0}^\infty \frac{y^{p}}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation}
Now, applying the operator $y\frac{d}{dy}$ $n$ times and taking the result at $y=2$ gives
\begin{equation}
\left.\left[ y\frac{d}{dy}\right]^n\frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt\right|_{y=2}=\sum_{p=0}^\infty \frac{p^{n}2^p}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation}
The function can be evaluated as
\begin{equation}
\frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\frac{4}{y}-8\frac{\sqrt{1-\frac{y}{4}}}{y^{3/2}}\arcsin\left( \frac{\sqrt{y}}{2} \right)
\end{equation}
The above result can be simplified by taking $z=\sqrt{y}/2$:
\begin{equation}
\left.\left[\frac{1}{2} z\frac{d}{dz}\right]^n
\left[ \frac{1}{z^2}-\frac{\sqrt{1-z^2}}{z^3}\arcsin z\right]
\right|_{z=\sqrt{2}/2}=\sum_{p=0}^\infty \frac{p^{n}2^p}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation}
As $\arcsin \left(\sqrt{2}/2\right)=\pi/4$ and since the successive applications of the operator on $z^{-2}$ and $z^{-3}\left( 1-z^2 \right)^k$ ($k$ is an integer) at $z=1/\sqrt{2}$ give rational results, we expect
\begin{equation}
F(n)=a_n+b_n\pi
\end{equation}
where $a_n$ and $b_n$ are rational. I could not find any simple expression for these coefficients however. First few values are $F(n)=2-\pi/2,-3+\pi,5-3\pi/2,-7+5\pi/2,13-3\pi,-7+17\pi/2, 93+27\pi/2\cdots$ for $n=0,1,2,3,4,5,6\cdots$. They do not correspond to the propose formula as remarked by @Mariusz Iwaniuk in a comment.
With $z=\exp(t/2)$, we have $1/2zd/dz=d/dt$, and we can build the generating function for the $F(n)$:
\begin{equation}
\sum_{n=0}^\infty F(n)\frac{t^n}{n!}=\phi\left( t-\ln2 \right)
\end{equation}
where
\begin{equation}
\phi\left( t-\ln 2 \right)=2e^{-3t/2}\left[ e^{t/2}-\sqrt{2-e^{t}}\arcsin\left(\frac{ e^{t/2}}{\sqrt{2}} \right)\right]
\end{equation}
| {
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About the equivalence relation on $\mathbb{Z}\times\mathbb{Z}\,$ s.t. $(a,b)\sim(c,d)\,$ if $\,2^{a^2+d^2}\equiv 2^{b^2+c^2} (\text{mod} \, 5)$ Let $\sim$ be an equivalence relation on $\mathbb{Z}\times\mathbb{Z}\,\,\,\,\,$ s.t. $\,\,\,\,\,\forall \,\,\,(a,b), (c,d)\in\mathbb{Z}\times\mathbb{Z}$:
$(a,b)\sim(c,d)\,\,\,\,\,$ if $\,\,\,\,\,\,\,2^{a^2+d^2}\equiv 2^{b^2+c^2} (\text{mod} \,\, 5)$
Show that if $f: \mathbb{Z}\times\mathbb{Z} \to \mathbb{Z_4}\,\,\,$ is defined by $\,\,\,f((a,b))=(a^2-b^2)+4\mathbb{Z},\,\,\,\, \forall \,\,\,\,(a,b)\in\mathbb{Z}\times\mathbb{Z}$ we have: $\,\,\,\,\,\,\sim \,\,\,\,= \,\,\,\,\sim_f$ $\,\,\,\,\,\,\,\,$-$\,\,\,\,$This point is quite easy using the definitions and the Fermat's little theorem.
The second point, which gives me hard times, is: determine $Im(f)$ and $\frac{\mathbb{Z}\times\mathbb{Z}}{\sim}$,$\,\,\,$ any suggestion?
| Since $\gcd(2,5)=1$ and $2^4 \equiv 1 \pmod{5}$, we have
$$2^{a^2+d^2} \equiv 2^{b^2+c^2} \pmod{5} \implies a^2+d^2 \equiv b^2+c^2 \pmod{4}\implies a^2-b^2 \equiv c^2-d^2 \pmod{4}.$$
Actually all these implications are reversible. So
$$2^{a^2+d^2} \equiv 2^{b^2+c^2} \pmod{5} \iff a^2-b^2 \equiv c^2-d^2 \pmod{4}.$$
This shows that the equivalence relation $\sim$ and the one generated by $f$ are the same.
For the image of $f$: consider the following:
\begin{align*}
f(0,0) & \equiv 0 \pmod{4}\\
f(1,0) & \equiv 1 \pmod{4}\\
f(0,1) & \equiv 3 \pmod{4}.
\end{align*}
Claim: Range of $f$ is $\{[0]_4,[1]_4,[3]_4\}$, i.e. $[2]_4 \not\in \text{Range}_f$.
If $a^2-b^2 \equiv 2 \pmod{4}$, then $a^2 \equiv b^2+2 \pmod{4}$. But $b^2+2 \equiv 2,3 \pmod{4}$. However, $\mod 4$, the only squares are $0,1$. Thus this is not possible.
Since the equivalence relation $\sim$ and the one generated by $f$ are the same, therefore the quotient set $\Bbb{Z} \times \Bbb{Z}/\sim$ will be (in a sense) isomorphic to the range of $f$. So
$$\Bbb{Z} \times \Bbb{Z}/\sim =\{[(0,0)], [(1,0)],[(0,1)]\}$$
| {
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Rolling five $6$-sided dice, where (as an example) $3$, $2$ and $1$ faces of each die are equivalent Let's say we have a $6$ sided die. $3$ of the sides have the value $A$, $2$ have the value $B$ and $1$ has the value $C$.
If we take $5$ of those, roll them together and look at the possible values, how would I go about finding the probability for specific combinations? Examples of combinations where X denotes any value could be:
*
*$AAXXX$
*$AAAXX$
*$AABBX$
*$ABCXX$
*$4$ of a kind
Is there a general way to figure the probability of obtaining a combination from a roll of $5$ dice? Both for a die face distribution $3A2B1C$ and possibly other alternate distributions like $3A1B1C1D$.
Thank you for any help you can give. I'm sorry if the question is a bit bare bones on my end but I was a bit unsure on exactly how to approach this.
| Someone else said use the multinomial distribution, but to be a little more clear:
If you have values $v_1,v_2,\ldots, v_k$ with multiplicity numbers $m_1,m_2,\ldots, m_k$ with $\sum_{i=1}^k m_i = M$, then the probability that a single fair die will roll value $v_i$ is $\dfrac{m_i}{M}$.
So, if you roll $n$ dice and you want to know the probability of getting $a_1\cdot v_1, a_2\cdot v_2,\ldots, a_k\cdot v_k$, it is given by:
$$\dbinom{n}{a_1,a_2,\cdots a_k}\prod_{i=1}^k\left(\dfrac{m_i}{M}\right)^{a_i}$$
In your example where you have $A\cdot 3, B\cdot 2, C\cdot 1$, the probability of rolling $A$ is $\dfrac{3}{3+2+1} = \dfrac{1}{2}$, the probability of rolling $B$ is $\dfrac{2}{3+2+1} = \dfrac{1}{3}$, and the probability of rolling $C$ is $\dfrac{1}{3+2+1} = \dfrac{1}{6}$.
Since you are rolling five dice, the probability of getting, say, $3\cdot A, 2\cdot B$ would be:
$$\dbinom{5}{3,2,0}\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{3}\right)^2\left(\dfrac{1}{6}\right)^0 = \dfrac{5!}{3!2!0!}\left(\dfrac{1}{8}\right)\left(\dfrac{1}{9}\right) = \dfrac{5}{36}$$
Extrapolating over all possible outcomes:
$$\begin{array}{c|c}\text{Outcome} & \text{Probability} \\ \hline \{A\cdot 5\} & \dfrac{1}{32} \\ \{A\cdot 4, B\cdot 1\} & \dfrac{5}{48} \\ \{A\cdot 4, C\cdot 1\} & \dfrac{5}{96} \\ \{A\cdot 3, B\cdot 2\} & \dfrac{5}{36} \\ \{A\cdot 3,B\cdot 1, C\cdot 1\} & \dfrac{5}{36} \\ \{A\cdot 3, C\cdot 2\} & \dfrac{5}{144} \\ \{A\cdot 2, B\cdot 3\} & \dfrac{5}{54} \\ \{A\cdot 2,B\cdot 2, C\cdot 1\} & \dfrac{5}{36} \\ \{A\cdot 2, B\cdot 1, C\cdot 2\} & \dfrac{5}{72} \\ \{A\cdot 2, C\cdot 3\} & \dfrac{5}{432} \\ \{A\cdot 1, B\cdot 4\} & \dfrac{5}{162} \\ \{A\cdot 1, B\cdot 3, C\cdot 1\} & \dfrac{5}{81} \\ \{A\cdot 1, B\cdot 2, C\cdot 2\} & \dfrac{5}{108} \\ \{A\cdot 1, B\cdot 1, C\cdot 3\} & \dfrac{5}{324} \\ \{A\cdot 1,C\cdot 4\} & \dfrac{5}{2592} \\ \{B\cdot 5\} & \dfrac{1}{243} \\ \{B\cdot 4, C\cdot 1\} & \dfrac{5}{486} \\ \{B\cdot 3, C\cdot 2\} & \dfrac{5}{486} \\ \{B\cdot 2, C\cdot 3\} & \dfrac{5}{972} \\ \{B\cdot 1, C\cdot 4\} & \dfrac{5}{3888} \\ \{C\cdot 5\} & \dfrac{1}{7776}\end{array}$$
Note: If you add up the probabilities for every possible outcome I listed, you will get 1 (100% probability that you wind up with one of those outcomes). This is known as the Law of Total Probability and shows that I did correctly describe the entire probability space of rolling five dice.
Since each outcome is disjoint, for any case you are trying to build (like at least two A's), find all outcomes that have at least two A's, and add their probabilities. For 4-of-a-kind, find every outcome with exactly 4 of a kind and add up their probabilities.
Also note: I strictly interpreted your question to be a question of dice rolled in any order. Rolling AAABB is the same as rolling ABBAA is the same as rolling ABABA, etc. in order.
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greatest common divisor of two elements Find all possible values of GCD(4n + 4, 6n + 3) for naturals n
and prove that there are no others
3·(4n + 4) - 2·(6n + 3) = 6,
whence the desired GCD is a divisor 6.
But 6n + 3 is odd, so only 1 and 3 remain.
n=1 and n=2 are examples for GCD=1 and GCD=3
is the solution correct ?
any other way to solve this ?
| Your way is correct.
Other way.
$\gcd(4n+4, 6n+3) = \gcd(4n+4, (6n+3) - (4n+4)) =$
$\gcd (4n+4, 2n -1) = \gcd(4n+4 - 2(2n-1), 2n-1)=$
$\gcd (6, 2n- 1) = $
... Now two things should be apparent. $2n-1$ is odd and $6$ is even so the prime factor $2$ of $6$ will not be a factor of $2n-1$. And Lemma: if $\gcd(j,b) = 1$ then $\gcd(j*a, b) = \gcd(a,b)$. That can be easily proven many ways.
So $\gcd(2*3, 2n-1) = \gcd(3,2n-1)$. Which is equal to $3$ if $3|2n-1$ which can happen if $2n-1 \equiv 0 \pmod 3$ or $n\equiv 2 \pmod 3$. Or is equal to $1$ if $3\not \mid 2n-1$ which can happen if $n\equiv 0, 1 \pmod 3$.
And another way:
$\gcd(4n+4, 6n+3) = \gcd(4(n+1), 3(2n+1)=$.
... as $3(2n+1)$ is odd....
$\gcd(n+1, 3(2n+1))$.
Now $\gcd(n+1, 2n+1) = \gcd(n+1, (2n+1)-(n+1) = \gcd(n+1, n) = \gcd(n+1 - n, n) = \gcd(1, n) = 1$.
So... $\gcd(n+1, 3(2n+1)) = \gcd(n+1, 3)$.
Which is $3$ if $3|n+1$ and is $1$ if not.
Perhaps we can retrofit this as
$\gcd(3,n+1) = \{1,3\}$
$\gcd(2n+1, n+1) = 1$ so
$\gcd(3(2n+1), n+1) = \gcd(3,n+1)$.
$\gcd(3(2n+1), 2) = 1$ so
$\gcd(3(2n+1), 2^2(n+1)) = \gcd(3,n+1)$.
All comes down to "casting out" relatively prime factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does this equation always work? $\sqrt{x \times (x+2)+1} = x + 1$ for non-negative $x$, and $=|x|-1$ for negative $x$ When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:
Non-Negative:
$$\forall x \in \Bbb R_0^+: \sqrt{x \times (x+2)+1} = x + 1$$
Negative:
$$\forall x \in \Bbb R^-: \sqrt{x \times (x+2)+1} = |x| - 1$$
| Note that you have a perfect square trinomial.
$$x(x+2)+1 = x^2+2x+1$$
You probably know that by binomial expansion, $(a\pm b)^2 = a^2\pm 2ab+b^2$.
$$(a\pm b)^2 = (a\pm b)(a\pm b) = a^2\pm ab\pm ba +b^2 = a^2\pm2ab+b^2$$
Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get
$$x^2+2(x)(1)+1^2 = (x+1)^2$$
So you have$\sqrt{(x+1)^2}$. You also have $\sqrt{a^2} = \vert a\vert$, so
$$\sqrt{(x+1)^2} = \vert x+1\vert = \begin{cases} \ x+1; \quad x \geq -1 \\ -x-1 = \vert x\vert-1; \quad x < -1\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
a problem on complex numbers Let $w\neq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=\cos(\frac{2\pi}{13})+i\sin(\frac{2\pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?
| Step 1: the equation you want is $(z-a)(z-b)=0$. Expand the product and you get $$z^2-(a+b)z+ab=0$$
Step 2: Use $w^{13}=1$, so $w^{-1}=w^{13}w^{-1}=w^{12}$ similarly, for all negative powers $$w^{-n}=w^{13-n}$$
Step 3: $$a+b=w+w^3+w^4+w^9+w^{10}+w^{12}+w^2+w^5+w^6+w^7+w^8+w^{11}=\\=\frac{w^{13}-1}{w-1}-1=-1$$
Step 4: To find $ab$, go to the trigonometric representation, and notice $$w^n+w^{-n}=2\cos\frac{2\pi n}{13}$$
Edit: After some manipulations, and using $w^{n+13}=w^n$, I've got $$ab=3(w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=\\-3+3(1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8+w^9+w^{10}+w^{11}+w^{12})=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$ Let x and a be real numbers > 0. Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$
My idea is that I'm going to use $a>b \iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a} \iff (\frac{1}{2}(x+\frac{a}{x}))^2 = \frac{1}{4}(x+\frac{a}{x})(x+\frac{a}{x})= \frac{1}{4}(x^2+2a+\frac{a^2}{x^2}) \ge \sqrt{a}^2=a$
And from that we get:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge \frac{1}{4}a^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a \iff \frac{1}{4}a^2x^2 +\frac{1}{2}ax^2+\frac{a^2}{4} \ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
| From here:
$$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a,$$
move the $a$ to the left side to yield
$$\frac{1}{4}x^2 - \frac{1}{2}a+\frac{a^2}{4x^2} \ge 0.$$
The left side is a perfect square:
$$\left(\frac{x}{2} - \frac{a}{2x}\right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of matrix of submatrices Can you check my solution to:
Task
Having two matrices $X,Y\in\mathbb{R}^{n,n}$ where $x,y\in\mathbb{R}$ and matrices are defined as
$
X=\begin{bmatrix}
x & 0 &0 & \dots & 0 \\
x & x & 0& \dots & 0\\
x & x & x& \dots & 0\\
\vdots&\vdots&\vdots&\vdots&\vdots \\
x & x & x & \dots & x \\
\end{bmatrix}
$ and $
Y=\begin{bmatrix}
0 & \dots & 0 &0 & y \\
0 & \dots & 0 & y & 0\\
0 & \dots & y & 0& 0\\
\vdots&\vdots&\vdots&\vdots&\vdots \\
y & \dots & 0 & 0 & 0 \\
\end{bmatrix}
$
find the $\det_{2n}\begin{bmatrix}
xI_n & Y \\
-Y & X \\
\end{bmatrix}
$.
Solution:
We can observe that for matrices $A,B,C,D\in\mathbb{R}^{n,n}$ when matrix $A$ is invertible then
$\begin{bmatrix}
A & B \\
C & D \\
\end{bmatrix}=\begin{bmatrix}
A & 0 \\
C & I_{n} \\
\end{bmatrix}\cdot\begin{bmatrix}
I_n & A^{-1}B \\
0 & D-CA^{-1}B \\
\end{bmatrix}
$
so
$\det_{2n}\begin{bmatrix}
A & B \\
C & D \\
\end{bmatrix}=\det_{n}(A) \cdot\det_{n}(D-CA^{-1}B)$
We see that $xI_n$ is invertible matrix when $x\neq 0$ and then we state that
$\det_{2n}\begin{bmatrix}
xI_n & Y \\
-Y & X \\
\end{bmatrix}=\det_{n}(xI_n) \cdot\det_{n}(X+Y(xI_n)^{-1}Y)
$
and now let's observe that $\det_{n}(xI_n)=x^n$ and
$
X+Y(xI_n)^{-1}Y=X+\frac{1}{x}YI_n^{-1}Y=X+\frac{1}{x}YI_{n}Y
=X+\frac{1}{x}Y^2=X+\frac{y^2}{x}I_n
$
so $\det_n (X+Y(xI_n)^{-1}Y)=\det_{n}\left(X+\frac{y^2}{x}I_n\right) = (x+\frac{y^2}{x})^n$
and in the end we can write that $\det_{2n}\begin{bmatrix}
xI_n & Y \\
-Y & X \\
\end{bmatrix}=x^n (x+\frac{y^2}{x})^n = (x^2+y^2)^n
$ when $x\neq 0$.
