Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
how to prove $\displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = .... $ How to prove $$ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{\sin^2 \theta}{\sin^2 \left( \frac{k\pi }{2n+1} \right ) } \right ) $$
So far, I manage to prove $ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{2n}\left(1 - \frac{\sin \theta}{\sin \left( \frac{k\pi }{2n+1} \right ) } \right ) $ though I am not sure I am aright.
| First consider the following Lemma
Lemma :
$$ \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}} $$
Proof : Note that,
$ \displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=n+1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right)$
$ \displaystyle = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\dfrac{(n+k)\pi}{2n+1}\right) $
$ \displaystyle = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \ \left(\because \prod_{k=1}^{n} f(k) = \prod_{k=1}^{n} f(n+1-k) \right) $
$ \displaystyle = \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) $
But,
$ \displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}} $ (For my proof of this, see here )
$ \displaystyle \implies \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}} $
Now, let $ \displaystyle \text{P} = (2n+1)\sin(\theta) \prod_{k=1}^n \left(1 -\dfrac{\sin^2(\theta)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right) $
$ \displaystyle = (2n+1)\sin(\theta) \dfrac{ \displaystyle \prod_{k=1}^n \left( \sin^2\left(\frac{k\pi}{2n+1}\right) - \sin^2(\theta) \right)}{ \displaystyle \prod_{k=1}^{n} \sin^2\left(\frac{k\pi}{2n+1}\right) } $
$ \displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^n \left( \cos^2(\theta) - \cos^2\left(\frac{k\pi}{2n+1}\right) \right) $ (Using the Lemma)
$ \displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) + \cos \left(\frac{k\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) $
$ \displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(2n+1 - k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \ \left(\because \cos (\pi -x) = -\cos x \right) $
$ \displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(n + k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \ \left(\because \prod_{k=1}^{n} f(k) = \prod_{k=1}^{n} f(n+1-k) \right) $
$ \displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=n+1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{ k \pi}{2n+1}\right) \right) \right) $
$ \displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) $
Also,
$ \displaystyle U_{n} (x) = 2^{n} \prod_{k=1}^{n} \left(x - \cos \left(\frac{k\pi}{n+1}\right) \right) $
where $ \displaystyle U_{n} (x)$ denotes the Chebyshev Polynomial of the Second kind.
$ \displaystyle \implies \text{P} = 2^{2n} \sin(\theta) \cdot 2^{-2n} \cdot U_{2n} (\cos \theta) $
$ \displaystyle = \sin ((2n+1) \theta) \ \left(\because U_{n} (\cos \theta) = \dfrac{\sin ((n+1) \theta)}{\sin \theta} \right) \quad \square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
epsilon-delta proof of $\lim_{x \to 4} \sqrt{x} = 2$ Prove $$\quad \lim_{x\to4}\sqrt{x}=2 $$
using the precise definition of limits. (Epsilon-Delta)
I think I proved this problem but when I look at the textbook to compare the proofs
they are quite different and I don't get how the book worked it all out.
So I am going to post mine for you to check if it's correct and the one from
the textbook to ask some questions.
$\\[10pt]$
My rough work: $\;$
Pick $\epsilon > 0$, then there exists $\delta>0$ such that
$\quad 0 < |x - 4| < \delta \quad \Rightarrow \quad |\sqrt{x} - 2| < \epsilon$
Establish a connection between $|x - 4|$ and $|\sqrt{x} - 2|$
$|\sqrt{x} - 2| \cdot \frac{|\sqrt{x} + 2|}{|\sqrt{x} + 2|} =
\frac{|x - 4|}{\sqrt{x} + 2}$, $\;$ Pick $\delta = 4$
$|x-4| < 4 \ \Rightarrow\ 0 < \sqrt{x} < \sqrt{8} \ \Rightarrow\ 2 < \sqrt{x} + 2 < \sqrt{8} + 2 \ \Rightarrow\ \frac{1}{\sqrt{8} + 2} < \frac{1}{\sqrt{x} + 2} < \frac{1}{2}$
This implies $\frac{|x - 4|}{\sqrt{x} + 2} < \frac{1}{2} \cdot |x-4| < \epsilon \ \Rightarrow\ |x-4|< 2 \cdot \epsilon$
$\\[20pt]$
My proof: $\;\delta = min\{4,2\epsilon\}$ and assume that $\ 0 < |x - 4| < \delta \ \Rightarrow \ |\sqrt{x} - 2| < \epsilon$
$\frac{|x - 4|}{\sqrt{x} + 2} < \frac{1}{2} \cdot |x-4| < \frac{1}{2} \cdot 2\epsilon < \epsilon \quad QED\quad $ Corrct?
$\\[20pt]$
From the book: $\;$To be able to form $\sqrt{x}$, we need to have $x \ge 0$.
To ensure this, we must have $\delta \le 4$. With $x \ge 0$, we can form $\sqrt{x}$ and write
$|x - 4| = |\sqrt{x}+2||\sqrt{x}-2|$.
Since $|\sqrt{x} + 2| \ge 2 > 1$, it follows that $\quad \leftarrow\;$ This is what I don't understand.
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $ where did "$\ge 2 > 1$" come from?
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $ the book doesn't explain
$|\sqrt{x}-2| \le |x - 4| \qquad \qquad \qquad \qquad \ \ \ \, \leftarrow$ How can I get this from the above inequality?
This last inequality suggests that we can simply set $\delta \le \epsilon \quad \leftarrow$ I got $2\epsilon$, how come it only has $\epsilon$?
Any help would be greatly appreciated.
| Answer to your where did $\geq2 > 1$ comes from:
It comes from the fact $\sqrt{x} \geq 0$, so $|\sqrt{x} + 2| \geq |0 + 2| = 2$ and of course $2 > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
solving $\sqrt{3-\sqrt{3+x}}=x$. Can we solve the following equation in $\mathbb{R}$ without expanding it into a fourth degree equation :
$$ \sqrt{3-\sqrt{3+x}} = x.$$
squaring both sides and squaring again is the only thing I could done, If you have any other idea just post hints.
| First, it must be $\,3+x\ge 0\Longleftrightarrow x\ge -3\,$ , for the inner-most square root to be defined on the real field.
Then it also must be
$$3-\sqrt{3+x}\ge 0\Longrightarrow9\ge 3+x\Longrightarrow x\le 6$$
for the outer square root to be defined, thus our definition domain is $\,-3\le x \le 6\,$ . Now directly, by successive squaring:
$$x=\sqrt{3-\sqrt{3+x}}\Longrightarrow x^2=3-\sqrt{3+x}\Longrightarrow x^4-6x^2+9=3+x\Longrightarrow$$
$$x^4-6x^2-x+6=0\Longleftrightarrow x^2(x^2-6)-(x-6)=0$$
We get by inspection the zero $\,x=\,$ , so
$$x^4-6x^2-x+6=(x-1)(x^3+x^2-5x-6)$$
Checking for rational solutions to the above we have that $\,x=-2\,$ is a zero, too, so
$$x^4-6x^2-x+6=(x-1)(x+2)(x^2-x-3)$$
Finally, the solutions to that quadratic are
$$x_{1,2}=\frac{1\pm\,\sqrt{13}}{2}=\begin{cases}\frac{1+\sqrt{13}}{2}\cong 2.3\\{}\\\frac{1-\sqrt{13}}{2}\cong -1.3\end{cases}$$
So the roots of the quartic are $\,-2\,,\,-1.3\,,\,1\,,\,2.3\,$ , but since we squared the original irrational equation to get the quartic, we now must check each of the above solutions in the original equation; we get at once that the negative ones must be dropped as we've a square root in one of the sides.
A quick calculator check also rules out the root $\,2.3\,$ of the quartic, so the only real solution is $\,x=1\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/311856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Quadratic equation $x^2-ax+2a = 0$ with Integers Roots Find Integer value of $a$ for which both roots of the equation $x^2-ax+2a = 0$ are also integers
My Try:: $\displaystyle x = \frac{a \pm \sqrt{a^2-8a}}{2}$. Now if the Roots are Integer . Then $a^2-8a=k^2$ where $k\in \mathbb{Z}$
Now How can I calculate after that., Thanks
| As you say, the question comes down to knowing when $a^2 - 8a - k^2 = 0$ for $k \in \mathbb{Z}$. Solving for $a$, $a = 4 \pm \sqrt{16 + k^2}$. As you can see, only for $k = 0$ or $k=3$ does this produce an integer solution for $a$ and so $a = 8$, $a = 0$, $a=9$ and $a=-1$ are possible.
The reason why those were the only $k$ values possible is because after a while, the difference between $k^2$ and $(k+1)^2$ is $2k+1$ and so when $k > 8$, $2k + 1$ becomes bigger than $16$. So we only need to test $k$ up to $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\sum\limits_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$? There is post here asking for $$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$
And the answer is $18-24\ln 2$.
It is easy to elvaluate that $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1+2+\cdots+n}=2$ and it is not difficult to justify the convergence of $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$ for all $k>1$.
Here comes my question: How to evaluate $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^3+2^3+\cdots+n^3}$ or in general, what is $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$ if $k$ is a positive integer.
| I guess the way is pretty much the same, as there are always formulas to evaluate
$$\sum_{i=1}^n i^k, $$
For $k=3$ the result of the sum above is
$$\frac{4}{3} (-9+\pi^2)$$
For higher terms you will probably not always get a nice result, but for $k=5$ you have $$ 4\left(15-\pi^2 + 2\sqrt{3} \pi \tan\left(\frac{\sqrt{3}\pi}{2}\right)\right)$$
I will calculate the $k=3$ case.
As $$\sum_{i=1}^n i^3 = \frac{1}{4} n^2 (1+n)^2$$ we have
\begin{align*}
\sum_{n=1}^\infty \frac{1}{\sum_{i=1}^n n^3 } &= 4 \sum_{n=1}^\infty \frac{1}{m^2(1+m)^2}\\
&=\sum_{n=1}^\infty \frac{1}{n^2} - \frac{2}{n} + \frac{2}{n+1} + \frac{1}{(n+1)^2}\\
&=4\sum_{n=1}^\infty \frac{1}{n^2} +4\sum_{n=1}^\infty \frac{1}{(1+n)^2} - 8 \cdot \sum_{n=1}^\infty \frac{1}{n} -\frac{1}{n+1}\\
&=4\left(\frac{\pi^2}{6} +\frac{\pi^2}{6}-1-2\right)\\
&=\frac{4}{3}(\pi^2-9)\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
| You are likely expected to integrate by parts (twice)
$$ \begin{eqnarray}
\int \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} \mathrm{d} x
\\
&=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} \mathrm{d} x
\\
&=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2} - \frac{\sin^2(x)}{x^2} \right) \mathrm{d}x
\end{eqnarray}
$$
where the last equality used
$$\begin{eqnarray}
3 \cos^2(x) \sin^2(x) - \sin^4(x) &=& 3 \cos^2(x) \sin^2(x) - \sin^2(x) (1-\cos^2(x)) \\ &=& \left(2 \sin(x) \cos(x) \right)^2 - \sin^2(x) = \sin^2(2x) - \sin^2(x)
\end{eqnarray}
$$
Now
$$\begin{eqnarray}
\int_0^\infty \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(2x)}{x^2} \mathrm{d} x - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{4}{3} \int_0^\infty \frac{\sin^2(y)}{y^2} \mathrm{d} y - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=&
\frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x = \frac{\pi}{3}
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 6,
"answer_id": 3
} |
If $v$ is an eigenvalue of $A$ and $c$ is an eigenvalue of $B$, must $vc$ be an eigenvalue of $AB$? Let $v$ be an eigenvalue of $A$ and $c$ be an eigenvalue of $B$. Is the product of $v$ and $c$ equal to an eigenvalue of $AB$?
| $1$ is an eigenvalue of $
\begin{pmatrix}
1 & 0\\
0 & 2\\
\end{pmatrix},$
$3$ is an eigenvalue of $
\begin{pmatrix}
2 & 0\\
0 & 3\\
\end{pmatrix}$
but $3$ is not an eigenvalue of $
\begin{pmatrix}
1 & 0\\
0 & 2\\
\end{pmatrix}
\begin{pmatrix}
2 & 0\\
0 & 3\\
\end{pmatrix}=
\begin{pmatrix}
2 & 0\\
0 & 6\\
\end{pmatrix}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Exponential equation, $(3+2\sqrt2)^x+1=6(\sqrt2+1)^x$ http://i.stack.imgur.com/YP2Ha.png
$$(3+2\sqrt2)^x+1=6(\sqrt2+1)^x \qquad\qquad x\in\mathbb{R}$$
I managed to find one of the solutions (x=2), but I got stuck. I would really appreciate a step by step solution. Thanks in advance :)
| $(\sqrt2+1)^2=3+2\sqrt2$
If $\sqrt2+1=y, y^{2x}+1=6{y^x}\implies y^{2x}-6y^x+1=0, y^x=\frac{6\pm\sqrt{6^2-4}}2=3\pm2\sqrt2$
Now, $3+2\sqrt2=y^2\implies 3-2\sqrt2=\frac1{3+2\sqrt2}=y^{-2}$
If $y^x=3+2\sqrt2,y^x=y^2\implies x=2$ as $y=\sqrt2+1\ne0,\pm1$
Similarly for $y^x=3-2\sqrt2,y^x=y^{-2}\implies x=-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
An irregular 6 faced dice An irregular 6 faced dice is such that the probability that it gives 3 even numbers in 5 throws is twice the probability that it gives 2 even numbers in 5 throws. How many sets of exactly 5 trials can be expected to give no even number out of 2500 sets?
(1) Nearly 5
(2) Nearly 10
(3) Nearly 15
(4) Nearly 20
| Let the probability of an even number be $p$. Thus the probability of an odd number is $1-p$.
The probability of $3$ even in $5$ throws is $\binom{5}{3}p^3(1-p)^2$. The probability of $2$ even in $5$ throws is $\binom{5}{2}p^2(1-p)^3$.
We are told that $\binom{5}{3}p^3(1-p)^2=2\binom{5}{2}p^2(1-p)^3$. Solve for $p$. There are the solutions $p=0$ and $p=1$, which we are presumably expected to discard, although there is no mathematical reason to do so. In addition, we have the solution $p=\frac{2}{3}$. So the probability that a single toss results in a number that is not even is $\frac{1}{3}$.
The probability of no even in $5$ trials is therefore $\frac{1}{3^5}$.
Now in $2500$ sets of $5$ trials, the expected number of times we have no even is $2500\cdot\frac{1}{3^5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/325754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Frequency response We consider the response $x(t)$ when the frequency of the input y(t) varies.
$$x''+bx'+kx=ky(t), y(t)=\cos \omega t$$
1) If $k=3.24$, what is the natural ($b=0$) frequency $\omega_n$?
I think we have $\omega_n=\frac{\sqrt{3.24}}{2\pi}$
2) If $k=3.24$ and $b=0.8$, what is the amplitude gain from the input $y(t)=\cos \omega_n t$?
I need to report the ratio of the amplitude of x to amplitude y.
Thank you
| Homogeneous equation is
$$
x''+bx+kx=0
$$
Characteristic equation of which
$$
\lambda^2+b\lambda+k=0
$$
which has solutions
$$
\lambda_{1,2}=-\frac b2 \pm \frac{\sqrt{b^2-4k}}2
$$
To have solution only in reals and keep order of ODE, you have two options.
First if $b^2-4k > 0$ you have two real solutions
$$
\lambda_{1,2} = \frac b2 \pm \tilde \lambda
$$
where $\tilde \lambda = \frac{\sqrt{b^2-4k}}2$ and general solutions $x_h = e^{-\frac b2t}\left (C_1 \cosh \tilde \lambda t + C_2 \sinh \tilde \lambda t \right)$. It's called Overdamped harmonic oscillator. This solution actually doesn't do any oscillations and exponentially dies out.
Second option, if you have $b^2-4k < 0$, so you have two complex solutions
$$
\lambda_{1,2} = -\frac b2 \pm i \tilde \lambda
$$
where $\tilde \lambda = \frac {\sqrt{4k-b^2}}2$ and general solution $x_h = e^{-\frac b2t} \left ( C_1 \cos \tilde \lambda t + C_2 \sin \tilde \lambda t\right)$. Concept of natural frequency or any frequency at all makes sense for this option. It can be defined as $\omega_0$ in
$$
\omega = \frac {\sqrt{4k-b^2}}2 = \sqrt{\frac {4k-b^2}4} = \sqrt{k-\frac {b^2}4} = \sqrt{\omega_0^2-\chi^2}
$$
so $\omega_0 = \sqrt k$ and $\chi = \frac b2$. It's also called undumped frequency. Latter serves as damping parameter. Even it's called frequency, it's angular frequency. As for the frequency (which is measured in Hertz), it's $\nu = \frac \omega {2\pi}$
When $k = 3.24$ and $b = 0$, natural or undumped angular frequency is $\sqrt {3.24} = 1.8$ and $\nu = \frac {1.8}{2\pi}$, so your answer is correct.
As for the inhomogeneous ODE, looking at the form of forcing term, you can guess the solution as $x_p = A\cos \omega t + B \sin \omega t$ so
$$
x_p' = \omega(-A \sin \omega t+B\cos \omega t) \\
x_p'' = -\omega^2 (A \cos \omega t + B \sin \omega t)
$$
After substitution to the ODE
$$
-\omega^2 (A\cos \omega t + B \sin \omega t) + b \omega (-A\sin \omega t+B\cos \omega t) + k(A\cos \omega t + B\sin \omega t) = k\cos \omega t \\
\left [(k-\omega^2)A+bB\omega \right ]\cos \omega t + \left [ (k-\omega^2)B-bA\omega\right ] \sin \omega t = k\cos \omega t
$$
After matching coefficient in front of sines and cosines
$$
(k-\omega^2)A+bB\omega = k \\
(k-\omega^2)B-bA\omega = 0
$$
so one can find $A$ and $B$
$$
A = \frac {k\left(\omega^2-k\right)}{(\omega^2-k)^2+b^2\omega^2} \\
B = \frac {kb \omega}{(\omega^2-k)^2+b^2\omega^2}
$$
and particular solution will be
$$
x_p = \frac {k\left(\omega^2-k\right)}{(\omega^2-k)^2+b^2\omega^2} \cos \omega t + \frac {kb \omega}{(\omega^2-k)^2+b^2\omega^2} \sin \omega t = \\
= \frac k{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \left ( \frac {\omega^2-k}{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \cos \omega t + \frac {b \omega}{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \sin \omega t \right )
$$
since
$$
\left ( \frac {\omega^2-k}{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \right)^2 + \left (\frac {b \omega}{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \right )^2 = 1
$$
you can subsitute
$$
\cos \delta = \frac {\omega^2-k}{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \\
\sin \delta = \frac {b \omega}{\sqrt{(\omega^2-k)^2+b^2\omega^2}}
$$
and particular solution will be
$$
x_p = \frac k{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \left ( \cos \delta \cos \omega t + \sin \delta \sin \omega t \right ) = \frac k{\sqrt{(\omega^2-k)^2+b^2\omega^2}} \cos \left ( \omega t - \delta\right)
$$
Amplitude ratio is
$$
AR = \frac k{\sqrt{(\omega^2-k)^2+b^2\omega^2}}
$$
Just substitute your values to get corresponding function of $\omega$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/326881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the appropriate base $$213_x=139_{10}$$
$$21_x=1021_4$$
How would I solve this
| $$213_x=139_{10}$$
$$2\cdot x^2+1\cdot x^1+3\cdot x^0=1\cdot 10^2+3\cdot 10^1+9\cdot10^0$$
$$2x^2+x+3=139$$
$$2x^2+x-136=0\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{1+8\cdot 136}}{4}=\frac{-1\pm33}{4}$$
$$x_1=8$$ the base $$x_2=-17/2$$
is not a correct base see comments below.
similarly
$$21_x=1021_4$$
$$2x+1=4^3+2\cdot 4 +1=73\Rightarrow x=36$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/327517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx$? How to find the follwing integral's value ?
$$\int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx$$
Actually, I don't know it can be represented as closed form.
| Here is another approach based on power series and beta function
$$ \int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx = \frac{2}{\pi}\,\int_{0}^{1}\!{\frac{\cos \left( \frac{\pi x }{2} \right) }{\sqrt {1-
{x}^{2}}}}{dx}$$
$$ = \frac{2}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k}}{(2k)!}\int_{0}^{1}\frac{x^{2k}}{\sqrt{1-x^2}}dx $$
$$ = \frac{2}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k} }{(2k)!}\frac{1}{2}\beta\left(k+\frac{1}{2},\frac{1}{2}\right)$$
$$ = \frac{1}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k} }{(2k)!}{\frac {\Gamma\left(\frac{1}{2}\right) \Gamma\left( k+\frac{1}{2} \right) }{\Gamma \left( k+1 \right) }}$$
$$ = \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty}\frac{(-1)^k \left(\frac{\pi}{2}\right)^{2k} }{(2k)!}{\frac { \Gamma\left( k+\frac{1}{2} \right) }{\Gamma \left( k+1 \right) }}$$
$$ = \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty}\frac{(-1)^k \left(\frac{\pi}{2}\right)^{2k} }{{\frac {{2}^{2k}\Gamma \left( k +1\right) \Gamma
\left( k+1/2 \right)}{\sqrt{\pi}}}}{\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma\left(k+1\right)}} $$
$$ = {{J_{0}}\left(\frac{\pi}{2} \right)}, $$
where
$$ (2k)! = \frac {{2}^{2k}\Gamma \left(k+1\right) \Gamma\left( k+1/2 \right) }{\sqrt {\pi }} ,$$
and $J_{\alpha}(x)$ is the Bessel function
$$ J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
Consider an equilateral triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Let $P'$, $P'_1$ and $P'_2$ be the reflection of the Point $P$ along $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$
Then we have
$${\triangle APB} \cong {\triangle AP'_2B}$$
$${\triangle APC} \cong {\triangle AP'_1B}$$
$${\triangle BPC} \cong {\triangle BP'C}$$
Thus we can safely say,
The area of the hexagon $AP'_1CP'BP'_2 = 2. {\triangle ABC}$
${\triangle AP'_2B}$, ${\triangle AP'_1B}$ and ${\triangle BP'C}$ are isosceles with an apex angle of $120^o$ and composed of $30^o$ and $60^o$ right angle triangle
Thus we have
$$\overline {P'_1P'_2} = 12\sqrt3$$
$$\overline {P'P'_2} = 13\sqrt3$$
$$\overline {P'_1P'} = 5\sqrt3$$
which implies $\angle P'P'_1P'_2 = 90^o$
So the area of the hexagon $AP'_1CP'BP'_2$
$$= \triangle AP'_1P'_2 + \triangle P'P'_1C + \triangle P'P'_2B + \triangle P'P'_1P'_2$$
$$= 12^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + \frac{1}{2}(5\sqrt3)(12\sqrt3)$$
$$=169\frac{\sqrt3}{2} + 90$$
$${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then
$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
$$\frac{\sqrt3}{4}a^2 = 169\frac{\sqrt3}{4} + 45$$
$$a^2 = 169 + 45\frac{4}{\sqrt3}$$
$$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$
Simplifying
$$a = \sqrt{169 + 60\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Transforming matrix-equation to overdetermined minimum problem i have broken down my problem to plainmath and could really use some help.
