Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$f(x)=x^3-ax^2-ax+1,\;\;|x_1|+|x_2|+|x_3|=3$
Find all $a\in \mathbb R$ such that the zero-points of the function
$$f(x)=x^3-ax^2-ax+1$$ Satisfy the equation: $$|x_1|+|x_2|+|x_3|=3$$
My work:
$$
\begin{split}
f(x) &= x^3-ax^2-ax+1 \\
&= x^3+1-ax(x+1)\\
&= (x+1)(x^2-x+1-ax) \\
&=(x+1)(x^2-(a+1)x+1) \\
x_1 &= -1
\end{split}
$$
By Viéta's formulae:
$$
\begin{split}
x_2 + x_3 &= a+1 \\
x_2 \cdot x_3 &= 1\implies x_2\;\&\;x_3>0 \lor x_2\;\&\;x_3<0\\
a &= 1\implies x_2=x_3=-1 \leftarrow \Delta=0 \\
&\text{square of a binomial, double solution}
\end{split}
$$
$$$$
Then it really is:
for $x_1=x_2=x_3=-1$
$$|x_1|+|x_2|+|x_3|=3$$
Is this correct?
| $ x_1 = -1 $, we have
$$ |x_2| + |x_3| = 2 $$
now, if $ x_2, x_3 $ real Number
can be four solution.
*
*$ x_2 > 0, x_3 > 0 $
$x_2 + x_3 = a+1$
$a =1$
*
*$ x_2 < 0, x_3 < 0 $
$x_2 + x_3 = -(a+1)$
$a =-3$
*
*$ x_2 < 0, x_3 > 0 $ or $ x_2 > 0, x_3 < 0 $
$ x_2 - x_3 = +- \sqrt{(a+1)^2 - 4} $
If $x_2 = x_3$
So,
$ (a+1)^2 - 4 = 0 $
$ a = -3 or a = 1$
if not ? $ a < -3 or a > 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $(x-1)^3$ divides $p(x)+1$ and $(x+1)^3$ divides $p(x)-1$. Find a polynomial $p(x)$ of degree 5. Let $(x-1)^3$ divides $p(x)+1$ and $(x+1)^3$ divides $p(x)-1$. Find a polynomial $p(x)$ of degree 5.
Here's what I have tried:—
As $(x-1)^3$ divides $p(x)+1$,
$p(1)+1=0$, $p(1)=-1$
$p(-1)=1$
Letting $p(x)=a_5x^{5} + a_4x^{4} + a_3x^{3} + a_2x^{2} + a_1x + a_0$
I don't know how to proceed any further.
I think the divisibility rule is gonna be used somewhere.
| Let $p(x)$ be of degree 5. Since $(x-1)^3|(p(x)+1)$ and $(x+1)^3|(p(x)-1)$ there exists two polynomials $f(x)=ax^2+bx+c$ and $g(x)=a_1 x^2+b_1 x +c_1$ such that
$$ (x-1)^3f(x)-1 = p(x) = (x+1)^3g(x)+1$$
$$ (x-1)^3f(x)- (x+1)^3g(x)-2 = 0$$
$$(x-1)^3(ax^2+bx+c)- (x+1)^3(a_1 x^2+b_1 x +c_1)-2 =0 \enspace \cdots(\ast) $$
In equation $(\ast)$, put $x=0$, we get
$-c-c_2 -2 =0$
Take $x=1$, $-8(a_1+b_1+c_1)-2=0$.
Take $x=-1$, $-8(-a-b+c)-2=0$
Similarily, take $x=\pm2$, and $x=\pm 3$. Finally, we will obtain $6$ linear equations in $6$ variables $a,b,c,a_1,b_1,c_1$ which can be solved by Gaussian elimination method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant of the matrix $(\omega^i \theta^j)_{i,j = 0,1,2}$ Given that $\omega=\frac{-1+\sqrt{-3}}{2}$ is a complex cube root of 1, $\theta$ is a real cube root of 2 and $\omega^{2}+\omega+1=0$, I am trying to maipulate the matrix
\begin{pmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{pmatrix}
to end up at
\begin{pmatrix}
1 & \theta &\theta^{2} \\
1 & \omega \theta & \omega^{2}\theta^{2} \\
1 & \omega^{2}\theta & \omega \theta^{2}
\end{pmatrix}
However I am struggling to come up with a way of manipulating the matrix in order to end up here.
| $$\begin{vmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{vmatrix}
=\omega\,\omega^2\theta\,\theta^2\begin{vmatrix}
1 & 1&1\\
1&1& 1 \\
1 & 1 & 1
\end{vmatrix}
$$
The values of $\theta$ and $\omega$ don't matter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even. Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even.
My solution:
$$
\begin{align}
f(-x)&=\ln\left ( \frac{-x+1}{-x-1} \right )\\
&=\ln \left ( \frac{-(x-1)}{-(x+1)}\right )\\
&=\ln\left ( \frac{x-1}{x+1} \right )\\
&=\ln (x-1)-\ln (x+1)\\
&=-(\ln (x+1)-\ln (x-1))\\
&=-\ln \left ( \frac{x+1}{x-1} \right ) = -f(x)
\end{align}
$$
It seems that function is odd. However, according to WolframAlpha it's neither odd nor even.
| While your conclusion is fine, there are still some problems with your computations, where some cautions is needed.
By definition, an even or odd function must have a symmetric domain of definition, so the first thing is to make this domain explicit: $\mathcal{D} := (-\infty, 1) \cup (1, +\infty)$.
Then we want to prove that $f$ is odd. In other words, we have to show that, for $x \in \mathcal{D}$,
$$f(-x) = -f(x).$$
If $x >1$, we have $x+1>0$ and $x-1>0$, so we can indeed write $f(x) = \ln (x+1)-\ln(x-1)$. However, for $x<1$, we have $x+1<0$ and $x-1<0$, so the expression $\ln (x+1)-\ln(x-1)$ is meaningless. Hence, the computation you gave is only valid for $x>1$. This is enough to prove your claim, but you need to make this explicit.
Another solution is to avoid splitting the logarithm altogether. For any $x\in\mathcal{D}$,
$$f(-x) = \ln \left(\frac{x-1}{x+1} \right) = \ln \left(\frac{1}{\left(\frac{x+1}{x-1}\right)} \right) = -\ln \left(\frac{x+1}{x-1} \right) = -f(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)
| Hint: Recall that $(a-b)(a^2+ab+b^2) = a^3-b^3$
Multiplying $$x^4 + x^2 + 1 = 0$$ with $x^2-1$ we get $$x^6-1=0$$ so $x^6=1$ and now is easy to finish...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
limit and summation I have to calculate $$ \lim_{m\rightarrow\infty}\sum_{n=0}^{\infty}\frac{1}{m^n} \sum_{k=0}^{\infty} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$
We know that $$ \sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=\frac{1}{4^n}\binom{2n}{n}=\frac1\pi\int_0^1\frac{x^n\,\mathrm{d}x}{\sqrt{x(1-x)}}$$
so
$$ \sum_{k=0}^{\infty} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=\lim_{n\rightarrow\infty} \sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=\lim_{n\rightarrow\infty}\frac{1}{4^n}\binom{2n}{n}=0$$
and
$$ \sum_{n=0}^{\infty}\frac{1}{m^n} \sum_{k=0}^{\infty} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k = \sum_{n=0}^{\infty}0=0$$
Am I right?
| For third limit, you could also have used that, by the Wallis product,
$$\binom{2 n}{n} \sim \frac {4^n} {\sqrt{\pi n}}$$ making
$$\frac{1}{4^n}\binom{2n}{n}\sim \frac {1} {\sqrt{\pi n}}$$
Remember also that there exist tight bounds for the central binomial coefficient
$$\frac {4^n} {\sqrt{4 n}} \leq \binom{2 n}{n} \leq\frac {4^n} {\sqrt{3 n+1}}\qquad \forall n \geq 1 $$ and, even more accurate
$$\binom{2 n}{n}\sim \frac {4^n} {\sqrt{\pi n}} \left(1-\frac {c_n} n \right)\qquad \text{where} \qquad \frac 19 < c_n <\frac 18 \qquad \forall n \geq 1$$
Using the very first approximation,we then have
$$\sum_{n=0}^{\infty}\frac{1}{m^n\sqrt{\pi n}}=\frac{1}{\sqrt{\pi }}\text{Li}_{\frac{1}{2}}\left(\frac{1}{m}\right)=\frac{1}{\sqrt{\pi }}\sum_{k=1}^\infty \frac 1 {\sqrt{k} \,m^k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does the beta distribution become U shaped when $\alpha$ and $\beta$ <1? In the Beta distribution (used to model Bernoulli probabilities), the $\alpha$ and $\beta$ parameters can be interpreted as the number of heads$+1$ and the number of tails$+1$ seen. So, if they were both $2$, it would lean towards the coin being fair and have a maximum at $0.5$. If they are both $20$, the distribution would become even surer we're dealing with a fair coin and peak even more at $p=0.5$.
What I don't get is its behavior when $\alpha$ and $\beta$ both become $<1$.
In that case, it becomes U-shaped and the density peaks at $p=0$ and $p=1$. Meaning the coin is likely to be two-sided. I know there is an intuition for this since I think I had an idea about it a long time ago. However, I've been trying to recollect all day and can't piece it together. Does anyone have an intuition?
| The Beta distribution is
$\dfrac{x^{a-1}(1-x)^{b-1}}{B(a, b)}
$
so the shape depends only on
$f(x)
=x^{a-1}(1-x)^{b-1}
$.
$f'(x)
=(a - 1) x^{a - 2} (1 - x)^{b - 1} - (b - 1) x^{a - 1} (1 - x)^{b - 2}\\
=x^{a - 2} (1 - x)^{b - 2}((a - 1) (1 - x) - (b - 1) x)
$
so $f'(x) = 0$
when
$(a - 1) (1 - x)
= (b - 1) x
$
or
$a-1
=x(b-1+a-1)
=x(a+b-2)
$
or
$x
=\dfrac{a-1}{a+b-2}
$.
Note that
$1-x
=\dfrac{b-1}{a+b-2}
$.
$\begin{array}\\
f''(x)
&=x^{a - 3} (1 - x)^{b - 3} ( x^2 (a + b - 3) (a + b - 2) - 2 (a - 1) (a + b - 3)x+a^2 - 3 a + 2)\\
&=x^{a - 3} (1 - x)^{b - 3} ( x^2 (a + b - 3) (a + b - 2) - 2 (a - 1) (a + b - 3)x+(a-1)(a-2))\\
&=x^{a - 3} (1 - x)^{b - 3} g(x)\\
\end{array}
$
where
$g(x)
= x^2 (a + b - 3) (a + b - 2) - 2 (a - 1) (a + b - 3)x+(a-1)(a-2)
$
(according to Wolfy).
If $a < 1$ and $b < 1$
then
$(a + b - 3) (a + b - 2)
\gt 0
$
so g(x) is u-shaped
and
$x^{a - 3} (1 - x)^{b - 3}
$
is u-shaped,
so their product
is u-shaped.
$g(0)
=(a-1)(a-2)
\gt 0$
and
$g(1)
=(b-1)(b-2)
\gt 0
$.
Also,
the discriminant of $g(x)$ is
$\begin{array}\\
d
&=(2 (a - 1) (a + b - 3))^2
-4(a-1)(a-2)(a + b - 3) (a + b - 2)\\
&=4(a-1)(a+b-3)( (a - 1) (a + b - 3)
-(a-2) (a + b - 2))\\
&=4(a-1)(a+b-3)( (a - 1) (a + b - 3)
-(a-2) (a + b - 2))\\
&=4(a-1)(a+b-3)(b-1)
\qquad\text{(quite surprisingly, to me}\\
\end{array}
$
If $a < 1$ and $b < 1$
then $d < 0$
(since all three terms
are negative),
so $g(x)$
has no real roots
so is always positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Evaluate $\iint_{[0,1]^2}\frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy$ I'm trying to prove $$I=\iint_{[0,1]^2}\frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy=\frac83(\sqrt2-1)$$ in an elementary way.
What I've tried
I'm not satisfied with my following method.
Integrating w.r.t. $x$:
$$I=\int_0^1\left[-\frac{20}3 \left(y^{3/2}-\sqrt{y^3+1}\right)-y^{-3/2}\, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{y^3}\right)\right]dy$$
Integrating w.r.t. $y$:
$$I=\int_0^1\left(10 \sqrt x-x^{-3/2}\right) \, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{x^3}\right)dx$$
Although I know how to handle some integrals of kind $\int_0^1x^\alpha(1-x)^\beta {}_pF_q(x)dx$ and how to cancel some "adjacent" hypergeometric function values, but concerning the answer and the integrand is so simple, I strongly believe that there are simple substitutions , IBPs, Feymann's tricks or something similar that can directly solve the integral. Any alternative solution will be appreciated.
| $$ I=\int_0^1\int_0^1 \frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy=\int_0^1\int_0^1 \frac{10y^2-1}{\sqrt{x^3+y^3}}dxdy$$
$$\Rightarrow I=5\int_0^1\int_0^1 \frac{x^2+y^2}{\sqrt{x^3+y^3}}dxdy-\int_0^1\int_0^1 \frac{1}{\sqrt{x^3+y^3}}dxdy$$
$$=10\int_0^1 \int_0^y \frac{x^2+y^2}{\sqrt{x^3+y^3}}dxdy-2\int_0^1\int_0^y \frac{1}{\sqrt{x^3+y^3}}dxdy$$
$$\overset{x=ty}=10\int_0^1 y \sqrt y\int_0^1\frac{t^2+1}{\sqrt{t^3+1}}dtdy-2\int_0^1\frac{1}{\sqrt y}\int_0^1 \frac{1}{\sqrt{t^3+1}}dtdy$$
$$=4\int_0^1 \frac{t^2+1}{\sqrt{t^3+1}}dt-4\int_0^1 \frac{1}{\sqrt{t^3+1}}dt=4\int_0^1 \frac{t^2}{\sqrt{t^3+1}}dt$$
$$=\frac83\sqrt{t^3+1}\bigg|_0^1=\frac83\left(\sqrt 2-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495004",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find coefficient of $x^{10}$ in $x^2(x^2-3x^3-1)^6$ I'm trying to solve the following problem (final exams preparation):
Find coefficient of $x^{10}$ in $x^2(x^2-3x^3-1)^6$.
The steps I have taken so far:
$x^2(x^2-3x^3-1)^6 = $
$=x^2\sum \binom{6}{n1,n2,n3}(x)^{2n_1}(-3)^{n_2}(x)^{3n_2}(-1)^{n_3}$
$n_1+n_2+n_3=6$
$2n_1+3n_2=8$
$\Rightarrow$ $n_1=1 \wedge n_2=2$
$\Rightarrow$ $n_1=4 \wedge n_2=0$
Are my steps correct? If so, how do I continue?
Thanks
| I'll translate the work you did into English.
So the coefficient of the $x^{10}$ term in $x^2(x^2-3x^3-1)^6$ is the same as the coefficient of the $x^8$ term in $(x^2-3x^3-1)^6$. To figure out which terms in that trinomial power contribute that that term, we effectively want to find the non-negative solutions to $2a+3b+c=8$ and $a+b+c=6$. As you note, the solutions to that are $(a,b,c)\in\{(4,0,2),(1,2,3)\}$.
Our last step is appealing to the multinomial formula to find those two terms. They are
$$\frac{6!}{4!2!}(x^2)^4(-3x^3)^0(-1)^2=15x^8\\\frac{6!}{3!2!}(x^2)^1(-3x^3)^2(-1)^3=90\cdot x^2\cdot9x^6\cdot(-1)=-540x^8$$
which lead to a total coefficient of $15-540=-525$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) =
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.
| You may continue as follows,
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
$$=2^{n - m} \cdot \dfrac{\left(\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}\right)
\left(\cos \dfrac{11m \pi}{6} - i \sin \dfrac{11m \pi}{6}\right)}
{\cos^2 \dfrac{11 m \pi}{6} + \sin^2 \dfrac{11 m \pi}{6}}$$
Then, set the the imaginary part of the numerator to zero,
$$I=\sin \dfrac{n \pi}{6}\cos \dfrac{11m \pi}{6} - \cos \dfrac{n \pi}{6}\sin \dfrac{11m \pi}{6} = -\sin\dfrac{(11m -n)\pi }{6}=\sin\dfrac{(m +n)\pi }{6}=0 $$
which leads to $\dfrac{(m+n)\pi }{6}= k\pi$. Thus, the relationship between $m$ and $n$ is
$$m+n=6k$$
with $k=0,1,2,...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Numbers 'poetics' 1 and 2019
One number is called 'poetic' when it can be represented of only one form how $2^a+2^b+2^c$ such that $a\ge b \ge c\ge 0.$ How many poetic numbers are between $1$ and $2019?$
Attempt: I separate in cases, I find that every $2^k$ with $k\ge2 , 2^k + 1$ with $k\ge1$ and every case with $a>b>c$
My final answer was $184$. Check that please .
$a\ge b\ge c\ge 0$
Numbers of form $2^k, k\ge 2$ and $2^k+1,k\ge 1$ are also unique re presentable
$2^k=2^{k-1}+2^{k-2}+2^{k-2}$
$2^{k}+1=2^{k-1}+2^{k-1}+2^0$
| We have $a\ge b\ge c\ge0$. But there is also a requirement that the representation is unique. So for example: 6 does not count because it is $4+1+1$ and also $2+2+2$.
Indeed, we have
(1): $2^a+2^a+2^b=2^{a+1}+2^{b-1}+2^{b-1}$ for $b\ge1$ and
(2): $2^a+2^b+2^b=2^{a-1}+2^{a-1}+2^{b+1}$ for $a>b+1$. [For $a=b+1$ the two expressions are identical]
We have $2^{11}=2048$, so $a\le10$ and if $a=10$, then we need $b<10$. Also if $a=10,b=9$, then $c<9$.
So for $a=10$ we can have $b=9$ and then $0\le c\le 8$, giving 9 possibilities. If $b=8$ then (2) requires $0\le c<8$ giving 8 possibilities. And so on, down to $b=1$ for which we must have $c=0$. So in total $a=10$ gives $1+2+\dots+9=45$.
For $a=9$ and $b=9$ only $c=0$ is allowed because of (1). For $b=8$ we can have $0\le c\le 8$ (9 poss). For $b=7$ we can have $0\le c\le 6$ (because (2) rules out $c=7$), giving 7 poss. For $b=6$ we have 6 poss, and so on down to 1 poss for $b=1$. So in total $a=9$ gives $(1+2+\dots+7)+9+1=38$.
Similarly for $a=8$ we get $1+2+\dots+6+9=30$, for $a=7$ we get $1+2+\dots+5+8=23$. For $a=6$ we get $1+2+\dots+4+7=17$. For $a=5$ we get $1+2+3+6=12$. For $a=4$ we get $1+2+5=8$, for $a=3$ we get $1+4=5$. For $a=2$ we get 3 (check: $9=4+4+1,8=4+2+2,7=4+2+1$). For $a=1$ we get 2 and for $a=0$ we get 1.
So grand total $45+38+30+23+17+12+8+5+3+2+1=184$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please?
| For the second one $$\frac{\log(x^2+x-1)}{\log(x^3+x-1)}$$ using what Lord Shark the Unknown suggested, it becomes
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}$$ Use the Taylor expansion of $\log(1+\epsilon)$ with $\epsilon=(3t+t^2)$ for the numerator and $\epsilon=(4t+3 t^2+t^3)$ for the denominator and use the binomial expansion or Taylor series to get
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}=\frac{3 t-\frac{7 }{2}t^2+O\left(t^3\right) } {4 t-5 t^2+O\left(t^3\right) }$$ Now, long division to get
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}=\frac{3}{4}+\frac{t}{16}+O\left(t^2\right)$$ which shows the limit and also how it is approached.
Just by curiosity, use your pocket calculator with $t=\frac 1 {10}$ (quite far away from $0$). You should get
$$\frac{\log \left(\frac{131}{100}\right)}{\log \left(\frac{1431}{1000}\right)}\approx 0.75348$$ while the above truncated expansion gives $\frac{121}{160}\approx 0.75625$.
| {
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How many solutions does the expression $3^x(3^x+1)+2=|3^x-1|+|3^x+2|$ have? I can solve this by opening every single modulus, doing alternate combinations of $\pm$ and checking whether every $3^x$ is $>0$ or not. But that consumes a lot of time. Is there is a shorter way to solve such questions.
The answer is 1
| $\because\forall x\in\mathbb R,3^x>0,\therefore 3^x+2>0$,so
$$\begin{align}
3^x(3^x+1)+2 & =|3^x-1|+|3^x+2|\\
\Longleftrightarrow \quad\,\,\,\, 3^{2x}+3^x+2 & =|3^x-1|+3^x+2\\
\Longleftrightarrow \qquad\qquad\quad 3^{2x} & =|3^x-1|
\end{align}$$
And when $3^x>1$,it’s
$$3^{2x}>3^{2x}-1=(3^x+1)(3^x-1)>3^x-1$$
so it haven’t solution when $3^x>1$,you can just find the solution in the case: $3^x\leqslant 1$
| {
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Taking the gradient of $f(\mathbf{x}) = \frac{1}{2}\|\mathbf{A} \mathbf{x} - \mathbf{b}\|_2^2$ Section 4.5 of the textbook Deep Learning by Goodfellow, Bengio, and Courville, says that the gradient of
$$f(\mathbf{x}) = \dfrac{1}{2}\|\mathbf{A} \mathbf{x} - \mathbf{b}\|_2^2$$
is
$$\nabla_{\mathbf{x}} f(\mathbf{x}) = \mathbf{A}^T (\mathbf{A}\mathbf{x} - \mathbf{b}) = \mathbf{A}^T \mathbf{A} \mathbf{x} - \mathbf{A}^T \mathbf{b}$$
My understanding is that $f(\mathbf{x}) = \dfrac{1}{2}\|\mathbf{A} \mathbf{x} - \mathbf{b}\|_2^2$ is the square of the Euclidean norm. So we have that
$$\begin{align} f(\mathbf{x}) = \dfrac{1}{2}\|\mathbf{A} \mathbf{x} - \mathbf{b}\|_2^2 &= \dfrac{1}{2} \left( \sqrt{(\mathbf{A} \mathbf{x} - \mathbf{b})^2} \right)^2 \\ &= \dfrac{1}{2} (\mathbf{A} \mathbf{x} - \mathbf{b})^2 \\ &= \dfrac{1}{2} (\mathbf{A} \mathbf{x} - \mathbf{b})(\mathbf{A} \mathbf{x} - \mathbf{b}) \\ &= \dfrac{1}{2} [ (\mathbf{A}\mathbf{x})(\mathbf{A} \mathbf{x}) - (\mathbf{A} \mathbf{x})\mathbf{b} - (\mathbf{A} \mathbf{x})\mathbf{b} + \mathbf{b}^2 ] \ \ \text{(Since matrix multiplication is distributive.)} \\ &= \dfrac{1}{2} [(\mathbf{A} \mathbf{x})^2 - 2(\mathbf{A} \mathbf{x})\mathbf{b} + \mathbf{b}^2] \ \ \text{(Note: Matrix multiplication is not commutative.)} \end{align}$$
It's at this point that I realised that, since we're working with matrices, I'm not really sure how to take the gradient of this. Taking the gradient of $f(\mathbf{x})$ with respect to $\mathbf{x}$, we get something like
$$\nabla_{\mathbf{x}} f(\mathbf{x}) = \dfrac{1}{2} [2 (\mathbf{A} \mathbf{x}) \mathbf{A}] - \dfrac{1}{2}[2(\mathbf{A} \mathbf{A} \mathbf{x})\mathbf{b}]$$
So what is the reasoning that leads us to get $\nabla_{\mathbf{x}} f(\mathbf{x}) = \mathbf{A}^T (\mathbf{A}\mathbf{x} - \mathbf{b}) = \mathbf{A}^T \mathbf{A} \mathbf{x} - \mathbf{A}^T \mathbf{b}$? Where did the transposed matrices come from?
