Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried:
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$ divides through both $xy$ and $x^2 + y^2$
This then allows me to introduce another equation $1=ax+by$ for some $a, b.$
But I then get stuck ...
| Suppose $xy|x^2+y^2$. Then $xy|x^2+y^2+2xy=(x+y)^2$. But if $\gcd(x,y)=1$, then also
$$
1=\gcd(x,x+y)=\gcd(y,x+y)=\gcd(xy,x+y)=\gcd(xy,(x+y)^2)
$$
from which it then follows that $xy=\pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Computing a gaussian integral involving both real and imaginary coefficients in a stochastic system I am stuck with the integral:
$$\int_{-\infty}^{\infty} \frac{\exp[-a(x-b)^2]}{1+cx^2} dx$$ where $a,c$ are real and $b$ is purely imaginary.
I tried to solve it by contour integration but the integral along the semicircular edge of the contour does not end up as 0. So the method doesnt work.
I dont have any idea about other ways to solve this. I am thinking about expanding the gaussian in terms of its Taylor series and integrating the respective terms. Still it is becoming complicated.
Does anyone have any idea about this?
| Since $b$ is purely imaginary, this is a Fourier Transform.
Assuming $a>0$, $c>0$ ,and $\Re(b) = 0$; make the substitution $- \pi s = \Im (b)$ or equivalently $-i\pi s =b$ or $\pi s = ib$
$$\begin{align*}\displaystyle & \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x-b\right)^2\right]}{1+cx^2} dx\\
\\
&= \int_{-\infty}^{\infty} \frac{\exp\left[-a\left(x+i\pi s\right)^2\right]}{1+cx^2} dx\\
\\
&= \int_{-\infty}^{\infty} \frac{\exp\left[-ax^2+a(\pi s)^2-2\pi i axs\right]}{cx^2+1} dx\\
\\
&= \frac{a^2}{c}e^{a(\pi s)^2}\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{a}(ax)^2}}{(ax)^2+\frac{a^2}{c}}e^{-2\pi i (ax)s} dx\\
\\
&= \frac{a}{c}e^{a(\pi s)^2}\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{a}y^2}}{y^2+\frac{a^2}{c}}e^{-2\pi iys} dy\\
\\
&= \frac{a}{c}e^{a(\pi s)^2}(2\pi)^2\frac{1}{2\left(\frac{2\pi a}{\sqrt{c}}\right)}\int_{-\infty}^{\infty} \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\space e^{-\pi^2\left(\frac{y}{\pi\sqrt{a}}\right)^2}\space e^{-2\pi iys} dy\\
\\
&= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\mathscr{F}\left\{ \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\space e^{-\pi\left(\frac{y}{\sqrt{\pi a}}\right)^2}\right\} \\
\\
&= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\left[\mathscr{F}\left\{ \frac{2\left(\frac{2\pi a}{\sqrt{c}}\right)}{(2\pi y)^2+\left(\frac{2\pi a}{\sqrt{c}}\right)^2}\right\} * \mathscr{F}\left\{ e^{-\pi\left(\frac{y}{\sqrt{\pi a}}\right)^2}\right\} \right]\\
\\
&= \frac{\pi}{\sqrt{c}}e^{a(\pi s)^2}\left[e^{-\frac{2 a}{\sqrt{c}}|\pi s|} * \sqrt{\pi a} e^{-a\left(\pi s\right)^2}\right] \\
\\
&= \pi \sqrt{\frac{\pi a}{c}}e^{a(\pi s)^2}\int_{-\infty}^{\infty}e^{-\frac{2 a}{\sqrt{c}}|\pi \tau|} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau \\
\\
&= \pi \sqrt{\frac{\pi a}{c}}e^{a(\pi s)^2}\left[\int_{-\infty}^{0}e^{\frac{2 a}{\sqrt{c}}\pi \tau} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau +\int_{0}^{\infty}e^{-\frac{2 a}{\sqrt{c}}\pi \tau} e^{-a\left(\pi s -\pi \tau\right)^2}\space d\tau \right]\\
\\
&= \pi \sqrt{\frac{\pi a}{c}}\left(\int_{-\infty}^{0}\exp\left[-a\left([\pi\tau]^2-2\left[\pi s+\frac{1}{\sqrt{c}}\right]\pi \tau\right)\right]\space d\tau +\int_{0}^{\infty}\exp\left[-a\left([\pi\tau]^2-2\left[\pi s-\frac{1}{\sqrt{c}}\right]\pi \tau\right)\right]\space d\tau \right)\\
\\
&= \pi \sqrt{\frac{\pi a}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\int_{-\infty}^{0}\exp\left[-a\left(\pi\tau-\left[\pi s+\frac{1}{\sqrt{c}}\right]\right)^2\right]\space d\tau +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\int_{0}^{\infty}\exp\left[-a\left(\pi\tau-\left[\pi s-\frac{1}{\sqrt{c}}\right]\right)^2\right]\space d\tau \right)\\
\\
&= \pi \sqrt{\frac{\pi a}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\frac{1}{\pi\sqrt{a}}\int_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)}e^{-u^2}\space du +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\frac{1}{\pi\sqrt{a}}\int_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}e^{-u^2}\space du\right)\\
\\
&= \sqrt{\frac{\pi}{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\int_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)}e^{-u^2}\space du +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\int_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}e^{-u^2}\space du\right)\\
\\
&= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\mathrm{erf}(u)\biggr{|}_{-\infty}^{-\sqrt{a}\left(\pi s+\frac{1}{\sqrt{c}}\right)} +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\mathrm{erf}(u)\biggr{|}_{-\sqrt{a}\left(\pi s-\frac{1}{\sqrt{c}}\right)}^{\infty}\right)\\
\\
&= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{a\left(\pi s+\frac{1}{\sqrt{c}}\right)^2}\left[\mathrm{erf}\left(-\sqrt{a}\left[\pi s+\frac{1}{\sqrt{c}}\right]\right)+1\right] +e^{a\left(\pi s-\frac{1}{\sqrt{c}}\right)^2}\left[1 -\mathrm{erf}\left(-\sqrt{a}\left[\pi s-\frac{1}{\sqrt{c}}\right]\right)\right]\right)\\
\\
&= \frac{\pi}{2}\frac{1}{\sqrt{c}}\left(e^{\left[-\sqrt{a}\left(ib+\frac{1}{\sqrt{c}}\right)\right]^2}\left[\mathrm{erf}\left(-\sqrt{a}\left[ib+\frac{1}{\sqrt{c}}\right]\right)+1\right] +e^{\left[-\sqrt{a}\left(ib-\frac{1}{\sqrt{c}}\right)\right]^2}\left[1 -\mathrm{erf}\left(-\sqrt{a}\left[ib-\frac{1}{\sqrt{c}}\right]\right)\right]\right)\\
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$u_{n+1} = a u_n +b u_{n-1} +c$ Is there a closed form formula for the general term of a recurrence sequence satisfying $$u_{n+1} = a u_n +b u_{n-1} +c$$ The sequence I am interested in satisfies $u_{n+1} = 8 u_n - 3 u_{n-1} -4$.
| This is a linear difference equation so can be stated as
$$
u = u_h + u_p
$$
with
$$
u_h(n+1)-au_h(n)-bu_h(n-1) = 0\\
u_p(n+1)-au_p(n)-bu_p(n-1) = c\\
$$
now making $u_h(n) = \phi^n$ and substituting we have
$$
\phi^n\left(\phi-a-b\phi^{-1}\right) = 0
$$
so
$$
\phi = \frac{1}{2} \left(a\pm\sqrt{a^2+4 b}\right)
$$
and
$$
u_h(n) = \frac{C_1}{2^n} \left(a-\sqrt{a^2+4 b}\right)^n+\frac{C_2}{2^n} \left(a+\sqrt{a^2+4 b}\right)^n
$$
and also
$$
u_p(n) = \frac{c}{1-a-b}
$$
so finally
$$
u(n) = \frac{C_1}{2^n} \left(a-\sqrt{a^2+4 b}\right)^n+\frac{C_2}{2^n} \left(a+\sqrt{a^2+4 b}\right)^n + \frac{c}{1-a-b}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.
| Suppose $z + \frac{1}{z} = e^{it}$ for some $t \in \mathbb{R}$. Then solving for $z$ gives
$$z = \frac{e^{it} \pm \sqrt{e^{2it} - 4}}{2} = e^{it} \cdot \frac{1 \pm \sqrt{1 - 4 e^{-2it}}}{2}.$$
Therefore,
$$ |z| = \frac{1}{2} \left| 1 \pm \sqrt{1 - 4 e^{-2it}} \right| \le \frac{1}{2} \left( |1| + |\sqrt{1 - 4 e^{-2it}}|\right) = \frac{1}{2} \left( 1 + \sqrt{1 - 4 e^{-2it}} \right)\le \\
\frac{1}{2} \left( 1 + \sqrt{|1| + |4 e^{-2it}|}\right) = \frac{1 + \sqrt{5}}{2}.$$
On the other hand, if we set $t := \frac{\pi}{2}$ so that $e^{-2it} = -1$ then we will get equality. Therefore, the maximum value of $|z|$ is $\frac{1+\sqrt{5}}{2}$, which is achieved for $z = \frac{1+\sqrt{5}}{2} i$. (In fact, a slight refinement of the argument will show that equality holds, i.e. the maximum is achieved, if and only if $z = \pm \frac{1+\sqrt{5}}{2} i$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
| Note that if $y=0$ then we get $-\frac 1x=0$ on second equation, so the system has no solutions.
So we can assume $y\neq 0$, set $\ x=ty\ $ and substitute.
We get $\begin{cases}t^3y^2+ty^2+3t-1=3t^2y+3y \\t^2y^2+y^2-t-3=0\end{cases}$
And this allow us to isolate $y^2=\dfrac{t+3}{t^2+1}$
Substituting in first equation rewritten $ty^2(t^2+1)+3t-1=3y(t^2+1)$
gives $t(t+3)+3t-1=3y(t^2+1)\iff 3y=\dfrac{t^2+6t-1}{t^2+1}$
Now be identifying $9y^2$ we get $(t^2+6t-1)^2=9(t+3)(t^2+1)$
We can remark $t=-1$ is a root $6^2=9(2)(2)$ else just graph it, and notice $-1$ and $2$ seems to be roots (so try to factor these terms).
Indeed the factorization is $$(t+1)(t-2)(t^2+4t+13)=0$$
Reporting in initial equations gives us the solutions:
*
*$t=2$ and $x=2y$ leads to $\begin{cases}2y+\frac 1y=3\\y=\frac 1y\end{cases}\iff \begin{cases}x=2\\y=1\end{cases}$
*$t=-1$ and $x=-y$ leads to $\begin{cases}-y-\frac 2y=3\\y=\frac 1y\end{cases}\iff \begin{cases}x=1\\y=-1\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsure how to start:
\begin{equation}
G_m = \sum_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right),
\end{equation}
where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
\begin{equation}
\sum_{j = 0}^{m - 1} c_j = 0
\end{equation}
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
\begin{equation}
\int_{0}^{1} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\frac{\pi}{\sin\left(\frac{\pi}{n} \right)}- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{2}\right)\right]
\end{equation}
Or for any positive upper bound $x$:
\begin{align}
I_n(x) &= \int_{0}^{x} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{x^n + 1}\right)\right]
\end{align}
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m \in \mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
\begin{align}
x^{2m} + 1 = 0 \rightarrow x^{2m} = e^{\pi i}
\end{align}
By De Moivre's formula, we observe that:
\begin{align}
x = \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) \mbox{ for } j = 0\dots 2m - 1,
\end{align}
which we can express as the set
\begin{align}
S &= \Bigg\{ \exp\left({\frac{\pi + 2\pi \cdot 0}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi \cdot 1}{2m} i} \right),\dots,\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 2)}{2m} i} \right)\\
&\qquad\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 1)}{2m} i} \right)\Bigg\},
\end{align}
which can be expressed as the set of $2$-tuples
\begin{align}
S &= \left\{ \left( \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi(2m - 1 - j )}{2m} i} \right)\right)\: \bigg|\: j = 0 \dots m - 1\right\}\\
& = \left\{ (z_j, c\left(z_j\right)\:|\: j = 0 \dots m - 1 \right\}
\end{align}
From here, we can factor $x^{2m} + 1$ into the form
\begin{align}
x^{2m} + 1 &= \prod_{r \in S} \left(x + r_j\right)\left(x + c(r_j)\right) \\
&= \prod_{i = 0}^{m - 1} \left(x^2 + \left(r_j + c(r_j)\right)x + r_j c(r_j)\right) \\
&= \prod_{i = 0}^{m - 1} \left(x^2 + 2\Re\left(r_j\right)x + \left|r_j \right|^2\right)
\end{align}
For our case here $\left|r_j \right|^2 = 1$ and $\Re\left(r_j\right) = \cos\left(\frac{\pi + 2\pi j}{2m} \right)= \cos\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) = c_j$
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &= \int_0^1 \log\left(\prod_{r \in S} \left(x^2 + 2c_jx+ \left|r_j \right|^2\right)\right)\\
&= \sum_{j = 0}^{m - 1} \int_0^1 \log\left(x^2 + 2c_jx + 1 \right)\\
&= \sum_{j = 0}^{m - 1} \left[2\sqrt{1 - c_j^2}\arctan\left(\frac{x + c_j}{\sqrt{1 - c_j^2}}\right) + \left(x + c_j\right)\log\left(x^2 + 2c_jx + 1\right) - 2x \right]_0^1 \\
&= \sum_{j = 0}^{m - 1} \left[ 2\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)c_j + \left(\log(2) - 2\right) + \left(c_j + 1\right)\log\left(c_j + 1\right) \right] \\
&= 2\sum_{j = 0}^{m - 1}\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
Thus,
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &=\sum_{j = 0}^{m - 1}c_j\sin\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
| Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^{(n)}$ as last time, we apply the $\log\prod_{i}a_i=\sum_i\log a_i$ property to see that
$$\log(1+x^n)=\log\prod_{k=1}^{n}(x-r_k^{(n)})=\sum_{k=1}^{n}\log(x-r_k^{(n)})$$
So
$$I_n=\int_0^1\log(1+x^n)\mathrm dx=\sum_{k=1}^{n}\int_0^1\log(x-r_k^{(n)})\mathrm dx$$
This last integral boils down to
$$\begin{align}
\int_0^1\log(x-a)\mathrm dx=&a\log\frac{a}{1+a}+\log(1-a)-1\\
=&\log\frac{a^a(1-a)}{e(1+a)^a}
\end{align}$$
So
$$I_n=\sum_{r\in S_n}\log\frac{r^r(1-r)}{e(1+r)^r}$$
And you know how I love product representations, so we again use $\log\prod_{i}a_i=\sum_i\log a_i$ to see that
$$
I_n=\log\prod_{r\in S_n}\frac{r^r(1-r)}{e(1+r)^r}\\
\prod_{r\in S_n}\frac{r^r(1-r)}{(1+r)^r}=\exp(n+I_n)
$$
Which I just think is really neat.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
A Series For the Golden Ratio
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?
Some background and motivation:
Wikipedia only provides one series for the golden ratio - see also the link in the comment by @Zacky. I became curious if I could construct another series for the Golden Ratio based on a slight modification to a known series representation of the $\sqrt{2}.$ At first I considered $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{(2n)!!}$$; which series can be accelerated via an Euler transform to yield $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{2^{3n+1}(n!)^2}$$ This last series became the impetus to try and and get to the Golden ratio. Through trial and error I stumbled upon
$$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$
| First of all note that
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$
Lets rewrite your sum as the following
$$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\binom{2n}n\left(\frac1{5^3}\right)^n=\frac15\frac1{\sqrt{1-\left(\frac4{5^3}\right)}}=\frac{\sqrt 5}{11}$$
And therefore you can correctly conclude that
$$\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac12+\frac{11}2\frac{\sqrt 5}{11}=\frac{1+\sqrt 5}2$$
$$\therefore~\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\phi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Find $ \sum_{n=1}^{\infty} \frac{1}{n(n+2)} $ I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
\begin{equation}
\sum_{n=1}^{\infty} \frac{1}{n(n+2)}
\end{equation}
First I set $ \frac{1}{n(n+2)}= \frac{A}{n}+\frac{B}{n+2}$ and solve to get $A=\frac{1}{2}$ and $B=-\frac{1}{2}$.
So I'm finding $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{2n+4})$
and proceed with the partial sum expansion.
At the end I am left with $S_N=\frac{1}{2}+\frac{1}{4}-\frac{1}{2N+4}$ by cancelling terms in between and then $\lim_{N \to \infty}S_N=\frac{3}{4}$.
Am I doing it correctly? Thank you for your time.
| $$\sum^{\infty}_{n=1}\frac{1}{n(n+2)} = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\frac{1}{n}-\frac{1}{n+2}\bigg]$$
$$ = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)+\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)\bigg]$$
$$ = \frac{1}{2}\bigg[\frac{1}{1}+\frac{1}{2}\bigg] = \frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to divide ${2k^3+3k^2+k-2j^3+3j^2-j}$ with $(k+1-j)$? The question I had was calculating $$\displaystyle\frac{1}{k+1-j}\sum_{i=j}^k i^2$$
Because I didn't know how to do a variable change, I did
$$\frac{1}{k+1-j}\sum_{i=j}^k i^2 = \frac{1}{k+1-j}\left(\sum_{i=1}^k i^2 - \sum_{i=1}^{j-1} i^2\right) = \frac{2k^3+3k^2+k-2j^3+3j^2-j}{6(k+1-j)}$$
The solution given used variable change and was able to cancel out $(k+1-j)$ very easily. Finally the answer was $$\frac{2k^2 + 2j^2 + 2kj - j + k}{6}$$ I have verified that these two solutions are the same by multiplying the second one with $(k+1-j).$
My question is:
How do you divide an expression like ${(2k^3+3k^2+k-2j^3+3j^2-j)}$ with $(k+1-j)$? Is it even worth it in this case to get the most simplified solution?
| Try to write the polynomial in $k$ as a polynomial in $m=k+1$.
$2k^3+3k^2+k=\big[2(k+1)^3-2-6k-6k^2\big]+\big[3(k+1)^2-3-6k\big]+(k+1)-1\\=2(k+1)^3+3(k+1)^2+(k+1)-6(1+2k+k^2)\\=2m^3-3m^2+m$
Therefore, $2m^3-3m^2+m-2j^3+3j^2-j=2(m^3-j^3)-3(m^2-j^2)+m-j\\=(m-j)\big[2(m^2+j^2+mj)-3(m+j)+1\big]$
Divide by $k+1-j=m-j$ and back-substitute $m$ to get,
$$2(m^2+j^2+mj)-3(m+j)+1=2k^2+2j^2+2kj+k-j$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the solution for $\int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx$ I used the rule :$$\int_{0}^{a}f(x) = \int_{0}^{a}f(a-x)\,dx$$
And got :$$\int_{0}^{\pi/2}\frac{\sin^2x}{\sin^2x+4\cos^2x}\,dx$$
Then, I added both the question and the above integrand, and I got :$$\int_{0}^{\pi/2}\frac{dx}{5}$$
Solving the above, I finally got the answer as $$\frac{\pi}{20}$$
But, according to my textbook, the answer is $$\frac{\pi}{6}$$
Where did I go wrong?
| So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.
In fact when you add the two terms you found :
$$A =\frac{\cos^2x}{\cos^2x+4\sin^2x}$$
$$B =\frac{\sin^2x}{\sin^2x+4\cos^2x}$$
$$\implies A + B = \frac {3 \cos(x)^4 + 3 \sin(x)^4+1}{(3 \sin(x)^2 +1) \cdot (3 \cos(x)^2 + 1)} $$
Here is an attempt to solve the exercice:
$$I = \int_{0}^{\pi/2}\frac{\cos^2x}{\cos^2x+4\sin^2x}dx = \int_{0}^{\pi/2}\frac{1}{1+4\tan^2x}dx $$
we can do this because the interval of integration permits it.
Then we change the variable of integration ; the substitution is the following :
$$ u = \tan x$$
we get :
$$I =\int_{0}^{\infty}\frac{1}{(1 +4u^2)(1 + u^2)}du = \int_{0}^{\infty} \frac {4}{3(4u^2 + 1)} - \frac {1}{3(u^2+1)} du $$
from there we can easily conclude.
In fact you just have to know this formula :
$$\int \frac 1 {x^2 + a^2} = \frac 1 {|a|} \arctan \left(\frac x a \right) + C $$
Where $C$ is the constant of integration.
And thus we get :
$$\int_{0}^{\infty} \frac {4}{3(4u^2 + 1)} du = \frac 2 3 \arctan(2u)\bigg\rvert_0^{\infty} = \pi / 3$$
$$\int_{0}^{\infty} \frac {1}{3(u^2+1)} du = \frac 1 3 \arctan u \bigg\rvert_0^{\infty} = \pi/6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Let $x>0$ , $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Then find limit
For $x>0$, let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. Then
$$
\lim_{x\to0^+}x\left(\left\lfloor\frac{1}{x}\right\rfloor
+\left\lfloor\frac{2}{x}\right\rfloor+\dots
+\left\lfloor\frac{10}{x}\right\rfloor\right)=\_\_\_
$$
I know , $\lfloor x\rfloor= x $ if $x\in Z$ &
$n$ if $x\notin Z$ and $n \in Z$ and $ n<x<n+1$.
I don't know how to proceed further !!
| $\frac{1}{x} -1< \lfloor \frac{1}{x}\rfloor<\frac{1}{x}$
multiplying by x we get
$\implies$x[$\frac{1}{x} -1]< x\lfloor \frac{1}{x}\rfloor<x[\frac{1}{x}]$
this gives $1-x< \lfloor \frac{1}{x}\rfloor<1$
taking limit x $\to 0^{+}$ by squeez theorem ,we get
$ lim_{x\to 0^{+}} x\lfloor \frac{1}{x}\rfloor=1$
similarly,
$\frac{2}{x} -1< \lfloor \frac{2}{x}\rfloor<\frac{2}{x}$
multiplying by x we get
$\implies$x[$\frac{2}{x} -1]< x\lfloor \frac{2}{x}\rfloor<x[\frac{2}{x}]$
this gives $2-x< \lfloor \frac{2}{x}\rfloor<2$
taking limit x $\to 0^{+}$ by squeez theorem ,we get
$ lim_{x\to 0^{+}} x\lfloor \frac{2}{x}\rfloor=2$
so this give value of complete expression =55.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve linear equation system with Gauss I have the following matrix and have to see if it has solutions depending on $a$. My solution:
$M=
\left[ {\begin{array}{cc}
a & a^2 &| &1 \\
-1 & -1& | & -a \\
1 & a & | & a
\end{array} } \right]
$
My attemp was:
Changing first with third line
$=
\left[ {\begin{array}{cc}
1 & a & | & a \\
-1 & -1& | & -a \\
a & a^2 &| &1 \\
\end{array} } \right]
$
Add the first row to the second one
$=
\left[ {\begin{array}{cc}
1 & a & | & a \\
0 & a-1& | & 0 \\
a & a^2 &| &1 \\
\end{array} } \right]
$
Add the $-a$ of the first line to the third line
$=
\left[ {\begin{array}{cc}
1 & a & | & a \\
0 & a-1& | & 0 \\
0 & 0 &| &1-a^2 \\
\end{array} } \right]
$
From $0=1-a^2$there we can obtain that the LES has a solution if $a=+/-1$
Is this a valid solution to the problem and is there a free variable?
| \begin{align}
M=
\left[\begin{array}{cc|c}
a & a^2 & 1 \\
-1 & -1 & -a \\
1 & a & a
\end{array}\right]
&\to
\left[\begin{array}{cc|c}
1 & a & a \\
-1 & -1 & -a \\
a & a^2 & 1
\end{array}\right]
&&R_1\leftrightarrow R_3
\\[4px]&\to
\left[\begin{array}{cc|c}
1 & a & a \\
0 & a-1 & 0 \\
0 & 0 & 1-a^2
\end{array}\right]
&&\begin{aligned}R_2&\gets R_2+R_1\\R_3&\gets R_3-aR_1\end{aligned}
\end{align}
Your work was pretty good.