In the case $x=0$ we see that
$\det_{2n}\begin{bmatrix}
xI_n & Y \\
-Y & X \\
\end{bmatrix}=\det_{2n}\begin{bmatrix}
0 & Y \\
-Y & 0 \\
\end{bmatrix} =(-1)^n \det_{2n}\begin{bmatrix}
Y & 0 \\
0 & -Y \\
\end{bmatrix} =(-1)^n \cdot y^n \cdot (-y)^n = (-1)^n\cdot (-1)^n \cdot y^{2n} = y^{2n}$
Can you maybe find any simpler solution to this task?
| Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $\det\pmatrix{0&Y\\ -Y&0}=\det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $\pmatrix{0&Y\\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$ I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2\sqrt{2}$ thus:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$ = $\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$ = $\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$ = $\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$ = $\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$ = $\frac{28\sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $\sqrt{2}$
This is as far as I can get. I don't know if $\frac{28\sqrt{2}-16}{14}-\sqrt{2}$ is still correct or close to the solution. How can I arrive at $\frac{-5\sqrt{2}-6}{7}$?
| You were doing fine until the place where you tried to expand
$(2\sqrt2 - 4)(4 + \sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
\begin{align}
(2\sqrt2 - 4)(4 + \sqrt2) &= (2\sqrt2 - 4)4 + (2\sqrt2 - 4)\sqrt2 \\
&= (8\sqrt2 - 16) + (4 - 4\sqrt2) \\
&= 4\sqrt2 - 12.
\end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$\frac{4\sqrt2 - 12}{14}.$
And finally you'll want to change the $-\sqrt2$ so that you have two fractions with a common denominator and can finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$) I need to simplify \begin{equation}
\frac{x^{5}}{x^{2}+1}
\end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
\begin{equation}
x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+...
\end{equation}
| The idea of polynomial division is like integer division. With integer division of $\frac nd$, we want integer $q,r$ so that $n=qd+r$ and $r\lt d$. With polynomial division of $\frac nd$, we want polynomial $q,r$ so that $n=qd+r$ and $\deg(r)\lt \deg(d)$.
$$
\require{enclose}
\begin{array}{rl}
&\phantom{)\,}\color{#C00}{x^3}\color{#090}{-x}\\[-4pt]
x^2+1\!\!\!\!\!&\enclose{longdiv}{x^5\qquad}\\[-4pt]
&\phantom{)\,}\underline{\color{#C00}{x^5+x^3}}\\[-2pt]
&\phantom{)\,x^5}{}-x^3\\[-4pt]
&\phantom{)\,x^5}\underline{\color{#090}{{}-x^3-x}}\\[-4pt]
&\phantom{)\,x^5{}-x^3-{}}x\\[-4pt]
\end{array}
$$
So we get a quotient of $x^3-x$ and a remainder of $x$, which allows us to write both
$$
\overbrace{\quad\,x^5\quad\,}^n=\overbrace{\left(x^3-x\right)}^q\overbrace{\left(x^2+1\right)}^d+\overbrace{\vphantom{x^5}\quad\;x\quad\;}^r
$$
and
$$
\frac{x^5}{x^2+1}=x^3-x+\frac{x}{x^2+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$
$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=\pm \sqrt{\frac{2}{3}}$
But $x \ne 0$ since $xy=-1$
$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$
Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$
Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$
Is there any other approach?
| From $xy=-1$
$$ S = \frac{4}{4-x^2} + \frac{9}{9-y^2} = 1 + \frac{35}{25 - \left(3x - \frac{2}{x}\right)^2}\geq \frac{12}{5}$$
Equality hold when $$3x=\frac{2}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$ =1 Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$= 1
My approach is , First I try remove xy term from the equation, to convert the given equation in the standard equation of ellipse and find the value of $a$ and $b$. For this , I use the concept of rotation of axis, using this concept first I find $\tan2\theta = -2$ then I find the value of $\sin \theta$ and $\cos \theta$. But the problem is the the value of $\sin \theta$ and $\cos\theta$ are complex and when I try to solve using the concept of rotation of axis but equation become too much complex and also take a lots of time to solve it. So I please tell me another approach or method to solve this question.
| The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r \cos \theta, y= r\sin \theta$
Substituting in the equation we get $r^2 = \dfrac{1}{2 \cos^2 \theta+6 \sin \theta \cos \theta+ 5 \sin^2 \theta} = \dfrac{2}{7+6\sin 2 \theta - 3 \cos 2 \theta} \le \dfrac{7+3 \sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = \dfrac{7+3 \sqrt 5}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Nigerian Olympiad Suppose $a,b,c,d$ are integers satisfying $ab + cd = 44,ad - bc = 9.$
Find the minimum possible value of $a² + b² + c² + d².$
| $$(ab+cd)^2=44^2\implies a^2b^2+2abcd+c^2d^2=44^2$$
Similarly:
$$a^2d^2-2abcd+c^2b^2=9^2$$
So:
$$a^2b^2+a^2d^2+c^2d^2+c^2b^2=44^2+9^2=2017\implies(a^2+c^2)(b^2+d^2)=2017$$
$2017$ is prime so one of $a^2+c^2$ or $b^2+d^2$ is $1$ and the other is $2017$. Assume $a^2+c^2=1$. Then one of $a$ and $c$ is $0$ and the other is $\pm 1$. If $a$ is $0$, then $d=\pm 44$. Then, $b^2=2017-d^2=9^2$.
So $$a^2+b^2+c^2+d^2=0^2+9^2+(\pm 1)^2+44^2=2018$$
You can exchange $a^2+c^2$ and $b^2+d^2$ w/o l.o.g. (and $a$ and $c$ or $b$ and $d$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Solving a Cauchy problem, differential equation I have the following Cauchy problem
\begin{cases} y'(x) + \frac{1}{x^2-1}y(x) = \sqrt{x+1} \\ y(0) = 0 \end{cases}
I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= \frac{1}{x^2-1}$ :
$$\int A(x)dx=\int \frac{1}{x^2-1}dx= \frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)+c $$
then I obtain : $$e^{A(x)}=e^{\frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)}=\Big(\frac{|x-1|}{|x+1|}\Big)^{\frac{1}{2}}=\sqrt{\frac{|x-1|}{|x+1|} }$$
I have attempted to solve it in this way:
$$ \sqrt{\frac{|x-1|}{|x+1|} }\cdot y'(x) + \frac{1}{x^2-1}\cdot \sqrt{\frac{|x-1|}{|x+1|} }y = \sqrt{x+1}\cdot\sqrt{\frac{|x-1|}{|x+1|} }$$
$$\sqrt{\frac{|x-1|}{|x+1|} }*y(x) =\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
$$y(x) =\Big(\sqrt{\frac{|x-1|}{|x+1|} }\Big)^{-1}\cdot\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.
| Computing $$\mu(x)=e^{\int\frac{1}{x^2-1}dx}=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ then you will get
$$\int\frac{d}{dx}\left(\frac{\sqrt{1-x}y(x)}{\sqrt{x+1}}\right)=\int\sqrt{1-x}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$, then find $\alpha$, $\beta$ in $(0,\frac\pi2)$
If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$,
where $\displaystyle \alpha, \beta \in \bigg(0,\frac{\pi}{2}\bigg)$. Then value of $\alpha,\beta$ are
Try: I am trying to convert it into sum of square of quantity
like $$(\cos^2 \alpha)^2+(2\sin^2 \beta)^2-2\cdot \cos^2 \alpha \cdot 2\sin^2 \beta-4\sqrt{2}\cos \alpha \sin \beta +2+4\cos^2 \alpha \cdot \sin^2 \beta$$
$$(\cos^2 \alpha--2\sin^2 \beta)^2-4\sqrt{2}\cos \alpha \sin \beta+4\cos^2 \alpha \cdot \sin^2 \beta$$
Now i did not know how to solve it, could some help me
| If you just call $x=\cos(\alpha)$ and $y=\sin(\beta)$ you get the 4th degree polynomial equation $x^4-4\sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction proof for stirling of first kind. I have to show that for every stirling number of the first kind $\forall n \geq 2 : s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $\frac{1}{24}2(2-1)(2-2)(3*2-1) = \frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $\forall n \in \mathbb{N} $ with $ n \geq 2$
that $s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n \rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = \frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $\frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$\frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n \choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?
| $s_{n+1,n-1} = \frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} \tag1$.
$s_{n,n-1} = \frac{n(n-1)}{2} \tag 2$.
Substitute this in the prior equation we get
$s_{n+1,n-1} = \frac{1}{24}n(n-1)(n-2)(3n-1)+\frac{n^2(n-1)}{2} $.
which simplifies to
$s_{n+1,n-1} = \frac{1}{24}(n+1)n(n-1)(3n+2)\tag 3 $.
Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.
| {
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How to see $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$? I was reading an example where the purpose was to compute a certain Galois group. Along the way, the writer says : note $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$. But how do I note this? I understand you can factorize by $x^2-1$, since when I draw on the unit circle I see that $-1$ and $+1$ are roots. But for the rest?
Edit :
I see you can then factor $(x^2-1)(x^4+x^2+1)$ and than substitute $x^2=y$ and solve quadratic equation but can you actually see the solution visually?
| $x^6-1=(x^3+1)(x^3-1)\\x^3-1=(x-1)(x^2+x+1)\\x^3+1=(x+1)(x^2-x+1)$ might help.
| {
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Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers. Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers.
WLOG, let a$\leqq b\leqq c$,
| If any two of $a,b,c$ are equal, then wlog. $a=b$. As $\frac{b+1}{a}=1+\frac1a$ is an integer, we conclude $a=b=1$. The remaining conditions are that $\frac{c+1}{1}$ and $\frac 2c$ are integers, which lead us to the solutions
$$(1,1,1),\qquad (1,1,2) $$
(and cyclic permutations of the latter).
So assume $a,b,c $ are pairwise different. By cyclic permutation, we may assume wlog that $a<b<c$ or that $a>b>c$.
In the first case, $0<\frac{a+1}{b}\le \frac bb=1$ and hence $a+1=b$. Likewise, $b+1=c$. Then the last integer is $\frac{c+1}a=\frac{a+3}a=1+\frac 3a$ and we must have $a=1$ or $a=3$, whic gives us the solutions
$$(1,2,3),\qquad (3,4,5) $$
(and cyclic permutations).
In the case $a>b>c$, we instead have that $0<\frac{c+1}{a}\le \frac{c+1}{c+2}<1$, not an integer. So this case does not produce additional solutions.
| {
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Expectation and variance of the number of elements of a random non-empty set selected from a finite power set
Let $S$ denote a finite set of cardinality $|S| = N$. Select randomly a non-empty subset of $S$. Let $X$ indicate the number of items belonging to this subset.
(a) Describe the probability mass function of $X$ and compute $\textbf{E}(X)$ and $\textbf{Var}(X)$.
(b) Prove that $\displaystyle\lim_{N\rightarrow\infty}\frac{\textbf{E}(X)}{N} = \frac{1}{2}$ and $\displaystyle\lim_{N\rightarrow\infty}\frac{\textbf{V}(X)}{N} = \frac{1}{4}$.
MY ATTEMPT
(a) In the first place, we first notice there are $C(N,x)$ subsets which has $x$ elements, where $1\leq x\leq N$. Therefore we conclude that
\begin{align*}
f_{X}(x) = \frac{1}{2^{N} - 1}{N\choose x}
\end{align*}
From then on we are able to determine $\textbf{E}(X)$ and $\textbf{Var}(X)$. Precisely speaking, we have
\begin{align*}
\sum_{x=1}^{N}{N\choose x}x & = \sum_{x=1}^{N}\frac{N!}{(x-1)!(N-x)!} = \sum_{x=0}^{N-1}\frac{N!}{x!(N-x-1)!}\\
& = N\sum_{x=0}^{N-1}\frac{(N-1)!}{x!(N-x-1)!} = N\sum_{x=0}^{N-1}{N-1\choose x} = N2^{N-1}
\end{align*}
according to the binomial theorem. From whence we conclude that
$$\textbf{E}(X) = \frac{N2^{N-1}}{2^{N}-1}$$
Further, we do also have the following identity
\begin{align*}
\sum_{x=1}^{N}{N\choose x}x^{2} & = \sum_{x=1}^{N}\left[\frac{N!}{(x-1)!(N-x)!}\right]x = \sum_{x=0}^{N-1}\left[\frac{N!}{x!(N-x-1)!}\right](x+1)\\
& = \sum_{x=1}^{N-1}\frac{N!}{(x-1)!(N-x-1)!} + \sum_{x=0}^{N-1}\frac{N!}{x!(N-x-1)!}\\
& = N(N-1)\sum_{x=0}^{N-2}\frac{(N-2)!}{x!(N-x-2)!} + N\sum_{x=0}^{N-1}\frac{(N-1)!}{x!(N-x-1)!}\\
& = N(N-1)\sum_{x=0}^{N-2}{N\choose x} + N\sum_{x=0}^{N-1}{N-1\choose x} = N(N-1)2^{N-2} + N2^{N-1}
\end{align*}
Therefore
$$\textbf{E}(X^{2}) = \frac{N(N-1)2^{N-2} + N2^{N-1}}{2^{N}-1}$$
Finally, we have the variance:
\begin{align*}
\frac{\textbf{Var}(X)}{N} & = \frac{\textbf{E}(X^{2}) - \textbf{E}(X)^{2}}{N}
\end{align*}
(b) As requested, we obtain the desired result
\begin{align*}
\lim_{N\rightarrow\infty}\frac{\textbf{E}(X)}{N} = \lim_{N\rightarrow\infty}\frac{2^{N-1}}{2^{N}-1} = \frac{1}{2}
\end{align*}
However, the same does not apply to the variance when I make use of the results obtained. Am I committing some conceptual mistake? I would be equally grateful if someone provided me a quicker or smarter way to tackle this problem.
EDIT
I edited my answer as suggested by Did. Nonetheless, I am still thinking there is something wrong with my solution. Could someone double-check my calculations? Thanks in advance.
| Seem probably simpler if you look at it this way: Say $S$ has $n$ elements. If $Y$ is the number of elements of a randomly selected element of the power set, including the empty set as a possibility, then $Y$ has the same distribution as $X_1+\dots+X_n$, where the $X_j$ are iid with $P(X_1=0)=P(X_1=1)=1/2$.
| {
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Probability of $2$ pairs on $5$ dice We throw $5$ dice. What is the probability of getting $2$ pairs ?
My solution says it is $$\frac{6\cdot 5\cdot 4\cdot 5! }{6^5\cdot 2\cdot 2!\cdot 2!},$$ where as for me it's $$\frac{6\cdot 5\cdot 4\cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
I do as follows: Throwing $5$ dice is the same thing as throwing one die $5$ times. What we want is $AABBC$.
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $\frac{5!}{2!2!}$. At the end, I get
$$\frac{6\cdot 5\cdot 4 \cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
What's wrong in my argument?
| You have to divide your result by $2$ because you are not interested in the order of the two different pairs: $11336$ is the same of $33116$.
We have $\binom{6}{2}$ ways to choose the values of the two couples (say $A$ and $B$ with $A<B$) and $4$ ways to choose the single value (say $C$). The number of permutations of "word" $AABBC$ is $5!/2!/2!$. Therefore the probability is
$$\frac{\binom{6}{2}\cdot 4 \cdot \frac{5!}{2! 2!}}{6^5}=\frac{6\cdot 5\cdot 4\cdot 5! }{6^5\cdot 2\cdot 2!\cdot 2!}$$
| {
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Creating an integral to represent the volume of the intersection of two balls in cartesian coordinates The question states:
Let $A$ be the intersection of the balls
$x^2+y^2+z^2\leq 9$ and $x^2+y^2+(z-8)^2\leq 49$
I am asked to just set up the iterated triple integral that represents the volume of $A$ in cartesian coordinates.
What I am able to determine so far is that for the equation
$x^2+y^2+z^2\leq 9 $:
$-\sqrt{9-x^2-y^2}\leq z \leq \sqrt{9-x^2-y^2}$
$-\sqrt{9-x^2}\leq y \leq \sqrt{9-x^2}$
$-3\leq x \leq 3$
If I were to set up this integral it would be:
$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{-\sqrt{9-x^2-y^2}}^{\sqrt{9-x^2-y^2}} dzdydx$
But I don't know how I'm supposed to set up the intersection of the two balls?
I was thinking of setting the two equations equal to each other so that
$x^2+y^2\leq 9-z^2$ and
$x^2+y^2\leq 49-(z-8)^2$
so then I have $9-z^2=49-(z-8)^2$
Solving for $z$ I get $z=3/2$ but I don't know what to do with this information.
| $\def\bbr{\mathbf{R}}$It is simpler to exploit the symmetries. From the picture below, you can see clearly that the intersection is a normal domain in $\bbr^3$, which means that you can write it as
$$
A=\{(x,y,z)\mid (x,y)\in D, f(x,y)\le z\le g(x,y)\}
$$
where $D$ is some two-dimensional region on the $xy$-plane. You can then set up the volume as
$$
V=\iint_D\int_{f(x,y)}^{g(x,y)}1\;dz dA
$$
and reduce it further to an iterated integral by working on the region $D$.
The ball $x^2+y^2+z^2\le 9$ can be written as:
$$
E_1=\{(x,y,z)\mid (x,y)\in D_1,-\sqrt{9-x^2-y^2}\le z\le\sqrt{9-x^2-y^2}\}
$$
where $D_1=\{(x,y)\mid x^2+y^2\le 9\}$ is a 2D region.
