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proving the identity $b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$ for $b \in \mathbb{N}$, $b > 2$, by counting a set in different ways I'm trying to prove that $\forall b \in \mathbb{N}, b>2,$ $b^{3} = 6\binom{b} {3} +6 \binom{b}{2} +b$ without just using algebra. The idea I've thought about is that there are $b^{3}$ ways to choose a triple of numbers from $\{1,\ldots, b\}$, and there are obviously $b$ ways to choose one element, the issue is I'm not sure how the $\binom{b}{3}$ and $\binom{b}{2}$ terms contribute. Any help would be greatly appreciated.
|
Slight different approach (Case 2 below)!
Consider $b=3$ for simplicity.
There are $b^3=27$ ways to form a $3$-digit number with the digits $1,2,3$.
Case 1: All three digits are the same:
$${b\choose 1}={3\choose 1}=3 \Rightarrow 111,222,333.$$
Case 2: Two digits are the same:
$${3\choose 2} \Rightarrow \color{red}{**\times},\color{green}{*\times*},\color{blue}{\times**} \quad \text{(* - same, $\times$ - different)}\\
{3\choose 2}{b\choose 1}={3\choose 2}{3\choose 1} \Rightarrow \color{red}{11\times,22\times,33\times},\color{green}{1\times1,2\times2,3\times3},\color{blue}{\times11,\times22,\times33}\\
{3\choose 2}{b\choose 1}{b-1\choose 1}={3\choose 2}{3\choose 1}{2\choose 1} \Rightarrow \color{red}{112,113,221,223,331,332},\color{green}{\cdots},\color{blue}{\cdots}.$$
Case 3: All three digits are different:
$${b\choose 3}\cdot 3!={3\choose 3}\cdot 3! \Rightarrow 123,132,213,231,312,321.$$
Hence:
$$\begin{align}b^3&={b\choose 3}\cdot 3!+{3\choose 2}{b\choose 1}{b-1\choose 1}+{b\choose 1}=\\
&=6\cdot {b\choose 3}+{3\choose 2}{b\choose 2}{2\choose 1}+{b\choose 1}=\\
&=6\cdot {b\choose 3}+6\cdot {b\choose 2}+b.\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$ $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$
So, getting rid of the denominators I got to:
$(x+1)(x+5) + x(x+5) < 2x(x+1)$
$\rightarrow x < \frac{-5}{9}$
And also keeping in mind we can't have $-1,-5$ in the solution set, I got that the solution set was:
$\left(-\infty ,-5\right)\cup \left(-5,-1\right)\cup \left(-1,-\frac{5}{9}\right)$
But this is not correct. Help appreciated!
|
To clear the fractions, you need to multiply by $|x(x+1)(x+5)|$ or $[x(x+1)(x+5)]^2.$ In these cases you can be sure that the inequality is preserved, since they are never negative.
|
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|
Prove that $2x^3+3x^2+x$ is always divisible by 6 if x is an integer. Prove that $2x^3+3x^2+x$ is always divisible by $6$ if $x$ is an integer.
I started by factoring the expression:
$x(2x^2+3x+1)=x(2x+1)(x+1)$
However I wasn’t sure how to progress from here to prove that it is always divisible by 6. Any ideas?
|
You are almost done
$$x(2x^2+3x+1)=x(2x+1)(x+1)=x(2x-2+3)(x+1)
\\=2(x-1)x(x+1)+3x(x+1)$$
Now, among any three consecutive integers one is a multiple of 3, and among two consecutive integers one is even.
|
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|
Find the maximal value of $c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$ in terms of $c^3 + a^3 = m$ where $a, b$ and $c$ are positives.
Let $a, b$ and $c$ be positive real numbers such that $c^3 + a^3 = m$ and $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0.$$ Find the maximal value of $$c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$$ in terms of $m$.
We have that $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0$$
$$\iff (b^3 - c^3)(a^3 - b^3) \ge 0 \implies b^3(c^3 + a^3 - b^3) \ge (ca)^3 \iff m - b^3 \ge \left(\frac{ca}{b}\right)^3$$
It is evident that $$c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a} = \frac{m - b^3 + 3b + 2}{c + a} \ge \frac{(ca)^3 + 3b^4 + 2b^3}{b^3(c + a)}$$
How should I continue?
|
Constraint
$$((b^3 - abc) + ab(a + b) - c(a^2 + b^2))((abc - b^3) - bc(b + c) + a(b^2 + c^2)) \le 0,$$
or
$$(b^3-a^3)(b^3 - c^3)\ge 0,$$
is equivalent to
$$b^2-(a+c)b+ac\ge 0\tag1.$$
The goal function
$$f(a,b,c)=c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$$
under the conditions
$$a^3 + c^3 = m\tag2$$
can be presented in the form of
$$f(a,b,c)=\dfrac1m(c^2-ca+a^2)(m-(b^3-3b-2)).\tag3$$
$\color{brown}{\textbf{At the edges}}.$
$$\max f(0,b,c) = \max f(a,b,0)
= \begin{cases}
\dfrac{m+4}{\sqrt[3]m},\quad\text{if}\quad m\in(0,1]\\
\dfrac{3\sqrt[3]m+2}{\sqrt[3]m},\quad\text{if}\quad m\in\left(1,3\sqrt3\right]\\
\dfrac{m+2}{\sqrt[3]m},\quad\text{if}\quad m\in\left(3\sqrt3,\infty\right).
\end{cases}\tag4$$
The greatest value of
$$g(a,c)=c^2-ca+a^2$$
under the condition $(2)$ can be achieved at the edges of the area or in the inner stationary points.
At the edges,
$$g_e = g\left(0,\sqrt[3]{m}\right) = g\left(0,\sqrt[3]{m}\right) = \sqrt[3]{m^2}.\tag5$$
The inner stationary points
of $g(a,c)$ under the condition $(2),$ in accordance with Lagrange multupliers method, corresponds with the stationary points of
$$G(a,c,\lambda) = a^2-ac+c^2+\lambda(m-a^3-c^3),$$
which can be found from the system $G'_a = G'_c = G'(\lambda) = 0,$ or
$$
\begin{cases}
2a-c-3\lambda a^2=0\\
-a+2c-3\lambda c^2 = 0\\
a^3+c^3=m
\end{cases}\ \Rightarrow
\begin{cases}
(2a-c)c^2=(-a+2c)a^2\\
a^3+c^3=m
\end{cases}\ \Rightarrow
\begin{cases}
(a-c)(a^2-ac+c^2)=0\\
a^3+c^3=m,
\end{cases}
$$
with the solution
$$g_s=g\left(\sqrt[3]{\dfrac m2},\sqrt[3]{\dfrac m2}\right)=\sqrt[3]{\dfrac {m^2}4}.\tag6$$
The global maximum is
$$\max f(a,0,c)=\dfrac{m+2}{\sqrt[3]m}\max(g_e,g_s) = \dfrac{m+2}{\sqrt m}\ \text{ at }\
\dbinom ac\in\left\{\dbinom{\sqrt[3]m}0,\dbinom0{\sqrt[3]m}\right\}.\tag6$$
$\color{brown}{\textbf{The inner stationary points}}$
of $f(a,b,c)$ under the conditions $(1),(2),$ in accordance with Lagrange multupliers method, corresponds with the stationary points of
$$F(a,b,c,\lambda,\mu) = (c^2-ca+a^2)(m-a^3+3a+2)+\lambda(m-a^3-c^3)+\mu(b^2-(a+c)b+ac),$$
which can be found from the system
$$F'_a = F'_b = F'_c = F'(\lambda) = F'_{\mu} = 0,$$
and can be solved similarly.
If the inner maxima are less than on the edges then $(4)$ defines the the global maxima.
|
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|
Knowing $\sin(45^\circ)={\sqrt2\over2}$, how do I find $\sin(135^\circ)$ and $\cos(135^\circ)$?
I just know about $$\sin(45^\circ)={\sqrt2\over2}$$
and I want understand how to find $\sin(135^\circ)$ and $\cos(135^\circ)$
Thank you.
|
The cosine and sine of a directed angle with vertex at the origin and initial side on the positive $x$-axis are, respectively, the $x$- and $y$-coordinates of the points where the terminal side of the angle intersects the unit circle $x^2 + y^2 = 1$.
We are given the sine of a first-quadrant angle. Since $(\cos\theta, \sin\theta)$ is a point on the unit circle,
$$\cos^2\theta + \sin^2\theta = 1$$
Therefore, we can solve for $\cos(45^\circ)$.
\begin{align*}
\cos^2(45^\circ) + \sin^2(45^\circ) & = 1\\
\cos^2(45^\circ) & = 1 - \sin^2(45^\circ)\\
\cos^2(45^\circ) & = 1 - \left(\frac{\sqrt{2}}{2}\right)^2\\
\cos^2(45^\circ) & = 1 - \frac{2}{4}\\
\cos^2(45^\circ) & = 1 - \frac{1}{2}\\
\cos^2(45^\circ) & = \frac{1}{2}\\
|\cos(45^\circ)| & = \sqrt{\frac{1}{2}}\\
\end{align*}
Since $45^\circ$ is a first-quadrant angle, the $x$-coordinate of the point where the terminal side of the angle intersects the unit circle is positive. Therefore, its cosine is positive. Hence,
\begin{align*}
\cos(45^\circ) & = \sqrt{\frac{1}{2}}\\
& = \frac{1}{\sqrt{2}}\\
& = \frac{\sqrt{2}}{2}
\end{align*}
Since $90^\circ < 135^\circ < 180^\circ$, $135^\circ$ is a second-quadrant angle. To determine its sine and cosine, we can use symmetry.
Since the sine of an angle is the $y$-coordinate of the point where its terminal side intersects the unit circle, two angles that intersect the unit circle at points with the same $y$-coordinate have the same sine. Hence,
$$\sin(\pi - \theta) = \sin\theta$$
Moreover, two angles with opposite $y$-coordinates have opposite sines. Hence,
\begin{align*}
\sin(\pi + \theta) & = -\sin\theta\\
\sin(-\theta) & = -\sin\theta
\end{align*}
Since any two coterminal angles have the same sine,
$$\sin(\theta + 2k\pi) = \sin\theta, k \in \mathbb{Z}$$
Therefore, $\sin\theta = \sin\varphi$ if
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or
$$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$
If we write $\sin(\pi - \theta) = \sin(\theta)$ in terms of degrees, we obtain
$$\sin(180^\circ - \theta) = \sin(\theta)$$
Since $135^\circ = 180^\circ - 45^\circ$, we obtain
$$\sin(135^\circ) = \sin(180^\circ - 45^\circ) = \sin(45^\circ)$$
Since the cosine of an angle is the $x$-coordinate of the point where its terminal side intersect the unit circle, two angles that intersect the unit circle at points with the same $x$-coordinate have the same cosine. Hence,
$$\cos(-\theta) = \cos\theta$$
Moreover, two angles with opposite $x$-coordinates have opposite cosines. Hence,
\begin{align*}
\cos(\pi - \theta) & = -\cos\theta\\
\cos(\pi + \theta) & = -\cos\theta
\end{align*}
Since any two coterminal angles have the same cosine,
$$\cos(\theta + 2k\pi) = \cos\theta, k \in \mathbb{Z}$$
Therefore, $\cos\theta = \cos\varphi$ if
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or
$$\varphi = -\theta + 2k\pi, k \in \mathbb{Z}$$
If we write $\cos(\pi - \theta) = -\cos\theta$ in terms of degrees, we obtain
$$\cos(180^\circ - \theta) = -\cos\theta$$
Thus,
$$\cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ)$$
|
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|
How do I prove the limit $\frac{\sin(xy)}{\sqrt{x^2 + y^2}}$ as (x, y) approaches (0, 0) using $\epsilon - \delta$ I know that I can convert this limit to polar coordinates and solve the limit, but I want to see how I would do it using the $\epsilon - \delta$ definition of a limit.
This is my work so far:
We know that $$\left|{\frac{\sin(xy)}{\sqrt{x^2 + y^2}}} - 0\right| < \epsilon$$ and $$
\left| \sqrt{x^2 + y^2} \right| < \delta $$
Then, $$
\begin{align}
\left|\sin(xy)\right| &< \epsilon \left|\sqrt{x^2 + y^2}\right| \\
\frac{\left|\sin(xy)\right|}{\epsilon} &< \left|\sqrt{x^2 + y^2}\right|
\end{align}
$$
I am stuck here, as normally I would get an expression that matches $\delta$, but here the signs are switched.
|
Let $(x,y)\not =(0,0)$.
$|\sin (xy)| \le |xy|;$ $x^2+y^2 \ge |xy|$;
$0\le |\dfrac{\sin (xy)}{\sqrt{x^2+y^2}}| \le\dfrac{|xy|}{\sqrt{x^2+y^2}}\le$
$ \dfrac{x^2+y^2}{\sqrt{x^2+y^2}}= \sqrt{x^2+y^2}.$
Choose $\delta =\epsilon$.
|
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|
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
I think that $(0,0)$ , $(1,0)$ and $(0,-1)$ are the only solutions to the above equation, but I'm unable to prove it.
I tried all sorts of things like working $\mod 9$ (but there are just too many cases ), a little bit of algebraic manipulations, tried to determine the parity of $x$ and $y$ etc. But they were to no avail for me.
I tried working modulo $9$ because $a^{3}\equiv 0,1$ or $-1 \pmod 9$.
The manipulations done by me were as follows:-
$x^2 + y^2 =(x-y)^3$ implies that by adding and subtracting $2xy$ on LHS we can rewrite the above equation as $(x-y)^2 +2xy=(x-y)^3$ . This can be rewritten as $2xy=(x-y)^3 -(x-y)^2$. This is all I could achieve here. One thing I did here was substitute $x-y=a$ and $x=a+y$ and rewrite the last equation as $2y^2 +2ay+a^2 -a^3=0$ and then I tried to find the roots of this quadratic in $y$ but this didn't work for me(I think there is something wrong with this approach, do tell me if you see it).That is all I could do.
Another question I would like to ask is do there exist integers $a,b$ and $c$, with none of them equal to zero, which satisfy $a^2 + b^2=c^3$ ? Thank you .
|
To extend your existing approach, multiply by $2$ to obtain :$$0=4y^2+4ay+2a^2-2a^3=(2y+a)^2+a^2(1-2a)$$
To obtain a factorisation with integer coefficients you need $2a-1=b^2$. For convenience, multiply through by $4$ to get $$0=(4y+2a)^2+4a^2(1-2a)=(4y+b^2+1)^2-(b^2+1)^2b^2$$ And the roots are $$4y=-(b^2+1)\pm b(b^2+1)=-(1\pm b)(1+b^2)$$
Now $b$ is odd, so the right-hand side is the product of two even numbers, and any odd value of $b$ will lead to a solution.
For example $b=3$ gives $x=10, y=5$.
|
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|
Find the maximum of the value $f(x)=\frac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$
Let $n$ be a given positive integer, find the maximum of the value of
$$f(x)=\dfrac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$$ where $x \le \dfrac{n^2}{4}$ and $x \in \mathbb N^{+}.$
If the $x$ is real positive number, the problem also not easy, because
$$f'(x)=-\dfrac{n^4}{x^2}+4(n^2+3)-8x=0$$
This cubic equation is not easy to handle. But if $x$ be positive integer, maybe have other method can solve it.
|
$n=1$ makes no sense, while for $n=2$ the only allowed value is $x=2$. Thus assume $n>2$.
Using Descartes' rule of signs on $x^2f'(x)=-8x^3+4(n^2+3)x^2-n^4$ we see that $f'(x)$ has zero or two positive roots. Next, $f'(1) = -n^4+4 n^2+4<0$ while $f'\left(\frac{n^2}{4}\right) = 2 \left(n^2-2\right)>0$, so that $f$ has two positive roots. Since $\lim_{x\to\infty} f(x) = -\infty$, the only possibility is that $f'$ has one root between $1$ and $\frac{n^2}{4}$, which must be a minimum, and another root, which must be a maximum, larger than $\frac{n^2}{4}$.
Finally, $f(1) = n^4-n^2+8$ and $f\left(\frac{n^2}{4}\right) = \frac{3 n^4}{4}+2 n^2$, and it's not hard to see that $f(1)\ge f\left(\frac{n^2}{4}\right)$. Thus the maximum occurs for $x=1$.
|
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|
Solving $\cos \frac{165}{2}$ What are the steps to solve $$\frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4}$$ into $$\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
Please explain. I got them from derivation of $$\cos(82.5^\circ)$$ in 2 different ways.
$$\cos(60^\circ+22.5^\circ)$$ and $$\cos\frac{(90^\circ+75^\circ)}{2}$$
Thanks in advance
|
If we are allowed to solve analytically then
\begin{align}2\sqrt{2}(2\sqrt{2}-1-\sqrt3)&=
4\sqrt{4}-2\sqrt{2}-2\sqrt{6}\\&=
8-2(\sqrt{2}+\sqrt{6})\\&=
8-2\left(2\sqrt{2+\sqrt{3}}\right)\\&=
8-4\sqrt{2+\sqrt{3}}\\&=
4\left(2-\sqrt{2+\sqrt{3}}\right)
\end{align}
so that
$$\frac{\sqrt{4\left(2-\sqrt{2+\sqrt{3}}\right)}}{4}=\frac{2\sqrt{2-\sqrt{2+\sqrt{3}}}}{4}=\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
|
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|
"Altered" Alternating Series Diverges or Converges? Consider the series
$$
1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
which alternates between a block of positives and a block of negatives, with the block sizes,
$$
1, 1, 2, 3, 4, 5, 6, \dots
$$
and so on.
Does this series converge or diverge?
|
Let: $$
S=1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
For a given sequence of additions assume a constant denominator, specifically the smallest in the group. For a series of subtractions pick a common constant denominator, the largest in the group. This guarantees a maximum effect of additions and a minimum effect of subtractions, guaranteeing greater partial sums than the original series.
Doing this we get:
$$S' = 1 -\frac{1}{2}+\frac{2}{3}-\frac{3}{7}+\frac{4}{8}-\frac{5}{17}+\frac{6}{18}...$$
Suppose the numerator is $2k$ for some integer $k$. Then the denominator is $\frac{5k^2-5k+6}{2}$. This term is positive. The term directly to the left is negative with numerator $2k-1$ and denominator $\frac{5k^2-5k+4}{2}$.
So:
$$S'=1 + \sum_{k=1}^\infty \frac{4k}{5k^2-5k+6}-\frac{4k-2}{5k^2-5k+4}$$
This preserves the order of terms.
If the two fractions are added, we get a second degree polynomial in the numerator and a 4th degree polynomial in the denominator for a terms on the order of $O(1/k^2)$.
So $S'$ converges by comparison to $1/k^2$ and $S$ converges by comparison with $S'$.
|
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|
Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
|
Mod $(2n^3+5)$, you have that $2n^3\equiv-5$. Now suppose $(2n^3+5)$ divides $n^4+n+1$. Then mod $(2n^3+5)$:
$$
\begin{align}
n^4&\equiv-n-1\\
2n^4&\equiv-2n-2\\
n(2n^3)&\equiv-2n-2\\
-5n&\equiv-2n-2\\
-3n&\equiv-2\\
2-3n&\equiv0
\end{align}
$$
So $2n^3+5$ must divide $2-3n$. You can then test values of $n$ to arrive at the inequality.
|
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|
Integral From Electricity and Magnetism In attempting to calculate the magnetic field due to a current loop at an arbitrary point, I ran into the following integrals
\begin{align}
&\int_0^{2\pi} \frac{z\cos t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\\&\int_0^{2\pi} \frac{z\sin t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\\&\int_0^{2\pi} \frac{1 - y\sin t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\end{align}
I've set $x = 0$ without loss of generality (there is rotational symmetry about the $z$ axis). The integrals are the $x$, $y$, and $z$ components of the magnetic field respectively with all physical constants set to $1$. Are these well-known integrals? They remind me of elliptic integrals, so I don't really expect a closed-form solution. It's often claimed that a current loop is a magnetic dipole, so, in principle, these integrals should look like a dipole field for large $y$ and $z$. Is there an easy way to see this?
|
I will work with integrals without normalization. Write $a$ for the diameter of the current loop. Then the first integral is simply
\begin{align*}
\int_{0}^{2\pi} \frac{a z \cos t}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t
&= 0.
\end{align*}
The other two integrals involve elliptic functions, but its asymptotic expansion far away from the origin can be computed easily as well. To this end, write $r = \sqrt{y^2 + z^2}$ for the distance from the origin. Then from the Taylor expansion, we have
\begin{align*}
\frac{1}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}}
&= \frac{1}{r^{3}} + \frac{3a y \sin t}{r^5} + \mathcal{O}\left(\frac{a^2}{r^5}\right)
\end{align*}
for $a \ll r$. Using this, we get
\begin{align*}
&\int_{0}^{2\pi} \frac{a z \sin t}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t \\
&\quad= \int_{0}^{2\pi} \left( \frac{1}{r^{3}} + \frac{3a y \sin t}{r^5} + \mathcal{O}\left(\frac{a^2}{r^5}\right) \right) (a z \sin t) \, \mathrm{d}t \\
&\quad = \frac{3a^2 \pi y z}{r^5} + \mathcal{O}\left(\frac{a^3}{r^4}\right)
\end{align*}
and
\begin{align*}
&\int_{0}^{2\pi} \frac{a(a - y \sin t)}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t \\
&\quad= \int_{0}^{2\pi} \left( \frac{1}{r^{3}} + \frac{3a y \sin t}{r^5} + \mathcal{O}\left(\frac{a^2}{r^5}\right) \right) a (a - y \sin t) \, \mathrm{d}t \\
&\quad= \frac{3a^2\pi z^2}{r^5} - \frac{\pi a^2}{r^3} + \mathcal{O}\left(\frac{a^3}{r^4}\right).
