Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdot \:x_2}{2}\right)^2$
$=\:x_1\cdot \:x_2\le \:\frac{\left(x_1\cdot \:\:x_2\right)^2}{4}\:$
$=\:\:x_1\cdot \:\:x_2\le \:\:\frac{{x_1}^2+2x_1x_2+{x_2}^2}{4}$
$=4\left(x_1\cdot x_2\right)\:\le {x_1}^2+2x_1x_2+{x_2}^2$
$=4\left(x_1\cdot \:x_2\right)\:-2x_1x_2\le \:{x_1}^2+{x_2}^2$
Substituting in the values for $x_1$ and $x_2$ we get:
$4\left(\frac{a+\sqrt{a^2-4a+4}}{2}\cdot \:\frac{a-\sqrt{a^2-4a+4}}{2}\right)\:-2\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)\le \:\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)^2+\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)^2$
$= 4\left(a-1\right)\:-\left(2a-2\right)\le \:a^2-2a+2$
$ = 0\le a^2-4a+4$
It seems as if I walked in circles through this process, can anyone help? Thanks in advance!
|
$x_1^2+x_2^2=(x_1+x_2)^2-2(x_1x_2)=a^2-2(a-1)$
The last equility is Vieta or if you wish $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+(x_1x_2).$
So the min is when $a=1$ (because of usual calculus).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4357090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Can we re-write Newton's Binomial formula as a power series in $\ r\ $ without any problems? Newton's Generalised Binomial theorem states that if $\ x\ $ and $\ y\ $ are real numbers with $\ \vert x \vert > \vert y \vert\ (\text{note that } \left\vert \frac{y}{x} \right\vert < 1),\ $ and $\ r\ $ is any complex number, one has
$$ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ .$$
If we restrict $\ r\ $ to be a real number, and view this as a power series in $\ r\ $ whilst preserving the order of terms, then:
*
*The constant term is $\ x^r.$
*The linear coefficient, i.e. the coefficient of $\ r,\ $ whilst preserving the order of terms in Newton's formula, is
$$\frac{1}{1} x^{r-1} y^1 - \frac{1}{2} x^{r-2} y^2 + \frac{1}{3} x^{r-3} y^3 - \frac{1}{4} x^{r-2} y^2 + \ldots = x^r\left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right)$$
I noticed that this is equal to $\ x^r \ln \left( 1 + \frac{y}{x} \right),\ $ although I'm not sure this fact is relevant.
*
*The quadratic coefficient, i.e. the coefficient of $\ r^2,\ $ again whilst preserving the order of terms in Newton's formula, is
$$ x^r\left( \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right)$$
The coefficients of $\ \left(\frac{y}{x}\right)^n\ $ are Stirling numbers of the first kind divided by factorials.
Now I'm sure all these series converge, by a little bit of work (which I'm working on...) and then the Alternating Series Test and/or/ Ratio Test, since $\ \left\vert \frac{y}{x} \right\vert < 1.\ $ So each coefficient of $\ r^k\ $ exists.
My main question is: is it true that
$$ (x+y)^r = x^r + x^r\left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + x^r\left( \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots\ ?$$
This can be re-written more nicely as:
$$ \left(1+\left(\frac{y}{x}\right) \right)^r = g(r):= 1 + \left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + \left( \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots$$
which is clearly a power series in $\ r.$
I'm not even sure if $\ g(r)\ $ exists for all values of $\ r\ $ let alone if it is equal to $\ \left(1+\left(\frac{y}{x}\right) \right)^r.$
I'm not sure if the Riemann Series Theorem has anything to say about this, since this is technically not a simple rearrangement of the terms in Newton's formula: more specifically, we have not permuted the terms...
|
$$ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ .$$
Dividing through by the constant $\ x^r,\ $ we see that the above series converges if and only if
$$ \left(1+\frac{y}{x}\right)^r = \sum_{k=0}^{\infty} \binom{r}{k} \left(\frac{y}{x}\right)^k = \underbrace{1}_{a_0} + \underbrace{\frac{r}{1!}}_{a_1} \left(\frac{y}{x}\right)^1 + \underbrace{\frac{r(r-1)}{2!} }_{a_2} \left(\frac{y}{x}\right)^2 + \underbrace{\frac{r(r-1)(r-2)}{3!} }_{a_3} \left(\frac{y}{x}\right)^3 + \ldots $$
converges, and we know that this Newton's Generalised Binomial Expansion converges for all real $\ r\in\mathbb{R}.$
However, this does not immediately imply the series of expanded brackets terms
$$ \underbrace{1}_{b_0} + \underbrace{\frac{r}{1!}}_{b_1}\left(\frac{y}{x}\right)^1 + \underbrace{- \frac{r}{2!} }_{b_2} \left(\frac{y}{x}\right)^2 + \underbrace{\frac{r^2}{2!} }_{b_3} \left(\frac{y}{x}\right)^2 + \underbrace{\frac{2r}{3!} }_{b_4} \left(\frac{y}{x}\right)^3 + \underbrace{- \frac{3r^2}{3!} }_{b_5} \left(\frac{y}{x}\right)^3 + \underbrace{\frac{r^3}{3!} }_{b_6} \left(\frac{y}{x}\right)^3 + \ldots$$
is equal to $\ \left(\ 1+\frac{y}{x}\right)^r:\ $ we have not yet even established if $\ \displaystyle\sum_k b_k \left(\frac{y}{x}\right)^k\ $ converges.
We proceed by showing that $\ \displaystyle\sum_k b_k \left(\frac{y}{x}\right)^k\ $ converges absolutely, as follows.
$$ \underbrace{1}_{c_0} + \underbrace{ \frac{ \vert r \vert }{1!} }_{c_1} \left\vert \frac{y}{x}\right\vert^1 + \underbrace{ \frac{ \vert r \vert ( \vert r \vert + 1 ) }{2!} }_{c_2} \left\vert\frac{y}{x}\right\vert^2 + \underbrace{ \frac{ \vert r \vert ( \vert r \vert + 1 )( \vert r \vert + 2 ) }{3!} }_{c_3} \left\vert\frac{y}{x}\right\vert^3 + \ldots $$
is a series of positive terms; this series converges by the ratio test, because
$$ \left \vert \frac{c_{k+1}\left\vert\frac{y}{x}\right\vert^{k+1} }{c_{k} \left\vert\frac{y}{x}\right\vert^k} \right \vert = \left\vert \frac{\left \vert r \right \vert + k}{k+1} \right\vert \left\vert \frac{y}{x} \right\vert \overset{k \to \infty}{\to} \vert 1 \vert \left\vert \frac{y}{x} \right\vert = \left\vert \frac{y}{x} \right\vert < 1. $$
So, splitting $\ c_k\ $ in the previous series up into individual terms by expanding brackets, we get:
$$ 1 + \frac{ \vert r \vert }{1!} \left\vert \frac{y}{x}\right\vert^1 + \frac{ \vert r \vert }{2!} \left\vert \frac{y}{x}\right\vert^2 + \frac{ \vert r \vert ^2 }{2!} \left\vert \frac{y}{x}\right\vert^2 + \frac{ 2 \vert r \vert }{3!} \left\vert\frac{y}{x}\right\vert^3 + \frac{ 3 \vert r \vert ^2 }{3!} \left\vert\frac{y}{x}\right\vert^3 + \frac{ \vert r \vert ^3 }{3!} \left\vert\frac{y}{x}\right\vert^3 + \ldots,\qquad (1) $$
which is a series of positive terms whose limit of partial sums must be equal to $\ \displaystyle\sum_{k=0}^{\infty}\ c_{k} \left\vert\frac{y}{x}\right\vert^k,\ $ by the Monotone Convergence Theorem. We can allow any of the terms in $\ (1)\ $ to be negative and it will still converge. In other words, $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ converges absolutely.
The first consequence of this is that $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ = \sum_{k=0}^{\infty} \binom{r}{k} \left(\frac{y}{x}\right)^k,\ $ for all values of $\ r\in\mathbb{R},\ $ because simply grouping together terms in the series $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ without changing the order of terms, doesn't change the fact that it converges, nor the value it converges to.
Secondly, by Fubini's theorem for infinite series (and not Riemann Series Theorem), the fact that $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ converges absolutely means that $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ is indeed equal to
$$ g(r):= 1 + \left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + \left( \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots\ $$
for all values of $\ r\in\mathbb{R}.$
Thus we finally have our result:
$$\ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ = \sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ = g(r):= 1 + \left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + \left( \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots $$
for all values of $\ r\in\mathbb{R}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4360296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Integral $\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx$ I am trying to compute the integral
$$\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx \tag{1}$$
Context: Originally I was trying to prove the following result:
$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2\binom{2n}{n}}=\frac83\beta(2)-\frac{\pi}3\ln(2+\sqrt{3})\tag{2}$$
Where $\beta(2)$ is the Catalan´s constant
To this end I started with the well known result
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{4^n x^{2n+1}}{(2n+1)\binom{2n}{n}} \tag{3}$$
Dividing both sides of $(3)$ by $x$ and integrating from $0$ to $1/2$ we obtain
$$\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx=\frac12\sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}} \tag{4}$$
So the task reduces to compute the integral in $(4)$. Therefore
$$
\begin{aligned}
\sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}}&=2\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx\\
&=2\int_0^{\pi/6}\frac{x}{\sin(x)}\,dx &(x \to \sin(x))\\
&=2\int_0^{\pi/2}\frac{x}{\sin(x)}\,dx-2\int_{\pi/6}^{\pi/2}\frac{x}{\sin(x)}\,dx\\
&=4\beta(2)-2\int_{\pi/6}^{\pi/2}\frac{x}{\sin(x)}\,dx\\
&=4\beta(2)-2\int_{0}^{\pi/3}\frac{\left(\frac{\pi}{2}-x\right)}{\cos(x)}\,dx & (x \to \frac{\pi}{2}-x)\\
&=4\beta(2)-\pi\int_{0}^{\pi/3}\sec(x)\,dx+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\
&=4\beta(2)-\pi\ln\left(\sec(x)+\tan(x) \right)\Big|_0^{\pi/3}+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}\frac{x}{e^{ix}+e^{-ix}}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}\frac{xe^{-ix}}{1+e^{-2ix}}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}xe^{-ix}\sum_{k=0}^\infty(-1)^ke^{-2ikx}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\sum_{k=0}^\infty(-1)^k\int_{0}^{\pi/3}xe^{-ix(2k+1)}\,dx\\
\end{aligned}
$$
The integral in the last line is $(1)$. I integrated by parts, but ended up with some nasty series not very promising.
|
$$I=\int_0^\frac{\pi}{3}\frac{x}{\cos(x)}\,dx $$
Making the substitution $x=\frac{\pi}{2}-t$
$$I=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{dt}{\sin t}-\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{t\,dt}{\sin t}=I_1-I_2$$
Next, we make the substitution $\frac{dt}{\sin t}=d\big(\ln(\tan\frac{t}{2})\big)$
$$I_1=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2}d\big(\ln(\tan\frac{t}{2})\big)=-\frac{\pi}{2}\ln(\tan\frac{\pi}{12})$$
Using $\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$ and $-\ln(2-\sqrt 3)=\ln(2+\sqrt 3)$
$$I_1=\frac{\pi}{2}\ln(2+\sqrt3)$$
$$I_2=\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{t\,dt}{\sin t}=t\ln(\tan\frac{t}{2})\Big|_\frac{\pi}{6}^\frac{\pi}{2}-\int_\frac{\pi}{6}^\frac{\pi}{2}\ln(\tan\frac{t}{2})dt$$
$$=\frac{\pi}{6}\ln(2+\sqrt 3)-\int_0^\frac{\pi}{2}\ln(\tan\frac{t}{2})dt+\int_0^\frac{\pi}{6}\ln(\tan\frac{t}{2})dt$$
$$=\frac{\pi}{6}\ln(2+\sqrt 3)-2\int_0^\frac{\pi}{4}\ln(\tan x)dx+2\int_0^\frac{\pi}{12}\ln(\tan x)dx$$
In the second term we make the substitution $\tan x=t$, and the third term was evaluated here
($G=\beta(2)$ - Catalan's constant)
$$I_2=\frac{\pi}{6}\ln(2+\sqrt 3)-2\int_0^1\frac{\ln t}{1+t^2}dt -\frac{4}{3}G=\frac{\pi}{6}\ln(2+\sqrt 3)+2G-\frac{4}{3}G$$
$$I_2=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac{2}{3}G$$
$$I=I_1-I_2=\frac{\pi}{3}\ln(2+\sqrt 3)-\frac{2}{3}G$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4360729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Show that $f(x) = x^3 + 3\sin x + 2\cos x$ is one-to-one. How would I show that $f(x) = x^3 + 3\sin x + 2\cos x$ is one to one?
Showing that the function is strictly increasing seemed to be the way to go, but then I need to show that $f'(x) = 3x^2 + 3\cos x - 2\sin x > 0$. I graphed the derivative function and it is indeed strictly positive. Thoughts?
|
To show that $f(x) = x^3 + 3 \sin x + 2 \cos x $ is one-to-one, it is enough to show that $ f'(x) = 3 x^2 + 3 \cos x - 2 \sin x > 0 $ for all $ x $. Notice that $ (3 \cos x - 2 \sin x) \ge -2 $ for all $ x $. So, whenever $ 3 x^2 > 2 $, $ f'(x) $ must be positive. Note that for $ x \le -\pi/2 $ or $ x \ge \pi/2 $, $ 3 x^2 \ge 3 (\pi/2)^2 > 3 (3/2)^2 > 2 $. Hence, $ f'(x) > 0 $ for $ x \le -\pi/2 $ or $ x \ge \pi/2 $. Next, for $ -\pi/2 < x < 0 $, all three terms in the expression of $f'(x)$ are positive, and so $f'(x)$ is also positive.
Finally, consider the case $ 0 \le x \le \pi/2 $. Two ways of establishing $f'(x) > 0$ is described below: one requires a calculator, the other does not.
Using calculator: Note that within $ 0 \le x \le \pi/2 $, $(3 \cos x - 2 \sin x) \le 0$ for $ \tan^{-1} (3/2) \le x \le \pi/2 $. Using a calculator (or your favorite application or programming language), verify that $3 (\tan^{-1} (3/2))^2 > 2$. Since $3 x^2 \ge 3 (\tan^{-1} (3/2))^2 > 2 $ for $ \tan^{-1} (3/2) \le x \le \pi/2 $, we have $ f'(x) > 0 $ for $ 0 \le x \le \pi/2 $.
Without using calculator: This method would not require any calculator, but only knowledge of the values of $\sin x$ and $\cos x$ for $x = (\pi/4), (\pi/3)$, which follows from basic geometry. Like before, it is sufficient to show that $ f'(x) > 0 $ for $ \tan^{-1} (3/2) \le x \le \pi/2 $. Since $\tan^{-1} x$ is a strictly increasing function in $ 0 \le x \le \pi/2 $, it follows that $ \pi/4 = \tan^{-1} 1 < \tan^{-1} (3/2) < \tan^{-1} \sqrt{3} = \pi/3 $. Note that $ 3 x^2 \ge 3 (\pi/3)^2 > 3 $ for $\pi/3 \le x \le \pi/2$, and hence $f'(x) > 0$ for $\pi/3 \le x \le \pi/2$. So, it is sufficient to show that $f'(x) > 0$ for $\pi/4 \le x \le \pi/3$ to complete the proof. Note that for $\pi/4 \le x \le \pi/3$, $3x^2$ is an increasing function, while $(3 \cos x - 2 \sin x)$ is a decreasing function. Therefore, for $\pi/4 \le x \le \pi/3$, $f'(x) \ge$ $ 3 (\pi/4)^2 + (3 \cos (\pi/3) - 2 \sin (\pi/3))$ $> 3 (3/4)^2 + ((3/2) - \sqrt{3}) = (51/16) - \sqrt{3} > 3 - \sqrt{3} > 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4361050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Determinant of a special block matrix $\big(\begin{smallmatrix}A&B\\B&A\end{smallmatrix}\big)$ Let be $ A,B\in \mathbb{K}^{n,n} $ arbitrary matrices. Then
$$ \det\Bigg(\underbrace{\begin{pmatrix}A&B\\B&A\end{pmatrix}}_{=:L}\Bigg)=\det(A+B)\cdot \det(A-B) $$
My idea: I consider
$$ \begin{align}&\det(A+B)\cdot \det(A-B)\\[10pt]&=\det(A+B)\cdot 1\cdot \det(A-B)\cdot 1\\[10pt]&=\Big(\det(A+B)\cdot \det(I_n)\Big)\cdot \Big(\det(A-B)\cdot \det(I_n)\Big)\\[10pt]&=\det\left(\begin{pmatrix}A+B&0\\0&I_n\end{pmatrix}\right)\cdot \det\left(\begin{pmatrix}I_n&0\\0&A-B\end{pmatrix}\right)\\[10pt]&=\det\Bigg(\underbrace{\begin{pmatrix}A+B&0\\0&A-B\end{pmatrix}}_{=:M}\Bigg) \end{align}$$
So far so good. No I tried to find a matrix $X\in \mathbb{K}^{n,n}$ such that $ L=X^{-1}\cdot M\cdot X $ but I couldn't.
|
You can use this decomposition:
$$
\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1} \begin{pmatrix} A &B \\ B &A \end{pmatrix} \begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix} = \begin{pmatrix} A+B &0 \\ 0 &A-B \end{pmatrix}
$$ You can see the motivation for this by thinking of $A, B$ as real numbers and $I_n$ as $1$.
The only thing the careful might ask is whether $\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1}$ makes sense and indeed it does. You can check that $\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1} = \frac{1}{2}\begin{pmatrix} I_n &I_n \\ -I_n &I_n \end{pmatrix}$ by manually multiplying it with $\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}$ to get the $2n \times 2n$ identity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4361322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How many ways are there to prove that $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G?$ In my post, I had found the values of the couple of integrals,
$$
\int_{0}^{\frac{\pi}{4}} x \tan x d x=-\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(1)}
$$
and
$$
\int_{0}^{\frac{\pi}{4}} x \cot x d x=\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(2)}
$$
I just wonder what happens when I combine these 2 results and consequently find a wonderful result
$$
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G.
$$
Adding (1) and (2) yields $$
G=\int_{0}^{\frac{\pi}{4}} x \tan x d x+ \int_{0}^{\frac{\pi}{4}} x \cot x d x$$
$$
\begin{aligned}
\because x \tan x+x \cot x &=x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right) \\
&=\frac{x}{\cos x \sin x} \\
&=\frac{2 x}{\sin (2 x)}
\end{aligned}
$$
$$
\therefore \int_{0}^{\frac{\pi}{4}} x \tan x d x+\int_{0}^{\frac{\pi}{4}} x \cot x d x=\int_{0}^{\frac{\pi}{4}} \frac{2 x}{\sin (2 x)} d x
$$Letting $x\mapsto \frac{x}{2}$ yields
$$
G=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x
$$
Now we can conclude that
$$
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G.
$$
I am curious whether there are more elegant proofs.
Your comments and alternate solutions are warmly welcome.
|
Inspired by Mr Svyatoslav,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x &=4 \int_{0}^{\frac{\pi}{2}} x d\left(\ln \left(\tan \frac{x}{2}\right)\right) \\
&=4\left[x \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-4 \int_{0}^{\frac{\pi}{2}} \ln \left(\tan \frac{x}{2}\right) d x \\\\
&=-2 \int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x
\end{aligned}
$$
By my post
$$
\int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x=-G,$$
we can conclude that $$
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4361625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
Here is my initial question:
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ lead to an improvement to the upper bound $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$?
MOTIVATION
Here is a proof that $I(n^2) \leq 2 - \frac{5}{3q}$ holds in general.
Suppose to the contrary that $I(n^2) > 2 - \frac{5}{3q}$ is true.
Note that
$$I(n^2) = \frac{2}{I(q^k)} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg),$$
so that we have
$$I(n^2) > 2 - \frac{5}{3q} \iff 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) > 2 - \frac{5}{3q} \iff \frac{5}{3q} > 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) \iff 5q^{k+1} - 5 > 6q^{k+1} - 6q \iff 0 > q^{k+1} - 6q + 5,$$
which then implies that $k=1$. (Otherwise, if $k > 1$, we have
$$0 > q^{k+1} - 6q + 5 \geq q^6 - 6q + 5,$$
since $k \equiv 1 \pmod 4$, contradicting $q \geq 5$.) Now, since $k=1$, we get
$$0 > q^2 - 6q + 5 = (q - 1)(q - 5),$$
which implies that $1 < q < 5$. This contradicts $q \geq 5$. This concludes the proof.
Now, let $Q = 2 - \frac{5}{3q}$.
Since $I(q^k) < I(n^2)$, then we obtain
$$I(q^k) < I(n^2) \leq Q \iff (I(q^k) - Q)(I(n^2) - Q) \geq 0$$
$$\iff 2 + Q^2 = I(q^k)I(n^2) + Q^2 \geq Q(I(q^k) + I(n^2) \iff I(q^k) + I(n^2) \leq \frac{2}{Q} + Q.$$
But $\frac{2}{Q} + Q$ can be rewritten as
$$\dfrac{2}{Q} + Q = \dfrac{2}{2 - \dfrac{5}{3q}} + \Bigg(2 - \dfrac{5}{3q}\Bigg) = 3 - \Bigg(\dfrac{5}{3q} - \dfrac{5}{6q-5}\Bigg) = 3 - \dfrac{5(3q - 5)}{3q(6q - 5)} = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$
Let
$$f(q) = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$
Then the derivative
$$f'(q) = \dfrac{5}{3q^2} - \dfrac{30}{(6q - 5)^2}$$
is positive for $q \geq 5$. This means that $f$ is an increasing function of $q$, which implies that
$$I(q^k) + I(n^2) \leq \dfrac{2}{Q} + Q < \lim_{q \rightarrow \infty}{f(q)} = 3.$$
FINAL QUESTION
Can we do better? If that is not possible, can you explain why?
|
This is a partial answer.
Note that
$$I(n^2) = \dfrac{2}{I(q^k)} \leq \dfrac{2}{I(q)} = \dfrac{2q}{q+1},$$
since $k \equiv 1 \pmod 4$ implies that $k \geq 1$.
Following exactly the same steps in the original post, we obtain
$$I(q^k) < I(n^2) \leq \dfrac{2q}{q+1}.$$
After working out the details, we get
$$I(q^k) + I(n^2) \leq 3 - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg).$$
Set $Q' = 3 - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg)$.
Let us compute the difference $Q' - \Bigg(\dfrac{2}{Q} + Q\Bigg)$ and determine its sign for $q \geq 5$. We obtain
$$Q' - \Bigg(\dfrac{2}{Q} + Q\Bigg) = \dfrac{5(3q - 5)}{3q(6q - 5)} - \Bigg(\dfrac{q - 1}{q(q + 1)}\Bigg) = -\dfrac{(3q - 8)(q - 5)}{3q(q + 1)(6q - 5)} \leq 0,$$
with equality occurring if and only if $q = 5$.
This shows that $Q'$ is a better upper bound for $I(q^k) + I(n^2)$ than $\dfrac{2}{Q} + Q$.
In other words, $\dfrac{2}{Q} + Q$ is a trivial upper bound for $I(q^k) + I(n^2)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4362475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful trick to evaluate the integral.
Noting that
$$I(1):= \int_{0}^{\infty} \frac{d x}{x^{4}-x^{2}+1} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{2}}{x^{4}-x^{2}+1}
$$
Combining them yields
\begin{aligned}
I(1)&=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\\&= \frac{1}{2}\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&= \frac{1}{2}\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
&= \frac{1}{2}\tan ^{-1}\left(x-\frac{1}{x}\right)_{0}^{\infty} \\
&= \frac{\pi}{2}
\end{aligned}
Later, I started to investigate the integrands with higher powers.
Similarly,
$$
I(2):= \int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{2}} \stackrel{x \mapsto \frac{1}{x}}{=}\int_{0}^{\infty} \frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}} d x
$$
By division, we decomposed $x^6$ and obtain $$
\frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}}=\frac{x^{2}+1}{x^{4}-x^{2}+1}-\frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}
$$$$
I(2)=\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x-\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}dx
$$
We can now conclude that $$I(2)=I(1)=\frac{\pi}{2} $$
My Question:
How about the integral $$\displaystyle I_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}$$ for any integer $n\geq 3$?
|
Note
$$
\int_0^\infty \frac1{(x^4-x^2+1)^{n+1}}dx=\frac{(-1)^{n}}{n!} \frac{d^{n} J(a)}{da^{n}}\bigg|_{a=1}
$$
where
$$J(a)=\int_0^\infty \frac{dx}{x^4-x^2+a}=\frac1{2\sqrt a}
\int_0^\infty \frac{d(x-\frac{\sqrt a}x)}{(x-\frac{\sqrt a}x)^2+2\sqrt a-1}
= \frac\pi{2\sqrt{a(2\sqrt a-1)} }
$$
Then
\begin{align}
& \int_0^\infty \frac1{x^4-x^2+1}dx= J(1)=\frac\pi2\\
&\int_0^\infty \frac1{(x^4-x^2+1)^2}dx= -\frac{dJ(a)}{da}\bigg|_{a=1}=\frac\pi2\\
&\int_0^\infty \frac1{(x^4-x^2+1)^3}dx=\frac12\frac{d^2J(a)}{da^2}\bigg|_{a=1}=\frac{9\pi}{16}\\
&\int_0^\infty \frac1{(x^4-x^2+1)^4}dx=-\frac16\frac{d^3 J(a)}{da^3}\bigg|_{a=1}=\frac{21\pi}{32}\\
&\int_0^\infty \frac1{(x^4-x^2+1)^5}dx=\frac1{24} \frac{d^4 J(a)}{da^4}\bigg|_{a=1}=\frac{25\pi}{32}\\
&\int_0^\infty \frac1{(x^4-x^2+1)^6}dx=-\frac1{120} \frac{d^5 J(a)}{da^5}\bigg|_{a=1}=\frac{483\pi}{512}\\
&\hspace{5mm}\cdots\hspace{2mm}\cdots
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4367988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
}
|
Approximating factorial using identity $\frac{1}{x}!\frac{2}{x}!\cdots\cdot\frac{x}{x}!=\frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$ I created a function that describes the product of the inverse multiples of a factorial
$$ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdots\frac{x-1}{x}!\cdot\frac{x}{x}!$$
for some reasons i thought this function might be useful, thats why i'm posting an incomplete explanation to the best of my knowledge but after some rigorous calculation, i was able to express $m(x)$ as a formula
$$ m(x) = \frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$$
it turns out that the graph of $m(x)$ is a very simple one, makes me suggest it would be easy to interpolate and create a super factorial approximation from it
$$ {x}! = x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot m(x)\cdot\sqrt{{2\pi}^{-x}}$$
and it resembles striling's approximation
$${x}! \approx x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot e^{-x} $$
$$
m(1)=1, m(2)=\frac{1}{2}!, m(3)=\frac{4\pi}{3^{5/2}}, m(4)=\frac{3\pi^{3/2}}{2^{9/2}}, m(5)=\frac{96\pi^2}{5^{9/2}}, m(6)=\frac{45\pi^{5/2}}{3^{13/2}} , m(7)=\cdots
$$
now my question, can you help with the approximation, does the function gives any more information
|
The Gauss multiplication formula states that
$$
\prod\limits_{k = 0}^{n - 1} {\Gamma\! \left( {z + \frac{k}{n}} \right)} = (2\pi )^{(n - 1)/2} n^{1/2 - nz} \Gamma (nz)
$$
for $n\geq 1$ and for all complex $z$ for which both sides are defined. Taking $z=1$ and using $r!=\Gamma(1+r)$ gives
$$
\prod\limits_{k = 1}^n {\left( {\frac{k}{n}} \right)!} = \prod\limits_{k = 0}^{n - 1} {\left( {\frac{k}{n}} \right)!} = \frac{{(2\pi )^{(n - 1)/2} n!}}{{n^n \sqrt n }}.
$$
Thus your result is a special case of the Gauss multiplication formula. An asymptotic expansion for $m$ coming from the Stirling series is
$$
m(x) \sim \left( {\frac{{\sqrt {2\pi } }}{e}} \right)^x \left( {1 + \frac{1}{{12x}} + \frac{1}{{288x^2 }} - \frac{{139}}{{51840x^3 }} - \cdots } \right)
$$
as $x\to +\infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4371422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Show using induction that for every $n\in \mathbb{N}$ the number $P_n=2^{2^{n+1}}+2^{2^n}+1$ is divisible by $21$. Show using induction that for every $n\in \mathbb{N}$ the number $P_n=2^{2^{n+1}}+2^{2^n}+1$ is divisible by $21$.