Basis: I have an image. In this image I have several UV-XYZ pairs. So i know the 3d position of serveral Pixels.
Given the following equations from 3d to 2d space.
$ Z_{c} \begin{pmatrix} U \\ V \\ 1 \end{pmatrix} = \begin{pmatrix} f & 0 & px \\ 0 & f & py \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} r_{1} & r_{2} & r_{3} & t_{1}\\ r_{4} & r_{5} & r_{6} & t_{2}\\ r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix}
\begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix} $
f, px, py are known constants. Many $UV \to XYZ$ pairs are known. I want to get $r_{1} - r_{9}, t_{1}, t_{2}$ and $t_{3}$
I can transform to:
$ \begin{pmatrix} f & 0 & px \\ 0 & f & py \\ 0 & 0 & 1 \end{pmatrix} ^{-1} \begin{pmatrix} U \\ V \\ 1 \end{pmatrix} = \frac 1 Z_{c} \begin{pmatrix} r_{1} & r_{2} & r_{3} & t_{1}\\ r_{4} & r_{5} & r_{6} & t_{2}\\ r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix}
\begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix} $
The $Z_{c}$ is the Z-Coordinate of the point AFTER the multiplication with the unknown transformation matrix.
$ Z_{c} = \begin{pmatrix} r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix} \begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix} $ This leads to these equations for U and V: $ U_{k} = \frac {r_{1}X + r_{2}Y + r_{3}Z + t_{1}} {r_{7}X + r_{8}Y + r_{9}Z + t_{3}} V_{k} = \frac {r_{4}X + r_{5}Y + r_{6}Z + t_{2}} {r_{7}X + r_{8}Y + r_{9}Z + t_{3}} $
The question in the end is quite simple. I need to transform this equation into sth., so I can apply 50-100 $UV \to XYZ$ pairs and get the camera position and rotation. Without the $Z_{c}$ it is pretty easy to transform into sth. like this:
$ \begin{pmatrix} U_{1} + V_{2} + 1 \\
U_{4} + V_{5} + 1 \\
\vdots \end{pmatrix} =
\begin{pmatrix} X_{1} & Y_{1} & Z_{1} & 1 & X_{2} & Y_{2} & Z_{2} & 1 & X_ {3} & Y_{3} & Z_{3} & 1 \\
X_{4} & Y_{4} & Z_{4} & 1 & X_{5} & Y_{5} & Z_{5} & 1 & X_ {6} & Y_{6} & Z_{6} & 1 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{pmatrix}
\begin{pmatrix} r_{1} \\ r_{2} \\ r_{3} \\ t_{1} \\ r_{4} \\ r_{5} \\ r_{6} \\ t_{2} \\ r_{7} \\ r_{8} \\ r_{9} \\ t_{3} \end{pmatrix} $
I hope you get the idea. I dont know if its correct, but it seems like it. So this is a classical overdetermined linear equation (if you add 100 Rows), which I could have solved with a QR-Decomposition (i hope its called like that). But I can't apply this idea to the new Problem with $Z_{c}$. Leaves me clueless.
| Let me do some rearranging and substituting to give you a more concise formula:
Your very first formula with the substitution
$$Z_{c} = \begin{pmatrix} r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix} \begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix}$$
gives
$$ \begin{pmatrix} U \\ V \\ 1 \end{pmatrix} \begin{pmatrix} r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix} \begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix} = \begin{pmatrix} f & 0 & px \\ 0 & f & py \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} r_{1} & r_{2} & r_{3} & t_{1}\\ r_{4} & r_{5} & r_{6} & t_{2}\\ r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix}
\begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix}$$
and now we have
$$ \left[\begin{pmatrix} f & 0 & px \\ 0 & f & py \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} r_{1} & r_{2} & r_{3} & t_{1}\\ r_{4} & r_{5} & r_{6} & t_{2}\\ r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix}-\begin{pmatrix}0 & 0 & U \\ 0 & 0 & V \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} r_{1} & r_{2} & r_{3} & t_{1}\\ r_{4} & r_{5} & r_{6} & t_{2}\\ r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix} \right]
\begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix} = 0$$
or
$$ \begin{pmatrix} f & 0 & px - U \\ 0 & f & py - V \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} r_{1} & r_{2} & r_{3} & t_{1}\\ r_{4} & r_{5} & r_{6} & t_{2}\\ r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix}
\begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix} = 0$$
$$ \begin{pmatrix} f & 0 & px - U \\ 0 & f & py - V \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} r_{1}X + r_{2}Y + r_{3}Z + t_{1}\\ r_{4}X + r_{5}Y + r_{6}Z + t_{2}\\ r_{7}X + r_{8}Y + r_{9}Z + t_{3} \end{pmatrix}
= 0$$
This gives the two equations
$$ fXr_{1} + fYr_{2} + fZr_{3} + ft_{1}+ (px - U)Xr_{7} + (px - U)Yr_{8} + (px - U)Zr_9 + (px - U)t_{3}
= 0$$
$$ fXr_{4} + fYr_{5} + fZr_{6} + ft_{2}+ (py - V)Xr_{7} + (py - V)Yr_{8} + (py - V)Zr_9 + (py - V)t_{3}
= 0$$
I tried to format into matrix times the r and t vector, but the equation was too wide to fit. This formulation allows for the separation into unknowns though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sum nx^n$ I am trying to find a closed form solution for $\sum_{n\ge0} nx^n\text{, where }\lvert x \rvert<1$.
This solution makes sense to me:
$\sum_{n\ge0} x^n=(1-x)^{-1} \\
\frac{d}{d x} \sum_{n\ge0} x^n = \frac{d}{d x} (1-x)^{-1} \\
\sum_{n\ge0} nx^{n-1} = (1-x)^{-2} \\
x \sum_{n\ge0} n x^{n-1} = x(1-x)^{-2} \\
\sum_{n\ge0} nx^n=\frac x{(1-x)^2}$
However, a book I am reading used the following method:
$$\sum_{n\ge0}nx^n=\sum_{n\ge0}x\frac d{dx}x^n=
x\frac d{dx}\sum\limits_{n\ge0}x^n=x\frac d{dx}\frac1{1-x}=\frac x{(1-x)^2}$$
This seems closely related to the solution I described above, but I am having difficulty understanding it. Can someone explain the method being used here?
| Here's a solution that use neither differentiation nor integraion. Since this question is marked as the duplicate target of How do I compute $\sum_{k=1}^{\infty} k \cdot p^k$, which no longer accepts new solution, I'm posting mine for fun.
Use summation by parts
$$S_N = \sum_{n=0}^N a_nb_n = a_N B_N - \sum_{n=0}^{N-1} B_n(a_{n+1}-a_n),$$
where $B_n = \sum_{k=0}^n b_k$ with $a_n = n$ and $b_n = x^n$ for all $n \in \Bbb{N}\cup\{0\}$ and $\lvert x\rvert<1$.
Use geometric sum formula to calculate $S_N$.
\begin{align}
B_n &= \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x} \\
a_{n+1}-a_n &= 1 \\
S_N &= N \cdot \frac{1-x^{N+1}}{1-x} - \sum_{n=0}^{N-1} \frac{1-x^{n+1}}{1-x} \\
&= \frac{1}{1-x} \left[(N- Nx^{N+1}) - \sum_{n=0}^{N-1} (1-x^{n+1}) \right] \\
&= \frac{1}{1-x} \left(- Nx^{N+1} + x\sum_{n=0}^{N-1} x^n\right) \\
&= \frac{1}{1-x} \left(- Nx^{N+1} + x \cdot \frac{1-x^N}{1-x} \right) \\
&= \frac{x}{(1-x)^2} - \frac{N(1+x)x^{N+1} + x^{N+1}}{(1-x)^2} \label{eqn} \tag{$\star$}
\end{align}
Take $N\to+\infty$ to kill the second term in \eqref{eqn}.
Hence $$\boxed{S = \lim_{N\to+\infty} S_N = \frac{x}{(1-x)^2}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Find a general control and then show that this could have been achieved at x2 Determine the general form of $u_0, u_1 ~\text{and} ~ u_2$ if a system of difference equations of the form
$$x_{n+1} = Ax_n + Bu_n,$$
where:
$$A = \begin{pmatrix}
3 & 2 & 2 \\
-1 & 0 & -1 \\
0 & 0 & 1
\end{pmatrix}$$
and:
$$B = \begin{pmatrix}
0 & 0 \\
0 & 1 \\
1 & 0
\end{pmatrix}$$
is to be controlled for $x_0 = 0 ~ to ~ x_3 = [2, 1, 2]^T$ .
Show this target could have been achieved at $x_2$
Solution
So far I have caculated the controlability matrix to be
$$
C
=\begin{pmatrix}
0&0&2&2&6&6\\
0&1&-1&0&-3&-2\\
1&0&1&0&1&0
\end{pmatrix}.
$$
Thus the system is controlable
Now putting Cv=x3 i have the 3 equations
$$
2c+2d+6e+6f=2\\
b-c-3e-2f=1\\
c+e+f=2\\
$$
which i have then put into augmented matrix row echleon form which i have found to be
$$
a-d-2e-3f=1\\
b+d+f=2\\
c+d+3e+3e=1\\
$$
How do I now solve with so many unknowns? also can you please check my working so far is correct. many thanks
| First of all we have forall $n\in N:$ $u_{n}\in R^{2}$.
So let
$$u_{0}=(u_{0,1},u_{0,2})^{t}$$ be given as well as $x_{0}=0$.
Then we get
$$
x_{1}=Ax_{0}+Bu_{0}=0+(0,u_{0,2},u_{0,1})^{t}$$
Again plugging this into the recursion we obtain
$$
x_{2}=Ax_{1}+Bu_{1}=A(0,u_{0,2},u_{0,1})^{t}+B(u_{1,1},u_{1,2}).$$
This has by assumption to be $x_{3}$ so you get $4$ variables to determine.
The system is underdetermined, but this should be no problem since you only have to solve the linear equation
$$x_{3}=A(0,u_{0,2},u_{0,1})^{t}+B(u_{1,1},u_{1,2})$$
with respect to $u_{0,1},u_{0,2},u_{1,1},u_{1,2}$.
An update to your problem. The Kalman matrix $C$ is correct. Now we want to find the vector $u:=(u_{0,1},u_{0,2},u_{1,1},u_{1,2},u_{2,1},u_{2,2})^{t}\in R^{6}$ such that
$$Cu=x_{3}.$$
We can write this in matrix vector representation as
$$
\begin{pmatrix}
0&0&2&2&6&6&|2\\
0&1&-1&0&-3&-2&|1\\
1&0&1&0&1&0&|2
\end{pmatrix}
$$
By changing rows and dividing by 2
$$
\begin{pmatrix}
1&0&1&0&1&0&|2\\
0&1&-1&0&-3&-2&|1\\
0&0&1&1&3&3&|1\\
\end{pmatrix}
$$
This system is in echelon form. So we first determine one specifical solution, for instance
$$u^{*}=\begin{pmatrix}
2\\1\\0\\1\\0\\0\end{pmatrix}
$$.
Next we need the KERNEL of the matrix, that is the elements
$$\operatorname{ker}(C):=\{x\in R^{6}: Cx=0\}.$$ In matrix vector representation this is
$$
\begin{pmatrix}
1&0&1&0&1&0&|0\\
0&1&-1&0&-3&-2&|0\\
0&0&1&1&3&3&|0\\
\end{pmatrix}
$$
Since this matrix has $\operatorname{Rank}(C)=3$, we have to find $3$ vectors, linear independent and satisfying the kernel condition: The vectors
$$v_{1}=\begin{pmatrix}
2\\0\\-3\\0\\1\\0
\end{pmatrix}
\quad
v_{2}=\begin{pmatrix}
0\\2\\0\\-3\\0\\1\end{pmatrix}
\quad
v_{3}=\begin{pmatrix}
1\\-1\\1\\-1\\-2\\2\end{pmatrix}
$$
do the job, so they generate the kernel of $C$. Altogether we have as solution
$$
u\in\{u^{*}+\lambda v_{1}+\mu v_{2}+\sigma v_{3}, \lambda,\mu,\sigma\in R\}.
$$
So if you pick a control vector $u$ in this set, you control the state to $x_{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
finding the real values of $x$ such that : $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
| Elaborating some on what @Fredrik Meyer suggested, one can get:
$$\begin{align*}x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}\hspace{5pt}&\Rightarrow\hspace{5pt} x^2=2+\sqrt{2-\sqrt{2+x}}\hspace{5pt}\Rightarrow\hspace{5pt} x^4-4x^2+4=2-\sqrt{2+x}\\
&\Rightarrow\hspace{5pt}x^8+16x^4+4-8x^6+4x^4-16x^2=2+x\\
&\Rightarrow\hspace{5pt} x^8-8x^6+20x^4-16x^2-x+2=0
\end{align*}$$
The last polynomial can be factored by noticing that it vanishes at $x=2$ and at $x=-1$ into $(x-2) (x+1) (x^3-3 x+1) (x^3+x^2-2 x-1)=0$ (using WA in the end).
Now you can find many roots - but beware: not all of them solve the initial problem: each transition above gives additional assumptions on $x$: first, $x\geq 0$. Then $x^2-2\geq0$ and after that $x^4-4x^2+2\leq0$. Do you see why?
All of those together imply that $\sqrt2\leq x\leq\sqrt{2+\sqrt2}$.
Using some real analysis (or WA :)), one can show that there exists a unique root of that polynomial on $(\sqrt2,\sqrt{2+\sqrt2})$.
NB: This is not the cleanest approach and not very elegant, but it works. I'm almost sure that I have seen much nicer way to solve it, but I can't think of one now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$ Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$
Im not sure should I use primitive root or quadratic residue.
For primitive root, $U_{33} = \{1,2,4,5,7,8,10,13,14,16,17,19,20,23,25,26,28,29,31,32\}$
Thus $\phi_{(3\cdot11)}=20=2^2\times{5}$
But it takes way too much time to test if each of them is primitive root or not...
But Im not quite familiar with the method of Quadratic Residue.
We have $x^8\equiv5\pmod3$ and $x^8\equiv 5 \pmod {11}$
so.... follows the rule of $x^2\equiv q \pmod n$,
we have $(x^4)^2 \equiv 5 \pmod 3$ and $(x^4)^2 \equiv 5 \pmod {11}$
and then how do I continue with it??
| Hint: Consider the equation modulo $3$. What are the solutions to $x^8 \equiv 5 \mod 3$?
Alternatively, you could reduce the above to $(x^4)^2 \equiv -1 \mod 3$, which means that for a solution, you need that $-1$ is a quadratic residue modulo $3$ and $x^4$ is its 'square root'. Is $-1$ a quadratic residue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that :
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ .
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $
What to do next?
| Hint:
From the identity : $\cos A= 1- 2 \sin^2(\frac{A}{2})$
$\sin^2(\frac{A}{2})=1-\cos A$
$1- \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2bc-b^2-c^2+a^2}{2bc}$
$= \dfrac{a^2-(b-c)^2}{2bc}=\dfrac{(a+b-c)(a-b+c)}{2bc}$
$a+b-c=\dfrac{(s-b)}{2}$, $a-b+c=\dfrac{(s-c)}{2}$
$2 \sin^2\dfrac{A}{2}=\dfrac{4(s-b)(s-c)}{2bc}$
$\sin^2(\dfrac{A}{2})=\dfrac{(s-b)(s-c)}{bc}$
Similarly you get $\cos^2(\dfrac{A}{2})$.
$\tan^2 \dfrac{A}{2} = \dfrac{(s-a)(s-c)}{(s)(s-a)}$
Add all such functions, you get:
$\dfrac{(s-b)^2(s-c)^2+(s-c)^2(s-a)^2+(s-a)^2(s-b)^2}{(s)(s-a)(s-b)(s-c)}$
To make the calculation simple, use $b+c-a=x$, $c+a-b=y$ and $a+b-c=z$.
$(s-a)=\dfrac{x}{2}$
$(s-b)=\dfrac{y}{2}$
$(s-c)=\dfrac{z}{2}$
$s=\dfrac{x+y+z}{2}$
Now it becomes algebraic expression. You can find the maximum of it by AM-GM inequality!
$\dfrac{(xy)^2+(yz)^2+(zx)^2}{(x+y+z)(xyz)} \ge \dfrac{3(xyz)^{4/3}}{(x+y+z)(xyz)}$
$\dfrac{3(xyz)^{1/3}}{(x+y+z)} \ge1$ (AM-GM)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Distance: sphere related question Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere ?
How do we do here as I have four options as answer :
(a) $3+\dfrac{\sqrt{30}}{2}$
(b) $3+\dfrac{\sqrt{69}}{3}$
(c) $3+\dfrac{\sqrt{123}}{4}$
(d) $\dfrac{52}{9}$
| Let the centers of your 3 spheres of radius 1 be $C_1, C_2, C_3$. Let the center of the sphere of radius 2 be $C_4$. These 4 points forms a tetrahedron with equilateral triangle base of length 2 ( $C_1C_2 = C_2C_3=C_3C_1=2$) , and a side length of 3 ($C_4C_1 = C_4C_2 = C_4C_3 = 3$). Let $O$ be the center of the equilateral base, then $OC_1 = \frac {2}{3} \sqrt{3} $.
Consider right triangle $OC_1 C_4$, we get that $OC_4 = \sqrt{ C_1C_4 ^2 - O C_1 ^2 } = \sqrt{ 9 - \frac {4}{3} } = \sqrt{ \frac {23}{3} } = \frac {\sqrt{69} }{3}$.
Hence, the distance from the plane to the top is $ 1 + \frac { \sqrt{69} }{3} + 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/341776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
logarithm problem - four tuple How many distinct four tuple (a,b,c,d) of rational numbers are there with $a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7=2005$
Can we proceed like this :
Using $\log a +\log b = \log(ab)$ and $m\log a = \log a^m$
$\Rightarrow \log_{10}2^a \cdot 3^b \cdot 5^c \cdot 7^d = 2005$
Please guide how to proceed further..
| You have
$$
2^a\cdot 3^b\cdot 5^c\cdot 7^d = 10^{2005} = 2^{2005}\cdot 5^{2005}.
$$
So this works if $a=c=2005$ and $b=d=0$.
To see if other solutions exist, observe that you'd get
$$
2^{a-2005}\cdot 5^{c-2005} = 3^{-b}\cdot 7^{-d}.
$$
All the exponents are rational, so we have
$$
2^{n_1/n_2}\cdot 5^{n_3/n_4} = 3^{n_5/n_6}\cdot 7^{n_7/n_8}.
$$
Raising both sides to the power $n_2 n_4 n_6 n_8$, we get all integer exponents. By uniqueness of prime factorizations, that can happen only if all of the integer exponenents are $0$. Therefore there can be no other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/343811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Infinite Factorization Power Series of $\sin(x)$ My teacher has given us a rather long problem and the last part is stumping me. How would one go about factoring the power series of sin(x)? Where:
p(x) = x - $\frac{x^{3}}{3!}$ + $\frac{x^{5}}{5!}$ - $\frac{x^{7}}{7!}$ ...
Better expressed as:
p(x) = $\sum_1^\infty (-1)^{n+1} * \frac{x^{2n-1}}{(2n-1)!}$
The exact phrasing of the last question is:
Use the roots to give an infinite factorization of p(x).
Where in a previous step of the problem, I have already figured out the roots to be -$\pi$, 0, and $\pi$.
| I show here the method used by Euler.
*
*Initial Identities:
Euler showed that
\begin{eqnarray}
\frac{\sin z}{z} = \prod_{k=1}^{\infty} \left ( 1 - \frac{z}{k^2 \, \pi^2} \right )
\end{eqnarray}
He started by considering his formula
\begin{eqnarray}
e^z = \lim_{n \to \infty} \left ( 1 + \frac{z}{n} \right )^n,
\end{eqnarray}
and the substitution of $z$ by $\mathrm{i} z$, so
\begin{eqnarray}
e^{\mathrm{i} z} = \lim_{n \to \infty}
\left ( 1 + \frac{\mathrm{i} z}{n} \right )^n
\end{eqnarray}
Then from using this in the identity,
\begin{eqnarray}
\sin z = \frac{e^{i z} - e^{-i z}}{2 \mathrm{i}}
\end{eqnarray}
he found
\begin{eqnarray}
\sin z =
\lim_{n \to \infty} \frac{ \left ( 1 + \frac{iz}{n} \right )^n
- \left ( 1 - \frac{i z}{n} \right )^n }{2 \mathrm{i}}.