I would greatly appreciate it if people would please take the time to clarify this.
| We must take the derivative with finesse, and that means we use the chain rule. Note that $f = g \circ h$, where $h(x) = Ax-b$ and $g(u) = (1/2) \|u\|^2$. The derivatives of $h$ and $g$ are $h'(x) = A$ and $g'(u) = u^T$. So by the chain rule
$$
f'(x) = g'(h(x)) h'(x) = (Ax-b)^T A.
$$
The gradient of $f$ is
$$
\nabla f(x) = f'(x)^T = A^T(Ax-b).
$$
| {
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Why is this the nearest integer? I was trying to prove that $(N+\sqrt{N^2-1})^k$, where k is a positive integer, differs from the integer nearest to it by less than $(2N-\frac{1}{2})^{-k}$. Note: N is an integer greater than 1.
So, I tried to look for the answer of the question, which I have taken it from an exam paper.
It said that:
We let T=$(N+\sqrt{N^2-1})^K +(N+\sqrt{N^2-1})^{-k}$= $(N+\sqrt{N^2-1})^k + (N-\sqrt{N^2-1})^k=2(N^k + kC2 N^{K-2}(N^2-1)+...$ which is clear it is an integer.
We know that:
$(N-\frac{1}{2})^2 =N^2-N+\frac{1}{4}=\frac{5}{4}-N$ This is <0 when N > 1, since N is an integer.
So,
$N-\frac{1}{2}<\sqrt{N^2-1}$.
$2N-\frac{1}{2} < N+\sqrt{N^2-1}$
$(2N-\frac{1}{2})^{-k}>(N+\sqrt{N^2-1}^{-k}$
Let $|T-(N+\sqrt{N^2-1})^{-k}$| we would be able to prove the question.
However, my question is that, we need to have the nearest integer, so, the integer that the answer used is
$(N+\sqrt{N^2-1})^k + (N + \sqrt{N^2-1})^{-k}$.
Why is this the nearest integer?
Thank you so much for your reply.
| By induction on $k$, we have
$$ (N\pm \sqrt{N^2-1})^k=a\pm b\sqrt{N^2-1}$$
with $a,b\in\Bbb Z$.
And of course
$(N-\frac12)^2=N^2-N+\frac14<N^2-1 $ implies
$$ 2N-\frac12<N+\sqrt{N^2-1}<2N.$$
Now from
$$ (N+\sqrt{N^2-1})^k(N-\sqrt{N^2-1})^k=((N+\sqrt{N^2-1})(N-\sqrt{N^2-1}))^k=1^k=1,$$
we conclude that $(N+\sqrt{N^2-1})^k=a+b\sqrt{N^2-1}$ differs from the integer $2a$ by
$$a-b\sqrt{N^2-1}=\frac1{(N+\sqrt{N^2-1})^k}<\frac1{(2N-\frac12)^k}.$$
| {
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Suppose that $z\in\mathbb C$ with $|z^2+1|\le 1$. How to prove $|z+1|\geq\frac12$. Let $z\in\mathbb C$ with $|z^2+1|\le 1$. I want to prove $|z+1|\geq\frac12$.
I noticed that $|z^2+1|\le 1$ means that $z$ lies in a cassini oval. I tried with the substitution $z=r\exp(i\theta)$ where $r\geq0, \theta\in[0,2\pi]$. The constraint now is $\frac{r^2}2\le-\cos(2\theta)$ and we need to prove $$\sqrt{r^2 \sin^2(\theta) + (r \cos(\theta) + 1)^2}=\sqrt{r^2+2r\cos\theta+1}\geq\frac12.$$
How can I do that?
| It suffices to prove that
$$|z+1| < \tfrac{1}{2} \quad \Longrightarrow \quad |z^2+1| > 1.$$
To this end, let $z = a + \mathrm{i}b$.
From $|z+1| < \frac{1}{2}$, we have
$(a+1)^2 + b^2 < \frac{1}{4}$ which results in
$a^2 - b^2 > a^2 + (a+1)^2 - \frac{1}{4} = 2(a+\frac{1}{2})^2 + \frac{1}{4} > 0$.
Then, we have
$|z^2 + 1| \ge \mathrm{Re}(z^2+1) = a^2 - b^2 + 1 > 1$. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| The formula for the tangent of a difference says
$$
\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a \tan b}.
$$
If $p = \tan a$ and $q=\tan b$ and $a = \tan^{-1}p$ and $b=\tan^{-1} q$ then we have
$$
\tan(a-b) = \frac{p-q}{1+pq}
$$
so
$$
a-b = \tan^{-1} \frac{p-q}{1+pq}
$$
or in other words
$$
\tan^{-1} p - \tan^{-1} q = \tan^{-1}\frac{p-q}{1+pq}.
$$
So plug in the numbers you specified:
$$
\tan^{-1} \frac {11} x - \tan^{-1} \frac 1 x = \tan^{-1} \frac{\frac{11}x - \frac 1 x}{1 + \left(\frac{11} x \cdot \frac 1 x\right)}
$$
$$
= \tan^{-1} \frac{10x}{x^2+11}
$$
Thus
$$
30^\circ = \tan^{-1} \frac{10x}{x^2+11}
$$
$$
\frac{10x}{x^2+11} = \tan 30^\circ = \frac{\sqrt 3}3
$$
$$
30x = \sqrt 3\, x^2 + 11\sqrt 3.
$$
This is a quadratic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Gaussian integral over a surface I have to solve the following integral:
$$
I(s)=\int_{S(s)} \frac{1}{2\pi^{3/2}}e^{-\frac{x^2+y^2+2z^2}{2}}dxdydz
$$
where $S(s)$ is the surface defined by $s=\sqrt{(x-y)^2+4z^2}$.
I parametrised $S(s$) using:
\begin{cases}
x = \frac{\xi+s\cos(\theta)}{2} \\
y = \frac{\xi-s\cos(\theta)}{2} \\
z = \frac{s\sin(\theta)}{2}
\end{cases}
with $-\infty<\xi<+\infty$ and $ 0 \leq \theta\leq2\pi$, for which $S(s)$ the element of area is equal to:
$$
dS=\frac{s}{2\sqrt{2}}\sqrt{2-\cos^2(\theta)}
$$
So the integral became:
$$
I(s)=\frac{1}{2\pi^{3/2}}\frac{s}{2\sqrt{2}}\int_{-\infty}^{+\infty}e^{-\frac{\xi^2+s^2}{4}}d\xi\int_{0}^{+2\pi}\sqrt{2-\cos^2(\theta)}d\theta
$$
but the later integral is not elementary.
On the other hand, using delta function seems to resolve all. Writing
$$
I(s)=\int_{-\infty}^{+\infty}\frac{1}{2\pi^{3/2}}e^{-\frac{x^2+y^2+2z^2}{2}}\delta\left(s-\sqrt{(x-y)^2+4z^2}\right)dxdydz
$$
and using the change of variables
\begin{cases}
x = \frac{\xi+r\cos(\theta)}{2} \\
y = \frac{\xi-r\cos(\theta)}{2} \\
z = \frac{r\sin(\theta)}{2}
\end{cases}
(the Jacobian is equal to $\frac{-r}{4}$) the integral becomes
$$
I(s)=\frac{1}{8\pi^{3/2}}\int_{0}^{+\infty}r\delta(s-r)dr\int_{0}^{+2\pi}d\theta\int_{-\infty}^{+\infty}e^{-\frac{\xi^2+r^2}{4}}d\xi=\frac{s}{2}e^{-\frac{s^2}{4}}
$$
So, what is wrong with my first attempt?
Thanks
| Your first attempt is entirely correct and the value of the integral is indeed
\begin{align}
I(s) &= \frac{1}{2\pi^{3/2}}\frac{s}{2\sqrt{2}}\int \limits_{-\infty}^\infty \mathrm{e}^{-\frac{\xi^2+s^2}{4}} \mathrm{d} \xi \int \limits_0^{2\pi}\sqrt{2-\cos^2(\theta)} \, \mathrm{d} \theta = \frac{s \mathrm{e}^{-s^2 /4}}{2\pi} \int \limits_0^{2\pi} \sqrt{1 - \frac{1}{2} \cos^2(\theta)} \, \mathrm{d} \theta \\
&= \frac{2}{\pi} \operatorname{E}\left(\frac{1}{\sqrt{2}}\right)s \mathrm{e}^{-s^2 /4}
\end{align}
with the complete elliptic integral of the second kind $\operatorname{E}$.
The mistake lies in your second approach. As discussed here, an additional factor must be taken into account when expressing a surface integral in terms of the delta function. Let $g(x,y,z) = \sqrt{(x-y)^2 + 4 z^2}$. Then the correct expression for your integral is
$$ I(s) = \int \limits_{\mathbb{R}^3} \frac{1}{2\pi^{3/2}} \mathrm{e}^{-\frac{x^2+y^2+2z^2}{2}} \delta(s - g(x,y,z)) \left|\nabla g(x,y,z)\right|\, \mathrm{d} x \, \mathrm{d}y \, \mathrm{d}z $$
and the gradient term turns into the missing square root after the change of variables.
| {
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Find area of ellipse $5x^2 -6xy +5y^2=8$
Find the area of ellipse whose equation in the $xy$- plane is given by $5x^2 -6xy +5y^2=8$
My attempt : I know that area of ellipse $ = \pi a b$ ,where $a$ is semi-major axis and $b$ is semi minor axis
Now if we make matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmatrix}$ Here eigenvalue $\lambda_1= 8 ,\lambda_2=2$
That is area of ellipse$ = \pi \frac{1}{\sqrt\lambda_2} \frac{1}{\sqrt\lambda_1}= \pi \frac{1}{2\sqrt 2}\frac{1}{\sqrt2}$
Is its correct ?
| Given $$5x^2-6xy+5y^2=8$$
Curve represent ellipse whose center is at $(0,0)$ and it is symmetrical about $y=x$ and $y=-x$ line
Put $y=x$ in $5x^2-6xy+5y^2=8$, getting $=\pm \sqrt{2}.$
So point of intersection of line and ellipse is $(\sqrt{2},\sqrt{2})$ or $(-\sqrt{2},-\sqrt{2})$
Which is notning but distance length of semi major axis i e $a = 2$
Put $y=-x$ in $5x^2-6xy+5y^2=8$, getting $\displaystyle =\pm \frac{1}{\sqrt{2}}.$
So point of intersection of line and ellipse is $\displaystyle \bigg(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\bigg)$ or $\displaystyle \bigg(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\bigg)$
Which is notning but distance length of semi minor axis i e $b=1$
So we have eccentricity $$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$$
And area of ellipse is $$\pi ab=2\pi$$
| {
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Inverse with Fermat, Modulo I wonder about an equation given as a solution to the following task:
Calculate the multiplicative inverse $$5^{−1} \pmod {13}$$
The solution ends in this equation:
$$
5^{11} \equiv 5^{10} \cdot 5 \equiv −1\cdot 5 \equiv 8 \pmod {13}
$$
and the check:
$$
5 \cdot 8 \equiv 40 \equiv 1 \pmod{13}
$$
1) What is the inverse here now?
2) Why is $5^{10} \equiv -1 \pmod {13}$?
As far as I understood Fermat the rule is given as:
$$
x^{(p−1)} \equiv 1 \pmod p
$$
which won't come into account here, since in this case
$p = 13$
and not ideal for calculating $5^{11}$.
If I would use $x^{p−1}$ it would be $5^{12}$ in this case, so I'd still need to divide by $5^1$ to get to the result of $5^{11}$.
I am so confused although it should be really easy.
| Question 2: Why is $5^{10}\equiv - \pmod {13}$.
Well....Because it is.
$5^2 \equiv 25\equiv -1\pmod {13}$ so $5^{10}\equiv (-1)^5\equiv -1 \pmod {13}$.
The real question is why are we trying to find $5^{10}$?
$5^{11}*5 \equiv 5^{12} \equiv 1 \pmod {13}$ so $5^{-1}\equiv 5^{11}$ and..... well, I guess the author just figured $5^{10}$ could be easily calculated by subsequent squaring. $5^{11} \equiv 5*5^{10}\equiv 5*(5^2)^5$ is an easy way do to it.
We could, just to be different figure $5^{11} = (5^3)^3*5^2\equiv 125^3*(-1)\equiv -8^3\equiv (-8)^3 \equiv 5^3 \equiv 125 \equiv 8\pmod {13}$ but that would be unnecessarily complicated.
Or we could just do increasing powers and see when we return to $1$. $5^1 \equiv 5$ and $5^2 \equiv -1$ so $5^3\equiv -1*5 \equiv 8$ and $5^4\equiv 1$. So $5^{4} =1$ and $5^{-1} \equiv 5^3 \equiv 8$
To be honest.... I do not know why the author assumed that it'd be "obvious" that $5^{10} \equiv -1$ but not that it'd be "obvious" that $5^{11}\equiv 8$.
But one way or another..... $5^{11}*5 \equiv 1\pmod {13}$ so $5^{-1}$ is $5^{11}$ whatever that is. And roll up your sleeves and do it... $5^{11}\equiv 8\pmod {13}$ by one method or another.
Question 1:
What is the inverse? Well, since $5*8\equiv 1$ it is $8$
| {
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Prove inequality of series sum by induction How should i prove by induction that:
$ 1 + 2^{(-2)} + 3^{-2} + ...+n^{(-2)} \le 2-n^{-1} $
I haven't had inequalities with sum of series at the class about induction so i can't really wrap my head around it.
| Induction is induction.
Base case: $1^{-2} = 1 \le 2- 1^{-1} = 1$.
Induction case:
If $1 + 2^{-2}+ 3^{-2} + ....... + n^{-2} \le 2-n^{-1}$ then
$1 + 2^{-2}+ 3^{-2} + ....... + n^{-2} + (n+1)^{-2} \le 2-n^{-1} + (n+1)^{-2}$.
So all we have to do is prove $ 2-n^{-1} + (n+1)^{-2} \le 2- (n+1)^{-1}$.
Can we do that?
All we have to do is calculate that
$2 - \frac 1n + \frac 1{(n+1)^2} =$
$2-(\frac 1n - \frac 1{(n+1)^2}) =$
$2 - (\frac {(n+1)^2- n}{n(n+1)^2} )=$
$2 - (\frac {n^2 +n + 1}{n(n+1)^2} )<$
$2 - (\frac {n^2 + n}{n(n+1)^2}) =$
$2- \frac {n+1}{(n+1)^2} =$
$2-\frac 1{n+1}= 2-(n+1)^{-1}$.
And that's it.
So $1 + 2^{-2}+ 3^{-2} + ....... + n^{-2} + (n+1)^{-2} \le 2-n^{-1} + (n+1)^{-2}< 2-(n+1)^{-1}$
| {
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Represent $f(x)=\ln x$ as power series, in powers of $(x-4)$ I was requested to represent $f(x)=\ln x$ as a power series, in powers of $(x-4)$. I only started studying power series today, and though I was able to represent different functions as power series I can't seem to work this one out, specially with the condition that the series must be in powers of $(x-4)$.
Any ideas on how to solve this problem?
| There are a few ways to go about this.
The most direct way.
$f(x) = f(x-a) + f'(x-a) (x-a) + \frac {f''(x-a)}{2!} (x-a)^2 + \cdots + \frac {f^{(n)}(x-a)}{n!} (x-a)^n + \cdots$
I hope that this looks familiar to you.
We set $a = 4$
$f(4) = \ln 4\\
f'(x) = \frac {1}{x}, f'(4) = \frac {1}{4}\\
f''(x) = -\frac {1}{x^2}, f''(4) = -\frac {1}{4^2}\\
f^{(n)}(x) = (-1)^{n-1} \frac {1}{x^n}, f^{(n)}(4) = (-1)^{n-1} \frac {(n-1)!}{4^n}$
And you can verify this last line via induction.
$f(x) = \ln 4 + \sum_\limits{n=1}^{\infty} (-1)^{n-1}\frac {(x-a)^n}{n4^n}$
And there is the tricky way to do this.
$f'(x) = \frac {1}{x}\\
f'(x) = \frac {1}{4 + (x-4)}\\
f'(x) = \frac 14 \frac {1}{1 + \frac {x-4}{4}}$
This can be thought of as the sum of a geometric series.
i.e $\frac {1}{1 - a} = \sum_\limits{n=0}^\infty a^n$ if $|a|< 1$
$f'(x) = \frac 14 \sum_\limits{n=0}^\infty (-1)^n \left(\frac {x-4}{4}\right)^n$
$f(x) - f(4) = \int \sum_\limits{n=0}^\infty (-1)^n \left(\frac {x-4}{4}\right)^n\ \frac {dx}{4}\\
= \sum_\limits{n=0}^\infty (-1)^n \frac {1}{(n+1)}\left(\frac {x-4}{4}\right)^{n+1}
\\ f(x) - \ln 4 = \sum_\limits{n=1}^\infty (-1)^{n-1} \frac {(x-4)^n}{n4^n}$
And still, the series will only converge if $|\frac {x-4}{4}| < 1$
And, if you have been working with power series for a while you might say.
I know $\ln (1-x) = -\sum_\limits{n=1}^\infty \frac {x^n}{n}\\
\ln (1+x) = \sum_\limits{n=1}^\infty (-1)^n \frac{x^n}{n}$
From doing exercises like the one above.
$\ln x = \sum_\limits{n=1}^\infty (-1)^n\frac {(x-1)^n}{n}$ substituting $(x-1)$ in for $x$ in the second equation above.
To center at 4.
$\ln x = \ln (4\frac {x}{4}) = \ln 4 + \ln \frac {x}{4}$
And substitute into what we had for $\ln x$
$\ln x = \ln 4 + \sum_\limits{n=1}^\infty (-1)^n\frac {(\frac{x}{4}-1)^n}{n}\\
\ln 4 + \sum_\limits{n=1}^\infty (-1)^n\frac {(x-1)^n}{n4^n}\\
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can we choose 4 different numbers from the set ${1,2,3,...,8,9,10}$ so that no two numbers are next to each other? I did this question using PIE and I'm confused as to why I'm not getting the right answer.
My approach:
Use complementary counting. There are $\binom{10}{4}$ ways to choose 4 different numbers.
I then subtracted $9\cdot\binom{8}{2}$ because there are $9$ ways to choose the pair of numbers and then $\binom{8}{2}$ ways to choose the last two numbers.
I then added $8\cdot\binom{8}{1}$ because I subtracted this case twice and thus have to add it in once.
I then subtracted $7$.
I got a final answer of $7$, but the correct answer is $35$. What did I do wrong?
| It looks like you applied the Inclusion-Exclusion Principle to cases with two consecutive, three consecutive, and four consecutive numbers. However, you should instead apply the Inclusion-Exclusion Principle to pairs of consecutive numbers.
There are $\binom{10}{4}$ ways to choose four numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
A pair of consecutive numbers: Your count is correct. The smaller of the two consecutive numbers must occur in one of the first nine positions. Choosing the smaller also determines the larger. The remaining two numbers can be selected in
$\binom{8}{2}$ ways, so there are $$\binom{9}{1}\binom{8}{2}$$ such selections.
Two pairs of consecutive numbers: This can occur in two ways. The pairs can overlap, or they are disjoint.
Two overlapping pairs: This means that three consecutive numbers are selected. Since the smallest of these three consecutive numbers must occur in one of the first eight positions. That leaves seven choices for the remaining number. Hence, there are $$\binom{8}{1}\binom{7}{1}$$ such selections.
Two disjoint pairs: We have eight available positions, two for the pairs and six for the other six numbers. Choose two of the eight positions for the pairs. Doing so determines the pairs. For instance, if we choose the third and fifth positions, then the pairs are $3, 4$ and $6, 7$.
$$1, 2, \boxed{3, 4}, 5, \boxed{6, 7}, 8, 9, 10$$
Hence, there are $$\binom{8}{2}$$ such selections.
Three pairs: Since we are only selecting four numbers, this can only occur if we have four consecutive numbers. The smallest of these numbers can be selected in seven ways. Hence, there are $$\binom{7}{1}$$ such selections.
By the Inclusion-Exclusion Principle, the number of ways four numbers can be selected from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ such that no two consecutive numbers are selected is
$$\binom{10}{4} - \binom{9}{1}\binom{8}{2} + \binom{8}{1}\binom{7}{1} + \binom{8}{2} - \binom{7}{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
| Yet another method in case you're interested: first square both sides
$$y^2=\frac{3-x^2}{(3+x)^2}$$
Then differentiate both sides:
$$2y\frac{dy}{dx}=\frac{(3+x)^2(-2x)-(3-x^2)2(3+x)}{(3+x)^4}$$
$$=-\frac{6(x+1)}{(3+x)^3}$$
Then divide both sides by $2y$ to get the result.
| {
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"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b^2\sin^2 t}}$$
Now let $$p=\cos(2t)$$
then we get a single variable function as:
$$h(p)=\frac{1}{\sqrt{2}}\frac{(a+b)+p(a-b)}{\sqrt{a^2+b^2+p(a^2-b^2)}}$$
where $p \in [-1, 1]$
Now we can find critical point and find minimum.
Is there a better approach, i tried lagrange multipliers but very tedious
| Note that
\begin{eqnarray*}
f(x,y) = \frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}} \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{ \sqrt{ab}}{a+b} \left( \sqrt{ \frac{a^2x^2+b^2y^2}{ab}} + \sqrt{ \frac{ab}{a^2x^2+b^2y^2}} \right)
\end{eqnarray*}
\begin{eqnarray*}= \frac{ \sqrt{ab}}{a+b} \left( \left( \sqrt[4]{ \frac{a^2x^2+b^2y^2}{ab}} - \sqrt[4]{ \frac{ab}{a^2x^2+b^2y^2}} \right)^2+2 \right).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Value of $f(f(x) – 2y) = 2x – 3y + f(f(y) – x)$ Let $f : \mathbb{R} \rightarrow \mathbb{R} $ be a polynomial function satisfying
$$f(f(x) – 2y) = 2x – 3y + f(f(y) – x),$$
where $x, y \in \mathbb{R}$. Find the value of $$f(21) – f(14).$$
| You are given
$$f(f(x) – 2y) = 2x – 3y + f(f(y) – x) \tag{1}\label{eq1A}$$
Since $x,y \in \mathbb{R}$, you can let $x = \frac{3y}{2}$ in \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
f\left(f\left(\frac{3y}{2}\right) - 2y\right) & = 2\left(\frac{3y}{2}\right) - 3y + f\left(f\left(y\right) - \frac{3y}{2}\right) \\
f\left(f\left(\frac{3y}{2}\right) - 2y\right) & = f\left(f\left(y\right) - \frac{3y}{2}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note $f$ can't be the constant function since \eqref{eq1A} then wouldn't hold for all $x$ and $y$. Actually, as WhatsUp's comment states, $f$ must be linear. To see this, consider the degree of $f$ to be some $n \gt 1$. Then $f(x) - 2y$ has an $x$ term of degree $n$ and $f(f(x) - 2y)$ has an $x$ term of degree $n^2$. However, on the right side, $f(f(y) - x)$ would only have a highest degree term of $x$ being $n$. Since $n^2 \gt n$, it would not be possible for \eqref{eq1A} to always hold.