So the first step is to check for $n=1$ $$P_1=2^{2^2}+2^2+1=16+5=21,$$ which is divisible by $21$, so the statement holds for $n=1$. Let for some $n=k\ge1,P_k=2^{2^{k+1}}+2^{2^k}+1$ is divisible by $21$. Now we should try to prove that the statement also holds for $n=k+1$, or show that $P_{k+1}=2^{2^{k+2}}+2^{2^{k+1}}+1$ is divisible by $21$. We can write $P_{k+1}$ as $$P_{k+1}=2^{2\cdot2^{k+1}}+2^{2\cdot2^k}+1$$ How do we use the induction hypothesis?
|
Let $A_n=2^{2^n}$. Then $P_n=2^{2^{n+1}}+2^{2^n}+1=A_n^2+A_n+1$ and...
$$P_{n+1}=2^{2^{n+2}}+2^{2^{n+1}}+1=2^{4\cdot2^n}+2^{2\cdot2^n}+1=A_n^4+A_n^2+1\\=(A_n^2+A_n+1)(A_n^2-A_n+1)=P_n(A_n^2-A_n+1)$$
And $P_1=21$. So... the answer is clear! If $21\mid P_n$, then automatically $21\mid P_{n+1}$ holds!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4371727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If x and y are real numbers such that $4x^2+2xy+9y^2=100$ then what are all possible values of $x^2+2xy+3y^2$ If $x$ and $y$ are real numbers and $4x^2+2xy+9y^2=100$ then what are all possible values of $x^2+2xy+3y^2$.
The first thing I did with this equation was to try and factor the first equation, but that didn't work so I tried to simplify it into a standard elliptical equation
$$\left(\frac{x}{5}\right)^2+\frac{xy}{50}+\left(\frac{3y}{10}\right)^2=1$$
Then I noticed that the numerator of the $x$ coefficient and the coefficient lined up but i couldn't get the denominators to cancel out.
Now I'm stuck and don't know how I should progress. Any help would be appreciated.
|
From $4 x^2 + 2 x y + 9 y^2 = 100 $, by defining $ r= [x, y]^T $ we can write
$ r^T Q r = 100 $
where $ Q = \begin{bmatrix} 4 && 1 \\ 1 && 9 \end{bmatrix} $
Diagonlizing $Q$, we can factor it into $ Q = R D R^T $
We get
$D = \begin{bmatrix} \frac{1}{2} ( 13 - \sqrt{29} ) && 0 \\ 0 && \frac{1}{2} ( 13 + \sqrt{29}) \end{bmatrix} $
And,
$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$
where $\theta = \frac{1}{2} \tan^{-1}(- \dfrac{2}{ 5} )$
So now we have the following
$ r^T R D R^T r = 100 $
Define $z = R^T r $ , then $ r = R z $
From which we know that,
$ D_{11} z_1^2 + D_{22} z_2^2= 100 $
Thus,
$ z_1 = \dfrac{10}{\sqrt{D_{11}}} \cos \theta $
$ z_2 = \dfrac{10}{\sqrt{D_{22}}} \sin \theta $
where $\theta \in \mathbb{R}$ is a parameter.
Now, $ r = R z = 10 R D^{-1/2} u $ where $ u = [ \cos \theta , \sin \theta ] ^T $
And the function that we have is
$ f(x, y) = x^2 + 2 x y + 3 y^2 = r^T Q_f r $
where $Q_f = \begin{bmatrix} 1 && 1 \\ 1 && 3 \end{bmatrix} $
Plugging $r$ from above,
$ f(u) = 100 u^T D^{-1/2} R^T Q_f R D^{-1/2} u $
Since $u$ is a unit vector, the range of values of $f$ is $[100 \lambda_{min}(Q_1), 100 \lambda_{max}(Q_1)] $ , where
$Q_1 = D^{-1/2} R^T Q_f R D^{-1/2} $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4376993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Can the value $\lim \frac{\sin x}{x} = 1$ be used even if $x$ is not approaching $0$? I was going through a solved example that involved calculating the coefficient $a$ for a piecewise function to be continuous at $x = 0$:
$f(x) = a \sin{\frac{\pi}{2} (x + 1)}$ if $x \le 0$
$f(x) = \frac{\tan{x} - \sin{x}}{x^3}$ if $x \gt 0$
For calculating the left-hand limit, the example proceeds by multiplying both the numerator and denominator by $\frac{\pi}{2} (x + 1)$ to get the $\sin x / x$ form and ends up with $\frac{a \pi}{2}$. Is this still applicable even when $x$ is not approaching $0$? Why can’t I just substitute $x = 0$ for the function since $0$ is included its domain and it reduces to $\frac{\sin{\pi / 2}}{\pi / 2}$, which is defined?
|
Let $f(x)$ a function such that $f(x)\to0$ when $x\to x_0\,\in\,\bar{\mathbb{R}}$, then:
$$\sin(f(x))\,\,\sim\,\,f(x)$$
This results can be shown using composite function limit theorem.
In your case, we are to study the continuity of:
$$f(x)=\begin{cases}
a \sin{\frac{\pi}{2} (x + 1)} & \text{ if } x\leq 0 \\
\frac{\tan{x} - \sin{x}}{x^3}& \text{ if } x > 0
\end{cases}$$
We must have:
$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)=f(0)$$
So:
$$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}a \sin{\frac{\pi}{2} (x + 1)}=a$$
Because when $x\to 0$, we have $\frac{\pi}{2} (x + 1)\to \frac{\pi}{2}$ and $\sin\left(\frac{\pi}{2}\right)=1$.
Also:
$$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{\tan{x} - \sin{x}}{x^3}=\lim_{x\to 0^+}\frac{\sin(x)}{x}\cdot\frac{\cos(x)-1}{\sin^2(x)}\,\,\sim\,\,\lim_{x\to 0^+}1\cdot\frac{-\frac{1}{2}x^2}{x^2}=-\frac{1}{2}$$
And:
$$f(0)=a$$
So, $f(x)$ is continous in $x=0$ if and only if $a=-\frac{1}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4377698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find two 2 permutations of the numbers of set $\{\frac 11,\frac 12,\dots,\frac 1n\}$ with certain properties. The problem is: Let $n>1$ be a natural numbers, $(a_1,a_2,\dots,a_n)$ and $(b_1,b_2,…,b_n)$ are $2$ permutations of the numbers in set $\{\frac 11,\frac 12,\dots,\frac 1n\}$, satisfying $a_i+b_i\ge a_j+b_j$ for all $i<j$. Does there exist $2$ permutations in which $a_i$ is not equal to $b_i$ for all $i$ and $\frac{a_1+b_1}{a_n+b_n}$ is an integer? I try for small $n$ like $3$ or $4$ and it seems to be there is no such permutation to satisfy. Can you guys help me?
|
Consider the counterexample:
\begin{align}
(a_1,a_2,\ldots,a_{12}) &= \bigg(\frac{1}{1},\frac{1}{4},\frac{1}{2},\frac{1}{3}, \frac{1}{5}, \frac{1}{9}, \frac{1}{6}, \frac{1}{10}, \frac{1}{7}, \frac{1}{11}, \frac{1}{8}, \frac{1}{12} \bigg) \\
(b_1,b_2,\ldots,b_{12}) &= \bigg(\frac{1}{4},\frac{1}{1},\frac{1}{3},\frac{1}{2}, \frac{1}{9}, \frac{1}{5}, \frac{1}{10}, \frac{1}{6}, \frac{1}{11}, \frac{1}{7}, \frac{1}{12}, \frac{1}{8} \bigg)
\end{align}
It is easy to see that $a_i+b_i \geqslant a_j+b_j$ for all $i<j$. Moreover:
$$\frac{a_1+b_1}{a_n+b_n} = \frac{\frac{1}{1}+\frac{1}{4}}{\frac{1}{8}+\frac{1}{12}} = \frac{\frac{5}{4}}{\frac{5}{24}} = 6$$
So what's the intuition behind this construction? We want to make sure that the numerator of $a_n+b_n$ is as small as possible to maximize our chance of getting $\frac{a_1+b_1}{a_n+b_n}$ to be an integer. Assuming one of $a_n$ and $b_n$ is $\frac{1}{n}$, we could try making the other as $\frac{1}{n/2}$. However, this will interfere with our condition that $a_i+b_i \geqslant a_j+b_j$ for all $i < j$ (I'll let you figure out why!). Thus, we could instead try keeping $\frac{1}{n}$ and $\frac{1}{2n/3}$. Then:
$$a_n+b_n = \frac{1}{n} + \frac{1}{2n/3} = \frac{5}{2n}$$
Now, we need to make sure that $5$ divides the numerator of $a_1+b_1$. It is easy to see that one of $a_1$ and $b_1$ must be $1$, so we can try taking the other as $\frac{1}{4}$. Then, we have:
$$\frac{a_1+b_1}{a_n+b_n} = \frac{\frac{1}{1}+\frac{1}{4}}{\frac{1}{2n/3}+\frac{1}{n}} = \frac{\frac{5}{4}}{\frac{5}{2n}} = \frac{n}{2}$$
We need to make sure that $\frac{n}{2}$ and $\frac{2n}{3}$ are integers, so $6 \mid n$. This strategy won't work for $n=6$ because $\frac{2n}{3} = 4$. For $n=12$, we can use the pairing trick done in the counterexample provided above to achieve the required.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4379227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations and I have a rather complicated question question that I could not write as the sum of squares:
$$ 0\leq 6x^2y^2z^2 +19(x^3y^3+y^3z^3+z^3x^3)+27(x^4yz+xy^4z+xyz^4)-24(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3).$$
Thanks for any ideas that might help me clarify the issue.
|
Another solution based on what you have already obtained and AM-GM:
It is easy to check that;
$$x^2y^2z^2+x^4yz+x^3y^3 \geq3x^3y^2z.$$
So we need to show that;
$$17(x^3y^3+x^3z^3+z^3y^3)+25(x^4yz+xy^4z+xyz^4)\geq21(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Now realize that we simply get;
$$x^3y^3+x^3y^3+x^3y^3+x^4yz+x^4yz+xyz^4\geq6x^3y^2z,$$
by adding all the similar inequalities together, we will have;
$$(x^3y^3+x^3z^3+z^3y^3)+(x^4yz+xy^4z+xyz^4)\geq(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Hence, it is just needed to prove that;
$$8(x^4yz+xy^4z+xyz^4)\geq4(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Equivalently, we should show that;
$$2xyz(x^3+y^3+z^3)\geq xyz(xy(x+y)+xz(x+z)+yz(y+z)).$$
And this is almost obvious since $x^3+y^3\geq xy(x+y).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4381204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
What is the decomposition into even and odd function of a function equal to $x$ when $x \geq 0$ and equal to zero when $x <0$? Every function $f$ with domain in $\mathbb{R}$ can be written
$$f=E+O$$
where $E$ is an even function and $O$ is an odd function.
Proof
Assume $f(x) = E(x) + O(x)$.
Then
$$f(-x)=E(-x) + O(-x)=E(x)-O(x)$$
Therefore, given a function $f$,
$$f(x) = E(x) + O(x)$$
$$f(-x)=E(x)-O(x)$$
represent a system of two equations in two unknowns.
We can solve for
$$E(x) = \frac{f(x) + f(-x)}{2}$$
$$O(x) = \frac{f(x) - f(-x)}{2}$$
This concludes the proof.
Now consider a function
$$f(x) = \begin{cases} 0\text{ if } x < 0 \\
x\text{ if } x \geq 0
\end{cases}$$
What do $E(x)$ and $O(x)$ look like?
$$E(x) = \begin{cases} \frac{0+x}{2}=\frac{x}{2}\text{ if } x<0 \\
\frac{x+0}{2}=\frac{x}{2}\text{ if } x \geq 0
\end{cases}$$
Isn't the function $E(x)=\frac{x}{2}$ odd?
Similarly, we reach
$$O(x) = \begin{cases} \frac{0-x}{2}=-\frac{x}{2}\text{ if } x<0 \\
\frac{x-0}{2}=\frac{x}{2}\text{ if } x \geq 0
\end{cases}$$
Isn't $O(x)$ even?
I must be missing something very silly here.
|
True, I felt a bit weird about it too. I should have made the proof something like as follows. Let $f$ be any function. Let $E(x)=\frac{f(x)+f(-x)}{2}$ and let $O(x)=\frac{f(x)-f(-x)}{2}$. Then, I easily show that $E$ is even, $O$ is odd, and $f=E+O$. Thus we can say $f=E+O$ with $E$ even and $O$ odd, for any function $f$.
Yes, exactly. You'd originally (in the Question) proved the required theorem's converse instead. These two parts together show that every function has a unique decomposition into a pair of even and odd functions.
$$E(x) = \begin{cases} \frac{0+x}{2}=\frac{x}{2}\text{ if } x<0 \\
\frac{x+0}{2}=\frac{x}{2}\text{ if } x \geq 0
\end{cases}$$ Isn't the function $E(x)=\frac{x}{2}$ odd?
Correction: $$E(x) = \begin{cases} \frac{0+(-x)}{2}=-\frac{x}{2}\text{ if } x<0 \\
\frac{x+0}{2}=\frac{x}{2}\text{ if } x \geq 0
\end{cases}.$$
$$O(x) = \begin{cases} \frac{0-x}{2}=-\frac{x}{2}\text{ if } x<0 \\
\frac{x-0}{2}=\frac{x}{2}\text{ if } x \geq 0
\end{cases}$$
Correction: $$O(x) = \begin{cases} \frac{0-(-x)}{2}=\frac{x}{2}\text{ if } x<0 \\
\frac{x-0}{2}=\frac{x}{2}\text{ if } x \geq 0
\end{cases}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4381416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Simplify $\sum n(n+1)$ According to WolframAlpha,
$$\sum_{n=1}^k n(n+1)=\frac{1}{3}k(k+1)(k+2)$$
and it is easy to verify this if we use induction.
However, I would like to know how one can actually come up with this, other than by thinking about how to force the terms to cancel out.
I tried: Since $k(k+1)/2=1+2+\cdots+k$,
$$\begin{align}
\sum_{n=1}^k n(n+1) &= 2\bigg(1(k)+2(k-1)+3(k-2)+\cdots+k(1)\bigg)\\
& = 2\sum_{n=1}^k n(k-n+1)\\
\sum_{n=1}^k n(n+1) &= 2\bigg(1(k)+2(k-1)+3(k-2)+\cdots+k(k-(k-1))\bigg)\\
& = 2\bigg( (1+2+3+\cdots+k)k-(0\cdot1+1\cdot2+2\cdot3+\cdots+(k-1)k) \bigg)\\
& = k^2(k+1)-2\sum_{n=1}^{k-1} n(n+1)\\
& = \sum_{n=1}^k (-2)^{k-n}n^2(n+1)
\end{align}$$
I feel like I'm only making it worse..
|
Note that $n(n+1) = 2\bigg(\frac{n(n+1)}{2!}\bigg)=2\binom{n+1}{2}$. Then the sum becomes ....
$$\begin{align}
S(k) &= \sum_{n=1}^k{n(n+1)}\\
&= 2\sum_{n=1}^k{\binom{n+1}{2}}\\
(1) &= 2\binom{k+2}{3}\\
&= 2\cdot\frac{k(k+1)(k+2)}{3!}\\
&= \frac{k(k+1)(k+2)}{3}\\
\end{align}$$
Where $(1)$ is just the hockey-stick identity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4382770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Computing explicit Riesz projection Consider the matrix:
$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $
which has eigenvalue $1$ with algebraic multiplicity $2$ and geometric multiplicity $1$.
I am trying to explictly construct the Riesz projection
\begin{align*}
P = - \frac{1}{2 \pi i} \int_{\gamma} \frac{1}{A - z} d \gamma
\end{align*}
where $\gamma$ is any curve enclosing the spectrum $\{1\}$. How to compute it?
|
It seems easiest to pick the contour $\gamma: \lbrack 0, 2 \pi \rbrack \to \mathbb{C} $ defined by
\begin{align*}
\gamma( \theta) = 1+ e^{i \theta}
\end{align*}
which is the circle of unit radius around $1$ and $\gamma'(\theta) = i e^{i \theta} $.
Now, to evaluate the contour integral
\begin{align*}
- \frac{1}{2 \pi i} \int_{0}^{2\pi} \frac{i e^{i \theta}}{ \begin{pmatrix} - e^{i \theta} & 1 \\ 0 & - e^{i \theta} \end{pmatrix}} d \theta
& = - \frac{1}{2 \pi i} \int_{0}^{2\pi} \frac{i e^{i \theta}}{e^{2i\theta}} \begin{pmatrix} - e^{i \theta} & - 1 \\ 0 & - e^{i \theta} \end{pmatrix} d \theta
= \frac{1}{2 \pi } \int_{0}^{2\pi} \begin{pmatrix} 1 & e^{-i \theta} \\ 0 & 1 \end{pmatrix} d \theta \\
& = \frac{1}{2 \pi } \begin{pmatrix} \int_{0}^{2\pi} 1 d\theta & \int_{0}^{2\pi} e^{-i \theta} d \theta \\ 0 & \int_{0}^{2\pi} 1 d \theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.
\end{align*}
Which is a projection onto the entire $\mathbb{C}^2$ as expected.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4382900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.
First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $2$ that isn't the multiple of $4$. Since the problem is asking for a minimal case that satisfies these conditions, $n+1$ should be the multiple of $4$, $n+2$ should be the multiple of $3$, and $n+3$ should be the multiple of $2$, thus giving $4 \cdot 3^2 \cdot 2^3 = 288.$
However, I am unsure as to whether or not this answer is correct, as testing cases has consistently yielded higher GCDs than this.
|
Following the comment of lulu, look at numbers of the form $n=19+24k$.
For $k=0$, $$(n+1)(n+2)^2(n+3)^3(n+4)^4=20\cdot 21^2\cdot 22^3\cdot 23^4$$ which has the factorization $2^5\cdot 3^2\cdot 5\cdot 7^2\cdot 11^3\cdot 23^4$ and is not divisible by any number containing more than five factors of $2$ and two factors of $3$
For $k=1$, the product $44\cdot 45^2 \cdot 46^3\cdot 47^4$ has no factors of $7$, so $7$ cannot be a factor of the divisor you seek.
For $k=2$, the product $68\cdot 69^2 \cdot 70^3\cdot 71^4$ has no factors of $11$, so $11$ cannot be a factor of the divisor you seek.
For $k=4$, the product $116\cdot 117^2 \cdot 118^3\cdot 119^4$ has no factors of $5$ or $23$, so neither $5$ nor $23$ can be a factor of the divisor you seek.
So the only factor of $20\cdot 21^2\cdot 22^3\cdot 23^4$ that can and must appear in the divisor you seek is $2^53^2$.
It's hard to see that in comparing two examples at a time, as many of them will have a $5$ or a $7$, as well as the occasional larger prime factor in common.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4385677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Distributing 6 marbles among 3 containers What's the probability of all containers being non-empty?
Interpretation: All configurations are equiprobable.
Each container can contain min. 0 and max. 6
marbles.
Total number of configurations is implied by both the condition of equiprobability
and the saturation condition: $3^{6}$.
First I place one of marbles into the first container
and so on till each container has exactly one marble.
That is $6\cdot 5 \cdot 4$ cases.
There remain 3 marbles.
I can distribute them like this: |** |** |** | => $3!$
or this: |**** |* |* | => $3$
or like this: |*** |** |* | => $2 \cdot \binom{3}{2} \binom{3}{2}$
Therefore the number of favorable configurations is
$6\cdot 5 \cdot 4 \cdot \left[3 + 3! + 2 \cdot \binom{3}{2} \binom{3}{2}\right]$
The probabilty we seek:
$\frac{1}{3^{6}}\cdot 6\cdot 5 \cdot 4 \cdot \left[3 + 3! + 2 \cdot \binom{3}{2} \binom{3}{2}\right]$
|
Your numerator is larger than your denominator, so your result cannot represent a probability. It is best to avoid choosing representatives of a set, then selecting additional members of a set since that leads to over counting.
Method 1: There are three choices for each marble, so there are $3^6$ ways to distribute the marbles.
For the favorable cases, we need to subtract those configurations in which there are one or more empty containers.
There are three ways to select a container to be left empty and $2^6$ ways to distribute the marbles to the remaining two containers.
However, if we subtract this amount from $3^6$, we will have subtracted each case in which there are two empty containers twice, once for each way of designating one of the empty containers as the empty containers. We only wish to subtract these three cases once, so we must add them to the total. Hence, the number of favorable cases is
$$3^6 - \binom{3}{1}2^6 + \binom{3}{2}1^6$$
by the Inclusion-Exclusion Principle.
Therefore, the probability that all containers are non-empty is
$$\Pr(\text{all containers are non-empty}) = \frac{3^6 - \dbinom{3}{1}2^6 + \dbinom{3}{2}1^6}{3^6}$$
Method 2: Alternatively, observe that there are three ways to express $6$ as a sum of exactly three positive integers:
\begin{align*}
6 & = 4 + 1 + 1\\
& = 3 + 2 + 1\\
& = 2 + 2 + 2
\end{align*}
Four marbles in one container, with one each in the other two containers: There are three ways to select the container which receives four marbles, $\binom{6}{4}$ ways to select which four of the six marbles are placed in that container, and two ways to distribute the remaining marbles to the remaining two containers. Hence, there are
$$\binom{3}{1}\binom{6}{4}\binom{2}{1}\binom{1}{1}$$
such distributions.
Three marbles in one container, two marbles in another container, and one marble in the other container: There are three ways to select the container which receives three marbles, $\binom{6}{3}$ ways to select which three of the six marbles that container receives, two ways to select the container which receives two marbles, $\binom{3}{2}$ ways to select which two of the remaining three marbles are placed in that container, and one way to place the remaining marble in the remaining container. Hence, there are
$$\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}\binom{1}{1} = 3!\binom{6}{3}\binom{2}{1}\binom{1}{1}$$
such distributions.
Two marbles are placed in each of the three containers: There are $\binom{6}{2}$ ways to place two of the six marbles in the first container, $\binom{4}{2}$ ways to place two of the remaining four marbles in the second container, and one way to place both of the remaining marbles in the third container. Hence, there are
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
such configuration.
Hence,
$$\Pr(\text{all containers are non-empty}) = \frac{\dbinom{3}{1}\dbinom{6}{4}\dbinom{2}{1}\dbinom{1}{1} + 3!\dbinom{6}{3}\dbinom{3}{2}\dbinom{1}{1} + \dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}{3^6}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4385983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Maximize $\lambda$ over $a^2 + b^2 + c^2 + d^2 \ge ab + \lambda bc + cd$ and $a,b,c,d\geq 0$. Find the largest real number $\lambda$ such that
$$a^2 + b^2 + c^2 + d^2 \ge ab + \lambda bc + cd$$ for all nonnegative real numbers $a,$ $b,$ $c,$ $d.$
I tried using AM-GM on like $a^2+\dfrac{b^2}{4}\geq ab$, and I tried combining the results, but it didn't get me anywhere. Could someone give me some guidance on how to proceed?
Thanks in advance!!!
|
You're very much on the right track. You can use
\begin{align*}
a^2+b^2+c^2+d^2
&=\left(a^2+\frac{b^2}4\right)+\left(\frac{c^2}4+d^2\right)+\frac{3(b^2+c^2)}4\\
&\geq ab+cd+\frac{3\cdot 2bc}4
\end{align*}
to get that $\lambda=3/2$ works. Now, you just need to determine if any larger $\lambda$ will work. Can you look at the equality cases of the above process to show that any larger $\lambda$ will fail?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4392427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Given diagonals in a parallelogram. Find sides. Given a parallelogram with $d_1 = AC = 26$ cm, $d_2 = BD = 18$ cm and $\sin \displaystyle \angle AOD = \frac{12}{13}$.
Find $AB = a$ and $AD = b$.
What I did in order to solve it
*
*Using the formula for the area
$S = \frac{d_1d_2\sin \displaystyle \Phi}{2} = \frac{26 \times 18 \times 12/13}{2} = 216$
*
*Using Heron's formula for $\triangle BOC$ to find $b$
Defining $OC = a = 26 / 2 = 13$, $OB = b = 18 / 2 = 9$, $BC = c$
$p = \frac{a + b +c}{2} = \frac{13 + 9 + c}{2} = \frac{22 + c}{2}$
$S = \sqrt{p(p-a)(p-b)(p-c)}$
I believe when I find $b$, I will able to find $a$ too, but kinda stuck at that point.
|
$\cos AOD =\pm \sqrt{1-\sin^2 AOD}=\pm \sqrt{1-(12/13)^2}=\pm 5/13$.
Since $AOD < 90^\circ$, $\cos AOD = 5/13$
So $b^2=AO^2+OD^2-2 AO\cdot OD \cos AOD = 13^2+9^2-2\cdot 13\cdot 9\cdot (5/13)=160$.
Since $\sin DOC=\sin(180^\circ-AOD)=\sin AOD=12/13$, we also have $\cos DOC=\pm 5/13$, but this time we choose the minus sign since $DOC>90^\circ$, so $\cos DOC = -5/13$
So $a^2=DO^2+OC^2-2 DO\cdot OC \cos DOC = 9^2+13^2-2\cdot 9\cdot 13\cdot (-5/13)=340$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4395766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order and prove that $\frac{1}{\pi}\arctan(\frac{4}{3})$ is irrational.
by contradiction $\exists n$ s.t. $(\frac{3}{5}+\frac{4}{5}i)^n=1$
$(3+4i)^n=5^n$
when $n=2$, $(3+4i)^2=3+4i\pmod5$
stuck how prove that $(3+4i)^n$=$3+4i\pmod5$ use induction? if yes how in this case?
for second part of the question again by contradiction.
$\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$
$\phi=\arctan(\frac{4}{3})=\frac{\pi m}{n}$ how continue from here?
|
For the second part of the question: O.P. assumed that $\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$ is rational. Then, $\cos(\frac{m}{n}\pi)=\frac{3}{5}$.
Let $U_{n-1}(x)$ be the Chebyshev polynomial of second kind of order $n-1$. Then $U_{n-1}(\frac{3}{5})=\frac{\sin m\pi)}{\sin(\frac{m}{n}\pi)}=0.$ We will show that this is impossible giving us a contradiction.
From https://en.wikipedia.org/wiki/Chebyshev_polynomials, I got
$$U_{n}(x)=\frac{(x+\sqrt{x^2-1})^n-(x-\sqrt{x^2-1})^n}{2\sqrt{x^2-1}}.$$
I am not sure if this is for all $x$. I hope so. Then, when we put $x=\frac{3}{5}$ we get
$$(3+4i)^{2n}=5^{2n}$$
which is not possible due to the answer of S. Eberhard.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4398533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to solve a difference equation $p(a,b)=[p(a-1,b)+p(a,b-1)]/2$ from the gambler's ruin problem with several boundary conditions Recently I read a probability book and it is going to deduce pascal distribution from the famous gambler's ruin problem,where person A need to win for $a$ times or person B to win for $b$ times to end up games.
$p(a,b)$ refers to in such situation the probability that A wins the final game.
The writer give a difference equation:$$p(a,b)=[p(a-1,b)+p(a,b-1)]/2 \quad \tag{1}$$ and it boundary conditions:
$$p(0,b)=1,\ p(a,0)=0,\ p(a,a)=1/2$$
I understand the probability meaning behind the equation and boundary conditions.But I have no idea how to solve a difference linear equation with two variables,the result is $$p(a,b)=\sum_{i=a}^{a+b-1}\binom{a+b-1}{i}(\frac{1}{2})^{a+b-1}$$
I have tried using generating function $G(x,y)=\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}p(a,b)x^ay^b$,
\begin{align*}
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}p(a,b)x^ay^b&=G(x,y)-\sum_{a=0}^{\infty}p(a,0)x^a-\sum_{b=0}^{\infty}p(0,b)y^b+p(0,0) \quad \tag{2}\\
&=G(x,y)-\sum_{b=0}^{\infty}y^b+\frac{1}{2}\\
&=G(x,y)-\frac{1}{1-y}+\frac{1}{2}\\
\end{align*}
then I use (1) and (2) as LHS, the RHS :
\begin{align*}
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}p(a-1,b)x^ay^b&=\sum_{a=0}^{\infty}\sum_{b=1}^{\infty}p(a,b)x^{a+1}y^b\\
&=x[G(x,y)-\sum_{a=0}^{\infty}p(a,0)x^a]\\
&=xG(x,y)
\end{align*}
and
\begin{align*}
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}p(a,b-1)x^ay^b&=\sum_{a=1}^{\infty}\sum_{b=0}^{\infty}p(a,b)x^{a}y^{b+1}\\
&=y[G(x,y)-\sum_{b=0}^{\infty}p(0,b)y^b]\\
&=yG(x,y)-\frac{y}{1-y}
\end{align*}
I solve the $G(x,y)$ is:
$$G(x,y)=\frac{1}{(y-1)(x+y-2)}$$
But it's still hard to expand $G(x,y)$ to series to get the $p(a,b)$.
|
Boundary conditions. If we assume $p(a,0)=0$ and $p(0,b)=1$ for $a,b>0$, together with the equation $2p(a,b)=p(a-1,b)+p(a,b-1)$ as in $(1)$, then we have $p(a,b)+p(b,a)=1$ for $(a,b)\neq(0,0)$, shown using induction. Thus, $p(a,a)=1/2$ holds for $a>0$; it need not be given. Assume $p(0,0)=q$ is also given (it doesn't play any significant role however).