Similarly, the ball $x^2+y^2+(z-8)^2\le 49$ can be written as
$$
E_2=\{(x,y,z)\mid (x,y)\in D_2,8-\sqrt{49-x^2-y^2}\le z\le 8+\sqrt{49-x^2-y^2}\}
$$
where $D_2=\{(x,y)\mid x^2+y^2\le 49\}$. So
$$
E_1\cap E_2=\{(x,y,z)\mid (x,y)\in D, 8-\sqrt{49-x^2-y^2}\le z\le\sqrt{9-x^2-y^2} \}\tag{1}
$$
for $D=\{x^2+y^2\le R^2\}$ with some real number $R$. To find out $R$, observe that the intersection curve (see picture below) of the two spheres is given by the equations:
$$
\begin{cases}
x^2+y^2+z^2= 9 \\
x^2+y^2+(z-8)^2= 49
\end{cases}
$$
which is equivalent to
$$
\begin{cases}
x^2+y^2+z^2= 9 \\
9-z^2+(z-8)^2= 49
\end{cases}
$$
which is
$$
\begin{cases}
x^2+y^2+z^2= 9 \\
z=\frac{3}{2}
\end{cases}
$$
So $R = \frac{3\sqrt{3}}{2}$. Now you can write the volume as
$$
V=\iint_D\int_{8-\sqrt{49-x^2-y^2}}^{\sqrt{9-x^2-y^2}}1\cdot dz dA
=\int_{-R}^{R}\int_{-\sqrt{R^2-y^2}}^{\sqrt{R^2-y^2}}\int_{8-\sqrt{49-x^2-y^2}}^{\sqrt{9-x^2-y^2}}1\cdot dzdxdy
$$
| {
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What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\right)\,$ independently, but that didn't go well for me.
*Multiplying and dividing by $\cos^2x$ or $\sin^2x$.
*Expressing $\cos^2x$ as $1-\sin^2x$ and splitting the integral, and I was stuck with $\int \left(\frac{\sin^3x}{\sin x - \cos x}\right)\, dx$ which I rewrote as $\int \frac{\csc^2x}{\csc^4x (1-\cot x) } dx,\,$ and tried a whole range of substitutions only to fail.
*I tried to substitute $\frac{1}{ \sin x - \cos x}$, $\frac{\sin x}{ \sin x - \cos x}$, $\frac{\cos x \sin x}{ \sin x - \cos x}$ and $\frac{\cos^2x \sin x}{\sin x - \cos x},$ independently, none of which seemed to work out.
*I expressed the denominator as $\sin\left(\frac{\pi}{4}-x\right)$ and tried multiplying and dividing by $\sin\left(\frac{\pi}{4}+x\right)$, and carried out some substitutions. Then, I repeated the same with $\cos\left(\frac{\pi}{4}+x\right)$. Neither of them worked.
| Hint:
Let $\dfrac\pi4-x=y$
$\sin x-\cos x=\sqrt2\sin y$
$\sin x=\dfrac{\cos y-\sin y}{\sqrt2}$
$2\cos^2x=1+\cos2x=1+2\sin y\cos y$
$$\dfrac{(\cos y-\sin y)(1+2\sin y\cos y)}{\sin y}=2\cos^2y-2\sin y\cos y+\cot y-1=\cos2y-\sin2y+\cot y$$
| {
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A Problem related to the Cauchy Distribution The following problem is from the book, "Introduction to Probability" by
Hoel, Port and Stone. My answer does not match the back of the book. What did I do wrong?
Thanks,
Bob
Problem:
Let $X$ have a Cauchy density. Find the upper quartile for $X$.
Answer:
Recall the density function for the Cauchy distribution is:
\begin{align*}
f(x) &= \frac{1}{x^2+1} \\
\end{align*}
Let $a$ be the number we seek.
\begin{align*}
F(x) &= \tan^{-1}{x} \\
\int_{-\infty}^{a} \frac{1}{x^2+1} \,\, dx &= \frac{3}{4} \\
\tan^{-1} \Big|_{-\infty}^{a} &= \frac{3}{4} \\
\tan^{-1}(a) - \frac{-\pi}{2} &= \frac{3}{4} \\
\tan^{-1}(a) + \frac{\pi}{2} &= \frac{3}{4} \\
\tan^{-1}(a) &= \frac{3}{4} - \frac{\pi}{2} \\
\end{align*}
The books's answer is $1$. What did I do wrong?
I ran the following command in R: pcauchy(1) . R returned 0.75 therefore,
I believe the book is right.
| You forgot the factor $\frac{1}{\pi}$ in the probability density function. Correct density function is
$$
f(x) =\frac{1}{\pi(1+x^2)}
$$ so that $\int_{-\infty}^\infty f(x) dx =1$. Your problem is to find $a$ such that
$$
\frac{1}{\pi}\int_{-\infty}^a \frac{dx}{1+x^2}=\frac{\arctan a+\frac{\pi}{2}}{\pi}=\frac{3}{4}
$$ or equivalently
$$
\arctan a=\frac{\pi}{4}.
$$ This gives $a=\tan \frac{\pi}{4}=1$.
| {
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Geometry with circle and triangle. Any other solutions(advice) are welcome.
What I am asking is this because I am studying mathematics through feedback process of my solution and learning new solutions.
Please, release Hold on.
When will the hold be resolved? I want to see your new solutions(or advices).
$\overline{AE}$ is a diameter.
and $\overline{AB}=1,\;\;\; \overline{AC}=2$
and $\overline{BD}:\overline{CD}=3:2$
Find the area of $\triangle ABC$ .
|
Here is a completely different soluiton based on trigonometry.
From triangle $ABD$:
$$\frac{3x}{\sin\alpha}=\frac{l}{\sin(90^\circ-\beta)}\tag{1}$$
From triangle $ADC$:
$$\frac{2x}{\sin\beta}=\frac{l}{\sin(90^\circ-\alpha)}\tag{2}$$
From triangle $ABC$:
$$\frac{1}{\sin(90^\circ-\alpha)}=\frac{2}{\sin(90^\circ-\beta)}\tag{3}$$
Rewrite (1),(2)(3):
$$\frac{3x}{\sin\alpha}=\frac{l}{\cos\beta}\tag{4}$$
$$\frac{2x}{\sin\beta}=\frac{l}{\cos\alpha}\tag{5}$$
$$\cos\beta=2\cos\alpha\tag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$\frac{3\sin\beta}{2\sin\alpha}=\frac{\cos\alpha}{\cos\beta}$$
$$3\sin\beta\cos\beta=2\sin\alpha\cos\alpha\tag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3\sin\beta=\sin\alpha\tag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $\alpha$:
$$9\sin^2\beta+\frac14\cos^2\beta=1$$
$$36\sin^2\beta+cos^2\beta=4$$
$$35\sin^2\beta=3$$
The rest is straightforward:
$$\sin\beta=\frac{\sqrt{3}}{\sqrt{35}}, \ \ \cos\beta=\frac{4\sqrt{2}}{\sqrt{35}}$$
From (8):
$$\sin\alpha=\frac{3\sqrt{3}}{\sqrt{35}}, \ \ \cos\alpha=\frac{2\sqrt{2}}{\sqrt{35}}$$
Now:
$$S=\frac12 AB\cdot h=\frac12 \cdot 1 \cdot 2 \sin(\alpha+\beta)=\sin(\alpha+\beta)$$
$$S=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{2\sqrt{6}}{5}$$
| {
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$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. What could be a proof for it?
| If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.
Then, we have \begin{align}\left\lfloor\frac{a-b}2\right\rfloor+\left\lceil\frac{a+b}2\right\rceil &= \left\lfloor\frac{(2m+1)-(2n+1)}2\right\rfloor+ \left\lceil\frac{(2m+1)+(2n+1)}2\right\rceil\\
&=\left\lfloor\frac{2m-2n}2\right\rfloor+ \left\lceil\frac{2m+2n+2}2\right\rceil\\
&=\lfloor m-n\rfloor + \lceil m+n+1\rceil\\
&= m-n + m+n+1\tag{$*$}\\
&= 2m+1\\
&=a\end{align}
We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).
If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.
So we have \begin{align}\left\lfloor\frac{a-b}2\right\rfloor+\left\lceil\frac{a+b}2\right\rceil &= \left\lfloor\frac{(2m)-(2n+1)}2\right\rfloor+ \left\lceil\frac{(2m)+(2n+1)}2\right\rceil\\
&=\left\lfloor\frac{2m-2n-1}2\right\rfloor+ \left\lceil\frac{2m+2n+1}2\right\rceil\\
&=\left\lfloor m-n-\frac 12\right\rfloor + \left\lceil m+n+\frac12\right\rceil\\
&= m-n -1+ m+n+1\tag{$\dagger$}\\
&= 2m\\
&=a\end{align}
Here we get to $(\dagger)$ because we round the first half down and the second half up to their respective nearest integers
| {
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If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| Hint:
Clearly, $\sin x=0\iff\cos^2x=?,$
$\cos x=0\iff\sin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
| {
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} |
Calculate $\lim\limits_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{6}{k(k+1)(k+3)}$ $$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{6}{k(k+1)(k+3)}$$
I tried to simplify the sum and I got $\frac{2}{k}-\frac{3}{k+1}+\frac{1}{k+3}$ but I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$\frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=\frac{(k+3)}{k(k+1)(k+2)(k+3)}-\frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.
| Hint
$$\frac{2}{k}-\frac{3}{k+1}+\frac{1}{k+3}=\frac{2}{k}-\frac{2}{k+1}-\frac{1}{k+1}+\frac{1}{k+3}=2\left(\frac{1}{k}-\frac{1}{k+1} \right)-\left(\frac{1}{k+1}-\frac{1}{k+3} \right)$$
| {
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"url": "https://math.stackexchange.com/questions/3089290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \\
0 = 0 \qquad \text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$\begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \\
x^2+xy+y^2 &= 3(x+y) \end{aligned}\\
\begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \\
x^2+x(y-3)+y(y-3) &=0 \\
x^2 + (y-3)(x+y) &= 0 \end{aligned}$$
What now? Am I on the right track?
| $x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $x\ne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=\frac{n+\sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $n\le 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show vector is approximately an eigenvector of matrix, thus find eigenvalue Say we have matrix $\mathbf{A}$
$$
\mathbf{A}=\begin{pmatrix}
-3&2&0\\
4&-6&2\\
0&1&-1
\end{pmatrix}
$$
We now must show that $\mathbf{v}=\begin{pmatrix}-1.34&-0.8&1\end{pmatrix}^T$ is an approximate eigenvector for $\mathbf{A}$ to 2 decimal places, and find the corresponding eigenvalue. We know what $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$ for some vector. So multiplying $\mathbf{A}$ by $\mathbf{v}$ we get
$$
\begin{pmatrix}
-3&2&0\\
4&-6&2\\
0&1&-1
\end{pmatrix}
\begin{pmatrix}
-1.34\\
-0.8\\
1
\end{pmatrix}=
\begin{pmatrix}
2.42\\
1.44\\
-1.8
\end{pmatrix}
$$
So we need to find eigenvalue $\lambda$ such that
$$
\begin{pmatrix}
2.42\\
1.44\\
-1.8
\end{pmatrix}
=\lambda
\begin{pmatrix}
-1.34\\
-0.8\\
1
\end{pmatrix}
$$
Say $\lambda=-1.8$, this gives us
$$
-1.8
\begin{pmatrix}
-1.34\\
-0.8\\
1
\end{pmatrix}=
\begin{pmatrix}
2.412\\
1.44\\
-1.8\\
\end{pmatrix}
$$
Now $\lambda=-1.8$ gives a good approximation for the eigenvector, but we're getting 2.41 instead of 2.42 for the first value of the eigenvector (to 2 decimal places). Is this enough to say that $\mathbf{v}$ is an approximate eigenvector for $\mathbf{A}$? or am I missing something in my method?
| Your method has a minor issue.
If you want numbers to match to a certain precision, you should use inequalities, e.g., solve
$$|-1.34\lambda-2.42|<0.01,$$
$$|-0.8\lambda-1.44|<0.01,$$
$$|1\cdot\lambda-(-1.8)|<0.01.$$
If you want it to match exactly to 2 decimal places, you may need to shrink these margins (they are not equivalent to matching to 2 decimal places). Or you can add some more inequalities, since if they are going to match to 2 decimal places, we'll need
$$-1.34\lambda \ge 2.42,$$
$$-0.8\lambda \ge 1.44,$$
$$1\cdot\lambda \le -1.8,$$
or else we could get something like $-1.34\lambda = 2.419$ or $1\cdot \lambda = -1.795$.
One example of such a value of $\lambda$ is $\lambda = -1.806$, which yields
$$\lambda \begin{pmatrix} -1.34 \\ -0.8 \\ 1 \end{pmatrix} = \begin{pmatrix}2.42004\\ 1.4448\\ -1.806\end{pmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
| This integral appeared on AoPS a while ago, it might be that it was on MSE too. Anyway I will quote what I did on AoPS.
$$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx=\int_0^1 \frac{1-x^3 }{1-x^5}dx$$ Recall the geometric series: $\displaystyle{\frac{1}{1-x^z}=\sum_{n=0}^\infty x^{nz}, \, |x|<1 } $. Applying it here: $$I=\int_0^1 (1-x^3) \left(\sum_{n=0}^\infty x^{5n}\right)dx=\sum_{n=0}^\infty \int_0^1 \left(x^{5n}-x^{5n+3}\right)dx$$ I don't know how to show that we can swap the sum and the integral here. $$I=\left(\frac{x^{5n+1}} {5n+1} - \frac{x^{5n+4}} {5n+4}\right) \bigg|_0^1 =\sum_{n=0}^\infty\left(\frac{1} {5n+1} - \frac{1 } {5n+4}\right) =\frac15 \sum_{n=0}^\infty\left(\frac{1} {n+\frac15} - \frac{1 } {n+\frac45}\right) $$
We have by the series formula of digamma function that:
$$\psi(z)-\psi(s)=\left(-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{z+n}\right)-\left(-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{s+n}\right)=\sum_{n=1}^\infty\left(\frac{1}{s+n}-\frac{1}{z+n}\right) $$ So $\displaystyle{ I=\frac15\left(\psi\left(\frac45\right) - \psi\left(\frac15\right)\right) } $. Also using the reflection formula:$$\psi(1-z)-\psi(z)=\pi\cot(\pi z)$$ We have: $$I=\frac{\pi} {5}\cot\left(\frac{\pi}{5}\right)=\frac{\pi} {5}\sqrt{1 +\frac{2} {\sqrt 5}}$$ see also: https://en.m.wikipedia.org/wiki/Digamma_function and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
New solution and quite elementary. Let:
$$x=\frac{1-t}{1+t}\rightarrow dx=-\frac{2}{(1+t)^2}dt$$
$$\Rightarrow I=2\int_0^1 \frac{x^2+3}{x^4+10x^2+5}dx$$
And by performing partial fractions we get:
$$I=\frac{\sqrt 5+1}{\sqrt 5}\int_0^1 \frac{dx}{x^2+2\sqrt 5+5}+\frac{\sqrt 5-1}{\sqrt 5}\int_0^1 \frac{dx}{x^2-2\sqrt 5+5}$$
And now there are two simple integrals left and some algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
} |
Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.
| This is handled in the same manner as in this answer.
Split the numerator like $$1-\cos x+\cos x(1-\cos 2x\cos 3x\dots\cos nx) $$ and the desired limit is equal to $$\lim_{x\to 0}\frac {1-\cos x} {x^2}+\lim_{x\to 0}\cos x\cdot \frac{1-\cos 2x\dots\cos nx} {x^2}$$ which is same as $$\frac {1} {2}+\lim_{x\to 0} \frac{1-\cos 2x\dots\cos nx} {x^2}$$ Applying same technique we see that the above is equal to $$\frac{1}{2}+\frac {2^2} {2}+\lim_{x\to 0} \frac{1-\cos 3x\dots\cos nx} {x^2}$$ and continuing in same fashion we see that the desired limit is equal to $$\frac{1^2+2^2+\dots+n^2}{2}=\frac{n(n+1)(2n+1)}{12}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating limit of definite integral I need to calculate:
$$
\lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}\, dt
$$ The result should be $1$.
Is there a quicker way than calculating the primitive function?
I thought about seperating to $\int_0^{x+1} -\int_0^x$ but still can't think of the solution.
| Use the estimate:
$$\frac{t-10}{t+10}<\frac{t^2+1}{t^2+20t+8}<\frac t{t+10}, \ t>1 \Rightarrow \\
\int_x^{x+1} \frac{t-10}{t+10}dt<I(x)<\int_x^{x+1} \frac t{t+10}dt \Rightarrow \\
1-20\ln \frac{x+11}{x+10}<I(x)<1-10\ln \frac{x+11}{x+10} \Rightarrow \\
\lim_{x\to\infty} \left(1-20\ln \frac{x+11}{x+10}\right)\le \lim_{x\to\infty} I(x) \le \lim_{x\to\infty} \left(1-10\ln \frac{x+11}{x+10}\right) \Rightarrow \\
\lim_{x\to\infty} I(x)=\lim_{x\to\infty} \int_x^{x+1}\frac{t^2+1}{t^2+20t+8}dt=1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$.
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$
*
*make exponent postive
$2x\frac{1}{\sqrt{4-x}} - 3\sqrt{4-x} = 0$
*
*square everything
$4x^2\frac{1}{4-x} - 9(4-x) =0$
*
*combine into one fraction
$\frac{1-9(4-x)[4x^2(4-x)]}{4x^2(4-x)} = 0$
*
*remove like terms
$1-9(4-x)=0$
$1-36-9x = 0$
$x = \frac{-35}{9}$
| If $$\frac{2x}{\sqrt{4-x}}-3\sqrt{4-x}=0$$
Then
$$\frac{2x}{\sqrt{4-x}}=\frac{3\sqrt{4-x}}{1}$$
If we cross multiply we get that
$$2x=3(4-x)=12-3x$$
You can go from here....
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find expected value of a random rectangle in a square grid I've tried to solve the following problem. Could you please tell if I did it right.
Problem. Given a n x n square grid. We take a random rectangle so that every ractangle is equally likely.
Find expected value of area of this rectangle.
Solution.