Since $f$ must be linear, then
$$f(y) = ay + b \tag{3}\label{eq3A}$$
for some real $a \neq 0$ and $b$. Let the arguments to the outer $f$ functions in \eqref{eq2A} be $g(y)$ and $h(y)$ to get using \eqref{eq3A} that $ag(y) + b = ah(y) + b \implies a(g(y) - h(y)) = 0 \implies g(y) = h(y)$. Thus, you get
$$f\left(\frac{3y}{2}\right) - 2y = f(y) - \frac{3y}{2} \tag{4}\label{eq4A}$$
Using $y = 14$ in \eqref{eq4A} gives
$$f(21) - 28 = f(14) - 21 \implies f(21) - f(14) = 7 \tag{5}\label{eq5A}$$
Alternatively, you can use \eqref{eq3A} in \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
f(ax + b - 2y) & = 2x - 3y + f(ay + b - x) \\
a(ax + b - 2y) + b & = 2x - 3y + a(ay + b - x) + b \\
a^2x + ab - 2ay & = 2x - 3y + a^2y + ab - ax \\
a^2x - 2ay & = 2x - 3y + a^2y - ax \\
(a^2 + a - 2)x & = (a^2 + 2a - 3)y
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
For this to be true for all real $x$ and $y$ requires $a^2 + a - 2 = a^2 + 2a - 3 = 0 \implies a = 1$. Using $a = 1$ in \eqref{eq3A} gives
$$f(21) - f(14) = (21 + b) - (14 + b) = 7 \tag{7}\label{eq7A}$$
This matches what was obtained in \eqref{eq5A}.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Infinity Summation problem (ANSWERED) Show that $$\sum_{n=3}^∞ 96n^4+156n^3+33n^2+54n+3 = \frac{100-π^3}{87}$$
Only way I know how to solve this is to put it in my graphing calculator, but that isn't much help in this situation.
This is a question I am solving on my own that I have discovered. If you send the answer and work it would not assist me cheating or anything malicious.
| (Big) hint: you can rewrite
$$
\frac{1}{4n^4+16n^3+23n^2+14n+3}= 2\left( \frac{1}{2n+1}-\frac{1}{2n+3}\right) - \frac{1}{(n+1)^2}
$$
and now, you have the sum of a telescoping series and a convergent series (whose sum is known).
That is, we can compute the sum of the telescopic series as
$$
\sum_{n=1}^N 2\left( \frac{1}{2n+1}-\frac{1}{2n+3}\right)
= 2\left( \frac{1}{2\cdot 1+1}-\frac{1}{2N+3}\right) \xrightarrow[N\to\infty]{} 2\left( \frac{1}{2\cdot 1+1}-0\right) = \frac{2}{3}
$$
while the last term will give
$$
\sum_{n=1}^\infty \frac{1}{(n+1)^2}
=\sum_{n=2}^\infty \frac{1}{n^2}
=\sum_{n=1}^\infty \frac{1}{n^2} - 1 = \frac{\pi^2}{6}-1
$$
so that
$$
\sum_{n=1}^\infty \frac{1}{4n^4+16n^3+23n^2+14n+3}
= \frac{2}{3}- (\frac{\pi^2}{6}-1) = \boxed{\frac{10-\pi^2}{6}\,.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all primes $p \geq 5$ such that $6^p \cdot (p - 4)! + 10^{3p}$ is divisible by $p$
Find all primes $p \geq 5$ such that $6^p \cdot (p - 4)! + 10^{3p}$ is divisible by $p$
I've tried this : First check $(p - 4)!:$
\begin{align*}
(p - 1)! &\equiv -1 \text{(mod p)}\tag{by Wilson's Theorem} \\
(p - 1)(p - 2)(p - 3)(p - 4)! &\equiv -1 \text{(mod p)} \\
(-1)(-2)(-3)(p - 4)! &\equiv -1 \text{(mod p)} \\
6(p - 4)! &\equiv 1
\end{align*}
By Fermat's Little Theorem, $6^{p - 1} \equiv 1$ (mod p), since $5 \nmid 6$ and the next primes are all greater than 6, so no prime $p \geq 5$ can divide 6. Then
\begin{align*}
6^{p - 1}6(p - 4)! &\equiv 1 \text{(mod p)} \\
6^p(p - 4)! &\equiv 1 \text{(mod p)}
\end{align*}
Now how would I apply Fermat's Little Theorem on $10^{3p}$? I've tried writing it as $(10^p)^3$, but since $5\mid 10$, it doesnt work. Would I discard the case when $p = 5$ to be able to use the theorem?
| HINT.-After you have $6^p(p-4)!\equiv 1\pmod p$ you need
$$1+(10^p)^3\equiv1+10^3=1001=7\cdot11\cdot13\equiv0$$ then you have three solutions $7,11$ and $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b$ in an arbitrary field Let $F$ be an arbitrary field, I need to prove that $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b \in F$
The difficulty that I am currently facing is that since $F$ is an arbitrary field, $x^2 - x + 1$ might be irreducible over $F$ and so I cannot solve it by factoring $x^2 - x + 1$. I also try to use the division algorithm to prove it using contradiction by assuming $x^2 - x + 1$ don't divide it, and hopefully something would happen to the remainder but this leads to a dead end as well. So any help would be appreciated.
| $$f(x)=x^4(x^6-x^3+1)+(ax+b)$$
Now reduce $f(x)$ by $x^2=x-1$ to have remainder $R(x)$ as
$$\implies R(x)=(x-1)^2[(x-1)^3-x(x-1)+1]+(ax+b)$$
$$\implies R(x)=(x-1-2x+1)[(x-1)(x^2-2x+1-x)+1])+ax+b$$
$$\implies R(x)=(x-1)^2[(x-1)(-2x)+1]+ax+b$$
$$\implies R(x)=(x-1)^2[-2(x-1)+2x+1]+ax+b$$
$$\implies R)x)=3(x-1)^2+ax+b \implies R(x)=-3x+ax+b$$
$$\implies R(x)=(a-3)x+b=0\implies a=3,b=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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ordered pair of $(p,q)$ in $20!$
A rational number given in the form
$\displaystyle \frac{p}{q},\;\;p,q\in \mathbb{Z}^{+}\;,\frac{p}{q}\in(0,1)$ and $p,q$ are coprime to each other.If $pq=20!.$ Then number of ordered pair of $p,q$ are
what i try
$20!=2^{18}\cdot 3^{8}\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$
Let $p=2^{a_{1}}\cdot 3^{a_{2}}\cdot 5^{a_{3}}\cdot 7^{a_{4}}\cdot 11^{a_{5}}\cdot 13^{a_{6}}\cdot 17^{a_{7}}\cdot 19^{a_{8}}$
$q=2^{b_{1}}\cdot 3^{b_{2}}\cdot 5^{b_{3}}\cdot 7^{b_{4}}\cdot 11^{b_{5}}\cdot 13^{b_{6}}\cdot 17^{b_{7}}\cdot 19^{b_{8}}$
$0\leq a_{1}\leq 18,0\leq a_{2}\leq 8,0\leq a_{3}\leq 4,0\leq a_{4}\leq 2,0\leq a_{5}\leq 1$
$0\leq a_{6}\leq 1,0\leq a_{7}\leq 1,0\leq a_{8}\leq 1$
and $0\leq b_{1}\leq 18,0\leq b_{2}\leq 8,0\leq b_{3}\leq 4,0\leq b_{4}\leq 2,0\leq b_{5}\leq 1$
$0\leq b_{6}\leq 1,0\leq b_{7}\leq 1,0\leq b_{8}\leq 1$
How do i solve it Help me please
| That $\gcd(p,q)=1$ means each prime power factor of $20!$ must either be wholly in $p$ or wholly in $q$. The fraction $\frac pq$ can never equal $1$, and if it is greater than $1$ then $\frac qp$ is less than $1$ and vice versa.
There are $2^8$ ways to assign the prime powers to either $p$ or $q$, so the number of fractions less than $1$ is half that, or $128$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Prove the following inequality: $\int_{0}^{\frac{\pi }{2}}\frac{sin(x)}{\sqrt{9-sin^{4}(x)}}dx\geq \frac{1}{3}$
Prove the following inequality: $$ \int_{0}^{\frac{\pi}{2}}\frac{\sin
x}{\sqrt{9-\sin^{4}x}}\ \mathrm dx\geq\frac{1}{3}. $$
I am thinking of replacing the equation with $\int_{0}^{\frac{\pi }{2}}\frac{\sin x}{9-\sin^{4} x }dx\geq \frac{1}{3}$, however I am stuck at this point.
Do you have any suggestions?
| \begin{align*}\int_{0}^{\pi/2} \frac{\sin{x}}{\sqrt{9-\sin^4{x}}}dx &\geq \int_{0}^{\pi/2}\frac{\sin{x}}{\sqrt{9}}dx\\& = \frac{1}{3}\int_{0}^{\pi/2} \sin{x}dx \\&=\frac{1}{3}[-\cos{\pi/2}-(-\cos{0})]\\&=\frac{1}{3}[0+1]\\&=\frac{1}{3}
\end{align*}
The inequality is because $\sqrt{9-\sin^4{x}}\leq \sqrt{9}$ due to positivity of $\sin^4{x}$. So ... $$\frac{1}{\sqrt{9-\sin^4{x}}}\geq \frac{1}{\sqrt{9}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535654",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=2-2=0$$
But on the right hand I have inrcrased $\sqrt{x^2-2x}$ rather then deceased
Another attempt:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
| Tips:
Try to find and sum up the two limits:
$$\lim_{x\to\infty} \sqrt[3]{x^3-2x}-x \text{ and }\lim_{x\to\infty}x-\sqrt{x^2-2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Laplace transform problem help I'm stuck with this Laplace transform problem.
Using Laplace transform, solve initial problem:
$$
\ y''' - y = 1
\\ y(0) = y'(0) = 1
\\ y''(0) = 0
$$
And from that we would get:
$$
\ s^3Y - s^2 - s - Y = \frac{1}{s}
$$
I've got it until the point I get:
$$
\ Y = \frac{s^3 + s^2 + 1}{s(s-1)(s^2 + s + 1)} = \frac{s^3 + s^2 + 1}{s(s^3 - 1)}
$$
and now I'm not sure what to do next.
I know that I should get something like:
$$
\ Y = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s^2 + s + 1}
$$
But I'm not able to get that, and if I got that I'm not sure what to do with fraction that has denominator like this:
$$
\ \frac{C}{s^2 + s + 1}
$$
That also makes me curios if I have overlooked on something and that term should actually cancel with something? (As it looks very similar to the numerator).
Any help and explanation of the situation would be of great help and use!
To summarize: I'm stuck at one point in solving this Laplace transform problem and my question would be how to solve it.
| Are you familiar with the general partial fraction method? With a quadratic (order 2) polynomial in the denominator, you should write a general linear (order 1) polynomial in the numerator. Then putting things over a common denominator,
$$\frac{A}{s} + \frac{B}{s-1} + \frac{Cs + D}{s^2+s+1} = \frac{(A+B+C)s^3+(B-C+D)s^2+(B-D)s-A}{s(s-1)(s^2+s+1)}$$
from which $A=-1$ and $B=C=D=1$. (Alternatively, you could factorize $s^2+s+1$ over the complex numbers.)
Then the inverse Laplace transform of
$$\frac{1}{s^2+s+1} = \frac{1}{\left(s+\frac{1}{2}\right)^2 + \frac{3}{4}}$$
is $e^{-x/2}$ multiplying the inverse Laplace transform of $$\frac{1}{s^2 + \frac{3}{4}}$$
which is itself $2/\sqrt{3} \times \sin \left(\sqrt{3}x/2\right)$.
Then multiplying by $s$ is the same as differentiating by $x$.
Note that the complex method (factorizing the quadratic with complex roots) is maybe more intuitive and has less arbitrary rule-following. It gives
$$
\frac{s+1}{s^2+s+1}
= \frac{\frac{1}{2} + \frac{i}{2\sqrt{3}}}{s+\frac{1}{2} + \frac{i\sqrt{3}}{2}}
+\frac{\frac{1}{2} - \frac{i}{2\sqrt{3}}}{s+\frac{1}{2} - \frac{i\sqrt{3}}{2}}
= {\rm Re} \left[ \frac{1 + \frac{i}{\sqrt{3}}}{s+\frac{1}{2} + \frac{i\sqrt{3}}{2}} \right]
$$
whose inverse Laplace transform is
$$
{\rm Re} \left[ \left(1 + \frac{i}{\sqrt{3}}\right) e^{-\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)x} \right] = e^{-x/2} \cos\left(\frac{\sqrt{3}}{2}x\right) + \frac{1}{\sqrt{3}}e^{-x/2} \sin\left(\frac{\sqrt{3}}{2}x\right)
$$
| {
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Integral $\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx$
Evaluate the integral $$P=\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx.$$
Context:
I started trying to evaluate the integral $$J=\int_0^\infty \frac{\arctan(x)^2}{x^2+x+1}dx,$$
and the integral $P$ is part of the process. At first, I tried $x\mapsto 1/x$, but it just ended up showing that $$J=\frac{\pi^2}{4}\int_0^\infty \frac{dx}{x^2+x+1}-\pi\int_0^\infty \frac{\arctan x}{x^2+x+1}dx+J,$$
which is of no use. Next, I tried integration by parts, using
$$\int\frac{dx}{x^2+x+1}=\frac{2}{\sqrt3}\arctan\frac{2x+1}{\sqrt3},$$
so that
$$J=\frac{\pi^3}{4\sqrt3}-\frac{4}{\sqrt3}\int_0^\infty\arctan(x)\arctan\left(\tfrac1{\sqrt3}+\tfrac{2}{\sqrt3}x\right)\frac{dx}{1+x^2}.$$
Then with $x\mapsto \tan x$ we have $$J=\frac{\pi^3}{4\sqrt3}-\frac{4}{\sqrt3}P.$$
Theoretically, integration by parts if possible from this point, as Wolfram provides an awful closed form for the anti-derivative of $\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)$, but I do not think this is really that realistic of an approach. Is there a better way to evaluate the integral $P$?
| $$J=\int_0^\infty \frac{\arctan^2 x}{1+x+x^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2}\frac{t^2}{1+\sin t\cos t}dt\overset{2t=\frac{\pi}{2}-x}=\frac14\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\left(\frac{\pi}{2}-x\right)^2}{2+\cos x}dx$$
$$=\frac12\int_0^\frac{\pi}{2}\frac{\frac{\pi^2}{4}+x^2}{2+\cos x}dx=\frac{\pi^3}{24\sqrt 3}+\frac12\int_0^\frac{\pi}{2}\frac{x^2}{2+\cos x}dx$$
Now we will use the following Fourier series (refer to this thread):
$$\frac{1}{2+\cos x}=\frac{1}{\sqrt 3}+\frac{2}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\cos(nx)$$
$$\Rightarrow J=\frac{\pi^3}{16\sqrt 3}+\frac{1}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\int_0^\frac{\pi}{2}x^2\cos(nx)dx$$
$$=\frac{\pi^3}{16\sqrt 3}+\frac{\pi^2}{4\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n}$$
$$+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\cos\left(\frac{n\pi}{2}\right)}{n^2}-\frac{2}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n^3}$$
$$\small =\frac{\pi^3}{16\sqrt 3}-\frac{\pi^2}{4\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{2n+1}+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n}}{(2n)^2}+\frac{2}{\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{(2n+1)^3}$$
$$=\boxed{\frac{\pi^3}{24\sqrt 3}+\frac{\pi}{4\sqrt 3}\operatorname{Li}_2\left(-\left(2-\sqrt 3\right)^2\right)+\frac{2}{\sqrt 3}\operatorname{Ti}_3\left(2-\sqrt 3\right)}$$
$$\text{where }\operatorname{Li}_k(x)=\sum\limits_{n=1}^\infty \frac{x^n}{n^k},\ \operatorname{Ti}_k(x)=\Im\operatorname{Li}_k(ix)=\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$ Given
$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$
I calculated $a_1$ to $a_5$
$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
which made me think of $\sin/\cos$. So I divided each by $2$, calculated $\arcsin(x)$ and got $$\left\{\frac{\pi }{4},\frac{\pi }{8},\frac{3 \pi }{16},\frac{5 \pi }{32},\frac{11 \pi }{64}\right\}$$
I found that this could be a formula
$$\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}$$
So $$\sin \left(\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}\right)$$ is $a_n/2$, but this way seems not natural. Is there a more natural way?
| Let $x_n$ for $n\in\Bbb N$ be the sequence defined by
\begin{align}
x_0&=\frac\pi 2&
x_n&=\frac{\pi-x_{n-1}}2
\end{align}
and $a_n=2\cos(x_n)$.
Then $0\leq x_n\leq\frac\pi 2$, hence $0\leq a_n\leq 2$.
Moreover:
\begin{align}
a_n^2
&=4\cos^2(x_n)\\
&=2\cos(2x_n)+2\\
&=2\cos(\pi-x_{n-1})+2\\
&=-2\cos(x_{n-1})+2\\
&=-a_{n-1}+2
\end{align}
so that $a_n=\sqrt{2-a_{n-1}}$.
On the other hand $x_n=\frac\pi 3+(-2)^{-n}(x_0-\frac\pi 3)$, hence
$$a_n=2\cos\Bigl(\frac\pi 3+(-2)^{-n}\frac\pi 6\Bigr)$$
gives an explicit formula for $a_n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int{\frac{\sin^3(x)}{2\cos^2(x)-3\sin^2(x)}}dx$ I want to integrate the following expression:
$$\int{\frac{\sin^3(x)}{2\cos^2(x)-3\sin^2(x)}}dx$$
I tried using a $t=\tan(\frac{x}{2})$ substitution but the terms were not cancelled. Then I tried using $1=\cos^2(x)+\sin^2(x)$ but couldn't get anywhere. I hope some of you can help, thanks.
| Hint:
Recall that $$\sin^2(x) = 1 - \cos^2(x)$$
Therefore,
$$\int\dfrac{\sin^3(x)}{2\cos^2(x) - 3\sin^2(x)}\,\mathrm dx\equiv\int\dfrac{1 - \cos^2(x)}{5\cos^2(x) - 3}\sin(x)\,\mathrm dx$$
Let $u = \cos(x)\implies\mathrm du = -\sin(x)\mathrm dx$. So,
$$\int\dfrac{1 - \cos^2(x)}{5\cos^2(x) - 3}\sin(x)\,\mathrm dx\equiv\int\dfrac{u^2 - 1}{5u^2 - 3}\,\mathrm du = \dfrac15\int\mathrm du - \dfrac25\int\dfrac1{5u^2 - 3}\,\mathrm du$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$
Prove that
$$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$
According to Wolfram the above holds. Could someone show me the steps for this?
| HINT
First note that
$$\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac{(n+1)-1}{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}-\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}$$
Now recall that
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n}=\frac1{\sqrt{1-x}},~~~\text{for }|x|<1$$
The task boils down to an exercise in integration. Can you take it from here?
For the interested reader I will add the complete solution now. As given, note that for the first sum we have the following
$$\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}=\sum_{n\geq0}\binom{2n}n\frac1{4^n}\int_0^1x^n{\rm d}x=\int_0^1\frac{{\rm d}x}{\sqrt{1-x}}=[-2\sqrt{1-x}]_0^1=2$$
This also gives us the following identity
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n(n+1)}=\int\frac{{\rm d}x}{\sqrt{1-x}}=-2\sqrt{1-x}+c$$
Take $x=0$ to see that $c=2$. Dividing by $x$ and integrating again over $[0;1]$ gives
\begin{align*}
\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}&=\int_0^1\frac{2-2\sqrt{1-x}}x{\rm d}x\\
&=2\int_0^1\frac{1-\sqrt{1-x}}x{\rm d}x\\
&=2\int_0^1\frac{1-\sqrt x}{1-x}{\rm d}x\\
&=2\int_0^1\frac1{1+\sqrt x}{\rm d}x\\
&=2\sum_{n\geq0}(-1)^n\int_0^1x^{\frac n2}{\rm d}x\\
&=2\sum_{n\geq0}\frac{(-1)^n}{\frac n2+1}\\
&=4\sum_{n\geq0}\frac{(-1)^n}{n+2}\\
&=4\left[\sum_{n\geq1}\frac{(-1)^n}n+1\right]\\
&=4-4\log2
\end{align*}
The result follows.
$$\therefore~\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}~=~4\log2-2$$
| {
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"url": "https://math.stackexchange.com/questions/3543054",
"timestamp": "2023-03-29T00:00:00",
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Is there any simpler way to find a remainder in multiple divisions? I got a question as follows.
$3x-5$ is the remainder when unknown $f(x)$ is divided by $x^2-x+1$
that has relatively complicated roots. Find the remainder when $f(x)$
is divided by $(x^2-x+1)(x-1)$. Express your answer in terms of
unknown $f(1)$, $x^2-x+1$, and $3x-5$.
Attempt
As the divisor is a cubic, then the remainder is at most a quadratic $a x^2 + bx +c$. Let $x_1$ and $x_2$ be the complicated root of $x^2-x+1$.
Now I have
\begin{align}
3x_1-5 &= a x_1^2 + bx_1 + c\\
3x_2-5 &= a x_2^2 + b x_2 +c \\
f(1) &= a + b +c
\end{align}
Finding $a$, $b$ and $c$ seems to be extremely tedious for me.
Question
Is there any simpler method to find the remainder in question?
Edit:
I have edited the quoted problem. The answer can be in terms of the divisor $x^2-x+1$ as well as the remainder $3x-5$.
| A possible way is as follows:
*
*$f(x) = (x^2-x+1)q(x) + 3x-5$
*$q(x) = (x-1)r(x) + c$
Hence,
$$f(x) = (x^2-x+1)((x-1)r(x) + c) + 3x -5$$ $$= (x^2-x+1)(x-1)r(x) + c(x^2-x+1) + 3x-5$$
Now you can find $c$:
$$f(1) =c -2 \Leftrightarrow c=2+f(1)$$
So, you get
$$f(x) = (x^2-x+1)(x-1)r(x) + \color{blue}{(2+f(1))(x^2-x+1) + 3x-5}$$
I leave it up to you to collect like terms of the remainder as it suits you.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please
| Hint:
If $x\le-1$ or $x\ge5$ then $-3x^2+12+4<0$, so its square root is not real;
that limits the possibilities to a few you could check.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})} = \frac {1-1/7^{4x} }{1+ 1/21^{4x}} $$
$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \dfrac {\lim\limits_{ x \to \infty } 1-1/7^{4x} }{ \lim\limits_{ x \to \infty } 1+ 1/21^{4x} }$$
but that’s defiantly wrong
| Put the constant outside of the limit:
$$
\begin {align}
l=&\lim\limits_{ x \to \infty }\color{blue}{\frac 1 3} \frac {7^{2x} +7^{-2x} }{ (7^{2x} - 7^{-2x})} \\
l=&\color{blue}{ \frac 1 3}\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 7^{2x} - 7^{-2x}} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{7^{2x}}( 1+ 7^{-4x}) }{ \color{blue}{7^{2x}} (1 - 7^{-4x})} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{} 1+ 7^{-4x} }{ \color{blue}{} 1 - 7^{-4x}} \\
l=&\frac 1 3
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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No positive real such that $\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$
Let $a>0$. Prove $\nexists x\in\mathbb R^+$ s.t.
$$\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$$
I know that $$k\in\mathbb Z\implies\left(\forall x\in\mathbb R\right) \lfloor x+k\rfloor=\lfloor x\rfloor + k.$$
I can say that $\left\lfloor\frac{144}{x+a}-1\right\rfloor=\left\lfloor\frac{144}{x+a}\right\rfloor-1$.
How should I proceed after this? Does this help in any way?
| Hint: The floor function, $\lfloor x\rfloor$, is bounded by $x-1< \lfloor x\rfloor\le x$ so you have $\text{RHS}\le \left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1<\text{LHS}$. By considering their (single) intersection, consider how you can form an inequality with $\left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1$, then check whether this intersection is a solution and find an inequality relating all four terms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Changing index in summation This question seems simple but it's been eating me for 20 mins. on a book I seen something like this :
$$
\sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1}\frac{1}{6} = \frac{1}{6}\left(\frac{5}{6}\right)^{7}\sum_{j=0}^{\infty}\left(\frac{5}{6}\right)^{j}
$$
Could anyone tell me why the above step is valid, I understand the $\frac{1}{6}$ part but not the $\left(\frac{5}{6}\right)^{7}$ part. It is from a probability book so conceptually I know this gives the right answer. please help, thank you very much
| If in doubt, write the terms out explicitly ... lets leave the $1/6$ out
\begin{eqnarray*}
\sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1} &=& \left(\frac{5}{6}\right)^{7} + \left(\frac{5}{6}\right)^{8} + \left(\frac{5}{6}\right)^{9} + \cdots \\
&=& \left(\frac{5}{6}\right)^{7} \left(1 + \frac{5}{6} + \left(\frac{5}{6}\right)^{2} + \cdots \right) \\
&=& \left(\frac{5}{6}\right)^{7} \sum_{j=0}^{\infty} \left(\frac{5}{6}\right)^{j}. \\
\end{eqnarray*}
| {
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Proof of $n\ge 6\implies 2n−8\le n^2−8n+ 16$ using induction.