Generating function. Note that $G(x,0)=\sum_{a=0}^\infty p(a,0)x^a=p(0,0)=q$ should not depend on $x$, while your computed result still does. You seem to assume $q=1/2$ initially, but $q=0$ in
\begin{align*}
\sum_{a,b>0}p(a-1,b)x^ay^b=x\big(G(x,y)-G(x,0)\big)&=x\big(G(x,y)-q\big),
\\
\sum_{a,b>0}p(a,b-1)x^ay^b=y\big(G(x,y)-G(0,y)\big)&=y\left(G(x,y)-q-\frac{y}{1-y}\right).
\end{align*}
So, the corrected equation is $$G(x,y)=q+\frac{y}{1-y}+\frac{x}{2}\big(G(x,y)-q\big)+\frac{y}{2}\left(G(x,y)-q-\frac{y}{1-y}\right),$$ the solution of which is $$G(x,y)=q+\frac{y(2-y)}{(1-y)(2-x-y)}.$$ As expected, $p(a,b)$ don't depend on $q$ for $(a,b)\neq(0,0)$. Assume $q=0$ from now on.
Getting an answer. This can be done in many ways, with results looking differently.
The stated one may be obtained using $\sum_{n\geqslant 0}\binom{n+m}{n}z^n=(1-z)^{-m-1}$:
\begin{align*}
G(x,y)&=\frac{2y}{2-y}\left(1-\frac{x}{2-y}\right)^{-1}\left(1-\frac{y}{2-y}\right)^{-1}
\\&=\frac{2y}{2-y}\sum_{m,n\geqslant 0}\left(\frac{x}{2-y}\right)^m\left(\frac{y}{2-y}\right)^n
\\&=y\sum_{m,n\geqslant 0}x^m y^n 2^{-m-n}(1-y/2)^{-m-n-1}
\\&=y\sum_{m,n,k\geqslant 0}x^m y^n 2^{-m-n}\binom{m+n+k}{k}(y/2)^k
\\\color{gray}{[n+k=s]}&=y\sum_{m,s\geqslant 0}x^m y^s 2^{-m-s}\sum_{k=0}^s\binom{m+s}{k},
\end{align*}
so that $2^{a+b}p(a,b+1)=\sum_{k=0}^{b}\binom{a+b}{k}=\sum_{k=0}^{b}\binom{a+b}{a+b-k}=\sum_{k=a}^{a+b}\binom{a+b}{k}$ as expected.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4399086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Galois group of splitting field of $x^4-16x^2+4$ isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$ I've already shown irreducible. I know I need to get intermediate fields since they correspond to subgroups (which I can use for isomorphisms). But how do I get the intermediate fields here?
edit: I'm thinking perhaps there is some way to show the order of the extension is $4$, and the only groups of order $4$ are the Klein-$4$ group and cyclic group and eliminating from there.
edit 2:
Treating $x^4-16x^2+4$ as a quadratic in $x^2$, applying the quadratic formula gives $$\frac{16\pm \sqrt{16^2-4\cdot 4}}{2}=\frac{16\pm 4\sqrt{15}}{2}=8\pm 2\sqrt{15}=(\sqrt{3}\pm\sqrt{5})^2.$$ and this is squared so $\pm(\sqrt{3}\pm\sqrt{5})$ gives the roots and there are four of them. Now what's next..
|
$\dfrac{1}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{5}+\sqrt{3}}{2}$.
Therefore, any root generates all three other roots. Therefore, the splitting field is of a degree-$4$ extension of $\mathbb{Q}$.
Now, since you want three distinct intermediate subfields of degree $2$,
$(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})=2\sqrt{5}$
$(\sqrt{5}+\sqrt{3})-(\sqrt{5}-\sqrt{3})=2\sqrt{3}$
$(\sqrt{5}+\sqrt{3})^2=8+ 2\sqrt{15}$
so you can use $\sqrt{5},\sqrt{3},\sqrt{15}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4400642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Continued fraction for $\sqrt 2$ I discovered this continued fraction for $\sqrt 2$ but could not find any sources in which it appeared. It goes as follows:
\begin{equation}
\sqrt{2} = 1 + \frac{1 + \frac{1 + \frac{1 + \dots}{3 + \dots}}{3 + \frac{1 + \dots}{3 + \dots}}}{3 + \frac{1 + \frac{1 + \dots}{3 + \dots}}{3 + \frac{1 + \dots}{3 + \dots}}}
\end{equation}
I can show that it approaches $\sqrt{2}$ via a computer program, but see no feasible way to prove it.
|
Let $x = \frac {1 + \frac {1 + \cdots}{3+\cdots}}{3 + \frac {1 + \cdots}{3+\cdots}}$
$x = \frac {1+x}{3+x}\\
x^2+ 3x = 1+x\\
x^2 + 2x - 1 = 0\\
x = -1 \pm \sqrt {2}$
We know that $x>0$ so we can discard the negative value.
$1 + x = \sqrt 2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4405010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$ Question
Evaluate $$\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$$
Here is my work (and where I stopped)
Solution found on Internet
(I didn’t understand it at all to be honest)
|
You should get $Li_{2}(x)$ function
https://mathworld.wolfram.com/Dilogarithm.html
$\int{\frac{ln(x)-x}{x^2-1}dx} = \int{\frac{ln{(x)}}{x^2-1}}-\int{\frac{x}{x^2-1}dx}\\
=-\int{\frac{ln{(x)}}{1-x^2}}dx - \frac{1}{2}\int{\frac{2x}{x^2-1}dx}\\
=-\int{\frac{ln{(x)}}{(1-x)(1+x)}}dx - \frac{1}{2}\ln{|x^2-1|} \\
-\frac{1}{2}\int{\frac{2ln{(x)}}{(1-x)(1+x)}dx} - \frac{1}{2}\ln{|x^2-1|}\\
-\frac{1}{2}\int{\frac{ln{(x)}((1-x)+(1+x))}{(1-x)(1+x)}dx} - \frac{1}{2}\ln{|x^2-1|}\\
-\frac{1}{2}\left(\int{\frac{ln{(x)}}{1+x}dx}+\int{\frac{ln{(x)}}{1-x}dx}\right) - \frac{1}{2}\ln{|x^2-1|}\\
-\frac{1}{2}\left(\ln(x+1)\ln(x) - \int{\frac{\ln(x+1)}{x}dx}+\int{\frac{\ln{(x)}}{1-x}dx}\right) - \frac{1}{2}\ln{|x^2-1|}
$
This two integrals can be expressed using $Li_{2}(x)$ function
For integral $\int{\frac{ln{(1+x)}}{x}dx}$ we can use substitution
$u = -x $
For integral $\int{\frac{ln{(x)}}{1-x}dx}$ we can use substitution
$v = 1-x$ and we will get $Li_{2}(x)$ function
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4405201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solutions of Triangles inequalities - Duplicate A,B,C are angles of a triangle we are supposed to prove that $sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$ $\leq$ $\frac{1}{8}$. I used trigonometric ratios of half angles which would give $\frac{(s-a)(s-b)(s-c)}{abc}$. How can I proceed after this step?
Any help would be appreciated ( I need the solution where angles are made in terms of sides and then the inequality is simplified)
|
As @Esgeriath's answer mentioned, the correct inequality should be
$$
\sin \frac{A}{2} \; \sin \frac{B}{2} \; \sin \frac{C}{2} \le \frac{1}{8}.
$$
That answer used multivariable calculus methods to calculate the extremum of the left-hand side.
Here is an elementary proof of this inequality.
You have shown that
$$
\sin \frac A2 \; \sin \frac B2 \; \sin \frac C2
= \frac{(s - a)(s - b)(s - c)}{abc}
$$
where $s = \frac{1}{2}(a + b + c)$. Recall that the area of the triangle is
$$
T = \sqrt{s (s - a)(s - b)(s - c)}
$$
by Heron's formula, the inradius is
$$
r = \frac{T}{s},
$$
and the circumradius is
$$
R = \frac{abc}{4T},
$$
so
$$
\frac{(s - a)(s - b)(s - c)}{abc}
= \frac{T^2}{4 s R T} = \frac{r}{4R}.
$$
But $R \ge 2r$ by Euler's inequality, with equality if and only if the triangle is equilateral, so
$$
\sin \frac A2 \; \sin \frac B2 \; \sin \frac C2 \le \frac{1}{8}
$$
where equality holds if and only if $A = B = C = 60^\circ$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4408491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Existence of $\sin\alpha=\frac{b}{\sqrt{a^2+b^2}}; \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$ Let $a,b\in\mathbb{R}$ such that $(a,b)\neq(0,0)$. Let $x$ be a real number.
We make a function such that $A(x)= a\cos(x)+b\sin(x)$. I need to show the existence of a number $\alpha\in\mathbb{R}$ such that:
$$\sin\alpha=\frac{b}{\sqrt{a^2+b^2}}; \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$$
I thought of using the fact that the sum of the square of the two numbers is equal to one is enough to deduce the existence but I feel like I am wrong. Do you guys have any hints?
|
Note that $\sin(x)$ is continuous in $x$, and that there is an $x'$ such that $\sin(x')=1$, and there is an $y$ such that $\sin(y)=-1$. So for any $c \in [-1,1]$, there is an $x$ such that $\sin(x)=c$. Now, for any such $a,b$, note that $\frac{b}{\sqrt{a^2+b^2}}$ is in $[-1,1]$, so there indeed
is an $\alpha'$ such that $\sin(\alpha')= \frac{b}{\sqrt{a^2+b^2}}$.
Now, we are almost done. Indeed, the equation $\sin^2(\alpha')+\cos^2(\alpha')=1$ must hold, so from this it follows that $\cos(\alpha')$ is
$\frac{\pm a}{\sqrt{a^2+b^2}}$, where $\alpha'$ is as in the previous pragraph. If on the one hand $\alpha'$ is such that $\cos(\alpha') = \frac{a}{\sqrt{a^2+b^2}}$, thenset $\alpha=\alpha'$ and we are done.
If on the other hand $\alpha'$ is such that $\cos(\alpha') = \frac{-a}{\sqrt{a^2+b^2}}$, then set $\alpha = \frac{\pi}{2}-\alpha$; then $\sin(\alpha)=\sin(\alpha')$ $=\frac{b}{\sqrt{a^2+b^2}}$, and $\cos(\alpha)=-\cos(\alpha') = -\frac{-a}{\sqrt{a^2+b^2}} = \frac{a}{\sqrt{a^2+b^2}}$, and so now we are done here as well.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4409503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Find $\exp\bigg[t\left ( \begin{matrix} -1& 1 \\ 0 & -1 \\ \end{matrix} \right )\bigg]$ by definition Find $\exp\bigg[t\left ( \begin{matrix}
-1& 1 \\
0 & -1 \\
\end{matrix} \right )\bigg]$ by definition
Denote $A=\left ( \begin{matrix}
-1& 1 \\
0 & -1 \\
\end{matrix} \right )$
I find out that
$$A^n=\left ( \begin{matrix}
-1& n \\
0 & -1 \\
\end{matrix} \right ) \forall n\in\mathbb{N} \text{ such that } n\mod 2 =1$$
$$A^n=\left ( \begin{matrix}
1& -n \\
0 & 1 \\
\end{matrix} \right ) \forall n\in\mathbb{N} \text{ such that } n\mod 2 =0$$
Denote $k=2n,j=2n+1$.
$e^{tA}=\sum_{k=1}^\infty \frac{t^k}{k!}A^k + \sum_{j=1}^\infty \frac{t^j}{j!}A^j$
I got a bit confused how to find $e^{tA}$ by definition
Thanks !
|
From your insight, it may be more helpful to rewrite the powers of $A$ (you made a slight error) as
$$A^n = (-1)^n \begin{pmatrix}
1 & -n \\ 0 & 1 \end{pmatrix}$$
So,
$$\sum_{n=0}^{\infty} \frac{t^n}{n!}A^n = \sum_{n=0}^\infty \frac{(-t)^n}{n!} \begin{pmatrix}
1 & -n \\ 0 & 1 \end{pmatrix} = \sum_{n=0}^\infty \begin{pmatrix}
\frac{(-t)^n}{n!} & -\frac{(-t)^n}{(n-1)!} \\ 0 & \frac{(-t)^n}{n!} \end{pmatrix}$$
Then you can compute these sums component-wise using the Taylor expansion for $e$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4410167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to integrate $\frac{\cos2x}{ \sin x + \sin 3x}$? I thought of using trig identities to get rid of $\cos2x$ and $\sin3x$, then use Weierstrass substitution, but I got myself into big trouble as the expression got too complicated.
Is there a simpler way to tackle this problem?
|
First, note that
$$\sin(x) + \sin(3x) = 4 \sin(x) \cos^2(x)$$
by the uncommonly-used triple angle identity for sine and the Pythagorean identity:
$$\begin{align*}
\sin(3x)
&= 3 \sin(x) - 4 \sin^3(x) \\
&= \sin(x) \Big( 3 - 4 \sin^2(x) \Big) \\
&= \sin(x) \Big( 3 - 4 \Big( 1 - \cos^2(x) \Big) \Big) \\
&= \sin(x) \Big( 3 - 4 + 4 \cos^2(x) \Big) \\
&= \sin(x) \Big( 4 \cos^2(x) - 1 \Big) \\
&= 4 \sin(x) \cos^2(x) - \sin(x)
\end{align*} $$
Then
$$\mathcal{I} := \int \frac{\cos(2x)}{\sin(x) + \sin(3x)} \, dx = \frac 1 4 \int \frac{\cos(2x)}{\sin(x)\cos^2(x)} \, dx$$
Then using a double-angle formula for cosine,
$$\cos(2x) = 2 \cos^2(x) - 1$$
we get
$$\mathcal{I} = \frac 1 2 \int \csc(x) \, dx - \frac 1 4 \int \frac{1}{\sin(x) \cos^2(x)} \, dx$$
The former is well-known. The second integral may be rewritten with basic trigonometry identities (turn it into $\csc(x) \sec^2(x)$, use Pythagoras, use linearity) and solved easily.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4411141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Maximizing $\sum_{i=1}^{10} \cos(3x_i)$, for real $x_i$ such that $\sum_{i=1}^{10} \cos(x_i)=0$
Determine the greatest possible value of $\sum_{i=1}^{10} \cos(3x_i)$ for real $x_1,x_2,\cdots,x_{10}$, where $\sum_{i=1}^{10} \cos(x_i)=0$.
My approach was to somehow get a equation solely in terms of one of the $\cos x_i$ and then differentiating and getting the maximum according to the range of cos which is $[-1,1]$. For achieving this, I first inductively proved that
$$a^3+b^3 +c^3 +\cdots+n^3= 3\cdot(\text{sum of product of terms taking three at a time})$$ but that just makes the expression to be that we need to maximize the value of
$$12\cdot(\text{sum of product of terms taking three at a time})$$ But now how do we get an equation in just one variable?
|
We apply Lagrange multipliers.
Let $f(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos 3x_i$ be the objective function and $g(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos x_i=0$ be the constraint.
For each $1\leq i\leq 10$, we have by the triple angle formula,
$\lambda \sin x_i = 3(3\sin x_i - 4\sin^3 x_i)$
which gives
$\sin x_i = 0$ or $\lambda-9=-12\sin^2 x_i$.
If $\sin x_i = 0$ then the corresponding $\cos 3x_i$ is maximized when $\cos x_i = 1$.
If $\sin x_i \neq 0$ then all remaining $\sin^2 x_i$ must be identical. Then all remaining $\cos^2 x_i$ are also identical. Let $\mu=|\cos x_i|$. Then $0\leq \mu\leq 1$.
Let $q$ be the number of indices with $\cos x_i = -\mu$, let $n$ be the number of indices with $\cos x_i = \mu$, and $m$ be the number of indices with $\cos x_i=1$.
Then necessarily, $q+n+m=10$ and the constraint becomes
$$
m+(n-q)\mu=0.
$$
The objective function becomes
$$
\begin{align}
m+ & n(4\mu^3-3\mu) + q(4(-\mu)^3 - 3(-\mu)) \\
&= m+ 4(n-q) \mu^3 - 3(n-q)\mu\\
&=4m+4(n-q)\mu \mu^2\\
&=4m(1-\mu^2)\\
&=4m\left( 1-\frac{m^2}{(q-n)^2}\right).\end{align}
$$
This is maximized with $q=7$, $n=0$, $m=3$. When this happens, we have $7$ of $x_1, \ldots, x_{10}$ satisfying $\cos x_i= -\frac 37$ and the other $3$ of them have $\cos x_i=1$. The maximum is
$$
12\left(1-\frac{9}{49}\right)=\frac{480}{49}\approx 9.7959.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4414200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solving recursive equation without Master's Theorem Cheers, I am asked to find the time complexity of a recursive algorithm, which splits the starting problem on $\sqrt{n}$ subproblems, of size $\sqrt{n}$ each, and then combines the solution in linear time.
My try for creating the equation is: $T(n) = \sqrt{n}T(\sqrt{n}) + n$, but I don't know if that is correct, or I have to do something like: $T(n) = \sqrt{n}T(\frac{n}{\sqrt{n}}) + n$. Other than that, I don't know how to proceed with either of them. Master's theorem can't be applied of course, but I am having trouble finding other ways, such as substitution. Can anyone provide some help? Thanks a lot.
|
$$T(n) = n ^ {\frac{1}{2}}T(n ^ {\frac{1}{2}}) + n$$
$$T(n ^ {\frac{1}{2}}) = n ^ {\frac{1}{4}}T(n ^ {\frac{1}{4}}) + n ^ {\frac{1}{2}}$$
Substitute:
$$T(n) = n ^ {\frac{1}{2}}n ^ {\frac{1}{4}}T(n ^ {\frac{1}{4}}) + n + n$$
Now,
$$T(n ^ {\frac{1}{4}}) = n ^ {\frac{1}{8}}T(n ^ {\frac{1}{8}}) + n ^ {\frac{1}{4}}$$
Substitute again,
$$T(n) = n ^ {\frac{1}{2}}n ^ {\frac{1}{4}}n ^ {\frac{1}{8}}T(n ^ {\frac{1}{8}}) + n + n + n$$
Notice the pattern:
$$T(n) = n ^ {\frac{1}{2}}n ^ {\frac{1}{4}}n ^ {\frac{1}{8}}\cdots T(1) + n + n + \cdots + n$$
Since $T(1) = 1$ and this repeats about $\operatorname{log_2}({\operatorname{log}_2(n)})$ times:
$$T(n) \approx n^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2 ^ {\operatorname{log_2}({\operatorname{log}_2(n)})}}} + n \operatorname{log_2}({\operatorname{log}_2(n)})$$
$$T(n) \approx n + n \operatorname{log} \operatorname{log} n$$
Thus, your algorithm works in $O(n\operatorname{log} \operatorname{log} n)$ time.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4416132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Proving the identity of $\sum_{k = 0}^n{4n \choose 4k} = 2^{4n - 2} + (-1)^n2^{2n - 1}$ combinatorially I want to prove the following identity combinatorially:
$$\sum_{k = 0}^n{4n \choose 4k} = 2^{4n - 2} + (-1)^n2^{2n - 1}$$
Here's my attempt so far: The left hand side is counting the number of teams among $4n$ people where the size of the team is a multiple of 4. Now let $A$ and $B$ be the last two people among these $4n$ people. Now let $T$ be a team such that $|T| = 4k$. Then we have 4 cases:
*
*$A, B \in T$
*$A, B \notin T$
*Only $A \in T$
*Only $B \in T$
Now I will construct all teams among the $4n - 2$ people after removing $A$ and $B$ (I don't care about their size) and convert them to a team of size $4k$ by adding a subset of $\{A, B\}$ to them. Clearly there are $2^{4n - 2}$ teams that don't contain $A$ or $B$. Let $T$ be one of these teams. Then we have the following cases:
*
*$|T| = 4k$. Teams of this kind correspond to teams of size $4k$ that don't contain $A$ and $B$.
*$|T| = 4k + 1$. Teams of this kind will just be discarded.
*$|T| = 4k + 2$. Teams of this kind correspond to teams of size $4k'$ that contain both $A$ and $B$. We will just add $A$ and $B$ to $T$.
*$|T| = 4k + 3$. Teams of this kind will be converted to teams of size $k'$ that contain either $A$ or $B$. Then for each team of this kind, we should add $1$ to $2^{4n - 2}$ because we have 2 cases here. Either add $A$ or add $B$.
Based on these cases, we will have $2^{4n - 2} - \sum_{i = 0}^{n - 1}{4n - 2 \choose 4i + 1} + \sum_{i = 0}^{n - 2}{4n - 2 \choose 4i + 3}$. If I've made no mistakes, then I should show that $- \sum_{i = 0}^{n - 1}{4n - 2 \choose 4i + 1} + \sum_{i = 0}^{n - 2}{4n - 2 \choose 4i + 3} = (-1)^n2^{2n - 1}$ but I don't know how to do this.
|
Here's an indirect approach that works.
First prove combinatorially that
$$(x-1)(1+x+x^2+\dots+x^{n-1}) = x^n - 1 \quad \text{for $n \ge 1$},$$
which is Identity 216 in Proofs That Really Count: The Art of Combinatorial Proof.
Then take $n=4$ and $x=i^k$ to yield
$$\frac{(i^k)^0+(i^k)^1+(i^k)^2+(i^k)^3}{4} = \begin{cases}1 & \text{if $4 \mid k$} \\ 0 & \text{otherwise}\end{cases}$$
and so
$$\sum_{k \ge 0}^n a_{4k} = \sum_{k \ge 0} \frac{1+i^k+(-1)^k+(-i)^k}{4} a_k.$$
Now take $a_k = \binom{4n}{k}$ and apply the binomial theorem, which has a clear combinatorial proof, to obtain
\begin{align}
\sum_{k \ge 0}^n \binom{4n}{4k}
&= \sum_{k \ge 0} \frac{1+i^k+(-1)^k+(-i)^k}{4} \binom{4n}{k} \\
&= \frac{1}{4} \sum_{k \ge 0} \binom{4n}{k}
+ \frac{1}{4} \sum_{k \ge 0} i^k \binom{4n}{k}
+ \frac{1}{4} \sum_{k \ge 0} (-1)^k \binom{4n}{k}
+ \frac{1}{4} \sum_{k \ge 0} (-i)^k \binom{4n}{k} \\
&= \frac{(1+1)^{4n} + (1+i)^{4n} + (1-1)^{4n} + (1-i)^{4n}}{4} \\
&= \frac{2^{4n} + (-4)^n + 0 + (-4)^n}{4} \quad \text{for $n>0$} \\
&= 2^{4n-2} + (-1)^n 2^{2n-1}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4417155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Is it hard to tackle the integral $\int_{0}^{\infty} \frac{x^{2}}{\left(1+x^{4}\right)^{2}} d x?$ Putting $ \displaystyle=4, \alpha=2, n=2 $ in my post,
$$\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}dx=\frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)$$
we can conclude that
$$
\begin{aligned}
I &=\frac{\pi}{4(2-1) !} \csc \frac{3 \pi}{4}\left(1-\frac{3}{4}\right) \\
&=\frac{\sqrt{2} \pi}{16}
\end{aligned}
$$
Question:
Is there any other method?
Your suggestion and alternative are warmly welcome.
|
Letting $x\mapsto \frac{1}{x} $ yields
$$
I=\int_{0}^{\infty} \frac{x^{4}}{\left(x^{4}+1\right)^{2}} d x
$$
Adding them together gives $$
\begin{aligned}
2 I &=\int_{0}^{\infty} \frac{x^{2}+x^{4}}{\left(x^{4}+1\right)^{2}} d x \\
&=\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}\right)^{2}} d x \\
&=\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^{2}+2\right]^{2}}
\end{aligned}
$$
Letting $x-\frac{1}{x}=\sqrt{2} \tan \theta$, we have
$$
I=\frac{\sqrt{2}}{8} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} \theta d \theta= \frac{\sqrt{2} \pi}{16}
$$
We can conclude that
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4419223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Is this nice-looking inequality actually trivial?
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz)≥2(x+y+z)^2 $$
I tried to use the inequality
$$x^2+y^2+z^2≥xy+yz+xz$$
Then I got,
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz) ≥2(xy+yz+xz)^2≥2(x+y+z)^2\implies xy+yz+xz≥x+y+z$$
Which is correct, I think. Because, the degree of the polynomial $xy+yz+xz$ is greater than the degree of the polynomial $x+y+z$. But, I am not sure.
Is this inequality trivial and is my solution correct?
|
Your solution is wrong, a value of a polynomial of smaller degree can be bigger than the corresponding value of a polynomial of bigger degree. For example $x+5$ has degree $1$, $x^9$ has degree $9$ but $1+5>1^9$.
Here is a correct solution. I would be glad to see a simpler solution, but I do not see a substantially different idea.
Denote $a=xy+yz+xz, b=x+y+z$. Your inequality is the same as $a(b^2-a)\ge 2b^2$ or $ab^2-a^2\ge 2b^2$ or $(a-2)b^2-a^2\ge 0$. Note that by $AM>GM, x^2+y^2+z^2\ge 3$, also $a\ge 3$.
Then $b^2=x^2+y^2+z^2+2a\ge 3+2a$. Hence your inequality will follow from $(a-2)(3+2a)-a^2\ge 0$ or $a^2-6-a\ge 0$. Since $a>0$, the latter inequality is equivalent to $a\ge 3$ which we have already established.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4421129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
ab+bc+ca=3 then $\sum\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}$ Let $a,b,c≥0:ab+bc+ca=3$. Prove that:
$$\frac{\sqrt{a+3}}{a+\sqrt{bc}}+\frac{\sqrt{b+3}}{b+\sqrt{ca}}+\frac{\sqrt{c+3}}{c+\sqrt{ab}}\ge\frac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}$$
This is really tough problem. I tried to use AM-GM as: $$2\sqrt{a+3}=2\sqrt{\frac{a+ab+bc+ca}{a+\sqrt{bc}}(a+\sqrt{bc})}\le\frac{a(a+b+c+1)}{a+\sqrt{bc}}+2\sqrt{bc}$$
But it seems useless. "sqrt" is a big ostacle. Is there any better idea? Thanks for all.
|
*Partial solution?
I tried to find related term due to AM-GM: $$2\sqrt{a+3}=2\sqrt{\frac{a+ab+bc+ca}{a+\sqrt{bc}}(a+\sqrt{bc})}\le\frac{a(a+b+c+1)}{a+\sqrt{bc}}+2\sqrt{bc}$$
It means that: $$\frac{a(a+b+c+1)}{a+\sqrt{bc}}\ge\frac{2a(b+c+1)}{\sqrt{a+3}+\sqrt{bc}}\implies \frac{\sqrt{a+3}+\sqrt{bc}}{a+\sqrt{bc}}\ge\frac{2(b+c+1)}{a+b+c+1}$$
Or: $$\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{b+c+1-a}{a+b+c+1}+\frac{a}{a+\sqrt{bc}}$$
Maybe the rest is easy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4425033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finding $(a,b)\in\mathbb{N}^2$ such that $\dfrac{a^2+b^2+1}{a+b} \in \mathbb{N}$. A pair $(a,b)\in\mathbb{N}^2$ is called good if $a < b$ and
$$\frac{a^2+b^2+1}{a+b}\in\mathbb{N}.$$
I think I've shown that there are infinitely many good pairs. However, the family of good pairs that I've found is not all of them, and the argument is not as elementary as I'd like.