Now you have to distinguish the cases $a=1$, $a=-1$, $a\ne\pm1$
Case $a=1$
The matrix becomes
$$
\left[\begin{array}{cc|c}
1 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
$$
and the system has infinitely many solutions
$$
\begin{bmatrix} 1-h \\ h \end{bmatrix}=
\begin{bmatrix} 1 \\ 0 \end{bmatrix} + h\begin{bmatrix} -1 \\ 1 \end{bmatrix}
$$
Case $a=-1$
The matrix becomes
$$
\left[\begin{array}{cc|c}
1 & -1 & -1 \\
0 & 2 & 0 \\
0 & 0 & 0
\end{array}\right]
\to
\left[\begin{array}{cc|c}
1 & 0 & -1 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array}\right]
$$
and the system has a single solution $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$.
Case $a\ne\pm1$
No solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Prove that $\int_a^b (f'(x))^2dx - h\sum_{i=0}^{n-1} \left(\frac{f(x_{i+1})-f(x_i)}{h}\right)^2 = O(h^2) $ Prove that:
$\int_a^b (f'(x))^2dx - h\sum_{i=0}^{n-1} {(\frac{f(x_{i+1})-f(x_i)}{h})}^2 = O(h^2) $
Where $h=\frac{b-a}{n} , x_k=a+kh$ and $f \in C^\infty[a,b]$
First I've tried to estimate the following: $\int_{x_i}^{x_{i+1}} (f'(x))^2dx - h{\left(\frac{f(x_{i+1})-f(x_i)}{h}\right)}^2$
By Taylor series we know:
$f'(x)=\frac{f(x+h)-f(x)}{h}-\frac{h}{2}f''(\alpha)$
(Where $ x\leq\alpha\leq x+h$)
So $\int_{x_i}^{x_{i+1}} (f'(x))^2dx = \int_{x_i}^{x_{i+1}} \left(\frac{f(x+h)-f(x)}{h}-\frac{h}{2}f''(\alpha)\right)^2dx$
Then I got stuck
| We have
\begin{align*}
A(x) &\overset{\text{def}}{=} \int_a^b [f'(x)]^2dx - \sum_{k=0}^{n-1}\left(\frac{f(x_{k+1}) - f(x_k)}{h}\right)^2h \\
&= \sum_{k=0}^{n-1}\left\{\color{red}{\int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - \color{blue}{\left(\frac{f(x_{k+1}) - f(x_k)}{h}\right)^2h}\right\}
\end{align*}
Now, for $x \in [x_k, x_{k+1}]$,
\begin{align*}
[f'(x)]^2 = [f'(x_k)]^2 + 2f'(x_k)f''(x_k)(x - x_k) + \{[f''(x_k)]^2 + f'(x_k)f'''(x_k)\}(x - x_k)^2 + O(h^3)
\end{align*}
and so
\begin{align*}
\color{red}{\int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} &= \color{green}{[f'(x_k)]^2h} + \color{orange}{2f'(x_k)f''(x_k)\cdot\frac{1}{2}h^2} + \{[f''(x_k)]^2 + f'(x_k)f'''(x_k)\}\cdot\frac{1}{3}h^3 + O(h^4) \\
&=\color{green}{[f'(x_k)]^2h} + \color{orange}{2f'(x_k)f''(x_k)\cdot\frac{1}{2}h^2} + O(h^3)
\end{align*}
Next,
\begin{align*}
f(x_{k+1}) = f(x_k) + hf'(x_k) + \frac{h^2}{2}f''(x_k) + O(h^3)
\end{align*}
and so
\begin{align*}
\color{blue}{\left(\frac{f(x_{k+1}) - f(x_k)}{h}\right)^2h} &= \left(f'(x_k) + \frac{h}{2}f''(x_k) + O(h^2)\right)^2 h \\
&= \color{green}{[f'(x_k)]^2h} + \color{orange}{f'(x_k)f''(x_k)h^2}+O(h^3)
\end{align*}
Therefore,
\begin{align*}
\color{red}{\int_{x_k}^{x_{k+1}}[f'(x)]^2 dx} - \color{blue}{\left(\frac{f(x_{k+1}) - f(x_k)}{h}\right)^2h} = O(h^3)
\end{align*}
Finally,
\begin{align*}
A(x) = \sum_{k=0}^{n-1}O(h^3) = O(h^2)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
trigonometric integral with tangent $$\int\biggl( \sqrt[3]{\frac{\sin x+1}{\cos x}}+\sqrt[3]\frac{\sin x-1}{\cos x}\biggr)\frac{1}{\cos^2x}\,dx ,\;x\in\Bigl(-\frac{\pi}{2},\frac{\pi}{2}\Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $\frac{1}{\cos^2x}$ is the derivative of tangent, but I don't know how to do it using the substitution $t=\tan x$ because there will be $\frac{1}{\cos x}$ inside and I think there is another method.
| I will use integration by parts twice. First we write
\begin{align}
I &= \int \left (\sqrt[3]{\frac{\sin x + 1}{\cos x}} + \sqrt[3]{\frac{\sin x - 1}{\cos x}} \right ) \sec^2 x \, dx\\
&= \int \big{(}\sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x} \big{)} \sec^2 x \, dx.
\end{align}
If we let
$$f(x) = \sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x}$$
and
$$g(x) = \sqrt[3]{\tan x + \sec x} - \sqrt[3]{\tan x - \sec x},$$
then obsereve that
$$f'(x) = \frac{1}{3} \sec x \cdot g(x) \quad \text{and} \quad g'(x) = \frac{1}{3} \sec x \cdot f(x).$$
So for the integral we have
\begin{align}
I &= \int \sec^2 x \cdot f(x) \, dx\\
&= f(x) \cdot \tan x - \int \tan x \cdot f'(x) \, dx \qquad \text{(by parts)}\\
&= f(x) \cdot \tan x - \frac{1}{3} \int \sec x \tan x \cdot g(x) \, dx \\
&= f(x) \cdot \tan x - \frac{1}{3} \sec x \cdot g(x) + \frac{1}{3} \int \sec x \cdot g'(x) \, dx \quad \text{(by parts)}\\
&= f(x) \cdot \tan x - \frac{1}{3} \sec x \cdot g(x) + \frac{1}{9} \int \sec^2 x \cdot f(x) \, dx\\
&= f(x) \cdot \tan x - \frac{1}{3} \sec x \cdot g(x) + \frac{1}{9} I,
\end{align}
giving
$$I = \frac{9}{8} \tan x \cdot f(x) - \frac{3}{8} \sec x \cdot g(x) + C$$
or
\begin{align}
I &= \frac{9}{8} \tan x \big{(} \sqrt[3]{\tan x + \sec x} + \sqrt[3]{\tan x - \sec x} \big{)}\\
& \qquad - \frac{3}{8} \sec x \big{(} \sqrt[3]{\tan x + \sec x} - \sqrt[3]{\tan x - \sec x} \big{)} + C.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.
How many solutions does
$$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$
have between $-90^\circ$ and $90^\circ$?
I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt{3}}{2}\right).$$
Since $a$ is between $-90^\circ$ and $90^\circ$, then $2a$ is between $-180^\circ$ and $180^\circ$. The RHS can be $60^\circ$, $120^\circ$, $-240^\circ$, and $-300^\circ$. Only $60^\circ$ and $120^\circ$ fit the criteria, but the answer is 4 solutions.
Where did I go wrong?
| You are right. Here is the image of the function using google.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $
I made this far:
$$
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
| You may proceed as follows transforming the limit into a derivative of a function at $0$.
First set $n = \frac{1}{t}$ and consider $t \to 0^+$:
\begin{eqnarray*} n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
& \stackrel{n=\frac{1}{t}}{=} & \frac{\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^4}+1}}-\sqrt{2}\frac{1}{t}}{t^3}\\
& = & \frac{\sqrt{1+\sqrt{1+t^4}}-\sqrt{2}}{t^4}\\
& \stackrel{y=t^4}{=} & \frac{\sqrt{1+\sqrt{1+y}}-\sqrt{2}}{y}\\
& \stackrel{y\to 0^+}{\longrightarrow} & f'(0) = \left.\frac{1}{4\sqrt{1+\sqrt{1+y}}\cdot \sqrt{1+y}} \right|_{y=0} = \boxed{\frac{1}{4\sqrt{2}}}\mbox{ for } f(y) =\sqrt{1+\sqrt{1+y}}\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
| Hint: If $a+b=c+d$ and $ab=cd$ then $\{a,b\}=\{c,d\}$. Use the identity
$$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)^3-3xy(x+y).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
Attempt
$$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx \\ =\int_0^\infty\frac{\operatorname{arcsinh}(2x)}{1+x^2} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{4+\sinh^2x} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{3+\cosh^2x} \, dx\\
=2\int_0^\infty\sum_{n=0}^\infty x(-3)^n\cosh^{-2n-1}(x) \, dx$$
I failed to integrate $x\cosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
| On the path of Kemono Chen...
\begin{align}J&=\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx\end{align}
Perform the change of variable $y=\operatorname{arcsinh}(2\tan x)$,
\begin{align}J&=\int_0^{+\infty}\frac{2x\cosh x}{4+\sinh^2 x}\,dx\\
&=\int_0^{+\infty}\frac{4x\left(\text{e}^{x}+\text{e}^{-x}\right)}{14+\text{e}^{2x}+\text{e}^{-2x}}\,dx\\
&=\int_0^{+\infty}\frac{4x\text{e}^{-x}\left(\text{e}^{2x}+1\right)}{14+\text{e}^{2x}+\text{e}^{-2x}}\,dx\\
\end{align}
Perform the change of variable $y=\text{e}^{-x}$,
\begin{align}J&=-\int_0^1 \frac{4\ln x\left(1+\frac{1}{x^2}\right)}{14+x^2+\frac{1}{x^2}}\\
&=-\int_0^1 \frac{4\ln x\left(1+x^2\right)}{x^4+14x^2+1}\\
&=\left[-\arctan\left(\frac{4x}{1-x^2}\right)\ln x\right]_0^1+\int_0^1 \frac{\arctan\left(\frac{4x}{1-x^2}\right)}{x}\,dx\\
&=\int_0^1 \frac{\arctan\left(\frac{4x}{1-x^2}\right)}{x}\,dx\\
&=\int_0^1 \frac{\arctan\left(\left(2+\sqrt{3}\right)x\right)}{x}\,dx+\int_0^1 \frac{\arctan\left(\left(2-\sqrt{3}\right)x\right)}{x}\,dx\\
\end{align}
In the first integral perform the change of variable $y=\left(2+\sqrt{3}\right)x$,
In the second integral perform the change of variable $y=\left(2-\sqrt{3}\right)x$,
\begin{align}J&=\int_0^{2+\sqrt{3}}\frac{\arctan x}{x}\,dx+\int_0^{2-\sqrt{3}}\frac{\arctan x}{x}\,dx\\
&=\Big[\arctan x\ln x\Big]_0^{2+\sqrt{3}}-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx+\Big[\arctan x\ln x\Big]_0^{2-\sqrt{3}}-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{5\pi}{12}\ln\left(2+\sqrt{3}\right)-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx+\frac{\pi}{12}\ln\left(2-\sqrt{3}\right)-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx
\end{align}
In the first integral perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\int_{2-\sqrt{3}}^{+\infty}\frac{\ln x}{1+x^2}\,dx-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\int_0^{+\infty}\frac{\ln x}{1+x^2}\,dx-2\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
\end{align}
Perform the change of variable $y=\tan x$,
\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\int_0^{\frac{\pi}{12}}\ln\left(\tan x\right)\,dx\\
\end{align}
It is well known that,
\begin{align}
\int_0^{\frac{\pi}{12}}\ln\left(\tan x\right)\,dx=-\frac{2}{3}\text{G}
\end{align}
(see: Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )
Thus,
\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\times -\frac{2}{3}\text{G}\\
&=\boxed{\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\frac{4}{3}\text{G}}
\end{align}
NB:
Observe that,
\begin{align}2-\sqrt{3}&=\frac{1}{2+\sqrt{3}}\\
\ln\left(2-\sqrt{3}\right)&=-\ln\left(2+\sqrt{3}\right)\\
\int_0^\infty \frac{\ln x}{1+x^2}\,dx&=0
\end{align}
(perform the change of variable $y=\dfrac{1}{x}$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 2
} |
general solution for double integral of square root of quadratic polynomials I wonder if there is a closed-form solution for double integral of square root of quadratic polynomials. Such as
$$
\int _0 ^1 \int _0 ^1 \sqrt{a \cdot x^2 + b \cdot y^2 + c \cdot x \cdot y + d \cdot x+e\cdot y +1 } \ dx dy
$$
I have tried to solve
$$
\int _0 ^1 \sqrt{a \cdot x^2 + b \cdot y^2 + c \cdot x \cdot y + d \cdot x+e\cdot y +1 } \ dx
$$
and this has a closed form of solution, but is very complicated that makes it nearlly impossible to get the integral further on y.
So I wonder if a closed form solution really exists, does anyone have some ideas?
| MAPLE got:
$\int _0^1\int_0^1\sqrt{a\cdot x^2+b\cdot y^2+c\cdot x\cdot y+d\cdot x+e\cdot y +1 }\ dx\ dy$
$$=\frac{1}{8}\left(-4\,ab\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)+4\,ab\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)-4\,ae\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)+4\,ae\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)+{c}^{2}\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)-{c}^{2}\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)+2\,cd\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)-2\,cd\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)-{d}^{2}\ln\left({\frac{d+2\,\sqrt{a}}{\sqrt{a}}}\right)+{d}^{2}\ln\left({\frac{2\,a+d+2\,\sqrt{a+d+1}\sqrt{a}}{\sqrt{a}}}\right)+{d}^{2}\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)-{d}^{2}\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)+4\,a\ln\left({\frac{d+2\,\sqrt{a}}{\sqrt{a}}}\right)-4\,a\ln\left({\frac{2\,a+d+2\,\sqrt{a+d+1}\sqrt{a}}{\sqrt{a}}}\right)-4\,a\ln\left({\frac{c+d+2\,\sqrt{b+e+1}\sqrt{a}}{\sqrt{a}}}\right)+4\,a\ln\left({\frac{2\,a+c+d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}}{\sqrt{a}}}\right)+4\,\sqrt{a+b+c+d+e+1}{a}^{3/2}-4\,\sqrt{a+d+1}{a}^{3/2}+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}c-2\,\sqrt{b+e+1}\sqrt{a}c+2\,\sqrt{a}d+2\,\sqrt{a+b+c+d+e+1}\sqrt{a}d-2\,\sqrt{b+e+1}\sqrt{a}d-2\,\sqrt{a+d+1}\sqrt{a}d\right){a}^{-3/2}$$
| {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
For I the incenter in △ABC, if AB+IC=AC+IB, then △ABC is isosceles.
Let $I$ be the center of the inscribed circle in $\triangle ABC$. Prove that if $$AB+IC=AC+IB$$ then the triangle is isosceles!
| Another way.
In the standard notation we have
$$c+\frac{r}{\sin\frac{\gamma}{2}}=b+\frac{r}{\sin\frac{\beta}{2}}$$ or
$$c+\frac{\frac{2S}{a+b+c}}{\sqrt{\frac{1-\frac{a^2+b^2-c^2}{2ab}}{2}}}=b+\frac{\frac{2S}{a+b+c}}{\sqrt{\frac{1-\frac{a^2+c^2-b^2}{2ac}}{2}}}$$ or
$$b-c=\frac{4S\sqrt{ab}}{(a+b+c)\sqrt{(a+c-b)(b+c-a)}}-\frac{4S\sqrt{ac}}{(a+b+c)\sqrt{(a+b-c)(b+c-a)}}$$ or
$$b-c=\frac{4S\sqrt{a}}{(a+b+c)\sqrt{b+c-a}}\left(\frac{\sqrt{b}}{\sqrt{a+c-b}}-\frac{\sqrt{c}}{\sqrt{a+b-c}}\right)$$ or
$$b-c=\frac{\sqrt{a}\left(\sqrt{b(a+b-c)}-\sqrt{c(a+c-b)}\right)}{\sqrt{a+b+c}}$$ or
$$b-c=\frac{\sqrt{a}\left(b(a+b-c)-c(a+c-b)\right)}{\sqrt{a+b+c}\left(\sqrt{b(a+b-c)}+\sqrt{c(a+c-b)}\right)}$$ or
$$b-c=\frac{\sqrt{a}(b-c)(a+b+c)}{\sqrt{a+b+c}\left(\sqrt{b(a+b-c)}+\sqrt{c(a+c-b)}\right)},$$ which gives $b=c$ or
$$\sqrt{b(a+b-c)}+\sqrt{c(a+c-b)}=\sqrt{a(a+b+c)}$$ or
$$2\sqrt{bc}=\sqrt{(a+b-c)(a+c-b)}$$ or
$$(b+c)^2=a^2,$$ which is impossible.
Id est, $b=c$ and we are done!
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\sum \frac {A_r}{(x-a-\sqrt ∆)^r} + \sum \frac {B_r}{(x-a+\sqrt ∆)^r}
$$
Putting $-\frac {1}{2\sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
\par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$\kappa \log \left( \frac {x-a-\sqrt ∆}{x-a+\sqrt ∆}\right) + \frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n \binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} \binom {2n-r-1}{n-1} D^{2n-r}$$
| Let $b\neq a^2$,
$$S(n)=\int \frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=\frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-\left( 2 A n-C-3 A\right) \, {{x}^{2}}-\left( \left( 2 B-2 A a\right) n+\left( 2 C+4 A\right) a-2 B\right) x\\+2 B a n-\left( -C-A\right) b-2 B a$$
$$A=\frac{1}{2 \left( b-{{a}^{2}}\right) \, \left( n-1\right) },B=-\frac{a}{2 \left( b-{{a}^{2}}\right) \, \left( n-1\right) },\\C=\frac{2 n-3}{2 \left( b-{{a}^{2}}\right) \, \left( n-1\right) }$$
Then
$$S(n)=\frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
\frac{2n-3}{2(n-1)(b-a^2)}S(n-1), \; n>1$$
$$S(1)=\int \frac {dx}{x^2-2ax+b}$$
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Extracting two balls with the same color vs distinct color I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(\text{distinct color})=P(\text{first white})P(\text{second black/first white})+P(\text{first black})P(\text{second white/first black})$$
$$=\frac{a}{a+b}\frac{b}{a+b-1}+\frac{b}{a+b}\frac{a}{a+b-1}=\frac{2ab}{(a+b)(a+b-1)}$$
$$P(\text{same color})=P(\text{first white})P(\text{second white/first white})+P(\text{first black})P(\text{second black/first black})$$
$$=\frac{a}{a+b}\frac{a-1}{a+b-1}+\frac{b}{a+b}\frac{b-1}{a+b-1}=\frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which is more probably to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true?
$$a^2+b^2-(a+b)< 2ab \equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer..
| Total number of possible extractions is $\binom{a+b}{2}$
Number of ways to extract:
*
*two white balls: $\binom{a}{2}$
*two black balls: $\binom{b}{2}$
*one of each color: $ab$
Probabilities can be calculated as $$\frac{\binom{a}{2}+\binom{b}{2}}{\binom{a+b}{2}}\text{ for same color and }\frac{ab}{\binom{a+b}{2}}\text{ for different colors}$$
Since you only want to know which has the higher probability, you can simply compare $\binom{a}{2}+\binom{b}{2}$ with $ab$ to see which is greater. Dependent on the values of $a$ and $b$, this could go either way.
Here's a chart detailing which outcome is more likely for $a,b\le30$. A plus (+) indicates that one ball of each color is more likely, a minus (-) indicates that two balls of one color is more likely, and the equal (=) indicates that the probability is split 50-50. Notice that they are equal when $a$ and $b$ are consectutive triangular numbers (1,3,6,10,15...).
Generally, when there are approximately equal numbers of each color, then drawing one ball of each color is more likely.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 + + = - - - - - - - - - - - - - - - - - - - - - - - - - - -
2 + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -
3 = + + + + = - - - - - - - - - - - - - - - - - - - - - - - -
4 - + + + + + + - - - - - - - - - - - - - - - - - - - - - - -
5 - - + + + + + + - - - - - - - - - - - - - - - - - - - - - -
6 - - = + + + + + + = - - - - - - - - - - - - - - - - - - - -
7 - - - + + + + + + + + - - - - - - - - - - - - - - - - - - -
8 - - - - + + + + + + + + - - - - - - - - - - - - - - - - - -
9 - - - - - + + + + + + + + - - - - - - - - - - - - - - - - -
10 - - - - - = + + + + + + + + = - - - - - - - - - - - - - - -
11 - - - - - - + + + + + + + + + + - - - - - - - - - - - - - -
12 - - - - - - - + + + + + + + + + + - - - - - - - - - - - - -
13 - - - - - - - - + + + + + + + + + + - - - - - - - - - - - -
14 - - - - - - - - - + + + + + + + + + + - - - - - - - - - - -
15 - - - - - - - - - = + + + + + + + + + + = - - - - - - - - -
16 - - - - - - - - - - + + + + + + + + + + + + - - - - - - - -
17 - - - - - - - - - - - + + + + + + + + + + + + - - - - - - -
18 - - - - - - - - - - - - + + + + + + + + + + + + - - - - - -
19 - - - - - - - - - - - - - + + + + + + + + + + + + - - - - -
20 - - - - - - - - - - - - - - + + + + + + + + + + + + - - - -
21 - - - - - - - - - - - - - - = + + + + + + + + + + + + = - -
22 - - - - - - - - - - - - - - - + + + + + + + + + + + + + + -
23 - - - - - - - - - - - - - - - - + + + + + + + + + + + + + +
24 - - - - - - - - - - - - - - - - - + + + + + + + + + + + + +
25 - - - - - - - - - - - - - - - - - - + + + + + + + + + + + +
26 - - - - - - - - - - - - - - - - - - - + + + + + + + + + + +
27 - - - - - - - - - - - - - - - - - - - - + + + + + + + + + +
28 - - - - - - - - - - - - - - - - - - - - = + + + + + + + + +
29 - - - - - - - - - - - - - - - - - - - - - + + + + + + + + +
30 - - - - - - - - - - - - - - - - - - - - - - + + + + + + + +
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{\pi/2}\cos^{2n+1}(x)dx$ How can I compute this Integral for integer $n$ from $0$ to $\pi/2$
$$\int_{0}^{\pi/2}\cos^{2n+1}(x)dx?$$
| Keywords: Wallis integrals.