The number of all rectangles is $\sum\limits_{i=1}^{n}{i^2} = \frac{n(n+1)(2n+1)}{6}$. Let $f((i,j))$ - number of rectangles which consist of square in i-th row and j-th column. $f((i,j)) = i(n-i+1)j(n-j+1)$. Let $X_{i,j} = 1$ if random rectangle consists of square in i-th row and j-th column and $X_{i,j} = 0$ otherwise. Expected value of area of a random rectangle is $E(\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{n}{X_{i,j}}}) = \sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{n}{\frac{f((i,j))}{\frac{n(n+1)(2n+1)}{6}}}}$
$= \sum\limits_{i=1}^{n}{\frac{1}{\frac{n(n+1)(2n+1)}{6}}\sum\limits_{j=1}^{n}{i(n-i+1)j(n-j+1)}} = \sum\limits_{i=1}^{n}{i(n-i+1)\frac{6}{n(n+1)(2n+1)}\sum\limits_{j=1}^{n}{j(n-j+1)}}$
$=\sum\limits_{i=1}^{n}{i(n-i+1)\frac{6}{n(n+1)(2n+1)}((n+1)\sum\limits_{j=1}^{n}{j} - \sum\limits_{j=1}^{n}{j^2})} = \sum\limits_{i=1}^{n}{i(n-i+1)\frac{6}{n(n+1)(2n+1)}(\frac{(n+1)(1+n)}{2} - \frac{n(n+1)(2n+1)}{6})}$
$=\frac{6}{n(n+1)(2n+1)}(\frac{(n+1)(1+n)}{2} - \frac{n(n+1)(2n+1)}{6})\sum\limits_{i=1}^{n}{i(n-i+1)} = \frac{6}{n(n+1)(2n+1)}(\frac{(n+1)(1+n)}{2} - \frac{n(n+1)(2n+1)}{6})^2$
| The expected area is the expected base times the expected height. If, for example, $n=10$, there are $10$ bases of length $1$, and $9$ of length $2$, on up to $1$ of length $n$: $\sum_{k=1}^n k$ in all. Thus the expected base length is $$b = \frac{\sum_{k=1}^n k ( n+1-k)}{\sum_{k=1}^n k}$$ and the expected height length is the same. So the expected area is $b^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate using trigonometric substitution. Am I on the right path? I have been trying to solve:
$$\int \frac{\sqrt{x^2-9}}{x^3} dx$$
I am letting $ x = 3\sec \theta$ and so $dx = 3 \sec \theta \tan \theta$
So then I have:
$$\int \frac{\sqrt{9\sec^2 \theta - 9}}{27 \sec^3 \theta} dx$$
$$\int \frac{\sqrt{9(\sec^2 \theta - 1)}}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ \int \frac{3\tan \theta}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ = \int \frac{9 \tan ^2 \theta}{27 \sec ^2 \theta} d \theta$$
$$ = \int \frac{\tan ^2 \theta}{3 \sec ^2 \theta} d \theta$$
$$ \int \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \frac{\cos^2 \theta}{3} d \theta$$
$$ \int \frac{\sin^2 \theta}{3} d \theta$$
$$\frac{1}{3} \int \sin^2 \theta d \theta$$
Am I on the right track?
| $$I=\int\frac{\sqrt{x^2-9}}{x^3}dx=\int\frac{3\sqrt{(x/3)^2-1}}{x^3}dx$$
and we know that $\cosh^2\theta-1=\sinh^2\theta$
$$x=3\cosh(y)$$
$$dx=3\sinh(y)dy$$
$$I=9\int\frac{\sqrt{\cosh^2(y)-1}}{27\cosh^3(y)}\sinh(y)dy=\frac13\int\frac{\sinh^2(y)}{\cosh^3(y)}dy=\frac13\int\text{sech}(y)-\text{sech}^3(y)dy$$
a reduction formula can then be used
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factorise $x^n + 1$ Is there a way to factorise $x^n + 1$
I thought of doing it like this: $$x^n +1 = (x+1)(x^{n-1} - x^{n-2} + \cdots - x + 1)$$ but can't seem to get anywhere using this method.
| If $n$ is even then the graph of $y=x^n+1$ is always above the $x$ axis so $x^n+1=0$ has no real roots.
If $n$ is odd then the graph of $y=x^n+1$ crosses the $x$ axis at $x=-1$ and cannot cross the $x$ axis anywhere else because $x^n+1$ is monotonically increasing. So $x^n+1=0$ has a single real root at $x=-1$, and so $x+1$ is a factor of $x^n+1$. In fact, for odd $n$:
$x^n+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}-x^{n-4}+\dots-x+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the value of $m+n$ if $\frac{m}{n}$ is the radius of the smallest of the three circles?
Circles of radii 5, 5, 8 and $\frac{m}{n} $(the smallest circle) are
mutually externally tangent to all circles, where $m$ and $n$ are
relatively prime positive integers. Find $m + n$.
Source: Bangladesh Math Olympiad 2017 Junior category.
I can not figure the radius of the smallest circle. Is there any formula for an internal point in a isosceles triangle which can help me to solve this math?
| Let $x$ be a radius of the little circle.
Thus, by Heron we can get an area of the left and of the right triangle:
$$\sqrt{(x+13)x\cdot8\cdot5}=\sqrt{40x(x+13)}.$$
An area of the lower triangle it's
$$\frac{10\sqrt{(x+5)^2-5^2}}{2}=5\sqrt{x^2+10x)}.$$
The area of the full triangle it's
$$\frac{10\sqrt{13^2-5^2}}{2}=60.$$
Thus, $$5\sqrt{x^2+10x}+2\sqrt{40x(x+13)}=60$$ or
$$\sqrt{5(x^2+10x)}+\sqrt{32(x^2+13x)}-12\sqrt5=0.$$
Can you end it now?
I got $$x=\frac{8}{9}.$$
Actually, the expression $\sqrt{5(x^2+10x)}+\sqrt{32(x^2+13x)}$ increases, which says it's enough to substitute $x=\frac{8}{9}.$
| {
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A sum of series problem with alternating sign of terms I came across a problem that requires me to find the sum of a series. The term of the series $T_n$ is given by
$$T_n = (-1)^{\frac{n(n+1)}2}n^2$$
Sum till $4n$ terms is to be found.
Writing down the first few terms:
$$-(1)^2-(2)^2+3^2+4^2-(5)^2-(6)^2+7^2+8^2+ \ldots$$
This can be arranged into a sum of difference between squares:
$$(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\ldots$$
$$=2\times(4+6+12+14+20+\ldots)$$
$$=4\times(2+3+6+7+10\ldots)$$
In the last step, the expression inside the parentheses contains $n$ terms.
I am unable to proceed from here. How do I find the sum of the series inside the parentheses? A hint would be much appreciated.
This is a question from Resonance DLPD Algebra, Page #30, Question #2.
| If the number of terms is multiple of$4,$ $$\sum_{r=0}^n(-(4r+1)^2-(4r+2)^2+(4r+3)^2+(4r+4)^2)=2\sum(8r+4+8r+6)=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute remainder of division of $P(x)$ by $x^2 -3x+2$?
The remainder of division of $P(x)$ by $x^2−1$ is $2x+1$, and the remainder of division of the same polynomial by $x^2−4$ is $x+4$. Compute the remainder of division of $P(x)$ by $x^2−3x+2$.
I will translate these into math equations
$$P(x) = (x^2-1)Q(x)+ 2x+1$$
$$P(x) = (x^2 -4)R(x)+x+4$$
And let
$$f(x) = P(x) $$
We're asked to find the remainder when this polynomial is divided by $x^2 -3x+2$. So, there are two equations, which is why I'm confused with what to use in the equation $f(x) = P(x)$. What am I missing here?
Regards
| HINT:
$$\frac{P(x)}{x^2-1}=Q(x)+\frac{2x+1}{x^2-1}\implies\frac{P(x)}{x-1}=[(x+1)Q(x)+2]+\frac 3{x-1}\\\frac{P(x)}{x^2-4}=R(x)+\frac{x+4}{x^2-4}\implies\frac{P(x)}{x-2}=[(x+2)R(x)+1]+\frac6{x-2}\\\boxed{\frac{P(x)}{x-2}-\frac{P(x)}{x-1}=\frac{(x-1)P(x)-(x-2)P(x)}{x^2-3x+2}=\frac{P(x)}{x^2-3x+2}}$$
| {
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Triple integral : volume I need to calculate the volume between $x^2+y^2\le z^2-1$ and $2x^2+y^2+z^2\le 2$.
So It's a hyperboloid of two-sheets intersected with an ellipsoid.
their intersection leads to: $3x^2+2y^2=1$ which is an ellipse.
Using Cylindrical coordinates : $(x,y,z)=(\frac{1}{\sqrt{3}}r\cos\theta,\frac{1}{\sqrt{2}}r\sin\theta,z)$
Jacobian $|J|=\frac{\sqrt{6}}{6}r$
$\sqrt{1+x^2+y^2} \le z\le \sqrt{2-2x^2-y^2}$
$$
Volume =2\int_{0}^{2\pi}\int_{0}^{1}\int_{\sqrt{1+\frac{1}{3}r^2\cos^2\theta+\frac{1}{2}r^2\sin^2\theta}}^{\sqrt{2-\frac{2}{3}r^2\cos^2\theta-\frac{1}{2}r^2\sin^2\theta}}\frac{\sqrt{6}}{6}rdzdrd\theta?
$$
If it is correct Is there a simpler way to get the volume ?
| The thought process was correct.
Hint: the integrand should be $\frac{\sqrt{6}}{6}r^2$.
Not $\frac{\sqrt{6}}{6}r$, since the Jacobian is $\frac{\sqrt{6}}{6}r$ for the change of variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111332",
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"source": "stackexchange",
"question_score": "2",
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Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3}$
I wrote down a few cases:
$X_1 = \sin\frac{\pi}{3}-4\cos\frac{\pi}{3}= \frac{\sqrt3-4}{2} $
$X_2 = \sin\frac{2\pi}{3}-4\cos\frac{2\pi}{3}= \frac{\sqrt3+4}{2} $
$X_3 = \sin\frac{3\pi}{3}-4\cos\frac{3\pi}{3}=4 $
$X_4 = \sin\frac{4\pi}{3}-4\cos\frac{4\pi}{3}= \frac{4-\sqrt3}{2} $
$X_5 = \sin\frac{5\pi}{3}-4\cos\frac{5\pi}{3}= \frac{4-\sqrt3}{2} $
$X_6 = \sin\frac{6\pi}{3}-4\cos\frac{6\pi}{3}= -4 $
So as found above, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.
Let's check whether its convergent or not:
The sequence is bounded as stated above so lets check if its decreasing or increasing.
$X_n \geq X_{n+1}$
$\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3} \geq\sin\frac{(n+1)\pi}{3}-4\cos\frac{(n+1)\pi}{3}$
$\sin\frac{n\pi}{3} - \sin\frac{(n+1)\pi}{3} \geq -4\cos\frac{(n+1)\pi}{3} + 4\cos\frac{n\pi}{3}$
I used trigonometrical identity for $\sin\alpha+\sin\beta$ and $\cos\alpha-\cos\beta$ :
$-\cos\frac{\pi(2n+1)}{6} \geq 4\sin\frac{\pi(2n+1)}{6}$
What should I do next? I am stuck here.
Thanks, and sorry if I made mistakes.
| If the sequence is convergent, then all of its subsequences should
converge to the same limit.
Consider the subsequence defined by $n_{k}=3k$:
$$
X_{n_{k}}=\sin(k\pi)-4\cos(k\pi)=-4\cos(k\pi)=\begin{cases}
-4 & \text{if }k\text{ is even}\\
+4 & \text{if }k\text{ is odd}.
\end{cases}
$$
This subsequence oscillates between the values of $-4$ and $+4$.
What does that tell you?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 5,
"answer_id": 3
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Is there any natural number triangle whose inscribed circle's radius is $1$ except length $(a,b,c)=(3,4,5)$?
Is there any natural number triangle that inscribed circle's radius is $1$ except length $(a,b,c)=(3,4,5)$?
I found that there are no right triangle except $(3,4,5)$.
Thm. There are only one natural number right triangle that inscribed circle's radius is 1.
Proof)WLOG, $0<a<b<c$. Than $$a^2+b^2=c^2$$ $$a+b+c=ab$$. Because it is right triangle, $$a-1+b-1=c$$ .So, $$b=\frac{2(a-1)}{(a-2)}$$
*
*If $a$ is even, $b$ is not a natural number.Because if $a=2k$ ,$b=\frac{2k-1}{k-1}=2+\frac{1}{k-1}$. Only integer solution in $k=2$.
*If $a$ is odd, $a=2k+1$ ,$a-1$is even but $a-2$ is odd. So $b$ is not integer except $a-2=1$.
I want to know general case about this. Such triangle is exist or not exist. I think It doesn't exist.
I think about it and I got one. Check It for me wrong or not.
We could make $$(a,b,c) =(x+y,y+z,z+x)$$ and by Heron's formula radius $$1= \frac{\sqrt{xyz}}{\sqrt{x+y+z}}$$ so $$x+y+z=xyz$$ and we knows only solution of this equation is $(x,y,z)=(3,2,1)$.
| Say $a\leq b\leq c$. If $s=(a+b+c)/2$ then $$r={S\over s}$$ where $S$ is area. By Heron formula we have
$$S=s \implies \boxed{(s-a)(s-b)(s-c)=s}$$
then $$s-a\mid s\implies s-a\mid a \implies a = k(s-a)$$
so $$k(b+c)=a(2+k)$$ Since $b+c\geq 2a$ we get $$2ka\leq a(2+k)\implies k\leq 2$$
*
*$k= 1$ we have $b+c=3a$ and $s=2a$, so $$(2a-b)(2a-c)=2\implies -2a^2+bc=2$$
since $c= 2a-b$ we get
$$(2a-b)(b-a)=2$$
here we have 2 possibilities:
If $b-a = 1$ and $2a-b = 2$ we get $a=3$ and $b= 4$ so $c= 5$
If $b-a = 2$ and $2a-b = 1$ we get $a=3$ and $b= 5$ so $c= 4$
*
*$k=2$ then $b+c=2a$ so $a=b=c$, so $(s-a)^3=s$ so $a^3/8 = 3a/2$ so $a^2 = 12$ which is impossibile.
So the only such triangle has sides $3,4,5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116044",
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"question_score": "3",
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conversion of Binomial identity into series sum Prove that $$\binom{n}{1}(1-x)-\frac{1}{2}\binom{n}{2}(1-x)^2+\frac{1}{3}\binom{n}{3}(1-x)^3+\cdots \cdots +(-1)^{n-1}\frac{1}{n}(1-x)^n$$
$$=(1-x)+\frac{1}{2}(1-x^2)+\frac{1}{3}(1-x^3)+\cdots +\frac{1}{n}(1-x^n)$$
what i try
$$\bigg[1-(1-x)\bigg]^n=\binom{n}{0}-\binom{n}{1}(1-x)+\binom{n}{2}(1-x)^2-\cdots +(-1)^n\binom{n}{n}(1-x)^n$$
Integrate with respect to $x$
$$\frac{x^{n+1}}{n+1}-1=-\binom{n}{0}(1-x)+\binom{n}{1}\frac{(1-x)^2}{2}-\binom{n}{2}\frac{(1-x)^3}{3}+\cdots +(-1)^{n-1}\frac{(1-x)^{n+1}}{n+1}$$
How do I solve it? Help me, please!
| We put
$$S_n(x) = \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1}
(1-x)^p.$$
Working first with the coefficient on $[x^q]$ where $1\le q\le n$
we see that it is
$$\sum_{p=q}^n (-1)^{p+1} \frac{1}{p}
{n\choose p} {p\choose q} (-1)^q.$$
Now
$${n\choose p} {p\choose q} =
\frac{n!}{(n-p)! \times q! \times (p-q)!}
= {n\choose q} {n-q\choose n-p}$$
so we find
$$ (-1)^q {n\choose q}
\sum_{p=q}^n (-1)^{p+1} \frac{1}{p}
{n-q\choose n-p}
\\ = (-1)^q {n\choose q}
\sum_{p=0}^{n-q} (-1)^{p+q+1} \frac{1}{p+q}
{n-q\choose n-p-q}
\\ = {n\choose q}
\sum_{p=0}^{n-q} (-1)^{p+1} \frac{1}{p+q}
{n-q\choose p}.$$
Introducing
$$f(z) =
\frac{(-1)^{n-q+1}
(n-q)!}{z+q} \prod_{k=0}^{n-q} \frac{1}{z-k}$$
we have
$$\sum_{p=0}^{n-q} \mathrm{Res}_{z=p} f(z)
= \sum_{p=0}^{n-q}
\frac{(-1)^{n-q+1} (n-q)!}{p+q}
\prod_{k=0}^{p-1} \frac{1}{p-k}
\prod_{k=p+1}^{n-q} \frac{1}{p-k}
\\ = \sum_{p=0}^{n-q}
\frac{(-1)^{n-q+1} (n-q)!}{p+q}
\frac{1}{p!} (-1)^{n-q-p} \frac{1}{(n-q-p)!}
\\ = \sum_{p=0}^{n-q} (-1)^{p+1} \frac{1}{p+q}
{n-q\choose p}.$$
This is the target sum omitting the binomial coefficient in front. Now
the residue at infinity of $f(z)$ is clearly zero and hence the sum
must be (residues sum to zero)
$$- \mathrm{Res}_{z=-q} f(z) =
- (-1)^{n-q+1}
(n-q)! \prod_{k=0}^{n-q} \frac{1}{-q-k}
\\ = - (n-q)! \prod_{k=0}^{n-q} \frac{1}{q+k}
= -(n-q)! \frac{(q-1)!}{n!}.$$
Restoring the binomial coefficient in front we thus have
$$[x^q] S_n(x) = -{n\choose q} (n-q)! \frac{(q-1)!}{n!}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
[x^q] S_n(x) = - \frac{1}{q},}$$
as claimed. Comntinuing with the constant coefficient we find
$$[x^0] S_n(x) =
\sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1}
[x^0] (1-x)^p
= \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1}.$$
Using the same technique as before we introduce
$$g(z) =
\frac{(-1)^{n+1} n!}{z} \prod_{k=0}^{n} \frac{1}{z-k}$$
We get for
$$\sum_{p=1}^n \mathrm{Res}_{z=p} g(z)
= \sum_{p=1}^n
\frac{(-1)^{n+1} n!}{p}
\prod_{k=0}^{p-1} \frac{1}{p-k}
\prod_{k=p+1}^{n} \frac{1}{p-k}
\\ = \sum_{p=1}^n
\frac{(-1)^{n+1} n!}{p}
\frac{1}{p!} (-1)^{n-p} \frac{1}{(n-p)!}
\\ = \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1}.$$
This is the target sum. Now the residue at infinity is zero
so this sum must be equal to
$$- \mathrm{Res}_{z=0} g(z)
= - \mathrm{Res}_{z=0}
\frac{(-1)^{n+1} n!}{z^2} \prod_{k=1}^{n} \frac{1}{z-k}
\\ = (-1)^{n} n!