Observation: we can check small values to show that good pairs do indeed exist, e.g. $(1,2),\, (4,7),\,(5,12),\, (11,16)$.
To construct an infinite family of good pairs, suppose we have
$$\frac{a^2+b^2+1}{a+b}=k\in\mathbb{N},$$
then $a^2+b^2+1 - k(a+b)=0$, and completing the square for $a$ and $b$ gives the equation for a circle:
$$\left(a-\frac{k}{2}\right)^2+\left(b-\frac{k}{2}\right)^2 = \frac{k^2}{2}-1$$
$$\implies a = \frac{k}{2} + \sqrt{\frac{k^2}{2}-1-\left(b-\frac{k}{2}\right)^2}.$$
Let $k=2j$ for some $j\in\mathbb{N}$ to eliminate the fractions, obtaining $a=j+\sqrt{j^2-1-b^2+2jb}$. In order to have $a\in\mathbb{N}$, we must have $j^2-1-b^2+2bj=n^2$ for some $n\in\mathbb{N}\cup\{0\}$. We can view this as a quadratic in $b$:
$$b^2-2jb+(n^2+1-j^2)=0.$$
Since we require $b\in\mathbb{N}$, the discriminant $\Delta$ must be of the form $\Delta = 4m^2$ for some $m\in\mathbb{N}\cup\{0\}$. Thus we have the following expressions for $a$ and $b$:
$$b=\dfrac{2j\pm\sqrt{\Delta}}{2} = j+m, \qquad a = j + \sqrt{n^2} = j+n.$$
Considering the discriminant:
$$\Delta = (-2j)^2 - 4(n^2+1-j^2) = 4m^2$$
$$\implies 2j^2-1 = m^2+n^2.$$
If $a<b$ we must have $n<m$. In particular, we can assume $n=0$ without restricting $m$. Then we have $m^2-2j^2=-1$, and so we have a correspondence between certain good pairs $(a,b)$ and elements of norm $-1$ in $\mathbb{Q}(\sqrt{2})$. For $x+y\sqrt{2}\in\mathbb{Z}[\sqrt{2}]$, define $N(x+y\sqrt{2})=x^2-2y^2$, and observe that this function is multiplicative. We have $N(1\pm\sqrt{2})=-1$, and thus $N((1\pm\sqrt{2})^{2r-1})=(-1)^{2r-1}=-1$ for any $r\in\mathbb{N}$. Let $m+j\sqrt{2} = (1+\sqrt{2})^{2r-1}$ and so $m-j\sqrt{2} = (1-\sqrt{2})^{2r-1}$. We can combine these expressions to obtain the following general formulas for values of $m$ and $j$:
$$m_r = \frac{(1+\sqrt{2})^{2r-1}+(1-\sqrt{2})^{2r-1}}{2},\qquad j_r = \frac{(1+\sqrt{2})^{2r-1}-(1-\sqrt{2})^{2r-1}}{2\sqrt{2}}.$$
Then we have $a=j+n=j$ and $b=m+j$, thus we have an infinite family of good pairs $(a_r,b_r)$ for $r\in\mathbb{N}$ given by:
$$a_r = \frac{(1+\sqrt{2})^{2r-1}-(1-\sqrt{2})^{2r-1}}{2\sqrt{2}},\qquad b_r = \frac{(1+\sqrt{2})^{2r}-(1-\sqrt{2})^{2r}}{2\sqrt{2}}.$$
The first few values of $r$ gives the good pairs $(1,2),\,(5,12),\,(29,70),\,(169,408),(985,2378)$ with corresponding values of $k=2,10,58,338, 1970$.
I seek advice both on this proof itself (i.e. does it work?) and also in seeking a more elementary argument (i.e. without requiring knowledge about $\mathbb{Q}(\sqrt{2})$ and the like) - it doesn't necessarily need to be constructive, in fact it would be interesting to know if there exists a more concise argument to that effect.
Regarding finding good pairs themselves, comparing the list constructed in my solution to thta of the observation, it's clear that this solution 'misses' some pairs - and this comes a no surprise since we disregard the cases that $k$ is odd, and $n>0$.
If $k=2j-1$, then we can use a similar method to show $j^2-j-b^2+2jb-b=n^2-n$ for some $n\in\mathbb{N}$, but I haven't been able to make any progress from this. The case for $k$ even and $n>0$ is essentially asking "when can $2j^2+1$ be written as the sum of two squares", which I'm not sure quite how to answer - using known results about sums of two square seems difficult as $2j^2+1$ has no general factorisation.
Are there totally different methods that can be used to construct more families of good pairs? And what about any other 'structure' to good pairs? It seems interesting that units in $\mathbb{Z}[\sqrt{2}]$ give rise to the solutions, can a more general statement be made?
|
$a^2-b^2$ = $(a-b)(a+b)$ is divisible by a+b, so $a^2+b^2+1$ is divisible by a+b if and only if $2a^2+1$ is.
To find all solutions with a, b >= 1:
Iterate through a = 0, 1, 2, …
Let z = 2a^2+1
Determine all divisors d of z.
For each d >= a, (a, d-a) is a solution.
Since z is divisible by z, we have the obvious b = z - a for each a. For example, a = 5, z = 51, b = 46, and $5^2 + 46^2 + 1 = 2142 = 51 \cdot 42$. Other divisors of z are 1, 3 and 17, leading to be = 17-5 = 12, and $5^2 + 12^2 + 1 = 170 is divisible by a+b = 17.
The values z have on average fewer divisors than random integer of the same size, since they are never divisible by 2, 5, 7, 11, or 13.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4429307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
}
|
Solving $(x^2+4x+3)\sqrt{x+2}+(x^2+9x+18)\sqrt{x+7}\geq x^3+10x^2+33x+36$ I am trying but I cannot solve the following inequation:
$(x^2+4x+3)\sqrt{x+2}+(x^2+9x+18)\sqrt{x+7}\geq x^3+10x^2+33x+36$
My attemption.
The inequation is equvalent to
$$
(x+1)(x+3)\sqrt{x+2}+(x+3)(x+6)\sqrt{x+7}\geq (x+3)^2(x+4)
$$
Since $x+2\geq 0$, dividing both sides by $x+3>0$, we have
$$
(x+1)\sqrt{x+2}+(x+6)\sqrt{x+7}\geq (x+3)(x+4)
$$
|
Let us solve
$$(x+1)\sqrt{x+2}+(x+6)\sqrt{x+7}\geq (x+3)(x+4).$$
We have
\begin{align*}
\mathrm{LHS} - \mathrm{RHS}
&= (x + 1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x + 7} - 3)\\[5pt]
&\qquad + 2(x + 1) + 3(x + 6) - (x + 3)(x + 4)\\[5pt]
&= (x + 1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x + 7} - 3) + (x + 4)(2 - x)\\[5pt]
&= (x + 1)\frac{x - 2}{\sqrt{x + 2} + 2} + (x + 6)\frac{x - 2}{\sqrt{x + 7} + 3} + (x + 4)(2 - x).
\end{align*}
If $-2 \le x \le 2$, we have
\begin{align*}
\mathrm{LHS} - \mathrm{RHS}
&= (2 - x)\left(-\frac{x + 1}{\sqrt{x + 2} + 2} - \frac{x + 6}{\sqrt{x + 7} + 3} + x + 4\right)\\
&= (2 - x)\left(\frac{1}{\sqrt{x + 2} + 2} - \frac{x + 2}{\sqrt{x + 2} + 2} - \frac{x + 6}{\sqrt{x + 7} + 3} + x + 4\right)\\
&\ge (2 - x)\left(0 - \frac{x + 2}{0 + 2} - \frac{x + 6}{0 + 3} + x + 4\right)\\
&= (2 - x)(x/6 + 1)\\
&\ge 0.
\end{align*}
If $x > 2$, we have
\begin{align*}
\mathrm{LHS} - \mathrm{RHS}
&= (x - 2)\left(\frac{x + 1}{\sqrt{x + 2} + 2} + \frac{x + 6}{\sqrt{x + 7} + 3} - x - 4\right)\\
&\le (x - 2)\left(\frac{x + 1}{0 + 2} + \frac{x + 6}{0 + 3} - x - 4\right)\\
&= (x - 2)(-x/6 - 3/2)\\
&< 0.
\end{align*}
Thus, the solution to the inequality is $-2 \le x \le 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4429521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finding the conditions for $\mathbf{A}×(\mathbf{B}×\mathbf{C}) = (\mathbf{A}×\mathbf{B})×\mathbf{C}$. I wanted to find the conditions in which $\mathbf{A}×(\mathbf{B}×\mathbf{C})$ is equal to $(\mathbf{A}×\mathbf{B})×\mathbf{C}$.
On solving $\mathbf{A}×(\mathbf{B}×\mathbf{C})-(\mathbf{A}×\mathbf{B})×\mathbf{C}=\mathbf{0}$, I got
$$
\mathbf{A}(\mathbf{B}\cdot\mathbf{C})- \mathbf{C}(\mathbf{A}\cdot\mathbf{B})=\mathbf{0}\tag{1}
$$
In the solution manual, it's given that this is possible if and only if either $\mathbf{A}$ is parallel to $\mathbf{C}$ or $\mathbf{B}$ is perpendicular to $\mathbf{A}$ and $\mathbf{C}$.
But as per what I have learnt, this means that $(1)$ will be $\mathbf{0}$ only when both terms are in it are $\mathbf{0}$. But here why don't we consider the equality of those two terms?
I am beginning vector analysis, So it will be really helpful if someone could help me out.
|
The Jacobi identity is
$$
\mathbf{A}\times(\mathbf{B}\times\mathbf{C})+
\mathbf{B}\times(\mathbf{C}\times\mathbf{A})+
\mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{0}
$$
Using anticommutativity, we get
$$
\mathbf{A}\times(\mathbf{B}\times\mathbf{C})-
(\mathbf{A}\times\mathbf{B})\times\mathbf{C}=
-\mathbf{B}\times(\mathbf{C}\times\mathbf{A})
$$
If $\mathbf{A}$ is parallel to $\mathbf{C}$, then $\mathbf{C}\times\mathbf{A}=\mathbf{0}$. If $\mathbf{B}$ is perpendicular to both $\mathbf{A}$ and $\mathbf{C}$, then it is parallel to $\mathbf{C}\times\mathbf{A}$ and therefore $\mathbf{B}\times(\mathbf{C}\times\mathbf{A})=\mathbf{0}$.
What about the converse? Suppose the two triple products are equal and that $\mathbf{A}$ is not parallel to $\mathbf{C}$. Then, in order for $\mathbf{B}\times(\mathbf{C}\times\mathbf{A})=\mathbf{0}$ we need that $\mathbf{B}$ is parallel to $\mathbf{C}\times\mathbf{A}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4430693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Calculating a matrix-exponential Let A be the following matrix.
$$\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$$
I have to calculate $e^A$.
My idea was to diagonalize A because then $e^A = Pe^DP^-1$ if $A = PDP^-1$.
But A cannot be diagonalized since 1 is a double eigenvalue and therefore A does not have 2 linearly independent eigenvectors.
How else can I calculate $e^A$?
Thank you!
|
Note that $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$.
So $e^A = \sum_{k=0}^{\infty} \frac{1}{k!}A^k = \sum_{k=0}^{\infty} \frac{1}{k!} \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \sum_{k=0}^{\infty} \frac{1}{k!} & \sum_{k=0}^{\infty} \frac{k}{k!} \\ 0 & \sum_{k=0}^{\infty} \frac{1}{k!} \end{pmatrix} = \begin{pmatrix} e & e \\ 0 & e \end{pmatrix}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4432299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit
*
*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our results for both odd and even multiples $n$ of $x$, we can conclude that
$$
\lim _{n \rightarrow \infty} \displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(nx)+\cos ^{6}(nx)\right] \ln (1+\tan x) d x =\frac{5 \pi\ln 2}{64}
$$
*As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below as an answer:
$$
I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)\right] \ln (1+\tan x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}
$$
In order to evaluate the even case
$$\int_{0}^{\frac{\pi}{4}}\left[\sin^{6}(2 n x)+\cos^{6}(2 nx)\right] \ln (1+\tan x) d x $$
we first simplify
$\displaystyle \begin{aligned}\sin ^{6}(2 n x)+\cos ^{6}(2 n x) =& {\left[\sin ^{2}(2 n x)+\cos ^{2}(2 n x)\right]\left[\sin ^{4}(2 n x)-\sin ^{2}(2 n x) \cos ^{2}(2 n x)\right) } \\&\left.+\cos ^{4}(2 n x)\right] \\=& 1-3 \sin ^{2}(2 n x) \cos ^{2}(2 n x) \\=& 1-\frac{3}{4} \sin ^{2}(4 n x) \\=& 1-\frac{3}{8}(1-\cos 8 n x) \\=& \frac{1}{8}(5+3 \cos (8nx))\end{aligned} \tag*{} $
To get rid of the natural logarithm, a simple substitution transforms the integral into
$\begin{aligned}I &=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln (1+\tan x) d x \\& \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\&=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(\frac{2}{1+\tan x}\right) d x \\&=\frac{1}{8} \ln 2 \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x) )d x-I \\I &=\frac{\ln 2}{16} \int_{0}^{\frac{\pi}{4}}(5+3 \cos 8 n x) d x\\&=\frac{\ln 2}{16}\left[5 x+\frac{3}{8 n} \sin (8 n x)\right]_0^{\frac{\pi}{4} }\\ &=\frac{5 \pi}{64} \ln 2\end{aligned} \tag*{} $
My Question:
How can we deal with the odd one
$$\displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x ?$$
Can you help?
|
Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)
$$
As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below:
$$
I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)\right] \ln (1+\tan x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}
$$
Proof:
Letting $x\mapsto \frac{\pi}{4}-x $ yields
$$
\begin{aligned}
I(m, n) &=\int_{0}^{\frac{\pi}{4}}\left[\sin ^{2 m}(2 x)+\cos ^{2 m}(2 x)\right) \ln \left(\frac{2}{1+\tan x}\right]d x \\
&=\ln 2 \int_{0}^{\frac{\pi}{4}}\left[\sin^{2m} (2 n x)+\cos ^{2 m}(2 n x) \right] d x-I(m, n)\\ I(m, n)&= \frac{\ln 2}{2} \int_{0}^{\frac{\pi}{4}}\left[\sin^{2m} (2n x)+\cos ^{2 m}(2nx)\right]d x\\& = \frac{\ln 2}{4} \int_{0}^{\frac{\pi}{2}}\left[\sin^{2m} (n x)+\cos ^{2 m}(nx)\right]d x\\&\stackrel{nx\mapsto x} {=}\frac{\ln 2}{2} \cdot \frac{1}{n} \int_{0}^{\frac{n \pi}{2}} \sin ^{2 m} x d x\\&= \frac{\ln 2}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2 m} x d x
\end{aligned}
$$
Using Wallis Formula, we can conclude that $$
\boxed{I(m, n)=\frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}},
$$
which is independent of $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$
Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.
I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2),$$
and $3 \le a+b+c$ (using $\frac{9}{a+b+c+6} \le \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$). But I didn't know how to deal with the $\sqrt{a}+\sqrt{b}+\sqrt{c}$.
Can anyone help me? Thank you so much.
|
Not beautiful solution.
WLOG $a\leq b\leq c$. Suppose $a>1$, then $b>1$, $c>1$, $ab+bc+ca+abc > 4$. Therefore $a\leq 1$.
Let $\sqrt{b}+\sqrt{c}=d$, $\sqrt{bc}=e$. Then $bc=e^2$, $d^2=b+c+2e$, $b+c=d^2-2e$. $(\sqrt{b}-\sqrt{c})^2=b+c-2e=d^2-4e\geq 0$, therefore $d^2\geq 4e$.
$$ab+bc+ca+abc=4\Rightarrow a(b+c+bc)+bc=4 \Rightarrow a(d^2-2e+e^2)+e^2=4$$
$$d^2\geq 4e \Rightarrow a(4e-2e+e^2)+e^2=\leq 4 \Rightarrow (a+1)e^2+2ae-4\leq 0$$
$$(a+1)e^2+2ae-4=(a+1)e^2+2(a+1)e-2e-4=((a+1)e-2)(e+2)$$
$$(a+1)e^2+2ae-4\leq 0\Rightarrow (a+1)e-2\leq 0 \Rightarrow e\leq\frac2{a+1}$$
$$a(d^2-2e+e^2)+e^2=4\Rightarrow ad^2=4+2ae-(a+1)e^2=4+\frac{a^2}{a+1}-(a+1)\left(e-\frac{a}{a+1}\right)^2$$
$$0\leq e \leq \frac2{a+1}, a\leq 1\Rightarrow \left|e-\frac{a}{a+1}\right|\leq \frac{2-a}{a+1}\Rightarrow ad^2\geq 4+\frac{a^2}{a+1}-(a+1)\left(\frac{2-a}{a+1}\right)^2\Rightarrow$$
$$ad^2\geq \frac{8a}{a+1} \Rightarrow d^2\geq \frac8{a+1}\Rightarrow d\geq\frac{2\sqrt{2}}{\sqrt{a+1}}$$
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{a}+d\geq \sqrt{a}+\frac{2\sqrt{2}}{\sqrt{a+1}}$$
Let $f=\sqrt{a}+\frac{2\sqrt{2}}{\sqrt{a+1}}$. Then $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq f$.
$$f^2=a+\frac{8}{a+1}+\frac{4\sqrt{2a}}{\sqrt{a+1}}=a+8-\frac{8a}{a+1}+\frac{4\sqrt{2a}}{\sqrt{a+1}}-1+1=8+a+1-\left(2\sqrt{\frac{2a}{a+1}}-1\right)^2$$
$$a\leq 1 \Rightarrow a+a\leq a+1\Rightarrow 2\sqrt{\frac{2a}{a+1}}\leq 2\Rightarrow \left|2\sqrt{\frac{2a}{a+1}}-1\right|\leq 1\Rightarrow$$
$$f^2\geq 8+a+1-1=8+a\Rightarrow \sqrt{a}+\sqrt{b}+\sqrt{c}\geq f\geq \sqrt{8+a}\geq\sqrt{8}=2\sqrt{2}$$
Minimum is obtained at $a=0$ and $b=c$, because it requires $8+a=8$ and $\left|e-\frac{a}{a+1}\right|=\frac{2-a}{a+1}\Rightarrow e=\frac{2}{a+1}\Rightarrow d^2=4e\Rightarrow (\sqrt{b}-\sqrt{c})^2=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4436752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
}
|
Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $p^3+q^3+3pq-1=0$
A sequence of numbers $t_0,t_1,t_2,...$ satisfies $$t_{n+2} = pt_{n+1} + qt_n, ~~~n\geq0$$
where $p$ and $q$ are real. Throughout this question, $x,y,z$ are non-zero real numbers.
Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $$p^3+q^3+3pq-1=0~~~~~~(*)$$
Deduce that either either $p+q+1=0$ or $(p-q)^2+(p+1)^2+(q+1)^2=0$. Hence show that either $x=y=z$ or $x+y+z=0$
I have attempted this question, but despite arriving at (*), I don't believe I did it in the correct way.
Here are my workings:
When $n+2 = 3p, 3p+1, 3p+2$ we get three equations.
$$\ z=py+qx\\ x=pz+qy \\
y=px+qz$$
Adding all three equations, we get, $$x+y+z = p(x+y+z) + q(x+y+z) \\ \therefore x+y+z \not=0 \implies p+q=1$$
I then cubed both sides of this equation resulting in $p^3+q^3+3p^2+3pq^2=1 \implies p^3+q^3+3pq(p+q)-1=0 \implies p^3+q^3+3pq-1=0$
then we can proceed to factor this expression to get $p+q=1$ or the sum of the squared terms.
I suspect this approach is incorrect, but not entirely sure why it's incorrect. I would appreciate an explanation on the matter.
|
For a shortcut, note that the characteristic polynomial of the linear recurrence is $\,x^2-px-q\,$, but the additional conditions also require $\,t_{n+3}=t_n\,$ with characteristic polynomial $\,x^3-1\,$. It follows that they must have a common root, so $\,0 = \text{res}(x^3-1, x^2 - p x - q)=-p^3 - 3 p q - q^3 + 1\,$.
Now back to OP's actual question.
I suspect this approach is incorrect, but not entirely sure why it's incorrect.
It is incorrect because of the assumption $x+y+z \ne 0$ made earlier, which is not given in the problem. This does not render the attempt useless, but leaves it incomplete. Going back to what's being asked:
$$
\begin{align}
p^3+q^3+3pq-1=0
\\ \iff \quad\quad\quad\quad\; (p + q - 1) (p^2 - p q + p + q^2 + q + 1) = 0
\\ \iff \quad\quad (p + q - 1)\big((p-q)^2+(p+1)^2+(q+1)^2\big) = 0\tag{*}
\end{align}
$$
You proved that if $\,x+y+z \ne 0\,$ then $\,p+q-1=0\,$, which is sufficient for $\,(*)\,$ to hold. However, what remains to still be proved is that if $\,x+y+z = 0\,$ then either factor of $\,(*)\,$ is zero.
For that, substitute $\,z=-x-y\,$ $\, \iff t_2 = -t_1 - t_0\,$ in the recurrence relations for $\,n=0, 1\,$:
$$
\begin{cases}
\begin{align}
-t_1 - t_0 = t_2 &= pt_1 +q t_0
\\ \iff \quad\quad (p+1)t_1+(q+1)t_0 &= 0 \tag{1}
\\ t_0 = t_3 &= pt_2 +q t_1 = p (pt_1 +q t_0 ) + q t_1
\\ \iff \quad (p^2+q)t_1+(pq-1)t_0 &= 0 \tag{2}
\end{align}
\end{cases}
$$
Regarding $\,(1),(2)\,$ as a homogeneous linear system in $\,t_0,t_1\,$ which is known to have non-trivial solutions since $\,x,y,z \ne 0\,$, then it follows its determinant must be zero, proving $\,(*)\,$:
$$
0 =
\begin{vmatrix}
p+1 & q+1
\\ p^2 + q & pq -1
\end{vmatrix}
= -\big(p^2 - p q + p + q^2 + q + 1\big)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4436910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Lower bound for $I(a,b)=\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \label{elliptic integral 2}$ I am studiyng this paper which states that the integral
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \tag{1}$$
satisfies the inequality
\begin{align}
\frac{\pi}{2a} \leq I(a,b) \leq \frac{\pi}{2b} \qquad \text{for} \,\, a \ge b \tag{2}
\end{align}
The right hand side of $(2)$ is easy to establish using Cauchy-Schwarz inequality
\begin{equation}
\left(\int_{a}^{b} f(x) g(x) \mathrm{d} x\right)^{2} \leq \int_{a}^{b}(f(x))^{2} \mathrm{~d} x \int_{a}^{b}(g(x))^{2} \mathrm{~d} x \nonumber
\end{equation}
\begin{align}
\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} &\leq \left(\int_0^\infty \left(\frac{1}{\sqrt{a^2+x^2}}\right)^2\,dx\right)^{1/2}\left(\int_0^\infty \left(\frac{1}{\sqrt{b^2+x^2}}\right)^2\,dx\right)^{1/2}\\
&=\left(\int_0^\infty \frac{1}{a^2+x^2}\,dx\right)^{1/2}\left(\int_0^\infty \frac{1}{b^2+x^2}\,dx\right)^{1/2}\\
&=\left(\frac{\pi}{2a}\right)^{1/2}\left(\frac{\pi}{2b}\right)^{1/2}\\
&=\frac{\pi}{2\sqrt{ab}}\\
&\leq \frac{\pi}{2\sqrt{b \cdot b}}\\
&= \frac{\pi}{2b} \qquad (\text{since} \,\ b\leq a)
\end{align}
Now, to prove the L.H.S. of $(2)$ I am unsure on how to proceed.
Edit
Following the hint below we obtain:
\begin{gather*}
\int_0^\infty\frac{dx}{a^2+x^2} \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \leq \int_0^\infty \frac{dx}{b^2+x^2}\\
\frac{1}{a} \int_0^\infty\frac{dx}{1+x^2} \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \leq \frac{1}{b}\int_0^\infty \frac{dx}{1+x^2}\\
\frac{1}{a} \arctan(x)\Big|_0^\infty \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \leq \frac{1}{b}\arctan(x)\Big|_0^\infty\\
\frac{\pi}{2a} \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \leq \frac{\pi}{2b}
\end{gather*}
|
hint We have
$$\frac{1}{\sqrt{b^2+x^2}\sqrt{b^2+x^2}} \ge \frac1{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}
\ge \frac{1}{\sqrt{a^2+x^2}\sqrt{a^2+x^2}} .$$
Therefore...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4438552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to calculate this limit $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $? $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $
I tried L'Hopital, but it gets complicated.
I tried also this:
$$ \lim\limits_{x\rightarrow0}\frac{1-(1+x^2)\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
$$ \lim\limits_{x\rightarrow0}\frac{\sin^2{x}-x^2\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
, but I don't know what to do from here.
|
I shall use this idea of yours:$$\lim_{x\to0}\frac{1-\sqrt{1+x^2}\cos(x)}{x^4}=\lim_{x\to0}\frac{1-(1+x^2)\cos^2(x)}{x^4\left(1+\sqrt{1+x^2}\cos(x)\right)}.$$Note that$$\lim_{x\to0}\left(1+\sqrt{1+x^2}\cos(x)\right)=2.$$So, what remains is to compute\begin{align}\lim_{x\to0}\frac{1-(1+x^2)\cos^2(x)}{x^4}&=\lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}\\&=\lim_{x\to0}\frac{\sin(x)-x\cos(x)}{x^3}\times\lim_{x\to0}\frac{\sin(x)+x\cos(x)}x\\&=\frac13\times2\\&=\frac23,\end{align}and therefore your limit is equal to $\frac13$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4444063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Prove that for $\alpha \leq2$ the sum $\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha$ is divergent, and if $\alpha \geq 4$ the sum is convergent. Prove that for $\alpha \leq2$ the sum $\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha$ is divergent, and if $\alpha \geq 4$ the sum is convergent.
Attempt:
$$\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha = \sum_{n=1}^\infty ((\sqrt{n+1}-\sqrt{n})\cdot\dfrac{(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})})^\alpha =$$
$$\sum_{n=1}^\infty (\dfrac{1}{\sqrt{n+1}+\sqrt{n}})^\alpha$$
Therefore, we will show for every $\alpha \leq 2$ these will going to make the sum diverge.
Easy to see that when $\alpha < 0 $ we get that:
$$\lim_{n\rightarrow \infty}(\dfrac{1}{\sqrt{n+1}+\sqrt{n}})^\alpha \neq 0$$
If $\alpha =1 $ then we get:
$$\sum_{n=1}^\infty (\dfrac{1}{\sqrt{n+1}+\sqrt{n}})$$
So, we can see that the sum is always positive so we can use the limit test of the sums,
Let $b_n = \sum_{n=1}^\infty (\dfrac{1}{\sqrt{n+1}+\sqrt{n}})$ and let $a_n = \sum_{n=1}^\infty (\dfrac{1}{\sqrt{n}})$
And we get $\lim_{n\rightarrow \infty}(\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n}}) =\lim_{n\rightarrow \infty}(\dfrac{\sqrt{n+1}}{\sqrt{n}}+1)=$ $\lim_{n\rightarrow \infty}(\sqrt{1+\dfrac{1}{n}}+1) = 2$
Therefore it diverges too.
For $\alpha=2$, I get struggled with that and I didn't succeed in proving it and how I can prove that converges for all $\alpha \geq 4$.
Thanks!
|
You could write $\sqrt{n+1} - \sqrt{n} = \sqrt{n}\left(\sqrt{1 + \frac{1}{n}} - 1\right)$ and then note that for $n\geqslant 1$,
$$
\frac{\sqrt 2 -1}{n} \leqslant \left( \sqrt{1+\frac{1}{n}} -1 \right) \leqslant\frac{1}{n}
$$
where the right hand inequality is derived from the basic property $x \geq \sqrt x$ when $x \geqslant 1$ and the left inequality from $\sqrt{x} \geqslant 1 + (\sqrt2 - 1)(x-1)$ when $1 \leqslant x \leqslant 2$ (picture the graph of $\sqrt x$ and its chord between $x=1$ and $x=2$).