\end{align*}
Now by writing $\mathbf{r} = y \hat{\mathbf{y}} + z\hat{\mathbf{z}}$ for the the displacement and $\mathbf{m}=I \pi a^2 \hat{\mathbf{z}}$ for the magnetic moment (where $I$ is the current on the loop), the magnetic field $\mathbf{B}$ takes the following asymptotic form
\begin{align*}
\mathbf{B}
&= \frac{\mu_0 I}{4\pi} \left[ \frac{3a^2 \pi y z}{r^5} \hat{\mathbf{y}} + \left( \frac{3a^2\pi z^2}{r^5} - \frac{\pi a^2}{r^3} \right) \hat{\mathbf{z}} + \mathcal{O}\left(\frac{a^3}{r^4}\right) \right] \\
&= \frac{\mu_0}{4\pi} \left[ \frac{3\mathbf{r} (\mathbf{m} \cdot \mathbf{r})}{r^5} - \frac{\mathbf{m}}{r^3} \right] + \mathcal{O}\left(\frac{a^3}{r^4}\right)
\end{align*}
The leading term is exactly the formula described in Wikipedia article.
|
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|
$f(x)$ is a quadratic function such that $f(0) = 1$ and $\int {f(x) \over x^2(x + 1)^3} dx $ is a rational function. Find $f(x)$.
$f(x)$ is a quadratic function such that $f(0) = 1$ and $$\int {f(x) \over x^2(x + 1)^3 } dx $$ is a rational function. Find $f(x)$.
Now first of all we don't even know what the degree of the numerator and denominator of the rational function is, so I suppose differentiating both sides is not an option or will become tedious task and since $f(x)$ is also obviously not given, so we can also not integrate then how to go about solving this question.
|
Let $$f(x)=Z(x+1)(2x+1)~~~~(1)$$ where $Z$ is a constant then $$I=\int \frac{f(x)dx}{x^2(x+1)^3}=Z\int \frac{2x+1}{(x(x+1))^2} dx= \frac{-Z}{x^2+x}+C.$$
Choose $Z=1$ in (1) to have $f(0)=1$.
Explaination: Since $f(0)=1$, $x$ or $x^2$ cannot be ia factor of $f(x)$. Try to have $(x+1)$ as a factor of $f(x)$ so that your integrand reduces. You may let the other factor be $(Kx+L)$. Then $$I=\int\frac{Kx+L}{(x^2+x)^2}dx.$$
Now, $K=2Z, L=Z$ is the obvious choice for $I$ to be algebraic and rational.
Else,
A Hint
You may do it by assuming $f(x)=Ax^2+Bx+C$,
Then by partial fractions $$I=\int\frac{Ax^2+Bx+C}{x^2(x+1)^3}=\int \left(\frac{P}{x}+\frac{Q}{x^2}+\frac{R}{x+1}+\frac{S}{(x+1)^2}+\frac{T}{(x+1)^3}\right) dx$$
For RHS to be rational and Algebraic $P=0=R$, this is to kill $\log x$ and $\log (x+1)$ on the RHS. Then you can choose sich values of $A,B,C$. Also you are given that $C=1$.
|
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|
Maximum value of $(1 − x)(2 − y)^ 2 (x + y)$ Additional Info
$x < 1, y<2, x+y>0$
I tried doing AM-GM (since all of the terms are positive.
$(1 − x)(2 − y)^ 2 (x + y) = (2-2x)(2-y)(2-y)(2x+2y)*1/4$
this way we can cancel the x and y's
thus the maximum value is
$((2-2x+2-y+2-y+2x+2y+1/4)/5)^5$
but the answer is said to be 1(?) where $x =0,$ and $y = 1$
Can someone tell me what I did wrong.
|
In your AM-GM the equality occurs for
$$2-2x=2-y=2x+2y=\frac{1}{4},$$ which is impossible.
I think, it's better to make the following.
By AM-GM
$$(1-x)(2-y)^2(x+y)=4(1-x)\left(1-\frac{y}{2}\right)^2(x+y)\leq$$
$$\leq4\left(\frac{1-x+2\left(1-\frac{y}{2}\right)+x+y}{4}\right)^4=\frac{81}{64}.$$
The equality occurs for $$1-x=1-\frac{y}{2}=x+y$$ or
$$(x,y)=\left(\frac{1}{4},\frac{1}{2}\right),$$ which says that we got a maximal value.
|
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|
Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.
|
Note that $$z=\sqrt{\frac{3 \pm \sqrt{5}}{2}}= \pm \left(\frac{1\pm \sqrt{5}}{2} \right).$$ So $$z_1=2\frac{1+\sqrt{5}}{4} =2 \cos 36^0$$ Hence the other three roots.
|
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|
Find the digits of a number Consider the following statement!
$\overline{ABCD}+\overline{EFG}=8768$, and $\overline{ABC}+\overline{DEFG}=6005$.
If $A,\,B,\,C,\,D,\,E,\,F,$ and $G$ are different numbers, Then, what is $\overline{ABCD}$?
My idea is :
$$\begin{aligned}
&\overline{ABCD}+\overline{EFG}=8768\\
&1000(A)+100(B+E)+10(C+F)+(D+G)=8768\\
&\begin{cases}
&A=8\\
&B+E=7\\
&C+F=6\\
&D+G=8
\end{cases}
\end{aligned}
$$
But i don't think that this is the best idea. Cz when i tried to solve this, i got the wrong numbers that doesn't satisfied the equation.
Please help.
|
When you have $X+Y\le 9+9$ the most you can carry is $1$.
And if you have $X+Y+1\le 9+9+1$ the most you can carry is $1$.
So you can never carry more than one. Now if you have $X+Y\to 0$ that means $X+Y$ or $X+Y + 1=10$ and you do carry $1$ to the next column.
So
$\overline{ABC} + \overline{DEFG} = 6005$
has $A+E\to 0$ so we carry and $D+1 = 6$ and $D=5$.
So
$\overline{ABCD}+\overline{EFG}= \overline{ABC}5+\overline{EFG}=8768$
$5 + G = 8$ or $5+G =18$ but $5+G \le 14$ so $G= 3$.
$\overline{ABC} + 5\overline{EF}3 = 6005$
$C+3 = 15$ or $C+3 = 5$ but $C+3 \le 12$ so $C+3 =5$ so $C=2$.
$\overline{ABCD}+\overline{EFG}= \overline{AB}25+\overline{EF}3=8768$
So $F+2=6$ (obviously not $16$) so $F = 4$
$\overline{AB}2 + 5\overline{E}43 = 6005$
So $B+4=10$ and $B=6$
$\overline{A}625+\overline{E}43=8768$
$6+E = 7$ so $E=1$
And $\overline A625 +143=8768$ and $\overline{A}62 + 5143 = 6005$ so
$A=8$ and $A+1+1 =10$. So $A=8$.
So $\overline{ABCD} = 8625$.
|
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|
Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$ It's a charming problem :
Let $a,b,c>0$ such that $a+b+c=1$ then we have :
$$\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$$
I know the identity :
Let $a,b,c>0$ such that $a+b+c=1$ then we have :
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=1.5$$
But I think it's not relevant here .
I try also majorization with the inequality :
Let $a\geq b\geq c>0$ such that $a+b+c=1$ then we have :
$$\exp\Big(\frac{-2}{3}\Big)a\geq \frac{a}{\exp(a+b)}$$
Second line of the majorization :
$$\exp\Big(\frac{-2}{3}\Big)^2ab\geq \frac{a}{\exp(a+b)}\frac{b}{\exp(b+c)}$$
Third line of the majorization :
$$\exp\Big(\frac{-2}{3}\Big)^3abc\geq \frac{a}{\exp(a+b)}\frac{b}{\exp(b+c)}\frac{c}{\exp(c+a)}$$
The lines are easy to check with the condition remains to apply Karamata's inequality and we are done . Unfortunately the second line fails .
My question :
Have you a proof ?
Thanks a lot for sharing your time and knowledge .
|
Fact 1: $\mathrm{e}^x \le \frac{2}{3}x^2 + x + 1, \quad \forall x \le \frac{2}{3}$.
(The proof is given at the end.)
The desired inequality is written as
$$a\mathrm{e}^{2/3-a-b} + b\mathrm{e}^{2/3-b-c} + c\mathrm{e}^{2/3-c-a} \le 1.$$
Let $f(x) \triangleq \frac{2}{3}x^2 + x + 1$. By Fact 1, it suffices to prove that
$$a f(2/3-a-b) + bf(2/3-b-c) + cf(2/3-c-a) \le 1.$$
With the substitutions $a = \frac{u}{u+v+w}, b=\frac{v}{u+v+w}, c = \frac{w}{u+v+w}$ for $u, v, w>0$, after clearing the denominators, it suffices to prove that
$$7u^3-12u^2v+6u^2w+6uv^2-3uvw-12uw^2+7v^3-12v^2w+6vw^2+7w^3\ge 0$$
or
$$(u^3+v^3+w^3-3uvw) + 6(u^3+v^3+w^3) - 12(u^2v+v^2w+w^2u) + 6(uv^2+vw^2+wu^2) \ge 0$$
which is obvious by using AM-GM (e.g.,$u^3 + uv^2 \ge 2u^2v$). We are done.
$\phantom{2}$
Proof of Fact 1: Let $g(x) = \ln (\frac{2}{3}x^2 + x + 1) - x$. We have $g'(x) = \frac{x(1-2x)}{2x^2+3x+3}$.
Thus, $g(x)$ is strictly decreasing on $(-\infty, 0)$, strictly increasing on $(0, 1/2)$,
strictly decreasing on $(1/2, \infty)$. Note also that $g(0) = 0$ and $g(\frac{2}{3}) > 0$. The desired result follows.
|
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Find $x,y,z$ for the given conditions $$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, which means $a^3+b^3+c^3=3abc$, but didn't get a breakthrough.
I am stuck here, please help me.
|
Just wanted to see. Here is how it looks if you make the Hessian matrix congruent to a diagonal matrix, the one that gives
$$ \frac{1}{4} \left(4x - 5 y - 3 z \right)^2 + \frac{3}{4} \left( 5y - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
is $ Q^T D Q = H $
The theorem is that the form does factor, allowing complex coefficients, if and only if the discriminant is zero. In this case, we see that $\det H = 0.$
As you saw, setting to zero gives us $5y=3z,$ after which $4x = 5y + 5y = 10y,$ giving $2x=5y=3z$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 5 }{ 4 } & 1 & 0 \\
\frac{ 3 }{ 2 } & \frac{ 3 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
8 & - 10 & - 6 \\
- 10 & 50 & - 15 \\
- 6 & - 15 & 18 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 5 }{ 4 } & \frac{ 3 }{ 2 } \\
0 & 1 & \frac{ 3 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
8 & 0 & 0 \\
0 & \frac{ 75 }{ 2 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 5 }{ 4 } & 1 & 0 \\
- \frac{ 3 }{ 4 } & - \frac{ 3 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
8 & 0 & 0 \\
0 & \frac{ 75 }{ 2 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 5 }{ 4 } & - \frac{ 3 }{ 4 } \\
0 & 1 & - \frac{ 3 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
8 & - 10 & - 6 \\
- 10 & 50 & - 15 \\
- 6 & - 15 & 18 \\
\end{array}
\right)
$$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
$$ 4 \left(x - \frac{5}{4} y - \frac{3}{4} z \right)^2 + \frac{75}{4} \left( y - \frac{3}{5} z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
$$ \frac{1}{4} \left(4x - 5 y - 3 z \right)^2 + \frac{3}{4} \left( 5y - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
|
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|
Proof by contradiction algebra Let n be an integer. Prove that if $n^2+2n-1$ is even, then $n$ is odd.
This is what i have tried so far
If $n$ is even then $n = 2k$
And $n^2+2n-1$ is odd
$$(2k)^2+2(2k)-1 = 4k^2+4k-1$$
|
If $n^2 + 2n - 1$ is even, then
$(n + 1)^2 = n^2 + 2n + 1 = (n^2 + 2n - 1) + 2 \tag 1$
is also even; now, the square of an even is even, since
$n = 2k \Longrightarrow n^2 = 4k^2 = 2(2k^2) \Longrightarrow 2 \mid n^2; \tag 2$
likewise, the square of an odd is odd:
$n = 2k + 1 \Longrightarrow n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1; \tag 3$
it follows that $n + 1$ is even, and hence
$n = (n + 1) - 1 \tag 4$
is odd; this may more formally seen by writing
$n + 1 = 2k$
$\Longrightarrow n = 2k - 1 = (2k -2) + 1 = 2(k - 1) + 1, \; \text{odd}. \tag 5$
$OE\Delta$.
The contradiction in the above proof is rather subtly buried, and not explicitly mentioned; it occurs when we infer $n + 1$ is even from $(n + 1)^2$ is even; there we tacitly assume $n + 1$ is odd which, in light of (2) and (3), forces $(n + 1)^2$ odd, contradicting the demonstrated even-ness of $(n + 1)^2$.
|
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|
$1^3+2^3+...+n^3= $? How I can find formula $1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$ . We can verify with ınduction and I know how I can prove it. How we find what is the formula...
I tried like this
$1^3+2^3+...+n^3= An^4+ Bn^3+Cn^2+Dn+E$
after a long process i found it this way
$A=-299/54,$
$B=-4/9,$
$C=107/54,$
$D=4.$
And then I tried how can show that
But I didn't find Maybe The value of a,b,c,d is wrong
Can I get proof of this formula in this way or another way
if we didn't know the formula, how would we get this formula
|
$0=E, 1=A+B+C+D+E, 1^3+2^3=9=16A+8B+4C+2D+E,$
$ 1^3+2^3+3^3=36=81A+27B+9C+3D+E, $
and $1^3+2^3+3^3+4^3=100=256A+64B+16C+4D+E \implies$
$ E=0, 7=14A+6B+2C, 33=78A+24B+6C, $ and $96=252A+60B+12C \implies$
$12=36A+6B$ and $54=168A+24B\implies$
$24A=6\implies A=\dfrac14$. Can you solve now for $B, C, $ and $D$?
|
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|
Algebraic proof of a combinatoric question (Combinatoric proof is given) I had a IMO training about double counting. Then, there is a problem which I hope there is a combinatoric proof. Here comes the problem:
For every positive integer $n$, let $f\left(n\right)$ be the number of all positive integers with exactly $2n$ digits, each having exactly $n$ of digits equal to $1$ and the other equal to $2$. Let $g\left(n\right)$ be the number of all positive integers with exactly $n$ digits, each of its digits can only be $1,2,3$ or $4$ and the number of $1$'s equals the number of $2$'s. Prove that $f\left(n\right)=g\left(n\right)$.
It is obvious to see that $f\left(n\right)=\binom{2n}{n}$, and $g\left(n\right)=\sum_{k\le\lfloor\frac{n}{2}\rfloor}\binom{n}{2k}\binom{2k}{k}2^{n-2k}$. However, it is hard to prove this in an algebraic way. I hope there are someone to prove it by algebraic way. Thank you!
Combinatoric proof
We can establish a one-to-one correspondence between $f\left(n\right)$ and $g\left(n\right)$. Let $F\left(n\right)$ be the set of all positive integers with exactly $2n$ digits, each having exactly $n$ of digits equal to $1$ and the other equal to $2$. Also, let $G\left(n\right)$ be the set of all positive integers with exactly $n$ digits, each of its digits can only be $1,2,3$ or $4$ and the number of $1$'s equals the number of $2$'s. Then, we can do this operation for all numbers in $F\left(n\right)$: For every two digits of the numbers in $F\left(n\right)$, $$\begin{cases}11\Rightarrow 1\\22\Rightarrow 2\\12\Rightarrow 3\\21\Rightarrow 4\end{cases}$$ Then all the numbers will change into a set which is totally same as $G\left(n\right)$, as we find that the difference between the number of $1$'s and $2$'s doesn't change at all. Therefore, we make a one-to-one correspondence between $F\left(n\right)$ and $G\left(n\right)$.
|
Here is the generating function approach:
\begin{align}
\sum_{n=0}^\infty \left(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\binom{2k}{k} 2^{n-2k}\right) z^n
&= \sum_{k=0}^\infty \binom{2k}{k} 2^{-2k} \sum_{n=2k}^\infty \binom{n}{2k}(2z)^n \\
&= \sum_{k=0}^\infty \binom{2k}{k} 2^{-2k} \frac{(2z)^{2k}}{(1-2z)^{2k+1}} \\
&= \frac{1}{1-2z} \sum_{k=0}^\infty \binom{2k}{k} \left[\left(\frac{z}{1-2z}\right)^2\right]^k \\
&= \frac{1}{1-2z} \cdot \frac{1}{\sqrt{1-4(z/(1-2z))^2}} \\
&= \frac{1}{\sqrt{1-4z}} \\
&= \sum_{n=0}^\infty \binom{2n}{n}z^n.
\end{align}
Hence $$\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\binom{2k}{k} 2^{n-2k} = \binom{2n}{n}$$ for all $n$.
|
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|
Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
|
Here is a systematic procedure. Let $\tan t=x$,
$$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx=\int_0^{\pi/4}\tan^4t(1-\tan t)^4dt$$
$$=I(8) -4I(7)+6I(6)-4I(5)+I(4)\tag{1}$$
where $I(n)=\int_0^{\pi/4}\tan^nt\>dt$ and it has the recursive relationship $I(n)=\frac{1}{n-1}-I(n-2)$.
Now, use the recursive equation to get
$$4I(7)+4I(5)=\frac 23$$
$$I(8)+6I(6)+I(4)=\frac87-4I(4)$$
$$I(4) = -\frac23 +\frac\pi4$$
Then, the integral evaluates to,
$$I=\frac{8}{7}-4(-\frac23 +\frac\pi4) -\frac{2}{3}= \frac{22}{7}-\pi$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
Hence,
\begin{equation*}
\lim_{n\to\infty} \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1} = \frac{0+\infty}{0+1} = \infty.
\end{equation*}
Thus, $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ diverges.
Is this correct? Thanks.
|
Yes. :)
|
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|
Given 2 red beads, 2 blue beads, 1 yellow bead, and 1 green bead, how many different necklaces can be made? I need to write a Matlab code to determine the answer (which was given as 16) and I need to utilize loops to remove flips and circular shifts
|
We use the Polya Enumeration Theorem (PET).
The cycle index $Z(D_6)$ of the dihedral group $D_6$ is given by
$$Z(D_6) = \frac{1}{12}
\left(\sum_{d|6} \varphi(d) a_d^{6/d}
+ 3 a_2^3 + 3 a_1^2 a_2^2\right).$$
This is
$$\frac{1}{12}
\left(a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6
+ 3 a_2^3 + 3 a_1^2 a_2^2\right)
\\ = \frac{1}{12}
\left(a_1^6 + 2 a_3^2 + 2 a_6
+ 4 a_2^3 + 3 a_1^2 a_2^2\right).$$
We seek
$$[R^2 B^2 Y G] Z(D_6; R+B+Y+G).$$
The only contribution comes from
$$\frac{1}{12}
\left(a_1^6 + 3 a_1^2 a_2^2\right).$$
This is because with the Polya substitution the terms $2 a_3^2 + 2 a_6
+ 4 a_2^3$ produce powers of three, six, and two, exclusively.
Continuing, we get
$$\frac{1}{12}
[R^2 B^2 Y G] (R+B+Y+G)^6
\\ + \frac{1}{4} [R^2 B^2 Y G]
(R+B+Y+G)^2 (R^2+B^2+Y^2+G^2)^2.$$
This is
$$\frac{1}{12} {6\choose 2,2,1,1}
+ \frac{1}{4} \times 2 \times 2 = 16.$$
BTW the convention at the OEIS is to use the term necklace for
cyclic symmetries and bracelet for dihedral ones.
|
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|
Integrate $\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$ for $\epsilon>0$ If I can solve this integral below in analytical form, then I will be able to provide the time dependant orbit for a two-body problem,
$$\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$$
for $\epsilon>0$.
|
Observe that,
$$\left(\frac{\sin\theta}{1+\epsilon\cos{\theta}} \right)'=
\frac{\cos\theta+\epsilon}{(1+\epsilon\cos{\theta})^2}
= \frac1\epsilon\left[\frac{1}{1+\epsilon\cos{\theta}}+\frac{\epsilon^2-1}{(1+\epsilon\cos{\theta})^2} \right]$$
and decompose the integral
$$I=\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$$
$$=\frac{1}{\epsilon^2-1}\left[\int_0^x d\left(\frac{\epsilon\sin\theta}{1+\epsilon\cos{\theta}}\right)-\int_0^x \frac{d\theta}{1+\epsilon\cos{\theta}}\right]$$
$$=\frac{\epsilon}{\epsilon^2-1}\frac{\sin x}{1+\epsilon\cos{x}}-\frac{1}{\epsilon^2-1}\int_0^x \frac{d\theta}{1+\epsilon\cos{\theta}}$$
The second integral is better known
$$\int_0^x \frac{d\theta}{1+\epsilon\cos{\theta}}
=\int_0^x \frac{ 2d(\tan\frac{\theta}{2}) }{(1-\epsilon)\tan^2\frac{\theta}{2}+(1+\epsilon)} = \frac{2}{\sqrt{1-\epsilon^2}} \tan^{-1} \left(\sqrt{\frac{1-\epsilon}{1+\epsilon}} \tan\frac x2 \right) $$
Thus,
$$I=-\frac{\epsilon}{1-\epsilon^2}\frac{\sin x}{1+\epsilon\cos{x}}+\frac{2}{{(1-\epsilon^2})^{3/2}} \tan^{-1} \left(\sqrt{\frac{1-\epsilon}{1+\epsilon}} \tan\frac x2 \right) $$
|
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|
Trying to solve a recurrence relation by using generating functions: $a_n=3a_{n-1} + a_{n-2}$ I'm trying to solve the recurrence relation below by using generating function:
\begin{equation}
a_n=\begin{cases}
0, & \text{if $n<0$}\\
2, & \text{if $n=0$}\\
1, & \text{if $n=1$}\\
3a_{n-1} + a_{n-2}, & \text{otherwise}.
\end{cases}
\end{equation}
The first thing I did was make the recurrence relation valid for all $n$ by using a kronecker delta:
$a_0 = 3.(0) + 0 + 2.(\delta_{n,0}) = 2$
$a_1 = 3.(2) + 0 - 5.(\delta_{n,1}) = 1$
The result I got was:
$$a_n = 3a_{n-1} + a_{n-2} + 2\delta_{n,0} - 5\delta_{n,1}$$
Multiplying by $x^n$:
$$a_n . x^n = 3a_{n-1} . x^n + a_{n-2} . x^n + 2\delta_{n,0} . x^n - 5\delta_{n,1} . x^n$$
Summing up both sides:
$$\sum_{n\geq0} a_n . x^n = \sum_{n\geq0}3a_{n-1} . x^n + \sum_{n\geq0}a_{n-2} . x^n + \sum_{n\geq0}2\delta_{n,0} . x^n - \sum_{n\geq0}5\delta_{n,1} . x^n$$
And making $F(x) = \sum_{n\geq0} a_n . x^n$, I got:
$$F(x) = 3xF(x) + x^2F(x) + 2 - 5x$$
which is:
$$F(x) = \frac{2 - 5x}{1-3x-x^2}$$
So far so good but from here on I can't find a way to calculate the $a_n$
I've heard it has something to do with partials fractions but I'm a newbie in this subject and I have no idea how to follow through.