\left.\left(
\prod_{k=1}^{n} \frac{1}{z-k}\right)'\right|_{z=0}
\\ = (-1)^{n} n!
\left.\left(
\prod_{k=1}^{n} \frac{1}{z-k} \sum_{k=1}^n \frac{1}{k-z}
\right)\right|_{z=0}
= (-1)^n n!
\times \frac{(-1)^n}{n!} \sum_{k=1}^n \frac{1}{k}.$$
We have shown that (with harmonic numbers)
$$\bbox[5px,border:2px solid #00A000]{
[x^0] S_n(x) = \sum_{k=1}^n \frac{1}{k} = H_n,}$$
which concludes the argument. If desired we may write
this as
$$S_n(x) = \sum_{k=1}^n \frac{1}{k}
- \sum_{q=1}^n \frac{1}{q} x^q$$
or
$$\bbox[5px,border:2px solid #00A000]{
S_n(x) = \sum_{q=1}^n \frac{1}{q} (1 - x^q).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find $a,b,c\in\mathbb{C}$, which minimize the value of the integral $\int_{-1}^1 |x^3-a-bx-cx^2|^2dx.$ We're working in the Hilbert Space $L^2([-1,1])$, we have $f(x)=x^3$ and $g(x)=cx^2 + bx + a$ with inner product
$$\langle f , g \rangle = \int_{-1}^1 |x^3-a-bx-cx^2|^2dx.$$
How does one go about finding the minimum?
| This is a simple least squares problem. You are looking for the projection of $f(x)=x^3$ on the subspace $S = span\{1,x,x^2\}$. The formula you presented is for $\|f-g\|^2$, not for $\langle f,g\rangle$. The coefficients $a,b,c$ will be the solutions of the linear system
$$
\begin{pmatrix}
\langle 1,1\rangle & \langle 1,x\rangle & \langle 1,x^2\rangle\\
\langle x,1\rangle & \langle x,x\rangle & \langle x,x^2\rangle\\
\langle x^2,1\rangle & \langle x^2,x\rangle & \langle x^2, x^2\rangle
\end{pmatrix}
\begin{pmatrix}a \\ b\\c\end{pmatrix}=
\begin{pmatrix} \langle f,1\rangle \\\langle f,x\rangle \\ \langle f,x^2 \rangle\end{pmatrix}\Leftrightarrow
$$
$$
\left(
\begin{array}{ccc}
2 & 0 & \frac{2}{3} \\
0 & \frac{2}{3} & 0 \\
\frac{2}{3} & 0 & \frac{2}{5} \\
\end{array}
\right)\begin{pmatrix}a \\ b\\c\end{pmatrix}=\begin{pmatrix}0 \\ 2/5\\0\end{pmatrix}.
$$
Solving the system you get $g(x)=\frac 35 x$. Please note that solving this system is the same as computing the stationary points of
$$
Q(a,b,c) = \int_{-1}^1 |f(x)-a-bx-cx^2|^2 dx.
$$
If you a use an orthogonal base of $S$ things will be even simpler.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
kangaroo maths competition How many three-digit positive integers $ABC$ exist, such that $(A + B)^c$ is a three-digit integer and an integer power of $2$?
Note: An integer power of $2$ is a number in the form $2^k$
, where $k$ is an integer.
(A) $15$
(B) $16$
(C) $18$
(D) $20$
(E) $21$
| Here is complete list of $21$ solutions.
*
*$2^7=128:$
*
*$2^7=(1+1)^7\implies ABC=117$
*$2^7=(2+0)^7\implies ABC=207$
*$2^8=4^4=16^2=256:$
*
*$2^8=(1+1)^8\implies ABC=118$
*$2^8=(2+0)^8\implies ABC=208$
*$4^4=(1+3)^4\implies ABC=134$
*$4^4=(2+2)^4\implies ABC=224$
*$4^4=(3+1)^4\implies ABC=314$
*$4^4=(4+0)^4\implies ABC=404$
*$16^2=(7+9)^2\implies ABC=792$
*$16^2=(8+8)^2\implies ABC=882$
*$16^2=(9+7)^2\implies ABC=972$
*$2^9=8^3=512:$
*
*$2^9=(1+1)^9\implies ABC=119$
*$2^9=(2+0)^9\implies ABC=209$
*$8^3=(1+7)^3\implies ABC=173$
*$8^3=(2+6)^3\implies ABC=263$
*$8^3=(3+5)^3\implies ABC=353$
*$8^3=(4+4)^3\implies ABC=443$
*$8^3=(5+3)^3\implies ABC=533$
*$8^3=(6+2)^3\implies ABC=623$
*$8^3=(7+1)^3\implies ABC=713$
*$8^3=(8+0)^3\implies ABC=803$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Every real matrix $A$ is the linear combination of $4$ orthogonal matrices Question:
Prove that every matrix $A\in M_n(\mathbb R)$ is the linear combination of $4$ orthogonal matrices $X, Y, Z, W$ , i.e. $A=aX+bY+cZ+dW$ for some $a,b,c,d\in\mathbb R$.
This problem is taken from a forum and this is my paraphrase. It is not obviously true. But I think the proof must invoke the singular-value decomposition (SVD) of a real matrix, but it's unclear to me what the next step is. Any idea is appreciated. Many thanks.
| In view of SVD, we may assume that $A\neq 0$ and $A = \operatorname{diag}(\lambda_1, \cdots, \lambda_n)$ such that $\lambda_i$'s are non-negative, and $\lambda_n$ is the largest among $\lambda_i$'s.
Under this assumption, we have $\lambda_n > 0$. Now write $\lambda_{2i-1} = a_i - b_i$ and $\lambda_{2i} = a_i + b_i$. In case $n$ is odd, we also set $a_{(n+1)/2} = 2\lambda_n$ and $b_{(n+1)/2} = \lambda_n$. This gives
$$ A = \operatorname{diag}(a_1, a_1, a_2, a_2, \cdots) - \operatorname{diag}(b_1, -b_1, b_2, -b_2, \cdots). $$
Now we notice that $|a_i| \leq 2\lambda_n$ and $|b_i| \leq \lambda_n$, aod so, there exist $\alpha_i$ and $\beta_i$ such that $a_i = 2\lambda_n \cos\alpha_i$ and $b_i = \lambda_n \cos\beta_i$. Now the trick is to consider matrices
$$ R_{\theta} = \begin{pmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix} \qquad \text{and} \qquad S_{\theta} = \begin{pmatrix} \cos \theta & \sin\theta \\ \sin\theta & -\cos \theta \end{pmatrix}. $$
Then both $R_\theta$ and $S_\theta$ are orthogonal. Moreover, we have
$$R_\theta + R_{-\theta} = (2 \cos\theta) \operatorname{diag}(1, 1) \qquad \text{and} \qquad S_\theta + S_{-\theta} = (2 \cos\theta) \operatorname{diag}(1, -1).$$
Now using this, we may write
$$ \operatorname{diag}(a_1, a_1, a_2, a_2, \cdots) = \lambda_n \begin{pmatrix} R_{\alpha_1} & & \\ & R_{\alpha_2} & \\ & & \ddots \end{pmatrix} + \lambda_n \begin{pmatrix} R_{-\alpha_1} & & \\ & R_{-\alpha_2} & \\ & & \ddots \end{pmatrix} $$
and
$$ \operatorname{diag}(b_1, -b_1, b_2, -b_2, \cdots) = \frac{\lambda_n}{2} \begin{pmatrix} S_{\beta_1} & & \\ & S_{\beta_2} & \\ & & \ddots \end{pmatrix} + \frac{\lambda_n}{2} \begin{pmatrix} S_{-\beta_1} & & \\ & S_{-\beta_2} & \\ & & \ddots \end{pmatrix}. $$
When $n$ is odd, the last block in the diagonal in each block matrix is the $1\times 1$ matrix with the entry $1$. Now all these four block matrices are orthogonal, and therefore the desired claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}=\int \frac{\sqrt{x+1}}{x}dx-\log\left (x\right )$, so I only have to deal with $\int \frac{\sqrt{x+1}}{x}dx$.
Setting $u=\sqrt{x+1}$, we obtain $du=\frac{1}{2\sqrt{x+1}}dx$, therefore $dx=2udu$ and $x=u^2-1$, so
$\int\frac{\sqrt{x+1}}{x}dx=\int\frac{2u^2}{u^2-1}du=\int \left(2+\frac{2}{u^2-1}\right)du=2\sqrt{x+1}+\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=$
$=2\sqrt{x+1}+\log \left (\sqrt{x+1}-1\right )-\log \left (\sqrt{x+1}+1\right )=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )$
But then $\int \frac{dx}{\sqrt{x+1}+1}=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )-\log\left (x\right )=2\sqrt{x+1}+\log\left ( \frac{x+2-2\sqrt{x+1}}{x^2} \right )$, which is not what I am supposed to obtain. The actual answer is $2\sqrt{x+1}-2\log\left ( 1+\sqrt{x+1} \right )$.
Where is my mistake?
(I already know a correct way to solve it, I just want to know where I am committing a mistake).
| But you have been right all along!
Notice that
$$\log\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} - \log x = \log\frac{\sqrt{x+1}-1}{x(\sqrt{x+1}+1)} = \log \frac{1}{(\sqrt{x+1}+1)^2} = -2\log(\sqrt{x+1}+1)$$
by your very first step.
Also, I would suggest that you use $\int\frac{dx}{x} = \log|x|$ (with the absolute value sign, to extend the domain).
What I would also suggest is to use an online function grapher if you are not sure that your solution is correct. If two functions have identical graphs or if they differ by a constant value, then both are correct solutions of the integral.
| {
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"url": "https://math.stackexchange.com/questions/3132934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Derivative of a product of trig functions: $3\sin(x)\cot(x)$ I am having a lot of trouble understanding on how to find the derivative of $3\sin x\cot x$.
I end up with $-3\cos x/\sin^2 x$
| You need to use the product rule:
$$
(3\sin{x}\cot{x})'=\\
3(\sin{x})'\cot{x}+3\sin{x}(\cot{x})'=\\
3\cos{x}\cot{x}+3\sin{x}(-\csc^2{x})=\\
3\cos{x}\cot{x}-3\sin{x}\csc^2{x}=\\
3\cos{x}\frac{\cos{x}}{\sin{x}}-3\sin{x}\frac{1}{\sin^2{x}}=\\
3\frac{\cos^2{x}}{\sin{x}}-3\frac{1}{\sin{x}}=\\
3\frac{1}{{\sin{x}}}(\cos^2{x}-1)=\\
3\frac{1}{{\sin{x}}}(1-\sin^2{x}-1)=-3\sin{x}.
$$
A much simpler way to do this, as was mentioned in the comments section, is to notice that $3\sin{x}\cot{x}=3\sin{x}\frac{\cos{x}}{\sin{x}}=3\cos{x}$ which is trivial to differentiate:
$$
(3\cos{x})'=3(\cos{x})'=3(-\sin{x})=-3\sin{x}.
$$
| {
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"url": "https://math.stackexchange.com/questions/3133189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does this pattern of summing polygonal numbers to get a square repeat indefinitely? I am using the table of polygonal numbers on this site:
http://oeis.org/wiki/Polygonal_numbers
The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.
Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
$$1+1+1+1=4=2^2$$
$$3+4+5+6+7=25=5^2$$
$$6+9+12+15+18+21=81=9^2$$
$$10+16+22+28+34+40+46=196=14^2$$
...
The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.
1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)
2-Do we know why we have such a pattern?
Edit 03-05-2019
Following the suggestion of Eleven-Eleven, I looked for other patterns similar to the one above. I found one that is even simpler. This time we skip the triangular numbers when we calculate the sum. We start with the squares and sum up enough terms to get another square.
$$1=1^2=T_{1}^2$$
$$4+5=3^2=9=T_{2}^2$$
$$9+12+15=6^2=36=T_{3}^2$$
$$16+22+28+34=10^2=100=T_{4}^2$$
$$25+35+45+55+65=15^2=225=T_{5}^2$$
We see the same pattern as above. The square with index $(n+1)$ requires the addition of one more term than the square with index $n$. The number of elements to sum up to get the square $T_{n}^2$ is simply $n$.
Can this be stated as the following theorem?
The square of a triangular number $T_{n}$ can be expressed as the sum of $n$ polygonal numbers excluding the triangular number itself.
| I'm adding another answer because I wanted to try and use the formula for polygonal numbers as given here in an attempt to better answer the original question by the OP. In both cases above we exploited the arithmetical sequence properties and only used Triangular numbers. Can we get the same result strictly using the polygonal number formula
$$P(s,n)=\frac{n^2(s-2)-n(s-4)}{2}$$
where $n$ represents the sequence number and $s$ represents the number of sides in a polygonal number. So, lets look at an example. In the second row, the OP has that
$$3+4+5+6+7=25=(T_2+2)^2$$
Well, $3,4,5,6,$ and $7$ are the second triangular, square, pentagonal, hexagonal, and heptagonal numbers, so $n=2$ and $s$ is indexed... therefore, we have
\begin{eqnarray*}\sum_{k=1}^5{P(k+2,2)}&=&\sum_{k=1}^5{\left[\frac{4((k+2)-2)-2((k+2)-4)}{2}\right]}\\&=&\sum_{k=1}^5{\left[\frac{4k-2(k-2)}{2}\right]}\\&=&\sum_{k=1}^5\left(k+2\right)\\&=&\sum_{k=1}^5{k}\\&=&T_5
\end{eqnarray*}
Now, this can be generalized. The claim is
$$\sum_{k=1}^{m+3}{P((k+2),m)}=\left(T_m+m\right)^2$$
Proof:
\begin{eqnarray*}\sum_{k=1}^{m+3}{P((k+2),m)}&=&\sum_{k=1}^{m+3}\left[\frac{m^2((k+2)-2)-m((k+2)-4)}{2}\right]\\&=&\sum_{k=1}^{m+3}\left[\frac{m^2k-mk+2m}{2}\right]\\&=&\frac{m^2}{2}\sum_{k=1}^{m+3}{k}-\frac{m}{2}\sum_{k=1}^{m+3}{k}+m\sum_{k=1}^{m+3}1\\&=&\frac{(m-1)m}{2}\sum_{k=1}^{m+3}k+m(m+3)\\&=&\frac{(m-1)m}{2}\frac{(m+3)(m+4)}{2}+\frac{4m(m+3)}{4}\\&=&\frac{m(m+3)}{4}\left[(m-1)(m+4)+4\right]\\&=&\frac{m(m+3)}{4}\left(m^2+3m\right)\\&=&\frac{m^2(m+3)^2}{2^2}\\&=&\left[\frac{m(m+3)}{2}\right]^2\\&=&\left[\frac{m(m+1+2)}{2}\right]^2\\&=&\left[\frac{m(m+1)+2m}{2}\right]^2\\&=&\left(T_m+m\right)^2
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| The easiest way to see this : Consider the number of heads we have in the first $8$ coins.
*
*If this number is even, we need a tail, we have probability $\frac{1}{2}$
*If this number is odd, we need a head, we have probability $\frac{1}{2}$
Hence no matter what the $8$ coins delivered, we have probability $\frac{1}{2}$ , which is the answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 5
} |
Getting different answers when using product rule and limit substitution than I do with quotient rule I'm trying to differentiate $$y = \frac{x+1}{x-1}.$$
Using quotient rule:
http://prntscr.com/mt90yc
Using product:
http://prntscr.com/mt91dd
Using limit definition:
http://prntscr.com/mt91ih
I get $-2$ for product and limit, but I get $\frac{-2}{x^{2}-2x+1}$ using quotient.
| It means that you're doing something wrong.
$$
\left(\frac{x+1}{x-1}\right)'=
\frac{(x+1)'(x-1)-(x+1)(x-1)'}{(x-1)^2}=\\
\frac{x-1-(x+1)}{(x-1)^2}=
\frac{x-1-x-1}{(x-1)^2}=-\frac{2}{(x-1)^2}
$$
$$
[(x+1)(x-1)^{-1}]'=\\
(x+1)'(x-1)^{-1}+(x+1)[(x-1)^{-1}]'=\\
(x-1)^{-1}+(x+1)(-1)(x-1)^{-2}=\\
\frac{1}{x-1}-\frac{x+1}{(x-1)^2}=
\frac{x-1}{(x-1)^2}-\frac{x+1}{(x-1)^2}=\\
\frac{x-1-x-1}{(x-1)^2}=-\frac{2}{(x-1)^2}
$$
$$
\lim\limits_{\Delta x \rightarrow 0}\frac{\frac{x+\Delta x+1}{x+\Delta x-1}-\frac{x+1}{x-1}}{\Delta x}=
\lim\limits_{\Delta x \rightarrow 0}\frac{\frac{(x+\Delta x+1)(x-1)-(x+1)(x+\Delta x-1)}{(x+\Delta x-1)(x-1)}}{\Delta x}=\\
\lim\limits_{\Delta x \rightarrow 0}\frac{x^2-x+x\Delta x-\Delta x+x-1-(x^2+x\Delta x-x+x+\Delta x-1)}{\Delta x(x+\Delta x-1)(x-1)}=\\
\lim\limits_{\Delta x \rightarrow 0}\frac{x^2+x\Delta x-\Delta x-1-(x^2+x\Delta x+\Delta x-1)}{\Delta x(x+\Delta x-1)(x-1)}=
\lim\limits_{\Delta x \rightarrow 0}\frac{x^2+x\Delta x-\Delta x-1-x^2-x\Delta x-\Delta x+1}{\Delta x(x+\Delta x-1)(x-1)}=\\
\lim\limits_{\Delta x \rightarrow 0}\frac{-2\Delta x}{\Delta x(x+\Delta x-1)(x-1)}=
\lim\limits_{\Delta x \rightarrow 0}\frac{-2}{(x+\Delta x-1)(x-1)}=\\
\frac{-2}{(x+0-1)(x-1)}=-\frac{2}{(x-1)^2}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find value of $\frac{\sum AA_1 \cos\left(\frac{A}{2}\right)}{\sum \sin A}$ Triangle $\Delta ABC$ is inscribed in a circle of radius one unit. If the internal angle bisectors of angles $\angle A, \angle B,\angle C$ meets the circle at the points $A_1,B_1,C_1$ respectively. Find value of $$S=\frac{\sum AA_1 \cos\left(\frac{A}{2}\right)}{\sum \sin A}$$
My try:
Letting $BC=a, AB=c, AC=b$
We have $$AD=\frac{2bc}{b+c}\cos\left(\frac{A}{2}\right)$$
Hence $$AA_1=AD+DA_1=\frac{2bc}{b+c}\cos\left(\frac{A}{2}\right)+DA_1$$
But how to find $DA_1$?
| Note that $A_1,$ $B_1,$ $C_1$ are the midpoints of $\overset{\huge\frown}{BC},$ $\overset{\huge\frown}{CA},$ $\overset{\huge\frown}{AB}$ respectively.