I have to do a proof by induction for this theorem:
For each $n\in \mathbb N$ such that $n\ge 6$ we have $$2n−8\le n^2−8n+ 16$$
Is this possible, or should I do a different type of proof. I am confused how to prove that $1\in S$ when $n\ge 6$, so I am thinking maybe I need to first prove $6\in S$ and then handle $n>6 $?
Thank you!
| Base case: Let $n=6$. Then the LHS is $4$ whereas the RHS is $(6-4)(6-4)=4$, so $2n-8\le n^2-8n+16=(n-4)^2$ is true for $n=6$.
Induction Hypothesis: Assume that for some fixed $k\ge 6$ we have $$2k-8\le (k-4)^2.\tag{$I$}$$
When $n=k+1$: Suppose $n=k+1$. Then
$$\begin{align}
2(k+1)-8&=2k-6\\
&=(2k-8)+2 \\
&\le (k-4)^2+2\quad \text{(by }(I)\text{)}\\
&=k^2-8k+16+2\\
&=k^2-6k+9- (2k-9)\\
&\le k^2-6k+9\quad \text{(for }k\ge 6\text{)}\\
&=((k+1)-4)^2\\
&=(k+1)^2-8(k+1)+16,
\end{align}$$
which is the same as $(I)$ but with $k$ replaced by $k+1$, so if it's true for $k$, it is true for $k+1$.
Conclusion: By induction on $n$ for all integers $n\ge 6$, we have $$2n-8\le n^2-8n+16.$$
| {
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Proving $(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$
If $\alpha \in \left(0, \frac{\pi}{2}\right)$, prove that:
$$(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$$
I know $\sin \alpha$ and $\cos \alpha$ are positive over $\alpha \in \left(0, \frac{\pi}{2}\right)$, so with $a=\sin \alpha,\ b=\cos \alpha$, $a>0,\ b>0$ and $a^2+b^2=1$, the inequality is
$$(a^2+ab)^a(b^2+ab)^b \leq 1$$
$$\Leftarrow a^ab^b(a+b)^{a+b} \leq 1$$
and here I don't know how to prove this inequality.
| The inequality is equivalent to: $$(1-\cos^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(1-\sin^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$$
By Bernoulli Inequality, $$(1-\cos^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(1-\sin^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\\\ \\ \leq \left(1+(-\cos^2 \alpha+\sin\alpha \cos \alpha){\sin \alpha}\right) \left( 1+(-\sin^2 \alpha+\sin \alpha \cos \alpha ){\cos \alpha}\right) \\\\ =\left(1-(\sin \alpha \cos \alpha )(\cos \alpha - \sin \alpha )\right) \left( 1+(\sin \alpha \cos \alpha )(\cos \alpha - \sin \alpha )\right)\\=1-[(\sin \alpha \cos \alpha )(\cos \alpha - \sin \alpha )]^2\\ \leq 1$$
| {
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For $\triangle ABC$, prove $\sum_{cyc}\frac{1}{4+\cos A\cos(B-C)}\geq\frac23$
For the triangle $\triangle ABC$, prove the inquality,
$$\frac1{4+\cos A\cos(B-C)}+\frac1{4+\cos B\cos(C-A)}+\frac1{4+\cos C\cos(A-B)}\ge \frac23$$
where $A$, $B$ and $C$ are the vertex angles.
I had trouble tackling it. Have treated it as a minimization problem, subject to the constraint $A+B+C=\pi$. In principle, it could be solved with the Lagrange multiplier. However, the required algebraic work seems to be prohibitive. Wonder whether there is a more elegant path to show it.
| First, we use the trig identity $ \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) $:
$$\frac1{4 + \cos A \cos(B-C)}+\frac1{4 +\cos B \cos(C-A)} + \frac1{4 + \cos C \cos(A-B)} \ge \frac23$$
$$ \Leftrightarrow \frac1{4 + \frac{1}{2} (\cos(A+B-C) + \cos(A-B+C))} + \frac1{4 + \frac{1}{2} (\cos(B+C-A) + \cos(B-C+A))} + \frac1{4 + \frac{1}{2} (\cos(C+A-B) + \cos(C-A+B))} \ge \frac23 $$
Since $A + B + C = \pi$, we get:
$$ \frac1{4 + \frac{1}{2} (\cos(\pi-2C) + \cos(\pi-2B))} + \frac1{4 + \frac{1}{2} (\cos(\pi-2A) + \cos(\pi-2C))} + \frac1{4 + \frac{1}{2} (\cos(\pi-2B) + \cos(\pi-2A))} \ge \frac23 $$
Using the trig identity $ \cos(\pi-x) = - \cos x$:
$$ \frac1{4 - \frac{1}{2} (\cos(2C) + \cos(2B))} + \frac1{4 - \frac{1}{2} (\cos(2A) + \cos(2C))} + \frac1{4 - \frac{1}{2} (\cos(2B) + \cos(2A))} \ge \frac23 $$
Applying the inequality $ \frac1a + \frac1b + \frac1c \ge \frac9{a+b+c}$:
$$ \frac1{4 - \frac{1}{2} (\cos(2C) + \cos(2B))} + \frac1{4 - \frac{1}{2} (\cos(2A) + \cos(2C))} + \frac1{4 - \frac{1}{2} (\cos(2B) + \cos(2A))} $$
$$ \ge \frac9{12 - (\cos(2A) + \cos(2B) + \cos(2C))} $$
Finally, use the inequality $ \cos(2A) + \cos(2B) + \cos(2C) \ge - \frac{3}{2}$ (which you can find the proof in here: Prove that $\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}$ for angles of a triangle):
$$ \frac{9}{12 - (\cos(2A) + \cos(2B) + \cos(2C))} \geq \frac{9}{12 - (- \frac{3}{2})} = \frac23 $$
The inequality holds when
$$A = B = C = \frac{\pi}{3} $$
or ABC is an equilateral triangle
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"language": "en",
"url": "https://math.stackexchange.com/questions/3553471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of $\sqrt{4x^2+x+7}+2x$ I've been working on this problem for a while now, but I can't solve it
$$\lim\limits_{x\to-\infty}{\sqrt{4x^2+x+7}+2x}$$
I've tried multiplying by
$$\frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}$$
but I didn't get it. Am I missing something really obvious? Can someone help me with this question?
| $$
\begin{aligned}
\lim_{x\to-\infty}\left(\sqrt{4x^2+x+7}+2x\right) &= \lim_{x\to-\infty}\frac{4x^2+x+7-4x^2}{\sqrt{4x^2+x+7}-2x}\\
& =\lim_{x\to-\infty}\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\
&=\lim_{x\to-\infty}\frac{-1-\frac{7}{x}}{\sqrt{4+\frac{1}{x}+\frac{7}{x^2}}+2}\\
&=-\frac{1}{4}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
A random sphere containing the center of the unit cube Inspired by a Putnam problem, I came up with the following question:
A point in randomly chosen in the unit cube, a sphere is then created using the random point as the center such that the sphere must be contained inside the cube (In other words, the largest sphere that fits). What is the probability that the center of the cube is contained inside the sphere created?
No real idea how to approach this one but thought some of you might find this interesting.
| Here's the solution, confirming the previous answers, using symmetry as I suggested. I'll reorder my variables to conform with previous discussions. So we're going to consider the cube $[-1,1]^3$ and restrict to centers of the sphere in the pyramid $0\le x\le y\le z\le 1$. This means that the face $z=1$ will be the closest. The origin will be (on or) inside such a sphere if and only if $x^2+y^2+z^2\le (1-z)^2$, i.e., $2z\le 1-(x^2+y^2)$.
Over what region in the $xy$-plane does our region project? Since $x\le y\le z$, we must have $2y\le 2z\le 1-(x^2+y^2)$, which means $x^2+y^2+2y\le 1$, or $x^2+(y+1)^2\le 2$. This results in the portion of $0\le x\le y$ lying inside the circle $x^2+(y+1)^2\le 2$. Note that $0\le x\le \dfrac{\sqrt3-1}2$.
Setting up the triple integral, the volume we desire is
$$\int_0^{\frac{\sqrt3-1}2}\int_x^{\sqrt{2-x^2}-1}\int_y^{\frac12(1-x^2-y^2)} dz\,dy\,dx,$$
and since we're comparing to the volume of the full pyramid, which is $1/6$, we take $6$ times this answer.
(We can also set this up nicely in polar coordinates:
$$\int_0^{\pi/4}\int_0^{\sqrt{\cos^2\theta+1}-\cos\theta}\int_{r\sin\theta}^{\frac12(1-r^2)}\,r\,dz\,dr\,d\theta.)$$
The integral, multiplied by $6$, turns into
\begin{align*}
\int_0&^{\frac{\sqrt3-1}2} \big({-}5+4\sqrt{2-x^2}+3x^2-2x^2\sqrt{2-x^2}-(3x-3x^2-4x^3)\big)dx\\ &= \int_0^{\frac{\sqrt3-1}2} \big({-}5-3x+4x^3+4\sqrt{2-x^2}+6x^2-2x^2\sqrt{2-x^2}\big)dx \\ &= \frac54-\frac98\sqrt3+\frac{\pi}4.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Show that $n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$
Hence or otherwise show that for all positive integers $n$
$$
n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}
$$
This is related to another part of this question where I proved $\ln \left(\frac{4-t}{t}\right) \geq 2-t$ for $0<t \leq 2$ by integrating $\frac{1}{s}+\frac{1}{4-s} \geq 1$(which is true for $0<s<4$) over $[t, 2] .$ However, I am not able to use this to prove that
$$n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$$ for all positive integers. Any help would be appreciated.
| Divide by $n$ both sides, you need to prove: $\ln\left(1+x\right) > \dfrac{2x}{2+x}= 2-\dfrac{4}{2+x}, 0 < x \le 1, x = \dfrac{1}{n}$. Consider $f(x) = \ln(1+x) +\dfrac{4}{2+x}-2, 0 < x \le 1\implies f'(x) = \dfrac{1}{1+x} - \dfrac{4}{(2+x)^2} = \dfrac{(x+2)^2-4(x+1)}{(1+x)(2+x)^2}= \dfrac{x^2+4x+4-4x-4}{(1+x)(2+x)^2}= \dfrac{x^2}{(1+x)(2+x)^2}> 0\implies f(x) > f(0) = 0\implies \ln(1+x) > 2 - \dfrac{4}{2+x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the indefinite integral $\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$ $$\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$$
My Attempt:
$$I = \int \frac{\sin^2\frac\theta2\cdot\sin\frac\theta2\cos\frac\theta2}{\cos^2\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}}
d\theta \\ = \frac12 \int \frac{(1-\cos\theta)\sin\theta}{(1+\cos\theta)\sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}}
d\theta$$
$$\text{let }\cos\theta = t \implies -\sin\theta d\theta = dt$$
$$I = \frac12 \int \frac{t-1}{(t+1) \sqrt{ t^3+t^2+t } } dt $$
I am not sure how to proceed further from here. Any hints/solutions on how to resolve the cubic expression under the square root?
| Set $\sqrt t=y,t=y^2,dt=2y\ dy$
$$I =\int\dfrac{(y^2-1)2y}{(y^2+1)y\sqrt{y^2(y^4+y^2+1)}}dy$$
$$=2\int\dfrac{1-1/y^2}{(y+1/y)\sqrt{y^2+1+1/y^2}}dy$$
Set $\int(1-1/y^2)dy =z$
Then $\sqrt{z^2+1}=u\implies z^2+1=u^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 0
} |
Is there another method ? of this summation $\sum_{k=0}^{n}{n \choose k}\cos kx$ Calculate or simplify $$\sum_{k=0}^{n}{n \choose k}\cos (kx)$$
by using complex method.
Let $A=\sum_{k=0}^{n}{n \choose k}\cos(kx)$ and $B=\sum_{k=0}^{n}{n \choose k}\ i \sin(kx)$ , then
$$\begin{align*}
A+iB &= \sum_{k=0}^{n}{n \choose k}\cos(kx)+\sum_{k=0}^{n}{n \choose k}\ i \sin(kx) \\
&= \sum_{k=0}^{n}{n \choose k}\left(\cos(kx)+i \sin(kx)\right) \\
&=\sum_{k=0}^{n}{n \choose k}e^{ikx}\end{align*}$$
By Binomial coefficient $(x+y)^{n}=\sum_{k=0}^{n}{n \choose k}x^{n-k}y^{k}$, so
$$A+iB=(1+e^{ix})^{n},$$
then
$$Re(A+iB)=Re(1+e^{ix})^{n},$$
thus
$$A=Re\left \{ e^{\frac{nix}{2}} (e^{\frac{-ix}{2}}+e^{\frac{ix}{2}})^n\right \}=Re\left ( e^{\frac{nix}{2}}2^{n}\cos^{n}\frac{x}{2} \right ).$$
So
$$\sum_{k=0}^{n}{n \choose k}\cos(kx)=2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{nx}{2}\right)$$
| Here's a induction proof. I think the proof using complex numbers is better, but the next proof illustrates a useful method.
We'll prove the following in parallel
$$\sum_k\binom{n}{k}\cos(k\ x) = 2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)$$
$$\sum_k\binom{n}{k}\sin(k\ x) = 2^n\cos^n\left(\frac{x}{2}\right)\sin\left(\frac{n\ x}{2}\right)$$
For $n=0$ the equalities are obvious.
For the inductive step we have
\begin{align}
&\sum_k\binom{n+1}{k}\cos(k\ x) =
\sum_k\binom{n}{k}\cos(k\ x)+ \sum_k\binom{n}{k-1}\cos(k\ x)\\
&=\sum_k\binom{n}{k}\cos(k\ x)+ \sum_k\binom{n}{k}\cos((k+1)\ x)\\
&=\sum_k\binom{n}{k}\cos(k\ x)+ \cos(x)\sum_k\binom{n}{k}\cos(k\ x)- \sin(x)\sum_k\binom{n}{k}\sin(k\ x)\\
&\stackrel{IH}= 2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)
+\cos(x)2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)
-\sin(x)2^n\cos^n\left(\frac{x}{2}\right)\sin\left(\frac{n\ x}{2}\right)\\
&= 2^n\cos^n\left(\frac{x}{2}\right)\left(\cos\left(\frac{n\ x}{2}\right)+\cos(x)\cos\left(\frac{n\ x}{2}\right)-\sin(x)\sin\left(\frac{n\ x}{2}\right)\right)\\
&= 2^n\cos^n\left(\frac{x}{2}\right)\left(\cos\left(\frac{n\ x}{2}\right)+\cos\left(\frac{(n+2)\ x}{2}\right)\right)\\
&= 2^n\cos^n\left(\frac{x}{2}\right)2\cos\left(\frac{x}{2}\right)\cos\left(\frac{(n+1)\ x}{2}\right) = 2^{n+1}\cos^{n+1}\left(\frac{x}{2}\right)\cos\left(\frac{(n+1)\ x}{2}\right)
\end{align}
where we used $\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)$ and $\cos(a)+\cos(b) = 2 \cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$.
A similar calculation using $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$, $\sin(a)+\sin(b) = 2 \sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$ and the IH proves that $$\sum_k\binom{n+1}{k}\sin(k\ x) = 2^{n+1}\cos^{n+1}\left(\frac{x}{2}\right)\sin\left(\frac{(n+1)\ x}{2}\right)$$
and that completes the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(2 - x)} = \frac{3}{2} + \sum_{n=1}^∞ \frac{x^n5}{ 2^{n+1}}$
Unfortunately I forgot to write down how he did it. How do I get that solution and is mine necessarily incorrect/incomplete?
| $$S=\frac{x+3}{2-x} = \frac{x-2+5}{2-x}=-1+\frac{5}{2-x}$$
$$S=-1+\frac 52\frac{1}{1-\dfrac x2}$$
$$S=-1+\frac 52\sum_{n=0}^{\infty}\left (\dfrac x 2\right )^n$$
Change the indice of the sum:
$$S=-1+\frac 52+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
Finally:
$$S=\frac 32+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
$$S=\frac 32+ 5\sum_{n=1}^{\infty}\dfrac {x^n}{ 2^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge.
I'm not allowed to use the formula $a+b+c=s$ where 's' is perimeter.
My Try
Using sine rule I was able to simplify,
\begin{align}\text{L. H. S.}& =\frac{(\sin A+\sin B+ \sin C)^2}{\sin^2A+\sin^2B+\sin^2C} \\
&=\frac{4\cos^2\dfrac C2\Bigl(\dfrac{\cos(A-B)}{2}+\sin\frac C2\Bigr)^2}{\sin^2A+\sin^2B+\sin^2C}
\end{align}
I'm unable to simplify further, Please give me hint. Thanks in advance.
| For a triangle ABCz, we have denominator of LHS as
$$D=(a^2+b^2+c^2)=[(b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)]$$ $$=[2bc \cos A+2ca \cos B+ 2ab \cos C]=2abc[\cos A/a+\cos B/b+\cos C/c]=\frac{abc}{R}[\cot A+\cot B+\cot C].~~~~(1)$$ In the last step we have used $ a/\sin A=b/\sin B=c/\sin C=2R.$
Next take the numerator of RHS as
$$N=(a+b+c)^2=4R^2[\sin A+ \sin B+\sin C]^2=64R^2[\cos^2(A/2) \cos^2(B/2) \cos^2(C/2)]$$
$$\implies N=\frac{16R^3 (\sin A \sin B \sin C) (64R^2[\cos^2(A/2) \cos^2(B/2)\cos^2(C/2)}{16R^3 (\sin A \sin B \sin C)}$$
$$\implies \frac{4(2abc)}{R}\frac{\cos^2(A/2) \cos^2(B/2)\cos^2(C/2)}{8(\sin(A/2) \cos(A/2)(\sin (B/2) \cos (B/2)(\sin (A/2) \cos (A/2)}$$
$$\implies N=\frac{abc}{R}[\cot(A/2) \cot(B/2) \cot(C/2)]=\frac{abc}{R}[\cot(A/2)+\cot(B/2)+\cot(C/2)]~~~~(2)$$
From (1) and (2) the required result follows.
Lastly we have used $$\cot(A/2)\cot(B/2)\cot(C/2)=\cot(A/2)+\cot(B/2)+\cot(C/2)]$$
in the triangle ABC.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{2},x_2=-1-\sqrt{2},x_3=1-2i,x_4=1+2i$. And after long computations the polynomial is $x^6+20x^2-144$. Are there clever way to find it?
| By Vieta's formulae, we have $x_1 + x_2 + x_3 + x_4 = 0$, $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = 0$, $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -12$, and $x_1x_2x_3x_4 = 5$. We can now calculate \begin{align*}(x_1+x_2)+(x_1+x_3)+(x_1+x_4)+(x_2+x_3)+(x_2+x_4)+(x_3+x_4) &= 3(x_1+x_2+x_3+x_4) \\ &= 0\end{align*} Similarly, \begin{align*}&(x_1+x_2)(x_1+x_3) + (x_1+x_2)(x_1+x_4) + \dotsb + (x_2+x_4)(x_3+x_4) \\ &= 3\left(x_1^2+x_2^2+x_3^2+x_4^2\right)+8\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\right) \\ &= 3\left[(x_1+x_2+x_3+x_4)^2-2(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)\right] + 8(0) \\ &= 3(0^2-2(0))+8(0) \\ &= 0\end{align*} and so on. Once we have computed all the symmetric polynomials, we can then use Vieta's formulae again to form an equation with the desired roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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D-operator-methods Solve the following differential equation:
$$(D^2-2D+1)y=x^2e^{3x}$$
I found the $C.F.=(c_1+c_2x)e^x$
$$\begin{align}
P.I. & =\frac{x^2e^{3x}}{(D^2-2D+1)}\\
& =e^{3x}\frac{x^2}{(D+3)^2-2(D+3)+1}\\
& =e^{3x}\frac{x^2}{D^2+6D+9-2D-6+1}\\
& =e^{3x}\frac{x^2}{D^2+4D+4}\\
& =\frac{e^{3x}}{4}x^2\left(1+\frac{D^2+4D}{4}\right)^{-1}\\
& =\frac{e^{3x}}{4}x^2\left(1-\frac{D^2+4D}{4}+........\right)\\
& =\frac{e^{3x}}{4}\left(x^2-\frac{2+8x}{4}\right)\\
& =\frac{e^{3x}}{4}\left(x^2-2x-\frac{1}{2}\right)\\
\end{align}$$
But in my book the P.I. is:
$$P.I.=\frac{1}{4}e^{3x}\left(x^2-2x+\frac{3}{2}\right)$$
I can't find my fault, please check this
| Your solution is missing a second order derivative term at this step
$$\frac{e^{3x}}{4}x^2\left(1-\frac{D^2+4D}{4}+........\right)$$
repeating the steps where $\text{C.F.}$ represents the homogeneous solution and $\text{P.I.}$ the particular solution
$$\begin{align}
\text{P.I.} & =\frac{x^2e^{3x}}{(D^2-2D+1)}\\
& =e^{3x}\left(\frac{1}{(D+3)^2-2(D+3)+1}\right)x^2\\
& =e^{3x}\left(\frac{1}{D^2+4D+4}\right)x^2\\
& =e^{3x}\left(\frac{1}{4\left[1+\dfrac{4D}{4}+\dfrac{D^2}{4}\right]}\right)x^2\\
& =\frac{e^{3x}}{4}\left(1+\left[D+\frac{D^2}{4}\right]\right)^{-1}x^2\\
& =\frac{e^{3x}}{4}\left(1-\left[D+\frac{D^2}{4}\right]+\left[D+\frac{D^2}{4}\right]^2\right)x^2\\
& =\frac{e^{3x}}{4}\left(1-D-\frac{D^2}{4}+D^2\right)x^2\quad \text{(we drop higher order derivatives)}\\
& =\frac{e^{3x}}{4}\left(1-D+\frac{3D^2}{4}\right)x^2\\
& =\frac{e^{3x}}{4}\left(x^2-2x+\frac{3}{2}\right)\\
\end{align}$$
which forms the general solution
$$y=y_c+y_p=\text{C.F.}+\text{P.I.}=c_1e^x+c_2xe^x+\frac{e^{3x}}{4}x^2-\frac{e^{3x}}{2}x+\frac{3e^{3x}}{8}$$
| {
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"url": "https://math.stackexchange.com/questions/3575097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Evaluating integral using residue calculus Evaluating the following using residue calculus:
$$I = \int_0^{2\pi} {sin^4\theta}d\theta$$
I have simplified to:
$$\oint_{\lvert z \rvert = 1} \frac{(z^2-1)^4}{16iz^5}$$
z= 0 is the isolated singular point here.
Let
$$g(z) = (z^2 -1)^4 $$ and
$$h(z) = 16iz^5$$
$h(z)$ has a zero of order 5 at $z = 0$
$g(z)$ and $h(z)$ are analytic at $z = 0$, $h(0) = 0$, $g(0)\neq 0$
Thus $f(z)$ has a pole of order 5 at$z=0$. Then use the formula:
$$Res[f(z), z_0] = \lim_{z\to 0} \frac{1}{(N-1)!}*\frac{d^(N-1)}{dz^(N-1)}[(z-z_0)^Nf(z)]$$
However, taking the 4th degree derivative of $g(z)$ over $h(z) is super messy. What did I do wrong and is there a way to do it cleaner?