For $k\geq 0$ integer, define
\begin{align}J_k=\int_0^{\frac{\pi}{2}}\cos^{2k+1} x\,dx\end{align}
Perform integration by parts, ($(\sin x)^\prime=\cos x$ )
\begin{align}J_{k+1}&= \int_0^{\frac{\pi}{2}}\cos^{2k+3} x\,dx\\
&=\Big[\sin x\cos^{2k+2} x\Big]_0^{\frac{\pi}{2}}+(2k+2)\int_0^{\frac{\pi}{2}}\sin^ 2x\cos^{2k+1} x\,dx\\
&=(2k+2)\int_0^{\frac{\pi}{2}}\sin^ 2x\cos^{2k+1} x\,dx\\
&=(2k+2)\int_0^{\frac{\pi}{2}}\left(1-\cos^ 2x\right)\cos^{2k+1} x\,dx\\
&=(2k+2)\left(J_{k}-J_{k+1}\right)
\end{align}
Therefore,
\begin{align}J_{k+1}&=\frac{2k+2}{2k+3}J_k\end{align}
Observe that,
\begin{align}J_0&=\int_0^{\frac{\pi}{2}} \cos x\,dx\\
&=\Big[\sin x\Big]_0^{\frac{\pi}{2}}\\
&=1
\end{align}
Addendum:
One obtains:
\begin{align}\boxed{J_k=\frac{2^kk!}{\big(2(k-1)+3\big)\big(2(k-2)+3\big)...\times 3}}\end{align}
| {
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"answer_count": 5,
"answer_id": 0
} |
Evaluate $\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $ $$\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $$
$$1 \cdot 3 \cdot 5 \cdots (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n+2)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} = \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} $$
But the follow up gives me $\infty$. How to approach this type of exercises?
The result should be $1/2$
| Observe that
$$
\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot3\cdot 5\cdots (2n+1)}\right)
$$
Hence
$$\sum_{n=1}^\infty\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\sum_{n=1}^\infty\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot 3\cdot 5\cdots (2n+1)}\right)=\frac{1}{2}\left(1-\frac{1}{1\cdot3}+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot 5}-\frac{1}{1\cdot3\cdot 5}+\cdots\right)=\frac{1}{2}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = \sqrt{1-x^2}, -1 \leq x \leq 1$ Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = \sqrt{1-x^2}, -1 \leq x \leq 1$
Solution attempt:
$f(x, y) = y$, along $y = \sqrt{1-x^2}, -1 \leq x \leq 1$
$\vec{r}(t) = (t, \sqrt{1-t^2}), -1 \leq t \leq 1$
$\vec{r}'(t) = (1, \frac{-2t}{2\sqrt{1-t^2}}) = (1, \frac{-t}{\sqrt{1-t^2}})$
$||\vec{r}'(t)|| = \sqrt{1+\frac{t^2}{1-t^2}} = \frac{1}{\sqrt{1-t^2}}$
Show $\int f(x, y) dt = \int_{-1}^{1} \sqrt{1-t^2}||\vec{r}'(t)|| dt = \int_{-1}^{1} \sqrt{1-t^2}\frac{1}{\sqrt{1-t^2}} dt = 1+1 = 2$
therefore the value of the required path integral is 2
is this correct?
| You can also parameterise the curve $x=cost$, $y=sint$
$$r'(t) = <-sint,cost>$$
$$||r'(t)|| = 1$$
$$F = sin t$$
$$I = \int_{0}^{\pi} sint dt = -cost|_{0}^{\pi} = 2$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhere. so what can I do?
| Here are two other takes.
Take 1
Let $\alpha = \sqrt[3]{3}+\sqrt[3]{9}$. Then $\alpha^3 = 9 \alpha + 12$.
By the rational root theorem, if $\alpha$ is rational, then $\alpha$ is an integer.
Now $1 < \sqrt[3]{3} < 2 $ and $2 < \sqrt[3]{9} < 3 $, and so $3 < \alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $\alpha$ is not an integer and so $\alpha$ is irrational.
Take 2
We have $\alpha = \beta+\beta^2$, where $\beta=\sqrt[3]{3}$.
If $\alpha$ were rational, then $\beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $\beta$ as a root is $x^3-3$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculating the determinant of a matrix $n\times n$ I was trying to calculate $detD_{1,1}$ when
$$ D=\begin{pmatrix}0 & -1 & -1 & \ldots & \ldots & -1 & -1 & -1\\
0 & 2 & -1 & \ldots & \ldots & 0 & 0 & 0\\
0 & 0 & 2 & \ldots & \ldots & -1 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \ldots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & \ldots & 2 & -1 & 0\\
0 & 0 & 0 & \ldots & \ldots & 0 & 2 & -1\\
0 & -1 & 0 & \ldots & \ldots & 0 & 0 & 2
\end{pmatrix}$$
I got to the following determinant of a matrix:
$$ \begin{vmatrix}0 & -1 & 0 & \ldots & \ldots & 0 & 0 & 4\\
0 & 0 & -1 & \ldots & \ldots & 0 & 0 & 8\\
0 & 0 & 0 & \ldots & \ldots & -1 & 0 & 16\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \ldots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & \ldots & 0 & -1 & 2^{n-1}\\
0 & 0 & 0 & \ldots & \ldots & 0 & 2 & -1\\
-1 & 0 & 0 & \ldots & \ldots & 0 & 0 & 2
\end{vmatrix}$$
But now I'm stuck. How can I continue from here? If there was no $-1$ at the bottom I could just multiplate the diagonal.
| Move the bottom row of the matrix to the top (permuting the rows cyclically) to find
$$
\begin{vmatrix}0 & -1 & 0 & \ldots & \ldots & 0 & 0 & 4\\
0 & 0 & -1 & \ldots & \ldots & 0 & 0 & 8\\
0 & 0 & 0 & \ldots & \ldots & -1 & 0 & 16\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \ldots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & \ldots & 0 & -1 & 2^{n-1}\\
0 & 0 & 0 & \ldots & \ldots & 0 & 2 & -1\\
-1 & 0 & 0 & \ldots & \ldots & 0 & 0 & 2
\end{vmatrix} = \\
(-1)^{n-1}
\begin{vmatrix}
-1 & 0 & 0 & \ldots & \ldots & 0 & 0 & 2\\
0 & -1 & 0 & \ldots & \ldots & 0 & 0 & 4\\
0 & 0 & -1 & \ldots & \ldots & 0 & 0 & 8\\
0 & 0 & 0 & \ldots & \ldots & -1 & 0 & 16\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \ldots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & \ldots & 0 & -1 & 2^{n-1}\\
0 & 0 & 0 & \ldots & \ldots & 0 & 2 & -1
\end{vmatrix}
$$
Because this matrix is block upper-triangular, we can compute
$$
\begin{vmatrix}
-1 & 0 & 0 & \ldots & \ldots & 0 & 0 & 2\\
0 & -1 & 0 & \ldots & \ldots & 0 & 0 & 4\\
0 & 0 & -1 & \ldots & \ldots & 0 & 0 & 8\\
0 & 0 & 0 & \ldots & \ldots & -1 & 0 & 16\\
\vdots & \vdots & \vdots & \ddots & \ldots & \vdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \ldots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & \ldots & 0 & -1 & 2^{n-1}\\
0 & 0 & 0 & \ldots & \ldots & 0 & 2 & -1
\end{vmatrix} = (-1)^{n-2} \det \begin{vmatrix}-1&2^{n-1}\\2&-1\end{vmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying compound fraction not producing answer provided by book I am working on a problem in a textbook(Precalculus Mathematics for Calculus, by James Stewart) and the answer in the back of the book for the problem(1.4 #67) is -xy but I am not getting that. Here is the problem:
\begin{align}
\frac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{x^2}-\frac{1}{y^2}}
\end{align}
Here is how I worked it out:
\begin{align}
\frac{\frac{x}{y}-\frac{y}{x}}{\frac{1}{x^2}-\frac{1}{y^2}}=\frac{\frac{x^2-y^2}{xy}}{\frac{y^2-x^2}{x^2y^2}}=\frac{x^2-y^2}{xy}*\frac{x^2y^2}{y^2-x^2}
=\frac{xy(x^2-y^2)}{y^2-x^2}=\frac{(xy)(x+y)(x-y)}{(y+x)(y-x)}=\frac{xy(x-y)}{y-x}=\frac{x^2y-xy^2}{y-x}
\end{align}
Can someone please explain what I am doing wrong here?
| You were almost there. $\frac{xy(x-y)}{y-x}$(your second last one)$=-xy$
| {
"language": "en",
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How to change the variables in $\frac{1}{(z+1)^{n+1}}=\frac{(-1)^n}{n!}\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right)$? We know that
$$\frac{1}{(z+1)^{n+1}}=\frac{(-1)^n}{n!}\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right).$$
If we put $z=e^t$ how can we change the variables above equation?
I know that $z\frac{d}{dz}=\frac{d}{dt}$. So how can we write $\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right)$?
Is $z^n \frac{d^n}{dz^n}=\frac{d^n}{dt^n}$?
| In general, if $f(z)$ is $n$ times differentiable, then we have
$$\left(z\frac{d}{dz}\right)^n\{f(z)\}=\sum_{k=1}^n \begin{Bmatrix} n \\ k \end{Bmatrix}z^k\frac{d^k\,f(z)}{dz^k}$$
where $\begin{Bmatrix} n \\ k \end{Bmatrix}=\frac1{k!}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}\,j^n$ are the Stirling Numbers of the Second Kind.
For $f(z)=\frac1{z+1}$, we find
$$\begin{align}
\left(z\frac{d}{dz}\right)^n\left(\frac{1}{1+z}\right)=\sum_{k=1}^n \begin{Bmatrix} n \\ k \end{Bmatrix}z^k\frac{(-1)^k\,k!}{(1+z)^{k+1}}
\end{align}$$
Letting $z=e^t$, we find that
$$\begin{align}
\frac{d^n}{dt^n}\left(\frac{1}{1+e^t}\right)&=\frac{1}{1+e^t}\sum_{k=1}^n \begin{Bmatrix} n \\ k \end{Bmatrix}(-1)^k\,k! \left(\frac{e^t}{1+e^t}\right)^k\\\\
&=\frac1{1+e^t}\sum_{k=1}^n\sum_{j=0}^k (-1)^j\binom{k}{j}j^n\left(\frac{1}{1+e^{-t}}\right)^k
\end{align}$$
| {
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"answer_count": 1,
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} |
Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
| Notice that $$\frac{x^3-x}{x^2+xy+x+y} = \frac{(x+1)(x^2-x)}{(x+1)(x+y)} =\frac{x^2-x}{x+y}$$
P.S.: You cannot simplify the $x$'s without considering all terms of the numerator and the denominator.
| {
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} |
Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$
Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.
We have
$$\begin{align}
\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2n^2-2-2n^2-3}{2(2n^2+3)}\right| \\
&= \left|\frac{-5}{2(2n^2+3)}\right|\\
&= \frac{5}{2(2n^2+3)} \\
&<\frac{5}{4n^2},
\end{align}$$
$$\frac{5}{4n^2}<\epsilon \iff \frac{1}{n^2}<\frac{4 \epsilon}{5} \iff n>\sqrt{\frac{5}{4\epsilon}}$$
We choose $n_0=\left[\sqrt{\frac{5}{4\epsilon}} \right]+1$, Then $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$.
Let $(x_n)=\frac{1}{\ln(n+1)}$ for $n \in \mathbb{N}$.
a) Use the definition of the limit to show that $\lim(x_n)=0$.
$|\frac{1}{\ln(n+1)}-0|=\frac{1}{\ln(n+1)}<\epsilon \Leftrightarrow ln(n+1) > \epsilon \Leftrightarrow n> e^{\epsilon} -1$
We choose $n_0=\left[ e^\epsilon -1 \right]+1$, Then $\lim(x_n)=0$.
b) Find specific value of $n_0 (\epsilon)$ as required in definition of limit for $\epsilon=\frac{1}{2}$.
$n_0=\left[\sqrt{e}-1\right]+1$
Is that true, please?
| Your answer is correct. However it seems you may have overcomplicated it.
\begin{align}
\lim_{n\to \infty}\left(\frac{n^2-1}{2n^2+3}\right) & = \lim_{n\to \infty}\left(\frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}\right)\\
\end{align}
Now use the fact that $\lim_{n\to \infty}\left(\frac{1}{n^k}\right)=0$, where $k$ is any positive integer.
Hence
\begin{align}
\lim_{n\to \infty}\left(\frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}\right) & = \lim_{n\to \infty}\left(\frac{1-0}{2+0}\right)\\
&= \frac{1}{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the $f$ of $g$ where $f(x) =x^2+2x+1$ and $g(x)=2x-3$ According to my textbook the $f$ of $g$ where $f(x) =x^2+2x+1$ and $g(x)=2x-3$ is $4x^2-8x+2$, but I get $4x^2-8x+4$.
What am I doing wrong?
My Steps:
$(2x-3)^2 + 2(2x-3)+1$
$(2x-3)(2x-3)+4x-6+1$
$4x^2-12x+9+4x-6+1$
$4x^2-8x+4$
| A simpler approach to your calculation is to factor $f$ first:
$$f(x) = x^2 + 2x + 1 = (x+1)^2.$$
Then, $$f(g(x)) = (g(x)+1)^2 = ((2x-3)+1)^2 = (2x-2)^2 = 2^2(x-1)^2 = 4x^2 - 8x + 4.$$
To check that the book's solution cannot be correct, it suffices to choose $x = 0$, which leads to $g(0) = -3$, and $$f(g(0)) = f(-3) = (-3)^2 + 2(-3) + 1 = 9 - 6 + 1 = 4.$$ Your computation is consistent with this result, since $$4(0)^2 - 8(0) + 4 = 4,$$ but the book's solution is not, since it gives $2$ when $x = 0$.
| {
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"url": "https://math.stackexchange.com/questions/3099037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A sum of Fibonacci numbers Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k}$$
How can I evaluate the following sums to show this identity (1) is correct?
$$\sum_{n=0}^{\infty}(-1)^n\frac{n^2}{{2n \choose n}(2n+1)}=A$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{{2n \choose n}(2n+1)}=B$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{1}{{2n \choose n}(2n+1)}=C$$
| Here is another answer, in the same spirit as that of @dan_fulea:
Both sides of $$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k+1}$$ can be considered as sequences defined by the same second order linear recurrence for the index $k$. Then, for them to be equal for all $k$, it suffices that they coincide at the first values $0$ and $1$ of the index $k$.
In other words, we just need to show that
\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&= F_1=1
\end{align}
and
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n^2+9n+4}{2n+1}\frac{1}{{2n \choose n}}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n} &=F_2=1
\end{align}
where we have introduced the Catalan number $C_n=\frac{1}{n+1} \binom{2n}{n}$.
But it is easy to verify, by induction, that
\begin{align}\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&=\frac{{2m+2 \choose m+1}+(-1)^m}{{2m+2 \choose m+1}}\\
\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n}&=\frac{C_{m+1}+(-1)^m}{C_{m+1}}
\end{align}
and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) }} {dx} \\
&= e^{x\ln(1+1/x)} \left(\ln(1+\frac{1}{x}) - \frac{1}{1+x} \right) >0
\end{align}
| $$
\frac{d}{dx} \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) = \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) \left( \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1} \right) > 0
$$
for $x > 1$, so that the function is strictly increasing on $(1, \infty)$. Indeed,
$$
\ln\left(1 + \frac{1}{x}\right) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} \mp \cdots,
$$
whereas
$$
\frac{1}{x + 1} = \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} \mp \cdots.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem to have understood what is done there. Could I get your assistance in order to understand it? Perhaps there's better way of factoring.
Regards
| An easier way is to note that $x^4-2x^3+x-2$ has the simple roots $x=-1$ and $x=2$, which are obvious candidates by the rational root theorem, so your polynomial is divisible by
$$(x+1)(x-2)=x^2-x-2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show, that the sequence $ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $ converges. I need to show, that the following sequence converges.
I think I can somehow do it by Riemann Integral, but I cannot figure out a way to extract $ \frac{1}{n} $ from it. I also cannot find two sequences which could let me show that it converges by the sandwich theorem. How to approach it?
$$ a_n = \left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{9}\right)\cdot\ldots\cdot\left(1+\frac{1}{3^n}\right) $$
| If you are ok with using GM-AM (inequality between geometric and arithmetic mean) and with using the well known limit
*
*$\left(1+ \frac{x}{n}\right)^n\stackrel{n\to \infty}{\longrightarrow}e^x$
then you can reason directly as follows:
\begin{eqnarray*} \prod_{k=1}^n \left(1+\frac{1}{3^k}\right)
& \leq & \left(\frac{\sum_{k=1}^n \left(1+\frac{1}{3^k}\right)
}{n} \right)^n\\
& = & \left(\frac{n + \sum_{k=1}^n \frac{1}{3^k}}{n} \right)^n\\
& \leq & \left(1 + \frac{\sum_{k=1}^{\infty} \frac{1}{3^k}}{n} \right)^n \\
& = & \left(1 + \frac{\frac{1}{2}}{n} \right)^n \\
& \stackrel{n \to \infty}{\longrightarrow} & \sqrt{e} \\
\end{eqnarray*}
Since $a_n$ is obviously increasing and according to above calculation also bounded, it follows that $a_n$ is also convergent.
| {
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Finding all real $x$ such that $1+\sum_{j=1}^n\sin{\frac{j\pi x}{n+1}} = 0$, where $n=18$.
The task is to find all $ x \in \mathbb R $ such that
$$ 1 + \sum_{j=1}^n \sin{\frac{j\pi x}{n + 1}} = 0, \qquad n = 18 $$
What I have tried
Using the following formulas:
$$1. \sin{x} + \sin{y} = 2\sin{\frac{x + y}{2}}\cos{\frac{x - y}{2}} $$
$$2. \cos{x} + \cos{y} = 2\cos{\frac{x + y}{2}}\cos{\frac{x - y}{2}} $$
$$3. \sin{x}\sin{y} = \frac{1}{2} (\cos{(x - y)} - \cos{(x + y)}) $$
First approach
$$ 1 + \sin{\frac{\pi x}{19}} + \sin{\frac{18\pi x}{19}} + \sin{\frac{2 \pi x}{19}} + \sin{\frac{17\pi x}{19}} + ... + \sin{\frac{9 \pi x}{19}} + \sin{\frac{10 \pi x}{19}} = 0 $$
Using formula 1:
$$ 1 + 2\sin{\frac{\pi x}{2}} (\cos{\frac{17\pi x}{19}} + \cos{\frac{15\pi x}{19}} + ... + \cos{\frac{\pi x}{19}})$$
Skipping some steps, we get:
$$ 1 + 2\sin{\frac{\pi x}{2}}\cos{\frac{9\pi x}{38}}(2\cos{\frac{3\pi x}{38} + 1})(2\cos{\frac{\pi x}{38}} + 1) = 0 $$
From here I couldn't find a way to go further
Second approach
Multiply both sides by $ \sin{\frac{\pi x}{38}} $:
$$ \sin{\frac{\pi x}{38}} (1 + \sin{\frac{\pi x}{19}} + \sin{\frac{2\pi x}{19}} + \sin{\frac{3 \pi x}{19}} + ... + \sin{\frac{17 \pi x}{19}} + \sin{\frac{18 \pi x}{19}}) = 0 $$
Using formula number 3
$$ \sin{\frac{\pi x}{38}} + \frac{1}{2}(\cos{\frac{\pi x}{38}} - \cos{\frac{3\pi x}{38}} + \cos{\frac{3\pi x}{38}} - \cos{\frac{5\pi x}{38}} + ... + \cos{\frac{35\pi x}{38}} - \cos{\frac{37\pi x}{38}}) = 0$$
$$ \sin{\frac{\pi x}{38}} - \sin{\frac{\pi x}{2}}\sin{\frac{18\pi x}{38}} = 0 $$
From here I couldn't find any way to go further
Any hints will be appreciated!
EDIT:
The upper-most formula had a mistake, but now it is up to date.
PS
The problem is supposed to be solvable assuming you have high-school level math (i.e only simple trigonometric formulas are allowed, no complex numbers, no derivatives for the sine and cosine functions).
Also, this equation on Wolfram Alpha shows the following solution:
$ x = \frac{19}{2}(4n - 1), n \in \mathbb Z $
Update 2: Using @Doug M's answer, I managed to conclude that ($ \sin{\frac{\pi x}{19}} = 1 $ and $ \cos{\frac{\pi x}{19}} = 0 $), which equalates to $ \frac{\pi x}{19} = \frac{\pi}{2} + 2\pi n, n \in \mathbb Z $ is one of the solutions to the equation
| Multiply the series by $\frac {\sin \frac {\pi x}{19}}{\sin \frac {\pi x}{19}}$
Then use
$\sin A\sin B = \frac 12 \cos (A-B) - \frac 12 \cos (A+B)$
And the series will telescope.
$1+\sum_\limits{j=1}^{18} \frac {\sin \frac{j\pi}{19}\sin \frac {\pi x}{19}}{\sin\frac {\pi x}{19}}$
$1+\sum_\limits{j=1}^{18} \left(\frac {\cos \frac{(j-1)\pi}{19}}{2\sin\frac {\pi x}{19}} - \frac {\cos \frac{(j+1)\pi}{19}}{2\sin\frac {\pi x}{19}}\right)$
$1+\frac{\cos 0 + \cos \frac {\pi x}{19} - \cos \frac {18\pi x}{19} - \cos \pi x}{2\sin \frac {\pi x}{19}} = 0$
let $y = \frac {\pi x}{19}$
$2\sin y + 1 + \cos y - \cos 18y - \cos 19y = 0$
(multiplying through by $\sin y$ has introduced an artificial solution at $\sin y = 0$)
$y = \frac {3\pi}{2} + 2k\pi$ is a solution, implying $x = \frac {57}{2}$
As for the others....