Then,
$$
\frac{(\sqrt 2- 1 )^\alpha}{n^{\alpha/2}} \leqslant (\sqrt{n+1}-\sqrt{n})^\alpha \leqslant \frac{1}{n^{\alpha/2}}.
$$
Thus your sum converges if and only if $\sum n^{-\alpha/2}$ converges, which is the case if and only if $\alpha > 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4447211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Show that $\cot\left(\frac{\pi}{4}+\beta\right)+\frac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$ Show that $$\cot\left(\dfrac{\pi}{4}+\beta\right)+\dfrac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$$
I'm supposed to solve this problem only with sum and difference formulas (identities).
So the LHS is $$\dfrac{\cot\dfrac{\pi}{4}\cot\beta-1}{\cot\dfrac{\pi}{4}+\cot\beta}+\dfrac{1+\cot\beta}{1-\cot\beta}=\dfrac{\cot\beta-1}{1+\cot\beta}+\dfrac{1+\cot\beta}{1-\cot\beta}=\dfrac{4\cot\beta}{1-\cot^2\beta}$$ I also tried to work with $\sin\beta$ and $\cos\beta$ and arrived at $$\dfrac{4\sin\beta\cos\beta}{\sin^2\beta-\cos^2\beta}$$ I don't see how to get $-2\tan2\beta$ from here (even with other identities).
|
$$\frac{4 \sin \beta \cos \beta}{\sin ^{2} \beta-\cos ^{2} \beta} = \frac{2 \sin2\beta}{-(\cos ^{2} \beta-\sin ^{2} \beta)}=-2\frac{\sin 2\beta}{\cos 2\beta}=-2 \tan(2\beta)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4447522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find $c$ for $P(\frac{5}{12}-c \le Y \le \frac{5}{12}+c)=\frac{1}{2}$ Find $c$ for $P(\frac{5}{12}-c \le Y \le \frac{5}{12}+c)=\frac{1}{2}$
We also have that
$$f_Y(y) = k\sum_{i=0}^{\infty}y^i, y\in(1/3, 1/2) \\ \implies k\int_{1/3}^{1/2}\frac{1}{1-y}dy \implies k=\frac{1}{\log(\frac{4}{3})}$$
What I have tried:
I had thought to apply chebyshev's inequality, in this case $\sigma=1$, so we have
$P(|Y-\frac{5}{12}|\le c)=\frac{1}{2} \\ \frac{1}{2} = 1-\frac{1}{c^2} \implies c = \sqrt{2} \\
\implies P(|Y-\frac{5}{12}| \le \sqrt{2})=\frac{1}{2}$
However, a part of me thinks I need to use the standard normal distribution inequality to figure this out instead.
Something like
$$P\left(\frac{\frac{5}{12}-c}{1} \le Y \le \frac{\frac{5}{12}+c}{1} \right)=\frac{1}{2}$$
and so we have to show that
$$\Phi(0) = \frac{1}{2} \\ \Phi(\frac{5}{12}+c)-\Phi(\frac{5}{12}-c)=\frac{1}{12} \\ \implies \Phi(\frac{5}{12}+c)-\left\{1-\Phi(c-\frac{5}{12}) \right\}\\ \text{we did that }\Phi(-z)=1-\Phi(z) \\ \implies \Phi(\frac{5}{12}+c) + \Phi(c-\frac{5}{12})-1=\frac{1}{2}$$
If I plug in the value of $\sqrt{2}$ like in the previous inequality, then I do not get the value of $\frac{1}{2}$. How do I approach this correctly with the chebyshev or standard normal?
|
If the distribution is not "normal" then why would you think of applying tricks from normal distribution? . Secondly it is Chebycheff's Inequality. So applying it won't yield you any "equality" for $c$.
$$P(\frac{5}{12}-c<Y<\frac{5}{12}+c)=k\int_{\frac{5}{12}-c}^{\frac{5}{12}+c}\frac{1}{1-y}\,dy = k\ln\bigg(\frac{\frac{7}{12}+c}{\frac{7}{12}-c}\bigg)$$.
This means that $$k\ln\bigg(\frac{\frac{7}{12}+c}{\frac{7}{12}-c}\bigg)=\frac{1}{2}$$
So $$\bigg(\frac{\frac{7}{12}+c}{\frac{7}{12}-c}\bigg)=e^{\frac{1}{2k}}$$.
Now solve for $c$.
It would be wise of you to go back a step and re-read the definitions of the Probability Density function and why integrating over a set gives the probability of the random variable lying in that set.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4449026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove $|1 - (1+\frac{z-1}{n})^m| \leq 1$ where $|z|\leq 1$ and $mHow to prove
$$\Big|1 - \big(1+\frac{z-1}{n}\big)^m\Big| \leq 1$$
where $z\in\mathbb{C}$, $|z|\leq 1$ and $m,n\in\mathbb{N}$ and $m<n$.
I have run numerical experiments and believe the inequality is correct.
|
By Maximum modulus principle, the maximum of $\displaystyle |1-(1+\frac{z-1}{n})^m|$ cannot be reached when $|z| < 1$. So we can suppose $|z|=1$.
Let $z=e^{i\theta}$, and $\displaystyle w = 1 + \frac{z-1}{n} = \rho e^{i \varphi}$. Then:
$$
\begin{align}
|1-(1+\frac{z-1}{n})^m | &= |1-w^m| \\
&=|1-w|\cdot|1+w+\cdots+w^{m-1}| \\
&\leq |1-w|\cdot (1+|w|+\cdots+|w|^{m-1}) \\
&\leq |1-w|\cdot (1+|w|+\cdots+|w|^{n-2}) \\
&= \sqrt{1 -2\cos \varphi \cdot \rho+\rho^2} \cdot(1+\rho+\cdots \rho^{n-2})
\end{align}
$$
Since
$
\displaystyle |\rho e^{i\varphi}-(1-\frac{1}{n})|= |\frac{e^{i\theta}}{n}|=\frac{1}{n}
$,
$$
\rho^2+(1-\frac{1}{n})^2-2\rho\cdot (1-\frac{1}{n})\cos\varphi=\frac{1}{n^2}
$$
. Hence
$$
2\cos \varphi \cdot \rho=\frac{\rho^2+(1-\frac{2}{n})}{1-\frac{1}{n}}
$$
. Hence
$$
\begin{align}
\sqrt{1 -2\cos \varphi \cdot \rho+\rho^2} &= \sqrt{1+\rho^2-\frac{\rho^2+(1-\frac{2}{n})}{1-\frac{1}{n}}} \\
&=\sqrt{\frac{1-\rho^2}{n-1}}
\end{align}
$$
So if we can prove
$$
(1-\rho^2)(1+\rho+\cdots+\rho^{n-2})^2 \leq n-1
$$, we can prove the original statement. Here we know that $\displaystyle \rho \in [1-\frac{2}{n}, 1]$.
Let $t^2 = 1-\rho$, then
$$
\begin{align}
(1-\rho^2)(1+\rho+\cdots+\rho^{n-2})^2 &= (2-t^2)t^2 \frac{(1-(1-t^2)^{n-1})^2}{t^4} \\
&\leq 2\frac{(1-(1-t^2)^{n-1})^2}{t^2}
\end{align}
$$
Hence if we can prove that
$$
1 - \sqrt{\frac{n}{2}}t \leq (1-t^2)^{n}
$$, we can prove the original problem.
But I cannot prove this statement. Plotting some graph for small $n$, I believe it's true for all $n \geq 4$ and $t \in [0, 0]$. I asked a new question about this statement: Prove that $1 - \sqrt{\frac{n}{2}}\cdot t \leq (1-t^2)^n$ for $n \geq 4$ and $t \in [0, 1]$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4451523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to prove the product of first n consecutive odd numbers is a square less than another square? I have observed for first few values of consecutive odd numbers, the result is always of the form:
$m^2 - n^2$, where $m$ and $n$ are two distinct positive integers. That is:
$$1\cdot 3\cdot 5 \cdot 7\cdots (2k-1) = m^2 - n^2$$
For example:
$
1\cdot 3 = 3 = 2^2 - 1^2 \\
1\cdot 3\cdot 5 = 15 = 4^2 - 1^2 \\
1\cdot 3\cdot 5\cdot 7 = 105 = 11^2 - 4^2 \\
1\cdot 3\cdot 5\cdot 7\cdot 9 = 945 = 31^2 - 4^2 \\
\vdots
$
But not sure, how to prove it.
Here is an attempt using induction:
Let it be true for some value of k, that is:
$1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1) = m^2 - n^2$
Then when $k$ takes the value of $k+1$, we have
$$\begin{align}
1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1)\cdot(2k + 1)
&= (m^2 - n^2)\cdot(2k + 1)\\
&= (m^2 - n^2)\cdot {(k+1)^2 - k^2}
\end{align}$$
and got stuck here.
Can you please suggest to proceed further or an altogether a different way of proving so or prove me wrong.
Thanks in advance.
|
You might note that the difference sequence of the sequence of squares is exactly the odd numbers. Here's the sequence of squares:
$$0, 1, 4, 9, 16, 25, 36, \ldots$$
Here's the sequence of the differences:
$$1, 3, 5, 7, 9, 11, \ldots$$
That means every odd number is the difference of two consecutive squares. And your number is odd.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4452548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find volume of solid bounded by given surfaces. $z=a+x,z=-a-x,x^2+y^2=a^2$ Find volume of solid bounded by given surfaces.
$$z=a+x, \qquad z=-a-x, \qquad x^2+y^2=a^2$$
This is the solid. We can find volume of solid that has positive $z$ value and multiply by $2$. And for finding volume of that solid I thought doing double integral where domain is circle and function is that ellipse. $\int_{-a}^a dy \int_{-a}^a f(x,y)dx$. Where $f(x,y)$ is intersection of plane $z=a+x$ and $x^2+y^2=a^2$. Is my thinking right? When I want to find intersection I substitute $x=z-a$ into $x^2+y^2=a^2$ I get $z=\frac{a+\sqrt{a^2-y^2}}{2}$ which when I graph does not look like ellipse. Can you help?
|
Intersection of planes is a line given by, $z = a + x = - a - x \implies x = - a, z = 0$
At intersection with the cylinder, $(-a)^2 + y^2 = a^2 \implies y = 0$
So, both planes and the cylinder intersect at a single point $(-a, 0, 0)$. That means the projection of the region in xy-plane is circle $x^2 + y^2 \leq a^2, z = 0$
The integral to find volume is given by,
$ \displaystyle V = \iint_{x^2 + y^2 \leq a^2} 2 |a + x| ~ dx ~ dy$
Note that if $a \lt 0, a + x $ is always negative (or zero) over the region and if $a \gt 0, a + x$ is always non-negative over the region.
Now as $x$ is an odd function, $[f(-x) = - f(x)]$, and the region is symmetric to yz plane, its integral over $x \gt 0$ will cancel out the integral over $x \lt 0$.
So, $ \displaystyle V = 2 |a| \iint_{x^2 + y^2 \leq a^2} dx ~ dy = 2 \pi |a|^3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4454804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Let $a_n$ be a sequence such that $a_1 >0$ and for all $n$ $a_{n+1}=a_n+ \frac{1}{a_n^3}$ then $\lim _\limits {n \to \infty}a_n=\infty$
Let $a_n$ be a sequence such that $a_1 >0$ and for all $n$ $a_{n+1}=a_n+ \frac{1}{a_n^3}$ then $\lim _\limits {n \to \infty}a_n=\infty$
I am not sure how to formally prove this , the book says the statement is true.
I just tried to gather some information from what is given I do not know if it is mostly needed for example according to $a_{n+1}=a_n+ \frac{1}{a_n^3}$ we get that $a_{n+1}-a_n=\frac{1}{a_n^3}$ so the sequence is increasing
also we know that the limit of $a_n$ is equal to the limit of a moved sequence (sorry if this is the incorrect word) so $\lim _\limits {n \to \infty}a_n=\lim _\limits {n \to \infty}a_{n+1}$
and the only way to get this it is if I can get $\lim _\limits {n \to \infty}\frac {1}{a_n^3}=0$ and that is when $\lim _\limits {n \to \infty}a_n= \infty$
This was my approach , according to all that we get that $\lim _\limits {n \to \infty}a_{n+1}= \lim _\limits {n \to \infty}a_n+ \frac{1}{a_n^3}$ since $\lim _\limits {n \to \infty}a_n= \infty$
the result will be $\lim _\limits {n \to \infty}a_{n+1}= \lim _\limits {n \to \infty}a_n=\infty$
Is my way correct? is there a different approach?
Thank you for any tips and help
|
Squaring both sides,
\begin{align*}
a_{n+1}^2 = a_n^2 + 2\frac{a_n}{a_n^3} + \frac{1}{a_n^6} > a_n^2 + 2\frac{1}{a_n^2}
\end{align*}
and squaring again,
\begin{align*}
a_{n+1}^4 > a_n^4 + 4 + \frac{4}{a_n^4} > a_n^4 + 4
\end{align*}
and so
\begin{align*}
a_{n+1}^4 = a_1^4 + \sum_{k=1}^{n}(a_{k+1}^4 - a_k^4) > 4n + a_1^4 \rightarrow \infty
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4454993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to do the following trigonometric simplification: $ \frac{1- \cos (3\alpha) }{1- \cos (\alpha)} = (1 + 2\cos (\alpha)^2) $ This is probably a trivial question but I don't understand it. Somehow I can't seem to understand how to simplify this expression:
$$ \frac{1- \cos (3\alpha) }{1- \cos (\alpha)} = (1 + 2\cos (\alpha))^2 $$
|
I prefer using capital $A$, so bare with me.
We will prove that $$1-\cos3A=(1-\cos A)(1+2\cos A)^2$$
First we will show that $$\cos3A=4\cos^3A-3\cos A$$
Indeed, we know that $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$
Thus, for $x=2A, y=A$ we obtain: $$(*):\cos 3A=\cos 2A\cos A-\sin 2A\sin A$$
but from $(*)$ we see that: $\cos 2A=\cos^2A-\sin^2A$
and we know that $\sin 2A=2\sin A\cos A$
So,$$\cos 3A=(\cos^2A-\sin^2A)\cos A-2\sin^2A\cos A\\=(2\cos^2A-1)\cos A-2(1-\cos^2)\cos A\\= 4\cos^3A-3\cos A$$
Now we have $$1-\cos3A=1-4\cos^3A+3\cos A$$
But $$(1-\cos A)(1+2\cos A)^2=\\(1-\cos A)(\cos^2A+2\cos A+1)=\\1-4\cos^3A+3\cos A$$
and we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4456088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$ is equal to $1$.
I'm looking for other approaches/ideas to factorize the expression.
|
You can observe that, if $x=-y$ or (not "and") $x=-z$ or (not "and") $y=z$ , then the expression equals to zero. Thus, we have
$$xy(x+y)+yz(y-z)-xz(x+z)=A(x + y) (x + z) (y - z)$$
Here, $A$ must be a real number, not an expression. Because, it is enough to see that the coefficient of $x^2y$ equals to $1$. This implies, $A=1$.
Write,
$$A=(x+z)y^2+y(x^2-z^2)-xz(x+z)\\
\Delta_y=(x-z)^2(x+z)^2+4xz(x+z)^2=(x+z)^2(x+z)^2=(x+z)^4$$
Then we have,
$$y_{1,2}=\frac{(z^2-x^2)±(x+z)^2}{2(x+z)}$$
Can you take from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4458244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\iff\tan^2\alpha+\tan\alpha-1=0$$ This equation has solutions $\left(\tan\alpha\right)_{1,2}=\dfrac{-1\pm\sqrt5}{2}$ but as $\alpha\in(0^\circ;45^\circ)\Rightarrow$ $\tan\alpha=\dfrac{\sqrt5-1}{2}$. Now $\sin\alpha=\dfrac{\sqrt5-1}{2}\cos\alpha$ and plugging into $\sin^2\alpha+\cos^2\alpha=1$ got me at $\cos^2\alpha=\dfrac{2}{5-\sqrt5}$
|
Hint:
Writing $A$ for $\alpha$
$$2C=2\sin3A\cos A=\sin4A+\sin2A$$
Now Weierstrass substitution to find $\sin4A$
$$\tan2A=2\implies\dfrac{\sin2A}2=\dfrac{\cos2A}1=\pm\sqrt{\dfrac1{1^2+2^2}}$$
As $0^\circ\le2A\le90^\circ, \sin2A\ge0$
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4460616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Group theory, when are elements conjugate to each other? In my book "Group Theory and Chemistry" by Bishop, it is written that if $$R = Q^{-1}PQ$$ $R$ is the transform of $P$ by $Q.$ But isn't $R$ not just equal to $P (Q^{-1} Q = E )$?
If I transform $P$ by $Q$ wouldn't I have $R = PQ$?
|
I will give two examples where this fails. The first is the smallest possible group that is non-abelian; that is, where the elements do not necessarily commute, so there exist $a,b$ in the group such that $ab\neq ba$. The second is usually one of the first examples one studies where commutativity fails.
Consider the group $D_3$ of all rotational and reflectional symmetries of an equilateral triangle.
Label the vertices of the triangle (pointing up) $1$, $2$, and $3$, going clockwise from the top.
Let $P$ be the clockwise rotation, sending $1$ to $2$, $2$ to $3$, and $3$ to $1$. Then $P^{-1}$ is the counter-clockwise rotation, sending, in the original configuration, $1$ to $3$, $3$ to $2$, and $2$ to $1$.
Let $Q$ be the vertical reflection, fixing $1$ and sending $2$ to $3$ and $3$ to $2$. Then clearly $Q^{-1}$ is $Q$.
Define concatenation $AB$ of symmetries $A,B$ as doing $B$ then $A$.
We have
$$Q^{-1}PQ=P^{-1},$$
whereas
$$PQ^{-1}Q=P.$$
Consider $GL_2(\Bbb R)$, the group of invertible $2\times 2$ matrices with entries in $\Bbb R$.
Consider
$$Q=\begin{pmatrix}
0 & 1\\
1 & -1
\end{pmatrix}.$$
Then
$$Q^{-1}=\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}.$$
Let
$$P=\begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix}.$$
Then
$$\begin{align}
Q^{-1}PQ&=\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & -1
\end{pmatrix}
\\
&=\begin{pmatrix}
2 & 0\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & -1
\end{pmatrix}
\\
&=\begin{pmatrix}
0 & 2\\
0 & 1
\end{pmatrix},
\end{align}$$
whereas
$$PQ^{-1}Q=P.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4465517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
How many real solutions are there for the equation $x^{3}-300x=3000$ How many real solutions are there for the equation $$x^{3}-300x=3000$$
Attempt
There is a solution here I found, without using any algebraic factorization: https://youtu.be/y845hT7aYAQ
and this seems to be most efficient.
But how about one that uses algebraic factorization, or without using derivative?
One method maybe:
$$(x+a)(x+b)(x+c)=x^{3}-300x-3000$$
Then solve $a,b,c$. LHS is
$$ x^{3} + (a+b+c)x^{2} + (ab+ac+bc)x +abc$$
So we have
$$a+b+c=0$$
$$ab+ac+bc=-300$$
$$abc=-3000$$
Using 3rd equation on the 2nd:
$$ -3000(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})=-300$$
Or $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=\frac{1}{10}$$
Then $a=-(b+c)$ so
$$(b+c)bc=3000$$
By symmetry we also have
$$(a+c)ac=3000$$
$$(b+a)ab=3000$$
|
By Descartes rule of signs, we have one positive root. Zero isn’t a root. Now suppose there are negative roots, then using $x \mapsto -x$, these are positive roots of $x^3+3000=300x$.
However by AM-GM,
$x^3+2000+1000> 300\sqrt[3]2\,x>300x$, hence there are no negative roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4466146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
find an equation satisfied by the three bisectors of a right triangle let $a,b,c$ be the sides of a generic triangle and $\alpha, \beta, \gamma$ the bisectors of the respective opposite angle.
It is well know that, for example, $\alpha$ divides the side $a$ in two segments, $x$ and $a-x$, and that we have $(a-x)c=bx$, in addition $\alpha$ satisfies the equation $\alpha^2=cb-x(a-x)$, thus we can eliminate $x$ obtaining $(b^2+2bc+c^2)\alpha^2=-a^2bc+b^3c+2b^2c^2+bc^3$.
Hence I have three equations involving only the sides and the bisectors, assuming that such triangle is right-angled (let $a$ be the hypotenuse), can we find other relations in order to (computationally) eliminate $a, b, c$ and get an equation in the only $\alpha, \beta, \gamma$? My attempts have not been successful so far.
|
(Too long for a comment.)
In the case of a right triangle with $\,b^2+c^2=a^2\,$ the bisector formulas simplify to:
*
*$\require{cancel}(b+c)^2\alpha^2 = bc\left((b+c)^2-a^2\right)=bc\left(\cancel{b^2+c^2}+2bc-\cancel{a^2}\right)=2b^2c^2\,$;
*$(c+a)^2\beta^2=ca\left(c^2+2ac+\underbrace{a^2-b^2}_{=\,c^2}\right)=2ac^2(c+a) \implies (c+a)\beta^2=2ac^2\,$.
This gives the system of equations:
$$
\begin{cases}
\begin{align}
(b+c)^2\alpha^2 - 2b^2c^2 &= 0
\\ (c+a)\beta^2 - 2ac^2 &= 0
\\ (a+b)\gamma^2 - 2ab^2 &= 0
\end{align}
\end{cases}
$$
Together with $\,b^2 + c^2 = a^2\,$, these are $\,4\,$ algebraic equations, between which it is technically possible to eliminate $\,a,b,c\,$ using polynomial resultants in order to get a relation in $\,\alpha,\beta,\gamma\,$ alone.
Variable $\,a\,$ can be eliminated by hand from the last two equations, by isolating the terms in $\,a\,$, squaring, then substituting $\,a^2=b^2+c^2\,$:
$$
\begin{align}
(c+a)\beta^2 - 2ac^2 = 0 \;\;\implies\;\; a(2c^2-\beta^2)=c\beta^2 \;\;\implies\;\; (b^2+c^2)(2c^2-\beta^2)^2 = c^2\beta^4
\end{align}
$$
This leaves three equations with $\,b,c\,$ remaining to be eliminated between them:
$$
\begin{cases}
\begin{align}
p(b,c) &= (b+c)^2\alpha^2 - 2b^2c^2 &= 0
\\ q(b,c) &= (b^2+c^2)(2c^2-\beta^2)^2 - c^2\beta^4 &= 0
\\ r(b,c) &=(b^2+c^2)(2b^2-\gamma^2)^2 - b^2\gamma^4 &= 0
\end{align}
\end{cases}
$$
The final relation between $\,\alpha,\beta,\gamma\,$ can be derived as $\,\text{res}_b\big(\text{res}_c(p,q), \text{res}_c(p,r)\big)=0\,$, which WA calculates to be a polynomial of total degree $\,120\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4468226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
How to solve long integration by partial fraction decomposition problems faster? Some problems are just too time consuming for short exam times what is the fastest way to solve problems like this one for example $$\int \frac{5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}dx$$
|
There are a few answers in here already, but I hope mine can contribute something. Except for very simple cases, there's no fastest partial fractions decomposition method. You have to decide which method(s) is(are) best for the case at hand. Although the algebraic method (matching coefficients) always works, the cover-up method (for distinct real roots) and the derivative method (for dealing if roots of multiplicity greater than one) are very useful to have in mind. As we'll see with my take on your example, a combination of methods is often warranted.
First we set up our partial fractions and eliminate the denominators to get
$$
\begin{split}
5x^4 - 21x^3 + 40x^2 - 37x + 14 &= A(x^2 - 2x + 2)^2 + (Bx + C)(x-2)(x^2 - 2x + 2)\\
&\phantom= +(Dx + E)(x-2).
\end{split}\tag1 \label{1}
$$
To solve for $A$, there's not much you can do other than direct evaluation by setting $x=2$, which will give you
$$
4A = 12 \Longrightarrow A = 3.
$$
Knowing $A$, $B$ can be easily calculated by noticing that the right-hand side of \eqref{1} can be expanded as
$$
5x^4 + \mathcal{O}(x^3) = (3 + B)x^4 + \mathcal{O}(x^3), \tag2 \label{2}
$$
where the big-O notation is used to represent terms proportional to $x^3$ at the most. Dividing both sides of \eqref{2} by $x^4$ and taking the limit as $x\to\infty$
$$
\begin{split}
\lim_{x\to\infty} 5 + {\mathcal{O}(x^3) \over x^4} &= \lim_{x\to\infty} 3 + B + {\mathcal{O}(x^3) \over x^4}\\
B &= 2.
\end{split}
$$
There are no clever ways to solve directly for $C$, $D$ or $E$, but if we expand the RHS of \eqref{1} we can find the last three unknowns by inspection. Substituting the known values of $A$ and $B$, we have
$$
- x^3 + 4x^2 -5x + 2 = C(x^3 -4x^2 + 6x - 4) + D(x^2 - 2x) + E(x - 2) \tag{3} \label{3}
$$
and it's obvious that $C = -1$.
Replacing $C$ by its value in \eqref{3} yields
$$
x - 2 = D(x^2 - 2x) + E(x - 2)
$$
from where it's evident that $D = 0$ and $E = 1$.
Finally
$$
{5x^4 - 21x^3 + 40x^2 - 37x + 14 \over (x-2)(x^2 - 2x + 2)^2} = {3 \over x-2} + {2x - 1 \over x^2 - 2x + 2} + {1 \over (x^2 - 2x + 2)^2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4469257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Doubt regarding validity of answer in a mod equation question
$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$
Find the set of values of $x$.
The answer given is $[0,\frac12]\cup [2,3]$.
What I don't get is how is the solution a range? I'm getting the solution as $0,\frac 12, 2 ,3, \frac 25$. That makes sense to me as the first expression is always positive, $$\because x^2-2x+1+1=(x-1)^2+1> 0$$ and the other two expressions give two cases which on being solved gives 4 solutions in total as it's a quadratic.
$$x^2-2x+2-|(x-2)(2x-1)|-|(x-0)(x-3)|=0$$
Using wavy curve on the two expressions inside mod, we get that they are negative for $x\in (\frac 12,2)$ and $x\in [0,3]$ respectively. So, we can conclude that if the first expression is negative then the second expression is negative too as $[\frac 12,2]$ is in $[0,3]$. So we can get 3 cases:
Both are positive, or both are negative, or the first one is positive and the second one is negative.
Solving for each gives the solutions which I said I've received earlier in the post.
So where is the range coming from? Am I interpreting the mod operator wrong? Or is my book wrong?
|
As you've mentioned, $|x^2-2x+2| = |(x-1)^2+1|$ is always positive.
So $|x^2-2x+2| =x^2-2x+2$.
$|2x^2-5x+2| = \begin{cases}2x^2-5x+2 & \text{for } x \in (-\infty,0.5] \cup [2,\infty)\\ -(2x^{2}-5x+2) &\text{for } x \in [0.5,2]\end{cases}$
So,
$|x^2-2x+2| - |2x^2-5x+2| = \begin{cases}-x^2+3x &\text{for } x \in (-\infty,0.5]\cup [2,\infty)\\3x^2-7x+4&\text{for } x \in [0.5,2]\end{cases}$
and,
$|x^2-3x| = \begin{cases}x^2-3x&\text{for } x \in (-\infty,0]\cup [3,\infty) \\ -(x^2-3x) &\text{for } x \in [0,3]\end{cases}$
Now segregating a little bit,
$|x^2-2x+2| - |2x^2-5x+2| = \begin{cases}-x^2+3x &\text{for } x \in (-\infty,0]\cup[0,0.5]\cup [2,3] \cup[3,\infty)\\3x^2-7x+4&\text{for } x \in [0.5,2]\end{cases}$
and,
$|x^2-3x| = \begin{cases}x^2-3x&\text{for } x \in (-\infty,0]\cup [3,\infty) \\ -x^2+3x &\text{for } x \in [0,0.5]\cup[0.5,2]\cup[2,3]\end{cases}$
So, the common range for which, $|x^2-2x+2| - |2x^2-5x+2|=|x^2-3x|$ is $[0,0.5] \cup [2,3]$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4470083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Find $X/1430$ when $X=(^{10}C_1)^2+2(^{10}C_2)^2+3(^{10}C_3)^2+ ...+10(^{10}C_{10})^2$ Let $X=(^{10}C_1)^2+2(^{10}C_2)^2+3(^{10}C_3)^2+ ...+10(^{10}C_{10})^2$, then what's the value of $X\over1430$?