\end{eqnarray}
The key idea of Euler here was the factorization of this
binomial as presented in the next section.
*
*Factorization using roots of unity
Assume $n= 2 p + 1$. The roots of the equation
\begin{eqnarray}
T^n -1 = 0
\end{eqnarray}
are
\begin{eqnarray}
T_r = e^{\frac{2 \pi r \mathrm{i}}{n}}, \quad (r=0, \pm 1, \pm2, \cdots
\pm p ).
\end{eqnarray}
One of the roots is $T_0=1$, while the other roots come in conjugate
pairs, leading to the second degree polynomial
\begin{eqnarray}
(T - e^{\frac{2 \, \pi \mathrm{i}}{n}})
(T - e^{-\frac{2 \, \pi \mathrm{i}}{n}} )
= T^2 - 2 \, T \cos \left ( \frac{2 r \pi}{n} \right ) + 1
\quad r=1,2, \cdots p,
\end{eqnarray}
so the polynomial $T^n -1$ can be factored out as
\begin{eqnarray}
T^n -1 = (T-1) \prod_{k=1}^{p} \left [ T^2 - 2 T \cos \left (
\frac{2 k \pi}{n} \right ) + 1 \right ] \quad (1) .
\end{eqnarray}
Next we make the substitution $T=X/Y$ and clear fractions to get
\begin{eqnarray}
X^n - Y^n = (X-Y) \prod_{k=1}^{p}
\left [ X^2 - 2 X Y \cos \left (
\frac{2 \, k \pi}{n} \right ) + Y^2 \right ]
\end{eqnarray}
We now we do one more substitution. This is
\begin{eqnarray}
X = 1 + \frac{iz}{n}, \quad \quad Y = 1 - \frac{iz}{n}
\end{eqnarray}
and after defining
\begin{eqnarray}
q_n(z) = \frac{ \left ( 1 + \frac{iz}{n} \right )^n
- \left ( 1 - \frac{i z}{n} \right )^n }{2 \mathrm{i} z}.
\end{eqnarray}
and using $X-Y= 2 \mathrm{i} z/n$, we find
\begin{eqnarray}
\label{qnz}
q_n(z) &=& \frac{1}{n} \prod_{k=1}^{p} \left [
\left ( 1 + \frac{iz}{n} \right )^2 -
2 \left ( 1 + \frac{iz}{n} \right ) \left ( 1 - \frac{iz}{n} \right )
\cos \left ( \frac{2 k \pi}{n} \right ) +
\left ( 1 - \frac{iz}{n} \right )^2 \right ] \nonumber \\
&=& \frac{1}{n} \prod_{k=1}^{p} \left (
2 - \frac{2 z^2}{n^2} - 2 \left ( 1 + \frac{z^2}{n^2} \right )
\cos \frac{2 k \pi}{n} \right ) \nonumber \\
&=& \frac{1}{n} \prod_{k=1}^{p} 2 \left (
1 - \frac{ z^2}{n^2} - \left ( 1 + \frac{z^2}{n^2} \right )
\cos \frac{2 k \pi}{n} \right ) \nonumber \\
&=& \frac{1}{n} \prod_{k=1}^{p} 2 \left [
\left ( 1 - \cos \frac{2 k \pi}{n} \right )
- \frac{z^2}{n^2} \left ( 1 + \cos \frac{2 k \pi}{n} \right ) \right ] \nonumber \\
&=& \frac{1}{n} \prod_{k=1}^{p} 2
\left ( 1 - \cos \frac{2 k \pi}{n} \right ) \left [
1 - \frac{z^2}{n^2}
\frac{ 1 + \cos \frac{2 k \pi}{n}} {1 - \cos \frac{2 k \pi}{n}} \right ] \quad (2)
\end{eqnarray}
Now, from equation (1) we see that
\begin{eqnarray}
\frac{T^n -1}{T-1} = \prod_{k=1}^{p} \left [ T^2 - 2 T \cos \left (
\frac{2 k \pi}{n} \right ) + 1 \right ].
\end{eqnarray}
On the other hand we know the trivial identity
\begin{eqnarray}
\frac{T^n -1}{T-1} = 1 + T + T^2 + \cdots + T^{n-1}
\end{eqnarray}
Therefore
\begin{eqnarray}
\lim_{T \to 1}\frac{T^n -1}{T-1} = \prod_{k=1}^{p} \left [ 2 - 2 \cos \left (
\frac{2 k \pi}{n} \right ) \right ] = n
\end{eqnarray}
so equation~\ref{qnz} simplifies to
\begin{eqnarray}
q_n(z) &=& \prod_{k=1}^{p} \left (
1 - \frac{z^2}{n^2} \;
\frac{ 1 + \cos \frac{2 k \pi}{n}} {1 - \cos \frac{2 k \pi}{n^2}} \right ) \quad (3) .
\end{eqnarray}
We know that
\begin{eqnarray}
\lim_{n \to \infty} q_n(z) = \frac{\sin z}{z}.
\end{eqnarray}
We now should find the limit as $n \to \infty$ in equation (3).
Euler assumed that the limit could go inside the product sign. We have the following
calculus identities (easily derived from Taylor expansion of the cosine function)
\begin{eqnarray}
\label{identu0}
\lim_{u \to 0} \frac{1 - \cos a u}{a^2 u^2} &=& \frac{1}{2} \nonumber \\
\lim_{u \to 0} \left ( 1 + \cos a \, u \right ) &=& 2 .
\end{eqnarray}
By choosing $a= 2 k \pi$ and $u=1/n$ in the first identity above we find
\begin{eqnarray}
\lim_{n \to \infty} n^2 \, \left ( 1 - \cos \frac{2 k \pi}{n} \right ) &=& 2 k^2 \pi^2.
\end{eqnarray}
so
\begin{eqnarray}
\label{qnz3}
\lim_{n \to \infty }q_n(z) =
\prod_{k=1}^{p} \left ( 1 - \frac{2 z^2}{2 k^2 \pi^2} \right ) =
\prod_{k=1}^{p} \left ( 1 - \frac{z^2}{k^2 \pi^2} \right ).
\end{eqnarray}
Therefore
\begin{eqnarray}
\frac{\sin z}{z} = \prod_{k=1}^{\infty} \left ( 1 - \frac{z^2}{k^2 \pi^2} \right ) \quad (4).
\end{eqnarray}
*
*Discussion
The materialization of this idea took Euler him more than 10 years of work.
Euler thought about the infinite series as extensions of polynomials where
the roots can be used to factorize them. If this is true, then it is natural
to think that
\begin{eqnarray}
\sin z = z \left ( 1- \frac{z}{\pi} \right ) \left ( 1 - \frac{z}{2 \pi} \right )
\cdots \left ( 1-\frac{z}{k \pi} \right ) \cdots \nonumber \\
\left ( 1- \frac{z}{-\pi} \right ) \left ( 1 - \frac{z}{-2 \pi} \right )
\cdots \left ( 1-\frac{z}{-k \pi} \right ) \cdots
\end{eqnarray}
where $\sin z$ vanishes at each $z= \pm k \, \pi$, $k=0,1, \cdots $. Then by grouping
each factor with that where its root has the opposite sign, he found
\begin{eqnarray}
\sin z = z \left ( 1- \frac{z^2}{\pi^2} \right ) \left ( 1 - \frac{z^2}{2^2 \pi^2} \right )
\cdots \left ( 1-\frac{z^2}{k^2 \pi^2} \right ) \cdots \nonumber \\
\end{eqnarray}
which coincides with (4).
However he got many objections. For example
*
*How does he knows that the order of the factorization
does not matter in the infinite.
*Are those the only complex roots of $\sin z$?,
*$\sin s/s$ and $e^s \sin s/s$ have the same roots, but clearly can not be factored
with the same formula.
Euler was encouraged by the fact that his formula checked out fine for some particular
values found before by Newton and Leibniz's. Also he evaluated the Zeta function
$\zeta(2)=\pi^2/6$ and matched with good accuracy.
The point of convergence is not addressed here. Is there a radius of convergence
(as in infinite series)?
In the proof presented here, still we can not guarantee that the limit can be interchanged
with the $\prod$ sign. That is, is the limit of the product equal to the product of the limits?
That this could be done was proved at a much later time by Weierstrass who generalized
Euler's result with his theorem, that entire functions can be represented by a product
involving their zeroes and in addition, every sequence tending to infinity has an
associated entire functions with zeroes at precisely the points of that sequence
(see Wikerstrass factorization theorem'').
Reference: https://books.google.com/books?id=CYyKTREGYd0C
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Exponential and logarithmic series: Find the sum of $2^2 + 3^2/ 2!+4^2/3!+...$ to infinity
Find the sum of the following series:
$ 2^2 + 3^2/2! + 4^2/3! + ...$ to infinity
The answer is given as $5e$ but I got it as $5e+1$
$T_n = 1/(n-2)! +3/(n-1)! + 1/(n)! $ for $n \ge 2$
and $T_1+T_2 + ... $ to infinity = $4 + e + 3(e-1) +(e-2)$
$=5e-1$
Can you teell me which is correct and how?
| So, $$T_n=\frac{(n+1)^2}{n!}$$
Let $(n+1)^2=n(n-1)+Bn+C$
$\implies n^2+2n+1=n^2+n(B-1)+C$
$\implies B-1=2,B=3, C=1$
So, $$T_n=\frac{(n+1)^2}{n!}=\frac1{(n-2)!}+3\frac1{(n-1)!}+\frac1{n!}$$
Putting $n=0,1,2,3,\cdots$
$$T_0=\frac1{0!}$$
$$T_1=0+3\frac1{0!}+\frac1{1!}$$
$$T_2=\frac1{0!}+3\frac1{1!}+\frac1{2!}$$
$$T_3=\frac1{1!}+3\frac1{2!}+\frac1{3!}$$
$$\cdots$$
$$\text{So, }\sum_{0\le r<\infty}\frac{(n+1)^2}{n!}=e+3e+e$$
$$\text{So, }\sum_{1\le r<\infty}\frac{(n+1)^2}{n!}=e+3e+e-\frac{(0+1)^2}{0!}=5e-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Dot Product Intuition I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ).
For example, suppose I have vector $\mathbf{a} = (a_1,a_2)$ and vector $\mathbf{b}=(b_1,b_2)$. What's the physical or geometrical meaning that
$$a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos(\theta)\;?$$
Why is multiplying $|\mathbf{b}|$ times $|\mathbf{a}|$ in direction of $\mathbf{b}$ the same as multiplying the first and second components of $\mathbf{a}$ and $\mathbf{b}$ and summing ?
I know this relationship comes out when we use the law of cosines to prove, but even then i cant get a intuition in this relationship.
This image clarifies my doubt:
Thanks
| By the scalar product (synonymously, the dot product) of two vectors $\mathbf{A}$ and $\mathbf{B}$, denoted by $\mathbf{A} \cdot\mathbf{B}$, we mean the quantity:
\begin{equation}
\mathbf{A} \cdot\mathbf{B} = |\mathbf{A}| |\mathbf{B}| cos(\mathbf{A},\mathbf{B})
\end{equation}
i.e., the product of the magnitudes of the vectors times the cosine of the angle between them. It follows from the above relationship that the scalar product of $\mathbf{A}$ and $\mathbf{B}$ equals the magnitude of $\mathbf{A}$ times the projection of $\mathbf{B}$ onto the direction of $\mathbf{A}$. As a matter of fact, the projection of $\mathbf{A}$ onto $\mathbf{A}$, denoted by $\mathbf{A_B}$ is length of the segment cut from $\mathbf{B}$ by the planes drawn through the end points of $\mathbf{A}$ perpendicular to $\mathbf{B}$, taken with the plus sign if the direction from the projection (onto $\mathbf{B}$) of the initial point of $\mathbf{A}$ to the projection of the end point of $\mathbf{B}$ coincides with the positive direction of $\mathbf{B}$, and with the minus sign otherwise. We can also conclude that:
\begin{equation}
\mathbf{A} \cdot\mathbf{B} = \mathbf{A_B} \cdot \mathbf{B} =\mathbf{A} \cdot \mathbf{B_A}
\end{equation}
Thus Scalar multiplication of two vectors is commutative. Given a system of rectangular coordinates ${x^1,x^2,x^3}$ let $\mathbf{i_1,i_2,i_3}$ the corresponding basis vectors. Then any vector \mathbf{A} can be represented in the form:
\begin{equation}
\mathbf{A} = A_1 \mathbf{i_1}+A_2 \mathbf{i_2}+A_3 \mathbf{i_3}
\end{equation}
Since the vectors $\mathbf{i_1,i_2,i_3}$ are orthonormal we have
\begin{equation}
\mathbf{i_j} \cdot \mathbf{i_k}=\delta_{j,k}
\end{equation}
Then
\begin{equation}
\mathbf{A} \cdot \mathbf{i_1} = (A_1 \mathbf{i_1}+A_2 \mathbf{i_2}+A_3 \mathbf{i_3}) \cdot \mathbf{i_1} = A_1
\end{equation}
In the similar way we have $\mathbf{A} \cdot \mathbf{i_2}=A_2 $ and $\mathbf{A} \cdot \mathbf{i_3}=A_3 $. In other words, $A_1$, $A_2$ and $A_3$ are the projections of the vector $\mathbf{A}$ onto the coordinate axes. Wae can also write:
\begin{equation}
\mathbf{A} = (\mathbf{A} \cdot \mathbf{i_1}) \cdot \mathbf{i_1} +(\mathbf{A} \cdot \mathbf{i_2}) \cdot \mathbf{i_2} + (\mathbf{A} \cdot \mathbf{i_3}) \cdot \mathbf{i_3}
\end{equation}
The scalar product of two vectors A and B can easily be expressed in
terms of their components:
\begin{equation}
\mathbf{A} \cdot \mathbf{B} = (A_1 \mathbf{i_1}+A_2 \mathbf{i_2}+A_3 \mathbf{i_3}) \cdot (B_1 \mathbf{i_1}+B_2 \mathbf{i_2}+B_3 \mathbf{i_3})
\end{equation}
And when using the orthonormality condition:
\begin{equation}
\mathbf{A} \cdot \mathbf{B} = A_1 B_1 + A_2 B_2 +A_3 B_3
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 14,
"answer_id": 11
} |
Proving $x$ is divisible by $20$ I need to prove that $x$ divisible by $20$ if and only if $x=0\pmod4$ and $x=0 \pmod 5$
proving that if $x=0 \pmod 4$ and $x=0 \pmod 5$ than $x$ divisible by $20$ is by the Chinese theorem (am I right??)
But the other way - I dont understand why its not enough to be divisible only by one of them ($5$ OR $4$)?
P.S. How does it help me prove $(7^n+4*2^n+8^n-3^n)|20$?? How do I open this?
| Hint $\rm\ \ 4,5\mid n\:\Rightarrow\: \dfrac{n}4,\dfrac{n}5\in\Bbb Z\:\Rightarrow\: \dfrac{n}4-\dfrac{n}5 = \dfrac{n}{20}\in\Bbb Z\:\Rightarrow\:20\mid n$
Edit $\ $ Regarding the new question in your edit, the above applies as follows
$\rm mod\ 4\!:\ f(n) = \color{#C00}{7^n}\!+4\cdot2^n\!+8^n\!-3^n\equiv \color{#C00}{3^n}+\,0\,+\,0\,-\,3^n\equiv 0$
$\rm mod\ 5\!:\ f(n) = \color{#C00}{7^n}\!+4\cdot2^n\!+\color{#0A0}{8^n}\!-3^n\equiv \color{#C00}{2^n}\!-2^n+ \color{#0A0}{3^n}-3^n\equiv 0$
Thus $\rm\:4,5\mid f(n)\:\Rightarrow\:20\mid f(n).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
functional equation problem in competition: $ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $
Find all $ f : \mathbb Q \to \mathbb Q $ such that $ f ( 1 ) = 2 $ and
$$ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $$
for all $ x , y \in \mathbb Q $.
thank you very much!
| Put $y=1$. Than we have
$$f(x)=f(x)\cdot f(1)- f(x+1)+1$$
Hence
$$f(x)=2f(x)- f(x+1)+1$$
Hence
$$f(x+1)=f(x)+1$$.
So we have $f(0)=1$.
Now we say $x=-y$, we have
$$f(-x^2)=f(x)f(-x)$$
When we have
$$0=f(-1)=f(-2\cdot \frac{1}{2}) = f(-2) f(\frac{1}{2})- f(-\frac{3}{2}) +1$$
So we have
$$0= -1\cdot f(\frac{1}{2}) - f(\frac{1}{2}) +3$$
Hence
$$0= -1(2\cdot f(\frac{1}{2})-3) $$
Hence $f(\frac{1}{2})=1$
You need more help?
In general we have for $x\in \mathbb{Z}$
\begin{align*}
0&=f(-1)\\
&=f(-x\cdot \frac{1}{x}) \\
&=f(-x) \cdot f(\frac{1}{x}) - f(-x+\frac{1}{x})+1\\
&= (-x+1)\cdot f( \frac{1}{x}) - f(\frac{1}{x}) +x\\
&= -x \cdot( f(\frac{1}{x}) -1)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the dimensions of a cuboid given the volume, surface area, and diagonal
Let the volume, surface area and length of the diagonal of a cuboid be as follows: $144$, $192$, $13$. Find the dimensions.
My trial: $$lbh = 144$$
$$2(lb+bh+lh) = 192 \implies lb+bh+lh = 96$$
$$l^2 + b^2 + h^2 = 13^2 = 169.$$
As $(l+b+h)^2 = l^2 + b^2 + h^2 + 2(lb+bh+lh) = 169 + 192 = 361$. So, $l+b+h = 19$.
Then?
NB: Sorry to all for posting my 1st question in a wrong manner. Thanks for helping me.
| $l,b$ and $h$ are the roots of cubic euqation equation :
$x^3+bx^2+cx+d=0$
$lbh= \dfrac{-d}{1}$
$lb+bh+hl=\dfrac{c}{1}$
$l+b+h= \dfrac{-b}{1}$
Plug in the values and use Vieta's formula.
Aliter:
$lbh=144$
$l^2+b^2+h^2=169$
$169 \equiv 1 \mod 4$
$l^2 \equiv 0 \mod 4, b^2 \equiv 0 \mod 4 $ and $h^2 \equiv 1 \mod 4$
$ \implies$ $$l=2k$$
$$b=2q$$
$$h=2m-1$$
$lbh=144$
One of the three numbers are odd. Odd divisors of $144$ are $3$ and $1$, if $h=1$, $l+b \neq 19$ (Why?). Therefore, $h=3$.
$l^2+b^2=160$ ,$lb=48$ and $l+b=16$
$(l-b)^2=l^2+b^2-2lh= 64 \implies l-b=8$
We get $l=12$, $b=4$ and $h=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/352022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{1}{2}1-\left(a_1+\frac{a_2}{2}+\cdots+\frac{a_n}{2^{n-1}}\right)$
Let $n>1$ be a positive integer and $\frac{1}{2}<a_{j}<1$ for $j=1,2,\ldots,n$. Show that $$(1-a_{1})(1-a_{2})\cdots (1-a_{n})>1-\left(a_{1}+\frac{a_{2}}{2}+\cdots+\frac{a_{n}}{2^{n-1}}\right).$$
I have no ideas.
| Let $b_i=1-a_i$, so that $0<b_i<\frac{1}{2}$. The inequality becomes:
$$b_1b_2 \ldots b_n>1-((1-b_1)+\frac{1-b_2}{2}+\ldots+\frac{1-b_n}{2^{n-1}})$$
$$1-\frac{1}{2^{n-1}}+b_1b_2 \ldots b_n>b_1+\frac{b_2}{2}+\ldots+\frac{b_n}{2^{n-1}}$$
We first prove that $\frac{1}{2}+b_1b_2>b_1+\frac{b_2}{2}$. This is equivalent to $(\frac{1}{2}-b_1)(1-b_2)$, which is clearly true.
Now, we induct on $n \geq 2$ to prove the given inequality.
When $n=2$, we have already proven it above.
Suppose it holds for $n=k \geq 2$, and consider $k+1$. Using the induction hypothesis, (since $0<b_kb_{k+1}<\frac{1}{2}$.)
\begin{align}
1-\frac{1}{2^{k}}+b_1b_2 \ldots b_{k+1} & =\frac{1}{2^k}+(1-\frac{1}{2^{k-1}}+b_1b_2 \ldots b_{k-1}(b_kb_{k+1})) \\
& >\frac{1}{2^k}+b_1+\frac{b_2}{2}+\ldots +\frac{b_{k-1}}{2^{k-2}}+\frac{b_kb_{k+1}}{2^{k-1}}
\end{align}
Using the $n=2$ base case,
$$\frac{1}{2^k}+\frac{b_kb_{k+1}}{2^{k-1}}=\frac{1}{2^{k-1}}(\frac{1}{2}+b_kb_{k+1})>\frac{1}{2^{k-1}}(b_k+\frac{b_{k+1}}{2})$$
Combining gives the desired inequality for $n=k+1$.
We are thus done by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving for three unknowns given three linear equations Have to admit, I forgot how to do simple 9th grade algebra problems. I need to know how to do this stuff for an exam tomorrow. I am doing recurrences and the only thing I need to do is find $a, b$, and $c$ from the following
$$\begin{align*}
a + b + c &= 3\\
4a + 2b+ c&= 6\\
9a + 3b + c& = 13
\end{align*}$$
Can someone refresh my memory on how to solve for $a, b$ and $c$
?
| Just use Gaussian elimination.
$a+b+c=3$
$4a+2b+c=6$
$9a+3b+c=13$
Subtracting equation 1 from equation 2 and equation 3:
$a+b+c=3$
$3a+b+0c=3$
$8a+2b+0c=10$
Subtract 2 times equation 2 from equation 3.