If we expand the $\cos 19 y, \cos 18 y$ terms we could turn this into a function of powers of $cos y$ and $sin y$ but we still have a 19 (or 17 degree when we factor out the two identified factors) degree polynomial to factor.
I put it into wolfram alpha and none of the other solutions are particularly nice.
https://www.wolframalpha.com/input/?i=sin+y++%2B+1+%2B+cos+y+-+cos+18y+-+cos+19y+%3D+0
This is as far as I know how to take it.
| {
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"timestamp": "2023-03-29T00:00:00",
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An algebra question based on sequences and series The following question was asked in JEE Advanced 2014 Paper I:
A pack contains $n$ cards numbered sequentially from $1$ to $n$. Two consecutive cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the number on the smaller card is $k$, then what is the value of $k-20$?
I could construct only one equation from this information:
$$\frac{n(n+1)}2-2k-1=1224$$
How should I proceed from here to solve this problem?
| Let the two removed cards be $k$ and $k + 1$.
$1 + 2 + .... n = \frac {n(n+1)}2,$ and removing $k$ and $k+1$ and adding them up we get $\frac {n(n+1)}2 - k -(k+1) =\frac {n(n+1)}2 - 2k-1 = 1224,$
so $k = \frac {n(n+1)}4 - \frac {1225}2,$ which means $\frac {n(n+1)}2$ is odd and that
$\frac {n(n+1)}2 > 1225$.
So $n(n+1) > 2450 = 2500 -50 = 50(50-1) = 49*50$.
So $n > 49$.
We also have $n > k = \frac {n(n+1)}4 - \frac {1225}2$
$4n > n^2 + n - 2450 > 0$
$2450 > n^2 - 3n,$ so
$2450+ \frac 94 > n^2 - 3n + \frac 92$
$\sqrt{2452.25}\approx 49.52 > n-\frac 32$
$51.02 > 51\ge n$
So $n = 50$ or $51$. But we must have $\frac {n(n+1)}2$ be odd so $n= 50$ and
$k =\frac {50(51)}4 = \frac {1225}2 = \frac {25*51 - 1225}2=25$.
And $k -20 = 5.$
And indeed, $(1+ ....+ 24) + (27+ ...+ 50) = (1+ .... + 50) - 25 - 26 = \frac {50*51}2 - 51 = 24*51 = 1224$.
.....
Argh. Stupid errors trying to solve equations in one's head!
| {
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Find the particular solution of the differential equation $\sqrt{x} + \sqrt{y}y' =0$, with $y(1) = 9$.
I get:
$\sqrt{y}dy = -\sqrt{x}dx \Rightarrow \frac{2}{3}y^{\frac{3}{2}} = -\frac{2}{3}x^{\frac{3}{2}}+C$.
What should I do from here?
| Multiply through by $3/2$ and rename $\frac32 C = c$
$$ y^{3/2} = c - x^{3/2} $$
The initial condition $y(1)=9$ gives $c = 28$. Solving for $y$ gives
$$ y^{3/2} = 28 - x^{3/2} \implies y = (28-x^{3/2})^{2/3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding limit of a (Laurent?) series I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.
| As $1+2+\ldots +n = \frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = \sum_{n=1}^{\infty} \frac{1}{\frac{k}{2}(k+1)} = 2 \sum_{k=1}^{\infty} \frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2\sum_{k=1}^{n} \frac{1}{k} - \frac{1}{k+1} = 2(1 - \frac{1}{n+1})$. Now taking limit when $n \rightarrow \infty$ you get that $S=2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$ = a - b? I'm given the complex rational expression
$$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$$
and asked to simplify. The solution provided is $a - b$; however I get $\frac{a^2 - b^2}{a+b}$.
My working:
Numerator first:
$$\frac{a}{b} - \frac{b}{a}$$
Least common denominator is $ab$.
Multiplying each part to get the LCD in the denominator on both sides I get:
$$\frac{a}{b} \cdot \frac{a}{a} - \frac{b}{a} \cdot \frac{b}{b}$$
$$= \frac{a^2 - b^2}{ab}$$
Multiplying this expression by the reciprocal in the original problem:
$$\frac{a^2 - b^2}{ab} \cdot \frac{ab}{a+b}$$
$$= \frac{ab(a^2-b^2)}{ab(a+b)}$$
Cancel out common factor $ab$:
$$\frac{a^2 - b^2}{a + b}$$
Where did I go wrong and how can I arrive at $a - b$?
| $$a^2-b^2=(a-b)(a+b)$$ If $a+b\ne 0$, you can divide both the numerator and denominator by it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple...
Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x^3 - 4x^2 + x + 6)$. Find the value of $|a + b + c|$.
My attempt :
$$x^2 + ax + b = (x + 1)Y$$
$$x^2 + bx + c = (x + 1)Z$$
From here onwards I do not how to continue. Please help. Thank you!
| Using GCD $\times$ LCM = Product, we get
$$(x+1)(x^3 - 4x^2 + x + 6) = (x^2+ax+b)(x^2+bx+c)$$
Now it remains to equate coefficients on both sides to get $(a, b, c) = (-1, -2, -3)$
| {
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u-sub vs trig-sub are giving different answers for $\int\frac{x+4}{x^2+2x+5}dx$ When I complete the square in the denominator and solve using u-sub, I can get the right answer:
$$\int\frac{x+4}{x^2+2x+5}dx$$
$$\int\frac{x+4}{x^2+2x+1+4}dx$$
$$\int\frac{(x+1)+4}{(x+1)^2+4}dx$$
$$u=x+1$$
$$\int\frac{u+3}{u^2+4}du$$
$$I_1=\int\frac{u}{u^2+4}du+I_2=\int\frac{3}{u^2+4}du$$
$$I_1=(w=u^2+4, dw=2udu)$$
$$I_1=\frac{1}{2}\int\frac{dw}{w}$$
$w = u^2+4,$
$w = (x+1)^2+4,$ so:
$$I_1=\frac{1}{2}\ln|x^2+2x+5|$$
$$I_2=3\int\frac{1}{u^2+4}du$$
$u = 2\tan\theta$
$du = 2\sec^2\theta d\theta$
$$I_2=3\int\frac{2\sec^2\theta}{(2\tan\theta)^2+4}d\theta$$
$$I_2=3\int\frac{2\sec^2\theta}{4\tan^2\theta+4}d\theta$$
$$I_2=3\int\frac{2\sec^2\theta}{4\sec^2\theta}d\theta$$
$$I_2=\frac{3}{2}\int d\theta$$
$$I_2=\frac{3\theta}{2}$$
At this point, I have to turn the integral back into terms of x, so I made the right triangle like normal:
Now solving the integrals:
$$\frac{1}{2}\ln|x^2+2x+5|+\frac{3\theta}{2}$$
$$\frac{1}{2}\ln|x^2+2x+5|+\frac{3\arctan(\frac{x+1}{2})}{2}+C$$
This obviously is the correct answer, however, when I try to solve this with trig-sub (which is the first thing that came to my mind when I looked at the problem, hence my frustration) I am getting a similar, albeit incorrect answer:
$$I_1=\int\frac{u}{u^2+4}du+I_2=\int\frac{3}{u^2+4}du$$
$u = 2\tan\theta$
$du = 2\sec^2\theta d\theta$
$$I_1=\int\frac{2\tan\theta}{4\tan^2\theta+4}\cdot\frac{2\sec^2\theta}{1}d\theta$$
$$I_1=\int\frac{2\tan\theta}{4\sec^2\theta}\cdot\frac{2\sec^2\theta}{1}d\theta$$
$$I_1=\int\frac{2\tan\theta}{2}d\theta$$
$$I_1=\int \tan\theta$$
$$I_2=3\int\frac{2\sec^2\theta}{4\sec^2\theta}d\theta$$
$$I_2=3\int\frac{1}{2}d\theta$$
$$I_2=\frac{3\theta}{2}$$
$$\int \tan\theta d\theta+\frac{3\theta}{2}$$
$$-\ln|\cos\theta|+\frac{3\theta}{2}$$
Now I put it back into terms of x like I did when solving it using u-sub:
$$-\ln\left|\frac{2}{u^2+4}\right|+\frac{3}{2}\arctan\frac{x+1}{2}$$
Since $u=x+1$:
$$-\ln\left|\frac{2}{(x+1)^2+4}\right|+\frac{3}{2}\arctan(\frac{x+1}{2})+C$$
But this is obviously wrong, since it looks like the $ln$ should have a $\frac{1}{2}$ in front of it, so something must be wrong with my trig-sub on $I_1$? I know it's a lot to read but I just wanted to put it step by step to see if there's some dumb algebraic mistake I made. If anyone can help, thanks a ton in advance.
| You plugged in the wrong expression for $\cos \theta$, it should be $2/\sqrt{u^2+4}$. Notice that the Pythagorean theorem tells you that the length of the hypotenuse should be $\sqrt{x^2+2x+5}$.
Note the minus sign outside of the logarithm. $$-\ln\left|\dfrac{2}{\sqrt{u^2+4}} \right|=\ln\left|\dfrac{\sqrt{u^2+4}}{2}\right|$$
Since $\sqrt{f(x)}$ can be written as $(f(x))^{1/2}$. Therefore by the logarithm properties.
*
*$\ln \left|\sqrt{u^2+4}\right|=1/2\cdot\ln\left|u^2+4\right|$
*$1/2\cdot\ln\left|(u^2+4)/\sqrt{2}\right|=1/2\cdot\ln\left|u^2+4\right|-1/2\cdot\ln\sqrt{2}$ which gets absorbed in the constant $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Undo this transformation I have two variables $x$, $y$ and calculate the following:
$a = \frac{x}{\sqrt{x^2+y^2}}$, $b = \frac{y}{\sqrt{x^2+y^2}}$
Using $a$ and $b$ is there a way I can derive my original $x$ and $y$?
| By squaring we get $$a^2=\frac{x^2}{x^2+y^2},b^2=\frac{y^2}{x^2+y^2}$$ and we get
$$a^2+b^2=\frac{x^2+y^2}{x^2+y^2}=1$$
and you can not compute $x$ or $y$.
| {
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"url": "https://math.stackexchange.com/questions/3123879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$. The question is exactly as in the title:
$$\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$$
It's been so long since the last time I tried to calculate something like this. I first thought about polar coordinates but that doesn't go well with the domain of integration.
What kind of substitution do we need for this kind of problem? I am sorry if similar problem has been asked, I just couldn't manage to find it.
| Let
$$ F(c)=\int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{1/2}}dxdy $$
and then
$$ F'(c)=-c\int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{3/2}}dxdy $$
or
$$ \int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{3/2}}dxdy=-\frac{F'(c)}{c}. $$
Since
$$ \int\frac {1}{(x^2 + c^2)^{1/2}}dx=\ln(x+\sqrt{x^2+c^2})$$
one has
\begin{eqnarray*}
F(c)&=&\int_{-a}^a\ln(x+\sqrt{x^2+y^2+c^2})\bigg|_{x=-b}^{x=b}dy\\
&=&\int_{-a}^a\bigg[\ln(b+\sqrt{b^2+y^2+c^2})-\ln(-b+\sqrt{b^2+y^2+c^2})\bigg]dy.
\end{eqnarray*}
It is easy to see
$$ F'(c)=-2\int_{-a}^a\frac{bc}{(c^2+y^2)\sqrt{b^2+c^2+y^2}}dy=-2\arctan\bigg(\frac{by}{c\sqrt{b^2+c^2+y^2}}\bigg) \bigg|_{-a}^a=-4\arctan\bigg(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\bigg)$$
and hence
$$ \int_{-a}^a\int_{-b}^b \frac {1}{(x^2 + y^2 +c^2)^{3/2}}dxdy=-\frac{F'(c)}{c}=\frac{4}{c}\arctan\bigg(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\bigg). $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x+2(x-1)^2 \ln(1-x)+3}{(1-x)^2}$$
$$f'''(x) = \dfrac{2x^2-5x+1}{(x-1)^3}$$
$$f^{(n)}(x) = \ ?$$
| Use the following.
$$f'''(x)=\frac{2x^2-5x+1}{(x-1)^3}=\frac{2x^2-4x+2-x+1-2}{(x-1)^3}=\frac{2}{x-1}-\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
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finding the mean and variance descriptive statistics Suppose the following data are obtained by recording $X$, the number of customers that arrive at an automatic banking machine during $15$ successive one-minute
time intervals.
Q) Record the mean and variance.
mean is
$u_{X} = \sum_{x=1}^{15} x f_{X}(x) = 1.67$
using the data from below:
$f_{X}(0) = 4/15, f_{X}(1) = 3/15, f_{X}(2) = 4/15, f_{X}(3) = 2/15, f_{X}(4) = 2/15$
variance is
$\sigma^2_{X} = \sum_{x=1}^{15} x f_{X}(x) - 1.67^2 = 1.88$
the mean is the same as the solution, but the variance they got was $1.952$. What am I doing wrong?
| Your method finds the population variance where
$$\begin{align*}
\mathsf{Var}(X)
&=\mathsf E(X^2)-\mathsf E(X)^2\\\\
&=0^2\left(\frac{4}{15}\right)+1^2\left(\frac{3}{15}\right)+2^2\left(\frac{4}{15}\right)+3^2\left(\frac{2}{15}\right)+4^2\left(\frac{2}{15}\right)-\left(\frac{25}{15}\right)^2\\\\
&=4.6-\left(\frac{25}{15}\right)^2\\\\
&\approx1.882
\end{align*}$$
The sample variance, which is desired, is defined as
$$s^2=\frac{1}{n-1}\cdot\sum_{i=1}^n (X_i-\bar{X})^2$$
so we get
$$\begin{align*}
s^2
&=\frac{1}{14}\left(\left(2-\frac{25}{15}\right)^2+\cdots+\left(4-\frac{25}{15}\right)^2\right)\\\\
&\approx 1.952
\end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor $x^5-x+15$ It's possible to factor $x^5-x+15$. WolframAlpha gives the answer of: $$(x^2+x+3)(x^3-x^2-2x+5)$$
According to the wikipedia article on quintic functions, the general form $x^5-x+a$ is factorable only when $a=±15$, $±22440$, or $±2759640$.
Question: How would one factor such an expression? For me, it seems close to impossible by hand.
Bonus question: Why do those specific values of $a$ make the expression factorable? If there's no simple answer, is there a paper/further reading about it?
| One can factor $$f(x) := x^5 - x \pm 15$$ manually (over $\Bbb Z$) without too much fuss.
First, if $f$ has a linear factor, it has a rational root and (because $f$ is monic) any rational root must be an integer. But $f(0) \equiv f(1) \equiv 1 \pmod 2$, so $f$ has no root modulo $2$ and hence no integer root and hence no linear factor. (Alternatively, we can show this with the Rational Root Theorem; see the bottom of this answer.)
Thus, if $f$ factors, it must factor as a product of a cubic and a quadratic, that is, (where we denote $\Lambda = \pm 15$)
$$x^5 - x + \Lambda = (x^3 + A x^2 + B x + C) (x^2 + D x + E)$$
for some integers $A, B, C, D, E$. Distributing the right-hand side and comparing coefficients gives various (at most) quadratic conditions on those integers:
\begin{align}
A + D &= 0 \\
A D + B + E &= 0 \\
A E + B D + C &= 0 \\
B E + C D &= - 1 \\
C E &= \Lambda \\
\end{align}
We can quickly reduce this system: The first equation gives $D = -A$, so the second equation gives $E = A^2 - B$, and then the third equation becomes $C = -A^3 + 2 A B$. Substituting leaves the system
\begin{align*}
A^4 - A^2 B - B^2 &= -1 \\
A (A^2 - 2 B) (A^2 - B) &= \Lambda .
\end{align*}
Since $A, A^2 - 2 B, A^2 - B$ are all integers and the prime factorization of $|\Lambda| = 15$ is $3 \cdot 5$, there are only a small number of combinations to check (in fact, since $15$ is a product of two primes, one of the three factors must be $\pm 1$), and we can quickly recover the factorization mentioned in the question.
Remark For readers unfamiliar with reducing polynomial equations modulo primes, we can conclude that $f$ has no linear factor with just a little more work, using the Rational Root Theorem, which in this case implies that the only possible rational roots are $\pm 1, \pm 3, \pm 5, \pm 15$; substituting shows that none of these are roots. One can even avoid most of this work: The derivative of the polynomial is $f'(x) = 5 x^4 - 1$, so $f$ is increasing where $|x| \geq 1$, but $f(-3) = -69, f(-1) = 13, f(1) = 15$, from which we can conclude none of the eight candidates are roots.
| {
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In equilateral ABC , Proving : $3(a^4 + b^4 + c^4 + s^4 ) = (a^2 + b^2 + c^2 + s^2 )^2$ My question is that:
In equilateral $\triangle ABC$, $AB= s$
There is a point P in same plane , the distance to three vertices are $a, b, c $
then
$$ 3(a^4 + b^4 + c^4 + s^4 ) = (a^2 + b^2 + c^2 + s^2 )^2 $$
Is this equation true? If it is true How can I prove?
| This is not a solution, although the equality seems to be true. One way to approach the problem might be to work with the points themselves as opposed to distances.
We can put your triangle on a coordinate plane and say that Point A is at $(-\frac{s}{2} , 0)$, Point B is at $(\frac{s}{2}, 0)$, and Point C is at $(0, \frac{s\sqrt{3}}{2})$.
Then, if we say Point P is at $(x, y)$, we have that
$$a = \sqrt{\Big(x+\frac{s}{2}\Big)^2 +y^2}$$
$$b = \sqrt{\Big(x-\frac{s}{2}\Big)^2 +y^2}$$
$$c = \sqrt{x^2 +\Big(y-\frac{s\sqrt{3}}{2}\Big)^2}$$
Now, foiling out the RHS of what you want to prove, we get that:
$$(a^2 + b^2 + c^2 + s^2)^2 = (a^4 +b^4 + c^4 + s^4) + 2a^2b^2 + 2a^2c^2 + 2a^2s^2 + 2b^2c^2 + 2c^2s^2 + 2b^2s^2$$
So, in other words, you question is equivalent to showing that:
$$a^4 + b^4 + c^4 + s^4 = a^2b^2 +a^2c^2 +a^2s^2 + b^2c^2 +c^2s^2 + b^2s^2 $$
Using the definitions of $a, b$ and $c$ from above might be a method of showing the right relationships to prove this is true. For example, $a$ and $b$ clearly have a close relationship as $a^2 - b^2 = 2sx$.
| {
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Finding the vertex, axis, focus, directrix, and latus rectum of the parabola $\sqrt{x/a}+\sqrt{y/b}=1$
Find the vertex, axis, focus, directrix, and length of latus rectum of the parabola
$$\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$
Try: Curve
$$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$$
represents a parabola.
While drawing the diagram, I am getting that the parabola touches $x$ axis at $A(a,0)$ and touches $y$ axis at $(0,b)$
Also, the tangents at points $A$ and $B$ intersect each other at $(0,0)$ at an angle of $90^\circ$.
I do not know how can I solve this problem. Could someone help me to solve it? Thanks.
| The given equation $\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$
can be simplified as
$$\displaystyle \sqrt{\frac{x}{a}} = 1 - \sqrt{\frac{y}{b}}$$
Squarring both sides,
$$\displaystyle \frac{x}{a}= 1 - 2\sqrt{\frac{y}{b}} + \frac{y}{b}$$
$$\displaystyle \frac{x}{a}-1+ \frac{y}{b}=- 2\sqrt{\frac{y}{b}}$$
Squarring again,
$$\displaystyle \left(\frac{x}{a}-1+ \frac{y}{b}\right)^2=4\frac{y}{b}$$
$$\displaystyle \left(X\right)^2=4Y$$
Where
$$X=\frac{x}{a}-1+ \frac{y}{b}$$
and $$Y=\frac{y}{b}$$
Now compare with the standard form, apply the formulae and find the required properties.
Hope this helps you.
| {
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What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtain
\begin{equation}
x^2+(1-x^2)+(1+x)+(1-x)
>4 (v^2)^{\frac{1}{4}}
\implies \frac{9}{16} \ge v.
\end{equation}
Therefore the max value is $\frac{9}{16}$.
Method 2
$$
v=x (1-x^2)
\implies 2v^2= 2x^2 (1-x^2)^2
$$
With the AM-GM-inequality
$$2x^2+(1-x^2)+(1-x^2)
>3 (2v^2)^{\frac{1}{3}}
\implies \frac{2}{(27)^{\frac{1}{2}}} > v.
$$
| Notice the A.M/G.M theorem says:
$a_1 + a_2 + .. + a_n \ge n\sqrt[n]{a_1a_2...a_n}$. And that $a_1 + a_2 + .. + a_n = n\sqrt[n]{a_1a_2...a_n}$ if and ONLY IF $a_1 = a_2 = .... = a_n$.
Your calculations for the inequalities were correct. But you ignored the requirements for equality.
So by your calculations.
$v \le \frac 9{16}$ ... which is true.
And $v = \frac 9{16}$ if and only if $x^2 = 1 -x^2 = 1+x = 1-x$. .... which never happens. As that is never possible we know that $v$ doesn't ever equal $\frac 9{16}$ . But we do know that $v < \frac 9{16}$ always. So this is an upper bound of $v$.
But it need not be a maximum. If a maximum does exist it is less than $\frac 9{16}$.
That's .... all true.
Also by your calculations.
$v \le \frac {2}{27^{\frac 12}} = \frac 2{3\sqrt 3} < \frac 9{16}$.
And $v = \frac 2{3\sqrt 3}$ if and only if $2x^2 = 1-x^2 =1-x^2$ or if and only if $x =\pm \frac 1{\sqrt 3}$.
As that is possible if $x = \frac 1{\sqrt 3}$. So we have $v \le \frac 2{3\sqrt 3}$ with equality holding if and only if $x = \frac 1{\sqrt 3}$.
So that is the maximum value.
$v \le \frac 2{3\sqrt 3} < \frac 9{16}$.
| {
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What would be the best way to memorize the 10 by 10 multiplication table? Hear me out before you start downvoting please. I have a learning disability so no matter how hard I try I can’t memorize the table. Please give some tips/hints on how to memorize the table. Thanks in advance.
| Without more context on why you're having difficulty, we can't specifically tailor an answer to you.