Does anyone can help me to finish the calculation?
Thanks in advance.
|
I'm going to rewrite the problem as $$a_{n+2} = 3a_{n+1} + a_n; \; a_0 = 2,\; a_1=1.$$ Multiply by $x^n$, sum, and let $A(x) = \sum_{n\ge 0}a_nx^n$. So,
$$\sum_{n\ge 0}a_{n+2}x^n = 3\sum_{n\ge 0}a_{n+1}x^n + \sum_{n\ge 0}a_n x^n,$$
and we can see that $\sum_{n\ge 0}a_{n+2}x^n = a_2 + a_3x + \cdots = (1/x^2)(A(x)-a_0-a_1x)$ and similarly $\sum_{n\ge 0}a_{n+1}x^n = a_1 + a_2x + \cdots = (1/x)(A(x)-a_0)$, so we obtain
$$\frac{1}{x^2}(A(x) - a_0 - a_1x) = \frac{3}{x} (A(x)-a_0 ) + A(x).$$
Substituting for $a_0$ and $a_1$ and solving for $A(x)$ yields your $F(x)$, ie $$A(x) = \frac{2-5x}{1-3x-x^2}.$$
Now the for the partial fraction decomposition, which is not so ugly if we keep our heads straight. Note that $1-3x-x^2$ has roots $\alpha_1 = -\frac{3}{2} - \frac{\sqrt{13}}{2}$ and $\alpha_2 = -\frac{3}{2} + \frac{\sqrt{13}}{2}$ and we want $$\frac{1}{1-3x-x^2} = \frac{1}{(x-\alpha_1)(x-\alpha_2)} = \frac{A}{(x-\alpha_1)} + \frac{B}{(x-\alpha_2)}.$$ Using this equation and solving for $A$ and $B$ yields
\begin{align}
\frac{1}{1-3x-x^2} &= \frac{1}{(\alpha_1-\alpha_2)(x-\alpha_1)} + \frac{1}{(\alpha_2-\alpha_1)(x-\alpha_2)}\\
&=\frac{1}{-\sqrt{13}(x-\alpha_1)} + \frac{1}{\sqrt{13}(x-\alpha_2)}
\end{align}
So we have
\begin{align}
A(x) &= (2-5x)\left(\frac{1}{-\sqrt{13}(x-\alpha_1)} + \frac{1}{\sqrt{13}(x-\alpha_2)} \right)\\
&=(2-5x) \left( \frac{1}{\alpha_1\sqrt{13}} \cdot \frac{1}{1-(x/\alpha_1)} + \frac{1}{-\alpha_2\sqrt{13}} \cdot \frac{1}{1-(x/\alpha_2)} \right)\\
&= (2-5x) \left( \frac{1}{\alpha_1\sqrt{13}} \sum_{n\ge 0} \left(\frac{1}{\alpha_1} \right)^n x^n + \frac{1}{-\alpha_2\sqrt{13}} \sum_{n\ge 0} \left(\frac{1}{\alpha_2} \right)^n x^n \right)\\
&= \frac{(2-5x)}{\sqrt{13}} \left(\sum_{n\ge 0} \left[ \left(\frac{1}{\alpha_1} \right)^{n+1} - \left(\frac{1}{\alpha_2} \right)^{n+1}\right] x^n \right)
\end{align}
Can you take it from here?
|
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|
If $a$, $b$, $c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are all integers, then $|a|=|b|=|c|$
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.
Well this is what I have done so far:
From the fact that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer, we get
$$abc \mid ab^2 + bc^2 + ca^2 \tag{1}$$
In the same way, we also have
$$abc \mid a^2b + b^2c + c^2a \tag{2}$$
so
$$\begin{align}
abc \mid a(ab^2 + bc^2 + ca^2) - b(a^2b + b^2c + c^2a) &\Rightarrow abc \mid c(a^3-b^3) \tag{3}\\
&\Rightarrow ab \mid a^3-b^3 \tag{4}\\
&\Rightarrow ab \mid b^3(a^3-b^3) \tag{5}
\end{align}$$ and so $a \mid b^6$. In the same way, we can also get $b \mid c^6$ and $c \mid a^6$.
But what should I do after this?
Any help is surely appreciated! Thanks!
|
I think what you've actually shown is that $a \mid b^5$, $b \mid c^5$ and $c \mid a^5$.
From this
you can argue as follows. Suppose $p$ is a prime factor of $a$. Then $a \mid b^5$ implies that
$p \mid b$, and similarly $b \mid c^5$ now implies that $p \mid c$.
Thus $p$ divides all of $a$, $b$, and $c$. Similarly, any prime factor of $b$ or $c$ divides all of $a$, $b$, and $c$.
So in the above situation, you can replace $a$ by $A = a/p$, $b$ by $B = b/p$, and $c$ by $C = c/p$ for any prime factor of $a,b,$ or $c$, and $A$, $B$, and $C$ will satisfy the conditions of the original problem.
Now repeat the above as many times as needed, as long as there are any prime factors left. In the end you are left with your numbers being $\pm 1$. Looking backwards this means $|a| = |b| = |c|$ as desired.
|
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|
$f(x+1) + f(x-1) = x^2$ ; $f(x+4) + f(x-4) = 2\sin x$ , then $f(x) =$? Given $f$ is a complex valued function satisfying $$f(x+1) + f(x-1) = x^2 \\
f(x+4) + f(x-4) = 2\sin x$$
what is $f(x)$ ?
Here, only for the first part MathWolfram alpha is showing $f(x)$ to be of type $c_1(i)^x + c_2(-i)^x + \dfrac{x^2 - 1}{2}$ but how are they introducing "$i$"? I used a generating function but was still unable to get those terms containing $i$ . Please help me with this problem.
|
The two conditions can be rewritten as
$$f(x+2)+f(x)=(x+1)^2$$
$$f(x+8)+f(x)=2\sin(x+4)$$
Iterating the first one gives
$$f(x+4)+f(x+2)=(x+3)^2$$
$$f(x+6)+f(x+4)=(x+5)^2$$
$$f(x+8)+f(x+6)=(x+7)^2$$
It follows
\begin{align}
2\sin(x+4) &= f(x+8)+f(x)\\
&= (x+7)^2-f(x+6)+f(x)\\
&= (x+7)^2-(x+5)^2+f(x+4)+f(x)\\
&= (x+7)^2-(x+5)^2+(x+3)^2-f(x+2)+f(x)\\
&= (x+7)^2-(x+5)^2+(x+3)^2-(x+1)^2+f(x)+f(x)\\
\end{align}
so
\begin{align}
f(x) &= \frac{2\sin(x+4)+(x+1)^2-(x+3)^2+(x+5)^2-(x+7)^2}{2}\\
&= \sin(x+4)-4x-16
\end{align}
|
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|
Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=0}^n \dfrac{\sqrt{n}}{n+k^2}(n=1,2,\cdots)$ I tried to change it into a Riemann sum but failed, since
\begin{align*}
\lim_{n \to \infty}\sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^n \frac{\sqrt{n}}{1+(k/\sqrt{n})^2}
,\end{align*}
which is not a standard form. Maybe, it need apply the squeeze theorem, but how to evaluate the bound.
By the way, WA gives its result
\begin{align*}
\lim_{n \to \infty}\sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\frac{\pi}{2}.
\end{align*}
|
Integral bounds do the job: $$\int_k^{k+1} \frac{1}{n+t^2} dt \leq \frac{1}{n+k^2}\leq \int_{k-1}^{k} \frac{1}{n+t^2} dt$$
Summing the left-hand side inequalities for $k\in \{0,\ldots,n\}$ and the right-hand side inequalities for $k\in \{1,\ldots,n\}$ yields
$$\int_0^{n+1}\frac{1}{n+t^2} dt \leq \sum_{k=0}^n \frac{1}{n+k^2} \leq \frac 1n + \int_{0}^{n} \frac{1}{n+t^2}$$
that is
$$\frac{1}{\sqrt{n}}\arctan\left(\frac{n+1}{\sqrt n}\right)\leq \sum_{k=0}^n \frac{1}{n+k^2} \leq \frac 1n + \frac{1}{\sqrt{n}}\arctan\left(\frac{n}{\sqrt n}\right)$$
Multiplying by $\sqrt n$ and squeezing yields
$$\lim_n \sqrt n \sum_{k=0}^n \frac{1}{n+k^2} = \lim_n \arctan\left(\frac{n}{\sqrt n}\right) = \frac \pi 2$$
|
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|
Find $n$ for a $\sum_{x=1}^n \left[(x+1)^3-x^3\right]$ $$ \sum_{x=1}^n \left[(x+1)^3-x^3\right]$$
This is my sum, I tried simplfifying and got $3x^2+3x+1$ but Im stuck on how to resolve the sum for $n$.
|
Note that
$$\sum_{x=1}^n \left[(x+1)^3-x^3\right] =\color{red}{2^3}-1^2+\color{red}{3^3}-\color{red}{2^3}+\ldots+\color{red}{n^3}-\color{red}{(n-1)^3}+(n+1)^3-\color{red}{n^3}$$
|
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|
What is the coefficient of x^2 in the expansion of (x+2)^4(x+3)^5 I'm missing something here.
I've calculated the $x^2$ coefficient of $(x+2)^4$ as 24 with constant term 16. And $x^2$ term coefficient of $(x+3)^5$ as 270 with constant term 243.
if I'm correct here then the answer should be $(16*270)+(24*243)$? but this does not appear to be the case.
any assistance appreciated.
|
Using the binomial theorem we have:
$$(x+2)^4(x+3)^5 = \left( \sum_{k=0}^{4}\binom{4}{k}x^k2^{4-k}\right) \left(\sum_{j=0}^{5}\binom{5}{j}x^j3^{5-j} \right).$$
We get $x^2$ by taking an $x$ term from each sum or an $x^2$ and $x^0$ term from one or the other sum. So the coefficient of $x^2$ is $$\overbrace{\binom{4}{1}2^3\binom{5}{1}3^4}^{x\cdot x} + \overbrace{2^4\binom{5}{2}3^3}^{x^0\cdot x^2} + \overbrace{\binom{4}{2}2^23^5}^{x^2\cdot x^0}$$
|
{
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|
Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $ Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $
My attempt is as follows:-
$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$
$$\left(\sin\theta-\cos\theta\right)^2+\sin\theta\left(\sin\theta-\cos\theta\right)=0$$
$$\left(\sin\theta-\cos\theta\right)(2\sin\theta-\cos\theta)=0$$
$$\tan\theta=1 \text { or } \tan\theta=\frac{1}{2}$$
$$\theta=n\pi+\frac{\pi}{4}$$
For $\tan\theta=\dfrac{1}{2} \text { how to make use of given condition } \tan71 ^{\circ}34^{\prime}=3$
|
Use $3=\frac{1+\frac12}{1-1\cdot\frac12}=\tan(45^\circ+\theta)$.
|
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|
How to show this inequality for complex numbers? Let $|y|<|z|$ then,
$$\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|\leq K |z|^{-1}|y|$$
for some constant $K.$
I want to use some Taylor expansion argument, but I am not sure if it would work for complex numbers. Any ideas will be much appreciated.
Edit: The expression should be equivalent to the one below:
$$\begin{align}
\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|&= \left| \frac{y/z+1}{|y/z+1|} - 1\right| \\
&\leq |1-y^2/z^2-1|\leq |z|^{-2}|y|^2\leq |z|^{-1}|y|
\end{align}$$
|
Let $y=\rho e^{i\phi}$ and $z=re^{i\theta}$. Then $|y|<|z|\implies\rho<r$ and $|z|^{-1}|y|=\rho/r$. Since \begin{align}|y+z|&=|(\rho\cos\phi+r\cos\theta)+i(\rho\sin\phi+r\sin\theta)|\\&=\sqrt{\rho^2\cos^2\phi+2\rho r\cos\theta\cos\phi+r^2\cos^2\theta+\rho^2\sin^2\phi+2\rho r\sin\theta\sin\phi+r^2\sin^2\theta}\\&=\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}\end{align} we have \begin{align}\small\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|&=\small\left|\frac{\rho\cos\phi+r\cos\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}+i\frac{\rho\sin\phi+r\sin\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}-\frac{re^{i\theta}}r\right|\\&=\small\sqrt{\left(\frac{\rho\cos\phi+r\cos\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}-\cos\theta\right)^2+\left(\frac{\rho\sin\phi+r\sin\theta}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}-\sin\theta\right)^2}\\&=\small\sqrt{\frac{(\rho\cos\phi+r\cos\theta)^2+(\rho\sin\phi+r\sin\theta)^2}{\rho^2+2\rho r\cos(\theta-\phi)+r^2}-\frac{2(\cos\theta(\rho\cos\phi+r\cos\theta)+\sin\theta(\rho\sin\phi+r\sin\theta))}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}+1}\\&=\sqrt{2-\frac{2(\rho\cos(\theta-\phi)+r)}{\sqrt{\rho^2+2\rho r\cos(\theta-\phi)+r^2}}}.\end{align} It can be shown that for $a,b,c,d\in\Bbb R^+$, the function $$f(x)=\frac{ax+b}{\sqrt{c+dx}}\implies\min f(x)=\frac{2\sqrt{a(bd-ac)}}d$$ where it exists at $x=(bd-2ac)/ad$. Substituting $a=2\rho$, $b=2r$, $c=\rho^2+r^2$, $d=2\rho r$ and $x=\cos(\theta-\phi)$, this gives \begin{align}\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|&\le\sqrt{2-\frac{\sqrt{2\rho(4\rho r^2-2\rho^3-2\rho r^2)}}{2\rho r}}=\sqrt{2-2\sqrt{1-\left(\frac\rho r\right)^2}}\le\sqrt2\cdot\frac\rho r.\end{align} The result follows with $K=\sqrt2$.
|
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|
using epsilon delta definition to proof limits
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
let $$f(x)=\frac{2x+3}{x-1}$$ then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_f \left(
0 < \left|x-1 \right|=x-1<\delta\Longrightarrow \large\left|\frac{2x+3}{x-1}\right|\right)>M$
$$M<\left|\frac{2x+3}{x-1}\right|=\frac{\left|2x+3\right|}{x-1}$$
take $\delta\le1$ implies:$$M<\frac{\left|2x+3\right|}{x-1}<\frac{7}{x-1}$$$$x-1<\frac{7}{M}$$
hence $$\delta\le\min\left\{1,\left(\frac{7}{M}\right)\right\}$$
is it true?
$$\lim_{\large x \to 1} \frac{\left(-1\right)^{\left[x\right]}}{x-1}=-∞$$
let $$g(x)=\frac{\left(-1\right)^{\left[x\right]}}{x-1}$$ ($\left[x\right]$ is floor function )
then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_g \left(
0 < \left|x-1 \right|<\delta\Longrightarrow \large\frac{\large \left(-1\right)^{\left[\large x\right]}}{x-1} <-M \right)$
take $\delta\le1$ implies:$0<x<2$, here is the problem, what is exactly $\left[x\right]$?, I tried another $\delta$ less than $1$, but still I have the problem.
|
The computations you show here are usefull, but they aren't what should appear in your proof, eg let's prove
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
We have to show that $\forall M>0, \ \exists \delta > 0, \ (x > 1 \textrm{ and } x-1 < \delta) \Longrightarrow \frac{2x+3}{x-1} >M$.
Take $M>0$. Your computation suggests that we can take $\delta_M := \min \{1;\frac{7}{M}\}$.
Indeed, if $1<x<1+\delta_M$, then $\frac{2x+3}{x-1} > (2x+3) \cdot \frac{M}{7}$, and knowing that $0<2x+3<7$ isn't usefull. What would be usefull here is knowing that $2x+3$ is bigger than 7, but that isn't the case. I suggest that you take $\delta_M := \min \{1;\frac{5}{M}\}$ since $2x+3 > 5$.
The floor function is defined as follow : $\lfloor x \rfloor$ is the biggest integer smaller or equal than x. In particular,
$$\lfloor x \rfloor = \left\{ \begin{matrix} 0 & \textrm{ if } 0\leqslant x <1 \\ 1 & \textrm{ if } 1\leqslant x < 2 \end{matrix} \right.$$
For the second limit, you'll hence have to distinguish the case where $0\leqslant x<1$ and the case where $1 \leqslant x < 2$. Anyway, taking $\delta_M := \frac{1}{M}$ works for any $M>1$:
Take $x$ such that $|x-1|<\frac{1}{M}$. Then $0<x<2$. Distinguish two cases :
*
*$1 < x <2$, then $(-1)^{\lfloor x \rfloor} = -1$ and $\frac{1}{M}>x-1>0$ hence $\frac{(-1)^{\lfloor x \rfloor}}{x-1} < -M$
*$0 < x <1$, then $(-1)^{\lfloor x \rfloor} = 1$ and $-\frac{1}{M}<x-1<0$ hence $\frac{(-1)^{\lfloor x \rfloor}}{x-1} < -M$
|
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|
Cosine of a $2 \times 2$ non-diagonalisable matrix Given
$$A = \begin{bmatrix}
\pi-1 & 1\\
-1 & \pi+1
\end{bmatrix}$$
I need to calculate its cosine, $\cos(A)$. Typically, I use diagonalisation to approach this type of problems:
$$\cos(A) = P \cos(D) P^{-1}$$
However, in this problem, the eigenvalues of the matrix are equal: $\lambda_1=\pi, \lambda_2=\pi$. Thus, there are no two linearly independent vectors and the method will not work.
Are there any alternative approaches?
Besides MacLaurin series expansion of $\cos(A)$, which does not work either since $A$ does not turn into a zero matrix at some point when multiplied by itself $n$ times.
|
When you do the Jordan decomposition, you get $A = SJS^{-1}$ with
$$
S = \begin{pmatrix}1&1\\1&2\end{pmatrix}\quad\text{and}\quad J = \begin{pmatrix}\pi&1\\0&\pi\end{pmatrix}.
$$
You find that
$$
J^n = \begin{pmatrix}\pi^n&n\pi^{n-1}\\0&\pi^n\end{pmatrix}.
$$
Since $A^n = SJ^nS^{-1}$, this implies
\begin{align}
\cos(A)
&= \sum_{n=0}^\infty \frac{(-1)^nA^{2n}}{(2n)!} = S\left(\sum_{n=0}^\infty \frac{(-1)^nJ^{2n}}{(2n)!}\right)S^{-1}\\
&= S\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\begin{pmatrix}\pi^{2n}&2n\pi^{2n-1}\\0&\pi^{2n}\end{pmatrix}\right)S^{-1}\\
&= S\begin{pmatrix}\cos(\pi)&\sum_{n=0}^\infty\frac{2n\pi^{2n-1}}{(2n)!}\\0&\cos(\pi)\end{pmatrix}S^{-1} = S\begin{pmatrix}-1&\sum_{n=1}^\infty\frac{(-1)^n\pi^{2n-1}}{(2n-1)!}\\0&-1\end{pmatrix}S^{-1}\\
&= S\begin{pmatrix}-1&-\sin(\pi)\\0&-1\end{pmatrix}S^{-1} = -I_2.
\end{align}
|
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|
Solve expression with unknown variables I am stuck in this next problem:
Given $x-y = 7$ and $xy = 5$, without solving for the values of $x$ and $y$, evaluate $x^2 + y^2$.
How can I solve this?
|
First expand $(x-y)^2$ to give
\begin{align}(x-y)^2&=x^2+y^2-2xy\end{align}
We can then substitute in the values we have been given and solve the resulting equation
\begin{align}7^2&=x^2+y^2-2\times 5\\
49&=x^2+y^2-10\\
x^2+y^2&=59\end{align}
Most of the time questions like this involve you spotting an expansion of one term which involves the other term, in this case we note that squaring $x-y$ will give us an $xy$ term and some $x^2$ and $y^2$ terms so this is probably a good place to start looking.
|
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|
Computing double integral $\ \iint \sqrt{4-r^2}r \ dr d\theta $ I am trying to solve the following integral:
$$\ 2 \int_0^\pi \int_0^{2\sin\theta} \sqrt{4-r^2} \ r \ dr d\theta $$
My attempt:
$$\ 2 \int_0^\pi \int_0^{2\sin\theta} \sqrt{4-r^2} \ r \ dr \ d\theta \stackrel{t = 4-r^2}{=} -\int_0^\pi \int t^{\frac{1}{2}} \ dt d\theta = -\frac{2}{3} \int_0^\pi t^{\frac{3}{2}} \ d\theta = -\frac{2}{3} \int_0^\pi \left[(4-r^2)^{3/2}\right]_0^{2\sin\theta} \ d \theta= -\frac{2}{3} \int_0^\pi (4 - 4 \sin^2\theta)^{3/2} - 8 \ d\theta = - \frac{2}{3}\int_0^\pi 8(1-\sin^2\theta)^{3/2} - 8 \ d \theta $$
because $\ (1-\sin^2\theta) = \cos^2\theta$ and $\ (\cos^2\theta)^{3/2} = \cos^3\theta $
$$\ = -\frac{16}{3} \int_0^\pi \cos^3\theta - 1 \ d\theta = \frac{16\pi}{3} $$
but I'm wrong somewhere along the way because the answer should be $\ \frac{16}{9} (3\pi -4) $ and with the limits I set I get the right answer using Wolfram integral calculator.
|
The error is due to assuming $(x^2)^{3/2}=x^3$ as opposed to $|x^3|$. This can be solved by noticing that $\cos^3x=|\cos^3x|$ for $x\in[0,\pi/2]$ and $-\cos^3x=|\cos^3x|$ for $x\in[\pi/2,\pi]$. Thus the integral should become $$2\times\left(-\frac{16}{3} \int_0^{\pi/2} \cos^3\theta - 1 \ d\theta\right)=\frac{16(3\pi-4)}9.$$
|
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|
Inverse of polynomial over a finite field The question is the following:
Can you deduce if ${2x + 1}$ is invertible in $\mathbb{Z}_3[x]/(x^2 + 2x + 2)$? In case of a positive answer, give its inverse.