$a+b+c=3$
$3a+b+0c=3$
$2a+0b+0c=4$
We can see that $a=2$
Substitute this into the other equations
$2+b+c=3$
$3(2)+b+0c=3$
$a=2$
We can now see $b=-3$ Substitute this into equation 1.
$2+(-3)+c=3$
$b=-3$
$a=2$
Which shows that $c=4$
Testing these by putting them into the original equations.
$2+-3+4=3$
$4(2)+2(-3)+4=6$
$9(2)+3(-3)+4=13$
And we see they solve the system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $
Evaluate
$$\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2
} \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $$
The question look's like there is a nice method to do it, but I can't figure out. Can someone provide some hint or answer?.
| $\displaystyle{\large%
\int_{0}^{1}
\left(x^{5} + x^{4} + x^{2}\right)\,
\sqrt {4x^{3} + 5x^2 + 10\;}\;{\rm d}x\quad:{\Huge ?}}$
Following @jdh8 ( $\color{#0000ff}{\mbox{There is a missing}\; \color{#ff0000}{\large 3/2}\
\mbox{factor in his formula}}$ ).
$P \equiv a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$
\begin{align}
\left(x^{5} + x^{4} + x^{2}\right)
&=
\left(4x^{3} + 5x^{2} + 10 \right)P'
+
\left(18x^{2} + 15x\right)P
\\[3mm]
{x^{4} + x^{3} + x \over 18x + 15}
&=
{4x^{3} + 5x^{2} + 10 \over 18x^{2} + 15x}\;P'
+
P
\end{align}
Take the limit $x \to 0$. In order to save a divergence we'll get $a_{1} = 0$ and
it follows that $a_{0} = 0$. Then
$$
{x^{4} + x^{3} + x \over 18x + 15}
=
{4x^{3} + 5x^{2} + 10 \over 18x + 15}\;\left(3a_{3} x + 2a_{2}\right)
+
\left[a_{3}x^{3} + a_{2}x^{2}\right]
$$
Again, take the limit $x \to 0$ and we get $a_{2} = 0$. The last expression is reduced to
$$
{x^{3} + x^{2} + 1 \over 18x + 15}
=
{4x^{3} + 5x^{2} + 10 \over 18x + 15}\;\left(3a_{3}\right)
+
\left[a_{3}x^{2}\right]
$$
One more time
$$
x \to 0
\quad\Longrightarrow\quad
{1 \over 15} = {10 \over 15}\left(3a_{3}\right)
\quad\Longrightarrow\quad
a_{3} = {1 \over 30}
\quad\Longrightarrow\quad
\color{#ff0000}{\large P = {x^{3} \over 30}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$ Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$
I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).
| Splitting the series into 2 yields
$$
\begin{aligned}
\sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} &=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}-\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}} \\
&=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}+\sum_{n=-1}^{\infty} \frac{1}{(-4 n-3)^{3}} \\
&=\frac{1}{64}\left[\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}+\sum_{n=-1}^{-\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}\right] \\
&=\frac{1}{64} \sum_{n=-\infty}^{\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}
\end{aligned}
$$
Using the theorem $$\pi \cot (\pi z)=\sum_{n=-\infty}^{\infty} \frac{1}{n+z} ,$$
where $z \not\in \mathbb{Z}.$
Differentiating both sides w.r.t $z$ twice yields
$$
2 \pi^{3}\left(\cot ^{2} (\pi z)+1\right) \cot (\pi z)=\sum_{n=-\infty}^{\infty} \frac{2}{(n+z)^{3}}
$$
Putting $z=\frac{1}{4}$ yields
$$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} = \frac{1}{64} \cdot \frac{1}{2}\left[2 \pi^{3}\left(\cot ^{2}\left(\frac{\pi}{4}\right)+1\right) \cot \frac{\pi}{4}\right]=\frac{\pi^{3}}{32}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 5
} |
integral with $\log\left(\frac{x+1}{x-1}\right)$ I encountered a tough integral and I am wondering if anyone has any ideas on how to evaluate it.
$$\displaystyle \int_{0}^{\infty}\frac{x}{x^{2}+a^{2}}\log\left(\left|\frac{x+1}{x-1}\right|\right)dx=\pi\tan^{-1}(1/a), \;\ a>1$$
I tried breaking up the log and differentiating w.r.t 'a', but did not make considerable progress. Is it possible to do this one with residues?.
I attempted to write it as $$\displaystyle \frac{1}{a^{2}}\int_{0}^{\infty}\frac{x}{1+(x/a)^{2}}\log(x+1)dx-\frac{1}{a^{2}}\int_{0}^{\infty}\frac{x}{1+(x/a)^{2}}\log(x-1)dx$$
Now, make the sub $t=x/a, \;\ dx=adt$
$\displaystyle \int_{0}^{\infty}\frac{t}{1+t^{2}}\log(1+at)dt-\int_{0}^{\infty}\frac{t}{1+t^{2}}\log(at-1)dt$
Differentiate w.r.t 'a' gives:
$$\displaystyle \int_{0}^{\infty}\frac{t^{2}}{(at+1)(t^{2}+1)}dt-\int_{0}^{\infty}\frac{t^{2}}{(at-1)(t^{2}+1)}dt$$
But, this may not be a good idea because of convergence issues.
Does anyone have a good idea of how to approach this one?.
Thanks and take care.
| This is a challenging integral with unexpected twists and turns in its evaluation. Ultimately, though, it all works out.
We begin by integrating by parts, something we normally have no business doing, but the infinities cancel:
$$\int_0^{\infty} dx \frac{x}{x^2+a^2}\log{\left(\frac{x+1}{x-1}\right)} = \\ \lim_{N \rightarrow \infty}\left[\frac{1}{2} \log{(x^2+a^2)}\log{\left(\frac{x+1}{x-1}\right)} \right]_0^N + \frac{1}{2} \int_0^{\infty} dx \log{(x^2+a^2)}\left(\frac{1}{x-1}-\frac{1}{x+1} \right)$$
Now, the limit in the first term on the RHS of the above equation converges to $-i \pi \, \log{a}$. The second term, the integral, has a singularity about which we may deform the interval of integration; the contribution of this deformation is
$$\lim_{\epsilon \rightarrow 0} i \frac{1}{2} \epsilon \int_0^1 d\phi e^{i \phi} \frac{\log{(1+a^2)}}{\epsilon e^{i \phi}} = i \frac{\pi}{2} \log{(1+a^2)}$$
Thus the integral is now equal to
$$PV \int_0^{\infty} dx \frac{\log{(x^2+a^2)}}{x^2-1} + i \frac{\pi}{2} \log{\left ( 1+ \frac{1}{a^2} \right )}$$
To evaluate this, let $I(a)$ be the real part of the above expression (the integral), and differentiate $I(a)$ with respect to $a$ (which I assume is a valid step, I will not prove it here) and get
$$\frac{\partial I}{\partial a} = 2 a \, PV \int_0^{\infty} \frac{dx}{x^2+a^2} \frac{1}{x^2-1} = a PV \int_{-\infty}^{\infty} \frac{dx}{x^2+a^2} \frac{1}{x^2-1}$$
Note that this integral is equal to
$$a \oint_C \frac{dz}{z^2+a^2} \frac{1}{z^2-1}$$
where $C$ is a semicircular arc in the upper half plane, just above the real axis (so we can ignore the poles at $z= \pm 1$, as the integral is a Cauchy principal value). This integral is equal to $i 2 \pi$ times the sum of the residues at the poles inside $C$. The only pole inside $C$ is at $z=ia$, so the integral is
$$a PV \int_{-\infty}^{\infty} \frac{dx}{x^2+a^2} \frac{1}{x^2-1} = a \, i 2 \pi \frac{1}{i 2 a} \frac{1}{(-a^2-1)} = -\frac{\pi}{a^2+1}$$
Integrating with respect to $a$, we find that
$$I(a) = K - \pi \arctan{a}$$
where $K$ is a constant of integration. We determine $K$ as being equal to $I(0)$. So the problem now reduces to evaluating
$$I(0) =2 \int_0^{\infty} dx \frac{\log{x}}{x^2-1}$$
Note that the singularity at $x=1$ is removable in this integral and therefore we do not need to use a Cauchy principal value. We evaluate this integral by once again appealing to the residue theorem, but this time, we consider
$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1}$$
where $C'$ is a keyhole contour with respect to the positive real axis. By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C'$ increases without bound, we get
$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$
This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C'$. As the only pole is at $z=-1$, we see that
$$\begin{align}\oint_{C'} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi \frac{\pi^2}{2}\end{align}$$
Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$. The Cauchy principal value is zero:
$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} - \int_0^{1-\epsilon} \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$
The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:
$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$
so that
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} - i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$
Therefore
$$K = 2 \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{2}$$
and
$$I(a) = \frac{\pi^2}{2} - \pi \arctan{a} = \pi \left ( \frac{\pi}{2} - \arctan{a}\right ) = \pi \arctan{\frac{1}{a}}$$
and
$$\int_0^{\infty} dx \frac{x}{x^2+a^2}\log{\left(\frac{x+1}{x-1}\right)} = \pi \arctan{\frac{1}{a}}+ i \frac{\pi}{2} \log{\left ( 1+ \frac{1}{a^2} \right )}$$
Note the imaginary part, which differs from the original problem statement. By using the absolute values, this imaginary part goes away and the stated result is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Finding a solution to the recurrence relation $a_n = 5a_{n−2} − 4a_{n−4}$ Find the solution to $$a_n = 5a_{n−2} − 4a_{n−4}$$ with $$a_0 = 3$$
$$a_1 = 2$$ $$a_2 = 6$$ $$a_3 = 8$$
My answer: Observe that the degree of recurrence is 4. Hence, the characteristic equation is: $x^4 - 5x^2 +4 = 0$. Solving for $y=x^2$ we get $y_1=1$, $y_2 = 4$. Hence, the solutions are $x_1 = -1$, $x_2 = 1$, $x_3 = -2$, $x_4 = 2$. This implies that $a_n = \alpha_1(-1)^n + \alpha_2 + \alpha_3 (-2)^n + \alpha_4(2^n)$. It remains to solve the initial conditions for $\alpha_i-s : \alpha_1 = 1, \alpha_2 = 1, \alpha_3 = 0, \alpha_4 = 1$.
Where should I go from here to ensure the answer is fully answered and complete? Thank you!
| Just for kicks, use generating functions on this one. Define $A(z) = \sum_{n \ge 0} a_n z^n$. Your recurrence is:
$$
a_{n + 4} = 5 a_{n + 2} - 4 a_n \quad a_0 = 3, a_1 = 2, a_2 = 6, a_3 = 8
$$
Using properties of ordinary generating afunctions:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2 - a_3 z^3}{z^4}
= 5 \frac{A(z) - a_0 - a_1 z}{z^2} - 4 A(z)
$$
Solving for $A(z)$, written as partial fractions:
$$
A(z) = \frac{1}{1 - z} + \frac{1}{1 + z} + \frac{1}{1 - 2 z}
$$
Expanding the geometric series:
$$
a_n = 1 + (-1)^n + 2^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Need help solving an Integral Equation Need help solving:
$$
f(x) = x + \lambda \int_{0}^{1}y(x+y)f(y)dy
$$
keeping terms through $\lambda^{2}$,
(a) by using the Fredholm method
(b) by using the Neumann method
| Ok, I guess I'll try to help with the Neumann method. (Hopefully it's not too late for you.) First, note that if $\lambda$ was $0$, then $f(x) = x$. If both $\lambda$ and $x$ were small, then we'd hope that $f(x) \approx x$. So let's plug $f(x)\approx x$ into the left-hand side of your equation, to get a new approximation $f_{1}(x)$ to $f(x)$:
\begin{align*}
f_{1}(x)
&= x+\lambda \int_{0}^{1} y(x+y)y\, \mathrm{d}y\\
&= x+\lambda\left(\frac{1}{4}+\frac{x}{3}\right) \\
&= \left(1+\frac{\lambda}{3}\right)x + \frac{\lambda}{4}
\end{align*}
Ok, so now let's plug $f_{1}(x)$ into the rhs of your equation to get a better approximation $f_{2}$:
\begin{align*}
f_{2}(x)
&= x+\lambda \int_{0}^{1} y(x+y)f_{1}(y)\, \mathrm{d}y\\
&= x+\lambda \int_{0}^{1} y(x+y)\left[ \left(1+\frac{\lambda}{3}\right)x + \frac{\lambda}{4} \right] \, \mathrm{d}y \\
&=x+\lambda \left[\frac{1}{4}+\frac{x}{3} + \frac{\lambda}{6} + \frac{17 \lambda}{72}x \right] \\
&=\frac{\lambda}{4} + \frac{\lambda^{2}}{6}+ \left(1+\frac{\lambda}{3}+ \frac{17\lambda^{2}}{72} \right)x
\end{align*}
Ok, so we have terms of order $\lambda^{2}$; let's make sure this doesn't change when we refine the approximation:
\begin{align*}
f_{3}(x)
&= x + \lambda \int_{0}^{1} y(x+y)\left[\frac{\lambda}{4} + \frac{\lambda^{2}}{6}+ \left(1+\frac{\lambda}{3}+ \frac{17\lambda^{2}}{72} \right)y \right] \, \mathrm{d}y\\
&=
x+\lambda\left[\frac{11 \lambda ^2}{96}+\frac{\lambda }{6}+\frac{35 \lambda ^2 x}{216}+\frac{17 \lambda x}{72}+\frac{x}{3}+\frac{1}{4} \right] \\
&=\frac{\lambda}{4} +\frac{\lambda^{2}}{6} + \left(1+ \frac{\lambda}{3} + \frac{17\lambda^{2}}{72} \right)x +\mathcal{O}(\lambda^{3}) \quad \text{as } \lambda \to 0.
\end{align*}
So we're good. I'm going to venture a guess that you can obtain the exact solution by assuming $f(x) = c_{0} + c_{1}x$ and then solving for $c_{0}$ and $c_{1}$ in terms of $\lambda$. Perhaps that will help with the Neumann method. When I carried out this procedure, I obtained
\begin{align*}
f(x) = -\frac{18\lambda}{-72+48\lambda +\lambda^{2}}+ \frac{24(\lambda-3)}{-72+48\lambda + \lambda^{2}} x
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Writing a series using Sigma notation How do I write $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}?$$
I have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.
| Notice that $2=\frac {2x^0}1$, so you want $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\frac {2x^0}1+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\sum_{i=1}^5\frac {(i+1)x^{i-1}}{2^{i-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Graph the polynomial $f(x)=x^4-3x^3-4x^2+3x-2$ using a graphing calculator Graph the polynomial $f(x)=x^4-3x^3-4x^2+3x-2$ using a graphing calculator?
a) The Range
b) The Real Zeros
c) The y-intercept
d) relative minimum and Relative Maximum
e) The interval where $f(x)\le0$
| With the help of the calculator is easy to establish:
$$
\begin{array}{c|ccccc} a & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline \\ f(a) & 115 & 16 & -5 & -2 &-5 & -20 & -29 & 10\end{array}
$$
This suggests the minimum is close to $-29$. In fact it is $-29.4136$. For very large $|x|$, the fourth power will dominate other terms, therefore the polynomial grows boundlessly for large $|x|$. Thus the range is $(-29.4136, \infty)$.
The tabulation indicates that real zeros are in the intervals $(-2,-1)$ and $(3,4)$. In fact, using calculator (or W|A) you could establish that
$$
x^4 -3 x^3 - 4x^2+3x-2 = \left(x-\frac{3}{4} \right)^4 - \frac{59}{8} \left(x-\frac{75}{236} \right)^2 - \frac{23737}{15104}
$$
which shows that these are the only real roots, and the polynomial is negative in between them. Approximate values of these roots are $-1.5163$ and $3.86814$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
prove a sequence is bounded above. I want to prove that $x_n=\left(1-\frac{1}{n}\right)^n$ is bounded above.
I know that $b_n=\left(1+\frac{1}{n}\right)^n$ is bounded above, because $b_n\rightarrow \mathcal e$. Indeed, it can be proved that $b_n<3$. Since $b_{n-1}=\frac{n^{n-1}}{(n-1)^{n-1}}$, $x_n=\frac{(n-1)^n}{n^n}=\frac{n-1}{n}.\frac{1}{b_{n-1}}$. So, $\frac{1}{b_{n-1}}>\frac13$, and $(x_n)$ is not bounded above. I commited a mistake, but where?
| If $n\ge 1$, you have $0 \le 1-\frac{1}{n} \le 1+\frac{1}{n}$. Hence $(1-\frac{1}{n})^n \le (1+\frac{1}{n})^n$.
Indeed, $(1-\frac{1}{n}) (1+\frac{1}{n}) = 1-\frac{1}{n^2} \le 1$. So you have $(1-\frac{1}{n})^n (1+\frac{1}{n})^n \le 1$, or $(1-\frac{1}{n})^n \le \frac{1}{(1+\frac{1}{n})^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\arccos$ within $\sin$ How to calculate the following integral:
$\int^{R}_{0}[2 \cos^{-1}(\frac{r}{2R}) -\sin(2 \cos^{-1}(\frac{r}{2R}) ) ] dr$.
This is a part of a complex formula.
| You can use substitution to solve this problem. Clearly the substitution is $u=\cos^{-1}\left(\frac{r}{2R}\right)$ and hence $\cos u=\frac{r}{2R}$. Then $r=0\to u=\frac{\pi}{2},r=R\to u=\frac{\pi}{3}$ and $dr=-2R\sin u$. Thus
\begin{eqnarray*}
I&=&-2R\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}(2u-\sin(2u))\sin udu\\
&=&2R\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(2u\sin u-\sin(2u)\sin u)du
\end{eqnarray*}
Now
\begin{eqnarray*}
\int(2u\sin u-\sin(2u)\sin u)du&=&-2\int ud\cos u-2\int\sin^2u\cos udu\\
&=&-2\int ud\cos u-2\int\sin^2ud\sin u\\
&=&-2u\cos u-\frac{2}{3}\sin^3u+2\int\cos udu\\
&=&-2u\cos u-\frac{2}{3}\sin^3u+2\sin u+C.
\end{eqnarray*}
So
\begin{eqnarray*}
I&=&\left.2R(-2u\cos u-\frac{2}{3}\sin^3u+2\sin u)\right|_{\frac{\pi}{3}}^{\frac{\pi}{2}}
&=&2R\cdot\frac{1}{12}(16-9\sqrt{3}+4\pi)\\
&=&\frac{R}{6}(16-9\sqrt{3}+4\pi).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the last digit of $3^{729}$ I'm practicing for my algebra exam but I stumbled on a question I don't know how to solve.
Let $N = 3^{729}$. What is the last digit of $N$?
The example answer says
Since $\gcd(3, 10) = 1$, check that
$3^4 = 81 = 1 \pmod {10}$:
Now, $729 = 182 \times 4 + 1,$ so we get we get (that might be a typo or missed a step)
$3^{729} = 3 \pmod {10}$.
Can anybody help me with this question? Thanks so much!
| $3^0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 1\equiv3^0\mod 10$
$3^1\equiv 3\times 1\equiv 3\equiv3^1\mod 10$
$3^2\equiv 3\times3\equiv 9\equiv3^2\mod 10$
$3^3\equiv 3\times9\equiv 7\equiv3^3\mod 10$
$3^4\equiv 3\times7\equiv 1\equiv3^0\mod 10$
$3^5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^1\mod 10$
$3^6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^2\mod 10$
$3^7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^3\mod 10$
$...$
$3^{728}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^0\mod 10$
$3^{729}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^1\mod 10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 3
} |
Prove that $3\sum\limits_{i=0}^k\binom{n}{3i}\leq2^n+2$ If $n\in \mathbb{Z^+}$ and $k$ is the largest integer for which $3k\leq n$, then is it true that $\sum_{i=0}^k\binom{n}{3i}\leq \frac{1}{3}(2^n+2)$?
My work: We can break this into two cases: $n=3k+1$ and $n=3k+2$.If $n=3k+1$, then we need to prove that $$\sum_{i=0}^k\binom{3k+1}{3k}\leq \frac{1}{3}(2^{3k+1}+2)$$ The first thing which strikes me is the fact that $2^{3k+1}=\sum_{i=0}^{3k+1}\binom{3k+1}{i}$ and the inequality can now be written after simplification as $$2\sum_{i=0}^k\binom{3k+1}{3i}\leq 2+\sum_{i=0}^k\binom{3k+1}{3i+1}+\sum_{i=0}^k\binom{3k+1}{3i+2}$$ but it does not seem to be helpful.Besides, there is another case($n=3k+2$) left to be tackled as well. Any ideas or hints will be useful.
| [This is essentially the same as Did's, but explicitly avoids the discrete Fourier Transform.]
We use the binomial theorem, which states that
$$ (x + 1 ) ^ n = \sum_{i=0}^n { n \choose i} x^ i $$
Since we want just those values where $i$ is a multiple of 3, we will use the fact that for $\omega \neq 1$ a cube root of 3, $\omega^2 + \omega + 1 = 0$, $ (\omega^2)^2 + \omega^2 + 1 = 0$, $ (\omega^3)^2 + \omega^3 + 1 = 3$.