"Memorizing" the times table isn't the most fruitful endeavor, but instead you should try and memorize the patterns that pop up. For example, $$\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\2 & & & & 10 & & & & & \\ 3 & & & & 15 & & & & & \\4 & & & & 20 & & & & & \\5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50\\6 & & & & 30 & & & & & \\7 & & & & 35 & & & & & \\8 & & & & 40 & & & & & \\9 & & & & 45 & & & & & \\10 & & & & 50 & & & & & \\\end{matrix}$$
For one, the table is symmetric, meaning you can cut down how much you need to memorize quite a bit. Take for example the times table written out for $5$. Notice how each number ends in either $5$ or $0$. If we look at the entries for multiplying by $10$, $$\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\2 & & & & & & & & & 20\\ 3 & & & & & & & & & 30\\4 & & & & & & & & & 40\\5 & & & & & & & & & 50\\6 & & & & & & & & & 60\\7 & & & & & & & & & 70\\8 & & & & & & & & & 80\\9 & & & & & & & & & 90\\10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100\\\end{matrix}$$ you basically take the numbers and add a zero. If we look at the times table for $2$: $$\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20\\ 3 & 6 & & & & & & & & \\4 & 8& & & & & & & & \\5 & 10 & & & & & & & & \\6 & 12 & & & & & & & & \\7 & 14 & & & & & & & & \\8 & 16 & & & & & & & & \\9 & 18 & & & & & & & & \\10 & 20 & & & & & & & & \\\end{matrix}$$ each product ends in $0$, $2$, $4$, $6$, or $8$.
Personally, if I'm asked to fill out the times table, these are the kinds of patterns I employ. However, again without more information on what you're struggling with, tailoring an answer is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Image of a region bounded by lines $x-y <2$ and $x+y>2$, under mapping $w=1/z$ I got $z = x+iy$ and $w=u+iv=\frac1 z = \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$
$u(x,y) = \frac{x}{x^2+y^2}$ and $v(x,y) = -\frac{y}{x^2+y^2}$
also $x(u,v) = \frac{u}{u^2+v^2}$ and $\;y(u,v) = -\frac{v}{u^2+v^2}$
$x-y<2 \implies \dfrac{u+v}{u^2+v^2} <2$
$x+y<2 \implies\dfrac{u-v}{u^2+v^2} > 2$
How to proceed from here? can't really figure out the region in $w$-plane.
| Answering my own question. I would like input/correction form any complex-analysis expert out there.
$\frac{u+v}{u^2+v^2} <2$ when compared to the standard equation of circle:
$x^2+y^2+\frac{B}{A}x+\frac{C}{A}y+D = 0$, gives a region in uv- plane which is outside the circle centered at (0.25,0.25) and radius=$\sqrt{2}/4$
Similarly $\frac{u-v}{u^2+v^2} >2$ is the region inside the circle centered at (0.25,-0.25) and radius=$\sqrt{2}/4$
The required region is the intersection of these two regions, which would lie inside the second circle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Units in this ring Consider the ring $ R = \mathbb{Z}[\frac{1 + \sqrt{-15}}{2}]$, then what are the units of this ring?
My try;
Let $\alpha \in R$. Then $\alpha = a + b(\frac{1 + \sqrt{-15}}{2})$, for some $a,b \in \mathbb{Z}$.
Now, $\alpha$ is a unity iff $N(\alpha) = 1$ iff $(a + b(\frac{1 + \sqrt{-15}}{2}))(a - b(\frac{1 + \sqrt{-15}}{2})) = 1$ iff $a^2 - b^2 (\frac{1 + \sqrt{-15}}{2})^2 = 1$.
Now, from here, can I simply state that $b = 0$ since we need to get rid of the complex part of the LHS? Then deduce that $\alpha = 1$ or $\alpha = -1$?
| In general $\alpha$ is a unit in $R$ if and only if $N(\alpha)$ is a unit in $\Bbb{Z}$. You haven't considered the case $N(\alpha)=-1$.
Second, the norm of $\alpha=a+b\left(\tfrac{1 + \sqrt{-15}}{2}\right)$ is not equal to
$$N(\alpha)=\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a - b\left(\frac{1 + \sqrt{-15}}{2}\right)\right).$$
Instead, it is given by taking the product over the conjugates of $a+b\left(\tfrac{1 + \sqrt{-15}}{2}\right)$, which gives
$$N(\alpha)=\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a+ b\left(\frac{1 - \sqrt{-15}}{2}\right)\right),\tag{1}$$
because $\frac{1-\sqrt{-15}}{2}$ is the other root of the minimal polynomial of $\frac{1+\sqrt{-15}}{2}$.
Proceeding from $(1)$ you want to find all integral solutions to
$$\left(a + b\left(\frac{1 + \sqrt{-15}}{2}\right)\right)\left(a+ b\left(\frac{1 - \sqrt{-15}}{2}\right)\right)=\pm1.$$
Expanding the left hand side of the equation yields an equation over the integers, and its solutions correspond to the units.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$
Evaluate the indefinite integral
$$\int \frac{x^2 \,dx}{a x^5 + b},$$
for real parameters $a, b \neq 0$.
No apparent substitutions simplify the expression (if the exponent of $x$ in the denominator were an integral multiple of $3$, the form of the integrand would suggest the substitution $u = x^3$, $du = 3 x^2 \,dx$, but the exponent is not). Applying integration by parts with $u = \frac{1}{a x^5 + b}$, $dv = x^2 dx$ is straightforward, but it produces an integrand with a much larger degree in the denominator and so appears only to make the situation worse. Applying integration by parts instead with $dw = \frac{x^k dx}{a x^5 + b}$ results in integrating $\frac{x^k dx}{a x^5 + b}$, which, except when $k \equiv 4 \pmod 5$ (which doesn't appear immediately fruitful), doesn't appear much easier than the given integral.
| Old post, but the original author bumped it so...
We want to scale the problem so we can concentrate on
$$\int\frac{x^2dx}{x^5+1}=\sum_{n=0}^4\int\frac{A_ndx}{x-x_n}=\sum_{n=0}^4A_n\ln(x-x_n)+C_1$$
Here $x_n^5=-1=e^{i(2n+1)\pi}$ so $x_n=e^{\frac{i(2n+1)\pi}5}$,
$$A_n=\lim_{x\rightarrow x_n}\frac{(x-x_n)x^2}{x^5+1}=\frac{x_n^2}{5x_n^4}=-\frac15x_n^3$$
$x_0^3=x_1=\cos\frac{3\pi}5+i\sin\frac{3\pi}5$; $x_1^3=x_4=x_0^*$; $x_2^3=x_2=-1$; $x_3^3=x_0=\cos\frac{\pi}5+i\sin\frac{\pi}5$; $x_4^3=x_3=x_1^*$, $x-x_n=r_ne^{i\theta_n}$ where
$$r_n=|x-x_n|=\sqrt{|x-x_n|^2}=\sqrt{x^2-2x\cos\frac{(2n+1)\pi}5+1}$$
$$\theta_n=\operatorname{atan2}\left(-\sin\frac{(2n+1)\pi}5,x-\cos\frac{(2n+1)\pi}5\right)=\tan^{-1}\left(\frac{x-\cos\frac{(2n+1)\pi}5}{\sin\frac{(2n+1)\pi}5}\right)-\frac{\pi}2$$
So
$$\begin{align}\int\frac{x^2dx}{x^5+1}&=-\frac15\left\{\cos\frac{3\pi}5\ln\left(x^2-2x\cos\frac{\pi}5+1\right)-2\sin\frac{3\pi}5\tan^{-1}\left(\frac{x-\cos\frac{\pi}5}{\sin\frac{\pi}5}\right)\right.\\
&\quad\left.+\cos\frac{\pi}5\ln\left(x^2-2x\cos\frac{3\pi}5+1\right)+2\sin\frac{\pi}5\tan^{-1}\left(\frac{x-\cos\frac{3\pi}5}{\sin\frac{3\pi}5}\right)\right.\\
&\quad\left.-\ln|x+1|\right\}+C\\
&=-\frac15\left\{\frac{1-\sqrt5}4\ln\left(x^2-\frac{1+\sqrt5}2x+1\right)-\frac{\sqrt{10+2\sqrt5}}2\tan^{-1}\left(\frac{4x-\sqrt5-1}{\sqrt{10-2\sqrt5}}\right)\right.\\
&\quad\left.+\frac{1+\sqrt5}4\ln\left(x^2+\frac{\sqrt5-1}2x+1\right)+\frac{\sqrt{10-2\sqrt5}}2\tan^{-1}\left(\frac{4x+\sqrt5-1}{\sqrt{10+2\sqrt5}}\right)\right.\\
&\quad\left.-\ln|x+1|\right\}+C\\
&=-\frac15\left\{-\frac1{2\phi}\ln\left(x^2-\phi x+1\right)-5^{1/4}\phi^{1/2}\tan^{-1}\left(\frac{2x-\phi}{5^{1/4}\phi^{-1/2}}\right)\right.\\
&\quad\left.+\frac{\phi}2\ln\left(x^2+\frac1{\phi}x+1\right)+5^{1/4}\phi^{-1/2}\tan^{-1}\left(\frac{2x+\frac1{\phi}}{5^{1/4}\phi^{1/2}}\right)-\ln|x+1|\right\}+C\end{align}$$
So if we organize our arithmetic reasonably, the operation isn't all that bad.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c- b)$$
If $c\neq {\rm mid}\{a, b, c\}$, the inequality is obviously true!
If $c={\rm mid}\{a, b, c\}$, we have $(a- c)(c- b)= 0\Leftrightarrow c= \dfrac{c^2+ ab}{a+ b}$. I tried to prove that:
$$f(c)- f(\frac{c^2+ ab}{a+ b})= (a- c)(c- b)F\geqq 0$$
where $f(c)= 3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$ but without success! I found this inequality by using discriminant and some coefficient skills. Thank you so much
| The task is homogenius. Let WLOG
$$a+b+c=2,\quad a,b,c \in(0,1),\tag1$$
$$f(a,b,c) = 3\big(a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\big) - b(a+b-c)(a-c)(c-b).\tag2$$
Using of the substiutions
$$a=x,\quad b= 1-xy,\quad c=1-x+xy,\quad x,y\in(0,1),\tag3$$
provides the conditions $(1)$ and allows to get
$\quad f\big(x,1-xy,1-x+xy\big) = 2xg(x,y),$
where
\begin{cases}
g(x,y) = 3(1-y) + (-10+9y+y^2)x + (11-12y+y^2+3y^3)x^2\\[4pt]
+ (-3+4y-2y^2-y^3-y^4)x^3\\[4pt]
g(0,y) = 3(1-y)\\[4pt]
g(1,y) = 1-2y+2y^3-y^4 = (1-y)^3(1+y)\\[4pt]
g(x,0) = 3-10x+11x^2-3x^3 = (1-x)^3 + 2(1-2x)^2 + x\\[4pt]
g(x,1) = 2x^2-2x^3.\tag4
\end{cases}
From $(4)$ should $g(x,y)\ge 0$ at the edges of the area.
On the other hand, at the inner stationary points
$$4g(x,y) = 4g(x,y) - g'_x(x,y) = 12(1-y) + 3(-10+9y+y^2)x + 2(11-12y+y^2+3y^3)x^2,$$
with the discriminant
\begin{align}
&D(y) = 9(-10+9y+y^2)^2 - 96(1-y)(11-12y+y^2+3y^3)\\[4pt]
& = -3 (1-y)(52-144y+89y^2+99y^3)\\[4pt]
& = -3 (1-y)\big(52(1-y)^3 + 12y(1-3y)^2 + 5y^2+43y^3\big) < 0.
\end{align}
Therefore, $g(x,y) \ge 0.$
$\color{brown}{\textbf{Is proved.}}$
$\color{green}{\textbf{Notes about the areas.}}$
The area $c=\operatorname{med}(a,b,c),\quad c\in\big[\min(a,b),\max(a,b)\big],$ corresponds with the area
$$y\in\left(\min\left(\frac12,2-\dfrac1x\right),\max\left(\frac12,2-\dfrac1x\right)\right)$$
(the plot of the area bounds see below).
However, applied universal approach allows to avoid such detalization.
| {
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"url": "https://math.stackexchange.com/questions/3138409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
*
*If $n$ is odd
$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv -1 \mod 8$
$7^n +1 \equiv 0 \mod 8$
Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.
*
*If $n$ is even
$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv 1 \mod 8$
$7^n +1 \equiv 2 \mod 8$
Therefore, the remainder of the division of $7^n+1$ is $2$.
Is that true, please?
| Note that
$\forall n \in \Bbb N, \; 7^n + 1 = (8 - 1)^n + 1 = \displaystyle \sum_0^n \dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + 1 = \sum_0^{n - 1} \dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + (-1)^n + 1 , \tag 1$
via the binomial theorem; when $n$ is odd this yields
$7^n + 1 = \displaystyle 8\sum_0^{n - 1} \dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k , \tag 2$
whence
$8 \mid 7^n + 1, \; \text{odd} \; n; \tag 3$
for even $n$ we obtain
$7^n + 1 = \displaystyle 8\sum_0^{n - 1} \dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k + 2; \tag 4$
we note that
$0 \le 2 < 8; \tag 5$
that is, the remainder of $7^n + 1$ when divided by $8$ is $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$ =
$\int \tan^{\frac{-5}{2}}\theta \sec\theta$
Here is where I get stuck...I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere...What is my next move from here? Did I even start this problem correctly? I can't tell :(
Update with more work after initial answers:
$\int \frac{\cos\theta}{\sin^2\theta}$ $u = \sin\theta, du = \cos\theta d\theta$
I found $\sin^{-1}\theta = \frac{\sqrt{x^2-1}}{x}$
$= \int \frac{du}{u^2} = \frac{1}{ \frac{1}{3}u^3} =
\frac{1}{3\sin^3\theta}
= 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3$
| $$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}} =
\int \frac{\tan\theta\sec\theta d\theta}{\underbrace{(\tan^2\theta)^{\frac{3}{2}}}_{\tan^3\theta}} =
\int \frac{\sec\theta d\theta}{\tan^2\theta}=
-\frac1{\sin \theta}=-\frac{x}{\sqrt{x^2-1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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How do find the fourth derivative of $e^{x^2}$ So using chain rule for the first derivative:
$$f' = 2x\cdot e^{x^2}$$
then using product rule for the $$f'' = 2x(2xe^{x^2}) + e^{x^2}\cdot 2$$
Is there an easy way to get the third? Does the second deriative simplify to:
$$4x^2e^{x^2} + 2e^{x^2} = (4x^2 + 2)(e^{x^2})$$
Can I then take chain rule again to get $f^{(3)}$?
| Use Maclaurin expansion: $y=e^{x^2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$:
$$\begin{align}y'&=\sum_{n=1}^{\infty} \frac{2nx^{2n-1}}{n!}\\
y'' &=\sum_{n=1}^{\infty} \frac{2n(2n-1)x^{2n-2}}{n!}\\
y''' &=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)x^{2n-3}}{n!}\\
y^{(4)}&=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)(2n-3)x^{2n-4}}{n!}=\\
&=4\sum_{n=2}^{\infty} \frac{(2n-1)(2n-3)x^{2(n-2)}}{(n-2)!}=\\
&=4\sum_{n=0}^{\infty} \frac{(2n+3)(2n+1)x^{2n}}{n!}=\\
&=16\sum_{n=1}^{\infty} \frac{(n-1+1)x^{2n}}{(n-1)!}+32\sum_{n=1}^{\infty} \frac{x^{2n}}{(n-1)!}+12\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}=\\
&=(16x^4+48x^2+12)e^{x^2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Trig Subsitution When There's No Square Root I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar \int_a^\infty \frac{dx}{(r^2+x^2)^{(3/2)}}$$
Anyway, so far, I have that:
$$x = r\tan \theta$$
$$dx = r\sec^2 \theta$$
$$\sqrt {(r^2+x^2)} = r\sec\theta$$
The triangle I based the above values on:
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (\sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $\lim \limits_{b \to \infty}$ in front of every line please.
\begin{align}
&= Ar \int_a^b \frac{r\sec^2\theta}{(r\sec\theta)^3}d\theta \\
&= Ar \int_a^b \frac{r\sec^2\theta}{r^3\sec^6\theta}d\theta \\
&= \frac{A}{r} \int_a^b \frac{1}{\sec^4\theta}d\theta \\
&= \frac{A}{r} \int_a^b \cos^4\theta d\theta \\
&= \frac{A}{r} \int_a^b (\cos^2\theta)^2 d\theta \\
&= \frac{A}{r} \int_a^b \left[\ \frac12 1+\cos(2\theta))\ \right]^2d\theta \\
&= \frac{A}{4r} \int_a^b 1 + 2\cos(2\theta) + \cos^2(2\theta)\ d\theta \\
&= \frac{A}{4r} \int_a^b 1 + 2\cos(2\theta)\ d\theta \quad+\quad \frac{A}{4r} \int_a^b \cos^2(2\theta)\ d\theta
\end{align}
And from there it gets really messed up and I end up with a weird semi-final answer of $$\frac{A}{4r}[2\theta+\sin(2\theta)] + \frac{A}{32r} [4\theta+\sin(4\theta)]$$ which is wrong after I make substitutions.
I already know that the final answer is $\dfrac{A}{r}\left(1-\dfrac{a}{\sqrt{r^2+a^2}}\right)$, but I really want to understand this.
| You are doing $(r\sec\theta)^3=r^6\sec^6\theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
\frac{A}{r}\int_{a/r}^{\infty}\frac{1}{(1+u^2)^{3/2}}\,du
$$
Now let's concentrate on the antiderivative
$$
\int\frac{1}{(1+u^2)^{3/2}}\,du=
\int\frac{1+u^2-u^2}{(1+u^2)^{3/2}}\,du=
\int\frac{1}{(1+u^2)^{1/2}}\,du-\int\frac{u^2}{(1+u^2)^{3/2}}\,du
$$
Do the second term by parts
$$
\int u\frac{u}{(1+u^2)^{3/2}}\,du=
-\frac{u}{(1+u^2)^{1/2}}+\int\frac{1}{(1+u^2)^{1/2}}\,du
$$
See what happens?
$$
\int\frac{1}{(1+u^2)^{3/2}}\,du=\frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.
Now
$$
\left[\frac{u}{(1+u^2)^{1/2}}\right]_{a/r}^{\infty}=1-\frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-\frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
\frac{A}{r}\left(1-\frac{a}{\sqrt{r^2+a^2}}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:
Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tangent from the vertices $A$, $B$, and $C$ respectively to the incircle. (So $a = y + z$, $b = x + z$, and $c = x + y$.)
Then the area of the triangle is $\boldsymbol{xy}$.
I can prove this algebraically,$^1$ but is there a picture proof of this fact?
What I have in mind is that we cut up the triangle $ABC$ into finitely many pieces, and rearrange them into a rectangle with sides $x$ and $y$.
$^1$ Using $x = \frac{b+c-a}{2}$ and $y = \frac{a+c-b}{2}$, we get $xy = \frac14\left(c - (a-b)\right)\left(c + (a-b)\right) = \frac14\left(c^2 - a^2 - b^2 + 2ab\right)$. From $c^2 = a^2 + b^2$ this reduces to $\frac14 (2ab) = \frac12 ab$, which is the area of the triangle.
| Here is an answer to @6005's comment "Maybe there is a way to also show $(x-z)(y-z)=z^2$ pictorally".
Referring to the pic below, it's immediate from the construction that, comparing it with the cut up area, A, of the triangle, the shaded area, $S$ equals $A-2z^2$. Then, $W = (x-z)(y-z) = 2A -S -A = 2z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{n\to\infty} \prod\limits_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $ I want to calculate $$\lim\limits_{n\to\infty} \prod_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $$
Taking logarithms, it's enough to find
$$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right).$$
Since $\lim\limits_{n\to\infty} \tan{\frac{x}{n^2}} = 0$, we can combine the Taylor series near $0$ of $\log(1+x)$ with the taylor series of $\tan{x}$ near $0$ to obtain the limit $e^\frac{1}{2}$.
My question is: is there any nicer way of evaluating this limit?
| Too long for a comment.
As I wrote in comments, composition of Taylor series is not only good for the limit but also allows quick and reasonable approximation of the partial product
$$P_n= \prod_{k=1}^{n} \left( 1 + \tan \left(\frac{k}{n^2}\right) \right)$$ Doing what you did taking logarithms and Taylor expansion, we have
$$\tan \left(\frac{k}{n^2}\right)=\frac{k}{n^2}+\frac{k^3}{3 n^6}+O\left(\frac{1}{n^{10}}\right)$$ making
$$\log \left(1+\tan \left(\frac{k}{n^2}\right)\right)=\frac{k}{n^2}-\frac{k^2}{2 n^4}+\frac{2 k^3}{3 n^6}+O\left(\frac{1}{n^{8}}\right)$$
$$\log(P_n)=\frac{1}{2}+\frac{1}{3 n}-\frac{1}{12 n^2}+\frac{2}{15 n^3}+O\left(\frac{1}{n^{4}}\right)$$ Continuing with Taylor, using $P_n=e^{\log(P_n)}$
$$P_n=\sqrt e \left(1+\frac{1}{3 n}-\frac{1}{36 n^2}+O\left(\frac{1}{n^{3}}\right) \right)$$
Computing for $n=10$, the exact result is $1.70341$ while the above approximation gives $\frac{3719 \sqrt{e}}{3600}\approx 1.70322$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$.
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
in $[-2\pi , 2\pi]$.
What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin x }{x}$$
I am getting somewhere from here, but I don't know how to continue. Please help!
The answer is 1
| $\forall x~(f'(x)=3x^2+2x+2+\cos x \gt 0)$ (the polynomial term is always at least $5/3$), so note that $f(-2 \pi) \lt 0, f(2 \pi) \gt 0$ and conclude that there is exactly $1$ root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of the variables scares me.
Here what I've done.
Using the AM-GM inequality and the Schur's inequality, we have that
$$a^3 + b^3 + c^3 \ge 3abc \implies P \ge \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$$
$$\ge \dfrac{9}{2}[ab(a + b) + bc(b + c) + ca(c + a)] = \dfrac{9}{2}[ab(2 - c) + bc(2 - a) + ca(2 - b)]$$
$$\ge \dfrac{9}{2}[2(ab + bc + ca) - 3abc] \ge \dfrac{27}{2}[2\sqrt[\frac{3}{2}]{abc} - abc]$$
Let $abc = m \implies m \le \left(\dfrac{a + b + c}{3}\right)^3 = \dfrac{8}{27}$
The problem becomes
Find the minimum value of $P' = 2\sqrt[\frac{3}{2}]{m} - m$ when $ 0 < m \le \dfrac{8}{27}$.
which is invalid because there isn't a minimum with the given condition.
| First, using the identity
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca),$$
we have
$$a^3 + b^3 + c^3 - 3abc = 8 - 6(ab + bc + ca). \tag{1}$$
Second, using three degree Schur
$$a^3 + b^3 + c^3 + 3abc \ge ab(a + b) + bc(b + c) + ca(c + a),$$
we have
$$a^3 + b^3 + c^3 + 6abc
\ge ab(a + b) + bc(b + c) + ca(c + a) + 3abc$$
or (using $ab(a + b) + abc = ab(a + b + c)$ etc.)
$$a^3 + b^3 + c^3 + 6abc \ge (a + b + c)(ab + bc + ca)
= 2(ab + bc + ca). \tag{2}$$
Third, using (1) and (2), we have
\begin{align*}
4(a^3 + b^3 + c^3) + 15abc
&= 3(a^3 + b^3 + c^3 + 6abc) + (a^3 + b^3 + c^3 - 3abc) \\
&\ge 3 \cdot 2(ab + bc + ca) + 8 - 6(ab + bc + ca)\\
&= 8.