Edit:
So I expanded:
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{(5-1)!}*\frac{d^4}{dz^4}[(z-0)^5*\frac{(z^2-1)^4}{16iz^5}]$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{4!}*\frac{d^4}{dz^4}[z^5*\frac{(z^2-1)^4}{16iz^5}]$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{384i}*\frac{d^4}{dz^4}[z^8-4z^6+6z^4-4z^2+1]$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{384i}*\frac{d^3}{dz^3}[8z^7-24z^5+24z^3-8z]$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{384i}*\frac{d^2}{dz^2}[56z^6-120z^4+72z^2-8]$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{384i}*\frac{d}{dz}[336z^5-480z^3+144z]$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{1}{384i}*(1680z^4-1440z^2+144)$$
$$Res[f(z), 0] = \lim_{z\to 0} \frac{144}{384i} = \frac{3}{8i}$$
So the result is $\frac{3\pi}{4}$
Appeared I have solved it typing this. Lol.... SMH
| $\oint_{|z| = 1} \frac {(z^2-1)^4}{16i z^5} \ dz$
Rather than differentiating (which is legit, but not necessary) find the Laurent series. Which just means expanding the binomial.
$\frac {1}{16i z^5} - \frac {4}{16i z^3} + \frac {6}{16i z} - \frac {4}{16i} z +\frac {1}{16i} z^3$
$\frac {6}{16i z}$ is the only term you care about to calculate the residual.
$(\frac 2\pi i) \frac {6}{16i z} = \frac {3\pi}{4}$
| {
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"answer_id": 0
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Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for the proof:
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
$$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
Let's go:
$$2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) = 2\cos(\frac{x}{2}+ \frac{y}{2})\cos(\frac{x}{2} - \frac{y}{2}) = $$
$$ = 2(\cos(\frac{x}{2})\cos(\frac{y}{2}) - \sin(\frac{x}{2})\sin(\frac{y}{2}))(\cos(\frac{x}{2})\cos(\frac{y}{2}) + \sin(\frac{x}{2})\sin(\frac{y}{2})) = $$
$$ = 2(\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - \sin^2(\frac{x}{2})\sin^2(\frac{y}{2})) = $$
$$ = 2\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - 2\sin^2(\frac{x}{2})\sin^2(\frac{y}{2})$$
Now I tried, I believe, almost every possible replacement by deriving from $\cos^2(x) + \sin^2(x) = 1$ and sadly nothing worked.
| Hint: $ x \rightarrow (x+y)/2 $ and $y \rightarrow (x-y)/2$ in the second & third equations. Now add them together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there a formula for expanding these kind of things (what kind of expression is it, by the way?)?
This is my solution:
$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$
$= \displaystyle\frac{[(\sqrt5+2)^3+(\sqrt5-2)^3][(\sqrt5+2)^3-(\sqrt5-2)^3]}{8\sqrt5}$
$=\displaystyle\frac{(\sqrt5+2+\sqrt5-2)[(\sqrt5+2)^2\color{red}{+}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2](\sqrt5+2-\sqrt5+2)[(\sqrt5+2)^2\color{red}{-}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2]}{8\sqrt5}$
$=\displaystyle\frac{[2\sqrt5(5+4\sqrt5+4+\color{red}{5-4}+5-4\sqrt5+4][4(5+4\sqrt5+4\color{red}{-(5-4)}+(5-4\sqrt5+4)]}{8\sqrt5}$
$=\displaystyle\frac{2584\sqrt5}{8\sqrt5}$
$=323$
Because of the multiplication, I still got the same answer as given in the book. However, is the book or I correct in terms of the positive/negative signs(in red)?
| Alternatively, let $a=9+4\sqrt5$ and $b=9-4\sqrt5$. Then, $a+b=18$, $ab=1$ and,
$$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}
=\frac{a^3 - b^3}{8\sqrt5}=\frac{(a-b)(a^2 +ab+b^2)}{8\sqrt5}$$
$$=\frac{\sqrt{(a+b)^2-4ab}\>[(a+b)^2 -ab)]}{8\sqrt5}
=\frac{\sqrt{18^2-4}\>(18^2 -1)}{8\sqrt5}=323$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Calculate the variance for the probability density function $f(x) = x/4$ for $1 \le x \le 3$ I am trying to use this formula:
$$V(X) = E(x^2)-E^2(x)$$
I already calculated $E(X) = \frac{13}{6}$.
I tried to calculate $E(x^2)$:
$$E(x) = \int_1^3 xf(x) dx \Leftrightarrow E(x^2) = \int_1^3x^2f(x^2) = ...$$ and so on. The final answer I got was $\approx 7.4$. It's different from my professor's solution because when he calculated $E(x^2)$ he did:
$$\int_1^3 x^2 f(x) dx = \int_1^3 x^2\frac{x}{4}dx$$
Did he make a mistake? It looks like he did but I want to make sure.
| I think you have made a mistake at the calculation. The variance is $$Var(X)=E(X^2)-[E(X)]^2=\int_1^3 x^2\cdot f(x)\, dx-\left(\int_1^3 x\cdot f(x) \, dx\right)^2$$
$$\int_1^3 x^2\cdot \frac{x}4\, dx-\left(\int_1^3 x\cdot \frac{x}4 \, dx\right)^2=\int_1^3 \frac{x^3}4\, dx-\left(\int_1^3 \frac{x^2}4 \, dx\right)^2$$
$$=\left[\frac{x^4}{16}\right]_1^3-\left(\left[\frac{x^3}{12}\right]_1^3\right)^2=\frac{3^4}{16}-\frac{1^4}{16}-\left(\frac{3^3}{12}-\frac{1^3}{12}\right)^2$$
$$=5-\left(\frac{26}{12}\right)^2=\frac{11}{36}=0.30\overline 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{A} \\
B = 2 + \dfrac{1}{B}$
Converting to quadratic equations:
$A^2 - A - 1 = 0 \\
B^2 -2B - 1 = 0$
Resulting to:
$2A = 1 + \sqrt{5} > 1 + \sqrt{2} = B$
My Question is:
Are there any more ways to solve this type of problem?
| An alternative approach
$$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} < 1 +\dfrac{1}{1} =2$$
$$B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}} > 2$$
showing $$A < B$$
Thus
$$2A = 2 +\dfrac{2}{A} > 2 +\dfrac{2}{2} =3$$
$$B = 2 +\dfrac{1}{B} < 2 + \dfrac12 $$
showing $$2A > B$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{5}\cdot\frac{4}{6}\cdot\frac{6}{7}+
\frac{2}{6}\cdot\frac{4}{7}\cdot\frac{6}{8}\cdot\frac{8}{9}+\cdots+\frac{2^nn!}{(2n)!\div (n+1)!}$$
I don't know what to do next...
| The product can be simplified to:
$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=
\sum_{j=1}^{\infty} \frac{j!(j+1)! 2^j}{(2j+1)!}=
\sum_{j=1}^{\infty} \frac{2^j}{\binom{2j+1}{j}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$
My question. Does there exist a closed-form expression of $I(a)$?
Some comments. Since $a-2<a+\cos(x)+\cos(y)<a+2$ and $\int_0^\pi \int_0^\pi \sin^2(x)\sin^2(y)\ \mathrm{d}x\, \mathrm{d}y=\frac{\pi^2}{4}$, we have the following bounds
$$
\frac{\pi^2}{4(a+2)} < I(a) < \frac{\pi^2}{4(a-2)},
$$
however I didn't manage to find an exact expression for $I(a)$. Any help is welcome!
| With CAS help:
$$\int _0^{\pi }\int _0^{\pi }\frac{\sin ^2(x) \sin ^2(y)}{a+\cos (x)+\cos (y)}dydx=\\\mathcal{L}_q\left[\int _0^{\pi }\int _0^{\pi }\mathcal{L}_a^{-1}\left[\frac{\sin ^2(x) \sin ^2(y)}{a+\cos (x)+\cos
(y)}\right](q)dydx\right](a)=\\\mathcal{L}_q\left[\int_0^{\pi } \left(\int_0^{\pi } e^{-q (\cos (x)+\cos (y))} \sin ^2(x) \sin ^2(y) \, dx\right) \, dy\right](a)=\\\mathcal{L}_q\left[\int_0^{\pi }
\frac{e^{-q \cos (y)} \pi I_1(q) \sin ^2(y)}{q} \, dy\right](a)=\\\mathcal{L}_q\left[\frac{\pi ^2 I_1(q){}^2}{q^2}\right](a)=\\\frac{a \pi ^2}{2}-\frac{2}{3} a \pi E\left(\frac{4}{a^2}\right)-\frac{1}{6}
a^3 \pi E\left(\frac{4}{a^2}\right)-\frac{2}{3} a \pi K\left(\frac{4}{a^2}\right)+\frac{1}{6} a^3 \pi K\left(\frac{4}{a^2}\right)=\\\frac{\pi ^2 \, _3F_2\left(\frac{1}{2},1,\frac{3}{2};2,3;\frac{4}{a^2}\right)}{4 a}$$
for: $a>2$
$$\frac{1}{6} \pi \left(-2 \left(a^2-4\right) K\left(\frac{a^2}{4}\right)-2 \left(a^2+4\right) E\left(\frac{a^2}{4}\right)+3 \pi a\right)$$
for: $a<2$
where: $K$,$E$ gives the elliptic integral of the first kind and second kind.
Mathematica code:
HoldForm[Integrate[(Sin[x]^2*Sin[y]^2)/(a + Cos[x] + Cos[y]), {x, 0, Pi}, {y, 0, Pi}] == (a \[Pi]^2)/2 - 2/3 a \[Pi] EllipticE[4/a^2] - 1/6 a^3 \[Pi] EllipticE[4/a^2] - 2/3 a \[Pi] EllipticK[4/a^2] + 1/6 a^3 \[Pi] EllipticK[4/a^2] == Pi^2/(4 a)*HypergeometricPFQ[{1/2, 1, 3/2}, {2, 3}, 4/a^2]] // TeXForm
Plot a solution:
f[a_?NumericQ] := NIntegrate[(Sin[x]^2*Sin[y]^2)/(a + Cos[x] + Cos[y]), {x, 0, Pi}, {y,0, Pi}];
g[a_] := (a \[Pi]^2)/2 - 2/3 a \[Pi] EllipticE[4/a^2] -
1/6 a^3 \[Pi] EllipticE[4/a^2] - 2/3 a \[Pi] EllipticK[4/a^2] +
1/6 a^3 \[Pi] EllipticK[4/a^2]; Plot[{f[a], g[a]}, {a, 2, 20},
PlotStyle -> {Red, {Dashed, Black}}, PlotLabels -> {"integral", "Analytic solution"}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that gets me back to where I started.
$\frac{1}{(x^2+2x+3)^2}$
What am I doing wrong?
| Hint:
Notice that $\left(x^2 + 2x + 3\right)^2\equiv\left((x + 1)^2 + 2\right)^2$. That is,
$$\int\frac1{\left(x^2 + 2x + 3\right)^2}\,\mathrm dx\equiv\int\frac1{\left((x + 1)^2 + 2\right)^2}\,\mathrm dx.$$
Let $u = x + 1\implies\mathrm du = \mathrm dx$. So,
$$\int\frac1{\left((x + 1)^2 + 2\right)^2}\,\mathrm dx\equiv\int\frac1{\left(u^2 + 2\right)^2}\,\mathrm du.$$
Now, apply an appropriate reduction formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+xy$$
$$2axy=c+xy$$
Also, the summation will be
$$a(1^2+2^2+3^2...10^2)+b(1+2+3+4...+10)+10c$$
$$=385a+55b+10c$$
$$375a+45b+30$$
I don’t know what to do with the x and y terms. How do I proceed?
| From $a+b+c=3$, we have $$f(1)=a+b+c=3$$
and from $f(x+y)=f(x)+f(y)+xy$, we have
$$f(x+1)=f(x)+f(1)+x=f(x)+x+3$$
therefore
$$\sum_{n=1}^{10}f(n+1)=\sum_{n=1}^{10} (f(n)+n+3)$$
or $$\sum_{n=2}^{11}f(n)=\sum_{n=1}^{10}f(n)+\sum_{n=1}^{10}(n+3)$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the general $n \times n$ form of the divisibility matrix? Background + Motivation
I had the following idea of using digits as vectors. Let us have say I have a three digit number $a$ in the basis $\lambda$:
$$ a = a_0 + a_1 \lambda + a_2 \lambda^2$$
Now, we want to convert $\lambda$ coefficients to $\lambda+1$ coefficients:
$$ a = a_0 -a_1 +a_2 + (a_1 - 2a_2) (\lambda +1) + a_2(\lambda +1)^2 $$
We note if $a_0 -a_1 +a_2$ is divisible by $\lambda+1$ then so is $a$. And since we are only interested checking if the number is divisible, we introduce the divisibility matrix (for $3$ digits):
$$
(\lambda+1) |a \implies (\lambda+ 1)|
\begin{pmatrix}
1 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 & -1 & 1 \\
0 & 1 & -2 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
a_0 \\
a_1 \\
a_2
\end{pmatrix}
$$
Similarly if the $a$ is divisible by $\lambda+2$
$$
(\lambda+2) | a \implies (\lambda+ 2) |
\begin{pmatrix}
1 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 & -1 & 1 \\
0 & 1 & -2 \\
0 & 0 & 1
\end{pmatrix}^2
\begin{pmatrix}
a_0 \\
a_1 \\
a_2
\end{pmatrix}
$$
Example
Let $\lambda = 10$ and $a=121$
Then we verify $1-2+1 = 0$ which is indeed divisible by $10 + 1 = 11$
Question
We have only done this for a $3$ digit number. What is general form of the divisibility matrix for a $n$ digit number?
| We make use of the following:
$$a = a_0 + a_1 \lambda + a_2 \lambda^2 + \dots $$
Or with $\lambda + 1$ coefficients:
$$ a = b_0 + b_1 (\lambda +1) + b_2 (\lambda +1)^2 + \dots$$
We define $\lambda +1 = \beta $ and combine the above $2$ equations:
$$ a_0 + a_1(\beta -1) + a_2 (\beta -1)^2 + \dots = b_0 + b_1 \beta + b_2\beta^2 +\dots $$
Setting $\beta = 0$:
$$b_0 = a_0 -a_1 +a_2 -a_3 + \dots = \sum_{i=0}^\infty (-1)^i a_i $$
Differentiating and setting $\beta = 0$ again:
$$ b_1 = a_1 - 2 a_2 + 3 a_3 - 4 a_4 + \dots = \sum_{i=1}^\infty a_i ( -1)^{i+1} i$$
Differentiating and setting $\beta = 0$ again:
$$ b_2 = \frac{2!}{2! 0!} a_2 - \frac{3!}{1!2!} a_3 + \frac{4!}{2! 2!} a_4 - \frac{5!}{3! 2!} a_4 + \dots = \sum_{i=2}^\infty a_i(-1)^{i} \text{ }{ }^i C_2 $$
Hence, in general:
$$ b_k = \sum_{i=k}^\infty a_i (-1)^{i-k} \text{ }{ }^i C_k $$
with $k \neq 0$
Now, we can construct an $n \times n$ divisibility matrix:
$$D_{jk} = \begin{cases}
0 & j < k \\
(-1)^k & j=0 \\
(-1)^{j-k} \text{ }{ }^j C_k & \text{else}
\end{cases}
$$
To write it some terms explicitly:
$$ D =
\begin{pmatrix}
1 & -1 & 1 & -1 & 1 &\dots \\
0 & 1 & -2 & 3 & -4 & \dots \\
0 & 0 & 1 & -3 & 6 & \dots \\
\vdots \\
0 & 0 & \dots & & &1 &
\end{pmatrix}$$
Interestingly each column is related to the binomial tree.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\
=n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$
I also tried proof by induction. Assume that it's true for n=k, so that
$$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\
n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\
(1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$
Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:
$$(n+2)^{2/3} -(n+1)^{2/3}\\
=(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\
<(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\
=\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$
But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!
| For $f(x) = x^{\frac{2}{3}}$, then by the Mean Value Theorem,
$(n+1)^{\frac{2}{3}}-n^{\frac{2}{3}}=\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c) = \frac{2}{3}c^{-\frac{1}{3}}$ for some $c \in (n,n+1)$.
Since $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3(^3\sqrt{x})}$ is decreasing, then $c > n \implies f'(c) < f'(n) = \frac{2}{3}n^{-\frac{1}{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$
$f(2+\sqrt{\frac{1}{3}})<0$ & $ f(2-\sqrt{\frac{1}{3}})<0$
It is Local Minimum at $2+\sqrt{\frac{1}{3}}$ and Local Maximum at $2-\sqrt{\frac{1}{3}}$
By hit and trial I got $f(5)=0$ viz a=5
Given abc=30, therefore bc=6.
Hence the answer is $\frac{6}{5}$ which is correct.
My only concern is to find the real value without using any HIT and TRIAL.
| Rewrite
$$(x – 1)(x – 2)(x – 3) = 24$$
as
$$x^3-6x^2+11x-30=0$$
which factorizes as
$$(x-5)(x^2-x+6)=0$$
Thus,
$$\frac{bc}a= \frac 65$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Identity for Catalan numbers Assume $C_{n}=\frac{1}{n+1} \binom{2n}{n}$ the Catalan numbers. I want to prove the following identity with generating functions.
$$C_{n+1}=\sum\limits_{k=0}^{\lfloor \frac{n}{2}\rfloor} \binom{n}{2k}2^{n-2k}C_k$$
I know that the generating function for the Catalan numbers is $C(x)=\frac{1-\sqrt{1-4x}}{2x}$ and I can prove the following $C_{n+1}=\sum\limits_{k=0}^{n}C_kC_{n-k}$. I tried to use those two results to derive the identity above but did not succeed so far. I'm not really sure if I approach this problem the right way, especially because I don't know how to obtain the floor function for the upper bound of the sum indexing.
| We may also use
$$C_n = {2n\choose n} - {2n\choose n+1}.$$
getting for the RHS of the sum
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} 2^{n-2k} {2k\choose k}
- \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} 2^{n-2k} {2k\choose k+1}.$$
Now for the first piece we have
$${n\choose 2k} {2k\choose k} =
\frac{n!}{(n-2k)! \times k! \times k!}
= {n\choose k} {n-k\choose k}.$$
This yields
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose k} 2^{n-2k} {n-k\choose n-2k}
\\ = [z^n] (1+z)^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose k} 2^{n-2k} \frac{z^{2k}}{(1+z)^k}.$$
The coefficient extractor enforces the range and we get
$$[z^n] (1+z)^n \sum_{k\ge 0}
{n\choose k} 2^{n-2k} \frac{z^{2k}}{(1+z)^k}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{z^{2}}{4(1+z)}\right)^n
\\ = \frac{1}{2^n} [z^n] (z+2)^{2n}
= {2n\choose n}.$$
Working with the second piece we find for $k\ge 1$
$${n\choose 2k} {2k\choose k+1} =
\frac{n!}{(n-2k)! \times (k-1)! \times (k+1)!}
= {n\choose k-1} {n-k+1 \choose k+1}.$$
This yields (the term at $k=0$ is zero)
$$\sum_{k=1}^{\lfloor n/2 \rfloor}
{n\choose k-1} 2^{n-2k} {n-k+1\choose n-2k}
\\ = [z^n] (1+z)^{n+1} \sum_{k=1}^{\lfloor n/2 \rfloor}
{n\choose k-1} 2^{n-2k} \frac{z^{2k}}{(1+z)^k}
\\ = \frac{1}{4} [z^n] (1+z)^{n+1} \frac{z^2}{1+z}
\sum_{k=0}^{\lfloor n/2 \rfloor - 1}
{n\choose k} 2^{n-2k} \frac{z^{2k}}{(1+z)^k}
\\ = \frac{1}{4} [z^{n-2}] (1+z)^{n}
\sum_{k=0}^{\lfloor (n-2)/2 \rfloor}
{n\choose k} 2^{n-2k} \frac{z^{2k}}{(1+z)^k}
.$$
The coefficient extractor once more enforces the range and we get
$$\frac{1}{4} [z^{n-2}] (1+z)^{n}
\sum_{k\ge 0}
{n\choose k} 2^{n-2k} \frac{z^{2k}}{(1+z)^k}
\\ = 2^{n-2} [z^{n-2}] (1+z)^n
\left(1+\frac{z^{2}}{4(1+z)}\right)^n
\\ = \frac{1}{2^{n+2}} [z^{n-2}] (z+2)^{2n}
= {2n\choose n+2}.$$
Collecting the two pieces we find
$$\left(\frac{(n+1)^2}{(2n+2)(2n+1)}
- \frac{(n+1)n(n-1)}{(2n+2)(2n+1)(n+2)}\right)
{2n+2\choose n+1}
\\ = \left(\frac{(n+2)(n+1)^2}{(2n+2)(2n+1)(n+2)}
- \frac{(n+1)n(n-1)}{(2n+2)(2n+1)(n+2)}\right)
{2n+2\choose n+1}.$$
Now
$$(n+2)(n+1)^2-(n+1)n(n-1) = (2n+2)(2n+1)$$
so we indeed obtain
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n+2} {2n+2\choose n+1} = C_{n+1}}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.
My approach: Given that $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$ Let us make the substitution $x^n=t$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$
Now since $0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$
This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$
So, as you can see, I am trying to solve the question using Sandwich theorem.
Can someone help me to proceed after this?
Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?
| you can use the Binomial series
$${{\left( 1+{{x}^{2}} \right)}^{-1/2}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right){{x}^{2k}}}$$
$${n{I}_{n}}=n\int_{0}^{1}{{{x}^{n}}{{\left( 1+{{x}^{2}} \right)}^{-1/2}}dx}=n\sum\nolimits_{k=0}^{\infty }{\left\{ \left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right)\int_{0}^{1}{{{x}^{2k+n}}dx} \right\}}=\sum\nolimits_{k=0}^{\infty }{\frac{n\left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right)}{\left( 2k+n+1 \right)}}$$
$$\underset{n\to \infty }{\mathop{\lim }}\,{n{I}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=0}^{\infty }{\frac{n\left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right)}{\left( 2k+n+1 \right)}}=\sum\nolimits_{k=0}^{\infty }{\underset{n\to \infty }{\mathop{\lim }}\,\frac{n\left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right)}{\left( 2k+n+1 \right)}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right)}={{\left( 1+1 \right)}^{-1/2}}=\frac{1}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Summation of binomial coefficients with indices having difference of 4
Let $S$ $=$ $\binom{n}3 $ $+$ $\binom{n}7$ $+$ $\binom{n}{11} $ $.....$
Find $S$.
This question can be solved by taking the difference of the binomial series of $\frac{-(1+x)^n - (1-x)^n}{2}$ and $\frac{(1+ix)^n - (1-ix)^n}{2i}$ , where $i^2$ $=$ $-1$ and substituting $x$ $=$ $1$ and dividing the resulting value by 2.
However, I wonder if the value of $S$ can be evaluated using calculus ? Please help me with this thought.
| Let $$S=\sum_{k=0}^{n} {n \choose 4k+3}$$
$$(1+x)^n=1+{n \choose 1} x+ {n \choose 2} x^2+ {n \choose 3}x^3+ {n \choose 4} x^4+...+{n \choose n}x^n~~~(1)$$
Let $z^4=1$ its roots are $1,a,a^2,a^3 (1,i,-1,-i)$, with $1+a+a^2+a^3=0$
Put $x=w$ in (1), then
putting $x=1,a,a^2,a^3(1,i,-1,-i)$ in (1) we get
$$2^n=\sum_{k=0}^{n} {n \choose k},~ (1+a)^n=\sum_{k=0}^{n} {n \choose k} a^k, ~(1+a^2)^n=\sum_{k=0}^n {n \choose k} a^{2k},$$ $$~(1+a^3)^n=\sum_{k=0}^{n} {n \choose k} a^{3k}$$
Multiplying these identities by $1, a, a^2, a^3$ respectively and adding them we get
$$2^n+a(1+a)^n+a^2(1+a^2)^n+a^3(1+a^3)= \sum_{k=0}^n (1+a^{k+1}+a^{2k+2}+a^{3k+3})
{n \choose k}~~~(2)$$
The term in the parenthesis in RHS of (2) is 1 only for$ k=3,7,11,15,...$, otherwise it vanishes.