Hence,
$\begin{array} {l l l l l l l l }
(1 + 1)^ n & = {n \choose 0} \times 1^0 & + {n \choose 1 } \times 1^1 & + {n\choose 2} \times 1^2 & + {n\choose 3} \times 1^3& + \ldots &+ {n \choose n} \times 1^n\\
(\omega + 1)^n & = {n \choose 0} \times \omega^0 & + {n \choose 1 } \times \omega^1 & + {n\choose 2} \times \omega^2 & + {n\choose 3} \times \omega^3 & + \ldots & + {n \choose n} \times \omega^n \\
(\omega^2 + 1)^n & = {n \choose 0} \times (\omega^2) ^0 & + {n \choose 1 } \times (\omega^2) ^1 & + {n\choose 2} \times (\omega^2)^2 & + {n\choose 3} \times (\omega^2)^3 & + \ldots & + {n \choose n} \times (\omega^2)^n \\
\hline\\
& {n \choose 0} \times 3 & + {n\choose 1 } \times 0 & + {n\choose 2} \times 0 & + {n\choose 3} \times 3 & + \ldots & + \ldots\\
\end{array}$
The last line is arrived at by adding up the previous 3 equations by the terms in the columns, and using the property of the cube root of unity. Hence, we see that
$$ \sum_{i=0}^{\lfloor\frac{n}{3} \rfloor} { n \choose 3i} = \frac{ 2^n + (1+\omega)^n + (1+\omega^2)^n} { 3} = \frac{ 2^n + (-\omega)^{2n} + (-\omega)^n}{3} $$
Hence, we can arrive at your inequality since $|\omega| = 1$.
This can further be simplified to the form that Thomas mentioned.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:
*
*$a_0 = 1$,
*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.
It looks like this:
*
*$a_0 = 1$
*$a_1 = {1 \choose 2} + 3$
*$a_2 = {2 \choose 2} + 3{1 \choose 2} + 9$
*$a_3 = {3 \choose 2} + 3{2 \choose 2} + 9{1 \choose 2} + 27$
*$a_4 = {4 \choose 2} + 3{3 \choose 2} + 9{2 \choose 2} + 27 {1 \choose 2} + 81$
I know what the generating function of the sequence $3 ^n = (1, 3, 9, 27, 81, \dots)$ is, as well as what the generating functions for some sequences of combinatorial numbers are, but how do I split the sequence up into these pieces I know?
(The problem is those combinatorial numbers "move right" every time. If they were growing left-to-right along with their coefficients, it would be much easier. And there is no constant difference between $a_i$ and $a_{i + 1}$.)
| As I said in my comment, you have:
$$\displaystyle a_n=3^n+\sum_{k=0}^{n-1}3^k{n-k \choose 2}$$
You can rewrite this as:
$$\displaystyle a_n=3^n+\sum_{k=1}^{n}3^{n-k}{k \choose 2}=3^n+3^n\sum_{k=1}^{n}3^{-k}{k \choose 2}=3^n\left(1+\sum_{k=1}^{n}\left(\frac{1}{3}\right)^{k}{k \choose 2}\right)$$
Also you have:
$$\displaystyle\sum_{k=1}^\infty \left(\frac{1}{3}\right)^{k}{k \choose 2}=\sum_{k=0}^\infty \left(\frac{1}{3}\right)^{k}{k \choose 2}=\dfrac{\left(\frac{1}{3}\right)^{2}}{\left(1-\frac{1}{3}\right)^{3}}=\frac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Stuck on Epsilon proof.. Using the $\epsilon-M $ definition of the limit, calculate
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$
Working so far:
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$
Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$
$$\left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$$
$$\left|\frac{3x^2+7-3(x^2+x+8)}{x^2+x+8} \right|<\epsilon$$
$$\left|\frac{-3x-17}{x^2+x+8} \right|<\epsilon$$
And now I'm stuck.. Any help would be great, thanks.
| Let $x>\max\{\sqrt{17},\frac{3}{\epsilon-1}\}$. We have $x>0$. This gives us
$$\left|\frac{-3x-17}{x^2+x+8} \right|= \left|\frac{3x+17}{x^2+x+8} \right|=\frac{3x+17}{x^2+x+8}<\frac{3x+17}{x^2}<\frac{3}{x}+1<\epsilon$$
This is incomplete as it assumes $\epsilon\neq 1$. When $\epsilon=1$ you can solve a quadratic and find which $x$'s work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
(4x^2+2kx-5)/(x+2) remainder is 3 find value of k? 2 methods - first is long division by $(x+2)$, 2nd is to use remainder theorem
let $f(x) = 4x^2+2kx-5$ and $g(x) = x+2$
to find the remainder of $\frac{f(x)}{g(x)}$ where $g(x) = (x+c)$ we need to evaluate $f(-c)$
$f(-2) = 4(-2)^2+2k(-2) -5$
Because the remainder is 3, we know that $f(-2)=3$
so $$16 - 4k -5 = 3$$
$$4k = 8$$
$$k=2$$
is that correct?
| $4x^2+2kx-5=(x+2)2x+x(2k-4)-5$
$=(x+2)4x+(x+2)(2k-8)-2(2k-8)-5$
$4x^2+2kx-5=(x+2)(2x+2k-8)+11-4k$
Alternatively, putting $x=-2,4(-2)^2+2k(-2)-5=11-4k$ which we have obtained by the long division
So, $11-4k=3\implies k=2$
So, both of the methods are correct. In fact, they are not independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$ To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$
$\frac{1}{x^2(x+a)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x+a)}+\frac{D}{(x+a)^2}$
One method of finding the values of the constants A,B,C & D is as follows.
$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1$
$Ax(x^2+a^2+2ax)+B(x^2+2ax+a^2)+C(x^3+ax^2)+Dx^2=1$
$A(x^3+a^2x+2ax^2)+B(x^2+2ax+a^2)+C(x^3+ax^2)+Dx^2=1$
$x^3(A+C)+x^2(2Aa+B+Ca+D)+x(Aa^2+2Ba)+Ba^2=1$
Equating coeffecients of $x^3,x^2,x^1,x^0$
$A+C=0$; $2Aa+B+Ca+D=0$; $Aa^2+2Ba=0$; $Ba^2=1$
$B=\frac{1}{a^2}$
$A=\frac{-2}{a^3}$
$C=\frac{2}{a^3}$
$D=\frac{1}{a^2}$
Is there a shorter method to find the values of A,B,C & D ?
| In this equation:
$$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1
$$
one little thing you can do is to evaluate at $x = -a$ to get $0 +0 +0 +D(-a)^2 = 1$ to get $D$. After that can evaluate at $x=0$ to get $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Expand the trinomial $(x+y+z)^4$ using the Multinomial Theorem
Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $\binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $\binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomial Theorem states:
$(x_1+\cdots+x_r)^n = \sum\limits_{n_1+...+n_r = n}\binom{n}{n_1,...,n_r}\cdot x_1^{n_1} \cdots x_r^{n_r}$.
The sum runs over all possible partitions $n_1,...,n_r$ such that $n_1+...+n_r=n$
There are only $3$ sets of numbers to get $n=4$:
1. $(4,0,0) \implies$ Here there are $3$ objects but $2$ are the same. So there are $3!/2! = 3$ permutations
2. $(3,1,0) \implies 3! = 6$ Permutations
3. $(2,2,0) \implies 3!/2! = 3$
$3+6+3 = 12$.
So for example, for item one you would have:
$\binom{4}{4,0,0} x^4 y^0 z^0 + \binom{4}{0,4,0} x^0 y^4 z^0 + \binom{4}{0,0,4} x^0 y^0 z^4$
For item $2\\$:
$(3,0,1), (3,1,0), \\
(1,3,0), (0,3,1), \\
(1,0,3), (0,1,3)$
For item $3\\$:
$(2,2,0), (0,2,2,), (2,0,2)$
What are the other $3$ terms? Thank you!
| Perhaps the other three terms that you are missing correspond to (1,1,2)?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
It is an easy question about integral,but I need your help. How to compute this integral?
$$ \int^{\pi}_{0} \frac{\sin(nx)\cos\left ( \frac{x}{2} \right )}{\sin \left ( \frac{x}{2} \right ) } \, dx$$
I need your help.
| Hint:
$\int_0^{\pi}\dfrac{\sin nx . \cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}=\int_0^{\pi}\dfrac{\sin n(\pi -x) . \sin \dfrac{x}{2}}{\cos\dfrac{x}{2}}=1$
$\sin (n \pi-nx)=\sin n{\pi} \cos nx-\cos{n \pi}\sin{nx}= (-1)^{n+1}\sin nx$,
$2I=\int_0^{\pi}\dfrac{\sin nx}{\cos \dfrac{x}{2} \cdot \sin \dfrac{x}{2}}$, when $n \in$ Even
$2I= \int_0^ \pi \dfrac{\sin nx \cdot \cos x}{\cos \dfrac{x}{2} \cdot \sin \dfrac{x}{2}}$, when $n \in$ Odd. Since we know $\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}=\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| Let $x=\sqrt[3]{2+\frac {10}{9}\sqrt 3}, y=\sqrt[3]{2-\frac {10}{9}\sqrt 3}$. Then
$$
x^3+y^3=4,\quad xy=\frac{2}{3}.
$$
The rest is yours.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 4
} |
Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. What is the closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$? We can use Fourier series of $e^{-bx}$ ($|x|<\pi$) to evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{n^2+b^2}$. But this one, it seems to me it is tough to get the closed form.
| This one can also be done with the standard technique of using the $\pi \cot(\pi z)$ multiplier, integrating $$f(z) = \frac{\pi \cot(\pi z)}{(z-a)^2+b^2}$$ along a circle using the same technique as here.
We integrate $f(z)$ along a circle of radius $R$ with $R$ going to infinity and the integral disappears in the limit so that the residues sum to zero (actually a square with vertices $$(\pm(N+1/2),\pm(N+1/2))$$ $N$ a positive integer is easier to handle computationally). The poles of $f(z)$ other than at the integers are at $$z_{0,1} = a\pm ib$$ and the residues are
$$\operatorname{Res}(f(z); z=z_0) =
\left.\frac{\pi \cot(\pi z)}{2(z-a)}\right|_{z=z_0}
= \frac{\pi\cot(\pi(a+bi))}{2bi}
= \frac{\pi}{2bi} i \frac{e^{i\pi(a+bi)}+ e^{-i\pi(a+bi)}}{e^{i\pi(a+bi)}- e^{-i\pi(a+bi)}}$$
and
$$\operatorname{Res}(f(z); z=z_1) =
\left.\frac{\pi \cot(\pi z)}{2(z-a)}\right|_{z=z_1}
= \frac{\pi\cot(\pi(a-bi))}{-2bi}
= \frac{\pi}{-2bi} i \frac{e^{i\pi(a-bi)}+ e^{-i\pi(a-bi)}}{e^{i\pi(a-bi)}- e^{-i\pi(a-bi)}}.$$
Now put $x=e^{i\pi a}$ and $y=e^{-\pi b}.$
The first residue becomes
$$\operatorname{Res}(f(z); z=z_0) =
\frac{\pi}{2b}\frac{xy+1/x/y}{xy-1/x/y}$$
and the second residue is
$$\operatorname{Res}(f(z); z=z_1) =
-\frac{\pi}{2b}\frac{x/y+y/x}{x/y-y/x}.$$
Adding the two contributions and simplifying with Euler's formula yields
$$-\frac{\pi x^2 (y^4-1)}{b (x^2y^2-1)(x^2-y^2)} =
-\frac{\pi (y^2-1/y^2)}{b (x^2-1/y^2)(1-y^2/x^2)} \\=
-\frac{\pi (y^2-1/y^2)}{b (x^2+1/x^2-y^2-1/y^2)} =
\frac{\pi}{b} \frac{\sinh(2\pi b)}{\cos(2\pi a)-\cosh(2\pi b)}.$$
Now with $S$ being our sum we have by the Cauchy Residue Theorem that
$$ S + \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cos(2\pi a)-\cosh(2\pi b)} = 0$$
so that finally
$$ S = \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)}.$$
This MSE link contains a similar computation, this one including the computation of the relevant bounds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 4,
"answer_id": 2
} |
Determine the $n$-th power of matrix Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1 \\ }$$ May you help me with the answer since the book states : $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n & 3\cdot4^n & 2+4^n \\ }$$ while mine misses the ${1\over 6}$. So who is right?
$$$$Thanks in advance for any help you are able to provide.
| I get for the symbolic form depending on n and the symbolic eigenvalues $\alpha,\beta,\gamma$ (where we insert later the known values $\small \alpha=0,\beta=1,\gamma=4$)
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
4 \cdot \beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & -4\cdot\beta^n+3\cdot\alpha^n+\gamma^n & \\
-2\cdot\beta^n+2\cdot\gamma^n & 3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n-3\cdot\alpha^n+\gamma^n \\
-2\cdot\beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n+3\cdot\alpha^n+\gamma^n
\end{array}
$$
and replacing with the actual values for $\small \alpha,\beta,\gamma$ we get
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
4 +2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & -4 +3\cdot 0^n+4^n & \\
-2+2\cdot 4^n & 3\cdot 0^n+3\cdot 4^n & 2-3\cdot 0^n+4^n \\
-2+2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & 2+3\cdot 0^n+4^n
\end{array}
$$
For $n \in \mathbb N \text{ and } n \gt 0$ this reduces to
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
4 +2\cdot 4^n & 3\cdot 4^n & -4+4^n & \\
-2+2\cdot 4^n & 3\cdot 4^n & 2+4^n \\
-2+2\cdot 4^n & 3\cdot 4^n & 2+4^n
\end{array}
$$
$ \qquad \qquad $ which matches the solution of the book.
For $n=0$ this gives with $0^0=1$
$$ A^0 = \frac16 \small \begin{array} {|r|r|r|r}
4 +2 & -3+3 & -4 +3+1 & \\
-2+2 & 3+3 & 2-3+1 \\
-2+2 & -3+3 & 2+3+1
\end{array} = \frac16 \small \begin{array} {|r|r|r|r}
6 & 0 & 0 & \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array} = I
$$
(So the textbook-solution should be commented with the required $0^0$-definition)
I used the diagonalization-procedure of Pari/GP for this and to get a textform for this answer:
M = [2,2,0;1,2,1;1,2,1]
tmpM=mateigen(M)
tmpW=tmpM^-1
tmpD=tmpW*M*tmpM
MPow = tmpM*matdiagonal(['a,'b,'c])*tmpW
\\ on text-level we have later to replace a by \alpha^n and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Weird Identities with Scalar Product & Transpose: $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^T$, $\vec{a}^T \cdot \vec{b} = \vec{b}^T \cdot \vec{a} $? Let's say I have a row vector $\vec{a}$ and a column vector $\vec{b}$:
\begin{align}
\vec{a}= \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad \vec{b} = \begin{pmatrix}1\\2\\3\end{pmatrix}
\end{align}
Now if I want to calculate a scalar product, it is easy for $\vec{a} \cdot \vec{b}= \begin{pmatrix}4 & 5 & 6\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}4\cdot1+ 5\cdot 2 + 6\cdot 3\end{pmatrix} = 32.$
but it is not as easy for $\vec{b}\cdot \vec{a}$ where the scalar product is undefined:
\begin{align}
\vec{b}\cdot \vec{a} = \begin{pmatrix}1\\2\\3\end{pmatrix} \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad{\scriptsize\text{Not defined!}}
\end{align}
So if I take a transpose of both of the vectors I get:
\begin{align}
\vec{b}{\hspace{0.4mm}}^\mathsf{T}\cdot \vec{a}{}^\mathsf{T} = \begin{pmatrix}1&2&3\end{pmatrix} \begin{pmatrix}4 \\ 5 \\ 6\end{pmatrix} = \begin{pmatrix}1\cdot4 + 2 \cdot 5 + 3 \cdot 6\end{pmatrix} = 32
\end{align}
So I get a relation that $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^{T} = 32$. But why do I find a different equation on Wikipedia which says: $(\vec{a}\cdot\vec{b} ){}^\mathsf{T}= \vec{b}{}^\mathsf{T} \cdot \vec{a}{\hspace{0.4mm}}^\mathsf{T}$?
EDIT: If i do it like @Samuel says i define original vectors differently:
\begin{align}
\vec{a}= \begin{pmatrix} 4\\ 5 \\ 6 \end{pmatrix} \qquad \vec{b}= \begin{pmatrix} 1\\ 2 \\ 3 \end{pmatrix}
\end{align}
And now i try to calculate $\vec{a} \cdot \vec{b}$ which is not defined. I can calculate ${\vec{a}}^T \vec{b}$ though:
\begin{align}
{\vec{a}}^T \cdot \vec{b}= \begin{pmatrix} 4& 5 &6 \end{pmatrix} \begin{pmatrix} 1\\ 2 \\ 3 \end{pmatrix} = 4\cdot 1 + 5\cdot 2 + 6 \cdot 3 = 32
\end{align}
Now i try to calculate scalar product $\vec{b} \cdot \vec{a}$ and neither this one is defined. But i can calculate ${\vec{b}}^T\cdot \vec{a}$ though:
\begin{align}
{\vec{b}}^T \cdot \vec{a}= \begin{pmatrix} 1& 2 &3 \end{pmatrix} \begin{pmatrix} 4\\ 5 \\ 6 \end{pmatrix} = 1\cdot 4 + 2\cdot 5 + 3 \cdot 6 = 32
\end{align}
So now i can say ${\vec{a}}^T \cdot \vec{b} = {\vec{b}}^T \cdot \vec{a} $. So this now is a lot wierder and i cant find it on Wikipedia.
| Vectors have a notation, they can be written only as a column matrix. In case of multiplying two such vectors, matrix multiplication is not defined as both would be 3x1 matrices.
\begin{gather}
\vec{u} = \left[ \begin{array}{c}
u_1 \\
u_2 \\
u_3 \\
\end{array} \right] & \vec{v} = \left[ \begin{array}{c}
v_1 \\
v_2 \\
v_3 \\
\end{array} \right]
\end{gather}
Then the following is not defined -
\begin{gather}
\vec{u}\cdot \vec{v} = \left[ \begin{array}{c}
u_1 \\
u_2 \\
u_3 \\
\end{array} \right] \cdot \left[ \begin{array}{c}
v_1 \\
v_2 \\
v_3 \\
\end{array} \right]
\end{gather}
A dyadic multiplication is defined, which gives rise to a column or a row matrix. It is as follows:
\begin{gather}
\vec{u} \otimes \vec{v} = \left[ \begin{array}{ccc}
u_1\cdot v_1 & u_1 \cdot v_2 & u_1 \cdot v_3\\
u_2\cdot v_1 & u_2 \cdot v_2 & u_2 \cdot v_3\\
u_3\cdot v_1 & u_3 \cdot v_2 & u_3 \cdot v_3\\
\end{array} \right]
\end{gather}
It is therefore
\begin{gather}
\vec{u} \otimes \vec{v} = (\vec{v} \otimes \vec{u})^T
\end{gather}
In terms of vector inner product same may be written as
\begin{gather}
\vec{u} \otimes \vec{v} = \vec{u} \cdot \vec{v}^T
\end{gather}
OR
Valid vector inner product is therefore,
\begin{gather}
\vec{u}\cdot \vec{v} = \left[ \begin{array}{c}
u_1 \\
u_2 \\
u_3 \\
\end{array} \right] \cdot \left[ \begin{array}{ccc}
v_1 & v_2 & v_3 \\
\end{array} \right]
\end{gather}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration of a rational function from +/- infinity I am trying to calculate the integral
$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$
where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?
| Why don't you want to simplify it by putting the term under the integral in the following manner:
$$\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}=\frac{\alpha +x}{\beta (c (x-a)^2+1)}-\frac{\gamma }{\delta +x}$$
with $$\alpha =\frac{4 a^2 c+a b^2 c+1}{b^2 c}$$
$$\beta =\frac{4 a^2 c+4 a b^2 c+b^4 c+1}{b^2 c}$$
$$\gamma =\frac{b^2}{4 a^2 c+4 a b^2 c+b^4 c+1}$$
$$\delta =a+b^2$$
which are just constants.
So
$$
\begin{eqnarray*}
\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx=\frac{\alpha}{\beta c}\int_{-\infty}^{\infty} \frac{1}{ (x-a)^2+\frac{1}{c}}dx&+&\\+\frac{1}{\beta c}\int_{-\infty}^{\infty} \frac{x}{ (x-a)^2+\frac{1}{c}}dx &-&\\-\gamma\int_{-\infty}^{\infty} \frac{1}{\delta +x}dx
\end{eqnarray*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why is this derivative incorrect? We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$
I would like to know why my approach is incorrect:
$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$
$$ = \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$$
$$ = \dfrac {2 \sin(x) - \sin(2x) \cdot \cos(x)}{\cos^3(2x) \cdot \sin^2(x)}$$
p.s. - To avoid confusion ; I wanted to get rid of the $\tan$. I'm sure there is a shorter method than this but I don't want it; I just want to know why this is wrong.
| Third line is: $ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)\cdot\cos^2(2x)}{\cos^2(2x) \cdot \sin^2(x)}$
instead of $\\\\ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$
$$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$$
Please help me in this integral, I've tried some substitutions, but nothing work.
Thanks in advance!
| First complete the square:
$$
x^2 + x + 2 = \left( x^2 + x + \tfrac{1}{4} \right) + \tfrac{7}{4} = \left( x + \tfrac{1}{2} \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2.
$$
Now, make the (inverse) trigonometric substitution:
$$
\tan t = \frac{x + \tfrac{1}{2}}{\tfrac{\sqrt{7}}{2}}.
$$
This choice of ratio is motivated by the sum of square expression above. As a consequence, we have
$$
x^2 + x + 2 = \left( \tfrac{\sqrt{7}}{2} \tan t \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2 = \tfrac{7}{4} \left( \tan^2 t + 1 \right) = \tfrac{7}{4} \sec^2 t.
$$
Now, the differential is
$$
dx = \tfrac{\sqrt{7}}{2} \sec^2 t \, dt
$$
and the limits of integration become
$$
\begin{align}
x &= 0 &\Longleftrightarrow \quad t(0) &= \arctan \tfrac{1}{\sqrt{7}} \\
x &= R &\Longleftrightarrow \quad t(R) &= \arctan \tfrac{2R + 1}{\sqrt{7}}.