\end{align*}
Also, when $a = b = c = 2/3$, we have $4(a^3 + b^3 + c^3) + 15abc = 8$.
Thus, the minimum of $4(a^3 + b^3 + c^3) + 15abc$ is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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} |
Prove that $\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$ Prove that $$\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$$
I have tried to solve this on my own, and I want to check my solution.
My steps:
Set $x=\left\lfloor \frac {n+3}{2}\right\rfloor $
then, for an integer x
$$ \frac {n+3}{2}=x+\epsilon , 0 \le \epsilon \lt 1$$
$$ n=2x+2 \epsilon-3 $$
We substitute in the right hand side and get
$$ \left\lceil \frac {2x+2\epsilon -1}{2} \right\rceil = \left\lceil x + \epsilon - \frac {1}{2}\right\rceil $$
Now, sine x is an integer we can write the last statement as $$ x +\left\lceil \epsilon - \frac {1}{2}\right\rceil $$
Now , I played around with the inequality. I have subtracted a half from all the sides so I got
$$ -\frac {1}{2} \le \epsilon -\frac {1}{2} \lt \frac {1}{2} $$
I have drawn the number line and found that the inequality turns to
$$ -1 \lt \epsilon -\frac {1}{2} \le 0 $$
(Since we are dealing with integers)
Now, we see that $\left\lceil \epsilon - \frac {1}{2}\right\rceil = 0$
So, we proved that$$ x +\left\lceil \epsilon - \frac {1}{2}\right\rceil = x $$
Which the same as the left hand side.
Is my procedure correct??
This is the solution that I have found
| You can pull integers out of floor/ceiling, so that setting $n=2k$ or $n=2k+1$ and pulling $k+1$
$$\left\lfloor \frac {2k+3}{2}\right \rfloor=\left\lceil \frac {2k+2}{2}\right\rceil$$ is equivalent to
$$\left\lfloor \frac {1}{2}\right \rfloor=\left\lceil \frac {0}{2}\right\rceil,$$
and
$$\left\lfloor \frac {2k+1+3}{2}\right \rfloor=\left\lceil \frac {2k+1+2}{2}\right\rceil$$ is equivalent to
$$\left\lfloor \frac {2}{2}\right \rfloor=\left\lceil \frac {1}{2}\right\rceil.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the surface area of two solids of revolution I have a couple of questions that are similar in nature:
1) I am trying to find the surface area of this when I rotate it around the x-axis. I have $y = \sqrt{5-x}$ when $3 \leq x \leq 5$
Say $dy/dx = \frac{1}{2}(5-x)^{\frac{-1}{2}}$. Then $(dy/dx)^2 = \frac{1}{4}(5-x)^{-1}$ and thus
$$SA = 2 \pi \int_3^5 \sqrt{5-x} * \sqrt{1 + \frac{1}{4} (5-x)^{-1}}$$
I want to find the exact surface area. Where do I go from here?
EDIT: Can I do this?
$$\begin{align}
SA &= 2 \pi \int_3^5 \sqrt{(5-x) + \frac{1}{4}}dx\\
&= 2 \pi \int_3^5 \sqrt{(5-x) + \frac{1}{4}}dx\\
&= 2 \pi \int_3^5 \sqrt{\frac{21}{4} - x} dx\\
&= 2 \pi \int_3^5 \sqrt{\frac{21}{4} - \frac{4x}{4}} dx\\
&= \pi \int_3^5 \sqrt{21 - 4x} dx
\end{align}$$
Am I on the right track?
2) In this case, I have $y^2 = x + 1$ when $0 \leq x \leq 3$
So $y = \sqrt{x+1}$ and so $$\frac{dy}{dx} = \frac{1}{2}(x+1)^{\frac{-1}{2}}$$ and $$\left( \frac{dy}{dx} \right)^2 = \frac{1}{4}(x+1)^{-1}$$
So
$$SA = 2 \pi \int_0^3 \sqrt{x+1} \sqrt{1 + \frac{1}{4 (x+1)}}$$
EDIT
Am I on the right track:
$$SA = 2 \pi * \int_0^3 \sqrt{x+1 + \frac{1}{4}}$$
$$SA = 2 \pi * \int_0^3 \sqrt{x+\frac{5}{4}}$$
and $u = x + \frac{5}{4}$ and $du = dx$
so
$$2 \pi \int_0^3 \sqrt{x + \frac{5}{4}}dx$$
$$2 \pi \int_{\frac{5}{4}}^{\frac{17}{4}} \sqrt{u}du$$
$$2 \pi \frac{2}{3}u^{\frac{2}{3}}\biggr]_{\frac{5}{4}}^{\frac{17}{4}}$$
$$\frac{4 \pi}{3} \bigg(\frac{17}{4} \sqrt{\frac{17}{4}} - \frac{5}{4} \sqrt{\frac{5}{4}}\bigg)$$
| For the first question, notice:
$$\begin{align}
\sqrt{5-x} * \sqrt{1 + \frac{1}{4} (5-x)^{-1}} &= \sqrt{(5-x)\left( 1 + \frac{1}{4} (5-x)^{-1} \right)} \\
&= \sqrt{(5-x)+(1/4)} \\
&= \sqrt{-x+ \frac{21}{4}}
\end{align}$$
Thus the integral becomes
$$SA = 2\pi \int_3^5 \sqrt{-x+ \frac{21}{4}} dx$$
A simple $u$-substitution where $u$ equals what is under the radicand should make the rest of the integral easy to calculate.
EDIT: To address the edit you made to the question OP, yes, that method is also a valid way of approaching this. You will still have to make the $u$-substitution of $u=21-4x$ there, though, so in my opinion just doing it right off the bat results in less computation. Matter of opinion though.
The same principle follows for your second question, yielding
$$SA = 2\pi \int_0^3 \sqrt{x + \frac 5 4} dx$$
And of course, a substitution where $u$ is the radicand makes the rest a fairly simple calculation.
For positive real numbers $a,b$, it is helpful to note that
$$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$$
This is the property we made use of here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $f(yf(x)+x/y)=xyf(x^2+y^2)$ over the reals
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(1)=1$ and for all real numbers $x$ and $y$ with $y \neq 0$, $$f\Bigg (yf(x)+\frac{x}{y}\Bigg)=xyf(x^2+y^2)$$
This seems quite hard. $f(x)=\begin{cases}\frac{1}{x}, x\neq 0 \\ 0, x=0 \end{cases}$ works by inspection, as does $f(x)=0$ (though I'm not sure if this is legitimate. If we set $y=1$, we have $f(f(x)+x)=xf(x^2+1)$. If we set $x=1$, we have $f(\frac{y^2+1}{y})=yf(y^2+1)$. which seems to imply $f(f(x)+x)=f(\frac{x^2+1}{x})$. If I could show that $f$ is injective I could go further, but I'm stuck - subbing in other values doesn't really seem to lead anywhere either.
I believe this problem came from an Olympiad camp.
| Let $f:\textbf{R}\rightarrow\textbf{R}$, such that $f(1)=1$ and
$$
f\left(yf(x)+\frac{x}{y}\right)=xyf(x^2+y^2)\textrm{, }\forall (x,y)\in\textbf{R}\times\textbf{R}^{*}\tag 1
$$
Set $x=0$, $y=1$ in (1), then easily $f(0)=0$ (the case $f(0)\neq 0$ is trivial). For $y=1$ in (1) we get
$$
xf(x^2+1)=f(f(x)+x)\textrm{, }x\neq 0.\tag 2
$$
Also using the symmetry we get
$$
f\left(yf(x)+\frac{x}{y}\right)=f\left(xf(y)+\frac{y}{x}\right)\textrm{, }x,y\neq 0
$$
For $y=1$, we get
$$
f\left(x+\frac{1}{x}\right)=f(f(x)+x)\textrm{, }x\neq 0\tag 3
$$
Assume that $h(x,y)$ is a surface (function) such that
$$
f(y+h(x,y))=f(x).\tag 4
$$
Note.
One choise of $h(x,y)$ is $h(x,y)=-y+x$ but may be there others. For example if $f(x)=x^2-x+1$, then $h(x,y)=x-y$ or $h(x,y)=1-x-y$.
We assume here that
$$
f(x)=h\left(x+\frac{1}{x},x\right)\tag{4.1}
$$
From (4) with $x\rightarrow x+\frac{1}{x}$ we get
$$
f\left(x+\frac{1}{x}\right)=f\left(y+h\left(x+\frac{1}{x},y\right)\right).
$$
Hence for $y=x$ in the above identity we get
$$
f\left(x+\frac{1}{x}\right)=f\left(x+h\left(x+\frac{1}{x},x\right)\right)=f(x+f(x))
$$
Hence we can get (3). From (2) and (3) we get also
$$
xf(x^2+1)=f\left(x+\frac{1}{x}\right)\tag 6
$$
Setting $x\rightarrow x^{-1}$ in (6)
$$
\frac{1}{x}f\left(1+\frac{1}{x^2}\right)=f\left(x+\frac{1}{x}\right)=xf(x^2+1).\tag 7
$$
Hence if we set $x^2\rightarrow x>0$ in (7), then
$$
f\left(1+\frac{1}{x}\right)=xf(x+1)\textrm{, for all }x>0.\tag 8
$$
Set also $x=y-1>0$ in (8), then
$$
f\left(\frac{y}{y-1}\right)=(y-1)f(y)\textrm{, }y>1.
$$
With $y=1/w>1$ we get
$$
\frac{1}{1-w}f\left(\frac{1}{1-w}\right)=\frac{1}{w}f\left(\frac{1}{w}\right).
$$
Hence if $0<w<1$ and $g(w):=\frac{1}{w}f\left(\frac{1}{w}\right)$, then
$$
f(x)=x^{-1}g\left(x^{-1}\right)\textrm{, where }x>1\textrm{ and }g(1-w)=g(w)\textrm{, }0<w<1\tag 9
$$
The solution (9) satisfies (8),(7),(6), but for to holds (3) we get plus a new functional equation for $f(x)$ and this is (4.1).
Hence the general solution is
$$
f(x)=h\left(x+\frac{1}{x},x\right)\tag{11}
$$
with
$$
\frac{1}{x-1}h\left(x-1+\frac{1}{x-1},\frac{1}{x-1}\right)=\frac{1}{x}h\left(x+\frac{1}{x},\frac{1}{x}\right)\tag{12}
$$
and $h(x,y)$ solution of
$$
f(y+h(x,y))=f(x).\tag{13}
$$
An obvious solution of $(13)$ is $h(x,y)=x-y$, but it may exist and other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Is it possible to express accuracy knowing sensitivity and specificity?
I know the value of the sensitivity Se and the value of the specificity of Sp, they are equal to 78.65 and 90.00, respectively. I know nothing but this. Can I somehow of the equations, which in the photo express the value of the accuracy Ac?
| Consider
$$X = \frac{A}{A+B}, \ Y = \frac{C}{C+D}, \ Z = \frac{A+C}{A+B+C+D}$$
Then $X$ and $Y$ don't determine $Z$. Counterexample:
$$A=1, B = 2, C = 3, D = 9 \Rightarrow X = \frac{1}{3}, Y = \frac{1}{4}, Z = \frac{4}{15}$$
$$A=3, B = 6, C = 1, D = 3 \Rightarrow X = \frac{1}{3}, Y = \frac{1}{4}, Z = \frac{4}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the following system of equations - (3)
Solve the following system of equations:
$$\large
\left\{
\begin{align*}
3x^2 + xy - 4x + 2y - 2 = 0\\
x(x + 1) + y(y + 1) = 4
\end{align*}
\right.
$$
I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.
| The resultant of $3\,{x}^{2}-xy-4\,x+2\,y-2$ and $
x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10\,{x}^{4}-24\,{x}^{3}-10\,{x}^{2}+42\,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots,
$x \approx -1.287147510$ and $x \approx 0.2049816008$, which correspond to
$y \approx -2.469872787$ and $y \approx 1.500750095$ respectively.
EDIT: For the corrected question, the resultant of $3\,{x}^{2}+xy-4\,x+2\,y-2$ and $x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the minimum value of $\left(\frac{n + p}{mx} + \frac{p + m}{ny} + \frac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$. Going back a few more years and you can find more and more interesting problems over the years as time turns back. I am still surprised at how easy this competition has become. Then I come across this problem, which goes by the following.
Given positive variables $x$, $y$ and $z$ and positive parameters $a, b, c$.
Find the minimum value of $\left(\dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}\right)\sqrt{yz + zx + xy}$.
It was simple, yet difficult. I wished to find a solution without using Lagrange multipliers but found no results. I would be grateful if you have a solution like so.
Well...
Perhaps delete what I had said 288 days ago. Let's start this all over again.
I can't leave this question to go waste. I shouldn't have overgeneralised this inequality.
Here is the correct inequality.
Given positive $x, y, z$ and distinct parameters $m, n, p > 0$. Calculate the minimum value of $$\large \left(\frac{n + p}{mx} + \frac{p + m}{ny} + \frac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$$
I have provided a solution below and I would be greatly appreciated if anyone could come up with a better solution than mine.
I apologise for the misunderstanding.
| Let $a = mx, b = ny, c = pz$. We have that $\dfrac{n + p}{mx} + \dfrac{p + m}{ny} + \dfrac{m + n}{pz}$
$$ = \frac{n + p}{a} + \frac{p + m}{b} + \frac{m + n}{c} = \left(\frac{1}{b} + \frac{1}{c}\right) \cdot m + \left(\frac{1}{c} + \frac{1}{a}\right) \cdot n + \left(\frac{1}{a} + \frac{1}{b}\right) \cdot p$$
$$ = \frac{b + c}{bc} \cdot m + \frac{c + a}{ca} \cdot n + \frac{a + b}{ab} \cdot p$$
$$ \implies \left(\dfrac{n + p}{mx} + \dfrac{p + m}{ny} + \dfrac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$$
$$ = \left(\frac{b + c}{bc} \cdot m + \frac{c + a}{ca} \cdot n + \frac{a + b}{ab} \cdot p\right) \cdot \sqrt{\frac{bc}{np} + \frac{ca}{pm} + \frac{ab}{mn}}$$
$$ \ge \sqrt{3\left(\frac{bc}{np} + \frac{ca}{pm} + \frac{ab}{mn}\right)\left[\frac{np(c + a)(a + b)}{bca^2} + \frac{pm(a + b)(b + c)}{cab^2} + \frac{mn(b + c)(c + a)}{abc^2}\right]}$$
$$ \ge \sqrt{3} \cdot \left[\frac{\sqrt{(c + a)(a + b)}}{a} + \frac{\sqrt{(a + b)(b + c)}}{b} + \frac{\sqrt{(b + c)(c + a)}}{c}\right]$$
$$ \ge \sqrt{3} \cdot \left(\frac{\sqrt b + \sqrt c}{\sqrt a} + \frac{\sqrt c + \sqrt a}{\sqrt b} + \frac{\sqrt a + \sqrt b}{\sqrt c}\right) \ge 3\sqrt 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reasoning about inequalities involving floor functions I am working on the beginning of an inductive argument and I wanted to confirm that my base case is sound.
Let $f(x) = \lfloor x\rfloor - \left\lfloor\dfrac{x}{2}\right\rfloor$ where is $x$ is a positive real.
Let $u,v$ be positive integers such that $u \ge 7$ and $v \ge 2$
I am trying to show that the following base case is true:
$$f(u^2+u) - f(u^2) - f\left(\frac{u^2+u}{v}\right) + f\left(\frac{u^2}{v}\right) > 0$$
Here is my argument for the base case:
(1) Let $a_1, a_2, a_3$ be nonnegative integers such that $a_1 < v$, $a_2 < v$, $a_3 < 2$ and:
$\left\lfloor\dfrac{u^2+u}{v}\right\rfloor = \dfrac{u^2+u-a_1}{v},\left\lfloor\dfrac{u^2}{v}\right\rfloor = \dfrac{u^2-a_2}{v},\left\lfloor\dfrac{u^2}{2}\right\rfloor = \dfrac{u^2-a_3}{2},$
(2) Let $b_1,b_2$ be nonnegative numbers with each $b_i < 2v$ and $b_i=a_i$ or $b_i = a_i+v$ such that:
$\left\lfloor\dfrac{u^2+u}{2v}\right\rfloor = \dfrac{u^2+u-b_1}{2v},\left\lfloor\dfrac{u^2}{2v}\right\rfloor = \dfrac{u^2-b_2}{2v}$
(3) Since $u^2+u$ is even, combining all terms, I get:
$f(u^2+u) - f(u^2) - f\left(\frac{u^2+u}{v}\right) + f\left(\frac{u^2}{v}\right) = $
$u - \dfrac{u^2+u}{2} + \dfrac{u^2-a_3}{2}-\dfrac{u^2+u-a_1}{v} + \dfrac{u^2+u-b_1}{2v} + \dfrac{u^2-a_2}{v} - \dfrac{u^2-b_2}{2v}$
$= \dfrac{u-a_3}{2} - \dfrac{u^2+u - 2a_1 + b_1}{2v} + \dfrac{u^2-2a_2 + b_2}{2v}$
$= \dfrac{u-a_3}{2} - \dfrac{u-2a_1+b_1+2a_2-b_2}{2v}$
$=\dfrac{(v-1)u -va_3 + 2a_1 - b_1 -2a_2 + b_2}{2v}$
(4) $\dfrac{va_3 - 2a_1 + b_1 + 2a_2 - b_2}{2v} < \dfrac{3}{2}$ since:
*
*$\dfrac{va_3}{2v} \le \dfrac{v}{2v} = \dfrac{1}{2}$
*$\dfrac{b_1 - 2a_1}{2v} \le \dfrac{(v + a_1) - 2a_1}{2v} \le \dfrac{v}{2v} = \dfrac{1}{2}$
*$\dfrac{2a_2 - b_2}{2v} \le \dfrac{2a_2 - (a_2)}{2v} = \dfrac{a_2}{2v} < \dfrac{v}{2v} = \dfrac{1}{2}$
(5) So that:
$f(u^2+u) - f(u^2) - f\left(\frac{u^2+u}{v}\right) + f\left(\frac{u^2}{v}\right) =$
$=\dfrac{(v-1)u -va_3 + 2a_1 - b_1 -2a_2 + b_2}{2v} > \dfrac{(v-1)u}{2v} - \dfrac{3}{2}$
(6) Since $u > 6$ and $v \ge 2$:
$\left(\dfrac{v-1}{v}\right)\dfrac{u}{2} \ge \left(\dfrac{1}{2}\right)\dfrac{u}{2} > \dfrac{6}{4}$
which means that:
$f(u^2+u) - f(u^2) - f\left(\frac{u^2+u}{v}\right) + f\left(\frac{u^2}{v}\right) > \dfrac{3}{2} - \dfrac{3}{2} = 0$
Is my reasoning correct? Did I make a mistake? Is any step in my argument unclear?
| I have gone through all of your steps fairly carefully and didn't see any mistakes. As for being clear, it took me a bit of time to figure out some of your lines in $(3)$ as you sometimes did several steps in one line. However, this is a minor point. Also, I've seen quite a few math proofs which are considerably harder to follow & figure out than what you provided.
If you're interested, the following is somewhat similar to what you did, but it takes a somewhat higher level view. First, you have
$$f(x) = \lfloor x\rfloor - \left\lfloor\dfrac{x}{2}\right\rfloor \tag{1}\label{eq1}$$
where is $x$ is a positive real. You are trying to show that
$$f(u^2+u) - f(u^2) - f\left(\frac{u^2+u}{v}\right) + f\left(\frac{u^2}{v}\right) > 0 \tag{2}\label{eq2}$$
is true for all positive integers $u,v$ with $u \ge 7$ and $v \ge 2$. Consider a positive integer $a$. If $a$ is even, i.e., $a = 2b$ for an integer $b$, then $f(a) =
\lfloor 2b\rfloor - \left\lfloor\dfrac{2b}{2}\right\rfloor = 2b - b = b = \dfrac{a}{2}$. If $a$ is odd, i.e., $a = 2b + 1$ instead, then $f(a) = \lfloor 2b + 1\rfloor - \left\lfloor\dfrac{2b + 1}{2}\right\rfloor = 2b + 1 - b = b + 1 = \dfrac{a + 1}{2}$. Using this with the first $2$ terms of \eqref{eq2}, with $u^2 + u$ always being even as you've already noted, then for $u$ being even
$$f(u^2+u) - f(u^2) = \frac{u^2 + u}{2} - \frac{u^2}{2} = \frac{u}{2} \tag{3}\label{eq3}$$
while if $u$ is odd, then
$$f(u^2+u) - f(u^2) = \frac{u^2 + u}{2} - \frac{u^2 + 1}{2} = \frac{u - 1}{2} \tag{4}\label{eq4}$$
The third & fourth terms of \eqref{eq2} involve $f\left(\dfrac{c}{v}\right)$ for some integer $c$. Consider $c = k(2v) + r$ for some integer $k$, plus $0 \le r \le 2v - 1$. First, if $0 \le r \lt v$, then
$$f\left(\frac{c}{v}\right) = \left\lfloor \frac{k(2v) + r}{v}\right\rfloor - \left\lfloor\frac{k(2v) + r}{2v}\right\rfloor = 2k - k = k \tag{5}\label{eq5}$$
Next, if $v \le r \lt 2v$, then
$$f\left(\frac{c}{v}\right) = \left\lfloor \frac{k(2v) + r}{v}\right\rfloor - \left\lfloor\frac{k(2v) + r}{2v}\right\rfloor = 2k + 1 - k = k + 1 \tag{6}\label{eq6}$$
Next, consider a lower limit of
$$\dfrac{c}{2v} - \dfrac{1}{2} = k + \dfrac{r}{2v} - \dfrac{1}{2} \tag{7}\label{eq7}$$
If $0 \le r \lt v$, then \eqref{eq7} is $\lt k$ and, thus, the value of \eqref{eq5}. If $v \le r \lt 2v$, then \eqref{eq7} is $\lt k + 1$ and, thus, the value of \eqref{eq6}. Thus, in either case,
$$f\left(\dfrac{c}{v}\right) \gt \dfrac{c}{2v} - \dfrac{1}{2} \tag{8}\label{eq8}$$
For an upper limit, let
$$\dfrac{c}{2v} + \dfrac{1}{2} = k + \dfrac{r}{2v} + \dfrac{1}{2} \tag{9}\label{eq9}$$
If $0 \le r \lt v$, then \eqref{eq9} is $\gt k$ and, thus, \eqref{eq5}. For $v \le r \lt 2v$, then \eqref{eq8} is $\ge k + 1$ and, thus, \eqref{eq6}. Thus, in either case,
$$\dfrac{c}{2v} + \dfrac{1}{2} \ge f\left(\dfrac{c}{v}\right) \tag{10}\label{eq10}$$
The minimum value of the third & fourth terms of \eqref{eq2} would be with the minimum value of the fourth term less the maximum value of the third term. Thus, using the limits given by \eqref{eq8} and \eqref{eq10}, then
$$f\left(\frac{u^2}{v}\right) - f\left(\frac{u^2+u}{v}\right) \gt \left(\frac{u^2}{2v} - \frac{1}{2}\right) - \left(\frac{u^2 + u}{2v} + \frac{1}{2}\right) = -\frac{u}{2v} - 1 \tag{11}\label{eq11}$$
Note for $v \ge 2$ that $\dfrac{v-1}{v} \ge \dfrac{1}{2}$. Thus, if $u$ is even, then from \eqref{eq3},
$$\frac{u}{2} - \frac{u}{2v} - 1 = \frac{u(v-1)}{2v} - 1 \ge \frac{8}{2(2)} - 1 = 1 \tag{12}\label{eq12}$$
If $u$ is odd, then from \eqref{eq4},
$$\frac{u - 1}{2} - \frac{u}{2v} - 1 = \frac{u(v-1)}{2v} - \frac{3}{2} \ge \frac{7}{2(2)} - \frac{3}{2} = \frac{1}{4} \tag{13}\label{eq13}$$
Since \eqref{eq12} and \eqref{eq13} are the minimum values of the LHS of \eqref{eq2}, this shows it is $\gt 0$ in either case, as requested to be proven.