Following the Wilson's theorem, for $K[X]/(f)$, any polynomial of degree $1 \leq deg < n$ will admit an inverse of degreee $1 \leq deg \leq n$ mod $f$. Since $n = 3$ because I'm working on $\mathbb{Z}_3$ and $deg = 2$, I guess $2x+1$ is irreductible in this case.
Once I want to find the inverse I've been following this post. The table using Euclidean algorithm is the following:
\begin{array}{r|r|r|r}
& & (x+\frac{3}{2})/2 & (2x+3)/5 \\\hline
1 & 0 & 1 & -(2x+3)/5\\\hline
0 & 1 & -\big(\dfrac{x}{2} + \dfrac{3}{4}\big) & \dfrac{x^2}{5} + \dfrac{3x}{5} + \dfrac{29}{20}\\\hline
x^2+2x+2 & 2x+1 & 5/4 & 0
\end{array}
So, finding the lineal combination I obtain this result:
$\dfrac{5}{4} = 1\times(x^2 +2x +2) - \big(\dfrac{x}{2}+\dfrac{3}{4}\big)\times(2x+1)\xrightarrow{}\dfrac{5}{4} = f(x) - \big(\dfrac{x}{2}+\dfrac{3}{4}\big)g(x)$,
Here is where I get stuck. I let 1 on the left side so I have the following result:
$1 = \bigg(\dfrac{4}{5}\bigg)(x^2+2x+2)- \bigg(\dfrac{2x}{5}+\dfrac{3}{5}\bigg)(2x+1)$
But once I arrive here I don't know how to get the value of the inverse.
Can anyone help me? Thank you very much.
My result, which I'm not absolutely sure is that the inverse of $2x+1$ is:
$-\big(\dfrac{2x}{5}+\dfrac{3}{5}\big)$ mod $x^2+2x+2$.
In the remote case it's correct, is there any way to get the final value instead of letting it in function of mod $f$?
Thank you again.
Bernat
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Note that $-\frac15 = 1$ in $\Bbb Z_3$, and $3 = 0$, so your inverse candidate is equal to $2x$. And we can just check whether this is in fact an inverse, using that $3 = 0$ and $x^2 = x+1$:
$$
2x\cdot (2x+1) = 4x^2 + 2x = 4(x+1) + 2x\\
= 6x + 4 = 1
$$
So yes, that is indeed the inverse you're looking for.
|
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Is my Proof for the following problem correct? Prove that $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$ If $a,b,c$ are positive real numbers prove that $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
I state the following:
Multiplying both sides by $abc$ yields $$(abc)(a+b+c)?{a^4+b^4+c^4}$$
By AM-GM for $abc$ we have $$abc\leq\left(\frac{a+b+c}{3}\right)^3\Rightarrow\left(\frac{a+b+c}{3}\right)^3(a+b+c)?{a^4+b^4+c^4}$$
This can be rewritten as the following:
$$\left(\frac{a+b+c}{3}\right)^4\leq\frac{a^4+b^4+c^4}{3}$$
which holds due to the power mean inequality.
I was also thinking that it may be proven by a double application of the Chebyshev inequality as well
To begin, hold that ${a}\ge{b}\ge{c}$ observe that $$\left[\begin{matrix} a^3 & b^3 & c^3 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{matrix}\right]=a^2+b^2+c^2\leq\left[\begin{matrix} a^3 & b^3 & c^3 \\ \frac{1}{b} & \frac{1}{c} & \frac{1}{a}\end{matrix}\right]=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}$$
Thus we can hold confidently even if we assume equality that $$\left[\begin{matrix} {a^2} & {b^2} & c^2 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{matrix}\right]=a+b+c\leq\left[\begin{matrix} \frac{a^3}
{b} & \frac{b^3}{c} & \frac{a^3}{a} \\ \frac{1}{c} & \frac{1}{a} & \frac{1}{b}\end{matrix}\right]=\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
Please provide feedback if you can.
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Using the inequality $$x^2+y^2+z^2\geq xy+yz+zx$$ twice we get
$$a^4+b^4+c^4\geq (ab)^2+(bc)^2+(ca)^2\geq abbc+abca+bcca=a^2bc+ab^2c+abc^2=abc(a+b+c)$$
|
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|
Prove that if p and q are both prime numbers, with p > q > 2, then $p^4− q^4$ is divisible by 16 Prove that if p and q are both prime numbers, with p > q > 2, then $p^4 − q^4$ is divisible by 16.
This is my attempt so far:
Since p and q are both prime numbers greater than 2, then they must be odd and hence can be written:
$p=2m+1$, for some integer n,
And $q=2n+1$, for some integer m.
Consider, $p^4-q^4$ $\implies$ $(p^2+q^2)(p+q)(p-q)$
$p-q=2m+1-(2n+1)=2(m-n),$ hence is a multiple of 2.
$p+q=2m+1+2n+1=2(m+n+1),$ hence is a multiple of 2.
$p^2+q^2=(2m+1)^2+(2n+1)^2=$
$2(2m^2+2m+2n^2+2n+1),$ hence is a multiple of 2.
However, this is where I am stuck. Doesn’t this just prove the statement is a multiple of 8, not 16? Could anyone help with this proof, or maybe suggest an alternative method. Thanks!
|
You're off to a great start.
Rewrite $$p^4-q^4=(p^2+q^2)(p+q)(p-q).$$
Regarding $p^2+q^2$:
Since $p$ and $q$ are both odd, they can be written as $p=2m+1$ and $q=2n+1$. Then we have
\begin{align}
p^2+q^2&=(2m+1)^2+(2n+1)^2\\
&=4m^2+4m+1+4n^2+4n+1\\
&=2(2m^2+2m+2n^2+2n+1)
\end{align}
This tells us that $2|p^2+q^2$ (and is exactly what you have).
Regarding $(p+q)(p-q)$:
Odd numbers are either $1\pmod 4$ or $3\pmod 4$. We have three cases:
Case 1: let $p\equiv1\pmod4$ and $q\equiv1\pmod 4$. Then we have $p+q\equiv 2\pmod 4$ and $p-q\equiv 0\pmod 4 \implies 4| p-q$.
Case 2: WLOG, let $p\equiv3\pmod4$ and $q\equiv1\pmod4$. Then we have $p+q\equiv0\pmod 4\implies 4|p+q$ and $p-q\equiv2\pmod4$.
Case 3: let $p\equiv3\pmod4$ and $q\equiv3\pmod4$. Then we have $p+q\equiv2\pmod4$ and $p-q\equiv0\pmod4\implies 4|p-q$.
In each of these three cases, while both $p+q$ and $p-q$ are both divisible by $2$, always only one of them is divisible by $4$. As such, $8|p^2-q^2$ for odd $p$ and $q$.
Together, as we have $2|p^2+q^2$ and $8|p^2-q^2$, we have $16|(p^2+q^2)(p^2-q^2)\implies 16|p^4-q^4$ as desired.
|
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|
Determine the set of all values of $ x \in [0, 2 \pi] $
Determine the set of all values of $ x \in [0, 2 \pi] $ that simultaneously satisfy $ \frac {2 \sin ^ 2 x + \sin x-1} {\cos x-1} < 0 $ and $ \tan x + \sqrt {3} < (1+ \sqrt {3} \cot x) \cot x $
My ''solution'':
$\frac{2{{\sin }^{2}}x+\sin x-1}{\cos x-1}<0\Rightarrow 2{{\sin }^{2}}x+\sin x-1>0\Rightarrow \sin x\in \left( \frac{1}{2},1 \right]$ and $\tan x+\sqrt{3}<(1+\sqrt{3}\cot x)\cot x\Leftrightarrow \tan x+\sqrt{3}<\frac{\tan x+\sqrt{3}}{\tan x}\cot x\Leftrightarrow \tan x+\sqrt{3}<\left( \tan x+\sqrt{3} \right)\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$
Am I right? Is there any better approach?
|
The first inequality,
$$ \frac {2 \sin ^ 2 x + \sin x-1} {\cos x-1}=\frac{(2\sin x -1)(\sin x+1)}{ \cos x-1}< 0 $$
Then,
$\sin x+1>0,\>\>\>\cos x-1<0 \implies \sin x > \frac12\implies x\in (\frac \pi6,\frac{5\pi}6)\tag{1}$
The second inequality,
$$\tan x+\sqrt{3}<\left( \tan x+\sqrt{3} \right)\cot^2x$$
Case 1.
$\tan x+\sqrt{3}>0,\>\>\>\cot^2x>1 \implies
x\in (0,\frac\pi4),\>(\frac{3\pi}4,\frac{5\pi}4),\>(\frac{7\pi}4,2\pi)\tag{2}$
Case 2.
$\tan x+\sqrt{3}<0,\>\>\>\cot^2x<1 \implies
x\in (\frac{\pi}2,\frac{2\pi}3),\>(\frac{3\pi}2,\frac{5\pi}3)\tag{3}$
Combine (2) and (3),
$x \in (0,\frac\pi4),(\frac\pi2,\frac{2\pi}3),(\frac{3\pi}4,\frac{5\pi}4),(\frac{3\pi}2,\frac{5\pi}3),(\frac{7\pi}4,2\pi)\tag{4}$
Thus, from (1) and (4), determine the set of all values
$$x \in (\frac\pi6,\frac\pi4),(\frac\pi2,\frac{2\pi}3),(\frac{3\pi}4,\frac{5\pi}6)$$
|
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|
Proving a Difficult Definite Integral in One Variable Let $t > 0 $, $N \in \mathbb{N}$, $q \geq 1 + \frac{1}{N}$.
Also let $r > 0 $, $r' = \frac{r}{r-1}$, such that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$.
We wish to prove the following integral equation:
$ \displaystyle \int^{t}_{0} \large (t-s)^{ -\frac{N}{2}(1 - \frac{1}{r}) - \frac{1}{2} } (t+s)^{ - \frac{N}{2} (q - \frac{1}{r'}) } \text{d}s = \large C t^{ \frac{1}{2} - \frac{N}{2} q } $,
where $C > 0$ is a constant, possibly just $1$.
I'm afraid I do not even know where to begin with this monster. Integral calculators have not been of much help. I presume it is a key point that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$, as the negative of this figure appears as the exponent of $(t-s)$.
Any hints/suggestions as to how I can calculate this integral are much appreciated.
EDIT
After substituting $s = tx$, we arrive at the integral:
$ \large t^{\frac{1}{2} - \frac{N}{2}q } \int^{1}_{0} (1-x)^{-\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} } (1+x)^{ -\frac{N}{2} (q - \frac{1}{r'}) } \text{d}x $.
Thus, it remains only to show that
$\int^{1}_{0} (1-x)^{-\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} } (1+x)^{ -\frac{N}{2} (q - \frac{1}{r'}) } \text{d}x = C$.
From our setting of $r$, as explained above, we have the following bounds on each of the exponents, which seem to be important for the existence of this integeral:
$ \large -1 < -\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} < 0 $
$ \large -\frac{1}{2} -\frac{N}{2}q < -\frac{N}{2} (q - \frac{1}{r'}) < \frac{1}{2} -\frac{N}{2}q $
In particular, by our definition of $q$, both exponents are always negative.
|
To show that the integral $$\int_0^1(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}\,dx$$ converges, note that $1\leq 1+x\leq 2$ in $(0,1)$, therefore the term $(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}$ is bounded above by a constant $C$. Therefore, \begin{align*}\int_0^1\left|(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}\right|\,dx&\leq C\int_0^1(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}\,dx\\ &=C\int_0^1y^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}\,dy,\end{align*} and the last integral converges, since the exponent of $y$ is greater than $-1$.
|
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Prove $n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$ by induction To prove:
$n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$
I assume the statement was derived from:
$n+2(n-1)+3(n-2)+...+(n-1)(n-(n-2))+n(n-(n-1))={n(n+1)(n+2)\over 6}$
For $n=k+1$
$(k+1)+2(k)+3(k-1)+...+2(k)+(k+1)={(k+1)(k+2)(k+3)\over 6}$
How do you use substitution from $n=k$, to prove $n=k+1$ when there are no duplicate terms?
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Hint: You can just extract $1$ from each product to use the induction hypothesis:
$$ \begin{aligned}
\phantom{=}&1\cdot(k+1) + 2\cdot k + 3\cdot(k-1) + \cdots + k\cdot 2 + (k+1) \\
=&(1\cdot k + \color{red}{1}) + (2\cdot (k-1) + \color{red}{2}) + \cdots + (k\cdot 1 + \color{red}{k}) + \color{red}{k+1}\\
=&\frac{(k-1)k(k+1)}6 + \color{red}{\frac{(k+1)(k+2)}2}
\end{aligned}
$$
|
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Tough complex numbers problem Let $c$ be a complex number. Suppose there exist distinct complex numbers $r$, $s$, and $t$ such that for every complex number $z$, we have
$$(z - r)(z - s)(z - t) = (z - cr)(z - cs)(z - ct).$$
Compute the number of distinct possible values of $c$.
OK, so I thought $1, \omega,$ and $\omega^2$ were the only answers, but apparently I'm missing something. Any help?
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Let $p(z)$ denote the cubic on the left-hand side; the right-hand side is then $c^3p(z/c)$. Write $p(z)=z^3+Az^2+Bz+C$ so$$z^3+Az^2+Bz+C\equiv z^3+cAz^2+c^2Bz+c^3C\\\implies(c-1)A=(c^2-1)B=(c^3-1)C=0.$$If $r,\,s,\,t$ are all nonzero, $C\ne0$ so $c^3=1$ as per your reasoning. For $c=\exp\frac{\pm 2\pi i}{3}$ to work, we need $A=B=0$, so $p(z)=z^3+C$, with three distinct roots as required, provided $C\ne0$.
Suppose without loss of generality $r=0$ so $s+t=-A,\,B=st\ne0,\,C=0$, and $c^2=1$ so $c=\pm1$. Only one of these values for $c$ are new to us, namely $c=-1$, whence $A=0$ and $t=-s$. Indeed, $z(z-s)(z+s)=z(z+s)(z-s)$ works for any $s\ne0$.
So, there are four possible values of $c$.
|
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In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\left(\dfrac{B}{2}\right)=\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{C}{2}\right)$$
$$2\sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$$
$$2\sqrt{\dfrac{(s-a)(s-c)(s-b)}{s(s-b)^2}}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-a)^2}}+\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-c)^2}}$$
$$\dfrac{2}{s-b}=\dfrac{1}{s-a}+\dfrac{1}{s-c}$$
$$\dfrac{2}{s-b}=\dfrac{s-c+s-a}{(s-a)(s-c)}$$
$$\dfrac{2}{s-b}=\dfrac{b}{(s-a)(s-c)}$$
$$2\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-c\right)=b\left(\dfrac{a+b+c}{2}-b\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$b^2-a^2-c^2+2ac=ba+bc-b^2$$
$$2b^2-a^2-c^2+2ac-ba-bc=0\tag{1}$$
$$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$$
$$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$
$$\cos A+\cos C=\dfrac{ab^2+ac^2-a^3+a^2c+b^2c-c^3}{2abc}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac^2-a^3+a^2c-c^3}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac(a+c)-(a+c)(a^2+c^2-ac)}{b}}{2ac}$$
Using equation $(1)$, $2ac-a^2-c^2=ba+bc-2b^2$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{(a+c)(ba+bc-2b^2)}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+(a+c)(a+c-2b)}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+a^2+c^2+2ac-2ba-2bc}{2ac}$$
$$\cos A+\cos C=\dfrac{a^2+c^2+2ac-ab-bc}{2ac}$$
Using equation $(1)$, $2ac-ba-bc=a^2+c^2-2b^2$
$$\cos A+\cos C=\dfrac{a^2+c^2+a^2+c^2-2b^2}{2ac}$$
$$\cos A+\cos C=\dfrac{2a^2+2c^2-2b^2}{2ac}$$
$$\cos A+\cos C=2\cdot\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A+\cos C=2\cos B$$
Is there any nice way to solve this question, mine goes very long. I tried various methods but this was the only way I was able to prove the required result.
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For $A\ne B,A+B+C=\pi$
using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$f(B,A)=\dfrac{\tan\dfrac B2-\tan\dfrac A2}{\cos B-\cos A}=-\dfrac1{2\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2}$$
By symmetry, $$f(B,A)=f(C,B)$$
Can you take it from here?
Similarly we can establish $$\dfrac{\cot\dfrac B2-\cot\dfrac A2}{\sin B-\sin C}=-\dfrac1{2\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}$$ so that a similar problem can be proved
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|
Baldi - Stochastic Calculus - Exercise about construsction of Stratonovich integral I need to prove that $$ \lim_n \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) \rightarrow \frac{1}{2}W_1^2 - \frac{1}{2}$$ in $L^2$.
where $W$ is a standard Wiener process.
So I started computing
$$E[( \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) + \frac{1}{2}(1-W_1^2) )^2]$$
and I split have that, after expanding the square:
$$E[( \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ))^2]=\frac{1}{2}$$
by independence the Gaussian distribution of independent increments.
Then, the double-product term $$\frac{1}{2} E[(1-W_1^2) \cdot \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )) ] = 0 $$ since it's a standard Wiener process and by independence of increments
The last term to compute is
$$ \frac{1}{4} E[(1-W_1^2)^2)] = \frac{1}{4} E[1+W_1^4 - 2 W_1^2 ] = \frac{1}{4} (1+3-2)=\frac{1}{2} $$
using that $E[W_1^2] = 1$ and $E[W_1^4] = 3$.
But in this way the sum is $1$ and not $0$... so what am I missing?
|
You expectation numbers are incorrect, you could see that in your proof, your solution is independent of $n$...
First notice that
$$ W_1^2= \sum_{i=0}^{2^n -1} W(\frac{i+1}{2^n})^2- W(\frac{i}{2^n})^2$$
$$ W_1^2= \sum_{i=0}^{2^n -1}\left( W(\frac{i+1}{2^n})+ W(\frac{i}{2^n})\right)\left( W(\frac{i+1}{2^n})- W(\frac{i}{2^n})\right)$$
Therefore, by factorizing, we have
$$\sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) + \frac{1}{2}(1-W_1^2) =\frac{1}{2}\left[1-\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]$$
You want to calculate
$$E\left[\frac{1}{4}\left[1-\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]^2\right]$$
The variance of a Brownian motion increment is known, we have
$$E\left[\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right] =\sum_{i=0}^{2^n -1}\frac{1}{2^n }=1$$
Finally,
$$E\left[\left[\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]^2\right]=E\left[\sum_{i,j=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2 (W(\frac{j+1}{2^n}) - W(\frac{j}{2^n}) )^2\right]$$
or
$$\sum_{i,j=0, i \ne j}^{2^n -1} E\left[( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2 (W(\frac{j+1}{2^n}) - W(\frac{j}{2^n}) )^2\right]+\sum_{i=0}^{2^n -1} E\left[( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^4\right]$$
Increments are independent , and you know the 4th moment of a Brownian motion, you can conclude when you make $n$ goes to infinity.
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For which integer values of $n$ does there exist an integer $m$ such that $n^{3} - m^{2} = -23$?
For which integer values of $n$ does there exist an integer $m$ such that $n^{3} - m^{2} = -23$?
I'm having a lot of trouble with this one, any help would be appreciated :)
So far, I've seen that if the expression were a perfect square we would have:
$n^{3} + 23 = x^{2}$
For some integer $x$. From this I've deduced that $n^{3}$ must be congruent to $x^{2}$ modulo $23$, however, I'm not sure how to proceed.
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Suppose that $n^3+23=m^2$ for some integers $n,m$. Observe that
$$(n+3)(n^2-3n+9)=n^3+27=m^2+4.$$
If $m$ is odd, then $n$ is even. Since $m^2\equiv 1\pmod{8}$, we must get $n^3\equiv 1-23\equiv 2\pmod{8}$. However, this is impossible as $n^3\equiv 0\pmod{8}$ for every even integer $n$. Thus $m$ is even.
Since $m$ is even, say $m=2k$, we get that $n$ is odd. Because $m^2\equiv 0\pmod{4}$, we obtain $n^3\equiv -23\equiv 1\pmod{4}$. Thus $n\equiv 1\pmod{4}$. That is,
$$k^2+1=\frac{m^2+4}{4}=\frac{n+3}{4}\left(n^2-3n+9\right).$$
Since $n\equiv 1\pmod{4}$, we must have
$$n^2-3n+9\equiv 1-3+9\equiv 3\pmod{4}.$$
Because $n^2-3n+9>0$, we conclude that a prime natural number $p\equiv 3\pmod{4}$ must divide $n^2-3n+9$. Thus, $k^2+1$ is divisible by $p$, so that $-1$ is a quadratic residue modulo $p$, which implies that $p\equiv 1\pmod{4}$. This is a contradiction.
|
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|
Identity about Fibonacci numbers If I note $(F_N)_N = \{0,1,1,2,3,5,...\}$ the Fibonacci sequence, I have proved the identity
$$ \forall N \geqslant 0,\,F_{N}^2 = F_{N} + 2\,\sum_{k=0}^{N-3} F_{k+1}\,F_{k+2}\,F_{N-k-2}. $$
This relation can be obtained by induction and by using the equality
$$F_{n+2}^2 =(F_{n+1}+F_n)^2=F^2_{n+1}+F^2_n+2F_{n+1}F_n.$$
Indeed we have
\begin{eqnarray*}
F_{n+2}^2 &=& F^2_{n+1}+F^2_n+2F_{n+1}F_n \\
\\
&=& F_{n+1} + 2\,\sum_{k=0}^{n-2} F_{k+1}\,F_{k+2}\,F_{n-k-1} + F_{n} + 2\,\sum_{k=0}^{n-3} F_{k+1}\,F_{k+2}\,F_{n-k-2} + 2\,F_{n+1}F_n \\
\\
&=& F_{n+2} + 2\,\sum_{k=0}^{n-3} F_{k+1}\,F_{k+2}\,F_{n-k} + 2\,F_{n-1}F_n + 2\,F_{n+1}F_n \\
\\
&=& F_{n+2} + 2\,\sum_{k=0}^{n-1} F_{k+1}\,F_{k+2}\,F_{n-k}.
\end{eqnarray*}
Question : I would like to know if anyone had ever seen a similar equality please. Thank you in advance.
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Define, for $n\ge3$, $\displaystyle S_n:=\sum_{k=0}^{n-3}F_{k+1}F_{k+2}F_{n-k-2}$. Then, the focus of your identity, $S_n=\frac{F_n^2-F_n}{2}$, is really on $S_n$ since the other terms are simpler.