Therefore, considering $\triangle A_1BC,$ $A_1B=A_1C$. Now we can use law of cosines in $\triangle AA_1B$ and $\triangle AA_1C$ to get,
$$(A_1B)^2=(AA_1)^2+c^2-2\cdot AA_1\cdot c\cos\frac A2$$
$$(A_1C)^2=(AA_1)^2+b^2-2\cdot AA_1\cdot b\cos\frac A2$$
Using these two, we can easily get $$AA_1\cos\left(\frac A2\right)=\frac{(b+c)}2$$
Also from law of sines we have $$\frac{\sin A}a=\frac{\sin B}b=\frac{\sin C}c=2R$$
where $R$ is the radius of the circumcircle.
Hence, $$\frac{\sum AA_1\cos\left(\frac A2\right)}{\sum\sin A}\equiv R\cdot\frac{\sum(b+c)}{\sum a}$$
Can you take it from here?
| {
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"answer_count": 1,
"answer_id": 0
} |
Solving $\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$
$$\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$$
I tried much of elementary methods to solve above integral but is not advancing.
Any methods from elementary to advanced are appreciated.
| Here is a way to simplify @Olivier Oloa's answer for the integral that is written in terms of the Gauss hypergeometric function.
Starting from (4) from a previous answer I gave here, it was shown that
$$_2F_1 \left (1, \frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right ) = \frac{2^{4/3}}{\sqrt{\pi}} \Gamma \left (\frac{4}{3} \right ) \Gamma \left (\frac{7}{6} \right ).$$
Thus
\begin{align}
I &= \frac{3 \sqrt{\pi}}{2^2} \cdot \frac{2^{4/3}}{\sqrt{\pi}} \cdot \frac{\Gamma (\frac{4}{3}) \Gamma (\frac{7}{6}) \Gamma (\frac{2}{3})}{\Gamma (\frac{1}{6})}\\
&= \frac{1}{2^{2/3} 6} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\
&= \frac{1}{2^{2/3} 6} \cdot \frac{\pi}{\sin (\frac{\pi}{3})}\\
&= \frac{\pi}{2^{2/3} 3^{3/2}},
\end{align}
giving the required simplification.
Actually, the reason for this coincidence is, as I have just discovered, unsurprising. As @Zacky correctly observes in the comments section of the linked question, the integral considered here is, to within a numerical factor, just that considered in the linked question, namely
$$\int_0^1 \frac{x^{2/3}}{\sqrt[3]{1 - x} (1 - x + x^2)} \, dx = 2^{5/3} \int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\sin x}}{4 - \sin^2 x} \, dx.$$
| {
"language": "en",
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"answer_count": 2,
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} |
Solve $\frac{dx}{dt}=3x-5y$ $\frac{dy}{dt}=5x-3y$ Given that $x(0)=1$ and $y(0)=1$
Solve
$$\frac{dx}{dt}=3x-5y$$ $$\frac{dy}{dt}=5x-3y$$ Given that $x(0)=1$ and $y(0)=1$
So I've got to the stage where i have two general solutions and I now need the particular solutions, but I don't know how to find one of the variables and both of the general solutions for $x$ and $y$ are the same which makes me think I've made a mistake. My workings are below any help would be great.
$$\frac{dx}{dt}=3x-5y$$
$$\therefore \frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 5\frac{dy}{dt}$$
$$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} -5 \left(5x-3y \right) = 3\frac{dx}{dt} -25x+15y$$
$$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} -25x+15 \left(\frac 3 5 x - \frac 1 5 \frac{dx}{dt}\right)$$
$$\frac{d^2x}{dt^2} +16x=0$$
$$\lambda ^2 +16=0 $$
$$\therefore \lambda = 4i$$
$$x=A\sin (4t) + B\cos(4t)$$
I then did the same for $y$ and obtained the same differential equation of $$\frac{d^2y}{dt^2} +16y=0$$ and thus $y=A\sin (4t) + B\cos(4t)$
Now using the initial conditions of $x(0)=1$ and $y(0)=1$ i found that $B=1$ but how do I find $A$ and won't it just be the same for both equations?
| There are several ways to solve this system. Here is yet another approach using eigenvalue/vector methods from linear algebra.
Let's re-write the system as:
$$ \frac{d\vec{r}}{dt} = \begin{pmatrix}3 &-5\\5 &-3 \end{pmatrix}\vec{r}(t) = \pmb{A}\,\vec{r}(t)$$
where $\vec{r}(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatrix}$. Then, solve the characteristic equation,
$$ \det(\pmb{A}-\lambda\pmb{I}) = 0 \Rightarrow \begin{vmatrix}3-\lambda &-5\\5 &-3-\lambda \end{vmatrix} = 0 \Rightarrow \lambda^2 +16 = 0 \Rightarrow \lambda = \pm \,4i$$
We find eigenvalues of $\pm \,4i$. Choosing $\lambda = 4i $ we find an associated eigenvector solving
$$(\pmb{A}-\lambda\pmb{I})\vec{v} = (\pmb{A}-4i\pmb{I})\vec{v}=\vec{0} \Rightarrow \begin{pmatrix}3-4i &-5\\5 &-3-4i \end{pmatrix}\vec{v} = \begin{pmatrix}0\\0\end{pmatrix}$$
where after some algebra $\vec{v} = \begin{pmatrix}5 \\ 3-4i\end{pmatrix}$. Looking at the real and imaginary parts of $\vec{v}\, e^{4it}$ will give use two linearly independent solutions to our system. With some algebra we find,
$$\vec{v}\, e^{4it} = \begin{pmatrix}5 \\ 3-4i\end{pmatrix}(\cos 4t +i\sin 4t) = \begin{pmatrix}5\cos 4t \\ 3\cos 4t + 4\sin 4t\end{pmatrix} +i \begin{pmatrix}5\sin 4t \\ 3\sin 4t -4\cos 4t\end{pmatrix} $$and thus our linear independent solutions are $\vec{r_1}(t) = \begin{pmatrix}5\cos 4t \\ 3\cos 4t + 4\sin 4t\end{pmatrix}$ and $\vec{r_2}(t) =\begin{pmatrix}5\sin 4t \\ 3\sin 4t -4\cos 4t\end{pmatrix} $.
With these we now seek a solution to the system of the form $\vec{r}(t) = C_1\vec{r_1}(t)+C_2\vec{r_2}(t) $ where $C_1$ and $C_2$ are constants that solve the initial value problem $x(0)=1$ and $y(0)=1$, or matching our matrix/vector form $\vec{r}(0)=\begin{pmatrix}1\\1\end{pmatrix}$. This gives,
$$\vec{r}(0) = C_1\begin{pmatrix}5\\3\end{pmatrix} +C_2\begin{pmatrix}0\\-4\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix} \Rightarrow \begin{pmatrix}5 &0\\3 &-4 \end{pmatrix}\begin{pmatrix}C_1\\C_2\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix}$$ and solving these linear equations gives $C_1 = \frac{1}{5}, C_2 = \frac{-1}{10}$. Finally substituting these into the expression for $\vec{r}(t)$ gives
$$ \vec{r}(t) = \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\frac{1}{5}\begin{pmatrix}5\cos 4t \\ 3\cos 4t + 4\sin 4t\end{pmatrix} - \frac{1}{10}\begin{pmatrix}5\sin 4t \\ 3\sin 4t -4\cos 4t\end{pmatrix} = \begin{pmatrix}\cos 4t - \frac{1}{2}\sin 4t\\\cos 4t + \frac{1}{2}\sin 4t\end{pmatrix}.$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to use a conditional statement in an infinite series to approach a specific limit If I were to sum the infinite series:
$$\sum_{i=1}^\infty \frac{1}{2^n} = \frac{1}{2^1}+\frac{1}{2^2}...\frac{1}{2^\infty}$$
but as soon as the previous partial sum
$\sum_{i=1}^\infty \frac{1}{2^{n-1}}$ becomes greater than 0.9 I subtract $\frac{1}{2^n}$ instead of adding it, forcing the series to approach 0.9.
What would would be the simplest form to write that, possibly using Iverson brackets? I'm not sure how to deal with the negative value though.
$$\sum_{i=1}^\infty \frac{1}{2^n}.[\sum_{i=1}^\infty \frac{1}{2^{n-1}} < 0.9]$$
or is this valid?
$$\sum_{i=1}^\infty f(n)=\begin {cases} \frac{1}{2^n} & \sum_{i=1}^\infty \frac{1}{2^{n-1}} < 0.9\\
\frac{-1}{2^n} & \sum_{i=1}^\infty \frac{1}{2^{n-1}} > 0.9 \end {cases}$$
or maybe:
$$\sum_{i=1}^\infty f(n)=\begin {cases} \frac{1}{2^n} & \sum_{i=1}^{n-1} \frac{1}{2^n} < 0.9\\
\frac{-1}{2^n} & \sum_{i=1}^{n-1} \frac{1}{2^n} > 0.9 \end {cases}$$
| Claim:
The sign of the $k$-th term is precisely $(-1)^{\lfloor\tfrac{k+1}{2}\rfloor}$, excep for $k=1$ and $k=2$. To be precise, setting
$$f(1):=\frac12,
\qquad
f(2):=\frac14,
\qquad\text{ and }\qquad
f(k):=\frac{(-1)^{\lfloor\tfrac{k+1}{2}\rfloor}}{2^k}
\quad
\text{ for all }k>2,$$
yields the desired sum. This means that for all $m\geq1$ the partial sum $S_m:=\sum_{k=1}^mf(k)$ satisfies
$$\begin{array}{ccc}
S_m<0.9\quad&\implies&\quad f(m+1)>0,\\
S_m>0.9\quad&\implies&\quad f(m+1)<0.\tag{1}
\end{array}$$
Proof:
This is clear for $m=1$ and $m=2$. To simplify $(1)$ note that $S_m\neq0.9$ for all $m$ because the denominator of $S_m$ is a power of $2$ whereas $0.9=\frac{3}{10}$. So $(1)$ is equivalent to
$$S_m<0.9\qquad\iff\qquad f(m+1)>0.$$
Moreover, the sign of $f(m+1)$ can be expressed simply in terms of $m$ because
$$f(m+1)>0
\quad\iff\quad
(-1)^{\lfloor\tfrac{m+2}{2}\rfloor}>0
\quad\iff\quad
m+2\equiv2,3\pmod{4}.$$
For $m>2$, to get rid of the first two irregular terms in the partial sum $S_m$, set $T_m:=\sum_{k=3}^mf(k)$. Then $T_m=S_m-\tfrac34$ and $(1)$ now simplifies to
$$T_m>\frac{3}{20}\qquad\iff\qquad m\equiv0,1\pmod{4}.$$
Writing out the first few summands of $T_m$ shows that if $m$ is even we can group the summands into pairs to get a simple geometric sum, for which it is easy to find a closed form:
\begin{eqnarray*}
T_{2n}&=&\left(\frac{1}{8}+\frac{1}{16}\right)
-\left(\frac{1}{32}+\frac{1}{64}\right)
+\left(\frac{1}{128}+\frac{1}{256}\right)
-\cdots\\
&=&\sum_{k=2}^n\left(-\frac{1}{4}\right)^k\left(2+1\right)
=3\sum_{k=2}^n\left(-\frac{1}{4}\right)^k\\
&=&3\frac{\left(-\tfrac{1}{4}\right)^2-\left(-\tfrac{1}{4}\right)^{n+1}}{1-\left(-\tfrac14\right)}
=\frac{3}{20}\left(1-\left(-\frac{1}{4}\right)^{n-1}\right)\\
&=&\frac{3}{20}-\frac{3}{20}\left(-\frac{1}{4}\right)^{n-1}\tag{2}
\end{eqnarray*}
In particular, this last expression shows that $T_{2n}>\frac{3}{20}$ if $n$ is even and $T_{2n}<\frac{3}{20}$ if $n$ is odd. This proves $(1)$ for all even $m$. For odd $m$, we simply add one summand to $(2)$ to get
\begin{eqnarray*}
T_{2n+1}&=&
\frac{3}{20}+\frac35\left(-\frac{1}{4}\right)^n+\frac{(-1)^{n+1}}{2^{2n+1}}\\
&=&\frac{3}{20}+(-1)^n\frac35\left(\frac{1}{4}\right)^n-\frac12\left(\frac{1}{4}\right)^n\\
&=&\frac{3}{20}+\left((-1)^n\frac35-\frac12\right)\left(\frac{1}{4}\right)^n,
\end{eqnarray*}
which shows that $T_{2n+1}>\tfrac{3}{20}$ if $n$ is even and $T_{2n+1}<\tfrac{3}{20}$ if $n$ is odd. This proves $(1)$ for all odd $m$, and hence for all $m$.
| {
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"url": "https://math.stackexchange.com/questions/3139360",
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"source": "stackexchange",
"question_score": "1",
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Solve the inequality $x^2 - 3 > 0$ For the inequality $x^2 - 3 > 0$, we have
\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}
Therefore,
\begin{align} x > -\sqrt 3 \end{align}
and
\begin{align} x > \sqrt 3 \end{align}
But, this is clearly wrong as we should get $x < -\sqrt 3$ and $x > \sqrt 3$ as the two intervals. What have I done wrong?
| HINT
Remember $ab > 0$ if $a,b > 0$ or $a,b < 0$
| {
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Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic equation as irreducible with no repeating factors.
$\int \frac{x+4}{x^2 + 2x + 5}$
= $\frac {Ax + B} {x^2 + 2x + 5} $
= $Ax+B(x^2 + 2x + 5)$
= $Ax^3 + 2Ax^2 + 5Ax + Bx^2 + 2Bx + 5B$
= $Ax^3 + (2A + 2B)x^2 + (5A + 2B)x + 5B$
However I get stuck trying to solve my system of equations. This leads me to believe that I did the partial fractions improperly.
$\begin{bmatrix}
1 & 0 & 0 \\
2 & 2 & 0 \\
5 & 2 & 1 \\
0 & 5 & 4 \\
\end{bmatrix}$
| The denominator has no real roots, which means you'll try to rewrite as follows ($A,B\in\mathbb{R}$):
$$\frac{x+4}{x^2 + 2x + 5} = A\underbrace{\frac{\left(x^2 + 2x + 5\right)'}{x^2 + 2x + 5}}_{\to \ln}+\underbrace{\frac{B}{x^2 + 2x + 5}}_{\to \arctan}$$
where $\left(x^2 + 2x + 5\right)'=2x+2$, so this comes down to finding $A$ and $B$ such that:
$$x+4=A\left(2x+2\right)+B$$
Once you have $A$ and $B$, you split the integral in two (easy) parts. Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143709",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluating $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$ without expansions in limits
Evaluate $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$
One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{\log_e (1+x)^{1/x}} = e^{\frac{1}{x} (x-\frac{x^2}{2} -\frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $\frac{11e}{24}$ as an answer.
Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!
| Solution without expansions by the L'Hospital's rule only:
$$\lim_{x\rightarrow0}\frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}\left(\frac{\ln(1+x)}{x}\right)'+\frac{1}{2}e}{2x}=$$
$$=\lim_{x\rightarrow0}\frac{\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)}{x^2}+\frac{1}{2}e}{2x}=\lim_{x\rightarrow0}\frac{\left(\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)}{x^2}\right)'}{2}$$ because
$$\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)}{x^2}=e\lim_{x\rightarrow0}\frac{x-(1+x)\ln(1+x)}{x^2+x^3}=$$
$$=e\lim_{x\rightarrow0}\frac{1-\ln(1+x)-1}{2x+3x^2}=-\frac{e}{2}$$ and we can continue:
$$ \lim_{x\rightarrow0}\frac{\left(\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)}{x^2}\right)'}{2}=$$
$$=\frac{1}{2}\lim_{x\rightarrow0}\left(\frac{\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)^2}{x^2}+(1+x)^{\frac{1}{x}}\left(\frac{1}{(1+x)^2}-\frac{1}{1+x}\right)}{x^2}-\frac{2}{x^3}(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)\right)=$$
$$=\frac{e}{2}\lim_{x\rightarrow}\left(\frac{\frac{\left(\frac{x}{1+x}-\ln(1+x)\right)^2}{x^2}-\frac{x}{(1+x)^2}}{x^2}-\frac{2}{x^3}\left(\frac{x}{1+x}-\ln(1+x)\right)\right)=$$
$$=\frac{e}{2}\lim_{x\rightarrow0}\left(-\frac{3x+1}{x^2(1+x)^2}+\frac{2\ln(1+x)}{(1+x)x^2}+\frac{\ln^2(1+x)}{x^4}\right)=$$
$$=\frac{e}{2}\lim_{x\rightarrow0}\frac{(1+x)^2\ln^2(1+x)+(2x^3+2x^2)\ln(1+x)-3x^3-x^2}{x^4}=...=\frac{11e}{24}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
differentiating each term
$$\frac{d}{dx} (11x^5)=11\frac{d}{dy}(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4] = 4yx^3+x^4$$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4] = y^5 + 5xy^4$$
finding $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{-x}{y} = \frac{-[4yx^3+x^4] + [y^5+5xy^4]}{55x^4}$$
According to the website I'm using, "WebWork", this is wrong.
| differentiating each term
$$\frac{d}{dx} (11x^5)=11(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4\color{red}{\frac{dy}{dx}}] $$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4\color{red}{\frac{dy}{dx}}] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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} |
Solve the recurrence $b_1 = 2$ for $b_n = 3b_{n-1} + 5$
Solve the recurrence $b_1 = 2$ for $$b_n = 3b_{n-1} + 5$$
I've tried solving this problem using iteration, but the formula I get in the end is wrong. It is not a closed formula since there's still recurrence. I think my error is starts from the last line below. I'm not sure how to fix it. The formula I get in the end is:
$2(3^{n-1})+ 5((3^n - 1) / 3)$
which I got by setting $k = n -1$ and using $\sum_{k=0}^{n-1}3^k\ = (3^n-1)/2$
\begin{align*}
b_n&=3b_{n-1}+5\\
&=3(3b_{n-2}+5)+5\\
&=3^2b_{n-2}+3\cdot5+5\\
&=3^2(3b_{n-3}+5)+3\cdot5+5\\
&=3^3b_{n-3}+3^2\cdot5-3\cdot5+5\\
&\;\vdots\\
&=3^kb_{n-k}+3^{k-1}\cdot5+3^{k-2}\cdot5+\ldots+3\cdot5+5\\
\end{align*}
| Put $k =n-1$ and you get,
\begin{align}b_n &= 3^{n-1}b_1 + 3^{n-2} \cdot 5 + 3^{n-3} \cdot 5 + \dots +5 \\
&= 2 \cdot 3^{n-1} + 5(1 + 3 + 3^2 + \dots + 3^{n-2}) \\
&= 2\cdot 3^{n-1} + 5\left( \frac{3^{n-1}-1}{2}\right)\\
&= \frac{3^{n+1}-5}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$? This is a problem that I stumbled upon in one of my books.
Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):
$
\rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\
\rightarrow R\cos x \cos \alpha - R\sin x \sin \alpha = 5\cos x - \sin x \\
\rightarrow ➊\hspace{0.25cm}5 = \sqrt{26}\cos \alpha \\
\rightarrow ➋\hspace{0.25cm}{-}1 = \sqrt{26}\sin \alpha \\
$
So here are my question(s): $\\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $\cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
| This is just my way of saying what everyone else has already said.
\begin{align}
R &=\sqrt{1^2+5^2}=\sqrt{26} \\
\cos \alpha &= \dfrac{5}{\sqrt{26}} \\
\sin \alpha &= \dfrac{1}{\sqrt{26}} \\
\alpha &\approx 11.31^\circ \\
\hline
5\cos x - 1 \sin x &= 4 \\
\cos x \cos \alpha - \sin x \sin \alpha &= \dfrac{4}{\sqrt{26}} \\
\cos(x + \alpha) &= \dfrac{4}{\sqrt{26}} \\
x + 11.31^\circ &\approx \pm 38.33^\circ + n 360^\circ \\
\hline
x &\approx 38.33^\circ - 11.31^\circ = 27.02^\circ \\
x &\approx -38.33^\circ - 11.31^\circ + 360^\circ = 310.36^\circ
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.
Any ideas?
| Prove instead that, for $0<x<1$,
$$
e^x<1+\frac{1}{\frac{1}{x}-1}=1+\frac{x}{1-x}=\frac{1}{1-x}
$$
which is the same as proving that, for $0<x<1$,
$$
e^{1-x}<\frac{1}{x}
$$
that is, $e^x>ex$. Now differentiating is easier, isn't it? If $f(x)=e^x-ex$,
$$
f'(x)=e^x-e
$$
Thus $f'$ is negative for $0<x<1$. Since $f(1)=0$, the inequality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all polynomials such that $p(x^2-2x)=p(x-2)^2$
Find all polynomials $p\in \mathbb{C}[x]$ such that $$p(x^2-2x)=p(x-2)^2$$
We can not say anything specific about the degree since both sides are of the degree $2n$.
Also by copering the coeficients we see that the leading coefficent must be $1$.
Setting $$x^2-2x = x-2\implies x\in\{1,2\}$$
If $x= 2$ we get $p(0)=p(0)^2$ so $p(0)\in\{0,1\}$.
If $x= 1$ we get $p(-1)=p(-1)^2$ so $p(-1)\in\{0,1\}$.
Now it sems there is a lot of options. Also if $p$ is linear we have $$a(x^2-2x)+b= (ax-2a+b)^2 =a^2x^2+2a(b-2a)x+(b-2a)^2$$
so $a=a^2$ and $b=(b-2a)^2$ and $-2a=2a(b-2a)$ so $a=1$ and $b=1$ thus $p(x)=x+1$. But how to finish in general?
| EDIT: (I made a change of variable $x-1\mapsto x$ as suggested by @Hw Chu.) We need to solve $\displaystyle p((x-1)^2-1)=p(x-1-1)^2$ or $p(x^2-1)=p(x-1)^2$ equivalently. Let $$\displaystyle q(x) =p(x-1)= cx^k\prod_{1\le i\le N\\a_i\ne 0} (x-a_i)^{n_i},$$ where $c\ne 0$ and $a_j\ne a_k$ for $j\ne k$. Plugging this into the equation $q(x^2)=q(x)^2$, we obtain
$$
cx^{2k}\prod_{i=1}^N(x^2-a_i)^{n_i} =c^2x^{2k}\prod_{i=1}^N (x-a_i)^{2n_i},\\
\prod_{i=1}^N(x+\sqrt{a_i})^{n_i}(x-\sqrt{a_i})^{n_i}=c\prod_{i=1}^N (x-a_i)^{2n_i}.
$$ Now, we find that the LHS has $2N$ linear factors because $$
x+\sqrt{a_j}\ne x-\sqrt{a_j},\\
x\pm\sqrt{a_j} \ne x\pm\sqrt{a_k}
$$ for all $j\ne k$. Since the RHS has only $N$ linear factors, it follows $N=0$ and $c=1$. Thus, non-trivial solutions are $p(x)=q(x+1)=(x+1)^k$ for some $k\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $f(2019)$ in definite integration If $\displaystyle f(n)=\int^{1}_{0}(1+x+x^2+\cdots +x^{n-1})(1+3x+5x^2+\cdots +(2n-1)x^{n-1})dx$. Then $f(2019)$ is
What I tried:
$$1+x+x^2+\cdots +x^{n-1}=\frac{1-x^n}{1-x}$$
and $$1+3x+5x^2+\cdots +(2n-1)x^{n-1}=\frac{1}{1-x}+\frac{2(1-x^n)}{1-x}-\frac{(2n-1)x^n}{1-x}$$
How do I solve it? Help me, please.
| By making change of variable $x=t^2$, we obtain
$$\begin{align*}
f(n)&=2\int_0^1 t(1+t^2+\cdots +t^{2n-2})(1+3t^2+5t^4+\cdots +(2n-1)t^{2n-2})dt.
\end{align*}$$ Further making substitution $u=t+t^3+\cdots +t^{2n-1}$, we get
$$
f(n)=2\int_0^n u\ du=\left[u^2\right]^n_0=n^2.
$$ So $f(2019)=2019^2$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Entropy of the upper and lower bits of a square number Consider a uniformly random number $x<2^n$. Let $H_{\star}(n)$ denote the (base-$2$ Shannon) entropy of the first $n$ bits of $x^2$, and let $H^{\star}(n)$ denote the entropy of the rest of the bits of $x^2$. In other words, writing $x^2 = q2^n + r$ with $r<2^n$, $H_{\star}(n)$ denotes the entropy of $r$ while $H^{\star}(n)$ denotes the entropy of $q$.
From numerical experiments up to $n=25$ it seems as though
$$
H_{\star}(n)=\left\{
\begin{array}{@{}ll@{}}
n + 2^{(4 - n)/2} - 3, & \text{if}\ n\equiv0\mod2 \\
n + 3\cdot2^{(1 - n)/2} - 3, & \text{if}\ n\equiv1\mod2
\end{array}\right.
$$
I don't have an exact expression for $H^{\star}(n)$, but it seems as though (for $n>1$), $H^{\star}(n)>n-\frac{3}{4}$, and $H^{\star}(n)$ approaches $n-\frac{3}{4}$ as $n$ gets larger.
Unfortunately I'm totally at a loss about how to prove these statements in general, and my experimentation program can't go higher because of the exponential blow-up in memory use. Do these results hold in general, and if so does anyone know how to prove them? Also is there a more exact expression for $H^{\star}(n)$?
| With help from Peter Taylor's comment, I was able to prove the expression for $H_{\star}(n)$. I believe the approximation for $H^{\star}(n)$ could also be proven by noting that for $x>2^{n/2}$, $\sqrt{x^2-2^{n-1}}\leqslant \sqrt{q}$ or $\sqrt{q}\leqslant\sqrt{x^2+2^{n-1}}$, and then expanding these bounds as Taylor series to show that most of the higher bits of $\sqrt{q}$ are the same as the higher bits of $x$, hence the entropy of $q$ must be nearly the same as the entropy of $x$. Unfortunately the argument seems to get pretty messy so I have yet to formalize it.
Let $(x,y)$ denote the greatest common divisor of $x$ and $y$. The proof of the expression for $H_{\star}(n)$ relies on the following proposition.
Proposition. For two numbers $x,y$ with $(x,2^n)=2^m$, $x^2\equiv y^2 \mod 2^n$ iff $x\equiv\pm y\mod 2^{n-\min\{m+1,\lfloor n/2\rfloor\}}$.
proof. First suppose $(x,2^n)=2^m$ and $x\equiv\pm y\mod 2^{n-\min\{m+1,\lfloor n/2\rfloor\}}$, so that we can write $x=i2^m$ and $y = x+ k2^{n-\min\{m+1,\lfloor n/2\rfloor\}}$. Then
\begin{align}
y^2 &= \left(x+ k2^{n-\min\{m+1,\lfloor n/2\rfloor\}}\right)^2\\
&= x^2 + 2kx2^{n-\min\{m+1,\lfloor n/2\rfloor\}} + k^22^{2\left(n-\min\{m+1,\lfloor n/2\rfloor\}\right)}\\
&= x^2 + kj2^{n+m+1-\min\{m+1,\lfloor n/2\rfloor\}} + k^22^{n+\left(n-\min\{2m+2,2\lfloor n/2\rfloor\}\right)}\\
&\equiv x^2 \mod 2^n.
\end{align}
Now suppose $(x,2^n)=2^m$ and $x^2\equiv y^2 \mod 2^n$. Then $2^n|(x+y)(x-y)$. Certainly it must be the case that $2^{\lceil n/2 \rceil}=2^{n-\lfloor n/2 \rfloor}$ divides one of these terms. Now suppose $2^{m+2}$ divides both of these terms. Then we would also have $2^{m+2}|(x+y)+(x-y)=2x$, so $2^{m+1}|x$ contradicting our assumptions. Thus $2^{n-(m+1)}$ must divide either $(x+y)$ or $(x-y)$. Since $2^{n-\min\{m+1,\lfloor n/2\rfloor\}}$ equals either $2^{n-(m+1)}$ or $2^{n-\lfloor n/2 \rfloor}$, we in particular have $2^{n-\min\{m+1,\lfloor n/2\rfloor\}}$ divides $(x+y)$ or $(x-y)$, i.e. $x=\pm y\mod 2^{n-\min\{m+1,\lfloor n/2\rfloor\}}$.$$\tag*{$\blacksquare$}$$
Now back to the original problem.
\begin{align}
H_{\star}(n) &= -\frac{1}{2^n}\sum_{x=0}^{2^n-1}\log_2\left(\frac{\#\{x:x^2\equiv y^2\mod 2^n\}}{2^n}\right)\\
&=-\frac{1}{2^n}\sum_{x=0}^{2^n-1}\log_2\left(2^{\min\{\log_2((x,2^n))+1,\lfloor n/2\rfloor\}-n}\cdot\left[x\equiv 0\mod 2^{n-\min\{\log_2((x,2^n))+1,\lfloor n/2\rfloor\}}?\,1:2\right]\right)\\
&=-\frac{1}{2^n}\sum_{x=0}^{2^n-1}\log_2\left(2^{\min\{\log_2((x,2^n))+2,\lfloor n/2\rfloor\}-n}\right)\\
&=\frac{1}{2^n}\sum_{x=0}^{2^n-1}\left(n-\min\{\log_2((x,2^n))+2,\lfloor n/2\rfloor\}\right)\\
&=n-\frac{1}{2^n}\sum_{x=0}^{2^n-1}\min\{\log_2((x,2^n))+2,\lfloor n/2\rfloor\}.
\end{align}
Since the number of values less than $2^n$ for which $2^m|x$ is exactly $2^{n-m}$, the sum restricted to the $x$'s such that $(x,2^n)=2^m$ with $m+2\leqslant \lfloor n/2\rfloor$ is just $(m+2)(n^{n-m}-2^{n-m-1})=(m+2)2^{n-m-1}$. Thus we get
\begin{align}
H_{\star}(n) &=n-\sum_{m=0}^{\lfloor n/2\rfloor-2}\frac{m+2}{2^{m+1}}-\lfloor n/2\rfloor2^{-\lfloor n/2\rfloor+1}.
\end{align}
From there a simple computation suffices to show the formula I gave originally is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of the sum using integral $\lim\limits_{n\rightarrow\infty}\sum_{k = 1}^{n} \frac{1}{(k+n)\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}$. I can find it using integral: $\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{k = 1}^{n} \frac{1}{(1 + \frac{k}{n})\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}$, but how get rid of $n$ and $\frac{k}{n^2}$ in sqrt? Gan you give me a hint, please?
| We have
\begin{align}
\frac 1{\sqrt{1 + n\ln\left({1+\frac{k}{n^2}}\right)}}
&=\left(1 + n\ln\left({1+\frac{k}{n^2}}\right)\right)^{-1/2}\\
&=\left(1 + \frac kn+O\left(\frac{k^2}{n^3}\right)\right)^{-1/2}\\
&=\frac 1{\sqrt{1 + \frac kn}}+\frac 1nO\left(\frac{k}{n}\right)^2
\end{align}
Consequently,
\begin{align}
\sum_{k = 1}^{n} \frac{1}{(k+n)\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}
&=\frac 1n\sum_{k = 1}^{n} \frac{1}{(1+\frac kn)\sqrt{1 + \frac kn}}+\frac 1n\sum_{k = 1}^{n} \frac{O\left(\frac{k^2}{n^3}\right)}{(1+\frac kn)\sqrt{1 + \frac kn}}\\
&=\frac 1n\sum_{k = 1}^{n}\left(1+\frac kn\right)^{-3/2}+\frac 1{n^2}\sum_{k = 1}^{n} O\left(\frac{k^2}{n^2}\right)\left(1+\frac kn\right)^{-3/2}
\end{align}
and the last sum vanish as $n\to\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Yet another difficult logarithmic integral This question is a follow-up to MSE#3142989.
Two seemingly innocent hypergeometric series ($\phantom{}_3 F_2$)
$$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{(-1)^n}{2n+1}\qquad \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2^n(2n+1)}$$
can be reduced to the logarithmic integrals
$$ \int_{0}^{1}\frac{\text{arctanh}(\sqrt{x})}{\sqrt{(1-x)(2-x)}}\,dx,\qquad\int_{0}^{1}\frac{\frac{1-x}{1+x}\log(x)}{\sqrt{x(1+x^2)}}\,dx\tag{A,B}$$
by FL-expansions and Fourier series, or directly by semi-integration by parts.
The issue is that neither (A) or (B) seem to be manageable through standard substitutions, the help of computer algebra systems or prayers to special deities, so I hope human beings are able to provide better insights. A promising substitution for (A) is $x=\frac{3-\cosh z}{2}$.
| Long Comment: Notes on evaluating $I_A$:
I first found the equivalent integral
$$I_A=2 \int_0^1 \frac{\sinh ^{-1}(x)}{x \sqrt{1-x^2}} \, dx\tag{1}$$
Integrating (1) by parts I then found
$$I_A=2 \int_0^1 \frac{\log \left(\frac{\sqrt{1-x^2}+1}{x}\right)}{\sqrt{x^2+1}} \, dx\tag{2}$$
Then since
$$\frac{1}{\sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\, x^{2 n}$$
we obtain using (2)
$$I_A=2 \sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\, \int_0^1 x^{2 n}\log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \, dx $$
Then using Mathematica
$$\int_0^1 x^{2 n} \log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \, dx=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{3}{2}\right)}{(2 n+1)^2\, \Gamma (n+1)}$$
$$I_A=2 \sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\,\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{3}{2}\right)}{(2 n+1)^2\, \Gamma (n+1)}=\pi \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};-1\right)$$
UPDATE 23/03/2019
In thinking about Jack D'Aurizio's comment below: I think the relationship between the integrals and Euler sums can perhaps be found by proving and then making use of (3) and (4)
$$\int_0^1 x^n \tanh ^{-1}\left(\sqrt{x}\right) \, dx=\frac{1}{n+1}\sum _{k=1}^{n+1} \frac{1}{2 k-1}\tag{3}$$
$$\int_0^1 x^{2n} \tanh ^{-1}\left(\sqrt{x}\right) \, dx=\frac{1}{2n+1}\sum _{k=1}^{2n+1} \frac{1}{2 k-1}\tag{4}$$
for Integrals A and B respectively.
In regard to integral A this looks difficult because it involves simplifying something like:
$$I_A=\frac{1}{\sqrt{2}} \int_0^1 \left(\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^n}{4^n}\right) \left(\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^n}{8^n} \right) \tanh ^{-1}\left(\sqrt{x}\right) \,dx$$
I get stuck trying to simplify the Cauchy Product.