Note this approach doesn't make the calculations very much, if any, simpler in your particular case. I believe the main advantage of the approach is that if instead of having $2$ terms each of an integer & fraction, you had many more terms (e.g., $4$, $6$, $8$ or even more), then this method would make dealing with all of the values generally shorter & easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$
$$3(\tan ^2 y-3)=3\tan ^2-1$$
$$9=1$$
The solution provided by the book $x=n\pi + \pi/4$ fits, so why did i get $9=1$?
| Another way to avoid confusion:
$$\dfrac31=\dfrac{\sin(x+15^\circ)\cos(x-15^\circ)}{\cos(x+15^\circ)\sin(x-15^\circ)}$$
Apply Componendo et Dividendo
$$\dfrac{3+1}{3-1}=\dfrac{\sin2x}{\sin30^\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Consecutive integers divisible by prime powers There's a similar question on here already but I don't think the answers are applicable in this case.
Question: Find the smallest three consecutive integers for which the first integer is divisible by the square of a prime; the second integer by the cube of a prime; and the third integer by the fourth power of a prime.
My attempt: Let the three consecutive integers be a, b and c. Then
$a\equiv 0\pmod{d^2}$
$a+1\equiv 0\pmod{e^3}$
$a+2\equiv 0\pmod{f^4}$
where $d,e,f$ are prime. This is equivalent to
$a\equiv 0\pmod{d^2}$
$a\equiv -1\pmod{e^3}$
$a\equiv -2\pmod{f^4}$
Then after applying the Chinese Remainder Theorem,
$e^3f^4a\equiv 1\pmod{d^2}$
$d^2f^4a\equiv 1\pmod{e^3}$
$d^2e^3a\equiv 1\pmod{f^4}$
I don't think this is getting me anywhere. Is this sort of general approach going to work or am I wasting my time?
| Hmm...
Well, by CRT $n \equiv 1 \pmod {p^2}$
$n \equiv 0 \pmod {q^3}$
$n \equiv-1 \pmod {r^4}$ will have unique solutions for any the primes $p, q, r$.
If I blanketly choose $r=2; q=3;p=5$ I will get.
$n\equiv 1 \pmod {25}$ and $n\equiv 0 \pmod 27$. $n= 27j = 1 + 25k = 1-2k + 27k$. If $k=14$ we have $j= 13$. This gives us $n\equiv 351 \pmod {675}$.
$n \equiv 351 \pmod{675}$ and $n\equiv -1 \pmod {16}$ gives us
$351+ 675m\equiv -1 + 3m \equiv -1 \pmod{16}$.
so $n = 351$ gives us $n-1 = 14*5^2; n = 13*3^2; n+1 =44*2^4$.
Is that the least possible?
Well $3^4< 352 <5^4$ so $r=2,3$ if we are to find any lower.
And $7^3< 351 < 11^3$ so $q=3,5,7$ if we are to find any lower.
And $17^2 < 351 < 19^2$ so $p = 7,11,13,17$ if we are to find any lower.
If worse comes to worse we can solve them all.... Or we could write a program to check all numbers less than $351$.
I doubt there is any lower but I'm not betting anything valuable on it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x - 3)^2 \implies a + b = \dfrac{a^2 - b^2}{a + b} = \dfrac{2x - 1}{(x - 3)^2}$.
$\left\{
\begin{align}
a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 106x + 80}{2(x - 3)^2}\\
b = \dfrac{(a + b) - (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 110x + 82}{2(x - 3)^2}
\end{align}
\right.$
| $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9 \implies (\sqrt{x + 2} - 2) - (\sqrt{3 - x} - 1) = x^2 - 6x + 8$$
$$ \implies (x - 2)\left(\frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1}\right) = (x - 2)(x - 4)$$
$$\implies \left[ \begin{align*} x - 2 &= 0\\ \frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1} &= x - 4 \end{align*} \right.$$
For the second equality, we have that $-2 \le x \le 3$ and $x \ge 4$ as the needed condition for the left and right side of the equality. So there're not any solutions for the second equality.
That means $x - 2 = 0 \implies x = 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Where does this proof of convergence fail? Given the series, $$\sum_{n=1}^{\infty} (-1)^{n}\frac{n}{n+1}$$
I know we can immediately conclude that it is obviously divergent by the divergence test. But I want to know where exactly am I going wrong in the following 'proof' as I have just started learning about convergence and divergence.
Here is what I did :
\begin{align}\sum_{n=1}^\infty (-1)^n \frac{n}{n+1} & = \left(-\frac{1}{2} + \frac{2}{3}\right)+\left(-\frac{3}{4} + \frac{4}{5}\right)+\left(-\frac{5}{6} + \frac{6}{7}\right)+\dots \\
& = \frac{1}{6}+\frac{1}{20}+\frac{1}{42} + \dots\\
& = \frac{1}{2\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{6\cdot 7}+\dots \\
& = \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)}-\left(\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\frac{1}{7\cdot 8}+\dots\right) \\
& = \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)}- \sum_{n=1}^\infty\frac{1}{(2n+1)(2n+2)}\\
& = \sum_{n=1}^\infty\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(2n+1)(2n+2)}\right)\\
& =3\sum_{n=1}^{\infty} \frac{n}{(n+2)(2n+1)(2n+2)},\end{align}
and this last series obviously converges by the comparison test.
I am able to represent this divergent sum as the difference of two convergent sums. Where is my flaw?
| Let
$s_m
=\sum_{n=1}^{} (-1)^{n}\frac{n}{n+1}
$.
Then
$\begin{array}\\
s_{2m}
&=\sum_{n=1}^{2m} (-1)^{n}\frac{n}{n+1}\\
&=\sum_{n=1}^{m} ((-1)^{2n-1}\frac{2n-1}{2n-1+1}+(-1)^{2n}\frac{2n}{2n+1})\\
&=\sum_{n=1}^{m} (-\frac{2n-1}{2n}+\frac{2n}{2n+1})\\
&=\sum_{n=1}^{m} \frac{-(2n-1)(2n+1)+4n^2}{2n(2n+1)}\\
&=\sum_{n=1}^{m} \frac{-(4n^2-1)+4n^2}{2n(2n+1)}\\
&=\sum_{n=1}^{m} \frac{1}{2n(2n+1)}\\
\text{and}\\
s_{2m+1}
&=s_{2m}+(-1)^{2m+1}\frac{2m+1}{2m+2}\\
&=s_{2m}-\frac{2m+1}{2m+2}\\
&=s_{2m}-(1-\frac{1}{2m+2})\\
\end{array}
$
Therefore
the even terms converge to
$\sum_{n=1}^{\infty} \frac{1}{2n(2n+1)}
$
and the odd terms converge to
$-1+\sum_{n=1}^{\infty} \frac{1}{2n(2n+1)}
$.
Therefore the series
does not converge.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$ I need to find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$. I am getting $\frac{518}{13}$ but this is wrong. Here is my work.
Using the mean value theorem $f'(c)=\frac{f(b) - f(a)}{b - a}$
$f(b) = 2(-5)^3-6(-5)^2-90(-5)+6 = -250-150+450+6 = 56$
$f(a) = 2(8)^3-6(8)^2-90(8)+6 = 1024-384-66 = 574$
$\frac{56-574}{-5-8}=\frac{518}{13}$
| Your error is a simple mistake in computing $f(8)$. To be precise, you write that
$$2(8)^3-6(8)^2-90(8)+6 = 1024-384-66,$$
where indeed $2\times8^3=1024$ and $6\times8^2=384$. You seem to have computed $9\times8$ instead of $90\times8$.
Apart from this, your work is entirely correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$
Dont know how to proceed from here
| Upon applying: $\ fg\bmod fh\, =\, f(g\bmod h) =\, $ mod Dstributive Law with $\,z = x^2$
$\begin{align}
&\ \,1+\cdots+z^{11}\,\bmod\, (1+\cdots+z^{5})(x+1)\\[.3em]
=\ &(1+\cdots+z^{5})\big(1+\color{#c00}{z^6}\,\bmod x+1 \big)\\[.3em]
=\ &(1+\cdots+z^{5})\big( 1 + \color{#c00}1\big)\, \ {\rm by}\ \color{#c00}{z^6}\equiv (x^2)^6\equiv \color{#c00}{1},\, \ {\rm by}\ \ x\equiv -1\!\!\!\pmod{\!x+1}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Fourier Series for $\cos^3(x)$ Consider the familiar trigonometric identity: $\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$
Show that the identity above can be interpreted as Fourier series expansion.
so we know that cos is periodic between $\pi$ and $-\pi$ and $\cos$ is an even function, therefore, $\cos^3$ is even.
so we need to compute $a_0$ ( the integral of $f(x)$ and it will equal $0$) and $a_n$ ( the integral from $\pi$ to $-\pi$ of $\cos^3(x) \cos(nx)$ )
how to compute $a_0$
thanks
| Recall: from a familiar trigonometric identity,
$$\cos^3(x) = \cos(x)\color{blue}{\cos^2(x)} = \cos(x)\color{blue}{(1 - \sin^2(x))}$$
Thus,
$$\int \cos^3(x)dx = \int (1-\sin^2(x))\cos(x)dx$$
Make the $u$-substitution $u = \sin(x), du = \cos(x)dx$ and you should be able to find the result easily. You should get
$$\int \cos^3(x)dx = \sin(x) - \frac 1 3 \sin^3(x) + C$$
Apply the fundamental theorem of calculus on the bounds $0,\pi$. Since $\sin(n\pi) \equiv 0 \; \forall n \in \Bbb Z$,
$$\int_0^\pi \cos^3(x)dx = \sin(0) - \frac 1 3 \sin^3(0) - \left( \sin(\pi) - \frac 1 3 \sin^3(\pi) \right) = 0$$
and thus, $a_0 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving for coefficients in polynomial I have the following equation:
$$8\eta_c-6x\eta_c+\frac{x^3}{2}\eta_c=2.5\eta_c+C_{01}x^0+C_{11}x+C_{21}x^2+C_{31}x^3$$
If I solve for these coefficients, I get:
$$C_{01}=5.5\eta_c$$
$$C_{11}=-6\eta_c $$
$$C_{21}=0$$
$$C_{31}=1/2\eta_c$$
If I change my equation from (x) to (x-1) like this, why do I get that the coefficients are the same? Wouldn't this imply that $x = x-1$?
My derivation with $x-1$ is below.
$$\eta_c\left(8-6(x-1)+\frac{(x-1)^3}{2}\right)=2.5\eta_c+C_{01}(x-1)^0+C_{11}(x-1)^1+C_{21}(x-1)^2+C_{31}(x-1)^3$$
When I expand the LHS:
$$\eta_c\left(8-6x+6+\frac{x^3}{2}-\frac{3x^2}{2}+\frac{3x}{2}-\frac{1}{2}\right)=2.5\eta_c+C_{01}+C_{11}(x-1)+C_{21}(x-1)^2+C_{31}(x-1)^3$$
Collecting terms on the LHS:
$$\eta_c\left(13.5-4.5x+\frac{x^3}{2}-\frac{3x^2}{2}\right)=2.5\eta_c+C_{01}+C_{11}(x-1)+C_{21}(x-1)^2+C_{31}(x-1)^3$$
Now if I expand the RHS:
$$\eta_c\left(13.5-4.5x+\frac{x^3}{2}-\frac{3x^2}{2}\right)=2.5\eta_c+C_{01}+C_{11}x-C_{11}+C_{21}x^2-2C_{21}x+C_{21}+C_{31}x^3-3C_{31}x^2+3C_{31}x-C_{31}$$
Now collecting terms on the RHS:
$$\eta_c\left(13.5-4.5x-\frac{3x^2}{2}+\frac{x^3}{2}\right)=(2.5\eta_c+C_{01}-C_{11}+C_{21}-C_{31})+x(C_{11}-2C_{21}+3C_{31})+x^2(C_{21}-3C_{31})+x^3(C_{31})$$
Then I solve for all coefficients and get the same as before.
| Rearranging the original equation gives
$$(C_{01}-5.5\eta_c)+(C_{11}+6\eta_c)x+C_{21}x^2+(C_{31}-\frac{\eta_c}{2})x^3=0$$
Now if we replace $x$ with any function $f(x)$ for example $f(x)=x-1$ we get
$$(C_{01}-5.5\eta_c)+(C_{11}+6\eta_c)f(x)+C_{21}f^2(x)+(C_{31}-\frac{\eta_c}{2})f^3(x)=0$$
Now this equation is always true when
$$C_{01}=5.5\eta_c$$
$$C_{11}=-6\eta_c$$
$$C_{21}=0$$
$$C_{31}=\frac{\eta_c}{2}$$
because both sides of the equation will then be equal to zero;
$$(5.5\eta_c-5.5\eta_c)+(-6\eta_c+6\eta_c)f(x)+(0)f^2(x)+(\frac{\eta_c}{2}-\frac{\eta_c}{2})f^3(x)=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
A nice Nesbitt inequality from a strange inequality Given $a,\,b,\,c> 0$$,$ prove that$:$
$$\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$$
See$:$ $\lceil$ https://artofproblemsolving.com/community/c6h354642p1923888 $\rfloor$
The only way I tried is Buffalo Way but the coefficient of$:$ $\text{coef}[\,c^{\,4}\,],\,\text{coef}[\,c^{\,3}\,],\,\text{coef}[\,c^{\,2}\,],\,\text{coef}[\,c\,],\,\text{coef}[\,c^{\,0}\,]$ aren$'$t same non$-$negative$,$ maybe I$'$m wrong because the equality occurs with $a= b$$.$ So I ask$,$ hope to see the best way$!$ Good luck everybody$!$
| We have: $$f\left(a,b,c\right)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{63}{5}\left[\frac{2c^2}{(a+b\, )^2}-\frac{c}{a+b}\right]\ge 0$$
Let $t=\dfrac{a+b}2>0$. We hope that $f(a,b,c)\ge f(t;t;c)$. Indeed,
$$f(a,b,c)-f\left(t,t,c\right)=\frac{\left(a-b\right)^2\left(a+b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b+2c\right)}\ge 0$$
So we only prove $f(t,t,c)\ge0$
Or $$63c^3+5c^2t-58ct^2+20t^3\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A probability question: Poor Alex
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money can be expressed as an irreducible fraction p/q.
What I did was--
WLOG let us assume the correct number be 1. Total number of outcomes can be calculated this way
(0,1)(0,2)(0,3)(0,4)(0,5)(0,6)(0,7)(0,8)(0,9)
(1)
(2,0)(2,1)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)
(3,0)(3,1)(3,2)(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)
(4,0)(4,1)(4,2)(4,3)(4,5)(4,6)(4,7)(4,8)(4,9)
(5,0)(5,1)(5,2)(5,3)(5,4)(5,6)(5,7)(5,8)(5,9)
(6,0)(5,1)(5,3)(5,4)(5,6)(5,7)(5,7)(5,8)(5,9)
(7,0)(7,1)(7,2)(7,3)(7,4)(7,5)(7,6)(7,8)(7,9)
(8,0)(8,1)(8,2)(8,3)(8,4)(8,5)(8,6)(8,7)(8,9)
(9,0)(9,1)(9,2)(9,3)(9,4)(9,5)(9,6)(9,7)(9,8)
So total number of outcomes is 82 and total number of favorable outcomes is 10 so my answer is 10/82 or 5/41. But it is not correct. Please help me understand this.....
| Alex tries the first number. The probability to get it wrong is $9/10$. If he does, he chooses the second number. The probability that he get the second number wrong is $8/9$ (assuming that he does not try to dial the same number expecting a different result). Then the total probability that he gets the number wrong is $$\frac 9{10}\frac89=\frac8{10}$$
That means that he get's it right with probability $2/10=20\%$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$ sin^2 x = \frac{1}{2}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sin x= \lvert\frac{1}{\sqrt2}\rvert$$
$$ \sin x= \frac{\sqrt2}{2} ; \sin x= -\frac{\sqrt2}{2}$$
So if I look for all the values of $x$ that solve the above I should get not only $ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°) but also $ x= \frac{3\pi}{4}$ (135°) ; $ x= \frac{7 \pi}{4}$ (315°).
What am I doing wrong?
| $\cos x=\sin x=\cos\left(\frac{\pi}{2}-x\right)$ thus $x=\frac{\pi}{2}-x \bmod [2\pi]$ or $x=-\frac{\pi}{2}+x \bmod [2\pi]$. The first equation has $\pi/4$ as solution and the second does not provide a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| Fixed Point Iteration
This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $\begin{bmatrix} x\\ y\end{bmatrix}$ on the left hand side.
Define
$f\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} (36-2y)/3 \\ (64-5x)/4\end{bmatrix}$
Start with an intial guess of $\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 0\\ 0\end{bmatrix}$
The result is $f\begin{bmatrix} 0\\ 0\end{bmatrix}=\begin{bmatrix} 12\\ 16\end{bmatrix}$
Now plug that back into f
The result is $f\begin{bmatrix} 12\\ 6\end{bmatrix}=\begin{bmatrix} 4/3\\ 1\end{bmatrix}$
Keep plugging the result back in. After 100 iterations you have:
$\begin{bmatrix} 7.9991\\ 5.9993\end{bmatrix}$
Here is a graph of the progression of the iteration:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 9
} |
Determinant with trigonometric functions
If
$$
\begin{vmatrix}
\sin 2x & \cos^2x & \cos 4x \\
\cos^2x & \cos2x & \sin^2x \\
\cos^4x & \sin^2x & \sin 2x \\
\end{vmatrix} = a_0 + a_1\sin x + a_2\sin^2x +\cdots+ a_n \sin^n x
$$
Then what is the value of $a_0$?
How do I solve this?
Thank you so much!!
| As noted in the comments, let $x=0$. Then we have
$$
\begin{vmatrix}
\sin 2x & \cos^2x & \cos 4x \\
\cos^2x & \cos2x & \sin^2x \\
\cos^4x & \sin^2x & \sin 2x \\
\end{vmatrix} = a_0 + a_1\sin x + a_2\sin^2x +\cdots+ a_n \sin^n x \\
\iff
\begin{vmatrix}
0 & 1 & 1 \\
1 & 1 & 0 \\
1 & 0 & 0 \\
\end{vmatrix} = a_0 + a_1 \cdot 0 + a_2\cdot 0 +\cdots+ a_n \cdot 0 = a_0
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.
| This answer is almost in the same spirit as @Quang Hoang's, but I hope this answer will add something. Let $$
P(z) = (z-a)(z-b)(z-c)=z^3-\sigma_1 z^2+\sigma_2 z-\sigma_3
$$ where $\sigma_1=a+b+c$, $\sigma_2=ab+bc+ca$ and $\sigma_3=abc$ by Vieta's formula. Note that for $z\in \{a,b,c\}$,
$$
z^{n+3} =\sigma_1 z^{n+2}-\sigma_2 z^{n+1}+\sigma_3 z^n,
$$ hence by summing over $z\in \{a,b,c\}$, we get recurrence relation
$$
s_{n+3}= \sigma_1 s_{n+2}-\sigma_2 s_{n+1}+\sigma_3 s_n
$$ for $s_n = a^n+b^n+c^n$. Given the data, it can be easily noted that $$\sigma_1=1 ,\quad \sigma_2 =\frac 12 \left((a+b+c)^2-(a^2+b^2+c^2)\right)=-\frac 12.$$ And by plugging $n=0$, we obtain
$$
3=1\cdot 2+\frac 12\cdot 1 +\sigma_3 s_0=2.5 + \sigma_3s_0,
$$ so $\sigma_3=abc\ne 0$ and $s_0=a^0+b^0+c^0=3$. This gives $\sigma_3=\frac 1 6$, implying that
$$
s_{n+3}=s_{n+2}+\frac 12 s_{n+1}+\frac 1 6 s_{n},\quad \forall n\ge 0.
$$ Now $s_4 =\frac {25}{6}$ and $s_5=6$ follows from the initial data $(s_3,s_2,s_1)=(3,2,1)$.
Note : The theory of homogeneous linear difference equations is behind it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Numerical Methods: calculate $b/a$ without division
Calculate $b/a$ in a calculator that only adds, subtracts and multiplies.
This problem is in the textbook for my numerical methods class. Obviously you can calculate it by
$$ \frac{1}{a} + \frac{1}{a} +\frac{1}{a} + ... \frac{1}{a} + = \frac{b}{a} $$
So the difficulty is in calculating $1/a$ without division. I would put my work here except I didn't have any idea on how to solve the problem.
| Do you know how long division by hand works? Subtract the largest multiple of $a$ from $b$.
That will be the quotient (value before the decimal point). Let the difference (reminder) be $d$. Then put a decimal point and consider $10 \cdot d$. Repeat the procedure to as many decimal values needed or until you get a repeated $d$.