Searching for the first few values of $S_n$ in the OEIS gives gives us sequence A191797, i.e. $S_n=\binom{F_n}{2}=\frac{2F_n^2-F_{n+4}+3F_{n+1}}{4}=T_{F_{n}-1}=\frac{F_n(F_n-1)}{2}$, where $\binom{n}{k}$ is the binomial coefficient and $T_n$ is the $n$th triangular number. This shows half your identity.
This should give you more places to look to see if the identity has been previously published. I've not found any reference to the sum of products formula, so it may be original.
We may prove the identity as follows
$$\require{cancel}
\begin{align}
S_n-S_{n-1}
&=\sum_{k=0}^{n-3}F_{k+1}F_{k+2}F_{n-k-2}-\sum_{k=0}^{n-4}F_{k+1}F_{k+2}F_{n-1-k-2}
\\
&=F_{n-2}F_{n-1}F_1+F_{n-3}F_{n-2}F_{2}+\sum_{k=0}^{n-5}F_{k+1}F_{k+2}F_{n-k-2}\ldots
\\&\quad\ldots-\left(F_{n-3}F_{n-2}F_{1}+\sum_{k=0}^{n-5}F_{k+1}F_{k+2}F_{n-k-3}\right)
\\
&=F_{n-2}F_{n-1}+\cancel{F_{n-3}F_{n-2}-F_{n-3}F_{n-2}}+\sum_{k=0}^{n-5}\left(F_{k+1}F_{k+2}F_{n-k-2}-F_{k+1}F_{k+2}F_{n-k-3}\right)
\\
&=F_{n-2}F_{n-1}+\sum_{k=0}^{n-5}F_{k+1}F_{k+2}\left[\underbrace{F_{n-k-2}}_{F_{n-k-3}+F_{n-k-4}}-F_{n-k-3}\right]
\\
&=F_{n-2}F_{n-1}+\sum_{k=0}^{(n-2)-3}F_{k+1}F_{k+2}\left(F_{(n-2)-k-2}\right)
\\
&=F_{n-2}F_{n-1}+S_{n-2}
\end{align}
$$
With $S_3=F_1F_2F_{3-0-2}=1$ and $S_4=F_1F_2F_{4-0-2}+F_2F_3F_{4-1-2}=3$, this recurrence relation gives us an alternate definition for $S_n$. Therefore, the aim is to inductively prove that, for $n\ge 3$, $S_n=\frac{F_n(F_n-1)}{2}$. For the two base cases, $S_3=1=\frac{F_3(F_3-1)}{2}$ and $S_4=3=\frac{F_4(F_4-1)}{2}$. Then, assuming $S_k=\frac{F_k(F_k-1)}{2}$ and $S_{k+1}=\frac{F_{k+1}(F_{k+1}-1)}{2}$, we have
$$\begin{align}S_{k+2}
&=F_{k}F_{k+1}+S_{k}+S_{k+1}
\\
&=F_{k}F_{k+1}+\frac{F_k(F_k-1)}{2}+\frac{F_{k+1}(F_{k+1}-1)}{2}
\\
&=\frac12\left(2F_{k}F_{k+1}+{F_k^2-F_k}+{F_{k+1}^2-F_{k+1}}\right)
\\
&=\frac12\left(2F_{k}F_{k+1}+\left[F_{k-1}F_{k+1}-(-1)^k\right]+\left[F_{k}F_{k+2}-(-1)^{k+1}\right]-F_{k+2}\right)\tag{*}
\\
&=\frac{1}{2}\left(F_{k+1}\left(2F_{k}+F_{k-1}\right)+F_{k}F_{k+2}-F_{k+2}\right)
\\
&=\frac{1}{2}\left(F_{k+1}\left(F_{k}+F_{k+1}\right)+F_{k}F_{k+2}-F_{k+2}\right)
\\
&=\frac{1}{2}\left(F_{k+1}F_{k+2}+F_{k}F_{k+2}-F_{k+2}\right)
\\
&=\frac{F_{k+2}(F_{k+2}-1)}{2}
\end{align}
$$
This completes the proof. $(*)$ uses Cassini's identity, $F_{n-1}F_{n+1}-F_n^2=(-1)^n$.
|
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|
Solution attempt $xuu_x+yuu_y=u^2-1$ Solve $$
\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}
$$
I have got using Lagrange method:
$$F\left(\frac{x}{y}\right)=\frac{x^2}{u^2-1}$$
Applying $u(x,x^2)=x^3$:
$$u^2=\frac{y^6-x^6}{x^2y^2}+1$$
But plug in it to the PDE show that there is a mistake
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Calling $v = u^2$ we have
$$
\frac 12 xv_x +\frac 12 y v_y = v-1
$$
with solution
$$
v = x^2\phi\left(\frac yx\right)+1
$$
now
$$
v(x,x^2) = x^6\Rightarrow \phi\left(\frac yx\right) = \frac{y^6-x^6}{x^4y^2}
$$
and
$$
v(x,y) = x^2\left(\frac{y^6-x^6}{x^4y^2}\right)+1
$$
and finally
$$
u(x,y) = \sqrt{\frac{y^6-x^6}{x^2y^2}+1}
$$
|
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|
Rewriting a double summation. If I am given this function: $$f(x) = \sum_{i = 1}^{\infty} \frac{1}{i^x}$$
Is there a way to rewrite:
$$g(x) = \sum_{ j = 1}^{\infty} \sum_{i = j}^{\infty} \frac{1}{(i \cdot j)^x}$$
In terms of f(x). By simple arthimatic I know that $f(x)^2 = 2 \cdot g(x) - f(2x)$.
The question is if for a more general way to rewrite the summation. If $g(x)$ contained 3 summation with $i,j,k$ how would you rewrite it in terms of $f(x)$.
|
Let
\begin{eqnarray*}
f_1(x) &=& \sum_{i \geq 1} \frac{1}{i^x} \\
f_2(x) &=& \sum_{i >j \geq 1} \frac{1}{(ij) ^x} \\
f_3(x) &=& \sum_{i>j>k \geq 1} \frac{1}{(ijk)^x}. \\
\end{eqnarray*}
In your question you have calculated
\begin{eqnarray*}
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) ^2 &=& \sum_{i \geq 1} \frac{1}{i^{2x} } + 2 \sum_{i >j \geq 1} \frac{1}{(ij) ^x} \\
(f_1(x))^2 &=& f_1(2x) +2 f_2(x).
\end{eqnarray*}
Calculating similarly
\begin{eqnarray*}
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) ^3 = \sum_{i \geq 1} \frac{1}{i^{3x} } + 3 \sum_{i >j \geq 1} \frac{1}{(i^2j) ^x} + 3 \sum_{i >j \geq 1} \frac{1}{(ij^2) ^x} + 6 \sum_{i >j>k \geq 1} \frac{1}{(ijk) ^x} \\
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) \left( 1 + \frac{1}{2^{2x}} + \frac{1}{3^{2x}} + \cdots \right) = \sum_{i \geq 1} \frac{1}{i^{3x} } + \sum_{i >j \geq 1} \frac{1}{(i^2j) ^x} + \sum_{i >j \geq 1} \frac{1}{(ij^2) ^x} . \\
\end{eqnarray*}
Multiply the second equation by $3$ and subtract it from the first gives
\begin{eqnarray*}
(f_1(x))^3 -3f_1(x) f_1(2x) = -2f_1(3x) +6 f_3(x).
\end{eqnarray*}
|
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|
Understanding the equation $|x+1|=x^2 -1$ I want to understand the equation
$$|x+1|=x^2 -1$$
$$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$
$$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \Rightarrow x^2 + x = 0 \Rightarrow x_3= 0 \text{, (but doesn't fulfill } x+1 < 0), x_4 = -1$$
$$\Rightarrow L = \{2,-1\}$$
What I don't get is how $x+1 < 0 \Rightarrow x^2 + x = 0$. How do we get $x^2 + x = 0$?
|
Suppose that $x$ satisfies
$$(*) \quad |x+1|=x^2 -1.$$
Then $|x+1|=(x -1)(x+1).$ Hence $|x+1|=|x-1| \cdot |x+1|.$
It is clear that $x=-1$ is a solution of $(*)$. Now we assume that $x \ne -1.$ Then we get $|x-1|=1.$ The last equation has the solutions $x=0$ and $x=2$. But only $x=2$ is a solutions of $(*)$.
Consequence: $(*)$ holds $ \iff x \in \{-1,2\}.$
|
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|
What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
]1
Solution:
Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\times\pi r^2=\dfrac{5\pi r^2}{2}$
The radius $R$ of the big circle is equal to $r$ plus the radius of the circumscribed circle of equailateral triangle ABC, whose side is $2r$.
The radius of the circumscribed circle of an equilateral triangle is the length of the sides divided by $\sqrt{3}$. Since the side here measures $2r$, the radius of the circumscribed circle is $\dfrac{2r}{\sqrt{3}}$.
So we have $R = r+\dfrac{2r}{\sqrt{3}}$
The area of the big circle is $\pi \times R^2$, which here is equal to $(r+\dfrac{2r}{\sqrt{3}})^2$
which, when expanded, gives
Big Circle area = $\dfrac{\pi r^2(7+4\sqrt{3})}{3}$
To obtain the shaded fraction, we need to divide the area of the colored region by the area of the big circle:
Shaded fraction = $\dfrac{\dfrac{5\pi r^2}{2}}{\dfrac{\pi r^2(7+4\sqrt{3})}{3}}$
Shaded fraction = $\dfrac{5\pi r^2}{2} \times \dfrac{3}{\pi r^2(7+4\sqrt{3})}$
Shaded fraction = $\dfrac{15}{2(7+4\sqrt{3})} \simeq 53.847 \% $
I think it's wrong. In the drawing the smaller circles are not tangent to the largest
|
Using your original drawing. Let the radius of the smaller circle $r=1$. The radius of the big circle is: $AD+AG=\frac{1}{\cos 30°}+\cos 30°=\frac{7\sqrt 3}{6}$. Thus, the ratio of the colored area to the area of the big circle is $$\frac{5 \over 2}{49 \over 12}=\frac{30}{49}$$
|
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$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=d$. Also notice that $(a+b+c+d)^{2}$ must be odd and only have three factors: $1, p, p^{2}$.
$$ p^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2(ab+ac+ad + bc+bd + cd) $$
$$ = 2(c^{2}+d^{2}+ac+ad++bc+bd) + (ab + 3cd) $$
$$ =2(c^{2}+d^{2}+ac+ad++bc+bd + cd) + (ab + cd) $$
So $(ab+cd)$ must be odd.
Now if $a+b+c+d$ is prime $>2$ then either 3 of them is odd and 1 is even, or 3 of them is even and one is odd.
$WLOG$, let $a,b,c$ be even and $d$ is odd, then $a^{2} + b^{2}+ ab$ is even and $c^{2} + d^{2}+ cd$ is odd, so we can't have 3 even and 1 odd.
But it is possible for 3 odd and 1 even.
|
We work in the principal ideal domain $\mathbb Z [\omega]$, where $\omega = \frac{-1 + \sqrt3 i}{2}$ (which we view as an element of $\mathbb C$).
Without loss of generality, assume that $a \geq b$ and $c \geq d$, and that we don't have equality at the same time.
Write the equation as $(a + b \omega)(a + b\overline\omega) = (c + d \omega)(c + d\overline\omega)$. Let $u$ be the greatest common divisor of $a + b\omega$ and $c + d\omega$. It is well defined up to multiplication by a sixth root of unity, i.e. $\pm 1, \pm \omega, \pm \omega^2$.
Write $a + b\omega = u v$ and $c + d\omega = u v'$, with $v, v'$ coprime. We then have $v\overline v = v' \overline {v'}$, which implies (since $v, v'$ coprime) that $v \mid \overline{v'}$ and $v'\mid \overline v$. But taking complex conjugation gives $\overline{v'}\mid v$, hence $v' = \varepsilon v$ for some sixth root of unity $\varepsilon$.
We therefore have $a + b\omega = uv$, $c + d\omega = u\overline v \varepsilon$. By replacing $u, v, \varepsilon$ with $u\delta^{-1}, v\delta, \varepsilon\delta^2$ for some sixth root of unity $\delta$, we may assume without loss of generality that $\varepsilon = \pm 1$.
Case 1: $\varepsilon = 1$.
We have $a + b\omega = uv$. Taking complex conjugation gives $a + b\overline \omega = \overline u\overline v$. Solving this linear equation, we get $a + b = -\omega u v - \overline \omega \overline u \overline v$.
Similarly, we get $c + d = -\omega u \overline v - \overline \omega \overline u v$.
Hence in the end $a + b + c + d = -(\omega u + \overline {\omega u})(v + \overline v)$.
Since both $(\omega u + \overline {\omega u})$ and $v + \overline v$ are integers, if the product is prime, then one of them must be $\pm 1$.
Now observe the fact that, for any element $x \in \mathbb Z[\omega]$, $x + \overline x = \pm 1$ will imply that $\arg x$ lies in either $[60^\circ, 120^\circ]$ or $[-120^\circ, -60^\circ]$.
It follows that $v + \overline v$ cannot be $\pm 1$. This is because both $a + b\omega$ and $c + d\omega$ lives in the region $\{z \in \mathbb C: 0 < \arg z \leq 60^\circ\}$, hence the number $v/\overline v$, being their quotient, must have $\arg$ in the range $(-60^\circ, 60^\circ)$.
But $\omega u + \overline{\omega u}$ cannot be $\pm 1$ either. The reasoning is similar: since $a + b\omega$ and $c + d\omega$ cannot both attain $60^\circ$ of $\arg$, the number $u^2 v \overline v$, being their product, must have $\arg$ in the range$(0^\circ, 120^\circ)$. This means $u$ must have $\arg$ in the range $(0^\circ, 60^\circ)$ or $(-180^\circ, -120^\circ)$, hence $\omega u$ has $\arg$ in the range $(120^\circ, 180^\circ)$ or $(-60^\circ, 0^\circ)$.
This completes the proof that $a + b + c + d$ cannot be a prime number in the case $\varepsilon = 1$.
Case 2: $\varepsilon = -1$.
Same as above, we work out the formula $a + b + c + d = -(v - \overline v)(\omega u - \overline{\omega u})$.
This case is now significantly easier, since the product must be a multiple of $3$. So if it is a prime, then we have $a + b + c + d = 3$ and it's already impossible.
|
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Find the value of $x$ in the figure
Attempt: $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$
$x= 6^\circ$
(The answer is $6^\circ$)
How to ensure there is no other solution?
|
Let $t = e^{ix}$. We then have $\sin nx = \frac{1}{2i}(t^n - t^{-n})$.
Expanding everything in the identity $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$, we have an equation: $$t^{20} - t^{18} - t^{16} + t^{12} + t^8 - t^4 - t^2 + 1 = 0,$$ which after factorization becomes: $$(t - 1)^2 (t + 1)^2 (t^{16} + t^{14} - t^{10} - t^8 - t^6 + t^2 + 1) = 0.$$
Of course, $t = \pm 1$ are not solutions to our problem. Thus $t$ is a root of the polynomial $t^{16} + t^{14} - t^{10} - t^8 - t^6 + t^2 + 1$, which happens to be exactly the $60$-th cyclotomic polynomial $\Phi_{60}(t)$.
Therefore $t$ is one of the $60$-th primitive roots of unity, and hence $x$ is equal to $6k^\circ$ for some $k$ prime to $60$.
But from the graph, we should have $5x < 180^\circ$, hence $k < 6$. This only leaves the possibility $k = 1$, or $x = 6^\circ$.
|
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prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle
Such that : $a+b+c=1$ Then
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$
My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$
Also :
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don't know how to complete it,any help is appreciated !
|
Using homogeneization and Ravi's substitution, the problem boils down to showing that
$$2(a+b+c)(a^2+b^2+c^2)+8abc \leq (a+b+c)^3 $$
holds for any triple $(a,b,c)$ of side lengths of a triangle, i.e. that
$$4(A+B+C)((B+C)^2+(A+C)^2+(A+B)^2)+8(A+B)(A+C)(B+C) \leq 8(A+B+C)^3 $$
holds for any triple $(A,B,C)$ of positive numbers. The previous inequality simplifies into
$$ (A+B+C)(A^2+B^2+C^2+AB+AC+BC)+(A^2 B+B^2 A+A^2 C+C^2 A+B^2 C+C^2 B+2ABC) \leq (A+B+C)^3$$
which is equivalent to the trivial $-ABC\leq 0$.
|
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Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$
I tried showing the equation, but my attempts did not get the result.
I already showed that
$$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})=\frac18$$ $$\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=-\frac12$$
Using in particular the trigonometric formulas:
$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ , $\sin(\pi+\alpha)=-\sin(\alpha)$ , $\sin(\pi-\alpha)=\sin(\alpha)$ , $\cos(\pi-\alpha)=-\cos(\alpha)$ , $\sin(-\alpha)=-\sin(\alpha)$ , and $2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)$
Any hint would be helpful. Thanks in advance.
|
Use the identity $\cos(a-b)+\cos(a+b) = 2\cos a\cos b$,
$$I=\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7} $$
$$=\frac12\left(\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}\right)
+\frac12\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}\right)
+\frac12\left(\cos\frac{2\pi}{7}+\cos\frac{10\pi}{7}\right) $$
Recognize $\cos\frac{8\pi}{7} = \cos\frac{6\pi}{7}$ and $\cos\frac{10\pi}{7} = \cos\frac{4\pi}{7}$ to simplify $I$,
$$I=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12$$
where the result you already obtained for the sum is used.
|
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Solve a system by putting new variables
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$$
The first step is to determine the domain: $\begin{array}{|l} x \ne 0 \\ y \ne 0 \end{array}$
We can simplify the first equation of the system, and we get: $\begin{array}{|l} \dfrac{x^2+y^2}{xy}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$
$\begin{array}{|l} \dfrac{25}{xy}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} 7xy=625 \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} xy=\dfrac{625}{7} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} x=\dfrac{625}{7y} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} x=\dfrac{625}{7y} \\ y^4-25y^2+\dfrac{390625}{49}=0 \end{array}$
The equation: $$y^4-25y^2+\dfrac{390625}{49}=0$$ has no solutions, so the whole system does not have a solution. In the text of the problem is said I should solve by "putting a new variable". I don't know how this method is called in English, and I would be grateful if you tell me. Let me give you a basic example of a system that can be solved using this method:
$$\begin{array}{|l} (x+2y)^2-(y-2x)^2=168 \\ (x+2y)^2+(y-2x)^2=12 \end{array}$$
Let $$\begin{array}{|l} (x+2y)^2=a \\ (y-2x)^2=b \end{array}...$$
|
let $b=\frac{x}{y}$ (defined because $x\ne 0,y\ne 0$). Then by the first equation:
$b+\frac{1}{b}=\frac{7}{25}$
$\iff (b\ne0)$
$b^2-\frac{7}{25}b+1=0$
$\iff$
False (no real solutions)
so there are no solutions
|
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Existence of $\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$
Prove that existence of limit $$\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$$ in $\mathbb{R}$.
Attempt.
Let $A_k$ be the $k$-th term of the sequence. Then $(A_k)$ is increasing. If we proved that it is upper bounded, then we would have convergence. But the estimation:
$$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x\leqslant \frac12\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\sin^2x}}\,\mathrm{d}x=+\infty,$$
although corrent, seems to overestimate the integral.
Thanks for the help.
|
How about this?
For fixed $x \in (0,\pi/2)$, the integrand
$$
\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}
$$
increases as $k$ increases. The integrand is nonnegative. So by the monotone convergence theorem,
$$
\lim_{k\to\infty}\int_0^{\pi/2}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\;dx \\=
\int_0^{\pi/2}\lim_{k\to\infty}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\;dx \\=
\int_0^{\pi/2}\;\frac{\sin^2x}{2\sqrt{1-\sin^2x}}\;dx = +\infty
$$
|
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Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$ \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} = \frac { 2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{n^2+1} \sim \frac { 2^{\ln n }}{n^2+1} \rightarrow 0$$
Is it right?
I have another exercize that ends similarly with
$$ \frac { 10^{\ln n }}{n^2+1} \rightarrow 0$$
But the book says that the result is $+\infty$.
|
This is right, because it depends what is in your base. Consider
$ (e^2)^{\ln(n)} = (e^{\ln(n)})^2 = n^2 $
Since $10 > e^2$ or rather $\ln(10) = a > 2$
You have: $10^{\ln(n)}$ is equivalent of $ n^a $ and therefore
$ \frac { 10^{\ln(n)}}{n^2+1} \rightarrow +\infty$
In your initial example $2 < e < e^2$ so the limit is zero.
|
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|
Proving inequality $(a^2 + b^2)^3 \ge 32(a^3 + b^3)(ab - a - b)$
If $a, b \in \mathbb R$ and $a + b \geq 0$, then prove that $$(a^2 + b^2)^3 \geq 32(a^3 + b^3)(ab - a - b)$$
Since $a + b ≥ 0$, we can apply A.M.-G.M. inequality, I tried to apply the inequality, but wasn't able to reach a conclusive decision. How can I solve it using A.M.-G.M. inequality, or by any other way which is much easier than prior method.
|
We need to prove that:
$$(a^2+b^2)^3+32(a+b)^2(a^2-ab+b^2)\geq32(a+b)(a^2-ab+b^2)ab$$ and since $a+b\geq0,$ it's enough to prove our inequality for $ab\geq0,$ which gives that $a$ and $b$ are non-negatives.
Now, by AM-GM
$$(a^2+b^2)^3+32(a+b)^2(a^2-ab+b^2)\geq2\sqrt{(a^2+b^2)^3\cdot32(a+b)^2(a^2-ab+b^2)}.$$
Thus, it's enough to prove that:
$$2\sqrt{(a^2+b^2)^3\cdot32(a+b)^2(a^2-ab+b^2)}\geq32(a+b)(a^2-ab+b^2)ab$$ or
$$(a^2+b^2)^3\geq8(a^2-ab+b^2)a^2b^2.$$
Let $a^2+b^2=2kab$.