I don't even know where to begin on this question. All solutions I've seen on various sites start by writing this as a summation and simplifying, and eventually bring it into this form:
$$=10\sum_{r=1}^{10}C^{10}_{r}*^{9}C_{r-1}$$. Until this point, I understand, but after this, I dont understand at all.
|
Here is a variation using the binomial theorem and a bit of algebra. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. This way we can write for instance
\begin{align*}
\binom{r}{k}=[x^k](1+x)^r\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{X}&\color{blue}{=\sum_{r=1}^{10}r\binom{10}{r}^2}=10\sum_{r=1}\binom{10}{r}\binom{9}{r-1}\tag{2}\\
&=10\sum_{r=1}^{10}\binom{10}{r}\binom{9}{10-r}\tag{3}\\
&=10\sum_{r=1}^{10}\binom{10}{r}[x^{10-r}](1+x)^9\tag{4}\\
&=10[x^{10}](1+x)^9\sum_{r=1}^{10}\binom{10}{r}x^r\tag{5}\\
&=10[x^{10}](1+x)^9\left((1+x)^{10}-1\right)\tag{6}\\
&=10[x^{10}](1+x)^{19}\tag{7}\\
&\;\;\color{blue}{=10\binom{19}{10}}\tag{8}
\end{align*}
We finally calculate $\color{blue}{X}=\frac{10}{1\,430}\binom{19}{10}\color{blue}{=646}$.
Comment:
*
*In (2) we use the binomial identity $\binom{r}{k}=\frac{r}{k}\binom{r-1}{k-1}$.
*In (3) we use the binomial identity $\binom{r}{k}=\binom{r}{r-k}$.
*In (4) we apply the coefficient of operator according to (1).
*In (5) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and factor out terms which do not depend on the summation index $r$.
*In (6) we apply the binomial theorem.
*In (7) we observe $[x^{10}](1+x)^{9}=0$.
*in (8) we select the coefficient of $x^{10}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4471784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\textrm{ and }$ $b+c$ are positive real numbers.
First of all, let’s ‘kill’ the term $\sin x$ and the square of $\cos x$ by identity and double angle formula.
$\displaystyle \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x =& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b) \cos ^{2} x+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b)\left(\frac{1+\cos 2 x}{2}\right)+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[\frac{a-b}{2} \cos 2 x+\left(b+c+\frac{a-b}{2}\right)\right] d x \\\stackrel{2x\mapsto x}{=} & \frac{1}{2} \int_{0}^{\pi} \ln \left[\frac{a-b}{2} \cos x+\left(\frac{a+b+2 c}{2}\right)\right] d x\end{aligned}\tag*{} $
By my post, I found, by Feynman’s Integration Technique, that
$\displaystyle \int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right),\tag*{} $
where $\left|\frac{c}{b}\right|\geq 1$.
$\displaystyle I=\frac{\pi}{2} \ln \left[\frac{\frac{a+b+2 c}{2}+\sqrt{\left(\frac{a+b+2 c}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}}}{2}\right]\tag*{} $
Simplifying gives the result
$\displaystyle \begin{aligned}I&=\frac{\pi}{2} \ln \left[\frac{a+b+2 c+2 \sqrt{(a+c)(b+c)}]}{4}\right] \\&=\frac{\pi}{2} \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)^{2} \\&=\pi \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)\end{aligned}\tag*{} $
|
Letting $t=\tan x$ transform the integral into
$$
\begin{aligned}
I &= 2 \int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x+\int_{0}^{\frac{\pi}{2}} \frac{\ln \left((a+c)+(b+c) t^{2}\right)}{1+t^{2}}dt\\&=-\pi\ln 2+\int_{0}^{\frac{\pi}{2} } \frac{\ln \left((a+c)+(b+c) t^{2}\right)}{1+t^{2}} d t
\end{aligned}
$$
By my post
$$
\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x=\pi \ln (\sqrt{a}+\sqrt{c})
$$
Replacing $a$ by $a+c$ and $c$ by $b+c$, we get
$$
I=\pi \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4472100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
If $a,b,c\in\mathbb{R},$ such that $a\geq 2$, $b\geq 5$, $c\geq 5$ and $2a^2+b^2+c^2=69$. Prove that $12a+13b+11c \geq 155.$ So,I am trying to prove the following inequality without using the Lagrange multipliers:
If $a,b,c\in\mathbb{R},$ such that $a\geq 2$, $b\geq 5$, $c\geq 5$ and $2a^2+b^2+c^2=69$. Prove that
$$12a+13b+11c \geq 155.$$
My try:
$\bullet a\geq 2, b\geq 5, c\geq 5,2a^2+b^2+c^2=69 \rightarrow 2 \le a<4,5 \le b,c \le 6.$
But,I don't know how to proceed further.
Any help would be highly appreciated.Thank you!
|
Let $a = 2 + x, b = 5 + y, c = 5 + z$. Then $x, y, z \ge 0$.
We have $2a^2 + b^2 + c^2 - 69
= 2x^2 + y^2 + z^2 + 8x + 10y + 10z - 11 = 0$.
We want to prove that $12x + 13y + 11z \ge 11$.
We split into two cases:
(1) If $x > 1$ or $y > 1$ or $z > 1$, we have $12x + 13y + 11z \ge 11$.
(2) If $x, y, z \in [0, 1]$, we have
$$2x^2 + y^2 + z^2 + 8x + 10y + 10z
\le 2x + y + z + 8x + 10y + 10z
= 10x + 11y + 11 z $$
which results in $10x + 11y + 11z \ge 11$.
Thus, we have $12x + 13y + 11z
\ge 10x + 11y + 11 z \ge 11$.
We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4472975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How can a cylinder be rotated at any angle $\theta$ about the origin using cartesian coordinates? How can a cylinder be rotated at any angle $\theta$ at origin using cartesian coordinates?
What would be the equation for a cylinder (first figure) rotated about the origin?
|
The equation of a cylinder in the upright position (i.e. with its axis along the vertical $z$ axis, is given by
$ \mathbf{p}^T Q \mathbf{p} = r^2 $
where $\mathbf{p} = [x, y, z]^T $ , and $r$ is the radius of the cylinder, and the $3 \times 3$ matrix is given by
$ Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} $
Now, if we rotate the axis of this center about an rotation axis $\mathbf{u}$ that passes through the origin by an angle $\theta$, then the rotation matrix $R$ will be
$ R = \mathbf{u u}^T + (I - \mathbf{u u}^T ) \cos(\theta) + S_u \sin(\theta) $
$S_u = \begin{bmatrix} 0 && - u_z && u_y \\ u_z&& 0 && - u_x \\ -u_y && u_x && 0 \end{bmatrix} $
The above formula is called the Rodrigues' rotation matrix formula.
The image of a point on the cylinder $\mathbf{p}$ is the point $\mathbf{p'}$ given by
$ \mathbf{p'} = R \mathbf{p} $
From this, we have
$\mathbf{p} = R^{-1} \mathbf{p'} = R^T \mathbf{p'} $
Substitute this in the equation of the cylinder, gives you,
$ \mathbf{p'}^T R Q R^T \mathbf{p'} = r^2 $
And this is the desired equation of the rotated cylinder.
For example, if the cylinder is rotate about the $y$ axis by $45^\circ$, then the rotation matrix will be
$ R = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && 0 && 1 \\ 0 && \sqrt{2} && 0 \\ -1 && 0 && 1 \end{bmatrix} $
The matrix we want is $ Q' = R Q R^T$, and equals,
$ Q' = R Q R^T = \frac{1}{2}\begin{bmatrix} 1 && 0 && 1 \\ 0 && \sqrt{2} && 0 \\ -1 && 0 && 1 \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} \begin{bmatrix} 1 && 0 && -1 \\ 0 && \sqrt{2} && 0 \\ 1 && 0 && 1 \end{bmatrix} $
Multiplying the first two matrix from the left
$Q' = \dfrac{1}{2} \begin{bmatrix} 1 && 0 && 0 \\ 0 && \sqrt{2} && 0 \\ - 1 && 0 && 0 \end{bmatrix} \begin{bmatrix} 1 && 0 && -1 \\ 0 && \sqrt{2} && 0 \\ 1 && 0 && 1 \end{bmatrix} $
Multiplying these two matrices, gives us $Q'$
$Q' = \dfrac{1}{2} \begin{bmatrix} 1 && 0 && - 1 \\ 0 && 2 && 0 \\ -1 && 0 && 1 \end{bmatrix} $
Now the equation of the rotated cylinder is
$ \mathbf{p'}^T Q' \mathbf{p'} = r^2 $
And since $\mathbf{p'} = [x, y, z]^T $, then the equation is
$ x^2 + 2 y^2 + z^2 - 2 x z = 2 r^2 $
EDIT: However, there is an easier way to write the equation of a rotated cylinder, using its rotated axis. Note that
$ Q = (I - \mathbf{kk}^T ) $
where $ \mathbf{k} $ is the unit vector along the $z$ axis ($\mathbf{k} = [0, 0, 1]^T$). Therefore, the matrix for the rotated cylinder as given above is
$ Q' = R Q R^T = R (I - \mathbf{k k}^T ) R^T = I - \mathbf{a a}^T $
where $ \mathbf{a} = R \mathbf{k} $ is the rotated axis unit vector, i.e. it is the vector resulting from applying the rotation matrix to the original unit axis vector $k$.
So, now for our example, instead of all that derivation, just compute the rotated cylinder axis unit vector. Since we're rotating the cylinder about the $y$ axis by $45^\circ$, then the new axis is
$ \mathbf{a} = R \mathbf{k} =\dfrac{1}{\sqrt{2}} [1, 0, 1]^T $
Hence,
$Q' = I - \mathbf{a a}^T = I - \dfrac{1}{2} \begin{bmatrix} 1 && 0&& 1 \\ 0 && 0 && 0 \\ 1 && 0 && 1 \end{bmatrix} = \dfrac{1}{2} \begin{bmatrix} 1 && 0 && -1 \\ 0 && 2 && 0 \\ -1 && 0 && 1 \end{bmatrix} $
which is what we got earlier. And the same equation in $x,y,z$ follows.
EDIT 2
As another example (a general one), suppose you rotated the vector $\mathbf{k}$ using the rotation matrix $R$, or you know the orientation of the final axis $\mathbf{a}$ in space, then you can express this vector is cylindrical coordinates as follows
$ \mathbf{a} = \begin{bmatrix} \sin(\theta) \cos(\phi) \\ \sin(\theta) \sin(\phi) \\ \cos(\theta) \end{bmatrix}$
Then the matrix $Q'$ is
$ Q' = I - \mathbf{a a}^T = \begin{bmatrix} a && d && e \\ d && b && f \\ e && f && c \end{bmatrix} $
And the equation of the cylinder will be
$ a x^2 + b y^2 + c y^2 + 2 d x y + 2 e x z + 2 f y z = r^2 $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4474103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
For all $n,k$ in positive integers, there exists $m_1,...,m_k$ such that the following holds Let $n,k$ be two positive integers. Prove that there exists $m_1,...,m_k$ in the positive integers such that
$$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$
Here is my attempt
We're going to construct a solution for the $m_i$'s, but rather than presenting it right away I will take time to go through the motivation behind it.
Assume that $\exists m_1,...,m_k$ such that $$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$
And consider $$\begin{align}
\prod_{i=1}^{k+1}\left(1+\frac{1}{m_i}\right)&=\left(1+\frac{1}{m_{k+1}}\right)\prod_{i=1}^k\left(1+\frac{1}{m_i}\right) \\ &= \left(1+\frac{1}{m_{k+1}}\right)\left(1+\frac{2^k-1}{n}\right) \\ &= 1+\frac{2^{k+1}-1}{n}
\end{align} $$
Where the last step follows from the assumption. Hence $$\frac{1}{m_{k+1}}=\frac{2^k}{2^k+n-1} \text{ or } m_i=\frac{2^{i-1}+n-1}{2^{i-1}}$$
Now we have to prove this solution actually works. In other words, $$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{2^{i-1}}{2^{i-1}+n-1}\right)=\prod_{i=1}^k\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right)$$
but expanding the RHS is not easy so let's go ahead and assume that the above is true $\forall k\le s$ we'll show that it's true for $k=s+1$
$$\begin{align}\prod_{i=1}^{s+1}\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right)&=\frac{2^{s+1}+n-1}{2^{s}+n-1}\prod_{i=1}^{s}\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right) \\&= \frac{2^{s+1}+n-1}{2^{s}+n-1}\left(1+\frac{2^s-1}{n}\right) \\&= 1+\frac{2^{s+1}-1}{n}\end{align} $$
Now my question is: Are the $m_i$'s integers? Since $$m_i=\frac{2^{i-1}+n-1}{2^{i-1}}$$
Then we should have $2^{i-1}\mid n-1$ but that's not true $\forall n,i$?
|
The $m_i$'s in your construction are not necessarily integers.
To illustrate the problem, consider the simple case of $k = 2$. You want to write $\frac{n + 3}n$ as a product $\frac{m_1 + 1}{m_1} \cdot \frac{m_2 + 1}{m_2}$.
Your strategy goes for $m_1 = n$ and $m_2 = \frac{n + 1}2$. However we see that $\frac{n + 1}2$ is not necessarily an integer.
In fact, the construction above works when $n$ is odd. For $n$ even, we can use instead $m_1 = \frac n 2$ and $m_2 = n + 2$.
This indicates how the solution would look like: you should separate the two cases of even/odd $n$.
We prove by induction on $k$. For $k = 1$, simply choosing $m_1 = n$ works. Now assume that the result is true for $k$ and we prove it for $k + 1$.
Thus we want to write $\frac{n + 2^{k + 1} - 1}n$ as a product $\prod_{i = 1}^{k + 1} \frac{m_i + 1}{m_i}$. As above, we will consider the parity of $n$.
If $n$ is odd, then we can write $$\frac{\frac{n + 1}2 + 2^k - 1}{\frac{n + 1}2} = \prod_{i = 1}^k \frac{m_i + 1}{m_i}$$ by induction hypothesis applied to $\frac{n + 1}2$. Choosing $m_{k + 1} = n$ gives us the willing identity.
If $n$ is even then we can write $$\frac{\frac n 2 + 2^k - 1}{\frac n 2} = \prod_{i = 1}^k \frac{m_i + 1}{m_i}$$ by induction hypothesis applied to $\frac n 2$. Choosing $m_{k + 1} = n + 2^{k + 1} - 2$ gives us the willing identity.
This finishes the induction step and hence the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4475634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Find expression of $c_n$, where $c_n = a_n + b_n$ Given the recurrence relation $a_{n+2} = 3a_{n+1} + 6a_n$ and $b_{n+2} = b_{n+1} + b_n$ I am supposed to find an expression of the recurrence relation for $c_n := a_n + b_n$. I tried to find some form of linear dependence to obtain a recurrence relation for $c_n$ but this did not let me finish.
Appreciate your help!
|
If I define the generating functions
$$ f(x) = \sum_{n=0}^\infty a_n x^n $$
$$ g(x) = \sum_{n=0}^\infty b_n x^n $$
then
$$ \begin{align*}
(6x^2 + 3x - 1) f(x) &= \sum_{n=0}^\infty (6 a_n x^{n+2} + 3 a_n x^{n+1} - a_n x^n)\\
&= -a_0 + (3 a_0 - a_1) x + \sum_{n=2}^\infty (6 a_{n-2} + 3 a_{n-1} - a_n) x^n \\
&= -a_0 + (3 a_0 - a_1) x
\end{align*} $$
$$ f(x) = \frac{A + Bx}{6x^2 + 3x - 1} $$
$$ \begin{align*}
(x^2+x-1) g(x) &= \sum_{n=0}^\infty (b_n x^{n+2} + b_n x^{n+1} - b_n x^n) \\
&= -b_0 + (b_0-b_1)x + \sum_{n=2}^\infty(b_{n-2}+b_{n-1}-b_n)x^n \\
&= -b_0 + (b_0-b_1)x
\end{align*} $$
$$ g(x) = \frac{C + Dx}{x^2+x-1} $$
The generating function for $c_n$ is $f+g$:
$$ \sum_{n=0}^\infty c_n x^n = \sum_{n=0}^\infty (a_n+b_n) x^n = f(x) + g(x) $$
And that sum is a fraction of polynomials, where the numerator $P(x)$ has degree at most $3$, but its exact form doesn't matter for this purpose:
$$ f(x) + g(x) = \frac{P(x)}{(6x^2+3x-1)(x^2+x-1)} = \frac{P(x)}{6x^4 + 9x^3 - 4x^2 - 4x + 1} $$
$$ \begin{align*}
P(x) &= (6x^4+9x^3-4x^2-4x+1)(f(x)+g(x)) \\
&= \sum_{n=0}^\infty (6c_n x^{n+4} + 9c_n x^{n+3} - 4c_n x^{n+2} - 4c_n x^{n+1} + c_n x^n) \\
&= Q(x) + \sum_{n=4}^\infty (6c_{n-4} + 9c_{n-3} -4c_{n-2} - 4c_{n-1} + c_n) x^n
\end{align*} $$
where $Q(x)$ is another polynomial of degree at most $3$ of "left over" terms. The equality forces $P(x) = Q(x)$ and all coefficients in the infinite sum must be zero.
So shifting the index, we get the recurrence relation for all $n \geq 0$:
$$ c_{n+4} = 4c_{n+3} + 4c_{n+2} - 9c_{n+1} - 6c_n $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4475812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Evaluate $\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$
$$\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$$
I completed the square:
$\int \dfrac{dx}{(1+x)\sqrt{2-(x-1)^2}}$
And then substituted $\sqrt 2\sin θ = x-1$ which gives
$\int \dfrac{dθ}{\sqrt2 \sinθ + 2}$
But now I'm stuck. Can someone please help?
|
If you have a rational function involving trigonometric functions, try a Weierstrass substitution.
$$t=\tan\left(\frac {\theta}2\right)\qquad\mathrm d\theta=\frac {2}{1+t^2}\,\mathrm dt\qquad\sin\theta=\frac {2t}{1+t^2}$$
Hence
$$\begin{align*}\int\frac {\mathrm d\theta}{2+\sqrt{2}\sin\theta} & =\int\frac {\mathrm dt}{t^2+t\sqrt{2}+1}\\ & =\int\frac {\mathrm dt}{\left(t+\frac 1{\sqrt{2}}\right)^2+\frac 12}\end{align*}$$
Can you complete the rest?
Another Approach
I also want to bring to light another possible substitution that, as far as I'm aware, isn't normally taught in a school setting (at least I didn't get taught this).
If you're daring enough and complacent with a large amount of algebraic manipulation up-front, then another possible solution would be to enforce an Euler Substitution of the Second Kind.
The substitution states that for an integral whose integrand is a purely rational function (i.e., no trigonometric terms)
$$\int R\left(x, \sqrt{ax^2+bx+c}\right)\,\mathrm dx$$
Where we impose the condition $c>0$, then an Euler substitution of the Second Kind gives
$$\sqrt{ax^2+bx+c}=xt+\sqrt{c}\qquad\qquad x=\frac {2t\sqrt{c}-b}{a-t^2}$$
In this case, since we have $c>0$, then our substitution is simply
$$\sqrt{1+2x-x^2}=xt+1\qquad\qquad x=\frac {2(1-t)}{1+t^2}$$
With this given substitution for $x$, we have that
$$x+1=\frac {3-2t+t^2}{1+t^2}\qquad\sqrt{1+2x-x^2}=\frac {1+2t-t^2}{1+t^2}\qquad\mathrm dx=\frac {2(t^2-2t-1)}{(1+t^2)^2}\,\mathrm dt$$
Right from the get-go, we can see that a lot of terms will cancel, which is the impressive nature of Euler Substitutions
$$\begin{align*}\int\frac {\mathrm dx}{(1+x)\sqrt{1+2x-x^2}} & =-2\int\frac {1+2t-t^2}{(1+t^2)^2}\frac {1+t^2}{t^2-2t+3}\frac {1+t^2}{1+2t-t^2}\,\mathrm dt\\ & =-2\int\frac {\mathrm dt}{(t-1)^2+2}\\ & =\sqrt{2}\arctan\left(\frac {1-t}{\sqrt2}\right)+C\end{align*}$$
Substituting our expression back in, then we get an impressive result
$$\int\frac {\mathrm dx}{(1+x)\sqrt{1+2x-x^2}}=\sqrt{2}\arctan\left(\frac {1+x-\sqrt{1+2x-x^2}}{x\sqrt2}\right)+C$$
Confirmed by Wolfram Alpha
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4478268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Find minimum of $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$ Let $a, b,c$ be the lengths of the sides of a triangle such that $a+b+c=2$, find the minimum value of $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$$
I don't have many ideas for this problem, my attemps:
The minimum value is $2\sqrt{2}+\dfrac{4}{3}$, the equality occurs for $a=b=c=\dfrac{2}{3}$
By Cauchy–Schwarz inequality, we have: $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge \dfrac{1}{\sqrt{2}}(a+b+b+c+c+a)=2\sqrt{2}$$
Thus, it suffices to prove that $$ab+bc+ca\ge\dfrac{4}{3}$$
This cannot be true because $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}=\dfrac{4}{3}$
Does anyone have any ideas, please give me a hint
|
My second proof:
Denote the expression by $f$.
Clearly, we have $a, b, c < 1$. We have
$$\sqrt{a^2 + b^2} - \frac{a + b}{\sqrt 2}
= \frac{\frac{(a - b)^2}{2}}{\sqrt{a^2 + b^2} + \frac{a + b}{\sqrt 2}}
\ge \frac{\frac{(a - b)^2}{2}}{\sqrt{1^2 + 1^2} + \frac{1 + 1}{\sqrt 2}} \ge \frac{(a - b)^2}{6}$$
which results in
$$\sqrt{a^2 + b^2} \ge \frac{a + b}{\sqrt 2} + \frac{(a - b)^2}{6}.$$
Thus, we have
\begin{align*}
f &\ge \frac{a + b + b + c + c + a}{\sqrt 2} + \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{6} + ab + bc + ca\\
&= \sqrt 2 (a + b + c) + \frac13 (a + b + c)^2\\
&= 2\sqrt 2 + \frac43.
\end{align*}
Also, when $a = b = c = 2/3$, we have $f = 2\sqrt{2} + 4/3$.
Thus, the minimum of $f$ is $2\sqrt 2 + \frac43$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4479540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Proving inequality using root of unity Let $\omega$ be a complex cube root of unity. It can be shown that if $a,b,c \in \mathbb{R}$, then $$(a+b+c)(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)= a^3+b^3+c^3-3abc$$
I was wondering if this could be extended to prove the AM/GM inequality for $n=3$. All that needs to be done is to prove that the left expression is non-negative, but I cannot find a way.
|
Leading the proof of $n=3$-AM-GM Ineq. from:
$$(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=a^3+b^3+c^3-3abc.$$
Enough to show that: $(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \ge 0.$
Let's make all of the conditions of equality the same in the proof of the above.
\begin{align}
\newcommand{w}{\omega}
& \text{WLOG in RHS, let } a\ge b\ge c.\\
&(a+b+c)(a+b\w+c\w^2)(a+b\w^2+c\w) \\
=&(a+b+c)\Big((a-c)+(b-c)\w\color{red}{+(c\w^2+c\w+c)}\Big)\Big((a-b)+(c-b)\w\color{red}{+(b\w^2+b\w+b)}\Big) \\
=&(a+b+c)\Big((a-c)+(b-c)\w\Big)\Big((a-b)+(c-b)\w\Big) \\
=&(a+b+c)\Big((a-b)(a-c)+(a-b)(b-c)\w+(a-c)(c-b)\w-(b-c)^2\w^2\Big) \\
=&(a+b+c)\Big((a-b)(a-c)+(a-b+c-a)(b-c)\w+(b-c)^2(\w+1)\Big) \\
=&(a+b+c)\Big((a-b)(a-c)-(b-c)^2\w+(b-c)^2\w+(b-c)^2\Big) \\
=&(a+b+c)\Big((a-b)(a-c)+(b-c)^2\Big) \\
\ge&(a+b+c)\Big((a-a)(a-a)+(b-b)^2 \Big) \\
= & 0.
\end{align}
Condition of Equality: $b=a, c=a, c=b \Rightarrow a=b=c.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4481472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluating $\lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$
Problem statement:
If
$$ A = \lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}, $$
Find $A^2$.
Solution:1)
\begin{align*}
\prod_{k=0}^{n} \binom{n}{k}
&= \prod_{k=1}^{n} \frac{n!}{k!(n-k)!}
= \frac{(n!)^{n+1}}{(1! \cdot 2!\cdots n!)^2} \\
&= \prod_{k=1}^{n} (n+1-k)^{n+1-2k} \\
&= \prod_{k=1}^{n} \left(\frac{n+1-k}{n+1}\right)^{n+1-2k},
\end{align*}
since $\sum_{k=1}^{n} (n+1-2k) = 0$. Taking log and limit, we get
\begin{align*}
& \frac{1}{n}\sum_{k=1}^{n}\left( 1 - \frac{2k}{n+1} \right) \ln\left( 1 - \frac{k}{n+1} \right) \\
&\to \int_{0}^{1} (1 - 2x)\log(1-x) \, \mathrm{d}x
= \frac{1}{2}.
\end{align*}
Source: FITJEE AITS 2020 FT-8 Paper 1 of JEE Advanced question 53
Could anybody explain the solution?
I tried taking log both sides to bring the power down but got stuck on evaluating $$\sum_{r=0}^n \ln\binom{n}{r}$$ further I think the answer should be $e$ instead of $0.5$ which is given.
|
We have the limit
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log \binom{n}{r}}{n(n+1)} $$
Applying Stolz Cesaro, the limit is equivalent to
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log \binom{n}{r}-\sum_{r=0}^{n+1} \log \binom{n+1}{r}}{n(n+1)-(n+1)(n+2)} $$
Since $\frac{\binom{n}{r}}{\binom{n+1}{r}}=1-\frac{r}{n+1}$, the limit simplifies to
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log(1-\frac{r}{n+1})}{-2(n+1)} $$
This is now a standard Riemann sum, which upon simplifying yields $\boxed{\frac12}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4483861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Proving $\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$ using $\epsilon$-$\delta$ definition I want to prove that $$\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$$ but I am not sure if my proof is valid because I did some algebraic manipulation. Can you please verify my proof?
Given $ \epsilon \gt 0 $, Choose $ \delta = \min\{3, \sqrt{2\epsilon}\} $
Suppose $ 0 \lt x - 2 \lt \delta $
Check:
$$\begin{align}
\left|\frac{x - 2}{\sqrt{x^{2} - 4}} + 1 - 1\right| &= \left| \frac{x - 2}{\sqrt{x^{2} - 4}}\right|\\
&= \frac{x - 2}{\sqrt{x^{2} - 4}} \\
&= \frac{x - 2}{\sqrt{x + 2}\sqrt{x - 2}} \\
&= \frac{\left(x - 2\right)^{1}}{\sqrt{x + 2}\cdot\left(x - 2\right)^{0.5}} \\
&= \frac{\sqrt{x - 2}}{\sqrt{x + 2}}
\end{align}$$
$$0 \lt x - 2 \lt 3\quad\Rightarrow\quad 4 \lt x + 2 \lt 7
\quad\Rightarrow\quad 2 \lt \sqrt{x + 2} \lt \sqrt{7}$$
$\Rightarrow$
$$\frac{\sqrt{x - 2}}{\sqrt{x + 2}} \lt \frac{\sqrt{\delta}}{2} \le \epsilon\tag*{$\blacksquare$}$$
|
HINT
A slightly different approach. Suppose that $0 < x - 2 < \delta_{\varepsilon}$. Then it results that
\begin{align*}
\left|\frac{x - 2}{\sqrt{x^{2} - 4}}\right| = \frac{\sqrt{(x - 2)^{2}}}{\sqrt{(x - 2)(x + 2)}} = \frac{\sqrt{x - 2}}{\sqrt{x + 2}} < \frac{\sqrt{x - 2}}{2} < \frac{\sqrt{\delta_{\varepsilon}}}{2} := \varepsilon
\end{align*}
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4484513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find the image of the parabola $x = \frac{9}{4} − \frac{y^2}{9}$ under the principal square root mapping $w = z^{1/2}$ with $z \in \mathbb{C}$. Find the image of the parabola $x = \frac{9}{4} − \frac{y^2}{9}$ under the principal square root
mapping $w = z^{1/2}$ with $z \in \mathbb{C}$.
Definition of principal square root mapping: $z^{1/2} = \sqrt{|z|}e^{i\operatorname{Arg}(z)/2}$.