Eg: Consider $\frac{25}{7}$
$$25 = \underline3\cdot 7 + 4 \tag{3.}$$
$$\color{red}{3}\cdot 10 = 30 = \underline4\cdot 7 + 2 \tag{3.4}$$
$$2 \cdot 10 = 20 = \underline2\cdot 7 + 6 \tag{3.42}$$
$$6 \cdot 10 = 60 = \underline8\cdot 7 + 4 \tag{3.428}$$
$$4 \cdot 10 = 40 = \underline5\cdot 7 + 5 \tag{3.4285}$$
$$5 \cdot 10 = 50 = \underline7\cdot 7 + 1 \tag{3.42857}$$
$$1 \cdot 10 = 10 = \underline1\cdot 7 + \color{red}{3} \tag{3.428571}$$
The last reminder $3$ is a repeated one and the repetition starts at the decimal starting with $4$, so the value is $3.\overline{428571}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{1-2y}\cdot\frac{1}{(1-\frac{1}{4y^2})^2}=$$ $$=-\frac{1}{2}\sum(2y)^n\sum(\frac{1}{4y^2})^n$$That is why I have $$f^{(3)}=3!\frac{-1}{2}\cdot2\cdot\frac{1}{4}=\frac{1}{4}$$Why is it not correct solution?
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{f}\pars{x} & \equiv
{\pars{1 + x}^{4} \over \pars{1 + 2x}^{3}\pars{1 - 2x}^{2}}
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
{1 + 4x + 6x^{2} \over
\pars{1 + 6x + 12x^{2}}\pars{1 - 4x + 4x^{2}}}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
{1 + 4x + 6x^{2} \over
1 - 4x + 4x^{2} + 6x - 24x^{2} + 12x^{2}}
\\[5mm] & =
\pars{1 + 4x + 6x^{2}}\pars{1 + 2x - 8x^{2}}^{-1}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
\pars{1 + 4x + 6x^{2}}\bracks{1 - \pars{2x - 8x^{2}} + \pars{2x - 8x^{2}}^{2}}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
\pars{1 + 4x + 6x^{2}}\pars{1 - 2x + 8x^{2} + 4x^{2}}
\\[5mm] & =
\pars{1 + 4x + 6x^{2}}\pars{1 - 2x + 12x^{2}}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
1 - 2x + 12x^{2} + 4x - 8x^{2} + 6x^{2} = 1 + 2x + 10x^{2}
\end{align}
$$
\bbx{\mrm{f}\pars{x} \equiv
{\pars{1 + x}^{4} \over \pars{1 + 2x}^{3}\pars{1 - 2x}^{2}}
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
\bbox[10px,#ffd]{1 + 2x + 10x^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the median in a triangle with trigonometry In a triangle $ABC$, $AB=7$, $AC=4$ and $\angle CAB=50º$. Let $M$ be the midpoint of $BC$. Determinate $AM$.
My try
I applied law of cosines $3$ times, first to find $BC$, then I let $\angle BCA=\alpha$ and $\angle ABC=130-\alpha$, and applied law of cosines in triangle $ACM$ and $ABM$, but it didn't work.
Any hints?
| The Law of Cosines tells us that
$$CB^2=7^2+4^2-2\cdot7\cdot 4\cdot\cos50=65-56\cos50$$
Finally, in virtue of Apollonius’s theorem, we obtain that
\begin{align*}
AM^2&=-\frac{BC^2}{4}+\frac{AC^2}{2}+\frac{AB^2}{2}\\
&=\frac{56\cos50-65+2\cdot 4^2+2\cdot 7^2}{4}\\
&=\frac{56\cos50+65}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right| < \left|\frac{-n -25}{2n^3+2n+8}\right| = \left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right|$$
Since $|-x| = |x|$:
$$\left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right| = \left|\frac{n +25}{2n^3+2n+8}\right|<\varepsilon $$
This is the part which I have trouble with. Here, I always end up with my minimum-required index to be smaller than zero, which seems peculiar to me:
$$\left|\frac{n +25}{2n^3+2n+8}\right| < |n+25| = n+ 25 < \varepsilon = 1/2$$
From here, I will get that my $N$ will be less than zero, which means that all the elements of the sequence are inside of my given epsilon environment, but I know that's not true since $a_1 = 5/6 < 3 - \varepsilon $, so what I did do wrong? Is it because I completely removed the denominator? If so, why does that break the inequality?
| Don't worry about $\epsilon$.
Instead,
try to get the difference
in a simple form
by assuming $n$
is as large as you need.
Then getting $n$
is much simpler.
Using your calculations:
$\begin{array}\\
\left|\dfrac{6n^3+5n-1}{2n^3+2n+8}-3\right|
&= \left|\dfrac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right|\\
&= \left|\dfrac{-n -25}{2n^3+2n+8}\right|\\
&= \left|\dfrac{n +25}{2n^3+2n+8}\right|\\
&\le \left|\dfrac{2n}{2n^3}\right|
\qquad\text{if } n \ge 25\\
&= \left|\dfrac{1}{n^2}\right|\\
&\lt \epsilon
\qquad\text{if } n \gt \frac1{\sqrt{\epsilon}}\\
\end{array}
$
Therefore the difference
is within $\epsilon$
if
$n \gt \max(25, \frac1{\sqrt{\epsilon}})$.
You don't have to get
the best possible $n$ -
just showing one exists
is good enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k+ n }}{k!} (n+a)^k$ I am trying to find the following limit:
\begin{align}
\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k+ n }}{k!} (n+a)^k
\end{align}
for some fixed $a>0$.
Things that tired.
We can come up with the following bound:
\begin{align}
\lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k+ n }}{k!} (n+a)^k \le \lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \left( \frac{\sqrt{k }}{k!} (n+a)^k + \frac{\sqrt{ n }}{k!} (n+a)^k\right)\\
\le \lim_{ n \to \infty} \frac{e^{-n}}{\sqrt{n}} \sum_{k=0}^\infty \frac{\sqrt{k }}{k!} (n+a)^k + e^{a}
\end{align}
| Defining $p(k, \lambda) = \frac{\lambda^k}{k!}e^{-\lambda}$, then
\begin{align*}
f(k, n) = \sqrt{1 + \frac{k}{n}}e^a \frac{(n+a)^k}{k!}e^{-(n+a)} = \sqrt{1 + \frac{k}{n}}e^ap(k, n+a)
\end{align*}
So the summation is equivalent to, for a random variable $X \sim \text{Pois}(n+a)$,
\begin{align*}
e^{a}\mathbb{E}\left[\sqrt{1 + \frac{X}{n}}\right]
\end{align*}
Since $X/n \rightarrow 1$ almost surely by the Strong Law of Large Numbers and $\mathbb{E}[\frac{X}{n}] = 1 + \frac{a}{n} \le 1 + a < \infty$, dominated convergence allows us to exchange $\mathbb{E}$ and $\lim$, and so
\begin{align*}
\lim_{n\rightarrow \infty} e^{a}\mathbb{E}\left[\sqrt{1 + \frac{X}{n}}\right] = e^{a}\mathbb{E}\left[\sqrt{1 + \lim_{n\rightarrow \infty}\frac{X}{n}}\right] = e^{a} \sqrt{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Show with induction that $\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$ Show with induction that
$\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$
n = 1
$LHS = \frac{1}{2}$
$RHS = 6 - \frac{1+4+6}{2} = \frac{1}{2}$
n = p
$LHS_{p} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + ... \frac{p^{2}}{2^{p}}$
$RHS_{p} = 6 - \frac{P^2+4p+6}{2^{p}}$
n = p + 1
$LHS_{p+1} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + ... \frac{p^{2}}{2^{p}} + \frac{(p+1)^{2}}{2^{(p+1)}}$
$RHS_{p+1} = 6 - \frac{(p+1)^2+4(p+1)+6}{2^{(p+1)}}$
Show that, $RHS_{p+1} = RHS + \frac{(p+1)^{2}}{2^{(p+1)}}$
$6 - \frac{(p+1)^2-4(p+1)+6}{2^{(p+1)}} = 6 - \frac{P^2+4p+6}{2^{p}} + \frac{(p+1)^{2}}{2^{(p+1)}}$
| Hint: The left-hand side is given by $$\frac{(k+1)^2}{2^{k+1}}-\frac{k^2-4k+6}{2^k}=-2^{-k-1} \left(k^2-10 k+11\right)$$
and the right-hand side: $$-{\frac {{k}^{2}-2\,k+3}{{2}^{k+1}}}$$ so your formula is not true.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof for range of average We can prove that average of two numbers $a,b$ where $a<b$ will be between $a$ and $b$ as follows
$a < b$ $a + a < a + b$a < $\dfrac{a + b}{2 }$
$a < b$ $a + b < b + b$ $\dfrac{a + b}{2 }< b$
Thus $a < \dfrac{a + b}{2 }< b$
Similarly, is it true always and can be proved, that average of three numbers, $a,b,c$ where $a<b<c$ will be between $a$ and $b$ if $b-a$ > $c-b$
| Yes, it is always true.
From what you've already got, we have
$$a<b<\frac{b+c}{2}<c\implies a<\frac{b+c}{2}$$
$$\implies 2a<b+c\implies 3a<a+b+c\implies a<\frac{a+b+c}{3}$$
Also, we have
$$b-a>c-b\implies a+c\lt 2b\implies a+b+c\lt 3b\implies \frac{a+b+c}{3}\lt b$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A trigonometry question from STEP examination Show that if at least one of the four angles A ± B ± C is a multiple of π, then
$$\sin^4A + \sin^4 B + \sin^4 C − 2 \sin^2 B \sin^2 C − 2 \sin^2 C \sin^2 A
− 2 \sin^2 A \sin^2 B + 4 \sin^2 A \sin^2 B \sin^2 C = 0$$
I want to start with proving $\sin(A+B+C)$ or $(\sin(A)+\sin(B)+\sin(C))^2$, however, I failed in both cases.
| Hint:
First of all writing $\sin A=a$ etc.,
$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$
$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$
Now if $A+B+C=\pi$
by this $\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$
and by this $\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos\dfrac C2$
Use $\sin2x=2\sin x\cos x$
We shall same expressions in some order if $A\pm B\pm C=\pi$
| {
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Finding the determinant of a tridiagonal matrix
$$\begin{vmatrix}x&1&0&0&⋯\\-n&x-2&2&0&⋯\\0&-(n-1)&x-4&3&⋯\\⋮&⋱&⋱&⋱&⋮\\0&⋯&-2&x-2(n-1)&n\\0&0&⋯&-1&x-2n\end{vmatrix}_{(n+1)×(n+1)}$$Find the value of the above determinant.
This problem comes from an advanced algebra book. I want to solve it with elementary transformation knowledge. I have been trying to solve it for a long time.
| Edit. Call your matrix $A$ and let $L$ be the zero-indexed Pascal matrix defined by
$$
l_{ij}=\begin{cases}
\binom{n-j}{i}&\text{ when }\ 0\le i\le n-j\le n,\\
0&\text{ otherwise}.
\end{cases}
$$
It is known that $(L^{-1})_{ij}=(-1)^{i+j}l_{ij}$. E.g. when $n=5$,
$$
L=\pmatrix{1&0&0&0&0&0\\ 5&1&0&0&0&0\\ 10&4&1&0&0&0\\ 10&6&3&1&0&0\\ 5&4&3&2&1&0\\ 1&1&1&1&1&1},
\ L^{-1}=\pmatrix{1&0&0&0&0&0\\ -5&1&0&0&0&0\\ 10&-4&1&0&0&0\\ -10&6&-3&1&0&0\\ 5&-4&3&-2&1&0\\ -1&1&-1&1&-1&1}.
$$
One may verify that $A-(x-n)I_{n+1}=L^{-1}NL$ for some nilpotent matrix $N$. More specifically, one may verify that $L(A-(x-n)I_{n+1})=NL$ where
$$
N=\pmatrix{0&1\\ &0&2\\ &&\ddots&\ddots\\ &&&\ddots&n\\ &&&&0}.
$$
In other words, if $V$ is the vector space of polynomials in $y$ of degrees $\le n$ and $D,g,g^{-1}:V\to V$ are the linear operators
\begin{aligned}
D(p)(y)&=p'(y),\\
g(p)(y)&=(1+y)^np\left(\frac{y}{1+y}\right),\\
g^{-1}(p)(y)&=(1-y)^np\left(\frac{y}{1-y}\right),
\end{aligned}
then $A-(x-n)I_{n+1}$ is the matrix representation of the linear map
$f=g^{-1}\circ D\circ g$ with respect to the ordered basis $(1,y,y^2,\ldots,y^n)$. (More explicitly, $f(p)(y)=n(1-y)p(y)+(1-y)^2p'(y)$, but this formula is unimportant here.) Since $D^n=0$, $f^n=g^{-1}\circ D^n\circ g$ is also zero. Hence $f$ and $A-(x-n)I_{n+1}$ are nilpotent and in turn $\det A=(x-n)^{n+1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show that $ab+bc+ca\le \frac34$ Let $a,b$ and $c$ be positive real numbers such that $(a+b)(b+c)(c+a) = 1$ , hen show that $$ab+bc+ca\le \frac34$$
I believe I need to use AM-GM inequality and use the fact $(a+b)(b+c)(c+a) = 1$ Using AM-GM in $a+b,b+c$ & $c+a$, I get $a+b+c\ge \frac32$. Any hint will be thankful.
| Consider the polynomial $$
\begin{align}p(x) &= (x-a)(x-b)(x-c)\\
&= x^3 - Ax^2 + Bx - C\end{align}\quad\text{ where }\quad
\begin{cases} A = a + b + c \\B = ab+bc+ca \\C = abc\end{cases}$$
Notice
$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C$$
We have
$$(a+b)(b+c)(c+a) = 1 \iff BA - C = 1 \iff B = \frac{1+C}{A}$$
Apply AM $\ge$ GM to the three pairs $(a,b), (b,c), (c,a)$ and combine the result, we obtain
$$8C = (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca}) \le (a+b)(b+c)(c+a) = 1$$
Apply AM $\ge$ GM again to the 3-tuple $(a+b,b+c,c+a)$, we find
$$\frac23 A = \frac13((a+b) + (b+c) + (c+a)) \ge \sqrt[3]{(a+b)(b+c)(c+a)} = 1$$
These mean $C \le \frac18$ and $A \ge \frac32$. Combine them, we can deduce
$$ab+bc+ca = B = \frac{1+C}{A} \le \frac{1 + \frac18}{\frac32} = \frac34$$
| {
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Finding floor of reciprocal sum
Evaluation of
$$\bigg \lfloor \frac{1}{\sqrt[3]{1}}+\frac{1}{\sqrt[3]{2^2}}+\frac{1}{\sqrt[3]{3^2}}+\cdots +\frac{1}{\sqrt[3]{(1000)^2}}\bigg\rfloor$$
Where $\lfloor x\rfloor $ is the floor of $x$
Try: It seems like we can solve it using Telescopic sums and that the sum lies between $2$ Telescopic sums, but could not figure out how to solve it.
Could someone help me to solve it? Thanks.
| Note that $\displaystyle \sum_{k=1}^{1000}\frac{3}{\sqrt[3]{(k+1)^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{k^2}}<\sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}<1+\sum_{k=2}^{1000}\frac{3}{\sqrt[3]{(k-1)^2}+\sqrt[3]{k(k-1)}+\sqrt[3]{k^2}}$.
$\displaystyle \sum_{k=1}^{1000}\frac{3}{\sqrt[3]{(k+1)^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{k^2}}=\sum_{k=1}^{1000}\frac{3(\sqrt[3]{k+1}-\sqrt[3]{k})}{(k+1)-k}=3(\sqrt[3]{1001}-1)>27$
$\displaystyle 1+\sum_{k=2}^{1000}\frac{3}{\sqrt[3]{(k-1)^2}+\sqrt[3]{k(k-1)}+\sqrt[3]{k^2}}=1+\sum_{k=2}^{1000}\frac{3(\sqrt[3]{k}-\sqrt[3]{k-1})}{(k-1)-k}=1+3(\sqrt[3]{1000}-1)=28$
So, $\displaystyle \left\lfloor \sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}\right\rfloor=27$.
| {
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"url": "https://math.stackexchange.com/questions/3205926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|x²-2x| + |x-4| > |x²-3x+4| , How do I solve for all real x? How do I solve this for all real x?
|x²-2x| + |x-4| > |x²-3x+4|
Looking at the question it is clear that it states |a| + |b| > |a-b|. How to proceed?
| First, note that $x^2-3x+4$ is always positive because discriminant $D=(-3)^2-4\times4<0$.
Case 1: $x\in(-\infty, 0]$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=-(x-4)$ and the inequality becomes:
$$x^2-2x-(x-4)>x^2-3x+4$$
$$x^2-3x+4>x^2-3x+4$$
...which is never true. So the ineqality has no solution in the given range.
Case 2: $x\in(0, 2]$
In this range: $|x^2-2x|=-(x^2-2x)$, $|x-4|=-(x-4)$ and the inequality becomes:
$$-(x^2-2x)-(x-4)>x^2-3x+4$$
$$-x^2+x+4 > x^2-3x+4$$
$$0 > 2x^2-4x$$
$$0 > x^2-2x=x(x-2)$$
...and this is obviously true for $x\in(0,2)$. Combined with the initial condition that $x\in(0,2]$ this gives one part of the solution: $x\in(0,2)$
Case 3: $x\in(2, 4]$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=-(x-4)$ and the inequality becomes:
$$x^2-2x-(x-4)>x^2-3x+4$$
$$x^2-3x+4>x^2-3x+4$$
...which is never true.
Case 4: $x\in(4, +\infty)$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=x-4$ and the inequality becomes:
$$x^2-2x+x-4>x^2-3x+4$$
$$-x-4>-3x+4$$
$$2x-8>0$$
$$x>4$$
So all values of $x$ in the given range are also solutions to the inequality.
If you summarize all four cases you get the final result:
$$\boxed{x\in(0,2)\cup(4,+\infty)}\tag{1}$$
The solution can be confirmed by looking at the plot of the function:
$$f(x)=|x^2-2x| + |x-4| - |x^2-3x+4|$$
...which is positive for values of $x$ defined by (1).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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One of $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$ is divisible by $8$ if $a,b,c∈\mathbb{Z}^+$ Given that $a,b,c$ are there distinct positive integers, prove that among $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$, there is at least one that is divisible by $8$.
I have no clue what to do and cannot even create cases as these are three random whole numbers. What do I do? Please help me guys!!!
Even a hint is graciously accepted!!!
| The three numbers are cyclic permutations of $ab(a^4-b^4)$. Therefore, at least one of the products must be formed from either $2$ even numbers or $2$ odd numbers.
If $a$ and $b$ are both even, then so is $(a^4-b^4)$, so $ab(a^4-b^4)$ is the product of three even numbers, which must be divisible by $8$.
If $a$ and $b$ are both odd, then $a^4 \equiv b^4 \pmod{8}$ (because any odd number is a square root of $1$ (mod $8$)), so $8|(a^4-b^4)$ and therefore $8|ab(a^4-b^4).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Optimization problem in a single variable. $$f(x) = x^3 - 3x $$
Find the maximum value of the function $f(x)$ on the set of real numbers satisfying
$$x^4 + 36 \le 13x^2$$
I have only learned how to solve such question when the constraints are in an equality rather than an inequality through the Lagrangian method. General method to solve such problems with inequality constraints would also be extremely helpful.
Thanks.
| $$\begin{align}
x^4+36 &\le 13x^2 \\
x^4-13x^2+36 &\le 0\\
x^4-9x^2-4x^2+36 &\le 0\\
x^2(x^2-9)-4(x^2-9) &\le 0\\
(x^2-2^2)(x^2-3^2) &\le 0\\
(x+3)(x+2)(x-2)(x-3) &\le 0\\
\end{align}$$
This inequality is only satisfied when $2 \le x \le 3$ or $-3 \le x \le -2$.
$f(x)=x^3-3x=x(x^2-3)$ so $f(x)$ is an odd function. Also, it is an increasing function so $f(3)$ is the maximum value of $f(x)$ in this domain.
We can verify this by observing that:
$$\begin{align}
f(-3) &= -18\\
f(-2) &= -2\\
f(2) &= 2\\
f(3) &= 18\\
\end{align}$$
General method:
$\cdot$ Find the domain.
$\cdot$ Find the gradient by differentiating the function.
$\cdot$ Equate gradient to $0$ and find the values of $x$.
$\cdot$ Differentiate the gradient and calculate $f''(x)$.
$\cdot$ If $f''(x) \gt 0$ it is a minimum.
$\ \ $If $f''(x) \lt 0$ it is a maximum.
$\ \ $If $f''(x) = 0$ it is a point of inflection.
| {
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$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$, $\lim_{n\to \infty}a_n$ is Given $a_1,a_2,n\in \mathbb N$
$$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$$
Then $\lim_{n\to \infty}a_n$ is
(A) $2(a_2-a_1)+a_1e^{-1}$
(B) $2(a_1-a_2)e^{-1}+a_2$
(C) $2(a_1-a_2)e^{-1}+a_1$
(D) $2(a_2-a_1)e^{-1}+a_1$
My attempt,
$a_1,a_2\in \mathbb N$
$$a_3=(1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1}$$
$$a_4=(1-\frac{1}{4})a_{3}+\frac{1}{4}a_{2}$$ $$=(1-\frac{1}{4})((1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1})+\frac{1}{4}a_{2}$$ $$=(1-\frac{1}{4})(1-\frac{1}{3})a_{2}+(1-\frac{1}{4})\frac{1}{3}a_{1}$$
$$a_5=(1-\frac{1}{5})a_{4}+\frac{1}{5}a_{3}$$ $$=(1-\frac{1}{5})((1-\frac{1}{4})(1-\frac{1}{3})a_{2}+(1-\frac{1}{4})\frac{1}{3}a_{1})+\frac{1}{5}((1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1})$$
But I am not able to conclude from here, I couldn't able to generalise anything from here.
| Let $f(x) = \sum_{n\geq 1} a_n x^n$. Then
\begin{align*}
(1-x) f'(x)
&= a_1 + (2a_2 - a_1) x + x \sum_{n \geq 1} ((n+2) a_{n+2} - (n+1) a_{n+1}) x^n \\
&= a_1 + (2a_2 - a_1) x + x \sum_{n \geq 1} a_n x^n \\
&= a_1 + (2a_2 - a_1) x + x f(x).
\end{align*}
This gives a first-order linear ODE, and solving this together with the initial condition $f(0) = 0$ gives
$$ f(x) = \frac{(2a_2 - a_1)x + 2(a_2 - a_1) (e^{-x} - 1)}{1-x}. $$
Comparing the coefficient of $x^n$ of both sides, we get
$$ a_n = 2a_2 - a_1 + 2(a_2 - a_1) \sum_{k=1}^{n} \frac{(-1)^{k}}{k!}. $$
As $n\to\infty$, this converges to
$$ \lim_{n\to\infty} a_n = 2a_2 - a_1 + 2(a_2 - a_1)(e^{-1} - 1) = 2(a_2 - a_1)e^{-1} + a_1. $$
| {
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