Thus, by AM-GM again $k\geq1$ and we need to prove that
$$(2k)^3\geq8(2k-1)$$ or
$$k^3-2k+1\geq0$$ or
$$k^3-k^2+k^2-k-k+1\geq0$$ or $$(k-1)(k^2+k-1)\geq0,$$ which is true for $k\geq1.$
|
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|
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
How do I solve it? Help me please.
|
$$\frac{\sin x}{x}\simeq 1-\frac{1}{6}x^2+\frac{1}{120}x^4+O(x^6)$$
$$\left(\frac{\sin x}{x}\right)^n\simeq 1-\frac{n}{6}x^2+\frac{n(5n-2)}{360}x^4+O(x^6)$$
$$\frac{\left(\frac{\sin x}{x}\right)^n}{1-\left(\frac{\sin x}{x}\right)^n}\simeq \frac{6}{n}x^{-2}-\frac{5n+2}{10n}+O(x^2)$$
$$\boxed{f(x)=\frac{x^n\sin^n x}{x^n-sin^nx}\simeq\frac{6}{n}x^{n-2}-\frac{5n+2}{10n}x^n+O(x^{n+2})}$$
If $\quad n<2\qquad$ $f(x)\sim\frac{6}{n}\frac{1}{x^{2-n}}\to\infty$
If $\quad n=2\qquad f(x)\sim 3-\frac35 x^2\to 3$
If $\quad n>2\qquad f(x)\sim\frac{6}{n}x^{n-2}\to 0 $
|
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How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?
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Consider :
$r>0$, $0\le \theta \lt 2π$.
Then obviously:
$r^4 \ge r\sin (2\theta) r^3 \sin^2 (\theta)$(Why?)
$r^4 \ge 2r \sin (\theta) \cos (\theta) r^3 \sin^2 (\theta)$
And with $x=r\cos (\theta)$, $y=r \sin (\theta)$:
$(x^2+y^2)^2\ge 2 xy^3.$
|
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|
$A (3,1)$ reflected to $y=2x$ what is area of AOA'? $y_1 = 2x$ = line of reflection
$x_1,y_1 = A(3,1) =$ reflected point
General formation of a line
$y = mx + k$
$Ax + By + C = 0$
Finding a line perpendicular to line of reflection
$m_2 = \frac{-1}{2}$
$y_2 = \frac{-1}{2}x + \frac{5}{2}$
Intersection of reflection line and line perpendicular to it
$y_1$ dan $y_2$
$4x = -x + 5$
$x = 1, y = 2$
Finding A'
Gradient from A' and 1,2 also -1/2
$\frac{y_3 - 2}{x_3 - 1} = \frac{-1}{2}$
$-2y_3 + 3 = x_3$
Distance 1,2 and 3,1 = $\sqrt 5$
Distance from 1,2 to A' also $\sqrt 5$
${(x_3 - 1)}^2+ {(y_3 - 2)}^2 = 5$
${(-2y_3 + 3)}^2+ {(y_3 - 2)}^2 = 5$
After find A' then find |OA'| |OA| and finf area. But is there less complicated way to find area? Am i missing something?
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Note that $|OA| = \sqrt{3^2+1}=\sqrt{10}$ and the angle $\theta$ between $y=2x$ and OA is given by
$$\tan\theta = \tan(\theta_2-\theta_1)
= \frac{\tan\theta_2 - \tan\theta_1}{1+\tan\theta_2\cdot \tan\theta_2 }
=\frac{2-\frac13}{1+2\cdot\frac13} = 1$$
which yields $\theta = 45^\circ$. Then, AOA' is an isosceles right triangle whose area is
$$A = \frac12 |OA|^2 = 5 $$
|
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|
Evaluate $\iint |xy|\,dx\,dy$ using polar coordinates If $R$ is the region bounded by $x^2+4y^2 \ge 1$ and $x^2+y^2 \le 1$. Then find the integral $$I=\iint_R |xy|\,dx\,dy.$$
I tried using Cartesian system and got the answer as $\frac{3}{8}$ using symmetry. Can we do this in Polar coordinates?
|
First, use symmetry to only do the integral in the first quadrant (this is allowed because both the integrand and the region of integration share the same symmetry):
$$\iint_R |xy|\: dA = 4 \iint_{R_1} |xy|\:dA$$
Next, we'll have to figure out the bounds. The upper bounds aren't difficult, but the lower ones are given by
$$r^2(\cos^2\theta+4\sin^2\theta) = 1 \implies \begin{cases} r = \frac{1}{\sqrt{1+3\sin^2\theta}} \\ \theta = \sin^{-1}\left(\frac{\sqrt{1-r^2}}{r\sqrt{3}}\right) \\ \end{cases} $$
which gives us two choices of integrals, either $dr$ first:
$$4\int_0^{\frac{\pi}{2}} \int_{\frac{1}{\sqrt{1+3\sin^2\theta}}}^1 r^3\sin\theta\cos\theta \: dr \: d\theta = \int_0^{\frac{\pi}{2}} \sin\theta \cos\theta \left(1- \frac{1}{(1+3\sin^2\theta)^2}\right)\: d\theta $$
or doing $d\theta$ first:
$$4\int_{\frac{1}{2}}^1 \int_{\sin^{-1}\left(\frac{\sqrt{1-r^2}}{r\sqrt{3}}\right)}^{\frac{\pi}{2}} r^3 \sin\theta \cos\theta \: d\theta \: dr = \int_{\frac{1}{2}}^1 2r^3\left(1 - \frac{1-r^2}{3r^2} \right)\: dr$$
This is one of the few times where doing the angular integral first is easier than the other way around. Finishing, we get
$$ \int_{\frac{1}{2}}^1 \frac{8}{3}r^3 - \frac{2}{3}r \: dr = \frac{2}{3}r^4 - \frac{1}{3}r^2\Bigr|_{\frac{1}{2}}^1 = \frac{3}{8}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Derivative of $\sec^{-1}x$ and integral of $\frac{1}{x\sqrt{x^2-1}}$ My attempt is as follows:-
*
*Derivative of $\sec^{-1}x$
Let $\theta=\sec^{-1}x,$ where $\theta\in [0,\pi]-{\dfrac{\pi}{2}}.$
$$\sec\theta=x.$$
Differentiating both sides with respect to $x:$
$$\sec\theta\cdot\tan\theta\cdot\dfrac{\mathrm d\theta}{\mathrm dx}=1\\
\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{1}{x\sqrt{x^2-1}}.$$
As $\sec^{-1}x$ is a strictly increasing function, its derivative should be positive, hence we write $x$ as $|x|$ to ensure that $\dfrac{\mathrm d\theta}{\mathrm dx}$ will not be negative if $x$ is negative. But I wonder why I didn't get $\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{1}{|x|\sqrt{x^2-1}}$ in the above calculation?
*Integral of $\dfrac{1}{x\sqrt{x^2-1}}$
Case $1:x>0$
Then the integral is definitely $\sec^{-1}x.$
Case $2: x<0$
Then the integral is $-\sec^{-1}x.$
But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$
Shouldn't $\displaystyle\int\frac{1}{|x|\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C\:?$ What am I missing here?
|
But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$
This is indeed wrong, since differentiating the equation at $x=-5$ gives $\displaystyle\frac{\sqrt6}{60}=-\frac{\sqrt6}{60}.$
Case $1:x>0$
the integral is definitely $\sec^{-1}x.$
Case $2: x<0$
the integral is $-\sec^{-1}x.$
Indeed, $\displaystyle\frac1{x\sqrt{x^2-1}}$ always has an antiderivative $$\frac x{|x|}\sec^{-1}x.\tag1$$ bjorn93 has also pointed out that it always has an antiderivative $$\sec^{-1}|x|.\tag{2}$$
Using the substitution $\displaystyle u=\frac1x$ and noting that
$\sec^{-1}x+\operatorname{cosec}^{-1}x\equiv\frac{\pi}2,$ we can also obtain \begin{align}\int\frac{\mathrm dx}{x\sqrt{x^2-1}}
&= \int\frac{-|u|}{u\sqrt{1-u^2}}\,\mathrm du\\
&= \begin{cases} \sin^{-1} u+C_1, &-1<u<0;\\
-\sin^{-1} u+C_2, &0<u<1\end{cases}\\
&= \begin{cases} \operatorname{cosec}^{-1}x+C_1, &x<-1;\\
-\operatorname{cosec}^{-1}x+C_2, &x>1\end{cases}\\
&= \begin{cases} \operatorname{cosec}^{-1}x+C_1, &x<-1;\\
\operatorname{sec}^{-1}x-\frac\pi2+C_2, &x>1\end{cases}\\
&= \begin{cases} \operatorname{cosec}^{-1}x+C, &x<-1;\\
\operatorname{sec}^{-1}x+D, &x>1.\tag3\end{cases}\end{align}
$(1),(2),(3)$ are consistent with one another because for $x<-1$ the expressions differ from one another by $\frac{\pi}2$ while for $x>1$ the expressions are identically equal. (Remember, each integration domain overlaps with just one of these intervals. Also, it is worth noting that when $C_1\ne C_2,$ differentiating the above result still returns the given integrand.)
P.S. It can be similarly derived that on $(-1,0)\cup(0,1)$ $$-\operatorname{sech}^{-1}|x| \text{ is an antiderivative of } \frac1{x\sqrt{1-x^2}},$$ and on $\mathbb R{\setminus}\{0\}$ $$-\operatorname{cosech}^{-1}|x| \text{ is an antiderivative of } \frac1{x\sqrt{1+x^2}}.$$
|
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|
Let $x_1,x_2 \in \mathbb{R}$ be the roots of the equation $x^2+px+q=0$
Let $x_1,x_2 \in \mathbb{R}$ be the roots of the equation
$x^2+px+q=0$. Find $p$ and $q$ if it is known that $x_1+1$ and $x_2+1$
are the roots of the equation $x^2-p^2x+pq=0$.
The roots of $x^2+px+q=0$ satisfy $$x_1+x_2=-\dfrac{b}{a}=-p,x_1x_2=\dfrac{c}{a}=q.$$ Also the roots of $x^2-p^2x+pq=0$ satisfy $$x_1+x_2=x_1+1+x_2+1=x_1+x_2+2=-\dfrac{b}{a}=p^2,x_1x_2=(x_1+1)(x_2+1)=\dfrac{c}{a}=pq.$$ Is it confusing I am using $x_1$ and $x_2$ for the roots of both equations? How can I improve what I've written? What to do next?
|
Another way :
Like If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
let $x_k+1=y_k; k=1,2$
$\implies x_k=y_k-1$
As $x_k$ is a root of $$x^2+px+q=0$$
$$(y_k-1)^2+p(y_k-1)+q=0$$
|
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|
Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2a}
$$
I think I am getting stuck here, is there a better substitution that I can chose so that the integral becomes more simple to evaluate ?
Solution as per my reference: $\dfrac{2(\sqrt{x}-1)}{\sqrt{1-x}}$
Note: I'd prefer to choose a substitution which does not make use of partial fractions, as there seems to be 4 terms for the substitution $\sqrt{x}=y\implies \frac{dx}{2\sqrt{x}}=dy$.
$$
I=\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}\\
=\int\frac{2dy}{y(1+y)(1-y^2)}=\int\frac{2dy}{y(1+y)^2(1-y)}\\
\frac{2}{y(1+y)^2(1-y)}=\frac{A}{y}+\frac{B}{1-y}+\frac{C}{1+y}+\frac{D}{(1+y)^2}
$$
|
Following @J.G's subbing
$\sqrt{x}=y$ then $y=\frac{1-u}{1+u}$
$$I\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{2}{y(1+y)(1-y^2)}dy=\frac12\int\frac{(1+u)^2}{u(u-1)}du$$
$$=-\frac12\int\left(\frac1u-\frac4{u-1}-1\right)du$$
|
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|
Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\in\Bbb Z$. Therefore $R$ is reflexive.
Now suppose $x^2\equiv y^2\pmod {4}$. This means $4\mid (x^2-y^2)$ and so $x^2-y^2=4a$, for some $a\in\Bbb Z$. Multiplying by $-1$ we have $-1(x^2-y^2=4a)\\\rightarrow -x^2+y^2=-4a\\\rightarrow y^2-x^2=4(-a)$
so $4\mid(y^2-x^2)$ and $y^2\equiv x^2\pmod{4}$. This shows that $R$ is symmetric.
Now we assume that $x^2\equiv y^2\pmod{4}$ and $y^2\equiv z^2\pmod{4}$. This means $4\mid(x^2-y^2)$ and $4\mid(y^2-z^2)$. Then we have $x^2-y^2=4a$ and $y^2-z^2=4b$ for some $a,b\in\Bbb Z$. Rearranging we get $x^2=4a+y^2$ and $z^2=y^2-4b$.
Then
$\begin{align*}x^2-z^2&=(4a+y^2)-(y^2-4b)\\&=4a+4b\\&=4(a+b)\end{align*}$
This shows that $4\mid(x^2-z^2)$ and $x^2\equiv z^2\pmod{4}$, therefore $R$ is transitive.
$\blacksquare$
Please forgive my rushed formatting, just wondering if my arguments here work and if this is a valid proof. Any feedback is appreciated, thanks!
|
This is straight forward and I think you did ok. The transitivity follows immediately from transitivity of congruence. In fact all three follow from the fact that congruence mod $n$ is indeed an equivalence relation.
|
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|
Evaluate $\int \frac{\cos^2x}{1+\tan x}dx$ Evaluate $\int \dfrac{\cos^2x}{1+\tan x}dx$
Here are my various unsuccessful attempts:-
Attempt $1$:
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
$$\ln(1+t)=y$$
$$\dfrac{1}{1+t}=\dfrac{dy}{dt}$$
$$\int \dfrac{dy}{\left(1+(e^y-1)^2\right)^2}$$
$$\int \dfrac{dy}{(1+e^{2y}+1-2e^y)^2}$$
$$\int \dfrac{dy}{(e^{2y}+2-2e^y)^2}$$
$$\int \dfrac{dy}{e^4y+4+4e^{2y}+4e^{2y}-4e^y(e^{2y}+2)}$$
$$\int \dfrac{dy}{e^4y-4e^{3y}+8e^{2y}-8e^y+4}$$
From here I gave up on this method.
Attempt $2$:
$$\int \dfrac{\cos^3x}{\cos x+\sin x}$$
$$\int\dfrac{3\cos x+\cos 3x}{4(\cos x+\sin x)}$$
$$\dfrac{1}{4}\left(3\cdot\int\dfrac{\cos x}{\cos x+\sin x}+\int \dfrac{\cos 3x}{\cos x+\sin x}\right)$$
Solving the first integral
$$\cos x=A(\cos x+\sin x)+B(-\sin x+\cos x)+C$$
$$A+B=1$$
$$A-B=0$$
$$A=\dfrac{1}{2},B=\dfrac{1}{2}$$
$$\dfrac{1}{4}\left(\dfrac{3}{2}\left(x+\ln|\sin x+\cos x|\right)+\int \dfrac{\cos 3x}{\cos x+\sin x}\right)$$
I was not understanding how to proceed for second integration.
Attempt $3$:
For @mvpq
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
Trying to write expression as
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A}{1+t}+\dfrac{B}{1+t^2}+\dfrac{C}{(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A(1+t^2)^2+B(1+t^2)+C(1+t)}{(1+t)(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=(A+B+C)+Ct+(B+2A)t^2+At^4$$
$$A+B+C=1\tag{1}$$
$$A=0$$
$$A+2B=0$$
$$B=0$$
$$C=0$$
But $A+B+C=1$, hence no solution, so cannot be solved by partial fractions.
Attempt $4$:
For @heropup
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
Trying to write expression as
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A}{1+t}+\dfrac{Bt+C}{1+t^2}+\dfrac{Dt+E}{(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A(1+t^2)^2+(Bt+C)(1+t^2)(1+t)+(Dt+E)(1+t)}{(1+t)(1+t^2)^2}$$
$$1=A(1+t^2)^2+(Bt+C)(1+t^2)(1+t)+(Dt+E)(1+t)$$
Placing $t=-1$ in the equation
$$4A=1, A=\dfrac{1}{4}$$
$$1=(Bt+C)(1+t+t^2+t^3)+(Dt+Dt^2+E+Et)$$
$$1=Bt+Bt^2+Bt^3+Bt^4+C+Ct+Ct^2+Ct^3+Dt+Dt^2+E+Et$$
$$1=Bt^4+(B+C)t^3+(B+C+D)t^2+(B+C+D+E)t+(C+E)$$
$$B=0$$
$$B+C=0$$
$$C=0$$
$$B+C+D=0$$
$$D=0$$
$$B+C+D+E=0$$
$$E=0$$
$$C+E=1$$
$$E=1$$
This is contradiction right?
Any hints?
|
To find $A,B,C,D,E$, you should have:
$$A(1+t^2)^2+(Bt+C)(1+t)(1+t^2)+(Dt+E)(1+t)=1$$
Then equate coefficients.
Writing the integrand in partial fractions, we have
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}=\int\left(\frac{1}{4(1+t)}+\frac{1-t}{4(1+t^2)}+\frac{1-t}{2(1+t^2)^2}\right)\,dt$$
$$=\frac{1}{4}\log(1+t)+\frac{1}{4}\arctan t-\frac{1}{8}\log(1+t^2)+\frac{1}{2}\int\frac{1-t}{(1+t^2)^2}\,dt$$
To evaluate the last integral, split it up:
$$\int\frac{-t}{(1+t^2)^2}\,dt=\frac{1}{2(1+t^2)}$$
and for the integral below, substitute $t=\tan u$:
$$\int\frac{1}{(1+t^2)^2}\,dt=\int\frac{\sec^2 u}{\sec^4u}\,du=\int \cos^2u\,du=\ldots$$
|
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|
Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In this new video Dr. Loh claims to discover a new way of solving the quadratic equation! How? It is the same as the above formula, by using the quadratic formula, the only thing I see different, is he rewrite it in the above form!
Can someone please explain to me how this is a new way?
|
Let us illustrate by example. Consider the equation
\begin{align}
x^2-2019x-2020 = (x-r_1)(x-r_2) = x^2-(r_1+r_2)x+r_1r_2=0.
\end{align}
The key observation is that the roots $r_1, r_2$ adds up to $2019$, which means the average of $r_1$ and $r_2$ is $\frac{2019}{2}$. Hence the roots have the form $r_\pm = \frac{2019}{2}\pm z$ for some $z$.
Next, it follows
\begin{align}
\frac{(2019)^2}{4}-z^2 = r_1r_2 = -2020 \ \ \implies \ \ z = \sqrt{\frac{(2019)^2}{4}+2020}.
\end{align}
Thus, we have
\begin{align}
r = \frac{2019}{2}\pm \sqrt{\frac{(2019)^2}{4}+2020}.
\end{align}
The "new" aspect is that the derivation avoids completing the square.
Here's another example
\begin{align}
x^2+212323x+24434 = 0
\end{align}
Again, the roots should have the form
\begin{align}
r_\pm = -\frac{212323}{2}\pm z \ \ \implies \ \ \frac{(212323)^2}{4}- z^2= 24434
\end{align}
which means
\begin{align}
z = \sqrt{\frac{(212323)^2}{4}-24434} \ \ \implies \ \ r_\pm = -\frac{212323}{2}\pm \sqrt{\frac{(212323)^2}{4}-24434}.
\end{align}
|
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How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:
*
*3 + 2
*3 + 1 + 1
*2 + 2 + 1
*2 + 1 + 1 + 1
*1 + 1 + 1 + 1 + 1
Is there any way to find out the solution with any two numbers without calculating every one of them?
|
It is a simple function call. In Mathematica, for example:
IntegerPartitions[5, {1, 5}, {1, 2, 3}]
(*
{{3, 2}, {3, 1, 1}, {2, 2, 1}, {2, 1, 1, 1}, {1, 1, 1, 1, 1}}
*)
If you merely seek the number of such partitions:
Length@IntegerPartitions[5, {1, 5}, {1, 2, 3}]
(* 5 *)
[This function asks for all integer partitions of the number 5, of length 1 through 5, using only the numbers 1, 2 and 3. Then take the "length" of this list, i.e., the number of partitions.]
Another example:
IntegerPartitions[10, {1, 10}, {1, 2, 3}]
(*
{{3, 3, 3, 1}, {3, 3, 2, 2}, {3, 3, 2, 1, 1}, {3, 3, 1, 1, 1, 1}, {3, 2, 2, 2, 1}, {3, 2, 2, 1, 1, 1}, {3, 2, 1, 1, 1, 1, 1}, {3, 1, 1, 1, 1, 1, 1, 1}, {2, 2, 2, 2, 2}, {2, 2, 2, 2, 1, 1}, {2, 2, 2, 1, 1, 1, 1}, {2, 2, 1, 1, 1, 1, 1, 1}, {2, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}
*)
Length@IntegerPartitions[10, {1, 10}, {1, 2, 3}]
(* 14 *)
|
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How I can show that $\prod_{i=1}^{n}\frac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\frac{1}{x}-\frac{1}{x}\left(\frac{2}{3}\right)^n+1$? How I can show that the following equality is true:
$$\prod_{i=1}^{n}\dfrac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\dfrac{1}{x}-\dfrac{1}{x}\left(\dfrac{2}{3}\right)^n+1\,?$$
|
Here we have a telescoping product.
We obtain for $n\geq 1$:
\begin{align*}
\color{blue}{\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^j(x+1)-3\cdot 2^{j-1}}}
&=\frac{1}{3^n}\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^{j-1}(x+1)-2^{j-1}}\tag{1}\\
&=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=1}^{n}\left(3^{j-1}(x+1)-2^{j-1}\right)}\\
&=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=0}^{n-1}\left(3^{j}(x+1)-2^{j}\right)}\tag{1}\\
&=\frac{1}{3^n}\,\frac{3^n(x+1)-2^n}{(x+1)-1}\tag{2}\\
&=\frac{3^nx+3^n-2^n}{3^nx}\\
&\,\,\color{blue}{=1+\frac{1}{x}-\left(\frac{2}{3}\right)^n\frac{1}{x}}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we factor out $\frac{1}{3^n}$.
*In (2) we shift the index in the product of the denominator by one to start with $j=0$.
*In (3) we use the telescopic property of the product and cancel equal factors in numerator and denominator.
|
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|
How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$ $$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$
This number looks like $\zeta(5)$ value.
We expand the terms
$$\int_0^1\frac{\frac{\pi^2}{6}}{1-x}\cdot \ln^2(x) \, \mathrm dx-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x} \cdot \ln^2(x) \, \mathrm dx$$
$$2\zeta(2)\zeta(3)-\int_0^1 \frac{\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$$
this last integral it is very complicate to compute...
Can any user please help to show that whether this value is equal to $\zeta(5)$ or not?
|
Here is an alternative approach through the harmonic sum thicket.