The proof may be easier in regard to integral B since
$$\frac{1}{\sqrt{1-x^2}}=\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^{2 n}} {4^n} $$
giving
$$I_B=\sqrt{2} \int_0^1\frac{ \tanh ^{-1}\left(\sqrt{x}\right)}{\sqrt{(1-x) (x+1)}}\; dx =\sqrt{2}\; \sum _{n=0}^{\infty } \frac{\binom{2 n}{n} \sum _{k=1}^{2 n+1} \frac{1}{2 k-1}}{4^n (2 n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
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integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$ I am looking for other methods to find the general integral
$$C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2x}$$
To do so, I first preformed $u=x-\pi$:
$$C(a,b)=\int_{-\pi}^{\pi}\frac{xdx}{a+b\cos^2x}+\pi\int_{-\pi}^{\pi}\frac{dx}{a+b\cos^2x}$$
The sub $x\mapsto -x$ provides
$$\int_{-\pi}^\pi \frac{xdx}{a+b\cos^2x}=0$$
and with symmetry,
$$C(a,b)=2\pi\int_0^\pi \frac{dx}{a+b\cos^2x}$$
Then we use $t=\tan(x/2)$:
$$C(a,b)=4\pi\int_0^\infty \frac1{a+b\left[\frac{t^2-1}{t^2+1}\right]^2}\frac{dt}{t^2+1}$$
$$C(a,b)=\frac{4\pi}{a+b}\int_0^\infty \frac{x^2+1}{x^4+2\frac{a-b}{a+b}x^2+1}dx$$
Then we consider
$$N_s(k)=\int_0^\infty\frac{x^{2s}}{x^4+2kx^2+1}dx$$
Then $x\mapsto 1/x$ gives
$$N_s(k)=N_{1-s}(k)$$
and with $s=0$:
$$C(a,b)=\frac{8\pi}{a+b}\int_0^\infty \frac{dx}{x^4+2kx^2+1}\qquad k=\frac{a-b}{a+b}$$
Then with
$$x^4+(2-c^2)x^2+1=(x^2+cx+1)(x^2-cx+1)$$
we have that
$$\begin{align}
u(c)=N_0\left(\frac{2-c^2}2\right)=&\frac1{4c}\int_0^\infty\frac{2x+c}{x^2+cx+1}dx-\frac1{4c}\int_0^\infty\frac{2x-c}{x^2-cx+1}dx\\
&+\frac14\int_0^\infty\frac{dx}{x^2+cx+1}+\frac14\int_0^\infty\frac{dx}{x^2-cx+1}
\end{align}$$
The first two integrals vanish and we have
$$u(c)=\frac14I(1,c,1)+\frac14I(1,-c,1)$$
where
$$I(a,b,c)=\int_0^\infty\frac{dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\left[\frac\pi2-\arctan\frac{b}{\sqrt{4ac-b^2}}\right]$$
So
$$4u(c)=\frac2{\sqrt{4-c^2}}\left[\frac\pi2-\arctan\frac{c}{\sqrt{4-c^2}}+\frac\pi2-\arctan\frac{-c}{\sqrt{4-c^2}}\right]$$
$$u(c)=\frac{\pi}{2\sqrt{4-c^2}}$$
$$N_0(a)=\frac\pi{2\sqrt{2+2a}}$$
And then
$$C(a,b)=\frac{8\pi}{a+b}N_0\left(\frac{a-b}{a+b}\right)$$
$$C(a,b)=\frac{4\pi^2}{\sqrt{a^2+ab}}$$
How else can you prove this? Have fun ;)
As it turns out, we may be missing a factor of $1/2$, so we may actually have
$$C(a,b)=\frac{2\pi^2}{\sqrt{a^2+ab}}$$
an issue discussed in the comment section of @Song's answer.
| Note that $\cos^2(x)=\frac{1+\cos(2x)}{2}$. Therefore,
$$\begin{align}
C(a,b)&=2\pi\int_0^\pi \frac1{a+b\cos^2(x)}\,dx\\\\
&=2\pi \int_0^\pi \frac{2}{2a+b+b\cos(2x)}\,dx\\\\
&=2\pi \int_0^{2\pi}\frac{1}{2a+b+b\cos(x)}\,dx\tag1\\\\
&=4\pi \int_0^\pi \frac{1}{2a+b+b\cos(x)}\,dx\tag2
\end{align}$$
We can proceed by either (i) letting $z=e^{ix}$ in $(1)$ and using contour integration or (ii) letting $t=\tan(x/2)$ in $(2)$.
METHODOLOGY $1$: CONTOUR INTEGRATION
Letting $z=e^{ix}$ in $(1)$ reveals
$$\begin{align}
C(a,b)&=2\pi \oint_{|z|=1} \frac{1}{2a+b +b\left(\frac{z+z^{-1}}{2}\right)}\,\frac1{iz}\,dz\\\\
&=\frac{4\pi}{ib}\oint_{|z|=1}\frac{1}{z^2+2(1+2a/b)z+1}\,dz\\\\
&=2\pi i \left(\frac{4\pi}{ib}\right)\text{Res}\left(\frac{1}{z^2+2(1+2a/b)z+1}, z=-(1+2a/b)+\frac2b\,\sqrt{a(a+b)}\right) \\\\
&=\frac{8\pi^2}{b}\left(\frac{1}{4\frac{\sqrt{a(a+b)}}{b}}\right)\\\\
&=\frac{2\pi^2}{\sqrt{a(a+b)}}
\end{align}$$
METHODOLOGY $2$: Tangent Half-Angle Substitution
Letting $t=\tan(x/2)$ in $(2)$, we obtain
$$\begin{align}
C(a,b)&=4\pi \int_0^\pi \frac{1}{2a+b+b\cos(x)}\,dx\\\\
&=4\pi \int_0^\infty \frac{1}{2a+b+b\frac{1-t^2}{1+t^2}}\,\frac{2}{1+t^2}\,dt\\\\
&=\frac{4\pi}{ a}\int_0^\infty \frac{1}{t^2+(1+b/a)}\,dt\\\\
&=\frac{4\pi}{a}\frac{\pi/2}{\sqrt{1+b/a}}\\\\
&=\frac{2\pi^2}{\sqrt{a(a+b)}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing $ \lim _{n \rightarrow \infty}\left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^{n}\right]$
$$
\lim _{n \rightarrow \infty}\left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^{n}\right] \text { equals }\_\_\_\_
$$
I tried to expand the term in power using binomial theorem but still could not obtain the limit.
| One has
$$n - \frac{n}{e} \left( 1 + \frac{1}{n}\right)^n = n - \frac{n}{e} \exp \left( n \ln \left( 1 + \frac{1}{n}\right)\right) =n - \frac{n}{e} \exp \left( n \left(\frac{1}{n}- \frac{1}{2n^2} + o \left( \frac{1}{n^2}\right)\right)\right) $$
so
$$n - \frac{n}{e} \left( 1 + \frac{1}{n}\right)^n = n - \frac{n}{e} \exp \left( 1 -\frac{1}{2n} + o \left( \frac{1}{n}\right)\right) = n - n \left( 1 - \frac{1}{2n} + o \left( \frac{1}{n}\right)\right) $$
so
$$n - \frac{n}{e} \left( 1 + \frac{1}{n}\right)^n = \frac{1}{2} + o(1).$$
Therefore the limit is
$$\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the area of $y = 12-x^2, y=x^2-6$ I found the x-intercepts via my graphing calculator to be $(-3,3)$ and $(3,3)$ and from it I formed the Area equation,
$$A = \int_{-3}^{3} (12-x^2 - x^2 - 6)dx = \int_{-3}^{3}(-2x^2+6)dx \\
= -\frac{2}{3}x^3 + 6x \bigg]_{-3}^{3}$$
$$A = \bigg(3\frac{2}{3}\cdot 3^3 + 6(3) \bigg) - \bigg(-\frac{2}{3} \cdot -3^3 + 6(-3)\bigg) \\
= 0 - 0$$
I calculated the integral and solved for the definite integral, but I got 0....I did the calculations correctly, but the answer doesn't seem to make sense for what the graphs look like.
Did I set the problem up wrong? or something?
Edit: I think I need to split the integral bounds up to $\int_{-3}^{0}$ and $\int_{0}^{3}$ to split the positive/negative area portions
| $$A = \int_{-3}^{3} (12-x^2 - (x^2 - 6))dx = \int_{-3}^{3}(-2x^2+18)dx
= -\frac{2}{3}x^3 + 18x \bigg]_{-3}^{3}$$
$$A = \bigg(-\frac{2}{3}\cdot 3^3 + 18(3) \bigg) - \bigg(-\frac{2}{3} \cdot -3^3 + 18(-3)\bigg)=36-(-36)=72.$$
You forgot the second set of parentheses inside the first integral. You don't need to split up the limits of integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$. I want to show:
If $c^2+8 \equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
| Hint:
As $c^2\equiv-8\pmod p,$
$$c(c+1)=c^2+c\equiv c-8\pmod p$$
$c(c+1)(c-8)\equiv?$
I believe this is how the problem naturally came into being .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How can I block diagonalise this matrix? I have this matrix:
$$A = \left(
\begin{array}{cccc}
0 & 0 & 0 & -1 \\
1 & 0 & 0 & -1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & -1 \\
\end{array}
\right)$$
and I would like to show that it can be block diagonalised into :
$$ B = \left(
\begin{array}{cccc}
\cos 2\pi/5 & -\sin 2\pi/5 & 0 & 0 \\
\sin 2\pi/5 & \cos 2\pi/5 & 0 & 0 \\
0 & 0 & \cos 4\pi/5 & -\sin 4\pi/5 \\
0 & 0 & \sin 4\pi/5 & \cos 4\pi/5\\
\end{array}
\right) = \left(
\begin{array}{cccc}
\frac{1}{4} \left(-1+\sqrt{5}\right) & -\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} & 0 & 0 \\
\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} & \frac{1}{4} \left(-1+\sqrt{5}\right) & 0 & 0 \\
0 & 0 & \frac{1}{4} \left(-1-\sqrt{5}\right) & -\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} \\
0 & 0 & \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} & \frac{1}{4} \left(-1-\sqrt{5}\right) \\
\end{array}
\right)$$
What is the general procedure?
| it is already a companion matrix (both that form and its transpose are used, depends on circumstances). The characteristic polynomial is $$x^4 + x^3 + x^2 + x + 1 = \frac{x^5-1}{x-1}$$
With four distinct eigenvalues (complex) it diagonalizes. Next, you need to figure out how to take a specific diagonal matrix, with complex $\alpha$ such that $|\alpha| = 1,$
$$
\left(
\begin{array}{rr}
\alpha & 0 \\
0 & \bar{\alpha}
\end{array}
\right)
$$
and send it back to one of your two by two blocks with sine and cosine.
To reverse the job: with some real angle $\theta,$ exactly how do you diagonalize
$$
\left(
\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}
\right) \; \; ? \;
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can someone shed some light on this inequality? I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$\frac{a_{n+1}}{a_n}>\left (1-\frac{1}{n+1}\right ) \left (\frac{n+1}{n}\right)$$
where does the equation in the first and second parenthesis come from?
Ok, I have another relating question:
why $$\frac{a_{n+1}}{a_n}> (1+\frac{1}{n})$$ ( The expression of third line.
| So, we have
$$\frac{a_{n+1}}{a_n} = \left(1 - \frac{1}{(n+1)^2}\right)^{n+1}\left(\frac{n+1}{n}\right).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$\left(1 - \frac{1}{(n+1)^2}\right)^{n+1} > 1 + (n+1)\left(\frac{-1}{(n+1)^2}\right) = 1 - \frac{1}{n+1}.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$\frac{a_{n+1}}{a_n} = \left(1 - \frac{1}{(n+1)^2}\right)^{n+1}\left(\frac{n+1}{n}\right) > \left(1-\frac{1}{n+1}\right)\left(\frac{n+1}{n}\right).$$
Finally, we multiply out the RHS of the inequality
$$\frac{a_{n+1}}{a_n} > \left(1-\frac{1}{n+1}\right)\left(\frac{n+1}{n}\right) = \frac{n+1}{n} - \frac{1}{n} = 1.$$
So, we have
$$\frac{a_{n+1}}{a_n} > 1 \implies a_{n+1} > a_n,$$
which means that $\{a_n\}$ is an increasing sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$ This is the last indefinite integral I am attempting before proceeding to definite integrals study. Any ideas how to solve it?
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$$
My attempt: (not the longest one... :P)
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx = \int \frac{\sqrt{(x-2)^3}}{x+1} dx = \left(\left(u=x-2,du=dx \right)\right) =\int\frac{\sqrt{u^3}}{u+3}du$$
But that substitution does not really make it simpler for me. If the expression inside the numerator's root was of the form $ax^2 + bx + c$ then I could apply one of Euler's substitution. But it's polynomial of third power.
How to deal with it?
| Consider
$$I = \int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx = \int \frac{\sqrt{(x-2)^3}}{x+1} dx$$
Let $u = x -2 \implies du = dx $
$$I = \int\frac{u^\frac{3}{2}}{u + 3}du$$
Let $u = 3\tan^2t \implies du = 6\tan t \sec^2t dt$
$$I = 6\sqrt{27}\int\frac{\tan^4t\sec^2t}{3\sec^2t}dt$$
$$ = 6\sqrt{3}\int\tan^4t \ dt$$
$$ = 6\sqrt{3}\int\tan^2t(\sec^2t - 1)\ dt = 6\sqrt{3}\frac{\tan^3t}{3} - 6\sqrt{3}\int\tan^2 t\ dt$$
$$ = 2\sqrt{3}\tan^3t - 6\sqrt{3}\int(\sec^2t - 1)\ dt$$
$$ = 2\sqrt3\tan^3t - 6\sqrt{3}\tan t + 6\sqrt3t$$
$$I = 2\sqrt3(\tan^3t - 3\tan t + 3t)$$
Now $3\tan^2t = u = x-2 \implies \tan t = \left(\frac{x-2}{3}\right)^\frac{1}{2}$
$$I = 2\sqrt3 \left(\frac{x-2}{3}\right)^\frac{1}{2} \left(\frac{x-2}{3} - 3 \right) + 6\sqrt3\arctan\left(\sqrt{\frac{x-2}{3}}\right)$$
$$ I = \boxed{\frac{2}{3}(x-11)\sqrt{x-2} + 6\sqrt3\arctan\left(\sqrt{\frac{x-2}{3}}\right)}$$
Note: Wolfram Mathematica gives the same result as one of the alternate forms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq 3( \sec^6 x \csc^6 x \sec^6 x\csc^6 x)^\frac{1}{3} $$
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x \geq 3( \sec x \csc)^4 $$
$$
\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq
\frac{3 * 2^4}{\sin ^4 2x} $$
Clearly least value is 48, but something is wrong here, as the answer is 80, if I use other methods.
| Others explain why $48$ is correct as a a lower bound but may not be the sharp lower bound.
One way to get a lower bound of $80$ involves using the fact that each term is a cubed quantity. Start with the decomposition
$\sec^6 x + \csc^6 x + \sec^6 x\csc^6 x=A+B$
$A=\sec^6 x + \csc^6 x$
$B=\sec^6 x\csc^6 x$
Factor $A$ as a sum of cubes:
$A=(\sec^2 x + \csc^2 x)(\sec^4 x - \sec^2 x\csc^2 x + \csc^4 x)$
$A=\dfrac{(\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x\sin^2 x + \sin^4 x)}{\cos^6 x\sin^6 x}$
Plugging in $\cos^2 x +\sin^2 x =1$ and $\cos^4 x +2 \cos^2 x\sin^2 x + \sin^4 x=(\cos^2 x +\sin^2 x)^2=1$:
$A=\dfrac{1 - 3\cos^2 x\sin^2 x}{\cos^6 x\sin^6 x}$
We have $(\cos x-\sin x)^2=1-2\cos x\sin x\ge 0$ forcing $|\cos x\sin x|\le 1/2$. Thereby
$A\ge (1 - 3/4)×(64)=16$
For $B$, simply render
$B=\dfrac{1}{\cos^6 x\sin^6 x}\ge 64$
where again we have put in $|\cos x\sin x|\le 1/2$.
Then
$A+B\ge 16+64=80$.
This bound may be proven sharp by putting in $x=\pi/4$, or by noting that the separate bounds on $A$ and $B$ both become sharp when $|\cos x|=|\sin x|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Simultaneous P.D.E. problem
Solve the following differential equation:
$$ \frac{dx}{y^2(x-y)} = \frac{dy}{x^2(x-y)} = \frac{dz}{z(x^2 + y^2)} $$
I have so far got
$x^2dx = y^2dy $ which implies that $(x-y)(x^2 +xy + y^2) = a$ ...(1)
Then $$\frac{dx+ dy}{(x-y)} = \frac{dz}{z}$$
Using the equation (1), I get
$\frac{1}{a} (x^2+xy+y^2)(dx+dy) = \frac{dz}{z}$
How do I proceed from this point? I could have solved this question easily if the third fraction in the given P.D.E. were
$\frac{dz}{z(x^2 - y^2)}$ but I need help for this problem. Please help me out.
| $$ \frac{dx}{y^2(x-y)} = \frac{dy}{x^2(x-y)} = \frac{dz}{z(x^2 + y^2)} $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)\frac{\partial z}{\partial x}+x^2(x-y)\frac{\partial z}{\partial y}=(x^2+y^2)z$$
A first characteristic equation comes from $\frac{dx}{y^2(x-y)} = \frac{dy}{x^2(x-y)} $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $\frac{dx}{y^2(x-y)} = \frac{dz}{z(x^2 + y^2)}$
$$\frac{dz}{z}=\frac{x^2+y^2}{y^2(x-y)}dx$$
with $y=(c_1+x^3)^{1/3}$
$$\frac{dz}{z}=\frac{x^2+ (c_1+x^3)^{2/3} }{(c_1+x^3)^{2/3}(x-(c_1-x^3)^{1/3})}dx$$
$$z=c_2\exp\left(\int \frac{x^2+ (c_1+x^3)^{2/3} }{(c_1+x^3)^{2/3}(x-(c_1+x^3)^{1/3})}dx \right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=\int \frac{x^2+ (c_1+x^3)^{2/3} }{(c_1+x^3)^{2/3}(x-(c_1+x^3)^{1/3})}dx $$
and a second characteristic equation would be :
$$z\,\exp(-I\left(c_1,x \right) )=c_2$$
$$z\,\exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=\exp(I(y^3-x^3,x) ) F\left(y^3-x^3 \right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
| {
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"url": "https://math.stackexchange.com/questions/3171667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.