Arg is the principal argument of $z$, so $-\pi<\operatorname{Arg}(z) \le \pi$
I've tried plugging in the expression $\frac{9}{4} − \frac{y^2}{9}+yi$ into the principal square root mapping, but can't seem to get anywhere.
|
Suppose $w=(u,v)$ is in the image of the parabola under the map principal branch of square root of $z$.
Then $\exists z=(x, y) $ on the parabola such that $√z=w$
Then $z=w^2$ implies
$x+iy=u^2-v^2+2iuv$
Hence $x=u^2-v^2 , y=2uv$
$(x, y) $ lies on the parabola $x=\frac{9}{4}-\frac{y^2}{9}$ implies
$u^2-v^2=\frac{9}{4}-\frac{4u^2v^2}{9}$
On simplification, we get
$36u^2-36v^2+16u^2v^2=81$
Hence image of the parabola is the set $\{(u,v):36u^2-36v^2+16u^2v^2=81\}$
$$\text{ Parabola} \space x=\frac{9}{4}-\frac{y^2}{9}$$
After solving, we get a pair of straight lines $u=\pm\frac{3}{2}$
$$\text{ Pair of straight lines} \space u=\pm\frac{3}{2}$$
For any point $w$ in the image, we have $-\frac{\pi}{2}<\operatorname{Arg} (w) \le\frac{\pi}{2}$.
Hence image of the parabola under the principal branch of $z^{\frac{1}{2}} $ is the line $u=\frac{3}{2}$
Image credit: GeoGebra
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4486587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Calculation of Residue. Let $c=\cos \dfrac{\pi}{5}, f(z)=\dfrac{z^2-2cz+1}{z^4-z^3+z^2-z+1}$.
$e^{\frac{3\pi}{5}i}$ is one of the pole of $f$. (This is because $f$ can be written as $f(z)=\frac{(z+1)(z^2-2cz+1)}{z^5+1}$.)
Then, calculate the Residue of $f$ at $e^{\frac{3\pi}{5}i}=:a$.
I calculated using the formula of Residue, but the calculation is complicated and I don't know how I should proceed.
\begin{align}
\mathrm{Res}(f,a)
&=\displaystyle\lim_{z\to a} (z-a)f(z)\\
&=\lim_{z\to a} \dfrac{(z-a)(z^2-2cz+1)}{z^4-z^3+z^2-z+1}\\
&=\lim_{z\to a}\dfrac{z^2-2cz+1+(z-a)(2z-2c)}{4z^3-3z^2+2z-1}\\
&=\dfrac{a^2-2ca+1}{4a^3-3a^2+2a-1}.
\end{align}
I have to simplify this, but I don't know how I can do.
I think I have to use some technical method. Thanks for any idea.
|
We can considerably simplify the residue calculation when expanding numerator and denominator with $z+1$. This way we can effectively get rid of the denominator. We consider
\begin{align*}
f(z)=\frac{(z+1)\left(z^2-2cz+1\right)}{z^5+1}
\end{align*}
and obtain
\begin{align*}
\mathrm{Res}&(f,a)
=\displaystyle\lim_{z\to a} (z-a)f(z)\\
&\color{blue}{=\lim_{z\to a}}\color{blue}{\frac{(z-a)(z+1)\left(z^2-2cz+1\right)}{z^5+1}}\\
&=\lim_{z\to a}\frac{\left(z^2+(1-a)z-a\right)\left(z^2-2cz+1\right)}{z^5+1}\\
&=\lim_{z\to a}\frac{\left(2z+(1-a)\right)\left(z^2-2cz+1\right)+\left(z^2+(1-z)z-a\right)(2z-2c)}{5z^4}\tag{1}\\
&=\frac{(a+1)\left(a^2-2ca+1\right)}{5a^4}\tag{2}\\
&=-\frac{1}{5}a(a+1)\left(a^2-2ca+1\right)\tag{3}\\
&=-\frac{1}{5}\left(a^4+(1-2c)a^3+(1-2c)a^2+a\right)\\
&=-\frac{1}{5}\left(a+a^4\right)-\frac{1}{5}\left(1-2c\right)\left(a^2+a^3\right)\\
&\color{blue}{=-\frac{1}{5}\left(a-\frac{1}{a}\right)-\frac{1}{5}(1-2c)\left(a^2-\frac{1}{a^2}\right)}\tag{4}
\end{align*}
Since
\begin{align*}
a-\frac{1}{a}&=\exp\left(3 i\pi/5\right)-\exp\left(-3 i\pi/5\right)=2i\sin(3\pi/5)\\
a^2-\frac{1}{a^2}&=-\exp\left(i\pi/5\right)+\exp\left(-i\pi/5\right)=-2i\sin(\pi/5)\\
\end{align*}
and noting that
\begin{align*}
c=\cos(\pi/5)=\frac{1}{4}\left(1+\sqrt{5}\right)
\end{align*}
it shouldn't be too hard to finish the calculation.
Comment:
*
*In (1) we apply L'Hôpital's rule.
*In (2) we calculate the limit noting the right-hand term of the numerator is zero.
*In (3) we use $a\cdot a^4=-1$.
*In (4) we again use properties of the units: $a\cdot a^4=a^2\cdot a^3=-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4489465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
According to the figure, what is the area of the triangle $BCD$ of type $x$?
$Q:$ In a circle $\omega \ $ with a center $A$, $E\in \omega \ $ is the tangent point, $[BA]\bot[AC] \ $, $\angle ABD= \angle DCB \ $, $\angle DBC= \angle DCA \ $, $D\in\omega \ $, $|EC|=x \ $ What is the area of the triangle $BCD$ of type $x$?
$\text{Try on me:}$ I drew $[AD]$ and named it $\angle BAD=\theta$ and $|BD|=a, |DC|=b$. I also knew that $\alpha+ \beta =45^{\circ}$. I first found $\frac{\sin{\beta}}{\sin{\alpha}}=\frac{b}{a}$ by applying the sine theorem in the triangles $\triangle ABD$ and $\triangle DCA$. Then $\tan{\theta}=\frac{a^2}{b^2}$ came out of the trigonometric ceva theorem. I have drawn a right triangle between the point $D$ and the right part of $\triangle DCA$ by selecting the constant of the ratio $k$. By trying a little, I got the following two equivalences, $r$ is the half-diameter of the circle: $$2r^2+2x^2=\frac{1}{2k^2},$$ $$a^4+b^4+\sqrt{2}ab=\frac{3}{4k^2}.$$ I tried to make a few more calculations and make observations by choosing $r=1$, but I couldn't get anything. Can you help? Any help will be greatly appreciated.
|
First of all note that $\angle ABC=\alpha+\beta=\angle ACB$, that means triangle ABC isosceles, so :
$\angle ABC=\angle ACB=45^o\Rightarrow \alpha=\beta=22.5^o$
Triangle AEC is right angled and we have:
$AC^2=r^2+x^2\Rightarrow AC=\sqrt{r^2+x^2}$
in triangle ABC:
$BC=AC\sqrt 2=\sqrt{2(r^2+x^2)}$
$AH=AC \sin 45^o=\frac{\sqrt {(r^2+x^2)}}{\sqrt 2}=HC$
$DH=HC tan 22.5^o=\frac{BC}2 tan 22.5^o=\frac{\sqrt{2(r^2+x^2)}}2 tan 22.5$
Area of DBC is:
$S_{DBC}= DH\times HC=\frac{\sqrt{2(r^2+x^2)}} 2 tan 22.5^o\times \frac{\sqrt {(r^2+x^2)}}{\sqrt 2}=\frac12(r^2+x^2)tan 22.5^o$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4491587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.
By multiplying both sides by $(x+1)^2$ we get,
$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$
$$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$
But I can't factor the third degree polynomial.
|
To solve $4x^3+12x^2+15x+5=0$, start by applying the rational root theorem. In this case, potential rational roots are $\pm \lbrace \frac{1}{4}, \frac{1}{2}, 1, \frac{5}{4}, \frac{5}{2}, 5 \rbrace$. Trying them out, we see that $x=\frac{-1}{2}$ satsfies the equation. So $x + \frac{1}{2}$ (or for the sake of avoiding fractions, $2x + 1$) is a factor.
$$4x^3+12x^2+15x+5 = (2x + 1)(ax^2 + bx + c)$$
The quadratic term works out to $2x^2 + 5x + 5$. By the quadratic formula, $x = \frac{-5 \pm i\sqrt{15}}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4493459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Minimum of a function defined by a sum I have the following function, for a fixed integer $n\in \mathbb N$, $\displaystyle\varphi(x) = \sum_{k=0}^{n} \frac{(-1)^k}{x-k}$, for $x\neq l$ for all $l\in\{0,\ldots,n\}$. I am suspecting that $\min\limits_{x\in (0,n)} \left|\varphi(x)\right|\ge 2$ and want to prove it. So my question is
*
*Is it true what I am suspecting for all $n\in\mathbb N$?
*If not what values of $n$ this is true?
I checked for $n=1$ and I proved easily that the minimum is larger than $4$. Can anyone help me with this in general case?
|
I think I have found an answer for the even case (when $n$ is even). Indeed, let $n=2p$ and since $\varphi(x) = -\varphi(2p-x)$ it is enough to study the following case:
*
*$x\in(2l-1, 2l)$ for $l\in\{1,\ldots,p\}$
\begin{align}
\left|\varphi(x)\right| &= \left|\sum_{k=0}^{2p} \frac{(-1)^k}{x-k}\right|\\
&= \left|\frac{1}x + \sum\limits_{k=1}^{p} \frac{1}{(x-2k)(x-2k+1)}\right|\\
&= \left|\frac{1}{(x-2l)(x-2l+1)} + \frac1x + \sum\limits_{k=1}^{l-1} \frac{1}{(x-2k)(x-2k+1)} + \sum\limits_{k=l+1}^{p} \frac{1}{(x-2k)(x-2k+1)}\right|\\
&\ge \left|\frac{1}{(x-2l)(x-2l+1)}\right| - \frac1x - \sum\limits_{k=1}^{l-1} \frac{1}{(x-2k)(x-2k+1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(x-2k)(x-2k+1)}\\
&\ge 4 - \frac1{2l-1} - \sum\limits_{k=1}^{l-1} \frac{1}{(2l-1-2k)(2l-1-2k+1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(2l-2k)(2l-2k+1)}\\
&= 4 - \frac1{2l-1}-\sum\limits_{k=1}^{l-1} \frac{1}{(2l-2k)(2l-2k-1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(2l-2k)(2l-2k+1)}\\
&= 4 - \frac1{2l} - \frac{1}{2l(2l-1)}-\sum\limits_{k=1}^{l-1} \frac{1}{2k(2k-1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(2(k-l))(2(k-l)-1)}\\
&\ge 4 - \frac12 - \sum\limits_{k=1}^l \frac{1}{2k(2k-1)} - \sum\limits_{k=1}^{p-l}\\
&\ge 4 - \frac12 - 2\sum_{k=1}^{\infty}\frac{1}{2k(2k-1)}\\
&= 4 - \frac12 - 2\ln 2\ge 2.
\end{align}
It seems that the odd case can be done by the same method.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4499350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $()^4$.
I tried to rewrite $8x^4=(2x)^4\times\frac{1}{2}$, but It wasn't helpful :
$$\left(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\right)((2x)^4+(3y)^4+(4z)^4)\ge (x+y+z)^4$$
So, is there any nice transform to apply C-S inequality?
or should I apply Lagrange Multiplier?
|
Let $\,u = 2x, v=3y, w=4z\,$ then the problem is to minimize $u^4 / 2 + v^4 / 3 + w^4 / 4$ under the constraint $u / 2 + v / 3 + w / 4 = 13/4$. By the weighted power means inequality:
$$
\sqrt[4]{\frac{u^4 / 2 + v^4 / 3 + w^4 / 4}{1/2+1/3+1/4}} \;\ge\; \frac{|u| / 2 + |v| / 3 + |w| / 4}{1/2+1/3+1/4} \;\ge\; \frac{13/4}{13/12} = 3
$$
The minimum is attained for $\,u=v=w=3\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4499469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
}
|
Proving that $\frac {(n+1)n(n-1)...(n-i)}{i+1}$ can be used as the summ of $k^h$ I have previously proved that $$\sum_{k=i}^{n} k(k-1)(k-2)\cdots(k-i+1) = \frac{(n+1)\cdot n\cdot (n-1)\cdots(n-i+1)}{i+1}$$
From here I am supposed to show that the formula above can be used to compute the sum of $k^h$, for example, $\sum_{k=1}^{n}k^2$ or $\sum_{k=1}^{n}k^3$, starting with $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$
I cannot figure out the connection between these and do not know where to start...
|
It is convenient to use the falling factorial notation
\begin{align*}
k^{\underline{i}}:=k(k-1)\cdots (k-i+1)
\end{align*}
With this notation OPs identity can be written as
\begin{align*}
\sum_{k=i}^nk^{\underline{i}}=\frac{(n+1)^{\underline{i+1}}}{i+1}
\end{align*}
and since summands with $1\leq k<i$ are zero we can also write
\begin{align*}
\color{blue}{S_i(n):=\sum_{k=1}^nk^{\underline{i}}=\frac{(n+1)^{\underline{i+1}}}{i+1}}\tag{1}
\end{align*}
We can now iteratively use $S_i(n), i=1,2,3,\ldots $ to calculate
\begin{align*}
\sum_{k=1}^nk^i
\end{align*}
Case $i=1$:
Since $k=k^{\underline{1}}$ we obtain from (1)
\begin{align*}
\color{blue}{\sum_{k=1}^n k}=S_1(n)=\frac{(n+1)^{\underline{2}}}{2}
\color{blue}{=\frac{1}{2}\left(n+1\right)n}\tag{2}
\end{align*}
Case $i=2$:
Since $k^2=k(k-1)+k=k^{\underline{2}}+k^{\underline{1}}$ we obtain from (1) and (2)
\begin{align*}
\color{blue}{\sum_{k=1}^n k^2}&=S_2(n)+S_ 1(n)=\frac{(n+1)^{\underline{3}}}{3}+\frac{(n+1)^{\underline{2}}}{2}\\
&=\frac{(n+1)n(n-1)}{3}+\frac{(n+1)n}{2}\\
&\,\,\color{blue}{=\frac{1}{6}\left(2n^2+3n^2+n\right)}
\end{align*}
This way we can iteratively calculate $\sum_{k=1}^n k^i, i=3,4,\ldots$. Here it is useful to know that
\begin{align*}
k^i=\sum_{j=0}^i\begin{Bmatrix} i \\ j \end{Bmatrix} k^{\underline{j}}
\end{align*}
with $\begin{Bmatrix} i \\ j \end{Bmatrix}$ the Stirling numbers of the second kind.
Note: The sum formula (1) is a discrete analogon to
\begin{align*}
\int_{0}^xt^i\,dt=\frac{x^{i+1}}{i+1}
\end{align*}
which is a main theme in this answer .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4500318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to find this two-variables limit of $f$ as $(x,y)$ approaches to $(0,0)$? I am trying to evaluate the limit $$\lim_{(x,y) \to (0,0)}\frac{2x^6-y^{10}}{x^2-y^3}$$
I have seen that both the reiterated limits are zero. Moreover, I have evaluated the limit by different curves such as $y=x^2$ and the limit equals zero as well.
However, the polar coordinates $$\lim_{r \to 0}\frac{r^4(2\cos^6(\theta)-r^4\sin^{10}(\theta))}{\cos^2(\theta)-r\sin^3(\theta)}$$ make me think that the limit does not exist because it depends on the $\cos^2\theta$ factor, but I don't know how to prove it or which curve to use.
|
Consider the sequence
$(\frac{1}{ n},0)\to(0,0)$
Then $f(\frac{1}{n},0)=\frac{1}{n^4}\to 0 $
Again the sequence $(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})\to(0,0)$
But $\begin{align}f(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})&=\frac{\frac{2(n^2+1)^3}{n^6}-\frac{1}{n^3\sqrt[3]n{}}}{\frac{1}{n^3}}\\&=2(1+\frac{1}{n^2}) ^3-\frac{1}{\sqrt[3]n}
\end{align}$
Hence $f(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})\to 2$
Hence $\lim_{(x,y) \to (0,0)}\frac{2x^6-y^{10}}{x^2-y^3}$ doesn't exists.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4501815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
How to find the roots of Martrix Norm?
Problem: Compute $||A||$ where $A = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$
Attempt:
$$
\begin{align*}
\det(A^TA-λI_2) &=
\det \left( \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}
- \begin{pmatrix} λ & 0 \\ 0 & λ\end{pmatrix} \right) \\
&= \det \left( \begin{pmatrix} 1 & 1 \\ 1 & 2\end{pmatrix}
- \begin{pmatrix} λ & 0 \\ 0 & λ\end{pmatrix} \right) \\
&= \det \left( \begin{pmatrix} 1-λ & 1 \\ 1 & 2-λ\end{pmatrix} \right) \\
&= (1-λ)(2-λ)-1(1) \\
&= λ^2-3λ+2-1 \\
&= λ^2-3λ+1
\end{align*}$$
The book, skipped the solution and has the answer
"which has roots $λ=\frac{3}{2}\pm\frac{1}{2}\sqrt{5}$. Therefore the larger of the two eigenvalues of $A^TA$ is $λ=\frac{3}{2}\pm\frac{1}{2}\sqrt{5}$, giving $||A||=\sqrt{\frac{3}{2}\pm\frac{1}{2}\sqrt{5}}$."
I don't get how it got that answer. I tried getting the eigenvalues of $A$, but it's $λ=1$.
|
Your calculation of the determinant is correct. To find the roots, just use the quadratic formula,
$$ax^2 + bx +c =0 \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here,
$$\begin{align*}
&\lambda^2 - 3 \lambda + 1 = 0 \\
&\implies \lambda = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9-4}}{2} =\frac{3 \pm \sqrt 5}{2}
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4502683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
|
We want to find the remainder from the division:$$\frac{x^{20} + x^{19} + \ldots + x^{11} + x^{10}}{x^3 + x}$$
Now note that $$ (x^3 + x)(x^{n+1} + x^n ) = x^{n+4} + x^{n+3} + x^{n+2} + x^{n+1}, $$
so that
$$ \frac{x^{20} + x^{19} + \ldots + x^{11} + x^{10}}{x^3 + x} = x^{17} + x^{16} + x^{13} + x^{12} + \frac{x^{12} + x^{11} + x^{10}}{x^3 + x} $$
and using long division you can show that the remainder from the final term is $-x.$
Peter's method is more efficient though.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
}
|
How does one algebraically find the total number of intersections between $\sin(x)$ and a linear equation? Take the function $f(x) = \dfrac{x}{8}$ and $\sin(x)$. By plotting, we can see that the graphs intersect at 7 distinct points, indicated by the grey dots. However, how would one approach such a problem algebraically? There seems to be no useful way to isolate the $x$ variable in a manner that would let us solve the problem.
|
At $x=\dfrac{5\pi}{2}$, compare the y-values of $y=\dfrac{x}{8}$ and $y=\sin{x}$:
$\dfrac{\frac{5\pi}{2}}{8}<\dfrac{5(3.2)}{16}=\sin{\frac{5\pi}{2}}$. So there are at least $7$ intersections.
If $x=3\pi$ then $\dfrac{x}{8}=\dfrac{3\pi}{8}>1$, and $1$ is the maximum value of $\sin{x}$. So for $x\ge 3\pi$ there are no intersections.
Therefore there are exactly $7$ intersections.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4506023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta}$$
Adding two versions together yields
$$
\begin{aligned}
2 I &=2 a \int_{0}^{\pi} \frac{d \theta}{a^{2}-b^{2} \cos ^{2} \theta} \\
&=4 a\int_0^{\frac{\pi}{2}} \frac{\sec ^{2} \theta}{a^{2} \sec ^{2} \theta-b^{2}} d \theta \quad \textrm{( By symmetry )}\\
&=4 a{\int_{0}^{\infty}} \frac{d t}{\left(a^{2}-b^{2}\right)+a^{2} t^{2}} \\
&=\frac{4}{\sqrt{a^{2}-b^{2}}}\left[\tan^{-1} \left(\frac{at}{\sqrt{a^{2}-b^{2}}}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\frac{\pi}{\sqrt{a^{2}-b^{2}}}}
$$
Are there any other methods to deal with the integral?
|
By Weierstrass substitution
Letting $t=\tan \frac{\theta}{2}$ transforms the integral
$$
\begin{aligned}
I&=\int_{0}^{\pi} \frac{1}{a+\frac{b\left(1-t^{2}\right)}{1+t^{2}}} \frac{2 d t}{1+t^{2}}\\
&=2 \int_{0}^{\infty} \frac{d t}{(a+b)+(a-b) t^{2}}\\
&=\frac{2}{\sqrt{a-b} \sqrt{a+b}} \left[\tan ^{-1}\left(\frac{\sqrt{a-b}t}{\sqrt{a+b}}\right)\right]_{0}^{\infty}\\
&=\frac{\pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4508099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
}
|
Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$
Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$
$$\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
My work:
Let$$S(n)=\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
By theorem of triviality, if any of $x_i$'s are $0$ the inequality is certainly true. So assume all numbers are $\gt0$
$S(1)$ says ${x_1}^3\le {x_1}^3$ which is certainly true.
$S(2)$ says $({x_1}^2+{x_2}^2)x_1x_2\le\frac18(x_1+x_2)^4$ which reduces to $0\le(x_1-x_2)^4$ which is certainly true.
Assume $S(k)$ is true. Now we just needs to prove that $S(k+1)$ is true. But I'm having a hard time in doing that.
Any help is greatly appreciated. Or is there any better method than induction$?$
|
theorem :
Let $x_1,x_2,x_3,y_1,y_2,y_3\in(0,\infty)$ then if we have :
$$x_1+x_2+x_3\geq y_1+y_2+y_3$$
And for $i\neq j,1\leq i\leq 3,1\leq j\leq 3$:
$$|x_i-x_j|\leq |y_i-y_j|$$
Then we have :
$$x_1x_2x_3\geq y_1y_2y_3$$
Case $n=3$ :
Let $y_1=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{xa},y_2=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{xb},y_3=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{ab}$
Let : $x_1=x_2=x_3=\left(a+b+x\right)/3$
We have the easy inequalities :
$$0\geq\left(3-\frac{6\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)}{3\left(a+b+x\right)}\right)^{\frac{1}{2}}\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)-\left(a+b+x\right)\geq \left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)-\left(a+b+x\right)$$
Remains to apply the theorem to get the case $n=3$ with a refinement (bonus)
The general case is similar .Also think to Maclaurin's inequalities .
Some details :
A classical idea is using the Bernoulli's inequality as we have with $a,b,c>0$ :
$$-2\left(2-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)+3\left(2-\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(a+b+c\right)}\right)\leq \left(3-\frac{6\left(\sqrt{ca}+\sqrt{cb}+\sqrt{ab}\right)}{3\left(a+b+c\right)}\right)^{3}-\left(3-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)^{2}$$
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4508477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
}
|
Find the limit of $\frac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$ Find the limit $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$$
For the numerator we have $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac{1}{3}\right)^{n+1}}{1-\frac13}=\dfrac32-\dfrac12\dfrac{1}{3^n}.$
By analogy, $1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}=\dfrac54-\dfrac14.\dfrac{1}{5^n}$
So we have $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}=\lim_{n\to\infty}{\dfrac{\frac32-\frac12.\frac{1}{3^n}}{\frac54-\frac14.\frac{1}{5^n}}}=\lim_{n\to\infty}{\left(\dfrac{3^{n+1}-1}{2.3^n}\cdot\dfrac{4.5^n}{5^{n+1}-1}\right)}$$ What am I supposed to do next?
My initial mistake was that I thought $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\left(\frac13\right)^n}{1-\frac13}$. How could we actually see (and show) that the terms in the sum(s) are not $n$, but $n+1$?
|
Sum of IGP is $$S_\infty=1+r+r^2+r^3+\ldots=\frac{1}{1-r}.$$
So the required limit $$=\frac{1/(1-1/3)}{1/(1-1/5)}=\frac{6}{5}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4509654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Proof for lemma for existence of Taylor's Polynomial There is lemma that says, for every $a \in \mathbb{R}$, for every polynomial $P_n(x)$ of degree at most $n$ is possible to write in this form:
$$P_n(x) = \sum_{k=0}^{n}c_{k}(x - a)^k$$
where $c_k = \frac{P_n^{(k)}(a)}{k!}$.
There is proof using induction:
$$P_{n+1}(x) = \sum_{k=0}^{n + 1}c_{k}x^{k} = c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\sum_{k=0}^{n}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} + \sum_{k=0}^{n}c_{k}x^{k}$$
I can understand right side is equal to the left side:
$$c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\sum_{k=0}^{n}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\left(\sum_{k=0}^{n + 1}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} - {n + 1 \choose n + 1}x^{n + 1}(-a)^{n + 1 - n - 1}\right) + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\left((x - a)^{n + 1} - x^{n + 1}\right) + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}x^{n + 1} + \sum_{k=0}^{n}c_{k}x^{k} = \sum_{k=0}^{n + 1}c_{k}x^{k}$$
But can't understand proof it's also equal to Taylor's Polynomial. The closest I got is:
$$c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\sum_{k=0}^{n}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}(x - a)^{n + 1} + \sum_{k=0}^{n}c_{k}x^{k} - c_{n + 1}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} = \\ = c_{n + 1}(x - a)^{n + 1} + \sum_{k=0}^{n}c_{k}(x - a)^k\left(\frac{x_k}{(x - a)^{k}} - \frac{c_{n + 1}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k}}{c_{k}(x - a)^{k}}\right)$$
Source: https://www2.karlin.mff.cuni.cz/~stanekj/vyukaLS2021/prednaska_MAII_2021_06_02.pdf - page 5
Thanks
|
I write the proof step by step. I hope that it would be helpful.
If the induction hypothesis would be true for $n$, then for $n + 1$ we can write:
\begin{align*}
P_{n+1}(x) = \sum_{k = 0}^{n + 1} c_kx^k = c_{n+1}\color{blue}{x^{n+1}} + \sum_{k = 0}^{n}c_kx^k
\end{align*}
But $x^{n + 1} = (x - a)^{n+1} - \sum_{k = 0}^n \binom{n + 1}{k}(-a)^{n + 1 - k}x^k$, so we have:
\begin{align*}
P_{n + 1}(x) &= c_{n + 1}\color{blue}{\left((x - a)^{n + 1} - \sum_{k = 0}^n \binom{n + 1}{k} (-a)^{n + 1 - k}x^k\right)} + \sum_{k = 0}^n c_kx^k\\
&= c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\color{magenta}{\sum_{k = 0}^n} \binom{n + 1}{k} (-a)^{n + 1 - k}x^k + \color{magenta}{\sum_{k = 0}^n} c_kx^k\\
&= c_{n + 1}(x - a)^{n + 1} + \color{magenta}{\sum_{k = 0}^n} \left(c_kx^k - \binom{n + 1}{k}(-a)^{n + 1 - k}c_{n + 1}x^k\right)\\
&= c_{n + 1}(x - a)^{n + 1} + \color{red}{\sum_{k = 0}^n \left(c_k - \binom{n + 1}{k}(-a)^{n + 1 - k}c_{n + 1}\right)x^k}\\
&= c_{n + 1}(x - a)^{n + 1} + \color{red}{Q_n(x)} \tag{$1$}
\end{align*}
Because $Q_n(x)$ is a polynomial of degree at most $n$, from the induction hypothesis we can write
\begin{align*}
Q_n(x) = \sum_{k = 0}^n \frac{Q_n^{(k)}(a)}{k!}(x - a)^k \tag{$2$}
\end{align*}
On the other hand, because $Q_n(x) = P_{n + 1}(x) - c_{n + 1}(x - a)^{n + 1}$ from $(1)$, by differentiating we have
\begin{align*}
\frac{Q_n^{(k)}(x)}{k!} &= \frac{P_{n + 1}^{(k)}(x)}{k!} - c_{n + 1}\frac{\left((x - a)^{n + 1}\right)^{(k)}}{k!}\\
&= \frac{P_{n + 1}^{(k)}(x)}{k!} - c_{n + 1}\frac{(n + 1)n\cdots(n + 1 - k)(x - a)^{n - k + 1}}{k!}
\end{align*}
By putting $x = a$ in the above equation, we get
\begin{align*}
\frac{Q_n^{(k)}(a)}{k!} = \frac{P_{n + 1}^{(k)}(a)}{k!} \tag{$3$}
\end{align*}
Also, it's obvious that $c_{n + 1} = \frac{P_{n + 1}^{(n + 1)}(a)}{(n + 1)!}$. By using $(3)$ and $(2)$ in $(1)$ we get the result:
\begin{align*}
P_{n + 1}(x) &= c_{n + 1}(x - a)^{n + 1} + Q_n(x)\\
&= \frac{P_{n + 1}^{(n + 1)}(a)}{(n + 1)!}(x - a)^{n + 1} + \sum_{k = 0}^n \frac{P_{n + 1}^{(k)}(a)}{k!}(x - a)^k\\
&= \sum_{k = 0}^{n + 1} \frac{P_{n + 1}^{(k)}(a)}{k!}(x - a)^k
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4511339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43$$ Or in general, where there are $n$ variables can we say that their sum (in the same format as above) is no less than $\frac{n}{n-1}$$?$
All the variables used here are positive real numbers.