\end{align}
$$
Now, substitute:
$$
\int_0^R \frac{dx}{(x^2 + x + 2)^3} = \int_{t(0)}^{t(R)} \frac{\tfrac{\sqrt{7}}{2} \sec^2 t \, dt}{\left( \tfrac{7}{4} \sec^2 t \right)^3} = \left( \tfrac{2}{\sqrt{7}} \right)^5 \int_{t(0)}^{t(R)} \cos^4 t \, dt.
$$
Can you finish it from here, using power reducing identities?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x^2+x+1$ is the cube of a prime. Please help me find all natural numbers $x$ so that $x^2+x+1$ is the cube of a prime number.(Used in here)
| This is never a complete solution, but larger than the size that a comment can afford
If $x^2+x+1=p^3$ where $p$ is prime.
Clearly, $p\ne2$
As $p$ divides $x^2+x+1,p$ will divide $(x-1)(x^2+x+1)=x^3-1$
$\implies x^3\equiv1\pmod p\implies ord_px=3\implies 3$ divides $p-1$
So, prime $p=3q+1$ for some natural number $q$
As, $p>2$ i.e., odd, $p$ can be written as $6r+1$ for some natural number $r$
Now, $x^2+x=p^3-1=(6r+1)^3-1=18r(12r^2+6r+1)$
Observe that $(18r,12r^2+6r+1)=1$
As $(x,x+1)=1,$
$x,x+1$ can be $18r,12r^2+6r+1$
$\implies 12r^2+6r+1-(18r)=x+1-x=1$
$\implies 12r^2-12r=0\implies r=1$ as $r>0$
Now, we can test for other obvious combinations unless we can factorize $12r^2+6r+1$
$x=2r,x+1=9(12r^2+6r+1)$
$x=9r,x+1=2(12r^2+6r+1)$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\cot^2 x$? How do you find $\int \cot^2 x \, dx$? Please keep this at a calc AB level. Thanks!
| Since the integrand $$\cot^2 x=\frac{\cos^2 x}{\sin^2 x}$$ is a rational fraction of $\sin x,\, \cos x$, you could use a universal standard substitution called the Weirstrasse substitution
$$
\begin{equation*}
\tan \frac{x }{2}=t,\qquad x =2\arctan t,\qquad dx =\frac{2}{1+t^{2}}dt
\end{equation*},
$$
which converts the integrand into a rational function of $t$ whose evaluation is by partial fractions expansion. From the double-angle formulas and the identity $\cos ^{2}\frac{x}{2}+\sin ^{2}\frac{x}{2}=1$ we get:
$$
\begin{eqnarray*}
\cos x &=&\cos \left(2\cdot\frac{x}{2}\right)=\cos ^{2}\frac{x}{2}-\sin ^{2}\frac{x}{2}=\frac{\frac{\cos ^{2}
\frac{x}{2}-\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}
\frac{x}{2}+\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{1-\tan ^{2}
\frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}=\frac{1-t^2}{1+t^2}, \\
&& \\
\sin x &=&\sin \left(2\cdot\frac{x}{2}\right)=2\sin \frac{x}{2}\cos \frac{x}{2}=\frac{\frac{2\sin \frac{x}{2}
\cos \frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}\frac{x}{2}+\sin ^{2}
\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{2\tan \frac{x}{2}}{1+\tan ^{2}
\frac{x}{2}}=\frac{2t}{1+t^2}.
\end{eqnarray*}
$$
Then
$$
\begin{eqnarray*}
\int \cot ^{2}xdx &=&\int \frac{\left( t^{2}-1\right) ^{2}}{2t^{2}\left(
1+t^{2}\right) }\,dt=\int \frac{1}{2}+\frac{1}{2t^{2}}-\frac{2}{1+t^{2}}dt \\
&=&\frac{1}{2}t-\frac{1}{2t}-2\arctan t+C \\
&=&\frac{1}{2}\tan \frac{x}{2}-\frac{1}{2\tan \frac{x}{2}}+C \\
&=&-\cot x+C,
\end{eqnarray*}
$$
because
$$
\begin{equation*}
\cot x=\frac{\cot ^{2}\frac{x}{2}-1}{2\cot \frac{x}{2}}=\frac{1}{2}\cot
\frac{x}{2}-\frac{1}{2\cot \frac{x}{2}}=\frac{1}{2\tan \frac{x}{2}}-\frac{1}{2}\tan \frac{x}{2}.
\end{equation*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| $$\frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - \sqrt{3b}$$
$$5+2\sqrt{3}=(a-\sqrt{3b})(7+\sqrt{3})$$
$(a-\sqrt{3b})(7+\sqrt{3})=7a+a\sqrt{3}-7\sqrt{3b}-3\sqrt{b}$
Now we have two equotation for two unknown $a$ and $b$:
$5 = 7a-3\sqrt{b}$
and$2\sqrt{3}=(a-\sqrt{b})\sqrt{3} \implies 2=a-\sqrt{b}$.
Now
$\sqrt{b}=\frac{7a-5}{3}=a-2$
$7a-5=3a-6 \implies a = -\frac{1}{4}$
$\sqrt{b} = a - 2 = -\frac{1}{4} - \frac{8}{4}=-\frac{9}{4} \implies b = \frac{81}{16}$
So,
$$a = -\frac{1}{4}, b = \frac{81}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving one function is greater than another How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?
My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0$ for all $x$. However, $q \neq 0,1$ so $f(x) > g(x)$ for all $x$. However, I'm looking for something more mathematically rigorous. Any suggestions?
| $\begin{align}
f(x)-g(x)
&=(x+1)^2-4qx\\
&=x^2+2x+1-4qx\\
&=x^2+2(1-2q)x+1\\
&=x^2+2(1-2q)+(1-2q)^2-(1-2q)^2+1\\
&=(x+1-2q)^2+1-(1-2q)^2\\
\end{align}
$.
Since $0 <q < 1$,
$-1 < 1-2q < 1$
so $0 < (1-2q)^2 < 1$
so $1 > 1-(1-2q)^2 > 0$
so (finally)
$f(x) > g(x)$.
Another way to get this final inequality is
$1-(1-2q)^2
= (1-(1-2q))(1+(1-2q))
= 2q(2-2q)
= 4q(1-q)
$
and both $1$ and $1-q$ are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
What I have done so far is:
$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$
And the solving for variables,
$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$
$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$
$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$
$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$
Am I going about this correctly? Or am I just doing this completely incorrect?
| How familiar are you with basic linear algebra? As @response has posted, you have numerous algebra mistakes. However, as for the method used to solve the problem, I think it's worth noting that you can easily transform this system to one more familiar looking and then use standard technique of Gaussian elimination. That is, rewrite the system as follows:
$
\begin{align}
\begin{bmatrix}
1 & -1 & 0 & 0\\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 3 \\
2 & 0 & 0 & -4 \\
\end{bmatrix}
\end{align}
$
$
\begin{align}
\begin{bmatrix}
a \\
b \\
c \\
d \\
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 \\
1\\
7 \\
6
\end{bmatrix}
\end{align}
$
And then solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/398802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is there any easy way to simplify the following term? $[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ]
+6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$
After the simplification , the result will be:
$3pq \cdot (p^2q + pq^2+2p^2-2pq+2q^2)$
I have tried to simplify it by using the binomial expansion for the power terms and by multiplying , adding, subtracting. it takes a lot of time.
Is there any alternative easy way to simplify the following term?
| Well... This bizzare thing $$[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ]
+6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$$
Can be split like : $T_1-T_2-3T_2+3T_3+3T_3-3T_4$ , Where $T_i$ denotes the $i^{\text{th}}$ term.
$[p (p+1) (p+2) (p+3) (p+q)^3] - [ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ] -3[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ]
+3[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)]+ 3[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$
Now take the common terms out,
$$p(p+1)(p+2)(p+q)^2[(p+3)(p+q)-p(p+q+3)] \\ 3p^2(p+1)(p+q)(p+q+3)[p(p+q+2)-(p+3)(p+q)] \\ + 3p^3(p+q+2)(p+q+3)[(p+1)(p+q)-p(p+q+1)]$$
Consider only the terms inside the square brackets for now. It's the best to just expand them as it is
$$ (p+3)(p+q)-p(p+q+3) = p^2 +pq +3p+3q-p^2 -pq-3p =3q \\
p(p+q+2)-(p+3)(p+q)= p^2+pq+2p -p^2 -pq-2p-2q= -2q \\
(p+1)(p+q)-p(p+q+1)= p^2+pq+p+q-p^2-pq-p =q$$
Now the equation becomes:
$$3pq(p+1)(p+2)(p+q)^2-6p^2q(p+1)(p+q)(p+q+3) +3p^3q(p+q+2)(p+q+3)$$
Again, take the common temrs out, i.e. $3pq$
$$3pq[(p+1)(p+2)(p+q)^2-2p(p+1)(p+q)(p+q+3) +p^2(p+q+2)(p+q+3)$$
Again, split the middle term and common terms out:
$$3pq[(p+1)(p+q)[(p+2)(p+q)-p(p+q+3)]+p^2(p+q+3)[p(p+q+2)-(p+1)(p+q)]]$$
Again, expand the terms in inner square brackets:
$$(p+2)(p+q)-p(p+q+3)=p^2+2p+2q+pq-p^2-pq-3p= 2q-p \\
p(p+q+2)-(p+1)(p+q)= p^2+pq+2p-p^2-pq-p-q=p-q
$$
The equation becomes:
$$3pq[(p+1)(p+q)(2q-p)+p^2(p-q)(p+q+3)]$$
I don't think it can be simplified more, so it's best to expand that...
$$=3pq[2p^2q+2pq^2+2pq+2q^2-p^3-p^2q-p^2-pq
+p^3+p^2q+3p^2-p^2q-pq^2-3pq]$$
Which is indeed equal to
$$3pq[p^2q+pq^2-2pq+2p^2+2q^2]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/399871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side
Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
I got $$s + 13 = 4 + 4\sqrt{7-s}+ 7 – s$$
I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$
I’m stuck at this point. Does anyone have an idea how to solve this equation?
| Clearly $7\ge s\ge-13\iff7-(-3)\ge s+3\ge-13-(-3)$
WLOG $s+3=10\cos 2y$ where $0\le2y\le\pi$
$\implies\sqrt{s+13}=\sqrt{10(1+\cos2y)}=2\sqrt5\cos y,\sqrt{7-s}=2\sqrt5\sin y$
$\implies2\sqrt5\cos y-2\sqrt5\sin y=2$
Squaring we get $$20(1-\sin2y)=4\iff\sin2y=\dfrac45\implies\cos2y=\pm\sqrt{1-\sin^22y}=\cdots$$
Check which values of $s=10\cos2y-3$ satisfies the given equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Solving Simple Mixed Fraction problem? How do you wrap your head around mixed fraction, does anyone knows how to figure out, can someone give me an example how it can be solved?
| Here is an example.
$$2 \frac{3}{11}=\frac{2 \cdot 11}{11}+\frac3{11}=\frac{22}{11}+\frac3{11}=\frac{25}{11}$$
Or conversely,
$$\frac{11}3=\frac{11-3 \cdot 3}{3}+\frac{3 \cdot 3}{3}=\frac{11-9}{3}+\frac{9}{3}=\frac{2}{3}+3=3 \frac{2}3$$.
In general,
$$x+\frac{y}{z}=\frac{x \cdot z}{z}+\frac{y}{z}=\frac{x \cdot z +y}z$$ and vice versa.
In most cases you'll encounter $y<z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the given integral I have got 2 questions I can't seem to get the same answer to as the book.
31) $\displaystyle \int \sin(\sqrt x)\,\mathrm{d}x = -2\sqrt x \cos(\sqrt x)+2\sin(\sqrt x )+c$.
I just get the integral to be $\frac{1}{2}\cos(\frac{1}{x})+c$
and
36) $\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} = -\frac{1}{4}x +\frac{1}{8} \tan^{-1}\left(\frac{2}{x}\right)+c$
This one has me all over the place. The best I can come to is setting $x = 2\tan x$ but thats about it.
| Hints:
For the first one, take: $ t = \sqrt{x} $ and you get $$ 2\int{t \sin(t) dx} $$
For the second:
$$
\begin{align}
\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} &= \dfrac{1}{4} \displaystyle\int \frac{(x^2 + 4 - x^2) \ \mathrm{d}x}{x^2(4+x^2)} \\
&= \frac{1}{4} \left[ \displaystyle\int \frac{\mathrm{d}x}{x^2} - \displaystyle\int \frac{\mathrm{d}x}{x^2 + 4} \right]
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can all points in the plane be represented like this? Solving a task regaring affine geometry, I've come across a problem:
Is it true that, for every point $(x,y)\in \mathbb{R}^2$, there exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that
$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$
I don't have any idea how to prove this or find a counterexample. All hints appreciated!
| Let's play around and see what happens.
I write $a$ for $\alpha$,
$c$ for $\cos \alpha$,
and $s$ for $\sin \alpha$
cause I'm lazy.
We have
$x = t c + s$
and
$y = t s + c$.
$x^2+y^2 = t^2 c^2+2tcs + s^2 + t^2 s^2 + 2tcs + c^2
= t^2+1+4tsc$.
$xy = t^2 cs + tc^2+t s^2 + sc
= cs(t^2+1)+t$.
We can solve for $cs$ in each of these.
$cs = (x^2+y^2-t^2-1)/(4t)$
and
$cs = (xy-t)/(t^2+1)$
so
$(x^2+y^2-t^2-1)/(4t) = (xy-t)/(t^2+1)$.
Now we can solve for $t$.
$4t(xy-t) = (t^2+1)(x^2+y^2-t^2-1)$
or
$4txy-4t^2 = (t^2+1)(x^2+y^2) - (t^2+1)^2
= t^2(x^2+y^2)+(x^2+y^2)-t^4-2t^2-1
$
so
$t^4-t^2(x^2+y^2+2)+4txy+1=0$.
This is a quartic, unfortunately,
and I don't see an obvious root.
We can also do this:
$xs-yc = (tcs + s^2)-(tsc+c^2)
= s^2-c^2
= 2s^2 - 1
$,
so
$yc = xs-2s^2+1$.
Squaring this,
and using $c^2 = 1-s^2$,
we get another quartic for $s$.
In other words,
we can get equations for
$t$
and $\sin \alpha$,
but they are quartics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
addition with a variable (mod) Given $2+x \equiv 7 \pmod 3$.
$2 + 0 = 2$
$2 + 1 = 3$
$2 + 2 = 4$
.
.
.
$2 + 5 = 7$
so, the answer will be $x = 5, 8, 11, 14, 17,\dots$
Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$
| We are given that:
$(2+x) = 7 (\mod 3)$
Thus, $2+x-7 = 3k$ where $k$ is an integer.
Simplifying, yields:
$x-5 = 3k$
Therefore, $x\in (5,8,..)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Whether the given series are convergent or divergent: Whether the given series is absolutely convergent, conditionally convergent or divergent:
$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.
$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.
$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$
Am i right?
| For (b), $\frac{n}{n+2} \to 1$,
so the series is absolutely
fairly violently divergent.
The sum of consecutive even-odd terms is
$\frac{2n}{2n+2} - \frac{2n+1}{2n+3}
= \frac{2n(2n+3)-(2n+2)(2n+1)}{(2n+2)(2n+3)}
= \frac{4n^2+6n-(4n^2+3n+2)}{(2n+2)(2n+3)}
= \frac{3n-2}{(2n+2)(2n+3)}
$,
and the sum of consecutive odd-even terms is
$-\frac{2n+1}{2n+3}+\frac{2n+2}{2n+4} -
= \frac{-(2n+1)(2n+4)+(2n+2)(2n+3)-}{(2n+3)(2n+4)}
= \frac{-(4n^2+7n+4)+(4n^2+6n+6)}{(2n+3)(2n+4)}
= \frac{-n+2}{(2n+3)(2n+4)}
$,
so the series is not even conditionally convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Indefinite integral $\int \left(\frac{\arctan x}{\arctan x - x}\right)^3 \mathrm{dx}$ My imagination doesn't help me with
$$\int \left(\frac{\arctan x}{\arctan x - x}\right)^3 \mathrm{dx}$$
What tools should I use? W|A doesn't help either.
| Hint:
$\because0<\dfrac{\tan^{-1}x}{x}\leq1$ $\forall x\in\mathbb{R}$
$\therefore\int\biggl(\dfrac{\tan^{-1}x}{\tan^{-1}x-x}\biggr)^3~dx$
$=\int\biggl(\dfrac{\dfrac{\tan^{-1}x}{x}}{\dfrac{\tan^{-1}x}{x}-1}\biggr)^3~dx$
$=-\int\dfrac{\dfrac{(\tan^{-1}x)^3}{x^3}}{\left(1-\dfrac{\tan^{-1}x}{x}\right)^3}dx$
$=-\int\dfrac{(\tan^{-1}x)^3}{x^3}\sum\limits_{n=0}^\infty\dfrac{(n+2)(n+1)(\tan^{-1}x)^n}{2x^n}dx$
$=-\int\sum\limits_{n=0}^\infty\dfrac{(n+2)(n+1)(\tan^{-1}x)^{n+3}}{2x^{n+3}}dx$
$=-\int\sum\limits_{n=3}^\infty\dfrac{(n-1)(n-2)(\tan^{-1}x)^n}{2x^n}dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ While playing around with Mathematica, I found that
$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$
Please help me prove this result.
| We can take $\ 2ab= a^2+b^2-(a-b)^2$ to get:
$$I= \frac12 \int_0^1 \frac{\ln^2 (1-x)}{1+x}dx+\frac12\int_0^1\frac{\ln^2(1+x)}{1+x}dx-\frac12 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx$$
By letting $\frac{1-x}{1+x}=t$ and expanding into power series the last one we get:
$$I=\frac12 J +\frac{\ln^3(1+x)}{6}\bigg|_0^1 -\frac12 \int_0^1 \frac{\ln^2 t}{1+t}dt=\frac12 J +\frac{\ln^3 2}{6}-\frac34\zeta(3) $$
$$J=\int_0^1 \frac{\ln^2(1-x)}{1+x}dx\overset{1-x\to x}=\frac12\int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx=\frac12 \sum_{n=0}^\infty \frac{1}{2^n} \int_0^1 x^{n}\ln^2 xdx$$
$$=\sum_{n=0}^\infty \frac{1}{2^n}\frac{1}{(n+1)^3}=2\operatorname{Li}_3 \left(\frac12\right)\Rightarrow \boxed{I=\operatorname{Li}_3 \left(\frac12\right)+\frac{\ln^3 2}{6}-\frac34\zeta(3)}$$
Of course one can rewrite the trilogarithm's value as seen from $(17)$ in this link, but this form is also valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 9,
"answer_id": 1
} |
Prove inequality $(x+y+z-2xyz)^2 \le 2$
Problem: Prove inequality $(x+y+z-2xyz)^2 \le 2\ (1)$ with
$x^2+y^2+z^2 = 1 \land x,y,z \in \mathbb R$
I tried expand $LHS$ and have:
$$(1)\iff 1 - 2 (xy+yz+xz) + 4 xyz(x+y+z)-(2xyz)^2 \ge 0$$
Denote: $xy = a, yz = b,xz=c \implies (1) \iff1-2\sum a+ 4 \sum ab - 2abc \ge0$
But stuck. Please help me.
| Expanding the $LHS$ we have
\begin{align}1+ 2(xy+ xz+ yz) - 4xyz(x+y+z) + 4x^2y^2z^2 \leq 3- 4xyz(x+y+z) + 4x^2y^2z^2
\end{align}
since $(xy+ xz+ yz) \leq (x^2+y^2+z^2)=1$ by Cauchy-Schwarz inequality. Then
\begin{align} 3- 4xyz(x+y+z) + 4x^2y^2z^2 &= 3- 4x^2y^2z^2(1/xy+1/yz+1/xz) + 4x^2y^2z^2\\& \leq 3- 32 x^2y^2z^2
\end{align}
since $\frac{3}{(1/xy+1/yz+1/xz)}\leq \frac{(x^2+y^2+z^2)}{3}\leq \frac{1}{3}$ by harmonic and arithmetic means inequality and the previous estimation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
| HINT:
SO, the equation whose roots are $\cos\frac{r\pi}{12}$ where $r=3,9$ is
$$\left(c-\cos \frac{3\pi}{12}\right)\left(c-\cos \frac{9\pi}{12}\right)=0$$
$$\implies \left(c-\frac1{\sqrt2}\right)\left(c+\frac1{\sqrt2}\right)=0\text{ as }\cos \frac{9\pi}{12}=\cos\left(\pi- \frac{3\pi}{12}\right)=-\cos\frac{3\pi}{12}$$
$$\implies 2c^2-1=0$$
SO, the equation whose roots are $\cos\frac{r\pi}{12}$ where $r=1,5,7,11$ will be
$$\frac{32c^6-48c^4+18c^2-1}{2c^2-1}=0$$
Now, apply Vieta's Formulas 1,2,3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/409396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\le\frac{2}{\sqrt{1+\sqrt{xy}}}$ Let $x,y>0$ and $xy\le 1$. Show that
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}\le\dfrac{2}{\sqrt{1+\sqrt{xy}}}.$$
This inequality have same follow methods?
I saw this.