So $$S=2^n+(1+i)^n+i^2(1+i^2)^n+i^3(1+i^3)^n= 2^ni[(1+i)^n-(1-i)^n]$$ $$=2^n-2^{n/2}~ 2\sin n\pi/4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$ Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality :
$$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$
Where :
$s : \operatorname{Semiperimeter}$
$R : \operatorname{circumradius}$
I know the relation :
$$m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$$
But when I multiple together I dont get simple formulas!
So, I need help finding a solution.
Thanks!
|
Note that the triangles ABD and EDC are similar. Then,
$$\frac{AD}{BD}=\frac{CD}{ED}\implies \frac{m_a}{\frac a2}=\frac{\frac a2}{AE-m_a}
\implies m_a^2 -AE\cdot m_a + \frac {a^2}4=0$$
which, since $AE \le 2R$ and $a=2R\sin A$, leads to
$$m_a =\frac12(AE+\sqrt{AE^2-a^2})\le \frac12\left[2R+\sqrt{(2R)^2-(2R\sin A)^2}\right] =2R\cos^2 \frac A2$$
Likewise, $m_b\le 2R\cos^2 \frac B2$ and $m_c\le 2R\cos^2 \frac C2$. Together, we have
$$\begin{align}
m_a m_bm_c & \le \frac12R^3\left( 4\cos\frac A2\cos \frac B2\cos \frac C2\right)^2 \\
& = \frac12R^3\left( 2 \cos\frac A2 \left(\cos \frac {B+C}2+\cos \frac {B-C}2 \right)\right)^2 \\
& = \frac12R^3\left( 2 \cos\frac A2 \sin\frac A2+2 \sin\frac {B+C}2\cos \frac {B-C}2 \right)^2 \\
& = \frac12R^3\left( \sin A + \sin B + \sin C \right)^2 \\
& = \frac12R^3 \left( \frac a{2R} + \frac b{2R} + \frac c{2R}\right)^2 \\
& = \frac12R \left( \frac{a + b+ c}2 \right)^2 \\
\end{align}$$
Thus,
$$m_a m_bm_c \le \frac12Rs^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Parameters of the Mandart inellipse What are the parameters of the Mandart ellipse
in terms of its associated triangle?
It looks that information on the Mandart ellipse
is less known than its Steiner counterpart.
Easily found references
wiki,
MathWorld
look incomplete,
so the provided answer collects the missing parts together.
|
The Mandart inellipse of a triangle
is an ellipse inscribed within the triangle,
tangent to its sides
at the contact points
$X_a,X_b,X_c$
of its excircles.
Its parameters
are expressed in terms of the coordinates of vertices $A,B,C$,
side lengths $a,b,c$,
semiperimeter $\rho=\tfrac12\,(a+b+c)$,
inradius $r$ and
circumradius $R$
of the associated $\triangle ABC$.
The Mandart inellipse is a generalized Steiner inellipse
with parameters
\begin{align}
u&=
\frac{(b-c)^2-a^2}{(b-c)^2-a\,(-a+2b+2c)}
,\\
v&=
\frac{(c-a)^2-b^2}{(c-a)^2-b\,(-b+2c+2a)}
,\\
w&=
\frac{(a-b)^2-c^2}{(a-b)^2-c\,(-c+2a+2b)}
\end{align}
or
\begin{align}
u&=1-\frac{a\,(\rho-a)}{r\,(r+4\,R)}
,\\
v&=1-\frac{b\,(\rho-b)}{r\,(r+4\,R)}
,\\
w&=1-\frac{c\,(\rho-c)}{r\,(r+4\,R)}
.
\end{align}
The foci are the roots of quadratic equation
\begin{align}
z^2-((1-u)A+(1-v)B+(1-w)C)\,z+u\,B\,C+v\,C\,A+w\,A\,B
&=0
,
\end{align}
\begin{align}
F_{1,2}&=
\tfrac12\,((1-u)\,A+(1-v)\,B+(1-w)\,C)
\\
&\pm
\sqrt{\tfrac14\,((1-u)\,A+(1-v)\,B+(1-w)\,C)^2-u\,B\,C-v\,C\,A-w\,A\,B}
.
\end{align}
The center of the Mandart inellipse is the mittenpunkt of the triangle,
\begin{align}
M&=\tfrac12\,((1-u)\,A+(1-v)\,B+(1-w)\,C)
.
\end{align}
The distance between the foci
\begin{align}
|F_1F_2|&=\frac{\rho}{1+\displaystyle\frac r{4R}}\,\sqrt[4]{1-\frac{2\,r}R}
.
\end{align}
Its area is found as
\begin{align}
S_{M}&=\frac{\pi((a+b-c)(b+c-a)(c+a-b))}{(2(ab+bc+ca)-(a^2+b^2+c^2))^{3/2}}\,S_{ABC}
\\
&=\frac{\pi\,\rho^2\,r^3}{(r^2+4\,r\,R)^{3/2}}
=\frac{\pi\,\rho\,r^2}{(r^2+4\,r\,R)^{3/2}}
\,S_{ABC}
=\frac{\pi\,r}{(r^2+4\,r\,R)^{3/2}}
\,S_{ABC}^2
,
\end{align}
Major and minor semi-axes are
\begin{align}
s_a&=\frac{\rho}{4+\displaystyle\frac rR}\,\sqrt{3-\Big(1+\frac rR\Big)^2
+2\,\sqrt{1-\frac{2\,r}R}}
,\\
s_b&=\frac{\rho}{4+\displaystyle\frac rR}\,\sqrt{3-\Big(1+\frac rR\Big)^2
-2\,\sqrt{1-\frac{2\,r}R}}
.
\end{align}
Edit:
Eccentricity of the Mandart inellipse is completely defined by
the ratio $V=\frac rR$:
\begin{align}
e^2=1-\frac{s_b^2}{s_a^2}
&=
\frac{4\Big(4V-2+\sqrt{1-2V}(3-(V+1)^2)\Big)}
{V^3(4+V)}
.
\end{align}
And vise versa, given the eccentricity $e$ of
the Mandart inellipse, the ratio $V=\frac rR$
is found as
a solution of the quartic
\begin{align}
V^4+4V^3+2qV-q&=0
,\quad
q=\tfrac{16}{e^4}\,(1-e^2)
,
\end{align}
\begin{align}
V=\frac rR&=
-1-\tfrac12\,\sqrt{3+t}+\tfrac12\,\sqrt{9-t+\frac{16+4q}{\sqrt{3+t}}}
,\quad \text{where}\quad t=(\sqrt[3]{2q}-1)^2
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suggestions for $ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $? I'm trying to evaluate $$
\lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2}
$$
over the domain $x>0$, $y>0$.
============
My attempt:
$f(x,x^2)\to +\infty$; so if the limit exists it must be $+\infty$.
I tried to evaluate the limits "near" $(x,x)$ where, I thought, there's may be some problems:
$f(x, x-x^2)\to +\infty$.
Then I convinced myself the limit could be $+\infty$:
since $f(x,y)>0$ over the domain, I had to find such $g(x,y)$ that:
1. $f(x,y) \ge g(x,y)$
2. $ \lim_{(x,y)\to (0,0)} g(x,y)\to +\infty $
$$
f(x,y)=\frac{x-\sqrt{xy}}{x^2-y^2}=\frac{x-\sqrt{xy}+y-y}{x^2-y^2}=\frac{x-\sqrt{xy}+y}{x^2-y^2}-\frac{y}{x^2-y^2}=\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2}
$$
Where the last step follows by $(x-\sqrt{xy}+y) \ge 0$ with $x>0$, $y>0$. $$
\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2} = \frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2}.
$$ From $\left[3\left(x-\sqrt{xy}+y\right)^2\right] \ge \left[x+xy+y^2\right]$, for every $(x,y)$ with $(x>y)$: $$
\frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(x^2-y^2)}+\frac{y}{y^2-x^2}.
$$ From here I observated that $ \left[\lim_{(x,y)\to (0,0)} g(x,y)\to +\infty \right]$ and eventually $ \left[\lim_{(x,y)\to (0,0)} f(x,y)\to +\infty \right]$ for $(x>y)$.
I thought that for $(y>x)$, the inequality was formally equivalent when I replace $(x)$ with $(y)$ and viceversa: $$
\frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(y^2-x^2)}+\frac{x}{x^2-y^2}.
$$
However I could see, through an online grapher, that it is false!!
So I remained without any chance to conclude the limit.
============
Is there anybody who knows why the last inequality isn't correct?
And also, has anybody some hints to evaluate the limit?
| If the limit exists, then it must be equal to the limit along any line, for example, $y=x/4$.
In that case,
$$ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} =\lim_{x\downarrow 0} \frac{8}{15}\frac{x}{x^2}=\frac{8}{15}\lim_{x\downarrow 0}\frac{1}{x}=\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n. If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.
I did and I got
$$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?
| In the first factor the multiplicity of the solution $x=2\vee x=-2$ is $3$, so here we have $6$ solutions.
In the second case the solution $x=-1$ has $n$ multiplicity, while in third factor there are solutions: $x=-2 \vee x=-3$ with multiplicity $m$.
We have to have:
$$3n+2m=12$$
So, every $(m,n)\in N$ such that $3n+2m=12$ are correct. In particular this is a diphantine equation with solutions:
$$m=6 \land n=0 \lor m=3 \land n=2 \lor m=0 \land n=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $u''(x)-\frac{u'(x)^2}{u(x)}+\frac{u'(x)}{x}=u(x)^2$ I am looking to solve the following ODE:
$$u''(x)-\frac{u'(x)^2}{u(x)}+\frac{u'(x)}{x}=u(x)^2.$$
As far as I can see, none of the common methods to solve ODEs (separation of variables, integrating factor, Laplace transform, etc.) yield anything useful.
Can I have some guidance on how to solve this equation?
EDIT: I have, with the help of computer algebra systems, found the solution to this equation to be
$$u(x)=\frac{\mathrm{e}^{AB}A^4x^{A-2}}{\left(x^A-\frac{A^2}2\mathrm{e}^{AB}\right)^2}$$
for arbitrary constants $A$ and $B$. However, I am still lost on how one might arrive at this solution. Any ideas?
| $\def\d{\mathrm{d}}$From the original equation $\dfrac{\d^2 u}{\d x^2} - \dfrac{1}{u} \left( \dfrac{\d u}{\d x} \right)^2 + \dfrac{1}{x} \dfrac{\d u}{\d x} = u^2$, letting $v = \dfrac{1}{ux^2}$ as @ClaudeLeibovici does yields$$
x^2 v \frac{\d^2 v}{\d x^2} - x^2 \left( \frac{\d v}{\d x} \right)^2 + xv \frac{\d v}{\d x} + v = 0,
$$
and letting $y = \ln x$ yields\begin{align*}
v \frac{\d^2 v}{\d y^2} - \left( \frac{\d v}{\d y} \right)^2 + v = 0.\tag{1}
\end{align*}
Denoting $w = \dfrac{\d v}{\d y}$, (1) is equivalent to the following autonomous system:$$
\begin{cases}
\dfrac{\d v}{\d y} = w\\
\dfrac{\d w}{\d y} = \dfrac{w^2}{v} - 1
\end{cases},
$$
and\begin{align*}
\frac{\d w}{\d v} = \frac{1}{w} \left( \dfrac{w^2}{v} - 1 \right) \Longrightarrow (v - w^2) \,\d v + vw \,\d w = 0.\tag{2}
\end{align*}
The integrating factor of (2) is $\dfrac{1}{v^3}$, thus the solution to (2) is $\dfrac{w^2 - 2v}{2v^2} = C_1$, where $C_1$ is a constant. Now,$$
\frac{\d v}{\d y} = w = \pm\sqrt{\smash[b]{2C_1 v^2 + 2v}},
$$
which can be solved explicitly (with a lot of calculation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all units of $\Bbb Q[x] / (x^2-1)$ Find all units of $\Bbb Q[x] / (x^2-1)$.
I just tried to find them by simply putting $(ax+b)(cx+d)=1$. But it was too complicated to find the undetermined coefficients. I cannot come up with anything related in case the ideal is not irreducible
Could you give me a few hints...?
| First, $b+(x^2-1)$ is a unit for all $0\ne b\in\mathbb{Q}$.
Also note that if for $a\ne0$, $ax+b+(x^2-1)$ is a unit, then so is $x+\frac{b}{a}+(x^2-1)$. So we need to check elements of the form $x+b+(x^2-1)$.
Now:
$$(x+b)(cx+d)=cx^2+dx+bcx+bd=c+bd+(d+bc)x=1$$
i.e. $d=-bc$ and $c+bd=1$ which gives
$$c-b^2c=1\implies c=\frac{1}{1-b^2}\quad\text{and}\quad d=\frac{-b}{1-b^2}$$
Therefore units of $\mathbb{Q}[x]/(x^2-1)$ are of the form $a(x+b)+(x^2-1)$ with $a\ne0$ and $b\ne\pm1$. The inverse is given by
$$(a(x+b)+(x^2-1))^{-1}=\frac{1}{a}\frac{1}{1-b^2}(x-b)+(x^2-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given a polynomial with roots $a, b, c, d, e$, find the polynomial whose roots are $abc, abd, abe, ...$ Let $p(x)=x^5-4x^4+3x^3-2x^2+5x+1$ and say $a, b, c, d, e$ are the roots of $p$. Find the polynomial whose roots are $abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde$.
By Viete's theorem we just need to find the values of the elementary symmetric functions corresponding to the ten roots. But each such function is a symmetric function of $a, b, c, d, e$, and hence can be written as a polynomial in the $5$ elementary symmetric functions coming from $a, b, c, d, e$, whose values are the coefficients of $p$. Thus it is possible to compute the coefficients of the desired polynomial without explicitly finding the values of $a, b, c, d, e$.
However, this will require $10$ different arduous computations. Is there a nifty way to do this?
| $p(x) = x^5-4x^4+3x^3-2x^2+5x+1$
$p$ has $5$ roots donated by $a$, $b$, $c$, $d$ and $e$
The elementary symmetric functions of the roots are
$a+b+c+d+e = 4$
$de+ce+be+ae+cd+bd+ad+bc+ac+ab = 3$
$cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$bcde+acde+abde+abce+abcd = 5$
$abcde = -1$
Let $z = abc$, Computing the elementary symmetric functions of $z$ which are symmetric functions in $a,b,c,d,e$ and expressing them in terms of the elementary symmetric functions of $x$
Writing out the conjugates of $z$ shows it's a polynomial of degree $10$
$(z-abc)(z-abd)(z-acd)(z-bcd)(z-abe)(z-ace)(z-bce)(z-ade)(z-bde)(z-cde)$
Expand to express the elementary symmetric functions of $z$
$z^{10}-s_1z^9+s_2z^8-s_3z^7+s_4z^6-s_5z^5+s_6z^4-s_7z^3+s_8z^2-s_9z+s_{10} = 0$
$s_1 = cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$s_2 = {.............}$
This process is large, requires tremendous calculations so I'll skip the details
$s_8 =
(abcde)^4(cde^2+bde^2+ade^2+bce^2+ace^2+abe^2+cd^2e+bd^2e+ad^2e+c^2de+b^2de+a^2de+bc^2e+ac^2e+b^2ce+a^2ce+ab^2e+a^2be+bcd^2+acd^2+abd^2+bc^2d+ac^2d+b^2cd+a^2cd+ab^2d+a^2bd+abc^2+ab^2c+a^2bc +3( bcde+acde+abde+abce+abcd ) )$
$s_9 = (abcde)^5(de+ce+be+ae+cd+bd+ad+bc+ac+ab) = (-1)^53 = -3$
$s_{10} = (abcde)^6 = 1$
Therefore our polynomial in $z$ is
$z^{12}-2z^9+19z^8-112z^7+82z^6+97z^5-15z^4+58z^3+3z^2+3z+1 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straightforward.
| On the path of Dennis Orton...
\begin{align}J&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\\
&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx+\int _0^1\frac{\ln \left(t^2-t+1\right)}{1-t}\:dt\\
&\overset{x=1-t}=2\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx\\
&=2\left(\int _0^1\frac{\ln \left(\frac{1+x^3}{1+x}\right)}{x}\:dx\right)\\
&=2\left(\int _0^1\frac{\ln \left(1+t^3\right)}{t}\:dt-\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\right)\\
&\overset{x=t^3}=\frac{2}{3}\int _0^1\frac{\ln \left(1+x\right)}{x}\:dt-2\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\\
&=-\frac{4}{3}\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\\
&=-\frac{4}{3}\left(\int_0^1\frac{\ln \left(1-t^2\right)}{t}\:dt-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\right)\\
&\overset{x=t^2}=-\frac{4}{3}\left(\frac{1}{2} \int _0^1\frac{\ln \left(1-x\right)}{x}\:dx-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\right)\\
&=\frac{2}{3}\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\\
&=\frac{2}{3}\left(-\int_0^1 \left(\sum_{n=1}^\infty\frac{x^{n-1}}{n}\right)\,dx\right)\\
&=-\frac{2}{3}\sum_{n=1}^\infty\left(\int_0^1 \frac{x^{n-1}}{n}\,dx\right)\\
&=-\frac{2}{3}\sum_{n=1}^\infty\frac{1}{n^2}\\
&=-\frac{2}{3}\times\frac{\pi^2}{6}\\
&=\boxed{-\frac{\pi^2}{9}}
\end{align}
NB: I assume $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}$
PS:
Sorry, i didn't see the soluton of Ali Shather
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Finding a closed form for $\sum_{k=0}^n (k^3-k-3)$ using generating functions I have been given a few hints. first of all I've been told to use the formula (for general series $a_n,b_n$):
$$\sum_{n=0}^\infty\left(\sum_{k=0}^n a_k b_{n-k}\right)x^n=\left(\sum_{i=0}^\infty a_i x^i\right)\left(\sum_{i=0}^\infty b_i x^i \right)$$
The direction I'm going with is simplifying the generating function of the series $\sum_{n=0}^\infty\left(\sum_{k=0}^n k^3-k-3 \right)x^n$ such that we set $a_k=k^3-k-3,b_k=1$ and thus get:
$$\left(\sum_{n=0}^\infty \left(k^3-k-3\right)x^n\right)\left(\sum_{n=0}^\infty x^n\right)$$
After simplifying A lot I finally got to the form $\frac{3x^{3}-3x^{2}+9x-3}{\left(-x+1\right)^{5}}$. I know this is not the intended way to do it because I have no idea how to get the coefficients from this. any help would be apperciated.
| \begin{align}
&\frac{3x^3-3x^2+9x-3}{(1-x)^5}\\
&=(3x^3-3x^2+9x-3)\sum_{n=0}^\infty \binom{n+4}{4}x^n\\
&=\sum_{n=0}^\infty (3x^3-3x^2+9x-3) \binom{n+4}{4}x^n\\
&= 3 \sum_{n=0}^\infty \binom{n+4}{4}x^{n+3}-3\sum_{n=0}^\infty \binom{n+4}{4}x^{n+2}+9 \sum_{n=0}^\infty \binom{n+4}{4}x^{n+1}-3\sum_{n=0}^\infty \binom{n+4}{4} x^n\\
&= 3 \sum_{n=3}^\infty \binom{n+1}{4}x^n-3\sum_{n=2}^\infty \binom{n+2}{4}x^n+9 \sum_{n=1}^\infty \binom{n+3}{4}x^n-3\sum_{n=0}^\infty \binom{n+4}{4} x^n\\
&= 3 \sum_{n=0}^\infty \binom{n+1}{4}x^n -3\sum_{n=0}^\infty \binom{n+2}{4}x^n+9 \sum_{n=0}^\infty \binom{n+3}{4}x^n-3\sum_{n=0}^\infty \binom{n+4}{4} x^n\\
&= \sum_{n=0}^\infty\left(3\binom{n+1}{4} -3\binom{n+2}{4}+9 \binom{n+3}{4}-3 \binom{n+4}{4}\right) x^n\\
&=\sum_{n=0}^\infty \left(\frac{n^4}{4} + \frac{n^3}{2} - \frac{n^2}{4} - \frac{7 n}{2} - 3\right) x^n,
\end{align}
which yields
$$\sum_{k=0}^n (k^3-k-3) = \frac{n^4}{4} + \frac{n^3}{2} - \frac{n^2}{4} - \frac{7 n}{2} - 3$$
Here's a more direct snake-oil approach from start to finish, using partial fraction decomposition.
\begin{align}
&\sum_{n=0}^\infty \left(\sum_{k=0}^n (k^3-k-3)\right)x^n\\
&=\sum_{k=0}^\infty (k^3-k-3) \sum_{n=k}^\infty x^n\\
&=\sum_{k=0}^\infty (k^3-k-3) \frac{x^k}{1-x}\\
&=\frac{1}{1-x}\sum_{k=0}^\infty (k^3-k-3)x^k \\
&=\frac{1}{1-x}\left(\sum_{k=0}^\infty k^3 x^k-\sum_{k=0}^\infty k x^k-3\sum_{k=0}^\infty x^k\right) \\
&=\frac{1}{1-x}\left(\frac{x(1+4x+x^2)}{(1-x)^4}-\frac{x}{(1-x)^2}-\frac{3}{1-x}\right) \\
&=\frac{x(1+4x+x^2)}{(1-x)^5}-\frac{x}{(1-x)^3}-\frac{3}{(1-x)^2} \\
&= \frac{-3}{(1-x)^2} + \frac{6}{(1-x)^3} - \frac{12}{(1-x)^4} + \frac{6}{(1-x)^5}\\
&= \sum_{n=0}^\infty \left(-3\binom{n+1}{1} + 6\binom{n+2}{2} - 12\binom{n+3}{3} + 6\binom{n+4}{4}\right)x^n\\
&=\sum_{n=0}^\infty \left(\frac{n^4}{4} + \frac{n^3}{2} - \frac{n^2}{4} - \frac{7 n}{2} - 3\right) x^n
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of equations in a neighborhood of the origin, $$x' = y+y^2 - 2xy + x^2$$ $$y'=x+y^2 - 2xy + x^2.$$
$\textbf{Solution:}$ Subtracting $x' = y+y^2 - 2xy + x^2$ from $y'=x+y^2 - 2xy + x^2$ gives us $y'-x' = x-y$ implies the following $$ x'+x=y'+y. \hspace{35pt} (1)$$ Integrating $x' = y+y^2 - 2xy + x^2$ with respect to $y$ and $y'=x+y^2 - 2xy + x^2$ with respect to $x$ gives us $$x = \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y \hspace{35pt} (2)$$ $$y=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3}. \hspace{35pt} (3)$$
Applying (2) and (3) to (1) gives us $$\frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y +y^2 -2xy+x^2$$ $$=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x + y^2 -2xy + x^2$$ $$\implies \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y = \frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x. \hspace{35pt}(4)$$
So, equation (4) denotes a function which if we replace $y$ with $x$ equation will be the same throughout. Thus, if a function of the form $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} - t^3 + t^2 + t \text{ where } t = x, y$$ implies $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} + t. \hspace{35pt} (5)$$
So equations (1), (4), and (5) define the stable and unstable points around the origin. So, $f'(x,y) >0$ as $$f'(x,y) = t + t^2 + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}.$$ Therefore, it will be unstable and we are done.
| The stable/unstable manifolds $W^s$/$W^u$ of the equilibrium point $0$ are the manifolds tangent to the stable/unstable eigenspace $E^s$/$E^u$ at $0$ (with the same dimension as the corresponding eigenspace) s.t. the trajectories starting in $W^s$/$W^u$ converge to $0$ when $t \to \pm \infty$.