My thoughts:
I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.
|
Applying Cauchy-Schwars inequality, we have
$$\left(\overbrace{\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}}^A\right)\left(x(y+z)+y(x+z)+z(y+x)\right)\ge(x+y+z)^2\implies A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}$$
Then let's apply AM-GM inequality:
$$
\begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\ y^2+z^2\ge 2yz\end{cases}
$$
\begin{align}&2(x^2+y^2+z^2)\ge2(xy+yz+xz)\\
\implies &x^2+y^2+z^2 \ge xy+yz+xz\\
\implies &(x+y+z)^2\ge 3(xy+yz+xz)\end{align}
Hence, we obtain:
$$A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}≥\frac{3(xy+yz+xz)}{2(xy+yz+xz)}=\frac 32$$
Now, let's apply Cauchy-Schwarz again:
$$\left(\overbrace{\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}}^A\right)\left(w(x+y+z)+x(w+y+z)+y(x+w+z)+z(x+y+w)\right)\implies A\ge \frac{(x+y+z+w)^2}{2(xy+xz+xw+yz+yw+zw)}$$
Let's apply AM-GM inequality again:
$$
\begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\x^2+w^2\ge 2xw\\ y^2+z^2\ge2yz\\y^2+w^2\ge 2yw\\z^2+w^2\ge 2zw\end{cases}
$$
\begin{align}&3(x^2+y^2+z^2+w^2)\ge 2(xy+xz+xw+yz+yw+zw)\\
\implies &x^2+y^2+z^2+w^2 \ge \frac 23 (xy+xz+xw+yz+yw+zw)\\
\implies &(x+y+z+w)^2\ge \frac 83 (xy+xz+xw+yz+yw+zw)\end{align}
Finally, we have
$$A\ge \frac{(x+y+z+w)^2}{2(xy+xz+xw+yz+yw+zw)}\implies A\ge \frac 86=\frac 43$$
Now, can you generalize this method using a simple (just a few steps) combinatorics?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4511844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
}
|
Can anyone explain this process of solving? (Differentiation) I'm at differentiation of algebraic functions. There's an example in the module that I couldn't quite get how it led to that.
$y=\frac {(x+1)^3}{x^2}$
It's solved by using a combination of quotient and power rules. I'll enumerate how it's solved.
$(1) y′= \frac{(x^2(3(x+1)^2))–(x+1)^32x}{x^4}$
$(2) y′= \frac{x(x+1)^2 (3x–2(x+1))}{x^4}$
$(3) y′= \frac {(x+1)^2 (3x-2x-2)}{x^3}$
$(4) y′= \frac {(x+1)^2 (x-2)}{x^3}$
How did $(2)$ came to be? Why was it done like that? I just can't wrap my head around it. It just looked like it skipped a couple steps (at least to me). Can anyone help me with this?
|
If it gets too messy, you can always use logarithmic differentiation. Take the natural logarithm of both sides to get:
$$\ln y = \ln((x+1)^3) - \ln(x^2)$$
$$\ln y = 3 \ln(x+1) - 2 \ln x$$
and so using the chain rule, we have that $\frac{d}{dx} (\ln y)$, where $y$ is a function of $x$), is $\frac{1}{y} \cdot y' = \frac{y'}{y}$:
$$\frac{y'}{y} = \frac{3}{x+1} - \frac{2}{x} = \frac{3x - 2(x+1)}{x(x+1)} = \frac{x-2}{x(x+1)}$$
so using what $y$ is, we have that:
$$y' = \frac{dy}{dx} = \frac{x-2}{x(x+1)} \frac{(x+1)^3}{x^2} = \frac{(x-2)(x+1)^2}{x^3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4512453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Squeeze theorem to prove that the sequence $a_n= \frac{3^n}{n!}$ is convergent? I want to know if the following proof of the convergence of a sequence is correct.
Proof. Let $a_n= \frac{3^n}{n!}$. Firstly, it is trivial to see that $a_n \geq 0$ for all $n \in \mathbb{N}$. Secondly, see that
$$a_n = \frac{3^n}{n!}= \frac{3 \times 3\times3 \times \cdot\cdot\cdot \times 3 \times 3 \times 3}{n \times (n-1) \times (n-2) \times \cdot \cdot \cdot 3 \times 2 \times 1} = \frac{3\times3\times3\times3}{n\times3\times2\times1} \times \frac{3 \times 3 \times...\times 3}{(n-1) (n-2)\cdot\cdot\cdot \times 4}$$
$$= \frac{3^4}{6n} \times \Big(\frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4}\Big)$$
Because $0 < \frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4} <1$ we know $a_n=\frac{3^4}{6n} \times \Big(\frac{3}{(n-1)}\times \frac{3}{n-2}\times...\times \frac{3}{4}\Big) \leq \frac{3^4}{6n}$. Using the fact that $a_n \geq 0$ we conclude that, since
$$0 \leq a_n \leq \frac{3^4}{6n}$$
and $0 \to 0, \frac{3^4}{6n} \to 0$ when $n \to 0$,
$$\lim_{n\to\infty}a_n=0$$
I skipped some trivial steps (e.g., showing that $\frac{3^4}{n}$ tends to $0$ if $n \to \infty$) because I assume theorems that prove such properties. I am mostly concerned about whether my manipulation of the factorial expression is correct and if the squeeze theorem is properly applied.
Thanks in advance!
|
Observe that: $0<a_{n+1} = \dfrac{3}{n+1}a_n < \dfrac{3}{n+1}\cdot \dfrac{3}{2}= \dfrac{9}{2n+2}$ for $n \ge 2$. Thus the squeeze theorem can be applied to conclude that $a_{n+1} \to 0$ as claimed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4514209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Solving the equation $f(x) = \int_0 ^x \sqrt{4-2f(t)} dt$ A continuous function $f(x)$ satisfies the following.
For all reals $x \le b$, $f(x)=a(x-b)^2+c$.
For all reals, $f(x) = \int_0 ^x \sqrt{4-2f(t)} dt$.
If $\int_0^6 f(x) dx = \frac{q}{p}$, where $p,q$ are relatively prime positive integers, find $p+q$
My approach is as follow $f\left( x \right) = \int\limits_0^x {\sqrt {4 - 2f\left( t \right)} dt} $
$f'\left( x \right) = \sqrt {4 - 2f\left( x \right)} $
$\frac{d}{{dx}}f\left( x \right) = \sqrt {4 - 2f\left( x \right)} $
$\frac{{d\left( {f\left( x \right)} \right)}}{{\sqrt {4 - 2f\left( x \right)} }} = dx$
Not able to proceed from here
|
We have
$$\frac{f'(x)}{\sqrt{4-2f(x)}}=1.$$
From here
$$-\sqrt{4-2f(x)}=x+c$$
hence
$$f(x) = 2-\frac12(x+c)^2.$$
As $f(0)=0$ we have $c=\pm2$. (As the vertex is $(\pm2,2)$ the condition $4-2f(x)\ge0$ is satisfied for all $x$.) Now integrate.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4516639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove $0\le ab+bc+ca-abc\le2$ A question asks
Let $a,b,c$ be non-negative reals such that $$a^2+b^2+c^2+abc=4$$
Prove that$$0\le ab+bc+ca-abc\le2$$
The lower bound can be proven as follows:
If $a,b,c>1$, then $$a^2+b^2+c^2+abc>4$$
This implies that at least one of $a,b,c$ is less than or equal to $1$, and we can assume WLOG that $a\le1$. Then,
$$ab+bc+ca-abc=a(b+c)+bc(1-a)\ge0$$
For the upper bound, I had the idea of treating $a^2+b^2+c^2+abc=4$ as a quadratic in $a$ but don't see any way to proceed from there. Please only give me hints!
|
$a^2+b^2+c^2+abc=4, a, b, c:$ non-negative. Prove that $0\leq ab+bc+ca-abc \leq 4.$
\begin{align}
&a, b, c: \text{non-negative.} \\
& \text{You found that }a \leq 1 \text{ or } b \leq 1 \text{ or } c \leq 1.\\
\Rightarrow \; & 1 \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \\
\therefore \; & abc \leq ab+bc+ca, 0 \leq ab+bc+ca-abc. \\
\ \\
& (a-b)^2+(b-c)^2+(c-a)^2 \geq 0. \\
\Rightarrow \; & a^2+b^2+c^2-ab-bc-ca \geq 0. \\
\Rightarrow \; & a^2+b^2+c^2+abc-ab-bc-ca-abc \geq 0. \\
\Rightarrow \; & 4 = a^2+b^2+c^2+abc \geq ab+bc+ca+abc \geq ab+bc+ca-abc.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4519609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Solving $\frac{(\sec x+\tan x)^2-(\sec 2x+\tan x)^2}{\sin 2x-\sin x}=2$ How can we solve for the value of $x$ in the range $[-\pi,\pi]$?
$$\frac{(\sec x+\tan x)^2-(\sec 2x+\tan x)^2}{\sin 2x-\sin x}=2$$
All i could do is write everything in terms of the elementary function $\sin$ and $\cos$ which gave me the equation
$$\cos^2 2x-\cos^2 x+2\sin x\cos^2 2x-\sin 2x\cos 2x=2\cos^2 x\cos^2 2x\sin x(2\cos x-1)$$
which doesn't seem any good to me. Please enlighten me with a clean solution to this problem.
|
If there is no typo, it is more than ugly.
Assuming that the denominator cannot be zero, that is to say $x\neq 0$, $x\neq \pm\frac \pi3$ and $x\neq \pm\pi$, using the tangent half-angle substitution $x=2\tan^{-1}(t)$, the lhs write
$$-\frac{ t \left(t^2-3\right) \left(t^2+1\right)^3 \left(2 t^5-3 t^4-12 t^3-2 t^2+2
t+1\right) } {(t-1)^2 (t+1)^2 \left(t^2-2 t-1\right)^2 \left(t^2+2 t-1\right)^2 \left(3 t^2-1\right) }$$ and we face the "nice" equation
$$5 t^{14}-3 t^{13}-55 t^{12}-2 t^{11}+193 t^{10}+19 t^9-307 t^8+36 t^7+367 t^6+19
t^5-85 t^4-2 t^3+11 t^2-3 t-1=0$$ which does not factor and does not show rational roots.
Cheating a little, there are only four real roots corresponding to
$$t_1=-0.211814 \quad \quad t_2=0.487927\quad\quad t_3=1.81774\quad\quad t_4=3.04026$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4519849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How many ways are there to use powers of $2$, each at most $3$ times, and sum up to $100$?
For $i=0,1,2,\ldots$, there're $3$ weights with mass $2^i$ gram(s). How many ways are there to weigh $100$ grams with them?
To solve with normal methods for counting doesn't seem possible. Although any way to select weights can be generated by starting from $100=2^6+2^5+2^2$ and replacing $2^{k+1}$ with $2^k+2^k$ numerous times, whether a replacement like that can take place depends on other replacements already made. For example, if there're already $2$ weights with mass $2^3$ used, you cannot replace $2^4$ with $2^3+2^3$ any more.
On the other hand, generating functions may help. I constructed the following function:$$\begin{aligned}f(x)=&\prod_{t=0}^6\left(1+x^{2^i}+x^{2\times2^i}+x^{3\times2^i}\right)\\=&(1+x+x^2+x^3)(1+x^2+x^4+x^6)\cdots(1+x^{64}+x^{128}+x^{192}).
\end{aligned}$$and we need to find the coeffitient of term $x^{100}$. How to compute this?
$\tiny\text{Why isn't there a tag for enumeration problems?}$
|
Your generating function $f(x)$ might as well be infinite:
$$\prod_{n=0}^{\infty}\left(1+x^{2^n}+x^{2(2^n)}+x^{3(2^n)}\right)$$
Now split this over even $n$ and odd $n$:
$$\prod_{n=0}^{\infty}\left(1+x^{4^n}+x^{2(4^n)}+x^{3(4^n)}\right)\prod_{n=0}^{\infty}\left(1+x^{2(4^n)}+x^{4(4^n)}+x^{6(4^n)}\right)$$
The left product is $\left(1+x+x^2+x^3\right)\left(1+x^4+x^8+x^{12}\right)\left(1+x^{16}+x^{32}+x^{48}\right)\cdots$. This is a factorization of $\left(1+x+x^2+x^3+\cdots\right)$.
Similarly the right product is $\left(1+x^2+x^4+x^6\right)\left(1+x^8+x^{16}+x^{24}\right)\left(1+x^{32}+x^{64}+x^{96}\right)\cdots$ and this is a factorization of $\left(1+x^2+x^4+x^6+\cdots\right)$.
So the generating function is
$$\left(1+x+x^2+x^3+\cdots\right)\left(1+x^2+x^4+x^6+\cdots\right)$$
Now you can see when this is multiplied out, what is the coefficient of $x^m$. If $m$ is even, you have $1\cdot x^m+x^2\cdot x^{m-2}+\cdots+x^m\cdot1=\frac{m+2}{2}x^m$.
If $m$ is odd, you have $x\cdot x^{m-1}+x^3\cdot x^{m-3}+\cdots +x^m\cdot1=\frac{m+1}{2}x^m$.
In the case of $m=100$, the coefficient is $\frac{102}{2}=51$.
Note that a new interpretation of the generating function $\left(1+x+x^2+x^3+\cdots\right)\left(1+x^2+x^4+x^6+\cdots\right)$ is that you can use $0$ or more $1$-weights, and $0$ or more $2$-weights. What is the combinatorial connection between that interpretation and the original description?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4520752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
For $0For $0<t\leq 1$ show
$$ \text{ln}(t)\geq \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
On a side note, I've been taking this summer class on inequalities. The professor told us that his main source are all sorts of math olympiads. Given that I'm gonna take Calc 3 next semester and Real Analysis afterwards, are those types of inequalities really as essential as my prof suggests? As you can tell, I'm not really capable of proving most of them by myself yet...
|
My try .
here (Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$) Michael Rozenberg show :
let $0<x\leq 1$ :
$$\ln{x}\geq(x-1)\sqrt[3]{\frac{2}{x^2+x}}$$
Now we need to show :
$$(x-1)\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{x-1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$
or :
$$\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$
I'm stuck here .Hope someone can achieve this .
Edit :
In fact using Wolfram Alpha we can conclude quicly see this logarithmic derivative link
Just some remarks for another proof :
We have the inequality :
$$\sqrt{\frac{\left(2x^{2}+5x+2\right)}{x}}\geq \frac{3}{2}\left(\sqrt{\frac{1}{x^{a}}}+\sqrt{x^{a}}\right)$$
Where :
$$a=\frac{2\sqrt{2}}{3}$$
So we need to show :
$$\left(\frac{3}{2}\left(\sqrt{\frac{1}{x^{a}}}+\sqrt{x^{a}}\right)+1\right)\left(x-1\right)\cdot\frac{1}{\left(2x+2\right)}-\ln\left(x\right)\leq 0$$
Or $x\to x^{\frac{2}{a}}$ :
$$\left(\frac{3}{2}\left(\frac{1}{x}+x\right)+1\right)\left(x^{\frac{2}{a}}-1\right)\cdot\frac{1}{\left(2x^{\frac{2}{a}}+2\right)}-\frac{\ln\left(x\right)2}{a}\leq 0$$
Now we substitute $\ln(x)=y$ and then we can use derivative .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4523983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Sum of $\sum_0^\infty \frac1{n^4+a^4}$ How to find the following series as pretty closed form by $a$?
$$S=\sum_0^\infty \frac1{n^4+a^4}$$
I first considered applying Herglotz trick, simply because the expressions are similar. So I changed it like this...
$$2S-a^{-4}=\sum_{-\infty}^\infty \frac1{n^4+a^4}$$
However, the attempt failed to find such an appropriate function like $\pi\cot\pi z$ in this post.
Next I found this post and used Fourier transform in a similar way, and the result was a nightmare!
How on earth can I calculate the value of this series?
|
If you write
$$n^4+a^4=(n-\alpha)(n-\beta)(n-\gamma)(n-\delta)$$
$$\alpha=-\frac{(1+i) a}{\sqrt{2}}\qquad \beta=\frac{(1+i) a}{\sqrt{2}}\qquad \gamma=-\frac{(1-i) a}{\sqrt{2}} \qquad \delta=\frac{(1-i) a}{\sqrt{2}}$$ Using partial fraction
$$\frac 1{n^4+a^4}=\frac{1}{(\alpha-\beta) (\alpha-\gamma) (\alpha-\delta) (x-\alpha)}+\frac{1}{(\beta-\alpha) (\beta-\gamma) (\beta-\delta) (x-\beta)}+$$ $$\frac{1}{(\gamma-\alpha) (\gamma-\beta) (\gamma-\delta) (x-\gamma)}+\frac{1}{(\delta-\alpha) (\delta-\beta) (\delta-\gamma) (x-\delta)}$$
Now,consider the partial sum
$$S_p(\epsilon)=\sum_{n=0}^p \frac 1{n-\epsilon}=\psi (p+1-\epsilon )-\psi (-\epsilon )$$ Compute all sums and use the asymptotics of the digamma function for large $p$ to obtain for large $p$
$$S_p=\sum_{n=0}^p \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi a\right)+\sin \left(\sqrt{2} \pi a\right)}{\cosh
\left(\sqrt{2} \pi a\right)-\cos \left(\sqrt{2} \pi a\right)}-\frac{1}{3 p^3}+O\left(\frac{1}{p^4}\right)$$
$$\color{red}{S_\infty=\sum_{n=0}^\infty \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi a\right)+\sin \left(\sqrt{2} \pi a\right)}{\cosh
\left(\sqrt{2} \pi a\right)-\cos \left(\sqrt{2} \pi a\right)}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4524326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
|
One more (why not?):
$$ \cos \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ -\frac{\sqrt2}{2} ·(\sin \alpha \ + \ \cos \alpha) $$ $$ \Rightarrow \ \ \sec \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{-\sqrt2}{(\sin \alpha \ + \ \cos \alpha)} \ \ ; $$
$$ \tan^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \sec^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ - \ 1 \ \ = \ \ \frac{2}{(\sin \alpha \ + \ \cos \alpha)^2} \ - \ 1 $$
$$ = \ \ \frac{2 \ - \ \sin^2 \alpha \ - \ 2 \sin \alpha \cos \alpha \ - \ \cos^2 \alpha}{\sin^2 \alpha \ + \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha} \ \ = \ \ \frac{1 \ - \ 2 \sin \alpha \cos \alpha }{1 \ + \ 2 \sin \alpha \cos \alpha } \ \ = \ \ \frac{1 \ - \ \sin (2\alpha) }{1 \ + \ \sin (2\alpha) } \ \ . $$
Your approach would be a "time-reversal" of this, but I don't think it's at all obvious that you would want to replace $ \ \sin^2 \alpha + \cos^2 \alpha \ $ with $ \ 2 - \sin^2 \alpha - \cos^2 \alpha \ $ in your numerator, and then work toward a "secant-squared" expression. (eMathHelp probably has the best way to continue from where you left off.)
ADDENDUM --
We may also apply the "angle-addition" formula for tangent:
$$ \tan \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{\tan \alpha \ + \ \tan \frac{3 \pi}{4}}{1 \ - \ \tan \alpha · \tan \frac{3 \pi}{4}} \ \ = \ \ \frac{\tan \alpha \ + \ (-1)}{1 \ - \ \tan \alpha · (-1)} $$
$$ \Rightarrow \ \ \tan^2 \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{ \tan^2 \alpha \ - \ 2 · \tan \alpha \ + \ 1 }{\tan^2 \alpha \ + \ 2 · \tan \alpha \ + \ 1 } $$
[multiplying the numerator and denominator by $ \ \cos^2 \alpha \ \ ] $
$$ = \ \ \frac{ \sin^2 \alpha \ - \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha }{\sin^2 \alpha \ + \ 2 \sin \alpha \cos \alpha \ + \ \cos^2 \alpha } \ \ , $$
and so to the left-side expression.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
}
|
Six-sided and four-sided dice question contradiction Suppose we have two fair dice, with six sides (numbered 1 to 6) and with four sides (numbered 1 to 4). Suppose we pick a die at random and throw it, the result is announced to be 2 and the die is discarded. Then we pick the remaining die and throw it. What is the expectation of the second throw?
On the one hand, us getting 2 on the first throw, does not provide us with information on whether the first die was six-sided or four-sided, so we could argue that by total expectation the expectation of the second throw is just the average of expectations:
$$
E(X)=(3.5)\frac{1}{2} + (2.5)\frac{1}{2}=3
$$
On the other hand, we know that we got 2 on the first throw, so the expectation of the second throw is:
$$
E(X)=(1)\frac{2}{9} + (2)\frac{1}{9} + (3)\frac{2}{9} + (4)\frac{2}{9} + (5)\frac{1}{9} + (6)\frac{1}{9}=3.22
$$
It seems to me that there is a contradiction?
|
Let $A$ represent the event we rolled the four-sided die in the first throw. Then $A^c$ represents the event we rolled a six-sided die in the first throw. Remember the second throw we are rolling the other die.
Let $B$ represent the event the outcome of the first roll is a $2$.
We are told $\Pr(A)=\Pr(A^c)=\frac{1}{2}$. We are told by the fact the dice are fair that $\Pr(B\mid A)=\frac{1}{4}$ and $\Pr(B\mid A^c)=\frac{1}{6}$.
Let $X$ be the random variable representing the outcome of the second die throw.
We are tasked with calculating $E[X\mid B]$.
For this, we look at $1\cdot \Pr(X=1\mid B)+2\cdot \Pr(X=2\mid B)+\dots+6\cdot \Pr(X=6\mid B)$
Let us examine $\Pr(X=1\mid B)$ more closely. We have $\Pr(X=1\mid B) = \dfrac{\Pr(X=1\cap B)}{\Pr(B)} = \dfrac{\Pr(X=1\cap A\cap B)+\Pr(X=1\cap A^c\cap B)}{\Pr(B\cap A)+\Pr(B\cap A^c)}$
$=\dfrac{\Pr(A)\Pr(X=1\cap B\mid A)+\Pr(A^c)\Pr(X=1\cap B\mid A^c)}{\Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)}=\dfrac{\Pr(A)\Pr(B\mid A)\Pr(X=1\mid B\cap A)+\Pr(A^c)\Pr(B\mid A^c)\Pr(X=1\mid B\cap A^c)}{\Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)}$
Recognize that $\Pr(X=1\mid A\cap B) = \Pr(X=1\mid A)$. Now... these we can actually find from the problem statement.
$=\dfrac{\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{6}+\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{4}}{\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{6}} =\dfrac{1}{5}$
Similarly we calculate for $X=2, X=3, X=4$. In the case of $X=5$ note that $\Pr(X=5\mid A^c)=0$ and not $\frac{1}{6}$ since there does not exist a face numbered $5$ on the four-sided die. That yields an answer of $\frac{1}{10}$ for those.
This gives a final calculation of our expectation as being:
$$1\cdot \frac{1}{5}+2\cdot \frac{1}{5}+3\cdot\frac{1}{5}+4\cdot\frac{1}{5}+5\cdot\frac{1}{10}+6\cdot\frac{1}{10}=3.1$$
Note, a priori I would not have expected each face of the two dice to be equally likely results of the second throw given a particular result of the first throw and so would have rejected an answer that attempts to assume that fact. This analysis does show that this happens to be the case here. Remember that this is the exception, not the norm. Be very careful about trying to use $\Pr(A)=\frac{|A|}{|S|}$ as this is many times blatantly incorrect to do. There are two outcomes to a lottery, you win or you lose. You don't win with probability $\frac{1}{2}$ however.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4532753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Noncommutative Ring with only Left Identity Does there exist a noncommutative ring $R$ without an identity and an element $e\in R$ such that $ex = x$ for all $x\in R$? (i.e. $xe \neq x$ for some $x\in R$).
|
$R=\left\{ \left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\in Mat_2(\mathbb Q) \right\} $ with the usual operations.
It satisfy the axioms of group, it is distributive, and a direct computation proves that the product is well defined in $R$, therefore is also associative:
$\left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\cdot \left[\begin{array}{ccc}c & d\\c & d\end{array}\right] = \frac 1 1 \left[\begin{array}{ccc}c(a+b) & d(a+b)\\c(a+b) & d(a+b)\end{array}\right] = (a+b)\left[\begin{array}{ccc}c & d\\c & d\end{array}\right]$
Note that the usual identity is not in $R$.
The matrix $\frac 1 2\left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right]$ is easily proven to be a left identity but not a right one:
$\frac 1 2\left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right]\cdot \left[\begin{array}{ccc}a & b\\a & b\end{array}\right] = \frac 1 1 \left[\begin{array}{ccc}a+a & b+b\\a+a & b+b\end{array}\right] = \left[\begin{array}{ccc}a & b\\a & b\end{array}\right]$
$\frac 1 2\left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\cdot \left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right] = \frac 1 1 \left[\begin{array}{ccc}a+b & a+b\\a+b & a+b\end{array}\right] = \frac {a+b} 2\left[\begin{array}{ccc}1 & 1\\1 & 1\end{array}\right]$
It is enought to take $a\neq b$ to see that these are different.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4532925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$using implicit differentiation
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ using implicit differentiation
My workings:
$y=\tan^{-1} x$,
$x= \tan y$
$\frac{d}{dx} (x) = \frac{d}{dx} \tan y$
$1 = \sec^2 y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$
How do I carry on from here?
|
$$x=\tan y\implies x^2=\tan^2y$$$$\implies 1+x^2=1+\tan^2y=\sec^2y$$$$\implies \frac{dy}{dx}=\frac{1}{\sec^2y}=\frac{1}{1+x^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4535215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Probability that first coin flip was heads given that 7 of 10 flips were heads? Let's say we flip a coin 10 times and we get heads 7 times. What is the probability that the first flip was heads? (My naïve estimation would be 7/10 but that's just intuition, trying to prove that).
|
I would use the binomial distribution to proof it, which is not very time consuming. Let $A$ be the event, that the first flip is head. And $B$ is the event that we get 7 times head in 10 coin flips. Then it is asked for $P(A|B)=\frac{P(A\cap B)}{P(B)}$.
$P(A\cap B)$ is the probability that we get head at the first flip and 6 heads in the remaining 9 flips. The probability is $0.5\cdot \binom{9}{6}\cdot 0.5^6\cdot 0.5^3$. And $P(B)=\binom{10}{7}\cdot 0.5^7\cdot 0.5^3$. Then the fraction is
$$P(A|B)=\frac{0.5\cdot \binom{9}{6}\cdot 0.5^6\cdot 0.5^3}{\ \ \ \quad \binom{10}{7}\cdot 0.5^7\cdot 0.5^3}=\frac{ \binom{9}{6}\cdot \require{cancel} \cancel{0.5^7}\cdot \require{cancel} \bcancel{0.5^3}}{\binom{10}{7}\cdot \require{cancel} \cancel{0.5^7}\cdot \require{cancel} \bcancel{0.5^3}}=\frac{ \binom{9}{6}}{\binom{10}{7}}$$
Using $\binom{n}{k}=\binom{n}{n-k}$ and $\binom{m}{r}=\frac{m\cdot (m-1)\cdot (m-2)\cdot \ldots \cdot (m-r+1)}{1\cdot 2\ \cdot \ldots \cdot r} (*)$ we obtain
$$\frac{\binom{9}{6}}{\binom{10}{7}}=\frac{ \binom{9}{3}}{\binom{10}{3}}=\frac{9\cdot 8\cdot 7}{1\cdot 2\cdot 3}\cdot \frac{1\cdot 2\cdot 3}{10\cdot 9\cdot 8}=\dots$$
Some steps you can omit, if you wish.
(*) The key point at this formula is that the number of factors are $r$ for both, the numerator and the denominator. At the numerator it starts with $m$ and then decreases. At the denominator it starts with $1$ and then increases.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4539796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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.