Let $x,y>0, xy\le 1$.
$$\dfrac{1}{1+x}+\dfrac{1}{1+y}\le\dfrac{2}{1+\sqrt{xy}},$$
because we have
\begin{align}
\dfrac{1}{1+x}+\dfrac{1}{1+y}-\dfrac{2}{1+\sqrt{xy}}&=\left(\dfrac{1}{1+x}-\dfrac{1}{1+\sqrt{xy}}\right)+\left(\dfrac{1}{1+y}-\dfrac{1}{1+\sqrt{xy}}\right)\\
&=\dfrac{(\sqrt{x}-\sqrt{y})^2(\sqrt{xy}-1)}{(1+x)(1+y)(1+\sqrt{xy})}\le 0
\end{align}
Thank you everyone, or have other nice methods?
| As you show
$$\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}},$$
it follows that
$$(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}})^2\le 2 (\frac{1}{1+x}+\frac{1}{1+y})\le \frac{4}{1+\sqrt{xy}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/409936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Finding Maximum Under Constraint Suppose $a$,$b$,$c$ satisfy $a+b+c=1$ and $a$,$b$,$c\in [0,1]$
Find the maximum value of $(a-b)(b-c)(c-a)$
| Another way is not to use calculus. WLOG, let $a=$MAX{$a,b,c$},then if we want max, it must >0, $→ a>c>b$, rewrite it to $\left(a-b\right)\left(c-b\right)\left(a-c\right)$, since $b\ge0$, so $0<a-b\le a,0<c-b\le c → \left(a-b\right)\left(c-b\right)\left(a-c\right)\le ac\left(a-c\right)$, then $b=0$ hold max,since $b=0$ will make $a,c$ much bigger also which make ac(a-c) bigger also, so we can sure $b=0$ is the condition ,ie, $a+c=1$ ,let $a=\cos ^2 x,c=\sin ^2 x$, we have:
$ac\left(a-c\right)=\left(\cos x *\sin x\right)^2\left(\cos ^2x-\sin ^2x\right)=\dfrac{\sin ^2\left( 2x\right)* \cos 2x}{4}=\dfrac{\left(1-\cos \left(4x\right)\right)\cos \left(2x\right)}{8}=\dfrac{\sqrt{\dfrac{\left(1+\cos 4x\right)\left(1-\cos 4x\right)^2}{2}}}{8}=\dfrac{\sqrt{\left(2+2\cos 4x\right)\left(1-\cos 4x\right)\left(1-\cos 4x\right)}}{16}\le \dfrac{\sqrt{\left(\dfrac{4}{3}\right)^3}}{16}=\dfrac{\sqrt{3}}{18}$ (AM-GM), the max will be hold when $1-\cos 4x=2+2coos4x$, ie.
$\cos 4x=-\dfrac{1}{3}→ \cos 2x=\sqrt{\dfrac{1-1/3}{2}}=\dfrac{\sqrt{3}}{3} → \cos ^2x=\dfrac{1+\dfrac{\sqrt{3}}{3}}{2}=\dfrac{3+\sqrt{3}}{6}=a, c=\dfrac{3-\sqrt{3}}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
Given $x,y,z\geq0$ and $xy+yz+zx=1$. Show that $\displaystyle\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
I've tried many things but all failed. The only thing I know is that the equality holds when $x=y=1, z=0$.
Please help. Thank you.
| $z=\dfrac{1-xy}{x+y} \ge 0 \to xy\le 1$
LHS$=\sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x+y}{1+x^2}}+\sqrt{\dfrac{x+y}{1+y^2}}=f(x,y)$
$f'_{x}=\dfrac{1-x^2-2xy}{(x^2+1) \sqrt{x^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(y^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<1>
$f'_{y}=\dfrac{1-y^2-2xy}{(y^2+1) \sqrt{y^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(x^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<2>
<1>$*(1+y^2)$-<2>$*(1+x^2)$:
$(y^2-x^2)\left(2(x+y)(1-xy)+\dfrac{(1+x^2)(1+y^2)}{\sqrt{1+x^2}{1+y^2}}\right)=0$
so we have $x=y$ only. put in <1>, we get: $4x-4x^3=(x^2+1)^{\frac{3}{2}}$,ie:
$15x^6-35x^4+13x^2-1=0 \to (3x^2-1)(5x^4-10x^2+1)=0$, we get two roots:
$x_1=\dfrac{1}{\sqrt{3}}, x_2=\sqrt{1-\dfrac{2}{\sqrt{5}}}$
$f(x_1)=2.79,f(x_2)=2.77$,but we note $f(x,y)$ when $x ,y \to \infty, f(x,y) \to 0$,so we have to check the bound, which is $xy=1$.
$f(x,y)=\sqrt{x}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{\dfrac{1}{x}+x}}$
let $m=\sqrt{x}+\dfrac{1}{\sqrt{x}} \ge2$
$f(m)=m+\dfrac{1}{\sqrt{m^2-2}}$, when $m\ge 2$, $f(m)$ is mono increasing, so the minimum will be got when $m=2 \to x=y=1, f(1)=2+\dfrac{1}{\sqrt{2}}< 2.77$, so the minimum is $2+\dfrac{1}{\sqrt{2}}$.
To prove $f(m)=m+\dfrac{1}{\sqrt{m^2-2}}$ is mono increasing function,
$f'(m)=1-\dfrac{m}{(m^2-2)^{3/2}}=1-\dfrac{1}{\sqrt{m^2-2}\left(m-\dfrac{2}{m}\right)}>0$ when $m\ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$ using contour integration The question is to solve the integral using concepts of contour integrals:
$$\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$$
| Remark: the integrand $P_r(t)$ is known as the Poisson kernel and the result follows immediately from a term-by-term integration of the normally converging (when $0\leq r<1$) series $P_r(t)=\sum_{n\in\mathbb{Z}}r^{|n|}e^{in t}$ over $[0,2\pi]$. But since you want contour integration...
Hint: when $0<r<1$ use Cauchy's integral formula given that
$$
\frac{1-r^2}{1+r^2-2r\cos t}=\mbox{Re}\left( \frac{1+re^{it}}{1-re^{it}}\right)=\mbox{Re}\left( \frac{1+z}{1-z}\right)\qquad z=re^{it}
$$
and factor $r^2$ out in the numerator and the denominator to reduce to the latter when $r>1$, which yields $-1$.
Details upon request: when $0<r<1$, the function$f(z)=\frac{1+z}{1-z}$ is holomorphic on the open unit disk, which contains $\gamma$ the circle of radius $r$ and center $0$. So by Cauchy's integral formula
$$1=f(0)=\frac{1}{2i\pi}\int_\gamma \frac{f(z)}{z}dz=\frac{1}{2i\pi}\int_0^{2\pi} \frac{1+re^{it}}{1-re^{it}}\cdot \frac{ire^{it}}{re^{it}}dt=\frac{1}{2\pi}\int_0^{2\pi} \frac{1+re^{it}}{1-re^{it}}dt$$whence
$$1=\mbox{Re}\left(\frac{1}{2\pi}\int_0^{2\pi} \frac{1+re^{it}}{1-re^{it}}dt\right)=\frac{1}{2\pi}\int_0^{2\pi}\mbox{Re}\left( \frac{1+re^{it}}{1-re^{it}}\right)dt=\frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1+r^2-2r\cos t}dt.$$
When $r>1$, we have
$$
\frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1+r^2-2r\cos t}dt=\frac{1}{2\pi}\int_0^{2\pi}\frac{-r^2(1-r^{-2})}{r^2(1+r^{-2}-2r^{-1}\cos t)}dt=-1
$$
by application of the above to $0<r^{-1}<1$.
For $r=0$, we find $1$. And for $r=1$, we get $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluting $\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} \operatorname dx$ $$\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx$$
My try:: $\displaystyle \int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx = \int_{1}^{2}\frac{\tan^{-1}x}{\tan^{-1}(x-1)-\tan^{-1}(x-2)}dx$
Now How can i solve after that.
plz help me
Thanks
| Observe that the quadratic $ x^2 -3x +3 $ has discriminant $$b^2-4ac=(-3)^2-4(1)(3)=9-12=-3$$ Hence no real solutions, which means it never crosses the $x$-axis, and since it opens up, this means it is always positive. Using this, and the following formula for $\arctan \frac{1}{u}$ $$\arctan \frac{1}{u}=\frac{1}{2}\pi - \arctan u, u>0 $$ We can rewrite the denominator of our integral as $$\frac{1}{2}\pi - \arctan (x^2-3x+3)$$ having used $u=x^2-3x+3$ to apply the formula. We now have $$\int_1^2 \frac{\arctan x}{\frac{1}{2} \pi - \arctan (x^2-3x+3)}dx$$ This is as far as I got, and wolfram alpha didn't find a closed-form solution for that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
How find $a$ such that $x^2-\sqrt{a-x}=a$ has exactly two real solutions Consider the equation
$$ x^2-\sqrt{a-x}=a.$$ I wish to determine the values of $a$ for which the above equation has exactly two real solutions (for $x$).
My idea:
$$a-x=(x^2-a)^2=x^4-2ax^2+a^2\Longrightarrow f(x)=x^4-2ax^2+x+a^2-a=0$$ and we must have
$$a-x\ge 0$$
$$x^2= a+\sqrt{a-x}\ge a,$$
so $f(x)=x^4-2ax^2+x+a^2-a=0$ has only real solution $x$, and this solution $x^2\ge a, x\le a.$ But can I use this to find the possible values of $a$?
then we have $f''(x)=0 ?$
$$\Longrightarrow f'(x)=4x^3-4ax+1,\Longrightarrow f''(x)=12x^2-4a=0$$
then $$12x^2=4a\Longrightarrow a=3x^2>0$$
so $$ a^2\le \sqrt{\dfrac{a}{3}}\le a$$
$$\Longrightarrow \dfrac{1}{3}\le a\le \sqrt[3]{\dfrac{1}{3}}$$
my reslut is true?and I think this problem have other nice methods.
I hope someone can write the final results，because I don't know the correct result.Thank you everyone
| Hint:$x^2-a=\sqrt{(a-x)}\ge 0$.So we have $x\le a\le x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge 6.$$
Without loss of generality we can let $x=\max{\{x,y,z\}}$
Proof: case 1
$x\ge y\ge z$
we can easily prove
$$\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge\sqrt{y^2+x+2}+\sqrt{z^2+z+2}$$
and
$$\sqrt{x^2+y+2}+\sqrt{y^2+x+2}\ge\sqrt{x^2+x+2}+\sqrt{y^2+y+2}$$
so we have
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge \sqrt{x^2+x+2}+\sqrt{y^2+y+2}+\sqrt{z^2+z+2}.$$
Then use
$$\sqrt{x^2+x+2}\ge\dfrac{3}{4}x+\dfrac{5}{4}$$
$$\sqrt{y^2+y+2}\ge\dfrac{3}{4}y+\dfrac{5}{4}$$
$$\sqrt{z^2+z+2}\ge\dfrac{3}{4}z+\dfrac{5}{4}$$
to get the result. Whereas the case 2 when $x\ge z\ge y$ can be proved using the same methods.
Now,I have another idea: using Holder inequality
we have
$$\left(\sum\sqrt{x^2+yz+2}\right)^2\left(\sum\dfrac{x^2+2yz+9}{x^2+yz+2}\right)\ge 36^3$$
$$\Longleftrightarrow \sum\dfrac{x^2+2yz+9}{x^2+yz+2}\le 1296$$
and the following link has some discussion about this problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538230
and
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752&p=3097872#p3097872
and Vasc gave the hint:
$$\sum\sqrt{8(a^2+bc+2)}\ge \sum\sqrt{(3a+b+c)^2+7}\ge 12\sqrt 2$$
How prove this hint?Thank you everyone.
and my other idea is as follows:
let $a=\min(a,b,c)$
we can prove
$$\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq \sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\tag{1}$$
\begin{align*}
&\sqrt{a^2+bc+2}+\sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8} \tag{2}
\end{align*}
Summing up
\begin{align*}
&\sum_{cyc}{\sqrt{a^2+bc+2}}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8}\\
=&\sqrt{a^2+\frac{(3-a)^2}{4}+2}+\sqrt{(3-a)^2+2a(3-a)+8}
\end{align*}
By the way: someone said $(1)$ is wrong? why? can anyone give an example? And hopefully someone can use this method to prove this inequality? Thank you very much!
| Here is my sketch, mostly to convince that alternative approach (i.e. inequalities instead of search for minimum) really may be possible. At the other hand, useful inequalities have "differential" nature (that is, I think they may be most easily proven by differentiation, but there may be other ways to proof).
Let us denote $a=x^2+yz, b=y^2+zx, c=z^2+xy$; then
$$a+b+c=x^2+yz+y^2+zx+z^2+xy=x^2+y^2+z^2+\frac{1}{2}((x+y+z)^2-(x^2+y^2+z^2))=$$
$$=x^2+y^2+z^2+4.5-\frac{1}{2}(x^2+y^2+z^2)\ge\frac{1}{2}3+4.5=6$$
Starting from reasonably looking inequality $\sqrt{u+2}+\sqrt{v+2}+\sqrt{w+2}\ge\sqrt{u+v+w+18}$ (where $u,v,w\ge 0$) we let $u=a,v=b,w=c$ and conclude that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge\sqrt{24}=\frac{6}{\sqrt{3/2}}$$
Starting from another inequality $f(u,v,w)\equiv\sqrt{u+2}+\sqrt{v+2}+\sqrt{w+2}\ge g(u,v,w)\equiv\sqrt{3(u+v+w)+18} $ (where $u,v,w\ge 0$ and $u\approx v\approx w$) we let $u=a,v=b,w=c$ and conclude that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge\sqrt{36}=6$$
Last inequality was made so that $f(2,2,2)=g(2,2,2)$ and $f'_u(2,2,2)=g'_u(2,2,2)$, $f'_v(2,2,2)=g'_v(2,2,2)$, $f'_w(2,2,2)=g'_w(2,2,2)$ (and yes, we need to analyze $f''$ vs. $g''$ too). Last inequality DOES NOT hold when, say, $u=7,v=0,w=0$ (5.83 vs. 6.25) so one probably need another similar inequality to cover, say, cases like $u\approx3v\approx3w$
Note, that it may be possible that I shall not convert the sketch into real proof.
I do not want to say that this alternative approach (that closely mimics differentiation) is the only possible. There may be possible other approaches, for example, based on the following lemma:
Let $U\ge u \ge \delta\ge 0$, then $\sqrt{U}+\sqrt{u}\ge\sqrt{U+\delta}+\sqrt{u-\delta}$
Such approach may be much more elegant, but I still had no success with it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
} |
Determinant of symmetric matrix Given the following matrix, is there a way to compute the determinant other than using laplace till there're $3\times3$ determinants?
\begin{pmatrix}
2 & 1 &1 &1&1 \\
1 & 2 & 1& 1 &1\\
1& 1 & 2 & 1 &1\\
1&1 &1 &2&1\\
1&1&1&1&-2
\end{pmatrix}
| Look at the matrix $$A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & -3 \end{pmatrix}.$$ It has rank two and its nonzero eigenvalues have sum $1$ (the trace) and product $$\det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} = -16$$
So the characteristic polynomial is $t^3 (t^2 - t - 16)$. Evaluate this at $t = -1$ and you get $$\det(-I - A) = (-1)^5 \det \begin{pmatrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & -2 \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
Thanks!
| This limit problem seems to cooperate rather nicely with the "conjugate factor" method. Multiply the numerator and denominator by the "conjugate" of the numerator to obtain
$$\lim_{x \rightarrow 1} \ \frac{\sqrt{x^2+3}-2}{x^2 - 1} \ \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2} $$
$$ = \ \lim_{x \rightarrow 1} \frac{(x^2+3) \ - \ 4 }{(x^2 - 1) \ \cdot \ (\sqrt{x^2+3}+2) } = \ \lim_{x \rightarrow 1} \frac{x^2 \ - \ 1 }{(x^2 - 1) \ \cdot \ (\sqrt{x^2+3}+2) } $$
$$= \ \lim_{x \rightarrow 1} \frac{ 1 }{ \sqrt{x^2+3}+2 } \ = \ \frac{1}{4} \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to solve this system of non-linear equations My knowledge of algebra is still insufficient to solve this problem. Any help in solving the system of equations would be greatly appreciated.
$$
xy(x+y)=30\\
x^3+y^3=35
$$
| Note that
$$
(x+y)^3
= x^3 + 3x^2y + 3xy^2 + y^3 = x^3+y^3 + 3xy(x+y)
= 35 + 3 \cdot 30 = 125
$$
and therefore $x+y = \sqrt[3]{125} = 5$ and $xy = \frac{30}{x+y} = 6$.
Thus $y = 5-x$ and $6 = xy = x(5-x)$, and so $x^2-5x+6=0$, which implies $x \in \{2,3\}$ and so $y \in \{3,2\}$. Thus, either $(x,y)=(2,3)$ or $(x,y) = (3,2)$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/422129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$
| We have
$$\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}=\frac{\sin x^2\cos\frac{1}{x}+\sin\frac{1}{x}\cos x^2-\sin\frac{1}{x}}{x}$$
and since $\sin x^2\sim_0 x^2$ and $\left|\cos \frac{1}{x}\right|\leq 1$ and since $\cos x^2\sim_0 1-\frac{x^4}{2}$ we can see easily that the given limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$ If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}$$ then how I proceed. Please help.
| We have by Titu's Lemma, for any $x,y$ reals and $a,b >0$
$ \dfrac{x^2}{a}+\dfrac{y^2}{b} \ge \dfrac{(x+y)^2}{a+b}$
$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$ is quite direct. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Linear Independence by row space.
Let $(1,1,-1), (2,1,0)$ and $(-1,0,1)$ be vectors, show if they are independent.
I wrote each vector on the rows of the matrix $A$.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}$
Then I put $A$ on an echelon form.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}\overset{L2-2L1} {\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ -1 & 0 & 1 \end{pmatrix}\overset{L3+L1} {\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 1 & 0 \end{pmatrix}\overset{L3+L2}{\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 0 & 2 \end{pmatrix}$
So, $r(A)=3$. If $\space r(A)=3$ that is equal to the number of rows$(n)$ of $A$, then I can conclued that the $3$ vectors are independent.
If $r(A)<n$ , then the set of the $3$ vectors would be dependent. Is this right? Thanks.
| Edit: The row operations you performed and the row echelon matrix are indeed correct.
You're correct that since $r = 3 = n$ and hence, the vectors are linearly independent.
Note, since the echelon form of the matrix is now in upper triangular form, we can "read off" the determinant of the matrix in echelon form, which is the same as the determinant of the matrix we started with: why? (Because the only elementary row operations used here are those that did not change the value of the determinant of the original matrix.)
$$\det(A) = \left|\begin{matrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{matrix}\right| = \left|\begin{matrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 0 & 2 \end{matrix}\right| = (1)(-1)(2) = -2 \neq 0$$
Since the $\det A \neq 0$, we know the rows (and columns) of $A$ are linearly independent.
And you are correct that when $r\lt n$, the vectors represented by the rows are then linearly dependent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the general equation of the ellipse that is not in the origin and rotated by an angle? I have the equation not in the center, i.e.
$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$$
But what will be the equation once it is rotated?
| If you came here looking for how
$$c_0x^2 + c_1y^2 + c_2xy + c_3x + c_4y + c_5 = 0$$
relates to $h, k, a, b, A$ its the following:
\begin{eqnarray}
c_0&=&\frac{\cos^2(A)}{a^2} + \frac{\sin^2(A)}{b^2}\\
c_1&=&\frac{\sin^2(A)}{a^2} + \frac{\cos^2(A)}{b^2}\\
c_2&=&\frac{\sin(2A)}{a^2} - \frac{\sin(2A)}{b^2}\\
c_3&=&-\frac{2 h \cos^2(A)}{a^2} - \frac{k\sin(2A)}{a^2} - \frac{2 h \sin^2(A)}{b^2} + \frac{k \sin(2A)}{b^2}\\
c_4&=&-\frac{h \sin(2A)}{a^2} - \frac{2k\sin^2(A)}{a^2} + \frac{h \sin(2A)}{b^2} - \frac{2k \cos^2(A)}{b^2}\\
\end{eqnarray}
$$c_5 = \frac{h^2 \cos^2(A)}{a^2} + \frac{h k \sin(2A)}{a^2} + \frac{k^2 \sin^2(A)}{a^2} + \frac{h^2 \sin^2(A)}{b^2} - \frac{ h k \sin(2A)}{b^2} + \frac{k^2 \cos^2(A)}{b^2} - 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 6,
"answer_id": 1
} |
A constrained extremum problem Find the maximum possible value of $$A = a^{333} + b^{333}+c^{333}$$ subject to the constraints $$a+b+c=0$$ and $$a^2+b^2+c^2=1,$$
where $a,b,c\in \mathbb{R}$
Thank you for helping me.
| $a+b+c=0$, so at least two of $a,b,c$ have same sign, WLOG, let $bc \ge 0 $
$A=a^{333}+b^{333}+c^{333} = (-b-c)^{333}+b^{333}+c^{333}=- \sum_{i=1} ^{332} C_i b^{333-i}c^i$,
We want $A$ is max, so b,c must be negative, ,let $x=-b,y=-c $, $ \to x \ge 0,y \ge 0, xy \le \dfrac{1}{6}, x+y=\sqrt{\dfrac{1}{2}+xy} $
$ A=(x+y)^{333}-(x^{333}+y^{333}) \le \left(\dfrac{1}{2}+xy \right)^{\frac{333}{2}}-2(xy)^{\frac{333}{2}} $
$f(t)=\left(\dfrac{1}{2}+t \right)^k-2t^k$, we will prove $f(t)$ is mono increasing function when $ 0 \le t \le \dfrac{1}{6} , k >3$:
$f'(t)=k\left[ \left(\dfrac{1}{2}+t \right)^{k-1}-2t^{k-1} \right]$, $f'(t)>0 \iff \left(\dfrac{1}{2}+t \right)^{k-1}-2t^{k-1} >0 \iff \left(\dfrac{1}{2}+t \right)^{k-1}>2t^{k-1} \iff \left(\dfrac{1}{2}+t \right) > 2^{\frac{1}{k-1}}t \iff \dfrac{1}{2}> (p-1)t , p=2^{\frac{1}{k-1}} , 1<p<2, $
so last one is true, $f'(t)>0 \implies A_{max}=\left(\dfrac{1}{2}+\dfrac{1}{6} \right)^{\frac{333}{2}}-2(\dfrac{1}{6})^{\frac{333}{2}}=\left(\dfrac{4}{6} \right)^{\frac{333}{2}}-2(\dfrac{1}{6})^{\frac{333}{2}}=\dfrac{2^{333}-2}{6^{166}\sqrt{6}}$, max will be hold when $x=y,xy=\dfrac{1}{6} \to x=y=\dfrac{1}{\sqrt{6}} \to b=c=-\dfrac{1}{\sqrt{6}}, a= \dfrac{2}{\sqrt{6}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that $d_n=2^n+3^n$, where $d_n = 5d_{n-1}-6d_{n-2}$ I have one more induction question.