Starting with Euler's reflexion formula for the dilogarithm function, namely
$$\operatorname{Li}_2 (x) + \operatorname{Li}_2 (1 - x) = \zeta (2) - \ln x \ln (1- x),$$
your integral $I$ can be rewritten as
$$I = \int_0^1 \frac{\ln^3 x \ln (1 - x)}{1 - x} \, dx + \int_0^1 \frac{\ln^2 x \operatorname{Li}_2 (x)}{1 - x} \, dx = I_1 + I_2.$$
For the first integral, from the generating function for the harmonic numbers $H_n$ we have
$$\frac{\ln (1 - x)}{1 - x} = - \sum_{n = 1}^\infty H_n x^n.$$
Thus
\begin{align}
I_1 &= -\sum_{n = 1}^\infty H_n \int_0^1 x^n \ln^3 x \, dx\\
&= -\sum_{n = 1}^\infty H_n \frac{d^3}{ds^3} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\
&= -\sum_{n = 1}^\infty H_n \frac{d^3}{ds^3} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= 6 \sum_{n = 1}^\infty \frac{H_n}{(n + 1)^4} = 6 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^4},
\end{align}
after reindexing $n \mapsto n - 1$. Since
$$H_n = H_{n - 1} + \frac{1}{n},$$
this leads to
$$I_1 = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} - 6 \sum_{n = 1}^\infty \frac{1}{n^2} = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} - 6 \zeta (5).$$
For the second integral, from the generating function for the generalised harmonic numbers of order two $H^{(2)}_n$ we have
$$\frac{\operatorname{Li}_2 (x)}{1 - x} = \sum_{n = 1}^\infty H^{(2)}_n x^n.$$
Thus
\begin{align}
I_2 &= \sum_{n = 1}^\infty H^{(2)}_n \int_0^1 x^n \ln^2 x \, dx\\
&= \sum_{n = 1}^\infty H^{(2)}_n \frac{d^2}{ds^2} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\
&= \sum_{n = 1}^\infty H^{(2)}_n \frac{d^2}{ds^2} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= 2\sum_{n = 1}^\infty \frac{H^{(2)}_n}{(n + 1)^3} = 2 \sum_{n = 2}^\infty \frac{H^{(2)}_{n-1}}{n^3},
\end{align}
after reindexing $n \mapsto n - 1$. Since
$$H^{(2)}_n = H^{(2)}_{n - 1} + \frac{1}{n^2},$$
we have
$$I_2 = 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 2 \sum_{n = 1}^\infty \frac{1}{n^5} = 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 2 \zeta (5).$$
Returning to our integral, we have
$$I = 6 \sum_{n = 1}^\infty \frac{H_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} - 8 \zeta (5).$$
Since
$$\sum_{n = 1}^\infty \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3),$$
and
$$\sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^3} = 3 \zeta (2) \zeta (3) - \frac{9}{2} \zeta (5),$$
it is immedate that $I = \zeta (5)$, as desired.
|
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|
Evaluate $\int\frac{dx}{(1+x^4)^{1/4}}$
Solve
$$
\int\frac{dx}{(1+x^4)^{1/4}}
$$
Set $t=\log x\implies x=e^t\implies dt=\dfrac{dx}{x}$
$$
\int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{\dfrac{1}{x}dx}{\big(\dfrac{1}{x^4}+1\big)^{1/4}}=\int\frac{dt}{(e^{-4t}+1)^{1/4}}
$$
Set $e^{-4t}+1=y\implies-4e^{-4t}dt=dy$
$$
I=\int\frac{e^{-4t}\,dt}{e^{-4t}(e^{-4t}+1)^{1/4}} = \frac{-1}{4}\int\frac{dy}{(y-1) y^{1/4}}
$$
The solution given in my reference is $$\frac{1}{2} \left(\dfrac{1}{2}\log\dfrac{1+z}{1-z}-\tan^{-1}z\right),\quad z=\frac{(1+x^4)^{1/4}}{x}$$
How do I see what substitution to choose in order to find such a solution ?
|
Once you have
$$I=\int\frac{dx}{(x-1)x^{1/4}},$$
take $x=u^4$ so that $dx=4u^3du$. We have $$I=4\int\frac{u^2du}{u^4-1}.$$
This is
$$\begin{align}
\tfrac14I&=\int\frac{u^2du}{(u^2+1)(u^2-1)}\\
&=\frac12\int\frac{du}{u^2+1}+\frac14\int\frac{du}{u-1}-\frac14\int\frac{du}{u+1}.
\end{align}$$
The rest is easy from there.
|
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|
integrate $\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx $ How can we integrate $$\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx ~~?$$Is it integrable function over $\mathbb R$?, Can we use the fact $\int_{-\infty}^\infty e^{ax-x^2}dx = \sqrt{\pi}e^{\frac{a^2}{4}}$
|
You can use the similar fact that
(i) $\int_{-\infty}^\infty e^{ax-bx^2}dx = \frac{\sqrt{\pi}}{\sqrt{b}}e^{\frac{a^2}{4b}}$
The expression in the exponent is:
$x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2} = x-\frac{x^2+1.9x + 0.95^2}{2\cdot (0.35)^2} = (1 + \frac{1.9x}{2\cdot (0.35)^2})x -\frac{1}{2\cdot (0.35)^2}x^2 + \frac{0.95^2}{2\cdot (0.35)^2}$
You can take
$e^{\frac{0.95^2}{2\cdot (0.35)^2}}$
Out of the integral to get
$\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx =
e^{\frac{0.95^2}{2\cdot (0.35)^2}} \int_{-\infty}^\infty e^{(1 + \frac{1.9x}{2\cdot (0.35)^2})x -\frac{1}{2\cdot (0.35)^2}x^2}dx$
And apply (i) from here.
|
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|
Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2-x^2] $$
|
Once you gave us the answer, the factoring became easy. You just have to look for the factor $1+y$
$$1 - y-x^2-y^2-yx^2+y^3= (1+y^3)-(y+y^2)-x^2(1+y)$$
$$=(1+y)(1-y+y^2)-y(1+y)-x^2(1+y)=(1+y)(1-y+y^2-y-x^2)$$
$$=(1+y)[(1-y)^2-x^2]$$
|
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|
Solve the irrational radical equation Solve equation:$$\frac{x + \sqrt{3}}{\sqrt{x} + \sqrt{x + \sqrt{3}}} + \frac{x - \sqrt{3}}{\sqrt{x} - \sqrt{x - \sqrt{3}}} = \sqrt{x}$$
answer: $S=\{2\}$
Attemp:
$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}}=\sqrt{x}\Leftrightarrow \frac{\left( \sqrt{x}-\sqrt{x+\sqrt{3}} \right)\left( x+\sqrt{3} \right)}{-\sqrt{3}}+\frac{\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right)\left( x-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}$$
$$\frac{2\sqrt{x}\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}\Leftrightarrow 2\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)=\sqrt{3}\Leftrightarrow \sqrt{x}\sqrt{x+\sqrt{3}}=3\sqrt{3}$$
|
You have a mistake in your algebra (see the previous answer) but rationalizing the denominator is a good idea. Note that the domain is $x\geq \sqrt 3$. After you do that and multiply both sides by $\sqrt{3}$, you obtain
$$x\sqrt{x-\sqrt{3}}-\sqrt{3}\sqrt{x-\sqrt{3}}+x\sqrt{x+\sqrt{3}}+\sqrt{3}\sqrt{x+\sqrt{3}}=3\sqrt 3\sqrt x $$ which can be rewritten as
$$(x-\sqrt3)^{3/2}+(x+\sqrt3)^{3/2}=3\sqrt 3\sqrt x $$
Then, square both sides (since they are positive, squaring produces an equivalent equation):
$$(x-\sqrt3)^{3}+(x+\sqrt3)^{3}+2(x^2-3)^{3/2}=27x \Leftrightarrow \\
2x^3+2(x^2-3)^{3/2}=9x $$
If you take the derivative of $f(x)=2x^3+2(x^2-3)^{3/2}-9x$ and use that $x\geq \sqrt 3$, it follows that $f'(x)>0$. This means that $f(x)$ is strictly increasing on the interval $x\geq \sqrt 3$, and the equation has no more than $1$ root. We can check $x=2$ is a root, so that's the final answer.
|
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|
Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please.
|
Let $(p,q,r,s) = (1-a,a-b,b-c,c)$, we have $p+q+r+s = 1$.
We are given
$$p^2 + q^2 + r^2 + s^2 = \frac14$$
This leads to
$$\begin{align} &\;\left(p - \frac14\right)^2 +
\left(q - \frac14\right)^2 +
\left(r - \frac14\right)^2 +
\left(s - \frac14\right)^2\\
= &\; (p^2+q^2+r^2+s^2) - \frac12(p+q+r+s) + \frac14\\
= &\; \frac14 - \frac12 + \frac14\\
= &\; 0
\end{align}
$$
As a result, $$p = q = r = s = \frac14\quad\implies\quad (a,b,c) = \left(\frac34,\frac12,\frac14\right)$$
|
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|
Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \mod (3)$
Hence this equation has no integer solution. Similar result comes out when we consider remainder of nomials on 4.
Is this solution correct?
|
I think you are trying to say:
If you assume $\gcd(y,3) =1$ (Why are you assuming this? What if $\gcd(y,3) = 3$?) or in other words if $3\not\mid y$ then $y\equiv 1 \pmod 3$ or $y\equiv 2 \pmod 3$.
ANd therefore $y^2\equiv 1 \pmod 3$ or $y^2\equiv 4\equiv 1\pmod 3$. and so $4y^2 \equiv 1\pmod 3$. I'm not sure why you didn't eliminate $4y^2 \equiv 2\pmod 3$.
And $3x^2 \equiv 0\pmod 3$.
Then you said "The common remainder between $3x^2$ and $4y^2$ is $0$" and I can not figure out what you mean and I can't think of any interpretation that would be compatible with you discovery that $3x^2 \not \equiv 4y^2$.
You claim $3x^2 - 4y^2 \equiv 0\pmod 3$ which is completely wrong as $4y^2 \equiv 1\pmod 3$ we have $3x^2 - 4y^2 \equiv -1 \equiv 2 \pmod 3$.
Which is enough to show $13\equiv 1\not \equiv 2\pmod 3$. So there is not integer solution.
IF we assume $\gcd(3,y)=1$. But you didn't explain why we are assuming $\gcd(3,y) = 1$.
Which you never considered.
Better to simply note:
$3x^2 - 4y^2 \equiv \begin{cases}3*0^2\equiv 0& x\equiv 0 \\3*1^2\equiv3\equiv 0 &x\equiv 1\pmod 3\\3*2^2\equiv 3*4\equiv 0&x\equiv 2\pmod 3\end{cases}-\begin{cases}4*0^2\equiv 0& y\equiv 0\pmod 3\\4*1^2\equiv 1&y\equiv 1\pmod 3\\4*2^2\equiv 16\equiv 1&y\equiv 2\pmod 3\end{cases}\pmod 3\equiv$
$0 - \begin{cases}0& \text{if }y\equiv 0\pmod 3\\1&\text{otherwise}\end{cases}\equiv$
$\begin{cases}0\\-1\end{cases}\equiv \begin{cases}0\\2\end{cases}\pmod 3$
$\not \equiv 1 \equiv 13 \pmod 3$.
And it's enough to prove it for just modulo $3$.
FWIW if you tried modulo $4$ you'd get.
$3x^2 - 4y^2 \equiv 3x^2\equiv 3*\begin{cases}0^2\\1^2\\2^2\\3^2\end{cases}\equiv$
$\begin{cases}0\\3\\0\\3\end{cases}\not \equiv 1\equiv 13\pmod 4$.
But there was utterly no reason to do those as we were done.
|
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|
Integration of a particular rational expression I am trying to solve the following integration, where $a,b,c,d,e$ and $f$ are constants:
$$I=\int\frac{x^4+ax^3+bx^2+cx+d}{x^3(x^3+ex+f)}dx$$
I tried to solve the integral using the following two methods, but both seemed to be very much complicated:
Method 1:
Using partial fraction decomposition by calculating the three roots of the denominator. Among the three roots, one root is real and the other two roots are complex (both are complex conjugates of each other). However, the roots are too much complicated and are as follows:
Method 2:
I tried to expand the denominator using binomial series as follows:
$$I=\int\frac{\frac{1}{x^2}+\frac{a}{x^3}+\frac{b}{x^4}+\frac{c}{x^5}+\frac{d}{x^6}}{1+\frac{e}{x^2}+\frac{f}{x^3}}dx$$
Then writing $\epsilon=\frac{e}{x^2}+\frac{f}{x^3}$, the above integral becomes
$$I=\int\left(\frac{1}{x^2}+\frac{a}{x^3}+\frac{b}{x^4}+\frac{c}{x^5}+\frac{d}{x^6}\right)\left(1+\epsilon\right)^{-1}dx$$
For a quite good approximation, it is required to expand the binomial series up to $\epsilon^{11}$, i.e. $40$ terms in the expression for the integral and this is too much cumbersome.
Even Mathematica expresses the result as a conditional expression and it is required to provide the range of the constants $a,b,c,d,e$ and $f$ to obtain the exact expression.
QUESTION:
Is it possible to solve the integration analytically using any other
suitable method? If no such methods are possible, do there exist any
approximation technique that might work?
|
Taking into account what you wrote about the roots of the denominator, I should write
$$\frac{x^4+ax^3+bx^2+cx+d}{x^3(x^3+ex+f)}=\frac{x^4+ax^3+bx^2+cx+d}{x^3(x-r)(x^2+s x+t)}$$Now, partial fraction decomposition would give
$$-\frac{d}{r t }\frac 1 {x^2}+\frac{d r s-d t-c r t}{r^2 t^2 }\frac 1 {x}+\frac{a r^3+b r^2+c r+d+r^4}{r^2 \left(r^2+r s+t\right)}\frac 1 {x-r}+\frac{A+B x}{t^2 \left(r^2+r s+t\right) \left(x^2+s x+t\right)}$$ where
$$A=-d r s^2 - d s^3 + d r t + 2 d s t + c r s t + c s^2 t - c t^2 -
b r t^2 - b s t^2 + a t^3 + r t^3$$
$$B=-a r t^2-b t^2+c r t+c s t-d r s-d s^2+d t+r s t^2+t^3$$ This makes the integral to be quite simple.
|
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|
Chain rule partial derivative
Find the partial derivatives $\partial Z / \partial u $ and $\partial Z / \partial \nu$ for the following function:
$$
Z(x,y,z) = 2x^3 - 3xy^2 + 0.75~yu - 5u^2\quad \text{where}~x = \sqrt{u+\nu}~\text{and}~y=\nu^2
$$
No matter how hard I try I couldn't solve it and couldn't find anything similar on internet either. Could you take a look at it?
|
We calculate $\frac{\partial Z}{\partial u}$ in two ways.
First variant: We consider
\begin{align*}
Z(x(u,v),y(u,v),z(u,v))=2x^3-3xy^2+\frac{3}{4}yz-5z^2
\end{align*}
where
\begin{align*}
x=\sqrt{u+v},\qquad y=v^2,\qquad z=u
\end{align*}
According to the chain rule we obtain
\begin{align*}
\color{blue}{\frac{\partial Z}{\partial u}}
&=\frac{\partial Z}{\partial x}\,\frac{\partial x}{\partial u}
+\frac{\partial Z}{\partial y}\,\frac{\partial y}{\partial u}
+\frac{\partial Z}{\partial z}\,\frac{\partial z}{\partial u}\\
&=\left(6x^2-3y^2\right)\frac{1}{2}\left(u+v\right)^{-\frac{1}{2}}+
\left(-6xy+\frac{3}{4}u\right)\cdot0+\left(\frac{3}{4}y-10z\right)\cdot 1\\
&=\left(6(u+v)-3v^4\right)\frac{1}{2}\left(u+v\right)^{-\frac{1}{2}}+\left(\frac{3}{4}v^2-10u\right)\\
&\,\,\color{blue}{=3\left(u+v\right)^{\frac{1}{2}}-\frac{3}{2}v^4\left(u+v\right)^{-\frac{1}{2}}+\frac{3}{4}v^2-10u}\tag{1}
\end{align*}
Second variant: We consider
\begin{align*}
Z(u,v)&=\left. 2x^3-3xy^2+\frac{3}{4}yz-5z^2\right|_{x=\sqrt{u+v}, y=v^2, z=u}\\
&=2\left(u+v\right)^{\frac{3}{2}}-3\left(u+v\right)^{\frac{1}{2}}v^4+\frac{3}{4}v^2u-5u^2\tag{2}
\end{align*}
We obtain from (2)
\begin{align*}
\color{blue}{\frac{\partial Z}{\partial u}}
&=\frac{\partial}{\partial u}\left(2\left(u+v\right)^{\frac{3}{2}}-3\left(u+v\right)^{\frac{1}{2}}v^4+\frac{3}{4}v^2u-5u^2\right)\\
&\,\,\color{blue}{=3\left(u+v\right)^{\frac{1}{2}}-\frac{3}{2}v^4\left(u+v\right)^{-\frac{1}{2}}+\frac{3}{4}v^2-10u}\tag{3}
\end{align*}
We observe (1) and (3) are equal as expected. The partial derivative $\frac{\partial Z}{\partial v}$ can be calculated similarly.
|
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|
$50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$
$50\cos^2 x + 5\cos x = 6\sin^2 x$
Find $\tan x$
I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$
I then solve this to get $\cos x = \dfrac27, -\dfrac38$
Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\dfrac{\sqrt{55}}{3}$
Are these $\pm$ signs correct? My reasoning came for a CAST diagram, can you confirm if this is correct and if not, why?
|
Using the Weierstrass transformation, the equation rationalizes to
$$50\left(\frac{1-t^2}{1+t^2}\right)^2+5\frac{1-t^2}{1+t^2}=6\left(\frac{2t}{1+t^2}\right)^2$$
and gives, by solving the biquadratic
$$45t^4-100t^2+55=24t^2,$$
$$t^2=\frac 59,\frac{11}5.$$
The tangent follows, by
$$\pm\frac{2\dfrac{\sqrt5}3}{1-\dfrac 59},\pm\frac{2\sqrt{\dfrac{11}5}}{1-\dfrac{11}5}.$$
Here you don't have to worry about angles nor quadrants.
|
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|
Can we solve for $a$ and $b$ only knowing the value of $z$? We have the following equation for $z$ > 0
$ a^2b+ab^2 = z $
Only the value of $z$ is known. How can we solve this equation to get the values of $a$ and $b$?
Simplifying,
$ab(a+b) = z $
eg. $z=84$
then $a = 3$, $b=4$
Any approach or suggestions or trial and error method ?
|
If $a,b,z \in \mathbb{R}$ then there are an infinite number of solutions.
If we assume $a=b$ then we have $2a^3=z \Rightarrow a=b=\root 3 \of {\frac z 2}$.
But if we assume $2a=b$ then we have $6a^3=z \Rightarrow a=\root 3 \of {\frac z 6}, b=2\root 3 \of {\frac z 6}$.
And, in general, if $ka=b$ then $(k+k^2)a^3=z \Rightarrow a=\root 3 \of {\frac z {k+k^2}}, b=k\root 3 \of {\frac z {k+k^2}}$.
On the other hand if $a,b$ are constrained to be integers then you can have zero, one, two or an infinite number of solutions, depending on the value of $z$.
|
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|
Is $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$ compact? Is the following set compact? How can I show it? $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$
Clearly it is closed as it contains its boundary, but I can not show that it is bounded..
|
Substituting, we find that $x^2+y^2-x-y = (x-\frac{1}{2})^2 - \frac{1}{4} + (y-\frac{1}{2})^2 - \frac{1}{4} = \frac{3}{4}$, ie $(x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 = \frac{1}{4}$, so the projection of $C$ on the z-plane is a circle, which is bounded as $|x|\le K$, $|y|\le K$ for some constant $K$.
Now, going back to the first equation we have that $|z| = |1-x-y| \le |1| + |x| + |y| \le 1+2K$, so $z$ is bounded as well.
We conclude that $C$ is bounded, and since you already know how to show that it is closed, we conclude that it is compact.
|
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|
If $\frac ab=\frac bc=\frac cd$, then $\frac ad=\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}$
If
$$\frac ab=\frac bc=\frac cd$$
then prove that
$$\frac ad=\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}$$
I don't want a full solution, just a little hint on how to start solving are welcome
|
Note
$$\frac{d^2}{a^2}\cdot {a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}
=\frac{a^3+\frac{b^2c^2}{a^2}+ac^2}{\frac{b^4c}{d^2}+d^2+b^2c}
=\frac{a^3+d^2+ac^2}{a^3+d^2+ac^2}=1$$
Thus,
$$\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}=\sqrt{\frac{a^2}{d^2}}=\frac ad$$
|
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|
Does there exist a power of 2 whose last 100 digits are composed only of the digits 1 or 2? Does there exist a power of 2 whose last 100 digits are composed only of the digits 1 or 2? If so, what techniques could be used to prove its existence?
|
Let $q_n$ for $n>0$ be the recurring sequence of natural numbers defined by:
\begin{align}
&q_1=1&&q_{k+1}=\frac{q_k+5^k(2-q_k\bmod 2)}2
\end{align}
then for every $k>0$ the number $a_k=2^kq_k$ satisfy:
*
*$0<a_k<10^k$;
*the decimal digits of $a_k$ belongs to $\{1,2\}$;
*there exists $m\geq k$ such that $2^m\equiv a_k\pmod{10^k}$.
While (1) and (2) are easily proved by induction, to prove (3), note that $5\nmid q_k$, for otherwise $10\mid a_k$ but this would implies that the last digit of $a_k$ is zero - a contradiction.
Since $2$ is a primitive root modulo $5^k$, there exists $n>0$ such that
$$2^n\equiv q_k\pmod{5^k}$$
Then multiplying by $2^k$, we get
$$2^{n+k}\equiv a_k\pmod{10^k}$$
To generate the recurrence above: assume $2^m\equiv a_k\pmod{10^k}$ with $m\geq k$ and $a_k=2^kq_k$ and let $2^n\equiv 10^kd+a_k\pmod{10^{k+1}}$ with $d\in\{1,2\}$ and $n\geq k+1$.
Dividing by $2^k$ we get $2^{n-k}\equiv 5^kd+q_k\pmod{2\cdot 5^{k+1}}$, hence reducing modulo $2$, we get $d\equiv -q_k\pmod 2$.