After the change of variables $(x, y) = (\zeta - \xi, \zeta + \xi)$, the system becomes
$$\begin {aligned}
\dot \xi &= -\xi \\
\dot \zeta &= \zeta + 4 \xi^2.
\end {aligned}$$
Find $E^s$ and $E^u$ first. Check that $W^u$ coincides with $E^u$. To find $W^s$, start with a series approximation. Substitute $\zeta = A \xi^\alpha$ into the equations and eliminate $\dot \xi$. The result is
$$A (\alpha + 1) \xi^\alpha + 4 \xi^2 = 0.$$
Equating the powers of $\xi$ gives $\alpha$ and equating the coefficients gives $A$. Verify that this happens to give the exact equation for $W^s$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
All integer values of $\frac{a^2+b^2+1}{ab-1}$ Determine all possible values of $\frac{a^2+b^2+1}{ab-1}$ where $a,b$ are positive integers.
I am quite certain one should use a Vieta jumping argument, but I cannot complete it.
Let $\frac{a^2+b^2+1}{ab-1} = k \in \mathbb{Z}$ with $a+b$ being minimal. Clearly $k > 0$ and we have $a^2 - kab + b^2 + k + 1 = 0$. Without loss of generality $a\geq b$. We see that $x_1 = a$ and $x_2 = kb-a = \frac{b^2+k+1}{a}$ are the roots of $x^2 - kbx + b^2 + k + 1 = 0$. The expression $kb-a$ shows they are both integers and $\frac{b^2+k+1}{a}$ they are both positive. To reach a contradiction, it suffices to have $b + \frac{b^2+k+1}{a} < a + b$, i.e. $b^2 + k + 1 < a^2$. Hmm, if we can perhaps somehow assure $k\leq a$, we will just consider pairs $(a,b)$ with $a\neq b$ and $a+b$ minimal $-$ that's how we can rule out the case $a\neq b$ and the rest will be very easy. So any idea how to complete this argument, if it indeed works?
Any help appreciated!
| If the ratio is $n,$ then we have $x^2 - nxy + y^2 + (n+1) = 0,$ with both $x,y$ positive integers.
Now, if $2y > nx,$ then we have $nx - y < y,$ which means that "jumping" the $y$ leads us to a smaller point in terms of $x+y.$ Same thing, if $2x > ny,$ jumping $x$ decreases $x+y.$ So, Hurwitz defined a "ground" solution ( Grundlösung) to be one with $2y \leq nx$ and $2x \leq ny.$ The picture becomes very simple, the part of the hyperbola arc between two slanted lines.
If there are any integer points, Vieta Jumping takes us to an integer point on the hyperbolic arc with $x \geq \frac{2}{n} y$ and $y \geq \frac{2}{n} x.$ There are such points for $n=3,$ at point $(2,2),$ then $n=6,$ points $(1,2)$ and $(2,1).$ That is it.
The solutions with $n=3$ are pairs $(x_{n+1}, x_n)$ from the sequence
$$ 2, 4, 10, 26, 68, 178, 466, ... $$
where $x_{n+2} = 3 x_{n+1} - x_n.$
The solutions with $n=6$ are pairs $(x_{n+2}, x_n)$ from the sequence
$$ 1, 1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552,... $$
where $x_{n+4} = 6 x_{n+2} - x_n.$
Ummm for any $n \geq 7,$ the hyperbola crosses the line $y = 1$ with $1<x<2.$ Also crosses the line $x = 1$ with $1<y<2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding Max/Min of a function in three variables, subject to two constraints
Find the maximizers or minimizers of
$$f(x_1,x_2,x_3) = \frac{1}{x^2_1 + x^2_2 + x^2_3}$$
with the given constraints
$$h_1(x_1,x_2,x_3) = 1 - x^2_1 - 2x^2_2 - 3x^2_3 = 0$$
and
$$h_2(x_1,x_2,x_3) = x_1 + 2x_2 + x_3 = 0.$$
My solution:
I tried to solve using Lagrange multipliers: thus coming up with these 3 equations, that we want to solve for $\lambda_1$ and $\lambda_2$
$$\begin{cases}
\dfrac{-2x_1}{(x^2_1 + x^2_2 + x^2_3)^2} = -2x_1 \lambda_1 + \lambda_2 \\
\dfrac{-2x_2}{(x^2_1 + x^2_2 + x^2_3)^2} = -4x_2 \lambda_1 + 2\lambda_2 \\
\dfrac{-2x_3}{(x^2_1 + x^2_2 + x^2_3)^2} = -6x_3 \lambda_1 + \lambda_2
\end{cases}$$
But trying to solve this - is just insane. I cannot produce any results at all. There must be a trick here or something? Perhaps another way of solving.
| Hint.
You found
$$
\begin{cases}
\dfrac{-2x_1}{(x^2_1 + x^2_2 + x^2_3)^2} = -2x_1 \lambda_1 + \lambda_2 \\
\dfrac{-2x_2}{(x^2_1 + x^2_2 + x^2_3)^2} = -4x_2 \lambda_1 + 2\lambda_2 \\
\dfrac{-2x_3}{(x^2_1 + x^2_2 + x^2_3)^2} = -6x_3 \lambda_1 + \lambda_2
\end{cases}
$$
Calling
$$
\mu_1 = \lambda_1(x^2_1 + x^2_2 + x^2_3)^2\\
\mu_2 = \lambda_2(x^2_1 + x^2_2 + x^2_3)^2
$$
you can follow with
$$
\begin{cases}
-2x_1= -2x_1 \mu_1 + \mu_2 \\
-2x_2 = -4x_2 \mu_1 + 2\mu_2 \\
-2x_3 = -6x_3 \mu_1 + \mu_2
\end{cases}
$$
now solving for $x_1,x_2,x_3,\mu_1$
$$
\begin{cases}
-2x_1= -2x_1 \mu_1 + \mu_2 \\
-2x_2 = -4x_2 \mu_1 + 2\mu_2 \\
-2x_3 = -6x_3 \mu_1 + \mu_2 \\
x_1+2x_2+x_3=0
\end{cases}
$$
we obtain
$$
\left[
\begin{array}{cccc}
x_1 & x_2 & x_3 & \mu_1\\
\frac{1}{2} \left(4+\sqrt{6}\right) \mu _2 & \left(1-\sqrt{6}\right) \mu _2 & \frac{1}{2} \left(-8+3
\sqrt{6}\right) \mu _2 & -\frac{1}{10} \left(6+\sqrt{6}\right) \\
\frac{1}{2} \left(4-\sqrt{6}\right) \mu _2 & \left(1+\sqrt{6}\right) \mu _2 & -\frac{1}{2} \left(8+3
\sqrt{6}\right) \mu _2 & \frac{1}{10} \left(-6+\sqrt{6}\right) \\\end{array}
\right]
$$
and substituting into $1 - x^2_1 - 2x^2_2 - 3x^2_3 = 0$ we obtain finally
$$
\cases{-250 \left(54+19 \sqrt{6}\right) \mu _2^2+342 \sqrt{6}+847=0\\
250 \left(19 \sqrt{6}-54\right) \mu _2^2-342 \sqrt{6}+847=0
}
$$
and thus, the solutions for the stationary points $x_1^*,x_2^*,x_3^*$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it valid to use operations on both sides before inequality is proven? As part of a bigger proof, I am trying to prove the inequality:
$$\frac{ab}{a^2 + b^2}< \frac{1}{2}$$
Is the following proving method correct?
$$\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}$$
$$2ab < (a^2 + b^2)$$
$$0 <(a-b)^2$$
$a \neq b$ thus $a - b \neq 0$ and every number squared is not negative thus $(a-b)^2$ is positive thus bigger than zero
My concern with this proof is that I am not sure if I am allowed to do operations on both sides before inequality is proven.
| What you've shown is $$\frac{ab}{a^2 + b^2} < \frac{1}{2} \implies 0 < (a-b)^2$$
But what you are (apparently) required to show is $$0 < (a-b)^2 \implies \frac{ab}{a^2 + b^2} < \frac{1}{2}$$
So the correct proof is basically the reverse of what you've written, namely:
Suppose $a ≠ b$. Then
$$\begin{align} 0 < (a-b)^2 &\implies 2ab < (a^2 + b^2)\\
& \implies\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}
\end{align}$$
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x+3)^{-\frac{2}{3}}+\frac{1}{3}(x-27)\cdot -\frac{2}{3}(x+3)^{-\frac{5}{3}}+\frac{1}{3}(x+3)^{-\frac{2}{3}} \\
&= \frac{2}{3}(x+3)^{-\frac{2}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}(x-27)\\
%
f'''(x) &= -\frac{2^2}{3^3}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2\cdot 5}{3^3}(x+3)^{-\frac{8}{3}}
\end{align}
This is very nasty, please help me to solve in some easy way.
| It is clear that,
$$(x+3)^{\frac{1}{3}} = 30^{\frac{1}{3}}[1+\frac{(x-27)}{30}]^{\frac{1}{3}}.$$
Therefore,
$$(x+3)^{\frac{1}{3}} = 30^{\frac{1}{3}}\sum_{k=0}^{\infty}\frac{\Gamma(\frac{4}{3})}{\Gamma(k+1)\Gamma(\frac{4}{3}-k)}\frac{(x-27)^{k}}{30^{k}},$$
and,
$$f(x) = 30^{\frac{1}{3}}\sum_{k=0}^{\infty}\frac{\Gamma(\frac{4}{3})}{\Gamma(k+1)\Gamma(\frac{4}{3}-k)}\frac{(x-27)^{k+1}}{30^{k}}.$$
Now it is easy to see that:
$f^{80}(27) = 30^{\frac{1}{3}}\frac{\Gamma(\frac{4}{3})}{\Gamma(80)\Gamma(\frac{-233}{3})}\frac{\Gamma (81)}{30^{79}}=30^{\frac{1}{3}}\frac{\Gamma(\frac{4}{3})}{\Gamma(\frac{-233}{3})}\frac{80}{30^{79}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}dx$ as partial fraction solved using matrix equation in order to solve the integral $$\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}\mathrm dx,$$
the expression inside the integral can be expressed as
$$(2x^5-2x^4+2x^3+3/2x^4-2x^3-x^2+1)= x+(A/(x-1))+(B/(x-1)^2)+(Cx+D/2x^2+2x+1)$$
from here I have been ask to set up the system of linear simultaneous equations that are needed to be solved to calculate the integral, by utilising MX=Z, where M is the coefficient matrix, X is the solution vector containing the coefficients, and Z is the RHS of the matrix equation
I have attempt to do this by factoring the LHS denominator to the form of
$$(x-1)^2(2x^2+2x+1)$$ and then multiplied both sides by this giving the resulting equation
$$2x^5-2x^4+2x^3+3 = 2x^5-2x^4+2x^3+3+A(x-1)(2x^2+2x+1)+B(2x^2+2x+1)+CxD(x-1)^2$$
if I subtract $(2x^5-2x^4+2x^3+3)$ from LHS I am left with
$$LHS = 3x^3-x+3$$
I have tried expanding out the rest of the RHS and then collected like terms and to try and set up four equations
$3=2A+CD$
$0=2B-2CD$
$-1=-A+2B+CD$
$3=-A+B$
however, there is no solution.
any help would be appreciated
| Hint. This is an exercise in simplifying the integrand.
First divide to obtain $$x+\frac{3x^3-x+3}{2x^4-2x^3-x^2+1}.$$ We now focus on the second part, and we immediately recognise that the denominator vanishes when $x=1.$ Hence $x-1$ is a factor of the denominator. Dividing out, we can express the fraction as $$\frac{3x^3-x+3}{(x-1)(2x^3-x-1)},$$ and again noticing that the cubic below has a factor $x-1$ we easily obtain $$\frac{3x^3-x+3}{(x-1)^2(2x^2+2x+1)},$$ or fully factored, $$\frac32\frac{x^3-x/3+1}{(x-1)^2\left(x^2+x+\frac12\right)}.$$ I shall now ignore the constant factor henceforth.
We then write the last fraction as $$\frac{\left(x^2+x+\frac12\right)-\left(x^2+x+\frac12\right)+x^3-x/3+1}{(x-1)^2\left(x^2+x+\frac12\right)}=\frac{\left(x^2+x+\frac12\right)}{(x-1)^2\left(x^2+x+\frac12\right)}+\frac{-\left(x^2+x+\frac12\right)+x^3-x/3+1}{(x-1)^2\left(x^2+x+\frac12\right)},$$ which simplifies to become $$\frac{1}{x-1}+\frac56\frac{x+9/5}{(x-1)^2\left(x^2+x+\frac12\right)}.$$ Ignoring the first part again, we focus on the second and write it as (also ignoring the constant $5/6$) $$\frac{x-1+1+9/5}{(x-1)^2\left(x^2+x+\frac12\right)}=\frac{x-1}{(x-1)^2\left(x^2+x+\frac12\right)}+\frac{1+9/5}{(x-1)^2\left(x^2+x+\frac12\right)},$$ which simplifies to become $$\frac{1}{(x-1)\left(x^2+x+\frac12\right)}+\frac{14}{5}\frac{1}{(x-1)^2\left(x^2+x+\frac12\right)}.$$
You may now deal with the last two fractions, as usual.
| {
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How many non-negative integer solutions are there for $a+b+c+d=25$ if $a\geq 1, b\geq 2,c\leq 6,d\leq 14$
How many non-negative integer solutions are there for $a+b+c+d=25$ if $a\geq 1, b\geq 2,c\leq 6,d\leq 14$
So first I let $x= a-1$, $y=b-2$ and get:
$x+y+c+d=22$
And if all are non-negative I get that there are ${n+k-1\choose k-1} ={25\choose 3}$ solutions
Then I need to subtract the solutions where $c\geq 7,d\geq 15$
Let $z= c-7$ then $x+y+z+d=15$ and there are ${18\choose 3}$ non negative solutions
Let $w= d-15$ then $x+y+c+w=7$ and there are ${10\choose 3}$ non negative solutions.
And combining $z= c-7, w=d-15$ gives $x+y+z+w=0$, which will have only $1$ solution
So there are ${25\choose 3}-{18\choose 3}-{10\choose 3}-1$ solutions.
Does this seem correct?
| Your solution is almost correct.
You subtract the case $"c\geq 7\; and\;d\geq 15"$ once in the case $c\geq 7$ and once in the case $d\geq 15$. So, you need to add the $1$ at the end.
Here for reconfirmation of the result the same calculation using generating functions:
$$[x^{22}]\frac{1}{1-x}\cdot\frac{1}{1-x}\cdot\frac{1-x^7}{1-x}\cdot\frac{1-x^{15}}{1-x} = [x^{22}]\frac{1-x^7-x^{15}+x^{22}}{(1-x)^4}$$
$$=[x^{22}](1-x^7-x^{15}+x^{22})\sum_{n \geq 0}\binom{n+3}3x^n$$
$$=\binom{25}3 - \binom{18}3-\binom{10}3 + \binom{3}3$$
| {
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Synthetic Proof for a Geometry problem A while back, this question was asked on MSE:
Find the length of $CE$
In fact, allow me to phrase the problem in a slightly different manner:
In quadrilateral $ABCD$, $AB=6$, $\angle{ABC}=90°$, $\angle{BCD}=45°$ and $\angle{CAD}=2\angle{ACB}$. If $DE$ is perpendicular to $AC$ with $E$ on side $BC$, prove that the length of $CE=12$.
I have managed to prove the above result, but was unable to avoid the use of some trigonometry and algebraic manipulations.
My solution is as follows:
Let $M$ be the point of intersection of line segments $AC$ and $DE$, and let $H$ be the foot of the perpendicular from $M$ to line segment $EC$. Also, let $BC=x$, $CE=a$. Finally, let $\angle ACB =\theta, \angle CAD = 2\theta, \angle ACD=45^{\circ}-\theta$.
By Pythagoras' Theorem, $AC=\sqrt{AB^2+BC^2}=\sqrt{36+x^2}$. Clearly, $\triangle{CME} \sim \triangle{CBA} \Rightarrow \frac{CM}{CE}=\frac{BC}{AC} \Rightarrow CM=CE \cdot \frac{BC}{AC}=\frac{ax}{\sqrt{36+x^2}}$.
Thus $AM=AC-MC=\sqrt{36+x^2}- \frac{ax}{\sqrt{36+x^2}}=\frac{36+x^2-ax}{\sqrt{36+x^2}} \Rightarrow \frac{CM}{AM} = \frac{ax}{36+x^2-ax}$. Now, $\tan(2\theta)=\frac{MD}{MA}, \tan(45^{\circ}-\theta)=\frac{MD}{MC} \Rightarrow \frac{\tan(2\theta)}{\tan(45^{\circ}-\theta)}=\frac{MC}{MA}=\frac{ax}{36+x^2-ax}$.
On the other hand, $\tan(\theta)=\frac{AB}{BC}=\frac{6}{x} \Rightarrow \tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}=\frac{2 \cdot \frac{6}{x}}{1-\frac{36}{x^2}}=\frac{12x}{x^2-36} $. Also, $\tan(45^{\circ}-\theta)=\frac{\tan(45^{\circ})-\tan(\theta)}{1+\tan(45^{\circ})\tan(\theta)}=\frac{1-\tan(\theta)}{1+\tan(\theta)}=\frac{1-\frac{6}{x}}{1+\frac{6}{x}}=\frac{x-6}{x+6} \Rightarrow \frac{\tan(2\theta)}{\tan(45^{\circ}-\theta)} = \frac{12x}{(x-6)^2}$.
Thus, we have $\frac{12x}{(x-6)^2}=\frac{ax}{36+x^2-ax} \Rightarrow a= (36+x^2-ax) \cdot \frac{12}{(x-6)^2} \Rightarrow a[1+\frac{12x}{(x-6)^2}]= 12 \cdot \frac{36+x^2}{(x-6)^2} \Rightarrow a \cdot \frac{x^2+36}{(x-6)^2} = 12 \cdot \frac{36+x^2}{(x-6)^2} \Rightarrow a=12$.
But this solution is, admittedly, rather tedious. Thus, I wonder if there exists a synthetic solution by any chance?
| Let $F$ on $CD$ so that $AF\parallel BC$ and $H$ on $AC$ so that $FH\perp CD$.
Since $AF\parallel BC$, it follows that
$$\angle CAF = \angle ACB = \frac{\angle CAD}{2}.$$
Thus $\angle CAF = \angle DAF$. Also since $AF\parallel BC$,
$$\angle AFD = \angle BCD = 45^\circ.$$
Therefore $\angle AFD = \angle AFH$. It follows that $D$ and $H$ are symmetry along $AF$, and that $DH\perp CE$. Since $CH\perp DE$, $H$ is the orthocenter of $\triangle CDE$, and so $E, F, H$ are colinear.
Finally, $\triangle EFC$ has $\angle EFC=90^\circ, \angle ECF = 45^\circ$, so $CE$ is twice the distance from $F$ to $CE$ and is $12$.
| {
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Find general solution for $\tan{x} +\tan{2x} +\tan{3x}= 0$ My approach :-
$\tan{(x+2x)}= \tan{3x}$
$\tan{x}+\tan{2x} = \tan{3x}(1-\tan{x}.\tan{2x})$
Putting this in my equation ,
$\tan{3x}.(2-\tan{x}tan{2x}) =0$
Now either
$\tan{3x}= 0$ or $\tan{x}\tan{2x}=2$
Hence we get either
$x=\frac{n\pi}{3}$ or $x= n\pi ± \arctan{\frac{1}{√2}}.$
However , my textbook says that the solution is
$x=n\pi$ or $x= n\pi \pm \frac{\pi}{3}$
Why is there a disparity in both the solutions? If they are equivalent , how?
I apologize in advance as I'm using latex for the first time. Thank You
| Remember
$$\tan (2 x)=\frac{2 \tan (x)}{1-\tan ^2(x)}$$
and
$$\tan(3x)=\frac{\tan (x) \left(\tan ^2(x)-3\right)}{3 \tan ^2(x)-1}$$
So the given equation can be written as
$$\frac{2 \tan (x)}{1-\tan ^2(x)}+\frac{\left(\tan ^2(x)-3\right) \tan (x)}{3 \tan ^2(x)-1}+\tan (x)=0$$
Adding the fractions
$$\frac{2 \tan (x) \left(\tan ^2(x)-3\right) \left(2 \tan ^2(x)-1\right)}{\left(\tan ^2(x)-1\right) \left(3 \tan ^2(x)-1\right)}=0$$
which is verified when
$$\tan(x)=0\lor \tan ^2(x)-3=0\lor 2 \tan ^2(x)-1=0$$
that is, $\forall k\in\mathbb{Z}$
$$x=k\pi\lor x=\pm \frac{\pi}{3}+k\pi \lor x=\pm\arctan\left(\sqrt{\frac{1}{2}}\right)+k\pi$$
| {
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Domain of $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$ $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$
I'm stuck in this one because this can be rewritten as:
$$f(x,y) = \ln(16-x^2-y^2) + \ln(x^2+y^2-4)$$
Yet, the domain of the given function is $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\} \lor \{(16-x^2-y^2<0)\land (x^2+y^2-4<0)\}$. But the domain of the rewritten on is only $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\}$. Which one is the correct one and why does this happen?
| In real analysis $\ln(ab)=ln(a)+ln(b)$ is true iff $a>0$ and $b>0$, because in other case the logarithms don't make sense.
| {
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Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Question: Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Solution: Let $f:\mathbb{N}\to\mathbb{Z}$ be such that $$f(n)=n^3+an^2+bn+c, \forall n\in\mathbb{N}.$$
Also assume for the sake of contradiction that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$.
We have $f(1)=1+a+b+c, f(2)=8+4a+2b+c, f(3)=27+9a+3b+c$ and $f(4)=64+16a+4b+c$.
Now since $f(4)$ is a perfect square $\implies f(4)\equiv 0,1\pmod 4\implies c\equiv 0,1\pmod 4.$
First let that $c\equiv 0 \pmod 4$. Then $f(2)\equiv 0\pmod 4\implies 2b\equiv 0\pmod 4\implies b\equiv 0,2\pmod 4.$
Also we have $f(1)\equiv 0 \pmod 4$. Now we have $b+c\equiv 0,2 \pmod 4\implies 1+b+c\equiv -1,1\pmod 4.$ Thus we have $a\equiv -1,1\pmod 4$.
Also $f(3)\equiv 0\pmod 4$.
Now we have $f(3)-2f(2)+f(1)\equiv 0 \pmod 4\implies 12+2a\equiv 0\pmod 4\implies 2a \equiv 0\pmod 4 \implies a\equiv 0,2\pmod 4.$ But we have $a\equiv -1,1\pmod 4$, which is a contradiction. Thus it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ when $c\equiv 0 \pmod 4$.
A similar analysis for $c\equiv 1\pmod 4$ will lead to a contradiction. Thus it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ when $c\equiv 1\pmod 4$.
Hence it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ in any case, i.e., $\exists n\in\mathbb{N}$ such that $f(n)$ is not a perfect square.
Is there any better way to solve this problem?
| Let's put:
$$f(n)=n^3+an^2+bn+c$$
Suppose the opposite: that for some $a,b,c$ function $f(n)$ is always a perfect square for every $n$.
It means that, for example:
$$f(n-1)=(n-1)^3+a(n-1)^2+b(n-1)+c=p^2\tag{1}$$
$$f(n+1)=(n+1)^3+a(n+1)^2+b(n+1)+c=q^2\tag{2}$$
...with $p,q$ being integers. Now subtract (1) from (2) and you get:
$$f(n+1)-f(n-1)=6n^2+2+4an+2b=q^2-p^2$$
$$2(3n^2+1+2an+b)=(q-p)(q+p)$$
Obviously, $p,q$ must be either both odd or both even. In both cases the RHS is divisible by 4. It follows that:
$$2\mid3n^2+b+(2an+1)\tag{3}$$
It is obvious that $2an+1$ is always odd so from (3) it follows that for all $n$:
$$2\nmid 3n^2+b\tag{4}$$
But this is impossible: for odd $b$ take any odd value of $n$ and (4) does not hold. If $b$ is even take any even value of $n$ and (4) does not hold. So for some values of $n$ (4) and consecutively (3) cannot be true and our assumptions that $f(n-1)$ and $f(n+1)$ are both perfect squares does not hold.
| {
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Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$.