$d_0 =2 $ $d_1=5$
let $d_n=5d_{n-1} - 6d_{n-2}$
Prove that $d_n=2^n+3^n$
| Assuming $d_0=2$ and $d_1=5$ (which also starts the induction, incidentally):
let $n \geq 2$, and suppose the property holds for all $0\leq k< n$. Then
$$
\begin{align*}
d_{n} &= 5d_{n-1} - 6d_{n-2} \\
&= 5\left(2^{n-1}+3^{n-1}\right) - 6\left(2^{n-2}+3^{n-2}\right) \qquad\qquad\text{(induction hypothesis)} \\
&= 5\cdot 2^{n-1}+5\cdot 3^{n-1} - 3\cdot 2^{n-1}- 2\cdot 3^{n-1} \\
&= 2\cdot 2^{n-1}+3\cdot 3^{n-1} \\
&= 2^{n}+3^{n}
\end{align*}
$$
so by induction, the property holds for all $n\in\mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $w^2 + x^2 + y^2 = z^2$, then $z$ is even if and only if $w$, $x$, and $y$ are even I'm trying to go through the MIT opencourseware Mathematics for Computer Science (6.042J). I've been stumped for half a day trying to figure it out. Something isn't clicking, and I could use some help.
Here is the problem:
Suppose that $w^2 + x^2 + y^2 = z^2$, where $w$, $x$, $y$, and $z$ always denote positive integers.
(Hint: It may be helpful to represent even integers as $2i$ and odd integers as $2j + 1$, where $i$ and $j$ are integers)
Prove the proposition: $z$ is even if and only if $w$, $x$, and $y$ are even. Do this by considering all the cases of $w$, $x$, $y$ being odd or even.
| First of all, the sum of a set of integers will be even iff the number of odd summands is even(may be $0$).
So, here either none or two of $w,x,y$ are odd.
If none of them is odd, we are done.
If two of them are odd, say $w=2W,x=2X+1,y=2Y+1$ where $W,X,Y$ are integers.
$w^2+x^2+y^2=(2W)^2+(2X+1)^2+(2Y+1)^2=4(W^2+X^2+X+Y^2+Y)+2\equiv2\pmod 4$
But $(\pm1)^2\equiv1, (\pm2)^2\equiv0\pmod 4$
So, there can be no integer $z,$ such that $z^2\equiv2\pmod 4$
Conversely, if $w,x,y$ are even $z^2=w^2+x^2+y^2\equiv0\pmod 4\implies z$ is even
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
| Forget your last term and Just set $x = 1/n$ in the second
The first term of the series is $x/2 - x/3=x/6.$
Since $x=1/n,$ the result is just $1/6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Calculate residue at essential singularity I know you can calculate a residue at an essential singularity by just writing down the Laurent series and look at the coefficient of the $z^{-1}$ term, but what can you do if this isn't so easy?
For instance (a friend came up with this function): what is the residue at $z = 0$ of the function $\dfrac{\sin\left(\dfrac{1}{z}\right)}{z-3}$?
The Laurent series of the sine is $\displaystyle \frac{1}{z} - \frac{1}{6z^{3}} + \frac{1}{120z^{5}} - \cdots + \cdots$
but if you divide by $(z-3)$, you get $\displaystyle \frac{1}{z(z-3)} - \frac{1}{6(z-3)z^{3}} + \frac{1}{120(z-3)z^{5}}+\cdots$
Now the series isn't a series solely "around" $z$! How to proceed further? Or shouldn't you try to write down the Laurent series?
Many thanks.
| We can look at the power series $$\frac{1}{\frac{1}{z} - 3} = \frac{z}{1 - 3z} = z + 3z^2 + ...$$ and $$\sin(z) = z - \frac{1}{6}z^3 + ...$$ so $$\frac{\sin(z)}{\frac{1}{z} - 3} = \Big(z + 3z^2 + ...\Big)\Big(z - \frac{1}{6}z^3 +...\Big) = z^2 + 3z^3 + ...$$ and $$\frac{\sin(\frac{1}{z})}{z - 3} = \frac{1}{z^2} + \frac{3}{z^3} +...$$ has residue $0$ at $z = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int \frac{\mathrm dz}{z^3 \sqrt{z^2 - 4}}$ $$\int \frac{dz}{z^3 \sqrt{z^2 - 4}}$$
$z = 4\sec\theta$
$dz = 4\sec\theta \tan d\theta$
$$\int \frac{\sec\theta \tan\theta}{4^3 \sec^3 \theta \tan \theta}$$
$$ \frac{1}{4^3} \int \frac{d \theta}{\sec^2 \theta}$$
I am again stuck, I have no idea how to proceed.
| Putting $z^2-4=u^2, zdz=udu$
$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{udu}{(u^2+4)^2u}=\frac{du}{(u^2+2^2)^2}$$
This can be addressed using Trigonometric identities $u=2\tan\theta$ like my other answer
Alternatively, $$\text{Let }I_n=\int \frac {du}{(u^2+a^2)^n}$$
$$=\int du\cdot\frac1{(u^2+a^2)^n}-\int\left(\int du \frac{d\frac1{(u^2+a^2)^n}}{du}\right)du $$
$$=\frac u{(u^2+a^2)^n}+2n\int \frac{u^2}{(u^2+a^2)^{n+1}}du$$
$$=\frac u{(u^2+a^2)^n}+2n\left(\int \frac1{(u^2+a^2)^n}-a^2\int \frac1{(u^2+a^2)^{n+1}} du\right)$$
$$\implies I_n=\frac u{(u^2+a^2)^n}+2n\left(I_n-a^2I_{n+1}\right)$$
$$\implies 2a^2n\cdot I_{n+1}=\frac u{(u^2+a^2)^n}+(2n-1)I_n $$
$$n=1\implies 2a^2 \int \frac {du}{(u^2+a^2)^2}=\frac u{(u^2+a^2)}+\int \frac {du}{(u^2+a^2)}=\frac u{(u^2+a^2)}+\frac1a\cdot \arctan \frac ua+K$$
$$a=2\implies 8 \int \frac {du}{(u^2+4)^2}=\frac u{(u^2+4)}+\int \frac {du}{(u^2+4)}=\frac u{(u^2+4)}+\frac12\cdot\arctan \frac u2+K$$
Now, $u=\sqrt{z^2-4}, u^2+4=z^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why such function does not exist? I could not prove the following:
A function $f \in \mathscr{C}^2([0, \pi])$, such that $$f(0) = f(\pi) = 0,\\
\int_0^{\pi} (f'(x))^2dx = 1,\\
\text{and }\int_0^{\pi} (f(x))^2dx = 2$$
Then such function does not exist.
I think that I have to use the Rayleigh quotient and have a contradiction for the eigenvalue $\frac{1}{2}$. Thanks in advance.
| This was meant as a comment to O.L.'s answer, but it was too long:
Here is an array of the four values of $(a,b)$ for $f(x)=(a+bx)x(\pi-x)$ that satisfy the given conditions:
$$
\left(
\begin{array}{cc}
\frac{\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}-\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi
^5} , -\frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \\[6pt]
\frac{\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}+\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi
^5} , -\frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \\[6pt]
\frac{-\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}-\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi
^5} , \frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \\[6pt]
\frac{-\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}+\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi
^5} , \frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}}
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Given $B \in M_{n\times n}(\mathbb R)$ is invertible and $B^2+B^4+B^7 = I$, find an expression for $B^{-1}$ in terms of only $B$. Given $B \in M_{n\times n}(\mathbb R)$ is invertible and $B^2+B^4+B^7=I$, find an expression for $B^{-1}$ in terms of only $B$.
I don't know where to start. Thanks in advance.
| Since $B^2 + B^4 + B^7 = I$, we have $B(B + B^3 + B^6) = I$; thus $B^{-1} = B + B^3 + B^6$. Note also that since $B^7 + B^4 + B^2 - I = 0$, any eigenvalue $\lambda$ of $B$ must satisfy $\lambda^7 + \lambda^4 + \lambda^2 - 1 =0$, hence cannot be $0$; thus the hypothesis that $B$ is invertable is extraneous, given that $B$ satisfies $B^2 + B^4 + B^7 = I$. Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find the limits $\lim\limits_{h\rightarrow 0} \frac{e^{-h}}{-h}$ and $\lim\limits_{h\rightarrow 0} \frac{|\cos h-1|}{h}$? How to work around to find the limit for these functions :
*
*$$\lim_{h\rightarrow 0} \frac{e^{-h}}{-h}$$
*$$\lim_{h\rightarrow 0} \frac{|\cos h-1|}{h}$$
For the second one i think that the limit doesn't exist.
| We know that $e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ Note that for $|h|<1$ and sufficiently small,
$$
\frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1
\leq
e^{-h}
\leq
\frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1+\frac{1}{2!}(-h)^2
$$
which is the same as
$$
1-h
\leq
e^{-h}
\leq
1-h+\frac{1}{2}h^2
$$
implies
$$
1-\frac{1}{h}
\geq
\frac{e^{-h}}{-h}
\geq
1-\frac{1}{h}-\frac{1}{2}h
$$
Make $h\to 0$. What you get? Now the other limit. Note that
\begin{align}
\frac{|\cos(h)-1|}{h}
=
&
2\frac{\left|\cos\left(2\dfrac{h}{2}\right)-1\right|}{\dfrac{h}{2}}
\\
=
&
2\frac{\left|1-2\sin^2\left(\dfrac{h}{2}\right)-1\right|}{\left(\dfrac{h}{2}\right)}
\\
=
&
4\frac{\left|\sin^2\left(\dfrac{h}{2}\right)\right|}{\left(\dfrac{h}{2}\right)}
\\
=
&
4\left(\dfrac{h}{2}\right)\frac{\sin^2\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)^2}
\end{align}
Make $h\to 0$. What you get?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Wolfram double solution to $\int{x \cdot \sin^2(x) dx}$ I calculated this integral :
$$\int{x \cdot \sin^2(x) dx}$$
By parts, knowing that $\int{\sin^2(x) dx} = \frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x) +c$. So I can consider $\sin^2(x)$ a derivative of $\frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x)$, and I get this result:
$$\frac{1}{4} \cdot x^2 - \frac{x}{4} \cdot \sin(2x) + \frac{1}{4} \cdot \sin^2(x) +c$$
I get the confirm on wolfram if I try to compute the derivative of $\frac{1}{4} \cdot x^2 - \frac{x}{4} \cdot \sin(2x) + \frac{1}{4} \cdot \sin^2(x) +c$, but here if I try to compute the integral of $\int{x \cdot \sin^2(x) dx}$ I get this result:
$$\frac{x^2}{4} -\frac{1}{4} \cdot x \cdot \sin(2x) -\frac{1}{8} \cdot \cos(2x) $$
But $\frac{1}{8} \cdot \cos(2x)$ isn't equal to $\frac{1}{4} \cdot \sin^2(x)$, which is the correct result?
| By the double angle identity for cosine, they differ by a constant. Add a constant of integration $d$ to your answer, and the two answers will be equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b^2+c^2 &\geq ab+bc+ca\\
2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\
\left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0
\end{align*}
$$
Which is obviously true.
However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:
$$
\begin{align*}
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\
a+b+c &\geq a+b+c
\end{align*}
$$
Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?
Edit: Please don't help me with the original inequality, I'll figure it out.
| (@HaiDangel told me.
https://diendantoanhoc.net/topic/182934-a3-b3-c3geqq-a2b-b2c-c2a/?p=731023)
A stronger version: Let $a, b, c$ be real numbers with $a + b \ge 0, b + c \ge 0$ and $c+a\ge 0$. Prove that
$$a^3 + b^3 + c^3 \ge a^2b + b^2c + c^2a.$$
I have an SOS expression:
\begin{align}
&a^3 + b^3 + c^3 - a^2b - b^2c - c^2a \\
=\ & \frac{(a^2+b^2-2c^2)^2 + 3(a^2-b^2)^2
+ \sum_{\mathrm{cyc}} 4(a+b)(c+a)(a-b)^2}{8(a+b+c)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Trigonometric problem : Eliminate $\theta$ and $\phi$ from the relation and find relation between p and q Question :
Eliminate $\theta$ and $\phi$ from the relation
$$\begin{align}
p \cot^2\theta + q \cot^2\phi &= 1 &(1)\\
p \cos^2\theta + q \cos^2\phi &= 1 &(2)\\
p \sin\theta &= q\sin\phi &(3)
\end{align}$$
Also find relation between $p$ and $q$.
I have tried different ways but unable to eliminate $\theta$ and $\phi$.
One method: If I subtract equations $(1)$ and $(2)$ then I got:
$$p\frac{\cos^4\theta}{\sin^2\theta} +q \frac{\cos^4\phi}{\sin^2\phi}=0$$
Please guide how to solve this.. thanks..
| HINT:
$$p\sin\theta=q\sin\phi\implies p^2\sin^2\theta=q^2\sin^2\phi$$
$$\implies p^2\cos^2\theta-q^2\cos^2\phi=p^2-q^2\ \ \ \ (4)$$
Use $(2),(4)$ to solve for $\cos^2\phi,\cos^2\theta$
If $\cos^2\phi=y, \cot^2\phi=\frac{\cos^2\phi}{1-\cos^2\phi}=\frac y{1-y}$
Put the values of $\cot^2\phi,\cot^2\theta$ in $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I write a sum of cosines as a product of sines? I am trying to prove that
$$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle.
I tried up to the stage of
$$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$
but how do I proceed from here?
| Recall the identities
$$
\begin{align}
2\sin(x)\sin(y)&=\cos(x-y)-\cos(x+y)\tag{1}\\
2\sin(x)\cos(y)&=\sin(x-y)+\sin(x+y)\tag{2}
\end{align}
$$
First use $(1)$ on $2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)$, then $(2)$ on the results. Finally, use $A+B+C=\pi$.
$$
\begin{align}
&4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\\
&=\color{#C00000}{2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)}
\color{#00A000}{-2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A+B}{2}\right)}\\
&=\color{#C00000}{\sin\left(\frac{C+B-A}{2}\right)+\sin\left(\frac{C+A-B}{2}\right)}
\color{#00A000}{+\sin\left(\frac{A+B-C}{2}\right)-\sin\left(\frac{A+B+C}{2}\right)}\\[6pt]
&=\cos(A)+\cos(B)+\cos(C)-1
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove more general identity:
$$
\begin{array}{|c|}
\hline \\
\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2 \\
\hline
\end{array}
, \qquad \mbox{ where } \ n \ \mbox{ is odd.}$$
| Whenever I see a problem like this, I think Chebyshev polynomials.
The Chebyshev polynomials of the first kind, $T_m(x)$, are defined so that:
$$T_m(\cos \theta) = \cos m\theta$$
$T_m$ is an $m$th degree polynomial, and the roots of $T_m(x)-1$ are exactly $\cos 2\pi k/m$ for $k=0,1,\dots,m-1$.
When $m=2n$ is even, we can write $$T_{2n}(x)=1+C\prod_{k=0}^{n-1}\left(x^2-\cos^2\frac{k\pi}{n}\right)$$ for some constant $C$. The occurrence of those $\cos^2\frac{k\pi}{n}$ suggested this might be a good approach.
Now, given a polynomial $p(x)=a_mx^m + a_{m-1}x^{m-1}\dots +a_0$, with non-zero roots $r_0,\dots, r_{m-1}$, we have formulas:
$$\sum \frac{1}{r_k} = -\frac{a_1}{a_0}$$
And:
$$\sum_{i< j} \frac{1}{r_ir_j} = \frac{a_2}{a_0}$$
So $$\sum \frac{1}{r_k^2} = \left(-\frac{a_1}{a_0}\right)^2 - 2\frac{a_2}{a_0}$$
Now let $p(x)=T_{2n}(x)-1$. The roots of $p$ are $r_k=\cos \frac{k\pi}{n}$ for $k=0,\dots,2n-1$, and the roots are non-zero since $n$ is odd. We know that $p(x)$ is even, so $a_1=0$. Finally, we know that $\sum_{k=0}^{2n-1} \frac{1}{r_k}^2$ is twice the sum that you are looking for.
So your sum is now reduced to finding $-\frac{a_2}{a_0}$ where the $a_0,a_2$ are coefficients of $p(x)$. Since $p(0)=T_{2n}(\cos \pi/2)-1 = \cos n\pi - 1 = -2$, so we know that $a_0=-2$. So $\sum r_k^{-2} = a_2$. (Again, we use $n$ odd here.)
So we need to prove that $a_2=2n^2$.
Now, $a_2=\frac{1}{2}T_{2n}^{''}(0)$. Let $f(x)=\cos 2nx = T_{2n}(\cos x)$. Differentiating we get:
$$-2n\sin 2nx = -T_{2n}^{'}(\cos x)\sin x$$
Differentiating both sides again:
$$-4n^2\cos 2nx = T_{2n}^{''}(\cos x)(\sin^2 x) - T_{2n}^{'}(\cos x)\cos x$$
Putting in $x=\pi/2$, then $\cos x=0$, $\sin x=1$, and $\cos 2nx=-1$. Therefore, we get $$4n^2 = T_n^{''}(0)\\a_n=\frac{1}{2}T_{2n}^{''}(0)=2n^2$$
and therefore your sum is $n^2$.
Note: Any symmetric rational function with rational coefficients of $\{\cos 2\pi k/n\mid k=0,\dots,n-1\}$ will be rational by this argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 4
} |
Partial fractions for $\frac{t+1}{2\sqrt{t}(t-1)}$ How do I use partial fractions for the expression $\dfrac{t+1}{2\sqrt{t}(t-1)}$? Because I have to find the integral of it... Thank you
| Your given integrand isn't a rational function (yet) to use partial fractions, we must first obtain a ratio of polynomials. We can do this by substituting $u = \sqrt t$.
$$u = \sqrt t \implies t = u^2 \implies dt = 2u\,du$$
Now, substituting the above into our original integral gives us:
$$\int \frac{t+1}{2\sqrt{t}(t-1)}\,dt = \int \frac{u^2 + 1}{2u(u^2 - 1)}\,(2u\,du) =\int \dfrac {u^2 + 1}{u^2 - 1}\,du$$
Now, polynomial division, followed by "partial fractions" gives us: $$\int \left(1 + \dfrac {1}{u^2 - 1}\right)\,du = \int \left(1 + \dfrac {1}{(u-1)(u+1)}\right)\,du = \int \left(1 + \dfrac{A}{u - 1} + \dfrac B{u + 1} \right) \,du$$
Now we solve for $A, B$:
$A(u+1) + B(u - 1) = 2 \iff Au + A + Bu - B = 2 \iff (A + B)u + (A - B) = 2$
$A + B = 0$
$A - B = 2$
Adding the equations gives us $2A = 2 \iff A = 1 \implies B = -1$ and we'll have a result of the form $$u + A\ln|u-1| + B\ln |u+1| + \text{Constant}$$ $$ = \sqrt t +\ln|\sqrt t - 1| - \ln |\sqrt t + 1| + \text{Constant} $$
$$ = \sqrt t+ \ln\left|\dfrac{\sqrt t - 1}{\sqrt t+1}\right| + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
| $a=\dfrac{b+c}{bc-1}>0 \to bc>1 \implies $ two of ${a,b,c} >1 $, otherwise ,one will be negative.
WOLG, let $a>1,b>1$. if $c\le 1, \implies (a-1)(b-1)(c-1) \le 0$ so the possible max value will be got when $c>1$.
$x=a-1>0,y=b-1>0,z=c-1>0, x+y+z+3=xyz+x+y+z+xy+yz+xz+1 \to xyz+xy+yz+xz=2$
$xy+yz+xz \ge 3(xyz)^{\frac{2}{3}} .\ \ u=(xyz)^{\frac{1}{3}}>0 \to u^3+3u^2 \le2 \iff u^3+3u^2-2 \le 0 \iff (u+1)(u^2+2u-2) \le 0 \to (u+1+\sqrt{3})(u+1-\sqrt{3})\le 0 \iff u \le \sqrt{3}-1 \implies xyz \le (\sqrt{3}-1)^3 \implies [(a-1)(b-1)(c-1)]_{max}=(\sqrt{3}-1)^3$
the "=" will hold when $x=y=z=\sqrt{3}-1 \implies a=b=c=\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/445293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to solve $x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$ solve the equalition
$$x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$$
I have seen some methods, for this problem.
my idea:
let $\sqrt[3]{x^3+x+1}=y,$ then
$\sqrt[3]{y^3+y+1}+y=-x-3$
and following have nice methods? thank you everyone.
| Let $y=\sqrt[3]{x^3+x+1}$ and $z=\sqrt[3]{y^3+y+1}$, then we translate the original equation into$$x+y+z+3=0$$ by following the procedure in the original post. We also have, by direct calculation:
$$z^3=y^3+y+1=x^3+x+1+y+1$$ so that $$z^3+z+1=x^3+(x+y+z+3)=x^3$$ So we have the suggestive equations$$y^3=x^3+x+1$$$$z^3=y^3+y+1$$$$x^3=z^3+z+1$$
From which we would immediately guess $x=y=z$ and $=-1$ follows immediately.
Suppose the three values are not equal and wlog that $y\gt x$, with $x$ being the least value, then (using the first of the three equations) $$(y-x)(x^2+xy+y^2)=x+1$$ The left-hand side is the product of two positive factors* so we have $x\gt -1$ and ($x$ being the least value, and values being unequal) $x+y+z\gt -3$, which is a contradiction.
For * note $$x^2+xy+y^2=(x+\frac y2)^2+\frac {3y^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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.