Since $d\in\{1,2\}$, we have $d=2-q_k\bmod 2$.
Moreover,
$$2^n\equiv 2^k(5^kd+q_k)=2^{k+1}q_{k+1}\pmod{10^{k+1}}$$
where $q_{k+1}=\frac{q_k+5^kd}2$.
|
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|
How do I evaluate the integral of the square root of a quartic equation? I'm currently trying to evaluate the integral
$$\int^1_0 \text{d}t\sqrt{(1-t^2)(1-k^2t^2)}$$
where $k\in(0,1)$. Is it the case that this can be expressed in terms of elliptic integrals? I'm struggling to see how this can be done as the elliptic integral that looks closest to this expression is
$$\int\frac{\text{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}$$
and I'm not sure how to relate the two.
|
Rewriting the integrand and integrating by parts we find
\begin{align}f(k) &\equiv \int \limits_0^1 \sqrt{(1-x^2)(1-k^2 x^2)} \, \mathrm{d} x = \int \limits_0^1 \frac{\sqrt{1-x^2}}{x} x \sqrt{1-k^2 x^2} \, \mathrm{d} x \\
&= \left[\frac{\sqrt{1-x^2}}{x} \frac{1-(1-k^2 x^2)^{3/2}}{3 k^2}\right]_{x=0}^{x=1} - \int \limits_0^1 \left(-\frac{1}{x^2 \sqrt{1-x^2}}\right) \frac{1-(1-k^2 x^2)^{3/2}}{3 k^2} \, \mathrm{d} x \\
&= \frac{1}{3k^2} \int \limits_0^1 \frac{1-(1-k^2 x^2)^{3/2}}{x^2 \sqrt{1-x^2}} \, \mathrm{d} x \, .
\end{align}
Mathematica can do the remaining integral. In order to get the correct result by hand, we rewrite the integrand again (in a less than obvious way, admittedly):
$$ f(k) = \frac{1}{3k^2} \int \limits_0^1 \left[\frac{(1+k^2)\sqrt{1-k^2 x^2}}{\sqrt{1-x^2}} - \frac{(1-k^2)}{\sqrt{(1-x^2)(1-k^2 x^2)}} - \frac{1 - k^2 x^4 -\sqrt{1-k^2 x^2}}{x^2 \sqrt{(1-x^2)(1-k^2 x^2)}} \right]\mathrm{d} x \, . $$
The first to terms lead to elliptic integrals and the third one has a surprisingly simple antiderivative, so we obtain
\begin{align}
f(k) &= \frac{1}{3k^2}\left[(1+k^2)\operatorname{E}(k) - (1-k^2) \operatorname{K}(k) - \left[\frac{\sqrt{1-x^2}\left(1-\sqrt{1-k^2x^2}\right)}{x}\right]_{x=0}^{x=1}\right] \\
&= \frac{(1+k^2)\operatorname{E}(k) - (1-k^2) \operatorname{K}(k)}{3k^2} \, .
\end{align}
|
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|
Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there some clever way to show this other than to add the fractions together by brute-force? For example, is there some way to group terms together and say something like "These terms sum to more than $\frac{1}{3}$, these terms sum to more than $\frac{1}{2}$, and these terms sum to larger than $\frac{1}{6}$, so the whole thing sums to more than $1$"?
|
For positive, unequal $a$ and $b$:
$\dfrac1a+\dfrac1b=\dfrac{a+b}{ab}>\dfrac4{a+b}$
because $(a+b)^2>4ab$ (the difference between these is $(a-b)^2$). So,
$\dfrac15+\dfrac17>\dfrac4{12}=\dfrac13$
$\dfrac19+\dfrac1{11}>\dfrac4{20}=\dfrac1{5}$
$\dfrac18+\dfrac1{12}>\dfrac4{20}=\dfrac1{5}$
When these inequalities are put into the given sum the claimed bound follows.
|
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|
Compute $\sigma$, given $\sigma^{11}$ in $S_{10}.$
Let $\sigma \in S_{10}$ with $\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)$. How to compute $\sigma$?
I tried:
$\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
Now I tried to find two permutations such that
$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ & & & & & & & & & \end{pmatrix}\cdot \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ & & & & & & & & & \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
But it's not working.
How to find $\sigma$ here?
|
Let $\tau= (314)(529687a)$ (with $a=10$) & so $\tau^{21}=e$ & $\sigma^{21}=e$. We have $ \tau= \sigma^{11}$. Squaring gives $\sigma=\tau^2$. Which is easy to compute.
|
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|
Methods for evaluating $\sum_{n=1}^\infty \frac1{a+(n-1)n}$ I am interested in methods for evaluating the sum $$\sum_{n=1}^\infty \frac1{a+(n-1)n}.$$
Indeed I will give my own answer below using the Residue Theorem.
Please feel free to post other methods for the evaluation, such as Maclaurin series, methods from harmonic/fourier analysis, ...
Related question: Continuity of $f(x) = \sum_{n=0}^{\infty} \frac{1}{n(n+1)+x}$.
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Let $s$ and $t$ to be the roots of $n^2-n+a=0$. So
$$\frac{1}{a+n(n-1)}=\frac{1}{(n-s)(n-t)}=\frac{1}{s-t}\left(\frac 1{n-s}-\frac 1{n-t} \right)$$ Recalling that
$$\sum_{n-1}^p \frac 1{n-x}=\psi(p-x+1)-\psi (1-x)$$ we have that, if
$$S={\sqrt{1-4 a}}\sum_{n-1}^p \frac{1}{a+n(n-1)}$$
$$S=\psi \left(\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a}\right)-\psi
\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4 a}\right)+$$ $$\psi \left(p+\frac{1}{2}-\frac{1}{2} \sqrt{1-4 a}\right)-\psi \left(p+\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a}\right)$$ Now, using the reflection formula for the digamma function
$$\psi \left(\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a}\right)-\psi
\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4 a}\right)=\pi \tan \left(\frac{1}{2} \pi \sqrt{1-4 a}\right)$$ Expanding the remaining terms as series for large values of $p$, then
$$\sum_{n-1}^p \frac{1}{a+n(n-1)}=\frac {\pi \tan \left(\frac{1}{2} \pi \sqrt{1-4 a}\right) } {\sqrt{1-4 a}}-\frac{1}{p}+\frac{a}{3 p^3}+O\left(\frac{1}{p^5}\right)$$
|
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What is the theory behind a simple pattern involving binomial coefficients? These binomial equations are all true.
$\binom{7}{4} = \binom{4}{4} + 3[ \binom{4}{3} + \binom{4}{2}] + \binom{4}{1}$
$\binom{8}{4} = \binom{5}{4} + 3[ \binom{5}{3} + \binom{5}{2}] + \binom{5}{1}$
$\binom{9}{4} = \binom{6}{4} + 3[ \binom{6}{3} + \binom{6}{2}] + \binom{6}{1}$
$\dots$
Is this the result of some general binomial identity or theory?
My work
I stumbled upon this while working with a recurrence relation in $(m,n)$ with $m,n \in \Bbb Z^+$
that I was trying to put into closed form.
|
By repeatedly applying Pascal's Theorem:
$$\begin{align*}
\binom{n}{4} &=\binom{n-1}{4}+\binom{n-1}{3}
\\ &=\left(\binom{n-2}{4}+\binom{n-2}{3}\right)+\left(\binom{n-2}{3}+\binom{n-2}{2}\right)
\\ &=\left(\left(\binom{n-3}{4}+\binom{n-3}{3}\right)+\left(\binom{n-3}{3}+\binom{n-3}{2}\right)\right)+\left(\left(\binom{n-3}{3}+\binom{n-3}{2}\right)+\left(\binom{n-3}{2}+\binom{n-3}{1}\right)\right)
\\ &= \binom{n-3}{4}+3\binom{n-3}{3}+3\binom{n-3}{2}+\binom{n-3}{1}
\end{align*}
$$
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Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and
$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$
$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$
$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
From this we can conclude that
$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
Is this solution correct?
|
Your answer is extremely close to the correct derivation. The error occurs when you write
$$\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=\sum_{\color{red}{n=0}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
where the last equality should have an index starting from $n=1$. The correct derivation is
\begin{align}\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)&=\sum_{\color{blue}{n=1}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}\\&=\sum_{\color{blue}{n=0}}^\infty \frac{\color{blue}{(n+1)}}{2^{\color{blue}{(n+1)+1}}} (x+2)^{\color{blue}{(n+1)-1}}\\&=\sum_{n=0}^\infty \frac{(n+1)}{2^{n+2}} (x+2)^{n}\end{align}
You could also find the power series through giobrach's answer, which might be the most straightforward technique.
|
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Given a polynomial $P$ with integer coefficients, find $P(2)$ Question:
$P(x)$ be a polynomial with non-negative integer coefficients such that $P(0)=33$ , $P(1)=40$ and $P(9)=60000$.
Find $P(2)$.
My attempt:
Suppose $P(x) = a_nx^n + \dotsb + a_1x + a_0$.
Now, $P(0) = 33$ implies $a_0 = 33$.
$P(1)=40$ implies that the sum of the coefficients other than constant term is $7$. Also,
$$P(9)-P(0)=3(2^5\times5^4-11)=3^2a_{1}+3^4a_{2}+\dotsb+a_n3^{2n}.$$
Dividing both sides by $3^3$ we can easily infer $a_{1}=3$. After solving it further, I ended up with
$$a_{2}+3a_{3}+\dotsb a_{n}3^{2n-3}=740,$$
but I can't find $P(2)$.
There are few more coefficients remaining other than $a_{1}$ whose sum is $4$, and I think casework can give a solution... like assuming consecutive $4$ coefficients $1$, $1$, $1$, $1$; $1$, $2$, $1$ etc., but I don't want to go that way.
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The polynomial can be at most of degree $4$ because
$9^5=59409$
so
$p(x)=33+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$
and
$a_1+a_2+a_3+a_4+a_5=7$
It is clear that $a_5=1$ otherwise
if $a_5=0$ then $p(9)<60000$
while if $a_5>1$ then $a_59^5>60000$
If $a_4\neq 0$ then $p(9)>60000$ so $a_4=0$. If $a_3=0$ we have
$a_1+a_2=6$
$9a_1+81a_2=918$
that it has not solutions
So We must suppose $a_3\neq 0$ and We must compute
$a_1+a_2+a_3=6$
$9a_1+81a_2+729a_3=918$
This means $a_3=1$ and so
$a_1+a_2=5$
So is clear that We must have $a_1=3$ and $a_2=2$. So the polinomial is
$p(x)=33+3x+2x^2+x^3+x^5$
|
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|
Combination of problem having identical objects without using generating function approach
A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is
My Attempt
Number of ways of partitioning $n$ identical objects into $r$ distinct groups, if empty groups are allowed is:${}^{n+r-1}C_{r-1}$.
Here, $n=2m$ and $r=4$
Req. number of ways = ${}^{n+r-1}C_{r-1}-$[#{more than m marks for paper 1}+#{more than m marks for paper 2}+#{more than m marks for paper 3}+#{more than m marks for paper 4}]=$${}^{2m+4-1}C_{4-1}-[{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}]={}^{2m+4-1}C_{4-1}-4.{}^{m+4-1}C_{4-1}\\
={}^{2m+3}C_{3}-4.{}^{m+3}C_{3}=\frac{(2m+3)(2m+2)(2m+1)}{3.2}-4\frac{(m+3)(m+2)(m+1)}{3.2}\\
=\frac{m+1}{3}.[4m^2+8m+3-2m^2-10m-12]=\frac{(m+1)(2m^2-2m-9)}{3}
$$
But my reference gives the solution $\dfrac{(m+1)(2m^2+4m+3)}{3}$, so what is going wrong here ?
Note: I am trying to solve this problem without using the generating function approach
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We have to count the number of solutions of
$$x_1+x_2+x_3+x_4=2m$$
where $0\leq x_i\leq m$ are integers for $i=1,2,3,4$.
By Stars and Bars, if we consider just the condition $x_i\geq 0$ we have
$\binom{2m+3}{3}$ solutions. On the other hand, we can have at most one variable $x_i\geq m+1$. If $i=1$, we have to count the number of solutions of
$$y_1+x_2+x_3+x_4=2m-(m+1)=m-1$$
where $y_1:=x_1-(m+1)\geq 0$, $x_i\geq 0$ for $i=2,3,4$. Such solutions, again by Stars and Bars, are $\binom{(m-1)+3}{3}$. By symmetry, the same number turns out for $i=2,3,4$. Hence the answer is
$$\binom{2m+3}{3}-4\binom{m+2}{3}=\frac{(m+1)(2m^2+4m+3)}{3}.$$
|
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Algebraic and Trigonometric expression is $>0$ for all real $x$
Prove that $$2x^2\sin x+2x\cos x+2x^2+1$$ is always positive for all real $x$.
From completing the square method
Write $1$ as $\sin^2 x+\cos^2 x$
$$x^4+2x^2\sin x+\sin^2 x+x^2+2x\cos x+\cos^2 x+x^2-x^4$$
$$(x^2+\sin x)^2+(x+\cos x)^2+x^2(1-x^2)$$
$\bullet\; $ In $|x|\leq 1$ our expression is $>0$.
Is there is any way to prove expression $>0$ for $|x|>1$?
Thanks in advance.
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By completing the square,
$$2(\sin x+1)x^2+2\cos x\,x+1\\
=2(\sin x+1)\left(x+\frac{\cos x}{2(\sin x+1)}\right)^2+1-\frac{\cos^2x}{2(\sin x+1)}\\
=2(\sin x+1)\left(\left(x+\frac{\cos x}{2(\sin x+1)}\right)^2+\frac{\sin x+1}2\right).$$
Obviously, $\sin x+1\ge0$. In case of equality, $\sin x=-1\implies \cos x=0$ and the expression reduces to $1$.
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Let $X,Y$ be independent normally distributed random variables. Find the density of $\frac{X^2}{Y^2+X^2}$ Let $X,Y$ be independent standard normally distributed random variables and $X,Y\neq 0$. Find the density of $\frac{X^2}{Y^2+X^2}$
I was given the tip of first calculating the density of $(X^2,Y^2)$ and then calculate the density of $(\frac{X^2}{Y^2+X^2},Y^2+X^2)$
When I follow the tips: I know that $X^2\sim\Gamma(\frac{1}{2},\frac{1}{2})$ and $Y$ too. Furthermore, $X^2$ and $Y^2$ are still independent. Therefore the density $f_{(X^2,Y^2)}(x,y)$ can be written as $f_{X^2}(x)f_{Y^2}(y)$ where $f_{X^2}$ and $f_{Y^2}$ are the density functions of $X^2$ and $Y^2$
My next idea, with the tip above in mind is to consider a map $\varphi: (x,y) \mapsto (\frac{x}{x+y},x+y)$
It then follows that $(\frac{X^2}{Y^2+X^2},Y^2+X^2)=\varphi(X^2,Y^2)$
and $f_{\frac{X^2}{Y^2+X^2},Y^2+X^2}(a,b) = f_{\varphi(X^2,Y^2)}(a,b)
$
Note that $\varphi^{-1}: (a,b)\mapsto (ba,b-ba)$ and thus $\det D \varphi^{-1}(a,b)=\det\begin{pmatrix} b & a \\
-b & 1-a
\end{pmatrix}=b(1-a)+ab\implies \det D \varphi^{-1}(\frac{X^2}{Y^2+X^2},Y^2+X^2)=(Y^2+X^2)(1-\frac{X^2}{Y^2+X^2})+(\frac{X^2}{Y^2+X^2})(Y^2+X^2)=Y^2+X^2$
And hence $P_{(\frac{X^2}{Y^2+X^2},Y^2+X^2)}(A)=\int_A f_{(\frac{X^2}{Y^2+X^2},Y^2+X^2)}(x,y)\,dx\,dy = \int_{\varphi^{-1}(A)}f_{(X,Y)}(x,y)\times (X^2+Y^2)\,dx\,dy = \int_{\varphi^{-1}(A)} f_X(x)\times f_Y(y)\times (X^2+Y^2)\,dx\,dy$
Where do I go from here?
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I’m on my phone so I can’t type it all out but X^2 and Y^2 are independent chi-squared r.v.s with one degree of freedom each.
These chi squared rvs are equivalent to Gamma(1/2,2) in distribution.
By a known fact about gamma random variables, if X and Y are independent gammas with the same location but different shape parameters a and b, X/X+Y is Beta(a,b). In this case we end up with Beta(1/2,1/2) as our final answer.
=====
Additional proof of independence of X/(X+Y) and X+Y along with derivation of distributions for gamma r.v.s X,Y:
$X,Y\sim \Gamma(\lambda_1,a),\Gamma(\lambda_2,a)$ and independent.
Let $,=/(+),+$ so $=,(1−)=$ and the jacobian $∣∣∂(,)/∂(,)∣∣=|([[,][−,1−]])|=||=$. The last equality is true since $,>0$.
Then $_{,}(,)=_{,}((,),(,)) \propto ^{−}^{_1−1}^{−}^{_2−1}=^{−(+)}^{_1−1}^{_2−1}.$
Simplifying further, we get
$$e^{-av}(uv)^{\lambda_1-1}((1-u)v)^{\lambda_2-1}v = e^{-av}v^{\lambda_1+\lambda_2-1}\times u^{\lambda_1-1}(1-u)^{\lambda_2-1}.$$
Thus $V\sim \Gamma(\lambda_1+\lambda_2,a)$ and $U=X/(X+Y)\sim Beta(\lambda_1,\lambda_2)$ and are independent.
|
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Computation of: $\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)$
Evaluate:
$$\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)\;n\in\mathbb N$$
My attempt:
Using the manual limit:
$$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$$
$$\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)$$
$$=\lim_{n\to\infty}\left(\sum_{i=1}^n\frac{\ln\left(1+\frac{1}{n^2+i}\right)}{\frac{1}{n^2+i}}\cdot\frac{1}{n(n^2+i)}\right)=0$$
Is this correct?
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You can squeeze it as follows:
$$n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+n}\right)\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+k}\right)\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2}\right)$$
Hence,
$$\underbrace{n^2\ln\left(1+\frac{1}{n^2+n}\right)}_{\stackrel{n\to\infty}{\longrightarrow}1}\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+k}\right)\leq \underbrace{n^2\ln\left(1+\frac{1}{n^2}\right)}_{\stackrel{n\to\infty}{\longrightarrow}1}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3535379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integrate $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Evaluate
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I have tried substitution of $\sin x$ as well as $\cos x$ but it is not giving an answer.
Do not understand if there is a formula for this or not.
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Let $x=\tan ^{-1}(t)$ to make
$$I=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx=\int \frac{dt}{t^{3/2}+t^2+\frac{1}{\sqrt{t}}+1}$$ Now, $t=u^2$
$$I=\int \frac{2 u^2}{u^5+u^4+u+1}\,du=\int \frac{du}{u+1}-\int \frac{u^3-u^2-u+1}{u^4+1}$$ For the second integral, use the roots of $u^4+1=0$ and partial fraction decomposition to face again a series of integrals
$$J_k=\int \frac {du}{u+k}=\log(u+k)$$ where $k$ is a complex number.
Recombine everything together to get
$$I=-\frac{1}{2} \tan ^{-1}\left(\frac{1}{u^2}\right)+\frac{1}{2 \sqrt{2}}\log \left(\frac{u^2-\sqrt{2} u+1}{u^2+\sqrt{2} u+1}\right)+\log \left(\frac{u+1}{\sqrt[4]{u^4+1}}\right)$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3535688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$ Given the inequality $$1+1.2 + 1.2^2 + 1.2 ^3 +\cdots + 1.2^n < \frac{N}{16}?$$
I need the value of $n$, or just an approximation. $N$ is known.
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Using the formula for the partial sum of a geometric series, rewrite this as
$$
\frac {1.2^{n+1}-1}{1.2-1} < \frac N{16} \implies
5(1.2^{n+1}-1) < \frac N{16} \implies
1.2^{n+1} < 1 + \frac N{80}.
$$
If we take a logarithm, we can solve this exactly to get
$$
n < \frac{\log(1 + \frac N{80})}{\log(1.2)} - 1.
$$
For a numerical approximation, we can note that $1.2^{n+1} \geq 1 + (0.2) \cdot (n+1)$, so that $n$ will necessarily satisfy the inequality if
$$
1 + \frac{n+1}{5} < 1 + \frac{N}{80} \implies n < \frac{N}{16} - 1.
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3538408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the radius of three equal circles, with radius $r$, inscribed in a rectangle. How can we find the radius of the small circles with the same given radius $r$. Three equal circles, with radius $r$, are inscribed in a rectangle in a way only one of them touches the others two, as the figure indicates. The circles centres form an isosceles triangle, is there any theorem can help in finding the value of $r$? Thanks for your help.
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If the white curve is a circle the you have three points of the big circle (assume symmetry: Big circle has $(-8,0),(0,6),(8,0)$) so the equation (and radius) can be figured.
Big Circle is center at $(0, -A)$ with radius $R$ and $8^2 + A^2 = R^2$ and $0^2 + (A+8)^2 = R^2$.
The three little circles have equations $(a_i - x)^2 + (b_i-y)^2 = r^2$. $a_1=-a_3;b_1=b_3; a_2 = 0$. circle 1: has the three points $(-w,0), (-v,z);$ (which satisfies $v^2 + (z+A)^2 = R^2$) and $(-m,k)$.
Circle 2: has three point $(-m,k),(m,k)$ and $(0,6)$. The point $(-m,k)$ shared between circle 1 and circle two is tangent and colinear to the centers.
And circle 3 has three points $(w,0),(v,z),(m,k)$.The point $(m,k)$ shared between circle 1 and circle two is tangent and colinear to the centers.
So solve the following equations.
Big Circle:
$(8-0)^2 + (0+A)^2 = R^2$
$(0-0)^2 + (6 + A)^2 = R^2$
$(m-0)^2 + (k+A)^2 = R^2$
$(v-0)^2 + (z+A)62 = R^2$
Circle 2:
$(m-0)^2 + (k-b_2)^2 = (0-0)^2 + (6-b_2)^2 = r^2$.
Circle 3:
$(a_3-w)^2 + (b_3-0)^2 = (a_3 - v)^2 + (b_3 - z)^2=(a_3-m)^2 + (b_3-k)^2 = r^2$
Colinear tangent of circle 2 and circle 3
$m = \frac {a_3}2$ and $k = \frac {b_2 + b_3}2$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3539093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.