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.
| Call the roots $\alpha, \alpha', \alpha''$ with $\alpha' = \alpha^2 + 2\alpha - 4$.
$$p(x) = x^3 + 2x^2 - 5x + 1 = (x - \alpha)(x - \alpha')(x - \alpha'') = q(x)(x - \alpha'')$$
with $$q(x) = x^2 + (-\alpha - 3 \alpha + 4) x + (\alpha^3 + 2 \alpha^2 - 4\alpha) = x^2 + (-\alpha - 3 \alpha + 4) x + (\alpha - 1)$$
(replacing $\alpha^2$ with $-2\alpha + 4$)
We can now compute the polynomial long division $p(x)/q(x)$ in $\mathbb Z(\alpha)$ to get $x - \alpha''$:
$$p(x) - x q(x) = (\alpha^2 + 3 \alpha - 2) x^2 + (-\alpha - 4) x + 1$$
$$p(x) - x q(x) - (\alpha^2 + 3 \alpha - 2) q(x) = 0$$
so $$p(x)/q(x) = x - (- \alpha^2 - 3 \alpha + 2).$$
| {
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Integer within certain interval. I have to show that there is a unique integer within the interval $[a,b]$ where
$a = -\frac{3}{2} + \sqrt{\frac{9}{4} + 2(n+1)}$
and
$b = -\frac{1}{2} + \sqrt{\frac{1}{4} + 2(n+1)}$
as well as $n \in \mathbb{N}$.
Now the uniqueness is easy by showing that there can be one integer within this interval at maximum. This is done by calculating $(b-a)$ and finding that $(b-a)<1$.
But how do I prove that there indeed exists one?
| Assume the contrary - $\nexists$ an integer lying in $[a,b]$. Then, $\exists\ m \in \mathbb{N}$ such that $[a,b] \subset [m,m+1]$.
So,
\begin{align}
m &< a \\
\implies m+\frac{3}{2} &< \sqrt{\frac{9}{4} + 2(n+1)} \\
\implies m^2 + 3m &< 2n+ 2
\end{align}
and
\begin{align}
b &< m+1 \\
\implies \sqrt{\frac{1}{4} + 2(n+1)} &< m+\frac{3}{2} \\
\implies 2n +2 &< m^2 + 3m + 2 \\
\implies 2n &<
m^2 + 3m\end{align}
Hence, we must have $m(m+3) = 2n+1$. But, $m(m+3)$ is an even number, while $2n+1$ is odd.
This is a contradiction, and so the result is proved.
| {
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Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{align}
I feel I exhausted all identities/properties of binomials without success. Mathematica says it is true, but how to show it?
| $\def\peq{\mathrel{\phantom{=}}{}}$For the first identity, because for $0 \leqslant j \leqslant n - 1$,\begin{align*}
&\peq \binom{n}{j} j (1 - x)^{n - j} x^{j - 1} - \binom{n}{j + 1} (j + 1) (1 - x)^{n - j - 1} x^j\\
&= \frac{n!}{(j - 1)!\, (n - j)!} (1 - x)^{n - j} x^{j - 1} - \frac{n!}{j!\, (n - j - 1)!} (1 - x)^{n - j - 1} x^j\\
&= \frac{n!}{j!\, (n - j)!} (1 - x)^{n - j - 1} x^{j - 1} (j(1 - x) - (n - j)x)\\
&= \binom{n}{j} (1 - x)^{n - j - 1} x^{j - 1} (j - nx),
\end{align*}
so\begin{align*}
&\peq \sum_{j = n - k}^n \binom{n}{j} (1 - x)^{n - j - 1} x^{j - 1} (j - nx)\\
&= nx^{n - 1} + \sum_{j = n - k}^{n - 1} \left( \binom{n}{j} j (1 - x)^{n - j} x^{j - 1} - \binom{n}{j + 1} (j + 1) (1 - x)^{n - j - 1} x^j \right)\\
&= nx^{n - 1} + \left( \binom{n}{n - k} (n - k) (1 - x)^k x^{n - k - 1} - nx^{n - 1} \right)\\
&= \binom{n}{n - k} (n - k) (1 - x)^k x^{n - k - 1}.
\end{align*}
For the second identity, note that\begin{gather*}
\frac{1}{n} \binom{n - 1}{k} \binom{n}{n - k} (n - k) = \frac{1}{n} · \frac{(n - 1)!}{k!\, (n - k - 1)!} · \frac{n!}{k!\, (n - k)!} · (n - k)\\
= \frac{(n - 1)!}{k!\, (n - k - 1)!} · \frac{(n - 1)!}{k!\, (n - k - 1)!} = \left( \binom{n - 1}{k} \right)^2,
\end{gather*}
thus dividing by $x^{n - 1}$ on both sides of the second identity and denoting $m = n - 1$, $t = -\dfrac{1}{x}$, it is equivalent to prove that\begin{gather*}
\sum_{k = 0}^m \left( \binom{m}{k} \right)^2 (t + 1)^k = \sum_{k = 0}^m \binom{m}{k} \binom{m + k}{k} t^{m - k}.\tag{2$'$}
\end{gather*}
Consider the polynomial $(s + t + 1)^m (s + 1)^m$. Since\begin{gather*}
(s + t + 1)^m (s + 1)^m = \left( \sum_{k = 0}^m \binom{m}{k} (t + 1)^k s^{m - k} \right) \left( \sum_{k = 0}^m \binom{m}{k} s^k \right),\\
(s + t + 1)^m (s + 1)^m = (s + 1)^m \sum_{k = 0}^m \binom{m}{k} (s + 1)^k t^{m - k} = \sum_{k = 0}^m \binom{m}{k} t^{m - k} (s + 1)^{m + k},
\end{gather*}
then\begin{gather*}
\left( \sum_{k = 0}^m \binom{m}{k} (t + 1)^k s^{m - k} \right) \left( \sum_{k = 0}^m \binom{m}{k} s^k \right) = \sum_{k = 0}^m \binom{m}{k} t^{m - k} (s + 1)^{m + k}.\tag{3}
\end{gather*}
Treating $t$ as a constant and $s$ as a variable for the moment, the coefficient of $s^m$ of the LHS of (3) is$$
\sum_{k = 0}^m \binom{m}{k} (t + 1)^k · \binom{m}{k} = \sum_{k = 0}^m \left( \binom{m}{k} \right)^2 (t + 1)^k,
$$
and that of the RHS of (3) is$$
\sum_{k = 0}^m \binom{m}{k} t^{m - k} · \binom{m + k}{m} = \sum_{k = 0}^m \binom{m}{k} \binom{m + k}{k} t^{m - k}.
$$
Therefore (2$'$) is true.
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Rational function integration (not sure wich technique to use) I have the integral $$\int \frac{x^2 - 11x -2}{(x-2)(x+2)(2x+1)} dx $$
From what I have learned in class, it seems that I need to use polynomial long division
is that the correct attitude for that problem? and if yes when I try to devide I have problem because the deg of $x^2 - 11x -2$$\ <$ $(x-2)(x+2)(2x+1)$ but on the other hand I can't use the "Partial fraction decomposition" technique here (I probably wrong)
I would like to get some help here
Thank you kindly
| Basically, it can be solved by using partial fraction decomposition :
\begin{equation}
\begin{array}{c}
=\int\left(\frac{71}{21(2 x+11)}-\frac{6}{7(x+2)}-\frac{1}{3(x-2)}\right) \mathrm{d} x \\
=\frac{71}{21} \int \frac{1}{2 x+11} \mathrm{d} x-\frac{6}{7} \int \frac{1}{x+2} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x-2} \mathrm{d} x \\
\int \frac{1}{2 x+11} \mathrm{d} x \\
\text { Substitute } u=2 x+11 \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=2 \text { (steps) } \longrightarrow \mathrm{d} x=\frac{1}{2} \mathrm{d} u \\
=\frac{1}{2} \int \frac{1}{u} \mathrm{d} u
\end{array}
\end{equation}
Plug in solved integrals:
$\frac{1}{2} \int \frac{1}{u} \mathrm{d} u$
$$
=\frac{\ln (u)}{2}
$$
Undo substitution $u=2 x+11$
$$
=\frac{\ln (2 x+11)}{2}
$$
Now solving:
$$
\begin{array}{c}
\int \frac{1}{x+2} \mathrm{d} x \\
\text { Substitute } u=x+2 \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=1_{(\text {steps })} \longrightarrow \mathrm{d} x=\mathrm{d} u:
\end{array}
$$
$=\int \frac{1}{u} \mathrm{d} u$
Use previous result:
$=\ln (u)$
Undo substitution $u=x+2$ $=\ln (x+2)$
\begin{equation}
\begin{array}{l}
\frac{71}{21} \int \frac{1}{2 x+11} \mathrm{d} x-\frac{6}{7} \int \frac{1}{x+2} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x-2} \mathrm{d} x \\
\quad=\frac{71 \ln (2 x+11)}{42}-\frac{6 \ln (x+2)}{7}-\frac{\ln (x-2)}{3}
\end{array}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ I=\iint\limits_S (x^5+z)\ dy\ dz\ \ \text{where}\ S\ \text{is an inner side of a hemisphere}\ x^2+y^2+z^2=R^2,\ z\leqslant 0 $ Evaluate $I$:
$$
I=\iint\limits_S (x^5+z)\ dy\ dz\ \ \text{where}\ S\ \text{is an inner side of a hemisphere}\ x^2+y^2+z^2=R^2,\ z\leqslant 0
$$
My attempt:
$$
\begin{aligned}
&z=-\sqrt{R^2-x^2-y^2}\Rightarrow
\begin{cases}
z'_x=\frac{x}{\sqrt{R^2-x^2-y^2}}\\
z'_y=\frac{y}{\sqrt{R^2-x^2-y^2}}
\end{cases}\\
&I=\iint\limits_{D(x,y)}\langle(x^5+z, 0, 0), (z'_x, z'_y,-1)\rangle\ dx\ dy=\iint\limits_{D(x,y)}\frac{x^6+xz}{\sqrt{R^2-x^2-y^2}}\ dx\ dy=\\
&=\iint\limits_{D(x,y)}\left(\frac{x^6}{\sqrt{R^2-x^2-y^2}}-x\right)\ dx\ dy=\int\limits_0^{2\pi}d\varphi\int\limits_0^R\left(\frac{r^6\cos^6\varphi}{\sqrt{R^2-r^2}}-r\cos\varphi\right)r\ dr=\\
&=\int\limits_0^{2\pi}\cos^6\varphi\ d\varphi\int\limits_0^R\frac{r^6}{\sqrt{R^2-r^2}}\ dr
\end{aligned}
$$
And then I got stuck because of the last integral. Perhaps I made a mistake somewhere in the beginning.
The answer should be the following:
$$
I=-\frac{2\pi R^7}{7}
$$
Could someone help me to solve this problem? I would appreciate it.
| You forgot to multiply by another factor of $r$. The integral should be
$$
\int_0^R \frac{r^7}{\sqrt{R^2 - r^2}} dr.
$$
Take $r = R\sin u$ then you obtain
$$
\int_0^R \frac{r^7}{\sqrt{R^2 - r^2}} dr \;\; =\;\; \int_0^{\frac{\pi}{2}} \frac{R^7\sin^7u}{R\sqrt{1-\sin^2u}}R\cos udu \;\; =\;\; \int_0^{\frac{\pi}{2}} R^7\sin^7udu.
$$
Expand $\sin^7u = \sin u\left (1-\cos^2u\right )^3$, and make the substitution $w = \cos u$. Then your integral becomes
$$
-\int_1^0 R^7\left (1-w^2 \right )^3dw \;\; =\;\; R^7\int_0^1\left (1 - 3w^2 + 3w^4 - w^6\right )dw \;\; = \;\; R^7\left (1 - 1 + \frac{3}{5} - \frac{1}{7}\right ) \;\; =\;\; \frac{16R^7}{35}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
n-th power of a matrix using the division of polynomials. Consider the matrix
$$
A=\begin{pmatrix}
0 & 0 & 0\\
-2 & 1 & -1\\
2 & 0 & 2
\end{pmatrix}
$$
*
*Calculate $A^3-3A^2+2A$.
*What is the remainder of the division of the polynomial $X^n$ by the polynomial $X^3-3X^2+2X$.
*Calculate $A^n$ for every natural number $n$.
I was solving the following problem and I was stuck in it. For part 1) the answer was the zero matrix. In part 2) I use the usual division and i get the following
$$
X^n=X^{n-3}(X^3-3X^2+2X)+3X^{n-1}-2X^{n-2}.
$$
When I pass to part 3) and using part 1) and 2), we obtain
$$
A^n=3A^{n-1}-2A^{n-2}.
$$
Using the fact that $A^3-3A^2+2A=O_{3\times 3}$. but if I use this answer for calculating $A^2$ the answer is not correct, so I think $A^n$ obtained is not correct. Now, one can use the diagonalization of the matrix $A$ and obtain
$$
A^n=\begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatrix}
$$
Can you help me in proving part 2 (if not correct) and part 3 without using the diagonalization method.
| Write
$$x^n = q(x)(x^3-3x^2+2x) + r(x) = q(x)x(x-1)(x-2)+r(x)$$
for polynomials $q,r \in \Bbb{R}[x]$ where $\deg r \le 2$. Plugging in $x = 0,1,2$ gives
$$r(0) = 0, \quad r(1)=1, \quad r(2)=2^n$$
so $r(x) = (2^{n-1}-1)x^2+(-2^{n-1}+2)x$. Now we get
$$A^n = q(A)(A^3-3A^2+2A) + r(A) = r(A) = (2^{n-1}-1)A^2+(-2^{n-1}+2)A$$
which yields precisely your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Pairs $(a,b)\in F^2$ such that $a^6+b^6=1$ where $F$ is a finite field of 25 elements Let $F$ be a field of 25 elements, and consider the group $G$ of all $2\times 2$ matrices $A$ with entries in $F$ satisfying $A_5A=I$, where $I$ is the identity matrix and
$A_5=\begin{pmatrix} a^5 & c^5 \\ b^5 & d^5 \end{pmatrix}$ if $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
I am asked to find the number of pairs $(a,b)\in F^2$ such that $a^6+b^6=1$, and conclude that $|G|=720$.
Note that $(a+b)^6=(a+b)^5(a+b)=(a^5+b^5)(a+b)=a^6+b^6+a^5b+ab^5$, so it suffices to show find the number of pairs $(a,b)$ such that $a^5b+ab^5=-1$. How do I have to proceed next?
| One needs to recognise that $a^6$ is the norm map from $F=\Bbb F_{25}$ to the field
$\Bbb F_5$ of order $5$, so $a^6\in\{0,1,2,3,4\}$. $a^6=0$ only has the solution $0$
but for $c\in\{1,2,3,4\}$ $a^6=c$ has six solutions in $F$. We need to count
the solutions of $(a^6,b^6)=(c,d)$ where $(c,d)$ is one of the pairs $(0,1)$, $(1,0)$,
$(2,4)$, $(3,3)$ and $(4,2)$. There are six solutions for each of the first two pairs
and thirty-six for the other three. So overall $a^6+b^6=1$ has $120$ solutions
in $F$.
Once you have done that you observe that $(c,d)$ must be "orthogonal" to $(a^5,b^5)$ i.e., $ca^5+db^5=0$. The solutions to this form a one-dimensional
subspace of $F^2$ and one needs to show exactly six of these $(c,d)$
also satisfy $c^6+d^6=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_n \frac{1}{n}\int_{1/n^2}^{+\infty} \frac{\tanh(\frac{\sqrt{x}}{n})}{x^2\sqrt{(n+1)x}}\ dx$ Let
$$f(x)=\frac{\tanh(\frac{\sqrt{x}}{n})}{x^2\sqrt{(n+1)x}} \ \ \ \text{and} \ \ f_n=f\chi_{[ 1/n^2,+\infty)}$$
I want to compute $$\lim_n \frac{1}{n}\int_{1/n^2}^{+\infty}f \ dx$$
My solution is the following:
Since $$\tanh(\frac{\sqrt{x}}{n})\leq \tanh(\sqrt{x}) \ \ \text{and} \ \ x^2\sqrt{(n+1)x} \geq x^2\sqrt{2x} \ \ \ \text{for} \ \ n\geq1$$
we have $$f_n\leq \frac{\tanh(\sqrt{x})}{x^2\sqrt{2x}} \ \ \text{for} \ \ x\in[1/n^2,1] \ \ n\geq 1$$
and $$f_n \leq \frac{1}{x^2} \ \ \text{for} \ x\in[1,+\infty)$$
Then by dominated convergence th. we have that the limit is equal to zero.
The solution given in my notes is much more involved than mine, so I guess I'm missing something..
Question
What is wrong with my solution?
| Change variables with $$ y = \frac{\sqrt{x}}{n} \Rightarrow dy = \frac{dx}{2n\sqrt{x}}. $$
Also, $$ \sqrt{x} = ny \Rightarrow x^2 = n^4y^4. $$
When $ x = \frac{1}{n} $ we get that $ y = \sqrt{1 + \frac{1}{n^2}} \to 1.$
Finally, observe that $$ \int_1^{\infty} \frac{tanh(y)}{y^5} < + \infty$$ hence $$ \lim_{n \to \infty} \frac{1}{n} \int_{\frac{1}{n^2}}^{\infty} \frac{tanh\left( \frac{\sqrt{x}}{n}\right)}{x^2\sqrt{2x}} dx = 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the closed path of given system of ordinary differential equation Consider the system of ordinary differential equations:
$$\begin{cases}\frac{dx}{dt}=4x^3y^2-x^5y^4\\
\frac{dy}{dt}=x^4y^5+2x^2y^3\end{cases}$$
Then for this system there exist
$1).$ A closed path in $\left \{(x,y) \in \mathbb{R^2}|x^2+y^2 \leq 5 \right \}$
$2).$ A closed path in $\left \{(x,y) \in \mathbb{R^2}|5<x^2+y^2 \leq 10 \right \}$
$3). $ A closed path in $\left \{(x,y) \in \mathbb{R^2}|x^2+y^2 >10 \right \}$
$4). $ No closed Path in $\mathbb{R^2}$
solution i tried- I first find out the $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{x^2y^3+2y}{4x-x^3y^2}$$
it will become
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0\;\;\;\;\;\;\;\
....................1$$
which is of from $$f_1(xy)ydx+f_2(xy)xdy$$
after that i find the $I.F$ of $1$ which comes out $$\frac{-1}{6xy}$$
now by multiplying this with $1$ i get $$\frac{1}{6} \left ( xy^2+\frac{2}{x} \right ) dx-\frac{1}{6} \left ( \frac{4}{y}+x^2y \right )dy=0$$
after solving this i get answer $$\frac{x^2y^2}{12}-\frac{1}{3}\log (\frac{x}{y^2})=c$$
but there is noting related to given option ,where i am making mistake please help
Thank you
| Starting from this line in your answer:
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0$$
$$-x^2y^2(ydx+xdy)-2ydx+4xdy=0$$
$$-x^2y^2d(xy)-2ydx+4xdy=0$$
Divide by $xy$:
$$-xyd(xy)-2\dfrac {dx}x+4\dfrac {dy}y=0$$
$$-(xy)^2-4\ln x+8\ln y=0$$
$$-(xy)^2+4\ln \dfrac {y^2}{x}=C$$
Or if you prefer:
$$(xy)^2+4\ln \dfrac x{y^2}=C_1$$
Our answers are different. There is a sign difference. This may explain AVK's comment:
There is a mistake somewhere in your solution. The picture of the phase plane of the original system does not coincide with the curves that you obtained.
Your mistake is here:
$$\frac{1}{6} \left ( xy^2+\frac{2}{x} \right ) dx-\frac{1}{6} \left ( \frac{4}{y}\color {red}{+x^2y} \right )dy=0$$
The term in red should be negative from the precedent line:
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0$$
You multiply by $-\dfrac 1 {6xy}$:
$$\dfrac 1 {6xy}(x^2y^2+2)ydx-\dfrac 1 {6xy}(4 \color{blue}{-x^2y^2})xdy=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Taylor series approximation of Gaussian $Q$ function , $Q(x) = \frac{1}{{\sqrt {2\pi } }}\int_x^\infty {{e^{ - \frac{{{v^2}}}{2}}}dv}$ $ \require{newcommand}\newcommand{\Erfc}{\operatorname{Erfc}}$
I am trying to find a Taylor series expansion for the Gaussian $Q$ function. I have seen that error function $\Erfc(x)$ is an approximation of $Q(x)$ (Is my assumption correct?).
$\Erfc(x)$ has a Taylor approximation. Is it possible to construct such for $Q(x)$ function? Is there any condition for that? Thank you.
| You can obtain a Taylor series as follows:
\begin{align*}
Q(x) & = \frac{1}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - t^2 /2} dt} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {e^{ - t^2 /2} dt} = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {e^{ - t^2 /2} dt}
\\ &
= \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}\left( { - \frac{{t^2 }}{2}} \right)^n } dt} = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}\left( { - \frac{1}{2}} \right)^n \frac{{x^{2n + 1} }}{{2n + 1}}} .
\end{align*}
Addendum:
A different series expansion is
$$
Q(x) = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}e^{ - x^2 /2} \sum\limits_{n = 0}^\infty {\frac{{x^{2n + 1} }}{{1 \cdot 3 \cdots (2n + 1)}}} .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose that $x,y,z$ are real numbers whose moduli are not equal to $1$.)
I discovered this problem from chapter 3 of Prelude to Mathematics by W. W. Sawyer. Sawyer thought that this problem arose from the sine law: let $a,b,c$ be respectively the lengths of the edges opposite to three vertices $A,B,C$ of a triangle. Define $x=\cos A$ and define $y,z$ analogously. Now equalities $(1)-(3)$ simply relate $a,b$ and $c$ to each other by the cosines of the angles and $(4)$ is just a rewrite of the sine law
$$
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.
$$
However, the algebraic version $(4)$ looks more general. For example, it does not state that $a,b,c$ must be positive or that they must satisfy the triangle inequality.
Sawyer wrote that this isn't a hard problem, but he didn't provide any solution. I can prove $(4)$ using linear algebra. Suppose that $(a,b,c)\ne(0,0,0)$ (otherwise $(4)$ is obvious). Rewrite $(1)-(3)$ in the form of $M\mathbf a=0$:
$$\begin{bmatrix}-1&z&y\\ z&-1&x\\ y&x&-1\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=0.$$
Since $x^2,y^2,z^2\ne1$, $M$ has rank $2$ and $D=\operatorname{adj}(M)$ has rank $1$. Hence all columns of $D$ are parallel to $(a,b,c)^T$ and $\frac{d_{11}}{d_{21}}=\frac{d_{12}}{d_{22}}=\frac{a}{b}$. Since $M$ is symmetric, $D$ is symmetric too. Therefore $\frac{1-x^2}{1-y^2}=\frac{d_{11}}{d_{22}}=\frac{d_{11}d_{12}}{d_{21}d_{22}}=\frac{a^2}{b^2}$, i.e. $\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.
As this problem comes from Hall and Knight's book, I think there should be a more elementary solution. Any ideas?
| By (1) and (3), $a=ay^2 + bxy +bz.$ Thus, $a(1-y^2)=b(xy+z)$ so that $$a^2(1-y^2)=ab(xy+z).$$
In a similar way, we derive from (2) and (3) that
$$b^2(1-x^2)=ab(xy+z).$$
Thus, the left sides of the two displayed equations are equal, yielding the first equality in (4). By symmetry, we're done.
IOW replace $(a,c)$ by $(c,a)$ and $(x,z)$ by $(z,x)$ above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.